高等微積分
Advanced Calculus
Ching-hsiao Arthur Cheng 鄭經斅
Copyright
本 Lecture Note 著作權屬於鄭經斅所有，並授權國立清華大學數學系於 104 學年度高等
微積分課程授課教師依實際上課需要修改。未經授權之改寫及大量複製、販售，本人保
留法律追訴權。
Contents
0 Introduction - Sets and Functions
i
0.1
Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
i
0.2
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iii
0.3
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
1 The Real Line and Euclidean Space
1.1
1
Ordered Fields and the Number Systems . . . . . . . . . . . . . . . . . . . .
1
1.1.1
Fields and partial orders . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1.2
The natural numbers, the integers, and the rational numbers . . . . .
8
1.1.3
Countability
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
Completeness and the Real Number System . . . . . . . . . . . . . . . . . .
14
1.2.1
Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
1.2.2
Monotone sequence property and completeness . . . . . . . . . . . .
19
1.3
Least Upper Bounds and Greatest Lower Bounds . . . . . . . . . . . . . . .
26
1.4
Cauchy Sequences
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
1.5
Cluster Points and Limit Inferior, Limit Superior . . . . . . . . . . . . . . .
34
1.6
Euclidean Spaces and Vector Spaces . . . . . . . . . . . . . . . . . . . . . .
41
1.7
Normed Vector Spaces, Inner Product Spaces and Metric Spaces . . . . . . .
43
1.8
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
1.2
2 Point-Set Topology of Metric spaces
55
2.1
Open Sets and the Interior of Sets . . . . . . . . . . . . . . . . . . . . . . . .
55
2.2
Closed Sets, the Closure of Sets, and the Boundary of Sets . . . . . . . . . .
60
2.3
Sequences and Completeness（完備性） . . . . . . . . . . . . . . . . . . . .
67
2.4
Series of real numbers and vectors . . . . . . . . . . . . . . . . . . . . . . . .
73
ii
2.5
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Compact and Connected Sets
3.1
76
83
Compactness（緊緻性） . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
3.1.1
The Heine-Borel theorem . . . . . . . . . . . . . . . . . . . . . . . .
94
3.1.2
The nested set property . . . . . . . . . . . . . . . . . . . . . . . . .
95
3.2
Connectedness（連通性） . . . . . . . . . . . . . . . . . . . . . . . . . . . .
96
3.3
Subspace Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
3.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99
4 Continuous Maps
104
4.1
Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
4.2
Operations on Continuous Maps . . . . . . . . . . . . . . . . . . . . . . . . . 108
4.3
Images of Compact Sets under Continuous Maps . . . . . . . . . . . . . . . 110
4.4
Images of Connected and Path Connected Sets under Continuous Maps . . . 114
4.5
Uniform Continuity（均勻連續） . . . . . . . . . . . . . . . . . . . . . . . . 118
4.6
Differentiation of Functions of One Variable . . . . . . . . . . . . . . . . . . 123
4.7
Integration of Functions of One Variable . . . . . . . . . . . . . . . . . . . . 128
4.8
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
5 Uniform Convergence and the Space of Continuous Functions
155
5.1
Pointwise and Uniform Convergence（逐點收斂與均勻收斂） . . . . . . . . 155
5.2
Series of Functions and The Weierstrass M -Test . . . . . . . . . . . . . . . . 164
5.3
Integration and Differentiation of Series
5.4
The Space of Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . 171
5.5
The Arzelà-Ascoli Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
. . . . . . . . . . . . . . . . . . . . 167
5.5.1
Equi-continuous family of functions . . . . . . . . . . . . . . . . . . . 174
5.5.2
Compact sets in C (K; V) . . . . . . . . . . . . . . . . . . . . . . . . 178
5.6
The Stone-Weierstrass Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 181
5.7
The Contraction Mapping Principle（收縮映射原理）and its Applications . 188
5.7.1
5.8
The existence and uniqueness of the solution to ODEs . . . . . . . . 191
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
iii
6 Differentiation of Maps
205
6.1 Bounded Linear Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
6.1.1
The matrix representation of linear maps between finite dimensional
normed spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
6.2
6.3
Definition of Derivatives and the Jacobian Matrices . . . . . . . . . . . . . . 210
Conditions for Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . 215
6.4
Properties of Differentiable Functions . . . . . . . . . . . . . . . . . . . . . . 219
6.4.1
Continuity of differentiable maps . . . . . . . . . . . . . . . . . . . . 219
6.4.2
6.4.3
The product rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
The chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
6.4.4
The mean value theorem . . . . . . . . . . . . . . . . . . . . . . . . . 224
6.5
Directional Derivatives and Gradient Vectors . . . . . . . . . . . . . . . . . . 226
6.6
6.7
Higher Order Derivatives of Functions . . . . . . . . . . . . . . . . . . . . . 230
Taylor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
6.8
Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
6.9
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
7 The Inverse and the Implicit Function Theorems
257
7.1
7.2
The Inverse Function Theorem（反函數定理） . . . . . . . . . . . . . . . . 257
The Implicit Function Theorem（隱函數定理） . . . . . . . . . . . . . . . . 264
7.3
The Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
7.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
8 Integration of Functions of Several Variables
275
8.1 Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275
8.2
Conditions for Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . 279
8.3
The Lebesgue Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
8.3.1
8.3.2
Volume and Sets of Measure Zero . . . . . . . . . . . . . . . . . . . . 283
The Lebesgue Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 287
8.4
Properties of the Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293
8.5
The Fubini Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300
8.6
8.7
Change of Variables Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 310
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321
Chapter 0
Introduction - Sets and Functions
0.1 Sets
Definition 0.1. A set is a collection of objects called elements or members of the set.
To denote a set, we make a complete list tx1 , x2 , ¨ ¨ ¨ , xN u or use the notation
␣
(
x : P (x)
or
␣ ˇ
(
x ˇ P (x) ,
where the sentence P (x) describes the property that defines the set. A set A is said to be a
subset of S if every member of A is also a member of S. We write x P A (or A contains x)
if x is a member of A, and write A Ď S (or S includes A) if A is a subset of S. The empty
set, denoted H, is the set with no member.
Definition 0.2. Let S be a given set, and A Ď S, B Ď S. The set A Y B, called the union
of A and B, consists of members belonging to set A or set B. Let A1 , A2 , ¨ ¨ ¨ be sets. The
8
Ť
set
Ai = tx | x P Ai for some iu is the union of A1 , A2 , ¨ ¨ ¨ . The set A X B, called the
i=1
intersection of A and B, consists of members belonging to both set A and set B. Let
8
Ş
Ai = tx | x P Ai for all iu is the intersection of A1 , A2 , ¨ ¨ ¨ .
A1 , A2 , ¨ ¨ ¨ be sets. The set
i=1
Remark 0.3. Let F be the collection of some subsets in S. Sometimes we also write the
Ť
union of sets in F as
A; that is,
APF
ď
ˇ
␣
(
A = x P S ˇDA P F Q x P A
APF
Similarly,
Ş
ˇ
(
A = x P S ˇ @A P F Q x P A is the intersection of sets in F .
␣
APF
i
ii
CHAPTER 0. Introduction - Sets and Functions
Example 0.4. Let A = t1, 2, 3, 4, 5u, B = t1, 3, 7u, S = t1, 2, 3, ¨ ¨ ¨ u, and F = tA, Bu.
Ť
Ş
Then
E ” A Y B = t1, 2, 3, 4, 5, 7u, and
E ” A X B = t1, 3u.
EPF
EPF
Definition 0.5. Let S be a given set, and A Ď S, B Ď S. The complement of A relative
to B, denoted BzA, is the set consisting of members of B that are not members of A. When
the universal set S under consideration is fixed, the complement of A relative to S or simply
the complement of A, is denoted by AA , or SzA.
Theorem 0.6. (De Morgan’s Law)
1. Bz
8
Ť
Ai =
i=1
2. Bz
8
Ş
i=1
Ai =
8
Ş
(BzAi ) or Bz
Ť
i=1
APF
8
Ť
Ş
(BzAi ) or Bz
i=1
A=
Ş
(BzA).
APF
A=
APF
Ť
(BzA).
APF
Proof. By definition,
x P Bz
8
ď
i=1
Ai ô x P B but x R
8
ď
Ai ô x P B and x R Ai for all i
i=1
ô x P BzAi for all i ô x P
8
č
(BzAi )
i=1
The proof of the second identity is similar, and is left as an exercise.
˝
Definition 0.7. An ordered pair (a, b) is an object formed from two objects a and b,
where a is called the first coordinate and b the second coordinate. Two ordered pairs
are equal whenever their corresponding coordinates are the same. An ordered n-tuples
(a1 , a2 , ¨ ¨ ¨ , an ) is an object formed from n objects a1 , a2 , ¨ ¨ ¨ , an , where for each j, aj is
called the j-th coordinate. Two n-tuples (a1 , a2 , ¨ ¨ ¨ , an ), (c1 , c2 , ¨ ¨ ¨ , cn ) are equal if aj = cj
for all j P t1, ¨ ¨ ¨ , nu.
Definition 0.8. Given sets A and B, the Cartesian product A ˆ B of A and B is the
ˇ
␣
(
set of all ordered pairs (a, b) with a P A and b P B, A ˆ B = (a, b) ˇ a P A and b P B . The
Cartesian of three or more sets are defined similarly.
Example 0.9. Let A = t1, 3, 5u and B = t‹, ˛u. Then
␣
(
A ˆ B = (1, ‹), (3, ‹), (5, ‹), (1, ˛), (3, ˛), (5, ˛) .
§0.2 Functions
iii
Example 0.10. Let A = [2, 7] and B = [1, 4]. The Cartesian product of A and B is the
square plotted below:
y
B
O
x
A
Figure 1: The Cartesian product [2, 7] ˆ [1, 4]
0.2 Functions
Definition 0.11. Let S and T be given sets. A function f : S Ñ T consists of two sets
S and T together with a “rule” that assigns to each x P S a special element of T denoted
by f (x). One writes x ÞÑ f (x) to denote that x is mapped to the element f (x). S is called
the domain (定義域）of f , and T is called the target or co-domain of f . The range
ˇ
␣
(
(值域）of f or the image of f , is the subset of T defined by f (S) = f (x) ˇ x P S .
f
T
x
S
f (x)
f (S)
Definition 0.12. A function f : S Ñ T is called one-to-one（一對一）, injective or
an injection if x1 ‰ x2 ñ f (x1 ) ‰ f (x2 ) (which is equivalent to that f (x1 ) = f (x2 ) ñ
x1 = x2 ). A function f : S Ñ T is called onto（映成）, surjective or an surjection
if @ y P T, D x P S, Q f (x) = y (that is, f (S) = T ). A function f : S Ñ T is called an
bijection if it is one-to-one and onto.
Remark 0.13 (映成函數的反敘述). If f : S Ñ T is not onto, then D y P T , Q @ x P S,
f (x) ‰ y. 一般來說，若有一個的數學的敘述 @ statement A, D statement B Q statement C
成立，那麼它的相反敘述的寫法為: D statement A, Q @ statement B, statement C 不成立。
簡單的記法：1. @ Ø D 2. D P Q Q Ø Q @ P ∼ Q.
iv
CHAPTER 0. Introduction - Sets and Functions
ˇ
␣
(
Definition 0.14. For f : S Ñ T , A Ď S, we call f (A) = f (x) ˇ x P A the image of A
ˇ
␣
(
under f . For B Ď T , we call f ´1 (B) = x P S ˇ f (x) P B the pre-image of B under f .
Example 0.15. f : R Ñ R, f (x) = x2 , B = [´1, 4] Ď T , f ´1 (B) = [´2, 2].
y
4
x
´2
´1 2
Figure 2: The preimage f ´1 ([´1, 4]) is [´2, 2] if f (x) = x2
Proposition 0.16. Let f : S Ñ T be a function, C1 ,C2 Ď T and D1 , D2 Ď S.
(a) f ´1 (C1 Y C2 ) = f ´1 (C1 ) Y f ´1 (C2 ).
(b) f (D1 Y D2 ) = f (D1 ) Y f (D2 ).
(c) f ´1 (C1 X C2 ) = f ´1 (C1 ) X f ´1 (C2 ).
(d) f (D1 X D2 ) Ď f (D1 ) X f (D2 ).
(e) f ´1 (f (D1 )) Ě D1 (“=” if f is one-to-one).
(f) f (f ´1 (C1 )) Ď C1 (“=” if C1 Ď f (S)).
§0.3 Exercises
v
Proof. We only prove (c) and (d), and the proof of the other statements are left as an
exercise.
(c) We first show that f ´1 (C1 X C2 ) Ď f ´1 (C1 ) X f ´1 (C2 ). Suppose that x P f ´1 (C1 X C2 ).
Then f (x) P C1 X C2 . Therefore, f (x) P C1 and f (x) P C2 , or equivalently, x P f ´1 (C1 )
and x P f ´1 (C2 ); thus x P f ´1 (C1 ) X f ´1 (C2 ).
Next, we show that f ´1 (C1 ) X f ´1 (C2 ) Ď f ´1 (C1 X C2 ). Suppose that x P f ´1 (C1 ) X
f ´1 (C2 ). Then x P f ´1 (C1 ) and x P f ´1 (C2 ) which suggests that f (x) P C1 and
f (x) P C2 ; thus f (x) P C1 X C2 or equivalently, x P f ´1 (C1 X C2 ).
(d) Suppose that y P f (D1 X D2 ). Then D x P D1 X D2 such that y = f (x). As a
consequence, y P f (D1 ) and y P f (D2 ) which implies that y P f (D1 ) X f (D2 ).
˝
Example 0.17. We note it might happen that f (D1 X D2 ) Ĺ f (D1 ) X f (D2 ). Take D1 =
[´1, 0] and D2 = [0, 1], and define f : S = R Ñ T = R to be f (x) = x2 . Then f (D1 ) =
f ([´1, 0]) = [0, 1] and f (D2 ) = f ([0, 1]) = [0, 1]. However,
f (D1 X D2 ) = f (t0u) = t0u Ĺ [0, 1] = f (D1 ) X f (D2 ) .
0.3 Exercises
§0.1 Sets
Problem 0.1. Let A, B, C be given sets. Show that
1. (A Y B) X C = (A X C) Y (B X C).
2. (A X B) Y C = (A Y C) X (B Y C).
§0.2 Functions
Problem 0.2. Let S and T be given sets, A Ď S, B Ď T , and f : S Ñ T . Show that
1. f (f ´1 (B)) Ď B, and f (f ´1 (B)) = B if B Ď f (S).
2. f ´1 (f (A)) Ě A, and f ´1 (f (A)) = A if f : S Ñ T is one-to-one.
Problem 0.3. Let A and B be two non-empty sets and f : A Ñ B. Show that
vi
CHAPTER 0. Introduction - Sets and Functions
1. f ´1 (D1 YD2 ) = f ´1 (D1 )Yf ´1 (D2 ), f ´1 (D1 XD2 ) = f ´1 (D1 )Xf ´1 (D2 ), f ´1 (D1 zD2 ) =
f ´1 (D1 )zf ´1 (D2 ) for all D1 , D2 Ď B.
2. f (C1 Y C2 ) = f (C1 ) Y f (C2 ), f (C1 X C2 ) Ď f (C1 ) X f (C2 ) for all C1 , C2 Ď A.
Problem 0.4. Let A and B be two non-empty sets and f : A Ñ B. Show that the following
three statements are equivalent; that is, show that each one of the following statements
implies the other four.
1. f is one-to-one.
2. For every y in B, the set f ´1 (tyu) contains at most one point.
3. f ´1 (f (C)) = C for all C Ď A.
4. f (C1 X C2 ) = f (C1 ) X f (C2 ) for all subsets C1 and C2 of A.
5. f (C2 zC1 ) = f (C2 ) ´ f (C1 ) for all C1 Ď C2 Ď A.
Chapter 1
The Real Line and Euclidean Space
1.1 Ordered Fields and the Number Systems
1.1.1 Fields and partial orders
Definition 1.1. A set F is said to be a field (體) if there are two operations + and ¨ such
that
1. x + y P F, x ¨ y P F if x, y P F. (封閉性)
2. x + y = y + x for all x, y P F. (commutativity, 加法的交換性)
3. (x + y) + z = x + (y + z) for all x, y, z P F. (associativity, 加法的結合性)
4. There exists 0 P F, called 加法單位元素, such that x + 0 = x for all x P F. (the
existence of zero)
5. For every x P F, there exists y P F (usually y is denoted by ´x and is called x 的加
法反元素) such that x + y = 0. One writes x ´ y ” x + (´y).
6. x ¨ y = y ¨ x for all x, y P F. (乘法的交換性)
7. (x ¨ y) ¨ z = x ¨ (y ¨ z) for all x, y, z P F. (乘法的結合性)
8. There exists 1 P F, called 乘法單位元素, such that x ¨ 1 = x for all x P F. (the
existence of unity)
9. For every x P F, x ‰ 0, there exists y P F (usually y is denoted by x´1 and is called
x 的乘法反元素) such that x ¨ y = 1. One writes x ¨ y ” x ¨ x´1 = 1.
1
2
CHAPTER 1. The Real Line and Euclidean Space
10. x ¨ (y + z) = x ¨ y + x ¨ z for all x, y, z P F. (distributive law, 分配律)
11. 0 ‰ 1.
Remark 1.2. Let x and y be both multiplicative inverse（乘法反元素）of a number a in
(F, +, ¨). Then
x¨a=1
(x ¨ a) ¨ y = 1 ¨ y = y
ñ
x ¨ 1 = x ¨ (a ¨ y) = y ;
ñ
thus x = y. In other words, the multiplicative inverse of a number is unique.
Remark 1.3. A set F satisfying properties 1 to 10 with 0 = 1 consists of only one member:
By distributive law, x¨0 = x¨(0+0) = x¨0+x¨0; thus ´(x¨0)+(x¨0) = ´(x¨0)+(x¨0)+(x¨0)
which implies that x ¨ 0 = 0. Therefore, if 0 = 1, then x = x ¨ 1 = x ¨ 0 = 0 for all x P F.
Hence, the set F consists only one element 0.
(
)
Remark 1.4. If x P F, then (1 + (´1) ¨ x = 0 which implies that x + (´1) ¨ x = 0.
Therefore, (´1) ¨ x = ´x + x + (´1) ¨ x = ´x + 0 = ´x.
! ˇ
)
qˇ
Example 1.5. Let Q =
ˇ p ‰ 0, p, q P Z : integers . Then Q is a field. (Check all the
p
properties from 1 to 11).
ˇ
␣
(
Example 1.6. Let N = n P Z ˇ n ą 0 . Then N is not a field because there is no zero.
Example 1.7. Let F = ta, b, cu with the operations + and ¨ defined by
+ a b c
a a b c
b b c a
c c a b
¨
a
b
c
a
a
a
a
b
a
b
c
c
a
.
c
b
Then F is a field because of the following: Properties 1, 2, 3, 6, 7 are obvious.
Property 4: D “0” Q x + “0” = x for all x P F. In fact, “0” = a.
Property 5: @ x P F, D y P F Q x + y = 0, here b = ´c, c = ´b.
Property 8: D “1” Q x ¨ “1” = x for all x P F. In fact, “1” = b (so Property 11 holds since
a ‰ b).
Property 9: @ x ‰ 0, P F, D z P F Q x ¨ z = 1, here z = x.
The validity of Property 10 is left as an exercise.
§1.1 Ordered Fields and the Number Systems
3
Example 1.8. Let (F, +, ¨) be a field. Then (x ´ y)(x + y) = x2 ´ y 2 for all x, y P F. In
fact,
(x ´ y)(x + y) = (x ´ y) ¨ x + (x ´ y) ¨ y
(by 分配律)
= x ¨ (x ´ y) + y ¨ (x ´ y)
(by 乘法交換律)
= x ¨ x + x ¨ (´y) + y ¨ x + y ¨ (´y)
(by 分配律)
= x2 ´ x ¨ y + x ¨ y ´ y 2
(by Remark 1.4 and 乘法交換律)
= x2 + 0 ´ y 2
(by Property 5)
= x2 ´ y 2
(by Property 4).
Definition 1.9. A partial order over a set P is a binary relation ď which is reflexive,
anti-symmetric and transitive (滿足遞移律), in the sense that
1. x ď x for all x P P (reflexivity).
2. x ď y and y ď x ñ x = y (anti-symmetry).
3. x ď y and y ď z ñ x ď z (transitivity).
A set with a partial order is called a partially ordered set.
Example 1.10. Let S be a given set, and 2S be the power set of S; that is,
␣ ˇ
(
P = 2S = A ˇ A Ď S = the collection of all subsets of S .
We define ď as Ě. Then
1. A Ě A (reflexivity).
2. A Ě B and B Ě A ñ A = B (anti-symmetry).
3. A Ě B and B Ě C ñ A Ě C (transitivity).
Hence, Ě is a partial order over 2S (or equivalently, (2S , Ě) is a partially ordered set).
Similarly, Ď on 2S is also a partial order.
Definition 1.11. Let (P, ď) be a partially ordered set. Two elements x, y P P are said to
be comparable if either x ď y or y ď x.
Definition 1.12. A partial order under which every pair of elements is comparable is called
a total order or linear order.
4
CHAPTER 1. The Real Line and Euclidean Space
Definition 1.13. An ordered field is a totally ordered field (F, +, ¨, ď) satisfying that
1. If x ď y, then x + z ď y + z for all z P F (compatibility of ď and +).
2. If 0 ď x and 0 ď y, then 0 ď x ¨ y (compatibility of ď and ¨).
Example 1.14. (Q, +, ¨, ě) is a totally ordered field, but is not an ordered field (since
Property 2 in Definition 1.13 is violated). On the other hand, (Q, +, ¨, ď) is an ordered
field.
From now on, the total order ď of an ordered field will be denoted by ď.
Definition 1.15. In an ordered field (F, +, ¨, ď), the binary relations ă, ě and ą are
defined by:
1. x ă y if x ď y and x ‰ y.
2. x ě y if y ď x.
3. x ą y if y ă x.
Adopting the definition above, it is not immediately clear that x ­ď y ô x ą y. However,
this is indeed the case, and to be more precise we have the following
Proposition 1.16. (Law of Trichotomy, 三一律) If x and y are elements of an ordered
field (F, +, ¨, ď), then exactly one of the relations x ă y, x = y or y ă x holds.
Proof. Since F is a totally ordered field, x and y are comparable. Therefore, either x ď y
or y ď x. Assume that x ď y.
1. If x = y, then x ­ă y and x ­ą y.
2. If x ‰ y, then x ă y. If it also holds that x ą y, then x ě y; thus by the property
of anit-symmetry of an order, we must have x = y, a contradiction. Therefore, it can
only be that x ă y.
The proof for the case y ď x is similar, and is left as an exercise.
Proposition 1.17. Let (F, +, ¨, ď) be an ordered field, and a, b, x, y, z P F.
1. If a + x = a, then x = 0.
If a ¨ x = a and a ‰ 0, then x = 1.
˝
§1.1 Ordered Fields and the Number Systems
2. If a + x = 0, then x = ´a.
If a ¨ x = 1 and a ‰ 0, then x = a´1 .
3. If x ¨ y = 0, then x = 0 or y = 0.
4. If x ď y ă z or x ă y ď z, then x ă z (the transitivity of ă).
5. If a ă b, then a + x ă b + x (the compatibility of ă and +).
If 0 ă a and 0 ă b, then 0 ă a ¨ b (the compatibility of ă and ¨).
6. If a + x = b + x, then a = b.
If a + x ď (ă) b + x, then a ď (ă) b.
If a ¨ x = b ¨ x and x ‰ 0, then a = b.
If a ¨ x ď (ă) b ¨ x and x ą 0, then a ď (ă) b.
7. 0 ¨ x = 0.
8. ´(´x) = x.
9. ´x = (´1) ¨ x.
10. If x ‰ 0, then x´1 ‰ 0 and (x´1 )´1 = x.
11. If x ‰ 0 and y ‰ 0, then x ¨ y ‰ 0 and (x ¨ y)´1 = x´1 ¨ y ´1 .
12. If x ď (ă) y and 0 ď (ă) z, then x ¨ z ď (ă) y ¨ z.
If x ď (ă) y and 0 ě (ą) z, then x ¨ z ě (ą) y ¨ z.
13. If x ď (ă) 0 and y ď (ă) 0, then x ¨ y ě (ą) 0.
If x ď (ă) 0 and y ě (ą) 0, then x ¨ y ď (ă) 0.
14. 0 ă 1 and ´1 ă 0.
15. x ¨ x ” x2 ě 0.
16. If x ą 0, then x´1 ą 0. If x ă 0, then x´1 ă 0.
Proof. 1. (´a) + a + x = (´a) + a = 0 ñ x = 0.
(a´1 ) ¨ a ¨ x = (a´1 ) ¨ a = 1 ñ x = 1.
5
6
CHAPTER 1. The Real Line and Euclidean Space
2. (´a) + a + x = (´a) + 0 = ´a ñ x = ´a.
(a´1 ) ¨ a ¨ x = (a´1 ) ¨ 1 = a´1 ñ x = a´1 .
3. Assume that x ‰ 0, then x´1 ¨ x ¨ y = x´1 ¨ 0 = 0 ñ y = 0.
Assume that y ‰ 0, then x ¨ y ¨ y ´1 = 0 ¨ y ´1 = 0 ñ x = 0.
4 and 5 are Left as an exercise.
6. a + 0 = a + x + (´x) = b + x + (´x) = b + 0 ñ a = b.
a + 0 = a + x + (´x) ď b + x + (´x) = b + 0 ñ a ď b (compatibility of ď and +).
a ¨ x ¨ x´1 = b ¨ x ¨ x´1 ñ a = b.
Suppose the contrary that b ă a. Then 0 = b + (´b) ď a + (´b). Since x ą 0, x ě 0;
thus
(
)
0 ď a + (´b) ¨ x = a ¨ x + (´b) ¨ x .
As a consequence, b ¨ x = 0 + b ¨ x ď a ¨ x + (´b) ¨ x + b ¨ x = a ¨ x. By assumption, we
must have a ¨ x = b ¨ x or (a ´ b) ¨ x = 0. Using 3, x = 0 (since a ‰ b), a contradiction.
7. See Remark 1.3.
8. (´x) + (´(´x)) = 0 = (´x) + x ñ x = ´(´x).
9. See Remark 1.4.
10. Assume x´1 = 0, 1 = x ¨ x´1 = x ¨ 0 = 0, a contradiction. Therefore, x´1 ‰ 0; thus
(x´1 )´1 ¨ x´1 = 1 = x ¨ x´1 ñ (x´1 )´1 = x (by 4).
11. That x ¨ y = 0 cannot be true since it is against Property 3, so x ¨ y ‰ 0. Moreover,
(x ¨ y)´1 (x ¨ y) = 1 = 1 ¨ 1 = (x ¨ x´1 ) ¨ (y ¨ y ´1 ) = (x´1 ¨ y ´1 ) ¨ (x ¨ y) ;
thus (x ¨ y)´1 = x´1 ¨ y ´1 (by 4).
12. If x ď (ă) y, then 0 = x + (´x) ď (ă) y + (´x). Since 0 ď (ă) z, by the compatibility
of ď (ă) and ¨ we must have 0 ď (ă) (y + (´x)) ¨ z = y ¨ z + (´x) ¨ z. Therefore, by
the compatibility of ď (ă) and +, x ¨ z = 0 + x ¨ z ď (ă) y ¨ z + (´x) ¨ z + x ¨ z = y ¨ z.
The second statement can be proved in a similar fashion.
13. Left as an exercise.
§1.1 Ordered Fields and the Number Systems
7
14. If 1 ď 0, then compatibility of ď and + implies that 0 ď ´1. By the compatibility of
ď and ¨, using 6 and 7 we find that 0 ď (´1) ¨ (´1) = ´(´1) = 1; thus we conclude
that 1 = 0, a contradiction. As a consequence, 0 ă 1; thus the compatibility of ă and
+ implies that ´1 ă 0.
15. Left as an exercise.
16. If x ą 0 but x´1 ď 0, then 1 = x ¨ x´1 ď x ¨ 0 = 0, a contradiction.
˝
Proposition 1.18. Let (F, +, ¨, ď) be an ordered field, and x, y P F.
1. If 0 ď x ă y, then x2 ă y 2 .
2. If 0 ď x, y and x2 ă y 2 , then x ă y.
Proof. 1. By definition of “<”, 0 ď x ď y and x ‰ y. Using 12 of Proposition 1.17,
x2 ď y ¨ x ă y ¨ y = y 2 .
By the transitivity of ă, we conclude that x2 ă y 2 .
2. Note that x ‰ y, for if not, then x2 ´ y 2 = 0 which contradicts to the assumption
x2 ă y 2 . Assume that y ă x, then 1 implies that y 2 ă x2 , a contradiction.
˝
Remark 1.19. Proposition 1.18 can be summarized as follows: if x, y ě 0, then
x ă y ô x2 ă y 2 .
Moreover, Example 1.8, Proposition 1.17 and Proposition 1.18 together suggest that if
x, y ě 0, then x ď y if and only if x2 ď y 2 .
Definition 1.20. The magnitude or the absolute value of x, denoted |x|, is defined as
"
x
if x ě 0 ,
|x| =
´x if x ă 0 .
Proposition 1.21. Let (F, +, ¨, ď) be an ordered field. Then
1. |x| ě 0 for all x P F.
2. |x| = 0 if and only if x = 0.
8
CHAPTER 1. The Real Line and Euclidean Space
3. ´|x| ď x ď |x| for all x P F.
4. |x ¨ y| = |x| ¨ |y| for all x, y P F.
5. |x + y| ď |x| + |y| for all x, y P F (triangle inequality, 三角不等式).
ˇ
ˇ
6. ˇ|x| ´ |y|ˇ ď |x ´ y| for all x, y P F.
Proof. Left as an exercise.
˝
Proposition 1.22. Define d(x, y) = |x ´ y|. Then
1. d(x, y) ě 0 for all x, y P F.
2. d(x, y) = 0 if and only if x = y.
3. d(x, y) = d(y, x) for all x, y P F.
4. d(x, y) ď d(x, z) + d(z, y) for all x, y, z P F (triangle inequality, 三角不等式).
Proof. Left as an exercise.
˝
Remark 1.23. d(x, y) is the “distance” of two elements x, y P F.
y
d(x, y)
x
d(z, y)
d(x, z)
z
Figure 1.1: An illustration of why 4 of Proposition 1.22 is called the triangle inequality.
1.1.2 The natural numbers, the integers, and the rational numbers
Definition 1.24. Let (F, +, ¨, ď) be an ordered field. The natural number system,
denoted by N, is the collection of all the numbers 1, 1 + 1, 1 + 1 + 1, 1 + 1 + ¨ ¨ ¨ + 1 and etc. in
F. We write 2 ” 1 + 1, 3 ” 2 + 1, and n ” 1looooooomooooooon
+ 1 + ¨ ¨ ¨ + 1. In other words, N = t1, 2, 3, ¨ ¨ ¨ u.
(n times)
The integer number system, denoted by Z, is the set Z = t¨ ¨ ¨ , ´3, ´2, ´1, 0, 1, 2, 3, ¨ ¨ ¨ u.
§1.1 Ordered Fields and the Number Systems
9
Principle of mathematical induction (Peano axiom, 皮亞諾公設):
If S is a subset of N Y t0u (or N) such that 0 P S (or 1 P S) and k + 1 P S if k P S, then
S = N Y t0u (or S = N).
Example 1.25. Prove
n
ř
k=
k=1
n(n + 1)
.
2
(‹)
ˇ ř
!
)
n(n + 1)
ˇ n
Proof. Let S = n P N ˇ
k=
（把所有滿足 (‹) 的 n 收集起來）. Then
2
k=1
1. If n = 1,
1
ř
k=
k=1
1ˆ2
= 1.
2
2. Assume that m P S, then
m+1
ÿ
k=
k=1
m
ÿ
k + (m + 1) =
k=1
m(m + 1)
(m + 1)(m + 2)
+ (m + 1) =
2
2
which implies that m + 1 P S.
By mathematical induction, we have S = N.
1
˝
1
Example 1.26. Prove that n ă for all n P N.
2
n
ˇ
!
)
1
1
ˇ
Proof. Let S = n P N ˇ n ă
. We show S = N by mathematical induction as follows:
2
(i) 1 P S ô
n
1
1
ă .
2
1
(ii) If n P S, then
1
2n+1
=
1 1
1 1
1
1
¨ ă ¨ =
ď
.
n
2 2
n 2
n+n
n+1
which implies that n + 1 P S.
By mathematical induction, we have S = N.
˝
Let (F, +, ¨, ď) be an ordered field. By the property of being a field, for any non-zero
n P N, there exists a unique multiplicative inverse n´1 . This inverse is usually denoted by
1
m
. We also use
to denote m ¨ n´1 . Giving this notation, we have the following
n
n
Definition 1.27. Let (F, +, ¨, ď) be an order field. The rational number system, denoted by Q, is the collection of all numbers of the form
is,
q
with p, q P Z and p ‰ 0; that
p
ˇ
)
q
ˇ
Q = x P F ˇ x = , p, q P Z, p ‰ 0 .
!
p
10
CHAPTER 1. The Real Line and Euclidean Space
Definition 1.28. An order field (F, +, ¨, ď) is said to have the Archimedean property
if @ x P F, D n P Z Q x ă n.
Theorem 1.29. Q has the Archimedean property.
Proof. If x ď 0, we take n = 1. Otherwise if 0 ă x =
and it is obvious that
q
ď q ă q + 1 = n.
p
q
with p, q P N, we take n = q + 1
p
˝
Definition 1.30. A well-ordered relation on a set S is a total order on S with the property
that every non-empty subset of S has a least (smallest) element in this ordering.
Proposition 1.31 (Well-Ordered Property of N). If S Ď N and S ‰ H, then S has a
smallest element; that is, D s0 P S Q @ x P S, s0 ď x.
Proof. Assume the contrary that there exists a non-empty set S Ď N such that S does not
ˇ
␣
(
have the smallest element. Define T = NzS, and T0 = n P N ˇ t1, 2, ¨ ¨ ¨ , nu Ď T . Then we
have T0 Ď T . Also note that 1 R S for otherwise 1 is the smallest element in S, so 1 P T
(thus 1 P T0 ).
Assume k P T0 . Since t1, 2, ¨ ¨ ¨ , ku Ď T , 1, 2, ¨ ¨ ¨ k R S. If k + 1 P S, then k + 1 is the
smallest element in S. Since we assume that S does not have the smallest element, k +1 R S;
thus k + 1 P T ñ k + 1 P T0 .
Therefore, by mathematical induction we conclude that T0 = N; thus T = N (since
T0 Ď T ) which further implies that S = H (since T = NzS). This contradicts to the
assumption S ‰ H.
˝
1.1.3 Countability
Definition 1.32. A set S is called denumerable or countably infinite（無窮可數的）if
S can be put into one-to-one correspondence with N; that is, S is denumerable if and only
if D f : N Ñ S which is one-to-one and onto. A set is called countable（可數的）if S is
either finite or denumerable.
1´1
1´1
onto
onto
Remark 1.33. If f : NÝÝÑS, then f ´1 : S ÝÝÑN. Therefore,
1´1
1´1
onto
onto
S is denumerable ô D f : NÝÝÑS ô D g = f ´1 : S ÝÝÑN.
␣
(
f can be thought as a rule of counting/labeling elements in S since S = f (1), f (2), ¨ ¨ ¨ .
§1.1 Ordered Fields and the Number Systems
11
1´1
Example 1.34. N is countable since f : NÝÝÑN with f (x) = x, @ n P N.
onto
Example 1.35. Z is countable. f : Z Ñ N with f (x) =
k
f(k)
−3
7
−2
5
−1
3
0
1
1
2
$
&
1
2x
%
´2x + 1
2
4
if x = 0
if x ą 0 .
if x ă 0
3
6
Figure 1.2: An illustration of how elements in Z are labeled
Example 1.36. The set N ˆ N =
␣
ˇ
(
(a, b) ˇ a, b P N is countable. In fact, two ways of
mapping are shown in the figures below.
y
y
5 11
5
4
7
12
3
4
8
2
2
1
1
1
4
13
3
5
9 14
2
3
6 10 15
1
2
3
4
5
x
10 11 12 13
9
8
7 14
2
3
6 15
1
4
5 16 17
1
2
3
4
5
x
Figure 1.3: The illustration of two ways of labeling elements in N ˆ N
Proposition 1.37. Let S be a non-empty set. The following three statements are equivalent:
(a) S is countable;
(b) there exists a surjection f : N Ñ S;
(c) there exists an injection f : S Ñ N.
Proof. “(a) ñ (b)” First suppose that S = tx1 , ¨ ¨ ¨ , xn u is finite. Define f : N Ñ S by
"
xk if k ă n ,
f (k) =
xn if k ě n .
12
CHAPTER 1. The Real Line and Euclidean Space
Then f : N Ñ S is a surjection. Now suppose that S is denumerable. Then by
1´1
definition of countability, there exists f : NÝÝÑS.
onto
“(a) ð (b)” W.L.O.G. (without loss of generality, 不失一般性) we assume that S is an
infinite set. Let k1 = 1. Since #(S) = 8, S1 ” Sztf (k1 )u ‰ H; thus N1 ” f ´1 (S1 ) is
a non-empty subset of N. By the well-ordered property of N (Proposition 1.31), N1 has
a smallest element denoted by k2 . Since #(S) = 8, S2 = Sztf (k1 ), f (k2 )u ‰ H; thus
N2 ” f ´1 (S2 ) is a non-empty subset of N and possesses a smallest element denoted by
k3 . We continue this process and obtain a set tk1 , k2 , ¨ ¨ ¨ u Ď N, where k1 ă k2 ă ¨ ¨ ¨ ,
and kj is the smallest element of Nj´1 ” f ´1 (Sztf (k1 ), f (k2 ), ¨ ¨ ¨ , f (kj´1 )u).
Claim: f : tk1 , k2 , ¨ ¨ ¨ u Ñ S is one-to-one and onto.
␣
(
Proof of claim: The injectivity of f is due to that f (kj ) R f (k1 ), f (k2 ), ¨ ¨ ¨ , f (kj´1 )
for all j ě 2. For surjectivity, assume that there is s P S such that s R f (tk1 , k2 , ¨ ¨ ¨ u).
Since f : N Ñ S is onto, f ´1 (tsu) is a non-empty subset of N; thus possesses a smallest
element k. Since s R f (tk1 , k2 , ¨ ¨ ¨ u), there exists ℓ P N such that kℓ ă k ă kℓ+1 . As a
consequence, there exists k P Nℓ such that k ă kℓ+1 which contradicts to the fact that
kℓ+1 is the smallest element of Nℓ .
Define g : N Ñ tk1 , k2 , ¨ ¨ ¨ u by g(j) = kj . Then g : N Ñ tk1 , k2 , ¨ ¨ ¨ u is one-to-one and
1´1
onto; thus h = g ˝ f : NÝÝÑS.
onto
“(a) ñ (c)” If S = tx1 , ¨ ¨ ¨ , xn u is finite, we simply let f : S Ñ N be f (xn ) = n. Then f is
1´1
clearly an injection. If S is denumerable, by definition there exists g : NÝÝÑS which
suggests that f = g ´1 : S Ñ N is an injection.
onto
1´1
“(a) ð (c)” Let f : S Ñ N be an injection. If f is also surjective, then f : S ÝÝÑN which
onto
implies that S is denumerable. Now suppose that f (S) Ĺ N. Since S is non-empty,
there exists s P S. Let g : N Ñ S be defined by
"
g(n) =
f ´1 (n) if n P f (S) ,
s
if n R f (S) .
Then clearly g : N Ñ S is surjective; thus the equivalence between (a) and (b) implies
that S is countable.
Theorem 1.38. Any non-empty subset of a countable set is countable.
˝
§1.1 Ordered Fields and the Number Systems
13
Proof. Let S be a countable set, and A be a non-empty subset of S. Since S is countable,
by by Proposition 1.37 there exists a surjection f : N Ñ S. On the other hand, since A is a
non-empty subset of S, there exists a P A. Define
"
x if x P A ,
g(x) =
a if x R A .
Then h = g ˝ f : N Ñ A is a surjection, and Proposition 1.37 suggests that A is countable.
˝
Example 1.39. The set N ˆ N is countable since the map f : N ˆ N Ñ N defined by
f ((m, n)) = 2m 3n is an injection.
Theorem 1.40. The union of denumerable denumerable sets is denumerable（無窮可數個
無窮可數集的聯集是無窮可數的）. In other words, if F is a denumerable collection of
Ť
denumerable sets, then
A is denumerable.
APF
ˇ
Proof. Let F = Ai ˇ i P N, Ai is denumerableu be an indexed family of denumerable
8
Ť
sets, and define A =
Ai . Since Ai is denumerable, Ai = txi1 , xi2 , xi3 , ¨ ¨ ¨ u. Then
i=1
␣ ˇ
(
A = xij ˇ i, j P N . Let f : NˆN Ñ A be defined by f ((i, j)) = xij . Then f : NˆN Ñ A is a
surjection. Moreover, Example 1.39 implies that there exists a bijection g : N Ñ N ˆ N; thus
␣
h = f ˝ g : N Ñ A is a surjection which, by Proposition 1.37, implies that A is countable.
Since A1 Ď A, A is infinite; thus A is denumerable.
˝
Corollary 1.41. The union of countable countable sets is countable（可數個可數集的聯集
是可數的）.
Proof. By adding empty sets into the family or adding N into a finite set if necessary, we
find that the union of countable countable sets is a subset of the union of denumerable
denumerable sets. By Theorem 1.38, we find that the union of countable countable sets is
countable.
˝
Example 1.42. Z ˆ Z is countable.
ˇ
␣
(
Proof. For i P Z, let Ai = (i, j) ˇ j P Z . By Example 1.35, Ai is countable for all i P Z.
Ť
Since Z ˆ Z = iPZ Ai which is countable union of countable sets, Theorem 1.40 implies
that Z ˆ Z is countable.
˝
14
CHAPTER 1. The Real Line and Euclidean Space
Theorem 1.43. Q is countable.
Proof. Define
$
q
’
(p, q), if x ą 0, x = , gcd(p, q) = 1, p ą 0.
’
&
p
(0, 0), if x = 0.
f (x) =
’
q
’
% (p, ´q), if x ă 0, x = ´ , gcd(p, q) = 1, p ą 0.
p
1´1
Then f : Q Ñ Z ˆ Z is one-to-one; thus f : QÝÝÑf (Q). Since Z ˆ Z is countable, its
onto
1´1
non-empty subset f (Q) is also countable. As a consequence, there exists g : f (Q)ÝÝÑN;
onto
1´1
thus h = g ˝ f : QÝÝÑN.
˝
onto
1.2 Completeness and the Real Number System
1.2.1 Sequences
Definition 1.44. A sequence in a set S is a function f : N Ñ S (not necessary one-to-one
or onto). The values of f are called the terms of the sequence.
Remark 1.45. A sequence in S is a countable list of elements in S arranged in a particular
␣
(8
order, and is usually denoted by f (n) n=1 or txn u8
n=1 with xn = f (n).
Definition 1.46. Let F be an ordered field. A sequence txn u8
n=1 Ď F is said to be convergent if there exists x P F such that for every ε ą 0,
ˇ
␣
(
# n P N ˇ xn R (x ´ ε, x + ε) ă 8 .
Such an x is called a limit of the sequence. In notation,
txn u8
n=1 Ď F is convergent ô D x P F Q @ ε ą 0, #tn P N | xn R (x ´ ε, x + ε)u ă 8.
8
If x is a limit of txn u8
n=1 , we say txn un=1 converges to x and write lim xn = x or xn Ñ x as
nÑ8
n Ñ 8. If no such x exists we say that txn u8
n=1 diverges or
lim xn does not exist.
Remark 1.47. The number N may depend on ε, and smaller ε usually requires larger N .
nÑ8
In the definition above, it could happen that there are two different limits of a convergent
sequence. In fact, this is never the case because of the following
§1.2 Completeness and the Real Number System
15
Proposition 1.48. If txn u8
n=1 is a sequence in an ordered field F, and xn Ñ x and xn Ñ y
as n Ñ 8, then x = y. (The uniqueness of the limit).
Proof. Assume the contrary that x ‰ y. W.L.O.G. we may assume that x ă y, and let
ε=
y´x
ą 0. Define
2
ˇ
␣
(
A1 = n P N ˇ xn R (x ´ ε, x + ε)
and
ˇ
␣
(
A2 = n P N ˇ xn R (y ´ ε, y + ε) .
Then by the definition of the convergence of sequences, #A1 ă 8 and #A2 ă 8. Let
N1 = max A1 , N2 = max A2 and N = maxtN1 , N2 u. Since A1 , A2 are finite, N ă 8. On the
other hand, N + 1 R A1 Y A2 which implies that xN +1 P (x ´ ε, x + ε) X (y ´ ε, y + ε) = H,
a contradiction.
˝
Example 1.49. Let xn =
(´1)n
. We show that txn u8
n=1 converges to 0. By definition,
n+1
ˇ
␣
(
we need to show for every ε ą 0 the set Aε = n P N ˇ xn R (´ε, ε) is finite. Note that
ˇ
␣
(
Aε = n P N ˇ |xn | ě ε ; thus
ˇ
!
ˇ
Aε = n P N ˇ
Therefore, #Aϵ =
[1]
ε
1
ěε
n+1
)
ˇ
!
)
1
ˇ
= n P Nˇn ď ´ 1 .
ε
´ 1 ă 8 which implies that txn u8
n=1 converges to 0.
Example 1.50. The sequence tyn u8
n=1 given by yn =
3 + (´1)n
diverges. To see this, we
2
have to show that any real number x cannot be the limit of tyn u8
n=1 .
Let y be given and ε =
1
. Then (y ´ ε, y + ε) at most contains one integer. Since yn
2␣
(
␣
(
only takes value 1 or 2 and # n P N | yn = 1 = # n P N | yn = 2 = 8, we find that
ˇ
␣
(
# n P N ˇ yn R (y ´ ε, y + ε) = 8
which implies tyn u8
n=1 cannot converges to y.
Example 1.51. A permutation of a non-empty set A is a one-to-one function from A
onto A. Let π : N Ñ N be a permutation of N, and txn u8
n=1 be a convergent sequence in
␣
(8
an ordered field F. Then xπ(n) n=1 is also convergent since if x is the limit of txn u8
n=1 and
ε ą 0,
ˇ
ˇ
␣
(
␣
(
# n P N ˇ xπ(n) R (x ´ ε, x + ε) = # n P N ˇ xn R (x ´ ε, x + ε) ă 8 .
16
CHAPTER 1. The Real Line and Euclidean Space
Proposition 1.52. Let F be an ordered field, txn u8
n=1 Ď F be a sequence, and x P F. Then
lim xn = x if and only if for every ε ą 0, there exists N ą 0 such that |xn ´ x| ă ε whenever
nÑ8
n ě N . In notation,
lim xn = x
nÑ8
@ ε ą 0, D N ą 0 Q n ě N ñ |xn ´ x| ă ε .
ô
ˇ
␣
(
8
Proof. “ñ” Let ε ą 0 be given, and Aϵ = n P N ˇ xn R (x ´ ε, x + ε) . Since txn un=1
converges to x, k ” #Aϵ ă 8 . Suppose that n1 ă n2 ă ¨ ¨ ¨ ă nk belongs to Aϵ . Let
N = nk + 1. Then if n ě N , n R Aϵ which implies that if n ě N , xn P (x ´ ε, x + ε)
or equivalently,
|xn ´ x| ă ε whenever n ě N .
ε
x1
(
x4 xN
0
ε
x
)
x5 x3
x2
xn for n ≥ N = N0 + 1
Figure 1.4: Let N0 be the largest index of those xn ’s outside (x ´ ε, x + ε). Then xn P
(x ´ ε, x + ε) whenever n ě N = N0 + 1.
“ð” Let ε ą 0 be given. Then for some N ą 0, if n ě N , we have |xn ´ x| ă ε or
equivalently, if n ě N , xn P (x ´ ε, x + ε). This implies that
ˇ
␣
(
# n P N ˇ xn R (x ´ ε, x + ε) ă N ă 8 .
˝
Remark 1.53. A sequence txn u8
n=1 Ď F diverges if (and only if)
@ x P F, D ε ą 0 Q #tn P N | xn R (x ´ ε, x + ε)u = 8
which is equivalent to that
@ x P F, D ε ą 0 Q tn P N | xn R (x ´ ε, x + ε)u = tn1 ă n2 ă ¨ ¨ ¨ ă nj ă ¨ ¨ ¨ u .
Therefore, txn u8
n=1 diverges if (and only if)
@ x P F, D ε ą 0 Q @ N ą 0, D n ě N such that |xn ´ x| ě ε .
Example 1.54. Now we use the ε-N argument as the definition of the convergence of
sequences to re-establish the convergence of sequences in Example 1.49, 1.50 and 1.51.
§1.2 Completeness and the Real Number System
17
[1]
(´1)n
Example 1.49 - revisit: Let ε ą 0 be given, and xn =
. Let N =
+ 1. Since
ε
n+1
[1] 1
1
1
ą ´ 1, if n ě N we must have n ą ´ 1; thus if n ě N ,
ă ε. Therefore,
ε
ε
ε
n+1
whenever n ě N
|xn ´ 0| ă ε
which implies that txn u8
n=1 converges to 0.
1
Example 1.50 - revisit: Let y be given, ε = , and N P N. Define
2
"
N + 1 if |yN ´ y| ă ε ,
n=
N + 2 if |yN ´ y| ě ε .
Then n ě N . Moreover, if |yN ´ y| ă ε, then |yn ´ y| ě |yn ´ yN | ´ |yN ´ y| ą 1 ´ ε = ε,
while if |yN ´ y| ě ε then clearly |yn ´ y| ě ε. Therefore,
@ y P F, D ε ą 0 Q @ N ą 0, D n ą N Q |yn ´ y| ě ε .
Example 1.51 - revisit: Now suppose that txn u8
n=1 is a convergent sequence with limit x,
and ε ą 0 be given. Then there exists N1 ą 0 such that if n ě N1 , we have |xn ´x| ă ε.
␣
(
Let N = max π ´1 (1), π ´1 (2), ¨ ¨ ¨ , π ´1 (N1 ) + 1. Then if n ě N , π(n) ě N1 which
implies that
ˇ
ˇ
ˇxπ(n) ´ xˇ ă ε whenever n ě N .
Therefore, lim xπ(n) = x.
nÑ8
From the example above, we notice that proving the convergence using the ε-N argument
seems more complicated; however, it is a necessary evil so we encourage the readers to major
it.
Lemma 1.55 (Sandwich). If lim xn = L, lim yn = L, tzn u8
n=1 is a sequence such that
nÑ8
xn ď zn ď yn , then lim zn = L.
nÑ8
nÑ8
Proof. Let ε ą 0 be given. Since lim xn = L and lim yn = L, by definition
nÑ8
nÑ8
D N1 ą 0 Q L ´ ε ă xn ă L + ε whenever n ě N1
and
D N2 ą 0 Q L ´ ε ă yn ă L + ε whenever n ě N2 .
Let N = maxtN1 , N2 u. Then for n ě N , L ´ ε ă xn ď zn ď yn ă L + ε; thus lim zn = L. ˝
nÑ8
18
CHAPTER 1. The Real Line and Euclidean Space
Proposition 1.56. If a ď xn ď b and lim xn = x, then a ď x ď b.
nÑ8
Proof. Assume the contrary that x R [a, b]. If x ă a, let ε = a ´ x ą 0. Since lim xn = x,
nÑ8
D N ą 0 Q xn P (x ´ ε, x + ε) whenever n ě N . Therefore, xn ă a for all n ě N , a
contradiction. So a ď x.
We can prove x ď b in a similar way, and the proof is left as an exercise.
˝
Corollary 1.57. If a ă xn ă b and lim xn = x, then a ď x ď b.
nÑ8
Definition 1.58. Let txn u8
n=1 be a sequence in an order field F.
1. txn u8
n=1 is said to be bounded（有界的）if there exists M ą 0 such that |xn | ď M
for all n P N.
2. txn u8
n=1 is said to be bounded from above（有上界）if there exists B P F, called
an upper bound of the sequence, such that xn ď B for all n P N.
3. txn u8
n=1 is said to be bounded from below（有下界）if there exists A P F, called a
lower bound of the sequence, such that A ď xn for all n P N.
Proposition 1.59. A convergent sequence is bounded（數列收斂必有界）.
Proof. Let txn u8
n=1 be a convergent sequence with limit x. Then there exists N ą 0 such
that
xn P (x ´ 1, x + 1)
@ n ě N.
␣
(
Let M = max |x1 |, |x2 |, ¨ ¨ ¨ , |xN ´1 |, |x| + 1 . Then |xn | ď M for all n P N.
Theorem 1.60. Suppose that xn Ñ x and yn Ñ y as n Ñ 8, λ is a constant. Then
1. xn ˘ yn Ñ x ˘ y as n Ñ 8.
2. λ ¨ xn Ñ λ ¨ x as n Ñ 8.
3. xn ¨ yn Ñ x ¨ y as n Ñ 8.
4. If yn , y ‰ 0, then
x
xn
Ñ as n Ñ 8.
yn
y
Proof. The proof of 1 and 2 are left as an exercise.
˝
§1.2 Completeness and the Real Number System
19
3. Since xn Ñ x and yn Ñ y as n Ñ 8, by Proposition 1.59 D M ą 0 Q |xn | ď M and
|yn | ď M . Let ε ą 0 be given. Moreover,
D N1 ą 0 Q |xn ´ x| ă
ε
@ n ě N1
2M
D N2 ą 0 Q |yn ´ y| ă
ε
@ n ě N2 .
2M
and
Define N = maxtN1 , N2 u. Then for all n ě N ,
|xn ¨ yn ´ x ¨ y| = |xn ¨ yn ´ xn ¨ y + xn ¨ y ´ x ¨ y| ď |xn ¨ (yn ´ y)| + |y ¨ (xn ´ x)|
ε
ε
ď M ¨ |yn ´ y| + M ¨ |xn ´ x| ă M ¨
+M ¨
= ε.
2M
2M
1
if yn , y ‰ 0 (because of 3). Since lim yn = y,
nÑ8
y
|y|
|y|
for all n ě N1 . Therefore, |y| ´ |yn | ă
for all n ě N1
D N1 ą 0 Q |yn ´ y| ă
2
2
|y|
which further implies that |yn | ą
for all n ě N1 .
2
|y|2
Let ε ą 0 be given. Since lim yn = y, D N2 ą 0 Q |yn ´ y| ă
ε for all n ě N2 .
nÑ8
2
4. It suffices to show that lim
1
nÑ8 yn
=
Define N = maxtN1 , N2 u. Then for all n ě N ,
ˇ1
|y|2
1 2
1 ˇˇ |yn ´ y|
ˇ
ă
ε¨
= ε.
ˇ ´ ˇ=
yn y
|yn ||y|
2
|y| |y|
˝
1.2.2 Monotone sequence property and completeness
Definition 1.61. A sequence txn u8
n=1 is said to be increasing/non-decreasing, decreasing/non-increasing, strictly increasing and strictly decreasing if xn ď xn+1 ,
xn ě xn+1 , xn ă xn+1 and xn ą xn+1 @ n P N, respectively. A sequence is called (strictly)
monotone if it is either (strictly) increasing or (strictly) decreasing.
Definition 1.62. An ordered field F is said to satisfy the (strictly) monotone sequence
property if every bounded (strictly) monotone sequence converges to a limit in F.
Remark 1.63. An equivalent definition of the monotone sequence property is that every
monotone increasing sequence bounded above converges; that is, if each sequence txn u8
n=1 Ď
F satisfying
(i) xn ď xn+1 for all n P N,
20
CHAPTER 1. The Real Line and Euclidean Space
(ii) D M P F Q @ n P N : xn ď M ,
is convergent, then we say F satisfies the monotone sequence property.
Example 1.64. (Q, +, ¨, ď) is an ordered field.
Question: Is there any bounded monotone sequence in Q that does not converge to a limit
in Q?
Answer: Yes! Consider the sequence
1
2
x1 = ,
x2 =
1
1
2+
2
,
1
x3 =
2+
,
1
2+
¨¨¨ ,
xn+1 =
1
.
2 + xn
1
2
Then txn u8
n=1 is a monotone decreasing sequence in Q. If lim xn = x, then Theorem 1.60
nÑ8 ?
1
from which we conclude that x = ´1 + 2. Since x R Q, txn u8
implies that x =
n=1
2+x
does not converge (to a limit) in Q. In other words, Q does not have the monotone sequence
property.
Proposition 1.65. An ordered field satisfying the monotone sequence property has the
Archimedean property; that is, if F is an ordered field satisfying the monotone sequence
property, then @ x P F, D n P N Q x ă n.
Proof. Assume the contrary that there exists x P F such that n ď x for all n P N. Let
xn = n. Then txn u8
n=1 is increasing and bounded above. By the monotone sequence property
p P F such that xn Ñ x
p as n Ñ 8; thus D N ą 0 such that
of F, there exists x
p| ă
|xn ´ x
1
4
1
4
@n ě N .
1
4
p| ă , |N + 1 ´ x
p| ă ; thus
In particular, |N ´ x
p| + |p
1 = |N + 1 ´ N | ď |N + 1 ´ x
x ´ N| ă
1 1
1
+ = ,
4 4
2
a contradiction.
˝
Example 1.66. Let (F, +, ¨, ď) be an ordered field satisfying the monotone sequence propN
erty, and y P F be a given positive number (that is, y ą 0). Define xn = nn , where Nn is
2
( Nn )2
( Nn + 1 )2
2
the largest integer such that xn ď y; that is, n ď y but
ą y (for example, if
n
y = 2, then x1 =
2
5
11
2
,
x
=
,
x
=
,
¨
¨
¨
).
Then
2
3
21
22
23
2
§1.2 Completeness and the Real Number System
21
1. xn is bounded above: since x2n ď y ď 2y + y 2 + 1 = (y + 1)2 , by the non-negativity of
xn and y and Remark 1.19 we must have 0 ď xn ď y + 1.
2. xn is increasing: by the definition of Nn ,
Nn2 ď 22n ¨ y ñ 4 ¨ Nn2 ď 22n+2 ¨ y = 22(n+1) ¨ y ñ
N
2N
( 2Nn )2
2n+1
ď y ñ 2Nn ď Nn+1 .
N
n
n+1
Therefore, xn = nn = n+1
ď n+1
= xn+1 . Since F satisfies the monotone sequence
2
2
2
property, D x P F Q xn Ñ x as n Ñ 8. By Theorem 1.60, x2n Ñ x2 , and by Proposition
1.56, x2 ď y.
Now we show x2 = y. To this end observe that
(
1 )2 ( N
1 ) ( Nn + 1 )2
xn + n = nn + n =
ą y;
n
2
2
2
2
(
1 )2
1
thus x2n ď y ď xn + n . By the Archimedean propery of F (Proposition 1.65), lim n = 0;
nÑ8 2
2
(
1 )2
2
2
thus Theorem 1.60 implies that x = lim xn = lim xn + n = y. Note that Proposition
nÑ8
nÑ8
2
1.18 implies that such an x is unique if x ą 0.
In general, one can define the n-th root of non-negative number y in an ordered field
satisfying the monotone sequence property. The construction of the n-th root of y P F is
left as an exercise.
Definition 1.67. For n P N, the n-th root of a non-negative number y in an ordered field
satisfying the monotone sequence property is the unique non-negative number x satisfying
?
xn = y. One writes y 1/n or n y to denote n-th root of y.
Definition 1.68. An ordered field F is said to be complete (完備) (have the completeness
property, 具備完備性) if it satisfies the monotone sequence property.
Remark 1.69. In an ordered field, completeness ô monotone sequence property (在 ordered field 裡，完備性 = 數列單調有界必收斂 = 數列遞增有上界必收斂). Moreover,
1. A complete ordered field is “Archimedean” (Proposition 1.65).
2. For n P N, the n-th root of a non-negative number in a complete ordered field is
well-defined (Definition 1.67).
Proposition 1.70. Let (F, +, ¨, ď) be an ordered field. Then F satisfies the monotone
sequence property if and only if F satisfies the strictly monotone sequence property.
22
CHAPTER 1. The Real Line and Euclidean Space
Proof. The “only if” part is trivial, so we only prove the “if” part. Let txn u8
n=1 be a bounded
increasing sequence in F. If txn u8
n=1 has finite number of values; that is,
ˇ
␣
(
# n P N ˇ xn ă xn+1 ă 8 ,
then there exists N P N such that xn = xN for all n ě N which implies that txn u8
n=1
converges to xN . Now suppose that
ˇ
␣
(
# n P N ˇ xn ă xn+1 = 8 .
Then there exists an infinite set tn1 , n2 , ¨ ¨ ¨ u Ď N such that xnk ‰ xnk+1 for all k P N. Let
yk = xnk . Since F satisfies the strictly monotone sequence property, yk Ñ y as k Ñ 8 for
8
some x P F. However, it is easy to see that the sequence txn un=1
also converges to y since
txn u8
n=1 is monotone increasing.
˝
Theorem 1.71. There is a “unique” complete ordered field, called the real number system
R.
Remark 1.72. Uniqueness means if F is any other complete ordered field (F, ‘, d, ď),
then there exists an field isomorphism ϕ : R Ñ F; that is, ϕ : R Ñ F is one-to-one and
onto, and satisfies that
1. ϕ(x + y) = ϕ(x) ‘ ϕ(y) for all x, y P R.
2. ϕ(x ¨ y) = ϕ(x) d ϕ(y) for all x, y P R.
3. x ď y ñ ϕ(x) ď ϕ(y) for all x, y P R.
Sketch of proof of Theorem 1.71. Let S be the collection of all bounded increasing sequences
in Q in which all terms in every sequence have the same sign; that is,
ˇ
!
ˇ
S = txn u8
n=1 ˇ xn P Q for all n P N, xj ¨ xk ě 0 for all k, j P N,
)
is
increasing
and
bounded
above
.
and txn u8
n=1
8
8
Define on S an equivalence relation „: txn u8
n=1 „ tyn un=1 if every upper bound of txn un=1 is
ˇ
␣[
(
]
ˇ txn u8
also an upper bound of tyn u8
txn u8
n=1 , and vice versa. Let R =
n=1
n=1 P S be the set
of equivalence class of S (the existence of such a set relies on the axiom of choice). We define
]
]
[
[
8
8
8
on R, +, ¨, ď as follows: if r = txn u8
n=1 and s = tyn un=1 (where txn un=1 , tyn un=1 P S),
then
§1.2 Completeness and the Real Number System
[
]
1. r + s = txn + yn u8
n=1 ;
]
$ [
8
tx
¨
y
u
’
n
n
n=1
’
’
& ´((´r) ¨ s)
(
)
2. r ¨ s =
’
´ r ¨ (´s)
’
’
%
(´r) ¨ (´s)
23
if r, s ě 0 ,
if r ă 0 and s ą 0 ,
if r ą 0 and s ă 0 ,
if r, s ă 0 ;
8
3. r ď s if every upper bound of tyn u8
n=1 is also an upper bound for txn un=1 .
One needs to verify that R is an ordered field, and this part is left as an exercise (or see
Remark 1.73 for some part of the verification).
␣ (8
]
[␣ (8 ] [
Claim 1: If xnk k=1 is a subsequence of txn u8
xnk k=1 = txn u8
n=1 , then
n=1 .
[
] [
]
8
Claim 2: If txn u8
ă
ty
u
n
n=1
n=1 , then for some N P N, xn ă yN for all n ě N .
The proofs of the claims above are not difficult and are left as an exercise.
Now we show the completeness of R by showing that R satisfies the strictly monotone
sequence property (Proposition 1.70). Let trk u8
k=1 be a bounded, strictly increasing sequence
[
]
8
in R+ . Write rk = txk,n u8
n=1 , where xk,n ď xk,n+1 for all k, n P N. Since trk uk=1 is bounded
in R, there is M P Q such that xk,n ď M for all k, n P N. Moreover, since rk ă rk+1 for all
k P N, by claims above we can assume that xk,n ă xk+1,1 for all k, n P N; thus
xk,n ă xℓ,m
@ ℓ ą k and n, m P N.
(‹)
8
Therefore, txn,n u8
n=1 is bounded and monotone increasing, so txn,n un=1 P S. Define r =
[
]
txn,n u8
n=1 . Then r P R, and
(i) r is an upper bound of trk u8
k=1 : Suppose the contrary that there exists M P Q such
that xn,n ď M for all n P N but xk,ℓ ą M for some k, ℓ P N.
(a) If k ě ℓ, then xk,k ě xk,ℓ ą M since txk,ℓ u8
ℓ=1 is increasing.
(b) If k ă ℓ, then xℓ,ℓ ą xk,ℓ ą M because of (‹).
In either case we conclude that M cannot be an upper bound of r, a contradiction.
(ii) r ´ ε is not an upper bound of trk u8
k=1 for all ε ą 0: Suppose the contrary that
8
r ´ ε is an upper bound of trk u8
k=1 . Write ε = tεk uk=1 , and W.L.O.G. we can assume
that there exists δ P Q such that εk ě 2δ ą 0 for all k P N. Then for all (fixed) k P N,
[
] [
] [
] [
]
8
8
8
txk,ℓ + δu8
ℓ=1 ă txk,ℓ + 2δuℓ=1 ď txk,ℓ + εk uℓ=1 ď txℓ,ℓ uℓ=1 .
24
CHAPTER 1. The Real Line and Euclidean Space
Let N1 = 1. By claim 2, for each k P N there exists Nk+1 P N such that Nk+1 ě Nk
and xNk ,ℓ + δ ă xNk+1 ,Nk+1 for all ℓ ě Nk+1 . On the other hand,
xNk+1 ,Nk+1 ě xNk ,Nk+1 + δ ě xNk ,Nk + δ ě ¨ ¨ ¨ ě x1,1 + kδ
which implies that txℓ,ℓ u8
ℓ=1 is not bounded, a contradiction.
As a consequence, r is the least upper bound of trk u8
k=1 .
˝
From now on R is the complete ordered field containing Q, Z, N.
Remark 1.73 (The existence of additive inverse of real numbers). Suppose that a bounded
8
increasing sequence txn u8
n=1 is not equivalent to any rational “number” tqun=1 for any q P Q,
then there exists a decreasing sequence tyn u8
n=1 such that xn ´ yn Ñ 0 as n Ñ 8. Such
tyn u8
n=1 can be obtained by choosing yn to be the smallest upper bound of the form
k
,
2n
where k P Z. By deleting terms if necessary, we can assume that all yn1 s have the same sign.
[
]
8
Then t´yn u8
n=1 is a bounded increasing sequence, and t´yn un=1 is the additive inverse of
[
]
txn u8
n=1 .
a
?
?
Example 1.74. In R, define xn inductively by x1 = 0, x2 = 2, x3 = 2 + 2, ¨ ¨ ¨ , xn+1 =
?
2 + xn . It is easy to see that txn u8
n=1 satisfies xn ě 0 for all n P N.
1. xn ď 2 for all n P N (boundedness): First of all, x1 ď 2. Assume that xn ď 2. Then
?
?
xn+1 = 2 + xn ď 2 + 2 = 2. By mathematical induction, xn ď 2 for all n P N.
2. xn ď xn+1 (monotonicity): Since xn ´ 2 ď 0 and xn + 1 ě 0, (xn ´ 2) ¨ (xn + 1) ď 0.
Expanding the product, we obtain that x2n ď xn + 2 = x2n+1 which implies that
xn ď xn+1 .
3. xn Ñ 2 as n Ñ 8 (convergence): Since txn u8
n=1 is a bounded monotone sequence in R,
lim xn = x for some x P R. Note that then xn+1 Ñ x as n Ñ 8. Since x2n+1 = xn + 2,
by Theorem 1.60 we must have x2 = x + 2. Then (x ´ 2)(x + 1) = 0 which implies
nÑ8
x = 2 or x = ´1 (failed). Therefore, txn u8
n=1 converges to 2.
Theorem 1.75. The interval (0, 1) in R is uncountable (不可數).
§1.2 Completeness and the Real Number System
25
Proof. Assume the contrary that there exists f : N Ñ (0, 1) which is one-to-one and onto.
Write f (k) in decimal expansion (十進位展開); that is,
f (1) = 0.d11 d21 d31 ¨ ¨ ¨
f (2) = 0.d12 d22 d32 ¨ ¨ ¨
..
..
.
.
f (k) = 0.d1k d2k d3k ¨ ¨ ¨
..
...
.
Here we note that repeated 9’s are chosen by preference over terminating decimals; that is,
for example, we write
1
1
= 0.249999 ¨ ¨ ¨ instead of = 0.250000 ¨ ¨ ¨ .
4
4
Let x P (0, 1) be such that x = 0.d1 d2 ¨ ¨ ¨ , where
"
dk =
5 if dkk ‰ 5 ,
7 if dkk = 5 .
（建構一個 x 使其小數點下第 k 位數與 f (k) 的小數點下第 k 位數不相等）. Then x ‰ f (k)
for all k P N, a contradiction; thus (0, 1) is uncountable.
˝
Corollary 1.76. R is uncountable.
Proposition 1.77. Q is dense (稠密) in R; that is, if x, y P R and x ă y, then D r P Q Q
x ă r ă y.
1
Ñ 0 as n Ñ 8 (by the Archimedean property of R, Proposition 1.65), there
n
ˇ1
ˇ
exists N ą 0 such that ˇ ´ 0ˇ ă y ´ x for all n ě N .
n
! ˇ
)
k ˇ
Claim:
ˇ k P Z X (x, y) ‰ H.
N
)
! ˇ
ℓ
k ˇ
Proof of claim: Suppose the contrary that
ď x and
ˇ k P Z X (x, y) = H. Then
N
N
ℓ+1
1
ě y for some ℓ P Z, while this fact will imply that y ´ x ď , a contradiction.
˝
N
N
Proof. Since
Remark 1.78. The denseness of Q in R can be rephrased as follows: if x P R and ε ą 0,
then D r P Q Q |x ´ r| ă ε.
( •
x´ε r
•
x
)
x+ε
26
CHAPTER 1. The Real Line and Euclidean Space
Corollary 1.79. The collection of irrational numbers QA ” RzQ is dense in R; that is, if
x, y P R and x ă y, D c P QA Q x ă c ă y.
Proof. Let x, y P R with x ă y. By Proposition 1.77 there exists r P Q, r ‰ 0 such that
?
x
y
? ă r ă ? . Let c = 2r. Then c P QA and x ă c ă y.
˝
2
2
Example 1.80. The harmonic sequence
x1 = 1
x2 = 1 +
..
.
xn = 1 +
..
.
1
2
..
.
n
ÿ
1
1
1 1
+ + ¨¨¨ + =
2 3
n k=1 k
..
.
is (monotone) increasing but not bounded above.
Proof. That the sequence is increasing is trivial. For the unboundedness, we observe that
1 1 1 1 1 1 1
1
+ + + + + + + ¨¨¨ + n
2 3 4 5 6 7 8
2
2n´1
1 1 1 1 1 1 1
ě 1 + + + + + + + + ¨¨¨ + n
2 4 4 8 8 8 8
2
1 1
1
n
= 1 + + + ¨¨¨ + = 1 +
2 2
2
2
x2n = 1 +
which is not bounded above (沒有上界).
˝
1.3 Least Upper Bounds and Greatest Lower Bounds
Definition 1.81. Let H ‰ S Ď R. A number M P R is called an upper bound (上界)
for S if x ď M for all x P S, and a number m P R is called a lower bound (下界) for S if
x ě m for all x P S. If there is an upper bound for S, then S is said to be bounded from
above, while if there is a lower bound for S, then S is said to be bounded from below. A
number b P R is called a least upper bound (最小上界) of S if
1. b is an upper bound for S, and
§1.3 Least Upper Bounds and Greatest Lower Bounds
27
2. if M is an upper bound for S, then M ě b.
A number a is called a greatest lower bound (最大下界) of S if
1. a is a lower bound for S, and
2. if m is a lower bound for S, then m ď a.
• (
m
a lower bound for S
)
S
•
M
an upper bound for S
If S is not bounded above, the least upper bound of S is set to be 8, while if S is not
bounded below, the greatest lower bound of S is set to be ´8. The least upper bound of
S is also called the supremum of S and is usually denoted by lubS or sup S, and “the”
greatest lower bound of S is also called the infimum of S, and is usually denoted by glbS
or inf S. If S = H, then sup S = ´8, inf S = 8.
Example 1.82. Let S = (0, 1). Then sup S = 1, inf S = 0.
Example 1.83. Let f : R Ñ R given by
"
1 ´ x2 if x ‰ 0,
f (x) =
0
if x = 0.
Define
S = tf (x) | x P Ru,
␣
1(
T = x P R | f (x) ą .
4
We can get S = (´8, 1), so sup(S) = 1, inf(S) = ´8.
?
?
?
?
(
1
3
3 ) (
3)
3
2
Solve 1´x = ñ x = ˘ , then we can get T = ´ , 0 Y 0,
, so sup(T ) =
,
?
3
inf(T ) = ´ .
2
4
2
2
2
2
Remark 1.84. The least upper bound and the greatest lower bound of S need not be a
member of S.
Remark 1.85. The reason for defining sup H = ´8 and inf H = 8 is as follows: if
H ‰ A Ď B, then sup A ď sup B and inf A ě inf B.
(
(
inf B inf A
A
)
sup A
B
)
sup B
28
CHAPTER 1. The Real Line and Euclidean Space
Since H is a subset of any other sets, we shall have sup H is smaller then any real number,
and inf H is greater than any real number. However, this “definition” would destroy the
property that inf A ď sup A.
The “definition” of sup H and inf H is purely artificial. One can also define sup H = 8
and inf H = ´8.
Definition 1.86. An open interval in R is of the form (a, b) which consists of all x P R Q
a ă x ă b. A closed interval in R is of the form [a, b] which consists of all x P R Q a ď
x ď b.
Proposition 1.87. Let S Ď R be non-empty. Then
1. b = sup S P R if and only if
(a) b is an upper bound of S.
(b) @ ε ą 0, D x P S Q x ą b ´ ε.
2. a = inf S P R if and only if
(a) a is a lower bound of S.
(b) @ ε ą 0, D x P S Q x ă a + ε.
Proof. “ñ” (a) is part of the definition of being a least upper bound.
(b) If M is an upper bound of S, then we must have M ě b; thus b ´ ε is not an upper
bound of S. Therefore, D x P S Q x ą b ´ ε.
“ð” First, we show that b is an upper bound for S. If not, there exists x P S such that
b ă x. Let ε = x ´ s ą 0. Then we do not have (i) since x P S but x ­ă s + ε.
Next, we show that if M is an upper bound of S, then M ě b. Assume the contrary.
Then D M such that M is an upper bound of S but M ă b. Let ε = b ´ M , then there
is no x P S Q x ą b ´ ε. ÑÐ
˝
So far it is not clear that whether the least upper bound or the greatest lower bound
for a subset S Ď R exists or not. The following theorem provides the existence of the least
upper bound or the greatest lower bound of a set S provided that S has certain properties.
Theorem 1.88. In R, the following two properties hold:
§1.3 Least Upper Bounds and Greatest Lower Bounds
29
1. Least upper bound property (L.U.B.P.):
Let S be a non-empty set in R that has an upper bound (or is bounded from above),
then S has a least upper bound. （非空集合有上界，則有最小上界）
2. Greatest lower bound property:
Let S be a non-empty set in R that has a lower bound (or is bounded from below), then
S has a greatest lower bound. (非空集合有下界，則有最大下界)
Proof. We only prove the least upper bound property since the proof of the greatest lower
bound property is similar.
Let H ‰ S Ď R be given. Let x0 be the smallest integer such that x0 is an upper bound
N1
, where N1 is the largest integer such that x2 is still an upper bound
10
N
of S. We continue this process, and define xn = xn´1 ´ nn , where Nn is the largest integer
10
of S. Let x1 = x0 ´
such that xn is an upper bound of S.（事實上，xn 就是十進位下小數點以下只有 n 位的
小數裡面，S 的上界中最小的那個數）
(
S
)
x3 x2
x1
Note that in the process of constructing txn u8
n=1 , Nn is always non-negative which im8
plies that txn u8
n=1 is decreasing. Moreover, any a P S is a lower bound of txn un=1 . By
completeness of R, txn u8
n=1 converges. Assume that xn Ñ x as n Ñ 8.
Claim: x = sup S (ô 1. x is an upper bound of S. 2. @ ε ą 0, D s P S Q s ą x ´ ε).
1. Assume the contrary that x is not an upper bound of S. Then D s P S Q s ą x. Since
xn Ñ x as n Ñ 8, D N ą 0 Q |xn ´ x| ă s ´ x for all n ě N ; thus
2x ´ s ă xn ă s
něN.
Therefore, xn cannot be an upper bound of S for all n ě N , a contradiction.
2. Assume the contrary that D ε ą 0 Q @ s P S, s ă x ´ ε. Choose k P N such that
εą
1
. Then
10k
xk´1 ´
Nk + 1
1
= xk ´ k ě x ´ ε ą s
k
10
10
which suggests that Nk is not the largest integer such that xk´1 ´
Nk
is still an upper
10k
30
CHAPTER 1. The Real Line and Euclidean Space
bound, a contradiction.
˝
Proposition 1.89. Suppose that H ‰ A Ď B Ď R. Then inf B ď inf A ď sup A ď sup B.
Proof. We proceed as follows.
1. sup A ď sup B: Let b = sup B, then @x P B, x ď b. Since A Ď B, then @x P A, x ď b;
hence b is also an upper bound for A. Since sup A is the least upper bound for A and
b is an upper bound for A, then sup A ď b = sup B.
2. It is similar to prove inf B ď inf A.
3. It is trivially true that inf A ď sup A.
˝
Theorem 1.90. Let (F, +, ¨, ď) be an ordered field such that F has the least upper bound
property, then F is complete.
Proof. We would like to show that any increasing bounded sequence converges. Let txn u8
n=1
be increasing and bounded above (by M ).
x = sup S
x−ε
x1
x2
x3 x4
(
M
s = xN
Define S = tx1 , x2 , ¨ ¨ ¨ , xn , ¨ ¨ ¨ u. Then S is non-empty and has an upper bound; thus by
the assumption that F satisfies the least upper bound property, sup S ” x exists.
1. x is an upper bound of S ñ xn ď x for all n P N.
2. By Proposition 1.87, @ ε ą 0, D s P S Q s ą x ´ ε . Note that s = xN for some N P N.
Since txn u8
n=1 is increasing, xN ď xn ď x for all n ě N . Therefore, if n ě N ,
x ´ ε ă xN ď xn ď x ă x + ε
which implies that |xn ´ x| ă ε if n ě N .
˝
Example 1.91. Q is not complete. Let S = tx1 = 3, x2 = 3.1, x3 = 3.14, ¨ ¨ ¨ u. Then S has
4 as an upper bound, but S has no least upper bound (in Q).
Remark 1.92. The two theorems above suggest that in an ordered field, completeness ô
the least upper bound property.
§1.4 Cauchy Sequences
31
1.4 Cauchy Sequences
So far the only criteria that we learn (from previous sections) for the convergence of a
sequence in an ordered field is that a bounded monotone sequence in R converges. Are there
any other criteria for the convergence of a sequence in an ordered field? By Proposition 1.48,
we know that if a sequence txn u8
n=1 in an ordered field F converges, then
ˇ
␣
(
D! x P F Q @ ε ą 0, # n P N ˇ xn R (x ´ ε, x + ε) ă 8.
We would like to investigate if the following much weaker statement
ˇ
␣
(
@ ε ą 0, D (a limit candidate) y P F Q # n P N ˇ xn R (y ´ ε, y + ε) ă 8
(‹)
leads to the convergence of a sequence. Note that statement (‹) is equivalent to statement
(‹‹) in the following
Definition 1.93. A sequence txn u8
n=1 in an ordered field is said to be Cauchy if
@ ε ą 0, D N ą 0 Q |xn ´ xm | ă ε whenever n, m ě N .
(‹‹)
Remark 1.94. (‹) 這個敘述的中心思想是：給定一正值 ε, 我們都能找到一個長度是 2ε
的區間使得落在此區間外的 xn 只有有限個。因為當對每個長度我們都能找到這樣的區間
時，才有機會找到 txn u8
n=1 的極限（極限若真的存在的話，那麼這個極限一定落在所有這
樣的區間之內）；要是連這樣的區間都找不到，就不可能會收斂了。
Example 1.95. In Q, x1 = 3, x2 = 3.1, x3 = 3.14, x4 = 3.141, ¨ ¨ ¨ . Then txn u8
n=1 is a
Cauchy sequence, but is not convergent. Therefore, a Cauchy sequence may not converge.
Proposition 1.96. Every convergent sequence is Cauchy.
Proof. Let txn u8
n=1 be a convergent sequence with limit x. For any ε ą 0, D N ą 0 Q
ε
|xn ´ x| ă if n ě N . Then by triangle inequality, if n, m ě N ,
2
|xn ´ xm | ď |xn ´ x| + |x ´ xm | ă
thus txn u8
n=1 is Cauchy.
Lemma 1.97. Every Cauchy sequence is bounded.
ε ε
+ = ε;
2 2
˝
32
CHAPTER 1. The Real Line and Euclidean Space
Proof. Let txn u8
n=1 be Cauchy. D N ą 0 Q |xn ´ xm | ă 1 for all n, m ě N . In particular,
|xn ´ xN | ă 1 if n ě N or equivalently,
xN ´ 1 ă xn ă xN + 1
@n ě N .
␣
(
Let M = max |x1 |, |x2 |, ¨ ¨ ¨ , |xN ´1 |, |xN | + 1 . Then |xn | ď M for all n P N.
˝
Definition 1.98. A sequence tyj u8
j=1 is called a subsequence (子 數 列) of a sequence
txn u8
n=1 if there exists an increasing function f : N Ñ N such that yj = xf (j) . In this case,
we often write f (j) = nj and yj = xnj .
In other words, a subsequence is a sequence that can be derived from another sequence
by deleting some elements without changing the order of remaining elements. Let f : N Ñ R
8
be a sequence an xn = f (n). A subsequence txnj u8
j=1 of txn un=1 is the image of an infinite
subset tn1 , n2 , ¨ ¨ ¨ u of N under the map f .
x1
x2 x3
xn 1 xn2
Example 1.99. Let txn u8
n=1 =
␣
x5 x8 x6 x4
xn 5 xn 3
x7
xn4
␣ 1 1 2 11
(
(
1 1 1 2 11
, ¨ ¨ ¨ , and tyn u8
, , , ,¨¨¨ .
n=1 =
2 7 3 3 8
2 7 3 8
1, , , , ,
8
Then tyn u8
n=1 can be viewed as a subsequence of txn un=1 by the relation yj = xnj ; that
is, y1 = x2 , y2 = x3 , y3 = x5 , y4 = x6 , and etc. The sequence txnj u8
j=1 is obtained by
deleting x1 and x4 (and maybe more) from the original sequence txn u8
n=1 . However, if
␣
(
1
11
8
tzn u8
, , 1, ¨ ¨ ¨ , then tzn u8
n=1 =
n=1 is not a subsequence of txn un=1 (but only a subset)
7 8
of txn u8
n=1 because the order is changed.
Theorem 1.100 (Bolzano-Weierstrass property). Every bounded sequence in R has a convergent subsequence; that is, every bounded sequence in R has a subsequence that converges
to a limit in R.
Proof. Let txn u8
n=1 be a bounded sequence satisfying |xn | ď M for all n P N. Divide
[´M, M ] into two intervals [´M, 0], [0, M ], and denote one of the two intervals containing
ˇ
␣
(
infinitely many xn as [a1 , b1 ]; that is, # n P N ˇ xn P [a1 , b1 ] = 8. Divide [a1 , b1 ] into two
[ a + b1 ] [ a1 + b1 ]
,
intervals a1 , 1
, b1 , and denote one of the two intervals containing infinitely
2
2
many xn as [a2 , b2 ]. We continue this process, and obtain a sequence of intervals [ak , bk ] such
that #tn P N | xn P [ak , bk ]u = 8.
§1.4 Cauchy Sequences
33
Let xn1 be an element belonging to [a1 , b1 ]. Since #tn P N | xn P [a1 , b1 ]u = 8, we can
choose n2 ą n1 such that xn2 P [a2 , b2 ], and for the same reason we can choose n3 ą n2 such
that xn3 P [a3 , b3 ]. We continue this process and obtain xnk P [ak , bk ] with nk ą nk´1 .
xn 2
a3
[
[
[
−M
O
a1
a2
b3
]
]
]
M
b1
b2
xn1
xn3
8
Since [ak , bk ] Ě [ak+1 , bk+1 ] for all k P N, we find that tak u8
k=1 is increasing and tbk uk=1
is decreasing. Moreover, ak ď M , bk ě ´M . As a consequence, by the monotone sequence
property, ak converges to a and bk converges to b.
On the other hand, we observe that bk ´ ak =
M
M
. Then b ´ a = lim k´1 = 0; thus
2k´1
kÑ8 2
a = b. Since ak ď xnk ď bk , by Sandwich lemma lim xnk = a = b P R.
kÑ8
˝
Lemma 1.101. If a subsequence of a Cauchy sequence is convergent, then this Cauchy
sequence also converges.
8
Proof. Let txn u8
n=1 be a Cauchy sequence with a convergent subsequence txnj uj=1 . Assume
lim xnj = x. Then @ ε ą 0,
jÑ8
ε
2
ε
D N ą 0 Q |xn ´ xm | ă
2
D K ą 0 Q |xnj ´ x| ă
if j ě K, and
if n, m ě N.
Choose j ě maxtK, N u. Then nj ě N ; thus if n ě N ,
|xn ´ x| ď |xn ´ xnj | + |xnj ´ x| ă
ε ε
+ = ε.
2 2
˝
Theorem 1.102. Every Cauchy sequence in R is convergent.
Theorem 1.103. Suppose that F is an ordered field with Archimedean property and every
Cauchy sequence converges. Then F is complete.
Proof. Suppose the contrary that there is a bounded increasing sequence txn u8
n=1 that does
not converge to a limit in F. By assumption, txn u8
n=1 cannot be Cauchy; thus
D ε ą 0 Q @ N ą 0 D n, m ě N Q |xn ´ xm | ě ε .
34
CHAPTER 1. The Real Line and Euclidean Space
Let N = 1, D n2 ą n1 ě 1 Q |xn1 ´ xn2 | ě ε. Let N = n2 + 1, D n4 ą n3 ě n2 + 1 Q
|xn3 ´ xn4 | ě ε. We continue this process and obtain a sequence txnj u8
j=1 satisfying
ˇ
ˇ
ˇx n
ˇ ě ε @ k P N.
´
x
n
2k´1
2k
≥ε
xn 1
≥ε
≥ε ≥ε
xn 8
xn 2 xn3 xn 4 xn5
xn6 xn7
␣ (8
Claim: xnj j=1 is unbounded (thus a contradiction to the boundedness of txn u8
n=1 ).
Proof of claim: Assume the contrary that there exists M P F such that xnj ď M for all
j P N. Since xn2k ě x1 + kε for all k P N, we must have
kď
M ´ x1
ε
@k P N
which violates the Archimedean property, a contradiction.
˝
Remark 1.104. In an ordered field with Archimedean property, Completeness ô Cauchy
completeness (Every Cauchy sequence converges).
Example 1.105. xn P R, |xn ´ xn+1 | ă
1
2n+1
@ n P N.
Claim: txn u8
n=1 is Cauchy. Given ε ą 0, choose N ą 0 Q
1
ă ε. Then if N ď n ă m,
2N
|xn ´ xm | ď |xn ´ xn+1 | + |xn+1 ´ xm |
ď |xn ´ xn+1 | + |xn+1 ´ xn+2 | + |xn+2 ´ xm |
ď ¨¨¨
ď |xn ´ xn+1 | + |xn+1 ´ xn+2 | + ¨ ¨ ¨ + |xm´1 ´ xm |
1
1
1
ď n+1 + n+2 + ¨ ¨ ¨ + m
2
2
2
1
1
ď n ď N ă ε;
2
2
thus txn u8
n=1 is Cauchy in R. This implies that the sequence is convergent.
1.5 Cluster Points and Limit Inferior, Limit Superior
Definition 1.106. A point x is called a cluster point of a sequence txn u8
n=1 if
ˇ
␣
(
@ ε ą 0, # n P N ˇ xn P (x ´ ε, x + ε) = 8 .
§1.5 Cluster Points and Limit Inferior, Limit Superior
35
Example 1.107. Let xn = (´1)n . Then 1 and ´1 are the only two cluster points of txn u8
n=1 .
1
n
Example 1.108. Let xn = (´1)n + .
Claim: 1 and ´1 are cluster points of txn u8
n=1 .
Let ε ą 0 be given. We observe that
ˇ
ˇ
( ␣
(
1
n P N ˇ xn P (1 ´ ε, 1 + ε) Ě n P N ˇ n is even, ă ε ;
n
ˇ
␣
(
thus # n P N ˇ xn P (1 ´ ε, 1 + ε) = 8. Similarly, ´1 is a cluster point.
Claim: @ a ‰ ˘1, a is not a cluster point of txn u8
n=1 (reasoning in the following proposition).
␣
Proposition 1.109. Let txn u8
n=1 Ď R and x P R.
1. x is a cluster point of txn u8
n=1 if and only if @ ε ą 0, N ą 0, D n ě N Q |xn ´ x| ă ε.
8
2. x is a cluster point of txn u8
n=1 if and only if there exists a subsequence txnj uj=1 of
txn u8
n=1 converges to x.
3. xn Ñ x as n Ñ 8 if and only if every proper subsequence of txn u8
n=1 converges to x.
4. xn Ñ x as n Ñ 8 if and only if txn u8
n=1 is bounded and x is the only cluster point of
txn u8
n=1 .
5. xn Ñ x as n Ñ 8 if and only if every proper subsequence of txn u8
n=1 has a further
subsequence that converges to x.
Proof. We only prove 1-4, and the proof of 5 is left as an exercise.
1. (ñ) Let ε ą 0 be given. Since there are infinitely many n1 s with |xn ´ x| ă ε, for any
fixed N P N, there are only finite number of the indices that are smaller than N . So
there must be some n ě N with |xn ´ x| ă ε.
(ð) Let ε ą 0 be given. Pick n1 ě 1 Q |xn1 ´ x| ă ε, then pick n2 ě n1 + 1
Q |xn2 ´ x| ă ε. We continue this process and obtain a subsequence txnj u8
j=1 satisfying
ˇ
␣
(
|xn ´ x| ă ε for all j P N. Then n P N ˇ xn P (x ´ ε, x + ε) Ě tn1 , n2 , ¨ ¨ ¨ u.
j
1
2
2. (ñ) By 1, we can pick n1 ě 1 Q |xn1 ´ x| ă 1 and pick n2 ě n1 + 1 Q |xn2 ´ x| ă .
In general, we can pick nk ě nk´1 + 1 Q |xnk ´ x| ă
x´
1
1
ă xnk ă x +
k
k
1
for all k ě 2. Then
k
@ k P N.
36
CHAPTER 1. The Real Line and Euclidean Space
By Sandwich lemma, lim xnk = x.
kÑ8
ˇ
␣
(
(ð) @ ε ą 0, D J ą 0 Q |xnj ´ x| ă ε if j ě J. Then n P N ˇ xn P (x ´ ε, x + ε) Ě
tnJ , nJ+1 , ¨ ¨ ¨ u.
8
3. (ñ) Let txnj u8
j=1 be a subsequence of a convergent sequence txn un=1 and lim xn = x.
nÑ8
Then @ ε ą 0, D N ą 0 Q |xn ´ x| ă ε for all n ě N . Since nj Ñ 8 as j Ñ 8, D J ą 0
Q nj ě N ; thus |xnj ´ x| ă ε whenever j ě J.
(ð) Assume the contrary that xn Ñ̂ x as n Ñ 8. Then
D ε ą 0 Q @ N ą 0, D n ě N Q |xn ´ x| ě ε.
Let n1 ě 1 such that |xn1 ´ x| ě ε, and n2 ě n1 + 1 such that |xn2 ´ x| ě ε. In general,
we can chose nk ě nk´1 such that |xnk ´x| ě ε for all k ě 2. The subsequence txnj u8
j=1
clearly does not converge to x, a contradiction.
4. (ñ) This direction is a direct consequence of Proposition 1.48 and 1.59.
(ð) Suppose that txn un=1 is a bounded sequence in R and has x as the only cluster
point but txn u8
n=1 does not converge to x. Then
ˇ
␣
(
D ε ą 0 Q # n P N ˇ xn R (x ´ ε, x + ε) = 8 .
ˇ
␣
(
Write n P N ˇ xn R (x ´ ε, x + ε) = tn1 , n2 , ¨ ¨ ¨ , nk , ¨ ¨ ¨ u. Then we find a subse␣ (8
␣ (8
quence xnk k=1 lying outside (x ´ ε, x + ε). Since xnk k=1 is bounded, the BolzanoWeierstrass property (Theorem 1.100) suggests that there exists a convergent subse␣
(8
quence xnkj j=1 with limit y. Since xnkj R (x ´ ε, x + ε), y R [x ´ ε, x + ε]; thus y ‰ x.
On the other hand, 2 suggests that y is a cluster point of txn u8
n=1 , a contradiction to
8
the assumption that x is the only cluster point of txn un=1 .
˝
Definition 1.110. A sequence txn u8
n=1 is said to diverge to infinity if @ M ą 0, D N ą 0
Q xn ą M whenever n ě N . It is said to diverge to negative infinity if t´xn u8
n=1
diverge to infinity. We use lim xn = 8 or ´8 to denote that txn u8
n=1 diverges to infinity
nÑ8
8
.
or negative infinity, and call 8 or ´8 the limit of txn un=1
Definition 1.111. The extended real number system, denoted by R˚ , is the number
system R Y t8, ´8u, where 8 and ´8 are two symbols satisfying ´8 ă x ă 8 for all
x P R.
§1.5 Cluster Points and Limit Inferior, Limit Superior
37
Remark 1.112. 1. R˚ is not a field since 8 and ´8 do not have multiplicative inverse.
2. The definition of the least upper bound of a set can be simplified as follows: Let
S Ď R˚ be a set (not necessary non-empty set). A number b P R˚ is said to be the
least upper bound of S if
(a) b is an upper bound of S (that is, s ď b for all s P S);
(b) If M P R˚ is an upper bound of S, then b ď M .
No further discussion (such as S = H or S is not bounded above) has to be made.
The greatest lower bound can be defined in a similar fashion.
3. Any sets in R˚ has a least upper bound and a greatest lower bound in R˚ , even the
empty set and unbounded set.
4. Proposition 1.87 can be rephrased as follows: Let S Ď R˚ . Then b = sup S P R if and
only if
(a) b is an upper bound of S;
(b) @ ε ą 0, D s P S Q s ą b ´ ε.
Note that b P R is crucial since there is no s P R˚ such that s ą 8 ´ ε = 8. The
greatest lower bound counterpart can be made in a similar fashion.
5. In light of Proposition 1.109 and Definition 1.110, we can redefine cluster points of a
real sequence as follows: A number x P R˚ is said to be a cluster point of a sequence
␣ (8
txn u8
n=1 Ď R if there exists a subsequence xnj j=1 such that lim xnj = x. Note that
jÑ8
now we can talk about if 8 or ´8 is a cluster points of a real sequence.
In the rest of the section, one is allowed to find the least upper bound and the greatest
lower bound of a subset in R˚ .
Definition 1.113. Let txn u8
n=1 be a sequence in R.
1. The limit superior of txn u8
n=1 , denoted by lim sup xn or lim xn , is the infimum of
nÑ8
nÑ8
!
␣ ˇ
()8
the sequence sup xn ˇ n ě k
.
k=1
2.
The limit inferior of txn u8
n=1 , denoted by lim inf xn or lim xn , is the supremum of
nÑ8
nÑ8
! ␣ ˇ
()8
the sequence
inf xn ˇ n ě k
.
k=1
38
CHAPTER 1. The Real Line and Euclidean Space
␣ ˇ
(
Remark 1.114. Let sup xn denote the number sup xn ˇ n ě k and inf xn denote the
něk
␣ ˇ
(něk
ˇ
number inf xn n ě k . Then the limit superior and the limit inferior can be written as
lim sup xn = inf sup xn
kě1 něk
nÑ8
and
lim inf xn = sup inf xn .
nÑ8
kě1 něk
Remark 1.115. Let txn u8
n=1 be a sequence in R, and yk = sup xn and zk = inf xn . Then
něk
něk
8
8
tyk uk=1 is a decreasing sequence, and tzk uk=1 is an increasing sequence. Therefore, the limit
8
of tyk u8
k=1 and the limit of tzk uk=1 both “exist” in the sense of Definition 1.46 and 1.110. In
8
8
fact, the limit of tyk u8
k=1 is the infimum of tyk uk=1 , and the limit of tzk uk=1 is the supremum
of tzk u8
k=1 . In other words,
lim sup xn = inf sup xn
and
lim sup xn = lim sup xn
and
kÑ8 něk
kě1 něk
lim inf xn = sup inf xn ;
kÑ8 něk
kě1 něk
thus
kÑ8 něk
nÑ8
lim inf xn = lim inf xn .
nÑ8
kÑ8 něk
8
= t1, 0, ´1, 1, 0, ´1, 1, 0, ´1, ¨ ¨ ¨ u. Then
Example 1.116. Let txn un=1
yk = sup xn = 1 ñ lim sup xn = 1.
něk
nÑ8
zk = inf xn = ´1 ñ lim inf xn = ´1.
něk
Example 1.117. Let xn =
nÑ8
1
. Then
n
1
ñ lim sup xn = 0.
k
nÑ8
něk
zk = inf xn = 0 ñ lim inf xn = 0.
yk = sup xn =
něk
"
Example 1.118. Let xn =
nÑ8
0 if n is even
; that is, txn u8
n=1 = t1, 0, 3, 0, 5, ¨ ¨ ¨ u. Then
n if n is odd
yk = sup xn = 8 ñ lim sup xn = 8.
něk
nÑ8
zk = inf xn = 0 ñ lim inf xn = 0.
něk
nÑ8
§1.5 Cluster Points and Limit Inferior, Limit Superior
Example 1.119. Let xn =
39
$
1
’
if n = 4k + 1,
1+
’
’
n
’
’
’
’
1
’
& ´1 ´
if n = 4k + 2,
n
1
’
’
1´
if n = 4k + 3,
’
’
n
’
’
’
’
% ´1 + 1 if n = 4k.
n
1
1
yk = sup xn = 1 + , zk = inf xn = ´1 ´ . lim sup xn = 1. lim inf xn = ´1.
nÑ8
něk
⃝
⃝ nÑ8
něk
Proposition 1.120. Let txn u8
n=1 be a sequence in R. Then
lim sup ´xn = ´ lim inf xn
nÑ8
nÑ8
and
lim inf ´xn = ´ lim sup xn .
nÑ8
nÑ8
Proof. By the fact that sup ´xn = ´ inf xn ,
něk
něk
lim sup ´xn = lim sup(´xn ) = lim
nÑ8
kÑ8 něk
kÑ8
(
)
´ inf xn = ´ lim inf xn = ´ lim inf xn .
něk
kÑ8 něk
nÑ8
The second identity holds simply by replacing xn by ´xn in the first identity.
Proposition 1.121. Let txn u8
n=1 be a sequence in R. Then
1. a = lim inf xn P R if and only if
nÑ8
(a) @ ε ą 0, D N ą 0 such that a ´ ε ă xn whenever n ě N ; that is,
ˇ
␣
(
@ ε ą 0, # n P N ˇ xn ď a ´ ε ă 8 ,
and
(b) @ ε ą 0 and N ą 0, D n ě N such that xn ă a + ε; that is,
ˇ
␣
(
@ ε ą 0, # n P N ˇ xn ă a + ε = 8 .
2. b = lim sup xn P R if and only if
nÑ8
(a) @ ε ą 0, D N ą 0 such that b + ε ą xn whenever n ě N ; that is,
ˇ
␣
(
@ ε ą 0, # n P N ˇ xn ě b + ε ă 8 ,
and
˝
40
CHAPTER 1. The Real Line and Euclidean Space
(b) @ ε ą 0 and N ą 0, D n ě N such that xn ą b ´ ε; that is,
ˇ
␣
(
@ ε ą 0, # n P N ˇ xn ą b ´ ε = 8 .
Proof. We only prove 1 since the proof of 2 is similar. Let zk = inf xn , and
něk
sup zk = lim zk = a P R˚ .
kě1
kÑ8
We show that a P R if and only if 1-(a) and 1-(b). Nevertheless, by Proposition 1.87 (or
Remark 1.112), a P R if and only if
(i) a is an upper bound of tzk u8
k=1 .
(ii) @ ε ą 0, D N P N Q zN ą a ´ ε.
We justify the equivalency between 1-(a) and (ii), as well as the equivalency between 1-(b)
and (i) as follows:
(i) a is an upper bound of tzk u8
k=1 ô a ě zk for all k P N ô @ ε ą 0, a + ε ą zk for all
k P N ô @ ε ą 0 and k P N, a + ε ą inf xn ô @ ε ą 0 and k P N, a + ε is not a lower
něk
bound of txn u8
něk ô @ ε ą 0 and k P N, D n ě k Q a + ε ą xn ô 1-(b).
(ii) @ ε ą 0, D N P N Q zN ą a ´ ε ô @ ε ą 0, D N ą 0 Q inf xn ą a ´ ε ô @ ε ą 0,
něN
D N ą 0 such that a ´ ε is a lower bound of txN , xN +1 , ¨ ¨ ¨ u ô @ ε ą 0, D N ą 0 such
that a ´ ε ď xn for all n ě N ô @ ε ą 0, D N ą 0 such that a ´ ε ă xn for all n ě N
˝
ô 1-(a).
Remark 1.122. By Proposition 1.121, if a = lim inf xn P R, then
nÑ8
ˇ
␣
(
@ ε ą 0, # n P N ˇ xn P (a ´ ε, a + ε) = 8
which suggests that a is a cluster point of txn u8
n=1 . Moreover, 1-(a) of Proposition 1.121
implies that no other cluster points can be smaller than a. In other words, if a = lim inf xn P
nÑ8
R, then a is the smallest cluster point of txn u8
n=1 . Similarly, b is the largest cluster point of
txn u8
n=1 if b = lim sup xn P R.
nÑ8
Theorem 1.123. Let txn u8
n=1 be a sequence in R. Then
1. lim inf xn ď lim sup xn .
nÑ8
nÑ8
§1.6 Euclidean Spaces and Vector Spaces
41
2. If txn u8
n=1 is bounded above by M , then lim sup xn ď M .
nÑ8
3. If txn u8
n=1 is bounded below by m, then lim inf xn ě m.
nÑ8
4. lim sup xn = 8 if and only if txn u8
n=1 is not bounded above.
nÑ8
5. lim inf xn = ´8 if and only if txn u8
n=1 is not bounded below.
nÑ8
6. If x is a cluster point of txn u8
n=1 , then lim inf xn ď x ď lim sup xn .
nÑ8
nÑ8
7. If a = lim inf xn is finite, then a is a cluster point.
nÑ8
8. If b = lim sup xn is finite, then b is a cluster point.
nÑ8
9. If txn u8
n=1 converges to x in R if and only if lim inf xn = lim sup xn = x P R.
nÑ8
nÑ8
Proof. Left as an exercise.
˝
Remark 1.124. Using the definition of cluster points of a sequence in Remark 1.112, Remark 1.122 and Theorem 1.123 together imply that the limit superior/inferior of a sequence
is the largest/smallest cluster point of that sequence.
Example 1.125. Let S = Q X [0, 1]. Then S is countable since it is a subset of a countable
1´1
set Q. Therefore, D f : NÝÝÑS or equivalently S = tq1 , q2 , ¨ ¨ ¨ , qn , ¨ ¨ ¨ u. The collection of
onto
all cluster points of tqn u8
n=1 is [0, 1] since Q X [0, 1] is dense in [0, 1].
1.6 Euclidean Spaces and Vector Spaces
Definition 1.126. Euclidean n-space, denoted by Rn , consists of all ordered n-tuples of
real numbers. Symbolically,
␣ ˇ
(
Rn = x ˇ x = (x1 , x2 , ¨ ¨ ¨ , xn ), xi P R .
Elements of Rn are generally denoted by single letters that stand for n-tuples such as
x = (x1 , x2 , ¨ ¨ ¨ , xn ), and speak of x as a “point” in Rn .
Definition 1.127. A real vector space V is a set of elements called vectors, with given
operations of vector addition + : V ˆ V Ñ V and scalar multiplication ¨ : R ˆ V Ñ V such
that
42
CHAPTER 1. The Real Line and Euclidean Space
1. v + w = w + v for all v, w P V.
2. (v + w) + u = v + (u + w) for all u, v, w P V.
3. D 0, the zero vector, Q v + 0 = v for all v P V.
4. @ v P V, D w P V Q v + w = 0.
5. λ ¨ (v + w) = λ ¨ v + λ ¨ w for all λ P R and v, w P V.
6. (λ + µ) ¨ v = λ ¨ v + µ ¨ v for all λ, µ P R and v P V.
7. (λ ¨ µ) ¨ v = λ ¨ (µ ¨ v) for all λ, µ P R and v P V.
8. 1 ¨ v = v for all v P V.
Example 1.128. Let the vector addition and scalar multiplication on Rn be defined by
x + y = (x1 + y1 , ¨ ¨ ¨ , xn + yn ) if
x = (x1 , ¨ ¨ ¨ , xn ), y = (y1 , ¨ ¨ ¨ , yn )
and
λ ¨ x = (λx1 , ¨ ¨ ¨ , λxn ) if
λ P R, x = (x1 , ¨ ¨ ¨ , xn ) .
Then Rn is a real vector space.
␣
(
Example 1.129. Let M ” n ˆ m matrix with entries in R . Define
A + B ” [aij + bij ],
λ ¨ A ” [λ ¨ aij ] if
λ P R, A = [aij ], B = [bij ] P M .
Then M is a real vector space.
Definition 1.130. W is called a subspace of a real vector space V if
1. W is a subset of V.
2. (W, +, ¨), with vector addition and scalar multiplication in V, is a real vector space.
Example 1.131. V = R3 , W = R2 ˆ t0u ” t(x, y, 0)|x, y P Ru. W is a subspace of V.
Lemma 1.132. If W is a subset of a real vector space V, then W is a subspace if and only
if λ ¨ v + µ ¨ w P W, @ λ, µ P R, v, w P W.
§1.6 Euclidean Spaces and Vector Spaces
43
Remark 1.133. “n” is called the dimension of Rn .
There are n linearly independent vectors e1 = (1, 0, ¨ ¨ ¨ , 0), e2 = (0, 1, 0, ¨ ¨ ¨ , 0), ¨ ¨ ¨ , en =
(0, 0, ¨ ¨ ¨ , 0, 1), but if v1 , v2 , ¨ ¨ ¨ , vn+1 are (n + 1) vectors in Rn , D λ1 , ¨ ¨ ¨ , λn+1 P R, Q λ1 v1 +
¨ ¨ ¨ + λn+1 vn+1 = 0, (λ1 , ¨ ¨ ¨ , λn+1 ) ‰ (0, ¨ ¨ ¨ , 0).
Definition 1.134. A subset H Ď Rn is called a hyperplane if H is (n ´ 1)-dimensional
subspace of Rn . An affine hyperplane is a set x+H ” tx+y | y P Hu for some hyperplane
H.
1.7 Normed Vector Spaces, Inner Product Spaces and
Metric Spaces
Definition 1.135. A nomed vector space (V, } ¨ }) is a real vector space V associated
with a function } ¨ } : V Ñ R such that
(a) }x} ě 0 for all x P V.
(b) }x} = 0 if and only if x = 0.
(c) }λ ¨ x} = |λ| ¨ }x} for all λ P R and x P V.
(d) }x + y} ď }x} + }y} for all x, y P V.
A function } ¨ } satisfies (a)-(d) is called a norm on V.
Example 1.136. Let V = R , and }x}2 ”
n
n
(ÿ
x2i
) 12
if x = (x1 , x2 , ¨ ¨ ¨ , xn ). Then } ¨ }2 is
i=1
a norm, called 2-norm, on Rn . It suffices to show that (d) in Definition 1.135 holds. Let
x = (x1 , x2 , ¨ ¨ ¨ , xn ) and y = (y1 , y2 , ¨ ¨ ¨ , yn ). Then
(}x + y}2 )2 =
n
ÿ
(xi + yi )2 =
n
ÿ
i=1
(x2i + 2xi yi + yi2 ) =
i=1
2
2
ď }x}2 + }y}2 + 2}x}2 }y}2
2
= (}x}2 + }y}2 ) ;
thus }x + y}2 ď }x}2 + }y}2 .
n
ÿ
i=1
x2i +
n
ÿ
yi2 + 2
i=1
(By Cauchy’s inequality)
n
ÿ
i=1
xi yi
44
CHAPTER 1. The Real Line and Euclidean Space
Example 1.137. Let V = Rn , and define
$
n
(ÿ
)1
’
p p
&
|xi |
if 1 ď p ă 8,
}x}p ”
i=1
’
␣
(
%
max |x1 |, ¨ ¨ ¨ , |xn | if p = 8,
for all x = (x1 , x2 , ¨ ¨ ¨ , xn ) P Rn .
Then } ¨ }p is a norm, called p-norm, on Rn . Property (d) in Definition 1.135; that is,
}x + y}p ď }x}p + }y}p , is left as an exercise.
␣
(
Example 1.138. Let Mnˆm ” "n ˆ m matrix with entries in R , and we remind the readRm Ñ Rn
ers that if A P Mnˆm , then A :
. Define
x ÞÑ Ax
}Ax}p
x‰0 }x}p
@ A P Mnˆm ;
}A}p = sup }Ax}p = sup
}x}p =1
)
! }Ax} ˇ
p ˇ
m
that is, }A}p is the least upper bound of the set
ˇ x ‰ 0, x P R . Therefore,
}x}p
}Ax}p
ď }A}p @ x ‰ 0; thus
}x}p
}Ax}p ď }A}p }x}p
@ x P Rm .
Consider the case p = 1, p = 2 and p = 8 respectively.
1. p = 2: Let (¨, ¨)Rk denote the inner product in Euclidean space Rk . Then
}Ax}22 = (Ax, Ax)Rn = (x, AT Ax)Rm = (x, P ΛP T x)Rm = (P T x, ΛP T x)Rn ,
in which we use the fact that AT A is symmetric; thus diagonalizable by an orthonormal
matrix P (that is, AT A = P ΛP T , P T P = I, Λ is a diagonal matrix). Therefore,
sup }Ax}22 = sup (P T x, ΛP T x) = sup (y, Λy) (Let y = P T x, then }y}2 = 1)
}x}2 =1
}x}2 =1
}y}2 =1
= sup (λ1 y12 + λ2 y22 + ¨ ¨ ¨ + λn yn2 )
}y}2 =1
␣
(
= max λ1 , ¨ ¨ ¨ , λn = maximum eigenvalue of AT A
a
maximum eigenvalue of AT A.
#
+
m
m
m
ÿ
ÿ
ÿ
2. p = 8: }A}8 = sup }Ax}8 = max
|a1j |,
|a2j |, ¨ ¨ ¨
|anj | .
which implies that }A}2 =
}x}8 =1
j=1
j=1
j=1
§1.6 Euclidean Spaces and Vector Spaces
45
[ ]
Reason: Let x = (x1 , x2 , ¨ ¨ ¨ , xn )T and A = aij nˆm . Then


a11 x1 + ¨ ¨ ¨ + a1m xm
 a21 x1 + ¨ ¨ ¨ + a2m xm 


Ax = 

..


.
an1 x1 + ¨ ¨ ¨ + anm xm
Assume max
1ďiďn
m
ÿ
|aij | =
j=1
m
ÿ
|akj | for some 1 ď k ď n. Let
j=1
x = (sgn(ak1 ), sgn(ak2 ), ¨ ¨ ¨ , sgn(akn )) .
m
ř
Then }x}8 = 1, and }Ax}8 =
|akj |.
j=1
On the other hand, if }x}8 = 1, then
|ai1 x1 + ai2 x2 + ¨ ¨ ¨ aim xm | ď
m
ÿ
#
|aij | ď max
j=1
#
thus }A}8 = max
m
ř
|a1j |,
j=1
m
ř
|a2j |, ¨ ¨ ¨
j=1
m
ÿ
|a1j |,
j=1
m
ř
m
ÿ
j=1
|a2j |, ¨ ¨ ¨
m
ÿ
+
|anj | ;
j=1
+
|anj | . In other words, }A}8 is the largest
j=1
sum of the absolute value of row entries.
#
+
n
n
n
ÿ
ÿ
ÿ
3. p = 1: }A}1 = max
|ai1 |, |ai2 |, ¨ ¨ ¨ , |aim | .
i=1
i=1
i=1
Example 1.139. Let C be the collection of all continuous real-valued functions on the
interval [0, 1]; that is,
ˇ
␣
(
C = f : [0, 1] Ñ R ˇ f is continuous on [0, 1] .
For each f P C , we define
}f }p =
$ [ż 1
] p1
’
’
&
|f (x)|p dx
if 1 ď p ă 8 ,
0
’
’
%
max |f (x)|
xP[0,1]
if p = 8 .
The function } ¨ }p : C Ñ R is a norm on C (Minkowski’s inequality).
(
)
Definition 1.140. An inner product space V, x¨, ¨y is a real vector space V associated
with a function x¨, ¨y : V ˆ V Ñ R such that
46
CHAPTER 1. The Real Line and Euclidean Space
(1) xx, xy ě 0, @ x P V.
(2) xx, xy = 0 if and only if x = 0.
(3) xx, y + zy = xx, yy + xx, zy for all x, y, z P V.
(4) xλx, yy = λxx, yy for all λ P R and x, y P V.
(5) xx, yy = xy, xy for all x, y P V.
A symmetric bilinear form x¨, ¨y satisfies (1)-(5) is called an inner product on V.
Example 1.141. Let (¨, ¨) : Rn ˆ Rn Ñ R be defined by
(x, y) =
n
ÿ
xi y i
@ x = (x1 , ¨ ¨ ¨ , xn ), y = (y1 , ¨ ¨ ¨ , yn ) .
i=1
Then (¨, ¨) is an inner product on Rn .
Example 1.142. Let C be defined as in Example 1.139. Define
ż1
xf, gy =
f (x)g(x)dx .
0
Then x¨, ¨y : C ˆ C Ñ R satisfies all the properties that an inner product has. Note that
xf, f y = }f }22 .
Proposition 1.143. If x¨, ¨y is an inner product on a real vector space V. Then
1. xλv + µw, uy = λxv, uy + µxw, uy for all u, v, w P V.
2. xu, λv + µwy = λxu, vy + µxu, wy for all u, v, w P V.
3. xv, λwy = λxv, wy for all v, w P V.
4. x0, wy = xw, 0y = 0 for all w P V.
Theorem 1.144. The inner product x¨, ¨y on a real vector space induces a norm } ¨ } given
a
by }x} = xx, xy and satisfies the Cauchy-Schwarz inequality
ˇ
ˇ
ˇxx, yyˇ ď }x} ¨ }y}
@ x, y P V .
(1.7.1)
§1.6 Euclidean Spaces and Vector Spaces
47
Proof. First, we observe that for all x, y P V fixed, we must have
0 ď xλx + y, λx + yy = }x}2 λ2 + 2xx, yyλ + }y}2
for all λ P R. Therefore,
xx, yy2 ´ }x}2 ¨ }y}2 ď 0
which implies (1.7.1).
It should be clear that (a)-(c) in Definition 1.135 are satisfied. To show that } ¨ } satisfies
the triangle inequality, by (1.7.1) we find that
(
)2
}x} + }y} ´ }x + y}2 = }x}2 + 2}x}}y} + }y}2 ´ xx + y, x + yy
(
)
= 2 }x}}y} ´ xx, yy ě 0 ;
thus the triangle inequality is also valid.
˝
Corollary 1.145. Let f, g : [0, 1] Ñ R be continuous. Then
ˇż 1
ˇ (ż 1
) 12 (ż 1
) 12
ˇ
ˇ
2
2
ˇ f (x)g(x)dxˇ ď
|f (x)| dx
|g(x)| dx .
ˇ
ˇ
0
0
0
Definition 1.146. A metric space (M, d) is a set M associated with a function d :
M ˆ M Ñ R such that
(i) d(x, y) ě 0 for all x, y P M .
(ii) d(x, y) = 0 if and only if x = y.
(iii) d(x, y) = d(y, x) for all x, y P M .
(iv) d(x, y) ď d(x, z) + d(z, y) for all x, y, z P M .
A function d satisfies (i)-(iv) is called a metric on M .
Example 1.147 (Discrete metric). Let M be a non-empty set, and d0 : M ˆ M Ñ R be
defined by
"
d0 (x, y) =
0 if x = y ,
1 if x ‰ y .
Then d0 is a metric on M , and we call d0 the discrete metric.
48
CHAPTER 1. The Real Line and Euclidean Space
Example 1.148 (Bounded metric). Let (M, d) be a metric space. Define ρ : M ˆ M Ñ R
by
ρ(x, y) =
d(x, y)
.
1 + d(x, y)
Then ρ is also a metric on M .
Proposition 1.149. If (V, } ¨ }) is a normed vector space, then the function d : V ˆ V Ñ R
defined by d(x, y) = }x ´ y} is a metric on V. In other words, (V, d) is a metric space, and
we usually write (V, } ¨ }) as the metric space.
1.8 Exercises
§1.1 Ordered Fields and the Number Systems
Problem 1.1. Let (F, +, ¨, ď) be an ordered field, and a, b, c, d P F.
1. Show that if a ď b and c ď d, then a + c ď b + d.
2. Show that if a ď b and c ă d, then a + c ă b + d.
Problem 1.2. Let S be a non-empty subset of N and satisfy that
1. 1, 2 P S.
2. if m and m + 1 P S, then m + 2 P S.
Show that S = N.
§1.2 Completeness and the Real Number System
Problem 1.3. Let F be an ordered field with Archimedean property, and x, y P F. Show
that x ď y if and only if @ ε ą 0, x ă y + ε.
Problem 1.4. Fix y ą 1. Complete the following.
1. Define y 1/n properly. (Hint: see how we define
?
y in class).
2. Show that y n ´ 1 ą n(y ´ 1) for all n P Nzt1u; thus y ´ 1 ą n(y 1/n ´ 1).
3. If t ą 1 and n ą (y ´ 1)/(t ´ 1), then y 1/n ă t.
§1.8 Exercises
49
4. Show that lim y 1/n = 1 as n Ñ 8.
nÑ8
Problem 1.5. Complete the following.
1. Let x ě 0 be a real number such that for any ε ą 0, x ď ε. Show that x = 0.
2. Let S = (0, 1). Show that for each ε ą 0 there exists an x P S such that x ă ε.
§1.3 Least Upper Bounds and Greatest Lower Bounds
Problem 1.6. Let A be a non-empty set of R which is bounded below. Define the set ´A
ˇ
␣
(
by ´A ” ´ x P R ˇ x P A . Prove that
inf A = ´ sup(´A) .
Problem 1.7. Let A, B be non-empty subset of R. Define A + B = tx + y | x P A, y P Bu.
Justify if the following statements are true or false by providing a proof for the true statement
and giving a counter-example for the false ones.
1. sup(A + B) = sup A + sup B.
2. inf(A + B) = inf A + inf B.
3. sup(A X B) ď mintsup A, sup Bu.
4. sup(A X B) = mintsup A, sup Bu.
5. sup(A Y B) ě maxtsup A, sup Bu.
6. sup(A Y B) = maxtsup A, sup Bu.
Problem 1.8. Let S Ď R be bounded below and non-empty. Show that
ˇ
␣
(
inf S = sup x P R ˇ x is a lower bound for S .
Problem 1.9. Let A, B be two sets, and f : A ˆ B Ñ R be a function. Show that
sup
(
)
(
)
f (x, y) = sup sup f (x, y) = sup sup f (x, y) .
(x,y)PAˆB
yPB
xPA
xPA
yPB
Problem 1.10. Fix b ą 1.
1. Show the law of exponents holds (for rational exponents); that is, show that
50
CHAPTER 1. The Real Line and Euclidean Space
(a) if r, s in Q, then br+s = br ¨ bs .
(b) if r, s in Q, then br¨s = (br )s .
ˇ
␣
(
2. For x P R, let B(x) = bt P R ˇ t P Q, t ď x . Show that br = sup B(r) if r P Q.
Therefore, it makes sense to define bx = sup B(x) for x P R. Show that the law of
exponents (for real exponents) are also valid.
3. Let y ą 0 be given. Using 4 of Problem 1.4 to show that if u, v P R such that bu ă y
and bv ą y, then bu+1/n ă y and bv´1/n ą y for sufficiently large n.
4. Let y ą 0 be given, and A be the set of all w such that bw ă y. Show that x = sup A
satisfies bx = y.
5. Prove that if x1 , x2 are two real numbers satisfying bx1 = bx2 , then x1 = x2 .
The number x satisfying bx = y is called the logarithm of y to the base b, and is denoted by
logb y.
§1.4 Cauchy Sequences
Problem 1.11. Let a P R. Define an through the iterated relation
an = a2n´1 ´ an´1 + 1
@ n ą 1, a1 = a .
For what a is the sequence tan u8
n=1 (1) monotone? (2) bounded? (3) convergent? Compute
the limit in the case of convergence.
Problem 1.12. Let F be an ordered field, and txn u8
n=1 be a sequence in F. Show that
8
txn un=1 is Cauchy if and only if
ˇ
␣
(
@ ε ą 0, D y P F Q # n P N ˇ xn R (y ´ ε, y + ε) ă 8 .
8
Problem 1.13. Let tan u8
n=1 and txn un=1 be two sequences in R, and define Sk =
k
ř
an (so
n=1
8
tSk u8
k=1 is also a sequence). Suppose that |xn ´ xn+1 | ă an for all n P N. Show that txn un=1
converges if tSk u8
k=1 converges.
Problem 1.14. Let f : R Ñ R be a function so that |f (x) ´ f (y)| ď
|x ´ y|
. Pick an
2
arbitrary x1 P R, and define xk+1 = f (xk ) for all k P N. Show that txn u8
n=1 is a Cauchy
sequence.
§1.8 Exercises
51
8
Problem 1.15. Suppose that txn u8
n=1 and tyn un=1 are two Cauchy sequence in R. Show
␣
(8
that the sequence |xn ´ yn | n=1 converges.
§1.5 Cluster Points and Limit Inferior, Limit Superior
8
Problem 1.16. Let txn u8
n=1 and tyn un=1 be sequences in R. Prove the following inequalities:
lim inf xn + lim inf yn ď lim inf(xn + yn ) ď lim inf xn + lim sup yn
nÑ8
nÑ8
nÑ8
nÑ8
nÑ8
ď lim sup(xn + yn ) ď lim sup xn + lim sup yn ;
nÑ8
nÑ8
nÑ8
(
)(
)
(
)(
)
lim inf |xn | lim inf |yn | ď lim inf |xn yn | ď lim inf |xn | lim sup |yn |
nÑ8
nÑ8
nÑ8
nÑ8
(
)( nÑ8
)
ď lim sup |xn yn | ď lim sup |xn | lim sup |yn | .
nÑ8
nÑ8
nÑ8
Give examples showing that the equalities are generally not true.
Problem 1.17. Prove that
lim inf
nÑ8
a
a
|xn+1 |
|xn+1 |
ď lim inf n |xn | ď lim sup n |xn | ď lim sup
.
nÑ8
|xn |
|xn |
nÑ8
nÑ8
a
Give examples to show that the equalities are not true in general. Is it true that lim n |xn |
nÑ8
|x
|
exists implies that lim n+1 also exists?
nÑ8 |xn |
Problem 1.18. Find the following limits.
1?
n
n! ,
nÑ8 n
lim
1a
n
(n + 1)(n + 2) ¨ ¨ ¨ (2n) .
nÑ8 n
lim
Problem 1.19. Given the following sets consisting of elements of some sequence of real
numbers. Find their sup and inf, and also the limsup and liminf of the sequence.
ˇ
(
cos m ˇ m = 0, 1, 2, ¨ ¨ ¨ .
1.
␣
2.
ˇ
␣a
(
m
| sin m| ˇ m = 1, 2, ¨ ¨ ¨ .
3.
␣
(1 +
(
1
mπ ˇˇ
) sin
m = 1, 2, ¨ ¨ ¨ .
m
6
Hint: For 1, first show that for all irrational α, the set
ˇ
␣
(
S = x P [0, 1] ˇ x = kα (mod 1) for some k P N
52
CHAPTER 1. The Real Line and Euclidean Space
is dense in [0, 1]; that is, for all y P [0, 1] and ε ą 0, there exists x P S X (y ´ ε, y + ε). Then
choose α =
1
to conclude that
2π
ˇ
␣
(
T = x P [0, 2π] ˇ x = k (mod 2π) for some k P N
is dense in [0, 2π]. To prove that S is dense in [0, 1], you might want to consider the following
set
ˇ
␣
(
Sk = x P [0, 1] ˇ x = ℓα (mod 1) for some 1 ď ℓ ď k + 1
Note that there must be two points in Sk whose distance is less than
(the multiples of) the difference of these two points?
1
. What happened to
k
§1.6 Euclidean Spaces and Vector Spaces
Problem 1.20. Show that the p-norm on Euclidean space Rn given by
$
n
(ÿ
) p1
’
&
|xi |p
if 1 ď p ă 8 ,
}x}p ”
x = (x1 , ¨ ¨ ¨ , xn )
i=1
’
␣
(
%
max |x1 |, ¨ ¨ ¨ , |xn |
if p = 8 ,
is indeed a norm.
§1.7 Normed Vector Spaces, Inner Product Spaces and Metric Spaces
Problem 1.21. Let Mnˆm be the collection of all n ˆ m matrices with real entries as in
Example 1.138. Define a function } ¨ } : M Ñ R by
}Ax}2
,
}x}2
xPRm
}A} = sup
x‰0
here we recall that } ¨ }2 is the 2-norm on Euclidean space given by
}x}2 =
k
(ÿ
x2i
)1/2
if
x = (x1 , ¨ ¨ ¨ , xk ) P Rk .
i=1
Show that
ˇ
(
␣
1. }A} = sup }Ax}2 = inf M P R ˇ }Ax}2 ď M }x}2 @ x P Rm .
xPRm
}x}2 =1
2. }Ax}2 ď }A}}x}2 for all x P Rm .
§1.8 Exercises
53
3. } ¨ } defines a norm on Mnˆm .
4. Let tAk u8
k=1 Ď Mnˆm . Show that lim }Ak } = 0 if and only if each entry of Ak conkÑ8
[ (k) ]
verges to 0. In other words, by writing Ak = aij 1ďiďn,1ďjďm , show that lim }Ak } = 0
kÑ8
(k)
if and only if lim aij = 0 for all 1 ď i ď m, 1 ď j ď n. In particular, Ak Ñ A in the
kÑ8
sense that }Ak ´ A} Ñ 0 as k Ñ 8 if and only if the (i, j)-th entry of Ak converges to
(i, j)-th entry of A for all 1 ď i ď n and 1 ď j ď m.
Problem 1.22. Let (V, +, ¨, x¨, ¨y) be an inner product space, and define }v} = xv, vy1/2 for
all v P V. Show that
1. 2}x}2 + 2}y}2 = }x + y}2 + }x ´ y}2 (parallelogram law).
2. }x + y}}x ´ y} ď }x}2 + }y}2 .
3. 4 xx, yy = }x + y}2 ´ }x ´ y}2 (polarization identity).
Can the p-norm } ¨ }p on Rn be induced from any inner product (on Rn ) for p ‰ 2?
Problem 1.23. Let (X, } ¨ }X ), (Y, } ¨ }Y ), (Z, } ¨ }Z ) be three normed vector spaces such
that X, Y Ď Z and
}x}Z ď C}x}X
@x P X
and
}y}Z ď C}y}Y
@y P Y .
Define
ˇ
␣
(
E = a P Z ˇ }a}E ” maxt}a}X , }a}Y u ă 8
and
ˇ
␣
F = a P Z ˇ }a}F ” a=x+y
inf (}x}X + }y}Y ) ă 8u .
xPX,yPY
Show that (E, } ¨ }E ) and (F, } ¨ }F ) are also normed vector spaces, and E = X X Y . The
space F is usually denoted by X + Y .
Problem 1.24 (True or False). Determine whether the following statements are true or
false. If it is true, prove it. Otherwise, give a counter-example.
1. Given two sets A and B. Then AˆB is countable if and only if A and B are countable.
2. Let txn u8
n=1 Ď R be a sequence and lim sup xn = x. Then sup xn = x.
nÑ8
nPN
54
CHAPTER 1. The Real Line and Euclidean Space
ˇ
␣
(
3. The set (x, y) P R2 ˇ x + y P Q is countable.
4. Let txn u8
n=1 Ď R be a sequence such that |xn ´ xn+1 | ď
R.
1
. Then txn u8
n=1 converges in
n
5. If a bounded sequence txn u8
n=1 in R satisfies xn+1 ´ ϵn ď xn for n P N, where
8
ř
ϵn
n=1
is an absolute convergent series; that is, the partial sum
k
ř
|ϵn | converges as k Ñ 8,
n=1
then txn u8
n=1 converges.
6. Let π : N Ñ N be one-to-one and onto (such map π is called a rearrangement), and
␣
(8
txn u8
n=1 is a convergent sequence. Then xπ(n) n=1 is also convergent.
7. Let A Ď R satisfy
sup
!ÿ
ˇ
)
ˇ
|b| ˇ B is a non-empty finite subsets of A ă 8 .
bPB
ˇ
␣
Then x P A ˇ x ‰ 0u is countable.
8. Any rearrangement of the series
n Ñ 8.
8
ř
xn diverges if and only if xn does not tend to 0 as
n=1
9. If txn u8
n=1 is a sequence of distinct non-zero real numbers such that lim xn = 0, then
nÑ8
ˇ
(
␣
the set mxn ˇ m P Z, n P N is dense in R.
Chapter 2
Point-Set Topology of Metric spaces
2.1 Open Sets and the Interior of Sets
Definition 2.1. Let (M, d) be a metric space. For each x P M and ε ą 0, the set
ˇ
␣
(
D(x, ε) = y P M ˇ d(x, y) ă ε
is called the ε-disk (ε-ball) about x or the disk/ball centered at x with radius ε.
M
x•
ε
Figure 2.1: The ε-ball about x in a metric space
Example 2.2. (R2 , } ¨ }p ) is a normed vector space. Consider x = 0, ε = 1 and p = 1, p = 2
and p = 8 respectively.
1. p = 1: }x}1 = |x1 | + |x2 |, d(x, y) = |x1 ´ y1 | + |x2 ´ y2 |.
2. p = 2: }x}2 =
a
a
x21 + x22 , d(x, y) = (x1 ´ y1 )2 + (x2 ´ y2 )2 .
␣
(
␣
(
3. p = 8: }x}8 = max |x1 |, |x2 | , d(x, y) = max |x1 ´ y1 |, |x2 ´ y2 | .
55
56
CHAPTER 2. Point-Set Topology of Metric spaces
p=1
p=2
1
1
´1
p=8
1
1
´1
´1
1
1
´1
´1
´1
Figure 2.2: The 1-ball about 0 in R2 with different p
Example 2.3. Let (M, d) be a metric space with discrete metric; that is,
#
1 if x ‰ y,
d(x, y) =
0 if x = y.
#
txu if 0 ă ε ď 1,
Then D(x, ε) =
M if ε ą 1.
Definition 2.4. Let (M, d) be a metric space. A set U Ď M is said to be open (in M ) if
@ x P U, D ε ą 0 Q D(x, ε) Ď U .
ˇ
␣
(
Example 2.5. The set A = (x, y) P R2 ˇ 0 ă x ă 1 is open: given (x, y) P A, take
ε = mint1 ´ x, xu, then D(x, ε) Ď A.
ˇ
␣
(
Example 2.6. A = (x, y) P R2 ˇ 0 ă x ď 1 is not open: let u = (1, 0), then @ ε ą 0 Q
(
(
ε )
ε )
D(u, ε) Ę A (since 1 + , 0 P D(u, ε) but 1 + , 0 R A).
2
2
Example 2.7. M ” (a, b) ˆ [c, d] Y tpu, p R [a, b] ˆ [c, d], and for two points (x1 , y1 ) and
(
) a
(x2 , y2 ) in M , we define the metric as d (x1 , y1 ), (x2 , y2 ) = (x1 ´ x2 )2 + (y1 ´ y2 )2 . Then
␣b ´ a
(
(a + b )
D(p, ε) = tpu if ε ! 1. Let q =
, d , ε ă min
, d ´ c . Then D(q, ε) is the
2
shaded region in red color shown in the figure below.
q
d
D(q, ε)
2
p
D(p, ε) = {p}
c
(
a
)
b
§2.1 Open Sets and the Interior of Sets
57
Proposition 2.8. Let (M, d) be a metric space. Then every ε-disk is open.
Proof. Let D(x, ε) be an ε-disk. We would like to show that @ y P D(x, ε), D δ ą 0 Q
D(y, δ) Ď D(x, ε). Let δ = ε ´ d(x, y) ą 0. Then if z P D(y, δ), we have
d(z, x) ď d(z, y) + d(x, y) ă δ + d(x, y) = ε ;
thus z P D(x, ε).
˝
Proposition 2.9. Let (M, d) be a metric space.
1. The intersection of finitely many open sets is open.
2. The union of arbitrary family of open sets is open.
3. The empty set H and the universal set M are open.
Proof. 1. Let U1 , U2 , ¨ ¨ ¨ , Uk be open sets in M , and U ”
k
Ş
Ui . If y P U , then y P Ui for
i=1
all 1 ď i ď k. Since Ui is open, D δi ą 0 Q D(y, δi ) Ď Ui . Let δ = mintδ1 , ¨ ¨ ¨ , δk u.
Claim: D(y, δ) Ď U .
Proof of claim: Let z P D(y, δ). Then d(y, z) ă δ ď δi if i = 1, 2, ¨ ¨ ¨ , k.
k
Ş
ñ z P D(y, δi ) @ i = 1, 2, ¨ ¨ ¨ , k ñ z P Ui if i = 1, 2, ¨ ¨ ¨ , k ñ z P
Ui ” U .
i=1
ˇ
(
Ť
Uα . If
2. Let F = Uα ˇ Uα open in M , α P I be a family of open sets, and U ”
␣
αPI
y P U , then y P Uβ for some β P I. Since Uβ is open, D δ ą 0 Q D(y, δ) Ď Uβ ; thus
Ť
D(y, δ) Ď
Uα ” U .
αPI
3. H is trivially an open set. Moreover, if y P M , then D(y, 1) Ď M (by definition).
˝
Corollary 2.10. Let (M, d0 ) be a metric space with discrete metric. Then every subset of
M is open.
( 1)
Ť ( 1)
is an open set in M . If A Ď M , A ‰ H, then A =
D y,
Proof. @ y P M, tyu = D y,
2
yPA
which suggests that A is open since it is an arbitrary union of open sets.
Remark 2.11. Infinite intersection of open sets need not be open:
8
( 1 1)
Ş
1. Take An = ´ , , then
An = t0u which is not open.
n n
n=1
2
˝
58
CHAPTER 2. Point-Set Topology of Metric spaces
8
Ş
1
1
2. Let Uk = (´2 ´ , 2 + ) Ď R. Then A =
Uk Ě [´2, 2]. Moreover, if x R [´2, 2],
k
k
k=1
(
1
x´2
1
´x ´ 2 )
then D k P N Q x R Uk If x ą 2,
ă
. If x ă ´2,
ă
. Therefore,
k
2
k
2
8
Ş
Uk = [´2, 2].
k=1
Example 2.12. Let A Ď Rn be open, and B Ď Rn . Then A + B = ta + b | a P A, b P Bu is
open.
Proof. Let y P A + B. Then y = a + b for some a P A, b P B. Since A is open, D δ ą 0 Q
D(a, δ) Ď A.
Claim: D(y, δ) Ď A + B.
Proof of claim: Let z P D(y, δ). Then }z ´ y}2 ă δ. Since z = b + (z ´ b), if we can show
that z ´ b P A, then z P A + B. Nevertheless, we have
}(z ´ b) ´ a}2 = }z ´ a ´ b}2 = }z ´ y}x ă δ
which implies that z ´ b P D(a, δ) Ď A.
˝
Definition 2.13. Let (M, d) be a metric space, and A Ď M be a subset of M . A point
x P A is called an interior point of A if D ε ą 0 Q D(x, ε) Ď A. The interior of A is the
collection of all interior points of A, and is denoted by int(A) or Å.
␣ 1 1
1
Example 2.14. Let M = R with d(x, y) = |x´y|, and A = [0, 1), B = 1, , , ¨ ¨ ¨ , , ¨ ¨ ¨ uY
2 3
n
␣ 1 (8
t0u =
Y t0u. Then Å = (0.1) and B̊ = H since
n=1
n
1. If x P (0, 1), then D ε ą 0 Q D(x, ε) Ď (0, 1) Ď A.
2. 0 is not an interior point since (´ε, ε) X [0, 1)A ‰ ϕ @ ε ą 0.
Remark 2.15. Å might be empty.
Theorem 2.16. Let (M, d) be a metric space, and A Ď M be a subset of M . The interior of
A is the largest open set contained in A. In other words, if U Ď A is open, then U Ď int(A).
Proof. Let z P U . Since U is open, D δ ą 0 Q D(z, δ) Ď U Ď A ñ z P Å ñ U Ď Å.
Ť
To show that Å is open, we observe that Å =
D(x, εx ), where εx ą 0 is chosen so that
xPÅ
D(x, εx ) Ď Å if x P Å, for the following reason:
§2.2 Closed Sets, the Closure of Sets, and the Boundary of Sets
59
1. “Ď”: trivial.
2. “Ě”: Let y P
Ť
D(x, εx ) ñ D x P Å Q y P D(x, εx ). Then if δ = εx ´ d(x, y),
xPÅ
˝
D(y, δ) Ď D(x, εx ) Ď A ñ y P A.
˝
Theorem 2.17. Let (M, d) be a metric space. A set A Ď M is open if and only if A = Å.
Example 2.18. Let (M, d) be a metric space, and A and B be two subsets of M .
1. int(A) Y int(B) Ď int(A Y B).
Proof. Let x P int(A) Y int(B). W.L.O.G. Assume x P int(A), then D r ą 0 such that
D(x, r) Ď A. Therefore, x P D(x, r) Ď A Y B, so x P int(A Y B).
˝
2. Strict containment might happen because of the following example:
Take A = [0, 1], B = [1, 2], then int(A) = (0, 1), int(B) = (1, 2).
Sine A Y B = [0, 2], int(A Y B) = (0, 2); however, int(A) Y int(B) = (0, 2)zt1u.
Hence, int(A) Y int(B) ‰ int(A Y B).
Another example is stated as follows: Let A = Q X [0, 1] and B = QA X [0, 1]. Then
(0, 1) = int([0, 1]) = int(A Y B) Ľ int(A) Y int(B) = H.
ˇ
(␣
Example 2.19. In a metric space (M, d), it is not always true that int y P M ˇ d(x, y) ď
ˇ
() ␣
(
R = y P M ˇ d(x, y) ă R . To see this, we consider the discrete metric
"
d0 (x, y) =
1 if x ‰ y,
0 if x = 0.
Let R = 1, and fix x0 P M ‰ H. Then
␣
ˇ
ˇ
(
()
(␣
y P M ˇ d0 (y, x0 ) ď 1 = M ñ int y P M ˇ d0 (y, x0 ) ď 1 = int(M ) = M .
ˇ
␣
(
Now y P M ˇ d0 (y, x0 ) ă 1 = tx0 u. As long as M has more than one point, we have
ˇ
()
(␣
int y P M ˇ d0 (y, x0 ) ď 1 = M ‰ tx0 u = D(x0 , 1).
60
CHAPTER 2. Point-Set Topology of Metric spaces
2.2 Closed Sets, the Closure of Sets, and the Boundary
of Sets
Definition 2.20. Let (M, d) be a metric space. A set F Ď M is said to be closed if
F A = M zF is open. In other words,
F is closed ô @ x P F A , D ε ą 0 Q D(x, ε) Ď F A .
Example 2.21. The set [0, 1] Ď R is closed, and the set (0, 1] Ď R is not open and not
closed.
Example 2.22. Let S = t(x, y) | x2 + y 2 ď 1u. Take z P R2 zS, then D(z, }z}2 ´ 1) Ď R2 zS.
As a consequence, R2 zS is open; thus S is closed.
ˇ
(
␣
Example 2.23. Let S = (x, y) ˇ 0 ă x ď 1, 0 ď y ď 1 . Since R2 zS is not open, S is not
closed.
Proposition 2.24. Any point in a metric space is closed; that is, if (M, d) is a metric space
and A = txu for some x P M , then A is closed.
1
2
Proof. We show that M ztxu is open. Let y P M ztxu. Pick r = d(x, y) ą 0.
Claim: D(y, r) Ď M ztxu.
y•
r=
1
d(x, y)
2
•x
1
2
Proof of claim: Let z P D(y, r). Then d(z, y) ă r = d(x, y). Then
1
1
d(z, x) ě d(x, y) ´ d(y, z) ě d(x, y) ´ d(x, y) = d(x, y) ą 0 ñ z ‰ x .
2
2
Proposition 2.25. Let (M, d) be a metric space.
1. The union of finitely many closed sets is closed.
2. The intersection of arbitrary family of closed sets is closed.
3. The universal set M and the empty set H are closed.
˝
§2.2 Closed Sets, the Closure of Sets, and the Boundary of Sets
Proof. 1. Let F1 , ¨ ¨ ¨ , Fk be closed sets, and F =
k
Ť
61
Fj . Then by De Morgan’s law,
j=1
A
F = M zF = M z
k
ď
Fj =
j=1
k
č
(M zFj ) =
j=1
Since Fj is closed, Fj A is open. By Proposition 2.9,
k
č
Fj A .
j=1
k
Ş
Fj A is open.
j=1
(
␣ ˇ
Ş
2. Let F = Fα ˇ Fα closed in M , α P I be a family of closed sets, and F ”
Fα .
αPI
Then by De Morgan’s law,
F A = Mz
č
Fα =
αPI
ď
(M zFα ) =
αPI
ď
FαA
αPI
␣
(
A
which suggests that F A is the union of open sets Fα αPI . By Proposition 2.9 we
conclude that F A is open or equivalently, F is closed.
3. M A = H, HA = M are both open.
˝
Corollary 2.26. Any set consists of finitely many points of a metric space is closed.
8
[
Ť
1
1]
Example 2.27. Let Fk = ´ 2 + , 2 ´
Ď R. Then B =
Fk Ď (´2, 2). Moreover, if
k
k
k=1
(
x+2
1
2´x
1
. If x ą 0, ă
). Therefore,
x P (´2, 2), then D k ą 0, Q x P Fk If x ď 0, ă
k
2
k
2
8
Ť
Fk = (´2, 2). This example suggests that an arbitrary union of closed sets might not be
k=1
closed.
Example 2.28. Let (M, d) be a metric space, and A = ty1 , y2 , . . . , yk u Ď M . Define
k
Ť
B = tx P M | d(x, yi ) ď 1 for some yi P Au = tx P M | d(x, yi ) ď 1u. Then B is closed.
i=1
Proof. It suffices to show Bi = tx P M | d(x, yi ) ď 1u is closed for i = 1, 2, . . . , k since
n
Ť
B=
Fi .. Take z P M zBi (if M zBi = H, then Bi = M and Bi is closed). Let N = tu P
i=1
M | d(u, z) ă d(z, yi ) ´ 1u.
Claim: N Ď M zBi ; that is, M zBi is open.
Proof of claim: Take u P N and compute d(u, yi ) ě d(yi , z) ´ d(u, z) ą d(z, yi ) ´ (d(z, yi ) ´
1) = 1. Hence u R Bi ñ u P M zBi . So N Ď M zBi .
˝
62
CHAPTER 2. Point-Set Topology of Metric spaces
Example 2.29. Let (M, d) be a metric space, A Ď M be closed, and B Ď M be finite
(#(B) ă 8). Then A + B is closed.
Proof. Left as an exercise.
˝
Definition 2.30. Let (M, d) be a metric space, and A Ď M .
1. A point x P M is called an accumulation point of A if @ ε ą 0, D(x, ε) contains
points in A other than x; that is, @ ε ą 0, D(x, ε) X (Aztxu) = (D(x, ε)ztxu) X A ‰ H.
2. A point x P M is called a limit point of A if @ ε ą 0, D(x, ε) contains points in A;
that is, @ ε ą 0, D(x, ε) X A ‰ H.
3. A point x P A is called an isolated point (孤立點) if D ε ą 0 Q D(x, ε) X A = txu.
4. The derived set of A is the collection of all accumulation points of A, and is denoted
by A1 .
s
5. The collection of all limit points of A is denoted by A.
Remark 2.31. 1. An accumulation point of A needs not to be in A.
2. If A = txu (that is, a single point), then A has no accumulation point; that is, A1 = H.
3. Accumulation points are called cluster points in some books.
s
4. If x P A1 , then x is a limit point of A. In other words, A1 Ď A.
s
5. If x P A, then x is a limit point of A. In other words, A Ď A.
s = [0, 1].
Example 2.32. Let A = (0, 1) Ď R, then A1 = [0, 1] and A
Example 2.33. Let A = (0, 1) Y t2u Ď R. Then
1. for any x P [0, 1], x P A1 ;
2. 2 R A1 , but 2 is a limit point of A;
3. if x R [0, 1] Y t2u then x R A1 .
Therefore, A1 = [0, 1]. Note that sup A = 2; thus sup A might not belong to A1 .
§2.2 Closed Sets, the Closure of Sets, and the Boundary of Sets
63
Example 2.34. Let A = txn u8
n=1 Ď R consists of a bounded sequence of distinct points.
Then A1 ‰ H.
Proof. By Bolzano-Weierstrass property (Theorem 1.100), A has a convergent subsequence
␣ (8
xnj j=1 converging to x P R.
Claim: x P A1 .
ˇ
ˇ
Proof of claim: @ ε ą 0, D K P N Q ˇxnj ´ xˇ ă ε for j ě K. Moreover, xnj P A.
˝
ˇ
␣
(
Example 2.35. In a metric space (M, d), let B(x, r) = y P M ˇ d(x, y) ď r . Is it true
that B(x, r) Ď D(x, r)1 ; that is, every point of B(x, r) is an accumulation point of D(x, r)?
Answer: No, take a metric space with discrete metric
"
1 if x ‰ y,
d0 (x, y) =
0 if x = 0.
and M has more than one point. We have D(x, 1) = txu, then D(x, 1)1 = H. Also,
B(x, 1) = M Ę H = D(x, 1)1 .
Proposition 2.36. If A Ď B, then A1 Ď B 1 .
Proof. Let x P A1 . Then @ ε ą 0, D y P A, y ‰ x Q y P D(x, ε) X A. Since A Ď B, y P B.
Therefore, @ ε ą 0, D y P B, y ‰ x Q y P D(x, ε) X B ô x P B 1 .
˝
Example 2.37. Let A be a subset of Rn . An interior point of A is an accumulation point
of A (Å Ď A1 if A Ď Rn ).
Proof. If x P Å, then D r ą 0, Q D(x, r) Ď A. Let ε ą 0 be given.
1. ε ě r, D(x, ε) X (Aztxu) Ě D(x, r) X (Aztxu) ‰ H.
2. ε ă r, D(x, ε) Ď D(x, r) Ď A ñ D(x, ε) X (Aztxu) ‰ H.
Then for all ε ą 0, D(x, ε) X (Aztxu) ‰ H ñ x P A1 .
˝
Theorem 2.38. Let (M, d) be a metric space and A Ď M , then A is closed if and only if
s
A = A.（一個集合是閉集合若且唯若該集合包含了它所有的
limit points）
Proof. A is closed ô @ y P AA , D r ą 0 Q D(y, r) Ď AA (or D(y, r) X A = H).
s (or y P A
sA ).
ô @ y P AA , y R A
s then y P A.
ô if y P A,
˝
64
CHAPTER 2. Point-Set Topology of Metric spaces
(
)
s = AYA1 = (AzA1 )YA1 .
Theorem 2.39. Let (M, d) be a metric space and A Ď M . Then A
s ô @ ε ą 0, D(x, ε) X A ‰ H.
Proof. By definition, x P A
s
ñ If x P AzA,
then @ ε ą 0, D(x, ε) X (Aztxu) ‰ H.
s
ñ If x P AzA,
then x P A1 .
s Ď A1 ñ A
s Ď A Y A1 . On the other hand, we also have (1) A Ď A
s and (2)
Therefore, AzA
s thus A Y A1 Ď A.
s
A1 Ď A;
˝
s Ď B.
s In
Corollary 2.40. Let (M, d) be a metric space, and A Ď B Ď M . Then A
s Ď B.
particular, if A Ď B and B is closed, then A
Proposition 2.41. Let (M, d) be a metric space, and A Ď M . Then AzA1 is the collection
of all isolated points of A.
Proof. Let x P AzA1 . Then x P A, but D ε ą 0 Q D(x, ε) X (Aztxu) = H. Therefore,
D(x, ε) X A = txu which implies that x is an isolated point.
˝
Theorem 2.42. Let (M, d) be a metric space, and A Ď M . Then A1 is closed; that is,
@ y R A1 , D r ą 0 Q D(y, r) X A1 = H.
Proof. Let y R A1 . Then D ε ą 0 Q D(y, ε) X (Aztyu) = (D(y, ε)ztyu) X A = H. Then
(
)A
A Ď D(y, ε)ztyu .
(
)A
Since D(y, ε)ztyu = D(y, ε) X tyuA is open, D(y, ε)ztyu is closed; thus Corollary 2.40
implies that
(
)
s Ď D(y, ε)ztyu A
A
or equivalently,
s X D(y, ε)ztyu = H .
A
s = A Y A1 , the equality above suggests that
Since A
A1 X D(y, ε)ztyu = H ;
thus the fact that y R A1 implies that D(y, ε) X A1 = H.
˝
Definition 2.43. Let (M, d) be a metric space and A Ď M . The closure of A is the
intersection of closed sets containing A, and is denoted by cl(A). In other word, cl(A) =
Ş
F (thus cl(A) is the smallest closed set containing A).
F closed.
AĎF
§2.2 Closed Sets, the Closure of Sets, and the Boundary of Sets
65
Proposition 2.44. Let (M, d) be a metric space, and A Ď M .
1. A Ď cl(A) (x P A ñ if F Ě A is closed, then x P F ).
2. A is closed if and only if A = cl(A).
s A Y A1 ).
Proposition 2.45. Let (M, d) be a metric space, and A Ď M . Then cl(A) = A(=
s Ď cl(A).
Proof. Since A Ď cl(A) and cl(A) is closed, Corollary 2.40 implies that A
s then D r ą 0 Q D(x, r) X A = H or in other words,
On the other hand, if x R A Y A1 = A,
A Ď D(x, r)A . By the definition of the closure of sets, cl(A) Ď D(x, r)A or equivalently,
s
D(x, r) Ď cl(A)A ; thus x R cl(A). Therefore, cl(A) Ď A.
˝
Example 2.46. Let A = [0.1) Y t2u Ď R. Find cl(A).
Answer: A1 = [0.1], cl(A) = A Y A1 = [0, 1] Y t2u.
?
Example 2.47. cl(A X B) = cl(A) X cl(B).
Answer: No. Take A = [0, 1], B = (1, 2]. Since A is closed, then cl(A) = A. Since cl(B) =
[1, 2], A X B = H. So cl(A X B) = H ‰ t1u = cl(A) X cl(B); thus cl(A X B) Ř cl(A) X cl(B).
Example 2.48. In a metric space (M, d),
ˇ
␣
(
x P cl(A) if and only if d(x, A) ” inf d(x, y) ˇ y P A = 0.
Proof. “ð” Suppose d(x, A) = 0. If x P A, then x P AYA1 = cl(A). If x R A, since d(x, A) =
0, @ ε ą 0 D y P A Q d(x, y) ă d(x, A) + ε = ε. In other words, (D(x, ε)ztxu) X A ‰ H.
Therefore, x P A1 ; thus x P A Y A1 = cl(A).
s = cl(A), @ ε ą 0, D(x, ε) X A ‰ H. In other words,
“ñ” Suppose x P cl(A). Since A
@ ε ą 0, D y P A Q d(x, y) ă ε.
Therefore, d(x, A) ă ε for all ε ą 0 which implies that d(x, A) = 0.
␣1 ˇ
(
ˇ n = 1, 2, ¨ ¨ ¨ . Find cl(A).
n
(
␣1 ˇ
1
ˇ n = 1, 2, ¨ ¨ ¨ Y t0u.
Answer: A = t0u ñ cl(A) = A Y A1 =
Example 2.49. A =
n
ˇ
(
Example 2.50. A = (x, y) ˇ x P Q . Find cl(A).
␣
Answer: A1 = R2 ñ cl(A) = R2 .
˝
66
CHAPTER 2. Point-Set Topology of Metric spaces
Definition 2.51. Let (M, d) be a metric space. A subset A Ď M is said to be dense（稠
密）in another subset B Ď M if A Ď B Ď cl(A).
Example 2.52. The rational numbers Q is dense in the real number system R.
Definition 2.53. Let (M, d) be a metric space, and A Ď M . The boundary of A, denoted
ĎA (BA = A
ĎA ).
s and A
sXA
by bd(A) or BA, is the intersection of A
Remark 2.54. 1. BA is closed since the closure of a set is closed.
2. By the definition of limit points of a set, we find that x P BA ô @ ε ą 0, D(x, ε) X A ‰
H and D(x, ε) X AA ‰ H.
3. BA = B(AA ).
s Å.
Proposition 2.55. Let (M, d) be a metric space, and A Ď M. Then BA = Az
Proof. If x P BA, then @ ε ą 0, D(x, ε) X AA ‰ H. Therefore, x R Å which implies that
s Å.
BA Ď Az
s Å, then @ ε ą 0, D(x, ε) Ę A. As a consequence, @ ε ą
On the other hand, if x P Az
ĎA and this further implies that x P A
ĎA = BA.
sXA
0, D(x, ε) X AA ‰ H; thus x P A
˝
Example 2.56. Let M = R, d(x, y) = |x ´ y|, and A = [0, 1] X Q. Then
1. A1 = [0, 1].
(
1
1
r P A, r + P A, r + Ñ r ñ r P A1 .
n
n
If s P [0, 1] X Q . D sn P A, sn Ñ s ñ s P A1 .
A
)
If t R [0, 1], Dε ą 0 Q D(t, ε) X [0, 1] = H ñ t R A1 .
s = [0, 1](= A Y A1 ).
2. A
3. Å = H.
4. BA = [0, 1].
Example 2.57. Let (M, d) be a metric space with discrete metric, and A Ď M . Recall that
every point is an open set.
1. A is open.
2. A is also closed since AA is open.
s = A.
5. cl(A) = A
3. Å = A.
4. A1 = H.
6. BA = H.
Remark 2.58. If A Ď B, then BA Ę BB. For example, let A = Q X [0, 1] and B = [0, 1].
Then A Ď B but BA = [0, 1], BB = t0, 1u.
§2.3 Sequences and Completeness（完備性）
67
Example 2.59. BA Ę A1 : take A = t0u, then A1 = H, BA = t0u.
Example 2.60. It is not always true that BA = B(int(A)). For example, take A = [0, 1] Y
t2u, then BA = t0, 1, 2u, int(A) = (0, 1), B(int(A)) = t0, 1u, so BA ‰ B(int(A)).
Example 2.61. Let (M, d) be a metric space, and A, B Ď M . Then
B(A Y B) Ď BA Y BB
and
B(A X B) Ď BA Y BB
since
x P B(A Y B) ô @ r ą 0, D(x, r) X (A Y B) ‰ H and D(x, r) X (AA X B A ) ‰ H
ñ @ r ą 0, D(x, r) X AA ‰ H, D(x, r) X B A ‰ H, and one of the following
holds: D(x, r) X A ‰ H or D(x, r) X B ‰ H
ĎA or x P B
ĎA ,
sXA
sXB
ñxPA
and with AA , B A replacing A, B in the inclusion we just arrive,
B(A X B) = B(A X B)A = B(AA Y B A ) Ď BAA Y BB A = BA Y BB .
2.3 Sequences and Completeness（完備性）
Definition 2.62. Let (M, d) be a metric space. A sequence in (M, d) is a function f : N Ñ
␣
(8
M , and is denoted by f (n) n=1 . Write xn for f (n). A sequence txn u8
n=1 in M is said to
converge to x if
@ ε ą 0, D N ą 0 Q d(xn , x) ă ε whenever n ě N.
ˇ
␣
(
ô @ ε ą 0, # n P N ˇ d(xn , x) ě ε ă 8.
ˇ
␣
(
ô @ ε ą 0, # n P N ˇ xn R D(x, ε) ă 8.
As Definition 1.46, one writes lim xn = x or xn Ñ x as n Ñ 8 to denote that the sequence
nÑ8
txn u8
n=1 converges to x.
Remark 2.63. Let (M, d) be a metric space, A Ď M be a subset.
x is a limit point of A ô @ ε ą 0, D(x, ε) X A ‰ H.
1
n
1
ô @ n ą 0, D xn P A, d(xn , x) ă .
n
ô @ n ą 0, D xn P A, xn P D(x, ).
ô D txn u8
n=1 Ď A Q xn Ñ x as n Ñ 8
68
CHAPTER 2. Point-Set Topology of Metric spaces
and
y is an accumulation point of A ô @ ε ą 0, D(y, ε) X (Aztyu) ‰ H.
1
n
1
ô @ n ą 0, D yn ‰ y, yn P A, d(yn , y) ă .
n
ô @ n ą 0, D yn ‰ y, yn P A, yn P D(y, ).
ô D tyn u8
n=1 Ď Aztyu Q yn Ñ y as n Ñ 8.
s ô If txn u8 Ď A and xn Ñ x as n Ñ 8,
Remark 2.64. A is closed ô A = cl(A) = A
n=1
then x P A.
Remark 2.65. The sequence txk u8
k=1 does not converge to x as k Ñ 8 if
ˇ
␣
(
D ε ą 0 Q @ N ą 0, D k ě N Q d(xk , x) ě ε ô D ε ą 0 Q # n P N ˇ d(xn , x) ě ε = 8
ˇ
␣
(
ô D ε ą 0 Q # n P N ˇ xn R D(x, ε) = 8.
Proposition 2.66. In Rn , a sequence of vectors converges if and only if every component
of the vectors converges. In other words, in Rn
Componentwise convergence ô Convergence.
(1)
(2)
(n)
n
Proof. Let tvk u8
k=1 , vk = (vk , vk , ¨ ¨ ¨ , vk ), be a sequence of vectors in R .
“ñ” Suppose vk Ñ v = (v (1) , ¨ ¨ ¨ , v (n) ) as k Ñ 8. Then
@ ε ą 0, D N ą 0 Q }vk ´ v}2 ă ε whenever k ě N ;
thus if k ě N ,
(i)
|vk ´ v (i) | ď }vk ´ v}2 =
b
(1)
(n)
(vk ´ v (1) )2 + ¨ ¨ ¨ + (vk ´ v (n) )2 ă ε.
(i)
“ð” Assume that vk Ñ ui as k Ñ 8 for i = 1, 2, ¨ ¨ ¨ , n. Then
ε
(i)
@ ε ą 0, D Ni ą 0, Q |vk ´ ui | ă ? whenever k ě Ni .
n
Let N = maxtN1 , N2 , ¨ ¨ ¨ , Nn u. Then if k ě N ,
c
b
ε2
ε2
(1)
(n)
}vk ´ u}2 = (vk ´ u1 )2 + ¨ ¨ ¨ + (vk ´ un )2 ă
+ ¨¨¨ +
= ε.
n
n
˝
§2.3 Sequences and Completeness（完備性）
Example 2.67. Let vk =
c
69
(1 1 )
, 2 P R2 . Then vk Ñ (0, 0) as k Ñ 8 since
k k
1
1
1?
( ´ 0)2 + ( 2 ´ 0)2 = 2 k 2 + 1 Ñ 0 as k Ñ 8 .
k
k
k
8
Proposition 2.68. Suppose that tvk u8
k=1 and twk uk=1 are sequences of vectors in a normed
space (V, } ¨ }), λk is a sequence in R, and vk Ñ v, wk Ñ w in V, λk Ñ λ in R as k Ñ 8.
Then
1. vk + wk Ñ v + w as k Ñ 8.
2. λk vk Ñ λv as k Ñ 8.
3.
1
1
vk Ñ v as k Ñ 8 if λk ‰ 0, λ ‰ 0.
λk
λ
Proposition 2.69. Let (M, d) be a metric space.
1. A set A Ď M is closed if and only if every convergent sequence txk u8
k=1 Ď A converges
to a limit in A.
s if and only if there is a sequence txk u8 Ď A, Q xk Ñ x as k Ñ 8
2. x P A
k=1
s Let txk u8 Ď A be a
Proof. “ñ” Since A is closed, Theorem 2.38 implies that A = A.
k=1
convergent sequence with limit x. Then
@ ε ą 0, D N ą 0 Q d(xk , x) ă ε whenever k ě N.
Therefore,
@ ε ą 0, D(x, ε) X A Ě txk u8
k=N ‰ H
s A).
which implies that x P A(=
“ð” Assume the contrary that A is not closed. Then
D x P AA Q @ ε ą 0, D(x, ε) Ę AA .
Let ε =
( 1)
1
, xn P D x,
X A. Then txn u8
n=1 Ď A and xn Ñ x as n Ñ 8; thus we
n
n
obtain a sequence txn u8
n=1 which converges to a point x R A, a contradiction.
˝
70
CHAPTER 2. Point-Set Topology of Metric spaces
Example 2.70. Suppose txk u Ď Rn is such that (i) }xk } ď 1 (ii) xk Ñ x as k Ñ 8.
Question 1: }x} ď 1?
Question 2: Can ď be replaced by ă; that is, is it true that }xk } ă 1, xk Ñ x as k Ñ 8,
then }x} ă 1?
ˇ
␣
(
Answer to Question 1: Yes, consider B(0, 1) = x P Rn ˇ }x} ď 1 . Then B is closed
since if x P B A , D ε = }x}2 ´ 1 ą 0 Q D(x, ε) Ď B A . Since txk u8
k=1 Ď B and xk Ñ x as
k Ñ 8, by Proposition 2.69 x P B; thus }x} ď 1.
On the other hand, we can obtain the inequality above by the triangle inequality:
}x}2 ď }xk ´ x}2 + }xk }2 ď }xk ´ x}2 + 1 @ k ą 0 ñ }x}2 ď lim }xk ´ x}2 + 1 = 1.
kÑ8
1
Answer to Question 2: No. For example, consider the case n = 1, and take xk = 1 ´ .
n
Then |xk | ă 1 and xk Ñ x = 1 as k Ñ 8. However, |x| = 1 ­ă 1.
Definition 2.71. A point x in a metric space is said to be a cluster point of a sequence
txn u8
n=1 if
ˇ
␣
(
@ ε ą 0, # n P N ˇ xn P D(x, ε) = 8 .
Proposition 2.72. If txn u8
n=1 is a sequence in a metric space (M, d), then
1. x is a cluster point of txn u8
n=1 if and only if @ ε ą 0 and N ą 0, D n ě N Q d(xn , x) ă ε.
8
2. x is a cluster point of txn u8
n=1 if and only if D txnj uj=1 Q xnj Ñ x as j Ñ 8.
3. xn Ñ x as n Ñ 8 if and only if every subsequence of txn u8
n=1 converges to x.
4. xn Ñ x as n Ñ 8 if and only if every proper subsequence of txn u8
n=1 has a further
subsequence that converges to x.
Proof. See the proof of Proposition 1.109 by changing | ¨ ´ ¨ | to d(¨, ¨).
˝
Theorem 2.73. The collection of cluster points of a sequence is closed.
8
Proof. Let txk u8
k=1 Ď M be a sequence, and A be the collection of cluster points of txk uk=1 .
If y P AA , then y is not a cluster point of txk u8
k=1 ; thus
ˇ
␣
(
D ε ą 0 Q # n P N ˇ xn P D(y, ε) ă 8 .
If z P D(y, ε), let r = ε ´ d(y, z) ą 0, then D(z, r) Ď D(y, ε) (Check!). As a consequence,
ˇ
␣
(
# n P N ˇ xn P D(z, r) ă 8.
§2.3 Sequences and Completeness（完備性）
71
ε
z
y
ε ´ d(y, z)
Figure 2.3: D(z, ε ´ d(y, z)) Ď D(y, ε) if z P D(y, ε)
Therefore, z P AA which implies that D(y, ε) Ď AA ; thus A is closed.
˝
Next we talk about the completeness of a metric space. Recall that the completeness
of an order field is defined by the monotone sequence property (or the least upper bound
property) which relies on the concept of order, so we cannot define the completeness of a
metric space via these two properties. On the other hand, Theorem 1.103 suggests that
when the concept of order is out of scope, the convergence of all Cauchy sequences seems
a good replacement for completeness. This is in fact how we define the completeness of
general metric spaces. To be more precise, we start with the following
Definition 2.74. Let (M, d) be a metric space. A sequence txk u8
k=1 Ď M is said to be
Cauchy if
@ ε ą 0, D N ą 0 Q d(xn , xm ) ă ε whenever n, m ě N.
Definition 2.75. A metric space (M, d) is said to be complete if every Cauchy sequence
in M converges to a limit in M .
Definition 2.76. A Banach space is a complete normed vector space.
Definition 2.77. A sequence txk u8
k=1 in a normed space (V, } ¨ }) is said to be bounded if
D B ą 0 Q }xk } ď B @ k P N.
Definition 2.78. A sequence txk u8
k=1 in a metric space (M, d) is said to be bounded if
D x0 P M and B ą 0 Q d(xk , x0 ) ď B @ k P N.
Remark 2.79. Adopting the definition of boundedness in a metric space, a sequence txk u8
k=1
in a nomed space (V, } ¨ }) is bounded if
D x0 P V and B ą 0 Q }xk ´ x0 } ď B @ k P N;
r Therefore, Definition 2.78 implies Definition 2.77.
thus }xk } ď }x0 } + B ” B.
72
CHAPTER 2. Point-Set Topology of Metric spaces
Proposition 2.80. A convergent sequence in (M, d) is bounded.
Proof. Let txk u8
k=1 be a convergent sequence in M with limit x0 . Then
@ ε ą 0, D N ą 0 Q d(xk , x0 ) ă ε whenever k ě N .
␣
(
Let C = max d(x1 , x0 ), d(x2 , x0 ), ¨ ¨ ¨ , d(xN ´1 , x0 ), ε + 1. Then d(xk , x0 ) ď C @ k P N.
˝
Proposition 2.81.
1. Every convergent sequence in (M, d) is Cauchy.
2. If a subsequence of Cauchy sequence converges, then this Cauchy sequence also converges.
Proof. See the proof of Proposition 1.96 and Lemma 1.101 by changing | ¨ | to d(¨, ¨).
˝
(
Theorem 2.82. A sequence in Rn converges if and only if the sequence is Cauchy because
)
?
(i)
(i)
of that max |vk ´ ui | ď }vk ´ u}2 ď n max |vk ´ ui | .
1ďiďn
1ďiďn
Theorem 2.83. Let (M, d) be a complete metric space, and N Ď M be a closed subset.
Then (N, d) is complete (完備空間中之閉集合是完備的).
Proof. Let txk u8
k=1 Ď N be Cauchy sequence. Then
@ ε ą 0, D N0 ą 0 Q d(xn , xm ) ă ε if n, m ě N0 .
Therefore, txk u8
k=1 is Cauchy in (M, d). By completeness of (M, d), D x P M Q xk Ñ x as
k Ñ 8. Note that x P N since N is closed.
˝
Theorem 2.84. Let (M, d) be a metric space, and A is dense subset of M ; that is, A Ď
s If every Cauchy sequence in A converges in M , then (M, d) is complete.
M Ď A.
be a Cauchy sequence in M . Since A is dense in M , for each n P N there
Proof. Let txn u8
␣ (n) (8 n=1
(n)
exists xk k=1 such that xk Ñ xn as k Ñ 8, and for each j P N, there exists N (j) ą 0
such that
( (n)
) 1
d xk , x n ă
j
@ k ě N (j) .
(k)
Let yk = xN (k) . Then
( (k)
)
(
1 1
(ℓ)
d(yk , yℓ ) ď d xN (k) , xk + d(xk , xℓ ) + d xℓ , xN (ℓ) ) ă + + d(xk , xℓ )
k ℓ
§2.4 Series of Real Numbers and Vectors
73
8
which implies that tyn u8
n=1 is a Cauchy sequence (for lim d(yk , yℓ ) = 0). Since tyn un=1 Ď A,
k,ℓÑ8
by assumption it converges to some point x P M ; thus for a given ε ą 0, there exists K ą 0
such that
( (n)
)
ε
d xN (n) , x = d(yn , x) ă
@n ě K .
2
␣
(
ε
1
Choose J ą 0 such that ă . Then if n ě max K, N (J) ,
J
2
(
( (n)
1 ε
(n)
d(xn , x) ď d xn , xN (n) ) + d xN (n) , x) ă + ă ε .
J 2
˝
2.4 Series of real numbers and vectors
Definition 2.85. Let (V, } ¨ }) be a normed space. A series
said to converge to S P V if the partial sum Sn =
n
ř
8
ř
xk , where txk u8
k=1 Ď V, is
k=1
xk converges to S, and one writes
k=1
S=
8
ř
xk if this is the case.
k=1
Theorem 2.86. Let (V, } ¨ }) be a complete normed space (called Banach space). A series
8
ř
xk converges if and only if
k=1
@ ε ą 0, D N ą 0 Q }xk + xk+1 + ¨ ¨ ¨ + xk+p } ă ε if k ě N, p ě 0.
Proof. Let Sn =
n
ř
xk be partial sum of
k=1
8
ř
xk . Then
k=1
8
tSn u8
n=1 converges in V ô tSn un=1 is Cauchy
ô @ ε ą 0, D N ą 0 Q }Sn ´ Sm } ă ε if n, m ě N
ô @ ε ą 0, D N ą 0 Q }xn+1 + xn+2 + ¨ ¨ ¨ + xm } ă ε if m ą n ě N
˝
ô @ ε ą 0, D N ą 0 Q }xk + xk+1 + ¨ ¨ ¨ + xk+p } ă ε if k ě N + 1, p ě 0.
8
ř
Corollary 2.87. If
xk converges, then }xk } Ñ 0 as k Ñ 8, and if }xk } Ñ̂ 0 as k Ñ 8,
k=1
then
8
ř
xk diverges.
k=1
Proof. Take p = 0 in Theorem 2.86.
˝
8
8
ř
ř
Definition 2.88. A series
xk is said to converge absolutely if
}xk } converges in
k=1
k=1
R. A series that is convergent but not absolutely convergent is said to be conditionally
convergent.
74
CHAPTER 2. Point-Set Topology of Metric spaces
8 (´1)k
ř
Example 2.89.
k=1
k
is conditionally convergent.
8
8
ř
ř
xk converges absolutely, then
xk
Theorem 2.90. In a complete normed space, if
k=1
k=1
converges.
Proof. If
8
ř
xk converges absolutely, then Sn =
k=1
n
ř
}xk } converges in R. Then
k=1
ˇ
ˇ
@ ε ą 0, D N ą 0 Q ˇ}xk } + }xk+1 } + ¨ ¨ ¨ + }xk+p }ˇ ă ε if k ě N, p ě 0.
Therefore, if k ě N, p ě 0,
}xk + xk+1 + ¨ ¨ ¨ + xk+p } ď }xk } + ¨ ¨ ¨ + }xk+p } ă ε .
Theorem 2.91. 1. Geometric series:
(a) If |r| ă 1, then
8
ř
rk converges absolutely to
k=1
(b) If |r| ą 1, then
8
ř
r
.
1´r
rk does not converge (diverge).
k=1
2. Comparison test:
(a) If
8
ř
k=1
(b) If
8
ř
8
ř
ak converges, ak ě 0, and 0 ď bk ď ak , then
bk converges.
k=1
ak diverges, ak ě 0, and ak ď bk , then
k=1
8
ř
bk diverges.
k=1
3. p-series:
8 1
ř
converges if p ą 1 and diverges if p ď 1.
p
k=1 k
4. Root test:
8
a
ř
(a) If lim sup k |xk | ă 1, then
xk converges absolutely.
kÑ8
k=1
8
a
ř
(b) If lim sup k |xk | ą 1, then
xk diverges.
kÑ8
k=1
5. Ratio and comparison test:
8
8
ř
ř
Let
ak and
bk be series, and bk ą 0 for all k P N.
k=1
k=1
˝
§2.4 Series of Real Numbers and Vectors
(a) lim sup
kÑ8
(b) lim inf
kÑ8
75
8
8
ř
ř
|ak |
ă 8,
bk is convergent, then
ak converges absolutely.
bk
k=1
k=1
8
8
ř
ř
ak
ą 0,
bk is divergent, then
ak diverges.
bk
k=1
k=1
6. Integral test:
If f is continuous, non-negative, and monotone decreasing on [1, 8), then
converges if and only if the improper integral
8
ř
f (k)
k=1
ż8
f (x)dx ă 8.
1
7. Alternative series:
8
ÿ
(´1)k ak is convergent if ak ě 0, ak Œ 0 (that is, ak ě ak+1 , ak Ñ 0 as k Ñ 8).
k=1
Remark 2.92. By Problem 1.17,
lim inf
kÑ8
a
a
|xk+1 |
|xk+1 |
ď lim inf k |xk | ď lim sup k |xk | ď lim sup
.
kÑ8
|xk |
|xk |
kÑ8
kÑ8
As a consequence, by the root test we obtain
1. if lim sup
kÑ8
2. if lim inf
kÑ8
8
ř
|xk+1 |
xk converges absolutely, and
ă 1, the series
|xk |
k=1
8
ř
|xk+1 |
xk diverges.
ą 1, the series
|xk |
k=1
This is called the ratio test.
Example 2.93. Let
xk =
$
’
’
&
1
2
1
’
’
%
that is, txk u8
k=1 =
1. lim inf
kÑ8
!
k
32
if k is odd,
if k is even,
1 1 1 1 1 1
, , , , , , ¨ ¨ ¨ , be a sequence in R. Then
2 3 4 9 8 27
)
|xk+1 |
= 0;
|xk |
a
1
2. lim inf k |xk | = ? ;
kÑ8
k+1
2
3
76
CHAPTER 2. Point-Set Topology of Metric spaces
a
1
3. lim sup k |xk | = ? ;
2
kÑ8
4. lim sup
kÑ8
Therefore,
|xk+1 |
= 8.
|xk |
8
ř
xk converges absolutely.
k=1
2.5 Exercises
§2.1 Open Sets and the Interior of Sets
Problem 2.1. Show that every open set in R is the union of at most countable collection
of disjoint open intervals; that is, if U Ď R is open, then
U=
ď
(ak , bk ) ,
kPI
where I is countable, and (ak , bk ) X (aℓ , bℓ ) = H if k ‰ ℓ.
Problem 2.2. Let (M, d) be a metric space, and A Ď M . An open cover of A is a collection
of open sets whose union contains A; that is, tUi uiPI is called an open cover of A if
1. Ui is open for all i P I.
2. A Ď
Ť
Ui .
iPI
Show that
␣
(8
1. if (ak , bk ) k=1 is an open cover of [a, b] Ď R, then there exists N ą 0 such that
N
Ť
(ak , bk ) Ě [a, b].
k=1
2. Using Exercise 2.1 to show that if tUk u8
k=1 is an open cover of [a, b], then there exists
N
Ť
N ą 0 such that
Uk Ě [a, b].
k=1
Problem 2.3. Let A and B be subsets of a metric space (M, d). Show that
1. int(int(A)) = int(A).
2. int(A X B) = int(A) X int(B).
§2.5 Exercises
77
§2.2 Closed Sets, the Closure of Sets, and the Boundary of Sets
s is
Problem 2.4. Let (M, d) be a metric space, and A Ď M . Show (by definition) that A
closed .
1
s
Problem 2.5. Let (M, d) be a metric space, and A Ď M . Show that A1 = Az(AzA
). In
other words, the derived set consists of all limit points that are not isolated points. Also
s 1 = AzA1 .
show that AzA
Problem 2.6. Let A Ď Rn . Define the sequence of sets A(m) as follows: A(0) = A and
A(m+1) = the derived set of A(m) for m P N. Do the following problems.
1. Prove that each A(m) for m P N is a closed set; thus A(1) Ě A(2) Ě ¨ ¨ ¨ .
2. Show that if there exists some m P N such that A(m) is a countable set, then A is
countable.
3. For any given m P N, is there a set A such that A(m) ‰ H but A(m+1) = H.
4. Let A be uncountable. Then each A(m) is an uncountable set. Is it possible that
8
Ş
A(m) = H?
m=1
5. Let A =
!
1
1ˇ
+ ˇ m ´ 1 ą k(k ´ 1), m, k P N . Find A(1) , A(2) and A(3) .
m
k
ˇ
)
Problem 2.7. Let A and B be subsets of a metric space (M, d). Show that
1. cl(cl(A)) = cl(A).
2. cl(A Y B) = cl(A) Y cl(B).
3. cl(A X B) Ď cl(A) X cl(B). Find examples of that cl(A X B) Ĺ cl(A) X cl(B).
Problem 2.8. Let (M, d) be a metric space, and A Ď M be a subset. Show that
(
) (
)
BA = A X cl(M zA) Y cl(A)zA .
Problem 2.9. Let A and B be subsets of a metric space (M, d). Show that
1. BA = B(M zA).
78
CHAPTER 2. Point-Set Topology of Metric spaces
2. B(BA) Ď B(A). Find examples of that B(BA) Ĺ BA. Also show that B(BA) = BA if A
is closed.
3. B(A Y B) Ď BA Y BB Ď B(A Y B) Y A Y B. Find examples of that equalities do not
hold.
4. If cl(A) X cl(B) = H, then B(A Y B) = BA Y BB.
5. B(B(BA)) = B(BA).
Problem 2.10. Let (M, d) be a metric space, and A Ď M be a subset. Determine which
of the following statements are true.
1. intA = AzBA.
2. cl(A) = M zint(M zA).
3. int(cl(A)) = int(A).
4. cl(int(A)) = A.
5. B(cl(A)) = BA.
6. If A is open, then BA Ď M zA.
7. If A is open, then A = cl(A)zBA. How about if A is not open?
Problem 2.11. Let (M, d) be a metric space. A set A Ď M is said to be perfect if A = A1 .
The Cantor set is constructed by the following procedure: let E0 = [0, 1]. Remove the
(1 2)
segment , , and let E1 be the union of the intervals
3 3
[
0,
1] [2 ]
, ,1 .
3
3
Remove the middle thirds of these intervals, and let E2 be the union of the intervals
[ 1] [2 3] [6 7] [8 ]
0, , , , , , , 1 .
9
9 9
9 9
9
Continuing in this way, we obtain a sequence of closed set Ek such that
(a) E1 Ě E2 Ě E2 Ě ¨ ¨ ¨ ;
§2.5 Exercises
79
(b) En is the union of 2n intervals, each of length 3´n .
The set C =
8
Ş
En is called the Cantor set.
n=1
1. Show that C is a perfect set; that is, C = C 1 .
2. Show that C is uncountable.
3. Find int(C).
Problem 2.12. In a metric space (M, d), if subsets satisfy A Ď S Ď cl(A), then A is said
to be dense in S. For example, Q is dense in R.
1. Show that if A is dense in S and if S is dense in T , then A is dense in T .
2. Show that if A is dense in S and B Ď S is open, then B Ď cl(A X B).
§2.3 Sequences and Completeness
Problem 2.13. Show that
1. Every convergent sequence in a metric space is a Cauchy sequence.
2. If a subsequence of a Cauchy sequence converges to x, then the sequence converges to
x.
3. x is a cluster point of txk u8
k=1 if and only if @ ε ą 0 and N ą 0, D k ą N with
d(xk , x) ă ε.
4. x is a cluster point of txk u8
k=1 if and only if there is a subsequence converging to x.
5. xk Ñ x as k Ñ 8 if and only if every subsequence of txk u8
k=1 converges to x.
6. xk Ñ x as k Ñ 8 if and only if every proper subsequence of txk u8
k=1 has a further
subsequence that converges to x.
Problem 2.14. Let (M, d) be a metric space, and N Ď M . Show that if (N, d) is complete,
then N is closed.
Remark: Theorem 2.83 states that if (M, d) is a complete metric space and N is a closed
subset of M , then (N, d) is complete. This problem gives a reverse statement.
80
CHAPTER 2. Point-Set Topology of Metric spaces
$
n+1
’
’
if n is odd ,
&
2n
Problem 2.15. Let an be defined by an =
Compute the value
n
’
’
if
n
is
even
.
% 3n
?
?
a
a
of lim inf n an , lim sup n an , lim inf n+1 and lim sup n+1 , and conclude that whether the
nÑ8
8
ř
an
nÑ8
nÑ8
an
nÑ8
series
an is convergent or not.
?
?
Hint: You can use lim n n = lim n n + 1 = 1 without proving it.
n=1
nÑ8
nÑ8
1
Problem 2.16. Let α P R, α ą . Discuss the absolute convergence or the conditional
3
8
ÿ
(´1)k
convergence of the series
.
α
k
k=2
k + (´1)
Problem 2.17. Determine whether the following series converge or not. Also test for their
absolute convergence.
1.
8
ÿ
sin(n´α ), α ą 0;
n=1
2.
8
ÿ
log(n + 1) ´ log n
arctan n2
n=1
3.
8
ÿ
a(a + 1) ¨ ¨ ¨ (a + n ´ 1)b(b + 1) ¨ ¨ ¨ (b + n ´ 1)
1 ¨ 2 ¨ ¨ ¨ n ¨ c(c + 1) ¨ ¨ ¨ (c + n ´ 1)
n=1
4.
8
ÿ
(´1)n (
n=1
5.
;
8
ÿ
n=2
n+1
?
1+
;
1
1 )
;
+ ¨¨¨ +
3
2n + 1
(´1)n
;
n + (´1)n
The a, b, c in (3) are not negative integers.
Problem 2.18. Let tan u8
n=1 Ď R ba a sequence. A series
8
ř
bn is said to be a rearrangement
n=1
8
ř
of the series
an if there exists a rearrangement π of N; that is, π : N Ñ N is bijective,
n=1
such that bn = aπ(n) . Show that if
8
ř
an converges absolutely, then any rearrangement of
n=1
the series
8
ř
n=1
an converges and has the value
8
ř
n=1
an .
§2.5 Exercises
81
§3.1 Compactness
Problem 2.19. Let (M, d) be a metric space.
1. Show that the union of a finite number of compact subsets of M is compact.
2. Show that the intersection of an arbitrary collection of compact subsets of M is compact.
Problem 2.20. A metric space (M, d) is said to be separable if there is a countable subset
A which is dense in M . Show that every compact set is separable.
Problem 2.21. Given tak u8
k=1 Ď R a bounded sequence. Define
ˇ
␣
␣ (8
(
A = x P R ˇ there exists a subsequence akj j=1 such that lim akj = x .
jÑ8
Show that A is a non-empty compact set in R. Furthermore , lim sup ak = sup A and
lim inf ak = inf A.
kÑ8
kÑ8
Problem 2.22. Let (M, d) be a compact metric space; that is, M itself is a compact set.
8
Ť
Fk ‰ H.
If tFk u8
k=1 is a sequence of closed sets such that int(Fk ) = H, then M z
k=1
Problem 2.23. Let d : R2 ˆ R2 Ñ R be defined by
#
|x1 ´ y1 |
if x2 = y2 ,
d(x, y) =
where x = (x1 , x2 ) and y = (y1 , y2 ).
|x1 ´ y1 | + |x2 ´ y2 | + 1 if x2 ‰ y2 .
1. Show that d is a metric on R2 . In other words, (R2 , d) is a metric space.
2. Find D(x, r) with r ă 1, r = 1 and r ą 1.
3. Show that the set tcu ˆ [a, b] Ď (R2 , d) is closed and bounded.
4. Examine whether the set tcu ˆ [a, b] Ď (R2 , d) is compact or not.
Problem 2.24. Let (M, d) be a complete metric space, and A Ď M be totally bounded.
Show that cl(A) is compact.
Problem 2.25. Let txk u8
k=1 be a convergent sequence in a metric space, and xk Ñ x as
k Ñ 8. Show that the set A ” tx1 , x2 , ¨ ¨ ¨ , u Y txu is compact by
82
CHAPTER 2. Point-Set Topology of Metric spaces
1. showing that A is sequentially compact; and
2. showing that every open cover of A has a finite subcover; and
3. showing that A is totally bounded and complete.
Problem 2.26. Let Y be the collection of all sequences tyk u8
k=1 Ď R such that
8
ř
|yk |2 ă 8.
k=1
In other words,
Y =
␣
ˇ
tyk u8
k=1 yk P R for all k P N,
ˇ
8
ÿ
(
|yk |2 ă 8 .
k=1
Define } ¨ } : Y Ñ R by
8
)1
›
› (ÿ
2 2
›tyk u8
›
=
|y
|
.
k
k=1
k=1
1. Show that } ¨ } is a norm on Y . The normed space (Y, } ¨ }) usually is denoted by ℓ2 .
2. Show that } ¨ } is induced by an inner product.
3. Show that (Y, } ¨ }) is complete.
ˇ
␣
(
4. Let B = y P Y ˇ }y} ď 1 . Is E compact or not?
Problem 2.27 (True or False). Determine whether the following statements are true or
false. If it is true, prove it. Otherwise, give a counter-example.
1. Every open set in a metric space is a countable union of closed sets.
2. Let A Ď R be bounded from above, then sup A P A1 .
3. An infinite union of distinct closed sets cannot be closed.
4. An interior point of a subset A of a metric space (M, d) is an accumulation point of
that set.
5. Let (M, d) be a metric space, and A Ď M . Then (A1 )1 = A1 .
6. There exists a metric space in which some unbounded Cauchy sequence exists.
7. Every metric defined in Rn is induced from some “norm” in Rn .
Chapter 3
Compact and Connected Sets
3.1 Compactness（緊緻性）
Definition 3.1. Let (M, d) be a metric space. A subset K Ď M is called sequentially
compact if every sequence in K has a subsequence that converges to a point in K.
Example 3.2. Any closed and bounded set in (R, | ¨ |) is sequentially compact.
8
Proof. Let txk u8
k=1 be a sequence in a closed and bounded set S. Then txk uk=1 is also
␣ (8
bounded; thus by Bolzano-Weierstrass property of R, there exists a subsequence xkj j=1
converging to a point x P R. Since S is closed, x P S; thus S is sequentially compact.
˝
Proposition 3.3. Let (M, d) be a metric space, and K Ď M be sequentially compact. Then
K is closed and bounded.
Proof. For closedness, assume that txk u8
Ď K and xk Ñ x as k Ñ 8. By the definition of
␣ k=1
(8
sequential compactness, there exists xkj j=1 converging to a point y P K. By Proposition
2.72, x = y; thus x P K.
For boundedness, assume the contrary that @ (x0 , B) P M ˆ R+ , there exists y P K such
that d(x0 , y) ą B. In particular, there exists
xk P K, d(xk , x0 ) ą 1 + d(xk´1 , x0 )
83
@ k P N.
84
CHAPTER 2. Point-Set Topology of Metric spaces
1
•x1
•
x0
•x2
1
•x3
Then any subsequence of txk u8
k=1 cannot be Cauchy since d(xk , xℓ ) ą 1 for all k, ℓ P N; thus
txk u8
k=1 has no convergent subsequence, a contradiction.
˝
Remark 3.4. Example 3.2 and Proposition 3.3 together suggest that in (R, | ¨ |),
sequentially compact ô closed and bounded .
Corollary 3.5. If K Ď R is sequentially compact, then inf K P K and sup K P K.
Proof. By Proposition 3.3, K must be closed and bounded. Therefore, inf K P R. Then
1
for each n P N, there exists xn P K such that inf K ď xn ă inf K + . Since txn u8
n=1 is
n
a bounded sequence in R, the Bolzano-Weierstrass theorem (Theorem 1.100) implies that
␣ (8
there is a subsequence xnk k=1 and x P R such that lim xnk = x. Note that x = inf K,
kÑ8
and by the closedness of K, x P K. The proof of sup K P K is similar.
˝
Definition 3.6. Let (M, d) be a metric space, and A Ď M . A cover of A is a collection of
␣ (
sets Uα αPI whose union contains A; that is,
ď
AĎ
Uα .
αPI
It is an open cover of A if Uα is open for all α P I. A subcover of a given cover is a
␣
␣ (
sub-collection Uα uαPJ of Uα αPI whose union also contains A; that is,
ď
AĎ
Uα , J Ď I .
αPJ
It is a finite subcover if #J ă 8.
Definition 3.7. Let (M, d) be a metric space. A subset K Ď M is called compact if every
open cover of K possesses a finite subcover; that is, K Ď M is compact if
ď
␣ (
@ open cover Uα αPI of K, D J Ď I, #J ă 8 Q K Ď
Uα .
αPJ
§3.1 Compactness
85
Example 3.8. Consider R ˆ t0u in the normed space (R2 , } ¨ }2 ).
␣ (
)(
D (x, 0), 1 xPR is an open cover of R ˆ t0u; that is,
ď (
)
R ˆ t0u Ď
D (x, 0), 1 .
For x P R, then
xPR
(
)
D (x, 0), 1
•
(x, 0)
Figure 3.1: An open cover of the x-axis
However, there is no finite subcover; thus R ˆ t0u is not compact.
(1 )
Example 3.9. Consider (0, 1] in the normed space (R, | ¨ |). Let Ik = , 2 . Then tIk u8
k=1
k
is an open cover of (0, 1]; that is,
(0, 1] Ď
8
ď
(1
k=1
k
)
,2 .
However, there is no finite subcover since
ď (1 )
1
R
,2 .
N +1
k
N
k=1
Therefore, (0, 1] is not compact.
Lemma 3.10. Let (M, d) be a metric space, and K Ď M be compact. Then K is closed.
In other words, compact subsets of metric spaces are closed.
Proof. Suppose the contrary that D txk u8
k=1 Ď K, xk Ñ x as k Ñ 8, but x R K. For y P K,
define the open ball Uy by
( 1
)
Uy = D y, d(x, y) .
2
␣ (
Ť
Then Uy yPK is an open cover of K; that is, K Ď
Uy . Since K is compact, there exist
yPK
ty1 , ¨ ¨ ¨ , yn u Ď K such that
KĎ
n
ď
i=1
U yi =
( 1
)
D yi , d(x, yi ) .
2
i=1
n
ď
86
CHAPTER 2. Point-Set Topology of Metric spaces
Let r =
␣
(
1
min d(x, y1 ), ¨ ¨ ¨ , d(x, yn ) ą 0. Then if d(x, z) ă r,
2
1
1
d(z, yi ) ě d(x, yi ) ´ d(x, z) ą d(x, yi ) ´ r ą d(x, yi ) ´ d(x, yi ) = d(x, yi )
2
2
which implies that D(x, r) X Uyi = H for all i = 1, ¨ ¨ ¨ , n.
U y3
U y1
1
d(x, y1 )
2
y3
y1
1
d(x, y3 )
2
x
r
1
d(x, y2 )
2
y2
U y2
On the other hand, since xk Ñ x as k Ñ 8, D N ą 0 such that
d(xk , x) ă r
@k ě N .
In particular, xN P D(x, r) X K; thus xN R Uyi for all i = 1, ¨ ¨ ¨ , n, which contradicts to
␣ (n
that Uyi i=1 is a cover of K.
˝
Lemma 3.11. Let (M, d) be a metric space, and K Ď M be compact. If F Ď K is closed,
then F is compact. In other words, closed subsets of compact sets are compact.
␣ (
␣ (
Proof. Let Uα αPI be an open cover of F . Then Uα αPI Y tF A u is an open cover of K;
thus possessing a finite subcover of K. Therefore, we must have
KĎ
n
ď
U αi Y F A
i=1
for some αi P I. In particular, F Ď
n
Ť
Uαi Y F A , so F Ď
i=1
n
Ť
U αi .
˝
i=1
Definition 3.12. Let (M, d) be a metric space. A subset A Ď M is called totally bounded
if for each r ą 0, there exists tx1 , ¨ ¨ ¨ , xN u Ď M such that
AĎ
N
ď
i=1
D(xi , r) .
§3.1 Compactness
87
Proposition 3.13. Let (M, d) be a metric space, and A Ď M be totally bounded. Then A
is bounded. In other words, totally bounded sets are bounded.
N
Ť
Proof. By total boundedness, there exists ty1 , ¨ ¨ ¨ , yN u Ď M such that A Ď
D(yi , 1). Let
i=1
␣
(
x0 = y1 and R = max d(x0 , y2 ), ¨ ¨ ¨ , d(x0 , yN ) + 1. Then if z P A, z P D(yj , 1) for some
j = 1, ¨ ¨ ¨ , N , and
d(z, x0 ) ď d(z, yj ) + d(yj , x0 ) ă 1 + d(x0 , yj ) ď R
which implies that A Ď D(x0 , R). Therefore, A is bounded.
˝
Example 3.14. In a general metric space (M, d), a bounded set might not be totally
bounded. For example, consider the metric space (M, d) with the discrete metric, and
A Ď M be a set having infinitely many points. Then A is bounded since A Ď D(x, 2) for
any x P M ; however, A is not totally bounded since A cannot be covered by finitely many
1
2
balls with radius .
Example 3.15. Every bounded set in (Rn , }¨}2 ) is totally bounded (Check!). In particular,
the set t1u ˆ [1, 2] in (R2 , } ¨ }) is totally bounded.
On the other hand, let d : R2 ˆ R2 Ñ R be defined by
#
|x1 ´ y1 |
if x2 = y2 ,
d(x, y) =
where x = (x1 , x2 ) and y = (y1 , y2 ).
|x1 ´ y1 | + |x2 ´ y2 | + 1 if x2 ‰ y2 .
Then (R2 , d) is also a metric space (exercise). The set t1u ˆ [1, 2] is not totally bounded. In
1
2
fact, consider open ball with radius :
( 1)
1
1
y P D x,
ô d(x, y) ă ô |x1 ´ y1 | ă and x2 = y2
2
2
2
(
1
1)
ô y 1 P x1 ´ , x 1 +
and x2 = y2 .
2
In other words,
2
( 1) (
1
1)
D x,
= x1 ´ , x 1 +
ˆ tx2 u ;
2
2
2
1
2
thus one cannot cover t1u ˆ [1, 2] by the union of finitely many balls with radius .
Proposition 3.16. Let (M, d) be a metric space, and T Ď M be totally bounded. If S Ď T ,
then S is totally bounded. In other words, subsets of totally bounded sets are totally bounded.
88
CHAPTER 2. Point-Set Topology of Metric spaces
Proof. Let r ą 0 be given. By the total boundedness of T , there exists tx1 , ¨ ¨ ¨ , xN u Ď M
such that
SĎT Ď
N
ď
D(xi , r) .
˝
i=1
Proposition 3.17. Let (M, d) be a metric space, and A Ď M . Then A is totally bounded
N
Ť
if and only if @ r ą 0, Dty1 , ¨ ¨ ¨ , yN u Ď A such that A Ď
D(yi , r).
i=1
Proof. It suffices to show the “only if” part. Let r ą 0 be given. Since A is totally bounded,
D ty1 , ¨ ¨ ¨ , yN u Ď M Q A Ď
N
ď
( r)
D yi , .
i=1
2
( r)
W.L.O.G., we may assume that for each i = 1, ¨ ¨ ¨ , N , D yi ,
X A ‰ H. Then for each
2
( r)
i = 1, ¨ ¨ ¨ , N , there exists xi P D yi ,
X A which suggests that
2
AĎ
N
ď
i=1
(
r) ď
D yi ,
Ď
D(xi , r)
2
N
i=1
( r)
since D yi ,
Ď D(xi , r) for all i = 1, ¨ ¨ ¨ , N .
˝
2
Lemma 3.18. Let (M, d) be a metric space, and K Ď M . If K is either compact or
sequentially compact, then K is totally bounded..
␣
(
Proof. Suppose first that K is compact. Let r ą 0 be given, then D(x, r) xPK is an open
cover of K. Since K is compact, there exists a finite subcover; thus D tx1 , ¨ ¨ ¨ , xN u Ď K
such that
KĎ
N
ď
D(xi , r) .
i=1
Therefore, K is totally bounded.
Now we assume that K is sequentially compact. Suppose the contrary that there is an
n
Ť
r ą 0 such that any finite set ty1 , ¨ ¨ ¨ , yn u Ď K, K Ę
D(yi , r). This implies that we can
i=1
choose a sequence txk u8
k=1 Ď K such that
xk+1 P Kz
k
ď
D(xi , r) .
i=1
Then txk u8
k=1 is a sequence in K without convergent subsequence since d(xk , xℓ ) ą r for all
k, ℓ P N.
˝
§3.1 Compactness
89
Theorem 3.19. Let (M, d) be a metric space, and K Ď M . Then the following three
statements are equivalent:
1. K is compact.
2. K is sequentially compact.
3. K is totally bounded and (K, d) is complete.
Proof. We show that 1 ñ 3 ñ 2 ñ 1 to conclude the theorem.
“1 ñ 3”: By Lemma 3.18, it suffices to show the completeness of (K, d). Let txk u8
k=1 be a
8
Cauchy sequence in K. Suppose that txk uk=1 does not converge in K. Then
␣
(
@ y P K, D δy ą 0 Q # k P N | xk P D(y, δy ) ă 8
(3.1.1)
for otherwise there is a subsequence of txk u8
k=1 that converges to x which will suggests
␣
(
the convergence of the Cauchy sequence. The collection D(y, δy ) yPK then is an open
␣ (
)(N
cover of K; thus possesses a finite subcover D yi , δyi i=1 . In particular, txk u8
k=1 Ď
N
(
)
Ť
D yi , δxi or
i=1
N
ď
ˇ
(
ˇ
# k P N xk P
D(yi , δy ) = 8
␣
i
i=1
which contradicts to (3.1.1).
“3 ñ 2”: The proof of this step is similar to the proof of the Bolzano-Weierstrass Theorem
in R (Theorem 1.100) that we proceed as follows. Let txk u8
k=1 be a sequence in T0 ” K.
␣ (1)
(1) (
Since K is totally bound, there exist y1 , ¨ ¨ ¨ yN1 Ď K such that
T0 ” K Ď
N1
ď
(1)
D(yi , 1) .
i=1
(1)
One of these D(yi , 1)’s must contain infinitely many xk ’s; that is, D 1 ď ℓ1 ď N1 such
ˇ
␣
(
(1)
(1)
that # k P N ˇ xk P D(yℓ1 , 1) = 8 . Define T1 = K X D(yℓ1 , 1). Then T1 is also
␣ (2)
(2) (
totally bounded by Proposition 3.16, so there exist y1 , ¨ ¨ ¨ , yN2 Ď T1 such that
T1 Ď
N2
ď
i=1
( (2) 1 )
D yi , .
2
90
CHAPTER 2. Point-Set Topology of Metric spaces
ˇ
␣
( (2) 1 )(
Suppose that # k P N ˇ xk P D yℓ2 ,
= 8 for some 1 ď ℓ2 ď N2 . Define T2 =
2
( (2) 1 )
T1 X D yℓ2 , . We continue this process, and obtain that for all n P N,
2
␣ (n)
(n) (
(1) D y1 , ¨ ¨ ¨ , yNn Ď Tn´1 such that
Tn´1 Ď
Nn
ď
( (n) 1 )
D yi ,
.
i=1
(n)
(2) Tn = Tn´1 X D(yℓn ,
n
1)
, where 1 ď ℓn ď Nn is chosen so that
n
ˇ
␣
( (n) 1 )(
# k P N ˇ xk P D yℓn ,
= 8.
(3.1.2)
n
ˇ
ˇ
␣
␣
( (1) (
( (j) 1 )(
such that
Pick an k1 P k P N ˇ xk P D yℓ1 , 1) , and kj P k P N ˇ xk P D yℓj ,
j
kj ą kj´1 for all j ě 2. We note such kj always exists because of (3.1.2). Then
␣ (8
xkj j=1 is a subsequence of txk u8
k=1 , and xkj P Tj Ď K for all j P N.
␣ (8
Claim: xkj j=1 is a Cauchy sequence.
1
ε
Proof of claim: Let ε ą 0 be given, and N ą 0 be large enough so that
ă . Since
N
2
( (N ) 1 )
if j ě N , we must have xkj P D yℓN ,
, we conclude that if n, m ě N , by triangle
inequality
N
(
)
(
(
1
1
(N ) )
(N ) )
+
ă ε.
d xkn , xkm ď d xkn , yℓN + d xkm , yℓN ă
N
N
␣ (8
Since (K, d) is complete, the Cauchy sequence xkj j=1 converges to a point in K.
␣ (
“2 ñ 1”: Let Uα αPI be an open cover of K.
Claim: there exists r ą 0 such that for each x P K, D(x, r) Ď Uα for some α P I.
Proof of claim: Suppose the contrary that for all k ą 0, there exists xk P K such
(
1)
Ę Uα for all α P I. Then txk u8
that D xk ,
k=1 is a sequence in K; thus by the
k
␣ (8
assumption of sequential compactness, there exists a subsequence xkj j=1 converging
in K. Suppose that xkj Ñ x as j Ñ 8, and x P Uβ for some β P I. Then
(1) there is r ą 0 such that D(x, r) Ď Uβ since Uβ is open.
(2) there exists N ą 0 such that d(xkj , x) ă
r
for all j ě N .
2
§3.1 Compactness
91
Choose j ě N such that
(
1)
1
r
ă . Then D xkj ,
Ď D(x, r) Ď Uβ , a contradiction.
kj
2
kj
By Lemma 3.18, there exists tx1 , ¨ ¨ ¨ , xN u Ď K such that K Ď
N
Ť
D(xi , r). For each
i=1
1 ď i ď N , the claim above implies that there exists αi P I such that D(xi , r) Ď Uαi .
N
N
Ť
Ť
Then
D(xi , r) Ď
Uαi which suggests that
i=1
i=1
KĎ
N
ď
U αi .
˝
i=1
Remark 3.20.
1. The equivalency between 1 and 2 is sometimes called the Bolzano-Weistrass Theorem.
2. A number r ą 0 satisfying the claim in the step “2 ñ 1” is called a Lebesgue number
␣ (
for the cover Uα αPI . The supremum of all such r is called the Lebesgue number
␣ (
for the cover Uα αPI .
Alternative Proof of Theorem 3.19. In this proof we show that 1 ñ 2 ñ 3 ñ 1 to conclude
the theorem.
“1 ñ 2”: Assume the contrary that K is not sequentially compact. Then there is a sequence txk u8
k=1 Ď K that does not have a convergent subsequence with a limit in K.
Therefore, for each x P K, there exists δx ą 0 such that
ˇ
␣
(
# k P N ˇ xk P D(x, δx ) ă 8
for otherwise x is a cluster point of txk u8
k=1 so Proposition 2.72 guarantees the existence
␣
(
8
of a subsequence of txk uk=1 converging to x. Since D(x, δx ) xPK is an open cover of
K, by the compactness of K there exists ty1 , ¨ ¨ ¨ , yN u Ď K such that
txk u8
k=1 Ď K Ď
N
ď
(
)
D y i , δ yi
i=1
ˇ
␣
(
)(
while this is impossible since # k P N ˇ xk P D yi , δyi ă 8 for all i = 1, ¨ ¨ ¨ N .
“2 ñ 3”: By Lemma 3.18, it suffices to show that (K, d) is complete. Let txk u8
k=1 Ď K be
␣ (8
a Cauchy sequence. By sequential compactness of K, there is a subsequence xkj j=1
converging to a point x P K. By Proposition 2.81, txk u8
k=1 also converges to x; thus
every Cauchy sequence in (K, d) converges to a point in K.
92
CHAPTER 2. Point-Set Topology of Metric spaces
“3 ñ 1”: We first prove the following
Claim: If tVα uαPI is an open cover of a totally bounded set A such that there is no
finite subcover, then for all r ą 0, there exists x P A such that A X D(x, r) does not
admit a finite subcover.
Proof of claim: Let r ą 0 be given. Since A is totally bounded, by Proposition 3.17
N
Ť
there exists ta1 , ¨ ¨ ¨ , aN u Ď A such that A Ď
D(aj , r). If for each j = 1, ¨ ¨ ¨ , N ,
j=1
A X D(aj , r) can be covered by finitely many Vα ’s, then A itself can be covered by
finitely many Vα ’s, a contradiction. Therefore, at least one AXD(aj , r) does not admit
a finite subcover.
Now assume the contrary that there exists an open cover tUα uαPI of K such that
there is no finite subcover. Let εn = 2´n . Since K is totally bounded, by the claim
there exists x1 P K such that K X D(x1 , ε1 ) which does not admit a finite subcover.
By Proposition 3.16, K X D(x1 , ε1 ) is totally bounded, so there must be an x2 P
K X D(x1 , ε1 ) such that K X D(x1 , ε1 ) X D(x2 , ε2 ) cannot be covered by the union
of finitely many Uα . We continuous this process, and obtain a sequence txk u8
k=1 such
that
(1) xk+1 P K X
k
Ş
D(xi , εi ) (which implies that d(xk+1 , xk ) ă εk );
i=1
(2) K X
k
Ş
D(xi , εi ) cannot be covered by the union of finitely many Uα .
i=1
Then similar to Example 1.105, we find that txk u8
k=1 is a Cauchy sequence in (K, d).
By the completeness of K, xk Ñ x as k Ñ 8 for some x P K.
Since tUα uαPI is an open cover of K, x P Uβ for some β P I. Since Uβ is open,
D r ą 0 such that D(x, r) Ď Uβ . For this particular r, there exists N ą 0 such that
r
r
d(xk , x) ă . Therefore, if k ě N such that εk ă ,
2
2
D(xk , εk ) Ď D(x, r) Ď Uβ
which contradicts to (2).
˝
Example 3.21. Let (M, d) be a metric space, and txk u8
k=1 be a convergent sequence with
limit x. Let A = tx1 , x2 , ¨ ¨ ¨ u Y txu. Then A is compact.
§3.1 Compactness
93
Definition 3.22. Let (M, d) be a metric space. A subset A Ď M is called pre-compact
s is compact. Let U Ď M be an open set, a subset A of U is said to be compactly
if A
s Ď U.
contained in U, denoted by AĂĂU, if A is pre-compact and A
Example 3.23. Let (M, d) be a complete metric space, and A Ď M be totally bounded.
s is compact. In other words, in a complete metric space, totally bounded sets are
Then A
pre-compact.
(Hint: Use the total boundedness equivalence to show compactness.)
Definition 3.24. Let (M, d) be a metric space, and A Ď M . A collection of closed sets
tFα uαPI is said to have the finite intersection property for the set A if the intersection
of any finite number of Fα with A is non-empty; that is, tFα uαPI has the finite intersection
property for A if
AX
č
Fα ‰ H for all J Ď I and #J ă 8.
αPJ
Theorem 3.25. Let (M, d) be a metric space, and K Ď M . The K is compact if and only
if every collection of closed sets with the finite intersection property for K has non-empty
intersection with K; that is,
KX
č
Fα ‰ H for all tFα uαPI having the finite intersection property for K.
αPI
Proof. It can be proved by contradiction, and is left as an exercise.
˝
[
1]
Example 3.26. Let A = (0, 1) Ď R, and Kj = ´ 1, . Take Kj1 , Kj2 , ¨ ¨ ¨ , Kjn , where
j
8
n
[
Ş
Ş
1]
j1 ă j2 ă ¨ ¨ ¨ ă jn . Then
Kjℓ X A = ´ 1,
X (0, 1) ‰ H. However x P
Kj ô
ℓ=1
8
Ş
jn
j=1
8
Ş
1
Kj = [´1, 0]; thus
Kj X A = H. Therefore, (0, 1) is not
´1 ď x ď for all j P N. So
j
j=1
j=1
compact.
Example 3.27. Let X be the collection of all bounded real sequences; that is,
ˇ
␣
(
ˇ
X = txk u8
k=1 Ď R for some M ą 0, |xk | ď M for all k .
›
›
›
The number sup |xk | ” supt|x1 |, |x2 |, ¨ ¨ ¨ , |xk |, ¨ ¨ ¨ u ă 8 is denoted by ›txk u8
k=1 . For examkě1
ple, if xk =
›
›
(´1)k
›
, then ›txk u8
k=1 = 1. Then (X, } ¨ }) is a complete normed space (left as
k
94
CHAPTER 2. Point-Set Topology of Metric spaces
an exercise). Define
ˇ
␣
1(
ˇ
A = txk u8
,
k=1 P X |xk | ď
k
ˇ
␣
(
ˇ
B = txk u8
k=1 P X xk Ñ 0 as k Ñ 8 ,
ˇ
(
␣
8
ˇ
P
X
the
sequence
tx
u
converges
,
C = txk u8
k
k=1
k=1
ˇ
␣
(
ˇ
D = txk u8
(the unit sphere in (X, } ¨ })).
k=1 P X sup |xk | = 1
kě1
The closedness of A (which implies the completeness of (A, } ¨ })) is left as an exercise. We
show that A is totally bounded.
Let r ą 0 be given. Then D N ą 0 Q
ˇ
!
ˇ
E = txk u8
k=1 ˇ x1 =
1
ă r. Define
N
i1
i2
iN ´1
, x2 =
, ¨ ¨ ¨ , xN ´1 =
for some
N +1
N +1
N +1
)
i1 , ¨ ¨ ¨ , iN ´1 = ´N , ´N + 1, ¨ ¨ ¨ , N ´ 1, N , and xk = 0 if k ě N + 1 .
Then
1. #E ă 8. In fact, #E = (2N + 1)N ´1 ă 8.
2. A Ď
Ť
txk u8
k=1 PE
(
1)
D txk u8
Ď
k=1 ,
N
Ť
txk u8
k=1 PE
(
)
D txk u8
k=1 , r .
Therefore, A is totally bounded.
On the other hand, B and C are not compact since they are not bounded; thus not
totally bounded by Proposition 3.13. D is bounded but not totally bounded. In fact, D
1
cannot be covered by the union of finitely many balls with radius since each ball with
2 )8
!␣
1
(k) (8
radius contains at most one of the points from the subset xj j=1
Ď D, where for
2
k=1
each k
(k)
txj u8
0, ¨ ¨ ¨ , 0, 1, 0, ¨ ¨ ¨ u ;
j=1 = t looomooon
(k ´ 1) terms
(k)
that is, xj = δkj , the kronecker delta.
3.1.1 The Heine-Borel theorem
Theorem 3.28. In the Euclidean space (Rn , } ¨ }2 ), a subset K is compact if and only if it
is closed and bounded.
§3.1 Compactness
95
Proof. By Proposition 3.13 and Theorem 3.19, it is clear that K is closed and bounded if
K is compact (in any metric space). It remains to show the direction “ð”. Nevertheless,
by Theorem 2.83 closed subsets of a complete metric space must be complete, so it suffices
to show that a bounded set in (Rn , } ¨ }2 ) is totally bounded.
Let r ą 0 be given. By the boundedness of K, for some
M ą 0 we have }x}2 ď M for
?
nM
ă r, and define
all x P K; thus K Ď [´M, M ]n . Choose N ą 0 so that
N
ˇ
!(
␣
()
M i1
M in ) ˇ
E=
,¨¨¨ ,
ˇ i1 , i2 , ¨ ¨ ¨ , in P ´ N, ´N + 1, ¨ ¨ ¨ , N ´ 1, N .
N
N
Then #E = (2N + 1)n ă 8, and
K Ď [´M, M ]n Ď
ď
D(x, r) .
˝
xPE
n
Alternative Proof of “ð”. Let txk u8
k=1 Ď K be a sequence. Since K Ď R , we can write
(1)
(2)
(n)
(j)
xk = (xk , xk , ¨ ¨ ¨ , xk ) P Rn . Since K is bounded, then all the sequence txk u8
k=1 ,
(j)
j = 1, 2, ¨ ¨ ¨ , n, are bounded; that is, ´Mj ď xk ď Mj for all k P N. Applying the
(1)
Bolzano-Weierstrass property (Theorem 1.100) to the sequence txk u8
k=1 , we obtain a se␣ (2) (8
␣ (2) (8
␣ (1) (8
(1)
(1)
quence xkj j=1 with xkj Ñ y as j Ñ 8. Now xkj j=1 has a subsequence xkj ℓ=1
ℓ
(2)
converges, say xkj Ñ y (2) as ℓ Ñ 8.
ℓ
Continuing in this way, we obtain a subsequence of txk u8
k=1 that converges to y =
(y (1) , y (2) , ¨ ¨ ¨ , y (n) ). Since K is close, y P K; thus K is sequentially compact which is
equivalent to the compactness of K.
˝
Corollary 3.29. A bounded set A in the Euclidean space (Rn , } ¨ }2 ) is pre-compact. In
n
particular, if txk u8
k=1 is a bounded sequence in R , there exists a convergent subsequence
␣ (8
xkj j=1 (the sentence in blue color is again called the Bolzano-Weierstrass theorem).
␣ 1
(
1
Example 3.30. Let A = t0u Y 1, , ¨ ¨ ¨ , , ¨ ¨ ¨ . Then A is compact in (R, | ¨ |).
2
n
Example 3.31. Let A = [0, 1] Y (2, 3] Ď (R, | ¨ |). Since A is not closed, A is not compact.
3.1.2 The nested set property
Theorem 3.32. Let tKn u8
n=1 be a sequence of non-empty compact sets in a metric space
8
Ş
(M, d) such that Kn Ě Kn+1 for all n P N. Then there is at least one point in
Kn ; that
n=1
is,
8
č
n=1
Kn ‰ H .
96
CHAPTER 2. Point-Set Topology of Metric spaces
8
8
8
)A
(Ş
Ş
Ť
Proof. Assume the contrary that
Kn = H. Then
KnA =
Kn = M . Since KnA
n=1
n=1
n=1
␣ (8
is open, KnA n=1 is an open cover of K1 ; thus by compactness of K1 , there exists J Ď N,
#J ă 8 such that
K1 Ď
ď
KnA =
nPJ
Therefore, K1 X
Ş
(č
Kn
)A
.
nPJ
Kn = H which implies that Kmax J = H, a contradiction.
˝
nPJ
Alternative Proof. By assumption, tKn u8
n=2 has the finite intersection property for K1 . Since
K1 is compact, by Theorem 3.25,
8
č
K1 X
Kn ‰ H .
˝
n=2
Corollary 3.33. Let tUk u8
k=1 be a collection of open sets in a metric space (M, d) such that
8
Ť
Uk Ď Uk+1 for all k P N and UkA is compact. Then
Uk ‰ M .
k=1
Proof. This is proved by letting Kn = UnA , and applying Theorem 3.32.
˝
Remark 3.34. If the compactness is removed from the condition, then the intersection
might be empty. Suppose that the metric space under consideration is (R, | ¨ |).
( 1)
1. If the closedness condition is removed, then Uk = 0,
has empty intersection.
k
2. If the boundedness condition is removed, then Fk = [k, 8) has empty intersection.
3.2 Connectedness（連通性）
Definition 3.35. Let (M, d) be a metric space, and A Ď M . Two non-empty open sets U
and V are said to separate A if
1. A X U X V = H ;
2. A X U ‰ H ;
3. A X V ‰ H ;
4. A Ď U Y V .
We say that A is disconnected or separated if such separation exists, and A is connected
if no such separation exists.
Proposition 3.36. Let (M, d) be a metric space. A subset A Ď M is disconnected if and
s2 = A
s1 X A2 = H for some non-empty A1 and A2 .
only if A = A1 Y A2 with A1 X A
§3.3 Subspace Topology
97
Proof. “ñ” Suppose that there exist U, V non-empty open sets such that 1-4 in Definition
3.35 hold. Let A1 = A X U and A2 = A X V. By 1, A1 Ď V A ; thus by the definition of
s1 Ď V A . This implies that A
s1 X A2 = H. Similarly, A
s2 X A1 = H.
the closure of sets, A
sA and V = A
sA be two open sets. Then V X A1 = U X A2 = H; thus
“ð” Let U = A
2
1
A X U X V = (A1 Y A2 ) X U X V = (A1 X U) X V = U X (A1 X V) = H .
Moreover, 2-4 in Definition 3.35 also hold since A1 Ď U and A2 Ď V.
˝
Corollary 3.37. Let (M, d) be a metric space. Suppose that a subset A Ď M is connected,
s2 = A
s1 X A2 = H. Then A1 or A2 is empty.
and A = A1 Y A2 , where A1 X A
Theorem 3.38. A subset A of the Euclidean space (R, | ¨ |) is connected if and only if it
has the property that if x, y P A and x ă z ă y, then z P A.
Proof. “ñ” Suppose that there exist x, y P A, x ă z ă y but z R A. Then A = A1 Y A2 ,
where
A1 = A X (´8, z)
and A2 = A X (z, 8) .
s1 XA2 = A1 X A
s2 = H;
Since x P A1 and y P A2 , A1 and A2 are non-empty. Moreover, A
thus by Proposition 3.36, A is disconnected, a contradiction.
“ð” Suppose that A is not connected. Then there exist non-empty sets A1 and A2 such
s1 X A2 = A1 X A
s2 = H. Pick x P A1 and y P A2 . W.L.O.G.,
that A = A1 Y A2 with A
we may assume that x ă y. Define z = sup(A1 X [x, y]) .
s1 .
Claim: z P A
Proof of claim: By definition, for any n ą 0 there exists xn P A1 X [x, y] such that
1
s1 .
z ´ ă xn ď z. Therefore, xn Ñ z as n Ñ 8 which implies that z P A
n
s1 , z R A2 . In particular, x ď z ă y.
Since z P A
(a) If z R A1 , then x ă z ă y and z R A, a contradiction.
s2 ; thus D r ą 0 such that (z ´ r, z + r) Ď A
sA . Then for
(b) If z P A1 , then z R A
2
all z1 P (z, z + r), z ă z1 ă y and z1 R A2 . Then x ă z1 ă y and z1 R A, a
contradiction.
˝
98
CHAPTER 2. Point-Set Topology of Metric spaces
3.3 Subspace Topology
Let (M, d) be a metric space, and N Ď M be a subset. Then (N, d) is a metric space, and
the topology of (N, d) is called the subspace topology of (N, d).
Remark 3.39. The topology of a metric is the collection of all open sets of that metric
space.
Proposition 3.40. Let (M, d) be a metric space, and N Ď M . A subset V Ď N is open in
(N, d) if and only if V = U X N for some open set U in (M, d).
Proof. “ñ” Let V Ď N be open in (N, d). Then @ x P V, D rx ą 0 such that
ˇ
␣
(
DN (x, rx ) ” y P N ˇ d(x, y) ă rx Ď V .
Ť
In particular, V =
DN (x, rx ). Note that DN (x, r) = D(x, r) X N ; thus if U =
xPV
Ť
D(x, rx ), then U is open in (M, d), and
xPV
V=
ď
D(x, rx ) X N = U X N .
xPV
“ð” Suppose that V = U X N for some open set U in (M, d). Let x P V. Then x P U; thus
D r ą 0 such that D(x, r) Ď U. Therefore,
ˇ
␣
(
DN (x, r) ” y P N ˇ d(x, y) ă r = D(x, r) X N Ď U X N = V ;
hence V is open in (N, d).
˝
Corollary 3.41. Let (M, d) be a metric space, and N Ď M . Let (M, d) be a metric space,
and N Ď M . A subset E Ď N is closed in (N, d) if and only if E = F X N for some closed
set F in (M, d).
Definition 3.42. Let (M, d) be a metric space, and N Ď M . A subset A is said to be
open
open
closed relative to N if A X N is closed in the metric space (N, d).
compact
compact
Theorem 3.43. Let (M, d) be a metric space, and K Ď N Ď M . Then K is compact in
(M, d) if and only if K is compact in (N, d).
§3.4 Exercises
99
Proof. “ñ” Let tVα uαPI be an open cover of K in (N, d). By Proposition 3.40, there are
open sets Uα in (M, d) such that Vα = Uα X N for all α P I. Then tUα uαPI is also an
open cover of K; thus possesses a finite subcover; that is, D J Ď I, #J ă 8 such that
Ť
KĎ
Uα which, together with the fact that K Ď N , implies that
αPJ
KĎ
(ď
)
ď
ď
(Uα X N ) =
Vα .
Uα X N =
αPJ
αPJ
αPJ
“ð” Let tUα uαPI be an open cover of K in (M, d). Letting Vα = Uα X N , by Proposition
3.40 we find that tVα uαPI is an open cover of K in (N, d). Since K is compact in
Ť
(N, d), there exists J Ď I, #J ă 8 such that K Ď
Vα ; thus
αPJ
KĎ
ď
Uα .
˝
αPJ
Remark 3.44. Another way to look at Theorem 3.43 is using the sequential compactness
equivalence. Let txk u8
k=1 Ď K be a sequence. By sequential compactness of K in either
␣ (8
(M, d) or (N, d), there exists xkj j=1 and x P K such that xkj Ñ x as j Ñ 8. As long as
the metric d used in different space are identical, the concept of convergence of a sequence
are the same; thus compactness in (M, d) or (N, d) are the same.
Example 3.45. Let (M, d) be (R, | ¨ |), and N = Q. Consider the set F = [0, 1] X Q. By
Corollary 3.41 F is closed in (Q, | ¨ |). However, F is not compact in (Q, | ¨ |) since F is not
complete. We can also apply Theorem 3.43 to see this: if F Ď Q is compact in (Q, | ¨ |),
then F is compact in (R, | ¨ |) which is clearly not the case since F is not closed in (R, | ¨ |).
Remark 3.46. Let (M, d) be a metric space. By Proposition 3.36 a subset A Ď M is
disconnected if and only if there exist two subsets U1 , U2 of A, open relative to A, such that
s1 and U2 = AzA
s2 , where
A = U1 Y U2 and U1 X U2 = H (one choice of (U1 , U2 ) is U1 = AzA
A1 and A2 are given by Proposition 3.36). Note that U1 and U2 are also closed relative to
A.
Given the observation above, if A is a connected set and E is a subset of A such that E
is closed and open relative to A, then E = H or E = A.
3.4 Exercises
§3.1 Compactness
100
CHAPTER 2. Point-Set Topology of Metric spaces
Problem 3.1. Let (M, d) be a metric space.
1. Show that the union of a finite number of compact subsets of M is compact.
2. Show that the intersection of an arbitrary collection of compact subsets of M is compact.
Problem 3.2. A metric space (M, d) is said to be separable if there is a countable subset
A which is dense in M . Show that every compact set is separable.
Problem 3.3. Given tak u8
k=1 Ď R a bounded sequence. Define
ˇ
␣
␣ (8
(
A = x P R ˇ there exists a subsequence akj j=1 such that lim akj = x .
jÑ8
Show that A is a non-empty compact set in R. Furthermore , lim sup ak = sup A and
lim inf ak = inf A.
kÑ8
kÑ8
Problem 3.4. Let (M, d) be a compact metric space; that is, M itself is a compact set. If
8
Ť
Fk ‰ H.
tFk u8
k=1 is a sequence of closed sets such that int(Fk ) = H, then M z
k=1
Problem 3.5. Let d : R2 ˆ R2 Ñ R be defined by
#
|x1 ´ y1 |
if x2 = y2 ,
d(x, y) =
where x = (x1 , x2 ) and y = (y1 , y2 ).
|x1 ´ y1 | + |x2 ´ y2 | + 1 if x2 ‰ y2 .
1. Show that d is a metric on R2 . In other words, (R2 , d) is a metric space.
2. Find D(x, r) with r ă 1, r = 1 and r ą 1.
3. Show that the set tcu ˆ [a, b] Ď (R2 , d) is closed and bounded.
4. Examine whether the set tcu ˆ [a, b] Ď (R2 , d) is compact or not.
Problem 3.6. Let (M, d) be a complete metric space, and A Ď M be totally bounded.
Show that cl(A) is compact.
Problem 3.7. Let txk u8
k=1 be a convergent sequence in a metric space, and xk Ñ x as
k Ñ 8. Show that the set A ” tx1 , x2 , ¨ ¨ ¨ , u Y txu is compact by
1. showing that A is sequentially compact; and
§3.4 Exercises
101
2. showing that every open cover of A has a finite subcover; and
3. showing that A is totally bounded and complete.
Problem 3.8. Let Y be the collection of all sequences tyk u8
k=1 Ď R such that
8
ř
|yk |2 ă 8.
k=1
In other words,
8
ÿ
ˇ
␣
(
ˇ
Y = tyk u8
y
P
R
for
all
k
P
N,
|yk |2 ă 8 .
k
k=1
k=1
Define } ¨ } : Y Ñ R by
8
)1
›
› (ÿ
2 2
›tyk u8
›
=
.
|y
|
k
k=1
k=1
1. Show that } ¨ } is a norm on Y . The normed space (Y, } ¨ }) usually is denoted by ℓ2 .
2. Show that } ¨ } is induced by an inner product.
3. Show that (Y, } ¨ }) is complete.
ˇ
␣
(
4. Let B = y P Y ˇ }y} ď 1 . Is E compact or not?
Problem 3.9. Let A, B be two non-empty subsets in Rn . Define
ˇ
␣
(
d(A, B) = inf }x ´ y}2 ˇ x P A, y P B
to be the distance between A and B. When A = txu is a point, we write d(A, B) as d(x, B).
ˇ
(
␣
(1) Prove that d(A, B) = inf d(x, B) ˇ x P A .
ˇ
ˇ
(2) Show that ˇd(x1 , B) ´ d(x2 , B)ˇ ď }x1 ´ x2 }2 for all x1 , x2 P Rn .
ˇ
␣
(
(3) Define Bε = x P Rn ˇ d(x, B) ă ε be the collection of all points whose distance from
Ş
B is less than ε. Show that Bε is open and
Bε = cl(B).
εą0
(4) If A is compact, show that there exists x P A such that d(A, B) = d(x, B).
(5) If A is closed and B is compact, show that there exists x P A and y P B such that
d(A, B) = d(x, y).
(6) If A and B are both closed, does the conclusion of (5) hold?
102
CHAPTER 2. Point-Set Topology of Metric spaces
Problem 3.10. Let K(n) denote the collection of all non-empty compact sets in Rn . Define
the Hausdorff distance of K1 , K2 P K(n) by
H
d (K1 , K2 ) = max
!
)
sup d(x, K1 ), sup d(x, K2 ) ,
xPK1
xPK2
in which d(x, K) is the distance between x and K given in Problem 3.9. Show that (K(n), dH )
is a metric space.
␣
(
Problem 3.11. Let M = (x, y) P R2 | x2 + y 2 ď 1 with the standard metric } ¨ }2 . Show
that A Ď M is compact if and only if A is closed.
Problem 3.12. 1. Let txk u8
k=1 Ď R be a sequence in (R, | ¨ |) that converges to x and let
8
Ş
Ďk . Is this true in any metric space?
Ak = txk , xk+1 , ¨ ¨ ¨ u. Show that txu =
A
k=1
2. Suppose that tKj u8
j=1 is a sequence of comapct non-empty sets satisfying the nested
set property; that is, Kj Ě Kj+1 , and diameter(Kj ) Ñ 0 as j Ñ 8, where
ˇ
␣
(
diameter(Kj ) = sup d(x, y) ˇ x, y P Kj .
Show that there is exactly one point in
8
Ş
Kj .
j=1
§3.2 Connectedness
Problem 3.13. Let (M, d) be a metric space, and A Ď M . Show that A is disconnected
(not connected) if and only if there exist non-empty closed set F1 and F2 such that
1. A X F1 X F2 = H ;
2. A X F1 ‰ H ;
3. A X F2 ‰ H ;
4. A Ď F1 Y F2 .
s
Problem 3.14. Prove that if A is connected in a metric space (M, d) and A Ď B Ď A,
then B is connected.
Problem 3.15. Let (M, d) be a metric space, and A Ď M be a subset. Suppose that A is
connected and contain more than one point. Show that A Ď A1 .
Problem 3.16. Show that the Cantor set C defined in Problem 2.11 is totally disconnected;
that is, if x, y P C, and x ‰ y, then x P U and y P V for some open sets U, V separate C.
§3.4 Exercises
103
Problem 3.17. Let Fk be a nest of connected compact sets (that is, Fk+1 Ď Fk and Fk
8
Ş
is connected for all k P N). Show that
Fk is connected. Give an example to show that
k=1
compactness is an essential condition and we cannot just assume that Fk is a nest of closed
connected sets.
Problem 3.18. Let tAk u8
k=1 be a family of connected subsets of M , and suppose that A
is a connected subset of M such that Ak X A ‰ H for all k P N. Show that the union
(Ť
)
Ak Y A is also connected.
kPN
Problem 3.19. Let A, B Ď M and A is connected. Suppose that A X B ‰ H and A X B A ‰
H. Show that A X BB ‰ H.
Problem 3.20. Given (M, d) a metric space and A Ď M a non-empty subset. A maximal
connected subset of A is called a connected component of A.
1. Let a P A. Show that there is a unique connected components of A containing a.
2. Show that any two distinct connected components of A are disjoint. Therefore, A is
the disjoint union of its connected components.
3. Show that every connected component of A is a closed subset of A.
4. If A is open, prove that every connected component of A is also open. Therefore,
when M = Rn , show that A has at most countable infinite connected components.
5. Find the connected components of the set of rational numbers or the set of irrational
numbers in R.
Problem 3.21 (True or False). Determine whether the following statements are true or
false. If it is true, prove it. Otherwise, give a counter-example.
1. There exists a non-zero dimensional normed vector space in which some compact nonzero dimensional linear subspace exists.
2. There exists a set A Ď (0, 1] which is compact in (0, 1] (in the sense of subspace
topology), but A is not compact in R.
3. Let A Ď Rn be a non-empty set. Then a subset B of A is compact in A if and only if
B is closed and bounded in A.
Chapter 4
Continuous Maps
4.1 Continuity
Definition 4.1. Let (M, d) and (N, ρ) be two metric spaces, A Ď M and f : A Ñ N be a
map. For a given x0 P A1 , we say that b P N is the limit of f at x0 , written
lim f (x) = b or f (x) Ñ b as x Ñ x0 ,
xÑx0
␣
(8
if for every sequence txk u8
k=1 Ď Aztx0 u converging to x0 , the sequence f (xk ) k=1 converges
to b.
Proposition 4.2. Let (M, d) and (N, ρ) be two metric spaces, A Ď M and f : A Ñ N be a
map. Then lim f (x) = b if and only if
xÑx0
@ ε ą 0, D δ = δ(x0 , ε) ą 0 Q ρ(f (x), b) ă ε whenever 0 ă d(x, x0 ) ă δ and x P A .
Proof. “ñ” Assume the contrary that D ε ą 0 such that for all δ ą 0, there exists xδ P A
with
0 ă d(xδ , x0 ) ă δ
and ρ(f (xδ ), b) ě ε .
1
k
In particular, letting δ = , we can find txk u8
k=1 Ď Aztx0 u such that
0 ă d(xk , x0 ) ă
1
k
and ρ(f (xk ), b) ě ε .
Then xk Ñ x0 as k Ñ 8 but f (xk ) Ñ̂ b as k Ñ 8, a contradiction.
104
§4.1 Continuity
105
“ð” Let txk u8
k=1 Ď Aztx0 u be such that xk Ñ x0 as k Ñ 8, and ε ą 0 be given. By
assumption,
D δ = δ(x0 , ε) ą 0 Q ρ(f (x), b) ă ε whenever 0 ă d(x, x0 ) ă δ and x P A .
Since xk Ñ x0 as k Ñ 8, D N ą 0 Q d(xk , x0 ) ă δ if k ě N . Therefore,
ρ(f (xk ), b) ă ε
@k ě N
which suggests that lim f (xk ) = b.
˝
kÑ8
Remark 4.3. Let (M, d) = (N, ρ) = (R, | ¨ |), A = (a, b), and f : A Ñ N . We write
lim f (x) and lim´ f (x) for the limit lim f (x) and lim f (x), respectively, if the later exist.
xÑa+
xÑb
Following this notation, we have
xÑa
xÑb
lim f (x) = L ô @ ε ą 0, D δ ą 0 Q |f (x) ´ L| ă ε if 0 ă x ´ a ă δ and x P (a, b) ,
xÑa+
lim f (x) = L ô @ ε ą 0, D δ ą 0 Q |f (x) ´ L| ă ε if 0 ă b ´ x ă δ and x P (a, b) .
xÑb´
Definition 4.4. Let (M, d) and (N, ρ) be two metric spaces, A Ď M , and f : A Ñ N
be a map. For a given x0 P A, f is said to be continuous at x0 if either x0 P AzA1 or
lim f (x) = f (x0 ).
xÑx0
Example 4.5. The identity map f :
Rn Ñ Rn
is continuous at each point of Rn .
x ÞÑ x
Example 4.6. The function f : (0, 8) Ñ R defined by f (x) =
point of (0, 8).
1
is continuous at each
x
Proposition 4.7. Let (M, d) and (N, ρ) be two metric spaces, A Ď M , and f : A Ñ N be
a map. Then f is continuous at x0 P A if and only if
@ ε ą 0, D δ = δ(x0 , ε) ą 0 Q ρ(f (x), f (x0 )) ă ε whenever x P D(x0 , δ) X A .
Proof. Case 1: If x0 P A1 , then f is continuous at x0 if and only if
@ ε ą 0, D δ = δ(x0 , ε) ą 0 Q ρ(f (x), f (x0 )) ă ε whenever x P D(x0 , δ) X Aztx0 u .
Since ρ(f (x0 ), f (x0 )) = 0 ă ε, we find that the statement above is equivalent to that
@ ε ą 0, D δ = δ(x0 , ε) ą 0 Q ρ(f (x), f (x0 )) ă ε whenever x P D(x0 , δ) X A .
106
CHAPTER 4. Continuous Maps
Case 2: Let x0 P AzA1 .
“ñ” then D δ ą 0 such that D(x0 , δ) X A = tx0 u. Therefore, for this particular δ, we
must have
ρ(f (x), f (x0 )) = 0 ă ε
whenever x P D(x0 , δ) X A .
“ð” We note that if x0 P AzA1 , f is defined to be continuous at x0 . In other words,
f is continuous at each isolated point.
˝
Remark 4.8. We remark here that Proposition 4.7 suggests that f is continuous at x0 P A
if and only if
@ ε ą 0, D δ ą 0 Q f (D(x0 , δ) X A) Ď D(f (x0 ), ε) .
Remark 4.9. In general the number δ in Proposition 4.7 also depends on the function f .
For a function f : A Ñ R which is continuous at x0 P A, let δ(f, x0 , ε) denote the largest
(
)
δ ą 0 such that if x P D(x0 , δ) X A, then ρ f (x), f (x0 ) ă ε. In other words,
ˇ (
␣
(
)
δ(f, x0 , ε) = sup δ ą 0 ˇ ρ f (x), f (x0 ) ă ε if x P D(x0 , δ) X A .
This number provides another way for the understanding of the uniform continuity (in
Section 4.5) and the equi-continuity (in Section 5.5). See Remark 4.51 and Remark 5.51 for
further details.
Definition 4.10. Let (M, d) and (N, ρ) be metric spaces, and A Ď M . A map f : A Ñ N
is said to be continuous on the set B Ď A if f is continuous at each point of B.
Theorem 4.11. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a map.
Then the following assertions are equivalent:
§4.1 Continuity
107
1. f is continuous on A.
2. For each open set V Ď N , f ´1 (V) Ď A is open relative to A; that is, f ´1 (V) = U X A
for some U open in M .
3. For each closed set E Ď N , f ´1 (E) Ď A is closed relative to A; that is, f ´1 (E) = F XA
for some F closed in M .
Proof. It should be clear that 2 ô 3 (left as an exercise); thus we show that 1 ô 2. Before
proceeding, we recall that B Ď f ´1 (f (B)) for all B Ď A and f (f ´1 (B)) Ď B for all B Ď N .
“1 ñ 2” Let a P f ´1 (V). Then f (a) P V. Since V is open in (N, ρ), D εf (a) ą 0 such that
D(f (a), εf (a) ) Ď V . By continuity of f (and Remark 4.8), there exists δa ą 0 such that
(
)
f (D(a, δa ) X A) Ď D f (a), εf (a) .
Therefore, by Proposition 0.16, for each a P f ´1 (V), D δa ą 0 such that
(
)
( (
))
D(a, δa ) X A Ď f ´1 f (D(a, δa ) X A) Ď f ´1 D f (a), εf (a) Ď f ´1 (V) .
Let U =
Ť
(4.1.1)
D(a, δa ). Then U is open (since it is the union of arbitrarily many
aPf ´1 (V)
open balls), and
(a) U Ě f ´1 (V) since U contains every center of balls whose union forms U;
(b) U X A Ď f ´1 (V) by (4.1.1).
Therefore, U X A = f ´1 (V).
“2 ñ 1” Let a P A and ε ą 0 be given. Define V = D(f (a), ε). By assumption there exists
U open in (M, d) such that f ´1 (V) = U X A . Since a P f ´1 (V), a P U; thus by the
openness of U, D δ ą 0 such that D(a, δ) Ď U. Therefore, by Proposition 0.16 we have
f (D(a, δ) X A) Ď f (U X A) = f (f ´1 (V)) Ď V = D(f (a), ε)
which suggests that f is continuous at a for all a P A; thus f is continuous on A. ˝
ˇ
␣
(
Example 4.12. Let f : Rn Ñ Rm be continuous. Then x P Rn ˇ }f (x)}2 ă 1 is open since
␣
ˇ
(
(
)
x P Rn ˇ }f (x)}2 ă 1 = f ´1 D(0, 1) .
108
CHAPTER 4. Continuous Maps
Remark 4.13. For a function f of two variable or more, it is important to distinguish the
continuity of f and the continuity in each variable (by holding all other variables fixed). For
example, let f : R2 Ñ R be defined by
"
1 if either x = 0 or y = 0,
f (x, y) =
0 if x ‰ 0 and y ‰ 0.
Observe that f (0, 0) = 1, but f is not continuous at (0, 0). In fact, for any δ ą 0, f (x, y) = 0
for infinitely many values of (x, y) P D((0, 0), δ); that is, |f (x, y) ´ f (0, 0)| = 1 for such
values. However if we consider the function g(x) = f (x, 0) = 1 or the function h(y) =
f (0, y) = 1, then g, h are continuous. Note also that
lim f (x, y) does not exists but
(x,y)Ñ(0,0)
lim (lim f (x, y)) = lim 0 = 0.
xÑ0 yÑ0
xÑ0
4.2 Operations on Continuous Maps
Definition 4.14. Let (M, d) be a metric space, (V, } ¨ }) be a (real) normed space, A Ď M ,
and f, g : A Ñ V be maps, h : A Ñ R be a function. The maps f + g, f ´ g and hf ,
mapping from A to V, are defined by
The map
(f + g)(x) = f (x) + g(x)
@x P A,
(f ´ g)(x) = f (x) ´ g(x)
@x P A,
(hf )(x) = h(x)f (x)
@x P A.
f
: Aztx P A | h(x) = 0u Ñ V is defined by
h
(f )
f (x)
(x) =
h
h(x)
@ x P Aztx P A | h(x) = 0u .
Proposition 4.15. Let (M, d) be a metric space, (V, } ¨ }) be a (real) normed space, A Ď M ,
and f, g : A Ñ V be maps, h : A Ñ R be a function. Suppose that x0 P A1 , and lim f (x) = a,
xÑx0
lim g(x) = b, lim h(x) = c. Then
xÑx0
xÑx0
lim (f + g)(x) = a + b ,
xÑx0
lim (f ´ g)(x) = a ´ b ,
xÑx0
lim (hf )(x) = ca ,
xÑx0
(f ) a
=
xÑx0 h
c
lim
if c ‰ 0 .
§4.2 Operations on Continuous Maps
109
Corollary 4.16. Let (M, d) be a metric space, (V, } ¨ }) be a (real) normed space, A Ď M ,
and f, g : A Ñ V be maps, h : A Ñ R be a function. Suppose that f, g, h are continuous at
f
x0 P A. Then the maps f + g, f ´ g and hf are continuous at x0 , and is continuous at
h
x0 if h(x0 ) ‰ 0.
Corollary 4.17. Let (M, d) be a metric space, (V, } ¨ }) be a (real) normed space, A Ď M ,
and f, g : A Ñ V be continuous maps, h : A Ñ R be a continuous function. Then the maps
f + g, f ´ g and hf are continuous on A, and
f
is continuous on Aztx P A | h(x) = 0u.
h
Definition 4.18. Let (M, d), (N, ρ) and (P, r) be metric space, A Ď M , B Ď N , and
f : A Ñ N , g : B Ñ P be maps such that f (A) Ď B. The composite function g ˝ f : A Ñ P
is the map defined by
(
)
(g ˝ f )(x) = g f (x)
@x P A.
Figure 4.1: The composition of functions
Theorem 4.19. Let (M, d), (N, ρ) and (P, r) be metric space, A Ď M , B Ď N , and
f : A Ñ N , g : B Ñ P be maps such that f (A) Ď B. Suppose that f is continuous at x0 ,
and g is continuous at f (x0 ). Then the composite function g ˝ f : A Ñ P is continuous at
x0 .
Proof. Let ε ą 0 be given. Since g is continuous at f (x0 ), D r ą 0 such that
(
)
g(D(f (x0 ), r) X B) Ď D (g ˝ f )(x0 ), ε .
Since f is continuous at x0 , D δ ą 0 such that
(
)
f (D(x0 , δ) X A) Ď D f (x0 ), r .
(
)
Since f (A) Ď B, f (D(x0 , δ) X A) Ď D f (x0 ), r X B; thus
(
)
(g ˝ f )(D(x0 , δ) X A) Ď g(D(f (x0 ), r) X B) Ď D (g ˝ f )(x0 ), ε .
˝
110
CHAPTER 4. Continuous Maps
Corollary 4.20. Let (M, d), (N, ρ) and (P, r) be metric space, A Ď M , B Ď N , and
f : A Ñ N , g : B Ñ P be continuous maps such that f (A) Ď B. Then the composite
function g ˝ f : A Ñ P is continuous on A.
Alternative Proof of Corollary 4.20. Let W be an open set in (P, r). By Theorem 4.11,
there exists V open in (N, ρ) such that g ´1 (W) = V X B. Since V is open in (N, ρ), by
Theorem 4.11 again there exists U open in (M, d) such that f ´1 (V) = U X A. Then
(
)
(g ˝ f )´1 (W) = f ´1 g ´1 (W) = f ´1 (V X B) = f ´1 (V) X f ´1 (B) = U X A X f ´1 (B) ,
while the fact that f (A) Ď B further suggests that
(g ˝ f )´1 (W) = U X A .
Therefore, by Theorem 4.11 we find that (g ˝ f ) is continuous on A.
˝
4.3 Images of Compact Sets under Continuous Maps
Theorem 4.21. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a
continuous map.
1. If K Ď A is compact, then f (K) is compact in (N, ρ).
2. Moreover, if (N, ρ) = (R, | ¨ |), then there exist x0 , x1 P K such that
ˇ
␣
(
f (x0 ) = inf f (K) = inf f (x) ˇ x P K
ˇ
␣
(
and f (x1 ) = sup f (K) = sup f (x) ˇ x P K .
Proof. 1. Let tVα uαPI be an open cover of f (K). Since Vα is open, by Theorem 4.11 there
Ť
exists Uα open in (M, d) such that f ´1 (Vα ) = Uα X A. Since f (K) Ď
Vα ,
αPI
K Ď f ´1 (f (K)) Ď
ď
f ´1 (Vα ) = A X
αPI
ď
Uα
αPI
which implies that tUα uαPI is an open cover of K. Therefore,
D J Ď I, #J ă 8 Q K Ď A X
ď
αPJ
thus f (K) Ď
Ť
αPJ
f (f ´1 (Vα )) Ď
Ť
αPJ
Vα .
Uα =
ď
αPJ
f ´1 (Vα ) ;
§4.3 Images of Compact Sets under Continuous Maps
111
2. By 1, f (K) is compact; thus sequentially compact. Corollary 3.5 then implies that
inf f (K) P f (K) and sup f (K) P f (K).
˝
8
Alternative Proof of Part 1. Let tyn u8
n=1 be a sequence in f (K). Then there exists txn un=1 Ď
K such that yn = f (xn ). Since K is sequentially compact, there exists a convergent subsequence txnk u8
k=1 with limit x P K. Let y = f (x) P f (K). By the continuity of f ,
(
)
(
)
lim ρ ynk , y = lim ρ f (xnk ), f (x) = 0
kÑ8
kÑ8
␣
(8
which implies that the sequence ynk k=1 converges to y P f (K). Therefore, f (K) is sequentially compact.
˝
Corollary 4.22 (The Extreme Value Theorem（極值定理）). Let f : [a, b] Ñ R be continuous. Then f attains its maximum and minimum in [a, b]; that is, there are x0 P [a, b] and
x1 P [a, b] such that
ˇ
␣
(
f (x0 ) = inf f (x) ˇ x P [a, b]
ˇ
␣
(
and f (x1 ) = sup f (x) ˇ x P [a, b] .
(4.3.1)
Proof. The Heine-Borel Theorem suggests that [a, b] is a compact set in R; thus Theorem
4.21 implies that f ([a, b]) must be compact in R. By the Heine-Borel Theorem again f ([a, b])
is closed and bounded, so
inf f ([a, b]) P f ([a, b]) and
sup f ([a, b]) P f ([a, b])
which further imply (4.3.1).
˝
ˇ
␣
Remark 4.23. If f attains its maximum (or minimum) on a set B, we use max f (x) ˇ x P
ˇ
ˇ
ˇ
((
␣
()
␣
((
␣
()
B or min f (x) ˇ x P B to denote sup f (x) ˇ x P B or inf f (x) ˇ x P B . Therefore,
(4.3.1) can be rewritten as
ˇ
␣
(
f (x0 ) = min f (x) ˇ x P [a, b]
and
ˇ
␣
(
f (x1 ) = max f (x) ˇ x P [a, b] .
Example 4.24. Two norms } ¨ } and ~ ¨ ~ on a real vector space V are called equivalent if
there are positive constants C1 and C2 such that
C1 }x} ď ~x~ ď C2 }x}
@x P V .
We note that equivalent norms on a vector space V induce the same topology; that is, if } ¨ }
and ~ ¨ ~ are equivalent norms on V, then U is open in the normed space (V, } ¨ }) if and
112
CHAPTER 4. Continuous Maps
only if U is open in the normed space (V, ~ ¨ ~). In fact, let U be an open set in (V, } ¨ }).
Then for any x P U, there exists r ą 0 such that
ˇ
␣
(
D}¨} (x, r) ” y P V ˇ }x ´ y} ă r Ď U .
ˇ
␣
(
Let δ = C1 r. Then if z P D~¨~ (x, δ) ” y P V ˇ ~x ´ y~ ă δ ,
}x ´ z} ď
1
1
~x ´ z~ ă
¨ C1 r = r
C1
C1
which implies that D~¨~ (x, δ) Ď D}¨} (x, r) Ď U. Therefore, U is open in (V, ~ ¨ ~). Similarly,
if U is open in (V, ~ ¨ ~), then the inequality ~x~ ď C2 }x} suggests that U is open in (V, }¨}).
Claim: Any two norms on Rn are equivalent.
Proof of claim: It suffices to show that any norm } ¨ } on Rn is equivalent to the two-norm
} ¨ }2 (check). Let tek unk=1 be the standard basis of Rn ; that is,
ek = ( 0,
¨ ¨ ¨ , 0 , 1, 0, ¨ ¨ ¨ , 0) .
looomooon
(k ´ 1) zeros
Every x P Rn can be written as x =
n
ř
xi ei , and }x}2 =
norms and the Cauchy-Schwarz inequality,
}x} ď
d
|xi |}ei } ď }x}2
thus letting C2 =
n
ÿ
}ei }2 ;
(4.3.2)
i=1
i=1
cn
ř
|xi |2 . By the definition of
i=1
i=1
n
ÿ
cn
ř
}ei }2 we have }x} ď C2 }x}2 .
i=1
On the other hand, define f : Rn Ñ R by
n
›ÿ
›
›
›
f (x) = }x} = › xi ei › .
i=1
Because of (4.3.2), f is continuous on Rn . In fact, for x, y P Rn ,
ˇ
ˇ ˇ
ˇ
ˇf (x) ´ f (y)ˇ = ˇ}x} ´ }y}ˇ ď }x ´ y} ď C2 }x ´ y}2
ˇ
␣
(
which guarantees the continuity of f on Rn . Let Sn´1 = x P Rn ˇ }x}2 = 1 . Then Sn´1 is a
compact set in (Rn , } ¨ }2 ) (since it is closed and bounded); thus by Theorem 4.21 f attains
§4.3 Images of Compact Sets under Continuous Maps
113
its minimum on Sn´1 at some point a = (a1 , ¨ ¨ ¨ , an ). Moreover, f (a) ą 0 (since if f (a) = 0,
a = 0 R Sn´1 ). Then for all x P Rn ,
x
P Sn´1 ; thus
}x}2
f
( x )
}x}2
ě f (a) .
The inequality above further implies that f (a)}x}2 ď f (x) = }x}; thus letting C1 = f (a) we
have C1 }x}2 ď }x}.
Remark 4.25.
1. Let f : R Ñ R be defined by f (x) = 0. Then f is continuous. Note that t0u Ď R is
compact (7 closed and bounded), but f ´1 (t0u) = R is not compact.
2. Let f : R Ñ R be defined by f (x) = x2 . Then f is continuous. Note that C = t1u is
connected, but f ´1 (C) = t1, ´1u is not connected.
Remark 4.26.
1. If K is not compact, then Theorem 4.21 is not true. Consider the following counter
example: K = (0, 1), f : K Ñ R defined by f (x) =
1
. Then f (K) is unbounded.
x
2. If f is not continuous, then Theorem 4.21 is not true either.
(a) Counter example 1: f : K = [0, 1] Ñ R defined by
$
& 1 if x ‰ 0,
x
f (x, y) =
% 0 if x = 0.
Then f (K) is unbounded ñ E x1 P K Q f (x1 ) = sup f (K).
(b) Counter example 2: f : [0, 1] Ñ R by
"
f (x, y) =
x if x ‰ 1,
0 if x = 1.
Then there is no x1 P [0, 1] such that f (x1 ) = sup f (x) = 1.
xP[0,1]
Example 4.27 (An example show that x0 , x1 in Theorem 4.21 are not unique). Let f :
[´2, 2] Ñ R be defined by f (x) = (x2 ´ 1)2 .
1. Critical point: f 1 (x) = 2(x2 ´ 1) ¨ 2x = 0 ô x = 0, ˘1.
114
CHAPTER 4. Continuous Maps
2. Comparison: f (0) = 1, f (1) = f (´1) = 0, f (2) = f (´2) = 9. Then
f (2) = f (´2) = sup f (x) and
f (1) = f (´1) =
inf f (x) .
xP[´2,2]
xP[´2,2]
Corollary 4.28. Let (M, d) be a metric space, K Ď M be a compact set, and f : K Ñ R
be continuous. Then the set
␣
ˇ
(
x P K ˇ f (x) is the maximum of f on K
is a non-empty compact set.
Proof. Let M = sup f (K). Then the set defined above is f ´1 (tMu), and
1. f ´1 (tMu) is non-empty by Theorem 4.21;
2. f ´1 (tMu) is closed since tMu is a closed set in (R, | ¨ |) and f is continuous on K.
Lemma 3.11 suggests that f ´1 (tMu) is compact.
˝
4.4 Images of Connected and Path Connected Sets under Continuous Maps
Definition 4.29. Let (M, d) be a metric space. A subset A Ď M is said to be path
connected if for every x, y P A, there exists a continuous map φ : [0, 1] Ñ A such that
φ(0) = x and φ(1) = y.
x
A
y
Figure 4.2: Path connected sets
Example 4.30. A set A in a vector space V is called convex if for all x, y P A, the line
segment joining x and y, denoted by xy, lies in A. Then a convex set in a normed space is
path connected. In fact, for x, y P A, define φ(t) = ty + (1 ´ t)x. Then
§4.4 Images of Connected and Path Connected Sets under Continuous Maps
115
1. φ : [0, 1] Ñ xy Ď A, φ(0) = x, φ(1) = y;
2. φ : [0, 1] Ñ A is continuous.
xy = φ([0, 1])
•
x
t •
1´t •
y
A
Figure 4.3: Convex sets
Example 4.31. A set S in a vector space V is called star-shaped if there exists p P S
such that for any q P S, the line segment joining p and q lies in S. A star-shaped set in a
normed space is path connected. In fact, for x, y P S, define
$
[ 1]
&
2tp + (1 ´ 2t)x
if t P 0, ,
φ(t) =
[ 2]
% (2t ´ 1)y + (2 ´ 2t)p if t P 1 , 1 .
2
Then
1. φ : [0, 1] Ñ xp Y py Ď S, φ(0) = x, φ(1) = y;
2. φ : [0, 1] Ñ A is continuous.
Theorem 4.32. Let (M, d) be a metric space, and A Ď M . If A is path connected, then A
is connected.
Proof. Assume the contrary that there are two open sets V1 and V2 such that
1. A X V1 X V2 = H;
2. A X V1 ‰ H;
3. A X V2 ‰ H;
4. A Ď V1 Y V2 .
Since A is path connected, for x P A X V1 and y P A X V2 , there exists φ : [0, 1] Ñ A such
that φ(0) = x and φ(1) = y. By Theorem 4.11, there exist U1 and U2 open in (R, | ¨ |) such
that φ´1 (V1 ) = U1 X [0, 1] and φ´1 (V2 ) = U2 X [0, 1]. Therefore,
[0, 1] = φ´1 (A) Ď φ´1 (V1 ) Y φ´1 (V2 ) Ď U1 Y U2 .
Since 0 P U1 , 1 P U2 , and [0, 1] X U1 X U2 = φ´1 (A X V1 X V2 ) = H, we conclude that [0, 1]
is disconnected, a contradiction.
˝
116
CHAPTER 4. Continuous Maps
Example 4.33. Let A =
␣(
(
1 ) ˇˇ
x, sin
x P (0, 1] Y (t0u ˆ [´1, 1]) . Then A is connected in
x
(R2 , } ¨ }2 ), but A is not path connected.
To see this, we assume the contrary that A is path connected such that there is a
␣(
(
1)ˇ
continuous function φ : [0, 1] Ñ A such that φ(0) = (x0 , y0 ) P x, sin ˇ x P (0, 1) and
x
ˇ
(
␣
φ(1) = (0, 0) P 0ˆ[´1, 1]. Let t0 = inf t P [0, 1] ˇ φ(t) P 0ˆ[´1, 1] . In other words, at t = t0
the path touches 0 ˆ [´1, 1] for the “first time”. By the continuity of φ, φ(t0 ) P 0 ˆ [´1, 1].
␣(
(
1)ˇ
Since φ(0) R 0 ˆ [´1, 1], φ([0, t0 )) Ď x, sin ˇ x P (0, 1) .
x
1a 2
Let 0 ă ε ă
x0 + y02 . Then there exists δ ą 0 such that if |t ´ t0 | ă δ, |φ(t) ´ φ(t0 )| ă
2
ε.
Theorem 4.34. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a
continuous map.
1. If C Ď A is connected, then f (C) is connected in (N, ρ).
2. If C Ď A is path connected, then f (C) is path connected in (N, ρ).
Proof. 1. Suppose that there are two open sets V1 and V2 in (N, ρ) such that
(a) f (C) X V1 X V2 = H; (b) f (C) X V1 ‰ H; (c) f (C) X V2 ‰ H; (d) f (C) Ď V1 Y V2 .
By Theorem 4.11, there are U1 and U2 open in (M, d) such that f ´1 (V1 ) = U1 X A and
f ´1 (V2 ) = U2 X A. By (d),
C Ď f ´1 (f (C)) Ď f ´1 (V1 ) Y f ´1 (V2 ) = (U1 Y U2 ) X A Ď U1 Y U2 .
Moreover, by (a) we find that
C X U1 X U2 = C X (U1 X A) X (U2 X A) = C X f ´1 (V1 ) X f ´1 (V2 )
Ď f ´1 (f (C) X V1 X V2 ) = H
which implies C X U1 X U2 = H. Finally, (b) implies that for some x P C, f (x) P V1 .
Therefore, x P f ´1 (V1 ) = U1 X A which suggests that x P U1 ; thus C X U1 ‰ H.
Similarly, C X U2 ‰ H. Therefore, C is disconnected which is a contradiction.
2. Let y1 , y2 P f (C). Then D x1 , x2 P C such that f (x1 ) = y1 and f (x2 ) = y2 . Since C is
path connected, D r : [0, 1] Ñ C such that r is continuous on [0, 1] and r(0) = x1 and
r(1) = x2 . Let φ : [0, 1] Ñ f (C) be defined by φ = f ˝ r. By Corollary 4.20 φ is
continuous on [0, 1], and φ(0) = y1 and φ(1) = y2 .
˝
§4.5 Uniform Continuity
117
Corollary 4.35 (The Intermediate Value Theorem（中間值定理）). Let f : [a, b] Ñ R be
continuous. If f (a) ‰ f (b), then for all d in between f (a) and f (b), there exists c P (a, b)
such that f (c) = d.
Proof. The closed interval [a, b] is connected by Theorem 3.38, so Theorem 4.34 implies that
f ([a, b]) must be connected in R. By Theorem 3.38 again, if d is in between f (a) and f (b),
then d belongs to f ([a, b]). Therefore, for some c P (a, b) we have f (c) = d.
˝
Example 4.36. Let f : [0, 1] Ñ [0, 1] be continuous. Then D x0 P [0, 1] Q f (x0 ) = x0 .
Proof. Let g(x) = x ´ f (x). Then
1. g(0) = 0 or g(1) = 0 ñ x0 = 0 or 1.
2. g(0) ‰ 0 or g(1) ‰ 0 ñ g(0) ă 0 and g(1) ą 0. Since g : [0, 1] Ñ R is continuous,
D x0 P [0, 1] Q g(x0 ) = 0 ñ D x0 P (0, 1) Q f (x0 ) = x0 .
˝
Remark 4.37. Such an x0 in Example 4.36 is called a fixed-point of f .
Example 4.38. Let f : [1, 2] Ñ [0, 3] be continuous, and f (1) = 0 and f (2) = 3. Then
D x0 P [1, 2] Q f (x0 ) = x0 .
Proof. Let g(x) = x ´ f (x). Then g : [1, 2] Ñ R is continuous. Moreover,
g(1) = 1 ´ f (1) = 1,
g(2) = 2 ´ f (2) = ´1 ;
thus D x0 P (1, 2) Q g(x0 ) = 0.
˝
Example 4.39. Let p be a cubic polynomial; that is, p(x) = a3 x3 + a2 x2 + a1 x + a0 for some
a0 , a1 , a2 P R and a3 ‰ 0. Then p has a real root x0 (that is, D x0 P R such that p(x0 ) = 0).
Proof. Note that p is obviously continuous and R is connected. Write
(
a
a
a )
p(x) = a3 x3 1 + 2 + 1 2 + 0 3 .
a3 x
a3 x
a3 x
α
Now lim
= 0 if n ą 0 and β ‰ 0, so
xÑ˘8 βxn
(
a
a
a )
lim 1 + 2 + 1 2 + 0 3 = 1 .
xÑ˘8
a3 x
a3 x
a3 x
Moreover,
3
lim ax =
xÑ8
"
8
if a ą 0,
´8 if a ă 0.
Suppose that a ą 0. Then lim ax3 = 8 and lim ax3 = ´8 ñ D x, y P R Q p(x) ă 0 ă
xÑ8
xÑ´8
p(y). By Corollary 4.35 D r P R Q p(r) = 0. The case that a ă 0 is similar.
˝
118
CHAPTER 4. Continuous Maps
4.5 Uniform Continuity（均勻連續）
Definition 4.40. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be
a map. For a set B Ď A, f is said to be uniformly continuous on B if for any
8
two sequences txn u8
n=1 , tyn un=1 Ď B with the property that lim d(xn , yn ) = 0, one has
nÑ8
(
)
lim ρ f (xn ), f (yn ) = 0.
nÑ8
Proposition 4.41. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a
map. If f is uniformly continuous on A, then f is continuous on A.
Proof. Let x0 P A X A1 , and txk u8
k=1 Ď A be a sequence such that xk Ñ x0 as k Ñ 8.
Let tyk u8
k=1 be a constant sequence with value x0 ; that is, yk = x0 for all k P N. Then
tyk u8
k=1 Ď A and d(xk , yk ) Ñ 0 as k Ñ 8. By the uniform continuity of f on A,
(
)
(
)
lim ρ f (xk ), f (x0 ) = lim ρ f (xk ), f (yk ) = 0
kÑ8
kÑ8
which implies that f is continuous on x0 .
˝
Example 4.42. Let f : [0, 1] Ñ R be the Dirichlet function; that is,
"
0 if x P Q ,
f (x) =
1 if x P QA .
and B = Q X [0, 1]. Then f is continuous nowhere in [0, 1], but f is uniformly continuous
on B. However, the proposition above guarantees that if f is uniformly continuous on A,
then f must be continuous on A (Check why the proof of Proposition 4.41 does not go
through if B is a proper subset of A).
Example 4.43. The function f (x) = |x| is uniformly continuous on R.
Proof. By the triangle inequality,
ˇ
ˇ ˇ
ˇ
ˇf (x) ´ f (y)ˇ = ˇ|x| ´ |y|ˇ ď |x ´ y| ;
8
thus if txn u8
n=1 and tyn un=1 are sequences in R and lim |xn ´ yn | = 0, by the Sandwich
nÑ8
ˇ
ˇ
lemma we must have lim ˇf (xn ) ´ f (yn )ˇ = 0.
˝
nÑ8
1
Example 4.44. The function f : (0, 8) Ñ R defined by f (x) = is uniformly continuous
x
on [a, 8) for all a ą 0. However, it is not uniformly continuous on (0, 8).
§4.5 Uniform Continuity
119
8
Proof. Let txn u8
n=1 and tyn un=1 be sequences in [a, 8) such that lim |xn ´ yn | = 0. Then
nÑ8
ˇ1
|xn ´ yn |
|xn ´ yn |
1ˇ
|f (xn ) ´ f (yn )| = ˇ ´ ˇ =
ď
Ñ 0 as
xn
yn
|xn yn |
a2
nÑ8
which implies that f is uniformly continuous on [a, 8) if a ą 0. However, by choosing
xn =
1
1
and yn = , we find that
n
2n
|xn ´ yn | =
1
2n
but |f (xn ) ´ f (yn )| = n ě 1 ;
thus f cannot be uniformly continuous on (0, 8).
˝
Remark 4.45. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : B Ď A Ñ N be a
map. Then the following four statements are equivalent:
(1) f is not uniformly continuous on B.
(
)
8
(2) D txn u8
n=1 , tyn un=1 Ď B Q lim d(xn , yn ) = 0 and lim sup ρ f (xn ), f (yn ) ą 0.
nÑ8
nÑ8
(
)
8
(3) D txn u8
n=1 , tyn un=1 Ď B Q lim d(xn , yn ) = 0 and lim ρ f (xn ), f (yn ) ą 0.
nÑ8
(4) D ε ą 0 Q @ n ą 0, D xn , yn P B and d(xn , yn ) ă
nÑ8
(
)
1
Q ρ f (xn ), f (yn ) ě ε.
n
Example 4.46. Let f : R Ñ R defined by f (x) = x2 . Then f is continuous in R but not
uniformly continuous on R. Let ε = 1, xn = n, and yn = n +
1
,
2n
ˇ
ˇ ˇ
ˇ ˇ
ˇ
ˇf (xn ) ´ f (yn )ˇ = ˇn2 ´ (n + 1 )2 ˇ = ˇn2 ´ n2 ´ 1 ´ 1 ˇ ą 1 @ n ą 0 .
2n
4n2
Example 4.47. The function f (x) = sin(x2 ) is not uniform continuous on R.
?
Proof. Let ε = 1, xn = 2n π +
?
?
?
π
π
and yn = 2n π ´
. Then
8n
8n
ˇ
ˇ ˇ
(
)
( 2
π
π )ˇˇ
π
ˇ sin(x2n ) ´ sin(yn2 )ˇ = ˇˇ sin 4n2 π + π + π
´ sin 4n π ´ +
;
ˇ = 2 cos
2
2
2 64n
2 64n
64n2
ˇ
ˇ
˝
thus if n is large enough, ˇ sin(x2n ) ´ sin(yn2 )ˇ ě 1.
1
Example 4.48. The function f : (0, 1) Ñ R defined by f (x) = sin is not uniformly
x
continuous.
120
CHAPTER 4. Continuous Maps
(
(
π )´1
π )´1
Proof. Let ε = 1, xn = 2nπ +
and yn = 2nπ ´
. Then
2
2
ˇ
ˇ
ˇ sin 1 ´ sin 1 ˇ = 2 ,
xn
while |xn ´ yn | =
π
2
4n2 π 2 ´ π4
=
1
(4n2 ´ 14 )π
yn
ď
1
for all n P N.
n
˝
Theorem 4.49. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a map.
For a set B Ď A, f is uniformly continuous on B if and only if
(
)
@ ε ą 0, D δ ą 0 Q ρ f (x), f (y) ă ε whenever d(x, y) ă δ and x, y P B .
Proof. “ð” Suppose the contrary that f is not uniformly continuous on B. Then there are
8
two sequences txn u8
n=1 , tyn un=1 in B such that
lim d(xn , yn ) = 0
kÑ8
Let ε =
(
)
lim sup ρ f (xn ), f (yn ) ą 0 .
but
nÑ8
(
)
1
lim sup ρ f (xn ), f (yn ) . Then by the definition of the limit and the limit
2 nÑ8
superior (or Proposition 1.121) we conclude that there exist subsequences txnk u8
k=1
such
that
and tynk u8
k=1
(
)
(
)
ρ f (xnk ), f (ynk ) ě lim sup ρ f (xn ), f (yn ) ´ ε = ε ą 0
nÑ8
while lim d(xnk , ynk ) = 0, a contradiction.
kÑ8
“ñ” Suppose the contrary that there exists ε ą 0 such that for all δ =
two points xn and yn P B such that
d(xn , yn ) ă
1
n
but
1
ą 0, there exist
n
(
)
ρ f (xn ), f (yn ) ě ε .
8
These points form two sequences txn u8
n=1 , tyn un=1 in B such that lim d(xn , yn ) = 0,
nÑ8
(
)
while the limit of ρ f (xn ), f (yn ) , if exists, does not converges to zero as n Ñ 8. As
a consequence, f is not uniformly continuous on B, a contradiction.
˝
Remark 4.50. The theorem above provides another way (the blue color part) of defining
the uniform continuity of a function over a subset of its domain. Moreover, according to
this alternative definition, if f : A Ñ N is uniformly continuous on B Ď A, then
)
( ε)
( ( δ)
X B Ď D c,
for some c P N ;
@ ε ą 0, D δ ą 0 Q @ b P M, f D b,
2
2
§4.5 Uniform Continuity
121
that is, the diameter of the image, under f , of subsets of B whose diameter is not greater
than δ is not greater than ε（在 B 中直徑不超過 δ 的子集合被函數 f 映過去之後，在對
應域中的直徑不會超過 ϵ）.
Remark 4.51. In terms of the number δ(f, x, ε) defined in Remark 4.9, the uniform continuity of a function f : A Ñ N is equivalent to that
δf (ε) ” inf δ(f, x, ε) ą 0
@ε ą 0.
xPA
The function δf (¨) is the inverse of the modulus of continuity of (a uniform
continuous) function f .
Theorem 4.52. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a map.
If K Ď A is compact and f is continuous on K, then f is uniformly continuous on K.
Proof. Let ε ą 0 be given. Since f is continuous on K,
(
) ε
@ a P K, D δ = δ(a) ą 0 Q ρ f (x), f (a) ă whenever x P D(a, δ) X A .
2
! (
)
δ(a) )
Then D a,
is an open cover of K; thus
2
aPK
D ta1 , ¨ ¨ ¨ , aN u Ď K Q K Ď
N
ď
( δ)
D ai , i ,
i=1
2
1
mintδ1 , ¨ ¨ ¨ , δN u. Then δ ą 0, and if x1 , x2 P K and d(x1 , x2 ) ă
2
( δ )
δ, there must be j = 1, ¨ ¨ ¨ , N such that x1 , x2 P B(aj , δj ). In fact, since x1 P D aj , j for
2
where δi = δ(ai ). Let δ =
some j = 1, ¨ ¨ ¨ , N , then
d(x2 , aj ) ď d(x1 , x2 ) + d(x1 , aj ) ă δ +
δj
ă δj .
2
Therefore, x1 , x2 P D(aj , δj ) X A for some j = 1, ¨ ¨ ¨ , N ; thus
(
)
(
)
(
) ε ε
ρ f (x1 ), f (x2 ) ď ρ f (x1 ), f (aj ) + ρ f (x2 ), f (aj ) ă + = ε .
2 2
˝
Alternative proof. Assume the contrary that f is not uniformly continuous on K. Then ((3)
8
of Remark 4.45 implies that) there are sequences txn u8
n=1 and tyn un=1 in K such that
lim d(xn , yn ) = 0
nÑ8
but
(
)
lim ρ f (xn ), f (yn ) ą 0 .
nÑ8
122
CHAPTER 4. Continuous Maps
8
Since K is (sequentially) compact, there exist convergent subsequences txnk u8
k=1 and tynk uk=1
with limits x, y P K. On the other hand, lim d(xn , yn ) = 0, we must have x = y; thus by
nÑ8
the continuity of f (on K),
(
)
(
)
(
)
0 = ρ f (x), f (x) = lim ρ f (xnk ), f (ynk ) = lim ρ f (xn ), f (yn ) ą 0 ,
kÑ8
nÑ8
a contradiction.
˝
Lemma 4.53. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be uniformly
␣
(8
continuous. If txk u8
k=1 Ď A is a Cauchy sequence, so is f (xk ) k=1 .
8
Proof. Let txk uk=1
be a Cauchy sequence in (M, d), and ε ą 0 be given. Since f : A Ñ N
is uniformly continuous,
(
)
D δ ą 0 Q ρ f (x), f (y) ă ε whenever d(x, y) ă δ and x, y P A .
For this particular δ, D N ą 0 Q d(xk , xℓ ) ă δ if k, ℓ ě N . Therefore,
)
ρ(f (xk ), f (xℓ ) ă ε if k, ℓ ě N .
˝
Corollary 4.54. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be
uniformly continuous. If N is complete, then f has a unique extension to a continuous
s that is, D g : A
s Ñ N such that
function on A;
s
(1) g is uniformly continuous on A;
(2) g(x) = f (x) for all x P A;
s Ñ N is a continuous map satisfying (1) and (2), then h = g.
(3) if h : A
s
Proof. Let x P AzA.
Then D txk u8
Ď A such that xk Ñ x as k Ñ 8. Since txk u8
k=1
␣
(k=1
8
is Cauchy, by Lemma 4.53 f (xk ) k=1 is a Cauchy sequence in (N, ρ); thus is convergent.
Moreover, if tzk u8
k=1 Ď A is another sequence converging to x, we must have d(xk , zk ) Ñ 0
␣
(8
␣
(8
as k Ñ 8; thus ρ(f (xk ), f (zk )) Ñ 0 as k Ñ 8, so the limit of f (xk ) k=1 and f (zk ) k=1
must be the same.
s Ñ N by
Define g : A
#
f (x)
if x P A ,
g(x) =
s
lim f (xk ) if x P AzA,
and txk u8
k=1 Ď A converging to x as k Ñ 8 .
kÑ8
§4.6 Differentiation of Functions of One Variable
123
Then the argument above shows that g is well-defined, and (2), (3) hold.
Let ε ą 0 be given. Since f : A Ñ N is uniformly continuous,
(
) ε
D δ ą 0 Q ρ f (x), f (y) ă whenever d(x, y) ă 2δ and x, y P A .
3
s such that d(x, y) ă δ. Let txk u8 , tyk u8 Ď A be sequences
Suppose that x, y P A
k=1
k=1
converging to x and y, respectively. Then D N ą 0 such that
d(xk , x) ă
(
) ε (
) ε
δ
δ
, d(yk , y) ă and ρ f (xk ), g(x) ă , ρ f (yk ), g(y) ă
2
2
3
3
@k ě N .
In particular, due to the triangle inequality,
d(xN , yN ) ď d(xN , x) + d(x, y) + d(y, yN ) ă
δ
δ
+ δ + = 2δ ;
2
2
(
) ε
thus ρ f (xN ), f (yN ) ă . As a consequence,
3
(
)
(
)
(
)
(
) ε ε ε
ρ g(x), g(y) ď ρ g(x), f (xN ) + ρ f (xN ), f (yN ) + ρ f (yN ), f (y) ă + + = ε . ˝
3 3 3
4.6 Differentiation of Functions of One Variable
Definition 4.55. A function f : (a, b) Ñ R is said to be differentiable at x0 if there
exists a number m such that
f (x) ´ f (x0 ) ´ m(x ´ x0 )
= 0.
xÑx0
x ´ x0
lim
The (unique) number m is usually denoted by f 1 (x0 ), and is called the derivative of f at
x0 .
Remark 4.56. The derivative of f at x0 can be computed by
f (x) ´ f (x0 )
.
xÑx0
x ´ x0
f 1 (x0 ) = lim
Remark 4.57. By the definition of the limit of functions, f : (a, b) Ñ R is differentiable at
x0 P (a, b) if and only if there exists m P R, denoted by f 1 (x0 ), such that
ˇ
ˇ
@ ε ą 0, D δ ą 0 Q ˇf (x) ´ f (x0 ) ´ f 1 (x0 )(x ´ x0 )ˇ ď ε|x ´ x0 | if |x ´ x0 | ă δ .
Definition 4.58. A function f : (a, b) Ñ R is said to be differentiable (on (a, b)) if f is
differentiable at each x0 P (a, b).
124
CHAPTER 4. Continuous Maps
Proposition 4.59. Suppose that a function f : (a, b) Ñ R is differentiable at x0 . Then f
is continuous at x0 .
Proof. For x ‰ x0 , f (x) ´ f (x0 ) =
f (x) ´ f (x0 )
¨ (x ´ x0 ); thus Proposition 4.15 implies that
x ´ x0
(
)
f (x) ´ f (x0 )
lim f (x) ´ f (x0 ) = lim
¨ lim (x ´ x0 ) = f 1 (x0 ) ¨ 0 = 0 .
˝
xÑx0
xÑx0
xÑx0
x ´ x0
Theorem 4.60. Suppose that functions f, g : (a, b) Ñ R are differentiable at x0 , and k P R
is a constant. Then
1. (kf )1 (x0 ) = kf 1 (x0 ).
2. (f ˘ g)1 (x0 ) = f 1 (x0 ) ˘ g 1 (x0 ).
3. (f g)1 (x0 ) = f 1 (x0 )g(x0 ) + f (x0 )g 1 (x0 ).
4.
( f )1
g
(x0 ) =
f 1 (x0 )g(x0 ) ´ f (x0 )g 1 (x0 )
if g(x0 ) ‰ 0.
g(x0 )2
Theorem 4.61 (Chain Rule). Suppose that a function f : (a, b) Ñ R is differentiable at x0 ,
and g : (c, d) Ñ R is differentiable at y0 = f (x0 ) P (c, d). Then g ˝ f is differentiable at x0 ,
and
(g ˝ f )1 (x0 ) = g 1 (f (x0 ))f 1 (x0 ) .
Proof. Let ε ą 0 be given. Since f : (a, b) Ñ R is differentiable at x0 and g : (c, d) Ñ R is
differentiable at y0 = f (x0 ),
!
ˇ
ˇ
1
ˇ
ˇ
D δ1 ą 0 Q f (x) ´ f (x0 ) ´ f (x0 )(x ´ x0 ) ď min 1,
ε
|x ´ x0 | if |x ´ x0 | ă δ1
2(1 + |g 1 (y0 )|)
)
and
ˇ
ˇ
D δ2 ą 0 Q ˇg(y) ´ g(y0 ) ´ g 1 (y0 )(y ´ y0 )ˇ ď
ε|y ´ y0 |
if |y ´ y0 | ă δ2 .
2(1 + |f 1 (x0 )|)
Moreover, by Proposition 4.59 f is continuous at x0 ; thus
ˇ
ˇ
D δ3 ą 0 Q ˇf (x) ´ f (x0 )ˇ ă δ2 if |x ´ x0 | ă δ3 and x P (a, b) .
Let δ = mintδ1 , δ3 u, and denote f (x) by y. Then if |x ´ x0 | ă δ, we have |y ´ y0 | ă δ2 and
ˇ
ˇ ˇ
ˇ
ˇ(g ˝ f )(x) ´ (g ˝ f )(x0 ) ´ g 1 (y0 )f 1 (x0 )(x ´ x0 )ˇ = ˇg(y) ´ g(y0 ) ´ g 1 (y0 )f 1 (x0 )(x ´ x0 )ˇ
ˇ
ˇ
= ˇg(y) ´ g(y0 ) ´ g 1 (y0 )(y ´ y0 ) + g 1 (y0 )(f (x) ´ f (x0 ) ´ f 1 (x0 )(x ´ x0 )ˇ
ˇ
ˇ
ˇ ε|x ´ x0 |
εˇf (x) ´ f (x0 )ˇ ˇˇ 1
ˇ
ď
+
g
(y
)
0
2(1 + |f 1 (x0 )|)
2(1 + |g 1 (y0 )|)
(
) ε
ε
1
x
ď
|x
´
|
+
|f
(x
)||x
´
x
|
+ |x ´ x0 | = ε|x ´ x0 | .
0
0
0
2(1 + |f 1 (x0 )|)
2
§4.6 Differentiation of Functions of One Variable
125
By Remark 4.57, g ˝ f is differentiable at x0 with derivative g 1 (f (x0 ))f 1 (x0 ).
˝
Proposition 4.62. If f : (a, b) Ñ R is differentiable at x0 P (a, b) and f attains a local
minimum or maximum at x0 , then f 1 (x0 ) = 0.
Proof. W.L.O.G. we assume that f attains its local minimum at x0 . Then f (x) ´ f (x0 ) ě 0
for all x P I, where I is an open interval containing x0 . Therefore,
f (x) ´ f (x0 )
f (x) ´ f (x0 )
= lim´
ď0
xÑx0
x ´ x0
x ´ x0
xÑx0
f 1 (x0 ) = lim
and
f (x) ´ f (x0 )
f (x) ´ f (x0 )
= lim+
ě 0.
xÑx0
x ´ x0
x ´ x0
xÑx0
f 1 (x0 ) = lim
As a consequence, f 1 (x0 ) = 0.
˝
Theorem 4.63 (Rolle). Suppose that a function f : [a, b] Ñ R is continuous, and is
differentiable on (a, b). If f (a) = f (b), then D c P (a, b) such that f 1 (c) = 0.
Proof. By the Extreme Value Theorem, there exists x0 and x1 in [a, b] such that
f (x0 ) = min f ([a, b]) and
f (x1 ) = max f ([a, b]) .
Case 1. f (x0 ) = f (x1 ), then f is constant on [a, b]; thus f 1 (x) = 0 for all x P (a, b).
Case 2. One of f (x0 ) and f (x1 ) is different from f (a). W.L.O.G. we may assume that
f (x0 ) ‰ f (a). Then x0 P (a, b), and f attains its global minimum at x0 . By Proposition
4.62, f 1 (x0 ) = 0.
˝
Theorem 4.64 (Cauchy’s Mean Value Theorem). Suppose that functions f, g : [a, b] Ñ R
are continuous, and f, g : (a, b) Ñ R are differentiable. If g(a) ‰ g(b) and g 1 (x) ‰ 0 for all
x P (a, b), then there exists c P (a, b) such that
f 1 (c)
f (b) ´ f (a)
=
.
1
g (c)
g(b) ´ g(a)
Proof. Consider the function
(
)(
) (
)(
)
h(x) ” f (x) ´ f (a) g(b) ´ g(a) ´ f (b) ´ f (a) g(x) ´ g(a) .
Then h : [a, b] Ñ R is continuous, and is differentiable on (a, b). Moreover, h(b) = h(a) = 0.
By Rolle’s theorem, there exists c P (a, b) such that
(
) (
)
h1 (c) = f 1 (c) g(b) ´ g(a) ´ f (b) ´ f (a) g 1 (c) = 0 .
˝
126
CHAPTER 4. Continuous Maps
Corollary 4.65 (Mean Value Theorem). Suppose that a function f : [a, b] Ñ R is continuous, and f : (a, b) Ñ R is differentiable. Then there exists c P (a, b) such that
f (b) ´ f (a)
.
b´a
Proof. Apply the Cauchy Mean Value Theorem for the case that g(x) = x.
f 1 (c) =
˝
Corollary 4.66. Suppose that a function f : [a, b] Ñ R is continuous and f 1 (x) = 0 for all
x P (a, b). Then f is constant.
Proof. Let x P (a, b) be given. By Mean Value Theorem, there exists c P (a, x) such that
f (x) ´ f (a) = f 1 (c)(x ´ a) = 0 .
Therefore, f (x) = f (a); thus for all x P (a, b), f (x) = f (a). Now by continuity, f (b) =
lim f (x) = f (a).
˝
xÑb´
Corollary 4.67 (L’Hôspital’s rule). Let f, g : (a, b) Ñ R be differentiable functions. Suppose
f 1 (x)
xÑx0 g 1 (x)
that for some x0 P (a, b), f (x0 ) = g(x0 ) = 0, g 1 (x) ‰ 0 for all x ‰ x0 , and the limit lim
exists. Then the limit lim
f (x)
xÑx0 g(x)
also exists, and
f (x)
f 1 (x)
= lim 1
.
xÑx0 g (x)
xÑx0 g(x)
lim
Proof. We first note that g(x) ‰ g(x0 ) for all x ‰ x0 since if not, the Mean Value Theorem
implies that the existence of c in between x and x0 such that g 1 (c) = 0 which contradicts
to the condition that g 1 (x) ‰ 0 for all x ‰ x0 . By Cauchy’s Mean Value Theorem, for all
x P (a, b) and x ‰ x0 , there exists ξ = ξ(x) in between x and x0 such that
f (x)
f (x) ´ f (x0 )
f 1 (ξ)
=
= 1
g(x)
g(x) ´ g(x0 )
g (ξ)
Since ξ Ñ x0 as x Ñ x0 , we have
f 1 (ξ)
f 1 (x)
f (x)
= lim 1
= lim 1
.
xÑx0 g (x)
xÑx0 g(x)
ξÑx0 g (ξ)
lim
˝
Theorem 4.68 (Taylor). Suppose that for some k P N, f : (a, b) Ñ R be (k + 1)-times
differentiable and c P (a, b). Then for all x P (a, b), there exists d in between c and x such
that
f (x) =
k
ÿ
f (j) (c)
j=0
j!
(x ´ c)j +
where f (j) denotes the j-th derivative of f .
f (k+1) (d)
(x ´ c)(k+1) ,
(k + 1)!
§4.6 Differentiation of Functions of One Variable
Proof. Let g(x) = f (x) ´
k f (j) (c)
ř
j=0
j!
127
(x ´ c)j , and h(x) = (x ´ c)k+1 . Then for 1 ď j ď k,
g (j) (c) = h(j) (c) = 0 ;
thus by the Cauchy mean value theorem (Theorem 4.64), there exists ξ1 in between x and
c, ξ2 in between ξ1 and c, ¨ ¨ ¨ , ξk+1 in between ξk and c such that
g(x)
g(x) ´ g(c)
g 1 (ξ1 )
g 1 (ξ1 ) ´ g 1 (c)
g 2 (ξ2 )
=
= 1
= 1
=
= ¨¨¨
h2 (ξ2 )
h(x)
h(x) ´ h(c)
h (ξ1 )
h (ξ1 ) ´ h1 (c)
g (k) (ξk )
g (k) (ξk ) ´ g (k) (c)
g (k+1) (ξk+1 )
f (k+1) (ξk+1 )
= (k)
= (k)
=
=
.
h (ξk )
h (ξk ) ´ h(k) (c)
h(k+1) (ξk )
(k + 1)!
Letting d = ξk+1 we conclude the theorem.
˝
Example 4.69. A function f : [a, b] Ñ R is said to be Lipschitz continuous if D M ą 0
such that
|f (x1 ) ´ f (x2 )| ď M |x1 ´ x2 |
@ x1 , x2 P [a, b] .
If the derivative of a differentiable function f : (a, b) Ñ R is bounded; that is, D M ą 0
Q |f 1 (x)| ď M for all x P (a, b), then the Mean Value Theorem suggests that f is Lipschitz
continuous. A Lipschitz continuous function must be uniformly continuous.
increasing
decreasing
Definition 4.70. A function f : (a, b) Ñ R is said to be
(on (a, b))
strictly increasing
strictly decreasing
ď
ě
if f (x1 )
f (x2 ) if a ă x1 ă x2 ă b. f is said to be monotone if f is either increasing
ă
ą
or decreasing on (a, b), and strictly monotone if f is either strictly increasing or strictly
decreasing.
Theorem 4.71. Suppose that f : (a, b) Ñ R is differentiable.
1. f is increasing on (a, b) if and only if f 1 (x) ě 0 for all x P (a, b).
2. f is decreasing on (a, b) if and only if f 1 (x) ď 0 for all x P (a, b).
3. If f 1 (x) ą 0 for all x P (a, b), then f is strictly increasing.
128
CHAPTER 4. Continuous Maps
4. If f 1 (x) ă 0 for all x P (a, b), then f is strictly decreasing.
Theorem 4.72 (Inverse Function Theorem). Let f : (a, b) Ñ R be differentiable, and f 1
be sign-definite; that is, f 1 (x) ą 0 for all x P (a, b) or f 1 (x) ă 0 for all x P (a, b). Then
f : (a, b) Ñ f ((a, b)) is a bijection, and f ´1 , the inverse function of f , is differentiable on
f ((a, b)), and
1
(f ´1 )1 (f (x)) = 1
@ x P (a, b) .
(4.6.1)
f (x)
Proof. W.L.O.G. we assume that f 1 (x) ą 0 for all x P (a, b). By Theorem 4.71 f is strictly
increasing; thus f ´1 exists.
Claim: f ´1 : f ((a, b)) Ñ (a, b) is continuous.
Proof of claim: Let y0 = f (x0 ) P f ((a, b)), and ε ą 0 be given. Then f ((x0 ´ ε, x0 + ε)) =
(
)
f (x0 ´ ε), f (x0 + ε) since f is continuous on (a, b) and (x0 ´ ε, x0 + ε) is connected. Let
(
δ = mintf (x0 ) ´ f (x0 ´ ε), f (x0 + ε) ´ f (x0 ) . Then δ ą 0, and
(
)
(y0 ´ δ, y0 + δ) = f (x0 ) ´ δ, f (x0 ) + δ Ď f ((x0 ´ ε, x0 + ε)) ;
thus by the injectivity of f ,
f ´1 ((y0 ´ δ, y0 + δ)) Ď f ´1 (f ((x0 ´ ε, x0 + ε))) = (x0 ´ ε, x0 + ε) = (f ´1 (y0 ) ´ ε, f ´1 (y0 ) + ε) .
The inclusion above implies that f ´1 is continuous at y0 .
Writing y = f (x) and x = f ´1 (y). Then if y0 = f (x0 ) P f ((a, b)),
f ´1 (y) ´ f ´1 (y0 )
x ´ x0
=
.
y ´ y0
f (x) ´ f (x0 )
Since f ´1 is continuous on f ((a, b)), x Ñ x0 as y Ñ y0 ; thus
f ´1 (y) ´ f ´1 (y0 )
1
x ´ x0
= lim
= 1
yÑy0
xÑx
y ´ y0
f (x0 )
0 f (x) ´ f (x0 )
lim
which implies that f ´1 is differentiable at y0 .
˝
4.7 Integration of Functions of One Variable
Definition 4.73. Let A Ď R be a bounded subset. A collection P of finitely many points
tx0 , x1 , ¨ ¨ ¨ , xn u is called a partition of A if inf A = x0 ă x1 ă ¨ ¨ ¨ ă xn´1 ă xn = sup A.
The mesh size of the partition P, denoted by }P}, is defined by
ˇ
␣
(
}P} = max xk ´ xk´1 ˇ k = 1, ¨ ¨ ¨ , n .
§4.7 Integration of Functions of One Variable
129
Definition 4.74. Let A Ď R be a bounded subset, and f : A Ñ R be a bounded function.
For any partition P = tx0 , x1 , ¨ ¨ ¨ , xn u of A, the upper sum and the lower sum of f
with respect to the partition P, denoted by U (f, P) and L(f, P) respectively, are numbers
defined by
U (f, P) =
n
ÿ
sup
fs(x)(xk ´ xk´1 ) =
k=1 xP[xk´1 ,xk ]
L(f, P) =
n
ÿ
k=1
inf
xP[xk´1 ,xk ]
n´1
ÿ
sup
fs(x)(xk+1 ´ xk ) ,
k=0 xP[xk ,xk+1 ]
fs(x)(xk ´ xk´1 ) =
n´1
ÿ
k=0
inf
xP[xk ,xk+1 ]
fs(x)(xk+1 ´ xk ) ,
where fs is an extension of f given by
"
fs(x) =
f (x) x P A ,
0
x R A.
(4.7.1)
The two numbers
ż
ˇ
␣
(
f (x)dx ” inf U (f, P) ˇ P is a partition of A ,
A
and
ż
ˇ
␣
(
f (x)dx ” sup L(f, P) ˇ P is a partition of A
A
are called the upper integral and lower integral of f over A, respective. The function
f is said to be Riemann (Darboux) integrable (over A) if
in this case, we express the upper and lower integral as
ż
ż
ż
f (x)dx =
A
f (x)dx, and
A
f (x)dx, called the integral of f
A
over A. The upper integral, the lower integral, and the integral of f over [a, b] sometimes
are also denoted by
żb
żb
f (x)dx,
a
Example 4.75.
f : [0, 1] Ñ R by
żb
a
f (x)dx, and
a
f (x)dx and
żb
f (x)dx.
a
żb
f (x)dx are not always the same. For example, define
a
"
f (x) =
1 if x P [0, 1]zQ,
0 if x P [0, 1] X Q.
Let P = t0 = x0 ă x1 ă ¨ ¨ ¨ ă xn = 1u be any partition on [0, 1]. Then for any k =
130
CHAPTER 4. Continuous Maps
0, 1, ¨ ¨ ¨ , n ´ 1,
sup
f (x) = 1 and
U (f, P) =
n´1
ÿ
inf
f (x) = 0; thus
xP[xk ,xk+1 ]
xP[xk ,xk+1 ]
sup
f (x)(xk ´ xk´1 ) =
k=0 xP[xk ,xk+1 ]
n
ÿ
(xk ´ xk´1 )
k=0
= (x1 ´ x0 ) + (x2 ´ x1 ) + ¨ ¨ ¨ + (xn ´ xn´1 ) = xn ´ x0 = 1 ´ 0 = 1
and
L(f, P) =
n
ÿ
0(xi ´ xi´1 ) = 0 .
i=1
As a consequence,
ż1
ż01
ˇ
␣
(
f (x)dx = inf U (f, P) ˇ P is a partition on [0, 1] = 1 ,
ˇ
␣
(
f (x)dx = sup L(f, P) ˇ P is a partition on [0, 1] = 0 ;
0
hence f is not Riemann integrable over [0, 1].
Example 4.76. Suppose f : [a, b] Ñ R is integrable and f ě 0 on [a, b], then
Reason: Since f ě 0 on [a, b] ñ
sup
żb
f (x)dx ě 0.
a
f (x) ě 0 for k = 0, 1, . . . , n ´ 1. Therefore,
xP[xk ,xk+1 ]
U (f, P) ě 0 for all partition P on [a, b], so
żb
żb
ˇ
(
f (x)dx =
f (x)dx = inftU (f, P) ˇ P is a partition on [a, b] ě 0 .
a
a
Definition 4.77. A partition P 1 of a bounded set A Ď R is said to be a refinement of
another partition P if P Ď P 1 .
Proposition 4.78. Let A Ď R be a bounded subset, and f : A Ñ R be a bounded function.
If P and P 1 are partitions of A and P 1 is a refinement of P, then
L(f, P) ď L(f, P 1 ) ď U (f, P 1 ) ď U (f, P) .
Proof. Let fs be the extension of f given by (4.7.1). Suppose that P = tx0 , x1 , ¨ ¨ ¨ , xn u, P 1 =
ty0 , y1 , ¨ ¨ ¨ , ym u, and P Ď P 1 . For any fixed k = 0, 1, ¨ ¨ ¨ , n ´ 1, either P 1 X (xk , xk+1 ) = H
or P 1 X (xk , xk+1 ) ‰ H.
1. If P 1 X (xk , xk+1 ) = H, then xk = yℓ and xk+1 = yℓ+1 for some ℓ. Therefore,
sup
xP[xk ,xk+1 ]
fs(x)(xk+1 ´ xk ) =
sup
xP[yℓ ,yℓ+1 ]
fs(x)(yℓ+1 ´ yℓ ).
§4.7 Integration of Functions of One Variable
131
2. If P 1 X (xk , xk+1 ) = tyℓ+1 , yℓ+2 , ¨ ¨ ¨ , yℓ+p u, then xk = yℓ and xk+1 = yℓ+p+1 . Therefore,
p+1
ÿ
sup
i=1 xP[yℓ+i´1 ,yℓ+i ]
fs(x)(yℓ+1 ´ yℓ )
xP[yℓ ,yℓ+1 ]
sup
+
sup
fs(x)(yℓ+i ´ yℓ+i´1 ) =
sup
fs(x)(yℓ+2 ´ yℓ+1 ) + ¨ ¨ ¨ +
xP[yℓ+1 ,yℓ+2 ]
xP[yℓ+p ,yℓ+p+1 ]
sup
ď
sup
fs(x)(yℓ+1 ´ yℓ ) +
xP[xk ,xk+1 ]
fs(x)(yℓ+2 ´ yℓ+1 ) + ¨ ¨ ¨
xP[xk ,xk+1 ]
sup
+
fs(x)(yℓ+p+1 ´ yℓ+p )
sup
fs(x)(yℓ+p+1 ´ yℓ+p ) =
xP[xk ,xk+1 ]
fs(x)(xk+1 ´ xk ) .
xP[xk ,xk+1 ]
In either case,
ÿ
sup
[yℓ´1 ,yℓ ]Ď[xk ,xk+1
] xP[yℓ´1 ,yℓ ]
fs(x)(yℓ ´ yℓ´1 ) ď
sup
fs(x)(xk+1 ´ xk ) .
xP[xk ,xk+1 ]
As a consequence,
U (f, P ) =
1
m´1
ÿ
sup
fs(x)(yℓ+1 ´ yℓ ) =
ℓ=0 xP[yℓ ,yℓ+1 ]
ď
n´1
ÿ
sup
n´1
ÿ
ÿ
fs(x)(yℓ ´ yℓ´1 )
k=0 [yℓ´1 ,yℓ ]Ď[xk ,xk+1 ]
fs(x)(xk+1 ´ xk ) = U (f, P) .
k=0 xP[xk ,xk+1 ]
Similarly, L(f, P) ď L(f, P 1 ); thus the fact that L(f, P 1 ) ď U (f, P 1 ) concludes the proposition.
˝
Corollary 4.79. Let f : [a, b] Ñ R be a function bounded by M ; that is, |f (x)| ď M for all
a ď x ď b. Then for all partitions P1 and P2 of [a, b],
żb
żb
´M (b ´ a) ď L(f, P1 ) ď
f (x)dx ď
f (x)dx ď U (f, P2 ) ď M (b ´ a) .
a
Proof. It suffices to show that
żb
a
żb
f (x)dx ď
a
f (x)dx. By the definition of infimum and
a
s and P
r such that
supremum, for any given ε ą 0, D partitions P
żb
żb
żb
żb
ε
ε
s
r
f (x)dx ´ ă L(f, P) ď
f (x)dx and
f (x)dx ď U (f, P) ă
f (x)dx + .
2
2
a
a
a
a
s Y P.
r Then P is a refinement of both P
s and P;
r thus
Let P = P
żb
żb
ε
ε
s
r
f (x)dx ´ ă L(f, P) ď L(f, P) ď U (f, P) ď U (f, P) ă
f (x)dx + .
2
2
a
a
132
CHAPTER 4. Continuous Maps
Since ε ą 0 is given arbitrarily, we must have
żb
żb
f (x)dx ď
a
f (x)dx.
˝
a
Proposition 4.80 (Riemann’s condition). Let A Ď R be a bounded set, and f : A Ñ R be
a bounded function. Then f is Riemann integrable over A if and only if
@ ε ą 0, D a partition P of A Q U (f, P) ´ L(f, P) ă ε .
Proof. “ñ” Let ε ą 0 be given. Since f is integrable over A,
inf
P: Partition of A
U (f, P) =
sup
P: Partition of A
L(f, P) =
ż
f (x)dx ;
A
thus there exist P1 and P2 , partitions of A, such that
ż
ż
ż
ε
ε
f (x)dx ´ ă L(f, P1 ) ď
f (x)dx ď U (f, P2 ) ă
f (x)dx + .
2
2
A
A
A
Let P = P1 Y P2 . Then P is a refinement of P1 and P2 ; thus
ż
ż
ε
f (x)dx
f (x)dx ´ ă L(f, P1 ) ď L(f, P) ď
2
A
A
ż
ε
ď U (f, P) ď U (f, P2 ) ă
f (x)dx +
2
A
which implies that U (f, P) ´ L(f, P) ă ε.
“ð” We note that for any partition P of A,
ż
ż
f (x)dx ď
f (x)dx ď U (f, P) ;
L(f, P) ď
A
A
so we have that for all partition P of A,
ż
ż
f (x)dx ´
f (x)dx ă U (f, P) ´ L(f, P) .
A
A
Let ε ą 0 be given. By choosing P so that U (f, P) ´ L(f, P) ă ε, we conclude that
ż
ż
f (x)dx ´
f (x)dx ă ε .
A
Since ε ą 0 is given arbitrarily,
over A.
A
ż
ż
f (x)dx =
A
A
f (x)dx; thus f is Riemann integrable
˝
§4.7 Integration of Functions of One Variable
133
Proposition 4.81. Suppose that f, g : [a, b] Ñ R are Riemann integrable, and k P R. Then
żb
1. kf is Riemann integrable, and
żb
(kf )(x)dx = k
f (x)dx.
a
a
2. f ˘ g are Riemann integrable, and
żb
żb
(f ˘ g)(x)dx =
f (x)dx ˘
a
a
żb
3. If f ď g for all x P [a, b], then
żb
g(x)dx.
a
żb
f (x)dx ď
a
g(x)dx.
a
4. If f is also Riemann integrable over [b, c], then f is Riemann integrable over [a, c],
and
żc
żb
f (x)dx =
a
żc
f (x)dx +
a
(4.7.2)
f (x)dx .
b
ˇż b
ˇ żb
ˇ
ˇ
5. The function |f | is also Riemann integrable, and ˇ f (x)dxˇ ď |f (x)|dx.
a
a
Proof. 1. Case 1. k ě 0. We note that
inf
inf
(kf )(x) = k
xP[xi´1 ,xi ]
f (x) and
sup (kf )(x) = k
xP[xi´1 ,xi ]
xP[xi´1 ,xi ]
sup
f (x) .
xP[xi´1 ,xi ]
Then
L(kf, P) =
=
n
ÿ
i=1
n
ÿ
inf
xP[xi´1 ,xi ]
(kf )(x)(xi ´ xi´1 )
inf
k
xP[xi´1 ,xi ]
i=1
f (x)(xi ´ xi´1 ) = kL(f, P) .
Similarly, U (kf, P) = kU (f, P) for every partition P. So
żb
sup
(kf )(x)dx =
P: Partition of [a, b]
a
L(kf, P) = k
f (x)dx = k
a
Similarly,
f (x)dx .
a
żb
żb
L(f, P)
żb
żb
=k
sup
P: Partition of [a, b]
(kf )(x)dx = k
f (x)dx. Hence kf is integrable and
a
a
żb
żb
(kf )(x)dx =
a
żb
(kf )(x)dx = k
a
żb
f (x)dx = k
a
f (x)dx .
a
134
CHAPTER 4. Continuous Maps
Case 2. k ă 0. We have
inf
(kf )(x) = k
xP[xi´1 ,xi ]
sup
f (x) and
xP[xi´1 ,xi ]
sup (kf )(x) = k
inf
f (x) .
xP[xi´1 ,xi ]
xP[xi´1 ,xi ]
Then L(kf, P) = kU (f, P) and U (kf, P) = kL(f, P); thus
żb
sup
(kf )(x)dx =
P: Partition of [a, b]
a
=k
Similarly,
inf
L(kf, P) =
P: Partition of [a, b]
żb
żb
(kf )(x)dx = k
a
sup
P: Partition of [a, b]
U (f, P) = k
kU (f, P)
żb
żb
f (x)dx = k
a
f (x)dx .
a
f (x)dx. Hence kf is Riemann integrable over [a, b] and
a
żb
żb
(kf )(x)dx =
a
żb
(kf )(x)dx = k
a
żb
f (x)dx = k
a
f (x)dx .
a
2. We prove the case of summation. For ant partition P, we have
L(f + g, P) =
ě
n
ÿ
i=1
n
ÿ
i=1
inf
xP[xi´1 ,xi ]
inf
xP[xi´1 ,xi ]
(f + g)(x)(xi ´ xi´1 )
f (x)(xi ´ xi´1 ) +
n
ÿ
i=1
inf
xP[xi´1 ,xi ]
g(x)(xi ´ xi´1 )
= L(f, P) + L(g, P) .
Similarly, U (f + g, P) ď U (f, P) + U (g, P). Therefore,
L(f, P) + L(g, P) ď L(f + g, P) ď U (f + g, P) ď U (f, P) + U (g, P) .
Let ε ą 0 be given. By Proposition 4.80, D P1 , P2 partitions of [a, b] such that
U (f, P1 ) ´ L(f, P1 ) ă
ε
2
and
U (g, P2 ) ´ L(g, P2 ) ă
ε
.
2
Let P = P1 Y P2 . By (4.7.3),
U (f + g, P) ´ L(f + g, P) ď (U (f, P) + U (g, P)) ´ (L(f, P) + L(g, P))
= (U (f, P) ´ L(f, P)) + (U (g, P) ´ L(g, P))
ď (U (f, P1 ) ´ L(f, P1 )) + (U (g, P2 ) ´ L(g, P2 )) ă
ε ε
+ = ε.
2 2
(4.7.3)
§4.7 Integration of Functions of One Variable
135
By Proposition 4.80, f + g is Riemann integrable over [a, b].
To see
żb
żb
(f + g)(x)dx =
żb
f (x)dx +
a
a
g(x)dx, we note that by Proposition 4.78,
a
ε
U (f, P) ď L(f, P) + U (f, P1 ) ´ L(f, P1 ) ă L(f, P) +
2
żb
żb
ε
ε
ď
f (x)dx + =
f (x)dx +
2
2
a
a
and similarly, U (g, P) ă
żb
ε
2
g(x)dx + . Therefore, by (4.7.3),
a
żb
żb
(f + g)(x)dx ď U (f + g, P)
żb
żb
ď U (f, P) + U (g, P) ă
f (x)dx + g(x)dx + ε .
(f + g)(x)dx =
a
a
a
(4.7.4)
a
On the other hand,
L(f, P) ą U (f, P) ´
ε
ě
2
żb
ε
ě
2
żb
and
L(g, P) ą U (g, P) ´
f (x)dx ´
ε
2
g(x)dx ´
ε
;
2
a
a
hence by (4.7.3),
żb
żb
(f + g)(x)dx =
(f + g)(x)dx ě L(f + g, P) ě L(f, P) + L(g, P)
a
a
żb
żb
f (x)dx +
ą
a
(4.7.5)
g(x)dx ´ ε .
a
By (4.7.4) and (4.7.5),
żb
żb
żb
żb
żb
f (x)dx + g(x)dx ´ ε ă (f + g)(x)dx ă
f (x)dx + g(x)dx + ε .
a
a
Since ε ą 0 is arbitrary,
a
a
żb
żb
żb
g(x)dx.
f (x)dx +
(f + g)(x)dx =
a
a
a
a
3. Let P = ta = x0 ă x1 ă ¨ ¨ ¨ ă xn = bu be a partition of [a, b]. Define
mi (f ) =
inf
xP[xi´1 ,xi ]
f (x)
and mi (g) =
inf
xP[xi´1 ,xi ]
g(x) .
136
CHAPTER 4. Continuous Maps
Since f (x) ď g(x) on [a, b], mi (f ) ď mi (g). As a consequence, for any partition P,
n
ÿ
L(f, P) =
mi (f )(xi ´ xi´1 ) ď
i=1
n
ÿ
mi (g)(xi ´ xi´1 ) = L(g, P) ;
i=1
thus taking the infimum over all partition P,
żb
żb
f (x)dx =
a
f (x)dx = sup L(f, P) ď sup L(g, P) =
P
a
żb
żb
g(x)dx =
P
a
g(x)dx .
a
4. Let ε ą 0 be given. Since f is Riemann integrable of [a, b] and [b, c], there exist a
partition P1 over [a, b] and a partition P2 of [b, c] such that
U (f, P1 ) ´ L(f, P1 ) ă
ε
2
and
U (f, P2 ) ´ L(f, P2 ) ă
ε
.
2
Let P = P1 Y P2 . Then P is a partition of [a, c] such that
U (f, P) ´ L(f, P) = U (f, P1 ) + U (f, P2 ) ´ L(f, P1 ) ´ L(f, P2 ) ă ε .
Therefore, Proposition 4.80 suggests that f is Riemann integrable over [a, c].
Now we show that
żc
żb
f (x)dx =
a
we let
żc
f (x)dx +
a
żc
A=
żb
f (x)dx,
a
f (x)dx. To simplify the notation,
b
B=
żc
f (x)dx,
C=
a
f (x)dx .
b
Let ε ą 0 be given. Then D partition P = tx0 , x1 , ¨ ¨ ¨ , xn u of [a, c] such that
A ď U (f, P) ă A + ε .
Let P 1 = P Y tbu. Then P 1 is a refinement of P. Moreover,
U (f, P 1 ) = U (f, P1 ) + U (f, P2 ) ,
where P1 = P 1 X [a, b] and P2 = P 1 X [b, c] are partitions of [a, b] and [b, c] whose union
is P. Therefore,
B + C ď U (f, P1 ) + U (f, P2 ) = U (f, P 1 ) ď U (f, P) ă A + ε .
On the other hand, D partition P1 of [a, b] and partition P2 of [b, c] such that
ε
ε
and C ď U (f, P2 ) ă C + .
B ď U (f, P1 ) ă B +
2
2
Let P = P1 Y P2 . Then P is a partition of [a, c]. Therefore,
A ď U (f, P) = U (f, P1 ) + U (f, P2 ) ă B + C + ε .
Therefore, @ ε ą 0, B + C ă A + ε and A ă B + C + ε; thus A = B + C.
§4.7 Integration of Functions of One Variable
137
5. Note that for any interval [α, β],
sup |f (x)| ´ inf |f (x)| ď sup f (x) ´ inf f (x) ;
xP[α,β]
xP[α,β]
(Check!)
xP[α,β]
xP[α,β]
thus for any partition P of [a, b],
U (|f |, P) ´ L(|f |, P) ď U (f, P) ´ L(f, P) .
Therefore, Proposition 4.80 suggests that |f | is Riemann integrable over [a, b]. Moreover, since ´|f (x)| ď f (x) ď |f (x)| for all x P [a, b], by 3 we have
żb
żb
żb
´ |f (x)|dx ď
f (x)dx ď
|f (x)|dx .
˝
a
a
a
Remark 4.82. The proof of 4 in Proposition 4.81 in fact also shows that if a ă b ă c, then
żc
żb
żc
f (x)dx =
f (x)dx +
f (x)dx .
a
a
b
Similar proof also suggests that
żc
żb
żc
f (x)dx =
f (x)dx +
f (x)dx .
a
a
Remark 4.83. If a ă b, we let the number
b
ża
f (x)dx denote the number ´
b
(4.7.2) holds for all a, b, c P R.
żb
f (x)dx. Then
a
Example 4.84. Let f : [0, 1] Ñ R be defined by
"
1 if x P Q,
f (x) =
´1 if x P QA .
Then f (x) is not Riemann integrable over [0, 1] since U (f, P ) = 1 and L(f, P ) = ´1.
However |f (x)| ” 1, thus |f | is Riemann integrable. In other words, if |f | is integrable, we
cannot know whether f is integrable or not.
Theorem 4.85. If f : [a, b] Ñ R is continuous, then f is Riemann integrable.
Proof. Let ε ą 0 be given. Theorem 4.52 suggests that
D δ ą 0 Q |f (x) ´ f (y)| ă
ε
whenever |x ´ y| ă δ and x, y P [a, b] .
2(b ´ a)
138
CHAPTER 4. Continuous Maps
Let P be a partition with mesh size less than δ. Then
U (f, P) ´ L(f, P) =
n
ÿ
(
k=1
sup
xP[xk´1 ,xk ]
f (x) ´
inf
xP[xk´1 ,xk ]
)
f (x) (xk ´ xk´1 )
n
ÿ
ε
ε
ď
(xk ´ xk´1 ) =
(xn ´ x0 ) ă ε ;
2(b ´ a) k=1
2(b ´ a)
thus by Proposition 4.80 f is Riemann integrable over [a, b].
˝
Corollary 4.86. If f : (a, b) Ñ R is continuous and f is bounded on [a, b], then f is
Riemann integrable over [a, b].
[
ε
ε ]
Proof. Let |f (x)| ď M for all x P [a, b], and ε ą 0 be given. Since f : a +
, b´
ÑR
8M
8M
is continuous, by Theorem 4.85 f is Riemann integrable; thus
[
ε
ε
ε ]
D P 1 : partition of a +
,b ´
Q U (f, P 1 ) ´ L(f, P 1 ) ă .
8M
8M
2
Let P = P 1 Y ta, bu. Then
U (f, P) ´ L(f, P)
(
ă
sup f (x) ´
ε
xP[a,a+ 8M
]
ď 2M ¨
inf ε f (x)
xP[a,a+ 8M ]
) ε
) ε
ε (
+ +
sup f (x) ´ infε f (x)
ε
xP[b´ 8M ,b]
8M
2
8M
xP[b´ 8M
,b]
ε
ε
ε
+ + 2M ¨
= ε;
8M
2
8M
thus Proposition 4.80 suggests that f is Riemann integrable over [a, b].
˝
Corollary 4.87. If f : [a, b] Ñ R is bounded and is continuous at all but finitely many
points of [a, b], then f is Riemann integral.
Proof. Let tc1 , ¨ ¨ ¨ , cN u be the collection of all discontinuities of f in (a, b) such that c1 ă
c2 ă ¨ ¨ ¨ ă cN . Let a = c0 and b = cN +1 . Then for all k = 0, 1, ¨ ¨ ¨ , N , f : (ck , ck+1 ) is
continuous and f : [ck , ck+1 ] is bounded; thus f is Riemann integrable by Corollary 4.87.
Finally, 4 of Proposition 4.81 suggests that f is Riemann integrable over [a, b].
˝
Theorem 4.88. Any increasing or decreasing function on [a, b] is Riemann integrable.
Proof. Let f : [a, b] Ñ R be a monotone function, and ε ą 0 be given. W.L.O.G. we may
assume that f (b) ‰ f (a). Let P = tx0 , x1 , ¨ ¨ ¨ , xn u be a partition of [a, b] with mesh size
§4.7 Integration of Functions of One Variable
less than
139
ε
. Then
|f (b) ´ f (a)|
U (f, P) ´ L(f, P) =
ă
n
ÿ
(
k=1
n
ÿ
sup
f (x) ´
xP[xk´1 ,xk ]
ˇ
ˇ
ˇf (xk ) ´ f (xk´1 )ˇ
k=1
inf
xP[xk´1 ,xk ]
)
f (x) (xk ´ xk´1 )
ˇ
ˇ
ε
ε
= ˇf (b) ´ f (a)ˇ
= ε;
|f (b) ´ f (a)|
|f (b) ´ f (a)|
thus Proposition 4.80 suggests that f is Riemann integrable over [a, b].
˝
Definition 4.89. A continuous function F : [a.b] Ñ R is called an anti-derivative（反導
函數）of f : [a, b] Ñ R if F is differentiable on (a, b) and F 1 (x) = f (x) for all x P (a, b).
Theorem 4.90 (Fundamental Theorem of Calculus（微積分基本定理）). Let f : [a, b] Ñ R
be continuous. Then f has an anti-derivative F , and
żb
f (x)dx = F (b) ´ F (a) .
a
Moreover, if G is any other anti-derivative of f , we also have
żb
f (x)dx = G(b) ´ G(a).
a
Proof. Define F (x) =
żx
f (y)dy, where the integral of f over [a, x] is well-defined because
a
of continuity of f on [a, x]. We first show that F is differentiable on (a, b).
Let x0 P (a, b) and ε ą 0 be given. Since [a, b] is compact,
ˇ
ˇ ε
D δ1 ą 0 Q ˇf (x) ´ f (y)ˇ ă whenever |x ´ y| ă δ1 and x, y P [a, b] .
2
Let δ = mintδ1 , x0 ´ a, b ´ x0 u. By 4 of Proposition 4.81, if x, x0 P (a, b),
ż x0
żx
żx
f (y)dy = F (x) ´ F (x0 ) ;
f (y)dy =
f (y)dy ´
x0
a
a
thus if 0 ă |x ´ x0 | ă δ,
ˇ F (x) ´ F (x )
ˇ ˇ 1 żx
ˇ ˇ 1 żx(
) ˇˇ
ˇ
ˇ ˇ
ˇ ˇ
0
´ f (x0 )ˇ = ˇ
f (y)dx ´ f (x0 )ˇ = ˇ
f (y) ´ f (x0 ) dy ˇ
ˇ
x ´ x0
x ´ x 0 x0
x ´ x 0 x0
ż maxtx0 ,xu
ż maxtx0 ,xu
ˇ
ˇ
1
1
ε
ˇ
ˇ
dy ă ε .
ď
f (y) ´ f (x0 ) dy ď
|x ´ x0 | mintx0 ,xu
|x ´ x0 | mintx0 ,xu 2
F (x) ´ F (x0 )
= f (x0 ) for all x0 P (a, b), so F 1 (x) = f (x) for all x P (a, b).
xÑx0
x ´ x0
Therefore, lim
140
CHAPTER 4. Continuous Maps
Next we show that F is continuous at x = a and x = b. This is simply because of the
boundedness of f on [a, b] which suggests that
żx
ˇż x
ˇ
ˇ
ˇ
ˇ
ˇ
lim sup ˇF (x) ´ F (a)ˇ = lim sup ˇ
f (t)dtˇ ď max |f (x)| ¨ lim sup
1dt = 0
xÑa+
xÑa+
xP[a,b]
a
xÑa+
a
and
żb
ˇż b
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
lim sup F (x) ´ F (b) = lim sup ˇ f (t)dtˇ ď max |f (x)| ¨ lim sup 1dt = 0 .
xÑb´
xÑb´
xP[a,b]
x
xÑb´
x
Therefore, F is an anti-derivative of f .
Now suppose that G is another anti-derivative of f . Then (G ´ F )1 (x) = 0 for all
x P (a, b). By Corollary 4.66, (G ´ F )(x) = (G ´ F )(a) for all x P [a, b]; thus G(b) ´ G(a) =
F (b) ´ F (a).
˝
Example 4.91. If f is only integrable but not continuous, then the function
żx
F (x) =
f (t)dt
a
is not necessarily differentiable. For example, consider
"
0 if 0 ď x ď 1,
f (x) =
1 if 1 ă x ď 2.
Then
"
F (x) =
0
if 0 ď x ď 1,
x ´ 1 if 1 ď x ď 2.
so F is continuous on [0, 2] but not differentiable at x = 1.
Theorem 4.92. Let f : [a, b] Ñ R be differentiable. If f 1 is Riemann integrable over [a, b],
then
żb
f 1 (x)dx = f (b) ´ f (a).
a
Proof. Let P = tx0 , x1 , ¨ ¨ ¨ xn u be a partition of [a, b]. Since f : [a, b] Ñ R is differentiable,
by the Mean Value Theorem there exists tξ1 , ¨ ¨ ¨ , ξn u with the property that xk ă ξk+1 ă
xk+1 for all k = 0, 1 ¨ ¨ ¨ , n ´ 1 such that
f 1 (ξk+1 )(xk+1 ´ xk ) = f (xk+1 ) ´ f (xk )
@ k = 0, 1, ¨ ¨ ¨ , n ´ 1 .
Therefore,
n´1
ÿ
k=0
inf
xP[xk ,xk+1 ]
1
f (x)(xk+1 ´ xk ) ď
n´1
ÿ
k=0
1
f (ξk+1 )(xk+1 ´ xk ) ď
n´1
ÿ
sup
k=0 xP[xk ,xk+1 ]
f 1 (x)(xk+1 ´ xk ) .
§4.7 Integration of Functions of One Variable
Since
n´1
ř
f 1 (ξk+1 )(xk+1 ´ xk ) =
k=0
141
n´1
ř(
)
f (xk+1 ) ´ f (xk ) = f (b) ´ f (a), the inequality above
k=0
implies that
L(f 1 , P) ď f (b) ´ f (a) ď U (f 1 , P) for all partitions P of [a, b] ;
thus by the definition of the upper and the lower integrals,
żb
żb
1
f (x)dx ď f (b) ´ f (a) ď
f 1 (x)dx .
a
a
We then conclude the theorem by the identity
żb
żb
1
f (x)dx =
a
1
żb
f (x)dx =
a
f 1 (x)dx
a
since f 1 is Riemann integrable.
˝
Definition 4.93. Let P = tx0 , x1 , ¨ ¨ ¨ , xn u be a partition of a bounded set A Ď R. A
collection of points tξ1 , ¨ ¨ ¨ , ξn u is called a sample set for the partition P if ξk P [xk´1 , xk ]
for all k = 1, ¨ ¨ ¨ , n.
Let f : A Ñ R be a bounded function with extension fs given by (4.7.1). A Riemann
sum of f for the partition P = ta = x0 ă x1 ă ¨ ¨ ¨ ă xn = bu of A is a sum which takes the
form
n´1
ÿ
fs(ξk )(xk+1 ´ xk ) ,
k=0
where the set Ξ = tξ0 , ξ1 , ¨ ¨ ¨ , ξn´1 u is a sample set for P.
Theorem 4.94 (Darboux). Let f : A Ñ R be a bounded function with extension fs given by
(4.7.1). Then f is Riemann integrable over A if and only if there exists I P R such that for
every given ε ą 0, there exists δ ą 0 such that if P is a partition of A satisfying }P} ă δ,
then any Riemann sum of f for the partition P lies in the interval (I ´ ε, I + ε). In other
words, f is Riemann integrable over A if and only if for every given ε ą 0, there exists δ ą 0
such that there exists I P R such that
ˇ
ˇ n´1
ˇ
ˇÿ s
f
(ξ
)(x
´
x
)
´
I
ˇăε
ˇ
k+1
k+1
k
(4.7.6)
k=0
whenever P = tx0 , x1 , ¨ ¨ ¨ , xn u is a partition of A satisfying }P} ă δ and tξ1 , ξ2 , ¨ ¨ ¨ , ξN u is
a sample set for P.
142
CHAPTER 4. Continuous Maps
Proof. “ð” Suppose the right-hand side statement is true. Let ε ą 0 be given. Then there
exists δ ą 0 such that if P is a partition of A satisfying }P} ă δ, then for all sets of
sample points tξ1 , ¨ ¨ ¨ , ξn u with respect to P, we must have
ˇ n´1
ˇ ε
ˇÿ s
ˇ
f (ξk+1 )(xk+1 ´ xk ) ´ I ˇ ă .
ˇ
4
k=0
Let P = tx0 , x1 , ¨ ¨ ¨ , xn u be a partition of A with }P} ă δ. Choose sets of sample
points tξ1 , ¨ ¨ ¨ , ξn u and tη1 , ¨ ¨ ¨ , ηn u with respect to P such that
(a)
sup
fs(x) ´
ε
ă fs(ξk+1 ) ď sup fs(x);
4(xn ´ x0 )
xP[xk ,xk+1 ]
fs(x) +
ε
ą fs(ηk+1 ) ě
inf fs(x).
4(xn ´ x0 )
xP[xk ,xk+1 ]
xP[xk ,xk+1 ]
(b)
inf
xP[xk ,xk+1 ]
Then
U (f, P) =
=
n´1
ÿ
sup
fs(x)(xk+1 ´ xk ) ă
k=0 xP[xk ,xk+1 ]
n´1
ÿ
n´1
ÿ
[
fs(ξk+1 ) +
k=0
]
ε
(xk+1 ´ xk )
4(xn ´ x0 )
n´1
fs(ξk+1 )(xk+1 ´ xk ) +
k=0
ÿ
ε ε
ε
ε
(xk+1 ´ xk ) ă I + + = I +
4(xn ´ x0 )
4 4
2
k=0
and
L(f, P) =
=
n´1
ÿ
k=0
n´1
ÿ
inf
xP[xk ,xk+1 ]
fs(x)(xk+1 ´ xk ) ą
n´1
ÿ
[
fs(ηk+1 ) ´
k=0
]
ε
(xk+1 ´ xk )
4(xn ´ x0 )
n´1
fs(ηk+1 )(xk+1 ´ xk ) ´
k=0
As a consequence, I ´
ÿ
ε
ε ε
ε
(xk+1 ´ xk ) ą I ´ ´ = I ´ .
4(xn ´ x0 )
4 4
2
k=0
ε
ε
ă L(f, P) ď U (f, P) ă I + ; thus U (f, P) ´ L(f, P) ă ε.
2
2
ż
“ñ” Let ε ą 0 be given, and I =
fs(x)dx. Since f is Riemann integrable over A, there
A
ε
2
exists a partition P1 = ty0 , y1 , ¨ ¨ ¨ , ym u of A such that U (f, P1 ) ´ L(f, P1 ) ă . Define
!
δ = min |y1 ´ y0 |, |y2 ´ y1 |, ¨ ¨ ¨ , |ym ´ ym´1 |,
ε
(
) .
4m sup f (A) ´ inf f (A) + 1
)
If P = tx0 , x1 , ¨ ¨ ¨ , xn u is a partition of A with }P} ă δ, then at most 2m intervals
of the form [xk , xk+1 ] contains one of these yj ’s, and each such interval [xk , xk+1 ] can
only contain one of these yj ’s. Let P 1 = P Y P1 .
§4.7 Integration of Functions of One Variable
143
ε
2
Claim: U (f, P) ´ U (f, P 1 ) ă .
Proof of claim: We note that
U (f, P) =
n´1
ÿ
sup
fs(x)(xk+1 ´ xk )
k=0 xP[xk ,xk+1 ]
ÿ
sup
0ďkďn´1 with
P1 X[xk ,xk+1 ]=H
xP[xk ,xk+1 ]
=
ÿ
sup
0ďkďn´1 with
P1 X[xk ,xk+1 ]‰H
xP[xk ,xk+1 ]
fs(x)(xk+1 ´ xk ) +
fs(x)(xk+1 ´ xk )
and
ÿ
sup
0ďkďn´1 with
P1 X[xk ,xk+1 ]=H
xP[xk ,xk+1 ]
U (f, P 1 ) =
[
ÿ
+
0ďkďn´1 with
P1 X[xk ,xk+1 ]=yj
fs(x)(xk+1 ´ xk )
sup fs(x)(yj ´ xk ) +
xP[xk ,yj ]
sup
]
fs(x)(xk+1 ´ yj ) .
xP[yj ,xk+1 ]
Therefore,
(
)
U (f, P) ´ U (f, P 1 ) ď sup f (A) ´ inf f (A)
ÿ
(xk+1 ´ xk )
0ďkďn´1 with
P1 X[xk ,xk+1 ]‰H
(
)
ε
ă 2m sup f (A) ´ inf f (A) δ ď .
2
On the other hand, the inequality U (f, P1 ) ´ L(f, P1 ) ă
U (f, P1 ) ´ I ă
ε
implies that
2
ε
.
2
As a consequence,
U (f, P) ´ I ď U (f, P) ´ I + U (f, P1 ) ´ U (f, P 1 ) ă ε .
Therefore, for any sample set tξ1 , ¨ ¨ ¨ , ξn u with respect to P,
n´1
ÿ
fs(ξk+1 )(xk+1 ´ xk ) ď U (f, P) ă I + ε .
k=0
Similar argument can be used to show that
n´1
ÿ
fs(ξk+1 )(xk+1 ´ xk ) ě L(f, P) ą I ´ ε ;
k=0
thus (4.7.6) is established.
˝
144
CHAPTER 4. Continuous Maps
Theorem 4.95 (Change of Variable Formula). Let g : [a, b] Ñ R be a one-to-one continuously differentiable function, and f : g([a, b]) Ñ R be Riemann integrable. Then (f ˝ g)g 1 is
also Riemann integrable, and
ż
żb
f (y) dy =
g([a,b])
f (g(x))|g 1 (x)| dx .
a
Proof. We only prove the case that f is continuous on g([a, b]), and the general case is
covered by Theorem 8.65 (which will be proved in detail).
W.L.O.G. we can assume that g 1 (x) ě 0 for all x P [a, b] so that g([a, b]) = [g(a), g(b)].
Let F be an anti-derivative of f . Then F is differentiable, and the chain rule suggests that
d
(F ˝ g)(x) = (F 1 ˝ g)(x)g 1 (x) = (f ˝ g)(x)g 1 (x) .
dx
Therefore, the fundamental theorem of Calculus implies that
żb
ż g(b)
ż
d
f (y)dy = F (g(b)) ´ F (g(a)) =
f (y)dy =
(F ˝ g)(x)dx
a dx
g(a)
g([a,b])
żb
= (f ˝ g)(x)g 1 (x)dx .
a
4.8 Exercises
§4.1 Continuity
Started from this section, for all n P N Rn always denotes the normed space (Rn , } ¨ }2 ).
Problem 4.1. Use whatever methods you know to find the following limits:
?
(?
)
1
1. lim+ (1 + sin 2x) x ;
2. lim
1 + x + x2 ´ 1 ´ x + x2 ;
xÑ´8
xÑ0
πx
3. lim (2 ´ x)sec 2 ;
xÑ1
4. lim x
xÑ8
(
( x )x )
5. lim x e´1 ´
;
xÑ8
(π
6. lim
x+1
xÑ8
2
)
x
;
x +1
´ sin´1 ? 2
( ax ´ 1 ) x1
x(a ´ 1)
, where a ą 0 and a ‰ 1.
Problem 4.2. Complete the following.
1. Find a function f : R2 Ñ R such that
lim lim f (x, y) and
xÑ0 yÑ0
exist but are not equal.
lim lim f (x, y)
yÑ0 xÑ0
˝
§4.8 Exercises
145
2. Find a function f : R2 Ñ R such that the two limits above exist and are equal but f
is not continuous.
3. Find a function f : R2 Ñ R that is continuous on every line through the origin but is
not continuous.
Problem 4.3. Complete the following.
R2 Ñ R
is continuous.
(x, y) ÞÑ x
ˇ
␣
(
2. Show that if U Ď R is open, then A = (x, y) P R2 ˇ x P U is open.
1. Show that the projection map f :
3. Give an example of a continuous function f : R Ñ R and an open set U Ď R such
that f (U) is not open.
Problem 4.4. Show that f : A Ñ Rm , where A Ď Rn , is continuous if and only if for every
B Ď A,
f (cl(B) X A) Ď cl(f (B)) .
Problem 4.5. Let } ¨ } be a norm on Rn , and f : Rn Ñ R be defined by f (x) = }x}. Show
that f is continuous on (Rn , } ¨ }2 ).
Hint: Show that |f (x) ´ f (y)| ď C}x ´ y}2 for some fixed constant C ą 0.
Problem 4.6. Let T : Rn Ñ Rm satisfy T (x + y) = T (x) + T (y) for all x, y P Rn .
1. Show that T (rx) = rT (x) for all r P Q and x P Rn .
2. Suppose that T is continuous on Rn . Show that T is linear; that is, T (cx + y) =
cT (x) + T (y) for all c P R, x, y P Rn .
3. Suppose that T is continuous at some point x0 in Rn . Show that T is continuous on
Rn .
4. Suppose that T is bounded on some open subset of Rn . Show that T is continuous on
Rn .
5. Suppose that T is bounded from above (or below) on some open subset of Rn . Show
that T is continuous on Rn .
146
CHAPTER 4. Continuous Maps
6. Construct a T : R Ñ R which is discontinuous at every point of R, but T (x + y) =
T (x) + T (y) for all x, y P R.
Problem 4.7. Let (M, d) be a metric space, A Ď M , and f : A Ñ R. For a P A1 , define
ˇ
␣
(
lim inf f (x) = lim+ inf f (x) ˇ x P D(a, r) X Aztau ,
xÑa
rÑ0
ˇ
␣
(
lim sup f (x) = lim+ sup f (x) ˇ x P D(a, r) X Aztau .
rÑ0
xÑa
Complete the following.
1. Show that both lim inf f (x) and lim sup f (x) exist (which may be ˘8), and
xÑa
xÑa
lim inf f (x) ď lim sup f (x) .
xÑa
xÑa
8
8
Furthermore, there exist sequences txn u8
n=1 , tyn un=1 Ď Aztau such that txn un=1 and
tyn u8
n=1 both converge to a, and
lim f (xn ) = lim inf f (x)
nÑ8
xÑa
and
lim f (yn ) = lim sup f (x) .
nÑ8
xÑa
2. Let txn u8
n=1 Ď Aztau be a convergent sequence with limit a. Show that
lim inf f (x) ď lim inf f (xn ) ď lim sup f (yn ) ď lim sup f (x) .
xÑa
nÑ8
nÑ8
xÑa
3. Show that lim f (x) = ℓ if and only if
xÑa
lim inf f (x) = lim sup f (x) = ℓ .
xÑa
xÑa
4. Show that lim inf f (x) = ℓ P R if and only if the following two conditions hold:
xÑa
(a) for all ε ą 0, there exists δ ą 0 such that ℓ ´ ε ă f (x) for all x P D(a, δ) X Aztau;
(b) for all ε ą 0 and δ ą 0, there exists x P D(a, δ) X Aztau such that f (x) ă ℓ + ε.
Formulate a similar criterion for limsup and for the case that ℓ = ˘8.
5. Compute the liminf and limsup of the following functions at any point of R.
$
& 0 if x P QA ,
(a) f (x) =
1
q
%
if x = with (p, q) = 1, q ą 0, p ‰ 0 .
p
p
§4.8 Exercises
147
"
(b) f (x) =
x if x P Q ,
´x if x P QA .
Problem 4.8. Let (M, d) be a metric space, and A Ď M . A function f : A Ñ R is called
lim inf f (x) ě f (a) ,
lower semi-continuous
xÑa
at a P A if
and is called lower/upper
upper semi-continuous
lim sup f (x) ď f (a) ,
xÑa
semi-continuous on A if f is lower/uppser semi-continuous at a for all a P A.
1. Show that if f : A Ñ R is lower semi-continuous on A, then f ´1 ((´8, r]) is closed
relative to A. Also show that if f : A Ñ R is upper semi-continuous on A, then
f ´1 ([r, 8)) is closed relative to A.
2. Show that f is lower semi-continuous at a if and only if for all convergent sequences
8
txn u8
n=1 Ď A and trn un=1 Ď R satisfying f (xn ) ď rn for all n P N, we have
(
)
f lim xn ď lim rn .
nÑ8
nÑ8
3. Let tfα uαPI be a family of lower semi-continuous functions on A. Prove that f (x) =
sup fα (x) is lower semi-continuous on A.
αPI
4. Let f : A Ñ R be given. Define
f ˚ (x) = lim sup f (y)
and
f˚ (x) = lim inf f (y) .
yÑx
yÑx
Show that f ˚ is upper semi-continuous and f˚ is lower semi-continuous, and f˚ (x) ď
f (x) ď f ˚ (x) for all x P A. Moreover, if g is a lowe semi-continuous function on A
such that g(x) ď f (x) for all x P A, then g ď f˚ .
§4.2 Operations on Continuous Maps
Problem 4.9.
Problem 4.10.
§4.3 Images of Compact Sets under Continuous Maps
Problem 4.11. Complete the following.
1. Show that if f : Rn Ñ Rm is continuous, and B Ď Rn is bounded, then f (B) is
bounded.
148
CHAPTER 4. Continuous Maps
2. If f : R Ñ R is continuous and K Ď R is compact, is f ´1 (K) necessarily compact?
3. If f : R Ñ R is continuous and C Ď R is connected, is f ´1 (C) necessarily connected?
Problem 4.12. Consider a compact set K Ď Rn and let f : K Ñ Rm be continuous and
one-to-one. Show that the inverse function f ´1 : f (K) Ñ K is continuous. How about if K
is not compact but connected?
Problem 4.13. Let (M, d) be a metric space, K Ď M be compact, and f : K Ñ R be lower
semi-continuous (see Problem 4.8 for the definition). Show that f attains its minimum on
K.
§4.4 Images of Connected and Path Connected Sets under Continuous Maps
Problem 4.14. Let D Ď Rn be an open connected set, where n ą 1. If a, b and c are any
three points in D, show that there is a path in G which connects a and b without passing
through c. In particular, this shows that D is path connected and D is not homeomorphic
to any subset of R.
Problem 4.15.
§4.5 Uniform Continuity
Problem 4.16. Check if the following functions on uniformly continuous.
1. f : (0, 8) Ñ R defined by f (x) = sin log x.
1
x
2. f : (0, 1) Ñ R defined by f (x) = x sin .
3. f : (0, 8) Ñ R defined by f (x) =
?
x.
4. f : R Ñ R defined by f (x) = cos(x2 ).
5. f : R Ñ R defined by f (x) = cos3 x.
6. f : R Ñ R defined by f (x) = x sin x.
sin(xa )
Problem 4.17. Find all positive numbers a and b such that the function f (x) =
is
1 + xb
uniformly continuous on [0, 8).
§4.8 Exercises
149
Problem 4.18. Find all positive numbers a and b such that the function f (x, y) = |x|a |y|b
is uniformly continuous on R2 .
Problem 4.19. Let f : Rn Ñ Rm be continuous, and lim f (x) = b exists for some b P Rm .
|x|Ñ8
Show that f is uniformly continuous on Rn .
Problem 4.20. Suppose that f : Rn Ñ Rm is uniformly continuous. Show that there exists
a ą 0 and b ą 0 such that }f (x)}Rm ď a}x}Rn + b.
Problem 4.21. Let f (x) =
q(x)
be a rational function define on R, where p and q are two
p(x)
polynomials. Show that f is uniformly continuous on R if and only if the degree of q is not
more than the degree of p plus 1.
Problem 4.22. Suppose that f : R Ñ R is a continuous periodic function; that is, D p ą 0
such that f (x + p) = f (x) for all x P R (and f is continuous). Show that f is uniformly
continuous on R.
Problem 4.23. Let (a, b) Ď R be an open interval, and f : (a, b) Ñ Rm be a function.
Show that the following three statements are equivalent.
1. f is uniformly continuous on (a, b).
2. f is continuous on (a, b), and both limits lim+ f (x) and lim´ f (x) exist.
xÑa
xÑb
ˇ
ˇ
ˇ
ˇ
ˇ f (x) ´ f (y) ˇ
ˇ
ˇ
3. For all ε ą 0, there exists N ą 0 such that f (x) ´ f (y) ă ε whenever ˇ
ˇą
x´y
N.
Problem 4.24. Suppose that f : [a, b] Ñ R is Hölder continuous with exponent α;
that is, there exist M ą 0 and α P (0, 1] such that
|f (x1 ) ´ f (x2 )| ď M |x1 ´ x2 |α
@ x1 , x2 P [a, b] .
Show that f is uniformly continuous on [a, b]. Show that f : [0, 8) Ñ R defined by
?
1
f (x) = x is Hölder continuous with exponent .
2
Problem 4.25. A function f : A ˆ B Ñ Rm , where A Ď R and B Ď Rp , is said to be
separately continuous if for each x0 P A, the map g(y) = f (x0 , y) is continuous and for
150
CHAPTER 4. Continuous Maps
y0 P B, h(x) = f (x, y0 ) is continuous. f is said to be continuous on A uniformly with
respect to B if
›
›
@ ε ą 0, D δ ą 0 Q ›f (x, y) ´ f (x0 , y)›2 ă ε whenever }x ´ x0 }2 ă δ and y P B .
Show that if f is separately continuous and is continuous on A uniformly with respect to
B, then f is continuous on A ˆ B.
Problem 4.26. Let (M, d) be a metric space, A Ď M , and f, g : A Ñ R be uniformly
continuous on A. Show that if f and g are bounded, then f g is uniformly continuous on A.
Does the conclusion still hold if f or g is not bounded?
§4.6 Differentiation of Functions of One Variable
Problem 4.27. Show that f : (a, b) Ñ R is differentiable at x0 P (a, b) if and only if there
exists m P R, denoted by f 1 (x0 ), such that
ˇ
ˇ
@ ε ą 0, D δ ą 0 Q ˇf (x) ´ f (x0 ) ´ f 1 (x0 )(x ´ x0 )ˇ ď ε|x ´ x0 | whenever |x ´ x0 | ă δ .
Problem 4.28. Suppose that f, g : R Ñ R are differentiable, and f ě 0. Find
d
f (x)g(x) .
dx
Problem 4.29. Suppose α and β are real numbers, β ą 0 and f : [´1, 1] Ñ R is defined
by
"
f (x) =
xα sin(x´β ) if x ‰ 0 ,
0
if x = 0 .
Prove the following statements.
1. f is continuous if and only if α ą 0.
2. f 1 (0) exists if and only if α ą 1.
3. f 1 is bounded if and only if α ě 1 + β.
4. f 1 is continuous if and only if α ą 1 + β.
5. f 2 (0) exists if and only if α ą 2 + β.
6. f 2 is bounded if and only if α ě 2 + 2β.
7. f 2 is continuous if and only if α ą 2 + 2β.
§4.8 Exercises
151
Problem 4.30 (The inverse statement of the chain rule). Let f : (a, b) Ñ R be continuous
and g : (c, d) Ñ R be differentiable at y0 = f (x0 ) P (c, d). Show that if (g˝f ) is differentiable
at x0 and g 1 (y0 ) ‰ 0, then f is differentiable at x0 .
Problem 4.31. Let f : R Ñ R be a polynomial, and f has a double root at a and b. Show
that f 1 (x) has at least three roots in [a, b].
Problem 4.32. Let f : R Ñ R be differentiable. Assume that for all x P R, 0 ď f 1 (x) ď
f (x). Show that g(x) = e´x f (x) is decreasing. If f vanishes at some point, conclude that f
is zero.
Problem 4.33. Let f : R Ñ R be twice differentiable. Suppose that f (x + h) ´ f (x) =
hf 1 (x + θh) for all x, h P R, where θ is independent of h. Show that f is a quadratic
polynomial.
Problem 4.34. Let f be a differentiable function defined on some interval I of R. Prove
that f 1 maps connected subsets of I into connected set; that is, f 1 has the intermediate
value property.
Problem 4.35. Let f : R Ñ R be a polynomial, and f has a double root at a and b. Show
that f 1 (x) has at least three roots in [a, b].
Problem 4.36. Let f : [´1, 1] Ñ R be a function such that x2 + f (x)2 = 1 for all |x| ď 1.
␣ ˇ
(
Define C = x ˇ |x| ď 1, f is continuous at x . Show that C contains at least 2 points and
C X (´1, 1) is an open set. Hence if f is continuous at more than 2 points, it is continuous
at uncountably many points.
Problem 4.37. Let f, g : R Ñ R be differentiable functions. Suppose that lim f (x) =
xÑ8
f 1 (x)
lim g(x) = 0, g 1 (x) ‰ 0 for all x P R, and the limit lim 1
exists. Show that the limit
xÑ8
xÑ8 g (x)
f (x)
lim
also exists, and
xÑ8 g(x)
f (x)
f 1 (x)
= lim 1
.
xÑ8 g(x)
xÑ8 g (x)
lim
Problem 4.38. Let f, g : (a, b) Ñ R be differentiable functions. Show that if lim+ f (x) =
xÑa
f 1 (x)
f (x)
lim+ g(x) = 8, g 1 (x) ‰ 0 for all x P (a, b), and the limit lim+ 1
exists, then lim+
xÑa
xÑa g (x)
xÑa g(x)
152
CHAPTER 4. Continuous Maps
exists and
lim+
xÑa
Hint: Let L = lim+
xÑa
f (x)
f 1 (x)
= lim+ 1
.
g(x) xÑa g (x)
(‹)
f 1 (x)
and ϵ ą 0 be given. Choose c P (a, b) such that
g 1 (x)
ˇ f 1 (x)
ˇ ϵ
ˇ
ˇ
´ Lˇ ă
ˇ 1
g (x)
2
@a ă x ă c.
Then for a ă x ă c, the Cauchy mean value theorem implies that for some ξ P (x, c) such
that
f (x) ´ f (c)
f 1 (ξ)
= 1
.
g(x) ´ g(c)
g (ξ)
Show that there exists δ ą 0 such that a + δ ă c and
ˇ f (x) ´ f (c) f (x) ˇ ϵ
ˇ
ˇ
´
@ x P (a, a + δ)
ˇ
ˇă
g(x) ´ g(c)
g(x)
2
and then conclude (‹).
Problem 4.39. Let f : (a, b) Ñ R be k-times differentiable, and c P (a, b). Let hk : (a, b) Ñ
R be given by
hk (x) = f (x) ´
k
ÿ
f (j) (c)
j=0
j!
(x ´ c)j .
hk (x)
= 0.
xÑc (x ´ c)k
Show that lim
Problem 4.40. Two metric spaces (M, d) and (N, ρ) are called homeomorphics if there
exists a continous map f : M Ñ N , called a homeomorphism between M and N , such
that f is one-to-one and onto, and its inverse f ´1 is also continuous. Homeomorphic metric
spaces have the same topological properties. In the following problems, (M, d) and (N, ρ)
are two metric spaces.
1. Suppose that M is compact, and f : M Ñ N is one-to-one and onto. Show that f is
a homeomorphism between M and N .
2. Suppose that f is a homeomorphism between M and N . Show that the restriction of
f to any subset A Ď M establishes a homeomorphism between A and f (A).
3. Determine which of the following pairs of metric spaces is homeomorphic.
§4.8 Exercises
153
(a) M = (a, b] Ď R and N = R.
(b) M is an open ball in Rn and N = Rn .
(c) M = R and N = Rn .
(d) M = [0, 1] ˆ [0, 1] Ď R2 and N = [0, 1] Ď R.
ˇ
␣
(
(e) M = (x, y) P R2 ˇ x2 + y 2 = 1 and N = [0, 1] Ď R.
ˇ
ˇ
␣
(
␣
(
(f) M = (x, y) P R2 ˇ x2 + y 2 = 1 and N = (x, y) P R2 ˇ x2 + xy + y 2 = 1 .
(g) M = R2 and N = R3 .
4. Let I Ď R be an interval and f : I Ñ R be a one-to-one continuous function. Show
that f must be strictly monotonic in I and f is a homeomorphism between I and
f (I).
If I Ď Rn for n ą 1 and f : I Ñ Rn is continuous and one-to-one, can we still assert
that f is homeomorphism between I and f (I)?
§4.7 Integration of Functions of One Variable
Problem 4.41. Let f : [a, b] Ñ R be a bounded function, and Pn denote the division of
[a, b] into 2n equal sub-intervals. Show that f is Riemann integrable over [a, b] if and only if
lim U (f, Pn ) = lim L(f, Pn ) .
nÑ8
nÑ8
Problem 4.42. Let f, g : [a, b] Ñ R be functions, where g is continuous, and f be nonnegative, bounded, Riemann integrable over [a, b]. Show that
1. f g is Riemann integrable.
2. D x0 P (a, b) such that
żb
żb
f (x)g(x)dx = g(x0 )
a
f (x)dx .
a
Problem 4.43. Let f : [a, b] Ñ R be differentiable and assume that f 1 is Riemann integrable. Prove that
żb
f 1 (x) dx = f (b) ´ f (a).
a
Hint: Use the Mean Value Theorem.
154
CHAPTER 4. Continuous Maps
Problem 4.44. Suppose that f : [a, b] Ñ R is Riemann integrable, m ď f (x) ď M for all
x P [a, b], and φ : [m, M ] Ñ R is continuous. Show that φ ˝ f is Riemann integrable on [a, b].
Problem 4.45 (True or False). Determine whether the following statements are true or
false. If it is true, prove it. Otherwise, give a counter-example.
1. Let f : R2 zt(0, 0)u Ñ R satisfy lim f (x, axn ) = 0 for all a P R, n P N and lim f (0, y) =
xÑ0
0. Then
lim
yÑ0
f (x, y) = 0.
(x,y)Ñ(0,0)
2. There exists a function f : R Ñ R which is continuous only at three points of R.
ˇ
␣
(
3. Let f : R Ñ R. Then f is continuous on R if and only if its graph (x, f (x)) ˇ x P R
is closed in R2 .
4. Let I1 and I2 be open intervals in R. Then f : I1 Ñ I2 is a diffeomorphism if and only
if f is differentiable and f 1 (x) ‰ 0 for all x P I1 .
5. Let f : [a, b] Ñ R be a function. If f 2 is Riemann integrable, then f is Riemann
integrable.
?
6. Let f : [a, b] Ñ R be a function. If f is Riemann integrable, then 3 f is Riemann
integrable.
7. Let f (x) = sin
1
be defined on (0, 1]. Then no matter how we define f (0), f is always
x
Riemann integrable on [0, 1].
Chapter 5
Uniform Convergence and the Space
of Continuous Functions
5.1 Pointwise and Uniform Convergence（逐點收斂與均
勻收斂）
Definition 5.1. Let (M, d) and (N, ρ) be two metric spaces, A Ď M be a set, and fk : A Ñ
N be functions for k = 1, 2, ¨ ¨ ¨ . The sequence of functions tfk u8
k=1 is said to converge
␣
(8
pointwise if fk (a) k=1 converges for all a P A. In other words, tfk u8
k=1 converges pointwise
if there exists a function f : A Ñ N such that
(
)
lim ρ fk (a), f (a) = 0
kÑ8
@ a P A.
In this case, tfk u8
k=1 is said to converge pointwise to f and is denoted by fk Ñ f p.w..
Let B Ď A be a subset. The sequence of functions tfk u8
k=1 is said to converge uniformly on B if there exists f : B Ñ N such that
(
)
lim sup ρ fk (x), f (x) = 0 .
kÑ8 xPB
In this case, tfk u8
k=1 is said to converge uniformly to f on B (or converge to f uniformly on
B). In other words, tfk u8
k=1 converges uniformly to f on B if for every ε ą 0, D N ą 0 such
that
(
)
ρ fk (x), f (x) ă ε
@ k ě N and x P B .
155
156
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Example 5.2. Let fk , f : [0, 1] Ñ R be given by
$
1
’
&
0
if ď x ď 1 ,
k
fk (x) =
and
’
% ´kx + 1 if 0 ď x ă 1 .
"
f (x) =
0 if x P (0, 1],
1 if x = 0.
k
Then tfk u8
k=1 converges pointwise to f on [0, 1]. To see this, fix x P [0, 1].
1. Case x ‰ 0: Let ε ą 0 be given, take N ą
1
1
ô
ă x. If k ě N ,
x
N
ˇ
ˇ ˇ
ˇ
ˇfk (x) ´ f (x)ˇ = ˇfk (x) ´ 0ˇ = |0 ´ 0| ă ε .
ˇ
ˇ
2. Case x = 0: For any ε ą 0, k = 1, 2, 3, . . . , ˇfk (0) ´ f (0)ˇ = |1 ´ 1| = 0 ă ε.
However, tfk u8
k=1 does not converge uniformly to f on [0, 1] because
ˇ
ˇ
ˇ
ˇ
sup ˇfk (x) ´ f (x)ˇ = 1 ñ lim sup ˇfk (x) ´ f (x)ˇ = 1 ‰ 0 .
kÑ8 xP[0,1]
xP[0,1]
Example 5.3. Let fk : [0, 1] Ñ R be given by fk (x) = xk . Then for each a P [0, 1), fk (a) Ñ 0
"
0 if x P [0, 1) ,
as k Ñ 8, while if a = 1, fk (a) = 1 for all k. Therefore, if f (x) =
then
1 if x = 1 ,
fk Ñ f p.w.. However, since
ˇ
ˇ
ˇ
ˇ
sup ˇfk (x) ´ f (x)ˇ = sup ˇfk (x)ˇ = 1 ,
xP[0,1)
xP[0,1]
we must have
ˇ
ˇ
lim sup ˇfk (x) ´ f (x)ˇ = 1 ‰ 0 .
kÑ8 xP[0,1]
Therefore, tfk u8
k=1 does not converge uniformly to f on [0, 1].
On the other hand, if 0 ă a ă 1, then
ˇ
ˇ
sup ˇfk (x) ´ f (x)ˇ ď ak ;
xP[0,a]
thus by the Sandwich lemma,
ˇ
ˇ
lim sup ˇfk (x) ´ f (x)ˇ = 0 .
kÑ8 xP[0,a]
Therefore, tfk u8
k=1 converges to uniformly f on [0, a] if 0 ă a ă 1.
§5.1 Pointwise and Uniform Convergence
157
Example 5.4. Let fk : R Ñ R be given by fk (x) =
sin x
1
. Then for each x P R, |fk (x)| ď
k
k
which converges to 0 as k Ñ 8. By the Sandwich lemma,
ˇ
ˇ
lim ˇfk (x)ˇ = 0
kÑ8
@x P R.
ˇ
ˇ
ˇ
ˇ
1
Therefore, fk Ñ 0 p.w.. Moreover, since sup ˇfk (x)ˇ ď , lim sup ˇfk (x)ˇ = 0 . Therefore,
tfk u8
k=1 converges uniformly to 0 on R.
k
xPR
kÑ8 xPR
Proposition 5.5. Let (M, d) and (N, ρ) be two metric spaces, A Ď M be a set, and
fk , f : A Ñ N be functions for k = 1, 2, ¨ ¨ ¨ . If tfk u8
k=1 converges uniformly to f on A, then
tfk u8
k=1 converges pointwise to f .
Proof. For each a P A,
(
)
(
)
ρ fk (a), f (a) ď sup ρ fk (x), f (x) ;
xPA
thus the Sandwich lemma shows that
(
)
lim ρ fk (a), f (a) = 0
kÑ8
since tfk u8
k=1 converges uniformly to f on A.
˝
Proposition 5.6 (Cauchy criterion for uniform convergence). Let (M, d) and (N, ρ) be two
metric spaces, A Ď M be a set, and fk : A Ñ N be a sequence of functions. Suppose that
(N, ρ) is complete. Then tfk u8
k=1 converges uniformly on B Ď A if and only if for every
ε ą 0, D N ą 0 such that
(
)
ρ fk (x), fℓ (x) ă ε
@ k, ℓ ě N and x P B .
Proof. “ñ” Suppose that tfk u8
k=1 converges uniformly to f on B. Let ε ą 0 be given. Then
D N ą 0 such that
(
) ε
ρ fk (x), f (x) ă
2
@ k ě N and x P B .
Then if k, ℓ ě N and x P B,
(
)
(
)
) ε ε
ρ fk (x), fℓ (x) ď ρ fk (x), f (x) + ρ(f (x), fℓ (x) ă + = ε .
2 2
158
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
␣
(8
“ð” Let b P B. By assumption, fk (b) k=1 is a Cauchy sequence in (N, ρ); thus is conver␣
(8
gent. Let f (b) denote the limit of fk (b) k=1 . Then tfk u8
k=1 converges pointwise to f
on B. We claim that the convergence is indeed uniform on B.
Let ε ą 0 be given. Then D N ą 0 such that
(
) ε
ρ fk (x), fℓ (x) ă
2
@ k, ℓ ě N and x P B .
Moreover, for each x P B there exists Nx ą 0 such that
(
) ε
ρ fℓ (x), f (x) ă
2
@ ℓ ě Nx .
Then for all k ě N and x P B,
(
)
(
)
(
) ε ε
ρ fk (x), f (x) ď ρ fk (x), fℓ (x) + ρ fℓ (x), f (x) ă + = ε
2 2
in which we choose ℓ ě maxtN, Nx u to conclude the inequality.
˝
Theorem 5.7. Let (M, d) and (N, ρ) be two metric spaces, A Ď M be a set, and fk : A Ñ N
be a sequence of continuous functions converging to f : A Ñ N uniformly on A. Then f is
continuous on A; that is,
lim f (x) = lim lim fk (x) = lim lim fk (x) = f (a) .
xÑa
xÑa kÑ8
kÑ8 xÑa
Proof. Let a P A and ε ą 0 be given. Since tfk u8
k=1 converges uniformly to f on A, D N ą 0
such that
(
) ε
ρ fk (x), f (x) ă
3
By the continuity of fN , D δ ą 0 such that
@ k ě N and x P A .
(
) ε
ρ fN (x), fN (a) ă whenever x P D(a, δ) X A .
3
Therefore, if x P D(a, δ) X A, by the triangle inequality
(
)
(
)
(
)
(
)
ρ f (x), f (a) ď ρ f (x), fN (x) + ρ fN (x), fN (a) + ρ fN (a), f (a)
ε ε ε
ă + + = ε;
3 3 3
thus f is continuous at a.
Example 5.8. Let fk : [0, 2] Ñ R be given by fk (x) =
˝
xk
. Then
1 + xk
§5.1 Pointwise and Uniform Convergence
159
1. For each a P [0, 1), fk (a) Ñ 0 as k Ñ 8;
2. For each a P (1, 2], fk (a) Ñ 1 as k Ñ 8;
1
3. If a = 1, then fk (a) = for all k.
2
$
’
& 0 if x P [0, 1) ,
1
8
Let f (x) =
Then tfk u8
if x = 1 ,
k=1 converges pointwise to f . However, tfk uk=1
’
% 2
1 if x P (1, 2] .
does not converge uniformly to f on [0, 2] since fk are continuous functions for all k P N
but f is not.
Remark 5.9. The uniform limit of sequence of continuous function might not be uniformly
1
continuous. For example, let A = (0, 1) and fk (x) = for all k P N. Then tfk u8
k=1 converges
x
1
uniformly to f (x) = , but the limit function is not uniformly continuous on A.
x
Theorem 5.10. Let I Ď R be a finite interval, fk : I Ñ R be a sequence of differentiable
␣
(8
functions, and g : I Ñ R be a function. Suppose that fk (a) k=1 converges for some a P I,
8
and tfk1 uk=1
converges uniformly to g on I. Then
1. tfk u8
k=1 converges uniformly to some function f on I.
2. The limit function f is differentiable on I, and f 1 (x) = g(x) for all x P I; that is,
d
d
fk (x) =
lim fk (x) = f 1 (x) .
kÑ8 dx
kÑ8
dx kÑ8
␣
(8
␣
(8
Proof. 1. Let ε ą 0 be given. Since fk (a) k=1 converges to f (a), fk (a) k=1 is a Cauchy
lim fk1 (x) = lim
sequence. Therefore, D N1 ą 0 such that
ˇ
ˇ
ˇfk (a) ´ fℓ (a)ˇ ă ε
@ k, ℓ ě N1 .
2
By the uniform convergence of tfk1 u8
k=1 on I and Proposition 5.6, D N2 ą 0 such that
ˇ 1
ˇ
ˇfk (x) ´ fℓ1 (x)ˇ ă ε
@ k, ℓ ě N2 and x P I ,
2|I|
where |I| is the length of the interval.
Let N = maxtN1 , N2 u. By the mean value theorem, for all k, ℓ ě N and x P I,
there exists ξ in between x and a such that
ˇ
ˇ
ˇfk (x) ´ fℓ (x) ´ fk (a) + fℓ (a)ˇ = |fk1 (ξ) ´ fℓ1 (ξ)||x ´ a| ă ε|x ´ a| ď ε ;
2|I|
2
160
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
thus for all k, ℓ ě N and x P I,
ˇ
ˇ ˇ
ˇ
ˇfk (x) ´ fℓ (x)ˇ ď ˇfk (a) ´ fℓ (a)ˇ + ε ă ε + ε = ε.
2
2 2
Therefore, Proposition 5.6 implies that tfk u8
k=1 converges uniformly on I.
2. Suppose that the uniform limit of tfk u8
k=1 is f . Let x P I be a fixed point, and define
$
& fk (t) ´ fk (x) if t P I, t ‰ x ,
t´x
ϕk (t) =
%
fk1 (x)
if t = x ,
$
& f (t) ´ f (x) if t P I, t ‰ x ,
t´x
and ϕ(t) =
%
g(x)
if t = x .
Then ϕk is continuous on I for all k P N, and tϕk u8
k=1 converges pointwise to ϕ.
Claim: tϕk u8
k=1 converges uniformly to ϕ on I.
Proof of claim: Let ε ą 0 be given. Since tfk1 u8
k=1 converges uniformly on I, there
exists N ą 0 such that
ˇ
ˇ
sup ˇfk1 (t) ´ fℓ1 (t)ˇ ă ε
@ k, ℓ ě N .
tPI
Since
ˇ
$ ˇˇ
fk (t) ´ fk (x) ´ fℓ (t) + fℓ (x)ˇ
&
ˇ
ˇ
if t ‰ x, t P I ,
ˇϕk (t) ´ ϕℓ (t)ˇ =
|t ´ x|
ˇ 1
ˇ
%
ˇf (x) ´ f 1 (x)ˇ
if t = x ,
k
ℓ
by the mean value theorem we obtain that
ˇ
ˇ
ˇ
ˇ
ˇϕk (t) ´ ϕℓ (t)ˇ ď sup ˇfk1 (s) ´ fℓ1 (s)ˇ ă ε
@ k, ℓ ě N and t P I .
sPI
Finally, by Theorem 5.7, ϕ is continuous on I; thus
f 1 (x) = lim ϕ(t) = ϕ(x) = g(x) .
tÑx
˝
Example 5.11. Assume that fk : I Ñ R is differentiable for all k P N, and tfk1 u8
k=1 converges
uniformly to g on I. Then tfk u8
k=1 might NOT converge. For example, consider fk (x) = k.
Then fk1 ” 0 but tfk u8
k=1 does not converge.
Example 5.12. Assume that fk : I Ñ R is differentiable for all k P N, and tfk u8
k=1
converges uniformly to f on I. Then f might NOT be differentiable. In fact, there are
§5.1 Pointwise and Uniform Convergence
161
differentiable functions fk : [a, b] Ñ R such that fk converges uniformly to f on [a, b] but f
is not differentiable. For example, consider
$
k 2
’
&
x
2
fk (x) =
ˇ ˇ
’
% ˇx ˇ ´ 1
2k
ˇ ˇ 1
if ˇxˇ ď ,
k
ˇ ˇ
1
if ď ˇxˇ ď 1.
k
Observe that fk (´x) = fk (x), so it suffices to consider x ě 0.
1. Let f (x) = |x|. Then fk Ñ f uniformly:
!
ˇ
ˇ
ˇ
ˇ)
ˇ
ˇ
ˇ
ˇ
sup ˇfk (x) ´ f (x)ˇ = sup ˇfk (x) ´ xˇ = max sup ˇfk (x) ´ xˇ, sup ˇfk (x) ´ xˇ
xP[´1,1]
= max
!
xP[ k1 ,1]
xP[0, k1 ]
xP[0,1]
ˇ kx2
ˇ
ˇ
ˇ)
1
sup ˇ
´ xˇ, sup ˇx ´
´ xˇ
xP[0, k1 ]
2
2k
xP[ k1 ,1]
ˇ kx2 ˇ ˇ ˇ k 1 2 1
ˇ + ˇxˇ ď ( ) + = 3 Ñ 0 as k Ñ 8 .
ď sup ˇ
2
2 k
k
2k
xP[0, 1 ]
k
1
2. To see if fk are differentiable, it suffices to show fk1 ( ) exists.
k
$ (
)
1
1
1
’
&
+
h
´
´
if h ą 0
1
1
f
(
+
h)
´
f
(
)
1
1
k
k
k
2k
2k
k
k
fk1 ( ) = lim
= lim
hÑ0
hÑ0 h ’
k
h
% k ( 1 + h)2 ´ 1
if h ă 0
2 k
2k
#
if h ą 0
1 h
= lim
= 1.
k 2
hÑ0 h
h + h if h ă 0
2
Example 5.13. Assume that fk : [´1, 1] Ñ R be given by
$
0
if x P [´1, 0] ,
’
’
’
’
2
’
( 1]
’
k 2
’
’
x
if x P 0, ,
&
2
k
fk (x) =
2
(
)
(
2 2
1 2]
k
’
’
1
´
x
´
if
x
P
, ,
’
’
2
k
k k
’
’
’
(2 ]
’
%
1
if x P , 1 .
k
$
0
if x P [´1, 0] ,
’
’
’
’
( 1]
’
’
’
if x P 0, ,
k2x
&
k
1 8
Then fk1 (x) =
)
(
(
1 2 ] and tfk uk=1 converges pointwise to 0 but not
2
2
’
if
x
P
,
´k
x
´
,
’
’
k
k k
’
’
’
(
’
2 ]
%
0
if x P , 1 ,
k
162
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
uniformly on [´1, 1]. We note that tfk u8
k=1 converges to a discontinuous function
"
f (x) =
0 if x P [´1, 0] ,
1 if x P (0, 1] ,
so the convergence of tfk u8
k=1 cannot be uniform on [´1, 1].
Example 5.14. Suppose fk : [0, 1] Ñ R are differentiable on (0, 1) and fk converges uniformly to f on [0, 1] for some f : [0, 1] Ñ R. Does fk1 converge uniformly?
Answer: No! Take fk =
sin(k 2 x)
, k = 1, 2, ¨ ¨ ¨ , then fk Ñ 0 uniformly on [0, 1] since
k
ˇ
ˇ
ˇ sin(k 2 x) ˇ 1
ˇ
ˇ
ˇ ď ñ lim sup ˇfk (x) ´ 0ˇ = 0 .
sup ˇfk (x) ´ 0ˇ = sup ˇ
kÑ8 xP[0,1]
k
k
xP[0,1]
xP[0,1]
Now fk1 (x) = k cos(k 2 x) and fk1 (0) = k Ñ 8 as k Ñ 8.
Example 5.15. There are differentiable functions fk : [a, b] Ñ R such that fk converges
uniformly to f on [a, b] but lim fk1 ‰ ( lim fk )1 . For example, take fk (x) =
[´1, 1]. Then fk1 (x) =
kÑ8
1 ´ k 2 x2
(1 + k 2 x2 )2
kÑ8
x
on
1 + k 2 x2
.
ˇ
x
1
´ 0ˇ = lim
= 0, fk converges uniformly to 0 on [´1, 1].
2
2
kÑ8 2k
kÑ8 xP[´1,1] 1 + k x
ˇ
1. Since lim sup ˇ
2. ( lim fk (x))1 = 01 = 0.
kÑ8
3.
1 ´ k 2 x2
lim fk1 (x) = lim
=
kÑ8
kÑ8 (1 + k 2 x2 )2
"
1 if x = 0,
ˇ ˇ
Note that fk1 does not converge
0 if x ‰ 0, ˇxˇ ă 1.
uniformly.
Theorem 5.16. Let fk : [a, b] Ñ R be a sequence of Riemann integrable functions which
converges uniformly to f on [a, b]. Then f is Riemann integrable, and
lim
żb
żb
kÑ8 a
fk (x)dx =
lim fk (x)dx =
a kÑ8
żb
f (x)dx .
(5.1.1)
a
Proof. Let ε ą 0 be given. Since tfk u8
k=1 converges uniformly to f on [a, b], D N ą 0 such
that
ˇ
ˇ
ε
ˇfk (x) ´ f (x)ˇ ă
@ k ě N and x P [a, b] .
(5.1.2)
4(b ´ a)
§5.2 Series of Functions and The Weierstrass M -test
163
Since fN is Riemann integrable on [a, b], by Riemann’s condition there exists a partition P
of [a, b] such that
U (fN , P) ´ L(fN , P) ă
ε
.
2
Using (4.7.3), we find that
U (f, P) ´ L(f, P) = U (f ´ fN + fN , P) ´ L(f ´ fN + fN , P)
ď U (f ´ fN , P) + U (fN , P) ´ L(f ´ fN , P) ´ L(fN , P)
ε
ε
ď
(b ´ a) +
(b ´ a) + U (fN , P) ´ L(fN , P)
4(b ´ a)
4(b ´ a)
ε ε ε
ă + + = ε;
4 4 2
thus by Riemann’s condition f is Riemann integrable on [a, b].
Now, if k ě N , (5.1.2) implies that
żb
żb
ˇ ˇż b(
ˇż b
ˇ
ˇ
) ˇˇ
ˇ ˇ
ˇ
ˇfk (x) ´ f (x)ˇdx
fk (x) ´ f (x) dxˇ ď
ˇ fk (x)dx ´ f (x)dxˇ = ˇ
a
a
a
ε
ε
ď
(b ´ a) = ă ε
4(b ´ a)
4
a
which shows (5.1.1).
˝
Example 5.17. Let tqk u8
k=1 be the rational numbers in [0, 1], and
"
fk (x) =
0 if x P tq1 , q2 , ¨ ¨ ¨ , qk u ,
1 otherwise .
Then fk converges pointwise to the Dirichlet function
"
f (x) =
0 if x P Q X [0, 1] ,
1 if x P [0, 1]zQ .
However, tfk u8
k=1 does not converge uniformly to f since fk are Riemann integrable on [0, 1]
for all k P N but f is not.
Example 5.18. Let fk : [0, 1] Ñ R be functions given in Example 5.13, and let gk = fk1 .
Then tgk u8
k=1 converges pointwise to 0, but
ż1
0
gk (x)dx = 1 for all k P N.
164
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
5.2 Series of Functions and The Weierstrass M -Test
Definition 5.19. Let (M, d) be a metric space, (V, } ¨ }) be a norm space, A Ď M be a
8
ř
gk converges pointwise if
subset, and gk , g : A Ñ V be functions. We say that the series
k=1
the sequence of partitial sum tsn u8
n=1 given by
sn =
n
ÿ
gk
k=1
converges pointwise. We use
pointwise to g. We say that
8
ř
k=1
8
ř
gk = g p.w. to denote that the series
8
ř
gk converges
k=1
gk converges uniformly on B Ď A if tsn u8
n=1 converges
k=1
uniformly on B.
Example 5.20. Consider the geometric series
8
ř
xk . The partial sum sn is given by
k=0
$
& 1 ´ xn+1 if x ‰ 1 ,
1´x
sn (x) =
%
n+1
if x = 1 .
Then
1.
8
ř
xk converges pointwise to g(x) =
k=0
2.
8
ř
1
in (´1, 1).
1´x
xk does not converge pointwise in (´8, ´1] Y [1, 8).
k=0
3.
8
ř
xk converges uniformly on (´a, a) if 0 ă a ă 1 since
k=0
|a|n+1
|x|n+1
ď
Ñ 0 as n Ñ 8.
1´a
xP(´a,a) 1 ´ x
sup |sn (x) ´ g(x)| = sup
xP(´a,a)
4.
8
ř
k=0
xk does not converge uniformly on (´1, 1) since sup |sn (x) ´ g(x)| = 8.
xP(´1,1)
The following two corollaries are direct consequences of Proposition 5.6 and Theorem
5.7.
§5.2 Series of Functions and The Weierstrass M -test
165
Corollary 5.21. Let (M, d) be a metric space, (V, } ¨ }) be a complete normed vector space,
8
ř
A Ď M be a subset, and gk : A Ñ V be functions. Then
gk converges uniformly on A if
k=1
and only if
n
› ÿ
›
@ ε ą 0, D N ą 0 Q ›
gk (x)› ă ε
@ n ą m ě N and x P A .
k=m+1
Corollary 5.22. Let (M, d) be a metric space, (V, } ¨ }) be a normed vector space, A Ď M
8
ř
be a subset, and gk , g : A Ñ V be functions. If gk : A Ñ V are continuous and
gk (x)
k=1
converges to g uniformly on A, then g is continuous.
Theorem 5.23. Let f : (a, b) Ñ R be an infinitely differentiable functions; that is, f (k) (x)
exists for all k P N and x P (a, b). Let c P (a, b) and suppose that for some 0 ă h ă 8,
ˇ (k) ˇ
ˇf (x)ˇ ď M for all x P (c ´ h, c + h) Ď (a, b). Then
f (x) =
8
ÿ
f (k) (c)
k=0
k!
(x ´ c)k
@ x P (c ´ h, c + h) .
Proof. First, we claim that
f (x) =
n
ÿ
f (k) (c)
k=0
k!
k
n
(x ´ c) + (´1)
żx
(y ´ x)n (n+1)
f
(y)dy
n!
c
@ x P (a, b) .
(5.2.1)
By the fundamental theorem or Calculus (Theorem 4.90) it is clear that (5.2.1) holds for
n = 0. Suppose that (5.2.1) holds for n = m. Then
m
ˇy=x ż x (y ´ x)m+1
[
]
m+1
ÿ
f (k) (c)
ˇ
k
m (y ´ x)
(m+1)
(x ´ c) + (´1)
f
(y)ˇ
´
f (m+2) (y)dy
f (x) =
k!
(m + 1)!
(m + 1)!
y=c
c
k=0
żx
m+1
ÿ f (k) (c)
(y ´ x)m+1 (m+2)
k
m+1
=
(x ´ c) + (´1)
f
(y)dy
k!
(m + 1)!
c
k=0
which implies that (5.2.1) also holds for n = m + 1. By induction (5.2.1) holds for all n P N.
n f (k) (c)
ř
Letting sn (x) =
(x ´ c)k , then if x P (c ´ h, c + h),
k=0
k!
ż
ˇ hn+1
ˇ
ˇ ˇ x hn
ˇ
ˇsn (x) ´ f (x)ˇ ď ˇˇ
M dy ˇ ď
M.
n!
c n!
ˇ hn+1 ˇ
hn+1
ˇ
ˇ
M = 0, D N ą 0 such that ˇ
ˇM ă ε if n ě N . As a
nÑ8 n!
n!
Let ε ą 0 be given. Since lim
consequence, if n ě N ,
ˇ
ˇ
ˇsn (x) ´ f (x)ˇ ă ε whenever n ě N .
˝
166
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
8
ř
Example 5.24. The series
(´1)k
k=0
subset of R.
x2k+1
converges to sin x uniformly on any bounded
(2k + 1)!
Theorem 5.25 (Weierstrass M -test). Let (M, d) be a metric space, (V, } ¨ }) be a complete
normed vector space, A Ď M be a subset, and gk : A Ñ V be a sequence of functions. Suppose
8
ř
that there exists Mk ą 0 such that sup }gk (x)} ď Mk for all k P N and
Mk converges.
xPA
Then
8
ř
k=1
gk converges uniformly and absolutely (that is,
k=1
8
ř
}gk } converges uniformly) on A.
k=1
n
ř
Proof. We show that the partial sum sn =
gk satisfies the Cauchy criterion. Let ε ą 0
k=1
be given. Since
N ą 0 such that
8
ř
Mk converges (which means
k=1
n
ř
Mk converges as n Ñ 8), there exists
k=1
n
ÿ
n
ˇ ÿ
ˇ
ˇ
ˇ
Mk = ˇ
Mk ˇ ă ε @ n ą m ě N .
k=m+1
k=m+1
Therefore,
n
n
n
›
› ÿ
ÿ
ÿ
›
›
›
›
›gk (x)› ď
Mk ă ε
gk (x)› ď
›
k=m+1
k=m+1
@ n ą m ě N and x P A .
k=m+1
Apply Proposition 5.6 to the sequence tsn u8
n=1 , we conclude the theorem.
˝
Theorem 5.7 and 5.25 together imply the following
Corollary 5.26. Let (M, d) be a metric space, (V, } ¨ }) be a complete normed vector space,
A Ď M be a subset, and gk : A Ñ V be a sequence of continuous functions. Suppose that
8
ř
Mk converges. Then
there exists Mk ą 0 such that sup }gk (x)} ď Mk for all k P N and
xPA
8
ř
k=1
gk is continuous on A.
k=1
Example 5.27. Consider the series f (x) =
8 ( k )2
ÿ
x
k=0
Moreover,
lim sup
kÑ8
k!
. For all x P [´R, R],
( x k )2
k!
ď
R2k
.
(k!)2
R2(k+1) / R2k
R2
=
lim
sup
= 0;
2
((k + 1)!)2 (k!)2
kÑ8 (k + 1)
thus the ratio test and the Weierstrass M -test imply that the series
8 ( k )2
ÿ
x
k=0
k!
converges
uniformly on [´R, R]. Theorem 5.7 then shows that f is continuous on [´R, R]. Since R is
arbitrary, we find that f is continuous on R.
§5.3 Integration and Differentiation of Series
167
Example 5.28. Let tak u8
k=0 be a bounded sequence. Then
uous function.
Example 5.29. Consider the function f (x) =
8
ÿ
ak
k!
k=0
xk converges to a contin-
8 cos(2k + 1)x
π
4 ř
´
. We can in fact show
2
π k=0 (2k + 1)2
(much later) that f (x) = |x| for all x P [´π, π], and by the Weierstrass M -test it is easy to
see that the convergence is uniform on R.
y
y = |x|
y = f0 (x)
y = f1 (x)
y = f2 (x)
π
´π/2
π/2
O
π
x
Figure 5.1: The graph of some partial sums
5.3 Integration and Differentiation of Series
The following two theorems are direct consequences of Theorem 5.10 and 5.16.
Theorem 5.30. Let gk : [a, b] Ñ R be a sequence of Riemann integrable functions. If
converges uniformly on [a, b], then
żbÿ
8
8
ř
gk
k=1
gk (x)dx =
a k=1
8 żb
ÿ
gk (x)dx .
k=1 a
Theorem 5.31. Let gk : (a, b) Ñ R be a sequence of differentiable functions. Suppose that
8
8
ř
ř
gk converges for some c P (a, b), and
gk1 converges uniformly on (a, b). Then
k=1
k=1
8
ÿ
k=1
8
gk1 (x) =
d ÿ
gk (x) .
dx k=1
Definition 5.32. A series is called a power series about c or centered at c if it is of
8
ř
the form
ak (x ´ c)k for some sequence tak u8
k=0 Ď R (or C) and c P R (or C).
k=0
168
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Proposition 5.33. If a power series centered at c is convergent at some point b ‰ c, then
the power series converges pointwise on D(c, |b´c|), and converges uniformly on any compact
subsets of D(c, |b ´ c|).
Proof. Since the series
8
ř
ak (b ´ c)k converges, |ak ||b ´ c|k Ñ 0 as k Ñ 8; thus there exists
k=0
M ą 0 such that |ak ||b ´ c|k ď M for all k.
1. x P D(c, |b ´ c|), the series
8
ř
ak (x ´ c)k converges absolutely since
k=0
8
ÿ
|ak (x ´ c)k | ď
k=0
8
ÿ
|ak ||x ´ c|k =
k=0
8
ÿ
8
|ak ||b ´ c|k
k=0
ÿ
|x ´ c|k
ď
M
|b ´ c|k
k=0
( |x ´ c| )k
|b ´ c|
which converges (because of the geometric series test or ratio test).
2. Let K Ď D(c, |b ´ c|) be a compact set. Then
ˇ
␣
(
dist(K, BD(c, |b ´ c|)) ” inf |x ´ y| ˇ x P K, |y ´ c| = |b ´ c| ą 0 .
Define r =
|b ´ c| ´ dist(K, BD(c, |b ´ c|))
. Then 0 ď r ă 1, and |x ´ c| ď r|b ´ c| for all
|b ´ c|
x P K. Therefore, |ak (x ´ c)k | ď M rk if x P K; thus the Weierstrass M -test implies
8
ř
˝
that the series
ak (x ´ c)k converges uniformly on K.
k=0
By the proposition above, we immediately conclude that the collection of all x at which
the power series converges must be connected and symmetric; thus is a disc or a point. This
observation induce the following
Definition 5.34. A non-negative number R is called the radius of convergence of the
8
ř
power series
ak (x ´ c)k if the series converges for all x P D(c, R) but diverges if x R
k=0
D(c, R). In other words,
8
ˇ ÿ
␣
(
R = sup r ě 0 ˇ
ak (x ´ c)k converges in D(c, R) .
k=0
The interval of convergence or convergence interval of a power series is the collection
of all x at which the power series converges.
Remark 5.35. A power series converges pointwise on its interval of convergence.
§5.3 Integration and Differentiation of Series
Theorem 5.36. Let tak u8
k=0 Ď C, c P C,
169
8
ř
ak (x ´ c)k be a power series with radius of
k=0
convergence R ą 0, and K Ď D(c, R) be a compact set. Then
1. The power series
8
ř
ak (x ´ c)k converges uniformly on K.
k=0
2. The power series
uniformly on K.
8
ř
(k + 1)ak+1 (x ´ c)k converges pointwise on D(c, R), and converges
k=0
Proof. 1. It is simply a restatement of Proposition 5.33.
2. By 1, it suffices to show that the power series
8
ř
(k+1)ak+1 (x´c)k converges pointwise
k=0
on D(c, R). Clearly the series converges at x = c. Let x P D(c, R) and x ‰ c. Since
|x ´ c| ă R, there exists b P D(c, R) such that
|b ´ c| =
Then if r =
8
ÿ
R + |x ´ c|
.
2
|x ´ c|
, 0 ă r ă 1 and
|b ´ c|
(k + 1)|ak+1 ||x ´ c|k ď
k=0
8
ÿ
(k + 1)|ak+1 ||b ´ c|k
k=0
( |x ´ c| )k
|b ´ c|
ďM
8
ÿ
(k + 1)rk
k=0
8
ř
for some M ą 0. Note that the ratio test implies that the series
(k + 1)rk converges
k=0
8
ř
k
if 0 ă r ă 1; thus the power series
(k + 1)|ak+1 ||x ´ c| converges by the comparison
k=0
test.
˝
8
ř
Corollary 5.37. Let tak u8
Ď
R
and
c
P
R,
and
ak (x ´ c)k be a power series with
k=0
k=0
8
ř
k
radius of convergence R ą 0. Then
ak (x ´ c) is differentiable in (c ´ R, c + R) and is
k=0
Riemann integrable over any closed intervals [α, β] Ď (c ´ R, c + R). Moreover,
8
8
ÿ
d ÿ
k
ak (x ´ c) =
kak (x ´ c)k´1
dx k=0
k=1
and
żβ ÿ
8
α k=0
k
ak (x ´ c) dx =
8
ÿ
k=0
@ x P (c ´ R, c + R)
żβ
ak
α
(x ´ c)k dx .
170
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Example 5.38. Let tak u8
k=0 be a bounded sequence. Then
8
8
8
ÿ
ak
d ( ÿ ak k ) ÿ
ak+1 k
k´1
x =
x
=
x .
dx k=0 k!
(k ´ 1)!
k!
k=1
k=0
żt
Example 5.39. We show
ex dx = et ´ 1 as follows. By Theorem 5.23, ex =
0
8 xk
ř
k=0 k!
and
the convergence is uniform on any bounded sets of R; thus Corollary 5.37 implies that
żt
żtÿ
8
8 żt k
8
8 k
ÿ
ÿ
ÿ
xk
x
tk+1
t
x
e dx =
dx =
dx =
=
= et ´ 1 .
k!
k!
(k
+
1)!
k!
0
0 k=0
k=0 0
k=0
k=1
d
dx
Example 5.40.
(ř
8 xk )
k=1 k
8
ř
=
xk´1 =
k=1
8
ř
xk for all x P (´1, 1); thus
k=0
1
d ( ÿ xk )
=
dx k=1 k
1´x
8
@ x P (´1, 1) .
As a consequence,
8 k
ÿ
t
k
k=1
d ( ÿ xk )
dx = ´ log(1 ´ t)
0 dx k=1 k
8
żt
=
@ t P (´1, 1) .
(5.3.1)
Using the alternating series test, it is clear that the left-hand side of (5.3.1) converges at
t = ´1. What is the value of
´
8
ÿ
(´1)k
k
k=1
d
Consider the partial sum
dx
=1´
(ř
n xk )
sides over [´1, 0],
k=1
k
=
1 1 1 1 1
+ ´ + ´ + ¨¨¨?
2 3 4 5 6
n´1
ř
xk =
k=0
1 ´ xn
1
xn
=
´
. Integrating both
1´x
1´x
1´x
ż0
n
ˇÿ
ˇ ż 0 |x|n
(´1)k
1
ˇ
ˇ
+ log 2ˇ ď
dx ď
(´x)n dx =
Ñ 0 as n Ñ 8;
ˇ
k
n+1
´1 1 ´ x
´1
k=1
thus
1´
1 1 1 1 1
+ ´ + ´ + ¨ ¨ ¨ = log 2 .
2 3 4 5 6
In other words,
8 k
ÿ
t
k=1
k
= ´ log(1 ´ t)
@ t P [´1, 1) .
§5.4 The space of Continuous Functions
Example 5.41. It is clear that
171
8
8
ř
ř
1
2 k
(´x
)
=
(´1)k x2k for all x P (´1, 1). So if
=
1 + x2
k=0
k=0
x P (´1, 1),
tan
´1
żx
dt
x=
=
2
0 1+t
8
ÿ
żxÿ
8
k 2k
(´1) t dt =
0 k=0
8 żx
ÿ
(´1)k t2k dt
k=0 0
k
(´1) 2k+1 ˇˇt=x
x3 x5 x7
=x´
=
t
+
´
+ ¨¨¨ .
ˇ
2k + 1
3
5
7
t=0
k=0
The right-hand side of the identity above converges at x = 1. What is the value of
8
ÿ
(´1)k
k=0
2k + 1
=1´
1 1 1
+ ´ + ¨ ¨ ¨?
3 5 7
Mimic the previous example, we consider
żx
żx
żx
dt
1 ´ (´t2 )n+1
(´t2 )n+1
´1
tan x =
=
dt +
dt
2
1 + t2
1 + t2
0 1+t
0
0
żxÿ
żx
n
(´t2 )n+1
k 2k
dt
=
(´1) t dt +
1 + t2
0 k=0
0
żx
żx
n żx
n
ÿ
ÿ
(´t2 )n+1
(´1)k 2k+1
(´t2 )n+1
k 2k
=
(´1) t dt +
dt
=
x
+
dt ;
2
2
1
+
t
2k
+
1
1
+
t
0
0
0
k=0
k=0
thus plugging x = 1,
ż 1 2(n+1)
ż1
n
ˇ
ÿ
(´1)k ˇˇ
1
t
ˇ
´1
2(n+1)
dt
ď
t
dt
=
Ñ 0 as n Ñ 8.
ˇ tan 1 ´
ˇď
2
2k + 1
2n + 3
0 1+t
0
k=0
Therefore,
1´
1 1 1
π
+ ´ + ¨ ¨ ¨ = tan´1 1 = .
3 5 7
4
5.4 The Space of Continuous Functions
Definition 5.42. Let (M, d) be a metric space, (V, } ¨ }) be a normed vector space, and
A Ď M be a subset. We define C (A; V) as the collection of all continuous functions on A
with value in V; that is,
ˇ
␣
(
C (A; V) = f : A Ñ V ˇ f is continuous on A .
172
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Let Cb (A; V) be the subspace of C (A; V) which consists of all bounded continuous functions
on A; that is,
ˇ
␣
(
Cb (A; V) = f P C (A; V) ˇ f is bounded .
Every f P Cb (A; V) is associated with a non-negative real number }f }8 given by
ˇ
␣
(
}f }8 = sup }f (x)} ˇ x P A = sup }f (x)}
xPA
The number }f }8 is called the sup-norm of f .
Proposition 5.43. Let (M, d) be a metric space, (V, } ¨ }) be a normed vector space, A Ď M
be a subset.
1. C (A; V) and Cb (A; V) are vector spaces.
(
)
2. Cb (A; V), } ¨ }8 is a normed vector space.
3. If K Ď M is compact, then C (K; V) = Cb (K; V).
Proof. 1 and 2 are trivial, and 3 is concluded by Theorem 4.21.
˝
Remark 5.44. In general } ¨ }8 is not a “norm” on C (A; V). For example, the function
f (x) =
1
belongs to C ((0, 1); R) and }f }8 = 8. Note that to be a norm }f }8 has to take
x
values in R, and 8 R R.
Proposition 5.45. Let (M, d) be a metric space, (V, } ¨ }) be a normed vector space, A Ď M
be a subset, and fk P Cb (A; V) for all k P N. Then tfk u8
k=1 converges uniformly on A if and
(
)
only if tfk u8
k=1 converges in Cb (A; V), } ¨ }8 .
(
)
Proof. (ð) Suppose that tfk u8
k=1 converges in Cb (A; V), } ¨ }8 . Then there exists f P
(
)
Cb (A; V), } ¨ }8 such that lim }fk ´ f }8 = 0 , and by the definition of the sup-norm,
kÑ8
lim sup }fk (x) ´ f (x)} = 0 .
kÑ8 xPA
Therefore, tfk u8
k=1 converges to f uniformly on A.
(ñ) Suppose that tfk u8
k=1 converges uniformly on A. Then there exists a function f : A Ñ
V such that
lim sup }fk (x) ´ f (x)} = 0 .
kÑ8 xPA
§5.4 The space of Continuous Functions
173
By the definition of the sup-norm, it suffices to show that f P Cb (A; V) in order to
conclude the proposition. By Theorem 5.7, we obtain that f P C (A; V). Moreover,
the uniform convergence implies that there exists N ą 0 such that
}fk (x) ´ f (x)} ă 1
@ k ě N and x P A .
In particular, the boundedness of fN provides M ą 0 such that }fN (x)} ď M for all
x P A; thus
}f (x)} ď }fN (x)} + }f (x) ´ fN (x)} ď M + 1
@x P A.
This implies that f is bounded; thus f P Cb (A; V).
˝
Theorem 5.46. Let (M, d) be a metric space, (V, }¨}) be a normed vector space, and A Ď M
(
)
be a subset. If (V, } ¨ }) is complete, so is Cb (A; V), } ¨ }8 .
(
)
Proof. Let tfk u8
k=1 be a Cauchy sequence in Cb (A; V), } ¨ }8 . Then
@ ε ą 0, D N ą 0 Q }fk ´ fℓ }8 ă ε if k, ℓ ě N .
By the definition of the sup-norm, the statement above implies that
@ ε ą 0, D N ą 0 Q }fk (x) ´ fℓ (x)} ă ε if k, ℓ ě N and x P A
8
which implies that tfk u8
k=1 satisfies the Cauchy criterion. By Proposition 5.6, tfk uk=1 con(
verges uniformly on A, and Proposition 5.45 shows that tfk u8
k=1 converges in Cb (A; V), } ¨
)
}8 .
˝
ˇ
␣
(
Example 5.47. The set B = f P C ([0, 1]; R) ˇ f (x) ą 0 for all x P [0, 1] is open in
(
)
C ([0, 1]; R), } ¨ }8 .
Reason: Let f P B be given. Since [0, 1] is compact and f is continuous, by the extreme
f (x )
0
. Now
value theorem there exists x0 P [0, 1] so that inf f (x) = f (x0 ) ą 0. Take ε =
2
xP[0,1]
ˇ
ˇ
f (x0 )
if g is such that }g ´ f }8 = sup ˇg(x) ´ f (x)ˇ ă ε =
, we have for any y P [0, 1],
xP[0,1]
2
ˇ
ˇ
ˇ
ˇ
ˇg(y) ´ f (y)ˇ ď sup ˇg(x) ´ f (x)ˇ ă f (x0 )
2
xP[0,1]
f (x0 )
f (x0 )
ď g(y) ď f (y) +
2
2
f (x0 )
f (x0 )
f (x0 )
ñ g(y) ě f (y) ´
ě f (x0 ) ´
=
ą 0.
2
2
2
ñ f (y) ´
174
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Therefore, g P B; thus D(f, ε) Ď B.
y
y = f (x)+ε
y = f (x)
y = f (x)´ε
x
O
Figure 5.2: g P D(f, ε) if the graph of g lies in between the two red dash lines
Example 5.48. Find the closure of B given in the previous example.
ˇ
␣
(
Proof. Claim: cl(B) = f P C ([0, 1], R) ˇ f (x) ě 0 .
ˇ
␣
(
Proof of claim: We show @ f P f P C ([0, 1], R) ˇ f (x) ě 0 , D fk P B Q }fk ´ f }8 Ñ 0 as
1
k
k Ñ 8. Take fk (x) = f (x) + , then fk P B (7 fk (x) ą 0), and
ˇ
ˇ
1
1
}fk ´ f }8 = sup ˇfk (x) ´ f (x)ˇ ď sup = Ñ 0 as k Ñ 8 .
k
xP[0,1]
xP[0,1] k
˝
5.5 The Arzelà-Ascoli Theorem
在這一節中，我們將研究一般情況下，連續函數列的逐點收斂與均勻收斂之間的具體差
異為何。更具體地說，我們希望能找到一個條件，使得逐點收斂的連續函數列，其均勻
收斂性等價於該條件成立。這個條件，刻劃了均勻收斂與逐點收斂的真實差異，而這個
特別的條件，也將提供額外（且有效）的判斷法，幫助我們判斷在連續函數空間裡面的集
合是否緊緻。
5.5.1 Equi-continuous family of functions
The first part of this section is devoted to the investigation of the difference between the
pointwise convergence and the uniform convergence of sequence of continuous functions.
Definition 5.49. Let (M, d) be a metric space, (V, } ¨ }) be a normed vector space, and
A Ď M be a subset. A subset B Ď Cb (A; V) is said to be equi-continuous（等度連續）if
@ ε ą 0, D δ ą 0 Q }f (x1 ) ´ f (x2 )} ă ε
whenever d(x1 , x2 ) ă δ, x1 , x2 P A, and f P B .
§5.5 The Arzelà-Ascoli Theorem
175
Remark 5.50. 1. If B Ď Cb (A; V) is equi-continuous, and C is a subset of B, then C is
also equi-continuous.
2. In an equi-continuous set of functions B, every f P B is uniformly continuous.
Remark 5.51. For a uniformly continuous function f , let δf (ε) (we have defined this
number in Remark 4.51) denote the largest δ that can be used in the definition of the
uniform continuity; that is, δf (ε) has the property that
}f (x) ´ f (y)} ă ε whenever d(x, y) ă δ, x, y P A
ô
0 ă δ ď δf (ε) .
Suppose that every element in B Ď Cb (A; V) is uniformly continuous on A. Then B is
equi-continuous if and only if inf δf (ε) ą 0.
f PB
ˇ
(
Example 5.52. Let B = f P Cb ((0, 1); V) ˇ |f 1 (x)| ď 1 for all x P (0, 1) . Then B is equicontinuous (by choosing δ = ϵ for any given ϵ, and applying the mean value theorem).
␣
Example 5.53. Let fk : [0, 1] Ñ R be a sequence of functions given by
$
1
’
’
kx
if 0 ď x ď ,
’
’
k
’
&
1
2
fk (x) =
2 ´ kx if ď x ď ,
’
k
k
’
’
’
2
’
%
0
if x ě ,
k
and B = tfk u8
k=1 . Then B is not equi-continuous since the largest δ for each k is
converges to 0.
ε
which
k
Lemma 5.54. Let (M, d) be a metric space, (V, } ¨ }) be a normed vector space, and K Ď M
be a compact subset. If B Ď C (K; V) is pre-compact, then B is equi-continuous.
Proof. Suppose the contrary that B is not equi-continuous. Then D ε ą 0 such that
@ k P N, D xk , yk P K and fk P B Q d(xk , yk ) ă
1
but }fk (xk ) ´ fk (yk )} ě ε .
k
(
)
Since B is pre-compact in C (K; V), } ¨ }8 and K is compact in (M, d), there exists a
␣ (8
␣ (8
subsequence fkj j=1 and txkj u8
j=1 such that fkj j=1 converges uniformly to some function
(
)
8
f P C (K; V), } ¨ }8 and txkj u8
j=1 converges to some a P K. We must also have tykj uj=1
converges to a since d(xkj , ykj ) ă
1
.
kj
176
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Since f is continuous at a,
D δ ą 0 Q }f (x) ´ f (a)} ă
ε
5
if x P D(a, δ) X K.
␣ (8
Moreover, since fkj j=1 converges to f uniformly on K and xkj , ykj Ñ a as j Ñ 8, D N ą 0
such that
}fkj (x) ´ f (x)} ă
ε
5
if j ě N and x P K
and
d(xkj , a) ă δ
and d(ykj , a) ă δ
if j ě N .
As a consequence, for all j ě N ,
ε ď }fkj (xkj ) ´ fkj (ykj )} ď }fkj (xkj ) ´ f (xkj )} + }f (xkj ) ´ f (a)}
+ }f (ykj ) ´ f (a)} + }f (ykj ) ´ fkj (ykj )} ă
4ε
5
which is a contradiction.
˝
Alternative proof of Lemma 5.54. Suppose the contrary that B is not equi-continuous. Then
D ε ą 0 such that
1
but }fk (xk ) ´ fk (yk )} ě ε .
k
␣ (8
(
)
Since B is pre-compact in C (K; V), } ¨ }8 , there exists a subsequence fkj j=1 converges
␣ (8
(
)
to some function f in C (K; V), } ¨ }8 . By Proposition 5.45, fkj j=1 converges uniformly
to f on K; thus there exists N1 ą 0 such that
@ k P N, D xk , yk P K and fk P B Q d(xk , yk ) ă
›
›
›fk (x) ´ f (x)› ă ε
j
4
@ j ě N1 and x P K .
Since f P C (K; V), by Theorem 4.52, f is uniformly continuous on K; thus
D δ ą 0 Q }f (x) ´ f (y)} ă
ε
4
if d(x, y) ă δ and x, y P K .
␣
(
[1]
Let N = max N1 ,
+ 1 , and j ě N . Then d(xkj , ykj ) ă δ and this further implies that
δ
ε ď }fkj (xkj) ´ fkj (ykj)} ď }fkj (xkj) ´ f (xkj)} + }f (xkj) ´ f (ykj)} + }f (ykj) ´ fkj (ykj)} ă
a contradiction.
3ε
,
4
˝
§5.5 The Arzelà-Ascoli Theorem
177
Corollary 5.55. Let (M, d) be a metric space, (V, }¨}) be a normed vector space, and K Ď M
8
be a compact subset. If tfk u8
k=1 converges uniformly on K, then tfk uk=1 is equi-continuous.
Example 5.56. Corollary 5.55 fails to hold if the compactness of K is removed. For
example, let tfk u8
k=1 be a sequence of identical functions fk (x) =
1
on (0, 1). Then tfk u8
k=1
x
converges uniformly on (0, 1) but tfk u8
k=1 is not equi-continuous since none of fk is uniformly
continuous on (0, 1) which violates Remark 5.50.
8
We have just shown that if tfk u8
k=1 converges uniformly on a compact set K, then tfk uk=1
must be equi-continuous. The inverse statement, on the other hand, cannot be true. For
8
example, taking tfk u8
k=1 to be a sequence of constant functions fk (x) = k. Then tfk uk=1
obviously does not converge, not even any subsequence. Therefore, we would like to study
under what additional conditions, equi-continuity of a sequence of functions (defined on
a compact set K) indeed converges uniformly. The following lemma is an answer to the
question.
Lemma 5.57. Let (M, d) be a metric space, (V, } ¨ }) be a Banach space, K Ď M be a
8
compact set, and tfk u8
k=1 Ď C (K; V) be a equi-continuous sequence of functions. If tfk uk=1
converges pointwise on a dense subset E of K (that is, E Ď K Ď cl(E)), then tfk u8
k=1
converges uniformly on K.
Proof. Let ε ą 0 be given. By the equi-continuity of tfk u8
k=1 ,
D δ ą 0 Q }fk (x) ´ fk (y)} ă
ε
3
if d(x, y) ă δ, x, y P K and k P N .
Since K is compact, K is totally bounded; thus
Dty1 , ¨ ¨ ¨ , ym u Ď K Q K Ď
m
ď
j=1
( δ)
D yj , .
2
δ
By the denseness of E in K, for each j = 1, ¨ ¨ ¨ , m, there exists zj P E such that d(zj , yj ) ă .
2
m
( δ)
Ť
Ď D(zj , δ); thus K Ď
D(zj , δ). Since tfk u8
Moreover, D yj ,
k=1 converges pointwise on
2
j=1
E, tfk (zj )u8
k=1 converges as k Ñ 8 for all j = 1, ¨ ¨ ¨ , m. Therefore,
D Nj ą 0 Q }fk (zj ) ´ fℓ (zj )} ă
ε
3
@ k, ℓ ě Nj .
Let N = maxtN1 , ¨ ¨ ¨ , Nm u, then
}fk (zj ) ´ fℓ (zj )} ă
ε
3
@ k, ℓ ě N and j = 1, ¨ ¨ ¨ , m .
178
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Now we are in the position of concluding the lemma. If x P K, there exists zj P E such
that d(x, zj ) ă δ; thus if we further assume that k, ℓ ě N ,
}fk (x) ´ fℓ (x)} ď }fk (x) ´ fk (zj )} + }fk (zj ) ´ fℓ (zj )} + }fℓ (zj ) ´ fℓ (x)} ă ε .
By Proposition 5.6, tfk u8
k=1 converges uniformly on K.
˝
Remark 5.58. Corollary 5.55 and Lemma 5.57 imply that “a sequence tfk u8
k=1 Ď C (K; V)
8
converges uniformly on K if and only if tfk uk=1 is equi-continuous and pointwise convergent
(on a dense subset of K)”.
5.5.2 Compact sets in C (K; V)
The next subject in this section is to obtain a (useful) criterion of determining the compactness (or pre-compactness) of a subset B Ď C (K; V) which guarantees the existence of a
␣ (8
(
)
convergent subsequence fkj j=1 of a given sequence tfk u8
k=1 Ď B in C (K; V), } ¨ }8 .
Lemma 5.59 (Cantor’s Diagonal Process). Let E be a countable set, (V, } ¨ }) be a Banach
␣
(8
space, and fk : E Ñ V be a sequence of functions. Suppose that for each x P E, fk (x) k=1
is pre-compact in V. Then there exists a subsequence of tfk u8
k=1 that converges pointwise on
E.
Proof. Since E is countable, E = txℓ u8
ℓ=1 .
␣
(8
␣ (8
1. Since fk (x1 ) k=1 is pre-compact in (V, } ¨ }), there exists a subsequence fkj j=1 such
␣
(8
that fkj (x1 ) j=1 converges in (V, } ¨ }).
␣
(8
␣
(8
␣
(8
2. Since fk (x2 ) k=1 is pre-compact in (V, } ¨ }), the sequence fkj (x2 ) j=1 Ď fk (x2 ) k=1
(8
␣
has a convergent subsequence fkjℓ (x2 ) ℓ=1 .
Continuing this process, we obtain a sequence of sequences S1 , S2 , ¨ ¨ ¨ such that
1. Sk consists of a subsequence of tfk u8
k=1 which converges at xk , and
2. Sk Ě Sk+1 for all k P N.
8
Let gk be the k-th element of Sk . Then the sequence tgk u8
k=1 is a subsequence of tfk uk=1
and tgk u8
˝
k=1 converges at each point of E.
§5.5 The Arzelà-Ascoli Theorem
179
␣
(8
The condition that “ fk (x) k=1 is pre-compact in V for each x P E” in Lemma 5.59
motivates the following
Definition 5.60. Let (M, d) be a metric space, (V, } ¨ }) be a normed vector space, and
compact
A Ď M be a subset. A subset B Ď Cb (A; V) is said to be pointwise pre-compact if the
bounded
compact
ˇ
␣
(
set Bx ” f (x) ˇ f P B is pre-compact in (V, } ¨ }) for all x P A.
bounded
Example 5.61. Let fk : [0, 1] Ñ R be given in Example 5.53, and B = tfk u8
k=1 . Then B
is pointwise compact: for each x P [0, 1], Bx is a finite set since if fk (0) = 0 for all k P N,
while if x ą 0, fk (x) = 0 for all k large enough which implies that #Bx ă 8.
是時候可以來看 C (K; V) 裡面的 compact sets 有什麼等價條件了。首先我們先看何
時 B Ď C (K; V) 是 compact set。給定一個函數列 tfk u8
k=1 Ď B，我們想知道能不能找到
␣ (8
一個在 sup-norm 下收斂的 subsequence fkj j=1（即 sequentially compact）。由 Diagonal
Process (Lemma 5.59) 知，我們得在 K 中找一個稠密的子集合 E 使得 tfk u8
k=1 在 E 上
是 pointwise pre-compact（這個部份只保證了可以找到 subsequence 逐點收斂），然後加
上 Lemma 5.57 的幫助，馬上知道加上 equi-continuity 的條件之後，逐點收斂會變均勻收
斂。因此，很自然地我們會要求 B 滿足 pointwise pre-compact 還有 equi-continuous 這兩
個條件來證出 B 是 C (K; V) 中的 compact set。而在一個 compact set K 中能不能找到
一個稠密子集合則是由下面這個 Lemma 所提供。
Lemma 5.62. A compact set K in a metric space (M, d) is separable; that is, there exists
a countable subset E of K such that cl(E) = K.
Proof. Since K is compact, K is totally bounded; thus @ n P N, D En Ď K such that
#En ă 8
and K Ď
( 1)
D y,
.
n
yPE
ď
n
Let E =
8
Ť
En . Then E is countable by Theorem 1.40. We claim that cl(E) = K.
n=1
To see this, first by the definition of the closure of a set, cl(E) Ď K (since K is closed).
( 1)
( 1)
Ť ( 1)
Let x P K. Since K Ď
D y, , x P D y,
for some y P En . Therefore, D x, XE ‰
yPEn
n
n
s = cl(E).
H for all n P N. This implies that x P E
n
˝
180
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Theorem 5.63. Let (M, d) be a metric space, (V, } ¨ }) be a Banach space, K Ď M be a
compact set, and B Ď C (K; V) be equi-continuous and pointwise pre-compact. Then B is
(
)
pre-compact in C (K; V), } ¨ }8 .
Proof. We show that every sequence tfk u8
k=1 in B has a convergent subsequence. Since K is
compact, there is a countable dense subset E of K (Lemma 5.62), and the diagonal process
␣ (8
(Lemma 5.59) implies that there exists fkj j=1 that converges pointwise on E. Since E is
␣ (8
␣ (8
dense in K, by Lemma 5.57 fkj j=1 converges uniformly on K; thus fkj j=1 converges in
(
)
C (K; V), } ¨ }8 by Proposition 5.45.
˝
Remark 5.64. Lemma 5.54 and Theorem 5.63 imply that “a set B Ď C (K; V) is precompact if and only if B is equi-continuous and pointwise pre-compact”. (That B is precompact implies that B is pointwise pre-compact is left as an exercise).
Corollary 5.65. Let (M, d) be a metric space, and K Ď M be a compact set. Assume that
B Ď C (K; R) is equi-continuous and pointwise bounded on K. Then every sequence in B
has a uniformly convergent subsequence.
␣
(8
Proof. By the Bolzano-Weierstrass theorem the boundedness of fk (x) k=1 implies that
␣
(8
fk (x) k=1 is pre-compact for all x P E. Therefore, we can apply Theorem 5.63 under the
setting (V, } ¨ }) = (R, | ¨ |) to conclude the corollary.
˝
The following theorem provides how compact sets look like in C (K; V).
Theorem 5.66 (The Arzelà-Ascoli Theorem). Let (M, d) be a metric space, (V, } ¨ }) be
a Banach space, K Ď M be a compact set, and B Ď C (K; V). Then B is compact in
(
)
C (K; V), } ¨ }8 if and only if B is closed, equi-continuous, and pointwise compact.
Proof. “ð” This direction is conclude by Theorem 5.63 and the fact that B is closed.
“ñ” By Lemma 3.10 and Lemma 5.54, it suffices to shows that B is pointwise compact.
␣
(8
Let x P K and fk (x) k=1 be a sequence in Bx . Since B is compact, there exists a
␣ (8
subsequence fkj j=1 that converges uniformly to some function f P B. In particular,
␣
(8
␣
(8
fkj (x) j=1 converges to f (x) P Bx . In other words, we find a subsequence fkj (x) j=1
␣
(8
of fk (x) k=1 that converges to a point in Bx . This implies that Bx is sequentially
˝
compact; thus Bx is compact.
Example 5.67. Let fk : [0, 1] Ñ R be a sequence of functions such that
§5.6 The Stone-Weierstrass Theorem
(1) |fk (x)| ď M1 for all k P N and x P [0, 1];
181
(2) |fk1 (x)| ď M2 for all k P N and x P [0, 1].
Then tfk u8
k=1 is clearly pointwise bounded. Moreover, by the mean value theorem
ˇ
ˇ
ˇfk (x) ´ fk (y)ˇ ď M2 |x ´ y|
@ x, y P [0, 1], k P N
which implies that tfk u8
k=1 is equi-continuous. Therefore, by Corollary 5.65 there exists a
␣ (8
subsequence fkj j=1 that converges uniformly on [0, 1].
Question: If assumption (1) of Example 5.67 is omitted, can tfk u8
k=1 still have a convergent
subsequence?
Answer: No! Take fk (x) = k, then tfk u8
k=1 does not have a convergent subsequence (note
that fk is continuous and fk1 (x) = 0; that is, Assumption (2) is fulfilled).
Example 5.68. We show that Assumption (1) of Example 5.67 can be replaced by fk (0) = 0
for all k P N.
Proof. (a) If fn (0) = 0, then by the mean value theorem we have for all x P (0, 1] and k P N,
fk (x) ´ fk (0) = fk1 (ck )(x ´ 0). Then Assumption (2) of Example 5.67 implies that
ˇ
ˇ ˇ
ˇˇ ˇ
ˇfk (x) ´ fk (0)ˇ = ˇfk1 (ck )ˇˇxˇ ď M2 |x| ď M2
which shows that tfk u8
k=1 is uniformly bounded by M2 .
(b) tfk u8
k=1 are equi-continuous (same proof as in Example 5.67).
˝
5.6 The Stone-Weierstrass Theorem
Theorem 5.69 (Weierstrass). Let f : [0, 1] Ñ R be continuous and let ε ą 0 be given. Then
there is a polynomial p : [0, 1] Ñ R such that }f ´ p}8 ă ε. In other words, the collection of
(
)
all polynomials is dense in the space C ([0, 1]; R), } ¨ }8 .
Proof. Let rk (x) = Ckn xk (1 ´ x)n´k . By looking at the partial derivatives with respect to x
n
ř
of the identity (x + y)n =
Ckn xk y n´k , we find that
k=0
1.
n
ř
rk (x) = 1; 2.
n
ř
krk (x) = nx; 3.
k(k ´ 1)rk (x) = n(n ´ 1)x2 .
k=0
k=0
k=0
n
ř
As a consequence,
n
ÿ
n
ÿ
[
]
(k ´ nx) rk (x) =
k(k ´ 1) + (1 ´ 2nx)k + n2 x2 rk (x) = nx(1 ´ x) .
k=0
2
k=0
182
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Since f : [0, 1] Ñ R is continuous on a compact [0, 1], f is uniformly continuous on [0, 1] (by
Theorem 4.52); thus
ˇ
ˇ ε
D δ ą 0 Q ˇf (x) ´ f (y)ˇ ă
2
Consider the Bernstein polynomial pn (x) =
if |x ´ y| ă δ, x, y P [0, 1] .
n
(k)
ř
rk (x). Note that
f
n
k=0
n (
n ˇ
ˇ ÿ
ˇ
ˇ ˇÿ
( k ))
( k )ˇˇ
ˇ
ˇ
ˇf (x) ´ pn (x)ˇ = ˇˇ
f (x) ´ f
rk (x)ˇ ď
f
(x)
´
f
ˇrk (x)
ˇ
n
k=0
(
ď
ÿ
)ˇ
( k )ˇˇ
ˇ
ˇf (x) ´ f
ˇrk (x)
ÿ
+
|k´nx|ăδn
k=0
n
n
|k´nx|ěδn
ÿ (k ´ nx)2
ε
ă + 2}f }8
rk (x)
2
(k ´ nx)2
|k´nx|ěδn
n
ε 2}f }8 ÿ
ε 2}f }8
ε }f }8
ď + 2 2
(k ´ nx)2 rk (x) ď +
x(1 ´ x) ď +
.
2
2
n δ k=0
2
nδ
2 2nδ 2
Choose N large enough such that
}f }8
ε
ă . Then for all n ě N ,
2N δ 2
2
ˇ
ˇ
}f ´ pn }8 = sup ˇf (x) ´ pn (x)ˇ ă ε .
˝
xP[0,1]
Remark 5.70. A polynomial of the form pn (x) =
n
ř
βk rk (x) is called a Bernstein poly-
k=0
nomial of degree n, and the coefficients βk are called Bernstein coefficients.
y
O
x
Figure 5.3: Using a Bernstein polynomial of degree 350 (the red curve) to approximate a
“saw-tooth” function (the blue curve)
(
)
Corollary 5.71. The collection of polynomials on [a, b] is dense in C ([a, b]; R), } ¨ }8 .
§5.6 The Stone-Weierstrass Theorem
183
(
)
Proof. We note that g P C ([a, b]; R) if and only if f (x) = g x(b ´ a) + a P C ([0, 1]; R); thus
ˇ
ˇ
ˇ
)ˇ
ˇf (x) ´ p(x)ˇ ă ε @ x P [0, 1] ô ˇˇg(x) ´ p( x ´ a ˇˇ ă ε @ x P [a, b] .
˝
b´a
Example 5.72.
Question: Let f P C ([0, 1], R) be such that }pn ´ f }8 Ñ 0 as n Ñ 8; that is, tpn u8
n=1
converges uniformly to f on [0, 1], where pn P P([0, 1]). Is f differentiable?
Answer: No! Take any continuous but not differentiable function f (for example, let
ˇ
1ˇ
f (x) = ˇx ´ ˇ). By Theorem 5.69, D pn : polynomial Q }pn ´ f }8 Ñ 0 as n Ñ 8.
2
Definition 5.73. Let (M, d) be a metric space, and E Ď M be a subset. A family A of
functions defined on E is called an algebra if
1. f + g P A for all f, g P A;
2. f ¨ g P A for all f, g P A;
3. αf P A for all f P A and α P R.
In other words, A is an algebra if A is closed under addition, multiplication, and scalar
multiplication.
Example 5.74. A function g : [a, b] Ñ R is called simple if we can divide up [a, b] into
sub-intervals on which g is constant except perhaps at the end-points. In other words, g is
called simple if there is a partition P = tx0 , x1 , ¨ ¨ ¨ , xN u of [a, b] such that
g(x) = g
( xi´1 + xi )
2
if x P (xi´1 , xi ) .
Then the collection of all simple functions is an algebra.
Proposition 5.75. Let (M, d) be a metric space, and A Ď M be a subset. If A Ď Cb (A; R)
is an algebra, then cl(A) is also an algebra.
8
8
Proof. Let f, g P cl(A). Then D tfk u8
k=1 , tgk uk=1 Ď A such that tfk uk=1 converges uniformly
to f on A, and tgk u8
k=1 converges uniformly to g on A. Since A is an algebra, fk + gk , fk ¨ gk
and αfk belong to A for all k P N. As a consequence, the uniform limit of fk + gk , fk ¨ gk
and αfk belong to cl(A) which implies that f + g, f ¨ g and αf belong to cl(A). Therefore,
cl(A) is an algebra.
˝
184
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Definition 5.76. Let (M, d) be a metric space, and A Ď M be a subset. A family F of
functions defined on A is said to
1. separate points on A if for all x, y P A and x ‰ y, there exists f P F such that
f (x) ‰ f (y).
2. vanish at no point of A if for each x P A there is f P F such that f (x) ‰ 0.
Example 5.77. Let P([a, b]) denote the collection of polynomials defined on [a, b] is an
algebra. Moreover, P([a, b]) separates points on [a, b] since p(x) = x does the separation,
and P([a, b]) vanishes at no point of [a, b].
Example 5.78. Let Peven ([a, b]) denote the collection of all polynomials p(x) of the form
p(x) =
n
ÿ
ak x2k = an x2n + an´1 x2n´2 + ¨ ¨ ¨ + a0 .
k=0
Then Peven ([a, b]) is an algebra. Moreover, Peven ([a, b]) vanishes at no point of [a, b] since the
constant functions are polynomials (since constant functions belongs to P([a, b]). However,
if ab ă 0, Peven ([a, b]) does not separate points on [a, b]. On the other hand, if ab ě 0, then
Peven ([a, b]) separates points on [a, b] since p(x) = x2 does the job.
Lemma 5.79. Let (M, d) be a metric space, and A Ď M be a subset. Suppose that A is an
algebra of functions defined on A, A separates points on A, and A vanishes at no point of
A. Then for all x1 , x2 P A, x1 ‰ x2 , and c1 , c2 P R (c1 , c2 could be the same), there exists
f P A such that f (x1 ) = c1 and f (x2 ) = c2 .
Proof. Since A separates points on A, D g P A such that g(x1 ) ‰ g(x2 ), and since A vanishes
at no point of A, D h, k P A such that h(x1 ) ‰ 0 and k(x2 ) ‰ 0. Then
[
]
[
]
g(x) ´ g(x2 ) h(x)
g(x) ´ g(x1 ) k(x)
]
]
f (x) = c1 [
+ c2 [
g(x1 ) ´ g(x2 ) h(x1 )
g(x2 ) ´ g(x1 ) k(x2 )
has the desired property.
˝
Theorem 5.80 (Stone). Let (M, d) be a metric space, K Ď M be a compact set, and
A Ď C (K; R) satisfying
1. A is an algebra.
2. A vanishes at no point of K.
3. A separates points on K.
§5.6 The Stone-Weierstrass Theorem
185
Then A is dense in C (K; R).
Example 5.81. Let K = [´1, 1] ˆ [´1, 1] Ď R2 . Consider the set P(K) of all polynomials
p(x, y) in two variables (x, y) P K. Then P(K) is dense in C (K; R).
Reason: Since K is compact, and P(K) is definitely an algebra and the constant function
p(x, y) = 1 P P(K) vanishes at no point of K, it suffices to show that P(K) separates
points. Let (a1 , b1 ) and (a2 , b2 ) be two different points in K. Then the polynomial
p(x, y) = (x ´ a1 )2 + (y ´ b1 )2
has the property that p(a1 , b1 ) ‰ p(a2 , b2 ). Therefore, P(K) separates points in K,
Proof of Theorem 5.80. We divide the proof into the following four steps:
s then |f | P A.
s
Step 1: We claim that if f P A,
Proof of claim: Let M = sup |f (x)|, and ε ą 0 be given. By Corollary 5.71, for every
xPK
ˇ
ˇ
ε ą 0 there is a polynomial p(y) such that ˇp(y) ´ |y|ˇ ă ε for all y P [´M, M]. Since
A is an algebra, by Proposition 5.75 cl(A) is also an algebra; thus g ” p(f ) P cl(A) if
f P cl(A). Nevertheless,
ˇ
ˇ
ˇg(x) ´ |f (x)|ˇ ă ε
@x P K
s
which shows that |f | P A.
Step 2: Let the functions maxtf, gu and mintf, gu be defined by
␣
(
maxtf, gu(x) = max f (x), g(x) ,
␣
(
mintf, gu(x) = min f (x), g(x) .
|f ´ g|
f +g
maxtf1 , ¨ ¨ ¨ , fn u P As and
mintf1 , ¨ ¨ ¨ , fn u P As .
f +g
|f ´ g|
Since maxtf, gu =
+
and mintf, gu =
´
, we find that if
2
2
2
2
s
s
s
s
f, g P A, then maxtf, gu P A and mintf, gu P A. As a consequence, if f1 , ¨ ¨ ¨ , fn P A,
Step 3: We claim that for any given f P C (K; R), a P K and ε ą 0, there exists a function
ga P As such that
ga (a) = f (a)
and ga (x) ą f (x) ´ ε
@x P K .
(5.6.1)
186
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Proof of claim: Since A separates points on K and A vanishes at no point of K, so
s Therefore, Lemma 5.79 implies that for every b P K with b ‰ a, there exists
does A.
hb P As such that hb (a) = f (a) and hb (b) = f (b). Note that every function in As is
continuous (by Theorem 5.7); thus the continuity of hb provides δ = δb ą 0 such that
[ (
)
(
)]
hb (x) ą f (x) ´ ε
@ x P D b, δb Y D a, δb X K .
(
)
(
)
Ť
Let Ub = D b, δb Y D a, δb . Then Ub is open. Since K Ď
Ub and K is combPK
b‰a
n
Ť
pact, there exists a finite set tb1 , ¨ ¨ ¨ , bn u Ď K such that K Ď
Ubj . Define
j=1
␣
(
s and ga (a) = f (a). Moreover, if x P K, x P Ub
ga = max hb1 , ¨ ¨ ¨ hbn . Then ga P A,
j
for some j; thus
ga (x) ě hbj (x) ą f (x) ´ ε
which implies that g satisfies (5.6.1).
Step 4: Let f P C (K; R) and ε ą 0 be given. For any a P K, let ga P As be a function
provided in Step 3 satisfying
ga (a) = f (a) and
ga (x) ą f (x) ´
ε
2
@x P K .
(5.6.2)
By the continuity of ga , there exists δ = δa ą 0 such that
ga (x) ă f (x) +
ε
2
(5.6.3)
@ x P D(a, δa ) X K .
Similar to Step 3, D ta1 , ¨ ¨ ¨ , am u Ď K such that
m
ď
)
(
KĎ
D aj , δaj .
(5.6.4)
j=1
␣
(
s and (5.6.2) shows that
Define h = min ga1 , ¨ ¨ ¨ , gam . Then h P A,
h(x) ą f (x) ´
ε
2
@ x P K.
Moreover, similar to Step 3 (5.6.3) and (5.6.4) imply that
h(x) ă f (x) +
ε
2
@ x P K.
s there exists p P A such that
On the other hand, since h P A,
ˇ
ˇ
ˇp(x) ´ h(x)ˇ ă ε
@x P K ;
2
thus
ˇ
ˇ ˇ
ˇ ˇ
ˇ
ˇp(x) ´ f (x)ˇ ď ˇp(x) ´ h(x)ˇ + ˇh(x) ´ f (x)ˇ ă ε
@x P K
§5.6 The Stone-Weierstrass Theorem
187
which concludes the theorem.
˝
ˇ
!
)
n
ř
ˇ
Example 5.82. Consider Peven ([0, 1]) = p(x) =
ak x2k ˇ ak P R (see Example 5.78).
k=0
Then A = Peven ([0, 1]) satisfies all the conditions in the Stone theorem, so Peven ([0, 1]) is
dense in C ([0, 1]; R).
On the other hand, if K = [´1, 1], then Peven ([´1, 1]) does not separate points on K
since if p P Peven ([´1, 1]), p(x) = p(´x); thus the Stone theorem cannot be applied to
conclude the denseness of Peven ([´1, 1]) in C ([´, 1]; R). In fact, Peven ([´1, 1]) is not dense
in C ([´1, 1]; R) since polynomials in Peven ([´1, 1]) are all even functions and f (x) = x
cannot be approximated by even functions.
Corollary 5.83. Let C (T) be the collection of all 2π-periodic continuous functions, and
Pn (T) be the collection of all trigonometric polynomials of degree n; that is,
Pn (T) =
!
n
ÿ
c0
ˇ
+
ck cos kx + sk sin kx ˇ tck unk=0 , tsk unk=1 Ď R .
2
ˇ
)
k=1
Let P(T) =
8
Ť
Pn (T). Then P(T) is dense in C (T). In other words, if f P C (T) and
n=0
ε ą 0 is given, there exists p P P(T) such that
ˇ
ˇ
ˇf (x) ´ p(x)ˇ ă ε
@x P R.
Proof. We note that C (T) can be viewed as the collection of all continuous functions defined
on the unit circle S1 in the sense that every f P C (T) corresponds to a unique F P C (S1 ; R)
such that f (x) = F (cos x, sin x), and vice versa. Since S1 Ď [´1, 1] ˆ [´1, 1] is compact,
Example 5.81 provides that P(S1 ), the collection of all polynomials defined on S1 , is an
algebra that separates points of S1 and vanishes at no point on S1 . The Stone-Weierstrass
Theorem then implies that there exists P P P(S1 ) such that
ˇ
ˇ
ˇF (x, y) ´ P (x, y)ˇ ă ε
@ (x, y) P S1 (that is, x2 + y 2 = 1).
Let p(x) = P (cos x, sin x). Note that
n
cos x =
( eix + e´ix )n
2
=
n
ÿ
1
n
ÿ
1
2
2n
C n eikx e´i(n´k)x =
n k
k=0
k=0
Ckn ei(2k´n)x
n
) ÿ
1 n(
1 n
=
Ck cos(2k ´ n)x + i sin(2k ´ n)x =
C cos(2k ´ n)x P Pn (T) ,
n
2
2n k
k=0
k=0
n
ÿ
188
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
and similarly, sinm x P Pm (T). Therefore, if P (x, y) =
n
ř
ak,ℓ xk y ℓ , then P (cos x, sin x) P
k,ℓ=0
P2n (T) because of the identities
]
1[
cos(θ ´ φ) + cos(θ + φ) ,
2
]
1[
sin θ cos φ =
sin(θ + φ) + sin(θ ´ φ) ,
2
]
1[
cos(θ ´ φ) ´ cos(θ + φ) .
sin θ sin φ =
2
cos θ cos φ =
As a consequence, we conclude that
ˇ
ˇ ˇ
ˇ
ˇf (x) ´ p(x)ˇ = ˇF (cos x, sin x) ´ P (cos x, sin x)ˇ ă ε
@x P R.
˝
5.7 The Contraction Mapping Principle（收縮映射原
理）and its Applications
Definition 5.84. Let (M, d) be a metric space, and Φ : M Ñ M be a mapping. Φ is said
to be a contraction mapping if there exists a constant k P [0, 1) such that
(
)
d Φ(x), Φ(y) ď kd(x, y)
@ x, y P M .
Remark 5.85. A contraction mapping must be (uniformly) continuous.
Reason: Given ε ą 0, take δ =
if d(x, y) ă δ, then
ε
, where k is set as in the definition of contraction. Now
k
(
)
ε
d Φ(x), Φ(y) ď kd(x, y) ă k ¨ = ε .
k
Example 5.86. For what r ă 1 do we have f : [0, r] Ñ [0, r] where f (x) = x2 a contraction?
Answer: By the mean value theorem, f (x) ´ f (y) = f 1 (c)(x ´ y), c between x, y; thus
ˇ
ˇ
ˇ ˇ
ˇf (x) ´ f (y)ˇ = ˇf 1 (c)ˇ|x ´ y| = 2c|x ´ y| ď 2r|x ´ y| .
1
2
Hence for all r ă , the map f : [0, r] Ñ [0, r] is a contraction where f (x) = x2 .
1
On the other hand, we show that f cannot be a contraction if r = . Suppose the
ˇ 2 ˇ
ˇ
[ 1] ˇ 2
2ˇ
ˇ
contrary that there exists k P [0, 1) such that for all x, y P 0, , x ´ y ď k ˇx ´ y ˇ. Then
2
ˇ
ˇ
ˇx2 ´ y 2 ˇ
ˇ
ˇ
sup
ˇx ´ y ˇ ď k ă 1 .
1
x‰y,x,yP[0, 2 ]
§5.7 The Contraction Mapping Principle and its Applications
189
[ 1]
1
1
´ , n = 1, 2, ¨ ¨ ¨ , x, yn P 0, . So
2
n
2
ˇ 2
ˇ
2
ˇ x ´ yn ˇ
ˇ
ˇ
(1 1 1 )
ˇ = lim ˇx + yn ˇ = lim
lim ˇˇ
+ ´
= 1.
nÑ8 x ´ yn ˇ
nÑ8
nÑ8 2
2 n
ˇ 2
ˇ
ˇx ´ y 2 ˇ
ă 1 is not possible.
This means
sup
x‰y,x,yP[0, 1 ] |x ´ y|
1
2
But we can take x = , yn =
2
Definition 5.87. Let (M, d) be a metric space, and Φ : M Ñ M be a mapping. A point
x0 P M is called a fixed-point for Φ if Φ(x0 ) = x0 .
Example 5.88. Let Φ : R Ñ R be given by Φ(x) =
is also a fixed-point.
x2 + 2
. Then 1 is a fixed-point, and 2
3
Theorem 5.89 (Contraction Mapping Principle). Let (M, d) be a complete metric space,
and Φ : M Ñ M be a contraction mapping. Then Φ has a unique fixed-point.
Proof. Let x0 P M , and define xn+1 = Φ(xn ) for all n P N Y t0u. Then
(
)
d(xn+1 , xn ) = d Φ(xn ), Φ(xn´1 ) ď kd(xn , xn´1 ) ď k n d(x1 , x0 ) ;
thus if n ą m,
d(xn , xm ) ď d(xm , xm+1 ) + d(xm+1 , xm+2 ) + ¨ ¨ ¨ + d(xn´1 , xn )
ď (k m + k m+1 + ¨ ¨ ¨ + k n´1 )d(x1 , x0 )
km
d(x1 , x0 ) .
ď k (1 + k + k + ¨ ¨ ¨ )d(x1 , x0 ) =
1´k
m
2
(5.7.1)
km
d(x1 , x0 ) = 0; thus
mÑ8 1 ´ k
Since k P [0, 1), lim
@ ε ą 0, D N ą 0 Q d(xn , xm ) ă ε @ n, m ě N .
In other words, txn u8
n=1 is a Cauchy sequence. Since (M, d) is complete, xn Ñ x as n Ñ 8
for some x P M . Finally, since Φ(xn ) = xn+1 for all n P N, by the continuity of Φ we obtain
that
Φ(x) = lim Φ(xn ) = lim xn+1 = x
nÑ8
nÑ8
which guarantees the existence of a fixed-point.
Suppose that for some x, y P M , Φ(x) = x and Φ(y) = y. Then
(
)
d(x, y) = d Φ(x), Φ(y) ď kd(x, y)
which implies that d(x, y) = 0 or x = y. Therefore, the fixed-point of Φ is unique.
˝
190
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Remark 5.90. The proof of the contraction mapping principle also provides an iterative
way, xk+1 = Φ(xk ), of finding the fixed-point of a contraction mapping Φ. Using (5.7.1), the
convergence rate of txm u8
m=1 to the fixed-point x is measured by
d(xm , x) = lim d(xm , xn ) ď
nÑ8
km
d(x1 , x0 ) .
1´k
Therefore, the smaller the contraction constant k, the faster the convergence.
Remark 5.91. Theorem 5.89 sometimes is also called the Banach fixed-point theorem.
Example 5.92. The condition k ă 1 in Theorem 5.89 is necessary. For example, let M = R,
ˇ
ˇ ˇ
ˇ
d(x, y) = |x ´ y|, and Φ : R Ñ R be given by Φ(x) = x + 1. Then ˇΦ(x) ´ Φ(y)ˇ = ˇx ´ y ˇ.
Suppose x‹ is a fixed-point of Φ. Then x‹ = Φ(x‹ ) = x‹ + 1 which leads to a contradiction
that 0 = 1.
1
x
Example 5.93. Let Φ : [1, 8) Ñ [1, 8) be given by Φ(x) = x + . Then if x ‰ y,
ˇ ˇ
ˇ
ˇ ˇ
(
)ˇ
ˇΦ(x) ´ Φ(y)ˇ = ˇx ´ y + 1 ´ 1 ˇ = ˇ(x ´ y) 1 ´ 1 ˇ ă |x ´ y| .
x y
xy
However, there is no fixed-point of Φ.
Example 5.94 (The secant method). Suppose that f is continuously differentiable, f 1 (x) ą
0 for all x P [a, b] and f (a)f (b) ă 0. By the intermediate value theorem there must be a
(unique) zero of f . How do we find this zero?
Assume that sup f 1 (x) ă 8. Let
xP[a,b]
M = max
!
sup f 1 (x), ´
xP[a,b]
be a positive constant, and consider Φ(x) = x ´
f (a) f (b)
,
b´a b´a
)
+1
f (x)
. Then by the mean value theorem,
M
1
1
ˇ
ˇ ˇ
ˇ (
)
ˇΦ(x) ´ Φ(y)ˇ = ˇ(x ´ y)(1 ´ f (ξ) )ˇ ď 1 ´ minξP[a,b] f (ξ) |x ´ y| ď k|x ´ y| ,
M
M
where k P [0, 1) is a fixed constant. Moreover, Φ1 (x) = 1 ´
f 1 (x)
ą 0; thus Φ is strictly
M
increasing. Since the choice of M implies that a ă Φ(a) ă Φ(b) ă b; thus Φ : [a, b] Ñ [a, b].
Therefore, the contraction mapping principle implies that one can find the fixed-point of Φ
(which is the zero of f ) using the iterative scheme xk+1 = Φ(xk ) (by picking any arbitrary
initial guess x0 P [a, b]).
§5.7 The Contraction Mapping Principle and its Applications
191
5.7.1 The existence and uniqueness of the solution to ODEs
In this sub-section we are concerned with if there is a solution to the initial value problem
of ordinary differential equation:
(
)
x1 (t) = f x(t), t
@ t P [t0 , t0 + ∆t],
(5.7.2a)
(5.7.2b)
x(t0 ) = x0 ,
where x : [t0 , t0 + ∆t] Ñ Rn and f : Rn ˆ [t0 , t0 + ∆t] Ñ Rn are vector-valued functions,
and x0 P Rn is a vector. Another question we would like to answer is “if (5.7.2) indeed has
a solution, is the solution unique?”
Theorem 5.95 (Fundamental Theorem of ODE). Suppose that for some r ą 0, f :
D(x0 , r) ˆ [t0 , T ] Ñ Rn is continuous and is Lipschitz in the spatial variable; that is,
›
›
DK ą 0 Q ›f (x, t) ´ f (y, t)›2 ď K}x ´ y}2
@ x, y P D(x0 , r) and t P [t0 , T ] .
Then there exists 0 ă ∆t ď T ´ t0 such that there exists a unique solution to (5.7.2).
Proof. For any x P C ([t0 , T ]; Rn ), define
żt
Φ(x)(t) = x0 +
(
)
f x(s), s ds .
t0
We note that if x(t) is a solution to (5.7.2), then x is a fixed point of Φ (for t P [t0 , t0 + ∆t]).
Therefore, the problem of finding a solution to (5.7.2) transforms to a problem of finding a
fixed-point of Φ.
To guarantee the existence of a unique fixed-point, we appeal to the contraction mapping
principle. To be able to apply the contraction mapping principle, we need to specify the
metric space (M, d). Let
!
∆t = min T ´ t0 ,
and define
r
1 )
,
,
Kr + 2}f (x0 , ¨)}8 2K
!
)ˇ
(
r)
n ˇ
M = x P C [t0 , t0 + ∆t]; R ˇ }x ´ x0 }8 ď
2
(
)
with the metric induced by the sup-norm } ¨ }8 of C [t0 , t0 + ∆t]; Rn . Then
(5.7.3)
192
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
1. We first show that Φ : M Ñ M . To see this, we observe that
żt
›ż t (
›ż t [ (
›
›
›
) ›
)
]
›
›Φ(x) ´ x0 › = ›› f x(s), s ds›› = ››
f x(s), s ´ f (x0 , s) ds +
f (x0 , s)ds›
8
8
t0
ż t0 +∆t
ď
t0
t0
t0
› (
›
)
›f x(s), s ´ f (x0 , s)› ds +
2
ż t0 +∆t
t0
8
›
›
›f (x0 , s)› ds
2
ż t0 +∆t
›
›
ďK
}x(s) ´ x0 }2 ds + ∆t›f (x0 , ¨)›8
[t0
›
› ]
ď ∆t K}x ´ x0 }8 + ›f (x0 , ¨)›8 ;
r
2
thus if x P M , (5.7.3) implies that }Φ(x) ´ x0 }8 ď .
›
›
2. Next we show that Φ is a contraction mapping. To see this, we compute ›Φ(x)´Φ(y)›8
for x, y P M and find that
›ż t [ (
›
›
)
(
)] ››
›Φ(x) ´ Φ(y)› ď ››
f
x(s),
s
´
f
y(s),
s
ds›
8
8
t0
ż t0 +∆t
1
ď
K}x(s) ´ y(s)}2 ds ď K∆t}x ´ y}8 ď }x ´ y}8 ;
2
t0
thus Φ : M Ñ M is a contraction mapping.
3. Finally we show that (M, d) is complete. It suffices to show that M is a closed subset
of C ([t0 , t0 + ∆t]; Rn ). Let txk u8
k=1 be a uniformly convergent sequence with limit x.
r
for all t P [t0 , t0 + ∆t], passing k to the limit we find that
2
r
r
}x(t) ´ x0 }2 ď for all t P [t0 , t0 + ∆t] which implies that }x ´ x0 }8 ď ; thus x P M .
2
2
Since }xk (t) ´ x0 }2 ď
Therefore, by the contraction mapping principle, there exists a unique fixed point x P M
which implies that there exists a unique solution to (5.7.2).
˝
Example 5.96. Let
#
xc (t) =
0
if 0 ď t ă c ,
1
(t ´ c)2 if t ě c .
4
1
Then for all c ą 0, xc (t) is a solution to x1 (t) = x(t) 2 for all t ą 0 with initial value x(0) = 0.
?
The reason for not having unique solution is that if f (x, t) = x, f : D(0, r) ˆ R Ñ R is
not Lipschitz in the spatial variabvle for all r ą 0. In other words, for all r, K ą 0, there
ˇ
exists x, y P D(0, r) satisfying ˇf (x) ´ f (y)| ě K|x ´ y| .
§5.7 The Contraction Mapping Principle and its Applications
193
Example 5.97. Findż a function x(t) satisfying x1 (t) = x(t) with initial value x(0) = 1.
t
Define Φ(x)(t) = 1 +
x(s)ds, x0 (t) = 1 and xk+1 (t) = Φ(xk )(t). Then
0
żt
żt
x1 (t) = 1 +
x0 (s)ds = 1 + t ñ x2 (t) = 1 +
ż0 t
ñ x3 (t) = 1 +
x1 (s)ds = 1 + t +
0
2
x2 (s)ds = 1 + t +
0
t2
2
3
t
t
+
2
3¨2
ñ ¨¨¨¨¨¨
ñ By induction, we have xk (t) = 1 + t +
which converges to x(t) =
8 tk
ř
k=0 k!
t2
tk
+ ¨¨¨ +
2
k!
= et .
Example 5.98. Findż a function x(t) satisfying x1 (t) = tx(t) with initial value x(0) = 3.
t
Define Φ(x)(t) = 3 +
sx(s)ds, x0 (t) = 3 and xk+1 (t) = Φ(xk )(t). Then
0
żt
żt
3t2
3t2
3t4
x1 (t) = 3 + 3sds = 3 +
ñ x2 (t) = 3 + sx1 (s)ds = 3 +
+
2
2
2¨4
0
ż0 t
2
4
6
3t
3t
3t
+
+
.
ñ x3 (t) = 3 + sx2 (s)ds = 3 +
2
2¨4 2¨4¨6
0
We can conjecture and prove that
xk (t) = 3 +
thus xk (t) Ñ x(t) = 3 + 3
3t2
3t4
3t2k
+
+ ¨¨¨ +
;
2
2¨4
2 ¨ 4 ¨ ¨ ¨ (2k)
t2k
. To see what x(t) is, we observe that
k=1 2 ¨ 4 ¨ ¨ ¨ (2k)
8
ř
(2 k
8
8
ÿ
ÿ
( t2 )
t /2)
t2k
t2k
1+
=
=
= exp
;
2 ¨ 4 ¨ ¨ ¨ (2k) k=0 2k k! k=0 k!
2
k=1
8
ÿ
thus the solution is x(t) = 3 exp
( t2 )
2
.
Remark 5.99. In the iterative process above of solving ODE, the iterative relation
xk+1 (t) = x0 +
żt (
)
f xk (s), s ds
t0
is called the Picard iteration.
194
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Example 5.100. Is there a solution to the Fredholm equation
żb
x(t) = λ K(t, s)x(s)ds + φ(t) ?
(5.7.4)
a
Define Φ : C ([a, b]; R) Ñ C ([a, b]; R) by
żb
Φ(x)(t) = λ
K(t, s)x(s)ds + φ(t) .
a
Then if K : [a, b] ˆ [a, b] Ñ R is continuous, and φ : [a, b] Ñ R is continuous, Φ(x) P
C ([a, b]; R) as long as x P C ([a, b]; R). Moreover,
ż
ˇ
ˇ ˇ b
(
) ˇ
ˇΦ(x)(t) ´ Φ(y)(t)ˇ ď ˇˇλ K(t, s) x(s) ´ y(s) dsˇˇ ď |λ|}K}8 |b ´ a|}x ´ y}8 ;
a
thus if |λ|}K}8 |b ´ a| ă 1, Φ is a contraction mapping. As a consequence, if
1. K : [a, b] ˆ [a, b] Ñ R is continuous;
2. φ : [a, b] Ñ R is continuous;
3. |λ|}K}8 |b ´ a| ă 1,
there exists a unique function x(t) satisfying (5.7.4).
5.8 Exercises
§5.1 Pointwise and Uniform Convergence
Problem 5.1. Let (M, d) be a metric space, A Ď M , and fk : A Ñ R be a sequence
of functions (not necessary continuous) which converges uniformly on A. Suppose that
a P cl(A) and
lim fk (x) = Ak
xÑa
exists for all k P N. Show that tAk u8
k=1 converges, and
lim lim fk (x) = lim lim fk (x) .
xÑa kÑ8
kÑ8 xÑa
Problem 5.2. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and fk : A Ñ N be
uniformly continuous functions, and tfk u8
k=1 converges uniformly to f : A Ñ N on A. Show
that f is uniformly continuous on A.
§5.8 Exercises
195
Problem 5.3. Complete the following.
(a) Suppose that fk , f, g : [0, 8) Ñ R are functions such that
1. @ R ą 0, fk and g are Riemann integrable on [0, R];
2. |fk (x)| ď g(x) for all k P N and x P [0, 8);
3. @ R ą 0, tfk u8
k=1 converges to f uniformly on [0, R];
4.
ż8
g(x)dx ” lim
RÑ8
0
Show that lim
kÑ8
żR
g(x)dx ă 8.
0
ż8
ż8
fk (x)dx =
0
f (x)dx; that is,
0
lim lim
żR
kÑ8 RÑ8 0
"
(b) Let fk (x) be given by fk (x) =
fk (x)dx = lim lim
żR
RÑ8 kÑ8 0
fk (x)dx .
1 if k ´ 1 ď x ă k ,
Find the (pointwise) limit f of
0 otherwise.
the sequence tfk u8
k=1 , and check whether
lim
kÑ8
explain why one can or cannot apply (a).
ż8
ż8
fk (x)dx =
0
x
(c) Let fk : [0, 8) Ñ R be given by fk (x) =
. Find lim
1 + kx4
kÑ8
f (x)dx or not. Briefly
0
ż8
fk (x)dx.
0
§5.2 The Weierstrass M -Test
Problem 5.4. Show that the series
8
ÿ
(´1)k
k=1
x2 + k
k2
converges uniformly on every bounded interval.
Problem 5.5. Consider the function
f (x) =
8
ÿ
1
.
1 + k2x
k=1
On what intervals does it converge uniformly? On what intervals does it fail to converge
uniformly? Is f continuous wherever the series converges? If f bounded?
196
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Problem 5.6. Determine which of the following real series
uniformly). Check the continuity of the limit in each case.
"
0
if x ď k ,
1. gk (x) =
(´1)k if x ą k .
$
1
’
& 2 if |x| ď k ,
k
2. gk (x) =
’
% 1 if |x| ą k .
2
8
ř
gk converge (pointwise or
k=1
x
3. gk (x) =
( (´1)k )
?
cos(kx) on R.
k
4. gk (x) = xk on (0, 1).
§5.3 Integration and Differentiation of Series
Problem 5.7. In the following series of functions defined on R, find its domain of convergence (classify it into domain of absolute and conditional convergence). If the series is a
power series, find its radius of convergence. Also discuss whether the series is uniformly
convergent in every compact subsets of its domain of convergence. Determine which series
can be differentiated or integrated term by term in its domain of convergence.
8
ř
(1)
x
, α ě 0, β ą 0;
α + k β x2
k
k=1
(2)
8 1 ?
ř
1 ´ x2k ;
k
(3)
k=1 2
8 1 ¨ 3 ¨ ¨ ¨ (2k ´ 1) (
ř
1
1)
1 + + ¨ ¨ ¨ + x2k ;
k=1
(4)
(5)
2 ¨ 4 ¨ ¨ ¨ (2k)
2
k
(´1)k´1
xk! ;
k=1 k log(k + 1)
8
ř
8
ř
ak xk , where tak u8
k=1 is defined by the recursive relation ak = 3ak´1 ´ 2ak´2 for
k=1
k ě 2, and a0 = 1, a1 = 2.
Also find the sum of the series in (5).
§5.8 Exercises
197
Problem 5.8. In this problem we investigate the differentiability of a complex power series.
8
8
ř
d ř
This requires a new proof of
ak x k =
kak xk´1 instead of making use of Theorem
dx k=0
5.10.
k=1
8
ř
Let tak u8
ak xk be a (real) power series with
k=0 Ď R be a real sequence, and f (x) =
k=0
n
ř
radius of convergence R ą 0. Let sn (x) =
ak xk be the n-th partial sum, Rn (x) =
k=0
8
ř
k´1
f (x) ´ sn (x), and g(x) =
kak x . For x, x0 P (´ρ, ρ) Ĺ (´R, R), where x ‰ x0 , write
k=1
(
) Rn (x) ´ Rn (x0 )
sn (x) ´ sn (x0 )
f (x) ´ f (x0 )
´ g(x) =
´ sn1 (x0 ) + sn1 (x0 ) ´ g(x0 ) +
.
x ´ x0
x ´ x0
x ´ x0
1. Show that
8
ˇ R (x) ´ R (x ) ˇ
ÿ
ˇ n
n 0 ˇ
k|ak |ρk´1 ,
ˇ
ˇď
x ´ x0
k=n+1
f (x) ´ f (x0 )
= g(x0 ).
xÑx0
x ´ x0
and use the inequality above to show that lim
2. Generalize the conclusion to complex power series; that is, show that if tak u8
k=0 Ď C
8
ř
ak z k has radius of convergence R ą 0, then
and the power series
k=0
8
8
ÿ
d ÿ
k
ak z =
kak z k´1
dz k=0
k=1
Assume that you have known
@ |z| ă R .
n
n
ř
d ř
ak z k =
kak z k´1 for all n P N Y t0u (this can
dz k=0
k=1
be proved using the definition of differentiability of functions with values in normed
vector spaces provided in Chapter 6).
Problem 5.9. Suppose that the series
8
ř
n=0
an = 0, and f (x) =
Show that f is continuous at x = 1 by complete the following.
8
ř
an xn for ´1 ă x ď 1.
n=0
1. Write sn = a0 + a1 + ¨ ¨ ¨ + an and sn (x) = a0 + a1 x + ¨ ¨ ¨ + an xn . Show that
sn (x) = (1 ´ x)(s0 + s1 x + ¨ ¨ ¨ + sn´1 xn´1 ) + sn xn
and f (x) = (1 ´ x)
8
ř
s n xn .
n=0
2. Using the representation of f from above to conclude that lim´ f (x) = 0.
xÑ1
198
3. What if
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
8
ř
an is convergent but not zero?
n=0
Problem 5.10. Construct the function g(x) by letting g(x) = |x| if x P
[
extending g so that it becomes periodic (with period 1). Define
f (x) =
8
ÿ
g(4k´1 x)
k=1
4k´1
1 1]
and
2 2
´ ,
.
1. Use the Weierstrass M -test to show that f is continuous on R.
2. Prove that f is differentiable at no point.
(So there exists a continuous which is nowhere differentiable!)
Hint: Google Blancmange function!
§5.4 The Space of Continuous Functions
Problem 5.11. Let δ : (C ([0, 1]; R), } ¨ }8 ) Ñ R be defined by δ(f ) = f (0). Show that δ is
linear and continuous.
Problem 5.12. Let (M, d) be a metric space, and K Ď M be a compact subset.
ˇ
␣
(
1. Show that the set U = f P C (K; R) ˇ a ă f (x) ă b for all x P K is open in
(
)
C (K; R), } ¨ }8 for all a, b P R.
ˇ
␣
(
2. Show that the set F = f P C (K; R) ˇ a ď f (x) ď b for all x P K is closed in
(
)
C (K; R), } ¨ }8 for all a, b P R.
3. Let A Ď M be a subset, not necessarily compact. Prove or disprove that the set
ˇ
␣
(
(
)
B = f P Cb (A; R) ˇ f (x) ą 0 for all x P A is open in Cb (A; R), } ¨ }8 .
§5.5 The Arzelà-Ascoli Theorem
Problem 5.13. Which of the following set B of continuous functions are equi-continuous
s compact in M ?
in the metric space M ? Are the closure B
ˇ
␣
(
1. B = sin kx ˇ k = 0, 1, 2, ¨ ¨ ¨ , M = C ([0, π]; R).
ˇ
?
␣
(
2. B = sin x + 4k 2 π 2 ˇ k = 0, 1, 2, ¨ ¨ ¨ , M = Cb ([0, 8); R).
§5.8 Exercises
3. B =
!
199
x2
ˇ
ˇ k = 0, 1, 2, ¨ ¨ ¨ , M = C ([0, 1]; R).
x2 + (1 ´ kx)2
ˇ
)
ˇ
(
␣
4. B = (k + 1)xk (1 ´ x) ˇ k P N , M = C ([0, 1]; R).
Problem 5.14. Let (M, d) be a metric space, (V, } ¨ }) be a normed space, and A Ď M be
a subset. Suppose that B Ď Cb (A; V) be equi-continuous. Prove or disprove that cl(B) is
equi-continuous.
Problem 5.15. Let fk : [a, b] Ñ R be a sequence of differentiable functions such that fk (a)
is bounded and |fk1 (x)| ď M for all x P [a, b] and k P N. Show that tfk u8
k=1 contains an
uniformly convergent subsequence. Must the limit function differentiable?
Problem 5.16. Let C 0,α ([0, 1]; R) denote the “space”
ˇ
!
)
|f (x) ´ f (y)|
ˇ
C 0,α ([0, 1]; R) ” f P C ([0, 1]; R) ˇ sup
ă
8
,
|x ´ y|α
x,yP[0,1]
where α P (0, 1]. For each f P C 0,α ([0, 1]; R), define
|f (x) ´ f (y)|
.
|x ´ y|α
x,yP[0,1]
}f }C 0,α = sup |f (x)| + sup
xP[0,1]
x‰y
(
)
1. Show that C 0,α ([0, 1]; R), } ¨ }C 0,α is a complete normed space.
ˇ
␣
(
2. Show that the set B = f P C ([0, 1]; R) ˇ }f }C 0,α ă 1 is equi-continuous.
(
)
3. Show that cl(B) is compact in C ([0, 1]; R), } ¨ }8 .
Problem 5.17. Given f : R Ñ R a continuous periodic function of period 1; that is,
f (x + 1) = f (x) for all x P R, and x1 , ¨ ¨ ¨ , xm P [0, 1] arbitrary m points, define a new
function
I(f ; x1 , ¨ ¨ ¨ , xm )(x) =
)
1(
f (x + x1 ) + ¨ ¨ ¨ f (x + xm )
m
@x P R.
Prove that the set
ˇ
␣
(
B = I(f ; x1 , ¨ ¨ ¨ , xm ) ˇ x1 , ¨ ¨ ¨ , xm P [0, 1], m P N
is uniformly bounded
and equi-continuous in the space C ([0, 1]; R). Moreover, show that the
!ż 1
)
derived set B 1 =
f (x)dx ; that is, the derived set of B consists of one single function
0
which is a constant function y =
ż1
f (x)dx.
0
200
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
Problem 5.18. Let (M, d) be a metric space, (V, } ¨ }) be a Banach space, K Ď M be
compact, and tfk u8
k=1 Ď C (K; V) be a sequence of continuous functions. Suppose that
8
for all x P K, if txk u8
k=1 , tyk uk=1 Ď K and lim xk = lim yk = x, the limits lim fk (xk ) and
kÑ8
kÑ8
kÑ8
8
lim fk (yk ) exist and are identical. Show that tfk uk=1 converges uniformly on K. How about
kÑ8
if K is not compact?
Problem 5.19. Assume that tfk u8
k=1 is a sequence of monotone increasing functions on R
with 0 ď fk (x) ď 1 for all x and k P N.
␣ (8
1. Show that there is a subsequence fkj j=1 which converges pointwise to a function f
(This is usually called the Helly selection theorem).
␣ (8
2. If the limit f is continuous, show that fkj j=1 converges uniformly to f on any
compact set of R.
§5.6 The Stone-Weierstrass Theorem
Problem 5.20. Define B to be the set of all even functions in the space C ([´1, 1]; R); that
is, f P B if and only if f is continuous on [´1, 1] and f (x) = f (´x) for all x P [´1, 1].
Prove that B is closed but not dense in C ([´1, 1]; R). Hence show that even polynomials
are dense in B, but not in C ([´1, 1]; R).
Problem 5.21. Let f : [0, 1] Ñ R be a continuous function.
1. Suppose that
ż1
f (x)xn dx = 0
@ n P N Y t0u .
0
Show that f = 0 on [0, 1].
2. Suppose that for some m P N,
ż1
f (x)xn dx = 0
@ n P t0, 1, ¨ ¨ ¨ , mu .
0
Show that f (x) = 0 has at least m distinct real roots around which f (x) change signs.
Problem 5.22. Let f : [0, 1] Ñ R be continuous. Show that
ż1
lim
f (x) sin(nx) dx = 0 .
nÑ8 0
§5.8 Exercises
201
Problem 5.23. Put p0 = 0 and define
x2 ´ p2k (x)
pk+1 (x) = pk (x) +
@ k P N Y t0u .
2
Show that tpk u8
k=1 converges uniformly to |x| on [´1, 1].
Hint: Use the identity
]
[
][
|x| + pk (x)
|x| ´ pk+1 (x) = |x| ´ pk (x) 1 ´
2
to prove that 0 ď pk (x) ď pk+1 (x) ď |x| if |x| ď 1, and that
)
(
2
|x| k
ă
|x| ´ pk (x) ď |x| 1 ´
2
k+1
if |x| ď 1.
Problem 5.24. Let f : [0, 1] Ñ R be continuous and ε ą 0. Prove that there is a simple
function g (defined in Example 5.74) such that }f ´ g}8 ă ε.
Problem 5.25. Suppose that pn is a sequence of polynomials converging uniformly to f on
[0, 1] and f is not a polynomial. Prove that the degrees of pn are not bounded.
Hint: An N th-degree polynomial p is uniquely determined by its values at N + 1 points
x0 , ¨ ¨ ¨ , xN via Lagrange’s interpolation formula
N
ÿ
p(xk )
,
πk (xk )
k=0
ś
(x ´ xj ).
where πk (x) = (x ´ x0 )(x ´ x1 ) ¨ ¨ ¨ (x ´ xN )/(x ´ xk ) =
p(x) =
πk (x)
1ďjďN
j‰k
Problem 5.26. Consider the set of all functions on [0, 1] of the form
h(x) =
n
ÿ
aj ebj x ,
j=1
where aj , bj P R. Is this set dense in C ([0, 1]; R)?
§5.7 The Contraction Mapping Principle and its Applications
Problem 5.27. Suppose that f : [a, b] Ñ R is twice continuous differentiable; that is,
f 1 , f 2 : [a, b] Ñ R are continuous, and f (a) ă 0 = f (c) ă f (b), and f 1 (x) ‰ 0 for all
x P [a, b]. Consider the function
Φ(x) = x ´
f (x)
.
f 1 (x)
202
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
1. Show that Φ : [a, b] Ñ R satisfies
|Φ(x) ´ Φ(y)| ď k|x ´ y|
@ x, y P [a, b]
for some k P [0, 1) if |b ´ a| are small enough.
2. Suppose that f 2 (x) ą 0 for all x P [a, b]. Show that there exists a ď r
a ă c such that
Φ : [r
a, b] Ñ [r
a, b].
3. Under the condition of 2, show that if x0 P [r
a, b], then the iteration
xk+1 = Φ(xk ) @ k P N Y t0u
provides a convergent sequence txk u8
k=1 with limit c.
(The iteration scheme above of finding the zero c of f is called the Newton method.)
Problem 5.28. Let (M, d) be a complete metric space, and f : M Ñ M . Define fk =
f ˝ f ˝ ¨ ¨ ¨ ˝ f , here the composition was taken for k ´ 1 times. Assume that there exists a
sequence tαk u8
k=1 Ď R such that
1. αk Ñ 0 as k Ñ 8.
(
)
2. d fk (x), fk (y) ď αk d(x, y) for all k P N, x, y P M .
Show that f has a unique fixed-point.
Problem 5.29. Let (M, d) be a metric space, and K Ď M be a compact set.
(1) Given f : M Ñ M a continuous map, define fk = f ˝ f ˝ ¨ ¨ ¨ ˝ f (as in the previous
problem) to the the k-th iterate of f . Prove that if fk has a unique fixed-point x0 ,
then f (x0 ) = x0 .
(
)
(2) Let f : K Ñ K be continuous and d f (x), f (y) ă d(x, y) for all x, y P K. Show that
f has a unique fixed-point in K. Show that the conclusion is false if K is not compact.
ˇ
ˇ
(3) Let K = [0, 1] be a closed interval in (2) and ˇf (x) ´ f (y)ˇ ď |x ´ y| for all x, y P K.
Given any x1 P [0, 1], define a sequence txk u8
k=1 by
xk+1 =
)
1(
xk + f (xk )
2
Show that txk u8
k=1 converges to a fixed-point of f .
@k ą 1.
§5.8 Exercises
203
(
)
Problem 5.30. Let (M, d) be a metric space, and f : M Ñ M be such that d f (x), f (y) ă
d(x, y) for all x, y P M , x ‰ y.
1. Fix x0 P M . Let xn+1 = f (xn ), and cn = d(xn , xn+1 ). Show that tcn u8
n=1 is a decreasing
sequence; thus c = lim cn exists.
nÑ8
␣ (8
2. Assume that there is a subsequence xnj j=1 of txn u8
n=1 such that xnj Ñ x as j Ñ 8.
Show that
(
)
(
)
c = d x, f (x) = d f (x), f (f (x)) .
and deduce that x is a fixed-point of f .
3. Suppose further that M is compact. Show that the sequence txn u8
n=1 itself converges
to x.
Problem 5.31. Find an upper bound on r ą 0 such that the mapping T : C ([0, r]; R) Ñ
C ([0, r]; R) defined by
żx
tf (t) dt
T (f )(x) = 1 + 3
0
is a contraction mapping. what is its fixed-point?
Problem 5.32. Let A = [a, b] ˆ [a, b] be a closed square in R2 , M = C ([a, b]; R), and
K : A Ñ R be a continuous function. For f P C ([a, b]; R), define
żb
T (f )(x) =
K(x, y)f (y)dy
@ x P [a, b] .
a
(1) Show that T (f ) P M for all f P M , and T : M Ñ M is Lipschitz continuous. Find a
Lipschitz constant for T .
ˇ
␣
(
(2) If B Ď M is a bounded subset of M , show that the image T (B) = T (f ) ˇ f P B is
uniformly bounded and equi-continuous.
1
(3) If the norm }K}8 ă
, show that T is a contraction mapping. What is its fixedb´a
point?
(4) If K satisfies the assumption in (3), show that the mapping S : M Ñ M defined by
S(f ) = f ´ T (f ) is a homeomorphism.
204
CHAPTER 5. Uniform Convergence and the Space of Continuous Functions
(5) Let a = 0, b = 1, and K(x, y) =
1 x+y´1
e
. Show that K satisfies the assumption in
4
(3). Given g P M , find f P M such that S(f ) = g.
Problem 5.33. Let A = [a, b]ˆ[a, b] be a closed square in R2 , and K : A Ñ R be continuous
on A. Define
ż
b
T (f )(x) =
K(x, y)g(y)dy ,
a
where f is a real-valued function defined on [a, b] such that the integral makes sense. For
a family of functions F consisting of f such that T (f ) is well-defined and |f (y)| ď M for
all y P [a, b], let G = T (F). Show that each sequence of G contains a uniformly convergent
subsequence.
Problem 5.34 (True or False). Determine whether the following statements are true or
false. If it is true, prove it. Otherwise, give a counter-example.
1. Let fn : [a, b] Ñ R be an uniformly convergent sequence of continuous functions. Then
the sequence of the indefinite integrals gn (x) defined by
żx
fn (t) dt
@ x P [a, b]
gn (x) =
a
converges uniformly to a continuously differentiable function.
2. Let fn : [0, 1] Ñ R be a equi-continuous sequence of functions such that the sequence
␣ 1 (8
fn ( ) n=1 is bounded in R. Then tfn u8
n=1 contains a convergent subsequence.
2
3.
4.
Chapter 6
Differentiation of Maps
6.1 Bounded Linear Maps
Definition 6.1. A map L from a vector space X into a vector space Y is said to be linear
if L(cx1 + x2 ) = cL(x1 ) + L(x2 ) for all x1 , x2 P X and c P R. We often write Lx instead of
L(x), and the collection of all linear maps from X to Y is denoted by L (X, Y ).
Suppose further that X and Y are normed spaces equipped with norms } ¨ }X and } ¨ }Y ,
respectively. A linear map L : X Ñ Y is said to be bounded if
sup }Lx}Y ă 8 .
}x}X =1
The collection of all bounded linear maps from X to Y is denoted by B(X, Y ), and the
number sup }Lx}Y is often denoted by }L}B(X,Y ) .
}x}X =1
Example 6.2. Let L : Rn Ñ Rm be given by Lx = Ax, where A is an m ˆ n matrix. Then
Example 1.138 shows that }L}B(Rn ,Rm ) is the square root of the largest eigenvalue of AT A
which is certainly a finite number. Therefore, any linear transformation from Rn to Rm is
bounded.
Proposition 6.3. Let (X, } ¨ }X ) and (Y, } ¨ }Y ) be normed spaces, and L P B(X, Y ). Then
ˇ
␣
(
}Lx}Y
= inf M ą 0 ˇ }Lx}Y ď M }x}X .
x‰0 }x}X
}L}B(X,Y ) = sup
In particular, the first equality implies that
}Lx}Y ď }L}B(X,Y ) }x}X
205
@x P X .
206
CHAPTER 6. Differentiation
Proposition 6.4. Let (X, } ¨ }X ) and (Y, } ¨ }Y ) be normed spaces, and L P L (X, Y ). Then
L is continuous on X if and only if L P B(X, Y ).
Proof. “ñ” Since L is continuous at 0 P X, there exists δ ą 0 such that
}Lx}Y = }Lx ´ L0}Y ă 1 if }x}X ă δ .
› ( δ )›
›δ ›
Then ›L x ›Y ď 1 if › x›X ă δ; thus by the properties of norm,
2
2
}Lx}Y ď
Therefore, sup }Lx}Y ď
}x}X =1
2
δ
if }x}X ă 2 .
2
which implies that L P B(X, Y ).
δ
“ð” If L P B(X, Y ), then M = }L}B(X,Y ) ă 8, and
}Lx1 ´ Lx2 }Y = }L(x1 ´ x2 )}Y ď M }x1 ´ x2 }X
which shows that L is uniformly continuous on X.
˝
(
)
Proposition 6.5. Let (X, }¨}X ) and (Y, }¨}Y ) be normed spaces. Then B(X, Y ), }¨}B(X,Y )
(
)
is a normed space. Moreover, if (Y, } ¨ }Y ) is a Banach space, so is B(X, Y ), } ¨ }B(X,Y ) .
(
)
Proof. That B(X, Y ), } ¨ }B(X,Y ) is a normed space is left as an exercise. Now suppose
(
)
that Y, } ¨ }Y is a Banach space. Let tLk u8
k=1 Ď B(X, Y ) be a Cauchy sequence. Then by
Proposition 6.3, for each x P X we have
}Lk x ´ Lℓ x}Y = }(Lk ´ Lℓ )x}Y ď }Lk ´ Lℓ }B(X,Y ) }x}X Ñ 0
as k, ℓ Ñ 8 .
Therefore, tLk xu8
k=1 is a Cauchy sequence in Y ; thus convergent. Suppose that lim Lk x = y.
kÑ8
We then establish a map x ÞÑ y which we denoted by L; that is, Lx = y. Then L is linear
since if x1 , x2 P X and c P R,
(
)
L(cx1 + x2 ) = lim Lk (cx1 + x2 ) = lim cLk x1 + Lk x2 = cLx1 + Lx2 .
kÑ8
kÑ8
Moreover, since tLk u8
k=1 is a Cauchy sequence, D M ą 0 such that }Lk }B(X,Y ) ď M for all
k P N. If ε ą 0 is given, for each x P X there exists N = Nx ą 0 such that
}Lk x ´ Lx}Y ă ε
@ k ě Nx .
§6.1 Bounded Linear Maps
207
Therefore, for k ě Nx ,
}Lx}Y ă }Lk x}Y + ε ď }Lk }B(X,Y ) }x}X + ε ď M }x}X + ε
which implies that sup }Lx}Y ď M + ε; thus L P B(X, Y ).
}x}X =1
Finally, we show that lim }Lk ´ L}B(X,Y ) = 0. Let x P X and ε ą 0 be given. Since
kÑ8
ε
8
tLk uk=1 is a Cauchy sequence, there exists N ą 0 such that }Lk ´ Lℓ }B(X,Y ) ă if k, ℓ ě N .
2
Then if k ě N ,
ε
}Lk x ´ Lx}Y = lim }Lk x ´ Lℓ x}Y ď lim sup }Lk ´ Lℓ }B(X,Y ) }x}X ď }x}X
ℓÑ8
2
ℓÑ8
which shows that }Lk ´ L}B(X,Y ) ă ε if k ě N .
˝
Proposition 6.6. Let (X, } ¨ }X ), (Y, } ¨ }Y ), (Z, } ¨ }Z ) be normed spaces, and L P B(X, Y ),
K P B(Y, Z). Then K ˝ L P B(X, Z), and
}K ˝ L}B(X,Z) ď }K}B(Y,Z) }L}B(X,Y ) .
We often write K ˝ L as KL if K and L are linear.
Proof. By the properties of the norm of a bounded linear map,
}K ˝ L(x)}Z = }K(Lx)}Z ď }K}B(Y,Z) }Lx}Y ď }K}B(Y,Z) }L}B(X,Y ) }x}X .
˝
From now on, when the domain X and the target Y of a linear map L is clear, we use
}L} instead of }L}B(X,Y ) to simplify the notation.
Theorem 6.7. Let (X, } ¨ }X ) and (Y, } ¨ }Y ) be normed spaces, and X be finite dimensional.
Then every linear map from X to Y is bounded; that is, L (X, Y ) = B(X, Y ).
Proof. Suppose that dim(X) = n. Let tek unk=1 Ď X be a linearly independent set of vectors.
From Example 4.24, every two norms on X are equivalent; thus we only focus on the norm
} ¨ }2 on X induced by the inner product
(
)
ei , ej X = δij
@ i = 1, ¨ ¨ ¨ n.
Since tek unk=1 is a linear independent set of vectors, every x P X can be expressed as a
unique linear combination of ek ’s; that is, for all x P X, D c1 = c1 (x), ¨ ¨ ¨ , cn = cn (x) P R
such that
x = c1 e1 + ¨ ¨ ¨ + cn en .
208
CHAPTER 6. Differentiation
These coefficients ck ’s in fact are determined by ck = (x, ek )X , and, by Example 4.24 and
the Cauchy-Schwarz inequality, satisfy
ˇ
ˇ
ˇck (x)ˇ ď }x}2 }ek }2 ď C}x}X .
As a consequence, if L is a linear map from X to Y , then
›
›
}Lx}Y = ›L(c1 (x)e1 + ¨ ¨ ¨ cn (x)en )›Y ď |c1 (x)|}Le1 }Y + ¨ ¨ ¨ |cn (x)|}Len }Y
␣
(
ď nC}x}X max }Le1 }Y , ¨ ¨ ¨ }Len }Y ď M }x}X
for some constant M ą 0; thus }L}B(X,Y ) ď M ă 8 which shows that L P B(X, Y ).
˝
Theorem 6.8. Let GL(n) be the set of all invertible linear maps on Rn ; that is,
ˇ
␣
(
GL(n) = L P L (Rn , Rn ) ˇ L is one-to-one (and onto) .
1. If L P GL(n) and K P B(Rn , Rn ) satisfying }K ´ L}}L´1 } ă 1 , then K P GL(n).
2. GL(n) is an open set of B(Rn , Rn ).
3. The mapping L ÞÑ L´1 is continuous on GL(n).
Proof. 1. Let }L´1 } =
1
and }K ´ L} = β. Then β ă α; thus for every x P Rn ,
α
α}x}Rn = α}L´1 Lx}Rn ď α}L´1 }}Lx}Rn = }Lx}Rn ď }(L ´ K)x}Rn + }Kx}Rn
ď β}x}Rn + }Kx}Rn .
As a consequence, (α ´ β)}x}Rn ď }Kx}Rn and this implies that K : Rn Ñ Rn is
one-to-one hence invertible.
(
1
1 )
2. By 1, we find that if }K ´ L} ă ´1 , then K P GL(n). Then D L, ´1 Ď GL(n)
}L
}L
}
}
if L P GL(n). Therefore, GL(n) is open.
3. Let L P GL(n) and ε ą 0 be given. Choose δ = min
!
1
ε
. If }K´L} ă δ,
,
´1
2}L } 2}L´1 }2
)
then K P GL(n). Since L´1 ´ K ´1 = K ´1 (K ´ L)L´1 , we find that if }K ´ L} ă δ,
1
}K ´1 } ´ }L´1 } ď }K ´1 ´ L´1 } ď }K ´1 }}K ´ L}}L´1 } ă }K ´1 }
2
which implies that }K ´1 } ă 2}L´1 }. Therefore, if }K ´ L} ă δ,
}L´1 ´ K ´1 } ď }K ´1 }}K ´ L}}L´1 } ă 2}L´1 }2 δ ă ε .
˝
§6.2 Definition of Derivatives and the Jacobian Matrices
209
Remark 6.9. There is another way to see that GL(n) is open in B(Rn , Rn ). Let M(n) be
the collection of n ˆ n real matrices, and } ¨ }2 be the matrix norm introduced in Example
1.138. Also define } ¨ } : M(n) Ñ R by
ˇ
(
␣
}A} = max |aij | ˇ A = [aij ]1 ď i, j ď n .
Then }¨} is also a norm on M(n). Since M(n) is finite dimensional (in fact, dim M(n) = n2 ),
by Example 4.24 } ¨ } and } ¨ }2 are equivalent norms on M(n); that is, there exists C, c ą 0
such that
@ A P M(n) .
c}A} ď }A}2 ď C}A}
Let tAk u8
k=1 Ď M(n) be a sequence of n ˆ n real matrices. The equivalence between } ¨ } and
}¨}2 implies that Ak Ñ A in M(n) if and only if each entry of Ak converges to corresponding
entry of A. Therefore, the determinant function is continuous on M(n). In other words,
@ A P M(n) .
lim det(Ak ) = det(A)
Ak ÑA
Since GL(n) can be viewed as the collection of n ˆ n matrices with non-zero determinant;
that is,
ˇ
␣
(
GL(n) = A P M(n) ˇ det(A) ‰ 0 ,
by the continuity of the determinant function and Theorem 4.11, we conclude that GL(n)
is open in M(n).
6.1.1 The matrix representation of linear maps between finite dimensional normed spaces
Let (X, } ¨ }X ) and (Y, } ¨ }Y ) be two finite dimensional normed spaces. Suppose that B =
tek un and Br = trek um are basis of X and Y , respectively. Then every x P X, and y P Y ,
k=1
k=1
there exists unique vectors (c1 , ¨ ¨ ¨ , cn ) P Rn and (d1 , ¨ ¨ ¨ , dm ) P Rm such that
and y = d1re1 + ¨ ¨ ¨ + dmrem .
x = c1 e1 + ¨ ¨ ¨ + cn en
We write [x]B for the column vector [c1 , ¨ ¨ ¨ , cn ]T and [y]Br for the column vector [d1 , ¨ ¨ ¨ , dm ]T .
Then for each L P L (X, Y ), the matrix
representation of L ]with respect to basis B and
[
r denoted by [L] r, is the matrix [Le1 ] r ... [Le2 ] r ... ¨ ¨ ¨ ... [Len ] r . The matrix [L] r has the
B,
B,B
B
B
property that
[Lx]Br = [L]B,Br[x]B .
B
B,B
210
CHAPTER 6. Differentiation
6.2 Definition of Derivatives and the Jacobian Matrices
Definition 6.10. Let (X, }¨}X ) and (Y, }¨}Y ) be two normed spaces. A map f : A Ď X Ñ Y
is said to be differentiable at x0 P A if there is a bounded linear map in B(X, Y ), denoted
by (Df )(x0 ) and called the derivative of f at x0 , such that
›
›
›f (x) ´ f (x0 ) ´ (Df )(x0 )(x ´ x0 )›
Y
lim
= 0,
xÑx0
}x ´ x0 }X
xPA
where (Df )(x0 )(x ´ x0 ) denotes the value of the linear map (Df )(x0 ) applied to the vector
x ´ x0 P X (so (Df )(x0 )(x ´ x0 ) P Y ). In other words, f is differentiable at x0 P A if there
exists L P B(X, Y ) such that
@ ε ą 0, D δ ą 0 Q }f (x) ´ f (x0 ) ´ L(x ´ x0 )}Y ď ε}x ´ x0 }X whenever x P D(x0 , δ) X A .
If f is differentiable at each point of A, we say that f is differentiable on A.
Remark 6.11. Suppose that f : A Ñ Y is differentiable on A, then Df itself is a map from
A to B(X, Y ). For each x P A, Df (x) is a linear map, but Df in general is not linear in x.
Example 6.12. Let f : (0, 8) Ñ R be given by f (x) =
1
. Then f is differentiable at any
x
x0 P (0, 8) since (Df )(x0 ) : R Ñ R is the linear map given by
(Df )(x0 )(x) = ´
1
¨ x.
x0 2
To see this, we observe that
lim
ˇ1
ˇ
ˇ ´ 1 ´ ´1 (x ´ x0 )ˇ
x
xÑx0
x0
x0 2
|x ´ x0 |
= lim
xÑx0
ˇ x0 2 ´ xx0 + x2 ´ xx0 ˇ
ˇ
ˇ
xx0 2
|x ´ x0 |
x0 2 ´ 2xx0 + x2
= lim
xÑx0
xx0 2 |x ´ x0 |
|x ´ x0 |
= 0.
= lim
xÑx0 xx0 2
Remark 6.13. Let f : (a, b) Ñ R be “differentiable” at x0 P (a, b) in the sense of Definition
4.55. The “derivative” f 1 (x0 ) and the derivative (Df )(x0 ) is related by (Df )(x0 )(h) =
f 1 (x0 )h since
lim
xÑx0
ˇ
ˇ
ˇf (x) ´ f (x0 ) ´ f 1 (x0 )(x ´ x0 )ˇ
|x ´ x0 |
= 0.
§6.2 Definition of Derivatives and the Jacobian Matrices
211
Example 6.14. View (C, | ¨ |) as a normed vector space (over field C). Then f : C Ñ R
given by f (z) = |z|2 is differentiable at z0 = 0 and (Df )(0) = 0 since
ˇ 2
ˇ
ˇ|z| ´ |0|2 ´ 0 ¨ (z ´ 0)ˇ
lim
= lim |z| = 0 .
zÑ0
zÑ0
|z|
However, f is not differentiable at any z0 ‰ 0. In fact, in Exercise Problem 6.3 one is asked
to show that if f : C Ñ R is differentiable at z0 , then (Df )(z0 ) = 0. Therefore, if f is
differentiable at z0 ‰ 0, then
ˇ 2
ˇ
ˇ|z| ´ |z0 |2 ´ 0 ¨ (z ´ z0 )ˇ
|z ´ z0 |
ˇ
ˇ ˇ
ˇ
ˇ z ¨ zs ´ z0 ¨ zs0 ˇ ˇ (z ´ z0 ) ¨ zs0 + z ¨ (z ´ z0 ) ˇ
=ˇ
ˇ=ˇ
ˇ
z ´ z0
z ´ z0
ˇ
ˇ ˇ
ˇ
z ¨ (z ´ z0 ) ˇ
z ¨ (z ´ z0 ) ˇ
ˇ
ˇ
= ˇzs0 +
ˇ = ˇzs0 + z ´ z0 + 0
ˇ
z ´ z0
z ´ z0
and the limit of the right-hand side as z approaches z0 does not exist since lim
zÑz0
z0 ¨ (z ´ z0 )
z ´ z0
does not exist (by the fact that the limit as z approaches z0 from the horizontal and vertical
directions are different).
On the other hand, the function g : R2 Ñ R given by g(x, y) ” f (x + iy) = x2 + y 2 is
differentiable everywhere and (Dg)(a, b)v = 2av1 + 2bv2 for all v = (v1 , v2 ) P R2 . To see
this,
(a + h)2 + (b + k)2 ´ (a2 + b2 ) ´ (2ah + 2bk)
h2 + k 2
?
?
=
Ñ 0 as (h, k) Ñ (0, 0)
h2 + k 2
h2 + k 2
›
› 2
›(x + y 2 ) ´ (a2 + b2 ) ´ (Dg)(a, b)(x ´ a, y ´ b)› 2
which implies that
R
lim
}(x ´ a, y ´ b)}R2
(x,y)Ñ(a,b)
= 0.
Example 6.15. Let (X, } ¨ }X ) and (Y, } ¨ }Y ) be two normed spaces. Then every bounded
linear map L : X Ñ Y is differentiable. In fact, (DL)(x0 ) = L for all x0 P X since
}Lx ´ Lx0 ´ L(x ´ x0 )}Y
= 0.
xÑx0
}x ´ x0 }X
lim
Example 6.16. Let f : GL(n) Ñ GL(n) be given by f (L) = L´1 , where GL(n) is defined in
Theorem 6.8. Then f is differentiable at any “point” L P GL(n) with derivative (Df )(K) P
B(GL(n), GL(n)) given by (Df )(L)(K) = ´L´1 KL´1 for all K P GL(n). The proof is left
as an exercise.
Theorem 6.17. Let (X, } ¨ }X ), (Y, } ¨ }Y ) be normed vector spaces, U Ď X be an open set,
and f : U Ñ Y be differentiable at x0 P U. Then (Df )(x0 ) is uniquely determined by f .
212
CHAPTER 6. Differentiation
Proof. Suppose L1 , L2 P B(X, Y ) are derivatives of f at x0 . Let ε ą 0 be given and e P X be
a unit vector; that is, }e}X = 1. Since U is open, there exists r ą 0 such that D(x0 , r) Ď U.
By Definition 6.10, there exists 0 ă δ ă r such that
ε
}f (x) ´ f (x0 ) ´ L1 (x ´ x0 )}Y
ă
}x ´ x0 }X
2
and
}f (x) ´ f (x0 ) ´ L2 (x ´ x0 )}Y
ε
ă
}x ´ x0 }X
2
if 0 ă }x ´ x0 }X ă δ. Letting x = x0 + λe with 0 ă |λ| ă δ, we have
1
}L1 (x ´ x0 ) ´ L2 (x ´ x0 )}Y
|λ|
›
›
› )
1 (››
ď
f (x) ´ f (x0 ) ´ L1 (x ´ x0 )›Y + ›f (x) ´ f (x0 ) ´ L2 (x ´ x2 )›Y
|λ|
›
›
›
›
›f (x) ´ f (x0 ) ´ L1 (x ´ x0 )›
›f (x) ´ f (x0 ) ´ L2 (x ´ x0 )›
Y
Y
=
+
}x ´ x0 }X
}x ´ x0 }X
ε ε
ă + = ε.
2 2
}L1 e ´ L2 e}Y =
Since ε ą 0 is arbitrary, we conclude that L1 e = L2 e for all unit vectors e P X which
( x )
( x )
guarantees that L1 = L2 (since if x ‰ 0, L1 x = }x}X L1
= }x}X L2
= L2 x). ˝
}x}X
}x}X
Example 6.18. (Df )(x0 ) may not be unique if the domain of f is not open. For example,
ˇ
␣
(
let A = (x, y) ˇ 0 ď x ď 1, y = 0 be a subset of R2 , and f : A Ñ R be given by f (x, y) = 0.
Fix x0 = (a, 0) P A, then both of the linear maps
and L2 (x, y) = ay
L1 (x, y) = 0
@ (x, y) P R2
satisfy Definition 6.10 since
ˇ
ˇ
ˇ
ˇ
ˇf (x, 0) ´ f (a, 0) ´ L2 (x ´ a, 0)ˇ
ˇf (x, 0) ´ f (a, 0) ´ L1 (x ´ a, 0)ˇ
›
›
›
›
lim
= lim
= 0.
›(x, 0) ´ (a, 0)› 2
›(x, 0) ´ (a, 0)› 2
(x,0)Ñ(a,0)
(x,0)Ñ(a,0)
R
R
Remark 6.19. Let U Ď R be an open set and suppose that f : U Ñ R is differentiable
on U. Then Df : U Ñ B(Rn , Rm ). Treating Df as a map from U to the normed space
(
)
B(Rn , Rm ), } ¨ }B(Rn ,Rm ) , and suppose that Df is also differentiable on U. Then the
n
m
derivative of Df , denoted by D2 f , is a map from U to B(Rn , B(Rn , Rm )). In other words,
for each a P U, (D2 f )(a) P B(Rn , B(Rn , Rm )) satisfying
›
›
›(Df )(x) ´ (Df )(a) ´ (D2 f )(a)(x ´ a)› n m
B(R ,R )
lim
= 0,
xÑa
}x ´ a}Rn
here (D2 f )(a) is bounded linear map from Rn to B(Rn , Rm ); thus (D2 f )(a)(x ´ a) P
B(Rn , Rm ).
§6.2 Definition of Derivatives and the Jacobian Matrices
213
Definition 6.20. Let tek unk=1 be the standard basis of Rn , U Ď Rn be an open set, a P U
and f : U Ñ R be a function. The partial derivative of f at a in the direction ej , denoted
by
Bf
(a), is the limit
Bxj
lim
hÑ0
f (a + hej ) ´ f (a)
h
if it exists. In other words, if a = (a1 , ¨ ¨ ¨ , an ), then
f (a1 , ¨ ¨ ¨ , aj´1 , aj + h, aj+1 , ¨ ¨ ¨ , an ) ´ f (a1 , ¨ ¨ ¨ , an )
Bf
(a) = lim
.
hÑ0
Bxj
h
Theorem 6.21. Suppose U Ď Rn is an open set and f : U Ñ Rm is differentiable at a P U.
Then the partial derivatives
Bfi
(a) exists for all i = 1, ¨ ¨ ¨ m and j = 1, ¨ ¨ ¨ n, and the matrix
Bxj
representation of the linear map Df (a) with respect to the standard basis of Rn and Rm is
given by


Bf1
Bf1
(a) ¨ ¨ ¨
(a)
 Bx1

Bxn


[
]
[
] 
Bfi

.
.
.
.
.
.
(a) .
(Df )(a) = 
 or (Df )(a) ij =
.
.
.


Bxj
 Bf

Bfm
m
(a) ¨ ¨ ¨
(a)
Bx1
Bxn
Proof. Since U is open and a P U, there exists r ą 0 such that D(a, r) Ď U. By the
differentiability of f at a, there is L P B(Rn , Rm ) such that for any given ε ą 0, there exists
0 ă δ ă r such that
}f (x) ´ f (a) ´ L(x ´ a)}Rm ď ε}x ´ a}Rn whenever x P D(a, δ) .
In particular, for each i = 1, ¨ ¨ ¨ , m,
ˇ f (a + he ) ´ f (a)
ˇ › f (a + he ) ´ f (a)
›
ˇ i
ˇ ›
›
j
i
j
´ (Lej )i ˇ ď ›
´ Lej › ď ε
ˇ
h
h
Rm
@ 0 ă |h| ă δ, h P R ,
where (Lej )i denotes the i-th component of Lej in the standard basis. As a consequence,
for each i = 1, ¨ ¨ ¨ , m,
fi (a + hej ) ´ fi (a)
= (Lej )i exists
hÑ0
h
lim
and by definition, we must have (Lej )i =
Bf
Bfi
(a). Therefore, Lij = i (a).
Bxj
Bxj
˝
214
CHAPTER 6. Differentiation
Definition 6.22. Let U Ď Rn be an open set, and f : U Ñ Rm . The matrix




Bf1
Bf1
Bf1
Bf1
¨¨¨
(x) ¨ ¨ ¨
(x)
 Bx1
 Bx1

Bxn 
Bxn






 ..

.
.
.
.
.
.
.
.
.
.
(Jf )(x) ”  .
 (x) ” 

.
.
.
.
.




 Bf



Bfm
Bfm
Bfm
m
¨¨¨
(x) ¨ ¨ ¨
(x)
Bx1
Bxn
Bx1
Bxn
is called the Jacobian matrix of f at x (if each entry exists). If n = m, the determinant
of (Jf )(x) is called the Jacobian of f at x.
Remark 6.23. A function f might not be differential even if the Jacobian matrix Jf exists;
however, if f is differentiable at x0 , then (Df )(x) can be represented by (Jf )(x); that is,
[
]
(Df )(x) = (Jf )(x).
Example 6.24. Let f : R2 Ñ R3 be given by f (x1 , x2 ) = (x21 , x31 x2 , x41 x22 ). Suppose that f
is differentiable at x = (x1 , x2 ), then


2x1
0
[
]
x31  .
(Df )(x) =  3x21 x2
3 2
4x1 x2 2x41 x2
Example 6.25. Let f : R2 Ñ R be given by
# xy
f (x, y) =
x2 + y 2
if (x, y) = (0, 0) .
0
Then
if (x, y) ‰ (0, 0) ,
[
] [
]
Bf
Bf
(0, 0) =
(0, 0) = 0; thus if f is differentiable at (0, 0), then (Df )(0, 0) = 0 0 .
Bx
By
However,
[ ]
ˇ
a
[
] x ˇˇ
|xy|
|xy|
ˇ
=
x2 + y 2 ;
ˇf (x, y) ´ f (0, 0) ´ 0 0
ˇ= 2
3
y
x + y2
(x2 + y 2 ) 2
thus f is not differentiable at (0, 0) since
|xy|
3
(x2 + y 2 ) 2
cannot be arbitrarily small even if x2 +y 2
is small.
Example 6.26. Let f : R2 Ñ R be given by
$
& x if y = 0 ,
y if x = 0 ,
f (x, y) =
%
1 otherwise .
§6.3 Conditions for Differentiability
Then
215
Bf
f (h, 0) ´ f (0, 0)
h
Bf
(0, 0) = lim
= lim = 1. Similarly,
(0, 0) = 1; thus if f is
Bx
h
By
hÑ0
hÑ0 h
[
] [
]
differentiable at (0, 0), then (Df )(0, 0) = 1 1 . However,
[ ]
ˇ
ˇ
[
] x ˇˇ ˇ
ˇ
ˇf (x, y) ´ f (0, 0) ´ 1 1
ˇ = ˇf (x, y) ´ (x + y)ˇ ;
y
thus if xy ‰ 0,
ˇ
ˇ
ˇf (x, y) ´ (x + y)ˇ = |1 ´ x ´ y| Ñ̂ 0 as (x, y) Ñ (0, 0), xy ‰ 0.
Therefore, f is not differentiable at (0, 0).
6.3 Conditions for Differentiability
Proposition 6.27. Let U Ď Rn be open, a P U, and f = (f1 , ¨ ¨ ¨ , fm ) : U Ñ Rm . Then
f is differentiable at a if and only if fi is differentiable at a for all i = 1, ¨ ¨ ¨ , m. In other
words, for vector-valued functions defined on an open subset of Rn ,
Componentwise differentiable ô Differentiable.
Proof. “ñ” Let (Df )(a) be the Jacobian matrix of f at a. Then
›
›
@ ε ą 0, D δ ą 0 Q ›f (x) ´ f (a) ´ (Df )(a)(x ´ a)›Rm ď ε}x ´ a}Rn if }x ´ a}Rn ă δ .
m
n
Let tej um
j=1 be the standard basis of R , and Li P L (R , R) be given by Li (h) =
n
eT
i [(Df )(a)]h. Then Li P B(R , R) by Theorem 6.7, and if }x ´ a}Rn ă δ,
ˇ
ˇ ˇ (
)ˇ
ˇfi (x) ´ fi (a) ´ Li (x ´ a)ˇ = ˇei ¨ f (x) ´ f (a) ´ (Df )(a)(x ´ a) ˇ
›
›
ď ›f (x) ´ f (a) ´ (Df )(a)(x ´ a)›Rm ď ε}x ´ a}Rn ;
thus fi is differentiable at a with derivatives Li .
“ð” Suppose that fi : U Ñ R is differentiable at a for each i = 1, ¨ ¨ ¨ , m. Then there exists
Li P B(Rn , R) such that
ˇ
ˇ
ε
@ ε ą 0, D δi ą 0 Q ˇfi (x) ´ fi (a) ´ Li (x ´ a)ˇ ď }x ´ a}Rn if }x ´ a}Rn ă δi .
m
n
m
Let L P L (R , R ) be given by Lx = (L1 x, L2 x, ¨ ¨ ¨ , Lm x) P Rm if x P Rn . Then
L P B(Rn , Rm ) by Theorem 6.7, and
m
ÿ
›
›
ˇ
ˇ
›f (x) ´ f (a) ´ L(x ´ a)› m ď
ˇfi (x) ´ fi (a) ´ Li (x ´ a)ˇ ď ε}x ´ a}Rn
R
i=1
␣
(
if }x ´ a}Rn ă δ = min δ1 , ¨ ¨ ¨ , δm .
˝
216
CHAPTER 6. Differentiation
Theorem 6.28. Let U Ď Rn be open, a P U, and f : U Ñ R. If
1. the Jacobian matrix of f exists in a neighborhood of a, and
2. at least (n ´ 1) entries of the Jacobian matrix of f are continuous at a,
then f is differentiable at a.
Bf
Bf Bf
,
, ¨¨¨,
are continuous at a. Let tej unj=1
Bx1 Bx2
Bxn´1
Bf
is continuous at a for i =
be the standard basis of Rn , and ε ą 0 be given. Since
Bxi
Proof. W.L.O.G. we can assume that
1, ¨ ¨ ¨ , n ´ 1,
ˇ
ˇ
Bf
ε
ˇ Bf
ˇ
D δi ą 0 Q ˇ (x) ´
(a)ˇ ă ? whenever }x ´ a}Rn ă δi .
Bxi
n
Bxi
On the other hand, by the definition of the partial derivatives,
ˇ
ˇ
Bf
ε
ˇ f (a + hen ) ´ f (a)
ˇ
D δn ą 0 Q ˇ
´
(a)ˇ ă ? whenever 0 ă |h| ă δn .
h
␣
Bxn
n
(
Let k = x ´ a and δ = min δ1 , ¨ ¨ ¨ , δn . Then
ˇ
[ Bf
]ˇ
Bf
ˇ
ˇ
(a)(x1 ´ a1 ) + ¨ ¨ ¨ +
(a)(xn ´ an ) ˇ
ˇf (x) ´ f (a) ´
Bx1
Bxn
ˇ
ˇ
Bf
Bf
ˇ
ˇ
= ˇf (a + k) ´ f (a) ´
(a)k1 ´ ¨ ¨ ¨ ´
(a)kn ˇ
Bx1
Bxn
ˇ
ˇ
Bf
Bf
ˇ
ˇ
= ˇf (a1 + k1 , ¨ ¨ ¨ , an + kn ) ´ f (a1 , ¨ ¨ ¨ , an ) ´
(a)k1 ´ ¨ ¨ ¨ ´
(a)kn ˇ
Bx1
Bxn
ˇ
ˇ
Bf
ˇ
ˇ
ď ˇf (a1 + k1 , ¨ ¨ ¨ , an + kn ) ´ f (a1 , a2 + k2 , ¨ ¨ ¨ , an + kn ) ´
(a)k1 ˇ
Bx1
ˇ
ˇ
Bf
ˇ
ˇ
(a)k2 ˇ
+ ˇf (a1 , a2 + k2 , ¨ ¨ ¨ , an + kn ) ´ f (a1 , a2 , a3 + k3 , ¨ ¨ ¨ , an + kn ) ´
Bx2
ˇ
ˇ
Bf
ˇ
ˇ
(a)kn ˇ .
+ ¨ ¨ ¨ + ˇf (a1 , ¨ ¨ ¨ , an´1 , an + kn ) ´ f (a1 , ¨ ¨ ¨ , an ) ´
Bxn
By the mean value theorem,
f (a1 , ¨ ¨ ¨ , aj´1 , aj + kj , ¨ ¨ ¨ , an + kn ) ´ f (a1 , ¨ ¨ ¨ , aj , aj+1 + kj+1 , ¨ ¨ ¨ , an + kn )
= kj
Bf
(a1 , ¨ ¨ ¨ , aj´1 , aj + θj kj , aj+1 + kj+1 , ¨ ¨ ¨ , an + kn )
Bxj
for some 0 ă θj ă 1; thus for j = 1, ¨ ¨ ¨ , n ´ 1, if }x ´ a}Rn = }k}Rn ă δ,
ˇ
ˇ
Bf
ˇ
ˇ
(a)kj ˇ
ˇf (a1 , ¨ ¨ ¨ , aj´1 , aj + kj , ¨ ¨ ¨ , an + kn ) ´ f (a1 , ¨ ¨ ¨ , aj , aj+1 + kj+1 , ¨ ¨ ¨ , an + kn ) ´
Bxj
ˇ
ˇ
ε
Bf
ˇ Bf
ˇ
=ˇ
(a1 , ¨ ¨ ¨ , aj´1 , aj + θj kj , aj+1 + kj+1 , ¨ ¨ ¨ , an + kn ) ´
(a)ˇ|kj | ď ? |kj | .
Bxj
Bxj
n
§6.3 Conditions for Differentiability
217
Moreover, if }x ´ a}Rn ă δ, then |kn | ď }k}Rn = }x ´ a}Rn ă δ ď δn ; thus
ˇ
ˇ
Bf
ε
ˇ
ˇ
(a)kn ˇ ď ? |kn | .
ˇf (a1 , ¨ ¨ ¨ , an´1 , an + kn ) ´ f (a1 , ¨ ¨ ¨ , an ) ´
n
Bxn
As a consequence, if }x ´ a}Rn ă δ, by Cauchy’s inequality,
ˇ
[ Bf
]ˇ
Bf
ˇ
ˇ
(a)(x1 ´ a1 ) + ¨ ¨ ¨ +
(a)(xn ´ an ) ˇ
ˇf (x) ´ f (a) ´
Bx1
n
ε ÿ
ď?
n
Bxn
|kj | ď ε}k}Rn = ε}x ´ a}Rn
j=1
which implies that f is differentiable at a.
Remark 6.29. When two or more components of the Jacobian matrix
[ Bf
Bx1
¨¨¨
]
Bf
Bxn
˝
of a
scalar function f are discontinuous at a point x0 P U, in general f is not differentiable at x0 .
For example, both components of the Jacobian matrix of the functions given in Example
6.25, 6.26, 6.41 are discontinuous at (0, 0), and these functions are not differentiable at
(0, 0).
ˇ
␣
(
Example 6.30. Let U = R2 z (x, 0) P R2 ˇ x ě 0 , and f : U Ñ R be given by
$
x
cos´1 a
if y ą 0 ,
’
’
2
’
x + y2
’
&
π
if y = 0 ,
f (x, y) = arg(x + iy) =
’
’
x
’
’
if y ă 0 .
% 2π ´ cos´1 a
x2 + y 2
Then
$
& ´
Bf
(x, y) =
%
Bx
Since
y
2
x + y2
if y ‰ 0 ,
0
if y = 0 ,
and
$
x
’
& x2 + y 2 if y ‰ 0 ,
Bf
(x, y) =
’
By
%
1
x
if y = 0 .
Bf
Bf
and
are both continuous on U, f is differentiable on U.
Bx
By
Definition 6.31. Let U Ď Rn be open, and f : U Ñ Rm be differentiable on U. f is
said to be continuously differentiable on U if Df : U Ñ B(Rn , Rm ) is continuous on
U. The collection of all continuously differentiable mappings from U to Rm is denoted by
C 1 (U; Rm ). The collection of all bounded differentiable functions from U to Rm whose
derivative is continuous and bounded is denoted by Cb1 (U; Rm ). In other words,
ˇ
␣
(
C 1 (U; Rm ) = f : U Ñ Rm is differentiable on U ˇ Df : U Ñ B(Rn , Rm ) is continuous
218
CHAPTER 6. Differentiation
and
Cb1 (U; Rm ) =
ˇ
!
)
1
m ˇ
f P C (U; R ) ˇ sup |f (x)| + sup }Df (x)}B(Rn ,Rm ) ă 8 .
xPU
xPU
Corollary 6.32. Let U Ď Rn be open, and f : U Ñ Rm . Then f P C 1 (U; Rm ) if and only if
the partial derivatives
Bfi
exist and are continuous on U for i = 1, ¨ ¨ ¨ , m and j = 1, ¨ ¨ ¨ , n.
Bxj
Proof. Note that for any matrix A = [aij ]mˆn , }A}B(Rn ,Rm ) ď
ř
|aij | ď nm}A}; thus
i,j
m ÿ
n ˇ
ˇ
ÿ
Bfi
ˇ Bfi
ˇ
ď
(x) ´
(x0 )ˇ
ˇ
B(Rn ,Rm )
›
›
›(Df )(x) ´ (Df )(x0 )›
i=1 j=1
Bxj
Bxj
›
›
ď nm›(Df )(x) ´ (Df )(x0 )›B(Rn ,Rm ) .
Therefore, the continuity of Df is equivalent to the continuity of the partial derivatives
for all i, j. The corollary is then concluded by Proposition 6.27 and Theorem 6.28.
Bfi
Bxj
˝
Example 6.33. If f : R Ñ R is differentiable at x0 , must f 1 be continuous at x0 ? In other
words, is it always true that lim f 1 (x) = f 1 (x0 )?
xÑx0
Answer: No! For example, take
$
& x2 sin 1 if x ‰ 0,
x
f (x) =
%
0
if x = 0.
Then f is differentiable at x = 0 since the limit
2
1
h sin
f (0 + h) ´ f (0)
h = lim h sin 1 = 0
lim
= lim
hÑ0
hÑ0
hÑ0
h
h
h
exists. Therefore,
$
& 2x sin 1 ´ cos 1 if x ‰ 0,
1
x
x
f (x) =
%
0
if x = 0.
However, lim f 1 (x) does not exist.
xÑ0
Proposition 6.34. Let U Ď Rn be open. Given f P Cb1 (U; Rm ), define
m ÿ
n
[
ÿ
ˇ]
ˇ Bfi
ˇ
}f }Cb1 (U ;Rm ) = sup |f (x)| +
(x)ˇ .
xPU
(
)
Then Cb1 (U; Rm ), } ¨ }Cb1 (U ;Rm ) is a Banach space.
i=1 j=1
Bxj
§6.4 Properties of Differentiable Functions
219
Proof. Left as an exercise.
˝
Theorem 6.35. Let U Ď Rn be open, K Ď U be compact, and f : U Ñ R be of class C 1 .
Then for each ε ą 0, there exists δ ą 0 such that
ˇ
ˇ
ˇf (y) ´ f (x) ´ (Df )(x)(y ´ x)ˇ ď ε}y ´ x}Rn if }y ´ x}Rn ă δ and x, y P K .
Proof. Define g : U ˆ U Ñ R by
ˇ
$ ˇˇ
& f (y) ´ f (x) ´ (Df )(x)(y ´ x)ˇ if y ‰ x ,
}y ´ x}Rn
g(x, y) =
%
0
if y = x .
Since f is of class C 1 , g is continuous on U ˆ U. In fact, it is clear that g is continuous at
(x, y) if x ‰ y, while the mean value theorem implies that f (w) ´ f (z) = (Df )(ξ)(w ´ z)
for some ξ on the line segment joining w and z; thus
ˇ
ˇ
ˇf (w) ´ f (z) ´ (Df )(z)(w ´ z)ˇ
lim sup
}w ´ z}Rn
(z,w)Ñ(x,x)
z‰w
ˇ(
ˇ
)
ˇ (Df )(ξ) ´ (Df )(z) (w ´ z)ˇ
›
›
= lim sup
ď lim sup ›(Df )(ξ) ´ (Df )(z)›B(Rn ,R) = 0 .
}w ´ z}Rn
(z,w)Ñ(x,x)
(z,w)Ñ(x,x)
z‰w
z‰w
Now by the compactness of K ˆ K, for each given ε ą 0, there exists δ ą 0 such that
|g(z, w) ´ g(x, y)| ă ε
if }(z, w) ´ (x, y)}R2n ă δ and x, y, z, w P K .
In particular, with (z, w) = (x, x) we find that |g(x, y)| ă ε if }x ´ y}Rn ă δ; thus
ˇ
ˇ
ˇf (y) ´ f (x) ´ (Df )(x)(y ´ x)ˇ
ăε
if 0 ă }x ´ y}Rn ă δ , x, y P K .
}y ´ x}Rn
˝
Corollary 6.36. Let U Ď Rn be open, K Ď U be compact, and f : U Ñ Rm be of class C 1 .
Then for each ε ą 0, there exists δ ą 0 such that
›
›
›f (y) ´ f (x) ´ (Df )(x)(y ´ x)› m ď ε}y ´ x}Rn
R
if }y ´ x}Rn ă δ and x, y P K .
6.4 Properties of Differentiable Functions
6.4.1 Continuity of differentiable maps
Theorem 6.37. Let (X, } ¨ }X ) and (Y, } ¨ }Y ) be normed spaces, U Ď X be open, and
f : U Ñ Y be differentiable at x0 P U. Then f is continuous at x0 .
220
CHAPTER 6. Differentiation
Proof. Since f is differentiable at x0 , there exists L P B(X, Y ) such that
›
›
D δ1 ą 0 Q ›f (x) ´ f (x0 ) ´ L(x ´ x0 )›Y ď }x ´ x0 }X @ x P D(x0 , δ1 ) .
As a consequence,
›
›
(
)
›f (x) ´ f (x0 )› ď }L} + 1 }x ´ x0 }X @ x P D(x0 , δ1 ) .
Y
!
)
ε
For a given ε ą 0, let δ = min δ1 ,
. Then δ ą 0, and if x P D(x0 , δ),
(6.4.1)
2(}L} + 1)
›
›
›f (x) ´ f (x0 )› ď ε ă ε .
˝
Y
2
Remark 6.38. In fact, if f is differentiable at x0 , then f satisfies the “local Lipschitz
property”; that is,
D M = M (x0 ) ą 0 and δ = δ(x0 ) ą 0 Q if }x´x0 }X ă δ, then }f (x)´f (x0 )}Y ď M }x´x0 }X
since we can choose M = }L} + 1 and δ = δ1 (see (6.4.1)).
Example 6.39. Let f : R2 Ñ R be given in Example 6.25. We have shown that f is not
differentiable at (0, 0). In fact, f is not even continuous at (0, 0) since when approaching
the origin along the straight line x2 = mx1 ,
lim
(x1 ,mx1 )Ñ(0,0)
mx21
m2
=
‰ f (0, 0) if m ‰ 0 .
x1 Ñ0 (m2 + 1)x2
m2 + 1
1
f (x1 , mx1 ) = lim
Example 6.40. Let f : R2 Ñ R be given in Example 6.26. Then f is not continuous at
(0, 0); thus not differentiable at (0, 0).
Example 6.41. Let f : R2 Ñ R be given by
$
x3
&
if (x, y) ‰ (0, 0) ,
x2 + y 2
f (x, y) =
%
0
if (x, y) = (0, 0) .
Then fx (0, 0) = 1 and fy (0, 0) = 0. However,
[ ]
ˇ
[
] x ˇˇ
ˇ
ˇf (x, y) ´ f (0, 0) ´ 1 0
ˇ
y
|x|y 2
a
=
3 Ñ̂ 0 as (x, y) Ñ (0, 0).
x2 + y 2
(x2 + y 2 ) 2
Therefore, f is not differentiable at (0, 0). On the other hand, f is continuous at (0, 0) since
ˇ
ˇ ˇ
ˇ
ˇf (x, y) ´ f (0, 0)ˇ = ˇf (x, y)ˇ ď |x| Ñ 0 as (x, y) Ñ (0, 0).
§6.4 Properties of Differentiable Functions
221
6.4.2 The product rules
Proposition 6.42. Let (X, } ¨ }X ), (Y, } ¨ }Y ) be normed vector spaces, U Ď X be open, and
f : U Ñ Y and g : U Ñ F be differentiable at x0 P U, where F is the scalar field associated
with the vector space Y . Then gf : U Ñ Y is differentiable at x0 , and
D(gf )(x0 )(v) = g(x0 )(Df )(x0 )(v) + (Dg)(x0 )(v)f (x0 ) .
Moreover, if g(x0 ) ‰ 0, then
is given by
(6.4.2)
f
f
: U Ñ Y is also differentiable at x0 , and D( )(x0 ) : X Ñ Y
g
g
(
)
(f )
g(x0 ) (Df )(x0 )(v) ´ (Dg)(x0 )(v)f (x0 )
D
(x0 )(v) =
.
g
g 2 (x0 )
(6.4.3)
Proof. We only prove (6.4.2), and (6.4.3) is left as an exercise.
Define Av = g(x0 )(Df )(x0 )(v) + (Dg)(x0 )(v)f (x0 ). Then A P B(X, Y ). Moreover,
(
)
(gf )(x) ´ (gf )(x0 ) ´ A(x ´ x0 ) = g(x0 ) f (x) ´ f (x0 ) ´ (Df )(x0 )(x ´ x0 )
(
)
+ g(x) ´ g(x0 ) ´ (Dg)(x0 )(x ´ x0 ) f (x)
(
)(
)
+ (Dg)(x0 )(x ´ x0 ) f (x) ´ f (x0 ) .
Since (Dg)(x0 ) P B(X, F), }(Dg)(x0 )}B(X,F) ă 8; thus using the inequality
ˇ
ˇ ›
›
ˇ(Dg)(x0 )(x ´ x0 )ˇ ď ›(Dg)(x0 )›
B(X,F)
}x ´ x0 }X
and the continuity of f at x0 (due to Theorem 6.37), we find that
ˇ ˇˇ(Dg)(x )(x ´ x )ˇˇ ›
› ˇˇ
›
›
›
›
ˇ
0
0 ›
lim ˇ
f (x) ´ f (x0 )›Y ˇ ď lim ›(Dg)(x0 )›B(X,F) ›f (x) ´ f (x0 )›Y = 0 .
xÑx0
}x ´ x0 }X
xÑx0
As a consequence,
›
›
›(gf )(x) ´ (gf )(x0 ) ´ A(x ´ x0 )›
Y
lim
xÑx0
}x ´ x0 }X
›
›
›f (x) ´ f (x0 ) ´ (Df )(x0 )(x ´ x0 )›
ˇ
ˇ
Y
ď ˇg(x0 )ˇ lim
xÑx0
}x ´ x0 }X
]
[ ˇˇg(x) ´ g(x ) ´ (Dg)(x )(x ´ x )ˇˇ
0
0
0
}f (x)}Y
+ lim
xÑx0
}x ´ x0 }X
[ ˇˇ(Dg)(x )(x ´ x )ˇˇ ›
› ]
0
0 ›
+ lim
f (x) ´ f (x0 )›Y = 0
xÑx0
}x ´ x0 }X
which implies that gf is differentiable at x0 with derivative D(gf )(x0 ) given by (6.4.2).
˝
222
CHAPTER 6. Differentiation
6.4.3 The chain rule
Theorem 6.43. Let (X, } ¨ }X ), (Y, } ¨ }Y ), (Z, } ¨ }Z ) be normed vector spaces, U Ď X and
V Ď Y be open sets. Suppose that f : U Ñ Y is differentiable at x0 P U, f (U) Ď V, and
g : V Ñ Z is differentiable at f (x0 ). Then the map F = g ˝ f : U Ñ Z defined by
(
)
F (x) = g f (x)
@x P U
is differentiable at x0 , and
(
)(
)
(DF )(x0 )(h) = (Dg) f (x0 ) (Df )(x0 )(h)
@h P X .
In particular, if X = Rn , Y = Rm and Z = Rℓ , then
(
m
ÿ
) Bf
)
Bgi (
(DF )(x0 ) ij =
f (x0 ) k (x0 ) .
k=1
Byk
Bxj
Proof. To simplify the notation, let y0 = f (x0 ), A = (Df )(x0 ) P B(X, Y ), and B =
(Dg)(y0 ) P B(Y, Z). Let ε ą 0 be given. By the differentiability of f and g at x0 and y0 ,
there exists δ1 , δ2 ą 0 such that if }x ´ x0 }X ă δ1 and }y ´ y0 }Y ă δ2 , we have
␣
(
ε
}f (x) ´ f (x0 ) ´ A(x ´ x0 )}Y ď min 1,
}x ´ x0 }X ,
2(}B} + 1)
ε
}g(y) ´ g(y0 ) ´ B(y ´ y0 )}Z ď
}y ´ y0 }Y .
2(}A} + 1)
Define
u(h) = f (x0 + h) ´ f (x0 ) ´ Ah
@ }h}X ă δ1 ,
v(k) = g(y0 + k) ´ g(y0 ) ´ Bk
@ }k}Y ă δ2 .
Then if }h}X ă δ1 and }k}Y ă δ2 ,
}u(h)}Y ď }h}X ,
}u(h)}Y ď
ε
}h}X
2(}B} + 1)
and
}v(k)}Z ď
ε
}k}Y .
2(}A} + 1)
Let k = f (x0 + h) ´ f (x0 ) = Ah + u(h). Then lim k = 0; thus there exists δ3 ą 0 such that
hÑ0
}k}Y ă δ2
whenever }h}X ă δ3 .
Since
(
)
F (x0 + h) ´ F (x0 ) = g(y0 + k) ´ g(y0 ) = Bk + v(k) = B Ah + u(h) + v(k)
= BAh + Bu(h) + v(k) ,
§6.4 Properties of Differentiable Functions
223
we conclude that if }h}X ă δ = mintδ1 , δ3 u,
ε
}F (x0 + h) ´ F (x0 ) ´ BAh}Z ď }Bu(h)}Z + }v(k)}Z ď }B}}u(h)}Y +
}k}Y
2(}A} + 1)
(
) ε
ε
ε
ε
}A}}h}X + }u(h)}Y ď }h}X + }h}X = ε}h}X
ď }h}X +
2
2(}A} + 1)
2
2
which implies that F is differentiable at x0 and (DF )(x0 ) = BA.
˝
Example 6.44. Consider the polar coordinate x = r cos θ, y = r sin θ. Then every function
f : R2 Ñ R is associated with a function F : [0, 8) ˆ [0, 2π) Ñ R satisfying
F (r, θ) = f (r cos θ, r sin θ) .
Suppose that f is differentiable. Then F is differentiable, and the chain rule implies that


Bx Bx
]
[
][
] [ Bf Bf ]
[
cos θ ´r sin θ
Bf Bf
BF BF
 Br Bθ 
=
.

=
Bx By
Bx By
By By
Br Bθ
sin θ r cos θ
Br
Bθ
Therefore, we arrive at the following form of chain rule
Bx B
By B
B
=
+
Br
Br Bx Br By
and
B
Bx B
By B
=
+
Bθ
Bθ Bx Bθ By
which is commonly seen in Calculus textbook.
(
)
2
Example 6.45. Let f : R Ñ
R
and
F
:
R
Ñ
R
be
differentiable,
and
F
x,
f
(x)
= 0 and
(
)
Fx x, f (x)
BF
BF
BF
) , where Fx =
‰ 0. Then f 1 (x) = ´ (
and Fy =
.
By
Bx
By
Fy x, f (x)
(
)
Example 6.46. Let γ : (0, 1) Ñ Rn and f : Rn Ñ R be differentiable. Let F (t) = f γ(t) .
(
)
Then F 1 (t) = (Df ) γ(t) γ 1 (t).
Example 6.47. Let f (u, v, w) = u2 v + wv 2 and g(x, y) = (xy, sin x, ex ). Let h = f ˝ g :
R2 Ñ R. Find
Bh
.
Bx
Way I: Compute
Bh
directly: Since
Bx
h(x, y) = f (g(x, y)) = f (xy, sin x, ex ) = x2 y 2 sin x + ex sin2 x ,
we have
Bh
= 2xy 2 sin x + x2 y 2 cos x + ex sin2 x + 2ex cos x .
Bx
224
CHAPTER 6. Differentiation
Way II: Use the chain rule:
Bh
Bf Bg1 Bf Bg2
Bf Bg3
=
+
+
= 2uv ¨ y + (u2 + 2wv) ¨ cos x + v 2 ¨ ex
Bx
Bu Bx
Bv Bx
Bw Bx
= 2xy 2 sin x + (x2 y 2 + 2ex sin x) cos x + ex sin2 x.
Example 6.48. Let F (x, y) = f (x2 + y 2 ), f : R Ñ R, F : R2 Ñ R. Show that x
BF
BF
=y .
By
Bx
Proof: Let g(x, y) = x2 + y 2 , g : R2 Ñ R, then F (x, y) = (f ˝ g)(x, y). By the chain rule,
[
]
[
]
[
]
BF BF
Bg Bg
1
= f (g(x, y)) ¨
= f 1 (g(x, y)) 2x 2y
Bx
By
Bx
By
which implies that
BF
= 2xf 1 (g(x, y)),
Bx
So y
BF
= 2yf 1 (g(x, y)) .
By
BF
BF
= f 1 (g(x, y))2xy = x .
Bx
By
6.4.4 The mean value theorem
Theorem 6.49. Let U Ď Rn be open, and f : U Ñ Rm with f = (f1 , ¨ ¨ ¨ , fm ). Suppose that
f is differentiable on U and the line segment joining x and y lies in U. Then there exist
points c1 , ¨ ¨ ¨ , cm on that segment such that
fi (y) ´ fi (x) = (Dfi )(ci )(y ´ x)
@ i = 1, ¨ ¨ ¨ , m.
Moreover, if U is convex and sup }(Df )(x)}B(Rn ,Rm ) ď M , then
xPU
}f (x) ´ f (y)}Rm ď M }x ´ y}Rn
@ x, y P U .
Proof. Let γ : [0, 1] Ñ Rn be given by γ(t) = (1 ´ t)x + ty. Then by Theorem 6.43, for each
i = 1, ¨ ¨ ¨ , m, (fi ˝ γ) : [0, 1] Ñ R is differentiable on (0, 1); thus the mean value theorem
(Corollary 4.65) implies that there exists ti P (0, 1) such that
(
)
fi (y) ´ fi (x) = (fi ˝ γ)(1) ´ (fi ˝ γ)(0) = (fi ˝ γ)1 (ti ) = (Dfi )(ci ) γ 1 (ti ) ,
where ci = γ(ti ). On the other hand, γ 1 (ti ) = y ´ x.
Let g(t) = (f ˝ γ)(t). Then the chain rule implies that g 1 (t) = (Df )(γ(t))(y ´ x); thus
}g 1 (t)}Rm ď }(Df )(γ(t))}B(Rn ,Rm ) }y ´ x}Rm ď M }x ´ y}Rn .
§6.4 Properties of Differentiable Functions
225
(
)
Define h(t) = g(1) ´ g(0) ¨ g(t). Then h : [0, 1] Ñ R is differentiable; thus by the mean
value theorem (Corollary 4.65) we find that there exists ξ P (0, 1) such that
(
)
h(1) ´ h(0) = h 1 (ξ) = g(1) ´ g(0) ¨ g 1 (ξ) ;
thus by the fact that g(0) = f (x) and g(1) = f (y),
}f (x) ´ f (y)}2Rm = h(1) ´ h(0) ď }g(1) ´ g(0)}Rm }g 1 (ξ)}Rm
ď M }f (x) ´ f (y)}Rm }x ´ y}Rn
which concludes the theorem.
˝
Example 6.50. Let f : [0, 1] Ñ R2 be given by f (t) = (t2 , t3 ). Then there is no s P (0, 1)
such that
(1, 1) = f (1) ´ f (0) = f 1 (s)(1 ´ 0) = f 1 (s)
since f 1 (s) = (2s, 3s2 ) ‰ (1, 1) for all s P (0, 1).
Example 6.51. Let f : R Ñ R2 be given by f (x) = (cos x, sin x). Then f (2π) ´ f (0) =
(0, 0); however, f 1 (x) = (´ sin x, cos x) which cannot be a zero vector.
Example 6.52. Let f be given in Example 6.30, and U be a small neighborhood of the
curve
ˇ
␣
( ␣
(
C = (x, y) ˇ x2 + y 2 = 1, x ď 0 Y (x, ˘1) | 0 ď x ď 1 .
Then
f (1, ´1) ´ f (1, 1) =
3π
.
2
On the other hand,
(Df )(x, y)(0, ´2) =
which can never be
(x, y) in U validates
[ ´y
x2 + y 2
[
]
x ] 0
2x
=´ 2
x2 + y 2 ´2
x + y2
ˇ 2x ˇ
3π
ˇ ď 3 if (x, y) P U while 3π ą 3. Therefore, no point
since ˇ 2
2
2
x +y
2
(
)
(Df )(x, y) (1, ´1) ´ (1, 1) = f (1, ´1) ´ f (1, 1) .
226
CHAPTER 6. Differentiation
Example 6.53. Suppose that A Ď Rn is an open convex set, and f : A Ñ Rm is differentiable and Df (x) = 0 for all x P A. Then f is a constant; that is, for some α P Rm we have
f (x) = α for all x P A.
Reason: Since A is convex, then the Mean Value Theorem can be applied to any x, y P A
such that fi (x)´fi (y) = Dfi (ci )(x´y) = 0 (7 Dfi = 0) for i = 1, 2, ¨ ¨ ¨ , m; thus f (x) = f (y)
for any x, y P A. Let α = f (x) P Rm , then we reach the conclusion.
Example 6.54. Let f : [0, 8) Ñ R be continuous and be differentiable on (0, 8). Suppose
that f (0) = 0 and f 1 (x) is non-decreasing (that is, if x ă y, then f 1 (x) ď f 1 (y)). Show that
g : (0, 8) Ñ R, g(x) =
f (x)
is also non-decreasing.
x
Proof: It suffices to prove g 1 (x) ě 0. Since f is differentiable on (0, 8), then g is differentiable
on (0, 8), and g 1 (x) =
xf 1 (x) ´ f (x)
. Hence
x2
g 1 (x) ě 0 ô xf 1 (x) ě f (x) .
Let x ą 0 be fixed. Applying the Mean Value Theorem to f we find that
D c P (0, x) Q f (x) = f (x) ´ f (0) = f 1 (c)(x ´ 0) ď xf 1 (x) .
6.5 Directional Derivatives and Gradient Vectors
Definition 6.55. Let f be real-valued and defined on a neighborhood of x0 P Rn , and let
v P Rn be a unit vector. Then
dˇ
f (x0 + tv) ´ f (x0 )
(Dv f )(x0 ) ” ˇ f (x0 + tv) = lim
tÑ0
dt t=0
t
ˇ
is called the directional derivative（方向導數）of f at x0 in the direction v.
Remark 6.56. Let tej unj=1 be the standard basis of Rn . Then the partial derivative
Bf
(x0 )
Bxj
(if it exists) is the directional derivative of f at x0 in the direction ej .
Theorem 6.57. Let U Ď Rn be open, and f : U Ñ R be differentiable at x0 . Then the
directional derivative of f at x0 in the direction v is (Df )(x0 )(v).
Proof. Let ε ą 0 be given. Since f is differentiable at x0 , there exists δ ą 0 such that
ˇ
ˇ
ˇf (x) ´ f (x0 ) ´ (Df )(x0 )(x ´ x0 )ˇ ď ε }x ´ x0 }Rn whenever }x ´ x0 }Rn ă δ .
2
§6.5 Directional Derivatives and Gradient Vectors
227
In particular, if x = x0 + tv with v being a unit vector in Rn and 0 ă |t| ă δ, then
ˇ f (x + tv) ´ f (x )
ˇ ˇˇf (x + tv) ´ f (x ) ´ (Df )(x )(tv)ˇˇ
0
0
0
ˇ
ˇ
0
0
´ (Df )(x0 )(v)ˇ =
ˇ
t
|t|
ˇ
ˇ
ˇf (x) ´ f (x0 ) ´ (Df )(x0 )(x ´ x0 )ˇ
ε
=
ď ă ε;
|t|
2
thus (Dv f )(x0 ) = (Df )(x0 )(v).
˝
Remark 6.58. When v P Rn but 0 ă }v}Rn ‰ 1, we let v =
v
. Then the direction
}v}Rn
derivatives of a function f : U Ď Rn Ñ R at a P U in the direction v is
f (a + tv) ´ f (a)
.
tÑ0
t
(Dv f )(a) = lim
Making a change of variable s =
t
. Then
}v}Rn
(Df )(x0 )(v) = }v}Rn (Df )(x0 )(v) = }v}Rn lim
tÑ0
f (a + sv) ´ f (a)
f (a + tv) ´ f (a)
= lim
.
sÑ0
t
s
We sometimes also call the value (Df )(x0 )(v) the “directional derivative” of f in the “direction” v.
Example 6.59. The existence of directional derivatives of a function f at x0 in all directions
does not guarantee the differentiability of f at x0 . For example, let f : R2 Ñ R be given as
in Example 6.41, and v = (v1 , v2 ) P R2 be a unit vector. Then
(Dv f )(0) = lim
tÑ0
f (tv1 , tv2 ) ´ f (0, 0)
= v31 .
t
However, f is not differentiable at (0, 0). We also note that in this example, (Dv f )(0) ‰
[
]
[ ]
Bf
Bf
(0, 0)
(0, 0) = 1 0 is the Jacobian matrix of f at (0, 0).
(Jf )(0)v, where (Jf )(0) =
Bx
By
Example 6.60. The existence of directional derivatives of a function f at x0 in all directions
does not even guarantee the continuity of f at x0 . For example, let f : R2 Ñ R be given by
$
& xy 2
if (x, y) ‰ (0, 0) ,
x2 + y 4
f (x, y) =
%
0
if (x, y) = (0, 0) ,
and v = (v1 , v2 ) P R2 be a unit vector. Then if v1 ‰ 0,
v2
t3 v1 v2
f (tv1 , tv2 ) ´ f (0, 0)
= lim 2 2 24 4 = 2
tÑ0 t(t v1 + t v2 )
tÑ0
t
v1
(Dv f )(0) = lim
228
CHAPTER 6. Differentiation
while if v1 = 0,
f (tv1 , tv2 ) ´ f (0, 0)
= 0.
t
However, f is not continuous at (0, 0) since if (x, y) approaches (0, 0) along the curve x = my 2
(Dv f )(0) = lim
tÑ0
with m ‰ 0, we have
m
my 4
= 2
2
4
4
yÑ0 m y + y
m +1
lim f (my 2 , y) = lim
yÑ0
which depends on m. Therefore, f is not continuous at (0, 0).
Example 6.61. Here comes another example showing that a function having directional
derivative in all directions might not be continuous. Let f : R2 Ñ R be given by
#
f (x, y) =
xy
x + y2
if x + y 2 ‰ 0 ,
0
if x + y 2 = 0 ,
and v = (v1 , v2 ) P R2 be a unit vector. Then if v1 ‰ 0,
t2 v1 v2
f (tv1 , tv2 ) ´ f (0, 0)
= lim
= v2
(Dv f )(0) = lim
tÑ0 t(tv1 + t2 v2
tÑ0
t
2)
while if v1 = 0,
f (tv1 , tv2 ) ´ f (0, 0)
= 0.
tÑ0
t
However, f is not continuous at (0, 0) since if (x, y) approaches (0, 0) along the polar curve
(Dv f )(0) = lim
θ(r) =
π
+ sin´1 (r ´ mr2 )
2
0 ă r ! 1,
we have
lim
(x,y)Ñ(0,0)
x=r cos θ(r),y=r sin θ(r)
f (x, y) = lim+
rÑ0
= lim+
rÑ0
r(´r + mr2 ) sin θ(r)
r2 cos θ(r) sin θ(r)
=
lim
r2 sin2 θ(r) + r cos θ(r) rÑ0+ r sin2 θ(r) ´ r + mr2
(´r + mr2 ) sin θ(r)
´1
=
2
m
sin θ(r) ´ 1 + mr
which depends on m. Therefore, f is not continuous at (0, 0).
Definition 6.62. Let U Ď Rn be an open set. The derivative of a scalar function f : U Ñ R
is called the gradient of f and is denoted by gradf or ∇f .
§6.5 Directional Derivatives and Gradient Vectors
229
Let U Ď Rn be an open set, a P U and f : U Ñ R be a real-valued function. Suppose
that f P C 1 (U; R) and (∇f )(a) ‰ 0. Then
we can assume that
Bf
(a) ‰ 0 for some 1 ď k ď n. W.L.O.G.,
Bxk
Bf
(a) ‰ 0. By the implicit function theorem, there exists an open
Bxn
neighborhood V Ď Rn´1 of (a1 , ¨ ¨ ¨ , an´1 ) and an open neighborhood W Ď R of an , as well as
ˇ
␣
(
a C 1 -function φ : V Ñ R such that in a neighborhood of a the level set x P U ˇ f (x) = f (a)
can be represented by xn = φ(x1 , ¨ ¨ ¨ , xn´1 ); that is,
(
)
f x1 , ¨ ¨ ¨ , xn´1 , φ(x1 , ¨ ¨ ¨ , xn´1 ) = f (a)
@ (x1 , ¨ ¨ ¨ , xn´1 ) P V .
Moreover,
(
)
fxj x1 , ¨ ¨ ¨ , xn´1 , φ(x1 , ¨ ¨ ¨ , xn´1 )
)
φxj (x1 , ¨ ¨ ¨ , xn´1 ) = ´ (
fxn x1 , ¨ ¨ ¨ , xn´1 , φ(x1 , ¨ ¨ ¨ , xn´1 )
Consider the collection of vectors tvj un´1
j=1 given by
ˇ
)
B ˇ (
vj =
x1 , ¨ ¨ ¨ , xn´1 , φ(x1 , ¨ ¨ ¨ , xn´1 )
ˇ
Bxj x=a
@ (x1 , ¨ ¨ ¨ , xn´1 ) P V .
(x1 , ¨ ¨ ¨ , xn´1 ) P V .
Then vj1 s are tangent vectors of the level surface. If tej unj=1 is the standard basis of Rn , then
(
(
)
fx (a) )
.
vj = ej + 0, ¨ ¨ ¨ , 0, φxj (a1 , ¨ ¨ ¨ , an´1 ) = ej ´ 0, ¨ ¨ ¨ , 0, j
fxn (a)
Therefore, the gradient vector (∇f )(a) is perpendicular to vj for all 1 ď j ď n ´ 1 which
conclude the following
Proposition 6.63. Let U Ď Rn be open and f P C 1 (U; R); that is, f : U Ñ R is contin(∇f )(x )
0
uously differentiable. Then if (∇f )(x0 ) ‰ 0, the vector
is the unit normal to
}(∇f
)(x
)}
0 Rn
ˇ
␣
(
the level set x P U ˇ f (x) = f (x0 ) at x0 .
ˇ
␣
(
Example 6.64. Find the normal to S = (x, y, z) ˇ x2 + y 2 + z 2 = 3 at (1, 1, 1) P S.
Solution: Take f (x, y, z) = x2 + y 2 + z 2 ´ 3. Then (∇f )(x, y, z) = (2x, 2y, 2z); thus
(∇f )(1, 1, 1) = (2, 2, 2) is normal to S at (1, 1, 1).
Example 6.65. Consider the surface
ˇ
␣
(
S = (x, y, z) P R3 ˇ x2 ´ y 2 + xyz = 1 .
Find the tangent plane of S at (1, 0, 1).
230
CHAPTER 6. Differentiation
Solution: Let f (x, y, z) = x2 ´ y 2 + xyz. Then
␣
(
S = (x, y, z) P R3 | f (x, y, z) = f (1, 0, 1) ;
that is, S is a level set of f . Since (∇f )(1, 0, 1) = (2, 1, 0) ‰ (0, 0, 0), (2, 1, 0) is normal to
S at (1, 0, 1); thus the tangent plane of S at (1, 0, 1) is 2(x ´ 1) + y = 0.
Proposition 6.66. Let f : Rn Ñ R be differentiable. Then ˘
˝
∇f
is the direction in
}∇f }Rn
which the function f increases/decreases most rapidly（最速上升／下降方向）.
Proof. Let x0 P Rn be given. Suppose that f increases most rapidly in the direction v,
then (Dv f )(x0 ) = sup (Dw f )(x0 ). Since f is differentiable, (Dw f )(x0 ) = (Df )(x0 )(w) =
}w}Rn =1
(∇f )(x0 ) ¨ w which is maximized in the direction
(∇f )(x0 )
.
}(∇f )(x0 )}Rn
˝
Example 6.67. Let f : R3 Ñ R be given by f (x, y, z) = x2 y sin z. Find the direction of
the greatest rate of change at (3, 2, 0).
Solution: We compute the gradient of f at (3, 2, 0) as follows:
(∇f )(3, 2, 0) =
( Bf
Bx
(3, 2, 0),
)
Bf
Bf
(3, 2, 0), (3, 2, 0)
By
Bz
ˇ
= (2xy sin z, x2 sin z, x2 y cos z)ˇ(x,y,z)=(3,,2,0) = (0, 0, 18).
Therefore, the direction of the greatest rate of change of f at (3, 2, 0) is (0, 0, 1).
6.6 Higher Order Derivatives of Functions
Let U Ď Rn be open, and f : U Ñ Rm is differentiable. By Proposition 6.5, the space
(
)
B(Rn , Rm ), } ¨ }B(Rn ,Rm ) is a normed space (in fact, it is a Banach space), so it is legitimate
to ask if Df : U Ñ B(Rn , Rm ) is differentiable or not. If Df is differentiable at x0 , we
call f twice differentiable at x0 , and denote the twice derivative of f at x0 as (D2 f )(x0 ). If
(
)
Df is differentiable on U, then D2 f : U Ñ B Rn , B(Rn , Rm ) . Similar, we can talk about
three times differentiability of a function if it is twice differentiable. In general, we have the
following
Definition 6.68. Let (X, } ¨ }X ) and (Y, } ¨ }Y ) be normed spaces, and U Ď X be open. A
function f : U Ñ Y is said to be twice differentiable at a P U if
§6.6 Higher Order Derivatives of Functions
231
1. f is (once) differentiable in a neighborhood of a;
(
)
2. there exists L2 P B X, B(X, Y ) , usually denoted by (D2 f )(a) and called the second
derivative of f at a, such that
›
›
›(Df )(x) ´ (Df )(a) ´ L2 (x ´ a)›
B(X,Y )
lim
= 0.
xÑa
}x ´ a}X
For any two vectors u, v P X, (D2 f )(a)(v) P B(X, Y ) and (D2 f )(a)(v)(u) P Y . The vector
(D2 f )(a)(v)(u) is usually denoted by (D2 f )(a)(u, v).
In general, a function f is said to be k-times differentiable at a P U if
1. f is (k ´ 1)-times differentiable in a neighborhood of a;
2. there exists Lk P B(X,
B(X, ¨ ¨ ¨ , B(X , Y lo
) ¨omo
¨ ¨ o))n , usually denoted by (Dk f )(a) and
loooooooooomoooooooooon
k copies of “X”
k copies of “)”
called the k-th derivative of f at a, such that
› k´1
›
›(D f )(x) ´ (Dk´1 f )(a) ´ Lk (x ´ a)›
B(X,B(X,¨¨¨ ,B(X,Y )¨¨¨ ))
lim
= 0.
xÑa
}x ´ a}X
For any k vectors u(1) , ¨ ¨ ¨ u(k) P X, the vector (Dk f )(a)(u(1) , ¨ ¨ ¨ , u(k) ) is defined as the
vector
(Dk f )(a)(u(k) )(u(k´1) ) ¨ ¨ ¨ (u(1) ) .
Example 6.69. Let (X, } ¨ }X ) and (Y, } ¨ }Y ) be two normed spaces, and f (x) = Lx for
some L P B(X, Y ). From Example 6.15, (Df )(x0 ) = L for all x0 P X; thus (D2 f )(x0 ) = 0
since Df : U P B(X, Y ) is a “constant” map. In fact, one can also conclude from
lim
›
›
›(Df )(x) ´ (Df )(x0 ) ´ 0(x ´ x0 )›
B(X,Y )
xÑx0
}x ´ x0 }X
=0
that (D2 f )(x0 ) = 0 for all x0 P X.
Remark 6.70. We focus on what (Dk f )(a)(uk )(¨ ¨ ¨ )(u1 ) means in this remark. We first
look at the case that f is twice differentiable at a. With x = a + tv for v P X with }v}X = 1
in the definition, we find that
lim
tÑ0
›
›
›(Df )(a + tv) ´ (Df )(a) ´ t(D2 f )(a)(v)›
B(X,Y )
|t|
= 0.
232
CHAPTER 6. Differentiation
Since (Df )(a + tv) ´ (Df )(a) ´ t(D2 f )(a)(v) P B(X, Y ), for all u P X with }u}X = 1 we
have
lim
tÑ0
›
›
›(Df )(a + tv)(u) ´ (Df )(a)(u) ´ t(D2 f )(a)(v)(u)›
Y
|t|
›[
] ›
› (Df )(a + tv) ´ (Df )(a) ´ t(D2 f )(a)(v) (u)›
Y
= lim
tÑ0
|t|
›
›
›(Df )(a + tv) ´ (Df )(a) ´ t(D2 f )(a)(v)›
B(X,Y )
ď lim
|t|
tÑ0
= 0.
On the other hand, by the definition of the direction derivative,
[ f (a + tv + su) ´ f (a + tv) f (a + su) ´ f (a) ]
(Df )(a + tv)(u) ´ (Df )(a)(u) = lim
´
;
sÑ0
s
s
thus the limit above implies that
f (a + tv + su) ´ f (a + tv) ´ f (a + su) + f (a)
tÑ0 sÑ0
st
f (a + tv + su) ´ f (a + tv)
f (a + su) ´ f (a)
lim
´ lim
sÑ0
s
s
= lim sÑ0
tÑ0
t
(D2 f )(a)(v)(u) = lim lim
= Dv (Du f )(a) .
Therefore, (D2 f )(a)(v)(u) is obtained by first differentiating f near a in the u-direction,
then differentiating (Df ) at a in the v-direction.
In general, (Dk f )(a)(uk ) ¨ ¨ ¨ (u1 ) is obtained by first differentiating f near a in the u1 direction, then differentiating (Df ) near a in the u2 -direction, and so on, and finally differentiating (Dk´1 f ) at a in the uk -direction.
Remark 6.71. Since (D2 f )(a) P B(X, B(X, Y )), if v1 , v2 P X and c P R, we have
(D2 f )(a)(cv1 + v2 ) = c(D2 f )(a)(v1 ) + (D2 f )(a)(v2 ) (treated as “vectors” in B(X, Y )); thus
(D2 f )(a)(cv1 + v2 )(u) = c(D2 f )(a)(v1 )(u) + (D2 f )(a)(v2 )(u)
@ u, v1 , v2 P X .
On the other hand, since (D2 f )(a)(v) P B(X, Y ),
(D2 f )(a)(v)(cu1 + u2 ) = c(D2 f )(a)(v)(u1 ) + (D2 f )(a)(v)(u2 )
@ u1 , u 2 , v P X .
Therefore, (D2 f )(a)(v)(u) is linear in both u and v variables. A map with such kind of
property is called a bilinear map (meaning 2-linear). In particular, (D2 f )(a) : X ˆ X Ñ Y
is a bilinear map.
§6.6 Higher Order Derivatives of Functions
233
In general, the vector (Dk f )(a)(u(1) , ¨ ¨ ¨ , u(k) ) is linear in u(1) , ¨ ¨ ¨ , u(k) ; that is,
(Dk f )(a)(u(1) , ¨ ¨ ¨ , u(i´1) , αv + βw, u(i+1) , ¨ ¨ ¨ , u(k) )
= α(Dk f )(a)(u(1) , ¨ ¨ ¨ , u(i´1) , v, u(i+1) , ¨ ¨ ¨ , u(k) )
+ β(Dk f )(a)(u(1) , ¨ ¨ ¨ , u(i´1) , w, u(i+1) , ¨ ¨ ¨ , u(k) )
for all v, w P X, α, β P R, and i = 1, ¨ ¨ ¨ , n. Such kind of map which is linear in each
component when the other k ´ 1 components are fixed is called k-linear.
␣
(
Consider the case that X is finite dimensional with dim(X) = n, e1 , e2 , . . . , en is a basis
of X, and Y = R. Then (D2 f )(a) : X ˆ X Ñ Y is a bilinear form (here the term “form”
means that Y = R). A bilinear form B : X ˆ X Ñ R can be represented as follows: Let
n
n
ř
ř
aij = B(ei , ej ) P R for i, j = 1, 2, ¨ ¨ ¨ , n. Given x, y P Rn , write u =
ui ei and v =
vj ej .
i=1
j=1
Then by the bilinearity of B,
 
v1
a11 ¨ ¨ ¨ a1n
] .
..   ..  .
...
un  ..
.  . 
vn
an1 ¨ ¨ ¨ ann

B(u, v) = B
n
(ÿ
ui ei ,
i=1
n
ÿ
n
ÿ
)
[
vj ej =
ui vj aij = u1 ¨ ¨ ¨
j=1
i,j=1
Therefore, if f : U Ď Rn Ñ R is twice differentiable at a, then the bilinear form (D2 f )(a)
can be represented as

[
(D2 f )(a)(u, v) = u1 ¨ ¨ ¨

(D2 f )(e1 , e1 ) ¨ ¨ ¨ (D2 f )(a)(e1 , en )  v 
  .1 
]
.
.
.

  ..  .
.
.
.
un 
.
.
.

vn
2
2
(D f )(en , e1 ) ¨ ¨ ¨ (D f )(a)(en , en )
The following proposition is an analogy of Proposition 6.27. The proof is similar to the
one of Proposition 6.27, and is left as an exercise.
Proposition 6.72. Let U Ď Rn be open, x0 P U, and f = (f1 , ¨ ¨ ¨ , fm ) : U Ñ Rm . Then
f is k-times differentiable at x0 if and only if fi is k-times differentiable at x0 for all
i = 1, ¨ ¨ ¨ , m.
Due to the proposition above, when talking about the higher-order differentiability of
f : U Ď Rn Ñ Rm and a point x0 P U, from now on we only focus on the case m = 1.
Example 6.73. In this example, we focus on what the second derivative (D2 f )(a) of a
function f is, or in particular, what (D2 f )(a)(ei , ej ) (which appears in the Remark 6.71) is,
if X = R2 .
234
CHAPTER 6. Differentiation
Let f : R2 Ñ R be differentiable, then
[
]
[
] [
]
Bf
Bf
(x, y)
(x, y) .
(Df )(x, y) = fx (x, y) fy (x, y) =
Bx
By
Suppose that f is twice differentiable at (a, b), and let L2 = (D2 f )(a, b). Then
›
(
)›
›(Df )(x, y) ´ (Df )(a, b) ´ L2 (x ´ a, y ´ b) ›
B(R2 ,R)
a
lim
=0
(x,y)Ñ(a,b)
(x ´ a)2 + (y ´ b)2
or equivalently,
›[
] [
] [ (
)]›
› fx (x, y) fy (x, y) ´ fx (a, b) fy (a, b) ´ L2 (x ´ a, y ´ b) ›
B(R2 ,R)
a
= 0,
lim
(x,y)Ñ(a,b)
(x ´ a)2 + (y ´ b)2
[ (
)]
(
where L2 (x ´ a, y ´ b) denotes the matrix representation of the linear map L2 (x ´ a, y ´
)
b) P B(R2 , R). In particular, we must have
]
›[
[
]››
› fx (x, b) ´ fx (a, b) fy (x, b) ´ fy (a, b)
lim ›
´ L2 e1 ›
=0
x´a
xÑa
and
›[
› f (a, y) ´ fx (a, b)
lim › x
y´b
yÑb
x´a
B(R2 ,R)
]
[
]›
fy (a, y) ´ fy (a, b)
´ L2 e2 ›
= 0.
y´b
B(R2 ,R)
›
Using the notation of second partial derivatives, we find that
[
] [
]
L2 e1 = fxx (a, b) fyx (a, b)
where fxy = (fx )y =
(
B Bf
By Bx
)
and
and fyx = (fy )x =
[
] [
]
L2 e2 = fxy (a, b) fyy (a, b) ,
(
)
B Bf
. Therefore, if v = v1 e1 + v2 e2 ,
Bx By
[
] [
]
[L2 v] = L2 (v1 e1 + v2 e2 ) = v1 fxx (a, b) + v2 fxy (a, b) v1 fyx (a, b) + v2 fyy (a, b) .
Symbolically, we can write
[ ] [[
]
L2 = fxx (a, b) fyx (a, b)
so that
[
[
(6.6.1)
]]
fxy (a, b) fyy (a, b)
[ ] [
[ ]
] [ ] v1
[
] [
] ] v1
fxy (a, b) fyy (a, b)
L2 (v1 e1 + v2 e2 ) = L2
= fxx (a, b) fyx (a, b)
v2
v2
[
]
[
]
= v1 fxx (a, b) fyx (a, b) + v2 fxy (a, b) fyy (a, b) .
§6.6 Higher Order Derivatives of Functions
235
For two vectors u and v, what does (D2 f )(a, b)(v)(u) or (D2 f )(a, b)(u, v) mean? To see
this, let u = u1 e1 + u2 e2 and v = v1 e1 + v2 e2 . Then
[ ]
[ 2
] [ 2
][ ] [
] u1
(D f )(a, b)(v)(u) = (D f )(a, b)(v) u = L2 (v1 e1 + v2 e2 )
u2
[ ]
[ ]
[
] u1
[
] u1
= v1 fxx (a, b) fyx (a, b)
+ v2 fxy (a, b) fyy (a, b)
u2
u2
[
][ ]
[
] fxx (a, b) fyx (a, b) u1
= v1 v2
.
fxy (a, b) fyy (a, b) u2
Therefore, (D2 f )(a, b)(e1 , e1 ) = fxx (a, b), (D2 f )(a, b)(e1 , e2 ) = fxy (a, b), (D2 f )(a, b)(e2 , e1 ) =
fyx (a, b) and (D2 f )(a, b)(e2 , e2 ) = fyy (a, b).
On the other hand, we can identify B(R2 ; R) as R2 (every 1ˆ2 matrix is a “row” vector),
[ ]T
and treat g ” Df : R2 Ñ R2 as a vector-valued function. By Theorem 6.21 (Dg)(a, b)
can be represented as a 2 ˆ 2 matrix given by
[
]
[
]
fxx (a, b) fxy (a, b)
(Dg)(a, b) =
.
fyx (a, b) fyy (a, b)
We note that the representation above means
]›
][
] [
] [
›[
fxx (a, b) fxy (a, b) x ´ a ›
fx (a, b)
› fx (x, y)
´
´
›
›
fyx (a, b) fyy (a, b) y ´ b R2
fy (a, b)
fy (x, y)
a
= 0.
lim
(x,y)Ñ(a,b)
(x ´ a)2 + (y ´ b)2
The equality above is equivalent to that
]
[
›[
] [
] [
] fxx (a, b) fyx (a, b) ››
›
›
› (Df )(x, y) ´ (Df )(a, b) ´ x ´ a y ´ b
fxy (a, b) fyy (a, b) R2
a
lim
=0
(x,y)Ñ(a,b)
(x ´ a)2 + (y ´ b)2
According to the equality above, L2 = (D2 f )(a, b) should be defined by
[
] ([
] [ ])T
[
] [
] fxx (a, b) fyx (a, b)
fxx (a, b) fxy (a, b) v1
L2 (v1 e1 + v2 e2 ) = v1 v2
=
fxy (a, b) fyy (a, b)
fyx (a, b) fyy (a, b) v2
which agrees with what (6.6.1) provides.
Proposition 6.74. Let U Ď Rn be open, and f : U Ñ R. Suppose that f is k-times
differentiable at a. Then for k vectors u(1) , ¨ ¨ ¨ , u(k) P Rn ,
n
ÿ
Bk f
(k)
(1) (2)
k
(1)
(k)
(a)uj1 uj2 ¨ ¨ ¨ ujk
(D f )(a)(u , ¨ ¨ ¨ , u ) =
Bx
Bx
¨
¨
¨
Bx
jk
jk´1
j1
j1 ,¨¨¨ ,jk =1
n
(
(
( Bf ) ))
ÿ
B
B
B
(k)
(1) (2)
=
¨¨¨
¨ ¨ ¨ (a)uj1 uj2 ¨ ¨ ¨ ujk ,
j1 ,¨¨¨ ,jk =1
Bxjk Bxjk´1
Bxj2 Bxj1
236
CHAPTER 6. Differentiation
(i)
(i)
(i)
where u(i) = (u1 , u2 , ¨ ¨ ¨ , un ) for all i = 1, ¨ ¨ ¨ , k.（上標括號中的數字指所給定的 k 個向
量中的第幾個向量，下標指每一個固定向量的第幾個分量）
Proof. Let tej unj=1 be the standard basis of Rn . By Remark 6.71 (on multi-linearity), it
suffices to show that
(Dk f )(a)(ejk )(ejk´1 ) ¨ ¨ ¨ (ej2 )(ej1 ) = (Dk f )(a)(ej1 , ¨ ¨ ¨ , ejk ) =
Bk f
Bxjk Bxjk´1 ¨ ¨ ¨ Bxj1
(a) (6.6.2)
provided that f is k-times differentiable at a since if so, we must have
n
n
(ÿ
)
ÿ
(1)
(k)
(D f )(a)(u , ¨ ¨ ¨ , u ) = (D f )(a)
uj1 ej1 , ¨ ¨ ¨ ,
ujk ejk
k
(1)
(k)
k
=
=
n
ÿ
n
ÿ
j1 =1 j2 =1
n
ÿ
¨¨¨
j1 =1
n
ÿ
jk =1
(1) (2)
jk =1
Bk f
Bxjk Bxjk´1 ¨ ¨ ¨ Bxj1
j ,¨¨¨ ,j =1
1
(k)
(Dk f )(a)(ej1 , ¨ ¨ ¨ , ejk )uj1 uj2 ¨ ¨ ¨ ujk
(1) (2)
(k)
(a)uj1 uj2 ¨ ¨ ¨ ujk .
k
We prove the proposition by induction. Note that the case k = 1 is true because of Theorem
6.21. Next we assume that (6.6.2) holds true for k = ℓ if f is (ℓ ´ 1)-times differentiable in a
neighborhood of a and f is ℓ-times differentiable at a. Now we show that (6.6.2) also holds
true for k = ℓ + 1 if f is ℓ-times differentiable in a neighborhood of a, and f is (ℓ + 1)-times
differentiable at a. By the definition of (ℓ + 1)-times differentiability at a,
lim
xÑa
›
› ℓ
›(D f )(x) ´ (Dℓ f )(a) ´ (Dℓ+1 f )(a)(x ´ a)›
B(Rn ,B(Rn ,¨¨¨ ,B(Rn ,R)¨¨¨ ))
}x ´ a}Rn
= 0.
Since
ˇ
ˇ[
]
ˇ
ˇ
ˇ (Dℓ f )(x) ´ (Dℓ f )(a) ´ (Dℓ+1 f )(a)(x ´ a) (ejℓ ) ¨ ¨ ¨ (ej2 )(ej1 )ˇ
›
›[
]
›
›
ℓ
ℓ
ℓ+1
}ej1 }Rn
ď › (D f )(x) ´ (D f )(a) ´ (D f )(a)(x ´ a) (ejℓ ) ¨ ¨ ¨ (ej2 )›
B(Rn ,R)
› ℓ
›
ď ›(D f )(x) ´ (Dℓ f )(a) ´ (Dℓ+1 f )(a)(x ´ a)›B(Rn ,B(Rn ,¨¨¨ ,B(Rn ,R)¨¨¨ )) }ej1 }Rn ¨ ¨ ¨ }ejℓ }Rn
›
›
= ›(Dℓ f )(x) ´ (Dℓ f )(a) ´ (Dℓ+1 f )(a)(x ´ a)› n
,
n
n
B(R ,B(R ,¨¨¨ ,B(R ,R)¨¨¨ ))
§6.6 Higher Order Derivatives of Functions
237
using (6.6.2) (for the case k = ℓ) we conclude that
lim
ˇ
ˇ
ˇ
Bℓ f
Bxjℓ Bxjk´1 ¨ ¨ ¨ Bxj1
xÑa
= lim
xÑa
ď lim
(x) ´
Bℓ f
Bxjℓ Bxjk´1 ¨ ¨ ¨ Bxj1
ˇ
ˇ
(a) ´ (Dℓ+1 f )(a)(ej1 , ¨ ¨ ¨ , ejℓ , x ´ a)ˇ
}x ´ a}Rn
ˇ ℓ
ˇ
ˇ(D f )(x)(ej , ¨ ¨ ¨ , ej ) ´ (Dℓ f )(a)(ej , ¨ ¨ ¨ , ej ) ´ (Dℓ+1 f )(a)(x ´ a)(ej , ¨ ¨ ¨ , ej )ˇ
1
1
ℓ
1
ℓ
}x ´ a}Rn
› ℓ
›
›(D f )(x) ´ (Dℓ f )(a) ´ (Dℓ+1 f )(a)(x ´ a)›
B(Rn ,B(Rn ,¨¨¨ ,B(Rn ,R)¨¨¨ ))
}x ´ a}Rn
xÑa
ℓ
= 0.
In particular, if x = a+tejℓ+1 for some jℓ+1 = 1, ¨ ¨ ¨ , n, by the definition of partial derivatives
we conclude that
Bℓ f
(D
ℓ+1
f )(a)(ej1 , ¨ ¨ ¨ , ejℓ , ejℓ+1 ) = lim
Bxjℓ Bxjk´1 ¨ ¨ ¨ Bxj1
(a + tejℓ+1 ) ´
Bℓ f
Bxjℓ Bxjk´1 ¨ ¨ ¨ Bxj1
(a)
t
tÑ0
B ℓ+1 f
=
(a)
Bxjℓ+1 Bxjℓ Bxjk´1 ¨ ¨ ¨ Bxj1
which is (6.6.2) for the case k = ℓ + 1.
˝
Example 6.75. Let f : R2 Ñ R be given by f (x1 , x2 ) = x21 cos x2 , and u(1) = (2, 0),
u(2) = (1, 1), u(3) = (0, ´1). Suppose that f is three-times differentiable at a = (0, 0) (in
fact it is, but we have not talked about this yet). Then
3
(1)
(2)
(3)
(D f )(a)(u , u , u ) =
=
2
ÿ
i,j,k=1
B3f
2
ÿ
B3f
B3f
(1) (2) (3)
(2)
(a)ui uj uk =
(a) ¨ 2 ¨ uj ¨ (´1)
Bxk Bxj Bxi
Bx2 Bxj Bx1
Bx2 Bx21
j=1
(0, 0) ¨ 2 ¨ 1 ¨ (´1) +
B3f
Bx22 Bx1
(0, 0) ¨ 2 ¨ 1 ¨ (´1) = 0 .
Corollary 6.76. Let U Ď Rn be open, and f : U Ñ R be (k + 1)-times differentiable at a.
Then for u(1) , ¨ ¨ ¨ , u(k) , u(k+1) P Rn ,
(Dk+1 f )(a)(u(1) , ¨ ¨ ¨ , u(k) , u(k+1) ) =
n
ÿ
j=1
(k+1)
uj
B ˇ
ˇ (Dk f )(x)(u(1) , ¨ ¨ ¨ , u(k) ) .
Bxj x=a
ˇ
In other words, (using the terminology in Remark 6.58) (Dk+1 f )(a)(u(1) , ¨ ¨ ¨ , u(k) , u(k+1) ) is
the “directional derivative” of the function (Dk f )(¨)(u(1) , ¨ ¨ ¨ , u(k) ) at a in the “direction”
u(k+1) .
238
CHAPTER 6. Differentiation
Proof. By Proposition 6.74,
(D
k+1
(1)
(k)
f )(a)(u , ¨ ¨ ¨ , u , u
(k+1)
n
ÿ
)=
j1 ,¨¨¨ ,jk ,jk+1 =1
=
=
n
ÿ
jk+1 =1
n
ÿ
(k+1)
ujk+1
j1 ,¨¨¨ ,jk =1
=
B
(k+1)
ujk+1
B k+1 f
(1)
(k)
(a)uj1 ¨ ¨ ¨ ujk
Bxjk+1 Bxjk ¨ ¨ ¨ Bxj1
n
ÿ
ˇ
ˇ
ˇ
Bxjk+1 x=a
jk+1 =1
n
ÿ
n
ÿ
(k+1)
ujk+1
jk+1 =1
B
B k+1 f
(1)
(k) (k+1)
(a)uj1 ¨ ¨ ¨ ujk ujk+1
Bxjk+1 Bxjk ¨ ¨ ¨ Bxj1
ˇ
ˇ
ˇ
Bxjk+1 x=a
j1 ,¨¨¨ ,jk =1
Bk f
(1)
(k)
(x)uj1 ¨ ¨ ¨ ujk
Bxjk ¨ ¨ ¨ Bxj1
(Dk f )(x)(u(1) , ¨ ¨ ¨ , u(k) ) .
˝
Example 6.77. Let f : R2 Ñ R be twice differentiable at a = (a1 , a2 ) P R2 . Then the
proposition above suggests that for u = (u1 , u2 ), v = (v1 , v2 ) P R2 ,
2
2
2
ÿ
B2f
i,j=1
2
B f
(D f )(a)(v)(u) = (D f )(a)(u, v) =
=
B2f
(a)ui vj
Bxj Bxi
B2f
B2f
(a)u1 v2 +
(a)u2 v1 + 2 (a)u2 v2
(a)u1 v1 +
Bx2 Bx1
Bx1 Bx2
Bx21
Bx2
 2

2
B f
B f
(a)
(a) [ ]
 v1
[
]  Bx21
Bx2 Bx1

= u1 u2 
 B2f
 v2 .
B2f
(a)
2 (a)
Bx1 Bx2
Bx2
In general, if f : Rn Ñ R be twice differentiable at a = (a1 , ¨ ¨ ¨ , an ) P Rn . Then for
u = (u1 , ¨ ¨ ¨ , un ), v = (v1 , ¨ ¨ ¨ , vn ) P R2

2
[
(D f )(a)(v)(u) = u1 ¨ ¨ ¨
B2f
 Bx2 (a)
1

]
..
un 
.


2
 B f
Bx1 Bxn

¨¨¨
..
B2f
(a)  
Bxn Bx1
 v
..
.
.
(a) ¨ ¨ ¨
B2f
(a)
Bx2n
The bilinear form B : Rn ˆ Rn Ñ R given by
B(u, v) = (D2 f )(a)(v)(u)
@ u, v P Rn
  .1 
  ..  .

 v
 n
§6.6 Higher Order Derivatives of Functions
239
is called the Hessian of f , and is represented (in the matrix form) as an n ˆ n matrix by
 2

B f
B2f
(a)
¨¨¨
 Bx2 (a)
Bxn Bx1
1




..
..
...

.
.
.




 B2f

B2f
(a) ¨ ¨ ¨
(a)
2
Bx1 Bxn
If the second partial derivatives
Bxn
B2f
(a) of f at a exists for all i, j = 1, ¨ ¨ ¨ , n (here the
Bxj Bxi
twice differentiability of f at a is ignored), the matrix (on the right-hand side of equality)
above is also called the Hessian matrix of f at a.
Even though there is no reason to believe that (D2 f )(a)(u, v) = (D2 f )(a)(v, u) (since
the left-hand side means first differentiating f in u-direction and then differentiating Df
in v-direction, while the right-hand side means first differentiating f in v-direction then
differentiating Df in u-direction), it is still reasonable to ask whether (D2 f )(a) is symmetric
or not; that is, could it be true that (D2 f )(a)(u, v) = (D2 f )(a)(v, u) for all u, v P Rn ? When
f is twice differentiable at a, this is equivalent of asking (by plugging in u = ei and v = ej )
that whether or not
B2f
B2f
(a) =
(a) .
Bxj Bxi
Bxi Bxj
(6.6.3)
The following example provides a function f : R2 Ñ R such that (6.6.3) does not hold at
a = (0, 0). We remark that the function in the following example is not twice differentiable
at a even though the Hessian matrix of f at a can still be computed.
Example 6.78. Let f : R2 Ñ R be defined by
$
2
2
& xy(x ´ y ) if (x, y) ‰ (0, 0) ,
x2 + y 2
f (x, y) =
%
0
if (x, y) = (0, 0) .
Then
and
$ 4
2 3
5
& x y + 4x y ´ y if (x, y) ‰ (0, 0) ,
(x2 + y 2 )2
fx (x, y) =
%
0
if (x, y) = (0, 0) ,
$ 5
3 2
4
& x ´ 4x y ´ xy if (x, y) ‰ (0, 0) ,
(x2 + y 2 )2
fy (x, y) =
%
0
if (x, y) = (0, 0) ,
240
CHAPTER 6. Differentiation
It is clear that fx and fy are continuous on R2 ; thus f is differentiable on R2 . However,
fx (0, k) ´ fx (0, 0)
= ´1 ,
k
kÑ0
fxy (0, 0) = lim
while
fy (h, 0) ´ fy (0, 0)
= 1;
h
hÑ0
fyx (0, 0) = lim
thus the Hessian matrix of f at the origin is not symmetric.
Definition 6.79. A function is said to be of class C r if the first r derivatives exist and
are continuous. A function is said to be smooth or of class C 8 if it is of class C r for all
positive integer r.
Now we would like to answer the question of what kind of functions are k-times differentiable. Suppose that U Ď Rn is open and f : U Ñ R. Note that by the definition of
differentiability, f is k-times differentiable in U if and only if Dk´1 f is differentiable in U.
This would further imply that f is k-times differentiable in U if and only if Dk´2 f is twice
differentiable in U. Therefore, Proposition 6.27 and Corollary 6.32 imply that
f is k-times (continuously) differentiable in U
ô Df is (k ´ 1)-times (continuously) differentiable in U
[ Bf Bf
Bf ]
ô
,
,¨¨¨ ,
is (k ´ 1)-times (continuously) differentiable in U
Bx1 Bx2
Bxn
Bf
ô
is (k ´ 1)-times (continuously) differentiable in U for all 1 ď j1 ď n
Bxj1
Bf
ôD
is (k ´ 2)-times (continuously) differentiable in U for all 1 ď j1 ď n
Bxj1
[ B2f
B2f ]
ô
,¨¨¨ ,
is (k ´ 2)-times (continuously) differentiable in U
Bx1 Bxj1
Bxn Bxj1
for all 1 ď j1 ď n
ô
B2f
is (k ´ 2)-times (continuously) differentiable in U for all 1 ď j1 , j2 ď n .
Bxj2 Bxj1
Applying similar argument several times, we obtain the following theorem which is an analogy of Corollary 6.32.
Theorem 6.80. Let U Ñ Rn and f : U Ñ R.
Bk f
Bxjk Bxjk´1 ¨ ¨ ¨ Bxj1
Suppose that the partial derivative
exists in a neighborhood of a P U and is continuous at a for all j1 , ¨ ¨ ¨ , jk =
§6.6 Higher Order Derivatives of Functions
241
1, ¨ ¨ ¨ , n. Then f is k-times differentiable at a. Moreover, if
on U, then f is of class C k .
Bk f
Bxjk Bxjk´1 ¨ ¨ ¨ Bxj1
is continuous
Theorem 6.81. Let U Ď Rn be open, and f : U Ñ R. Suppose that the mixed partial
derivatives
Then
Bf Bf
B2f
B2f
,
,
,
exist in a neighborhood of a, and are continuous at a.
Bxi Bxj Bxj Bxi Bxj Bxi
B2f
B2f
(a) =
(a) .
Bxj Bxi
Bxi Bxj
(6.6.4)
Proof. Let S(a, h, k) = f (a + hei + kej ) ´ f (a + hei ) ´ f (a + kej ) + f (a), and define φ(x) =
f (x + hei ) ´ f (x) as well as ψ(x) = f (x + kej ) ´ f (x) for x in a neighborhood of a. Then
S(a, h, k) = φ(a + kej ) ´ φ(a) = ψ(a + hei ) ´ ψ(a); thus the mean value theorem implies that
there exists c on the line segment joining a and a + kej and d on the line segment joining a
and a + hei such that
( Bf
)
Bφ
Bf
(c) = k
(c + hei ) ´
(c) ,
Bxj
Bxj
Bxj
(
)
Bf
Bψ
Bf
S(a, h, k) = ψ(a + hei ) ´ ψ(a) = h (d) = h
(d + kej ) ´
(d) .
Bxi
Bxi
Bxi
S(a, h, k) = φ(a + kej ) ´ φ(a) = k
As a consequence, if h ‰ 0 ‰ k,
) S(a, h, k)
)
Bf
1 ( Bf
Bf
1 ( Bf
(d + kej ) ´
(d) =
=
(c + hei ) ´
(c)
k Bxi
Bxi
hk
h Bxj
Bxj
By the mean value theorem again, there exists c1 and d1 on the line segment joining c,
c + hei and d, d + kej , respectively, such that
B2f
B2f
(d1 ) =
(c1 ) .
Bxj Bxi
Bxi Bxj
The theorem is then concluded by the continuity of
B2f
B2f
and
at a, and c1 Ñ a
Bxi Bxj
Bxj Bxi
and d1 Ñ a as (h, k) Ñ (0, 0).
˝
Corollary 6.82. Let U Ď Rn be open, and f is of class C 2 . Then
(D2 f )(a)(u, v) = (D2 f )(a)(v, u)
@ a P U and u, v P Rn .
Remark 6.83. In view of Remark 6.70, (6.6.4) is the same as the following identity
f (a + hei + kej ) ´ f (a + hei ) ´ f (a + kej ) + f (a)
hÑ0 kÑ0
hk
f (a + hei + kej ) ´ f (a + hei ) ´ f (a + kej ) + f (a)
= lim lim
kÑ0 hÑ0
hk
lim lim
242
CHAPTER 6. Differentiation
which implies that the order of the two limits lim and lim can be interchanged without
hÑ0
kÑ0
changing the value of the limit (under certain conditions).
Example 6.84. Let f (x, y) = yx2 cos y 2 . Then
fxy (x, y) = (2xy cos y 2 )y = 2x cos y 2 ´ 2xy(2y) sin y 2 = 2x cos y 2 ´ 4xy 2 sin y 2 ,
fyx (x, y) = (x2 cos y 2 ´ yx2 (2y) sin y 2 )x = (x2 cos y 2 ´ 2x2 y 2 sin y 2 )x
= 2x cos y 2 ´ 4xy 2 sin y 2 = fxy (x, y) .
The following two theorems concern the C k -regularity of inverse functions and implicit
functions.
Theorem 6.85. Let D Ď Rn be open, f : D Ñ Rn be injective and be of class C k . If f ´1 ,
the inverse function of f , exists and is differentiable in f (D), then f ´1 is of class C k .
Proof. Let y0 P f (D). Then y0 = f (x0 ) for some x0 P D. Since f is differentiable at x0 and
f ´1 is differentiable at y0 , by the chain rule we must have
In = [D(f ˝ f ´1 )](y0 ) = [Df ](x0 )[Df ´1 ](y0 ) ,
where In is the n ˆ n identity matrix. Therefore, [Df ](x0 ) is invertible, and the inverse
function theorem implies that f ´1 is of class C 1 (in a neighborhood of y0 ).
We note that the map g : GL(n) Ñ GL(n) given by g(L) = L´1 is infinitely many times
differentiable; thus using the identity (from the inverse function theorem)
(
)´1 (
)
(Df ´1 )(y) = (Df )(x)
= g ˝ (Df ) ˝ f ´1 (y) ,
by the chain rule we find that if f P C k , then Df ´1 P C k´1 which is the same as saying
that f ´1 P C k .
˝
Theorem 6.86. Let D Ď Rn ˆ Rm be open, and F : D Ñ Rm be a function of class C k .
Suppose that for some open set U Ď Rm and some differentiable function f : U Ñ Rm ,
U ˆ f (U) Ď D and F (x, f (x)) = 0 for all x P U. Then f is of class C k .
Proof. (Not yet finished!!!)
˝
§6.7 Taylor’s Theorem
243
6.7 Taylor’s Theorem
Recall the Taylor theorem for functions of one variable that if f : (a, b) Ñ R be of class
C k+1 for some k P N and c P (a, b), then for all x P (a, b), there exists d in between c and x
such that
f (x) =
k
ÿ
f (j) (c)
j=0
j!
(x ´ c)j +
f (k+1) (d)
(x ´ c)k+1 ,
(k + 1)!
where f (j) (c) denotes the j-th derivative of f at c. In this section, we extend this result to
functions of several variables.
Theorem 6.87 (Taylor). Let U Ď Rn be open, and f : U Ñ R be (k+1)-times differentiable.
Suppose that x, a P U and the line segment joining x and a lies in U. Then there exists a
point c on the line segment joining x and a such that
f (x) ´ f (a) =
k
ÿ
1
j copies of x ´ a
hkkkkkkkkkikkkkkkkkkj
(D f )(a)(x ´ a, ¨ ¨ ¨ , x ´ a)
j
j!
1
+
(Dk+1 f )(c)(x
´ a, ¨ ¨ ¨ , x ´ a ) .
looooooooomooooooooon
(k + 1)!
j=1
(6.7.1)
(k + 1) copies of x ´ a
(
)
Proof. Let g(t) = f (1 ´ t)a + tx . Since xa Ď U and U is open, there exists δ ą 0 such
that (1 ´ t)a + tx P U for all t P (´δ, 1 + δ). By the chain rule, for t P (´δ, 1 + δ),
n
ÿ
(
)
)
Bf (
(1 ´ t)a + tx (xi ´ ai ) ;
g (t) = (Df ) (1 ´ t)a + tx (x ´ a) =
Bxi
i=1
1
thus for t P (´δ, 1 + δ), Proposition 6.74 shows that
2
g (t) =
)
(
)
B2f (
(1 ´ t)a + tx (xi ´ ai )(xj ´ aj ) = (D2 f ) (1 ´ t)a + tx (x ´ a, x ´ a).
Bxj Bxi
i,j=1
n
ÿ
By induction, we conclude that
(
)
g (j) (t) = (Dj f ) (1 ´ t)a + tx (x
´ a, ¨ ¨ ¨ , x ´ a) .
looooooooomooooooooon
j copies of x ´ a
By the fact that f is (k + 1)-times differentiable, g : (´δ, 1 + δ) Ñ R is (k + 1)-times
differentiable as well. Theorem 4.68 then implies that for some t0 P (0, 1),
g(1) ´ g(0) =
k
ÿ
g (j) (0)
j=1
j!
Letting c = (1 ´ t0 )a + t0 x, (6.7.2) implies (6.7.1).
+
g (k+1) (t0 )
.
(k + 1)!
(6.7.2)
˝
244
CHAPTER 6. Differentiation
Definition 6.88. Let U Ď Rn be open, and f : U Ñ R be k-times differentiable. The k-th
degree Taylor polynomial for f centered at a is the polynomial
k
ÿ
1
j!
j=0
(Dj f )(a)(x
´ a, ¨ ¨ ¨ , x ´ a).
looooooooomooooooooon
j copies x ´ a
Corollary 6.89. Let U Ď Rn be open, f : U Ñ R be (k + 1)-times differentiable, and define
the remainder
Rk (a, h) = f (a + h) ´
k
ÿ
1
j!
j=0
(Dj f )(a)(h, ¨ ¨ ¨ , h).
Rk (a, h)
= 0, or in notation, Rk (a, h) = O(}h}Rk n ) as h Ñ 0.
hÑ0 }h}k n
R
Then lim
Example 6.90. Let f (x, y) = ex cos y. Compute the fourth degree Taylor polynomial for
f centered at (0, 0).
Solution: We compute the zeroth, the first, the second, the third and the fourth mixed
derivatives of f at (0, 0) as follows:
f (0, 0) = 1 ,
fx (0, 0) = 1 ,
fy (0, 0) = 0 ,
fxx (0, 0) = 1 ,
fxy (0, 0) = fyx (0, 0) = 0 ,
fyy (0, 0) = ´1 ,
fxxx (0, 0) = 1 ,
fxxy (0, 0) = fxyx (0, 0) = fyxx (0, 0) = 0 ,
fyyy (0, 0) = 0 ,
fyyx (0, 0) = fyxy (0, 0) = fxyy (0, 0) = ´1 ,
and
fxxxx (0, 0) = 1 ,
fyyyy (0, 0) = 1 ,
fxxxy (0, 0) = fxxyx (0, 0) = fxyxx (0, 0) = fyxxx (0, 0) = 0 ,
fxyyy (0, 0) = fyxyy (0, 0) = fyyxy (0, 0) = fyyyx (0, 0) = 0 ,
fxxyy (0, 0) = fxyxy (0, 0) = fxyyx (0, 0) = fyxxy (0, 0)
= fyxyx (0, 0) = fyyxx (0, 0) = ´1 .
§6.8 Maxima and Minima
245
Then the fourth degree Taylor polynomial for f centered at (0, 0) is
]
1[
f (0, 0) + fx (0, 0)x + fy (0, 0)y + fxx (0, 0)x2 + 2fxy (0, 0)xy + fyy (0, 0)y 2
2
]
1[
+ fxxx (0, 0)x3 + 3fxxy (0, 0)x2 y + 3fxyy (0, 0)xy 2 + fyyy (0, 0)y 3
6
1[
+
fxxxx (0, 0)x4 + 4fxxxy (0, 0)x3 + 6fxxyy (0, 0)x2 y 2
24
]
3
4
+ 4fxyyy (0, 0)xy + fyyyy (0, 0)y
) 1(
)
)
1(
1( 4
= 1 + x + x2 ´ y 2 + x3 ´ 3xy 2 +
x ´ 6x2 y 2 + y 4 .
2
6
24
Observing that using the Taylor expansions
1
1
1
e x = 1 + x + x2 + x3 + x4 + ¨ ¨ ¨
2
6
24
and
1
1
cos y = 1 ´ y 2 + y 4 + ¨ ¨ ¨ ,
2
24
we can “formally” compute ex cos y by multiplying the two “polynomials” above and obtain
that
ex cos y “=” 1 + x +
) (1 4 1 2 2
) (1
1
1 )
1( 2
x ´ x y + y 2 + h.o.t. ;
x ´ y 2 + x3 ´ xy 2 +
2
6
2
24
4
24
where h.o.t. stands for the higher order terms which are terms with fifth or higher degree.
Definition 6.91. Let U Ď Rn be open. A function f : U Ñ R is said to be real analytic
8 1
ř
at a P U if f (x) =
(Dk f )(a)(x ´ a, ¨ ¨ ¨ , x ´ a) in a neighborhood of a.
k=0 k!
Example 6.92. Let f : R Ñ R be defined by
$
(
)
& exp ´ 1
if x ą 0 ,
|x|2
f (x) =
%
0
if x ă 0 .
Then f is of class C 8 , and f (k) (0) = 0 for all k P N. Therefore, f is not real analytic at 0.
6.8 Maxima and Minima
Definition 6.93. Let U Ď Rn be open, and f : U Ñ R.
1. If there is a neighborhood of x0 P U such that f (x0 ) is a maximum in this neighborhood, then x0 is called a local maximum point of f .
246
CHAPTER 6. Differentiation
2. If there is a neighborhood of x0 P U such that f (x0 ) is a minimum in this neighborhood,
then x0 is called a local minimum point of f .
3. A point is called an extreme point of f if it is either a local maximum point or a
local minimum point of f .
4. A point x0 is a critical point of f if f is differentiable at x0 and (Df )(x0 ) = 0; that
is, (Df )(x0 ) P B(Rn , R) is the trivial map (which sends every vector in Rn to zero
vector).
5. A point x0 is a saddle point of f if x0 is a critical point of f but not an extreme
point of f .
Theorem 6.94. Let U Ď Rn be open, f : U Ñ R be differentiable, and x0 P U is an extreme
point of f . Then x0 is a critical point of f .
Proof. Suppose the contrary that the linear map (Df )(x0 ) : Rn Ñ R is not the zero map;
that is, there exists u P Rn , u ‰ 0, such that (Df )(x0 )(u) = c ‰ 0 for some constant c P R.
W.L.O.G, we can assume that }u}Rn = 1 and c ą 0 (for otherwise change u to ´u). By the
differentiability of f ,
ˇ
ˇ c
D δ ą 0 Q ˇf (x0 + h) ´ f (x0 ) ´ (Df )(x0 )(h)ˇ ď }h}Rn
2
whenever }h}Rn ă δ .
Then for any 0 ă λ ă δ,
ˇ
ˇ
ˇf (x0 ˘ λu) ´ f (x0 ) ¯ λ(Df )(x0 )(u)ˇ ď λc .
2
Therefore, ´
λc
λc
ď f (x0 ˘ λu) ´ f (x0 ) ¯ λc ď
which further implies that
2
2
f (x0 ) ď f (x0 + λu) ´
λc
λc
ă f (x0 + λu) and f (x0 ) ě f (x0 ´ λu) +
ą f (x0 ´ λu)
2
2
for all λ ą 0 small enough. As a consequence, x0 cannot be a local extreme point of f , a
contradiction.
˝
Definition 6.95. If f : U Ñ R is of class C 2 , the Hessian of f at x0 is the bilinear
function Hx0 (f ) : Rn ˆ Rn Ñ R given by
Hx0 (f )(u, v) = (D2 f )(x0 )(u, v)
@ u, v P Rn .
§6.8 Maxima and Minima
247
The matrix representation of Hx0 (f )(¨, ¨) is given by
 2
B f
¨¨¨
 Bx2 (x0 )
1

[
] 
..
...
Hx0 (f ) = 
.


 B2f
(x0 ) ¨ ¨ ¨

B2f
(x0 )
Bxn Bx1

..
.
B2f
(x0 )
Bx2n
Bx1 Bxn





[
]
[
]
in the sense that Hx0 (f )(u, v) = [u]T Hx0 (f ) [v] = [v]T Hx0 (f ) [u].
positive definite
if
negative definite
ą
positive semi-definite
ě
B(u, u) 0 for all u ‰ 0, and is called
if B(u, u) 0 for all u P Rn .
ă
negative semi-definite
ď
Definition 6.96. A bilinear form B : Rn ˆ Rn Ñ R is called
Theorem 6.97. Let U Ď Rn be open, and f : U Ñ R be a function of class C 2 .
1. If x0 is a critical point of f such that the Hessian Hx0 (f ) is
has a local
negative
definite, then f
positive
maximum
point at x0 .
minimum
2. If f has a local
maximum
negative
point at x0 , then Hx0 (f ) is
semi-definite.
minimum
positive
Proof. 1. Suppose that Hx0 (f ) is negative definite.
Claim: There exists 0 ă λ ă 8 such that
@ u P Rn .
Hx0 (f )(u, u) ď ´λ}u}2Rn
(6.8.1)
Proof of claim: Since Hx0 (f )(u, u), viewed as a function of u, is continuous, by Theorem
4.21 λ = ´ max Hx0 (f )(u, u) exists and is positive. Then for all u P Rn with u ‰ 0,
}u}Rn =1
( u
Hx0 (f )
,
}u}Rn
u )
ď ´λ
}u}Rn
@ u P Rn , u ‰ 0 .
The inequality (6.8.1) follows from that the Hessian Hx0 (f ) is bilinear.
Since f P C 2 , there exists δ ą 0 such that D(x0 , δ) Ď U and
› 2
›
λ
›(D f )(x) ´ (D2 f )(x0 )› n
ď
B(R ,B(Rn ,R))
2
@ x P D(x0 , δ).
(6.8.2)
248
CHAPTER 6. Differentiation
Now since x0 is a critical point of f , (Df )(x0 ) = 0. As a consequence, by Taylor’s
theorem (Theorem 6.87), for any x P D(x0 , δ), we can find c = c(x) P xx0 such that
1
f (x) = f (x0 ) + (Df )(x0 )(x ´ x0 ) + (D2f )(c)(x ´ x0 , x ´ x0 )
2
]
1 2
1[
= f (x0 ) + (D f )(x0 )(x ´ x0 , x ´ x0 ) + (D2f )(c) ´ (D2f )(x0 ) (x ´ x0 , x ´ x0 )
2
2
ˇ[
ˇ
]
1
1
ˇ
ˇ
ď f (x0 ) ´ λ}x ´ x0 }2Rn + ˇ (D2f )(c) ´ (D2f )(x0 ) (x ´ x0 , x ´ x0 )ˇ
2
2
›
1›
1
ď f (x0 ) ´ λ}x ´ x0 }2Rn + ›(D2f )(c) ´ (D2f )(x0 )›B(Rn ;B(Rn ,R)) }x ´ x0 }2Rn .
2
2
Note that c = c(x) P D(x0 , δ) if x P D(x0 , δ); thus (6.8.2) implies that if x P D(x0 , δ),
1λ
λ
λ
}x ´ x0 }2Rn ď f (x0 ) ´ }x ´ x0 }2Rn .
f (x) ď f (x0 ) ´ }x ´ x0 }2Rn +
2
22
4
As a consequence, for all x P D(x0 , δ), f (x) ď f (x0 ) which validates that x0 is a local
maximum point of f .
2. Suppose the contrary that f has a local maximum point at x0 but for some u P Rn ,
Hx0 (f )(u, u) ą 0 .
W.L.O.G, we can assume that }u}Rn = 1. By Theorem 6.94, (Df )(x0 ) = 0; thus
Taylor’s Theorem implies that
[
]
1
1
f (x) = f (x0 ) + (D2 f )(c)(x ´ x0 , x ´ x0 ) = f (x0 ) + (x ´ x0 )T Hc (f ) (x ´ x0 ) .
2
2
Since x0 is a local maximum point of f , there exists δ ą 0 such that f (x) ď f (x0 ) for
all x P D(x0 , δ). As a consequence, for some c = c(x) P xx0 ,
[
]
[
]
(x ´ x0 )T Hc (f ) (x ´ x0 ) = 2 f (x) ´ f (x0 ) ď 0
@ x P D(x0 , δ) .
Let 0 ă t ă δ and x = x0 + tu. Then x P D(x0 , δ); thus
Hc (f )(u, u) ď 0
@ t P (0, δ) .
We note that c depends on t, and c Ñ x0 as t Ñ 0. Therefore, by the continuity of
H‚ (f ), passing t Ñ 0 in the inequality above we find that
Hx0 (f )(u, u) = lim Hc (f )(u, u) ď 0
tÑ0
which is a contradiction.
˝
§6.9 Exercises
249
Remark 6.98. Inequality (6.8.1) can also be obtained by studying the largest eigenvalue of
Hx0 (f ). Note that since f P C 2 , Hx0 (f ) is symmetric by Theorem 6.81. As a consequence,
there exists an orthonormal matrix O P GL(n) whose columns are (real) eigenvectors of
Hx0 (f )
]
[
Hx0 (f ) = OΛOT ,
where Λ is a diagonal matrix whose diagonal entries are eigenvalues of Hx0 (f ). Note that
by the orthonormality of O, every vector u P Rn satisfies }OT u}Rn = }u}Rn . Therefore,
Hx0 (f )(u, u) = uT OΛOT u = (OT u)T Λ(OT u) ď λ}OT u}2Rn = λ}u}2Rn ,
where λ is the largest eigenvalue of Λ.
[
]
Remark 6.99 (Sylvester’s criterion). To justify if a matrix Hx0 (f ) is positive/negative
definite, let


B2f
 Bx21

 .
∆k = 
 ..

 B2f
Bx1 Bxk
Then Hx0 (f ) is
¨¨¨
...
¨¨¨
B2f
Bxk Bx1 


 (x0 ) .


2
B f 
..
.
Bx2k
positive
det(∆k ) ą 0
definite if and only if
for all k = 1, ¨ ¨ ¨ , n.
negative
(´1)k det(∆k ) ą 0
6.9 Exercises
n
m
Problem 6.1. Let tTk u8
k=1 Ď B(R , R ) be a sequence of bounded linear maps from
Rn Ñ Rm . Prove that the following three statements are equivalent:
1. tTk u8
k=1 converges pointwise (to a function T );
2. lim }Tk ´ Tℓ }B(Rn ,Rm ) = 0;
k,ℓÑ8
n
3. tTk u8
k=1 converges uniformly (to T ) on every compact subsets of R .
Problem 6.2. Let P((0, 1)) Ď Cb ((0, 1); R) be the collection of all polynomials defined on
(0, 1).
1. Show that the operator
d
: P((0, 1)) Ñ Cb ((0, 1)) is linear.
dx
250
CHAPTER 6. Differentiation
2. Show that
that
(
)
(
)
d
: P((0, 1)), } ¨ }8 Ñ Cb ((0, 1)), } ¨ }8 is unbounded; that is, show
dx
sup }p1 }8 = 8 .
}p}8 =1
§6.2 Definition of Derivatives and the Jacobian Matrices
Problem 6.3. Show that if f : C Ñ R is differentiable at z0 , then (Df )(z0 ) = 0.
Hint: Show that B(C, R) = t0u.
Problem 6.4. Let f : R2 Ñ R be given by
"
0 if xy = 0 ,
f (x, y) =
1 if xy ‰ 0 .
Compute
Bf
Bf
(x, y) and
(x, y).
Bx
By
Problem 6.5. Investigate the differentiability of
$
xy
&a
if (x, y) ‰ (0, 0) ,
x2 + y 2
f (x, y) =
%
0
if (x, y) = (0, 0) .
Problem 6.6. Investigate the differentiability of
# xy
if x + y 2 ‰ 0 ,
2
x
+
y
f (x, y) =
0
if x + y 2 = 0 .
Problem 6.7. Define f : R2 Ñ R by
$
& (x2 + y 2 ) sin a 1
if (x, y) ‰ (0, 0) ,
x2 + y 2
f (x, y) =
%
0
if (x, y) = (0, 0) .
Discuss the differentiability of f . Find (∇f )(x, y) at points of differentiability.
Problem 6.8. Let r ą 0 and α ą 1. Suppose that f : D(0, r) Ñ R satisfies |f (x)| ď }x}α
for all x P D(0, r). Show that f is differentiable at 0. What happens if α = 1?
Problem 6.9. Suppose that f, g : R Ñ Rm are differentiable at a and there is a δ ą 0 such
that g(x) ‰ 0 for all 0 ă |x ´ a| ă δ. If f (a) = g(a) = 0 and (Dg)(a) ‰ 0, show that
}(Df )(a)}
}f (x)}
=
.
xÑa }g(x)}
}(Dg)(a)}
lim
§6.9 Exercises
251
Problem 6.10. Consider the map δ defined in Problem 5.11 in Chapter 5; that is, δ :
C ([0, 1]; R) Ñ R be defined by δ(f ) = f (0). Show that δ is differentiable. Find (Dδ)(f )
(for f P C ([0, 1]; R)).
Problem 6.11. Let f : GL(n) Ñ GL(n) be given by f (L) = L´1 . In class we have shown
that f is continuous on GL(n). Show that f is differentiable at each “point” (or more
precisely, linear map) of GL(n).
Hint: In order to show the differentiability of f at L P GL(n), we need to figure out what
(Df )(L) is. So we need to compute f (L + h) ´ f (L), where h P B(Rn , Rn ) is a “small”
linear map. Compute (L + h)´1 ´ L´1 and make a conjecture what (Df )(L) should be.
Problem 6.12. Let I : C ([0, 1]; R) Ñ R be defined by
ż1
I(f ) =
f (x)2 dx .
0
Show that I is differentiable at every “point” f P C ([0, 1]; R).
Hint: Figure out what (DI)(f ) is by computing I(f + h) ´ I(f ), where h P C ([0, 1]; R) is
a “small” continuous function.
Remark. A map from a space of functions such as C ([0, 1]; R) to a scalar field such as R
or C is usually called a functional. The derivative of a functional I is usually denoted by
δI instead of DI.
§6.3 Conditions for Differentiability
§6.4 Properties of Differentiable Functions
Problem 6.13. Let U Ď Rn be open, and f : U Ñ R. Suppose that the partial derivatives
Bf
Bf
,¨¨¨ ,
are bounded on U; that is, there exists a real number M ą 0 such that
Bx1
Bxn
ˇ
ˇ
ˇ Bf
ˇ
(x)ˇ ď M
ˇ
@ x P U and j = 1, ¨ ¨ ¨ , n .
Bxj
Show that f is continuous on U.
Hint: Mimic the proof of Theorem 6.28.
Problem 6.14. Let U Ď Rn be open, and f : U Ñ R. Show that f is differentiable at a P U
if and only if there exists a vector-valued function ε : U Ñ Rn such that
f (x) ´ f (a) ´
n
ÿ
Bf
j=1
and ε(x) Ñ 0 as x Ñ a.
Bxj
(a)(xj ´ aj ) = ε(x) ¨ (x ´ a)
252
CHAPTER 6. Differentiation
Problem 6.15. Verify the chain rule for
u(x, y, z) = xey ,
v(x, y, z) = yz sin x
and
f (u, v) = u2 + v sin u
(
)
with h(x, y, z) = f u(x, y, z), v(x, y, z) .
Problem 6.16. Let (r, θ, φ) be the spherical coordinate of R3 so that
x = r cos θ sin φ, y = r sin θ sin φ, z = r cos φ .
1. Find the Jacobian matrices of the map (x, y, z) ÞÑ (r, θ, φ) and the map (r, θ, φ) ÞÑ
(x, y, z).
2. Suppose that f (x, y, z) is a differential function in R3 . Express |∇f |2 in terms of the
spherical coordinate.
Problem 6.17. Let U Ď Rn be open and convex, and f : U Ñ Rm be differentiable on U.
Show that for each a, b P U and vector v P Rm , there exists c on the line segment joining a
and b such that
[
]
v ¨ f (b) ´ f (a) = v ¨ D(f )(c)(b ´ a) .
Problem 6.18. Let U Ď Rn be open, and for each 1 ď i, j ď n, aij : U Ñ R be differentiable
functions. Define A = [aij ] and J = det(A). Show that
(
BJ
BA )
= tr Adj(A)
Bxk
Bxk
@1 ď k ď n,
where for a square matrix M = [mij ], tr(M ) denotes the trace of M , Adj(M ) denotes the
adjoint matrix of M , and
Bmij
BM
denotes the matrix whose (i, j)-th entry is given by
.
Bxk
Bxk
Hint: Show that
ˇ Ba
ˇ ˇ
ˇ
ˇ
ˇ
Ba1n ˇ
ˇ 11 a ¨ ¨ ¨ a ˇ ˇ a Ba12 a ¨ ¨ ¨ a ˇ
ˇa ¨ ¨ ¨ a
ˇ Bx
ˇ 11 Bx
ˇ 11
ˇ
12
1n ˇ
13
1n ˇ
(n´1)1
Bxk ˇ
k
ˇ k
ˇ ˇ
ˇ
ˇ
ˇ Ba21
ˇ ˇ
ˇ
ˇ
ˇ
Ba22
ˇ
ˇ a21 ¨ ¨ ¨ a(n´1)2 Ba2n ˇ
a22 ¨ ¨ ¨ a2n ˇˇ ˇˇ a21
a23 ¨ ¨ ¨ a2n ˇˇ
BJ
ˇ
ˇ
Bxk
Bxk ˇ
= ˇ Bxk
ˇ+ˇ .
ˇ + ¨¨¨ + ˇ .
.. ˇˇ
.
.
.
Bxk
.. ˇ ˇ ..
..
ˇ ..
ˇ
ˇ ..
. ˇ
ˇ
ˇ ˇ
ˇ
ˇ
ˇ Ban1
ˇ ˇ
ˇ
ˇ
Ban1 ˇ
Ban2
an2 ¨ ¨ ¨ ann ˇ ˇan1
an3 ¨ ¨ ¨ ann ˇ
ˇ
ˇan1 ¨ ¨ ¨ a(n´1)n
ˇ
Bxk
Bxk
Bxk
§6.9 Exercises
253
and rewrite this identity in the form which is asked to prove. You can also show the differentiation formula by applying the chain rule to the composite function F ˝ g of maps
(
)
2
2
g : U Ñ Rn and F : Rn Ñ R defined by g(x) = a11 (x), a12 (x), ¨ ¨ ¨ , ann (x) and
(
)
BF
F (a11 , ¨ ¨ ¨ , ann ) = det [aij ] . Check first what
is.
Baij
Problem 6.19. Let ψ : Rn Ñ Rn be a differentiable function such that
B 2 ψk
B ( Bψk )
=
Bxi Bxj
Bxi Bxj
exists and is continuous in Rn for each 1 ď i, j, k ď n. Suppose that (Dψ)(x) P GL(n) for all
x P Rn , and define A = (Dψ)´1 (or in terms of their matrix representation, [A] = [Dψ]´1 ).
Let ψ = (ψ1 , ¨ ¨ ¨ , ψn ) and [A] = [aij ].
n
ř
1. Show that
k=1
aik
n Bψ
ř
Bψk
i
=
akj = δij , where δij is the Kronecker delta; that is,
Bxj
Bx
k
k=1
δij = 1 if i = j or δij = 0 if i ‰ j.
2. Show that for each 1 ď i, j, k ď n, aij : Rn Ñ R is differentiable, and
n
ÿ
Baij
B 2 ψr
=´
air
asj .
Bxk
Bxk Bxs
r,s=1
Problem 6.20. Let U Ď Rn be open and connected, and f : U Ñ R be a function such
that
Bf
(x) = 0 for all x P U. Show that f is constant in U.
Bxj
§6.5 Directional Derivatives and Gradient Vectors
Problem 6.21. Let
f (x, y) =
$ 3
& xy
if (x, y) ‰ (0, 0) ,
%
if (x, y) = (0, 0) .
x4 + y 2
0
Show that the directional derivative of f at the origin exists in all directions u, and
( Bf
)
Bf
(Du f )(0, 0) =
(0, 0), (0, 0) ¨ u .
Bx
By
§6.6 Higher Derivatives of Functions
π
2
Problem 6.22. Let f (x, y, z) = (x2 +1) cos(yz), and a = (0, , 1), u = (1, 0, 0), v = (0, 1, 0)
and w = (2, 0, 1).
1. Compute (Df )(a)(u).
254
CHAPTER 6. Differentiation
2. Compute (D2 f )(a)(v)(u).
3. Compute (D3 f )(a)(w)(v)(u).
Problem 6.23. 1. If f : A Ď Rn Ñ Rm and g : B Ď Rm Ñ Rℓ are twice differentiable and
f (A) Ď B, then for x0 P A, u, v P Rn , show that
D2 (g ˝ f )(x0 )(u, v)
(
)
(
)
= (D2 g)(f (x0 )) (Df )(x0 )(u), Df (x0 )(v) + (Dg)(f (x0 )) (D2 f )(x0 )(u, v) .
2. If p : Rn Ñ Rm is a linear map plus some constant; that is, p(x) = Lx + c for some
L P B(Rn , Rm ), and f : A Ď Rm Ñ Rs is k-times differentiable, prove that
(
)(
)
Dk (f ˝ p)(x0 )(u(1) , ¨ ¨ ¨ , u(k) ) = (Dk f ) p(x0 ) (Dp)(x0 )(u(1) ), ¨ ¨ ¨ , (Dp)(x0 )(u(k) .
§6.7 Taylor’s Theorem
Problem 6.24. Let f (x, y) be a real-valued function on R2 . Suppose that f is of class C 1
(that is, all first partial derivatives are continuous on R2 ) and
Show that
B2f
B2f
B2f
exists and
=
.
ByBx
BxBy
ByBx
B2f
exists and is continuous.
BxBy
Hint: Mimic the proof of Theorem 6.81.
Problem 6.25. Let f : Rn Ñ Rm be differentiable, and Df is a constant map in B(Rn , Rm );
that is, (Df )(x1 )(u) = (Df )(x2 )(u) for all x1 , x2 P Rn and u P Rn . Show that f is a linear
term plus a constant and that the linear part of f is the constant value of Df .
Problem 6.26. Let U Ď Rn be open, and f : U Ñ Rn be of class C 2 such that Df : U Ñ
B(Rn , Rn ) satisfies (Df )(x) P GL(n) for all x P U. Define J = det([Df ]) and A = [Df ]´1 .
With aij denoting the (i, j)-th entry of A, show the Piola identity
n
ÿ
B
i=1
Bxi
(Jaij )(x) = 0
@ 1 ď j ď n and x P U .
(6.9.1)
Is f continuous at (0, 0)? Is f differentiable at (0, 0)?
Problem 6.27. Let U Ď Rn be open, and f : U Ñ R be of class C k and (Dj f )(x0 ) = 0 for
j = 1, ¨ ¨ ¨ , k ´ 1, but (Dk f )(x0 )(u, u, ¨ ¨ ¨ , u) ă 0 for all u P Rn , u ‰ 0. Show that f has a
local maximum at x0 ; that is, D δ ą 0 such that
f (x) ď f (x0 )
@ x P D(x0 , δ) .
§6.9 Exercises
255
§6.8 Maxima and Minima
Problem 6.28. Let f (x, y) = x3 + x ´ 4xy + 2y 2 ,
1. Find all critical points of f .
(
(
))
2. Find the corresponding quadratic from Q(x, y, h, k) or (D2 f (x, y) (h, k), (h, k) at
these critical points, and determine which of them is positive definite.
3. Find all relative extrema and saddle points.
4. Find the maximal value of f on the set
ˇ
␣
(
A = (x, y) ˇ 0 ď x ď 1, 0 ď y ď 1, x + y ď 1 .
Problem 6.29. Let f : R2 Ñ R be given by
$
6 2
& x2 + y 2 ´ 2x2 y ´ 4x y
if (x, y) ‰ (0, 0) ,
(x4 + y 2 )2
f (x, y) =
%
0
if (x, y) = (0, 0) .
1. Show that f is continuous (at (0, 0)) by showing that for all (x, y) P R2 ,
4x4 y 2 ď (x4 + y 2 )2 .
2. For 0 ď θ ď 2π, ´8 ă t ă 8, define
gθ (t) = f (t cos θ, t sin θ) .
Show that each gθ has a strict local minimum at t = 0. In other words, the restriction
of f to each straight line through (0, 0) has a strict local minimum at (0, 0).
3. Show that (0, 0) is not a local minimum for f .
Problem 6.30 (True or False). Determine whether the following statements are true or
false. If it is true, prove it. Otherwise, give a counter-example.
1.
2.
256
CHAPTER 6. Differentiation
3.
4. Let U Ď Rn be open. Then f : U Ñ R is differentiable at a P U if and only if each
directional derivative (Du f )(a) exists and
(Du f )(a) =
n
ÿ
Bf
j=1
Bxj
(a)uj =
( Bf
)
Bf
(a), ¨ ¨ ¨ ,
(a) ¨ u
Bx1
Bxn
where u = (u1 , ¨ ¨ ¨ , un ) is a unit vector.
5. Let f : R2 Ñ R be of class C 1 . Assume that all second order partial derivatives of f
exist, then f is second times differentiable in R2 .
6. Let f be a function defined on R2 , and A be an invertible matrix. Define y = Ax for
x P Rn . Then f (y) is differentiable if and only if f (Ax) is differentiable as a function
of x.
7. Let f : [a, b] Ñ R2 be continuous and be differentiable on (a, b). If f (a) = f (b), then
there exists some c P (a, b) such that f 1 (c) = 0.
Chapter 7
The Inverse and the Implicit Function
Theorems
7.1 The Inverse Function Theorem（反函數定理）
反函數定理是用來探討一個函數的反函數是否存在的問題。只要一個函數不是一對一的，
一般來說都不能定義其反函數，例如三角函數中，正弦、餘弦及正切函數都是周期函
數，所以全域的反函數不存在。但是我們也知道有所謂的反三角函數 sin´1 （或 arcsin）,
cos´1 （或 arctan）及 tan´1 （或 arctan），這是因為我們限制了原三角函數的定義域使其
在新的定義域上是一對一的（因此反函數存在）。因此，要討論一個定在某一個（大範圍
的）定義域的函數的反函數，常常我們最多只能說反函數只在某一小塊區域上存在。
如何知道一個函數在一小塊區域上的反函數存在，我們首先該問的是在定義域是一維
（或是指單變數函數）的情況下發生什麼事？由一維的反函數定理 (Theorem 4.72) 我們知
道首先應該要保留的條件是類似於微分不為零的這個條件。但是在多變數函數之下，微
分不為零的條件該怎麼呈現，這是第一個問題。而當我們觀察 (4.6.1)，應該可以猜出在
多變數版本裡面所該對應到的條件，即是 (Df )(x) 這個 bounded linear map 的可逆性。
另外，假設 f P C 1 ，那麼由 Theorem 6.8 我們知道在一個點 x0 如果 (Df )(x0 ) 可逆的
話，那麼在一個鄰域裡 (Df )(x) 都可逆。所以下面這個反函數定理的條件中只有 (Df ) 在
一個點可逆這個條件，因為我們目前想先知道小區域的反函數存不存在。
Theorem 7.1 (Inverse Function Theorem). Let D Ď Rn be open, x0 P D, f : D Ñ Rn be
of class C 1 , and (Df )(x0 ) be invertible. Then there exist an open neighborhood U of x0 and
an open neighborhood V of f (x0 ) such that
1. f : U Ñ V is one-to-one and onto;
257
258
CHAPTER 7. Inverse and Implicit Function Theorems
2. The inverse function f ´1 : V Ñ U is of class C 1 ;
(
)´1
3. If x = f ´1 (y), then (Df ´1 )(y) = (Df )(x) ;
4. If f is of class C r for some r ą 1, so is f ´1 .
Proof. Assume that A = (Df )(x0 ). Then }A´1 }B(Rn ,Rn ) ‰ 0. Choose λ ą 0 such that
2λ}A´1 }B(Rn ,Rn ) = 1. Since f P C 1 , there exists δ ą 0 such that
›
›
›
›
›(Df )(x) ´ A› n n = ›(Df )(x) ´ (Df )(x0 )› n n ă λ whenever x P D(x0 , δ) X D .
B(R ,R )
B(R ,R )
By choosing δ even smaller if necessary, we can assume that D(x0 , δ) Ď D. Let U = D(x0 , δ).
Claim: f : U Ñ Rn is one-to-one (hence f : U Ñ f (U) is one-to-one and onto).
(
)
Proof of claim: For each y P Rn , define φy (x) = x + A´1 y ´ f (x) (and we note that every
fixed-point of φy corresponds to a solution to f (x) = y). Then
(
)
(Dφy )(x) = Id ´ A´1 (Df )(x) = A´1 A ´ (Df )(x) ,
where Id is the identity map on Rn . Therefore,
›
›
›
›
›(Dφy )(x)› n n ď }A´1 }B(Rn ,Rn ) ›A ´ (Df )(x)› n n ă 1
B(R ,R )
B(R ,R )
2
@ x P D(x0 , δ) .
By the mean value theorem (Theorem 6.49),
›
›
›φy (x1 ) ´ φy (x2 )› n ď 1 }x1 ´ x2 }Rn
R
2
@ x1 , x2 P D(x0 , δ), x1 ‰ x2 ;
(7.1.1)
thus at most one x satisfies φy (x) = x; that is, φy has at most one fixed-point. As a
consequence, f : D(x0 , δ) Ñ Rn is one-to-one.
Claim: The set V = f (U) is open.
Proof of claim: Let b P V. Then there is a P U with f (a) = b. Choose r ą 0 such that
D(a, r) Ď U. We observe that if y P D(b, λr), then
(
)
r
}φy (a) ´ a}Rn ď }A´1 y ´ f (a) }Rn ď }A´1 }B(Rn ,Rn ) }y ´ b}Rn ă λ}A´1 }B(Rn ,Rn ) r = ;
2
thus if y P D(b, λr) and x P D(a, r),
›
›
1
r
}φy (x) ´ a}Rn ď ›φy (x) ´ φy (a)›Rn + }φy (a) ´ a}Rn ă }x ´ a}Rn + ă r .
2
2
Therefore, if y P D(b, λr), then φy : D(a, r) Ñ D(a, r). By the continuity of φy ,
φy : D(a, r) Ñ D(a, r) .
§7.1 The Inverse Function Theorem
259
On the other hand, (7.1.1) implies that φy is a contraction mapping if y P D(b, λr); thus by
the contraction mapping principle 5.89 φy has a unique fixed-point x P D(a, r). As a result,
every y P D(b, λr) corresponds to a unique x P D(a, r) such that φy (x) = x or equivalently,
f (x) = y. Therefore,
(
)
D(b, λr) Ď f D(a, r) Ď f (U) = V .
Next we show that f ´1 : V Ñ U is differentiable. We note that if x P D(x0 , δ),
}(Df )(x0 ) ´ (Df )(x)}B(Rn ,Rn ) }A´1 }B(Rn ,Rn ) ă λ}A´1 }B(Rn ,Rn ) =
1
;
2
thus Theorem 6.8 implies that (Df )(x) is invertible if x P D(x0 , δ).
Let b P V and k P Rn such that b + k P V. Then there exists a unique a P U and
h = h(k) P Rn such that a + h P U, b = f (a) and b + k = f (a + h). By the mean value
theorem and (7.1.1),
›
›
›φy (a + h) ´ φy (a)› n ă 1 }h}Rn ;
R
2
thus the fact that f (a + h) ´ f (a) = k implies that
1
}h ´ A´1 k}Rn ă }h}Rn
2
which further shows that
1
1
}h}Rn ď }A´1 k}Rn ď }A´1 }B(Rn ,Rn ) }k}Rn ď
}k}Rn .
2
2λ
(7.1.2)
As a consequence, if k is such that b + k P V,
› ´1
›
(
) ›
(
) ›
›f (b + k) ´ f ´1 (b) ´ (Df )(a) ´1 k › n
›a + h ´ a ´ (Df )(a) ´1 k › n
R
R
=
}k}Rn
}k}Rn
›
›
›k ´ (Df )(a)(h)› n
›(
)´1 ›
R
ď › (Df )(a) ›B(Rn ,Rn )
}k}Rn
›
›
›
›(
)´1 ›
f (a + h) ´ f (a) ´ (Df )(a)(h)›Rn }h}Rn
ď › (Df )(a) ›B(Rn ,Rn )
}h}Rn
}k}Rn
›(
)´1 ›
›
›
› (Df )(a) › n n ›f (a + h) ´ f (a) ´ (Df )(a)(h)›
B(R ,R )
Rn
ď
.
λ
}h}Rn
Using (7.1.2), h Ñ 0 as k Ñ 0; thus passing k Ñ 0 on the left-hand side of the inequality
above, by the differentiability of f we conclude that
› ´1
(
) ›
›f (b + k) ´ f ´1 (b) ´ (Df )(a) ´1 k › n
R
= 0.
lim
kÑ0
n
}k}R
260
CHAPTER 7. Inverse and Implicit Function Theorems
This proves 3.
To see 4, we note that the map g : GL(n) Ñ GL(n) given by g(L) = L´1 is infinitely
many time differentiable; thus using the identity
(
)´1 (
)
(Df ´1 )(y) = (Df )(x)
= g ˝ (Df ) ˝ f ´1 (y) ,
by the chain rule we find that if f P C r , then Df ´1 P C r´1 which is the same as saying that
f ´1 P C r .
˝
Remark 7.2. Since f ´1 : V Ñ U is continuous, for any open subset W of U f (W) =
(f ´1 )´1 (W) is open relative to V, or f (W) = O X V for some open set O Ď Rn . In other
words, if U is an open neighborhood of x0 given by the inverse function theorem, then
f (W) is also open for all open subsets W of U. We call this property as f is a local open
mapping at x0 .
Remark 7.3. Since (Df )(x0 ) P B(Rn , Rn ), the condition that (Df )(x0 ) is invertible can
be replaced by that the determinant of the Jacobian matrix of f at x0 is not zero; that is,
det
([
])
(Df )(x0 ) ‰ 0 .
The determinant of the Jacobian matrix of f at x0 is called the Jacobian of f at x0 . The
Jacobian of f at x sometimes is denoted by Jf (x) or
B(f1 , ¨ ¨ ¨ , fn )
.
B(x1 , ¨ ¨ ¨ , xn )
Example 7.4. Let f : R Ñ R be given by
#
f (x) =
x + 2x2 sin
0
1
x
if x ‰ 0 ,
if x = 0 .
Let 0 P (a, b) for some (small) open interval (a, b). Since f 1 (x) = 1 ´ 2 cos
1
1
+ 4x sin for
x
x
x ‰ 0, f has infinitely many critical points in (a, b), and (for whatever reasons) these critical
points are local maximum points or local minimum points of f which implies that f is not
locally invertible even though we have f 1 (0) = 1 ‰ 0. One cannot apply the inverse function
theorem in this case since f is not C 1 .
Corollary 7.5. Let U Ď Rn be open, f : U Ñ Rn be of class C 1 , and (Df )(x) be invertible
for all x P U. Then f (W) is open for every open set W Ď U.
§7.1 The Inverse Function Theorem
261
在證明了小區域的（local）反函數定理 (Theorem 7.1) 之後，我們接下來要問的是全
域的（global）反函數在什麼條件之下會存在。如果照一維的反函數定理，我們會猜測是
不是只要 (Df )(x) 在整個區域都可逆就能得到在全域的反函數都存在。以下給個反例說
單單在這個條件之下，函數不一定會有一對一的性質。
Example 7.6. Let f : R2 Ñ R2 be given by
f (x, y) = (ex cos y, ex sin y) .
Then
[ x
]
e cos y ´ex sin y
[
]
(Df )(x, y) = x
.
e sin y ex cos y
It is easy to see that the Jacobian of f at any point is not zero (thus (Df )(x) is invertible for
all x P R2 ), and f is not globally one-to-one (thus the inverse of f does not exist globally)
since for example, f (x, y) = f (x, y + 2π).
要再加什麼條件進來才能得到反函數在全域都存在是個不容易的問題。在一維的情
況下，導數是 sign definite 就表示函數在全域是嚴格單調的。在高維度的情況，即使是
(Df )(x) 到處都可逆，仍然有很多情況可能發生（如上例）。下面這個定理（全域的反函
數存在定理），從某種角度來說並沒有真的加了什麼條件以確保全域的反函數存在，只是
多要求了在所考慮的區域邊界上函數是一對一的。這個條件在一維的情況之下是自動成
立的：因為如果一單變數函數的導數是 sign definite，那麼函數在邊界上必定是一對一的
（因為嚴格單調的關係）。
Theorem 7.7 (Global Existence of Inverse Function). Let D Ď Rn be open, f : D Ñ Rn be
of class C 1 , and (Df )(x) be invertible for all x P D. Suppose that K is a connected compact
subset of D, and f : BK Ñ Rn is one-to-one. Then f : K Ñ Rn is one-to-one.
ˇ
␣
(
Proof. Define E = x P K ˇ D y P K, y ‰ x Q f (x) = f (y) . Our goal is to show that E = H.
Claim 1: E is closed.
Proof of claim 1: Suppose the contrary that E is not closed. Then there exists txk u8
k=1 Ď E,
xk Ñ x as k Ñ 8 but x P KzE. Since xk P E, by the definition of E there exists yk P E
such that yk ‰ xk and f (xk ) = f (yk ). By the compactness of K, there exists a convergent
␣ (8
subsequence ykj j=1 of tyk u8
k=1 with limit y P K. Since x R E and f (xkj ) = f (ykj ) Ñ f (y)
as j Ñ 8, we must have x = y; thus ykj Ñ x as j Ñ 8.
262
CHAPTER 7. Inverse and Implicit Function Theorems
Since (Df )(x) is invertible, by the inverse function theorem there exists δ ą 0 such that
␣ (8
␣ (8
f : D(x, δ) Ñ Rn is one-to-one. By the convergence of sequences xkj j=1 and ykj j=1 ,
there exists N ą 0 such that
xkj , ykj P D(x, δ)
@j ě N .
(
( )
This implies that f : D(x, δ) Ñ Rn cannot be one-to-one since xkj ‰ ykj but f xkj =
( ))
f ykj , a contradiction. Therefore, E is closed.
Claim 2: E is open relative to K; that is, for every x P E, there exists an open set U such
that x P U and U X K Ď E.
Proof of claim 2:
1. x P BK X E: By the injectivity of f on BK, there exists x1 P E X int(K), x1 ‰ x,
such that f (x) = f (x1 ). Since (Df )(x) and (Df )(x1 ) are invertible, by the inverse
function theorem there exist open neighborhoods U1 of x and U2 of x1 , as well as open
neighborhoods V1 , V2 of f (x), such that f : U1 Ñ V1 and f : U2 Ñ V2 are both
one-to-one and onto. Since x ‰ x1 , W.L.O.G. we can assume that U2 Ď int(K) and
U1 X U2 = H. Since V1 X V2 is open, the continuity of f implies that f ´1 (V1 X V2 ) =
O X D for some open set O; thus
f : U1 X O X K Ñ V1 X V2 X f (K) is one-to-one ,
f : U2 X O X K Ñ V1 X V2 X f (K) is one-to-one and onto .
s P U X K corresponds to a unique x
r P U2 X O X K
Let U = U1 X O. Then every x
s‰x
r. Therefore, x
s P E, or
such that f (s
x) = f (r
x). Since U1 X U2 = H, we must have x
equivalently, U X K Ď E.
2. x P int(K) X E: It suffices to show that there exists r ą 0 such that B(x, r) Ď E.
Suppose
Now we show that E = H. Since K is connected, E is open relative to K and E is
closed, Remark 3.46 implies that E = K or E = H. Suppose the case that E = K. Let
x P BK Ď E. Then there exists y P E such that y ‰ x and f (x) = f (y). Since f : BK Ñ Rn
(
)
is one-to-one, y R BK. Therefore, we have shown that if E = K, then f (BK) Ď f int(K) .
By Theorem 4.21, the compactness of K implies that f (K) is compact; thus there is
b P Rn such that b R f (K). Consider the function φ : K Ñ R defined by
n
1
1ÿ
φ(x) = }f (x) ´ b}2Rn =
|fi (x) ´ bi |2 .
2
2 j=1
§7.1 The Inverse Function Theorem
263
Then φ is a continuous function on K; thus φ attains its maximum at x0 P K. Since
(
)
f (BK) Ď f int(K) , we can assume that x0 P int(K); thus Theorem 6.94 implies that
(Dφ)(x0 ) = 0. As a consequence,
[
]T [
]
(Df )(x0 ) f (x0 ) ´ b = 0 .
By the choice of b, f (x0 ) ´ b ‰ 0; thus we must have that (Df )(x0 ) is not invertible, a
contradiction.
˝
ˇ
␣
Example 7.8. Let f : R2 Ñ R2 be given as in Example 7.6, and D = (x, y) ˇ x P R, 0 ă
(
y ă 2π . Then f : D Ñ R2 is one-to-one. If K is a compact subset of D, then f : K Ñ R2
is also one-to-one (thus f : BK Ñ R2 must be one-to-one as well).
Corollary 7.9. Let D Ď Rn be a bounded open convex set, and f : D Ñ Rn be of class C 1
such that
s
1. f and Df are continuous on D;
2. the Jacobian det
([
])
s
(Df )(x) ‰ 0 for all x P D;
3. f : BD Ñ Rn is one-to-one.
s Ñ Rn is one-to-one. Moreover, f ´1 : f (D)
s Ñ Rn is continuous, and f ´1 :
Then f : D
f (D) Ñ D is of class C 1 .
Proof. We first claim that there exists a small ε ą 0 such that f : Dε Ñ Rn is one-to-one,
where
ˇ
␣
(
Dε ” x P D ˇ d(x, BD) ă ε .
Assume the contrary that for every k ą 0, there exists xk , yk P D such that
(a) xk ‰ yk ;
(b) d(xk , BΩ) ă
1
1
and d(yk , BΩ) ă ;
k
k
(c) f (xk ) = f (yk ).
8
Since txk u8
k=1 and tyk uk=1 are bounded (due to the boundedness of D), by the Bolzano␣ (8
␣ (8
Weierstrass Theorem (or Corollary 3.29) there exist xkj j=1 and ykj j=1 such that xkj Ñ
s and yk Ñ y P D.
s By (b), x, y P BD; thus the fact that f : BD Ñ Rn is one-to-one
xPD
j
implies that x = y. Therefore, xkj Ñ x and ykj Ñ x as j Ñ 8.
xkj ´ ykj
.
}xkj ´ ykj }Rn
n
Since tuj u8
j=1 is bounded in R , by the Bolzano-Weierstrass
␣ (8
Theorem again there is a convergent subsequence ujℓ ℓ=1 with limit u ‰ 0. Moreover, by
Let uj =
264
CHAPTER 7. Inverse and Implicit Function Theorems
the convexity of D, the mean value theorem implies that for each i = 1, ¨ ¨ ¨ , n, there exists
ciℓ on the line segment joining xkjℓ and ykjℓ such that
( )
0 = fi (xkjℓ ) ´ fi (ykjℓ ) = (Dfi )(ciℓ )(xkjℓ ´ ykjℓ ) = }xkjℓ ´ ykjℓ }Rn (Dfi )(ciℓ ) ujℓ
( )
which by (a) further shows that (Dfi )(ciℓ ) ujℓ = 0 for all i = 1, ¨ ¨ ¨ , n and ℓ P N. Since
ciℓ Ñ x as ℓ Ñ 8, passing ℓ Ñ 8 we conclude that (Dfi )(x)(u) = 0. This holds for each
(
)
i = 1, ¨ ¨ ¨ , n; thus (Df )(x)(u) = 0. Therefore, det [(Df )(x)] = 0, a contradiction.
Now suppose that there exists x, y P D such that f (x) = f (y). Choose a compact set
K Ď D such that x, y P K and BK Ď Dε (this can be done, for example, by choosing that
s δ for some small δ ą 0). Since f : Dε Ñ Rn is one-to-one, f : BK Ñ Rn is
K = DzD
one-to-one. By Theorem 7.7, f : K Ñ Rn is one-to-one. Then x = y; thus f : D Ñ Rn is
one-to-one.
s Ñ Rn is one-to-one. Assume the contrary that there exists
Next, we show that f : D
x P D and y P BD such that f (x) = f (y). By the inverse function theorem there exists open
neighborhood U of x and V of f (x) such that f : U Ñ V is one-to-one and onto. By choosing
U even smaller if necessary, we can assume that there exists tyk u8
k=1 Ď DzU and yk Ñ y
as k Ñ 8. By the continuity of f , f (yk ) Ñ f (y) as k Ñ 8. However, since f : D Ñ Rn
␣
(8
␣
(8
is one-to-one, f (yk ) k=1 R V; thus f (yk ) k=1 cannot converge to f (y) as k Ñ 8 (since
f (y) P V), a contradiction.
Finally, the inverse function theorem implies that f ´1 : f (D) Ñ D is of class C 1 , and
s follows from the fact that (f ´1 )´1 (F ) = f (F ) is closed in D
s
the continuity of f ´1 on f (D)
for all closed subset F of Rn .
˝
Remark 7.10. Suppose that D Ď Rn in Corollary 7.9 is open, bounded, connected but not
convex. The Whitney extension theorem (which is not covered in this text) implies that
s Then Theorem
there exists a function F : Rn Ñ Rn so that F = f and DF = Df on D.
s
7.7 can be applied to guarantee that F is one-to-one on D.
7.2 The Implicit Function Theorem（隱函數定理）
Theorem 7.11 (Implicit Function Theorem). Let D Ď Rn ˆ Rm be open, and F : D Ñ Rm
be a function of class C 1 . Suppose that for some (x0 , y0 ) P D, where x0 P Rn and y0 P Rm ,
§7.2 The Implicit Function Theorem
F (x0 , y0 ) = 0 and
265

BF1
 By1
[
] 

(Dy F )(x0 , y0 ) =  ...

 BF

¨¨¨
...
m
By1
¨¨¨
BF1
Bym 

..  (x , y )
. 
 0 0
BF 
m
Bym
is invertible. Then there exists an open neighborhood U Ď Rn of x0 , an open neighborhood
V Ď Rm of y0 , and f : U Ñ V such that
(
)
1. F x, f (x) = 0 for all x P U;
2. y0 = f (x0 );
(
)´1
(
)
3. (Df )(x) = ´ (Dy F )(x, f (x)) (Dx F ) x, f (x) for all x P U, where the matrix representation of Dx F (x, f (x)) P B(Rn , Rn ) is given by


BF1
BF1
¨¨¨
 Bx1
Bxn 


[
]  .

.
.
.
.
.
(x, y) .
(Dx F )(x, y) =  .
.
. 



 BF
BFm
m
¨¨¨
Bx1
Bxn
4. f is of class C 1 ;
5. If F is of class C r for some r ą 1, so is f .
Proof. Let z = (x, y) and w = (u, v), where x, u P Rn and y, v P Rm . Define G by
(
)
G(x, y) = x, F (x, y) , and write w = G(z). Then G : D Ñ Rn+m , and
[
]
In
0
[
]
(DG)(x, y) =
,
(Dx F )(x, y) (Dy F )(x, y)
where In is the n ˆ n identity matrix. We note that the Jacobian of G at (x0 , y0 ) is
(
)
det [(Dy F )(x0 , y0 )] which does not vanish since (Dy F )(x0 , y0 ) is invertible, so the inverse
function theorem implies that there exists open neighborhoods O of (x0 , y0 ) and W of
(
)
x0 , F (x0 , y0 ) = (x0 , 0) such that
(a) G : O Ñ W is one-to-one and onto;
(b) the inverse function G´1 : W Ñ O is of class C r ;
266
CHAPTER 7. Inverse and Implicit Function Theorems
(
) (
)´1
(c) (DG´1 ) x, F (x, y) = (DG)(x, y) .
By Remark 7.2, W.L.O.G. we can assume that O = U ˆ V, where U Ď Rn and V Ď Rm are
open, and x0 P U, y0 P V.
(
)
Write G´1 (u, v) = φ(u, v), ψ(u, v) , where φ : W Ñ U and ψ : W Ñ V. Then
(
) (
)
(u, v) = G φ(u, v), ψ(u, v) = φ(u, v), F (u, ψ(u, v))
(
)
which implies that φ(u, v) = u and v = F (u, ψ(u, v)). Let f (x) = ψ(x, 0). Then u, f (u) P
(
)
U ˆ V is the unique point satisfying F u, f (u) = 0 if u P U. Therefore, f : U Ñ V, and
(
)
F x, f (x) = 0
@x P U .
(
)
(
)
Since G(x0 , y0 ) = (x0 , 0) = G x0 , f (x0 ) , (x0 , y0 ), x0 , f (x0 ) P O, and G : O Ñ W is
one-to-one, we must have y0 = f (x0 ).
By (b) and (c), we have G´1 is of class C 1 , and
(
)´1
(DG´1 )(u, v) = (DG)(x, y)
.
As a consequence, ψ P C 1 , and
[
]
(Du φ)(u, v) (Dv φ)(u, v)
(Du ψ)(u, v) (Dv ψ)(u, v)
[
=
[
In
0
]´1
(Dx F )(x, y) (Dy F )(x, y)
]
In
0
=
(
)´1
(
)´1 .
´ (Dy F )(x, y) (Dx F )(x, y) (Dy F )(x, y)
Evaluating the equation above at v = 0, we conclude that
(
)´1
(
)
(Df )(u) = (Du ψ)(u, 0) = ´ (Dy F )(u, f (u)) (Dx F ) u, f (u)
which implies 3. We also note that 4 follows from (b) and 5 follows from 3.
˝
Alternative proof of Theorem 7.11 without applying the inverse function theorem. Let z =
(x, y), z0 = (x0 , y0 ), A = (Dx F )(x0 , y0 ) and B = (Dy F )(x0 , y0 ). Define
r(x, y) = F (x, y) ´ A(x ´ x0 ) ´ B(y ´ y0 ) .
Our goal is to solve the equation
0 = A(x ´ x0 ) + B(y ´ y0 ) + r(x, y)
§7.2 The Implicit Function Theorem
267
for y. By the invertibility of B, this is equivalent of finding a fixed-point of the map
[
]
Φx (y) = y0 ´ B ´1 A(x ´ x0 ) + r(x, y) .
Since r is of class C 1 and (Dr)(x0 , y0 ) = 0,
D δ ą 0 Q }(Dr)(x, y)}B(Rm ,Rm ) ă min
␣
1
4m}B ´1 }B(Rm ,Rm )
,
1 (
2m
@ z P D(z0 , δ) .
Therefore, the mean value theorem (Theorem 6.49) implies that
m
m
ÿ
›
›
ˇ
ˇ ÿ
ˇ
ˇ
›r(x, y) ´ r(x0 , y0 )› m ď
ˇri (x, y) ´ ri (x0 , y0 )ˇ =
ˇ(Dri )(ci )(z ´ z0 )ˇ
R
i=1
ď
i=1
}z ´ z0 }Rn+m
δ
ă
´1
´1
4}B }B(Rm ,Rm )
4}B }B(Rm ,Rm )
( δ)
( δ
Ď D(z0 , δ), and
for all z = (x, y) P D x0 , ) ˆ D y0 ,
2
2
›
›
›Φx (y1 ) ´ Φx (y2 )› m ď }B ´1 }B(Rm ,Rm ) }r(x, y1 ) ´ r(x, y2 )}Rm ă 1 }y1 ´ y2 }Rm
R
4
(7.2.1)
( δ)
( δ)
if x P D x0 , , y1 , y2 P D y0 ,
and y1 ‰ y2 . As a consequence, for each (fixed) x satisfying
2
2
)
!
δ
δ
δ
)
m ă
,
if
}y
´
y
}
we have
,
}x ´ x0 }Rn ă r ” min (
0
R
´1
4 1 + }A}B(Rn ,Rm ) }B
}B(Rm ,Rm ) 2
2
›
›
}Φx (y) ´ y0 }Rn ď }B ´1 }B(Rm ,Rm ) ›A(x ´ x0 ) + r(x, y)›Rm
[
] δ
ď }B ´1 }B(Rm ,Rm ) }A}B(Rn ,Rm ) }x ´ x0 }Rn + }r(x, y)}Rm ) ă .
2
(7.2.2)
ˇ
␣
δ(
. Then for each x P U ” D(x0 , r), (7.2.1) and (7.2.2)
Let M = y P Rm ˇ }y ´ y0 }Rm ď
2
imply that Φx : M Ñ M is a contraction mapping; thus there is a unique
fixed-point y P M .
?
(
3δ )
(the choice of this
Denote this unique fixed-point as f (x). Then f : U Ñ V ” D y0 ,
V guarantees that U ˆ V Ď D(z0 , δ)) and
2
(
)
(
)
(
)
F x, f (x) = A(x ´ x0 ) + B f (x) ´ y0 + r x, f (x) = 0 .
Moreover, since F (x, y) = 0 if and only if y is a fixed-point of Φx , and the contraction
mapping principle provides the uniqueness of the fixed-point if (x, y) P U ˆ M . Since
(x0 , y0 ), (x0 , f (x0 )) P U ˆ M , we must have y0 = f (x0 ).
268
CHAPTER 7. Inverse and Implicit Function Theorems
To see the differentiability of f , we first claim that f : U Ñ V is continuous. Since f (x)
is the fixed-point of Φx ,
(
)
f (x) = y0 ´ B ´1 A(x ´ x0 ) + r(x, f (x)) .
If x1 , x2 P U, then (x1 , f (x1 )), (x2 , f (x2 )) P D(z0 , δ); thus (7.2.1) implies that
›
›
›
›
›
›
›f (x1 ) ´ f (x2 )› m = ›B ´1 A(x1 ´ x2 )› m + ›r(x1 , f (x1 )) ´ r(x2 , f (x2 ))› m
R
R
R
b
1
ď }B ´1 A}B(Rn ,Rm ) }x1 ´ x2 }Rn +
}x1 ´ x2 }2Rn + }f (x1 ) ´ f (x2 )}2Rm
2
1
1
ď }B ´1 A}B(Rn ,Rm ) }x1 ´ x2 }Rn + }x1 ´ x2 }Rn + }f (x1 ) ´ f (x2 )}Rm .
2
2
Therefore,
›
›
(
)
›f (x1 ) ´ f (x2 )› m ď 2}B ´1 A}B(Rn ,Rm ) + 1 }x1 ´ x2 }Rn
R
(7.2.3)
which implies that f : U Ñ V is (Lipschitz) continuous.
r = (Dx F )(b), B
r =
Now let a P U and ε ą 0 be given. Define b = (a, f (a)), and A
(Dy F )(b). We would like to show that there exists δ1 ą 0 such that
›
›
r ´1 A(x
r ´ a)› m ď ε}x ´ a}Rn
›f (x) ´ f (a) + B
R
@ x P D(a, δ1 ) .
Since F P C 1 and the map L ÞÑ L´1 is continuous, there exists δ2 ą 0 such that
›
›
›(Dy F )(z)´1 (Dx F )(z) ´ (Dy F )(z0 )´1 (Dx F )(z0 )› n m ď ε
B(R ,R )
4
@ z P D(z0 , δ2 ) .
Moreover, since r P C 1 , there exists δ3 ą 0 such that
›
›
›r(z) ´ r(b) ´ (Dr)(b)(z ´ b)› m ď
R
ε
}z ´ b}Rn+m
@ z P D(b, δ3 )
}z ´ b}Rn+m
@ z P D(b, δ3 ) .
2}B ´1 }B(Rm ,Rm )
and
›
›
›(Dr)(z) ´ (Dr)(b)› n+m m ď
B(R
,R )
ε
2}B ´1 }B(Rm ,Rm )
δ3
δ δ
. Then if }x ´ a}Rn ă δ1 , using (7.2.3)
Choose δ1 = min 2 , 3 ,
´1
2 2 2(2}B A}B(Rn ,Rm ) + 1)
!
)
we find that
›
›
›(x, f (x)) ´ (a, f (a))› n+m ď }x ´ a}Rn + }f (x) ´ f (a)}Rm ă mintδ2 , δ3 u ;
R
§7.2 The Implicit Function Theorem
269
thus if }x ´ a}Rn ă δ1 ,
›
›
r ´1 A(x
r ´ a)› m
›f (x) ´ f (a) + B
R
›( ´1
)
(
)›
´1
r A
r ´ B A (x ´ a) + B ´1 r(x, f (x)) ´ r(a, f (a)) › m
=› B
R
›
› ´1
´1
r ´ B A› n m }x ´ a}Rn
r A
ď ›B
B(R ,R )
›
(
)›
´1
+ }B }B(Rm ,Rm ) ›r(x, f (x)) ´ r(a, f (a)) ´ (Dr)(b) x ´ a, f (x) ´ f (a) ›Rm
›
(
)›
+ }B ´1 }B(Rm ,Rm ) ›(Dr)(b) (x ´ a, f (x) ´ f (a) › m ď ε}x ´ a}Rn .
R
Therefore, f is differentiable on U, and
(
)´1
(
)
(Df )(x) = ´ (Dy F )(x, f (x)) (Dx F ) x, f (x)
@x P U .
(7.2.4)
Since F is of class C 1 and f is continuous on U, we find that Df is continuous; thus f is of
class C 1 .
˝
Example 7.12. Let F (x, y) = x2 + y 2 ´ 1.
1. If (x0 , y0 ) = (1, 0), then Fx (x0 , y0 ) = 2 ‰ 0; thus the implicit function theorem implies
that locally x can be expressed as a function of y.
2. If (x0 , y0 ) = (0, ´1), then Fy (x0 , y0 ) = ´2 ‰ 0; thus the implicit function theorem
implies that locally y can be expressed as a function of x.
3. If (x0 , y0 ) =
(
?
?
3)
1
´ ,
, then Fx (x0 , y0 ) = ´1 ‰ 0 and Fy (x0 , y0 ) = 3 ‰ 0; thus the
2 2
implicit function theorem implies that locally x can be expressed as a function of y
and locally y can be expressed as a function of x.
Example 7.13. Suppose that (x, y, u, v) satisfies the equation
"
xu + yv 2 = 0
xv 3 + y 2 u6 = 0
and (x0 , y0 , u0 , v0 ) = (1, ´1, 1, ´1). Let F (x, y, u, v) = (xu + yv 2 , xv 3 + y 2 u6 ). Then
F (x0 , y0 , u0 , v0 ) = 0.


BF1 BF1
]
[
1
1
 Bx
By 
is invertible,
1. Since (Dx,y F )(x0 , y0 , u0 , v0 ) =  BF BF  (x0 , y0 , u0 , v0 ) =
2
2
´1 ´2
Bx
By
locally (x, y) can be expressed in terms of u, v; that is, locally x = x(u, v) and y =
y(u, v).
270
CHAPTER 7. Inverse and Implicit Function Theorems

BF1
 By
2. Since (Dy,u F )(x0 , y0 , u0 , v0 ) =  BF
2
By

BF1
]
[
1 1
Bu 
is invertible,
(x , y , u , v ) =
BF2  0 0 0 0
´2 6
Bu
locally (y, u) can be expressed in terms of x, v.
Example 7.14. Let f : R3 Ñ R2 be given by
f (x, y, z) = (xey + yez , xez + zey ) .
Then f is of class C 1 , f (´1, 1, 1) = (0, 0) and
[ y
]
[
]
e xey + ez
yez
(Df )(x, y, z) = z
.
e
zey
xez + ey
[
]
0 e
Since (Dy,z f )(´1, 1, 1) =
is invertible, the implicit function theorem implies that the
e 0
system
" y
xe + yez = 0
xez + zey = 0
can be solved for y and z as continuously differentiable function of x for x near ´1 and (y, z)
near (1, 1). Furthermore, if we write (y, z) = g(x) for x near ´1, then
[
xey + ez
yez
g (x) =
zey
xez + ey
1
]´1 [ y ]
e
.
ez
7.3 The Lagrange Multipliers
In this section we are concerned with the optimization problem
“find the extreme value of function y = f (x) subject to the constraint g(x) = 0”, (7.3.1)
where f : Rn Ñ R is a real-value differentiable function, and g : Rn Ñ Rm is a vector-valued
␣ ˇ
(
C 1 -function for some m ă n. We assume that the feasible set x ˇ g(x) = 0 is non-empty
and does not contain isolated points (or the extreme value of f is not attained at isolated
points of the feasible set), and all the local extreme points of f do not belong to the feasible
set (otherwise we can ignore the constraint and find the local extreme value of f directly).
Suppose that f attains its extreme value, subject to the constraint g = 0, at point a,
and the Jacobian matrix of g at a has rank m (that is, [(Dg)(a)] has full rank). Without
§7.3 The Lagrange Multipliers
271
loss of generality we can assume that the square matrix


Bg1
Bg1
(a) ¨ ¨ ¨
(a)
 Bx1

Bxm
 .

.
...
 ..

.
.


 Bg

Bgm
m
(a) ¨ ¨ ¨
(a)
Bx1
Bxm
is invertible (thus has rank m). Then the implicit function theorem implies there exist an
open neighborhood V of (am+1 , ¨ ¨ ¨ , an ) and a C 1 -function φ : V Ñ Rm such that in a
neighborhood of a, the feasible set can be expressed as
(xm+1 , ¨ ¨ ¨ , xn ) P V .
(x1 , ¨ ¨ ¨ , xm ) = φ(xm+1 , ¨ ¨ ¨ , xn )
Under these settings, the original optimization problem (7.3.1) is transformed as “the func(
)
tion y = f φ(xm+1 , ¨ ¨ ¨ , xn ), xm+1 , ¨ ¨ ¨ , xn attains its extreme value at (am+1 , ¨ ¨ ¨ , an )”;
thus
ˇ
(
)
ˇ
f φ(xm+1 , ¨ ¨ ¨ , xn ), xm+1 , ¨ ¨ ¨ , xn = 0 ,
ˇ
Bxm+1 (xm+1 ,¨¨¨ ,xn )=(am+1 ,¨¨¨ ,an )
ˇ
(
)
B ˇ
f φ(xm+1 , ¨ ¨ ¨ , xn ), xm+1 , ¨ ¨ ¨ , xn = 0 ,
ˇ
B
Bxm+2 (xm+1 ,¨¨¨ ,xn )=(am+1 ,¨¨¨ ,an )
..
.
(
)
B ˇ
f φ(xm+1 , ¨ ¨ ¨ , xn ), xm+1 , ¨ ¨ ¨ , xn = 0 ,
ˇ
Bxn (xm+1 ,¨¨¨ ,xn )=(am+1 ,¨¨¨ ,an )
ˇ
or using the chain rule,
[
]
Bf
Bf
Bφ
Bf
(a) ¨ ¨ ¨
(a)
(am+1 , ¨ ¨ ¨ , an ) +
(a) = 0
Bx1
Bxm
Bxj
for m + 1 ď j ď n .
Bxj
Noting that the implicit function theorem implies that
´1  Bg

Bg1
Bg1
1
(a) ¨ ¨ ¨
(a) ¨ ¨ ¨
(a)
  Bxm+1
 Bx1
Bxm

 .
[
]
..
.. 
..

(Dφ)(am+1 , ¨ ¨ ¨ , an ) = ´ 
.
.
. 
 
 ..
  Bg
 Bg
Bg
m
m
m
(a) ¨ ¨ ¨
(a)
(a) ¨ ¨ ¨
Bx1
we find that for m + 1 ď j ď n,
Bxm
Bxm+1
¨¨¨
´1  Bg

Bg1
1
(a)
(a)
  Bxj

Bxm

Bg1
(a)
 Bx1

Bφ
(am+1 , ¨ ¨ ¨ , an ) = ´ 

Bxj

..
.
..
.
Bgm
(a) ¨ ¨ ¨
Bx1
..
.
Bgm
(a)
Bxm




 .
 ..

 Bg

.


m
(a)
Bxj
(7.3.2)

Bg1
(a)
Bxn

..
.
Bgm
(a)
Bxn

,


272
CHAPTER 7. Inverse and Implicit Function Theorems
Define

[
λ1 λ2 ¨ ¨ ¨
[
]
Bf
λm = ´
(a) ¨ ¨ ¨
Bx1
Bg1
(a)
]  Bx1
 .
Bf
(a) 
 ..
Bxm
 Bg
m
Bx1

¨¨¨
...
(a) ¨ ¨ ¨
´1
Bg1
(a)

Bxm




Bgm
(a)
..
.
.
Bxm
Then (7.3.2) implies that
Bf
Bg
Bg
Bg
(a) + λ1 1 (a) + λ2 2 (a) + ¨ ¨ ¨ + λm m (a) = 0
Bxj
Bxj
Bxj
Bxj
for m + 1 ď j ď n.
(7.3.3)
On the other hand, by the fact that


Bg1
(a)
 Bx1
 .
 ..

 Bg
m
Bx1
for 1 ď j ď m,
¨¨¨
..
.
(a) ¨ ¨ ¨
.. 
. 


Bgm
(a)
Bg1
(a)
 Bx1
m
Bx1
 .
 ..

 Bg
m
Bxm

 .
 ..

 Bg


´1
Bg1
Bg1
(a)
(a)
  Bx1
Bxm
¨¨¨
..
.
(a) ¨ ¨ ¨
Bx1
Bg1
(a)

Bxm
¨¨¨
..

 = Imˆm ,


Bgm
(a)
..
.
.
(a) ¨ ¨ ¨
Bxm
´1  Bg

Bg1
1
(a)
(a)
  Bxj

Bxm
..
.




Bgm
(a)
 .
 ..

 Bg
Bxm
Bxj
m

 = ej ,


(a)
m
where tej um
j=1 is the standard basis of R . Therefore, left multiplying the equation above
[
]
Bf
Bf
(a) ¨ ¨ ¨
(a) , we obtain that
by the row vector ´
Bx1
λ1
Bxm
Bg1
Bg
Bg
Bf
(a) + λ2 2 (a) + ¨ ¨ ¨ + λm m (a) = ´
(a)
Bxj
Bxj
Bxj
Bxj
for 1 ď j ď m.
Combining (7.3.3) and (7.3.4), we conclude that
Bf
Bg
Bg
Bg
(a) + λ1 1 (a) + λ2 2 (a) + ¨ ¨ ¨ + λm m (a) = 0
Bxj
Bxj
Bxj
Bxj
or equivalently,
[
] [
(Df )(a) + λ1 λ2 ¨ ¨ ¨
We then establish the following
λm
][
for 1 ď j ď n
]
(Dg)(a) = 0 .
(7.3.4)
§7.4 Exercises
273
Theorem 7.15 (Lagrange Multipliers). Let U Ď Rn be an open set, f : U Ñ R be differenˇ
␣
(
tiable, g : U Ñ Rm be of class C 1 for some m ă n, and S = x P U ˇ g(x) = 0 . Suppose that
[
]
S ‰ H and f |S , the restriction of f on S, attains its extreme value at a P S. If (Dg)(a)
has full rank, then there exist λ1 , λ2 , ¨ ¨ ¨ , λm P R such that
[
] [
(Df )(a) + λ1 λ2 ¨ ¨ ¨
λm
][
]
(Dg)(a) = 0 .
7.4 Exercises
§7.1 The Inverse Function Theorem
Problem 7.1. Prove Corollary 7.4; that is, show that if U Ď Rn is open, f : U Ñ Rn is
of class C 1 , and (Df )(x) is invertible for all x P U, then f (W) is open for every open set
W Ď U.
§7.2 The Implicit Function Theorem
Problem 7.2. Assume that one proves the implicit function theorem without applying the
inverse theorem. Show the inverse function using the implicit function theorem.
Problem 7.3. Suppose that F (x, y, z) = 0 is such that the functions z = f (x, y), x =
g(y, z), and y = h(z, x) all exist by the implicit function theorem. Show that fx ¨gy ¨hz = ´1.
Problem 7.4. Suppose that the implicit function theorem applies to F (x, y) = 0 so that
y = f (x). Find a formula for f 2 in terms of F and its partial derivatives. Similarly, suppose
that the implicit function theorem applies to F (x1 , x2 , y) = 0 so that y = f (x1 , x2 ). Find
formulas for fx1 x1 , fx1 x2 and fx2 x2 in terms of F and its partial derivatives.
Problem 7.5. Given F : R4 Ñ R2 and suppose that F (x, y) = 0, where x = (x1 , x2 ) and
y = (y1 , y2 ). State conditions which guarantees that the equation
By1 Bx2 By2 Bx2
+
=0
Bx1 By1 Bx1 By2
holds. Prove or justify your answer.
§7.3 The Lagranege Multipliers
Problem 7.6 (True or False). Determine whether the following statements are true or
false. If it is true, prove it. Otherwise, give a counter-example.
274
1.
2.
3.
CHAPTER 7. Inverse and Implicit Function Theorems
Chapter 8
Integration of Functions of Several
Variables
In this chapter, we focus on the integration of bounded functions on bounded subsets of Rn .
8.1 Integrable Functions
We start with a simpler case n = 2.
Definition 8.1. Let A Ď R2 be a bounded set. Define
ˇ
␣
(
a1 = inf x P R ˇ (x, y) P A for some y P R ,
ˇ
␣
(
b1 = sup x P R ˇ (x, y) P A for some y P R ,
ˇ
␣
(
a2 = inf y P R ˇ (x, y) P A for some x P R ,
ˇ
␣
(
b2 = sup y P R ˇ (x, y) P A for some x P R .
A collection of rectangles P is called a partition of A if there exists a partition Px of [a1 , b1 ]
and a partition Py of [a2 , b2 ],
␣
(
␣
(
Px = a1 = x0 ă x1 ă ¨ ¨ ¨ ă xn = b1
and Py = a2 = y0 ă y1 ă ¨ ¨ ¨ ă ym = b2 ,
such that
ˇ
␣
(
P = ∆ij ˇ ∆ij = [xi , xi+1 ] ˆ [yj , yj+1 ] for i = 0, 1, ¨ ¨ ¨ , n ´ 1 and j = 0, 1, ¨ ¨ ¨ , m ´ 1 .
The mesh size of the partition P and also called the norm of P, denoted by }P}, is defined
by
ˇ
!b
)
ˇ
}P} = max
(xi+1 ´ xi )2 + (yj+1 ´ yj )2 ˇ i = 0, 1, ¨ ¨ ¨ , n ´ 1, j = 0, 1, ¨ ¨ ¨ , m ´ 1 .
275
276
CHAPTER 8. Integration
The number
a
(xi+1 ´ xi )2 + (yj+1 ´ yj )2 is often denoted by diam(∆ij ), and is called the
diameter of ∆ij .
Similar to the integrability of f on a bounded subset of R, we have the following
Definition 8.2. Let A Ď R2 be a bounded set, and f : A Ñ R be a bounded function. For
ˇ
␣
(
any partition P = ∆ij ˇ ∆ij = [xi , xi+1 ] ˆ [yj , yj+1 ], i = 0, ¨ ¨ ¨ , n ´ 1, j = 0, ¨ ¨ ¨ , m ´ 1 , the
upper sum and the lower sum of f with respect to the partition P, denoted by U (f, P)
and L(f, P) respectively, are numbers defined by
U (f, P) =
ÿ
A
sup f (x, y)A(∆ij ) ,
0ďiďn´1 (x,y)P∆ij
0ďjďm´1
L(f, P) =
ÿ
inf
0ďiďn´1
0ďjďm´1
(x,y)P∆ij
A
f (x, y)A(∆ij ) ,
A
where A(∆ij ) = (xi+1 ´ xi )(yj+1 ´ yj ) is the area of the rectangle ∆ij , and f is an extension
of f , called the extension of f by zero outside A, given by
A
f (x) =
"
f (x) x P A ,
0
x R A.
The two numbers
ż
ˇ
␣
(
f (x, y)dA ” inf U (f, P) ˇ P is a partition of A
A
and
ż
ˇ
␣
(
f (x, y)dA ” sup L(f, P) ˇ P is a partition of A
A
are called the upper integral and lower integral of f over A, respectively. The function
f is said to be Riemann (Darboux) integrable (over A) if
and in this case, we express the upper and lower integral as
of f over A.
ż
ż
ż
f (x, y)dA =
A
f (x, y)dA,
A
f (x, y)dA, called the integral
A
In general, we can consider the integrability of a bounded function f defined on a
bounded set A Ď Rn as follows
§8.1 Integrable Functions
277
Definition 8.3. Let A Ď Rn be a bounded set. Define the numbers a1 , a2 , ¨ ¨ ¨ , an and
b1 , b2 , ¨ ¨ ¨ , bn by
ˇ
␣
(
ak = inf xk P R ˇ x = (x1 , ¨ ¨ ¨ , xn ) P A for some x1 , ¨ ¨ ¨ , xk´1 , xk+1 , ¨ ¨ ¨ , xn P R ,
ˇ
␣
(
bk = sup xk P R ˇ x = (x1 , ¨ ¨ ¨ , xn ) P A for some x1 , ¨ ¨ ¨ , xk´1 , xk+1 , ¨ ¨ ¨ , xn P R .
A collection of rectangles P is called a partition of A if there exists partitions P (k) of
␣
(
(k)
(k)
(k)
[ak , bk ], k = 1, ¨ ¨ ¨ , n, P (k) = ak = x0 ă x1 ă ¨ ¨ ¨ ă xNk = bk , such that
ˇ
!
ˇ
(1)
(1)
(2)
(2)
(n)
(n+1)
P = ∆i1 i2 ¨¨¨in ˇ ∆i1 i2 ¨¨¨in = [xi1 , xi1 +1 ] ˆ [xi2 , xi2 +1 ] ˆ ¨ ¨ ¨ ˆ [xin , xin+1 ],
)
ik = 0, 1, ¨ ¨ ¨ , Nk ´ 1, k = 1, ¨ ¨ ¨ , n .
The mesh size of the partition P, denoted by }P} and also called the norm of P, is defined
by
#g
f
}P} = max
ˇ
n
fÿ
e (x(k) ´ x(k) )2 ˇˇ ik = 0, 1, ¨ ¨ ¨ , Nk ´ 1, k = 1, ¨ ¨ ¨ , n
ik +1
ik
+
.
k=1
d
The number
n
ř
(k)
(k)
(xik +1 ´ xik )2 is often denoted by diam(∆i1 i2 ¨¨¨in ), and is called the di-
k=1
ameter of the rectangle ∆i1 i2 ¨¨¨in .
Definition 8.4. Let A Ď Rn be a bounded set, and f : A Ñ R be a bounded function. For
any partition
ˇ
!
ˇ
(n+1)
(n)
(2)
(2)
(1)
(1)
P = ∆i1 i2 ¨¨¨in ˇ ∆i1 i2 ¨¨¨in = [xi1 , xi1 +1 ] ˆ [xi2 , xi2 +1 ] ˆ ¨ ¨ ¨ ˆ [xin , xin+1 ],
)
ik = 0, 1, ¨ ¨ ¨ , Nk ´ 1, k = 1, ¨ ¨ ¨ , n ,
the upper sum and the lower sum of f with respect to the partition P, denoted by
U (f, P) and L(f, P) respectively, are numbers defined by
ÿ
A
U (f, P) =
sup f (x, y)ν(∆) ,
∆PP (x,y)P∆
L(f, P) =
ÿ
∆PP
A
inf f (x, y)ν(∆) ,
(x,y)P∆
where ν(∆) is the volume of the rectangle ∆ given by
(1)
(1)
(2)
(2)
(n)
(n)
ν(∆) = (xi1 +1 ´ xi1 )(xi2 +1 ´ xi2 ) ¨ ¨ ¨ (xin +1 ´ xin )
278
CHAPTER 8. Integration
(1)
(1)
(2)
(2)
(n)
A
(n)
if ∆ = [xi1 ´ xi1 +1 ] ˆ [xi2 ´ xi2 +1 ] ˆ ¨ ¨ ¨ ˆ [xin ´ xin +1 ], and f is the extension of f by
zero outside A given by
A
f (x) =
"
f (x) x P A ,
0
x R A.
(8.1.1)
The two numbers
ż
ˇ
␣
(
f (x) dx ” inf U (f, P) ˇ P is a partition of A ,
A
and
ż
ˇ
␣
(
f (x) dx ” sup L(f, P) ˇ P is a partition of A
A
are called the upper integral and lower integral of f over A, respective. The function f
is said to be Riemann (Darboux) integrable (over A) if
in this case, we express the upper and lower integral as
ż
ż
f (x) dx =
A
ż
f (x) dx, and
A
f (x) dx, called the integral of f
A
over A.
Definition 8.5. A partition P 1 of a bounded set A Ď Rn is said to be a refinement of
another partition P of A if for any ∆1 P P 1 , there is ∆ P P such that ∆1 Ď ∆. A partition
P of a bounded set A Ď Rn is said to be the common refinement of another partitions
P1 , P2 , ¨ ¨ ¨ , Pk of A if
1. P is a refinement of Pj for all 1 ď j ď k.
2. If P 1 is a refinement of Pj for all 1 ď j ď k, then P 1 is also a refinement of P.
In other words, P is a common refinement of P1 , P2 , ¨ ¨ ¨ , Pk if it is the coarsest refinement.
“+”
“=”
Figure 8.1: The common refinement of two partitions
Qualitatively speaking, P is a common refinement of P1 , P2 , ¨ ¨ ¨ , Pk if for each j =
1, ¨ ¨ ¨ n, the j-th component cj of the vertex (c1 , ¨ ¨ ¨ , cn ) of each rectangle ∆ P P belongs to
(j)
Pi
for some i = 1, ¨ ¨ ¨ , k.
Similar to Proposition 4.78 and Corollary 4.79, we have
§8.2 Conditions for Integrability
279
Proposition 8.6. Let A Ď Rn be a bounded subset, and f : A Ñ R be a bounded function.
If P and P 1 are partitions of A and P 1 is a refinement of P, then
L(f, P) ď L(f, P 1 ) ď U (f, P 1 ) ď U (f, P) .
Corollary 8.7. Let A Ď Rn be a bounded subset, and f : A Ñ R be a bounded function. If
P1 and P2 are partitions of A, then
ż
ż
f (x)dx ď
f (x)dx ď U (f, P2 ) .
L(f, P1 ) ď
A
A
8.2 Conditions for Integrability
In the following two sections, we discuss some equivalent conditions for Riemann integrability of bounded functions (over bounded sets). We recall that in Section 4.7 we have
talked about two equivalent conditions for Riemann integrability: the Riemann condition
(Proposition 4.80) and the Darboux theorem (Theorem 4.94). This section contributes to
the n-dimensional version of Riemann’s condition and Darboux theorem.
The proof of the following proposition is identical to the proof of Proposition 4.80.
Proposition 8.8 (Riemann’s condition). Let A Ď Rn be a bounded set, and f : A Ñ R be
a bounded function. Then f is Riemann integrable over A if and only if
@ ε ą 0, D a partition P of A Q U (f, P) ´ L(f, P) ă ε .
Definition 8.9. Let P = t∆1 , ∆2 , ¨ ¨ ¨ , ∆N u be a partition of a bounded set A Ď Rn . A
collection of N points tξ1 , ¨ ¨ ¨ , ξN u is called a sample set for the partition P if ξk P ∆k for
all k = 1, ¨ ¨ ¨ , N . Points in a sample set are called sample points for the partition P.
Let A Ď Rn be a bounded set, and f : A Ñ R be a bounded function. A Riemann
sum of f for the the partition P = t∆1 , ∆2 , ¨ ¨ ¨ , ∆N u of A is a sum which takes the form
N
ÿ
A
f (ξi )νn (∆k ) ,
k=1
where the set Ξ = tξ1 , ξ2 , ¨ ¨ ¨ , ξN u is a sample set for the partition P.
Similar to Theorem 4.94, the following theorem establishes the equivalence between
the Riemann condition and the Darboux integrals. The idea of the proof of the following
theorem are essentially identical to the proof of Theorem 4.94; however, the detail proof
requires a slight modification due to the fact that the dimension is bigger than one.
280
CHAPTER 8. Integration
Theorem 8.10 (Darboux). Let A Ď Rn be a bounded set, and f : A Ñ R be a bounded
function with extension f
A
given by (8.1.1). Then f is Riemann integrable over A if and
only if there exists I P R such that for every given ε ą 0, there exists δ ą 0 such that if P
is a partition of A satisfying }P} ă δ, then any Riemann sums for the partition P belongs
to the interval (I ´ ε, I + ε). In other words, f is Riemann integrable over A if and only if
there exists I P R such that for every given ε ą 0, there exists δ ą 0 such that
N
ˇÿ
ˇ
A
ˇ
ˇ
f
(ξ
)ν(∆
)
´
I
ˇ
ˇăε
k
k
(8.2.1)
k=1
(
whenever P = t∆1 , ¨ ¨ ¨ , ∆N is a partition of A satisfying }P} ă δ and tξ1 , ξ2 , ¨ ¨ ¨ , ξN u is a
sample set for P.
[ r r ]n
Proof. The boundedness of A guarantees that A Ď ´ ,
for some r ą 0. Let R =
2 2
[ r r ]n
.
´ ,
2 2
“ð” Suppose the right-hand side statement is true. Let ε ą 0 be given. Then there exists
(
δ ą 0 such that if P = t∆1 , ¨ ¨ ¨ , ∆N is a partition of A satisfying }P} ă δ, then for
all sets of sample points tξ1 , ¨ ¨ ¨ , ξN u for P, we must have
N
ˇ ε
ˇÿ
A
ˇ
ˇ
f (ξk )ν(∆k ) ´ I ˇ ă .
ˇ
4
k=1
Let P = t∆1 , ¨ ¨ ¨ , ∆N
(
be a partition of A with }P} ă δ. Choose two sample sets
tξ1 , ¨ ¨ ¨ , ξN u and tη1 , ¨ ¨ ¨ , ηN u for P such that
A
(a) sup f (x) ´
xP∆k
A
(b) inf f (x) +
xP∆k
A
A
ε
ă f (ξk ) ď sup f (x);
4ν(R)
xP∆k
A
A
ε
ą f (ηk ) ě inf f (x).
xP∆k
4ν(R)
Then
U (f, P) =
=
N
ÿ
A
sup f (x)ν(∆k ) ă
k=1 xP∆k
N
ÿ
N
ÿ
[ A
f (ξk ) +
k=1
N
ε
ε ε
ε ÿ
f (ξk )ν(∆k ) +
ν(∆k ) ă I + + = I +
4ν(R)
4 4
2
A
k=1
ε ]
ν(∆k )
4ν(R)
k=1
§8.2 Conditions for Integrability
281
and
L(f, P) =
=
N
ÿ
k=1
N
ÿ
A
inf f (x)ν(∆k ) ą
xP∆k
k=1
“ñ” Let I =
ż
ε ]
ν(∆k )
4ν(R)
N
A
f (ηk )ν(∆k ) ´
k=1
As a consequence, I ´
N
ÿ
[ A
f (ηk ) ´
ε ε
ε
ε ÿ
ν(∆k ) ą I ´ ´ = I ´ .
4ν(R)
4 4
2
k=1
ε
ε
ă L(f, P) ď U (f, P) ă I + ; thus U (f, P) ´ L(f, P) ă ε.
2
2
f (x)dx, and ε ą 0 be given. Since f is Riemann integrable over A, there
A
ε
(i)
exists a partition P1 of A such that U (f, P1 ) ´ L(f, P1 ) ă . Suppose that P1 =
2
␣ (i) (i)
(i) (
y0 , y1 , ¨ ¨ ¨ , ymi for 1 ď i ď n. With M denoting the number m1 + m2 + ¨ ¨ ¨ + mn ,
we define
δ=
(
ε
A
A
4rn´1 (M + n) sup f (R) ´ inf f (R) + 1
).
Then δ ą 0. Our goal is to show that if P is a partition of A with }P} ă δ and
tξ1 , ¨ ¨ ¨ , ξN u is a set of sample points for P, then (8.2.1) holds.
Assume that P = t∆1 , ∆2 , ¨ ¨ ¨ , ∆N u is a given partition of A with }P} ă δ.
Let P 1 be the common refinement of P and P1 . Write P 1 = t∆11 , ∆12 , ¨ ¨ ¨ , ∆1N 1 u and
(1)
(2)
(n)
∆k = ∆k ˆ ∆k ˆ ¨ ¨ ¨ ˆ ∆k
definition of the upper sum,
U (f, P) =
N
ÿ
1(1)
as well as ∆1k = ∆k
1(2)
1(n)
ˆ ∆k
ˆ ¨ ¨ ¨ ˆ ∆k . By the
ÿ
sup f (x)ν(∆k )
A
sup f (x)ν(∆k )
k=1 xP∆k
ÿ
=
y
A
sup f (x)ν(∆k ) +
xP∆k
1ďkďN with
(i)
(i)
R∆ for all i, j
j
k
A
xP∆k
1ďkďN with
(i)
(i)
y
P∆ for some i, j
j
k
and similarly,
U (f, P 1 ) =
ÿ
A
sup f (x)ν(∆1k ) +
xP∆k1
1ďkďN 1 with
(i)
1(i)
y
R∆
for all i, j
j
k
(i)
ÿ
xP∆1k
1ďkďN 1 with
(i)
1(i)
y
P∆
for some i, j
j
k
By the fact that ∆k P P 1 if yj R ∆k for all i, j, we must have
ÿ
ÿ
ν(∆k ) =
ν(∆1k ) .
y
1ďkďN with
(i)
(i)
P∆ for some i, j
j
k
1(i)
1ďkďN 1 with
(i)
1(i)
y
P∆
for some i, j
j
k
A
sup f (x)ν(∆1k ) .
282
CHAPTER 8. Integration
The equality above further implies that
ÿ
A
sup f (x)ν(∆k )´
U (f, P)´U (f, P 1 ) =
xP∆k
1ďkďN with
(i)
(i)
y
P∆ for some i, j
j
k
(
)
A
A
ď sup f (R) ´ inf f (R)
ÿ
A
sup f (x)ν(∆1k )
xP∆k1
1ďkďN 1 with
(i)
1(i)
y
P∆
for some i, j
j
k
ÿ
ν(∆k ) .
1ďkďN with
(i)
(i)
y
P∆ for some i, j
j
k
Moreover, for each fixed i, j,
ď
[ r r ]i´1 [ (i)
] [ r r ]n´i
(i)
ˆ yj ´ δ, yj + δ ˆ ´ ,
;
∆k Ď ´ ,
(i)
2 2
(i)
2 2
1ďkďN with yj P∆k
thus
ÿ
ν(∆k ) ď 2δrn´1
@ 1 ď i ď n, 1 ď j ď mi .
(i)
(i)
1ďkďN with yj P∆k
Therefore,
mi
n ÿ
(
)ÿ
A
A
U (f, P) ´ U (f, P 1 ) ď sup f (R) ´ inf f (R)
ÿ
ν(∆k )
i=1 j=0 1ďkďN with y (i) P∆(i)
j
k
(
mi
n ÿ
)ÿ
ď sup f (R) ´ inf f (R)
2δrn´1
A
A
i=1 j=0
ď 2δr
n´1
(
) ε
A
R
(m1 + m2 + ¨ ¨ ¨ + mn + n) sup f (R) ´ inf f (A) ă ,
2
and the fact that U (f, P1 ) ´ L(f, P1 ) ă
ε
shows that
2
U (f, P) ´ I ď U (f, P) ´ I + U (f, P1 ) ´ U (f, P1 )
ď U (f, P) ´ L(f, P1 ) + U (f, P1 ) ´ U (f, P 1 ) ă ε .
Therefore, for any sample set tξ1 , ¨ ¨ ¨ , ξN u for P,
N
ÿ
A
f (ξk )ν(∆k ) ď U (f, P) ă I + ε .
k=1
Similar argument can be used to show that
N
ÿ
A
f (ξk )ν(∆k ) ě L(f, P) ą I ´ ε
k=1
which concludes the Theorem.
˝
§8.3 Lebesgue’s Theorem
283
In Section 5.1, we show that if a sequence of Riemann integrable functions tfk u8
k=1
converges to a function f uniformly on [a, b], then f is also Riemann integrable over [a, b] and
the integral of the limit function is the same as the limit of the integrals (of the sequences).
This theorem can also be established if the domain A under consideration is a bounded
subset of Rn . In fact, the same proof used to establish Theorem 5.16 can be applied to
conclude the following
Theorem 8.11. Let A Ď Rn be a bounded set, and fk : A Ñ R be a sequence of Riemann
integrable functions over A such that tfk u8
k=1 converges uniformly to f on A. Then f is
Riemann integrable over A, and
lim
ż
kÑ8 A
ż
fk (x) dx =
f (x) dx .
(8.2.2)
A
From now on, we will simply use fs to denote the zero extension of f when the
domain outside which the zero extension is made is clear.
8.3 The Lebesgue Theorem
In this section, we talk about another equivalent condition of Riemann integrability, named
the Lebesgue theorem. The Lebesque theorem provides a more practical way to check the
Riemann integrability in the development of theory. To understand the Lebesgue theorem,
we need to talk about a new concept, sets of measure zero.
8.3.1 Volume and Sets of Measure Zero
Definition 8.12. Let A Ď Rn be a bounded set, and 1A (or χA ) be the characteristic
function of A defined by
"
1A (x) =
1 if x P A ,
0 otherwise .
A is said to have volume if 1Aż is Riemann integrable (over A), and the volume of A,
1A (x) dx. A is said to have volume zero or content
denoted by ν(A), is the number
zero if ν(A) = 0.
A
Remark 8.13. Not all bounded set has volume.
Proposition 8.14. Let A Ď Rn be bounded. Then the following three statements are
equivalent.
284
CHAPTER 8. Integration
(a) A has volume zero;
(b) for every ε ą 0, there exists finite open rectangles S1 , ¨ ¨ ¨ , SN whose sides are parallel
to the coordinate axes such that
AĎ
N
ď
Sk
and
k=1
N
ÿ
(8.3.1)
ν(Sk ) ă ε
k=1
(c) for every ε ą 0, there exists finite rectangles S1 , ¨ ¨ ¨ , SN such that (8.3.1) holds.
Proof. It suffices to show (a)ñ(b) and (c)ñ(a) since it is clear that (b)ñ(c).
“(a)ñ(b)” Since A has volume zero,
ż
1A (x) dx = 0; thus for any given ε ą 0, there exists
A
a partition P of A such that
U (1A , P) ă
ż
1A (x) dx +
A
"
Since sup 1A (x) =
xP∆
ε
ε
= .
2
2
1 if ∆ X A ‰ H ,
we must have
0 otherwise ,
ř
∆PP
∆XA‰H
ν(∆) ă
ε
. Now if ∆ P P
2
and ∆XA ‰ H, we can find an open rectangle l such that ∆ Ď l and ν(l) ă 2ν(∆).
N
N
Ť
ř
Let S1 , ¨ ¨ ¨ , SN be those open rectangles l. Then A Ď
Sk and
ν(Sk ) ă ε.
k=1
k=1
“(c)ñ(a)” W.L.O.G. we can assume that the ratio of the maximum length and minimum
length of sides of Sk is less than 2 for all k = 1, ¨ ¨ ¨ , N (otherwise we can divide Sk into
smaller rectangles so that each smaller rectangle satisfies this requirement). Then each
Sk can be covered by a closed rectangle lk whose sides are parallel to the coordinate
? n
axes with the property that ν(lk ) ď 2n´1 n ν(Sk ). Let P be a partition of A such
that for each ∆ P P with ∆ X A ‰ H, ∆ Ď lk for some k = 1, ¨ ¨ ¨ , N . Then
U (1A , P) =
ÿ
ν(∆) ď
∆PP
∆XA‰H
thus the upper integral
ż
N
ÿ
N
? nÿ
? n
ν(lk ) ď 2n´1 n
ν(Sk ) ă 2n´1 n ε ;
k=1
k=1
1A (x) dx = 0. Since the lower integral cannot be negative,
A
we must have
ż
A
1A (x) dx =
ż
1A (x) dx = 0 which shows that A has volume zero. ˝
A
Example 8.15. Each point in Rn has volume zero.
§8.3 Lebesgue’s Theorem
285
Example 8.16. The Cantor set (defined in Exercise Problem 2.11) has volume zero.
Definition 8.17. A set A Ď Rn (not necessarily bounded) is said to have measure
zero（測度為零）or be a set of measure zero（零測度集）if for every ε ą 0, there exist
8
(
)
Ť
countable many rectangles S1 , S2 , ¨ ¨ ¨ such that tSk u8
Sk
k=1 is a cover of A that is, A Ď
k=1
8
ř
and
ν(Sk ) ă ε.
k=1
Example 8.18. The real line R ˆ t0u on R2 has measure zero: for any given ε ą 0, let
[ ´ε
ε ]
Sk = [´k, k] ˆ k+3 , k+3 . Then
2
k 2
R ˆ t0u Ď
8
ď
Sk
k
and
k=1
8
ÿ
ν(Sk ) =
k=1
8
ÿ
2k ¨
k=1
2ε
2k+3 k
=
8
ÿ
ε
ε
=
ă ε.
k+1
2
2
k=1
Similarly, any hyperplane in Rn also has measure zero.
Proposition 8.19. Let A Ď Rn be a set of measure zero. If B Ď A, then B also has
measure zero.
Modifying the proof of Proposition 8.14, we can also conclude the following
Proposition 8.20. A set A Ď Rn has measure zero if and only if for every ε ą 0, there
exist countable many open rectangles S1 , S2 , ¨ ¨ ¨ whose sides are parallel to the coordinate
8
8
Ť
ř
Sk and
ν(Sk ) ă ε.
axes such that A Ď
k=1
k=1
Remark 8.21. If a set A has volume zero, then it has measure zero.
Proposition 8.22. Let K Ď Rn be a compact set of measure zero. Then K has volume
zero.
Proof. Let ε ą 0 be given. Then there are countable open rectangles S1 , S2 , ¨ ¨ ¨ such that
KĎ
8
ď
k=1
Sk
and
8
ÿ
ν(Sk ) ă ε .
k=1
Since tSk u8
k=1 is an open cover of K, by the compactness of K there exists N ą 0 such that
N
N
8
Ť
ř
ř
KĎ
Sk , while
ν(Sk ) ď
ν(Sk ) ă ε. As a consequence, K has volume zero.
˝
k=1
k=1
k=1
Since the boundary of a rectangle has measure zero, we also have the following
286
CHAPTER 8. Integration
Corollary 8.23. Let S Ď Rn be a bounded rectangle with positive volume. Then S is not a
set of measure zero.
8
Ť
Theorem 8.24. If A1 , A2 , ¨ ¨ ¨ are sets of measure zero in Rn , then
Ak has measure
k=1
zero.
Proof. Let ε ą 0 be given. Since A1k s are sets of measure zero, there exist countable
␣ (k) (8
rectangles Sj j=1 , such that
8
ď
Ak Ď
(k)
8
ÿ
and
Sj
j=1
(k)
ν(Sj ) ă
j=1
ε
@k P N.
2k+1
(k)
Consider the collection consisting of all Sj ’s. Since there are countable many rectangles in
this collection, we can label them as S1 , S2 , ¨ ¨ ¨ , and we have
8
ď
Ak Ď
8
ÿ
ν(Sℓ ) =
Therefore,
8 ÿ
8
ÿ
=
(k)
8
ď
Sℓ
ℓ=1
ν(Sj ) ď
k=1 j=1
k=1
8
Ť
(k)
Sj
k=1 j=1
k=1
and
8 ď
8
ď
8
ÿ
k=1
ε
2k+1
=
ε
ă ε.
2
Ak has measure zero.
˝
k=1
Corollary 8.25. The set of rational numbers in R has measure zero.
Theorem 8.26. Let A Ď Rn be bounded and B Ď Rm be a set of measure zero. Then A ˆ B
has measure zero in Rn+m .
Proof. Let ε ą 0 be given. Since A is bounded, there exist a bounded rectangle R such that
m
A Ď R. Since B has measure zero, there exist countable rectangles tSk u8
k=1 Ď R such that
BĎ
8
ď
Sk
k=1
Then A ˆ B Ď
8
Ť
and
8
ÿ
k=1
νm (Sk ) ă
ε
.
ν(R)
(R ˆ Sk ), and
k=1
8
ÿ
k=1
νn+m (R ˆ Sk ) =
8
ÿ
k=1
νn (R)νm (Sk ) = νn (R)
8
ÿ
νm (Sk ) ă ε .
k=1
Since R ˆ Sk is a rectangle for all k P N, we conclude that A ˆ B has measure zero.
˝
§8.3 Lebesgue’s Theorem
287
8.3.2 The Lebesgue Theorem
在之前我們提到了函數 Riemann 可積的兩個等價條件：Riemann’s condition 和 Darboux
定理。在這一節中，我們將引進函數是 Riemman 可積的另一個等價條件。這個等價條件
A
說的是一個函數 f 在 A 上是 Riemann 可積的若且唯若 f 的延拓 f （在函數可積分的定
義中有定義）的不連續點所構成的集合其測度為零。為了證明這個敘述，我們先對一個
函數的連續點做一個量化的刻劃。這個刻劃的方式，可以很容易用來檢驗一個函數在一
個點是否連續。
Definition 8.27. Let f : Rn Ñ R be a function. For any x P Rn , the oscillation of f at
x is the quantity
osc(f, x) ” inf
sup
δą0 x1 ,x2 PD(x,δ)
ˇ
ˇ
ˇf (x1 ) ´ f (x2 )ˇ .
我們注意到在上述定義中被取 infimum 的這個量 h(δ; x) ”
sup
ˇ
ˇ
ˇf (x1 ) ´ f (x2 )ˇ 是
x1 ,x2 PD(x,δ)
個 δ 的遞減函數（x 固定），而 osc(f, x) 則是 h(δ; x) 當 δ Ñ 0 時的極限。另外，我們也
注意到說 h(δ; x) 也可以寫成 sup f (y) ´ inf f (y).
yPD(x,δ)
yPD(x,δ)
以下的 Lemma 是關於如何檢驗一個函數在一個點是連續的。
Lemma 8.28. Let f : Rn Ñ R be a function, and x0 P Rn . Then f is continuous at x0 if
and only if osc(f, x0 ) = 0.
Proof. “ñ” Let ε ą 0 be given. Since f is continuous at x0 ,
ˇ
ˇ ε
D δ ą 0 Q ˇf (x) ´ f (x0 )ˇ ă
whenever x P D(x0 , δ).
3
In particular, for any x1 , x2 P D(x0 , δ),
ˇ
ˇ ˇ
ˇ ˇ
ˇ
ˇf (x1 ) ´ f (x2 )ˇ ď ˇf (x1 ) ´ f (x0 )ˇ + ˇf (x0 ) ´ f (x2 )ˇ ă 2ε ;
3
thus
sup
ˇ
ˇ
ˇf (x1 ) ´ f (x2 )ˇ ď 2ε which further implies that
3
x1 ,x2 PD(x0 ,δ)
0 ď osc(f, x0 ) ď
2ε
ă ε.
3
Since ε is given arbitrarily, osc(f, x0 ) = 0.
“ð” Let ε ą 0 be given. By the definition of infimum, there exists δ ą 0 such that
ˇ
ˇ
ˇf (x1 ) ´ f (x2 )ˇ ă ε .
sup
x1 ,x2 PD(x0 ,δ)
ˇ
ˇ
In particular, ˇf (x) ´ f (x0 )ˇ ď
sup
x1 ,x2 PD(x0 ,δ)
ˇ
ˇ
ˇf (x1 ) ´ f (x2 )ˇ ă ε for all x P D(x0 , δ).
˝
288
CHAPTER 8. Integration
␣
Lemma 8.29. Let f : Rn Ñ R be a function. Then for all ε ą 0, the set Dε = x P
ˇ
(
Rn ˇ osc(f, x) ě ε is closed.
Proof. Suppose that tyk u8
k=1 Ď Dε and yk Ñ y. Then for any δ ą 0, there exists N ą 0 such
that yk P D(y, δ) for all k ě N . Since D(y, δ) is open, for each k ě N there exists δk ą 0
such that D(yk , δk ) Ď D(y, δ); thus we find that
ˇ
ˇ
ˇf (x1 ) ´ f (x2 )ˇ ď
sup
x1 ,x2 PD(yk ,δk )
sup
ˇ
ˇ
ˇf (x1 ) ´ f (x2 )ˇ
@k ě N .
x1 ,x2 PD(y,δ)
The inequality above implies that osc(f, y) ě ε; thus y P Dε and Dε is closed.
˝
Theorem 8.30 (Lebesgue). Let A Ď Rn be bounded, f : A Ñ R be a bounded function, and
A
f be the extension of f by zero outside A; that is,
"
f (x) if x P A ,
A
f (x) =
0
otherwise .
A
Then f is Riemann integrable over A if and only if the collection of discontinuity of f is a
set of measure zero.
ˇ
ˇ
␣
(
␣
(
A
A
Proof. Let D = x P Rn ˇ osc(f , x) ą 0 and Dε = x P Rn ˇ osc(f , x) ě ε . We remark
8
Ť
D1 .
here that D =
k=1
k
“ñ” We show that D 1 has measure zero for all k P N (if so, then Theorem 8.24 implies
k
that D has measure zero).
Let k P N be fixed, and ε ą 0 be given. By Riemann’s condition there exists a partition
P of A such that
]
ÿ[
A
A
ε
sup f (x) ´ inf f (x) ν(∆) ă .
∆PP
xP∆
xP∆
k
Define
ˇ
␣
(
(1)
D 1 ” x P D 1 ˇ x P B∆ for some ∆ P P ,
k
k
ˇ
(
␣
(2)
D 1 ” x P D 1 ˇ x P int(∆) for some ∆ P P .
k
(1)
(2)
k
(1)
Then D 1 = D 1 Y D 1 . We note that D 1 has measure zero since it is contained in
k
k
k
k
Ť
(2)
B∆ while each B∆ has measure zero. Now we show that D 1 also has measure
∆PP
k
§8.3 Lebesgue’s Theorem
zero. Let C =
␣
289
ˇ
(
Ť
(2)
∆ P P ˇ int(∆) X D 1 ‰ H . Then D 1 Ď
∆ . Moreover, we
k
k
∆PC
1
also note that if ∆ P C, sup f (x) ´ inf f (x) ě . In fact, if ∆ P C, there exists
xP∆
k
xP∆
A
A
y P int(∆) X D 1 ; thus choosing δ ą 0 such that D(y, δ) Ď int(∆),
k
ˇ A
ˇ
A
A
A
sup f (x) ´ inf f (x) = sup ˇf (x1 ) ´ f (x2 )ˇ ě
xP∆
xP∆
x1 ,x2 P∆
sup
ˇ
ˇ A
ˇf (x1 ) ´ f A (x2 )ˇ
x1 ,x2 PD(y,δ)
ˇ
ˇ A
1
A
A
ě inf
sup ˇf (x1 ) ´ f (x2 )ˇ = osc(f , y) ě .
δą0 x1 ,x2 PD(y,δ)
k
As a consequence,
ÿ
1 ÿ
ν(∆) ď
k
∆PC
∆PP
which implies that
ř
[
]
A
A
ε
sup f (x) ´ inf f (x) ν(∆) = U (f, P) ´ L(f, P) ă
xP∆
k
xP∆
(2)
ν(∆) ă ε. In other words, we establish that D 1 has measure
k
∆PC
zero. Therefore, D has measure zero for all k P N; thus D has measure zero.
1
k
sĎ
“ð” Let R be a bounded closed rectangle with sides parallel to the coordinate axes and A
int(R), and ε ą 0 be given. Define ε1 =
ε
, where }f }8 = sup |f (x)|.
2}f }8 + ν(R) + 1
xPA
1. Since Dε1 is a subset of D, Proposition 8.19 implies that Dε1 has measure zero;
thus Proposition 8.20 provides open rectangles S1 , S2 , ¨ ¨ ¨ whose sides are parallel
8
8
ř
Ť
ν(Sk ) ă ε1 . In addition,
Sk , and
to the coordinate axes such that Dε1 Ď
k=1
k=1
we can assume that Sk Ď R for all k P N since Dε1 Ď R.
2. Since Dε1 Ď R is bounded, Lemma 8.29 implies that Dε1 is compact; thus Dε1 Ď
N
Ť
Sk for some N P N.
k=1
Ďk , and P1 be a partition of R satisfying
Let lk = S
(a) For each ∆ P P1 with ∆ X Dε1 ‰ H, ∆ Ď lk for some k = 1, ¨ ¨ ¨ , N .
(b) For each k = 1, ¨ ¨ ¨ , N , lk is the union of rectangles in P1 .
(c) Some collection of ∆ P P1 forms a partition P2 of A.
290
CHAPTER 8. Integration
R
R
SN
ñ
Dε1
S1
SN
Dε1
S1
A
A
Figure 8.2: Constructing partitions P1 and P2 from finite rectangles S1 , ¨ ¨ ¨ , SN
ˇ
␣
(
Rectangles in P1 fall into two families: C1 = ∆ P P1 ˇ ∆ Ď lk for some k = 1, ¨ ¨ ¨ , N ,
ˇ
␣
(
and C2 = ∆ P P1 ˇ ∆ Ę lk for all k = 1, ¨ ¨ ¨ , N . By the definition of the oscillation
function, for x R Dε1 we let δx ą 0 be such that
sup
A
f (y) ´
xPD(y,δy )
A
inf
xPD(y,δy )
ˇ A
ˇ
ˇf (x1 ) ´ f A (x2 )ˇ ă ε1 .
sup
f (y) =
x1 ,x2 PD(x,δx )
(
∆ is compact, there exists r ą 0 the Lebesgue number associated
∆PC2
)
Ť
with K and open cover
D(x, δx ) such that for each a P K, D(a, r) Ď D(y, δy ) for
Since K =
Ť
xPK
some y P K. Let P 1 be a refinement of P1 such that }P 1 } ă r. Then if ∆1 P P 1 satisfies
that ∆1 Ď ∆ for some ∆ P C2 , we must have ∆1 Ď D(y, δy ) for some y P K; thus
A
A
sup f (x) ´ inf1 f (x) ď sup
xP∆1
xP∆
A
f (y) ´
xPD(y,δy )
f (y)
xPD(y,δy )
ˇ A
ˇ
ˇf (x1 ) ´ f A (x2 )ˇ ă ε1
sup
=
A
inf
x1 ,x2 PD(y,δy )
ˇ
␣
(
if ∆1 Ď ∆ for some ∆ P C2 . Let P = ∆1 P P 1 ˇ ∆1 Ď ∆ for some ∆ P P2 . Then P is
a partition of A and
U (f, P) ´ L(f, P) =
(
ÿ
∆1 PP 1
∆1 Ď∆PC1
ď 2}f }8
ÿ
+
∆1 PP 1
∆1 Ď∆PC2
ÿ
∆1 PP 1
∆1 Ď∆PC1
ď 2}f }8
ÿ
)(
)
A
A
sup f (x) ´ inf1 f (x) ν(∆1 )
xP∆
xP∆1
ν(∆1 ) + ε1
ÿ
ν(∆1 )
∆1 PP 1
∆1 Ď∆PC2
ν(∆) + ε1 ν(R)
∆PPXC1
ď 2}f }8
N
ÿ
(
)
ν(Sk ) + ε1 ν(R) ă 2}f }8 + ν(R) ε1 ď ε ;
k=1
thus f is Riemann integrable over A by Riemann’s condition.
˝
§8.3 Lebesgue’s Theorem
291
Example 8.31. Let A = Q X [0, 1], and f : A Ñ R be the constant function f ” 1. Then
"
1 if x P Q X [0, 1] ,
fs(x) =
0 otherwise .
The collection of points of discontinuity of fs is [0, 1] which, by Corollary 8.23, cannot be a
set of measure zero; thus f is not Riemann integrable.
Another way to see that f is not Riemann integrable is U (f, P) = 1 and L(f, P) = 0 for
all partitions P of A.
Corollary 8.32. A bounded set A Ď Rn has volume if and only if the boundary of A has
measure zero.
Proof. It suffices to show that the collection of discontinuities of the function 1A (which is
A
the same as 1Ď
A ) is indeed BA.
1. If x0 R BA, then there exists δ ą 0 such that either D(x0 , δ) Ď A or D(x0 , δ) Ď AA ;
thus 1A is continuous at x0 R BA since 1A (x) is constant for all x P D(x0 , δ).
2. On the other hand, if x0 P BA, then there exists xk P A, yk P AA such that xk Ñ x0 and
yk Ñ x0 as k Ñ 8. This implies that 1A cannot be continuous at x0 since 1A (xk ) = 1
while 1A (yk ) = 0 for all k P N.
As a consequence, the collection of discontinuity of 1A is exactly BA, and the corollary
follows from Lebesgue’s theorem.
˝
Corollary 8.33. Let A Ď Rn be a bounded set with volume. A bounded function f : A Ñ R
with a finite or countable number of points of discontinuity is Riemann integrable over A.
ˇ
ˇ
␣
(
␣
(
Proof. We note that x P Rn ˇ osc(fs, x) ą 0 Ď BA Y x P A ˇ f is discontinuous at x . ˝
Remark 8.34. In addition to the set inclusion listed in the proof of Corollary 8.33, we also
have
␣
ˇ
ˇ
( ␣
(
x P A ˇ f is discontinuous at x Ď x P Rn ˇ osc(fs, x) ą 0 .
Therefore, if A Ď Rn is a bounded set with volume, then a bounded function f : A Ñ R is
Riemann integrable if and only if the collection of points of discontinuity of f has measure
zero.
Corollary 8.35. A bounded function is integrable over a compact set of measure zero.
292
CHAPTER 8. Integration
Proof. If f : K Ñ R is bounded, and K is a compact set of measure zero, then the collection
of discontinuities of fs is a subset of K.
˝
Corollary 8.36. Suppose that A, B Ď Rn are bounded sets with volume, and f : A Ñ R is
Riemann integrable over A. Then f is Riemann integrable over A X B.
Proof. By the inclusion
␣
ˇ
ˇ
( ␣
(
AXB
A
x P int(A X B) ˇ osc(f
, x) ą 0 Ď x P Rn ˇ osc(f , x) ą 0 ,
we find that
␣
ˇ
ˇ
(
␣
(
AXB
AXB
x P Rn ˇ osc(f
, x) ą 0 Ď B(A X B) Y x P int(A X B) ˇ osc(f
, x) ą 0
ˇ
␣
(
A
Ď BA Y BB Y x P Rn ˇ osc(f , x) ą 0 .
Since BA and BB both have measure zero, the integrability of f over A X B then follows
from the integrability of f over A and the Lebesgue Theorem.
˝
Remark 8.37. Suppose that A Ď Rn is a bounded set of measure zero. Even if f : A Ñ R
is continuous, f might not be Riemann integrable. For example, the function f given in
Example 8.31 is not Riemann integrable even though f is continuous on A.
Remark 8.38. When f : A Ñ R is Riemann integrable over A, it is not necessary that A
has volume. For example, the zero function is Riemann integrable over A = Q X [0, 1] even
though A does not has volume.
Corollary 8.39 (Lebesgue’s Differentiation Theorem, a simple version). Let A Ď Rn be a
bounded set with volume, and f : A Ñ R be bounded and Riemann integrable over A. Then
ż
1
lim
f (x) dx = f (x0 )
(8.3.2)
rÑ0 ν(D(x0 , r) X A) D(x ,r)XA
0
for almost every x0 P A; that is, the equality above does not hold only for x0 from a set of
measure zero.
Proof. Let ε ą 0 be given, and suppose that fs, the zero extension of f outside A, is
continuous at x0 . Then there exists δ ą 0 such that
ˇ
ˇ
ˇfs(x) ´ fs(x0 )ˇ ă ε
2
@ x P D(x0 , δ) X A .
§8.4 Properties Of The Integrals
293
Since BA has measure zero, by the fact that B(D(x0 , r) X A) Ď BD(x0 , r) Y BA we find that
B(D(x0 , r) X A) also has measure zero for all r ą 0. In other words, D(x0 , r) X A has volume.
Then if 0 ă r ă δ,
ˇ
ˇ
ˇ
ż
ˇ
1
ˇ
f (x) dx ´ f (x0 )ˇ
ν(D(x0 , r) X A) D(x0 ,r)XA
ż
ˇ
(
) ˇˇ
1
ˇ
=ˇ
fs(x) ´ fs(x0 ) dxˇ
ν(D(x0 , r) X A) D(x0 ,r)XA
ż
ˇ
ˇ
1
ˇfs(x) ´ fs(x0 )ˇ dx
ď
ν(D(x0 , r) X A) D(x0 ,r)XA
ż
ε
1
ε
ď
1 dx = ă ε .
2 ν(D(x0 , r) X A) D(x0 ,r)XA
2
This implies that (8.3.2) holds for all x0 at which fs is continuous. The theorem then follows
from the Lebesgue theorem.
˝
8.4 Properties of the Integrals
Proposition 8.40. Let A Ď Rn be bounded, and f, g : A Ñ R be bounded. Then
ż
(a) If B Ď A, then
(f 1B )(x) dx =
ż
A
(b)
ż
B
ż
(c) If c ě 0, then
g(x) dx ď
A
ż
ż
(d) If f ď g on A, then
f (x) dx and
A
ż
A
A
g(x) dx.
A
ż
ż
(cf )(x) dx = c
g(x) dx and
f (x) dx ď
A
ż
f (x) dx +
f (x) dx.
A
ż
f (x) dx.
ż
(f +g)(x) dx ď
A
(cf )(x) dx = c
ż
B
ż
(f +g)(x) dx ď
A
A
(f 1B )(x) dx =
A
ż
f (x) dx +
A
f (x) dx and
ż
A
ż
ż
f (x) dx ď
A
g(x) dx.
A
(e) If A has volume zero, then f is Riemann integrable over A, and
ż
f (x) dx = 0.
A
Proof. We only prove (a), (b), (c) and (e) since (d) are trivial.
(a) Let ε ą 0 be given. By the definition of the lower integral, there exist partition PA of
A and PB of B such that
ż
ÿ
A
(f 1B )(x) dx ´ ε ă L(f 1B , PA ) =
inf f 1B (x)ν(∆)
A
∆PPA
xP∆
294
CHAPTER 8. Integration
and
ż
f (x) dx ´
B
ÿ
ε
B
ă L(f, PB ) =
inf f (x)ν(∆) .
xP∆
2
∆PP
B
Let PA1 be a refinement of PA such that some collection of rectangles in PA1 forms a
B
partition of B. Denote this partition of B by PB1 . Since inf f (x) ď 0 if ∆ P PA1 zPB1 ,
xP∆
Proposition 8.6 implies that
ż
ÿ
A
inf f 1B (x)ν(∆)
(f 1B )(x) dx ´ ε ă L(f 1B , PA ) ď L(f 1B , PA1 ) =
A
(
ÿ
=
1 zP 1
∆PPA
B
ď
ÿ
1
∆PPB
ÿ )
+
inf f
xP∆
xP∆
B
inf f (x)ν(∆)
xP∆
1
∆PPB
B
1
∆PPA
ż
(x)ν(∆) = L(f, PB1 ) ď
f (x) dx .
B
rA be a partition of A such that PB Ď P
rA and
On the other hand, let P
ÿ
ε
,
ν(∆) ď
2(M + 1)
r
∆PPA zPB , ∆XB‰H
where M ą 0 is an upper bound of |f |. Then
ÿ
B
inf f (x)ν(∆) ě ´M
rA zPB , ∆XB‰H
∆PP
xP∆
ÿ
B
f (x)ν(∆) ě ´
rA zPB , ∆XB‰H
∆PP
ε
2
which further implies that
ż
ÿ
A
rA ) =
inf (f 1B (x)ν(∆)
(f 1B )(x) dx ě L(f 1B , P
A
=
( ÿ
∆PPB
rA
∆PP
xP∆
ÿ
+
rA zPB , ∆XB‰H
∆PP
B
inf f (x)ν(∆) ą
rA zPB , ∆XB‰H
∆PP
B
inf f (x)ν(∆)
xP∆
rA zPB , ∆XB=H
∆PP
ÿ
= L(f, PB ) +
)
ÿ
+
ż
f (x) dx ´ ε .
xP∆
B
Therefore, we establish that
ż
ż
ż
f (x) dx ´ ε ă
(f 1B )(x) dx ă
f (x) dx + ε .
B
A
B
Since ε ą 0 is given arbitrarily, we conclude that
ż
(f 1B )(x) dx =
A
argument can be applied to conclude that
ż
(f 1B )(x) dx =
A
ż
f (x) dx. Similar
B
ż
f (x) dx.
B
§8.4 Properties Of The Integrals
295
(b) Let ε ą 0 be given. By the definition of the lower integral, there exist partitions P1
and P2 of A such that
ż
ε
f (x) dx ´ ă L(f, P1 ) and
2
A
ż
g(x) dx ´
A
ε
ă L(g, P2 ) .
2
Let P be a common refinement of P1 and P2 . Then
ż
ż
f (x) dx + g(x) dx ´ ε ă L(f, P1 ) + L(f, P2 ) ď L(f, P) + L(g, P)
A
A
ÿ
ÿ
=
inf fs(x)ν(∆) +
inf gs(x)ν(∆)
∆PP
ÿ
ď
∆PP
xP∆
∆PP
xP∆
inf (fs + gs)(x)ν(∆) = L(f + g, P) ď
xP∆
ż
(f + g)(x) dx .
A
Since ε ą 0 is given arbitrarily, we conclude that
ż
ż
ż
f (x) dx +
g(x) dx ď
(f + g)(x) dx .
A
Similarly, we have
A
ż
A
ż
ż
(f + g)(x) dx ď
A
g(x) dx; thus (b) is established.
f (x) dx +
A
A
(c) It suffices to show the case c = ´1. For each k P N, there exist partitions Pk and Qk
of A such that
ż
1
´f (x) dx ´ ă L(´f, Pk ) ď
k
A
ż
´f (x) dx
A
and
ż
f (x) dx ď U (f, Qk ) ă
A
ż
f (x) dx +
A
1
.
k
Let Rk be the common refinement of Pk and Qk . Then
ż
ż
1
´f (x) dx ´ ă L(´f, Pk ) ď L(´f, Rk ) ď
´f (x) dx
k
A
A
and
ż
f (x) dx ď U (f, Rk ) ď U (f, Qk ) ă
A
ż
f (x) dx +
A
1
.
k
which implies that
lim U (f, Rk ) =
kÑ8
ż
f (x) dx
A
and
lim L(´f, Rk ) =
kÑ8
ż
´f (x) dx
A
296
CHAPTER 8. Integration
Since
A
ÿ
L(´f, Rk ) =
∆PRk
we conclude that
ÿ
inf (´f ) (x)ν(∆) = ´
xP∆
A
sup f (x)ν(∆) = U (f, Rk ) ,
∆PRk xP∆
ż
ż
´f (x) dx = ´
A
f (x) dx.
A
(e) Since f is bounded on A, there exist M ą 0 such that ´M ď f (x) ď M for all x P A.
f
ď 1A on A; thus (c) and (d) imply that
Therefore, ´1A ď
M
ż
ż
ż
ż
f (x)
1
0=
1A (x) dx =
1A (x) dx ě
dx =
f (x) dx
M A
A
A
A M
which implies that
that
ż
A
ż
ż
f (x) dx ď 0. Similarly,
A
´f (x) dx ď 0 which further implies
A
f (x) dx ě 0. Therefore, by Corollary 8.7 we conclude that
ż
ż
0ď
f (x) dx ď
f (x) dx ď 0
A
A
ż
˝
which implies that f is Riemann integrable over A and
f (x) dx = 0.
A
Remark 8.41. Let A Ď Rn be bounded and f, g : A Ñ R be bounded. Then (b) of
Proposition 8.40 also implies that
ż
ż
ż
ż
ż
ż
(f ´g)(x) dx ď
f (x) dx´ g(x) dx and
f (x) dx´ g(x) dx ď
(f ´g)(x) dx .
A
A
A
A
A
A
Corollary 8.42. Let A, B Ď R be bounded such that A X B has volume zero, and f :
A Y B Ñ R be bounded. Then
ż
ż
ż
ż
ż
ż
f (x) dx +
f (x) dx ď
f (x) dx ď
f (x) dx ď
f (x) dx +
f (x) dx .
n
A
B
AYB
AYB
A
B
Proof. Note that f 1A + f 1B = f 1AYB + f 1AXB on A Y B. Therefore, (a), (b) of Proposition
8.40 and Remark 8.41 implies that
ż
ż
ż
ż
ż
f (x) dx +
f (x) dx =
(f 1A )(x) dx +
(f 1B )(x) dx ď
(f 1A + f 1B )(x) dx
A
B
AYB
AYB
AYB
ż
(
)
=
f 1AYB ´ (´f 1AXB ) (x) dx
żAYB
ż
f 1AYB (x) dx ´
(´f 1AXB )(x) dx
ď
AYB
AYB
ż
ż
=
f (x) dx ´
(´f )(x) dx
AYB
AXB
§8.4 Properties Of The Integrals
297
which, with the help of Proposition 8.40 (e), further implies that
ż
ż
ż
f (x) dx +
f (x) dx ď
f (x) dx .
A
B
AYB
The case of the upper integral can be proved in a similar fashion.
˝
Having established Proposition 8.40, it is easy to see the following theorem (except (c)).
The proof is left as an exercise.
Theorem 8.43. Let A Ď Rn be bounded, c P R, and f, g : A Ñ R be Riemann integrable.
Then
ż
(a) f ˘ g is Riemann integrable, and
ż
(f ˘ g)(x) dx =
A
(b) cf is Riemann integrable, and
ż
f (x) dx ˘
A
ż
g(x) dx.
A
ż
(cf )(x) dx = c
A
f (x) dx.
A
ˇ ż
ˇż
ˇ
ˇ
|f (x)|dx.
(c) |f | is Riemann integrable, and ˇ f (x) dxˇ ď
A
A
(d) If f ď g, then
ż
ż
f (x) dx ď
A
g(x) dx.
A
ˇż
ˇ
ˇ
ˇ
(e) If A has volume and |f | ď M , then ˇ f (x) dxˇ ď M ν(A).
A
Theorem 8.44. Let A Ď R be bounded, and f : A Ñ R be a bounded integrable function.
n
1. If A has measure zero, then
ż
f (x) dx = 0.
A
2. If f (x) ě 0 for all x P A, and
measure zero.
ż
ˇ
␣
(
f (x) dx = 0, then the set x P A ˇ f (x) ‰ 0 has
A
Proof. 1. We show that L(f, P) ď 0 ď U (f, P) for all partitions P of A. Let P =
␣
(
∆1 , ¨ ¨ ¨ , ∆N be a partition of A. By Corollary 8.23, ∆k X AA ‰ H for k = 1, ¨ ¨ ¨ , N ;
thus we must have inf fs(x) ď 0 and sup fs(x) ě 0. As a consequence, if P is a
xP∆k
partition of A,
L(f, P) =
N
ÿ
k=1
thus
inf fs(x)ν(∆k ) ď 0 and U (f, P) =
xP∆k
ż
N
ÿ
sup fs(x)ν(∆k ) ě 0 ;
k=1 xP∆k
ż
f (x) dx ď 0 ď
A
xP∆k
A
f (x)dx. Since f is integrable over A,
ż
f (x) dx = 0.
A
298
CHAPTER 8. Integration
ˇ
␣
1(
2. Let Ak = x P A ˇ f (x) ě
. We claim that Ak has measure zero for all k P N.
k
ż
Let ε ą 0 be given. Since
f (x)dx = 0, there exists a partition P of A such that
A ˇ
␣
(
Ť
ε
U (f, P) ă . Let C = ∆ P P ˇ ∆ X Ak ‰ H . Then Ak Ď
∆, and
k
∆PC
ÿ
ÿ
1 ÿ
ε
ν(C) ď
sup fs(x)ν(∆) ď
sup fs(x)ν(∆) = U (f, P) ă
k ∆PC
k
∆PC xP∆
∆PP xP∆
which implies that
implies that A =
ř
ν(∆) ă ε. Therefore, Ak has measure zero; thus Theorem 8.24
∆PC
8
Ť
Ak also has measure zero.
˝
k=1
Remark 8.45. Combining Corollary 8.35 and Theorem 8.44, we conclude that the integral
of a bounded function over a compact set of measure zero is zero.
Remark 8.46. Let A = Q X [0, 1] and f : A Ñ R be the constant function f ” 1. We have
shown in Example 8.31 that f is not Riemann integrable. We note that A has no volume
since BA = [0, 1] which is not a set of measure zero. However, A has measure zero since it
consists of countable number of points.
1. Since f is continuous on A, the condition that A has volume in Corollary 8.33 cannot
be removed.
2. Since A has measure zero, the condition that f is Riemann integrable in Theorem 8.44
cannot be removed.
Theorem 8.47 (Mean Value Theorem for Integrals). Let A be a subset of Rn such that A
has volume and is compact and connected. Suppose that f : A Ñ R is continuous, then there
exists x0 P A such that
ż
f (x) dx = f (x0 )ν(A) .
A
1
The quantity
ν(A)
ż
f (x) dx is called the average of f over A.
A
Proof. Because of Theorem 8.44, it suffices to show the case that ν(A) ‰ 0. Let m =
min f (x) and M = max f (x). Then
xPA
xPA
m1A (x) ď f (x) ď M 1A (x) ;
§8.5 Fubini’s Theorem
299
thus (b) and (d) of Theorem 8.43 imply that
ż
ż
mν(A) =
ż
m1A (x) dx ď
A
M 1A (x) dx = M ν(A) .
f (x) dx ď
A
A
By the connectedness of A and continuity of f , Theorem 4.21 żand Theorem 3.38 implies
that f (A) = [m, M ]; thus by the fact that the quantity
1
ν(A)
must be x0 P A such that
1
f (x0 ) =
ν(A)
f (x) dx P [m, M ], there
A
ż
f (x) dx .
˝
A
Definition 8.48. Let A Ď Rn be a set and f : A Ñ R be a function. For B Ď A, the
ˇ
restriction of f to B is the function f ˇ : A Ñ R given by f |B = f 1B . In other words,
B
ˇ
f ˇ (x) =
"
B
f (x) if x P B ,
0
if x P AzB .
The following lemma is a direct consequence of Proposition 8.40 (a).
Lemma 8.49. Let A Ď Rn be bounded, and f : A Ñ R be a bounded function. Suppose that
ˇ
B Ď A, and f ˇ is Riemann integrable over A. Then f is Riemann integrable over B, and
B
ż
A
ˇ
f ˇ (x) dx =
B
ż
f (x) dx .
B
Theorem 8.50. Let A, B be bounded subsets of Rn be such that A X B has measure zero,
ˇ
ˇ
ˇ
and f : A Y B Ñ R be such that f ˇAXB , f ˇA and f ˇB are all Riemann integrable over A Y B.
Then f is integrable over A Y B, and
ż
ż
ż
f (x) dx =
AYB
f (x) dx +
A
f (x) dx .
B
Proof. Since 1AYB = 1A + 1B ´ 1AXB , we have
ˇ
ˇ
ˇ
f = f 1AYB = f ˇA + f ˇB ´ f ˇAXB ;
thus Theorem 8.43, Theorem 8.44 and Lemma 8.49 imply that
ż
ż
ż
ż
ż
ˇ
ˇ
f (x) dx =
f ˇA (x) dx +
f ˇB (x) dx =
f (x) dx + f (x) dx .
AYB
AYB
AYB
A
B
˝
300
CHAPTER 8. Integration
8.5 The Fubini Theorem
If f : [a, b] Ñ R is continuous, the fundamental theorem of Calculus (Theorem 4.90) can be
applied to computed the integral of f over [a, b]. In the following two sections, we focus on
how the integral of f over A Ď Rn , where n ě 2, can be computed if the integral exists. We
start with the special case n = 2.
Definition 8.51. Let S = [a, b] ˆ [c, d] be a rectangle in R2 , and f : S Ñ R be bounded.
For each fixed x P [a, b], the lower integral of the function f (x, ¨) : [c, d] Ñ R is denoted
by
żd
f (x, y) dy, and the upper integral of f (x, ¨) : [c, d] Ñ R is denoted by
c
żd
f (x, y) dy.
c
If for each x P [a, b] the upper integral and the lower integral of f (x, ¨) : [c, d] Ñ R are
żd
the same, we simply write
żb
żb
f (x, y) dx,
a
f (x, y) dy for the integrals of f (x, ¨) over [c, d]. The integrals
c
f (x, y) dx and
a
żb
f (x, y) dx are defined in a similar way.
a
Lemma 8.52. Let A = [a, b] ˆ [c, d] be a rectangle in R2 , and f : A Ñ R be bounded. Then
ż
ż b(ż d
ż b(ż d
ż
)
)
f (x, y) dA ď
f (x, y) dy dx ď
f (x, y) dy dx ď
f (x, y) dA (8.5.1)
A
a
and
ż
c
ż d(ż b
f (x, y) dA ď
A
c
a
c
A
ż d(ż b
ż
)
)
f (x, y) dx dy ď
f (x, y) dx dy ď
f (x, y) dA . (8.5.2)
a
c
a
A
Proof. It suffices to prove (8.5.1). Let ε ą 0 be given. Choose a partition
ˇ
␣
(
P = ∆ij ˇ ∆ij = [xi , xi+1 ] ˆ [yj , yj+1 ] for i = 0, 1, ¨ ¨ ¨ , n ´ 1 and j = 0, 1, ¨ ¨ ¨ , m ´ 1
of A such that L(f, P) ą
ż
f (x, y) dA ´ ε. Using (4.7.3) and Remark 4.82, we find that
A
ż b(ż d
)
f (x, y) dy dx =
a
c
n´1
ÿ ż xi+1 ( m´1
ÿ ż yj+1
i=0
ě
xi
yj
n´1
ÿ m´1
ÿ ż xi+1 ( ż yj+1
i=0 j=0
ě
j=0
n´1
ÿ m´1
ÿ
i=0 j=0
xi
inf
(x,y)P∆ij
)
f (x, y) dy dx
)
f (x, y) dy dx
yj
f (x, y)ν(∆ij ) = L(f, P) ą
ż
f (x, y) dA ´ ε .
A
§8.5 Fubini’s Theorem
301
Since ε ą 0 is given arbitrarily, we must have
)
ż b(ż d
ż
f (x, y) dy dx ě
a
Similarly,
ż b (ż d
A
)
ż
f (x, y) dA, so (8.5.1) is concluded.
f (x, y) dy dx ď
a
f (x, y) dA .
c
c
˝
A
Theorem 8.53 (Fubini’s Theorem, the case n = 2). Let A = [a, b] ˆ [c, d] be a rectangle in
R2 , and f : A Ñ R be Riemann integrable. Then
1. the functions
2. the functions
żd
żd
f (¨, y) dy and
c
c
żb
żb
f (x, ¨)dx and
a
f (¨, y) dy are Riemann integrable over [a, b];
f (x, ¨)dx are Riemann integrable over [c, d], and
a
3. The integral of f over A is the same as the iterated integrals; that is,
ż b(ż d
ż
f (x, y) dA =
f (x, y) dy dx =
A
a
c
c
a
ż d(ż b
=
Proof. It suffices to prove that
ż b(ż d
)
żd
a
c
c
a
)
f (x, y) dy dx
ż d(ż b
)
)
f (x, y) dx dy =
f (x, y) dx dy .
f (x, y) dy is Riemann integrable over [a, b] and
c
ż b(ż d
)
ż
f (x, y) dy dx =
a
Since
ż b (ż d
)
c
ż b (ż d
f (x, y) dy dx ď
a
c
a
c
f (x, y) dA ď
A
c
ż b(ż d
a
The integrability of
ż b(ż d
)
a
c
)
f (x, y) dy dx
c
ż
f (x, y) dy dx ď
ď
żd
)
f (x, y) dy dx ď
a
(8.5.3)
)
f (x, y) dy dx , Lemma 8.52 implies that
ż b(ż d
ż
f (x, y) dA .
A
f (x, y) dA .
A
f (x, y) dy and the validity of (8.5.3) are then concluded by the
c
integrability of f over A.
˝
302
CHAPTER 8. Integration
żbżd
Remark 8.54. To simplify the notation, sometimes we use
f (x, y) dydx to denote
ż b (ż d
) a c
the iterated integral the iterated integral
f (x, y) dy dx. Similar notation applies
a
c
to the upper and the lower integrals. For example, we also have
ż b (ż d
)
f (x, y) dy dx.
a
żb żd
f (x, y) dydx =
a
c
c
Remark 8.55. For each x P [a, b], define φ(x) =
żd
f (x, y) dy and ψ(x) =
żd
c
f (x, y) dy.
c
Then φ(x) ď ψ(x) for all x P [a, b], and the Fubini Theorem implies that
żb
[
]
ψ(x) ´ φ(x) dx = 0 .
a
ˇ
(
By Theorem 8.44, the set x P [a, b] ˇ ψ(x) ´ φ(x) ‰ 0 has measure zero. In other words,
␣
except on a set of measure zero, f (x, ¨) is Riemann integrable over [c, d] if f is Riemann
integrable over [a, b] ˆ [c, d]. This property can be rephrased as that “f (x, ¨) is Riemann
integrable over [c, d] for almost every x P [a, b] if f is Riemann integrable over the rectangle
[a, b] ˆ [c, d]”. Similarly, f (¨, y) is Riemann integrable for almost every y P [c, d] if f is
Riemann integrable over [a, b] ˆ [c, d].
Remark 8.56. The integrability of f does not guarantee that f (x, ¨) or f (¨, y) is Riemann
integrable. In fact, there exists a function f : [0, 1] ˆ [0, 1] Ñ R such that f is Riemann
integrable, f (¨, y) is Riemann integrable for each y P [0, 1], but f (x, ¨) is not Riemann
integrable for infinitely many x P [0, 1]. For example, let
$
& 0 if x = 0 or if x or y is irrational ,
f (x, y) =
1
q
%
if x, y P Q and x = with (p, q) = 1 .
p
p
Then
1. For each y P [0, 1], f (¨, y) żis continuous atżall irrational numbers. Therefore, f (¨, y) is
1
Riemann integrable, and
1
f (x, y) dx =
0
f (x, y) dx = 0.
0
2. For x = 0 or x R Q, f (x, ¨) is Riemann integrable, and
ż1
f (x, y) dy = 0.
0
§8.5 Fubini’s Theorem
3. If x =
303
q
with (p, q) = 1, f (x, ¨) is nowhere continuous in [0, 1]. In fact, for each
p
y0 P [0, 1],
lim f (x, y) =
yÑy0
yPQ
1
p
while
lim f (x, y) = 0 ;
yÑy0
yRQ
thus the limit of f (x, y) as y Ñ y0 does not exist. Therefore, the Lebesgue theorem
implies that f (x, ¨) is not Riemann integrable if x P Q X (0, 1]. On the other hand, for
q
x = with (p, q) = 1 we have
p
ż1
f (x, y) dy = 0
ż1
and
f (x, y) dy =
0
4. Define φ(x) =
ż1
0
f (x, y) dy and ψ(x) =
0
ż1
1
.
p
f (x, y) dy. Then 2 and 3 imply that φ
0
and ψ are Riemann integrable over [0, 1], and
ż1
ż1
φ(x)dx =
0
ψ(x)dx = 0.
0
5. For each a R Q X [0, 1] and b P [0, 1], f is continuous at (a, b). In fact, for any given
ε ą 0, there exists a prime number p such that
1
ă ε. Let
p
ˇ
!ˇ
)
ℓ ˇˇ ˇ
ˇ
δ = min a ´ ˇ 0 ď ℓ ď k ď p, k P N, ℓ P N Y t0u .
k
(
)
Then δ ą 0, and if (x, y) P D (a, b), δ X ([0, 1] ˆ [0, 1]), we have
ˇ
ˇ ˇ
ˇ
ˇf (x, y) ´ f (a, b)ˇ = ˇf (x, y)ˇ ă 1 ă ε ,
p
(
)
ℓ
where we use the fact that if (x, y) P D (a, b), δ and x P Q, then x = (in reduced
k
form) for some k ą p.
ˇ
␣
(
As a consequence, (a, b) P R2 ˇ fs is discontinuous at (a, b) Ď Q ˆ [0, 1]. Since
Q ˆ [0, 1] is a countable union of measure zero sets, it has measure zero; thus f is
Riemann integrable by the Lebesgue theorem. The Fubini theorem then implies that
ż1ż1
ż
f (x, y) dxdy = 0 .
f (x, y) dA =
[0,1]ˆ[0,1]
0
0
Remark 8.57. The integrability of f (x, ¨) and f (¨, y) does not guarantee the integrability of
f . In fact, there exists a bounded function f : [0, 1] ˆ [0, 1] Ñ R such that f (x, ¨) and f (¨, y)
304
CHAPTER 8. Integration
are both Riemann integrable over [0, 1], but f is not Riemann integrable over [0, 1] ˆ [0, 1].
For example, let
$
& 1 if (x, y) = ( k , ℓ ), 0 ă k, ℓ ă 2n odd numbers, n P N ,
2n 2n
f (x, y) =
%
0 otherwise .
Then for each x P [0, 1], f (x, ¨) only has finite number of discontinuities; thus f (x, ¨) is
Riemann integrable, and
ż
1
f (x, y) dy = 0 .
0
Similarly, f (¨, y) is Riemann integrable, and
ż1
f (x, y) dx = 0. As a consequence,
0
ż1ż1
ż1ż1
f (x, y) dydx =
0
f (x, y) dxdy = 0 .
0
0
0
However, note that f is nowhere continuous on [0, 1] ˆ [0, 1]; thus the Lebesgue theorem
implies that f is not Riemann integrable. One can also see this by the fact that U (f, P) = 1
and L(f, P) = 0 for all partition of [0, 1] ˆ [0, 1].
Corollary 8.58. 1. Let φ1 , φ2 : [a, b] Ñ R be continuous maps such that φ1 (x) ď φ2 (x)
ˇ
␣
(
for all x P [a, b], A = (x, y) ˇ a ď x ď b, φ1 (x) ď y ď φ2 (x) , and f : A Ñ R be
continuous. Then f is Riemann integrable over A, and
ż
ż b ( ż φ2 (x)
)
f (x, y) dA =
f (x, y) dy dx .
A
a
φ1 (x)
2. Let ψ1 , ψ2 : [c, d] Ñ R be continuous maps such that ψ1 (y) ď ψ2 (y) for all y P [c, d],
ˇ
␣
(
A = (x, y) ˇ c ď y ď d, ψ1 (y) ď x ď ψ2 (y) , and f : A Ñ R be continuous. Then f is
Riemann integrable over A, and
ż
ż d ( ż ψ2 (y)
)
f (x, y) dA =
f (x, y) dx dy .
A
c
ψ1 (y)
Proof. It suffices to prove 1. First we show that f is Riemann integrable over A. By
ˇ (
␣
(
)
Lebesgue’s theorem, it suffices to show that the set (x, y) P R2 ˇ osc fs, (x, y) ą 0 has
measure zero, where fs is the extension of f by zero outside A. Nevertheless, we note that
␣
ˇ (
(
)
[
]
[
]
(x, y) P R2 ˇ osc fs, (x, y) ą 0 Ď tau ˆ φ1 (a), φ2 (a) Y tbu ˆ φ1 (b), φ2 (b) Y
␣(
( ␣(
(
)ˇ
)ˇ
Y x, φ1 (x) ˇ x P [a, b] Y x, φ2 (x) ˇ x P [a, b] .
§8.5 Fubini’s Theorem
305
[
]
[
]
It is clear that tauˆ φ1 (a), φ2 (a) and tbuˆ φ1 (b), φ2 (b) have measure zero since they have
␣(
(
␣(
(
)ˇ
)ˇ
volume zero. Now we claim that the sets x, φ1 (x) ˇ x P [a, b] and x, φ2 (x) ˇ x P [a, b]
also have measure zero.
Let ε ą 0 be given. Since φ1 is continuous on a compact set [a, b], φ1 is uniformly
continuous on [a, b]; thus there exists δ ą 0 such that
ˇ
ˇ
ˇφ1 (x1 ) ´ φ1 (x2 )ˇ ă
ε
b´a
whenever |x1 ´ x2 | ă δ .
Let P = ta = x0 ă x1 ă ¨ ¨ ¨ ă xn´1 ă xn = bu be a partition of [a, b] such that |xi+1 ´xi | ă δ
[
]
] [
for all i = 0, ¨ ¨ ¨ , n ´ 1, and let ∆i = xi , xi+1 ˆ
min φ1 (x), max φ1 (x) . Then
xP[xi ,xi+1 ]
xP[xi ,xi+1 ]
ď
␣(
( n´1
)ˇ
x, φ1 (x) ˇ x P [a, b] Ď
∆i
i=0
and
n´1
ε
ε ÿ
(xi+1 ´ xi ) = ε .
(xi+1 ´ xi ) =
b
´
a
b
´
a
i=0
i=0
i=0
␣(
(
␣(
(
)ˇ
)ˇ
Therefore, x, φ1 (x) ˇ x P [a, b] has volume zero; thus x, φ1 (x) ˇ x P [a, b] has measure
␣(
(
␣
)ˇ
zero. Similarly, x, φ2 (x) ˇ x P [a, b] also has measure zero. By Theorem 8.24, (x, y) P
ˇ (
(
)
R2 ˇ osc fs, (x, y) ą 0 has measure zero; thus f is Riemann integrable over A.
n´1
ÿ
ν(∆i ) ă
n´1
ÿ
Let m = min φ1 (x), M = max φ2 (x), and S = [a, b] ˆ [m, M ]. Then A Ď S. By Lemma
xP[a,b]
xP[a,b]
8.49 and the Fubini Theorem,
ż b ( ż φ2 (x)
ż b(ż M
ż
ż
)
)
s
s
f (x, y) dy dx
f (x, y) dy dx =
f (x, y) dA = f (x, y) dA =
A
S
a
m
a
φ1 (x)
which concludes 1.
˝
ˇ
(
␣
Example 8.59. Let A = (x, y) P R2 ˇ 0 ď x ď 1, x ď y ď 1 , and f : A Ñ R be given by
f (x, y) = xy. Then Corollary 8.58 implies that
ż1(
ż 1(ż 1
ż 1 2 ˇy=1
ż
)
x x3 )
xy ˇ
1 1
1
f (x, y) dA =
xy dy dx =
´
dx = ´ = .
ˇ dx =
2 y=x
2
2
4 8
8
0
A
0
x
0
ˇ
␣
(
On the other hand, since A = (x, y) P R2 ˇ 0 ď y ď 1, 0 ď x ď y , we can also evaluate the
integral of f over A by
ż1 3
ż 1 2 ˇx=y
ż
ż 1(ż y
)
y
x yˇ
1
xy dx dy =
dy = .
xy dA =
ˇ dy =
2 x=0
8
0 2
0
0
0
A
306
CHAPTER 8. Integration
ˇ
␣
(
?
Example 8.60. Let A = (x, y) P R2 ˇ 0 ď x ď 1, x ď y ď 1 , and f : A Ñ R be given by
3
f (x, y) = ey . Then Corollary 8.58 implies that
)
y3
dy
dx .
e
?
ż 1(ż 1
ż
f (x, y) dA =
A
0
x
Since we do not know how to compute the inner integral, we look for another way of finding
ˇ
␣
(
the integral. Observing that A = (x, y) P R2 ˇ 0 ď y ď 1, 0 ď x ď y 2 , we have
ż 1 ( ż y2
ż
f (x, y) dA =
A
0
0
ż1
)
1 y3 ˇˇy=1 e ´ 1
2 y3
e dx dy =
y e dy = e ˇ
=
.
3
3
y=0
0
y3
Now we prove the general Fubini Theorem.
Theorem 8.61 (Fubini’s Theorem). Let A Ď Rn and B Ď Rm be rectangles, and f :
A ˆ B Ñ R be bounded. For x P Rn and y P Rm , write z = (x, y). Then
ż (ż
ż
f (z) dz ď
)
ż (ż
f (x, y)dy dx ď
AˆB
A
B
)
ż
f (x, y)dy dx ď
A
B
f (z) dz
(8.5.4)
f (z) dz .
(8.5.5)
AˆB
and
ż (ż
ż
f (z) dz ď
AˆB
B
ż (ż
ż
)
)
f (x, y)dx dy ď
f (x, y)dx dy ď
A
B
A
AˆB
In particular, if f : A ˆ B Ñ R is Riemann integrable, then
ż (ż
ż
)
f (z) dz =
AˆB
ż (ż
)
f (x, y)dy dx
f (x, y)dy dx =
A
B
B
A
ż (ż
=
A
B
B
A
ż (ż
)
)
f (x, y)dx dy =
f (x, y)dx dy .
Proof. It sufficesż to prove (8.5.4). Let ε ą 0 be given. Choose a partition P of A ˆ B such
f (z) dz ´ ε. Since P is a partition of A ˆ B, there exist partition Px
ˇ
␣
(
of A and partition Py of B such that P = ∆ = R ˆ S ˇ R P Px , S P Py . By Proposition
that L(f, P) ą
AˆB
§8.5 Fubini’s Theorem
307
8.40 and Corollary 8.42, we find that
ż (ż
ż
)
(ż
f (x, y) dy dx = Ť
1A (x) Ť
A
B
RPPx R
ě
ě
R
f
R
ÿ
ě
ÿ
f
AˆB
)
(x, y) dy dx
f
AˆB
(x,y)PRˆS
AˆB
inf f
(x,y)P∆
∆PP
)
(x, y) dy dx
S
inf
R PPx ,S PPy
AˆB
S
S PPy
ÿ ÿ ż (ż
R PPx S PPy
=
SPPy S
ÿ ż ( ÿ ż
R PPx
)
f (x, y)1B (y) dy dx
(x, y)νm (S)νn (R)
(x, y)νn+m (∆) = L(f, P) ą
ż
f (z)dz ´ ε .
AˆB
Since ε ą 0 is given arbitrarily, we conclude that
ż
ż (ż
)
f (z) dz ď
f (x, y)dx dy .
AˆB
Similarly,
ż (ż
A
B
ż
)
f (x, y)dy dx ď
B
A
f (z) dz; thus (8.5.4) is concluded.
˝
AˆB
Corollary 8.62. Let S Ď Rn be a bounded set with volume, φ1 , φ2 : S Ñ R be continuous
ˇ
␣
(
maps such that φ1 (x) ď φ2 (x) for all x P S, A = (x, y) P Rn ˆR ˇ x P S, φ1 (x) ď y ď φ2 (x) ,
and f : A Ñ R be continuous. Then f is Riemann integrable over A, and
ż ( ż φ2 (x)
ż
f (x, y) d(x, y) =
S
A
)
f (x, y) dy dx .
(8.5.6)
φ1 (x)
Proof. Since BA has measure zero, and f is continuous on A, Corollary 8.33 implies that f is
Riemann integrable over A. Let m = min φ1 (x) and M = max φ2 (x). Then A Ď S ˆ[m, M ];
xPS
xPS
thus Theorem 8.50 and the Fubini Theorem imply that
ż
ż
f (x, y) d(x, y) =
A
ż (ż M
A
f (x, y) d(x, y) =
Sˆ[m,M ]
ż (ż M
A
)
S
)
A
f (x, y) dy dx
m
f (x, y) dy dx .
=
S
m
A
Noting that [m, M ] has a boundary of volume zero in R, and for each x P S, f (x, ¨) is
A
continuous except perhaps at y = φ1 (x) and y = φ2 (x), Corollary 8.33 implies that f (x, ¨) is
308
CHAPTER 8. Integration
Riemann integrable over [m, M ] for each x P S. Therefore,
żM
żM
A
f (x, y) dy =
m
which further implies that
ż
ż (ż M
f (x, y) d(x, y) =
A
S
A
f (x, y) dy
m
)
A
f (x, y) dy dx .
(8.5.7)
m
ˇ
␣
(
A
For each fixed x P S, let Ax = y P R ˇ φ1 (x) ď y ď φ2 (x) . Then f (x, y) = f (x, y)1Ax (y)
A
for all (x, y) P S ˆ [m, M ] or equivalently, f (x, ¨) = f (x, ¨)|Ax for all x P S; thus Proposition
8.40 (a) implies that
żM
ż
A
f (x, y) dy =
m
ż φ2 (x)
f (x, y) dy =
f (x, y) dy
Ax
@x P S .
(8.5.8)
φ1 (x)
Combining (8.5.7) and (8.5.8), we conclude (8.5.6).
˝
ˇ
␣
Example 8.63. Let A Ď R3 be the set (x1 , x2 , x3 ) P R3 ˇ x1 ě 0, x2 ě 0, x3 ě 0, and x1 +
(
x2 + x3 ď 1 , and f : A Ñ R be given by f (x1 , x2 , x3 ) = (x1 + x2 + x3 )2 . Let S =
[0, 1] ˆ [0, 1] ˆ [0, 1], and fs : R3 Ñ R be the extension of f by zero outside A. Then
Corollary 8.33 implies that f is Riemann integrable (since BA has measure zero). Write
p1 = (x2 , x3 ), x
p2 = (x1 , x3 ) and x
p3 = (x1 , x2 ). Lemma 8.49 implies that
x
ż
ż
f (x)dx = fs(x)dx ,
A
S
and Theorem 8.61 implies that
ż
ż
s
f (x)dx =
S
)
fs(p
x3 dx3 .
x3 , x3 )dp
(ż
[0,1]
[0,1]ˆ[0,1]
ˇ
(
Let Ax3 = (x1 , x2 ) P R2 ˇ x1 ě 0, x2 ě 0, x1 + x2 ď 1 ´ x3 . Then for each x3 P [0, 1],
ż 1´x3 ( ż 1´x3 ´x2
ż
ż
)
s
f (p
x3 , x3 )dp
x3 =
f (p
x3 , x3 )dp
x3 =
f (x1 , x2 , x3 )dx1 dx2 .
␣
[0,1]ˆ[0,1]
Ax3
0
0
Computing the iterated integral, we find that
ż
ż 1 [ ż 1´x3 ( ż 1´x3 ´x2
) ]
2
f (x)dx =
(x1 + x2 + x3 ) dx1 dx2 dx3
A
0
0
0
ż 1 [ ż 1´x3
]
(x1 + x2 + x3 )3 ˇˇx1 =1´x3 ´x2
dx2 dx3
=
ˇ
3
x1 =0
ż01 [ ż01´x3 (
]
3)
1 (x2 + x3 )
=
´
dx2 dx3
3
3
ż01 ( 0
4)
1 x3 x3
1 1
1
15 ´ 10 + 1
1
=
´
+
dx3 = ´ +
=
=
.
4
3
12
4 6 60
60
10
0
§8.6 Change of Variables Formula
309
Example 8.64. In this example we compute the volume of the n-dimensional unit ball ωn .
By the Fubini theorem,
ż 1 ż ?1´x21
ωn =
´1
Note that the integral
´
´
1´x21
dxn ¨ ¨ ¨ dx1 .
?
1´x21 ´¨¨¨´x2n´1
n´1
1
¨¨¨
1´x21
?
´
ż1
ωn =
´
ż ?1´x2 ´¨¨¨´x2
1
?
¨¨¨
?
ż ?1´x2
ż ?1´x21 ´¨¨¨´x2n´1
n´1
1´x21 ´¨¨¨´x2n´1
2
ωn´1 (1 ´ x )
n´1
2
dxn ¨ ¨ ¨ dx2 is in fact ωn´1 (1´x21 ) 2 ; thus
żπ
2
dx = 2 ωn´1
cosn θdθ .
(8.5.9)
0
´1
Integrating by parts,
żπ
żπ
żπ
ˇθ= π
2
2
2
ˇ 2
n
n´1
n´1
cos θ dθ =
cos
θ d(sin θ) = cos
θ sin θˇ
+ (n ´ 1)
cosn´2 θ sin2 θ dθ
θ=0
0
0
0
żπ
2
cosn´2 θ(1 ´ cos2 θ) dθ
= (n ´ 1)
0
which implies that
żπ
n´1
cos θ dθ =
n
0
2
n
żπ
2
cosn´2 θ dθ .
0
As a consequence,
żπ
2
0
n
cos θ dθ =
$
żπ
(n ´ 1)(n ´ 3) ¨ ¨ ¨ 2 2
’
’
cos θ dθ
&
n(n ´ 2) ¨ ¨ ¨ 3
’
’
%
if n is odd ,
0
(n ´ 1)(n ´ 3) ¨ ¨ ¨ 1
n(n ´ 2) ¨ ¨ ¨ 2
żπ
2
dθ
if n is even ;
0
2ω
n´2
thus the recursive formula (8.5.9) implies that ωn =
π . Further computations shows
n
that
$
n´1
’
(2π) 2
’
’
ω1
if n is odd ,
&
n(n ´ 2) ¨ ¨ ¨ 3
ωn =
n´2
’
’
(2π) 2
’
%
ω2
if n is even .
n(n ´ 2) ¨ ¨ ¨ 4
ż8
Let Γ be the Gamma function defined by Γ(t) =
xt´1 e´x dx for t ą 0. Then Γ(x + 1) =
0
(1) ?
xΓ(x) for all x ą 0, Γ(1) = 1 and Γ
= π. By the fact that ω1 = 2 and ω2 = π, we can
express ωn as
2
n
π2
ωn = ( n+2 ) .
Γ 2
310
CHAPTER 8. Integration
8.6 Change of Variables Formula
Fubini theorem can be used to find the integral of a (Riemann integrable) function over a
rectangular domain if the iterated integrals can be evaluated. However, like the integral of
a function of one variable, in many cases we need to make use of several change of variables
in order to transform the integral to another integral that can be easily evaluated. In this
section, we establish the change of variables formula for the integral of functions of several
variables.
Theorem 8.65 (Change of Variables Formula). Let U Ď Rn be an open bounded set, and
g : U Ñ Rn be an one-to-one C 1 mapping with C 1 inverse; that is, g ´1 : g(U) Ñ U is
also continuously differentiable. Assume that the Jacobian of g, Jg = det([Dg]), does not
vanish in U, and E ĂĂU has volume. Then g(E) has volume. Moreover, if f : g(E) Ñ R is
bounded and integrable, then (f ˝ g)Jg is integrable over E, and
ż
ż
ż
ˇ
ˇ
ˇ
ˇ
ˇ B(g , ¨ ¨ ¨ , gn ) ˇ
f (y) dy = (f ˝ g)(x)ˇJg (x)ˇ dx = (f ˝ g)(x)ˇ 1
ˇ dx .
g(E)
E
E
B(x1 , ¨ ¨ ¨ , xn )
Remark 8.66. The condition that g has to be defined on a larger open set U can be
removed. In other words, E Ď U has volume is enough for the change of variable formula
to hold; however, we will not prove this more generalized version here.
The proof of the change of variables formula is separated into several steps, and we list
each step as a lemma.
First, we show that the map g in Theorem 8.65 has the property that g ´1 (Z) has measure
zero (or volume zero) if Z itself has measure zero (or volume zero). This establishes that if
A and B are not overlapping; that is, ν(A X B) = 0, then ν(g ´1 (A X B)) = 0.
Lemma 8.67. Let U Ď Rn be an open set, and ϕ : U Ñ Rn be Lipschitz continuous; that
is, there exists L ą 0 such that }ϕ(x) ´ ϕ(y)}Rn ď L}x ´ y}Rn for all x, y P U. Suppose
that Z Ď U is a set of measure zero (or a set of volume zero) and Zs Ď U. Then ϕ(A) has
measure zero (or volume zero).
Proof. We prove the case that Z has measure zero, and the proof for the case that Z has
volume zero is obtained by changing the countable sum/union to finite sum/union.
First we note that if S Ď U is a rectangle on which the ratio of the maximum length and
minimum length of sides is less than 2, then ϕ(S) Ď R for some n-cube R with side of length
§8.6 Change of Variables Formula
311
?
?
L nδ, where δ is the maximum length of sides of S. Therefore, ν(ϕ(S)) ď (2 nL)n ν(S).
Let ε ą 0 be given. Since Z has measure zero, there exists countable rectangles S1 , S2 , ¨ ¨ ¨
8
8
Ť
ř
ε
such that Z Ď
Sk and
ν(Sk ) ă ?
. Moreover, as in the proof of Proposition
n
k=1
(2 nL)
k=1
8.14 we can also assume that the ratio of the maximum length and minimum length of sides
8
8
Ť
ř
Rk and
ν(Rk ) ă ε for some rectangles
of Sk is less than 2 for all k P N; thus ϕ(Z) Ď
k=1
Rk ’s.
k=1
˝
Next, we prove that it suffices to show the change of variables formula for the case that
f is a constant and E is the pre-image of closed rectangle under g in order to establish the
full result.
Lemma 8.68. Let U Ď Rn be an open bounded set, and g : U Ñ Rn be an one-to-one C 1
mapping that has a C 1 inverse. Assume that the Jacobian of g, Jg = det([Dg]), does not
vanish in U, and
ż
|Jg (x)|dx
ν(R) =
for all closed rectangle R Ď g(U) .
(8.6.1)
g ´1 (R)
Then if E ĂĂU has volume and f : g(E) Ñ R is bounded and integrable, then (f ˝ g)|Jg | is
Riemann integrable over E, and
ż
ż
f (y) dy = (f ˝ g)(x)|Jg (x)|dx .
g(E)
E
Proof. Consider the extensions of f and (f ˝ g)|Jg | given by
f
g(E)
"
(x) =
f (x) if x P g(E) ,
E
and (f ˝ g)|Jg | (x) =
A
0
if x P g(E) ,
"
(f ˝ g)(x)|Jg |(x) if x P E ,
0
if x P E A .
ˇ g(E)
␣
(
By the integrability of f over g(E), the set y P Rn ˇ f
is discontinuous at y has measure
zero. Since
␣
ˇ
(
E
x P Rn ˇ (f ˝ g)|Jg | is discontinuous at x
ˇ
␣
(
Ď BE Y x P int(E) ˇ f is discontinuous at g(x)
ˇ
␣
(
= BE Y y P g(int(E)) ˇ f is discontinuous at y
ˇ g(E)
␣
(
Ď BE Y y P Rn ˇ f
is discontinuous at y ,
312
CHAPTER 8. Integration
ˇ
␣
(
E
we conclude that x P Rn ˇ (f ˝ g)|Jg | is discontinuous at x has measure zero. Therefore,
(f ˝ g)|Jg | is Riemann integrable over E. On the other hand, by the fact that
(f
g(E)
E
˝ g)|Jg | = (f ˝ g) |Jg | = (f ˝ g)|Jg |
the Lebesgue theorem also implies that (f
F Ď U since
(f ˝ g)|Jg |
E
g(E)
E
on
U,
˝ g)|Jg | is Riemann integrable over F if E Ď
F
E
= (f ˝ g)|Jg |
@F Ě E .
Moreover, it follows from Lemma 8.49 that
ż
(f
g(E)
ż
˝ g)(x)|Jg (x)|dx =
@E Ď F Ď U .
(f ˝ g)(x)|Jg (x)|dx
F
(8.6.2)
E
Since the Jacobian of g does not vanish in U, Remark 7.2 implies that g is an open
s is compact, there exists an open set V
mapping; thus g(U) is open. By the fact that g(E)
s Ď V ĂĂg(U). It then follows from g ´1 P C 1 (g(U)) and V
s Ď U that
in Rn such that g(E)
› ´1
›
there exists L ą 0 such that ›g (y1 ) ´ g ´1 (y2 )›Rn ď L}y1 ´ y2 }Rn for all y1 , y2 P V. In other
words, g ´1 is Lipschitz on V, and Lemma 8.67 implies that g ´1 (Z) has volume zero if Z Ď V
has volume zero.
Note that there exists δ ą 0 such that d(x, y) ą δ for all x P g(E) and y P V A . Let P be
a partition of g(E) such that }P} ă δ; that is, diam(∆) ă δ for all ∆ P P. Then ∆ Ď V if
∆ P P and ∆ X g(E) ‰ H. Since inf f
g(E)
yP∆
inf
(f
´1
(y) =
xPg
g(E)
(∆)
˝ g)(x) if ∆ Ď U, using (8.6.1)
we find that
L(f, P) =
ÿ
∆PP
∆Xg(E)‰H
=
ÿ
∆PP
∆Xg(E)‰H
g(E)
yP∆
ÿ
∆PP
∆Xg(E)‰H
ď
inf f
ÿ
(y)ν(∆) =
inf
(f
´1
∆PP
∆Xg(E)‰H
inf (f
g(E)
xPg ´1 (∆)
ż
(f
xPg
(∆)
g(E)
˝ g)(x)ν(∆)
ż
˝ g)(x)
|Jg (x)|dx
g ´1 (∆)
g(E)
˝ g)(x)|Jg (x)|dx .
g ´1 (∆)
Since each “face” of the rectangle ∆ P P has volume zero, Lemma 8.67 implies that g ´1 (∆)X
§8.6 Change of Variables Formula
313
g ´1 (∆1 ) has volume zero if ∆ X ∆1 has volume zero. Therefore, Corollary 8.42 shows that
ż
g(E)
L(f, P) ď Ť
(f ˝ g)(x)|Jg (x)|dx
ż ∆PP,∆Xg(E)‰H
g ´1 (∆)
=
(f
g(E)
˝ g)(x)|Jg (x)|dx
Ť
g ´1 ( ∆PP,∆Xg(E)‰H ∆)
ż
(f
=
g(E)
˝ g)(x)|Jg (x)|dx
Ť
g ´1 ( ∆PP,∆Xg(E)‰H ∆)
ż
=
(f ˝ g)(x)|Jg (x)| dx ,
E
where we have used the integrability of (f
g(E)
˝ g)|Jg | over the set g ´1 (
Ť
∆PP,∆Xg(E)‰H ∆)
(since this set is a super set of E) and (8.6.2) to conclude the last two equalities.
Similarly, by the fact that sup f
g(E)
U (f, P) =
sup (f
xPg
∆PP
∆Xg(E)‰H
ÿ
ě
∆PP
∆Xg(E)‰H
g(E)
(f
˝ g)(x) if ∆ Ď U, we obtain that
ż
|Jg (x)|dx
˝ g)(x)
´1 (∆)
ż
g(E)
xPg ´1 (∆)
yP∆
ÿ
sup (f
(y) =
g ´1 (∆)
g(E)
ż
(f
˝ g)(x)|Jg (x)|dx =
g ´1 (∆)
g(E)
˝ g)(x)|Jg (x)|dx
Ť
g ´1 ( ∆PP,∆Xg(E)‰H ∆)
ż
=
(f ˝ g)(x)|Jg (x)| dx .
E
The integrability of f over g(E) then implies that
ż
ż
f (y) dy =
g(E)
(f ˝ g)(x)|Jg (x)|dx. ˝
E
Since the differentiability of g implies that locally g is very closed to an affine map; that
is, g(x) « Lx+c for some L P B(Rn , Rn ) and c P Rn (in fact, g(x) « g(x0 )+(Dg)(x0 )(x´x0 )
in a neighborhood of x0 ), our next step is to establish (8.6.1) first for the case that g is an
affine map. Since the volume of a set remains unchanged under translation, W.L.O.G. we
can assume that g is linear.
Lemma 8.69. Let g P B(Rn , Rn ), and A Ď Rn be a set that has volume. Then g(A) has
volume, and
(
)
ν g(A) =
ż
ż
1dy =
g(A)
|Jg (x)|dx .
(8.6.3)
A
Remark 8.70. If g P B(Rn , Rn ), then g(x) = Lx for some n ˆ n matrix. In this case
Jg (x) = det(L) for all x P A; thus (8.6.3) is the same as that
ż
ż
(
)
ν L(A) =
1dy =
| det(L)|dx = | det(L)|ν(A) .
L(A)
A
(8.6.4)
314
CHAPTER 8. Integration
Therefore, in the following we prove (8.6.4) instead of (8.6.3).
Proof of Lemma 8.69. Since any n ˆ n matrices can be expressed as the product of elementary matrices, it suffices to prove the validity of the lemma for the case that L is an
elementary matrix.
Suppose first that A = [a1 , b1 ] ˆ ¨ ¨ ¨ ˆ [an , bn ] is a rectangle.
1. If L is an elementary matrix of the type


1 0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0
.. 
.

. 
 0 .. 0
 . .
.. 
 .. . . 1 . . .
. 



 ..
.
.
 .
0
0
0
1
. 

 Ð the i0 -th row
.. 
 ..
..
..
. 1
.
 .
. 
 .
.. 
.
.. 0
L=
0
. 
 ..

 .

.
.
.
..
.. 
.. 1
 ..


 ..
.. 
 .
Ð the j0 -th row
1
0
0
0
. 


.
...
... . 

1
. 
 0


..
 0
. 0 
0
0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0 1
Ò
Ò
the i0 -th column
the j0 -th column
then
L(A) = [a1 , b1 ] ˆ ¨ ¨ ¨ ˆ [ai0 ´1 , bi0 ´1 ] ˆ [aj0 , bj0 ] ˆ [ai0 +1 , bi0 +1 ] ˆ ¨ ¨ ¨ ˆ
ˆ[aj0 ´1 , bj0 ´1 ] ˆ [ai0 , bi0 ] ˆ [aj0 +1 , bj0 + 1] ˆ ¨ ¨ ¨ ˆ [an , bn ] ;
(
)
thus ν L(A) = ν(A) = | det(L)|ν(A).
2. If L is an elementary matrix of the type


1 0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0

.. 
 0 1
0
. 
 .
.. 
 .. . . . . . . . . .
. 


 .
.. 
 ..
0
1
0
. 



..  Ð the k -th row
L =  ...
0
0
c
0
. 


.. 
 ..
0
1
0
. 
 .
 .
. . . . . . . . . .. 
 ..
. 


 ..

 .
0
1 0 
0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0 1
§8.6 Change of Variables Formula
315
then
L(A) = [a1 , b1 ] ˆ ¨ ¨ ¨ ˆ [ak0 ´1 , bk0 ´1 ] ˆ [cak0 , cbk0 ] ˆ [ak0 +1 , bk0 +1 ] ˆ ¨ ¨ ¨ ˆ [an , bn ]
if c ě 0 or
L(A) = [a1 , b1 ] ˆ ¨ ¨ ¨ ˆ [ak0 ´1 , bk0 ´1 ] ˆ [cbk0 , cak0 ] ˆ [ak0 +1 , bk0 +1 ] ˆ ¨ ¨ ¨ ˆ [an , bn ]
(
)
if c ă 0. In either case, ν L(A) = |c|ν(A) = | det(L)|ν(A).
3. If L is an elementary matrix of the type


1 0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0
 0 1
0
0 


.
 . .. .. ..

.
.
.
c
0  Ð the i0 -th row
 .
 .

... ... ...
 ..

0


 .
.. 
.

0
1
0
. 
L= .

 ..
.. 
.. .. ..
.
.
.
 .
. 
 .

. . . . . . .. 
 ..
. . 
.
.

 .

 ..
0
1 0 
0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0 1
Ò
the j0 -th column
then L(A) is a parallelepiped
ˇ
(
␣
L(A) = (x1 , ¨ ¨ ¨ , xi0 ´1 , xi0 + cxj0 , xi0 +1 , ¨ ¨ ¨ , xn ) P Rn ˇ xi P [ai , bi ] @ 1 ď i ď n
ˇ
␣
= (x1 , ¨ ¨ ¨ , xi0 ´1 , yi0 , xi0 +1 , ¨ ¨ ¨ , xn ) P Rn ˇ ai0 + cxj0 ď yi0 ď bi0 + cxj0 ,
(
xi P [ai , bi ] @ i ‰ i0 ;
xi0 axis
xi0 axis
L
xj0 axis
xj0 axis
(x1 , ¨ ¨ ¨ , xi0 ´1 , xi0 +1 , ¨ ¨ ¨ , xn ) hyper-space
(x1 , ¨ ¨ ¨ , xi0 ´1 , xi0 +1 , ¨ ¨ ¨ , xn ) hyper-space
Figure 8.3: The image of a rectangle under a linear map induced by the elementary matrix
of the third type
316
CHAPTER 8. Integration
thus the Fubini theorem (or Corollary 8.62) implies that
ż
( ż bi0 +cxj0
)
ν(L(A)) =
1dyi0 dp
xi0 = ν(A) .
[a1 ,b1 ]ˆ¨¨¨ˆ[ai0 ´1 ,bi0 ´1 ]ˆ[ai0 +1 ,bi0 +1 ]ˆ¨¨¨ˆ[an ,bn ]
(
ai0 +cxj0
)
On the other hand, | det(L)| = 1, so ν L(A) = | det(L)|ν(A) is validated.
Therefore, (8.6.4) holds if A is a rectangle and L is an elementary matrix. An immediate
consequence of this observation is that if Z is a set of measure zero, so is L(Z).
Now suppose that A is an arbitrary set with volume, and L is an elementary matrix.
1. If det(L) = 0, L must be an elementary matrix of the second type (with c = 0), and
in this case,
L(A) Ď [´r, r] ˆ ¨ ¨ ¨ ˆ [´r, r] ˆ lo
[´ε,
omooε]
n ˆ ¨ ¨ ¨ ˆ [´r, r]
the k0 -th slot
for some r ą 0 sufficiently large and arbitrary ε ą 0. Therefore, L(A) has volume
(
)
zero; thus L(A) has volume and ν L(A) = | det(L)|ν(A).
2. Suppose that det(L) ‰ 0. Let ε ą 0 be given. Since A has volume, by Riemann’s
condition there exists a partition of A such that
U (1A , P) ´ L(1A , P) ă
ε
.
| det(L)|
Note that the inequality above also implies that
ε
ε
and ν(A) ´ L(1A , P) ă
.
| det(L)|
| det(L)|
ˇ
ˇ
␣
(
␣
(
Ť
∆
Let C1 = ∆ P P ˇ ∆ X A ‰ H and C2 = ∆ P P ˇ ∆ Ď A , and define R1 =
∆PC
1
Ť
and R2 =
∆. Then R2 Ď A Ď R1 . Moreover,
U (1A , P) ´ ν(A) ă
∆PC2
ÿ
ÿ
(
)
ν L(R1 ) =
ν(L(∆)) =
| det(L)|ν(∆) = | det(L)|U (1A , P)
∆PC1
∆PC1
ă | det(L)|ν(A) + ε
and
ÿ
ÿ
(
)
ν L(R2 ) =
ν(L(∆)) =
| det(L)|ν(∆) = | det(L)|L(1A , P)
∆PC2
∆PC2
ą | det(L)|ν(A) ´ ε .
§8.6 Change of Variables Formula
317
As a consequence, by the fact that L(R2 ) Ď L(A) Ď L(R1 ) we conclude that
ˇż
ˇ
ˇ
ż
1dx ´
L(A)
ˇ
(
)
(
)
(
)
ˇ
1dxˇ ď ν L(R1 ) ´ ν L(R2 ) = | det(L)| U (1A , P) ´ L(1A , P) ă ε .
L(A)
Since ε ą 0 is arbitrary, we find that
ż
ż
1dx which implies that 1L(A) is
1dx =
L(A)
L(A)
Riemann integrable, or equivalently, L(A) has volume.
On the other hand, observing that
(
)
(
)
(
)
| det(L)|ν(A) ´ ε ă ν L(R2 ) ď ν L(A) ď ν L(R1 ) ă | det(L)|ν(A) + ε ,
(
)
we conclude that ν L(A) = | det(L)|ν(A) again because ε ą 0 is arbitrary.
˝
Lemma 8.71. Let U Ď Rn be an open bounded set, and g : U Ñ Rn be an one-to-one C 1
mapping that has a C 1 inverse. Assume that the Jacobian of g, Jg = det([Dg]), does not
vanish in U. Then
ż
ν(R) =
|Jg (x)|dx
for all closed rectangle R Ď g(U) .
(8.6.1)
g ´1 (R)
Proof. First, we note that by the the compactness of R, there exist m ą 0 and Λ ą 0 such
that
|Jg (x)| ě m
›
›
and ›(Dg)(x)›B(Rn .Rn ) ď Λ
@ x P g ´1 (R) .
Let 0 ă ε ă 1 be given. By the compactness of g ´1 (R), (Theorem 4.52 implies that)
Jg : g ´1 (R) Ñ R is uniformly continuous; thus there exists δ1 ą 0 such that
ˇ
ˇ
ˇJg (x1 ) ´ Jg (x2 )ˇ ă mε
if }x1 ´ x2 }Rn ă δ1 and x1 , x2 P g ´1 (R) .
Since g ´1 is of class C 1 , the continuity of g ´1 and Corollary 6.36 guarantee that there exists
δ ą 0 such that if }y1 ´ y2 }Rn ă δ and y1 , y2 P R, we have
› ´1
›
›g (y1 ) ´ g ´1 (y2 )› n ă δ1
R
and
› ´1
›
›g (y2 ) ´ g ´1 (y1 ) ´ (Dg ´1 )(y1 )(y2 ´ y1 )› n ď ?ε }y1 ´ y2 }Rn .
R
2 nΛ
Let P be a partition of R with mesh size }P} ă δ and the ratio of the maximum length
and minimum length of sides of each ∆ is less than 2. For ∆ P P, let c∆ denote the center
318
CHAPTER 8. Integration
(
)
of ∆ for ∆ P P, and define A∆ = (Dg)(g ´1 (c∆ )) as well as h∆ (x) = A∆ x ´ g ´1 (c∆ ) + c∆ .
Then
ˇ
ˇ
ˇJg (g ´1 (y)) ´ Jg (g ´1 (c∆ ))ˇ ă mε
@y P ∆.
´1
Moreover, the inverse function theorem (Theorem 7.1) implies that A´1
∆ = (Dg )(c∆ ); thus
for y P ∆,
› (
›
›
)›
›(h ˝ g ´1 )(y) ´ y › n = ›A∆ g ´1 (y) ´ g ´1 (c∆ ) ´ (Dg ´1 )(c∆ )(y ´ c∆ ) › n
R
R
› ´1
›
´1
´1
›
ď }A∆ }B(Rn ,Rn ) g (y) ´ g (c∆ ) ´ (Dg )(c∆ )(y ´ c∆ )› n
R
´1
ď
ε}(Dg)(g (c∆ ))}B(Rn ,Rn )
ε
?
}y ´ c∆ }Rn ď ? diam(∆) .
2 nΛ
4 n
The inequality above implies that for all ∆ P P,
(
)
(1 ´ ε)n ν(∆) ď ν (h∆ ˝ g ´1 )(∆) ď (1 + ε)n ν(∆) .
Since Jh∆ = det(A∆ ) = Jg (g ´1 (c∆ )), Lemma 8.69 or (8.6.4) provides that
ż
ż
ˇ
ˇ
(
)
(
) (
)
ˇJg (x)ˇ dx ď
|Jg (g ´1 (c∆ ))| + mε dx = |Jg (g ´1 (c∆ ))| + mε ν g ´1 (∆)
g ´1 (∆)
g ´1 (∆)
(
)
´1
(
)
ν
(h
˝
g)
(∆)
∆
= |Jg (g ´1 (c∆ ))| + mε
ď (1 + ε)n+1 ν(∆) .
|Jg (g ´1 (c∆ ))|
A similar argument provides a lower bounded of the left-hand side, and we conclude that
ż
n+1
(1 ´ ε) ν(∆) ď
|Jg (x)|dx ď (1 + ε)n+1 ν(∆)
@∆ P P .
g ´1 (∆)
Summing over all ∆ P P, we find that
(1 ´ ε)
n+1
ν(R) ď
ÿż
∆PP
Identity (8.6.1) is then concluded since
is arbitrary.
|Jg (x)|dx ď (1 + ε)n+1 ν(R) .
g ´1 (∆)
ř ż
∆PP
ż
|Jg (x)|dx =
g ´1 (∆)
|Jg (x)|dx and ε P (0, 1)
g ´1 (R)
˝
Example 8.72. Let A be the triangular region with vertices (0, 0), (4, 0), (4, 2), and f :
A Ñ R be given by
f (x, y) = y
a
x ´ 2y .
§8.6 Change of Variables Formula
319
( u ´ v)
Let (u, v) = (x, x ´ 2y). Then (x, y) = g(u, v) = u,
; thus
2
ˇ
ˇ
ˇ1 0 ˇ
1
ˇ
ˇ
Jg (u, v) = ˇ 1
1ˇ = ´ .
ˇ ´ ˇ
2
2
2
Define E as the triangle with vertices (0, 0), (4, 0), (4, 4). Then A = g(E).
y
v
g
E
A
u
x
Figure 8.4: The image of E under g
Therefore,
ż
ż
(
)
1
f (x, y)d(x, y) =
f (x, y)d(x, y) =
f g(u, v) d(u, v)
2 E
g(E)
A
ż4żu
ż
?
1 4 [ 2 3 2 5 ]ˇˇv=u
1
=
(u ´ v) vdvdu =
uv 2 ´ v 2 ˇ du
4 0 0
4 0 3
5
v=0
ż4
ˇ
u=4
(
)
1
1
2 7ˇ
256
2 2 5
=
´ u 2 du =
ˆ u2 ˇ
=
.
4 0 3 5
15 7 u=0 105
ż
Example 8.73. Suppose that f : [0, 1] Ñ R is Riemann integrable and
ż1
(1´x)f (x) dx = 5
0
(note that the function g(x) = (1 ´ x)f (x) is Riemann integrable over [0, 1] because of the
Lebesgue theorem). We would like to evaluate the iterated integral
ż1żx
f (x ´ y) dydx.
0
0
It is nature to consider the change of variables (u, v) = (x ´ y, x) or (u, v) = (x ´ y, y).
Suppose the later case. Then (x, y) = g(u, v) = (u + v, v); thus
ˇ
ˇ
ˇ1 1 ˇ
ˇ = ´1 .
Jg (u, v) = ˇˇ
1 0ˇ
Moreover, the region of integration is the triangle A with vertices (0, 0), (1, 0), (1, 1), and
three sides y = 0, x = 1, x = y correspond to u = 0, u + v = 1 and v = 0. Therefore, if
320
CHAPTER 8. Integration
E denotes the triangle enclosed by u = 0, v = 0 and u + v = 1 on the (u, v)-plane, then
g(E) = A, and
ż1żx
ż
ż
f (x ´ y) dydx =
f (x ´ y)d(x, y) =
0
0
A
ż
=
f (x ´ y)d(x, y)
g(E)
(
)
f g1 (u, v) ´ g2 (u, v) |Jg (u, v)|d(u, v) =
żE1
=
ż 1 ż 1´u
f (u) dvdu
0
0
(1 ´ u)f (u) du = 5 .
0
Example 8.74 (Polar coordinates). In R2 , when the domain over which the integral is taken
is a disk D, a particular type of change of variables is sometimes very useful for the purpose
of evaluating the integral. Let (x, y) = (x0 + r cos θ, y0 + r sin θ) ” ψ(r, θ), where (x0 , y0 ) is
the center of D under consideration. If the radius of D is R, then D, up to removing a line
segment with length R, is the image of (0, R) ˆ (0, 2π) under ψ. Note that the Jacobian of
ψ is
ˇ
ˇ Bψ1
ˇ
ˇ
Jψ (r, θ) = ˇ Br
ˇ Bψ2
ˇ
Br
ˇ
ˇ
Bψ1 ˇ ˇ
ˇ ˇcos θ ´r sin θˇ
ˇ
ˇ
ˇ
Bθ
ˇ=ˇ
ˇ = r.
Bψ2 ˇ ˇ sin θ r cos θ ˇ
ˇ
Bθ
Therefore, if f : D Ñ R is Riemann integrable, then
ż
ż
ż
ˇ
ˇ
(f ˝ ψ)(r, θ)ˇJψ (r, θ)ˇ d(r, θ)
f (x, y) d(x, y) =
f (x, y) d(x, y) =
(0,R)ˆ(0,2π)
ψ((0,R)ˆ(0,2π))
D
ż
=
f (x0 + r cos θ, y0 + r sin θ) r d(r, θ) .
(0,R)ˆ(0,2π)
Example 8.75 (Cylindrical coordinates). In R3 , when the domain over which the integral
is taken is a cylinder C; that is, C = D ˆ [a, b] for some disk D and ´8 ă a ă b ă R, then
the change of variables
ψ(r, θ, z) = (x0 + r cos θ, y0 + r sin θ, z)
0 ă r ă R , 0 ă θ ă 2π , a ď z ď b ,
where (x0 , y0 ) is the center of D and R is the radisu of D, is sometimes very useful for
evaluating the integral. Since the Jacobian of ψ is
ˇ
ˇ
ˇ Bψ1 Bψ1 Bψ1 ˇ
ˇ
ˇ
ˇ ˇ
ˇ Br
Bθ
Bz ˇ ˇcos θ ´r sin θ 0ˇ
ˇ
ˇ
ˇ ˇ
ˇ
ˇ Bψ Bψ2 Bψ2 ˇ ˇ
Jψ (r, θ, z) = ˇ 2
ˇ = ˇ sin θ r cos θ 0ˇ = r ,
ˇ
ˇ Br
Bθ
Bz ˇ ˇ
ˇ
ˇ
ˇ ˇ 0
0
1
ˇ Bψ3 Bψ3 Bψ3 ˇ
ˇ
ˇ
Br
Bθ
Bz
§8.7 Exercises
321
we must have
ż
ż
f (x, y, z) d(x, y, z) =
f (x, y, z) d(x, y, z)
C
ψ((0,R)ˆ(0,2π)ˆ[a,b])
ż
ˇ
ˇ
=
(f ˝ ψ)(r, θ, z)ˇJψ (r, θ, z)ˇ d(r, θ, z)
(0,R)ˆ(0,2π)ˆ[a,b]
ż
=
f (x0 + r cos θ, y0 + r sin θ, z) r d(r, θ, z) .
(0,R)ˆ(0,2π)ˆ[a,b]
Example 8.76 (Spherical coordinates). In R3 , when the domain over which the integral is
taken is a ball B, the change of variables
ψ(ρ, θ, ϕ) = (x0 + ρ cos θ sin ϕ, y0 + ρ sin θ sin ϕ, z0 + ρ cos ϕ) 0 ă ρ ă R, 0 ă θ ă 2π, 0 ă ϕ ă π,
where (x0 , y0 , z0 ) is the center of B and R is the radius of B, is often used to evaluate the
integral a function over B. Since the Jacobian of ψ is
ˇ
ˇ
ˇ Bψ1 Bψ1 Bψ1 ˇ
ˇ
ˇ
ˇ Bρ
Bθ
Bϕ ˇ ˇˇcos θ sin ϕ ´ρ sin θ sin ϕ ρ cos θ cos ϕˇˇ
ˇ
ˇ
ˇ
ˇ Bψ2 Bψ2 Bψ2 ˇ ˇˇ
ˇ = ˇ sin θ sin ϕ ρ cos θ sin ϕ ρ sin θ cos ϕ ˇˇ
Jψ (ρ, θ, ϕ) = ˇˇ
Bθ
Bϕ ˇˇ ˇ
ˇ
ˇ Bρ
ˇ
ˇ Bψ3 Bψ3 Bψ3 ˇ ˇ cos ϕ
0
´ρ
sin
ϕ
ˇ
ˇ
ˇ Bρ
Bθ
Bϕ ˇ
= ´ρ2 cos2 θ sin3 ϕ ´ ρ2 sin2 θ sin ϕ cos2 ϕ ´ ρ2 cos2 θ sin ϕ cos2 ϕ ´ ρ2 sin2 θ sin3 ϕ
= ´ρ2 sin3 ϕ ´ ρ2 sin ϕ cos2 ϕ = ´ρ2 sin ϕ ,
if the radius of B is R, we must have
ż
ż
f (x, y, z) d(x, y, z) =
f (x, y, z) d(x, y, z)
B
ψ((0,R)ˆ(0,2π)ˆ(0,π))
ż
ˇ
ˇ
=
(f ˝ ψ)(ρ, θ, ϕ)ˇJψ (ρ, θ, ϕ)ˇ d(ρ, θ, ϕ)
(0,R)ˆ(0,2π)ˆ(0,π)
ż
f (x0 + ρ cos θ sin ϕ, y0 + ρ sin θ sin ϕ, z0 + ρ cos ϕ) ρ2 sin ϕ d(r, θ, z) .
=
(0,R)ˆ(0,2π)ˆ(0,π)
8.7 Exercises
§8.2 Conditions for Integrability
322
CHAPTER 8. Integration
Problem 8.1. Let f : [0, 1] ˆ [0, 1] Ñ R be a bounded function such that f (x, y) ď
f (x, z) if y ă z and f (x, y) ď f (t, z) if x ă t. In other words, f (x, ¨) and f (¨, y) are both
non-decreasing functions for fixed x, y P [0, 1]. Show that f is Riemann integrable over
[0, 1] ˆ [0, 1].
Problem 8.2. Let A Ď Rn be a bounded set, and fk : A Ñ R be a sequence of Riemann
integrable functions which converges uniformly to f on A. Show that f is Riemann integrable
over A, and
lim
ż
ż
kÑ8 A
fk (x) dx =
lim fk (x) dx =
A kÑ8
ż
f (x) dx .
A
§8.3 Lebesgue’s Theorem
Problem 8.3. Complete the following.
1. Show that if A is a set of volume zero, then A has measure zero. Is it true that if A
has measure zero, then A also has volume zero?
2. Let a, b P R and a ă b. Show that the interval [a, b] does not have measure zero (in
R).
3. Let A Ď [a, b] be a set of measure zero (in R). Show that [a, b]zA does not have
measure zero (in R).
4. Show that the Cantor set (defined in Exercise Problem 2.11) has volume zero.
8
Ť
Problem 8.4. Let A =
volume?
D
k=1
8 (1
(1 1 )
Ť
1 1
1)
, k =
´ k , + k be a subset of R. Does A have
k 2
k=1
k
2
k
2
Problem 8.5. Let f : [a, b] Ñ R be bounded and Riemann integrable. Show that the graph
of f has volume zero by considering the difference of the upper and lower sums of f .
Problem 8.6. Let A Ď żRn be an open bounded set with volume, and f : A Ñ R be
f (x) dx = 0 for all subsets B Ď A with volume, then f = 0.
continuous. Show that if
B
Problem 8.7. Prove the following statements.
1. The function f (x) = sin
1
is Riemann integrable over (0, 1).
x
§8.7 Exercises
323
2. Let f : [0, 1] Ñ R be given by
$
& 1 if x = q P Q, (p, q) = 1 ,
p
p
f (x) =
% 0 if x is irrational.
Then f is Riemann integrable over [0, 1]. Find
ż1
f (x)dx as well.
0
3. Let A Ď Rn be a bounded set, and f : A Ñ R is Riemann integrable. Then f k （f 的
k 次方）is integrable for all k P N.
Problem 8.8. Suppose that f : [a, b] Ñ R is Riemann integrable, and the set
żb
ˇ
(
ˇ
[a, b] f (x) ‰ 0 has measure zero. Show that
f (x) dx = 0.
␣
x P
a
§8.5 Fubini’s Theorem
Problem 8.9. Evaluate the iterated integral
ż1żx
0
2
(2y ´ y 2 ) 3 dydx.
0
Problem 8.10. Let A = [a, b] ˆ [c, d] be a rectangle in R2 , and f : A Ñ R be Riemann
integrable. Show that the sets
!
żs d
żs b
ˇ żd
ˇ żb
)
!
)
ˇ
ˇ
x P [a, b] ˇ
f (x, y)dy ‰ f (x, y)dy
and
y P [c, d] ˇ
f (x, y)dx ‰ f (x, y)dx
c
c
a
a
have measure zero (in R1 ).
Problem 8.11. Define a set S Ď [0, 1] ˆ [0, 1] by
ˇ
!(
)
p k)
ˇ
S=
,
P [0, 1] ˆ [0, 1] ˇ m, p, k P N , gcd(m, p) = 1 and 1 ď k ď m ´ 1 .
m m
Show that
ż 1(ż 1
0
0
)
1S (x, y) dy dx =
ż 1(ż 1
0
)
1S (x, y) dx dy = 0
0
but 1S is not Riemann integrable over [0, 1] ˆ [0, 1].
Problem 8.12. Let f : [0, 1] ˆ [0, 1] Ñ R be given by
$
22n
if (x, y) P [2´n , 2´n+1 ) ˆ [2´n , 2´n+1 ), n P N ,
’
&
´22n+1 if (x, y) P [2´n , 2´n+1 ) ˆ [2´n´1 , 2´n ), n P N ,
f (x, y) =
’
%
0
otherwise .
324
CHAPTER 8. Integration
1. Show that
ż1
f (x, y) dx = 0 for all y P [0, 1).
0
2. Show that
ż1
[ 1)
f (x, y) dy = 0 for all x P 0, .
2
0
3. Justify if the iterated (improper) integrals
ż1ż1
0
are identical.
Problem 8.13.
1. Draw the region corresponding to the integral
f (x, y)dxdy and
ż1ż1
f (x, y) dydx
0
0
ż 1 ( ż ex
0
0
)
(x + y) dy dx and evaluate.
1
2. Change the order of integration of the integral in 1 and check if the answer is unaltered.
§8.6 Change of Variables Formula
Problem 8.14. Prove Theorem 4.95 using Theorem 8.65.
ˇ
(
␣
Problem 8.15. Find the volume of the set (x, y, z) P R3 ˇ 0 ď x2 + y 2 + xy ď z 2 ď 4 .
Problem 8.16. Suppose that U Ď Rn is an nonempty open set, and f : U Ñ R is of class
C 1 such that Jf (x) ‰ 0 for all x P U. Show that
(
)
ν f (D(x0 , r))
(
) = Jf (x0 )
lim
rÑ0+ ν D(x0 , r)
@ x0 P U .
Problem 8.17. 1. Let A be the parallelogram with vertices (0, 0),
(1 1)
, . Evaluate the integral
(2
3
,´
1)
, (1, 0) and
3
3 3
ż
a
?
x ´ y x + 2y dA .
A
2. Let A be the parallelogram bounded by lines x = 3y, x = 1 + 3y, y = ´2x and
y = 1 ´ 2x. Evaluate the integral
ż a
3
2x2 ´ 5xy ´ 3y 2 dA .
A
3. Let A be the trapezoid with vertices (1, 1), (2, 2), (2, 0) and (4, 0). Evaluate the
integral
ż
e(y´x)/(y+x) dA .
A
§8.7 Exercises
325
Problem 8.18 (True or False). Determine whether the following statements are true or
false. If it is true, prove it. Otherwise, give a counter-example.
1. Let A Ď Rn be bounded, and f : A Ñ R be Riemann integrable. If P be a partition
of A, and m ď f (x) ď M for all x P A. Then mν(A) ď L(f, P) ď U (f, P) ď M ν(A).
s is countable, then A has volume zero.
2. Let A Ď Rn be a set of measure zero. If AzA
3. Let A Ď Rn be a closed rectangle and f, g : B Ñ R be Riemann integrable functions.
If there exists aż set Z Ď A żsuch that Z has measure zero and g(x) = f (x) for all
x P AzZ, then
f (x) dx =
A
g(x) dx.
A
4. Let A Ď Rn be a closed rectangle. Suppose that f and g are two bounded real-valued
functions defined on A such that f is continuous and g = f except on a set of measure
zero, then f and g are both Riemann integrable over A.
5. Let A, B Ď R be bounded, and f : A Ñ R and g : f (A) Ñ R be Riemann integrable.
Then g ˝ f is Riemann integrable over A.
6.
326