Spivak Calculo en Variedades Solucionario
April 6, 2017 | Author: Juarez Carlos | Category: N/A
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Calculus on Manifolds A Solution Manual for Spivak (1965)
Jianfei Shen School of Economics, The University of New South Wales
Sydney, Australia
2010
Contents
1
Functions on Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Norm and Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Subsets of Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Functions and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 1 6 9
2
Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Basic Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Implicit Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 13 18 26 34 38 40
3
Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Measure Zero and Content Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Fubini’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45 45 51 51
4
Integration on Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4.1 Algebraic Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
iii
1 FUNCTIONS ON EUCLIDEAN SPACE
1.1 Norm and Inner Product I Exercise 1 (1-1 ). Prove that kxk 6
Pn ˇˇ i ˇˇ iD1 ˇx ˇ.
Proof. Let x D x 1 ; : : : ; x n . Then
12 0 n n n ˇ ˇ ˇ X X Xˇˇ X 2 2 ˇ @ ˇˇx i ˇˇA D x i D kxk2 : xi C ˇx i x j ˇ > iD1
i D1
i¤j
i D1
Taking the square root of both sides gives the result.
t u
I Exercise 2 (1-2). When does equality hold in Theorem 1-1 (3) kx C y k 6 kxk C ky k ? ˇ ˇ Proof. We reprove that ˇhx; yiˇ 6 kxk ky k for every x; y 2 Rn . Obviously, if x D 0 or y D 0, then hx; yi D kxk ky k D 0. So we assume that x ¤ 0 and y ¤ 0. We first find some w 2 Rn and ˛ 2 R such that hw; ˛yi D 0. Write w D x ˛y . Then 0 D hw; ˛yi D hx ˛y; ˛yi D ˛ hx; yi ˛ 2 ky k2 implies that
ı ˛ D hx; yi ky k2 : Then
kxk2 D kwk2 C k˛y k2 > k˛y k2 D ˇ
hx; yi ky k
2 :
ˇ
Hence, ˇhx; yiˇ 6 kxk 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 . Since
kx C y k2 D hx C y; x C yi D kxk2 C ky k2 C 2 hx; yi 6 kxk2 C ky k2 C 2 kxk ky k 2 D kxk C ky k ;
1
2
CHAPTER 1
FUNCTIONS ON EUCLIDEAN SPACE
equality holds precisely when hx; yi D kxkjjyjj, i.e., when one is a nonnegative multiple of the other. t u
I Exercise 3 (1-3). Prove that kx
y k 6 kxk C ky k. When does equality hold?
Proof. By Theorem 1-1 (3) we have kx y k D kx C . y/k 6 kxk C k y k D kxk C ky k. The equality holds precisely when one vector is a non-positive multiple of the other. t u
ˇ I Exercise 4 (1-4). Prove that ˇ kxk
ˇ ky kˇ 6 kx
y k.
2 Pn P y k2 D yi D kxk2 C ky k2 2 niD1 xi yi > iD1 xi 2 kxk2 C ky k2 2 kxk ky k D kxk ky k . Taking the square root of both sides gives the result. t u Proof. We have kx
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 xk 6 kz y k C ky x k. 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. t u
I Exercise 6 (1-6). If f and g be integrable on Œa; b. ˇR ˇ b
a. Prove that ˇ
a
ˇ R 12 R 12 ˇ b b f gˇ 6 a f 2 a g2 .
b. If equality holds, must f D g for some 2 R? What if f and g are continuous? 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:
ˇ ˇ 0 11=2 0 11=2 ˇX ˇ X X ˇ ˇ ˇ f .xi / g .xi / xi ˇˇ 6 @ f .xi /2 xi A @ g .xi /2 xi A : ˇ ˇ i ˇ i i 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 is true. In fact, suppose that for each 2 R, there is an x 2 Œa; b with
SECTION 1.1
3
NORM AND INNER PRODUCT
2 g .x/ > 0. Then the inequality holds true in an open neighbor2 Rb hood of x since f and g are continuous. So a f g > 0 since the integrand is always non-negative and is positive on some subinterval of Œa; b. Rb Rb Rb Expanding out gives a f 2 2 a f g C 2 a g 2 > 0 for all . Since the quadratic has no solutions, it must be that its discriminant is negative.
f .x/
c. Let a D 0, b D n, f .x/ D xi and g .x/ D yi for all x 2 Œi 1; i/ for i D 1; : : : ; n. Then part (a) gives the inequality of Theorem 1-1 (2). Note, however, that the equality condition does not follow from (a). t u
I Exercise 7 (1-7). A linear transformation M W Rn ! Rn is called norm preserving if kM x k D kxk, and inner product preserving if hM x; M yi D hx; yi. 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:
hM x; M yi D
kM x C M yk2
D
M yk2
kM .x
y/k2
4 2
D
kM x
kM .x C y/k
4 x y k2 k
kx C y k2 4
D hx; yi : If M is inner product preserving, then one has by Theorem 1-1 (4):
p p hM x; M xi D hx; xi D kxk :
kM x k D
(b) Take any M x; M y 2 Rn with M x D M y . Then M x
0 D hM x
M y; M x
M yi D hx
y; x
M y D 0 and so
yi 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, M 1 2 L.Rn / exists. For every x; y 2 Rn , we have
kM and
D M Therefore, M
1
1
x; M
1
1
M M D k x
y D M M E
1
1
x
D kxk ;
x ;M M
1
y
D hx; yi :
is also norm preserving and inner product preserving.
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4
CHAPTER 1
FUNCTIONS ON EUCLIDEAN SPACE
I Exercise 8 (1-8). If x; y 2 Rn are the angle between x and y , non-zero, ı y denoted † .x; y/, is defined as arccos hx; yi kxk k 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 ¤ 0 we have † .Tx; Ty/ D † .x; y/. 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 Txi D i xi , prove that T is angle preserving if and only if all ji j are equal. c. What are all angle preserving T W Rn ! 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/ D arccos
hTx; Tyi kTx k kTyk
D arccos
hx; yi kxk ky k
D † .x; y/ :
(b) We first suppose that T is angle preserving. Since .x1 ; : : : ; xn / is a basis of Rn , all xi ’s are nonzero. Since
˝ ˛ ! ˛ ! i xi ; j xj
D arccos
D arccos kTxi k Txj ki xi k j xj ! ˝ ˛ i j xi ; xj ˇ ˇ D arccos ji j ˇj ˇ kxi k kxj k D † xi ; xj ; ˝
† Txi ; Txj
Txi ; Txj
it must be the case that
ˇ ˇ i j D ji j ˇj ˇ : Then i and j have the same signs.
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I Exercise 9 (1-9). If 0 6 < , let T W R2 ! R2 have the matrix ! cos sin AD : sin cos Show that T is angle preserving and if x ¤ 0, then † .x; Tx/ D . Proof. For every x; y 2 R2 , we have
T x; y D Therefore,
cos sin
!
sin cos
! x D y
! x cos C y sin : x sin C y cos
2
2
T x; y D x 2 C y 2 D x; y I
that is, T is norm preserving. Then it is angle preserving by Exercise 8 (a).
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 a2 C b 2 cos : Hence,
hx; Txi † .x; Tx/ D arccos kxk kTx k
D arccos
! a2 C b 2 cos D : a2 C b 2
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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
AD
˘ ˘
:: :
a11 :: : an1
a1m :: : anm
´
˝
Then
a1 :: : : an
˘
˛ a1 ; h :: ; : n ha ; hi
M h D Ah D
and so
0 1 n D n n E2 X X X 2 ai ; h 6 kai k khk D @ kM hk2 D kai k2 A khk2 ; i D1
i D1
iD1
p
that is,
0 B kM hk 6 @
p Let M D
n X
n X i D1
1 C kai kA khk :
kai k and we get the result.
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iD1
I Exercise 11 (1-11). If x; yq2 Rn and z; w 2 Rm , show that h.x; z/ ; .y; w/i D hx; yi C hz; wi and k.x; z/k D kxk2 C kzk2 . Proof. We have .x; z/ ; .y; w/ 2 RnCm . Then
h.x; z/ ; .y; w/i D
n X iD1
xi yi C
m X
zj wj D hx; yi C hz; wi ;
j D1
and
k.x; z/k2 D h.x; z/ ; .x; z/i D hx; xi C hz; zi D kxk2 C kzk2 :
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6
CHAPTER 1
FUNCTIONS ON EUCLIDEAN SPACE
I Exercise 12 (1-12 ). Let .Rn / denote the dual space of the vector space Rn . If x 2 Rn , define 'x 2 .Rn / by 'x .y/ D hx; yi. Define M W Rn ! .Rn / by M x D 'x . Show that M is a 1-1 linear transformation and conclude that every ' 2 .Rn / is 'x for a unique x 2 Rn . Proof. We first show M is linear. Take any x; y 2 Rn and a; b 2 R. Then M .ax C by/ D 'axCby D a'x C b'y D a M x C b M y; where the second equality holds since for every z 2 Rn ,
'axCby .z/ D hax C by; zi D a hx; zi C b hy; zi D a'x .z/ C b'y .z/ : To see M is 1-1, we need only to show that ıM D f0g, where ıM 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. t u
I Exercise 13 (1-13 ). If x; y 2 Rn , then x and y are called perpendicular (or orthogonal) if hx; yi D 0. If x and y are perpendicular, prove that kx C y k2 D kxk2 C ky k2 . Proof. If hx; yi D 0, we have
kx C y k2 D hx C y; x C yi D kxk2 C 2 hx; yi C ky k2 D kxk2 C ky k2 :
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1.2 Subsets of Euclidean Space I Exercise 14 (1-14 ). Simple. Omitted.
˚ I Exercise 15 (1-15). Prove that x 2 Rn W kx
ak < r is open.
Proof. For any y 2 x 2 Rn W kx ak < r µ B.aI r/, let " D r that B.yI "/ B.aI r/. Take any z 2 B.yI "/. Then
˚
ka; zk 6 ka; y k C ky; zk < ka; y k C " D r: I Exercise 16 (1-16). Simple. Omitted.
I Exercise 17 (1-17). Omitted.
ka; y k. We show
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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 .ai ; bi /, show that @A D Œ0; 1 X A. 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 ¤ ¿, which means that x 2 Ax. Hence, X D Ax, i.e., A is dense in X . t u
I Exercise 19 (1-19 ). If A is a closed set that contains every rational number r 2 Œ0; 1, show that Œ0; 1 A. 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 0; 1 2 A, the claim holds. t u
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 … A, ˚ and let Gm D y 2 Rn W kx y k > 1=m , m D 1; 2; : : :. If y 2 A, then x ¤ y ; hence, kx y k > 1=m for some m; therefore y 2 Gm (see Figure 1.1). Thus, S A 1 mD1 Gm , and by compactness we have a finite subcovering. Now observe that the Gm for an increasing sequence of sets: G1 G2 ; therefore, a finite union of some of the Gm is equal to the set with the highest index. Thus, K Gs for some s , and it follows that B.xI 1=s/ Ac . Therefore, Ac is open.
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
˚ O D . i; i/n W i 2 N be an open cover of A. Then there is a finite subcover . i1 ; i1 /n ; : : : ; . im ; im /n of A. Let i 0 D max fi1 ; : : : ; im g. Hence, A i 0 ; i 0 ; that is, A is bounded. t u
˚
I Exercise 21 (1-21 ). a. If A is closed and x … A, prove that there is a number d > 0 such that ky x k > d for all y 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 . c. Give a counterexample in R2 if A and B are closed but neither is compact. Proof. (a) A is closed implies that Ac is open. Since x 2 Ac , there exists an open ball B.xI d / with d > 0 such that x 2 B.xI d / Ac . Then ky x k > d for all y 2 A. (b) For every x 2 B , there exists dx > 0 such that x 2 B.xI dx =2/ Ac and ky x k > dx for all y 2 A. Then the family fB.xI dx =2/ W x 2 Bg is an open cover of B . Since B is compact, there is a finite set fx1 ; : : : ; xn g such that ˚ B.x1 I dx1 =2/; : : : ; B.xn I dxn =2/ covers B as well. Now let
˚ . d D min dx1 =2; : : : ; dxn =2 2: Then for any x 2 B , there is an open ball B.xi I xi =2/ containing x and ky di . Hence,
ky
x k > ky
xi k
kxi
x k > di
xi k >
di =2 D di =2 > d:
(c) See Figure 1.2.
0
Figure 1.2.
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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 . t u
Proof.
1.3 Functions and Continuity I Exercise 23 (1-23). If f W A ! Rm and a 2 A, show that limx!a f .x/ D b if and only if limx!a f i .x/ D b i for i D 1; : : : ; m. 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
f i .x/
ak < ı ,
" bi < p ; m
i D 1; : : : ; m:
Therefore, for every x 2 A which satisfy 0 < kx
p
kf .x/
bk D
m X f i .x/
bi
2
iD1
<
ak < ı ,
p
m X "2 =m D "I iD1
that is, limx!a f .x/ D b . Only if: Now suppose that limx!a f .x/ D b . Then for every number " > 0 there is a number ı > 0 such that kf .x/ bk < " for all x 2 A which satisfy 0 < kx ak < ı . But then for every i D 1; : : : ; m,
f i .x/ i.e. limx!a f i .x/ D b i .
b i 6 kf .x/
bk < "; t u
I Exercise 24 (1-24). Prove that f W A ! Rm is continuous at a if and only if 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/ D f .a/ if and only if limx!a f i .x/ D f i .a/ for every i D 1; : : : ; m; that is, if and only if f i is continuous at a for each i D 1; : : : ; m. t u
I Exercise 25 (1-25). Prove that a linear transformation T W Rn ! Rm is continuous.
10
CHAPTER 1
FUNCTIONS ON EUCLIDEAN SPACE
Proof. Take any a 2 Rn . Then, by Exercise 10 (1-10), there exists M > 0 such that Tx Ta D T .x a/ 6 M kx ak : Hence, for every " > 0, let ı D "=M . Then Tx Ta < " when x 2 Rn and 0 < kx ak < ı D "=M ; that is, limx!a Tx D Ta, and so T is continuous. t u
I Exercise 26 (1-26). Let A D
n
o x; y 2 R2 W x > 0 and 0 < y < x 2 .
a. Show that every straight line through .0; 0/ contains an interval around .0; 0/ which is in R2 X A. 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 gh W R ! R by gh .t / D f .th/. Show that each gh is continuous at 0, but f is not continuous at .0; 0/. Proof. (a) Let the line through .0; 0/ be y D ax . If a 6 0, then the whole line is in R2 X A. If a > 0, then ax intersects x 2 at a; a2 and .0; 0/ and nowhere else; see Figure 1.3. yD 2 x
y
yD
ax
A
x
0
Figure 1.3.
(b) We first show that f is not continuous at 0. Clearly, f .0/ D 0 since 0 … A. For every ı > 0, there exists x 2 A satisfying 0 < kxk < ı , but jf .x/ f .0/j D 1. We next show gh .t/ D f .th/ is continuous at 0 for every h 2 R2 . If h D 0, then g0 .t / D f .0/ D 0 and so is continuous. So we now assume that h ¤ 0. It is clear that
gh .0/ D f .0/ D 0: The result is now from (a) immediately.
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SECTION 1.3
11
FUNCTIONS AND CONTINUITY
˚ I Exercise 27 (1-27). Prove that x 2 Rn W kx the function f W Rn ! R with f .x/ D kx ak.
ak < r is open by considering
Proof. We first show that f is continuous. Take a point b 2 Rn . For any " > 0, let ı D ". Then for every x satisfying kx bk < ı , we have
jf .x/
ˇ f .b/j D ˇ kx
Hence, x 2 Rn W kx
˚
kb
ak
ak < r D f
ˇ akˇ 6 kx 1
ak
kb
a k 6 kx
. 1; r/ is open in Rn .
bk < ı D ": t u
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
x k for all y 2 A.
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I Exercise 29 (1-29). Simple. Omitted.
I Exercise 30 (1-30). Let f W Œa; b ! R be an increasing function. If x1 ; : : : ; xn 2 P Œa; b are distinct, show that niD1 o f; xi < f .b/ f .a/. Proof.
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2 DIFFERENTIATION
2.1 Basic Definitions I Exercise 31 (2-1 ). Prove that if f W Rn ! Rm is differentiable at a 2 Rn , then it is continuous at a. Proof. Let f be differentiable at a 2 Rn ; then there exists a linear map W Rn ! Rm such that lim
kf .a C h/
h!0
f .a/ khk
.h/k
D 0;
or equivalently,
f .a C h/
f .a/ D .h/ C r.h/;
(2.1)
where the remainder r.h/ satisfies
ı
lim kr.h/k khk D 0:
h!0
(2.2)
Let h ! 0 in (2.1). The error term r.h/ ! 0 by (2.2); the linear term .h/ aslo Pn tends to 0 because if h D i D1 hi ei , where .e1 ; : : : ; en / is the standard basis of P Rn , then by linearity we have .h/ D niD1 hi .ei /, and each term on the right tends to 0 as h ! 0. Hence,
lim f .a C h/
h!0
f .a/ D 0I
that is, limh!0 f .a C h/ D f .a/. Thus, f is continuous at a.
t u
I Exercise 32 (2-2). A function f W R2 ! R is independent of the second variable if for each x 2 R we have f .x; y1 / D f .x; y2 / for all y1 ; 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/ D f .x; 0/. Conversely, if f .x; y/ D g.x/, then f .x; y1 / D g.x/ D f .x; y2 /. If f is independent of the second variable, then
13
14
CHAPTER 2
lim
ˇ ˇf .a C h; b C k/
.h;k/!0
f .a; b/ k.h; k/k
ˇ g 0 .a/hˇ
D
lim
.h;k/!0
6 lim
DIFFERENTIATION
ˇ g.a/ g 0 .a/hˇ k.h; k/k ˇ g.a/ g 0 .a/hˇ jhj
ˇ ˇg.a C h/
ˇ ˇg.a C h/
h!0
D 0I hence, f 0 .a; b/ D .g 0 .a/; 0/.
t u
I Exercise 33 (2-3). Define when a function f W R2 ! R is independent of the first variable and find f 0 .a; b/ for such f . Which functions are independent of the first variable and also of the second variable? Proof. We have f 0 .a; b/ D .0; g 0 .b// with a similar argument as in Exercise 32. If f is independent of the first and second variable, then for any .x1 ; y1 /, .x2 ; y2 / 2 R2 , we have f .x1 ; y1 / D f .x2 ; y1 / D f .x2 ; y2 /; that is, f is constant. t u
I Exercise 34 (2-4). Let g be a continuous real-valued function on the unit ˚ circle x 2 R2 W kxk D 1 such that g.0; 1/ D g.1; 0/ D 0 and g. x/ D g.x/. Define f W R2 ! R by ı if x ¤ 0; kxk g x kxk f .x/ D 0 if x D 0:
˚
a. If x 2 R2 and h W R ! R is defined by h.t / D f .t x/, show that h is differentiable. 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,
h.t / D f .t x/ D t kxk g
tx t kxk
h i D kxk g x= kxk t D f .x/t I
finally, if x ¤ 0 and t < 0,
h.t / D f .t x/ D
t kxk g
tx t kxk
x= kxk h i D kxk g x= kxk t D
t kxk g
D f .x/t: Therefore, h.t / D f .x/t for every given x 2 R2 , and so is differentiable: Dh D h. (b) Since g.1; 0/ D 0 and g. x/ D g.x/, we have g. 1; 0/ D g. .1; 0// D g.1; 0/ D 0. If f is differentiable at .0; 0/, there exists a matrix .a; b/ such that Df .0; 0/.h; k/ D ah C bk . First consider any sequence .h; 0/ ! .0; 0/. Then
SECTION 2.1
15
BASIC DEFINITIONS
0 D lim
jf .h; 0/
f .0; 0/ jhj
h!0
ahj
D lim
h!0
D lim
ˇ ˇ ˇ jhj g h=jhj ; 0 jhj ˇ ˇ jhj g .˙1; 0/
ˇ ˇ ahˇ ˇ ahˇ
jhj
h!0
D jaj implies that a D 0. Next let us consider .0; k/ ! .0; 0/. Then
0 D lim
jf .0; k/
f .0; 0/ jkj
k!0
bkj
D lim
ˇ ˇ ˇ jkj g 0; k=jkj
ˇ ˇ bk ˇ
jkj
k!0
D jbj
forces that b D 0. Therefore, f 0 .0; 0/ D .0; 0/ and Df .0; 0/.x; y/ D 0. If g.x/ ¤ 0, then
lim
jf .x/
x!0
f .0/ kxk
0j
D lim
x!0
ˇ ˇˇ ˇ ˇ kxk g x= kxk ˇ kxk
ˇ ˇˇ ˇ D lim ˇg x= kxk ˇ ¤ 0; x!0
and so f is not differentiable. Of course, if g.x/ D 0, then f .x/ D 0 and is differentiable.
t u
I Exercise 35 (2-5). Let f W R2 ! R be defined by q x jyj x 2 C y 2 if .x; y/ ¤ 0; f .x; y/ D 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
f .x; y/ D q
x jyj x2 C y2
x jyj D D k.x; y/k k.x; y/k
x jyj : k.x; y/k k.x; y/k
(2.3)
If we let g W x 2 R2 W kxk D 1 ! R be defined as g.x; y/ D x jyj, then (2.3) can 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
xj yj D
xjyj D
f .x; y/I
that is, g satisfies all of the properties listed in Exercise 34. Since g.x/ ¤ 0 unless x D 0 or y D 0, we know that f is not differentiable at 0. A direct proof can be found in Berkovitz (2002, Section 1.11). t u
I Exercise 36 (2-6). Let f W R2 ! R be defined by f .x; y/ D f is not differentiable at .0; 0/.
p
jxyj. Show that
16
CHAPTER 2
DIFFERENTIATION
Proof. It is clear that lim
h!0
jf .h; 0/j jf .0; k/j D 0 D lim I k!0 jhj jkj
hence, if f is differentiable at .0; 0/, it must be that Df .0; 0/.x; y/ D 0 since derivative is unique if it exists. However, if we let h D k > 0, and take a sequence f.h; h/g ! .0; 0/, we have
jf .h; h/ f .0; 0/ lim .h;h/!.0;0/ k.h; h/k
0j
p 1 h2 D lim D p ¤ 0: .h;h/!.0;0/ k.h; h/k 2
Therefore, f is not differentiable.
t u
I Exercise 37 (2-7). Let f W R2 ! R be a function such that jf .x/j 6 kxk2 . Show that f is differentiable at 0. Proof. jf .0/j 6 k0k2 D 0 implies that f .0/ D 0. Since
jf .x/j jf .x/ f .0/j D lim 6 lim kxk D 0; x!0 kxk x!0 x!0 kxk lim
Df .0/.x; y/ D 0.
t u
I Exercise 38 (2-8). Let f W R ! R2 . Prove that f is differentiable at a 2 R if and only if f 1 and f 2 are, and that in this case ! 1 0 .f / .a/ f 0 .a/ D : .f 2 /0 .a/ 0
Proof. Suppose that f is differentiable at a with f .a/ D
! c1 . Then for i D c2
1; 2,
0 6 lim
ˇ ˇ i ˇf .a C h/
h!0
f i .a/ jhj
ˇ ˇ c i hˇ
kf .a C h/
6 lim
h!0
f .a/ jhj
Df .a/.h/k
D0
implies that f i is differentiable at a with .f i /0 .a/ D c i . Now suppose that both f 1 and f 2 are differentiable at a, then by Exercise 1,
ˇ
06
kf .a C h/
f .a/ jhj
Df .a/.h/k
6
2 ˇˇf i .a X iD1
C h/
f i .a/ jhj
! .f 1 /0 .a/ implies that f is differentiable at a with f .a/ D . .f 2 /0 .a/ 0
ˇ ˇ .f i /0 .a/ hˇ
t u
I Exercise 39 (2-9). Two functions f; g W R ! R are equal up to n-th order at a if
SECTION 2.1
17
BASIC DEFINITIONS
f .a C h/ g.a C h/ D 0: hn h!0 lim
a. Show that f is differentiable at a if and only if there is a function g of the form g.x/ D a0 C a1 .x a/ such that f and g are equal up to first order at a. b. If f 0 .a/; : : : ; f .n/ .a/ exist, show that f and the function g defined by
g.x/ D
n X f .i / .a/ i D0
iŠ
.x
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/ C f 0 .a/ h D 0; h
f .a C h/
h!0
so we can let g.x/ D f .a/ C f 0 .a/ .x a/. On the other hand, if there exists a function g.x/ D a0 C a1 .x lim
f .a C h/
g.a C h/ h
h!0
D lim
f .a C h/
a0
a1 h
h
h!0
a/ such that
D 0;
then a0 D f .a/, and so f is differentiable at a with f 0 .a/ D a1 . (b) By Taylor’s Theorem1 we rewrite f as
f .x/ D
n 1 X f .i / .a/ iD0
iŠ
.x
a/i C
f .n/ .y/ .x nŠ
a/n ;
where y is between a and x . Thus, lim
x!a
f .x/ g.x/ D lim x!a .x a/n D lim
f .n/ .y/ .x nŠ
f .n/ .y/
x!a
.n/
a/n f nŠ.a/ .x .x a/n
a/n
f .n/ .x/ nŠ
D 0:
t u
1
(Rudin, 1976, Theorem 5.15) Suppose f is a real function on Œa; b, n is a positive integer, f .n 1/ is continuous on Œa; b, f .n/ exists for every t 2 .a; b/. Let ˛ , ˇ be distinct points of Œa; b, and define n X1 f .k/ .˛/ P .t/ D .t ˛/k : kŠ kD0
Then there exists a point x between ˛ and ˇ such that
f .ˇ / D P .ˇ / C
f .n/ .x/ .ˇ nŠ
˛/n :
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 following: a. f .x; y; z/ D x y . 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/ . 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
. j. f .x; y/ D sin.xy/; sin x sin y ; x y . i. f .x; y/ D sin.xy/
Solution. Compare this with Exercise 47. y (a) We have f .x; y; z/ D x y D e ln x D e y ln x D exp B. 2 ln 1 /.x; y; z/. It follows from the Chain Rule that
h i 0 f 0 .a; b; c/ D exp0 . 2 ln 1 /.a; b; c/ 2 ln 1 .a; b; c/ h i D exp.b ln a/ .ln 1 /. 2 /0 C 2 .ln 1 /0 .a; b; c/ h i D ab 0; ln a; 0 C b=a; 0; 0 D ab 1 b ab ln a 0 : (b) By (a) and Theorem 2-3(3), we have
f 0 .a; b; c/ D
ab
1
0
b
ab ln a 0
! 0 : 1
(c) We have f .x; y/ D sin B. 1 sin. 2 //. Then, by the chain rule,
h i h i0 f 0 .a; b/ D sin0 . 1 sin. 2 //.a; b/ 1 sin. 2 / .a; b/ h i 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 : (d) Let g.y; z/ D sin.y sin z/. Then
SECTION 2.2
19
BASIC THEOREMS
f .x; y; z/ D sin x g.y; z/ D sin. 1 g. 2 ; 3 //: Hence,
f 0 .a; b; c/ D sin0 ag .b; c/ . 1 g. 2 ; 3 //0 .a; b; c/ i h D cos ag .b; c/ g .b; c/ . 1 /0 C ag 0 . 2 ; 3 / .a; b; c/ i h D cos ag .b; c/ g .b; c/ ; 0; 0 C ag 0 . 2 ; 3 /.a; b; c/ : It follows from (c) that
g 0 . 2 ; 3 /.a; b; c/ D 0
cos .b sin c/ sin c
b cos .b sin c/ cos c :
Therefore,
f 0 .a; b; c/ D cos a sin .b sin c/ sin .b sin c/
a cos .b sin c/ sin c
ab cos .b sin c/ cos c :
(e) Let g.x; y/ D x y . Then
f .x; y; z/ D x g.y;z/ D g x; g.y; z/ D g 1 ; g. 2 ; 3 / : Then
i h Df .a; b; c/ D Dg a; g .b; c/ B D 1 ; Dg. 2 ; 3 / .a; b; c/: By (a),
Dg a; g .b; c/ .x; y; z/ D ag.b;c/ g .b; c/ =a
ag.b;c/ ln a
0 1 x B C 0 @y A z
c
D
c ab b c x C ab ln a y; a
D 1 .a; b; c/.x; y; z/ D x; and
Dg. 2 ; 3 /.a; b; c/.x; y; z/ D Dg .b; c/ B D 2 ; D 3 .a; b; c/.x; y; z/ D
bc c y C b c ln b z: b
Hence,
c bc c c ab b c b c Df .a; b; c/.x; y; z/ D x C a ln a y C b ln b z ; a b and
20
CHAPTER 2
ı f 0 .a; b; c/ D ab c b c a
DIFFERENTIATION
c ab b c ln a ln b : D g x; y C z D g 1 ; 2 C 3 .
ı c ab b c c ln a b
(f) Let g.x; y/ D x y . Then f .x; y; z/ D x yCz Hence,
Df .a; b; c/.x; y; z/ D Dg .a; b C c/ B D 1 ; D 2 C D 3 .a; b; c/.x; y; z/ D Dg .a; b C c/ B x; y C z abCc .b C c/ x C abCc ln a y C z ; D a and
f 0 .a; b; c/ D
abCc .bCc/ a
abCc ln a
abCc ln a :
(g) Let g.x; y/ D x y . Then
f .x; y; z/ D .x C y/z D g x C y; z D g 1 C 2 ; 3 : Hence,
h i Df .a; b; c/.x; y; 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 .a C b/c c D .x C y/ C .a C b/c ln .a C b/ z; .a C b/ and
.aCb/c c .aCb/
.aCb/c c .aCb/
.a C b/c ln .a C b/ : (h) We have f .x; y/ D sin.xy/ D sin B 1 2 . Hence, h i f 0 .a; b/ 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/ f 0 .a; b; c/ D
D cos .ab/ .b; a/ D b cos .ab/ a cos .ab/ : (i) Straightforward. (j) By Theorem 2-3 (3), we have
1 0 sin.xy/ .a; b; c/ Bh C i0 C f 0 .a; b; c/ D B @ sin x sin y .a; b; c/A 0
Œx y 0 .a; b; c/
0
b cos .ab/ B D @cos .a sin b/ sin b ab 1 b
1 a cos .ab/ C a cos .a sin b/ cos b A : ab ln a
t u
I Exercise 41 (2-11). Find f 0 for the following (where g W R ! R is continuous):
SECTION 2.2
21
BASIC THEOREMS
a. f .x; y/ D
R xCy
b. f .x; y/ D
R xy
a a
g.
g.
c. f .x; y; z/ D
R sin
x sin.y sin z /
xy
g.
Rt
g . Then f .x; y/ D h B .1 C 2 / .x; y/, and so h i f 0 .a; b/ D h0 .a C b/ . 1 C 2 /0 .a; b/
Solution. (a) Let h.t / D
a
D g .a C b/ .1; 1/ D g .a C b/ g .a C b/ : (b) Let h.t / D Hence,
Rt
a g . Then f .x; y/ D
R xy a
h i g D h.xy/ D h B 1 2 .x; y/.
h i 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/ : (c) We can rewrite f .x; y; z/ as
Z
a
Z
sin.x sin.y sin z//
Z
Let .x; y; z/ D sin x sin.y sin z/ , k.x; y; z/ D Then f .x; y; z/ D k.x; y; z/ h.x; y; z/, and so
f 0 .a; b; c/ D k 0 .a; b; c/
Z
R .x;y;z/ a
xy
g:
g a
a
a
xy
sin.x sin.y sin z//
gD
gC
f .x; y; z/ D
g , and h.x; y; z/ D
R xy a
g.
h0 .a; b; c/:
It follows from Exercise 40 (d) that
k 0 .a; b; c/ D k 0 .a; b; c/ 0 .a; b; c/: The other parts are easy.
t u
I Exercise 42 (2-12). A function f W Rn Rm ! Rp 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
kf .h; k/k D 0: .h;k/!0 k.h; k/k lim
22
CHAPTER 2
DIFFERENTIATION
b. Prove that Df .a; b/.x; y/ D f .a; y/ C f .x; b/. c. Show that the formula for Dp.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 , respectively. Then for any x 2 Rn and y 2 Rm , we have
xD
n X
x i eni ;
and
yD
iD1
m X
j y j em :
j D1
Therefore,
1 0 1 0 n m n m X X X X jA jA x i eni ; y j em D f @x i eni ; y j em f .x; y/ D f @ i D1
i D1
j D1
D
j D1
n X m X
j f x i eni ; y j em
i D1 j D1
D
n X m X
j x i y j f eni ; em :
i D1 j D1
Then, by letting M D
P
i;j
f e i ; e j , we have n m
X ˇ ˇ
i j i j
Xˇ i j ˇ i j x y f e ; e n m 6 f e ; e kf .x; y/k D ˇx y ˇ n m
i;j
i;j " ˇ ˇ ˇ ˇ# ˇ iˇ ˇ ˇ 6 M max ˇx ˇ max ˇy j ˇ i
j
6 M kxk ky k : Hence,
M khk kkk kf .h; k/k 6 lim .h;k/!0 k.h; k/k .h;k/!0 k.h; k/k M khk kkk D lim .h;k/!0 X 2 2 hi C k j lim
p
i;j
D
M khk kkk : q .h;k/!0 khk2 C kkk2
˚
Now
khk kkk 6
lim
khk2 if kkk 6 khk kkk2
Hence khk kkk 6 khk2 C kkk2 , and so
if khk 6 kkk :
SECTION 2.2
23
BASIC THEOREMS
lim
.h;k/!0
M khk kkk 6 lim M q .h;k/!0 2 2 khk C kkk
q
khk2 C kkk2 D 0:
(b) We have
kf .a C h; b C k/
f .a; b/ f .a; k/ f .h; b/k .h;k/!0 k.h; k/k kf .a; b/ C f .a; k/ C f .h; b/ C f .h; k/ f .a; b/ D lim .h;k/!0 k.h; k/k kf .h; k/k D lim .h;k/!0 k.h; k/k lim
f .a; k/
f .h; b/k
D0 by (a); hence, Df .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. Hence, by (b), we have
Dp.a; b/.x; y/ D p a; y C p .x; b/ D ay C xb:
t u
I Exercise 43 (2-13). Define IP W Rn Rn ! R by IP.x; y/ D hx; yi. a. Find D .IP/ .a; b/ and .IP/0 .a; b/. b. If f; g W R ! Rn are differentiable and h W R ! R is defined by h.t / D hf .t /; g.t /i, show that
D E D E h0 .a/ D f 0 .a/T ; g.a/ C f .a/; g 0 .a/T : D
E
c. If f W R ! Rn is differentiable and kf .t /k D 1 for all t , show that f 0 .t /T ; f .t / D 0. d. Exhibit a differentiable function f W R ! R such that the function jf j defined by jf j .t / D jf .t /j is not differentiable. 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; yi C hx; bi D hb; xi C ha; yi ; and so .IP/0 .a; b/ D .b; a/. (b) Since h.t / D IP B f; g .t /, by the chain rule, we have
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/; g 0 .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
DIFFERENTIATION
h.t / D kf .t /k2 D 1 is constant, and so h0 .a/ D 0; that is,
D E D E D E 0 D f 0 .a/T ; f .a/ C f .a/; f 0 .a/T D 2 f 0 .a/T ; f .a/ ; D
E
and so f 0 .a/T ; f .a/ D 0. (d) Let f .t / D t . Then f is linear and so is differentiable: Df D t . However, lim
t !0C
jtj D 1; t
lim
t!0
jtj D t
1I
that is, jf j is not differentiable at 0.
t u
I Exercise 44 (2-14). Let Ei , i D 1; : : : ; k be Euclidean spaces of various dimensions. A function f W E1 Ek ! Rp is called multilinear if for each choice of xj 2 Ej , j ¤ i the function g W Ei ! Rp defined by g.x/ D f .x1 ; : : : ; xi 1 ; x; xi C1 ; : : : ; xk / is a linear transformation. a. If f is multilinear and i ¤ j , show that for h D .h1 ; : : : ; hk /, with h` 2 E` , we have
lim
f a1 ; : : : ; hi ; : : : ; hj ; : : : ; ak khk
h!0
D 0:
b. Prove that
Df .a1 ; : : : ; ak / .x1 ; : : : ; xk / D
k X
f .a1 ; : : : ; ai
1 ; xi ; aiC1 ; : : : ; ak / :
i D1
Proof. (a) To light notation, define
a
i j
´ a1 ; : : : ; ai
1 ; ai C1 ; : : : ; aj 1 ; aj C1 ; : : : ; ak
Let g W Ei Ej ! Rp be defined as g xi ; xj D f a and so
lim
g a
h!0
i j ; hi ; hj
khk
i j ; xi ; xj
:
. Then g is bilinear
g a i j ; hi ; hj
6 lim D0
h!0
hi ; hj
by Exercise 42 (a). (b) It follows from Exercise 42 (b) immediately.
t u
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 Rn . a. Prove that det W Rn Rn ! R is differentiable and
SECTION 2.2
25
BASIC THEOREMS
1 a1 B C B :: C B:C n X B C C D det .a1 ; : : : ; an / .x1 ; : : : ; xn / D det B B xi C : C B : i D1 B :: C @ A an 0
b. If aij W R ! R are differentiable and f .t / D det aij .t / , show that
0 a .t / B 11: :: B : : B : n X B 0 0 B f .t / D det B aj1 .t / B : j D1 :: B :: : @ an1 .t /
1 a1n .t / :: C C : C C aj0 n .t / C C: :: C : C A ann .t /
¤ 0 for all t and b1 ; : : : ; bn W R ! R are differentiable, let s1 ; : : : ; sn W R ! R be the functions such that s1 .t /; : : : ; sn .t / are the solutions of the equations n X aj i .t /sj .t / D bi .t /; i D 1; : : : ; n:
c. If det aij .t /
j D1
Show that si is differentiable and find si0 .t /. Proof. (a) It is easy to see that det W Rn Rn ! R is multilinear; hence, the conclusion follows from Exercise 44. (b) By (a) and the chain rule,
f 0 .t / D det
0
aij .t / B a10 .t /; : : : ; an0 .t / 0 1 a11 .t / a1n .t / B : :: C :: B : C : B : : C n X B 0 C 0 C D det B B aj1 .t / aj n .t / C : B :: C j D1 :: B ::: : : C @ A an1 .t / ann .t /
(c) Let
0
a11 .t / B : :: : ADB : @ : a1n .t /
1 an1 .t / :: C C : A; ann .t /
0
1 s1 .t / B : C : C sDB @ : A; sn .t /
Then
As D b;
0
1 b1 .t / B : C : C and b D B @ : A: bn .t /
26
CHAPTER 2
and so
si .t / D
DIFFERENTIATION
det .Bi / ; det .A/
where Bi is obtained from A by replacing the i -th column with the b. It follows from (b) that si .t / is differentiable. Define f .t / D det .A/ and gi .t / D det .Bi /. Then 1 0
a .t / B 11: :: B : : B : n X B 0 0 B f .t / D det B a1j .t / B : j D1 :: B :: : @ a1n .t /
an1 .t / :: C C : C C 0 anj .t / C C; :: C : C A ann .t /
and
0
a .t / B 11: :: B : : : n B X B B a0 .t / gi0 .t / D B 1j j D1 B :: B ::: : @ a1n .t /
ai
b1 .t / :: :: : : ai0 1;j .t / bj0 .t / :: :: : : ai 1;n .t / bn .t /
Therefore,
si0 .t / D
1;1 .t /
aiC1;1 .t / :: :: : : 0 aiC1;j .t / :: :: : : ai C1;n .t /
1 an1 .t / :: C C : C C 0 anj .t / C C: :: C : C A ann .t /
f 0 .t /gi0 .t / f .t /gi0 .t / : f 2 .t /
t u
I Exercise 46 (2-16). Suppose f W Rn ! Rn is differentiable and has a differenh i 1 0 . tiable inverse f 1 W Rn ! Rn . Show that f 1 .a/ D f 0 f 1 .a/ Proof. We have f B f 1 .x/ D x . On the one hand D f B f f B f 1 is linear; on the other hand,
D f Bf Therefore, Df
1
1
.a/ .x/ D Df f
h .a/ D Df f
1
i .a/
1
.a/ B Df
1
1
.a/ .x/ D x since
.a/ .x/:
1
.
2.3 Partial Derivatives I Exercise 47 (2-17). Find the partial derivatives of the following functions: a. f .x; y; z/ D x y . 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/ .
t u
SECTION 2.3
27
PARTIAL DERIVATIVES 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/
.
Solution. Compare this with Exercise 40. (a) D1 f .x; y; z/ D yx y
1
, D2 f .x; y; z/ D x y ln x , and D3 f .x; y; z/ D 0.
(b) D1 f .x; y; z/ D D2 f .x; y; z/ D 0, and D3 f .x; y; z/ D 1.
(c) D1 f .x; y/ D sin y cos x sin y , and D2 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 z y z ln y x y ln x .
z
1
z
, D2 f .x; y; z/ D x y zy z
(f) D1 f .x; y; z/ D y C z x yCz
1
1
ln x , and D3 f .x; y; z/ D
, and D2 f .x; y; z/D3 f .x; y; z/ D x yCz ln x .
(g) D1 f .x; y; z/ D D2 f .x; y; z/ D z.x C y/z D3 f .x; y; z/ D .x C y/z ln.x C y/.
1
, and
(h) D1 f .x; y/ D y cos.xy/, and D2 f .x; y/ D x cos.xy/.
cos 3
(i) D1 f .x; y/ D cos 3 sin.xy/
cos 3 D2 f .x; y/ D cos 3 sin.xy/
1
1
y cos.xy/, and
x cos.xy/.
t u
I Exercise 48 (2-18). Find the partial derivatives of the following functions (where g W R ! R is continuous): a. f .x; y/ D
R xCy a
g.
Rx
y g. R xy c. f .x; y/ D a g .
b. f .x; y/ D
d. f .x; y/ D
R .R y g / b
a
g.
Solution. (a) D1 f .x; y/ D D2 f .x; y/ D g.x C y/. (b) D1 f .x; y/ D g.x/, and D2 f .x; y/ D
g y .
(c) D1 f .x; y/ D yg.xy/, and D2 f .x; y/ D xg.xy/.
28
CHAPTER 2
R y b g .
(d) D1 f .x; y/ D 0, and D2 f .x; y/ D g y g
DIFFERENTIATION
t u
I Exercise 49 (2-19). If f .x; y/ D x x
xx
y
0
C ln x @arctan arctan arctan sin cos xy
!1 A ln.x C y/
find D2 f 1; y .
Solution. Putting x D 1 into f .x; y/, we get f 1; y D 1. Then D2 f 1; y D 0. t u
I Exercise 50 (2-20). Find the partial derivatives of f in terms of the derivatives of g and h if a. f .x; y/ 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/. Solution. (a) D1 f .x; y/ D g 0 .x/ h y , and D2 f .x; y/ D g.x/h0 y .
(b) D1 f .x; y/ D h y g.x/h.y /
g .x/, and D2 f .x; y/ D h0 y g.x/h.y / ln g.x/.
1 0
(c) D1 f .x; y/ D g 0 .x/, and D2 f .x; y/ D 0. (d) D1 f .x; y/ D 0, and D2 f .x; y/ D g 0 y .
(e) D1 f .x; y/ D D2 f .x; y/ D g 0 .x C y/.
t u
I Exercise 51 (2-21 ). Let g1 ; g2 W R2 ! R be continuous. Define f W R2 ! R by Z x Z y f .x; y/ D g1 .t; 0/ dt C g2 .x; t/ dt: 0
0
a. Show that D2 f .x; y/ D g2 .x; y/. b. How should f be defined so that D1 f .x; y/ D g1 .x; y/? c. Find a function f W R2 ! R such that D1 f .x; y/ D x and D2 f .x; y/ D y . Find one such that D1 f .x; y/ D y and D2 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
Z
Z
g1 t; y dt C
f .x; y/ D 0
y
g2 .a; t/ dt; 0
where t 2 R is a constant. (c) Let
f .x; y/ D x 2 C y 2 =2. f .x; y/ D xy .
t u
I Exercise 52 (2-22 ). If f W R2 ! R and D2 f D 0, show that f is independent of the second variable. If D1 f D D2 f D 0, show that f is constant. Proof. Fix any x 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 proved immediately. t u
˚ 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 D1 f D D2 f D 0, show that f is constant. b. Find a function f W A ! R such that D2 f D 0 but f is not independent of the second variable. Proof. (a) As in Figure 2.1, for any .a; b/; .c; d / 2 R2 , we have
f .a; b/ D f . 1; b/ D f . 1; d / D f .c; d / : (b) For example, we can let
˚
f .x; y/ D
0 x
if x < 0 or y < 0
t u
otherwise.
I Exercise 54 (2-24). Define f W R2 ! R by
˚
f .x; y/ D
2
2
xy xx2 Cyy 2
.x; y/ ¤ 0;
0
.x; y/ D 0:
a. Show that D2 f .x; 0/ D x for all x and D1 f 0; y D
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 D1;2 f .0; 0/ ¤ D2;1 f .0; 0/. Proof.
(a) We have
x .x 4 y 4 4x 2 y 2 /
.x; y/ ¤ 0;
2
.x 2 Cy 2 /
D2 f .x; y/ D
.x; y/ D 0;
0
and
y .y 4 x 4 4x 2 y 2 /
.x; y/ ¤ 0;
2
.x 2 Cy 2 /
D1 f .x; y/ D
.x; y/ D 0:
0
Hence, D2 f .x; 0/ D x and D1 f 0; y D
y.
(b) By (a), we have D1;2 f .0; 0/ D D2 D1 f 0; y
D1
.0/ D
1; but D2;1 f .0; 0/ D
D2 .x; 0/ .0/ D 1.
t u
I Exercise 55 (2-25 ). Define f W R ! R by
˚
f .x/ D
2
x
e
x¤0 x D 0:
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 ¤ 0 and
k 2 N, we show that f .k/ .x/ D p3k x 3
x
2
1
Step 1
Clearly, f 0 .x/ D 2x
Step 2
Suppose that f .k/ .x/ D p3k x
Step 3
Then by the chain rule,
e
e
x
2
1
e
x
. We do this by induction.
.
2
.
SECTION 2.3
31
PARTIAL DERIVATIVES
y
−2
0
−1
2 x
1
Figure 2.2.
h i0 f .kC1/ .x/ D f .k/ .x/ 2 0 D p3k x 1 x 2 e x C p3k x 1 2x 0 x 1 x 2 C p3k x 1 2x 3 e D p3k 2 D q3kC1 x 1 C q3kC3 x 1 e x 2 D p3.kC1/ x 1 e x ;
3
e
x
2
x
2
where q3kC1 and q3kC3 are polynomials. Therefore, f .x/ 2 C 1 for all x ¤ 0. It remains to show that f .k/ .x/ is defined and continuous at x D 0 for all k . Step 1
Obviously,
f 0 .0/ D lim
f .x/
f .0/
D lim
x
x!0
e
x!0
x
x
2
D lim 2x
3
x!0
e
x
2
D0
by L’Hôpital’s rule. Step 2
Suppose that f .k/ .0/ D 0.
Step 3
Then,
f .kC1/ .0/ D lim
f .k/ .x/
f .k/ .0/
x D lim p3kC1 x 1 e x!0 p3kC1 x 1 D lim : x!0 ex 2 x!0
x
2
Hence, if we use L’Hôpital’s rule 3k C 1 times, we get f .kC1/ .0/ D 0. A similar computation shows that f .k/ .x/ is continuous at x D 0.
I Exercise 56 (2-26 ). Let
˚
f .x/ D
e 0
.x 1/
2
e
.xC1/
2
x 2 . 1; 1/ ; x … . 1; 1/ :
t u
32
CHAPTER 2
DIFFERENTIATION
a. Show that f W R ! R is a C 1 function which is positive on . 1; 1/ and 0 elsewhere. b. Show that there is a C 1 function g W R ! Œ0; 1 such that g.x/ D 0 for x 6 0 and g.x/ D 1 for x > ". 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 " an
"; an C "
and zero elsewhere. d. If A Rn is open and C A is compact, show that there is a non-negative C 1 function f W A ! R such that f .x/ > 0 for x 2 C and f D 0 outside of some closed set contained in A. e. Show that we can choose such an f so that f W A ! Œ0; 1 and f .x/ D 1 for x 2 C. Proof. (a) If x 2 . 1; 1/, then x 1 ¤ 0 and x C 1 ¤ 0. It follows from Exercise 55 that 2 2 e .x 1/ 2 C 1 and e .xC1/ 2 C 1 . Then it is straightforward to check that f 2 C 1 . See Figure 2.3 y
1 x
0
−1
Figure 2.3.
(b) By letting z D x C 1, we derive a new function j W R ! R from f as follows:
˚
j .z/ D
e
.z 2/
2
e
z
2
z 2 .0; 2/ ; z … .0; 2/ :
0
By letting w D "z=2, we derive a function k W R ! R from j as follows:
˚
k .w/ D
e .2w=" 2/ 0
2
e .2w="/
2
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 2 C 1 , which is positive on .0; "/ and 0 elsewhere. Now let x
Z g.x/ D
,Z k .x/
0
Then g 2 C
1
"
k .x/ :
0
; it is 0 for x 6 0, increasing on .0; "/, and 1 for x > ".
(c) It follows from (a) immediately. (d) For every x 2 C , let Rx ´ . "; "/n be a rectangle containing x , and Rx is contained in A (we can pick such a rectangle since A is open and C A). Then fRx W x 2 C g is an open cover of C . Since C is compact, there exists ˚ fx1 ; : : : ; xm g C such that Rx1 ; : : : ; Rxm covers C . For every xi , i D 1; : : : ; n, we define a function gi W Rxi ! R as
gi .x/ D f
xi1
ai1
!
"
f
xin
ain "
;
where ai1 ; : : : ; ain 2 Rn is the middle point of Rxi . Finally, we define g W Rx1 [ [ Rxm ! R as follows:
g.x/ D
m X
gi .x/ :
i D1
Then g 2 C 1 ; it is positive on C , and 0 outside Rx1 [ [ Rxm . (e) Follows the hints.
t u
˚ I Exercise 57 (2-27). Define g; h W x 2 R2 W kxk 6 1 ! R3 by q 2 2 g.x; y/ D x; y; 1 x y ; q 2 2 1 x y : h.x; y/ D x; y; Show that the maximum of f on x 2 R3 W kxk D 1 is either the maximum of ˚ f B g or the maximum of f B h on x 2 R2 W kxk 6 1 .
˚
Proof. Let A ´ g .A/ [ h .A/.
˚
˚ x 2 R2 W kxk 6 1 and B ´ x 2 R3 W kxk D 1 . Then B D t u
34
CHAPTER 2
DIFFERENTIATION
2.4 Derivatives I Exercise 58 (2-28). Find expressions for the partial derivatives of the following functions: a. F .x; y/ D f g.x/k y ; g.x/ C h y . b. F .x; y; z/ D f
g.x C y/; h y C z .
c. F .x; y; z/ D f x y ; y z ; z x .
d. F .x; y/ D f x; g.x/; h.x; y/ . Proof.
(a) Letting a ´ g.x/k y ; g.x/ C h y , we have
D1 F .x; y/ D D1 f .a/ g 0 .x/ k y C D2 f .a/ g 0 .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 y
1
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 z z
D3 F .x; y; z/ D D2 f .a/ y ln y C D3 f .a/ xz
1
;
x 1
:
(d) Letting a ´ x; g.x/; h.x; y/, we have
D1 F .x; y/ D D1 f .a/ C D2 f .a/ g 0 .x/ C D3 f .a/ D1 h.x; y/ D2 F .x; y/ D D3 f .a/ D2 h.x; y/:
t u
I Exercise 59 (2-29). Let f W Rn ! R. For x 2 Rn , the limit lim
t !0
f .a C tx/ t
f .a/
;
if it exists, is denoted Dx f .a/, and called the directional derivative of f at a, in the direction x . a. Show that Dei f .a/ D Di f .a/. b. Show that D tx f .a/ D t Dx f .a/.
SECTION 2.4
35
DERIVATIVES
c. If f is differentiable at a, show that Dx f .a/ D Df .a/.x/ and therefore DxCy f .a/ D Dx f .˛/ C Dy f .a/. Proof. (a) For ei D .0; : : : ; 0; 1; 0; : : : ; 0/, we have
f .a C t ei / f .a/ t !0 t f .a1 ; : : : ; ai 1 ; ai C t; ai C1 ; : : : ; an / D lim t !0 t D Di f .a/
Dei f .a/ D lim
f .a/
by definition. (b) We have
D tx f .a/ D lim
s!0
f .a C stx/ s
f .a/
D lim t st !0
f .a C st x/ st
f .a/
D tDx f .a/:
(c) If f is differentiable at a, then for any x ¤ 0 we have
0 D lim
jf .a C t x/
t !0
D lim
jf .a C tx/
t !0
ˇ ˇ f .a C tx/ D lim ˇˇ t !0 t
f .a/ ktxk f .a/ jtj f .a/
and so
f .a C tx/ t The case of x D 0 is trivial. Therefore, Dx f .a/ D lim
Df .a/.t x/j t Df .a/.x/j ˇ ˇ Df .a/.x/ˇˇ f .a/
t!0
1 kxk 1 ; kxk
D Df .a/.x/:
DxCy f .a/ D Df .a/ .x C y/ D Df .a/.x/ C Df .a/ .y/ D Dx f .a/ C Dy f .a/:
t u
I Exercise 60 (2-30). Let f be defined as in Exercise 34. Show that Dx f .0; 0/ exists for all x , but if g ¤ 0, then DxCy f .0; 0/ ¤ Dx f .0; 0/ C Dy f .0; 0/ for all x; y . Proof. Take any x 2 R2 .
lim
t!0
f .tx/
f .0; 0/ t
jtj kxk g tx D lim
t!0
t
.
jt j kxk
:
Therefore, Dx f .0; 0/ exists for any x . Now let g ¤ 0; then, D.0;1/ f .0; 0/ D D.1;0/ f .0; 0/ D 0, but D.1;0/C.0;1/ f .0; 0/ D D.1;1/ f .0; 0/ ¤ 0. t u
36
CHAPTER 2
DIFFERENTIATION
I Exercise 61 (2-31). Let f W R2 ! R be defined as in Exercise 26. Show that Dx f .0; 0/ exists for all x , although f is not even continuous at .0; 0/. Proof. For any x 2 R2 , we have lim
f .t x/
f .0/ t
t !0
D lim
t !0
f .t x/ D0 t
by Exercise 26 (a).
t u
I Exercise 62 (2-32). a. Let f W R ! R be defined by
˚
f .x/ D
x 2 sin x1
x¤0
0
x D 0:
Show that f is differentiable at 0 but f 0 is not continuous at 0. b. Let f W R2 ! R be defined by
‚
x 2 C y 2 sin q
f .x; y/ D
1 2
x C y2
.x; y/ ¤ 0 .x; y/ D 0:
0
Show that f is differentiable at .0; 0/ but Di f is not continuous at .0; 0/. Proof. (a) We have lim
f .x/
f .0/ x
x!0
x 2 sin x1 1 D lim x sin D 0: x!0 x!0 x x
D lim
Hence, f 0 .0/ D 0. Further, for any x ¤ 0, we have
f 0 .x/ D 2x sin
1 x
cos
1 : x
It is clear that limx!0 f 0 .x/ does not exist. Therefore, f 0 is not continuous at 0. (b) Since
x 2 C y 2 sin q lim
.x;y/!.0;0/
q
1 2
x C y2
x2 C y2
lim
D
q
.x;y/!.0;0/
x 2 C y 2 sin q
1 x2 C y2
we know that f 0 .0; 0/ D .0; 0/. Now take any .x; y/ ¤ .0; 0/. Then
D1 f .x; y/ D 2x sin q
1 2
x Cy
2
2x cos q
1 2
x Cy
: 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 D2 f .
t u
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 a 1 ,
lim
ˇ ˇ ˇf a1 C h1 ; a
h!0
6 lim
1
ˇ ˇ D1 f .a/ h1 ˇ
f .a/
khk ˇ ˇ ˇf a1 C h1 ; a
1
h1 !0
f .a/ ˇ ˇ ˇh1 ˇ
ˇ ˇ D1 f .a/ h1 ˇ
D 0:
See aslo Apostol (1974, Theorem 12.11).
t u
I Exercise 64 (2-34). A function f W Rn ! R is homogeneous of degree m if f .t x/ D t m f .x/ for all x . If f is also differentiable, show that n X
x i Di f .x/ D mf .x/:
i D1
Proof. Let g.t / D f .tx/. Then, by Theorem 2-9,
g 0 .t / D
n X
Di f .tx/ x i :
(2.4)
i D1
On the other hand, g.t / D f .t x/ D t m f .x/; then
g 0 .t / D mt m
1
f .x/:
Combining (2.4) and (2.5), and letting t D 1, we then get the result.
(2.5)
t u
I Exercise 65 (2-35). If f W Rn ! R is differentiable and f .0/ D 0, prove that there exist gi W Rn ! R such that
38
CHAPTER 2
f .x/ D
n X
DIFFERENTIATION
x i gi .x/:
iD1
Proof. Let hx .t / D f .t x/. Then 1
Z
h0x .t / dt D hx .1/
0
hx .0/ D f .x/
f .0/ D f .x/:
Hence,
Z f .x/ D 0
1
h0x .t / dt D
Z
1
f 0 .tx/ dt D
0
1
Z 0
D D
n X iD1 n X
2 3 n X 4 xi Di f .t x/5 dt iD1
xi
1
Z
Di f .t x/ dt 0
x i gi .x/;
i D1
where gi .x/ D
R1 0
Di f .t x/ dt .
t u
2.5 Inverse Functions For this section, Rudin (1976, Section 9.3 and 9.4) is a good reference.
I Exercise 66 (2-36 ). Let A Rn be an open set and f W A ! Rn a continuously differentiable 1-1 function such that det f 0 .x/ ¤ 0 for all x . Show that f .A/ is an open set and f 1 W f .A/ ! A is differentiable. Show also that f .B/ is open for any open set B A. 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 Rn 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 differentiable, it follows that f 1 is differentiable at y ; since y is chosen arbitrary, it follows that f 1 W f .A/ ! A is differentiable. 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 t u
I Exercise 67 (2-37). a. Let f W R2 ! R be a continuously differentiable function. Show that f is not 1-1.
SECTION 2.5
39
INVERSE FUNCTIONS
b. Generalize this result to the case of a continuously differentiable function f W Rn ! Rm with m < n. Proof. (a) Let f 2 C 0 . Then both D1 f and D2 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 there is x0 ; y0 2 R2 such that D1 f x0 ; f0 ¤ 0. The continuity of D1 f implies that there is an open set A R2 containing x0 ; y0 such that D1 f .x/ ¤ 0 for all x 2 A. Define a function g W A ! R2 with
g.x; y/ D f .x; y/; y : Then for all .x; y/ 2 A,
! D1 f .x; y/ D2 f .x; y/ g .x; y/ D ; 0 1 0
and so det g 0 .x; y/ D D1 f .x; y/ ¤ 0; furthermore, g 2 C 0 .A/ and g is 1-1. Then by Exercise 66, we know that g .A/ is open. We now show that g .A/ cannot be open actually. Take a point f x0 ; y0 ; yz 2 g .A/ with y ¤ y0 . Then for any .x; y/ 2 A, we
must have
g.x; y/ D f .x; y/; y D f x0 ; y0 ; yz H) .x; y/ D x0 ; y0 I
that is, there is no .x; y/ 2 A such that g.x; y/ D f x0 ; y0 ; yz . This proves that f cannot be 1-1. (b) We can write f W Rn ! Rm as f D f 1 ; : : : ; f m , where f i W Rn ! R for every i D 1; : : : ; m. As in (a), there is a mapping, say, f 1 , a point a 2 Rn , and an open set A containing a such that D1 f 1 .x/ ¤ 0 for all x 2 A. Define g W A ! Rm as
g x1; x where x
1
1
D f .x/; x
1
;
´ x 2 ; : : : ; x n . Then as in (a), it follows that f cannot be 1-1.
t u
I Exercise 68 (2-38). a. If f W R ! 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/ ¤ 0 for all .x; y/ but f is not 1-1.
Proof.
40
CHAPTER 2
DIFFERENTIATION
(a) Suppose that f is not 1-1. Then there exist a; b 2 R with a < b such that f .a/ D f .b/. It follows from the mean-value theorem that there exists c 2 .a; b/ such that
f .a/ D f 0 .c/ .b
0 D f .b/
a/ ;
which implies that f 0 .c/ D 0. A contradiction. (b) We have
Dx e x cos y f .x; y/ D Dx e x sin y 0
D
e x cos y e x sin y
Dy e x cos y Dy e x sin y ! e x sin y : e x cos y
!
Then
det f 0 .x; y/ D e 2x cos2 y C sin2 y D e 2x ¤ 0:
However, f .x; y/ is not 1-1 since f .x; y/ D f x; y C 2k for all .x; y/ 2 R2 and k 2 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). t u
I Exercise 69 (2-39). Use the function f W R ! R defined by
˚
f .x/ D
x 2
C x 2 sin x1
0
x¤0 xD0
to show that continuity of the derivative cannot be eliminated from the hypothesis of Theorem 2-11. Proof. If x ¤ 0, then
f 0 .x/ D if x D 0, then
1 1 C 2x sin 2 x
cos
1 I x
h=2 C h2 sin 1= h 1 D : f .0/ D lim h 2 h!0 0
Hence, f 0 .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). Proof. Define f W R Rn ! Rn by
SECTION 2.6
41
IMPLICIT FUNCTIONS
0
Figure 2.6.
t u
f i .t; s/ D
n X
aj i .t /s j
bi .t /;
j D1
for i D 1; : : : ; n. Then
0
D2 f 1 .t; s/ B :: :: M´B : : @ n D2 f .t; s/
1 0 a11 .t / D1Cn f 1 .t; s/ C B : :: :: CDB : : : A @ : n a1n .t / D1Cn f .t; s/
1 an1 .t / :: C C : A; ann .t /
and so det .M/ ¤ 0. It follows from the Implicit Function Theorem that for each t 2 R, there is a unique s.t / 2 Rn such that f t; s.t / D 0, and s is differentiable. u t
I Exercise 71 (2-41). Let f W R R ! R be differentiable. For each x 2 R define gx W R ! R by gx 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 D2;2 f .x; y/ ¤ 0 for all .x; y/, show that c is differentiable and 0
c .x/ D
D2;1 f x; c .x/
: D2;2 f x; c .x/
b. Show that if c 0 .x/ D 0, then for some y we have
D2;1 f .x; y/ D 0; D2 f .x; y/ D 0: c. Let f .x; y/ D x y log y
y
y log x . Find
42
CHAPTER 2
" max
1=26x62
DIFFERENTIATION
# min f .x; y/ :
1=36y61
Proof. (a) For every x , we have gx0 y D D2 f .x; y/. Since for every x there is a unique y D c .x/ such that D2 f x; c .x/ D 0, the solution c .x/ is the same as obtained from the Implicit Function Theorem; hence, c .x/ is differentiable, and by differentiating D2 f x; c .x/ D 0 with respect to x , we have
D2;1 f x; c .x/ C D2;2 f x; c .x/ c 0 .x/ D 0I that is, 0
c .x/ D
D2;1 f x; c .x/ : D2;2 f x; c .x/
(b) It follows from (a) that if c 0 .x/ D 0, then D2;1 f x; c .x/ D 0. Hence, there exists some y D c .x/ such that D2;1 f .x; y/ D 0. Furthermore, by definition, D2 x; c .x/ D D2 f .x; y/ D 0.
(c) We have
D2 f .x; y/ 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
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/ D 1, and c.a/ D 1=3 for some a > 1=2 (see Figure 2.7). Therefore,
min f x; y D f x; y .x/ ;
1=36y61
where (see Figure 2.8)
SECTION 2.6
IMPLICIT FUNCTIONS
„
y .x/ D
43 if 1=2 6 x 6 a
1=3 c .x/ D x
1=x
if a < x 6 1 if 1 < x 6 2:
1
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 C ln 3 1 ln x: x 3 3 1=26x6a It is easy to see that x D 1=2, and so f x ; 1=3 D ln 4=3e =6.
max f x; 1=3 D
a 0, we have U f; P 0 L f; P 0 < "=2 and U g; P 00 L g; P 00 < "=2. Let Px refine both P 0 and P 00 . Then
U f; Px Hence,
" L f; Px < 2
and
U g; Px
" L g; Px < : 2
48
CHAPTER 3
U f C g; Px
INTEGRATION
L f C g; Px < ";
and so f C g is integrable. Now, by definition, for any " > 0, there exists a partition P (by using a R common refinement partition if necessary) such that A f < L f; P C "=2, R R R A g < L g; P C"=2, U f; P < A f C"=2, and U g; P < A g C"=2. Therefore,
Z
Z f C
Z
" < L f; P C L g; P 6 L f C g; P 6
g
A
A
f Cg
A
6 U f C g; P 6 U f; P C U g; P Z Z < f C g C ": A
Hence,
R
f Cg D
A
R A
f C
R A
A
g.
(c) First, suppose that c > 0. Then for any partition P and any subrectangle S , we have mS cf D cmS f and MS cf D cMS f . But then L cf; P D cL f; P and U cf; P D cU f; P . Since f is integrable, for any " > 0 there exists a partition P such that U f; P L f; P < "=c . Therefore, h i U cf; P L cf; P D c U f; P L f; P < "I that is, cf is integrable. Further,
" < cL f; P D L cf; P 6 c
Z c
f A
R
Z
cf 6 U cf; P D cU f; P A Z "
X
h
MS2 f
mS2 f
i
v .S2 /
S2 22 .S /
D U f S; P2
L f S; P2 I
that is, f S is integrable.
y d
c x 0
a
b Figure 3.2.
If: Now suppose that f S is integrable for each S . For each partition P 0 , let ˇ 0ˇ ˇP ˇ be the number of subrectangles induced by P 0 . Let PS be a partition such that " U f S; PS L f S; PS < : 2jP j Let P 0 be the partition of A obtained by taking the union of all the subsequences defining the partitions of the PS ; see Figure 3.2. Then there are
50
CHAPTER 3
INTEGRATION
refinements PS0 of PS whose rectangles are the set of all subrectangles of P 0 which are contained in S . Hence,
XZ
f S
"<
S
S
X
X L f S; PS 6 L f S; PS0 D L f; P 0
S
S
6 U f; P 0 X D U f S; PS0 S
6
X
U f S; PS
S
<
XZ
Therefore, f is integrable, and
R A
f D
P R S
S
f S C ": S
S
f S .
t u
I Exercise 76 (3-5). Let f; g W A ! R be integrable and suppose f 6 g . Show R R that A f 6 A g . Proof. Since f is integrable, the function f is integrable by Exercise 74 (c); R 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
R
AgC
R A
R f D Ag
R
A f ; hence,
R
Af 6
R A
g
f
D
R A gC
f
R
A g.
D t u
I Exercise 77 (3-6). If f W A ! R is integrable, show that jf j is integrable and ˇ R ˇR ˇ f ˇ 6 jf j. A A Proof. Let f C D max ff; 0g and f
f DfC
f
D max f f; 0g. Then 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 f C and f are integrable if f is. Further,
ˇZ ˇ ˇZ ˇ ˇ ˇ ˇ f ˇ Dˇ fC ˇ ˇ ˇ A
A
ˇ Z ˇˇ ˇˇZ ˇ C ˇ ˇ f ˇˇ f ˇ Dˇ f ZA ZA 6 fCC f A A Z C D f Cf ZA D jf j : A
I Exercise 78 (3-7). Let f W Œ0; 1 Œ0; 1 ! R be defined by
t u
SECTION 3.3
51
FUBINI’S THEOREM
„
f x; y D
0
x irrational
0
x rational, y irrational
1=q x rational, y D p=q is lowest terms. Show that f is integrable and
R Œ0;1Œ0;1
f D 0. t u
Proof.
3.2 Measure Zero and Content Zero I Exercise 79 (3-8). Prove that Œa1 ; b1 Œan ; bn does not have content 0 if ai < bi for each i . Proof. Similar to the Œa; b case.
t u
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.
t u
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.
t u
Proof.
3.3 Fubini’s Theorem I Exercise 82 (3-27). If f W Œa; b Œa; b ! R is continuous, show that Z
b
Z
y
Z
b
b
Z
f x; y dy dx:
f x; y dx dy D a
a
Proof. As illustrated in Figure 3.3,
a
x
52
CHAPTER 3
n
INTEGRATION
o
x; y 2 Œa; b2 W a 6 x 6 y and a 6 y 6 b n o D x; y 2 Œa; b2 W a 6 x 6 b and x 6 y 6 b :
C D
y y
D
x
y first b
C x first
a x 0
a
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 Z Z 1C x; y dx dy D 1C x; y dy dx D 0: Œ0;1
Œ0;1
Œ0;1
Œ0;1
Proof. There must be typos.
t u
I Exercise 84 (3-31). If A D Œa1 ; b1 Œan ; bn and f W A ! R is continuous, define F W A ! R by Z F .x/ D What is Di F .x/, for x 2 int.A/? Solution. Let c 2 int.A/. Then
f: Œa1 ;x 1 Œan ;x n
SECTION 3.3
53
FUBINI’S THEOREM
Di F .c/ D lim
F c i ; ci C h
F .c/
h
h!0
R
Œa1 ;c 1 Œai ;c i ChŒan ;c n f F .c/ h h!0 R c i Ch R f dxi 1 i 1 i C1 n ai Œa1 ;c Œai 1 ;x Œai C1 ;c Œan ;c D lim h h!0 R c i Ch R f dxi 1 i 1 i C1 n ci a ;c a ;c a ;c Œa ;c Œ 1 Œ i 1 Œ i C1 n D lim h h!0 Z D f x i ; ci : Œa1 ;c 1 Œai 1 ;c i 1 Œai C1 ;c i C1 Œan ;c n D lim
F .c/
t u
I Exercise 85 (3-32 ). Let f W Œa; b Œc; d ! R be continuous and suppose Rb D2 f is continuous. Define F y D a f x; y dx . Prove Leibnitz’s rule: F 0 y D Rb a D2 f x; y dx . Proof. We have
F yCh F y F y D lim h h!0 Rb Rb a f x; y C h dx a f x; y dx D lim h h!0 Z b f x; y C h f x; y dx: D lim h h!0 a 0
By DCT, we have 0
b
Z
F y D a
"
f x; y C h lim h h!0
f x; y
# dx
b
Z
D2 f x; y dx:
D
t u
a
I Exercise 86 (3-33). If f W Œa; b Œc; d ! R is continuous and D2 f is contin Rx uous, define F x; y D a f t; y dt . a. Find D1 F and D2 F . b. If G .x/ D
R g.x/ a
f .t; x/ dt , find G 0 .x/.
Solution.
(a) D1 F x; y D f x; y , and D2 F D
Rx a
D2 f t; y dt .
(b) It follows that G .x/ D F g .x/ ; x . Then
G 0 .x/ D g 0 .x/ D1 F g .x/ ; x C D2 F g .x/ ; x Z g.x/ D g 0 .x/ f g .x/ ; x C D2 f .t; x/ dt: a
t u
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 / D 1. What would the right side be if the factor .k C `/Š=kŠ`Š did not appear in the definition of ^?
˘
b. Show that 'i1 ^ ^ 'ik .v1 ; : : : ; vk / is the determinant of the k k minor of
v1 :: : vk
obtained by selecting columns i1 ; : : : ; ik .
Proof. (a) Since 'ij 2 T .Rn /, for every j D 1; : : : ; k , we have
kŠ Alt 'i1 ˝ ˝ 'ik .ei1 ; : : : ; eik / 1Š 1Š X D .sgn. //'i1 .e .i1 / / 'ik .e .ik / /
'i1 ^ ^ 'ik .ei1 ; : : : ; eik / D
2Sk
D 1: If the factor .k C `/Š=kŠ`Š did not appear in the definition of ^, then the solution would be 1=kŠ. (b)
t u
I Exercise 88 (4-9 ). Deduce the following properties of the cross product in R3 . e1 e1 D 0 a. e1 e2 D e3 e1 e3 D e2
e2 e1 D e3 e2 e2 D 0 e2 e3 D e1
e3 e1 D e2 e3 e2 D e1 e3 e3 D 0
Proof. 55
56
CHAPTER 4
INTEGRATION ON CHAINS
(a) We just do the first line.
e
1
hw; zi D e1 D 0 H) z D e1 e1 D 0;
w
e
2
hw; zi D e1 D w3 H) e2 e1 D e3 ;
w
e
3
hw; zi D e1 D w2 H) e3 e1 D e2 :
w t u
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]
57
Index
Directional derivative, 24
59
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