Analysis III

M.S (M&6-*0&65) S*21 %*&4 P&3*4-"II A1&/95.5 - III &-4+75-

Dear Students, We welcome you as a student of the Second year M.Sc degree course. This paper deals with the subject BANAL'"I"

material

for



IIIC

this

.

The

paper

learning will

be

supplemented by contact lectures. In this book the first five units deal with Real analysis and the last five units deal with Measure Theory. Learning

through

Education

mode,

the as

you

Distance are

all

aware, involves self learning and self

assess ment and in this regard you are expected to put in disciplined and dedicated effort. As

our

part,

we

assure

guidance and support. With best wishes,

of

our

"'LLAB$" M.Sc., Second Year

P)8-:  %II  A6)4@;1;  III

Unit

1: De fi ni ti on s an d ex is te nc e of Integral, Properties of the

th e

Integral,

In te gr at io n

an d

di ff er en ti at io n,

Integration

of

functions,

Rectifiable

vector

valued curves

(Chapter 6 : Sections 6,1 to 6.23)

Unit 2: Di sc us si on of th e ma in pr ob le m, uniform

convergence,

convergence

and

Uniform continuity,

uniform

convergence

int egr ati on,

uni for m

and

con ver gen ce

and Differentiation, Equi continuous families of functions, The Stone – Wei ers tra ss The ore m (Ch apt er 7: Sections 7.1 to 7.33)

Unit 3: Power series, The exponential and Logarithmic

functions,

The

trigonometric algebraic

Functions,

completeness

The

of

the

Compl ex fiel d, Four ier seri es, The Gamma

function

(Chapter

8:

Sections 8.1 to 8.22)

Unit 4: Linear

Transformations

Di ff er en tia tio n pr in ci pl e

-

–theorem

Th e

Th e

con tr ac tion

in ve rs e

(Chapter

fu nc ti on

9

relevant

sections)

Unit 5: The implicit fu nction theorem – T ra nk

theo re m

Derivatives

he

– Det er min ant s –

of

higher

order

–

Differentiation of integrals (Chapter 9 relevant sections)

Unit 6: Lebesgue

outer

measure –

Measurable sets – Regularity.

Unit 7: Meas ura ble fun cti ons – Bor el and Lebesgue measurability.

Unit 8: Integration

of

non-negative

functions – The general integral – Integration of series

Unit 9: Riemann and Lebesgue integrals – The four derivatives – Continuous non–differentia

ble functions.

Unit 10: Functions of bounded variations – Le bes gu e diff ere nti atio n theo re m – Differentiation

and

integration

–

The Lebesgue set

#-?< B773;:

1.

Principles

of

Mathematical

Analysis by Walter Rudin , Third Edition,

McGraw

International

Student

Hill

,

Edition,

1976 Chapters 6,7,8,9. 2.

Measure Theory and Integration by G. de Barra, Willey Eastern Ltd 2 edition 1991 Chapters 2,3 and 4

SCHEME OF LESSONS

ANALYSIS – III

".

#I#LE

N7

$6< 1

1.1:

Definitions

and

existence

of

the

1 Integral

2

1.2 : Properties of the Integral.

3

1.3 : Integration and differentiation

1.4

:

Integration

of

vector

valued

4 functions

5

1.5 : Rectifiable curves

$6< 2

6

2.1 : Discussion of the main problem

7

2.2: Uniform convergence

8

2. 3: Un if or m co nv er ge nc e an d co nt in ui ty

2.4:

uniform

convergence

and

convergence

and

9 integration

2.5:

uniform

10 Differentiation

11

2.6 : Equ i con tin uou s fam il ies of fun ct ion s

12

2. 7: Th e St on e – We ie rs tr as s Th eo re m

$61< 3

13

3.1: Power series

3. 2:

Th e ex po ne nt ia l an d Lo ga ri th mi c

14 functions

15

3. 3: Th e tr ig on om et ri c Fu nc ti on s

3.4: The algebraic completeness of the 16 Complex field

17

3.5: Fourier series

18

3.6: The Gamma function

$61< 4

19

4.1: Linear Transformations

20

4.2: Differentiation

21

4.3: The contraction principle.

22

4.4: The inverse function theorem

$61< 5

23

5.1: The implicit function theorem

24

5.2: The rank theorem

25

5.3: Determinants

26

5.4: Derivatives of Higher Order

27

5.5: Differentiation of integrals

$NI#-1

$6< "<:=+<=:

Section 1.1 : Definitions and existence of the Integral Section 1.2: Properties of the Integral. Section 1.3 : Integration and differentiation Section

1.4:

Integration

of

vector valued functions Section 1.5 : Rectifiable curves

I6<:7,=+<76

A satis facto ry discu ssion of the main concepts of analysis must be based on an accurately defined number

concept. In this unit we discuss the concept of Riemann-Stieltjes integral and its properties, some theorems of Integrati on of vector valued functions and Rectifiable curves.

SECION-1.1

DEFINIIONS &

E$ISENCE OF HE INEGRAL

D-.161<176:

Let [a,b] be a given interval, by a partition P of [a,b] we mean a finite set of points x 0 , x 1 , x 2 ,...,x n where a=x 0 ≤ x 1 ≤ x 2 ...≤x n = b We write .,n.

x i = x i – x i–1 for i = 1,2,..

Supp ose f is a boun ded real func tion defined on [a,b] corresponding to each partition P of [a,b] we put Mi = supf(x), ( xi-1 ≤ x ≤ xi), mi = inf f(x), ( xi-1 ≤ x ≤ xi), n

n

U(P,f) = ∑ MiΔ"i, L(P,f) = ∑ miΔ"i i

=1

i

=1

b

_

and finally put

∫fd" = inf U (P, f)

------------------(1)

a b

∫fd" =supL (P,f)

------------------(2)

¯ a

where the inf and sup are taken over all partitions P of [a,b]. The left members of (1) and (2) are called Upper and Lower Riemann integrals.If

the

upper

and

lower

integrals are equal, we say that f

is Riemann integrable on [a,b], we write f

F

common b

>

and we denote the

value

of

(1)

&

(2)

b

∫fd" or ∫f(")d" .This is the Riemann a

a

integral of f over [a,b]. Since f is bounded, there exist m and M such that m ≤ f(x) ≤ M (a≤x≤b). Hence, for every P, m(b – a) ≤

L(P,f)

≤ U(P,f) ≤ M(b – a), so that the numbers L(P,f) and U(P,f) form a bounded set. That is, the upper and lower integrals are defined for every bounded function f. D-.161<176:

Riemann-Stielties integral

Let

be monotonically increasing

function on [a,b]. Corresponding to each partition P of [a,b] we write Δα i =

(xi )–

(xi–1 ) for i = l,2,...,n.

Supp ose f is a boun ded real func tion on [a,b] we put n

n

U(P,f,α) = ∑ MiΔαi, L(P,f,α) = ∑ miΔαiand we define i

=1

i

=1

b

_

∫fd

= inf U(P,f,α)

-------------------(3)

a b

∫fd

= sup L(P,f,α)

-------------------(4)

¯ a

where inf and sup are taken over all partitions P of [a,b]. The left members of (3) and (4) are called the upper and lower Riemann-Stieltjes integral of f with respect to [a,b] and we write f

F >(

, over ).

If the left members of (3) and (4) are equal we say that f is RiemannStieltjes integrable with respect to

,

over [a,b]. !-5)3:

By taking

(x) = x, the Riemann

integral becomes a specia l ca se of Riemann-Stieltjes integral. D-.161<176:

We say that the partition P* is a refinement of P if P* Given two partitions P

P

P. 1

and P 2 we say

that P* is there common refinement if P* = P 1 K P 2 .

#0-7-5 1.1.1:

If P* is a_refinement of P then L(P,f,

) ≤ L(P*,f,

U(P,f,

).

) and U(P*,f,

) ≤

P77:

Let P = (a= x 0 , x 1 , x 2 ,..x n = b} be a partition of [a,b]. Let P* be a refinement of P. Then P* P P. Su ppose first that P* co ntains just one point more that P. Let

this

extra

point

be

x*

and

x i–1 ≤x*≤ x i where x i–1 and x i are two consecutive points of P. Put w1 = inf(f(x)), xi-1 ≤ x ≤ x* w2 = inf ( f(x)), x* ≤ x ≤ xi and mi = inf(f(x)), xi-1 ≤ x ≤ xi Now L(P*,f,α) − L(P,f,α)

=w1[α(x*) − α(xi-1)] + w2[α(xi) − α(xi-1)] =w1[α(x*) − α(xi-1)] + w2[α(xi) − α(x*)] -mi[α(xi) − α(xi-1) + α(x*) − α(x*)] = w1 − mi α(x*) − α xi-1

)[

(

≥0.

(

+ w2 − mi α xi − α(x*)

)] (

)[ ( )

]

(i.e)L(P*,f,α) − L(P,f,α) ≥ 0. L(P*,f,α) ≥ L(P,f,α) L(P*,f,α) ≤ L(P,f,α)

If P* contains k points more than P , we repeat this reasoning k times and arrive at a result L(P*,f

) ≥ L(P,f,

Similarly,

prove

U(P*,f,

we

) ≤ U(P,f,

can

that

).

#0-7-5 1.1.2: b

b

_

∫fdα ≤ ∫fdα ¯ a

a

P77:

Let P* be the common refinement of two partitions P 1 and P 2 .

).

Then

L(P 1 ,f,

U(P*,f,

)

≤

) ≤ U(P 2 ,f,

L(P*,f,

)

≤

).

Then L (P1,f,α) ≤ L(P,f,α) ≤ U(P*,f,α) ≤ U(P2,f,α) Hence L(P1, f,

------------(1) ) ≤ U(P2, f, ) If P 2 is fixed and supremum is taken

over all P 1 , (1) gives sup L(P 1 ,f, U(P 2 ,f,

)≤

).

sup L(P1, f,

) ≤ U(P2, f, )

b

∫fdα ≤ U(P , f,

(i.e)

2

)

-----------------(2)

¯ a

By taking inf over all P 2 in (2), we get b

∫fd

≤ infU(P2, f,

)

¯ a

b

L

b

∫fdα ≤ ∫fdα ¯ a

¯ a

#0-7-5 1.1.3:

( Necessary and Sufficient condition for Riemann-Stieltjes integrability)

f

F >(

) on [a,b] if and only if for

every

> 0, ther e exist a partition P

such that U(P,f,

) – L(P,f,

)<

P77.:

N-+-;;)@ C76,1<176: Let f F

>(

).

Then,

by

definition

b

_

b

∫fdα = ∫fd ¯

----------------(1)

a

a

b

Since

∫fd

is the supremum of

a ¯

L(P,f,

) over all partitions P , there

exist

a

partition

b

∫fdα < L(P , f, 1

¯ a

ε

)+ 2

P 1

such

that

b

_

Also , since

∫fd

is the infimum of

a

U(P,f,

) over all partitions P ,there

exist

a

partition

P 2

such

that

b

_

U(P 2 ,f,

)<

∫

fdα

+

ε 2

a

If P = P 1 K P 2 , then P is the common refinement of P 1 and P 2 Then, by definition of refinement, we have b

∫fdα < L(P,f,α) +

F

------------------(2)

2

¯ a b

_

and U(P,f,α) <

∫

F

fdα + 2

a

Adding (2) and (3), we get

------------ (3)

b

_

b

∫fdα + U(P,f,α) < L(P,f,α) + ∫fdα + ε ¯

a

a

By (1), we have U(P,f,α) < L(P,f,α) + ε

(i.e)U(P,f,α)-L(P,f,α)<ε "=..1+1-6< C76,1<176

Let U(P,f, For

) – L(P,f,

every

partition

we

have

_

∫fdα ≤ ∫fdα ≤ U(P,f,α) ¯

a

a

b _

P

.

b

b

L(P,f,α) ≤

)<

b _

∫fdα − ∫fdα ≤ U(P,f,α) − L(P,f,α) < ε a

a b

(i.e)

b

_

∫fdα − ∫fdα < ε ¯ a

a

This is true for

> 0, we have

b

_

b

∫fdα = ∫fdα. ¯

a

a

Hence f

F >(

) on [a,b].

#0-7-5 1.1.4:

a.

If U(P,f,

) – L(P,f,

) <

for some P and some inequality

holds

holds then the

for

every

refinement P. b.

If If U(P,f, for P = { x

) – L(P,f, 0,

)<

holds

x 1 x 2 ,..., x n } and

if s i ,t i are arbitrary points in [x i–1 ,x i ] then n

∑ |f(si) − f(ti)|Δαi < ε i=1

c.

If f

F >(

) and ti F [x i–1 ,x i ] then

n

b

∑ f(ti) Δαi − ∫fdα a

i=1

|

|

P77.:

a.

<ε

Let U(P,f,

) – L(P,f,

) <

some partition P and some

for > 0.

Let P* be the refinement of P. Then U(P*,f, L(P,f, L

) ≤ U(P,f,

) ≤ L(P*,f,

).

We have U(P*,f,

≤ U(P,f, L U(P*,f,

) and

) – L(P,f, ) – L(P*,f,

) – L(P*,f, )< )<

Hence the result holds good for every refinement of P. b.

Let s i ,t i he two arbitrary points in [x i–1 ,x i ].

)

Then |f(si) − f(ti)| ≤ Mi − mi n L

n

∑ |f(si) − f(ti)|Δαi ≤ ∑ (Mi − mi)Δαi i=1

i=1 n

n

≤ ∑ (MiΔαi) − ∑ (miΔαi) i=1

i=1

≤ U(P,f,α) − L(P,f,α) < ε n

(i.e)∑ |f(si) − f(ti)| Δαi < ε. i=1

Let t i F [xi-1, xi],

D

i=1,2,....,n

Then M i ≤ f(ti) ≤ mi n L

D

i=1,2,....,n

n

n

∑ miΔαi ≤ ∑ f(ti) Δαi ≤ ∑ MiΔαi i=1

i=1

i=1

n L

L(P,f,α) ≤ ∑ f(ti) Δαi ≤ U(P,f,α) ---------(1) i=1 b

_

b

c.

Also L (P,f,α) ≤

∫fdα ≤ ∫fdα ≤ U(P,f,α). a

¯ a b L

L(P,f,α) ≤

∫fd

≤ U(P,f,α)

------------(2)

a

From (1) and (2), we get

|

n

b

∑ f(ti)Δαi − ∫fdα i=1

a

|

≤ U(P,f,α) − L(P,f,α) < ε

#0-7- 1.1.5:

If f is continuous on [a,b] then f F >

(

) on [a,b].

P77:

Let

> 0 be given.

Choose

> 0 so that (

(b) –

(a))

< ε. Since f is continuous on [a,b] and [a,b]

is

compact,

f

is

uniformly

continuous. L

There exists a

F [a,b]

> 0 such that x,t

and

|x-t| < δ

B

| f ( x ) − f (t )| <

-------(1)

If P is any partition of [a,b] such that Δx i <

, D i, then (1) implies

Mi − mi < η n

L

n

U(P,f,α) − L(P,f,α) = ∑ MiΔαi − ∑ miΔαi i=1

i=1 n

=∑ (Mi-mi)Δαi i=1 n

< ∑ ηΔαi i=1

= η(Δα1 + Δα2 +...+

n

)

(

= η α(x1) − α(x0) + α(x2) − α(x1) +...+

(

= η α(xn) − α(x0)

(xn) − α(xn-1))

)

= η α(b) − α(a) <ε L

(

)

U(P,f,α) − L(P,f,α) < ε

By theorem 1.1.3, f

F >(

) on [a,b].

#0-7-5 1.1.6:

If f is monotonic on [a,b] and if is continuous on [a,b] then f F > ( ). P77:

Let

> 0 be given.

Double click this page to view clearly

Since

is continuous on [a,b], for

any positive integer n, choose a partition Δαi =

α(b) − α(a) , n

such

that

i=1,2,....,n

By hypothesis f is monotonic on [a,b]. Suppose f is monotonically increasing (the proof is analogo us in the other case). Then M i = f(x i ) and m i = f(x i–1 ). n L

U(P,f,α) − L(P,f,α) =

n

∑ MiΔαi − ∑ miΔαi i=1

i=1 n

=

(Mi − mi)Δαi ∑ i=1 n

(

=∑ f(xi) − f(xi-1)

)

α(b) − α(a) n

i=1

=

α(b) − α(a) f n

(xn) − f(x0)

=

α(b) − α(a) f n

(b) − f(a)

<

, if n is taken large enough.

By theorem 1.1.3, f

F >(

) on [a,b].

#0--5 1.1.7:

Suppose f is bounded on [a,b] , f has

only

finitely

discontinuity

on

many [a,b],

points and

of is

continuous at every point at which f is discontinuous. Then f

F >(

).

P:

Let

> 0 be given.

Put M = sup|f(x)|, let E be the set of points at which f is discontinuous. Since E is finite and

is continuous

at every point of E, we can cover E by finitely many disjoint intervals [u j ,v j ]

O

[a,b] such that the sum of

the corresponding differences

(v

j)

–

α(u j ) < Further more , we can place these intervals in such a way that every poi nt of E

J

(a,b) lies in the interior

of some [u j ,v j ]. Remove the segments (u [a,b].

The

compact.

remaining

Hence

f

is

j

,v j ) from

set

is

uniformly

continuous on K, and there exist 0 such that |f(s) – f(t)| <

K

s

if s

> F

K,

t F K, |s – t| <

0, x1, Now form a partition P = {x x 2 ,..x n } of [a,b] as follows. Each u j

occurs in P. Each v

j

occurs in P. No

point of any segment (u j v j ) occurs in P. If x i–1 is not one of the u j , then < δ.

xi

Note that M i – m i ≤ 2M for every i, and that M i – m i <

unless xi–1 is one

of the u j . Hence U(P,f, (a)] Since

) – L(P,f,

) ≤ [

(b) –

+ 2M > 0 is arbitrary, by theorem

1.1.6, f F > (

).

#0-7-5 1.1.8:

Suppose f

F >(

) on [a,b], m ≤ f

≤ M, ; is continuous on [m, M] and h(x) = ; f(x)) on [a,b]. Then he h >(

F

) on [a,b].

P77:

Choose

> 0.

Since ; is continuous on [m, M] and [m , M] is co mpact, continuous on [m, M]

; is uni for mly

L there exist

> 0 such that

<

and s,t F [m, M] and |s –1| <

B | ; (s) – ; (t)| <

Since f F > (

) on [a,b], there is a

partition P of [a,b] such that 2

U(P,f,α) − L(P,f,α) < δ .

------------------ (1)

Let M i = sup f (x), ,xi-1 ≤ x ≤ x i,

mi = inf f(x), xi-1 ≤x≤x

i,

Mi * = sup f(x), xi-1 ≤x≤x

i,

mi * = inf h(x), xi-1 ≤x≤x

i,

Divide the number., 1,2....,n in two classes A and B as follows, i F A if M i – m i <

and i F B if M i –

m i ≥ δ. For i F A our choice of

shows that

M* i – m i * < For i F B, M i *– m i *≤ 2K, where K = sup| ; |(t)|, m ≤ t ≤ M.

We have

∑ Δαi ≤ ∑ (Mi − mi)Δαi iFB

=

iFB

∑ MiΔαi − ∑ miΔαi iFB

iFB

<δ .(by(1)) Hence

∑ Δαi<

--------------------- (2)

iFB

U(P,f,α) − L(P,f,α) = ∑ (Mi * − mi * )Δαi + ∑ (Mi * − mi * )Δαi iFA

iFB

<∑ε Δαi + ∑2K iFA

i

iFB

<ε α(xn) − α(x0) + 2K

[

]

(by(2))

<ε[α(b) − α(a)] + 2K <ε[α(b) − α(a) + 2K] Since

>0 is arbitary,we see that hF >(α) on [a,b]

C'P $E"#ION":

1.

Suppose

increases on [a,b], a

≤ x 0 ≤ b,

is continuous at x

f(x 0 ) = 1, and f(x) = 0 if x ≠ x Prove that f fd

F >(

0, 0.

) and that ∫

=0


If f(x) = 0 for all irrationa l x, f(x)

2.

= 1 for all rational x, prove that f 

>(

)on [a,b] for any a < b.

SECION-1.2

PROPERIES OF

HE INEGRAL #0-7-5 1.2.1: a.

f1  >(α) and f2  >(α) on [ a,b], then f 1 + f2  >(α), cf  >(α) for b

every constant c,and

b

∫( f

1

+ f2) d

=

b

∫f

a

1

d

If f1(x) ≤ f2(x)on [ a,b],then

∫f

a

b 1

dα ≤

∫f dα. 2

a

If f  >(α) on [ a,b] and if a(α) on [a,c] c

and on[c,b], and

b

b

∫fdα+∫fdα=∫fdα a

d.

∫

= c fdα

a

a

c.

∫ cfd

+

a

b

b.

b

c

a

If f  >(α) on [a,b] and if | f(x)| ≤ M on [a,b],then b

|

∫fd

| ≤ M[α(b) − α(a)].

a

e.

If f  >(α1) and f  >(α2), then f  >(α1+α2) and b

b

b

∫fd(α +α )=∫fdα +∫fdα : 1

2

1

a

a

2

a

If f  >(α) and c is a positive constant,then f  >(cα) and b

b

∫fd(cα) = c ∫fdα a

a


P77.:

If f = f 1 + f 2 and P is any partition of [a,b], then we have L(P,f1, α) + L(P,f2, α) ≤ L(P,f,

Let If

) ≤ U(P,f,α) ≤ U(P,f1, α) + U(P,f2, α).! .................(1)

> 0 be given. f1F

>(

), then there exists a

partition P 1 such that U(P1,f1,α) − L(P1,f1, α) < ε. ---------(2) Also if f 2

F >(

), then there exists a

partition P 2 such that U(P2,f2,α) − L(P2,f2,α) <

-------(3 )

If P is the common refinement of P 1 and P2,then ( 2) implies U(P,f1,α) − L(P,f1,α) < U(P1,f1,α) − L(P1,f1,α)

(by theorem 1.1.1) <

-------------------(4)

and(3) implies U(P,f2, α) − L(P,f2,α) < U(P2,f2,α) − L(P2,f2,α)(by theorem 1.1.1) <

---------------------(5)

Adding ( 4) and ( 5), we get U(P,f1,α) + U(P,f2,α) − L(P,f1,α) − L(P,f2,α) < 2

----------(6)

Now U(P,f,α) − L(P,f,α) ≤ U(P,f1,α) + U(P,f2,α) − L(P,f1,α) − L(P,f2,α)(by(1)) < 2ε(by(6))

(i.e)U(P,f,α) − L(P,f,α) < 2ε By theorem 1.1.3 f F

>(α)

on [ a,b]


(i.e) f 1 +f 2 F

>(

) on [a,b].

Now for partition P, we have b

U(P,f1,α) <

∫f d 1

+

---------------(7)

+

---------------(8)

a b

U(P,f2,α) <

∫f d 2

a

Adding(7) and (8), we get b

U(P,f1,α) + U(P,f2,α) <

b

∫f d

∫

+

1

+ f2d

a

+

a b

=

b

∫f dα + ∫f d 1

a

2

+2

a b

By (1) U(P,f,α) ≤ U(P,f1,α) + U(P,f2,α) <

∫f dα + ∫f d 1

a b

b 2

+2

a

b

(i.e)U(P,f,α) < ∫f1dα + ∫f2d +2 a

a

b

But

∫fd

≤ U(P,f,α)

a

From the above two inequalities, we get b

b

∫fd a

<

b

∫f d 1

a

+

∫f d 2

+2

a


Since

>0 is arbitary,we conclude that

b

b

∫fd

b

∫f dα + ∫f d

≤

a

1

--------------------(10)

2

a

a

Replace by f1 and -f1 and f2 and -f2 in (10), we get b

b

∫fd

b

∫f dα + ∫f d

≤

a

1

--------------------(11)

2

a

a

From (10) and(11), we get b

b

∫fd

b

∫f dα + ∫f dα

≤

a

1

2

a

a

b

b

b

(i.e)∫(f1 + f2)dα = ∫f1dα + ∫f2dα a

a

a

(b) Let f 1 (x) ≤ f 2 (x) on [a,b]. Then f 2 (x) – f 1 (x) ≥ 0. Since

is monotonically increasing in

[a,b],

(b) >

(a). b

Then we have

∫ ( f ( x ) − f (x ) ) d 2

1

≥0

a b

b

(i.e)∫(f2 − f1)d a

≥0

b

B ∫f d 2

a

−

∫f d 1

≥0

a


b

b

B ∫f d 2

a

≥

b

b

∫f dα B ∫f dα ≤ ∫f dα 1

1

a

a

2

a

Similarly we can prove (c), (d) and (e). #0-7-5 1.2.2:

If f F > (

) and g

F >(

) on [a,b]

then i.

fg F > (

); b

ii.

b

|f| F >(α) and |∫fd | ≤∫|f| dα. a

a

P77:

2

If we take ; (t) = t , then by theorem 1.1.8, 2

f F >(α) B f F >(α)

-------------------(1)

f F >(α) and g F >(α) B f+g F >(α), f-g F >(α) (by theorem 1.2.1) B

2

(f+g)

2

F >(α), ( f-g) F >(α) ( by(1))


B

2

2

(f+g) + (f-g) F >(α) (by theorem 1.2.1)

B 4fg F >(α)

B

1 4

(4fg) F >(α) on [ a,b] (by therom 1.2.)

B fg F >(α) on [ a,b]

If we take ; (t) = | t |, then by theorem 1.1.8, f F >(

) B | f | F >(

). b

Choose c=±1,so that c

∫fd

≥ 0.

a b

∫fd

Then |

a

b

|=c

b

∫ fd =∫ a

a

b

∫|f|d

cfd ≤

, since cf ≤|f|.

a

D-.161<176:

The unit step function I is defined by

I (x ) =

{

0 ( x≤0 ) 1

(x>0)


#0-7-5 1.2.3:

If a < s < b, f is bounded on [a,b], f

is

continuous

at

s,

b

∫fd

and α(x) = I(x-s), then

= f(s).

a

P77:

Consider partitions P = {x

0 ,x 1 ,x 2 ,x 3

}, where x 0 = a and x 1 = s < x 2 < x 3 =b. Then U(P,g,α)=M1Δα1+M2Δα2+M3Δα3 [

]

[

]

[

= M1 α(x1) − α(x0) + M2 α(x2) − α(x1) + M3 α(x3) − α(x2)

[

]

[

]

]

[

= M1 I(x1 − s) − I(x0 − s) + M2 I(x2 − s) − I(x1 − s) + M3 I(x3 − s) − I(x2 − s)

]

= M1[0 − 0] + M2[1 − 0] + M3[1 − 1]

(by the definition of unit step function) =M2 ly

III L(P,f,α) = m2.

Since f is continuous at s, we see that M 2 and m 2 converges to f(s) as x (i.e.) U(P,f,

) and L(P,f,

to f(s) as x 2

A

2

A

s.

) converges

s.


b

∫fd

Therefore

= f(s)

a

#0-7-5 1.2.4:

Su ppose c n ≥ 0 for n = l,2,3,..., c n converges, {s n } is a sequence of distinct

points

in

(a,b),

and

∞

α(x) = ∑ cnI(x-sn) be continuous on n=1

[a,b]. Then

b

∞

∫fd

=∑ cnf(sn)

a

n=1

P77:

Since

cn

converges,

the

series

∞

∑ cnI(x-sn)

is also converges for

n=1

every x. (by Comparison test). Clearly = 0 and

(x) is monotonic. Also (b) =

(a)

c n (by the definition

of unit step function).

∞

>0 be given.Choose N so that∑ cn <

Let

.

---------(1)

n=N+1 N

Put

1

∞

(x) = ∑ cnI(x-sn) and

2

(x) = ∑ cnI(x-sn).

n=1

n=N+1

By the properties of the integral and by theorem 1.2.3,

b

N

fdα1 =

∫

cnf sn

-------------------- 2

( )

∑

a

n=1

( )

∞

Since

2

∞

(b) − α2(a) = ∑ cnI(b-sn) − ∑ cnI(a-sn) n=N+1

∞

n=N+1

∞

b

= ∑ cn − 0 = ∑ cn < ε.(by ( 1)), we have n=N+1

n=N+1

where

M

|∫ |

fdα2 ≤ M

,-----------(3)

(by

the

a

=

sup|f(x)|.

properties of integral) Since

|

b

= N

∫fd a

||

−∑ cnf(sn) = n=1

b

=

|∫ a

1

b

fdα1 +

∫ a

+ α 2 , we have

b

N

∫fd(α +α ) − ∑ c f(s )

N

1

2

n

n=1

||

fdα2 − ∑ cnf(sn) = n=1

n

a N

∑ cnf(sn) + n=1

| b

∫ a

N

|

fdα2 − ∑ cnf(sn) (by ( 2)) n=1


b

=

|∫ | |∫

fdα2 ≤ Mε(by(3))

a

b

(i.e)

fd

a

N

|

−∑ cnf(sn) ≤ Mε n=1

b

If we let N

A

∞,we get

∞

∫fd

= ∑ cnf(sn)

a

n=1

#0-7- 1.2.5:

Assume α' F

increase

>(

monotonically and

) on [a,b]. Let f be a bounded

real function on [a,b]. Then f if and only if f b

'

F >(

F > (α)

). In that case

b

∫fd

∫f(x) α '(x)dx

=

a

a

P77:

Let

> 0 be given.

Since

'F

>(

), by theorem 1.1.3,

there exist a partition P = {a = x 0 ,x 1 ,x 2 ,...,x n =b} such that U(P,α') − L(P,α') <

----------(1 )


By the mean value theorem,

i

= α (xi) − α(xi-1) = α (ti)Δxi, ------- (2)

where ti F [xi-1, xi] for i=1,2,.....n. n

If si F [xi-1, xi], then

'

'

∑ |α (si) − α (ti)| Δx i <

-----------(3) ( by ( 1) and theorem 1.1.4(b))

i=1

Put M=sup |f(x)|. n

Since

n

∑ f(si) Δαi = i=1

|

----------------------------- (4)

∑ f(si) α'(ti) Δxi(by(2)), we have i=1

n

n

||

∑ f(si) Δαi − ∑ f(si) αi(si)Δxi i=1

=

|

i=1 n

|

∑ f(si) (α '(ti) − α '(si))Δxi i=1

=

n

n

i=1

i=1

n

≤ ∑ |f(si)| |α '(si) − α '(ti)|Δxi i=1

= Mε ( by ( 3) and ( 4)) Inn particular

∑ f(si)Δαi ≤ U(P,f

') + M

,for all cho ices of i sF [xi-1, xi], so that

i=1

U(P,f,α) ≤ U(P,fα') + M

--------------------(5)

Similarly we can show that U(P,f Therefore | U(P,f,α) ≤ U(P,f

') ≤ U(P,f,α) + M

')| ≤ M

------------(6)

------------(7)

If P is replaced by any refinement, then (1) is true and hence (7) is Therefore also true.

|

b

b _

- f(x)α'(x)dx ∫fd ∫ ¯ a

a

|

∑ f(si) α '(ti)Δxi − ∑ f(si) α '(si)Δxi

|

≤M


b _

b _

∫

Since

>0 is arbitary, fd a

∫

= f(x)α'(x)dx a

for any bounded f. b

Similarly

b

∫fd

∫

= f(x)α'(x)dx for any bounded f.

¯ a

¯ a

b _

f F >(α)

b _

b

b

C ∫fd =∫fdα ∫f(x)α'(x)dx = ∫f(x)α'(x)dx C

a

¯ a

a

¯ a

b

C

fα ' F >(α) and

b

∫fd

=

a

∫f(x)α '(x)dx a

#0-7-5 1.2.6: (C0/- 7 %*4-;)

Suppose

is a Strictly increasing continuous function that maps an interval [A,B] onto [a,b]. Suppose a is monotonically increasing on [a,b] and f

F >(

) on [a,b]. Define

g on [A,B] by f(

(y) =

(

and

(y)), g(y) =

(y)) . Then Double click this page to view clearly

b

g F >(β) and

b

∫gd

∫fdα

=

a

a

P77.:

To each partition P={x

0,

x 1 , x 2 ,...,

x n } of [a,b] corresponds a partition Q={y 0 , y 1 , y 2 ,..., y n } of [A,B], so that x i =

(p(y i ). Al l partitions of

[A,B] are obtained in this way. Since the value s taken by f on [x i–1 ,x i are exactly the same as those taken by g on [y i–1 ,y i ], we see that U(Q,g, U(P,f,

), L(Q,g,

) = L(P,f,

)=

).

n

(U(Q,g,β) = ∑ Mi * Δβi where Mi *

= sup g(x), (xi-1 ≤ x ≤ xi)

i=1

= sup f(φ(y)) = sup f(x) = Mi and

(

)

(

i

= β(yi) − β(yi-1)

)

= α φ(yi) − α φ(yi-1) = α(yi) − α(yi-1) = Δαi Therefore U(Q,g,β) = U(P,f,α))


Since f

 >(

), P can be chosen so

that both U(P,f,

) and L(P,f,

) are

b

close to

∫fd

Hence U(Q,g,

) and

a

L(Q,g, theorem

) are also close and by 1.1.3,

g

b

g  >(β)and

∫gd



>(

) and

b

=

a

∫fd a

C'P $E"#ION": 1.

Pro ve theor em 1.2.1 (c) , (d) and (e).

SECION-1.3

INEGRAION

AND DIFFERENIAION

#0-7-5 1.3.1: Let f

 >(

) on [a,b]. For a ≤ x ≤ b, b

put F (x) =

∫f(t)dt a

Then F is continuous on [a,b]. Also if f is continuous at x 0 of [a,b], then F is differentiable at x 0 and F‘(x 0 ) = f(x 0 ). P77.:

Since f

F >(

) , f is bounded.

Then there exist a real number M such that | f(t) | ≤ M for a ≤ x ≤ b. y

If a ≤ x ≤ y ≤ b, then | F(y) − F(x)| =

x

|∫

f(t)dt-

a

a a

|

|∫ | x

y

x

f(t)dt ≤

=

|

∫f(t)dt+∫f(t)dt

=

y

∫f(t)dt

a

y

|

∫|f(t)|dt x

≤ M(y-x). Given

>0 we have|y-x| <

ε M

B

|F(y) − F(x)| < ε


Therefore F is uniformly continuous on [a,b]. B

F is continuous on [a,b].

Suppose f is continuous at a point x 0 . Given

> 0, choose

| f(t) – F(x 0 ) | <

> 0 such that

if |t – x0 |<

and

a ≤ t ≤ b. Therefore, if x 0 – +

|

0

0

and a ≤ s < t ≤ b

F(t)-F(s) t-s

|

− f(x0) =

|

t 1 t-s

∫(f(u) − f(x ))du 0

s

|

t

≤

1 t-s

f(u) − f(x0) du

∫( s

t

1

<ε t-s

∫du s

<ε Hence F'(x0) = f(x0).


)

#0-7-5 1.3.2: (#0-

F=6,5-6<4

<0-7-5

7

C4+=4=;) If f

F >(

) on [a,b] and if there is a

differentiable function F on [a,b] such

that

F’=

f

then

b

∫f(x) dx=F(b) − F(a) a

P77:

Let

> 0 be given.

Choose a partition P = {x 0 ≤ x 1 ≤ x 2 ....≤x n } of [a,b] so that U(P,f) – L(P,f) < By the Mean Value theorem, there are points t i

F

[x i–1 ,x i such that

F(x i ) – F(x i–1 ) = f(t i )Δx i for i = 1,2,...., n. n

Thus

f(t ) Δx ∑ i=1 i

i

= F (b ) − F (a ).

It follows from theorem 1.1.4 (c) that

|

n

b

∑ f(ti)Δxi − ∫fdα i=1

(i.e)

a

|

|

<ε

b

(F(b) − F(a)) − ∫fdα a

|

Since this is hold for every have b

∫f(x)dx = F(b) − F(a) a

#0-7-5 1.3.3: (I6<-/<76 *@ P<;)

<ε

> 0, we

Suppose F and G are differentiable F

function on [a,b], F’= f b

G'=g F > then

>

and

b

∫F (x)g(x)dx=F(b)G(b) − F(a)G(a) − ∫f(x)G(x)dx a

a

P77.:

Put H(x) = F(x)G(x),

DxF

[a,b].

Then H'(x) = F'(x)G(x) + F(x)G'(x) = f(x)G(x) + F(x)g(x) F’= f F > and G’= g F > B H’ F > on [a,b]. Let

us

apply

the

Fundamental

theorem of Calculus to the function H and its derivative H’, we get b

∫H'(x)dx=H(b) − H(a) a


b

∫(f(x)G(x) + F(x)g(x))dx=F(b)G(b) − F(a)G(a) a b

b

∫f(x)G(x)dx+∫F(x)g(x)dx = F(b)G(b) − F(a)G(a). a

a b

b

(i.e)∫F(x)g(x)dx=F(b)G(b) − F(a)G(a) − ∫f(x)G(x) dx. a

a

Hence the theorem. C'P $E"#ION": 1.

Suppose f ≥ b0, f is continuous on [a,b], and

∫f(x) dx=0 Pr ove

that

a

f(x) =0 for all x

S*(62 1.4

F [a,b].

I6*,&62 2+

8*(62 8&/7*) +7(625

D-.161<176: Let f 1 ,f 2 ,...,f k be a real functions on [a,b], and let


 = (f 1 ,f 2 ,...,f k ) be the corresponding mapping

of

[a,b]

3

into

! .

If

α

increases monotonically on [a,b], to say that f >

(

F >

(

) means that f j F

) for j = 1,2 ,.. .., k. If thi s is the

case we define b

b

fd

∫ a

=

(

b

f1 d

, ....,

∫

fkdα

∫

a

)

a

#0-7-5 1.4.1: (A67/=-

7

<0-

F=6,5-6<

<0-7-5 7 C+==; 7 >-+<7 >=-, =6+<76;) If  and F map [a,b] into on [a,b] and if

F ' =  then

b

∫f(t)dt =F(b) − F(a) a

3

! , if 

F >

P77.:

Let . = (f 1 ,f 2 ,... ,f k ) and

F

= (F 1

,F 2 ,... ,F k ).

=f B

F’

B

By

the

(F1’, F2’, ..., Fk’)

=

(f1, f2, ..., fk).

Fj ' = f j, j = 1, 2, ...., k.

Fundamental

theorem

Calculus, for real valued function f

of j,

b

we have

∫fj(!)d! = F (b) − F (a) j

j

a b

b

b

∫f (!)d!, ∫f (!)d!, ....., ∫fk(!)d!

(i.e)

1

(

a

2

a

a

)

=(F1(b) − F 1(a), F2(b) − F 2(a), ...., Fk(b) − F k(a)) b

(i.e.)∫f(!)d!=F(b) − F (a) a

#0-7-5 1.4.2:

If  maps [a,b] into !

3

and if 

F > (α)

for some mo notonically increasi ng function on [a,b] then |  | F > (α) and b

| | ∫fdα

b

≤

a

∫|f|dα a

P77:

Let  = (f 1 ,f 2 ,...,f k ). Then | 2

2



| =

2

(f 1 +f 2 +.. .+f k ) f F >(α) B

f1, f2, ...fk F >(α) on [a,b]

B f12, f22, ...., fk2 F >(α) on [a,b] B f12+f22 +...+f k2 F >(α) on [a,b]

B (f12+f22 +....+f B

k2

)

1/2

F >(α)

on [a,b]

|f| F >(α) on [a,b]


b

Put y=(y1, y2, ...yk), where yj =

∫f d

for j=1,2,....k.

j

a b

k

k

b

2

∫fd

Then y=

and|y| =

a

j

y .y ∑ j=1

j

=

y f dα ∑ ∫a j=1 j

b

=

j

b

∫∑ y f dα ≤ ∫|#||f| dα j j

a

a b

=|#|

∫|f| dα. a

If @ = 0, the inequality is trivial. If @ ≠ 0, then divide by | b

@ |, we get

b

b

|y| ≤ ∫|f|d . That is a

| | ∫fdD

≤

∫ | f| d α a

a

C'P $E"#ION": b

1.

Define

∫fd

if f:[a,b]

A

k

R

a


S*(62 1.5

R*(6+&'/* (748*

D-.161<176:

A continuous mapping [a,b] into ! If

3

is called a curve in

is 1 – 1 , then (a) =

of an interval

(b),

!

3

.

is called an arc.If

is said to be a closed

curve. We associate to each partition P = {x 0 , x 1 , X 2 ,..x n }of [a,b] and to each curve

on

[a,b],

the

number

n

I ( P,γ) =

γ(xi) − γ(xi-1) ∑ | | i=1

Here I (P,

) is the length of the

polygonal path with vertices at

(x

0 ),

γ(x 1 ),...,

(xn ). The length of

defined as If

I(

I(

) = sup

I (P,

is

).

) < ∞, we say that

is

rectifiable. #0-7- 1.5.1:

If

’is continuous on [a,b] , then

is

b

rectifiable and

I

(γ) = ∫|γ'(t)dt a

P77:

If a ≤ x i–1 ≤ x i ≤ b, then |

(x i ) –

x i–1 )| xi

=

Hence

|

I

xi

∫γ'(t)dt xi-1

|

≤

∫|γ'(t)| dt xi-1

b

(P,γ) ≤ ∫|γ'(t)| dt for every a

partition P of [a,b].

(

b

L

We have sup

I

(P,γ) ≤ ∫|γ'(t)| dt a

b

(i.e.) I (γ) ≤

∫|γ'(t)| dt

-----(1)

a

Now to prove the reverse inequality. Let

> 0 be given.

Since

‘ is continuous on [a,b], it is

uniformly continuous on [a,b]. > 0 such that | s – t | <

LE

B

| γ

‘(s) – γ ‘ (t)| < ε. Let P = P = {x 0 , x 1 , x 2 ,...,x n } be a partition of [a,b] with If x i–1 ≤t≤x –

i,

x i<

it follows that |

‘(xi )| <

(i.e.) |

‘(t)| – |

(i.e.) |

‘(t)| ≤ |

, Di

‘(x i )| < ‘(x i )| +

‘(t)

xi

Hence

xi

xi

∫|γ'(t)|dt ≤ ∫|γ'(x ) + ε| dt ≤ (|γ'(x )| + ε)∫dt i

xi-1

i

xi-1

xi-1

≤

(|γ'(x )| + ε)(x − x

≤

(|γ'(x )| + ε)Δx

i

i

i

|

)

i-1

i

|

≤ γ'(xi) Δxi +

ix

xi

≤

) |∫ ( | ) |∫ | |∫ ( | | | ∫( ) | | γ'(t) + γ'(xi) − γ'(t) dt +

ix

xi-1 xi

≤

xi

γ'(xi) − γ'(t) dt

γ'(t)dt +

xi-1

xi-1

xi

≤ γ(xi) − γ(xi-1) +

γ'(xi) − γ'(t) dt

xi-1

xi

| ∫

|

≤ γ(xi) − γ(xi-1) +

dt +

i

x

xi-1

|

|

|

|

≤ γ(xi) − γ(xi-1) + ε(xi + xi-1)εΔxi ≤ γ(xi) − γ(xi-1) + 2

ix

If we add this inequality for i = 1,2,.. .,n, n

we get

xi

n

n

∑ ∫|γ'(t)| dt ≤ ∑ |γ(xi) − γ(xi-1)|+2ε∑ Δxi i=1 xi-1

i=1

i=1

xn

(i.e.)∫|γ'(t)| dt ≤

I

(P,γ) + 2ε(xn − x0)

x0


b

L

∫|γ'(t)| dt ≤

I

(P,γ) + 2ε(b-a)

a

Since

> 0 is arbitrary,we have

b

∫|γ'(t)| dt ≤

I

(P,γ).

a b

Hence

∫|γ'(t)| dt ≤

I

(γ ).

-------------------- (2)

a b

From (1) and (2), we get I (γ) =

∫|γ'(t)| dt a

C'P $E"#ION": 1. Suppose

f is a bounded real 2

function on [a,b] , and f

F>

on

F >

[a,b] , Does it follow that f

? Does the answer change if we assume that f 2.

3

F >

?

Let P be the Cantor set. Let f be a boun ded real funct ion on [0,1] which

is

continuous

at

every

point outside P. Prove that f f >

on [0,1]. Double click this page to view clearly

2

F

$NI#-2

$61< "<:=+<=:

Sectio n 2.1: Discussion of the main problem Section

2.2:

Uniform

2.3:

Uniform

convergence Section

convergence and continuity Section

2.4:

uniform

convergence and integration Section

2.5:

uniform

convergence and Differentiation Section

2.6:

Equi

continuous

families of functions Section

2.7:

The

Weierstrass Theorem

Stone

–

I6<7,=+<176

In this unit we confine our attention to

complex

valued

functions,

although many of the theorems and proofs which fo llow extend without difficulty to vector-valued functions, and to mappings into general metric spaces. We discuss the concept of uniform convergence and continuity, differentiation and integration.

SECION-2.1

DISC!SSION OF

HE MAIN PROBLEM

D-.161<176:

Suppose

{f n }, n = 1,2,..,, is a

sequence of functions defined on a

set E, and suppose that the sequence of numbers {f n (x)} converges for every x

E. We can then define a

F

function f by f(x) = lim fn(x)( xFE ). -------------( 1 ) n

∞

A

{f n } converges to f point wise on

E if

(1) holds. Similarly, if

f

every

E, and if we define

x

F

n

(x) co nverges for

∞

f(x) = ∑ fn(x)( xFE ) the function f is n=1

called the sum of the series

∑ fn

To say that f is continuous at x means

lim f(t) = f(x) Hence to ask

Ax

t

whether the limit of a sequence of

continuous functions is continuous is the

same

as

to

ask

whether

lim x lim f(t) = lim t lim x f(t) n ∞ n ∞ A A

t

A

(i.e.)

A

whether the order in which limit processes

are

carried

out

is

immaterial. On the LHS , we first let n

A

∞ then t

then n

A

A

x; on the RHS t

A

x first,

∞.

We shall now show by means of several examples that limit processes ca nnot in genera l be interchanged without affecting the result.

E?)584- 1: For m sm,n =

= 1,2,...,n = 1,2,..., let

m m+n

Then

,

for

every 1 1+n/m

lim sm,n = lim m

A

∞

lim n

A

m

A

fixed

=1

n,

so

that

∞

lim sm,n = 1

∞ m

A

∞

On the other hand, for every fixed m

m, lim sm,n = n(m+1)= 0. n

A

that

∞

nlim∞ nlim∞ A

so

A

sm,n = 0.

E?)584- 2:

Le! f n(x)

=

x

2 2 n

(1+x ) ∞

(x real; n=0,1,2,...), ∞

x and consider fn(x) =

f ∑ n=0

n

2

(x) = ∑ 1+x2 n . n=0 ( )

------------- (1)

Since f n (0) = 0, we have f(0) = 0. For x ≠ 0, the last series in (1) is convergent with sum (?).


Hence f(x) =

{

(x=0)

0 1+x

2

, so that

( x≠0 )

a co nvergen t series of co ntinuous functions may have a discontinuous sum.

E?)584- 3: For m=1,2,...,put fm(x) = lim m

When m!x is an integer , f For all other values of x, f

A

For irrational x, f

∞

m (x) m (x)

= 1.

= 0.

fm(x)

Now let f(x ) = lim m

A

(cosm !

∞

m (x)

= 0 for every

m; hence f(x) = 0. For rational x, say x = p/q, where p and q are integers, we see that m!x

x)

2n

is an integer if m ≥ q, so that f(x) = 1.

lim (cosm!

Hence lim m

A

∞ n

A

x)

2n

=

∞

{

0

(x irrational )

1

(x rational )

(i.e.) an everywhere discontinuous limit function, which is not Riemannintegrable.

E?)58- 4: sin nx

Let f n(x) = √n f(x) = lim n

A

(x real;n=1,2,3,... ), and

fn(x) = 0

∞

Then f ‘(x) = 0, and f

n'

(x) = √n cos nx

, so that {f n ‘} does not converge to

f

‘.

For

instance

fn '(0) = √n

+ ∞ as n

A

A

∞ , whereas

f’(0) = 0.

E?)84- 5: 2

(

Let fn(x) = n x 1 − x

2 n

) (0 ≤ x ≤ 1,n=1,2,3....) --------(2)

For 0 ≤ x ≤ 1,we have lim fn(x) = 0 n

(

A

∞

α

Since,if p>0 and

is real,then lim n

∞

A

n

Since fn(0) = 0, we see that lim fn(x) = 0 ( 0 ≤ x ≤ 1). n

A

n

(1+p)

=0

)

------------ (3)

∞

1

Also

∫(

x 1−x

2 n

1

) dx= 2n+2 .

0 1

∫

L

n

2

fn(x)dx= 2n+2

A

0

If we replace n

2

+ ∞ as n

∞

A

by n in (2), (3) still

holds, but we have 1

lim n

A

1

∫f (x)dx= lim

∞ 0

n

n

A

∞

n 2n+2

=

1 2,

whereas

∫ 0

[

]

lim fn(x) dx=0. n

A

∞

L

The limit of the integral need not

be equal to the integral of the limit, even if both are finite. C'P $E"#ION": 1.

Give more examples for the limit of the integral need not be equal to the integral of the limit, even if both are finite.

SECION-2.2

!NIFORM

CON"ERGENCE

D-.161<176: A sequence of functions {f

}, n = n

1,2,3,....,

is

said

to

converge

uniformly on E to a function f if for eve ry

> 0 there is an integ er N such

that n ≥ N implies |f for all x

 E.---(1)

n (x)

– f(x)| ≤

Note 1: Every uniformly convergent sequence is point wise convergent. Note

2:

The

difference

between

uniform convergent and poin t wise convergent is : If {f n }converges po int wise on E, then there exists a function f such that, for every > 0 and for every x

F

E, there is an

integer N , depending on

and on

x, such that (1) holds if n ≥ N. If {f n } converges uniformly on E, it is possible , for each

> 0, to find one

integer N which will do for all x

F

Note 3: We say that the series

f

E. n

(x) converges uniformly on E if the sequence {s n }of partial sums defined n

by

∑ f i( x ) = s n ( x ) i=1

uniformly on E.

converges

C) =+ 0@ + < - 7 6 7  =6  7 5 +76>-/-6+-: #0-7-52.2.1 : The

sequence

of

functions

{f n },

defined on E, converges uniformly on E if and only if for every

> 0 there

is an integer N such that n ≥ N, m

≥

N, x F E implies |f n (x) – f m (x)| ≤ P77:

Suppose {f n } converges uniformly on E. Let f be the limit function and let 0.

>

Then there is an integer N such that n ≥ N, x

F

E implies

ε

|fn(x) − fm(x)| ≤ 2 Therefore |fn(x) − fm(x)| ≤ |fn(x) − f(x)| + |f(x) − fm(x)| ε

<2 +

Conversely,

ε 2

=

if n ≥ N,m ≥ N, xF E

suppose

the

Cauchy

condition holds. (i.e.) for every

> 0 there is an

integer N such that n ≥ N, m ≥ N, x F E implies

| f n (x ) − f m (x )| ≤

-----(1)

By a theorem, (f n (x)} converges, say to f(x) for every x ( since R is complete). Thus the sequence {f n } converges on E, to f.

We

have

to

prove

that

the

convergence is uniformly. Fix n, and let m Since f m (x)

A H in(1).

A

f(x) as m A ∞, this

gives |f n (x) – f(x)| ≤ N and every x L {f n }

for every n ≥

F E.

converges uniformly to f on

E.

#0-7-5 2.2.2:

Suppose lim fn(x) = f(x) ( x F E ) n

∞

A

. Put Mn = sup |fn(x) − f(x)| xFE

Then f n

A

f uniformly on E if and only

if Mn

A

0 as n

A

∞.

P77.:

Suppose f n

f uniformly on E.

A

By definition, for every > 0 there is an integer N such that n ≥ N implies |f n (x) – f(x)| ≤

for all x

F E.

Therefore sup |fn(x) − f(x)| ≤ ε xFE

(i.e.) M n ≤ (i.e) M n

A

if n ≥ N. 0 as n

A ∞.

Conversely, suppose M n

A

0 as n A

∞. Then, given > 0 there is an integer N such that n ≥ N B M n ≤ ε.

(i.e.) n ≥ N

B

B

sup |fn(x) − f(x)| ≤ ε xFE

|fn(x)

− f(x)| ≤

for every xF E.

Therefore fn

A

f uniformly on E.

&-1-;<);;

<0-7-5 76

=6175 +76>-/-6+-. #0-7-5 2.2.3: Suppose

{f n }is

a

sequence

of

functions defined on E, and suppose |f n (x)| ≤ M n (x F E, n = 1,2,3,....). Then if

fn convergenc e uniformly on E

Mn converges.

P77:

Suppose Then,

|

m

for

|

∑ fi(x) i=n

M n converges. arbitrary m

≤

∑ |fi(x)| ≤ ∑ Mi ≤ i=n

>

0,

m

( xF E )

i=n

(x F E) provided m and n are large enough.

(i.e.) there is an integer N such that n ≥ N, m ≥ N, x

E

implies

|f n (x)– f m (x)|≤ By theorem 2.2.1,

f

n

convergence

uniformly on E. C'P $E"#ION": 1. Prove

that

every

uniformly

convergent sequence of bounded functions is uniformly bounded.

SECION-2.3

!NIFORM

CON"ERGENCE AND CONIN!I%

#0-7-5 2.3.1: Suppose f n



f unifor mly on a set E in

a metric space.Let x be a limit point of E, and suppo se that

lim

tAx

fn(t) = An

(n

=

1,2,3,....)

.

Then

{A n }converges, lim

and

f t = lim A

n Ax ( )

t

n

A

n

∞

In other words, the conclusion is that lim

Ax

t

lim fn(t) = lim n

A

∞

n

A

lim

Ax

fn(t)

∞ t

P77.:

Let

> 0 be given.

Since {f n }converges uniformly on E, there exists N such that n ≥ N, m ≥ N,t

F

E|fn(t) − fm(t)| ≤

---------------------(1)

By hypothesis,for n=1,2,3,......, t lim x fn(t) = An A

Letting t

A

--------(2)

x in (1) and using (2), we

get n, m ≥ N implies |A

n–

Am| ≤ ε

Ther efore {A n }is a Cauchy sequence in the set of real numbers R.

Since R is complete, {A

n}

converges

to some point, say, A. |f(t) − A| ≤ |f(t) − fn(t) + fn(t) − An + An − A| ≤ |f(t) − fn(t)| + |fn(t) − An| + |An − A|

Since f n

A

----------------(3)

f uniformly on E, choose a

positive integer n such that ε

|f(t) − fn(t)| ≤ 3 , for all t F E Since An

-------------------- (4)

A,we have m ≥ N implies | An − A| ≤

A

Then,

for

this

n,

we

ε 3

--------- (5)

choose

a

neighborhood V of x such that ε

|f(t) − An| ≤ 3 if t F V J E,t ≠ x.

(

since

lim

t

Ax

fn(t) = An

Using (4),(5) and (6) in (3),we get | f(t) − A| ≤

ε 3

+

if t F V J E,t ≠ x

(i.e.) lim f(t) = A= lim An

Ax

t

(i.e) lim t

Ax

n

A

∞

lim fn(t) = lim n

A

∞

n

A

∞ t

lim

Ax

fn(t)


) ε 3

-----(6)

+

ε 3

=ε

#0-7-5 2.3.2:

(Corollary to theorem 2.3.1) If {f

n}

is a sequence of continuous functions on E, and if f

n

A

f uniformly on E,

then f is continuous on E. P77:

Since

{f n }

is

a

sequence

of

continuous functions on E, for every lim fn(t) = fn(x)

n, we have By

theorem

Ax

2.3.1,

lim fn(t) = lim

lim

t

tA x

n

A

∞

n

A

we lim

∞ t

(

Ax

have fn(t)

)

(i.e.) lim fn(t) = lim f n(x) = f(x) since fn f uniformly on E .

Ax

t

n

A

∞

A

(i.e) lim f(t) = f(x)

Ax

t

By definition of continuous function, f is continuous on E.


Note:The

converse

of

the

above

theorem is need not be true.

E?)584-: 2

2 n

f n (x) = n x(1 – x ) (0 ≤ x ≤ 1, n = 1,2,3,....)

#0-7-5 2.3.3:

Suppose K is compact, and a. {f n } is a sequence of continuous functions on K, b.

{f n }converges point wise to a continuous function f on K,

c.

f n (x) ≥ f n+1 (x) for all x 1,2,3,.....

Then f n

A

f uniformly on K.

P77:

Put g n = f n – f.

F

K, n=

Since f n and f are continuous, g n is also continuous. Since f n

A

f point wise , g n

A

0 point

wise. Also , since f n (x) ≥ f n+1 (x) for all x F K, f n (x) – f(x) ≥ f n+1 (x) – f(x), for all x F K g n (x) ≥ g n+1 (x) for all x We

have

to

prove

that

fn

F

K.

A

f

uniformly on K. (i.e.) to prove that g n

A

0 uniformly

on K. Let

> 0 be given.

Let K n = {x F K \ g n (x) ≥ (i.e.) K n = {x F K \ g n (x) (i.e.) K n = {x F K \ x F g n (i.e.) K n = g n

–1

([

,∞)).

} F

–1

[

([

,∞)}.

,∞))}.

Since g n is continuous and [

,∞) is

closed, K n is closed ,and hence K n is compact (since closed subsets are compact). Let x F K n+1 . Then g n+1 (x) ≥ Since g n (x) ≥ g n+1 (x) ≥

, xF K n .

Then K n R K n+1 D n. Fix x F K. Since g n (x) A 0 point wise, we see that x G K n if n is sufficiently large. Thus x

G J Kn. J

In other words,

K n is empty.

L K n is empty for some N. It follows

that

0≤g n (x)<

,D x F

Therefore | g n (x) –0| < and for all n ≥ N.

K& D n

≥

N.

for all xF K

(i.e.) g n (i.e.) f n

A

A

0 uniformly on K. f uniformly on K.

D-.161<176:

If X is a metric space,

C(X)

will

denote the set of all complex valued, continuous, bounded functions with domain X. C(X)

consists

of

all

complex

continuous functions on X if X is compact. F

C

We associate with each f (X) its supremum norm ||f|| = sup |f(x)|. xFE Since f is bounded, ||f|| < ∞. Also ||f|| = 0

C

f(x) = 0 for every x

FX

(i.e.) ||f|| = 0 If

h

=

f

C

+

f=0.

g,

then

|h(x)|

=

|f(x)+g(x)|≤ |f(x)|+|g(x)| ≤ ||f||+||g|| for all x F X Hence ||f+g|| = ||h|| ≤ ||f||+||g||. Also

C(X)

is a metric space with the

metric d(f,g) = ||f – g||. #0-7-5 2.3.4: C(X)

is a complete metric space.

P77:

Let {f n }be a Cauchy sequence in C (X). L

to. each

> 0, there exists a

positive integer N such that n, m ≥ N implies ||f n – f m || <

It follows that there is a function f with

domain

converges

X

to

which

uniformly,

(by

{f n } Cauchy

criterio n for uniform co nvergence). By theorem 2.3.2, f is

continuous.

Since f n is bounded and there is an n such that |f(x) – f

n (x)|

<1 for all

x F X, f is bounded. Thus f F

C(X).

Since f n

A

f uniformly on X, we have

||f– f n ||

A

0 as n

A

∞.

C'P $E"#ION" 1. If

{f n }

and

{g n }

converge

uniformly on a set E, prove that {f n + g n } converges uniformly on E.

2. If

{f n }

and

{g n }

converge

uniformly on a set E and if {f

n}

and

of

{g n }are

sequences

bounded functions, prove that {f n g n } converges uniformly on E.

SECION-2.4

!NIFORM

CON"ERGENCE AND INEGRAION

#0-7-5 2.4.1:

Let

be monotonically increasing on

[a,b]. Suppose f

n = (

).on [a,b], for

n = 1,2,3,...., and suppose f n uniformly b

on

[a,b].Then

f



,and

b

∫fd

=n lim ∞ A

a

∫f dα n

a

P77:

Put

n

= sup |f n (x) – f(x)|, the

supremum being taken over a ≤ x ≤ b.

f

(i.e) − εn ≤ f-fn ≤ εn.

----------------- (1)

(i.e.)fn − εn ≤ f ≤ fn + εn b

b _

b

∫( f

− εn) d

n

≤

∫ fd ≤∫ fd ≤∫(f

¯ a

¯ a

b

∫

a

b

(fn − εn) d

≤

∫

a

∫fd

0≤

−

a

b

∫

≤

∫( f

fd

≤

a

b

b

∫(f

≤

¯ a

)dα-∫(fn-εn)dα

n + εn

a

a

b

b

∫ (f

≤

n + εn

∫

= 2 εndα=2εn[α(b) − α(a)] -----(2)

− fn + εn)d

a

a

By theorem 2.2.2,

From(1), 0 ≤

∫fd

∫

− fdα

A

∫fd

=

∫fd

0 as n

A

∞

-----------(3)

).[a,b].

Using(3) in ( 1), we getεn[α(b) − α(a)] ≤

n

a

∫fd

b

a

∫

− fn d

|

∫

a

a

b

∫fd a

b

∫

− a fndα ≤

∫ε dα n

a

≤ n[α(b) − α(a)]

a

− fndα ≤ εn[α(b) − α(a)]

fd

b

b

a

b

b

∫ε dα ≤ ∫fd

b

Therefore

f uniformly on [ a,b]

¯ a

b

|∫

A

b

a

(i.e)

(

∞. fn

¯ a

b _

=(

A

b

a

Therefore

0 as n

n A

b _

(i.e.) f

)dα

n + εn

a

b

∫fd

)dα

n + εn

a

_ b

fd

¯ a b _

b _

A

0 as n

A

b

= lim n

A

∫f dα. n

∞ a


∞

)

C'P $E"#ION" 1.

If

fn

=(

).on

[a,b]

and

if

∞

f (x ) =

f ∑ n=1

(x)(a ≤ x ≤ b),

n

the

series converging uniformly on b

[a,b], then

∞

∫fd

=∑

a

n=1

SECION-2.5

b

∫f dα n

a

!NIFORM

CON"ERGENCE AND DIFFERENIAION

#0-7- 2.5.1: Suppose

{f n }

is

a

sequence

of

functions, differentiable on [a,b] and such

that

{f n (x 0 )}

converges

for

some point x 0 on [a,b]. If {f n '} converges uniformly on [a ,b], then {f n } converges unif ormly on [a,b],

to

a

function

f,

f '(x) = lim fn '(x)(a≤x≤b n

A

and

)

∞

P77.:

Let

> 0 be given.

Since

{f n (x 0 )}converges

point

x0

on

convergent

[a,b],

for and

sequence

is

some every

Cauchy

,

choose N such that n ≥ N,

|

|

m ≥ N,t F E implies fn(x0) − fm(x0) ≤

ε 2

------------------(1)

Also, since {f n '} converges uniformly on [a,b], say to f‘, we have ε |fn '(t) − fm '(t)| < 2(b-a) (a≤t≤b )

------------------(2)

Apply Mean Value theorem to the function f n – f m , we get


|(fn − fm)(x) − (fn − fm)(t)| = |fn(x) − fn(t) + fm(t)|

|

= (fn(x) − fn(t)) + (fm(x)-fm(t))

|

≤ |x-t||fn '(t) − fm '(t)| ( by MVT ) ε

<|x-t| 2(b-a) (by(2))

------------(3)

ε

≤ 2 .for any x and t on

[a,b], if n,m ≥ N ------------- (4)

|

Also | fn(x) − fm(x)| = fn(x) − fm(x) − fn(x0) + fm(x0) + fn(x0) − fm(x0)

|

||

|

≤ fn(x) − fm(x) − fn(x0) + fm(x0) fn(x0) − fm(x0)

Therefore

ε <2

ε 2

+

=

,for any x on [ a,b],if n,m ≥ N

|

(by(1) and ( 4))

{fn} converges uniformly on [a,b]

Let f (x) = lim fn(x)(a ≤ x ≤ b n

A

)

∞

Let us now fix a point x on [a,b] and ;n(t)

define ;(t)

=

f(t) − f(x) t-x

Allowing n

=

fn(t) − fn(x) t-x

for a ≤ t ≤ b,t ≠ x.

∞ in

A

;n(t),

we get lim n

A

;n(t) =

∞

lim n

A

fn(t) − fn(x) , t-x

∞

lim fn(t) − lim fn(x) n

A

∞

n

A

∞

lim x A

Also

t

and

lim

Ax

f(t) − f(x) t-x

=

t-x

;n(t) ; (t )

=

= lim t-x

t

= ;(t). ----------- (5)

fn(t) − fn(x) t-x lim t-x f(t) − f(x) t-x

= fn '(x) -----------(6)

= f '(x), for a ≤ t ≤ b,t ≠ x

-------------- (7)

The inequality ( 3)can be rewritten as

|

fn(x) − fn(t) − fm(x) + fm(t) t-x

(i.e)

|

fn(x) − f n(t) t-x

−

|<

ε 2(b-a)

fm(x) − f m(t) t-x

(i.e.)|;n(t) − ;m(t)| <

|<

ε 2(b-a)

ε . 2(b-a)


The above equation shows that { converges uniformly to

;

Apply theorem 2.3.1 to { ;n ( t )

lim lim t-x n

A

= lim lim

∞

n

A

∞ t-x

;n}

for t ≠ x. ; n },

we get

;n(t).

(i.e) lim ;n(t) = lim fn '(x) ( by(5) and (6)). t-x

L

n

A

∞

f '(x) = lim fn '(x) ( by(7)). n

A

∞

#0-7- 2.5.2:

There

exists

a

real

continuous

function on the real line which is nowhere differentiable. P77:

Define

;(x)

= | x|

( − 1 ≤ x ≤ 1) -------------(1)

Extend the definition of

; (x)

to all

real x by requiring that ;(x+2)

= ; (x )

-----------------------------(2)


Then,for all s and t,we have In particular,

|;(s) − ;(t)| = |s| − |t| ≤ |s-t|.

------(3)

1

is conti nuou s on R .

;

(?).

∞

Define f(x) = ∑ n=0

Since 0 ≤

3 4

n

()

n

(4 x)

;

≤ 1,we have | f(x)| =

;

|

∞

∑(

3 4

n=0

)

n

;

∞

≤∑ n=0

∞

n=0

() 3 4

∞

∑ n=0

By

∑ n=0

on

3 4

() 1

n

R .

∑( n=0

3 4

)| n

is a geometric series

3 4

()

;

n

(4 x)

By

3 4

< 1 and

n

theorem

∞

≤

n

with the common ratio hence

||

(4 x)

∞

n

∑

Since

3

(4)

n

converges in R

1

2.2.3,

series

conv erg es uni for mly

theorem

continuous on R

the

.

1

2.3.2,

f

.


is

Fix a real number x and a positive integer/m.

1

m=

± 2.4

-m

sign is so chosen that no m

between 4

m

x and 4

where

the

integ er lies

(x +

). This m

m

can be done , since | 4 4

m

|

(x +

m)

–

x |

m

|

= 4 δm = 4 Define

m

=

m

|δm| = 4

m

1

-m

|± 2 4 | = 4

n

m -m 1 4 2

=

1 2

n

(4 (x+δm)) − ;(4 x)

;

δm n

n

1

n

(

When n > m, 4 δm = 4 δm = 4 ± 2 .4

-m

) = ± 2 4n-m = ± 2 22(n-m) = 22(n-m) − 1 1

1

= even integer.

Therefore

(4

= 0. (i.e.)

n

∑ =

=

| |

n=0

;

(

n

m

∑ n=0

4

n=0

δm

; (4

n

x)

n

;

|

(4nx)

4

∞

||

()

=

∑ n=0

n

( )γ| 3 4

n

( ) γ |(since,γ = 0 when n > m) 3 4

3 m γm 4

()

3

)

n

n

n

m-1

≥

|

δm

4 x+δm − ∑

()

5m )) –

= 0 when n > m.

∞

n

3

(x +

f(x+δm) − f(x)

We conclude that ∞

n

−∑ n=0

3 4

m-1

n

( )γ

n

≥

3 m m 4 4

()

−∑ n=0

3 4

n

( ) γ (by (5)) n


m-1

≥3

m

n

− ∑3 ≥ 3

m

(

− 1+3+3

2

+....+3

m-1

)

n=0

≥3 ≥

(

m

−

m

(

3 +1 2

As m

A

m

3 −1 3 − 12

)≥3

m

−

(

m

3 −1 2

m

)≥

(

m

)

2.3 − 3 − 1 2

). ∞,δm

A

0

It follows that f is not differentiable at x.

SECION-2.6 EQ!ICONIN!O!S FAMILIES OF F!NCIONS D-.161<176:

Let {f n } be a sequence of functions defined on a set E. We say that {f

n}

is point wise bounded on E if the sequence {f n (x)}is bounded for every x  E, that is, if there exists a finitevalued function

; defined on E such


that

|f n (x)|

<

; (x)

(x F E, n =

1,2,3,....)

We say that {f n }is uniformly bounded on E if there exis ts a number M such that |f n (x)| < M (x

F E,

n = 1,2,3,....)

Note: If {f n }is uniformly bounded sequence of continuous functions on a compact set E, there need not exist a subsequence which converges point wise on E.

E?)584-: Let f n (x) = sin nx (0 ≤ x ≤ 2

,n =

1,2,3,....) Suppose

there

exists

{n k }such that {sin n

a

k x}

sequence converges,

for every x

[0,27

F

]. In that case,

we have

klim ∞ A

(sin nkx-sin nk+1x) = 0, (0 ≤ x ≤ 2 ) and hence 2

lim (sin nkx-sin nk+1x) = 0, (0 ≤ x ≤ 2 ) k

A

∞

By

Lebesgue's

theorem,

2π

lim k

A

∫(sin n x-sin n

k+1x

k

)

2

--------(1)

dx=0

∞ 0

But 2π

∫(sin n x-sin n

k+1x

k

)

2

dx = 2

, which contradicts(1)

0

Note:

Every

need

not

convergent

contains

a

sequence uniformly

convergent subsequence. For

example

fn(x) =

x 2

2

x + (1 − nx)

2

(0 ≤ x ≤ 1,n=1,2,3,.....)

Then | f n (x) | ≤ 1 so that {f uniformly bounded on [0,1].

n }is

Also lim fn(x) = 0, (0 ≤ x ≤ 1) n

A

∞

But f n (l/n) =1, (n = 1,2,3,....), so that no su bsequence ca n co nverge uniformly on [0,1]. D-.161<176:

A family @ of co mplex functions f defined on a set E in a metric space X is said to be equicontinuous on E if for every

> 0 there exists a

such that |f(x) – f(y)| < d(x,y) <

>0

whenever

, x F E, y F E, and f F @ .Here

d denotes the metric of X. Note:

Every

member

equicontinuous fa mily continuous.

is

of

an

uniformly

#0-7-5 2.6.1:

If {f n } is a point wise bounded sequence of complex functions on a countable set E, then {f n } has a sub seq uen ce {f nk } such that {f nk (x)} converges for every x

F

E .

P77:

Let {x i }, i = 1,2,3,..., be the points of E, arranged in a sequence. Since {f n (x 1 )}is bounded, there exists a subsequence , which we shall denote {f 1,k },

such

converges as k

that A

{f 1,k (x 1 )}

∞.

Let us now consider sequences S

1,

S 2 , S 3 ,....., which we represent by the array

S1: f1,1f1,2f1,3f1,4.... .... S2: f2,1f2,2f2,3f2,4.... .... S3:f3,1f3,2f3,3f3,4.... .... .... ........ ........ ............ and

which

have

the

following

properties: a.

S n is a sub seq uen ce of S n–1 , for n = 2,3,4,...... A

b.

{f n,k (x n )} converges as k

c.

The order in which the functions appear

is

the

same

in

∞.

each

sequ ence; i.e., if one fu nction precedes another in S 1 , they are in the same relation in every S

n,

until one or the other is deleted. Hence, when going from one row in the above array to the next belo w, fu nctions may move to the left but never to the right.

We now go down the diagonal of the arrays; i.e., we consider the sequence S: f 1,1 f 2,2 f 3,3 f 4,4 .... .... By

(c),

the

sequence

S

(except

possibly its first n – 1 terms) is a subsequence

of

1,2,3,......Hence

Sn, (b)

for

n

implies

{f n,n (x i )} converges as n every x i F E.

A

= that

∞, for

#0-7-5 2.6.2:

If K is a compact metric space, if f n F c (K) for n = 1,2,3,....,and if {f n }converges uniformly on K, then {f n } is , equicontinuous on K. P77:

Let

> 0 be given.

Since {f n }converges uniformly, there is an integer N such that n > N implies ||f n – f N || < definition of ?(X) in sec 2.3). Since

continuous

uniformly

– f i (y)| <

(refer

functions

continuous

sets, there is a

.

on

are

compact

> 0 such that |f

i (x)

if 1 ≤ i ≤ N and d(x,y) <

δ. If n > N and d(x,y) <

, it follows

that |fn(x) − fn(y)| ≤ |fn(x) − fN(x) + fN(x) − fN(y) + fN(y) − fn(y)| ≤ |fn(x) − fN(x)| + |fN(x) − fN(y)| + |fN(y) − fn(y)| ≤ < fn − fN < + |fN(x) − fN(y)| + < fN − fn < <

+

+

=3

Therefore {f n } is equicontinuous on K.

#0-7-5 2.6.3:

If K is a compact, if f = 1,2,3,....,and if {f

nF n}

?(K) for n

is point wise

bounded and equico ntin uous on K, then a. b.

{f n } is uniformly bounded on K, {f n }

contains

a

uniformly

convergent subsequence. P77:

a.

Let

> 0 be given.

Since {f n } is equicontinuous on K, there exis ts a

> 0 such tha t

x,y F K d(x,y) < δ B |fn(x) − fn(y)| <

,for all n.

-------------(1)

Since K is compact, there are finitely many points p

1,

p 2 ,.....,

p r in K such that to every x corresponds at least one p

F

i

K

with

d(x, p i ) < Since {f n } is point wise bounded, there exist M i < ∞ such that (6) and ( 7) implies that h(t) < f(t) + (i.e.)f(t) − (i.e.) −

< h (t) < f(t) +

(i.e)|h(t) − f(t)| <

(Ftk )

( tFK ).

(F tK)

(F tK)

|fn(pi)| < Mi for all n.

------------------- (2)

|

If M=max(M1, M2, .....Mr), then|fn(x)| = fn(x) − fn(pi) + fn(pi)

|

|

| |

≤ fn(x) − fn(pi) + fn(pi) <

|

+Mi.(by(2) & (3))

≤ε+M

Therefore,

{f n }

is

uniformly

bounded on K. b.

Let E be a subset of K.

countable

dense

By theorem 2.6.1, {f n } has a subsequence { f ni } such that {f ni (x)} converges for every x

F


E .

Put f ni = g i . We

shall

prove

that

{g i }

converges uniformly on K. Let

> 0 be given.

Since {f n } is equicontinuous on K, there exists a

> 0 such that

d(x,y) < δ B |fn(x) − fn(y)| <

, for all n.

----------------(4)

Let V(x,δ) = y F E/d(x,y) < δ .

{

}

Since E is dense in K and K is compact, there are finitely many points x 1 , x 2 ,....., x m in E such that

KO

V(x 1 ,

δ) K .... K V(x m , Since

{g i (x)}

δ) K V(x 2 ,

). converges

for

every x F E, there is an integer N such that


|gi(xs) − gj(xs)| <

If

whenever i ≥ N,j ≥ N,1 ≤ s ≤ m. --------( 5)

xFKO

δ) K .... K V(x m ,

V(x 1 ,

δ) K V(x 2 ,

), then x

F V(x s ,

) for some s (i.e.) d(x, x B

|gi(x s ) – g j (x s )| <

s)

< δ

for every

i. If i ≥ N, j ≥ N , |gi(x) − gj(x)| = |gi(x) − gj(xs) + gi(xs) − gj(xs) + gj(xs) − gj(x)|

|

| |

| |

|

≤ gi(x) − gi(xs) + gi(xs) − gj(xs) + gj(xs) − gj(x)

< Therefore,

{g i }

+

+

=3

converges

uniformly on K. (i.e.) {g i } co nverges uniformly on K (i.e.) { f ni } converges uniformly on K


(i.e.) {f n } contains a uniformly convergent subsequence. C'P $E"#ION": 1.

Suppose {f n }, {g n } are defined on E, and a.

Σf n has uniformly bo unded partial sums;

b.

gn

c.

g 1 (x)

A

0 uniformly on E; ≥

g 2 (x)

≥.....for every x Prove that on E.

F

≥

g 3 (x)

E.

f n g n converges uniformly

SECION-2.7

HE SONE :

#EIERSRASS HEOREM

#0--5 2.7.1:

If f is a continuous complex function on [a,b], there exists a sequence of polynomials P n su ch t hat lim "

= f(x) uniformly on [a,b]. If f

A

P n (x) ∞

is real,

the P n may be taken real. P:

We may assume, without loss of generality, that [a ,b] = [0,1]. We may also assume that f(0) = f(1) = 0. For if the theorem is proved for this case, consi der g(x) = f(x) – f(0) – x[f(1) – f(0)] (0 ≤ x ≤

1).

Here g(0) = g(1 ) = 0, and if g can be obtained as the limit of a uniformly converge nt sequence of polynomi als , it is clear that the same is

true for f,

since f – g is a polynomial. Furthermore, we define f(x) = 0 for x outside [0,1]. Then f is uniformly continuous on the whole line.

(

We put Qn(x) = cn 1 − x

2 n

)

(n = 1,2,3,....),

------ (1)

1

∫Q (x) dx=1(n=1,2,3,...) --------(2)

where cn is chosen so that

n

−1 1

Now

∫ (1 − x ) −1

1 / √n

1 2 n

2 n

∫ (1 − x )

dx=2

2 n

∫ (1 − x )

dx ≥ 2

−0

dx

0

1 / √n

∫(1 − nx ) dx(by binomial theorem) 2 n

≥2

0

3

[

≥ 2 x-

[

1

nx 3

=2 √n −

]

1 / √n 0 1

3

( √n )

n

3

]

[

1

1

− 0 = 2 √n − 3√n

]


=2

2

4

1

[ 3√ n ] = [ 3√ n ] ≥ √ n 1

Equation ( 2) implies that

1

∫ (

cn 1 − x

2 n

)

B

dx=1

−1

1

B 1>cn √n B cn <

−1

----------- (3)

√n

(

≤|x| ≤ 1

where Q n (x )

A

2 n

∫(1 − x ) dx=1

>0(1) and ( 3) implies that Qn(x) < √n 1 − δ

For any

L

cn

2 n

)

------------------ (4)

0 uniformly in

≤|x| ≤ 1.

1

Now set Pn(x) =

∫f(x+t)Q (t)dt n

---------------- (5)

−1 −x

P n (x ) =

1 −x

1

∫f(x+t)Q (t)dt+∫f(x+t)Q (t)dt+∫f(x+t)Q (t)dt n

n

−1

n

−x

1 −x

Put x + t = y. Then dx = dt & t = –1 B y = x –

1,t=

t=1–x B y=1,t=1

B

–x

Pn(x) =

y = 0,

y = x + 1.

0 L

B

1

x+1

∫f(y) Q (y-x)dy+∫f (y) Q (y-x)dy+∫f(y) Q (y-x)dy n

n

x-1

0

n

1

1

=0+

∫f (y) Q (y-x)dy+0(?) n

0

The

RHS

integral

is

clearly

a

polynomial in x. Thus {P n } is a sequence of polynomials whic h are real if f is real.


Therefore, given

> 0, we choose

> 0 such that |y – x| < ε

|f(y) − f(x)| < 2 . Let M= sup

x F [0.1]

implies

----------------- (6)

|f(x)| (since f is bounded on[0, 1])

--------(7)

If 0 ≤ x ≤ 1 1

|Pn(x) − f(x)| =

|∫ |∫

1

−1

n

−1

1

=

|

∫Q (t)dt (by(2) & (5))

f(x+t)Qn(t)dt-f(x)

|

(f(x+t) − f(x))Qn(t)dt

−1

1

≤

∫|(f(x+t) − f(x))Q (t)dt| n

−1

−δ

δ

∫|

≤

| ∫|(f(x+t) − f(x))Q (t)dt|

(f(x+t) − f(x))Qn(t)dt +

−1

n

−δ 1

+

∫|(f(x+t) − f(x))Q (t)dt| n

δ -δ

δ

∫(

≤

−1

(f(x+t) − f(x))

−δ

-δ

<

-δ

∫|

)

|f(x+t)| + |f(x)| Qn(t)dt+

∫(M+M) Q (t)dt Q (t)dt n

−1

1

∫(M+M) Qn(t)dt+∫(M+M)Qn(t)dt(by(7)) −1

δ -δ

δ 2

∫√ n ( 1 − δ )

≤ 2M

ε dt+ 2

−1

( (

2 n

< 4M√n 1 − δ ε 2

2

∫ Q (t)dt+2M∫√n(1 − δ )dt(by(4)) n

−δ 2 n

< 2M√n 1 − δ

<

1

) )

ε +2

(1)+2M√n(1 − δ

δ 2 n

) (1 − δ)(by(2))

ε

+ 2.

ε

+ 2 , for large enough n

=ε Therefore lim Pn(x) = f(x) uniformly on[a,b]. n

A

∞


n

Cor: For every interval [–a,a] there is a sequence of real polynomials P

n

such that P n (0) = 0 and such that lim Pn(x) = |x| uniformly on [–a,a] n

A

∞

P77.:

By the above theorem, there exists a sequence {P n *} of real polynomials which converges uniformly to |x| on [–a,a]. In particular, P A

n *(0)

A

0 as n

∞. The polynomials P n (x) = P n *(x)

– P n *(0) for n = 1,2,3,..have the desired properties. D-.16<176:

A

family

a

of

complex

functions

defined on a set E is said to be an algeb ra if (i) f+g

F

? (ii)fg

F

? (iii) cf

F

? for all f

F

?,

g F

? and for all

complex constants c, that is, if ? is closed under addition, multiplication, and scalar multiplication. For the algebra of real functions , we have to consider (iii) for all real c. If ? has the property that f

?

F

when ever f n F ? ( n = 1,2,3,....) and fn

A

f uniformly on E, then ? is said

to be uniformly closed. Let

?

be the set of all functions which

are limits of uniform ly convergent sequence of membe rs of ? . Then

?

is

called the uniform closure of ?. Theorem 2.7.2: Let

?

be the uniform

closure of an algebra ? of bounded

functions.

Then

?

is a uniformly

closed algebra. P77.:

If

f F?

and

uniformly

g

F?

, there

convergent

{f n },{g n }such that f n

A

exist

sequences f , gn

A

g

and f n F ? ,g n F ? . Since the functions are bounded, we have f n + g n A f + g, f n g n A fg, cf n A cf, where c is any constant, the convergence is uniform in each case. Hence f +g Therefore

F?

?

, fg

F?

and cf

is an algebra.

SinceB is the closur e of ? , (uniformly).

F?

?

is closed

Definition: Let ? be a family of functions on a set E. Then ? is said to separate points on E if to every pair of distinct point x corresponds a function f

1 ,x 2

F

F

E there

? , such

that f(x 1 ) ≠ f(x 2 ). If to each x function g

F

F

E there corresponds a

? , such that g(x) ≠ 0, we

say that ? Vanishes at no

point of E.

#0--5 2.7.3:

Suppose ? , is an algebra of functions on a set E, ? , separates points on E, and ? ,vanishes at no point of E. Suppose x 1 ,x 2 are distinct points of E, and c 1 ,c 2 are constants (real if ?, is a real algebra). Then ? , contains a

function f such that f(x 1 ) = c 1 , f(x 2 ) = c2. P77.:

Since ? , separates points on E and ? , vanishes at no point of E, we have g(x 1 ) ≠ g(x 2 ), h(x 1 ) ≠ 0, k(x 2 ) ≠ 0, where g,h,k

F

? ,.

Put u = gk – g(x 1 )k, v = gh – g(x 2 )h. Since g,h,k u,v

F

F

? , and ? ,is an algebra,

? , .

Also u(x 1 ) = g(x 1 )k – g(x 1 )k = 0 and v(x 2 )= g(x 2 )h – g(x 2 )h = 0, u(x 2 ) = g(x 2 )k – g(x 1 )k ≠ 0 & v(x 1 ) = g(x 1 )h – g(x 2 )h ≠ 0. Let f=

c1 v v(x1)

+

c2 u u(x2)

.

Then f(x1) = f(x 2 ) =

c1v(x2) v(x1)

c1v(x1) v(x1)

+

+

c2u(x2) u(x2)

c 2u(x1) u(x2)

= c1 + 0 = c 1 and

= 0 + c 2 = c2

"#ONE'" GENE!ALI(A#ION OF #HE &EIE!"#!A"" #HEO!EM

#0--5 2.7.4:

Let ? be an algebra of real continuou s functions on a compact set K. If ? separates

points

on

K

and

if

?

vanishes at no point of K, then the uniform closure

?

of ? consists of all

real continuous functions on K.

We shall divide the proof into four steps. "#EP 1:

If f

F ?

then |f|

F ?

P:77.

Let a=sup | f(x)|, ( xFK ). Let By

------------(1)

> 0 be given. the

corollary

to

the

Stone-

Weierstrass theorem, there exist real numbers

c 1 ,c 2 ,....,c n

such

that

n i

|

∑ i=1

c iy

F

− |y| <

|

,

y [-a,a].

-------(2)

Since ? is an algebra and f F ? B c i i

f ? for i = 1,2,... ,n. n

Hence

the

function

g=

c if ∑ i=1

i

is

a

member of ? . By (1) and (2), we have |g(x) |f(x)|| <

(x F K)

Since ?

?

, |f| F

is uniformly clos ed and g(x) ?

"#EP 2:

F

.

F?

If f

F?

and g

max(f,g) and min(f,g)

F?

, then

.

By max(f,g) we mean the function h defined by

h (x ) =

f(x) if f(x) ≥ g(x),

{

g (x )

and min(f,g) =

{

if f(x)
f(x) if f(x) < g(x), g(x) if f(x) ≥ g(x),

.

P77.:

Consider max(f,g) =

the f+g 2

and min(f,g) =

+

identities |f-g|

f+g 2

2

−

|f-g| 2

.

max

Since gF

?

?

is an algebra and f

F?

and

, we have

f+g,f-g F B. Also

f+g |f-g| 2 , 2 F B

Therefore max(f,g) and min(f,g)

F ?

. By

can

be

extended to any finite set functio ns, (i.e.) if f 1 ,f 2 ,.. .,f n

of

,

iteration,

then

the

result

max(f 1 ,f 2 ,...,f n ) F

?

F

?

and

min(f 1 ,f 2 ,.. .,f n ) F? "#EP

3:

Given a real function f,

continuous on K, a point x

F K,

> 0, there exists a function g

and x F?

such that g x (x) = f(x) and g x (t) > f(t) –

(t F K)

P77.:

By hypothesis

? O

? , and ? satisfies

the hypothesis of theorem 2.7.3, also

satisfies

the

hypothesis

?

of

theorem 2.7.3. Hence, for every y

F K,

we can find a

function h y F? such that h y (x)=f(x),h y =f(y)

(refer

theorem

2.7.3)-------(3) By the continuity of h an open set J

y,

y

there exists

containing y, such

that h y( t ) > f ( t ) −

( t F Jy)

-------------( 4)

Since K is compact, there is finite set of points y 1 ,y 2 ,...,y n such that

K

O

Jy1 K Jy2 K ...

K

Jyn. -------------( 5)

Put g x = max(h y1 ,h y2 ,...,h yn ). By step 2, g

F? .

By (3), (4) and (5), we have g x (x) = f(x) and g x (t) > f(t) – "#EP

4:

Given a real function f,

continuous on K, and exists a function h F – f(x)| <

?

> 0, there

such that |h(x)

. (xF K)

P77.:

Let us consider the functions g x , for eaph x F K, constructed in Step 3. By the continuity of g x , there exist open sets V x containing x, such that

g x( t ) < f( t ) +

( t F Vx) . -----------( 6)

Since K is compact, there exists a finite set of points x 1 ,X 2 ,...,x m such that K

O

Vx1 K Vx2 K ...

By Step 2, h

F?

K

Vxm. -----------( 7)

.

By Step 3, h(t) > f(t) –

(t

F K).

(6) and (7) implies that h(t) < f(t) + (t F K). (i.e.) f(t) – (i.e.) –

< h(t) < f(t) +

< h(t) – f(t) <

(i.e.) |h(t) – f(t)| <

(t

(t F K).

(tF K). F

K).

#0-- 2.7.5:

Suppose ? is a self-ad joint algebra of complex continuous functions on a compact set K, ? separates points on K and ? vanishes at no point of K. Then the uniform closure B of ? consists of all complex continuous functions on K. In other words ? is dense in C(X). P:

Let ? r be the set of all real functions on K which belong to ? If f

F

? and f =

¯ u + iv, with u,v real , then 2u = f + f , and since ? is a self-ad joint, we see that u F ? r . If x 1 ≠ x 2 , there exists f that f(x 1 ) = 1, f(x 2 ) = 0.

F

? such

Hence 0 = u(x 2 ) ≠ u(x 1 ) = 1. Therefore ? r separates points on K. If x F K, then g(x) ≠ 0 for some g F ? , and there is a complex number $ such that $ g(x) > 0. If f = $ g , f = u+iv, it follows that u(x) > 0. Hence ? R vanishes at no point of K. Therefore ? R sa tisfies the hypothesi s of theorem 2.7.4. It follows that every real continuous function on K lies in the uniform closure of ? R , and hence lies in

?

. If

f is a complex continuous function on K, f = u + iv, then u

F?

, vF

?

.

Hence f

F ?

.

C'P $E"#ION": 1.

Prove that the set C (X) of all complex valued, continuous, bounded functions with domain X, with d(f,g) = ||f – g|| is a metric space.

2. Distin guish

between uniformly convergent and point wise convergent.

$NI#-3

$61< "<:=+<=:

Section 3.1: Power series Section 3.2: The exponential and Logarithmic functions Sectio n 3.3: The trigonom etric Functions Section

3.4:

completeness

The of

the

algebraic Complex

Field Section 3.5: Fourier series Section function

3.6:

The

Gamma

I6<:7,=+<176

In this unit we shall derive some properties

of

functions which

are

represented by power series. We also discuss

the

concept

Exponential, Trigonometric

of

the

Logarithmic, functions

,

Gamma

function and Fourier series.

SECION-3.1

PO#ER SERIES

We have already discussed about the power serie s in Analy sis I in the first year. The form of the power series ∞

is f(x) = ∑ cnx

n

n=0 ∞

f(x) = ∑ cn(x-a) n=0

n

or more generally

These are ca lled analytic fu nctions. We shall discu ss the power series only for real values of x. Instead of circles

of

convergence

we

shall

encounter intervals of convergence. ∞

The series f (x) = ∑ cnx

n

converges

n=0

for all x in (– R ,R), for some R. ∞

If

the

series

f

(x) = ∑ cn(x-a)

n

n=0

converges for |x – a | < R, then f is said to be expanded in a power series about the point x = a. #0-7-5 3.1.1: ∞

Suppose

the

series

∑ cnx n=0

converges for |x| < R, and define

n

∞

f(x) = ∑ cnx

n

( |x | < R )

. Then

the

n=0 ∞ n

c n=0 ∑

series

on [– R +

nx

conve rges unifo rmly

,R s], no matter which

> 0 is chosen. The function f is continuous and differentiable in (– R ,R), and ∞

f '(x) = ∑ ncnx

n-1

( |x | < R )

n=0

P77.:

Let

> 0 be given.

For |x| ≤ R – |c n (R –

n

) |.

, we have |c

nx

n

| ≤

∞

∑ cn(R-ε)

Since

n

converges

n=0

abso lutely ( M every power series converges absolutely in the interior of its interval of convergenc

e , by the

root test) ∞

the

L

series

cnx

n

converges

n=0 ∑

uniformly on [ –R +

, R –

].(by

theorem 2.2.3) Since

lim n

A

n

√n=1 n

limsup n|cn| = limsup n ∞ n ∞

√

have

∞

n

A

we

A

cn| | √ ∞

Therefore the series f (x) = ∑ cnx n=0

n

∞

and the ser ies f ' (x) = ∑ ncnx

n-1

have

n=1

the same interval of convergence. ∞

Since f ' (x) = ∑ ncnx

n-1

is a power

n=1

series, it converges uniformly in [– R +

,R –

], for every

> 0 we

can apply theorem 2.5.1( for series instead

of

sequence),

(i.e.)

∞

f '(x) = ∑ ncnx

n-1

holds for |x|
n=1

holds if |x| ≤ R –

But, given any x such that |x| < R, we can find an > 0 such that |x|
This

shows

that

f ' (x) = ∑ ncnx n=1

n-1

holds for |x| < R. Cor: Under the hypothesis of the above theorem, f has derivatives of all orders in (– R ,R), which are given by ∞

f

( k)

(x) = ∑ n(n-1)....(n-k+1)cnx

n-k

n=k

In particular, f

(k)

(0) = k!c k , (k =

0,1,2,....). P77.:

By

the

above

theorem

∞ n

f(x) = ∑ cnx we get n=0 ∞

f '(x) = ∑ ncnx n=1

n-1

from

Apply theorem 3.1.1, to f ‘, we get Successively apply theorem 3.1.1 to f (3)

”,f

,...., we get ∞

(k )

f

(x) = ∑n(n-1)....(n-k+1)cnx

n-k

, -----------(1)

n=1

Putting x = 0 in (1), we get f

(k)

(0)=

k!c k , (k = 0,1,2,....).

A*-4'; #0-7-5 #0-7-5 3.1,2: Suppose

cn

converges.

Put

∞ n

f (k ) =

c n=1 ∑

nx

( −1
lim f(x) = ∑ cn

Then x

A1

n=1

P77:

Let s n = c 0 + c 1 + ....+ c n , s –1 = 0.

m

Then

m

∑ c nx

n

= ∑ (sn − sn-1)x

n=0

n

n=0 0

1

2

= (s0 − s − 1)x + (s1 − s0)x + (s2 − s1)x +...+ 1

1

2

2

= s01 + s1x − s0x + s2x − s1x ..... − sm-1x

m-1

(sm − sm-1)x

+ smx

m

m

= (1 − x)s0 + (1 − x)s1x1 + (1 − x)s2x2....(1 − x)sm-1xm-1 + smxm m-1 n

= (1-x)∑ snx + smx

m

n=0 ∞

For | x| < 1, let m

A

∞,we get

m n

∑ cnx = (1 − x)∑ snx n=0

n

n=0

∞

(i.e.)f(x) = (1 − x)∑ snx

n

n=0

Since

c n converges to s (say).

Then its partial sum sequence {s converges to s.

(i.e.) lim sn = s. n

Let

A

∞

> 0 be given. ε

Choose N so that n>N B |s-sn| < 2 . ---------- (1) ∞

By the geometric series test,we have

∑x

n

= 1 if | x|<1.

n=0


n}

∞ n (i.e)(1 − x)∑ x = 1 if | x| < 1. ------------ (2). n=0 ∞

Now to prove that

lim

x

A1

n

f(x) = ∑ c = s n=0

∞

∞

∞

|f(x)-s| = (1-x)∑ snxn − s = (1 − x)∑ snxn − s(1 − x)∑ snxn (by(2))

| |

n=0 ∞

n

= (1-x)∑ (sn-s)x n=0

|| |

n=0

n=0

∞ n

| |(

≤ ( 1 − x)∑ |sn-s| x

M

|x+y| ≤ |x| + |y|)

n=0 ∞

∞ n

n

≤ ( 1 − x)∑ |sn-s||x| + (1 − x)∑ |sn-s||x| n=0

n=N+1

n

∞ n

n

ε

≤ ( 1 − x)∑ |sn-s||x| + 2 (1 − x)∑ |x| (by(1)) n=0

n=N+1

N n

ε

< (1 − x)∑ |sn-s||x| + 2 .

---------------(3)

n=0

Choose

> 0 such that 1-x <

Then

(3) ε

becomes

ε

f(x) − s < 2+ 2=

|

|

(i.e.)|f(x) − s| <

if x>1∞

(i.e) lim f(x) = s =∑ cn x

A1

n=0


|

Note:If Σ a n ,Σ b n , Σc n , converge to A,B,C , and if c

n

= a 0 b n + a 1 b n–1

+.... + a n b 0 , then C = AB. #0-7-5 3.1.3:

Given a double sequence {a 1,2,3,...,

j

=

ij },

1,2,3,...,

∞

suppose that

∞

∑ |aij| = bi ( i=1,2,3,....) and

∑

i =

biconverges.Then

j=1

∞

∞

∞

∑ ∑ aij = ∑ ∑ aij. i=1 j=1

j=1 i=1

Let E be a countable set, consisting of the points x 0 ,x 1 ,x 2 ,...,x n , and suppose x n

A

x 0 as

A

∞. Define

P77.: ∞

Let

∑ |aij| = bi

(i=1,2,3,....)

-----------------(1)

j=1 ∞

fi(x0) = ∑ aij(i=1,2,3,...) ----------(2) j=1 n

fi(xn) = ∑ aij(i,n=1,2,3,...) ----------(3) j=1 ∞

g(x) = ∑ aijfi(x)( xFE )

---------- (4)

i=1


Now, (2) and (3), together with (1), we get

n

∞

fi(n) = ∑ aij j=1

(i.e.) x n

A

A

∑ aij = fi(x0) as n

A

∞

f i (x 0 ) as n

A

j=1

x 0 B f i (x n )

A

∞. This shows that each f

i

is continuous

at x 0 . ∞

Now | fi(x)| = ∑ |aij| = bi for x F E &i=1,2,3,.... j=1 ∞ L

g(x) = ∑ fi(x) converges uniformly. i=1

By theorem 2.3.1, g is continuous at x0.


∞

Therefore

∞

∞

∑ ∑ aij = ∑ fi(x0) (by(2)) i=1 j=1

i=1

=g(x0) ( by(4)) = lim g(xn) n

A

∞ ∞

= lim n

A

∑ fi(xn)(by(4))

∞ i=1 ∞

= lim n

A

∑ ∑ (aij)(by(3))

∞ i=1 j=1 ∞

= lim n

A

∑ ∑ (aij)(by(3)) n

∞ L

∞

∞

A

∞

∞ j=1 i=1

= lim n

n

∞

∞

∞

∑ ∑ (aij) = ∑ ∑ aij

∞ j=1 i=1

j=1 i=1

∞

∑ ∑ aij = ∑ ∑ aij i=1 j=1

j=1 i=1

#ALO!'" #HEO!EM

#0-7-5 3.1.4: ∞

Su ppose f (x) = ∑ cnx

n

, the series

n=0

converging in |x| < R. If –R< a < R, then f can be expanded in a power series about the point x = a which Double click this page to view clearly

converges in | x – a | < R – |a|, and ∞

f (x ) = ∑

f

(n)

(a )

(x-a) n!

n

(|x-a| < R-|a|)

n=0

P77.: ∞

We have f(x) = ∑ cnx

n

n=0 ∞

=∑ cn(x-a+a)

n

n=0 ∞

=

∑ cn((x-a)+a)

n

n=0 ∞

(

n

=∑ c n

∑

n=0

m=0

)

n n-m m a x-a) ( m

( )

Therefore f(x) can be extended in the form of power series about the point x = a. ∞

But

n

∑∑ n=0

m=0

∞

|( ) cn

|

n n n-m m a x-a) = ∑ |cn|(|x-a| + |a|) ( m n=0

which converges if |x – a| + |a| < (i.e.) if |x – a| < R –

|a|

R.

We know that, by corollary to the theorem 3.1.1, f(k)(0) = k!ck.

(k=0,1,2,....).

( k)

(i.e.)ck = ∞ L

(0)

f

f (x ) = ∑

k!

f

(n)

(k=0,1,2,...)

(a )

(x-a) n!

n

(|x-a| < R-|a|)

n=0

#0-7- 3.1.5:

Suppose the series

an x

n

and

bn x

n

converge in the segment S = (–R,R). Let E be the set of all x

F

S at which

∞

∑ anx n=0

∞ n

=

∑ bn x

n

n=0

If E has a limit point in S, then a n = b n for n = 0,1,2,3, ....

∞

Hence

∞ n

∑ a nx n=0

n

= ∑ bnx holds for all n=0

x F S. P77.:

Put

cn

=

an

–

bn

and

∞

f(x) = ∑ cnx

n

( xFS )

n=0

∞ n

= ∑ (an − bn)x = 0 for x F S.(by hypothesis) n=0

Let A be the set of all limit points of E in S. Let B be the set of all other points of S. By the definition of limit point, B is open. Let x 0 F A.

By

the

above

theorem,

∞

f(x) = ∑ dn(x-x0)

n

(|x-x0| < R-|x0|)

n=0

Now to prove that d n = 0 for all n. Suppose there exist a smallest nonnegative integer k such that d

k

≠ 0.

k

f(x) = (x-x0) g(x) |x-x0| < R-|x0| ,

(

)

∞

where g(x) = ∑ dn+k(x-x0)

m

m=0

Since g is continuous at x 0 , and g(x 0 ) = d k ≠ 0, there exists a

> 0 such

that g(x 0 ) = d k ≠ 0 if |x – x 0 | < k

Therefore f(x) = (x – x 0 ) g(x) ≠ 0 if 0 < |x – x 0 | <

, which is a

contradiction to the fact that x 0 is a limit point of E.

Therefore d n = 0 for all n. ∞ n

So f(x) = ∑ dn(x-x0) = 0 if | x-x0| < R-|x0| n=0

(i.e.) in a neighborhood of x

0.

Therefore A is open. (i.e.) A and B are disjoint open sets. (i.e.) A and B are separated. Since S = A

K

B and S is connected,

one of A and B must be empty. By hypothesis A is not empty. Hence B is empty and A = S. Since f is continuous in S, A O E. Thus E = S and

(n)

cn = 

f

(0 ) n!

= 0 (n=0,1,2,....)

an = bn for n=0,1,2,3,.... ∞

∑ a nx



∞ n

n

= ∑ bnx holds for all x  S

n=0

n=0

C'P $E"#ION":

1.

Define f(x) =

e

−1

2

/ x ( x≠0 )

.Prove that f has derivatives of all

(x=0)

0

{ order s at x = 0 and that f

(n)

(0) =

0 for n = 1,2,3,...

SECION-3.2

HE

E$PONENIAL AND LOGARIHMIC F!NCIONS

D-.161<176: ∞

n

z E(z) = ∑ n! n=0

Note

1:

By

the

ratio

test,

E(z)

converges for every complex z. Note

2:

(Addition ∞

n

z E(z)E(w) = ∑ n!

∞

m

w ∑ m!

n=0

m=0

∞

n

k

=∑ ∑ n=0

formula)

k=0

n-k

z w k!(n-k) !

∞

n

n=0

k=0

∞

(z+w) 1 n k n-k =∑ ∑ z w =∑ n! k n!

()

n

n=0

= E(z+w), where z and w are complex numbers.

Note 3: E(z)E(–z) = E(z – z) = E(0) = 1, where,z is a complex number. L

E(z) ≠ 0 for all z.

By the definition, E(x) > 0 if x > 0 and E(x) > 0 for all real x. Again by the definition, E(x) x

A

A

+∞ as

+∞ and E(x) A 0 as x A – ∞ along

the real axis.


Also 0 < x < y implies E(x) < E(y) and E(–y) < E(–x). Hence E is strictly increasing on the whole real axis. lim h=0

E(z+h) − E(z) h

= E(z)lim h=0

E( h) − 1 h

= E(z).1 = E(z)

By iteration of the addition formula gives E(z 1 + z 2 + ...+ z n ) = E(z 1 )E(z 2 ). . .E(z n ). If z 1 = z 2 = ...= z n = 1,then E(n) = n

e ,n= 1,2,3,...(

M

E(1) = e)

If p = n/m, where n and m are positive integers, then [E(p)]

m

= E(mp) = E(n) = e

E(p) = e rational).

p

and E(–p) = e

n

, so that

–p

(p>0, p

#0-7-5 3.2.1:

Let

e

x

be

defined

on

∞

xn n!

x

e = E (x ) = ∑ n=0

a.

x

e

is

continuous

and

differentiable for all x; x

x

b.

(e ) ’= e

;

c.

e is a strictly increasing function

x

of x, and e d.

(d) e

e.

e x

x

A

A

x+y

A

x

y

= e e ; A

+∞ and e

x

A

0 as

– ∞; n

x

> 0;

+∞ as x

lim x e

f.

x

−x

= 0 for every n.

+∞

P77:

For the proof of (a) to (e), refer Note 1,2,3.

∞

x

x

n

(f)By definition e = ∑ n! n=0 2

x!

=1+ 1 ! + > x (i.e.)e >

n

x x

(n+1) !

B e

−x

x

x

n

x

n+1

.... + n! + (n+1) ! + .....

n+1

(n+1) !

<

x 2! +

for x>0

(n+1)! n

x x

-x n

B e x <

(n+1) ! x

A 0 as x A +∞;

n -x (i.e.) lim x e = 0 for every n. x

A

+∞

Note:Since

E is strictly increasing 1

and differentiable on R , it has an inverse function L, which is also strictly increasing and differentiable and whose domain is E(R

1

), that is,

the set of all positive numbers, L is defined by E(L(y)) = y (y > 0) or L(E(x)) = x (x real). C'P $E"#ION": 1.

Find the following limits. lim

i. x

A0

e-(1 + x) x

1/x

n log n

1/n

− 1]

SECION-3.3

HE

ii.

lim n

A

∞

[n

RIGONOMERIC F!NCIONS

Let C(x) =

us 1 2

define 1

[E(ix) + E(-ix)] and S(x) = 2i [E(ix) − E(-ix)]

¯

¯ Since E z = E(z), C(x) and S(x) are

()

real for real x. Also E(ix) = C(x) + iS(X) (i.e.) C(x) and S(x) are real and imaginary parts of E(ix),respectively, if x is real.

¯

|E(ix)|2 = E(ix)E(ix) = E(ix)E(-ix) = 1, so that |E(ix)| = 1(x real) , so that |E(ix)| = 1 (x real).

Also C(0) = 1 and S(0) = 0 , and C'(x) = –S(x), S'(x) = C(x).


Now

to

positive

show real

that

there

number x

exist

such

a

that

C(x)=0. Suppose C(x) ≠ 0 for all x. Since C(0) = 1, we have C(x) > 0 for all x > 0. Since S(0) = 0 and S'(x) = C(x), we have S'(x) > 0 for all x > 0. L

S(x) is strictly increasing function.

If 0 < x < y then S(x) < S(y). #

y

(i.e.)S(x)(y-x) <

∫

S(t)dt <

x

∫ − C '(!)d! = -(C(t))

y x

= − [C(y) − C(x)]

"

= [C(x) – C(y)] 2

2

2

Since 1 = |E(ix)| = C (x) + S (x), 2

C (x) ≤ 1.


L

C(x) and – C(x) ≤ 1.

L

[C(x) – C(y)] ≤ 2.

(i.e.)S(x) (y – x) ≤ 2. Since S(x) > 0 this inequality cannot be true for large values of y , we get a contradiction. L E

a positive number x such that

C(x) = 0. Let

x0

be

the

smallest

positive

integer such that C(x 0 ) = 0.

Define the number Then

C(

by

=

2x 0 .

/2) = C(x 0 ) = 0

Since E(ix) = C(x) + iS(X) & |E(ix)| = 1, we have S(

/2) = ±1

Since C(x) > 0 in (0,

/2), S is an

increasing function in (0,

/2)

Therefore S( E(πi)=E

πi

/2) = 1. Thus

πi

πi

.

( πi2 ) = i

πi

( 2 + 2 )=E( 2 ).E( 2 )=i.i=-1

E(2πi)=E(πi+πi)=E(πi).E(πi) = ( − 1)( − 1) = 1. Also E(z+2πi)=E(z).E(2πi)=E(z)

#0-7-5 3.3.1: a.

The function E is periodic, with period 2

b.

The

i,

functions

C

periodic, with period 2

and

S

are

,

c.

If 0 < t < 2

, then E(it) ≠ 1,

d.

If z is a complex number with |z| = 1, there is a unique t in [0, 2 such that E(it) = z.

)

P77.:

a.

Since

E(z+2

i)

=

E(z),

periodic, with period 2 b.

C(x+2

)=

1 2

E

is

i.

[E(i(x+2 )) + E(-i(x+2 ))] 1 2

=

[E(ix+2

)i + E(-ix-i2π))]

1

= 2 [E(ix) + E(-ix)](Since E(2πi) = 1) =C(x)

Therefore C(x) is periodic with period 2 Similarly S(x) is periodic with period 2 c.

Suppose 0 < t <

/2 and E(it) =

x + iy, where x & y are real. L

0 < x < 1 & 0 < y < 1.

1 = |E(it)|

2

= x

E(4it) = (E(it)) – 6

2 2

y

+ y

4

2

4

+ y

2

. 4

= (x + iy) = x

+ 4ixy(x

2

2

– y )

4

If E(4it) is real, then x 2

, that is x Since x

2

½ and y

+ (1/2)

2

2

= 0

2

2

= 1,2x

2

= 1B x

2

=

= ½.

E(4it) = (1/2)

L

– y

= y .

+ y

2

2

2

2

– 6(l/2) (l/2)

2

+ 4ixy(0)

= 1/4 – 6/4 + 1/4 = – 4/4 = –1. L E(it) ≠ 1. d.

Choose z so that |z| = 1. Write z = x + iy, with x and y real. Suppose first that x ≥ 0 and y ≥ 0. On [0,

/2], C decreases from 1

to 0. Hence C(t) = x for some t F [0, 2].

/

Since C [0,

2

+ S

2

= 1 and S ≥ 0 on

/2], it follows that z = E(it).

If x < 0 and y ≥ 0, the preceding conditions are satisfied by –iz . Hence –iz = E(it) for some t F [0, /2], and since i = E(i obtain z = E(i(t+

/2), we

/2)).

Finally if y < 0, the preceding two cases show that –z = E(it) for some t F (0, –E(it) = E(i(t+

). Hence z =

)).

Suppose 0 ≤ t 1 < t 2 < 2π , E(it 2 )[E(it 1 )

–1

= E(i(t 2 – t 1 )) ≠

1( by (c)) Therefore there is a unique t in [0,2

) such that E(it) = z.

C'P $E"#ION": 1.

π

2

2

π

If 0
<

sinx x

<1

SECION-3.4

HE ALGEBRAIC

COMPLEENESS OF HE COMPLE$ FIELD

#0-7-5 3.4.1:

Suppose

a 0 ,a 1 ,...,a n

are

complex n

numbers, n ≥ 1, a n ≠ 0,P(z) =

∑

a kz

k

0

. Then P(z) = 0 for some complex number z. P77:

Without loss of generality we assume that a n = 1.

Put

= inf |P(z)| (z complex) .

If |z| = R, then |P(z)| = | a 0 + a 1 z + n

... + a n z |=|a n

|z | – |a n –1 z

n–1

0+

n

a 1 z + ... + z | ≥

| –.....– |a 0 |

n

≥ R – |a n–1 |R

n–1

n

= R [1 – |a n–1 |R A

– .... – |a 0 | –1

–.....– |a 0 |R

–n

]

∞ as R A ∞

Hence there exist R 0 such that |P(z)| >

if |z| > R 0 .

Since |P| is continuous on the closed disc with centre at 0 and radius R 0 , |P(z 0 )| = Claim:

for some z 0 . = 0.

Suppose

≠ 0.

Put Q(z) =

|

P(z+z0) P(z0)

|

Since P(z0) = L

P(z+z0) p(z0)

.then Q(0) =

P(z0) P(z0)

= 1.

=inf| P(z)|, P(z+z0) ≥ P(z0)

≥ 1.(i.e) Q(z) ≥ 1 for all z.

L

there is a smallest integer k, 1 ≤ k k

≤ n, such that Q(z) = 1 + b k z + a 1 z n

+.... + b n z , b k ≠ 0. By theorem 3.3.1(d), there is a real such that e

ikθ

b k = – |b k |. k ikθ

If r > 0 and |b k | < 1, |1 + b k r e

|

k

= 1 – r |b k | so that |Q(re

iθ

k

)| ≤ 1 – r {|b k | – r|b k+1 | –

.... – r

n–k

For

sufficiently

|b n |}

expression

in

small

braces

is

r,

the

positive.

Hence |Q(re

iθ

)| < 1, which is a contradiction

to Q(z) ≥ 1 for all z.

= 0, that is P(z 0 ) = 0.

L

SECION-3.5

FO!RIER SERIES

D-.161<176:

A trigonometric polynomial is a finite sum

of

the

form

N

f( x ) = a0 +

(a ∑ n=1

ncos

nx+bn sin nx)(x real)

where a 0 ,a 1 ,.. .,a N ,b 0 ,b 1 ,.. .,b N are complex C (x ) =

1 2

numbers,Since 1

[E(ix) + E(-ix)] and S(x) = 2! [E(ix) − E(-ix)], f(x)

can also be written in the form N

f (x ) =

cne ∑ −N

Note

inx

(x real)

1:

polynomial 2

Every is periodic

trigonometric with

period

Note2: If n is a nonzero integer,

e

inx

which is also has period 2 π 1 2π

inx

is the derivative of e

1

inx

∫e

dx =

-π

{

/in,

Hence

if n=0

0 if n=±1,±2,... N

If we multiply f

(x) = ∑ cne

inx

by e

imx

,

-N

where m is an integer, if we integrate the

π 1

cm= 2π

product,

∫ f( x ) e

we

get

-imx

dx for |m| ≤ N.

-π

If |m| > N, the above integral is ze

ro.

Note 3: The trigonometric polynomial N

f(x) = ∑ cne −N

= 0,1,...,N.

inx

is real iff c

-n

¯

= cn for n

Note 4: We define a trigonom etric series to be a series of the form ∞

∑ cn e

inx

(x real) the nth partial sum

−∞ N

of this series is defined to be

∑ cne

inx

−N

Note 5: If f is an integrable function on [–

,

], the numbers cm for all

integers m are called the Fourier coefficients

of

f,

and

the

series

∞

∑ cn e

inx

formed

with

these

the

Fourier

−∞

coefficients series of f.

is

called

D-.161<176: Let {

;n}

(n = 1,2,3,...) be a

sequ ence of complex fu nctions on [a,b], such that

¯

∞

∫

; n (x ); m (x )

dx=0( n≠m ) (n ≠ m).

−∞

Then {

;n

} is said to be an

orthogonal sy stem of functio ns on b

∫|

[a,b]. If, in addition

; n (x )

2

| dx = 1,

a

, for all n, {

;n

} is said to be

orthonormal.

E?)584-: The

fu nctions

(2π)

−

1 2

e

orthonormal system on [–

inx

,

form

an

].

Note: If { ; n } is orthonormal on [a,b]

and

if

¯

b

cn =

∫

f(t);n(t) dt (n=1,2,...) we call c n

a

the

nth

Fourier

relative to {

coefficients

of

f

; n }. ∞

We write f (x) N

∑ c ; (x) and call this n

n

1

series the Fourier series of f. #0-7-3.5.1:

Let{

;n}

be orthonormal on [a,b]. n

Let s n(x) = ∑ cm;m(x) be the nth m=1

partial sum of the Fourier series of n

f, and suppose t

(x) = ∑ γm;m(x)

n

m=1 b

.Then

∫|f-sn| a

b 2

dx ≤

∫|f-tn| a

2

dx

and

equality holds iff

=c m ,(m =

m

1,2,...,n). P77.:

Let ∫ denote the integral over [a,b], the sum from 1 to n.

∫

Then

¯

¯ ¯ γ ; =

∫∑

ft n =

f

m

m

∑

¯ ¯ γ f; = m

∫

m

¯∫

∑

¯ γ c

m m

(b# !he defini!in of cm). Now

2

∫|tn|

2

=

∫

¯

tntn =

b 2

∫|f-t | dx =∫(f-t )( n

n

a

b

∫ a

∫

¯

∫

∫

b

∫

∫

a

a

¯−

mγm

a

∫|f| dx-∑ |cm| a

m=1

b

∫(f¯f − ft¯ − t ¯f + t ¯t ) dx n

n

n n

a

∫t ¯t dx n n

2

b 2

∫|t | dx n

a

∑ c¯γ − ∑ γ¯ γ m m

b 2

∑

a

b

∫|f| dx-∑ c 2

n

b

¯ tn fdx+

a

b

=

m=1

)

2 ¯ ¯ |f| dx- ftndx- tn fdx+

a

=

m=1

b

ftndx-

a

b

=

∑

¯ ¯ f -tn dx=

a

¯ f fdx-

∑

γm;m =

;n

b

b

n

γm;m

(since { } is orthonormal)

= ∑ |γm|

=

∫

n

+

∑

m

2

|γm − cm|

m

----------(1)

,


¯¯ ∑ γm;m n

γm;m

m=1

which

is

evidently

minimized

iff

γ m − cm b

L

b 2dx

∫|f-s | n

≤

a

∫|f-t | n

2

dx

a

b

Put in

m

= cm(1), we get

b 2

2

∫|f-t | dx=∫|f| dx-∑ |c n

a b

Since

m

|

2

a

b 2

2

∫|f-t | dx ≥ 0,∫|f| dx-∑ |c n

a

m

|

2

≥0

a

b

L

2

∫|f| dx ≥ ∑ |c

|

2

m

a n

b

2

∫|s (x)| dx =∑ |c n

2

m

|

m=1

≤

2

∫|f(x)| dx a

#0-7-3.5.2: (B-; ;-4' ; 16-9 =41 <@) If { ; n } is

orthonormal

on

∞

∞

b 2

2

[a,b] and if f(x) N ∑ cn; (x), then ∑ |cn| ≤ ∫|f(x)| dx n

n=1

n=1

.In particu lar lim cn = 0. n

A

∞


a

P77.:

From the above theorem, we have

∫

n

2

2

|sn(x)| dx = ∑ |cm| ≤ m=1

b

2

∫|f(x)| dx. a

∞

Letting n

A

∞,we get

b

∑ | c n|

2

≤

2

∫|f(x)| dx

n=1

a

Also lim cn = 0 .( ? ) n

A

∞

Trigonometric

series:

We

shall

consider functions f that have period 2

and that are Riem ann-integrable

on [ – is

,

given

] The Fourier series of f .

by

∞

f(x) =

π

∑ c ne

inx

where c n =

1 2π

∫f(x)e

− ∞

-inx

dx

-π

, the Nth partia l sum of the Fourier series

of

f

is

given


by

N

SN(x) = SN(f;x) =

∑c e

inx

n

-N π 1 L 2π

N

∫|SN(x)|

2

π 2

1 2π

dx =∑ |cn| ≤

-π

n=-N

∫|f(x)|

2

dx

-π

The Dirichlet kernel D N(x) is defined by DN(X) = ∑ e

inx

=

sin(N + 1/2 )x sin(x / 2)

n = −N π

Since c n =

1 2π

N

∫f(x)e

-inx

dx,SN(f;x) =

-π N

=∑ −N

π

∫ f ( t) e

∑ cn e

=

-in(x-t)

1

dt= 2π

-π

-N

N

-int

dte

inx

-π

π

∫ f ( t) ∑ e -π

π

1 2π

∑ ∫ f ( t) e

−N

π

1 2π

N inx

-in(x-t)

-N

1

dt= 2π

∫ f ( t) D

(x-t)dt

N

-π

π 1 = 2π

∫f(x-t)D

(t)dt

N

-π

#0-7- 3.5.3:

If, for some x, there are constants >

0

and

M

|f(x+t) − f(x)| ≤ M|t| for all t

< F

∞such

that

(-δ,δ), then lim SN(f;x) = f(x) N

A

∞


P77.: Define g (t) =

f(x-t) − f(x)

for 0< |t| <

sin (t/2)

π

Now

1 2π

π

∫

1 DN(x)dx= 2π

N inx

∫∑ e

-π

-

, and put g (0) = 0

dx

n=-N

(by the definition of Dirichlet kernel ) N 1 2π

=

π

N

∑ ∫e

inx

dx=∑ e

n=-N -

(

1 2π

M

dx

n=-N

π

=1

inx

inx

∫e

{

1

dx=

-π

if n=0

0 n=±1, ± 2, .. .

)

π L

SN(f;x) − f(x) =

1 2π

π

∫f(x-t)D

(t)dt-f(x) =

N

-π

1 2π

∫

∫D

∫

π

1 (f(x-t)-f(x))DN(t)dt= 2π

∫g(t)sin(t/2)D

-π

=

(t)dt

N

-π

π 1 2π

(t)dt

N

-π

π

=

(t)dt-f(x).1

N

π 1

f(x-t)DN(t)dt-f(x). 2π

−π

1 2π

∫f(x-t)D -π

π

=

1 2π

π

∫g(t) sin(t/2)

sin(N+1/2)t sin(t/2)

1 dt= 2π

-π

∫g(t) sin(N+1/2)dt -π

π 1

= 2π

∫g(t) [sin Nt cos t/2+cos Nt sin t/2

]dt

-π π 1 = 2π

π

∫[g(t)cos t/2 ]sin Nt dt+ ∫[g(t)sin t/2 ]cos Nt dt 1 2π

-π

-π

By |f(x+t) – f(x)| ≤ M|t| for all t F (–

,

) and the definition of g(t),

g(t)cos(t/2)

and

g(t)sin(t/2)

bounded. Double click this page to view clearly

are

The last two integrals tend to 0 as N A

L

∞ (by theorem 3.5.2)

lim SN(f;x) = f(x)

N

A

∞

#0-7- 3.5.4:

If f is continuous (with period 2 if

) and

> 0, then there is a trigonometric

polynomial P such that |P(x) – f(x)| < for all real x. P77:

If we identify x and x + 2 regard the 2

, we may

-periodic functions on

R 1 as functions on the unit circle T, by means of the mapping x

A

ix

e . The

trigonometric polynomials , i.e., the N

functions of the form f

(x) = ∑ cne -N

inx

,

form a self-adjoint algebra ? , which separates point on T, and which vanishes at no point of T. Since T is compact, ? is dense in ?(T).

#0-7- 3.5.5: (P  ;-> 4 '; and

g

<0 -7 - )

are

functions

with

period

inx

, g(x) N

∑ γne

−∞

and

−∞

1 2π

Then lim ∞

∫|f(x)-S

2

N

(f;x)|

1 dx=0, 2π

-π

π 1 2π

,

inx

π

A

2

∞

∑ cne

N

f

Riemann-integrable

∞

f(x) N

Suppose

¯ ¯ ∫f(x) g(x) dx=∑ c γ , π

∞

-π

−∞

n n

∞ 2

∫|f(x)| dx=∑ |c |

2

n

-π

−∞

P77:

Let

us 2

{

use

the

π 1 2π

∫ | h (x )| -π

2

dx

}

notation

1 2


Let

> 0 be given.

Since f F> and f( ) = f(– ), there exists a continuous 2 -periodic function h with ||f – h||

2

< ε.

By the above theorem, there is a trigonometric polynomial P such that |h(x) – P(x)| <

< h-P <

=

{

2

2

=

}

ε 2π 2π

1 2

{

for all x. 1 2

π 1 2π

2

∫|h(x)-P(x)| dx -π

} { <

1 2

π 1 2π

2

∫ε dx -π

π

} {∫} 2

=

ε 2π

dx

-π

=ε

If P has degree N 0 , theorem 3.5.1 shows that ||h – S n (h)|| 2 < ||h – P|| 2 <

, for all N ≥ N 0 .


1 2

< S N(h) − SN(f) < 2 = < SN(h-f)2 < 2 ≤ < h-f < 2 < ε.(by theorem 3.5.1 )

Now < f-SN(f) < 2 = < f-h+h-SN(h) + SN(h) − SN(f) < 2 ≤ < f-h < 2 + < h-SN(h) < 2 + < S N(h) − SN(f) < 2 <

+

+

=3

L lim < f-SN(f) < 2 = 0 N

A

∞ 1 2

π

(i.e) lim N

A

{ {∫ 1 2π

∞

} }

2

∫|f(x)-S (f;x)| dx N

-π π

1 2π

(i.e) lim N

A

∞

|f(x)-SN(f;x)|

¯ ∑ c ∫e ¯ g(x) dx

π

∫

¯ 1 SN(f)gdx= 2π

∫

-π N

dx = 0

-π

π 1

Next, 2π

2

=0

N

-π

π

1 n 2π

inx

cne g(x)dx=

-N

inx

-π

¯

=∑ cnγn ---------------(1) n=-N

By Schwarz inequality,we have

|∫

∫

(i.e.)

|∫ ∫

¯ fg −

|

¯ SN(f)g ≤

∫

|f-SN(f)|

¯ fg −

¯ g ≤

||

{

¯ S N (f ) g

∫|f-SN(f)|

|

A

2

¯ g

2

| |}

1 2

A

0 as N

0 as N

A

π

L lim N

A

∫S (f)¯g = ∫f¯g N

∞ -π

π

(i.e.) lim N

A

∞

1 2π

π

∫S (f)¯g = ∫f¯g N

-π

1 2π

-π


A

∞

∞

N

By(1),

lim

N

A

π

π

¯ 1 ¯ ∑ cnγn = 2π ∫fg

∞ n=-N

-π ∞

¯ ¯ cnγn (i.e.) 2π -π fg = ∑ n=-∞ 1

∫

¯ f x γ ( ) ∫ g(x) dx = ∑ c ¯ π

1

(i.e.) 2π

∞

n n

-π

-∞

Put f = g in the above equation, we get 1 2π

π

∞

∫|f(x) | dx = ∑ |c | 2

2

n

-π

n=-∞

C'P $E"#ION":

1.

Supp ose 0 < ≤

, f(x) = 0 if

f(x+2

<

, f(x) = 1 if |x| < |x| ≤

, and

) = f(x) for all x.

a.Compute

the

Fourier

coefficients of f. ∞

2

sin (nδ) π-δ b. Conclude that ∑ = 2 n −∞


SECION 3.6

HE GAMMA

F!NCION

D-<7 ∞

For 0
x-1 -t

( x )∫ t

e dt

0

The integral converges for x



(0, ∞)

#0-7-5 3.6.1: a.

The functional equation x

(x) holds if 0 < x <

∞

(n+1) = n! for n = 1,2,....

b. c.

(x+1) =

log

is convex on (0, ∞).

P77: a.

Let 0 < x < ∞. ∞

Γ(x) =

∫t

x-1 -t

e dt

0 ∞

Γ(x+1) =

∫t

∞ x+1-1 -t

e dt=

∞ x -t

∫t e

0

dt=

0

=

∞ x-1 -t

( − t e )0 + ∫xt 0

-t

0

∞ x -t ∞

x

∫t d(e )

∫t

e dt=0+x

x-1 -t

e dt=xΓ(x)

0


By (a)Γ(n+1) = nΓ(n) = n(n-1)Γ(n-1) =n(n-1)(n-2)....(1)Γ(1) ∞

b.

∞

∫t

Γ(1) =

∫t e

1-1 -t

Let 1
0

1 1 p+ q=

1.

(

x p

∞ x

y

x

y

1

1

1

1

+ − 1 -t + − + −t + + ) = ∫t( p q ) e dt =∫t( p q ) ( p q )e ( p q )dt y q

0

0

∞

∞ x

t( p

−

1 p

y

1

t

t

x-1

t

y-1

t

) − ( q + q )e − p − q dt= t( p )e − p t( q )e − q dt

∫

∫

0

0

1/p

∞

{∫

t

x-1 -t

e dt

0

c.

dt=1

0

∞

≤

-t

dt=

Γ(n+1) = n(n+1)(n-2).....1 = n!

L

=

∫e

0 -t

e dt=

0

Now Γ

∞

∞

} {∫ .

t

y-1 -t

e dt

0

1/p

= {Γ(x)}

{ Γ (x )}

1/q

}

(by Holder's inequality)

1/q

Taking log on both sides,we get log Γ

1/p

1

1

=p .Then 1-

1/p

1

1

1

=1=q p

L

logΓ(λx+(1 − λ)y) ≤

L

log Γ is convex on(0, ∞)

log

1/q

+ log{Γ(x)}

(x) + qlog (y)

= plog Put

1/q

x y ( p + q ) ≤ log[{Γ(x)} {Γ(y)} ] = log{Γ(x)}

(x) + (1 − λ)log Γ(y)

#0-7-5 3.6.2:

If f is a positive function on (0, ∞) such that Double click this page to view clearly

a.

f(x+1) = xf(x)

b.

f(1)= 1

c.

log f is convex,

then f(x) =

(x).

P77.:

Since

satisfies (a), (b) and (c) , it is

enough to prove that f(x) is uniquely determined by (a), (b) and (c), for all x > 0. By (a), it is enough to do this for x F (0,1).

Put

=logf. Then f(x+1) = xf(x)

log f(x+1) = log x + log f(x) φ(x+1) = log x+ Since f(1) = 1, By (c) ,

(x). ----------(1) (1) = log f(1) =

is convex .

0.

B

Suppose 0 < x < 1, and n is a positive integer. Then f(n+1) = nf(n) = n(n – 1 )f(n – 1) = n! L

φ( n+1) = log f( n+1) = logn! -----( 2)

Consider the difference quotients of on

the

intervals

[n,n+1],[n+1,n+1+x], [n+1, n+2]. φ(n+1) − φ(n) n+1-n

log n ≤

≤

φ(n+1+x) − φ(n+1) n+1+x-n-1

φ(n+1+x) − φ(n+1) ≤ n+1+x-n-1

From(1), we have

≤

φ(n+2) − φ(n+1) n+2-n-1

log (n+1)(by(1)) ----------(3)

(n+x+1) = log (n+x) + φ(n+x)

=log(n+x) + log(n+x-1) + φ(n+x-1) =log(n+x) + (n+x-1) + φ(n+x-1) =....=φ(x) + log[x(x+1)...(x+n)]

(3) B logn ≤ B x log n ≤

φ(x) + log[x(x+1)...(x+n)] − log n! x

≤ log

(n+1)(by(2))

(x) + log[x(x+1)...(x+n)] − log n! ≤ x log (n+1)

Subtract xlog n,we get 0 ≤ φ(x) + log[x(x+1)...(x+n)] − log n!-x log n ≤ x log 0 ≤ φ(x) + log[x(x+1)...(x+n)] − log n!-x log n 0 ≤ φ(x) − log

[(

x

n!n x x+1)...(x+n)

(n+1) − x log n

x

≤ x ( log(n+1) − x log n

1 n

)

] ≤ x log ( ) n+1 n

(

=x log 1 +

A

0 as n

A

∞


)

(i.e.) φ(x) − log

[

x

n!n x(x+1)...(x+n)

]

0 as n

A

A

∞

x

(i.e.) φ(x) =

lim n

(i.e.)

A

n!n x(x+1)...(x+n)

log

∞

[

]

(x) is determined uniquely

(i.e.) log f(x) is

determined uniquely

B

f(x) is detennined uniquely

B

f(x) =

(x).

#0-7-5 3.6.3:

If

x

>

0

(1 − t )

y-1

and

1

∫

t

x-1

dt=

0

(This

integral

y

>

0,

then

called

beta

Γ (x )Γ (y ) Γ(x+y) is

so

function B(x,y)) P77:

By the definition of gamma function, we have

∞

Γ(x) =

∫t

x-1

y-1

(1-t)

dt

0 1

Now B (",

#)

=

∫t

"

−1

y-1

(1-t)

dt

0 1

Therefore B (1,y) =

1

∫

t

1− 1

y-1

(1-t)

y−1

∫(1-t)

dt=

0

(

dt= −

0

(1-t) y

y

1

)

=

0

1 y

Let p,q be real numbers such that 1 p

+

1 q

=1 1 x p

B(x,y) = B

+

x q,

x

y =

) ∫0

(

1

x

y-1 + −1 t( p q ) (1-t) dt

1

1

1

1

1

1

1

1

y + − + + − + =∫t( p q ) ( p q )(1-t) ( p q ) ( p q )dt x

x

1

1

0 1 y + − + + − + =∫t( p q ) ( p q )(1-t) ( p q ) ( p q )dt x

x

1

1

0 1

y

1

y

1

− − − − + =∫t( p p ) ( q q )(1-t)( p p )(1 − t)( q p )dt x

1

x

1

0 1 y-1

y-1

=∫t( p )t( q )(1-t)( p )(1 − t)( q )dt x-1

x-1

0


1

≤

{∫ (

t(

x-1 p

y-1

1/p

1

} {∫(

p

) (1 − t)( p ) dt

)

0

t(

x-1 p

y-1

}

q

1/q

) (1 − t)( p ) dt

0

)

(by Holder's inequality) 1/p

1

=

{

∫

t

x-1

( 1 − t)

dt

0 x p

(i.e.)B( +

x q,

)

y ≤ B(x,y)

1/q

1

} {

y-1

∫

t

x-1

( 1 − t)

y-1

dt

0

1/p

.B(x,y)

}

1/p

= B(x,y)

, B(x,y)

1/q

1/q

Taking log on both sides , we get log B

x

x

1

1

( p + q , y) ≤ p log B (x,y) + q log B (x,y) 1

1 y-1

x+1-1

Now B(x+1,y) = 1

=

t

∫ 0

(1-t)

t

∫ (1 − t )

x

x (1-t) (1 − t)

y-1

dt=

0

x

t

t

1

dv=

t

x

0

∫( 1 − t ) (1-t)

dt x+y-1

dt

0

( 1 − t ) , then du=x( 1 − t )

Let u=

∫

(1-t)

t

1

x

y-1

x

dt =

∫(1 − t)

x+y-1

dt,v=-

(1 − t )

x-1

(

1

(1 − t)

2

)dt and

x+y

x+y

0

L

(

B(x+1,y) =

−

(

x (1-t)x+y 1 t 1 −t x+y 0

)

)

1

+

∫

(1 − t)

=0+

x+y

0

1 x x+y

x+y

x

t 1-t

( )

x-1

1

( 1 − t)

1 x+y

t x-1 1 x 2 dt= x+y 1-t (1-t)

∫(1-t) ( ) 0

∫(1-t)

x+y-x+1-2 x-1

t

dt

0

1 x = x+y

∫(1-t)

y-1 x-1

t

dt =

x x+y B

(x,y)

0

(i.e.)B(x+1,y) = Let f(x) =

Γ (x+y) Γ(y)

x x+y B

(x,y). ------------ (1)

.B(x,y).


2 dt

Then f(x+1) = = =

Γ(x+1+y) Γ(y)

Γ(x+1+y) Γ(y)

.B(x+1,y)

x

. x+y B(x,y). ( by(1))

xΓ(x+y) (x+y)Γ(x+y) x . x+y B(x,y) = Γ(y) .B(x,y) = xf(x) Γ(y)

f(1) =

Γ(y+1)

.B(1,y). = Γ(y)

Γ(1+y) 1 Γ(y) y

=

yΓ(y) 1 Γ(y) y

=1

log f(x) = log (x+y) + log B (x,y) − log (y) Then f(x+1) = Γ(x+1+y) Γ(y)

= =

Γ(x+1+y) Γ(y)

.B(x+1,y)

x

. x+y B(x,y). ( by(1))

xΓ(x+y) (x+y)Γ(x+y) x . x+y B(x,y) = Γ(y) .B(x,y) = xf(x) Γ(y)

f(1) =

Γ(y+1)

.B(1,y). = Γ(y)

Γ(1+y) 1 Γ(y) y

=

yΓ(y) 1 Γ(y) y

=1

log f(x) = log (x+y) + log B (x,y) − log (y)

log f(x) = log log

(x+y) + log B(x,y) –

(y).

Since log

and log B

are convex, log

f is also convex.


By the above theorem f(x) =

(i.e.)

Γ(x+y) Γ( y )

(x).

.B(x,y) = Γ(x).

)Γ(y)) B(x,y) = ΓΓ((xx+y

L

1

(i.e.) ∫tx-1(1 − t)

y-1

Γ(x)Γ(y) dt= Γ(x+y)

0

"75-

The

C76;-9=-6+-; 2

substitution t = sin equation

in the above

turns

into

π/2

2

x-1

2

2

∫(sin θ) (1 − sin θ)

y-1

sin

cos

d

Γ(x)Γ(y)

= Γ(x+y)

0 π/2

(i.e.)2∫(sin2θ)

x-1

2

(1 − sin θ)

y-1

sin

cos

Γ(x)Γ(y)

d

= Γ(x+y)

0

π/2

Γ(1 / 2)Γ(1 / 2)

=2

Γ(1/2+1/2)

2 1 / 2) − 1

∫(sin θ) (

2(1 / 2) − 1

(cos )

0

(i.e.)

(

)

Γ (1 / 2) Γ(1)

π/2

2

=2

∫dθ=2(π/2)=π 0

Since

2

(1) = 1, we have(Γ(1 / 2))

Therefore

=π

(1 / 2) = √π.


dθ

∞ 2

Put t=s in Γ(x) =

∫t

x-1 -t

e dt,we get

0 ∞

Γ(x) =

∞ 2x-2

∫s

−s

e

2

2sds =2

∫s

0

2x-1

e

−s

2

ds

0

Put x = ½, we get ∞

√

∞

∫

=

(1 /2)=

2(1/2) − 1

s

e

− s

2

∫e

ds=

0 x−1

Also

(x) = 2 √ π

Γ

− s

2

ds.

0 1

( 2 ) Γ(

x+1 2

) (Verify).

C'P $E"#ION": 1. Prove

lim x

A

∞

the

Stirling's

Γ( x + 1 ) (x/e)

x

√2πx

formula

= 1.

(x) =

2

x−1

() ( Γ

x+1 2

Prove that

3.

If f(x) = 0 for all x in some

√π

Γ

1 2

2.

)

segment J, then prove that lim S N (f:x)

=

0

for

every

x

J.(Localization theorem) 4.

Prove t hat

lim x

every

A

x

−α

logx=0 for

+∞

>0.


F

$NI#-4

$61< "<:=+<=:

Section

4.1:

Linear

Transformations Section 4.2: Differentiation Section

4.3:

The

contraction

principle. Section 4.4: The inverse function Theorem

I6<:7,=+<176

In this unit we shall discuss about the set of vectors in Euclidean nspace

n

R ,

linear

transformation,

diff erentiation and the contraction

principle

and

finally

the

inverse

function theorem.

SECION-4.1

: LINEAR

RANSFORMAIONS

D-.161<176:

A non-empty set X space if x + y every x,y







R

n

is a vector

X and cx



X, for

X and for all scalar c.

D-.161<176:

If x 1 ,x 2 ,...,x k  R

n

and c 1 ,c 2 ,...,c k are

scalars then x 1 c 1 + x 2 c 2 +... + x k c k is called a linear combination of x ,x 2 ,... ,x k .

1

D-.161<176:

Let S

O

R

n

.The n the set of all linea r

combination of elements of S is called a linear span of S and it is denoted by L(S). N7<-:

L(S) is a vector space. D-.161<176:

Linear independent: A set consist of vector s x 1 ,x 2 ,.. .,x k is said to be linearly independent if x + x 2 c 2 +... + x k c k = 0 c k = 0 .Otherwise {x dependent.

B

1c1

c 1 = c 2 = ....

1 ,x 2 ,...,x k }

are

D-.161<176: If a vector space X contains an independent set of r vectors but contains no independent set of r+1 vectors , then X has dimension r and we write dim X = r.

D-.161<176: An independent subset of a vector space X which spans the space X is called a basis of X.

N7<-(1): Let R n be the set of all ordered ntuples

x

=

(x 1 ,x 2 ,...,x n )

where

x 1 ,x 2 ,...,x n are real numbers and the element of R

n

called points or vectors

. Define x + y = (x

1

+y 1 ,x 2 +y 2 ,...

,x n +y n ) where y = (y 1 ,y 2 ,...,y n ) and x=( αx

F

x1 ,αx 2 ,...,

x n ). Then x + y,

n

R . n

(ie) R

is closed under addition and

scalar multiplication.

L

R

n

is a vector space over the field

R.

N7<-(11): Let

e1

=

(1,0,0,...,0),

e 2

=

(0,1,0,...,0), .... e n = (0,0,0,...,1). If x

F

n

R , x = (x 1 ,x 2 ,...,x n ) , then x =

Σ x j e j . We shall call {e 1 ,e 2 ,...,e n } the standard basis of R

n

.

N7<-(111): If a subset S = {v

1 ,v 2 ,...,v k }

of X

contains the zero vector , then the set S is linear dependent. In particular if v 1 = 0, then 1 v 1 +0v 2 +.. .+0v k = 0 L

B 1≠

0.

The set S is linearl y dependent set.

N7<-(1>): Let S = {v 1 ,v 2 ,...,v k } be linearly dependent iff there exist a vector in S which is a linear combin ation of remaining vectors in S.

N7<-(>): The set {v} consist of single vectors is linearly independent iff the vector v ≠ 0.

N7<-(>): If the set S is linearly independent then any non-empty subset of S is also linearly independent.

D-6<76: If V is a vector space over a field F and if W O V then W is a subspace of V if W itself is a vector space over F.

#0-7-5 4.1.1: Let r be a positive integer. If a vector space X is spanned by a set of r vectors, then dim X ≤ r.

P77.:

Suppose dim X > r.

Let dim X = r + 1. Therefore there is a vector space X which co ntains an independent set Q = {y

1 ,y 2 ,...,y r+1 }

and which is

spanned by a set S 0 consisting r vectors {x 1 ,x 2 ,...,x r }. Suppo se 0 ≤ i < r and suppose a set S i has been constru cted which spans X and which consist of all y

j

with 1 ≤

j ≤ i plus a certain collection of r – i members of S

0

(In other words, S

say {x 1 ,x 2 ,...,x r–i } i

is obtained from

S 0 by replacing i of its elements by members of Q, without altering the

span.) Since S the span of S scalars

i i

spans X, y i+1 is in and hence there are

a 1 ,a 2 ,...,a i+1 ,b 1 ,b 2 ,...,b r–i

with a i+1 = 1, such that i+1

r−i

∑ ajyj + ∑ bkxk = 0. j=1

k=1

If all bk‘s were 0, the independence of Q would force all a

j ‘s

to be zero,

which is a contradiction. It follows that some x

kFSi

is a linea r

combination of the other members of S i K {y i+1 }. Let T i = S i K {y i+1 }. Remove this x k from T i and call the remaining set S i+1 .Then S i+1 spans

the same set as T

i,

namely X, (ie)

L(S i+1 ) = X, and S i has the properti es postulate d for S i with i + 1 in place of i. Starting with S i , we have constructed sets S 1 ,S 2 ,...,S r and S r

consist

of

y 1 ,y 2 ,...,y r and L(S r ) = X. But Q is linearly indep endent and hence y r+1 G L(S r ), which is a contradiction to dim X > r. L dim X ≤ r.

Cor: dim R

n

= n.

P77.:

Since {e 1 ,e 2 ,.. .,e n } spans R

n

, the

above theorem shows that dim R

n

≤

n.

Since

{e 1 ,e 2 ,..

independent, dim R

n

.,e n }

is

≥ n.

Therefore dim R n = n. #0-7-5 4.1.2:

Suppose X is a vector space and dim X = n. a.

A set E of n vectors in X spans X iff E is independent.

b.

X has a basis, and every basis consist of n vectors

c.

If 1 ≤ r ≤ n and {y 1 ,y 2 ,...,y r } is an independent set in X, then X

has

a

basis

containing

{y 1 ,y 2 ,...,y r }. P77:

a.

Suppose E = {x 1 ,x 2 ,...,x n }.

If E is independent then to prove that L(E) = X. Let y F X. Then

A

=

dependent. ( L

{x 1 ,x 2 ,...,x n ,y} M

is

dim X = n )

a vector in A which is a

E

linear combination of remaining vectors. Since E is independent, no

vector

in

combination

E

is

of

a

linear

preceding

vectors. L

y is a linear combination of

{x 1 ,x 2 ,...,x n }. L

y

F

L(E).

L

X

Q

L(E).

But L(E)

Q

X. Therefore X = L(E).

Conversely, let X = L(E).

Now

to

prove

that

E

is

independent. Suppose E is dependent. Then one of the elements, say, {x k } is a linear combination of preceding vectors. (i.e.) we can eliminate x k without changing the span of E. Hence E cannot span X, which is a contradiction to L(E) = X. Therefore E is independent. b.

Since dim X = n, X contains an independent set of n vectors. (i.e.)

E

=

{x 1 ,x 2 ,...,x n }

independent set in X. By (a), L(E) = X.

is

L

X has a basis consists of n

vectors. c.

Let {x 1 ,x 2 ,...,x n } be a basis of X. By hypothesis, {y 1 ,y 2 ,...,y r } is an independent set in X. L

he t

set

{y 1 ,y 2 ,...,y r ,x 1 ,x 2 ,...,x n }

spans

X and is dependent, since it contains more than n vectors. L

one of the X i ‘s is a linear

combination

of

the

other

members of S. If we remove this x

i

from S, the

remaining set still spans X. This can be repeated r times and leads to the basis of X which contains {y 1 ,y 2 ,...,y r }.

D-.161<176: A mapping A of a vector space X into a vector space Y is said to be 416 -) <) 6;.7 5) <176 if A(x 1 +x 2 ) = Ax 1 +Ax 2 , A(cx) = cA(x), for all x

1 ,x 2 ,x F X

and

all scalars c.

D-.161<176: Linear transformations of X into X is called linear operators on X. If A is a 416-) 78-)<7 on X which i. ii.

is one-to-one an d maps X onto X, we say that A is invertible.

–1

We define A x or A(A

–1

on X that, A

–1

(Ax) =

x) = x.

N7<-:

i.

If A is linear then A.0 = 0.

ii. If

the set of

n vectors say

{x 1 ,x 2 ,...,x n } is a basis of X then any element

x FX

representation

has of

a the

unique function

n

x = ∑ cixi and the linearity of A i=1

allows us to compute Ax from the vectors

Ax 1 ,Ax 2 ,

coordinates formula

...,Ax n

c 1 ,c 2 ,...

,c n

and

the

by

the

n

Ax = ∑ ciAxi i=1

#0-7-5 4.1.3:

A

linear

operator

A

on

a

finite-

dimensional vector space X is one-toone iff the range of A is all of X. P77:

Let {x 1 ,x 2 ,. . .,x n } be a basis of X. Let

{

= (A)

=

n

∑ ciAxi / xi

F

}

X =L

i=1

Now

=

(A)

{Ax/x F X}

=

=

({Ax1, Ax2, ....Axn})

L(Q)

where

Q

=

{Ax 1 ,Ax 2 ,...,Ax n }. Now to prove that A is one-to-one iff =

(A) = X.

(i.e.) to prove that A is one-to-one iff L(Q) = X. By theorem 4.1.2.(a), it is enough to prove that A is one-to-one iff Q is independent. Suppose A is one-to-one .

n

B

Let ∑ ciAxi = 0 i=1

A

(

n

∑ c ix i i=1

n

B

A

(

∑ c ix i i=1

)

=0

n

)

=A.0

B

∑ c ix i = 0 i=1

B c1 = c2 = .... cn = 0 (since {x1, x2, ...., xn} is independent)

L

Q is independent.

Conversely, let Q be independent and

(

A

n

∑ cixi i=1

)

=0

n

Then ∑ ciAxi = 0 i=1

B

c1 = c2 = ....cn = 0 (Q is independent)

Therefore Ax = 0 only if x = 0. Now, Ax = Ay

B

A(x – y) = 0 B x– y

= 0 B x = y.

Therefore A is one-to-one. D-.161<176:

Let L(X,Y) be the set of all linear transformations of the vector space X

into the vector space Y. If Y = X , then we write L(X,Y) = L(X).

D-.161<176:

If

A 1 ,A 2 F L(X,Y)

scalars,

define

and

if

c 1 ,c 2

are

c 1 A 1 +c 2 A 2

by

(c 1 A 1 +c 2 A 2 )x = c 1 A 1 x+c 2 A 2 x (x F X).

It is clear that c 1 A 1 +c 2 A 2 F L(X,Y). D-.161<176:

If X,Y,Z are vector spaces and if A,B F L(X,Y) we denote their product BA to be the composition of A and B : (BA)x = B(Ax) (x Then BA

F

L(X,Z).

F

X).

Note that BA need not be the same as AB, even if X = Y = Z

D-6<76:

For A

F

{|Ax|/x

n

m

n

with

L(R ,R ), define ||A|| = sup F

R

|x|≤1}

and

the

inequality |Ax| ≤ ||A|| |x| holds for all x

F

n

R . Also if

is such that |Ax| ≤ n

|x| for all x F R then ||A|| ≤

.

#0-7-5 4.1.4: a.

n

m

If A F L(R ,R ), then ||A|| < ∞ and A is a uniformly continuous mapping of R n into R m .

b.

If A,B

F

n

m

L(R ,R ) and c is a scalar

, then ||A+B|| ≤ ||A|| + ||B||, ||cA|| = |c| ||A||. With the

distance

between

A

and n

B

m

defined as ||A – B||, L(R ,R ) is a metric space. c.

n

m

n

m

If A F L(R ,R ) and B F L(R ,R ), then ||BA|| ≤ ||B|| ||A||.

P77.:

a.

Let

{e 1 ,e 2 ,...,e n }

standard basis in R x=

c

iei,

n

be

the

and suppose

|x| ≤ 1, so that |c i | ≤

1 for i = 1,2,...,n. Then |Ax| = |A( Ci

|| Ae i |≤

ci e i )|=|

ci Ae i |≤

| Aei | (since |c i | ≤

1). |Ax| ≤

|

| Ae i |< ∞.

Therefore Sup|Ax| < ∞. (i.e.) ||A|| < ∞.

Since |Ax – Ay| = |A(x – y)| ≤ ||A|| |x – y| where x,y Let

F

n

R .

> 0 be given.

Choose

Now | x − y | <

ε

= > 0. ||A|| B

| Ax − Ay | ≤

Therefore

A

ε

||A|| | x − y | < ||A|| ||A|| = ε

is

uniformly

continuous. b.

||A + B|| = sup{|(A+B)x|/x

F

R

n

with |x|≤1} =

sup{|(Ax+Bx)|/x F R

n

with

|x|≤1} ≤ su p {|Ax|+|B x|/x

F

R n with

|x|≤1} = sup{|Ax|/x sup{|Ax|/x

F

F

R

n

R

n

with |x|≤1} +

with |x|≤1}


= ||A|| + ||B||. Hence ||A + B|| ≤ ||A|| + ||B||. If c is a scalar, then ||cA|| = sup{|(cA)x|/x

FR

n

with |x|≤1} n

with

n

with

= sup{|c|| Ax|/x F R |x|≤1} = |c| sup{|Ax|/x F R |x|≤1} = |c| ||A|| n

m

Now to prove that L(R ,R ) is a metric space. Define d(A,B) = ||A – B||, then d(A,B) ≥ 0 d(A,B) = 0

C

||A – B|| = 0

C

A

= B. Also d(A,B) = ||A – B|| = ||B – A|| = d(B,A).

d(A,C) = ||A – C|| where C n

F

m

L(R ,R ) = ||A – B + B – C|| ≤ || A – B|| + ||B – C|| = d(A,B) + d(B,C) n

m

Therefore L(R ,R ) is a metric space. c.

F

Let A,B

L(R n ,R m ).

Now ||BA|| = sup{|(BA)x|/x R

n

F

with |x|≤1} = sup{|B(Ax)|/x

F

R

n

with

|x|≤ 1} ≤

|Ax|/x F R

n

sup{|Ax|/x F R

n

sup{||B||

with |x|≤ 1} =

||B||

with |x|≤ 1}

= ||B|| ||A||. #0-7- 4.1.5:

Let be the set of all invertible linear n operator on R .

a.

If

n

A F Ω,

A||.||A

–1

B F L(R ),

and

|| < 1, then B

||B

–

F Ω.

n b.

is an open subset of L(R ), –1 and the mapping A A A is continuous on

P77:

a.

Put ||A

–1

|| = 1/

and ||B – A||

= β. Then ||B – A||.||A < 1

B

Therefore

–1

< –

> 0,

|| < l

B

β/α

Now A

–1

–1

|x| =

(Ax)| ≤

| A

||A

= |Ax| (

L

–1

A(x)| =

|

|| |A(x)| –1

||A

|| = 1/

)

= |Ax – Bx + Bx| ≤ |Ax – Bx| + |Bx| = |(A–B)x| + |Bx| ≤ ||A–B|| |x| + |Bx| = ||B – A|||x| + |Bx|

=

= )

L

|x| –

|x| + |Bx|(

M

||B – A||

|x| ≤ |Bx| B (α –

)|x|≤ |Bx| |Bx| ≥ (

–

)|x| > 0 (M α – β

> 0) L

Bx ≠ 0.

Now Bx ≠ By

C

B(x–y) ≠0

x–y≠0

L

C

B is one-to-one.

Bx – By ≠ 0 C

C

x ≠ y.

By

theorem

4.1.3,

BF

.

This

holds for all B with ||B – A|| < and b.

n

is open subset of L(R

Replace x by B

–1

y in (

–

).

)|x|≤

|Bx|, we get (

–

= |y| (y

FR

–1

)| B y | ≤ |BB

n

| B –1 y |≤ |y|(

B

|| B

}≤(

≤

– A

–1

– A

1 α−β

As B

–1

–1

–1

A

(A – B)A

–1

= B

FR

n

A – B

–1

(A – B)A

–1

.

|| = ||B

–1

|| ||A – B|| ||A

β

A,

A

0.

–1

–1

–1

.α.

A

y|/y

–1

= B

|| ≤ || B

–1

) .

–1

B A ||B

–1

)–1 .

–

|| =sup{| B

–

Now B

y |

)

B

–1

–1

–1

||

–1

L

||B

L

AA A

– A –1

–1

|| as

A

0.

is continuous on

.

D-.161<176: M)<1+-;

Suppose

{x 1 ,x 2 ,...,x n }

and

{y 1 ,y 2 ,...,y m } are bases of vector spaces X and Y respectively. Th en every A

F

L(X,Y) determines a set of

numbers

a ij

uch s

that

m

Axj = ∑ aijyi

(1 ≤ j ≤ n )

i=1

It is co nvenient to visualize these numbers in a rectangular array of m rows and n columns, called an m by n matrix:

[A] =

[

a11

a12

S

a1n

a21

a22

S

a2n

S

S

S

S

S

amn

am1 am2

]

Observe that the co ordinates a ij of the vector Ax j appear in the jth column of [A]. The vectors Ax

are

j

called the column vectors of [A].

The

range of A is spanned by the column vectors of [A]. n

Ifx=

(

n

)

∑ cjxj thenAx=A ∑ cjxj j=1

m

=∑ i=1

[

n

]

∑ aijcj j=1

j=1

n

n

m

= ∑ cjAxj = ∑ cj∑ aijyj j=1

j=1

i=1

yi.

Next to prove that

| |A ||

≤

{∑ } aij2


1/2

Suppose

{x 1 ,x 2 ,...,x n }

and

{y 1 ,y 2 ,...,y m } are the standard bases of

R

n

and

m

R

def ini tio n Ax =

[ ]

∑ ∑a c

ij j

i

=

respectively.

∑ [a

By

yi

j

] yi

i1c1 + ai2c2 +....+a 1ncn

i

= [a11c1 + a12c2 +....+a

]y1 + [a21c1 + a22c2+....+a

1ncn

+ [am1c1 + am2c2 +...+a

=

(

n

) (

∑ a1jcj j=1

m 2

|Ax| = ∑ i=1

y1 +

[

n

∑ a2jcj j=1

n

2

]

∑ aijcj j=1

)

] yn.

mncn

y2 + .... +

m

≤

∑ i=1

] + y1....

2ncn

[

(

n

∑ anjcj

n

j=1

n

)

yn.

]

∑ aij2∑ cj2 (by Schwarz inequality) j=1

j=1

2

=

∑

aij2 |x|

i, j

Thus ||A|| ≤

{∑ }

1/2

aij2


C'P $E"#ION": 1.

If S is a non-empty subset of a vector space X, prove that the span of S is a vector space.

2.

Prove that BA is linear if A and B are linear transformatio ns. Prove also

that

A

–1

is

linear

and

invertible. 3.

Assume A  L(X,Y) and Ax = 0 only whe n x = 0. Prvoe that A is 1 – 1.

SECION-4.2

:

DIFFERENIAION

If f is a real valued function with domain (a,b)  R

1

and if x

 (a,b),

then f'(x) is usually defined to be the real number

f(x + h) − f(x) provi h

lim

h

A

ded that this

0

limit exists.

(i. e.) f'(x) = h

L

f(x + h) − f(x) h

lim

A

0

f(x+h) – f(x) = f’(x)h + r(h), where

the remainder r(h) is

small

in

lim

r(h) h

h

A

the

sense

that

= 0.

0

D-.161<176:

If f is a map from (a,b)

O

R

differentiable mapping and x then f'(x) is the

1

to R F

m

is

(a,b)

linear transformation

of R

1

to

A

m

that

f(x + h) − f(x) − f'(x)h h

lim

h

R

0

=0

satisfies

D-6<76:

Suppose E is an open set in R m

n

and

F

f maps E into R and x E if there n exist a linear transformation A of R into lim h

A

R

|f(x + h)

0

m

such

− f(x) − Ah|

= 0,

|h |

.....

that

( 1)

then f is differentiable at x and we write f'(x) = A. If f is differentiable at every x F E, we say that f is differentiable in E. #0-7-5 4.2.1:

Suppose E and f are as in the above definition , x

F

E, and (1) holds with A

= A 1 and with A = A 2 . Then A 1 = A 2 . P77:

Let B = A 1 – A 2 .

Now Bh = (A 1 – A 2 )h = A 1 h – A 2 h = f(x+h) – f(x) – f(x+h) + f(x) + A 1 h – A 2 h = [f(x+h) – f(x) – A 2 h] – [f(x+h) – f(x) – A 2 h] |Bh| ≤ |f(x+h) – f(x) – A 1 h| + |f(x+h) – f(x) – A 2 h| |B(h)| |h|

≤

|f(x + h)

− f(x) − A1h|

|h|

+

|f(x + h)

− f(x) − A2h|

|h|

A

as h

|B(h)| | h|

A

0ash

For fixed h

|B(th)| |th|

A

0ast

A

≠

0

A

0

0

0, it follows that A

0

.....(2)

The lineari ty of B shows that the

left

side of (2) is independent of t. n

Thus Bh = 0 for every h

F

R . Hence

B = 0. Therefore A 1 = A 2 .

(C016 =-) #0-7-5 4.2.2: Suppose E is an open set in R maps E into R

m

n

, f

, f is differentiable at

x 0 F E, g maps an open set containi

ng

k

f(E) into R , and g is differentiable at f(x 0 ). Then the mapping F of E into R

k

defined by F(x) = g(f(x)) is

differen tiable at x 0 , and F'(x 0 ) = g'(f(x 0 ))f’(x 0 ).

P77.:

Put y 0 = f(x 0 ), A = f’(x 0 ), B = g'(y 0 ), and define

.

u(h) = f(x 0 +h) – f(x 0 ) – Ah, v(k) = g(y 0 +k) – g(y 0 ) – Bk, for all hF R

n

and k F R

m

for which f(x 0 +h)

and g(y 0 +k) are defined. Since f is differentiable at x 0 , we have h

(i. e.)

L

A |u(h)|

lim

h

|f(x0 + h) − f(x0) − Ah|

lim

A

0

|h|

|h|

0

= 0. (i. e.)

(h) A 0 as h

ε(h) =0where

lim

h

A

A

=0

(h) =

0

| u(h)| | h|

0.

|||ly g is differentiable at x 0 , we have lim

k

A0

|g(y0 + k) − g(y0) − Bk| | k|

=0


(i. e.)

|v(k)|

lim

h

A

0

|k|

= 0. (i. e.) h

(k) A 0 as k

L

η(k) =0where

lim

A

A

( k) =

0

| v(k)| |k|

0.

Given h, put k = f(x 0 +h) – f(x 0 ). Then |k| = |f(x 0 +h) – f(x 0 )| = |Ah + u(h)| ≤ |Ah| + |u(h)| ≤ ||A|| |h| +

(h)|h|

(i.e.) |k| ≤ [||A|| + Now

F(x 0 +h)

–

(h)] |h|. F(x 0 )

–

BAh

= g(y 0 +k) – g(y 0 ) – Bah = Bk + v(k) – BAh = B(k – Ah) + v(k)

=

B(u(h)) + v(k). L

|F(x 0 +h) – F(x 0 ) – Bah| = |B(u(h))

+ v(k)| ≤ |B(u(h))| + |v(k)|


|F(x0 + h) − F(x0) − BAh | | h|

≤

||B|| ε(h)|h| + η(k)| [||A|| + ε(h)] |h|.

Since

(h)

as k

0, the RHS tends to

A

A

0 as h

A

0 and

(k)

A

0

Zero

(i. e.) h

L

|F(x0 + h) − F(x0) − BAh |

lim

A

|h|

0

F is differentiable at x

0

=0

and F'(x 0 )

= BA. (i.e.) F'(x 0 ) = g'(y 0 )f'(x 0 ) F'(x 0 ) = g'(f(x 0 )).f’(x 0 ). D-.161<176:

Consider a function f that maps an open

set

E

O

R

n

into

R

m

Let

{e 1 ,e 2 ,...,e n } and {u 1 ,u 2 ,...,u m } be


the standard bases of R

n

and R

respect ively. The components of f

m

are

real functions f 1 ,f 2 ,...,f m defined by m

f ( x ) = ∑ f i( x ) u i

(x

F

E)

or,

i=1

equivalently, by f i (x) = f(x).u i , 1≤i≤ m. For x

F

E, 1≤ i ≤ m, 1≤ j ≤ n, we

define

( D j f i( x ) ) =

lim

t

A0

fi(x+te

)

j

− fi(x)

t

provided

the limit exists, where D j f i is the deriva tive of f i with respect to X j , keeping the other variables fixed, we denote

∂ fi ∂ xj

in place of D j f i , and D j f i is

called the partial derivative.

#0-7-5 4.2.3:

Suppose f maps an open set E R

n

into R

m

O

,and f is differentiable

at a point x

F

E. Then the partial

derivatives (D j f i )(x) and

exist,

m

f'(x)ej = ∑ (Djfi)(x)ui

(1≤j≤n

)

i=1

P77:

Fix j. Since f is differentiable at x, f(x + te j ) – f(x) = f'(x)(te j ) + r(te j ) where |r(te j )|/t The lim

t

linearity

A0

f(x+te

)

j

t

A

of − f(x)

0 as t f’(x)

A

0.

shows

= f'(x)ej.

that

.....(1)

If we represent f interms of its components, then (1) becomes m

fi(x+te j) ∑ t tA0 lim

i

− f i (x )

ui = f'(x)ej.

=1

m

(i. e.) ∑ i=1

lim

t

A0

fi(x+te

)

j

− f i (x )

t

ui = f'(x)ej.

m

∑ (Djfi)(x)ui = f'(x)ej i=1

Therefore the partial deriva tives of (D j f i )(x) exist, and m

f'(x)ej = Note

:

( D f ) (x )u ( 1 ≤ j ≤ n ∑ i=1 j i

By

the

i

above

).

theorem,

m

f'(x)ej = ∑ (Djfi)(x)ui i=1

(1 ≤ j ≤ n ).

m L

m

f'(x)e1 = ∑ (D1fi)(x)ui, f'(x)e2 = ∑ (D2fi)(x)ui, ...., i=1

i=1

m

f'(x)ej =

Let

D f )(x)u . i=1 ( ∑ j i

[f’(x)]

i

be

the

matrix

that

represen t f’(x) with respect to the standard bases. Then

[f'(x)] =

[

(D1f1)(x)

S

( D n f 1 ) (x )

(D1f2)(x)

S

( D n f 2 ) (x )

S

S

S

( D 1 f m ) (x )

S

(Dnfm)(x)

#0-7-5 4.2.4:

]

Suppose f maps a convex open set EO R

n

into R

m

, f is differentiable in

E, and there is a real number M such that ||f'(x)|| ≤ M for every x

F

E.

Then |f(b) – f(a)| ≤ M|b – a| for all a E, b F E.

F

P77.:

Fix a F E and b Define tF R

1

F

E.

(t) = (1 – t)a + tb, for every

such that

(t)

F

E.

Since E is convex, 0 ≤ t ≤1 implies (t)F E. Put g(t) = f( g'(t) = f'(

(t)). (t)).

'(t) = f'(

(t)) (b –

a). |g'(t)| = | f’( f’(

(t)) (b – a) | ≤ ||

(t))|| |b – a| ≤ M |b – a|.

When t = 0, g(0) = f(

(0)) = f(a).

When t = 1, g(1) = f(

(1)) = f(b).

By mean value theorem for vectorvalued functions, we have |g(1) –g(0)|≤(1–0)g'(t)

(i.e.) | f(b) – f(a) | ≤ M (b

– a)

Cor: If, in addition, f’(x) = 0 then f is constant. P77.:

To

prove

this,

note

that

the

hypothesis of the theorem holds with M = 0.

A differentiable mapping f of an open set E

O

R

n

into R

m

is said to be

continuously differentiable in E, if f’is a

continuous n

mapping

of

E

into

m

L(R ,R ). (i.e.) for every x

F

0, corresponds a

E and to every

>

> 0 such that

|| f’(y) – f’(x)|| <

if y

F

E and |x –

y| < If this so, we also say that f is a ? mapping , or that f

F

? (E).

#0-7- 4.2.5:

Suppose f maps an open set E into R

m

. Then f

derivatives

F

D jfi

O

R

? ‘(E) iff the partial exist

and

are

n

con tin uou s on E for 1 ≤ i ≤ m, 1 ≤ j ≤ n. P77.:

Assume frist that f

F

? ‘( E ) .

By theorem 4.2.3, we have f’(x)e

j .u i

= (D j f i )(x)for all i and j and for all x F E. Now (D j f i (y) – (D j f i (x) = f’(y)e j .u i – f’(x)e j .u i =

(f'(y)

–

f’(x))e j .u i . Since

{e 1 ,e 2 .....e n }

{u 1 ,u 2 ,...,u m } bases of R

n

and

are

the

and standard

R

m

L

respectively, | u i | = | e j | = 1.

| (D j f i (y) – (D j f i )(x)| = |(f'(y) –

f'(x))e j .u i | = |(f'(y) – f’(x))e j | ≤ ||f’(y) – f’(x)|| |e

j|

= ||f’(y) – f’(x)|| L

D j f i is continuous.

Conversely let D j f i be continuous. Consider the case m = 1. (i.e.) f = f 1 . Fix x F E and

> 0.

Since E is an open set, there is an open ball S O E with center at x and radius r.

Since D j f i is continuous, r can be so chosen that ε

|(Djfi)(y) − (Djfi)(x)| < n

.....(1)

(y

F

S, 1 ≤ j ≤ n

)

Suppose n

h = ∑ hjej with |h|
0 =0,and

j=1

v k = h 1 e 1 +h 2 e 2 + .... + h k e k for 1 ≤ k ≤ n. Then f(x+h) – f(x) = f(x + v n ) – f(x + v0) = f(x + v 1 ) – f(x + v 0 ) + f(x + v 2 ) – f(x + v 1 )

+

f(x + v 3 ) – f(x + v 2 )

+......+ f(x + v n ) – f(x + v n–1 )

n

[

= ∑ f( x + v j ) − f(x + v j − 1 )

]

.....(2)

j=1

Since | v k | < r for 1 ≤ k ≤ n a nd since S is convex, the segments with end points f(x + v

j–1 )

and f(x + v

j)

lie in S. Since v j = h 1 e 1 +h 2 e 2 + .... + h j e j and v j–1 = h 1 e 1 +h 2 e 2 + .... + h j–1 e j–1 we have v j – v j–1 = h j e j

B

v j = v j–1 +

hjej, Apply Mean value theorem to the jth summand in (2), f(x + v j ) - f(x + v j–1 ) = [(x + v j ) – (x + v j–1 )] (f’(x + v j–1 +θ j h j e j ))

wher e 0 <

j < 1 and this diff ers from

| h|

h j D j f(x) by less than

F

n

, using (1)

f(x + v j ) – f(x + v j-1 ) = [v j – v j–1 ] D j f(x + v j–1 +θ j h j e j ) =

h j D j f(x +

v j–1 +θ j h j e j )

Substitute in (2), we get n

[

f(x + h) − f(x) = ∑ hjDj f(x + v j − 1 + θjhjej)

]

j=1 n

f(x + h) − f(x) −

∑ hj(Djf)(x) j=1

n

n

[

]

= ∑ hjDj f(x + v j − 1 + θjhjej) − j=1

|

j=1

n

f(x + h) − f(x) −

∑ hj(Djf)(x)

|

∑ hj(Djf)(x) j=1


=

=

| |

n

n

∑ h jD j[ f (x + v j − 1 + θ jh je j)]

−

j=1

|

∑ hj(Djf)(x) j=1

n

∑ j=1

[[

[

]]

]

hj Dj f(x + v j − 1 + θjhjej) − Dj[f(x)]

|

n

∑ |h j| | D j[ f (x + v j − 1 + θ jh je j) ]

≤

|

− Dj[f(x)]

j=1 n

<

ε j n h ∑| |

(by (1))

j=1

≤ |h| ε

L

L

|

n

f(x + h) − f(x) −

∑ hj(Djf)(x) j=1

|h|

<ε

f is differentiable at x. m

f'(x)ej = ∑ (Djfi)(x)ui

,

which

i=1

nothing but jth column vector


is

on [f'(x)], where [f'(x)] =

[

n

Ifh=

(D1f1)(x)

S

(Dnf1)(x)

(D1f2)(x)

S

(Dnf2)(x)

S

S

S

(D1fm)(x)

S

(Dnfm)(x)

n

n

j=1 n

j=1

j=1

m

= ∑ hj∑ (Djfi)(x)ui j=1

i=1 n

Since m = 1, f'(x)h = ∑ hj(Djf)(x) j=1

The matrix [f’(x)] consists of the row (D 1 f)(x).....(D n f)(x); and since D 1 f,.....,

(D n f)

functions on E,f

are

F

continuous

? ‘(E).

C'P $E"#ION": 1.

]

∑ hjej then f'(x)h=f' (x) ∑ hjej = ∑ f '(x)hjej

If f(0,0) = 0 and f(x, y) =

xy 2

x +y

2

if f(", #) ≠ (0, 0)


Prove

that

(D 1 f)(x,y)

and

(D 2 f)(x,y) exist at every point of R

2

, altho ugh f is not contin uous

at (0,0). 2.

Suppose that f is a real-valued function defined in an open set E



n

R , and that the partial

derivatives are bounded in E. Prove that f is continuous in E.

SECION-4.3

: HE

CONRACION PRINCIPLE.

D-.161<76:

Let X be a metric space, with metric d. If

maps X into X and if the re is a

rea l num ber c < 1 suc h tha t d( (y)) ≤ cd(x,y) for all x,y

F

(x) ,

X, then

is said to be a contraction of X into X.

#0-7-5 4.3.1: 5886/ <0-7-5)

(C76<+<76

If X is a complete metric space, and if is a contraction of X into X, then there exists one and only one x F X such that

(x) = x.

P77:

Let

x 0 F X.

Define

{x n }by

setting

φ(x n ) = x n+1 . First to prove that {x n } is a Cauchy sequence in X. Since

is a contraction, there exists

a number c such that 0 < c <1 such that d( x,y F X

(x),

(y)) < cd(x,y) for all

For n ≥ 1, we have d(x n+1 , x n ) = d( cd(

(x n ),

( x n–1 )) ≤ c d(x n , x n–1 )

(x n–1 ),

( x n–2 ))

=

≤

c.c

2

d(x n–1 ,x n–2 ) = c d(x n–1 ,x n–2 ) =.... n

d(x n+1 , x n ) ≤ c d(x 1 ,x 0 ) . if n < m, then it follows that d(x n , x m ) ≤ d(x n , x n+1 ) + d(x n+l , x n+2 ) +.....+ d(x m–1 , x m )

≤c n d(x 1 ,x 0 ) +.....+ c

m–1

+

c n+ 1d(x 1 ,x 0 )

d(x 1 ,x 0 ).

n

≤ c d[1 + c + c

2

+....+ c

m–n–1

]

d(x 1 ,x 0 ) < c n d[1 + c + c 2 +....] d(x 1 ,x 0 ) n

<

c 1−c

d(x1, x0)

Since c < 1, (c

n

)

A

0 as n

A

∞

Therefore, given

> 0, there exists a

positive integer n such that n

c 1−c

d(x1, x0) < ε

L

d(x n , X m ) < ε

L

{x n } is a Cauchy sequence in X.

Since X is complete metric space, {x n } A x. Since

is

contraction,

is

continuous. L

φ(x n )

A

(x).

(i. e.) φ(x) = lim φ(xn) = lim xn + 1 = x n

L

A

∞

n

A

∞

x is a fixed point of X.

Now to prove the uniqueness:

Suppose y

F

X such that y ≠ x and

(y) = y.

Now d(x,y) = d(

(x),

(y)) ≤ cd(x,y)

(i. e.) (1−c)d (x, y) ≤ 0 .....(1) Since c < 1, we have 1 – c > 0 and since d(x,y) ≥ 0, (1

–

c)d(x,y)

≥

0,

which

contradiction to (1). L

x is a unique fixed point of X,

is

a

SECION-4.4:

HE IN"ERSE

F!NCION HEOREM #HE IN%E!"E F$NC#ION #HEO!EM

Theorem 4.4.1:

Suppose f is a ? ‘mapping of an open set E  R

n

into

n

R , f’(a) is invertible for some a  E and b = f(a) then a.

there exist open sets U and V in R

n

such that a  U, b  V, f is one-

to-one on U, and f(U) = V; b.

if g is the inverse of f [which exists, by (a)], defined in V by g(f(x)) = x, x

P77:

a.

Put f’(a) = A.

 U,

then g



C ‘(V)

Given that f’(a) is invertible. L

A is invertible (i.e.) A

Choose

so that

1

= 2

–1

||

exists.

.

A −1

||

Since f’ is continuous at a , there exists an open ball U O E with center at ‘a’ such that ||f’(x) – f’(a)|| <

(i. e.) ||f'(x) − A|| <

.

Define a function

by,

+ A

–1

.....(*) (x) = x

(y – f(x)) where x F E and

n

yF R . Note that f(x) = y iff x is a fixed point of Let f(x) = y. Then A

–1

L

(y – y) = x + A

–1

x is a fixed point of

(x) = x + (0) = x.

Conversely, let

(x) = x. –1

Then x = x + A –1

A (y – f(x)) = 0 y = f(x). Now A

–1

'(x) = I + A A – A

–1

(y – f(x))

B

B

y–f(x) = 0

–1

f’(x) = A

B

(0 – f'(x)) =

–1

(A – f’(x))

||φ'(x)|| = ||A − 1 (A − f' (x))|| ≤ ||A − 1 || |A − f' (x)| < 2λ . = 2 . 1

1

..... (1)

By Mean value theorem for single variable, |

(x1 ) –

||

'(x)||, x

(x 2 )| = |x 1 –x 2 | F (x 2 ,

x1)

1

|φ(x1) − φ(x2)| < 2 |x1 − x2| where x1, x2 L

F

U ..... (2)

is a contraction.

By using fixed point theorem( theorem 4.3.1) , the function has atmost one fixed point in U


so that f(x) = y for at most one x F U. L

By theorem 4.1.3, f is one-to-

one in U. Put V = f(U). Let y 0 F V. Then y 0 = f(x 0 ) for some x 0 F U. Let B be an open ball with center at x 0 and radius r > 0, so small that its closure

¯

B lies in U.

Now to prove that V is open. (i.e.) to prove that |y – y 0 | B

<

λr

y F V.

Since

(x0 ) = x 0

+ A

–1

f(x 0 )), φ(x 0 ) – x 0 = A

–1

(y – f(x 0 ))

(y –

|φ(x0) − x0| = |A (y − f(x0))| ≤ ||A || |y − f(x0)| ≤ −1

<

1 2λ

r 2

λr =

−1

1 2λ

|y

− y0|

......(3)

¯ If x F B , then | –

(x0 ) +

(x) – x0 | = |

(x 0 ) – x 0 | ≤ |

φ(x 0 )| + | <

(x)

(x) –

(x 0 ) - x 0 |

1 r x − x0| + | 2 2

L

|

(x) – x 0 | < r. (x)F B(x 0 ,r).

Also if x1, x2

F

¯

|

|

B, then φ(x1) − φ(x2) <

1 2

| x1

(by (2))

− x2|

¯ L

is a contraction of

Since R

n

¯

B into B .

¯

is complete and B is

¯

closed, B is complete.( since any closed

subset

of

a

complete

metric space is complete).


By the above theorem, fixed point, x L L

y = f(")

F

F

has a

¯ B

¯

f(B)

O

f(U) = V

y F V.

(i.e.) y is an interior point of V. L

b.

y is open set.

Now to prove that g

F

Fix y F V and y + k

F

V. Then there

exist x

U so that

F

U, x + h

F

? ’ (V).

y = f(x); y + k = f(x + h) Now, A

–1

(x + h)–

(x) = x + h +

(y – f(x + h)) – x – A

–1

(y –

f(x)) =h+A

–1

(y – f(x

+ h) – y + f(x)) =h+A A

−1

−1

( f( x + h )

k .....( 4)

− f( x ) ) = h −

1 2

Now |φ(x + h) − φ(x)| ≤

B

1 2

B

A

1

k ≥

|

B |h |

By (*),

−1

|h − A

|

k ≤

1 2

| h|

−1 −1 | h| ≥ |h − A k| ≥ |h| − |A k| −1

|

| h| B

2

|

≤ 2 A

−1

| h | B 2 |A

−1

|

k ≤ 2

k ≥ |h|

|

||A || |k| ≤ −1

1 λ

| k|. ..... (5)

1 ||f'(x) − A|| < λ. B ||f'(x) − A|| < 2||A ||

B

−1

A

||

−1

|f'(x)

− A| <

|| |

1 2

< 1.

|

By theorem 4.1.5, f’(x) is invertible linear operator. (i.e.) f’(x) has an inverse, say T.

(i. e.) T = f'(1x) . Now g(y+k) – g(y) – Tk = g(f(x+h)) – g(f(x)) – T = x+h – x –


Tk = h – Tk = –T[k – h/ T] = –T[f (x+ h) – f(x) –hf ’(x )]

|g(y + k)

|

− g(y) T(k)| = − T[f(x + h) − f(x) − hf'(x)]

| |T || | f (x + h )

≤

|g(y + k)

≤

− f(x) − hf'(x)|

− g(y) − Tk|

|k|

| |T | |

|f(x + h)

|f(x + h) ≤

|

||T||

− f(x) − hf'(x)| λ(h)

− f(x) − hf'(x)|

|k|

(by (5))

Since f is differentiable, the RHS of the inequality tends to zero. Clearly, g is differentiable and g'(y) = t. But T was chosen to be the inverse of f (x) = f’(g(y)),


g'(y) = [f'(x)]

−1

[

= f'(g(y))

]

−1

.

..... (6)

Since g is a continuous mapping of V onto U and f’ is a continuous mappi of U into the set

of all invertible n

elements of L(R

ng

), and that inversion

is a continuous mapping of (by theorem 4.1.3) By (6) g

onto F

? ‘(V)

Theorem 4.4.2: If f is a ? ‘ mapping of an set B

O

R

n

into R

invertible for every x an open subset of R

n

F

n

and if f’(x) is

E, then f(W) is

for every open

set W O E. P77.:

Let y

F

element x W

O

f(W). Th en there exist an F

W such that f(x)= y Since

E, x F E.

By hypothesis, f’(x) is

invertible.

By the inversion function theorem, there exists an open set U and V in R

n

such that x F U and y F V, f is one-

to-one and f(U) = V. Since W is open and x selected so that U Therefore f(U) (i.e,) V

O

f(W).

But f(x)

F

V

O

O

F

W, U can be

W.

f(W).

(i.e,) y F V and V is open, there exists a neighborhood Nr(y) But V

O

O

V,

f(W). Ttherefore Nr(y)

f(W). Therefore f(W) is an open in R

n

,

O

C'P $E"#ION": 1.

Suppose that f is a differentiable real function in an open set E O

n

R , and that f has a local

maximum at a point x

F

E. Prove

that f’(x) = 0 2.

If f is a differentiable mapping of a connected open set E

O

R

n

m

R , and if f'(x) = 0. for every x F E, prove that f is constant in E.

into

$NI#-5

$61< "<:=+<=:

Section 5.1: The implicit function theorem Section 5.2: The rank theorem Section 5.3: Determinants. Section 5.4: higher order Sectio n 5.5:

Derivatives

of

Dif ferentiation of

integrals

I6<:7,=+<176

In this unit we shall discuss about the im plicit function theorem, the rank theorem,

determin ants, derivatives

of higher order and differentiation of integrals.

SECION-5.1

HE IMPLICI

F!NCION HEOREM

If f is a continuously differentiable real function in the plane, then the equation f(x,y) = 0 can be solved for y interms of x in a neighborhood of any point (a,b) for which f(a,b) = 0 and

∂f ∂y

≠ 0 . The preceding very

info rmal statem ent is the simplest case

of

the

so-called

“

implicit

function theorem”. N7<)<176:

and y = (y

If x = (x

1 ,x 2 ,...,x n )  R

1 ,y 2 ,...,y m )  R

m

n

, let us

write (x,y) for the point (or vector)

(x 1 ,x 2 ,...,x n , y 1 ,y 2 ,...,y m ) F R

Every A F L(R linear

n+m

.

linear n+m

transformation

n

, R ) can be split into two

transformations A x and A y ,

defined by A x h = A(h,0), A y k A(0,k) for any h

F

n

R , kF R

m

=

. Now to

show that A x F L(R n ), A y F L(R n+m , R n ) and A(h,k) = A x h + A y k. 877.: i.

If h 1 ,h 2

F

R

n

then Ax(h 1 + h 2 ) =

A(h 1 + h 2 ,0) = A[(h 1 ,0), (h 2 ,0)] = A[(h 1 ,0)] + A[(h 2 ,0)] =A x h 1 + A y k 2 .

ii.

A x (ch) = A(ch,0) = A[c(h,0)] = cA(h,0) = c A

xh

where c is a

scalar. A x is a linear. |||ly A y is a linear. iii.

A x h + A y k = A(h,0) + A(0,k) = A[(h,0) + (0,k)] = A(h,k)

Theorem 5.5.1: If A F L(R n+m , R n ) and if A x is invertible, then there corresponds to every k

F

R

m

a unique

n

h F R such tha t A(h,k) = 0. This h can be computed form k by the formula h = –(A x ) –1 A y k. P77.:

Given that A(h,k) = 0.

Since A(h,k) =A x h + A y k, A x h + A y k = 0. B A x h = –A y k.

Since A x is invertible, A x-1 exists. (i.e.) A x A x h = –(A x )

–1

–1

= Ax

–1

AX= I

B

Ax

–1

Ax

Ayk

B

h = –(A x )

–1

A y k.

Now to prove the uniquness. Suppose h 1 ,h 2 F R –(A x )

–1

n

such that h 1 =

A y k 1 ., h 2 = –(A x )

–1

Ayk2

then h 1 – h 2 = 0. Therefore h is unique.

(I841+1< F=6+<176 #0-7-) #0-7- 5.5.2:

Let f be a ? ‘-mapping of an open set n+m n E O R into R ,such that f(a,b) = 0 for some point (a,b)

F

E. Put A

= f’(a,b) and assume that A x

is

invertible. Then there exist open sets O

n+m

m

O

U R and W R , with (a,b) and b F W, having the property: To every y

F

F

U

W corresponds a unique

x such that (x,y)

F

U and f(x,y) = 0.

If this x is defined to be g(y), then g is a ? ‘–mapping of W into R a, f(g(Y),y) = 0 (y –(A x )

–1

F

n,

g(b) =

W), and g'(b) =

Ay.

[This g is called implicit function].

P77.:

Define F: E

A

R

n+m

(f(x,y),y) for every (x,y)

by F(x,y) = F

E

Then F is a ? ‘‘-mapping of E into R

n+m

.

Now to prove that F is differentiable at (x,y)

F

of E into R Let (x,y)

E. and F’ is continuous map n+m

F

. n

E and let h F R , k F R

such that (x+h,y+k)

F

m

E,

F(x+h,y+k) – F(x,y) (f(x+h,y+k),y+k) – (f(x,y),y) = (f(x+h,y+k) – f(x,y), k) = (f'(x,y)(h,k) + r(h,k), k+0),

=

where

|r(h, k)|

lim (h, k)

A

0

(h, k)

=0

F(x+h,y+k) – F(x,y) = (f’(x,y)(h,k) ,k) + (r(h,k), 0). Therefore F is differentiable at (x,y) F

E.

Let

> 0 be given.

Since

f’is

exists a

continuous

on

E,

there

> 0 such that

|| f’(x,y) – f’(u,v) || <

whenever

|(x,y) – (u,v)| < Now

F'(x,y)(h,k)

–

F'(u,v)(h,k)

(f’(x,y)(h,k) ,k) – (f’(u,v)(h,k),k)

=

= ((f’(x,y) – f’(u,v))(h,k),0)

|| F'(x,y)(h,k) – F'(u,v)(h,k) || ≤ || f’(x,y ) – f’(u,v )|| |h,k| < if |h,k|≤1. | F'(x,y)(h,k) – F'(u,v)(h,k) || < L

F is a ? ‘-mapping of an open set E n+m

into R

.

Next to prove that F'(a,b) is an invertible element of L(R Since f(a,b) = 0, we have

n+m

f(a+h,b +k)

– f(a,b) = f’(a,b)(h,k)+r(h,K), where

lim (h, k)

A

0

|r(h, k)| =0 |(h, k)|

).

L

f(a+h,b+k)

=

A(h,k)

+

r(h,k)

(since f(a,b) = 0)

Since F(a+h,b+k) – F(a,b) (f(a+h,b+k),b+k) – (f(a,b),b)

=

= (f(a+h,b+k),k)

= (A(h,k)+r(h,k),k) = (A(h,k),k) + (r(h,k),0)

L

F’is the linear operator on R

that maps (h,k) to (A(h,k),k). (i-e. ) F'(a,b)(h,k) = (A(h,k),k).

n+m

Suppose

F'(a,b)(h,k)

(A(h,k),k) = 0

B

(i.e.) A(h,0) = 0

=

Ax

=

0

–1

B

(A x h) = 0

then

A(h,k) = 0 and k =0

B

A x h = 0.

Since A x is invertible, A x

L

0

B

(A x

–1

–1

exists .

A x )h

=

B

Ih

h = 0.

L

F'(a,b)(h,k)

L

F'(a,b) is a one-to-one on L(R

Since R

n+m

=

0

B

(h,k) = (0,0). n+m

).

is a finite dimensional

vector space and by theorem 4.1.3, F'(a,b) is onto, we have F'(a,b) is invertible.

Now

apply

the

inverse

function

theorem to F.

It shows that there exist open sets U and V in R F(a,b)

F

n+m

, with (a,b)

F

U and

V such that F is a 1 – 1

mapping of U onto V.

Now F(a,b) = (f(a,b),b) = (0,b) Let W = {y Since (0,b)

F

F

R

m

/(0,y)

F

V}.

V, b F W.

If y F W then (0,y)

F

L

(0,y) = F(x,y), x,y

L

(0,y) = (f(x,y),y)

L

f(x,y) = 0.

V = F(U). F

U

F

V .

Now to prove the uniqueness. Suppose,

with

the

same

y,

that

(x’,y) F U and f(x’,y) = 0. F(x’,y)

=

(f(x’,y),y)

=

(0,y)

=

(f(x,y),y) = F(x,y) Since F is 1 – 1,F(x’,y) = F(x,y) (x’,y) = (x,y)

B

B

x’= x.

Next to prove the second part. Define g(y) = x, for y F W such that (g(y),y)

FU

and f(g(y),y) = 0.

(

)

Consider F(g(y), y) = f(g(y), y), y = (0, y)

.....(1)

Since F is 1 – 1 on U and F(U) = V, G = F

–1

exists.

By inverse function theorem, G F ? ‘(V).

(1) gives G(F(g(y),y)) = G(0,y) (g(y),y) = G(0,y) L

F

g F ? ‘(W) (or) g

? ’ (V)

F?

‘. –1

Next to prove g'(b) = –(Ax) Put

B

Ay.

(y) = (g(y),y).

Let k F R Now

m

such that y+k

(y+k) –

F

W.

(y) = (g(y+k),y+k) –

(g(y),y) = g(y),k)

(g(y+k)

–

=

(g'(y)k

+

r(k),k).

= (g'(y)k,k) + (r(k),0) L

Φ is differentiable and Φ'( y) k = ( g'( y)

k, k)

.....( 2)

Since

(y)

=

(g(y),y),

f(

(y))

=

f(g(y),y) = 0 in w. The chain rule therefore shows that f’(

(y))

'(y) = 0

When y = b,f'( Therefore and f'(

(b))

'(b) = 0.

(y) = (g(y),y) = (a,b)

(y)) = A.

Thus f'( a, b) Φ'( b) = 0. ( i. e.) AΦ'( b) = 0. .....( 3)

Consider A x g'(b)k+ A y k = A(g'(b)k,k) =A

'(b)k (by (2))

A x g'(b)k+A y k = 0 (by (3)) A x g'(b) +A y = 0 Axg'(b) = - A y . Since A x is invertible, A x

-1

exists.

-I

A x A x g'(b) = - (A x -1)Ay.

L

g'(b) = – (Ax

–1

)A y .

E?)584-: Take n =2 , m =3 and consider the 5

mapping f = (f 1 ,f 2 ) of R given by f 1 (x 1 ,x 2 ,y 1 ,y 2 ,y 3 ) = 2e + 3

x1

2

into R

+ x 2 y 1 – 4y 2

f 2 (x 1 ,x 2 ,y 1 ,y 2 ,y 3 ) = x 2 cos x 1 – 6x 1 + 2y 1 – y 3 .

If a = (0,1) and b = (3,2,7), then f(a,b) = 0. The matrix of the transformation A = f'(a,b) is given by

[

[A] = D 1 f1 D 2 f1 D 3 f1 D 4 f1 D 5 f1

=

[

2

D 1 f2 D 2 f2 D 3 f2 D 4 f2 D 5 f2

3 1

− 6 1 2

Hence [Ax] =

[

2 −6

− 4

0

0

− 1

3 1

]

]

]

and [Ay] =

[

1 2

−4 0

0 −1

]

We see that the column vectors of [A x ] are independent, A x is invertible and the implicit function theorem

.

asserts the existence of a –mapping g , defined in a neighborhood of (3,2,7), such that g(3,2,7) = (0,1) and f(g(y),y) = 0.

Since

−1

[(A ) ] = [A ] x

−1

g'(3,2,7 ) = − Ax

= −

1 20

[

= −

1 20

[

−1

=

x

1

|Ax|

Ay = −

1 20

1 − 6

− 4

3

6 + 4

− 24

− 2

−5

− 4

3

10

− 24

− 2

AdjA=

]

[

1

−3

6

2

]

1

− 3

1

−4

0

6

2

2

0

− 1

[

][ =

1 20

][ 1/4

1/5

− 1/2 6/5

] − 3 / 20 1 / 10

Interms of pa rtial derivativ es, the conclusion is that D1g1 = D1g2 =

1 4 −1 2

D2g1 =

1 5

D2g2 =

D3g1 = 6 5

D3g2 =

−3 20 1 10

at the point (3,2,7


)

]

C'P $E"#ION": 1.

Take n = m =1 in the implicit function theorem , and interpret the theorem graphically.

SECION-5.2

: HE RANK

HEOREM

D-.161<176: Suppose X and Y are vector spaces, and A F L(X,Y). The 6 = 4 4 ; 8 ) + - of A, ? (A) , is the set of all x F X at which Ax = 0. (i.e.) ? (A) = { x

F X/Ax

= 0}.

? (A) is a vector space in X. The )6/- of A, !-;=4< 1: X.

=

(A) = {Ax/x

F X}.

? (A) is a vector space in

Let x 1 ,x 2

F

? (A). Then Ax 1 = 0, Ax 2

= 0. Now A(x 1 +x 2 ) = Ax 1 + Ax 2 =0+0= 0. L

X 1 + X 2 F ? (A).

Also A(cx) = cAx = c0 = 0, where c is a scalar. L

cx F ? (A).

L

? (A) is a vector space in X.

!-;=4< 2 :

=

(A) is a vector space in

X. Let y 1 ,y 2

F =

(A). Then there exist

x 1 ,x 2 F X such that

y 1 = Ax 1, y 2 = Ax 2 . Now A(x 1 +x 2 ) = Ax 1 + Ax 2 = y 1 +y 2 . L

x1+ x2

If y

F

F =

(A).

. = (A), then there exists x F X

such that y = Ax cy = cAx = Acx L =

F=

(A)

(A) is a vector space in X.

D-.161<176:

The )6 of A is defined to be the dimension of

=

(i.e.) r(A) = dim

(A). =

(A).

!-;=4< 3:

Show that all invertible elements in n

L(R ) have rank ‘n’ and conversely.

Suppose A

F

(i.e.) A: R

n

n

L(R ) is invertible.

A

R

n

Since A is onto,

is 1 – 1 and onto.

=

n

(A) = R .

Therefore r(A) = dim

=

(A) = dim R

n

= n.

Conversely,let the rank of A

F

L(R n )

be n. (i.e.) r(A) = n. (i.e.) dim

=

(A)

Therefore

=

= R .

=

dim R

n

n

(i.e.) A is onto. Since R

n

is a finite dimensional vector

space , by theorem 4.1.3,

A is 1 – 1 Therefore A is invertible.

P72-+<176;: Let X be a vector space. An operator P F L(X) is said to be a projection in X if P

2

= P.

Example: Define P by P(x) = x , x F X. Let x 1 ,x 2

F

X. Then P(x 1 ) = x 1 , P(x 2 )

= x2. Now P(x 1 + x 2 ) = P(x 1 )+P(x 2 ) = x 1 + x 2 and P(cx) = cP(x) = cx. Therefore

P

transformation.

is

a

linear

2

Now P (x) = P(P(x)) = P(x). Therefore P is a projection. !-;=< 4:

If P is a projection in X then every x F X has a unique representation of the form x = x 1 + x 2 where x 1 F = (P), x 2 F ? (A). Since x 1 F = (P) , x 2 F ? (A), we have x 1 = P(x 1 ) and Px 2 = 0. x 2 =x–x

1 B

P(x 2 ) = P(x) – P(x 1 ) B

0 = x 1 – x 1, by putting P(x) = x 1 . Suppose x = x 3 + x 4 where x 3 F = (P) , x 4 F ? (A).. Then x 1 = Px = Px 3 + Px 4 = x 3 + 0 = x 3 and x 4 = x - x 3 = x - x 1 = x 2

Therefore the expression is unique. !-;=4< 5:

If X is a finite dimensional vector space and if X 1 is a vector space in X, then there is a polynomial P in X with =

(P) = X 1 .

Since X is finite dimensional and X 1 O X, dim X 1 is finite. If X 1 contains only 0, this is trivial, put Px = 0 for all x F X 1 . (P(x 1 +x 2 ) = 0 B P(x 1 )+ P(x 2 ) = 0+0 =0 and P(cx) = cP(x) = c0 = 0, 2

P (x) = P(P(x)) = P(0) = 0 = P(x).) Therefore = (P) = X 1 .

Assume dim X 1 = k > 0. Then it has a basis {u 1 ,u 2 ,...,u k } for X 1 and {u 1 ,u 2 ,.. .,u n ) for X. Define P by P(c 1 u 1 +c 2 u 2 +... .+c n u n ) = c 1 u 1 +c 2 u 2 +.. .+c k u k . Let

xF

X1.

Then

x

=

c 1 u 1 +c 2 u 2 +...+c k u k . Px

=

P[c l u 1 +c 2 u 2 +..

P[c l u l +c 2 u 2 +..

.+c k u k ]

.+c k u k +0.u k+l

+

.+0.u n ] = c l u l +c 2 u 2 +...+c k u k (i.e.) Px = x , for Therefore x But L =

=

F=

(P) O X 1 .

(P) = X 1 .

every x

=

x.

F

X 1.

(P). (i.e.) X 1 O= (P).

= ..

(#0- !63 #0-7-) #0-7- 5.2.1: Suppose

m,n,r

are

non-negative

integers with m ≥ r, n ≥ r. F is a ? ‘-mapping of an open set E R

m

O

R

n

into

and the derivative F'( x) has rank r

for every x

F

E. Fix a

F

E , put A = F'(a)

, let Y 1 be the range of A, and let P be a projection in R

m

whose range

is Y 1 .Let Y 2 be the null space of P. Then there are open sets U and V in R

n

,with a

F

U, U O E, and there is a 1

– 1 ? ‘-mapping H of V onto U (whose inverse is also of class ? ’ ) F(H(x)) = Ax +

such that

(Ax) (x

F

V)

where

is a ? ‘-mapping of the open

set A(V)

O

Y 1 into Y 2 .

P77.:

Case (i): Let r = 0. Then rank F’(x) = 0 for every x

F

E.

By definition dim {range of F'(x)} = 0. Range of F'(x) = {0}.

Therefore F'(x) = 0. ||F'(x)|| = 0. By

Mean

Value

theorem

||F(b)

–

F(a)|| ≤ |b – a| ||F'(x)|| = 0 L

||F(b) – F(a)||) = 0.

L

F(b) = F(a). F(x) is constant.

Let V = U. Define H(x) = x, for every x F V and

(0) = F(a).

Now F(H(x)) = F(x) = F(a) = 0x +

(0) =

(0x).

F(H(x)) = Ax + (Ax) where A = F'(a) = 0, since F(a) is constant. Case (ii): Let r > 0. Then rank F'(x) = r, for every x

F

E.

rank F'(a) = r, for every a L

rank A = r, since A =

F

E.

F'(a).

dim {range of A} = r. dim Y 1 = r. Y 1 has a basis containing r elements, say {y 1 ,y 2 ,... ,y r }. n

Choose Z i F R so that AZ i = Y i ,1≤i ≤ r.

Define

S:

Y1

R

A

n

by

S(c 1 y 1 +c 2 y 2 +...+c r y r )

=

c 1 z 1 +c 2 z 2 +...+c r z r , where c 1 ,c 2 ,...,c r are scalars. Clearly S is linear. L

ASy i = Az i B ASy i = y i , 1 ≤ i ≤ r.

L

ASy = y, if y

Define G: E

O

F

R

Y 1 . (or) AS = I.

n

A

R

n

by G(x) = x +

SP[F(x) – Ax]. Then

G(x+h)

–

G(x)

=

x

+h+

SP[F(x+h) – A(x+h)] – x – SP[F(x) – Ax] = h+ SP[F(x+h) – F(x) – Ax – Ah + Ax]

= h+ SP[F(x+h) – F(x) – Ah]

G(x+h) – G(x) – Dh = h+ SP[F(x+h) – F(x ) – A h] – h[I +SP [F' (x) –A], where D = I +SP[F'(x)–A = SP[F(x+h) – F(x) – F'(x)+h] G(x + h) − G(x) − Dh h

|G(x + h)

− G(x) − Dh|

|h|

When

h

A

≤

=

SP[F(x + h) − F(x) − F'(x) + h] h

||SP||

0,

|F(x + h)

the

− F(x) − F'(x) + h|

|h|

RHS

inequality tends to zero. Therefore LHS tends to zero.

of

the

L

G'(x) = D, for every x

F

E.

(i.e.) G'(x) = = I +SP[F'(x)–A]. G'(a) = I +SP[F'(a)–A] = I +SP[A–A] ( since F'(a) = A) = I.

L

G'(a) is an identity operator on R

n

.

Clearly G is a ? ’-mapping on E. Apply the inverse function theorem to G, there are open sets U and V in R ,with a G(a)

F

F

n

U,U O E.

V, G is 1 – 1 and G(U) = V.

G has an inverse H: V A U which is also bijection and H

F

? ’(V).

Next to prove that ASPA = A. Let x F R

n

. Then ASPA(x) ASP[A(x)]

= ASPy, where y = Ax,

= ASy = Iy = y = Ax. Therefore ASPA = A. Since AS = I, PA = A. Since G(x) = x + SP[F(x) – Ax], AG(x) = Ax + ASP[F(x) – Ax] = Ax + ASPF(x) – ASPAx = Ax + ASPF(x) – Ax = ASPF(x) = PF(x) ( since AS = I) Therefore AG(x) = PF(x), for every x F E.

In particular, AG(x) = PF(x) holds for x F U. (since U

O

E)

If we replace x by H(x), AG(H(x)) = PF(H(x)), for every x F U=V. Ax = PF(H(x)) Define every x P

(x) = F(H(x)) – Ax , for F

V.

(x) = PF(H(x)) – PAx = Ax – Ax =

0. L

(x)

Clearly

F

? (P) = Y 2 . is a ? ‘-mapping of V into

Y2. To complete the proof, we have to show that there is a ? ‘-mapping

of

A(V) into Y 2 satisfies

(Ax)

(x), for

every x F V.

Put

(x) = F(H(x)), for every x

F V.

'(x) = F'(H(x))H'(x) L

Rank

of

'(x)

=

Rank[F'(H(x))H'(x)] = r dim(range of '(x)) = r. dim M = r, where M = range of

'(x).

Since Y 1 = = (A), dim Y 1 = dim = (A)

=

rank A = r.

Also PF(H(x)) = Ax. Put (x) Ax B P P maps M into

'(x) = A =

(A) = Y 1 .

L

P is 1 – 1 and onto, since M and Y 1

have same dimensions

Suppose Ah = 0, where h = x 2 – x 1 F V P

'(x)h = 0 = P.0

Again

'(x)h = 0 .

B

(x) = F(H(x)) – Ax, x

F

V

'(x) = F'(H(x))H'(x) – A. '(x) = F'(H(x))H'(x) – A =

'(x)h –

Ah = 0, Define

g(t)

=

(x1 +th)

t F [0,1]. g'(t) =

'(x 1 +th)h = 0.

Therefore g(t) is constant.

where

g(0) = g(1). But g(0) = L

(x 1 ) and g(1) =

Ψ(x 1 ) =

Define

(x 2 ). on A(V)

(Ax) =

(x 2 ).

O

Y 1 such that

(x).

Next to prove that

F

? ‘(A(V)) ,

Let y 0 be a point in A(V). Then there exists x 0 F V such that y 0 =Ax 0 . Since

V

is

open

and

y0

has

a

neighborhood W in Y 1 such that X = X 0 + S(y – y 0 ) lies in V, Ax = Ax 0 + AS(y – y 0 ) = y 0 +y–y (since AS = I)

0

= y (y) =

L

(Ax) =

(x) =

(x 0 + S(y

– y 0 )) '(y) = L

φ

F

’(x 0 + S(y – y 0 )) S.

? ‘(A(V)).

C'P $E"#ION": 2

1.

For (x,y) ≠ (0,0), define f= (f1,f2)b y f1(x, y) = f2(x, y) =

xy 2

2

x +y

x − y 2

x +y

compute the rank

of f’(x,y) and find the range of f.

2.

n

m

Suppose A F L(R ,R ), let r be the rank of A. a.

Define S as in the proof of theorem 5.2.1. Show that R (S).

2

2

is a projection in R

n

whose

null space between and whose range is b.

=

(S).

Use (a) to show that dim N (A) + dim R (A)= n.

SECION-5.3

:

DEERMINANS.

D-.161<176:

If (j 1 ,j 2 ,...,j n ) is an ordered n-tuples of integers. Define s(j1, j2, ...., jn) = Π sgn (jq − jp) p
where sgn x = 1 if x > 0, sgn x = –1 if x < 0, sgn x = 0 if x = 0. Then s(j 1 ,j 2 ,..., j n ) = 1, – 1 , 0, and it

changes sign if any two of the j's are interchanged.

D-.161<176:

Let [A] be the matrix of the linear operator A in R

n

with standard basis

{e 1 ,e 2 ,...,en }. Let a(i,j) be the entry in the ith row and jth column of [A]. We define det

[A]

=

s(j 1 ,j 2 ,...,j n )a(1,j 1 )a(2,j 2 )....a(n,j n ) . the sum extends over all ordered ntuples of integers (j 1 ,j 2 ,...,j n ) with 1≤ j≤n.

E?)584-:

If [A] =

2 4 , then det A = [ ] 5 3

[ ]

∑

s(j1, j2)a(1, j1)a(2, j2)

= s(1,2)a(1,1)a(2,2)

+

s(2,1)a(1,2)a(2,1) =

1.2.3

+

(–1)4.5 = 6 – 20 = –14

Note : Let [A] be the matrix.

[A] =

[

a11 a12

S

a1j

S

a1n

a

S

a

S

a

S

S

S

S

S

anj

S

ann

a 21

S

22

S

an1 an2

2j

2n

]

n

The jth column vector xj = a1je1 + a2je1 +....+a

n

njen = ∑ aijei = ∑ a(i, j)ei j=1

Therefore .,x n ).

det[A]

=

det

i=1

(x 1 ,x 2 ,..

#0-7- 5.3.1: a.

nR

then det[A] = det (x

.,x n )

1 ,x 2 ,..

= 1. b.

det is a linear function of each of the column vectors x, if the others are held fixed.

c.

If [A] 1 is obtained from [A] by interchanging two columns, then det[A] 1 = –det[A].

d.

n

If I is the identity operator o

If [A] has two equal co lumns, then det[A] = 0.


P77.:

a.

If the matrix [A] = [I] then a a(i, j) =

{

1

if i = j

0

if i ≠ i

det[I] =

s(j 1 ,

(i,j)

j 2 ,... j n )a(1,

j 1 )a(2,j 2 )....a(n,j n ) = s( 1,2,... ,n)a( 1,1 )a(2,2).. ..a(n,n) = s(1,2,...,n).1 =

sgn(2

–

1)sgn(3

–

2).....sgn(n – (n – 1)) = 1. b.Since

s(j

1 ,j 2 ,...,j n )

=

Π sgn (jq − jp) where sgn(j q –

p
j p ), where sgn x = 1 if x > 0, sgn x = –1 if x < 0, sgn x = 0 if x = 0, s(j

1 ,j 2 ,..

.,j n )

=

0 if

any two of the j's are equal. Each of the remaining n! products in det

[A]

=

s(j 1 ,j 2 ,...j n )a(1,j 1 )a(2,j 2 )....a(n,j n )

contains exactly one factor from each column. Therefore det is a linear function of each of the column vectors x, if the others are held fixed. c.

In s(j 1 ,j 2 ,...,j n ) if two entries are interchanged, change

of

it sign.

effects

the

Therefore

det[A] 1 = –det[A]. d.

Suppose two columns of [A] are equal, then interchange the two columns, –det[A].

we

get

det[A] 1

=

But [A] = [A] 1 . Therefore det[A] = –det[A]. (i.e.) 2det[A] = 0. (i.e.) det[A] = 0. #0-7- 5.3.2:

If [A] and [B] are n by n matrices then det([B][A]) = det[B]det[A ]. P77:

Let x 1 ,x 2 ,.. .,x n be the columns of [A]. Define

B

(x1, x2, ....,xn) = ΔB[A] = det([B][A])

......(1)

(i.e.) Bx 1 ,Bx 2 ,...,Bx n are the column vector of [B][A]. (i. e.) ΔB(x1, x2, ...., xn) = det (Bx1, Bx2, ....., Bxn). .....(2)

By (2) and theorem 5.3.1,

b also

has the properties (b) to (d). n

Since xj = ∑ a(i, j)ei, ΔB[A] = ΔB i=1

[

n

∑ a(i, j)ei, x2, ..., xn i=1

]

n

= ∑ a(i, 1) ΔB[ei, x2, ..., xn]. i=1

Repeating this process with x

2 ,...

,x n

we obtain ΔB[A] =

∑ a(i , 1)a(i , 2)..........a(i , n) Δ (e , e , .....e ) 1

2

n

B

i1

i2

in

where the sum being extended over all ordered n-tuples (i 1 ,i 2 , .... i n ) with the condition 1 ≤ i r ≤ n . Now

B [e i ,x 2 ,...,x n

] = t(i 1 ,i 2 ,.... i n ).

Δ B (e 1 ,e 2 ,...,e n )


Since [B][I] = [B], det([B][I]) = det[B].

L

Δ B (e 1 ,e 2 ,...,e n ) det([B][I]) Δ b [A]

L

=

Δ B [I]

=

=

a(i1 ,1)a(i 2 ,2)....a(i n ,n)t(i 1 ,i 2 ,.... i n ).

Δ B (e 1 ,e 2 ,...,e n )

Δ B [A]

=

det([B][A])

=

a(i 1 ,1)a(i 2 ,2)....a(i n ,n)

t(i 1 ,i 2 ,.... i n ).det[B]- (1) Take [B] = [I], we get det[A] =

a(i 1 ,1)a(i 2 ,2)....a(i n ,n)

t(i 1 ,i 2 ,.... i n ).det[I] =

a(i 1 , 1 )a(i 2 ,2)....a(i n ,n)

t(i 1 ,i 2 ,.... i n ) ( since det[I] = 1) Substitute in (1), we get

det([B][A])

=

det[B]det[A]. #0-7- 5.3.3:

A linear operator A on is invertible iff det[A] ≠ 0. P77:

Suppose A is invertible. Then [A][A

–1

] = I.

By theorem 5.3.2, det[A]det[A det([A][A

–1

]) = det[I] = 1.

Therefore det[A] ≠ 0. Conversely , let det[A] ≠ 0. Now to prove that A is invertible.

–1

] =

Suppose A is not invertible. Then the column vectors x

1 ,x 2 ,...

.,x n

of [A] are dependent. Therefore there exists a vector x k such that x k + ∑ cjxj = 0 for some j≠k

scalars c j . If we replace x k by x k + c j x j , then the determinant of the matrix is unchanged. The same result is true if we replace x xk + ∑ cjxj

(i.e.

k

by

zero).

But

the

j≠k

determinant of a matrix contains a column of zeros is zero.

L

(

)

det [A] = det x1, x2, ....., xk + ∑ cjxj, ....xn = det(x1, x2, ...., 0, ...., xn) j≠k


         

  



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

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

 





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                     

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   

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

 

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

           



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 

derivatives D 1 f, D 2 f,..., D n f . If the function

D jf

are

themselves

differ enti able, then the secon d order partial derivative of f are defined by D ij f= D i D j f where i = l,2,...,n, j = l,2,...,n. If all these functions D

ij f

are

continuous in E then we say that f is of class ? ”in E (or) f Note:

If

the

F

? ”(E)

derivatives

are

continuous then D ij f = D ji f. #0-7- 5.4.1:

Suppose f is defined- in an open set 2

E O R , and D i f, D 21 f are exist at every point of E. Suppose Q

O

E is a

closed rectang le with sides parallel to

the coordinate axes, having (a,b) and (a+h,b+k) as opposite vertices (h ≠ 0, k ≠ 0). Put (f,Q) = f(a+h,b+k) – f(a+h,b) – f(a,b+k) + f(a,b). Then there is a point (x,y) in the interior of Q such that (f,Q) = hk(D 21 f)(x,y). P77.:

Let u(t) = f(t,b+k Given

) − f(t, b). .....(1)

(f,Q)

=

f(a+h,b+k)

–

f(a+h,b) – f(a,b+k) + f(a,b). = u(a+h) – u(a) = h u'(x) where x is a point lies between a and a+h.

Differe ntiate (1) with respect to t, we get

u'(t) = D 1 f(t,b+k) – D 1 f(t,b) L

u'(x) = D 1 f(x,b+k) – D 1 f(x,b)

L

(f,Q) = h u'(x) = h[D 1 f(x,b+k) –

D 1 f(x,b)] = hk(D 21 f)(x,y) where b < y < b+k. #0-7- 5.4.2:

Suppose f is defined in an open set 2

EO R

, and suppose D

1 f,

D 21 f and

D 2 f are exist at every point of E and D 21 f. is continuous at some point (a,b) and

F

E. Then D 21 f exists at (a,b)

(D 12 f)(a,b) = (D 21 f)(a,b).

P77.:

Suppose A = (D 21 f)(a,b). Given that (D 21 f.) is continuous at (a,b)

F

E.

Let

> 0 be given.

If Q is a rectangle contained in E such that |(D 21 f)(x,y) –(D 21 f)(a,b)|< (i.e.) |(D 21 f)(x,y) –A | < By

theorem

5.4.1,

hk(D 21 f)(x,y). L

L

Δ(f, Q) hk

|

= (D21f)(x, y).

Δ(f, Q) hk

|

− A <ε

(f,Q)

=

h

0, k

A

Δ(f, Q) hk 1

=

..... (1)

0

=

f (a+h,b+k

) − f(a+h,b ) − f(a,b+k ) + f(a, b) hk

f(a+h,b+k

hk 1 h

A

(f, Q) = f(a + h, b + k) − f(a + h, b) f(a, b + k) + f(a, b),

Since

=

Δ(f, Q) hk =A

lim

L

) − f(a+h,b ) − f(a,b+k ) − f(a, b)

(

[

[

f(a+h,b+k

) − f(a+h,b ) k

−

f(a,b+k

k

Keeping h fixed and allowing k

A

=

1 h

1 h

{[ A

A

f(a+h,b+k

lim

0, k

k

0

] [A −

f(a,b+k

lim

k

) − f(a, b) k

0

]}

) − (D2f)(a, b)]

Δ(f, Q) hk = A

) − f(a+h,b )

0

[(D2f)(a+h,b lim

h

0,

0

k

=

A

]

Δ(f, Q) hk

lim

k

)]

) − f(a, b)

1 h

lim h

A

0

[(D2f)(a+h,b

) − (D2f)(a, b)] = D12f(a, b)

From A = D 12 f(a,b) Therefore (D 21 f)(a,b) = (D 12 f)(a,b). C'P $E"#ION": 1.

Show that the existence of D

12 f

does not imply the existence of D 21 f. For example, let f(x,y) = g(x),

where

g

is

nowhere

differentiable. Double click this page to view clearly

SECION-5.5

:

DIFFERENIAION OF INEGRALS

#0-7- 5.5.1: "=887;(x,t) is defined for a ≤ x ≤ b, c

a.

≤ t ≤d; is

b.

an

increasing

function

on[a,b]; c. d.

φ'  R (

) for every t



[c,d];

c < s < d, and to every corresponds a |(D 2

> 0 such that

(x,t) –(D2

)(x,s)| <

all x  [a,b] and for all t s+

> 0



(s –

for ,

). b

Define f(t) =

∫φ(x, t) dα(x)

(c≤t≤d

a

Then (D 2 φ)

s



=

(

),f'(s) exists,

)

b

and f' (s) =

∫(D φ)(x, s) dα(x) 2

a

P77.:

Consider the difference quotients ψ(x, t) =

φ(x, t) − φ(x, s) t−s

for0<|t−s|<

By Mean vale theorem, (x,t) –

(x,s) = (t – s) D2

(x,u),

where t < u < s. L

φ(x, t) − φ(x, s) t−s

(i.e.)

= D2φ(x, u).

(x,t) = D 2

By (d), |(D 2

(x,u).

(x,t) –(D2

(x,s)| <

for all x F [a,b] and for all t s+

F

(s –

).

(i. e.) |ψ(x, t) − D2φ(x, s)| <

.....(1)

,

b

Define f(t) =

∫φ(x, t) dα (x). a

b

Then f(s) =

∫φ(x, s) dα (x). a

b f(t) − f(s) t−s

=

∫

φ(x, t) − φ(x, s) dα (x) t−s

a b

=

∫ψ(x, t) dα(x) a

From D2

(1),

(x,t)

converges

to

(x,s), t

Therefore on [a,b].

(x)

By theorem 2.4.1,

A

s

(D 2 φ)

uniformly

is monotonically

increasing on [a,b] and

fnF

=

(

).on [a,b], for n =

and suppose f n [a,b]. Then f b

A

1,2,3,....,

f uniformly on

F =

(

)on [a,b], and

(

) (since

b

∫fd

= lim n

a

A

∫f

∞ a

we have (D 2 φ) (

n dα

s

F =

t

F =

)) b

Since

f (t) − f(s) t−s

=

∫ψ(x, t) dα (x), a b

lim

t

As

f(t) − f(s) = t−s

A s ∫a

lim t

b

ψ(x, t) dα(x) =

lim ∫A a

t

ψ(x, t) dα (x) s

b

=

∫(D φ)(x, s) dα(x) (by (1)) 2

a b

herefore f'(s) =

∫(D φ)(x, s) dα (x) 2

a


E?)584-: Compute

the

integrals

∞ 2

f(t) =

∫e

− x

cos(xt) dx

cos(xt)dx

− ∞

and ∞

g(t) = −

∫

2

xe

− x

sin (xt) dx, where − ∞ < t < ∞.

− ∞

We claim that f is differentiable and f’(t) = g(t). If

>0,then = sin

=

sin

−

cos(α + β) − c os β −

sin

+ cos (α +β) − c os β

+ cos (α + β) − c os β

sin

= (α + β) sin

+sin

+cos (α + β) − ( β

+

sin

+sin sin

+cos )

α+β

=

1 β

∫( sin

− sin t ) dt

α

|

cos(α + β) − co s β

+sin

|

α+β

| = ∫( sin 1 β

α


|

− sin t ) dt

α+β 1

∫| sin

≤ |β|

− sin t | dt

α α+β

≤

1 |β|

∫| t −

| |dt|

α

(by Mean value theorem) 1

≤ | β|

[

(t− 2

2

)

]

α+β

1

[

≤ |β| (

α

1

+

−

2

1

2

]

)− 0 2

| |=

≤ |β| β = |β| β

(i. e.)

|

cos(α + β) − co s β

| β|

|≤

+sin

|β| 2

.....(1)

2

∞

Given that f(t) =

∫e

2

− x

cos(xt) dx

− ∞

∞

f(t + h) − f(t) =

∞

∫e

− x

2

cos(x(t + h)) dx −

∫e

− ∞

2

− x

cos(xt) dx

− ∞

∞ f(t + h) − f(t) h

=

1 h

∫e

− x

2

[cos(xt+xh )) − cos (xt)] dx

− ∞

Let

=xt,

f(t + h) − f(t) h

=xh. − g( t ) ∞

=

1 h

∫

∞ 2

e

− x

[

cos(xt+xh

]

))

− cos (xt) dx +

− ∞

∫xe

2

− x

sin(xt) dx

− ∞

∞

=

1 h

∫e

2

− x

[cos(xt+xh )) − cos (xt) +xhsin (xt)] dx

− ∞ ∞

=

1 h

∫e

2

− x

[cos(α + β) − c os

+

sin ] dx

− ∞

∞

|

f(t + h) − f(t) h

From (1)

|

|

− g(t) ≤

1 |h|

cos(α + β) − cos + β

∫e

− x

2

|cos(α + β) − c os

sin

|≤

sin | dx

|β| 2 2

(i. e.) |cos(α + β) − c os

+

− ∞

+

sin | ≤

|β| 2

2

< |β| .


∞ L

|

f(t + h) − f(t) h

|

1 |h|

− g(t) <

∫e

1

2

2

|β| dx

− ∞

∞

< |h|

− x

∞

∫

2

e

− x

2

∞ 2

2

2

2

− x − x 2 x dx |x| |h| dx = ∫e |x| dx = ∫e

− ∞

− ∞

When h

− ∞

0, f is differentiable at t

A

and f’(t) = g(t). Next to find f(t): ∞

f(t) =

∫

− x

e

2

cos(xt)dx.

− ∞ − x

Letu=e Then

L

du dx

2

,dv=cos

= − 2xe

f(t) =

[

2

e

− x

− x

sin(xt) t

2

(xt) dx. v=sin (xt) / t ∞

∞

]

+2 − ∞

∫

xe

− x

2

sin xt dx t

− ∞

∞ 2

=0+

t

− x

∫xe − ∞

2

sin xtdx = −

2 t

g(t)

(i. e.) tf(t) = − 2g (t) Since f’(t) = g(t), f satisfies the differential equation 2f’(t) + tf(t) = 0


I.F=e

∫ 2t dt = e

2

t 4

2

t

∫

Therefore e 4 f(t) =

0+c=c. ∞

∞ 2

Initially, when t = 0, f(0) =

e

− x

2

dx+2

∫

Letz=x

. Then

dz dx

− x

dx

∫

− ∞

2

e 0

= 2x.

dz (i. e.) dx = dz 2x = d(z) ∞

f(0) = 2

∞

∫e 0

L

1

−z

2 √z

dz =

∫z

∞ −1 2

e

−z

dz =

∫z

0

1 2

− 1

e

−z

dz = √π

0

c = √π. 2

t 4

The required solution is e f(t) = √π. 2

f(t) = √πe

−

t 4

.

C'P $E"#ION": 1.

For t ≥ 0, put

φ(x, t) =

and put

{

x

(0 ≤ x ≤ √ t )

− x + 2 √t

(√ t ≤ x ≤ 2 √ t )

0

otherwise

(x,t) = –

(x,|t|) if t <

0. Double click this page to view clearly

2

Show that and (D 2

is continuous on R ,

)(x,0) = 0 for all x. 1

Define f(t) =

∫φ(x, t) dx −1

Show that f(t) =tif

|t | <

1

f'(0) ≠

∫(D φ)(x, 0) dx 2

−1

1 4

Hence

Analysis III

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