Calculus on Manifolds - Jianfei Shen 行到水窮處 ... · Calculus on Manifolds A Solution Manual forSpivak(1965) Jianfei Shen School of Economics, The University of New South Wales

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

1.1 Norm and Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Subsets of Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Functions and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 Basic Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.4 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.5 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.6 Implicit Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.2 Measure Zero and Content Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.3 Fubini’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4 Integration on Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.1 Algebraic Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

iii

1FUNCTIONS ON EUCLIDEAN SPACE

1.1 Norm and Inner Product

I Exercise 1 (1-1�). Prove that kxk 6PniD1

ˇ̌̌xiˇ̌̌.

Proof. Let x D�x1; : : : ; xn

�. Then0@ nX

iD1

ˇ̌̌xiˇ̌̌1A2 D nX

iD1

�xi�2C

Xi¤j

ˇ̌̌xixj

ˇ̌̌>

nXiD1

�xi�2D kxk

2 :

Taking the square root of both sides gives the result. ut

I Exercise 2 (1-2). When does equality hold in Theorem 1-1 (3)�kx C yk 6 kxk C kyk

�?

Proof. We reprove thatˇ̌hx;yi

ˇ̌6 kxk � kyk for every x;y 2 Rn. Obviously, if

x D 0 or y D 0, then hx;yi D kxk � kyk 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 kyk2

implies that

˛ D hx;yiıkyk

2 :

Then

kxk2D kwk

2C k˛yk

2 > k˛yk2D

�hx;yi

kyk

�2:

Hence,ˇ̌hx;yi

ˇ̌6 kxk � kyk. 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 yk2D hx C y;x C yi D kxk

2C kyk

2C 2 hx;yi 6 kxk2 C kyk2 C 2 kxk � kyk

D�kxk C kyk

�2;

1

2 CHAPTER 1 FUNCTIONS ON EUCLIDEAN SPACE

equality holds precisely when hx;yi D kxk�jjyjj, i.e., when one is a nonnegative

multiple of the other. ut

I Exercise 3 (1-3). Prove that kx � yk 6 kxk C kyk. When does equality hold?

Proof. By Theorem 1-1 (3) we have kx � yk D kx C .�y/k 6 kxk C k�yk D

kxkC kyk. The equality holds precisely when one vector is a non-positive mul-

tiple of the other. ut

I Exercise 4 (1-4). Prove thatˇ̌kxk � kyk

ˇ̌6 kx � yk.

Proof. We have kx � yk2DPniD1

�xi � yi

�2D kxk

2C kyk

2� 2

PniD1 xiyi >

kxk2C kyk

2� 2 kxk kyk D

�kxk � kyk

�2. Taking the square root of both sides

gives the result. ut

I Exercise 5 (1-5). The quantity ky � xk is called the distance between x and y .

Prove and interpret geometrically the “triangle inequality”: kz � xk 6 kz � ykC

ky � xk.

Proof. The inequality follows from Theorem 1-1 (3):

kz � xk D k.z � y/C .y � x/k 6 kz � yk 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�.

a. Prove thatˇ̌̌R baf � g

ˇ̌̌6�R baf 2� 1

2

�

�R bag2� 1

2

.

b. If equality holds, must f D �g for some � 2 R? What if f and g are continu-

ous?

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:ˇ̌̌̌ˇ̌Xi

f .xi / g .xi /�xi

ˇ̌̌̌ˇ̌ 6

0@Xi

f .xi /2�xi

1A1=20@Xi

g .xi /2�xi

1A1=2 :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 NORM AND INNER PRODUCT 3�f .x/ � �g .x/

�2> 0. Then the inequality holds true in an open neighbor-

hood of x since f and g are continuous. SoR ba

�f � �g

�2> 0 since the in-

tegrand is always non-negative and is positive on some subinterval of Œa; b�.

Expanding out givesR baf 2 � 2�

R baf � g C �2

R bag2 > 0 for all �. Since the

quadratic has no solutions, it must be that its discriminant is negative.

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). ut

I Exercise 7 (1-7). A linear transformation M W Rn ! Rn is called norm pre-

serving if kM xk 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 DkM x CM yk

2� kM x �M yk

2

4

DkM .x C y/k2 � kM .x � y/k2

4

Dkx C yk

2� kx � yk

2

4

D hx;yi :

If M is inner product preserving, then one has by Theorem 1-1 (4):

kM xk DphM x;M xi D

phx;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 yi D hx � y;x � 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�1 xk D

M �M�1 x

� D kxk ;and D

M�1 x;M�1 yED

�M�M�1 x

�;M

�M�1 y

��D hx;yi :

Therefore, M�1 is also norm preserving and inner product preserving. ut


I Exercise 8 (1-8). If x;y 2 Rn are non-zero, the angle between x and y ,

denoted † .x;y/, is defined as arccos�hx;yi

ıkxk � kyk

�, 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

�ixi , prove that T is angle preserving if and only if all j�i 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

kTxk � kTyk

�D arccos

�hx;yi

kxk � kyk

�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

†�Txi ;Txj

�D arccos

˝Txi ;Txj

˛kTxik �

Txj !D arccos

˝�ixi ; �jxj

˛k�ixik �

�jxj !

D arccos

�i�j

˝xi ;xj

˛j�i j �

ˇ̌�jˇ̌� kxik � kxj k

!D †

�xi ;xj

�;

it must be the case that

�i�j Dj�i j �ˇ̌�jˇ̌:

Then �i and �j have the same signs. ut

I Exercise 9 (1-9). If 0 6 � < � , let T W R2 ! R2 have the matrix

A D

cos � sin �

� 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

cos � sin �

� sin � cos �

! x

y

!D

x cos � C y sin �

�x sin � C y cos �

!:

Therefore, T �x; y� 2 D x2 C y2 D �x; y� 2 Ithat is, T is norm preserving. Then it is angle preserving by Exercise 8 (a).

SECTION 1.1 NORM AND INNER PRODUCT 5

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 b2

�cos �:

Hence,

† .x;Tx/ D arccos

�hx;Txi

kxk � kTxk

�D arccos

�a2 C b2

�cos �

a2 C b2

!D �: 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 6M khk for h 2 Rm.

Proof. Let M’s matrix be

A D

�a11 � � � a1m:::

: : ::::

an1 � � � anm

˘

´

�a1

:::

an

˘

:

Then

M h D Ah D

� ˝a1;h

˛:::

han;hi

˘

;

and so

kM hk2D

nXiD1

Dai ;h

E26

nXiD1

�kaik � khk

�2D

0@ nXiD1

kaik2

1A � khk2 ;that is,

kM hk 6

0B@p

nXiD1

kaik

1CA � khk :

Let M D

pnXiD1

kaik and we get the result. ut

I Exercise 11 (1-11). If x;y 2 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 Dqkxk

2C kzk

2.

Proof. We have .x; z/ ; .y;w/ 2 RnCm. Then

h.x; z/ ; .y;w/i D

nXiD1

xiyi C

mXjD1

zjwj D hx;yi 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


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 aM x C bM 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. ut

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 yk2D

kxk2C kyk

2.

Proof. If hx;yi D 0, we have

kx C yk2D hx C y;x C yi D kxk

2C 2 hx;yi C kyk2 D kxk2 C kyk2 : ut

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�ka;yk. We show

that B.yI "/ � B.aI r/. Take any z 2 B.yI "/. Then

ka; zk 6 ka;yk C ky; zk < ka;yk C " D r: ut

I Exercise 16 (1-16). Simple. Omitted.

I Exercise 17 (1-17). Omitted.

SECTION 1.2 SUBSETS OF EUCLIDEAN SPACE 7

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.

Proof. Let X ´ Œ0; 1�. Obviously, A is open since A DSi .ai ; bi /. Then X X A

is closed in X and so X X A D X X A. Since @A D xA \ X X A D xA \ .X X A/, it

suffices to show that

X X A � xA: (1.1)

But (1.1) holds if and only if xA 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 xA. Hence, X D xA, i.e., A is dense in X . ut

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 xA D A. Since

0; 1 2 A, the claim holds. ut

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 � yk > 1=m

, m D 1; 2; : : :. If y 2 A, then x ¤ y ;

hence, kx � yk > 1=m for some m; therefore y 2 Gm (see Figure 1.1). Thus,

A �S1mD1Gm, 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.

1=m

xA

Figure 1.1. A compact set is closed

Let A be compact. We first show that A is bounded. Let


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; : : : ; img. Hence, A ��i 0; i 0

�; that is, A is bounded. ut

I Exercise 21 (1-21�).

a. If A is closed and x … A, prove that there is a number d > 0 such that

ky � xk > 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 � xk > 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 � xk > 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 � xk > 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; : : : ; xng such that˚B.x1I dx1

=2/; : : : ;B.xnI 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 � xik >di . Hence,

ky � xk > ky � xik � kxi � xk > di � di=2 D di=2 > d:

(c) See Figure 1.2.

0

Figure 1.2.ut

SECTION 1.3 FUNCTIONS AND CONTINUITY 9

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 � DB and D � U .

Proof. ut

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 fi .x/ D bi for i D 1; : : : ; m.

Proof. Let f W A! Rm and a 2 A.

If: Assume that limx!a fi .x/ D bi for i D 1; : : : ; m. Then for every "=

pm > 0,

there is a number ıi > 0 such that f i .x/ � bi < "=

pm for all x 2 A which

satisfy 0 < kx � ak < ıi , for every i D 1; : : : ; m. Put

ı D min fı1; : : : ; ımg :

Then for all x 2 A satisfying 0 < kx � ak < ı, f i .x/ � bi < "pm; i D 1; : : : ; m:

Therefore, for every x 2 A which satisfy 0 < kx � ak < ı,

kf .x/ � bk D

pmXiD1

�f i .x/ � bi

�2<

pmXiD1

�"2=m

�D "I

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/ � bi 6 kf .x/ � bk < ";i.e. limx!a f

i .x/ D bi . ut

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 fi .x/ D f i .a/ for every i D 1; : : : ; m; that is, if and only if f i is contin-

uous at a for each i D 1; : : : ; m. ut

I Exercise 25 (1-25). Prove that a linear transformation T W Rn ! Rm is con-

tinuous.


Proof. Take any a 2 Rn. Then, by Exercise 10 (1-10), there exists M > 0 such

that

Tx � Ta D T .x � a/ 6M 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. ut

I Exercise 26 (1-26). Let A Dn�x; y

�2 R2 W x > 0 and 0 < y < x2

o.

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 XA. If a > 0, then ax intersects x2 at�a; a2

�and .0; 0/ and nowhere else; see

Figure 1.3.

0 x

y

yDx2

yDax

A

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, butjf .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. ut

SECTION 1.3 FUNCTIONS AND CONTINUITY 11

I Exercise 27 (1-27). Prove that˚x 2 Rn W kx � ak < r

is open by considering

the function f W Rn ! R with f .x/ D kx � ak.

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/ � f .b/j Dˇ̌kx � ak � kb � ak

ˇ̌6 kx � ak � kb � ak 6 kx � bk < ı D ":

Hence,˚x 2 Rn W kx � ak < r

D f �1 .�1; r/ is open in Rn. ut

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

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

Œa; b� are distinct, show thatPniD1 o

�f; xi

�< f .b/ � f .a/.

Proof. ut

2DIFFERENTIATION

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

limh!0

kf .aC h/ � f .a/ � �.h/k

khkD 0;

or equivalently,

f .aC h/ � f .a/ D �.h/C r.h/; (2.1)

where the remainder r.h/ satisfies

limh!0kr.h/k

ıkhk D 0: (2.2)

Let h! 0 in (2.1). The error term r.h/! 0 by (2.2); the linear term �.h/ aslo

tends to 0 because if h DPniD1 hiei , where .e1; : : : ; en/ is the standard basis of

Rn, then by linearity we have �.h/ DPniD1 hi�.ei /, and each term on the right

tends to 0 as h! 0. Hence,

limh!0

�f .aC h/ � f .a/

�D 0I

that is, limh!0 f .aC 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 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 g0?

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 DIFFERENTIATION

lim.h;k/!0

ˇ̌f .aC h; b C k/ � f .a; b/ � g0.a/h

ˇ̌k.h; k/k

D lim.h;k/!0

ˇ̌g.aC h/ � g.a/ � g0.a/h

ˇ̌k.h; k/k

6 limh!0

ˇ̌g.aC h/ � g.a/ � g0.a/h

ˇ̌jhj

D 0I

hence, f 0.a; b/ D .g0.a/; 0/. ut

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; 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/,

.x2; y2/ 2 R2, we have f .x1; y1/ D f .x2; y1/ D f .x2; y2/; that is, f is con-

stant. ut

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

f .x/ D

˚kxk � g

�xıkxk

�if x ¤ 0;

0 if x D 0:

a. If x 2 R2 and h W R ! R is defined by h.t/ D f .tx/, show that h is differen-

tiable.

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 .tx/ D t kxk � g

�tx

t kxk

�D

hkxk � g

�x= kxk

�i� t D f .x/t I

finally, if x ¤ 0 and t < 0,

h.t/ D f .tx/ D �t kxk � g

�tx

�t kxk

�D �t kxk � g

��x= kxk

�D

hkxk � g

�x= kxk

�i� t

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 ahC bk. First consider any sequence .h; 0/! .0; 0/. Then

SECTION 2.1 BASIC DEFINITIONS 15

0 D limh!0

jf .h; 0/ � f .0; 0/ � ahj

jhjD limh!0

ˇ̌̌jhj � g

�h=jhj ; 0

�� ah

ˇ̌̌jhj

D limh!0

ˇ̌jhj � g .˙1; 0/ � ah

ˇ̌jhj

Djaj

implies that a D 0. Next let us consider .0; k/! .0; 0/. Then

0 D limk!0

jf .0; k/ � f .0; 0/ � bkj

jkjD limk!0

ˇ̌̌jkj � g

�0; k=jkj

�� bk

ˇ̌̌jkj

Djbj

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

limx!0

jf .x/ � f .0/ � 0j

kxkD lim

x!0

ˇ̌̌kxk � g

�x= kxk

�ˇ̌̌kxk

D limx!0

ˇ̌̌g�x= kxk

�ˇ̌̌¤ 0;

and so f is not differentiable.

Of course, if g.x/ D 0, then f .x/ D 0 and is differentiable. ut

I Exercise 35 (2-5). Let f W R2 ! R be defined by

f .x; y/ D

�x jyj

�qx2 C y2 if .x; y/ ¤ 0;

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/ Dx �jyjqx2 C y2

Dx �jyj

k.x; y/kD k.x; y/k �

�x

k.x; y/k�jyj

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). ut

I Exercise 36 (2-6). Let f W R2 ! R be defined by f .x; y/ Dpjxyj. Show that

f is not differentiable at .0; 0/.


Proof. It is clear that

limh!0

jf .h; 0/j

jhjD 0 D lim

k!0

jf .0; k/j

jkjI

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

lim.h;h/!.0;0/

jf .h; h/ � f .0; 0/ � 0j

k.h; h/kD lim.h;h/!.0;0/

ph2

k.h; h/kD

1p2¤ 0:

Therefore, f is not differentiable. ut

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

limx!0

jf .x/ � f .0/j

kxkD lim

x!0

jf .x/j

kxk6 lim

x!0kxk D 0;

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

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

c2

!. Then for i D

1; 2,

0 6 limh!0

ˇ̌̌f i .aC h/ � f i .a/ � ci � h

ˇ̌̌jhj

6 limh!0

kf .aC h/ � f .a/ � Df .a/.h/k

jhjD 0

implies that f i is differentiable at a with .f i /0.a/ D ci .

Now suppose that both f 1 and f 2 are differentiable at a, then by Exercise 1,

0 6kf .aC h/ � f .a/ � Df .a/.h/k

jhj6

2XiD1

ˇ̌̌f i .aC h/ � f i .a/ � .f i /0.a/ � h

ˇ̌̌jhj

implies that f is differentiable at a with f 0.a/ D

.f 1/0.a/

.f 2/0.a/

!. ut

I Exercise 39 (2-9). Two functions f; g W R! R are equal up to n-th order at a

if


limh!0

f .aC h/ � g.aC h/

hnD 0:

a. Show that f is differentiable at a if and only if there is a function g of the

form g.x/ D a0C 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

nXiD0

f .i/.a/

i Š.x � a/i

are equal up to n-th order at a.

Proof. (a) If f is differentiable at a, then by definition,

limh!0

f .aC h/ ��f .a/C f 0.a/ � h

�h

D 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 � a/ such that

limh!0

f .aC h/ � g.aC h/

hD limh!0

f .aC h/ � a0 � a1h

hD 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�1XiD0

f .i/.a/

i Š.x � a/i C

f .n/.y/

nŠ.x � a/n;

where y is between a and x. Thus,

limx!a

f .x/ � g.x/

.x � a/nD limx!a

f .n/.y/nŠ

.x � a/n � f .n/.a/nŠ

.x � a/n

.x � a/n

D limx!a

f .n/.y/ � f .n/.x/

nŠ

D 0: ut

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 ofŒa; b�, and define

P.t/ D

n�1XkD0

f .k/.˛/

kŠ.t � ˛/k :

Then there exists a point x between ˛ and ˇ such that

f .ˇ/ D P.ˇ/Cf .n/.x/

nŠ.ˇ � ˛/n:


2.2 Basic Theorems

I Exercise 40 (2-10). Use the theorems of this section to find f 0 for the follow-

ing:

a. f .x; y; z/ D xy .

b. f .x; y; z/ D .xy ; z/.

c. f .x; y/ D sin.x siny/.

d. f .x; y; z/ D sin�x sin.y sin z/

�.

e. f .x; y; z/ D xyz.

f. f .x; y; z/ D xyCz .

g. f .x; y; z/ D .x C y/z .

h. f .x; y/ D sin.xy/.

i. f .x; y/ D�sin.xy/

�cos3.

j. f .x; y/ D�sin.xy/; sin

�x siny

�; xy

�.

Solution. Compare this with Exercise 47.

(a) We have f .x; y; z/ D xy D elnxyD ey lnx D exp B.�2 � ln�1/.x; y; z/. It

follows from the Chain Rule that

f 0.a; b; c/ D exp0h.�2 ln�1/.a; b; c/

i�

��2 ln�1

�0.a; b; c/

D exp.b ln a/ �h.ln�1/.�2/0 C �2.ln�1/0

i.a; b; c/

D ab �h�0; ln a; 0

�C�b=a; 0; 0

�iD

�ab�1b ab ln a 0

�:

(b) By (a) and Theorem 2-3(3), we have

f 0.a; b; c/ D

ab�1b ab ln a 0

0 0 1

!:

(c) We have f .x; y/ D sin B.�1 sin.�2//. Then, by the chain rule,

f 0.a; b/ D sin0h.�1 sin.�2//.a; b/

i�

h�1 sin.�2/

i0.a; b/

D cos .a sin b/ �h.sin�2/.�1/0 C �1.sin�2/0

i.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 BASIC THEOREMS 19

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/

�� .�1g.�2; �3//0.a; b; c/

D cos�ag .b; c/

��

hg .b; c/ .�1/0 C ag0.�2; �3/

i.a; b; c/

D cos�ag .b; c/

��

h�g .b; c/ ; 0; 0

�C ag0.�2; �3/.a; b; c/

i:

It follows from (c) that

g0.�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 xy . Then

f .x; y; z/ D xg.y;z/ D g�x; g.y; z/

�D g

��1; g.�2; �3/

�:

Then

Df .a; b; c/ D Dg�a; g .b; c/

�B

hD�1;Dg.�2; �3/

i.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

�0B@xyz

1CADab

cbc

ax C

�ab

c

ln a�y;

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/

Dbcc

by C

�bc ln b

�z:

Hence,

Df .a; b; c/.x; y; z/ Dab

cbc

ax C

�ab

c

ln a� �bcc

by C

�bc ln b

�z

�;

and


f 0.a; b; c/ D�ab

c

bcıa ab

c

bcc ln aıb ab

cbc ln a ln b

�:

(f) Let g.x; y/ D xy . Then f .x; y; z/ D xyCz D g�x; y C z

�D g

��1; �2 C �3

�.

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

�DabCc .b C c/

ax C

�abCc ln a

� �y C z

�;

and

f 0.a; b; c/ D�abCc.bCc/

aabCc ln a abCc ln a

�:

(g) Let g.x; y/ D xy . Then

f .x; y; z/ D .x C y/z D g�x C y; z

�D g

��1 C �2; �3

�:

Hence,

Df .a; b; c/.x; y; z/ D Dg .aC b; c/ BhD�1 C D�2;D�3

i.a; b; c/.x; y; z/

D Dg .aC b; c/ B�x C y; z

�D.aC b/c c

.aC b/.x C y/C

�.aC b/c ln .aC b/

�z;

and

f 0.a; b; c/ D�.aCb/cc.aCb/

.aCb/cc.aCb/

.aC b/c ln .aC b/�:

(h) We have f .x; y/ D sin.xy/ D sin B��1�2

�. Hence,

f 0.a; b/ D .sin/0 .ab/ �hb.�1/0.a; b/C a.�2/0.a; b/

iD 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

f 0.a; b; c/ D

0BB@�sin.xy/

�0.a; b; c/h

sin�x siny

�i0.a; b; c/

Œxy �0 .a; b; c/

1CCAD

0B@ b � cos .ab/ a � cos .ab/

cos .a sin b/ � sin b a � cos .a sin b/ � cos b

ab�1b ab ln a

1CA : ut

I Exercise 41 (2-11). Find f 0 for the following (where g W R! R is continuous):


a. f .x; y/ DR xCya

g.

b. f .x; y/ DR xyag.

c. f .x; y; z/ DR sin

�x sin.y sin z/

�xy g.

Solution. (a) Let h.t/ DR tag. Then f .x; y/ D

�h B .�1 C �2/

�.x; y/, and so

f 0.a; b/ D h0 .aC b/ �h.�1 C �2/0.a; b/

iD g .aC b/ � .1; 1/

D

�g .aC b/ g .aC b/

�:

(b) Let h.t/ DR tag. Then f .x; y/ D

R xyag D h.xy/ D

hh B

��1 � �2

�i.x; y/.

Hence,

f 0.a; b/ D h0 .ab/ �hb � .�1/0.a; b/C a � .�2/0.a; b/

iD g .ab/ � .b; a/

D

�b � g .ab/ a � g .ab/

�:

(c) We can rewrite f .x; y; z/ as

f .x; y; z/ D

Z a

xy

g C

Z sin.x sin.y sin z//

a

g D

Z sin.x sin.y sin z//

a

g �

Z xy

a

g:

Let .x; y; z/ D sin�x sin.y sin z/

�, k.x; y; z/ D

R .x;y;z/a

g, and h.x; y; z/ DR xyag.

Then f .x; y; z/ D k.x; y; z/ � h.x; y; z/, and so

f 0.a; b; c/ D k0.a; b; c/ � h0.a; b; c/:

It follows from Exercise 40 (d) that

k0.a; b; c/ D k0� .a; b; c/

�� 0.a; b; c/:

The other parts are easy. ut

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

lim.h;k/!0

kf .h;k/k

k.h;k/kD 0:


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).

Proof. (a) Let�en1 ; : : : ; e

nn

�and

�em1 ; : : : ; e

mm

�be the stand bases for Rn and Rm,

respectively. Then for any x 2 Rn and y 2 Rm, we have

x D

nXiD1

xiein; and y D

mXjD1

yj ejm:

Therefore,

f .x;y/ D f

0@ nXiD1

xiein;

mXjD1

yj ejm

1A D nXiD1

f

0@xiein; mXjD1

yj ejm

1AD

nXiD1

mXjD1

f�xiein; y

j ejm

�

D

nXiD1

mXjD1

xiyjf�ein; e

jm

�:

Then, by letting M DPi;j

f �ein; ejm

� , we have

kf .x;y/k D

Xi;j

xiyjf�ein; e

jm

� 6Xi;j

ˇ̌̌xiyj

ˇ̌̌ f �ein; ejm

� 6M

"maxi

�ˇ̌̌xiˇ̌̌�

maxj

�ˇ̌̌yjˇ̌̌�#

6M kxk kyk :

Hence,

lim.h;k/!0

kf .h;k/k

k.h;k/k6 lim.h;k/!0

M khk kkk

k.h;k/k

D lim.h;k/!0

M khk kkkpXi;j

��hi�2C

�kj�2�

D lim.h;k/!0

M khk kkkqkhk

2C kkk

2

:

Now

khk kkk 6

˚khk

2 if kkk 6 khkkkk

2 if khk 6 kkk :

Hence khk kkk 6 khk2 C kkk2, and so


lim.h;k/!0

M khk kkkqkhk

2C kkk

2

6 lim.h;k/!0

M

qkhk

2C kkk

2D 0:

(b) We have

lim.h;k/!0

kf .aC h;bC k/ � f .a;b/ � f .a;k/ � f .h;b/k

k.h;k/k

D lim.h;k/!0

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

k.h;k/k

D lim.h;k/!0

kf .h;k/k

k.h;k/k

D 0

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: ut

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

h0.a/ DDf 0.a/T; g.a/

EC

Df .a/; g0.a/T

E:

c. If f W R! Rn is differentiable and kf .t/k D 1 for all t , show thatDf 0.t/T; f .t/

ED

0.

d. Exhibit a differentiable function f W R ! R such that the function jf j defined

by jf j .t/ Djf .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/; 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


h.t/ D kf .t/k2 D 1

is constant, and so h0.a/ D 0; that is,

0 DDf 0.a/T; f .a/

EC

Df .a/; f 0.a/T

ED 2

Df 0.a/T; f .a/

E;

and soDf 0.a/T; f .a/

ED 0.

(d) Let f .t/ D t . Then f is linear and so is differentiable: Df D t . However,

limt!0C

jt j

tD 1; lim

t!0�

jt j

tD �1I

that is, jf j is not differentiable at 0. ut

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;xiC1; : : : ;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

limh!0

f �a1; : : : ;hi ; : : : ;hj ; : : : ; ak� khk

D 0:

b. Prove that

Df .a1; : : : ; ak/ .x1; : : : ;xk/ DkXiD1

f .a1; : : : ; ai�1;xi ; aiC1; : : : ; ak/ :

Proof.

(a) To light notation, define

a�i�j ´�a1; : : : ; ai�1; aiC1; : : : ; aj�1; ajC1; : : : ; ak

�:

Let g W Ei�Ej ! Rp be defined as g�xi ;xj

�D f

�a�i�j ;xi ;xj

�. Then g is bilinear

and so

limh!0

g �a�i�j ;hi ;hj � khk

6 limh!0

g �a�i�j ;hi ;hj � �hi ;hj � D 0

by Exercise 42 (a).

(b) It follows from Exercise 42 (b) immediately. 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 Rn.

a. Prove that det W Rn � � � � � Rn ! R is differentiable and


D�det

�.a1; : : : ; an/ .x1; : : : ;xn/ D

nXiD1

det

0BBBBBBBB@

a1:::

xi:::

an

1CCCCCCCCA:

b. If aij W R! R are differentiable and f .t/ D det�aij .t/

�, show that

f 0.t/ D

nXjD1

det

0BBBBBBBB@

a11.t/ � � � a1n.t/:::

: : ::::

a0j1.t/ � � � a0jn.t/:::

: : ::::

an1.t/ � � � ann.t/

1CCCCCCCCA:

c. If det�aij .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 equationsnX

jD1

aj i .t/sj .t/ D bi .t/; i D 1; : : : ; n:

Show that si is differentiable and find s0i .t/.

Proof.

(a) It is easy to see that det W Rn � � � � � Rn ! R is multilinear; hence, the con-

clusion follows from Exercise 44.

(b) By (a) and the chain rule,

f 0.t/ D�det

�0 �aij .t/

�B�a01.t/; : : : ; a

0n.t/

�

D

nXjD1

det

0BBBBBBBB@

a11.t/ � � � a1n.t/:::

: : ::::

a0j1.t/ � � � a0jn.t/:::

: : ::::

an1.t/ � � � ann.t/

1CCCCCCCCA:

(c) Let

A D

[email protected]/ � � � an1.t/:::

: : ::::

a1n.t/ � � � ann.t/

1CCA ; s D

[email protected]/:::

sn.t/

1CCA ; and b D

[email protected]/:::

bn.t/

1CCA :Then

As D b;


and so

si .t/ Ddet .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

f 0.t/ D

nXjD1

det

0BBBBBBBB@

a11.t/ � � � an1.t/:::

: : ::::

a01j .t/ � � � a0nj .t/:::

: : ::::

a1n.t/ � � � ann.t/

1CCCCCCCCA;

and

g0i .t/ D

nXjD1

0BBBBBBBB@

a11.t/ � � � ai�1;1.t/ b1.t/ aiC1;1.t/ � � � an1.t/:::

: : ::::

::::::

: : ::::

a01j .t/ � � � a0i�1;j .t/ b0j .t/ a0iC1;j .t/ � � � a0nj .t/:::

: : ::::

::::::

: : ::::

a1n.t/ � � � ai�1;n.t/ bn.t/ aiC1;n.t/ � � � ann.t/

1CCCCCCCCA:

Therefore,

s0i .t/ Df 0.t/g0i .t/ � f .t/g

0i .t/

f 2.t/: ut

I Exercise 46 (2-16). Suppose f W Rn ! Rn is differentiable and has a differen-

tiable inverse f �1 W Rn ! Rn. Show that�f �1

�0.a/ D

hf 0�f �1.a/

�i�1.

Proof. We have f B f �1.x/ D x. On the one hand D�f B f �1

�.a/ .x/ D x since

f B f �1 is linear; on the other hand,

D�f B f �1

�.a/ .x/ D

�Df

�f �1 .a/

�B Df �1 .a/

�.x/:

Therefore, Df �1 .a/ DhDf

�f �1 .a/

�i�1. ut

2.3 Partial Derivatives

I Exercise 47 (2-17). Find the partial derivatives of the following functions:

a. f .x; y; z/ D xy .

b. f .x; y; z/ D z.

c. f .x; y/ D sin.x siny/.

d. f .x; y; z/ D sin�x sin.y sin z/

�.

SECTION 2.3 PARTIAL DERIVATIVES 27

e. f .x; y; z/ D xyz.

f. f .x; y; z/ D xyCz .

g. f .x; y; z/ D .x C y/z .

h. f .x; y/ D sin.xy/.

i. f .x; y/ D�sin.xy/

�cos3.

Solution. Compare this with Exercise 40.

(a) D1f .x; y; z/ D yxy�1, D2f .x; y; z/ D xy ln x, and D3f .x; y; z/ D 0.

(b) D1f .x; y; z/ D D2f .x; y; z/ D 0, and D3f .x; y; z/ D 1.

(c) D1f .x; y/ D�siny

�cos

�x siny

�, and D2f .x; y/ D x cosy cos

�x siny

�.

(d) D1f .x; y; z/ D sin.y sin z/ cos�x sin.y sin z/

�,

D2f .x; y; z/ D cos�x sin.y sin z/

�x cos.y sin z/ sin z, and

D3f .x; y; z/ D cos�x sin.y sin z/

�x cos.y sin z/y cos z.

(e) D1f .x; y; z/ D yzxyz�1, D2f .x; y; z/ D xy

zzyz�1 ln x, and D3f .x; y; z/ D

yz lny�xy

zln x

�.

(f) D1f .x; y; z/ D�y C z

�xyCz�1, and D2f .x; y; z/D3f .x; y; z/ D xyCz ln x.

(g) D1f .x; y; z/ D D2f .x; y; z/ D z.x C y/z�1, and

D3f .x; y; z/ D .x C y/z ln.x C y/.

(h) D1f .x; y/ D y cos.xy/, and D2f .x; y/ D x cos.xy/.

(i) D1f .x; y/ D cos 3�sin.xy/

�cos3�1y cos.xy/, and

D2f .x; y/ D cos 3�sin.xy/

�cos3�1x cos.xy/. ut

I Exercise 48 (2-18). Find the partial derivatives of the following functions

(where g W R! R is continuous):

a. f .x; y/ DR xCya

g.

b. f .x; y/ DR xyg.

c. f .x; y/ DR xyag.

d. f .x; y/ DR .R y

bg/

ag.

Solution.

(a) D1f .x; y/ D D2f .x; y/ D g.x C y/.

(b) D1f .x; y/ D g.x/, and D2f .x; y/ D �g�y�.

(c) D1f .x; y/ D yg.xy/, and D2f .x; y/ D xg.xy/.


(d) D1f .x; y/ D 0, and D2f .x; y/ D g�y�� g�R ybg�. ut

I Exercise 49 (2-19). If

f .x; y/ D xxxxy

C�ln x

�0@arctan

arctan

�arctan

�sin

�cos xy

�� ln.x C y/

��!1Afind D2f

�1; y

�.

Solution. Putting x D 1 into f .x; y/, we get f�1; y

�D 1. Then D2f

�1; y

�D

0. ut

I Exercise 50 (2-20). Find the partial derivatives of f in terms of the deriva-

tives 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) D1f .x; y/ D g0 .x/ h�y�, and D2f .x; y/ D g.x/h0

�y�.

(b) D1f .x; y/ D h�y�g.x/h.y/�1g0 .x/, and D2f .x; y/ D h0

�y�g.x/h.y/ lng.x/.

(c) D1f .x; y/ D g0 .x/, and D2f .x; y/ D 0.

(d) D1f .x; y/ D 0, and D2f .x; y/ D g0�y�.

(e) D1f .x; y/ D D2f .x; y/ D g0.x C y/. ut

I Exercise 51 (2-21�). Let g1; g2 W R2 ! R be continuous. Define f W R2 ! R by

f .x; y/ D

Z x

0

g1 .t; 0/ dt C

Z y

0

g2 .x; t/ dt:

a. Show that D2f .x; y/ D g2.x; y/.

b. How should f be defined so that D1f .x; y/ D g1.x; y/?

c. Find a function f W R2 ! R such that D1f .x; y/ D x and D2f .x; y/ D y. Find

one such that D1f .x; y/ D y and D2f .x; y/ D x.

Proof.


(a) D2f .x; y/ D 0C g2.x; y/ D g2.x; y/.

(b) We should let

f .x; y/ D

Z x

0

g1�t; y

�dt C

Z y

0

g2 .a; t/ dt;

where t 2 R is a constant.

(c) Let

� f .x; y/ D�x2 C y2

�=2.

� f .x; y/ D xy. ut

I Exercise 52 (2-22�). If f W R2 ! R and D2f D 0, show that f is independent

of the second variable. If D1f D D2f 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 D2f

�x; y�

� �y2 � y1

�D 0:

Hence, f�x; y1

�D f

�x; y2

�; that is, f is independent of y.

Similarly, if D1f D 0, then f is independent of x. The second claim is then

proved immediately. ut

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 D1f D D2f D 0, show that f is constant.

b. Find a function f W A! R such that D2f 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 if x < 0 or y < 0

x otherwise.ut

I Exercise 54 (2-24). Define f W R2 ! R by

f .x; y/ D

˚xy x

2�y2

x2Cy2 .x; y/ ¤ 0;

0 .x; y/ D 0:

a. Show that D2f .x; 0/ D x for all x and D1f�0; y

�D �y for all y.


.a; b/ .�1; b/

.�1; d/ .c; d/

x

y

Figure 2.1. f is constant

b. Show that D1;2f .0; 0/ ¤ D2;1f .0; 0/.

Proof.

(a) We have

D2f .x; y/ D

�x.x4�y4�4x2y2/

.x2Cy2/2 .x; y/ ¤ 0;

0 .x; y/ D 0;

and

D1f .x; y/ D

��y.y4�x4�4x2y2/

.x2Cy2/2 .x; y/ ¤ 0;

0 .x; y/ D 0:

Hence, D2f .x; 0/ D x and D1f�0; y

�D �y.

(b) By (a), we have D1;2f .0; 0/ D D2�D1f

�0; y

��.0/ D �1; but D2;1f .0; 0/ D

D1�D2 .x; 0/

�.0/ D 1. ut

I Exercise 55 (2-25�). Define f W R! R by

f .x/ D

˚e�x

�2x ¤ 0

0 x D 0:

Show that f is a C1 function, and f .i/ .0/ D 0 for all i .

Proof. Figure 2.2 depicts f .x/. We first show that f 2 C1.

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�1

�e�x

�2. We do this by induction.

Step 1 Clearly, f 0 .x/ D 2x�3e�x�2

.

Step 2 Suppose that f .k/ .x/ D p3k�x�1

�e�x

�2.

Step 3 Then by the chain rule,


0 x

y

−2 −1 1 2

Figure 2.2.

f .kC1/ .x/ Dhf .k/ .x/

i0D p03k

�x�1

��

��x�2

�� e�x

�2

C p3k

�x�1

�� 2x�3 � e�x

�2

D

�p03k

�x�1

��

��x�2

�C p3k

�x�1

�� 2x�3

�� e�x

�2

D

�q3kC1

�x�1

�C q3kC3

�x�1

�� e�x

�2

D p3.kC1/

�x�1

�� e�x

�2

;

where q3kC1 and q3kC3 are polynomials.

Therefore, f .x/ 2 C1 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 limx!0

f .x/ � f .0/

xD limx!0

e�x�2

xD limx!0

2x�3e�x�2

D 0

by L’Hôpital’s rule.

Step 2 Suppose that f .k/ .0/ D 0.

Step 3 Then,

f .kC1/ .0/ D limx!0

f .k/ .x/ � f .k/ .0/

x

D limx!0

p3kC1

�x�1

�e�x

�2

D limx!0

p3kC1�x�1

�ex�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. ut

I Exercise 56 (2-26�). Let

f .x/ D

˚e�.x�1/

�2

� e�.xC1/�2

x 2 .�1; 1/ ;

0 x … .�1; 1/ :


a. Show that f W R ! R is a C1 function which is positive on .�1; 1/ and 0

elsewhere.

b. Show that there is a C1 function g W R ! Œ0; 1� such that g.x/ D 0 for x 6 0and 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 C1 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

C1 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

e�.x�1/�2

2 C1 and e�.xC1/�2

2 C1. Then it is straightforward to check that

f 2 C1. See Figure 2.3

0 x

y

−1 1

Figure 2.3.

(b) By letting z D xC 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/ ;

0 z … .0; 2/ :

By letting w D "z=2, we derive a function k W R! R from j as follows:

k .w/ D

˚e�.2w="�2/

�2

� e�.2w="/�2

w 2 .0; "/ ;

0 w … .0; "/ :


0 x

y

1 2

j

0 x

yk

Figure 2.4.

It is easy to see that k 2 C1, which is positive on .0; "/ and 0 elsewhere. Now

let

g.x/ D

�Z x

0

k .x/

�,�Z "

0

k .x/

�:

Then g 2 C1; 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; : : : ; xmg � 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

x1i � a

1i

"

!� � � f

�xni � a

ni

"

�;

where�a1i ; : : : ; a

ni

�2 Rn is the middle point of Rxi

.

Finally, we define g W Rx1[ � � � [Rxm

! R as follows:

g.x/ D

mXiD1

gi .x/ :

Then g 2 C1; it is positive on C , and 0 outside Rx1[ � � � [Rxm

.

(e) Follows the hints. ut

I Exercise 57 (2-27). Define g; h W˚x 2 R2 W kxk 6 1

! R3 by

g.x; y/ D

�x; y;

q1 � x2 � y2

�;

h.x; y/ D

�x; y;�

q1 � x2 � y2

�:

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 ´˚x 2 R2 W kxk 6 1

and B ´

˚x 2 R3 W kxk D 1

. Then B D

g .A/ [ h .A/. ut


2.4 Derivatives

I Exercise 58 (2-28). Find expressions for the partial derivatives of the follow-

ing 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�xy ; yz ; zx

�.

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

D1F.x; y/ D D1f .a/ � g0 .x/ � k

�y�C D2f .a/ � g

0 .x/ ;

D2F.x; y/ D D1f .a/ � g.x/ � k0�y�C D1f .a/ � h

0�y�:

(b) Letting a´ g.x C y/; h�y C z

�, we have

D1F.x; y; z/ D D1f .a/ � g0.x C y/;

D2F.x; y; z/ D D1f .a/ � g0.x C y/C D2f .a/ � h

0�y C z

�;

D3F.x; y; z/ D D2f .a/ � h0�y C z

�:

(c) Letting a´ xy ; yz ; zx , we have

D1F.x; y; z/ D D1f .a/ � yxy�1C D3f .a/ � z

x ln z;

D2F.x; y; z/ D D1f .a/ � xy ln x C D2f .a/ � zy

z�1;

D3F.x; y; z/ D D2f .a/ � yz lny C D3f .a/ � xz

x�1:

(d) Letting a´ x; g.x/; h.x; y/, we have

D1F.x; y/ D D1f .a/C D2f .a/ � g0 .x/C D3f .a/ � D1h.x; y/

D2F.x; y/ D D3f .a/ � D2h.x; y/: ut

I Exercise 59 (2-29). Let f W Rn ! R. For x 2 Rn, the limit

limt!0

f .aC tx/ � f .a/

t;

if it exists, is denoted Dxf .a/, and called the directional derivative of f at a, in

the direction x.

a. Show that Deif .a/ D Dif .a/.

b. Show that Dtxf .a/ D tDxf .a/.

SECTION 2.4 DERIVATIVES 35

c. If f is differentiable at a, show that Dxf .a/ D Df .a/.x/ and therefore

DxCyf .a/ D Dxf .˛/C Dyf .a/.

Proof.

(a) For ei D .0; : : : ; 0; 1; 0; : : : ; 0/, we have

Deif .a/ D lim

t!0

f .aC tei / � f .a/

t

D limt!0

f .a1; : : : ; ai�1; ai C t; aiC1; : : : ; an/ � f .a/

t

D Dif .a/

by definition.

(b) We have

Dtxf .a/ D lims!0

f .aC stx/ � f .a/

sD limst!0

tf .aC stx/ � f .a/

stD tDxf .a/:

(c) If f is differentiable at a, then for any x ¤ 0 we have

0 D limt!0

jf .aC tx/ � f .a/ � Df .a/.tx/j

ktxk

D limt!0

jf .aC tx/ � f .a/ � t � Df .a/.x/j

jt j�1

kxk

D limt!0

ˇ̌̌̌f .aC tx/ � f .a/

t� Df .a/.x/

ˇ̌̌̌�1

kxk;

and so

Dxf .a/ D limt!0

f .aC tx/ � f .a/

tD Df .a/.x/:

The case of x D 0 is trivial. Therefore,

DxCyf .a/ D Df .a/ .x C y/

D Df .a/.x/C Df .a/ .y/

D Dxf .a/C Dyf .a/: ut

I Exercise 60 (2-30). Let f be defined as in Exercise 34. Show that Dxf .0; 0/

exists for all x, but if g ¤ 0, then DxCyf .0; 0/ ¤ Dxf .0; 0/ C Dyf .0; 0/ for all

x;y .

Proof. Take any x 2 R2.

limt!0

f .tx/ � f .0; 0/

tD lim

t!0

jt j � kxk � g

�tx.�jt j � kxk

��t

:

Therefore, Dxf .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. ut


I Exercise 61 (2-31). Let f W R2 ! R be defined as in Exercise 26. Show that

Dxf .0; 0/ exists for all x, although f is not even continuous at .0; 0/.

Proof. For any x 2 R2, we have

limt!0

f .tx/ � f .0/

tD lim

t!0

f .tx/

tD 0

by Exercise 26 (a). ut

I Exercise 62 (2-32).

a. Let f W R! R be defined by

f .x/ D

˚x2 sin 1

xx ¤ 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

f .x; y/ D

‚ �x2 C y2

�sin 1q

x2 C y2.x; y/ ¤ 0

0 .x; y/ D 0:

Show that f is differentiable at .0; 0/ but Dif is not continuous at .0; 0/.

Proof.

(a) We have

limx!0

f .x/ � f .0/

xD limx!0

x2 sin 1x

xD limx!0

x sin1

xD 0:

Hence, f 0 .0/ D 0. Further, for any x ¤ 0, we have

f 0 .x/ D 2x sin1

x� cos

1

x:

It is clear that limx!0 f0 .x/ does not exist. Therefore, f 0 is not continuous at

0.

(b) Since

lim.x;y/!.0;0/

�x2 C y2

�sin 1q

x2 C y2qx2 C y2

D lim.x;y/!.0;0/

qx2 C y2 sin

1qx2 C y2

D 0;

we know that f 0.0; 0/ D .0; 0/. Now take any .x; y/ ¤ .0; 0/. Then

D1f .x; y/ D 2x sin1q

x2 C y2� 2x cos

1qx2 C y2

:

SECTION 2.4 DERIVATIVES 37

0

Figure 2.5.

As in (a), limx!0 D1f .x; 0/ does not exist. Similarly for D2f . ut

I Exercise 63 (2-33). Show that the continuity of D1f 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,

limh!0

ˇ̌̌f�a1 C h1; a�1

�� f .a/ � D1f .a/ � h1

ˇ̌̌khk

6 limh1!0

ˇ̌̌f�a1 C h1; a�1

�� f .a/ � D1f .a/ � h1

ˇ̌̌ˇ̌h1ˇ̌ D 0:

See aslo Apostol (1974, Theorem 12.11). ut

I Exercise 64 (2-34). A function f W Rn ! R is homogeneous of degree m if

f .tx/ D tmf .x/ for all x. If f is also differentiable, show that

nXiD1

xiDif .x/ D mf .x/:

Proof. Let g.t/ D f .tx/. Then, by Theorem 2-9,

g0.t/ D

nXiD1

Dif .tx/ � xi : (2.4)

On the other hand, g.t/ D f .tx/ D tmf .x/; then

g0.t/ D mtm�1f .x/: (2.5)

Combining (2.4) and (2.5), and letting t D 1, we then get the result. ut

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


f .x/ D

nXiD1

xigi .x/:

Proof. Let hx.t/ D f .tx/. ThenZ 1

0

h0x.t/dt D hx .1/ � hx .0/ D f .x/ � f .0/ D f .x/:

Hence,

f .x/ D

Z 1

0

h0x.t/dt D

Z 1

0

f 0.tx/dt D

Z 1

0

24 nXiD1

xiDif .tx/

35 dt

D

nXiD1

xiZ 1

0

Dif .tx/dt

D

nXiD1

xigi .x/;

where gi .x/ DR 10

Dif .tx/dt . ut

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 differen-

tiable, 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��f�B

�0.x/

�¤ 0 for

all x 2 B � A, it follows that f .B/ is open. ut


a. Let f W R2 ! R be a continuously differentiable function. Show that f is not

1-1.

SECTION 2.5 INVERSE FUNCTIONS 39

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 D1f and D2f are continuous. Assume that f is 1-1;

then both D1f and D2f cannot not be constant and equal to 0. So suppose that

there is�x0; y0

�2 R2 such that D1f

�x0; f0

�¤ 0. The continuity of D1f implies

that there is an open set A � R2 containing�x0; y0

�such that D1f .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,

g0.x; y/ D

D1f .x; y/ D2f .x; y/

0 1

!;

and so det�g0.x; y/

�D D1f .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

�; zy�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

�; zy�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

�; zy�. 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 D1f 1.x/ ¤ 0 for all x 2 A. Define g W A! Rm as

g�x1;x�1

�D

�f .x/;x�1

�;

where x�1´�x2; : : : ; xn

�. Then as in (a), it follows that f cannot be 1-1. ut


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�ex cosy; ex siny

�. Show that det

�f 0.x; y/

�¤

0 for all .x; y/ but f is not 1-1.

Proof.


(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

0 D f .b/ � f .a/ D f 0 .c/ .b � a/ ;

which implies that f 0 .c/ D 0. A contradiction.

(b) We have

f 0.x; y/ D

Dxex cosy Dyex cosy

Dxex siny Dyex siny

!

D

ex cosy �ex siny

ex siny ex cosy

!:

Then

det�f 0.x; y/

�D e2x

�cos2 y C sin2 y

�D e2x ¤ 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). ut

I Exercise 69 (2-39). Use the function f W R! R defined by

f .x/ D

˚x2C x2 sin 1

xx ¤ 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/ D1

2C 2x sin

1

x� cos

1

xI

if x D 0, then

f 0 .0/ D limh!0

h=2C h2 sin�1=h

�h

D1

2:

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 IMPLICIT FUNCTIONS 41

0

Figure 2.6.ut

f i .t; s/ D

nXjD1

aj i .t/sj� bi .t/;

for i D 1; : : : ; n. Then

M´

0BB@D2f 1 .t; s/ � � � D1Cnf 1 .t; s/

:::: : :

:::

D2f n .t; s/ � � � D1Cnf n .t; s/

1CCA [email protected]/ � � � an1.t/:::

: : ::::

a1n.t/ � � � ann.t/

1CCA ;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. 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 gx�y�D f .x; y/. Suppose that for each x there is a unique y with

g0x�y�D 0; let c .x/ be this y.

a. If D2;2f .x; y/ ¤ 0 for all .x; y/, show that c is differentiable and

c0 .x/ D �D2;1f

�x; c .x/

�D2;2f

�x; c .x/

� :b. Show that if c0 .x/ D 0, then for some y we have

D2;1f .x; y/ D 0;

D2f .x; y/ D 0:

c. Let f .x; y/ D x�y logy � y

�� y log x. Find


max1=26x62

"min

1=36y61f .x; y/

#:

Proof.

(a) For every x, we have g0x�y�D D2f .x; y/. Since for every x there is a unique

y D c .x/ such that D2f�x; c .x/

�D 0, the solution c .x/ is the same as ob-

tained from the Implicit Function Theorem; hence, c .x/ is differentiable, and

by differentiating D2f�x; c .x/

�D 0 with respect to x, we have

D2;1f�x; c .x/

�C D2;2f

�x; c .x/

�� c0 .x/ D 0I

that is,

c0 .x/ D �D2;1f

�x; c .x/

�D2;2f

�x; c .x/

� :(b) It follows from (a) that if c0 .x/ D 0, then D2;1f

�x; c .x/

�D 0. Hence, there

exists some y D c .x/ such that D2;1f .x; y/ D 0. Furthermore, by definition,

D2�x; c .x/

�D D2f .x; y/ D 0.

(c) We have

D2f .x; y/ D x lny � ln x:

Let D2f .x; y/ D 0 we have y D c .x/ D x1=x . Also, D2;2f .x; y/ D x=y > 0 since

x; y > 0. Hence, for every fixed x 2�1=2; 2

�,

minyf .x; y/ D f

�x; c .x/

�:

0

x

y

0.5 1 1.5 2

0.5

1

1.5 c .x/

Figure 2.7.

It is easy to see that c0 .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,

min1=36y61

f�x; y

�D f

�x; y� .x/

�;

where (see Figure 2.8)

SECTION 2.6 IMPLICIT FUNCTIONS 43

y� .x/ D

„1=3 if 1=2 6 x 6 ac .x/ D x1=x if a < x 6 11 if 1 < x 6 2:

0

x

y

0.5 1 1.5 2

0.5

1

1.5

y� .x/

Figure 2.8.

1=2 6 x 6 a In this case, our problem is

max1=26x6a

f�x; 1=3

�D �

�1C ln 3

3

�x �

1

3ln x:

It is easy to see that x� D 1=2, and so f�x�; 1=3

�D ln

�4=3e

�=6.

a < x 6 1 In this case, our problem is

maxa<x61

f�x; x1=x

�D �x1C1=x :

It is easy to see that the maximum of f occurs at x� D a and y��x��D 1=3.

1 < x 6 2 In this case, our problem is

max1<x62

f .x; 1/ D �x � ln x:

The maximum of f occurs at x� D 1.

Now, as depicted in Figure 2.9, we have x� D 1=2, y� D 1=3, and f�x�; y�

�D

ln�4=3e

�=6. ut


0

xf�x; y� .x/

�0.5 1 1.5 2

−2.5

−2

−1.5

−1

−0.5

Figure 2.9.

3INTEGRATION

3.1 Basic Definitions

I Exercise 72 (3-1). Let f W Œ0; 1� � Œ0; 1�! R be defined by

f�x; y

�D

˚0 if 0 6 x < 1=21 if 1=2 6 x 6 1:

Show that f is integrable andRŒ0;1��Œ0;1�

f D 1=2.

Proof. Consider a partition P D .P1; P2/ with P1 D P2 D�0; 1=2; 1

�. Then

L�f; P

�D U

�f; P

�D 1=2. It follows from Theorem 3-3 (the Riemann condition)

that f is integrable andRŒ0;1��Œ0;1�

f D 1=2. ut

I Exercise 73 (3-2). Let f W A! R be integrable and let g D f except at finitely

many points. Show that g is integrable andRAf D

RAg.

Proof. Fix an " > 0. It follows from the Riemann condition that there is a

partition P of A such that

U�f; P

�� L

�f; P

�<"

2:

Let P 0 be a refinement of P such that:

� for every x 2 A with g .x/ ¤ f .x/, it belongs to 2n subrectangles of P 0, i.e.,

x is a corner of each subrectangle.

� for every subrectangle S of P 0,

v .S/ <"

2nC1d .u � `/;

where

45

46 CHAPTER 3 INTEGRATION

d Dˇ̌̌˚

x W f .x/ ¤ g .x/ˇ̌̌;

u D supx2A

fg .x/g � infx2Aff .x/g ;

` D infx2Afg .x/g � sup

x2A

ff .x/g :

x

Figure 3.1.

With such a choice of partition of A, we have

U�g; P 0

�� U

�f; P 0

�D

dXiD1

24 2nXjD1

hMSij

�g��MSij

�f�iv�Sij�35

6 d2nuv;

where v ´ supS2P 0 fv .S/g is the least upper bound of the volumes of the

subrectangles of P 0. Similarly,

L�g; P 0

�� L

�f; P 0

�D

dXiD1

24 2nXjD1

hmSij

�g��mSij

�f�iv�Sij�35

> d2n`v:

Therefore,

U�g; P 0

�� L

�g; P 0

�6hU�f; P 0

�C d2nuv

i�

hL�f; P 0

�C d2n`v

i6"

2C d2n .u � `/ v

D"

2C d2n .u � `/

"

2nC1d .u � `/

D "I

that is, g is integrable. It is easy to see now thatRAg D

RAf . ut

I Exercise 74 (3-3). Let f; g W A! R be integrable.

a. For any partition P of A and subrectangle S , show that mS�f�C mS

�g�6

mS�f C g

�and MS

�f C g

�6 MS

�f�C MS

�g�

and therefore L�f; P

�C

L�g; P

�6 L

�f C g; P

�and U

�f C g; P

�6 U

�f; P

�C U

�g; P

�.


b. Show that f C g is integrable andRA

�f C g

�DRAf C

RAg.

c. For any constant c, show thatRAcf D c

RAf .

Proof.

(a) We show that mS�f�C mS

�g�

is a lower bound ofn�f C g

�.x/ W x 2 S

o. It

is clear that mS�f�6 f .x/ and mS

�g�6 g .x/ for any x 2 S . Then for every

x 2 S we have

mS�f�CmS

�g�6 f .x/C g .x/ D

�f C g

�.x/ :

Hence, mS�f�CmS

�g�6 mS

�f C g

�.

Similarly, for every x 2 S we have MS

�f�> f .x/ and MS

�g�> g .x/;

hence,�f C g

�.x/ D f .x/ C g .x/ 6 MS

�f�C MS

�g�

and so MS

�f C g

�6

MS

�f�CMS

�g�.

Now for any partition P of A we have

L�f; P

�C L

�g; P

�D

XS2P

mS�f�v .S/C

XS2P

mS�g�

D

XS2P

hmS

�f�CmS

�g�iv .S/

6XS2P

mS�f C g

�v .S/

D L�f C g; P

�;

(3.1)

and

U�f; P

�C U

�g; P

�D

XS2P

MS

�f�v .S/C

XS2P

MS

�g�v .S/

D

XS2P

hMS

�f�CMS

�g�iv .S/

>XS2P

MS

�f C g

�v .S/

D U�f C g; P

�:

(3.2)

(b) It follows from (3.1) and (3.2) that for any partition P ,

U�f C g; P

�� L

�f C g; P

�6hU�f; P

�C U

�g; P

�i�

hL�f; P

�C L

�g; P

�iD

hU�f; P

�� L

�f; P

�iC

hU�g; P

�� L

�g; P

�i:

Since f and g are integrable, there exist P 0 and P 00 such that for any " > 0,

we have U�f; P 0

��L

�f; P 0

�< "=2 and U

�g; P 00

��L

�g; P 00

�< "=2. Let xP refine

both P 0 and P 00. Then

U�f; xP

�� L

�f; xP

�<"

2and U

�g; xP

�� L

�g; xP

�<"

2:

Hence,


U�f C g; xP

�� L

�f C g; xP

�< ";

and so f C g is integrable.

Now, by definition, for any " > 0, there exists a partition P (by using a

common refinement partition if necessary) such thatRAf < L

�f; P

�C "=2,R

Ag < L

�g; P

�C"=2, U

�f; P

�<RAf C"=2, and U

�g; P

�<RAgC"=2. Therefore,Z

A

f C

ZA

g � " < L�f; P

�C L

�g; P

�6 L

�f C g; P

�6ZA

�f C g

�6 U

�f C g; P

�6 U

�f; P

�C U

�g; P

�<

ZA

f C

ZA

g C ":

Hence,RA

�f C g

�DRAf C

RAg.

(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,

U�cf; P

�� L

�cf; P

�D c

hU�f; P

�� L

�f; P

�i< "I

that is, cf is integrable. Further,

c

ZA

f �"

c< cL

�f; P

�D L

�cf; P

�6ZA

cf 6 U�cf; P

�D cU

�f; P

�< c

ZA

f C"

c;

i.e.,RAcf D c

RAf .

Now let c < 0. Then for any partition P of A, we have mS�cf�D cMS

�f�

and MS

�cf�D cmS

�f�. Hence L

�cf; P

�D cU

�f; P

�and U

�cf; P

�D cL

�f; P

�.

Since f is integrable, for every " > 0, choose P such that U�f; P

�� L

�f; P

�<

�"=c. Then

U�cf; P

�� L

�cf; P

�D �c

hU�f; P

�� L

�f; P

�i< "I

that is, cf is integrable. Furthermore,

�c

ZA

f C"

c< �cL

�f; P

�D �U

�cf; P

�6 �

ZA

cf 6 �L�cf; P

�D �cL

�f; P

�< �c

ZA

f �"

c;

i.e.,RAcf D c

RAf . ut


I Exercise 75 (3-4). Let f W A! R and let P be a partition of A. Show that f is

integrable if and only if for each subrectangle S the function f�S is integrable,

and that in this caseRAf D

PS

RSf�S .

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 P1 of A

such that U�f; P1

�� L

�f; P1

�< " for any given " > 0. Let P2 be a common

refinement of P and P1. Then

U�f; P2

�� L

�f; P2

�6 U

�f; P1

�� L

�f; P1

�< ";

and there are rectangles˚S12 ; : : : ; S

n2

µ �2 .S/ with respect to P2, such that

S DSniD1 S

i2. Therefore,

U�f; P2

�� L

�f; P2

�D

XS2

hMS2

�f��mS2

�f�iv .S2/

>X

S22�2.S/

hMS2

�f��mS2

�f�iv .S2/

D U�f�S;P2

�� L

�f�S;P2

�I

that is, f�S is integrable.

0

x

y

a b

c

d

Figure 3.2.

If: Now suppose that f �S is integrable for each S . For each partition P 0, letˇ̌P 0ˇ̌

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 sub-

sequences defining the partitions of the PS ; see Figure 3.2. Then there are


refinements P 0S of PS whose rectangles are the set of all subrectangles of P 0

which are contained in S . Hence,XS

ZS

f�S � " <XS

L�f�S;PS

�6XS

L�f�S;P 0S

�D L

�f; P 0

�6 U

�f; P 0

�D

XS

U�f�S;P 0S

�6XS

U�f�S;PS

�<XS

ZS

f�S C ":

Therefore, f is integrable, andRAf D

PS

RSf�S . ut

I Exercise 76 (3-5). Let f; g W A ! R be integrable and suppose f 6 g. Show

thatRAf 6

RAg.

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 seeRA

�g � f

�> 0

since g > f . It follows from Exercise 74 thatRA

�g � f

�DRA

�g C

��f

��DR

Ag C

RA

��f

�DRAg �

RAf ; hence,

RAf 6

RAg. ut

I Exercise 77 (3-6). If f W A! R is integrable, show that jf j is integrable andˇ̌RAfˇ̌6RAjf j.

Proof. Let f C D max ff; 0g and f � D max f�f; 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

f C and f � are integrable if f is. Further,ˇ̌̌̌ZA

f

ˇ̌̌̌D

ˇ̌̌̌ZA

�f C � f �

�ˇ̌̌̌D

ˇ̌̌̌ZA

f C �

ZA

f �ˇ̌̌̌

6ZA

f C C

ZA

f �

D

ZA

�f C C f �

�D

ZA

jf j : ut

I Exercise 78 (3-7). Let f W Œ0; 1� � Œ0; 1�! R be defined by

SECTION 3.3 FUBINI’S THEOREM 51

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 andRŒ0;1��Œ0;1�

f D 0.

Proof. ut

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. ut


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


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 W Œa; b� � Œa; b�! R is continuous, show thatZ b

a

Z y

a

f�x; y

�dx dy D

Z b

a

Z b

x

f�x; y

�dy dx:

Proof. As illustrated in Figure 3.3,


C Dn�x; y

�2 Œa; b�2 W a 6 x 6 y and a 6 y 6 b

oD

n�x; y

�2 Œa; b�2 W a 6 x 6 b and x 6 y 6 b

o:

0

x

y

0

x

y

a b

a

byDx

C

y first

x first

Figure 3.3. Fubini’s Theoremut

I Exercise 83 (3-30). Let C be the set in Exercise 17. Show thatZŒ0;1�

ZŒ0;1�

1C�x; y

�dx

!dy D

ZŒ0;1�

ZŒ0;1�

1C�x; y

�dy

!dx D 0:

Proof. There must be typos. ut

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

F .x/ D

ZŒa1;x1��Œan;xn�

f:

What is DiF .x/, for x 2 int.A/?

Solution. Let c 2 int.A/. Then

SECTION 3.3 FUBINI’S THEOREM 53

DiF .c/ D limh!0

F�c�i ; ci C h

�� F .c/

h

D limh!0

RŒa1;c1��Œai ;c

iCh��Œan;cn� f � F .c/

h

D limh!0

R ciCh

ai

�RŒa1;c1��Œai�1;x

i�1��ŒaiC1;ciC1��Œan;cn� f

�dxi � F .c/

h

D limh!0

R ciCh

ci

�RŒa1;c1��Œai�1;c

i�1��ŒaiC1;ciC1��Œan;cn� f

�dxi

h

D

ZŒa1;c1��Œai�1;c

i�1��ŒaiC1;ciC1��Œan;cn�

f�x�i ; ci

�: ut

I Exercise 85 (3-32�). Let f W Œa; b� � Œc; d � ! R be continuous and suppose

D2f is continuous. Define F�y�DR baf�x; y

�dx. Prove Leibnitz’s rule: F 0

�y�DR b

aD2f

�x; y

�dx.

Proof. We have

F 0�y�D limh!0

F�y C h

�� F

�y�

h

D limh!0

R baf�x; y C h

�dx �

R baf�x; y

�dx

h

D limh!0

Z b

a

f�x; y C h

�� f

�x; y

�h

dx:

By DCT, we have

F 0�y�D

Z b

a

"limh!0

f�x; y C h

�� f

�x; y

�h

#dx

D

Z b

a

D2f�x; y

�dx: ut

I Exercise 86 (3-33). If f W Œa; b� � Œc; d �! R is continuous and D2f is contin-

uous, define F�x; y

�DR xaf�t; y

�dt .

a. Find D1F and D2F .

b. If G .x/ DR g.x/a

f .t; x/ dt , find G0 .x/.

Solution.

(a) D1F�x; y

�D f

�x; y

�, and D2F D

R xa

D2f�t; y

�dt .

(b) It follows that G .x/ D F�g .x/ ; x

�. Then

G0 .x/ D g0 .x/D1F�g .x/ ; x

�C D2F

�g .x/ ; x

�D g0 .x/ f

�g .x/ ; x

�C

Z g.x/

a

D2f .t; x/ dt: ut

4INTEGRATION ON CHAINS

4.1 Algebraic Preliminaries

I Exercise 87 (4-1�). Let e1; : : : ; en be the usual basis of Rn and let '1; : : : ; 'nbe 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

'i1 ^ � � � ^ 'ik .ei1 ; : : : ; eik / DkŠ

1Š � � � 1ŠAlt

�'i1 ˝ � � � ˝ 'ik

�.ei1 ; : : : ; eik /

D

X�2Sk

.sgn.�//'i1.e�.i1// � � �'ik .e�.ik//

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) ut

I Exercise 88 (4-9�). Deduce the following properties of the cross product in

R3.

a.e1 � e1 D 0 e2 � e1 D �e3 e3 � e1 D e2

e1 � e2 D e3 e2 � e2 D 0 e3 � e2 D �e1

e1 � e3 D �e2 e2 � e3 D e1 e3 � e3 D 0

Proof.

55

56 CHAPTER 4 INTEGRATION ON CHAINS

(a) We just do the first line.

hw; zi D

e1

e1

w

D 0 H) z D e1 � e1 D 0;

hw; zi D

e2

e1

w

D �w3 H) e2 � e1 D �e3;

hw; zi D

e3

e1

w

D w2 H) e3 � e1 D e2:

ut

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, Mono-

graphs 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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Calculus on Manifolds - Jianfei Shen 行到水窮處 ... · Calculus on Manifolds A Solution Manual forSpivak(1965) Jianfei Shen School of Economics, The University of New South Wales