calculus on manifolds - jianfei shen 行到水窮處 ... · calculus on manifolds a solution manual...
TRANSCRIPT
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
4 CHAPTER 1 FUNCTIONS ON EUCLIDEAN SPACE
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
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 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
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; : : : ; 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.
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/ 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/.
16 CHAPTER 2 DIFFERENTIATION
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
SECTION 2.1 BASIC DEFINITIONS 17
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:
18 CHAPTER 2 DIFFERENTIATION
2.2 Basic Theorems
I Exercise 40 (2-10). Use the theorems of this section to find f 0 for the follow-
ing:
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
20 CHAPTER 2 DIFFERENTIATION
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):
SECTION 2.2 BASIC THEOREMS 21
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:
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).
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
SECTION 2.2 BASIC THEOREMS 23
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
24 CHAPTER 2 DIFFERENTIATION
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
SECTION 2.2 BASIC THEOREMS 25
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
sn.t/
1CCA ; and b D
bn.t/
1CCA :Then
As D b;
26 CHAPTER 2 DIFFERENTIATION
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/.
28 CHAPTER 2 DIFFERENTIATION
(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.
SECTION 2.3 PARTIAL DERIVATIVES 29
(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.
30 CHAPTER 2 DIFFERENTIATION
.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,
SECTION 2.3 PARTIAL DERIVATIVES 31
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/ :
32 CHAPTER 2 DIFFERENTIATION
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; "/ :
SECTION 2.3 PARTIAL DERIVATIVES 33
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
34 CHAPTER 2 DIFFERENTIATION
2.4 Derivatives
I Exercise 58 (2-28). Find expressions for the partial derivatives of the follow-
ing functions:
a. F.x; y/ 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
36 CHAPTER 2 DIFFERENTIATION
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
38 CHAPTER 2 DIFFERENTIATION
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
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 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
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�ex cosy; ex siny
�. 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
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
42 CHAPTER 2 DIFFERENTIATION
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
44 CHAPTER 2 DIFFERENTIATION
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
�.
SECTION 3.1 BASIC DEFINITIONS 47
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,
48 CHAPTER 3 INTEGRATION
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
SECTION 3.1 BASIC DEFINITIONS 49
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
50 CHAPTER 3 INTEGRATION
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
I Exercise 80 (3-9).
a. Show that an unbounded set cannot have content 0.
b. Give an example of a closed set of measure 0 which does not have content 0.
Proof.
(a) Finite union of bounded sets is bounded.
(b) Z or N. ut
I Exercise 81 (3-10).
a. If C is a set of content 0, show that the boundary of C has content 0.
b. Give an example of a bounded set C of measure 0 such that the boundary of
C does not have measure 0.
Proof. ut
3.3 Fubini’s Theorem
I Exercise 82 (3-27). If f 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,
52 CHAPTER 3 INTEGRATION
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