Calculus on Manifolds A Solution Manual for Spivak for Spivak ( (1965 1965))
Jianfei Shen School of Economics, The University of New South Wales
Sydn Sydney ey,, Aust Austra rali lia a
2010 2010
Contents
1
Functions Functi ons on Euc Euclid lidean ean Spa Space ce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Norm and Inne Innerr Product Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Subs Subsets ets of of Euclid Euclidean ean Space Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Fun Functio ctions ns and and Continu Continuity ity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 1 6 9
2
Differen Diffe renti tiat atio ion n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Basi Basic c Defini Definition tions s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Basi Basic c Theor Theorems ems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Part Partial ial Deri Derivativ vatives es . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Der Deriva ivativ tives es . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Inve Inverse rse Func Function tions s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Impl Implicit icit Fun Functio ctions ns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 13 18 26 34 38 40
3
Integr Inte grat atio ion n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Basi Basic c Defini Definition tions s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Measu Measure re Zero Zero and Conte Content nt Zero Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Fubi Fubini’s ni’s Theor Theorem em . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45 45 51 51
4
Integra Integ rati tion on on Ch Chai ains ns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4.1 Algeb Algebraic raic Preli Prelimina minaries ries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
iii
1 FUNCTIONS ON EUCLIDEAN SPACE
1.1 Norm and Inner Product I Exercise 1 (1-1 ). Prove that x 6
k k
x 1 ; : : : ; x n . Then
0Xˇ @ ˇ
Proof. Let x
D
iD1
x
i
n iD1
xi .
ˇˇ1A D X C Xˇˇ ˇˇ X D k k 2
n
P ˇˇ ˇˇ
n
x
i
n
2
iD1
i
x x
j
xi
>
2
x
2
:
i D1
i¤j
Taking the square root of both sides gives the result.
I Exercise 2 (1-2). When does equality hold in Theorem 1-1 (3)
ut kx C y k 6 kxk C ky k ?
y for every x ; y Rn . Obviously, if 6 x y x 0. So we assume that x x 0 or y 0, then x ; y 0 and y 0. n 0. Write w x ˛ y . We first find some w R and ˛ R such that w; ˛ y Then 0 ˛ x; y ˛2 y 2 w; ˛ y x ˛y ; ˛ y Proof. We reprove that
D
D
ˇh iˇ x; y
k k k k h i D k kk k D 2 2 h
Dh
iDh
iD h
2
iD
¤
¤ D
i k k
implies that
D hx; y i ky k2 :
ı
˛ Then
2
x; y kxk D kwk C k˛y k > k˛y ky k : Hence, hx ; y i 6 kx k ky k. Particularly, the above display holds with equality if and only if kwk D 0, if and only if w D 0 , if and only if x ˛ y D 0 , if and only if x D ˛ y . 2
2
2
h i k D 2
ˇ ˇ
Since
kx C y k2 D hx C y ; x C y i D kxk2 C ky k2 C 2 hx; y i 6 kxk2 C ky k2 C 2 kxk ky k D kxk C ky k 2 ;
1
2
CHAPTER 1
FUNCTIONS ON EUCLIDEAN SPACE
equality holds precisely when x ; y multiple of the other.
i D kxkjjy jj, i.e., when one is a nonnegative ut I Exercise 3 (1-3). Prove that kx y k 6 kx k C ky k. When does equality hold? Proof. By Theorem 1-1 (3) we have kx y k D k x C .y /k 6 k x k C ky k D kxk Cky k. The equality holds precisely when one vector is a non-positive multiple of the other. ut I Exercise 4 (1-4). Prove that kx k ky k 6 kx y k. 2 n 2 Proof. We have k x y k D D kxk2 C ky k2 2 niD1 xi yi > i D1 xi yi kxk2 C ky k2 2 kxk ky k D kxk ky k 2. Taking the square root of both sides gives the result. ut I Exercise 5 (1-5). The quantity ky x k is called the distance between x and y . Prove and interpret geometrically the “triangle inequality”: kz x k 6 kz y kC ky xk. h
ˇ ˇ P
P
Proof. The inequality follows from Theorem 1-1 (3):
kz xk D k.z y / C .y x/k 6 kz y k C ky xk : Geometrically, if x , y , and z are the vertices of a triangle, then the inequality says that the length of a side is no larger than the sum of the lengths of the other two sides.
ut
I Exercise 6 (1-6). If f and g be integrable on Œa;b. 1 2
ˇˇR ˇˇ R R
a. Prove that
b a
f g 6
b. If equality holds, must f ous?
b a
f 2
b a
g2
1 2
.
D g for some 2 R ? What if f and g are continu-
c. Show that Theorem 1-1 (2) is a special case of (a). Proof.
a. Theorem 1-1 (2) implies the inequality of Riemann sums:
ˇˇX ˇ i
ˇˇ 0X ˇ @
f .xi / g . xi / xi 6
1=2
f .xi /2 xi
i
1 0X A @
1=2
g .xi /2 xi
i
1 A
:
Taking the limit as the mesh approaches 0, one gets the desired inequality. b. No. We could, for example, vary f at discrete points without changing the values of the integrals. If f and g are continuous, then the assertion R, there is an x is true. In fact, suppose that for each Œa;b with
2
2
SECTION 1.1
3
NORM AND INNER PRODUCT
g .x/ 2 > 0. Then the inequality holds true in an open neighbor2 b hood of x since f and g are continuous. So a f g > 0 since the in-
f .x/
R R
tegrand is always non-negative and is positive on some subinterval of Œa; b. b b b Expanding out gives a f 2 2 a f g 2 a g 2 > 0 for all . Since the quadratic has no solutions, it must be that its discriminant is negative.
R
R
C
c. Let a 0, b n, f .x/ xi and g .x/ yi for all x Œi 1; i / for i 1 ; : : : ; n. Then part (a) gives the inequality of Theorem 1-1 (2). Note, however, that the equality condition does not follow from (a).
D
D
D
D
2
D
ut
I Exercise 7 (1-7). A linear transformation M Rn Rn is called norm pre- x , and inner product preserving if M x ; M y serving if M x x; y .
W
k k D k k
! h
iDh
i
a. Prove that M is norm preserving if and only if M is inner product preserving. b. Prove that such a linear transformation M is 1-1 and M 1 is of the same sort. Proof.
(a) If M is norm preserving, then the polarization identity together with the linearity of M give: 2 2 hM x; M y i D kM x C M y k 4 kM x M y k M .x C y /k2 kM .x y /k2 k D
4 x
2
2 D kx C y k 4 k y k D hx; y i :
If M is inner product preserving, then one has by Theorem 1-1 (4):
p
p
kM xk D hM x; M xi D hx; xi D kxk : (b) Take any M x; M y 2 Rn with M x D M y . Then M x M y D 0 and so 0 D hM x M y ; M x M y i D hx y ; x y i I but the above equality forces x D y ; that is, M is 1-1. Since M 2 L.Rn / and M is injective, it is invertible; see Axler (1997, Theorem 3.21). Hence, M1 2 L.Rn / exists. For every x ; y 2 Rn , we have
kD ED
kM1 x and
D
M1 x ; M1 y
1
M M
x
D k k D h x ;
M M1 x ; M M1 y
x; y :
i
Therefore, M1 is also norm preserving and inner product preserving.
ut
4
CHAPTER 1
I Exercise 8 (1-8). If x ; y denoted
FUNCTIONS ON EUCLIDEAN SPACE
Rn are non-zero, the angle between x and y ,
2
ı
† .x; y /, is defined as arccos hx; y i kxk ky k
, which makes sense by
Theorem 1-1 (2). The linear transformation T is angle preserving if T is 1-1, and for x ; y .x ; y /. 0 we have .Tx ; Ty /
¤
†
D†
a. Prove that if T is norm preserving, then T is angle preserving. b. If there is a basis . x1 ; : : : ; xn / of Rn and numbers 1 ; : : : ; n such that T xi i xi , prove that T is angle preserving if and only if all i are equal.
j j
c. What are all angle preserving T Rn
W
D
! Rn?
Proof.
(a) If T is norm preserving, then T is inner product preserving by the previous exercise. Hence, for x ; y 0,
¤
Tx; Ty Ty Tx
h
i
† .Tx; Ty / D arccos k k k k
D
arccos
h
x; y
i
kxk ky k
D†
.x ; y / :
(b) We first suppose that T is angle preserving. Since . x1 ; : : : ; xn / is a basis of Rn , all x i ’s are nonzero. Since
†
Txi ; Txj
D
arccos
˝ ˛! D k k Txi ; Txj
Txj
Txi
˝ ˛! k k ˝ ˛ ! ˇˇ D j j k kk k D † ˇˇ j arccos
arccos
i xi ; j xj
i xi
j xj
i j xi ; xj xj xi j
i
xi ; xj ;
it must be the case that i j
Dji
Then i and j have the same signs.
j :
I Exercise 9 (1-9). If 0 6 < , let T R2
W
A
!
cos sin : sin cos
D
Show that T is angle preserving and if x
¤ 0, then † .x; Tx/ D .
R2 , we have
2 ! ! C D D C D C D I
Proof. For every x; y
T x; y Therefore,
! R2 have the matrix
cos sin sin cos
T x; y
2
x2
x y
y2
x cos x sin
x; y
!
y sin : y cos
2
that is, T is norm preserving. Then it is angle preserving by Exercise 8 (a).
ut
SECTION 1.1
5
NORM AND INNER PRODUCT
Let x
D .a;b/ ¤ 0. We first have hx; Txi D a .a cos C b sin / C b .a sin C b cos / D
Hence, x ; Tx x Tx
h
a2
D
i
a2
C b2
cos :
!
C b2 cos D : a2 C b 2
† .x; Tx/ D arccos k k k k arccos ut I Exercise 10 (1-10 ). If M W Rm ! Rn is a linear transformation, show that there is a number M such that kM hk 6 M khk for h 2 Rm . Proof. Let M’s matrix be
A
D
a11 :: : an1
::
˘ ˘ ˝ ˛˘
a1m :: : anm
:
Then Mh
´
a1 ; h :: : n a ;h
D Ah D
a1 :: : : an
h
;
i
and so n
2
0X 1 kk k D@ k k Ak k 0X 1 B@ k kCA k k
n
D E X X k k D
kM h
2
i
a ;h
6
iD1
kM hk 6 n
Let M
D
i D1
h
a
n
2
ai
iD1
that is,
pX
i
p
2
h
n
ai
h :
i D1
ut
I Exercise 11 (1-11). If x ; y
Rn and z; w
hx; y i C hz; wi and k.x; z/
x
2 q kD k k Ck k
Proof. We have . x; z/ ; .y ; w/
z
2
.
2 Rm, show that h.x; z/ ; .y ; w/i D
2 RnCm. Then n
X iD
h.x; z/ ; .y ; w/
;
i D1
kai k and we get the result. 2
2
iD1
m
X C
xi yi
j D1
zj wj
D hx; y i C hz; wi ;
and
k.x; z/k2 D h.x; z/ ; .x; z/i D hx; xi C hz; zi D kxk2 C kzk2 :
ut
6
CHAPTER 1
FUNCTIONS ON EUCLIDEAN SPACE
I Exercise 12 (1-12 ). Let .Rn / denote the dual space of the vector space Rn . If .Rn / by M x 'x . x Rn , define ' x .Rn / by 'x .y / x ; y . Define M Rn Show that M is a 1-1 linear transformation and conclude that every ' .Rn / is 'x for a unique x Rn .
2
2
Dh
W
i
!
2
2
D
2 Rn and a; b 2 R. Then M .ax C b y / D 'a Cb D a' C b' D a M x C b M y ; where the second equality holds since for every z 2 Rn , 'a Cb . z/ D hax C b y ; zi D a hx; zi C b hy ; zi D a' .z/ C b' . z/ : To see M is 1-1, we need only to show that ı M D f0g, where ı M is the null set of M. But this is clear and so M is 1-1. Since dim .Rn / D dim Rn , M is also onto. This proves the last claim. ut I Exercise 13 (1-13 ). If x ; y 2 Rn , then x and y are called perpendicular (or orthogonal) if hx ; y i D 0 . If x and y are perpendicular, prove that kx C y k2 D kxk2 C ky k2. Proof. If hx ; y i D 0, we have kx C y k2 D hx C y ; x C y i D kxk2 C 2 hx; y i C ky k2 D kxk2 C ky k2 : ut Proof. We first show M is linear. Take any x ; y x
x
x
y
y
x
y
y
1.2 Subsets of Euclidean Space I Exercise 14 (1-14 ). Simple. Omitted.
˚ 2
I Exercise 15 (1-15) . Prove that x
Rn
W kx ak < r is open. µ B.aI r/, let " D r ka; y k. We show
2 x 2 Rn W kx ak < r that B .y I "/ B.aI r/. Take any z 2 B.y I "/. Then ka; zk 6 ka; y k C ky ; zk < ka; y k C " D r: Proof. For any y
˚
I Exercise 16 (1-16) . Simple. Omitted.
I Exercise 17 (1-17) . Omitted.
ut
SECTION 1.2
7
SUBSETS OF EUCLIDEAN SPACE
I Exercise 18 (1-18) . If A Œ0; 1 is the union of open intervals .ai ; bi / such that each rational number in .0;1/ is contained in some .a i ; bi /, show that @A Œ0; 1 A.
D
X
S
Proof. Let X
´ Œ0;1. Obviously, A is open since A D i .ai ; bi /. Then X X A is closed in X and so X X A D X X A. Since @A D Ax \ X X A D Ax \ .X X A/, it suffices to show that x X X A A: (1.1) But (1.1) holds if and only if Ax D X . Now take any x 2 X and any open nhood U of x in X . Since Q is dense, there exists y 2 U . Since there exists some i such that y 2 .ai ; bi /, we know that U \ .ai ; bi / ¤ ¿ , which means that U \ A ¤ ¿ , x Hence, X D A,x i.e., A is dense in X . ut which means that x 2 A. I Exercise 19 (1-19 ). If A is a closed set that contains every rational number r Œ0; 1, show that Œ0; 1 A.
2
Proof. Take any r
2 .0; 1/ and any open interval r 2 I .0;1/. Then there exists q 2 Q \ .0; 1/ such that q 2 I . Since q 2 A, we know that r 2 Ax D A. Since ut 0; 1 2 A, the claim holds. I Exercise 20 (1-20). Prove the converse of Corollary 1-7: A compact subset of Rn is closed and bounded. Proof. To show A is closed, we prove that Ac is open. Assume that x
D ˚ 2
n
… A, 2 A, then x ¤ y ;
x y > 1=m , m and let Gm 1; 2 ; : : :. If y y R Gm (see Figure 1.1). Thus, hence, x y > 1=m for some m; therefore y 1 A mD1 Gm , and by compactness we have a finite subcovering. Now observe G2 that the Gm for an increasing sequence of sets: G1 ; therefore, a finite union of some of the G m is equal to the set with the highest index. Thus, K Gs for some s , and it follows that B .x 1=s/ Ac . Therefore, A c is open.
k S
k
W k k
D
2
I
A
1=m x
Figure 1 .1. A compact set is closed
Let A be compact. We first show that A is bounded. Let
8
CHAPTER 1
FUNCTIONS ON EUCLIDEAN SPACE
D .i;i/n W i 2 N
O
˚
˚
be an open cover of A. Then there is a finite subcover . i1 ; i1 /n ; : : : ; . im ; im /n of A. Let i 0 max i1 ; : : : ; im . Hence, A i 0 ; i 0 ; that is, A is bounded.
D
f
g
I Exercise 21 (1-21 ).
ut
a. If A is closed and x y x > d for all y
… A, prove that there is a number d > 0 such that 2 A. b. If A is closed, B is compact, and A \ B D ¿, prove that there is d > 0 such that ky x k > d for all y 2 A and x 2 B . k k
c. Give a counterexample in R 2 if A and B are closed but neither is compact. Proof.
(a) A is closed implies that A c is open. Since x A c , there exists an open ball B.x d / with d > 0 such that x B .x d / Ac . Then y x > d for all y A.
2
2 k k (b) For every x 2 B, there exists d x > 0 such that x 2 B .x I d x =2/ Ac and ky xk > d x for all y 2 A. Then the family fB.xI d x =2/ W x 2 B g is an open cover of B. Since B is compact, there is a finite set fx1 ; : : : ; xn g such that B .x1 I d x =2/;:::; B.xn I d x =2/ covers B as well. Now let I
˚
2
n
1
D min
d
Then for any x d i . Hence,
I
˚
d x1 = 2 ; : : : ; dx n =2
.
2:
2 B, there is an open ball B.xi I xi =2/ containing x and ky xi k >
ky xk > ky xi k kxi xk > d i d i =2 D d i =2 > d: (c) See Figure 1.2.
0
Figure 1 .2.
t u
SECTION 1.3
9
FUNCTIONS AND CONTINUITY
I Exercise 22 (1-22 ). If U is open and C U is compact, show that there is a compact set D such that C D B and D U .
Proof.
ut
1.3 Functions and Continuity Rm and a A , show that lim x!a f .x/ I Exercise 23 (1-23) . If f A and only if lim x!a f i .x/ b i for i 1 ; : : : ; m.
W ! 2 D b if D D Proof. Let f W A ! Rm and a 2 A. p If: Assume that limx!a f i .x/ D b i for i D 1; : : : ; m. Then for every "= m > 0, p there is a number ıi > 0 such that f i .x/ b i < "= m for all x 2 A which satisfy 0 < kx ak < ıi , for every i D 1 ; : : : ; m. Put ı D min fı1 ; : : : ; ım g : Then for all x 2 A satisfying 0 < kx ak < ı, " f i .x/ b i < p ; i D 1 ; : : : ; m : m
Therefore, for every x
2 A which satisfy 0 < kx ak < ı,
pX m
kf .x/ bk D that is, limx!a f .x/
D b.
f i .x/
bi
i D1
2
p
m
X D I <
"2 =m
"
iD1
Only if: Now suppose that limx!a f .x/ b. Then for every number " > 0 A which satisfy there is a number ı > 0 such that f .x/ b < " for all x 0 < x a < ı. But then for every i 1 ; : : : ; m,
D k
k D
k k
f i .x/
i.e. limx!a f i .x/
D bi .
bi
6 f .x/
k
2
bk < "; ut
W ! Rm is continuous at a if and only if
I Exercise 24 (1-24). Prove that f A each f i is.
Proof. By definition, f is continuous at a if and only if limx!a f .x/
D
f .a/; it follows from Exercise 23 that limx!a f .x/ f .a/ if and only if i i f .a/ for every i 1 ; : : : ; m; that is, if and only if f i is continlimx!a f .x/ uous at a for each i 1 ; : : : ; m.
D
D
D
D
ut
I Exercise 25 (1-25). Prove that a linear transformation T Rn tinuous.
W
! Rm is con-
10
CHAPTER 1
Proof. Take any a
that
FUNCTIONS ON EUCLIDEAN SPACE
2 Rn. Then, by Exercise 10 (1-10), there exists M
> 0 such
Tx
Ta D T .x a/ 6 M kx ak : Hence, for every " > 0, let ı D "=M . Then Tx T a < " when x 2 Rn and 0 < kx ak < ı D "=M ; that is, lim ! Tx D Ta, and so T is continuous. ut I Exercise 26 (1-26) . Let A D x; y 2 R2 W x > 0 and 0 < y < x 2 . x
a
n
o
a. Show that every straight line through .0;0/ contains an interval around .0;0/ which is in R 2 A.
X b. Define f W R2 ! R by f .x / D 0 if x … A and f .x / D 1 if x 2 A. For h 2 R2 define g W R ! R by g .t / D f .t h/. Show that each g is continuous at 0, but h
h
h
f is not continuous at .0;0/.
Proof.
(a) Let the line through .0;0/ be y ax. If a 6 0, then the whole line is in 2 2 R A. If a > 0, then ax intersects x at a; a2 and .0; 0/ and nowhere else; see Figure 1.3.
D
X
y
2
x
D
y
y
D
a x
A
x
0
Figure 1 .3.
(b) We first show that f is not continuous at 0 . Clearly, f .0/ 0 since 0 A. 1. For every ı > 0, there exists x A satisfying 0 < x < ı , but f .x / f .0/ 2 f .t h/ is continuous at 0 for every h R . If h We next show gh .t / 0, then g 0 .t / f .0/ 0 and so is continuous. So we now assume that h 0 . It is clear that gh .0/ f .0/ 0:
D
D
D
2
kk
D
The result is now from (a) immediately.
D
… jD j 2 D ¤
D
ut
SECTION 1.3
11
FUNCTIONS AND CONTINUITY
x I Exercise 27 (1-27) . Prove that x Rn the function f Rn x a . R with f .x /
W
˚ 2
W k ak < r Dk k
!
is open by considering
Proof. We first show that f is continuous. Take a point b
let ı
2 Rn. For any " > 0,
D ". Then for every x satisfying kx bk < ı, we have jf .x/ f .b/j D kx ak kb ak 6 kx ak kb ak 6 kx bk < ı D ": ut Hence, x 2 Rn W kx ak < r D f 1 .1; r/ is open in R n . I Exercise 28 (1-28) . If A Rn is not closed, show that there is a continuous function f W A ! R which is unbounded. Proof. Take any x 2 @A. Let f .y / D 1= ky xk for all y 2 A. ut
˚
ˇ
ˇ
I Exercise 29 (1-29) . Simple. Omitted.
I Exercise 30 (1-30). Let f Œa;b R be an increasing function. If x1 ; : : : ; xn n Œa;b are distinct, show that i D1 o f; xi < f .b/ f .a/.
W
Proof.
P !
2 ut
2 DIFFERENTIATION
2.1 Basic Definitions I Exercise 31 (2-1 ). Prove that if f Rn it is continuous at a .
W
! Rm is differentiable at a 2 Rn, then
2 Rn ; then there exists a linear map W Rn ! kf .a C h/ f .a/ .h/k D 0; lim ! khk
Proof. Let f be differentiable at a Rm such that
h
or equivalently,
0
f .a
C h/ f .a/ D .h/ C r.h/;
(2.1)
where the remainder r .h/ satisfies lim r.h/
h!0
ı
k khk D 0:
k
(2.2)
Let h 0 in (2.1). The error term r .h/ 0 by (2.2); the linear term .h/ aslo n tends to 0 because if h i D1 hi ei , where . e1 ; : : : ; en / is the standard basis of n n R , then by linearity we have . h/ i D1 hi .ei /, and each term on the right tends to 0 as h 0. Hence,
!
!
!
D P
D
lim f .a
h!0
that is, limh!0 f .a
P
C h/ f .a/ D 0I
C h/ D f .a/. Thus, f is continuous at a. ut I Exercise 32 (2-2). A function f W R2 ! R is independent of the second vari- able if for each x 2 R we have f .x;y1 / D f.x;y2 / for all y 1 ; y2 2 R. Show that f is independent of the second variable if and only if there is a function g W R ! R such that f .x;y/ D g.x/. What is f 0 .a;b/ in terms of g 0 ? Proof. The first assertion is trivial: if f is independent of the second variable,
we can let g be defined by g.x/ f.x;0/. Conversely, if f .x;y/ f.x;y1 / g.x/ f.x;y2 /. If f is independent of the second variable, then
D
D
D
D g.x/, then
13
14
CHAPTER 2
lim
.h;k/!0
ˇ
f .a
C h; b C k/ f.a;b/ g0.a/h D lim k.h;k/k .h;k/!
C h/ g.a/ g0.a/h k.h;k/k g.a C h/ g.a/ g 0 .a/h 6 lim jhj h!0 D 0I
ˇ
hence, f 0 .a;b/
DIFFERENTIATION
ˇ
0
ˇ
g.a
ˇ
D .g0.a/; 0/.
ˇ ut
R is independent of the I Exercise 33 (2-3). Define when a function f R2 0 first variable and find f .a;b/ for such f . Which functions are independent of the first variable and also of the second variable?
W
!
Proof. We have f 0 .a;b/
D .0; g0.b// with a similar argument as in Exercise 32.
If f is independent of the first and second variable, then for any .x1 ; y1 /, R2 , we have f .x1 ; y1 / .x2 ; y2 / f .x2 ; y1 / f .x2 ; y2 /; that is, f is constant.
2
D
D
ut
I Exercise 34 (2-4). Let g be a continuous real-valued function on the unit circle x R2 x 1 such that g.0; 1/ g.1; 0/ 0 and g. x / g.x /. 2 R by Define f R
˚ 2
W k kD !
W
f .x / R2 and h R a. If x tiable.
D
˚
D kxk g 0
D
ık k x
x
D
if x
¤ 0; if x D 0:
W ! R is defined by h.t/ D f .t x/, show that h is differen-
2
b. Show that f is not differentiable at .0;0/ unless g
D 0. Proof. (a) If x D 0 or t D 0, then h.t/ D f .0/ D 0; if x ¤ 0 and t > 0, D f .t x/ D t kxk g t kt xxk finally, if x ¤ 0 and t < 0,
D hk k k ki D
h.t/
h.t/
D f .t x/ D t kxk g
x
tx t x
g x= x
t
f .x /t
I
D k k k k kk h i D k k kk t x x
g
x= x
g x= x
t
D f .x/t:
Therefore, h.t/ f .x /t for every given x R2 , and so is differentiable: Dh h. 0 and g. x / g.x /, we have g. 1;0/ g. .1;0// (b) Since g.1; 0/ g.1; 0/ 0. If f is differentiable at .0;0/, there exists a matrix .a; b/ such that Df.0;0/.h;k/ ah bk. First consider any sequence .h; 0/ .0;0/. Then
D
D
D
D C
D
2
D
!
D D
SECTION 2.1
15
BASIC DEFINITIONS
0
jf.h;0/ f.0;0/ ahj D lim D h!0 lim jhj h!0
ˇˇ j j j j ˇˇ ˇ j j ˙j j ˇ h
g h= h ; 0
ah
h
h g . 1;0/ ah lim D h!0 jhj Djaj implies that a D 0. Next let us consider .0; k/ ! .0; 0/. Then jf.0;k/ f.0;0/ bk j D lim jkj g 0;k=jkj bk Djbj 0 D lim jkj jkj k!0 k!0 forces that b D 0. Therefore, f 0 .0; 0/ D .0; 0/ and D f.0;0/.x;y/ D 0. If g. x / ¤ 0,
ˇˇ
ˇˇ
then
jf .x/ f .0/ 0j D lim lim ! ! kxk
x
0
x
0
ˇˇ k k k kˇˇ ˇ ˇ D ˇ k k ˇ¤ kk
and so f is not differentiable. Of course, if g. x / 0, then f .x /
D
I Exercise 35 (2-5). Let f R2
W
f.x;y/
D
g x= x
lim g x = x
x
x!0
0;
D 0 and is differentiable.
ut
! R be defined by
q j j
x y 0
x
x2
C y2
if .x; y/
¤ 0; if .x; y/ D 0:
Show that f is a function of the kind considered in Exercise 34, so that f is not differentiable at .0; 0/. Proof. If .x; y/
¤ 0, we can rewrite f.x;y/ as x jy j x jy j x jy j : (2.3) D k.x;y/ D k f.x;y/ D .x;y/k k k.x;y/k k.x;y/k x2 C y2 If we let g W x 2 R2 W kx k D 1 ! R be defined as g.x;y/ D x jy j, then (2.3) can
˚
q
be rewritten as
f.x;y/
D k.x;y/k g..x; y/= k.x;y/k/:
It is easy to see that g.0; 1/
D g.1; 0/ D 0;
and g. x; y/
D xjyj D xjy j D f.x;y/I
that is, g satisfies all of the properties listed in Exercise 34. Since g.x / 0 0 or y 0, we know that f is not differentiable at 0 . A direct proof unless x can be found in Berkovitz (2002, Section 1.11).
D
¤
D
ut
I Exercise 36 (2-6) . Let f R2 f is not differentiable at .0;0/.
W
p
! R be defined by f .x;y/ D jxy j. Show that
16
CHAPTER 2
DIFFERENTIATION
Proof. It is clear that
jf.h;0/j D 0 D lim jf.0;k/j I jhj jkj h!0 k!0 lim
hence, if f is differentiable at .0; 0/, it must be that Df.0;0/.x;y/ 0 since k > 0, and take a derivative is unique if it exists. However, if we let h .0; 0/, we have sequence .h; h/
D
D
g! f.h;h/ f.0;0/ 0j h2 1 j D D p ¤ 0: lim lim k.h; h/k .h;h/!.0;0/ .h;h/!.0;0/ k.h; h/k 2 f
p
Therefore, f is not differentiable. I Exercise 37 (2-7). Let f R2 that f is differentiable at 0 .
W
ut
! R be a function such that jf .x/j 6 kxk2. Show
j k k2 D 0 implies that f .0/ D 0. Since jf .x/ f .0/j D lim jf .x/j 6 lim kxk D 0; lim ! ! kxk kxk ! Df .0/.x;y/ D 0. ut I Exercise 38 (2-8). Let f W R ! R2 . Prove that f is differentiable at a 2 R if Proof. f .0/ 6 0
j
x
0
x
0
x
0
and only if f 1 and f 2 are, and that in this case
f 0 .a/
D
!
.f 1 /0 .a/ : .f 2 /0 .a/
Proof. Suppose that f is differentiable at a with f 0 .a/
D
!
c1 . Then for i c2
D
1; 2,
ˇˇ
f i .a
ˇˇ
C h/ f i .a/ c i h jhj
kf .a C h/ f.a/ Df .a/.h/k D 0 jhj h!0 h!0 implies that f i is differentiable at a with .f i /0 .a/ D c i . 0 6 lim
6 lim
Now suppose that both f 1 and f 2 are differentiable at a, then by Exercise 1,
X ˇˇ 2
06
kf .a C h/ f.a/ Df .a/.h/k 6 jhj i D1
f i .a
implies that f is differentiable at a with f 0 .a/ I Exercise 39 (2-9). Two functions f; g R if
D
ˇˇ
C h/ f i .a/ .f i /0.a/ h jhj
!
.f 1 /0 .a/ . .f 2 /0 .a/
ut
W ! R are equal up to n-th order at a
SECTION 2.1
17
BASIC DEFINITIONS
lim
f .a
C h/ g.a C h/ D 0: hn
h!0
a. Show that f is differentiable at a if and only if there is a function g of the form g.x/ a0 a1 .x a/ such that f and g are equal up to first order at a .
D C
.a/ exist, show that f and the function g defined by b. If f 0 . a / ; : : : ; f .n/ n
g.x/
f .i / .a/ .x iŠ
X D iD0
a/i
are equal up to n -th order at a . Proof. (a) If f is differentiable at a, then by definition,
lim
f .a
h!0
C h/
f.a/ h
C f 0 .a/ h D 0;
so we can let g.x/ f.a/ f 0 .a/ .x a/. On the other hand, if there exists a function g.x/
D
C
D a0 C a1.x a/ such that f .a C h/ g.a C h/ D lim f .a C h/ a0 a1h D 0; lim h
h!0
h
h!0
f.a/, and so f is differentiable at a with f 0 .a/ then a 0 (b) By Taylor’s Theorem1 we rewrite f as
D
n1
f.x/
X D iD0
f .i / .a/ .x iŠ
i
a/ C
f .n/ .y/ .x nŠ
D a1 .
a/n;
where y is between a and x . Thus, f.x/ g.x/ lim x!a .x a/n
f .n/ .y/ .x nŠ
.n/
a/n f nŠ.a/ .x a/n D x!a lim .x a/n .n/ .n/ D lim f .y/ f .x/ nŠ
x!a
D 0:
ut
1
(Rudin, 1976, Theorem 5.15) Suppose f is a real function on Œa;b, n is a positive integer, is continuous on Œa;b, f .n/ exists for every t 2 .a;b/. Let ˛, ˇ be distinct points of f Œa;b, and define .n 1/
n1
P.t/ D
X
k D0
f .k/ .˛/ .t ˛/ k : kŠ
Then there exists a point x between ˛ and ˇ such that f.ˇ/ D P.ˇ/ C
f .n/ .x/ .ˇ ˛/n : nŠ
18
CHAPTER 2
DIFFERENTIATION
2.2 Basic Theorems I Exercise 40 (2-10) . Use the theorems of this section to find f 0 for the follow- ing:
D xy . b. f.x;y;z/ D .x y ; z/ . c. f.x;y/ D sin.x sin y/ . d. f.x;y;z/ D sin x sin.y sin z/ . e. f.x;y;z/ D x y . f. f.x;y;z/ D x yCz . g. f.x;y;z/ D .x C y/ z . h. f.x;y/ D sin.xy/. cos 3 . i. f.x;y/ D sin.xy/ j. f.x;y/ D sin.xy/; sin x sin y ; x y a. f.x;y;z/
z
.
Solution. Compare this with Exercise 47.
xy e ln x (a) We have f.x;y;z/ follows from the Chain Rule that
D
f 0 .a;b;c/
D
y
D ey ln x D exp B. 2 ln 1/.x;y;z/. It 0
h i h i h i
D exp0 . 2 ln 1/.a;b;c/ 2 ln 1 .a;b;c/ D exp.b ln a/ .ln 1/. 2/0 C 2.ln 1/0 .a;b;c/ D ab 0; ln a; 0 C b=a;0;0 D ab1b ab ln a 0 :
(b) By (a) and Theorem 2-3(3), we have f 0 .a;b;c/ (c) We have f .x;y/ f 0 .a;b/
(d) Let g.y; z/
D
ab1 b 0
ab ln a 0
!
0 : 1
D sin B. 1 sin. 2//. Then, by the chain rule,
h
ih
i
0
D sin0 . 1 sin. 2//.a; b/ 1 sin. 2/ .a;b/ D cos .a sin b/ .sin 2/. 1/0 C 1.sin 2/0 .a;b/ D cos .a sin b/ sin b .1; 0/ C a .0; cos b/ D cos .a sin b/ sin b a cos .a sin b/ cos b :
h
D sin.y sin z/. Then
i
SECTION 2.2
19
BASIC THEOREMS
D sin x g.y;z/ D sin. 1 g. 2; 3//:
f.x;y;z/ Hence, f 0 .a;b;c/
D sin0 D cos D cos
It follows from (c) that 0
2
3
g . ; /.a;b;c/ Therefore,
. 1 g. 2 ; 3 //0 .a;b;c/
h h
ag .b;c/
ag .b;c/ ag .b;c/
D
i
g.b;c/. 1 /0
C ag0. 2; 3/ .a;b;c/ g.b;c/;0;0 C ag0 . 2 ; 3 /.a;b;c/
:
b cos .b sin c/ cos c :
cos .b sin c/ sin c
0
i
f 0 .a;b;c/
D cos
a sin .b sin c/
sin .b sin c/
a cos .b sin c/ sin c
D xy . Then f.x;y;z/ D x g.y;z/ D g
(e) Let g.x; y/
Then Df.a;b;c/
D Dg
By (a),
x;g.y;z/
a;g.b;c/
Dg a;g.b;c/ .x;y;z/
D
a
Bh
g.b;c/
D
ab cos .b sin c/ cos c :
1
2
i
D 1 ; Dg. 2 ; 3 / .a;b;c/:
a
g.b;c/
c
D
g ;g. ; / :
g . b; c / =a
ab b c x a
3
C
c
ab ln a y;
D 1 .a;b;c/.x;y;z/
ln a
0
0B@ 1CA x y z
D x;
and Dg. 2 ; 3 /.a;b;c/.x;y;z/
B C
Dg.b;c/
D c D bbc y
Hence,
D 2 ; D 3 .a;b;c/.x;y;z/
b c ln b z:
c
Df.a;b;c/.x;y;z/
and
D
ab b c x a
C
a
bc
ln a
bc c y b
C
b c ln b z ;
20
CHAPTER 2
f 0 .a;b;c/ (f) Let g.x;y/ Hence,
D D C D C D C B D C B C C C D C D
D
ı bc c
a b
bc c
a
Df.a;b;c/.x;y;z/
and
ı
x yCz
1
g x; y
Dg.a;b
c/
D ; D
Dg.a;b
c/
x; y
abCc .b a
c/
2
abCc ln a
Hence,
h
and f .a;b;c/
D
.aCb/c c .aCb/
.aCb/c c .aCb/
D sin.xy/ D sin B
f 0 .a;b/
abCc ln a :
1
C 2; 3
:
i
h
.a
C b/
c
ln .a
C b/
1 2 . Hence,
:
i
D .sin/0 .ab/ b. 1/0.a;b/ C a. 2/0.a;b/ D cos .ab/ b .1;0/ C a.0; 1/ D cos .ab/ .b;a/ D b cos .ab / a cos .ab/ :
(i) Straightforward. (j) By Theorem 2-3 (3), we have 0
0 Bh i D@ 0 D B@
sin.xy/ .a;b;c/
f 0 .a;b;c/
z ;
D Dg .a C b;c/ B D 1 C D 2; D 3 .a;b;c/.x;y;z/ D Dg .a C b;c/ B x C y; z c D .a.aCCb/b/ c .x C y/ C .a C b/c ln .a C b/ z;
Df.a;b;c/.x;y;z/
(h) We have f .x;y/
.a;b;c/.x;y;z/
y
3
C 3 .
z
D xy . Then f.x;y;z/ D .x C y/ z D g x C y; z D g
(g) Let g.x; y/
g 1; 2
z
D
abCc ln a
x
abCc .bCc/ a
f 0 .a;b;c/
c
ab b c ln a ln b :
a b c ln a b
D x y . Then f.x;y;z/
0
DIFFERENTIATION
sin x sin y
0
1 CA
.a;b;c/
Œx y 0 .a;b;c/
b cos .ab/ cos .a sin b/ sin b ab1 b
a cos .ab/ a cos .a sin b/ cos b ab ln a
I Exercise 41 (2-11) . Find f 0 for the following (where g R
1 CA
:
ut
W ! R is continuous):
SECTION 2.2
BASIC THEOREMS
a. f.x;y/
R R R D
D b. f.x;y/ D c. f.x;y;z/
x Cy a xy a
21
g.
g.
sin x sin.y sin z / xy
Solution. (a) Let h.t/
D
t g. a
R
f 0 .a;b/
(b) Let h.t/ Hence,
g.
D h B .1 C 2/
h
i
D h0 .a C b/ . 1 C 2/0.a;b/ D g .a C b/ .1; 1/ D g .a C b/ g .a C b/ :
t a g.
Then f .x;y/
xy a
D R
.x;y/, and so
h i D B i 1 2
D R
Then f.x;y/
f 0 .a;b/
D h0 .ab/ b . 1/0.a;b/ C a . 2/0.a;b/ D g .ab/ .b;a/ D b g . ab/ a g .ab/ :
g
h
sin.x sin.y sin z/ /
a
Z Z D C D g
xy
sin.x sin.y sin z/ /
Z D D R
g
a
f 0 .a;b;c/
.x;y/.
.x;y;z/ a
xy
g
a
Let .x;y;z/ sin x sin.y sin z/ , k.x;y; z/ Then f.x; y;z/ k.x;y;z/ h.x;y;z/, and so
D
h
(c) We can rewrite f.x; y;z/ as f.x;y;z/
D h.xy/
Z
g:
a
g, and h.x;y; z/
D
xy a
R
g.
D k0.a;b;c/ h0.a;b;c/:
It follows from Exercise 40 (d) that k 0 .a;b;c/
D k0
The other parts are easy.
.a;b;c/
0 .a;b;c/:
ut I Exercise 42 (2-12). A function f W Rn Rm ! R p is bilinear if for x ; x1 ; x2 2 Rn , y ; y1 ; y2 2 Rm , and a 2 R we have f .ax ; y / D af .x ; y / D f .x ; ay / ; f .x1 C x2 ; y / D f .x1 ; y / C f .x2 ; y / ; f .x ; y1 C y2 / D f .x ; y1 / C f .x ; y2 / : a. Prove that if f is bilinear, then lim
.h;k/!0
kf .h; k/k D 0: k.h; k/k
22
CHAPTER 2
b. Prove that D f .a; b/.x ; y /
DIFFERENTIATION
D f .a; y / C f .x; b/.
c. Show that the formula for D p.a; b/ in Theorem 2-3 is a special case of (b). m Proof. (a) Let e1n ; : : : ; enn and e1m ; : : : ; em be the stand bases for Rn and Rm ,
2 X D
respectively. Then for any x
Rn and y
2 Rm, we have
n
m
x i eni ;
x
and y
iD1
X D
j y j em :
j D1
Therefore,
0X X 1 X 0 X 1 AD @ A D @ XX D XX D P D X Xˇ ˇ kD "ˇ ˇ ˇˇ ˇˇ ˇˇ ˇˇ# n
f .x ; y /
m
n
x i eni ;
f
iD1
j y j em
m
x i eni ;
f
j D1
iD1
j y j em
j D1
n
m
j f x i eni ; y j em
iD1 j D1 n
m
j x i y j f eni ; em :
iD1 j D1
Then, by letting M
i;j
j f eni ; em
, we have
j x i y j f eni ; em
kf .x; y /
x i y j
6
i;j
j f eni ; em
i;j
6 M max i
xi
max j
k kky k :
6 M x
Hence, lim
.h;k/!0
kf .h; k/k 6 lim M khkkkk k.h; k/k . ; /! k.h; k/k M khkkkk D . lim ; /! 2 hi C k j hk
0
hk
0
pX
i;j
D . lim ; /! hk
Now
khkkkk 6
˚
khk2 kkk2
k kkkk 6 khk2 C kkk2, and so
Hence h
0
M h
q k kk kkC k kk h
2
k
k
if k 6 h
: 2
k k k k if k hk 6 kkk :
2
y j
SECTION 2.2
23
BASIC THEOREMS
M h
lim
.h;k/!0
(b) We have
k kkkk 6 lim khk2 C kkk2 . ; /!
q
hk
0
q k k C k k D
M
2
h
k
2
0:
kf .a C h; b C k/ f .a; b/ f .a; k/ f .h; b/k . ; /! k.h; k/k kf .a; b/ C f .a; k/ C f .h; b/ C f .h; k/ f .a; b/ f .a; k/ f .h; b/k D . lim k.h; k/k ; /! kf .h; k/k D . lim k.h; k/k ; /! D0 by (a); hence, D f .a; b/.x ; y / D f .a; y / C f .x ; b/. (c) It is easy to check that p W R2 ! R defined by p.x;y/ D xy is bilinear. lim
hk
0
hk
0
hk
0
Hence, by (b), we have
Dp.a;b/.x;y/
D p a; y C p.x;b/ D ay C xb: I Exercise 43 (2-13) . Define IP W Rn Rn ! R by IP.x ; y / D hx ; y i.
ut
a. Find D .IP/ .a; b/ and .IP/0 .a; b/. Rn are differentiable and h R b. If f; g R f.t/;g.t/ , show that
h
W
i
W
!
h0 .a/
D
D
!
ECD
f 0 .a/T ;g.a/
R is defined by h.t/
D
E
f.a/;g0 .a/T :
D
E
W ! Rn is differentiable and kf.t/k D 1 for all t , show that f 0 .t/T ;f.t/ D
c. If f R 0.
d. Exhibit a differentiable function f R by f .t/ f.t/ is not differentiable.
j j Dj
W ! R such that the function jf j defined
j
Proof. (a) It is evident that IP is bilinear; hence, by Exercise 42 (b), we have D .IP/ .a; b/.x ; y /
D IP .a; y / C IP .x; b/ D ha; y i C hx; bi D hb; xi C ha; y i ;
and so .IP/0 .a; b/ .b; a/. (b) Since h.t/ IP f; g .t/, by the chain rule, we have
D D B
Dh.a/ .x/
˛ ˝
D D .IP/ f .a/;g.a/ B Df.a/.x/; Dg.a/ .x/ D hg.a/; Df.a/.x/i C hf .a/; Dg.a/ .x/i D g.a/; f 0 .a/ x C f.a/;g0.a/ x: (c) Let h.t/ D hf.t/;f.t/i with kf.t/k D 1 for all t 2 R. Then
˝
˛
24
CHAPTER 2
h.t/
DIFFERENTIATION
D kf.t/k2 D 1
is constant, and so h 0 .a/
D 0; that is, 0 D f 0 .a/T ;f.a/ C f.a/;f 0 .a/T D 2 f 0 .a/T ;f.a/
D
D
ED
and so f 0 .a/T ;f.a/ (d) Let f .t/
E D
E D
E
;
0.
D t . Then f is linear and so is differentiable: Df D t . However, jt j D 1; lim jt j D 1I lim t!0C
t
t!0
t
that is, f is not differentiable at 0.
ut
j j
I Exercise 44 (2-14). Let Ei , i 1 ; : : : ; k be Euclidean spaces of various dimensions. A function f E1 Ek Rp is called multilinear if for Ej , j Rp defined by g. x / each choice of xj i the function g Ei f .x1 ; : : : ; xi 1 ; x ; xiC1 ; : : : ; xk / is a linear transformation.
D W ¤
2
a. If f is multilinear and i have lim
h!0
!
W
!
D
¤ j , show that for h D .h1; : : : ; hk /, with h` 2 E`, we
f a1 ; : : : ; hi ; : : : ; hj ; : : : ; ak
khk
D 0:
b. Prove that k
Df .a1 ; : : : ; ak / .x1 ; : : : ; xk /
D
X
f .a1 ; : : : ; ai 1 ; xi ; aiC1 ; : : : ; ak / :
iD1
Proof.
(a) To light notation, define
´
ai j
W ! Rp be defined as g
Let g Ei Ej and so
g aij ; hi ; hj
lim
h!0
by Exercise 42 (a).
a1 ; : : : ; ai1 ; ai C1 ; : : : ; aj 1 ; aj C1 ; : : : ; ak :
khk
D xi ; xj
D
f aij ; xi ; xj . Then g is bilinear
g ai j ; hi ; hj
6 lim
h!0
hi ; hj
(b) It follows from Exercise 42 (b) immediately.
0
ut
I Exercise 45 (2-15) . Regard an n n matrix as a point in the n -fold product Rn Rn by considering each row as a member of R n .
a. Prove that det Rn
W
Rn ! R is differentiable and
SECTION 2.2
25
BASIC THEOREMS
0 BB X B D B@ D 1 CC CC A
a1 :: : n det xi :: iD1 : an
D det .a1 ; : : : ; an / .x1 ; : : : ; xn /
b. If a ij R
W ! R are differentiable and f .t/ 0
f .t/
1 CC CC A
:
det aij .t/ , show that
0 BB X B D B@
a11 .t/ :: : n 0 det aj1 .t/ :: j D1 : an1.t/
::
:
::
:
a1n .t/ :: : 0 aj n .t/ : :: : ann .t/
¤
R are differentiable, let c. If det aij .t/ 0 for all t and b1 ; : : : ; bn R R be the functions such that s1 . t / ; : : : ; s n .t/ are the solutions s1 ; : : : ; sn R of the equations
W !
W !
n
X
aj i .t/sj .t/
j D1
D bi .t /; i D 1 ; : : : ; n :
Show that s i is differentiable and find s i0 .t/. Proof.
(a) It is easy to see that det Rn clusion follows from Exercise 44.
W
Rn ! R is multilinear; hence, the con-
(b) By (a) and the chain rule,
f 0 .t/
0
a10 . t / ; : : : ; a0n .t/
B 0 BB X B D B@ D
det
aij .t/
a11 .t/ :: : n 0 det aj1 .t/ :: j D1 : an1.t/
(c) Let
A
Then
0 B D@
a11 .t/ :: : a1n .t/
::
:
1 CA
an1.t/ :: ; : ann .t/
s
As
0 B D@
::
:
::
:
1 CC CC A
a1n .t/ :: : 0 aj n .t/ : :: : ann .t/
1 CA
s1 .t/ :: ; : sn .t/
D b;
and
b
0 B D@
1 CA
b1 .t/ :: : : bn .t/
26
CHAPTER 2
and so si .t/
DIFFERENTIATION
.Bi / D det ; det .A/
where Bi is obtained from A by replacing the i -th column with the b. It follows from (b) that s i .t/ is differentiable. Define f .t/ det .A/ and g i .t/ det .Bi /. Then a11 .t/ an1.t/ :: :: :: : : : n 0 0 0 f .t/ anj .t/ ; det a1j .t/ :: :: j D1 :: : : : a1n .t/ ann .t/
D
0 BB X BB D B@
and
0 BB X D B BB@
a11 .t/ :: : 0 a1j .t/ :: : a1n .t/
n
gi0 .t/
j D1
Therefore,
::
:
::
:
si0 .t/
D
1 CC CC CA
ai 1;1.t/ b1 .t / ai C1;1 .t/ :: :: :: : : : 0 0 0 ai1;j .t / bj .t / ai C1;j .t/ :: :: :: : : : ai 1;n .t / bn .t / aiC1;n .t/ 0
0
::
:
::
:
1 CC CC CA
an1.t/ :: : 0 anj .t/ : :: : ann .t/
0
D f .t/gi .t/f 2.t/f.t/gi .t/ :
ut
I Exercise 46 (2-16) . Suppose f Rn
a differen- ! Rn is differentiable and has 1 0 tiable inverse f 1 W Rn ! Rn . Show that f 1 .a/ D f 0 f 1 .a/ .
W
B f 1.x/ D x. On the one hand D
Proof. We have f 1
f f
B
h i B B
is linear; on the other hand,
B D h i D
D f
Therefore, D f 1 .a/
f 1 .a/ .x /
Df f 1 .a/
Df f 1 .a/ 1
f f 1 .a/ .x /
D x since
Df 1 .a/ .x/:
.
2.3 Partial Derivatives I Exercise 47 (2-17) . Find the partial derivatives of the following functions:
D xy . b. f.x;y;z/ D z . c. f.x;y/ D sin.x sin y/ . d. f.x;y;z/ D sin x sin.y sin z/ . a. f.x;y;z/
ut
SECTION 2.3
27
PARTIAL DERIVATIVES z
D xy . f. f.x;y;z/ D x yCz . g. f.x;y;z/ D .x C y/ z . h. f.x;y/ D sin.xy/. cos 3 . i. f.x;y/ D sin.xy/ e. f.x;y;z/
Solution. Compare this with Exercise 40.
D yx y1, D 2f.x;y;z/ D xy ln x, and D 3f.x;y;z/ D 0. (b) D1 f.x;y;z/ D D2 f.x;y;z/ D 0, and D 3 f.x;y;z/ D 1. (c) D1 f.x;y/ D sin y cos x sin y , and D 2 f.x;y/ D x cos y cos x sin y . (d) D1 f.x;y;z/ D sin.y sin z/ cos x sin.y sin z/ , D2 f.x;y;z/ D cos x sin.y sin z/ x cos.y sin z/ sin z, and D3 f.x;y;z/ D cos x sin.y sin z/ x cos.y sin z/y cos z. (e) D1 f.x;y;z/ D y z x y 1 , D2 f.x;y;z/ D x y zy z1 ln x, and D3 f.x;y;z/ D (a) D1 f.x;y;z/
D C
z
z
z
y z ln y x y ln x . (f) D1 f.x;y;z/
z x yCz1 , and D 2 f.x;y;z/D3 f.x;y;z/
y
D D C y/z1, and D C C (h) D1 f.x;y/ D y cos.xy/, and D 2 f.x;y/ D x cos.xy/. cos 31 (i) D1 f.x;y/ D cos 3 sin.xy/ y cos.xy/, and cos 31 D2 f.x;y/ D cos 3 sin.xy/ x cos.xy/.
D xyCz ln x.
D2 f.x;y;z/ (g) D1 f.x;y;z/ z.x z D3 f.x;y;z/ .x y/ ln.x y/.
ut
I Exercise 48 (2-18) . Find the partial derivatives of the following functions R is continuous): (where g R
W ! x Cy a. f.x;y/ D a g. x b. f.x;y/ D y g . xy c. f.x;y/ D a g. R . g/ g. d. f.x;y/ D a Solution.
R R R R
y b
(a) D1 f.x;y/
D D2f.x;y/ D g.x C y/. (b) D1 f.x;y/ D g.x/, and D 2 f.x;y/ D g y . (c) D1 f.x;y/ D yg.xy/, and D 2 f.x;y/ D xg.xy/.
28
CHAPTER 2
(d) D1 f.x;y/
D 0, and D2f.x;y/ D g
I Exercise 49 (2-19) . If
f.x;y/
Dx
xx
DIFFERENTIATION
R y
g
y b
g .
ut
xy
0 C @ ln x
arctan arctan arctan sin cos xy
find D 2 f 1; y .
Solution. Putting x
0.
D 1 into f .x;y/, we get f
D 1; y
ln.x
C y/
!1A D
1. Then D2 f 1; y
ut
I Exercise 50 (2-20). Find the partial derivatives of f in terms of the deriva- tives of g and h if
D g.x/h y . b. f.x;y/ D g.x/h.y / . c. f.x;y/ D g.x/. d. f.x;y/ D g y . e. f.x;y/ D g.x C y/ . a. f.x;y/
Solution.
D g0 .x/h y , and D 2f.x;y/ D g.x/h0 y . (b) D1 f.x;y/ D h y g.x/h.y /1 g 0 .x/, and D 2 f.x;y/ D h0 y g.x/h.y / ln g.x/. (c) D1 f.x;y/ D g 0 .x/, and D 2 f.x;y/ D 0. (d) D1 f.x;y/ D 0, and D 2 f.x;y/ D g 0 y . (e) D1 f.x;y/ D D2 f.x;y/ D g0 .x C y/. ut I Exercise 51 (2-21 ). Let g 1 ; g2 W R2 ! R be continuous. Define f W R2 ! R by (a) D1 f.x;y/
x
f.x;y/
Z D
y
g1 .t;0/ dt
0
a. Show that D 2 f.x;y/
Z C
g2 .x;t / dt:
0
D g2.x;y/.
b. How should f be defined so that D 1 f.x;y/
D g1.x;y/? c. Find a function f W R2 ! R such that D1 f.x;y/ D x and D 2 f.x;y/ D y . Find one such that D 1 f.x;y/ D y and D 2 f.x;y/ D x . Proof.
SECTION 2.3
29
PARTIAL DERIVATIVES
(a) D2 f.x;y/
D 0 C g2.x;y/ D g2.x;y/.
(b) We should let x
f.x;y/
y
Z Z D C g1 t; y dt
0
where t
2 R is a constant.
g2 .a; t/ dt;
0
(c) Let
f.x;y/ D x2 C y 2 f.x;y/ D xy.
=2.
ut
R and D 2 f I Exercise 52 (2-22 ). If f R2 0, show that f is independent D2 f of the second variable. If D 1 f 0, show that f is constant.
W
!
D
D
D
2 R. By the mean-value theorem, for any y1; y2 2 R, there exists a point y 2 y1 ; y2 such that f x; y2 f x; y1 D D2 f x; y y2 y1 D 0: Hence, f x; y1 D f x; y2 ; that is, f is independent of y . Similarly, if D1 f D 0, then f is independent of x . The second claim is then ut proved immediately. I Exercise 53 (2-23 ). Let A D .x;y/ 2 R2 W x < 0 , or x > 0 and y ¤ 0 . a. If f W A ! R and D 1 f D D2 f D 0, show that f is constant. b. Find a function f W A ! R such that D 2 f D 0 but f is not independent of the Proof. Fix any x
˚
second variable.
Proof.
2 R2, we have f.a;b/ D f .1;b/ D f .1;d/ D f .c;d/:
(a) As in Figure 2.1, for any .a;b/;.c;d /
(b) For example, we can let
f.x;y/
D
˚ ˚
0
f.x;y/
a. Show that D 2 f .x;0/
D
tu
x otherwise.
I Exercise 54 (2-24) . Define f R2
W
if x < 0 or y < 0
! R by 2
2
y xy xx2 Cy 2 .x;y/
0
D x for all x and D 1f
¤ 0; .x; y/ D 0:
D 0; y
y for all y .
30
CHAPTER 2
DIFFERENTIATION
y .a;b/
. 1;b/
x
. 1;d/
.c;d/
Figure 2 .1. f is constant
b. Show that D 1;2f.0;0/
¤ D2;1 f.0;0/.
Proof.
(a) We have D2 f.x;y/
D
and D1 f.x;y/
Hence, D 2 f .x;0/
D
D x and D1f
(b) By (a), we have D1;2f.0;0/
D1 D2 .x;0/ .0/
D 1.
x .x 4 y 4 4x 2 y 2 /
2
.x2 Cy 2 /
.x;y/
¤ 0; .x; y/ D 0;
0 y .y 4 x 4 4x 2 y 2 /
2
.x2 Cy 2 /
.x;y/
¤ 0; .x; y/ D 0:
0
D D 0; y
y.
D2 D1 f 0; y
.0/
D 1; but D2;1 f.0;0/ D ut
I Exercise 55 (2-25 ). Define f R
W ! R by
f .x/
D
˚
e x
2
0
x
¤ 0 x D 0:
Show that f is a C 1 function, and f .i / .0/
D 0 for all i . Proof. Figure 2.2 depicts f .x/. We first show that f 2 C 1 .
Let pn y be a polynomial with degree n with respect to y. For x .k /
2 N, we show that f
k
.x/
D p3k
Step 1 Clearly, f 0 .x/ Step 2 Step 3
x
e
x 2
x 1 e x
¤ 0 and
. We do this by induction.
2
D 2x3ex . Suppose that f .k/ .x/ D p3k Then by the chain rule,
1
2
.
SECTION 2.3
31
PARTIAL DERIVATIVES
y
−2
−1
0
2 x
1
Figure 2 .2.
.kC1/
f
.x/
0
h i D C D D C D C D f .k / .x/
0 p3k x 1
x 2
0 p3k x 1
e x
x 2
q3kC1 x 1
2
p3k x 1
p3k x 1
q3kC3 x 1
p3.kC1/ x 1
e x
2
e x
2x 3
2x 3 e x
e x
2
2
2
;
where q 3kC1 and q 3kC3 are polynomials.
C 1 for all x 0. It remains to show that f .k / .x/ is Therefore, f .x/ defined and continuous at x 0 for all k.
2
¤
D
Step 1 Obviously, 0
f .0/
D x!0 lim
f .x/
2
f .0/ D lim ex D lim 2x3ex D 0 x!0 x!0 x x 2
by L’Hôpital’s rule. Step 2
Suppose that f .k/ .0/
D 0.
Step 3 Then,
f .k C1/ .0/
f .k / .x/
lim D x!0 D x!0 lim p3kC1
p3kC1 D x!0 lim ex
f .k / .0/ x
x 1 e x
x 1
2
2
:
C 1 times, we get f .k C1/ .0/ D 0. A similar computation shows that f .k/ .x/ is continuous at x D 0. Hence, if we use L’Hôpital’s rule 3k
I Exercise 56 (2-26 ). Let
f .x/
D
˚
2
e .x1/ 0
2
e.xC1/
x
2 .1;1/; x … .1;1/:
ut
32
CHAPTER 2
DIFFERENTIATION
R is a C 1 function which is positive on . 1;1/ and 0
a. Show that f R elsewhere.
W !
b. Show that there is a C 1 function g R and g.x/ 1 for x > ".
W ! Œ0;1 such that g.x/ D 0 for x 6 0
D
c. If a
2 Rn, define g W Rn ! R by g.x /
D f
!
x1
a1 f "
xn
an "
:
Show that g is a C 1 function which is positive on
a1
"; a1 C "
and zero elsewhere.
an
"; an C "
Rn is open and C A is compact, show that there is a non-negative d. If A 1 R such that f .x/ > 0 for x C function f A C and f 0 outside of some closed set contained in A .
W !
2
D
e. Show that we can choose such an f so that f A x C .
W ! Œ0;1 and f .x/ D 1 for
2
Proof.
(a) If x . 1; 1/, then x 1 0 and x 1 0. It follows from Exercise 55 that 2 2 e .x1/ C 1 and e .xC1/ C 1 . Then it is straightforward to check that f C 1 . See Figure 2.3
2
2 2
¤
C ¤
2
y
−1
1 x
0 Figure 2 .3.
(b) By letting z
D x C 1, we derive a new function j W R ! R from f as follows:
˚
2
e .z2/
ez
2
2 .0;2/; 0 z … .0;2/: By letting w D "z=2, we derive a function k W R ! R from j as follows: j .z/
k .w/
D
˚
D
2
e .2w="2/ 0
z
2
e.2w="/
w
2 .0;"/; w … .0;"/:
SECTION 2.3
33
PARTIAL DERIVATIVES
y
y j
0
k
2 x
1
x
0
Figure 2 .4.
It is easy to see that k let
2 C 1, which is positive on .0; "/ and 0 elsewhere. Now
,Z Z D x
g.x/
"
k .x/
k .x/ :
0
Then g
0
2 C 1; it is 0 for x 6 0, increasing on .0; "/, and 1 for x > ".
(c) It follows from (a) immediately. (d) For every x C , let Rx . ";"/n be a rectangle containing x, and Rx is A). contained in A (we can pick such a rectangle since A is open and C Then Rx x C is an open cover of C . Since C is compact, there exists x1 ; : : : ; xm C such that Rx1 ; : : : ; Rxm covers C . For every xi , i 1; : : : ; n, R as we define a function g i R xi
2
´
f W 2 g g
f
W
gi .x /
˚
!
D f
xi1
ai1 "
! f
D
xin
ain "
;
Rn is the middle point of R xi . where ai1 ; : : : ; ain R as follows: Finally, we define g R x1 Rxm
2
W
[ [
! m
g.x/
X D
gi .x/ :
i D1
Then g
2 C 1; it is positive on C , and 0 outside Rx [ [ Rx
m
1
.
(e) Follows the hints.
ut W x 2 R2 W kxk 6 1 ! R3 by
q D q D ˚˚ 2 W k k D B 2 W k k Wkk ´ ˚ 2 W k k D
I Exercise 57 (2-27) . Define g; h
˚
g.x;y/
x;y;
h.x;y/
x;y;
Show that the maximum of f on x f g or the maximum of f h on x
B
Proof. Let A
g .A/
[ h.A/.
´ x 2 R2
˚
x2
1
1
R3
R2
x 6 1 and B
y2 ;
x2
y2 :
x 1 is either the maximum of x 61 . x
R3
x
1 . Then B
D ut
34
CHAPTER 2
DIFFERENTIATION
2.4 Derivatives I Exercise 58 (2-28) . Find expressions for the partial derivatives of the follow- ing functions:
a. F.x;y/
C C C D D D C ´ D f
b. F.x;y;z/ c. F.x;y;z/ d. F.x;y/ Proof.
(a) Letting a
g.x/k y ;g.x/
f g.x
y/;h y
.
h y z
.
f x y ; y z ; z x .
f x;g.x/;h.x;y/ .
g.x/k y ;g.x/
h y , we have
D D1f .a/ g0 .x/ k y C D2f .a/ g0 .x/; D2 F.x;y/ D D1 f .a/ g.x/ k 0 y C D1 f .a/ h0 y : (b) Letting a ´ g.x C y/;h y C z , we have D1 F.x;y;z/ D D1 f .a/ g 0 .x C y/; D2 F.x;y;z/ D D1 f .a/ g 0 .x C y/ C D2 f .a/ h0 y C z ; D3 F.x;y;z/ D D2 f .a/ h0 y C z : (c) Letting a ´ x y ; y z ; z x , we have D1 F.x;y;z/ D D1 f .a/ yx y1 C D3 f .a/ z x ln z; D2 F.x;y;z/ D D1 f .a/ x y ln x C D2 f .a/ zy z1 ; z x1 D3 F.x;y;z/ D D2 f .a/ y ln y C D3 f .a/ xz : (d) Letting a ´ x;g.x/;h.x;y/, we have 0 D1 F.x;y/ D D1 f .a/ C D2 f .a/ g .x/ C D3 f .a/ D1 h.x;y/ D2 F.x;y/ D D3 f .a/ D2 h.x; y/: I Exercise 59 (2-29) . Let f W Rn ! R. For x 2 Rn , the limit f .a C t x / f .a/ lim ;
D1 F.x;y/
t!0
ut
t
if it exists, is denoted D x f .a/, and called the directional derivative of f at a , in the direction x .
a. Show that D ei f .a/ b. Show that D t x
D Di f .a/. f .a/ D t D f .a/. x
SECTION 2.4
35
DERIVATIVES
c. If f is differentiable at a, show that Dx f .a/ DxCy f .a/ Dx f .˛/ Dy f .a/.
D
C
D
Df .a/.x / and therefore
Proof.
(a) For e i
D . 0 ; : : : ; 0 ; 1 ; 0 ; : : : ; 0 /, we have f .a C t ei / f .a/ D f .a/ D lim ei
t !0
t f .a1 ; : : : ; ai 1 ; ai
C t; ai C1; : : : ; an/ f .a/
D lim t !0 D Di f .a/
t
by definition. (b) We have D t x f .a/
f .a C st x / f .a/ f .a C st x / f .a/ lim t D lim D D tD s!0 st !0 s st
x f .a/:
(c) If f is differentiable at a , then for any x
¤ 0 we have jf .a C t x/ f .a/ Df .a/.t x/j 0 D lim t! kt xk jf .a C t x/ f .a/ t Df .a/.x/j 1 D lim t !0 jt j kxk f .a C t x / f .a/ 1 Df .a/.x / D lim t !0 t kxk ; 0
ˇˇ
and so
Dx f .a/
The case of x
ˇˇ
f .a C t x / f .a/ D lim D Df .a/.x/: t !0 t
D 0 is trivial. Therefore, D C f .a/ D Df .a/ .x C y / D Df .a/.x/ C Df .a/ .y / D D f .a/ C D f .a/: x y
x
ut
y
I Exercise 60 (2-30) . Let f be defined as in Exercise 34. Show that Dx f.0;0/ Dx f.0;0/ Dy f.0;0/ for all exists for all x , but if g 0, then Dx Cy f.0;0/ x; y .
¤
Proof. Take any x
lim
t !0
¤
2 R2.
f .t x /
f.0;0/ D lim jt j kxk g t
t!0
.j j k k
Therefore, D x f.0;0/ exists for any x . Now let g 0; then, D.0;1/f.0;0/ D .1;0/f.0;0/ D.1;1/ f.0;0/ 0.
¤ ¤
C
D
tx t
t
x
:
D 0, but D.1;0/C.0;1/f.0;0/ D ut
36
CHAPTER 2
DIFFERENTIATION
R be defined as in Exercise 26 . Show that I Exercise 61 (2-31) . Let f R2 Dx f.0;0/ exists for all x , although f is not even continuous at .0; 0/.
W
Proof. For any x
!
2 R2, we have f .t x/
lim
f .0/ D lim f .t x/ D 0
t
t !0
t
t !0
by Exercise 26 (a).
ut
I Exercise 62 (2-32) .
a. Let f R
W ! R be defined by f .x/
D
˚
x 2 sin x1
x
¤ 0 x D 0:
0
Show that f is differentiable at 0 but f 0 is not continuous at 0 .
b. Let f R2
W
! R be defined by f.x;y/
D
‚
x2
C y2
0
1
q C sin
x
2
y
.x;y/
¤0
.x; y/
D 0:
2
Show that f is differentiable at .0;0/ but D i f is not continuous at .0; 0/. Proof.
(a) We have
lim
f .x/
x!0
Hence, f 0 .0/
f .0/ D lim x2 sin x1 D lim x sin 1 D 0: x!0 x!0 x x x
D 0. Further, for any x ¤ 0, we have f 0 .x/
D 2x sin 1x cos 1x :
It is clear that lim x!0 f 0 .x/ does not exist. Therefore, f 0 is not continuous at 0. (b) Since
lim
.x;y/!.0;0/
x2
C y2
1
q q C C sin
x
x
we know that f 0 .0; 0/
2
y
2
y2
2
lim D .x;y/!.0;0/
q C x2
y 2 sin
1
q C x
D .0; 0/. Now take any .x; y/ ¤ .0;0/. Then
D1 f.x;y/
D 2x sin
1
q C x
2
y
2
2x cos
1
q C x
2
: y
2
2
y
2
D 0;
SECTION 2.4
37
DERIVATIVES
0
Figure 2 .5.
As in (a), limx!0 D1 f .x;0/ does not exist. Similarly for D 2 f .
ut
I Exercise 63 (2-33) . Show that the continuity of D1 f j at a may be eliminated from the hypothesis of Theorem 2-8. Proof. It suffices to see that for the first term in the sum, we have, by letting
a2 ; : : : ; an
µ
a1 ,
lim
ˇˇ
f a1
h!0
ˇˇ
C h1; a1 f .a/ D1f .a/ h1 khk f a1 C h1 ; a1 f .a/ D1 f .a/ h1
6 lim
h1 !0
ˇˇ
See aslo Apostol (1974, Theorem 12.11).
ˇˇ h1
ˇˇ D
0:
ut
R is homogeneous of degree m if I Exercise 64 (2-34) . A function f Rn m f .t x / t f .x / for all x . If f is also differentiable, show that
W
D
!
n
X
x i Di f .x /
iD1
Proof. Let g.t/
D mf .x/:
D f .t x/. Then, by Theorem 2-9, n 0
g .t/
X D iD1
i
Di f .t x / x :
(2.4)
D f .t x/ D t mf .x/; then g 0 .t/ D mt m1 f .x /: (2.5) ut Combining (2.4) and (2.5), and letting t D 1, we then get the result. I Exercise 65 (2-35) . If f W Rn ! R is differentiable and f .0/ D 0, prove that there exist g i W Rn ! R such that On the other hand, g.t/
38
CHAPTER 2
DIFFERENTIATION
n
f .x /
X D
x i gi .x/:
i D1
Proof. Let h x .t/
D f .t x/. Then
1
Z 0
h0x .t/ dt
D h .1/ h .0/ D f .x/ f .0/ D f .x/: x
x
Hence, 1
f .x /
Z D 0
h0x .t/ dt
1
Z D 0
f 0 .t x / dt
2X Z D 4 Z X D X D 0
xi Di f .t x / dt
i D1
n
x
i D1 n
3 5
n
1
1
i
Di f .t x/ dt
0
x i gi .x /;
i D1
where g i .x/
D
1 Di f .t x / dt . 0
R
ut
2.5 Inverse Functions For this section, Rudin (1976, Section 9.3 and 9.4) is a good reference.
Rn a continuously I Exercise 66 (2-36 ). Let A Rn be an open set and f A differentiable 1-1 function such that det f 0 .x/ 0 for all x . Show that f .A/ is 1 an open set and f f .A/ A is differentiable. Show also that f .B/ is open for any open set B A.
W
!
¤
W !
Proof. For every y 2 f .A/, there exists x 2 A such that f .x / D y . Since f 2 C 0 .A/ and det f 0 .x / ¤ 0, it follows from the Inverse Function Theorem that there is an open set V A containing x and an open set W R n containing y such that W D f .V /. This proves that f .A/ is open. Since f W V ! W has a continuous inverse f 1 W W ! V which is differen-
tiable, it follows that f 1 is differentiable at y ; since y is chosen arbitrary, it A is differentiable. follows that f 1 f .A/
W
! Take any open set B A. Since f B 2 C 0 .B/ and det all x 2 B A, it follows that f .B/ is open.
¤ 0
f B .x /
0 for
ut
I Exercise 67 (2-37) .
a. Let f R2 1-1.
W
! R be a continuously differentiable function. Show that f is not
SECTION 2.5
39
INVERSE FUNCTIONS
b. Generalize this result to the case of a continuously differentiable function Rm with m < n . f Rn
W
!
Proof.
(a) Let f C 0 . Then both D 1 f and D 2 f are continuous. Assume that f is 1-1; then both D1 f and D2 f cannot not be constant and equal to 0. So suppose that 0. The continuity of D 1 f implies there is x0 ; y0 R 2 such that D 1 f x0 ; f 0 2 R containing x0 ; y0 such that D1 f .x / 0 for that there is an open set A all x A. R2 with Define a function g A
2
2
2
¤
W !
g.x;y/ Then for all .x; y/
2 A, g 0 .x;y/
D
D
¤
f.x;y/;y :
!
D1 f.x;y/
D2 f.x;y/
0
1
;
D1 f.x;y/ 0; furthermore, g C 0 .A/ and g is 1-1. Then and so det g0 .x;y/ by Exercise 66, we know that g .A/ is open. We now show that g .A/ cannot be open actually. g .A/ with y y 0 . Then for any .x; y/ A, we Take a point f x0 ; y0 ; y
D ¤ 2 z 2 ¤ 2 D z H) D I D 2 D z W ! W ! D
must have
g.x;y/
f.x;y/;y
that is, there is no .x; y/
f x0 ; y0 ; y
.x;y/
A such that g.x;y/
x0 ; y0
f x0 ; y0 ; y . This proves
that f cannot be 1-1.
Rm as f R for every (b) We can write f Rn f 1 ; : : : ; f m , where f i Rn 1 n i 1; : : : ; m. As in (a), there is a mapping, say, f , a point a R , and an open Rm as set A containing a such that D 1 f 1 .x / 0 for all x A. Define g A
D
1
g x ;x where x 1
´
1
¤ D
2
f .x /; x
1
2
W !
;
x 2 ; : : : ; x n . Then as in (a), it follows that f cannot be 1-1.
I Exercise 68 (2-38) .
ut
W ! R satisfies f 0 .a/ ¤ 0 for all a 2 R, show that f is 1-1 (on all of R ). b. Define f W R2 ! R2 by f.x;y/ D e x cos y; e x sin y . Show that det f 0 .x;y/ ¤ a. If f R
0 for all .x; y/ but f is not 1-1.
Proof.
40
CHAPTER 2
DIFFERENTIATION
R with a < b such that (a) Suppose that f is not 1-1. Then there exist a; b f.a/ f.b/. It follows from the mean-value theorem that there exists c .a;b/ such that 0 f.b/ f.a/ f 0 .c/.b a/ ;
2
D
D D which implies that f 0 .c/ D 0. A contradiction.
2
(b) We have
f 0 .x;y/
D D
Then
Dx e x cos y
Dy e x cos y
Dx e x sin y
Dy e x sin y
e x cos y e x sin y
D
0
det f .x;y/
e
2x
ex sin y e x cos y
2
2
cos y
C sin
y
!
!
:
D
e 2x
¤ 0:
However, f .x;y/ is not 1-1 since f .x;y/ f x; y 2k for all .x; y/ R2 and k N. This exercise shows that the non-singularity of Df on A implies that f is locally 1-1 at each point of A, but it does not imply that f is 1-1 on all of A. See Munkres (1991, p. 69).
2
D
C
2
ut
I Exercise 69 (2-39) . Use the function f R
W ! R defined by
f .x/
˚
x 1 2 D 2 C x sin x x ¤ 0 0 x D 0
to show that continuity of the derivative cannot be eliminated from the hypoth- esis of Theorem 2-11. Proof. If x
¤ 0, then f 0 .x/
if x
D 0, then
0
f .0/
D 12 C 2x sin x1 cos x1 I
D h!0 lim
0
h=2
C h2 sin h
1= h
D 12 :
Hence, f .x/ is not continuous at 0. It is easy to see that f is not injective for any neighborhood of 0 (see Figure 2.6).
2.6 Implicit Functions I Exercise 70 (2-40) . Use the implicit function theorem to re-do Exercise 45 (c).
W Rn ! Rn by
Proof. Define f R
SECTION 2.6
41
IMPLICIT FUNCTIONS
0
Figure 2 .6.
t u
n i
f .t; s/
X D
aj i .t/s j
j D1
for i
bi .t/;
D 1 ; : : : ; n. Then M
0 B ´@
D2 f 1 .t; s/
:: :
::
n
D2 f .t; s/
:
D1Cn f 1 .t; s/
1 0 CA D B@
a11 .t/ :: : a1n .t/
:: : D1Cn f n .t; s/
::
:
and so det .M/ 0. It follows from the Implicit Function Theorem that for each t 0, and s is differentiable. unique s .t/ Rn such that f t; s.t/
¤
1 CA
an1.t/ :: ; : ann .t/
2 R, there is a 2 ut I Exercise 71 (2-41) . Let f W R R ! R be differentiable. For each x 2 R define gx W R ! R by g x y D f.x;y/. Suppose that for each x there is a unique y with gx0 y D 0; let c .x/ be this y . a. If D 2;2 f.x;y/ ¤ 0 for all .x; y/, show that c is differentiable and
D
0
c .x/ b. Show that if c 0 .x/
c. Let f.x;y/
Dx
D2;1 f x; c .x/
D D
2;2 f
x; c .x/
D 0, then for some y we have D2;1 f.x;y/ D 0; D2 f.x;y/ D 0:
y y log x. Find
y log y
:
42
CHAPTER 2
max
1=26x 62
"
DIFFERENTIATION
#
min f.x;y/ :
1=36y 61
Proof. D 2 f.x;y/. Since for every x there is a unique (a) For every x , we have g x0 y y c .x/ such that D2 f x; c .x/ 0, the solution c .x/ is the same as obtained from the Implicit Function Theorem; hence, c .x/ is differentiable, and by differentiating D 2 f x; c .x/ 0 with respect to x, we have
D D D C
D
D2;1 f x; c .x/
that is,
c 0 .x/
D2;2 f x; c .x/
D DD2;1 f f 2;2
c 0 .x/
x; c .x/ x; c .x/
D 0I
:
(b) It follows from (a) that if c 0 .x/ 0, then D2;1 f x; c .x/ 0. Hence, there c .x/ such that D2;1 f.x;y/ 0. Furthermore, by definition, exists some y D2 x; c .x/ D2 f.x;y/ 0.
D D
D
D
D
D
(c) We have
D x ln y ln x: Let D2 f.x;y/ D 0 we have y D c .x/ D x 1=x . Also, D2;2 f.x;y/ D x=y > 0 since x;y > 0. Hence, for every fixed x 2 1=2;2 , min f.x;y/ D f x; c .x/ : y D2 f.x;y/
y
c . x /
1.5 1 0.5
x 0
0.5
1
1.5
2
Figure 2 .7.
It is easy to see that c 0 .x/ > 0 on 1=2;2 , c .1/ a > 1=2 (see Figure 2.7). Therefore,
D
min f x; y
1=36y 61
where (see Figure 2.8)
D 1, and c.a/ D 1=3 for some
f x; y .x/ ;
SECTION 2.6
43
IMPLICIT FUNCTIONS
y .x/
D
„
if 1=2 6 x 6 a
1=3
D x1=x
c .x/ 1
if a < x 6 1 if 1 < x 6 2:
y 1.5
y .x/
1 0.5
x 0
0.5
1
1.5
2
Figure 2 .8.
1=2 6 x
6
a
In this case, our problem is 1
max f x;1=3
1=26x 6a
It is easy to see that x a
6
1
ln 3 x 3
1 ln x: 3
D C D D D 1=2, and so f x ;1=3
ln 4=3e =6.
In this case, our problem is
max f x; x 1=x
a
x 1C1=x :
It is easy to see that the maximum of f occurs at x 1
6
2
In this case, our problem is max f .x;1/
1
The maximum of f occurs at x
x
D
1=3.
D x ln x:
D 1.
Now, as depicted in Figure 2.9, we have x ln 4=3e =6.
D a and y
D 1=2, y D 1=3, and f
x; y
D
ut
44
CHAPTE CHAPTER R2
f x; y .x/ 0
0.5
1
−0.5 −1 −1.5 −2 −2.5
Figure 2 .9.
x 1.5
2
DIFFER DIFFERENT ENTIAT IATION ION
3 INTEGRATION
3.1 Basic Definitions I Exercise 72 (3-1). Let f f Œ0;1
W W
f x; y
Show that f f is integrable and
Œ0; Œ0; 1 ! R be defined by
˚ D
R
0 if 0 0 6 x < 1=2 1 if 1=2 1=2 6 x 6 1:
Œ0;1 Œ0;1 Œ0;1 f
Proof. Consider a partition P
D 1=2.
.P 1 ; P 2 / with P 1 P 2 0;1=2;1 . Then L f; P U f; P 1=2. 1=2. It foll follow ows s from from Theo Theore rem m 3-3 3-3 (the (the Riem Rieman ann n cond condit itio ion) n) f 1=2. 1=2. that f that f is is integrable and Œ0;1 Œ0;1Œ0;1
D D R
D
D
D
R be integrable and let g I Exercise 73 (3-2) . Let f f A g many points. Show that g . g is integrable and A f A g
W W !
R D R
D
ut
D f except at finitely
Proof. Fix an " > 0. It follows from the Riemann condition that there is a
A such that partition P partition P of A
U f; P
L f; P <
" : 2
Let P Let P 0 be a refinement of P of P such such that:
for every x 2 A with g P 0 , i.e., with g .x / ¤ f .x/, it belongs to 2 n subrectangles of P x is a corner of each subrectangle.
S of P P 0 , for every subrectangle subrectangle S v .S/ <
2nC1 d
" .u
`/ ;
where
45
46
CHAPTE CHAPTER R3
ˇ D D ˇ˚ W
d
x f .x /
INTEG INTEGRAT RATION ION
ˇˇ
¤ g .x / ; u D sup fg .x /g inf ff .x /g ; 2A 2A ` D inf fg .x /g sup ff .x /g : 2A x
x
x
x2A
x
Figure 3 .1.
With such a choice of partition of A of A,, we have
2 3 h i X X D 4 5 d
0
2n
0
U g; P
U f; P
M Sij g
i D1
M Sij f
v S ij ij
j D j D1
6 d 2n uv;
where v supS 2P 0 v.S/ is the least upper bound of the volumes of the P 0 . Similarly, subrectangles of P
´
f
g
2 3 h i D X 4X 5 d
0
2n
0
L g; P
L f; P
mSij g
i D1
mSij f
v S ij ij
j D j D1
> d 2n `v:
Therefore,
h C
U g; P 0
L g; P 0 6 U f; P 0 " 2 " 2
d 2n uv
ih C i L f; P 0
C d 2n .u `/ v D C d 2n .u `/ 2nC1d ".u `/ D "I that is, g is, g is integrable. It is easy to see now that A g D A f . . I Exercise 74 (3-3). Let f; f; g W A ! R be integrable. 6
R R
d 2n `v
ut
C C C C C C C C
a. For any partition P of A and subrectangle S , show that mS f mS g 6 mS f g and M S f g 6 M S f M S g and therefore L f; P L g; P 6 L f g; P and U f g; P 6 U f; P U g; P .
C C
C C
SECTION 3.1
47
BASIC DEFINITIONS
R R C R R R n C W 2 o C 2 2 C C D C C C 2 C D C C C C C D X C X X h C i D X C D C C D X C X X h C i D X C D C C C h C i h C i h i h i D C x x x x x
b. Show that f
C g is integrable and A f C g D c. For any constant c , show that A cf D c A f .
A f
A g
.
Proof.
(a) We show that m S f
mS g is a lower bound of
f
is clear that mS f 6 f .x/ and m S g 6 g .x/ for any x x S we have mS f
mS g 6 f .x/
g .x/
f
g .x/ x
S . It
S . Then for every
g .x/:
mS g 6 mS f g . Hence, m S f S we have M S f > f .x/ and M S g > g.x/; Similarly, for every x f .x/ g .x/ 6 M S f M S g and so M S f g 6 hence, f g .x/ M S f M S g . Now for any partition P of A we have L f; P
L g; P
mS f v.S/
mS g
S2P
S2P
mS f
mS g
v.S/
(3.1)
S2P
mS f
6
g v.S/
S2P
L f
g; P ;
and
U f; P
U g; P
M S f v.S/
S2P
M S g v.S/
S2P
M S f
M S g
v.S/
(3.2)
S2P
M S f
>
g v.S/
S2P
U f
g; P :
(b) It follows from (3.1) and (3.2) that for any partition P ,
U f
g; P
L f
g; P 6 U f; P
U g; P
L f; P
L g; P
U f; P
L f; P
U g; P
L g; P
:
Since f and g are integrable, there exist P 0 and P 00 such that for any " > 0, L f; P 0 < "=2 and U g; P 00 L g; P 00 < "=2. Let P refine we have U f; P 0 both P 0 and P 00 . Then U f; P
Hence,
L f; P <
" 2
and U g; P
L g; P <
" : 2
48
CHAPTER 3
INTEGRATION
C x C x
U f
g; P
L f
g; P < ";
and so f g is integrable. Now, by definition, for any " > 0, there exists a partition P (by using a common refinement partition if necessary) such that A f < L f; P "=2, g < L g; P "=2, U f; P < A f "=2, and U g; P < A g "=2. Therefore, A
C
R R C R C R C C Z C Z C C Z C C C Z C Z C R C D R C R D D D D D h i I f
g
A
" < L f; P
L g; P 6 L f
g; P 6
A
f
g
A
6 U f
g; P
6 U f; P
<
U g; P
f
g
A
Hence,
f
A
g
A
f
A
":
A
g.
(c) First, suppose that c > 0. Then for any partition P and any subrectangle S , we have mS cf cm S f and M S cf cM S f . But then L cf; P cL f; P and U cf; P cU f; P . Since f is integrable, for any " > 0 there L f; P < "=c. Therefore, exists a partition P such that U f; P
U cf; P
L cf; P
c U f; P
L f; P < "
that is, cf is integrable. Further, c
Z f
A
" < cL f; P c
D Z L cf; P 6
D Z C
cf 6 U cf; P
A
cU f; P
" ; c
f
A
R
R
i.e., A cf c A f . Now let c < 0. Then for any partition P of A, we have mS cf c mS f . Hence L cf; P c U f; P and U cf; P and M S cf Since f is integrable, for every " > 0, choose P such that U f; P "=c. Then
D
D D D D D h i I U cf; P
L cf; P
c U f; P
cM S f cL f; P . L f; P <
L f; P < "
that is, cf is integrable. Furthermore,
Z C c
f
A
" < c
cL
D Z f; P
U cf; P 6
A
cf 6
L
D Z cf; P
cL f; P
<
c
f
A
i.e.,
R
A cf
Dc
R
A f .
" ; c
ut
SECTION 3.1
49
BASIC DEFINITIONS
R and let P be a partition of A . Show that f is I Exercise 75 (3-4). Let f A integrable if and only if for each subrectangle S the function f S is integrable, and that in this case A f S S f S .
W !
R D P R
Proof. Let P be a partition of A, and S be a subrectangle with respect to P .
Only if: Suppose that f is integrable. Then there exists a partition P 1 of A L f; P 1 < " for any given " > 0. Let P 2 be a common such that U f; P 1 refinement of P and P 1 . Then
˚ µ S D D X h i X h i I D U f; P 2
L f; P 2 6 U f; P 1
and there are rectangles S 21 ; : : : ; S2n n i S iD1 S 2 . Therefore, U f; P 2
L f; P 2
L f; P 1 < ";
2 .S / with
M S2 f
respect to P 2 , such that
mS2 f
v .S 2 /
S2
M S2 f
>
mS2 f
v .S 2 /
S2 22 .S/
U f S; P 2
L f S; P 2
that is, f S is integrable.
y d
c x a
0
b Figure 3 .2.
If: Now suppose that f S is integrable for each S . For each partition P 0 , let P 0 be the number of subrectangles induced by P 0 . Let P S be a partition such that " U f S; P S L f S; P S < : 2 P
ˇˇ
j j
Let P 0 be the partition of A obtained by taking the union of all the subsequences defining the partitions of the P S ; see Figure 3.2. Then there are
50
CHAPTER 3
INTEGRATION
refinements P S0 of P S whose rectangles are the set of all subrectangles of P 0 which are contained in S . Hence,
X Z S
f S
S
" <
X
X
L f S; P S0
L f S; P S 6
S
S
L f; P 0
D X D X X Z C 6 U f; P 0
U f S; P S0
S
U f S; P S
6
S
<
f S
S
Therefore, f is integrable, and
":
S
R D P R f
S
A
S
f S .
ut
I Exercise 76 (3-5). Let f; g A that A f 6 A g .
W ! R be integrable and suppose f 6 g. Show
R R
Proof. Since f is integrable, the function
f is integrable by Exercise 74 (c); then g f is integrable by Exercise 74 (b). It is easy to see A g f > 0 since g > f . It follows from Exercise 74 that A g f D A g C f D C A f D A g A f ; hence, A f 6 A g. ut A g I Exercise 77 (3-6). If f W A ! R is integrable, show that jf j is integrable and 6 A jf j. A f Proof. Let f C D max ff; 0g and f D max ff; 0g. Then f D f C f and jf j D f C C f : It is evident that for any partition P of A, both U f C ; P L f C ; P 6 U f; P L f; P and U f ; P L f ; P 6 U f; P L f; P ; hence, both
R R R R ˇR ˇ R ˇˇZ
R R R
R R
ˇˇ D ˇˇZ ˇˇ D ˇˇZ Z Z D Z D j j
f C and f are integrable if f is. Further,
A
f
Z Z C
f C
f
f C
A
A
A
f C
6
A
C
f
A
f
f
A
C f
f :
A
I Exercise 78 (3-7). Let f Œ0;1
W
ˇˇ
Œ0; 1 ! R be defined by
ut
SECTION 3.3
51
FUBINI’S THEOREM
f x; y
„ D
0
x irrational
0
x rational, y irrational
1=q
x rational, y
Show that f is integrable and Proof.
R
Œ0;1Œ0;1 f
D p=q is lowest terms.
D 0. ut
3.2 Measure Zero and Content Zero I Exercise 79 (3-8). Prove that Œa 1 ; b1 ai < bi for each i .
Œan; bn does not have content 0 if
Proof. Similar to the Œa; b case.
ut
I Exercise 80 (3-9).
a. Show that an unbounded set cannot have content 0 . b. Give an example of a closed set of measure 0 which does not have content 0 . Proof.
(a) Finite union of bounded sets is bounded. (b) Z or N .
ut
I Exercise 81 (3-10) .
a. If C is a set of content 0 , show that the boundary of C has content 0 . b. Give an example of a bounded set C of measure 0 such that the boundary of C does not have measure 0 . Proof.
ut
3.3 Fubini’s Theorem I Exercise 82 (3-27) . If f Œa;b
W
b
Œa;b ! R is continuous, show that
y
Z Z
b
b
Z Z D
f x; y dx dy
a
a
Proof. As illustrated in Figure 3.3,
f x; y dy dx:
a
x
52
CHAPTER 3
C
D D
n 2 n 2 x; y x; y
Œa;b2 a 6 x 6 y and a 6 y 6 b
W Œa;b2 W a 6 x 6 b and x 6 y 6 b
o o
INTEGRATION
:
y y first
x
y
D
b
C x first
a x a
0
b
Figure 3 .3. Fubini’s Theorem
t u
I Exercise 83 (3-30) . Let C be the set in Exercise 17 . Show that
Z Z ! D Z Z ! 1C x; y dx
Œ0;1
dy
1C x; y dy
Œ0;1
Œ0;1
Œ0;1
Proof. There must be typos.
I Exercise 84 (3-31). If A R by define F A
W !
Solution. Let c
D 0: ut
D Œa1; b1 Œan ; bn and f W A ! R is continuous,
F .x /
What is D i F .x /, for x
dx
2 int.A/?
2 int.A/. Then
Z D
Œa1 ;x1 Œan ;xn
f:
SECTION 3.3
R R R
F c i ; c i Di F .c /
D h!0 lim D
53
FUBINI’S THEOREM
C h F .c / h
Œa1 ;c 1 Œai ;ci ChŒan ;cn f lim h h!0
lim D h!0
D h!0 lim
Z D
Œa1
F .c/
c i Ch ai
c i Ch ci
;c 1
R R
Œa1
;c 1
Œai 1
;x i 1
Œai C1 h
;c i C1
Œan
;c n
f dxi
F .c/
Œa1 ;c1 Œai 1 ;ci 1 Œai C1 ;ci C1 Œan ;c n f dxi h
Œai 1
;c i 1
Œai C1
;c i C1
Œan
f x i ; c i : ;c n
ut
R be continuous and suppose I Exercise 85 (3-32 ). Let f Œa;b Œc; d b D2 f is continuous. Define F y . Prove Leibnitz’s rule: F 0 y a f x; y dx b a
W D R !
R
D2 f x; y dx .
D
Proof. We have
F 0 y
D
C R C R Z C F y
lim
h!0
lim D h!0
F y
h
b a
b a
h dx
f x; y
f x; y dx
h
b
D h!0 lim
h
f x; y
h
f x; y
h
a
dx:
By DCT, we have
D Z " C # Z D D R W ! D R D D R D C D Z D C b
0
lim
F y
f x; y
h
h
h!0
a
f x; y
dx
b
D2 f x; y dx:
a
I Exercise 86 (3-33) . If f Œa;b Œc;d x uous, define F x; y f t; y dt . a
ut
R is continuous and D 2 f is contin-
a. Find D 1 F and D 2 F . b. If G .x/
g .x/ f a
.t; x/ dt , find G 0 .x/ .
Solution.
(a) D1 F x; y
f x; y , and D 2 F
(b) It follows that G .x/
G 0 .x/
x D2 f a
t; y dt .
F g . x / ; x . Then
g0 .x/ D1 F g . x / ; x 0
D2 F g . x / ; x
g.x/
g .x/f g . x / ; x
D2 f .t;x/ dt:
a
ut
4 INTEGRATION ON CHAINS
4.1 Algebraic Preliminaries I Exercise 87 (4-1 ). Let e1 ; : : : ; en be the usual basis of Rn and let '1 ; : : : ; 'n be the dual basis.
a. Show that ' i1 'ik .ei1 ; : : : ; eik / 1 . What would the right side be if the factor .k `/Š=kŠ`Š did not appear in the definition of ?
C
^ ^
D
^
'ik .v1 ; : : : ; vk / is the determinant of the k b. Show that 'i1 v1 :: : obtained by selecting columns i 1 ; : : : ; ik . vk
^ ^
˘
k minor of
Proof.
(a) Since ' ij
'i1
2 T .Rn /, for every j D 1 ; : : : ; k, we have
^ ^ 'i
.ei1 ; : : : ; eik / k
D 1Š kŠ 1Š Alt 'i ˝ ˝ 'i .ei ; : : : ; ei D .sgn.//'i .e.i // 'i .e.i //
X
1
k
1
1
1
k
k
/
k
2Sk
D 1: If the factor .k `/Š=kŠ`Š did not appear in the definition of solution would be 1=kŠ.
^, then the
C
(b)
ut
I Exercise 88 (4-9 ). Deduce the following properties of the cross product in R3 .
e1 e1 a. e1 e2 e1 e3
D 0 D e3 D e2
e2 e1 e2 e2 e2 e3
D e3 D 0 D e1
e3 e1 e3 e2 e3 e3
D e2 D e1 D 0
Proof.
55
56
CHAPTER 4
INTEGRATION ON CHAINS
D H) D D D H) D D H) D
tu
(a) We just do the first line.
hw; z hw; z hw; z
iD iD iD
e1 e1 w e2 e1 w e3 e1 w
0
z
w3
w2
e1
e2
e3
e1
0;
e1
e1
e3 ;
e2 :
References
[1] Apostol, Tom M. (1974) Mathematical Analysis : Pearson Education, 2nd edition. [37] [2] Axler, Sheldon (1997) Linear Algebra Done Right , Undergraduate Texts in Mathematics, New York: Springer-Verlag, 2nd edition. [] [3] Berkovitz, Leonard D. (2002) Convexity and Optimization in Rn , Pure and Applied Mathematics: A Wiley-Interscience Series of Texts, Monographs and Tracts, New York: Wiley-Interscience. [15] [4] Munkres, James R. (1991) Analysis on Manifolds , Boulder, Colorado: Westview Press. [40] [5] Rudin, Walter (1976) Principles of Mathematical Analysis , New York: McGraw-Hill Companies, Inc. 3rd edition. [ 17, 38] [6] Spivak, Michael (1965) Calculus on Manifolds: A Modern Approach to Classical Theorems of Advanced Calculus , Boulder, Colorado: Westview Press. [i]
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