Conway, John B - Functions of One Complex Variables I (1).pdf

Graduate Texts in Mathematics

11

Editorial Board S. Axler F.W. Gehring K.A. Ribet

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John B. Conway

Functions of One Cotnplex Variable I Second Edition With 30 Illustrations

~ Springer

John B. Conway Mathematics Department University of Tennessee Knoxville, Tennessee 37996-1301 USA

Editorial Board S. Axler Mathematics Department San Francisco State University Sall Francisco, CA 94132 USA [email protected]

F.W. Gehring Mathematics Department East Hall University of Michigan Ann Arbor, MI 48109 USA [email protected]

K.A. Ribet Mathematics Department University of California, Berkeley Berkeley, CA 94720-3840 USA [email protected]

Mathematics Subject Classification (2000): 30-01 Library of Congress Cataloging-in-Publication Data Conway, John B. Functions of one complex variable I I John B. Conway.-2nd ed. p. cm. - (Graduate texts in mathematics; 11) Rev. corr. ed. of: Functions of one complex variable. 2nd ed. c1978. "Seventh corrected printing"-T.;. verso. Includes bibliographical references (p. - ) and index.

ISBN- 13: 978-0-387-94234-6 DOl: 10.1007/978-1-4612-6313-5

e_ISBN-13: 978-1-4612-6313-5

1. Functions of complex variables. I. Conway. John B. of one complex variable. II. Title. III. Series. QA331.7C68 1978 515'.93---{jc20 Printed on acid-free paper.

Functions 95-5931

© 1973, 1978 Springer Science+Business Media. Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, Inc., 233 Spring Street, New York. NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks. and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights.

Solkover reprint of the hardcover 15t Edition 1978 15 14 13 springeronline.com

To Ann

PREFACE FOR THE SECOND EDITION I have been very pleased with the success of my book. When it was apparent that the second printing was nearly sold out, Springer-Verlag asked me to prepare a list of corrections for a third printing. When I mentioned that r had some ideas for more substantial revisions, they reacted with characteristic enthusiasm. There are four major differences between the present edition and its predecessor. First, John Dixon's treatment of Cauchy's Theorem has been included. This has the advantage of providing a quick proof of the theorem in its full generality. Nevertheless, I have a strong attachment to the homotopic version that appeared in the first edition and have proved this form of Cauchy's Theorem as it was done there. This version is very geometric and quite easy to apply. Moreover, the notion of homotopy is needed for the later treatment of the monodromy theorem; hence, inclusion of this version yields benefits far in excess of the time needed to discuss it. Second. the proof of Runge's Theorem is new. The present proof is due to Sandy Grabiner and does not use "pole pushing". In a sense the "pole pushing" is buried in the concept of uniform approximation and some ideas from Banach algebras. Nevertheless, it should be emphasized that the proof is entirely elementary in that it relies only on the material presented in this text. Next, an Appendix B has been added. This appendix contains some bibliographical material and a guide for further reading. Finally, several additional exercises have been added. There are also minor changes that have been made. Several colleagues in the mathematical community have helped me greatly by providing constructive criticism and pointing out typographical errors. I wish to thank publicly Earl Berkson, Louis Brickman, James Deddens, Gerard Keough, G. K. Kristiansen, Andrew Lenard, John Mairhuber, Donald C. Meyers, Jeffrey Nunemacher, Robert Olin, Donald Perlis, John Plaster, Hans Sagan, Glenn Schober, David Stegenga, Richard Varga, James P. Williams, and Max Zorn. Finally, I wish to thank the staff at Springer-Verlag New York not only for their treatment of my book, but also for the publication of so many fine books on mathematics. In the present time of shrinking graduate enrollments and the consequent reluctance of so many publishers to print advanced texts and monographs, Springer-Verlag is making a contribution to our discipline by increasing its efforts to disseminate the recent developments in mathematics. John B. Conway VI

PREFACE This book is intended as a textbook for a first course in the theory of functions of one complex variable for students who are mathematically mature enough to understand and execute E - 8 arguments. The actual prerequisites for reading this book are quite minimal; not much more than a stiff course in basic calculus and a few facts about partial derivatives. The topics from advanced calculus that are used (e.g., Leibniz's rule for differentiating under the integral sign) are proved in detail. Complex Variables is a subject which has something for all mathematicians. In addition to having applications to other parts of analysis, it can rightly claim to be an ancestor of many areas of mathematics (e.g., homotopy theory, manifolds). This view of Complex Analysis as "An Introduction to Mathematics" has influenced the writing and selection of subject matter for this book. The other guiding principle followed is that all definitions, theorems, etc. should be clearly and precisely stated. Proofs are given with the student in mind. Most are presented in detail and when this is not the case the reader is told precisely what is missing and asked to fill in the gap as an exercise. The exercises are varied in their degree of difficulty. Some are meant to fix the ideas of the section in the reader's mind and some extend the theory or give applications to other parts of mathematics. (Occasionally, terminology is used in an exercise which is not defined-e.g., group, integral domain.) Chapters I through V and Sections VI.l and VI.2 are basic. It is possible to cover this material in a single semester only if a number of proofs are omitted. Except for the material at the beginning of Section VI.3 on convex functions, the rest of the book is independent of VI. 3 and VI.4. Chapter VII initiates the student in the consideration of functions as points in a metric space. The results of the first three sections of this chapter are used repeatedly in the remainder of the book. Sections four and five need no defense; moreover, the Weierstrass Factorization Theorem is necessary for Chapter XI. Section six is an application of the factorization theorem. The last two sections of Chapter VII are not needed in the rest of the book although they are a part of classical mathematics which no one should completely disregard. The remaining chapters are independent topics and may be covered in any order desired. Runge's Theorem is the inspiration for much of the theory of Function Algebras. The proof presented in section VIII. I is, however, the classical one involving "pole pushing". Section two applies Runge's Theorem to obtain a more general form of Cauchy's Theorem. The main results of sections three and four should be read by everyone, even if the proofs are not. Chapter IX studies analytic continuation and introduces the reader to analytic manifolds and covering spaces. Sections one through three can be considered as a unit and will give the reader a knowledge of analytic vii

viii

Preface continuation without necessitating his going through all of Chapter IX. Chapter X studies harmonic functions including a solution of the Dirichlet Problem and the introduction of Green's Function. If this can be called applied mathematics it is part of applied mathematics that everyone should know. Although they are independent, the last two chapters could have been combined into one entitled "Entire Functions". However, it is felt that Hadamard's Factorization Theorem and the Great Theorem of Picard are sufficiently different that each merits its own chapter. Also, neither result depends upon the other. With regard to Picard's Theorem it should be mentioned that another proof is available. The proof presented here uses only elementary arguments while the proof found in most other books uses the modular function. There are other topics that could have been covered. Some consideration was given to including chapters on some or all of the following: conformal mapping, functions on the disk, elliptic functions, applications of Hilbert space methods to complex functions. But the line had to be drawn somewhere and these topics were the victims. For those readers who would like to explore this material or to further investigate the topics covered in this book, the bibliography contains a number of appropriate entries. Most of the notation used is standard. The word "iff" is used in place of the phrase "if and only if", and the symbol. is used to indicate the end of a proof. When a function (other than a path) is being discussed, Latin letters are used for the domain and Greek letters are used for the range. This book evolved from classes taught at Indiana University. I would like to thank the Department of Mathematics for making its resources available to me during its preparation. I would especially like to thank the students in my classes; it was actually their reaction to my course in Complex Variables that made me decide to take the plunge and write a book. Particular thanks sho~ld go to Marsha Meredith for pointing out several mistakes in an early draft, to Stephen Berman for gathering the material for several exercises on algebra, and to Larry Curnutt for assisting me with the final corrections of the manuscript. I must also thank Ceil Sheehan for typing the final draft of the manuscript under unusual circumstances. Finally, I must thank my wife to whom this book is dedicated. Her encouragement was the most valuable assistance I received. John B. Conway

TABLE OF CONTENTS Preface for the Second Edition

VI

Preface

viii

Contents, Volume II

1.

The Complex Number System §l. §2. §3. §4. §5. §6.

n.

3 4 6

8

Definition and examples of metric spaces Connectedness. Sequences and completeness Compactness Continuity Uniform convergence

11 14

17

20

24 28

Power series Analytic functions Analytic functions as mapping, Mobius transformations

30 33 44

Complex Integration §1. §2. §3. §4. §5. §6.

§7. §8.

V.

1 1

Elementary Properties and Examples of Analytic Functions §1. §2. §3.

IV.

The real numbers The field of complex numbers The complex plane Polar representation and roots of complex numbers Lines and half planes in the complex plane The extended plane and its spherical representation

Metric Spaces and the Topology of IC §1. §2. §3. §4. §5. §6.

m.

xi

Riemann-Stieltjes integrals Power series representation of analytic functions Zeros of an analytic function The index of a closed curve Cauchy's Theorem and Integral Formula The homotopic version of Cauchy's Theorem and simple connectivity . Counting zeros; the Open Mapping Theorem Goursat's Theorem .

58

68 76 80 83 87 97 100

Singularities §l.

Classification of singularities ix

103

x

Table of Contents

§2. §3.

VI.

Residues The Argument Principle

The Maximum Modulus Theorem The Maximum Principle §2. Schwarz's Lemma §3. Convex functions and Hadamard's Three Circles Theorem §4. Phragmen-LindelOf Theorem

128 130 133 138

Compactness and Convergence in the Space of Analytic Functions §1. The space of continuous functions C(G,Q) §2. Spaces of analytic functions §3. Spaces of merom orphic functions . §4. The Riemann Mapping Theorem . §5. Weierstrass Factorization Theorem §6. Factorization of the sine function §7. The gamma function §8. The Riemann zeta function .

142 151 155 160 164 174 176 187

Runge's Theorem Runge's Theorem §2. Simple connectedness §3. Mittag-Leffler's Theorem

195 202 204

Analytic Continuation and Riemann Surfaces Schwarz Reflection Principle. §2. Analytic Continuation Along A Path §3. Monodromy Theorem §4. Topological Spaces and Neighborhood Systems §5. The Sheaf of Germs of Analytic Functions on an Open Set §6. Analytic Manifolds §7. Covering spaces

210 213 217 221 227 233 245

Harmonic Functions Basic Properties of harmonic functions §2. Harmonic functions on a disk §3. Subharmonic and superharmonic functions §4. The Dirichlet Problem §5. Green's Functions

252 256 263 269 275

§1.

VII.

VIII.

§1.

IX.

§1.

X.

112 123

§1.

Table of Contents

XI.

xi

Entire Functions Jensen's Formula The genus and order of an entire function Hadamard Factorization Theorem

§l. §2. §3.

XII.

280 282 287

The Range of an Analytic Function Bloch's Theorem The Little Picard Theorem Schottky's Theorem The Great Picard Theorem

§l. §2. §3. §4.

292

296 297 300

Appendix A: Calculus for Complex Valued Functions on an Interval

303

Appendix B: Suggestions for Further Study and Bibliographical Notes

307

References

311

Index

313

List of Symbols

317

TABLE OF CONTENTS, VOLUME II Preface 13.

Return to Basics §l. §2. §3. §4.

§S.

14.

Regions and Curves Derivatives and other recollections Harmonic conjugates and primitives Analytic arcs and the reflection principle Boundary values for bounded analytic functions

Conformal Equivalence For Simply Connected Regions §1.

§2. §3. §4.

§S.

Elementary properties and examples Crosscuts Prime Ends Impressions of a prime end Boundary values of Riemann maps

xii

Table of Contents

§6. §7.

15.

Conformal Equivalence For Finitely Connected Regions §I. §2. §3. §4. §5. §6. §7.

16.

The Area Theorem Disk mappings: the class S

Analysis on a finitely connected region Conformal equivalence with an analytic Jordan region Boundary values for a conformal equivalence between finitely connected Jordan regions Convergence of univalent functions Conformal equivalence with a circularly slit annulus Conformal equivalence with a circularly slit disk Conformal equivalence with a circular region

Analytic Covering Maps Results for abstract covering spaces §2. Analytic covering spaces §3. The modular function §4. Applications of the modular function §5. The existence of the universal analytic covering map §1.

17.

De Branges's Proof of the Bieberbach Conjecture §I. §2. §3. §4. §5. §6.

18.

Some Fundamental Concepts From Analysis § I.

§2. §3. §4.

§5. §6. §7.

19.

Subordination Loewner chains Loewner's differential equation The Milin Conjecture Some special functions The proof of de Branges's Theorem

Bergman spaces of analytic and harmonic functions Partitions of unity Convolution in Euclidean space Distributions The Cauchy transform An application: rational approximation Fourier series and Cesaro sums

Harmonic Functions Redux §1. Harmonic functions on the disk §2. Fatou's Theorem §3. Semicontinuous functions §4. Subharmonic functions §5. The logarithmic potential §6. An application: approximation by harmonic functions §7. The Dirichlet problem §8. Harmonic majorants §9. The Green function

Table of Contents

§1O. §ll.

Regular points for the Dirichlet problem The Dirichlet principle and Sobolev spaces

20.

Hardy Spaces on the Disk §l. Definitions and elementary properties §2. The Nevanlinna Class §3. Factorization of functions in the Nevanlinna class §4. The disk algebra §5. The invariant subspaces of HP §6. Szeg6's Theorem

21.

Potential Theory in the Plane §I. Harmonic measure §2. The sweep of a measure §3. The Robin constant §4. The Green potential §5. Polar sets §6. More on regular points §7. Logarithmic capacity: part 1 §8. Some applications and examples of logarithmic capacity §9. Removable singularities for functions in the Bergman space §1O. Logarithmic capacity: part 2 §ll. The transfinite diameter and logarithmic capacity §12. The refinement of a subharmonic function §13. The fine topology §14. Wiener's criterion for regular points

References

List of Symbols Index

xiii

Chapter 1 The Complex Number System §1. The real numbers We denote the set of all real numbers by IR. It is assumed that each reader is acquainted with the real number system and all its properties. In particular we assume a knowledge of the ordering of IR, the definitions and properties of the supremum and infimum (sup and inf), and the completeness of IR (every non-empty set in IR which is bounded above has a supremum). It is also assumed that every reader is familiar with sequential convergence in IR and with infinite series. Finally, no one should undertake a study of Complex Variables unless he has a thorough grounding in functions of one real variable. Although it has been traditional to study functions of several real variables before studying analytic function theory, this is not an essential prerequisite for this book. There will not be any occasion when the deep results of this an:a are needed.

§2. The field of complex numbers We define C, the complex numbers, to be the set of an ordered pairs (a, b) where a and b are real numbers and where addition and multiplication are defined by:

(a, b)+(c, d) = (a+c, b+d) (a, b) (c, d) = (ac-bd, bc+ad) It is easily checked that with these definitions C satisfies all the axioms for a field. That is, C satisfies the associative, commutative and distributive laws for addition and multiplication; (0,0) and (1,0) are identities for addition and multiplication respectively, and there are additive and multiplicative inverses for each nonzero element in C. We will write a for the complex number (a, 0). This mapping a --* (a, 0) defines a field isomorphism of IR into C so we may consider IR as a subset of C. If we put i = (0, 1) then (a, b) = a+bi. From this point on we abandon the ordered pair notation for complex numbers. Note that i 2 = - 1, so that the equation Z2 + 1 = 0 has a root in IC. In fact, for each z in C, Z2+ I = (z+i) (z-i). More generally, if z and ware complex numbers we obtain Z2+ W2

=

(z+iw) (z-iw) 1

2

The Complex Number System

By letting z and w be real numbers a and b we can obtain (with both a and b "# 0)

a~ib = a~~~2 = a2:b2 -

{a 2!b 2)

so that we have a formula for the reciprocal of a complex number. When we write z = a + ib (a, b E IR) we call a and b the real and imaginary parts of z and denote this by a = Re z, b = Im z. We conclude this section by introducing two operations on IC which are not field operations. If z = x + iy(x, y E IR) then we define Izl = (x 2+ y2)!- to be the absolute value of z and z = x - iy is the conjugate of z. Note that 2.1 In particular, if z "# 0 then

Izl2

Z

The following are basic properties of absolute values and conjugates whose verifications are left to the reader. 2.2

Rez = !(z+z)

and

(z+w) = z+w

2.3

Imz =

1

2i (z-z).

zw = zw.

and

Izwl = Izllwl· Izjwl = Izlllwl· Izl = Izl·

2.4 2.5 2.6

The reader should try to avoid expanding z and w into their real and imaginary parts when he tries to prove these last three. Rather, use (2.1), (2.2), and (2.3). Exercises 1. Find the real and imaginary parts of each of the following:

!.

z-~(aEIR)' z3. 3+5i. (-I+iJ3)3;

z ' z+a

"

7i+ 1 '

(-I- J 3)6 . (1J2+ i)n 2

i

;

In;

2

for

2::; n ::; 8.

2. Find the absolute value and conjugate of each of the following:

3-i

i

-2+i; -3; (2+i)(4+3i); J2+3i; i+3 ; (I + i)6; i 17.

The complex plane

3

3. Show that z is a real number if and only if Z = Z. 4. If z and ware complex numbers, prove the following equations: Iz+wl2

= Iz1 2 +2Re zw+ Iwlz.

IZ-WJ2 = Iz1 2 -2Re zw+ Jwi 2 • Iz+ wlz + Iz- wlz = 2(lz12 + IwI Z). 5. Use induction to prove that for z = ZI + .. . +zn; W = WIW2 ••• Wn: Iwl = IWII·· ·Iwnl; 2 = 21 +., .+2n; W = WI ... wn ' 6. Let R(z) be a rational function of z. Show that R(z) = R(i) if all the coefficients in R(z) are real.

§3. The complex plane From the definition of complex numbers it is dear that each z in C can be identified with the unique point (Re z, 1m z) in the plane [R2. The addition of complex numbers is exactly the addition law of the vector space [R2. If z and ware in C then draw the straight lines from z and w to = (0, 0». These form two sides of a parallelogram with 0, z and w as three vertices. The fourth vertex turns out to be z + w. Note also that Iz-wl is exactly the distance between z and w. With this in mind the last equation of Exercise 4 in the preceding section states the parallelogram law: The sum of the squares of the lengths of the sides of a parallelogram equals the sum of the squares of the lengths of its diagonals. A fundamental property of a distance function is that it satisfies the triangle inequality (see the next chapter). In this case this inequality becomes

°(

IZ1 - z 21 s IZI - Z31 + for complex numbers ZI, Z2, Z3' By using easy to see that we need only show 3.1

Iz+wl

s

IZ3-z21

ZI-Z2

Izl + Iwl (z, WE

=

(ZI-Z3)+(Z3-Z2),

it is

q.

To show this first observe that for any z in C,

3.2 Hence, Re (zw)

-IZI

s

s s

Rez

-izi Imz izwi = Jzllwl. Thus,

s s

Izl Izi

Iz+wl2 = Iz1 2 +2Re (zw) + Iwl 2

s

IzI2+2Izllwl+lwI2

= (izi + Iwl)2, from which (3.1) follows. (This is called the triangle inequality because, if we represent z and w in the plane, (3.1) says that the length of one side of the triangle [0, z, z -+ w] is less than the sum of the lengths of the other two sides. Or, the shortest distance between two points is a straight line.) On encounter-

4


ing an inequality one should always ask for necessary and sufficient conditions that equality obtains. From looking at a triangle and considering the geometrical significal'lce of (3.1) we are led to consider the condition z = tw for some t E IR, t ;:: O. (or IV = tz if IV = 0). It is clear that equality will occur when the two points are colinear with the origin. In fact, if we look at the proof of (3.1) we see that a necessary and sufficient condition for Iz + wi = Izi + Iw, is that Izwl = Re (zw). Equivalently, this is zw ;:: 0 (i.e., zw is a real number and is non negative). MultiplyinK this by w/w we get IwI 2 (z/w);:: Oifw # O. If t

1 )' 2 = z/w . = (~ 1\V12 I'wi (z/w)

then t ;:: 0 and z = two By induction we also get

3.3 Also useful is the inequality 3.4

Izl-Iwll

s:;

Iz-wl

Now that we have given a geometric interpretation of the absolute value let us see what taking a complex conjugate does to a point in the plane. This is also easy; in fact, Z is the point obtained by reflecting z across the x-axis (i.e., the real axis). Exercises 1. Prove (3.4) and give necessary and sufficient conditions for equality. 2. Show that equality occurs in (3.3) if and only if zdz/ ;:: 0 for any integers k and I, 1 s:; k, I s:; n, for which Zl # O. 3. Let a E rr~ and c > 0 be fixed. Describe the set of points z satisfying

Iz-al-Iz+al =

2c

for every possible choice of a and c. Now let a be any complex number and, using a rotation of the plane, describe the locus of points satisfying the above equation.

§4. Polar representation and roots of complex numbers Consider the point z = x + iy in the complex plane C. This point has polar coordinates Cr, 8): x = r cos 8, y = r sin e. Clearly r = Izi and () is the angle between the positive real axis and the line segment from 0 to Z. Notice that 8 plus any multiple of 211" can be substituted for 8 in the above equations. The angle () is called the argument ofz and is denoted by e = arg z. Because of the ambiguity of e, "arg" is not a function. We introduce the notation 4.1

cis 8 = cos 8+i sin 8.

Polar representation and roots of complex numbers

5

Let ZI = 1'1 cis 8 1 , Z2 = 1'2 cis 82 , Then Z1 Z2 = 1'11'2 cis 01 cis O2 = 1'11'2 [(cos 0 1 cos Oz-sin Ol sin (2)+i (sin 0 1 cos 02+sin 82 cos OJ)]. By the formulas for the sine and cosine of the sum of two angles we get

4.2 Alternately, arg(ZtZ2) = arg ZI + arg Z1' (What function of a real variable takes products into sums?) By induction we get for Z k = r k cis (j h' 1 ~ k ~ 11,

4.3 In particular, 4.4

Z" =

I'"

cis (ne),

for every integer n ;::: O. Moreover if z #- 0, z· [r - 1 cis (- 0)] = 1; so that (4.4) also holds for all integers n, positive, negative, and zero, if z #- O. As a special case of (4.4) we get de M oivre's formula: (cos O+i sin 8t = cos n8+i sin nO. We are now in a position to consider the following problem: For a given complex number a#-O and an integer n ;::: 2, can you find a number z satisfying zn = a? How many such z can you find? In light of (4.4) the solution is easy. Let a = lal cis ex; by (4.4), Z = lal ' /II cis (aln) fills the bill. ]

However this is not the only solution because z' = la1 1/11 cis - (cc+ 27T) also n satisfies (zT = a. In fact each of the numbers 1 (a + 27Tk), 0 ~ k ~ n-l, n

lal 1 / n cis -

4.5

is an nth root of a. By means of (4.4) we arrive at the following: for each non zero number a in IC there are n distinct nth roots of a; they are given by formula (4.5). Example Calculate the nth roots of unity. Since 1 = cis 0, (4.5) gives these roots as

27T

417

217

1, cis - , cis ~-, ... , cis - (n-l). n n n In particular, the cube roots of unity are

Exercises I. Find the sixth roots of unity.

6


2. Calculate the following: (a) the square roots of i (b) the cube roots of i (c) the square roots of ..)3 +3i 3. A primilive nth rool oj unity is a complex number a such that l,a,a 2, ••• ,a n - 1 are distinct nth roots of unity. Show that if a and bare primitive nth and mth roots of unity, respectively, then ab is a kth root of unity for some integer k. What is the smallest value of k? What can be said if a and bare nonprimitive roots of unity? 4. Use the binomial equation

(a+b)" = where

f (n)k a"-kbk,

k=O

(z) k!(nn~k)! ' =

and compare the real and imaginary parts of each side of de Moivre's formula to obtain the formulas: cos n(J

= cos" (J- (;)

sin

= (~)

n(J

cos·- 2 sin 2 (J+

COS"-l (J

sin

(~) COS"-4 (J sin4 (J- • ••

(J- (;) COS,,-3 (J

sin 3

(J+ ••.

. 21T t ' . h 1 lor an Integer n ~ 2. S ow that 1 +z+ ... +z"- = O. n 6. Show that «pCt) = cis t is a group homomorphism of the additive group IR onto the multiplicative group T = {z: Izl = I}. 7. If Z E C and Re(zn)~ 0 for every positive integer n, show that z is a non-negative real number.

5. Let z =

CIS -

§5. Lines and half planes in the complex plane Let L denote a straight line in C. From elementary analytic geometry, L is determined by a point in L and a direction vector. Thus if a is any point in Land b is its direction vector then L

Since b

=1=

= {z = a+tb: -

00

< t < oo}.

0 this gives, for z in L,

1m ( z-a) b = O. In fact if z is such that O=lm ( z-a) b

then 1=

-( z-a)

b

7

Lines and half planes in the complex plane

implies that z = a + tb, -

00

< t <

00.

That is

5.1 What is the locus of each of the sets

{z: Im(z~a) > o}, {z: (z~a) < o} 1m

?

As a first step in answering this question, observe that since b is a direction we may assume Ibl = 1. For the moment, let us consider the case where a = 0, and put Ho = {z: 1m (z/b) > O}, b = cis f3. If z = r cis () then z/b = r cis «()-f3). Thus, z is in Ho if and only if sin «()-f3) > 0; that is, when f3 < 8 < 7T+f3. Hence Ho is the half plane lying to the left of the line L if

we are "walking along L in the direction of b." If we put

then it is easy to see that Ha = a+Ho == {a+w: WE Ho}; that is, Ha is the translation of Ho by a. Hence, Ha is the half plane lying to the left of L. Similarly,

is the half plane on the right of L. Exercise 1. Let C be the circle {z:

Iz-cl

=

r}, r > 0; let a = c+r cis

(X

and put

8

Lp

=

{z:


1m

(z~a) =

o}

where b = cis {3. Find necessary and sufficient conditions in terms of {3 that Lp be tangent to C at a.

§6. The extended plane and its spherical representation Often in complex analysis we will be concerned with functions that become infinite as the variable approaches a given point. To discuss this situation we introduce the extended plane which is IC U {oo} == 1C00. We also wish to introduce a distance function on 1C00 in order to discuss continuity properties of functions assuming the value infinity. To accomplish this and to give a concrete picture of 1C00 we represent 1C00 as the unit sphere in 1R 3 ,

s=

{(XI,X2,X3)EIR3:xf+x~+x~

= I}.

Let N = (0,0, 1); that is, N is the north pole on S. Also, identify IC with {(Xl' X2' 0): Xl' X 2 E IR} so that IC cuts S along the equator. Now for each point z in IC consider the straight line in 1R3 through z and N. This intersects .'v

the sphere in exactly one point Z #- N. If /z/ > 1 then Z is in the northern hemisphere and if /z/ < I then Z is in the southern hemisphere; also, for /z/ = I, Z = z. What happens to Z as /z/ -+ oo? Clearly Z approaches N; hence, we identify N and the point 00 in 1C00. Thus 1C00 is represented as the sphere S. Let us explore this representation. Put z = X+ iy and let Z = (Xl' X2' X3) be the corresponding point on S. We will find equations expressing Xl' x 2, and X3 in terms of X and y. The line in 1R3 through z and N is given by {tN+(I-t)z: -00 < t < oo}, or by

6.1

{«I-t)x,(I-t)y,t):

-00

< t < oo}.

Hence, we can find the coordinates of Z if we can find the value of t a't

The extended plane and its spherical representation

which this line intersects S. If

1

9

is this value then

1 = (l_t)2x2+(1_t)2y2+t 2 = (l_t)2IzI2+t 2

From which we get 1-12

= (1-1)2IzI2.

Since t :/= 1 (z :/= 00) we arrive at

Thus

6.2 But this gives

1

- i(z - i)

z+i x =-I Iz12+ I

6.3

Iz12+ I

Iz 12 -1

x =---

IzI 2+1·

3

If the point Z is given (Z :/= N) and we wish to find z then by setting X3 and using (6.1), we arrive at

=

Xl +iX2 Z=---

6.4

I-X 3

Now let us define a distance function between points in the extended plane in the following manner: for z, z' in Coo define the distance from z to z', d(z, z'), to be the distance between the corresponding points Z and Z' in 1R3. If Z = (Xl' X2' X3) and Z' = (x;, X2' Xl) then

6.5 Using the fact that Z and Z' are on S, (6.5) gives

6.6 By using equation (6.3) we get

6.7

d(') 2Iz-z'l z, z = [(1 + Iz12) (1 + Iz'12W·'

In a similar manner we get for

Z

(' Z, Z

E

C)

in C

6.8 This correspondence between points of Sand C", is called the stereographic projection.


10

Exercises 1. Give the details in the derivation of (6.7) and (6.8). 2. For each of the following points in C, give the corresponding point of S: 0, 1 +i, 3+2i. 3. Which subsets of S correspond to the real and imaginary axes in C? 4. Let A be a circle lying in S. Then there is a unique plane P in 1R3 such that P (\ S = A. Recall from analytic geometry that

P

=

{(X t :X2,X3): XtfJl+X2fJ2+X3fJ3

= /}

where (/31' /32' /3 3) is a vector orthogonal to P and I is some real number. It can be assumed that + fii + fif = 1. Use this information to show that if A contains the point N then its stereo graphic projection on C is a straight line. Otherwise, A projects onto a circle in C. 5. Let Z and Z' be points on S corresponding to z and z' respectively. Let W be the point on S corresponding to z+z'. Find the coordinates of Win terms of the coordinates of Z and Z'.

fir

Chapter II Metric Spaces and the Topology of

C

§1. Definition and examples of metric spaces A metric space is a pair (X, d) where X is a set and d IS a function from X x X into IR, called a distance function or metric, which satisfies the following conditions for x, y, and z in X: 0

d(x,y)

~

d(x, y)

= 0 if and only if x = y

d(x, y)

= dey, x) (symmetry)

d(x, z) ::::; d(x, y)+d(y, z) (triangle inequality)

If x and r > 0 are fixed then define B(x;r) = {yeX: d(x,y) < r} H(x;r) = {yeX: d(x,y)::::; r}. B(x; r) and H(x; r) are called the open and closed balls, respectively, with center x and radius r.

Examples

1.1 Let X = IR or C and define d(z, w) = Jz-wJ. This makes both (IR, d) and (C, d) metric spaces. In fact, (C, d) will be the example of principal interest to us. If the reader has never encountered the concept of a metric space before this, he should continually keep (C, d) in mind during the study of this chapter. 1.2 Let (X, d) be a metric space and let Y eX; then (Y, d) is also a metric space. 1.3 Let X = C and define d(x+iy, a+ib) = Jx-aJ+Jy-bJ. Then (C, d) is a metric space. 1.4 Let X = C and define d(x+iy, a+ib) = max {Jx-aJ, Jy-bJ}. 1.5 Let X be any set and define d(x, y) = 0 if x = y and d(x, y) = I if x of. y. To show that the function d satisfies the triangle inequality one merely considers all possibilities of equality among x, y, and z. Notice here that B(x; E) consists only of the point x if E ::::; 1 and B(x; E) = X if E > 1. This metric space does not appear in the study of analytic function theory. 1.6 Let X = IRn and for x = (Xl> ••• , x n), y = (Yl' ... ,Yn) in IR n define d(x,y)

=

Lt 11

(X j - y )2r

12

Metric Spaces and the Topology of C

1.7 Let S be any set and denote by B(S) the set of all functions f: S such that

IlflL" == sup {If(s)l: s E S} <

~

C

00.

That is, B(S) consists of all complex valued functions whose range is contained inside some disk of finite radius. For f and g in B(S) define d(f; g) = Ilf-g;lw' We will show that d satisfies the triangle inequality. In fact if J, g, and h are in B(S) and s is any point in S then If(s)-g(s)1 = If(s)-h(s)+ h(s)-g(s)1 ::; If(s)-h(s)1 + Ih(s)-g(s)[ ::; Ilf-hil oo + Ilh-giloo" Thus, when the supremum is taken over all s in S, Ilf-gilro ::; IIf-h 1100 +IIh-gll oo , which is the triangle inequality for d.

1.8 Definition. For a metric space (X, d) a set G c X is open if for each x in G there is an E > 0 such that B(x; E) c G. Thus, a set in C is open if it has no "edge." For example, G = {z E C: a < Re z < b} is open; but {z: Re z < O} u {O} is not because B(O; E) is not contained in this set no matter how small we choose E. We denote the empty set, the set consisting of no elements, by D. 1.9 Proposition. Let (X, d) be a metric space; then: (a) The sets X and D are open; (b) If G 1 , . . • , Gn are open sets in X then so is (c)

If

n n

Gk ;

k~l

{G/ j E J} is a collection of open sets in X, J any indexing set, then G = U {G j : j E J} is also open.

n n

Proof The proof of (a) is a triviality. To prove (b) let x

E

G

=

Gk ; then

k~1

x E Gk for k = 1, ... , n. Thus, by the definition, for each k there is an Ek > 0 such that B(x; Ek) C Gk • But if E = min {El' E2, ••• , En} then for 1 ::; k ::; n B(x; E) c B(x; Ek) C Gk • Thus B(x; E) c G and G is open. The proof of (c) is left as an exercise for the reader. • There is another class of subsets of a metric space which are distinguished. These are the sets which contain all their "edge"; alternately, the sets whose complements have no "edge." 1.10 Definition. A set F c X is closed if its complement, X - F, is open. The following proposition is the complement of Proposition 1.9. The proof, whose execution is left to the reader, is accomplished by applying de Morgan's laws to the preceding proposition. 1.11 Proposition. Let (X, d) be a metric space. Then: (a) The sets X and D are closed; n (b) If F 1 , ••• , Fn are closed sets in X then so is U Fk ; k~l

(c) If {Fj : j E J} is any collection of closed sets in X, J any indexing set, then F = n {Fj : j E J} is also closed. The most common error made upon learning of open and closed sets is to interpret the definition of closed set to mean that if a set is not open it is

Definition and examples of metric spaces

13

closed. This, of course, is false as can be seen by looking at {z E C: Re z > O} u {O}; it is neither open nor closed.

n

1.12 Definition. Let A be a subset of X. Then the interior of A, int A, is the {F: F set U {G: G is open and G c A}. The closure of A, A-, is the set is closed and F :::> A}. Notice that int A may be empty and A - may be X. If A = {a+bi: a and b are rational numbers} then simultaneously A- = C and int A = O. By Propositions 1.9 and 1.11 we have that A - is closed and int A is open. The boundary of A is denoted by OA and defined by OA = An (X-A)-. 1.13 Proposition. Let A and B be subsets of a metric space (X, d). Then: (a) A is open if and only if A = int A; (b) A is closed if and only if A = A -; (c) int A = X-(X-A)-; A- -= X-int (X-A); OA = A--int A; (d) (A u B)- = A- u B-; (e) Xo E int A if and only if there is an € > 0 such that B(xo; €) C A; (f) Xo E A - if and only if for every € > 0, B(xo; €) n A :f. O.

Proof. The proofs of (a)-(e) are left to the reader. To prove (f) assume xoEA- = X-int (X-A); thus, xo¢int (X-A). By part (e), for every € > 0 B(xo; €) is not contained in X -A. That is, there is a point y E B(xo; €) which is not in X-A. Hence, yEB(xo; €)nA. Now suppose xo¢A- = X -int (X -A). Then Xo E int (X -A) and, by (e), there is an € > 0 such that B(xo; €) eX-A. That is, B(xo; €) n A = 0 so that Xo does not satisfy the condition . • Finally, one last definition of a distinguished type of set. 1.14 Definition. A subset A of a metric space X is dense if A - = X. The set of rational numbers 0 is dense in IR and {x+iy: x, yEO} is dense in Co

Exercises 1. Show that each of the examples of metric spaces given in (1.2)-(1.6) is, indeed, a metric space. Example (1.6) is the only one likely to give any difficulty. Also, describe B (x; r) for each of these examples. 2. Which of the following subsets of C are open and which are closed: (a) {z:lzl
14


that G (') Y is open in (Y, d). Conversely, show that if G1 C Y is open in (Y, d), there is an open set G c X such that G 1 = G (') Y. 9. Do Exercise 8 with "closed" in place of "open." 10. Prove Proposition 1.13. 11. Show that {cis k: k a non-negative integer} is dense in T = {z E C : Izl = I}. For which values of 8 is {cis(k8): k a non-negative integer} dense in T?

§2. Connectedness Let us start this section by giving an example. Let X = {z € c: Izl :::; I} Iz-31 < I} and give X the metric it inherits from Co (Hencefol'ward, whenever we consider subsets X of IR or C as metric spaces we will assume, unless stated to the contrary, that Xhas the inherited metric d(z, w) = Iz- wi.) Then the set A = {z: Izl :::; I} is simultaneously open and closed. It is closed because its complement in X, B = X -A = {z: Iz-31 < I} is open; A is open because if a E A then B(a; 1) c A. (Notice that it may not happen that {z € C: Iz-al < I} is contained in A-for example, if a = 1. But the definition of B(a; 1) is {z € X: Iz-al < I} and this is contained in A.) Similarly B is also both open and closed in X. This is an example of a non-connected space.

u {z:

2.1 Definition. A metric space (X, d) is connected if the only subsets of X which are both open and closed are D and X. If A c X then A is a connected subset of X if the metric space (A, d) is connected. An equivalent formulation of connectedness is to say that X is not connected if there are disjoint open sets A and B in X, neither of which is empty, such that X = A u B. In fact, if this condition holds then A = X - B is also closed. 2.2 Proposition. A set X c IR is connected

iff X

is an interval.

Proof Suppose X = [a, b], a and b elements of IR. Let A c X be an open subset of X such that a € A, and A # X. We will show that A cannot also be closed-and hence, X must be connected. Since A is open and a E A there is an e > 0 such that [a, a+e) c A. Let r = sup {e: [a, a+£) c A} Claim. [a,a+r) c A.lnfact,ifa:::; x < a+rthen,puttingh = a+r-x > 0, the definition of supremum implies there is an e with r - h < e < rand [a, a+£) c A. But a :::; x = a+(r-h) < a+e implies x E A and the claim is established. However, a + r If A; for if, on the contrary, a + rEA then, by the openness of A, there is a 8 > 0 with [a+r, a+r+8) c A. But this gives [a, a+r+8) c A, contradicting the definition of r. Now if A were also closed then a+r E B = X-A which is open. Hence we could find a 8 > 0 such that (a + r - 8, a + r] c B, contradicting the above claim. The proof that other types of intervals are connected is similar and it will be left as an exercise.

Connectedness

15

The proof of the converse is Exercise 1. • If wand z are in C then we denote the straight line segment from z to w by

[z, w]

=

{tw+{l-t)z: 0

A polygon from a to b is a set P Wk

=

Zk+l

for 1

~ k ~ n -

n

= U [Zk' k=1

1; or, P = [a,

~

t

~

I}

wk ] where

Z2'"

Zn'

Zt

= a,

w"

= band

b).

2.3 Theorem. An open set Gee is connected iff for any two points a, b in G there is a polygon from a to b lying entirely inside G. Proof Suppose that G satisfies this condition and let us assume that G is not connected. We will obtain a contradiction. From the definition, G = A U B where A and B are both open and closed, A n B = D, and neither A nor B is empty. Let a E A and bE B; by hypothesis there is a polygon P from a to b such that peG. Now a moment's thought will show that one of the segments making up P will have one end point in A and another in B. So we can assume that P = [a, b]. Define,

S = {s E [0,1]: sb+(I-s)a E A} T = {t

E

[0,1]: tb+{1-t)a

E

B}

Then S n T = D, S u T = [0, 1],0 E Sand 1 E T. However it can be shown that both Sand T are open (Exercise 2), contradicting the connectedness of [0, 1]. Thus, G must be connected. Now suppose that G is connected and fix a point a in G. To show how to construct a polygon (lying in G!) from a to a point b in G would be difficult. But we don't have to perform such a construction; we merely show that one exists. For a fixed a in G define A = {b E G: there is a polygon peG from a to b}. The plan is to show that A is simultaneously open and closed in G. Since = G and the theorem will be proved. To show that A is open let bE A and let P = [a, Z2"'" zn' b] be a polygon from a to b with PeG. Since G is open (this was not needed in the first half), there is an t: > 0 such that B(b; t:) c G. But if Z E B(b; t:) then [b, z] c B(b; t:) c G. Hence the polygon Q = P U [b, z] is inside G and goes from a to z. This shows that B(b; t:) c A, and so A is open. To show that A is closed suppose there is a point Z in G-A and let E > 0 be such that B(z; E) c G. If there is a point b in A n B(z; E) then, as above, we can construct a polygon from a to z. Thus we must have that B(z; E) n A = D, or B(z; E) c G-A. That is, G-A is open so that A is closed. • a E A and G is connected this will give that A

2.4 Corollary. If Gee is open and connected and a and b are points in G then there is a polygon P in G from a to b which is made up of line segments parallel to either the real or imaginary axis. Proof There are two ways of proving this corollary. One could obtain a

16


polygon in G from a to b and then modify each of its line segments so that a new polygon is obtained with the desired properties. However, this proof is more easily executed using compactness (see Exercise 5.7). Another proof can be obtained by modifying the proof of Theorem 2.3. Define the set A as in the proof of (2.3) but add the restriction that the polygon's segments are all parallel to one of the axes. The remainder of the proof will be valid with one exception. If z EO B(b; €) then [b, z] may not be parallel to an axis. But it is easy to see that if z = x+iy, b = p+iq then the polygon [b, p+iy] U [p+iy, z] c B(b; €) and has segments parallel to an axis. III It will now be shown that any set S in a metric space can be expressed, in a canonical way, as the union of connected pieces. 2.5 Definition. A subset D of a metric space X is a component of ,\:' if it is a maximal connected subset of X. That is, D is connected and there is no connected subset of X that properly contains D. If the reader examines the example at the beginning of this section he will notice that both A and B are components and, furthermore, these are the only components of X. For another example let X = {D, I, .1, t, ... }. Then clearly every component of X is a point and each point is a component. Notice that while the components

{M

are all open in X, the component {O}

is not. 2.6 Lemma. Let x 0 E X and let {D j: j E J} be a collection of connected subsets of X such that x 0 EO D j for each j in J. Then D = U {D / j E J} is connected.

Proof Let A be a subset of the metric space CD, d) which is both open and closed and suppose that A i= O. Then A n D j is open in (D j, d) for each j and it is also closed (Exercises l.8 and 1.9). Since D j is connected we get that either A n D j = or An D j = D j • Since A i= 0 there is at least one k such that A n Dk i= 0; hence, A n Dk = D k • In particular x 0 E A so that Xo EA n D j for every j. Thus An D j = D j , or D j C A, for each index). This gives that D = A, so that D is connected. III 2.7 Theorem. Let (X, d) be a metric space. Then: (a) Each Xo in X is contained in a component of X. (b) Distinct components of X are disjoint. Note that part (a) says that X is the union of its components.

Proof (a) Let!» be the collection of connected subsets of X which contain the given point Xo. Notice that {xo} E 9 so that!» i= O. Also notice that the hypotheses of the preceding lemma apply to the collection g. Hence C = U {D: D EO g} is connected and Xo E C. But C must be a component. In fact, if D is connected and C C D then x 0 EO D so that DE!»; but then DeC, so that C = D. Thus C is maximal and part (a) is proved. (b) Suppose C 1 and C z are components, C 1 i= C z , and suppose there is a point Xo in C 1 n C z . Again the lemma says that C 1 U C z is connected.

17

Sequences and completeness

Since both C 1 and C 2 are components, this gives C 1 = C 1 U C 2 = C z, a contradiction. II 2.8 Proposition. (a) If A c X is connected and A c Be A -, then B is connected. (b) If C is a component of X then C is closed. The proof is left as an exercise.

2.9 Theorem. Let G be open in IC; then the components (!f G are open and there are only a countable number of them. Proof. Let C be a component of G and let Xo c C. Since G is open there is an ( > 0 with B(xo; f) c G. By Lemma 2.6, B(xo; €) U C is connected and so it must be C. That is B( xo; () c C and C is, therefore. open. To see that the number of components is countable let S = {a+ ib: a and b are rational and a + bi E G}. Then S is countable and each component of G contains a point of S, so that the number of components is countable. II Exercises

1. The purpose of this exercise is to show that a connected subset of IR is an interval. (a) Show that a set A c IR is an interval iff for any two points a and b in A with a < b, the interval [a, b] c A. (b) Use part (a) to show that if a set A c IR is connected then it is an interval. 2. Show that the sets Sand T in the proof of Theorem 2.3 are open. 3. Which of the following su bsets X of IC are connected; if X is not connected, what are its components: (a) X = {z: Izl :s; I} U {z: IZ-21 < I}. (b) X =

[0,1)

U

{1+~: n ~

I}. (c) X = IC-(A

°

U

B) where A = [0, (0) and B =

{z = r cis 0: r = 0, :s; e:s; oo}? 4. Prove the following generalization of Lemma 2.6. If {D j: j E J} is a collection of connected subsets of X and if for each j and k in J we have D j n Dk -# D then D = U {D j : j E J} is connected. 5. Show that if F c X is closed and connected then for every pair of points a, b in F and each to > there are points Zo, ZI, . . • ,Zn in F with Zo = a, Zn = band d(zk _ 1, Zk) < to for 1 :s; k :s; n. I s the hypothesis that F be closed needed? If F is a set which satisfies this property then F is not necessarily connected, even if F is closed. Give an example to illustrate this.

°

§3. Sequences and completeness One of the most useful concepts in a metric space is that of a convergent sequence. Their central role in calculus is duplicated in the study of metric spaces and complex analysis. 3.1 Definition. If {Xl' x 2 ,

••• }

is a sequence in a metric space (X, d) then

18


{xn} converges to x-in symbols x = lim Xn or Xn -+ x-if for every" > 0 there is an integer N such that d(x, xn) < " whenever n ;:::: N. Alternately, x = lim XII if 0= lim d(x, XII)' If X = C then z = lim z" means that for each" > 0 there is an N such that Iz-z,,1 < Eo" when n ;:::: N. Many concepts in the theory of metric spaces can be phrased in terms of sequences. The following is an example.

3.2 Proposition. A set F c X is closed x = lim xn we have x E F.

iff for

each sequence {xn} in F with

Proof. Suppose F is closed and x = lim X Il where each X Il is in F. So for every E > 0, there is a point x ll in B(x; E); that is B(x; £) n F", D, so that x EO F~ = F by Proposition l.13(f). Now suppose F is not closed; so there is a point Xo in F~ which is not in F. By Proposition 1.13(f), for every Eo" > 0 we have B(x 0; €) n F # D-

In particular for every integer n there is a point d(xo,

X,I)

<1 which implies that

n

X,I

-+

Xo'

Since

XII

Xo

in

B(

X

0;

~) n F. Thus,

tt F, this says the con-

clition fails . •

3.3 Definition. If A c X then a point X in X is a limit point of A if there is a sequence {XII} of distinct points in A such that x = lim Xn' The reason for the word "distinct" in this definition can be illustrated by the following example. Let X = C and let A = [0, 1] u {i}; each point in [0, IJ is a limit point of A but i is not. We do not wish to call a point such as i a limit point; but if "distinct" were dropped from the definition we could taken XII = i for each i and have i = lim x.. 3.4 Proposition. (a) A set is closed iff it contains all its limit points. (b) If A c X then A~ = A u {x: X is a limit point of A}. The proof is left as an exercise. From real analysis we know that a basic property of [f\\: is that any sequence whose terms get closer together as n gets large, must be convergent. Such sequences are called Cauchy sequences. One of their attributes is that you know the limit will exist even though you can't produce it.

°

3.5 Definition. A sequence {xn} is called a Cauchy sequence if for every € > there is an integer N such that d(xn' xm) < € for all n, m ;:::: N. If (X, d) has the property that each Cauchy sequence has a limit in X then (X, d) is complete. 3.6 Proposition. C is complete. Proof If {xn+iYn} is a Cauchy sequence in C then {xn} and {Yn} are Cauchy sequences in [f\\:. Since [f\\: is complete, Xn --+ X and Yn -+ Y for points x, Y in [f\\:. It follows that X + iy = lim (xn + iYn), and so C is complete. III Consider Coo with its metric d (1. 6.7 and I. 6.8). Let z'" z be points in C;

19

Sequences and completeness

it can be shown that d(zn' z) -3> 0 if and only if IZn-zl -3> O. In spite of this, any sequence {zn} with lim IZnl = 00 is Cauchy in Coo, but, of course, is not Cauchy in C. If A c X we define the diameter of A by diam A = sup {d(x, y): x and yare in A}. 3.7 Cantor's Theorem. A metric space (X, d) is complete

iff for

any sequence

{Fn} of non-empty closed sets with F) ~ F2 ~ ... and diam Fn consists of a single point.

-3>

0,

n Fn 00

n~1

Proof Suppose (X, d) is complete and let {Fn} be a sequence of closed sets having the properties: (i) F) ~ F2 ~ ... and (ii) lim diam Fn = O. For each n, let Xn be an arbitrary point in Fn; if n, m ;::: N then Xn, Xm are in FN so that, by definition, d(xm xm) :<;; diam FN. By the hypothesis N can be chosen sufficiently large that diam FN < 10; this shows that {xn} is a Cauchy sequence. Since X is complete, Xo = lim Xn exists. Also, Xn is in FN for all n ;::: N

nFn = F. 00

since Fn c F N; hence, Xo is in FN for every N and this gives Xo

E

n~1

So F contains at least one point; if, also, y is in F then both Xo and yare in Fn for each n and this gives d(xo, y) :<;; diam Fn -3> O. Therefore d(xo, y) = 0, or Xo = y. Now let us show that X is complete if it satisfies the stated condition. Let {xn} be a Cauchy sequence in X and put Fn = {xn' Xn+ 1, .. . }-; then Fl ~ Fi ~ .... If 10 > 0, choose N such that d(xn, xm) < 10 for each n, m ;::: N; this gives that diam {xn' X n + 1> ••• } :<;; 10 for n ;:::: N and so diam Fn :<;; 10 for n ;::: N (Exercise 3). Thus diam Fn -3> 0 and, by hypothesis, there is a point Xo in X with {xo} = F) (J F2 (J • • • • In particular Xo is in Fn, and so d(xo, xn) :<;; diam Fn -3> O. Therefore, Xo = lim Xn. • There is a standard exercise associated with this theorem. It is to find a sequence of sets {Fn} in IR which satisfies two of the conditions: (a) each Fn is closed, (b) F) ~ F2 ~ ... , (c) diam Fn -3> 0; but which has F = F) (J F2 (J ••• either empty or consisting of more than one point. Everyone should get examples satisfying the possible combinations. 3.8 Proposition. Let (X,d) be a complete metric space and let Y eX. Then (Y, d) is a complete metric space iff Y is closed in X. Proof. It is left as an exercise to show that (Y,d) is complete whenever Y is a closed subset. Now assume (Y,d) to be complete; let Xo be a limit point of Y. Then there is a sequence {y n} of points in Y such that Xo = limyn' Hence {y n} is a Cauchy sequence (Exercise 5) and must converge to a point Yo in Y, since (Y,d) is complete. It follows that Yo=xo and so Y contains all its limit points. Hence Y is closed by Proposition 3.4. •

20


Exercises 1. Prove Proposition 3.4. 2. Furnish the details of the proof of Proposition 3.8. 3. Show that diam A = diam A - . 4. Let Zn' Zbe points in IC and let d be the metric on 1C00- Show that IZn - zi -+ 0 if and only if d(zn' z) -+ O. Also show that if IZnl -+ 00 then {zn} is Cauchy in IC",. (Must {zn} converge in ICC()?) 5. Show that every convergent sequence in (X, d) is a Cauchy sequence. 6. Give three examples of non complete metric spaces. 7. Put a metric don IR such that Ixn-xi -+ 0 if and only if d(xn' x) -+ 0, but that {xn} is a Cauchy sequence in (IR, d) when IXnl -+ 00. (Hint: Take inspiration from 1C 8. Suppose {xn} is a Cauchy sequence and {xnJ is a subsequence that is convergent. Show that {xn} must be convergent. 00 . )

§4. Compactness The concept of compactness is an extension of the benefits of finiteness to infinite sets. Most properties of compact sets are analogues of properties of finite sets which are quite trivial. For example, every sequence in a finite set has a convergent subsequence. This is quite trivial since there must be at least one point which is repeated an infinite number of times. However the same statement remains true if "finite" is replaced by "compact." 4.1 Definition. A subset K of a metric space X is compact if for every collection ~ of open sets in X with the property

4.2

Kc

U {G:GE~},

there is a finite number of sets G 1 , ••• , Gn in ~ such that K c G 1 U G 2 U ... U Gn • A collection of sets ~ satisfying (4.2) is called a cover of K; if each member of ~ is an open set it is called an open cover of K. Clearly the empty set and all finite sets are compact. An example of a non compact set is D = 2, 3, ... , then {G 2 , G 3 , finite subcover.

{z E IC: Izi ••• }

< I}. If G. =

{Z: Izi

< 1-

~} for n =

is an open cover of D for which there is no

4.3 Proposition. Let K be a compact subset of X; then: (a) K is closed; (b) ifF is closed and Fe K then F is compact. Proof To prove part (a) we will show that K

= K-. Let

Xo E

K-; by Pro-

position 1.13(0, B(xo; €) n K =f. D for each € > O. Let G. = X and suppose that

Xo

!# K. Then each Gn is open and K c

-B(xo;~)

U G. (because n 00

00

.-1

.-1

21

Compactness

B(XO;~) =

Kc UG n~1

n•

{xo}). Since

But G1 c G2

gives that B( x 0;

~) n

C

Kis compact there is an integer m such that so that K G = X-B(XO;~)' But this m ...

C

m

K = D, a contradiction. Thus K = K - .

To prove part (b) let f§ be an open cover of F. Then, since F is closed, {X-F} is an open cover of K. Let G 1 , ••. , Gn be sets in f§ such that K c G l U ... U Gn U (X -F). Clearly, Fe G 1 U ... U Gn and so F is compact. • If .'F is a collection of subsets of X we say that .'F has the finite intersection property (f.i.p.) if whenever {Fl' F 2, ... , Fn} c .'F, Fl n F2 n ... n Fn =1= D. An example of such a collection is {D - G2 , D - G 3 , ••• } where the sets Gn are as in the example preceding Proposition 4.3. f§ U

4.4 Proposition. A set K c X is compact iff every collection .'F of closed subsets of K with the fi.p. has {F: FE.'F} =1= D·

n

Proof Suppose K is compact and .'F is a collection of closed subsets of K having the f.i.p. Assume that {F: FE.'F} = D and let f§ = {X - F: FE.'F}. Then, U {X-F: FE.'F} = X {F: FE.'F} = X by the assumption; in particular, f§ is an open cover of K. Thus, there are F I , •.• ,

n

Fn E.'F such that K

n

c

U(X -

Fk) = X -

k~1

C

n

n Fk. n

But this gives that

nFk n

k~1

X - K, and since each Fk is a subset of K it must be that

contradicts the f.i.p. The proof of the converse is left as an exercise . •

k

n Fk n

k~1

=

D. This

~I

4.5 Corollary. Every compact metric space is complete.

Proof This follows easily by applying the above proposition and Theorem 3.7. •

4.6. Corollary.

If X is compact then every infinite set has a limit point in

X.

Proof Let S be an infinite subset of X and suppose S has no limit points. Let {ai' a 2 , ••• } be a sequence of distinct points in S; then Fn = {an, an + I, ..• } also has no limit points. But if a set has no limit points it contains all its limits points and must be closed! Thus, each Fn is closed and {Fn:

n Fn 00

n ~ I} has the f.i.p. However, since the points ai' a2 , = D, contradicting the above proposition. •

•••

are distinct,

n~ I

4.7 Definition. A metric space (X, d) is sequentially compact if every sequence in X has a convergent subsequence. It will be shown that compact and sequentially compact metric spaces are the same. To do this the following is needed. 4.8 Lebesgue's Covering Lemma.

If (X, d) is sequentially compact and

22


'1J is an open cover of X thell there is an set G in 'lJ with Bex; IE) c G.

E

> 0 such that if x is in X, there is a

Proof The proof is by contradiction; suppose that 'lJ is an open cover of X and no such f' > 0 can be found. In particular, for every integer n there is a

point x" in X such that B(t"n; \

1)' is not contained in any set Gin '1J. Since X

11

is sequentially compact there is a point Xo in X and a subsequence {xIIJ such that Xo = lim x"". Let Go E 'lJ such that Xo E' Go and choose f' > 0 such that B(xo; E) C Go. Now let N be such that d(xo, xnJ < E/2 for all 11k :.?:: N. Let 11, be any integer larger than both Nand 2/E, and let y E' Bexnk ; link)· Then d(x o, y) ~ d(xo, xnJ+d(Xllk' Y) < E/2+ I < E. That is, B(xn,; link) C B(xo; E) C GIl> contradicting the choice of X llk • II There are two ,common misinterpretations of Lebesgue's Covering Lemma; one implies that it says nothing and the other that it says too much. Since 'lJ is an open covering of X it follows that each x in X is contained in some G in 'lJ. Thus there is an E > 0 such that B(x; E) C G since G is open. The lemma. however, gives one E > 0 such that for any x, B(x; E) is contained in some member of 'lJ. The other misinterpretation is to believe that for the E > 0 obtained in the lemma. B(x; fc) is contained in each G in '!J such that xc G.

4.9. Theorem. Let eX, d) be a metric space; then the following are equimlent statements: (a) X is compact; (b) Every infinite set ill X has a limit point; (c) X is sequentially compact; (d) X is complete and for every E > 0 there are a finite number of points x I, • . . , xn in X such that

U B(x II

X =

k;

fc).

k= 1

(The property mentioned in Cd) is called total boundedness.) Proof That (a) implies (b) is the statement of Corollary 4.6. (b) implies (c): Let {XII} be a sequence in X and suppose, without loss of generality, that the points Xj, X2 • . . . are all distinct. By (b), the set {Xl' x 2 , ••• } has a limit point Xo' Thus there is a point x n , (B(xo; 1); similarly, there is an integer 112 > 111 with x n , E' B(x 0: 1/2). Continuing we get integers n! < 112 < ... , with X"k E B(xo; 11k). Thus, -'0 = lim x nk and X is sequentially compact. (c) implies (d): To see that X is complete let {xn} be a Cauchy sequence, apply the definition of sequential compactness, and appeal to Exercise 3.8. Now let € > 0 and fix Xl E' X. If X = B(x!; E) then we are done; otherwise choose X 2 EX -B(x!; E). Again, if X = B(x!; E) U B(x 2 ; E) we are done;

Compactness

23

if not, let x 3 E X - [B(x 1 ; E) sequence {xn} such that

e)]. If this process never stops we find a

U B(x 2;

X.+ I EX -

n

U B(Xk; 1:-).

k= I

But this implies that for n # m, d(xn, xm) ;;.:: e > O. Thus {xn} can have no convergent subsequence, contradicting (c). (d) implies (c): This part of the proof will use a variation of the "pigeon hole principle." This principle states that if you have more objects than you have receptacles then at least one receptacle must hold more than one object. Moreover, if you have an infinite number of points contained in a finite number of balls then one ball contains infinitely many points. So part (d) says that for every E > 0 and any infinite set in X, there is a point Y E X such that B(y; e) contains infinitely many points of this set. Let {x,,} be a sequence of distinct points. There is a point YI in X and a subsequence {x~t)} of {xn} such that {x~l)} c B(YI; I). Also, there is a point Yz in X and a subsequence {X~2)} of {x~l)} such that {x~)} c B(yz; t). Continuing, for each integer k ;;.:: 2 there is a point Yk in X and a subsequence {~k)} of {x~k-l)}suchthat {X~k)} c B(Yk; Ilk). LetFk = {x~k)}-;thendiamFk:5 21k

n Fk o:J

and Fl ~ F2 ~ .... By Theorem 3.7,

=

{x o }. We claim that X~k)--)oo

k~1

Xo (and {X~k)} is a subsequence of {x.}). In fact, Xo E Fk so that d(xo, 4k » :5 diam Fk :5 21k, and Xo = lim X~k). (c) implies (a): Let '!f be an open cover of X. The preceding lemma gives an E > 0 such that for every x E X there is a G in ~ with R(x; e) c G. Now

(c) also implies (d); hence there are points X =

n

U B(x

k;

k~1

Hence X

=

n

€). Now for 1 :5 k :5 n there is a set Gk

UG

k;

that is, {G 1 ,

••• ,

• , Xn

XI>" E

in X such that

'!f with B(xk ; e) c Gk •

Gn } is a finite subcover of '!f . •

k~1

4.10 Heine-Borel Theorem. A subset K oflR" (n ;;.:: 1) is compact

iff K is closed

and bounded. Proof. If K is compact then K is totally bounded by part (d) of the preceding theorem. It follows that K must be closed (Proposition 4.3); also, it is easy to show that a totally bounded set is also bounded. Now suppose that K is closed and bounded. Hence there are real numbers al, ... ,an and bl, ... ,bn such that KcF=[al,b;lX ... X [a",bnl. If it can be shown that F is compact then, because K is closed, it follows that K is compact (Proposition 4.3(b». Since Rn is complete and F is dosed it follows that F is complete. Hence, again using part (d) of the preceding theorem we need only show that F is totally bounded. This is easy although somewhat "messy" to write down. Let (>0; we now will write F as the union of n-dimensional rectangles each of diameter less than f. m

After doing this we will have FeU B (x,,; £) where each 1<=1

XI<

belongs to

24


one of the aforementioned rectangles. The execution of the details of this strategy is left to the reader (Exercise 3). • Exercises 1. Finish the proof of Proposition 4.4. 2. Let P = (Pl' ... ,Pn) and q = (ql' ... , qn) be points in ~. with Pk < qk for each k. Let R = [Ph ql] X ••• x [Pn, qnl and show that diam R

= d(p, q) =

3. Let F = [aJ> btl x ... x [an, bnl c

[t

~n

k=l

(qk_Pk)2]t.

and let

€

> 0; use Exercise 2 to

show that there are rectangles R l , ... , Rm such that F

=

m

U Rk and

k-I

diam

Rk < € for each k. If Xk E Rk then it follows that Rk c B(xk; f). 4. Show that the union of a finite number of compact sets is compact. 5. Let X be the set of all bounded sequences of complex numbers. That is, {xn } E X iff sup {lx.l: n ~ I} < 00. If x = {x.} and Y = {Yn}, define d(x, y) = sup {Ix. - Y.I : n ~ I}. Show that for each x in X and € > 0, B(x; €) is not totally bounded although it is complete. (Hint: you might have an easier time of it if you first show that you can assume x = (0, 0, ... ).) 6. Show that the closure of a totally bounded set is totally bounded.

§5. Continuity One of the most elementary properties of a function is continuity. The presence of continuity guarantees a certain degree of regularity and smoothness without which it is difficult to obtain any theory of functions on a metric space. Since the main subject of this book is the theory of functions of a complex variable which possess derivatives (and so are continuous), the study of continuity is basic.

° °

5.1 Definition. Let eX, d) and (0, p) be metric spaces and let f: X -+ be a function. If a E X and W EO, then limf(x) = w if for every € > there is a

°such that p(f(x), w) < whenever °< d(x, a) < S. The function f is continuous at the point a if limf(x) = f(a). Iffis continuous at each point of X-wl

S>

€

X-WI

X then f is a continuous function from X to 0.

5.2 Proposition. Let f: (X, d) -+ (0, p) be a function and a E X, Ct = f(a). The following are equivalent statements: (a) f is continuous at a; (b) For every € > O,f-l(B(Ct; €» contains a ball with center at a; (c) Ct = limf(x.) whenever a = lim x •.

The proof will be left as an exercise for the reader. That was the last proposition concerning continuity of a function at a

Continuity

25

point. From now on we will concern ourselves only with functions continuous on all of X. 5.3 Proposition. Let f: (X, d) ~ (0, p) be a function. The following are equivalent statements: (a) fis continuous; (b) If Ll is open in 0 then f-l(Ll) is open in X; (c) -If r is closed in 0 then f- 1 (r) is closed in X. Proof (a) implies (b): Let Ll be open in 0 and let x Ef-l(Ll). If w = f(x) then w is in Ll; by definition, there is an E > 0 with B(w ; E) C Ll. Since f is continuous, part (b) of the preceding proposition gives a 0 > 0 with B(x; 0) c f-l(B(w; E» c f-l(Ll). Hence, f-l(Ll) is open. (b) implies (c): If reO is closed then let Ll = o-r. By (b),J-l(Ll) = X-f-l(r) is open, so thatf-l(r) is closed. (c) implies (a): Suppose there is a point x in X at whichfis not continuous. Then there is an E > 0 and a sequence {xn } such that p(f(xn ), f(x» ~ E for every n while x = lim x n • Let r = O-B(f(x); E); then r is closed and each Xn is inf-l(f). Since (by (c»f-l(r) is closed we have x Ef-l(r). But this implies p(f(x), f(x» ~ E > 0, a contradiction . • The following type of result IS probably well understood by the reader and so the proof is left as an exercise.

5.4 Proposition. Let f and g be continuous functions from X into C and let

IX, (3 E IC. Then IXf+ (3g and fg are both continuous. Also, fIg is continuous provided g(x) of- 0 for every x in X. 5.5 Proposition. Letf: X ~ Yandg: Y ~Z be continuousfunctions. Then gof (where go f(x) = g(f(x») is a continuous function from X into Z. Proof If U is open in Z then g-l(U) is open in Y; hence,J-l(g-l(U» (gaf)-I(U) is open in X . •

=

5.6 Definition. A function f: (X, d) ~ (0, p) is uniformly continuous if for every E > 0 there is a 0 > 0 (depending only on E) such that p(f(x),J(y» < E whenever d(x, y) < o. We say thatfis a Lipschitzfunction if there is a constant M > 0 such that p(f(x),J(y» :s; Md(x, y) for all x and y in X. It is easy to see that every Lipschitz function is uniformly continuous. In fact, if E is given, take 0 = ElM. It is even easier to see that every uniformly continuous function is continuous. What are some examples of such functions? If X = 0 = IR then f(x) = x 2 is continuous but not uniformly continuous. If X = 0 = [0, 1] then f(x) = xt is uniformly continuous but is not a Lipschitz function. The following provides a wealthy supply of Lipschitz functions. Let A be a non-empty subset of X and x E X; define the distance from x to the set A, d(x, A), by d(x, A) = inf {d(x, a): a E A}. 5.7 Proposition. Let A be a non-empty subset of X; then: (a) d(x, A) = d(x, A -);

26


(b) d(x, A) = 0 iff x E A - ; (c) Id(x, A) - dey, A)I ::::; d(x, y) for all x, y in X. Proof (a) If A c B then it is clear from the definition that d(x, B) ::::; d(x, A). Hence, d(x, A -) ::::; d(x, A). On the other hand, if E > 0 there is a point y in A - such that d(x, A -) ~ d(x, y) - E/2. Also, there is a point a in A with dey, a) < E/2. But Id(x, y) - d(x, a)1 ::::; dey, a) < E/2 by the triangle inequality. In particular, d(x, y) > d(x, a)-E/2. This gives, d(x, A-) ~ d(x, a)-E ~ d(x, A) -E. Since E was arbitrary d(x, A -) ~ d(x, A), so that (a) is proved. (b) If x E A - then 0 = d(x, A -) = d(x, A). Now for any x in X there is a minimizing sequence {an} in A such that d(x, A) = lim d(x, an). So if d(x, A) = 0, lim d(x, an) = 0; that is, x = lim an and so x E A -. (c) For a in A d(x, a) ::::; d(x, y)+d(y, a). Hence, d(x, A) = inf {d(x, a): a E A} ::::; inf {d(x, y) + dey, a): a E A} = d(x, y) + dey, A). This gives d(x, A)dey, A) ::::; d(x, y). Similarly dey, A) - d(x, A) ::::; d(x, y) so the desired inequality follows . • Notice that part (c) of the proposition says that f: X --+ IR defined by f(x) = d(x, A) is a Lipschitz function. If we vary the set A we get a large supply of these functions. It is not true that the product of two uniformly continuous (Lipschitz) functions is again uniformly continuous (Lipschitz). For example, f(x) = x is Lipschitz butfIis not even uniformly continuous. However ifbothfand g are bounded then the conclusion is valid (see Exercise 3). Two of the most important properties of continuous functions are contained in the following result.

5.8 Theorem. Let f: (X, d) --+ (0, p) be a continuous function. (a) If X is compact then f(X) is a compact subset of O. (b) If X is connected then f(X) is a connected subset of O. Proof To prove (a) and (b) it may be supposed, without loss of generality, that f(X) = O. (a) Let {w n } be a sequence in 0; then there is, for each n ~ 1, a point Xn in X with Wn = f(x n ). Since X is compact there is a point x in X and a subsequence {xnJ such that x = lim x nk ' But if W = f(x), then the continuity off gives that W = lim w nk ; hence 0 is compact by Theorem 4.9. (b) Suppose ~ c 0 is both open and closed in 0 and that ~ # D. Then, because/eX) = 0, D # /-1(~); also,f-l(~) is both open and closed because / is continuous. By connectivity, /-1(~) = X and this gives 0 = ~. Thus, 0 is connected . •

5.9 Corollary. If f: X --+ 0 is continuous and K c X is either compact or connected in X then f(K) is compact or connected, respectively, in O.

5.10 Corollary. Iff: X

~ R is continuous and X is connected then f( X) is an interval. This foIIows from the characterization of connected subsets of IR as intervals.

Continuity

27

5.11 Intermediate Value Theorem. Iff: [a, b] --+ IR is continuous andf(a) :::; :::; feb) then there is a point x, a :::; x :::; b, with f(x) = g.

g

5.12 Corollary. Iff: X --+ IR is continuous and K c X is compact then there are points Xo and Yo in K withf(x o) = sup {f(x): x E K} andf(yo) = inf {f(x): XEK}. Proof If IX = sup {f(x): x E K} then IX is in f(K) because f(K) is closed and bounded in IR. Similarly f3 = inf {f(x): x E K} is in f(K) . • 5.13 Corollary. If K c X is compact and f: X are points x 0 and Yo in K with If(xo) I = sup {If(x)l: x

E

K} and If(Yo)1

--+

C is continuous then there

= inf {If(x) I: x E K}.

Proof This corollary follows from the preceding one because g(x) = If(x) I defines a continuous function from X into IR. 5.14 Corollary. If K is a non-empty compact subset of X and x is in X then there is a point y in K with d(x, y) = d(x, K). Proof Define f: X --+ IR by fey) = d(x, y). Then f is continuous and, by Corollary 5.12, assumes a minimum value on K. That is, there is a point y in K with fey) :::; fez) for every Z E K. This gives d(x, y) = d(x, K) . • The next two theorems are extremely important and will be used repeatedly throughout this book with no specific reference to the theorem numbers. 5.15. Theorem. Suppose f: X uniformly continuous.

--+ Q

is continuous and X is compact; then f is

Proof Let E > 0; we wish to find a 8 > 0 such that d(x, y) < 8 implies p(f(x), fey)) < E. Suppose there is no such 8; in particular, each 8 = lin will fail to work. Then for every n ~ I there are points Xn and Yn in X with d(xn> Yn) < lin but p(f(xn), f(Yn)) ~ E. Since X is compact there is a subsequence {xnJ and a point x E X with x = lim x nk . Claim. x = lim Ynk. In fact, d(x, Ynk) :::; d(x, xnJ+ link and this tends to zero as k goes to 00. But if w = f(x), w = lim f(xnJ = lim f(YnJ so that E :::;

p(f(xnJ,J(Ynk))

:::; p(f(xnJ, w)+p(w,J(ynJ)

and the right hand side of this inequality goes to zero. This is a contradiction and completes the proof. • 5.16. Definition. If A and B are non-empty subsets of X then define the distance from A to B, dCA, B), by dCA, B)

= inf {dCa, b): a E A, b E B}.

Notice that if B is the single-point set {x} then dCA, {x}) = d(x, A). If

28


A = {y} and B = {x} then d( {x}, {y}) = d(x, y). Also, if A n B of. [: then dCA, B) = 0, but we can have dCA, B) = 0 with A and B disjoint. The most popular type of example is to take A = {(x, 0): x E SIl} C IFR2 and B = {ex, eX): x E SIl}. Notice that A and B are both closed and disjoint and still dCA, B) = O.

5.17 Theorem. If A and B are non-empty di!Jjoint sets in X with B closed and A compact then dCA, B) > O. Proof DefineJ: ) ( + SIl byJ(x) = d(x, B). Since A n B = ::::J and B is closed, J(a) > 0 for each a in A. But since A is compact there is a point a in A such that 0 < J(a) = inf {f(x): x E A} = dCA, B). III

Exercises 1. 2. X 3.

Prove Proposition 5.2. Show that if J and g are uniformly continuous (Lipschitz) functions from into C then so is J+ g. We say that f: X --+ C is bounded if there is a constant M > 0 with IJ(x)! ::; M for all x in X. Show that if J and g are bounded uniformly continuous (Lipschitz) functions from X into C then so is f'5. 4. Is the composition of two uniformly continuous (Lipschitz) functions again uniformly continuous (Lipschitz)? 5. Supposef: X --+ n is uniformly continuolls; show that if {x n } is a Cauchy sequence in X then {l(xn)} is a Cauchy sequence in £2. Is this still true if we only assume that J is continuous? (Prove or give a counterexample.) 6. Recall the definition of a dense set (1.l4). Suppose that n is a complete metric space and that J: (D, d) .~)- en; p) is uniformly continuous, where D is dense in (X, d). Use Exercise 5 to show that there is a uniformly continuous function g: X -+ 0 with g(x) = J(x) for every x in D. 7. Let G be an open subset of C and let P be a polygon in G from a to b. Use Theorems 5.15 and 5.17 to show that there is a polygon Q c G from a to b which is composed of line segments which are parallel to either the real or imaginary axes. 8. Use Lebesgue's Covering Lemma (4.8) to give another proof of Theorem 5.15. 9. Prove the following converse to Exercise 2.5. Suppose (,\:', d) is a compact metric space having the property that for every E > 0 and for any points a, b in X, there are points Zo, ZI' ••. , Zn in X with Zo = a, z" = b, and d(zk-J' Zk) < E for 1 ~ k ::; n. Then (X, d) is connected. (Hint: Use Theorem 5.17.) 10. Let J and g be continuous functions from (X, d) to (D, p) and let D be a dense subset of X. Prove that if J(x) = g(x) for x in D then J = g. Use this to show that the function g obtained in Exercise 6 is unique. §6. Uniform convergence Let X be a set and (n, p) a metric space and suppose f j~, J2' ... are functions from X into n. The sequence {in} converges uniformly to J-written

29

Uniform convergence

f = u-lim fn-if for every € > 0 there is an integer N (depending on alone) such that p(f(x), fn(x» < € for all x in X, whenever n ~ N. Hence,

€

sup {p(f(x),fn(x»: x E X} ::; € whenever n ~ N. The first problem is this: If X is not just a set but a metric space and each j~ is continuous does it follow thatfis continuous? The answer is yes. 6.1 Theorem. Suppose f,,: (X, d) ---+ (0, p) is continuous for each n and that f = u -lim fn; then f is continuous. Proof Fix Xo in X and E > 0; we wish to find a 0 > 0 such that p(f(xo), f(x» < € when d(xo, x) < O. Sincef= u-limfm there is a functionf" with p(f(x), j~(x» < €/3 for all x in X. Since fn is continuous there is a 0 > 0 such that p(f,,(xo),f,,(x» < €/3 when d(xo, x) < O. Therefore, if d(xo, x) <0, p(f(x o),f(x» ::; p(f(xo), fn(x 0» + p(f,,(x 0), f,,(x» + p(f,,(x), f(x» < € • • Let us consider the special case where 0 = IC. If Un: X ---+ C, let f,,(x) = U1 (x) + ... + un(x). We say f(x) =

The series

ro

L

ro

L unCx) ifff(x)

= limfn(x) for each x in X. n=l un is uniformly convergent to f iff f = u -lim f".

I

6.2 Weierstrass M-Test. Let un: X

---+

C be a function such that Iun(x) I ::; Mn

for every x in X and suppose the constants satisfy uniformly convergent.

ro

L Mn

n=l

<

00.

Then

ro

L Un is I

Proof Letf,,(x) = Ul(X)+ ... +un(x). Then for n > m, 1f,,(x)-fm(x)1 = lum +1(x)+ .. . + un(x) I ::;

n

L

k=m+l

00

Mk for each x. Since

L Mk 1

converges, {fnCx)} is a Cauchy sequence in C. Thus there is a number g E C with g = lim fn(x). Define f(x) = g; this gives a function f: X ---+ C. Now 00

00

00

00

If(x)-f,,(x)1 = I L un(x) I ::; L lun(x)l::; L M k ; k=n+l k=n+l k=n+l

since L Mk is convergent, for any I

Mk < E whenever n n ~ N.•

~

€

> 0 there is an integer N such that

N. This gives If(x) - f,,(x) I <

E

00

L

k=n+l for all x in X when

Exercise 1. Let {/,,} be a sequence of uniformly continuous functions from (X, d) into (n, p) and suppose that f = u - limf" exists. Prove that f is uniformly continuous. If each In is a Lipschitz function with constant Mn and sup Mn < 00, show that I is a Lipschitz function. If sup Mn = 00, show that

I may fail to be Lipschitz.

Chapter III Elementary Properties and Examples of Analytic Functions §1. Power series

In this section the definition and basic properties of a power series will be given. The power series will then be used to give examples of analytic functions. Before doing this it is necessary to give some elementary facts on infinite series in C whose statements for infinite series in IR should be well 00

known to the reader. If an is in C for every n ~ 0 then the series L an m

n~O

converges to Z iff for every E > 0 there is an integer N such that I Lan - zl < E n=O whenever m ~ N. The series L an converges absolutely if L lanl converges.

L an

1.1 Proposition. If L an converges absolutely then

Proof Let

E

converges.

> 0 and put Zn = aO+a 1 + ... + an" Since 00

there is an integer N such that L lanl < n=N

IZm-zkl =

m

m

E.

lanl converges

Thus, if m > k ~ N, 00

L ani::::; L lanl::::; L n=k+l n=k+l n=N

I

L

lanl

<

E.

That is, {zn} is a Cauchy sequence and so there is a Z in C with Z = lim Zn. Hence Lan = z . • Also recall the definitions of limit inferior and superior of a sequence in IR. If {an} is a sequence in IR then define lim inf an = lim [inf {an> an+1 ,

••• }]

n~oo

lim sup an = lim [sup {an> an+1 ,

••• }]

n~oo

An alternate notation for lim inf an and lim sup an is lim an and lim an" If = inf {an> an + 1 , ••• } then {b n } is an increasing sequence of real numbers or {- oo}. Hence, lim inf an always exists although it may be ± 00. Similarly lim sup an always exists although it may be ± 00. A number of properties of lim inf and lim sup are included in the exercises of this section. 00 A power series about a is an infinite series of the form L a,,(z-a)n. One n=O of the easiest examples of a power series (and one of the most useful) is the

bn

geometric series

L zn. 00

n=O

For which values of z does this series converge and 30

31

Power series

when does it diverge? It is easy to see that I-z so that

n+ 1

1.2

l+z+ ... +zn

If

Izl

< 1 then 0

= (I-z) (I +z+ . .. +zn),

l-zn+ 1

= --l-z

= lim zn and so the geometric series is convergent with 1 = l-z·

00

L zn o

Izl > I then lim Izln = ro and the series diverges. Not only is this result an archetype for what happens to a general power series, but it can be used to explore the convergence properties of power series.

If

1.3 Theorem. For a given power series L anCz-a)n define the number R, o ::; R ::; 00, by n= 0
~ = lim sup lanll!n, then:

(a) if Iz-al (b) if Iz-al

< R, the series converges absolutely: > R, the terms of the series become unbounded and so the

series diverges; (c) if 0 < r < R then the series converges uniformly on {z: Iz - al ::; r}. Moreover, the number R is the only number having properties (a) and (b). Proof We may suppose that a

= O. If Izl < R there is an r with Izl < r < R.

Thus, there is an integer N such that

~}

But then

the tail end

lanl

<

lanl l !.

<

!r for all n ~

N (because

~ and so Ianz· I < (';IJ for all n ~ N.

~r

>

This says that

I a.znis dominated by the series L (~)n, and since ~ < 1 r r

n=N

the power series converges absolutely for each Izl < R. Now suppose r < R and choose p such that r < p < R. As above, let N be such that

lanl

<

~ for

all n

~

N. Then if

Izl ::;

r, lanznl ::;

GJ

and

(~) < 1. Hence the Weierstrass M-test gives that the power series converges Izl ::; To prove (b), let Izi

uniformly on {z:

r}. This proves parts (a) and (c). .

1

Izl

1

> r > R. Hence - < - ; r R from the definition of lim sup, this gives infinitely many integers n with

~<

> R and choose r with

la.ll!n. It follows that lanz·1 >

(';IJ

and, since

(';1)

> 1, these terms

become unbounded . • The number R is called the radius of convergence of the power series.

32

1.4 Proposition. If vergence R, then

L

Elementary Properties and Examples of Analytic Functions

an(z -a)" is a given power series with radius of con-

if this limit exists. Proof Again assume that a = 0 and let ex = lim la./a.+ 11, which we suppose to exist. Suppose that Izi < r < ex and find an integer N such that r < la./an+ti for all n::::: N. Let B= laNlr N ; then la,'HllrN+1 = laN+drr N < laNlr N = B; laN+2Ir N+2 = laN+zlrr N+1 < la N+1 !r N+1 < B;continuingweget

~ B for all n ::::: N. But then lanz" I = lanr"1 EJ: .~ BEJ: for all n ::::: N. ~ ~ Since Izi < r we get that L lanz"1 is dominated by a convergent series and ianr"!

00

n=1

hence converges. Since r < ex was arbitrary this gives that ex ~ R. On the other hand if Izl > r > ex, then lanl < rlan + 11 for all n larger than some integer N. As before, we get jan," 1 ::::: B = jaNrNI for n ::::: N. This

1 gives lanz"1 ::::: B 1 1;1 : which approaches and so R ~ ex. Thus R = ex. III Consider the series

f

n~O

00

as n does. Hence,

L a.z" diverges

z· ; by Proposition 1.4 we have that this series

n!

has radius of convergence 00. Hence it converges at every complex number and the convergence is uniform on each compact subset of C. Maintaining a parallel with calculus, we designate this series by

_

e" = exp z =

~

z"

n= 0

n.

L !'

the exponential series or junction. Recall the following proposition from the theory of infinite series (the proof will not be given).

1.5 Proposition. Let L an and L bn be two absolutely convergent series and put n

Cn

Then

Lc

n

=

L0 akbn-k'

k=

is absolutely convergent with sum

1.6 Proposition. Let L an(z-a)n and I bll(z-at be power series with radius oj convergence::::: r > O. Put n

Cn

=

I

k~'

()

akbn-k;

then both power series I (an+b n ) (z-a)" and I cn(z-a)" have radius of convergence ::::: r, and

Analytic functions

I

for

Iz-al

(an + bn) (z-ay

I

=

cn(z-aY =

33

[I an(z-ay+ I bn(z-ay] [I an(z-ay] [I bn(z-aY]

< r.

Proof We only give an outline of the proof. If 0 < s < r then for lan+bnl Izln ~ lanlsn+ Ibnls n < 00; 1e,,1 Izln ~ (I Ibnls n) < 00. From here the proof can easily be completed . •

we get

I

I

I

I

Izl

~ s,

(I lanls")

Exercises l. Prove Proposition 1.5. 2. Give the details of the proof of Proposition 1.6. 3. Prove that lim sup (a" + b,,) ~ lim sup a" + lim sup b" and lim inf (a" + b,,) 2. lim inf a" + lim inf b" for bounded sequences of real numbers {a,,} and {b ll } . 4. Show that lim inf a n ~ lim sup an for any sequence in ~. 5. If {an} is a convergent sequence in ~ and a = lim an' show that a = lim inf an = lim sup an. 6. Find the radius of convergence for each of the following power series:

I

00

(a)

a"zn, a

E

C; (b)

n~O

co

I

an2 z", a

I

00

E

n~O

C; (c)

I

00

knz", k an integer #-0; (d)

,,~O

zn!.

,,~O

7. Show that the radius of convergence of the power series

i

(-nI)n zn(n+l)

n=1

is 1, and discuss convergence for z = 1, -1, and i. (Hint: The nth coefficient of this series is not (-IY/n.) §2. Analytic functions In this section analytic functions are defined and some examples are given. It is also shown that the Cauchy-Riemann equations hold for the real and imaginary parts of an analytic function. 2.1 Definition. If G is an open set in C and f: at a point a in G if

G~C

then f is differentiable

a _+---,h)'-----=-P---'-(a--,-) lim f'---.:(ch~O

h

exists; the value of this limit is denoted by f'(a) and is called the derivative of . f at a. Iff is differentiable at each point of G we say that f is differentiable on G. Notice that if f is differentiable on G then f'(a) defines a function f': G -+ Co Iff' is continuous then we say thatfis continuously differentiable. Iff' is differentiable then f is twice differentiable; continuing, a differentiable function such that each successive derivative is again differentiable is called infinitely differentiable.

(Henceforward, all functions will be assumed to take their values in C unless it is stated to the contrary.)


34

The following was surely predicted by the reader. 2.2 Proposition. continuous at a.

If f: G -? IC is differentiable at a point a in G then f is

Proof In fact,

I.iml/(z) -

:~a

.

l(a)1 =

llim I/(z) - 1(~}J.l· [limlz - all Iz - al ::~a Z---"'U

=1/'(a)I·O=O.1III 2.3 Definition. A function I: G--.IC is analytic if I is continuously differentiable on G. It follows readily, as in calculus. that sums and products of functions analytic on G are analytic. Also, if I and g are analytic on G and G I is the set of points in G where g doesn't vanish, thenl/g is analytic on G I . Since constant functions and the function z are clearly analytic it follows that all rational functions are analytic on the complement of the set of zeros of the denominator. Moreover, the usual laws for differentiating sums, products, and quotients remain valid. 2.4 Chain Rule. Let I and g be analytic on G and Q respective(v and suppose I(G)c Q. Then gal is analytic on G and (gof)'(z) = g'(f(z) )/'(z) for all z in G.

Proof. Fix Zo in G and choose a positive number r such that B (zo; r) c G. We must show that if O
J(zo+ hn ) -J(zo)

Since lim[f(zo + hn ) - J(zo)] = 0 by (2.2) we have that limhn- I [ gof(zo + hn }

-

goJ(zo) ] = g'(f(zo»/'(zo)

*'

Case 2 J(zo)= J(zo+ hn ) for infinitely many values of n. Write {hn } as the union of two sequences {kn} and {In} where f (zo) J(zo+kn ) andf(zo)=J(zo+!n) for all n. SinceJis differentiable, f'(zo) = lim ',,-IU(ZO + In) - J(zo)] =0. Also liml,,-I[ goJ(zo + In) - goJ(zo)] = o. By Case I, lim k n- 1[ gof(zo + k n ) - goJ(zo)] = gl(f(ZO»/'(zo) = O. Therefore limh,,-I[ gof(zo + h,,) - goJ(zo)] =0= gl(f(ZO»/'(zo)' The general case easily follows from the preceding two.

II

In order to define the derivative, the function was assumed to be defined on an open set. If we say f is analytic on a set A and A is not open, we mean that f is analytic on an open set containing A.

35

Analytic functions

Perhaps the definition of analytic function has been anticlimatic to many readers. After seeing books written on analytic functions and year-long courses and seminars on the theory of analytic functions, one can excuse a certain degree of disappointment in discovering that the definition has already been encountered in calculus. Is this theory to be a simple generalization of calculus? The answer is a resounding no. To show how vastly different the two subjects are let us mention that we will show that a differentiable function is analytic. This is truly a remarkable result and one for which there is no analogue in the theory of functions of a real variable (e.g., consider x 2 sin

~) . Another equally remarkable result is that every analytic function

is infinitely differentiable and, furthermore, has a power series expansion about each point of its domain. How can such a humble hypothesis give such far-reaching results? One can get some indication of what produces this phenomenon if one considers the definition of derivative. In the complex variable case there are any infinity of directions in which a variable can approach a point a. In the real case, however, there are only two avenues of approach. Continuity of a function defined on ~, for example. can be discussed in terms of right and left continuity; this is far from the case for functions of a complex variable. So the statement that a function of a complex variable has a derivative is stronger than the same statement about a function of a real variable. Even more, if we consider a function f defined on Gee as a function of two real variables by putting g(x, y) = f(x + iy) for (x, y) E G, then requiring that g be Frechet differentiable will not ensure that f has a derivative in our sense. In an exercise we ask the reader to show thatf(z) = Izl2 has a derivative only at z = 0; but, g(x, y) = f(x + iy) = x 2 + y2 is Frechet differentiable. That differentiability implies analyticity is proved in Chapter IV; but right now we prove that power series are analytic functions. 00

L

2.5 Proposition. Let j(z) = an(z - at have radius of convergence R > Then: n;O (a) For each k ~ 1 the series

o.

ao

L n(n-l) ... (n-k+ l)an(z-a)n-k

2.6

n=k

has radius of convergence R; (b) The function f is infinitely difJerentiable on B(a; R) and, furthermore, J
L


36

wish to show that Rrule that lim log n

1

=

=

lim sup Il1a n l l /(n-I). Now it follows from l'H6pitai's

0, so that lim n 1/(.--

n-!ooon-l

!)

1. The result will follow from

=

n-+oo

Exercise 2 if it can be shown that lim sup lanll/(nl)=R- I . Let (R,)-l = lim sup la"II/(n-1); then R' is the radius of convergence of x

x;.

La,,'::"

La". I;:"·

1 =

1

Izi

Notice that

()

L lanz"l

< R' then

=

laol + 1;:1 Liane Iz"l

Izi ~ Rand z *- 0 then L la"z"l < + -1001 < 00, so that R ::s; R'. This Izi

proof of part (a).

(b) For

Izl

zLa,,+lz" + 0 0

00

and

<

00.

gives that R

=

=

hence if

This gives R' ::s; R. If

L 10", lz"1

1

=

G-I" L

n

n~l

akz k. Fix a point win BCO; R) and fix r with

Iwl

la"z"l

R' and completes the

L nanz""I, Sn(z) = L akz\ 00

< R put g(z)

La"z";

=

L 00

and RnCz) =

k~O

k~n+l

< r < R; we wish to show thatF(w) exists and is equal to g(w). To do this let 8 > 0 be arbitrary except

for the restriction that Rew; 8) c B(O; r). (We will further restrict 8 later in the proof.) Let z E B( w; S); then

2.8

jJ~)~f(w) z-w

_ g(w) =

[S.(Z) -sn(w) z-w

S~(W)J + [s~(w) -

g(w)]

+ [!!.~_~Z) - R_~\~!J z-)t!

Now

But

Hence, Rn(Z)-Rn(W)I:::; I z-w Since r < R,

L'" laklkr

k- 1

I

iaklkr k -

1

k=II+1

converges and so for any

E'

> 0 there is an integer

k~l

N J such that for n 2:: N J

< ~ IRn(Z)-Rn(W)1 z-w 3

(z

E

B(w; 8».

Also, lim s~(w) = g(w) so there is an integer N z such that Is~(w)-g(w)1 <

j

whenever n 2:: N 2 • Let n = the maximum of the two integers Nl and N z .

37

Analytic functions

Then we can choose 8 > 0 such that

_ ISn(Z)-Sn(W) z-w

S~(W)I <

-<

3

whenever 0 < [z - wi < 8. Putting these inequalities together with equation (2.8) we have that

-=[(~! If(Z)z-w

g(W)\ <

IE

for 0 < [z-w[ < 8. That is,!,(w) = g(w). (c) By a straightforward evaluation we get f(O) = f(O)(O) = aQ. Using (2.6) (for a = 0), we get f(k)(O) = k !ak and this gives formula (2.7). II

If the series L anCz -- a)n has radius of convergence R > 00

2.9 Corollary.

=

fez)

0 then

n~O

00

L an(z -

a)" is analytic in B(a; R).

n =0

Hence. expz =

L zn / n! is analytic in C. Before further examining the oc

n=O

exponential function and defining cosz and sinz. the following result must be proved. 2.10 Proposition. If G is open and connected and f:G -» C is differentiable with !'(z) = 0 for all z in G, then f is constant.

Proof Fix z 0 in G and let Wo = fez 0)' Put A = {z E G: fez) = wo}; we will show that A = G by showing that A is both open and closed in G. Let z E G and let {zn} c A be such that z = lim Zn' Since f(z.) = Wo for each n ~ 1 andfis continuous we getf(z) = Wo, or z E A. Thus, A is closed in G. Now fix a in A, and let IE > 0 be such that BCa; E) c G. If z E B(a; £), set g(t) = f(tz+(l-t)a), 0 ::; t ::; 1. Then g(t)-g(s) t-s

2.11 Thus. if we let

(-H

.

lIm

g(t)-g(s) (t-s)z+{s-t)a (t-s)z+(s-t)a t-s

we get (A.4(b), Appendix A)

g(t) - g(s)

1-5

t-s

,

--- = f (sz+(l-s)a)'(z-a) = O.

That is, g'(s) = 0 for 0 ::; s ::; 1, implying that g is a constant. Hence, fez) = g(l)=g(O) = f(a) = woo That is, BCa; IE) c A and A is also open. II Now differentiate fez) = eZ; we do this by Proposition 2.5. This gives that

!'(z)

=

~ ~ zn-l

n~ln!

=

i

_1_ zn-I

n~l(n-l)!

=

i

~ z" =f(z).

.~on!

Thus the complex exponential function has the same property as its real counterpart. That is d

2.12

--eZ=e" dz

38


Putg(z) = eZea-Zfor some fixed ain C; theng'(z) = eZea-z+eZ(_ea-z) = O. Hence g(z) = w for all z in C and some constant w. In particular, using eO = 1 we get w = g(O) = ea. Then eZe a- z = ea for all z. Thus ea +b = eae b for all a and b in IC. This also gives 1 = eZe- Z which implies that eZ # 0 for any z and e- = lje z • Returning to the power series expansion of et , since all the coefficients of this series are real we have exp z = exp z. In particular, for (J a real number we get le wl2 = eiOe- iO = eO = l. More generally, le z I2 = eZe' = ezH = exp (2 Re z). Thus, Z

2.13

iexp z! = exp (Re z).

We see, therefore, that e has the same properties that the real function eX has. Again by analogy with the real power series we define the functions cos z and sin z by the power series Z

cos

Z

= 1-

Z2

2!

Z4

z2n

4!

(2n)!

+- + ... + ( - 1 ) " - + _..

_3

sm z = z

-

3!

z5

+

5!

7

2n- 1

+ ... +(-1)"( -

) + ...

2n - 1 !

Each of the series has infinite radius of convergence and so cos z and sin z are analytic in IC. By using Proposition 2.5 we find that (cos z)' = - sin z and (sin z)' = cos z. By manipulating power series (which is justified since these series converge absolutely) 2.14

- z = 2i 1 (iz sm e _... e - iz)

cos z = J( e iz + e - iz)

This gives for z in C, cos 2 z + sin 2 z 2.15 In particular if we let z Hence, for z in IC

e iz =

=

=

1 and

cos z+ i sin z.

a real number

(J

in (2.15) we get e iO = cis

(J.

2.16 where B=argz. Since eX+IY=exe'Y we have lezl=exp(Rez) and arge z = Imz.

A functionfisperiodic with period c iff(z+c)=f(z) for all z in IC. If c is a period of e Z then eZ=ez+c=ezec implies that e C = 1. Since I =lecl= expRe(c), Re(c)=O. Thus c=i(} for some B in IR. But !=ec=eil1=cosB+ i sin () gives that the periods of e Z are the multiples of 2wi. Thus, if we divide the plane into infinitely many horizontal strips by the lines Imz = w(2k - I), k any integer, the exponential function behaves the same in each of these strips. This property of periodicity is one which is not present in the real exponential function. Notice that by examining complex functions we have demonstrated a relationship (2.15) between the exponential function and the trigonometric functions which was not expected from our knowledge of the real case.

Analytic functions

39

Now let us define log z. We could adopt the same procedure as before and let log z be the power series expansion of the real logarithm about some point. But this only gives log z in some disk. The method of defining the logarithm as the integral of t- 1 from 1 to x, x > 0, is a possibility, but proves to be risky and unsatisfying in the complex case. Also, since e" is not a one-one map as in the real case, log z cannot be defined as the inverse of e". We can, however, do something similar. We want to define log w so that it satisfies w = e" when z = log w. Now since eZ =f. 0 for any z we cannot define log O. Therefore, suppose e" = w and w =f. 0; if z = x+iy then Iwl = e" and y = arg w+211'k, for some k. Hence 2.17

{log Iwl +i(arg w+211'k): k is any integer}

is the solution set for e"

= w. (Note that log Iwl is the usual real logarithm.)

2.18 Definition. If G is an open connected set in C and /: G -+- C is a continuous function such that z = exp /(z) for all z in G then / is a branch 0/ the logarithm. Notice that 0, G. Suppose / is a given branch of the logarithm on the connected set G and suppose k is an integer. Let g(z) = /(z) +211'ki. Then exp g(z) = exp/(z) = z, so g is also a branch of the logarithm. Conversely. if/and g are both branches of log z then for each z in G,/(z) = g(z) +211'ki for some integer k, where k depends on z. Does the same k work for each z in G? The answer is

yes. In fact, if h(z)

1 .[/(z)-g(z)] = -2 11'1

then h is continuous on G and h(G)

C I, the integers. Since G is connected, h(G) must also be connected (Theorem II. 5.8). Hence there is a k in I with/(z) + 211'ki = g(z) for all z in G. This gives

2.19 Proposition. If G c C is open and connected and f is a branch of log z on G then the totality 0/ branches 0/ log z are the functions /(z) + 211'ki, k e Z. Now let us manufacture at least one branch of log z on some open connected set. Let G=C-{ZEIR:Z:S:O}; that is, "slit" the plane along the negative real axis. Clearly G is connected and each z in G can be uniquely represented by z = Izle" where -11' < 8 < 11'. For 8 in this range, define f(re " ) = log r+i8. We leave the proof of continuity to the reader (Exercise 9). It follows that/is a branch ofthe logarithm onG. Is/analytic? To answer this we first prove a general fact. 2.20 Proposition. Let G and n be open subsets of C. Suppose that /: G -+- C and g: n -+- C are continuous functions such that /(G) c nand g(f(z» = z for all z in G. If g is differentiable and g'(z) =f. 0, / is differentiable and

I'(z) =

g'~(Z»

40


If g is analytic,

f

is analytic.

C such that II # 0 and a+h g(f(a» and a+1I = g(f(a+h» impliesf(a) # f(a+h). Also Proof. Fix a in G and let

hE

g(f(a+

E

G. Hence a

=

h» - g(f(a»

------""-------------_.. _-

h

g(j{a+h»-g(f(a» ---. __

.----------------- - ---

f(a +h) -./(a)

---

. f(a+h)-f(a) --------h

Now the limit of the left hand side as h --+ 0 is, of course, 1; so the limit of the right hand side exists. Since lim [/(a+I1)-f(a)J = 0, h~O

. g(f(a + 11» - g(f(a» = g ,(j(a». . Itm ............................. f(a +h) - ./(a)

h·. 0

Hence we get that . l(a+17)-/(a)

I!m--·--··-

h~O h exists since g'(f(a» # 0, and I = g'(j(a»f'(a). Thus,f'(z) = [g'(f(z))J -1. If g is analytic then g' is continuous and this gives that f is analytic. III

2.21 Corollary. A branch of the logarithm function is analytic and its derivative is

Z-I.

We designate the particular branch of the logarithm defined above on

C - {z E ~: z< O} to be the principal branch of the logarithm. If we write log z as a function we will always take it to be the principal branch of the logarithm unless otherwise stated. Iff is a branch of the logarithm on an open connected set G and if b in C is fixed then deline g: G --+ C by g(z) = exp (bICz». If b is an integer, then g(z) = z/'. In this manner we define a branch of Zb, b in C, for an open connected set on which there is a branch of log z. If we write g(z) = Zb as a function we will always understand that Zb = exp (b log z) where log z is the principal branch of the logarithm; Zb is analytic since log z is. As is evident from the considerations just concluded, connectedness plays an important role in analytic function theory. For example, Proposition 2.W is false unless G is connected. This is analogous to the role played by intervals in calculus. Because of this it is convenient to introduce the term "region." A region is an open connected subset of the plane. This section concludes with a discussion of the Cauchy-Riemann equations. Let f: G --+ C be analytic and let u(x, y) = Re f(x + iy), vex, y) = 1m ./(x+iy) for x+iy in G. Let us evaluate the limit

f

, . I(z+h) - fez) (z) = hm -------h-O h

in two different ways. First let h and h real we get

--+

0 through real values of h. For h # 0

41

Analytic functions

f(z+h)-f(z) h

f(x+h+iy) -f(x+iy) h

= Letting h

-+

u(x+h, y)-u(x, y) h

. v(x+h, y)-v(x, y) h

+1

0 gives ou F(z) = ox (x, y)

2.22

Now let h h real,

-+

ov

+ i ox (x, y)

0 through purely imaginary values; that is, for h t= 0 and

f(z+ ih) - fez) ih

. u(x, y+h)-u(x, y) h

-l

+

vex, y+h)-v(x, y) h

Thus, 2.23

F(z)

=

ou (x, y) oy

-i -

ov (x, y) oy

+-

Equating the real and imaginary parts of (2.22) and (2.23) we get the Cauchy-Riemann equations ou ox

2.24

ov oy

-=-

and

ou oy

ov ox

-=--

Suppose that u and v have continuous second partial derivatives (we will eventually show that they are infinitely differentiable). Differentiating the Cauchy-Riemann equations again we get

Hence, 02U

2.25

o2U

-+--0 8x 2 oy2 - .

Any real-valued function with continuous second derivatives satisfying (2.25) is said to be harmonic. In a similar fashion, v is also harmonic. We will study harmonic functions in Chapter X. Let G be a region in the plane and let u and v be functions defined on G with continuous partial derivatives. Furthermore, suppose that u and v satisfy the Cauchy-Riemann equations. Ifj(z)=u(z)+iv(z) thenjcan be shown to be analytic in G. To see this, let z = x + ry E G and let B (z; r) c G. If h=s+it E B(O;r) then u(x+s, y+t)-u(x, y) = [u(x+s, y+t)-u(x, y+t)]+[u(x, y+t)-u(x, y)]

Applying the mean value theorem for the derivative of a function of one variable to each of these bracketed expressions, yields for each s+it in B(O; r) numbers SI and 11 such that ISll < lsi and Itt! < It I and 2.26

{

u(x+s, y+t) -u(x, y+t) = u..,(x+s 1 , y+t)s u(x, y+t)-u(x, y) = u,(x, y+( 1 )t


42

Letting
(2.26) gives that ((!(s,t) s _ t s+ir = s+lt [ux(x+sj,y+t)-ux(x.y)]+ s+it [uy(x,y+tj)-uy(x,y)]

But lsi ~ Is+itl, It I ~ is+iti, are continuous gives that

IS11

<

lsi, It11

< Itl, and the fact that

Ux

and uy

.
2.27 Hence

u(X+s, y+t)-u(x, y)

=

uxCx, y)s+uy(.x, y)t+
where cp satisfies (2.27). Similarly vex + s, y + t) - vex, y) = v.,(x, y)s --1-- Vy(x, y)t + .p(s, t)

where 2.28

.p satisfies . .p(s, t) lIm-- = 0

s+i(~O

s+it

Using the fact that u and v satisfy the Cauchy-Riemann equations it is easy to see that f(z+s+it)-f(z) -----:----s+zt

=

. ux(z)+wx(z)

+

In light of (2.27) and (2.28), f is differentiable and 1"(z) = ux(z) + iv.Jz). Since U x and Vx are continuous, 1" is continuous and f is analytic. These results are summarized as follows.

2.29. Theorem. Let u and v be real-valued functions defined on a region G and suppose that u and v have contilluous partial derivatives. Then f: G --;- IC defined by fez) = u(z) + iv(z) is analytic iff u and v satisfy the Cauchy-Riemann equations. Example. Is u(x, y) = log (Xl + y2)t harmonic on G = IC - {O}? The answer is yes! This could be shown by differentiating u to see that it satisfies (2.25). However, it can also be shown by observing that in a neighborhood of each point of G, u is the real part of an analytic function defined in that neighborhood. (Which function?) Another problem concerning harmonic functions which will be taken up in more detail in Section VIII. 3, is the following. Suppose G is a region in the plane and u: G --;- IR is harmonic. Does there exist a harmonic function v: G --;- IR such thatf = u+ iv is analytic in G? If such a function v exists it is called a harmonic conjugate of u. If VI and V2 are two harmonic conjugates of u then i(v j -v 2 ) = (u+ivj)-(U+iv2) is analytic on G and only takes on

Analytic functions

43

purely imaginary values. It follows that two harmonic conjugates of a harmonic function differ by a constant (see Exercise 14). Returning to the question of the existence of a harmonic conjugate, the above example u(z) = log Izl of a harmonic function on the region G = C{O} has no harmonic conjugate. Indeed, if it did then it would be possible to define an analytic branch of the logarithm on G and this cannot be done. (Exercise 21.) However, there are some regions for which every harmonic function has a conjugate. In particular, it will now be shown that this is the case when G is any disk or the whole plane. 2.30 Theorem. Let G be either the whole plane C or some open disk. u: G -7 IR is a harmonic function then u has a harmonic conjugate.

If

Proof To carry out the proof of this theorem, Leibniz's rule for differentiating under the integral sign is needed (this is stated and proved in Proposition IV. 2.1). Let G = B(O; R), 0 < R :-:; 00, and let u: G -7 IR be a harmonic function. The proof will be accomplished by finding a harmonic function v such that u and v satisfy the Cauchy-Riemann equations. So define

fo uix, t)dt+cp(x) y

vex, y) =

and determine cp so that Vx = -uy. Differentiating both sides of this equation with respect to x gives

f uxix, t) dt+tp'(x) y

vix, y) =

o

-f Uyy(x, t) dt+cp'(x) y

o

-u/x, y)+u/x, 0) + cp'(x)

So it must be that cp'(x) = - uy(x, 0). It is easily checked that u and

fo Ux(X, t)dt- f u/s, O)ds x

y

vex, y)

=

0

do satisfy the Cauchy-Riemann equations . • Where was the fact that G is a disk or C used? Why can't this method of proof be doctored sufficiently that it holds for general regions G? Where does the proof break down when G = C- {O} and u(z) = log Izl? Exercises 1. Show that f(z) = Izl2 = x 2 + y2 has a derivative only at the origin. 2. Prove that if bn, an are real and positive and 0 < b = lim bn < 00, a = lim sup an then ab = lim sup (a nbn). Does this remain true if the requirement of positivity is dropped? 3. Show that lim nl/n = 1.

44


4. Show that (cos z)' = - sin z and (sin z)' = cos z. 5. Derive formulas (2.14). 6. Describe the following sets: {z: eZ = i}, {z: eZ = -I}, {z: eZ = {z: cos z = O}, {z: sin z = O}. 7. Prove formulas for cos (z+w) and sin (z+w). sin z ... . 8. Define tan z = - - ; where IS thIS functIon defined and analytIc? cos z 9. Suppose that Zn' Z E G = IC - {z: z :=; O} and Zn = rne'·n, z = re i8 where -71" < 8, 8n < 71". Prove that if Zn -+ z then en -+ e and rn -+ r. 10. Prove the following generalization of Proposition 2.20. Let G and Q be open in IC and suppose f and h are functions defined on G, g : Q-+C and suppose that f( G) c Q. Suppose that g and h are analytic, g'(w)*O for any w, thatfis continuous, h is one-one, and that they satisfy h(z)=g(f(z» for Z in G. Show thatf is analytic. Give a formula for F(z). I J. Suppose that f: G -+ IC is a branch of the logarithm and that n is an integer. Prove that zn = exp (nf(z» for all z in G. 12. Show that the real part of the function z; is always positive. 13. Let G = C - {z E ~: Z :=; O} and let n be a positive integer. Find all analytic functions f: G ~ C such that z = (f(z»" for all z E G. 14. Suppose f: G ~ C is analytic and that G is connected. Show that if f( z) is real for all z in G then f is constant.

-n,

15. For

r > 0 let A = {w: w= exp (D where 0 < Izi < r} ;determine the

set A. 16. Find an open connected set G c IC and two continuous functions f and g defined on G such thatf(z)2 = g(Z)2 = 1-z2 for all z in G. Can you make G maximal? Are f and g analytic? 17. Give the principal branch of .jI-z. 18. Letf: G -+ IC and g: G -+ IC be branches of za and Zb respectively. Show thatfg is a branch of za+b andf/g is a branch of za-b. Suppose thatf(G) c G and g( G) c G and prove that both fog and g f are branches of zab. 19. Let G be a region and define G* = {z: Z E G}. Iff: G -+ IC is analytic prove that f*: G* -+ C, defined by /*(z) = fez), is also analytic. 20. Let z l' Z 2' ... , z" be complex numbers such that Re z k > 0 and Re(zl ... zk»O for l:=;k:=;n. Show that log(zt ... z,,)=logzt + ... + log z,,' where log z is the principal branch of the logarithm. If the restrictions on the z k are removed, does the formula remain valid? 21. Prove that there is no branch of the logarithm defined on G = C - {O}. (Hint: Suppose such a branch exists and compare this with the principal branch.) 0

§3. Analytic functions as mappings. Mobius transformations Consider the function defined by fez) = Z2. If z = x+iy and /L+iv=f(z) then /L = x 2 - y2, v = 2xy. Hence, the hyperbolas x 2 - y2 = c and 2xy = d are mapped by f into the straight lines /L = c, v = d. One interesting fact is

Analytic functions as mappings. Mobius transformations

45

that for c and d not zero, these hyperbolas intersect at right angles, just as their images do. This is not an isolated phenomenon and this property will be explored in general later in this section. Now examine what happens to the lines x = c and y = d. First consider x = c(y arbitrary) ;fmaps this line into fL = c 2 - y2 and v = 2cy. Eliminating y we get that x = c is mapped onto the parabola v2 = - 4C 2 (fL - ( 2 ). Similarly, 1 takes the line y = d onto the parabola v 2 = 4d 2 (fL + d 2 ). These parabolas intersect at (c 2 - d 2 , ± 21 cd!). It is relevant to point out that as c ~ 0 the parabola jJ2 = -4c 2 (1l - c 2 ) gets closer and closer to the negative real axis. This corresponds to the fact that the function z~ maps G = C - {z EO IR: z~o} onto {z: Rez>O}. Notice also that x=c and x= -c (and y = d. y = - d) are mapped on to the same parabolas. What happens to a circle centered at the origin? If z = re iB then I(z) = 2 2 r e ;8; thus, the circle of radius r about the origin is mapped onto the circle of radius r2 in a two to one fashion. Finally, what happens to the sector Sea, (3) = {z: a < arg z < {3}, for a < f3? It is easily seen that the image of Sea, (3) is the sector S(2a, 2(3). The restriction of 1 to S(a, (3) will be one-one exactly when f3 - a < 1T. The above discussion sheds some light on the nature of I( z) = Z 2 and, likewise, it is useful to study the mapping properties of other analytic functions. In the theory of analytic functions the following problem holds a paramount position: given two open connected sets G and Q, is there an analytic function I defined on G such that I( G) = Q? Besides being intrinsically interesting, the solution (or rather, the information about the existence of a solution) of this problem is very useful. 3.1. Definition. A path in a region Gee is a continuous function y: [a, b] _.-> G for some interval [a, bJ in IR. If y'(t) exists for each t in [a, b] and y': [a, b] -> C is continuous then y is a smooth path. Also y is piecewise smooth if there is a partition of [a, b], a = to < t[ < ... < t" = b, such that y is smooth on each subinterval [f}_I' 1 ~j ~ n. To say that a function y: la, b] -> C has a derivative y'(t) for each point t in [a, b] means that

tJ,

.

y{l+h)-y(t)

l~ _·······)-1- _... =Y'(l)

exists for a < t < b and that the right and left sided limits exist for t = a and t = b, respectively. This is, of course, equivalent to saying that Rey and 1m y have a derivative (see Appendix A). Suppose y:[a,b]-,G is a smooth path and that for some to in (a,b), y'(to) 7'" O. Then y has a tangent line at the point zo= yUo). This line goes through the point Zo in the direction of (the vector) y'(lo); or, the slope of the line is tan(argy'(to». If y, and Y2 are two smooth paths with y,«(,)= Y2(t2)=zO and y;(I,)7"'O, y;(l2)7"'0, then define the angle between the paths y, and Y2 at Zo to be


46

Suppose y is a smooth path in G andJ: G-C is analytic. Then (1= Joy is also a smooth path and (1'(t)=f'(y(t»y'(t). Let zo=y(to), and suppose that y'(to)*O and f'(zo)*O; then (1'(to)*O and arg(1'(to)=argJ'(zo)+ argy'(to). That is,

3.2

arg(1'(to) - arg y'( to) = argJ'(zo)·

Now let YI and Y2 be smooth paths with YI(t I)= yi t 2) = Zo and y;(tI)* 0* y;(t2); let (11 = JOYI and (12 = J O Y2. Also, suppose that the paths YI and Y2 are not tangent to each other at zo; that is, suppose y;UI)*y;(t2). Equation (3.2) gives

3.3 This says that given any two paths through z 0, f maps these paths onto two paths through Wo = f(zo) and, whenJ'(zo) 'f 0, the angles between the curves are preserved both in magnitude and direction. This summarizes as follows. 3.4 Theorem. Iff: G - IC is analytic then f preserves angles at each point Zo of G where J'(zo) 'f O. A functionf: G - IC which has the angle preserving property and also has . If(z) - f(a) I I1m z-a Iz-al

existing is called a conformal map. If f is analytic and J'(z) 'f 0 for any z then f is conformal. The converse of this statement is also true. If fez) = e Z then f is conformal throughout IC; let us look at the exponential function more closely. If z = c+iy where c is fixed thenf(z) = re iy for r = eC • That is, f maps the line x = c onto the circle with center at the origin and of radius eC • Also, f maps the line y = d onto the infinite ray {reid: 0 < r < co} .

--- ---.

."i

di

e'

-------

~-.-....:

-Tr'

----

We have already seen that e Z is one-one on any horizontal strip of width <2'11. Let G={z: -'I1

47

the circle {e(e ill : -17 < () < 17}, and the horizontal line y = d, -17 < d < 17, goes onto the ray making an angle d with the positive real axis. Notice that log z, the principal branch of the logarithm, does the opposite. It maps n onto the strip G, circles onto vertical segments in G, rays onto horizontal lines in G. The exploration of the mapping properties of cos z, sin z, and other analytic functions will be done in the exercises. We now proceed to an amazing class of mappings, the Mobius transformations. 3.5 Definition. A mapping of the form S(z) = az+db is called a linear fraeez+ tional transformation. If a, b, c, and d also satisfy ad-be 1= 0 then S(z) is called a Mobius transformation. dz-b If S is a Mobius transformation then S-I(Z) = satisfies -ez+a S(S -I(Z» = S -I(S(Z» = z; that is, S -I is the inverse mapping of S. If Sand T are both linear fractional transformations then it follows that SoT is also. Hence, the set of Mobius maps forms a group under composition. Unless otherwise stated, the only linear fractional transformations we will consider are Mobius transformations. Let S(z) = az+db ; if ,\ is any non-zero complex number, then ez+ S(z)

= ('\a)z+(Ab) . ('\e)z +(M)

That is, the coefficients a, b, e, d are not unique (see Exercise 20). We may also consider S as defined on Coo with S( (0) = ale and S( -die) = 00. (Notice that we cannot have a = 0 = e or d = 0 = e since either situation would contradict ad-be 1= 0.) Since S has an inverse it maps Coo onto Coo. If S(z) = z + a then S is called a translation; if S(z) = az with a > 0 then S is a dilation; if S(z) = eillz then it is a rotation; finally, if S(z) = liz it is the inversion. 3.6 Proposition. If S is a Mobius transformation then S is the composition of translations, dilations, and the inversion. (Of course, some of these may be missing.) Proof. First, suppose e = O. Hence S(z) = (ald)z+(bld) so if Sl(Z) = (ald)z, S2(Z) = z+(bld), then S2 0 Sl = S and we are done. (be-ad) Now let e 1= 0 and put Sl(Z) = z+d/e, S2(Z) = liz, S3(Z) = 2 z,

e

S4(Z) = z+a/e. Then S = S4 0 S3 0 S2° Sl' • What are the finite fixed points of S? That is, what are the points z satisfying S(z)=z. If z satisfies this condition then ez 2+{d-a)z-b = O.

48


Hence, a Mobius transformation can have at most two fixed points unless S(z)=z for all z. (Why?) Now let S be a Mobius transformation and let a, b, c be distinct points in Coo with IX = S(a), f3 = S(b), y = S(c). Suppose that T is another map with this property. Then T- 1 S has a, b, and c as fixed points and, therefore, T- 1 S = I = the identity. That is, S = T. Hence, a Mobius map is uniquely determined by its action on any three given points in Coo. Let Z2' Z3' Z4 be points in COO" Define S: Coo -+ Coo by 0

0

Z-Z3 Z-Z4

S(z)

= ~

S(z)

=--

S(z)

=--

if Zz

= 00;

Z2- Z4 if Z3 = Z-Z4

00',

Z-Z3 Z2 -Z3

00.

if Z4

=

In any case S(zz) = I, S(Z3) = 0, S(Z4) = tion having this property.

00

and S is the only transforma-

3.7 Definition. If Z1 E C DC then (z l' Z2' Z 3' Z4) (The cross ratio of ZI' Z2' Z3' and Z4) is the image of Zl under the unique Mobius transformation which takes z2 to 1, Z3 to 0, and Z4 to 00. For example: (Z2' Z2' Z3' Z4) = 1 and (z, 1,0, (0) = z. Also, if M is any Mobius map and Wb W3' 1V 4 are the points such that MW2 = I, MWj = 0, MW4 = 00 then Mz = (z, IVz, W3' 1V 4 ). 3.8 Proposition. formation then for any point

If Z2' Z3' Z4 are distinct points and T is any Mobius trans-

Z l'

Proof Let Sz = (z, ZZ, Z3' Z4); then S is a Mobius map. If M = ST- 1 then M(Tz z ) = I, M(Tz3) = 0, M(Tz 4) = 00; hence, ST-1z = (z, Tz 2, Tz 3, Tz 4) for all z in COO" In particular, if z = Tz I the desired result follows . • 3.9 Proposition. If z 2, Z3, Z4 are distinct points in Coo and W2, w 3, w4 are also distinct points of Coo, then there is one and only one Mobius transformation S such that SZz = W2, SZ3 = W3' SZ4 = W4 Proof Let Tz = (z, Z2' Z3' Z4), Mz = (z, wz, W3' w4) and put S = M-1T. Clearly S has the desired property. If R is another Mobius map with Rz j = W j for j = 2, 3,4 then R- 1 S has three fixed points (zz, Z3' and Z4)' Hence R -loS = I, or S = R . • It is well known from high school geometry that three points in the plane determine a circle. (Recall that a circle in Coo passing through 00 corresponds to a straight line in Co Hence there is no need to inject in the previous state0


49

ment the word "non-colinear". A straight line in the plane will be called a circle. The next result explains when four points lie on a circle. 3.10 Proposition. Let Z 3,

ZI' zz, Z3, Z4 be four distinct points in Coo. Then Z4) is a real number iff all four points lie on a circle.

(ZI,

Z2'

Proof Let S: Coo -+ Coo be defined by Sz = (z, Z2' Z3' Z4); then S-I(!R) = the set of z such that (z, Zz, Z3' Z4) is real. Hence, we will be finished if we can show that the image of !Roo under a Mobius transformation is a circle.

Let Sz

=

az

+b if + d'

---'

cz

Z =

x

E

!R and w

=

S-l(X) =F

00

then x

=

Sw

implies that S( w) = S( w ). That is, aw+b cw+d

dw+b cw+d

-----

Cross mUltiplying this gives 3.11

(ac-iic) Iwlz+(ad-bc)w+(bc-dii)w+(bd-bd)

If ac is real then ac-iic = 0; putting multiplying (3.11) by i gives 3.12

IX

= O.

= 2(ad-bc), f3 = i(bd-hd) and

0= 1m (lXw)-f3 = 1m (lXw-f3)

since f3 is real. That is, w lies on the line determined by (3.12) for fixed ex and f3. If ac is not real then (3.11) becomes IwI2+yw+yw-1)

for some constants y in C,

I)

in !R. Hence,

Iw+yl =

3.13

=0

A

where A=

ad-bcl (lyl2 +I)t = l-_ -_ ac-ac

> O.

Since y and A are independent of x and since (3.13) is the equation of a circle, the proof is finished . • 3.14 Theorem. A Mobius transformation takes circles onto circles. Proof Let r be any circle in Coo and let S be any Mobius transformation. Let Z2' Z3' Z4 be three distinct points on r and put Wj = SZj forj = 2,3,4. Then W2, W3, W4 determine a circle r'. We claim that S(r) = r'. In fact, for any Z in Coo

3.15 by Proposition 3.8. By the preceding proposition, if Z is on r then both sides of (3.15) are real. But this says that Sz E r' . • Now let rand r' be two circles in Coo and let Z2' Z3' Z4 E r; W2, W3, W4 E r'. Put Rz = (z, Z2, Z3' Z4), Sz = (z, W2' W3, w4)' Then T = S-l 0 R maps

50


I' onto 1". In fact, Tz j = follows that T(r) = 1".

for j = 2, 3, 4 and, as in the above proof, it

W j

3.16 Proposition. For any given circles I' and I" in CoO there is a Mobius transformation T such that T(r) = I". Furthermore we can specify that T take any three points on r onto any three points of I". if we do specify Tz j for j = 2, 3, 4 (distinct Z j in r) then T is unique. Proof The proof, except for the uniqueness statement, is given in the previous paragraph. The uniqueness part is a trivial exercise for the reader. II Now that we know that a Mobius map takes circles to circles, the next question is: What happens to the inside and the outside of these circles? To answer this we introduce some new concepts.

3.17 Definition. Let r be a circle through points Z2, z* in CO) are said to be symmetric with respect to I' if

Z3' Z4'

The points z,

3.18 As it stands, this definition not only depends on the circle but also on the points Z2' z 3' Z4-,It is left as an exercise for the reader to show that symmetry is independent of the points chosen (Exercise 11). Also, by Proposition 3.10 z is symmetric to itself with respect to r if and only if Z E r.

/

/

/

I

/

z !'

Let us investigate what it means for z and Z* to be symmetric. If r is a straight line then our linguistic prejudices lead us to believe that z and z* are symmetric with respect to r if the line through z and z* is perpendicular to rand z and z* are the same distance from r but on opposite sides of r. This is indeed the case. If r is a straight line then, choosing Z4 = 00, equation (3.18) becomes Z*-Z3 Z2- Z 3

This gives Iz*-z31

=

Z-Z3

2 2 -2 3

Iz - z31; since z3 was not specified, we have that z


and z* are equidistant from each point on

r.

51

Also

Z*-Z3 Z -Z3 Im--=Im-Z2 -Z3 Z2 -Z3

3- Z 3

-1m

Z2- Z 3

Hence, we have (unless Z E r) that Z and z* lie in different half planes determined by r. It now follows that [z, z*] is perpendicular to r. Now suppose that r = {z: lz-al = R} (0 < R < 00). Let Z2' Z3, Z4 be points in r; using (3.18) and Proposition 3.8 for a number of Mobius transformations gives

(z*, Z2, Z3' Z4)

=

(z, Z2' Z3, Z4)

=

(z-a, zz-a, Z3-a, Z4-a)

=

(z-a, z:~~, z:~~, ;:~~)

=

(_R2_, z-a

=

(~R2~. + a, z-a

zz-a, Z3- a, Z4- a) Z2,

Z3, Z4)

Hence, z* = a+ R2(Z -a) -1 or (z* -a)(z -a) = R2. From this itfollows that

R2

z*-a ...- - ~-- > 0 z-a - Iz-al 2 '

-.~

so that z* lies on the ray {a+t(z-a): 0 < t < oJ} from a through z. Using the fact that Iz - al lz* - al = R2 we can obtain z* from z (if z lies inside r) as in the figure below. That is: Let L be the ray from a through z. Construct

a--~----~------------~~z* z

a line P perpendicular to L at z and at the point where P intersects r construct the tangent to r. The point of intersection of this tangent with L is the point z*. Thus, the points a and Cf) are symmetric with respect to r. 3.19 Symmetry Principle. If a Mijbius transformation T takes a circle 1'1 onto the circle 1'2 then any pair of points symmetric with respect to 1'1 are mapped by T onto a pair of points symmetric with respect to r 2'

52


Proof Let =2, Z 3, respect to r 1 then

Z4 E

r 1;

it follows that if z and z* are symmetric with

(Tz*, Tz 2 , Tz 3 , Tz 4 ) = (z*,

=

Z2, Z3, Z4)

(Tz, Tz 2 , Tz 3 , Tz 4 )

by Proposition 3.8. Hence Tz* and Tz are symmetric with respect to r 2 • • Now we will discuss orientation for circles in Coo; this will enable us to distinguish between the "inside" and "outside" of a circle in Coo. Notice that on Coo (the sphere) there is no obvious choice for the inside and outside of a circle. 3.20 Definition. If r is a circle then an orientation for r is an ordered triple of points (ZI' Z2, Z3) such that each Zj is in r. Intuitively, these three points give a direction to r. That is we "go" from ZI to Z2 to Z3. If only two points were given, this would, of course, be ambiguous. az+b Let r = ~ and let z\> Z2' Z3 E~; also, put Tz = (z, ZI' Z2, Z3) = - - . ez+d Since T(~oo) = ~oo it follows that a, b, e, d can be chosen to be real numbers (see Exercise 8). Hence,

az+b Tz=-ez+d az+b lez+dl 2

•

(ez+d)

1

lez+dl 2 [aelzI 2 +bd+bez+adz]

Hence, 1m (z,

(ad-be)

ZI' Z2' Z3)

= lez+dl 2 1m z.

Thus, {z: 1m (z, z l' Z 2, z 3) < o} is either the upper or lower half plane depending on whether (ad-be) > 0 or (ad-be) < O. (Note that ad-be is the "determinant" of T.) Now let r be arbitrary, and suppose that z[, Z2' Z3 are on r; for any Mobius transformation S we have (by Proposition 3.8) {z: 1m (z,

Z[, Z2' Z3)

> O} = {z: 1m (Sz,

=

S-1

SZI' SZ2, SZ3)

{z: 1m (z,

> O}

SZI' SZ2, SZ3)

> O}

In particular, if S is chosen so that S maps r onto ~('" then {z: 1m (z, ZI' z 2, z 3) > O} is equal to S - 1 of either the upper or lower half plane. If (ZI' Z2' Z3) is an orientation of r then we define the right side of r (with respect to (ZI' Z2' Z3» to be {z: 1m (z,

ZI' Z2' Z3)

> O}.


53

Similarly, we define the left side of l' to be

{z: 1m (z,

ZI' Z2, Z3)

< O}.

The proof of the following theorem is left as an exercise.

3.21 Orientation Principle. Let 1'1 and 1'2 be two circles in C oc and let T be a Mobius tran.~formation such that T(1' 1) = 1'2' Let (Zl, Z2, Z3) be an orientation for r l ' Then T takes the right side and the left side of 1\ onto the right side and left side of r 2 Ivith respect to the orientation (Tz 1, Tz 2 , Tz 3)' Consider the orientation (l, 0, co) of R By the definition of the cross ratio, (z, 1, 0, (0) = z. Hence, the right side of IR with respect to (1, 0, (0) is the upper half plane. This fits our intuition that the right side lies on our right as we walk along IR from 1 to 0 to co. As an example consider the following problem: Find an analytic function f: G -+ IC, where G = {z: Re z > O}, such thatf(G) = D = {z: Izl < I}. We solve this problem by finding a Mobius transformation which takes the imaginary axis onto the unit circle and, by the Orientation Principle, takes G onto D (that is, we must choose this map carefully in order that it does not send G onto {z: Izl > 1 If we give the imaginary axis the orientation ( - i, 0, i) then {z: Re z > O} is on the right of this axis. In fact,

n.

2z

(z, -I, 0, i) = -~; Z-l

2z z+i

z-iz+i 2

= -- .

Iz-W

(!zI2+iz) I

Hence, {z: 1m (z, -I, 0, i) > O} = {z: 1m (iz) > O} = {z: Re z > O}. Giving r the orientation (-i, -1, i) we have that D lies on the right of r. Also, 2i z+ I . (z - i -1 i) = - ..~...~- . , , , i-I z-i If 2z. Sz = Z-l

and

Rz

( ;2i- ) =

1-1

(zz-z+

1) --.

then T = R - 1 S maps G onto D (and the imaginary axis onto f). By algebraic manipUlations we have

z-l z+1

Tz=-····

e"-I

Combining this with previous results we have that g(z) = ---1 maps

e-+

the infinite strip {z: 11m zl < 71"/2} onto the open unit disk D. (It is worth e"-1 mentioning that eX+! = tanh (z/2).)

S4


Let G 1, G 2 be open connected sets; to try to find an analytic functionJ such thatJ(G 1) = Gz we try to map both G 1 and G z onto the open unit disk. If this can be done, J can be obtained by taking the composition of one function with the inverse of the other. /L

f,

/

/

/

/

/

/

/

As an example, let G be the open set inside two circles r 1 and r 2, intersecting at points a and b (a t= b). Let L be the line passing through a and b and give L the orientation (00, a, b). Then Tz = (z, 00, a, b) =

(z -

a) z-b maps L onto the real axis (Too = 1, Ta = 0, Tb = 00). Since T must map circles onto circles, T maps fl and f2 onto circles through and 00. That is, T(f l ) and T(f2) are straight lines. By the use of orientation we have that T( G) = {w: - a < arg w < a} for some a > 0, or the complement of some such closed sector. By the use of an appropriate power of z and possibly a rotation we can map this wedge onto the right half plane. Now, composing with the map (z - l)(z + 1)-1 gives a map of G onto D = {z:lzl < I}.

°

Exercises 1. Find the image of {z: Re z < 0, 11m zl < 7T} under the exponential function. 2. Do exercise 1 for the set {z: 11m zl < 7T/2}. 3. Discuss the mapping properties of cos z and sin z. 4. Discuss the mapping properties of z· and zl/n for n ~ 2. (Hint: use polar coordinates. ) 5. Find the fixed points of a dilation, a translation and the inversion on Coo. 6. Evaluate the following cross ratios: (a) (7 + i, I, 0, 00) (b) (2, 1- i, 1, 1 + i) (c)(O, 1, i, -I)(d)(i-l, 00, l+i, 0). az+b . 7. If Tz = - - find Z2' Z3' Z4 (m terms of a, b, c, d) such that Tz = (z, cz+d


az+b B. If Tz = - - show that T(IR",) cz+d real numbers.

55

.

= IR", Iff we can choose a, b, c, d to be

az+b fi nd necessary an d su ffi' . 9. If Tz = --d' Clent cond'Itl0ns that T(r) = r cz+ where r is the unit circle {z: Izl = I}. 10. Let D = {z: Izl < I} and find all Mobius transformations T such that T(D) = D.

11. Show that the definition of symmetry (3.17) does not depend on the choice of points Z2' Z3' Z4' That is, show that if W2, W3, W4 are also in r then equation (3.1B) is satisfied iff (z*, W2, W3, w4) = (z, W2, W3, w4)' (Hint: Use Exercise B.) 12. Prove Theorem 3.4. 13. Give a discussion of the mappingf(z) = -!(z+ liz). 14. Suppose that one circle is contained inside another and that they are tangent at the point a. Let G be the region between the two circles and map G conformally onto the open unit disk. (Hint: first try (z - a)-I.) 15. Can you map the open unit disk conformally onto {z :0< Izl < l}? 16. Map G = C- {z: -1 ::; z ::; I} onto the open unit disk by an analytic function f Can f be one-one? 17. Let G be a region and suppose that f: G - C is analytic such that f( G) is a subset of a circle. Show that f is constant. z-ia lB. Let - 00 < a < b < 00 and put Mz = - . . Define the lines Ll = z-lb {z: 1m z = b}, L2 = {z: 1m z = a} and L3 = {z: Re z = O}. Determine which of the regions A, B, C, D, E, F in Figure 1, are mapped by M onto the regions U, V, W, X, Y, Z in Figure 2.

A

ib

D

u

E

B

w

o

z

x ia

C

F

Figure 1

Figure 2

19. Let a, b, and M be as in Exercise 18 and let log be the principal branch of the logarithm.

56


(a) Show that log (Mz) is defined for all z except z = ie, a :S: c :S: b; and if h(z) = 1m [log Mz] then 0 < h(z) < 71' for Re z > O. (b) Show that log (z-ie) is defined for Re z > 0 and any real number e; also prove that 11m log (z-ie)1 <

71'

2 if Re z

(c) Let h be as in (a) and prove that h(z) (d) Show that

f dt

=

> O. 1m [log (z-ia)-Iog (z-ib)J.

b

--. = z-/l

i[log (z-ib)-log (z-ia)]

a

(Hint: Use the Fundamental Theorem of Calculus.) (e) Combine (c) and Cd) to get that b

h(x+iy) =

fx

2

+(:_t)2

dt = arctan

a

e:

a ) -arctan

(Y:b)

(f) Interpret part (e) geometrically and show that for Rez > O. h(z) is the

angle depicted in the figure.

ih

ia

az+h

Otz+f3

and Tz = ---- ; show that S = T iff there is a non zero cz+d yz+" complex number A such that or: = Aa, f3 = Ab, y = Ae, " = Ad. 21. Let T be a Mobius transformation with fixed points ZI and Z2' If S is a Mobius transformation show that S -1 TS has fixed points S - I Z I and S - I Z 2' 22. (a) Show that a Mobius transformation has 0 and 00 as its only fixed points iff it is a dilation. but not the identity. 20. Let Sz

= --


57

(b) Show that a Mobius transformation has 00 as its only fixed point iff it is a translation, but not the identity. 23. Show that a Mobius transformation T satisfies T(O) = 00 and T( 00) = 0 iff Tz = az- 1 for some a in C. 24. Let T be a Mobius transformation, T #- the identity. Show that a Mobius transformation S commutes with T if Sand T have the same fixed points. (Hint: Use Exercises 21 and 22.) 25. Find all the abelian subgroups of the group of Mobius transformations. 26. (a) Let GL 2 (C) = all invertible 2 x 2 matrices with entries in C and let vIt be the group of Mobius transformations. Define 'P: GL 2 (C) - vIt by

'P(: :) = : : : : . Show that 'P is a group homomorphism of GL 2(C) onto vIt. Find the kernel of 'P. (b) Let SL 2 (C) be the subgroup of GL 2 (C) consisting of all matrices of determinant 1. Show that the image of SL 2 (C) under 'P is all of vIt. What part of the kernel of'P is in SL 2 (C)? 27. If f§ is a group and.JV is a subgroup then.IV is said to be a normal subgroup of f§ if S -1 TS E.JV whenever T E.IV and S E f§. f§ is a simple group if the only normal subgroups of f§ are {J} (/ = the identity of f§) and f§ itself. Prove that the group vIt of Mobius transformations is a simple group. 28. Discuss the mapping properties of (1- z)'. 29. For complex numbers a and fJ with laI 2+lfJI2=1

U",.p(z)

=

az -13 fJz+ii

and let

U=

{u",.p:l aI2 +lfJI 2 =1}.

(a) Show that U is a group under composition. (b) If SU2 is the set of all unitary matrices with determinant I, show that SU2 is a group under matrix multiplication and that for each A in SU2 there are unique complex numbers a and fJ with lal 2+ I fJ 12 = I and

A=( - fJ~ ~). a (c) Show that ( _;

:

}-~Ua.p is a homomorphism of the group SU2onto

U. What is its kernel? (d) If I E{ 0, I, ~, ... } let HI = all the polynomials of degree $. 2/. For ua.p=u in U define TJ/):Hr-~HI by (TJ I>j)(z)=(fJz+ii)2'J(U(Z». Show that Tu(l) is an invertible linear transformation on HI and U f--+ Tu(l) is an injective homomorphism of U into the group of invertible linear transformations of HI onto HI' 30. For Izl < I define fez) b\

t,

f(z)=ex p { -ilog[i(

:~;)r2}.

(a) Show that f maps D = {z : Izl < I} conformally onto an annulus G. (b) Find all Mobius transformations S(z) that map D onto D and such that f(S (z» = fez) when Izl < I.

Chapter IV Complex Integration In this chapter results are derived which are fundamental in the study of analytic functions. The theorems presented here constitute one of the pillars of Mathematics and have far ranging applications. §1. Riemann-Stieltjes integrals We will begin by defining the Riemann-Stieltjes integral in order to define the integral of a function along a path in C. The discussion of this integral is by no means complete, but is limited to those results essential to a cogent exposition of line integrals. 1.1 Definition. A function y: [a, b] -+ C, for [a, b] c IR, is of bounded variation if there is a constant M > 0 such that for any partition P = {a = to < t 1 < ... < tm = b} of [a, b]

v{y; P)

=

m

L k=l

ly(tk)-y{tk - 1)1 ~ M.

The total variation of y, V{y), is defined by V{y)

= sup {v(y;

P): P a partition of[a, b]}.

Clearly V{y) ~ M < 00. It is easily shown that y is of bounded variation if and only if Re y and 1m yare of bounded variation. If y is real valued and is non-decreasing then .y is of bounded variation and V(y) = y(b) -y(a). (Exercise I) Other examples will be given, but first let us give some easily deduced properties of these functions. 1.2 Proposition. Let y: [a, b] -+ C be of bounded variation. Then: (a) If P and Q are partitions of[a, b] and pc Q then v{y; P)

~ v(y; Q); (b) If a: [a, b] -+ C is also of bounded variation and
The proof is left to the reader. The next proposition gives a wealthy collection of functions of bounded variation. In actuality this is the set of functions which is of principal concern to us. 1.3 Proposition. If y: [a,b]-+C is piecewise smooth then y is of bounded variation and

f ly'(r)1 dt b

V(y) =

a

58

59

Riemann-Stieltjes integrals

Proof. Assume that y is smooth (the complete proof is easily deduced from this). Recall that when we say that y is smooth this means y' is continuous. Let P = { a = to < t I < ... < 1m = b }. Then, from the definition, m

v(y; P) =

L

k=l

Jy(tk)-y(tk - 1 )J

m

=

k~l I

f

t"

y'(t) dtJ

Ik-I

: ; JI f m

tk

ly'(!)J dt

'k-l

f ly'(t)1 dt b

=

a

Hence V(y) ::; J~ ly'(t)1 dt, so that y is of bounded variation. Since y' is continuous it is uniformly continuous; so if € > 0 is given we can choose 81 > 0 such that Is-tl < 81 implies that ly'(s)-y'(t)1 < E. Also, we may choose 82 > 0 such that if P = {a = to < 11 < .. .
where "k is any point in [tk-I> ~

J ly'(t)1 dt

::;

E

+

tkl. Hence

m

k1;l ly'h)1 Uk-fk-I)

a

f ly'(t)1 dt ::; e+£(b-a) + .1;lly(tk)-y(tk-l)1 m

b

a

=

£[1 +(b-a)J+v(y; P)

::; E[l + (b-a)]+ V(y).

Complex Integration

60

Letting

€

-+ 0

+,

gives

Jly'(t)J dt ::; VCy), b

a

which yields equality . • 1.4 Theorem. Let y: fa, b] -+ C be oj bounded variation and suppose that J: [a, b] -+ C is continuous. Then there is a complex number I slich that Jar every € > 0 there is a 0 > 0 such that when P = {to < tl < .. .
II - Jl J(Tk) [y(tk)-y(t,,-l)] I <

€

Jar whatever choice oj points Tb t k - 1 ::; Tk ::; tk. This number I is called the Riemann-Stieljes integral off with respect to y over [a. bland is designated by

JJdy = JJ(t) dy(t). b

/ =

b

a

a

Proof Since J is continuous it is uniformly continuous; thus, we can find (inductively) positive numbers 0 1 > 82 > 03 > ... such that if Is-tl < Om'

~. For each m :2: 1 let & m = the collection of all partitions P m of [a, bJ with [[PII < 8m ; so &1 ::> 9 2 ::> Finally define Fm to be the IJ(s) - J(t) I <

closure of the set:

1.5

Ltl J

(Tk)[y(tk)-y(t k-

1)]:PE&m and

tk- J

::;

Tk::; tk}'

The following are claimed to hold:

1.6

(

F!

::>

F2

::> •••

diam Fm ::;

~

m

and V(y)

If this is done then, by Cantor's Theorem (II. 3.7), there is exactly one complex number I such that / E Fm for every m :2: L Let us show that this will complete the proof. If E > 0 let m > (2/E) V(y); then E > (21m) V(y) :2: diam Fm. Since / E Fm , Fm c B(/; E). Thus, if 0 = Om the theorem is proved. Now to prove (1.6). The fact that FI ::> F2 ::> ••• follows trivially from 2 the fact that &1 ::> &2 ::> •••• To show that diam Fm ::; -- V(y) it suffices

m

2 to show that the diameter of the set in (1.5) is s: ---- V(y). This is done in two m stages, each of which is easy although the first is tedious. If P = {to < ... < In} is a partition we will denote by S (P) a sum of the form L!(Tk)[y(tk)-y(lk-l)] where Tk is any point with lk-l ::;Tk ::;lk'


61

the first step will be to show that if P c Q (and 1 hence Q E 9 m ) then IS(P) - S(Q)I < - V(y). We only give the proof for m the case where Q is obtained by adding one extra partition point to P. Let 1 ~ P ~ In and let f p- 1 < t* < tp; suppose that P U {t*} = Q. If tp 1 ~ a ~ t*, t* ~ a' ~ (p' and Fix

:2c 1 and let P E 9

In

S(Q)=

m;

L j(ak)[y(tk)-y(tk .. l)]+](O)[y(t*)-y(tp_l)] k#'p + ](0')[ y(tp } - y(t*)

J,

then, using the fact that 1](1")-}(0)1<

IS(P)- S(Q)I=

-.l m

for 11"-al
Ik,&p L [f(Tk)-f(O'k)] [y(tk)-y(tk-1)]+f(Tp) [y(tp)-y(tp_

1 )]

-f(O') [y(t*)-y{tp-l)]-f(O") [y(tp)-y(t*)]1 ~

-m1 k,&p L ly(tk)-y(tk-1)1+I[f(Tp)-f(0')] [y(t*)-y(t p_

1)]

+ [/( Tp) - f(O")] [yet p) - y(t*)] 1 ~

1

- L ly(t )-y(tkm k,&p k

1

)1

1

+ _····ly(t*)-y(tP-l)1 m

1

+ -ly(tp)-y(t*)1 m

~

1

- V(y)

m

For the second and final stage let P and R be any two partitions in & m. Then Q = PuR is a partition and contains both P and R. Using the first part

we get

IS(P)- S(R)I ::;IS(P) - S(Q)I + IS( Q) - S(R)I

It now follows that the diameter of (1.5) is

~~

~2 V(y). m

V(y) . •

m The next result follows from the definitions by a routine" -

a argument.

1.7 Proposition. Let f and g be continuous functions on [a, b] and let y and a be functions of bounded variation on [a, b]. Then for any complex scalars a and /3: (a) J~ (lXf+f3g) dy = IX J~fdy+f3 J~gdy (b) J:fd(lXy+f3O') = IXJ~fdy+f3 J~fdO'. The following is a very useful result in calculating these integrals. 1.8 Proposition. Let y: [a, b] _ C be of bounded variation and letf: [a, b] - C be continuous. If a = to < tl < ... < tn = b then

f fdy = ~ "'f_I fdy. b

"

"tk

k-l

62

Complex Integration

The proof is left as an exercise. As was mentioned before, we will mainly be concerned with those '1 which are piecewise smooth. The following theorem says that in this case we can find jfd'1 by the methods of integration learned in calculus.

1.9 Theorem. If y is piecewise smooth and f: [a,bj-)C is continuous then

.f

fdy=.C f(t)y'(l}dt.

Proof Again we only consider the case where y is smooth. Also, by looking at the real and imaginary parts of y, we reduce the proof to the case where '1([ a, b]) c Il\t Let e: > 0 and choose 8 > 0 such that if P = {a = to < ... < In = b} has II P II < 8 then

11 fdy -

1.10

J/(Tk) [Y(tk)-Y(f k- I)]\ <

lE

and

1.11 for any choice of Tk in [fk-j, tk]' If we apply the Mean Value Theorem for derivatives we get that there is a Tk in [tk-" fkJ with y'( Tk) = [y(t k) - y(tk- I)] (tk - [k-I) -!. (Note that the fact that Y is real valued is needed to apply the Mean Value Theorem.) Thus, n

L f(Tk) [y(tk)-y(t

k=1

k-

j )]

=

n

L f(Tk)y'(Tk) (tk-f._

I )·

k=1

Combining this with inequalities (1.10) and (1.11) gives

I[fdY

- /f(t)Y'(t) dt\ <

E.

Since € > 0 was arbitrary, this completes the proof of the theorem. III We have already defined a path as a continuous function y: [a, b] -'). C. If y: [a, b] -0>- C is a path then the set {y(t): a :::; t :::; b} is called the trace of y and is denoted by {y}. Notice that the trace of a path is always a compact set. y is a rectifiable path if y is a function of bounded variation. If P is a partition of [a, b] then vCy; P) is exactly the sum of lengths of line segments connecting points on the trace of y. To say that Y is rectifiable is to say that Y has finite length and its length is V(y). In particular, if y is piecewise smooth then Y is rectifiable and its length is J~ Iy'l dt. If y: [a, b] ----,.. C is a rectifiable path with {y} c E c C and f: E --+ C is a continuous function then f 0 y is a continuous function on [a, b]. With this in mind the following definition makes sense.

Riemann·Stieltjes integrals

63

1.12 Definition. If ,,: [a, b) ~ C is a rectifiable path and I is a function defined and continuous on the trace of" then the (line) integral 01I along" is

f1(,,(/» d,,(/). b

"

LI

This line integral is also denoted by = L/(z) dz. As an example let us take ,,: [0, 21T) ~ C to be ,,(t)

f(z) =

!

z

= e it

and define

for z "i' O. Now " is differentiable so, by Theorem 1.9 we have

Jy!z dz = J5

K

e-it(ie it ) dl

= 21Ti.

Using the same definition of 'I and letting m be any integer 2!: 0 gives exp (i(m+l)t) dl=iH'fI cos (m+l)t dtf~'fIsin{m+ 1)ldl=O. Now let a, bEe and put '1(1) = tb + (1 - t)a for 0:::; t:::; 1. Then Y'{t) = b - a, and using the Fundamental Theorem of Calculus we get that for

J zmdz=f5'f1eiml(ieil)dl=ig'fl

n 2!: 0,

Jy

zndz

= (b-a)

J~

[tb+(l-t)a]n dt

=-

1

(bn+l_an+ I ).

n+l There are more examples in the exercises, but now we will prove a certain "invariance" result which, besides being useful in computations, forms the basis for our definition of a curve. If,,: [a, b] -~ C is a rectifiable path and 'P: [c, d) ~ [a, b) is a continuous non· decreasing function whose image is all of [a, b) (i.e., 'P(c) = a and !P(d) = b) then" 0 'P: [c, d) ~ C is a path with the same trace as". Moreover, "0'P is rectifiable because if c = So < SI < ... < Sn = d then a = 'P(so) :::; 'P(SI) :::; ... :::; 'P(sn) = b is a partition of [a, b). Hence n

L

k=1

1"('P(Sk»-"('P(Sk-I»I:::; V(,,)

so that V(" 0 'P) :::; V(,,) < 00. So if I is continuous on {,,} JyO tp f is well defined.

= {"

0

'P} then

1.13 Proposition. II,,: [a, b) ~ C is a rectifiable path and 'P: [c, d) ~ [a, bj is a continuous non·decreasing lunction with !p(c) = a, 'P(d) = b; then lor any lunction I continuous on {y}

f1= f f

y

yo tp

Proof Let £ > 0 and choose 8 1 > 0 such that for {so < SI < ... < sn}, a partition of [c, d) with (SAl-SAl-I) < 81 , and Sk-I ~ (1k ~ S1 we have

Similarly choose 82 > 0 such that if {to < t I < ... < tn} is a partition of [a, b) with (t" - t"-I) < 82 and tit. -1 ~ 'Tit. ~ tAl, then

64

Complex Integration

1.15 But fP is uniformly continuous on [c, d]; hence there is a.5 > 0, which can be chosen with 8 < 8 b such that IfP(s)-fP(s')1 < 82 whenever Is-s'l < 8. So if {so < SI < ... < sn} is a partition of [c, dJ with (Sk-Sk-l) < .5 < 01 and tk = fP(Sk), then {to :$ tl :$"':$ In} is a partition of [a, b] with (tk-f.-l) < 82 , If Sk-l :$ Uk :$ Sk and 'Tk = fP(u k) then both (1.14) and (1.15) hold. Moreover, the right hand parts of these two differences are equal! It follows that

Since € > 0 was arbitrary, equality is proved . • We wish to define an equivalence relation on the collection of rectifiable paths so that each member of an equivalence class has the same trace and so that the line integral of a function continuous on this trace is the same for each path in the class. It would seem that we should define a and y to be equivalent if a = y fP for some function fP as above. However, this is not an equivalence relation! 0

1.16 Definition. Let u: [c, d]-+ C and y: [a, b] -+ C be rectifiable paths. The path u is equivalent to y if there is a function fP: [c, d] -+ [a, b] which is continuous, strictly increasing, and with fP(c) = a, fP(d) = b; such that a = yo fP. We call the function fP a change 0/ parameter. A curve is an equivalence class of paths. The trace of a curve is the trace of anyone of its members. If/is continuous on the trace of the curve then the integral of/ over the curve is the integral of/over any member of the curve. A curve is smooth (piecewise smooth) if and only if some one of its representatives is smooth (piecewise smooth). Henceforward, we will not make this distinction between a curve and its representative. In fact, expressions such as "let y be the unit circle traversed once in the counter-clockwise direction" will be used to indicate a curve. The reader is asked to trust that a result for curves which is, in fact, a result only about paths will not be stated. Let y: [a, b] -+ C be a rectifiable path and for a :$ t :$ b, let Iyl (t) be V(y; [a, tJ). That is,

Iyl (t) =

sup

{f

iy(tk)-y(tk-1)1: {to,··., t n } is a partition of [a,

k=1

{l}.

Clearly Iii (t) is increasing and so Iii: [a, bj-J> !R is of bounded variation. If / is continuous on {y} define

J/Idzl = J/(y(t)) diyl (t). b

y

a

If y is a rectifiable curve then denote by

-y

the curve defined by (-y) (t) =


65

y( - t) for - b ~ t ~ - a. Another notation for this is y - I. Also if c E IC let y+c denote the curve defined by (y+c) (t) = y(t)+c. The following

proposition gives many basic properties of the line integral. 1.17 Proposition. Let y be a rectifiable curve and suppose that I is a lunction continuous on {y}. Then:

L

I = - S- y I; (a) (b) IL/I ~ L I/lldzl ~ V(y) sup [1/(z)l: Z E {y}]; (c) If c E IC then JJ(z) dz = L+cf'(z-c) dz. The proof is left as an exercise. The next result is the analogue of the Fundamental Theorem of Calculus for line integrals. 1.18 Theorem. Let G be open in IC and let y be a rectifiable path in G with initial and end points a and f3 respectively. f: G --+ IC is a continuous lunction with a primitive F: G --+ IC, then

fr

f1= F(8) - F(a) y

(Recall that F is a primitive of I when F'

=

f)

The following useful fact will be needed in the proof of this theorem. 1.19 Lemma.IJ G is an open set in C, y : [a,bj--+G is a rectifiable path, and J: G --+ C is continuous Then Jor every £ > 0 there is a polygonal path f in G such that f(a)=y(a), f(b)=y(b), and IfyJ-IrJI
Proof. Case 1. Suppose G is an open disk. Since {y} is a compact set. d= dist({ y}, aG» O. It follows that if G = B (c; r) then {y} c B (c; p) where p = r - ~ d. The r~ason for passing to this smaller disk is that J is uniformly continuous in B(c;p)cG. Hence without loss of generality it can be assumed that J is uniformly continuous on G. Choose () > 0 such that IJ(z)-J(w)I
'I

IY(s) - y(t}1 < 8/2

1.19a

1.20 Define f: [a,b]--+C by

f(T)=

I tk-tk

_ 1

[(lk-T)y(tk_I)+(t-lk_I)y(tk)]

if tk _ 1 ~ t ~ t k · So on [lk I' ld, f(t) traces out the straight line segment from y(tk _ I) to y(td; that is, f is a polygonal path in G. From (1.I9a)

66

Complex Integration

1.21 Since

fr f = f!:i([(t»['(t)dl (f=

Jr

i

k=

I

it follows that

y(tk)-y(tk-I)f l, f(r(t»dt. tk - tk - I I, I

Using (1.20) we obtain

L /y(tk)-y(tk-I)/(tk-tk-I)-Ifl, n

:S{+

k= I

I,

/f(r(t»-f(Y(7'k»)/dt. I

Applying (1.21) to the integrand gives

If f - f/I:s{+{~ /y(tk)-y(tk _ I)/ :S{(I+ V(y»). y

I

k

=

I

The proof of Case I now follows.

Case II. G is arbitrary. Since {y} is compact there is a number r with 0< r0 such that /y(s)- y(t)1 < r when Is - II < 8. If P={to
Proof of Theorem 1.18. Case I. y: [a,bl~C is piecewise smooth. Then fi = ';f(y(t»y'(t)dl = f;F'(y(t»y'(t)dt = f;(Foy)'(t)dt = F(y(b»F(y(a»= F(f3)- F(a) by the Fundamental Theorem of Calculus. Case II The General Case. If f > 0 then Lemma 1.19 implies there is a polygonal path [ from a to 13 such that IIi -- fr f/ < L But [ is piecewise [F( 13)- F(a)]1 < {; smooth, so by Case I I rf= F( 13) - F(a). Hence If since (> 0 is arbitrary, the desired equality follows . • The use of Lemma 1.19 in the proof of Theorem 1.18 to pass from the piecewise smooth case to the rectifiable case is typical of many proofs of results on line integrals. We shall see applications of Lemma 1.19 in the future. A curve y: [a, b] ~ IC is said to be closed if yea) = y(b).

i-

1.22 Corollary. Let G, y, and f satisfy the same hypothesis as in Theorem 1.18. If y' is a closed curve then

Jf= 0 y

67


The Fundamental Theorem of Calculus says that each continuous function has a primitive. This is far from being true for functions of a complex variable. For example let fez) = Izl2 = x 2 + y2. If F is a primitive of f then F is analytic. So if F = U+iV then X 2+y2 = F'(x+iy). Now, using the Cauchy-Riemann equations,

oU oV 2 2 aU ov - = - = x + y and - = - = 0.

ex

aU

oy =

But -

oy

oy

°implies that U(x, y) = u(x) for some differentiable function u. aU

But this gives x 2 + y2 = -

ox =

see that 8).

ox

u'(x), a clear contradiction. Another way to

Izl2 does not have a primitive is to apply Theorem 1.18 (see Exercise

Exercises l. Let y: [a, b] ---i> IR be non decreasing. Show that y is of bounded variation and V(y) = y(b)-y(a). 2. Prove Proposition 1.2. 3. Prove Proposition 1.7. 4. Prove Proposition 1.8 (Use induction). 5. Let yet) = exp « -1 +i)t- 1 ) for 0 < t ::; 1 and yeO) = O. Show that y is a rectifiable path and find V(y). Give a rough sketch of the trace of y. 6. Show that if y; [a, b] -+ C is a Lipschitz function then y is of bounded variation. 1

C, defined by yet) = t+it sin - for t¥-O and t yeO) = 0, is a path but is not rectifiable. Sketch this path. 8. Let y and G be the two polygons [1, i] and [1, 1 + i, i]. Express y and G as paths and calculate Lfand Juf where f(z) = Iz12.

7. Show that y: [0, I]

9. Define y: [0, 27T]

-+

-+

C by yet) = exp (int) where n is some integer (positive,

negative, or zero). Show that

°: ;

f1'z~ dz = 21Tin.

10. Define yet) = eit for t ::; 21T and find I1' zn dz for every integer n. 11. Let y be the closed polygon [1 - i, 1 + i, - 1 + i, - 1 - i, 1 - i]. Find

f! "z

dz.

12. Let I(r) =

f

l'

that lim I(r) r->-oo

=

0.

ei%

-

z

dz where y: [0, 1T]

.

---i>

C is defined by yet) = re't. Show

13. Find I" z-t dz where: (a) y is the upper half of the unit circle from + 1 to - 1 : (b) " is the lower half of the unit circle from + 1 to - 1.

Complex Integration

68

14. Prove that if cp: [a, b] ~ [c, d] is continuous and cp(a) = c, cp(b) = d then cp is one-one iff cp is strictly increasing. 15. Show that the relation in Definition 1.16 is an equivalence relation. 16. Show that if y and a are equivalent rectifiable paths then V(y) = V(a). 17. Show that if y: [a, b] ~ C is a path then there is an equivalent path a: [0, 1] ~ C. 18. Prove Proposition 1.17. 19. Let yet) = 1 + e il for 0 ::; t::; 2'TT and find fy (Z2 - 1)-1 dz. 20. Let yet) = 2e il for -'TT::; t ::; 'TT and find fy (z2 - 1)-1 dz. 21. Show that if Fl and F2 are primitives for f: G ~ C and G is open and connected then there is a constant c such that FI (z) = c + F2 (z) for each z in G. 22. Let y be a closed rectifiable curve in an open set G and a $. G. Show that for n z 2, fy (z - a)-I! dz = O. 23. Prove the following integration by parts formula. Let f and g be analytic in G and let y be a rectifiable curve from a to b in G. Then

fyfg'

=

f(b)g(b) - f(a)g(a) - fJ'g·

§2. POHier series representation of analytic functions In this section we will see that a function f, analytic in an open set G, has a power series expansion about each point of G. In particular, an analytic function is infinitely differentiable. We begin by proving Leibniz's rule from Advanced Calculus. 2.1 Proposition. Let cp: [a, b] x [c, d] g: [C, d] ~ C by

~

C be a continuous function and define

f cp(s, t) ds. b

2.2

g(t) =

a

Then g is continuous. Moreover, if ocp exists and is a continuous function on

or

[a, b] x [c, d] then g is continuously differentiable and b

2.3

g'(t) =

fat

-ocp (s, t) ds.

a

Proof. The proof that g is continuous is left as an exercise. Notice that if we prove that g is differentiable with g' given by formula (2.3) then it will follow '· . . ocp. . H f rom t herst fi part t hat g IS contInUOUS SInce at IS contInUOus. ence, we

need only verify formula (2.3).

Power series representation of analytic functions

Fix a point to in [c, d] and let

€

69 OqJ > O. Denote - by qJ2; it follows

ot

that qJ2 must be uniformly continuous on [a, b] x [c, d]. Thus, there is a I) > 0 such that IqJzCs " t')-qJ2(S, t)1 < € whenever (S-S')2+(t-t')2 < 1)2. In particular

2.4 whenever It-tol < a ~ S ~ b,

I)

and a ~ S ~ b. This gives that for It-tol <

I)

and

2.5 But for a fixed S in [a, b] <1>(t) = qJ(s, t) - tqJ2(S, to) is a primitive of qJz{s, t)qJz{s, to). By combining the Fundamental Theorem of Calculus with inequality (2.5), it follows that IqJ(s, t)-qJ(s, to)-(t-tO)qJ2(S, to)1 ~

for any S when It-tol <

I).

€

It-tol

But from the definition of g this gives

g(t)-g(to) - Ib qJz{s, to) ds I ~ I~-.:::....:..-= t-t o a

E(b-a)

when 0 < It-tol < I)• • This result can be used to prove that 2"

f -.-ds = 21T e"S-z eis

if

Izl

< 1.

o

Actually, we will need this formula in the proof of the next proposition. e is Let qJ(s, t) = -.-s- for 0 ~ t ~ 1, 0 ~ s ~ 21T; (Note that qJ is continuously e" -tz differentiable because Izl < 1.) Hence get) = J~" qJ(S, t) ds is continuously differentiable. Also, g(O) = 21T; so if it can be shown that g is a constant, then 21T = g(1) and the desired result is obtained. Now

but for t fixed, <1>(s) = zi(eiS-tz)-l has <1>'(s) = -zi(e is -tz)-2(ie is)= ze is(e i'_tz)-2. Hence g'(t) = <1>(21T)-<1>(O) = 0, so g must be a constant.

70

Complex Integration

The next result, although very important, is transitory. We will see a much more general result than this-Cauchy's Integral Formula; a formula which is one of the essential facts of the theory. 2.6 Proposition. Let f: G ---* C be analytic and suppose B(a; r) c G(r > 0). If yet) = a + re il , 0 ::; t ::; 2rr, then 1 ff(W) fez) = ----; dw 2rrl w-z y

for

Iz-al

< r.

H

Proof By considering G1 =

(z-a): z E G} and the function g(z) = f(a+

rz) we see that, without loss of generality, it may be assumed that a = 0 and r = 1. That is we may assume that B(O; 1) c G.

Fix z,

Izi

< 1; it must be shown that fez) = -1. ff(W) - - dw

2m

w-z

1

2K

=LJ"f(~iS)eiS ds; 2rr elS-z o

that is, we want to show that 2"

f(eis)eis 0= f - . - - ds-2rrf(z) eiS-z

o

2.

=

iS f[ f(eiS)e e"-z

- .------ -

] fez) ds

o

We will apply Leibniz's rule by letting rp( s, t ) =

fez -+. tee;s - z))e is .

e"-z

-

f()

z,

for 0 :S t :s 1 and 0 :s s ::; 2'7T. Since Iz + (e z)1 = Iz(1 - t) + te"1 ::; 1, rp is well defined and is continuously differentiable. Let g( t) = (o2wrp(s, t) ds; so g has a continuous derivative. The proposition will be proved if it can be shown that g(l) = 0; this is done by showing that g(O) = 0 and that g is constant. To see that g(O) = 0 IS

----

71


compute:

fo cp(S, 0) ds 2"

g(O) =

2"

=

f [f(z)eiS eis_z -

fez)

]

ds

o 2"

f

ei.

= fez) - .•o

since we showed that

f

2"

=

e' -z

ds-2.".f(z)

0,

ei.

-.-- ds = 2.". prior to the statement of this pro-

o elS-z

position. To show that g is constant compute g'. By Leibniz's rule, g'(t) = J~" cpis, t) ds where

However, for 0 < t :;:; 1 we have that <1>(s) = -U-1f(z+t(eiS-z» is a primitive of cpis, t). So g'(t) = <1>(27T) -<1>(0) = 0 for 0 < t :;:; 1. Since g' is continuous we have g' = 0 and g must be a constant. • How is this result used to get the power series expansion? The answer is that we use a geometric series. Let Jz-aJ < r and suppose that w is on the circle Jw-aJ = r. Then

- 1- = - _ . w-z

w-a

1

1

z-aJ = (w-a) 1- [ - w-a

L (z-a)" w-a
n-O

since Jz-aJ < r = Jw-aJ. Now, multiplying both sides by [f(w)f2.".i] and integrating around the circle y: Jw-aJ = r, the left hand side yieldsf(z) by the preceding proposition. The right hand side becomes-what? To find the answer we must know that we can distribute the integral through the infinite sum. 2.7 Lemma. Let y be a rectifiable curve in C and suppose that Fn and Fare continuous functions on {y}. If F = u -lim Fn on {y} then

fF

Y

=

lim

fF

7

n•

72

Complex Integration

Proof Let E > 0; then there is an integer Nsuch that IFn(w)-F(w)1 < EIV(y) for all IV on {y} and n 2:: N. But this gives, by Proposition 1.17(b),

If F - f Fnl If (F-Fn)1 ~ f IF(w)-Fn(w)lldwl =

whenever n 2:: N.

III 00

2.8 Theorem. Let f be analytic in B(a; R); then fez) =

L an(z -

a)" for

n~O

I iz - al < R where an = - pl/l(a) and this series has radius of convergence 2:: R.

n!

Proof Let 0 < r < R so that BCa; r) then by Proposition 2.6, fez) =

But, since

Iz - al

1

c

B(a; R). If y(t)

.ffCW) ' ."~ dw for

2m

=

a+re it , 0 ~ t .:C~ 217,

1z-a[ < r.

w-z

< rand w is on the circle {y},

1.f1\:'~~~"~~ln ,~ ~ C=;llIJ '[ h were lV

=

"1/'( It')"I, max t,

I'lv-ai

i

1 = rf.

S'lIlce "Iz-al < I ,th e W'eIerstrass r

L

M-test gives that f( w) (z - a)"f(IV - a)" + I converges uniformly for {y}. By Lemma 2.7 and the discussion preceding it

IV

on

2.9 If we set

then an is independent of z, and so (2.9) is a power series which converges for Iz-al < r. It follows (Proposition III. 2.5) that an

=

-

1

n!

pn' l (a), so that the value of an

is independent of y; that is, it is independent of r. So w

2.10 for

fez)

Iz-al

=

I

n= 0

anCz-a)"

< r. Since r was chosen arbitrarily, r < R, we have that (2.10)


73

holds for Iz - al < R; giving that the radius of convergence of (2.10) must be at least R . •

2.11 Corollary. Iff: G -r C is analytic and a E G thenf(z) = Iz-al < R where R = d(a, fJG).

co

L a"(z-a)" for 0

Proof. Since R = d(a, fJG), B{a; R) c G so that f is analytic on B{a; R). The result now follows from the theorem. •

2.12 Corollary. Iff: G -r C is analytic then f is infinitely differentiable. 2.13 Corollary. Iff: G -r C is analytic and B(a; r) f(")(a) =

~

27Ti

f

c

G then

f{w) dw (w-ay+l

y

where y(t) =

a+re it ,

0 :::;; t :::;; 27T.

2.14 Cauchy's Estimate. Let f be analytic in B(a; R) and suppose If(z) 1 for all z in B{a; R). Then

:::;;

M

Proof. Since Corollary 2.13 applies with r < R, Proposition 1.17 implies that

Since r < R is arbitrary, the result follows by letting r -r R - . • We will conclude this section by proving a proposition which is a special case of a more general result which will be presented later in this chapter.

2.15 Proposition. Let f be analytic in the disk B(a; R) and suppose that y is a closed rectifiable curve in B( a; R). Then f has a primitive and hence

Jf= O. y

Proof. This is proved by showing that f has a primitive (Corollary 1.22). Now, by Theorem 2.8, f{z) = ~>"(z-aY for Iz-al < R. Let F{z) =

~ (~) {z-a)"+l

.. -0

n+l

= (z-a)

~ (~) (z-a)" .

~ n+l

Since lim (n+ 1)1/" = I, it follows that this power series has the same radius of convergence as L a,,(z-a)". Hence, F is defined on B(a; R). Moreover, F'{z) = f{z) for Iz-al < R . • Exercises 1. Show that the function defined by (2.2) is continuous. 2. Prove the following analogue of Leibniz's rule {this exercise will be

74

Complex Integration

frequently used in the later sections.) Let G be an open set and let y be a rectifiable curve in C. Suppose that cp: {y} x G --7- C is a continuous function and define g: G --7- C by

f cp(w, z) dw

g(z) =

y

then g is continuous. If

ocp exists for each (w,

oz

then g is analytic and g'(z)

=

f

ocp

-

OZ

z) in {y} x G and is continuous

(w, z) dw.

y

3. Suppose that y is a rectifiable curve in C and cp is defined and continuous on {y}. Use Exercise 2 to show that g(z) =

fcp(}>}

w-z

dw

y

is analytic on C - {y} and g(n)(z) = n!

f-. _ _~(~L (W-Z)"+I

dw.

y

4. (a) Prove Abel's Theorem: Let Ian (z - a)" have radius of convergence 1 and suppose that I an converges to A. Prove that

Jim La/'

=

r-+ 1-

A.

(Hint: Find a summation formula which is the analogue of integration by parts.) (b) Use Abel's Theorem to prove that log 2 = 1-1+t-... 5. Give the power series expansion of log z about z = i and find its radius of convergence. 6. Give the power series expansion of about z = I and find its radius of convergence. 7. Use the results of this section to evaluate the following integrals:

.Jz

(a)

(b)

(c)

f;:%

dz,

f z-a dz ,

f

Sin z 7dz,

o :::;

t :::;

211' and n 2: O.


7S

8. Use a Mobius transformation to show that Proposition 2.15 holds if the disk B(a; R) is replaced by a half plane. 9. Evaluate the following integrals: (a)

f · -. f f2 ~ z"

dz where n is a positive integer and yet)

= eit,O

SIS 21T;

7

(b)

dz 1

(z - 2)

y

(c)

n

where n is a positive integer and yet)

dz where yet) z +1

= 2e i', 0

=!- +e it , 0 S ts2'IT;

S t S 271. (Hint: expand

(Z2 + 1)-1 by

y

means of partial fractions);

(d)

f f

Sin z . -z- dz where yet) = e"', 0 S t S 271;

y

(e)

zltm

- - ) dz where yet)

(z-I

7

10. Evaluate

m

= I +te i', 0 s

f

z2+1 2 ) dz where yet) z(z +4

s

t

= re ll , 0

271.

S t S 271, for all possible

y

values of r, 0 < r < 2 and 2 < r < 00. II. Find the domain of analyticity of

fez) = also, show that tan fez)

= z (i.e., f

fez) =

(I

~.Iog I-Iz +~z) ;

21

is a branch of arctan z). Show that

2: (_l)k 2Z2k+ -k- for Izl < I +I co

1

k-O

(Hint: see Exercise III. 3.19.) 12. Show that

sec z

=1+

E L: ~ Z2k (2k)! co

k-l

for some constants E l , E 4 ,' ••• These numbers are called Euler's numbers. What is the radius of convergence of this series? Use the fact that I = cos z sec z to show that

E2n-Cn2:2)E2n-2 + Cn2:4) E2n- + ... +( 4

_1)n-l

C2n)E2 +( -It=O.

Evaluate E 2 , E4 , E6 , E 8 • (E10 = 50521 and Ell = 2702765).

Complex Integration

76

eZ-l 13. Find the series expansion of - - about zero and determine its radius z

z

= -:;:-

of convergence. Consider fez)

e"-l

fez) =

and let co

ak k z ~ ....... k~O

k!

be its power series expansion about zero. What is the radius of convergence? Show that

o=

ao

+

(

/ 11

+ 1)· 1

a 1 + ...

+

(n+ 1)

n / an"

Using the fact that f(z) +1Z is an even function show that ak = 0 for k odd and k > 1. The numbers B Zn = (-1)n- 1a 2" are called the Bernoulli numbers for 11 ;::0: 1. Calculate B 2 , B4 ,' .• BI o. 14. Find the power series expansion of tan z about z = 0, expressing the coefficients in terms of Bernoulli numbers. (Hint: use Exercise 13 and the formula cot 2z = 1 cot Z-1· tan z.) §3. Zeros of an analytic function If p(z) and q(z) are two polynomials then it is well known that p(z) = s (z)q(z) + r(z) where s(z) and r(z) are also polynomials and the degree of r(z) is less than the degree of q(z). In particular, if a is such that pea) = 0 then choose (z-a) for q(z). Hence, p(z) = (z-a)s(z)+r(z) and r(z) must be a constant polynomiaL But letting z = a gives 0 = pea) ~~ rea). Thus, p(z) = (z-a)s(z). If we also have that sea) = 0 we can factor (z-a) from s(z). Continuing we get p(z) = (z-a)m fez) where 1 ::: m ::: degree of p(z), and t(z) is a polynomial such that tea) # O. Also, degree t(z) = degree p(z)-m. 3.1 Definition. If f: G --7> C is analytic and a in G satisfies f(a) = 0 then a is a zero off of multiplicity m ;::0: 1 if there is an analytic function g: G --7> C such thatf(z) = (z-a)"'g(z) where g(a) # O. Returning to the discussion of polynomials, we have that the multiplicity of a zero of a polynomial must be less than the degree of the polynomial. If n "-7. the degree of the polynomial p(z) and ai' ... , ak are all the distinct zeros of p(z) thenp{z) = (z_al)m t •• • (z-ak)mks(z) where s(z) is a polynomial with no zeros. Now the Fundamental Theorem of Algebra says that a polynomial with no zeros is constant. Hence, jf we can prove this result we will have succeeded in completely factoring p(z) into the product of first degree polynomials. The reader might be pleasantly surprised to know that after many years of studying Mathematics he is right now on the threshold of proving the Fundamental Theorem of Algebra. But first it is necessary to prove a famous result about analytic functions. It is also convenient to introduce some new terminology.

Zeros of an analytic function 77 3.2 Definition. An entire function is a function which is defined and analytic in the whole complex plane Co (The term "integral function" is also used.) The following result follows from Theorem 2.8 and the fact that C contains BCO; R) for arbitrarily large R. 3.3 Proposition.

If f is

an entire function then f has a power series expansion 00

fez)

=

L

"= 0

anz"

with infinite radius of convergence.

In light of the preceding proposition, entire functions can be considered as polynomials of "infinite degree". So the question arises: can the theory of polynomials be generalized to entire functions? For example, can an entire function be factored? The answer to this is difficult and is postponed to Section VII. 5. Another property of polynomials is that no non constant polynomial is bounded. Indeed, if p(z) = z"+an_Iz n- I + ... +ao then limp(z) = lim z" [1+an_IZ-l+"'+aoz-n] = 00. The fact that this also z-+OO

z-+oo

holds for entire functions is an extremely useful result.

3.4 Liouville's Theorem. If I is a bounded entire lunction then I is constant. Proof Suppose I/(z) :0; M for all z in C. We will show that I'{z) = 0 for all z in Co To do this use Cauchy's Estimate (Corollary 2.14). Since I is analytic in any disk B(z; R) we have that il'(z)l :0; MfR. Since R was arbitrary, it follows that I'(z) = 0 for each z in Co • The reader should not be deceived into thinking that this theorem is insignificant because it has such a short proof. We have expended a great deal of effort building up machinery and increasing our knowledge of analytic functions. We have plowed, planted, and fertilized; we shouldn't be surprised if, occasionally, something is available for easy picking. Liouville's Theorem will be better appreciated in the following application. 1

II p(z) is a non constant polynomial then there is a complex number a with pea) = O.

3.5 Fundamental Theorem of Algebra.

Proof Supposep(z) 1= 0 for all z and let/(z) = [p(Z)J-l; thenlis an entire function. If p is not constant then, as was shown above, lim p(z) = 00;

so lim I(z) = O. In particular, there is a number R > 0 such that I/(z) 1 < 1 z-+oo

if Izl > R. Butlis continuous on BCO; R) so there is a constant M such that I/(z) 1:0; M for Izl :0; R. Hencelis bounded and, therefore, must be constant by Liouville's theorem. It follows that p must be constant, contradicting our assumption. III

If p(z) is a polynomial and aI' ... ,am are its zeros with a j having multiplicity k j thenp(z) = c(z-alll.· ·(z-am)kmlor some constant c and kl + ... +km is the degree 01 p.

3.6 Corollary.

Returning to the analogy between entire functions and polynomials, the

78

Complex Integration

reader should be warned that this cannot be taken too far. For example, if p is a polynomial and a E C then there is a number z with p(z) = a. In fact, this follows from the Fundamental Theorem of Algebra by considering the polynomial p(z)-a. However the exponential function fails to have this property since it does not assume the value zero. (Nevertheless, we are able to show that this is the worst that can happen. That is, a function analytic in C omits at most one value. This is known as Picard's Little Theorem and will be proved later.) Moreover, no one should begin to make an analogy between analytic functions in an open set G and a polynomial p defined on C; rather, you should only think of the polynomials as defined on G. F or example, let

+z) ,Izl < 1.

1 I(z) = cos ( ----

1-z

Notice that

1+ z

1- z

=

{z: Izl < I} onto G = {z: Re z > O}. The

! n1T -

2

}

are the points \ ------------: n is odd and positive ; so f has in• n1T --r- 2 finitely many zeros. However, as n ---> Xi the zeros approach 1 which is not in the domain of analyticity D. This is the story for the most general case. zeros of

f

maps D

3.7 Theorem. Let G be a connected open set and let f: G - IC be an analytic Junction. Then. the Jollowing are equivalent statements: (a) J=- 0; (b) there is a point a in G such that I(O)(a) = 0 Jor each n 2:: 0; (c) {z E G: J(z) = O} has a limit point in G.

Proof Clearly (a) implies both (b) and (c). (c) implies (b): Let a E G and a limit point of Z = {z E G: J(z) = O}, and let R > 0 be such that B(a; R) c G. Since a is a limit point of Z and J is continuous it follows that J(a) = O. Suppose there is an integer n 2:: 1 such that J(a) = /,(a) = ... = p.-Il(a) = 0 and J(o)(a) # O. Expanding J in power series about a gives that

L'" ak(z-a)k

J(z) =

k=n

for

Iz-al

< R. If g(z) =

ro

L ak(z-al-

n

k==n

then g is analytic in B(a; R), J(z) = (z-a)"g(z), and g(a) = an # O. Since g is analytic (and therefore continuous) in B(a; R) we can find an r, 0 < r < R, such that g(z) # 0 for Iz-al < r. But since a is a limit point of Z there is a point b with I(b) = 0 and 0 < Ib-aJ < r. This gives 0 = (b-a)"g(b) and so g(b) = 0, a contradiction. Hence no such integer n can be found; this proves part (b). (b) implies (a): Let A = {z E G: pn)(z) = 0 for all n 2:: O}. From the hypothesis of (b) we have that A # O. We will show that A is both open

Zeros of an analytic function

79

and closed in G; by the connectedness of G it will follow that A must be G and so f == O. To see that A is closed let Z E A - and let Zk be a sequence in A such that Z = lim Zk' Since each pHl is continuous it follows that f(n)(z) = Iimi'n l (zk) = O. So Z E A and A is closed. To see that A is open, let a E A and let R > 0 be such that B(a; R) c::: G. 1 Then fez) = ~ a.(z-a)" for Iz-al < R where a. = - pN)(a) = 0 for each

n!

n ~ O. Hencef(z) = 0 for all z in B(a; R) and, consequently, B(a; R) c::: A. Thus A is open and this completes the proof of the theorem. II 3.8 Corollary. Iff and g are analytic on a region G then f == g iff {z E G: fez) = g(z)} has a limit point in G. This follows by applying the preceding theorem to the analytic function f-g· 3.9 Corollary. Iff is analytic on an open connected set G and f is not identically zero then for each a in G with f(a) = 0 there is an integer n ~ 1 and an analytic function g: G -l> C such that g(a) f= 0 and fez) = (z -a)"g(z) for all z in G. That is, each zero off has finite multiplicity.

Proof. Let n be the largest integer (2: 1) such that /(k)( a) and define g(z)

=

(z - a)-"/(z) for z

*-

a and g(a)

=

=

1

0 for 0 ~ k < n

-/(n)(a). Then g

n!

is clearly analytic in G - {a}; to see that g is analytic in G it need only be shown to be analytic in a neighborhood of a. This is accomplished by using the method of the proof that (c) implies (b) in the theorem. II 3.10 Corollary. Iff: G -l> C is analytic and not constant, a E G, and f(a) = 0 then there is an R > 0 such that 8(a; R) c G andf(z) f= 0 for 0 < Iz - al < R.

Proof By the above theorem the zeros of f are isolated . • There is one instance where the analogy between polynomials and analytic functions works in reverse. That is, there is a property of analytic functions which is not so transparent for polynomials.

3.n Maximum Modulus Theorem.lfG is a region andf: G...,.. C is an analytic function. such that there is a point a in G with If(a) 1 ~ If(z) 1 for all z in G, then f is constant. Proof Let 8(a; r) c::: G, y(t) = a+re it for 0 ~ t ~ 217'; according to Proposition 2.6 f(a)

= ~ f f(w) dw w-a

2m

y

2x

= ~ff(a+re'l) dt 2?T

o

80

Complex Integration

Hence

If· 2.

!/(a)1

:s:

I/(a+relt)1 dt :::~ I/(a) I

27T

o

since I/(a+rei')1

:s:

I/(a)1 for all t. This gives that

°= J 2n

[1/(a)! -1/(a + re it )!] dt;

o

but since the integrand is non-negative it follows that I/(a)1 = I/(a + reit)1 for all t. Moreover, since r was arbitrary, we have that I maps any disk B(a; R) c G into the circle Izl = 10(1 where 0( = I(a). But this implies that I is constant on B(a; R) (Exercise III. 3.17). In particular I(z) = 01. for Iz-al < R. According to Corollary 3.8, 1== 01.. II According to the Maximum Modulus Theorem, a non-constant analytic function on a region cannot assume its maximum modulus; this fact is far from obvious even in the case of polynomials. The consequences of this theorem are far reaching; some of these, along with a closer examination of the Maximum Modulus Theorem, are presented in Chapter VI. (Actually, the reader at this point can proceed to Sections VI. I and VI. 2.)

Exercises L Let I be an entire function and suppose there is a constant M, an R > 0, and an integer n ~ I such that I/(z)1 :s: M Izl" for Izi > R. Show that I is a polynomial of degree :s: n. 2. Give an example to show that G must be assumed to be connected in Theorem 3.7. 3. Find all entire functions I such that I(x) = eX for x in JR. 4. Prove that ez + a = eZe" by applying Corollary 3.8. 5. Prove that cos (a + b) = cos a cos b - sin a sin b by applying Corollary 3.8. 6. Let G be a region and suppose that I: G ->- C is analytic and a E G such that I/(a) 1 :s: I/(z) I for all z in G. Show that either 1(0) = 0 or I is constant. 7. Give an elementary proof of the Maximum Modulus Theorem for polynomials. 8. Let G be a region and let f and g be analytic functions on G such that f(z)g(z) = 0 for all z in G. Show that either f == 0 or g == O. 9. Let U: C ~ R be a harmonic function such that U( z) ~ for all z in C; prove that U is constant. 10. Show that if f and g are analytic functions on a region G such that f g is analytic then either f is constant or g == O.

°

§ 4. The index of a closed curve

It is easy to see that L(z - a)-1 dz = 2nin if i'(t) = a + eZ"int, 0 The following result shows that this is not peculiar to the path y.

:s: t:s:

l.

Cauchy's Theorem

81

4.1 Proposition. If y: [0, I]_C is a closed rectifiable curve and a • {y} then

2~i f /'~a y

is an integer. Proof. An easy argument using Lemma 1.19 shows that it suffices to prove this proposition under the additional assumption that y is smooth. In this case define g: [0, 1] ~ C by 1 y'(s) g(t)= () ds

£ o

s -a

y

Hence, g(O) = 0 and g( I) = f y(z - a)-I dz. We also have that g'(t)=

for

y'(t) y(t)-a

O:5t:51.

But this gives

.!!... e - g ( y dt

a) = e - gy' - g' e - g ( y - a) =e- g [ y'-y'(y-a)-I(y-a)J

=0 So e-g(y-a) is the constant function e-g(O)(y(O)-a)=y(O)-a= e -8(1)( y( I) - a). Since y(O) = y(l) we have that e -g(1) = I or that g(I)= 2'ITik for some integer k . • 4.2 Definition. If y is a closed rectifiable curve in C then for a 1/ {y}

1 .j(z-a)-I dz n(y;a)= -2 'IT!

y

is called the index of y with respect to the point a. It is also sometimes called the winding number of y around a. Recall that if y: [0, I]_C is a curve, - y or y -\ is the curve defined by (-y)(I)=y(I-/) (this is actually a reparametrization of the original definition). Also if y and 0 are curves defined on [0, I] with y(l) = 0(0) then y+o is the curve (y+ 0 )(1)= y(21) if 0:5 1 :5 ~ and (y+ o)(t) =0(2t-l) if ~:5 t:5 1. The proof of the following proposition is left to the reader. 4.3 Proposition. If y and 0 are closed rectifiable curves having the same initial points then (a) n(y;a)= -n(-y;a) for every a* {y}; (b) n(y+o;a)=n(y;a)+n(o;a) for every a. {y}u{o}. Why is n(y; a) called the winding number of y about a? As was said before if y(t)=a+e 2 ,..;", for 0:5 1 S 1 then n(y; a)=n. In fact if Ib-al < 1 then n(y; b)=n and if Ib-al> 1 then n(y; b)=O. This can be shown directly or one can invoke Theorem 4.4 below. So at least in this case n(y; b) measures the number of times y wraps around b- with the minus sign indicating that the curve goes in the clockwise direction.

82

Complex Integration

The following discussion is intuitive and mathematically imprecise. Actually, with a little more sophistication this discussion can be corrected and gives insight into the Argument Principle (V.3). If y is smooth then -I y'(t) (z-a) dz= ( ) dt. 0 y t -a y Taking inspiration from calculus one is tempted to write f/ z - a) -1 dz = log[y(t) - a1l::6. Since y(l) = y(O), this would always give zero. The difficulty lies in the fact that y( t) - a is complex valued and unless y( t) - a lies in a region on which a branch of the logarithm can be defined, the above inspiration turns out to be only so much hot air. In fact if y wraps around the point a then we cannot define log( y( t) - a) since there is no analytic branch of the logarithm defined on C - {a}. Nevertheless there is a correct interpretation of the preceding discussion. If we think of log z = logl z I+ i arg z as defined then

11

f

f (z - a) - dz = log[ y(l) - a] -Iog[ y(O) - a] = 1

y

{iogi y ( I ) - a 1- log/ y (0) - a I} + i { arg [ y ( I ) - a ] - arg [ y (0) - a] }.

Since the difficulty in defining logz is in choosing the correct value of argz, we can think of the real part of the last expression as equal to zero. Since y(l)= y(O) it must be that even with the ambiguity in defining argz, arg[y(l)-a]-arg[y(O)-a] must equal an integral multiple of 2w, and furthermore this integer counts the number of times y wraps around a. Let y be a closed rectifiable curve and consider the open set G = C {y}. Since {y} is compact (z:/zl>R}cG for some sufficiently large R. This says that G has one, and only one, unbounded component. 4.4 Theorem. Let y be a closed rectifiable curve in C. Then n(y;a) is constant for a belonging to a component of G = C - { y }. A/so, n (y; a) = 0 for a belonging to the unbounded component of G.

Proof. Define f: G~C by f(a) = n( y; a). It will be shown that f is continuous. If this is done then it follows that f(D) is connected for each component D of G. But since f(G) is contained in the set of integers it follows that f(D) reduces to a single point. That is, f is constant on D. To show that f is continuous recall that the components of G are open (Theorem II. 2.9). Fix a in G and let r = d (a, (y». If Ia - bl < <5 < ~ r then If(a) - f(b)1

= 2~ I~[ (z- a)-I_(z I

= 2w

f

(a-b) (z-a)(z-b) dz

y

la-blj

:5~

b)-I] dzl

y

Idz/ Iz-allz-b/

Cauchy's Theorem

83

But for la-bl
Z

on {y) we have that Iz-al2:r>t' and

28

IZ--bl>tr. It follows that If(a)-f(b)I<-2 V(y). So if (>0 is given 1fr then, by choosing 5 to be smaller than ~ rand ('IT r 2() /2 V( y), we see that j must be continuous. (Also, see Exercise 2.3.) Now let U be the unbounded component of G. As was mentioned before the theorem there is an R > 0 such that U::> {z: Izi > R}. If { > 0 choose a with lal > Rand Iz - al > (21f£)-1V(y) uniformly for z on {y}; then In(y; a)1 < (. That is, n(y; a) ~ 0 as a --+ 00. Since n(y; a) is constant on U, it must be zero . •

Exercises 1. Prove Proposition 4.3. 2. Give an example of a closed rectifiable curve y in C such that for any integer k there is a point a $. {y} with n(y; a) = k. 3. Let p(z) be a polynomial of degree n and let R > 0 be sufficiently large so that p never vanishes in {z: Izl 2 R}. If yet) = Reir, 0 S t s 2n, show that

PI(z) f -.-dz y

p(z)

=

.

21f1n.

4. Fix w = re lll *0 and let y be a rectifiable path in C- {O} from I to w. Show that there is an integer k such that iyz - 1 dz = logr + iii + 21fik. § 5. Cauchy's Theorem and Integral Formula We have already proved Cauchy's Theorem for functions analytic in a disk: if G is an open disk then /J=O for any analytic functionj on G and any closed rectifiable curve y in G (Proposition 2.15). For which regions G does this result remalll valid? There are regions for which the result is false. For example, if G=C,-{O} andf(z)=z-I then y(t)=e il for OS IS 21f gives that 1 = 21fi. The difficulty with C- {O} is the presence of a hole (namely {O». In the next section it will be shown that 1/=0 for every analytic function f and every closed rectifiable curve '( in regions G that have no "holes." In the present section we adopt a different approach. Fix a region G and an analytic function f on G. Is there a condition on a closed rectifiable curve y such that I J = O? The answer is furnished by the index of y with respect to points outside G. Before presenting this result we need the following lemma. (This has already been seen in Exercise 2.3.)

J

5.1 Lemma. Let '( be a rectifiable curve and suppose cp is a junction defined and continuous on {y}. For each In 2: I let Fm(z) = 1yCP(w)(w - z)-m dw for z
Complex Integration

84

bounded there; and use the factorization. (

5.2

I )m - (

W-Z

I

w-a

)m = [ ~ W

=(z-a) [

Z -

W

Fm(z)-Fm(a) z-a

=

k=1

+

E

\m -

W-Z

k (

I

w-a

I (w-z)m-'(w-ai

)k _ 1

+ ...

G. z 1= a. I t follows from (5.2) that

fqJ(w)(w-a)-' y

(

I ] (w-z)(w-a)m

The details are left to the reader. Now fix a in G = C - {y) and let z 5.3

~a ] ~

I (w-z)m(w-a)

+

m

(w-z)m

dw + ... +

fqJ(w)(w-a)-m y

w-z

dw

Since a f {r}, qJ(w)(w-a)-k is continuous on {y} for each k. By the first part of this proof (the part left to the reader). each integral on the right hand side of (5.3) defines a continuous function of z,z in G. Hence letting z--->a. (5.3) gives that the limit exists and

F:" (a) =

I

y

qJ(W) dw + . . . + (w-a)m+1

I

y

qJ(W) dw (w-a)m+1

=mFm+l(a) . •

5.4 Cauchy's Integral Formula (First Version). Let G be an open subset oj the plane and J: G---> C an analytic Junction. IJ y is a closed rectifiable curve in G such that n(y; w)=O Jor all w in C - G, then for a in G- {y} I

n(y;a)J(a)= -2' 7rI

1 y

J(z) --dz. z-a

Proof. Define qJ: G X G--->C by qJ(z. w) = [fez) - J(w)l/(z - w) if z 1= wand qJ(z ,z) =F(z). It follows that qJ is continuous; and for each w in G, z--->qJ(z.w) is analytic (Exercise I). Let H={wEC:n(y;w)=O}. Since n(y; w) is a continuous integer-valued function of w, H is open. Moreover HuG = C by the hypothesis. Define g: C--->C by g(z) = f yqJ(Z, w)dw if z z)-~(w)dw if z E H. If z E GnH then

f qJ(z, w) dw

=

'Y

f

W -

Z

J(w) - - dw- J(z)n(y;z)27Ti yW-Z

=fJ(w)dW. yW-Z Hence g is a well-defined function.

G and g(z) = f y(w-

fJ(w) - J{z) dw 'Y

=

E

Cauchy's Theorem

8S

By Lemma 5.1 g is analytic on H and by an analogue of Leibniz's rule (for example, see Exercise 2.2) g is analytic on G; that is, g is an entire function. But Theorem 4.4 implies that H contains a neighborhood of 00 in Coo- Since f is bounded on {y} and lim (W-Z)-I=O uniformly for w in {y}, Z --+ 00

5.5

}~~ g(z)= }~~

f ~,~'~

dw=O.

y

In particular (5.5) implies lhere is an R >0 such that Ig(z)l:<:; 1 for Iz I ~ R. Since g is bounded on B (0; R) it follows that g is a bounded entire function. Hence g is constant by Liouville's Theorem. But then (5.5) says that g=O. That is, if Q E G- {y} then 0=

f f

y

=

Y

J(z)-f(a) Z-Q

dz

J(z) dz- f(a)f~·

Z-Q

y

Z-Q

This proves the theorem . • Often there is a need for a more general version of Cauchy's Integral Formula that involves more than one curve. For example in dealing with an annulus one needs a formula involving two curves. 5.6 Cauchy's Integral Formula (Second Version). Let G be an open subset of the plane and f: G ...... C an analytic function. If YI" .. 'Ym are closed rectifiable curves in G such that n(Yl; w) + ... + n(y",; w) = 0 for all w in C - G, then for Q in G - U ;~l {Yd J(a)

L n(Yk;a)= L A I m

m

=

k= I

I f J(z) -. --- dz. 2771 y,

Z -

a

Proof The proof follows the lines of Theorem 5.4. Define cp(z, w) as it was done there and let H = {W:n(rl; w) + ... + n(Ym; w) = O}. Now g(z) is defined as in the proof of (5.4) except that the sum of the integrals over ri' ... , I'm is used. The remaining details of the proof are left to the reader. III Though an easy corollary of the preceding theorem, the next result is very important in the development of the theory of analytic functions.

5.7 Cauchy's Theorem (First Version). Let G be an open subset of the plane and J: G~C an analytic Junction. IJ YI'"'' Ym are closed rectifiable curves in G such (hat n(YI: w)+'" + n(Ym: w)=OJor all w in C- G then

2. f J=O. m

J.;~

I

y,

Proof. Substitute f(z)(z - a) for J in 111eorem 5.6.111 Let G={z:RI
86

Complex Integration

n(y,; w)= I = - n(Y2; w); if Iwl ~ R2 then n(y,; w)= n(Y2; w)=O. So n(y,; w) + n(Y2; w)=O for all win C- G. 5.8 Theorem. Let G be an open subset of the plane and f: G~C an analytic function. If Y"""Y m are closed rectifiable curves in G such that n(y,;w) + ... +n(Ym;w)=Ofor all w in C-- G then for a in G-{y} and k~ I In m I Pk)(a).L n(y;a)=k! "" - .

. ;=,

}

L.J 2m }=,

f

y,

f(z)

( z-a )k+'

dz.

Proof. This follows immediately by differentiating both sides of the formula in Theorem 5.6 and applying Lemma 5.1 . •

5.9 Corollary. Let G be an open set andf: G ---C an analytic function. If y is a closed rectifiable curve in G such that n (y; w) = 0 for all w in C - G then for a in G - { y} k' pk)(a)n( y; a) = -'.

2m

f y

f(z) dz. (z-a)k+1

Cauchy's Theorem and Integral Formula is the basic result of complex analysis. With a result that is so fundamental to an entire theory it is usual in mathematics to seek the outer limits of the theorem's validity. Are there other functions that satisfy J~f = 0 for all closed curves y? The answer is no as the following converse to Cauchy's Theorem shows. A closed path T is said to be triangular if it is polygonal and has three sides. 5.10 Morera's Theorem. Let G be a region and let f: G~C be a continuous function such that I Tf = 0 for evel), triangular path T in G; then f is analytic in G.

Proof. First observe that f will be shown to be analytic if it can be proved thatf is analytic on each open disk contained in G. Hence, without loss of generality, we may assume G to be an open disk; suppose G= R(a; R). Use the hypothesis to prove that f has a primitive. For z in G define F(z)= Ira.:lf. Fix Zo in G; then for any point z in G the hypothesis gives that F(z) = I[a.:olf + I[z".:lf. Hence F(z)-F(zo) z-zo

=-l-f Z-Zo

j.

[Zo.:]

This gives

= _____ 1-

(z-zo)

f [zo.z]

[f(w)-f(zo)] dw.

Cauchy's Theorem

87

But by taking absolute values

I

F( z) - F( z ) I z-zo 0 -f(zo) :::;;sup{lf(w)-f(zo)I:WE[Z,Zo]}

which shows that . F(z)-F(ZO)_f(). I1m - Zo'

Z-+Zo

Z - Zo

Exercises I. Suppose f: G-+C is analytic and define cp: G X G-+C by cp(z, w)=[J(z) - f(w)](z - W)-I if z"* wand cp(z,z)= /,(z). Prove that cp is continuous and for each fixed w, z-+cp(z, w) is analytic. 2. Give the details of the proof of Theorem 5.6. 3. Let B± = B( ± 1; !), G = B(O; 3) - (B+ u B_). Let Yl' Y2' Y3 be curves whose traces are Iz-II= I, Iz+ 11= I, and Izl=2, respectively. Give YI'Y2' and Y3 orientations such that n(YI;w)+n(Y2;w)+n(Y3;w)=0 for all w in C-G. 4. Show that the Integral Formula follows from Cauchy's Theorem. 5. Let Y be a closed rectifiable curve in C and at!. {y}. Show that for n ~ 2 fy(z-a)-ndz=O. 6. Letf be analytic on D = B (0; I) and suppose If(z)1 : :; 1 for Izl < l. Show 1/,(0)1 :::;; I. 7. Let y( t) = 1+ e il for 0:::;; t :::;; 2'1T. Find f z ~ I dz for all positive integers n. 8. Let G be a region and suppose fn : G-+C is analytic for each n ~ I. Suppose that Un} converges uniformly to a functionf: G-+C Show thatf is analytic. 9. Show that if f: C-+C is a continuous function such that f is analytic off [ - I, I] then f is an entire function. 10. Use Cauchy's Integral Formula to prove the Cayley-Hamilton Theorem: If A is an n X n matrix over C and f (z) = det( z - A) is the characteristic polynomial of A then f(A)=O. (This exercise was taken from a paper by C. A. McCarthy, Amer. Math. Monthly, 82 (1975), 390-391).

y(

r

§6 The homotopic version of Cauchy's Theorem and simple COIfIU!ctivity

This section presents a condition on a closed curve Y such that f/=O for an analytic function. This condition is less general but more geometric than the winding number condition of Theorem 5.7. This condition is also used to introduce the concept of a simply connected region; in a simply connected region Cauchy's Theorem is valid for every analytic function and every closed rectifiable curve. Let us illustrate this condition by

Complex Integration

88

considering a closed rectifiable curve in a disk, a region where Cauchy's Theorem is always valid (Proposition 2.15). Let G = B (a; R) and let y : [0, ll~ G be a closed rectifiable curve. If 0:0;,:0; I and O:O;s:O;l, and we put z=ta+(l-t)y(s); then z lies on the straight line segment from a to yes). Hence, z must lie in G. Let y,(s)= fa +(I-t)y(s) for O:O;s:S1 and O:St:S 1. So, Yo=y and y, is the curve constantly equal to a; the curves y/ are somewhere in between. We were able to draw y down to a because there were no holes. If a point inside y were missing from G (imagine a stick protruding up from the disk with its base inside y), then as y shrinks it would get caught on the hole and could not go to the constant curve. 6.1 Definition. Let Yo, y, : [0, 1]~G be two closed rectifiable curves in a region G; then Yo is homotopic to y, in G if there is a continuous function

[;[0, l]x[O,

6.2

11~G

{

such that

f(S'O)=Yo(S) f(O,t)=f(l,t)

and r(s,I)=y,(s) (O:St:Sl)

(O:O;s:O;l)

So if we define y,: [0, IJ~G by y,(s)=f(s,t) then each y/ is a closed curve and they form a continuous family of curves which start at Yo and go to y,. Notice however that there is no requirement that each y, be rectifiable. In practice when Yo and y, are rectifiable (or smooth) each of the Yl will also be rectifiable (or smooth). If Yo is homotopic to y, in G write Yo-y,. Actually a notation such as yo-y,(G) should be used because of the role of G. If the range of f is not required to be in G then, as we shall see shortly, all curves would be homotopic. However, unless there is the possibility of confusion, we will only write YO-YI' lt is easy to show that "-" is an equivalence relation. Clearly any curve is homotopic to itself. If YO-YI and r: [0, 1] X [0, l]~G satisfies (6.2) then define A(s,t)=f(s,l-t) to see that Y1-YO' Finally, if Yo-y, and YI-Y2 with f satisfying (6.2) and A:[O. l]x[O, l]~G satisfying A(s,O)=

.\

r ()L-_ _ _ _ _ _......I

Cauchy's Theorem

89

)'1(S), A(s.I) = 'Y2(S). and A(O. t) = A(I. t) for all sand t; define ~: [0.1] x [0,1] .... G by

cI>(s,t) = {

[(S.2t)

A(s,2t-l)

iq s t s 1

Then cI> is continuous and shows that Yo - Y2. The reader can best understand the motivation for the definition of cI> by consulting the figure above, where [2 = [0,1] X [0,1]. 6.3 Definition. A set G is convex if given any two points a and b in G the line segment joining a and b, [a,b], lies entirely in G. The set G is star shaped if there is a point a in G such that for each z in G, the line segment [a,z] lies entirely in G. Clearly each convex set is star shaped but the

converse is just as clearly false. We will say that G is a- star shaped if [a,z]c G whenever z e G. If G is a- star shaped and z and ware points in G then [z,a, w] is a polygon in G connecting z and w. Hence, each star shaped set is connected. 6.4 Proposition. Let G be an open set which is a- star shaped. If Yo is the curve which is constantly equal to a then every closed rectifiable curve in G is homotopic to Yo. Proof. Let y\ be a closed rectifiable curve in G and put r(s,t)=ty\(s)+ (1- t)a. Because G is a- star shaped, r(s, t) e G for 0 S s, t sLIt is easy to see that r satisfies (6.2). • The situation in which a curve is homotopic to a constant curve is one that we will often encounter. Hence it is convenient to introduce some new terminology. 6.S Definition. If y IS a closed rectifiable curve in G then y is homotopic to zero (y-O) if y is homotopic to a constant curve. 6.6 Cauchy's 1beorem (Second Version). Iff: G-C is an analytic junction and y is a closed rectifiable curve in G such that y-O, then

!f=O. y

This version of Cauchy's Theorem would follow immediately from the first version if it could be shown that n(y;w)=O for all w in C- G whenever y-O. This can be done. A plausible argument proceeds as follows. Let y\ = Y and let Yo be a constant curve such that Y\-Yo. Let r satisfy (6.2) and define h(t)= n(y/; w), where y/(s)=r(s,t) for 0 S s, t S 1 and w is fixed in C - G. Now show that h is continuous on [0, I]. Since h is integer valued and h(O)=O it must be that h(t) ===0. In particular, n(y; w)=O for all w inC-G. The only point of difficulty with this argument is that for 0 < t < 1 it may be that y/ is not rectifiable. As was stated after Definition 6.1, in practice each of the curves y/ will not only be rectifiable but also smooth. So there is justification in making

w

~~~~

this assumption and providing the details to transform the preceding paragraph into a legitimate proof (Exercise 9). Indeed. in a course designed for physicists and engineers this is probably preferable. But this is nol desirable for the training of mathematicians. The statement of a theorem is not as important as its proof. Proofs are important in mathematics for several reasons. not the least of which is that a proof deepens our insight into the meaning of the theorems and gives a natural delineation of the extent of the theorem's validity. Most important for the education of a mathematician. it is essential to examine other proofs because they reveal methods. A good method is worth a thousand theorems. Not only is this statement valid as a value judgement. but also in a literal sense. An important method can be reused in other situations to obtain further results. With this in mind a complete proof of Theorem 6.6 will be presented. In fact. we will prove a somewhat more general fact since the proof of this new result necessitates only a little more effort than the proof that n(y:w)=O for w in C-C whenever y-O. In fact. the proof of the next result more clearly reveals the usefulness of the method.

6.7 Cauchy's lbeorem (Third Version). If Yo and Y1 are two closed rectifiable curces in C and YO-YI then

for ecoy fUllction f analytic

Of!

C.

Before proceeding let us consider a special case. Suppose r satisfies (6.2) and also suppose r has continuous second partial derivatives. Hence

throughout the square /2

=

(21'

e2 r

('sit

(tEs

[0, 1] x [0, 1]. Define 1

get) = ff(r(s, t»)

o then g(O) = derivative,

La f

and gel)

=

frI J.

1

g'(t)

f[f'Cr(s, t»)

=

o

(,E (s. t) ds; os

By Leibniz's rule g has a continuous

~~ ~r + fCres, t)) rc~J ds cs c{ etcs

But iT -;-cf -;:-b [ (f f) ...Ef] ~..... = (f , . I ) -;:os c't ('.I' ct 0

0

a2 r + for -;--. n

esct

;

91

Cauchy's Theorem

hence

or

or (0, t).

g'(t) = l(r(1, t» -- (1, t)-/(r(o, t» -

at

at

Since r(l, t) = r(0, t) for all 1 we get g'(t) = 0 for all t. So g is a constant; in particular Lol = Proof of Theorem 6.7. Let r: [2--+G satisfy (6.2). Since r is continuous and [2 is compact. r is uniformly continuous and r(l2) is a compact subset of G. Thus

LJ

= d(r(I2),

r

IC-G) > 0

and there is an integer n such that if (s-sY +(1- t')2 < 4/n 2 then

Ir(s, t)-r(s',

1')1

< r.

Let

and put J' k

'

for 0

~ j, k ~ n -

= [~ , j + n

n

I]

x

[~ , k + n

n

1]

1. Since the diameter of the square

~k is Iin , it follows

that f(~k) C B(Zjk; r). So if we let Pjk be the closed polygon [Zjk' Zj+1. k' Zj+1,k+1' Zj.k+1' Zjd; then, because disks are convex, ljk c B(Zjk; r). But from Proposition 2.15 it is known that

6.8 for any function 1 analytic in G. It can now be shown that Lof = LJ by going up the ladder we have constructed, one rung at a time. That is, let Qk be the closed polygon [ZO.k' Zl,k' .. , Znd· We will show that = fQoI = fQ.! = ... = fQJ = 1 (one rung at a time !). To see that Lof = f Qo 1 observe that if ait) = yo(t) for

LJ

L.

then uj+[Zj+I,o, Zjo] (the + indicating that uj is to be followed by the polygon) is a closed rectifiable curve in the disk B( Z jO; r) c G. Hence

Jf=aj

J

[Zj+"o,Z,o]

f=

J

[ZjO, Z, + .. 0]

Adding both sides of this equation for 0 ~ j < larly fyJ = fQJ

f

n yields fyJ = fQJ.

Simi-

92

Complex Integration

To see that fQ

k

f = fQ k+ I f use equation (6.8); this gives 0 =

6.9

n-i

L

ff

j=O Pj"

ZJ'

However, notice that the integral

f Pjk f

includes the integral over [Zj

+ I, k'

Zj+I,k+ll, which is the negative of the integral over [Zj+l,k+l' Zj+l,kl,

which is part of the integral ZO,k =

JPj+1..f

r(o,~)

Also, =

r(l,~)

= Zn,k

so that [ZO,k+I' Zo,kl = -[Zlk' Zl,k+d. Hence, taking these cancellations into consideration, equation (6.9) becomes

This completes the proof of the theorem . • The second version of Cauchy's Theorem immediately follows by letting YI be a constant path in (6.7). 6.10 Corollary, If Y is a closed rectifiable curve in G such that y-O then n(y;w)=Ofor all w in C- G.

The converse of the above corollary is not valid. That is, there is a closed rectifiable curve y in a region G such that n( Y; w) = 0 for all w in C - G but y is not homotopic to a constant curve (Exercise 8). If G is an open set and Yo and YI are closed rectifiable curves in G then n(Yo:a)=n(y\;a) for each a in C- G provided Yo-YI(G). Let Yo(t) = e 21TiI and YI(t)=e- 21Til for O:$; IS J. Then n(yo;O)= 1 and n(YI;O)= -I so that Yo and YI are not homotopic in C - {OJ. 6.11 Definition. If Yo, YI : [0, Il~G are two rectifiable curves in G such that Yo(O)=y\(O)=a and yo(l)=y\(l)=b then Yo and Y\ are fixed-end-point

Cauchy's Theorem

homotopic (FEP homotopic) if there is a continuous map that

6.12

f(s,O)

= Yo(s)

f(O, t) = a

f(s, 1)

= Yl(S)

f(I, t)

= b

r: I2~G

93

such

for 0 ::::; s, t ::::; 1. Again the relation of FEP homotopic is an equivalence relationship on curves from one given point to another (Exercise 3). Notice that if Yo and Yl are rectifiable curves from a to b then YO-Yl is a closed rectifiable curve. Suppose f satisfies (6.12) and define Y: [0, I] ---* G by yes) = Yo(3s) for 0 ::::; s ::::; !; yes) = b for t : : ; s ::::; t; and yes) = Yl (3 -3s) for t : : ; s ::::; 1. We will show that Y""" O. In fact, define A: 12 ---* G by A(s,t)=

I

f(3s(l- t), t) if 0::::; s ::::; ! f(1-t,3s-1+2t-3st) if t::::;s::::;i

t::::;

if

Yl«3-3s)(1-t»

s ::::; 1.

Although this formula may appear mysterious it can easily be understood by seeing that for a given value of t, A is the restriction of f to the boundary of the square [0, 1- t] x [t, 1] (see the figure). It is left to the reader to check that A shows Y"'" O.

U~-----------r--~

I-I

Hence, for f analytic on G the second version of Cauchy's Theorem gives

o=Jf=Jf-Jf. 1

10

1\

This is summarized in the following.

6.13 Independence of Path Theorem. If Yo and Y1 are two rectifiable curves in G from a to b and Yo and Y1 are FEP homotopic then

If= If Yo

y,

for any function f analytic in G. Those regions G for which the integral of an analytic function around a closed curve is always zero can be characterized.

6.14 Definition. An open set G is simply connected if G is connected and every closed curve in G is homotopic to zero.

94

Complex Integration

6.15 Cauchy's Theorem (Fourth Version). If G is simply connected then fJ =0 for every closed rectifiable curve and every analytic function f. Let us now take a few moments to digest the concept of simple connectedness. Clearly every star shaped region is simply connected. Also, examine the complement of the spiral r = e. That is, let G = C - {ee i6 : :$ e < oo}; then G is simply connected. In fact, it is easily seen that G is open and connected. If one argues in an intuitive way it is not difficult to become convinced that every curve in G is homotopic to zero. A rigorous proof will be postponed until we have proved the following: A region G is simply connected iff Coo - G, its complement in the extended plane, is connected in Coo. This will not be proved until Chapter VIII. If this criterion is applied to the region G above then G is simply connected since Coo - G consists of the spiral r = e and the point at infinity. Notice that for G = C - {O}, C - G = {o} is connected but Coo - G = {a, oo} is not. Also, the domain of the principal branch of the logarithm is simply connected. It was shown earlier in this chapter (Corollary 1.22) that if an analytic function f has a primitive in a region G then the integral off around every closed rectifiable curve in G is zero. The next result should not be too surprising in light of this.

°

6.16 Corollary. If G is simply connected and f: G~ C is analytic in G then f has a primitive in G. Proof. Fix a point a in G and let Yr, Y2 be any two rectifiable curves in G from a to a point z in G. (Since G is open and connected there is always a path from a to any other point of G.) Then, by Theorem 6.15, o = fY1 -yJ = fyJ - fyJ (where Yr - Y2 is the curve which goes from a to z along Yr and then back to a along - Y2)' Hence we can get a well defined function F: G ~ C by setting F( z) = fyi where Y is any rectifiable curve from a to z. We claim that F is a primitive of j. If Zo E G and r > 0 is such that B(zo; r) c G, then let y be a path from a to zoo For z in B(zo; r) let y. = y+[zo, z); that is, y. is the path y followed by the straight line segment from Zo to z. Hence

F(z)-F(zo) = _1_ Z-Zo (z-zo)

f

f.

[zo.z]

Now proceed as in the proof of Morera's Theorem to show that F(zo) = f(zo)· • Perhaps a somewhat less expected consequence of simple connectedness is the fact that a branch of log f(z) where f is analytic and never vanishes, can be defined on a simply connected region. Nevertheless this is a direct consequence of the preceding corollary. 6.17 Corollary. Let G be simply connected and let f: G~ C be an analytic function such that f(z) =1= 0 for any z in G. Then there is an analytic function

95

Cauchy's Theorem

g: G--'» C such that fez) = expg(z). If Zo such that g(zo) = woo Proof Since / never vanishes,

~

E

G and e

WO

=f(zo),

we may choose g

is analytic on G; so, by the preceding

corollary, it must have a primitive gl' If h(z) = exp g\(z) then h is analytic and never vanishes. So, ~ is analytic and its derivative is h(z)I'(z) --- h'(z)/(z) h(z) 2 "-"-----.-------------.----~

But h' = g~h so that hI' ~/h' = O. Hence flh is a constant c for aU z in G. That is fez) = c exp gl(Z) = exp [gl(Z)+C'] for some c'. By letting g(z) = gl(Z)+C' +2rrik for an appropriate k, g(zo} = Wo and the theorem is proved. III Let us emphasize that the hypothesis of simple connectedness is a topological one and this was used to obtain some basic results of analysis. Not only are these last three theorems (6.15, 6.16, and 6.17) consequences of simple connectivity, but they are equivalent to it. It will be shown in Chapter VIII that if a region G has the conclusion of each of these theorems satisfied for every function analytic on G, then G must be simply connected. We close this section with a definition. 6.18 Definition. If G is an open set then y is homologous to zero, in symbols y~O, if n(y;w)=O for all w in C~ G. Using this notation, Corollary 6.10 says that y-O implies y~O. This result appears in Algebraic Topology when it is shown that the first homology group of a space is isomorphic to the abclianization of the fundamental group. In fact, those familiar with homology theory will recognize in the proof of Theorem 6.7 the elements of simplicial approximation. Exercises 1. Let G be a region and let (11' (12: [0, 1] ~ G be the constant curves == a, (12(t) == b. Show that if y is a closed rectifiable curve in G and y - (11 then y - (12' (Hint: connect a and b by a curve_) 2. Show that if we remove the requirement "r(O, t) = r(l, t) for all t" from Definition 6.1 then the curve Yo(t) = e lltj " 0 ::; t ::; 1, is homotopic to (1j(t)

the constant curve Yl(t) == I in the region G = C- {OJ. 3. Let ~ = aU rectifiable curves in G joining a to b and show that Definition 6.l1 gives an equivalence relation on ~. 4. Let G = C - {O} and show that every closed curve in G is homotopic to a closed curve whose trace is contained in {z: Izl = I}.

96

5. Evaluate the integral

Complex Integration

fz 2d:

1 where yeO)

= 21cos 201

e for °: :; 0 :::; 2".. iB

y

6. Let yeO) = Oe iO for 0:::; {} :::; 217 and yeO) = 47T-0 for 217 :::; 0 :::; 47T. Evaluate

f-~

Z2 +71"2 •

y

7. Let fez) = [(z-J-i)'(z- I -ii)'(z-I-2)'(z-i-i)]-1 and let y be the polygon [0, 2, 2 + 2i, 2i, OJ. Find L f 8. Let G = C - {a, b}, a cp b, and let y be the curve in the figure below. ----------~------

'h

(a) Show that nCy; a) = n(y; b) = O. (b) Convince yourself that y is not homotopic to zero. (Notice that the word is "convince" and not "prove". Can you prove it?) Notice that this example shows that it is possible to have a closed curve y in a region such that nCy; z) = 0 for all z not in G without y being homotopic to zero. That is, the converse to Corollary 6.10 is false. 9. Let G be a region and let Yo and YI be two closed smooth curves in G. Suppose YO-YI and r satisfies (6.2). Also suppose that YI(S)=f(s,t) is smooth for each t. If IV E C - G define h(t) = n( Y1 : w) and show that h: [0, I]->Z is continuous.

10. Find all possible values of

f~ where Y is any closed rectifiable 1+ y

z2

curve in C not passing through ± i. II. Evaluate e Z - e- Z dz where Y is one of the curves depicted below.

f

y

Z4

(Justify your answer.)

(u)

(b)

97

Counting zeros; the Open Mapping Theorem

§7. Counting zeros,' the Open Mapping Theorem

In this section some applications of Cauchy's Integral Theorem are given. It is shown how to count the number of zeros inside a curve; also, using some information on the existence of roots of an analytic equation, it will be proved that a non-constant analytic function on a region maps open sets onto open sets. In section 3 it was shown that if an analytic function/had a zero at z = a we could write/(z) = (z-a)mg(z) where g is analytic and g(a) # O. Suppose G is a region and let/be analytic in G with zeros at a b ••• ,am' (Where some of the ak may be repeated according to the multiplicity of the zero.) So we can write/(z) = (z-a l ) (Z-a2)' .. (z-am)g(z) where g is analytic on G and g(z) # 0 for any z in G. Applying the formula for differentiating a product gives 7.1

for z # aI' ... , am' Now that this is done, the proof of the following theorem is straightforward.

7.2 Theorem. Let G be a region and let f be an analytic function on G with zeros a l , ... , am (repeated according to multiplicity). If y is a closed rectifiable curve in G which does not pass through any point a k and if y ::::: 0 in G then

1 -2. '17/

f -I'(z) f( ) dz y

Z

m

L n(y; a k=l

=

k )

Proof. If g( z) =1= 0 for any z in G then g'( z)/ g( z) is analytic in G; since y ::::: 0 in G, Cauchy's Theorem gives

f g'((Z)), dz g z

=

O. So, using (7.1) and

y

the definition of the index, the proof of the theorem is finished. II 7.3 Corollary. Let f, G, and y be as in the preceding theorem except that al, ... ,am are the points in G that satisfy the equationf(z)=a; then

f -I'(z)- dz = L n(y; ak) 2m /Cz)-Of. 1

m

-----:

y

. 1et us caIcu I ate A s an I'11 ustratIOn,

f

k=l

(2z+ - -1) - dz where y is the circle Z2+Z+ 1

y

Izl =

2. Since the denominator of the integrand factors into (z - WI) (z where WI and Wz are the non-real cubic roots of 1, Theorem 7.2 gives

f y

2z+ 1 dz = 4n-i. Z2+Z+ 1

W2 ),

98

Complex Integration

As another illustration. let '1 : [0. 11~G be a closed rectifiable curve in C. '10::::;0. Suppose that]is analytic in G. Then]o'1=a is a closed rectifiable curve in C (Exercise I). Suppose that a is some complex number with a ¢ { a} =]( (y}), and let us calculate n( a; a). We get

n(a; .:l()

I =~. 2m

f -dw-

w-ex

<1

= _1

f

F(z) dz 27Ti J(z)-ex y

m

=

L n(y; a

k)

k~l

where ak are the points in G withj(ak)=a. (To show the second equality above takes a little effort, although for '1 smooth it is easy. The details are left to the reader.) Note. It may be that there are infinitely many points in G that satisfy the equation ]( z) = a. However. from what we have proved, all limit points of this sequence must lie on the boundary of G. It follows that n('1; z) '* 0 for at most a finite number of solutions of fez) = a. (See Exercise 2.) Now if 13 in C - {a} belongs to the same component of 1[:- {a} as does a, then n(a;a}=n(a:f3); or.

where Zk (a) and zi( 13) are the points in G that satisfy ](z) = It and ](z) = 13 respectively. If we had chosen '1 so that n('1;zk(a»=1 for each k, we would have that]( G) contains the component of I[: - ]( {'1}) that contains a. We would also have some information about the number of solutions of ](z)= 13. This procedure is used to prove the following result which, in addition to being of interest in itself, will yield the Open Mapping Theorem as a consequence. 7.4. Theorem. Suppose] is analytic in B( a; R) and let a = ] ( a). I] f( z) - a has a zero of order m at z = a then there is an { > 0 and 8 > 0 such that for o < I~ - al < 8. the equationf(z) = ~ has exactly m simple roots in B(a; t). A simple root of fez) = t is a zero of fez) - t of multiplicity 1. Notice that this theorem says that f(B(a; E)) ::J B(a; 8). Also, the condition that f( z) - a have a zero of finite multiplicity guarantees that f is not constant.

Counting zeros; the Open Mapping Theorem

99

Proof of Theorem. Since the zeros of an analytic function are isolated we can choose € > 0 such that € < !R, fez) = ex has no solutions with 0 < Iz-al < 2€, and f'(z) # 0 if 0 < Iz-al < 2€. (If m ~ 2 then f'(a) = 0.) Let y(t) = a + t:eXp(27Tit), 0:5; t:5; 1, and put 0 =/oy. Now a fE. {a}; so there is a 8 > 0 such that B( a; 8) n {o} = D. Thus, B( a; 8) is contained in some component of C - {O'}; that is, la - fl < 8 implies nCo; a) p

=

nCo; f)

=

L

n(y; Zk(f». But since n(y; z) must be either zero or one,

k~l

we have that there are exactly m solutions to the equation I( z) = f inside B(a; E). Since f'(z) =F 0 for 0 < Iz - al < E, each of these roots (for f =F a) must be simple (Exercise 3). • 7.5 Open Mapping Theorem. Let G be a region and suppose that f is a non constant analytic function on G. Then for any open set V in G, f( V) is open. Proof If U eGis open, then we will have shown that f( V) is open if we can show that for each a in V there is as> 0 such that B(ex; S) c f(V), where ex = f(a). But only part of the strength of the preceding theorem is needed to find an € > 0 and as> 0 such that B(a; €) c V andf(B(a; €» ::.> B(ex; S) . • If X and n are metric spaces and f: X -* n has the property that f( V) is open in n whenever V is open in X, then f is called an open map. Iff is a one-one and onto map then we can define the inverse map f- l : n -* X by f-l(w) = x where f(x) = w. It follows that f- 1 is continuous exactly when f is open; in fact, for U c X, (f-l)-l(V) = f(V).

7.6 Corollary. Suppose f: G -* C is one-one, analytic and f(G) = n. Then f- l : n -* C is analytic and (f-l)'(w) = [f'(z)r 1 where w = fez). Proof By the Open Mapping Theorem, f- 1 is continuous and n is open. Since z = f- 1 (f(z» for each z E G, the result follows from Proposition III. 2.20. •

Exercises l. Show that if I: G --+ C is analytic and y is a rectifiable curve in G then loy is also a rectifiable curve. 2. Let G be open and suppose that y is a closed rectifiable curve in G such that y~o. Set r=d({y},aG) and H={z EC:n(y;z)=O}. (a) Show that {z: d(z, aG)< ir} c H. (b) Use part (a) to show that if I: G--+C is analytic then I(z)=a has at most a finite number of solutions z such that n(y;z)*O. 3. Letfbe analytic in B(a; R) and suppose thatf(a) = 0. Show that a is a zero of multiplicity m ifff(",-1)(a) = ... = f(a) = 0 andf(m)(a) # 0.

100

Complex Integration

4. Suppose that I: G -+ IC is analytic and one-one; show that f'(z) ¥ 0 for any z in G. 5. Let X and Q be metric spaces and suppose that I: X -+ Q is one-one and onto. Show that I is an open map iff I is a closed map. (A function I is a closed map if it takes closed sets onto closed sets.) 6. Let P: IC -+ ~ be defined by P(z) = Re z; show that P is an open map but is not a closed map. (Hint: Consider the set F = {z: 1m z = (Re Z)-l and Re z ¥ O}.) 7. Use Theorem 7.2 to give another proof of the Fundamental Theorem of Algebra.

§8. Goursat's Theorem Most modern books define an analytic function as one which is differentiable on an open set (not assuming the continuity of the derivative). In this section it is shown that this definition is the same as ours.

Goursat's Theorem. Let G be an open set and let I: G -+ IC be a differentiable lunction; then I is analytic on G. Proof. We need only show that l' is continuous on each open disk contained in G; so, we may assume that G is itself an open disk. It will be shown that f is analytic by an application of Morera's Theorem (S.lO). That is, we must show that Jrf=O for each triangular path Tin G. Let T = [a, b, c, a] and let ~ be the closed set formed by T and its inside. Notice that T = 8~. Now using the midpoints of the sides of ~ form four triangles ~l' ~2' ~3' ~4 inside ~ and, by giving the boundaries appropriate

-a

L-----~~------~------_.------~b

directions, we have that each 1j = 8.1

O~j

is a triangular path and 4

JI= T

.L

JI

J= 1 TJ

Among these four paths there is one, call it T(l), such that Ih(1)/1 ~ IhJI for j = 1, 2, 3, 4. Note that the length of each T j (perimeter of ~ )-denoted by t(T)-is !leT). Also diam T j = t diam T; finally, using (8.1)

101

Goursat's Theorem

Now perform the same process on T(1), getting a triangle T(2) with the analogous properties. By induction we get a sequence {T(n)} of closed triangular paths such that if fl.(n) is the inside of T(n) along with T(n) then;

8.1 8.3

8.4 8.5 These equations imply:

8.6

8.7 8.8

diam Since each

fl.(n)

= (t)"d where d = diam fl..

fl.(n)

is closed, (8.2) and (8.8) allow us to apply Cantor's

Theorem (11.3.7), and get that Let

n 00

fl.(n) consists of a single point Zo0

n=l

~

> 0; since I has a derivative at

Z0

B(zo; 8) c G and

we can find a 8 > 0 such that

I

I

/(Z)-/(Zo) - /'(Zo) < Z-Zo

~

whenever 0 < Iz-zol < 8. Alternately,

8.9 whenever Iz-zol < 8. Choose n such that diam fl.(n) = (t)nd < 8. Since Zo E fl.(n) this gives fl.(h) c B(zo; 8). Now Cauchy's Theorem implies that o= dz = Z dz. Hence

h(n)

h(n)

If II = If T(n)

[/(z)-/(zo)-/'(zo) (z-zo)] dzl

T(n)

~~

I

IZ-Zolldzl

T(n)

~ ~

[diam fl.(n)] [t(T(n»]

= ~dt(!}"

Complex Integration

102

But using (8.6) this gives

If II ~ 4 «;dtw n

n

= Edt.

T

Since

«;

was arbitrary and d and

t

are fixed,

JT I

= O. •

Chapter V Singularities In this chapter functions which are analytic in a punctured disk (an open disk with the center removed) are examined. From information about the behavior of the function near the center of the disk, a number of interesting and useful results will be derived. In particular, we will use these results to evaluate certain definite integrals over the real line which cannot be evaluated by the methods of calculus. §1. Classification of singularities This section begins by studying the best behaved singularity-the removable kind. 1.1 Definition. A function! has an isolated singularity at z = a if there is an R>O such that! is defined and analytic in B(a;R)-{a} but not in B (a; R). The point a is called a removable singularity if there is an analytic function g: B(a; R)-,;C such that g(z)= !(z) for 0< Iz - al < R. . sin z 1 1 . . . The functIons - - , - , and exp - all have Isolated smgularitles at z = O.

z

z

z

sin-z h as a removab ' I ' (see ExerCIse .) However, only Ie smgu anty 1 . It is left to z the reader to see that the other two functions do not have removable singularities. How can we determine when a singularity is removable? Since the function has an analytic extension to B(a; R), J = 0 for any closed curve in the punctured disk; but this may be difficult to apply. Also it must happen that lim J(z) exists. This is easier to verify, but a much weaker criterion is

J

available.

1.2 Theorem. IJ J has an isolated singularity at a then the point z = a is a removable singularity iff lim (z-a)J(z) = 0 Proof. Suppose J is analytic in {z: 0 < Iz-al < R}, and define g(z) = (z-a)J(z) for z #- a and g(a) = O. Suppose lim (z-a)J(z) = 0; then g is z->-a

clearly a continuous function. If we can show that g is analytic then it follows that a is a removable singularity. In fact, if g is analytic we have g(z) = (z-a)h(z) for some analytic function defined on B(a; R) because g(a) = 0 (IV. 3.9). But then h(z) andJ(z) must agree for 0 < Iz-al < R, so that a is, by definition, a removable singularity. 103

Singularities To show that g is analytic we apply Morera's Theorem. Let T be a triangle in B(a; R) and let Ll be the inside of T together with T. If art Ll then T"" 0 in {z: 0 < Jz-aJ < R} and so, hg = 0 by Cauchy's Theorem. If a is a vertex of T then we have T = [a, b, e, aJ. Let x E [a, bJ and y E [e, aJ and 104

c

aL-------~------------------------~b

form the triangle TI = [a, x, y, aJ. If P is the polygon [x, b, e, y, x] then h g = h, g+ g = h, g since P "" 0 in the punctured disk. Since g is continuous and g(a) = 0, for any E > 0 x and y can be chosen such that Jg(z)J ~ Elt for any z on T I , where t is the length of T. Hence Jh gl = Ih I gl ~ E; since E was arbitrary we have h g = O. If a E Ll and T = [x, y, z, x] then consider the triangles TI = [x, y, a, x], T2 = [y, z, a, y], T3 = [z, x, a, z]. From the preceding paragraph hj g = 0

Jp

x'OC:;;"--------------------------------..:::=..\y for j = 1,2,3 and so, hg = h,g+ hzg+ h3g = O. Since this exhausts all possibilities, g must be analytic by Morera's Theorem. Since the converse is obvious, the proof of the theorem is complete. • The preceding theorem points out another stark difference between functions of a real variable and functions of a complex variable. The function f(x) = IxI. x E JR, is not differentiable because it has a "corner" at x = O. Such a situation does not occur in the complex case. For a function to have an honest singularity (i.e., a non-removable one) the function must behave badly in the vicinity of the point. That is, either If(z) I becomes infinite as z nears the point (and does so at least as quickly as (z-a)-I), or If(z) I doesn't have any limit as z -+ a.

Classification of singularities

105

1.3 Definition. If z = a is an isolated singularity of f then a is a pole of f if lim If(z)1 = 00. That is, for any M > 0 there is a number € > 0 such that z--+a

If(z) 1 ?: M whenever 0 < Iz-al < E. If an isolated singularity is neither a pole nor a removable singularity it is called an essential singularity. It is easy to see that (z - a) -m has a pole at z = a for m ?: 1. Also, it is not difficult to see that although z = 0 is an isolated singularity of exp (Z-1), it is neither a pole nor a removable singularity; hence it is an essential singularity. Suppose thatfhas a pole at z = a; it follows that [f(Z)]-l has a removable singularity at z = a. Hence, h(z) = [f(zW 1 for z of. a and h(a) = 0 is analytic in B(a; R) for some R > O. However, since h(a) = 0 it follows by Corollary IV. 3.9 that h(z) = (z-a)mhj(z) for some analytic function hI with 111(a) of. Oandsomeintegerm ?: 1. But this gives that (z-a)"'!(z) = [h 1 (z)rI has a removable singularity at z = a. This is summarized as follows. 1.4 Proposition. If G is a region with a in G and iff is analytic on G - {a} with a pole at z = a then there is a positive integer m and an analytic function g: G -)- C such that

fez)

1.5

=

g(z) . (z-a)m

1.6 Definition. Iff has a pole at z = a and m is the smallest positive integer such that fez) (z - a)'" has a removable singularity at z = a then f has a pole of order m at z = a. Notice that if m is the order of the pole at z = a and g is chosen to satisfy (1.5) then g(a) of. O. (Why?) Letfhave a pole of order m at z = a and putf(z) = g(z) (z-a)-m. Since g is analytic in a disk B(a; R) it has a power series expansion about a. Let

L 00

g(z)

=

A m +A m - 1 (z-a)+'" +A 1 (z-a)m-l+(z-ar

ak(z-at·

k= 0

Hence

1.7 where g 1 is analytic in B(a; R) and Am of. O.

1.8 Definition. Iff has a pole of order m at z = a and f satisfies (1.7) then Am(z - a) -m + ... + A j (z - a) -1 is called the singular part of fat z = a. As an example consider a rational function r(z) = p(z)/q(z), where p(z) and q(z) are polynomials without common factors. That is, they have no common zeros; and consequently the poles of r(z) are exactly the zeros of q(z). The order of each pole of r(z) is the order of the zero of q(z). Suppose q(a) = 0 and let S(z) be the singular part of r(z) at a. Then r(z) - S(z) = r 1 (z) and r 1 (z) is a rational function whose poles are also poles of r(z). More-

106

Singularities

over, it is not difficult to see that the singular part of r I (z) at any of its poles is also the ~ingular part of r(z) at that pole. Using induction we arrive at the following: if a J , " , an are the poles of r(z) and S/z) is the singular part of r(z)atz = ajthen

r(z) =

1.9

n

I

S/z)+P(z)

j '" 1

where P(z) is a rational function without poles. But, by the Fundamental Theorem of Algebra, a rational function without poles is a polynomial! So P(z) is a polynomial and (1.9) is nothing else but the expansion of a rational function by partial fractions. Is this expansion by partial fractions (1.9) peculiar only to rational functions'? Certainly it is if we require P(z) in (1.9) to be a polynomial. But if we allow P(z) to be any analytic function in a region G, then (1.9) is valid for any function r(z) analytic in G except for a finite number of poles. Suppose we have a function f analytic in G except for infinitely many poles (e.g., f( z) = (cos z) -1); can we get an analogue of (1.9) where we replace the finite sum by an infinite sum? The answer to this is yes and is contained in Mittag-Leffler's Theorem which will be proved in Chapter VIII. There is an analogue of the singular part which is valid for essential singularities. Actually we will do more than this as we will investigate functions which are analytic in an annulus. But first, a few definitions. 1.10 Definition. If

{Zn:

complex numbers,

I

n

= 0, ± 1, ± 2, ... } is a doubly infinite sequence of

I

00

n= -

00

Zn

is absolutely convergent if both

00

00

are absolutely convergent. In this case

I

n= -

fun,ction on a set S for n = 0, ± 1, ... and

Zn 00

=

n=O

00

00

E

Z-n

n=l

n=O

I=-" + I

z". If

Un

is a

n=t

C(J

I

un{s) is absolutely convergent

-00

for each s

I

00

z" and

cc

:x;.

S, then the convergence is uniform over S if both

I

Un

and I

n~O

n~l

converge uniformly on S. The reason we are limiting ourselves to absolute convergence is that this is the type of convergence we will be most concerned with. One can define U- n

m

00

convergence of

I

-co

Zm

but the definition is not that the partial sums

I

Zn

n= -m

I -1 satisfies this criterion but it is clearly not a "",0 n series we wish to have convergent. On the other hand, if L z" is absolutely

converge. In fact, the series

00

-
convergent with sum z then it readily follows that z = lim

m

I 11=

-m

z".

If 0 :0; Rl < R2 :0; 00 and a is any complex number, define ann (a; R I , R 2 ) = {z: Rl < Iz-al < R 2 }. Notice that ann (a; 0, R 2 ) is a punctured disk.


107

1.11 Laurent Series Development. Let f be analytic in the annulus ann (a; R l' R z). Then

L ,,= Xl

fez)

=

a.(z-a)n OJ

where the conrergence is absolute and uniform orer ann (a; '1' '2)- if R j < < r2 < R z . Also the coefficients an are given by the formula

'1

1.12

where y is the circle Iz-al is unique.

=

r Jor any r, RI < r < R z . MoreOl;er, this series

I.::-al

Proof If RI < r 1 < '2 < R2 and YI, Y2 are the circles = '1' Iz-al = '2 respectively, then YI ,...., Yz in ann (a; R I , R2)' By Cauchy's Theorem we

L,

In

g = g. In have that for any function g analytic in ann (a; R I , R z), particular the integral appearing in (1.12) is independent of, so that for each integer n, an is a constant. Moreover,J2: B(a; R z ) --+ IC given by the formula

Jiz)

1.13

1

J(w) dw

=-.

w-z

27T1

,

where Iz - al < r2 , Rl < '2 < R 2 , is a well defined function. Also, by Lemma IV.5.1 J2 is analytic in B(a; '2)' Similarly, if G = {z: Iz - al > R j } then II : G --> C defined by

f

1.14

'I

'I

Iw-al=r,

J(w) dw , w-z

where Iz-al > and RI < < R z , is analytic in G. If RI < Iz-al < R2 let r 1 and r 2 be chosen so that RI < r 1 < Iz-al < '2 < R 2· Let YI(I) = a+r 1e it and Y2(t) = a+r 2e it , 0 :s: t :s: 27T. Also choose a straight line segment A going from a point on Yl radially to Y2 which misses z. Since Yl ,...., Y2 in ann (a; R 1 , R z ) we have that the closed curve

)'2

Singularities

108

Y = Y2 - A - Yl

+ A is

homotopic to zero. Also n(Y2; z) = 1 and n(Yl' z)

= 0 gives, by Cauchy's Integral Formula, that I(z)

1 fl(W) = -. -dw

w-z

2m

y

= _1

fl(W) dw _ _ 1 fl(W) dw

w-z

2~

w-z

2~

YZ

11

= 12(Z) +/1 (z). The plan now is to expand II and 12 in power series (/1 having negative powers of (z-a»; then adding them together will give the Laurent series development of I(z). Since 12 is analytic in the disk B(a; R 2 ) it has a power series expansion about a. Using Lemma IV. 5.1 to calculate 11n )(a), 00

1.15

12(Z)

= L a,,(z-at n=O

where the coefficients an are given by (1.12). Now define g(z) for

o<

!) ;

~

by g(z) = II(a + RI Z so Z = 0 is an isolated singularity. We claim that z = 0 is a removable singularity. In fact, if r > RI then let p(z) = d(z, C) where C is the circle {w: Iw-al = r}; also put M = max {1/(w)l: WE C}. Then for Iz-al > r Mr 1/1(Z) I :;:; p(z)' But lim p(z)

=

Izl <

00; so that lim g(z) = lim/l(a

::-0

Hence, if we define g(O)

,:-0

!) = o. Z

= 0 then g is analytic in B(O; g(z) =

1.16

+

IjR 1). Let

L Bn z" 00

n=1

be its power series expansion about O. It is easy to show that this gives 1.17

fl(z)

=

00

L

a_n(z-a)-n

n=1

where a_ n is defined by (1.12) (the details are to be furnished by the reader

L 00

in Exercise 3). Also, by the convergence properties of (1.15) and (1.17),

-00

a,,(z - at converges absolutely and uniformly on properly smaller annuli. The uniqueness of this expansion can be demonstrated by showing that if fez) =

00

L

n=

-00

a,,(z - at converges absolutely and uniformly on proper

annuli then the coefficients a" must be given by the formula (1.12) . •

109


We now use the Laurent Expansion to classify isolated singularities.

co

1.18 Corollary. Let z = a be an isolated singularity off and let fez) = L a,,(z-a)" be its Laurent Expansion in ann (a; 0, R). Then: -co (a) z = a is a removable singularity iff a" = 0 for n :s; -1; (b) :I = a is a pole of order m iff a_ m # 0 and a" = 0 for n :s; -(m+ 1); (c) z = a is an essential singularity iff a" # 0 for infinitely many negative integers n. Proof (a) If a" = 0 for n :s; -1 then let g(z) be defined in B(a; R) by co

g(z) = La,,(z-a)"; thus, g must be analytic and agrees withfin the punc-

,,-0

tured disk. The converse is equally as easy. (b) Suppose a" = 0 for n :s; + 1); then (z - a)mf(z) has a Laurent Expansion which has no negative powers of (z - a). By part (a), (z a)mf(z) has a removable singularity at z = a. Thus f has a pole of order m at z = a. The converse follows by retracing the steps in the preceding argument. (c) Since f has an essential singularity at z = a when it has neither a removable singularity nor a pole, part (c) follows from parts (a) and (b). • One can also classify isolated singularities by examining the equations

-em

1.19

lim Iz-aISlf(z)1 = 0

1.20

lim Iz-aISlf(z)1

=

00

where s is some real number. This is outlined in Exercises 7, 8, and 9; the reader is strongly encouraged to work through these exercises. The following gives the best information which can be proved at this time concerning essential singUlarities. We know thatfhas an essential singularity at z = a when lim If(z) I fails to exist ("existing" includes the possibility that z .....

the limit is infinity). This means that as z approaches a the values of fez) must wander through C. The next theorem says that not only do they wander, but, as z approaches a,f(z) comes arbitrarily close to every complex number. Actually, there is a result due to Picard that says that fez) assumes each complex value with at most one exception. However, this is not proved until Chapter XII.

1.21 Casorati-Weierstrass Theorem. Iff has an essential singularity at z thenfor every 8 > 0, {f[ann (a; 0, 8)]} - = c.

=a

Proof Suppose thatfis analytic in ann (a; 0, R); it must be shown that if c and E > 0 are given then for each 8 > 0 we can find a z with Iz-.al < 8 and If(z)-cl < E. Assume this to be false; that is, assume there is a c in C and E > 0 such that If(z)-cl ~ E for all z in G = ann (a; 0, 8). Thus lim Z-+(J

Iz-al-11f(z)-cl=oo, which implies that (z-a)-I(f(z)-c) has a pole at z=a. If m is the order of this pole then lim Iz-alm+1If(z)-cl=O. Z-+(J

no

Singularities

Hence Iz - al m + 11](z)1 $ Iz - al m + 11](z} - cl + Iz - al m + llel gives that lim Iz-alm+'I](z)I=O since m ~l. But, according to Theorem 1.2, this gives that ](z)(z - a)'" has a removable singularity at z = a. This contradicts the hypothesis and completes the proof of the theorem. • z~a

Exercises 1. Each of the following functions I has an isolated singularity at z = O. Determine its nature; if it is a removable singularity define 1(0) so that I is analytic at z = 0; if it is a pole find the singular part; if it is an essential singularity determine/({z: 0 < Izl < S}) for arbitrarily small values of 3. (a) fez)

sin z

= -; z

cos z-1 (e) fez) = - - - - ; z

(e) fez)

= log (~+ 1) z

;

= cos z ;

(d)f(z)

=

z

exp(z-l);

(f) fez) = cos

~1-1)

;

Z

Z2+ 1 (g) fez) = z(z-1) ; (i) fez)

(b) fez)

= z sin! ; z

(h) fez) = (l_e z )-l; (j) fez)

2. Give the partial fraction expansion of r(z)

1

= z" sin - . z

=

2

+1-------------2

(z +z+ l)(z-l) 3. Give the details of the derivation of (1.1 7) from (1.l6).

4. Letf(z)

•

=

1 ; give the Laurent Expansion of fez) in each of z(z-l) (z-2) the following annuli: (a) ann (0; 0, 1); (b) ann (0; 1,2); (c) ann (0; 2, co). 5. Show that fez) = tan z is analytic in C except for simple poles at

Z

=

:2 + n7T, for each integer n.

1T

Determine the singular part of I at each of

these poles. 6. If f: G --» C is analytic except for poles show that the poles of f cannot have a limit point in G. 7. Let f have an isolated singularity at z = a and suppose fez) ¢= o. Show that if either (1.19) or (1.20) holds for some sin IR then there is an integer m such that (1.19) holds if s > m and (1.20) holds if s < m. 8. Let/, a, and m be as in Exercise 7. Show: (a) m = 0 iff z = a is a removable singularity and/(a) # 0; (b) m < 0 iff z = a is a removable singularity andfhas a zero at z = a of order -m; (c) m > 0 iff z = a is a pole off of order m. 9. A function f has an essential singularity at z = a iff neither (U9) nor (1.20) holds for any real number s.

111


10. Suppose that f has an essential singularity at z = a. Prove the following strengthened version of the Casorati-Weierstrass Theorem. If C E C and E > 0 are given then for each 1) > 0 there is a number o£, Ie - (XI < E, such that fez) = IX has infinitely many solutions in B(a; 8). 11. Give the Laurent series development of fez) = exp

(D.

Can you

generalize this result? 12. (a) Let A E C and show that exp{!.>.(z for 0 < Izl <

00,

+D}

where for n an =

=

ao

+~an(z' +~)

0

~

"

~ f e~ cos ( cos nt dt o

(b) Similarly, show

{ ( I)}

expp z - ; for 0 < Izl <

00,

where bn

=

~

=

bo

+ ~ bn 00

(

z"

1t)

+ ---;n( _

f "

cos (nt-.>. sin t) dt.

o

13. Let R > 0 and G = {z: Izl > R}; a functionf: G ---')- C has a removable singularity, a pole, or an essential singularity at infinity if f(Z-I) has, respectively, a removable singularity, a pole, or an essential singularity at z = O. If f has a pole at 00 then the order of the pole is the order of the pole of f(Z-I) at z = O. (a) Prove that an entire function has a removable singularity at infinity iff it is a constant. (b) Prove that an entire function has a pole at infinity of order m iff it is a polynomial of degree m. (c) Characterize those rational functions which have a removable singularity at infinity. (d) Characterize those rational functions which have a pole of order m at infinity. 14. Let G = {z: 0 < lzl < I} and let f: G ---')- C be analytic. Suppose that y is a closed rectifiable curve in G such that nCy; a) = 0 for all a in C - G. What is J1f? Why? 15. Let f be analytic in G = {z: 0 < Iz-ai < r} except that there is a sequence of poles {an} in G with an ---')- a. Show that for any w in C there is a sequence {zn} in G with a = lim z" and w = lim f(zn).

112

Singularities

16. Determine the regions

III

which the functions

f (z) = (sin 1. )-, z

and

g(z)= Ia'(t-z)-Jdt are analytic. Do they have any isolated singularities? Do they have any singularities that are not isolated? 17. Letfbe analytic in the region G=ann(a;O,R). Show that if ff)f(x+ iy )/2 dx qy < 00 then f has a removable singularity at z = a. Suppose that p > 0 and f f)f(x+ iy)/P dxqy < 00; what can be said about the nature of the singularity at z = a? § 2. Residues

The inspiration behind this section is the desire for an answer to the following question: If f has an isolated singularity at z = a what are the possible values for f J when y is a closed curve homologous to zero and not passing through a? If the singularity is removable then clearly the integral will be zero. If z = a is a pole or an essential singularity the answer is not always zero but can be found with little difficulty. In fact, for some curves y, the answer is given by equation (1.12) with n=-1. 2.1 Definition. Let f have an isolated singularity at z

fez) =

= a and let

00

L

n= -

a.(z-a)" 00

be its Laurent Expansion about z = a. Then the residue of fat z = a is the coefficient a _ l ' Denote this by Res (/; a) = a _ l ' The following is a generalization of formula (l.l2) for n = -I. 2.2 Residue Theorem. Let f be analytic in the region G except for the isolated singularities ai' a 2 , ••. ,am' If y is a closed rectifiable curve in G which does not pass Ihrough any of the points a k and if y=:O in G then 1 m -2. f = n ( y; a k ) Res (j; a k ) • 71"1

J Y

L

k=l

Proof. Let mk = n( y; ak) for 1~ k :::; m, and choose positive numbers rJ, ... ,rm such that no two disks B(ak;rd intersect, none of them intersects {y}, and each disk is contained in G. (This can be done by induction and by using the fact that y does not pass through any of the singularities.) Let Yk (I) = ak + rk exp( - 271"imkt) for 0:::; t :::; I. Then for I '$ j :::; m m

n(y;a)+ ~ n(Yk;a)=O. k= J

m

n(y;a)+ ~ n(Yk;a)=O k= J

Residues

for all a not in G-{al, ... ,am }. Since Theorem IV.5.7 gives

J

113

is analytic

In

G-{al, ... ,am }

m

2.3

0=

fJ+ L f J. k= I

y

If J(z) =

I:

y,

bn(z - ad n is the Laurent expansion about z

-oc

series converges uniformly on aB(a k; rk )· Hence f

f (z b- f (z -

But

=

ak

then this

oc

a k )" = 0 if

11

"* -}

J = I: bnf (z

y,

- oc

- ak)"·

Yk

since (z - ad" has a primitive. Also

Yk

1

y,

a k )-1

= 2'ITin(Yk; ak)Res(f; ak)' Hence (2.3) implies the de-

sired result. I11III Remark. The condition in the Residue Theorem that J have only a finite number of isolated singularities was made to simplify the statement of the theorem and not because the theorem is invalid when J has infinitely many isolated singularities. In fact, if J has infinitely many singularities they can only accumulate on aGo (Why?) If r=d({y},JG) then the fact that y~O gives that n(y; a) =0 whenever d(a; aG) < tr. (See Exercise IV.7.2.) The Residue Theorem is a two edged sword; if you can calculate the residues of a function you can calculate certain line integrals and vice versa. Most often, however, it is used as a means to calculate line integrals. To use it in this way we will need a method of computing the residue of a function at a pole. Suppose 1 has a pole of order m ::::: I at z = a. Then g(z) = (z-ar/(z) has a removable singularity atz = aandg(a) # O.Letg(z) = bo+b1(z-a)+ ... be the power series expansion of g about z = a. It follows that for z near but not equal to a, bo bm - I ~ 'k I(z) = - ( ) .. + ... + (--) + 1... bm+k(z-a). z-a z-a k=O This equation gives the Laurent Expansion of 1 in a punctured disk about z = a. But then Res (f; a) = bm - 1 ; in particular, if z = a is a simple pole Res (f; a) = g(a) = lim (z-a)/(z). This is summarized as follows. z-+a

2.4 Proposition. Suppose 1 has a pole (z-a)m/(z); then

Res (f; a)

01 order m at z 1

= a and put g(z) =

g(m - I )(a). (m-l)! The remainder of this section will be devoted to calculating certain integrals by means of the Residue Theorem 2.5 Example. Show

=

114

Singularities

2

z If fez) = - 14 then! has as its poles the fourth roots of -l. These are

+z

exactly the numbers e iO where (J

rr 3rr 5rr

=-

4' 4 ' 4 '

7rr

and 4 .

Let

an =

exp({~ +(n-l)~J)

for n = 1, 2, 3, 4; then it is easily seen that each an is a simple pole off. Consequently, Res (f; al) = lim (z-al)!(z) = ai(al-a2)-1(al-a3)-1(al-a4)-1

(rri) 4 .

l-i

= 4~2 = ! exp Similarly Res (f; a 2 )

=

~~~i = t exp (-!rri).

Now let R > 1 and let y be the closed path which is the boundary of the upper half of the disk of radius R with center zero, traversed in the counter-clockwise direction. The Residue Theorem gives

-R -I

0

1 R

~f! = Res (f; a1)+Res (f; a2)

2m

1

-i

2~2

But, applying the definition of line integral,

This gives 2.6

Residues

115 4il For 0 S t S'TT, 1+ R 4e lies on the circle centered at 1 of radius R4; hence II + R 4e4i' l ~ R4-1. Therefore

2

and since ~ ~ 0 for all x in IR, it follows from (2.6) that l+x

f~ co

-

l+x

fR

dx = lim R .... oo

-R

00

x 24 dx l+x

7T

= ../2 2.7 Example. Show

f 00

Sin

xd _

--;- x-

o

7T

2·

ei % The functionJ(z) = - has a simple pole at z = O. If 0 < r < R let Y be the

z

closed curve that is depicted in the adjoining figure. It follows from Cauchy's

Theorem that 0

2.8

Jy f

=

o~ t

Breaking y into its pieces,

dx+ f'>+ T'>+ f'> -R

YR

y,

where YR and Yr are the semicircles from R to - Rand - r to r respectively. But

f

x I feiX-e-'iX dx --dx=-

R

R

Sin

,

x

2i

x

r

R

1 =2i

-r

Ie-dx+1 I -dx ex x 2; i

,

ix

"

-R

116

Singularities

Also

I

I n

e iZ ~dz = i

exp (iRe i9 ) de

o

YR

: ; oJlexp (i R eiB)1 de 1[

Jexp ( - R sin e) de 1t

o

By the methods of calculus we see that, for 0 > 0 sufficiently small, the largest possible value of exp (- R sin 8), with 0 ::; e ::; 7r - 0, is exp (- R sin 0). (Note that 0 does not depend on R if R is larger than 1.) This gives that

I~~:

J

,,- 8

+

dz ::; 20

exp ( - R sin 8) de

YR

5

< 20+11" exp (-R sin 0). If

E

t

> 0 is given then, choosing 0 <

there is an Ro such that exp ( - R

E,

E

sin 0) < - for ali R > Ro. Hence 3'IT lim R .... co

J

e iZ

-

dz

=

O.

Z

YR

eiz_l Since - - has a removable singularity at z = 0, there is a constant

z

iZ

M > 0 such that /e z

11::;

M for

f

eiZ -

Izl ::;

1. Hence,

I

- z - dz ::::; 7rrM;

Yr

that is,

But

I~

°

= lim

dz

= -'ITt

' .... 0

I

eiZ-1

- - - dz.

for each r so that -7ri = lim , .... 0

z

I

e iZ - dz z

Yr

So, if we let r -+ 0 and R -+

00

in (2.8)

117

Residues

Notice that this example did not use the Residue Theorem. In fact, it could have been presented after Cauchy's Theorem. It was saved until now because the methods used to evaluate this integral are the same as the methods used in applying the Residue Theorem. 2.9 Example. Show that for a > I,

f '"

o "0

If z = el then

z=

dO a + cos 0 =

'TT

.ja2 - I .

I

- and so

z

( + -I)

a+cos B = a+t(z+z} = a+-!- z Hence

f '"

o

z

=

z2+2az+1 . 2z

tf f 2",

dB a+cos B -

dB a+cos B

0

dz z2+2a z+ I

= -i

y

where y is the circle Izl = 1. But z2+2a z+ I = (z-or:) (z-{3) where or: = -a+(a 2-I)t, fJ = -a-(a 2 -I)t. Since a > 1 it follows that 1Or:1 1. By the Residue Theorem

f

dz

Z2

'TTi

+ 2a z + I = .ja2 -

I ;

y

by combining this with the above equation we arrive at

f

"

o

dB a + cos B =

'TT

.ja2 - I .

2.10 Example. Show that

f 00

o

log x dx 1+x2

=

O.

To solve this problem we do not use the principal branch of the logarithm. Instead define log z for z belonging to the region G = {Z

E

C: z "# 0 and

-2'TT

3'TT} ; < arg z <"2

118

if z = Izl

-~

oF 0 with

el9

Singularities

< fJ < 3;, let t(z) = log IzJ+i8. Let 0 < r < R

and let y be the same curve as in Example 2.7. Notice that t'(x) = log x for x > 0, and t(x) = log Ixl +7Ti for x < O. Hence,

f

1.11

f f R

t'(z) d -- Z = l+z2

7

"

log x d 'R f[lOg R+ifJ] 19.11l - - X+I e "u l+x2 I+R 2 e219 0

0

-r

+

. f[IOgr+ifJ] 19.1 log Ix l +7Ti d 1 2 X+1f 1 2 219 e ,,'8 +X +r e "

-R

Now the only pole of t(z) (1 + Z2) -1 inside y is at z = i; furthermore, this is a simple pole. By Proposition 2.4 the residue of t(z) (1 + Z2) -1 is [log Iii +tn-i]

7T

= 4' So,

f

t(z) dz = 7T 2 i l+z2 2

y

f

f

r

r

Also, R

R

-r

x dx logf I + x2

+

flOg Ixl + 7Ti dx = 2 1+ x 2

-R

r

Letting r _ 0+ and R _

00,

R

logx dX+7TI. -dx1+ x 2 1+ x 2

and using the fact that

(Exercise 2(f», it follows from (2.11) that 00

IOgX f - 12 dx = +x o

"

' . f[IOgr+ifJ] 19 1 r"'O+ lIm If 1 2 2111 e dfJ +r e

- t

0

lim iR

R ... oo

f

"[lOg R+ifJ]

o

1 R2

+

2i9

e

i8

e dfJ.

We now show that both of these limits are zero. If p > 0 then '" [log p

IP jo

1

+

+ iO]o

I

pilog pi j'"

. e' dO < p2e 2 'O 11 - p21

'lTpllog pi 11 - p21

0

dO

---::- + letting p - 0 + or p -

00,

+

P

j'11

11 - p21

0

P'IT 2

.

211 _ p21 '

the limit of this expression is zero.

0 dO

~

Residues

f

119

00

2.12 Example. Show that

o

x- dx = C

1+x

~ if 0 < sm 7rC

c < 1.

To evaluate this integral we must consider a branch of the function z-c. The point z = 0 is called a branch point of z-c, and the process used to

evaluate this integral is sometimes called integration around a branch point. Let G = {z: z i= 0 and 0 < arg z < 27r}; define a branch of the logarithm on G by putting t(re iO ) = log r + i{) where 0 < () < 21T. For z in G put I(z) = exp [- ct(z)]; so I is a branch of z- We now select an appropriate curve Y in G. Let 0 < r < 1 < R and let 8 > O. Let Ll be the line segment [r+8i,. R+8i]; YR the part of the circle [z[ = R from R+8i counterclockwise to R-8i; L2 the line segment [R-8i, r-8i]; and Y, the part of the circle [z[ = r from r - 8i clockwise to r + 8i. Put Y = Ll + YR + L2 + Yr. Since Y""'" 0 in G and Res (f(z) (l +Z)-l; -1) = I( -1) = e- inc , the Residue Theorem gives C.

d 2· f l(Z) 1+z z = me

2.13

-inc

.

Using the definition of a line integral

f

L,

fez} --dz= 1+ z

fR f(/+i8} dl r

1+ 1+ i8

.

Let g( 1,8) be defined on the compact set [r, R] X [0, ~ 'IT] by f(/+i8) I- I g(/,8)= I 1+I+i8 - 1+1 C

when 8 >0 and g(t,O)=O. Then g is continuous and hence uniformly continuous. If E:>O then there is a 80 such that if (/-t'f+(8-8,)2<8g then [g(t,8)-g(/',8')[
Singularities

120

and 0 < 80 , Thus

f

g ( t, 0 ) dt ::; f

R

r

for 8 < 00 , This implies that

f R I+t t-C

2.14

.

- - dt= hm

8~O+

r

f

J(z) L,I+z

- - dz.

Similarly. using the fact that I (i) = I (z) + 27Ti t-- dt- I'1m -e -bTicjR C

2.15

r

I+t

8~O+

f

J(z) - dz.

L,

I+z

Now the value of the integral in (2.13) does not depend on 8. Therefore, letting O~O+ and using (2.14) and (2.15) gives t -- d t- I'1m [ . - iwc - ( I - e - 2wic ) j R27Tle . 1+ t Il~O+

2.16

C

1 1

r

J (z-) -

y,

1+ z

J (z) 1 -.

+

Now if p>O and p=i=I and if Yp is the part of the circle yp2_8 2 +io to yp2_8 2 -i8 then

YR

1+ z

Izl=p

from

Since this estimate is independent of 8, (2.16) implies

27Tie-i7T('-(I-e-2Wi<')jR ~ dtl::;

1

r

1+ t

~27Tr+ ~27TR. II - rl II - R I

But as r~O+ and R~oo the right-hand side of this last inequality converges to zero. Hence

f co

. 27Ti e- mc = (l-e- 2',nc)

or,

o

f co

o

-t-

t-C d 27Ti e- i1tc - - t - ----::-:1+1 - l-e 2n;c

27Ti 7T sin 7TC

=--

C

l+t

dt· '

Residues

121

Exercises 1. Calculate the following integrals:

Jcos ;2- 1 dx co

(b)

f ft

(c)

o

cos 28d8 2 where a2 < 1 I-2acos 8+a

f ft

(d)

o

d8 8)2 where a > 1. (a+cos

2. Verify the following equations:

f(lOgI+x2 co

(b)

X)3

dx = 0

o

()f co

cos ax d _ 1T(a+ 1) e- a if a > 0; (1+X2)2 X 4

C

o

f

,,/2

(d)

o

d8 1T a+sin28 = 2[a(a+ I)]t '

co

(e)

log x f (I+x2)2 dx = -

if a > 0;

1T

4.;

o

f co

(g)

-co

eax

7T

- - dx = - - if 0 < a < 1; 1 +ex sina7T

2"

(h) flog sin 2 28d8

I log sin 8d8 = -% log 2. "

=

o

4

0

3. Find all possible values of f y expz - 1 dz where y is any closed curve not passing through z = O. 4. Suppose thatfhas a simple pole at z = a and let g be analytic in an open set containing a. Show that Res (fg; a) = g(a) Res (f; a). 5. Use Exercise 4 to show that if G is a region andfis analytic in G except for simple poles at aI' ... ,an; and if g is analytic in G then

f

1 . fg -2

1Tl

1

= Ln

k= I

n(y; ak) g(ak) Res (f; ak)

122

Singularities

for any closed rectifiable curve y not passing through ai' ... , an such that y:::;:: 0 in G. 6. Let y be the rectangular path [n --I--H-- ni, -11--t+ni, -n--t -ni, n +-t-ni, n+-t+ni] and evaluate the integral L7T(z+ar- z cot 7Tzdz for a # an integer. Show that lim 7T(Z --1-- a) - 2 cot 17zdz = 0 and, by using the first part, deduce that 11--+'"

L

7T 2

1

ro

sin2-;~ = n=~CX) (a+~)2 (Hint: Use the fact that for z = x + iy. Icoszl 2 = cos z X + sinh 2 y and Isinzl 2 = sin 2 x + sinh 2 y to show that Icotnzl :::;; 2 for zon y ifnis sufficiently large.) 7. Use Exercise 6 to deduce that 1

w

~ (211+ 1)2

8

8. Let y be the polygonal path defined in Exercise 6 and evaluate Sr ?TeZ 2 - a 2)-1 cot 7Tzdz for a =f. an integer. Show that lim 7T(z2 - a 2 ) -[ cot 7Tzdz = 0, and consequently Il--->W

L

7T

cot 'ITa = -1 a

+

2: -2a00

11=1

a 2 _n 2

for a =f. an integer. 9. Use methods similar to those of Exercises 6 and 8 to show that

for a =f. an integer. 10. Let y be the circle Izl Show that

and let m and n be non-negative integers.

(± 1)p(n+2p)! p! (n+p)! '

o

if m

= 2p+n,

p ?:: 0

otherwise

II. In Exercise 1.12, consider all and bn as functions of the parameter A and use Exercise 10 to compute power series expansions for anCA) and bnCA). (bnCA) is called a Bessel function.) 12. Let f be analytic in the plane except for isolated singularities at ai' a2, ... , am' Show that Res (f; 00) =

m

2: Res (f; a k=l

k )·

The Argument Principle

123

(Res(j; (0) is defined as the residue of -Z-1(Z-I) at z=O. Equivalently, Res(f; (0) = - - 12. If when y(t) = Re", 0::; I ::; 21)", for sufficiently large R.) 1)"1 .

What can you say if j has infinitely many isolated singularities? 13. LetJbe an entire function and let a, b E C such that lai < Rand Ibl < R. Ify(t) = Reit,O::; t:::; 27T,eva!uateL[(z-a)(z-b)]-IJ(z)dz. Use this result to give another proof of Liouville's Theorem.

§3 The Argument Principle Suppose thatJis analytic and has a zero of order mat z = a. SoJ(z) (z-a)mg(z) where g(a) of O. Hence

=

I'(z) m g'(z) -=-+-J(z) z-a g(z)

3.1

and g'lg is analytic near z = a since g(a) of O. Now suppose that J has a pole of order m at z = a; that is, J(z) = (z - a) -mg(z) where g is analytic and g(a) of O. This gives I'(z) .............. =_-m ........... J(z) Z-Q

3.2

g'(z) + ....... g(z)

and again g'!g is analytic near z = a. Also, to avoid the phrase "analytic except for poles" which may have already been used too frequently, we make the following standard definition.

3.3 Definition. If G is open andJis a function defined and analytic in G except for poles, then J is a meromorphic Junction on G. Suppose that J is a meromorphic function on G and define f: G --)- Coo by setting J(z) = co whenever z is a pole of I It easily follows that J is continuous from G into Coo (Exercise 4). This fact allows us to think of meromorphic functions as analytic functions with singularities for which we can remove the discontinuity of f, although we cannot remove the nondifferentiability of I 3.4 Argument Principle. Let J be meromorphic in G with poles PI' P2' ... ,Pm and zeros ZI' Z2, . . . ,Zn counted according to multiplicity. IJ y is a closed rectifiable curve in G with y:::::; 0 and not passing through PI' ... ,Pm ; ZI' ••• , Zn; then

3.5

I fI'CZ)

-.

J(z)

27T1

dz =

L n(y; Zk) k=1 n

-

L n(y;pJ. j=1 m

y

Prool By a repeated application of (3.1) and (3.2) I'(z)

J(z)

f

f.

_1__ _1_ k=lz-zk r 1Z - Pj

+ g'(z).. g(z)

124

Singularities

where g is analytic and never vanishes in G. Since this gives that g'lg is analytic, Cauchy's Theorem gives the result. III Why is' this called the "Argument Principle"? The answer to this is not completely obvious, but it is suggested by the fact that if we could define logf(z) then it would be a primitive for I'll Thus Theorem 3.4 would give that as z goes around y, logf(z) would change by 27riK where K is the integer on the right hand side of (3.5). Since 27riK is purely imaginary this would give that 1m log fez) = arg fez) changes by 27TK. Of course we can't define log fez) (indeed, if we could then L f'lf = 0 since f'lf has a primitive). However, we can put the discussion in the above paragraph on a solid logical basis. Since no zero or pole off lies on y there is a disk B(a; r), for each a in {y}, such that a branch of logf(z) can be defined on Bea; r) (simply select r sufficiently small that fez) #- or 00 in B(a; r». The balls form an open cover of {y}; and so, by Lebesgue's Covering Lemma, there is a positive number E > 0 such that for each a in {y} we can define a branch of log fez) on B(a; E). Using the uniform continuity of y (suppose that y is defined on [0, 1]), there is a partition 0 = to < t 1 < ... < tk = I such that y(t)EB(y(tj_l); E) for lj-l.:s; t.:s; tj and I .:s;j.:s; k. Let I j be a branch of log f defined on B(y(tj-I); E) for 1 .:s;j.:s; k. Also, since thej-th and U+ 1)-5t sphere both contain y(t) we can choose II, ... , I~ so that II (y(tl» = 12(y(tj»; t 2(y(t2) = (3(yt 2»;··· ; (k-I(y(tk-l» = tk(y(t k - 1 ) · If I' j is the path y restricted to [t j _ I, t J then, since tj = f'lf,

°

for 1 .:s; j .:s; k. Summing both sides of this equation the right hand side "telescopes" and we arrive at

where a = 1'(0) = 1'(1). That is, tk(a)-tl(a) = 27TiK. Because 2niK is purely imaginary we get 1m tkCa)-lm IICa) = 27TK. This makes precise our contention that as z traces out y, arg f{z) changes by 27T K. The proof of the following generalization is left to the reader (Exercise 1). 3.6 Theorem. Letfbe meromorphic in the region G with zeros Z1, Z2 •••• ' Zn and poles PI' ...• Pm counted according to multiplicity. If g is analytic in G and y is a closed rectifiable curve in G with y ~ 0 and not passing through any z i or Pj then 1 j~ f' . g - = 27T1 f

L g(zi)n(y; Zi) n

;=1

-

L m

j=1

g(p)n(y;p).

y

We already know that a onc-one analytic function f has an analytic inverse (IV. 7.6). It is a remarkable fact that Theorem 3.6 can be used to give


125

a formula for calculating this inverse. Suppose R > 0 and that I is analytic and one-one 011 B(a; R); let n = I[B(a; R)]. If Iz-al < Rand g = I(z) E n then I( w) - g has one, and only one, zero in B(a; R). If we choose g( w) == w, Theorem 3.6 gives

f

z = _1 wf'(w) dw 2'T1'i I(w)-g y

where y is the circle Iw-al = R. But z =I-I(g); this gives the following 3.7 Proposition. Let I be analytic on an open set containing B(a; R) and suppose that I is one-one on B(a; R). II n = I[B(a; R)] and y is the circle Iz-al = R thenl-l(w) is definedlor each w in n by thelormula

f

I-I(w) = _1 zf'(z) dz. 2'T1'i I(z)-w y

This section closes with Rouche's Theorem.

3.8 _Rouche's Theorem. Suppose J and g are meromorphic in a neighborhood

oj B (a; R) with no zeros or poles on the circle y = {z: Iz - al = R}. IJ Zf' Zg (Pf , Pg ) are the number oj zeros (poles) oj J and g inside y counted according to their multiplicities and if IJ(z) + g(z)1 < IJ(z)1 + 1 g(z)1 on y, then Proof. From the hypothesis

I IJ(z) I g(z) +1 < g(z) +1 IJ(z) on y. If A = J(z)/g(z) and if A is a positive real number then this inequality becomes A + 1 < A + 1, a contradiction. Hence the meromorphic function f / g maps y into n = C - [0, (0). If I is a branch of the logarithm on n then l(f(z)jg(z» is a well-defined primitive for (f/g)'(f/g)-l in a neighborhood of y. Thus

£ = 2~i f[ j - ~]

0=

2~i

y

(f/g)'(f/g)-l

y

=(Zf- Pf) -(Zg- Pg). •

This statement of Roucbe's Theorem was discovered by Irving Glicksberg (A mer. Math. Monthly, 83 (1976), 186-187). In the more classical statements of the theorem, f and g are assumed to satisfy the inequality

126

Singularities

1/+ gl < Igl on y. This weaker version often suffices in the applications as can be seen in the next paragraph. Rouche's Theorem can be used to give another proof of the Fundamental Theorem of Algebra. If p(z) = z"+a 1 z,,-1 + ... +a" then

p(z) a1 a" -=1+-+"'+z" z z" and this approaches 1 as z goes to infinity. So there is a sufficiently largt. number R with

/p;:) - 11 < 1

for Izl = R; that is, Ip(z)-z"1 < Izl" for Izl = R. Rouchc's Theorem says that p(z) must have n zeros inside Izl = R. We also mention that the use of a circle in Rouchc's Theorem was a convenience and not a necessity. Any closed rectifiable curve" with" """" 0 in G could have been used, although the conclusion would have been modified by the introduction of winding numbers.

Exercises I. Prove Theorem 3.6. 2. Suppose / is analytic on .8(0; I) and satisfies 1/(z)1 < I for Izl = I. Find the number of solutions (counting multiplicities) of the equation/(z)=zn where n is an integer lar~r than or equal to I. 3. Let / be analytic in B(O; R) with /(0)=0, /,(0)*0 and /(z)*O for 0< Izl S R. Put p=min{if(z)I:lzl = R} >0. Define g: B(O; p)_C by

g(w)= _I 2'1Ti

1 y

z/,(z) dz /(z)-w

where y is the circle Izl = R. Show that g is analytic and discuss the properties of g. 4. If / is meromorphic on G andj: G-C"" is defined by j(z) = 00 when z is a pole of / and j(z) = fez) otherwise. show that j is continuous. 5. Let f be meromorphic on the region G and not constant; show that neither the poles nor the zeros of f have a limit point in G. 6. Let G be a region and let B(G) denote the set of all analytic functions on G. (The letter" B" stands for holomorphic. Some authors call a differentiable function holomorphic and call functions analytic if they have a power series expansion about each point of their domain. Others reserve the term "analytic" for what many call the complete analytic function, which we will not describe here.) Show that B(G) is an integral domain; that is, B(G) is a commutative ring with no zero divisors. Show that M(G), the meromorphic functions on G, is a field. We have said that analytic functions are like polynomials; similarly, meromorphic functions are analogues of rational functions. The question


127

arises, is every meromorphic function on G the quotient of two analytic functions on G? Alternately, is M(G) the quotient field of H(G)? The answer is yes but some additional theory will be required before this answer can be proved. 7. State and prove a more general version of Rouche's Theorem for curves other than circles in G. 8. Is a non-constant meromorphic function on a region G an open mapping of G into C? Is it an open mapping of G into Coo ? 9. Let A > I and show that the equation A-z-e- z = 0 has exactly one solution in the half plane {z: Re z > O}. Show that this solution must be real. What happens to the solution as A---+ I ? 10. Let f be analytic in a neighborhood of D= ii(O; I). If If(z)1 < I for Izl = I, show that there is a unique z with Izi < I and f(z) = z. If If(z)1 ~ I for Izl = I, what can you say?

Chapter VI The Maximum Modulus Theorem This chapter continues the study of a property of analytic functions first seen in Theorem IV. 3.1 I. In the first section this theorem is presented again with a second proof, and other versions of it are also given. The remainder of the chapter is devoted to various extensions and applications of this maximum principle.

§l. The Maximum Principle Let n be any subset of IC and suppose a is in the interior of n. We can, therefore, choose a positive number p such that B(a; p) C [2; it readily follows that there is a point f in n with If I > 14 To state this another way, if a is a point in n with lal 2': I~i for each ~ in the set n then ex belongs to on.

1.1 Maximum Modulus Theorem---First Version. Iff is analytic in a region G and a is a point in G with If(a)[ 2': If(z)[ for all z in G then f must be a constant function. Proof Let 0 = f(G) and put a = f(a). From the hypothesis we have that IIXI 2': If I for each gin il; as in the discussion preceding the theorem IX is in 00 n n. In particular, the set .0 cannot be open (because then .0 n 8.0 = [J). Hence the Open Mapping Theorem (IV. 7.5) says thatfmust be constant. II

1.2 Maximum Modulus Theorem-Second Version. Let G be a bounded ope/! set in IC and suppose f is a continuous function on G- which is analytic in G. Then max {If(z)l: z E G-} = max {If(z)l: z E oG}. Proof Since G is bounded there is a point a E G-- such that I/(a)1 2': I/(z) I for all z in G-. Iflis a constant function the conclusion is trivial; iflis not constant then the result follows from Theorem 1.1. II Note that in Theorem 1.2 we did not assume that G is connected as in Theorem 1.1. Do you understand how Theorem 1.1 puts the finishing touches on the proof of 1.2? Or, could the assumption of connectedness in Theorem 1.1 be dropped? Let G = {z = x+iy: -t1T < y < 11T} and put/(z) = exp [exp z]. Then I is continuous on G- and analytic on G. If z E oG then z = x ± 11Ti so I/(z) I = lexp (± ieX)1 = 1. However, as x goes to infinity through the real numbers, I(x) -+ 00. This does not contradict the Maximum Modulus Theorem because G is not bounded. In light of the above example it is impossible to drop the assumption of the boundedness of G in Theorem 1.2; however, it can be replaced. The 128

The Maximum Principle

129

substitute is a growth condition on [f(z) [ as z approaches infinity. In fact, it is also possible to omit the condition that f be defined and continuous on G-. To do this, the following definitions are needed. 1.3 Definition. Iff: G -+ IR and a E G- or a = 00, then the limit superior of fez) as z approaches a, denoted by lim sup fez), is defined by

lim supf(z) = lim sup {fez): z r-+O+

z ...... a

E

G n B(a; r)}

(If a = 00, B(a; r) is the ball in the metric of COO') Similarly, the limit. inferior offez) as z approaches a, denoted by lim inf fez), is defined by z-+a

lim inff(z) = lim inf {fez): z r-+O+

z-+a

E

G n B(a; r)}.

It is easy to see that lim fez) exists and equals

lim inf fez).

IX

z-+a

iff

0:

= lim sup fez) =

If G c C then let 000G denote the boundary of G in Coo and call it the extended boundary of G. Clearly 000G = oG if G is bounded and 000G = oG U {oo} if G is unbounded. After these preliminaries the final version of the Maximum Modulus Theorem can be stated. 1.4 Maximum Modulus Theorem-Third Version. Let G be a region in C and f an analytic function on G. Suppose there is a constant M such that lim sup Ifez) [ :::; M for all a in 0ooG. Then Ifez) [ :::; M for all z in G. z-+a

Proof Let S > 0 be arbitrary and put H = {z E G: If(z) [ > M+S}. The theorem will be demonstrated if H is proved to be empty. Since If I is continuous, H is open. Since lim sup [f(z) [ :::; M for each z->a

a in 0ooG, there is a ball B(a; r) such that If(z) I < M+S for all z in G n B(a; r). Hence H- c G. Since this condition also holds if G is unbounded and a = 00, H must be bounded. Thus, H - is compact. So the second version of the Maximum Modulus Theorem applies. But for z in oH, If(z) [ = M+8 since H- c {z: If(z) I ~ M+S}; therefore, H = 0 or fis a constant. But the hypothesis implies that H = 0 if f is a constant. • Notice that in the example G = {z: [1m z[ < 1-1T}, fez) = exp (e f satisfies the condition lim sup [f(z) [ :::; 1 for all a in oG but not for a = 00. Z

),

z-+a

Exercises 1. Prove the following Minimum Principle. If f is a non-constant analytic function on a bounded open set G and is continuous on G-, then either f has a zero in G or If I assumes its minimum value on oG. (See Exercise IV. 3.6.) 2. Let G be a bounded region and suppose f is continuous on G- and

130

The Maximum Modulus Theorem

analytic on G. Show that if there is a constant c ~ 0 such that If(z) 1 = c for all z on the boundary of G then either f is a constant function or f has a zero in G. 3. (a) Letfbe entire and non-constant. For any positive real number c show that the closure of {z: /f(z)/ < c} is the set {z: If(z)1 :s; c}. (b) Let p be a non-constant polynomial and show that each component of {z:lp(z)1 < c} contains a zero of p. (Hint: Use Exercise 2.) (c) If p is a non-constant polynomial and c> 0 show that {z: Ip(z)1 = c} is the union of a finite number of closed paths. Discuss the behavior of these paths as c -> XJ. 4. Let 0 < r < R and put A = {z: r :s; Izl :s; R}. Show that there is a positive number £ > 0 such that for each polynomial p, sup {/p(Z)-Z-l/:

ZE

A} ~

E

This says that Z-1 is not the uniform limit of polynomials on A. 5. Let f be analytic on B(O; R) with If(z)1 :s; M for Izi :s; Rand If(O) I = a > O. Show that the number of zeros off in B(O; lR) is less than or equal to

IO~ 2 log ( ; ) .

Hint: If Z I'

... ,

z. are the zeros off in B(O; 1R), consider

the function

and note that g(O)

= f(O). (Notation:

{I ak = a 1a2 ... an.)

k~1

6. Suppose that bothfand g are analytic on B(O; R) with If(z)1 = /g(z)/ for Izi = R. Show that if neither f nor g vanishes in B(O; R) then there is a constant '\, 1,\1 = I, such thatf = '\g. 7. Let f be analytic in the disk BCO; R) and for 0 :s; r < R define A(r) = max {Re/(z): Izl = r}. Show that unless 1 is a constant, A(r) is a strictly increasing function of r. 8. Suppose G is a region,!: G -)- C is analytic, and M is a constant such that whenever z is on GroG and {zn} is a sequence in G with z = lim Zn we have lim sup /f(zn) I :s; M. Show that I/(z)1 :s; M, for each z in G. §2. Schwarz's Lemma

2.1 Schwarz's Lemma. Let D = {z: Izl < I} and suppose f is analytic on D with (a) If(z) 1:s; 1 for z in D, (b) f(O) = O.

Then 11'(0)1 :s; I and If(z)1 :s; Izlfor all z in the disk D. Moreover ifl1'(O») = 1 or if I/(z)1 = /zlJor some z #- 0 then there is a constant c, lei = 1, such that f(w) = cw for all w in D.

Schwarz's Lemma Proof. Define

g:

D~C by g(z)= f(z) z

for Z'FO and g(O) =1'(0); then

g is

analytic in D. Using the Maximum Modulus Theorem, Ig(z)l:s; r- for Izl $ rand 0 < r < l. Letting r approach 1 gives Ig(z)1 $ 1 for all z in D. That is, If(z)l:s; Izl and 11'(0)1 = Ig(O)1 :s; 1. If If(z)1 = Izi for some z in D, Z9"O, or 11'(0)1 = 1 then Igl assumes its maximum value inside D. Thus, again applying the Maximum Modulus Theorem, g(z)= e for some constant e with lei = l. This yields f(z)= ez and completes the proof of the theorem. II We will apply Schwarz's Lemma to characterize the conformal maps of the open unit disk onto itself. First we introduce a class of such maps. If lal < 1 define the Mobius transformation: 1

z-a fP,,(z) = - - _

l-az

Notice that fPa is analytic for Jzl < lal- 1 so that it is analytic in an open disk containing the closure of D = {z: Izl < 1}. Also, it is an easy matter to check that for

Izi < 1. Hence qJa maps D into itself in a one-one fashion. Let 8 be a real number; then

eiO -

I = e

iO -

a

ii

I

I

= 1 This says that qJa(i3D) = aD, and, from the preceding material, qJ(D) = D. These facts, and other pertinent information which can be easily checked, are summarized as follows. 2.2 Proposition. 1/ lal < 1 then fPa is a one-one map 0/ D = {z: Izi < I} onto itself; the inverse of fPa is fP-a' Furthermore, fP" maps 8D onto 8D, fPia) = 0, fP~(O) = l-laI 2 , andfP~(a) = (1-laI 2)-1. Let us see how these functions fP" can be used in applying Schwarz's Lemma. Suppose / is analytic on D with I/(z) I $ 1. Also, suppose lal < 1 and/(a) = IX (so IIXI < 1 unless/is constant). Among all functions/having these properties what is the maximum possible value of If'(a) I? To solve this problem let g = fP" 0 / 0 fP-.' Then g maps D into D and also satisfies g(O) = rp,,(/(a)) = fP,,(Q() = O. Thus we can apply Schwarz's Lemma to obtain that Ig'(O)1 $ 1. Now obtain an explicit formula for g'(O). Applying the chain rule

132


g'(O) = (qJl> f)'( qJ_,,(O))qJ'_JO) 0

= ('1'..

0

1)'(0)(1-101 2 )

= 'P~(a.)f'(0)

(1-101 2)

= 1- 10 12 1'(0). l-11X12

Thus,

11'(0)1

2.3

:5

l-11X12 .

l-lal 2

Moreover equality will occur exactly when Ig'(O)1 = I, or, by virtue of Schwarz's Lemma, when there is a constant c with lei = I and

2.4

f (z) = q; _ a (cCPa (z»)

Izl < 1. We are now ready to state and prove one of the main consequences of Schwarz's Lemma. Note that if lei = 1 and lal < 1 then f = eCPa defines a one-one analytic map of the open unit disk D onto itself. The next result says that the converse is also true. for

2.5 Theorem. Let f: D --'> D be a one-one analytic map of D onto itself and suppose f( a) = O. Then there is a complex number c with lei = 1 such that

f

=

CqJ".

Proof Since I is one-one and onto there is an analytic function g: D -+ D such that g{ I(z» = z for Izl < I. Applying inequality (2.3) to both I and g gives II'(a)1 :5 (I -Ial l ) - I and Ig'(O)! :5 I -lal 2 (since g(O) = a). But since

1= g'(O)I'(a), II'(a)[ = (1-laI 2)-I. Applying formula (2.4) we have that f = CqJ(1 for some e, lei = I. II

Exercises

1. Suppose If(z)1 :5 1 for Izl < 1 and f is a non-constant analytic function. By considering the function g: D --'> D defined by g(z)

I(z)-a

= I-a/(z)

where a = I(O}, prove that

I/(O)I-I~I :5 I/(z)1 :5 1/(0)1 +fl

1+ I/(O)llzl

1-

1/(0)1 Izi

for Izi < 1. 2. Does there exist an analytic function/: D---)o D with/(t) =

,?

i

andl'(t) =

133

Convex functions and Hadamard's Three Circles Theorem

3. Suppose f: D-+C satisfies Re f( z) ~ 0 for all z in D and suppose that f is analytic and not constant. (a) Show that Ref(z»O for all z in D. (b) By using an appropriate Mobius transformation, apply Schwarz's Lemma to prove that if f(O) = I then

I/(z) I ~ I + Izl

l-lzl

for Izi < 1. What can be said if1(0) =F I? (c) Show that if f(O) = I, f also satisfies I/(z)1

~ 1-izi 1+ Izi

.

(Hint: Use part (a». 4. Prove Caratheodory's Inequality whose statement is as follows: Letf be analytic on li(O; R) and let M(r)=max{lf(z)I:lzl=r}, A(r)= max{ Ref(z): Izl = r}; then for 0< r < R. if A (r) ~ 0,

M(r) ~ R+r [A(R) + 1/(0)1] R-r (Hint: First consider the case where 1(0) = 0 and examine the function

g(z) =/(Rz) [2A(R)+/(Rz>r 1 for Izl < 1.) 5. Letl be analytic in D = {z: Izl < I} and suppose that I/(z) I all z in D. (a) If I(zk) = 0 for 1 ~ k ~ n show that

~

M for

°

for Izl < 1. (b) If !(zk) = for 1 ~ k.:5; n, each zk"* 0, and 1(0) = Me ia (z l z2'···' zn)' find a formula for f. 6. Suppose I is analytic in some region containing B(O; 1) and II(z) I = 1 where Izl = 1. Find a formula for f (Hint: First consider the case where I has no zeros in B(O; 1).) 7. Suppose I is analytic in a region containing B(O; 1) and I/(z)1 = 1 when Izl = 1. Suppose that I has a zero at z = i(1 + i) and a double zero at z = 1· Can 1(0) = 1? 8. Is there an analytic function/ on B(O; 1) such that I/(z) I < 1 for Izl < 1, 1(0) = t, and 1'(0) = !? If so, find such an f Is it unique? §3. Conpex!unctiollS lind HtUlIImllrd'. T""ee Circle. Tlteorem In this section we will study convex functions and logarithmically convex functions and show that such functions appear in connection with the study of analytic functions.

134


3.1 Definition. If [a, b] is an interval in the real line, a functionf: [a, b] is convex if for any two points Xl and X2 in [a, b]

-+ ~

whenever 0 ::; t ::; 1. A subset A c IC is convex if whenever z and IV are in A, tz + (I - t)w is in A for 0 ::; t ::; 1; that is, A is convex when for any two points in A the line segment joining the two points is also in A. (See IV. 4.3 and IV. 4.4.) What is the r(,lation between convex functions and convex sets? The answer is that a function is convex if and only if the portion of the plane lying above the graph of the function is a convex set. 3.2 Proposition. A function f: [a, b]

-+ ~

is convex

iff the

set

A = {(x,y):a::; x::; b andf(x)::; y} is convex. Proof. Suppose f: [a, b] -+ ~ is a convex function and let (Xl' y,) and (Xz, Y2) be points in A. If 0 ::; t ::; 1 then, by the definition of convex function, f(tx 2 +(I-t)x,)::; (f(x 2)+(l-t)f(x,) ::; tYz+(l-t)y,. Thus t(X2' Yz)+ (I-t) (x" y,) = (tx 2 +(1-t)x" tYz+(l-t)y,) is in A; so A is convex. Suppose A is a convex set and let x" X 2 be two points in [a, b]. Then (tx 2 +(l-t)x" tf(x 2)+(I-t)f(x,» is in A if 0::; t::; 1 by virtue of its convexity. But the definition of A gives that f(tx2+(I-t)x,)::; (f(X2)+ (1- t)f(x ,); that is,f is convex . • The proof of the next proposition is left to the reader.

3.3 Proposition. (a) A function f: [a, b]

-+ ~

iff for

is convex

n

x" ... ,x" in [a, b] and real numbers t" ... , t" 2: 0 with

L tk

k='

any points

= 1,

fet, tkXk) ::; kt, td(Xk)'

(b) A set A

c

IC is convex

numbers t" ... , t" 2: 0 with

n

iff for

L tk

k~'

= 1,

any points z" ... ,z" in A and real n

L tkzk belongs to

k~'

A.

What are the virtues of convex functions and sets? We have already seen the convex sets used in connection with complex integration. Also, the fact that disks are convex sets has played a definite role, although this may not have been apparent since this fact is taken for granted. The use of convex functions may not be so familiar to the reader; however it should be. In the first course of calculus the fact (proved below) that f is convex when f" is non-negative is used to obtain a local minimum at a point to whenever !'(to) = O. Moreover, convex functions (and concave functions) are used to obtain inequalities. Iff: [a, b] -+ IR is convex then it follows from Proposition 3.2 that f(x) ::; max {f(a),f(b)} for all x in [a, b]. We now give a necessary condition for the convexity of a function.


135

3.4 Proposition. A differentiablefunctionf on [a, b] is convex ifff' is increasing. Proof. First assume that f is convex; to show that I' is increasing let a S x < y s b and suppose that 0 < t < 1. Since 0 < (l-t)x+ty-x = t(y-x), the definition of convexity gives that f«l-t)x+ty)-f(x) f(y)-f(x) S . t(y-x) y-x

Letting t --+ 0 gives that

3.5

f'(x) S fey) -f(x) . y-x

Similarly, using the fact that 0 > (l-t)x+ ty- y = (I-t) (x- y) and letting t --+ I gives that

3.6

f'(y) 2: fey) - f(x) . y-x

So, combining (3.5) and (3.6), we have thatf' is increasing. Now supposing thatf' is increasing and that x < u < y, apply the Mean Value Theorem for differentiation to find rand s· with x < r < u < s < y such that f'(r) = feu) - f(x) u-x

and f'(s)

= fCy) - feu)

.

y-u

Since f'(r) S f'(s) this gives that feu) - f(x) u-x

s

fey) - feu) y-u

whenever x < u < y. In particular by letting u = (1- t)x + ty where t < I, f(u)-f(x) < f(y)-f(u) . t(y-x) - (I-t)(y-x) , and hence (I - t) [feu) - f(x)] s t[f(y) - feu)].

o<

This shows that f must be convex . • In actuality we will mostly be concerned with functions which are not only convex, but which are logarithmically convex; that is, logf(x) is convex. Of course this assumes that f(x) > 0 for each x. It is easy to see that a logarithmically convex function is convex, but not conversely.

3.7 Theorem. Let a - IR! by

-0;

C is continuous and f is analytic in G.

If

we define M:

M(x) = sup {If(x+ iy)!: - w < y < w}, and If(z)1 < B for all z in G, then log M(x) is a convex fimction. Before proving this theorem, note that to say that 10gM(x) is convex means (Exercise 3) that for a ::s; x < u < y ::s; b, (y-x) log M(u) ::s; (y-u) log Al(x)+(u-x) log M(y) Taking the exponential of both sides gives

3.8 whenever a ::s; x < u < y ::s; b. Also, since log M(x) is convex we have that log M(x) is bounded by max {log M(a), log M(b)}. That is, for a ::s; x ::s; b

M(x) ::s; max {Mea), M(b)}. This gives the following.

3.9 Corollary. If f and G are as in Theorem 3.7 and f is not constant then !f(z)! < sup {If(w)!: WE cG} for all z in G. To prove Theorem 3.7 the following lemma is used. 3.10 Lemma. Let f and G be as in Theorem 3.7 alld further suppose that If(z)1 ::s; 1 for z on oG. Then If(z)1 ::s; I for all z in G.

Proof For each E > 0 let g,(z) = [l+E(z-a)]-l for each z in G-. Then for z = x+iy in GIg,(z)!::s; !Re[l+E(z-a)ll-l =

[I +E(x-a)r 1

::s; 1. So for z in cG if(z)g,(z) I ::s; I. Also, since f is bounded by B in G,

!f(z)g.(z) I ::s; BII +E(z-a)!-)

3.n

::s; BrE 11m zlr 1

So if R = {x+iy: a::s; x::s; b, Iyl ::s; Bid, inequality (3.11) gives If(z)g,(z) I ::s; 1 for z in oR. It follows from the Maximum Modulus Theorem that IfCz)g.(z) I ::s; 1 for z in R. But if 11m zl > BIE then (3.11) gives that IfCz)gcCz) I ::s; 1. Thus for all z in G.

If(z) I ::s; 11 +E(z-a)l· Letting

E

approach zero the desired result follows.

II

Proof of Theorem 3.7. First observe that to prove the theorem we need only establish


137

for a 0, A" = exp (z log A) is an entire function of z with no zeros. So g(z) defined by

g(z) = M(a)(b-zl!(b-a)M(b)(Z-a)!(b-a) is entire, never vanishes, and (because IAxl

= ARe.) for z =

x+iy

Ig(z) I = M(a)(b-x)!(b-a)M(b)(X-al!(b-a).

3.12

(It is assumed here that Mea) and M(b) -# 0. However, if either M(a) or

M(b) is zero then f == 0.) Since the expression on the right hand side of (3.12) is continuous for x in [a, b] and never vanishes, Igl- 1 must be bounded in G-. Also, Ig(a+iy)1 = M(a) and Ig(b+iy)1 = M(b) so that If(z)/g(z) 1 S; 1 for z in 8G; and fig satisfies the hypothesis of Lemma 3.10. Thus If(z) 1

S;

Ig(z)l, z

E

G.

Using (3.12) this gives for a < u < b

M(u) s; M(a)(b-u){(b-alM(b)(u-a)/(&-a) which is the desired conclusion. III Hadamard's Three Circles Theorem is an analogue of the preceding theorem for an annulus. Consider ann (0; R I , R 2 ) = A where 0 < R\ '< R z < 00. If G is the strip {x+iy: log RI < X < log R 2 } then the exponential function maps G onto A and 8G onto 8A. Using this fact one can prove the following from Theorem 3.7 (the details are left to the reader).

3.13 Hadamard's Three Circles Theorem. Let 0 < R 1 < R 2 < 00 and suppose f is analytic and not identically zero on ann (0; R j , R 2)' If R j < r < R 2' define M(r) = max {If(reIB)I: 0 s; () s; 2'lT}. Then for R j < '1 s; , s; '2 < R2 and,[ '2. iogr 2 -logr logr-Iogr, log M(r) S; log M(r t ) + ------------------- log M(r 2 ). iogr 2 -Iogr 1 log r 2 -10g r 1

*'

Another way of expressing Hadamard's Theorem is to say that log M(r) is a convex function of log r. Exercises 1. Let f: [a, b] -0- ~ and suppose that f(x) > 0 for all x and that f has a continuous second derivative. Show that f is logarithmically convex iff r(x)f(x) - [f'(x)] 2 ~ 0 for all x. 2. Show that iff: (a, b) -0- ~ is convex thenfis continuous. Does this remain true iffis defined on the closed interval [a, b]? 3. Show that a function f: Ca, b] -0- ~ is convex iff any of the following equivalent conditions is satisfied: (a) a

S;

x < u < y

S;

1)

f(U) U b gives det ( f(x) x 1 fey) y 1

~

0;

138


(b) a S x < u < y S b give/(u)-f(x) S f(y)-f(x) ; u-x y-x (c) a S x < u < y S b givesf(u)-f(x) S f(y)-f(u) . u-x y-u Interpret these conditions geometrically. 4. Supply the details in the proof of Hadamard's Three Circles Theorem. 5. Give necessary and sufficient conditions on the function f such that equality occurs in the conclusion of Hadamard's Three Circles Theorem. 6. Prove Hardy's Theorem : Iff is analytic on B(O; R) and not constant then I(r)

=~ 211'

f

2"

If(re i8 )ld8

o

is strictly increasing and log I(r) is a convex function of log r. Hint: If r 1 < r < r 2 find a continuous function tp: [0, 211'] --* C such that tp(8)f(re i8 ) = If(re i8 ) I and consider the function F(z) = f(ze i8)tp(8)d8. (Note that r is fixed, so rp may depend on r.) 7. Let f be analytic in ann (0; R I , R 2 ) and not identically zero; define

o<

g"

Iz(r) =

~ 211'

f

2"

If(re i8 )1 2 d8.

o

Show that log Iir) is a convex function of log r, Rl < r < R z .

§4. The Phragmen-LindeliJ! Theorem This section presents some results of E. Phragmlm and E. LindelOf (published in 1908) which extend the Maximum Principle by easing the requirement of bounded ness on the boundary. The Phragmen-LindelOf Theorem bears a relation to the Maximum Modulus Theorem which is analogous to the relationship of the following result to Liouville's theorem. If f is entire and If(z) I S 1+ Izlt then f is a constant function. (Prove it!) So it is not necessary to assume that an entire function is bounded in order to prove that it is constant; it is sufficient to assume that its growth as z --* 00 is restricted by 1+ Izlt. The PhragmenLindelof Theorem places a growth restriction on an analytic function f: G --* C as z nears a point on the extended boundary. Nevertheless, the conclusion, like that of the Maximum Modulus Theorem, is thatfis bounded. 4.1 Phragmen-LindelOf Theorem. Let G be a simply connected region and let f be an analytic function on G. Suppose there is an analytic function tp: G --* C which never vanishes and is bounded on G. If M is a constant and 000G = A u B such that: (a) for every a in A, lim suplf(z)1 z_a

s

M;

139

Phragmen-Linde1of Theorem

(b) for every b in B, and TJ > 0, lim sup If(z)1 l!p(z)I'I z"'b

then If(z)l:s; MforallzinG.

~

M;

Proof. Let 1
~ " for all z in G. Also because G is simply connected there is an analytic branch of log 0; and Ig(z)1 = 1p, F satisfies the hypothesis of Theorem 1.4. Thus W(z)1 ~ max(M, ,,-'1M) for all z in G. This gives

lJ(z)l:s; 1"/ O. Letting TJ - 0+ gives that lJ(z) I :s; M tor all z in G. •

4.2 Corollary. Let a

~

1- and put G

= {z:

largzl <

~}.

Suppose that f is analytic on G and there is a constant M such that lim sup z-+w

If(z) 1 :s; M for all w in BG. If there are positive constants P and b < a such that

4.3 for all z with Izl sufficiently large, then If(z) 1 :s; M for all z in G. Proof Let b < c < a and put 'P(z) = exp (-z~ for z in G. If z = rei', I/JI < Tr/2a, then Re ZC = r C cos c/J. So for z in G 1'P(z) 1 = exp (_r C cos c/J) when z = re i9 • Since c < a, cos c/J ~ P > 0 for all z in G. This gives that 'P is bounded on G. Also, if TJ > 0 and z = re i9 is sufficiently large,

If(z)II'P(z)l1J :s; P exp (r b-TJr c cos c/J) :s; P exp (r b -7]r c p) But rb_TJrcp=rC(rb-c_TJp). Since b
=0

Hence, f and 'P satisfy the hypothesis of the Phragmen-Lindelof Theorem so that If(z) 1 :s; M for each z in G. • Note that the size of the angle of the sector G is the only relevant fact in this corollary; its position is inconsequential. So if G is any sector of angle TrIa the conclusion remains valid.

140


4.4 Corollary. Let a

~

t, G

= {z:

and suppose that for every w in

largzl < ;:},

aG, lim sup If(z) I s M. Moreover, assume that .-+w

for every 8 > 0 there is a constant P (which may depend on 8) such that 4.5

for z in G and Izl sufficiently large. Then If(z) I S M for all z in G. Proof. Define F: G ~ C by F(z) = fez) exp (_€ZG) where £ > 0 is arbitrary. If x > 0 and 8 is chosen with 0 < 8 < £ then there is a constant P with

s P exp [(8 - £)x 00 in IR; so MI = sup {IF(x)l:

IF(x) I

G ).

But then IF(x)1 ~ 0 as x ~ < 00. Define M2 = max {M I, M} and

H+

=

{ZEG:O < argz < 7T/2a},

H_

=

{z

E

0 < x < oo}

G: 0 > arg z > -7T/2a};

then lim sup /f(z) I S M2 for all z in oH+ and oH_. Using hypothesis (4.5), .-+w

Corollary 4.2 gives /F(z)1 S M2 for all z in H + and H _ hence, IF(z)/ S M2 for all z in G. We claim that M2 = M. In fact, if M2 = MI > M then IFI assumes its maximum value in G at some point x, 0 < x < 00 (because IF(x)1 ~ 0 as x ~ 00 and lim sup If(x)1 = lim sup IF(x)1 S M < M I). This would give -,,-+0

-" ..... 0

that F is a constant by the Maximum Principle and so M = MI' Thus, M2 = M and IF(z) I S M for all z in G; that is,

If(z)1 S M exp (£ Re ZD) for all z in G; since M is independent of t, we can let £ ~ 0 and get /f(z) I S M for all z in G. • Let G = {z: z * 0 and largzl < wj2a} and let fez) = exp(zU) for z E G. Then f(z) = exp(lzlacos a O) where 0 = argz. So for z in oG, If(z)1 = I; but fez) is clearly unbounded in G. In fact, on any ray in G we have that If(z)1 -> 00. This shows that the growth condition (4.5) is very delicate and can't be improved. 1

1

Exercises I. In the statement of the Phragmen-Lindelof Theorem, the requirement that G be simply connected is not necessary. Extend Theorem 4.1 to regions G with the property that for each z in a'X)G there is a sphere V in Coo centered at z such that V Ii G is simply connected. Give some examples of regions that are not simply connected but have this property and some which don't.

141

The Phragmen-Lindelof Theorem

2. In Theorem 4.1 suppose there are bounded analytic functions 9),92"'" 9n on C that never vanish and doe C = A u B) u ... U En such that condition (a) is satisfied and condition (b) is also satisfied for each 9" and Ek • Prove that If(z)1 ~ M for all z in C. 3. Let C = {z: 1m zl < 11'} and suppose I: G -71[: is analytic and lim sup %->w

I/(z)] '.S M for win cG. Also, suppose A < 00 and a < 1 can be found such that I/(z)1 < exp [A exp (a IRe zl)] for all z in G. Show that I/(z)1 ~ M for all z in C. Examine exp (exp z) to see that this is the best possible growth condition. Can we take a = 1 above? 4. Let f: C -7 I[: be analytic and suppose M is a constant such that lim sup I/(z,,)I ~ M for each sequence {zn} in G which converges to a point in aooG. Show that I/(z)1 ~ M. (See Exercise 1.8). 5. Let f: G -7 I[: be analytic and suppose that G is bounded. Fix Z 0 in 8G and suppose that lim sup If(z)] ~ M for w in cC, w oF zoo Show that if z-~w

lim Iz-zol' I/(z)1 = 0 for every

€

:>

0 then I/(z)1

~

M for every z in aGo

z -+Zo

(Hint: If a¢. G, consider O} and let I: G -7 I[: be an analytic function with lim sup II(z) I :::; M for w in aG, and also suppose that for every E > 0, z-w

lim sup {exp( -E/r) Il(re i8 ) I : 181 < 1n} = O.

r--+oo

Show that I/(z) ~ Ai for all z in G. 7. Let G = {z: Re z > O} and let I: G-·;..I[: be analytic such that/(l) = 0 1

and such that lim sup

I/(z)1

~

M for

IV

in cG. Also, suppose that for every

z~w

8, 0 < 8 < I, there is a constant P such that Prove that

( Hint: Consider I(z)

G~;).)

Chapter VII Compactness and Convergence in the Space of Analytic Functions In this chapter a metric is put on the set of all fixed region G, and compactness and convergence discussed. Among the applications obtained is a Mapping Theorem. Actually some more general results are obtained study spaces of meromorphic functions.

analytic functions on a in this metric space is proof of the Riemann which enable us to also

§I. The space of continuous functiolls C(G,Q) In this chapter (n, d) will always denote a complete metric space. Although much of what is said does not need the completeness of fl, those results which hold the most interest are not true if (n, d) is not assumed to be complete. 1.1 Definition. If G is an open set in C and (l..!, d) is a complete metric space then designate by C(G, n) the set of all continuous functions from G to n. The set C(G, n) is never empty since it always contains the constant functions. However, it is possible that C(G, n) contains only the constant functions. For example, suppose that G is connected and f2 = N = {l, 2, ... }. Iff is in C(G, £1) then f(G) must be connected in n and, hence, must reduce to a point. However, our principal concern will be when n is either C or C". For these two choices of n, C(G, £1) has many non constant elements. In fact, each analytic function on G is in C(G, q and each meromorphic function on G is in C(G, C",) (see Exercise V. 3.4). To put a metric on C(G, [1) we must first prove a fact about open subsets of IC. The third part of the next proposition will not be used until Chapter

VIII. 1.2 Proposition. If G is open in C then there is a sequence {Kn} of compact

subsets of G such that G

=

U Kn. CD

Moreover, the sets Kn can be chosen to

n~l

satisfy the following conditions: (a) Kn c: int Kn+ 1; (b) K c: G and K compact implies K C Knfor some n; (c) Erery component of Ceo - Kn contains a component of C" - G. 142

The space of continuous functions C(G,O)

143

Proof. For each positive integer n let

K. = {z:

Izl ~

n}

n {z: d(z, C-G) ~ ~}

;

since Kn is clearly bounded and it is the intersection of two closed subsets of C, K. is compact. Also, the set {z:

Izl

< n+ I}

n {z: d(z, C-G)

>

_l_} n+l

is open, contains K., and is contained in K.+ l ' This gives that (a) is satisfied. Since it easily follows that G =

00

00

n~.

n~.

U K. we also get that G = U int K.;

so if

K is a compact subset of G the sets {int K.} form an open cover of K. This

gives part (b). To see part (c) note that the unbounded component of Coo -K. (:::> Coo - G) must contain 00 and must therefore contain the component of Coo - G which contains 00. Also the unbounded component contains {z: Izl > n}. So if D is a bounded component of C;, - K. it contains a point z with d(z, C - G) <

Iw - z I <

~.

But by definition this gives a point w in C - G with

n

~. But then z

E

B( w; ~) c

Coo - Kn; since disks are connected and

z is in the component D of Coo - K", B( w;

~)

c D. If D. is the component

of Ceo - G that contains w it follows that D) cD . • If G =

U Kn

where each Kn is compact and Kn c int K.+ 1 , define

n~1

Pn(/' g) = sup {d(f(z), g(z»: z

1.3

E

Kn}

for all functions f and g in C(G, Q). Also define p(/, g)

1.4

=

f

(t)"

n; 1

P.(/' ~L 1+ Pn(/' g)

L

since t(l + t) - 1 ~ 1 for all t ~ 0, the series in (1.4) is dominated by (t). and must converge. It will be shown that P is a metric for C(G, Q). To do this the following lemma, whose proof is left as an exercise, is needed. 1.5 Lemma.

If (S,

d) is a metric space then

Ji- (s, t)

= _ -:d(,--"-s,-..::t)_

1 +d(s, t)

is also a metric on S. A set is open in (S, d) iff it is open in (S, Ji-); a sequence is a Cauchy sequence in (S, d) iff it is a Cauchy sequence in (S, Ji-).

1.6 Proposition. (C(G, Q), p) is a metric space.

144

Compactness and Convergence

Proof It is clear that p(f, g) = p(g,f). Also, since each Pn satisfies the triangle inequality, the preceding lemma can be used to show that p satisfies the 00

triangle inequality. Finally, the fact that G = U Kn gives thatf = g whenever p(f, g) = O. • n=l The next lemma concerns subsets of C(G, Q) x C(G, Q) and is very useful because it gives insight into the behavior of the metric p. Those who know the appropriate definitions will recognize that this lemma says that two uniformities are equivalent. 1.7 Lemma. Let the metric p be defined as in (1.4). If € > 0 is given then there is a 8 > 0 and a compact set K c: G such that for f and gin C(G, Q), sup {d(f(z), g(z»: z

1.8

E

K} < 8 => p(f, g) <

€.

Conversely, if 8 > 0 and a compact set K are given, there is an that for f and gin C(G, Q), p(f, g) <

1.9

€

=> sup {d(f(z), g(z»: z

E

K} < 8.

€

ao

> 0 such

nt

0 is fixed let p be a positive integer such that L < 1€ n=p+1 . t and put K = Kp. Choose 0 > 0 such that 0 :s; t < 0 gIves 1 + t < -rE. Suppose Proof If

€>

fand g are functions in C(G, Q) that satisfy sup {d(f(z), g(z»: z E K} < o. Since Kn c: Kp = K for 1 :s n :s; p, Pn(f, g) < 0 for 1 :s; n :s; p. This gives Pn(f, g)

<

1E

1 +pif, g)

for 1 :s; n :s; p. Therefore p

p(f, g) <

00

L (W(!-") + n=p+l L (W n=l

< " That is, (1.8) is satisfied. Now suppose K and 0 are given. Since G = compact there is an integer p

~

1 such that K

00

U Kn

n=1

c:

€

.

.

U int Kn and K is

n

=

I

Kp; this gives

pif, g) ~ sup {d(f(z), g(z»: z

Let

00

=

E

K}

s I-s

1

> 0 be chosen so that 0 :s; s < 2P " Impltes - - < 0; then -1-

+1

< 2P €

implies t < O. So if p(f, g) < " then- pif, g) < 2P " and this gives pif, g) 1 +pif, g) < O. But this is exactly the statement contained in (1.9) . • 1.10 Proposition. (a) A set (!) c: (C(G, Q), p) is open iff for eachf in is a compact set K and a 0 > 0 such that (!) :::>

{g: d(f(z), g(z» < 0, z

E

K}

(!)

there

The space of continuous functions C(G,O)

145

(b) A sequence {fn} in (C( G, Q), p) converges to f iff {/,,} converges to f uniformly on all compact subsets of G. € > 0, (2 :::> {g: p(J, g) < €}. But now the first part of the preceding lemma says that there is as> 0 and a compact set K with the desired properties. Conversely, if (2 has the stated property and f E (2 then the second part of the lemma gives an € > 0 such that (2 :::> {g: p(J, g) < €}; this means that (2 is open. The proof of part (b) will be left to the reader. •

Proof If (2 is open and fE (2 then for some

1.11 Corollary. The collection of open sets is independent of the choice of the sets {Kn}. That is, ifG

=

co

U K~ where each K~ is compact and K~ c

int K~+l

n~1

and if I'- is the metric defined by the sets iff it is open in (C( G, Q), p).

{K~}

then a set is open in (C(G, Q), 1'-)

Proof This is a direct consequence of part (a) of the preceding proposition since the characterization of open sets does not depend on the choice of the sets {Kn } • • Henceforward, whenever we consider C(G, Q) as a metric space it will be assumed that the metric p is given by formula (1.4) for some sequence {Kn} of compact sets such that Kn c int Kn+l and G

=

co

U Kn. Actually, the

n~1

requirement that Kn c int Kn+ 1 can be dropped and the above results will remain valid. However, to show this requires some extra effort (e.g., the Baire Category Theorem) which, though interesting, would be a detour. Nothing done so far has used the assumption that Q is complete. However, if Q is not complete then C(G, Q) is not complete. In fact, if {wn } is a non-convergent Cauchy sequence in Q and /,,(z) = Wn for all z in G, then {fn} is a non-convergent Cauchy sequence in C(G, Q). However, we are assuming that Q is complete and this gives the following. 1.12 Proposition. C(G, Q) is a complete metric space. Proof Again utilize Lemma I.? Suppose{/,,} is a Cauchy sequence inC(G, Q). Then for each compact set KeG the restrictions of the functions In to K gives a Cauchy sequence in C(K, Q). That is, for every S > 0 there is an integer N such that

1.13

sup {d(/,,(z),Jm(z)): z

E

K} < S

for n, m ~ N. In particular {fn(z)} is a Cauchy sequence in Q; so there is a point f(z) in n such that f(z) = lim fn(z). This gives a function f: G --+ Q; it must be shown that f is continuous and p(/", f) --+ O. Let K be compact and fix 13 > 0; choose N so that (1.13) holds for n, m ~ N. If z is arbitrary in K but fixed then there is an integer m ~ N so that d(f(z), fm(z)) < 13. But then d(f(z),/,,(z)) < 213

146


for all n :2: N. Since N does not depend on z this gives sup {d(f(z),fn(z»:

Z E

K}f 0

as n -+ 00. That is, {fn} converges to f uniformly on every compact set in G. In particular, the convergence is uniform on all closed balls contained in G. This gives (Theorem I r. 6.1) that f is continuous at each point of G. Also, Proposition 1.10 (b) gives that p(j~. f) -,. O. • The next definition is derived from the classical origins of this subject. Actually it could have been omitted without interfering with the development of the chapter. However, even though there is virtue in maintaining a low ratio of definitions to theorems, the classical term is widely used and should be known by the reader.

1.14 Definition. A set :F c C(G, il) is normal if each sequence in .'j: has a subsequence which converges to a function f in C(G, £2). This of course looks like the definition of sequentially compact subsets, but the limit of the subsequence is not required to be in the set !IF. The next proof is left to the reader. 1.15 Proposition. A set :::F c C(G, £2) is lIormal if! its closure is compact. 1.16 Proposition. A set !IF c C(G, il) is normal iff for aery compact set KeG alld 8 > 0 there are functions fl, ....f,. in .::F such that for f in .ry; there is at least one k, 1 <::: k <::: II, with sup

{d(f(z),j~(::)):

::

E

K} < 8.

Proof Suppose :::F is normal and let K and 8 > 0 be given. By Lemma 1.7 there is an E > 0 such that (1.9) holds. But since !IF- is compact, .'!i' IS totally bounded (actually there are a few details to fill in here). So there are fl, ... , j~ in :::F such that .ry; c

n

U

k '" ]

But from the choice of

.'F

E n

c

this gives

U {f: d(f(z),J;..(z))

< ii,:: E K};

k=1

that is, ,'F satisfies the condition of the proposition. For the converse, suppose ,ry; has the stated property. Since it readily follows that !IF - also satisfies this condition, assume that ,ry; is closed. But since C(G, ~2) is complete !IF must be complete. And, again using Lemma 1.7, it readily follows that ~ is totally bounded. From Theorem II.4.9, ~ is compact and therefore normal. • This section concludes by presenting the Arzela-Ascoli Theorem. Although its proof is not overly complicated it is a deep result which has proved extremely useful in many areas of analysis. Before stating the theorem a few results of a more general nature are needed.

147

The space of continuous functions C(G,fl)

Let (X., d.) be a metric space for each n ;::: 1 and let X cartesian product. That is, X = {g = {xn}: Xn g = {xn} and '7 = {Yo} in X define d(g, 7])

1.17 1.18 Proposition.

CD

X. for each n ;::: I}. For

l+d.(xn,Yn)

n=1

Xm d), where d is defined by (1.17), is a metric space.

e = {X:}:'_I is in X TI Xn then e =

-+

• -1

Also,

TI Xn be their

n-l

= ~ H)'--- d.(xn' Yn) .

00

If

E

=

00

g = {xn} iff x~ ----;.. x.for each n .

if each (Xm dn) is compact then X is compact,

Proof The proof that d is a metric is left to the reader. Suppose d(gk, g)

--+

0;

SInce

we have that

, I1m k-oo

d"(x~, x.) - 0 k -. I +d.(xm x.)

This gives that x~ -+ x. for each n ;::: I. The proof of the converse is left to the reader. Now suppose that each (X., do) is compact. To show that (X, d) is compact it suffices to show that every sequence in X has a convergent subsequence; this is accomplished by the Cantor diagonalization process. Let gk = {~} E X for each k ;::: 1 and consider the sequence of the first entries of the that is, consider {xnt'_1 C XI' Since Xl is compact there is a point XI in Xl and a subsequence of {x~} which converges to it. We are now faced with a problem in notation. If this subsequence of {Xnt'-l is denoted by {x~j}j": 1 there is little confusion at this stage. However, the next step in the proof is to consider the corresponding subsequence of second entries {x~j} /':~! and take a subsequence of this. Furthermore, it is necessary to continue this process for all the entries. It is easy to see that this is opening up a notational Pandora's Box. However, there is an alternative. Denote the convergent subsequence of {x~} by {x~: kEN I }, where N 1 is an infinite subset of the positive integers N. Consider the sequence of second entries of {e: kEN! }. Then there is a point x 2 in X 2 and an infinite subset N 2 C N 1 such that lim {X~: kEN 2} = X2' (Notice that we still have lim {xt: kEN 2} = XI') Continuing this process gives a decreasing sequence of infinite subsets of N, N 1 ::J N 2::J ... ; and points x n in Xn such that

e;

1.19

lim {~: k

E

Nn }

= xn

{e

J }; we claim that Let kj be the }th integer in Nj and consider {xn} as} -+ 00. To show this it suffices to show that

1.20

tkj -+ g =

148


for each n ;::: 1. But since!\lJ c !\In for};::: n, {X~J:};::: n} is a subsequence of {X~: k E !\In}. So (1.20) follws from (1.19) . • The following definition plays a central role in the Arzela-Ascoli Theorem.

1.21 Definition. A set §" c C(G, Q) is equicontinuous at a point for every E > 0 there is as> 0 such that for Iz-zol < S, d(f(z),J(zo» <

Zo

in G iff

E

for every fin §". §" is equicontinuous over a set E c G iffor every E > 0 there is as> 0 such that for z and z' in E and Iz-z'l < S,

d(f(z),J(z'» <

E

for all / in §". Notice that if §" consists of a single function/then the statement that §" is equicontinuous at Zo is only the statement that/is continuous at zoo The important thing about equicontinuity is that the same S will work for all the functions in §". Also, for §" = {f} to be equicontinuous over E is to require that / is uniformly continuous on E. For a larger family §" to be equicontinuous there must be uniform uniform continuity. Because of this analogy with continuity and uniform continuity the following proposition should not come as a surprise. 1.22 Proposition. Suppose §" c C( G, Q) is equicontinuous at each point then §" is equicontinuous over each compact subset 0/ G.

Proof Let KeG be compact and fix a Sw > 0 such that

E

0/ G;

> O. Then for each w in K there is

d(f(w'),J(w» <

!E

for allfin§"whenever Iw-w'l <8w ' Now {B(w; 8w ): we K} forms an open cover of K; by Lebesgue's Covering Lemma (II. 4.8) there is a 8>0 such that for each z in K, B(z; 8) is contained in one of the sets of this cover. So if z and z' are in K and Iz-z'l <8 there is a w in K with z' e B(z; 8) c B(w; 8w )' That is, Iz-wl<8w and Iz'-wl<8w ' This gives d(f(z),f(w» and d(f(z'),f(w»

1.23 Arzela-Ascoli Theorem. A set §" two conditions are satisfied:

c

C(G, Q) is normal

iff the following

(a) Jor each z in G, {fez): fe §"} has compact closure in Q; (b) oF is equicontinuous at each point oj G.

Proof First assume that §" is normal. Notice that for each z in G the map of C(G, Q) -+ Q defined by J --+ J(z) is continuous; since ~- is compact its image is compact in n and (a) foHows. To show (b) fix a point Zo in G and let E > O. If R > 0 is chosen so that K = B(z 0; R) c G then K is compact

The space of continuous functions C(G,n)

149

and Proposition 1.16 implies there are functions II' ... , j~ in :IF such that for each / in :IF there is at least one J.. with sup {d(f(z),J..(z»): z

1.24

E

K} <

E

3

.

But since eachJ.. is continuous there is a 0, 0 < 0 < R, such that implies that

Iz-zol

< 0

d(J..Cz),J..(zo)
d(f(z)J(zo»)

Iz - z 01

< 0, IE:IF, and k is chosen so that

:s: d(f(z),J..(z)) + d(J..{z),J..(zo)) + d(J..(zo)J(zo» <

E

That is, ,'F is equicontinuous at zo' Now suppose:IF satisfies conditions (a) and (b); it must be shown that :IF is normal. Let {zn} be the sequence of all points in G with rational real and imaginary parts (so for z in G and S > 0 there is a Zn with iz - znl < 0). For each n ;::: 1 let Xn = {f(Zn):IE:IF}- en; from part (a), (X., d) is a compact metric space. Thus, by Proposition US,

n Xn is a compact metric space. For I in :IF define 1 in X by 00

X

=

n

~

1

1=

{f(Zj)J(Z2),·· .}.

Let {J..} be a sequence in :IF; so Ud is a sequence in the compact metric space X. Thus there is a in X and a subsequence of Uk} which converges to E. For the sake of convenient notation, assume that g = lim lk' Again from Proposition US,

e

1.25

lim J..(z,,) =

W"

k~co

where g = {w n }. It will be shown that {J..} converges to a function/in C(G, Q). By (1.25) this function / will have to satisfy fez,,) = W n • The importance of (1.25) is that it imposes control over the behavior of {J..} on a dense subset of G. We will use the fact that {J..} is equicontinuous to spread this control to the rest of G. To find the function/and show that {J..} converges to/it suffices to show that {J..} is a Cauchy sequence. So let K be a compact set in G and let E > 0; by Lemma 1.10(b) it suffices to find an integer J such that for k, j ;:::: J, 1.26

sup {d(J..(z)Jj(z): z

E

K} <

<'.

Since K is compact R = d(K, aG) > O. Let Kl = {z: d(z, K) :s: tR}; then Kl is compact and K c int K j C Kl C G. Since :IF is equicontinuous at each

150


point of G it is equicontinuous on K1 by Proposition 1.22. So choose I) < tR, such that

o<

I),

E

d(f(z),J(z'» < -

1.27

3

for all/in ~ whenever z and z' are in K1 with Iz-z'l < I). Now let D be the collection of points in {z.} which are also points in K 1 ; that is D

= {z.: z. E K 1 }

If z E K then there is a z. with Iz-z.1 < 0; but 0 < tR gives that d(z., K) < tR, or that z. E K 1 • Hence {B(w; 0): wED} is an open cover of K. Let w 1 , ••• , W. ED such that K c

•

U B(w i ; 0).

i= 1

Since lim fk(W;) exists for 1 sis n (by (1.25» there is an integer J such k--+oo

that for j, k

~

J

1.28

for i = 1, ... , n. Let z be an arbitrary point in K and let Wi be such that !wi-z! < k andj are larger than J then (1.27) and (1.28) give d(h.(z),!j(z» s d(h.(z),h.(w i»

<

I).

If

+ d(h.(w;), !j(w;)+ d(!j(Wi),!j(Z»

E.

Since z was arbitrary this establishes (1.26) . • Exercises

1. Prove Lemma 1.5 (Hint: Study the function f(t)

=

-1 t for t > -1.)

+t

2. Find the sets K. obtained in Proposition 1.2 for each of the following choices of G: (a) G is an open disk; (b) G is an open annulus; (c) G is the plane with n pairwise disjoint closed disks removed; (d) G is an infinite strip; (e) G = C-I. 3. Supply the omitted details in the proof of Proposition 1.18. 4. Let F be a subset of a metric space (X, d) such that F- is compact. Show that F is totally bounded. 5. Suppose {I.} is a sequence in C(G, Q) which converges tofand {zn} is a sequence in G which converges to a point z in G. Show lim I.(zn) = fez). 6. (Dini's Theorem) Consider C(G, IR) and suppose that {In} is a sequence in C(G, IR) which is monotonically increasing (i.e., fn(z) s 1.+1(Z) for all z in G) and lim/,,(z) = fez) for all z in G wherefE C(G, IR). Show that/" -+ f 7. Let {In} C C(G, Q) and suppose that {/,,} is equicontinuous at each point of G. If fE C(G, Q) and fez) = limfn(z) for each z then show that/" -+ f

151

Spaces of analytic functions 00

8. (a) Let f be analytic on B(O; R) and let fez)

J,.(z) =

n

L akzk , show that J,. ~ fin

q.

C(G;

k-O

= L anzn for Izl < n-O

R. If

(b) Let G = ann (0; 0, R) and letfbe analytic on G with Laurent series development fez) C(G; q.

=

L anzn. 00

=

Put J,.(z)

L ak zk "

and show that fn ~ fin

k=-oo

"=-00

§2. Spaces of analytic functions Let G be an open subset of the complex plane. If H (G) is the collection of analytic functions on G, we can consider H (G) as a subset of C(G,C). We use H(G) to denote the analytic functions on G rather than A (G) because it is a universal practice to let A (G) denote the collection of continuous functionsf:G-~C that are analytic in G. Thus A(G)'I: H (G). The letter H is used in reference to "analytic" because the word holomorphic is commonly used for analytic. Another term used in place of analytic is regular. The first question to ask about H(G) is: Is H(G) closed in C(G,C)? The next result answers this question positively and also says that the functionf~f' is continuous from H(G) into H(G). 2.1 Theorem./f{J,.} is a sequence in H(G) andfbelongs to C(G, In ~ f then f is analytic and f~k) ~ j
q

such that

Proof. We will show that f is analytic by applying Morera's Theorem (IV. 5.10). So let T be a triangle contained inside a disk D c G. Since T is compact, {fn} converges to f uniformly over T. Hence f Ti = lim f Tin = 0 since each In is analytic. Thus f must be analytic in every disk D c G; but this gives that f is analytic in G. To show that f: k) ~ f(k), let D = B(a; r) c G; then there is a number R > r such that B(a; R) C G. If" is the circle Iz-al = R then Cauchy's Integral Formula gives f:k)(Z)_j
for z in D. Using Cauchy'S Estimate, 2.2

If~k)(Z)-f(k)(z)1 ~ k!Mnk~1 (R-r)

for Iz-al

~ r,

whereMn = sup {/f,.(w)-f(w)/: /w-a/ = R}. Butsincef,.~J, limMn = O. Hence, it follows from (2.2) that f~k) ~ f(k) uniformly on R(a; r). Now if K is an arbitrary compact subset of G and 0 < r < d(K, oG) then there are

ai' ... , an in K such that K c

n

U B(a

j ;

j-I

r). Since

f~k) ~ j(k)

uniformly on

each B(aj ; r), the convergence is uniform on K. • We will always assume that the metric on H(G) is the metric which it

152


inherits as a subset of C(G, complete.

q.

The next result follows because C(G,

is

2.3 Corollary. II(G) is a complete metric space.

L fn(z) 00

2.4 Corollary. If fn: G -+ C is analytic and

converges uniformly on

n=l

compact sets to fez) then

L f~k)(Z). 00

Pk)(z) =

n= 1

It should be pointed out that the above theorem has no analogue in the theory of functions of a real variable. For example it is easy to convince oneself by drawing pictures that the absolute value function can be obtained as the uniform limit of a sequence of differentiable functions. Also, it can be shown (using a Theorem of Weierstrass) that a continuous nowhere differentiable function on [0, 1] is the limit of a sequence of polynomials. Surely this is the most emphatic contradiction of the corresponding theorem for Real Variables. A contradiction in another direction is furnished by the 1 following. Let i~(x) = x" for 0 ::s; x ::s; I. Then 0 = u -lim fn; however the n sequence of derivatives {f~} does not converge uniformly on [0, 1]. To further illustrate how special analytic functions are, let us examine a result of A. Hurwitz. As a consequence it follows that if i~ -+f and each fn never vanishes then either f == 0 or f never vanishes.

2.5 Hurwitz's Theorem. Let G be a region and suppose the sequence {In} in II(G) converges to f Iff¢' 0, B(a; R) c G, and fez) =1= 0 for Iz-al = R then there is an integer N such that for n 2 N, f and i~ have the same number of zeros in BCa; R). Proof Since fez) -#

°

for

o=

iz -al inf

= R,

{If(z)i: Iz-ai =

R} > 0.

But 1;, -> I uniformly on {z: Iz - al = R} so there is an integer N such that if n ? Nand Iz - al = R then 1~(z) 0 for Iz - al = Rand

'*

li( z) - In (z)1 <~ 8 < II( z)1 ::s; II( z)1 + lin (z)i· Hence Rouche's Theorem (V.3.8) implies that number of zeros in B(a; R).II

I

and

In

have the same

2.6 Corollary. Ii G is a region and {In} c 1I( G) converges to I in H( G) and each /" never vanishes on G then either 1== 0 or I never vanishes. In order to discuss normal families in H( G) the following terminology is needed.

Spaces of analytic functions

153

2.7 Definition. A set :F c H(G) is locally bounded if for each point a in G there are constants M and r > 0 such that for all I in :F,

]/(z)] :::; M, for ]z - a] < r. Alternately, :F is locally bounded if there is an r > 0 such that sup {]/(z)]: ]z-a] < r,JE:F} <

00.

That is, :F is locally bounded if about each point a in G there is a disk on which :F is uniformly bounded. This immediately extends to the requirement that :F be uniformly bounded on compact sets in G. 2.8 Lemma. A set :F in H(G) is locally bounded KeG there is a constant M such that

iff lor

each compact set

]/(z)] :::; M lor all I in :F and z in K. The proof is left to the reader.

2.9 Montel's Theorem. Alamity:F in H(G) is normal iff :F is locally bounded. Proof Suppose :F is normal but fails to be locally bounded; then there is a compact set KeG such that sup {]/(z)!: z E K,JE:F} = 00. That is, there is a sequence {In} in :F such that sup {]/n(z)]: z E K} 2: n. Since :F is normal there is a function I in H(G) and a subsequence {Ink} such that Ink -+ f But this gives that sup {]/nk(z) - I(z)]: z E K} -~ 0 as k -+ 00. If ]/(z)] :::; M for z in K, nk :::; sup {llnN)-/(z)]: z E K}+M;

since the right hand side converges to M, this is a contradiction. Now suppose :F is locally bounded; the Arzela-Ascoli Theorem (1.23) will be used to show that :F is normal. Since condition (a) of Theorem 1.23 is clearly satisfied, we must show that :Fis equicontinuous at each point of G. Fix a point a in G and ( > 0; from the hypothesis there is an r > 0 and M> 0 such that B(a; r) c G and ]/(z)] :::; M for all z in B(a; r) and for all 1 in .'F. Let ]z - a] < !r and 1 E:F; then using Cauchy's Formula with yet) = a + re il , 0 :::; t:::; 277, I/(a) - I(z)]

=---1

21T

\J' -----lew) (a-z)

(IV-a) (w-z)

dw I

y

2M :::; -]a-z] r Letting is < min

<

E"

{lr'4~ £}

for all/in:F . •

it follows that la-z] < is gives ]/(a)-/(z)]

154


2.10 Corollary. A set ff' c H(G) is compact iff it is closed and locafly bounded. Exercises I. Let f, 11,12' ... be elements of H(G) and show that f" ---* I iff for each closed rectifiable curve y in G,ln(z) ---* I(z) uniformly for z in {y}. 2. Let G be a region, let a E IR, and suppose that I: [a, 00] x G ---* C is a continuous function. Define the integral F(z) = S: I(t, z)dt to be uniformly convergent on compact subsets of G if lim S~ l(t, z)dt exists uniformly for z b-HO

in any compact subset of G. Suppose that this integral does converge uniformly on compact subsets of G and that for each t in (a, co), I(t, .) is analytic on G. Prove that F is analytic and 00

F(k)(Z) =

f ~kl(t,31 dt a

cz k

3. The proof of Montel's Theorem can be broken up into the following sequence of definitions and propositions: (a) Definition. A set ff' c C(G, C) is locally Lipschitz if for each a in G there are constants M and r > 0 such that I/(z)-/(a)1 :0; Mlz-al for alliin ff' and Iz-al < r. (b) If ff' c C(G, C) is locally Lipschitz then ff' is equicontinuous at each point of G. (c) If ofF c H(G) is locally bounded then ,CY; is locally Lipschitz. 4. Prove Vitali's Theorem: If G is a region and Un}C:.H(G) is locally bounded and f E H(G) that has the property that A = {z E G: limfn(z) = fez)} has a limit point in G then fn-+(. 5. Show that for a set §- c H(G) the following are equivalent conditions: (a) ff' is normal; (b) For every E > 0 there is a number c > 0 such that {c/: IE ff'} c B(O; E) (here B(O; E) is the ball in H(G) with center at 0 and radius E). 6. Show that if ff' c H(G) is normal then ff" = {f':/E ff'} is also normal. Is the converse true? Can you add something to the hypothesis that ff" is normal to insure that ff' is normal? 7. Supposeff' is normal in H(G) and Q is open in C such that/(G) c Q for every I in ff'. Show that if g is analytic on Q and is bounded on bounded sets then {g I: IEff'} is normal. 8. Let D = {z: Izl < I} and show that ff' c H(D) is normal iff there is a sequence {Mn} of positive constants such that lim sup I:./ Mn :0; 1 and if 0

I(z)

=

a:>

Lo anzn is in ff' then rani

:0;

Mn for all n.

9. Let D=B(O; I) and for O
for each r, O
155

Spaces of meromorphic functions

II. Suppose that Un} is a sequence in H(G), f is a.non-constant function, and.r.. -o.f in H(G). Let a E G and 0: = f(a); show that there is a sequence {an} in G such that: (i) a = lim an; (ii) fn(a.) = 0: for sufficiently large n. 12. Show that lim tan nz = - i uniformly for z in any compact subset of G = {z: 1m z < O}. 13. (a) Show that if fis analytic on an open set containing the disk B(a; R) then If(aW $

;~2

ff

2" R

o

If(a+re i6W rdrd6.

0

(b) Let G be a region and let M be a fixed positive constant. Let ~ be If(zW dxdy $ M. Show the family of all functions! in H (G) such that that ~ is normal. G

ff

§3. Spaces of meromorphic functions If G is a region and f is a meromorphic function on G, and if f (z) = 00 whenever z is a pole of f then f: G -> Coo is a continuous function (Exercise V. 3.4). If M(G) is the set of all meromorphic functions on G then consider M(G) as a subset of C(G, C",) and endow it with the metric of C(G, C",,). In this section this metric space will be discussed as H(G) was discussed in the previous section. Recall from Chapter I that the metric d is defined on Ceo as follows: for Zl and Z2 in C

and for z in C

Notice that for non zero complex numbers

Zt

and zz,

3.1 and for z

~

0

3.2

d(z, 0)

Also recall that if {Zn} d(z,z,,)_O then

Iz -

IS

=d

G' (0).

a sequence in C and Z E C that satisfies

znl-O.

Some facts about the relationship between the metric spaces C and Coo are summarized in the next proposition. In order to avoid confusion B(a; r) will be used to designate a ball in C and Boo(a; r) to designate a ball in Coo-

156


3.3 Proposition. (a) If a is in C and r > 0 then there is a number p > U such that BroCa; p) C B(a; r). (b) Conversely, if p > 0 is given and a E C then there is a number r > 0 such that B(a; r) c BocCa; p). (c) If p > 0 is given thell there is a compact set K c C such that Coo - K c B",(oo; p).

(d) Conversely, if a compact set K c C is given, there is a number p > 0 such that Boo(oo; p) c C",---K. The proof is left to the reader. The first observation is that M(G) is not complete. In fact if j,,(z) == Il then U~} is a Cauchy sequence in M(G). But {fn} converges to the function which is identically (fJ in C(G, Coo) and this is not meromorphic. However this is the worst that can happen. 3.4 Theorem. Let {f,,} be a sequence in M(G) and supposej" ~ fin C(G, C",J. Then either f is meromorphic or f == 00. each j" is analytic then either f is analytic or f == 00.

rr

Proof Suppose there is a point a in G withf(a) # XJ and put M = If(a)1 + 1. Using part (a) of Proposition 3.3 we can find a number p > 0 such that Bx(f(a); p) c B(f(a); M). But since j,,+ f there is an integer flo such that d(j,,(a)j(a» < 1p for all n ? 11 0 , Also UJIJb ... } is compact in C(G, Cx» so that it is equicontinuous. That is, there is an r > 0 such that Iz - al < r implies d(fn(z), .!n(a» < 1p. That gives that d(j,,(z), f(a» ::; p for Iz-al ::; r and for fl ? no. But by the choice of p, Ij,,(z) 1 ::; Ij,,(z) - f(a) I + If(a) 1 ::; 2M for all z in B(a; r) and n ? flo. But then (from the formula for the metric d) L

(1 + 4M2) Ij,,(z) -

fez)! ::; d(j,,(z),J(z»

for z in BCa; r) and n ? 110' Since d(j,,(z) , fez») ~ 0 uniformly for z in B(a; r), this gives that if~(z)-f(z)1 -+ 0 uniformly for z in BCa; r). Since the tail end of the sequence {j,,} is bounded on BCa; r), j" has no poles and must be analytic near z = a for Il ~ flo. It follows thatfis analytic in a disk about a. Now suppose there is a point a in G withfCa) = 00. For a function g in C(G, Coo) define! by g 00;

fn

and

----7

(g/1)-- (z)

= 00

(1) g

(z)

=L if g(z) # g(z)

0 or 00;

1

if g(z) = O. It follows that g

E

(1) g

(z) = 0 if g(z)

C(G, C w )' Also, since

I fin C(G, C"J it follows from formulas (3.1) and (3.2) that ;~

C(G, C",). Now each

1

function.f~

=

I

~f

in

is meromorphic on G; so the preceding

1 1 paragraph gives a number r > 0 and an integer no such that f and j" are

analytic on BCa; r) for

11 ~ flo

and

/"

~

1

f

uniformly on B(a; r). From


157

I

1

Hurwitz's Theorem (2.5) either j == 0 or j has isolated zeros in B(a; r). So if I ¢

00

then

j

¢ 0 and I must be merom orphic in BCa; r). Combining this

with the first part of the proof we have that I is meromorphic in G if I is not identically infinite. 1 If each is analytic then - has no zeros in B(a; r). It follows from

f:

I.

Corollary 2.6 to Hurwitz's Theorem that either; == 0 or

But since I(a) =

00

~

never vanishes.

1 we have that I has at least one zero; thus

f ==

B(a; r). Combining this with the first part of the proof we see that I

is analytic.

III

==

00

00

in

or I

3.5 Corollary. M(G) u {oo} is a complete metric space. 3.6 Corollary. H(G) U {oo} is closed in (,(G, C oc ). To discuss normality in M(G) one must introduce the quantity

2il'(z)1

1+11\;)1

2 '

for each meromorphic function f However if z is a pole off then the above expression is meaningless since I'(z) has no meaning. To rectify this take the limit of the above expression as z approaches the pole. To show that the limit exists let a be a pole off of order m 2': I; then

I(z) = g(z) +

Am_ + ... + . Al

(z-a)m

(z-a)

for z in some disk about a and g analytic in that disk. For z =ft a

Thus

211'(z)1

mAm _ 2 I (z _ a)m +l

1+ I/(zW

- ~I-'- ~m

AI,

+ ... + ~-;;-)2 _

(z-a)'"

Iz - al 2m

I

+ ... +1.1_ + g(Z)1 2 (z-a)

+ Al(Z - a)ml - g'(z)(z - a)m+ll + lAm +- ... +- Al(Z - a)ml +- g(z)(z - a)ml2

21z - aim-limA," +- ...

-----------------

g (z)

158


So if m :?: 2 lim z-'a

If m

=

~!.f'(=-2L

1+[f{z)[2

0

=

1 then

. 21f'(z)[ lIm --------z~a 1 + [f(zW

=

2

-----

[All

3.7 Definition. Iff is a meromorphic function on the region G then define f1-(f): G -----3>- [R by

whenever z is not a pole of J, and

. 2if'(z)1 f1-(f) (a) = h m - - z~a 1 + if(zW if a is a pole off It follows that f1-(f) E C(G, q. The reason for introducing f1-(f) is as follows: If f: G -----3>- Coo is meromorphic then for z close to z' we have that d(f(z), fez')) is approximated by f1-(f) (z) Iz-z'l. So if a bound can be obtained for f1-(f) thenfis a Lipschitz function. Iff belongs to a family of functions and /1-(f) is uniformly bounded for fin this family, then the family is a uniformly Lipschitz set of functions. This is made precise in the following proof. 3.8 Theorem. Afamity :F f E :F} is locally bounded.

c

M(G) is normal in C(G, C",) iffA:!F) == {f1-(f):

0;:. T hus:f/' = {'f,}' n. IS 1+n normal in C(G, CG(J and f1-(,SF) is locally bounded. However, .'F is not normal in M(G) since the sequence {fn} converges to the constantly infinite function which does not belong to M(G).

Note. If f,,(z)

=

nz for n :?: 1 then f1-(f,,) (z)

=

- - -21121-;'

z,-

Proof of Theorem 3.8. We will assume that f1-(:F) is locally bounded and prove that .SF is normal by applying the Arzela-Ascoli Theorem. Since C", is compact it suffices to show that ,SF is equicontinuous at each point of G. So let K be an arbitrary closed disk contained in G and let M be a constant with f1-(f) (z) ::; M for all z in K and all f in :F. Let z and z' be arbitrary points in K. Suppose neither z nor z' are poles of a fixed function f in :F and let c< > 0 be an arbitrary number. Choose points Wo = Z, WI, . . . , Wn = z' in K which satisfy the following conditions: 3.9 win

[W k - 1 ,

wd implies w is not a pole of /;

159


3.10

L

k=1

3.11

IWk-Wk-d

:s:

2Iz-z'l;

+ 1/(Wk_1)1 2 ) [(1 + 1/(wk)1 2 )(1 + 1/(Wk-1)lZ))' (1

I

3.12I/(Wk)-/(Wk-l) - /'(Wk-I)I < Wk-W k - I

IX,

-

1 < a,

1 ::s: k::s: n;

1:S: k :S: n.

To see that such points can be found select a polygonal path Pin K satisfying (3.9) and (3.10). Cover P by small disks in which conditions similar to (3.11) and (3.12) hold, choose a finite subcover, and then pick points W o, ••• , W II on P such that each segment [Wk - I , wk] lies in one of these disks. Then {wo, ... , wII } will satisfy all of these conditions. If 13k = [(I + I/(Wk-1W) (l + I/(wkW)]t then

2 d(f(Wk-l),f(Wk» II

d(/(z),f(z'») :S:

k-I

=

2 2 13k-1/(wk)-/(wk-I)1 II

*

k _I

:S: L.. -21 / (W k)-/(Wk- 1) - j"( Wk- 1)\1 Wk-Wk- 11 k -1 13k Wk - Wk- 1

+

2 2 13k-//'(Wk-I)//wk-w.-d II

Ic= 1

Using the fact that 211'(wk)1 :S: M(I + /1(w.W) and the conditions on wo, ••. , W II this becomes

2 M/w.-w._ I/ n

:S: (4IX+2IXM) Iz-z'/ +

k-I

:S: (4cx+2cxM+2M)/z-z'l Since 3.13

IX

> 0 was arbitrary this gives that if z and z' are not poles of.f then d(f(z),f(z'» :S: 2M/z-z'/.

Now suppose z' is a pole of 1 but z is not. If then it follows from (3.13) that

W

is in K and is not a pole

d(/(z), (0) :S: d(/(z),f(w»+d(f(w), (0)

:S: 2M/z-w/ + d(f(w) , (0).


160

Since it is possible to let w approach z' without wever being a pole of f (poles are isolated!), this gives thatf(lI') -lo- fez') = CI) and IZ-Wl-lo-Iz-zl Thus (3.13) holds if at most one of z and z' is a pole. But a similar procedure gives that (3.13) holds for all z and z' in K. So if K = B(a; r) and E > 0 are given then for [; < min {r, E/2M} we have that Iz - al < [; implies d(f(z) , J(a» < E, and [; is independent offin ff. This gives that ff is equicontinuous at each point a in G. The proof of the converse is left to the reader. • Exercises

1. Prove Proposition 3.3. 2. Show that if ff c M(G) is a normal family in C(G, Coo) then p,(ff) is locally bounded. §4. The Riemann Mapping Theorem

We wish to define an equivalence relation between regions in C. After doing this it will be shown that all proper simply connected regions in C are equivalent to the open disk D = {z: Izl < I}, and hence are equivalent to one another. 4.1 Definition. A region G 1 is coriformally equivalent to G z ifthereis an analytic function f: G 1 -lo- C such that l is one-one and l(G) = Gz• Clearly, this is an equivalence relation. It is immediate that C is not equivalent to any bounded region by Liouville's Theorem. Also it is easy to show from the definitions that if G 1 is simply connected and G 1 is equivalent to G z then G z must be simply connected. If l is the principal branch of the square root then f is one-one and shows that C - {z: z :s; o} is equivalent to the right half plane. 4.2 Riemann Mapping Theorem. Let G be a simply connected reginn which is not the whole plane and let a E G. Then there is a unique analytic function f: G -lo- C having the properties:

(a) f(a) = 0 andf'(a) > 0; (b) l is one-one; (c) f(G) = {z: Izl < I}. The proof that the function f is unique is rather easy. In fact, if g also has the properties off and D = {z: Iz 1< I} then fog - ) : D -lo- D is analytic, one-one, and onto. Also f 0 g-l(O) = lea) = 0 so Theorem VI. 2.5 implies there is a consta'1t c with lei = 1 and f g-l(Z) = cz for all z. But then fez) = cg(z) gives that 0 < f'(a) = cg'(a); since g'(a) > 0 it follows that c = 1, orf= g. To motivate the proof of the existence of J, consider the family ff of all analytic functions f having properties (a) and (b) and satisfying 1f(z) 1 < 1 for z in G. The idea is to choose a member of ff having property (c). Suppose 0

The Riemann Mapping Theorem

161

UKn = G and a 00

{Kn} is a sequence of compact subsets of G such that

E

Kn

n~l

for each n. Then {f(Kn)} is a sequence of compact subsets of D = {z: [z[ < I}. Also, as n becomes larger f(Kn) becomes larger and larger and tries to fill out the disk D. By choosing a function f in ff with the largest possible derivative at a, we choose the function which "starts out the fastest" at z = a. It thus has the best possible chance of finishing first; that is, of having 00

Uf(Kn) = D. n~l

Before carrying out this proof, it is necessary for future developments to point out that the only property of a simply connected region which will be used is the fact that every non-vanishing analytic function has an analytic square root. (Actually it will be proved in Theorem VIII. 2.2 that this property is equivalent to simple connectedness.) So the Riemann Mapping Theorem will be completely proved by proving the following. 4.3 Lemma. Let G be a region which is not the whole plane and such that every non-vanishing analytic function on G has an analytic square root. If a E G then there is an analytic function f on G sllch that: (a) f(a) = 0 andf'(a) > 0; (b) f is one-one; (c) f{G) = D = {z: [z[ < I}.

Proof Define ff by letting .SF

=

{I,

H(G):fis one-oneJ(a)

= O,f'(a) > O,f(G)

c D}

Since f(G) c D, sup {[f(z)!: z E G} :s:: 1 for f in ff; by MonteI's Theorem ff is normal if it is non-empty. So the first fact to be proved is

4.4 It will be shown that

4.5

.SF- = ff u {O}.

Once these facts are known the proof can be completed. Indeed, suppose (4.4) and (4.5) hold and consider the function f ---?- f'(a) of H(G) ---?- C. This is a continuous function (Theorem 2.1) and, since ff- is compact, there is an fin ff- with f'(a) 2: g'(a) for all g in ff. Because ff "# 0, (4.5) implies that fE:7. It remains to show that f(G) = D. Suppose WED such that w rtf(G). Then the function fCz)-w l-cvf(z) is analytic in G and never vanishes. By hypothesis there is an analytic function h: G -;.. :c such that

4.6

[h(z)J

2

f(z)-w 1-wf(z)

= ---

.

162


n

Since the Mobius transformation

~-w

l~=~i; maps D onto D, h(G) c D.

=

Define g: G -)0 IC by

g(z) = Ih'(a)[ ~~i~)-IJ(a) h'(a) I-h(a)h(z) Then g(G) c D, g(a) =c 0, and g is one-one (why?). Also g'(a)

= !!l'(~J . h'(a) [l_-:-jh(a)I~~] [1-lh(a)12]2

h'(a)

~~L

l-lh(aW

But [h(aW = I-wi = Iwl and differentiating (4.6) gives (sincef(a) = 0) that

2h(a)h'(a) = Therefore

rea) (l-lwI2).

rCa) (~-::J(UL2 . _J

g'(a) =

2y'l wl

l-Iwl

=

rea) (~JI~i)

>

rCa)

This gives that g is in :F and contradicts the choice of f Thus it must be thatf(G) = D. Now to establish (4.4) and (4.5). Since G i= IC, let bE IC-G and let g be a function analytic on G such that [g(Z)]2 = z-b. If Z1 and Z2 are points in G and g(Zj) = ±g(zz) then it follows that ZI = Z2' In particular, g is one-one. By the Open Mapping Theorem there is a number r > 0 such that

4.7

g(G)

::>

B(g(a); r)

So if there is a point z in G such that g(z) E B( - g(a); r) then r > Ig(z) + g(a) I = l-g(z)-g(a)l. According to (4.7) there is a IV in G with g(w) = -g(z); but the remarks preceding (4.7) show that }\' = z which gives g(z) = O. But then z-b = [g(Z)]2 = 0 implies b is in G, a contradiction. Hence 4.8

g(G) n

g:

1~+g(a)1

< r} = D.

Let U be the disk g: i~+g(a)i < r} = B( --g(a); r). There is a Mobius transformation T such that T(ICw-U-) = D. Let gj = Tog; then gj is analytic and gj(G) c D. If C( = gj(a) then let gz(z) = CF. gj(z); so we still have that g2(G) c D and gz is analytic, but we also have that g2(a) = O. Now it is a simple matter to find a complex number c, lei = 1, such that g3(Z) = cgzCz) has positive derivative at z = a and is, therefore, in ~. This establishes (4.4). 0

The Riemann Mapping Theorem

Suppose {in} is a sequence in ff and In and since f:(a) -i> f'(a) it follows that

4.9

...,.. fin

163

HeG). Clearly f(a)

=

0

f'(a) ::?:: O.

Let z 1 be an arbitrary element of G and put' = fez 1); let ,. = .f~(z 1.). Let z 2 E G, z 2 -.f. Z I and let K be a closed disk centered at z 2 such that z t ¢ K. Thenfn(z) -~" never vanishes on K sincef. is one-one. Butfl::) -~. -i> fez) - ~ uniformly on K, so Hurwitz's Theorem gives that fez) - ~ never vanishes on K or fez) == ~. If fez) == ~ on K then f is the constant function' throughout G; since.f(a) = 0 we have thatf(z) == O. Otherwise we get thatf(z2) -.f. f(zl) for Z2 -.f. z 1; that is,fis one-one. But if fis one-one thenf' can never vanish; so (4.9) implies thatf'(a) > 0 andfis in ff. This proves (4.S) and the proof of the lemma is complete. III 4.10 Corollary. Among the simply connected regions there are only two equivalence classes; one consisting of C alone and the other containing all the proper simply connected regions.

Exercises 1. Let G and 0 be open sets in the plane and let f: G -i> 0 be a continuous function which is one-one, onto, and such thatf- I : 0 -i> G is also continuous (a homeomorphism). Suppose {zn} is a sequence in G which converges to a point z in 8G; also suppose that w = lim f(z.) exists. Prove that WEan. 2. (a) Let G be a region, let a E G and suppose thatf: (G- {a}) -i> C is an analytic function such that f{G - {a}) = n is bounded. Show that f has a removable singularity at z = a. If f is one-one, show that f(a) E 80. (b) Show that there is no one-one analytic function which maps G = {z: < Izl < I} onto an annulus 0 = {z: r < Izl < R} where r > O. 3. Let G be a simply connected region which is not the whole plane and suppose that Z E G whenever z E G. Let a E G () IR and suppose that f: G -i> D = {z: [zl < I} is a one-one analytic function with/(a) = O,f'(a) >0 and /(G) = D. Let G+ = {z E G: Im z > O}. Show that /(G+) must lie entirely above or entirely below the real axis. 4. Find an analytic function f which maps {z: Izl < 1, Re z > O} onto B(O; I) in a one-one fashion. S. Let f be analytic on G = {z: Re z > O}, one-one, with Re fez) > 0 for all z in G, andf(a) = a for some real number a. Show that If'(a) 1 ::; 1. 6. Let G 1 and G 2 be simply connected regions neither of which is the whole plane. Let f be a one-one analytic mapping of G 1 onto G 2 • Let a E G 1 and put ex = f(a). Prove that for anyone-one analytic map h of G 1 into G 2 with Ii(a) = ex it follows that Ih'(a)1 ::; 1f'(a)l. Suppose Ii is not assumed to be one-one; what can be said? 7. Let G be a simply connected region and suppose that G is not the whole plane. Let .6. = {~: !~I < I} and suppose that f is an analytic, one-one map of G onto .6. with f(a) = 0 and f'(a) > 0 for some point a in G. Let g be any other analytic, one-one map of G onto .6. and express g in terms off

°

164


8. Let '" '2' R" R 2 , be positive numbers such that R,f" = R 2 f'2; show that ann (0; '10 R I ) and ann (0; '2' R 2 ) are conformally equivalent. 9. Show that there is an analytic function f defined on G = ann(O; 0, 1) such that f' never vanishes and f(G) = B(O; 1).

§ 5. The Weierstrass Factorization Theorem The notion of convergence in H(G) can be used to solve the following problem. Given a sequence {ak } in G which has no limit point in G and a sequence of integers {mk}. is there a function J which is analytic on G and such that the only zeros of J are at the points ak • with the multiplicity of the zero at ak equal to m k? The answer to the question is yes and the result is due to Weierstrass. If there were only a finite number of points, ai' ... , an then J(z) = (z-an)m, ... (z-an)m n would be the desired function. What happens if there are infinitely many points in this sequence? To answer this we must discuss the convergence of infinite products of numbers and functions. Clearly one should define an infinite product of numbers Zn (denoted

n zn) as the limit of the finite products. Observe, however, that if one of 00

by

n-I

the numbers Zn is zero, then the limit is zero, regardless of the behavior of the remaining terms of the sequence. This does not present a difficulty, but it shows that when zeros appear, the existence of an infinite product is trivial.

5.1 Definition. If {zn} is a sequence of complex numbers and if z exists, then z is the infinite p,oduct of the numbers

Zn

= lim

n n

Zk

k -I

and it is denoted by

TI 00

Suppose that no one of the numbers and is also not zero. Let Pn

=

Zn

is zero, and that

=

Zn

exists

n=1

n

TI

Z

Zk

for

11

k=1

C 1; then no Pn is zero and

~ = Zn' Since Z # 0 and Pn -+ Z we have that lim Zn = 1. So that except Pn-I for the cases where zero appears, a necessary condition for the convergence of an infinite product is that the n-th term must go to 1. On the other hand, note that for Zn = a for all nand lal < 1, = 0 although lim Zn = a # O. Because of the fact that the exponential of a sum is the product of the exponentials of the individual terms, it is possible to discuss the convergence of an infinite product (when zero is not involved) by discussing the convergence of the series L log Zn' where log is the principal branch of the logarithm. However, before this can be made meaningful the Zn must be

TIz.

165

Weierstrass Factorization Theorem

restricted so that log Zn is meaningful. If the product is to be non-zero, then Zn ~ I. So it is no restriction to suppose that Re zn > 0 for all n. Now n

suppose that the series

L log Zn converges. If Sn = L log Zk and Sn ~ S then

n Zk so that n Zn is convergent to Z = e' i= O. > 0 for all n ~ 1. Then n Zn converges to a non n

00

k~1

n~1

k~1

exp Sn ~ exp s. But exp Sn =

5.2 Proposition. Let Re Zn

00

n~1

00

zero number iff the series L log Zn converges. n~1

Proof Let Pn = (Zl·· ·zn), Z = re i9 , -7T < 8 :s; 7T, and t(Pn) = log IPnl+i8n where (}-7T < 8n :s; 8+7T. If Sn = log Zl + ... +log Zn then exp (sn) = Pn so that Sn = t(Pn) + 27Tikn for some integer k n. Now suppose that Pn ~ z. Then sn-sn-l = log zn~O; also t(Pn)-t(Pn-l)~O, Hence, (kn-kn-l)~O as n ~ 00. Since each k n is an integer this gives that there is an no and a k such that k m = k n = k for m, n ~ no. So Sn ~ t(z) + 27Tik; that is, the series L log Zn converges. Since the converse was proved above, this completes the proof. • Consider the power series expansion of log (I-t-z) about Z = 0:

10g(l+z) =

zn 2: (_l)n-l_ = n=l n 00

Z -

Z2 -

2

+ ... ,

which has radius of convergence 1. If Izi < 1 then 11 _IOg(;+Z)1 = l-!-z-tz 2 +···1 :s; -!-(Izi + IzI2 + ...)

Izi

= -!- l-lzl· If we further require Izi < -t then 11 - log (; +Z)I :s;

t.

This gives that for Izi < -t

5.3

-tizi :s; Ilog (1 + z)1 :s; 1-14

This will be used to prove the following result. 5.4 Proposition. Let Re zn > -1; then the series

absolutely iff the series L zn converges absolutely.

L

L

log (1 +zn) converges

Proof If IZnl converges then Zn ~ 0; so eventually IZnl < t. By (5.3) Ilog (1 +zn)1 is dominated by a convergent series, and it must converge Ilog (1 +zn)1 converges, then it follows that IZnl < -!also. If, conversely, for sufficiently large n (why?). Again (5.3) allows us to conclude that L IZnl converges . •

L

L

166


We wish to define the absolute convergence of an infinite product. The first temptation should be avoided. That is, we do not want to say that IT IZnl converges. Why? If IT IZnl converges it does not follow that IT Zn converges. In fact, let Zn = -1 for all n; then IZnl = 1 for all n so that n

IT IZnl converges to 1. However IT Zk is ± 1 depending on whether 11 is even k-I or odd, so that IT Zn does not converge. Thus, if absolute convergence is to

imply convergence, we must seek a different definition. On the basis of Proposition 5.2 the following definition is justified.

5.5 Definition. If Re Zn > 0 for all n then the infinite product IT Zn is said to converge absolutely if the series L log Zn converges absolutely. According to Proposition 5.2 and the fact that absolute convergence of a series implies convergence, we have that absolute convergence of a product implies the convergence of the product. Similarly, if a product converges absolutely then any rearrangement of the terms of the product results in a product which is still absolutely convergent. If we combine Propositions 5.2 and 5.4 with the definition, the following fundamental criterion for convergence of a infinite product is obtained. 5.6 Corollary. If Re Zn > 0 then the product the series

L (zn -

IT

Zn converges absolutely

iff

1) converges absolutely.

Although the preceding corollary gives a necessary and sufficient condition for the absolute convergence of an infinite product phrased in terms with which we are familiar, it does not give a method for evaluating infinite products in terms of the corresponding infinite series. To evaluate a particular product one must often resort to trickery. We now apply these results to the convergence of products of functions. A fundamental question to be answered is the following. Suppose {J,,} is a sequence of functions on a set X and J,,(x) -+ f(x) uniformly for x in X; when will exp (fn(x) -+ exp (f(x» uniformly for x in X? Below is a partial answer which is sufficient to meet our needs. 5.7 Lemma. Let X be a set and let f, f1' f2' ... be functions from X into C such that j~(x) -+ f(x) uniformly for x in X. If there is a constant a such that Re f(x) :0; a for all x in X then exp J,,(x) -+ exp f(x) uniformly for x in X. Proof If € > 0 is given then choose 0 > 0 such that Ie' - 11 < E e - a whenever Izl < o. Now choose no such that Ifn(x) - f(x) I < 0 for all x in X whenever

n ;;::

110'

Thus

a-a> lexp [fn(x)-f(x)]-II

=

lexp fn(x) - 1/ expf(x)

It follows that for any x in X and for n ;;:: no,

lexpJ,,(x)-expf(x)1 <

E

e- a lexpj(x)1

:0;

E• •

5.S Lemma. Let (X, d) be a compact metric space and let {gn} be a sequence


167

of continuous functions from X into IC such that and uniformly for x in X. Then the product

L gn(x) converges absolutely

n (1 + g.(x)) .=1 co

f(x) =

converges absolutely and uniformly for x in X. Also there is an integer no such that f(x) = 0 iff gn(x) = -1 for some n, 1 ~ n ~ no'

L

Proof Since g.(x) converges uniformly for x in X there is an integer no such that [g.(x) [ < t for all x in X and n > .no. This implies that Re [1 + g.(x)] > 0 and also, according to inequality (5.3), [log (l + gn{x» [ ~ t [gn(x) I for all n > no and x in X. Thus 00

hex) =

I

n =no

+1

log (1 + g.(x»)

converges uniformly for x in X. Since h is continuous and X is compact it follows that h must be bounded; in particular, there is a constant a such that Re hex) < a for all x in X. Thus, Lemma 5.7 applies and gives that

n 00

exp hex) =

n:;;no+ 1

(1 + gn(x»

converges uniformly for x in X. Finally, f(x) = [1 + g 1 (x)] .. . [1 + gnJx)] exp hex) and exp hex) #- 0 for any x in X. So iff(x) = 0 it must be that gn(x) = -1 for some n with 1 ~ n ~ no. III We now leave this general situation to discuss analytic functions. 5.9 Theorem. Let G be a region in IC and let Un} be a sequence in H(G) such that no fn is identically zero. If I. [J,,(z) -1] converges absolutely and uniformly 00

on compact subsets ofG then Ilfn(z) converges in H(G) to an analyticfunction n~1

f(z). If a is a zero off then a is a zero of only a finite number of the functions J", and the multiplicity of the zero off at a is the sum of the multiplicities of the zeros of the functions fn at a. Proof Since [j~(z) -1] converges uniformly and absolutely on compact subsets of G, it follows from the preceding lemma that fez) = nj~(z) converges uniformly and absolutely on compact subsets of G. That is, the infinite product converges in H(G). Suppose fCa) = 0 and let r > 0 be chosen such that B(a; r) c G. By hypothesis, I [J,,(z) -1] converges uniformly on BCa; r). According to Lemma 5.8 there is an integer n such that fCz) = fl Cz) .. .. f,.(z)g(z) where g does not vanish in B(a; r). The proof of the remainder of the theorem now follows. III Let us now return to a discussion of the original problem. If {an} is a sequence in a region G with no limit point in G (but possibly some point may be repeated in the sequence a finite number of times), consider the

I

168


functions (z - an). According to Theorem 5.9 if we can find functions gnCz) which are analytic on G, have no zeros in G, and are such that LI(z ---an)gn(z)---- 11 converges uniformly on compact subsets of G; thenf(z) = II(z - an)gn(z) is analytic and has its zeros only at the points Z = an' The safest way to guarantee that gn(z) never vanishes is to express it as gn(z) =exp hn(z) for some analytic function hn(z). In fact, if G is simply connected it follows that gn(z) must be of this form. The functions we are looking for were introduced by Weierstrass. 5.10 Definition. An elementary factor is one of the following functions Eiz) for p = 0, 1, ... :

Eo(z)

=

l-z,

Ep(z) = (1-z) exp (z

~ + ... + ~) ,p :2:

+

The function Eiz/a) has a simple zero at z

if

=

1.

a and no other zero, Also

bis a point in IC - G then Ep(az-b - b) has a simple zero at z

=

a and is

analytic in G. These functions will be used to manufacture analytic functions with prescribed zeros of prescribed multiplicity, but first an inequality must be proved which will enable us to apply Theorem 5.9 and obtain a convergent infinite product.

5.11 Lemma.

If Izl ::; 1

and p:2: 0 then II-Eiz)1 ::; IzlP+l,

Proof We may restrict our attention to the case where p :2: 1. For a fixed p let Eiz) = 1

GO

I akz k=l

+

k

be its power series expansion about z = O. By differentiating the power series as well as the original expression for Eiz) we obtain

I

OCJ

E-~(z) =

kakzk-l

k~1

= - zP exp

(z + .. , + ~)

Comparing the two expressions gives two pieces of information about the coefficients ak' First, a 1 = a z = , . , = a p = 0; second, since the coefficients of the expansion of exp

(z + ... + ~)

are all positive, ak

Thus, lakl = -ak for k :2: p+ 1; this gives

o=

Ell)

or 00

L

k~p+

I

la k = !

=

1

+

if;

I ak' k=p+l

::;

0 for k :2: P + I,


Hence, for

izl ::;

169

I,

IEp{z) -

11

=

I

00

L

akzkl

k =p+ 1

=lzIP+li L akzk-P-!I 00

k=p+l

L lakl 00

::; IzIP+ I = IzIP+!

k=p+l

which is the desired inequality. III Before solving the general problem of finding a function with prescribed zeros, the problem for the case where G = C will be solved. This is done for several reasons. In a later chapter on entire functions the specific information obtained when G is the whole plane is needed. Moreover, the proof of the general case, although similar to the proof for C, tends to obscure the rather simple idea behind the proof.

5.12 Theorem. Let {an} be a sequence in C such that lim la,,1 = 00 and an # 0 for all n 2:: 1. (This is not a sequence of distinct points; but, by hypothesis, no point is repeated an infinite number of times.) If {Pn} is any sequence of integers such that 5.13

i

n= 1

(~)pn+ 1 1"1

<

00

for all r > 0 then f(z)

=

n EPn (z/a.) 00

n=1

converges in H(C). The function I is an entire function with zeros only at the points an' II z 0 occurs in the sequence {an} exactly m times then f has a zero at z = z 0 of multiplicity m. Furthermore, if Pn = n -I then (5.13) will be satisfied. Proof Suppose there are integers Pn such that (5.13) is satisfied. Then, according to Lemma 5.1 I,

whenever Izl ::; rand r ::; lanl. For a fixed r > 0 there is an integer N such that lanl ;;:: r for all n 2:: N (because lim la"1 = (0). Thus for each r > 0 the series L II - EpJ z / an)1 is dominated by the convergent series (5.13) on the disk 11(0; r). This gives that [l-Ep.(z/a n )] converges absolutely in H(C).

L

n Ep.(z/a 00

By Theorem 5.9, the infinite product

n

~

I

n)

converges in H(C).

To show that {P.} can be found so that (5.13) holds for all r is a trivial

170


matter. For any r there is an integer N such that lalll > 2r for all n ~ N. This gives that

C~:I)

<

!

for all n

~

N; so if Pn

= n-l for all n, the tail

L

end of the series (5.13) is dominated by CD". Thus, (5.13) converges. 18 There is, of course, a great latitude in picking the integers PII' If Pn were bigger than /1- I we would have the same conclusion. However, there is an advantage in choosing the Pn as small as possible. After all, the smaller the integer Pn the more elementary the elementary factor Epn{z/an ). As is evident in considering the series (5.13), the size of the integers Pn depends on the rate at which {:anl} converges to infinity. This will be explored later in Chapter XI. 5.14 The Weierstrass Factorization Theorem. Let f be an entire function and let {an} be the non-zero zeros off repeated according to multiplicity; suppose f has a zero at z = 0 of order m ;;. 0 (a zero of order m =0 at z =0 means f(O) *0). Then there is an entire function g and a sequence of integers {Pn} such that

Proof According to the preceding theorem integers {Pn} can be chosen such that

has the same zeros as f with the same multiplicities. It follows that f(z)/h(z) has removable singularities at z = 0, a j , a z , .... Thus flh is an entire function and, furthermore, has no zeros. Since IC is simply connected there is an entire function g such that

fez) h(z) The result now follows.

=

ef/(Z)

18

5.15 Theorem. Let G be a region and let {a j } be a sequence oj distinct points in G with no limit point in G; and let {m be a sequence oj integers. Then there is an analytic function f defined on G whose only zeros are at the points a j ; furthermore, aj is a zero off of multiplicity mj'

J

Proof We begin by showing that it suffices to prove this theorem for the special case where there is a number R > 0 such that 5.16

{z:

Izl

> R}

C

G and

lajJ

~ R

for allj ~ 1.

f in R(G) = the multiplicity of the zero at z = a j ;

It must be shown that with this hypothesis there is a function

with the a/s as its only zeros and mj

171


and with the further property that

lim /(z)

5.17

z-+oo

= 1.

In fact, if such an f can always be found for a set satisfying (5.16), let G 1 be an arbitrary open set in C with {a j } a sequence of distinct point~ in G 1 with no limit point, and let {mj} be a sequence of integers. Now if B(a; r) is a disk in G 1 such that aj fE B(a; r) for all j:;:,: 1, consider the Mobius transformation Tz = (z - a)-I. Put G = T(G 1) \ {oo}; it is easy to see that G satisfies condition (5.16) where aj = Taj = (a j - a)-I. If there is a function f in H(G) with a zero at each aj of multiplicity mi' with no other zeros, and such that f satisfies (5.17); then g( z) = f(Tz) is analytic in G I - {a} with a removable singularity at z = a. Furthermore, g has the prescribed zero at each aj of multiplicity m j. So assume that G satisfies (5.16). Define a second sequence {z.} consisting of the points in {a j}, but such that each aj is repeated according to its multiplicity m j ' Now, for each n :;:,: 1 there is a point w" in C - G such that

Iw.-z.1

= d(z",

C-G).

Notice that the hypothesis (5.16) excludes the possibility that G = C unless the sequence {aj} were finite. In fact, if {aj} were finite the theorem could be easily proved so it suffices to assume that {a)} is infinite. Since laJi :s R for allj and {aj} has no limit point in G it follows that C-G is non-empty as well as compact. Also,

lim Iz. - w.1 Consider the functions

=

o.

E,,(Z.-W,,) ; z-w"

each has a simple zero at z = Zn' It must be shown that the infinite product of these functions converges in H(G). To do this let K be a compact subset of G so that d(C-G, K) > O. For any point z in K

:s Iz,,-w.1 [d(w", K)r IZn-Wnl z-w" :s

1

Iz.-w.1 [d(C-G, K)r 1

It follows that for any 8, 0 < 8 < 1, there is an integer N such that

<8 Iz.-w·1 z-w"

for all z in K and n :;:,: N. But then Lemma 5.11 gives that 5.18

IE.(;~;:) - 11 :s 8.+

1


172

for all z in K and n

~

N. But this gives that the series

converges uniformly and absolutely on K. According to Theorem 5.9

CEn(~~::)

j(z) =

converges in H(G), so thatjis an analytic function on G. Also, Theorem 5.9 implies that the points {aj} are the only zeros ofjand mj is the order of the zero at z = a j (because a j occurs m j times in the sequence {zn To show that limj(z) = I, let E > 0 be an arbitrary number and let Rl > R (Rl will

n.

Z->OO

be further specified shortly). If Izl

~ Rl

then, because

G c B(O; R),

[Zn-Wnl

-Iz-w n

IZnl : :; Rand Wn

E

11:-

2R
So, if we choose Rl > R so that 2R < 8(Rl -R) for some 8, 0 < 8 < 1, (5.18) holds for Iz I ~ Rl and for all n for all nand Izi 5.19

~ R1 ;

~

I. In particular, Re

En( Zn-Wn) z- Wn

so that

Ij(z)-11 = lex p

(~ log En (:n~::)) - 11

is a meaningful equation. On the other hand (5.3) and (5.18) give that

: :;

- - - 11 L 31- En (Zn-Wn) oo

n= 1

<

2

Z-JV

n

~ ~ 8n + 1

- ;S2

3 1}2 21-8

=---

for Izi

~ R1•

If we further restrict 8 so that leW-II <

E

whenever

> 0


173

then equation (5.19) gives that If(z)-II < limf(z) = 1..

€

whenever Izl ~ R I • That is,

Z->
One of the more interesting results that follows from the above theorem says (in algebraic terms) that M(G) is the quotient field of the integral domain H(G). Avoiding this language the result is as follows. 5.20 Corollary. Iff is a meromorphic function on an open set G then there are analytic functions g and h on G such that f = gjh. Proof Let {a j} be the poles of f and let m j be the order of the pole at a j. According to the preceding theorem there is an analytic function h with a zero of multiplicity m j at each z = a j and with no other zeros. Thus hf has removable singularities at each point a j. It follows that g = hf is analytic in G. •

Exercises

1. Show that

n (1 +zn) converges absolutely iff n (I + IZnD converges.

2. Prove that lim

log (1 +z)

z

z-->o

= 1.

3. Let f and g be analytic functions on a region G and show that there are analytic functions.fl' gt. and h on G such that.f(z) = h(zVt(z) and g(z) = h(z)g)(z) for all z in G; and.f) and gt have no common zeros. 4. (a) Let 0 < lal < 1 and Izl :5 r < 1; show that a+lalzII+r l(I-az)a :5 I-r

(b) Let {an} be a sequence of complex numbers with 0 < lanl < 1 and < 00. Show that the infinite product

L (I-Ianl)

converges in H(B(O; 1» and that IB(z)1 :5 1. What are the zeros of B? (B(z) is called a Blaschke Product.) (c) Find a sequence {an} in B(O; I) such that (I-lanD < 00 and every number e i8 is a limit point of {an}.

L

n 00

5. Discuss the convergence of the infinite product

n=1

6. Discuss the convergence of the infinite products 7. Show that

n 00

(

1-

1) = I-.

2"

Ip for p >

n

o.

n [1 + ~J and nil + ~I.

n=2 n 2 8. For which values of z do the products (I-zn) and (1 +z2n) converge? Is there an open set G such that the product converges uniformly on each compact subset of G? If so, give the largest such open set.

n

n

174


9. Use Theorem 5.15 to show there is an analytic function fan D = {z: G which properly contains D. 10. Suppose G is an open set and {J,,} is a sequence in H(G) such that /(z) = converges in H(G). (a) Show that

Iz I < I} which is not analytic on any region

llt..(z)

k~l ~k(Z) JJkt..(Z)] converges in H(G) and equals J'(z). (b) Assume that f is not the identically zero function and let K be a compact subset of G such thatf(z) # 0 for all z in K. Show that J'(z) = ~ f~(z) fez) J,,(z)

.:S

and the convergence is uniform over K. 11. A subset J of H(G), G a region, is an ideal iff: (i) f and g in J implies af+bgis in Jfor all complex numbers a and b; (ii)fin J andg any function in H(G) impliesfg is in J. J is called a proper ideal if J # (0) and J # H(G); J is a maximal ideal if J is a proper ideal and whenever J is an ideal with J c J then either J = J or J = H(G); J is a prime ideal if whenever f and g e H(G) and /g e J then either fe J or g e J. If fe H(G) let ~(.f) be the set of zeros of/counted according to their multiplicity. So ~«z-a)3) = {a, a, a}. If!/ c H(G) then ~(!/) = n {~(f):fe!/}, where the zeros are again counted according to their multiplicity. So if!/ = {(z-a)3 (z-b), (z-a)2} then ~(!/) = {a, a}. (a) Iff and g e H(G) then / divides g (in symbols, fIg) if there is an h in H(G) such that g = fh. Show that fIg iff ~(f) c ~(g). (b) If Yc H(G) and Y contains a non-zero function then f is a greatest common divisor of Y if: (i) fig for each g in Y and (ii) whenever hlg for each g in Y, hi/, In symbols, f = g.c.d.Y. Prove that f = g.c.d.Y. iff !Ir(f) = !Ir(Y) and show that each non-empty subset of H(G) has a g.c.d. (c) If A c G let J(A) = {fe H(G): ~(f) ::l A}. Show that J(A) is a closed ideal in H(G) and J(A) = (0) iff A has a limit point in G. (d) Let a E G and J = J( {a}). Show that J is a maximal ideal. (e) Show that every maximal ideal in H(G) is a prime ideal. (f) Give an example of an ideal which is not a prime ideal. 12. Find an entire function / such that fen + in) = 0 for every integer n (positive, negative or zero). Give the most elementary example possible (i.e., choose the Pn to be as small as possible). 13. Find an entire function/such that f(m + in) = 0 for all possible integers m, n. Find the most elementary solution possible.

§6. FIlctoriZlltion of the sine function In this section an application of the Weierstrass Factorization Theorem to sin 1TZ is given. If an infinite sum or product is followed by a prime

175

Factorization of the sine function

(apostrophe) (i.e., 2:' or TI'), then the sum or product is to be taken over all the indicated indices 11 except n = 0. For example, co

00

00

2:' an = n=1 L a_ n + "==1 Lan" n=-oo The zeros of sin

7TZ =

1 .. h . 21 (e mZ - e - InZ) are precisely t e Integers; moreover,

each zero is simple. Since

n~~ GY < for all r > 0, one can (5.13) choose fl. Factorization Theorem. Thus sin

7TZ =

[exp g(z)] z

=

00

1 for all n

111

the Weierstrass

J~I~ (1 - ~) ex!n;

or, because the terms of the infinite product can be rearranged,

6.1

sin

7TZ =

[exp g(z)] z

for some entire function g(z). If fez) = sin 'IT

D(1 -::) 7TZ

then, according to Theorem 2.1,

Fez)

cot 'lTZ = -.-fez) 1

=

2z ;2_n 2

00

+ -z + L, '"

g'(z)

n~l

and the convergence is uniform over compact subsets of the plane that contain no integers (actually, a small additional argument is necessary to justify this-see Exercise 5.10). But according to Exercise V. 2.8, 'IT

cot 'lTZ

1 =

Z

+

2z 2: .=1z-n 00

-2--2

for z not an integer. So it must be that g is a constant, say g(z) = a for all z. It follows from (6.1) that for 0 < Izi < 1 sin

TTZ =

7TZ

::

fr (1 _;z~).

Letting z approach zero gives that ea =

6.2

sin 'lTZ =

n

7T n= 1

'lTZ

'IT.

This gives the following:

1:':'r (1 _z:) n= 1

fl

and the convergence is uniform over compact subsets of C.

176


Exercises 1. Show that cos

D[

(in4~2I)2 J.

1-

TTZ =

2. Find a factorization for sinh z and cosh z. 3. Find a factorization of the function cos I \ • ,

4. Prove WallIs s formula:

7T

:2

xo n·

:.J -sin(

7TZ).

. 4

!

(2n)2

= n~ 1 (2n -1)(2~+ 1) .

§7. The gamma/unction Let G be an open set in the plane and let {/,.} be a sequence of analytic functions on G. If {/,.} converges in H(G) tofandfis not identically zero, then it easily follows that {In} converges to f in M(G). Since d(ZI, Z2) =

d(~, ~), where dis the spherical metric on Coo (see (3.1)), it follows that '" I

Z2

converges to! {~} /,. f verges uniformly to

in ll'I(G). It is an easy exercise to show that

.!

f

{~} /,.

con-

on any compact set K on which no /,. vanishes. (What

does Hurwitz's Theorem have to say about this situation?) Since, according to Theorem 5.12, the infinite product

converges in H(C) to an entire function which only has simple zeros at Z = - I, - 2, ... , the above discussion yields that

1"1 ro

7.1

11=

1

(

1

+~

)-1

e z/ n

converges on compact subsets of C - { - I, -- 2, ... } to a function with simple poles at z = -1, -2, .... 7.2 Definition. The gamma function, r(z), is the meromorphic function with simple poles at z = 0, -1, ... defined by

7.3

r(z) =

e~YZ(1(1

+

where y is a constant chosen so that r(l)

~rlez/n,

= I.

Oil

C

The gamma function

177

The first thing that must be done is to show that the constant y exists; this is an easy matter. Substituting z = 1 in (7.1) yields a finite number C

D + ~1)-1 00

=

(

1

e l/n

which is clearly positive. Let y = log c; it follows that with this choice of y, equation (7.3) for z = 1 gives r(l) = 1. This constant y is called Euler's constant and it satisfies

7.4

=

eY

D + ~1)-1 00

(

1

e lln

Since both sides of (7.4) involve only real positive numbers and the real logarithm is continuous, we may apply the logarithm function to both sides of (7.4) and obtain y

~ 10g[(1 + O-le

=

=

i [~ -

k~1

= lim

k

log (k+ 1)+log k]

.i: [~ k

" .... OOk=1

= lim [(I + " .... 00

log (k+ 1)+log k]

~2 +

Adding and subtracting log

11

fact that lim log ( 11 :

yields

7.5

y

1lk ]

1) = 0

... +

!)

n

-IOg(n+1)].

to each term of this sequence and using the

= lim [(1 + n .... oo

!+

2

... +

~)

11

- log

n] .

This last formula can be used to approximate y. Equation (7.5) is also used to derive another expression for r(z). From the definition of r(z) it follows that

0

- yz r(z) = _e_ lim " ( 1 + ~ z n .... ook=1 k - yz

= _e_ lim z

lim

n .... oo

)-1

ezlk

o_en

n .... oo k=1

k

zlk

z+k

I))

e-yzn' . exp ( z( 1 +1+ ... + z(z+ I) ... (z+n) n

178


However e-- YZ exP [(1+1+ ...

+~)zJ = nzexp[z( -y+I+± + ... + ~ -IOgn)].

So that the following is obtained 7.6 Gauss's Formula. For z #- 0, -1, ...

n' n- - fez) = lim --------------. "--->00 z(z+ 1) ... (z+n) Z

The formula of Gauss yields a simple derivation of the functional equation satisfied by the gamma function. 7.7 Functional Equation. For z #- 0, -1, ... f(z+ 1)

=

zf(z)

To obtain this important equation substitute z + 1 for z in (7.6); this gives n! n<+ 1

1'(z+ 1) = lim - - - - - - "--->0') (z+ I) ... (z+n+ I)

n_)

= zf(z)

since lim ( _ _ = 1. z+n+l Now consider 1'(z+2);we have 1'(z+2) = 1'«z+ 1)+ 1) = (z+1)1'(z+l) by the functional equation. A second application of (7.7) gives r(z+2) = z(z+ 1)I'(z). In fact, by reiterating this procedure

7.8

r(z+n)

=

z(z+ 1) ... (z+n-l)1'Cz)

for n a non negative integer and z #- 0, -I, .... In particular setting z = 1 gives that

7.9

I'(n+l)

= n!

That is, the I' function is analytic in the right half plane and agrees with the factorial function at the integers. We may therefore consider the gamma function as an extension of the factorial to the complex plane; alternately, if z #- - I, - 2, ... then letting z! = f(z + 1) is a justifiable definition of z!. As has been pointed out, I' has simple poles at z = 0, - 1, ... ; we wish to find the residue of l' at each of its poles. To do this recall from Proposition V. 2.4 that Res (1'; -n)

= lim (z+n)1'(z) z- -n

179

The gamma function

for each non-negative integer n. But from (7.8)

(z+n)r(z)

r(z+n+ 1) = ---'----'--z(z+ 1) ... (z+n-l)

So letting z approach - n gives that (-It

= --,

Res(r; -n)

7.10

n!

n

~

0.

According to Exercise 5.10 we can calculate r'jr by 1

r'(z)

7.11

r(z)

= -y -

~+

~

6

z n(n+z)

for z -# 0, -1, ... and convergence is uniform on every compact subset of C - {O, -I, ... }. It follows from Theorem 2.1 that to calculate the derivative of l"jr we may differentiate the series (7.11) term by term. Thus when z is not a negative integer

( r'(z»)' r(z) =

7.12

1 Z2

6 00

+

1 (n+z)2'

At this time the reader may well be asking when this process will stop. Will we calculate the second derivative of r'fr? The answer to this question is no. The answer to the implied question of why anyone would want to derive formulas (7.11) and (7.12) is that they allow us to characterize the gamma function in a particularly beautiful way. Notice that the definition of r(z) gives that rex) > if x > 0. Thus, log rex) is well defined for x > and, according to formula (7.12), the second derivative of log r(x) is always positive. According to Proposition VI. 3.4 this implies that the gamma function is logarithmically convex on (0, (0); that is, log rex) is convex there. It turns out that this property together with the functional equation and the fact that r(1) = 1 completely characterize the gamma function.

°

°

°

7.13 Bohr-Mollerup Theorem. Let f be a function defined on (0, 00) such that f(x) > for all x > O. Suppose that f has the following properties:

(a) logf(x) is a convex function; (b) f(x+ 1) = xf(x) for all x; (c) f(1) = 1. Then f(x)

=

rex) for all x.

Proof Begin by noting that since f has properties (b) and (c), the function also satisfies 7.14

f(x+n) = x(x+ 1) ... (x+n-l)f(x).

for every non-negative integer n. So if f(x) = rex) for 0 < x :::; 1, this

180


equation will give that f and f are everywhere identical. Let 0 < x ::; I and let n be an ~nteger larger than 2. From Exercise VI. 3.3 _Io-=-g___ f('-n_-_l)'-----_I--'og:::..::f---.:(n-- "-) < log f( x + n) - log f( n) < _lo..:::g.::-f(,-n_+_1)'-----_l--'og:::..::f---.:(n-'-.) (n-l)-n (x+n)-n (n+ I)-n Since (7.14) holds we have that f(m) = (m-I)! for every integer m Thus the above inequalities become

-log(n-2)!+log(n-I)!

logf(x+n)-Iog (n-I)! x

s

s

~

1.

logn!-Iog(n-I)!;

or

x log (n- I) ::; logf(x+n)-Iog (n-I)! ::; xlogn. Adding log (n -I)! to each side of this inequality and applying the exponential (exp is a monotone increasing function and therefore preserves inequalities) gives (n-lY(n-I)! ::;f(x+n)::; nX(n-I)! Applying (7.14) to calculate f(x+n) yields _~(n_-~l)~X~(n_-~I)_!__

_ ____ nX----'(~n_-_I~)!____ - x(x+I) ..• (x+n-I)

x(x+I) ... (x+n-I) -

______n_X_n_!

[x+n]

= x(x+ 1) ... (x+n) --n- . Since the term in the middle of this sandwich, f(x), does not involve the integer n and since the inequality holds for all integers n ~ 2, we may vary the integers on the left and right hand side independently of one another and preserve the inequality. In particular, n+ I may be substituted for non the left while allowing the right hand side to remain unchanged. This gives

n~ n! ----------s x(x+ I) ... (x+n)

for all n

f(x)

s

nX n! [x+n] x(x+ I) ... (x+n) 11

~ 2 and x in [0, I). Now take the limit as n --+ 00. Since lim (x:n)

= I, Gauss's formula implies that f(x) = f(x) for 0 < x ::; 1. The result now follows by applying (7.14) and the Functional Equation . • 7.15 Theorem.

If Re

z > 0 thell

fez) =

J'" e-tt

z- 1

dt

o

The integrand in 7.15 behaves badly at t = 0 and t = 00, so that the meaning of the above equation must be explicitly stated. Rather than give a formal definition of the convergence of an improper integral, the properties of this particular integral are derived in Lemma 7.16 below (see also Exercise 2.2).

181

The gamma function

7.16 Lemma. Let S = {z: a :5: Re z :5: A} where 0 < a < A <

00.

(a) For every E > 0 there is a 0 > 0 such that for all z in S

whenever 0 < a < f3 < O. (b) For every E > 0 there is a number

whenever

f3 >

>

a

such that for all z in S

K

K.

Proof. To prove (a) note that if 0 < t :5: 1 and z is in S then (Re z-I) log t :5: (a-I) log t; since e- I :5: 1, le- I t Z - 1 1 :5: t Re z-1 :5: t a - 1 •

So if 0 < a <

f3 < 1 then

f e- t

ft

fl

I

fl

Z -

1

dt :5:

a- 1

dt

"

"

for all z in S. If E > 0 then we can choose 0, 0 < 0 < 1, such that a- 1(f3a_ a a) < E for la-f31 < o. This proves part (a). To prove part (b) note that for z in Sand t ;::: 1, It Z - 1 1 :5: t A - 1 • Since A t - 1 exp (-It) is continuous on [I, 00) and converges to zero as t --+ 00, there is a constant c such that t A -1 exp ( - 1t) :5: c for all t ;::: 1. This gives that le- I t Z - 1 1 :5: ce--l: 1 for all z in Sand t ;::: 1. If f3 >

f

a

> 1 then

"

f

P

fl

e- I t z -

1

dt :5: c

e--!-I

a

= 2c(e--!-a- e --!-fl). Again, for any E > 0 there is a number K > 1 such that 12c(e--!-a- e--!-P)1 < E whenever a, f3 > K, giving part (b) . • The results of the preceding lemma embody exactly the concept of a uniformly convergent integral. In fact, if we consider the integrals

182


for 0 < IX < 1, then part (a) of Lemma 7.16 says that these integrals satisfy a Cauchy criterion as IX -+ O. That is, the difference between any two will be arbitrarily small if IX and f3 are taken sufficiently close to zero. A similar interpretation is available for the integrals

for

IX

> 1. The next proposition formalizes this discussion.

7.17 Proposition. If G

= {z: Re z > O} and

f e-tt n

/,,(z) =

Z- 1

dt

lin

for n ~ 1 and z in G, thell each J.. is analytic vergent ill B(G).

011

G and the sequence is

Proof. Think of/,,(z) as the integral of cp(t, z) = e - tt z segment

[~,

l

COIl-

along the straight line

11] and apply Exercise IV. 2.2 to conclude that J.. is analytic.

Now if K is a compact subset of G there are positive real numbers a and A such that K c {z: a :::; Re z :::; A}. ,Since

f e-tt

lin

fm(z)-J..(Z) =

z-

11m

l dt

+

f e-tt m

z- l

dt

n

for m > n, Lemma 7.16 and Lemma 1.7 imply that {J..} is a Cauchy sequence in B(G). But B(G) is complete (Corollary 2.3) so that {J..} must converge. • If f is the limit of the functions {J..} from the above proposition then define the integral to be this function. That is,

f e-tt
fez) =

7.18

z- l

dt, Re z > O.

o

To show that this functionf(z) is indeed the gamma function for Re z > 0 we only have to show that f(x) = rex) for x ~ 1. Since [1, 00) has limit points in the right half plane and both f and r are analytic then it follows thatfmust be r (Corollary IV. 3.8). Now observe that successive performing of integration by parts on (1- tjlltt,,-l yields

f( n

o

1 - ~). t X n

1

dt = _ _ _ Il_!n_"_ __ x(x+ 1) ... (x+n)

which converges to rex) as 11 -+ 00 by Gauss's formula. If we can show that the integral in this equation converges to SO' e-tt x - 1 dt = f(x) as 11 -+ 00 then

183

The gamma function

Theorem 7.15 is proved. This is indeed the case and it follows from the following lemma.

+

7.19 Lemma. (a) {( 1

(b)

Iff

~

0 then

~)"}

converges to e in H(C). Z

(1 -~)" :S:

~

e-'forall n

t.

Proof (a) Let K be a compact subset of the plane. Then lzj < K and 11 sufficiently large. It suffices to show that

lim /llog

n-+oo

(1 +=)

for all z in

z

=

11

11

uniformly for z in K by Lemma 5.7. Recall that

L (-It-

wk

00

logO+w) =

1

k

k=!

for

IlVl

< J. Let n > II

So 7.20

Izi

log

for all

( + -z) 1

n

( z) -

n log 1 + ~

z

z in

K; if

I Z2 2 n

= z -

=

z is

any point in K then

1 Z3 3n

+ - - 2 - . .. .

[I2(z) +:31(Z)2 -...] ; ~

z -

~

taking absolute values gives that

In log (1 +~) - z\:S: Izi ~~I~r-l :S: Iz/

~ I~r

k=l

Izl2

1

--_.-----

n I

-Iz/nl

R2 - n-R

<--

where R ~ Izi for all z in K. If n -:.. 00 then this difference goes to zero uniformly for z in K. (b) Now let t ~ 0 and substitute - t for z in (7.20) where t s: n. This gives 11

(

t) + t =

log 1 - -

n

-

t

L -l(t)k-l :s: 0 00

k=2k

11,

184

Thus II

log

(1 -

n: ;


-t;

and since exp is a monotone function part (b) is proved . • Proof of Theorem 7.15. Fix x > 1 and let £ > O. According to Lemma 7.16 (b) we can choose K > 0 such that r

f e-tt

7.21

x- 1

dt <

~

whenever r > K. Let II be any integer larger than defined in Proposition 7.17. Then

K

and letfn be the function

J[e- (1 + ~YJ t

-

tx -

1

dt

lin

Now by Lemma 7.19 (b) and Lemma 7.16 (a) 7.22

for sufficiently large II. Also, if II is sufficiently large, part (a) of the preceding lemma gives

for tin [0, K] where M = So

tx-

1 dt. Thus

7.23

Using Lemma 7.19 (b) and (7.21)

for

II

>

K.

If we combine this inequality with (7.22) and (7.23), we get

185

The gamma function

for n sufficiently large. That is

-X(-X-+-l-;-~~-~-(x-+-nJ

= lim [/"(X) = I(x) - rex).

This completes the proof of Theorem 7.15 . • As an application of Theorem 7.15 and the fact that 2) notice that

reD =

.;:;

(Exercise

.J; = fe-let dt. o

Performing a change of variables by putting t =

f e-

S2

gives

...;;. =

52

s- 1 (2s) ds

o

That is,

f
e

-52

.J;

d

s=2

o

This integral is often used in probability theory.

Exercises 1. Show that 0 < y < 1. (An approximation to y is .57722. It is unknown whether y is rational or irrational.) 2. Show that r(z) r(l-z) = 7T esc 7TZ for z not an integer. Deduce from this that ret) = 3. Show: r(2z) = 2 2z - l r(z) r(z+!). (Hint: Consider the function r(z) f(z+!) r(2z)-I.) 4. Show that log r(z) is defined for z in C -( - 00, 0] and that

.J;. .J;

log r(z) = -log z-yz -

f

n=1

[lOg

(I + -=) - -=]. n

n

5. Let/be analytic on the right half plane Re z > 0 and satisfy:/(I) = I, . /(z+n) /(z+ I) = z/(z), and lIm Zfi() = 1 for all z. Show that/ = r. n"'oo n n


186

6. Show that

r(z) =

00

2: --)_. + Je-tt z OJ

(-In

n=O

n!(z+n)

~

1

dt

°

1

for z of 0, -1, -2, ... (not for Re z > 7. Show that

Jsin (t2) dt = Jcos (t2) dt = -tJt1T. ex:

°

alone).

00

o

0

8. Let u > and v > 0 and express r(u) rev) as a double integral over the first quadrant of the plane. By changing to polar coordinates show that

J(cos &)2"-1 (sin 8)2,,-1 d8.

,,12

r(u) 1'(1')

=

2r(u+ l)

o

The function

Beu L:) ,

r(u) rev) I'(u+c)

= _._--

is called the beta jilllctioll. By changes of variables show that

ft I

B(u, v) =

u-

1 (1-t)"-1 dt

o

Can this be generalized to the case when u and v are complex numbers with positive real part? 9. Let CX. be the volume of the ball of radius one in [Rn (n ~ 1). Prove by induction and iterated integrals that

f (1_t 1

CX n

=

2cxn -

1

2 )<.-I)/2

dt

o

10. Show that Tr nl2

CX n

=

(n/2)l'(n/i)

where CX n is defined in problem 9. Show that if n 11. The Gaussian psi function is defined by 'Y(z) =

=

2k, k ~ I, then

CX.

= 1Tkjk!

1"(z)

1'&Y

(a) Show that 'I" is meromorphic in IC with simple poles at z and Res ('I"; -n) = -1 for n ~ O.

= 0, - I, ...

187

The Riemann zeta function

(b) Show that 1¥(1) = -yo 1 (c) Show that 1¥(z+ 1) -1¥(z) = - . z

(d) Show that 1¥(z)-1¥(I-z) = -7T cot 7TZ. (e) State and prove a characterization of 1¥ analogous to the BohrMollerup Theorem. §8. The Riemann zeta function Let z be a complex number and (z log 11)1 = exp (Re z log n). Thus n

L

Ik-zi =

k= 1

n

na

positive integer. Then

L

exp(-Rezlogk)

L

k- Re

In'l =

lexp

k= 1

=

11

,

k=1

So if Re z ;::: 1 + € then

L n

L n

Ik-zi

~

k= 1

k-(l+,>;

k= 1

that is, the series

converges uniformly and absolutely on {z: Re z ;::: 1 +€}. In particular, this series converges in H( {z: Re z > I}) to an analytic function ~(z). 8.1 Definition. The Riemann zeta function is defined for Re z > 1 by the equation ~(z)

=

00

L

n-'.

n=1

The zeta function, as well as the gamma function, has been the subject of an enormous amount of mathematical research since their introduction. The analysis of the zeta function has had a profound effect on number theory and this has, in turn, inspired more work on ~. In fact, one of the most famous unsolved problems in Mathematics is the location of the zeros of the zeta function. We wish to demonstrate a relationship between the zeta function and the gamma function. To do this we appeal to Theorem 7.15 and write

f e- t,-1 dt 00

r(z) =

t

o

188


for Re z > O. Performing a change of variable in this integral by letting t = nu gives

Je-ntt%-l dt; 00

r(z) = n%

o

that is

Je-ntt 00

Il- Z r(z)

=

z- 1

dt.

o

If Re z > 1 and we sum this equation over all positive n, then {(z)r(z) =

8.2

I

00

n-Zr(z)

n~l

We wish to show that this infinite sum can be taken inside the integral sign. But first, an analogue of Lemma 7.16. 8.3 Lemma. (a) Let S = {z: Re z ::::: a} where a > 1. If E: > 0 then there is a number a, 0 < a < 1, such that for all z in S

whenever b > fJ > CL > O. (b) Let S = {z: Re z ::; A} where -00 < A < a number K > 1 such that for all z in S

whenever

f3 >

IX

>

00.

IfE: > 0 then there is

K.

Proof (a) Since e t -1 ::::: t for all t ::::: 0 we have that for 0 < t ::; 1 and z in S

l(e t _I)-l t

Z-

1

1::;

ta -

2•

Since a > 1 the integral SA t a - 2 dt is finite so that acan be found to satisfy (a). (b) If t ::::: 1 and z is any point in S then, as in the proof of Lemma 7.16 (b), there is a constant c such that

Since ett(e t _1)-1 is integrable on [1, found . •

00)

the required number

K

can be


189

=

8.4 Corollary. (a) If S the integral

{z: a::::; Re z::::; A} where 1 < a < A <

00

then

I (et_l)-ltZ-l dt 00

o

converges uniformly on S. (b) If S = {z: Re z ::::; A} where -

I (e

00

< A <

00

then the integral

00

I) - 1 t Z -

t -

dt

1

I

converges uniformly on S.

8.5 Proposition. For Re z > 1

I (e'-1)-lt 00

~(z)r(z) =

1 dt

Z -

o

Proof According to the above corollary this integral is an analytic function in the region {z: Re z > I}. Thus, it suffices to show that ~(z)r(z) equals this integral for z = x > 1. From Lemma 8.3 there are numbers IX and {J, 0 < IX < {J < 00, such that: «

f(et-l)-lt X - 1 dt

<~,

o

f 00

(et_1)-lt X -

dt <

~.

/3

Since n

L

L 00

e- kt ::::;

k= 1

for all n

1

~

e- kt = (e t -1)-l

k= 1

1,

f ~f a.

~

dt <

~,

e- nt t x - 1 dt <

~.

e-ntt x -

1

o

00

fl

Using equation (8.2) yields

Is(x)r(x) -

[(e t -I) -1 tX-t dtl ::::;

- f (e _1)- t t

"

l

X-

1

dtl

E;

+

In~1

!

e-ntt x -

1

dt

190


I

But e - 01 converges to (e' -1) - I uniformly on [Ot, ,8], so that the right hand side is exactly E • • We wish to use Proposition 8.5 to extend the domain of definition of ~ to {;:: Re z > -I} (and eventually to all C). To do this, consider the Laurent expansion of (e Z - 1)-- 1 ; this is

1

I

~-=1

8.6 for some constants neighborhood of t

2: anz"

I

= z - 2+

00

n ::: 1

Thus [eel -1) - I - r IJ remains bounded in a O. But this implies that the integral

{J I ' {J2' • • • •

=

j'(_I__ ~) 1

e' -1

o

tZ-1

dt

t

converges uniformly on compact subsets of the right half plane {z: Re z > O} and therefore represents an analytic function there. Hence

_ (1 I) ~(z)f(z)= J -, - 1

8.7

~

e-I

o

t

" +j (I')

tz -

1

dt+(Z-l)-1

tZ -

1

-,--dt, e-l

1

and (using Corollary 8ACb» each of these summands, except (z -1) -I, is analytic in the right half plane. Thus one may define ~(z) for Re z > 0 by setting it equal to [r(z)] - 1 times the right hand side of (8.7). In this manner ~ is meromorphic in the right half plane with a simple pole at z = 1 n- I diverges) whose residue is 1. Now suppose 0 < Re z < I; then

(I

co

(Z-1)-1

Jt

= -

z-

2 dt.

I

Applying this to equation (8.7) gives

f(_,_l- - 1) co

8.8

~(z)f(z)

=

o

e -1

t

tz -

I

dt,

0< Re z < 1.

Again considering the Laurent expansion of (e Z -1) - 1 (8.6) we see that [ee' - 1) -1 - r 1+ -!l :::; ct for some constant c and all t in the unit interval [0, IJ. Thus the integral

is uniformly convergent on compact subsets of {z: Re z > -I}. Also, since lim t (}:- I~oc

_f

_1_)' 1 e -1 =

t

191


there is a constant c' such that

(~

1- )
This gives that the integral

f( 1

1)

00

-- -

-

e t -1

t

t ;:::: 1.

t%- 1

dt

1

converges uniformly on compact subsets of {z: Re z < I}. Using these last two integrals with equation (8.8) gives

1+ 2--1) f( e' 1 - 1

1+ f( -1-1 - -1) t:-

1

8.9

'(z)f{z)

=

-- -

t

o

00

t Z - 1 dt - -

2z

et -

t

1

dt

1

for 0 < Re z < 1. But since both integrals converge in the strip -1 < Re z < 1 (8.9) can be used to define ,(z) in {z: -1 < Re z < I}. What happens at z = O? Since the term (2z) -1 appears on the right hand side of (8.9) will , have a pole at z = O? The answer is no. To define ,(z) we must divide (8.9) by fez). When this happens the term in question becomes [2zf(Z)]-1 = [2f(z + 1 which is analytic at z = O. Thus, if , is so defined in the strip {z: - 1 < Re z < I} it is analytic there. If this is combined with (8.7), ,(z) is defined for Re z > -1 with a simple pole at z = 1. Now if -1 < Re z < 0 then

nr

f 00

t,-1 dt =

z

1

inserting this in (8.9) gives

1 1+ -I)" f( e -1 2 00

t(z)f(z) =

8.10

-

But

-.- t

o

1

et _ 1

1

+i

t

t .-1 dt

+ 1) i i1 (e;t=1 = i

'

-1 < Re z < O.

t

=

cot (Yt).

A straightforward computation with Exercise V. 2.8 gives

2

cot (-!it) = --:- - 4it It

for t

=1=

O. Thus

1 2'"= -2--22 + 4n

n

1

t

1T


192

Applying this to (8.10) gives
~(z)r(Z)=2f(~ ;S I

8.11

o

2

f

12

+4n 7T

2)t

Z

dl

= 2 L.. '"
tZ 2 4 2 2 dt t+n7T

n;lO

=2

L00 (27Tn)%-1 fOO-t /+1- dt %

n;l

o

f 00

= 2(27Ty- 1W-Z)

t%

-2-

o

t

+I

dt,

for -1 < Re z < O. (It is left to the reader to justify the interchanging of the sum and the integral.) Now for x a real number with -1 < x < 0, the change of variable s = t 2 gives (by Example V. 2.12) 00 00 tX I fsHX-IJ --dt = - --ds 8.12 t2 + 1 2 s+ 1

f

o

0

=

1

27T cosec [1-7T(l- x)] I

= 2 7T sec (t7TX). But Exercise 7.2 gives 1

r(x) =

r{l-x)

7T

sin TTX

=

r(l-x)

7T

[2 sin (-t7TX) cos (-t 7Tx)]

Combining this with (8.11) and (8.12) yields the following. 8.13 Riemann's Functional Equation. ~(z)

= 2(27Ty-l r(l-z)W -z) sin (!7TZ)

for - I < Re z < O. Actually this was shown for x real and in (-I, 0); but since both sides of (8.13) are analytic in the strip -I < Re z < 0, (8.13) follows. The same type of reasoning gives that (8.13) holds for -1 < Re z < 1 (what happens at z = O?). But we wish to do more than this. We notice that the right hand side of (8.13) is analytic in the left hand plane Re z < O. Thus, use (8.13) to extend the definition of ,(z) to Re z < O. We summarize what was done as follows.


193

8.14 Theorem. The zeta function can be defined to be meromorphic in the plane with only a simple pole at z = 1 and Res (L 1) = 1. For z t= 1 ~ satisfies Riemann's functional equation. Since f(l-z) has a pole at z = 1, 2, ... and since' is analytic at z = 2, 3, ... we know, from Riemann's functional equation, that 8.15

W -z) sin (l7rz)

= 0

for z = 2, 3, .... Furthermore, since the pole of r(l- z) at z = 2, 3, ... is simple, each of the zeros of (8.15) must be simple. Since sin (l7rz) = 0 whenever z is an even integer, W -z) = 0 for z = 3, 5, .... That is ~(z) = o for z = - 2, - 4, - 6, .... Similar reasoning gives that , has no other zeros outside the closed strip {z: 0 ::;; Re z ::;; I}. 8.16 Definition. The points z = - 2, - 4, ... are called the trivial zeros of' and the strip {z: 0 ::;; Re z ::;; I} is called the critical strip. We now are in a position to state one of the most celebrated open questions in all of Mathematics. Is the following true? The Riemann Hypothesis. If z is a zero of the zeta function in the critical strip then Re z = t. It is known that there are no zeros of, on the line Re z = I (and hence none on Re z = 0 by the functional equation) and there are an infinite number of zeros on the line Re z = t. But no one has been able to show that , has any zeros off the line Re z = t and no one has been able to show that all zeros must lie on the line. A positive resolution of the Riemann Hypothesis will have numerous beneficial effects on number theory. Perhaps the best way to realize the connection between the zeta function and number theory is to prove the following theorem.

8.17 Euler's Theorem. If Re z > 1 then

fr

~(z) = n~l (d-:z) Pn where {Pn} is the sequence of prime numbers. Proof First use the geometric series to find

I

~=

8.18

Pn

00

LPn- mz m=O

for all n ;::: 1. Now if n ;::: 1 and we take the product of the terms (1- pi Z) - I for I ::;; k ::;; n, then by the distributive law of multiplication and by (8.18), 8.19

TI n

k~1

where the integers n l'

11 2' ...

(

1

T=-=z Pk

)

Ln 00

=

j~1

j-

Z

are all the integers which can be factored as a


194

product of powers of the prime numbers PI' ... ,Pn alone. (The reason that no number nj - Z has a coefficient in this expansion other than 1 is that the factorization of nj into the product of primes is unique.) By letting n -+ 00 the result is achieved. •

Exercises 1. Let g(z) = z(z-I)7T-tz~(z)r(tz) and show that g is an entire function which satisfies the functional equation g(z) = g{l-z). 2. Use Theorem 8.17 to prove that L P; 1 = 00. Notice that this implies that there are an infinite number of primes. den) . 3. Prove that ~2(Z) = for Re z > 1, where den) IS the number of

2:"

divisors of n. 4. Prove that

nZ

n~l

~(z)~(z-l)

=

. 2:'" a(n) - , for Re z > 1, where a(n) IS the sum of n Z

n~l

the divisors of n. Uz-I)

5. Prove that - - - = ~(z)

2:"" cp(n) - for Re z > I, where cp(n) is the number of

n~l nZ

integers less than n and which are relatively prime to n. 1 ~ p.(n) . 6. Prove that = L. - for Re z > 1, where p.(n) IS defined as follows. ~(z) n~l nZ pftp~2 •.. p~m be the

Let n = factorization of n into a product of primes PI' •.• ,Pm and suppose that these primes are distinct. Let p.(l) = 1; if k I = ... = k m = I then let p.(n) = ( _l)m; otherwise let p.(n) = o. nz) ~ A(n) . 7. Prove that 1(z) = - L. 7 for Re z > 1, where A(n) = log P If n = n~l

pm for some prime P and m ;::: I; and A(n) = 0 otherwise. 8. (a) Let 7](z) = ~'(z)/~(z) for Re z > 1 and show that lim (z-zo)7](z) is z-+zo

always an integer for Re z 0 ;::: I. Characterize the point z 0 (in its relation to 0 in terms of the sign of this integer. (b) Show that for E > 0 Re 7] (1 +E+it) =

L 00

n:;;;; 1

A(n)n-(1+<) cos (t log n)

where A(n) is defined in Exercise 7. (c) Show that for all E > 0, 3Re 7] (1 +E)+4Re 7] (1 + E+ it) + Re 7] (1 +€+2it) ::; (d) Show that

~(z)

¥ 0 if Re z

=

1 (or 0).

o.

Chapter VIII Runge's Theorem In this chapter we will prove Runge's Theorem and investigate simple connectedness. Also proved is a theorem of Mittag-Leffler on the existence of meromorphic functions with prescribed poles and singular parts. §1. Runge's Theorem In Chapter IV we saw that an analytic function in an open disk is given by a power series. Furthermore, on proper subdisks the power series converges uniformly to the function. As a corollary to this result, an analytic function on a disk D is the limit in H(D) of a sequence of polynomials. We ask the question: Can this be generalized to arbitrary regions G? The answer is no. As one might expect the counter-example is furnished by G = {z: o < 1=1 < 2}. If {Pn(z)} is a sequence of polynomials which converges to an analytic functionj on G, and y is the circle Izl = 1 then JJ = lim JyPn = O. But Z-1 is in H(G) and SyZ-l 1= O. The fact that functions analytic on a disk are limits of polynomials is due to the fact that disks are simply connected. If G is a punctured disk then the Laurent series development shows that each analytic function on G is the uniform limit of rational functions whose poles lie outside G (in fact at the center of G). That is, each j in H(G) is the limit of a sequence of rational functions which also belong to H(G). This is what can be generalized to arbitrary regions, and it is part of the content of Runge's Theorem. We begin by proving a version of the Cauchy Integral Formula. Unlike the former version, however, the next proposition says that there exist curves such that the formula holds; not that the formula holds for every curve. 1.1 Proposition. Let K be a compact subset oj the region G; then there are straight line segments YI"'" Yn in G---- K such that Jor every Junction J in H(G),

j(z) =

~ _1. f£(~~ dw k=

12m

w-z

Yk

Jor all z in K. The line segments Jorm a Jinite number oj closed polygons. ProoJ. Observe that by enlarging K a little we may assume that K = (intKf· Let O
196

Runge's Theorem

(there are only a finite number of them because K is compact). Also let aRj be the boundary of Rj , I::; j ::; m, considered as a polygon with the counter-clockwise direction. If z E R j , 1 ::; j ::; m, then d(z, K) < J2S so that R j C G by the choice of S. Also, many of the sides of the rectangles R I , ••• , Rm will intersect. Suppose R j and Ri have a common side and let U j and U i be the line segments in oR j , and oRi respectively, such that R; n R j = {u j } = {uJ. From the direction given oR j and oR;, Uj and U i are directed in the opposite sense. So if g; is any continuous function on {u j},

J g; + Jg; Clj

=

O.

Uj

Let YI""'Yn be those directed line segments that constitute a side of exactly one of the Rj , I ::; j ::; m. Thus

1.2 for every continuous function g; on

m

UI oR

je,

j •

We claim that each Yk is in G - K. In fact, if one of the Yk intersects K, it is easy to see that there are two rectangles in the grid with Yk as a side and so both meet K. That is, Yk is the common side of two of the rectangles R I , ... , Rm and this contradicts the choice of Yk' If z belongs to K and is not on the boundary of any R j then
is continuous on

m

U oR

j

=I~. (J(W») 2m

w-z

for J in H(G). It follows from (1.2) that

j~l

1.3

But z belongs to the interior of exactly one R j ' If z rt R j' _I

27Ti

f [~'~

jV-z

dw = O' '

~Ri

and if z is in R j' this integral equals J(z) by Cauchy's Formula. Thus (1.3) becomes 1.4

J(z)

U oR

k=12m

w-z

Yk

m

whenever z E K -

1 fl(W) = L~ --: -- dw

j'

But both sides of (1.4) are continuous functions

j~l

on K (because each Yk misses K) and they agree on a dense subset of K. Thus, (1.4) holds for all z in K. The remainder of the proof follows . •

Runge's Theorem

197

This next lemma provides the first step in obtaining approximation by rational functions.

°

1.5 Lemma. Let y be a rectifiable curve and let K be a compact set such that K n {y} = O. IfJ is a continuous Junction on {y} and € > then there is a rational Junction R{z) having all its poles on {y} and such that

f

w-z dw-R{z) I < I J{W)

€

y

Jor all z in K. Proof Since K and {y} are disjoint there is a number r with 0 < r < d{K, {y}). If y is defined on [0, 1] then for s s, t s 1 and z in K

°

Iyf{(y)(t» t -z

s

Ji{(Y)(S»1 y s -z

~ IJ(Y(/»Y(S) - J(y(s»y(t) -z[f(y{/»

r

s

- J(y(s»]

I

r~ /I(Y{/»IIY{S)-y{t)1 + ~ IY(/)I keyes»~

- J{Y{/»I

°

+

~~I keyes»~ - J{y{t»1

There is a constant c > such that Izl s c for all z in K, ly(/) I s c and IJ(y(/» I s c for all I in [0, 1]. This gives that for all s and I in [0, 1] and z in K, (I) I-J(Y(/» y -z

Is

c -i ly(s)-y(t)1 r

J{y{s» -(-)y s -z

2c

+ 2IJ{y(s»-J(y(/»1 r

Since both y andJ 0 yare uniformly continuous on [0, 1], there is a partition {O = to < 11 < ... < tn = I} such that _ IJ{y(t» y(t)-z

1.6

I

J(y{t j» < - _ € y(I)-Z V(y)

for t j - 1 sIs t j , 1 s j s n, and z in K. Define R(z) to be the rational function R{z) =

n

L J(y{t j - 1»

j;l

[y(t)-y(t j _ 1)] [y(tj_1)-Zr 1

The poles of R(z) are yeO), y(tl), ... ,y(tn - 1). Using (1.6) yields that dW-R(Z)/ = / i If J(W) w-z

j;l

y

»]

fti[J(y(t» - [Cy(ti- t dy(t)1 y(t)-z y(tj_l)-z

0-1

f Ij

S

V~Y) ~

Ij-t

=€

for all z in K . •

dIYI(t)

198

Runge's Theorem

Before stating Runge's Theorem let us agree to say that a polynomial is a rational function with a pole at ::c. It is easy to see that a rational function whose only pole is at x is a polynomial.

1.7 Runge's Theorem. Let K he a compact suhset of C and let E be a subset oj C x - K that meets each component of C x ..... K. /j j is analytic in an open set containing K and E>O then there is a rational junctiun R(z) whose only poles lie in E and such that Ij(z)-R(z)l
jar all z in K. The proof that will be given here was obtained by S. Grabiner (Amer. Math. Monthly, 83 (1976), 807-808). For this proof we place the result in a different setting. On the space C (K, C) we define a distance function p by pUg) = sup{ Ij(z) - g(z)1 : z

E

K}

for j and g in C (K, C). I l is easy to see that p(jnJ)~O iff j = u -limjn on K. Hence C(K,C) is a complete metric space. So Runge's Theorem says that if j is analytic on a neighborhood of K and (> 0 then there is a rational function R (zl with poles in E such that r(j, R) < (. By taking I:: = 1/ n it is seen that we want to find a sequence of rational functions {Rn (z)} with poles in E such that r(j, Rn)----'>O; that is, such that j = u -lim Rn on K. Let B(E)=all functions j in C(K,C) such that there is a se4uence {Rn} of rational functions with poles in E such that j = u····1im Rn on K. Runge's Theorem states that if j is analytic in a neighborhood of K then flK, the restriction of jto K, is in B(E). 1.8 Lemma. B (E) is a closed subalgebra oj C (K, C) that contains every rational junction with a pole in E. To say that B(E) is an algebra is to say that iff and g are in B(E) and ex E C then rxf, j + g, and jg are in B (E). The proof of Lemma 1.8 is left to the reader.

1.9 Lemma. Let V and U be open subsets of C with V c U and dV /j H is a component of U and H n V7'=O then He V.

n U = O.

Proof. Let a E H n V and let G be the component of V such that a E G. Then HuG is connected (II.2.6) and contained in U. Since H is a component of U, C c fl. But dC c av and so the hypothesis of the lemma says that dC n H =0. This implies that H - C= H

n [(c -

) u DC]

C

=Hn(C-G

)

so that H-C is open in H. But G is open implies that H-G=Hn(cG) is closed in H. Since H is connected and G 7'= 0, H- G=O. That is, H=Gc V.JIII

199

Runge's Theorem

1.10 Lemma. If a e C - K then (z - a)-I e B (E).

Proof. Case 1. 00 , E. Let U=C-K and let V={a e C:(z-a)-I e B(E)}; so Ec V cU.

1.11

If a

E

Vand 1 b - al < dCa, K) then

bE

V.

The condition on b gives the existence of a number r, O
1.12

]-1

Hence Ib-allz -ai-I < r< 1 for all z in K gives that

[1- bz-a

a ]

1.13

-I

=

f (b -

a

n=O z-a

)n

converges uniformly on K by the Weierstrass M-test. If Qn(z)= ~ ~

(b-a)k --

then (z-a)-IQn(z) e B(E) since a e V and z-a B (E) is an algebra. Since B (E) is closed (1.12) and the uniform convergence of (1.13) imply that (Z-b)-' e B(E). That is, b e V. Note that (1.11) implies that V is open. If b e av then let {an} be a sequence in V with b = liman. Since b; V it follows from (1.11) that Ib-anl2: d(an,K). Letting n~oo gives (by 11.5.7) that O=d(b,K), or be K. Thus av (')U=o. If H is a component of U=C-K then H(')E'I=O, so H(') V'I=O. By Lemma 1.9, He V. But H was arbitrary so U c V, or V= U. k=O

Case 2. 00 e E. Let d= the metric on Coo' Choose ao in the unbounded component of C-K such that d(ao,oo)~dd(oo,K) and laol>2max{lzl:zeK}. Let Eo=(E - {oo })u{ao}; so Eo meets each component of Coo - K. If a e CK then Case I gives that (z-a)-I E B(Eo).1f we can show that (Z-aO)-1 eB(E) then it follows that B(Eo)cB(E) and so (z-a)-I eB(E) for each a in C - K. Now Izi aol:S; t for all z in K so

converges uniformly on K. So Qn(z)= -ao-'~k=o(zlaot is a polynomial and (z-ao)-'=u-limQn on K. Since Qn has its only pole at 00, Qn e B(E). Thus (z-ao)-I e B(E). • The Proof of Runge's Theorem. If f is analytic on an open set G and KeG then for each f > 0 Proposition 1.1 and Lemma 1.5 imply the existence of a rational function R(z) with poles in C-K such that If(z)-R(z)l
200

Runge's Theorem

all z in K. But Lemma 1.10 and the fact that B (E) is an algebra gives that R

B(E). III

E

1.14 Corollary. Let G be an open subset of the plane and let E be a subset of C x - G such that E meets every component of C G. Let R (G, E) be the set of rational functions with poles in E and consider R (G, E) as a subspace of H(G). Iff E H(G) then there is a sequence {Rn} in R(G,E) such that f= lim R". That is, R(G,E) is dense in H(G). 00 -

Proof. Let K be a compact subset of G and I: > 0; it must be show that there is an R in R(G, E) such that If(z) - R(z)1 < € for all z in K. According to Proposition vn.1.2 there is a compact set Kl such that K c Kl C G and each component of Coo - Kl contains a component of Coo - G. lienee, E meets each component of Coo - K j • The corollary now follows from Runge's Theorem. II1II The next corollary follows by letting E = {oo} and using the fact that a rational function whose only pole is at 00 is a polynomial. 1.15 Corollary. If G is an open subset of C such that C oc - G is connected then for each analytic function f on G there is a sequence of polynomials { Pn} such that f = lim p" in H(G). Corollary 1.14 can be strengthened a little by requiring only that Emeets each component of Coo - G. The reader is asked to do this (Exercise I). The condition that E - meet every component of lC eo - G cannot be relaxed. This can be seen by considering the punctured plane IC - {O} = G. So 1C 00 - G = {O, oo}. Suppose that for this case we could weaken Runge's Theorem by assuming that E consisted of 00 alone. Then for each integer n ;::: 1 we could find a polynomial Pn(z) such that

I ~ _PnCZ)! <1n

1.16

for

~

n then

Z

:s;

Izi

:s; n. Then Il-zPn(z)! :s;

jPn(z)i

1

= - IZPn(z) I :s; n

Izl n

:s; 1 for

~

n

1 - IZPn(z) -II

n

:s;

Izi 1

+n

:s; n. But if Izl

~

n

2 :s; - •

n

. 2 By the MaXImum Modulus Theorem, /Pn(Z) I :s; - for Izi :s; n. [n particular, n Pn(z) -+ 0 uniformly on Izi :s; 1. This clearly contradicts (1.16) and shows that E must be the set {a, Cf)}. Of course, the point in the above paragraph could have been made by appealing to what was said about this same example at the beginning of this section. However, this further exposition gives an introduction to a concept whose connection with Runge's Theorem is quite intimate. 1.21 Definition. Let K be a compact subset of the plane; the polynomially

Runge's Theorem

201

convex hull of K, denoted by K, is defined to be the set of all points w such

that for every polynomial p /p(w)/ :::; max {/p(z)/: z E K}.

That is, if the right hand side of this inequality is denoted by lip 11K , then

K=

{w:

/p(w)/ :::; IlpllKfor all polynomialsp}.

If K is an annulus then K is the disk obtained by filling in the interior hole. In fact, if K is any compact set the Maximum Modulus Theorem gives that K is obtained by filling in any" holes" that may exist in K.

Exercises

I. Prove Corollary 1.14 if it is only assumed that E - meets each componentofCoo-G. 2. Let G be the open unit disk B(O; I) and let K = {z: ! :::; /z/ :::; i-}. Show that there is a function I analytic on some open subset G! containi'l!: K which cannot be approximated on K by functions in H(G). Remarks. The next two problems are concerned with the following question. Given a compact set K contained in an open set G! c G, can functions in R(G!) be approximated on K by functions in R(G)? Exercise 2 says that for an arbitrary choice of K, G, and G! this is not true. Exercise 4 below gives criteria for a fixed K and G such that this can be done for any G 1 •

Exercise 3 is a lemma which is useful in proving Exercise 4. 3. Let K be a compact subset of the open set G and suppose that any bounded =1= O. Then every componentofCoo-K contains a component of Coo - G. 4. Let K be a compact subset of the open set G; then the following are equivalent: (a) If I is analytic in a neighborhood of K and EO > 0 then there is a g in R(G) with /I(z) - g(z) I < EO for all z in K; (b) If D is a bounded component of G - K then D - () 8G =1= 0; (c) If z is any point in G-K then there is a function/in R(G) with componentDofG-KhasD- () 8G

I/(z)/ > sup {/I(w)l: win K}. 5. Can you interpret part (c) of Exercise 4 in terms of K? 6. Let K be a compact subset of the region G and define KG = {z E G: /I(z)/ :::; II/IIK for all I in R(G)}. (a) Show that if Coo - G is connected then KG = K. (b) Show that d(K, C-G) = d(KG' C-G). (c) Show that RG c the convex hull of K == the intersection of all convex subsets of C which contain K.

202

Runge's Theorem

(d) If KG c G 1 c G and G1 is open then for every g in B(G 1 ) and € > 0 there is a functionfin B(G) such that /f(z)-g(z)/ < € for all z inKG' (Hint: see Exercise 4.) (e) KG = the union of K and all bounded components of G-K whose closure does not intersect aG. §2. Simple connectedness Recall that an open connected set G is simply connected if and only if every closed rectifiable curve in G is homotopic to zero. The purpose of this section is to prove some equivalent formulations of simple connectedness. 2.1 Definition. Let X and n be metric spaces; a homeomorphism between X and U is a continuous map f: X -* n which is one-one, onto, and such thatf-l: n -* X is also continuous. Notice that if f: X -* n is one-one, onto, and continuous then f is a homeomorphism if and only if f is open (or, equivalently, f is closed). If there is a homeomorphism between X and n then the metric spaces X and n are homeomorphic. We claim that C and D = {z: /z/ < I} are homeomorphic. In fact fez) = z(l + /zl) - I maps C onto D in a one-one fashion and its inverse, f-l(w) = w(l-Iwi)-I, is clearly continuous. Also, iffis a one-one analytic function on an open set G and n = f(G) then G and n are homeomorphic. Finally, all annuli are homeomorphic to the punctured plane. 2.2 Theorem. Let G be an open connected subset of C. Then the following are equivalent: (a) G is simply connected; (b) n(y; a) = 0 for every closed rectifiable curve y in G and every point a in C-G; (c) C",-G is connected; (d) For any f in B (G) there is a sequence of polynomials that converges

tofinB(G); (e) For anyf in H(G) and any closed rectifiable curve y in G, Srf = 0; (f) Every function f in B (G) has a primitive; (g) For any fin B(G) such that fez) #- 0 for all z in G there is afunction g in B(G) such that fez) = exp g(z); (h) For any fin B(G) such that fez) #- 0 for all z in G there is a function

ginB(G)suchthatf(z) = [g(Z)]2;

(i) G is homeomorphic to the unit disk; (j) If u: G -* R is harmonic then there is a harmonic function v: G such that f = u+iv is analytic on G.

Proof. The plan is to show that (a) = (b) = ... = (i) = (a) and (h) = (g). Many of these implications have already been done.

-*

=

R

(j)

Simple connectedness 203 (a) => (b) If y is a closed rectifiable curve in G and a is a point in the complement of G then (z - a) - 1 is analytic in G, and part (b) follows by

Cauchy's Theorem. (b) => (c) Suppose Ceo -G is not connected; then Ceo -G = Au B where A and B are disjoint, non-empty, closed subsets of Coo. Since 00 must be either in A or in B, suppose that 00 is in B; thus, A must be a compact subset of C (A is compact in Coo and doesn't contain (0). But then G 1 = G u A = Coo - B is an open set in C and contains A. According to Proposition 1.1 there are a finite number of polygons Yl' ."•. ,Ym in G 1 -A = G such that for every analytic function f on G 1

fez) =

i

~.J·f(W)

k=12m

w-z

dw

7k

for all z in A. In particular, if fez) == 1 then

1=

m

L n(Yk; z) k=l

for all z in A. Thus for any z in A there is at least one polygon Yk in G such that n(Yk; z) "# O. This contradicts (b). (c) => (d) See Corollary I.lS. (d) => (e) Let Y be a closed rectifiable curve in G, let f be an analytic function on G, and let {Pn} be a sequence of polynomials such thatf = lim Pn in B(G). Since each polynomial is analytic in C and Y '" 0 in C, LPn = 0 for every n. But {Pn} converges to f uniformly on {y} so that L f = lim LPn = o. (e) => (f) Fix a in G. From condition (e) it follows that there is a function F: G -+ C defined by letting F(z) = Lfwhere Y is any rectifiable curve in G from a to z. It follows that F' = f (see the proof of Corollary IV. 6.16). (f) => (g) Iff(z) "# 0 for all z in G then/'/fis analytic on G. Part (f) implies there is a function F such that F' = /,If It follows (see the proof of Corollary IV. 6.17) that there is an appropriate constant c such that g = F + c satisfies fez) = exp g(z) for all z in G. (g) => (b) This is trivial. (b) => (i) If G = C then the function z(1 + Izi) -1 was shown to be a homeomorphism immediately prior to this theorem. If G "# C then Lemma VII. 4.3 implies that there is an analytic mapping f of G onto D which is one-one. Such a map is a homeomorphism. (i) => (a) Let h: G -+ D = {z: Izl < I} be a homeomorphism and let Y be a closed curve in G (note that Y is not assumed to be rectifiable). Then a(s) = h(y(s» is a closed curve in D. Thus, there is a continuous function A: [2 -+ D such that A(s, 0) = a(s) for 0 :s; s :s; I, A(s, I) = 0 for 0 :s; s :s; 1 and A(O, t) = A(I, t) for 0 :s; t :s; 1. It follows that r = h- 1 0 A is a continuous map of [2 into G and demonstrates that Y is homotopic to the curve which is constantly equal to h- 1(0). The details are left to the reader. (b) => (j) Suppose that G "# C; then the Riemann Mapping Theorem implies there is an analytic function h on G such that h is one-one and h(G)

204

Runge's Theorem

= D. If u: G ---J. R is harmonic then U 1 = U h - I is a harmonic function on D. By Theorem III. 2.30 there is a harmonic function VI: D ---J. R such that 11 = U 1 + iv, is analytic on D. Let I = 11 h. Then I is analytic on G and u is the real part off Thus v = 1m I = VI 0 h is the sought after harmonic conjugate. Since Theorem m. 2.30 also applies to C, (j) follows from (h). (j) (g) Suppose I: G ---J. C is analytic and never vanishes, and let u = Re J, v = 1m f If V: G ---J. R is defined by Vex, y) = log I/(x+iy)1 = log [u(x, y)2 + vex, y)2]t then a computation shows that V is harmonic. Let V be a harmonic function on G such that g = V + iV is analytic on G and 0

0

=

let h(z) = exp g(z). Then h is analytic, never vanishes, and

I/(Z)I = h(z)

1 for all

z in G. That is,f/h is an analytic function whose range is not open. It follows that there is a constant c such that/(z) = c h(z) = c exp g(z) = exp [g(z) + CI]' Thus, g(z)+c l is a branch of log/(z). This completes the proof of the theorem. III This theorem constitutes an aesthetic peak in Mathematics. Notice that it says that a topological condition (simple connectedness) is equivalent to analytical conditions (e.g., the existence of harmonic conjugates and Cauchy's Theorem) as well as an algebraic condition (the existence of a square root) and other topological conditions. This certainly was not expected when simple connectedness was first defined. Nevertheless, the value of the theorem is somewhat limited to the fact that simple connectedness implies these nine properties. Although it is satisfying to have the converse of these implications, it is only the fact that the connectedness of Coo - G implies that G is simply connected which finds wide application. No one ever verifies one of the other properties in order to prove that G is simply connected. For an example consider the set G = C - {z = re ir : 0 :::; r < ro}; that is, G is the complement of the infinite spiral r = (), 0 :::; () < roo Then CO() - G is the spiral together with the point at infinity. Since this is connected, G is simply connected.

Exercise I. The set G={re il : ----ro
Consider the following problem: Let G be an open subset of C and let {ad be a sequence of distinct points in G such that {a,,} has no limit point in G. For each integer k ~ 1 consider the rational function

3.1

Mittag-Leffler's Theorem

205

where mk is some positive integer and A Ik, ••• , Amkk are arbitrary complex coefficients. Is there a merom orphic function f on G whose poles are exactly the points {ad and such that the singular part of fat Z = ak is Sk(Z)? The answer is yes and this is the content of Mittag-Leffler's Theorem.

3.2 Mittag-Lerner's Theorem. Let G be an open set, {ad a sequence of distinct points in G without a limit point in G, and let {Sk(Z)} be the sequence of rational functions giren by equation (3.1). Then there is a meromorphic function f on G whose pole.s are exactly the points {ak } and such that the singular part off at ak is SkCZ). Proof Although the details of this proof are somewhat cumbersome, the idea is simple. We use Runge's Theorem to find rational functions {RJz)}

with poles in Coo - G such that

Lt

Sk(Z) - Rk(Z)} is a Cauchy sequence in

MeG). The resulting limit is the sought after meromorphic function. (Actually we must do a little more than this.) Use Proposition VII. 1.2 to find compact subsets of G such that 00

G

=

U Km Kn

n=l

c int Kn+I'

and each component of Ceo - Kn contains a component of CoO - G. Since each Kn is compact and {a k } has no limit point in G, there are only a finite number of points ak in each Kn- Define the sets of integers In as follows:

for n :?: 2. Define functions J" by fn(z) =

L

Sk(Z)

kErn

for n :?: 1. Then In is rational and its poles are the points {ak : k E In} c Kn Kn _ }. (If In is empty let In = 0.) Since j~ has no poles in Kn _ I (for n:?: 2) it is analytic in a neighborhood of Kn _ I' According to Runge's Theorem there is a rational function Rn(z) with its poles in C x - G and that satisfies /J,,(z) - R.(z)/ < (!-)"

for all z in K n -

3.3

1•

We claim that fez) = fl(z)

+

L [J,,(z)-R'(z)] n=2 Cf)

is the desired meromorphic function. It must be shown that

f

is a mero-

206

Runge's Theorem

morphic function and that it has the desired properties. Start by showing that the series in (3.3) converges uniformly on every compact subset of G - {ak: k ~ I}. This will give thatfis analytic on G - {ak: k ~ I} and it will only remain to show that each ak is a pole with singular part SkCZ). So let K be a compact subset of G - {ak: k :::': I}; then K is a compact subset of G and, therefore, there is an integer N such that K c:: K N • If n :::': N then If.(z) - R,(z)i < for all z in K. That is, the series (3.3) is dominated on K by a convergent series of numbers; by the Weierstrass M-test (II. 6.2) the series (3.3) converges uniformly on K. Thus f is analytic on G - {a k : k :::': I}. Now consider a fixed integer k :::': 1; there is a number R > 0 such that laj-akl > R for j =I- k. Thusf(z) = Sk(Z)+g(z) for 0 < iz-akl < R, where g is analytic in B(ak ; R). Hence, z = ak is a pole off and Sk(Z) is its singular part. This completes the proof of the theorem. II Just as there is merit in choosing the integers Pn in Weierstrass's Theorem (VII. 5.2) as small as possible, there is merit in choosing the rational functions RnCz) in (3.3) to be as simple as possible. As an example let us calculate the simplest meromorphic function in the plane with a pole at every integer n.

at

The simplest singular part is (z - n) - 1 but

co

I

(z -n) -1 does not converge in

-- co

M(C). However (z_n)-1+(z+II)-1 = 2Z(Z2_ n 2)"1 and 1 Z

+ ~'_. ~z_ L.,

"=1

Z2

_n 2

does converge in M (C). The singular part of this function at Z = n is (z - n) -1. In fact, from Exercise V. 2.8 we have that this function is 7r cot 7TZ. Exercises 1. Let G be a region and let {an} and {b m} be two sequences of distinct points in G without limit points in G such that an =I- bn for all n, m. Let Sn(z) be a singular part at an and let Pm be a positive integer. Show that there is a meromorphic functionfon G whose only poles and zeros are {an} and {b m} respectively, the singular part at z = an in Sn(z), and z = b m is a zero of multiplicity Pm' 2. Let {an} be a sequence of points in the plane such that lani--+- co, and let {b n } be an arbitrary sequence of complex numbers. (a) Show that jf integers {k n } can be chosen such that 3.4 converges absolutely for all r > 0 then

3.5 converges in M(C) to a function/with poles at each point z

=

an'


kn

207

(b) Show that if lim sup Ibnl < 00 then (3.4) converges absolutely if n for all n. (c) Show that if there is an integer k such that the series

=

3.6 converges absolutely, then (3.4) converges absolutely if k n = k for all n. (d) Suppose there is an r > 0 such that Ian - ami ;::: r for all n # m. Show that L lanl- 3 < 00. In particular, if the sequence {b n } is bounded then the series (3.6) with k = 2 converges absolutely. (This is somewhat involved and the reader may prefer to prove part (f) directly since this is the only application.) (e) Show that if the series (3.5) converges in M(I[) to a meromorphic function f then

fez)

=

[b

b{ ( '7)

(

~ ----::- + -" ] + -"'- + ... + ---- )kn L

n= 1

Z

an

an

'7

an

an

-I}]

(f) Let wand w' be two complex numbers such that 1m (w'lw) # O. Using the previous parts of this exercise show that the series

where the sum is over all w = 2nw+2n'w for 1'1, n' = 0, ± I, ±2, ... but not w = 0, is convergent in Mel[) to a meromorphic function, with simple poles at the points 2nw+2n'w'. This function is called the Weierstrass zeta function. (g) Let &,0(Z) = - nz); fp is called the Weierstrass pe function. Show that gJ(z)

=

;12

+ L' ((~~W)2 -

:2)

where the sum is over the same w as in part (f). Also show that fp(z) = gJ(z+2nw+2n'w') for all integers nand 1'1'. That is, gJ is doubly periodic with periods 2w and 2w'. 3. This exercise shows how to deduce Weierstrass's Theorem for the plane (Theorem VII. 5.12) from Mittag-Leffler's Theorem. (a) Deduce from Exercises 2(a) and 2(b) that for any sequence {an} in C with lim an = 00 and an # 0 there is a sequence of integers {k n } such that h(z) =

(z) + ... + ---I (z-- )kn - J

~,[ --1 + -1 + -1 L

n=1

z-a n

an

an an

1

an an

is a meromorphic function on C with simple poles at aI' a 2, .. , . The remainder of the proof consists of showing that there is a functionf such that h = f'lf This function f will then have the appropriate zeros.

Runge's Theorem

208

(b) Let z be an arbitrary but fixed point in I[; - {aI' a z, ••• }. Show that if Y1 and yz are any rectifiable curves in 1[;- {aI, a2,"'} from 0 to z and his the function obtained in part (a), then there is an integer m such that

Jh - Jh 1'1

=

27Tim.

Y2

(c) Again let h be the meromorphic function from part (a). Prove that for z i= aI' a z, .•. and y any rectifiable curve in 1[;- {ai' a z , .• •},

fez) = exp

(f h)

defines an analytic function on I[; - {aI, a2, ... } with j'lf = h. (That is, the value of fez) is independent of the curve y and the resulting function f is analytic. (d) Suppose that z E {ai' a 2 , •• • }; show that z is a iemovable singularity of the function f defined in part (c). Furthermore, show that fez) = 0 and that the multiplicity of this zero equals the number of times that z appears in the sequence {a l , a z , ... }. (e) Show that

3.7

fez) =

n 00

n= I

(

1 - -z ) exp [ -z an an

+ 1- (Z)2 . . . . + ... + -1 (z- )knJ 2 an

k n an

Remark. We could have skipped parts (b), (c), and (d) and gone directly from (a) to (e). However this would have meant that we must show that (3.7) converges in H(C) and it could hardly be classified as a new proof. The steps outlined in parts (a) through (d) give a proof of Weierstrass's Theorem without introducing infinite products. 4. This exercise assumes a knowledge of the terminology and results of Exercise VII. 5.11. (a) Define two functionsfandg in H(G) to be relatively prime (in symbols, (I, g) = 1) if the only common divisors off and g are non-vanishing functions in H(G). Show that (I, g) = 1 iff !!t'(f) n !!t'(g) = O. (b) If (I, g) = 1, show that there are functionsfl' gl in H(G) such that ill + gg 1 = 1. (Hint: Show that there is a meromorphic function 'P on G such that II = 'Pg E H(G) and gi(i-fJ1)') (c) Let fl' ... ,f" E H(G) and g = g.c.d {II,' .. ,fn}. Show that there are functions 'Pl"'" 'Pn in H(G) such that g = 'Plf1 + .. '+'Pn/". (Hint: Use (b) and induction.) (d) If {.J'~J is a collection of ideals in H(G), show that .J' = ,i'a is

n a

also an ideal. If Y' c H(G) then let .J' = n {)"": )"" is an ideal of H(G) and Y' c ),,"}. Prove that .J' is the smallest ideal in H(G) that contains Y' and .J' = {'Pdl + ... +'P,,/": 'Pk E H(G),jk E Y' for 1 ~ k ~ n} . .J' is called the ideal generated by Y' and is denoted by .J' = (Y'). If Y' is finite then (Y') is called a finitely generated ideal. If .7 = {f} for a single function f then (f) is called a principal ideal.


209

(e) Show that every finitely generated ideal in H(G) is a principal ideal. (f) An ideal ..1' is called a fixed ideal if :2"(..1') D; otherwise it is caned a free ideal. Prove that if ..1'= (Y') then :2"(..1') = :2"(,sP) and that a proper principal ideal is fixed. (g) LetJiz) = sin (2-·z) for all n ~ 0 and let ..1' = ({ft,J2' ... D. Show that J is a fixed ideal in H{C) which is not a principal ideal. (h) Let J be a fixed ideal and prove that there is an J in H(G) with :!l'(f} = :!l'(..1') and J c (f). Also show that ..1' = (f) if ..1' is finitely generated. (i) Let Jt be a maximal ideal that is fixed. Show that there is a point a in G such that Jt = «z - a». (j) Let {an} be a sequence of distinct points in G with no limit point in G. Let J = {f E H(G): J(an ) = 0 for all but a finite number of the an}. Show that J is a proper free ideal in H(G). (k) ILf is a proper free ideal show that for any finite subset Y' off, :2"(Y') :F O. Use this to show that J can contain no polynomials. (1) Let ..1' be a proper free ideal; then f is a maximal ideal iff whenever g E H(G) and :2"(g) n :2"(/) D for all f in f then g E f. 5. Let G be a region and let {a.} be a sequence of distinct points in G with no limit point in G. For each integer n ~ 1 choose integers k n ~ 0 and constants A~k), 0 ::; k ::; k n • Show that there is an analytic function J on G such thatJ(k)(an) = k!A~k). (Hint: Let g be an analytic function on G with a zero at an of multiplicity k n + 1. Let h be a meromorphic function on G with poles at each an of order k n + I and with singular part Sn(z). Choose the Sn so thatJ = gh has the desired property.) 6. Find a meromorphic function f with poles of order 2 at 1, 12, 13, ... such that the residue at each pole is 0 and lim (z-.Jn)2J(z) = 1 for all n.

'*

'*

z-v'n

Chapter IX Analytic Continuation and Riemann Surfaces Consider the following problem. Letfbe an analytic function on a region G; when can f be extended to an analytic function fl on an open set G I which properly contains G? If G 1 is obtained by adjoining to G a disjoint open set so that G becomes a component of G 1 , f can be extended to G I by defining it in any way we wish on G 1 - G so long as the result is analytic. So to eliminate such trivial cases it is req uired that G 1 also be a region. Actually, this process has already been encountered. Recall that in the discussion of the Riemann zeta function (Section VII. 8) ~(z) was initially defined for Re Z > 1. Using various identities, principal among which was Riemann's functional equation, ~ was extended so that it was defined and analytic in IC - {I} with a simple pole at z = 1. That is, ~ was analytically continued from a smaller region to a larger one. Another example was in the discussion that followed the proof of the Argument Principle (V.3.4). There a meromorphic function f and a closed rectifiable curve y not passing through any zero or pole of f was given. If z = a is the initial poi!'t of y (and the final point), we put a disk Dl about a on which it was possible to define a branch tl of log f Continuing, we covered y by a finite number of disks D 1 , D z, . .. , Dn, where consecutive disks intersect and such that there is a branch tj of logf on D j ' Furthermore, the functions tj were chosen so that t/z) = tj-1(z) for z in D j - 1 n D j , 2 ::; j ::; n. The process analytically continues t1 to D 1 U D 2, then DIU D z U D 3 , and so on. However, an unfortunate thing (for this continuation) happened when the last disk Dn was reached. According to the Argument Principle it is distinctly possible that tn(z) i= t 1(z) for z in Dn n D j • In fact, t"(z)-t1(z) = 27TiK for some (possibly zero) integer K. This last example is a particularly fruitful one. This process of continuing a function along a path will be examined and a criterion will be derived which ensures that continuation around a closed curve results in the same function that begins the continuation. Also, the fact that continuation around a closed path can lead to a different function than the one started with, will introduce us to the concelJt of a Riemann Surface. This chapter begins with the Schwarz Reflection Principle which is more like the process used to continue the zeta function than the process of continuing along an arc. §1. Schwarz Reflection Principle If G is a region and G* = {z: Z E G} and if fis an analytic function on G, then f*: G* --+ IC defined by f*(z) = fez) is also analytic. Now suppose that 210

Schwarz Reflection Principle

211

G = G*; that is, G is symmetric with respect to the real axis. Then g(z) = fez) - fez) is analytic on G. Since G is connected it must be that G contains an open interval of the real line. Suppose f(x) is real for all x in G () IR; then g(x) == 0 for x in G () IR. But G () IR has a limit point in G so thatf(z) = fez) for all z in G. The fact that f must satisfy this equation is used to extend a function defined on G () {z: 1m z ~ O} to all of G. If G is a symmetric region (i.e., G = G*) then let G + = {z E G: 1m z > O}, G_ = {z E G: 1m z < O}, and Go = {z E G: 1m z = O}. 1.1 Schwarz Reflection Principle. Let G be a region such that G = G*. If f: G + u Go -+ C is a continuous function which is analytic on G + and if f(x) is real for x in Go, then there is an analytic function g: G -+ C such that g(z) = fez) for z in G + U Go. Proof For z in G _ define g(z) = fez) and for z in G + u Go let g(z) = fez). It is easy to see that g: G -+ C is continuous; it must be shown that g is analytic. It is trivial that g is analytic on G + U G _ so fix a point x 0 in Go and let R > 0 with B(xo; R) c G. It suffices to show that g is analytic on B(xo; R); to do this apply Morera's Theorem. Let T = [a, b, c, a] be a triangle in B(xo; R). To show that Jrg = 0 it is sufficient to show that Jpg = 0 a

-------?'------------~----G()

b~--------~G~--------whenever P is a triangle or a quadrilateral lying entirely in G + U Go or G _ u Go. In fact, this is easily seen by considering various pictures such as the one above. Therefore assume that T c G + u Go and [a, b) c Go. The proof of the other cases is similar and will be left to the reader. (See Exercise 1 for a general proposition which proves all these cases at once.) Let Ll designate T together with its inside; then g(z) = fez) for all z in Ll. By hypothesisfis continuous on G+ u Go and sofis uniformly continuous on Ll. So if E > 0 there is a 8 > 0 such that when z and z' Ell and \z-z'\ < 8 then If(z)-f(z')\ < E. Now choose 0: and (:3 on the line segments [c, a] and [b, c) respectively, so that \o:-a\ < 8 and 1(:3-bl < 8. Let T J = [0:, (:3, c, 0:] and Q = [a, b, (:3, 0:, a]. Then hf = JTJ+ JQf, but T J and its inside are contained in G + and f is analytic there; hence 1.2

Jf= Jf

T

Q

Analytic Continuation and Riemann Surfaces

212

0.

r------

------>.. B

- - - - a - L . . . . - - - - - - - - - - - - - ' - h- - 0 "

But if 0 :,;

t :,;

1 then i[t;3+(I -t)01]-[tb+(1-t)a]1 < S

so that !/(t/l+(l-t)01)-I(tb+(l-t)a)! <

If M

=

max {i/(z)l: z Ell} and

t=

E.

the perimeter of T then

J1/ = ICb - a) II(tb + (1 ... t)a)dt _. (;3····01) I j{t;3 + (1- t)01)dtl I

1.3

I

I 1+

[a,b]

1

0

[~,C(J

0

Ij, [f(tb+(l-t)a)-/(I(3+(l-t)01)dl I 1

:,; Ib-a!

l

o

1

+ I(b -

a) -

«(3 - a)IIJ(I(3 + (1- t)a)dtl o

:,; €lb-a!+M!Cb-.B)+(a-a)! :,; d+2M8 Also

I JII :,; Mia-a! :,; MS [Ct,

aJ

and

I IIlJ II:,; MS. [b,

Combining these last two inequalities with (1.2) and (l.3) gives that

IIII :,; d+4MS. T

Since it is possible to choose 3 < € and since E is arbitrary, it follows that h/= 0; thuslmust be analytic . • Can the reflection principle be generalized? For example, instead of requiring that G be a region which is symmetric with respect to the real axis, suppose that G is symmetric with respect to a circle. (Definition III. 3.17).

Analytic Continuation Along a Path

213

Can the reflection principle be formulated and proved in this setting? The answer is provided in the exercises below. Exercises 1. Let y be a simple closed rectifiable curve with the property that there is a point a such that for all z on y the line segment [a, z] intersects {y} only at z; i.e. [a, z] n {y} = {z}. Define a point w to be inside y if [a, w] n {y} = 0 and let G be the collection of all points that are inside y. (a) Show that G is a region and G- = G u {y}. (b) Letf: G- --J>- IC be a continuous function such thatfis analytic on G. Show that J = O. (c) Show that n(y; z) = ± 1 if z is inside y and n(y; z) = 0 if z
J

Remarks. It is not necessary to assume that y has such a point a as above; each part of this exercise remains true if y is only assumed to be a simple closed rectifiable curve. Of course, we must define what is meant by the inside of y. This is difficult to obtain. The fact that a simple closed curve divides the plane into two pieces (an inside and an outside) is the content of the Jordan Curve Theorem. This is a very deep result of topology. 2. Let G be a region in the plane that does not contain zero and let G* be the set of all points z such that there is a point w in G where z and ware symmetric with respect to the circle I~I = 1. (See III. 3.17.) (a) Show that G* = {z: (I/z) E G}. (b) Iff: G --J>- IC is analytic, define 1*: G* --J>- IC by f*(z) = fO/2). Show that f* is analytic. (c) Suppose that G = G* and f is an analytic function defined on G such that fez) is real for z in G with Izl = 1. Show that f = f*. (d) Formulate and prove a version of the Schwarz Reflection Principle where the circle 1fl = 1 replaces ~. Do the same thing for an arbitrary circle. 3. Let G, G +, G _, Go be as in the statement of the Schwarz Reflection Principle and letf: G+ u Go --J>-lC ao be a continuous function such thatfis rneromorphic on G +. Also suppose that for x ill Go f(x) E ~. Show that there is a merom orphic function g : G -+ IC 00 such that g(z) = f(z) for z in G + u Go. Is it possible to allow fto assume the value 00 on Go? §2. Analytic Continuation Along a Path Let us begin this section by recalling the definition of a function. We use the somewhat imprecise statement that a function is a triple (I, G, D) where G and n are sets andfis a "rule" which assigns to each element of G a unique element of D. Thus, for two functions to be the same not only must the rule be the same but the domains and the ranges must coincide. If we enlarge the range Q to a set D1 then (I, G, .nJ is a different function. However, this point should not be emphasized here; we do wish to emphasize that a change in the domain results in a new function. Indeed, the purpose of analytic continuation is to enlarge the domain. Thus, let G = {z: Re z > -1} and


214

fez) = log (l +z) for z in G, where log is the principal branch of the logarithm. Let D = B(O; 1) and let g(z)

=

.i (-__

1).-1

~

n for z in D. Then (J, G, C) =F (g, D, q even thoughf(z) = g(z) for all z in D. Nevertheless, it is desirable to recognize the relationship between f and g. This leads, therefore, to the concept of a germ of analytic functions. "=1

2.1 Definition. A function element is a pair U; G) where G is a region and fis an analytic function on G. For a given function element (J, G) define the germ of f at a to be the collection of all function elements (g, D) such that a E D and fez) = g(z) for all z in a neighborhood of a. Denote the germ by [fla. Notice that [fla is a collection of function elements and it is not a function element itself. Also (g, D) E [fla if and only if (J, G) E [gJa. (Verify!). It should also be emphasized that it makes no sense to talk of the equality of two germs [flo and [gh unless the points a and b are the same. For example, if (J, G) is a function element then it makes no sense to say that [fla = [f]b for two distinct points a and b in G. 2.2 Definition. Let y: [0, lJ --+ C be a path and suppose that for each t in [0, 1] there is a function element U;, D t ) such that: (a) yet) EDt; (b) for each t in [0, 1] there is a 0 > 0 such that yes) E Dr and

Is - tl < 0

implies

2.3 Then (fl' Dd is the analytic continuation of (fo. Do) along the path y; or, (fl' D j ) is obtained from (fa, Do) by analytic continuation along y. Observe that the sets Dr in the definition can be enlarged or diminished without affecting the fact that there is a continuation. This will be useful in the proofs below. Before proceeding, examine part (b) of this definition. Since y is a con tinuous function and y( t) is in the open set DI' it follows that there is a 8 > 0 such that y(s) c D f for Is - [I < 8. The important content of part (b) is that (2.3) is satisfied whenever Is - tl < (i. That is,h(z) = j,(z), whenever Is - II < (i and z belongs to the component of Ds n Dr that contains yes). Whether for a given curve and a given function element there is an analytic continuation along the curve can be a difficult question. Since no degree of generality can be achieved which justifies the effort, no existence theorems for analytic continuations will be proved. Each individual case will be considered by itself. Instead uniqueness theorems for continuations are proved. One such theorem is the Monodromy Theorem of the next section. This theorem gives a criterion by which one can tell when a continuation along two different curves connecting the same points results in the same function element.

Analytic Continuation Along a Path

215

The next proposition fixes a curve and shows that two different continuations along this curve of the same function element result in the same function element. Alternately, this result can be considered as an affirmative answer to the following question: Is it possible to define the concept of "the continuation of a germ along a curve?" 2.4 Proposition. Let y: [0, 1]-+ IC be a path from a to b and let {eft, D ,): o ~ t ~ I} and {(gr' B,): ~ t ~ I} be analytic continuations along y such that (fO]a = (g O]a' Then [fdb = [g th·

°

Proof This proposition will be proved by showing that the set T = {t E [0,1]: (f,L(,) = [g']Y(I)} is both open and closed in [0, 1]; since T is non-empty (0 E T) it will foHow that T = [0, 1] so that, in particular, 1 E T.

The easiest part of the proof is to show that T is open. So fix t in T and assume t i= 0 or 1. (If t = 1 the proof is complete; if t = then the argument about to be given will also show that [a,a + 0) c T for some I) > 0.) By the definition of analytic continuation there is a I) > such that for Is-tl < 0, yes) E Dr n B, and

°

°

2.5

{ffs]y(s) :

[1.]y(S)· [gs]y(S) - [g,]y(S)' Let II be a connected subset of Dr n B, which contains yes) and yet). But since t E T,J,(z) = g,(z) for all z in II. Hence [J,]r(s) = [g,]y(s) for aU yes) in D, n B,. So it follows from (2.5) that [fsL(s) = [gsL(s) whenever Is - tl < b. That is, (t - b, t + b) c Tand so Tis open. To show that T is closed let t be a limit point of T, and again choose I) > 0 so that yes) E D/ n B, and (2.5) is satisfied whenever Is- tl < o. Since t is a limit point of T there is a point s in T with Is - tl < b. Let G be a region such that y«t - b, t - b)) £; G £; D/ n B,; in particular, yes) E G. Thus, !s(z) = gs(z) for all z in G by the definition of T. But, according to (2.5), j~(z) = /,(z) and gs(z) = g,(z) for all z in G. So /,(z) = g,(z) for all z in G and, because G has a limit point in D, n B" this gives that [I.]y(l) = [gIL(t). That is, t E T and so T is closed. II 2.6 Definition. If y: [0, 1) --+ IC is a path from a to band {(J" D,): 0 ~ t ~ I} is an analytic continuation along y then the germ [fdb is the analytic continuation of [fola along y.

The preceding proposition implies that Definition 2.6 is unambiguous. As stated this definition seems to depend on the choice of the continuation {(/" Dr)}' However, Proposition 2.4 says that if {(g" B,)} is another continuation along y with [foJa = [golD then (flh = [glh. So in fact the definition does not depend on the choice of continuation. 2.7 Definition. If (f, G) is a function element then the complete analytic function obtained from (f, G) is the collection ff of all germs [gJo for which there is a point a in G and a path y from a to b such that (gh is the analytic continuation of [f]a along y.

216


A collection of germs :7 is called a complete analytic function if there is a function element (I, G) such that ff is the complete analytic function obtained from (I, G). Notice that the point a in the definition is immaterial; any point in G can be chosen since G is an open connected subset of IC (see II. 2.3). Also, if :7 is the complete analytic function associated with (I, G) then [flz E ff for all z in G. Although there is no ambiguity in the definition of a complete analytic function there is an incompleteness about it. Is it a function? We should refrain from calling an object a function unless it is indeed a function. To make ff a function one must manufacture a domain (the range will be q and show that:7 gives a "rule". This is easy. In a sense we let .fF be its own domain; more precisely, let

Sf

=

{(z, [flz): [f]z

E

:7}.

Define .fF: Sf -J> IC by ff(z, [fJz) = fez). In this way ff becomes an "honest" function. Nevertheless there is still a lingering dissatisfaction. To have a satisfying solution a structure will be imposed on j f which will make it possible [0 discuss the concept of analyticity for functions defined on 212. In this setting, the function ,'IF defined above becomes analytic; moreover, it reflects the behavior of each function element belonging to a germ that is in ff. The introduction of this structure is postponed until Section 5. Exercises 1. The collection {Do, D 1 , ••• ,Dn} of open disks is called a chain of disks if D j - 1 (\ D j ¥ for I ::; j ::; n. If {(ij' D j ): 0 ::; j ::; n} is a collection of function elements such that {Do, D l , ... , Dn} is a chain of disks andfj ___ I(z) = fiz) for z in D j - I n D j ' I ::; j ::; n; then {(fj, D): 0 ::; j ::; n} is called an analytic continuation along a chain of disks. We say that (I., Dn) is obtained by an analytic continuation of (fa, Do) along a chain of disks. (a) Let {(fj, D): 0 ::; j ::; n} be an analytic continuation along a chain of disks and let a and b be the centers of the disks Do and Dn respectively. Show that there is a path y from a to b and an analytic continuation {(go B t )}

=:

n

along y such that {y}

cUD j'

[fola = [g o]a> and

Lfnh =

[g Ih·

j~O

(b) Conversely, let {(/" D t ): 0::; t::; I} be an analytic continuation along a path y: [0, 11 -J> IC and let a = yeO), b = y(l). Show that there is an analytic continuation along a chain of disks {(g j ' B j): 0 ::; j ::; n} such that n

{y}

C

U Bj ,

[JoL

=

[gaL, and

U~h

=

[gnh·

j~O

2. Let Do

B(l; 1) and let fo be the restriction of the principal branch of Let yet) = exp (2-rrit) and aCt) = exp (41Tit) for 0 ::; t ::; 1. (a) Find an analytic continuation {(/" Dr): 0 ::; t ::; I} of (j~, Do) along y and show that [Jdl = [-fol1.

.Jz to Do.

=

Monodromy Theorem

217

(b) Find an analytic continuation {(g" B,): 0 :::; t :::; I} of (fo, Do) along u and show that [gl]l

= [gO]l'

3. Letfbe an entire function, Do = B(O; 1), and let y be a path from 0 to h. Show that if {(/" D,): 0 :::; t :::; I} is a continuation of (f, Do) along y then fl{Z) = fez) for all z in D l . (This exercise is rather easy; it is actually an exercise in the use of the terminology.) 4. Let y: [0, 1] ~ C be a path and let {(/" D,): 0 :::; t :::; I} be an analytic continuation along y. Show that {(f:, D,): 0 :::; t :::; I} is also a continuation along y. 5. Suppose y: [0, 1] ~ C is a closed path with yeO) = y(I) = a and let {(/" D,): 0 :::; t :::; I} be an analytic continuation along y such that [fl]a = [f~]a andfo ~ O. What can be said about (fo, Do)? 6. Let Do = B(I; 1) and let fo be the restriction to Do of the principal branch of the logarithm. For an integer n let yet) = exp (27Tint), 0 :::; t :::; 1. Find a continuation {(f" D,): 0 :::; t :::; I} along y of (fo, Do) and show that [fdl = [fo+27Tin]1' 7. Let y: [0, 1] ~ C be a path and let {(/" D,): 0 :::; t :::; I} be an analytic continuation along y. Suppose G is a region such that /,(D,) c G for all t, and suppose there is an analytic function h: G ~ C such that h{fo(z» = z for all z in Do. Show that h(f,(z» = z for all z in D, and for all t. Hint: Show that T = {t: h(/,(z» = z for all z in D,} is both open and closed in [0, 1]. 8. Let y: [0, 1] ~ C be a path with yeO) = 1 and yet) ~ 0 for any t. Suppose that {{/" D,): 0 :::; t :::; I} is an analytic continuation of fo(z) = log z. Show that each/, is a branch of the logarithm. §3. Monodromy Theorem Let a and b be two complex numbers and suppose y and a are two paths from a to b. Suppose {{It,D,)} and {{gI'B,)} are analytic continuations along y and a respectively, and also suppose that [fO]a = [gola' Does it follow that [fdb = [g,h? If y and a are the same path then Proposition 2.4 gives an affirmative answer. However, if y and a are distinct then the answer can be no. In fact, Exercises 2.2 and 2.6 furnish examples that illustrate the possibility that [fdb [g .lb' Since both of these examples involve curves that wind around the origin, the reader might believe that a sufficient condition for [fdb and [gdb to be equal can be couched in the language of homotopy. However, since all curves in the plane are homotopic the result would have to be phrased in terms of homotopy in a proper subregion of C. For the examples in Exercises 2.2 and 2.6, this sought after criterion must involve the homotopy of the curves in the punctured plane. This is indeed the case. The origin is discarded in the above examples because there is no germ [h]o centered at zero that belongs to the complete analytic function obtained from (fo, Do). If (f, D) is a function element and a E D then f has a power series expan-

*

218


sion at z = a. The first step in proving the Monodromy Theorem is to investigate the behavior of the radius of convergence for an analytic continuation along a curve.

3.1 Lemma. Let y: [0, 1] --.. IC be a path and let {(I" D/): 0 :s; t :s; I} be an analytic continuation along y. For 0 :s; t :s; 1 let R(t) be the radius of convergence of the power series expansion of1, about z = yet). Then either R(t) == 00 or R: [0, 1] --.. (0, (0) is continuous. Proof If R(t) = 00 for some value of t then it is possible to extendJ, to an entire function. It follows that fs(z) = J,(z) for all z in D. so that R(s) = 00 for each s in [0, 11; that is R(s) == 00. So suppose that R(t) < 00 for all t. Fix t in [0, 11 and let T = y(t); let

J,(z) =

co

L TnCZ-T)n n= 0

be the power series expansion of 1, about T. Now let Sl > 0 be such that < 8 1 implies that yes) E D t n B(T; R(t» and [fsly(s) = [J,ly(s)' Fix s with Is-tl < 8 1 and let a = yes). Now 1, can be extended to an analytic function on B(T; R(t». Sincefs agrees withJ, on a neighborhood of a,!. can be extended so that it is also analytic on B(T; R(t» U Ds. If I. has power series expansion

Is-tl

fs(z) =

00

L

n= 0

aiz-a)"

about z = a, then the radius of convergence R(s) must be at least as big as the distance from a to the circle Iz - TI = R(t): that is, R(s) ~ dCa, {z: Iz - 71 = Ret)}) = R(t) - IT - al· But this gives that R(t) - R(s) :s; Iy(t) - y(s)l. A similar argument gives that R(s) - R(t) :s; Iy(t) - y(s)l; hence IR(s)-R(t)1

$

Jy(t)-y(s)J

for Js- tj < 8 1 , Since y: [0, 1] --.. IC is continuous it follows that R must be continuous at t . • 3.2 Lemma. Let y: [0, 1]->- C be a path from a to b and let {(I" :s; I} be an analytic continuation along y. There is a number € > if a: [0, 1] --.. IC is any path from a to b with Iy(t) - a(t)1 < € for {(gt, B t): 0 $ t $ I} is any continuation along a with [g ola = [glh = [fdb'

Dr): 0 :s; t 0 such that all t, and if [fola; then

Proof For 0 $ t :s; 1 let R(t) be the radius of convergence of the power series expansion of ft about z = yet). It is left to the reader to show that if R(t) == 00 then any value of to will suffice. So suppose R(t) < co for all t. Since, by the preceding lemma, R is a continuous function and since R(t) > 0 for all t, R has a positive minimum value. Let

3.3

o<

€

<

t

min {R(t): 0 :s; t :s; I}

and suppose that a and {(gr' B,)} are as in the statement of this lemma.

Monodromy Theorem

219

Furthermore, suppose that Dr is a disk of radius R(t) about y(t). The truth of the conclusion will not be affected by this assumption (Why?), and the exposition will be greatly simplified by it. For the same reason it will also be assumed that each Br is a disk. Since la(t)-y(t)1 < E < R(t), a(t) E B t n D t for all t. Thus, it makes sense to ask whether gtCz) = J,(z) for all z in B t n D t. Indeed, to complete the proof we must show that this is precisely the case for t = 1. Define the set T = {t E [0, 1]: J,(z) = gt(z) for z in B t n D t }; and show that 1 E T. This is done by showing that T is a non-empty open and closed subset of [0, 1].

From the hypothesis of the lemma, 0 E T so that T oF O. To show Tis open fix I in T and choose S > 0 such that

3.4

Ins) - y(t)1 < { la(s) - a(t)1 <

E,

[J.]y(S) = [J,]y(s),

E,

[g.]u(S) = [gt]u(S)' and

a(s) E B t

whenever Is-tl < S. We will now show that Bs n B t n Ds n D t oF D for Is-tl < S; in fact, we will show that a(s) is in this intersection. If Is-tl < S then la(s)-y(s)1 < E < R(s) so that a(s) E D •. Also la(s) - y(t)1 :::;; la(s) - y(s) I+ Iy(s) - y(t) I < 2E < R(t)

by (3.3); so a(s) EDt. Since we already have that a(s) E Bs n B t by (3.4), a(s) E B. n B t n Ds n D t = G. Since t E T it follows thatJ,(z) = gt(z) for all z in G. Also, from (3.4)J.(z) = J,(z) and gs(z) = gt(z) for all z in G. Thus J.(z) = g.(z) for z in G; but since G has a limit point in Bs n Ds it must be that sET. That is, (t-S, t+S) c T and T is open. The proof that T is closed is similar and will be left to the reader. • 3.S Definition. Let (J, D) be a function element and let G be a region which contains D; then (f, D) admits unrestricted analytic continuation in G if for any path y in G with initial point in D there is an analytic continuation of (f, D) along y., If D = {z: Iz -11 < I} and f is the principal branch of .jz or log z then (f, D) admits unrestricted continuation in the punctured plane but not in the whole plane (see Exercise 2.7). It has been stated before that an existence theorem for analytic continuations will not be proved. In particular, if (f, D) is a function element and G is a region containing D, no criterion will be given which implies that (f, D) admits unrestricted continuation in G. The Monodromy Theorem assumes that G has this property and states a uniqueness criterion. 3.6 Monodromy Theorem. Let (f, D) be a function element and let G be a region containing D such that (f, D) admits unrestricted continuation in G.

220


Let a E D, bEG and let Yo and Y1 be paths in G from a to b; let {(fl' D, ): t :$ I} and {( g" B, ): 0 :$ t :$ I} be analytic continuations of (f, D) along Yo and Y1 respectively. If Yo and Y1 are FEP homotopic in G then

o :$

Proof Since Yo and YI are fixed-end-point homotopic in G there is a continuous function r: [0, 1J x [0, I] -+ G such that

r(t, 1) = YI(l)

ret,O) = YoU)

1'(1, u) = b

reO, u) = a

for all t and u in [0, 1]. Fix u, 0 ::; u ::; I and consider the path Yu, defined by yit) = r(t, u), from a to b. By hypothesis there is an analytic continuation

{(hr,u, D, ..): 0 ::; t ::; I} of (f, D) along YU' It follows from Proposition 2.4 that [g Jb = [h 1, Ih and lfdb = [hI, 01&· So it suffices to show that

[hl,olb

=

[hj,lh·

To do this introduce the set

u=

°

{u

E

[0,1]: [h1,uh

=

[hi, olb}'

and show that U is a non-empty open and closed subset of [0, IJ. Since E U, U O. To show that U is both open and closed we will establish the following.

*

°

3.7 Claim. For u in [0, 1J there is as> such that if [u -1'[ < S then [11 1. uh = [hl,vh. For a fixed u in [0, IJ apply Lemma 3.2 to find an E > 0 such that if a is any path from a to b with [Yu(t) -aCt)[ < € for all t, and if {(k " E,)} is any continuation of (f, D) along a, then

3.8 Now r is a uniformly continuous function, so there is as> 0 such that if [u - v[ < S then [YuC!} - yJt)[ = [r(t, u) - {'(t, v)[ < E for all t. Claim 3.7 now follows by applying (3.8). Suppose u E U and let 8 > 0 be the number given by Claim 3.7. By the definition of U, (u - 8, u + 8) c U; so U is open. If Ii E U- and 5 is again chosen as in (3.7) then there is a u in U such that lu - vi < 5. But by (3.7) [hl.uJb = [hl.vh; and since u E U, [hl.vh = [hl,ok Therefore [h1,ulb = [hl.OJb so that u E U, that is, U is closed. II The following corollary is the most important consequence of the Monodromy Theorem.

3.9 Corollary. Let (f, D) be a function element which admits unrestricted continuation in the simply connected region G. Then there is an analytic function F: G -+ IC such that F(z) = fez) for all z in D. Proof Fix a in D and let z be any point in G. If y is a path in G from a to z

Topological Spaces and Neighborhood Systems

221

and HI" Dr): 0 ~ t ~ I} is an analytic continuation of (f, D) along y then let F(z, y) = II (z). Since G is simply connected F(z, y) = F(:, 0) for any two paths y and a in G from a to z. Thus, Fez) = F(z, y) gives a well defined function F: G --+ C. To show that F is analytic let z F G and let y and {(f;, Dr)} be as above. A simple argument gives that F(w) = Ii (II') for w in a neighborhood of z (Verify!); so F must be analytic. II Exercises

1. Prove that the set T defined in the proof of Lemma 3.2 is closed. 2. Let (f, D) be a function element and let a E D. If y: [0, 1] --',> IC is a path with yeO) = a and y( 1) = b and {(I" Dr): 0 ~ t ~ I} is an analytic continuation of (f; D) along y, let R(t) be the radius of convergence of the power series expansion of fr at z = yet). (a) Show that R(t) is independent of the choice of continuation. That is, if a second continuation {(g" B t )} along y is given with [g oL = [lL and ret) is the radius of convergence of the power series expansion of g, about z = yet) then ret) = R(t) for all t. (b) Suppose that D = B(l; 1), I is the restriction of the principal branch of the logarithm to D, and y(t) = 1 +a t for 0 ~ t ~ 1 and a > O. Find R(t).

(c) Let (f, D) be as in part (b), let 0 < a < 1 and let yet) = (1·-- a I) exp (27Tit) for 0 ~ t ~ 1. Find R(t). (d) For each of the functions R(t) obtained in parts (b) and (c), find min {R(t): ~ t ~ I} as a function of a and examine the behavior of this function as a --+ CIJ or a --+ 0. 3. Let 1': [0, 1] x [0, 1] ....>- G be a continuous function such that 1'(0, 1/) = a, 1'(1, u) = b for all u. Let yu(t) = ret, u) and suppose that {Ut,u, D t ..,): ~ t ~ I} is an analytic continuation along y" such that [!O,,,]a = [Io,,,1o for all u and v in [0, 1]. Let R(t, u) be the radius of convergence of the power series expansion of.fr." about z = u). Show that either R(t, u) == CIJ or R: [0, 1] x [0, 1]+ (0, co) is a continuous function. 4. Use Exercise 3 to give a second proof of the Monodromy Theorem.

°

°

ret,

§4. Topological Spaces and Neighborhood Systems The notion of a topological space arises by abstracting one of the most important concepts in the theory of metric spaces-that of an open set. Recall that in Chapter II we were given a metric or distance function on a set X and this metric was used to define what is meant by an open set. In a topological space we are given a collection of subsets of a set X which are called open sets, but there is no metric available. After axiomatizing the properties of open sets, it will be our purpose to recreate as much of the theory of metric spaces as is possible.

4.1 Definition. A topological space is a pair (X, 3') where X is a set and .'1 is a collection of subsets of X having the following properties:

222


(a) U E:Yand XE:Y; (b) if U j ,

• •• ,

Un are in:Y then j

(c) if {U i : i

E

n U E:Y;

n

j

~J

J} is any collection of sets in /Y then

U U i is in :Y.

iEl

The collection of sets :Y is called a topology on X, and each member of :Y is called an open set. Notice that properties (a), (b), and (c) of this defimtion are the properties of open subsets of a metric space that were proved in Proposition II. 1.9. So if (X. d) is a metric space and c'T is the collection of all open subsets of X then (X"Y) is a topological space. When it is said that a topological space is an abstraction of a metric space, the reader should not get the impression that he is merely playing a game by discarding the metric. That is, no one should believe that there is a distance function in the background, but the reader is now required to prove theorems without resorting to it. This is quite false. There are topological spaces (X, :T) such that for no metric d on X is ,Y the collection of open sets obtained via d. We will see such an example shortly, but it is first necessary to further explore this concept of a topology. The statement "Let X be a topological space" is, of course, meaningless; a topological space consists of a topology :Y as well as a set X. However, this phase will be used when there is no possibility of confusion. 4.2 Definition. A subset F of a topological space X is closed if X - F is open. A point a in X is a limit point of a set A if for every open set U that contains a there is a point x in An U such that x cf- a. Many of the proofs of propositions in this section follow along the same lines as corresponding propositions in Chapter n. When this is the case the proof will be left to the reader. Such is the case with the following two propositions. 4.3 Proposition. Let eX, .r) be a topological space. Then: (a) [J and X are closed sets; (b) if F J , ••• , Fn are closed sets then FI U ... U Fn is closed; Fi is a closed set. (c) ({ {Fi: i E J} is a collection of closed sets then

n

f

E

I

4.4 Proposition. A subset of a topological space is closed iff it contains all its limit points. Now for an example of a topological space that is not a metric space. Let X = [0, I] = { [ : O:s; t:S; I} and let:Y consist of all sets U such that: (i) if 0 E U then X - U is either empty or a sequence of points in X; (ii) if 0 ¢ U then U can be any set. It is left to the reader to prove that eX, :Y) is a topological space. Some of the examples of open sets in this topology are: the set of all irrational numbers in X; the set of all irrational numbers together with zero. To see


223

that no metric can give the collection of open sets :Y, suppose that there is such a metric and obtain a contradiction. Suppose that d is a metric on X such that U E :Y iff for each .x in U there is an € > 0 such that B(x; €) = {y:d(x,y) < €} C U. Now let A = (0, 1);ifUE:YandOE Uthenthereisa point a in UnA, a i= 0 (in fact there is an infinity of such points). Hence, o is a limit point of A. It follows that there is a sequence {t n } in A such that d(tn' 0)-)-0. But if U= {XEX: x i= tn for any n} = X-{tl' fz, .. . }then o E U and U is open. So it must follow that tn E U for n sufficiently large; this is an obvious contradiction. Hence, no metric can be found. This example illustrates a technique that, although available for metric spaces, is of little use for general topological spaces: the convergence of sequences. It is possible to define "convergent sequence" in a topological space (Do it !), but this concept is not as intimately connected with the structure of a topological space as it is with a metric space. For example, it was shown above that a point can be a limit point of a set A but there is no sequence in A that converges to it. If a topological space (X, :Y) is such that a metric d on X can be found with the property that a set is in :Y iff it is open in (X, d), then (X, :Y) is said to be metrisable. There are many non-metrisable spaces. In addition to inventing non-metrisable topologies as was done above, it is possible to define processes for obtaining new topological spaces from old ones which will put metrisable spaces together to obtain non-metrisable ones. For example, the arbitrary cartesian product of topological spaces can be defined; in this case the product space is not metrisable unless there are only a countable number of coordinates and each coordinate space is itself metrisable. (See VII. 1.18.) Another example may be obtained as follows. Consider the unit interval 1= [0, 1]. Stick one copy of I onto another and we have a topological space which still "looks like" 1. For example, [1, 2] is a copy of I and if we stick it onto I we obtain [0, 2]. In fact, if we "stick" a finite number of closed intervals together another closed interval is obtained. What happens if a countable number of closed intervals are stuck together? The answer is that we obtain the infinite interval [0, (0). (If the intervals are stuck together on both sides then u-;g is obtained.) What happens if we put together an uncountable number of copies of J? The resulting space is called the long line. Locally (i.e., near each point) it looks like the real line. However, the long line is not metrisable. As a general rule of thumb, it may be said that if a process is used to obtain new spaces from old ones, a non-metrisable space will result if the process is used an uncountable number of times. For another example of a space that is non-metrisable let X be a set consisting of three points-say X = {a, b, c}. Let /:1 = {D, X, {a}, {b}, {a, b} }; it is easy to check that ff is a topology for X. To see that (X, /T) is not metrisable notice that the only open set containing c is the set X itself. There do not exist disjoint open sets U and V such that a E U and c E V. On the other hand if there was a metric d on X such that ..'T is the collection of open sets relative to this metric then it would be possible to find such

224


open sets (e.g., let U = B(a; t-) and V = B(c; €) where € < dCa, c». In other words, (X, 51 fails to be metrisable because .r does not have enough open sets to separate points. (Does the first example of a non-metrisable space also fail because of this deficiency?) The next definition hypothesizes enough open sets to eliminate this difficulty. 4.5 Definition. A topological space (X, .r) is said to be a Hausdorff space if for any two distinct points a and b in X there are disjoint open sets U and V such that a E U and b E V. Every metric space is a Hausdorff space. As we have already seen there are examples of topological spaces which are not Hausdorff spaces. Many authors include in the definition of a topological space the property of a Hausdorff space. This policy is easily defended since most of the examples of topological spaces which one encounters are, indeed, Hausdorff spaces. However there are also some fairly good arguments against this combining of concepts. The first argument is that mathematical pedagogy dictates that ideas should be separated so that they may be more fully understood. The second, and perhaps more substantial reason for not assuming all spaces to be Hausdorff, is that more examples of nDn-Hausdorff spaces are arising in a natural context. Even though there will be no non-Hausdorff space which will appear in this book, this separation of the two concepts will be maintained for a while longer. The next step in this development is the definition, in the setting of topological spaces, of certain concepts encountered in the theory of metric spaces and the stating of analogous propositions.

.r) is connected if the only nonempty subset of X which is both open and closed is the set X itself.

4.6 Definition. A topological space (X,

4.7 Proposition. Let (X, .r) be a topological space; then X = U {Ci : i E I} where each C i is a component of X (a maximal connected subset of X). Furthermore, distinct components of X are disjoint and each component is closed. 4.8 Definition. Let eX, .r) and (n, 9') be topological spaces. A function f: X -+ is continuous iff-l(~) E . r whenever ~ E 9'.

°

4.9 Proposition. Let (X, .r) and (0,9') be topological spaces and letf: X -+ be a function. Then the following are equivalent: (a) f is continuous; (b) if A is a closed subset ofn thenf-I(A) is a closed subset of X; (c) if a E X and if f(a) E ~ E 9' then there is a set U in . r such that a andf(U) c ~.

E

° U

4.10 Proposition. Let (X, 51 and (0, 9') be topological spaces and suppose that X is connected. Iff: X -+ n is a continuous function such that f(X) = 0, then n is connected. 4.11 Definition. A set K c X is compact if for every sub-collection (!) of

.r


U {U: U E (9} there are a n (9 such that K c U Uk'

such that K in

C

225

finite number of sets U 1,

••. ,

Un

k~1

4.12 Proposition. Let (X, :T) and (n, ,9") be topological spaces and suppose K is a compact subset of X. Iff: X --i>- n is a continuous function then f(K) is compact in n. If (X, d) is a metric space and Y c X then (Y, d) is also a metric space. Is there a way of making a subset of a topological space into a topological space? We could, of course, declare every subset of Y to be open an~ this would make Y into a topological space. But what is desired is a topology on Ywhich has some relationship to the topology on X; a natural topology on a subset of a topological space. If (X, :T) is a topological space and Y c X then define :T y

It is easy to check that

::r y

=

{Un Y: UE:T}.

is a topology on Y.

,r

4.13 Definition. If Y is a subset of a topological space (X, :T) then y is called the relative topology on Y. A subset W of Y is relatively open in Y if WE:T y; W is relatively closed in Y if Y - WE :T y. Whenever we speak of a subset of a topological space as a topological space it will be assumed that it has the relative topology unless the contrary is explicitly stated. 4.14 Proposition. Let (X, :T) be a topological space and let Y be a subset of X. (a) If X is compact and Y is a closed subset of X then (Y, :T y) is compact. (b) Y is a compact subset of X iff (Y, :T y) is a compact topological space. (c) If(X, :T) is a Hausdorff "'pace then (Y, :T y) is a Hausdorff space. (d) rr(X, :T) is a Hausdorff space and (Y, :T y) is compact then Y is a closed subset ofX. Proof The proofs of (a), (b), and (c) are left as exercises. To prove part (d) it suffices to show that for each point a in X - Y there is an open set U such that a E U and Un Y = O. So fix a point a in X - Y; for each point y in Y there are open sets Uyand Vy in X such that a E UY' Y E V y, and U y n Vy = o (because (X, ,r) is a Hausdorff space). Then {Vy n Y: y E Y} is a collection of sets in :T y which covers Y. So there are points yj, ... , Yn in Y such

that Y c

U(V)"

i~l

n Y) c

UV y,

i~l

=

V. Since a E U y, for each i, a E U = n

Up,; also U E:T. It is easily verified that Un V = U (U n V y ,) that Un Y = O. • ;; 1 The proof of this proposition yields a stronger result.

n

i~1

=

0 so

4.15 Corollary. Let (X, ,.71 be a Hausdorff space and let Y be a compact subset of X; then for each point a in X - Y there are open sets U and V in X such that a E U, Y c V and Un V = O. If we return to the consideration of metric spaces we can discover a new

Analytic Continuation and Riemann Surfaces 226 way to define a topology. The sequence of steps by which open sets are obtained in a metric space are as follows: the metric is given, then open balls are defined, then open sets are defined as those sets which contain a ball about each of their points. What we wish to mimic now is the introduction of balls; this is done by defining a neighborhood system on a set.

4.16 Definition. Let X be a set and suppose that for each point x in X there is a collection JVx of subsets of X having the following properties: (a) for each V in JVx , x E V; (b) if V and V E.AI", there is a Win JVx such that We V n V; (c) if V E JVx and V E JVy then for each z in V n V there is a Win JV;, such that W c V n V. Then the collection {JVx : x E X} is called a neighborhood system on X. If (X, d) is a metric space and if x E X then JVx = {B(x; €): € > O} gives a collection {u-V:: x E X} which is a neighborhood system. In fact, this was the prototype of the above definitions. Notice that condition (c) relates the neighborhood systems of different points. If only conditions (a) and (b) were satisfied, it would not follow that these neighborhoods would be open sets in the topology to be defined. For example, if X is a metric space and JVx is the collection of closed balls about x then {JVx : x E X} satisfies conditions (a) and (b) but not (c). Moreover, it is easy to verify that by letting x = y = z condition (b) can be deduced from (c) (Verify!). The next proposition relates neighborhood systems and topological spaces. 4.17 Proposition. (a) If (X, Y) is a topological space and JVx = {V E Y: x E U} then {.AI,,: x E X} is a neighborhood system on X. (b) If {JVx : x E X} is a neighborhood system on a set X then let Y = {V: x in V implies there is a V in .AI" such that V c V}. Then Y is a topology on X and JVx c Y for each x. (c) If(X, Y) is a topological space and {JVx : x E X} is defined as in part (a), then the topology obtained as in part (b) is again Y. (d) If {JVx : x E X} is a given neighborhood system and Y is the topology defined in part (b), then the neighborhood system obtained from Y contains {JVx : x E X}. That is, if V is one of the neighborhoods of x obtained from Y then there is a V in .AI" such that V c v.

Proof. The proof of parts (a), (c), and (d) are left to the reader. To prove part (b), first observe that both X and 0 are in Y(O E Y since the conditions are vacuously satisfied). Let VI' ... , Vn

E

Y and put V

=

nV n

j •

If x

E

V

j~l

then for each) there is a Vj in.Al" such that Vj

C

V j . It follows by induction

on part (b) of Definition 4.16 that there is a V in.Al" such that V c

n Vj . n

.i~1

Thus V c V and V must belong to Y. Since the union of a collection of

The Sheaf of Germs of Analytic Functions on an Open Set

227

sets in Y is clearly in Y, Y is a topology. Finally, fix x in and let U E JVx • If y E U it follows from part (c) of Definition 4.16 that there is a V in JVy such that V c U. Thus U E Y . • 4.18 Definition. If {,Ai'x: x E X} is a neighborhood system on X and Y is the topology defined in part (b) of Proposition 4.17, then Y is called the topology induced by the neighborhood system. 4.19 Corollary. If {JVx : x E X} is a neighborhood system on X and Y is the induced topology then (X, Y) is a Hausdorff space iff for any two distinct points x and y in X there is a set U in JY" and a set V in JVy such that Un V

= O.

Exercises 1. Prove the propositions which were stated in this section without proof. 2. Let (X, Y) and (0, sP) be topological spaces and let Y c X. Show that if f: X --+ a is a continuous function then the restriction off to Y is a continuous function of (Y, Y y) into (0, sP). 3. Let X and a be sets and let {JVx : x E X} and {.Aw: W EO} be neighborhood systems and let Y and sP be the induced topologies on X and a respecti vel y. (a) Show that a function f: X --+ a is continuous iff when x E X and w = f(x), for each Ll in .A(i) there is a U in JVx such thatf(U) c Ll. (b) Let X = a = C and let ~ = .Az = {B(z; E): E > O} for each z in Co Interpret part (a) of this exercise for this particular situation. 4. Adopt the notation of Exercise 3. Show that a function f: X --+ a is open iff for each x in X and U in JVx there is a set Ll in.Aw (where w = f(x» such that Ll c f( U). 5. Adopt the notation of Exercise 3. Let Y c X and define 'Ply = {Y n U: U EJVy } for each yin Y. Show that {'Ply: y E Y} is a neighborhood system for Yand the topology it induces on Y is Y y. 6. Adopt the notation of Exercise 3. For each point (x, w) in Xx a let 'PI(x,W)

= {UxLl: UEJY",LlE.Aw }

(a) Show that {ott(X,W): (x, w) EXXO} is a neighborhood system on Xx a and let flJ be the induced topology on Xx O. (b) If U E Y and Ll E sP, call the set Ux Ll an open rectangle. Prove that a set is in flJ iff it is the union of open rectangles. (c) Define PI: Xx a --+ X and P2: Xx a --+ a by PI(X, w) = x and P2(X, w) = w. Show that PI and P2 are open continuous maps. Furthermore if (Z, fJ€) is a topological space show that a functionf: (Z, fJ€) --+ (Xx 0, flJ) is continuous iff PI 0 f: Z --+ X and P2 0 f: Z --+ a are continuous.

§5. The Sheaf of Germs of Analytic Functions on an Open Set This section introduces a topological space which plays a vital role in complex analysis. In addition to the topological structure, an analytical

228


structure is also imposed on this space (in the next section). This will furnish the setting in which to study the complete analytic function as an analytic function. If (f, D) is a function element, recall that the germ of (f, D) at a point z = a in D is the collection of all function elements (g, B) such that a E B and g(z) = fez) for all z in the component of B n D that contains a. This germ is denoted by [fl •. 5.1 Definition. For an open set G in C let .'l'(G) = {(z,[f]z):zEG,fisanalyticatz}. Define a map p: .9"(G) -'>- C by p(z, [f]z) = z. Then the pair (.9"(G), p) is called the sheaf of germs of analytic functions on G. The map p is called the projection map; and for each z in G, p -I(Z) = P -I( {z}) is called the stalk or fiber over z. The set G is called the base space of the sheaf. Notice that for a point (z, [fL) to be in .9"(G), it is not necessary thatfbe defined on all of G; it is only required that f be analytic in a neighborhood of z. How do we picture this sheaf? (There are, of course, too many dimensions to form an accurate geometrical picture.) One way is to follow the agricultural terminology used in the definition. On top of each stalk there is a collection of germs; the stalks are tied together into a sheaf. A better feeling for .9"(G) can be obtained by examining the notation for points in .9"(G). When we consider a point (z, [f]z) in.9"( G), think of a function element (f, D) in the germ [f]. instead of the germ itself. For every point win D there is a point (w, [f]w) in .9"(G). Thus about (z, [f]z) there is a sheet or surface few; [f]w): wED}. In fact, .9"(G) is entirely made up of such sheets and they overlap in various ways. Alternately, we can think of .9"(G) as the union of graphs; each point (z, [fJ.) in .9"(G) corresponding to the point (z, fez»~ on the graph of (f, D). (The graph of (f, D) is a subset of C 2 or ~4.) Two function elements are equivalent at a point z if their graphs coincide near z. A topology will be defined on .9"(G) by defining a neighborhood system. For an open set D contained in G and a function f: D -'>- C analytic on D define

5.2

N(f, D) = {(z, [f]z): zED}.

That is, N(f, D) is defined for each function element (f, D). If we think of .9"(G) as a collection of sheets lying above G which are indexed by the germs, then N(f, D) is the part of that sheet indexed by f and which lies above D. 5.3 Theorem. For each point (a, [f].) in .9"(G) let .A(.,[f)a)

= {N(g, B): a E B and [g]. = [f].}.

then {.A(. ,[JJa): (a, [f].) E .9"( G)} is a neighborhood system on .9"( G) and the induced topology is Hausdorff. Furthermore. the induced topology makes the map p: .9"(G) -'>- G continuous. Proof. Fix (a, [f].) in .9"(G); since it is clear that condition (a) of Definition


229

4.16 holds, it remains to verify that condition (c) holds. (Condition (b) is a consequence of (c).) Let N(g!, B I ) and N(g2' B 2 ) EJY(a,[fla) and let

5.4 It is necessary to find a function element (k, W) such that N(k, W) EA''(b,[hl b ) and N(k, W) c N(gl, B 1 ) n N(gz, B2)' It follows from (5.4) that bE BI n B2 and [hh = [gdb = [gzh. If bE WeB! n B z and h is defined on W then N(h, W) c N(g!, B I ) n N(g2' Bl)' To show that the induced topology is Hausdorff, use Corollary 4.19. So let (a, [fl.) and (b, [g]b) be distinct points of 9'(G). We must find a neighborhood N(f, A) of (a, [f],,) and a neighborhood N(g, B) of (b, [g]b) such that N(f, A) n N(g, B) = O. How can it happen that (a, [fl.) =f. (b, [gh)? There are two possibilities. Either a =f. b or a = band [fla =f. [g]". If a =f. b then let A and B be disjoint disks about a and b respectively; it follows immediately that N(f, A) n N(g, B) = O. If a = b but [fla =f. [g]a, we must work a little harder (but not much). Since [fl. =f. [g]a there is a disk D = BCa; r) such that bothfand g are defined on D andf(z) =f. g(z} for 0 < Iz-al < r. (It may happen that f(a) = g(a) but this is inconsequential.)

Claim. N(f, D) n N(g, D) = O. In fact, if (z, [hl z) E N(f, D) n N(g, D} then zED, [h]z = [f]., and [hl: = [g]z. It follows thatfand g agree on a neighborhood of z, and this is a contradiction. Hence the induced topology is Hausdorff. Let U be an open subset of G; begin the proof that p: .9"'(G) -7 G is continuous by calculating p -I( U). Since p(z, [flz) = Z, p-l(U) = {(z, [flz): z

E

U}.

So if (z, [f]z) E P-I( U) and D is a disk about z on whichf is defined and such that D c U, N(f, D) c p U). It follows that p must be continuous (Exercise 4.3). I11III Consider what was done when we showed that the induced topology was Hausdorff. If a =f. b then (a, [fl.) and (b, [glb) were on different stalks p -lea) and p -l(b); so these distinct stalks were separated. In fact, if a E A, bE B and An B = 0 then p-l(A) n p-I(B) = D. If a = b then (a, lIJa) and (a, [g]a) lie on the same stalk p -lea). Since [f1o # [gL we were able to divide the stalk. That is, one germ was "higher up" on the stalk than the other. In the remainder of this section some of the properties of 9'(G) as a topological space are investigated. In particular, it will be of interest to characterize the components of 9'(G). However, we must first digress to study some additional topological concepts.

-Ie

5.5 Definition. Let (X, :Y) be a topological space. If Xo and Xl E X then an arc (or path) in X from X () to Xl is a continuous function y: [0, 1]-+ X such that yeO) = x 0 and y( I) = X l' The point x 0 is called the initial point of y and x 1 is called the final point or terminal point. The trace of y is the set {y} = {y(t): S t S; I}.

°

230


A subset A of X is said to be arcwise or path wise connected if for any two points Xo and Xl in A there is a path from Xo to Xl whose trace lies in A. The topological space (X, ff) is called locally arcwise or path wise connected if for each point X in X and each open set U which contains X there is an open arcwise connected set V such that X E V and V c U. For each X in X let ~ be the collection of all open arcwise connected subsets of X which contain x. Then X is locally arcwise connected iff {~: x E X} is a neighborhood system which induces the original topology on X. (Verify!) The proof of the following proposition is left to the reader. 5.6 Proposition. Let (X, ff) be a topological space. (a) If A is an arcwise connected subset of X then A is connected. (b) If X is locally arcwise connected then each component of X is an open

set. The converse to part (a) of the preceding proposition is not true. For example let

X =

{t+isin~

: t>

o}

U

{si: -1

~s~

1}.

Since X is the closure of a connected set it is itself connected. Howcrver, there is no arc from the point 1/7T to i which lies in X. X is also an example of a topological space which is connected but not locally arcwise connected. Suppose X is connected and locally arcwise connected; does it follow that X is arcwise connected? The answer is yes. In fact, this is an abstract version of a theorem which was proved about open connected subsets of the plane. Since disks in the plane are connected, it follows that open subsets of IC are locally arcwise connected. Recall that in Theorem II. 2.3 it was proved that for an open connected subset G of the plane, any two points in G can be joined by a polygon which lies in G. Hence, a partial generalization of this (the concept of a polygon in an abstract metric space is meaningless) is the following proposition whose proof is virtually identical to the proof of II. 2.3. 5.7 Proposition. If X is locally arcwise connected then an open connected subset of X is arcwise connected. We now return to the sheaf of germs of analytic functions on an open set G. 5.S Proposition. Let G be an open subset of the plane and let U be an open connected subset of G such that there is an analytic function f defined on U. Then N(j, U) is arcwise connected in Y(G).

Proof Let (a, [f]a) and (b, [fh) be two generic points in N(f, U); then a, b E U. Since U is a region there is a path y: [0, 1] --+ U from a to b. Define a: [0, 1] --+ N(j, U) by aCt)

= (y(t), [f]Y(I).

231


Clearly a(O) = (a, [f]a) and a(1) = (b, [fh); if it can be shown that a is continuous then a is the desired arc. Fix t in [0, 1] and let N(g, V) be a neighborhood of a(t). Then yet) E= V and [fL(t) = [gL(t). So there is a number r > 0 such that B(y(t); r) c Un V and fez) = g(z) for Iz-y(t)1 < r. Also, since y is continuous there is a 8 > 0 such that Iyes) -y(t)1 < r whenever Is- Ii < 8. Combining these last two facts gives that (t-8, t+8) c a- 1(N(g, V». It follows from Exercise 4.3 that a is continuous.

III

5.9 Corollary. 9"(G) is locally arcwise connected and the components of 9"(C) are open arc wise connected sets. Proof The first part of this corollary is a direct consequence of the preceding proposition. The second half follows from Proposition S.6(b) and Proposition 5.7. III In light of this last corollary, it is possible to gain insight into the nature of the components of 9"(G) by studying the curves in 9"(G).

5.10 Theorem. There is a path in 9"(G) from (a, [flo) to (b, [g]b) iff there is a path y in G from a to b such that [gh is the analytic continuation of[f]a along y. Proof Suppose that a: [0, 1] - ? ,51'(G) is a path with a(O) = (a, [flo), a(l) = (b, [gh). Then y = po a is a path in G from a to b. Since a(t) (o9"(G) for each t, there is a germ [ftly(!} such that

aCt) = (y(t), [It],(t».

°: :;

We claim that {[J,]y(t): t :::;; I} is the required continuation of [fla along y. Since [/la = [fo]a and [g]b = [/lh, it is only necessary to show that {[J,]y(t)} is a continuation. For each t let D t be a disk about z = yet) such that D t C G and J, is defined on D t • Fix tin [0, 1]; since NUt, D,) is a neighborhood of a(t) and a is continuous, there is a 8 > 0 such that (t-8, 1+8) c a- 1(N(J" D,».

That is, if Is - tl < 8 then (y(s), [j;')y(s» = a(s) E N(J" D,). But, by definition, this gives that yes) E= D t and [Is]'(s) = [j~]y(S); and this is precisely the condition needed to insure that {(ft, D ,): 0 :::; t :::; I} is a continuation along y (Definition 2.2). Now suppose that y is a curve in G from a to band (JJ,],(t): t :::;; I} is a continuation along y such that [fO]a = [fla and [fl]b = [g]b' Define a: [0, 1]->- Y(G) by aCt) = (y(t), [J,ly(t»; it is claimed that a is a path from (a, [fla) to (b, [gh). Since the initial and final points of a are the correct ones it is only necessary to show that a is continuous. Because the details of this argument consist in retracing the steps of the first half of this proof, their execution is left to the reader. III

°: :;

5.U Theorem. Let 9"(G) (ff

C(f

c 9"(G) and let (a, [fla)

E

C(f;

then

C(f

is a component of

232

ce =


reb, [glb): [glb is the continuation of[fla along some curve in G}.

ce

ce

Proof Suppose is a component of 9"(G); by Corollary 5.9 is an open arcwise connected subset of 9"(G). So by the preceding theorem, for each point (b, [gh) in [gh is the continuation of [fla along some curve in G. Conversely, if [glb is the continuation of [fla along some curve in G then (b, [gh) belongs to the component of 9"(G) which contains (a, [fJ.); that is, (b, [gh) E Now suppose that consists of all points (b, [gh) such that [glb is a continuation of [fla. Then is arcwise connected and hence is connected. If is the component of 9"(G) containing then by the first half of this proof it follows that = I' • Notice that the point (a, [f]a) in the statement of the preceding theorem has a transitory role; any point in the component will do. Fix a function element (f, D) and recall that the complete analytic function :F associated with (J, D) is the collection of all germs [glz which are analytic continuations of [fJ. for any a in D (see Definition 2.7). Let

ce,

ce.

cel

5.12

ce

ce

ce

ce

ce

G = {z: there is a germ [g]z in:F];

it follows that G is open. In fact, if Z E G then there is a germ [glz in :F and a disk B about z on which g is defined. Clearly BeG so that G is open. It follows from Theorem 5.11 that

5.13 is a component of 9"«(:) and that p(~) 5.14 Definition. Let :F defined in (5.13) and p pair (Jr, p) is called the (5.12) is called the base

= G. (It is also true that ~

c 9"(G).)

be a complete analytic function. If /J£ is the set is the projection map of the sheaf 9"«(:) then the Riemann Surface of :F. The open set G defined in space of ::;;.

5.15 Theorem. Let::;; be a complete analytic function with base space G and let (Jr, p) be its Riemann Surface. Theil p: P-£ ~ G is an open continuous map. Also, if(a, [f]a) is a point in [j£ then there is a neighborhood NC.f, D) of (a, [fla) such that p maps N(J, D) homeomorphically onto an open disk in the plane. Proof Consider ,'?/t as a component of 9"«(:). Since p: 9"«(:) ~ C is continuous it follows that p: ~ ~ G is continuous. To show that p is open it is sufficient to show that p(NU; U» is open for each (J, U) (Exercise 4.4). But p(N(J, U» = U which is open. If (a, [fla) E.~ let D be an open disk such that (J, D) E [fla. Then p: N(J, D) ~ D is an open, continuous, onto map. To show that p is a homeomorphism it only remains to show that p is one-one on N(J, D). But if (b, [fh) and (c, [fle) are distinct points of N(J, D) then b =1= c which gives that p is one-one. • Remarks. In its standard usage the Riemann surface of a complete analytic function consists not only of what is here called a Riemann Surface but also

Analytic Manifolds

233

some extra points-the branch points. These branch points correspond to singularities and permit a deeper analysis of the complete analytic function. Exercises 1. Define F: 9'«(;) -+ C by F(z, [f]z) = fez) and show that F is continuous. 2. Let :F be the complete analytic function obtained from the principal branch of the logarithm and let G = C- {O}. If D is an open subset of G andf: D -+ C is a branch of the logarithm show that [f]a E :F for all a in D. Conversely, if (J, D) is a function element such that [f]a E :F for some a in D, show thatf: D -+ C is a branch of the logarithm. (Hint: Use Exercise 2.8.) 3: Let G = C - {O}, let :F be the complete analytic function obtained from the principal branch of the logarithm, and let (3l, p) be the Riemann surface of:F (so that G is the base of 3l). Show that 3l is homeomorphic to the graph r = {(z, e z E G} considered as a subset of C x C. (Use the map h: 3l -+ r defined by h(z, [f).) = (f(z), z) and use Exercise 2.) State and prove an analogous result for branches of zl/n. 4. Consider the sheaf 9'«(;), let B = {z: Jz-IJ < I}, let t be the principal branch of the logarithm defined on B, and let tl (z) = t(z) + 21ri for all z in B. (a) Let D = {z: JzJ < I} and show that (t, [t]t) and (t, [ttlt) belong to the same component of p -I(D). (b) Find two disjoint open subsets of 9'«(;) each of which contains one of the points (t, [t1t) and (t, [ttlt). Z ):

§6. Analytic Manifolds In this section a structure will be defined on a topological space which, when it exists, enables us to define an analytic function on the space. Before making the necessary definitions it is instructive to consider a previously encountered example of such a structure. The extended plane Coo can be endowed with an analytic structure in a neighborhood of each of its points. If a E Coo and a =I- 00 then a finite neighborhood V of a is an open subset of the plane. If fP: V -+ C is the identity map, fP{z) = z, then fP gives a "coordinatization" of the neighborhood V. (Do not become confused over the preceding triviality. The introduction of the identity function fP seems an unnecessary nuisance. After all, V is an open subset of C so we know what it means to have a function analytic on V. Why bring up fP? The answer is that for the general definition it is necessary to consider pairs such as (V, fP); trivialities appear here because this is a trivial example.) If a = 00 then let Voo = {z: IzJ> l}u {oo} and define fPoo: Voo-+C by fPoo(z) = Z-I for z =I- 00, fPoo( 00) = o. So fPoo is a homeomorphism of V 00 onto the open disk B(O; 1). Hence to each point a in Coo a pair (Va' fPa) is attached such that Va is a neighborhood of a and fPa is a homeomorphism of Va onto an open subset of the plane. What happens if two of the sets Va and Vb intersect? Suppose for example that a =I- 00 and Va n V =I- O. Let Goo = B(O; 1) = fPoo( V 00) and let Ga = fPa(Va) (=Va)' Then fP;;;t(z) = Z-I for all z in Goo; thus fPa fP;;;;I(Z) = Z-I for all z in fPoclVoo n Va). Since 00 tVa, 0 tfPoo(Voo n Va) 00

0

234


so that 'fa () 'P-;' 1 is analytic on its domain. Similarly, if both a and bare finite then 'Po 'Pb- 1 is analytic on its domain. This is the crucial property. 0

6.1 Definition. Let X be a topological space; a coordinate patch on X is a pair ( U, 'f) where U is an open subset of X and 'f is a homeomorphism of U onto an open subset of the plane. If a E U then the coordinate patch (U, 'P) is said to contain a.

6.2 Definition. An analytic manifold is a pair (X, <1» where X is a Hausdorff connected topological space and is a collection of coordinate patches on X such that: (i) each point of X is contained in at least one member of <1>, and (ii) if (Ua, with Uan Ub*D then of coordinate patches is called an analytic structure on X. An analytic manifold is also called an analytic surface. Note immediately that 'Po" 'Pb- 1 is one-one since both 'Po and 'Pb are. Henceforward, for the sake of brevity, care will not be taken in mentioning the appropriate domain of a function such as 'Po 'Pb- 1. Next, it must be emphasized that the definition of an analytic manifold is tied to its analytic structure. It is possible to give the open disk two incompatible analytic structures (one of them the natural one). This is also the case with the torus which can be made into an analytic manifold in an uncountable number of incompatible ways. (Exercise 4.) But this investigation must be postponed until we have the notion of an isomorphism between analytic surfaces. With the collection of coordinate patches introduced prior to Definition 6.1, Coo becomes an analytic manifold. However, a closed disk is not a mani~ fold since the points on the boundary cannot be surrounded by a coordinate patch. Similarly, the union of two intersecting planes in iR( 3 is not an analytic manifold. A number of examples of analytic surfaces will be available after substantiating the following observations which are gathered into a proposition. 0

6.3 Proposition. (a) Suppose (X, v = { ( Un V,
E

}

then (V,
(Q, '1') is an analytic surface.

235

Analytic Manifolds

Proof. (a) This is a triviality. (b) It is clear that Q is connected, that 'It is a collection of coordinate

patches for Q, and that each point of Q is contained in at least one member of 'It. So let (U,cp) and (V,P.)E such that h(U)nh(V)'f'D. But then UnV#O since h(U)nh(V)=h(UnV). So, (qJoh-1)-lo(floh- 1)= (Cpoh-I)o(hop.-I)=cpop.-I which is analytic by the condition on <1>. Thus (Q, 'It) is an analytic manifold. • In virtue of part (a) of the preceding proposition the following assumption is made on all analytic structures that will be discussed: If (U,cp) E and V is an open subset of U then (V,cp) E <1>.

Of course, when an analytic structure is defined one does not bother to give all the coordinate patches, but only those that generate in the above manner. Proposition 6.3 also implies that any space that is homeomorphic to a region in the plane is an analytic surface. Hence a piece of paper with a crease in it is an analytic surface. Is this a shock to you? If the reader has ever seen an introduction to differentiable manifolds, he may be surprised at this. If X is a subset of 1R3 then X is a differentiable 2-manifold if each point in X is contained in a coordinate patch (U,cp) such that cp - I : cp( U)~ U C 1R3 has coordinate functions with continuous partial derivatives. That is, let G=cp(U) and cp-I(S,t)=(~(s,t), 1/(s,t), K(s,t» for all (s,t) in G. It is required that ~, 1/, and K be functions from G into IR with continuous partial derivatives. A folded piece of paper is not a differentiable 2-manifold. In fact, if (U,cp) is a patch that contains a point on the crease then cp - I has at least one non-differentiable coordinate function. Since analyticity is a stronger notion than differentiability this seems confusing, but the explanation is simple. If h is a homeomorphism of X onto the region G in the plane then an analytic structure is imposed on X via h. In fact, by considering X with this structure we are only considering G under a different guise. In the definition of a differentiable 2-manifold there is no differentiable structure "imposed" on X; the structure is restricted by conditions that it inherits as a subset of 1R3 (where there is already a differentiable structure). In a similar fashion the surface of a cube in 1R3 is an analytic surface since it is homeomorphic to Coco Suppose now that G is a region and f: G~C is an analytic function such that j'(Z)'f' for any Z in G. We wish to give the graph

°

f={(Z.!(Z»:ZEG}

an analytic structure. If p: r ~ G is defined by p(z, /(z» = z then p is a homeomorphism of r onto G; consequently r inherits the analytic structure of G as in 6.3 (b). But this does not use the analyticity of j, let alone the fact

236


r

that doesn't vanish. This is surely an uninteresting structure. So, to rephrase the problem: put an analytic structure on r that is "connected" to the analytic properties of f Fix Ci = (a,/(a)) in r. Since('(a) # 0 there is a disk D" about a such that Da c G and f is one-one on D". Let

6.4

6.5

CP.(z,/(z))

=

fez)

for each (z, fez)) in V •.

6.6 Proposition. Let G be a region in the plane and let f be an analytic function on G with non-vanishing dericalive. 1f l' is the graph off and
=

{( U.,

CPa):

Ci E

1', V. and CPa as in (6.4) and (6.5)}

then (1',
p

0

0

6.7 Theorem. 1f(2/2, p) is the Riemann surface of a complete al1alyticfunctiol1 andrtJ = {( U, p): U is open in ,rJ,f, p is one-one Oil U}, then C;.f, q) is all analytic manifold. Proof It follows from Theorem 5.15 that each point of dIt is contained in an open set on which p is one-one and that p is a homeomorphism there. Furthermore, if(U, p) and (V, p)Eq) and Un Vol. then po(p:U)-l is the identity map. (The notation pi U is used to denote the restriction of p to V.) Since dIt is connected it is an analytic surface . • 6.8 Definition. Let C"t', <1J) and (U, 'F) be analytic manifolds and let f: X ->- n be a continuous function; let a E X and a = f(a). The function f is analytic al a if for any patch (.\, .j;) in tj? which containsCi there is a patch (U, 'P) in
(i)f(U) c A; (ii) Ij; 0 f 0 'P - 1 is analytic on cp( U) c C.

Analytic Manifolds

237

The function is analytic on X if it is analytic at each point of X. The condjtion that (U, rp) can be found such that a E U and I( U) c A is a consequence of the continuity of{CProposition 4.9(c». The heart of the definition is the requirement that f I rp - 1 be an analytic function from rp( U) c C into iC. For two given analytic surfaces there may be many analytic functions from Olle to the other or there may be very few. Clearly every constant function is analytic; but there may be no other analytic functions. For example, if )( = C and n is a bounded region in the plane then Liouville's Theorem implies there are no non-constant analytic functions from X into n. Also, suppose thatf: C CXl --;.. C is an analytic function; thenf(C co ) is compact so that the restriction of f to C is a bounded analytic function on iC. Again, Liouville's Theorem says that each such f is a constant function. On 0

0

the other hand, if p is a polynomial and a E C then both p(z) and p(_l_)

z-a

are analytic functions from Coo to Coo. In fact, these are practically the only analytic functions from Coo to Coo (Exercise 7). If (X, iC. Before proving some of the basic properties of analytic functions on manifolds, one further example will be given. This example is stated as a theorem and justifies the terminology "complete analytic function". 0

6.9 Theorem. Let ,'F be a complete analytic function with Riemann Surface (9£, p). If ,'7: :~ --;.. C is defined by ff(z, [f]z) = fez) then .,'7 is an analytic function. Proof Fix ex = (a, [f]u) in:?ll and let D be a diSK about a on whichfis defined and analytic. Let U be the component of p -1 (D) which contains ex; so (U, p) is a coordinate patch. Let p -1 denote the inverse of p: U --;.. p( U). We must show that ff p -1 is analytic on pC U) c iC. But for z in peU), 0

,'70 p -1(Z) = ff(z,f(z»)

=

f(z);

that is, ,'70 P-1 = f which is analytic. III The next several results are generalizations of theorems about analytic functions defined on regions in the plane. 6.10 Proposition. S'uppose (X,
238

analytic fun ctions from X into thenf = g.

n. If {x

Analytic Continuation and Riemann Surfaces E

X:f(x) = g(x)} has a limit point in X

Proof Define the subset A of X by

A = {x:f and g agree on a neighborhood of x}. This set A will be shown to be non-empty and the reader will be required to prove that A is both open and closed in X. By hypothesis there is a point a in X such that for every neighborhood V of a there is a point x in V with x # a and f(x) = g(x). It is easy to conclude that f(a) = g(a) = CY:. If (A, o/J) E 'Y and CY: E A then there is a patch (V, tp) in <1> such that f( V) and g( V) are contained in A with both o/J 0 f 0 tp -! and o/J 0 g 0 tp - 1 analytic in a disk D about Zo = - C be an analytic function. If there is a point a E X and a neighborhood V of a such that If(a) ? If(x) for all x in V then f is a constant function. The proof is left to the reader. The Maximum Modulus Theorem allows us to generalize Liouville's Theorem in the following way. 1

1

6.13 Liouyille's Theorem. If eX, <1» is a compact analytic manifold then there is no non-constant analytic fUllction from X into C. 6.14 Open Mapping Theorem. Let (X, <1» and (n, 'Y) be analytic manifolds and let f: X -+ n be a non-constant analytic function. If V is an open subset of X then[C U) is open in it Proof Let V be an open subset of X and let Ct E f( U). If a E: V such that = f(a) and CA, I~) E '¥ which contains IX, then let (V, qJ) E <1> with a in V such thatf(V) c A and I~ 0 fo
6.15 Definition. If (X, <1» and (n, 'Y) are analytic manifolds, an isomorphism of X onto n is an analytic function f: X -+ n which is one-one and onto. If an isomorphism exists then eX, <1» is said to be isomorphic to (n, 'F). If f: X -+ n is an isomorphism then f is an open mapping by (6.14). It follows that f- 1: n -+ X is a homeomorphism. Is f- 1 also analytic? The answer is yes and the reader is asked to prove the next proposition. 6.16 Proposition. isomorphism.

If f:

X

-+

n is an isomorphism then f

-1:

n -+ X

is also an

Analytic Manifolds

239

Recall that the Riemann Mapping Theorem states that if G is a simply connected region in the plane and G =F C, then G is isomorphic to the open unit disk if both are considered as analytic surfaces. Also recall that Proposition 6.3(b) states that if (X, <1» is an analytic manifold and h: X -+ n is a homeomorphism of X onto the topological space n, then h induces an analytic structure on n; denote this structure by 0 h -1. Suppose n already has an analytic structure 'Y. It is easy to see that h - 1 = 'Y iff h is analytic; that is, iff h is an isomorphism from (X, <1» onto (n, 'Y). As an example let X = C and n = {z: Izl < I}, and define h: X·-+ n by 0

h(z)

z

=-11' 1+ z

Then h is a homeomorphism of C onto n, but it is clearly not analytic. So using h and the analytic structure on C, a structure can be induced on n which is completely unrelated to its natural structure. For example, with this induced structure n will have no bounded non-constant analytic functions. Consider the following situation: let G be a region in the plane and let f: G -+ C be an analytic function with non-vanishing derivative. Let (g, D) be a function element such that g(D) c G and f(g(z» = z for all z in D. (That is, g is a "local" inverse off) If §' is the complete analytic function obtained from (g, D) and (&l, p) is its Riemann surface, let us examine whether &l and the graph of f are isomorphic analytic manifolds. (The analytic structure for the graph of f was introduced in Proposition 6.6.) The answer to this question is yes provided that the domain of f is not restricted. To illustrate what can go wrong let G = {z: Izl < I} and letfbe the exponential function e Z • Then :IF consists of all germs of branches of the logarithm (Exercise 5.3); so &l is rather large and complicated. However, {(z, eZ ) : z E G} is a simple copy of the disk G. The difficulty arises because the domain of e Z has been artificially restricted. If instead G is the whole complex plane then §' and the graph of G are indeed isomorphic analytic surfaces (see Exercise 5.3). 6.17 Definition. Let f: G -+ C be an analytic function with non-vanishing derivative. If a E G and Q£ = f(a), let (g, D) be a function element such that Q£ E D and f(g(z» = z for all z in D. If :IF is the complete analytic function obtained from (g, D) then :IF is called the complete analytic function of local inverses for f

There are two questions that arise in connection with this definition. First, does the definition of the complete analytic function of local inverses for f depend on the choice of the function element? Could we have started with another local inverse and still have obtained the same complete analytic function? Second, does :IF contain the germ of every local inverse off? The answer to each of these questions is given in the following proposition; it is "yes". 6.18 Proposition. Let f be an analytic function with non-vanishing derivative on a region G. Let a and bEG, Q£ = f(a), f3 = feb); and let ~o and ~1 be disks


240

about a and (3 respectively such that there are analytic functions go: ~o ---+ IC and gl: n l ---+ IC with go(a) = a, gl({3) = b, f(gom) = ~ for all ~ in ~o, f(gIW) = ~for all ~ in ~I' Then there is a path a ill f(G) from a to {3 such that (gl' ~l) is the continuation of (go, ~o) along a. Prool Since G is connected there is a path yin G from a to b. For 0 :s t :s let D, be a disk about y(t) such that D, c G and on which f is one-one. Let a = f y and let ~, be a disk about a(t) such that n, c f(D,}. Finally, let g,: ~, ---+ IC be an analytic function such that 0

f(g,(~»

=

g,(a(t»

= yet)

~

for

in

~

~,

Claim. {(g" ~,)} is an analytic continuation along a. To show this fix t and let S be chosen so that yes) Ef-l(~,) n D, whenever Is-tl < 8. Now fix s y a

,...--

I

/

- ............. D,

"-

/

I /

__

--...

y/

\

\

"- \ ........

'-. .........

/

\

"\ \

y(s)/

_/

I

--

__

",.

..- . /

/

I

yet) '\

II

/\ / I '\ I

\

\1'I

//,

I //D,

I

,

/

/

/

/

/

I

/

/~-:..-::.. ....... /

/

I I" \ \

'\

a(1)

",

\',

.

\ -", \ I

- - 1(8)/ a(s)} / -~ /

I I I

-_/

I I

I

\

I \.

I

//I(D,)

.....................

......./

with IS-II < 8 and let B be a disk about yes) such that B c f-l(~,) n Ds n D,. So feB) is an open set containing a(s) = f(y(s» and feB) c feD,). By definition gsCf(z» = z for z in B; thus f(gsCO) = ~ for all ~ in feB). But feB) c ~, which gives that f(g,(O) = ~ for all ~ in feB). But for ~ in feB) both gsm and g,m are in f-l(~,) n Ds n D, and f is one-one here. Hence gsm = g,m for all ~ inf(B); alternately, [gs]a(s) = [g,L(s) whenever Is - 11 < S. This substantiates the claim . • Recall that if (3f, p) is the Riemann surface of a complete analytic function §, the symbol § is also used to denote the analytic function § : !J£ ---+ IC defined by §(z, [flz) = fez) (Theorem 6.9).

6.19 Proposition. Let f be an analytic function on a region G which has a nonvanishing derivative, let § be the complete analytic function of local inverses

Analytic Manifolds

241

off, and let (81, p) be the Riemann surface of ff'. f(G) then T(Z, [g),) = (g(z), z)

If ff'(81) =

G and p(81)

=

defines an isomorphism between the analytic manifold 8l and graph (f). Before proving this proposition we must show that under the same hypothesis each member of ff' is a local inverse off This provides a partial converse to Proposition 6.18. 6.20 Lemma. Let f, G, ff' and (81, p) be as in the preceding proposition. If [g]a E ff' then there is a disk D about a on which g is defined and such that f(g(z» = z for all z in D.

Proof Fix [g]a in ff' (so (a, [g]a) E 8l); then, by hypothesis, g(a) = ff'(a, [g]a) E G. If b = f(g(a» then there is a disk B about b and an analytic function h: B --+ C such that h(b) = g(a) and f(h(z» = z for all z in B. From Proposition 6.18, [hh E ff'. Also a = pea, [g]a) E p(81) = f(G). According to Theorem 5.11 there is a path yin f(G) from a to b such that [h]b is the continuation along y of [g]a. Let {(go D t ) } be a continuation along y with each D t a disk such that [goL = [g]a and [g l]b = [h]b. Define a subset T of [0, 1] by T

= {t:f(gtCz» = z for all z in D t }.

°

We want to show that T = [0, IJ. In fact, once this is proved it follows that E T so that (g, D) must be a local inverse off Since T contains 1 it is non-empty; it must be shown that T is both open and closed in [0, 1]. For any number t in [0, 1] let S > such that [gs]y(S) = [gt]y(s) whenever Is- 11 < S. If t E T then f(gs(z» = f(grCz» = z for all z in Ds n D t and Is - tl < S. It follows that (t - S, t + S) c T and so T is open. If t E T- then there is an s in T such that Is-tl < S. For z in Ds n D t we have that f(grCz» = f(gs(z» = t so that t E T. That is, T is closed . •

°

Proof of Proposition 6.19. It follows from the preceding lemma that (g(a), a) E graph (f) if (a, [g]a) E 8l. Thus, T does indeed map 8l into graph (f). Suppose (o:,f(o:» E graph (f) and a = f(o:). If (g, D) is a function element such that a E D, g(a) = 0:, and f(g(z» = z for all z in D, then [g]a E ff' and T(a, [g]a) = (0:, f(o:». That is T(~) = graph (f). To show that T is one-one let (a, [gla) and (b, [hh) E 8l such that T(a, [g]a) = T(b, [hh); that is, (g(a), a) = (h(b), b). Thus, a = band g(a) = h(b) = 0:. Moreover, Lemma 6.20 implies that 6.21

f(g(z» = f(h(z»

for z in a neighborhood of a. Butf'(o:) "# 0, so thatfis one-one in a neighborhood of ~ = 0:. It follows from (6.21) that [g]a = [h]~ and therefore, T is one-one. It remains to show that T is analytic. This is actually an easy argument once the question to be answered is made explicit. So fix (a, [gla) in 8l and put 0: = g(a). Let ~ be a disk about 0: on which f is one-one and let D be a disk about a on which g is defined and such that D c f(~). Let U =

m,


242

/

f I

/

/

;/

I

!

---" '--~

"-

'\

. (a, [gj,)

\

\

\

\.

"-

\

'-.._----,....

./

/

I

/

\ \ I

I I

p

E Ll} and define 'P: U --+ C by 'PCs, fCD) = flO; so (U, q;) is a coordinate patch on graph (1') (Proposition 6.6). Also N(f. D) = fez, [glz): zED} gives that (N(g, D), p) is a coordinate patch on 24 (Theorem 6.7) containing Ca, [gla), and satisfying r(N(g, D) c U. I-Ience to complete the proof it must be shown that 'P c r p - 1 is an analytic function on D. This is trivial. In fact, if ZED then

fm): ~

0

'P

<0

r

0

p-l(Z)

= 'P

= -

0

r(z, [g]z)

'P(g(z), z) ~,

that is, 'P r p - 1 is the identity function on D. III If G is the punctured plane and fez) = z" for some n or if G = C and fez) = e Z then the hypothesis of Proposition 6.19 is satisfied. Let us examine this a little more closely for the case where fez) = Z2, Z E G = C - {O}. So [j.f is isomorphic to the graph of Z2; let r = fez, Z2): z #- O}. Now r: ;j£ --+ I' is defined by rCa, [gla) = (g(a), a). Recall that .'F: 3f --+ C is defmed by .%(a, [gla) = g(a). For this case .% acts like the square root function. In other words, we have found a natural domain of definition of Z1. The corresponding function on f would project (z, Z2) to its first coordinate. Even though we have shown that [j.f (a very abstract object) is equivalent to a less abstract object (the graph f), this is still somewhat dissatisfying. After all, the graph is a subset of C 2 which is beyond our geometric intuition. 0

0

Analytic Manifolds

243 Plane I

Plane 2

Therefore, we would like to have a more geometric picture of the Riemann surface. Consider two copies of the plane that have been slit along the negative real axis. So imagine the planes as having two negative real axes and label them * and,....., as shown in the figure. The space X we will describe will be the union of the two planes but where the two *-axes are identified and the two ,.....,-axes are identified. So if a curve in Plane 1 approaches the *-axis and hits it at -x then it exits in Plane 2 at -x on the *-axis. For the point -Ion the *-axis a typical neighborhood would consist of a half disk about - 1 above the *-axis in Plane 1 and half a disk about -1 below the *-axis in Plane 2. This is a representation of the Riemann surface of local inverses for Z2 (the Riemann surface of z for short). To see this define a map k: X -+ !Jf as follows. If z is in Plane I, z not on the negative real axis, let h(z) = (z, [gL) where g is the principal branch of the square root. If z is in Plane 2 but not on the negative real axis let h(z) = (z, [g]z) where - g is the principal branch of the square root. It remains to define h(z) for z on the *-axis and the ,.....,-axis. This we leave to the reader along with the proof that the resulting function h is an isomorphism (the space X has a natural analytic structure). rn the case fez) = z" we can carry out the same construction, but here n copies of the plane are required. If fez) = e Z the same ideas are again employed but now it is necessary to use an infinite number of planes indexed by all the integers. In the case of the surface for Zl;", a curve which passes through the negative real axis of one plane exits through the negative real axis of the next one. If it is in the l1-th plane, then it exits through the negative axis of the first plane. For the surface of log z, a curve can continue hopping from one plane to the next and will never return to the plane where it started unless it "retraces its steps".

-J

Exercises I. Show that an analytic manifold is locally compact. That is, prove that if a E X and U is an open neighborhood of a then there is an open neighborhood V of a such that V - c U and V - is compact.

244


2. Which of the following are analytic manifolds? What is its analytic structure if it is a manifold? (a) A cone in 1R3. (b) {(Xl. X2, Xl) E 1R3: xr+x~ + x~ = 1 or + x~ > I and X 3 = O} 3. The following is a generalization of Proposition 6.3(b). Let (X, 41) be an analytic manifold, let S1 be a topological space, and suppose there is a continuous function h of X onto S1 that is locally one-one (that is, if X EX there is an open set U such that x E U and h is one-one on U). If (U, ~) E cP and h is one-one on U let Ll = h( U) and let .p: Ll -?- C be defined by .p( w) = ~ 0 (h/U)-I(w). Let 'F be the collection of all such pairs (Ll, .p). Prove that (0, 0/) is an analytic manifold and h is an analytic function from X to O. 4. Let T = {z: Izl = I} x {z: Izl = I}; then T is a torus. (This torus is homeomorphic to the usual hollow doughnut in 1R 3 .) If wand w' are complex numbers such that 1m (w/w') # 0 then wand w', considered as elements of the vector space Cover IR, are linearly independent. So each z in C can be uniquely represented as z = tw+t'w'; t, t' in R Define h: C -?- Tby h(tw+ tw') = (e 2nit , e2nit '). Show that h induces an analytic structure on T. (Use Exercise 3.) (b) If w, w' and ~, {' are two pairs of complex numbers such that 1m (w/w') # 0 and 1m WO # 0, define a(s~+s'n = (e 2l1is , e 2"IS') and -r(tw+t'w') = (e 2l1it , e2 "it'). Let G = {tw+t'w': 0 < t < 1,0 < (' < I} and o = {s~+s'{':O < s < 1,0 < s' < I}; show that both aand-rareone-one on G and 0 respectively. (Both G and 0 are the interiors of parallelograms.) If <1>r and <1>", are the analytic structures induced on T by -r and a respectively, and if the identity map of (T, <1>,) into (T, <1>a) is analytic then show that the function I: G -?- 0 defined by I = a - I a -r is analytic. (To say that the identity map of (T, <1>r) into CT, <1>",) is analytic is to say that <1>r and <1>" are equivalent structures.) (c) Let w = 1, w' = i, ~ = 1, = C( where 1m ex # 0; define a, T, G, n and I as in part (b). Show that <1>r and <1>" are equivalent analytic structures if and only if C( = i. (Hint: Use the Cauchy-Riemann equations.) (d) Can you generalize part (c)? Conjecture a generalization? 5. (a) Let I be a meromorphic function defined on C and suppose I has two independent periods wand w'. That is,j(z) = ICz+nw+n'w') for all z in C and all integers nand n', and 1m (w/w') + O. Using the notation of Exercise 4(a) show that there is an analytic function F: T -?- C", such that 1= F h. (For an example of a meromorphic function with two independent periods see Exercise VIII. 3.2(g).) (b) Prove that there is no non-constant entire function with two independent periods. 6. Show that an analytic surface is arcwise connected. 7. Suppose that I: COO ....;. Coo is an analytic function. (a) Show that either I == co or 1- I( co) is a finite set. (b) If I"" co, let ai' ... , an be the points in C where I takes on the value co. Show that there are polynomials Po, PI' ... ,Pn such that

xi

r

0

I(z) = PoCz) for z in IC.

+

~Pk (;~~J

Covering Spaces

245

(c) If f is one-one, show that either f(z) fez) =

~ + b (some a,

z-c

b, c in

=

az+b (some a, b in

q.

8. Furnish the details of the discussion of the surface for this section.

9. LetG = {z: -

~

q or

";1: at the

end of

< Rez < ;}anddefinef:G--+CbYf(Z) = sinz.Give

a discussion for f similar to the discussion of ";z at the end of this section.

§7. Covering spaces In this section the concept of a covering space will be introduced and some of its elementary properties will be deduced. One byproduct of this study is the fact that two closed curves in the punctured plane are homotopic iff they have the same winding number about the origin. Intuitively, a topological space X is a covering space for the topological space n if X can be wrapped around n in such a way that it can be easily unwrapped. What is meant by "wrapping" one space around another? This seems to indicate that we want a function from X onto U. To say that it must be easily unwrapped must mean that we can find an inverse for the function. 7.1 Definition. If n is a topological space then a covering space of n is a pair (X, p) where X is a connected topological space and p is a continuous function of X onto n such that: for each w in Q there is a neighborhood Ll of w such that each component of p -I(Ll) is open, and p maps each of these components homeomorphically onto Ll. Such an open set Ll is called fundamental and Ll is properly covered. Both (C, exp) and (C - {O}, z") are covering spaces of the punctured plane; also, each open disk in the punctured plane is fundamental. If r = {z: Izl = I} and p: IR --+ r is defined by pet) = exp (21Tit), then (IR, p) is a covering space of r. Every proper arc in r is fundamental. The following is a list of properties of covering spaces. Their proof is left to the reader. 7.2 Proposition. Let (X, p) be a covering space of n. (a) p is an open mapping of X onto n. (b) If x E X then there is an open neighborhood U of x on which p is a homeomorphism. (c) Every fundamental open set is connected. (d) If Q is locally arcl1'ise connected then so is X.

In light of part (b) of the preceding proposition it is natural to ask if every locally one-one function p of X onto n makes (X, p) a covering space of Q. The answer is no. For example, let X = {z E C: z oF 0,0 < arg z < 517/4}, p(z) = Z2, and let n = p(X) = C - {O}; then p is locally one-one. Let Llr = {t: It-ij < r} for any r, b < r < l. It is easy to see that p-l(Llr) consists

246


of two components. One of these (the one in the first quadrant) is mapped homeomorphically onto ~" while the other is not. In fact, g = i is not in the image of this second component. One of the most important properties of covering spaces is the fact that a curve in n can be lifted to a curve in X. 7.3 Definition. Let (X, p) be a covering space of n and let y be a path in D. A path y in X is called a lifting of y if po Y = y. A useful way of understanding what a lifting is is to consider the following diagrams. If I = [0, 1] and the path y is given then we have the following mapping diagram:

x p

/-_.--+

n"

y

To say that y can be lifted is to say that the diagram can be completed in such a way that it is a commutative diagram:

x

that is, one can go from one place in the diagram to another without being concerned about which path is taken. It is an important property of covering spaces that every path may be lifted when the base space is locally arcwise connected. Actually a stronger result which will be of use later can be prov.ed. 7.4 Theorem. Let (X, p) be a covering space of the space Q. ifF: [0, 1] x [0, 1] + Q is a continuous function with F(O, 0) = Wo and if x 0 is any point in X with p(xo) = Wo, then there is a unique continuous function 1': [0, 1] x [0, 1] -~ X such that 1'(0, 0) = Xo and po l' = F. Before giving the proof of this theorem let us state an important corollary. 7.5 Corollary. Let (X, p) be a covering space of the space i2. If y is a path in n with initial point Wo and p(x o) = Wo, then there is a unique lifting y of y with initial point xo.

Covering Spaces

247

Proof Define F: [0, 1] x [0, 1] -?- Q by F(s, t) = y(s); so F is continuous and F(O, 0) = woo According to Theorem 7.4 there is a unique function F: [0, 1] x [0, 1] ---,.. X such that F(O, 0) = Xo and po F = F. Let yes) = F(s, 0); then y has initial point Xo and is a lifting of y. To prove the uniqueness of y, suppose iJ is also a path in X with initial point Xo and which lifts y. Define K: [0, 1] x [0, IJ -?- X by K(s, t) = O'(s). Then K: (0, 0) = Xo and p 0 K(s, t) = poO'(s) = yes) = F(s, t). By the uniqueness part of Theorem 7.4, F = K. Thus yes) = F(s, 0) = K(s, 0) = O'(s) . • Proof of Theorem 7.4. Let {O = So < S1 < .,. < Sn = I} and {O = to < < ... < In = I} be partitions of[O, 1] such that for 0::; i,j::; n - 1, F([s" si+dx [tj'

tl

t j + 1J)

is contained in a fundamental open set !!ij in Q. (Verify that this can be done.) Now Wo = F(O, O)E!!OO' Let U oo be the component p-l(!!OO) which contains Xo. Since pi U ao is a homeomorphism of UOI) onto !!oo, it is possible to define F: [0, S1] x [0, ttJ -?- X by

- t) = (p IU oo )- 1 F(s,

0

F(s, t).

Now extend F to [0, S2] x [0, ttl as follows. F({sj} x [0, tID is connected (Why?) and, since F({sdx[O, tID c !!10, it is contained in p-l(!!lO)' Let U IO be the component of p -l(!!1 0) which contains F( {S1} X [0, t I])' Then PiUI0 is a homeomorphism; define F(s, t)

= (pIUlOr- 1 F(s, t) 0

for (s, t) in [SI' S2) x [0, td. This gives a continuous function F on [0, S2) x [0, t d. (The domain can be written as the union of two closed sets, on each of these sets F is continuous, and F agrees on their intersection; hence, F is continuous on their union.) Continuing this process leads to a continuous function F: [0, 1] x [0, IJ -?- X such that po F = F and F(O, 0) = Xo. Since at each stage of this construction the definition of F is unique (because p is a homeomorphism on each Ui)' it follows that F is unique. III . The next result is called the Monodromy Theorem. To distinguish this from the theorem of the same name which was obtained in §3, the present version is referred to as the "abstract" theorem. Later it will be shown how the original theorem can be deduced from this abstract one. 7.6 Abstract Monodromy Theorem. Let (X, p) be a covering space of Q and let y and u be two paths in 12 with the same initial and final points. Let y and {j be paths in X with fhe same initial points such that y and (j are liflings of y and u respectively. If y""" u (FEP) in 12 then y and a have the same final points and y ,....." a (FEP) in X. Note. Although we have not defined the concept of FEP homotopy between two curves in an arbitrary topological space, the definition is similar to that given for curves in a region of the plane (Definition IV. 6.11'). Proof Let

Wo

and

Wi

be the initial and final points, respectively, of y and

u.


248

Let x 0 E P-1(wO) such that 1'(0) = 0'(0) = x o' By hypothesis there is a continuous function F: [0, 1] x [0, 1] -)- n such that F(O, t) = wo, F(l, t) = WI' F(8, 0) = yes), and F(s, 1) = a(s) for all sand tin [0, 1]. According to Theorem 7.4 there is a unique continuous function 1': [0, 1] x [0, 1] ---7 X such that 1'(0, 0) = Xo and poF= F. Now F({O}x[O, 1]) = {wo} and poF= F implies that F({O}x[O, 1]) c p-!(wo). But each component of p -l(wo) consists of a single point (Exercise 4) and 1'( {O} x [0, 1]) is connected. Therefore F(O, t) = x 0 for all t. Similarly, there is a point x I such that F(l, t) = Xl for all t and p(x!) = WI' By the uniqueness of y and the fact that s ---7 F(s, 0) is a path with initial point Xo \vhich lifts y, it must be that YCs) = F(s, 0). Similarly, O'(s) = F(s, 1). Therefore y(l) = 0'(1) = XI = F(I, t) for all t, and l' demonstrates that y "-' O'(FEP) in X. III In order to show that the Monodromy Theorem can be deduced from the preceding version, it is necessary to first prove a lemma. This preliminary result is actually the Monodromy Theorem for a disk and the reader is asked to supply an elementary proof. 7.7 Lemma. Lei (g. A) be a function element and let B be a disk such that A c Band (g, A) admits unrestricted analytic continuation in B. flY is a closed curve in B with i'(O) = {'(I) = a in A and {(gr' At)} is an analytic continuation of (g, A) along y then [g oJa = [g da·

The next theorem will facilitate the deduction of the Monodromy Theorem from the Abstract Monodromy Theorem. Additionally, it has some interest by itself. 7.8 Theorem. Let (f; D) be a function element which admits unrestricted continuation in the region G, and let 95 be the component of the shea/(Y(G), p) thal contains (z 0, [fLo) for some z 0 in D. Then (95, p) is a covering space of G. Proof Let B be any disk such that BeG and let oll be a component of p - 1 (B) which is contained in 95. The proof will be completed by showing that p maps ol! homeomorphically onto B.

Fix (a, [gL) in q[; then 7.9 Claim. (z, [h]z) E Yl! iff z E Band [hl: is the continuation of [gla along some curve in B. In fact, if (z, [h]z) is such a point then there is a curve y in p - I (B) (c .'l'(B) from (a, [gla) to (z, [h]z) (Theorem 5.10). Thus (z, [h].) must belong to the same component of p -l(B) as does (a, [gJa); that is, (z, [h]z) E "If. For the converse, let (z, [II],) E 0l!; since q[ is pathwise connected (Proposition 5.7) there is a path in 0/1 from (a, [g]a) to (z, [111z)' But this implies that [hL is the continuation of [g]a along a path in p(q[) c B (Theorem 5.10). So claim 7.9 has been shown. Since (I; D) admits unrestricted continuation in G and [g]a is a continuation of [fLo (zo in D), it is a trivial matter to see that [g]a admits unrestricted continuation in B. In view of (7.9) this gives that pCff) = B. It only remains to prove that piol/ is one-one. This amounts to showing

Covering Spaces

249

that if (b, [hh) and (b, [kh) e Ol/ then [hJb = [kh. But since Ol/ is arcwise connected, this is exactly the conclusion of Lemma 7.7. • Let us retain the notation of the preceding theorem. Fix a in D and let y and a be paths in G from a to a point z = b. Suppose that {(fr, Dr)} and {(gr, Br)} are continuations of (J, D) along y and a respectively; so [/oJa = [goJa = [f]a' Now yet) = (y(t), [frJy(t) and aCt) = (a(t), [grJa(t» are paths in f'(l (see the proof of Theorem 5.10) with the same initial point (a, [fJa). Moreover p 0 y = y and p 0 a = a; so y and a are the unique liftings of y and a to f'(l. According to the Abstract Monodromy Theorem, if y ,...., a(FEP) in G then y and a have the same final point. That is, (b, [/tlb) = y(l) = a(l) = (b, [glh) so that [ftlb = [gtlb' This is precisely the conclusion of Theorem 3.6. For another application of the Abstract Monodromy Theorem we wish to prove that closed curves in the punctured plane are homotopic iff they have the same winding number about the origin. To do this let r = {z: Izl = I}; as we observed at the beginning of this section, if pet) = exp (21Tit) then (IR, p) is a covering space of r. If y is any rectifiable curve in C - {O} it is easy to see that y is homotopic (in C - {O}) to the curve a defined by aCt) = y(t)/Iy(t) I (Exercise IV. 6.4). So assume that /y(t)/ = I for all t. Similarly, we can assume that yeO) = 1. Let y be the unique curve in IR such that yeO) = 0 and yet) = exp (21Tiy(t». Since y is rectifiable it is easy to see that y is also rectifiable. Also n(y; 0)

fdZ

I = -2' -

z

TTl

y

1 fdy(t) 1

= 21Ti

o

yet)

1

= _1 fd exp (21Tiy(t» 21Ti

yet)

o

1

=

f dy(t); o

so

7.10 since yeO) =

n(y, 0)

= y(l)

o.

So if a is also a closed rectifiable curve with /a(t)/ == I, a(O) = 0 = a(l) and n(y; 0) = n(a, 0) = n then a(l) = y(l) = n, where a is the unique lifting of a to IR such that a(O) = O. Let F: [0, 1] x [0, 1] -+ r be defined by

F(s, t)

= exp {2ni[(l

- t)a(s)

+ tY(s)]}

then F(O, t) = 1 = F(I, t) (Why?) and F demonstrates that y""" a.


250

Notice that the rectifiability of y and a was only used to define the winding number of y and a about the origin. It is possible to extend the definition of winding number to non-rectifiable curves. 7.11 Definition. If y is any closed curve in C- {O} with yeO) = y(1) = 1, let YI(t) = y(t)!ly(t)1 and let y be the unique curve in IR such that yeO) = 0 and YI(t) = exp (27riy(t»; the winding number ofy about the origin is

n(y; 0) = Y(1).

In view of (7.10) this definition agrees with the former definition for rectifiable curves. 7.12 Theorem. Let y and a be two closed curves in C - {O} such that a(O) yeO) = 1; then y '" a in C- {O} iffn(y; 0) = n(a; 0).

=

Proof It was shown above that if n(y; 0) = n(a; 0) then Y'" a. Conversely, if y ,......, a then the Abstract Monodromy Theorem implies that the liftings y and (j such that yeO) = (j(0) = 0 have the same end point. That is, n(y; 0) = n(a; 0) . • Exercises 1. Suppose that (X, p) is a covering space of 0 and (0, 'IT) is a covering space of Y; prove that (X, 'IT p) is also a covering space of Y. 2. Let (X, p) and (Y, a) be covering spaces of 0 and A respectively. Define pxa: Xx Y~OxA by (pxa) (x, y) = (p(x), a(y» and show that (Xx Y, p x a) is a covering space of 0 x A. 3. Let (0,1/;) be an analytic manifold and let (X, p) be a covering space of O. Show that there is an analytic structure on X such that p is an analytic function from (X, <1» to (0, 1/;). 4. Let (X, p) be a covering space of 0 and let wED. Show that each component of p -1( w) consists of a single point and p -I( w) has no limit points in 0

X.

5. Let 0 be a pathwise connected space and let (X, p) be a covering space of O. If WI and W2 are points in 0, show that p -I( WI) and p -1( w2) have the same cardinality. (Hint: Let y be a path in 0 from WI to W2; if Xl E p-l(w l ) and y is the lifting of y with initial point Xl = y(O) , letf(xI) = Y(I). Show that fis a one-one map of p-I(WI) onto p-I(W2).) 6. In this exercise all spaces are regions in the plane. (a) Let (G,j) be a covering of 0 and suppose thatfis analytic; show that if 0 is simply connected then f is one-one. (Hint: If fez 1) = f(Z2) let y be a path in G from z I to z 2 and consider a certain analytic continuation along fo y; apply the Monodromy Theorem.) (b) Suppose that (GUfl) and (G 2'/2) are coverings of the region 0 such that both fl and f2 are analytic. Show that if GI is simply connected then there is an analytic function f: G I ~ G2 such that (G 1 , f) is a covering of G2 andfl = f2 f That is the diagram is commutative. 0

Covering Spaces

251

f

G,-----------.~

G2

\1 Q

(c) Let (G 1'/1)' (Gz,/z) and 0 be as in part (b) and, in addition, assume that both G 1 and G2 are simply connected. Show that there is a one-one analytic function mapping G 1 onto G z . 7. Let G and 0 be regions in the plane and suppose that f: G -+ 0 is an analytic function such that (G,f) is a covering space of O. Show that for every region 0 1 contained in 0 which is simply connected there is an analytic function g: 0 1 -+ G such thatf(gl{w» = w for all w in 0 1 • 8. What is a simply connected covering space of the figure eight? 9. Give two nonhomeomorphic covering spaces of the figure eight that are not simply connected. 10. Prove that the closed curve in Exercise IV.6.8 is not homotopic to zero in the doubly punctured plane.

Chapter X Harmonic Functions In this chapter harmonic functions will be studied and the Dirichlet Problem will be solved. The Dirichlet Problem consists in determining all regions G such that for any continuous function f: 8G -+ IR there is a continuous function u: G- -+ IR such that u(z) = f(z) for z in 8G and u is harmonic in G. Alternately, we are asked to determine all regions G such that Laplace's Equation is solvable with arbitrary boundary values. §1. Basic properties of harmonic functions We begin by recalling the following definition and giving some examples of harmonic functions. 1.1 Definition. If G is an open subset of C then a function u: G -+ IR is harmonic if u has continuous second partial derivatives and 82 u

82 u

+ -8y2 = 0. 8x 2 This equation is called LAPLACE'S EQUATION. We also review the following facts about harmonic functions. (1) (Theorem III.2.29) A function! on a region G is analytic iff Re! = u and 1m f = v are harmonic functions which satisfy the Cauchy-Riemann equations. (2) (Theorem VIII.2.2(j» A region G is simply connected iff for each harmonic function u on G there is a harmonic function v on G such that f = u+iv is analytic on G. 1.2 Definition. Iff: G -+ C is an analytic function then u = Re! and v = Imf are called harmonic conjugates. With this terminology, Theorem VIII.2.2(j) becomes the statement that every harmonic function on a simply connected region has a harmonic conjugate. If u is a harmonic function on G and D is a disk that is contained in G then there is a harmonic function v on D such that u + iv is analytic on D. In other words, each harmonic function has a harmonic conjugate locally. Finally note that if VI and V2 are both harmonic conjugates of u then i(VI-V2)=(u+ivl)-(U+iv2) is an analytic function whose range is contained in the imaginary axis; hence VI = V2 + c, for some constant c. 1.3 Proposition. If u: G

->

R is harmonic then u is infinitely differentiable.

Proof. Fix Zo = xo+iyo in G and let 0 be chosen such that B(zo;o) c G. 252

Basic properties of harmonic functions

253

Then u has a harmonic conjugate von B(zo;o). That is,f = u+iv is analytic and hence infinitely differentiable on B(z 0 ;0). It now follows that u is infinitely differentiable. • The preceding proposition gives a property that harmonic functions share with analytic functions. The next result is the analogue of the Cauchy Integral Formula. 1.4 Mean Value Theorem. If u: G ..... ~ is a harmonic function and B(a; r) is a closed disk contained in G, then 2n

u(a)

=

-~fu(a+rei8) dB

27T

o

Proof Let D be a disk such that BCa; r) cDc G and let f be an analytic function on D such that u = Ref It is easy to deduce from Cauchy's Integral Formula that 2n

f(a)

-~ff(a+rei8)dB. 27T

=

o

By taking the real part of each side of this equation we complete the proof. • In order to study this property of harmonic functions we isolate it. 1.5 Definition. A continuous function u: G ~ ~ has the Mean Value Property (MVP) if whenever B(a;r) c G 2n

u(a)

=

~fu(a+rei8)dB.

27T

o

In the following section it will be shown that any continuous function defined on a region that has the MVP must be .a harmonic function. One of the main tools used in showing this is the following analogue of the Maximum Modulus Theorem for harmonic functions. 1.6 Maximum Principle (First Version). Let G be a region and suppose that u is a continuous real t'alued function 011 G with the MVP. If there is a point a in G such that u(a) ::::: u(z) for all z ill G then u is a constant function.

Proof Let the set A be defined by A

= {z E G: u(z) = u(a)}.

Since u is continuous the set A is closed in G. If Zo E A let r be chosen such that BCz 0 ;r) c G. Suppose there is a point b in B(z 0 ;r) such that u(b) i= u(a); then u(b) < u(a). By continuity, u(z) < u(a) = u(zo) for all z in a neighborhood of b. In particular, if p = Iz 0- bl and b = z 0+ pe ii3 , 0 ::;; f1 < 27T then

254

Harmonic Functions

f3 E I and

there is a proper interval I of [0, 217] such that for all {} in I. Hence, by the MVP

u(z 0 + pew) < u(z 0)

2n

uCza) =

~fu(zo+pei8)d{} 211"

< u(zo),

o

a contradiction. So B(z 0 ;r) c A and A is also open. By the connectedness of G, A = G.II 1.7 Maximum Principle (Second Version). Let G be a region and let u and v be two bounded continuous real valued junctions on G that have the MVP. Ij jar each point a in the extended boundary a~p, limsupu(z)~ ::----.a

liminft:{z) ;:-----'f>il

then either u(z) 0 let Gb to the hypothesis,

o~ =

G n B(a;8). Then according

=

Iim[sup{u{z):zEGJ}-inf{v(z):ZEG~}] 54 0

lim [sup {u(z) : Z

E

G,d + sup{ - v(z): z E G,d]

a~o

~

lim sup {u(z)-v(z): Z

E

G,d.

6~O

So lim sup [u(z) - vez)]

~

0 for each a in 0ooG. So it is sufficient to prove the

theorem under the assumption that v(z) = 0 for all z in G. That is, assume

1.8

lim sup u(z) :s; 0

for all a in oooG and show that either u(z) O. Let f>O be chosen so that u(b»f: and let B={z E G:u(z)~t:}. If a E oooG then (1.8) implies there is a ()={)(a) such that u(z)<€ for all z in G (\ B(a; c5). Using the Lebesgue Covering Lemma, a c5 can be found that is independent of a. That is, there is a {j > 0 such that if z E G and d (z, iYoo G) < (j then u(z) < t:. Thus, Be {z

E

G: d(z, o,,,,G)

~

S}.

This gives that B is bounded in the plane; since B is clearly closed, It IS compact. So if B -# 0, there is a point Zo in B such that u(zo) ~ u(z) for all z in B. Since u(z) < € for z in G - B, this gives that u assumes a maximum value at a point in G. So u must be constant. But this constant must be u(zo) which is positive and this contradicts (1.8). II

Basic properties of harmonic functions

255

The following corollary is a useful special case of the Maximum Principle. 1.9 Corollary. Let G be a bounded region and suppose that w: G ... ~ IR is a continuous function that satisfies the MVP on G. If w(z}=O for all z in dG then w(z}=Ofor all z in G.

Proof First take w = u and v = 0 in Theorem 1.7. So w(z) < 0 for all z or w(z) == O. Now take w = v and II = 0 in (1.7); so either w(z) > 0 for ali z or w(z) == O. Since both of these hold, w == O. • Even though Theorem 1.7 is called the Maximum Principle, it is also a Minimum Principle. For the sake of completeness, a Minimum Principle corresponding to Theorem 1.6 is stated below. It can be proved either by appealing to (1.7) or by considering the function -u and appealing to (1.6).

1.10 Minimum Principle. Let G be a region and suppose that u is a continuous real valued function on G with the MYP. If there is a point a in G such that u(a) ::; u(z) for all z in G then u is a constant function. Exercises

1. Show that if u is harmonic then so are 2. If u is harmonic, show that f 3. Let p(x, y) =

Ux -

= U x - iUy

ou and ox

ou

uy = -;::-. oy

is analytic.

L" ak/xki for all x, y in IR.

k.l; 0

Show that p is harmonic iff: (a) k(k-l)ak,I-2 +1(l-!)ak-2,1 = 0 for 2 ::; k, I ::; n; (b) an-I,I = an,l = 0 for 2::; I ::; n; (c) a k ,n-l = ak,n = 0 for 2 ::; k ::; n.

4. Prove that a nonconstant harmonic function on a region is an open map. (Hint: Use the fact that the connected subsets of IR are intervals.) 5. Iffis analytic on G andf(z) #. 0 for any z show that u = log/fl is harmonic on G. 6. Let u be harmonic in G and suppose B(a ;R) c G. Show that u(a) =

7. For

Izl

;12 JJ

u(x, y) dx dy.

B(a;R)

< 1 let u(z)

=

1m

[G~~YJ

Show that u is harmonic and lim u(re i9 ) = 0 for all O. Does this violate Y-+l-

Theorem 1.7? Why? 8. Let u:G -'>- IR be a function with continuous second partial derivatives and define UCr, 0) = u(r cos 0, r sin 0). (a) Show that

256

So if 0 ¢ G then u is harmonic iff 8

(8U) + 88£]2U= o. 2

r iir r a;.

(b) Let u have the property that it depends only on Izl and not arg z. That is, u(z) = 9'(lzl). Show that u is harmonic iff u(z) = a log Izl +b for some constants a and b. 9. Let u:G --* IR be harmonic and let A = {z E G: u,,(z) = u,(z) = O}; that is, A is the set of zeros of the gradient of u. Can A have a limit point in G? 10. State and prove a Schwarz Reflection Principle for harmonic functions. II. Deduce the Maximum Principle for analytic functions from Theorem 1.6.

§2. Harmonic functions on a disk Before studying harmonic functions in the large it is necessary to study them locally. That is, we must study these functions on disks. The plan is to study harmonic functions on the open unit disk {z:lzl < I} and then interpret the results for arbitrary disks. Of basic importance is the Poisson kernel. 2.1 Definition. The function Pr(O)

=

L 00

n=

rlnleino,

-00

for 0 ::; r < 1 and - 00 < 0 < 00, is called the Poisson kernel. Let z == re iO , 0 ::; r < 1; then 1 +re iO -1--;0 = (1 +z)(1 +Z+Z2 + ...) -re

00

=

1 +2 2: r"e inO n=1

Hence, 00 1 +reiO) Re ( --iO = 1 +2 2: r" cos nO l-re n=1

=

1+

2: r"(einO+e-inO) 00

n= 1

Harmonic functions on a disk

1+re i6 l-re,e

Also - - . =

257

1+rei8_re-iO_r2 16 2

Il-re I

2.2

P(O) r

-

so that

(1

iB ) 1-r2 +re - Re 1-2r cos 8+r2 l-re i8

2.3 Proposition. The Poisson kernel satisfies the following:

(a)

~ 217

f "

P,(O) de = 1;

(b) Pr(8) > 0 for ai,

e, Pre -8) = P,(8), and P

r

is periodic in 8 with period

217; (c) Pr(O) < PreS) if 0 < S < 101 :::;; 17; (d) for each S > 0, lim p,(e) = 0 uniformly in e for 17 ;::: lei;:::

s.

Proof. (a) For a fixed value of r, 0 :::;; r < 1, the series (2.1) converges uniformly in O. So

-"

-"

(b) From equation (2.2), P.(O) = (l_r 2 )11_re i8 1- 2 > 0 since r < 1. The rest of (b) is an equally trivial consequence of (2.2). (c) Let 0 < S < e : :; 17 and define f: [S, e] -+ IR by f(t) = PrCt). Using (2.2), a routine calculation shows that F(t) < 0 so that f(S) > fee). (d) We must show that lim [sup{P,(8):S :::;; 101:::;; 17}] = 0

,. ...... 1-

But according to part (c), p,(e) :::;; P,(S) if 3 :::;; 181 :::;; 17; so it suffices to show that lim PreS) = O. But, again, this is a trivial consequence of equation (2.2).• ,.-.1-

Before going to the applications of the Poisson kernel, the reader should take time to consider the significance of Proposition 2.3. Think of P,(fJ) not as a function of rand e but as a family of functions of e, indexed by r. As r approaches 1 these functions converge to zero uniformly on any closed subinterval of [-17, 17] which does not contain e = 0 (part d). Nevertheless, part (a) is still valid. So as r approaches 1, the graph of P, becomes closer to the 8 axis for 8 away from zero but rises sharply near zero so that 2.3(a) is maintained. The next theorem states that the Dirichlet Problem can be solved for the unit disk.

2.4 Theorem. Let D = {z: Izi < I} and suppose that f: aD -+ IR is a continuous function. Then there is a continuous function u:D - ~ IR such that (a) u(z) = f(z) for z in aD; (b) u is harmonic in D.

258

Harmonic Functions

Moreover u is unique and is defined by the formula

2.5 for 0

u(re''0 ) $

r < I, 0

$

f)

$

"

ft)f(e't) ' dt = - IP,(f)27T

27T.

Proof Define u:D ~ IR by letting u(re iO ) be as in (2.5) if 0 $ r < 1 and letting u(e iO ) = f(e io ). Clearly u satisfies part (a); it remains to show that u is continuous on D- and harmonic in D. (i) u is harmonic in D. If 0 $ r < 1 then

, 1f" Re [1 +rei(O-t)] " "(O-t) f(e't) dt

u(re'o) = 27T

I-re'

1t

{ If

,[I

+rei(O-t)] } = Re 27T f(e") 1- rei(O-t) dt . -1t

{If " 1t

it

iO

= Re 27T fCe't) [eeit +re _ reiO ] dt } . So define g:D

~

IC by 1t

I ff(e't) ' [eit+z] eit _ z dt.

g(z) = 27T

-" Since u = Re g we need only show that g is analytic. But this is an easy consequence of Exercise IV.2.2. (ii) u is continuous on D-. Since u is harmonic on D it only remains to show that u is continuous at each point of the boundary of D. To accomplish this we make the following

2.6 Claim. Given IX in [ -7T, 7T] and E" > 0 there is a p, 0 < P < I, and an arc A of aD about e iot such that for p < r < 1 and e iO in A, [u(re iO ) - f(e iot)[ <

E"

Once claim 2.6 is proved the continuity of u at e iot is immediate since f is a continuous function. To avoid certain notational difficulties, the claim will only be proved for IX = O. (The general case can be obtained from this one by an argument which involves a rotation of the variables.) Since f is continuous at z = I there is a 0 > 0 such that 2.7

<

1

-E"

3


259

if 101 < o. Let M = max {If(eiB)I: 101 ::; 7T}; from Proposition 2.3 (d) there is a number p, 0 < P < 1, such that €

2.8

Pr(O) < 3M

for p < r < 1 and 101 ~ and p < r < 1,

u(reiB)-f(l) =

~ 27T

= -1

27T

"

f

oand 101

Let A be the arc {e iB : 101 < 10}. Then if e iB

"

f

E

A

P.(O-t)f(e it) dt-f(1)

-"

III
Ifltl ~

-to.

P.(O-t)[f(elt)-f(l)]dt+ -1 27T

f P (O-t)[f(e,t)-f(1)]dt. " r

1II:O:B

-is; so from (2.7) and (2.8) it follows that

::; 10thenlt-01 ~

lu(re iB )-f(1)1 ::;

~ €+2M(3~) =

€.

This proves Claim 2.6. Finally, to show that u is unique, suppose that v is a continuous function on D- which is harmonic on D and v(e iB ) = f(e iB ) for all O. Then u-v is harmonic in D and (u-v) (z) = 0 for all z in oD. It follows from Corollary 1.9 that u-v == 0. • 2.9 Corollary. IJ u : D then

~R

is a continuous Junction that is harmonic in D

u(re'"B) = - 1 27T

f "

"

Pr(O-t)u(e't) dt

-"

for 0

:$

r < 1 and all O. Moreover, u is the real part of the analytic function f(z)

1 7T

= -2

f" eit+z u(e't)" dt.

-"

-i/-

e-z

Proof The first part of the corollary is a direct consequence of the theorem. The second part follows from the fact thatfis an analytic function (Exercise IV.2.2) and formula (2.2) . • 2.10 Corollary. Let a E C, p > 0, and suppose h is a continuous real valued function on {z: Iz - al = p}; then there is a unique continuous function w:B(a; p) -+IR such that w is harmonic on B(a; p) and w(z) = h(z) for Iz-al = p.

Proof Considerf(e iB ) = h(a+ pe i8 ); thenfis continuous on oD. Ifu:D- -+ IR is a continuous function such that u is harmonic in D and u(e iB) = f(e i8 ) then

Harmonic Functions

260

it is an easy matter to show that w(z)

=

u(

z~a) is the desired function on

B(a; p) . • It is now possible to give the promised converse to the Mean Value

Theorem. 2.11 Theorem. If u: G -r III is a continuous function which has the MVP then u is harmonic. Proof Let a E G and choose p such that B(a; p) c G; it is sufficient to show that u is harmonic on B(a; p). But according to Corollary 2.10 there is a continuous function w: B(a; p) -rill which is harmonic in B(a; p) and w(a+pe i9 ) = u(a+pe i9 ) for all 8. Since u-w satisfies the MVP and (u-w)(z) = 0 for Iz-al = p, it follows from Corollary 1.9 that u == w in B(a; p); in particular, u must be harmonic. • To prove the above theorem we used Corollary 2.10, which concerns functions harmonic in an arbitrary disk. It is desirable to derive a formula for the Poisson kernel of an arbitrary disk; to do this one need only make a change of variables in the formula (2.2). If R > 0 then substituting rlR for r in the middle of (2.2) gives R2_r2 .. _ - R2 -2rR cos 8+r2

2.12

~ r < R and all B. So if u is continuous on B(a; R) and harmonic in B(a; R) then

for 0

2.13

u(a+re i9 )

I =-21T

f[ "

2

R -

2

R2_r2 ] (8) 2 u(a+Re it ) dt - t +r

r R cos

Now (2.12) can also be written R2 _

r2

IR - re i9 12

and R-r ~ IRe it -re i8 ~ R+r. Therefore 1

R-r

R2 _r2

R+r

--<--~--.

R+r - R2-2rR cos (8-t)+r 2

R-r

If u :2: 0 then equation (2.13) yields the following. 2.14 Harnack's Inequality. If u: B(a; R) -r III is continuous, harmonic in B(a; R), and u :2: 0 then for 0 ~ r < Rand allB R-r . R+r - - u(a) ~ u(a+re I8 ) ~ - - u(a) R+r R-r

Before proceeding, the reader is advised to review the relevant definitions and properties of the metric space C(G, Ill) (Section VII.l).


261

2.15 Definition. If G is an open subset of IC then Har( G) is the space of harmonic functions on G. Since Har( G) c C (G, IR) it is given the metric that it inherits from C(G,IR). 2.16 Harnack's Theorem. Let G be a region. (a) The metric space Har(G) is complete. (b) If {un} is a sequence in Har(G) such that U 1 ::; U 2 ::; ••• then either un(z) ---+ 00 uniformly on compact subsets oj G or {un} converges in Har(G) to a harmonic junction. Proof (a) To show that Har(G) is complete, it is sufficient to show that it is a closed subspace of C(G, IR). So let {un} be a sequence in Har(G) such that Un ---+ U in C(G, IR}. Then (Lemma IV.2.7) it follows that u has the MVP and so, by Theorem 2.11, u must be harmonic. Let u(z) (b) We may assume that U 1 2: 0 (if not, consider {un-u 1 = sup {u.(z): n 2: I} for each z in G. So for each z in G one of two possibilities occurs: u(z) = 00 or u(z) E IR and u.(z) ---+ u(z). Define A = {z E G: u(z) = co}

n.

B

= {ZEG:U(Z) < co};

then G = A u B and A n B = O. We will show that both A and B are open. If a E G, let R be chosen such that D(a; R) c G. By Harnack's inequality 2.17 for all Z in B(a; R) and all n 2: 1. If a E A then u.(a) ---+ 00 so that the left half of (2.17) gives that un(z) ---+ 00 for all Z in B(a; R). That is, B(a; R) c A and so A is open. In a similar fashion, if a E B then the right half of (2.17) gives that u(z) < co for Jz-aJ < R. That is B is open. Since G is connected, either A = G or B = G. Suppose A = G; that is U == 00. Again if D(a; R) c G and 0 0 and (2.17) gives that M un(a) :s; un(z) for Jz-aJ ::; p. Hence un(z) ---+ 00 uniformly for z in D(a; pl. In other words, we have shown that for each a in G there is a p > 0 such that un(z) ---+ 00 uniformly for Jz-aJ ::; p. From this it is easy to deduce that u.(z) ---+ 00 uniformly for z in any compact set. Now suppose B = G, or that u(z) < co for all z in G. If p < R, then for m :$; n Harnack's Inequality applied to the positive harmonic function U. - U m implies there is a constant C depending only on p and R such that O:s; u.(z) - um(z) :::;; C[u.(a) - um(a)]

for Iz - al :::;; p. Thus, {un(z)} is a uniformly Cauchy sequence on R(a; p). It follows that {un} is a Cauchy sequence in Har(G) and so, by part (a), must converge to a harmonic function. Since un(z) ---+ u(z), u is this harmonic function . • It is possible to give ·alternate proofs of Harnack's Theorem. One involves

262

Harmonic Functions

applying Dini's Theorem (Exercise VII.I.6). Another involves using the Monotone Convergence Theorem from measure theory to obtain that u has the MVP. However, both these approaches necessitate proving that the function u is continuous. This is rather easy to accomplish by appealing to (2.17) and the fact that uiz) -+ u(z) for all z; these facts imply that R-Iz-al R+ Iz-al I I u(a) I I u(a) ~ u(z) ~ R-z-a R+z-a Hence -2Iz-al 2lz-al u(a) ~ u(z)-u(a) ~ I I u(a); I I R+ z-a R- z-a or lu(z) - u(a)1 ~

2 Iz-al I I u(a) R- z-a

So as z -+ a, it is clear that u(z) -+ u(a).

Exercises 1. Let D = {z: Izi < I} and suppose thatf: D- -+ C is a continuous function such that both Refand Imfare harmonic. Show that

f(re i6 ) =

"

~f f(eit)P,(B-t) dt 21T -"

for all re i6 in D. Using Definition 2.1 show thatfis analytic on D iff

f" f(eit)eint dt = 0

-"

for all n ~ l. 2. In the statement of Theorem 2.4 suppose that f is piecewise continuous on aD. Is the conclusion of the theorem still valid? If not, what parts of the conclusion remain true? 3. Let D = {z:lzl < I}, T = aD = {z:lzl = I} (a) Show that if g:D- -+ C is a continuous function and g,:T -+ C is defined by g,{z) = g(rz) then g,(z) -+ g(z) uniformly for z in T as r -+ 1-. (b) If f:T -+ C is a continuous function define j:D- -+ C by](z) = fez) for z in Tand

j(re i8 ) =

2~

f "

-"

f(eit)Pr{fj - t) dt

(So Rej and Imj are harmonic in D). Define.i,.: T -C by .i,.(z) = j(rz). Show that for each r< I there is a sequence {Pn(z,Z)} of polynomials in z and i such thatpn(z,Z)-.i,.(z) uniformly for z in T. (Hint: Use Definition 2.1.)

Subharmonic and superharmonic functions

263

(c) Weierstrass approximation theorem for T. Iff: T ~C is a continuous function then there is a sequence {Pn(z,Z)} of polynomials in z and i such thatpn(z,z)~f(z) uniformly for z in T. (d) Suppose g:[O, l]~C is a continuous function such that g(O)=g(l). Use part (c) to show that there is a sequence {Pn} of polynomials such that Pn(t)~g(t) uniformly for t in [0, I]. (e) Weierstrass approximation theorem for [0,1]. If g: [0, I]~C is a continuous function then there is a sequence {Pn} of polynomials such that Pn(t)~g(t) uniformly for t in [0,1]. (Hint: Apply part (d) to the function get) + (1- t) g(l) + tg(O).) (f) Show that if the function g in part (e) is real valued then the polynomials can be chosen with real coefficients. 4. Let G be a simply connected region and let r be its closure in Coo; 000G = r-G. Suppose there is a· homeomorphism cp of r onto D-(D {z: /z/ < I}) such that cp is analytic on G. (a) Show that cp(G) = D and cp(oooG) = aD. (b) Show that if f: 000G ~ IR is a continuous function then there is a continuous function u: r ~ IR such that u(z) = fez) for z in 000G and u is harmonic in G. (c) Suppose that the function f in part (b) is not assumed to be continuous at 00. Show that there is a continuous function u:G- ~ IR such that u(z) = fez) for z in fJG and u is harmonic in G (see Exercise 2). 5. Let G be an open set, a E G, and Go = G - {a}. Suppose that u is a harmonic function on Go such that lim u(z) exists and is equal to A. Show that

*

z .... a

if U:G ~ IR is defined by/U(z) = u{z) for z a and U(a) = A then U is harmonic on G. 6. Let f: {z:Re z = O} ~ IR be a bounded continuous function and define u: {z:Re z>O} ~ IR by 00 . 1 xf(it) U(X+lY) = 2 ( )2 dt. 7T x + y-t -00 Show that u is a bounded harmonic function on the right half plane such that for c in 1R,J(ic) = lim u(z).

f

z-+ic

7. Let D= {z : /z/ < I} and suppose f: aD~1I! is continuous except for a jump discontinuity at z = I. Define u: D~IR by (2.5). Show that u is harmonic. Let v be a harmonic conjugate of u. What can you say about the behavior of vCr) as r~I -? What about v(re iO ) as r~I - and f}~O? §3. Subharmonic and superharmonic functions In order to solve the Dirichlet Problem generalizations of harmonic functions are introduced. According to Theorem 2.11, a function is harmonic exactly when it has the MVP. With this in mind, the choice of terminology in the next definition becomes appropriate.

264

Harmonic Functions

3.1 Definition. Let G be a region and let cp: G -+ IR be a continuous function. cp is a subharmonic function if whenever B(a; r) c G,

Lf

2"

cp(a) :s;

cp(a+re iB ) dB.

o

cp is a superharmonic function if whenever B(a; r)

c

G,

2"

cp(a)

~ 21T ~f cp(a+re

iB )

dB.

o

The first comment that should be made is that cp is super harmonic iff - cp is subharmonic. Because of this, only the results on subharmonic functions will be given and it will be left to the reader to state the analogous result for superharmonic functions. Nevertheless, we will often quote results on superharmonic functions as though they had been stated in detail. In the definition of a subharmonic function cp it is possible to assume only that cp is upper semi-continuous. However this would make it necessary to use the Lebesgue Integral in the definition instead of the Riemann Integral. So it is assumed that cp is continuous when cp is subharmonic even though there are certain technical advantages that accrue if only upper semicontinuity is assumed. Clearly every harmonic function is subharmonic as well as superharmonic. In fact, according to Theorem 2.11, u is harmonic iff u is both subharmonic and superharmonic. If CP! and CP2 are subharmonic then so is a!CP! + a 2CP2 ifa!, a2 ~ o. It is interesting to see which of the results on harmonic functions also hold for subharmonic functions. One of the most important of these is the Maximum Principle. 3.2 Maximum Principle (Third Version). Let G be a region and let cp: G -+ IR be a subharmonic function. If there is a point a in G with cp(a) ~ cp(z) for all z in G then cp is a constant function. The proof is the same as the proof of the first version of the Maximum Principle. (Notice that only the Minimum Principle holds for superharmonic functions.) The second version of the Maximum Principle can also be extended, but here both subharmonic and superharmonic functions must be used. 3.3 Maximum Principle (Fourth Version). Let G be a region and let cp and if; be bounded real valued functions defined on G such that cp is subharmonic and '" is superharmonic. If for each point a in a
then either cp(z) < if;(z)for all z in G or cp

= if; and cp is harmonic.


265

Again, the proof is identical to that of Theorem 1.7 and will not be repeated here. Notice that we have not excluded the possibility that a subharmonic function may assume a minimum value. Indeed, this does happen. For example, cp(x, Y) = x 2 + y2 is a subharmonic function and it assumes a minimum at the origin. This failure of the Minimum Principle is due to the fact that if u has the MVP then so does -u; however, if rp is subharmonic then -rp is never subharmonic unless it is harmonic. When we say that a function satisfies the Maximum Principle, we refer to the third version. That is, we suppose that it does not assume a maximum value in G unless it is constant.

3.4 Theorem. Let G be a region and rp: G - IR a continuous function. Then rp is subharmonic ifffor every region GI contained in G and every harmonic function UI on GI , rp-u l satisfies the Maximum Principle on G1 •

Proof Suppose that rp is subharmonic and G1 and Ul are as in the statement of the theorem. Then rp-Ul is clearly subharmonic and must satisfy the Maximum Principle. N ow suppose cp is continuous and has the stated property; let Ji( a, r) C G. According to Corollary 2.10 there is a continuous function u: Ji(a, r) ~ IR which is harmonic in B(a; r) and u(z) = cp(z) for Iz - al = r. By hypothesis, cp - u satisfies the Maximum Principle. But (cp - u)( z) 0 for Iz - al = r. So cp ~ u and

=

f -~ f

cp(a)~u(a)=

2"

21w

u(a+reiO)d(J

o

2n

- 2w

cp(a

+ reiO)

d(J.

o

Therefore rp is subharmonic. • 3.5 Corollary. Let G be a region and cp: G~ IR a continuous function; then cp is subharmonic iff for every bounded region G t such that Gt- c G and for every continuous function U t : G t- ~IR that is harmonic in Gt and satisfies cp(z):s ut(z) for z on aGt,cp(z)~ ut(z) for z in G t •

3.6 Corollary. Let G be a region and rpi and rp2 subharmonic functions on G; if rp(z) = max {rpl(Z), rp2(Z)} for each z in G then rp is a subharmonic function. Proof Let G1 be a region such that G1 c G and let U1 be a continuous function on G1- which is harmonic on G I with rp(z) :S u1(z) for all z in aG I • Then both rpl(Z) and rp2(Z) :S u1(z) on aGI' From Corollary 3.5 we get that rpl(Z) and rpiz) :S ul(z) for all z in G 1 • So cp(z) :S u1(z) for z in Gl> and, again by Corollary 3.5, rp is subharmonic. •

266 Harmonic Functions 3.7 Corollary. Let rp be a subharmonic function on a region G and let B(a; r) c G. Let rp' be the function defined on G by:

(i) rp'(z) = rp(z) if z E G-B(a; r); (ii) rp' is the continuous function on B(a; r) which is harmonic in B(a; r) and agrees with rp(z) for [z - a[ = r. Then rp' is subharmonic. The proof is left to the reader. As was mentioned at the beginning of this section, one of the purposes in studying subharmonic functions is that they enter into the solution of the Dirichlet Problem. Indeed, the fourth version of the Maximum Principle gives an insight into how this occurs. If G is a region and u: G- --+ IR is a continuous function (G- = the closure in 1C
3.8 u(z) = sup {rp(z): rp is subharmonic and lim suprp(z) :-::; u(a) for allain O
3.10 for each z in G. Conversely, if f is given and u is defined by (3.10) then u must be the solution of the Dirichlet Problem with boundary values f; that is, provided the Dirichlet Problem can be solved. In order to show that (3.10) is a solution two questions must be answered affirmatively. (a) Is u harmonic in G? (b) Does lim u(z) = f(a) for each a in 0ooG? z--+a


267

The first question can always be answered "Yes" and this is shown in the next theorem. The second question sometimes has a negative answer and an example will be given which demonstrates this. However, it is possible to impose geometrical restrictions on G which guarantee that the answer to (b) is always yes for any continuous functionf This will be done in the next section. 3.11 Theorem. Let G be a region and/:oooG -* ~ a continuous/unction; then u(z) = sup {cp(z):cP E &(J, G)} defines a harmonic/unction u on G. Proof Let //(a)/ ::; M for all a E 0ooG. The proof begins by noting that

3.12

cp(z) ::; M for all z in G, cP in &(J, G)

This follows because, by definition, lim sup cp(z) ::; M whenever cp E &(J, G); so (3.12) is a direct consequence of the Maximum Principle. Fix a in G and let R(a; r) c G. Then u(a) = sup {cp(a):cp E &(J, G)}; so there is a sequence {CPn} in &(f, G) such that u(a) = lim cpnCa). Let n = max {CPI' •.. ,CPn}; by Corollary 3.6 n is subharmonic. Let ~ be the subharmonic function on G such that ~(z) = n(z) for z in G-B(a; r) and ~ is harmonic on B(a; r) (Corollary 3.7). It is left to the reader to verify the following statements:

3.13

~ ~ ~+ I

3.14

CPn ::; n ::; ~

3.15

~

E

&(J, G).

Because of (3.15), ~(a) ::; u(a); from (3.14) and the choice of {CPn}, this gives that 3.16

u(a) = lim ~(a).

Moreover, statement (3.12) gives that ~ ::; M for all n; so using (3.13), Harnack's Theorem implies that there is a harmonic function U on B(a; r) such that UCz) = lim ~(z) uniformly for z in any proper subdisk of B(a; r). It follows from (3.15) and (3.16) that U ::; u and U(a) = u(a), respectively. Now let Zo E B(a; r) and let Nn} be a sequence in &(J, G) such that u(zo) = lim .fn(ZO). Let Xn = max {CPn, .fn}, Xn = max {Xl' ... , Xn}, and let X~ be the subharmonic function which agrees with Xn off B(a; r) and is harmonic in B(a; r). As above, this leads to a harmonic function U 0 on BCa; r) such that U o ::; u and Uo(zo) = u(zo). But n::; Xn so that ~::; X~. Hence U ::; U 0 ::; u and U(a) = U o(a) = u(a). Therefore U - U 0 is a negative harmonic function on BCa; r) and (U- Uo)(a) = O. By the Maximum Principle, U = U 0; so U(z 0) = u(z 0). Since z 0 was arbitrary, U = U in B(a; r). That is, U is harmonic on every disk contained in G. • 3.17 Definition. Let G be a region and let /: 0eeP

-* ~

be a continuous

268

Harmonic Functions

function. The harmonic function u obtained in the preceding theorem is called the Perron Function associated with f The next step in solving the Dirichlet Problem is to prove that for each point a in 000G lim u(z) exists and equalsj(a). As was mentioned earlier, this does not always hold. The following example illustrates this phenomenon. Let G = {z:O < Izl < I}, T= {z:lzl = I}; so oG = Tu {O}. Define I: oG --+ ~ by /(z) = 0 if z E T and 1(0) = 1. For 0 < E 0 for z in G, u.(z) = 0 for z in T, and u.(z) = I if Izl = Eo Suppose that v E 9(j, G); since III :$ I, v(z):::; 1 for all z in G. If R. = {z: E < \z\ < I} then lim sup v(z):::; u.(a) for all a in oR.; by the Maximum Principle, v(z) :$ u.(z) for all z in R.. Since € was arbitrary this gives that for each z in G, v(z) :$ lim u.(z) = O. Hence the Perron function associated with I is the identically zero function, and the Dirichlet Problem cannot be solved for the punctured disk. (Another proof of this is available by using Exercise 2.5 and the Maximum Principle.)

Exercises 1. Which of the following functions are subharmonic? superharmonic? harmonic? neither subharmonic nor superharmonic? (a) cp(x, y) = x 2 + y2; (b) cp(x, y) = x 2 _y2; (c) cp(x, y) = X2+y; (d) cp(x, y) = x 2 _y; (e) cp(x, y) = X+y2; (f) cp(x, y) = x_y2. 2. Let Subhar(G) and Superhar(G) denote, respectively, the sets of subharmonic and superharmonic functions on G. (a) Show that Subhar(G) and Superhar(G) are closed subsets of C(G; ~). (b) Does a version of Harnack's Theorem hold for subharmonic and superharmonic functions? 3. (This exercise is difficult.) If G is a region and if f: 8",p ----+ ~ is a continuous function let uf be the Perron Function associated with f. This defines a map T: C( 8(YeP; IR) ----+ Har( G) by TU) = uf' Prove: (a) T is linear (i.e., T(aJl + a 2f2) = alTUl) + a 2T(2»' (b) T is positive (i.e., ifj(a) :::: 0 for all a in oooG then T(f)(z) :::: 0 for all z in G). (c) T is continuous. Moreover, if {/,,} is a sequence in C(0ooG; ~) such thatf,. --+1 uniformly then T(f,.) --+ T(f) uniformly on G. (d) If the Dirichlet Problem can be solved for G then T is one-one. Is the converse true? 4. In the hypothesis of Theorem 3.11, suppose only that I is a bounded function on oooG; prove that the conclusion remains valid. (This is useful if G is an unbounded region and g is a bounded continuous function on oG. If we define I: o",G --+ IR by I(z) = g(z) for z in oG and I( 00) = 0 then the conclusion of Theorem 3.11 remains valid. Of course there is no reason to expect that the harmonic function will have predictable behavior near 00-

The Dirichlet Problem

269

we could have assigned any value to J( 00). However, the behavior near points of oG can be studied with hope of success.) 5. Show that the requirement that G1 is bounded in Corollary 3.5 is necessary. 6. If f: G ~ n is analytic and cp: n ~ R is subharmonic, show that cp f is subharmonic if f is one-one. What happens if f '( z) "* 0 for all z in G? 0

§4. The Dirichlet Problem

4.1 Definition. A region G is called a Dirichlet Region if the Dirichlet Problem can be solved for G. That is, G is a Dirichlet Region if for each continuous functionJ: o O. 4.2 Definition. Let G be a region and let a E o O} of functions such that: (a) t/I, is defined and superharmonic on G (a; r) with 0 $ t/I,(z) $ I; (b) lim rp,(z) = 0; (c) lim rp,(z) z-w

=

1 for win G ("\ {w:lw-al

= r}.

The following observation is useful: if~, is defined by letting~, = rp, on G(a; r) and ~,(z) = 1 for z in G-B(a; r), then~, is superharmonic. (Verify!) So the functions ~ "approach" the function which is one everywhere but z = a, where it is zero. The second observation which must be made is that if G is a Dirichlet Region then there is a barrier for G at each point of o O. Define rP,:G(a; r) -+ R by rP,(z)

= ~ min {u(z), c,}.1t is left to c,

the reader to

check that {rp,} is a barrier. The next result provides a converse to the above facts. 4.3 Theorem. Let G be a region and let a E aco G such that there is a barrier Jor at a. If J: 0coG -+ R is continuous and u is the Perron Function associated

(i

Harmonic Functions

270

withf then

lim u(z)

=

z-o

f(a)

Proof Let {",,: r > O} be a barrier for G at a and for convenience assume a # 00; also assume thatf(a) = 0 (otherwise consider the function f-fCa». Let € > 0 and choose 0 > 0 such that !f(w)! < E whenever WE 0aP and [w-a[ < 28; let", = "'~. Let rji:G -'>-IR be defined by rji(z) = ",(z) for z in G(a; 0) and rji(z) = 1 for z in G-B(a; 0). Then rji is superharmonic. If !f(w)! ::;; M for all win o",G, then -Mrji'-E is subharmonic. 4.4 Claim. - Mrji - € is in f!jJ(f, G). If WE 0ooG-B(a; 0) then lim sup [-Mrji(z)-e] = -M-£
rji(z) ~ 0, it follows that lim sup [-Mrji(Z)-E] ::;; -e for all win 0coG. In

particular, if w E 000G

(1

BCa; 0) then lim sup [-Mrji(Z)-E] ::;;

--to

< few) by

the choice of O. This substantiates Claim 4.4. Hence

4.5

·-Mrji(z)-e ::;; u(z)

for all z in G. A similar analysis yields lim inf [Mrji(z)+€] ~ lim sup
or, combining this with (4.5),

4.6

-Mrji(Z)-E ::;; u(z)

:0;

Mrji(Z)+E

for all z in G. But limrji(z) = lim ",(z) = 0; since e was arbitrary, (4.6) gives that z-a z~a lim u(z) =

°

=

f(a).

This completes the proof. II Notice that the purpose of the barrier was to construct the function rji which "squeezed" u down to zero. 4.7 Corollary. A region G is a Dirichlet Region iff there is a barrier for G at each point of 0ooG. The above corollary is not the solution to the problem of characterizing Dirichlet Regions. True, it gives a necessary and sufficient condition that a region be a Dirichlet Region and this condition is formally weaker than the definition. However, there are aesthetic and practical difficulties with Corollary 4.7. One difficulty is that the condition in (4.7) is not easily verified.


271

Another difficulty is that it is essentially the same type of condition as the definition; both hypothesize the existence of functions with prescribed boundary behavior. What is desired? Both tradition and aesthetics dictate that we strive for a topological-geometric condition on G which is necessary and sufficient that G be a Dirichlet Region. Such conditions are usually easy to verify for a given region, and the equivalence of a geometric property and an analytic one is the type of beauty after which most mathematicians strive. At the present time no such equivalence is known and we must be content with sufficient conditions. 4.8 Lemma. Let G be a region in C and let S be a closed connected subset of

Ceo - G such that 00 E Sand S n oooG = {a}. If Go is the component of - S that contains G then Go is a simp(v connected region in the plane.

C oc

Proof Let Go, G 1 , ••• be the components of Coo - S with G c Go; note that each Go is a region in Co If z E 0ooG. then G. u {z} is connected (Exercise II.2.1). Since Gn is a component it follows that 000Gn c S. By Lemma

U (G. u 00

n.2.6 Gn u S ( = G;; uS) is connected and, consequently, so is

S)

n~l

=

Coo - Go· In virtue of Theorem VIII.2.2(c), Go is simply connected. II

4.9 Theorem. Let G be a region in C and suppose that a E 000G such that the component ofC", - G which contains a does not reduce to a point. Then G has a barrier at a. Proof. Let S be the component of C cx) - G such that a E S. By considering an appropriate Mobius transformation if necessary, we may assume that a = 0 and 00 E S. Let Go be the component of C S which contains G. The preceding lemma gives that Go is simply connected; since 0 (/:. Go there is a branch t' of log z defined on Go. In particular t' is defined on G. For r> 0, let t'r(z) = t'(z) - log r for z in G(O; r). So -t'/G(O; r» is a subset of the right half plane. Now let Cr = G n {z: Izl = r}; then Cr is the union of at most a countable number of pairwise disjoint open arcs 'fk in {z: Izl = r}. But -t'r(Yk) = (ia k , i13 k ) = {it: a k < t < 13k} for k;;::: 1. Hence XJ

-

co

- t'r( Cr ) =

U (iak' iflk)

k=l

and these intervals are pairwise disjoint. Furthermore, the length of Yk is r(flk -
4.10

I (flk-ak) k=l

~ 27T.

Now if log is the principal branch of the logarithm then

.

(Z-i(Xk)

hk(z) = 1m log ---.z-lflk

272

Harmonic Functions

is harmonic in the right half plane and 0 < hk(z) < Exercise 1II.3.19). Moreover

for Re z > 0 (see

1r

4.11 if x > O. From (4.11) it follows that

"

f 00

L hkeX+iy)

:s;

k=l

X

2

x

+(y-t)

2

dt

= 7T.

_.- 00

Since each hk ;::: 0, Harnack's Theorem gives that h the right half plane. Hence 1 .fr(z) = .... h ( - t'r(z»

L hk is harmonic in 00

=

k= 1

1r

is harmonic in G(O; r). ft will be shown that {!/IT} is a barrier at O. Fix r > 0; then lim Re [- t',(z)] = + 00. So it suffices to show that h(z) -+0 as Re z -+

z-o

+ 00.

Using (4.11) and (4.10) it follows that for x > 0,

L 00

h(x+iy) =

hk(x+iy)

k=l

So, indeed, lim hex + iy) = 0 uniformly in y; this gives that lim x-+

+ co

To prove that lim rf;r(z) = 1 for w in G with prove that z-w 4.12

lim h(z) =

1r

z-.. ic

So fix k ;::: 1 and fix c in

(atk'

Iwl =

.frCz) = O.

z-+O

r, it is sufficient to

if atk < c < flk for some k

f3k)' The following will be proved.

4.13 Claim. There are numbers at and f3 such that at < atk < f3k < fl and if

u(z) = 1m log (:-~at), Z-latk

v(z)

=

1m log

G=~~k)


273

OCk < Y < 13k implies 0 ~ h(x+iy)-hk(X+iy) ~ u(x+iy) +v(x+iy). Once 4.13 is established, Equation 4.12 is proved as follows. From Exercise 111.3.19,

then x > 0 and

v(x+iy) = arctan (Y:f3k) -arctan (Y:f3} so if x+ iy ~ ic, c < 13k < 13, then v(x+ iy) ~ o. Similarly u(x+iy) ~ 0 as x+iy ~ ic, with oc < OCk < c. Hence, Claim 4.13 yields 4.14

lim [h(z)-hiz)] = O.

But

hk(X+iy) = arctan e~OCk) -arctan e:f3k). so lim hk(z) z-+ic

= ?T;

this combined with (4.14) implies Equation (4.12).

It remains to substantiate Claim 4.13; we argue geometrically. Recall (Exercise III.3.19) that hiz) is the angle in the figure. Consider all the intervals (ioc j' if3 j) lying above (iock' if3k) and translate them downward along the imaginal')' axis, keeping them above (iock' if3k) until their endpoints coincide and such that one of the endpoints coincides with if3k. Since r,(f3 J - oc j) ~ 2?T there is a number 13 < (13k + 2?T) such that each of the translated intervals lies in (if3k' if3). Now if z = x+iy, x > 0 and OCk < Y < 13k,

z

Harmonic Functions

274

then the angle h j(z) increases as the interval (icx j' ifl) is translated downward. Hence CXk < 1m z < flk implies

"[.h j(z)

4.15

:::; v(z),

j

where v is as in the statement of the claim and the sum is over all j such that cx j ;::.: (3k' By performing a similar upward translation of the intervals (icx j. i(3j) with (3j :::; CX k, there is a number cx < (cx k -21T) such that the translates lie in the interval (icx, icxk)' So if u is as in the claim and CXk < 1m z < (3k'

"[.h /z)

4.16

:::; u(z)

j

where the sum is over allj with flj :::; CX k. By combining (4.15) and (4.16) the claim is established . • 4.17 Corollary. Let G be a region such that no component of Coo - G reduces to a point; then G is a Dirichlet Region. 4.18 Corollary. A simply connected region is a Dirichlet Region. Proof If G #- C the result is clear since Coo - G has only one component. If G = C then the result is trivial. • Theorem 4.9 has no converse as the following example illustrates. Let 1 > rl > r2 > ... with rn -? 0; for each n let Yn be a proper closed arc of

the circle Izl = ro with length V(Yn)' Put G = B(O; 1) suppose that lim V(Yn)/r n = 2n. So Coo - G

00

[U {Yn} n=1 00

= {O} u U

n=l

{Yn}

U

U

{O}] and

{z:

Izl;;::': I}.

According to Theorem 4.9 there is a barrier at each point of oooG = oG with the possible exception of zero. We will show that there is also a barrier at zero. If rn-

m

1

> r > rn and if m > n, let Bm = B(O; r)-U {Yj}. Let h m be the j=n

continuous function on B;;' which is harmonic on Bm with hm(z)

Izl

=

rand hm(z)

=

0 for

z

in

m

U {y j}. Then

=

1 for

{h m} is a decreasing sequence of

j~n

positive harmonic functions on G(O; r); by Harnack's Theorem {h m } converges to a harmonic function h on G(O; r) which is also positive (Why?) Since lim h(z) = 1 for Iwl = r, we need only show that lim h(z) = O. Let k m z-w

z-o

be the harmonic function on B(O; rm) which is 0 on {Ym} and 1 on {z: Izl = rm} - {Ym} (this does not have continuous boundary values, only piecewise continuous boundary values which are sufficient-see Exercise 2.2). Then o :::; h :::; k m on B(O; rm) and

f

2n

km(O) =

~

21T

km(r mew) de

o

= ~ (21T _ V(Ym») 21T

rm

Green's Function

Since V(Ym] ~ 211, km(O) rm a barrier at zero.

275

~ 0; it follows that h(z) ~ 0 as z ~ O. Thus, G has

Exercises l. Let G = B(O; 1) and find a barrier for G at each point of the boundary. 2. Let G = C - (00, 0] and construct a barrier for each point of 0exp. 3. Let G be a region and a a point in 000G such that there is a harmonic function u: G ~ IR with lim u(z) = 0 and lim inf u(z) > 0 for all w in oooG, w -# a. Show that there is a barrier for G at a. 4. This exercise asks for an easier proof of a special case of Theorem 4.9. Let G be a bounded region and let a E oG such that there is a point b with [a, b] n G- = {a}. Show that G has a barrier at a. (Hint: Consider the transformation (z-a)(z-b)-I.)

§5. Green's Function In this section Green's Function is introduced and its existence is discussed. Green's Function plays a vital role in differential equations and other fields of analysis.

5.1 Definition. Let G be a region in the plane and let a E G. A Green's Function of G with singularity at a is a function ga : G - {a} ~ IR with the properties: (a) ga is harmonic in G- {a}; (b) G(z) = ga(z)+log Iz-al is harmonic in a disk about a; (c) lim ga(z) = 0 for each win 0ooG. For a given region G and a point a in G, ga need not exist. However, if it exists, it is ,unique. In fact if ha has the same properties, then, from (b), ha - ga is harmonic in G. But (c) implies that lim [ha(z) - ga(z)] = 0 for z-w

every win 8rJ:JG; so ha = ga by virtue of the Maximum Principle. A second observation is that a Green's Function is positive. In fact, ga is harmonic in G - {a} and lim ga(z) = + 00 since ga(z) + log Iz - al is z-+a

harmonic at z = a. By the Maximum Principle, ga( z) > 0 for all z in G - {a}. Given this observation it is easy to see that C has no Green's Function with a singularity at zero (or a singularity at any point, for that matter). In fact, suppose go is the Green's Function with singularity at zero and put g = -go; so g(z) < 0 for all z. We will show g must be a constant function, which is a contradiction. To do this, it is sufficient to show that if 0 -# Zl -# Z2 -# 0 then g(Z2) :; g(zl)' If £ > 0 then there is as> 0 such that

276

Harmonic Functions

Ig(Z)-g(Zl)1 < e if IZl -zi < 8; so g(z) < g(zl)+e if IZ-Zll < 8. Let r > IZl -z21 > 8; then

hr(z)

=

(8)

g(Zl) + e log -

IZ-Z11 log - r -

r

is harmonic in C - {Z l}' It is left to the reader to check that g(z) :::; hr(z) for Z on the boundary of the annulus A = {z:8 < jz-zll < r}. By the Maximum Principle, g(z) :::; hr(z) for Z in A; in particular, h.(z2) ;;:: g(zz). Letting r _ 00 we get r-+ '"

since e was arbitrary, g(Z2) :5 g(Zl) and g must be a constant function. When do Green's Functions exist?

5.2 Theorem. If G is a bounded Dirichlet Region then for each a in G there is a Green's Function on G with singularity at a. Proof Define f: 8G -IR by fez) = log .jz-aj, and let u:G- _IR be the unique continuous function which is harmonic on G and such that u(z)=f(z) for Z in oG. Then ga(z) = u(z)-log jz-al is easily seen to be the Green's Function . • This section will close with one last result which says that Green's Functions are conformal invariants.

5.3 Theorem. Let G and Q be regions such that there is a one-one analytic function f of G onto Q; let a E G and a. = f(a). If ga and y~ are the Green's Functions for G and Q with singularities a and a. respectively, then ga(z) = yif(z». Proof Let cp: G - IR be defined by cp = y~ f To show that cp = ga it is sufficient to show that cp has the properties of the Green's Function with singularity at z = a. Clearly cp is harmonic in G - {a}. If w EO", G then lim cp(z) = 0 will follow if it can be shown that lim cp(zn) = 0 for any sequence 0

z-+w

{zn} in G with Zn -+ w. But {f(zn)} is a sequence in Q and so there is a subsequence {znJ such thatf(znk) -+ w in Q- (closure in C"')' So 'Y~(f(znJ) -+ O. Since this happens for any convergent subsequence of {f(zn)} it follows that lim cp(zn} = lim yif(zn» = O. Hence lim cp(z) = 0 for every w in oooG. z-+w

Finally, taking the power series expansion off about z fez) = a.+A 1(z-a)+A 2(z-a)2+ ... ;

or f(z)-a.

=

(z-a)[Al +A 2 (z-a)+ .. .].

Hence

5.4

loglf(z) - a.J = 10gJz-aJ +h(z),

=

a,

Green's Function 277 where h(z) = log IAI +Aiz-a)+ ... is harmonic near z = a since AI =1= 0. Suppose that y~( w) = ~(w) -logl IV - 0:1 where ~ is a harmonic function on Q. Using (5.4) 1

'F{z) =

~{f{z»-loglf(z)-o:l

= [~(r{z)-h(z)]-loglz-al·

Since ~ f - h is harmonic near z z = a. • 0

=

a, 'F(z) + log Iz - al is harmonic near

Exercises 1. (a) Let G be a simply connected region, let a E G, and let f: G -+ D = {z: Izi < I} be a one-one analytic function such that f(G) = D and f{a) = 0. Show that the Green's Function on G with singularity at a is ga{z) = -log If(z)l· (b) Find the Green's Functions for each of the foHowing regions: (i) G = C-(oo, 0]; (ii) G = {z:Re z > o}; (iii) G = {z:o < 1m z < 21T}. 2. Let ga be the Green's Function on a region G with singularity at z = a. Prove that if!f is a positive superharmonic function on G - {a} with lim inf

[!f(z)+log Iz-all > - 00, then ga(z) .:s:; !fez) for z =1= a. 3. This exercise gives a proof of the Riemann Mapping Theorem where it is assumed that if G is a simply connected region, G =1= C, then: (i) Coo - G is connected, (ii) every harmonic function on G has a harmonic conjugate, (iii) if a (Z) for a real number 0:. (So fis analytic in G.) Prove that If(z)1 = exp ( - ga(z)) and that lim If(z) I = I for each w in 8G (Compare this with Exercise 1). Prove that for 0< r < I, Cr = {z:lf(z)1 = r} consists of a finite number of simple closed curves in G (see Exercise VI.I.3). Let Gr be a component of {z:lf{z)1 < r} and apply Rouche's Theorem to get that fez) = andf(z)-w o = 0, IWol < r, have the same number of solutions in Gr. Prove thatfis one-one on Gr. From here conclude thatf(G) = D = {z: Izi < I} and f'(a) > 0, for a suitable choice of 0:. (b) Let G be a simply connected region with G =1= C, but assume that G is unbounded and 0, 00 E 0ocP. Let t be a branch of log z on G, a E G, and ex = tea). Show that t is one-one on G and ((z) =1= ex+21Ti for any z in G. Prove that 'F(z) = [t(Z)-0:-21Tirl is a conformal map of G onto a bounded simply connected region in the plane. (Show that t omits all values in a neighborhood of 0: + 21Ti.) (c) Combine parts (a) and (b) to prove the Riemann Mapping Theorem. 4. (a) Let G be a region such that 8G = y is a simple continuously differen-

°

278

Harmonic Functions

tiable closed curve. If !:iJG ->- IR is continuous and g(z, a) Green's Function on G with singularity at a, show that

5.5

Il(a)

=

f!CZ)

=

gaCz) is the

~-Gn (z, a) ds

is a formula for the solution of the Dirichlet Problem with boundary values!; where

_~!I is the derivative of g in the direction of the outw:lfd normal to y and en

ds indicates that the integral is with respect to arc length. (Note: these concepts are not discussed in this book but the formula is sufficiently interesting so as to merit presentation.) (Hint: Apply Green's formula

fJ'~[WlV-VdUJdxdy = n

J'[

;,n -

(Iv

II,

Gn

GU] ds -u -;_. en

with n = G- {z: I,:-al s 8}, (3 < dCa, {y}), lie, h, Ii = gaCz) = g(z, a).) (b) Show that if G = {z: 121 < I} then (5.5) reduces to equation (2.5).

Chapter XI Entire Functions To begin this chapter, let us recall the Weierstrass Factorization Theorem for entire functions (VII. 5.14). Let f be an entire function with a zero of multiplicity m ~ 0 at z = 0; let {an} the zeros off, an of- 0, arranged so that a zero of multiplicity k is repeated in this sequence k times. Also assume that JaIl :::; la 2 1:::; .... If {Pn} is a sequence of integers such that 0.1 for every r > 0 then P(z) =

0.2

00

TI

Epn(z/an)

n=1

converges uniformly on compact subsets of the plane, where Ep(z) = (l-z)exp ( z

0.3 for P

~

+ 2 + ... + Z2

zP) p

1 and Eo(z) = l-z

Consequently 0.4

fez)

=

zm e 9 (Z) P(z)

where g is an entire function. An interesting line of investigation begins if we ask the questions: What properties off can be deduced if g and P are assumed to have certain "nice" properties? Can n:strictions be imposed on f which will imply that g and P have particular properties? The plan that will be adopted in answering these questions is to assume that g and P have certain characteristics, deduce the implied properties off, and then try to prove the converse of this implication. How to begin? Clearly the first restriction on g in thIs program is to suppose that it is a polynomial. It is equally clear that such an assumption must impose a growth condition on e 9 (Zj. A convenient assumption on Pis that all the integers Pn are equal. From equation (0.1), we see that this is to assume that there is an integer P ~ 1 such that 00

I lanl-

n;l

P

< 00;

that is, it is an assumption on the growth rate of the zeros off In the first 279

280

Entire Functions

section of this chapter Jensen's Formula is deduced. Jensen's Formula says that there is a relation between the growth rate of the zeros of I and the growth of M (r) = sup {1/(reie)l: 0 :5 8 :5 217} as r increases. In succeeding sections we study the growth of the zeros ofland the growth of M(r). Finally, the chapter culminates in the beautiful factorization theorem of Hadamard which shows an intimate relation between the growth rate of M(r) and these assumptions on g and P. §1. Jensen's Formula

If I is analytic in an open set containing B(O; r) and I doesn't vanish in B(O; 1') then log III is harmonic there. Hence it has the Mean Value Property eX. 1.4); that is

f ' 2n

1.1

log

1/(0)1 = --1

217

log i/(re''0 )1

de.

o

Suppose I has exactly one zero a = rei> on the circle Izi (z - a) -1 then (1.1) can be applied to g to obtain

f;. f [log I/(re )1-log ire

= r. If g(z) = I(z)

21t

log ig(O) I =

i8

iB -

rei'l]

de

o

Since log ig(O)1 = log I/(O)I-Iog r, it will follow that (Ll) remains valid where I has a single zero on iz' = r, if it can be shown that

j'

21t

2;1

' d8 log Ire'' e -re"1

= log r;

o

alternately, if it can be shown that 2"

flog II-e i8 1 d8 = O.

o

But this follows from the fact that

Jlog (sin2 28) dB = 2"

417 log 2

o

(Exercise V. 2.2(h)). So (1.1) remains valid if I has a single zero on Izl = r; by induction, (1.1) is valid as long as/has no zeros in B(G; r). The next step is to examine what happens if I has zeros inside B(O; r). In this case, log !/(z)1 is no longer harmonic so that the MVP is not present.

1.2 Jensen's Formula. Let I be an analytic lunction on a region containing

281

Jensen's Formula

S(O; r) and suppose that a1, ... , all are the zeros of fin B(O; r) repeated according to multiplicity. If f(O) 0 then

*"

log If(O) I =

-

L

~ log C;kl) +

2"

flOg If(reiB)1 dB.

o

Proof If Ibl < 1 then the map (z-b) (l_hZ)-1 takes the disk B(O; 1) onto itself and maps the boundary onto itself. Hence r 2(z- ak) r2-iikz maps B(O; r) onto itself and takes the boundary to the boundary. Therefore

-akz f(z) TI r(z-ak) n

F(z) =

r

2

-

k=1

is analytic in an open set containing B(O; r), has no zeros in B(O; r), and IF(z)1 = If(z) I for Izi = r. So (1.1) applies to F to give 2"

log IF(O) I =

2~ flOg If(rei6) I d8. o

However

F(O) = f(O)

D(-;J

so that Jensen's Formula results . • If the same methods are used but the MVP is replaced by Corollary X. 2.9, log If(z) I can be found for z oft ak' 1 :$ k :$ n. 1.3 Poisson-Jensen Formula. Let f be analytic in a region which contains B(O; r) and let ai' ... , an be the zeros off in B(O; r) repeated according to multiplicity. if Izl < r andf(z) oft 0 then

log If(z) I = -

-ii zj 1 f2" (reIB+z) 2:" log Ir2r(Z-ak k) + 2- Re -I-B- log If(re IB )I d8. 71" re -z

k=1

o

Exercises 1. In the hypothesis of Jensen's Formula, do not suppose that f(O) oft O. Show that if f has a zero at z = 0 of multiplicity m then

log

f
-

6 n

( r ) 1 f2" log la,,1 + 271" log o

If(rel~1 dB.

282

Entire Functions

2. Let f be an entire function, M(r) = sup {lfCrei")I: 0 ~ IJ :c: 21T}, nCr) = the number of zeros offin BCO; r) counted according to multiplicity. Suppose that j{O) = 1 and show that nCr) log 2 ~ log M(2r). 3. In Jensen's Formula do not suppose that/is analytic in a region containing B(O; r) but only that/is meromorphic with no pole at z = O. Evaluate 2n

I' 21T log If(re i8 )1 de.

J

D

4. (a) Using the notation of Exercise 2, prove that

I r

o

n(!) - dt = t

~ L

(r )'

Iog-,

k=l

lakl

where ai' ... ,an are the zeros of / in B(O; 1'). (b) Let / be meromorphic without a pole at z = 0 and let nCr) be the number of zeros of / in B(O; r) minus the number of poles (each counted according to multiplicity). Evaluate

I~l~t) r

dt.

o

5. Let D = B(O; 1) and suppose thatf: D .....)-1[ is an analytic function which is bounded. (a) If {an} are the non-zero zeros of/in D counted according to multiplicity, prove that I(1-la.l) < 00. (Hint: Use Proposition VII. 5.4). (b) If / has a zero at z = 0 of multiplicity m ~ 0, prove that fez) = zmB(z) exp (-g(z» where B is a Blaschke Product (Exercise VII. 5.4) and g is an analytic function on D with Re g (z) ~ ·-Iog M (M = sup {!f(z)l:

Izl

< 1

n·

§2. The genus and order of an entire function 2.1 Definition. Let/be an entire function with zeros {at. a 2 , ••• }, repeated a.::cording to multiplicity and arranged such that la 1 ~ la 2 1~ .... Then f is of finite rank if there is an integer p such that

I

00

2.2

lanl-(P+ 1) <

00.

n=J

Ifp is the smallest integer such that this occurs, thenfis said to be o(rank p; a function with only a finite number of zeros has rank O. A function is of infinite rank if it is not of finite rank.

The genus and order of an entire function

283

From equation (0.1) it is seen that iff has finite rankp then the canonical product P in (0.4) can be taken to be

nEp(z/a 00

2.3

P(z) =

n=1

n )·

Notice that iff is of finite rank and p is any integer larger than the rank of J, then (2.2) remains valid. So there is a second canonical product (2.3), and this shows that the factorization (0.4) off is not unique. However, if the product P is defined by (2.3) where p is the rank off then the factorization (0.4) is unique except that g may be replaced by g+21Tmi for any integer m.

2.4 Definition. Letfbe an entire function of rank p with zeros {ai, a2'" .}. Then the product defined in (2.3) is said to be in standard form for! Iff is understood then it will be said to be in standard form. 2.5 Definition. An entire function f has finite genus iff has finite rank and if

fez) = zm eg(z) P(z), where P is in standard form, and g is a polynomial. If p is the rank of f and q is the degree of the polynomial g, then fL = max (p, q) is called the genus of! Notice that the genus off is a well defined integer because once P is in standard form, then g is uniquely determined up to adding a multiple of 21Ti. In particular, the degree of g is determined. 2.6 Theorem. Let f be an entire function of genus fL. For each positive number IX there is a number ro such that for Izi > ro

If(z) I < exp (1XIzl"+ I) Proo! Since f is an entire function of genus fL fez) = zm eg(z)

n E,,(z/a 00

n ),

n=1

where g is a polynomial of degree :::;; fL. Notice that if Izi < 2.7

log IE,,(z) I

=

Re {log(l-z)+z+ .. . +Z"/fL}

=

1 +2 } I +I Re { - fL + 1 zIt - fL + 2 Zll - ...

:::;; Izl/+1

{_1_ +K fL+l fL+2

+ ... }

:::;; Izl,,+1(1+-t+(!)2+ ... )

= 2Izl"+1 Also

IE,.(z) I :::;; (1 + IzJ) exp (izi + ... + Izl"/fL), so that

1- then

log jE,.(z)1 :::;; log (1 + Izl) + Izi + ... + Izl"/fL·

Entire Functions

284

Hence,

r

z~n;,

1£/z)1 Izll'+ 1 -

log

0

So if A > 0 then there is a number R > 0 such that 2.8

But on {z: t ~ Izl ~ R} the function Izl-(I'+I) log I£I'(z) I is continuous except at z = + I, where it tends to - 00. Hence there is a constant B > 0 such that 2.9

Combining (2.7), (2.8), and (2.9) gives that log I£/z) I ~ Mlzll'+ 1

2.10

for all z in C, where M Since Ilan l-
=

max {2, A, B}. an integer N can be chosen so that

00,

But, using (2.10), 2.11

n;~1 log 1£I'(zja I ~ M n;~1 I~r+l ~ ~ Izll'+l. n)

Now notice that in the derivation of (2.8), A could be chosen as small as desired by taking R sufficiently large. So choose r 1 > 0 such that DC

log I£I'(z) I ~ 4N IzII'+1, for If r2 = max {la 1 1r 1 , la21 r 1 ,

••• ,

L log I£I'(z/a N

n)

laNI

Izi

> r 1•

r 1 } then

I ~ ~ Izll'+ 1 for Izi

> r 2.

n;1

Combining this with (2.11) gives that 2.12

log

IP(z)1 =

2: log co

1£I'(zjan) I .:s;

~lzll'+1

n;1

for

Izi

> r2. Since g is a polynomial of degree ~IL, .

11m %-00

Izl + Ig(z) I IZ 11'+1 = o.

m log

So there is an r3 > 0 such that m log Izl + Ig(z)1 < Together with (2.12) this yields log If(z) I < ex Izll'+1

t IX Izll'+1

for

Izl > r3.

The genus and order of an entire function

285

for Izl > ro = max {fl, r3}' By taking the exponential of both sides, the desired inequality, is obtained. III The preceding theorem says that by restricting the rate of growth of the zeros of the entire function fez) = zm exp g(z)P(z) and by requiring that g be a pofynomiai, then the growth of M(r) = max {If(re i8 )1: 0 :0; 8 :0; 27T} is dominated by exp (~lzIJi+l) for some 11- and any ~ > O. We wish to prove the converse to this result. 2.13 Definition. An entire function f is of finite order if there is a positive constant a and an ro > 0 such that If(z) 1< exp (lzla) for Izi > roo Iffis not of finite order then f is of infinite order. If f is of finite order then the number A = inf {a: If(z)1 < exp (lzl") for Izl sufficiently large} is called the order off Notice that if If(z) 1< exp (Izla) for Izl > ra > I and b > a then If(z)1 < exp (IzJh). The next proposition is an immediate consequence of this observation. 2.14 Proposition. Let f be an entire function of finite order A. If E > 0 then If(z) 1< exp (lzIH£) for all z with Izl sufficiently large; and a z can be found, with Izl as large as desired, such that If(z)1 ;?: exp (lzI;'-<). Although the definition of order seems a priori weaker than the conclusion of Theorem 2.6, they are, in fact, equivalent. The reader is asked to show this for himself in Exercise 3. So it is desirable to know if every function of finite order has finite genus (a converse of Theorem 2.6). That this is in fact the case is a result of Hadamard's Factorization Theorem, proved in the next section. The proof of the next proposition is left to the reader. 2.15 Proposition. Let f be an entire function of order A and let M(r) = max {If(z)l:

Izi =

r}; then

.

A = hmsup

log log M ( r )

logr

r~oc;

.

Consider the function fez) = exp (e then If(z) 1 = exp (Re e (e'" cos 8) if z = re i8 • Hence M(r) = exp (e'") and Z );

log log M(r) log r

--------------

Z)

= exp

r log r '

thus, f is of infinite order. On the other hand if g(z) = exp(z"), 11 ;?: 1, then Ig(z)1 = exp (Re zn) = exp (r" cos nO). Hence M(r) = exp (r") and so log log M(r) = n; log r thus g is of order n. For further examples see Exercise 7. Using this terminology, Theorem 2.6 can be rephrased as follows 2.16 Corollary. Iff is an entire function offinite genus 11- thenfis offinite order .1:0;11-+1.

286

Entire Functions

Exercises 1. Let I(z) = Ienz n be an entire function of finite genus fJ.; prove that

lim en(n!)I/(I'+ 1) = O. (I-lint: Use Cauchy's Estimate.) 2. Let II and 12 be entire functions of finite orders Al and A1 respectively. Show that 1=11 +12 has finite order A and ,\ :::: max (AI' A2)' Show that A = max (AI' A2 ) if Al # '\2 and give an example which shows that A < max (AI' '\2) withl # 0. 3. Suppose / is an entire function and A, B, a are positive constants such that there is a ro with 1/(z)1 ~ exp(Alzlli + B) for Izl > ro' Show that / is of finite order ~ a. 4. Prove that if I is an entire function of order ,\ then I' also has order A. 5. Let I(z) = Ienz" be an entire function and define the number by (X

0(

=

· 111 . f -log I1m n~oo

Ie",i

nlogn

(a) Show that if / has finite order then (X > O. (Hint: If the order of f is ,\ and f3 > ,\ show that le,,1 ~ r- 11 exp (rl1) for sufficiently large r, and find the minimum value of this expression.) (b) Suppose that < 0( < (J) and show that for any E > 0, E < ex, there is an integer p such that l/n < 11-(,-d for n > p. Conclude that for Izl = r > 1 there is a constant A such that

°

I/(z)1 < Ar P + n~

c:::~)n

(c) Let p be as in part (b) and Jet IV be the largest integer ::::(2r)I/(.-,). Take r sufficiently large so that IV > p and show that X

1/ = ,\, -

1

r

(n"') i

L

:V

'!l

L

= {J 'r

!

r

',II

(;~)

1

where B is a constant which does not depend on r. (d) Use parts (b) and (c) to show that if 0 < < (J) then I has finite order A and ,\ :::: 0( - ! . (e) Prove that I is of finite order iff > 0, and if I has order ,\ then A=ex- I . 6. Find the order of each of the following functions: (a) sin z; (b) cos z; (X

(X

I

CD

(c) cosh

z; (d)

n-D"zn where a > O. (Hint: For part (d) use Exercise 5.)

n=l

7. Let II and 12 be entire functions of finite order AI' A2 ; show that I = 11/2 has finite order A :S max (AI' '\2)' 8. Let {an} be a sequence of non-zero complex numbers. Let p = inf {a: I lanl- a < (J)}; the number p is called the exponent 01 eom-ergence of {an}.

Hadamard Factorization Theorem

287 (a) Iff is an entire function of rank p then the exponent of convergence p of the non-zero zeros off satisfies: p ::; p ::; p + 1. (b) If p' = the exponent of convergence of {an} then for every E" > 0, lanl-(P+E) < 00 and lanl-(p-E) = 00. (c) Letfbe an entire function of order,\ and let {ai, a2, ... } be the nonzero zeros of f counted according to multiplicity. If p is the exponent of convergence of {an} prove that p ::; ,\. (Hint: See the proof of (3.5) in the next section.)

L

L

n co

(d) Let P(z)

=

£/z/an) be a canonical product of rank p, and let p

n~1

be the exponent of convergence of {an}. Prove that the order of P is p. (Hint: If,\ is the order of P, p ::; '\; assume that lall ::; la 2 1::; ... and fix z, Izl > 1. Choose N such that lanl ::; 2 Izl if n ::; Nand lanl > 2 Izi if n ~ N + l. Treating the cases p 0

Prove that for Izi ~ 1, log 1£/z)1 < B of z. Use this to prove that

IzlP where B is a constant independent

for some constant C independent of z.) 9. Find the order of the following entire functions:

(a) fez) =

n (l-anz), co

D( :J

0 <

lal

< 1

n~1

(b) fez) =

1-

(c) fez) = [r(z)r l . §3. Hadamard Factorization Theorem In this section the converse of Corollary 2.16 is proved; that is each function of finite order has finite genus. Since a function of finite genus can be factored in a particularly pleasing way this gives a factorization theorem.

3.1 Lemma. Let f be a non-constant entire function of order A with f(O) #- 0 and let {a 1, a 2 , ••• } be the zeros off counted according to multiplicity and arrang~d so that la 11 ::; la 2 1 ::; . . . . If an integer p > ,,- 1 then

288

Entire Functions

Proof. Let n = nCr) = the number of zeros off in BCO; 1'); according to the Poisson-Jensen formula

. = log If(z)1

z Jog Ir---iikz k=! r(z-ak)

Ln. ,

I+

1

271"

f2n

o

i8 Re (re - , -+ z) log If(re''0 )1 dB re,e_ z

.

for Izl < r. Using Exercise 1 and Leibniz's rule for differentiating under an integral sign this gives

2"

+ for

Izl

~f

271"

2re iO (re i8 -z)-Zlog If(re i8 )1 dB

o

< rand z of- ai' ... , an" Differentiating p times yields:

21<

+(p+ I)!

~ J' 2re i8(re i8 _Z)-p-2 log If(re

217

i8 )

1

dB.

o

Now as I' --+ 00, n(r)>- 00 so that the result will follow if it can be shown that the last two summands in (3.2) tend to zero as r ~"" 00. To see that the second sum converges to zero let r > 2Jzl; then !akl ~ I' gives Ir2-(lkzl :2:!r 2 so that (Iakl Ir 2 -ak zl!)p+! ~ (2/r)P+!. Hence the second summand is dominated by nCr) (2/r)P+!. But it is an easy consequence of Jensen's Formula (see Exercise 1.2) that log 2n(r)::s; log M(2r). Sincefis of order A, for any E > 0 and r sufficiently large (log 2)n(r)r(p+1)::s; log[M(2r)]r-(P+!)

But p+ 1 > ,\ so that E may be chosen with ('\+E)-(P+ 1) < O. Hence nCr) (2/rY+ 1 --+ 0 as r --+ 00; that is, the second summand in (3.2) converges to zero. To show that the integral in (3.2) converges to zero notice that

r reiB(reiO_z)-P-Z dO

2"

o

=

0

since this integral is a multiple of the integral of (w-z)-p-2 around the circle Iwi = I' and this function has a primitive. So the value of the Integral in (3.2) remains unchanged if we substitute log Ifl-Iog M(r) for log If I·

289


So for 21z1 < r, the absolute value of the integral in (3.2) is dominated by 2"

3.3

(p+ I)! 2 P +3 r-(p+l)

~f

21T

[log M(r)-log If(re i9 )1] dB.

o

But according to Jensen's Formula, 2"

~ flog If(re i 9) I dB ;:::

21T

0

o

since f(O) = I. Also log M(r) :::; rH' for sufficiently large r so that (3.3) is dominated by (p+l)! 2 P +3 r H .-(p+l) As before, £ can be chosen so that rHo -(p+ I) ~ 0 as r ~ 00 • • Note that the preceding lemma implicitly assumes that f has infinitely many zeros. However, iff has only a finite number of zeros then the sum in Lemma 3. I becomes a finite sum and the lemma remains valid. 3.4 Hadamard's Factorization Theorem. order A then I has finite genus fL :::; A.

If f is an entire function of finite

Proof Let p be the largest integer less than or equal to A; so p :::; A < p + 1. The first step in the proof is to show that I has finite rank and that the rank is not larger than p. So let {ai' a 2 , ••• } be the zeros off counted according to multiplicity and arranged such that la II :::; la21 :::; .... It must be shown that

3.5

L la.r(p+ .;1 00

I)

<

00.

There is no loss in generality in assuming that 1(0) = 1. Indeed, if I has a zero at the origin of multiplicity m and M(r) = max {If(z)l: Izl = r} then for any £ > 0 and Izl = r log If(z)z-ml :::; log [M(r)r- m] :::; rH<-m log r :::; r'l.+ 2<

if r is sufficiently large. So f(z)z-m is an entire function of order A with no zero at the origin. Since multiplication by a scalar does not affect the order, the assumption that f(O) = 1 is justified. Let nCr) = the number of zeros of fin li(O; r). It follows (Exercise 1.2) that [log 2] nCr) ::; log M(2r). Sincefhas order A, log M(2r) ::; (2r)Ht. for any t: > 0 so that lim n(r)r-(A+<) = O. Hence nCr) :::; r A+< for sufficiently large r. Since lall ;~~I:::; ... , k:::; n(lakl) :::; lakI A+< for all k larger than some integer ko. Hence,

290

Entire Functions

for k > k o' So if E is chosen with It + E < p + 1 (recall that ;\ < p + 1) then iakl-(P+ I) is dominated by a convergent series; (3.5) now follows. Let fez) = 1"(::) exp (g(z)) where P is a canonical product in standard form. Hence for::: # al.

I

/,(z) f(:::)

, =

g (z)

+

P'(z) PC::)

U sing Lemma 3.1 gives that

However it is easy to show that

for z # aI' a 2 , .... Hence gtp+ I) == 0 and g must be a polynomial of degree sp. So the genus off s p s ,\. III As an application of Hadamard's Theorem a special case of Picard's Theorem can be proved. This theorem is proved in full generality in the next chapter.

3.6 Theorem. Let f be an entire function of finite order, then f assumes each complex number with one possible exception. Proof Suppose there are complex numbers IX and {3, IX # {3, such thatf(z) # IX and fez) # {3 for all z in C So f -IX is an entire function that never vanishes; hence there is an entire function g such that fez) -IX = exp g(z). Since f has finite order, so does I--- IX; by Hadamard's Theorem g must be a polynomial. But exp g(z) never assumes the value f3-rx and this means that g(z) never assumes the value log ({3 -IX), a contradiction to the Fundamental Theorem of Algebra. III One might ask how many times f assumes a given value IX. If g is a polynomial of degree n ?: 1, then every IX is assumed exactly n times. Howeverf = ea assumes each value (with the exception of zero) an infinite number times. Since the order of e g is n (see Exercise 2.5) the next result lends some confusion to this problem; the confusion will be alleviated in the next chapter.

3.7 Theorem. Let f be an entire junction of finite order .\ where ;\ is not an integer; then f has infinitely many zeros. Proof Suppose j has only a finite number of zeros {ai' az, ... , an} counted according to multiplicity. Then j(z) = e 9 (z)(z-a j ) • • • (z-a n ) for an entire function g. By Hadamard's Theorem, g is a polynomial of degree s:;;\. But it is easy to see that j and e a have the same order. Since the order of e 9 is the degree of g, ,\ must be an integer. This completes the proof. III

291


3.8 Corollary. Iff is an entire function of order A and A is not an integer then f assumes each complex value an infinite number of times. PrDoj If a. Ef(C), apply the preceding theorem to f-a. . •

Exercises

1. Let f be analytic in a region G and suppose that f is not identically zero. Let Go = G- {z: f(z) = O} and define h: Go ~ IR by h(z) = log If(z)l. 8h

Show that -

ax

8h

- i-

8y

=

f' -f on Go·

2. Refer to Exercise 2.7 and show that if ;\1 =f. ;\2 then ;\ = max (;\1' ;\2)' 3. (a) Let f and g be entire functions of finite order A and suppose that f(a.) = g(a.) for a sequence {a.} such that la,r(J·+l) = 00. Show that\

L

f=g·

(b) Use Exercise 2.8 to show that iff, g and {an} are as in part (a) with 00 for some € > 0 then f = g. (c) Find all entire functions f of finite order such that f(log n) = n. (d) Give an example of an entire function with zeros {log 2, log 3, ... } and no other zeros.

L la.I-(.I.+<) =

Chapter XII The Range of an Analytic Function In this chapter the range of an analytic function is investigated. A generic problem of this type is the following: Let y; be a family of analytic functions on a region G which satisfy some property P. What can be said aboutf(G) for eachfin y;? Are the setsf(G) uniformly big in some sense? Does there exist a ball B(a; r) such thatf(G) :::> B(a; r) for eachfin Y;? Needless to say, the answers to such questions depend on the property P that is used to define Y;. In fact there are a few theorems of this type that have already been encountered. For example, the Casorati-Weierstrass Theorem says that if G = {z: 0 < Iz - al < r} and Y; is the set of analytic functions on G with an essential singularity at z = a, then for each 0, 0 < 0 < r, and each f in Y; f(ann (a; 0; 0» is dense in C (Y. 1.21). Recall (Exercise Y. 1.13) that iffis entire and f(l/z) has a pole at z = 0, then f is a polynomial. So iff is not a polynomial thenf(l/z) has an essential singularity at z = o. So as a corollary to the Casorati-Weierstrass Theorem, fCC) is dense in C for each entire function (if f is a polynomial then fCC) = C). This chapter will culminate in the Great Picard Theorem that substantially improves the Casorati-Weierstrass Theorem. Indeed, it states that if f has an essential singularity at z = a then f(ann (a; 0; 0» is equal to the entire plane with possibly one point deleted. Moreover, f assumes each of the values in this punctured disk an infinite number of times. (See Exercise Y. 1.10.) As above, this yields that fCC) is also the whole plane, with one possible point deleted, whenever f is an entire function. This is known as the Little Picard Theorem. However, this latter result will be obtained independently. Before these theorems of Picard are proved, it is necessary to obtain further results about the range of an analytic function-which results are of interest in themselves. §1. Bloch's Theorem To fit the result referred to in the title of this section into the general questions posed in the introduction, let D = B(O; 1) and let Y; be the family of all functions f analytic on a region containing D - such that f(O) = 0 and 1'(0) = 1. How "big" can j(D) be? Put another way: because 1'(0) = 1 =f 0, f is not constant and so feD) is open. That is, feD) must contain a disk of positive radius. As a consequence of Bloch's Theorem, there is a positive constant B such thatf(G) contains a disk of radius Bfor eachfin Y;. 292

Bloch's Theorem

293

1.1 Lemma. Let f be analytic in D = {z: Izl < I} and suppose that f(O) = 0, 1'(0) = 1, and If(z) I :$; M for all z in D. Then M ~ 1 andf(D) =:> B(O; 116M). Proof Let 0 < r < 1 and fez) = z+a2z2+ ... ; according to Cauchy's Estimate /anl :$; Mlr n for n ~ 1. So I = fall :$; M. If /zl = (4M)-1 then 00

/f(z)/ ~ /zl ~

L n=2

(4M)-1 -

lanz n/

L
n=2

M(4M)-n

= (4M)-1_(l6M-4)-1 ~

(6M)-1

since M ~ 1. Suppose Iwl < (6M) -1; it will be shown that g(z) = fez) - w has a zero. In fact, for /z/ = (4M)-r, If(z)-g(z)1 = Iwl < (6M)-1 :$; If(z)l. So, by Roucht!'s Theorem, f and g have the same number of zeros in B(O; 1/4M). Sincef(O) = 0, g(zo) = 0 for some zo; hencef(D) =:> B(O; 116M). •

1.2 Lemma. Suppose g is analytic on B(O; R), g(O) = 0, /g'(O)/ = p. > 0, and /g(z)/ :$; M for all z, then geBCO; R» Proof {z: Iz/ to the of the

=:>

B

(0; ~;;2)

Let fez) = [Rg' (O)r J g(Rz) for Izl < 1; then f is analytic on D = < 1 },J(O) = O,J'(O) = 1, and If(z) I :$; M/p.R for all z in D. According preceding lemma, feD) =:> B(O; p.RI6M). If this is translated in terms original function g, the lemma is proved . •

1.3 Lemma. Let f be an analytic function on the disk B(a; r) such that /J'(z)-J'(a)/ < IJ'(a) Ifor all z in B(a; r), z of- a; thenfis one-one. Proof Suppose Zl and Z2 are points in B(a; r) and Zl of- Z2' If y is the line segment [Zl> Z2] then an application of the triangle inequality yields If(Zl)-f(Z2)/

= IIJ'(z)dzl y

~

IIJ'(a) y

~

dzl-II

[J'(z)-J'(a)] dzl

y

/J'(a)/I Z I-Z21 - I/J'(z)-J'(a)/Idz/. y

Using the hypothesis, this gives /f(Zl)-f(Z2)1 > 0 so thatf(zl) of-f(Z2) and is one-one . •

f

1.4 Bloch's Theorem. Let f be an analytic function on a region containing the closure of the disk D = {z: Iz/ < I} and satisfying f(O) = 0,1'(0) = 1. Then there is a disk S c: D on which f is one-one and such that f(S) contains a disk of radius 1/72.

294

The Range of an Analytic Function

Proof Let K(r) = max {11'(z)l: Izl = r} and let her) = (l-r)K(r). It is easy to see that h: [0, 1] -IR is continuous, h(O) = 1, h(l) = O. Let ro = sup {r: her) = I}; then hero) = 1, ro < 1, and her) < 1 if r > ro (Why?). Let a be chosen with lal = ro and II'(a) I = K(ro); then

1.5

II'(a)1

Now if Iz-al < t(1-ro) definition of r0 gives 1.6

=

Po,

(1-ro)-1.

=

Izl

< Hl+ro); since ro < t (l+ro), the

II'(z) I ::; K(t(1 + r 0» = h(t(1+ro» [l-t(l+ro)r 1 < [1--!-(1 +ro)]-1

= l/po for Iz-al < Po. Combining (1.5) and (1.6) gives

II'(z) - I'(a) 1

II'(z) 1

::;

+

II'(a) I

< 3/2po· According to Schwarz's Lemma, this implies that

II'(z)-I'(a)1 < for z in B(a; Po). Hence if z E S

=

31z2~al Po

B(a; !Po),

II'(z)-I'(a)1 <

2~ = II'(a) 1

Po By Lemma 1.4,f is one-one on S. It remains to show that f(S) contains a disk of radius 1/72. For this define g: B(O; tpo) - C by g(z) = f(z+a)-f(a) then g(O) = 0, Ig'(O)1 = /I'(a)/ = (2pO)-1. If Z E B(O; lPo) then the line segment y = [a, z+a] lies in S c B(a; Po). So by (1.6) /g(z) 1 =

If I'(w) dW\ 1

1

::; -/zl Po

<

t·

Applying Lemma l.2 gives that where

geBCO; tpo»

:::>

B(O; a)

1 72

Bloch's Theorem

295

If this is translated into a statement about J, it yields that f(S)

:J

B(f(a);

7~)'.

1.7 Corollary. Let f be an analytic function on a region containing B(O; R); 1 then f(B(O; R» contains a disk of radius 72 R/I' (0)/. Proof Apply Bloch's theorem to the function g(z) = [f(Rz) - f(O)]/RI' (0) (the result is trivial if 1'(0) = 0, so it may be assumed that 1'(0) i= 0) . • 1.8 Definition. Let ~ be the set of all functions f analytic on a region containing the closure of the disk D = {z: lzl < I} and satisfying f(O) = 0, 1'(0) = 1. For each f in ~ let f3(f) be the supremum of all numbers r such that there is a disk Sin D on whichfis one-one and such thatf(S) contains

a disk of radius r. (so f3(f) by

~ 712) . Bloch's constant is the number B defined

B = inf {f3(f):fE

According to Bloch's Theorem, B

~

~}.

1

72 . If one considers the function

fez) = z then clearly B:s; 1. However, better estimates than these are known. In fact, it is known that .43 :s; B :s; .47. Although the exact value of B remains unknown, it has been conjectured that

B

rG)rG~)

= --'---'---'-,'--"-

(1+)3)!rG)

A related constant is dehned as follows. 1.9 Definition. Let ~ be as in Definition 1.8. For eachfin ~ define I.(!) = sup {r:f(D) contains a disk of radius r}. Landau's constant L is defined by L = inf {..\(f):fE

~}.

Clearly L ~ B and it is easy to see that L :s; 1. Again the exact value of L is unknown but it can be proved that .50 :s; L :s; .56. In particular, L > B. 1.10 Proposition. Iff is analytic on a region containing the closure of the disk D = {z: Izl < I} andf(O) = 0,/'(0) = 1; thenf(D) contains a disk of radius L.

Proof The proof will be accomplished by showing that feD) contains a disk

of radius I. = A(f). For each n there is a point
feD). Now

Ef(D)

C

A-~)

f(D-) and this last set is compact. So there is a

296


point ex inf(D-) and a subsequence {ex n.} such that exn• ----)- ex. It is easy to see that we may assume that ex = lim ex n. If Iw-IX! < A, choose no such that !w-ex! < A-I/n o. There is an integer n l > no such that

Ian - 0:1 for n

~

n t • Hence !w-ex.!

1 < A --- - - Iw no :0;

0:1

!w-exl + lex-exn!

1 < A --

no 1

if n 2:: n l . That is WE B(ex n; A-lin) c feD). Since w was arbitrary it follows that B(ex; A) c feD) . •

1.11 Corollary. Let f be analytic on a region that contains B (0; R); then f(B(O; R» contains a disk of radius R!f'(O)!L.

Exercises 1. Examine the proof of Bloch's Theorem to prove that L ~ 1/24. 2. Suppose that in the statement of Bloch's Theorem it is only assumed that f is analytic on D. What conclusion can be drawn? (Hint: Consider the functionsJ.(z) = s-If(sz), 0 < s < L) Do the same for Proposition 1.10.

§2. The Little Picard Theorem The principal result of this section generalizes Theorem XL 3.6. However, before proceeding, a lemma is necessary.

2.1 Lemma. Let G be a simply connected region and suppose that f is an analytic function on G that does not assume the values 0 or 1. Then there is an analytic function g on G such that

fez) = -exp (bT cosh [2g(z)]) for z in G. Proof. Since f never vanishes there is a branch t of log f(z) defined on G; that is et = f. Let F(z) = (27Ti)-lt(Z); if F(a) = n for some integer n then lea) = exp (27Tin) = 1, which cannot happen. Hence F does not assume any integer values. Since F cannot assume the values 0 and I, it is possible to define H(z) = F(z) F(z)-1.

.J

.J

Now H(z) of- 0 for any z so that it is possible to define a branch g of logH on G. Hence cosh(2g)+ 1 = -t(e 2o+e- 2e )+ 1 = 1-(eO+e- 9)2 =

Schottky's Theorem

297

1 t(H + I/H)2 = 2F = -. 1Tl

t.

But this gives / = e t = exp (1Ti+1Ti cosh (2g)] =

-exp [rri cosh (2g)J. I11III Suppose / and g are as in the lemma, n is a positive integer, and m is any integer. If there is a point a in G with g(a) = ± log (v'n +.,;n=n +-!- im1T, then 2 cosh[2g(a)] = e 2g (a)+e- 2g (a) = e imn (y'n+,.[i!=T)±2+e -im"(v'n+ Jii"=I)+2 = {-om[(v'n+v'n-l)2 +(v'n-Jii-=l)2] = (-l)m[2(2n-l)]; or cosh [2g(a)] = (-l)m(2n-I). Therefore/(a) = -exp [( _l)m(2n-1)1Ti] and, since (2n -1) must be odd,j(a) = L Hence g cannot assu!lle any of the values

{± log (v'~+ v'~-I)+tim1T: n

2: 1, m = 0,

± 1, ... }.

These points form the vertices of a grid of rectangles in the plane. The height of an arbitrary rectangle is Itimrr-ti(m+ IH = !1T < v'3.

The width is log (v'n+ 1 +v'n)-Iog (v'n+v'n-1) > 0. Now cp(x) = log
2.3 Little Picard Theorem. IJ J is an entire Junction that omits two values J is a constant.

then

Proof. If I(z) "# a and I(z) "# b for all z then (f-a) (b-a)-l omits the values 0 and 1. So assume that/(z) "# 0 and/(z) "# 1 for all z. According to Lemma 2.2, this gives an entire function g such that g(e) contains no disk of radius 1. Moreover, if 1 is not a constant function then g is not constant so there is a point Zo with g'(zo) "# O. By considering g(z+zo) if necessary, it may be supposed that g'(O) "# O. But according to Corollary 1.11, geBCO; R) contains a disk of radius LRlg'(O)I. If R is chosen sufficiently large this gives that g(e) does contain a disk of radius I-a contradiction. So / must be constant, I11III

Exercises

I. Show that if J is a meromorphic function on C such that Coo - J(C) has at least three points then J is a constant. (Hint: What if 00 rt J«(:)?) 2. For each integer Il 2: 1 determine all meromorphic functions J and g on C with a pole at 00 such that In + gn = 1. §J'. Schottky's Theorem

Let 1 be a function defined on a simply connected region containing the disk 1) and suppose that 1 never assumes the valu«s 0 and 1. Let us examine the proof of Lemma 2.1. If t is any branch of log / let

nco;

298


I F=-I.

27Ti '

g = a branch of log H. There are two places in this scheme wm:re we are allowed a certain amount of latitude; namely, in picking the functions t and g which are branches of log f and log H, respectively. For the proof of Schottky's Theorem below specify these branches by requiring 3.1 0 ~ 1m teO) < 27T,

o~

3.2

1m g(O) < 27T.

3.3 Schottky's Theorem. For each a and Jj, U <. a < 00 and 0 ~ f3 ~ I, there is a constant C (a, f3) such thai if f is an analytic function on some simply connected region containing B (0; I) that omits the values 0 and I, and such that If(O)1 ~ a; then If(z)1 ~ C (a,p) for Izl ~ p.

Proof It is only necessary to prove this theorem for 2 is accomplished by looking at two cases.

~

IX <

00.

The proof

Case 1. Suppose 1- ~ If(O) I ~ IX. Recalling the functions F, H, and g in Lemma 2.1 (and rediscussed at the beginning of this section), (3.1) gives I IF(O) I = 27T Iiogi f(O) I+ i 1m t(O) I ~

Let COCIX)

3.4

I

I -loglX+l; 27T

= 27T log IX+ 1. Also

,,/F(O) ±~F(O)-II ~ I~F(O)I

= exp =

+ I~F(O)-II

0 log IF(O)I) + exp 0 log IF(O) -

11)

IF(O)ll + 1F(0) - III

~ CO(IX)t+[CO(IX)+lJ!

LetC 1(IX)

= CO(IX)t+[CotlX)+W. Now if IH(O)I ;::: 1 then (3.2) and (3.4) give Ig(O) I = Iiogi H(O) 1+ i 1m g(O)j IH(O)I +27T log C 1 (IX)+27T.

~ log ~

If IH(O)I < I then in a similar fashion.

Ig(O)j ~ -log IH(O)I + 27T

= log

CH~O)I) + 27T

= log I~ F(O) + ~'-F(-O)---ll + 27T ~

log C 1 (IX)+27T.

Schottky's Theorem

Let Cioc)

=

299

log C 1 (oc)+21T.

If lal < 1 then Corollary 1.11 implies that g(B(a; l-Ial) contains a disk of radius 3.5 L(I-lal) Ig'(a)l. On the other hand, Lemma 2.2 says that geBCO; 1» contains no disk of radius 1. Hence, the expression (3.5) must be less than 1; that is,

3.6

Ig'(a) I < [L(1-lal)]-1 for /al < 1.

If lal < 1, let,., be the line segment [0, a]; then Ig(a) I : :; Ig(O)1 + Ig(a)- g(O) I

:::;; Cz(oc)+lf g'(z) dzl y

:::;; Cz(oc) + /al max {lg'(z)l: z E [0, aJ} Using (3.6) this gives

Ig(a) I : :;

lal Cz(oc) + L(I-Ial)

If C 3 (oc, (:J) = Cioc)+(:J[L(I-(:J)r 1 then this gives Ig(z) I : :; C 3(OC, (:J)

if Izl :::;; (:J. Consequently if Izl :::;; (:J, I/(z) I = lexp [1Ti cosh 2g(z)]1

:::;; exp [1Tlcosh 2g(z)1J :::;; exp [1Te Zlg (z)l] :::;; exp [1Te ZC3 (12:.P)]; define C4 (oc, (:J) = exp {1T exp [2Cioc, (:J)]). Case 2. Suppose 0 < 1/(0)1 < t. In this case (1-f) satisfies the conditions of Case 1 so that 11-I(z) I : :; C4 (2, (:J) if Izl :::;; (:J. Hence I/(z) I : :; 1 + C4 (2, (:J). If we define qoc, (:J) = max {C4 (oc, (:J), 1 + C4 (2, (:J)}, the theorem is proved . • 3.7 Corollary. Let I be analytic on a simply connected region containing B(O; R) and suppose that I omits the values 0 and 1. If qoc, (:J) is the constant obtained in Schottky's Theorem and 1/(0)/ :::;; oc then /f(z)/ :::;; qoc, ~) for Iz/ :::;; (:JR. Proof. Consider the function I(Rz) , /z/ :::;; 1.. What Schottky's Theorem (and the Corollary that follows it) says is that a certain family of functions is uniformly bounded on proper subdisks of B(O; 1). By Montel's Theorem, it follows that this family is normal. It is this observation which will be of use in proving the Great Picard Theorem in the next section.


300

§4. The Great Picard Theorem The main tool used in the proof of Picard's Theorem is the following result. 4.1 Montel-Caratheodory Theorem. If ff is the family of all analytic functions on a region G that do not assume the values 0 and 1, then § is normal in C(G, C",,). Proof Fix a point Zo in G and define the families ~ and .Yf' by C§

= {jE§":

If(zo)1 ~ I},

.Yf' = {jE §": If(zo)1 ~ I}; so §" = ~ u .Yf'. It will be shown that ~ is normal in H(G) and that .Yf'is normal in C(G, Coo) (that 00 is a limit of a sequence in .Yf' is easily seen by considering constant functions). To show that ~ is normal in H(G), Montel's Theorem is invoked; that is, it is sufficient to show that ~ is locally bounded. If a is any point in G let y be a curve in G from Zo to a; let Do, D 1 , ••• , Dn be disks in G with centers zo, Z I' ... , Zn = a on {y} and such that Zk-I and Zk are in D k - I n Dk for 1 ~ k ~ n. Also assume that Dk c G for o ~ k ~ n. We now apply Schottky's Theorem to Do. It follows that there is a constant Co such that If(z) I ~ Co for;; in Do and f in ~. (If Do = B(zo; r) and R > r is such that B(zo; R) c G then, according to Corollary 3.7, If(z) I ~ C(I, 13) for Z in Do andfin C§ whenever 13 is chosen with r < f3R). In particular If(zl)1 ~ Co so that Schottky's Theorem gives that ~ is uniformly bounded by a constant CIon D 1 • Continuing, we have that ~ is uniformly bounded on Dn. Since a was arbitrary, this gives that C§ is locally bounded. By Montel's Theorem, C§ is normal in H(G). Now consider .Yf' = {f E §": If(zo)1 ~ I}. If f E .Yf' then Ilf is analytic on G because f never vanishes. Also Ilf never vanishes and never assumes the value 1; moreover 1(llf) (zo)1 ~ 1. Hence £ = {Ilf: f E .Yf'} c C§ and £ is normal in H(G). So if {jn} is a sequence in .Yf' there is a subsequence {f"k} and an analytic function h on G such that {l/f"J converges in B(G) to h. According to Corollary VII. 2.6 (Corollary to Hurwitz's Theorem), either h == 0 or h never vanishes. If h == 0 it is easy to see that fnk(z) -+ 00 uniformly on compact subsets of G. If h never vanishes then Ilh is analytic and it follows that /,.Jz) -+ IJh(z) uniformly on compact subsets of G . • 4.2 Great Picard Theorem. Suppose an analytic function f has an essential singularity at z = a. Then in each neighborhood of a f assumes ea.ch complex number, with one possible exception, an infinite number of times. Proof For the sake of simplicity suppose that f has an essential singularity

at

Z =

O. Suppose that there is an R such that there are two numbers not in

{fez): 0 < Izl < R}; we will obtain a contradiction. Again, we may suppose that fez) #- 0 and fez) #- 1 for 0 < Izi < R. Let G = B(O; R)- {O} and define flJ.: G -+ C by /,.(z) = J(zJn). So each!" is analytic and no fa assumes the value 0 or 1. According to the preceding theorem, {/,.} is a normal family in C(G, Coo).

The Great Picard Theorem

301

Let {.f..k } be a subsequence of {.f..} such that Ink -> rp uniformly on {z: /z/ = tR}, where rp is either analytic on G or rp == 00. If rp is analytic, let M = max {/rp(z)/: /z/ = tR}; then /f(Zjnk)/ = /.f..k(z)/ ~ Ilnk(z)-rp(z)1 + /rp(z)/ :;; 2M for nk sufficiently large and Izl = tR. Thus /f(z)/ :;; 2M for Iz/ = R/2nk and for sufficiently large nk. According to the Maximum Modulus Principle, f is uniformly bounded on concentric annuli about zero. This gives that f is bounded by 2M on a deleted neighborhood of zero and, so, z = must be a removable singularity. Therefore rp cannot be analytic and must be identically infinite. It is left to the reader to show that if rp == 00 then f must have a pole at zero. So at most one complex number is never assumed. If there is a complex number w which is assumed only a finite number of times then by taking a sufficiently small disk, we again arrive at a punctured disk in whichffails to assume two values . • An alternate phrasing of this theorem is the following.

°

4.3 Corollary. If f has an isolated singularity at z = a and if there are two complex numbers that are not assumed infinitely often by f then z = a is either a pole or a removable singularity. In the preceding chapter it was shown that an entire function of order .\, where .\ is not an integer, assumes each value infinitely often (Corollary XI.3.8). Functions of the form e g , for g a polynomial, assume each value infinitely often, although there is one excepted value-namely, zero. The Great Picard Theorem yields a general result along these lines (although an exceptional value is possible, so that the following result is not comparable with Corollary XI.3.8). 4.4 Corollary. If f is an entire function that is not a polynomial then f

assumes every complex number, with one possible exception, an infinite number of times.

°

Proof. Consider the function g(z)=f(l/z). Sincefis not a polynomial, g has an essential singularity at z = (Exercise V. l. 13). The result now follows from the Great Picard Theorem . • Notice that Corollary 4.4 is an improvement of the Little Picard Theorem. Exercises

l. Let f be analytic in G = B (0; R) - {o} and discuss all possible values of the integral _1 f'(z) dz 211i f(z)-a

f

1

where y is the circle /z/ = r < R and a is any complex number. If it is assumed that this integral takes on certain values for certain numbers a, does this imply anything about the nature of the singularity at z = 01 2. Show that if f is a one-one entire function then f(z) = az + b for some constants a and b, a*O. 3. Prove that the closure of the set .% in Theorem 4.1 equals .% together with the constant functions 00, 0, and 1.

Appendix A Calculus for Complex Valued Functions on an Interval In this Appendix we would like to indicate a few results for functions defined on an interval, but whose values are in C rather than IR. If f: [a, b] --+ C is a given function then one can easily study its calculus type properties by considering the real valued functions Re f and 1m f For example, the fact that for a complex number z = x+iy max (ixl, Iyi) ~ Izl =

.Jx 2+ y2

~ 2 max (lxi, Iyl),

easily allows us to show thatfis continuous iff Refand lmfare continuous. However we sometimes wish to have a property defined and explored directly in terms of f without resorting to the real and imaginary parts of f This is the case with the derivative off A.I Definition. A functionf: [a, b] _ C is differentiable at a point x in (a, b) if the limit l'

.lm

h-+O

f(x+h)-f(x) h

exists and is finite. The value of this limit is denoted by f'(x). For the points x = a or b we modify this definition by taking right or left sided limits. Iff is differentiable at each point of [a, b] then we say that f is a differentiable function and we obtain a new function f': [a, b] --+ C which is called the derivative off The next Proposition has a trivial proof which we leave to the reader. A.2 Proposition. Afunctionf: [a, b] --+ C is differentiable iff Ref and lmf are differentiable. Also, f'(x) = (Re f)'(x) + i(Im f)'(x) for all x in [a, b]. Of course it makes no sense to talk of the derivative of a complex valued function being positive; accordingly, the geometrical significance of the derivative of a real valued function has no analogue for complex valued fune-tions. However the reader is invited to play a game by assuming that Re f and 1m f have positive or negative derivatives, and then interpret these conditions for f One fact remains true for derivatives and this is the consequence of a vanishing derivative. A.3 Proposition. If a function f: [a, b] all x then f is a constant.

--+

303

C is differentiable and f'(x) = 0 for

304

Appendix A

Proof If f'(x) = 0 for all x then (Re f)'(x) = (1m f)'(x) = 0 for all x. It follows that Re f and 1m f are constant and, hence, so is f

One important theorem about the derivative of a real valued function which is not true for complex valued functions is the Mean Value Theorem. In fact, if f(x) = x 2 + ix 3 it is easy to show that f(b)-f(a)

= f'(c) (b-a)

for some point c in [a,b] only when a= b. One of the principal applications of the Mean Value Theorem for derivatives is the proof of the Chain Rule. In ligi;, of the discussion in the preceding paragraph one might well doubt the validity of the Chain Rule for complex valued functions. The Chain Rule tells us how to calculate the derivative of the composition of two differentiable functions; this leads to two different situations. First suppose that j: [a,b]~C is differentiable and let g: [c,d]~[a,b] also be differentiable. Then j(g(t»=Rej(g(t»+ i1mj( g(t»; from here the Chain Rule follows by applying the Chain Rule from Calculus. In the second case the result still holds. Let G be an open subset of C such that j([a,bJ) e.G and suppose h: G~C is analytic. We wish to show that hojis differentiable and calculate (h o])'. Since the proof of this Chain Rule follows the line of argument used to prove the Chain Rule for the composition of two analytic functions (Proposition 111.2.4) we will not repeat it here. We summarize this discussion in the following. A.4 Proposition. Let f: [a, b] --+ C be a differentiable function. (a) If g: [c, d] --+ [a, b] is differentiable then fog is differentiable and (f g)'(t) = f'(g(t»g'(t). (b) If G is an open subset of C containing f([a, b]) and h: G --+ IC is an analytic function then h 0 f is differentiable and (h 0 f)'(x) = h'(f(x»f'(x). To discuss integral calculus for complex valued functions we adopt a somewhat different approach. We define the integral in terms of the real and imaginary parts of the function. 0

A.S Definition. Iff: [a, b] of f over [a, b] by

--+

IC is a continuous function, we define the integral

Jf(x)dx = JRef(x)dx+i JImf(x)dx. b

a

b

a

b

a

If the reader wishes to see a direct development of the integral he need only work through Section IV. 2 of the text with yet) = t for all t. However, this hardly seems worthwhile. Besides the additivity of the integral the only result which interests us is the Fundamental Theorem of Calculus. Recall that if F: [a, b] --+ IC is a function and f = F' then F is called a primitive off.

Appendix A

305

A.6 Fundamental Theorem of Calculus. A continuous function f: [a, b] --* C has a primitive and any two primitives differ by a constant. If F is any primitive off then

f f(x) dx = F(b)-F(a) b

a

Proof If g and h are primitives of Refand Imfthen F of f The result now easily follows.

=

g+ih is a primitive

Appendix B Suggestions for Further Study and Bibliographical Notes GENERAL

The theory of analytic functions of one complex variable is a vast one. The books by Ahlfors [1], Caratheodory [9], Fuchs [17], Heins [24], Hille [26], Rudin [41], Saks and Zygmund [42], Sansone and Gerretsen [43], and Veech [46] treat some topics not covered in this book. In addition they contain material which further develops some of the topics discussed here. We have not touched upon the theory of functions of several complex variables. The book by Narasimhan [37] contains an elementary introduction to functions of several complex variables. Also Cartan [10] contains an introduction to the subject. The book by Whittaker and Watson [47] contains some very classical analysis and several bibliographical comments. Finally the two volume work by P6lya and Szego [39] should be looked at by every student. These books contain problems on analysis with the solutions in the back. CHAPTER III § I. A more thorough treatment of power series and infinite series in general can be seen in the book by Knopp [28]. §2. There are several ways to define an analytic function. Some books (for example, Ahlfors [I]) define a function to be analytic if it has a derivative at every point in an open subset of the plane. Other books (for example, Cartan [10]) define a function to be analytic in an open subset of the plane if at every point of this open set the function has a power series expansion. This latter approach has one advantage in that it is the standard way of defining a function to be analytic in several variables. In many ways the study of analytic function theory can be considered as the study of the logarithm function. This will become more evident in the remainder of the book. §3. More information concerning Mobius transformations can be obtained from the book by Caratbeodory [9].

CHAPTER IV

§3. See the paper by Burdick and Lesley [8] for more on uniqueness theorems. 307

308

Appendix B

§4 and §5. Cauchy's Theorem first appeared in the treatise (11]. However Cauchy's original statement was far different from the one that appears in this book. Cauchy proved his theorem under the assumption that the function] is differentiable and that the derivative is continuous on and inside a simple closed smooth curve. Goursat [20, 21] removed the assumption that l' is continuous but retained the assumption that the curve over which the integral is to be taken is a simple closed smooth curve. Pringsheim [40] introduced the method of proof that is used by many today. He first proved the theorem for triangles (the method used to prove Goursat's Theorem in §S) and then approximated the contours by polygons. The role of the winding number in Cauchy's theorem and the extension to a system of curves such that the sum of the winding numbers with respect to every point outside of the region of analyticity is zero, seems to have first been observed by Artin [4]. The proof of Theorem IV. 5.4 is due to Dixon [13]. CHAPTER V

For more examples of the use of residues to calculate integrals see the books by Lindelof [31] and Mitrinovic [35]. An interesting paper is the one by Boas and Friedman [7]. §3. The reference for Glicksberg's statement of Rouche's Theorem is [IS]. CHAPTER VI

The original paper of Ph ragmen and Lindelof [3S] is still worth reading. CHAPTER VII § 1. Much of the material in this section can be done in a more general setting. See for example the books by Kelley [27] and Dugundji [15]. §2. Montel's book [36] on normal families is well worth reading. In this treatise he explores several applications of the theory of normal families to problems in complex analysis. §4. There is a wealth of material on conformal mappings. Caratheodory [9] has some additional information. Also the books by Bergman [5], Goluzin [19], and Sansone and Gerretsen, vol. II [43] have extensive treatments. It is also possible to use Hilbert space techniques to construct conformal maps. See Bergman [5] and the second volume of Hille [26]. §7. For more information on the gamma function look at the book of Artin [3]. Also the paper [30] contains more information about the gamma function. An interesting survey article is one by Davis [12]. §S. The book by Edwards [16] gives a complete exposition of the Reimann zeta hypothesis from an historical point of view. This book

Appendix B

309

examines Riemann's original paper point by point and fully explicates his results. There is a result of Beurling (see the book by Donoghue [l4]) that gives an equivalent formulation of the Riemann zeta hypothesis in terms of functional analysis. CHAPTER VIII

The reference to Grabiner's treatment of Runge's Theorem is [22]. There are several other proofs of Runge's Theorem. One proof is by "pole pushing"; this was used in the earlier edition of this book and also appears in [42]. A proof using functional analysis appears in Rudin's book [41]. CHAPTER IX

§3. An interesting paper on the monodromy theorem is [45]. §6. The book by Springer [44] gives a readable introduction to the theory of Riemann surfaces. A more advanced treatment is [2]. §7. An excellent treatment of covering spaces can be found in Massey [33]. CHAPTER X

Further results on harmonic functions can be found in Helms [25]. This area of harmonic functions has been extended to functions of more than two variables. In addition the Dirichlet problem can be formulated and solved in this more general setting. CHAPTER XI

The theory of entire functions is one of the largest branches of analytic function theory. The book by Boas [6] is a standard reference. CHAPTER XII The book by Hayman [23] contains many generalizations of the theorems presented in this chapter. Also the paper by MacGregor [32] contains many applications and additional results. This paper also contains an interesting bibliography. There are essentially two ways of proving Picard's Theorem. The elementary approach used in this chapter is based on the treatment found in Landau's book [29]. The other treatment uses the modular function and can be found in [1] or [46J.

References

[1] [2] [3] [4] [5] [6] [7] [8] [9]

[10] [II]

[12] [I3] [I4] [IS] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] (29] [30] [31] (32] (33J [34] [35] [36] [j7] [38]

V. AHLFORS, Complex Analysis, McGraw-Hill Book Co. (1966). V. AHLFORS and L. SARlO, Riemann Surfaces, Princeton University Press (1960). AIlTIN, The Gamma Function, Holt, Reinhart, and Winston (1964). AIlTIN, Collected Papers, Addison-Wesley Publishing Co. (1965). BERGMAN, The Kernel Function and Conformal Mapping, American Mathematical Society (1970). R. P. BOAS, Entire Functions, Academic Press (1954). H. P. BOAS and E. FRIEDMAN, A simplification of certain contour integrals, Amer. Math. Monthly, 84 (1977), 467-468. D. BURDICK AND F. D. LESLEY, Some uniqueness theorems for analytic functions, Amer. Math. Monthly, 82 (1975), 152-155. C. CAllAmEoDOIlY, Theory of Functions (2 vols.), Chelsea Publishing Co. (1964). H. CAIlTAN, Elementary Theory of Analytic Functions of One qr Several CO"fJlex Variables, Addison-Wesley Publishing Co. (1963). A. CAUCHY, Memoire sur les Integrales Definies entre des Limites Imaginaires, Bure Freres (1825). PInuP J. DAVIS, Leonhard Euler's Integral: A historical profile of the gamma function, Amer. Math. Monthly 66 (1959), 849-869. JOHN D. DIXON, A brief proof of Cauchy's integral theorem, Proc. Amer. Math. Soc., 29 (1971), 625-<;26. W. F. DoNOGHUE, Distributions and Fourier Transforms, Academic Press (1969). J. DUGUNDn, Topology, Allyn and Bacon (1966). H. M. EowAIlDs, Riemann's Zeta Function, Academic Press (1974). W. H. J. FUCHS, Topics in the Theory of Functions of One Complex Variable, D. Van Nostrand Co. (1967). I. GUCKSBERG, A remark on Rouche's theorem, Amer. Math. Monthly, 83 (1976), 186. G. M. GoLUZIN, Geometric Theory of Functions of a Complex Variable, American Mathematical Society (1969). E. GoURSAT, Demonstration du TI!eoreme de Cauchy, Acta Math., 4 (1884), 197-200. E. GoURSAT, Sur la Definition General des Fonctions Analytiques d'apres Cauchy, Trans. Amer. Math. Soc., 1 (1900), 14-16. S. GIlABINEIl, A short proof of Runge's theorem, Amer. Math. Monthly, 83 (1976), 807-808. W. K. HAYMAN, Meromorphic Functions, Oxford at the Clarendon Press (1964). M. HBINS, Selected Topics in the Classical Theory of Functions of a CO"fJlex Variable, Holt, Reinhart, and Winston (1962). L. L. HELMS, Introduction to Potential Theory, Wiley-Interscience (1969). E. HILLE, Analytic Function Theory (2 vols.), Ginn and Co. (1959). J. L. KBu.Bv, General Topology, Springer-Verlag (1975). K. KNopp, Infinite sequences and series. Dover Publications (1956). E. LANDAU, Damellung und BegrUndung einiger neuerer Ergebnisse der Funktionentheorie, Springer-Verlag (1929). R. LBIPNIK AND R. OBERG, Subvex functions and Bohr's uniqueness theorem, Amer. Math. Monthly, 74 (1967) 1093-1094. E. LINom.OF, Le Calcui des Residues et ses Applications /0 tMor;e des Fonetions, Chelsea Publishing Co. (1947). T. H. MAcGllBGOIl, Geometric Problems in Complex Analysis, Amer. Math. Monthly 79 (1972), 447-468. W. MAssBY, Algebraic Topology: An Introduction, Springer-Verlag (1977). CJiA.tuJls A. McCAIlTIfY, The Cayley-Hamilton Theorem, Amer. Math. Monthly 82 (1974), 390-391. D. S. MITIlINOVIC, Calculus of Residues, P. Noordhoff (1966). P. MoNTEL, ~ons sur les series de polynomes, Gauthier-Villars (1910). R. NAIlASIMHAN. Several Complex Variables, The University of Chicago Press (1971). E. PHIlAGMEN AND E. LlNDm.OF, Sur une extension d'un principe chusique de I'lI1IIIIyse et

L. L. E. E. S.

a

sur quelques proprietes des fonctions manogeue.s dons Ie ooisinage d'1I1I point singulier,

Acta Math. 31 (1908),381-406.

311

312

References

[39) G. POLYA AND G. SZEGO, Problems and Theorems in Analysis (2 vols.), Spnnger-Veriag (vol. I, 1972; vol. 2, 1976). (40) A. PRINGSHEIM, {jber den Goursat'schen Beweis des Cauchy'schen Integralsatzes, Trans. Amer. Math. Soc., 2 (1901), 413-421. (41) W. RUDIN, Real and Complex Analysis, McGraw-Hill Book Co. (1966). (42) S. SAKS AND A. ZYGMUND, Analytic Functions, Monografie Matematyczne (1952). (43) G. SANSONE and J. GERRETSEN, Lectures on the Theory of Functions of a Complex Variable (2 vols.), P. Noordhoff (vol. I, 1960; vol. 2, 1969). (44) G. SPRINGER, Introduction to Riemilnn Surfaces, Addison-Wesley Publishing Co. (1957). (45) DAVID STYER and C. D. MINDA, The use of the monodromy theorem, Amer. Math. Monthly, 81 (1974),639--642. [46) W. A. VEECH, A Second Course in Complex Analysis, W. A. Benjamin, Inc. (1967). [47) E. WHITTAKER and G. N. WATSON, A Course of Modern Analysis, Cambridge University Press (1962).

INDEX Cayley-Hamilton Theorem 87 Chain of disks 216 chain rule 34 Change of parameter 64 closed ball 11 closed curve 66 closed map 99 closed set 12, 222 closure 13 compact 20, 224 complete analytic function 215, 216 - of local inverses 239 complete metric space 18 component 16 conformal mapping 46 conformally equivalent 160 conjugate 2 connected 14, 224 continuous function 24, 224 convergent sequence 18 convergent series 30, 106 convex function 134 convex set 89, 134 coordinate patch 234 cover 20 covering space 245 critical strip 193 cross ratio 48 curve 64

Abel's Theorem 74 absolute value 2 absolutely convergent product 166 absolutely convergent series 30, 106 analytic continuation along a path 214 - along a chain of disks 216 - of a germ 215 analytic function 34, 236 analytic manifold 234 analytic structure 234 analytic surface 234 angle between two paths 45 arc 229 arcwise connected 230 argument 4 argument principle 123 Arzela-Ascoli Theorem 148 barrier 269 base space 228, 232 Bernoulli numbers 76 Bessel function 122 beta function 186 Blaschke product 173 Bloch's constant 295 Bloch's Theorem 293 Bohr-Mollerup Theorem 179 boundary 13 bounded function 12 branch of the logarithm 39, 40

DeMoivre's Formula 5 dense 13 diameter of a set 19 differentiable function 33 dilation 47, 56 Dini's Theorem 150 Dirichlet problem 252, 269 Dirichlet region 269 distance between a point and a set 25 distance between two sets 27 distance function 11

Cantor's Theorem 19 Caratheodory's inequality 133 Casorati-Weierstrass Theorem 109, III

Cauchy's estimate 73 Cauchy's Integral Formula 84 Cauchy-Riemann equations 41 Cauchy sequence 18 Cauchy's Theorem - first version 85 - second version 89 - third version 90 - fourth version 94

elementary factor 168 empty set 12 313

314

Index

entire function 77 equicontinuous 148 equivalence of paths 64 essential singularity 105, III Euler's constant 75, 177, 185 Euler's Theorem 193 exponent of convergence 286 exponential function 32 exponential series 32 extended boundary 129 extended plane 8 FEP homotopic 93 f.i.p. 21 fiber 228 final point 229 finite genus 283 finite intersection property 21 finite order 285 finite rank 282 finitely generated ideal 208 fixed ideal 209 free ideal 209 function element 214 function of bounded variation 58 functional equation of the gamma function 178 of the zero function 192 fundamental set 245 Fundamental Theorem of Algebra 77, 100 gamma function 176 Gaussian psi function 186 Gauss's Formula 178 genus 283 geometric series 30 germ 214 Goursat's Theorem 100 Great Picard Theorem 300 greatest common divisor 174 Green's function 275 Hadamard's Factorization Theorem 289 Hadamard's Three Circles Theorem 137

Hardy's Theorem 138 harmonic conjugate 42, 252

harmonic function 41, 252 Harnack's Inequality 260 Harnack's Theorem 261, 268 Hausdorff space 224 Heine-Borel Theorem 23 holomorphic function 126, 151 homeomorphism 163, 202 homologous to zero 95 homomorphism 6 homotopic closed curves 88 homotopic to zero 89 Hurwitz's Theorem 152 ideal 174 ideal generated by Y 208 imaginary part 2 Independence of Path Theorem 93 index 81 infinite order 285 infinite product 164 infinite rank 282 initial point 229 integration around a branch point 119 interior 13 Intermediate Value Theorem 27 inversion 47 isolated singularity 103 isomorphism 238 Jensen's Formula 280 Jordan Curve Theorem 213 Landau's Constant 295 Laplace's Equation 252 Laurent Series Development 107 Lebesgue's Covering Lemma 21 left side of a circle 53 Leibniz's Rule 68, 73 lifting 246 limit inferior of a function 129 - of a sequence 30 limit point 18, 222 limit superior of a function 129 - of a sequence 30 line integral 63 linear fractional transformation 47 Liouville's Theorem 77, 123, 238 Lipschitz function 25 Little Picard Theorem 297

Index locally bounded 153 locally arcwise connected 230 locally Lipschitz 154 logarithmically convex function 135

.MVP 253, 260 maximal ideal 174 Maximum Modulus Theorem, - first version 79, 128, 238 - second version 128 - third version 129 Maximum Principle for harmonic functions - first version 253 - second version 254 - third version 264 - fourth version 264 mean value property 253, 260 Mean Value Theorem 253 meromorphic function 123 metric space II metrisable space 223' Minimum Principle 129,255 Mittag-Leffler'S Theorem 205 Mobius transformation 47 Monodromy Theorem 219,221,247 Montel-Caratheodory Theorem 300 Montel's Theorem 153 Morera's theorem 86 multiplicity of a zero 76 neighborhood system 226 normal set of functions 146 normal subgroup 57 open ball 11 open cover 20 open map 99 Open Mapping Theorem 99, 238 open set, 12, 222 order 285 orientation 52 Orientation Principle 53 parallelogram law 3 partial fractions 106 path 45, 229 pathwise connected 230

31S

periodic function 38 Perron family 266 Perron function 268 Phragmen-Lindelof Theorem 138 Picard's Great Theorem 300 Picard's Little Theorem 297 Poisson-Jensen Formula 281 Poisson Kernel 256 pole 105, 111 polygon 15 polynomially convex hull 200 power series 30 prime ideal 174 primitive 65, 94 principal branch of the loagritbm 40 principal ideal 208 projection map 228 proper ideal 174 properly covered 245 punctured disk 103 radius of convergence 31 rank 282 real part 2 rectifiable path 62 region 40 regular function 151 relative topology 225 relatively prime 208 removable singularity 103, III residue 112 Residue Theorem 112 Riemann's Functional Equation 192 Riemann Hypothesis 193 Riemann Mapping Theorem 160,277 Riemann Surface 232 Riemann Zeta Function 187 right side of a circle 53 roots of unity 5 rotation 47 Rouche's Theorem 125 Runge's Theorem 198 Schottky's Theorem 298 Schwarz Reflection Principle 211, 256 Schwarz's Lemma 130 sequentially compact 21 sheaf of germs 228 simple group 57

Index

316

simple root 98 simply connected 93, 202 singular part 105 singularity at infinity III stalk 228 standard form 283 star shaped set 89 stereographic projection 9 subharmonic function 264 superharmonic function 264 symmetric points 50 Symmetry Principle 52 terminal point 229 topological space 221 topology induced by a neighborhood system 227 . total boundedness 22 total variation 58 trace of a path 62, 229 translation 47

triangle inequality 3, 11 triangular path 86 trivial zeros 193 uniform convergence 28, 29 uniformly continuous 25 uniformly convergent integral 154 unrestricted analytic continuation 219 Vitali's Theorem 154 Wallis's Formula 176 Weierstrass Approximation Theorem 263 Weierstrass Factorization Theorem 170,207 Weierstrass M-test 29 Weierstrass Pe Function 207 Weierstrass Zeta Function 207 winding number 81, 250

LIST OF SYMBOLS C,IR

I

JzJ argz cisO

2 4 4 8 12 12 13 13 13

z,

Coo

0

B(S)

A~

intA ilA Q [z,wl limj(x)=w x_a

d(x,A) dCA, B) j=u-limjn liminfan lim supan GL 2(C) SLiC) SU2

v(y; P) V(y) f~/dy fJ fJldzl

n(y;a)

Yo -

Yl

y-O y",,=,O Res(j; a) 8",G liminfj(z) z~a

limsup/(z)

88 89 95 112 129 129 129

z~a

C(G,Q) p(j,g) H(G) M(G) JL(j) '£ (f), '£ (g) yeA)

13

15 24 25 27

L', II'

29 30 30 57 57 57 58 58 60 63

£ &J (z)

G* (j, G)

[fla ffy

N(j,D)

AI(a,rfla) (U,
64

81

317

142 143 151 155 158 174 174 175 201 207 210,213 214 214 225 228 228 234 234 261 266

Graduate Texts in Mathematics (co",inuetlfrom pag' Ii)

64 65 66 67 68 69 70 71 72 73 74 75 76 77

78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95

EDWARDS. Fourier Series. Vol. I. 2nd ed. WELLS. Differential Analysis on Complex Manifolds. 2nd ed. WATBRHOUSE. Introduction to Affine Group Schemes. SERRE. Local Fields. WEIDMANN. Linear Operators in Hilbert Spaces. LANG. Cyclotomic Fields ll. MASSEY. Singular Homology Theory. FARKAsIKRA. Riemann Surfaces. 2nd ed. STILLWELL. Classical Topology and Combinatorial Group Theory. 2nd ed. HUNGERFORD. Algebra. DAVENPORT. Multiplicative Number Theory. 3rd ed. HOCHSCHILD. Basic Theory of Algebraic Groups and Lie Algebras. IITAKA. Algebraic Geometry. HECKE. Lectures on the Theory of Algebraic Numbers. BURRIslSANKAPPANAVAR. A Course in Universal Algebra. WALTERS. An Introduction to Ergodic Theory. ROBINSON. A Course in the Theory of Groups. 2nd ed. FORSTER. Lectures on Riemann Surfaces. BOTTITu. Differential Forms in Algebraic Topology. WASHINGTON. Introduction to Cyclotomic Fields. 2nd ed. IRELAND/ROSEN. A Classical Introduction to Modem Number Theory. 2nd ed. EDWARDS. Fourier Series. Vol. ll. 2nd ed. VAN UNT. Introduction to Coding Theory. 2nded. BROWN. Cohomology of Groups. PIERCE. Associative Algebras. LANG. Introduction to Algebraic and Abelian Functions. 2nd ed. BRfllNDSTEO. An Introduction to Convex Polytopes. BEARDON. On the Geometry of Discrete Groups. DIESTEL. Sequences and Series in Banach Spaces. DuBROVIN/FoMENKO!NovIKOV. Modem Geometry-Methods and Applications. Part I. 2nd ed. WARNER. Foundations of Differentiable Manifolds and Lie Groups. SHIRYAEV. Probability. 2nd ed.

96 97 98 99 100 101 102 103 104 105 106 107 !O8 109 110 111 112 113 114 115 116 117 118 119 120 121 122

CONWAY. A Course in Functional Analysis. 2nded. KOSUTZ. Introduction to Elliptic Curves and Modular Forms. 2nd ed. BROCKER/TOM DIECK. Representations of Compact Lie Groups. GRovE/BENSON. Finite Reflection Groups. 2nded. BERG/CHRISTENsEN/REsSEL. Harmonic Analysis on Sernigroups: Theory of Positive Definite and Related Functions. EDWARDS. Galois Theory. VARADARAJAN. Lie Groups, Lie Algebras and Their Representations. LANG. Complex Analysis. 3rd ed. DUBROVINlFoMENKoINoVIKOV. Modem Geometry-Methods and Applications. Parlll. LANG. S~(R). Sn.VERMAN. The Arithmetic of Elliptic Curves. OLVER. Applications of Lie Groups to Differential Equations. 2nd ed. RANGE. Holomorphic Functions and Integral Representations in Several Complex Variables. LEHTO. Univalent Functions and Teichmiiller Spaces. LANG. Algebraic Number Theory. HUSEMOILER. Elliptic Curves. 2nd ed. LANG. Elliptic Functions. KARATZAslSHREVE. Brownian Motion and Stochastic Calculus. 2nd ed. KosUTZ. A Course in Number Theory and Cryptography. 2nd ed. BERGERIGOSTIAUX. Differential Geometry: Manifolds, Curves, and Surfaces. KELLEy/SRTNTVASAN. Measure and Integral. Vol. I. J.-P. SERRE. Algebraic Groups and Class Fields. PEDERSEN. Analysis Now. ROTMAN. An Introduction to Algebraic Topology. ZIEMER. Weakly Differentiable Functions: Sobolev Spaces and Functions of Bounded Variation. LANG. Cyclotomic Fields I and ll. Combined 2nd ed. REMMERT. Theory of Complex Functions. Readings in Mathematics

123 ESBINGHAUslHERMES et al. Numbers. Readings in Mathematics

124 DUBROVINlFoMENKoINoVIKOV. Modem Geometry-MethOds and Applications. Part ill

125 BERENSTBlNfGAY. Complex Variables: An Introduction. 126 BOREL. Linear Algebraic Groups. 2nd ed. 127 MASSEY. A Basic Course in Algebraic Topology. 128 RAUCH. Partial Differential Equations. 129 FuLTONfHARRIs. Representation Theory: A First Course. Readings in Mathematics 130 DODSONfPoSTON. Tensor Geometry. 131 LAM. A First Course in NoncoDnnutative Rings. 2nd ed. 132 BEARDON. Iteration of Rational Functions. 133 HARRIS. Algebraic Geometry: A First Course. 134 ROMAN. Coding and Information Theory. 135 ROMAN. Advanced Linear Algebra. 2nded. 136 AnKINsfWEINTRAUB. Algebra: An Approach via Module Theory. 137 AxLERlBoURDONfRAMEY. Harmonic Function Theory. 2nd ed. 138 COHEN. A Course in Computational Algebraic Number Theory. 139 BREDON. Topology and Geometry. 140 AUBIN. Optima and Equilibria. An Introduction to Nonlinear Analysis. 141 BECKERlWEISPFENNINGfKREOEL. Grabner Bases. A Computational Approach to Commutative Algebra. 142 LANG. Real and Functional Analysis. 3rd ed. 143 DooB. Measure Theory. 144 DENNISfFARB. Noncommutative Algebra. 145 VICK. Homology Theory. An Introduction to Algebraic Topology. 2nded. 146 BRIDGES. Computability: A Mathematical Sketchbook. 147 ROSENBERG. Algebraic K-Theory and Its Applications. 148 ROTMAN. An Introduction to the Theory of Groups. 4th ed. 149 RATCLIFFE. Foundations of Hyperbolic Manifolds. ISO EISENBUD. Commutative Algebra with a View Toward Algebraic Geometry. lSI SILVERMAN. Advanced Topics in the Arithmetic of Elliptic Curves. 152 ZIEGLER. Lectures on Polytopes. 153 FuLTON. Algebraic Topology: A First Course. 154 BROWNfPEARCY. An Introduction to Analysis. 155 KAssEL. Quantum Groups. 156 KECHRIS. Classical Descriptive Set Theory.

157 MALLIAVIN. Integration and Probability. 158 ROMAN. Field Theory. 159 CONWAY. Functions of One Complex Variable II. 160 LANG. Differential and Riemannian Manifolds. 161 BORWEINfERDELYI. Polynomials and Polynomial Inequalities. 162 ALPERlNfBELL.Groupsand Representations. 163 DIXONfMORTIMER. Permutation Groups. 164 NATHANSON. Additive Number Theory: The Classical Bases. 165 NATHANSON. Additive Number Theory: Inverse Problems and the Geometry of Sumsets. 166 SHARPE. Differential Geometry: Cartan's Generalization of Klein's Erlangen Program. 167 MORANDI. Field and Galois Theory. 168 EWALD. Combinatorial Convexity and Algebraic Geometry. 169 BHATIA. Matrix Analysis. 170 BREDON. Sheaf Theory. 2nd ed. 171 PETERSEN. Riemannian Geometry. 172 REMMERT. Classical Topics in Complex Function Theory. 173 DIESTEL. Graph Theory. 2nd ed. 174 BRIDGES. Foundations of Real and Abstract Analysis. 175 LICKORISH. An Introduction to Knot Theory. 176 LEE. Riemannian Manifolds. 177 NEWMAN. Analytic Number Theory. 178 CLARKEfLEDY AEvfSTERNfWOLENSKl. Nonsmooth Analysis and Control Theory. 179 DOUGLAS. Banach Algebra Techniques in Operator Theory. 2nd ed. 180 SRIVASTAVA. A Course on Borel Sets. 181 KREss. Numerical Analysis. 182· WALTER. Ordinary Differential Equations 183 MEGGINSON. An Introduction to Banach Space Theory. 184 BOLLOBAS. Modem Graph Theory. 185 CoXlLI1TLEfO'SHEA. Using Algebraic Geometry. 2nd ed. 186 RAMAKRISHNANNALENZA. Fourier Analysis on Number Fields. 187 HARRlsfMORRISON. Moduli of Curves. 188 GOLDBLATT. Lectures on the Hyperreals: An Introduction to Nonstandard Analysis. 189 LAM. Lectures on Modules and Rings. 190 EsMONDFlMURTY. Problems in Algebraic Number Theory. 2nd ed

191 LANG. Fundamentals of Differential Geometry. 192 HIRscH/LACOMBE. Elements of Functional Analysis. 193 COHEN. Advanced Topics in Computational Number Theory. 194 ENGEllNAGEL. One-Parameter Semigroups for Linear Evolution Equations. 195 NATHANSON. Elementary Methods in Number Theory. 196 OSBORNE. Basic Homological Algebra. 197 EISENBUD/HARRIs. The Geometry of Schemes. 198 ROBERT. A Course in p-adic Analysis. 199 HEDENMALMlKoRENBLUMlZHU. Theory of Bergman Spaces. 200 BAO/CHERNISHEN. An Introduction to Riemall.n-Finsler Geometry. 201 HINDRY/SILVERMAN. Diophantine Geometry: An Introduction. 202 LEE. Introduction to Topological Manifolds. 203 SAGAN. The Symmetric Group: Representations, Combinatorial Algorithms, and Symmetric Functions. 204 ESCOFIER. Galois Theory. 205 FEuxlHALPERINITHOMAS. Rational Homotopy Theory. 2nd ed. 206 MURTY. Problems in Analytic Number Theory. Readings in Mathematics 207 GODsILIROYLE. Algebraic Graph Theory. 208 CHENEY. Analysis for Applied Mathematics. 209 ARVESON. A Short Course on Spectral Theory. 210 ROSEN. Number Theory in Function Fields. 211 LANG. Algebra. Revised 3rd ed 212 MATOUSEK. Lectures on Discrete Geometry. 213 FRITzSCHElGRAUERT. From Holomorphic Funciions to Complex Manifolds. 214 JOST. Partial Differential Equations. 215 GOLDSCHMIDT. Algebraic Functions and Projective Curves. 216 D. SERRE. Matrices: Theory and Applications. 217 MARKER. Model Theory: An Introduction. 218 LEE. Introduction to Smooth Manifolds. 219 MACLACHLANIREID. The Arithmetic of Hyperbolic 3-Manifolds. 220 NESTRUEV. Smooth Manifolds and Observables. 221 GRONBAUM. Convex Polytopes. 2nd ed. 222 HALL. Lie Groups, Lie Algebras, and Representations: An Elementary Introduction.

223 VRETBLAD. Fourier Analysis and Its Applications. 224 WALSCHAP. Metric Structures in Differential Geometry. 225 BUMP: Lie Groups 226 ZHU. Spaces of Holomorphic Functions in the Unit Ball. 227 MILLERISTURMFELS. Combinatorial Commutative Algebra. 228 DIAMOND/SHURMAN. A First Course in Modular Forms. 229 EISENBUD. The Geometry of Syzygies. 230 STROOCK. An Introduction to Markov Processes. 231 BJORNERIBRENTI. Combinatorics of Coxeter Groups. 232 EVERESTIWARD. An Introduction to Number Theory.

Conway, John B - Functions of One Complex Variables I (1).pdf

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