The Real And Complex Number Systems

The Real And Complex Number Systems 

Integers 

1.1 Prove that there is no largest prime. 

Proof: Suppose p is the largest prime. Then p! + 1 is NOT a prime. So, 

there exists a prime q such that 

q |p! + 1 ⇒ q |1 

which is impossible. So, there is no largest prime. 

Remark: There are many and many proofs about it. The proof that we 

give comes from Archimedes 287-212 B. C. In addition, Euler Leonhard 

(1707-1783) find another method to show it. The method is important since 

it develops to study the theory of numbers by analytic method. The reader 

can see the book, An Introduction To The Theory Of Numbers by 

Loo-Keng Hua, pp 91-93. (Chinese Version) 

1.2 If n is a positive integer, prove the algebraic identity 

∑n−1 

a n − b n = (a − b) a k b n−1−k 

k=0 

Proof: It suffices to show that 

∑n−1 

x n − 1 = (x − 1) x k . 

k=0 

1

Consider the right hand side, we have 

∑n−1 

∑n−1 

∑n−1 

(x − 1) x k = x k+1 − 

k=0 

= 

k=0 

n∑ 

x k − 

k=1 

= x n − 1. 

k=0 

∑n−1 

x k 

k=0 

x k 

1.3 If 2 n − 1 is a prime, prove that n is prime. A prime of the form 

2 p − 1, where p is prime, is called a Mersenne prime. 

Proof: If n is not a prime, then say n = ab, where a > 1 and b > 1. So, 

we have 

∑b−1 

2 ab − 1 = (2 a − 1) (2 a ) k 

which is not a prime by Exercise 1.2. So, n must be a prime. 

Remark: The study of Mersenne prime is important; it is related 

with so called Perfect number. In addition, there are some OPEN problem 

about it. For example, is there infinitely many Mersenne nembers 

The reader can see the book, An Introduction To The Theory 

Of Numbers by Loo-Keng Hua, pp 13-15. (Chinese Version) 

1.4 If 2 n + 1 is a prime, prove that n is a power of 2. A prime of the 

form 2 2m + 1 is called a Fermat prime. Hint. Use exercise 1.2. 

Proof: If n is a not a power of 2, say n = ab, where b is an odd integer. 

So, 

2 a + 1 ∣ ∣ 2 ab + 1 

and 2 a + 1 < 2 ab + 1. It implies that 2 n + 1 is not a prime. So, n must be a 

power of 2. 

k=0 

Remark: (1) In the proof, we use the identity 

2n−2 

∑ 

x 2n−1 + 1 = (x + 1) (−1) k x k . 

k=0 

2

Proof: Consider 

2n−2 

∑ 

(x + 1) (−1) k x k = 

k=0 

= 

2n−2 

∑ 

k=0 

2n−1 

(−1) k x k+1 + 

2n−2 

∑ 

k=0 

2n−2 

(−1) k x k 

∑ 

∑ 

(−1) k+1 x k + (−1) k x k 

k=1 

= x 2n+1 + 1. 

k=0 

(2) The study of Fermat number is important; for the details the reader 

can see the book, An Introduction To The Theory Of Numbers by 

Loo-Keng Hua, pp 15. (Chinese Version) 

1.5 The Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, ... are defined by the recursion 

formula x n+1 = x n + x n−1 , with x 1 = x 2 = 1. Prove that (x n , x n+1 ) = 1 

and that x n = (a n − b n ) / (a − b) , where a and b are the roots of the quadratic 

equation x 2 − x − 1 = 0. 

So, 

Proof: Let d = g.c.d. (x n , x n+1 ) , then 

Continue the process, we finally have 

So, d = 1 since d is positive. 

Observe that 

and thus we consider 

i.e., consider 

If we let 

d |x n and d |x n+1 = x n + x n−1 . 

d |x n−1 . 

d |1 . 

x n+1 = x n + x n−1 , 

x n+1 = x n + x n−1 , 

x 2 = x + 1 with two roots, a and b. 

F n = (a n − b n ) / (a − b) , 

3

then it is clear that 

So, F n = x n for all n. 

F 1 = 1, F 2 = 1, and F n+1 = F n + F n−1 for n > 1. 

Remark: The study of the Fibonacci numbers is important; the reader 

can see the book, Fibonacci and Lucas Numbers with Applications 

by Koshy and Thomas. 

1.6 Prove that every nonempty set of positive integers contains a smallest 

member. This is called the well–ordering Principle. 

Proof: Given (φ ≠) S (⊆ N) , we prove that if S contains an integer 

k, then S contains the smallest member. We prove it by Mathematical 

Induction of second form as follows. 

As k = 1, it trivially holds. Assume that as k = 1, 2, ..., m holds, consider 

as k = m + 1 as follows. In order to show it, we consider two cases. 

(1) If there is a member s ∈ S such that s < m + 1, then by Induction 

hypothesis, we have proved it. 

(2) If every s ∈ S, s ≥ m + 1, then m + 1 is the smallest member. 

Hence, by Mathematical Induction, we complete it. 

Remark: We give a fundamental result to help the reader get more. We 

will prove the followings are equivalent: 

(A. Well–ordering Principle) every nonempty set of positive integers 

contains a smallest member. 

(B. Mathematical Induction of first form) Suppose that S (⊆ N) , 

if S satisfies that 

Then S = N. 

(1). 1 in S 

(2). As k ∈ S, then k + 1 ∈ S. 

(C. Mathematical Induction of second form) Suppose that S (⊆ N) , 

if S satisfies that 

(1). 1 in S 

(2). As 1, ..., k ∈ S, then k + 1 ∈ S. 

4

Then S = N. 

Proof: (A ⇒ B): If S ≠ N, then N − S ≠ φ. So, by (A), there exists 

the smallest integer w such that w ∈ N − S. Note that w > 1 by (1), so we 

consider w − 1 as follows. 

Since w − 1 /∈ N − S, we know that w − 1 ∈ S. By (2), we know that 

w ∈ S which contadicts to w ∈ N − S. Hence, S = N. 

(B ⇒ C): It is obvious. 

(C ⇒ A): We have proved it by this exercise. 

Rational and irrational numbers 

1.7 Find the rational number whose decimal expansion is 0.3344444444.... 

Proof: Let x = 0.3344444444..., then 

x = 3 10 + 3 

10 + 4 

2 10 + ... + 4 + .., where n ≥ 3 

3 10n = 33 

10 + 4 ( 

1 + 1 2 10 3 10 + ... + 1 ) 

10 + .. n 

= 33 

10 + 4 ( 1 

2 10 3 1 − 1 

10 

= 33 

10 + 4 

2 900 

= 301 

900 . 

) 

1.8 Prove that the decimal expansion of x will end in zeros (or in nines) 

if, and only if, x is a rational number whose denominator is of the form 2 n 5 m , 

where m and n are nonnegative integers. 

Proof: (⇐)Suppose that x = 

k , if n ≥ m, we have 

2 n 5 m 

k5 n−m 

2 n 5 n = 5n−m k 

10 n . 

So, the decimal expansion of x will end in zeros. Similarly for m ≥ n. 

(⇒)Suppose that the decimal expansion of x will end in zeros (or in 

nines). 

5

For case x = a 0 .a 1 a 2 · · · a n . Then 

∑ n 

k=0 

x = 

10n−k a k 

= 

10 n 

∑ n 

k=0 10n−k a k 

2 n 5 n . 

For case x = a 0 .a 1 a 2 · · · a n 999999 · · · . Then 

∑ n 

k=0 

x = 

10n−k a k 

+ 9 

9 

+ ... + + ... 

2 n 5 n 10n+1 10n+m ∑ n 

k=0 

= 

10n−k a k 

+ 9 ∑ ∞ 

10 −j 

2 n 5 n 10 n+1 j=0 

∑ n 

k=0 

= 

10n−k a k 

+ 1 

2 n 5 n 10 n 

= 1 + ∑ n 

k=0 10n−k a k 

. 

2 n 5 n 

So, in both case, we prove that x is a rational number whose denominator is 

of the form 2 n 5 m , where m and n are nonnegative integers. 

1.9 Prove that √ 2 + √ 3 is irrational. 

Proof: If √ 2 + √ 3 is rational, then consider 

(√ 

3 + 

√ 

2 

) (√ 

3 − 

√ 

2 

) 

= 1 

which implies that √ 3 − √ 2 is rational. Hence, √ 3 would be rational. It is 

impossible. So, √ 2 + √ 3 is irrational. 

Remark: (1) √ p is an irrational if p is a prime. 

Proof: If √ p ∈ Q, write √ p = a , where g.c.d. (a, b) = 1. Then 

b 

Write a = pq. So, 

By (*) and (*’), we get 

b 2 p = a 2 ⇒ p ∣ ∣ a 2 ⇒ p |a (*) 

b 2 p = p 2 q 2 ⇒ b 2 = pq 2 ⇒ p ∣ ∣ b 2 ⇒ p |b . (*’) 

p |g.c.d. (a, b) = 1 

which implies that p = 1, a contradiction. So, √ p is an irrational if p is a 

prime. 

6

Note: There are many and many methods to prove it. For example, the 

reader can see the book, An Introduction To The Theory Of Numbers 

by Loo-Keng Hua, pp 19-21. (Chinese Version) 

(2) Suppose a, b ∈ N. Prove that √ a+ √ b is rational if and only if, a = k 2 

and b = h 2 for some h, k ∈ N. 

Proof: (⇐) It is clear. 

(⇒) Consider 

( √a + 

√ 

b 

) ( √a − 

√ 

b 

) 

= a 2 − b 2 , 

then √ a ∈ Q and √ b ∈ Q. Then it is clear that a = h 2 and b = h 2 for some 

h, k ∈ N. 

1.10 If a, b, c, d are rational and if x is irrational, prove that (ax + b) / (cx + d) 

is usually irrational. When do exceptions occur 

Proof: We claim that (ax + b) / (cx + d) is rational if and only if ad = bc. 

(⇒)If (ax + b) / (cx + d) is rational, say (ax + b) / (cx + d) = q/p. We 

consider two cases as follows. 

(i) If q = 0, then ax + b = 0. If a ≠ 0, then x would be rational. So, a = 0 

and b = 0. Hence, we have 

ad = 0 = bc. 

(ii) If q ≠ 0, then (pa − qc) x+(pb − qd) = 0. If pa−qc ≠ 0, then x would 

be rational. So, pa − qc = 0 and pb − qd = 0. It implies that 

qcb = qad ⇒ ad = bc. 

(⇐)Suppose ad = bc. If a = 0, then b = 0 or c = 0. So, 

If a ≠ 0, then d = bc/a. So, 

{ 

ax + b 0 if a = 0 and b = 0 

cx + d = b 

. 

if a = 0 and c = 0 

ax + b 

cx + d = 

d 

ax + b 

cx + bc/a 

= 

a (ax + b) 

c (ax + b) = a c . 

Hence, we proved that if ad = bc, then (ax + b) / (cx + d) is rational. 

7

1.11 Given any real x > 0, prove that there is an irrational number 

between 0 and x. 

Proof: If x ∈ Q c , we choose y = x/2 ∈ Q c . Then 0 < y < x. If x ∈ Q, 

we choose y = x/ √ 2 ∈ Q, then 0 < y < x. 

Remark: (1) There are many and many proofs about it. We may prove 

it by the concept of Perfect set. The reader can see the book, Principles 

of Mathematical Analysis written by Walter Rudin, Theorem 2.43, 

pp 41. Also see the textbook, Exercise 3.25. 

(2) Given a and b ∈ R with a < b, there exists r ∈ Q c , and q ∈ Q such 

that a < r 

Proof: We show it by considering four cases. (i) a ∈ Q, b ∈ Q. (ii) 

a ∈ Q, b ∈ Q c . (iii) a ∈ Q c , b ∈ Q. (iv) a ∈ Q c , b ∈ Q c .( 

) 

1 − √ 1 

2 

b. 

(i) (a ∈ Q, b ∈ Q) Choose q = a+b 

2 

and r = √ 1 

2 

a + 

(ii) (a ∈ Q, b ∈ Q c ) Choose r = a+b 

2 

and let c = 1 

< b−a, then a+c := q. 

2 n 

(iii) (a ∈ Q c , b ∈ Q) Similarly for (iii). 

(iv) (a ∈ Q c , b ∈ Q c ) It suffices to show that there exists a rational 

number q ∈ (a, b) by (ii). Write 

Choose n large enough so that 

b = b 0 .b 1 b 2 · · · b n · ·· 

a < q = b 0 .b 1 b 2 · · · b n < b. 

(It works since b − q = 0.000..000b n+1 ... ≤ 1 

10 n ) 

1.12 If a/b < c/d with b > 0, d > 0, prove that (a + c) / (b + d) lies 

bwtween the two fractions a/b and c/d 

Proof: It only needs to conisder the substraction. So, we omit it. 

Remark: The result of this exercise is often used, so we suggest the 

reader keep it in mind. 

1.13 Let a and b be positive integers. Prove that √ 2 always lies between 

the two fractions a/b and (a + 2b) / (a + b) . Which fraction is closer to √ 2 

Proof: Suppose a/b ≤ √ 2, then a ≤ √ 2b. So, 

a + 2b 

a + b − √ (√ ) (√ ) 

2 − 1 2b − a 

2 = 

≥ 0. 

a + b 

8

In addition, 

(√ a 

) 

2 − − 

b 

( a + 2b 

a + b − √ ) 

2 = 2 √ 2 − 

( a 

b + a + 2b ) 

a + b 

= 2 √ 2 − a2 + 2ab + 2b 2 

ab + b 2 

1 

[( 

= 2 √ ) 

2 − 2 ab + 

ab + b 2 

≥ 

= 0. 

1 

ab + b 2 [ ( 

2 √ 2 − 2 

) 

a√ a + 2 

( 

2 √ ) ] 

2 − 2 b 2 − a 2 

( 

2 √ ) ( ) ] 2 

a 

2 − 2 √2 − a 2 

So, a+2b is closer to √ 2. 

a+b 

Similarly, we also have if a/b > √ 2, then a+2b 

a+b 

to √ 2 in this case. 

Remark: Note that 

a 

b < √ 2 < a + 2b 

a + b < 2b 

a 

< √ 2. Also, a+2b 

a+b 

by Exercise 12 and 13. 

is closer 

And we know that a+2b is closer to √ 2. We can use it to approximate √ 2. 

a+b 

Similarly for the case 

2b 

a < a + 2b 

a + b < √ 2 < a b . 

1.14 Prove that √ n − 1 + √ n + 1 is irrational for every integer n ≥ 1. 

Proof: Suppose that √ n − 1 + √ n + 1 is rational, and thus consider 

( √n 

+ 1 + 

√ 

n − 1 

) ( √n 

+ 1 − 

√ 

n − 1 

) 

= 2 

which implies that √ n + 1 − √ n − 1 is rational. Hence, √ n + 1 and √ n − 1 

are rational. So, n − 1 = k 2 and n + 1 = h 2 , where k and h are positive 

integer. It implies that 

h = 3 2 and k = 1 2 

which is absurb. So, √ n − 1 + √ n + 1 is irrational for every integer n ≥ 1. 

9

1.15 Given a real x and an integer N > 1, prove that there exist integers 

h and k with 0 < k ≤ N such that |kx − h| < 1/N. Hint. Consider the N +1 

numbers tx − [tx] for t = 0, 1, 2, ..., N and show that some pair differs by at 

most 1/N. 

Proof: Given N > 1, and thus consider tx − [tx] for t = 0, 1, 2, ..., N as 

follows. Since 

0 ≤ tx − [tx] := a t < 1, 

so there exists two numbers a i and a j where i ≠ j such that 

|a i − a j | < 1 N ⇒ |(i − j) x − p| < 1 , where p = [jx] − [ix] . 

N 

Hence, there exist integers h and k with 0 < k ≤ N such that |kx − h| < 1/N. 

1.16 If x is irrational prove that there are infinitely many rational numbers 

h/k with k > 0 such that |x − h/k| < 1/k 2 . Hint. Assume there are 

only a finite number h 1 /k 1 , ..., h r /k r and obtain a contradiction by applying 

Exercise 1.15 with N > 1/δ, where δ is the smallest of the numbers 

|x − h i /k i | . 

Proof: Assume there are only a finite number h 1 /k 1 , ..., h r /k r and let 

δ = min r i=1 |x − h i /k i | > 0 since x is irrational. Choose N > 1/δ, then by 

Exercise 1.15, we have 

∣ 

1 ∣∣∣ 

N < δ ≤ x − h k ∣ < 1 

kN 

which implies that 

1 

N < 1 

kN 

which is impossible. So, there are infinitely many rational numbers h/k with 

k > 0 such that |x − h/k| < 1/k 2 . 

Remark: (1) There is another proof by continued fractions. The 

reader can see the book, An Introduction To The Theory Of Numbers 

by Loo-Keng Hua, pp 270. (Chinese Version) 

(2) The exercise is useful to help us show the following lemma. {ar + b : a ∈ Z, b ∈ Z} , 

where r ∈ Q c is dense in R. It is equivalent to {ar : a ∈ Z} , where r ∈ Q c is 

dense in [0, 1] modulus 1. 

10

Proof: Say {ar + b : a ∈ Z, b ∈ Z} = S, and since r ∈ Q c , then by Exercise 

1.16, there are infinitely many rational numbers h/k with k > 0 such 

that |kr − h| < 1 . Consider (x − δ, x + δ) := I, where δ > 0, and thus choosing 

k 0 large enough so that 1/k 0 < δ. Define L = |k 0 r − h 0 | , then we have 

k 

sL ∈ I for some s ∈ Z. So, sL = (±) [(sk 0 ) r − (sh 0 )] ∈ S. That is, we have 

proved that S is dense in R. 

1.17 Let x be a positive rational number of the form 

x = 

n∑ 

k=1 

where each a k is nonnegative integer with a k ≤ k − 1 for k ≥ 2 and a n > 0. 

Let [x] denote the largest integer in x. Prove that a 1 = [x] , that a k = 

[k!x] − k [(k − 1)!x] for k = 2, ..., n, and that n is the smallest integer such 

that n!x is an integer. Conversely, show that every positive rational number 

x can be expressed in this form in one and only one way. 

and 

Proof: (⇒)First, 

[ ] 

n∑ a k 

[x] = a 1 + 

k! 

k=2 

[ n∑ 

] 

a k 

= a 1 + since a 1 ∈ N 

k! 

k=2 

n∑ a k 

n∑ 

= a 1 since 

k! ≤ k − 1 

k! 

k=2 

k=2 

Second, fixed k and consider 

k!x = k! 

n∑ 

j=1 

(k − 1)!x = (k − 1)! 

= 

k−1 

a j 

j! = k! ∑ 

n∑ 

j=1 

j=1 

a k 

k! , 

n∑ 

k=2 

1 

(k − 1)! − 1 k! = 1 − 1 n! < 1. 

a j 

j! + a k + k! 

k−1 

a j 

j! = (k − 1)! ∑ 

j=1 

n∑ 

j=k+1 

a j 

j! 

a j 

n∑ 

j! + (k − 1)! 

j=k 

a j 

j! . 

11

So, 

and 

[ 

∑k−1 

[k!x] = k! 

j=1 

∑k−1 

= k! 

j=1 

a j 

j! + a k + k! 

a j 

j! + a k since k! 

[ 

∑k−1 

k [(k − 1)!x] = k (k − 1)! 

j=1 

∑k−1 

= k (k − 1)! 

∑k−1 

= k! 

j=1 

a j 

j! 

j=1 

n∑ 

j=k+1 

n∑ 

a j 

j! 

j=k+1 

] 

a j 

j! < 1 

a j 

n∑ 

j! + (k − 1)! 

j=k 

] 

a j 

j! . 

a j 

n∑ 

j! since (k − 1)! 

j=k 

a j 

j! < 1 


a k = [k!x] − k [(k − 1)!x] for k = 2, ..., n. 

Last, in order to show that n is the smallest integer such that n!x is an 

integer. It is clear that 

n∑ a k 

n!x = n! 

k! ∈ Z. 

In addition, 

So, we have proved it. 

k=1 

(n − 1)!x = (n − 1)! 

n∑ 

k=1 

n−1 

∑ 

= (n − 1)! 

k=1 

/∈ Z since a n 

n 

12 

a k 

k! 

a k 

k! + a n 

n 

/∈ Z.

(⇐)It is clear since every a n is uniquely deermined. 

Upper bounds 

1.18 Show that the sup and the inf of a set are uniquely determined whenever 

they exists. 

Proof: Given a nonempty set S (⊆ R) , and assume sup S = a and 

sup S = b, we show a = b as follows. Suppose that a > b, and thus choose 

ε = a−b , then there exists a x ∈ S such that 

2 


b < a + b 

2 

= a − ε < x < a 

b < x 

which contradicts to b = sup S. Similarly for a < b. Hence, a = b. 

1.19 Find the sup and inf of each of the following sets of real numbers: 

(a) All numbers of the form 2 −p + 3 −q + 5 −r , where p, q, and r take on all 

positive integer values. 

Proof: Define S = {2 −p + 3 −q + 5 −r : p, q, r ∈ N}. Then it is clear that 

sup S = 1 + 1 + 1 , and inf S = 0. 

2 3 5 

(b) S = {x : 3x 2 − 10x + 3 < 0} 

Proof: Since 3x 2 − 10x + 3 = (x − 3) (3x − 1) , we know that S = ( 1 

3 , 3) . 

Hence, sup S = 3 and inf S = 1 3 . 

(c) S = {x : (x − a) (x − b) (x − c) (x − d) < 0} , where a 

Proof: It is clear that S = (a, b)∪(c, d) . Hence, sup S = d and inf S = a. 

1.20 Prove the comparison property for suprema (Theorem 1.16) 

Proof: Since s ≤ t for every s ∈ S and t ∈ T, fixed t 0 ∈ T, then s ≤ t 0 

for all s ∈ S. Hence, by Axiom 10, we know that sup S exists. In addition, 

it is clear sup S ≤ sup T. 

Remark: There is a useful result, we write it as a reference. Let S and T 

be two nonempty subsets of R. If S ⊆ T and sup T exists, then sup S exists 

and sup S ≤ sup T. 

13

Proof: Since sup T exists and S ⊆ T, we know that for every s ∈ S, we 

have 

s ≤ sup T. 

Hence, by Axiom 10, we have proved the existence of sup S. In addition, 

sup S ≤ sup T is trivial. 

1.21 Let A and B be two sets of positive numbers bounded above, and 

let a = sup A, b = sup B. Let C be the set of all products of the form xy, 

where x ∈ A and y ∈ B. Prove that ab = sup C. 

Proof: Given ε > 0, we want to find an element c ∈ C such that ab−ε < 

c. If we can show this, we have proved that sup C exists and equals ab. 

Since sup A = a > 0 and sup B = b > 0, we can choose n large enough 

such that a − ε/n > 0, b − ε/n > 0, and n > a + b. So, for this ε ′ = ε/n, 

there exists a ′ ∈ A and b ′ ∈ B such that 


a − ε ′ < a ′ and b − ε ′ 

ab − ε ′ (a + b − ε ′ ) < a ′ b ′ since a − ε ′ > 0 and b − ε ′ > 0 



ab − ε n (a + b) < a′ b ′ := c 

ab − ε < c. 

1.22 Given x > 0, and an integer k ≥ 2. Let a 0 denote the largest integer 

≤ x and, assumeing that a 0 , a 1 , ..., a n−1 have been defined, let a n denote the 

largest integer such that 

a 0 + a 1 

k + a 2 

k 2 + ... + a n 

k n ≤ x. 

Note: When k = 10 the integers a 0 , a 1 , ... are the digits in a decimal 

representation of x. For general k they provide a representation in 

the scale of k. 

(a) Prove that 0 ≤ a i ≤ k − 1 for each i = 1, 2, ... 

14

Proof: Choose a 0 = [x], and thus consider 

[kx − ka 0 ] := a 1 

then 

0 ≤ k (x − a 0 ) < k ⇒ 0 ≤ a 1 ≤ k − 1 

and 

a 0 + a 1 

k ≤ x ≤ a 0 + a 1 

k + 1 k . 

Continue the process, we then have 

0 ≤ a i ≤ k − 1 for each i = 1, 2, ... 

and 

a 0 + a 1 

k + a 2 

k 2 + ... + a n 

k n ≤ x < a 0 + a 1 

k + a 2 

k 2 + ... + a n 

k n + 1 

k n . (*) 

(b) Let r n = a 0 + a 1 k −1 + a 2 k −2 + ... + a n k −n and show that x is the sup 

of the set of rational numbers r 1 , r 2 , ... 

Proof: It is clear by (a)-(*). 

Inequality 

1.23 Prove Lagrange’s identity for real numbers: 

( n∑ 

) 2 ( n∑ 

) ( n∑ 

) 

a k b k = a 2 k b 2 k − 

∑ 

(a k b j − a j b k ) 2 . 

k=1 

k=1 k=1 1≤k

and 

( n∑ 

) ( n∑ 

) 

a k b k a k b k = ∑ 

a k b k a j b j = 

k=1 

k=1 

1≤k,j≤n 

So, 

( n∑ 

k=1 

a k b k 

) 2 

= 

= 

= 

( n∑ 

k=1 

( n∑ 

k=1 

( n∑ 

k=1 

a 2 k 

a 2 k 

a 2 k 

) ( n∑ 

k=1 

) ( n∑ 

k=1 

) ( n∑ 

k=1 

b 2 k 

b 2 k 

b 2 k 

) 

n∑ 

k=1 

a 2 kb 2 k + ∑ k≠j 

+ ∑ a k b k a j b j − ∑ a 2 kb 2 j 

k≠j 

k≠j 

) 

∑ 

+ 2 a k b k a j b j − 

∑ 

) 

− 

1≤k| ≤ ‖x‖ ‖y‖ by Remark (1). 

1.24 Prove that for arbitrary real a k , b k , c k we have 

( n∑ 

) 4 ( n∑ 

) ( n∑ 

) 2 ( n∑ 

) 

a k b k c k ≤ a 4 k b 2 k c 4 k . 

k=1 

k=1 k=1 k=1 

16

Proof: Use Cauchy-Schwarz inequality twice, we then have 

( n∑ 

) ⎡ 4 ( n∑ 

) ⎤ 2 

2 

a k b k c k = ⎣ a k b k c k 

⎦ 

k=1 

k=1 

( n∑ 

) 2 ( n∑ 

) 2 

≤ a 2 kc 2 k b 2 k 

k=1 

k=1 

( n∑ 

) 2 ( n∑ 

) ( n∑ 

≤ a 4 k c 4 k 

k=1 

k=1 k=1 

( n∑ 

) ( n∑ 

) 2 ( n∑ 

= a 4 k b 2 k 

k=1 k=1 k=1 

b 2 k 

c 4 k 

) 2 

) 

. 

1.25 Prove that Minkowski’s inequality: 

( n∑ 

k=1 

(a k + b k ) 2 ) 1/2 

≤ 

a 2 k) 1/2 

+ 

b 2 k) 1/2 

. 

( n∑ 

k=1 

( n∑ 

k=1 

This is the triangle inequality ‖a + b‖ ≤ ‖a‖+‖b‖ for n−dimensional vectors, 

where a = (a 1 , ..., a n ) , b = (b 1 , ..., b n ) and 

‖a‖ = 

( n∑ 

k=1 

a 2 k) 1/2 

. 


n∑ 

n∑ n∑ 

n∑ 

(a k + b k ) 2 = a 2 k + b 2 k + 2 a k b k 

k=1 

k=1 

k=1 

k=1 

( 

n∑ n∑ 

n∑ 

) 1/2 ( n∑ 1/2 

≤ a 2 k + b 2 k + 2 a 2 k bk) 2 by Cauchy-Schwarz inequality 

k=1 k=1 

k=1 

k=1 

⎡ 

1/2 ) ⎤ 1/2 

2 

= ⎣ ak) 2 + 

⎦ . 

( n∑ 

( n∑ 

k=1 

k=1 

b 2 k 

17

So, 

( n∑ 

k=1 

(a k + b k ) 2 ) 1/2 

≤ 

a 2 k) 1/2 

+ 

b 2 k) 1/2 

. 

( n∑ 

k=1 

( n∑ 

k=1 

1.26 If a 1 ≥ ... ≥ a n and b 1 ≥ ... ≥ b n , prove that 

( n∑ 

) ( n∑ 

) ( n∑ 

) 

a k b k ≤ n a k b k . 

k=1 k=1 

k=1 

Hint. ∑ 1≤j≤k≤n (a k − a j ) (b k − b j ) ≥ 0. 


0 ≤ ∑ 

(a k − a j ) (b k − b j ) = ∑ 

1≤j≤k≤n 


Since 

∑ 

1≤j≤k≤n 

∑ 

1≤j≤k≤n 

a k b j + a j b k = 

1≤j≤k≤n 

a k b j + a j b k ≤ 

= 

= 

∑ 

1≤j

In addition, 

∑ 

a k b k + a j b j 

1≤j≤k≤n 

= 

= n 

n∑ 

a k b k + na 1 b 1 + 

k=1 

n∑ 

a k b k + (n − 1) a 2 b 2 + ... + 

k=2 

n∑ 

a k b k + a 1 b 1 + a 2 b 2 + ... + a n b n 

k=1 

= (n + 1) 

n∑ 

a k b k 

k=1 

which implies that, by (**), 

( n∑ 

) ( n∑ 

) ( n∑ 

) 

a k b k ≤ n a k b k . 

k=1 k=1 

k=1 

n∑ 

k=n−1 

a k b k + 2a n−1 b n−1 + ∑ k=n 

a k b k 

Complex numbers 

1.27 Express the following complex numbers in the form a + bi. 

(a) (1 + i) 3 

Solution: (1 + i) 3 = 1 + 3i + 3i 2 + i 3 = 1 + 3i − 3 − i = −2 + 2i. 

(b) (2 + 3i) / (3 − 4i) 

Solution: 2+3i 

3−4i = (2+3i)(3+4i) 

(3−4i)(3+4i) = −6+17i 

25 

= −6 

25 + 17 

25 i. 

(c) i 5 + i 16 

Solution: i 5 + i 16 = i + 1. 

(d) 1 2 (1 + i) (1 + i−8 ) 

Solution: 1 2 (1 + i) (1 + i−8 ) = 1 + i. 

1.28 In each case, determine all real x and y which satisfy the given 

relation. 

19

(a) x + iy = |x − iy| 

Proof: Since |x − iy| ≥ 0, we have 

(b) x + iy = (x − iy) 2 

x ≥ 0 and y = 0. 

Proof: Since (x − iy) 2 = x 2 − (2xy) i − y 2 , we have 

x = x 2 − y 2 and y = −2xy. 

We consider tow cases: (i) y = 0 and (ii) y ≠ 0. 

(i) As y = 0 : x = 0 or 1. 

(ii) As y ≠ 0 : x = −1/2, and y = ± √ 3 

2 . 

(c) ∑ 100 

k=0 ik = x + iy 

Proof: Since ∑ 100 

k=0 ik = 1−i101 

1−i 

= 1−i 

1−i 

= 1, we have x = 1 and y = 0. 

1.29 If z = x+iy, x and y real, the complex conjugate of z is the complex 

number ¯z = x − iy. Prove that: 

(a) Conjugate of (z 1 + z 2 ) = ¯z 1 + ¯z 2 

Proof: Write z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 , then 

z 1 + z 2 = (x 1 + x 2 ) + i (y 1 + y 2 ) 

= (x 1 + x 2 ) − i (y 1 + y 2 ) 

= (x 1 − iy 1 ) + (x 2 − iy 2 ) 

= ¯z 1 + ¯z 2 . 

(b) z 1 z 2 = ¯z 1¯z 2 

Proof: Write z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 , then 

z 1 z 2 = (x 1 x 2 − y 1 y 2 ) + i (x 1 y 2 + x 2 y 1 ) 

= (x 1 x 2 − y 1 y 2 ) − i (x 1 y 2 + x 2 y 1 ) 

and 

¯z 1¯z 2 = (x 1 − iy 1 ) (x 2 − iy 2 ) 

= (x 1 x 2 − y 1 y 2 ) − i (x 1 y 2 + x 2 y 1 ) . 

20

So, z 1 z 2 = ¯z 1¯z 2 

(c) z¯z = |z| 2 

Proof: Write z = x + iy and thus 

z¯z = x 2 + y 2 = |z| 2 . 

(d) z + ¯z =twice the real part of z 

Proof: Write z = x + iy, then 

z + ¯z = 2x, 

twice the real part of z. 

(e) (z − ¯z) /i =twice the imaginary part of z 

Proof: Write z = x + iy, then 

z − ¯z 

i 

= 2y, 

twice the imaginary part of z. 

1.30 Describe geometrically the set of complex numbers z which satisfies 

each of the following conditions: 

(a) |z| = 1 

Solution: The unit circle centered at zero. 

(b) |z| < 1 

Solution: The open unit disk centered at zero. 

(c) |z| ≤ 1 

Solution: The closed unit disk centered at zero. 

(d) z + ¯z = 1 

Solution: Write z = x + iy, then z + ¯z = 1 means that x = 1/2. So, the 

set is the line x = 1/2. 

(e) z − ¯z = i 

21

Proof: Write z = x + iy, then z − ¯z = i means that y = 1/2. So, the set 

is the line y = 1/2. 

(f) z + ¯z = |z| 2 

Proof: Write z = x + iy, then 2x = x 2 + y 2 ⇔ (x − 1) 2 + y 2 = 1. So, the 

set is the unit circle centered at (1, 0) . 

1.31 Given three complex numbers z 1 , z 2 , z 3 such that |z 1 | = |z 2 | = |z 3 | = 

1 and z 1 + z 2 + z 3 = 0. Show that these numbers are vertices of an equilateral 

triangle inscribed in the unit circle with center at the origin. 

Proof: It is clear that three numbers are vertices of triangle inscribed in 

the unit circle with center at the origin. It remains to show that |z 1 − z 2 | = 

|z 2 − z 3 | = |z 3 − z 1 | . In addition, it suffices to show that 

Note that 

which is equivalent to 


|z 1 − z 2 | = |z 2 − z 3 | . 

|2z 1 + z 3 | = |2z 3 + z 1 | by z 1 + z 2 + z 3 = 0 

|2z 1 + z 3 | 2 = |2z 3 + z 1 | 2 

(2z 1 + z 3 ) (2¯z 1 + ¯z 3 ) = (2z 3 + z 1 ) (2¯z 3 + ¯z 1 ) 


|z 1 | = |z 3 | . 

1.32 If a and b are complex numbers, prove that: 

(a) |a − b| 2 ≤ ( 1 + |a| 2) ( 1 + |b| 2) 


( 

1 + |a| 

2 ) ( 1 + |b| 2) − |a − b| 2 = (1 + āa) ( 1 + ¯bb ) − (a − b) ( ā − ¯b ) 

= (1 + āb) ( 1 + a¯b ) 

= |1 + āb| 2 ≥ 0, 

22

so, |a − b| 2 ≤ ( 1 + |a| 2) ( 1 + |b| 2) 

(b) If a ≠ 0, then |a + b| = |a| + |b| if, and only if, b/a is real and 

nonnegative. 

Proof: (⇒)Since |a + b| = |a| + |b| , we have 




Then 

(⇐) Suppose that 

|a + b| 2 = (|a| + |b|) 2 

Re (āb) = |a| |b| = |ā| |b| 

b 

a = āb 

āa 

b 

a 

āb = |ā| |b| 

= 

|ā| |b| 

|a| 2 ≥ 0. 

= k, where k ≥ 0. 

|a + b| = |a + ka| = (1 + k) |a| = |a| + k |a| = |a| + |b| . 

1.33 If a and b are complex numbers, prove that 

|a − b| = |1 − āb| 

if, and only if, |a| = 1 or |b| = 1. For which a and b is the inequality 

|a − b| < |1 − āb| valid 

Proof: (⇔) Since 

|a − b| = |1 − āb| 

⇔ ( ā − ¯b ) (a − b) = (1 − āb) ( 1 − a¯b ) 

⇔ |a| 2 + |b| 2 = 1 + |a| 2 |b| 2 

⇔ ( |a| 2 − 1 ) ( |b| 2 − 1 ) = 0 

⇔ |a| 2 = 1 or |b| 2 = 1. 

23

By the preceding, it is easy to know that 

|a − b| < |1 − āb| ⇔ 0 < ( |a| 2 − 1 ) ( |b| 2 − 1 ) . 

So, |a − b| < |1 − āb| if, and only if, |a| > 1 and |b| > 1. (Or |a| < 1 and 

|b| < 1). 

1.34 If a and c are real constant, b complex, show that the equation 

az¯z + b¯z + ¯bz + c = 0 (a ≠ 0, z = x + iy) 

represents a circle in the x − y plane. 


[ ( 

z¯z − 

b ¯b ¯z − 

−a −a z + b ) ] b 

= 

−a −a 

−ac + |b|2 

a 2 , 

so, we have 

( )∣ b ∣∣∣ 

2 

∣ z − = 

−a 

−ac + |b|2 

a 2 . 

Hence, as |b| 2 − ac > 0, it is a circle. 

−ac+|b| 2 

a 2 

< 0, it is not a circle. 

Remark: The idea is easy from the fact 

We square both sides and thus 

|z − q| = r. 

z¯z − q¯z − ¯qz + ¯qq = r 2 . 

As −ac+|b|2 

a 2 = 0, it is a point. As 

1.35 Recall the definition of the inverse tangent: given a real number t, 

tan −1 (t) is the unique real number θ which satisfies the two conditions 

− π 2 < θ < +π , tan θ = t. 

2 

If z = x + iy, show that 

(a) arg (z) = tan ( ) −1 y 

x , if x > 0 

24

Proof: Note that in this text book, we say arg (z) is the principal argument 

of z, denoted by θ = arg z, where −π < θ ≤ π. 

So, as x > 0, arg z = tan −1 ( y 

x) 

. 

(b) arg (z) = tan −1 ( y 

x) 

+ π, if x < 0, y ≥ 0 

Proof: As x < 0, and y ≥ 0. The point (x, y) is lying on S = {(x, y) : x < 0, y ≥ 0} . 

Note that −π < arg z ≤ π, so we have arg (z) = tan −1 ( y 

x) 

+ π. 

(c) arg (z) = tan −1 ( y 

x) 

− π, if x < 0, y < 0 

Proof: Similarly for (b). So, we omit it. 

(d) arg (z) = π 2 if x = 0, y > 0; arg (z) = − π 2 

if x = 0, y < 0. 

Proof: It is obvious. 

1.36 Define the folowing ”pseudo-ordering” of the complex numbers: 

we say z 1 < z 2 if we have either 

(i) |z 1 | < |z 2 | or (ii) |z 1 | = |z 2 | and arg (z 1 ) < arg (z 2 ) . 

Which of Axioms 6,7,8,9 are satisfied by this relation 

Proof: (1) For axiom 6, we prove that it holds as follows. Given z 1 = 

r 1 e i arg(z 1) , and r 2 e i arg(z 2) , then if z 1 = z 2 , there is nothing to prove it. If 

z 1 ≠ z 2 , there are two possibilities: (a) r 1 ≠ r 2 , or (b) r 1 = r 2 and arg (z 1 ) ≠ 

arg (z 2 ) . So, it is clear that axiom 6 holds. 

(2) For axiom 7, we prove that it does not hold as follows. Given z 1 = 1 

and z 2 = −1, then it is clear that z 1 < z 2 since |z 1 | = |z 2 | = 1 and arg (z 1 ) = 

0 < arg (z 2 ) = π. However, let z 3 = −i, we have 

z 1 + z 3 = 1 − i > z 2 + z 3 = −1 − i 

since 

and 

|z 1 + z 3 | = |z 2 + z 3 | = √ 2 

arg (z 1 + z 3 ) = − π 4 > −3π 4 = arg (z 2 + z 3 ) . 

(3) For axiom 8, we prove that it holds as follows. If z 1 > 0 and z 2 > 0, 

then |z 1 | > 0 and |z 2 | > 0. Hence, z 1 z 2 > 0 by |z 1 z 2 | = |z 1 | |z 2 | > 0. 

(4) For axiom 9, we prove that it holds as follows. If z 1 > z 2 and z 2 > z 3 , 

we consider the following cases. Since z 1 > z 2 , we may have (a) |z 1 | > |z 2 | or 

(b) |z 1 | = |z 2 | and arg (z 1 ) < arg (z 2 ) . 

As |z 1 | > |z 2 | , it is clear that |z 1 | > |z 3 | . So, z 1 > z 3 . 

25

As |z 1 | = |z 2 | and arg (z 1 ) < arg (z 2 ) , we have arg (z 1 ) > arg (z 3 ) . So, 

z 1 > z 3 . 

1.37 Which of Axioms 6,7,8,9 are satisfied if the pseudo-ordering is 

defined as follows We say (x 1 , y 1 ) < (x 2 , y 2 ) if we have either (i) x 1 < x 2 or 

(ii) x 1 = x 2 and y 1 < y 2 . 

Proof: (1) For axiom 6, we prove that it holds as follows. Given x = 

(x 1 , y 1 ) and y = (x 2 , y 2 ) . If x = y, there is nothing to prove it. We consider 

x ≠ y : As x ≠ y, we have x 1 ≠ x 2 or y 1 ≠ y 2 . Both cases imply x < y or 

y < x. 

(2) For axiom 7, we prove that it holds as follows. Given x = (x 1 , y 1 ) , 

y = (x 2 , y 2 ) and z = (z 1 , z 3 ) . If x < y, then there are two possibilities: (a) 

x 1 < x 2 or (b) x 1 = x 2 and y 1 < y 2 . 

For case (a), it is clear that x 1 + z 1 < y 1 + z 1 . So, x + z < y + z. 

For case (b), it is clear that x 1 + z 1 = y 1 + z 1 and x 2 + z 2 < y 2 + z 2 . So, 

x + z < y + z. 

(3) For axiom 8, we prove that it does not hold as follows. Consider 

x = (1, 0) and y = (0, 1) , then it is clear that x > 0 and y > 0. However, 

xy = (0, 0) = 0. 

(4) For axiom 9, we prove that it holds as follows. Given x = (x 1 , y 1 ) , 

y = (x 2 , y 2 ) and z = (z 1 , z 3 ) . If x > y and y > z, then we consider the 

following cases. (a) x 1 > y 1 , or (b) x 1 = y 1 . 

For case (a), it is clear that x 1 > z 1 . So, x > z. 

For case (b), it is clear that x 2 > y 2 . So, x > z. 

1.38 State and prove a theorem analogous to Theorem 1.48, expressing 

arg (z 1 /z 2 ) in terms of arg (z 1 ) and arg (z 2 ) . 

Proof: Write z 1 = r 1 e i arg(z 1) and z 2 = r 2 e i arg(z 2) , then 

z 1 

z 2 

= r 1 

r 2 

e i[arg(z 1)−arg(z 2 )] . 

Hence, 

where 

( ) 

z1 

arg = arg (z 1 ) − arg (z 2 ) + 2πn (z 1 , z 2 ) , 

z 2 

⎧ 

⎨ 0 if − π < arg (z 1 ) − arg (z 2 ) ≤ π 

n (z 1 , z 2 ) = 1 if − 2π < arg (z 1 ) − arg (z 2 ) ≤ −π 

⎩ 

−1 if π < arg (z 1 ) − arg (z 2 ) < 2π 

. 

26

1.39 State and prove a theorem analogous to Theorem 1.54, expressing 

Log (z 1 /z 2 ) in terms of Log (z 1 ) and Log (z 2 ) . 

Proof: Write z 1 = r 1 e i arg(z 1) and z 2 = r 2 e i arg(z 2) , then 

Hence, 

∣ Log (z 1 /z 2 ) = log 

z 1 ∣∣∣ 

∣ + i arg 

z 2 

z 1 

z 2 

= r 1 

r 2 

e i[arg(z 1)−arg(z 2 )] . 

( 

z1 

z 2 

) 

= log |z 1 | − log |z 2 | + i [arg (z 1 ) − arg (z 2 ) + 2πn (z 1 , z 2 )] by xercise 1.38 

= Log (z 1 ) − Log (z 2 ) + i2πn (z 1 , z 2 ) . 

1.40 Prove that the nth roots of 1 (also called the nth roots of unity) 

are given by α, α 2 , ..., α n , where α = e 2πi/n , and show that the roots ≠ 1 

satisfy the equation 

1 + x + x 2 + ... + x n−1 = 0. 

Proof: By Theorem 1.51, we know that the roots of 1 are given by 

α, α 2 , ..., α n , where α = e 2πi/n . In addition, since 


x n = 1 ⇒ (x − 1) ( 1 + x + x 2 + ... + x n−1) = 0 

1 + x + x 2 + ... + x n−1 = 0 if x ≠ 1. 

So, all roots except 1 satisfy the equation 

1 + x + x 2 + ... + x n−1 = 0. 

1.41 (a) Prove that |z i | < e π for all complex z ≠ 0. 

Proof: Since 

z i = e iLog(z) = e − arg(z)+i log|z| , 

27

we have 

∣ ∣z i ∣ ∣ = e 

− arg(z) < e π 

by −π < arg (z) ≤ π. 

(b) Prove that there is no constant M > 0 such that |cos z| < M for all 

complex z. 

Proof: Write z = x + iy and thus, 

cos z = cos x cosh y − i sin x sinh y 


|cos x cosh y| ≤ |cos z| . 

Let x = 0 and y be real, then 

e y 

2 ≤ 1 ∣ e y + e −y∣ ∣ ≤ |cos z| . 

2 

So, there is no constant M > 0 such that |cos z| < M for all complex z. 

Remark: There is an important theorem related with this exercise. We 

state it as a reference. (Liouville’s Theorem) A bounded entire function 

is constant. The reader can see the book, Complex Analysis by Joseph 

Bak, and Donald J. Newman, pp 62-63. Liouville’s Theorem can 

be used to prove the much important theorem, Fundamental Theorem of 

Algebra. 

1.42 If w = u + iv (u, v real), show that 

z w = e u log|z|−v arg(z) e i[v log|z|+u arg(z)] . 

Proof: Write z w = e wLog(z) , and thus 

wLog (z) = (u + iv) (log |z| + i arg (z)) 

= [u log |z| − v arg (z)] + i [v log |z| + u arg (z)] . 

So, 

z w = e u log|z|−v arg(z) e i[v log|z|+u arg(z)] . 

28

1.43 (a) Prove that Log (z w ) = wLog z +2πin. 

Proof: Write w = u + iv, where u and v are real. Then 

Log (z w ) = log |z w | + i arg (z w ) 

= log [ e u log|z|−v arg(z)] + i [v log |z| + u arg (z)] + 2πin by Exercise1.42 

= u log |z| − v arg (z) + i [v log |z| + u arg (z)] + 2πin. 

On the other hand, 

wLogz + 2πin = (u + iv) (log |z| + i arg (z)) + 2πin 

= u log |z| − v arg (z) + i [v log |z| + u arg (z)] + 2πin. 

Hence, Log (z w ) = wLog z +2πin. 

Remark: There is another proof by considering 

e Log(zw) = z w = e wLog(z) 


Log (z w ) = wLogz + 2πin 

for some n ∈ Z. 

(b) Prove that (z w ) α = z wα e 2πinα , where n is an integer. 

Proof: By (a), we have 

(z w ) α = e αLog(zw) = e α(wLogz+2πin) = e αwLogz e 2πinα = z αw e 2πinα , 

where n is an integer. 

1.44 (i) If θ and a are real numbers, −π < θ ≤ π, prove that 

(cos θ + i sin θ) a = cos (aθ) + i sin (aθ) . 

Proof: Write cos θ + i sin θ = z, we then have 

(cos θ + i sin θ) a = z a = e aLogz = e a[log|eiθ |+i arg(e iθ )] = e 

iaθ 

= cos (aθ) + i sin (aθ) . 

29

Remark: Compare with the Exercise 1.43-(b). 

(ii) Show that, in general, the restriction −π < θ ≤ π is necessary in (i) 

by taking θ = −π, a = 1 2 . 

Proof: As θ = −π, and a = 1 , we have 

2 

( 

(−1) 1 1 

2 = e 2 Log(−1) = e π −π 

2 i = i ≠ −i = cos 

2 

) ( −π 

+ i sin 

2 

) 

. 

(iii) If a is an integer, show that the formula in (i) holds without any 

restriction on θ. In this case it is known as DeMorvre’s theorem. 

Proof: By Exercise 1.43, as a is an integer we have 

(z w ) a = z wa , 

where z w = e iθ . Then 

( 

e 

iθ ) a 

= e iθa = cos (aθ) + i sin (aθ) . 

1.45 Use DeMorvre’s theorem (Exercise 1.44) to derive the triginometric 

identities 

sin 3θ = 3 cos 2 θ sin θ − sin 3 θ 

cos 3θ = cos 3 θ − 3 cos θ sin 2 θ, 

valid for real θ. Are these valid when θ is complex 

Proof: By Exercise 1.44-(iii), we have for any real θ, 

By Binomial Theorem, we have 

(cos θ + i sin θ) 3 = cos (3θ) + i sin (3θ) . 

sin 3θ = 3 cos 2 θ sin θ − sin 3 θ 

and 

cos 3θ = cos 3 θ − 3 cos θ sin 2 θ. 

30

For complex θ, we show that it holds as follows. Note that sin z = eiz −e −iz 

2i 

and cos z = eiz +e −iz 

, we have 

2 

( ) e 

3 cos 2 z sin z − sin 3 iz + e −iz 2 ( ) ( ) e iz − e −iz e iz − e −iz 3 

z = 3 

− 

2 

2i 

2i 

( ) ( ) e 2zi + e −2zi + 2 e iz − e −iz 

= 3 

+ e3zi − 3e iz + 3e −iz − e −3zi 

4 

2i 

8i 

= 1 [ ( 

3 e 2zi + e −2zi + 2 ) ( e zi − e −zi) + ( e 3zi − 3e iz + 3e −iz − e −3zi)] 

8i 

= 1 [( 

3e 3zi + 3e iz − 3e −iz − 3e −3zi) + ( e 3zi − 3e iz + 3e −iz − e −3zi)] 

8i 

= 4 ( 

e 3zi − e −3zi) 

8i 

= 1 ( 

e 3zi − e −3zi) 

2i 

Similarly, we also have 

= sin 3z. 

cos 3 z − 3 cos z sin 2 z = cos 3z. 

1.46 Define tan z = sin z/ cos z and show that for z = x + iy, we have 

tan z = 

sin 2x + i sinh 2y 

cos 2x + cosh 2y . 

31

Proof: Since 

tan z = sin z sin (x + iy) sin x cosh y + i cos x sinh y 

= = 

cos z cos (x + iy) cos x cosh y − i sin x sinh y 

(sin x cosh y + i cos x sinh y) (cos x cosh y + i sin x sinh y) 

= 

(cos x cosh y − i sin x sinh y) (cos x cosh y + i sin x sinh y) 

( 

sin x cos x cosh 2 y − sin x cos x sinh 2 y ) + i ( sin 2 x cosh y sinh y + cos 2 x cosh y sinh y ) 

= 

(cos x cosh y) 2 − (i sin x sinh y) 2 

= sin x cos x ( cosh 2 y − sinh 2 y ) + i (cosh y sinh y) 

cos 2 x cosh 2 y + sin 2 x sinh 2 since sin 2 x + cos 2 x = 1 

y 

(sin x cos x) + i (cosh y sinh y) 

= 

cos 2 x + sinh 2 since cosh 2 y = 1 + sinh 2 y 

y 

1 

2 

= 


2 

cos 2 x + sinh 2 since 2 cosh y sinh y = sinh 2y and 2 sin x cos x = sin 2x 

y 


= 

2 cos 2 x + 2 sinh 2 y 


= 

2 cos 2 x − 1 + 2 sinh 2 y + 1 


= 

cos 2x + cosh 2y since cos 2x = 2 cos2 x − 1 and 2 sinh 2 y + 1 = cosh 2y. 

1.47 Let w be a given complex number. If w ≠ ±1, show that there exists 

two values of z = x + iy satisfying the conditions cos z = w and −π < x ≤ π. 

Find these values when w = i and when w = 2. 

Proof: Since cos z = eiz +e −iz 

, if we let e iz = u, then cos z = w implies 

2 

that 

w = u2 + 1 

⇒ u 2 − 2wu + 1 = 0 

2u 


So, by Theorem 1.51, 

(u − w) 2 = w 2 − 1 ≠ 0 since w ≠ ±1. 

e iz = u = w + ∣ w 2 − 1 ∣ 1/2 e iφ k 

, where φ k = arg (w2 − 1) 

2 

( 

) 

= w ± ∣ w 2 − 1 ∣ 1/2 e i arg(w 2 −1) 

2 

+ 2πk , k = 0, 1. 

2 

32

So, 

ix−y = i (x + iy) = iz = log 

∣ w ± ∣ w 2 − 1 ∣ ⎛ 

( 

)⎞ 

1/2 e i arg (w 2 −1) 

2 

∣ +i arg ⎝w ± ∣ w 2 − 1 ∣ 1/2 e i arg ( w 2 −1 ) 

2 

⎠ 

Hence, there exists two values of z = x+iy satisfying the conditions cos z = w 

and 

⎛ 

( 

)⎞ 

−π < x = arg ⎝w ± ∣ w 2 − 1 ∣ 1/2 e i arg(w 2 −1) 

2 

⎠ ≤ π. 

For w = i, we have 

( 

iz = log ∣ 1 ± √ ) 

2 i∣ + i arg 


z = arg 

(( 

1 ± √ ) ) 

2 i 

For w = 2, we have 

∣ 

iz = log ∣2 ± √ 3∣ + i arg 

(( 

1 ± √ ) ) 

2 i 

( 

− i log ∣ 1 ± √ ) 

2 i∣ . 

( 

2 ± √ ) 

3 


( 

z = arg 2 ± √ ) 

3 

∣ 

− i log ∣2 ± √ 3∣ . 

1.48 Prove Lagrange’s identity for complex numbers: 

∣ n∑ ∣∣∣∣ 

2 n∑ n∑ 

a k b k = |a k | 2 |b k | 2 − ∑ ( ) 2 

ak¯bj − ā j b k . 

∣ 

k=1 

k=1 

k=1 

1≤k

Proof: By Exercise 1.44 (i), we have 

sin nθ = 

[ n+1 

2 ] 

∑ 

k=1 

( n2k−1 

) 

sin 2k−1 θ cos n−(2k−1) θ 

⎧ 

⎫ 

⎪⎨ [ n+1 

2 

∑ 

] 

( 

= sin n θ 

n2k−1 

) 

⎪⎬ 

cot n−(2k−1) θ 

⎪⎩ 

⎪⎭ 

k=1 

= sin n θ { ( n 1) cot n−1 θ − ( n 3) cot n−3 θ + ( n 5) cot n−5 θ − +... } . 

(b) If 0 < θ < π/2, prove that 

sin (2m + 1) θ = sin 2m+1 θP m 

( 

cot 2 θ ) 

where P m is the polynomial of degree m given by 

P m (x) = ( ) 

2m+1 

1 x m − ( ) 

2m+1 

3 x m−1 + ( ) 

2m+1 

5 x m−2 − +... 

Use this to show that P m has zeros at the m distinct points x k = cot 2 {πk/ (2m + 1)} 

for k = 1, 2, ..., m. 

Proof: By (a), 

sin (2m + 1) θ 

= sin 2m+1 θ 

{ (2m+1 

1 

) ( 

cot 2 θ ) m 

− 

( 2m+1 

3 

= sin 2m+1 θP m 

( 

cot 2 θ ) , where P m (x) = 

) ( 

cot 2 θ ) m−1 ( 

+ 

2m+1 

) ( 

5 cot 2 θ ) } 

m−2 

− +... 

m+1 

∑ 

k=1 

( 2m+1 

) 

2k−1 x m+1−k . (*) 

In addition, by (*), sin (2m + 1) θ = 0 if, and only if, P m (cot 2 θ) = 0. Hence, 

P m has zeros at the m distinct points x k = cot 2 {πk/ (2m + 1)} for k = 

1, 2, ..., m. 

(c) Show that the sum of the zeros of P m is given by 

m∑ 

cot 2 πk 

2m + 1 

k=1 

34 

m (2m − 1) 

= , 

3

and the sum of their squares is given by 

m∑ 

cot 4 πk 

2m + 1 = m (2m − 1) (4m2 + 10m − 9) 

. 

45 

k=1 

Note. There identities can be used to prove that ∑ ∞ 

∑ n=1 n−2 = π 2 /6 and 

∞ 

n=1 n−4 = π 4 /90. (See Exercises 8.46 and 8.47.) 

Proof: By (b), we know that sum of the zeros of P m is given by 

( ( m∑ m∑ 

x k = cot 2 πk − 2m+1 

)) 

2m + 1 = − ( 3 m (2m − 1) 

2m+1 

) = . 

3 

k=1 

k=1 

And the sum of their squares is given by 

m∑ m∑ 

x 2 k = cot 4 πk 

2m + 1 

k=1 k=1 

( m 

) 2 ( 

∑ 

∑ 

= x k − 2 

k=1 

1 

1≤i

So, let z = 1, we obtain that 

n−1 

∏ ( 

n = 

) n−1 

∏ 

[( 

1 − e 

2πik/n 

= 1 − cos 2πk 

n 

k=1 

n−1 

∏ 

= 

k=1 

n−1 

∏ 

= 

k=1 

n−1 

∏ 

= 2 n−1 

( 

2 sin 2 πk 

n 

( 

2 sin πk 

n 

k=1 

n−1 

∏ 

= 2 n−1 

= 

[ 

k=1 

∏ 

2 n−1 n−1 

k=1 

n−1 

∏ 

= 2 n−1 

k=1 

) 

− i 

( 

sin πk 

n 

( 

sin πk 

n 

k=1 

) ( 

sin πk 

n 

( 

2 sin πk πk 

cos 

n n 

) 

) ( 

cos 

) 

e i( 3π 2 + πk 

n ) 

( 

sin πk ) ] e ∑ n−1 3π 

k=1 2 + πk 

n 

n 

( 

sin πk 

n 

) 

. 

) 

) 

πk 

− i cos 

n 

( 3π 

2 + πk ) 

+ i sin 

n 

( 

− i sin 2πk )] 

n 

( 3π 

2 + πk )) 

n 

36

Some Basic Notations Of Set Theory 

References 

There are some good books about set theory; we write them down. We 

wish the reader can get more. 

1. Set Theory and Related Topics by Seymour Lipschutz. 

2. Set Theory by Charles C. Pinter. 

3. Theory of sets by Kamke. 

4. Naive set by Halmos. 

2.1 Prove Theorem 2.2. Hint. (a, b) = (c, d) means {{a} , {a, b}} = 

{{c} , {c, d}} . Now appeal to the definition of set equality. 

Proof: (⇐) It is trivial. 

(⇒) Suppose that (a, b) = (c, d) , it means that {{a} , {a, b}} = {{c} , {c, d}} . 

It implies that 

{a} ∈ {{c} , {c, d}} and {a, b} ∈ {{c} , {c, d}} . 

So, if a ≠ c, then {a} = {c, d} . It implies that c ∈ {a} which is impossible. 

Hence, a = c. Similarly, we have b = d. 

2.2 Let S be a relation and let D (S) be its domain. The relation S is 

said to be 

(i) reflexive if a ∈ D (S) implies (a, a) ∈ S, 

(ii) symmetric if (a, b) ∈ S implies (b, a) ∈ S, 

(iii) transitive if (a, b) ∈ S and (b, c) ∈ S implies (a, c) ∈ S. 

A relation which is symmetric, reflexive, and transitive is called an equivalence 

relation. Determine which of these properties is possessed by S, if S 

is the set of all pairs of real numbers (x, y) such that 

(a) x ≤ y 

Proof: Write S = {(x, y) : x ≤ y} , then we check that (i) reflexive, (ii) 

symmetric, and (iii) transitive as follows. It is clear that D (S) = R. 

1

(i) Since x ≤ x, (x, x) ∈ S. That is, S is reflexive. 

(ii) If (x, y) ∈ S, i.e., x ≤ y, then y ≤ x. So, (y, x) ∈ S. That is, S is 

symmetric. 

(iii) If (x, y) ∈ S and (y, z) ∈ S, i.e., x ≤ y and y ≤ z, then x ≤ z. So, 

(x, z) ∈ S. That is, S is transitive. 

(b) x < y 

Proof: Write S = {(x, y) : x < y} , then we check that (i) reflexive, (ii) 


(i) It is clear that for any real x, we cannot have x < x. So, S is not 

reflexive. 

(ii) It is clear that for any real x, and y, we cannot have x < y and y < x 

at the same time. So, S is not symmetric. 

(iii) If (x, y) ∈ S and (y, z) ∈ S, then x < y and y < z. So, x < z wich 

implies (x, z) ∈ S. That is, S is transitive. 

(c) x < |y| 

Proof: Write S = {(x, y) : x < |y|} , then we check that (i) reflexive, (ii) 


(i) Since it is impossible for 0 < |0| , S is not reflexive. 

(ii) Since (−1, 2) ∈ S but (2, −1) /∈ S, S is not symmetric. 

(iii) Since (0, −1) ∈ S and (−1, 0) ∈ S, but (0, 0) /∈ S, S is not transitive. 

(d) x 2 + y 2 = 1 

Proof: Write S = {(x, y) : x 2 + y 2 = 1} , then we check that (i) reflexive, 

(ii) symmetric, and (iii) transitive as follows. It is clear that D (S) = [−1, 1] , 

an closed interval with endpoints, −1 and 1. 

(i) Since 1 ∈ D (S) , and it is impossible for (1, 1) ∈ S by 1 2 + 1 2 ≠ 1, S 

is not reflexive. 

(ii) If (x, y) ∈ S, then x 2 + y 2 = 1. So, (y, x) ∈ S. That is, S is symmetric. 

(iii) Since (1, 0) ∈ S and (0, 1) ∈ S, but (1, 1) /∈ S, S is not transitive. 

(e) x 2 + y 2 < 0 

Proof: Write S = {(x, y) : x 2 + y 2 < 1} = φ, then S automatically satisfies 

(i) reflexive, (ii) symmetric, and (iii) transitive. 

(f) x 2 + x = y 2 + y 

Proof: Write S = {(x, y) : x 2 + x = y 2 + y} = {(x, y) : (x − y) (x + y − 1) = 0} , 

then we check that (i) reflexive, (ii) symmetric, and (iii) transitive as follows. 

It is clear that D (S) = R. 

2

(i) If x ∈ R, it is clear that (x, x) ∈ S. So, S is reflexive. 

(ii) If (x, y) ∈ S, it is clear that (y, x) ∈ S. So, S is symmetric. 

(iii) If (x, y) ∈ S and (y, z) ∈ S, it is clear that (x, z) ∈ S. So, S is 

transitive. 

2.3 The following functions F and G are defined for all real x by the 

equations given. In each case where the composite function G ◦ F can be 

formed, give the domain of G◦F and a formula (or formulas) for (G ◦ F ) (x) . 

(a) F (x) = 1 − x, G (x) = x 2 + 2x 

Proof: Write 

G ◦ F (x) = G [F (x)] = G [1 − x] = (1 − x) 2 + 2 (1 − x) = x 2 − 4x + 3. 

It is clear that the domain of G ◦ F (x) is R. 

(b) F (x) = x + 5, G (x) = |x| /x if x ≠ 0, G (0) = 0. 

Proof: Write 

G ◦ F (x) = G [F (x)] = 

{ 

G (x + 5) = 

|x+5| 

if x ≠ −5. 

x+5 

0 if x = −5. 


{ { 2x, if 0 ≤ x ≤ 1 

x 

(c) F (x) = 

G (x) = 

2 , if 0 ≤ x ≤ 1 

1, otherwise, 

0, otherwise. 

Proof: Write 

⎧ 

⎨ 4x 2 if x ∈ [0, 1/2] 

G ◦ F (x) = G [F (x)] = 0 if x ∈ (1/2, 1] 

⎩ 

1 if x ∈ R − [0, 1] 


Find F (x) if G (x) and G [F (x)] are given as follows: 

(d) G (x) = x 3 , G [F (x)] = x 3 − 3x 2 + 3x − 1. 

Proof: With help of (x − 1) 3 = x 3 − 3x 2 + 3x − 1, it is easy to know 

that F (x) = 1 − x. In addition, there is not other function H (x) such that 

G [H (x)] = x 3 − 3x 2 + 3x − 1 since G (x) = x 3 is 1-1. 

(e) G (x) = 3 + x + x 2 , G [F (x)] = x 2 − 3x + 5. 

3 

.

Proof: Write G (x) = ( x + 1 2) 2 

+ 

11 

4 , then 

G [F (x)] = 

( 

F (x) + 1 ) 2 

+ 11 

2 4 = x2 − 3x + 5 



(2F (x) + 1) 2 = (2x − 3) 2 

F (x) = x − 2 or − x + 1. 

2.4 Given three functions F, G, H, what restrictions must be placed on 

their domains so that the following four composite functions can be defined 

G ◦ F, H ◦ G, H ◦ (G ◦ F ) , (H ◦ G) ◦ F. 

Proof: It is clear for answers, 

R (F ) ⊆ D (G) and R (G) ⊆ D (H) . 

Assuming that H ◦ (G ◦ F ) and (H ◦ G) ◦ F can be defined, prove that 

associative law: 

H ◦ (G ◦ F ) = (H ◦ G) ◦ F. 

Proof: Given any x ∈ D (F ) , then 

((H ◦ G) ◦ F ) (x) = (H ◦ G) (F (x)) 

= H (G (F (x))) 

= H ((G ◦ F ) (x)) 

= (H ◦ (G ◦ F )) (x) . 

So, H ◦ (G ◦ F ) = (H ◦ G) ◦ F. 

2.5 Prove the following set-theoretic identities for union and intersection: 

4

(a) A ∪ (B ∪ C) = (A ∪ B) ∪ C, A ∩ (B ∩ C) = (A ∩ B) ∩ C. 

Proof: For the part, A∪(B ∪ C) = (A ∪ B)∪C : Given x ∈ A∪(B ∪ C) , 

we have x ∈ A or x ∈ B ∪ C. That is, x ∈ A or x ∈ B or x ∈ C. Hence, 

x ∈ A∪B or x ∈ C. It implies x ∈ (A ∪ B)∪C. Similarly, if x ∈ (A ∪ B)∪C, 

then x ∈ A ∪ (B ∪ C) . Therefore, A ∪ (B ∪ C) = (A ∪ B) ∪ C. 

For the part, A∩(B ∩ C) = (A ∩ B)∩C : Given x ∈ A∩(B ∩ C) , we have 

x ∈ A and x ∈ B ∩C. That is, x ∈ A and x ∈ B and x ∈ C. Hence, x ∈ A∩B 

and x ∈ C. It implies x ∈ (A ∩ B) ∩ C. Similarly, if x ∈ (A ∩ B) ∩ C, then 

x ∈ A ∩ (B ∩ C) . Therefore, A ∩ (B ∩ C) = (A ∩ B) ∩ C. 

(b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) . 

Proof: Given x ∈ A ∩ (B ∪ C) , then x ∈ A and x ∈ B ∪ C. We consider 

two cases as follows. 

If x ∈ B, then x ∈ A ∩ B. So, x ∈ (A ∩ B) ∪ (A ∩ C) . 

If x ∈ C, then x ∈ A ∩ C. So, x ∈ (A ∩ B) ∪ (A ∩ C) . 

So, we have shown that 

A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) . (*) 

Conversely, given x ∈ (A ∩ B) ∪ (A ∩ C) , then x ∈ A ∩ B or x ∈ A ∩ C. 

We consider two cases as follows. 

If x ∈ A ∩ B, then x ∈ A ∩ (B ∪ C) . 

If x ∈ A ∩ C, then x ∈ A ∩ (B ∪ C) . 

So, we have shown that 

(A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C) . (**) 

By (*) and (**), we have proved it. 

(c) (A ∪ B) ∩ (A ∪ C) = A ∪ (B ∩ C) 

Proof: Given x ∈ (A ∪ B) ∩ (A ∪ C) , then x ∈ A ∪ B and x ∈ A ∪ C. 

We consider two cases as follows. 

If x ∈ A, then x ∈ A ∪ (B ∩ C) . 

If x /∈ A, then x ∈ B and x ∈ C. So, x ∈ B ∩ C. It implies that 

x ∈ A ∪ (B ∩ C) . 

Therefore, we have shown that 

(A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C) . (*) 

5

Conversely, if x ∈ A ∪ (B ∩ C) , then x ∈ A or x ∈ B ∩ C. We consider 

two cases as follows. 

If x ∈ A, then x ∈ (A ∪ B) ∩ (A ∪ C) . 

If x ∈ B ∩ C, then x ∈ A ∪ B and x ∈ A ∪ C. So, x ∈ (A ∪ B) ∩ (A ∪ C) . 

Therefore, we have shown that 

A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C) . (*) 


(d) (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A) = (A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) 

Proof: Given x ∈ (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A) , then 

x ∈ A ∪ B and x ∈ B ∪ C and x ∈ C ∪ A. (*) 

We consider the cases to show x ∈ (A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) as follows. 

For the case (x ∈ A): 

If x ∈ B, then x ∈ A ∩ B. 

If x /∈ B, then by (*), x ∈ C. So, x ∈ A ∩ C. 

Hence, in this case, we have proved that x ∈ (A ∩ B)∪(A ∩ C)∪(B ∩ C) . 

For the case (x /∈ A): 

If x ∈ B, then by (*), x ∈ C. So, x ∈ B ∩ C. 

If x /∈ B, then by (*), it is impossible. 

Hence, in this case, we have proved that x ∈ (A ∩ B)∪(A ∩ C)∪(B ∩ C) . 

From above, 

(A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A) ⊆ (A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) 


(A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A) . 

So, we have proved it. 

Remark: There is another proof, we write it as a reference. 


(A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A) 

= [(A ∪ B) ∩ (B ∪ C)] ∩ (C ∪ A) 

= [B ∪ (A ∩ C)] ∩ (C ∪ A) 

= [B ∩ (C ∪ A)] ∪ [(A ∩ C) ∩ (C ∪ A)] 

= [(B ∩ C) ∪ (B ∩ A)] ∪ (A ∩ C) 

= (A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) . 

6

(e) A ∩ (B − C) = (A ∩ B) − (A ∩ C) 

Proof: Given x ∈ A ∩ (B − C) , then x ∈ A and x ∈ B − C. So, x ∈ A 

and x ∈ B and x /∈ C. So, x ∈ A ∩ B and x /∈ C. Hence, 

x ∈ (A ∩ B) − C ⊆ (A ∩ B) − (A ∩ C) . (*) 

Conversely, given x ∈ (A ∩ B) − (A ∩ C) , then x ∈ A ∩ B and x /∈ A ∩ C. 

So, x ∈ A and x ∈ B and x /∈ C. So, x ∈ A and x ∈ B − C. Hence, 


(f) (A − C) ∩ (B − C) = (A ∩ B) − C 

x ∈ A ∩ (B − C) (**) 

Proof: Given x ∈ (A − C) ∩ (B − C) , then x ∈ A − C and x ∈ B − C. 

So, x ∈ A and x ∈ B and x /∈ C. So, x ∈ (A ∩ B) − C. Hence, 

(A − C) ∩ (B − C) ⊆ (A ∩ B) − C. (*) 

Conversely, given x ∈ (A ∩ B)−C, then x ∈ A and x ∈ B and x /∈ C. Hence, 

x ∈ A − C and x ∈ B − C. Hence, 


(A ∩ B) − C ⊆ (A − C) ∩ (B − C) . (**) 

(g) (A − B) ∪ B = A if, and only if, B ⊆ A 

Proof: (⇒) Suppose that (A − B) ∪ B = A, then it is clear that B ⊆ A. 

(⇐) Suppose that B ⊆ A, then given x ∈ A, we consider two cases. 

If x ∈ B, then x ∈ (A − B) ∪ B. 

If x /∈ B, then x ∈ A − B. Hence, x ∈ (A − B) ∪ B. 

From above, we have 

A ⊆ (A − B) ∪ B. 

In addition, it is obviously (A − B) ∪ B ⊆ A since A − B ⊆ A and B ⊆ A. 

2.6 Let f : S → T be a function. If A and B are arbitrary subsets of S, 

prove that 

f (A ∪ B) = f (A) ∪ f (B) and f (A ∩ B) ⊆ f (A) ∩ f (B) . 

7

Generalize to arbitrary unions and intersections. 

Proof: First, we prove f (A ∪ B) = f (A) ∪ f (B) as follows. Let y ∈ 

f (A ∪ B) , then y = f (a) or y = f (b) , where a ∈ A and b ∈ B. Hence, 

y ∈ f (A) ∪ f (B) . That is, 

f (A ∪ B) ⊆ f (A) ∪ f (B) . 

Conversely, if y ∈ f (A) ∪ f (B) , then y = f (a) or y = f (b) , where a ∈ A 

and b ∈ B. Hence, y ∈ f (A ∪ B) . That is, 

f (A) ∪ f (B) ⊆ f (A ∪ B) . 

So, we have proved that f (A ∪ B) = f (A) ∪ f (B) . 

For the part f (A ∩ B) ⊆ f (A) ∩ f (B) : Let y ∈ f (A ∩ B) , then y = 

f (x) , where x ∈ A∩B. Hence, y ∈ f (A) and y ∈ f (B) . That is, f (A ∩ B) ⊆ 

f (A) ∩ f (B) . 

For arbitrary unions and intersections, we have the following facts, and 

the proof is easy from above. So, we omit the detail. 

And 

f (∪ i∈I A i ) = ∪ i∈I f (A i ) , where I is an index set. 

f (∩ i∈I A i ) ⊆ ∩ i∈I f (A i ) , where I is an index set. 

Remark: We should note why the equality does NOT hold for the case 

of intersection. for example, consider A = {1, 2} and B = {1, 3} , where 

f (1) = 1 and f (2) = 2 and f (3) = 2. 

f (A ∩ B) = f ({1}) = {1} ⊆ {1, 2} ⊆ f ({1, 2}) ∩ f ({1, 3}) = f (A) ∩ f (B) . 

2.7 Let f : S → T be a function. If Y ⊆ T, we denote by f −1 (Y ) the 

largest subset of S which f maps into Y. That is, 

f −1 (Y ) = {x : x ∈ S and f (x) ∈ Y } . 

The set f −1 (Y ) is called the inverse image of Y under f. Prove that the 

following for arbitrary subsets X of S and Y of T. 

(a) X ⊆ f −1 [f (X)] 

8

Proof: Given x ∈ X, then f (x) ∈ f (X) . Hence, x ∈ f −1 [f (X)] by 

definition of the inverse image of f (X) under f. So, X ⊆ f −1 [f (X)] . 

Remark: The equality may not hold, for example, let f (x) = x 2 on R, 

and let X = [0, ∞), we have 

f −1 [f (X)] = f −1 [[0, ∞)] = R. 

(b) f (f −1 (Y )) ⊆ Y 

Proof: Given y ∈ f (f −1 (Y )) , then there exists a point x ∈ f −1 (Y ) 

such that f (x) = y. Since x ∈ f −1 (Y ) , we know that f (x) ∈ Y. Hence, 

y ∈ Y. So, f (f −1 (Y )) ⊆ Y 

Remark: The equality may not hold, for example, let f (x) = x 2 on R, 

and let Y = R, we have 

f ( f −1 (Y ) ) = f (R) = [0, ∞) ⊆ R. 

(c) f −1 [Y 1 ∪ Y 2 ] = f −1 (Y 1 ) ∪ f −1 (Y 2 ) 

Proof: Given x ∈ f −1 [Y 1 ∪ Y 2 ] , then f (x) ∈ Y 1 ∪ Y 2 . We consider two 

cases as follows. 

If f (x) ∈ Y 1 , then x ∈ f −1 (Y 1 ) . So, x ∈ f −1 (Y 1 ) ∪ f −1 (Y 2 ) . 

If f (x) /∈ Y 1 , i.e., f (x) ∈ Y 2 , then x ∈ f −1 (Y 2 ) . So, x ∈ f −1 (Y 1 ) ∪ 

f −1 (Y 2 ) . 

From above, we have proved that 

f −1 [Y 1 ∪ Y 2 ] ⊆ f −1 (Y 1 ) ∪ f −1 (Y 2 ) . (*) 

Conversely, since f −1 (Y 1 ) ⊆ f −1 [Y 1 ∪ Y 2 ] and f −1 (Y 2 ) ⊆ f −1 [Y 1 ∪ Y 2 ] , 

we have 

f −1 (Y 1 ) ∪ f −1 (Y 2 ) ⊆ f −1 [Y 1 ∪ Y 2 ] . (**) 

From (*) and (**), we have proved it. 

(d) f −1 [Y 1 ∩ Y 2 ] = f −1 (Y 1 ) ∩ f −1 (Y 2 ) 

Proof: Given x ∈ f −1 (Y 1 ) ∩ f −1 (Y 2 ) , then f (x) ∈ Y 1 and f (x) ∈ Y 2 . 

So, f (x) ∈ Y 1 ∩ Y 2 . Hence, x ∈ f −1 [Y 1 ∩ Y 2 ] . That is, we have proved that 

f −1 (Y 1 ) ∩ f −1 (Y 2 ) ⊆ f −1 [Y 1 ∩ Y 2 ] . (*) 

9

Conversely, since f −1 [Y 1 ∩ Y 2 ] ⊆ f −1 (Y 1 ) and f −1 [Y 1 ∩ Y 2 ] ⊆ f −1 (Y 2 ) , 

we have 

f −1 [Y 1 ∩ Y 2 ] ⊆ f −1 (Y 1 ) ∩ f −1 (Y 2 ) . (**) 


(e) f −1 (T − Y ) = S − f −1 (Y ) 

Proof: Given x ∈ f −1 (T − Y ) , then f (x) ∈ T − Y. So, f (x) /∈ Y. We 

want to show that x ∈ S − f −1 (Y ) . Suppose NOT, then x ∈ f −1 (Y ) which 

implies that f (x) ∈ Y. That is impossible. Hence, x ∈ S − f −1 (Y ) . So, we 

have 

f −1 (T − Y ) ⊆ S − f −1 (Y ) . (*) 

Conversely, given x ∈ S −f −1 (Y ) , then x /∈ f −1 (Y ) . So, f (x) /∈ Y. That 

is, f (x) ∈ T − Y. Hence, x ∈ f −1 (T − Y ) . So, we have 


S − f −1 (Y ) ⊆ f −1 (T − Y ) . (**) 

(f) Generalize (c) and (d) to arbitrary unions and intersections. 

Proof: We give the statement without proof since it is the same as (c) 

and (d). In general, we have 

and 

f −1 (∪ i∈I A i ) = ∪ i∈I f −1 (A i ) . 

f −1 (∩ i∈I A i ) = ∩ i∈I f −1 (A i ) . 

Remark: From above sayings and Exercise 2.6, we found that the 

inverse image f −1 and the operations of sets, such as intersection and union, 

can be exchanged. However, for a function, we only have the exchange of 

f and the operation of union. The reader also see the Exercise 2.9 to get 

more. 

2.8 Refer to Exercise 2.7. Prove that f [f −1 (Y )] = Y for every subset Y 

of T if, and only if, T = f (S) . 

Proof: (⇒) It is clear that f (S) ⊆ T. In order to show the equality, it 

suffices to show that T ⊆ f (S) . Consider f −1 (T ) ⊆ S, then we have 

f ( f −1 (T ) ) ⊆ f (S) . 

10

By hyppothesis, we get T ⊆ f (S) . 

(⇐) Suppose NOT, i.e., f [f −1 (Y )] is a proper subset of Y for some 

Y ⊆ T by Exercise 2.7 (b). Hence, there is a y ∈ Y such that y /∈ 

f [f −1 (Y )] . Since Y ⊆ f (S) = T, f (x) = y for some x ∈ S. It implies that 

x ∈ f −1 (Y ) . So, f (x) ∈ f [f −1 (Y )] which is impossible by the choice of y. 

Hence, f [f −1 (Y )] = Y for every subset Y of T. 

2.9 Let f : S → T be a function. Prove that the following statements 

are equivalent. 

(a) f is one-to-one on S. 

(b) f (A ∩ B) = f (A) ∩ f (B) for all subsets A, B of S. 

(c) f −1 [f (A)] = A for every subset A of S. 

(d) For all disjoint subsets A and B of S, the image f (A) and f (B) are 

disjoint. 

(e) For all subsets A and B of S with B ⊆ A, we have 

f (A − B) = f (A) − f (B) . 

Proof: (a) ⇒ (b) : Suppose that f is 1-1 on S. By Exercise 2.6, we 

have proved that f (A ∩ B) ⊆ f (A) ∩ f (B) for all A, B of S. In order to 

show the equality, it suffices to show that f (A) ∩ f (B) ⊆ f (A ∩ B) . 

Given y ∈ f (A) ∩ f (B) , then y = f (a) and y = f (b) where a ∈ A 

and b ∈ B. Since f is 1-1, we have a = b. That is, y ∈ f (A ∩ B) . So, 

f (A) ∩ f (B) ⊆ f (A ∩ B) . 

(b) ⇒ (c) : Suppose that f (A ∩ B) = f (A) ∩ f (B) for all subsets A, B 

of S. If A ≠ f −1 [f (A)] for some A of S, then by Exercise 2.7 (a), there is 

an element a /∈ A and a ∈ f −1 [f (A)] . Consider 

φ = f (A ∩ {a}) = f (A) ∩ f ({a}) by (b) (*) 

Since a ∈ f −1 [f (A)] , we have f (a) ∈ f (A) which contradicts to (*). Hence, 

no such a exists. That is, f −1 [f (A)] = A for every subset A of S. 

(c) ⇒ (d) : Suppose that f −1 [f (A)] = A for every subset A of S. If 

A ∩ B = φ, then Consider 

φ = A ∩ B 

= f −1 [f (A)] ∩ f −1 [f (B)] 

= f −1 (f (A) ∩ f (B)) by Exercise 2.7 (d) 

11

which implies that f (A) ∩ f (B) = φ. 

(d) ⇒ (e) : Suppose that for all disjoint subsets A and B of S, the image 

f (A) and f (B) are disjoint. If B ⊆ A, then since (A − B) ∩ B = φ, we have 


f (A − B) ∩ f (B) = φ 

f (A − B) ⊆ f (A) − f (B) . (**) 

Conversely, we consider if y ∈ f (A) − f (B) , then y = f (x) , where x ∈ A 

and x /∈ B. It implies that x ∈ A − B. So, y = f (x) ∈ f (A − B) . That is, 

By (**) and (***), we have proved it. 

f (A) − f (B) ⊆ f (A − B) . (***) 

(d) ⇒ (a) : Suppose that f (A − B) = f (A) − f (B) for all subsets A and 

B of S with B ⊆ A. If f (a) = f (b) , consider A = {a, b} and B = {b} , we 

have 

f (A − B) = φ 

which implies that A = B. That is, a = b. So, f is 1-1. 

2.10 Prove that if A˜B and B˜C, then A˜C. 

Proof: Since A˜B and B˜C, then there exists bijection f and g such 

that 

f : A → B and g : B → C. 

So, if we consider g ◦ f : A → C, then A˜C since g ◦ f is bijection. 

2.11 If {1, 2, ..., n} ˜ {1, 2, ..., m} , prove that m = n. 

Proof: Since {1, 2, ..., n} ˜ {1, 2, ..., m} , there exists a bijection function 

f : {1, 2, ..., n} → {1, 2, ..., m} . 

Since f is 1-1, then n ≤ m. Conversely, consider f −1 is 1-1 to get m ≤ n. So, 

m = n. 

2.12 If S is an infinite set, prove that S contains a countably infinite 

subset. Hint. Choose an element a 1 in S and consider S − {a 1 } . 

12

Proof: Since S is an infinite set, then choose a 1 in S and thus S − {a 1 } 

is still infinite. From this, we have S − {a 1 , .., a n } is infinite. So, we finally 

have 

{a 1 , ..., a n , ...} (⊆ S) 

which is countably infinite subset. 

2.13 Prove that every infinite set S contains a proper subset similar to S. 

Proof: By Exercise 2.12, we write S = ˜S ∪ {x 1 , ..., x n , ...} , where ˜S ∩ 

{x 1 , ..., x n , ...} = φ and try to show 

as follows. Define 

by 

˜S ∪ {x 2 , ..., x n , ...} ˜S 

f : ˜S ∪ {x 2 , ..., x n , ...} → S = ˜S ∪ {x 1 , ..., x n , ...} 

f (x) = 

{ 

x if x ∈ ˜S 

x i if x = x i+1 

. 

Then it is clear that f is 1-1 and onto. So, we have proved that every infinite 

set S contains a proper subset similar to S. 

Remark: In the proof, we may choose the map 

f : ˜S ∪ {x N+1 , ..., x n , ...} → S = ˜S ∪ {x 1 , ..., x n , ...} 

by 

f (x) = 

{ 

x if x ∈ ˜S 

x i if x = x i+N 

. 

2.14 If A is a countable set and B an uncountable set, prove that B − A 

is similar to B. 

Proof: In order to show it, we consider some cases as follows. (i) B∩A = 

φ (ii) B ∩ A is a finite set, and (iii) B ∩ A is an infinite set. 

For case (i), B − A = B. So, B − A is similar to B. 

For case (ii), it follows from the Remark in Exercise 2.13. 

For case (iii), note that B ∩ A is countable, and let C = B − A, we have 

B = C ∪ (B ∩ A) . We want to show that 

(B − A) ˜B ⇔ C˜C ∪ (B ∩ A) . 

13

By Exercise 2.12, we write C = ˜C ∪ D, where D is countably infinite and 

˜C ∩ D = φ. Hence, 

( ) 

˜C ∪ D 

C˜C ∪ (B ∩ A) ⇔ 

⇔ 

( 

˜C ∪ D 

) 

˜ 

[ ] 

˜ ˜C ∪ (D ∪ (B ∩ A)) 

( 

˜C ∪ D 

′) 

( ) ( ) 

where D ′ = D∪(B ∩ A) which is countably infinite. Since ˜C ∪ D ˜ ˜C ∪ D 

′ 

is clear, we have proved it. 

2.15 A real number is called algebraic if it is a root of an algebraic 

equation f (x) = 0, where f (x) = a 0 + a 1 x + ... + a n x n is a polynomial 

with integer coefficients. Prove that the set of all polynomials with integer 

coefficients is countable and deduce that the set of algebraic numbers is also 

countable. 

Proof: Given a positive integer N (≥ 2) , there are only finitely many 

eqautions with 

n∑ 

n + |a k | = N, where a k ∈ Z. (*) 

k=1 

Let S N = {f : f (x) = a 0 + a 1 x + ... + a n x n satisfies (*)} , then S N is a finite 

set. Hence, ∪ ∞ n=2S n is countable. Clearly, the set of all polynomials with 

integer coefficients is a subset of ∪ ∞ n=2S n . So, the set of all polynomials with 

integer coefficients is countable. In addition, a polynomial of degree k has at 

most k roots. Hence, the set of algebraic numbers is also countable. 

2.16 Let S be a finite set consisting of n elements and let T be the 

collection of all subsets of S. Show that T is a finite set and find the number 

of elements in T. 

Proof: Write S = {x 1 , ..., x n } , then T =the collection of all subsets of 

S. Clearly, T is a finite set with 2 n elements. 

2.17 Let R denote the set of real numbers and let S denote the set 

of all real-valued functions whose domain in R. Show that S and R are not 

equinumrous. Hint. Assume S˜R and let f be a one-to-one function such 

that f (R) = S. If a ∈ R, let g a = f (a) be the real-valued function in S which 

correspouds to real number a. Now define h by the equation h (x) = 1+g x (x) 

if x ∈ R, and show that h /∈ S. 

14

Proof: Assume S˜R and let f be a one-to-one function such that f (R) = 

S. If a ∈ R, let g a = f (a) be the real-valued function in S which correspouds 

to real number a. Define h by the equation h (x) = 1 + g x (x) if x ∈ R, then 


h = f (b) = g b 

h (b) := 1 + g b (b) = g b (b) 

which is impossible. So, S and R are not equinumrous. 

Remark: There is a similar exercise, we write it as a reference. The 

cardinal number of C [a, b] is 2 ℵ 0 

, where ℵ 0 = # (N) . 

Proof: First, # (R) = 2 ℵ 0 

≤ # (C [a, b]) by considering constant function. 

Second, we consider the map 

by 

f : C [a, b] → P (Q × Q) , the power set of Q × Q 

f (ϕ) = {(x, y) ∈ Q × Q : x ∈ [a, b] and y ≤ ϕ (x)} . 

Clearly, f is 1-1. It implies that # (C [a, b]) ≤ # (P (Q × Q)) = 2 ℵ 0 

. 

So, we have proved that # (C [a, b]) = 2 ℵ 0 

. 

Note: For notations, the reader can see the textbook, in Chapter 4, pp 

102. Also, see the book, Set Theory and Related Topics by Seymour 

Lipschutz, Chapter 9, pp 157-174. (Chinese Version) 

2.18 Let S be the collection of all sequences whose terms are the integers 

0 and 1. Show that S is uncountable. 

Proof: Let E be a countable subet of S, and let E consists of the sequences 

s 1 , .., s n , .... We construct a sequence s as follows. If the nth digit 

in s n is 1, we let the nth digit of s be 0, and vice versa. Then the sequence 

s differes from every member of E in at least one place; hence s /∈ E. But 

clearly s ∈ S, so that E is a proper subset of S. 

We have shown that every countable subset of S is a proper subset of S. 

It follows that S is uncountable (for otherwise S would be a proper subset 

of S, which is absurb). 

Remark: In this exercise, we have proved that R, the set of real numbers, 

is uncountable. It can be regarded as the Exercise 1.22 for k = 2. (Binary 

System). 

15

2.19 Show that the following sets are countable: 

(a) the set of circles in the complex plane having the ratiional radii and 

centers with rational coordinates. 

Proof: Write the set of circles in the complex plane having the ratiional 

radii and centers with rational coordinates, {C (x n ; q n ) : x n ∈ Q × Q and q n ∈ Q} := 

S. Clearly, S is countable. 

(b) any collection of disjoint intervals of positive length. 

Proof: Write the collection of disjoint intervals of positive length, {I : I is an interval of positive 

S. Use the reason in Exercise 2.21, we have proved that S is countable. 

2.20 Let f be a real-valued function defined for every x in the interval 

0 ≤ x ≤ 1. Suppose there is a positive number M having the following 

property: for every choice of a finite number of points x 1 , x 2 , ..., x n in the 

interval 0 ≤ x ≤ 1, the sum 

|f (x 1 ) + ... + f (x n )| ≤ M. 

Let S be the set of those x in 0 ≤ x ≤ 1 for which f (x) ≠ 0. Prove that S 

is countable. 

Proof: Let S n = {x ∈ [0, 1] : |f (x)| ≥ 1/n} , then S n is a finite set by 

hypothesis. In addition, S = ∪ ∞ n=1S n . So, S is countable. 

2.21 Find the fallacy in the following ”proof” that the set of all intervals 

of positive length is countable. 

Let {x 1 , x 2 , ...} denote the countable set of rational numbers and let I 

be any interval of positive length. Then I contains infinitely many rational 

points x n , but among these there will be one with smallest index n. Define 

a function F by means of the eqaution F (I) = n if x n is the rational number 

with smallest index in the interval I. This function establishes a one-to-one 

correspondence between the set of all intervals and a subset of the positive 

integers. Hence, the set of all intervals is countable. 

Proof: Note that F is not a one-to-one correspondence between the set 

of all intervals and a subset of the positive integers. So, this is not a proof. 

In fact, the set of all intervals of positive length is uncountable. 

Remark: Compare with Exercise 2.19, and the set of all intervals of 

positive length is uncountable is clear by considering {(0, x) : 0 < x < 1} . 

16

2.22 Let S denote the collection of all subsets of a given set T. Let f : S → 

R be a real-valued function defined on S. The function f is called additive 

if f (A ∪ B) = f (A) + f (B) whenever A and B are disjoint subsets of T. If 

f is additive, prove that for any two subsets A and B we have 

f (A ∪ B) = f (A) + f (B − A) 

and 

f (A ∪ B) = f (A) + f (B) − f (A ∩ B) . 

Proof: Since A ∩ (B − A) = φ and A ∪ B = A ∪ (B − A) , we have 

f (A ∪ B) = f (A ∪ (B − A)) = f (A) + f (B − A) . (*) 

In addition, since(B − A) ∩ (A ∩ B) = φ and B = (B − A) ∪ (A ∩ B) , we 

have 

f (B) = f ((B − A) ∪ (A ∩ B)) = f (B − A) + f (A ∩ B) 


By (*) and (**), we have proved that 

f (B − A) = f (B) − f (A ∩ B) (**) 

f (A ∪ B) = f (A) + f (B) − f (A ∩ B) . 

2.23 Refer to Exercise 2.22. Assume f is additive and assume also that 

the following relations hold for two particular subsets A and B of T : 

and 

and 

f (A ∪ B) = f (A ′ ) + f (B ′ ) − f (A ′ ) f (B ′ ) 

f (A ∩ B) = f (A) f (B) 

f (A) + f (B) ≠ f (T ) , 

where A ′ = T − A, B ′ = T − B. Prove that these relations determine f (T ) , 

and compute the value of f (T ) . 

17

Proof: Write 

f (T ) = f (A) + f (A ′ ) = f (B) + f (B ′ ) , 

then 

[f (T )] 2 = [f (A) + f (A ′ )] [f (B) + f (B ′ )] 

= f (A) f (B) + f (A) f (B ′ ) + f (A ′ ) f (B) + f (A ′ ) f (B ′ ) 

= f (A) f (B) + f (A) [f (T ) − f (B)] + [f (T ) − f (A)] f (B) + f (A ′ ) f (B ′ ) 

= [f (A) + f (B)] f (T ) − f (A) f (B) + f (A ′ ) f (B ′ ) 

= [f (A) + f (B)] f (T ) − f (A) f (B) + f (A ′ ) + f (B ′ ) − f (A ∪ B) 

= [f (A) + f (B)] f (T ) − f (A) f (B) + [f (T ) − f (A)] + [f (T ) − f (B)] 

− [f (A) + f (B) − f (A ∩ B)] 

= [f (A) + f (B) + 2] f (T ) − f (A) f (B) − 2 [f (A) + f (B)] + f (A ∩ B) 

= [f (A) + f (B) + 2] f (T ) − 2 [f (A) + f (B)] 


[f (T )] 2 − [f (A) + f (B) + 2] f (T ) + 2 [f (A) + f (B)] = 0 


x 2 − (a + 2) x + 2a = 0 ⇒ (x − a) (x − 2) = 0 

where a = f (A) + f (B) . So, x = 2 since x ≠ a by f (A) + f (B) ≠ f (T ) . 

18

Charpter 3 Elements of Point set Topology 

Open and closed sets in R 1 and R 2 

3.1 Prove that an open interval in R 1 is an open set and that a closed interval is a 

closed set. 

proof: 1. Let a, b be an open interval in R 1 , and let x a, b. Consider 

minx a, b x : L. Then we have Bx, L x L, x L a, b. Thatis,x is an 

interior point of a, b. Sincex is arbitrary, we have every point of a, b is interior. So, 

a, b is open in R 1 . 

2. Let a, b be a closed interval in R 1 , and let x be an adherent point of a, b. Wewant 

to show x a, b. Ifx a, b, then we have x a or x b. Consider x a, then 

Bx, a 2 

x a, b 3x 2 

a , x 2 

a a, b 

which contradicts the definition of an adherent point. Similarly for x b. 

Therefore, we have x a, b if x is an adherent point of a, b. Thatis,a, b contains 

its all adherent points. It implies that a, b is closed in R 1 . 

3.2 Determine all the accumulation points of the following sets in R 1 and decide 

whether the sets are open or closed (or neither). 

(a) All integers. 

Solution: Denote the set of all integers by Z. Letx Z, and consider 

Bx, x1 x S . So,Z has no accumulation points. 

2 

However, Bx, x1 S x . SoZ contains its all adherent points. It means that 

2 

Z is closed. Trivially, Z is not open since Bx, r is not contained in Z for all r 0. 

Remark: 1. Definition of an adherent point: Let S be a subset of R n ,andx a point in 

R n , x is not necessarily in S. Then x is said to be adherent to S if every n ball Bx 

contains at least one point of S. To be roughly, Bx S . 

2. Definition of an accumulation point: Let S be a subset of R n ,andx a point in R n , 

then x is called an accumulation point of S if every n ball Bx contains at least one point 

of S distinct from x. To be roughly, Bx x S . Thatis,x is an accumulation 

point if, and only if, x adheres to S x. Note that in this sense, 

Bx x S Bx S x. 

3. Definition of an isolated point: If x S, but x is not an accumulation point of S, then 

x is called an isolated point. 

4. Another solution for Z is closed: Since R Z nZ n, n 1, we know that R Z 

is open. So, Z is closed. 

5. In logics, if there does not exist any accumulation point of a set S, then S is 

automatically a closed set. 

(b) The interval a, b. 

solution: In order to find all accumulation points of a, b, we consider 2 cases as 

follows. 

1. a, b :Letx a, b, then Bx, r x a, b for any r 0. So, every 

point of a, b is an accumulation point. 

2. R 1 a, b , a b, : For points in b, and , a, it is easy to know 

that these points cannot be accumulation points since x b, or x , a, there

exists an n ball Bx, r x such that Bx, r x x a, b . For the point a, it is easy 

to know that Ba, r a a, b . That is, in this case, there is only one 

accumulation point a of a, b. 

So, from 1 and 2, we know that the set of the accumulation points of a, b is a, b. 

Since a a, b, we know that a, b cannot contain its all accumulation points. So, 

a, b is not closed. 

Since an n ball Bb, r is not contained in a, b for any r 0, we know that the point 

b is not interior to a, b. So,a, b is not open. 

(c) All numbers of the form 1/n, (n 1,2,3,.... 

Solution: Write the set 1/n : n 1,2,... 1, 1/2, 1/3, . . . , 1/n,... : S. 

Obviously, 0 is the only one accumulation point of S. So,S is not closed since S does not 

contain the accumulation point 0. Since 1 S, andB1, r is not contained in S for any 

r 0, S is not open. 

Remark: Every point of 1/n : n 1,2,3,... is isolated. 

(d) All rational numbers. 

Solutions: Denote all rational numbers by Q. It is trivially seen that the set of 

accumulation points is R 1 . 

So, Q is not closed. Consider x Q, anyn ball Bx is not contained in Q. Thatis,x 

is not an interior point of Q. In fact, every point of Q is not an interior point of Q. So, 

Q is not open. 

(e) All numbers of the form 2 n 5 m , (m, n 1,2,.... 

Solution: Write the set 

2 n 5 m : m, n 1, 2, . . m m1 1 2 5m , 1 4 5m ,..., 1 

2 

n 5m ,... : S 

1 2 1 5 , 1 2 1 5 ,..., 1 2 2 5 1 

m ,... 

1 4 1 5 , 1 4 1 5 ,..., 1 2 4 5 1 

m ,... 

........................................................ 

 

2 1 n 1 5 , 1 

2 

n 1 5 ,..., 1 2 2 

n 5 1 

m ... 

..................................................... 

So,wefindthatS 1 : n 1, 2, . . . 1 : m 1, 2, . . . 

2 n 5 

0. So,S is not 

m 

closed since it does not contain 0. Since 1 S, andB 1 , r is not contained in S for any 

2 

2 

r 0, S is not open. 

Remark: By (1)-(3), we can regard them as three sequences 

1 

2 m 5m , 1 5 m m 

and 1 

m1 4 m1 2 

n m 5m , respectively. 

m1 

And it means that for (1), the sequence 5 m m m1 

moves 1 . Similarly for others. So, it is 

2 

easy to see why 1 is an accumulation point of 1 2 2 5m m 

m1 . And thus get the set of all 

accumulation points of 2 n 5 m : m, n 1, 2, . . . 

(f) All numbers of the form 1 n 1/m, (m, n 1,2,.... 

Solution: Write the set of all numbers 1 n 1/m, (m, n 1, 2, . . . as 

1 m 1 m m 

1 1 

m1 m : S. 

m1 

1 

2 

3

And thus by the remark in (e), it is easy to know that S 1, 1. So,S is not closed 

since S S. Since2 S, andB2, r is not contained in S for any r 0, S is not open. 

(g) All numbers of the form 1/n 1/m, (m, n 1,2,.... 

Solution: Write the set of all numbers 1/n 1/m, (m, n 1, 2, . . . as 

1 1/m m m1 

1/2 1/m m m1 

...1/n 1/m m m1 

...: S. 

We find that S 1/n : n N 1/m : m N 0 1/n : n N 0. So,S is 

not closed since S S. Since1 S, andB1, r is not contained in S for any r 0, S is 

not open. 

(h) All numbers of the form 1 n /1 1/n, (n 1,2,.... 

Soluton: Write the set of all numbers 1 n /1 1/n, (n 1, 2, . . . as 

k 

1 

1 1 2k k1 

 

1 

1 1 

2k1 

k 

k1 

: S. 

We find that S 1, 1. So,S is not closed since S S. Since 1 S, andB 1 

2 2 

not contained in S for any r 0, S is not open. 

, r is 

3.3 The same as Exercise 3.2 for the following sets in R 2 . 

(a) All complex z such that |z| 1. 

Solution: Denote z C :|z| 1 by S. It is easy to know that S z C :|z| 1. 

So, S is not closed since S S. Letz S, then |z| 1. Consider Bz, |z|1 S, soevery 

2 

point of S is interior. That is, S is open. 

(b) All complex z such that |z| 1. 

Solution: Denote z C :|z| 1 by S. It is easy to know that S z C :|z| 1. 

So, S is closed since S S. Since1 S, andB1, r is not contained in S for any r 0, S 

is not open. 

(c) All complex numbers of the form 1/n i/m, (m, n 1, 2, . . . . 

Solution: Write the set of all complex numbers of the form 1/n i/m, 

(m, n 1, 2, . . . as 

1 m i m 

1 i m 

m1 2 m ... 1 

m1 

n m i m 

...: S. 

m1 

We know that S 1/n : n 1,2,... i/m : m 1, 2, . . . 0. So,S is not closed 

since S S. Since1 i S, andB1 i, r is not contained in S for any r 0, S is not 

open. 

(d) All points x, y such that x 2 y 2 1. 

Solution: Denote x, y : x 2 y 2 1 by S. We know that 

S x, y : x 2 y 2 1. So,S is not closed since S S. Letp x, y S, then 

x 2 y 2 1. Itiseasytofindthatr 0 such that Bp, r S. So,S is open. 

(e) All points x, y such that x 0. 

Solution: Write all points x, y such that x 0asx, y : x 0 : S. It is easy to 

know that S x, y : x 0. So,S is not closed since S S. Letx S, then it is easy 

to find r x 0 such that Bx, r x S. So,S is open. 

(f) All points x, y such that x 0. 

Solution: Write all points x, y such that x 0asx, y : x 0 : S. It is easy to

know that S x, y : x 0. So,S is closed since S S. Since0, 0 S, and 

B0, 0, r is not contained in S for any r 0, S is not open. 

3.4 Prove that every nonempty open set S in R 1 contains both rational and irratonal 

numbers. 

proof: Given a nonempty open set S in R 1 .Letx S, then there exists r 0 such that 

Bx, r S since S is open. And in R 1 , the open ball Bx, r x r, x r. Since any 

interval contains both rational and irrational numbers, we have S contains both rational and 

irrational numbers. 

3.5 Prove that the only set in R 1 which are both open and closed are the empty set 

and R 1 itself. Is a similar statement true for R 2 

Proof: Let S be the set in R 1 , and thus consider its complement T R 1 S. Then we 

have both S and T are open and closed. Suppose that S R 1 and S , we will show that 

it is impossible as follows. 

Since S R 1 ,andS , then T and T R 1 . Choose s 0 S and t 0 T, then we 

consider the new point s 0t 0 

which is in S or T since R S T. If s 0t 0 

S, wesay 

2 2 

s 0 t 0 

s 

2 1, otherwise, we say s 0t 0 

t 

2 1. 

Continue these steps, we finally have two sequences named s n S and t m T. 

In addition, the two sequences are convergent to the same point, say p by our construction. 

So, we get p S and p T since both S and T are closed. 

However, it leads us to get a contradiction since p S T . Hence S R 1 or 

S . 

Remark: 1. In the proof, the statement is true for R n . 

2. The construction is not strange for us since the process is called Bolzano Process. 

3. 6 Prove that every closed set in R 1 is the intersection of a countable collection of 

open sets. 

proof: Given a closed set S, and consider its complement R 1 S which is open. If 

R 1 S , there is nothing to prove. So, we can assume that R 1 S . 

Let x R 1 S, then x is an interior point of R 1 S. So, there exists an open interval 

a, b such that x a, b R 1 S. In order to show our statement, we choose a smaller 

interval a x , b x so that x a x , b x and a x , b x a, b R 1 S. Hence, we have 

R 1 S xR 1 S a x , b x 


S R 1 xR 1 S a x , b x 

xR 1 S R 1 a x , b x 

n n1 R 1 a n , b n (by Lindelof Convering Theorem). 

Remark: 1. There exists another proof by Representation Theorem for Open Sets on 

The Real Line. 

2. Note that it is true for that every closed set in R 1 is the intersection of a countable 

collection of closed sets. 

3. The proof is suitable for R n if the statement is that every closed set in R n is the 

intersection of a countable collection of open sets. All we need is to change intervals into 

disks.

3.7 Prove that a nonempty, bounded closed set S in R 1 is either a closed interval, or 

that S can be obtained from a closed interval by removing a countable disjoint collection of 

open intervals whose endpoints belong to S. 

proof: If S is an interval, then it is clear that S is a closed interval. Suppose that S is not 

an interval. Since S is bounded and closed, both sup S and inf S are in S. So,R 1 S 

inf S,supS S. Denote inf S,supS by I. Consider R 1 S is open, then by 

Representation Theorem for Open Sets on The Real Line, we have 

R 1 S m m1 I m 

I S 


S I m m1 I m . 

That is, S can be obtained from a closed interval by removing a countable disjoint 

collection of open intervals whose endpoints belong to S. 

Open and closed sets in R n 

3.8 Prove that open n balls and n dimensional open intervals are open sets in R n . 

proof: Given an open n ball Bx, r. Choose y Bx, r and thus consider 

By, d Bx, r, whered min|x y|, r |x y|. Then y is an interior point of Bx, r. 

Since y is arbitrary, we have all points of Bx, r are interior. So, the open n ball Bx, r is 

open. 

Given an n dimensional open interval a 1 , b 1 a 2 , b 2 ...a n , b n : I. Choose 

x x 1, x 2 ,...,x n I and thus consider r minin i1 r i ,wherer i minx i a i , b i x i . 

Then Bx, r I. Thatis,x is an interior point of I. Sincex is arbitrary, we have all points 

of I are interior. So, the n dimensional open interval I is open. 

3.9 Prove that the interior of a set in R n isopeninR n . 

Proof: Let x intS, then there exists r 0 such that Bx, r S. Choose any point of 

Bx, r, sayy. Then y is an interior point of Bx, r since Bx, r is open. So, there exists 

d 0 such that By, d Bx, r S. Soy is also an interior point of S. Sincey is 

arbitrary, we find that every point of Bx, r is interior to S. Thatis,Bx, r intS. Sincex 

is arbitrary, we have all points of intS are interior. So, intS is open. 

Remark: 1 It should be noted that S is open if, and only if S intS. 

2. intintS intS. 

3. If S T, then intS intT. 

3.10 If S R n , prove that intS is the union of all open subsets of R n which are 

contained in S. This is described by saying that intS is the largest open subset of S. 

proof: It suffices to show that intS AS A,whereA is open. To show the statement, 

we consider two steps as follows. 

1. Let x intS, then there exists r 0 such that Bx, r S. So, 

x Bx, r AS A.Thatis,intS AS A. 

2. Let x AS A, then x A for some open set A S. SinceA is open, x is an 

interior point of A. There exists r 0 such that Bx, r A S. Sox is an interior point 

of S, i.e., x intS. Thatis, AS A intS. 

From 1 and 2, we know that intS AS A,whereA is open. 

Let T be an open subset of S such that intS T. SinceintS AS A,whereA is open,

we have intS T AS A which implies intS T by intS AS A. Hence, intS is the 

largest open subset of S. 

3.11 If S and T are subsets of R n , prove that 

intS intT intS T and intS intT intS T. 

Proof: For the part intS intT intS T, we consider two steps as follows. 

1. Since intS S and intT T, wehaveintS intT S T which implies 

that Note that intS intT is open. 

intS intT intintS intT intS T. 

2. Since S T S and S T T, wehaveintS T intS and 

intS T intT. So, 

intS T intS intT. 

From 1 and 2, we know that intS intT intS T. 

For the part intS intT intS T, we consider intS S and intT T. So, 

intS intT S T 

which implies that Note that intS intT is open. 

intintS intT intS intT intS T. 

Remark: It is not necessary that intS intT intS T. For example, let S Q, 

and T Q c , then intS , andintT . However, intS T intR 1 R. 

3.12 Let S denote the derived set and S theclosureofasetSinR n . Prove that 

(a) S is closed in R n ; that is S S . 

proof: Let x be an adherent point of S . In order to show S is closed, it suffices to 

show that x is an accumulation point of S. Assume x is not an accumulation point of S, i.e., 

there exists d 0 such that 

Bx, d x S . * 

Since x adheres to S , then Bx, d S . So, there exists y Bx, d such that y is an 

accumulation point of S. Note that x y, by assumption. Choose a smaller radius d so that 

By, d Bx, d x and By, d S . 

It implies 

By, d S Bx, d x S by (*) 

which is absurb. So, x is an accumulation point of S. Thatis,S contains all its adherent 

points. Hence S is closed. 

(b) If S T, then S T . 

Proof: Let x S , then Bx, r x S for any r 0. It implies that 

Bx, r x T for any r 0sinceS T. Hence, x is an accumulation point of T. 

That is, x T .So,S T . 

(c) S T S T 

Proof: For the part S T S T , we show it by two steps. 

1. Since S S T and T S T, wehaveS S T and T S T by (b). 

So, 

S T S T 

2. Let x S T , then Bx, r x S T . Thatis,

Bx, r x S Bx, r x T . 

So, at least one of Bx, r x S and Bx, r x T is not empty. If 

Bx, r x S , then x S .AndifBx, r x T , then x T .So, 

S T S T . 

From 1 and 2, we have S T S T . 

Remark: Note that since S T S T ,wehaveclS T clS clT, 

where clS is the closure of S. 

(d) S S . 

Proof: Since S S S , then S S S S S S since S S by 

(a). 

(e) S is closed in R n . 

Proof: Since S S S by (d), then S cantains all its accumulation points. Hence, S 

is closed. 

Remark: There is another proof which is like (a). But it is too tedious to write. 

(f) S is the intersection of all closed subsets of R n containing S. That is, S is the 

smallest closed set containing S. 

Proof: It suffices to show that S AS A,whereA is closed. To show the statement, 

we consider two steps as follows. 

1. Since S is closed and S S, then AS A S. 

2. Let x S, then Bx, r S for any r 0. So, if A S, then 

Bx, r A for any r 0. It implies that x is an adherent point of A. Hence if A S, 

and A is closed, we have x A. Thatis,x AS A.So,S AS A. 

From 1 and 2, we have S AS A. 

Let S T S, whereT is closed. Then S AS A T. It leads us to get T S. 

That is, S is the smallest closed set containing S. 

Remark: In the exercise, there has something to remeber. We list them below. 

Remark 1. If S T, then S T . 

2. If S T, then S T. 

3. S S S . 

4. S is closed if, and only if S S. 

5. S is closed. 

6. S is the smallest closed set containing S. 

3.13 Let S and T be subsets of R n . Prove that clS T clS clT and that 

S clT clS T if S is open, where clS is the closure of S. 

Proof: Since S T S and S T T, then clS T clS and, 

clS T clT. So,clS T clS clT. 

Given an open set S , and let x S clT, then we have

1. x S and S is open. 

Bx, d S for some d 0. 

 

Bx, r S Bx, r if r d. 

Bx, r S Bx, d if r d. 

and 

2. x clT 

Bx, r T for any r 0. 

From 1 and 2, we know 

Bx, r S T Bx, r S T Bx, r T if r d. 

Bx, r S T Bx, r S T Bx, d T if r d. 

So, it means that x is an adherent point of S T. Thatis,x clS T. Hence, 

S clT clS T. 

Remark: It is not necessary that clS T clS clT. For example, S Q and 

T Q c , then clS T and clS clT R 1 . 

Note. The statements in Exercises 3.9 through 3.13 are true in any metric space. 

3.14 AsetSinR n is called convex if, for every pair of points x and y in S and every 

real satisfying 0 1, we have x 1 y S. Interpret this statement 

geometrically (in R 2 and R 3 and prove that 

(a) Every n ball in R n is convex. 

Proof: Given an n ball Bp, r, and let x, y Bp, r. Consider x 1 y, where 

0 1. 


x 1 y p x p 1 y p 

x p 1 y p 

r 1 r 

r. 

So, we have x 1 y Bp, r for 0 1. Hence, by the definition of convex, 

we know that every n ball in R n is convex. 

(b) Every n dimensional open interval is convex. 

Proof: Given an n dimensional open interval I a 1 , b 1 ...a n , b n .Letx, y I, 

and thus write x x 1 , x 2 ,...,x n and y y 1 , y 2 ,...y n . Consider 

x 1 y x 1 1 y 1 , x 2 1 y 2 ,...,x n 1 y n where 0 1. 


a i x i 1 y i b i ,wherei 1, 2, . . , n. 

So, we have x 1 y I for 0 1. Hence, by the definition of convex, we 

know that every n dimensional open interval is convex. 

(c) The interior of a convex is convex. 

Proof: Given a convex set S, and let x, y intS. Then there exists r 0 such that 

Bx, r S, andBy, r S. Consider x 1 y : p S, where 0 1, since S 

is convex.

Claim that Bp, r S as follows. 

Let q Bp, r, We want to find two special points x Bx, r, andy By, r such 

that q x 1 y. 

Since the three n balls Bx, r, By, r, andBp, r have the same radius. By 

parallelogram principle, we let x q x p, andy q y p, then 

x x q p r, andy y q p r. 

It implies that x Bx, r, andy By, r. In addition, 

x 1 y 

q x p 1 q y p 

q. 

Since x, y S, andS is convex, then q x 1 y S. It implies that Bp, r S 

since q is arbitrary. So, we have proved the claim. That is, for 0 1, 

x 1 y p intS if x, y intS, andS is convex. Hence, by the definition of 

convex, we know that the interior of a convex is convex. 

(d) The closure of a convex is convex. 

Proof: Given a convex set S, and let x, y S. Consider x 1 y : p, where 

0 1, and claim that p S, i.e., we want to show that Bp, r S . 

Suppose NOT, there exists r 0 such that 

Bp, r S . * 

Since x, y S, then Bx, r S and By, r S . And let x Bx, r S and 

2 2 2 

y By, r S. Consider 

2 

x 1 y p x 1 y x 1 y 

x x 1 y 1 y 

 

x x x x 

1 y 1 y 1 y 1 y 

x x 1 y y | |x y 

 

2 r | |x y 

r 

if we choose a suitable number , where 0 1. 

Hence, we have the point x 1 y Bp, r. Note that x, y S and S is convex, 

we have x 1 y S. It leads us to get a contradiction by (*). Hence, we have proved 

the claim. That is, for 0 1, x 1 y p S if x, y S. Hence, by the 

definition of convex, we know that the closure of a convex is convex. 

3.15 Let F be a collection of sets in R n , and let S AF A and T AF A. For each 

of the following statements, either give a proof or exhibit a counterexample. 

F. 

(a) If x is an accumulation point of T, then x is an accumulation point of each set A in 

Proof: Let x be an accumulation point of T, then Bx, r x T for any 

r 0. Note that for any A F, wehaveT A. Hence Bx, r x A for any 

r 0. That is, x is an accumulation point of A for any A F. 

The conclusion is that If x is an accumulation point of T AF A, then x is an 

accumulation point of each set A in F.

(b) If x is an accumulation point of S, then x is an accumulation point of at least one set 

AinF. 

Proof: No! For example, Let S R n ,andF be the collection of sets consisting of a 

single point x R n . Then it is trivially seen that S AF A.Andifx is an accumulation 

point of S, then x is not an accumulation point of each set A in F. 

3.16 Prove that the set S of rational numbers in the inerval 0, 1 cannot be 

expressed as the intersection of a countable collection of open sets. Hint: Write 

S x 1 , x 2 ,..., assume that S k k1 S k , where each S k is open, and construct a 

sequence Q n of closed intervals such that Q n1 Q n S n and such that x n Q n . 

Then use the Cantor intersection theorem to obtain a contradiction. 

Proof: We prove the statement by method of contradiction. Write S x 1 , x 2 ,..., and 

assume that S k k1 S k , where each S k is open. 

Since x 1 S 1 , there exists a bounded and open interval I 1 S 1 such that x 1 I 1 . 

Choose a closed interval Q 1 I 1 such that x 1 Q 1 .SinceQ 1 is an interval, it contains 

infinite rationals, call one of these, x 2 .Sincex 2 S 2 , there exists an open interval I 2 S 2 

and I 2 Q 1 . Choose a closed interval Q 2 I 2 such that x 2 Q 2 . Suppose Q n has been 

constructed so that 

1. Q n is a closed interval. 

2. Q n Q n1 S n1 . 

3. x n Q n . 

Since Q n is an interval, it contains infinite rationals, call one of these, x n1 .Since 

x n1 S n1 , there exists an open interval I n1 S n1 and I n1 Q n . Choose a closed 

interval Q n1 I n1 such that x n1 Q n1 .So,Q n1 satisfies our induction hypothesis, and 

the construction can process. 

Note that 

1. For all n, Q n is not empty. 

2. For all n, Q n is bounded since I 1 is bounded. 

3. Q n1 Q n . 

4. x n Q n . 

Then n n1 Q n by Cantor Intersection Theorem. 

Since Q n S n , n n1 Q n n n1 S n S. So, we have 

S n n1 Q n n n1 Q n 

which is absurb since S n n1 Q n by the fact x n Q n . Hence, we have proved that 

our assumption does not hold. That is, S the set of rational numbers in the inerval 0, 1 

cannot be expressed as the intersection of a countable collection of open sets. 

Remark: 1. Often, the property is described by saying Q is not an G set. 

2. It should be noted that Q c is an G set. 

3. For the famous Theorem called Cantor Intersection Theorem, the reader should 

see another classical text book, Principles of Mathematical Analysis written by Walter 

Rudin, Theorem 3.10 in page 53. 

4. For the method of proof, the reader should see another classical text book, Principles 

of Mathematical Analysis written by Walter Rudin, Theorem 2.43, in page 41.

Covering theorems in R n 

3.17 If S R n , prove that the collection of isolated points of S is countable. 

Proof: Denote the collection of isolated points of S by F. Letx F, there exists an 

n ball Bx, r x x S . Write Q n x 1 , x 2 ,..., then there are many numbers in 

Q n lying on Bx, r x x. We choose the smallest index, say m mx, and denote x by 

x m . 

So, F x m : m P, whereP N, a subset of positive integers. Hence, F is 

countable. 

3.18 Prove that the set of open disks in the xy plane with center x, x and radius 

x 0, x rational, is a countable covering of the set x, y : x 0, y 0. 

Proof: Denote the set of open disks in the xy plane with center x, x and radius x 0 

by S. Choose any point a, b, wherea 0, and b 0. We want to find an 2 ball 

Bx, x, x S which contains a, b. It suffices to find x Q such that 

x, x a, b x. Since 

x, x a, b x x, x a, b 2 x 2 x 2 2a bx a 2 b 2 0. 

Since x 2 2a bx a 2 b 2 x a b 2 2ab, we can choose a suitable rational 

number x such that x 2 2a bx a 2 b 2 0sincea 0, and b 0. Hence, for any 

point a, b, wherea 0, and b 0, we can find an 2 ball Bx, x, x S which 

contains a, b. 

That is, S is a countable covering of the set x, y : x 0, y 0. 

Remark: The reader should give a geometric appearance or draw a graph. 

3.19 The collection Fof open intervals of the form 1/n,2/n, where n 2, 3, . . . , is an 

open covering of the open interval 0, 1. Prove (without using Theorem 3.31) that no 

finite subcollection of F covers 0, 1. 

Proof: Write F as 1 ,1, 1 , 2 ,..., 1 2 3 3 n , 2 n ,.... Obviously, F is an open covering 

of 0, 1. Assume that there exists a finite subcollection of F covers 0, 1, and thus write 

them as F n 1 1 

1 

, m1 ,...., n 1 1 

k 

, mk . Choose p 0, 1 so that p min 1ik n 1 i 

. Then 

p n 1 1 

i 

, mi , where 1 i k. It contracdicts the fact F covers 0, 1. 

Remark: The reader should be noted that if we use Theorem 3.31, then we cannot get 

the correct proof. In other words, the author T. M. Apostol mistakes the statement. 

3.20 Give an example of a set S which is closed but not bounded and exhibit a 

coubtable open covering F such that no finite subset of F covers S. 

Solution: Let S R 1 , then R 1 is closed but not bounded. And let 

F n, n 2 : n Z, then F is a countable open covering of S. In additon, it is 

trivially seen that no finite subset of F covers S. 

3.21 Given a set S in R n with the property that for every x in S there is an n ball Bx 

such that Bx S is coubtable. Prove that S is countable. 

Proof: Note that F Bx : x S forms an open covering of S. SinceS R n , then 

there exists a countable subcover F F of S by Lindelof Covering Theorem. Write 

F Bx n : n N. Since 

S S nN Bx n nN S Bx n , 

and 

S Bx n is countable by hypothesis.

Then S is countable. 

Remark: The reader should be noted that exercise 3.21 is equivalent to exercise 3.23. 

3.22 Prove that a collection of disjoint open sets in R n is necessarily countable. Give an 

example of a collection of disjoint closed sets which is not countable. 

Proof: Let F be a collection of disjoint open sets in R n , and write Q n x 1 , x 2 ,.... 

Choose an open set S in F, then there exists an n ball By, r S. In this ball, there 

are infinite numbers in Q n . We choose the smallest index, say m my. Then we have 

F S m : m P N which is countable. 

For the example that a collection of disjoint closed sets which is not countable, we give 

it as follows. Let G x : x R n , then we complete it. 

3.23 Assume that S R n . A point x in R n is said to be condensation point of S if every 

n ball Bx has the property that Bx S is not countable. Prove that if S is not 

countable, then there exists a point x in S such that x is a condensation point of S. 

Proof: It is equivalent to exercise 3.21. 

Remark: Compare with two definitions on a condensation point and an accumulation 

point, it is easy to know that a condensation point is an accumulation point. However, am 

accumulation point is not a condensation point, for example, S 1/n : n N. Wehave 

0 is an accumulation point of S, but not a condensation point of S. 

3.24 Assume that S R n and assume that S is not countable. Let T denote the set of 

condensation points of S. Prove that 

(a) S T is countable. 

Proof: If S T is uncountable, then there exists a point x in S T such that xisa 

condensation point of S T by exercise 3.23. Obviously, x S is also a condensation 

point of S. It implies x T. So, we have x S T which is absurb since x S T. 

Remark: The reader should regard T as a special part of S, and the advantage of T 

helps us realize the uncountable set S R n . Compare with Cantor-Bendixon Theorem 

in exercise 3.25. 

(b) S T is not countable. 

Proof: Suppose S T is countable, then S S T S T is countable by (a) 

which is absurb. So, S T is not countable. 

(c) T is a closed set. 

Proof: Let x be an adherent point of T, then Bx, r T for any r 0. We want to 

show x T. That is to show x is a condensation point of S. Claim that Bx, r S is 

uncountable for any r 0. 

Suppose NOT, then there exists an n ball Bx, d S which is countable. Since x is an 

adherent point of T, then Bx, d T . Choose y Bx, d T so that By, Bx, d 

and By, S is uncountable. However, we get a contradiction since 

By, S is uncountable Bx, d S is countable . 

Hence, Bx, r S is uncountable for any r 0. That is, x T. SinceT contains its all 

adherent points, T is closed. 

(d) T contains no isolated points. 

Proof: Let x T, andifx is an isolated point of T, then there exists an n ball Bx, d 

such that Bx, d T x. On the other hand, x T means that Bx, d x S is

uncountable. Hence, by exercise 3.23, we know that there exists y Bx, d x S 

such that y is a condensation point of Bx, d x S. So,y is a condensation point of 

S. It implies y T. It is impossible since 

1. y x T. 

2. y Bx, d. 

3. Bx, d T x. 

Hence, x is not an isolated point of T, ifx T. Thatis,T contains no isolatd points. 

Remark: Use exercise 3.25, by (c) and (d) we know that T is perfect. 

Note that Exercise 3.23 is a special case of (b). 

3.25 AsetinR n is called perfect if S S, that is, if S is a closed set which contains 

no isolated points. Prove that every uncountable closed set F in R n canbeexpressedinthe 

form F A B, where A is perfect and B is countable (Cantor-Bendixon theorem). 

Hint. Use Exercise 3.24. 

Proof: Let F be a uncountable closed set in R n . Then by exercise 3.24, 

F F T F T, whereT is the set of condensation points of F. Note that since F is 

closed, T F by the fact, a condensation point is an accumulation point. Define 

F T A and F T B, then B is countable and A T is perfect. 

Remark: 1. The reader should see another classical text book, Principles of 

Mathematical Analysis written by Walter Rudin, Theorem 2.43, in page 41. Since the 

theorem is famous, we list it below. 

Theorem 2.43 Let P be a nonempty perfect set in R k . Then P is uncountable. 

Theorem Modefied 2.43 Let P be a nonempty perfect set in a complete separable 

metric space. Then P is uncountable. 

2. Let S has measure zero in R 1 . Prove that there is a nonempty perfect set P in R 1 such 

that P S . 

Proof: SinceS has measure zero, there exists a collection of open intervals I k such 

that 

S I k and |I k | 1. 

Consider its complement I k c which is closed with positive measure. Since the 

complement has a positive measure, we know that it is uncountable. Hence, by 

Cantor-Bendixon Theorem, we know that 

I k c A B, whereA is perfect and B is countable. 

So, let A P, wehaveproveit. 

Note: From the similar method, we can show that given any set S in R 1 with measure 

0 d , there is a non-empty perfect set P such that P S . In particular, S Q, 

S the set of algebraic numbers, and so on. In addition, even for cases in R k , it still holds. 

Metric Spaces 

3.26 In any metric space M, d prove that the empty set and the whole set M are both 

open and closed. 

proof: In order to show the statement, it suffices to show that M is open and closed 

since M M . Letx M, then for any r 0, B M x, r M. Thatis,x is an interior

point of M. Sincx is arbitrary, we know that every point of M is interior. So, M is open. 

Let x be an adherent point of M, it is clearly x M since we consider all points lie in 

M. Hence, M contains its all adherent points. It implies that M is closed. 

Remark: The reader should regard the statement as a common sense. 

3.27 Consider the following two metrics in R n : 

in 

d 1 x, y max 1in |x i y i |, d 2 x, y |xi i1 

y i |. 

In each of the following metric spaces prove that the ball Ba; r has the geometric 

appearance indicated: 

(a) In R 2 , d 1 , a square with sides parallel to the coordinate axes. 

Solution: It suffices to consider the case B0, 0,1. Letx x 1 , x 2 B0, 0,1, 

then we have 

|x 1 | 1, and |x 2 | 1. 

So, it means that the ball B0, 0,1 is a square with sides lying on the coordinate axes. 

Hence, we know that Ba; r is a square with sides parallel to the coordinate axes. 

(b) In R 2 , d 2 , a square with diagonals parallel to the axes. 

Solution: It suffices to consider the case B0, 0,1. Letx x 1 , x 2 B0, 0,1, 

then we have 

|x 1 x 2 | 1. 

So, it means that the ball B0, 0,1 is a square with diagonals lying on the coordinate 

axes. Hence, we know that Ba; r is a square with diagonals parallel to the coordinate 

axes. 

(c) A cube in R 3 , d 1 . 

Solution:It suffices to consider the case B0, 0, 0,1. Let 

x x 1 , x 2 , x 3 B0, 0, 0,1, then we have 

|x 1 | 1, |x 2 | 1, and |x 3 | 1. 

So, it means that the ball B0, 0, 0,1 is a cube with length 2. Hence, we know that 

Ba; r is a cube with length 2a. 

(d) An octahedron in R 3 , d 2 . 

Solution: It suffices to consider the case B0, 0, 0,1. Let 

x x 1 , x 2 , x 3 B0, 0, 0,1, then we have 

|x 1 x 2 x 3 | 1. 

It means that the ball B0, 0, 0,1 is an octahedron. Hence, Ba; r is an octahedron. 

Remark: The exercise tells us one thing that Ba; r may not be an n ball if we 

consider some different matrices. 

3.28 Let d 1 and d 2 be the metrics of Exercise 3.27 and let x y denote the usual 

Euclidean metric. Prove that the following inequalities for all x and y in R n : 

d 1 x, y x y d 2 x, y and d 2 x, y n x y nd 1 x, y. 

Proof: List the definitions of the three metrics, and compare with them as follows.

Then we have 

(a) 

(b) 

(c) 

(d) 

x y 

1. d 1 x, y max 1in |x i y i |. 

2. x y in i1 

x i y i 2 1/2 

. 

in 

3. d 2 x, y |xi i1 

y i |. 

d 1 x, y max 

1in |x i y i | 

 

 

in 

1/2 

x i y i 2 

i1 

in 

1/2 

x i y i 2 

i1 

in 

|x i y i | 

i1 

2 1/2 

in 

1/2 

n x y n x i y i 2 

i1 

d 2 x, y 2 

n n max |x i y i | 

1in 

d 1 x, y. 

in 

|x i y i | 

i1 

2 

max |x i y i | 2 1/2 

1in 

x y. 

in 

|x i y i | d 2 x, y. 

i1 

 

2 1/2 

in 

1/2 

n x i y i 2 

i1 

n max 

1in |x i y i | 

in 

x i y i 2 2|x i y i ||x j y j | 

i1 

1ijn 

in 

in 

x i y i 2 n 1 x i y i 2 by A. P. G. P. 

i1 

i1 

in 

n x i y i 2 

i1 

nx y 2 . 

So, 

d 2 x, y n x y. 

From (a)-(d), we have proved these inequalities. 

Remark: 1. Let M be a given set and suppose that M, d and M, d are metric spaces. 

We define the metrics d and d are equivalent if, and only if, there exist positive constants 

, such that 

dx, y dx, y dx, y. 

The concept is much important for us to consider the same set with different metrics. For

example, in this exercise, Since three metrics are equivalent, it is easy to know that 

R k , d 1 , R k , d 2 ,andR k , . are complete. (For definition of complete metric space, 

the reader can see this text book, page 74.) 

2. It should be noted that on a finite dimensional vector space X, any two norms are 

equivalent. 

3.29 If M, d is a metric space, define d x, y dx,y . Prove that 1dx,y d isalsoametric 

for M. Note that 0 d x, y 1 for all x, yinM. 

Proof: In order to show that d isametricforM, we consider the following four steps. 

(1) For x M, d x, x 0sincedx, x 0. 

(2) For x y, d x, y dx,y 0sincedx, y 0. 

1dx,y 

(3) For x, y M, d x, y dx,y dy,x d y, x 

1dx,y 1dy,x 

(4) For x, y, z M, 

d dx, y 

x, y 

1 dx, y 1 1 

1 dx, y 

1 1 since dx, y dx, z dz, y 

1 dx, z dz, y 

dx, z dz, y 

 

1 dx, z dz, y 

dx, z 

 

1 dx, z dz, y dz, y 

1 dx, z dz, y 

dx, z 

 

1 dx, z dz, y 

1 dz, y 

d x, z d z, y 

Hence, from (1)-(4), we know that d is also a metric for M. Obviously, 

0 d x, y 1 for all x, y in M. 

Remark: 1. The exercise tells us how to form a new metric from an old metric. Also, 

the reader should compare with exercise 3.37. This is another construction. 

2. Recall Discrete metric d, we find that given any set nonempty S, S, d is a metric 

space, and thus use the exercise, we get another metric space S, d , and so on. Hence, 

here is a common sense that given any nonempty set, we can use discrete metric to form 

many and many metric spaces. 

3.30 Prove that every finite subset of a metric space is closed. 

Proof: Let x be an adherent point of a finite subet S x i : i 1, 2, . . . , n of a metric 

space M, d. Then for any r 0, Bx, r S . Ifx S, then B M x, S where 

min 1ijn dx i , x j . It is impossible. Hence, x S. Thatis,S contains its all adherent 

points. So, S is closed. 

3.31 In a metric space M, d the closed ball of radius r 0 about a point a in M is the 

set B a; r x : dx, a r. 

(a) Prove that B a; r is a closed set. 

Proof: Let x M B a; r, then dx, a r. Consider Bx, , where dx,ar , 

2 

then if y Bx, , wehavedy, a dx, a dx, y dx, a dx,ar r. Hence, 

2 

Bx, M B a; r. That is, every point of M B a; r is interior. So, M B a; r is 

open, or equivalently, B a; r is a closed set.

(b) Give an example of a metric space in which B a; r is not the closure of the open 

ball Ba; r. 

Solution: Consider discrete metric space M, then we have let x M 

The closure of Ba;1 a 

and 

B a;1 M. 

Hence, if we let a is a proper subset of M, then B a;1 is not the closure of the open ball 

Ba;1. 

3.32 In a metric space M, if subsets satisfy A S A , where A is the closure of A, 

thenAissaidtobedense in S. For example, the set Q of rational numbers is dense in R. If 

A is dense in S and if S is dense in T, prove that A is dense in T. 

Proof: Since A is dense in S and S is dense in T, wehaveA S and S T. Then 

A T. Thatis,A is dense in T. 

3.33 Refer to exercise 3.32. A metric space M is said to be separable if there is a 

countable subset A which is dense in M. For example, R 1 is separable becasue the set Q of 

rational numbrs is a countable dense subset. Prove that every Euclidean space R k is 

separable. 

Proof: Since Q k is a countable subset of R k ,andQ k 

R k , then we know that R k is 

separable. 

3.34 Refer to exercise 3.33. Prove that the Lindelof covering theorem (Theorem 3.28) 

is valid in any separable metric space. 

Proof: Let M, d be a separable metric space. Then there exists a countable subset 

S x n : n N M which is dense in M. Given a set A M, and an open covering F 

of A. Write P Bx n , r m : x n S, r m Q. 

Claim that if x M, andG is an open set in M which contains x. Then 

x Bx n , r m G for some Bx n , r m P. 

Since x G, there exists Bx, r x G for some r x 0. Note that x clS since S is 

dense in M. Then, Bx, r x /2 S . So, if we choose x n Bx, r x /2 S and r m Q 

with r x /2 r m r x /3, then we have 

x Bx n , r m 

and 

Bx n , r m Bx, r x 

since if y Bx n , r m , then 

dy, x dy, x n dx n , x 

r m r x 

2 

r x 

3 r x 

2 

r x 

So, we have prvoed the claim x Bx n , r m Bx, r x G or some Bx n , r m P. 

Use the claim to show the statement as follows. Write A GF G, and let x A, then 

there is an open set G in F such that x G. By the claim, there is Bx n , r m : B nm in P 

such that x B nm G. There are, of course, infinitely many such B nm corresponding to 

each G, but we choose only one of these, for example, the one of smallest index, say 

q qx. Then we have x B qx G.

The set of all B qx obtained as x varies over all elements of A is a countable collection 

of open sets which covers A. To get a countable subcollection of F which covers A, we 

simply correlate to each set B qx one of the sets G of F which contained B qx . This 

complete the proof. 

3.35 Refer to exercise 3.32. If A is dense in S and B is open in S, prove that 

B clA B, where clA B means the closure of A B. 

Hint. Exercise 3.13. 

Proof: Since A is dense in S and B is open in S, A S and S B B. Then 

B S B 

A B, B is open in S 

clA B 

by exercise 3.13. 

3.36 Refer to exercise 3.32. If each of A and B is dense in S and if B is open in S, prove 

that A BisdenseinS. 

Proof: Since 

clA B, B is open 

clA B by exercise 3.13 

S B since A is dense in S 

B since B is open in S 

then 

clA B B 

which implies 

clA B S 

since B is dense in S. 

3.37 Given two metric spaces S 1 , d 1 and S 2 , d 2 , ametric for the Cartesian 

product S 1 S 2 can be constructed from d 1 d 2 in may ways. For example, if x x 1 , x 2 

and y y 1 , y 2 are in S 1 S 2 , let x, y d 1 x 1 , y 1 d 2 x 2 , y 2 . Prove that is a 

metric for S 1 S 2 and construct further examples. 

Proof: In order to show that isametricforS 1 S 2 , we consider the following four 

steps. 

(1) For x x 1 , x 2 S 1 S 2 , x, x d 1 x 1 , x 1 d 2 x 2 , x 2 0 0 0. 

(2) For x y, x, y d 1 x 1 , y 1 d 2 x 2 , y 2 0 since if x, y 0, then x 1 y 1 

and x 2 y 2 . 

(3) For x, y S 1 S 2 , 

x, y d 1 x 1 , y 1 d 2 x 2 , y 2 

d 1 y 1 , x 1 d 2 y 2 , x 2 

y, x. 

(4) For x, y, z S 1 S 2 , 

x, y d 1 x 1 , y 1 d 2 x 2 , y 2 

d 1 x 1 , z 1 d 1 z 1 , y 1 d 2 x 2 , z 2 d 2 z 2 , y 2 

d 1 x 1 , z 1 d 2 x 2 , z 2 d 1 z 1 , y 1 d 2 z 2 , y 2 

x, z z, y.

Hence from (1)-(4), we know that isametricforS 1 S 2 . 

For other metrics, we define 

1 x, y : d 1 x 1 , y 1 d 2 x 2 , y 2 for , 0. 

d 

2 x, y : d 1 x 1 , y 1 2 x 2 , y 2 

1 d 2 x 2 , y 2 

and so on. (The proof is similar with us by above exercises.) 

Compact subsets of a metric space 

3.38 Assume S T M. Then S is compact in M, d if, and only if, S is compact in 

the metric subspace T, d. 

Proof: Suppose that S is compact in M, d. LetF O : O is open in T be an 

open covering of S. SinceO is open in T, there exists the corresponding G which is open 

in M such that G T O . It is clear that G forms an open covering of S. Sothereis 

a finite subcovering G 1 ,...,G n of S since S is compact in M, d. Thatis,S kn k1 G k . 


S T S 

T kn k1 G k . 

kn k1 T G k 

kn k1 O k F. 

So, we find a fnite subcovering O 1 ,...,O n of S. Thatis,S is compact in T, d. 

Suppose that S is compact in T, d. LetG G : G is open in M be an open 

covering of S. SinceG T : O is open in T, the collection O forms an open 

covering of S. So, there is a finite subcovering O 1 ,...,O n of S since S is compact in 

T, d. Thatis,S kn k1 O k . It implis that 

S kn k1 O k kn k1 G k . 

So, we find a finite subcovering G 1 ,...,G n of S. Thatis,S is compact in M, d. 

Remark: The exercise tells us one thing that the property of compact is not changed, 

but we should note the property of being open may be changed. For example, in the 

2 dimensional Euclidean space, an open interval a, b is not open since a, b cannot 

contain any 2 ball. 

3.39 If S is a closed and T is compact, then S T is compact. 

Proof: Since T is compact, T is closed. We have S T is closed. Since S T T, by 

Theorem 3.39, we know that S T is compact. 

3.40 The intersection of an arbitrary collection of compact subsets of M is compact. 

Proof: Let F T : T is compacet in M , and thus consider TF 

T, whereF F. 

We have TF 

T is closed. Choose S F . then we have TF 

T S. Hence, by 

Theorem 3.39 TF 

T is compact. 

3.41 The union of a finite number of compact subsets of M is cmpact. 

Proof: Denote T k is a compact subset of M : k 1, 2, . . n by S. LetF be an open 

covering of kn k1 T k . If there does NOT exist a finite subcovering of kn k1 T k , then there 

does not exist a finite subcovering of T m for some T m S. SinceF is also an open 

covering of T m , it leads us to get T m is not compact which is absurb. Hence, if F is an open 

covering of kn k1 T k , then there exists a finite subcovering of kn k1 T k .So,kn k1 T k is

compact. 

3.42 Consider the metric space Q of rational numbers with the Euclidean metric of 

R 1 . Let S consists of all rational numbers in the open interval a, b, whereaandbare 

irrational. Then S is a closed and bounded subset of Q which is not compact. 

Proof: Obviously, S is bounded. Let x Q S, then x a, orx b. Ifx a, then 

B Q x, d x d, x d Q Q S, whered a x. Similarly, x b. Hence, x is an 

interior point of Q S. Thatis,Q S is open, or equivalently, S is closed. 

Remark: 1. The exercise tells us an counterexample about that in a metric space, a 

closed and bounded subset is not necessary to be compact. 

2. Here is another counterexample. Let M be an infinite set, and thus consider the 

metric space M, d with discrete metric d. ThenbythefactBx,1/2 x for any x M, 

we know that F Bx,1/2 : x M forms an open covering of M. It is clear that there 

does not exist a finite subcovering of M. Hence, M is not compact. 

3.In any metric space M, d, we have three equivalent conditions on compact which 

list them below. Let S M. 

(a) Given any open covering of S, there exists a finite subcovering of S. 

(b) Every infinite subset of S has an accumulation point in S. 

(c) S is totally bounded and complete. 

4. It should be note that if we consider the Euclidean spaceR n , d, wehavefour 

equivalent conditions on compact which list them below. Let S R n . 

Remark (a) Given any open covering of S, there exists a finite subcovering of S. 

(b) Every infinite subset of S has an accumulation point in S. 

(c) S is totally bounded and complete. 

(d) S is bounded and closed. 

5. The concept of compact is familar with us since it can be regarded as a extension of 

Bolzano Weierstrass Theorem. 

Miscellaneous properties of the interior and the boundary 

If A and B denote arbitrary subsets of a metric space M, prove that: 

3.43 intA M clM A. 

Proof: In order to show the statement, it suffices to show that M intA clM A. 

1. Let x M intA, we want to show that x clM A, i.e., 

Bx, r M A for all r 0. Suppose Bx, d M A for some d 0. Then 

Bx, d A which implies that x intA. It leads us to get a conradiction since 

x M intA. Hence, if x M intA, then x clM A. Thatis, 

M intA clM A. 

2. Let x clM A, we want to show that x M intA, i.e., x is not an interior 

point of A. Suppose x is an interior point of A, then Bx, d A for some d 0. However, 

since x clM A, then Bx, d M A . It leads us to get a conradiction since 

Bx, d A. Hence, if x clM A, then x M intA. Thatis,clM A M intA. 

From 1 and 2, we know that M intA clM A, or equvilantly, 

intA M clM A.

3.44 intM A M A . 

Proof: Let B M A, and by exercise 3.33, we know that 

M intB clM B 


intB M clM B 


intM A M clA. 

3.45 intintA intA. 

Proof: Since S is open if, and only if, S intS. Hence, Let S intA, wehavethe 

equality intintA intA. 

3.46 

(a) intn 

i1 A i n i1 intA i , where each A i M. 

Proof: We prove the equality by considering two steps. 

(1) Since n 

i1 

i 1, 2, . . . , n. Hence, intn 

i1 

A i A i for all i 1,2,...,n, then intn 

i1 

A i n i1 intA i . 

(2) Since intA i A i , then n 

i1 intA i n 

i1 

have 

n 

i1 

From (1) and (2), we know that intn 

i1 

intA i intn i1 A i . 

A i i1 

A i .Sincen 

i1 

n 

intA i . 

A i intA i for all 

intA i is open, we 

Remark: Note (2), we use the theorem, a finite intersection of an open sets is open. 

Hence, we ask whether an infinite intersection has the same conclusion or not. 

Unfortunately, the answer is NO! Just see (b) and (c) in this exercise. 

(b) int AF A AF intA, if F is an infinite collection of subsets of M. 

Proof: Since AF A A for all A F. Then int AF A intA for all A F. 

Hence, int AF A AF intA. 

(c) Give an example where eqaulity does not hold in (b). 

Proof: Let F 1 

n , 1 n : n N, then int AF A , and AF intA 0. So, 

we can see that in this case, int AF A is a proper subset of AF intA. Hence, the 

equality does not hold in (b). 

Remark: The key to find the counterexample, it is similar to find an example that an 

infinite intersection of opens set is not open. 

3.47 

(a) AF intA int AF A. 

Proof: Since intA A, AF intA AF A.Wehave AF intA int AF A 

since AF intA is open. 

(b) Give an example of a finite collection F in which equality does not hold in (a). 

Solution: Consider F Q, Q c , then we have intQ intQ c and 

intQ Q c intR 1 R 1 . Hence, intQ intQ c is a proper subset of 

intQ Q c R 1 . That is, the equality does not hold in (a). 

3.48

(a) intA if A is open or if A is closed in M. 

Proof: (1) Suppose that A is open. We prove it by the method of contradiction. Assume 

that intA , and thus choose 

x intA 

intclA clM A 

intclA M A 

intclA intM A since intS T intS intT. 

Since 

x intclA Bx, r 1 clA A A 

and 

x intM A Bx, r 2 M A A c * 

we choose r minr 1 , r 2 , then Bx, r A A A c A A c . However, 

x A and x A Bx, r A for this r. ** 

Hence, we get a contradiction since 

Bx, r A by (*) 

and 

Bx, r A by (**). 

That is, intA if A is open. 

(2) Suppose that A is closed, then we have M A is open. By (1), we have 

intM A . 

Note that 

M A clM A clM M A 

clM A clA 

A 

. Hence, intA if A is closed. 

(b) Give an example in which intA M. 

Solution: Let M R 1 ,andA Q, then 

A clA clM A clQ clQ c R 1 . Hence, we have R 1 intA M. 

3.49 If intA intB and if A is closed in M, then intA B . 

Proof: Assume that intA B , then choose x intA B, then there exists 

Bx, r A B for some r 0. In addition, since intA , we find that Bx, r A. 

Hence, Bx, r B A . It implies Bx, r M A . Choose 

y Bx, r M A, then we have 

y Bx, r By, 1 Bx, r, where 0 1 r 

and 

y M A By, 2 M A, forsome 2 0. 

Choose min 1 , 2 , then we have 

By, Bx, r M A 

A B A c 

B.

That is, intB which is absurb. Hence, we have intA B . 

3.50 Give an example in which intA intB but intA B M. 

Solution: Consider the Euclidean sapce R 1 , |. |. LetA Q, andB Q c , then 

intA intB but intA B R 1 . 

3.51 A clA clM A and A M A. 

Proof: By the definition of the boundary of a set, it is clear that 

A clA clM A. In addition, A clA clM A, and 

M A clM A clM M A clM A clA. Hence, we have 

A M A. 

Remark: It had better regard the exercise as a formula. 

3.52 If clA clB , then A B A B. 

Proof: We prove it by two steps. 

(1) Let x A B, then for all r 0, 

Bx, r A B Bx, r A Bx, r B 

and 

Bx, r A B c Bx, r A c B c * 

Note that at least one of Bx, r A and Bx, r B is not empty. Without loss of 

generality, we say Bx, r A . Then by (*), we have for all r 0, 

Bx, r A , andBx, r A c . 

That is, x A. Hence, we have proved A B A B. 

(2) Let x A B. Without loss of generality, we let x A. Then 

Bx, r A , andBx, r A c . 

Since Bx, r A , wehave 

Bx, r A B Bx, r A Bx, r B . ** 

Claim that Bx, r A B c Bx, r A c B c . Suppsoe NOT, it means that 

Bx, r A c B c . Then we have 

Bx, r A Bx, r clA 

and 

Bx, r B Bx, r clB. 

It implies that by hypothesis, Bx, r clA clB which is absurb. Hence, we have 

proved the claim. We have proved that 

Bx, r A B by(**). 

and 

Bx, r A B c . 

That is, x A B. Hence, we have proved A B A B. 

From (1) and (2), we have proved that A B A B. 

Supplement on a separable metric space 

Definition (Base) A collection V of open subsets of X is said to be a base for X if the 

following is true: For every x X and every open set G X such that x G, we 

have 

x V G for some .

In other words, every open set in X is the union of a subcollection of V . 

Theorem Every separable metric space has a countable base. 

Proof: Let M, d be a separable metric space with S x 1 ,...,x n ,... 

satisfying clS M. Consider a collection Bx i , 1 : i, k N, then given any 

k 

x M and x G, whereG is open in X, wehaveBx, G for some 0. 

Since S is dense in M, we know that there is a set Bx i , 1 for some i, k, such that 

k 

x Bx i , 1 Bx, G. So, we know that M has a countable base. 

k 

Corollary R k ,wherek N, has a countable base. 

Proof: SinceR k is separable, by Theorem 1, we know that R k has a countable 

base. 

Theorem Every compact metric space is separable. 

Proof: LetK, d be a compact metric space, and given a radius 1/n, wehave 

p 

K i1 B x n i ,1/n . 

Let S x n i : i, n N , then it is clear S is countable. In order to show that S is 

dense in K, givenx K, we want to show that x is an adherent point of S. 

Consider Bx, for any 0, there is a point x n i in S such that 

B x n i ,1/n Bx, since 1/n 0. Hence, we have shown that Bx, S . 

That is, x clS which implies that K clS. So, we finally have K is separable. 

Corollary Every compact metric space has a countable base. 

Proof: It is immediately from Theorem 1. 

Remark This corallary can be used to show that Arzela-Ascoli Theorem.

Limits And Continuity 

Limits of sequence 

4.1 Prove each of the following statements about sequences in C. 

(a) z n 0if|z| 1; z n diverges if |z| 1. 

Proof: For the part: z n 0if|z| 1. Given 0, we want to find that there exists a 

positive integer N such that as n N, wehave 

|z n 0| . 

Note that log|z| 0since|z| 1, hence if we choose a positive integer N log |z| 

1, 

then as n N, wehave 

|z n 0| . 

For the part: z n diverges if |z| 1. Assume that z n converges to L, thengiven 

1, there exists a positive integer N 1 such that as n N 1 ,wehave 

|z n L| 1 

|z| n 1 |L|. * 

However, note that log|z| 0since|z| 1, if we choose a positive integer 

N max log |z| 

1 |L| 1, N 1 , then we have 

|z| N 1 |L| 

which contradicts (*). Hence, z n diverges if |z| 1. 

Remark: 1. Given any complex number z C 0, lim n|z| 1/n 1. 

2. Keep lim n n! 1/n in mind. 

3. In fact, z n is unbounded if |z| 1. ( z n diverges if |z| 1. ) Since given 

M 1, and choose a positive integer N log |z| 

M 1, then |z| N M. 

(b) If z n 0andifc n is bounded, then c n z n 0. 

Proof: Since c n is bounded, say its bound M, i.e., |c n| M for all n N. In 

addition, since z n 0, given 0, there exists a positive integer N such that as n N, 

we have 

|z n 0| /M 

which implies that as n N, wehave 

|c n z n| M|z n| . 

That is, lim n c n z n 0. 

(c) z n /n! 0 for every complex z. 

Proof: Given a complex z, and thus find a positive integer N such that |z| N/2. 

Consider (let n N). 

z n 

z N 

z nN 

n! N! N 1N 2 n 

Hence, z n /n! 0 for every complex z. 

 

zN 

N! 

1 

2 

nN 

0asn . 

Remark: There is another proof by using the fact 

n1 

a n converges which implies 

z 

a n 0. Since n 

n1 

converges by ratio test for every complex z, then we have 

n!

z n /n! 0 for every complex z. 

(d) If a n n 2 2 n, then a n 0. 

Proof: Since 

0 a n n 2 2 n 2 

n 2 2 n 1 n for all n N, 

and lim n 1/n 0, we have a n 0asn by Sandwich Theorem. 

4.2 If a n2 a n1 a n /2 for all n 1, show that a n a 1 2a 2 /3. Hint: 

a n2 a n1 1 2 a n a n1 . 

Proof: Since a n2 a n1 a n /2 for all n 1, we have b n1 b n /2, where 

b n a n1 a n . So, we have 

n 

b n1 1 b1 0asn . * 

2 

Consider 


So we have 

n1 

a n2 a 2 b k 1 

2 n 

k2 

k1 

b n 

3a n1 

2 

b k 

a 1 2a 2 

2 

a n a 1 2a 2 /3 by (*). 

1 

2 

a n1 a 1 

. 

4.3 If 0 x 1 1andifx n1 1 1 x n for all n 1, prove that x n is a 

decreasing sequence with limit 0. Prove also that x n1 /x n 1 2 . 

Proof: Claim that 0 x n 1 for all n N. We prove the claim by Mathematical 

Induction. As n 1, there is nothing to prove. Suppose that n k holds, i.e., 0 x k 1, 

then as n k 1, we have 

0 x k1 1 1 x k 1 by induction hypothesis. 

So, by Mathematical Induction, we have proved the claim. Use the claim, and then we 

have 

x 

x n1 x n 1 x n 1 x n n x n 1 

1 x n 2 1 x n 0since0 x n 1. 

So, we know that the sequence x n is a decreasing sequence. Since 0 x n 1 for all 

n N, by Completeness of R, (Thatis,a monotonic sequence in R which is bounded is 

a convergent sequence.) Hence, we have proved that x n is a convergent sequence, 

denoted its limit by x. Note that since 

x n1 1 1 x n for all n N, 

we have x lim n x n1 lim n 1 1 x n 1 1 x which implies xx 1 0. 

Since x n is a decreasing sequence with 0 x n 1 for all n N, we finally have x 0. 

For proof of x n1 /x n 1 .Since 2 

1 1 x 

lim 

x0 x 1 2 

then we have 

*

x n1 

x n 

1 1 x n 

x n 

1 2 . 

Remark: In (*), it is the derivative of 1 1 x at the point x 0. Of course, we can 

prove (*) by L-Hospital Rule. 

4.4 Two sequences of positive integers a n and b n are defined recursively by taking 

a 1 b 1 1 and equating rational and irrational parts in the equation 

a n b n 2 a n1 b n1 2 2 for n 2. 

Prove that a n2 2b n2 1 for all n 2. Deduce that a n /b n 2 through values 2 ,and 

that 2b n /a n 2 through values 2 . 

Proof: Note a n b n 2 a n1 b n1 2 2 for n 2, we have 

a n a2 n1 2b2 n1 for n 2, and 

b n 2a n1 b n1 for n 2 

since if A, B, C, andD N with A B 2 C D 2 , then A C, andB D. 

Claim that a n2 2b n2 1 for all n 2. We prove the claim by Mathematical 

Induction. As n 2, we have by (*) 

a 22 2b 22 a 12 2b 12 2 22a 1 b 1 2 1 2 2 22 2 1. Suppose that as n k 2 

holds, i.e., a k2 2b k2 1, then as n k 1, we have by (*) 

a2 k1 2b2 k1 a k2 2b k2 2 22a k b k 2 

a k4 4b k4 4a k2 b2 

k 

a k2 2b k2 2 

1 by induction hypothesis. 

Hence, by Mathematical Induction, we have proved the claim. Note that a n2 2b n2 1for 

all n 2, we have 

* 

2 2 

a n 1 2 2 

b n bn 

and 

2 

2b n 

a n 2 2 2. 

a2 n 

a 

Hence, lim n 

1 

n b n 

2 by lim n b n 

0 from (*) through values 2 ,and 

2b 

lim n 

1 n a n 

2 by lim n an 

0 from (*) through values 2 . 

Remark: From (*), we know that a n and b n is increasing since a n N and 

b n N. That is, we have lim n a n , and lim n b n . 

4.5 A real sequence x n satisfies 7x n1 x n3 6forn 1. If x 1 1 , prove that the 

2 

sequence increases and find its limit. What happens if x 1 3 or if x 2 1 5 2 

Proof: Claim that if x 1 1 ,then0 x 2 n 1 for all n N. We prove the claim by 

Mathematical Induction. Asn 1, 0 x 1 1 1. Suppose that n k holds, i.e., 

2 

0 x k 1, then as n k 1, we have 

0 x k1 x k 3 6 

 

7 

1 7 

6 1 by induction hypothesis. 

Hence, we have prove the claim by Mathematical Induction. Since 

x 3 7x 6 x 3x 1x 2, then

x n1 x n x n 3 6 

x 

7 n 

x n 3 7x n 6 

7 

0since0 x n 1 for all n N. 

It means that the sequence x n (strictly) increasing. Since x n is bounded, by 

completeness of R, we know that he sequence x n is convergent, denote its limit by x. 

Since 

x 

x lim n 

x n1 lim n3 6 

n 

 

7 

x3 6 , 

7 

we find that x 3, 1, or 2. Since 0 x n 1 for all n N, we finally have x 1. 

Claim that if x 1 3 ,then1 x 2 n 2 for all n N. We prove the claim by 

Mathematical Induction. Asn 1, there is nothing to prove. Suppose n k holds, i.e., 

1 x k 2, then as n k 1, we have 

1 1 6 x k1 x k 3 6 

23 6 2. 

7 

7 7 

Hence, we have prove the claim by Mathematical Induction. Since 

x 3 7x 6 x 3x 1x 2, then 

x n1 x n x n 3 6 

x 

7 n 

x n 3 7x n 6 

7 

0since1 x n 2 for all n N. 

It means that the sequence x n (strictly) decreasing. Since x n is bounded, by 

completeness of R, we know that he sequence x n is convergent, denote its limit by x. 

Since 

x 

x lim n 

x n1 lim n3 6 

n 

 

7 

x3 6 , 

7 

we find that x 3, 1, or 2. Since 1 x n 2 for all n N, we finally have x 1. 

Claim that if x 1 5 , then x 2 n 5 for all n N. We prove the claim by 

2 

Mathematical Induction. Asn 1, there is nothing to prove. Suppose n k holds, i.e., 

x k 5 , then as n k 1, 

2 

x k1 x k 3 6 

5 2 3 6 

 

7 7 

173 

56 3 5 2 . 

Hence, we have proved the claim by Mathematical Induction. Ifx n was convergent, 

say its limit x. Then the possibilities for x 3, 1, or 2. However, x n 5 for all n N. 

2 

So, x n diverges if x 1 5 . 2 

Remark: Note that in the case x 1 5/2, we can show that x n is increasing by the 

same method. So, it implies that x n is unbounded. 

4.6 If |a n| 2and|a n2 a n1 | 1 8 |a 2 

n1 a n2 | for all n 1, prove that a n 

converges. 

Proof: Let a n1 a n b n , then we have |b n1 | 1 8 |b n||a n1 a n| 1 2 |b n|, since 

|a n| 2 for all n 1. So, we have |b n1 | 1 2 n |b 1 |. Consider (Let m n)

|a m a n| |a m a m1 a m1 a m2 ...a n1 a n | 

|b m1 | ...|b n| 

|b 1 | 1 

2 

m2 

... 1 

2 

n1 

, * 

then a n is a Cauchy sequence since 1 2 k converges. Hence, we know that a n is a 

convergent sequence. 

Remark: In this exercise, we use the very important theorem, every Cauchy sequence 

in the Euclidean space R k is convergent. 

4.7 In a metric space S, d, assume that x n x and y n y. Prove that 

dx n , y n dx, y. 

Proof: Since x n x and y n y, given 0, there exists a positive integer N such 

that as n N, wehave 

dx n , x /2 and dy n , y /2. 

Hence, as n N, wehave 

|dx n , y n dx, y| |dx n , x dy n , y| 

dx n , x dy n , y /2 /2 

. 

So, it means that dx n , y n dx, y. 

4.8 Prove that in a compact meric space S, d, every sequence in S has a subsequence 

which converges in S. This property also implies that S is compact but you are not required 

to prove this. (For a proof see either Reference 4.2 or 4.3.) 

Proof: Given a sequence x n S, and let T x 1 , x 2 ,.... If the range of T is finite, 

there is nothing to prove. So, we assume that the range of T is infinite. Since S is compact, 

and T S, wehaveT has a accumulation point x in S. So, there exists a point y n in T such 

that By n , x 1 n . It implies that y n x. Hence, we have proved that every sequence in S 

has a subsequence which converges in S. 

Remark: If every sequence in S has a subsequence which converges in S, then S is 

compact. We give a proof as follows. 

Proof: In order to show S is compact, it suffices to show that every infinite subset of S 

has an accumulation point in S. Given any infinite subset T of S, and thus we choose 

x n T (of course in S). By hypothesis, x n has a subsequence x kn which converges 

in S, say its limit x. From definition of limit of a sequence, we know that x is an 

accumulation of T. So,S is compact. 

4.9 Let A beasubsetofametricspaceS. IfA is complete, prove that A is closed. Prove 

that the converse also holds if S is complete. 

Proof: Let x be an accumulation point of A, then there exists a sequence x n such that 

x n x. Sincex n is convergent, we know that x n is a Cauchy sequence. And A is 

complete, we have x n converges to a point y A. By uniqueness, we know x y A. 

So, A contains its all accumulation points. That is, A is closed. 

Suppose that S is complete and A is closed in S. Given any Cauchy sequence 

x n A, we want to show x n is converges to a point in A. Trivially, x n is also a 

Cauchy sequence in S. SinceS is complete, we know that x n is convergent to a point x in 

S. By definition of limit of a sequence, it is easy to know that x is an adherent point of A. 

So, x A since A is closed. That is, every Cauchy sequence in A is convergent. So, A is

complete. 

Supplement 

1. Show that the sequence 

lim 

n 

2n!! 

2n 1!! 0. 

Proof: Let a and b be positive integers satisfying a b 1. Then we have 

a!b a!b! a b! ab!. * 

So, if we let fn 2n!, then we have, by (*) 

2n!! 

2n 1!! fn! 

fn2n 1! 1 

2n 1 0. 

Hence, we know that lim n 

2. Show that 

2n!! 

2n1!! 

0. 

a n 1 1 n 1 2 n 1 n n e 1/2 as n , 

where x means Gauss Symbol. 

Proof: Since 

x 1 2 x2 log1 x x, for all x 1, 1 

we have 

k n 

 

k1 

k 

n 1 2 

k 

n 

k n 

2 

log an log 1 n 

k 

k1 

k n 

 

k1 

Consider i 2 n i 1 2 , then by Sandwish Theorem, we know that 

lim n 

log a n 1/2 

which implies that a n e 1/2 as n . 

3. Show that n! 1/n n for all n N. ( n! 1/n as n .) 

Proof: We prove it by a special method following Gauss’ method. Consider 

n! 1 k n 

n n k 1 1 

and thus let fk : kn k 1, it is easy to show that fk f1 n for all 

k 1, 2, . . . , n. So, we have prove that 

n! 2 n n 


n! 1/n n . 

Remark: There are many and many method to show n! 1/n as n . We do not 

give a detail proofs about it. But We method it as follows as references. 

(a) By A. P. G. P. , we have 

k1 

n 1 

k 

n 1 

n! 

1/n 

k 

n

and use the fact if a n converges to a, thensois 

n 

a k k1 

n . 

(b) Use the fact, by Mathematical Induction, n! 1/n n/3 for all n. 

(c) Use the fact, A n /n! 0asn for any real A. 

(d) Consider pn n! 

n 

1/n , and thus taking log pn. 

n 

(e) Use the famuos formula, a n are positive for all n. 

lim inf a n1 

a n 

lim infa n 1/n lim supa n 1/n lim sup a n1 

a n 

and let a n n! 

n 

. n x 

(f) The radius of the power series k 

k0 

is . 

k! 

(g) Ue the fact, 1 1/n n e 1 1/n n1 , then en n e n n! en n1 e n . 

(h) More. 

Limits of functions 

Note. In Exercise 4.10 through 4.28, all functions are real valued. 

4.10 Let f be defined on an opne interval a, b and assume x a, b. Consider the 

two statements 

(a) lim h0 |fx h fx| 0; 

(b) lim h0 |fx h fx h| 0. 

Prove that (a) always implies (b), and give an example in which (b) holds but (a) does 

not. 

Proof: (a) Since 

lim|fx h fx| 0 lim|fx h fx| 0, 

h0 h0 

we consider 

|fx h fx h| 

|fx h fx fx fx h| 

|fx h fx| |fx fx h| 0ash 0. 

So, we have 

lim|fx h fx h| 0. 

h0 

(b) Let 

fx 

|x| if x 0, 

1ifx 0. 


lim|f0 h f0 h| 0, 

h0 

but 

lim|f0 h f0| lim||h| 1| 1. 

h0 h0 

So, (b) holds but (a) does not. 

Remark: In case (b), there is another example,

1/|x| if x 0, 

gx 

0ifx 0. 

The difference of two examples is that the limit of |g0 h g0| does not exist as h 

tends to 0. 

4.11 Let f be defined on R 2 .If 

lim 

x,ya,b 

fx, y L 

and if the one-dimensional lim xa fx, y and lim yb fx, y both exist, prove that 

lim xa 

lim 

yb 

fx, y 

lim 

yb 

lim xa 

fx, y L. 

Proof: Since lim x,ya,b fx, y L, thengiven 0, there exists a 0 such that as 

0 |x, y a, b| , wehave 

|fx, y L| /2, 

which implies 

lim|fx, y L| 

yb 

which implies 

limfx, y L 

yb 

/2 if 0 |x, y a, b| 

lim xa 

limfx, y L /2 if 0 |x, y a, b| . 

yb 

Hence, we have proved lim xa|lim yb x, y L| /2 . Since is arbitrary, we have 


lim xa 

limfx, y L 0 

yb 

lim xa 

limfx, y L 0. 

yb 

So, lim xa lim yb fx, y L. The proof of lim yb lim xa fx, y L is similar. 

Remark: 1. The exercise is much important since in mathematics, we would encounter 

many and many similar questions about the interchange of the order of limits. So, we 

should keep the exercise in mind. 

2. In the proof, we use the concept: |lim xa fx| 0if,andonlyiflim xa fx 0. 

3. The hypothesis fx, y L as x, y a, b tells us that every approach form these 

points x, y to the point a, b, fx, y approaches to L. Use this concept, and consider the 

special approach from points x, y to x, b and thus from x, b to a, b. Note that since 

lim yb fx, y exists, it means that we can regrad this special approach as one of approaches 

from these points x, y to the point a, b. So, it is natural to have the statement. 

4. The converse of statement is not necessarily true. For example, 

x y if x 0ory 0 

fx, y 

1 otherwise. 

Trivially, we have the limit of fx, y does not exist as x, y 0, 0. However,

lim 

y0 

fx, y 

0ifx 0, 

1ifx 0. 

and lim 

x0 

fx, y 

0ify 0, 

1ify 0. 

lim 

x0 

lim 

y0 

fx, y 

lim 

y0 

limfx, y 1. 

x0 

In each of the preceding examples, determine whether the following limits exist and 

evaluate those limits that do exist: 

lim 

x0 

limfx, y ; lim 

y0 y0 

limfx, y ; 

x0 

lim fx, y. 

x,y0,0 

Now consider the functions f defined on R 2 as follows: 

(a) fx, y x2 y 2 

if x, y 0, 0, f0, 0 0. 

x 2 y 2 

Proof: 1. Since(x 0) 

we have 

2. Since (y 0) 

we have 

x 

limfx, y lim 

2 y 2 

x0 x0 x 2 y 2 

lim 

y0 

x 

limfx, y lim 

2 y 2 

y0 y0 x 2 y 2 

lim 

x0 

 

limfx, y 1. 

x0 

 

limfx, y 1. 

y0 

y 2 

1 ify 0, 

y 2 1ify 0, 

x 2 

1ifx 0, 

x 2 1ifx 0, 

3. (x, y 0, 0) Letx r cos and y r sin , where 0 2, and note that 

x, y 0, 0 r 0. Then 

lim fx, y lim x 2 y 2 

x,y0,0 x,y0,0 x 2 y 2 

lim 

r0 

r 2 cos 2 sin 2 

r 2 

cos 2 sin 2 . 

So, if we choose and /2, we find the limit of fx, y does not exist as 

x, y 0, 0. 

Remark: 1. This case shows that 

1 lim 

x0 

lim 

y0 

fx, y 

lim 

y0 

lim 

x0 

fx, y 

1 

2. Obviously, the limit of fx, y does not exist as x, y 0, 0. Since if it was, then 

by (*), (**), and the preceding theorem, we know that 

which is absurb. 

lim 

x0 

lim 

y0 

fx, y 

lim 

y0 

lim 

x0 

fx, y 

**

xy 

(b) fx, y 

2 

if x, y 0, 0, f0, 0 0. 

xy 2 xy 2 

Proof: 1. Since (x 0) 

xy 2 

limfx, y lim 

0 for all y, 

x0 x0 

xy 2 x y 2 

we have 


we have 

lim 

y0 

limfx, y 0. 

x0 

xy 2 

limfx, y lim 

0 for all x, 

y0 y0 

xy 2 x y 2 

lim 

x0 

limfx, y 0. 

y0 



xy 2 

fx, y 

xy 2 x y 2 

So, 

r 4 cos 2 sin 2 

r 4 cos 2 sin 2 r 2 2r 2 cossin 

cos 2 sin 2 . 

cos 2 sin 2 12cossin 

r 2 

0ifr 0 

fx, y 

1 if /4 or 5/4. 

Hence, we know that the limit of fx, y does not exists as x, y 0, 0. 

(c) fx, y 1 x sinxy if x 0, f0, y y. 


limfx, y lim 1 

x0 x0 x sinxy y * 

we have 


we have 

limfx, y 

y0 

lim 

y0 

lim 

x0 

limfx, y 0. 

x0 

1 

lim y0 x sinxy 0ifx 0, 

lim y0 y 0ifx 0, 

limfx, y 0. 

y0 


x, y 0, 0 r 0. Then

fx, y 

1 

r cos sinr2 cossin if x r cos 0, 

r sin if x r cos 0. 

0ifr 0, 

 

0ifr 0. 

So, we know that lim x,y0,0 fx, y 0. 

Remark: In (*) and (**), we use the famuos limit, that is, 

lim sin x 

x0 x 1. 

There are some similar limits, we write them without proofs. 

(a) lim t t sin1/t 1. 

(b) lim x0 x sin1/x 0. 

(c) lim x0 ,ifb 0. 

(d) fx, y 

sinax 

sinbx 

a b 

x y sin1/x sin1/y if x 0andy 0, 

0 if x 0ory 0. 

** 


x y sin1/x sin1/y x sin1/x sin1/y y sin1/x sin1/y if y 0 

fx, y 

0ify 0 

we have if y 0, the limit fx, y does not exist as x 0, and if y 0, lim x0 fx, y 0. 

Hence, we have (x 0, y 0) 


lim 

y0 

lim 

x0 

fx, y 

does not exist. 

x y sin1/x sin1/y x sin1/x sin1/y y sin1/x sin1/y if x 0 

fx, y 

0ifx 0 

we have if x 0, the limit fx, y does not exist as y 0, and if x 0, lim y0 fx, y 0. 

Hence, we have (x 0, y 0) 

3. (x, y 0, 0) Consider 

we have 

lim 

x0 

|fx, y| 

lim 

y0 

fx, y 


|x y| if x 0andy 0, 

0ifx 0ory 0. 

lim fx, y 0. 

x,y0,0 

sinxsiny 

tan xtan y 

,iftanx tan y, 

(e) fx, y 

cos 3 x if tan x tan y. 

Proof: Since we consider the three approaches whose tend to 0, 0, we may assume 

that x, y /2, /2. and note that in this assumption, x y tan x tan y. Consider 

1. (x 0)

So, 

So, 

2. (y 0) 

lim 

x0 

fx, y 

lim 

y0 

fx, y 

sinxsiny 

lim x0 tan xtan y 

cosy if x y. 

1 if x y. 

lim 

y0 

limfx, y 1. 

x0 

sinxsiny 

lim y0 tan xtan y 

cosx if x y. 

cos 3 x if x y. 

lim 

x0 

limfx, y 1. 

y0 

3. Let x r cos and y r sin , where 0 2, and note that 


lim 

lim fx, y 

x,y0,0 

sinr cos sinr sin 

r0 if cos sin . 

tanr cos tanr sin 

lim r0 cos 3 r cos if cos sin . 

1ifcos sin , byL-Hospital Rule. 

 

1ifcos sin . 

So, we know that lim x,y0,0 fx, y 1. 

and 

then 

So, 

Remark: 1. There is another proof about (e)-(3). Consider 

x y 

sin x sin y 2cos sin 

x y 

2 2 

lim 

lim fx, y 

x,y0,0 

That is, lim x,y0,0 fx, y 1. 

tan x tan y sin x cosx sin y 

cosy , 

sin x sin y 

tan x tan y 

xy 

cos 

2 

cosxcosy 

cos xy . 

2 

x,y0,0 

cos 

xy 

cos xcos y 

2 

cos 

xy 

2 

lim x,y0,0 cos 3 x 1ifx y. 

1ifx y, 

2. In the process of proof, we use the concept that we write it as follows. Since its proof 

is easy, we omit it. If 

or 

L 

lim fx, y 

x,ya,b 

if x y 

L if x y

then we have 

lim fx, y L if x 0andy 0, 

x,ya,b L if x 0ory 0. 

lim fx, y L. 

x,ya,b 

4.12 If x 0, 1 prove that the following limit exists, 

lim 

m 

lim n cos2n m!x , 

and that its value is 0 or 1, according to whether x is irrational or rational. 

Hence, 

Proof: If x is rational, say x q/p, whereg. c. d. q, p 1, then p!x N. So, 

lim n 

cos 2n m!x 

lim 

m 

1ifm p, 

0ifm p. 

lim n cos2n m!x 1. 

If x is irrational, then m!x N for all m N. So, cos 2n m!x 1 for all irrational x. 

Hence, 

lim n 

cos 2n m!x 0 lim m 

Continuity of real-valued functions 

lim n 

cos 2n m!x 0. 

4.13 Let f be continuous on a, b and let fx 0 when x is rational. Prove that 

fx 0 for every x a, b. 

Proof: Given any irrational number x in a, b, and thus choose a sequence x n Q 

such that x n x as n . Note that fx n 0 for all n. Hence, 

0 lim n 

0 

lim n 

fx n 

f lim n 

x n by continuity of f at x 

fx. 

Since x is arbitrary, we have shown fx 0 for all x a, b. Thatis,f is constant 0. 

Remark: Here is another good exercise, we write it as a reference. Let f be continuous 

on R, andiffx fx 2 , then f is constant. 

Proof: Since fx f x 2 fx 2 fx, we know that f is an even function. So, 

in order to show f is constant on R, it suffices to show that f is constant on 0, . Given 

any x 0, , sincefx 2 fx for all x R, wehavefx 1/2n fx for all n. Hence, 

fx lim n 

fx 

lim n 

fx 1/2n 

f lim n 

x 1/2n by continuity of f at 1 

f1 since x 0. 

So, we have fx f1 : c for all x 0, . In addition, given a sequence 

x n 0, such that x n 0, then we have

c lim n 

c 

lim n 

fx n 

f lim n 

x n by continuity of f at 0 

f0 

From the preceding, we have proved that f is constant. 

4.14 Let f be continuous at the point a a 1 , a 2 ,...,a n R n . Keep a 2 , a 3 ,...,a n fixed 

and define a new function g of one real variable by the equation 

gx fx, a 2 ,...,a n . 

Prove that g is continuous at the point x a 1 . (This is sometimes stated as follows: A 

continuous function of n variables is continuous in each variable separately.) 

Proof: Given 0, there exists a 0 such that as y Ba; D, whereD is a 

domain of f, wehave 

|fy fa| . * 

So, as |x a 1 | , which implies |x, a 2 ,...,a n a 1 , a 2 ,...,a n | , wehave 

|gx ga 1 | |fx, a 2 ,...,a n fa 1 , a 2 ,...,a n | . 

Hence, we have proved g is continuous at x a 1 

Remark: Here is an important example like the exercise, we write it as follows. Let 

j : R n R n ,and j : x 1 , x 2 ,...,x n 0,.,x j ,..,0. Then j is continuous on R n for 

1 j n. Note that j is called a projection. Note that a projection P is sometimes 

defined as P 2 P. 

Proof: Given any point a R n ,and given 0, and choose , then as 

x Ba; , wehave 

| j x j a| |x j a j | x a for each 1 j n 

Hence, we prove that j x is continuous on R n for 1 j n. 

4.15 Show by an example that the converse of statement in Exercise 4.14 is not true in 

general. 

Proof: Let 

fx, y 

x y if x 0ory 0 

1otherwise. 

Define g 1 x fx,0 and g 2 y f0, y, then we have 

limg 1 x 0 g 1 0 

x0 

and 

limg 2 y 0 g 2 0. 

y0 

So, g 1 x and g 2 y are continuous at 0. However, f is not continuous at 0, 0 since 

limfx, x 1 0 f0, 0. 

x0 

Remark: 1. For continuity, if f is continuous at x a, then it is NOT necessary for us 

to have 

lim xa 

fx fa

this is because a can be an isolated point. However, if a is an accumulation point, we then 

have 

f is continuous at a if, and only if, lim xa 

fx fa. 

4.16 Let f, g, andh be defined on 0, 1 as follows: 

fx gx hx 0, whenever x is irrational; 

fx 1andgx x, whenever x is rational; 

hx 1/n, ifx is the rational number m/n (in lowest terms); 

h0 1. 

Prove that f is not continuous anywhere in 0, 1, that g is continuous only at x 0, and 

that h is continuous only at the irrational points in 0, 1. 

Proof: 1. Write 

0ifx R Q 0, 1, 

fx 

1ifx Q 0, 1. 

Given any x R Q 0, 1, andy Q 0, 1, and thus choose x n Q 0, 1 

such that x n x, andy n R Q 0, 1 such that y n y. Iff is continuous at x, 

then 

1 lim n 

fx n 

f lim n 


fx 

0 

which is absurb. And if f is continuous at y, then 

0 lim n 

fy n 

f lim n 

y n by continuity of f at y 

fy 

1 

which is absurb. Hence, f is not continuous on 0, 1. 

2. Write 

0ifx R Q 0, 1, 

gx 

x if x Q 0, 1. 

Given any x R Q 0, 1, and choose x n Q 0, 1 such that x n x. Then 

x 

lim n 

x n 

lim n 

gx n 

limg lim n 

x n by continuity of g at x 

gx 

0 

which is absurb since x is irrational. So, f is not continous on R Q 0, 1. 

Given any x Q 0, 1, and choose x n R Q 0, 1 such that x n x. Ifg is

continuous at x, then 

0 

lim n 

gx n 

g lim n 


gx 

x. 

So, the function g may be continuous at 0. In fact, g is continuous at 0 which prove as 

follows. Given 0, choose , as|x| , wehave 

|gx g0| |gx| |x| . So,g is continuous at 0. Hence, from the preceding, 

we know that g is continuous only at x 0. 

3. Write 

1ifx 0, 

hx 0ifx R Q 0, 1, 

1/n if x m/n, g. c. d. m, n 1. 

Consider a 0, 1 and given 0, there exists the largest positive integer N such that 

N 1/. LetT x : hx , then 

0, 1 x : hx 1 x : hx 1/2...x : hx 1/N if 1, 

T 

if 1. 

Note that T is at most a finite set, and then we can choose a 0 such that 

a , a acontains no points of T and a , a 0, 1. So,if 

x a , a a, wehavehx . It menas that 

lim xa 

hx 0. 

Hence, we know that h is continuous at x 0, 1 R Q. For two points x 1, and 

y 0, it is clear that h is not continuous at x 1, and not continuous at y 1bythe 

method mentioned in the exercise of part 1 and part 2. Hence, we have proved that h is 

continuous only at the irrational points in 0, 1. 

Remark: 1. Sometimes we call f Dirichlet function. 

2. Here is another proof about g, we write it down to make the reader get more. 

Proof: Write 

0ifx R Q 0, 1, 

gx 

x if x Q 0, 1. 

Given a 0, 1, andifg is continuous at a, then given 0 a, there exists a 0 

such that as x a , a 0, 1, wehave 

|gx ga| . 

If a R Q, choose 0 so that a Q. Then a a , a which 

implies |ga ga| |ga | a a. But it is impossible. 

If a Q, choose 0 so that a R Q. a a , a which 

implies |ga ga| |a| a a. But it is impossible. 

If a 0, given 0 and choose , thenas0 x , wehave 

|gx g0| |gx| |x| x . It means that g is continuous at 0. 

4.17 For each x 0, 1, letfx x if x is rational, and let fx 1 x if x is

irrational. Prove that: 

(a) ffx x for all x in 0, 1. 

Proof: If x is rational, then ffx fx x. Andifx is irrarional, so is 

1 x 0, 1. Then ffx f1 x 1 1 x x. Hence, ffx x for all x in 

0, 1. 

(b) fx f1 x 1 for all x in 0, 1. 

Proof: If x is rational, so is 1 x 0, 1. Then fx f1 x x 1 x 1. And 

if x is irrarional, so is 1 x 0, 1. Then fx f1 x 1 x 1 1 x 1. 

Hence, fx f1 x 1 for all x in 0, 1. 

(c) f is continuous only at the point x 1 . 2 

Proof: If f is continuous at x, then choose x n Q and y n Q c such that x n x, 

and y n x. Then we have, by continuity of f at x, 

fx f lim n 

x n 

lim n 

fx n lim n 

x n x 

and 

fx f lim n 

y n lim n 

fy n lim n 

1 y n 1 x. 

So, x 1/2 is the only possibility for f. Given 0, we want to find a 0 such that as 

x 1/2 ,1/2 0, 1, wehave 

|fx f1/2| |fx 1/2| . 

Choose 0 so that 1/2 ,1/2 0, 1, then as 

x 1/2 ,1/2 0, 1, wehave 

|fx 1/2| |x 1/2| if x Q, 

|fx 1/2| |1 x 1/2| |1/2 x| if x Q c . 

Hence, we have proved that f is continuous at x 1/2. 

(d) f assumes every value between 0 and 1. 

Proof: Given a 0, 1, wewanttofindx 0, 1 such that fx a. Ifa Q, then 

choose x a, wehavefx a a. Ifa R Q, then choose x 1 a R Q, we 

have fx 1 a 1 1 a a. 

Remark: The range of f on 0, 1 is 0, 1. In addition, f is an one-to-one mapping 

since if fx fy, then x y. (The proof is easy, just by definition of 1-1, so we omit it.) 

(e) fx y fx fy is rational for all x and y in 0, 1. 

Proof: We prove it by four steps. 

1. If x Q and y Q, then x y Q. So, 

fx y fx fy x y x y 0 Q. 

2. If x Q and y Q c , then x y Q c .So, 

fx y fx fy 1 x y x 1 y 2x Q. 

3. If x Q c and y Q, then x y Q c .So, 

fx y fx fy 1 x y 1 x y 2y Q. 

4. If x Q c and y Q c , then x y Q c or x y Q. So,

fx y fx fy 

1 x y 1 x 1 y 1 Q if x y Q c , 

x y 1 x 1 y 2 Q if x y Q. 

Remark: Here is an interesting question about functions. Let f : R 0, 1 R. Iff 

satisfies that 

fx f x 1 

x 1 x, 

then fx x3 x 2 1 

. 2xx1 

Proof: Let x x1 

x , then we have 2 x 1 

x1 ,and3 x x. So, 

fx f x 1 

x fx fx 1 x * 


fx f 2 x 1 x ** 

and 

f 2 x f 3 x f 2 x fx 1 2 x. *** 

So, by (*), (**), and (***), we finally have 

fx 1 2 1 x x 2 x 

x3 x 2 1 

2xx 1 . 

4.18 Let f be defined on R and assume that there exists at least one x 0 in R at which f 

is continuous. Suppose also that, for every x and y in R, f satisfies the equation 

fx y fx fy. 

Prove that there exists a constant a such that fx ax for all x. 

Proof: Let f be defined on R and assume that there exists at least one x 0 in R at which f 

is continuous. Suppose also that, for every x and y in R, f satisfies the equation 

fx y fx fy. 

Prove that there exists a constant a such that fx ax for all x. 

Proof: Suppose that f is continuous at x 0 , and given any r R. Since 

fx y fx fy for all x, then 

fx fy x 0 fr, wherey x r x 0 . 

Note that y x 0 x r, then 

lim xr 

fx lim xr 

fy x 0 fr 

yx0 

lim fy x 0 fr 

fr since f is continuous at x 0 . 

So, f is continuous at r. Sincer is arbitrary, we have f is continuous on R. Definef1 a, 

andthensincefx y fx fy, wehave 

f1 f 1 m .. m 1 mtimes 

mf 

1 m 

f 1 m f1 

m *

In addition, since f1 f1 by f0 0, we have 

f1 f 1 m .. m 

1 

mf 

1 m 

f 1 m 

mtimes 

f1 

m *’ 

Thus we have 

f m n f1/m ...1/m ntimes nf1/m m n f1 by (*) and (*’) ** 

So, given any x R, and thus choose a sequence x n Q with x n x. Then 

fx f lim n 

x n 

lim n 

fx n by continuity of f on R 

lim n 

x n f1 by (**) 

xf1 

ax. 

Remark: There is a similar statement. Suppose that fx y fxfy for all real x 

and y. 

(1) If f is differentiable and non-zero, prove that fx e cx ,wherec is a constant. 

Proof: Note that f0 1sincefx y fxfy and f is non-zero. Since f is 

differentiable, we define f 0 c. Consider 

fx h fx fh f0 

fx fxf 

h 

h 

0 cfx as h 0, 

we have for every x R, f x cfx. Hence, 

fx Ae cx . 

Since f0 1, we have A 1. Hence, fx e cx ,wherec is a constant. 

Note: (i) If for every x R, f x cfx, then fx Ae cx . 

Proof: Since f x cfx for every x, we have for every x, 

f x cfxe cx 0 e cx fx 0. 

We note that by Elementary Calculus, e cx fx is a constant function A. So,fx Ae cx 

for all real x. 

(ii) Suppose that fx y fxfy for all real x and y. Iffx 0 0forsomex 0 , then 

fx 0 for all x. 

Proof: Suppose NOT, then fa 0forsomea. However, 

0 fx 0 fx 0 a a fx 0 afa 0. 

Hence, fx 0 for all x. 

(iii) Suppose that fx y fxfy for all real x and y. Iff is differentiable at x 0 for 

some x 0 , then f is differentiable for all x. And thus, fx C R. 

Proof: Since

fx h fx 

h 

fx 0 h x x 0 fx 0 x x 0 

h 

fx x 0 fx 0 h fx 0 

fx x 0 f x 0 as h 0, 

h 

we have f x is differentiable and f x fx x 0 f x 0 for all x. And thus we have 

fx C R. 

(iv) Here is another proof by (iii) and Taylor Theorem with Remainder term R n x. 

Proof: Since f is differentiable, by (iii), we have f n x f 0 n fx for all x. 

Consider x r, r, then by Taylor Theorem with Remainder term R n x, 

n 

fx 

k0 


f k 0 

k! 

x k R n x, whereR n x : fn1 

n 1! xn1 , 0, x or x,0, 

|R n x| fn1 

n 1! xn1 

 

f 0 n1 f 

n 1! 

f 

 

0r n1 

n 1! 

0asn . 

Hence, we have for every x r, r 

 

fx 

k0 

f0 

x n1 

M, whereM max 

xr,r |fx| 

f 

k 

0 

k! 

 

 

k0 

x k 

f 0x k 

k! 

e cx ,wherec : f 0. 

Since r is arbitrary, we have proved that fx e cx for all x. 

(2) If f is continuous and non-zero, prove that fx e cx ,wherec is a constant. 

Proof: Since fx y fxfy, wehave 

0 f1 f 1 n ... 1 n 

n f 1 

ntimes n f 1 n f1 1/n * 

and (note that f1 f1 1 by f0 1, ) 

0 f1 f 1 n ... 1 n f n 

1 

ntimes n f 1 n f1 1/n *’ 

f m n f 1 n ... 1 n mtimes 

f 1 n 

m 

f1 

m n 

by (*) and (*’) ** 

So, given any x R, and thus choose a sequence x n Q with x n x. Then

lim 

r n0 

fx f lim n 

x n 

lim n 

fx n by continuity of f 

lim n 

f1 xn by (**) 

f1 x 

e cx , where log f1 c. 

Note: (i) We can prove (2) by the exercise as follows. Note that fx 0 for all x by 

the remark (1)-(ii) Consider the composite function gx log fx, then 

gx y log fx y log fxfy log fx log fy gx gy. Since log and f 

are continuous on R, its composite function g is continuous on R. Use the exercise, we 

have gx cx for some c. Therefore, fx e gx e cx . 

(ii) We can prove (2) by the remark (1) as follows. It suffices to show that this f is 

differentiable at 0 by remark (1) and (1)-(iii). Sincef m n f1 m n 

then for every real r, 

fr f1 r a 

by continuity of f. Note that lim r b r 

r0 r exists. Given any sequence r n 

with r n 0, and thus consider 

fr n f0 

1 

1 

r n 

f1rn r n 

f1rn rn 

r n 

exists, 

we have f is differentiable at x 0. So,byremark(1),wehavefx e cx . 

(3) Give an example such that f is not continuous on R. 

Solution: Consider gx y gx gy for all x, y. Then we have gq qg1, 

where q Q. ByZorn’s Lemma, we know that every vector space has a basis 

v : I. Note that v : I is an uncountable set, so there exists a convergent 

sequence s n v : I. Hence, S : v : I s n 

n1 

sn 

n 

n1 

is a new 

basis of R over Q. Givenx, y R, and we can find the same N such that 

N 

x 

k1 

N 

q k v k and y p k v k ,wherev k S 

k1 

Define the sume 

N 

x y : p k q k v k 

k1 

By uniqueness, we define gx to be the sum of coefficients, i.e., 

N 

gx : q k . 

k1 

Note that 

g 

s n 

1 for all n lim n 

g 

and 

s n 0asn 

Hence, g is not continuous at x 0 since if it was, then 

s n 1

1 lim n 

g 

s n 

g lim 

s n n by continuity of g at 0 

g0 

0 

which is absurb. Hence, g is not continuous on R by the exercise. To find such f, it suffices 

to consider fx e gx . 

Note: Such g (or f) isnot measurable by Lusin Theorem. 

4.19 Let f be continuous on a, b and define g as follows: ga fa and, for 

a x b, letgx be the maximum value of f in the subinterval a, x. Show that g is 

continuous on a, b. 

Proof: Define gx maxft : t a, x, and choose any point c a, b, wewant 

to show that g is continuous at c. Given 0, we want to find a 0 such that as 

x c , c a, b, wehave 

|gx gc| . 

Since f is continuous at x c, then there exists a 0 such that as 

x c , c a, b, wehave 

fc /2 fx fc /2. * 

Consider two cases as follows. 

(1) maxft : t a, c a, b fp 1 ,wherep 1 c . 

As x c , c a, b, wehavegx fp 1 and gc fp 1 . 

Hence, |gx gc| 0. 

(2) maxft : t a, c a, b fp 1 ,wherep 1 c . 

As x c , c a, b, wehaveby(*)fc /2 gx fc /2. 

Hence, |gx gc| . 

So, if we choose ,thenforx c , c a, b, 

|gx gc| by (1) and (2). 

Hence, gx is continuous at c. And since c is arbitrary, we have gx is continuous on 

a, b. 

Remark: It is the same result for minft : t a, x by the preceding method. 

4.20 Let f 1 ,...,f m be m real-valued functions defined on R n . Assume that each f k is 

continuous at the point a of S. Define a new function f as follows: For each x in S, fx is 

the largest of the m numbers f 1 x,...,f m x. Discuss the continuity of f at a. 

Proof: Assume that each f k is continuous at the point a of S, then we have f i f j and 

|f i f j | are continuous at a, where 1 i, j m. Sincemaxa, b ab|ab| , then 

2 

maxf 1 , f 2 is continuous at a since both f 1 f 2 and |f 1 f 2 | are continuous at a. Define 

fx maxf 1 ,...f m ,useMathematical Induction to show that fx is continuous at 

x a as follows. As m 2, we have proved it. Suppose m k holds, i.e., maxf 1 ,...f k is 

continuous at x a. Then as m k 1, we have 

maxf 1 ,...f k1 maxmaxf 1 ,...f k , f k1 

is continuous at x a by induction hypothesis. Hence, by Mathematical Induction, we 

have prove that f is continuos at x a. 

It is possible that f and g is not continuous on R whihc implies that maxf, g is 

continuous on R. For example, let fx 0ifx Q, andfx 1ifx Q c and gx 1

if x Q, andgx 0ifx Q c . 

Remark: It is the same rusult for minf 1 ,...f m since maxa, b mina, b a b. 

4.21 Let f : S R be continuous on an open set in R n , assume that p S, and assume 

that fp 0. Prove that there is an n ball Bp; r such that fx 0 for every x in the 

ball. 

Proof: Since p S is an open set in R n , there exists a 1 0 such that Bp, 1 S. 

Since fp 0, given fp 0, then there exists an n ball Bp; 

2 2 such that as 

x Bp; 2 S, wehave 

fp 

fp fx fp 3fp . 

2 

2 

Let min 1 , 2 , then as x Bp; , wehave 

fx fp 0. 

2 

Remark: The exercise tells us that under the assumption of continuity at p, we roughly 

have the same sign in a neighborhood of p, iffp 0 or fp 0. 

4.22 Let f be defined and continuous on a closed set S in R. Let 

A x : x S and fx 0 . 

Prove that A is a closed subset of R. 

Proof: Since A f 1 0, andf is continous on S, wehaveA is closed in S. And 

since S is closed in R, we finally have A is closed in R. 

Remark: 1. Roughly speaking, the property of being closed has Transitivity. Thatis, 

in M, d let S T M, ifS is closed in T, andT is closed in M, then S is closed in M. 

Proof: Let x be an adherent point of S in M, then B M x, r S for every r 0. 

Hence, B M x, r T for every r 0. It means that x is also an adherent point of T in 

M. SinceT is closed in M, we find that x T. Note that since B M x, r S for every 

r 0, we have (S T) 

B T x, r S B M x, r T S B M x, r S T B M x, r S . 

So, we have x is an adherent point of S in T. And since S is closed in T, wehavex S. 

Hence, we have proved that if x is an adherent point of S in M, then x S. Thatis,S is 

closed in M. 

Note: (1) Another proof of remark 1, since S is closed in T, there exists a closed subset 

U in Msuch that S U T, and since T is closed in M, wehaveS is closed in M. 

(2) There is a similar result, in M, d let S T M, ifS is open in T, andT is open 

in M, then S is open in M. (Leave to the reader.) 

2. Here is another statement like the exercise, but we should be cautioned. We write it 

as follows. Let f and g be continuous on S, d 1 into T, d 2 .LetA x : fx gx, 

show that A is closed in S. 

Proof: Let x be an accumulation point of A, then there exists a sequence x n A 

such that x n x. So, we have fx n gx n for all n. Hence, by continuity of f and g, we 

have 

fx f lim n 

x n lim n 

fx n lim n 

gx n g lim n 

x n gx. 

Hence, x A. Thatis,A contains its all adherent point. So, A is closed.

Note: In remark 2, we CANNOT use the relation 

fx gx 

since the difference "" are not necessarily defined on the metric space T, d 2 . 

4.23 Given a function f : R R, define two sets A and B in R 2 as follows: 

A x, y : y fx, 

B x, y : y fx. 

and Prove that f is continuous on R if, and only if, both A and B are open subsets of R 2 . 

Proof: () Suppose that f is continuous on R. Leta, b A, then fa b. Sincef is 

continuous at a, thengiven fab 0, there exists a 0 such that as 

2 

|x a| , wehave 

fa b 

fa fx fa . * 

2 

Consider x, y Ba, b; , then |x a| 2 |y b| 2 2 which implies that 

1. |x a| fx fa b by (*) and 

2 

2. |y b| y b b fa b . 

2 

Hence, we have fx y. Thatis,Ba, b; A. So,A is open since every point of A is 

interior. Similarly for B. 

() Suppose that A and B are open in R 2 . Trivially, a, fa /2 : p 1 A, and 

a, fa /2 : p 2 B. SinceA and B are open in R 2 , there exists a /2 0 such 

that 

Bp 1 , A and Bp 2 , B. 

Hence, if x, y Bp 1 , , then 

x a 2 y fa /2 2 2 and y fx. 

So, it implies that 

|x a| , |y fa /2| , andy fx. 

Hence, as |x a| , wehave 

y fa /2 

fa /2 y fx 

fa y fx 

fa fx. ** 

And if x, y Bp 2 , , then 

x a 2 y fa /2 2 2 and y fx. 

So, it implies that 

|x a| , |y fa /2| , andy fx. 

Hence, as |x a| , wehave 

fx y fa /2 fa . *** 

So, given 0, there exists a 0 such that as |x a| , we have by (**) and (***) 

fa fx fa . 

That is, f is continuous at a. Sincea is arbitrary, we know that f is continuous on R. 

4.24 Let f be defined and bounded on a compact interval S in R. IfT S, the

number 

f T supfx fy : x, y T 

is called the oscillation (or span) of f on T. Ifx S, the oscillation of f at x is defined to 

be the number 

f x lim 

h0 

 

fBx; h S. 

Prove that this limit always exists and that f x 0if,andonlyif,f is continuous at x. 

Proof: 1. Note that since f is bounded, say |fx| M for all x, wehave 

|fx fy| 2M for all x, y S. So, f T, the oscillation of f on any subset T of S, 

exists. In addition, we define gh f Bx; h S. Note that if T 1 T 2 S, wehave 

f T 1 f T 2 . Hence, the oscillation of f at x, f x lim h0 

gh g0 since g is 

an increasing function. That is, the limit of f Bx; h S always exists as h 0 . 

2. Suppose that f x 0, then given 0, there exists a 0 such that as 

h 0, , wehave 

|gh| gh f Bx; h S /2. 

That is, as h 0, , wehave 

supft fs : t, s Bx; h S /2 


/2 ft fx /2 as t x , x S. 

So, as t x , x S. wehave 

|ft fx| . 

That is, f is continuous at x. 

Suppose that f is continous at x, thengiven 0, there exists a 0 such that as 

t x , x S, wehave 

|ft fx| /3. 

So, as t, s x , x S, wehave 

|ft fs| |ft fx| |fx fs| /3 /3 2/3 


supt fs : t, s x , x S 2/3 . 

So, as h 0, , wehave 

f Bx; h S supt fs : t, s x , x S . 

Hence, the oscillation of f at x, f x 0. 

Remark: 1. The compactness of S is not used here, we will see the advantage of the 

oscillation of f in text book, Theorem 7.48, in page 171. (On Lebesgue’s Criterion for 

Riemann-Integrability.) 

2. One of advantage of the oscillation of f is to show the statement: Let f be defined on 

a, b Prove that a bounded f does NOT have the properties: 

f is continuous on Q a, b, and discontinuous on R Q a, b. 

Proof: Since f x 0if,andonlyif,f is continuous at x, we know that f r 0for 

r R Q a, b. DefineJ 1/n r : f r 1/n, then by hypothesis, we know that 

 

n1 J 1/n R Q a, b. It is easy to show that J 1/n is closed in a, b. Hence, 

intclJ 1/n intJ 1/n for all n N. It means that J 1/n is nowhere dense for all 

n N. Hence,

a, b 

n1 J 1/n Q a, b 

is of the firse category which is absurb since every complete metric space is of the second 

category. So, this f cannot exist. 

Note: 1 The Boundedness of f can be removed since we we can accept the concept 

0. 

2. (J 1/n is closed in a, b) Given an accumulation point x of J 1/n ,ifx J 1/n ,wehave 

f x 1/n. So, there exists a 1 ball Bx such that f Bx a, b 1/n. Thus, no 

points of Bx can belong to J 1/n , contradicting that x is an accumulation point of J 1/n . 

Hence, x J 1/n and J 1/n is closed. 

3. (Definition of a nowhere dense set) In a metric space M, d, letA be a subset of M, 

we say A is nowhere dense in M if, and only if A contains no balls of M, ( intA ). 

4. (Definition of a set of the first category and of the second category) AsetA in a 

metric space M is of the first category if, and only if, A is the union of a countable number 

of nowhere dense sets. A set B is of the second category if, and only if, B is not of the first 

category. 

5. (Theorem) A complete metric space is of the second category. 

We write another important theorem about a set of the second category below. 

(Baire Category Theorem) A nonempty open set in a complete metric space is of the 

second category. 

6. In the notes 3,4 and 5, the reader can see the reference, A First Course in Real 

Analysis written by M. H. Protter and C. B. Morrey, in pages 375-377. 

4.25 Let f be continuous on a compact interval a, b. Suppose that f has a local 

maximum at x 1 and a local maximum at x 2 . Show that there must be a third point between 

x 1 and x 2 where f has a local minimum. 

Note.Tosaythatf has a local maximum at x 1 means that there is an 1 ball Bx 1 such 

that fx fx 1 for all x in Bx 1 a, b. Local minimum is similarly defined. 

Proof: Let x 2 x 1 . Suppose NOT, i.e., no points on x 1 , x 2 can be a local minimum 

of f. Sincef is continuous on x 1 , x 2 , then inffx : x x 1 , x 2 fx 1 or fx 2 by 

hypothesis. We consider two cases as follows: 

(1) If inffx : x x 1 , x 2 fx 1 , then 

(i) fx has a local maximum at x 1 and 

(ii) fx fx 1 for all x x 1 , x 2 . 

By (i), there exists a 0 such that x x 1 , x 1 x 1 , x 2 ,wehave 

(iii) fx fx 1 . 

So, by (ii) and (iii), as x x 1 , x 1 , wehave 

fx fx 1 

which contradicts the hypothesis that no points on x 1 , x 2 can be a local minimum of f. 

(2) If inffx : x x 1 , x 2 fx 1 , it is similar, we omit it. 

Hence, from (1) and (2), we have there has a third point between x 1 and x 2 where f has 

a local minimum. 

4.26 Let f be a real-valued function, continuous on 0, 1, with the following property: 

For every real y, either there is no x in 0, 1 for which fx y or there is exactly one such 

x. Prove that f is strictly monotonic on 0, 1.

Proof: Since the hypothesis says that f is one-to-one, then by Theorem*, we know that f 

is trictly monotonic on 0, 1. 

Remark: (Theorem*) Under assumption of continuity on a compact interval, 1-1 is 

equivalent to being strictly monotonic. We will prove it in Exercise 4.62. 

4.27 Let f be a function defined on 0, 1 with the following property: For every real 

number y, either there is no x in 0, 1 for which fx y or there are exactly two values of 

x in 0, 1 for which fx y. 

(a) Prove that f cannot be continuous on 0, 1. 

Proof: Assume that f is continuous on 0, 1, and thus consider max x0,1 fx and 

min x0,1 fx. Then by hypothesis, there exist exactly two values a 1 a 2 0, 1 such 

that fa 1 fa 2 max x0,1 fx, and there exist exactly two values b 1 b 2 0, 1 

such that fb 1 fb 2 min x0,1 fx. 

Claim that a 1 0anda 2 1. Suppose NOT, then there exists at least one belonging 

to 0, 1. Without loss of generality, say a 1 0, 1. Sincef has maximum at a 1 0, 1 

and a 2 0, 1, we can find three points p 1 , p 2 ,andp 3 such that 

1. p 1 a 1 p 2 p 3 a 2 , 

2. fp 1 fa 1 , fp 2 fa 1 ,andfp 3 fa 2 . 

Since fa 1 fa 2 , we choose a real number r so that 

fp 1 r fa 1 r fq 1 ,whereq 1 p 1 , a 1 by continuity of f. 

fp 2 r fa 1 r fq 2 ,whereq 2 a 1 , p 2 by continuity of f. 

fp 3 r fa 2 r fq 3 ,whereq 3 p 3 , a 2 by continuity of f. 

which contradicts the hypothesis that for every real number y, there are exactly two values 

of x in 0, 1 for which fx y. Hence, we know that a 1 0anda 2 1. Similarly, we 

also have b 1 0andb 2 1. 

So, max x0,1 fx min x0,1 fx which implies that f is constant. It is impossible. 

Hence, such f does not exist. That is, f is not continuous on 0, 1. 

(b) Construct a function f which has the above property. 

Proof: Consider 0, 1 Q c 0, 1 Q 0, 1, and write 

Q 0, 1 x 1 , x 2 ,...,x n ,.... Define 

1. fx 2n1 fx 2n n, 

2. fx x if x 0, 1/2 Q c , 

3. fx 1 x if x 1/2, 1 Q c . 

Then if x y, then it is clear that fx fy. Thatis,f is well-defined. And from 

construction, we know that the function defined on 0, 1 with the following property: For 

every real number y, either there is no x in 0, 1 for which fx y or there are exactly 

two values of x in 0, 1 for which fx y. 

Remark: x : f is discontinuous at x 0, 1. Givena 0, 1. Note that since 

fx N for all x Q 0, 1 and Q is dense in R, forany1ball Ba; r Q 0, 1, 

there is always a rational number y Ba; r Q 0, 1 such that |fy fa| 1. 

(c) Prove that any function with this property has infinite many discontinuities on 

0, 1. 

Proof: In order to make the proof clear, property A of f means that

for every real number y, either there is no x in 0, 1 for which fx y or 

there are exactly two values of x in 0, 1 for which fx y 

Assume that there exist a finite many numbers of discontinuities of f, say these points 

x 1 ,...,x n . By property A, there exists a unique y i such that fx i fy i for 1 i n. 

Note that the number of the set 

S : x 1 ,...,x n y 1 ,...,y n x : fx f0, andfx f1 is even, say 2m 

We remove these points from S, and thus we have 2m 1 subintervals, say I j , 

1 j 2m 1. Consider the local extremum in every I j ,1 j 2m 1 and note that 

every subinterval I j ,1 j 2m 1, has at most finite many numbers of local extremum, 

say # t I j : fx is the local extremum t j j 

1 ,..,t pj p j . And by property A, 

there exists a unique s j j 

j 

k such that f t k f s k for 1 k p j . We again remove these 

points, and thus we have removed even number of points. And odd number of open 

intervals is left, call the odd number 2q 1. Note that since the function f is monotonic in 

every open interval left, R l ,1 l 2q 1, the image of f on these open interval is also an 

open interval. If R a R b , sayR a a 1 , a 1 and R b b 1 , b 2 with (without loss of 

generality) a 1 b 1 a 2 b 2 , then 

R a R b by property A. 

(Otherwise, b 1 is only point such that fx fb 1 , which contradicts property A.) Note 

that given any R a , there must has one and only one R b such that R a R b . However, we 

have 2q 1 open intervals is left, it is impossible. Hence, we know that f has infinite many 

discontinuities on 0, 1. 

4.28 In each case, give an example of a real-valued function f, continuous on S and 

such that fS T, or else explain why there can be no such f : 

(a) S 0, 1, T 0, 1. 

Solution: Let 

fx 

2x if x 0, 1/2, 

1ifx 1/2, 1. 

(b) S 0, 1, T 0, 1 1, 2. 

Solution: NO! Since a continuous functions sends a connected set to a connected set. 

However, in this case, S is connected and T is not connected. 

(c) S R 1 , T the set of rational numbers. 

Solution: NO! Since a continuous functions sends a connected set to a connected set. 

However, in this case, S is connected and T is not connected. 

(d) S 0, 1 2, 3, T 0, 1. 

Solution: Let 

(e) S 0, 1 0, 1, T R 2 . 

fx 

0ifx 0, 1, 

1ifx 2, 3. 

Solution: NO! Since a continuous functions sends a compact set to a compact set. 

However, in this case, S is compact and T is not compact.

(f) S 0, 1 0, 1, T 0, 1 0, 1. 

Solution: NO! Since a continuous functions sends a compact set to a compact set. 

However, in this case, S is compact and T is not compact. 

(g) S 0, 1 0, 1, T R 2 . 

Solution: Let 

fx, y cot x,coty. 

Remark: 1. There is some important theorems. We write them as follows. 

(Theorem A) Letf : S, d s T, d T be continuous. If X is a compact subset of S, 

then fX is a compact subset of T. 

(Theorem B) Letf : S, d s T, d T be continuous. If X is a connected subset of S, 

then fX is a connected subset of T. 

2. In (g), the key to the example is to find a continuous function f : 0, 1 R which is 

onto. 

Supplement on Continuity of real valued functions 

Exercise Suppose that fx : 0, R, is continuous with a fx b for all 

x 0, , and for any real y, either there is no x in 0, for which fx y or 

there are finitely many x in 0, for which fx y. Prove that lim x fx exists. 

Proof: For convenience, we say property A, it means that for any real y, either 

there is no x in 0, for which fx y or there are finitely many x in 0, for 

which fx y. 

We partition a, b into n subintervals. Then, by continuity and property A, asx 

is large enough, fx is lying in one and only one subinterval. Given 0, there 

exists N such that 2/N . For this N, we partition a, b into N subintervals, then 

there is a M 0 such that as x, y M 

|fx fy| 2/N . 

So, lim x fx exists. 

Exercise Suppose that fx : 0, 1 R is continunous with f0 f1 0. Prove that 

(a) there exist two points x 1 and x 2 such that as |x 1 x 2 | 1/n, wehave 

fx 1 fx 2 0 for all n. In this case, we call 1/n the length of horizontal strings. 

Proof: Define a new function gx fx 1 n fx : 0, 1 1 n . Claim that 

there exists p 0, 1 1 n such that gp 0. Suppose NOT, byIntermediate 

Value Theorem, without loss of generality, let gx 0, then 

g0 g 1 n ...g 1 1 n f1 0 

which is absurb. Hence, we know that there exists p 0, 1 1 n such that 

gp 0. That is, 

f p 1 n fp. 

So,wehave1/n as the length of horizontal strings. 

(b) Could you show that there exists 2/3 as the length of horizontal strings 

Proof: The horizontal strings does not exist, for example,

fx 

x, ifx 0, 1 4 

x 1 ,ifx 1 , 3 . 

2 4 4 

x 1, if x 3 ,1 4 

Exercise Suppose that fx : a, b R is a continuous and non-constant function. Prove 

that the function f cannot have any small periods. 

Proof: Say f is continuous at q a, b, and by hypothesis that f is 

non-constant, there is a point p a, b such that |fq fp| : M 0. Since f is 

continuous at q, thengiven M, there is a 0 such that as 

x q , q a, b, wehave 

|fx fq| M. * 

If f has any small periods, then in the set q , q a, b, there is a point 

r q , q a, b such that fr fp. It contradicts to (*). Hence, the 

function f cannot have any small periods. 

Remark 1. There is a function with any small periods. 

Solution:The example is Dirichlet function, 

0, if x Q 

fx 

c 

1, if x Q . 

Since fx q fx, for any rational q, we know that f has any small periods. 

2. Prove that there cannot have a non-constant continuous function which has 

two period p, andq such that q/p is irrational. 

Proof: Since q/p is irrational, there is a sequence qn 

p n 

Q such that 

q n 

p n 

q p 1 p 

 

p2 |pq n qp n| 

n 

p n 

0asn . 

So, f has any small periods, by this exercise, we know that this f cannot a 

non-constant continuous function. 

Note: The inequality is important; the reader should kepp it in mind. There are 

many ways to prove this inequality, we metion two methods without proofs. The 

reader can find the proofs in the following references. 

(1) An Introduction To The Theory Of Numbers written by G.H. Hardy and 

E.M. Wright, charpter X, pp 137-138. 

(2) In the text book, exercise 1.15 and 1.16, pp 26. 

3. Suppose that fx is differentiable on R prove that if f has any small periods, 

then f is constant. 

Proof: Givenc R, and consider 

fc p n fc 

p n 

0 for all n. 

where p n is a sequence of periods of function such that p n 0. Hence, by 

differentiability of f, we know that f c 0. Since c is arbitrary, we know that 

f x 0onR. Hence, f is constant. 

Continuity in metric spaces

In Exercises 4.29 through 4.33, we assume that f : S T is a function from one metric 

space S, d S to another T, d T . 

4.29 Prove that f is continuous on S if, and only if, 

f 1 intB intf 1 B for every subset B of T. 

Proof: () Suppose that f is continuous on S, and let B be a subset of T. Since 

intB B, wehavef 1 intB f 1 B. Note that f 1 intB is open since a pull back of 

an open set under a continuous function is open. Hence, we have 

intf 1 intB f 1 intB intf 1 B. 

That is, f 1 intB intf 1 B for every subset B of T. 

() Suppose that f 1 intB intf 1 B for every subset B of T. Given an open subset 

U T, i.e., intU U, sowehave 

f 1 U f 1 intU intf 1 U. 

In addition, intf 1 U f 1 U by the fact, for any set A, intA is a subset of A. So,f is 

continuous on S. 

4.30 Prove that f is continuous on S if, and only if, 

fclA clfA for every subset A of S. 

Proof: () Suppose that f is continuous on S, and let A be a subset of S. Since 

fA clfA, then A f 1 fA f 1 clfA. Note that f 1 clfA is closed 

since a pull back of a closed set under a continuous function is closed. Hence, we have 

clA clf 1 clfA f 1 clfA 


fclA ff 1 clfA clfA. 

() Suppose that fclA clfA for every subset A of S. Given a closed subset 

C T, and consider f 1 C as follows. Define f 1 C A, then 

fclf 1 C fclA 

clfA clff 1 C 

clC C since C is closed. 

So,wehaveby(fclA C) 

clA f 1 fclA f 1 C A 

which implies that A f 1 C is closed set. So, f is continuous on S. 

4.31 Prove that f is continuous on S if, and only if, f is continuous on every compact 

subset of S. Hint. If x n p in S, the set p, x 1 , x 2 ,... is compact. 

Proof: () Suppose that f is continuous on S, then it is clear that f is continuous on 

every compact subset of S. 

() Suppose that f is continuous on every compact subset of S, Givenp S, we 

consider two cases. 

(1) p is an isolated point of S, then f is automatically continuous at p. 

(2) p is not an isolated point of S, thatis,p is an accumulation point p of S, then there 

exists a sequence x n S with x n p. Note that the set p, x 1 , x 2 ,... is compact, so we 

know that f is continuous at p. Sincep is arbitrary, we know that f is continuous on S. 

Remark: If x n p in S, the set p, x 1 , x 2 ,... is compact. The fact is immediately

from the statement that every infinite subset p, x 1 , x 2 ,... of has an accumulation point in 

p, x 1 , x 2 ,.... 

4.32 A function f : S T is called a closed mapping on S if the image fA is closed 

in T for every closed subset A of S. Prove that f is continuous and closed on S if, and only 

if, 

fclA clfA for every subset A of S. 

Proof: () Suppose that f is continuous and closed on S, and let A be a subset of S. 

Since A clA, wehavefA fclA. So, we have 

clfA clfclA fclA since f is closed. * 

In addition, since fA clfA, wehaveA f 1 fA f 1 clfA. Note that 

f 1 clfA is closed since f is continuous. So, we have 

clA clf 1 clfA f 1 clfA 


fclA ff 1 clfA clfA. ** 

From (*) and (**), we know that fclA clfA for every subset A of S. 

() Suppose that fclA clfA for every subset A of S. Gvien a closed subset C 

of S, i.e., clC C, then we have 

fC fclC clfC. 

So, we have fC is closed. That is, f is closed. Given any closed subset B of T, i.e., 

clB B, we want to show that f 1 B is closed. Since f 1 B : A S, wehave 

fclf 1 B fclA clfA clff 1 B clB B 


fclf 1 B B clf 1 B f 1 fclf 1 B f 1 B. 

That is, we have clf 1 B f 1 B. So,f 1 B is closed. Hence, f is continuous on S. 

4.33 Give an example of a continuous f and a Cauchy sequence x n in some metric 

space S for which fx n is not a Cauchy sequence in T. 

Solution: Let S 0, 1, x n 1/n for all n N, andf 1/x : S R. Then it is clear 

that f is continous on S, andx n is a Cauchy sequence on S. In addition, Trivially, 

fx n n is not a Cauchy sequence. 

Remark: The reader may compare the exercise with the Exercise 4.54. 

4.34 Prove that the interval 1, 1 in R 1 is homeomorphic to R 1 . This shows that 

neither boundedness nor completeness is a topological property. 

Proof: Since fx tan x : 1, 1 R is bijection and continuous, and its 

2 

converse function f 1 x arctanx : R 1, 1. Hence, we know that f is a Topologic 

mapping. (Or say f is a homeomorphism). Hence, 1, 1 is homeomorphic to R 1 . 

Remark: A function f is called a bijection if, and only if, f is 1-1 and onto. 

4.35 Section 9.7 contains an example of a function f, continuous on 0, 1, with 

f0, 1 0, 1 0, 1. Prove that no such f can be one-to-one on 0, 1. 

Proof: By section 9.7, let f : 0, 1 0, 1 0, 1 be an onto and continuous function. 

If f is 1-1, then so is its converse function f 1 . Note that since f is a 1-1 and continous 

function defined on a compact set 0, 1, then its converse function f 1 is also a continous

function. Since f0, 1 0, 1 0, 1, we have the domain of f 1 is 0, 1 0, 1 which 

is connected. Choose a special point y 0, 1 0, 1 so that f 1 y : x 0, 1. 

Consider a continous function g f 1 | 0,10,1y , then 

g : 0, 1 0, 1 y 0, x x,1 which is continous. However, it is impossible 

since 0, 1 0, 1 y is connected but 0, x x,1 is not connected. So, such f cannot 

exist. 

Connectedness 

4.36 Prove that a metric space S is disconnected if, and only if there is a nonempty 

subset A of S, A S, which is both open and closed in S. 

Proof: () Suppose that S is disconnected, then there exist two subset A, B in S such 

that 

1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S. 

Note that since A, B are open in S ,wehaveA S B, B S A are closed in S. So,ifS 

is disconnected, then there is a nonempty subset A of S, A S, which is both open and 

closed in S. 

() Suppose that there is a nonempty subset A of S, A S, which is both open and 

closed in S. Then we have S A : B is nonempty and B is open in S. Hence, we have two 

sets A, B in S such that 


That is, S is disconnected. 

4.37 Prove that a metric space S is connected if, and only if the only subsets of S which 

are both open and closed in S are empty set and S itself. 

Proof: () Suppose that S is connected. If there exists a subset A of S such that 

1. A , 2.A is a proper subset of S, 3.A is open and closed in S, 

then let B S A, wehave 


It is impossible since S is connected. So, this A cannot exist. That is, the only subsets of S 

which are both open and closed in S are empty set and S itself. 

() Suppose that the only subsets of S which are both open and closed in S are empty 

set and S itself. If S is disconnected, then we have two sets A, B in S such that 


It contradicts the hypothesis that the only subsets of S which are both open and closed in S 

are empty set and S itself. 

Hence, we have proved that S is connected if, and only if the only subsets of S which 

are both open and closed in S are empty set and S itself. 

4.38 Prove that the only connected subsets of R are 

(a) the empty set, 

(b) sets consisting of a single point, and 

(c) intervals (open, closed, half-open, or infinite). 

Proof: Let S be a connected subset of R. Denote the symbol #A to be the number of 

elements in a set A. We consider three cases as follows. (a) #S 0, (b) #S 1, (c) 

#S 1. 

For case (a), it means that S , and for case (b), it means that S consists of a single 

point. It remains to consider the case (c). Note that since #S 1, we have inf S sup S.

Since S R, wehaveS inf S,supS. (Note that we accept that inf S or 

sup S .) If S is not an interval, then there exists x inf S,supS such that x S. 

(Otherwise, inf S,supS S which implies that S is an interval.) Then we have 

1. , x S : A is open in S 

2. x, S : B is open in S 

3. A B S. 

Claim that both A and B are not empty. Asume that A is empty, then every s S, wehave 

s x inf S. By the definition of infimum, it is impossible. So, A is not empty. Similarly 

for B. Hence, we have proved that S is disconnected, a contradiction. That is, S is an 

interval. 

Remark: 1. We note that any interval in R is connected. It is immediate from Exercise 

4.44. But we give another proof as follows. Suppose there exists an interval S is not 

connceted, then there exist two subsets A and B such that 


Since A and B , we choose a A and b B, and let a b. Consider 

c : supA a, b. 

Note that c clA A implies that c B. Hence, we have a c b. In addition, 

c B clB, then there exists a B S c; B . Choose 

d B S c; c , c S so that 

1. c d b and 2. d B. 

Then d A. (Otherwise, it contradicts c supA a, b. Note that 

d a, b S A B which implies that d A or d B. We reach a contradiction since 

d A and d B. Hence, we have proved that any interval in R is connected. 

2. Here is an application. Is there a continuous function f : R R such that 

fQ Q c ,andfQ c Q 

Ans: NO! If such f exists, then both fQ and fQ c are countable. Hence, fR is 

countable. In addition, fR is connected. Since fR contains rationals and irrationals, we 

know fR is an interval which implies that fR is uncountable, a cotradiction. Hence, such 

f does not exist. 

4.39 Let X be a connected subset of a metric space S. LetY be a subset of S such that 

X Y clX, whereclX is the closure of X. Prove that Y is also connected. In 

particular, this shows that clX is connected. 

Proof: Given a two valued function f on Y, we know that f is also a two valued 

function on X. Hence, f is constant on X, (without loss of generality) say f 0onX. 

Consider p Y X, it ,means that p is an accumulation point of X. Then there exists a 

dequence x n X such that x n p. Note that fx n 0 for all n. So, we have by 

continuity of f on Y, 

fp f lim n 

x n 

lim n 

fx n 0. 

Hence, we have f is constant 0 on Y. Thatis,Y is conneceted. In particular, clX is 

connected. 

Remark: Of course, we can use definition of a connected set to show the exercise. But, 

it is too tedious to write. However, it is a good practice to use definition to show it. The 

reader may give it a try as a challenge.

4.40 If x is a point in a metric space S, letUx be the component of S containing x. 

Prove that Ux is closed in S. 

Proof: Let p be an accumulation point of Ux. Letf be a two valued function defined 

on Ux p, then f is a two valud function defined on Ux. SinceUx is a component 

of S containing x, then Ux is connected. That is, f is constant on Ux, (without loss of 

generality) say f 0onUx. And since p is an accumulation point of Ux, there exists a 

sequence x n Ux such that x n p. Note that fx n 0 for all n. So, we have by 

continuity of f on Ux p, 

fp f lim n 

x n 

lim n 

fx n 0. 

So, Ux p is a connected set containing x. SinceUx is a component of S containing 

x, wehaveUx p Ux which implies that p Ux. Hence, Ux contains its all 

accumulation point. That is, Ux is closed in S. 

4.41 Let S be an open subset of R. By Theorem 3.11, S is the union of a countable 

disjoint collection of open intervals in R. Prove that each of these open intervals is a 

component of the metric subspace S. Explain why this does not contradict Exercise 4.40. 

Proof: By Theorem 3.11, S 

n1 I n ,whereI i is open in R and I i I j if i j. 

Assume that there exists a I m such that I m is not a component T of S. Then T I m is not 

empty. So, there exists x T I m and x I n for some n. Note that the component Ux is 

the union of all connected subsets containing x, then we have 

T I n Ux. * 

In addition, 

Ux T ** 

since T is a component containing x. Hence, by (*) and (**), we have I n T. So, 

I m I n T. SinceT is connected in R 1 , T itself is an interval. So, intT is still an interval 

which is open and containing I m I n . It contradicts the definition of component interval. 

Hence, each of these open intervals is a component of the metric subspace S. 

Since these open intervals is open relative to R, not S, this does not contradict Exercise 

4.40. 

4.42 Given a compact S in R m with the following property: For every pair of points a 

and b in S and for every 0 there exists a finite set of points x 0 , x 1 ,...,x n in S with 

x 0 a and x n b such that 

x k x k1 for k 1,2,..,n. 

Prove or disprove: S is connected. 

Proof: Suppose that S is disconnected, then there exist two subsets A and B such that 

1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S. * 

Since A and B , we choose a A, andb A and thus given 1, then by 

hypothesis, we can find two points a 1 A, andb 1 B such that a 1 b 1 1. For a 1 , 

and b 1 ,given 1/2, then by hypothesis, we can find two points a 2 A, andb 2 B 

such that a 2 b 2 1/2. Continuous the steps, we finally have two sequence a n A 

and b n B such that a n b n 1/n for all n. Sincea n A, andb n B, we 

have a n S and b n S by S A B. Hence, there exist two subsequence 

a nk A and b nk B such that a nk x, andb nk y, wherex, y S since S is 

compact. In addition, since A is closed in S, andB is closed in S, wehavex A and 

y B. On the other hand, since a n b n 1/n for all n, wehavex y. Thatis,

A B which contradicts (*)-3. Hence, we have prove that S is connected. 

Remark: We given another proof by the method of two valued function as follows. Let 

f be a two valued function defined on S, and choose any two points a, b S. If we can 

show that fa fb, we have proved that f is a constant which implies that S is 

connected. Since f is a continuous function defined on a compact set S, then f is uniformly 

on S. Thus, given 1 0, there exists a 0 such that as x y , x, y S, we 

have |fx fy| 1 fx fy. Hence, for this , there exists a finite set of 

points x 0 , x 1 ,...,x n in S with x 0 a and x n b such that 

x k x k1 for k 1,2,..,n. 

So, we have fa fx 0 fx 1 ... fx n fb. 

4.43 Prove that a metric space S is connected if, and only if, every nonempty proper 

subset of S has a nonempty boundary. 

Proof: () Suppose that S is connected, and if there exists a nonempty proper subset U 

of S such that U , thenletB clS U, wehave(defineclU A 

1. A . B since S U , 

2. A B clU clS U U S U S 

S A B, 

3. A B clU clS U U , 

and 

4. Both A and B are closed in S Both A and B are open in S. 

Hence, S is disconnected. That is, if S is connected, then every nonempty proper subset of 

S has a nonempty boundary. 

() Suppose that every nonempty proper subset of S has a nonempty boundary. If S is 

disconnected, then there exist two subsets A and B such that 

1. A, B are closed in S, 2.A and B , 3.A B , and 4. A B S. 

Then for this A, A is a nonempty proper subset of S with (clA A, andclB B) 

A clA clS A clA clB A B 

which contradicts the hypothesis that every nonempty proper subset of S has a nonempty 

boundary. So, S is connected. 

4.44. Prove that every convex subset of R n is connected. 

Proof: Given a convex subset S of R n , and since for any pair of points a, b, the set 

1 a b :0 1 : T S, i.e., g : 0, 1 T by g 1 a b is a 

continuous function such that g0 a, andg1 b. So,S is path-connected. It implies 

that S is connected. 

Remark: 1. In the exercise, it tells us that every n ball is connected. (In fact, every 

n ball is path-connected.) In particular, as n 1, any interval (open, closed, half-open, or 

infinite) in R is connected. For n 2, any disk (open, closed, or not) in R 2 is connected. 

2. Here is a good exercise on the fact that a path-connected set is connected. Given 

0, 1 0, 1 : S, andifT is a countable subset of S. Prove that S T is connected. (In 

fact, S T is path-connected.) 

Proof: Given any two points a and b in S T, then consider the vertical line L passing 

through the middle point a b/2. Let A x : x L S, and consider the lines form 

a to A, and from b to A. Note that A is uncountable, and two such lines (form a to A, and

from b to A) are disjoint. So, if every line contains a point of T, then it leads us to get T is 

uncountable. However, T is countable. So, it has some line (form a to A, and from b to A) 

is in S T. So, it means that S T is path-connected. So, S T is connected. 

4.45 Given a function f : R n R m which is 1-1 and continuous on R n .IfA is open 

and disconnected in R n , prove that fA is open and disconnected in fR n . 

Proof: Theexerciseiswrong. There is a counter-example. Let f : R R 2 

f 

cos 2x 

1x 

cos 2x 

1x 

2 

2 

2x 

,1 sin if x 0 

1x 2 

2x 

, 1 sin if x 0 

1x 2 

Remark: If we restrict n, m 1, the conclusion holds. That is, Let f : R R be 

continuous and 1-1. If A is open and disconnected, then so is fA. 

Proof: In order to show this, it suffices to show that f maps an open interval I to 

another open interval. Since f is continuous on I, andI is connected, fI is connected. It 

implies that fI is an interval. Trivially, there is no point x in I such that fx equals the 

endpoints of fI. Hence, we know that fI is an open interval. 

Supplement: Here are two exercises on Homeomorphism to make the reader get more 

and feel something. 

1. Let f : E R R. Ifx, fx : x E is compact, then f is uniformly continuous 

on E. 

Proof: Let x, fx : x E S, and thus define gx Ix x, fx : E S. 

Claim that g is continuous on E. Consider h : S E by hx, fx x. Trivially, h is 1-1, 

continuous on a compact set S. So, its inverse function g is 1-1 and continuous on a 

compact set E. The claim has proved. 

Since g is continuous on E, we know that f is continuous on a compact set E. Hence, f 

is uniformly continuous on E. 

Note: The question in Supplement 1, there has another proof by the method of 

contradiction, and use the property of compactness. We omit it. 

2. Let f : 0, 1 R. Ifx, fx : x 0, 1 is path-connected, then f is continuous 

on 0, 1. 

Proof: Let a 0, 1, then there is a compact interval a a 1 , a 2 0, 1. Claim 

that the set 

x, fx : x a 1 , a 2 : S is compact. 

Since S is path-connected, there is a continuous function g : 0, 1 S such that 

g0 a 1 , fa 1 and g1 a 2 , fa 2 . If we can show g0, 1 S, wehaveshown 

that S is compact. Consider h : S R by hx, fx x; h is clearly continuous on S. So, 

the composite function h g : 0, 1 R is also continuous. Note that h g0 a 1 ,and 

h g1 a 2 , and the range of h g is connected. So, a 1 , a 2 hg0, 1. Hence, 

g0, 1 S. We have proved the claim and by Supplement 1, we know that f is 

continuous at a. Sincea is arbitrary, we know that f is continuous on 0, 1. 

Note: The question in Supplement 2, there has another proof directly by definition of 

continuity. We omit the proof. 

4.46 Let A x, y :0 x 1, y sin 1/x, B x, y : y 0, 1 x 0, 

and let S A B. Prove that S is connected but not arcwise conneceted. (See Fig. 4.5,

Section 4.18.) 

Proof: Let f be a two valued function defined on S. SinceA, andB are connected in S, 

then we have 

fA a, andfB b, wherea, b 0, 1. 

Given a sequence x n A with x n 0, 0, then we have 

a lim n 

fx n f limx n by continuity of f at 0 

n0 

f0, 0 

b. 

So, we have f is a constant. That is, S is connected. 

Assume that S is arcwise connected, then there exists a continuous function 

g : 0, 1 S such that g0 0, 0 and g1 1, sin 1. Given 1/2, there exists a 

0 such that as |t| , wehave 

gt g0 gt 1/2. * 

1 

Let N be a positive integer so that 

2N 

Define two subsets U and V as follows: 

U 

V 

1 

, thus let ,0 : p and 1 

,0 : q. 

2N 2N1 

x, y : x p q 

2 

x, y : x p q 

2 

gq, p 

gq, p 

Then we have 

(1). U V gq, p, (2). U , sincep U and V , sinceq V, 

(3). U V by the given set A, and (*) 

Since x, y : x pq 

2 

gq, p. So, we have 

and x, y : x 

pq 

2 are open in R2 , then U and V are open in 

(4). U is open in gq, p and V is open in gq, p. 

From (1)-(4), we have gq, p is disconnected which is absurb since a connected subset 

under a continuous function is connected. So, such g cannot exist. It means that S is not 

arcwise connected. 

Remark: This exercise gives us an example to say that connectedness does not imply 

path-connectedness. And it is important example which is worth keeping in mind. 

4.47 Let F F 1 , F 2 ,... be a countable collection of connected compact sets in R n 

such that F k1 F k for each k 1. Prove that the intersection 

k1 F k is connected and 

closed. 

Proof: Since F k is compact for each k 1, F k is closed for each k 1. Hence, 

 

k1 F k : F is closed. Note that by Theorem 3.39, we know that F is compact. Assume 

that F is not connected. Then there are two subsets A and B with 

1.A , B . 2.A B . 3.A B F. 4.A, B are closed in F. 

Note that A, B are closed and disjoint in R n . By exercise 4.57, there exist U and V which 

are open and disjoint in R n such that A U, andB V. Claim that there exists F k such 

that F k U V. Suppose NOT, thenthereexistsx k F k U V. Without loss of 

generality, we may assume that x k F k1 . So, we have a sequence x k F 1 which 

implies that there exists a convergent subsequence x kn , say lim kn x kn x. Itis 

clear that x F k for all k since x is an accumulation point of each F k . So, we have

x F 

k1 F k A B U V 

which implies that x is an interior point of U V since U and V are open. So, 

Bx; U V for some 0, which contradicts to the choice of x k . Hence, we have 

proved that there exists F k such that F k U V. LetC U F k ,andD V F k , then 

we have 

1. C since A U and A F k ,andD since B V and B F k . 

2. C D U F k V F k U V . 

3. C D U F k V F k F k . 

4. C is open in F k and D is open in F k by C, D are open in R n . 

Hence, we have F k is disconnected which is absurb. So, we know that F 

k1 F k is 

connected. 

4.48 Let S be an open connected set in R n .LetT be a component of R n S. Prove 

that R n T is connected. 

Proof: If S is empty, there is nothing to proved. Hence, we assume that S is nonempty. 

Write R n S xR n S Ux, whereUx is a component of R n S. So, we have 

R n S xR n S Ux. 

Say T Up, forsomep. Then 

R n T S xR n ST Ux. 

Claim that clS Ux for all x R n S T. If we can show the claim, given 

a, b R n T, and a two valued function on R n T. Note that clS is also connected. We 

consider three cases. (1) a S, b Ux for some x. (2)a, b S. (3)a Ux, 

b Ux . 

For case (1), let c clS Ux, then there are s n S and u n Ux with 

s n c and u n c, then we have 

fa lim n 

fs n f lim n 

s n fc f lim n 

u n lim n 

fu n fb 

which implies that fa fb. 

For case (2), it is clear fa fb since S itself is connected. 

For case (3), we choose s S, and thus use case (1), we know that 

fa fs fb. 

By case (1)-(3), we have f is constant on R n T. Thatis,R n T is connected. 

It remains to show the claim. To show clS Ux for all x R n S T, i.e., to 

show that for all x R n S T, 

clS Ux S S Ux 

S Ux 

. 

Suppose NOT, i.e., for some x, S Ux which implies that Ux R n clS 

which is open. So, there is a component V of R n clS contains Ux, whereV is open by 

Theorem 4.44. However, R n clS R n S, sowehaveV is contained in Ux. 

Therefore, we have Ux V. Note that Ux R n S, andR n S is closed. So, 

clUx R n S. By definition of component, we have clUx Ux, whichis 

closed. So, we have proved that Ux V is open and closed. It implies that Ux R n or 

which is absurb. Hence, the claim has proved. 

4.49 Let S, d be a connected metric space which is not bounded. Prove that for every

a in S and every r 0, the set x : dx, a r is nonempty. 

Proof: Assume that x : dx, a r is empty. Denote two sets x : dx, a r by A 

and x : dx, a r by B. Then we have 

1. A since a A and B since S is unbounded, 

2. A B , 

3. A B S, 

4. A Ba; r is open in S, 

and consider B as follows. Since x : dx, a r is closed in S, B S x : dx, a r 

is open in S. So, we know that S is disconnected which is absurb. Hence, we know that the 

set x : dx, a r is nonempty. 

Supplement on a connected metric space 

Definition Two subsets A and B of a metric space X are said to be separated if both 

A clB and clA B . 

AsetE X is said to be connected if E is not a union of two nonempty separated 

sets. 

We now prove the definition of connected metric space is equivalent to this definiton 

as follows. 

Theorem A set E in a metric space X is connected if, and only if E is not the union of 

two nonempty disjoint subsets, each of which is open in E. 

Proof: () Suppose that E is the union of two nonempty disjoint subsets, each 

of which is open in E, denote two sets, U and V. Claim that 

U clV clU V . 

Suppose NOT, i.e., x U clV. Thatis,thereisa 0 such that 

B X x, E B E x, U and B X x, V 


B X x, V B X x, V E 

B X x, E V 

U V , 

a contradiction. So, we have U clV . Similarly for clU V . So,X is 

disconnected. That is, we have shown that if a set E in a metric space X is 

connected, then E is not the union of two nonempty disjoint subsets, each of which 

is open in E. 

() Suppose that E is disconnected, then E is a union of two nonempty 

separated sets, denoted E A B, whereA clB clA B . Claim that A 

and B are open in E. Suppose NOT, it means that there is a point x A which is 

not an interior point of A. So, for any ball B E x, r, there is a correspounding 

x r B, wherex r B E x, r. It implies that x clB which is absurb with 

A clB . So, we proved that A is open in E. Similarly, B is open in E. Hence, 

we have proved that if E is not the union of two nonempty disjoint subsets, each of 

whichisopeninE, then E in a metric space X is connected. 

Exercise Let A and B be connected sets in a metric space with A B not connected and 

suppose A B C 1 C 2 where clC 1 C 2 C 1 clC 2 . Show that 

B C 1 is connected.

Proof: Assume that B C 1 is disconnected, and thus we will prove that C 1 is 

disconnected. Consider, by clC 1 C 2 C 1 clC 2 , 

C 1 clC 2 A B C 1 clA B C 1 clB * 

and 

clC 1 C 2 A B clC 1 A B clC 1 B ** 

we know that at least one of (*) and (**) is nonempty by the hypothesis A is 

connected. In addition, by (*) and (**), we know that at leaset one of 

C 1 clB 

and 

clC 1 B 

is nonempty. So, we know that C 1 is disconnected by the hypothesis B is connected, 

and the concept of two valued function. 

From above sayings and hypothesis, we now have 

1. B is connected. 

2. C 1 is disconnected. 

3. B C 1 is disconnected. 

Let D be a component of B C 1 so that B D; we have, let 

B C 1 D E C 1 , 

D clE clD E 


clE A E , andclA E E . 

So,wehaveprovethatA is disconnected wich is absurb. Hence, we know that 

B C 1 is connected. 

Remark We prove that clA E E clE A E as follows. 

Proof: Since 

D clE , 

we obtain that 

clE A E 

clE D C 2 A B 

clE D C 2 B 

clE D C 2 since B D 

clE C 2 since D clE 

And since 

we obtain that 

clC 1 C 2 since E C 1 

. 

clD E ,

clA E E 

clD C 2 A B E 

clD C 2 B E 

clD C 2 E since B D 

clC 2 E since clD E 

clC 2 C 1 since E C 1 

. 

Exercise Prove that every connected metric space with at least two points is uncountable. 

Proof: LetX be a connected metric space with two points a and b, where 

a b. Define a set A r x : dx, a r and B r x : dx, a r. It is clear 

that both of sets are open and disjoint. Assume X is countable. Let 

da,b 

r , da,b , it guarantee that both of sets are non-empty. Since 

4 2 

da,b 

, da,b is uncountable, we know that there is a 0 such that 

4 2 

A B X. It implies that X is disconnected. So, we know that such X is 

countable. 

Uniform continuity 

4.50 Prove that a function which is uniformly continuous on S is also continuous on S. 

Proof: Let f be uniformly continuous on S, thengiven 0, there exists a 0 such 

that as dx, y , x and y in S, then we have 

dfx, fy . 

Fix y, called a. Thengiven 0, there exists a 0 such that as dx, a , x in S, 

then we have 

dfx, fa . 

That is, f is continuous at a. Sincea is arbitrary, we know that f is continuous on S. 

4.51 If fx x 2 for x in R, prove that f is not uniformly continuous on R. 

Proof: Assume that f is uniformly continuous on R, thengiven 1, there exists a 

0 such that as |x y| , wehave 

|fx fy| 1. 

Choose x y ,( 

2 

|x y| , then we have 

2 

|fx fy| y 1. 2 

When we choose y 1 , then 1 

2 

2 

1 

2 

1 

which is absurb. Hence, we know that f is not uniformly continuous on R. 

Remark: There are some similar questions written below. 

1. Here is a useful lemma to make sure that a function is uniformly continuous on 

a, b, but we need its differentiability. 

(Lemma) Letf : a, b R R be differentiable and |f x| M for all x a, b. 

Then f is uniformly continuous on a, b, wherea, b may be .

Proof: By Mean Value Theorem, wehave 

|fx fy| |f z||x y|, wherez x, y or y, x 

M|x y| by hypothesis. * 

Then given 0, there is a /M such that as |x y| , x, y a, b, wehave 

|fx fy| , by (*). 

Hence, we know that f is uniformly continuous on a, b. 

Note: A standard example is written in Remark 2. But in Remark 2, we still use 

definition of uniform continuity to practice what it says. 

2. sin x is uniformly continuous on R. 

Proof: Given 0, we want to find a 0 such that as |x y| , wehave 

Since sin x sin y 2cos xy 

2 

|sin x sin y| . 

xy 

sin , 

2 

|sin x| |x|, and|cosx| 1, we have 

|sin x sin y| |x y| 

So, if we choose , then as |x y| , it implies that 

|sin x sin y| . 

That is, sinx is uniformly continuous on R. 

Note: |sin x sin y| |x y| for all x, y R, can be proved by Mean Value Theorem 

as follows. 

proof: By Mean Value Theorem, sin x sin y sin z x y; itimpliesthat 

|sin x sin y| |x y|. 

3. sinx 2 is NOT uniformly continuous on R. 

Proof: Assume that sinx 2 is uniformly continuous on R. Thengiven 1, there is a 

0 such that as |x y| , wehave 

|sinx 2 siny 2 | 1. * 

Consider 

n 

2 n 2 

n n 0, 

4 n 

2 

and thus choose N 

 

 

1 

4 2 4 2 

N 2 

which implies 

N . 

So, choose x N 2 

and y N , thenby(*),wehave 

|fx fy| sin N sinN sin 

2 

1 1 

2 

which is absurb. So, sinx 2 is not uniformly continuous on R. 

4. x is uniformly continuous on 0, . 

Proof: Since x y |x y| for all x, y 0, , thengiven 0, there exists 

a 2 such that as |x y| , x, y 0, , wehave 

So, we know that 

x y |x y| . 

x is uniformly continuous on 0, .

Note: We have the following interesting results:. Prove that, for x 0, y 0, 

|x p y p | 

|x y| p if 0 p 1, 

p|x y|x p1 y p1 if 1 p . 

Proof: (As 0 p 1) Without loss of generality, let x y, consider 

fx x y p x p y p , then 

f x p x y p1 x p1 0, note that p 1 0. 

So, we have f is an increasing function defined on 0, for all given y 0. Hence, we 

have fx f0 0. So, 

x p y p x y p if x y 0 


|x p y p | |x y| p 

for x 0, y 0. 

Ps: The inequality, we can prove the case p 1/2 directly. Thus the inequality is not 

surprising for us. 

(As 1 p Without loss of generality, let x y, consider 

x p y p pz p1 x y, wherez y, x, byMean Value Theorem. 

which implies 

for x 0, y 0. 

px p1 x y, note that p 1 0, 

px p1 y p1 x y 

|x p y p | p|x y|x p1 y p1 

5. In general, we have 

x r is uniformly continuous on 0, , ifr 0, 1, 

is NOT uniformly continuous on 0, , ifr 1, 

and 

sinx r 

is uniformly continuous on 0, , ifr 0, 1, 

is NOT uniformly continuous on 0, , ifr 1. 

Proof: (x r )Asr 0, it means that x r is a constant function. So, it is obviuos. As 

r 0, 1, thengiven 0, there is a 1/r 0 such that as |x y| , x, y 0, , 

we have 

|x r y r | |x y| r by note in the exercise 

r 

. 

So, x r is uniformly continuous on 0, , ifr 0, 1. 

As r 1, assume that x r is uniformly continuous on 0, , thengiven 1 0, 

there exists a 0 such that as |x y| , x, y 0, , wehave 

|x r y r | 1. * 

By Mean Value Theorem, we have (let x y /2, y 0)

x r y r rz r1 x y 

ry r1 /2. 

So, if we choose y 2 1 

r1 

, then we have 

r 

x r y r 1 

which is absurb with (*). Hence, x r is not uniformly continuous on 0, . 

Ps: The reader should try to realize why x r is not uniformly continuous on 0, , for 

r 1. The ruin of non-uniform continuity comes from that x is large enough. At the same 

time, compare it with theorem that a continuous function defined on a compact set K is 

uniflormly continuous on K. 

(sin x r )Asr 0, it means that x r is a constant function. So, it is obviuos. As r 0, 1, 

given 0, there is a 1/r 0 such that as |x y| , x, y 0, , wehave 

|sin x r sin y r x 

| 2cos 

r y r 

sin 

xr y r 

2 

2 

|x r y r | 

|x y| r by the note in the Remark 4. 

r 

. 

So, sin x r is uniformly continuous on 0, , ifr 0, 1. 

As r 1, assume that sin x r is uniformly continuous on 0, , thengiven 1, there 

is a 0 such that as |x y| , x, y 0, , wehave 

|sin x r sin y r | 1. ** 

Consider a sequence n 2 1/r n 1/r , it is easy to show that the sequence tends to 

0asn . So, there exists a positive integer N such that |x y| , x n 2 1/r , 

y n 1/r . Then 

sin x r sin y r 1 

which contradicts (**). So, we know that sin x r is not uniformly continuous on 0, . 

Ps: For n 2 1/r n 1/r : x n 0asn 0, here is a short proof by using 

L-Hospital Rule. 

Proof: Write 

x n n 1/r 

n 

1/r 

2 

n 1/r 1 1 2n 

1/r 

1 

and thus consider the following limit 

 

1 1 

2n 1/r 1 

n 1/r

lim x 

1 1 2x 1/r 1 

x 1/r , 0 

0 

lim 1/r 

x 2 x 1 r 1 

1 

2x 

1 1 r 1 

by L-Hospital Rule. 

0. 

Hence x n 0asn . 

6. Here is a useful criterion for a function which is NOT uniformly continuous defined 

a subset A in a metric space. We say a function f is not uniformly continuous on a subset A 

in a metric space if, and only if, there exists 0 0, and two sequences x n and y n 

such that as 

lim n 

x n y n 0 


|fx n fy n | 0 for n is large enough. 

The criterion is directly from the definition on uniform continuity. So, we omit the 

proof. 

4.52 Assume that f is uniformly continuous on a bounded set S in R n . Prove that f 

must be bounded on S. 

Proof: Since f is uniformly continuous on a bounded set S in R n ,given 1, then 

there exists a 0 such that as x y , x, y S, wehave 

dfx, fy 1. 

Consider the closure of S, clS is closed and bounded. Hence clS is compact. Then for 

any open covering of clS, there is a finite subcover. That is, 

clS xclS Bx; /2, 

clS kn k1 Bx k ; /2, wherex k clS, 

S kn k1 Bx k ; /2, wherex k clS. 

Note that if Bx k ; /2 S for some k, then we remove this ball. So, we choose 

y k Bx k ; /2 S, 1 k n and thus we have 

Bx k ; /2 By k ; for 1 k n, 

since let z Bx k ; /2, 

z y k z x k x k y k /2 /2 . 

Hence, we have 

S kn k1 By k ; , wherey k S. 

Given x S, then there exists By k ; for some k such that x By k ; . So, 

dfx, fx k 1 fx Bfy k ;1 

Note that kn k1 Bfy k ;1 is bounded since every Bfy k ;1 is bounded. So, let B be a 

bounded ball so that kn k1 Bfy k ;1 B. Hence, we have every x S, fx B. Thatis, 

f is bounded. 

Remark: If we know that the codomain is complete, then we can reduce the above 

proof. See Exercise 4.55. 

4.53 Let f be a function defined on a set S in R n and assume that fS R m .Letg be 

defined on fS with value in R k , and let h denote the composite function defined by

hx gfx if x S. Iff is uniformly continuous on S and if g is uniformly continuous 

on fS, show that h is uniformly continuous on S. 

Proof: Given 0, we want to find a 0 such that as x y R 

n , x, y S, we 

have 

hx hy gfx gfy . 

For the same , sinceg is uniformly continuous on fS, then there exists a 0 such 

that as fx fy R 

m ,wehave 

gfx gfy . 

For this ,sincef is uniformly continuous on S, then there exists a 0 such that as 

x y R 

n , x, y S, wehave 

fx fy R 

m . 

So, given 0, there is a 0 such that as x y R 

n , x, y S, wehave 

hx hy . 

That is, h is uniformly continuous on S. 

Remark: It should be noted that (Assume that all functions written are continuous) 

(1) uniform continuity uniform continuity uniform continuity. 

(2) uniform continuity NOT uniform continuity 

(3) NOT uniform continuity uniform continuity 

(a) NOT uniform continuity, or 

(b) uniform continuity. 



(4) NOT uniform continuity NOT uniform continuity 



For (1), it is from the exercise. 

For (2), (a) let fx x, andgx x 2 , x R fgx fx 2 x 2 . 

(b) let fx x ,andgx x 2 , x 0, fgx fx 2 x. 

For (3), (a) let fx x 2 ,andgx x, x R fgx fx x 2 . 

(b) let fx x 2 ,andgx x , x 0, fgx f x x. 

For (4), (a) let fx x 2 ,andgx x 3 , x R fgx fx 3 x 6 . 

(b) let fx 1/x, andgx 1 , x 0, 1 fgx f 1 

x . 

x x 

Note. In (4), we have x r is not uniformly continuous on 0, 1, forr 0. Here is a 

proof. 

Proof: Let r 0, and assume that x r is not uniformly continuous on 0, 1. Given 

1, there is a 0 such that as |x y| , wehave 

|x r y r | 1. * 

Let x n 2/n, andy n 1/n. Then x n y n 1/n. Choose n large enough so that 1/n . 

So, we have

|x r y r | 2 n 

r 

1 n 

r 

r 

1 n |2r 1| , asn since r 0, 

which is absurb with (*). Hence, we know that x r is not uniformly continuous on 0, 1, for 

r 0. 

Ps: The reader should try to realize why x r is not uniformly continuous on 0, 1, for 

r 0. The ruin of non-uniform continuity comes from that x is small enough. 

4.54 Assume f : S T is uniformly continuous on S, whereS and T are metric 

spaces. If x n is any Cauchy sequence in S, prove that fx n is a Cauchy sequence in T. 

(Compare with Exercise 4.33.) 

Proof: Given 0, we want to find a positive integer N such that as n, m N, we 

have 

dfx n , fx m . 

For the same , sincef is uniformly continuous on S, then there is a 0 such that as 

dx, y , x, y S, wehave 

dfx, fy . 

For this , sincex n is a Cauchy sequence in S, then there is a positive integer N such 

that as n, m N, wehave 

dx n , x m . 

Hence, given 0, there is a postive integer N such that as n, m N, wehave 

dfx n , fx m . 

That is, fx n is a Cauchy sequence in T. 

Remark: The reader should compare with Exercise 4.33 and Exercise 4.55. 

4.55 Let f : S T be a function from a metric space S to another metric space T. 

Assume that f is uniformly continuous on a subset A of S and let T is complete. Prove that 

there is a unique extension of f to clA which is uniformly continuous on clA. 

Proof: Since clA A A , it suffices to consider the case x A A. Since 

x A A, then there is a sequence x n A with x n x. Note that this sequence is a 

Cauchy sequence, so we have by Exercise 4.54, fx n is a Cauchy sequence in T since f 

is uniformly on A. In addition, since T is complete, we know that fx n is a convergent 

sequence, say its limit L. Note that if there is another sequence x n A with x n x, 

then fx n is also a convergent sequence, say its limit L . Note that x n x n is still a 

Cauchy sequence. So, we have 

dL, L dL, fx n dfx n , fx n dfx n , L 0asn . 

So, L L . That is, it is well-defined for g : clA T by the following 

gx 

fx if x A, 

lim n fx n if x A A, wherex n x. 

So, the function g is a extension of f to clA. 

Claim that this g is uniformly continuous on clA. That is, given 0, we want to 

find a 0 such that as dx, y , x, y clA, wehave 

dgx, gy . 

Since f is uniformly continuous on A, for /3, there is a 0 such that as

dx, y , x, y A, wehave 

dfx, fy . 

Let x, y clA, and thus we have x n A with x n x, andy n A with y n y. 

Choose /3, then we have 

dx n , x /3 and dy n , y /3 as n N 1 

So, as dx, y /3, wehave(n N 1 ) 

dx n , y n dx n , x dx, y dy, y n /3 /3 /3 . 

Hence, we have as dx, y , (n N 1 ) 

dgx, gy dgx, fx n dfx n , fy n dfy n , fy 

* 

dgx, fx n dfy n , gy 

And since lim n fx n gx, and lim n fy n gy, we can choose N N 1 such that 

dgx, fx n and 

dfy n , gy . 

So, as dx, y , (n N) wehave 

dgx, gy 3 by (*). 

That is, g is uniformly on clA. 

It remains to show that g is a unique extension of f to clA which is uniformly 

continuous on clA. If there is another extension h of f to clA which is uniformly 

continuous on clA, thengivenx A A, we have, by continuity, (Say x n x) 

hx h lim n 

x n lim n 

hx n lim n 

fx n lim n 

gx n g lim n 

x n gx 

which implies that hx gx for all x A A. Hence, we have hx gx for all 

x clA. Thatis,g is a unique extension of f to clA which is uniformly continuous on 

clA. 

Remark: 1. We do not require that A is bounded, in fact, A is any non-empty set in a 

metric space. 

2. The exercise is a criterion for us to check that a given function is NOT uniformly 

continuous. For example, let f : 0, 1 R by fx 1/x. Sincef0 does not exist, we 

know that f is not uniformly continuous. The reader should feel that a uniformly continuous 

is sometimes regarded as a smooth function. So, it is not surprising for us to know the 

exercise. Similarly to check fx x 2 , x R, and so on. 

3. Here is an exercise to make us know that a uniformly continuous is a smooth 

function. Let f : R R be uniformly continuous, then there exist , 0 such that 

|fx| |x| . 

Proof: Since f is uniformly continuous on R, given 1, there is a 0 such that as 

|x y| , wehave 

|fx fy| 1. * 

Given any x R, then there is the positive integer N such that N |x| N 1. If 

x 0, we consider 

y 0 0, y 1 /2, y 2 ,...,y 2N1 N 2 , y 2N x. 

Then we have

N 

|fx f0| |fy 2k fy 2k1 | |fy 2k1 fy 2k2 | 

k1 

2N by (*) 


|fx| 2N |f0| 

2 1 |x| |f0| since |x| N 1 

 

2 |x| 2 |f0|. 

 

Similarly for x 0. So, we have proved that |fx| |x| for all x. 

4.56 In a metric space S, d, letA be a nonempty subset of S. Define a function 

f A : S R by the equation 

f A x infdx, y : y A 

for each x in S. The number f A x is called the distance from x to A. 

(a) Prove that f A is uniformly continuous on S. 

(b) Prove that clA x : x S and f A x 0 . 

Proof: (a) Given 0, we want to find a 0 such that as dx 1 , x 2 , x 1 , x 2 S, 

we have 

|f A x 1 f A x 2 | . 

Consider (x 1 , x 2 , y S) 

dx 1 , y dx 1 , x 2 dx 2 , y, anddx 2 , y dx 1 , x 2 dx 1 , y 

So, 

infdx 1 , y : y A dx 1 , x 2 infdx 2 , y : y A and 

infdx 2 , y : y A dx 1 , x 2 infdx 1 , y : y A 


f A x 1 f A x 2 dx 1 , x 2 and f A x 2 f A x 1 dx 1 , x 2 


|f A x 1 f A x 2 | dx 1 , x 2 . 

Hence, if we choose , thenwehaveasdx 1 , x 2 , x 1 , x 2 S, wehave 

|f A x 1 f A x 2 | . 

That is, f A is uniformly continuous on S. 

(b) Define K x : x S and f A x 0 , we want to show clA K. We prove it 

by two steps. 

() Letx clA, then Bx; r A for all r 0. Choose y k Bx;1/k A, then 

we have 

infdx, y : y A dx, y k 0ask . 

So, we have f A x infdx, y : y A 0. So, clA K. 

() Letx K, then f A x infdx, y : y A 0. That is, given any 0, there 

is an element y A such that dx, y . Thatis,y Bx; A. So,x is an adherent 

point of A. Thatis,x clA. So, we have K clA. 

From above saying, we know that clA x : x S and f A x 0 . 

Remark: 1. The function f A often appears in Analysis, so it is worth keeping it in mind.

In addition, part (b) comes from intuition. The reader may think it twice about distance 0. 

2. Here is a good exercise to pratice. The statement is that suppose that K and F are 

disjoint subsets in a metric space X, K is compact, F is closed. Prove that there exists a 

0 such that dp, q if p K, q F. Show that the conclusion is may fail for two 

disjoint closed sets if neither is compact. 

Proof: Suppose NOT, i.e., for any 0, there exist p K, andq F such that 

dp , q . Let 1/n, then there exist two sequence p n K, andq n F such 

that dp n , q n 1/n. Note that p n K, andK is compact, then there exists a 

subsequence p nk with lim nk p nk p K. Hence, we consider dp nk , q nk n 1 k 

to get a 

contradiction. Since 

dp nk , p dp, q nk dp nk , q nk n 1 , k 

then let n k , wehavelim nk q nk p. Thatis,p is an accumulation point of F which 

implies that p F. So, we get a contradiction since K F . That is, there exists a 

0 such that dp, q if p K, q F. 

We give an example to show that the conclusion does not hold. Let 

K x,0 : x R and F x,1/x : x 0, then K and F are closd. It is clear that 

such cannot be found. 

Note: Two disjoint closed sets may has the distance 0, however; if one of closed sets is 

compact, then we have a distance 0. The reader can think of them in R n , and note that 

a bounded and closed subsets in R n is compact. It is why the example is given. 

4.57 In a metric space S, d, letA and B be disjoint closed subsets of S. Prove that 

there exists disjoint open subsets U and V of S such that A U and B V. Hint. Let 

gx f A x f B x, in the notation of Exercise 4.56, and consider g 1 ,0 and 

g 1 0, . 

Proof: Let gx f A x f B x, then by Exercixe 4.56, we have gx is uniformly 

continuous on S. So,gx is continuous on S. Consider g 1 ,0 and g 1 0, , and 

note that A, B are disjoint and closed, then we have by part (b) in Exercise 4.56, 

gx 0ifx A and 

gx 0ifx B. 

So, we have A g 1 ,0 : U, andB g 1 0, : V. 

Discontinuities 

4.58 Locate and classify the discontinuities of the functions f defined on R 1 by the 

following equations: 

(a) fx sin x/x if x 0, f0 0. 

sinx 

Solution: f is continuous on R 0, and since lim x0 x 1, we know that f has a 

removable discontinuity at 0. 

(b) fx e 1/x if x 0, f0 0. 

Solution: f is continuous on R 0, and since lim x0 

e 1/x and lim x0 

e 1/x 0, 

we know that f has an irremovable discontinuity at 0. 

(c) fx e 1/x sin 1/x if x 0, f0 0. 

Solution: f is continuous on R 0, and since the limit fx does not exist as x 0, 

we know that f has a irremovable discontinuity at 0.

(d) fx 1/1 e 1/x if x 0, f0 0. 

Solution: f is continuous on R 0, and since lim x0 

e 1/x and lim x0 

e 1/x 0, 

we know that f has an irremovabel discontinuity at 0. In addition, f0 0and 

f0 1, we know that f has the lefthand jump at 0, f0 f0 1, and f is 

continuous from the right at 0. 

4.59 Locate the points in R 2 at which each of the functions in Exercise 4.11 is not 

continuous. 

(a) By Exercise 4.11, we know that fx, y is discontinuous at 0, 0, where 

fx, y x2 y 2 

if x, y 0, 0, andf0, 0 0. 

x 2 y2 Let gx, y x 2 y 2 ,andhx, y x 2 y 2 both defined on R 2 0, 0, we know that g 

and h are continuous on R 2 0, 0. Note that h 0onR 2 0, 0. Hence, f g/h is 

continuous on R 2 0, 0. 

(b) By Exercise 4.11, we know that fx, y is discontinuous at 0, 0, where 

xy 2 

fx, y 

if x, y 0, 0, andf0, 0 0. 

xy 2 2 

x y 

Let gx, y xy 2 ,andhx, y xy 2 x y 2 both defined on R 2 0, 0, we know 

that g and h are continuous on R 2 0, 0. Note that h 0onR 2 0, 0. Hence, 

f g/h is continuous on R 2 0, 0. 

(c) By Exercise 4.11, we know that fx, y is continuous at 0, 0, where 

fx, y 1 x sinxy if x 0, and f0, y y, 

since lim x,y0,0 fx, y 0 f0, 0. Letgx, y 1/x and hx, y sinxy both defined 

on R 2 0, 0, we know that g and h are continuous on R 2 0, 0. Note that h 0 

on R 2 0, 0. Hence, f g/h is continuous on R 2 0, 0. Hence, f is continuous on 

R 2 . 

(d) By Exercise 4.11, we know that fx, y is continuous at 0, 0, where 

x y sin1/x sin1/y if x 0andy 0, 

fx, y 

0 if x 0ory 0. 

since lim x,y0,0 fx, y 0 f0, 0. It is the same method as in Exercise 4.11, we know 

that f is discontinuous at x,0 for x 0andf is discontinuous at 0, y for y 0. And it is 

clearly that f is continuous at x, y, wherex 0andy 0. 

(e) By Exercise 4.11, Since 

we rewrite 

fx, y 

fx, y 

cos 

sinxsiny 

tan xtan y 

,iftanx tan y, 


xy 

cos xcos y 

2 

cos 

xy 

2 

if tan x tan y 


We consider x, y /2, /2 /2, /2, others are similar. Consider two cases (1) 

x y, and (2) x y, wehave 

(1) (x y) Since lim x,ya,a fx, y cos 3 a fa, a. Hence, we know that f is 

, 

.

continuous at a, a. 

(2) (x y) Sincex y, itimpliesthattanx tan y. Note that the denominator is not 0 

since x, y /2, /2 /2, /2. So, we know that f is continuous at a, b, a b. 

So, we know that f is continuous on /2, /2 /2, /2. 

Monotonic functions 

4.60 Let f be defined in the open interval a, b and assume that for each interior point x 

of a, b there exists a 1 ball Bx in which f is increasing. Prove that f is an increasing 

function throughout a, b. 

Proof: Suppose NOT, i.e., there exist p, q with p q such that fp fq. Consider 

p, q a, b, and since for each interior point x of a, b there exists a 1 ball Bx in 

which f is increasing. Then p, q xp,q Bx; x , (The choice of balls comes from the 

hypothesis). It implies that p, q n 

k1 Bx n ; n : B n . Note that if B i B j , we remove 

such B i and make one left. Without loss of generality, we assume that x 1 .. x n . 

.fp fx 1 ... fx n fq 

which is absurb. So, we know that f is an increasing function throughout a, b. 

4.61 Let f be continuous on a compact interval a, b and assume that f does not have a 

loacal maximum or a local minimum at any interior point. (See the note following Exercise 

4.25.) Prove that f must be monotonic on a, b. 

Proof: Since f is continuous on a, b, wehave 

max fx fp, wherep a, b and 

xa,b 

min fx fq, whereq a, b. 

xa,b 

So, we have p, q a, b by hypothesis that f does not have a local maximum or a local 

minimum at any interior point. Without loss of generality, we assume that p a, and 

q b. Claim that f is decreasing on a, b as follows. 

Suppose NOT, then there exist x, y a, b with x y such that fx fy. Consider 

x, y and by hyothesis, we know that f| x,y has the maximum at y, andf| a,y has the 

minimum at y. Then it implies that there exists By; x, y such that f is constant on 

By; x, y, which contradicts to the hypothesis. Hence, we have proved that f is 

decreasing on a, b. 

4.62 If f is one-to one and continuous on a, b, prove that f must be strictly 

monotonic on a, b. That is, prove that every topological mapping of a, b onto an 

interval c, d must be strictly monotonic. 

Proof: Since f is continuous on a, b, wehave 

max fx fp, wherep a, b and 

xa,b 

min fx fq, whereq a, b. 

xa,b 

Assume that p a, b, then there exists a 0 such that fy fp for all 

y p , p a, b. Choose y 1 x , x and y 2 x, x , thenwehaveby 

1-1, fy 1 fx and fy 2 fx. And thus choose r so that 

fy 1 r fx fz 1 r, wherez 1 y 1 , x by Intermediate Value Theorem, 

fy 2 r fx fz 2 r, wherez 2 x, y 2 by Intermediate Value Theorem, 

which contradicts to 1-1. So, we know that p a, b. Similarly, we have q a, b.

Without loss of generality, we assume that p a and q b. Claim that f is strictly 

decreasing on a, b. 

Suppose NOT, then there exist x, y a, b, with x y such that fx fy. ("" 

does not hold since f is 1-1.) Consider x, y and by above method, we know that f| x,y has 

the maximum at y, andf| a,y has the minimum at y. Then it implies that there exists 

By; x, y such that f is constant on By; x, y, which contradicts to 1-1. Hence, 

for any x y a, b, wehavefx fy. ("" does not hold since f is 1-1.) So, we 

have proved that f is strictly decreasing on a, b. 

Reamrk: 1. Here is another proof by Exercise 4.61. It suffices to show that 1-1 and 

continuity imply that f does not have a local maximum or a local minimum at any interior 

point. 

Proof: Suppose NOT, it means that f has a local extremum at some interior point x. 

Without loss of generality, we assume that f has a local minimum at the interior point x. 

Since x is an interior point of a, b, then there exists an open interval 

x , x a, b such that fy fx for all y x , x . Note that f is 1-1, so 

we have fy fx for all y x , x x. Choose y 1 x and 

y 2 x, x , then we have fy 1 fx and fy 2 fx. And thus choose r so that 

fy 1 r fx fp r, wherep y 1 , x by Intermediate Value Theorem, 

fy 2 r fx fq r, whereq x, y 2 by Intermediate Value Theorem, 

which contradicts to the hypothesis that f is 1-1. Hence, we have proved that 1-1 and 

continuity imply that f does not have a local maximum or a local minimum at any interior 

point. 

2. Under the assumption of continuity on a compact interval, one-to-one is 

equivalent to being strictly monotonic. 

Proof: By the exercise, we know that an one-to-one and continuous function defined 

on a compact interval implies that a strictly monotonic function. So, it remains to show that 

a strictly monotonic function implies that an one-to-one function. Without loss of 

generality, let f be increasing on a, b, then as fx fy, we must have x y since if 

x y, then fx fy and if x y, then fx fy. So, we have proved that a strictly 

monotonic function implies that an one-to-one function. Hence, we get that under the 

assumption of continuity on a compact interval, one-to-one is equivalent to being strictly 

monotonic. 

4.63 Let f be an increasing function defined on a, b and let x 1 ,..,x n be n points in 

the interior such that a x 1 x 2 ... x n b. 

n 

(a) Show that k1 

fx k fx k fb fa . 

Proof: Let a x 0 and b x n1 ;sincef is an increasing function defined on a, b, we 

know that both fx k and fx k exist for 1 k n. Assume that y k x k , x k1 , then 

we have fy k fx k and fx k1 fy k1 . Hence, 

n 

 

k1 

fx k fx k fy k fy k1 

n 

k1 

fy n fy 0 

fb fa . 

(b) Deduce from part (a) that the set of dicontinuities of f is countable.

Proof: Let D denote the set of dicontinuities of f. Consider 

D m x a, b : fx fx m 1 , then D 

m1 D m . Note that #D m , so 

we have D is countable. That is, the set of dicontinuities of f is countable. 

(c) Prove that f has points of continuity in every open subintervals of a, b. 

Proof: By (b), f has points of continuity in every open subintervals of a, b, since 

every open subinterval is uncountable. 

Remark: (1) Here is another proof about (b). Denote Q x 1 ,...,x n ,..., and let x be 

a point at which f is not continuous. Then we have fx fx 0. (If x is the end 

point, we consider fx fx 0orfx fx 0) So, we have an open interval I x 

such that I x fa, b fx. The interval I x contains infinite many rational numbers, 

we choose the smallest index, say m mx. Then the number of the set of discontinuities 

of f on a, b is a subset of N. Hence, the number of the set of discontinuities of f on a, b 

is countable. 

(2) There is a similar exercise; we write it as a reference. Let f be a real valued function 

defined on 0, 1. Suppose that there is a positive number M having the following 

condition: for every choice of a finite number of points x 1 ,..,x n in 0, 1, wehave 

n 

M i1 

x i M. Prove that S : x 0, 1 : fx 0 is countable. 

Proof: Consider S n x 0, 1 : |fx| 1/n, then it is clear that every S n is 

countable. Since S 

n1 S n , we know that S is countable. 

4.64 Give an example of a function f, defined and strictly increasing on a set S in R, 

such that f 1 is not continuous on fS. 

Solution: Let 

x if x 0, 1, 

fx 

1ifx 2. 

Then it is clear that f is strictly increasing on 0, 1, sof has the incerse function 

x if x 0, 1, 

f 1 x 

2ifx 1. 

which is not continuous on fS 0, 1. 

Remark: Compare with Exercise 4.65. 

4.65 Let f be strictly increasing on a subset S of R. Assume that the image fS has one 

of the following properties: (a) fS is open; (b) fS is connected; (c) fS is closed. Prove 

that f must be continuous on S. 

Proof: (a) Given a S, then fa fS. Given 0, wewantofinda 0 such 

that as x Ba; S, wehave|fx fa| . SincefS is open, then there exists 

Bfa, fS, where . 

Claim that there exists a 0 such that fBa; S Bfa, . Choose 

y 1 fa /2 and y 2 fa /2, then y 1 fx 1 and y 2 fx 2 ,wehave 

x 1 a x 2 since f is strictly increasing on S. Hence, for x x 1 , x 2 S, wehave 

fx 1 fx fx 2 since f is strictly increasing on S. So,fx Bfa, .Let 

mina x 1 , x 2 a, then Ba; S a , a S x 1 , x 2 S which implis 

that fBa; S Bfa, .( Bfa, ) 

Hence we have prove the claim, and the claim tells us that f is continuous at a. Sincea 

is arbitrary, we know that f is continuous on S.

(b) Note that since fS R, andfS is connected, we know that fS is an interval I. 

Given a S, then fa I. We discuss 2 cases as follows. (1) fa is an interior point of I. 

(2) fa is the endpoint of I. 

For case (1), it is similar to (a). We omit the proof. 

For case (2), it is similar to (a). We omit the proof. 

So, we have proved that f is continuous on S. 

(c) Given a S, then fa fS. SincefS is closed, we consider two cases. (1) fa 

is an isolated point and (2) fa is an accumulation point. 

For case (1), claim that a is an isolated point. Suppose NOT, there is a sequence 

 

x n S with x n a. Consider x n n1 

x : x n a x : x n a, and thus we 

may assume that x : x n a : a n is a infinite subset of x n 

n1 

.Sincef is 

monotonic, we have lim n fx n fa . SincefS is closed, we have fa fS. 

Therefore, there exists b fS such that fa fb fa. 

If fb fa, then b a since f is strictly increasing. But is contradicts to that fa is 

isolated. On the other hand, if fb fa, then b a since f is strictly increasing. In 

addition, fa n fa fb implies that a n b. But is contradicts to that a n a. 

Hence, we have proved that a is an isolated point. So, f is sutomatically continuous at 

a. 

For case (2), suppose that fa is an accumulation point. Then Bfa; fS and 

Bfa; has infinite many numbers of points in fS. Choose y 1 , y 2 Bfa; fS 

with y 1 y 2 , then fx 1 y 1 ,andfx 2 y 2 . And thus it is similar to (a), we omit the 

proof. 

So, we have proved that f is continuous on S by (1) and (2). 

Remark: In (b), when we say f is monotonic on a subset of R, its image is also in R. 

Supplement. 

It should be noted that the discontinuities of a monotonic function need not be isolated. 

In fact, given any countable subset E of a, b, which may even be dense, we can 

construct a function f, monotonic on a, b, discontinuous at every point of E, and at 

no other point of a, b. To show this, let the points of E be arranged in a sequence x n , 

n 1, 2, . . . Let c n be a sequence of positive numbers such that c n converges. Define 

fx c n a x b 

x nx 

Note: The summation is to be understood as follows: Sum over those indices n for whcih 

x n x. If there are no points x n to the left of x, the sum is empty; following the usual 

convention, we define it to be zero. Since absolute convergence, the order in which the 

terms are arranged is immaterial. 

Then fx is desired. 

The proof that we omit; the reader should see the book, Principles of Mathematical 

Analysis written by Walter Rudin, pp97. 

Metric space and fixed points 

4.66 Let BS denote the set of all real-valued functions which are defined and 

bounded on a nonempty set S. Iff BS, let 

f sup 

xS 

The number f is called the " sup norm " of f. 

|fx|.

(a) Provet that the formula df, g f g defines a metric d on BS. 

Proof: We prove that d isametriconBS as follows. 

(1) If df, g 0, i.e., f g sup xS |fx gx| 0 |fx gx| for all x S. 

So, we have f g on S. 

(2) If f g on S, then |fx gx| 0 for all x S. Thatis,f g 0 df, g. 

(3) Given f, g BS, then 

df, g f g 

sup|fx gx| 

xS 

sup|gx fx| 

xS 

g f 

dg, f. 

(4) Given f, g, h BS, thensince 

|fx gx| |fx hx| |hx gx|, 

we have 

|fx gx| 

sup 

xS 

|fx hx| sup|hx gx| 

f h h g 


f g sup|fx gx| f h h g. 

xS 

So,wehaveprovethatd isametriconBS. 

(b) Prove that the metric space BS, d is complete. Hint: If f n is a Cauchy 

sequence in BS, show that f n x is a Cauchy sequence of real numbers for each x in S. 

Proof: Let f n be a Cauchy sequence on BS, d, That is, given 0, there is a 

positive integer N such that as m, n N, wehave 

df, g f n f m sup|f n x f m x| . * 

xS 

So, for every point x S, the sequence f n x R is a Cauchy sequence. Hence, the 

sequence f n x is a convergent sequence, say its limit fx. It is clear that the function 

fx is well-defined. Let 1 in (*), then there is a positive integer N such that as 

m, n N, wehave 

|f n x f m x| 1, for all x S. ** 

Let m , andn N, wehaveby(**) 

|f N x fx| 1, for all x S 


|fx| 1 |f N x|, for all x S. 

Since |f N x| BS, say its bound M, and thus we have 

|fx| 1 M, for all x S 

which implies that fx is bounded. That is, fx BS, d. Hence, we have proved that 

BS, d is a complete metric space. 

Remark: 1. We do not require that S is bounded. 

2. The boundedness of a function f cannot be remove since sup norm of f is finite. 

xS

3. The sup norm of f, often appears and is important; the reader should keep it in mind. 

And we will encounter it when we discuss on sequences of functions. Also, see Exercise 

4.67. 

4. Here is an important theorem, the reader can see the definition of uniform 

convergence in the text book, page 221. 

4.67 Refer to Exercise 4.66 and let CS denote the subset of BS consisting of all 

funtions continuous and bounded on S, where now S is a metric space. 

(a) Prove that CS is a closed subset of BS. 

Proof: Let f be an adherent point of CS, then Bf; r CS for all r 0. So, 

there exists a sequence f n x such that f n f as n . So, given 0, there is a 

positive integer N such that as n N, wehave 

df n , f f n f sup|f n x fx| . 

xS 

So, we have 

|f N x fx| . for all x S. * 

Given s S, and note that f N x CS, so for this , there exists a 0 such that as 

|x s| , x, s S, wehave 

|f N x f N s| . ** 

We now prove that f is continuous at s as follows. Given 0, and let /3, then there 

is a 0 such that as |x s| , x, s S, wehave 

|fx fs| |fx f N x| |f N x f N s| |f N s fs| 

/3 /3 /3 by (*) and (**) 

. 

Hence, we know that f is continuous at s, and since s is arbitrary, we know that f is 

continuous on S. 

(b) Prove that the metric subspace CS is complete. 

Proof: By (a), we know that CS is complete since a closed subset of a complete 

metric space is complete. 

Remark: 1. In (b), we can see Exercise 4.9. 

2. The reader should see the text book in Charpter 9, and note that Theorem 9.2 and 

Theorem 9.3. 

4.68 Refer to the proof of the fixed points theorem (Theorem 4.48) for notation. 

(a) Prove that dp, p n dx, fx n /1 . 

Proof: The statement is that a contraction f of a complete metric space S has a unique 

fixed point p. Take any point x S, and consider the sequence of iterates: 

x, fx, ffx,... 

That is, define a sequence p n inductively as follows: 

p 0 x, p n1 fp n n 0, 1, 2, . . . 

We will prove that p n converges to a fixed point of f. First we show that p n is a 

Cauchy sequence. Since f is a contraction (dfx, fy dx, y, 0 1 for all 

x, y S), we have 

dp n1 , p n dfp n , fp n1 dp n , p n1 ,

so, by induction, we find 

dp n1 , p n n dp 1 , p 0 n dx, fx. 

Use the triangel inequality we find, for m n, 

m1 

dp m , p n dp k1 , p k 

kn 

m1 

dx, fx k 

kn 

dx, fx n m 

1 

dx, fx 

n 

1 . * 

Since n 0asn , we know that p n is a Cauchy sequence. And since S is 

complete, we have p n p S. The uniqueness is from the inequality, 

dfx, fy dx, y. 

From (*), we know that (let m ) 

dp, p n dx, fx n 

1 . 

This inequality, which is useful in numberical work, provides an estimate for the 

distance from p n to the fixed point p. An example is given in (b) 

(b) Take fx 1 x 2/x, S 1, . Prove that f is contraction of S with 

2 

contraction constant 1/2 and fixed point p 2 . Form the sequence p n starting wth 

x p 0 1 and show that p n 2 2 n . 

Proof: First, fx fy 1 x 2/x 1 y 2/y 1 yx 

x y 2 

2 2 2 xy , then we 

have 

|fx fy| 1 y x 

x y 2 

2 xy 

1 2 x y 1 xy 

2 

1 2 |x y| since 1 2 xy 1. 

So, f is a contraction of S with contraction constant 1/2. By Fixed Point Theorem, we 

know that there is a unique p such that fp p. Thatis, 

1 

2 p 2 p p p 2 .( 2 is not our choice since S 1, .) 

By (a), it is easy to know that 

p n 2 2 n . 

Remark: Here is a modefied Fixed Point Theorem: Let f be function defined on a 

complete metric space S. If there exists a N such that df N x f N y dx, y for all 

x, y S, where 0 1. Then f has a unique fixed point p S. 

Proof: Since f N is a contraction defined on a complete metric space, with the 

contraction constant , with 0 1, by Fixed Point Theorem, we know that there 

exists a unique point p S, such that

f N p p 

ff N p fp 

f N fp fp. 

That is, fp is also a fixed point of f N . By uniqueness, we know that fp p. In addition, 

if there is p S such that fp p . Then we have 

f 2 p fp p ,...,f N p .. p . Hence, we have p p .Thatis,f has a unique 

fixed point p S. 

4.69 Show by counterexample that the fixed-point theorem for contractions need not 

hold if either (a) the underlying metric space is not complete, or (b) the contraction 

constant 1. 

Solution: (a) Let f 1 1 x : 0, 1 R, then 2 |fx fy| 1 2 

|x y|. So,f is a 

contraction on 0, 1. However, it has no any fixed point since if it has, say this point p, we 

get 1 1 p p p 1 0, 1. 

2 

(b) Let f 1 x : 0, 1 R, then |fx fy| |x y|. So,f is a contraction with 

the contraction constant 1. However, it has no any fixed point since if it has, say this point 

p, weget1 p p 1 0, a contradiction. 

4.70 Let f : S S be a function from a complete metric space S, d into itself. 

Assume there is a real sequence a n which converges to 0 such that 

df n x, f n y n dx, y for all n 1 and all x, y in S, wheref n is the nth iterate of f; that 

is, 

f 1 x fx, f n1 x ff n x for n 1. 

Prove that f has a unique point. Hint. Apply the fixed point theorem to f m for a suitable m. 

Proof: Since a n 0, given 1/2, then there is a positive integer N such that as 

n N, wehave 

|a n| 1/2. 

Note that a n 0 for all n. Hence, we have 

df N x, f N y 1 dx, y for x, y in S. 

2 

That is, f N x is a contraction defined on a complete metric space, with the contraction 

constant 1/2. By Fixed Point Theorem, we know that there exists a unique point p S, 

such that 

f N p p 

ff N p fp 

f N fp fp. 

That is, fp is also a fixed point of f N . By uniqueness, we know that fp p. In addition, 

if there is p S such that fp p . Then we have 

f 2 p fp p ,...,f N p .. p . Hence, we have p p .Thatis,f has a unique 

fixed point p S. 

4.71 Let f : S S be a function from a metric space S, d into itself such that 

dfx, fy dx, y 

where x y. 

(a) Prove that f has at most one fixed point, and give an example of such an f with no 

fixed point.

Proof: If p and p are fixed points of f where p p , then by hypothesis, we have 

dp, p dfp, fp dp, p 

which is absurb. So, f has at most one fixed point. 

Let f : 0, 1/2 0, 1/2 by fx x 2 . Then we have 

|fx fy| |x 2 y 2 | |x y||x y| |x y|. 

However, f has no fixed point since if it had, say its fixed point p, then 

p 2 p p 1 0, 1/2 or p 0 0, 1/2. 

(b) If S is compact, prove that f has exactly one fixed point. Hint. Show that 

gx dx, fx attains its minimum on S. 

Proof: Let g dx, fx, and thus show that g is continuous on a compact set S as 

follows. Since 

dx, fx dx, y dy, fy dfy, fx 

dx, y dy, fy dx, y 

2dx, y dy, fy 

dx, fx dy, fy 2dx, y * 

and change the roles of x, andy, wehave 

dy, fy dx, fx 2dx, y ** 

Hence, by (*) and (**), we have 

|gx gy| |dx, fx dy, fy| 2dx, y for all x, y S. *** 

Given 0, there exists a /2 such that as dx, y , x, y S, wehave 

|gx gy| 2dx, y by (***). 

So, we have proved that g is uniformly continuous on S. 

So, consider min xS gx gp, p S. We show that gp 0 dp, fp. Suppose 

NOT, i.e., fp p. Consider 

df 2 p, fp dfp, p gp 

which contradicts to gp is the absolute maximum. Hence, gp 0 p fp. Thatis, 

f has a unique fixed point in S by (a). 

(c)Give an example with S compact in which f is not a contraction. 

Solution: Let S 0, 1/2, andf x 2 : S S. Then we have 

|x 2 y 2 | |x y||x y| |x y|. 

So, this f is not contraction. 

Remark: 1. In (b), the Choice of g is natural, since we want to get a fixed point. That 

is, fx x. Hence, we consider the function g dx, fx. 

2. Here is a exercise that makes us know more about Remark 1. Let f : 0, 1 0, 1 

be a continuous function, show that there is a point p such that fp p. 

Proof: Consider gx fx x, then g is a continuous function defined on 0, 1. 

Assume that there is no point p such that gp 0, that is, no such p so that fp p. So, 

by Intermediate Value Theorem, we know that gx 0 for all x 0, 1, orgx 0 

for all x 0, 1. Without loss of generality, suppose that gx 0 for all x 0, 1 which 

is absurb since g1 f1 1 0. Hence, we know that there is a point p such that 

fp p.

3. Here is another proof on (b). 

Proof: Given any point x S, and thus consider f n x S. Then there is a 

convergent subsequence f nk x, say its limit p, sinceS is compact. Consider 

dfp, p d f limf nk x , limf nk x 

k k 

and 

Note that 

d limff nk x, limf nk x by continuity of f at p 

k k 

limdf nk1 x, f nk x 1 

k 

df nk1 x, f nk x ... df 2 f nk1 x, ff nk1 x. 2 

limdf 2 f nk1 x, ff nk1 x 

k 

d limf 2 f nk1 x, limff nk1 x 

k k 

d f 2 limf nk1 x , f limf nk1 x by continuity of f 2 and f at p 

k k 

df 2 p, fp. 3 

So, by (1)-(3), we know that 

fp, fp df 2 p, fp p fp 

by hypothesis 

dfx, fy dx, y 

where x y. Hence, f has a unique fixed point p by (a) in Exercise. 

Note. 1.Ifx n x, andy n y, then dx n , y n dx, y. Thatis, 

lim n 

dx n , y n d lim n 

x n , lim n 

y n . 


dx n , y n dx n , x dx, y dy, y n and 

dx, y dx, x n dx n , y n dy n , y, 

then 

|dx n , y n dx, y| dx, x n dy, y n 0. 

So,wehaveproveit. 

2. The reader should compare the method with Exercise 4.72. 

4.72 Assume that f satisfies the condition in Exercise 4.71. If x S, letp 0 x, 

p n1 fp n ,andc n dp n , p n1 for n 0. 

(a) Prove that c n is a decreasing sequence, and let c limc n . 


c n1 c n dp n1 , p n2 dp n , p n1 

dfp n , fp n1 dp n , p n1 

dp n , p n1 dp n , p n1 

0, 

so c n is a decreasing sequence. And c n has a lower bound 0, by Completeness of R, 

we know that c n is a convergent sequence, say c limc n .

(b) Assume there is a subsequence p kn which converges to a point q in S. Prove that 

c dq, fq dfq, ffq. 

Deduce that q is a fixed point of f and that p n q. 

Proof: Since lim n p kn q, and lim n c n c, wehavelim n c kn c. So,we 

consider 

c lim n 

c kn 

lim n 

dp kn , p kn1 

lim n 

dp kn , fp kn 

dq, fq 

and 

dp kn , p kn1 dp kn1 , p kn ... dfp kn1 , f 2 p kn1 , 

we have 

c dq, fq lim n 

dfp kn1 , f 2 p kn1 dfq, f 2 q. * 

So, by (*) and hypoethesis 

dfx, fy dx, y 

where x y, we know that q fq c 0, in fact, this q is a unique fixed point. . 

In order to show that p n p, we consider (let m kn) 

dp m , q dp m , fq dp m1 , q .. dp kn , q 

So, given 0, there exists a positive integer N such that as n N, wehave 

dp kn , q . 

Hence, as m kN, wehave 

dp m , q . 

That is, p n p.

Derivatives 

Real-valued functions 

In each following exercise assume, where mecessary, a knowledge of the formulas for 

differentiating the elementary trigonometric, exponential, and logarithmic functions. 

5.1 Assume that f is said to satisfy a Lipschitz condition of order at c if there 

exists a positive number M (which may depend on c) and 1 ball Bc such that 

|fx fc| M|x c| 

whenever x Bc, x c. 

(a) Show that a function which satisfies a Lipschitz condition of order is continuous 

at c if 0, and has a derivative at c if 1. 

Proof: 1. As 0, given 0, there is a /M 1/ such that as 

x c , c Bc, wehave 

|fx fc| M|x c| M . 

So, we know that f is continuous at c. 

2. As 1, consider x Bc, andx c, wehave 

fx fc 

x c M|x c| 1 0asx c. 

So, we know that f has a derivative at c with f c 0. 

Remark: It should be note that (a) also holds if we consider the higher dimension. 

(b) Given an example of a function satisfying a Lipschitz condition of order 1 at c for 

which f c does not exist. 

Solution: Consider 

||x| |c|| |x c|, 

we know that |x| is a function satisfying a Lipschitz condition of order 1 at 0 for which 

f 0 does not exist. 

5.2 In each of the following cases, determine the intervals in which the function f is 

increasing or decreasing and find the maxima and minima (if any) in the set where each f is 

defined. 

(a) fx x 3 ax b, x R. 

Solution: Since f x 3x 2 a on R, we consider two cases: (i) a 0, and (ii) a 0. 

(i) As a 0, we know that f is increasig on R by f 0onR. In addition, if f has a 

local extremum at some point c, then f c 0. It implies that a 0andc 0. That is, 

fx x 3 b has a local extremum at 0. It is impossible since x 3 does not. So, we know 

that f has no maximum and minimum. 

(ii) As a 0, since f 3x 2 a 3 x a/3 x a/3 , we know that 

f x : 

, a/3 a/3 , a/3 a/3 , 

0 0 0 


fx : 

, a/3 a/3 , a/3 a/3 , 

 

. *

Hence, f is increasing on , a/3 and a/3 , , and decreasing on 

a/3 , a/3 . In addition, if f has a local extremum at some point c, then f c 0. 

It implies that c a/3 . With help of (*), we know that fx has a local maximum 

f a/3 and a local minimum f a/3 . 

(b) fx logx 2 9, |x| 3. 

Solution: Since f x 

2x , |x| 3, we know that 

x 2 9 


f x : 

, 3 3, 

0 0 

, 3 3, 

fx : 

. 

 

Hence, f is increasing on 3, , and decreasing on , 3. It is clear that f cannot have 

local extremum. 

(c) fx x 2/3 x 1 4 ,0 x 1. 

Solution: Since f x 2x13 

3x 1/3 


7x 1, 0 x 1, we know that 

f x : 

0, 1/7 1/7, 1 

0 0 

0, 1/7 1/7, 1 

fx : 

. ** 

 

Hence, we know that f is increasing on 0, 1/7, and decreasing on 1/7, 1. In addition, if f 

has a local extremum at some interior point c, then f c 0. It implies that c 1/7. With 

help of (**), we know that f has a local maximum f1/7, and two local minima f0, and 

f1. 

Remark: f has the absolute maximum f1/7, and the absolute minima 

f0 f1 0. 

(d) fx sin x/x if x 0, f0 1, 0 x /2. 

Sulotion: Since f x cosx xtan x as 0 x /2, and f 

x 2 

0 0, in addition, 

f x 0asx 0 by L-Hospital Rule, we know that 


f x : 

0, /2 

0 

0, /2 

fx : . (***) 

 

Hence, we know that f is decreasing on 0, /2. In addition, note that there is no interior 

point c such that f c 0. With help of (***), we know that f has local maximum f0, 

and local minimum f/2. 

Remark: 1. Here is a proof on f 0 :Since

lim sin x 1 

x0 x 

 

lim 

x0 

 

2sin x/2 2 

x 0, 

we know that f 0 0. 

2. f has the absolute maximum f0, and the absolute minimum f/2. 

5.3 Find a polynomial f of lowest possible degree such that 

fx 1 a 1 , fx 2 a 2 , f x 1 b 1 , f x 2 b 2 

where x 1 x 2 and a 1 , a 2 , b 1 , b 2 are given real numbers. 

Proof: It is easy to know that the lowest degree is at most 3 since there are 4 unknows. 

The degree is depends on the values of a 1 , a 2 , b 1 , b 2 . 

5.4 Define f as follows: fx e 1/x2 if x 0, f0 0. Show that 

(a) f is continuous for all x. 

Proof: In order to show f is continuous on R, it suffices to show f is continuous at 0. 

Since 

1/x 2 

limfx lim 

x0 x0 

1 e 0 f0, 

we know that f is continuous at 0. 

(b) f n is continuous for all x, andf n 0 0, (n 1,2,...) 

Proof: In order to show f n is continuous on R, it suffices to show f n is continuous at 

0. Note that 

px 

lim 

x e x 0, where px is any real polynomial. * 

Claim that for x 0, we have f n x e 1/x2 P 3n 1/x, whereP 3n t is a real polynomial of 

degree 3n for all n 1,2,.... As n 0, f 0 x fx e 1/x2 e 1/x2 P 0 1/x, where 

P 0 1/x is a constant function 1. So, as n 0, it holds. Suppose that n k holds, i.e., 

f k x e 1/x2 P 3k 1/x, whereP 3k t is a real polynomial of degree 3k. Consider 

n k 1, we have 

f k1 x f k x 

** 

e 1/x2 P 3k 1/x by induction hypothesis 

e 1/x2 2 1 x 

3 2 

P3k 

1 x 1 x P3k 

1 x . 

Since 2t 3 P 3k t t 2 P 3k t is a real polynomial of degree 3k 3, we define 

2t 3 P 3k t t 2 P 3k t P 3k3 t, and thus we have by (**) 

f k1 x e 1/x2 P 3k3 1/x. 

So, as n k 1, it holds. Therefore, by Mathematical Induction, we have proved the 

claim. 

Use the claim to show that f n 0 0, (n 1,2,...) as follows. As n 0, it is trivial 

by hypothesis. Suppose that n k holds, i.e., f k 0 0. Then as n k 1, we have

f k x f k 0 

x 0 

fk x 

x by induction hypothesis 

P 

e1/x2 3k 1/x 

x 

tP 3kt 

(let t 1/x) 

 

e t2 

tP 3k t 

e t 

e t 

e t2 0ast ( x 0) by (*). 

Hence, f k1 0 0. So, by Mathematical Induction, we have proved that f n 0 0, 

(n 1,2,...). 

Since 

lim 

x0 

f n x lim 

x0 

e 1/x2 P 3n 1/x 

lim 

x0 

lim t 

P 3n 1/x 

e 1/x2 

P 3n t 

e t 

0by(*) 

f n 0, 

we know that f n x is continuous at 0. 

Remark: 1. Here is a proof on (*). Let Px be a real polynomial of degree n, and 

choose an even number 2N n. We consider a Taylor Expansion with Remainder as 

follows. Since for any x, wehave 

2N1 

e x 1k! x k e x,0 

2N1 

2N 2! x2N2 1k! x k , 

k0 

k0 

then 

0 Px Px 

e x 

0asx 

2N1 1 

k! xk 

k0 

2N1 

since degPx n deg 1 

k0 k! xk 2N 1. By Sandwich Theorem, wehave 

proved 

Px 

lim 

x e x 0. 

2. Here is another proof on f n 0 0, (n 1,2,...). By Exercise 5.15, it suffices to 

show that lim x0 f n x 0. For the part, we have proved in this exercise. So, we omit the 

proof. Exercise 5.15 tells us that we need not make sure that the derivative of f at 0. The 

reader should compare with Exercise 5.15 and Exercise 5.5. 

3. In the future, we will encounter the exercise in Charpter 9. The Exercises tells us one 

important thing that the Taylor’s series about 0 generated by f converges everywhere 

on R, but it represents f only at the origin. 

5.5 Define f, g, andh as follows: f0 g0 h0 0 and, if x 0, 

fx sin1/x, gx x sin1/x, hx x 2 sin1/x. Show that 

(a) f x 1/x 2 cos1/x, ifx 0; f 0 does not exist. 

e t 

e t2

Proof: Trivially, f x 1/x 2 1 

cos1/x, ifx 0. Let x n 

2n 1 2 

consider 

fx n f0 

sin1/x n 

x n 0 x n 

2n 1 as n . 

2 

Hence, we know that f 0 does not exist. 

(b) g x sin1/x 1/x cos1/x, ifx 0; g 0 does not exist. 

, and thus 

Proof: Trivially, g 1 

x sin1/x 1/x cos1/x, ifx 0. Let x n ,and 

2n 1 2 

y n 1 , we know that 2n 

gx n g0 

sin 1 

x n 0 xn 

1 for all n 

and 

gy n g0 

sin 1 

y n 0 yn 

0 for all n. 

Hence, we know that g 0 does not exist. 

(c) h x 2x sin1/x cos1/x, ifx 0; h 0 0; lim x0 h x does not exist. 

Proof: Trivially, h x 2x sin1/x cos1/x, ifx 0. Consider 

hx h0 

|x sin1/x| |x| 0asx 0, 

x 0 

so we know that h 1 

0 0. In addition, let x n ,andy 

2n 1 n 1 ,wehave 

2n 

2 

h x n 2 

2n 1 and h y n 1 for all n. 

2 

Hence, we know that lim x0 h x does not exist. 

5.6 Derive Leibnitz’s formula for the nth derivative of the product h of two functions 

f and g : 

n 

h n kn f k g nk x, where kn n! 

k!n k! . 

k0 

Proof: We prove it by mathematical Induction. As n 1, it is clear since 

h f g g f. Suppose that n k holds, i.e., h k k 

j0 

jk f j g kj x. Consider 

n k 1, we have

h k1 h k 

k 

jk f j g kj x 

j0 

k 

jk f j g kj x 

j0 

k 

jk f j1 g kj f j g kj1 

j0 

 

 

 

k1 j0 

jk f j1 g kj f k1 g 0 

k 

j1 

jk f j g kj1 f 0 g k1 

 

k1 

 

j0 jk f j1 g kj f k1 g 0 

k1 

j0 

k 

j1 

f j1 g kj f 0 g k1 

k1 

 

j0 

k1 

 

j0 

jk 

k1 

j1 

k 

j1 

f j1 g kj f k1 g 0 f 0 g k1 

f j1 g kj f k1 g 0 f 0 g k1 

k1 

jk f j1 g kj . 

j0 

So, as n k 1, it holds. Hence, by Mathematical Induction, we have proved the 

Leibnitz formula. 

Remark: We use the famous formula called Pascal Theorem:n1 k1 kn n k1 , 

where 0 k n. 

5.7 Let f and g be two functions defined and having finite third-order derivatives f x 

and g x for all x in R. Iffxgx 1 for all x, show that the relations in (a), (b), (c), 

and (d) holds at those points where the denominators are not zero: 

(a) f x/fx g x/gx 0. 

Proof: Since fxgx 1 for all x, wehavef g g f 0 for all x. By hypothesis, we 

have 

f g g f 

0 for those points where the denominators are not zero 

fg 


f x/fx g x/gx 0. 

(b) f x/f x 2f x/fx g x/g x 0. 

Proof: Since f g g f 0 for all x, wehavef g g f f g 2f g g f 0. By 

hypothesis, we have

(c) f x 

f x 

3 f xg x 

fxg x 

3 f x 

fx 

0 f g 2f g g f 

f g 

f 

f 

f 

f 

g x 

g x 

2 g 

g 

2 f 

f 

0. 

g 

g 

g 

g 

f 

f 

by (a). 

Proof: By (b), we have f g 2f g g f 0 f g 3f g 3f g fg .By 

hypothesis, we have 

0 f g 3f g 3f g fg 

f g 

(d) f x 

f x 

3 2 

f x 

f x 

f 

f 

3 f g 

f g 3 g 

g fg 

f g 

f 

f 3f g 

f g 

f 

f 

2 

 

g x 

g x 

Proof: By(c),wehave f 

f 

f 

f 

we know that f 

f 

f x 

f x 

3 2 

f x 

f x 

f g 

fg 

g 

 

 

1 2 

1 2 

3 f 

f 

3 2 

3 f g 

fg 

g x 

g x 

g 

3 f 

g f 

f 

f 

f 

f 

3 g 2 

2 

g x 

3 g x 2 

f 

f 

f 

f 

f 

f 

 

 

f 2 

 

f 

 

3g 1 g g f 

f g 

g 

g 

f 2 

 

g 2 

f g 

g x 

g x 

2 

. 

g 

g 

2 

. 

f g 

fg 

f 

f 

g 

g 

g 2 

g 

by (a). 

.Since 

g 

g 

, 

f 

f 

 

g 

g 


by (b) 

Note. The expression which appears on the left side of (d) is called the Schwarzian 

derivative of f at x. 

(e) Show that f and g have the same Schwarzian derivative if 

gx afx b/cfx d, wheread bc 0. 

Hint. If c 0, write af b/cf d a/c bc ad/ccf d, and apply part 

(d). 

Proof: If c 0, we have g a f b . So, we have 

d d

g x 

g x 3 2 

a 

d f x 

a 

d f x 3 2 

f x 

f x 3 2 

g x 

g x 

2 

a 

d f x 

a 

d f x 

f x 

f x 

So, f and g have the same Schwarzian derivative. 

If c 0, write g af b/cf d a/c bc ad/ccf d, then 

cg a 1 cf d 1sincead bc 0. 

LetG cg a, andF 

1 

bcad 


F 

F 3 2 

bc ad 

cf d, then GF 1. It implies that by (d), 

F 

F 3 2 

F 

F 

2 

. 

2 

G 

G 3 2 

F 2 c 

 

bcad f 

3 F c 

 

bcad f 2 

f 

f 3 2 

G 

G 3 2 

cg 

cg 3 2 

f 2 

f 

G 2 

G 

2 

G 2 

G 

 

 

cg 2 

cg 

c 

2 

bcad f 

c 

bcad f 

g 

3 g 2 

. 

g 2 g 

So, f and g have the same Schwarzian derivative. 

5.8 Let f 1 , f 2 , g 1 , g 2 be functions having derivatives in a, b. DefineF by means of 

the determinant 

Fx 

f 1 x f 2 x 

g 1 x g 2 x 

,ifx a, b. 

(a) Show that F x exists for each x in a, b and that 

F x 

f 1 x f 2 x 

g 1 x g 2 x 

 

f 1 x f 2 x 

g 1 x g 2 x 

. 

f 1 x f 2 x 

Proof: SinceFx 

f 1 g 2 f 2 g 1 ,wehave 

g 1 x g 2 x 

F f 1 g 2 f 1 g 2 f 2 g 1 f 2 g 

1 

f 1 g 2 f 2 g 1 f 1 g 2 f 2 g 1 

f 

1 x f 2 x f 1 x f 2 x 

 

g 1 x g 2 x g 1 x g 2 x 

.

then 

(b) State and prove a more general result for nth order determinants. 

Proof: Claim that if 

F x 

f 11 f 12 ... f 

1n 

Fx 

f 21 f 22 ... f 2n 

 

... ... ... ... 

f n1 f n2 ... f nn 

f 11 f 12 ... f 1n 

f 21 f 22 ... f 2n 

, 

... ... ... ... 

f n1 f n2 ... f nn 

f 11 f 12 ... f 1n 

f 21 f 22 ... f 

2n 

... 

... ... ... ... 

f n1 f n2 ... f nn 

f 11 f 12 ... f 1n 

f 21 f 22 ... f 2n 

... ... ... ... 

f 

n1 

f n2 ... f 

nn 

We prove it by Mahematial Induction. As n 2, it has proved in (a). Suppose that n k 

holds, consider n k 1, 

f k11 f k12 ... f k1k1 

 

f 11 f 12 ... f 1k1 

f 21 f 22 ... f 2k1 

... ... ... ... 

 

f 12 ... f 1k1 

f 11 ... f 1k 

1 k11 f k11 ... ... ... ..1 k1k1 f k1k1 ... ... ... 

f k2 ... f kk1 f k1 ... f kk 

. . . . (The reader can write it down by induction hypothesis). 

Hence, by Mathematical Induction, wehaveprovedit. 

Remark: The reader should keep it in mind since it is useful in Analysis. For example, 

we have the following Theorem. 

(Theorem) Suppose that f, g, andh are continuous on a, b, and differentiable on 

a, b. Then there is a a, b such that 

Proof: Let 

f g h 

fa ga ha 

fb gb hb 

Fx 

fx gx hx 

fa ga ha 

fb gb hb 

then it is clear that Fx is continuous on a, b and differentiable on a, b since the 

operations on determinant involving addition, substraction, and multiplication without 

division. Consider 

Fa Fb 0, 

then by Rolle’s Theorem, we know that 

0. 

, 

.


F 0, where a, b, 

f g h 

fa ga ha 

fb gb hb 

0. * 

(Application- Generalized Mean Value Theorem) Suppose that f and g are 

continuous on a, b, and differentiable on a, b. Then there is a a, b such that 

fb fag f gb ga. 

Proof: Let hx 1, and thus by (*), we have 



 

f g 

fb gb 

f g 0 

fa ga 1 

fb gb 1 

 

0, 

f g 

fa ga 

fb fag f gb ga. 

Note: Use the similar method, we can show Mean Value Theorem by letting 

gx x, andhx 1. And from this viewpoint, we know that Rolle’s Theorem, Mean 

Value Theorem, and Generalized Mean Value Theorem are equivalent. 

5.9 Given n functions f 1 ,...,f n , each having nth order derivatives in a, b. A 

function W, called the Wronskian of f 1 ,...,f n , is defined as follows: For each x in a, b, 

Wx is the value of the determinant of order n whose element in the kth row and mth 

column is f k1 m x, wherek 1, 2, . . , n and m 1,2,...,n. [The expression f 0 m x is 

written for f m x.] 

(a) Show that W x can be obtained by replacing the last row of the determinant 

defining Wx by the nth derivatives f n 1 x,...,f n n x. 

Proof: Write 

Wx 

f 1 f 2 ... f n 

f 1 f 2 ... f 

n 

... ... ... ... 

f 1 

n1 

f 2 

n1 

... f n 

n1 

and note that if any two rows are the same, its determinant is 0; hence, by Exercise 5.8-(b), 

we know that 

, 

0

W x 

f 1 f 2 ... f n 

... ... ... ... 

f 1 

n2 

f 1 

n 

f 2 

n2 

f 2 

n 

... f n 

n2 

... f n 

n 

. 

(b) Assuming the existence of n constants c 1 ,...,c n , not all zero, such that 

c 1 f 1 x ...c n f n x 0 for every x in a, b, show that Wx 0 for each x in a, b. 

Proof: Since c 1 f 1 x ...c n f n x 0 for every x in a, b, wherec 1 ,...,c n , not all 

zero. Without loss of generality, we may assume c 1 0, we know that 

c 1 f k 1 x ...c n f k n x 0 for every x in a, b, where 0 k n. Hence, we have 

Wx 

f 1 f 2 ... f n 

f 1 f 2 ... f 

n 

... ... ... ... 

f 1 

n1 

f 2 

n1 

... f n 

n1 

0 

since the first column is a linear combination of other columns. 

Note. A set of functions satisfing such a relation is said to be a linearly dependent set 

on a, b. 

(c) The vanishing of the Wronskian throughout a, b is necessary, but not sufficient. 

for linear dependence of f 1 ,...,f n . Show that in the case of two functions, if the Wronskian 

vanishes throughout a, b and if one of the functions does not vanish in a, b, then they 

form a linearly dependent set in a, b. 

Proof: Let f and g be continuous and differentiable on a, b. Suppose that fx 0for 

all x a, b. Since the Wronskian of f and g is 0, for all x a, b, wehave 

fg f g 0 for all x a, b. * 

Since fx 0 for all x a, b, wehaveby(*), 

fg f g 

0 g 

0 for all x a, b. 

f 2 f 

Hence, there is a constant c such that g cf for all x a, b. Hence, f, g forms a 

linearly dependent set. 

Remark: This exercise in (b) is a impotant theorem on O.D.E. We often write (b) in 

other form as follows. 

(Theorem) Letf 1 ,..,f n be continuous and differentiable on an interval I. If 

Wf 1 ,...,f n t 0 0forsomet 0 I, then f 1 ,..,f n is linearly independent on I 

Note: If f 1 ,..,f n is linearly independent on I, ItisNOT necessary that 

Wf 1 ,...,f n t 0 0forsomet 0 I. For example, ft t 2 |t|, andgt t 3 . It is easy to 

check f, g is linearly independent on 1, 1. AndWf, gt 0 for all t 1, 1. 

Supplement on Chain Rule and Inverse Function Theorem. 

The following theorem is called chain rule, it is well-known that let f be defined on an 

open interval S, letg be defined on fS, and consider the composite function g f defined 

on S by the equation

g fx gfx. 

Assume that there is a point c in S such that fc is an interior point of fS. Iff is 

differentiable at c and g is differentiable at fc, then g f is differentiable at c, andwe 

have 

g f c g fcf c. 

We do not give a proof, in fact, the proof can be found in this text book. We will give 

another Theorem called The Converse of Chain Rule as follows. 

(TheConverseofChainRule) Suppose that f, g and u are related so that 

fx gux. Ifux is continuous at x 0 , f x 0 exists, g ux 0 exists and not zero. 

Then u x 0 is defined and we have 

f x 0 g ux 0 u x 0 . 

Proof: Sincef x 0 exists, and g ux 0 exists, then 

fx fx 0 f x 0 x x 0 o|x x 0 | * 

and 

gux gux 0 g ux 0 ux ux 0 o|ux ux 0 |. ** 

Since fx gux, andfx 0 gux 0 , by (*) and (**), we know that 

f 

ux ux 0 

x 0 

g ux 0 x x 0 o|x x 0 | o|ux ux 0 |. *** 

Note that since ux is continuous at x 0 , we know that o|ux ux 0 | 0asx x 0 . 

So, (***) means that u x 0 is defined and we have 

f x 0 g ux 0 u x 0 . 

Remark: The condition that g ux 0 is not zero is essential, for example, gx 1on 

1, 1 and ux |x|, wherex 0 0. 

(Inverse Function Theroem) Suppose that f is continuous, strictly monotonic function 

which has an open interval I for domain and has range J. (It implies that 

fgx x gfx on its corresponding domain.) Assume that x 0 is a point of J such 

that f gx 0 is defined and is different from zero. Then g x 0 exists, and we have 

g x 0 1 

f gx 0 . 

Proof: Itisaresultofthe converse of chain rule note that 

fgx x. 

Mean Value Theorem 

5.10 Given a function defined and having a finite derivative in a, b and such that 

lim xb 

fx . Prove that lim xb 

f x either fails to exist or is infinite. 

Proof: Suppose NOT, we have the existence of lim xb 

f x, denoted the limit by L. 

So, given 1, there is a 0 such that as x b , b we have 

|f x| |L| 1. * 

Consider x, a b , b with x a, then we have by (*) and Mean Value Theorem, 

|fx fa| |f x a| where a, x 

|L| 1|x a| 

which implies that

|fx| |fa| |L| 1 

which contradicts to lim xb 

fx . 

Hence, lim xb 

f x either fails to exist or is infinite. 

5.11 Show that the formula in the Mean Value Theorem can be written as follows: 

fx h fx 

f 

h 

x h, 

where 0 1. 

Proof: (Mean Value Theorem) Letf and g be continuous on a, b and differentiable 

on a, b. Then there exists a a, b such that fb fa f b a. Note that 

a b a, where 0 1. So, we have proved the exercise. 

Determine as a function of x and h, and keep x 0 fixed, and find lim h0 in each 

case. 

(a) fx x 2 . 


fx h fx 

h 


x h2 x 2 

h 

2x h 2x h f x h 

1/2. 

Hence, we know that lim h0 1/2. 

(b) fx x 3 . 


fx h fx 

x h3 x 3 

3x 

h 

h 

2 3xh h 2 3x h 2 f x h 


3x 9x2 9xh 3h 2 

3h 

Since 0 1, we consider two cases. (i) x 0, (ii) x 0. 

(i) As x 0, since 

0 3x 9x2 9xh 3h 2 

1, 

3h 

we have 

 

3x 9x 2 9xh3h 2 

3h 

if h 0, and h is sufficiently close to 0, 

3x 9x 2 9xh3h 2 

if h 0, and h is sufficiently close to 0. 

3h 

Hence, we know that lim h0 1/2 by L-Hospital Rule. 

(ii) As x 0, we have 

 

3x 9x 2 9xh3h 2 

3h 

if h 0, and h is sufficiently close to 0, 

3x 9x 2 9xh3h 2 

if h 0, and h is sufficiently close to 0. 

3h 

Hence, we know that lim h0 1/2 by L-Hospital Rule. 

From (i) and (ii), we know that as x 0, we have lim h0 1/2.

Remark: For x 0, we can show that lim h0 3 3 

as follows. 

Proof: Since 

we have 

 

0 3h2 

3h 

1, 

3h 2 

3h 

3 h 

3h 

3 3 

if h 0, 

3h 2 

3 h 3 if h 0. 

3h 3h 3 

Hence, we know that lim h0 3 . 3 

(c) fx e x . 



fx h fx 

h 

exh e x 

h 

e xh f x h 

log eh 1 

h 

. 

h 

Hence, we know that lim h0 1/2 since 

log eh 1 

h 

lim lim 

h0 h0 h 

lim e h h e h 1 by L-Hospital Rule. 

h0 he h 1 

Note that e h 1 h h2 

2 oh2 

lim 

h0 

1 

1/2. 

(d) fx log x, x 0. 


fx h fx 

h 


h o1 

2 

1 h oh 

2 

 

log1 h x 

h 

h 

x log1 h x 

h 

x log1 h x . 

1 

x h 

Since log1 t t t2 2 

ot2 ,wehave

h h 

x x 1 

lim lim 

h 2 x 2 o h x 2 

h0 h0 h h 

x x 1 h 2 x 2 o h x 2 

lim 

h0 

1 

lim 

2 h x 2 o h x 2 

h x 2 1 2 h x 3 o h x 3 

o1 

2 

1 1 h 2 x o h x 

h0 

1 

1/2. 

5.12 Take fx 3x 4 2x 3 x 2 1andgx 4x 3 3x 2 2x in Theorem 5.20. Show 

that f x/g x is never equal to the quotient f1 f0/g1 g0 if 0 x 1. How 

do you reconcile this with the equation 

fb fa 

gb ga 

obtainable from Theorem 5.20 when n 1 

Solution: Note that 

12x 2 6x 2 12 x 1 

4 

11 

48 

f x 1 

g x 1 , a x 1 b, 

x 1 

4 

11 

48 

, where (0 1 4 11 

48 1). 

So, when we consider 

f1 f0 

g1 g0 0 

and 

f x 12x 3 6x 2 2x xg x x12x 2 6x 2, 

we CANNOT write f x/g x x. Otherwise, it leads us to get a contradiction. 

Remark: It should be careful when we use Generalized Mean Value Theorem, we 

had better not write the above form unless we know that the denominator is not zero. 

5.13 In each of the following special cases of Theorem 5.20, take n 1, c a, x b, 

and show that x 1 a b/2. 

(a) fx sin x, gx cosx; 

Proof: Since, by Theorem 5.20, 

sin a sin b sinx 1 2cos a b sin 

2 

a 2 

b sinx 1 

cosa cosbcosx 1 

2sin a 2 

b sin a 2 

b cosx 1 , 

we find that if we choose x 1 a b/2, then both are equal. 

(b) fx e x , gx e x . 

Proof: Since, by Theorem 5.20, 

e a e b e x 1 e 

a 

e b e x 1 , 

we find that if we choose x 1 a b/2, then both are equal. 

Can you find a general class of such pairs of functions f and g for which x 1 will always 

be a b/2 and such that both examples (a) and (b) are in this class

Proof: Look at the Generalized Mean Value Theorem, we try to get something from 

the equality. 

fa fbg a 2 

b ga gbf a 2 

b , * 

if fx, andgx satisfy following two conditions, 

(i) f x gx and g x fx 

and 

(ii) fa fb f a 2 

b ga gb g a 2 

b , 

then we have the equality (*). 

5.14 Given a function f defined and having a finite derivative f in the half-open 

interval 0 x 1 and such that |f x| 1. Define a n f1/n for n 1, 2, 3, . . . , and 

show that lim n a n exists. 

Hint. Cauchy condition. 

Proof: Consider n m, and by Mean Value Theorem, 

|a n a m| |f1/n f1/m| |f p| 1 n m 1 1 n m 

1 

then a n is a Cauchy sequence since 1/n is a Cauchy sequence. Hence, we know that 

lim n a n exists. 

5.15 Assume that f has a finite derivative at each point of the open interval a, b. 

Assume also that lim xc f x exists and is finite for some interior point c. Prove that the 

value of this limit must be f c. 

Proof: It can be proved by Exercise 5.16; we omit it. 

5.16 Let f be continuous on a, b with a finite derivative f everywhere in a, b, 

expect possibly at c. If lim xc f x exists and has the value A, show that f c must also 

exist and has the value A. 

Proof: Consider, for x c, 

fx fc 

x c f where x, c or c, x by Mean Value Theorem, * 

since lim xc f x exists, given 0, there is a 0 such that as x c , c c, 

we have 

A f x A . 

So, if we choose x c , c c in (*), we then have 

fx fc 

A x c f A . 

That is, f c exists and equals A. 

Remark: (1) Here is another proof by L-Hospital Rule. Since it is so obvious that we 

omit the proof. 

(2) We should be noted that Exercise 5.16 implies Exercise 5.15. Both methods 

mentioned in Exercise 5.16 are suitable for Exercise 5.15. 

5.17 Let f be continuous on 0, 1, f0 0, f x defined for each x in 0, 1. Prove 

that if f is an increasing function on 0, 1, then so is too is the function g defined by the 

equation gx fx/x. 

Proof: Sincef is an increasing function on 0, 1, we know that, for any x 0, 1

f x fx 

x 

So, let x y, wehave 

f x 

fx f0 

x 0 

f x f 0where 0, x. * 

gx gy g zx y, wherey z x 

f zz fz 

x y 

0by(*) 

which implies that g is an increasing function on 0, 1. 

5.18 Assume f has a finite derivative in a, b and is continuous on a, b with 

fa fb 0. Prove that for every real there is some c in a, b such that 

f c fc. 

Hint. Apply Rolle’s Theorem to gxfx for a suitable g depending on . 

Proof: Consider gx fxe x , then by Rolle’s Theorem, 

ga gb g ca b, wherec a, b 

0 


f c fc. 

z 2 

Remark: (1) The finding of an auxiliary function usually comes from the equation that 

we consider. We will give some questions around this to get more. 

(2)There are some questions about finding auxiliary functions; we write it as follows. 

(i) Show that e e . 

Proof: (STUDY) Since log x is a strictly increasing on 0, , in order to show 

e e , it suffices to show that 

log e log e log e e log 


log e 

Consider fx logx 

x 

e 

: e, , wehave 

f x 1 log x 

x 2 

log 

. 

0wherex e, . 

So, we know that fx is strictly decreasing on e, . Hence, 

e e . 

(ii) Show that e x 1 x for all x R. 

loge 

e 

log 

 

.Thatis, 

Proof: By Taylor Theorem with Remainder Term, we know that 

e x 1 x ec 

2 x2 ,forsomec. 

So, we finally have e x 1 x for all x R. 

Note: (a) The method in (ii) tells us one thing, we can give a theorem as follows. Let 

f C 2n1 a, b, andf 2n x exists and f 2n x 0ona, b. Then we have 

2n1 

fx f 

k 

a 

. 

k! 

k0

Proof: By Generalized Mean Value Theorem, we complete it. 

(b) There are many proofs about that e x 1 x for all x R. We list them as a 

reference. 

(b-1) Let fx e x 1 x, and thus consider the extremum. 

(b-2) Use Mean Value Theorem. 

(b-3) Since e x 1 0forx 0ande x 1 0forx 0, we then have 

 

0 

x 

et 1dt 0and 

x 

0 

et 1dt 0. 

So, e x 1 x for all x R. 

(iii) Let f be continuous function on a, b, and differentiabel on a, b. Prove that there 

exists a c a, b such that 

f fc fa 

c . 

b c 

Proof: (STUDY) Since f c fcfa , we consider f 

bc 

cb c fc fa. 

Hence, we choose gx fx fab x, then by Rolle’s Theorem, 

ga gb g ca b where c a, b 

which implies thatf c fcfa . 

bc 

(iv) Let f be a polynomial of degree n, iff 0onR, then we have 

f f ..f n 0onR. 

Proof: Let gx f f ..f n , then we have 

g g f 0onR since f is a polynomial of degree n. * 

Consider hx gxe x , then h x g x gxe x 0onR by (*). It means that h 

is a decreasing function on R. Since lim x hx 0bythefactg is still a polynomial, 

then hx 0onR. Thatis,gx 0onR. 

(v) Suppose that f is continuous on a, b, fa 0 fb, and 

x 2 f x 4xf x 2fx 0 for all x a, b. Prove that fx 0ona, b. 

Proof: (STUDY) Since x 2 f x 4xf x 2fx x 2 fx by Leibnitz Rule, let 

gx x 2 fx, then claim that gx 0ona, b. 

Suppose NOT, there is a point p a, b such that gp 0. Note that since fa 0, 

and fb 0, So, gx has an absolute maximum at c a, b. Hence, we have g c 0. 

By Taylor Theorem with Remainder term, wehave 

gx gc g cx c g 

x c 2 ,where x, c or c, x 

2! 

gc since g c 0, and g x 0 for all x a, b 

0sincegc is absolute maximum. 

So, 

x 2 fx c 2 fc 0 ** 

which is absurb since let x a in (**). 

(vi) Suppose that f is continuous and differentiable on 0, , and 

lim x f x fx 0, show that lim x fx 0. 

Proof: Since lim x f x fx 0, then given 0, there is M 0 such that as 

x M, wehave

f x fx . 

So, as x M, wehave 

e x e M e M fM e x 

e x fx 

e x e x e M e M fM . 

If we let e x e M e M fM gx, ande x e M e M fM hx, then we have 

g x e x fx h x 

and 

gM e M fM hM. 

Hence, for x M, 

e x e M e M fM gx 

e x fx 

hx e x e M e M fM 

It implies that, for x M, 

e x e M e M fM fx e x e M e M fM 


lim fx 0since is arbitrary. 

x 

Note: Intheprocessofproof,weusetheresultonMean Value Theorem. Letf, g, and 

h be continuous on a, b and differentiable on a, b. Suppose fa ga ha and 

f x g x h x on a, b. Show that fx gx hx on a, b. 

Proof: By Mean Value theorem, wehave 

gx fx ga fa gx fx 

g c f c, wherec a, x. 

0 by hypothesis. 

So, fx gx on a, b. Similarly for gx hx on a, b. Hence, fx gx hx 

on a, b. 

(vii) Let fx a 1 sin x ...a n sin nx, wherea i are real for i 1, 2, . . n. Suppose that 

|fx| |x| for all real x. Prove that |a 1 ..na n| 1. 

Proof: Let x 0, and by Mean Value Theorem, we have 

|fx f0| |fx| |a 1 sin x ...a n sin nx| 

|f cx|, wherec 0, x 

|a 1 cosc ...na n cosncx| 

|x| by hypothesis. 

So, 

|a 1 cosc ...na n cosnc| 1 

Note that as x 0 ,wehavec 0 ; hence, |a 1 ..na n| 1. 

Note: Here are another type: 

(a) |sin 2 x sin 2 y| |x y| for all x, y. 

(b) |tan x tan y| |x y| for all x, y , . 2 2 

(viii) Let f : R R be differentiable with f x c for all x, wherec 0. Show that

there is a point p such that fp 0. 

Proof: ByMean Vaule Theorem, we have 

fx f0 f x 1 x f0 cx if x 0 

f0 f x 2 x f0 cx if x 0. 

So, as x large enough, we have fx 0andasx is smalle enough, we have fx 0. Since 

f is differentiable on R, it is continuous on R. Hence, by Intermediate Value Theorem, 

we know that there is a point p such that fp 0. 

(3) Here is another type about integral, but it is worth learning. Compare with (2)-(vii). 

If 

c 0 c 1 

... c n 

2 n 1 0, where c i are real constants for i 1, 2, . . n. 

Prove that c 0 ...c n x n has at least one real root between 0 and 1. 

Proof: Suppose NOT, i.e., (i) fx : c 0 ...c n x n 0 for all x 0, 1 or (ii) 

fx 0 for all x 0, 1. 

In case (i), consider 

1 

0 fxdx c0 c 1 

... c n 

0 2 n 1 0 

which is absurb. Similarly for case (ii). 

So, we know that c 0 ...c n x n has at least one real root between 0 and 1. 

5.19. Assume f is continuous on a, b and has a finite second derivative f in the 

open interval a, b. Assume that the line segment joining the points A a, fa and 

B b, fb intersects the graph of f in a third point P different from A and B. Prove that 

f c 0forsomec in a, b. 

Proof: Consider a straight line equation, called gx fa fbfa x a. Then 

ba 

hx : fx gx, we knwo that there are three point x a, p and b such that 

ha hp hb 0. 

So, by Mean Value Theorem twice, we know that there is a point c a, b such that 

h c 0 

which implies that f c 0sinceg is a polynomial of degree at least 1. 

5.20 If f has a finite third derivative f in a, b and if 

fa f a fb f b 0, 

prove that f c 0forsomec in a, b. 

Proof: Sincefa fb 0, we have f p 0wherep a, b by Rolle’s 

Theorem. Since f a f p 0, we have f q 1 0whereq 1 a, p and since 

f p f b 0, we have f q 2 0whereq 2 p, b by Rolle’s Theorem. Since 

f q 1 f q 2 0, we have f c 0wherec q 1 , q 2 by Rolle’s Theorem. 

5.21 Assume f is nonnegative and has a finite third derivative f in the open interval 

0, 1. Iffx 0 for at least two values of x in 0, 1, prove that f c 0forsomec in 

0, 1. 

Proof: Since fx 0 for at least two values of x in 0, 1, sayfa fb 0, where 

a, b 0, 1. By Rolle’s Theorem, we have f p 0wherep a, b. Note that f is 

nonnegative and differentiable on 0, 1, so both fa and fb are local minima, where a 

and b are interior to a, b. Hence, f a f b 0.

Since f a f p 0, we have f q 1 0whereq 1 a, p and since 

f p f b 0, we have f q 2 0whereq 2 p, b by Rolle’s Theorem. Since 

f q 1 f q 2 0, we have f c 0wherec q 1 , q 2 by Rolle’s Theorem. 

5.22 Assume f has a finite derivative in some interval a, . 

(a) If fx 1andf x c as x , prove that c 0. 

Proof: Consider fx 1 fx f y where y x, x 1 by Mean Value Theorem, 

since 

lim fx 1 

x 


limfx 1 fx 0 

x 

which implies that (x y ) 

lim 

x f y 0 y 

lim fy 

Since f x c as x , we know that c 0. 

Remark: (i) There is a similar exercise; we write it as follows. If fx L and 

f x c as x , prove that c 0. 

Proof: By the same method metioned in (a), we complete it. 

(ii) The exercise tells that the function is smooth; its first derivative is smooth too. 

(b) If f x 1asx , prove that fx/x 1asx . 

Proof: Given 0, we want to find M 0 such that as x M 

fx 

x 1 . 

Since f x 1asx , thengiven 3 ,thereisM 0 such that as x M ,we 

have 

|f x 1| 

3 |f x| 1 

3 

 

By Taylor Theorem with Remainder Term, 

fx fM f x M 

fx x fM f 1x f M , 

then for x M , 

fx 

x 1 fM 

x |f 1| f M 

x 

 

fM 

x 3 1 3 

Choose M 0 such that as x M M ,wehave 

fM 

x 

3 M 

and x 

 

M 

x by (*) 

* 

** 

/3 

1 3 . *** 

Combine (**) with (***), we have proved that given 0, there is a M 0 such that as 

x M, wehave 

fx 

x 1 . 

That is, lim x 

fx 

x 1.

Remark: If we can make sure that fx as x , we can use L-Hopital Rule. 

We give another proof as follows. It suffices to show that fx as x . 

Proof: Since f x 1asx , thengiven 1, there is M 0 such that as 

x M, wehave 

|f x| 1 1 2. 

Consider 

fx fM f x M 

by Taylor Theorem with Remainder Term, then 

lim fx since x f x is bounded for x M. 

(c) If f x 0asx , prove that fx/x 0asx . 

Proof: The method metioned in (b). We omit the proof. 

Remark: (i) There is a similar exercise; we write it as follows. If f x L as x , 

prove that fx/x L as x . The proof is mentioned in (b), so we omit it. 

(ii) It should be careful that we CANNOT use L-Hospital Rule since we may not have 

the fact fx as x . Hence, L-Hospital Rule cannot be used here. For example, f 

is a constant function. 

5.23 Let h be a fixed positive number. Show that there is no function f satisfying the 

following three conditions: f x exists for x 0, f 0 0, f x h for x 0. 

Proof: It is called Intermediate Value Theorem for Derivatives. (Sometimes, we 

also call this theorem Darboux.) See the text book in Theorem 5.16. 

(Supplement) 1. Suppose that a R, andf is a twice-differentiable real function on 

a, . LetM 0 , M 1 ,andM 2 are the least upper bound of |fx|, |f x|, and|f x|, 

respectively, on a, . Prove that M 12 4M 0 M 2 . 

Proof: Consider Taylor’s Theorem with Remainder Term, 

fa 2h fa f a2h f 

2h 2 ,whereh 0. 

2! 

then we have 

f a 

2h 1 fa 2h fa f h 


Since gh : M 0 

h 

|f a| M 0 

h hM 2 M 1 M 0 

h hM 2. * 

hM 2 has an absolute maximum at , hence by (*), we know that 

M 12 4M 0 M 2 . 

Remark: 

2. Suppose that f is a twice-differentiable real function on 0, , andf is bounded on 

0, , andfx 0asx . Prove that f x 0asx . 

Proof: Since M 12 4M 0 M 2 in Supplement 1, wehaveproveit. 

3. Suppose that f is real, three times differentiable on 1, 1, such that f1 0, 

f0 0, f1 1, and f 0 0. Prove that f 3 x 3forsomex 1, 1. 

Proof: Consider Taylor’s Theorem with Remainder Term, 

M 0 

M 2

fx f0 f 0x f 0 

x 

2! 

2 f3 c 

x 

3! 

3 ,wherec x,0 or 0, x, 

Then let x 1, and subtract one from another, we get 

f 3 c 1 f 3 c 2 6, where c 1 and c 2 in 1, 1. 

So,wehaveprovef 3 x 3forsomex 1, 1. 

5.24 If h 0andiff x exists (and is finite) for every x in a h, a h, andiff is 

continuous on a h, a h, show that we have: 

(a) 

fa h fa h 

f 

h 

a h f a h,0 1; 

Proof: Let gh fa h fa h, then by Mean Vaule Theorem, wehave 

gh g0 gh 

g hh, where 0 1 

f a h f a hh 


fa h fa h 

f 

h 

a h f a h,0 1. 

(b) 

fa h 2fa fa h 

f 

h 

a h f a h,0 1. 

Proof: Let gh fa h 2fa fa h, then by Mean Vaule Theorem, wehave 

gh g0 gh 

g hh, where 0 1 

f a h f a hh 


fa h 2fa fa h 

f 

h 

a h f a h,0 1. 

(c) If f a exists, show that. 

f fa h 2fa fa h 

a lim 

h0 h 2 

Proof: Since 

fa h 2fa fa h 

lim 

h0 h 2 

f 

lim 

a h f a h 

by L-Hospital Rule 

h0 2h 

lim 

h0 

f a h f a 

2h 

1 2 2f a since f a exists. 

f a. 

f a f a h 

2h 

Remark: There is another proof by using Generalized Mean Value theorem. 

Proof: Let g 1 h fa h 2fa fa h and g 2 h h 2 , then by Generalized 

Mean Value theorem, we have


Hence, 

g 1 h g 1 0g 2 h g 1 hg 2 h g 2 0 

fa h 2fa fa h 

h 2 

f a h f a h 

2h 

fa h 2fa fa h 

lim 

h0 h 2 

f 

lim 

a h f a h 

h0 2h 

f a since f a exists. 

(d) Give an example where the limit of the quotient in (c) exists but where f a does 

not exist. 

Solution: (STUDY) Note that in the proof of (c) by using L-Hospital Rule. We know 

that |x| is not differentiable at x 0, and |x| satisfies that 

|0 h| |0 h| 

f 

lim 

0 lim 

0 h f 0 h 

h0 2h 

h0 2h 

So,letustrytofindafunctionf so that f x |x|. So, consider its integral, we know that 

x 2 

2 

fx 

if x 0 

x2 if x 0 

2 

Hence, we complete it. 

Remark: (i) There is a related statement; we write it as follows. Suppose that f defined 

on a, b and has a derivative at c a, b. Ifx n a, c and y n c, b with such 

that x n y n 0asn . Then we have 

f fy 

c lim n fx n 

n y n x n 

. 

Proof: Sincef c exists, we have 

fy n fc f cy n c oy n c * 

and 

fx n fc f cx n c ox n c. *’ 

If we combine (*) and (*’), we have 

Note that 

we have 

fy n fx n 

y n x n 

y n c 

y n x n 

lim n 

lim n 

0. 

which implies that, by (**) 

f c oy n c 

y n x n 

1and 

oy n c 

y n x n 

oy n c 

y n c 

x n c 

y n x n 

1 for all n, 

ox n c 

y n x n 

y n c 

y n x n 

ox n c 

x n c 

ox n c 

y n x n 

. ** 

x n c 

y n x n

f fy 

c lim n fx n 

n y n x n 

. 

(ii) There is a good exercise; we write it as follows. Let f C 2 a, b, andc a, b. 

For small |h| such that c h a, b, write 

fc h fc f c hhh 

where 0 1. Show that if f c 0, then lim h0 h 1/2. 

Proof: Since f C 2 a, b, byTaylor Theorem with Remainder Term, we have 

fc h fc f ch f 

h 

2! 

2 ,where c, h or h, c 

f c hhh by hypothesis. 

So, 

f c hh f c 

h 

h f 

2! 

and let h 0, we have c by continuity of f at c. Hence, 

limh 1/2 since f c 0. 

h0 

Note: We can modify our statement as follows. Let f be defined on a, b, and 

c a, b. For small |h| such that c h a, b, write 

fc h fc f c hhh 

where 0 1. Show that if f c 0, and x x for x a h, a h, then 

lim h0 h 1/2. 

Proof: Use the exercise (c), we have 

f fc h 2fc fc h 

c lim 

h0 h 2 

f 

lim 

c hh f c hh 

by hypothesis 

h0 h 

f 

lim 

c hh f c hh 

2h since x x for x a h, a h. 

h0 2hh 

Since f c 0, we finally have lim h0 h 1/2. 

5.25 Let f have a finite derivatiive in a, b and assume that c a, b. Consider the 

following condition: For every 0, there exists a 1 ball Bc; , whose radius 

depends only on and not on c, such that if x Bc; , andx c, then 

fx fc 

x c f c . 

Show that f is continuous on a, b if this condition holds throughout a, b. 

Proof: Given 0, we want to find a 0 such that as dx, y , x, y a, b, we 

have 

|f x f y| . 

Choose any point y a, b, and thus by hypothesis, given /2, there is a 1 ball 

By; , whose radius depends only on and not on y, such that if x By; , and 

x y, then, 

fx fy 

x y f y /2 . * 

Note that y Bx, , so, we also have 

,

fx fy 

x y f x /2 *’ 

Combine (*) with (*’), we have 

|f x f y| . 

Hence, we have proved f is continuous on a, b. 

Remark: (i) The open interval can be changed into a closed interval; it just need to 

consider its endpoints. That is, f is continuous on a, b if this condition holds throughout 

a, b. The proof is similar, so we omit it. 

(ii) The converse of statement in the exercise is alos true. We write it as follows. Let f 

be continuous on a, b, and 0. Prove that there exists a 0 such that 

fx fc 

x c f c 

whenever 0 |x c| , a x, c, b. 

Proof: Given 0, we want to find a 0 such that 

fx fc 

x c f c 

whenever 0 |x c| , a x, c b. Sincef is continuous on a, b, we know that f is 

uniformly continuous on a, b. That is, given 0, there is a 0 such that as 

dx, y , wehave 

|f x f y| . * 

Consider dx, c , x a, b, thenby(*),wehave 

fx fc 

x c f c |f x f c| by Mean Value Theorem 

where dx , x . So, we complete it. 

Note: This could be expressed by saying that f is uniformly differentiable on a, b if f 

is continuous on a, b. 

5.26 Assume f has a finite derivative in a, b and is continuous on a, b, with 

a fx b for all x in a, b and |f x| 1 for all x in a, b. Prove that f has a 

unique fixed point in a, b. 

Proof: Given any x, y a, b, thus, by Mean Value Theorem, wehave 

|fx fy| |f z||x y| |x y| by hypothesis. 

So, we know that f is a contraction on a complete metric space a, b. So,f has a unique 

fixed point in a, b. 

5.27 Give an example of a pair of functions f and g having a finite derivatives in 0, 1, 

such that 

but such that lim x0 

f x 

g x 

lim 

x0 

fx 

gx 0, 

does not exist, choosing g so that g x is never zero. 

Proof: Letfx sin1/x and gx 1/x. Then it is trivial for that g x is never zero. 

In addition, we have 

fx 

f 

lim 0, and lim 

x 


x0 gx x0 g x

Remark: In this exercise, it tells us that the converse of L-Hospital Rule is NOT 

necessary true. Here is a good exercise very like L-Hospital Rule, but it does not! We 

write it as follows. 

Suppose that f a and g a exist with g a 0, and fa ga 0. Prove that 


lim xa 

fx 

gx lim xa 

 

lim xa 

fx 

gx 

fx fa 

gx ga lim xa 

f a 

g a . 

fx fa/x a 

gx ga/x a 

f a 

g a since f a and g a exist with g a 0. 

Note: (i) It should be noticed that we CANNOT use L-Hospital Rule since the 

statement tells that f and g have a derivative at a, we do not make sure of the situation of 

other points. 

(ii) This holds also for complex functions. Let us recall the proof of L-Hospital Rule, 

we need use the order field R; however, C is not an order field. Hence, L-Hospital Rule 

does not hold for C. Infact,noordercanbedefinedinthecomplexfieldsincei 2 1. 

Supplement on L-Hospital Rule 

We do not give a proof about the following fact. The reader may see the book named A 

First Course in Real Analysis written by Protter and Morrey, Charpter 4, pp 88-91. 

Theorem ( 0 ) Let f and g be continuous and differentiable on a, b with 0 g 0ona, b. 

If 

then 

lim 

xa 

lim 

xa 

fx 0 lim gx 0and 

* 

 

xa f x 

g x L, 

lim 

xa 

fx 

gx L. 

Remark: 1. The size of the interval a, b is of no importance; it suffices to 

have g 0ona, a , forsome 0. 

2. (*) is a sufficient condition, not a necessary condition. For example, 

fx x 2 ,andgx sin 1/x both defined on 0, 1. 

3. We have some similar results: x a ; x a; x 1/x 0 ; 

x 1/x 0 . 

Theorem ( ) Let f and g be continuous and differentiable on a, b with g 0on 

a, b. If 

then 

lim 

xa 

lim 

xa 

fx lim gx and 

* 

 

xa f x 

g x L,

lim 

xa 

fx 

gx L. 

Remark: 1. The proof is skilled, and it needs an algebraic identity. 

2. We have some similar results: x a ; x a; x 1/x 0 ; 

x 1/x 0 . 

3. (*) is a sufficient condition, not a necessary condition. For example, 

fx x sin x, andgx x. 

Theorem (O. Stolz) Suppose that y n , andy n is increasing. If 

lim 

x n x n1 

n y n y n1 

L, (or ) 

then 

lim 

x n 

n y n 

L. (or ) 

Remark: 1. The proof is skilled, and it needs an algebraic identity. 

2. The difference between Theorem 2 and Theorem 3 is that x is a continuous 

varibale but x n is not. 

Theorem (Taylor Theorem with Remainder) Suppose that f is a real function defined on 

a, b. Iff n x is continuous on a, b, and differentiable on a, b, then (let 

x, c a, b, with x c) there is a x, interior to the interval joining x and c such 

that 

fx P f x fn1 x 

n 1! x cn1 , 

where 

n 

f 

P f x : 

k c 

x c k . 

k! 

k0 

Remark: 1. As n 1, it is exactly Mean Value Theorem. 

2. The part 

f n1 x 

n 1! x cn1 : R n x 

is called the remainder term. 

3. There are some types about remainder term. (Lagrange, Cauchy, Berstein, 

etc.) 

Lagrange 

R n x fn1 x 

n 1! x cn1 . 

Cauchy 

Berstein 

R n x fn1 c x c 

n! 

R n x 

1 n x c n1 , where 0 1. 

1 

n! c 

x 

x t n f n1 tdt

5.28 Prove the following theorem: 

Let f and g be two functions having finite nth derivatives in a, b. For some interior 

point c in a, b, assume that fc f c ... f n1 c 0, and that 

gc g c ... g n1 c 0, but that g n x is never zero in a, b. Show that 

lim xc 

fx 

gx 

fn c 

g n c . 

NOTE. f n and g n are not assumed to be continuous at c. 

Hint. Let 

Fx fx x cn2 f n c 

, 

n 2! 

define G similarly, and apply Theorem 5.20 to the functions F and G. 

Proof: Let 

Fx fx 

fn c 

x cn2 

n 2! 

and 

Gx gx 

gn c 

x cn2 

n 2! 

then inductively, 

F k x f k f 

x 

n c 

x cn2k 

n 2 k! 

and note that 

F k c 0 for all k 0, 1, . . , n 3, and F n2 c f n c. 

Similarly for G. Hence, by Theorem 5.20, wehave 

n2 

Fx 

k0 

F 

k 

k! x ck G n1 x 1 F n1 x 1 Gx 

k0 

n2 

G 

k 

k! 

x c k 

where x 1 between x and c, which implies that 

fxg n1 x 1 f n1 x 1 gx. * 

Note that since g n is never zero on a, b; it implies that there exists a 0 such that 

every g k is never zero in c , c c, wherek 0, 1, 2. . . , n. Hence, we have, by 

(*), 

lim xc 

fx 

gx lim xc 

lim 

x1 c 

lim 

x1 c 

 

f n1 x 1 

g n1 x 1 

f n1 x 1 f n1 c 

g n1 x 1 g n1 c since x c x 1 c 

f n1 x 1 f n1 c/x 1 c 

g n1 x 1 g n1 c/x 1 c 

fn c 

g n c since fn exists and g n exists 0 on a, b. 

Remark: (1) The hint is not correct from text book. The reader should find the 

difference between them. 

(2) Here ia another proof by L-Hospital Rule and Remark in Exercise 5.27. 

Proof: Since g n is never zero on a, b, it implies that there exists a 0 such that

every g k is never zero in c , c c, wherek 0, 1, 2. . . , n. So, we can apply 

n 1 times L-Hospital Rule methoned in Supplement, and thus get 

fx 

lim xc gx lim f n1 x 

xc 

g n1 x 

lim xc 

lim xc 

 

f n1 x f n1 c 

g n1 x g n1 c 

f n1 x f n1 c/x c 

g n1 x g n1 c/x c 

fn c 

g n c since fn exists and g n exists 0 on a, b. 

5.29 Show that the formula in Taylor’s theorem can also be written as follows: 

n1 

fx 

k0 

f 

k 

c 

k! 

x c k x cx x 1 n1 

n 1! 

f n x 1 , 

where x 1 is interior to the interval joining x and c. Let1 x x 1 /x c. Show that 

0 1 and deduce the following form of the remainder term (due to Cauchy): 

1 n1 x c n 

f 

n 1! 

n x 1 c. 

Hint. Take Gt t in the proof of Theorem 5.20. 

Proof: Let 

and note that 

n1 

Ft 

k0 

f 

k 

t 

k! 

x t k ,andGt t, 

F t 

fn t 

x tn1 

n 1! 

then by Generalized Mean Value Theorem, wehave 

Fx FcG x 1 Gx GcF x 1 


n1 

fx 

k0 

f 

k 

c 

k! 

So,wehaveprovethat 

where 

x c k 

fn x 1 

n 1! x x 1 n 

fn x 1 c 

n 1! 

n1 

fx 

k0 

f 

k 

c 

k! 

x c n 1 n ,wherex 1 x 1 c. 

x c k R n1 x, 

R n1 x fn x 1 c 

x c n 1 n ,wherex 

n 1! 

1 x 1 c 

is called a Cauchy Remainder. 

Supplement on some questions. 

1. Let f be continuous on 0, 1 and differentiable on 0, 1. Suppose that f0 0and

|f x| |fx| for x 0, 1. Prove that f is constant. 

Proof: Given any x 1 0, 1, by Mean Value Theorem and hypothesis, we know that 

|fx 1 f0| x 1 |f x 2 | x 1 |fx 2 |, wherex 2 0, x 1 . 

So, we have 

|fx 1 | x 1 x n|fx n1 | Mx 1 x n ,wherex n1 0, x n ,andM sup |fx| 

xa,b 

Since Mx 1 x n 0, as n , we finally have fx 1 0. Since x 1 is arbirary, we find 

that fx 0on0, 1. 

2. Suppose that g is real function defined on R, with bounded derivative, say |g | M. 

Fix 0, and define fx x gx. Show that f is 1-1 if is small enough. (It implies 

that f is strictly monotonic.) 

Proof: Suppose that fx fy, i.e., x gx y gy which implies that 

|y x| |gy gx| M|y x| by Mean Value Theorem, and hypothesis. 

So, as is small enough, we have x y. Thatis,f is 1-1. 

Supplement on Convex Function. 

Definition(Convex Function) Let f be defined on an interval I, and given 0 1, 

we say that f is a convex function if for any two points x, y I, 

fx 1 y fx 1 fy. 

For example, x 2 is a convex function on R. Sometimes, the reader may see another weak 

definition of convex function in case 1/2. We will show that under continuity, two 

definitions are equivalent. In addition, it should be noted that a convex function is not 

necessarily continuous since we may give a jump on a continuous convex function on its 

boundary points, for example, fx x is a continuous convex function on 0, 1, and 

define a function g as follows: 

gx x, ifx 0, 1 and g1 g0 2. 

The function g is not continuous but convex. Note that if f is convex, we call f is concave, 

vice versa. Note that every increasing convex function of a convex function is convex. (For 

example, if f is convex, so is e f . ) It is clear only by definition. 

Theorem(Equivalence) Under continuity, two definitions are equivalent. 

Proof: It suffices to consider if 

f 

x y 

2 

 

fx fy 

2 

then 

fx 1 y fx 1 fy for all 0 1. 

Since (*) holds, then by Mathematical Induction, itiseasytoshowthat 

f 

x 1 ...x 2 

n 

2 n fx 1 ...fx 2 

n 

2 n . 

Claim that 

f 

x 1 ...x n 

n fx 1 ...fx n 

n for all n N. ** 

Using Reverse Induction, letx n x 1...x n1 

, then 

n1 

*

f 

x 1 ...x n 

n 

f 

x 1 ...x n1 

n x n 

fx n 

fx 1 ...fx n 

n by induction hypothesis. 

So, we have 

f 

x 1 ...x n1 

fx 1 ...fx n1 

. 

n 1 

n 1 

Hence, we have proved (**). Given a rational number m/n 0, 1, where 

g. c. d. m, n 1; we choose x : x 1 ... x m ,andy : x m1 ... x n , thenby(**),we 

finally have 

f mx n my 

n n mfx n mfy 

n n m n fx 1 m n 

fy. *** 

Given 0, 1, then there is a sequence q n Q such that q n as n . Then 

by continuity and (***), we get 

fx 1 y fx 1 fy. 

Remark: The Reverse Induction is that let S N and S has two properties:(1) For 

every k 0, 2 k S and (2) k S and k 1 N, then k 1 S. Then S N. 

(Lemma) Letf be a convex function on a, b, then f is bounded. 

Proof: LetM maxfa, fb, then every point z I, write z a 1 b, we 

have 

fz fa 1 b fa 1 fb M. 

In addition, we may write z ab t, wheret is chosen so that z runs through a, b. So, 

2 

we have 


2f 

f 

a b 

2 

a b 

2 

1 2 f 

f 

a b 

2 

a b 

2 

t 1 2 f a b 

2 

t f a b 

2 

t 

t 

fz 


2f a b M : m fz. 

2 

Hence, we have proved that f is bounded above by M and bounded below by m. 

(Theorem) Iff : I R is convex, then f satisfies a Lipschitz condition on any closed 

interval a, b intI. In addition, f is absolutely continuous on a, b and continuous on 

intI. 

Proof: We choose 0 so that a , b intI. By preceding lemma, we 

know that f is bounded, say m fx M on a , b . Given any two points x, andy, 

with a x y b We consider an auxiliary point z y , and a suitable 

then y z 1 x. So, 

fy fz 1 x fz 1 fx fz fx fx 


fy fx M m y x 

Change roles of x and y, we finally have 

 

M m. 

yx 

yx ,

|fy fx| K|y x|, whereK M m . 

That is, f satisfies a Lipschitz condition on any closed interval a, b. 

We call that f is absolutely continuous on a, b if given any 0, there is a 0 

n 

such that for any collection of a i , b i i1 

of disjoint open intervals of a, b with 

n 

b i a i , wehave 

k1 

n 

|fb i fa i | . 

k1 

Clearly, the choice /K meets this requirement. Finally, the continuity of f on intI is 

obvious. 

(Theorem) Letf be a differentiable real function defined on a, b. Prove that f is 

convex if and only if f is monotonically increasing. 

Proof: () Suppose f is convex, and given x y, we want to show that f x f y. 

Choose s and t such that x u s y, then it is clear that we have 

fu fx 

u x 

fs fu 

s u 

Let s y , wehaveby(*) 

fu fx 

u x f y 

fy fs 

y s . * 

which implies that, let u x 

f x f y. 

() Suppose that f is monotonically increasing, it suffices to consider 1/2, if 

x y, then 

fx fy 

2 

f 

x y 

2 

 

fx f 

xy 

2 

fy f 

xy 

2 

2 

x y 

4 f 1 f 2 , where 1 2 

0. 

Similarly for x y, and there is nothing to prove x y. Hence, we know that f is convex. 

(Corollary 1) Assume next that f x exists for every x a, b, and prove that f is 

convex if and only if f x 0 for all x a, b. 

Proof: () Suppose that f is convex, we have shown that f is monotonically increasing. 

So, we know that f x 0 for all x a, b. 

() Suppose that f x 0 for all x a, b, it implies that f is monotonically 

increasing. So, we know that f is convex. 

(Corollary 2) Let0 , then we have 

|y 1 | ...|y n| 

n 

1/ 

 

|y 1 | ...|y n| 

n 

Proof: Letp 1, and since x p pp 1x p2 0 for all x 0, we know that 

fx x p is convex. So, we have (let p ) 

by 

x 1 ...x n 

n 

1/ 

/ 

 

x 1 / ...x n 

/ 

n *

f 

x 1 ...x n 

n fx 1 ...fx n 

n . 

Choose x i |y i | ,wherei 1, 2, . . , n. Thenby(*),wehave 

|y 1 | ...|y n| 

n 

1/ 

 

|y 1 | ...|y n| 

n 

1/ 

. 

(Corollary 3) Define 

M r y 

|y 1 | r ...|y n| r 

n 

1/r 

,wherer 0. 

Then M r y is a monotonic function of r on 0, . In particular, we have 

M 1 y M 2 y, 

that is, 

|y 1 | ...|y n| 

n 

 

|y 1 | 2 ...|y n| 2 

n 

Proof: It is clear by Corollary 2. 

(Corollary 4) By definition of M r y in Corollary 3, wehave 

lim M ry |y 1 | |y 

r0 n| 1/n : M 0 y 

 

and 

lim r 

M r y max|y 1 |,...,|y n| : M y 

Proof: 1. Since M r y 

Theorem, wehave 

log 

So, 

2. As r 0, we have 

|y 1| r ...|y n| r 

n 0 

r 0 

|y 1 | r ...|y n| r 

n 

max|y 1 |,...,|y n| r 

n 

1/2 

. 

1/r 

, taking log and thus by Mean Value 

1 n 

n |yi i1 

| r log|y i | 

1 n 

, where 0 r r. 

n |yi i1 

| r 

log 

lim M ry lim e 

r0 r0 

 

1 n 

lim e 

r0 

 

i1 

e n 

1/r 

n 

y 

r 

1 ...|yn| r 

n 

r 

n 

i1 

1 

n 

log y i 

n 

i1 

|y 1 | |y n| 1/n . 

y i 

r log y i 

y i 

r 

M r y max|y 1 |,...,|y n| r 1/r 

which implies that, by Sandwich Theorem, 

lim r 

M r y max|y 1 |,...,|y n| 

since lim r 1 n 1/r 1. 

(Inequality 1) Letf be convex on a, b, and let c a, b. Define

lx fa 

then fx lx for all x c, b. 

fc fa 

c a 

x a, 

Proof: Consider x c, d, then c xa xc a ca 

xa x, wehave 

fc x a c fa c x a fx 


fc fa 

fx fa c a x a lx. 

(Inequality 2) Letf be a convex function defined on a, b. Leta s t u b, 

then we have 

ft fs 

t s 

 

fu fs 

u s 

 

fu ft 

u t 

Proof: By definition of convex, we know that 

fu fs 

fx fs u s x s, x s, u * 

and by inequality 1, we know that 

ft fs 

fs 

t s 

x s fx, x t, u. ** 

So, as x t, u, by (*) and (**), we finally have 

ft fs fu fs 

t s 

u s . 

Similarly, we have 

Hence, we have 

ft fs 

t s 

fu fs 

u s 

 

 

fu fs 

u s 

fu ft 

u t 

 

. 

fu ft 

u t 

Remark: Using abvoe method, it is easy to verify that if f is a convex function on a, b, 

then f x and f x exist for all x a, b. In addition, if x y, wherex, y a, b, then 

we have 

f x f x f y f y. 

That is, f x and f x are increasing on a, b. We omit the proof. 

(Exercise 1) Letfx be convex on a, b, and assume that f is differentiable at 

c a, b, wehave 

lx fc f cx c fx. 

That is, the equation of tangent line is below fx if the equation of tangent line exists. 

Proof: Sincef is differentiable at c a, b, we write the equation of tangent line at c, 

lx fc f cx c. 

Define 

fs fc 

ms s c where a s c and mt 

then it is clear that 

ms f c mt 


. 

. 

ft fc 

t c 

where b t c,

lx fc f cx c fx. 

(Exercise 2) Letf : R R be convex. If f is bounded above, then f is a constant 

function. 

Proof: Suppose that f is not constant, say fa fb, wherea b. Iffb fa, we 

consider 

fx fb fb fa 

,wherex b 

x b b a 

which implies that as x b, 

fb fa 

fx x b fb as x 

b a 

And if fb fa, we consider 

fx fa fb fa 

x a ,wherex a 

b a 

which implies that as x a, 

fb fa 

fx x a fa as x . 

b a 

So, we obtain that f is not bouded above. So, f must be a constant function. 

(Exercise 3) Note that e x is convex on R. Use this to show that A. P. G. P. 

Proof: Sincee x e x 0onR, we know that e x is convex. So, 

ex 1 ...e 

x n 

n ,wherex i R, i 1, 2, . . . , n. 

So, let e x i y i 0, for i 1, 2, . . . , n. Then 

e x 1 ...xn 

n 

Vector-Valued functions 

y 1 y n 1/n y 1 ...y n 

n . 

5.30 If a vector valued function f is differentiable at c, prove that 

f c lim 1 fc h fc. 

h0 h 

Conversely, if this limit exists, prove that f is differentiable at c. 

Proof: Write f f 1 ,...,f n : S R R n , and let c be an interior point of S. Then if 

f is differentiable at c, each f k is differentiable at c. Hence, 

lim 1 fc h fc 

h0 h 

f 

lim 1 c h f 1 c 

,..., f nc h f n c 

h0 h 

h 

f 

lim 1 c h f 1 c f 

,...,lim n c h f n c 

h0 h 

h0 h 

f 1 c,...,f n c 

f c. 

Conversly, it is obvious by above. 

Remark: We give a summary about this. Let f be a vector valued function defined on 

S. Write f : S R n R m , c is a interior point. 

f f 1 ,...,f n is differentiable at c each f k is differentiable at c, * 

and

f f 1 ,...,f n is continuous at c each f k is continuous at c. 

Note: The set S can be a subset in R n , the definition of differentiation in higher 

dimensional space makes (*) holds. The reader can see textbook, Charpter 12. 

5.31 A vector-valued function f is differentiable at each point of a, b and has constant 

norm f. Prove that ft f t 0ona, b. 

Proof: Sincef, f f 2 is constant on a, b, wehavef, f 0ona, b. It implies 

that 2f, f 0ona, b. Thatis,ft f t 0ona, b. 

Remark: The proof of f, g f , g f, g is easy from definition of differentiation. 

So, we omit it. 

5.32 A vector-valued function f is never zero and has a derivative f which exists and is 

continuous on R. If there is a real function such that f t tft for all t, prove that 

there is a positive real function u and a constant vector c such that ft utc for all t. 

Proof: Since f t tft for all t, wehave 

f 1 t,...,f n t f t tft tf 1 t,...,tf n t 


f i t 

t since f is never zero. * 

fit 

Note that f i t 

f i 

is a continuous function from R to R for each i 1,2,...,n, sincef is 

t 

continuous on R, we have, by (*) 

f 

1 t 

ax f 1 t dt x 

tdt fi t f ia 

e t for i 1, 2, . . . , n. 

a 

e a 

So, we finally have 

ft f 1 t,...,f n t 

where ut e t and c 

e t 

utc 

f 1 a 

e a 

f 1 a fna 

,..., . 

a 

e e a 

,..., f na 

e a 

Supplement on Mean Value Theorem in higher dimensional space. 

In the future, we will learn so called Mean Value Theorem in higher dimensional 

space from the text book in Charpter 12. We give a similar result as supplement. 

Suppose that f is continuous mapping of a, b into R n and f is differentiable in a, b. 

Then there exists x a, b such that 

fb fa b af x. 

Proof: Letz fb fa, and define x fx z which is a real valued function 

defined on a, b. It is clear that x is continuous on a, b and differentiable on a, b. 

So, by Mean Value Theorem, we know that 

b a xb a, wherex a, b 


|b a| | xb a| 

fb fa xb a by Cauchy-Schwarz inequality.

So, we have 

Partial derivatives 

fb fa b af x. 

5.33 Consider the function f defined on R 2 by the following formulas: 

xy 

fx, y if x, y 0, 0 f0, 0 0. 

x 2 y2 Prove that the partial derivatives D 1 fx, y and D 2 fx, y exist for every x, y in R 2 and 

evaluate these derivatives explicitly in terms of x and y. Also, show that f is not continuous 

at 0, 0. 

Proof: It is clear that for all x, y 0, 0, wehave 

D 1 fx, y y y2 x 2 

x 2 y 2 and D 2fx, y x x2 y 2 

2 x 2 y 2 . 2 

For x, y 0, 0, wehave 

fx,0 f0, 0 

D 1 f0, 0 lim 

0. 

x0 x 0 


D 2 f0, 0 0. 

In addition, let y x and y 2x, wehave 

limfx, x 1/2 limfx,2x 2/5. 

x0 x0 

Hence, f is not continuous at 0, 0. 

Remark: The existence of all partial derivatives does not make sure the continuity of f. 

The trouble with partial derivatives is that they treat a function of several variables as a 

function of one variable at a time. 

5.34 Let f be defined on R 2 as follows: 

fx, y y x2 y 2 

if x,,y 0, 0, f0, 0 0. 

x 2 y2 Compute the first- and second-order partial derivatives of f at the origin, when they exist. 

Proof: For x, y 0, 0, it is clear that we have 

4xy 

D 1 fx, y 

3 

x 2 y 2 and D 2fx, y x4 4x 2 y 2 y 4 

2 x 2 y 2 2 

and for x, y 0, 0, wehave 

fx,0 f0, 0 

f0, y f0, 0 

D 1 f0, 0 lim 

0, D 

x0 x 0 

2 f0, 0 lim 

1. 

y0 y 0 

Hence, 

and 

D 1,2 f0, 0 lim 

x0 

D 1 fx,0 D 1 f0, 0 

0, 

x 0 

D 2 fx,0 D 2 f0, 0 

lim 2 

x 0 

x0 x does not exist, 

D 1 f0, y D 1 f0, 0 

0, 

y 0 

D 1,1 f0, 0 lim 

x0 

D 2,1 f0, 0 lim 

y0

D 2,2 f0, 0 lim 

y0 

D 2 f0, y D 2 f0, 0 

y 0 

lim 

y0 

0 y 0. 

Remark: We do not give a detail computation, but here are answers. Leave to the 

reader as a practice. For x, y 0, 0, wehave 

D 1,1 fx, y 4y3 y 2 3x 2 

x 2 y 2 3 , 

and 

complex-valued functions 

D 1,2 fx, y 4xy2 3x 2 y 2 

x 2 y 2 3 , 

D 2,1 fx, y 4xy2 3x 2 y 2 

x 2 y 2 3 , 

D 2,2 fx, y 4x2 yy 2 3x 2 

x 2 y 2 3 . 

5.35 Let S be an open set in C and let S be the set of complex conjugates z, where 

z S. Iff is defined on S, defineg on S as follows: gz fz, the complex conjugate 

of fz. Iff is differentiable at c, prove that g is differentiable at c and that g c f c. 

Proof: Sincec S, we know that c is an interior point. Thus, it is clear that c is also an 

interior point of S . Note that we have 

the conjugate of 

fz fc 

z c 

z 

f fc 

z c 

gz gc 

 

by gz fz. 

z c 

Note that z c z c, so we know that if f is differentiable at c, prove that g is 

differentiable at c and that g c f c. 

5.36 (i) In each of the following examples write f u iv and find explicit formulas 

for ux, y and vx, y : ( These functions are to be defined as indicated in Charpter 1.) 

(a) fz sin z, 

Solution: Since e iz cosz i sin z, we know that 

sin z 1 2 ey e y sin x ie y e y cosx 

from sin z eiz e iz 

2i 

and 

. So, we have 

ux, y ey e y sin x 

2 

vx, y ey e y cosx 

. 

2 

(b) fz cosz, 

Solution: Sincee iz cosz i sin z, we know that 

cosz 1 2 ey e y cosx e y e y sin x 

from cos z eiz e iz 

2 

. So, we have

ux, y ey e y cosx 

2 

and 

vx, y ey e y sin x 

. 

2 

(c) fz |z|, 

Solution: Since|z| x 2 y 2 1/2 , we know that 

ux, y x 2 y 2 1/2 

and 

vx, y 0. 

(d) fz z, 

Solution: Since z x iy, we know that 

ux, y x 

and 

vx, y y. 

(e) fz arg z, (z 0), 

Solution: Since argz R, we know that 

ux, y argx iy 

and 

vx, y 0. 

(f) fz Log z, (z 0), 

Solution: Since Log z log|z| i argz, we know that 

ux, y logx 2 y 2 1/2 

and 

vx, y argx iy. 

and 

(g) fz e z2 , 

Solution: Since e z2 

e x2 y 2 i2xy 

, we know that 

ux, y e x2 y 2 cos2xy 

vx, y e x2 y 2 sin2xy. 

(h) fz z ,( complex, z 0). 

Solution: Since z e Log z , thenwehave(let 1 i 2 ) 

z e 1i 2 log|z|i arg z 

e 1 log|z| 2 arg zi 2 log|z| 1 arg z 

. 

So, we know that 

ux, y e 1 log|z| 2 arg z 

cos 2 log|z| 1 arg z 

e 1 logx 2 y 2 1/2 2 argxiy 

cos 2 logx 2 y 2 1/2 1 argx iy

and 

vx, y e 1 log|z| 2 arg z 

sin 2 log|z| 1 arg z 

e 1 logx 2 y 2 1/2 2 argxiy 

sin 2 logx 2 y 2 1/2 1 argx iy . 

(ii) Show that u and v satisfy the Cauchy -Riemanns equation for the following values 

of z :Allz in (a), (b), (g); no z in (c), (d), (e); all z except real z 0in(f),(h). 

So, 

and 

Proof: (a)sinz u iv, where 

ux, y ey e y sin x 

2 

u x v y ey e y cosx 

2 

and vx, y ey e y cosx 

2 

for all z x iy 

u y v x ey e y sin x 

for all z x iy. 

2 

(b) cos z u iv, where 

ux, y ey e y cosx 

and vx, y ey e y sin x 

2 

2 

So, 

u x v y ey e y sin x 


2 

and 

u y v x ey e y cosx 


2 

(c) |z| u iv, where 

ux, y x 2 y 2 1/2 and vx, y 0. 

So, 

u x xx 2 y 2 1/2 v y 0ifx 0, y 0. 

and 

u y yx 2 y 2 1/2 v x 0ifx 0, y 0. 

So, we know that no z makes Cauchy-Riemann equations hold. 

(d) z u iv, where 

ux, y x and vx, y y. 

So, 

u x 1 1 v y . 

So, we know that no z makes Cauchy-Riemann equations hold. 

(e) arg z u iv, where 

ux, y argx 2 y 2 1/2 and vx, y 0. 

Note that 

. 

.

ux, y 

(1) arctany/x, ifx 0, y R 

(2) /2, if x 0, y 0 

(3) arctany/x , ifx 0, y 0 

(4) arctany/x , ifx 0, y 0 

(5) /2, if x 0, y 0. 

and 

v x v y 0. 

So, we know that by (1)-(5), for x, y 0, 0 

y 

u x 

x 2 y 2 

and for x, y x, y : x 0, y 0, wehave 

u y x 

x 2 y . 2 

Hence, we know that no z makes Cauchy-Riemann equations hold. 

Remark: We can give the conclusion as follows: 

y 

arg z x 

for x, y 0, 0 

x 2 y 2 

and 

arg z y 

x for x, y x, y : x 0, y 0. 

x 2 y2 (f) Log z u iv, where 

ux, y logx 2 y 2 1/2 and vx, y argx 2 y 2 1/2 . 

Since 

u x x 

x 2 y and u y 

2 y 

x 2 y 2 

and 

y 

v x 

x 2 y for x, y 0, 0 and v 2 y x for x, y x, y : x 0, y 0, 

x 2 y2 we know that all z except real z 0 make Cauchy-Riemann equations hold. 

Remark: Log z is differentiable on C x, y : x 0, y 0 since Cauchy-Riemann 

equations along with continuity of u x iv x ,andu y iv y . 

(g) e z2 u iv, where 

ux, y e x2 y 2 cos2xy and vx, y e x2 y 2 sin2xy. 

So, 

u x v y 2e x2 y 2 xcos2xy ysin 2xy for all z x iy. 

and 

u y v x 2e x2 y 2 ycos2xy xsin 2xy for all z x iy. 

Hence, we know that all z make Cauchy-Riemann equations hold. 

(h) Since z e Log z ,ande z is differentiable on C, we know that, by the remark of (f), 

we know that z is differentiable for all z except real z 0. So, we know that all z except 

real z 0 make Cauchy-Riemann equations hold. 

( In part (h), the Cauchy-Riemann equations hold for all z if is a nonnegative integer,

and they hold for all z 0if is a negative integer.) 

Solution: It is clear from definition of differentiability. 

(iii) Compute the derivative f z in (a), (b), (f), (g), (h), assuming it exists. 

Solution: Sincef z u x iv x , if it exists. So, we know all results by (ii). 

5.37 Write f u iv and assume that f has a derivative at each point of an open disk D 

centered at 0, 0. Ifau 2 bv 2 is constant on D for some real a and b, not both 0. Prove 

that f is constant on D. 

Proof: Letau 2 bv 2 be constant on D. We consider three cases as follows. 

1. As a 0, b 0, then we have 

v 2 is constant on D 


vv x 0. 

If v 0onD, it is clear that f is constant. 

If v 0onD, that is v x 0onD. So, we still have f is contant. 

2. As a 0, b 0, then it is similar. We omit it. 

3. As a 0, b 0, Taking partial derivatives we find 

auu x bvv x 0onD. 1 

and 

auu y bvv y 0onD. 

By Cauchy-Riemann equations the second equation can be written as we have 

auv x bvu x 0onD. 2 

We consider 1v x 2u x and 1u x 2v x , then we have 

bvv x2 u x2 0 3 

and 

auv x2 u x2 0 4 

which imply that 

au 2 bv 2 v x2 u x2 0. 5 

If au 2 bv 2 c, constant on D, wherec 0, then v x2 u x2 0. So, f is constant. 

If au 2 bv 2 c, constant on D, wherec 0, then if there exists x, y such that 

v x2 u x2 0, then by (3) and (4), ux, y vx, y 0. By continuity of v x2 u x2 , we know 

that there exists an open region S D such that u v 0onS. Hence, by Uniqueness 

Theorem, we know that f is constant. 

Remark: In complex theory, the Uniqueness theorem is fundamental and important. 

The reader can see this from the book named Complex Analysis by Joseph Bak and 

Donald J. Newman.

Functions of Bounded Variation and Rectifiable Curves 

Functions of bounded variation 

6.1 Determine which of the follwoing functions are of bounded variation on 0, 1. 

(a) fx x 2 sin1/x if x 0, f0 0. 

(b) fx x sin1/x if x 0, f0 0. 

Proof: (a) Since 

f x 2x sin1/x cos1/x for x 0, 1 and f 0 0, 

we know that f x is bounded on 0, 1, in fact, |f x| 3on0, 1. Hence, f is of 

bounded variation on 0, 1. 

1 

(b) First, we choose n 1beanevenintegersothat 1, and thus consider a 

2 n1 

partition P 0 x 0 , x 1 1 

we have 

2 

n2 

 

k1 

, x 2 1 

2 2 

,..., x n 1 

n 2 

|f k | 2 2 

n 

k1 

, x n1 

1 

n1 2 

1/k . 

, x n2 1 , then 

Since 1/k diverges to , we know that f is not of bounded variation on 0, 1. 

6.2 A function f, defined on a, b, is said to satisfy a uniform Lipschitz condition of 

order 0ona, b if there exists a constant M 0 such that |fx fy| M|x y| for 

all x and y in a, b. (Compare with Exercise 5.1.) 

(a) If f is such a function, show that 1 implies f is constant on a, b, whereas 

1 implies f is of bounded variation a, b. 

Proof: As 1, we consider, for x y, wherex, y a, b, 

|fx fy| 

0 M|x y| 1 . 

|x y| 

Hence, f x exists on a, b, and we have f x 0ona, b. So, we know that f is 

constant. 

As 1, consider any partition P a x 0 , x 1 ,..., x n b, wehave 

n 

 

k1 

n 

|f k | M |x k1 x k | Mb a. 

k1 

That is, f is of bounded variation on a, b. 

(b) Give an example of a function f satisfying a uniform Lipschitz condition of order 

1ona, b such that f is not of bounded variation on a, b. 

Proof: First, note that x satisfies uniform Lipschitz condition of order , where 

1 

0 1. Choosing 1 such that 1 and let M k1 

since the series 

k 

converges. So, we have 1 

1 1 

. 

M k1 k 

Define a function f as follows. We partition 0, 1 into infinitely many subsintervals. 

Consider 

x 0 0, x 1 x 0 

M 1 1 1 , x 2 x 1 

M 1 1 2 ,..., x n x n1 

M 1 1 n ,.... 

And in every subinterval x i , x i1 ,wherei 0,1,...., we define

fx x x i x i1 

2 

then f is a continuous function and is not bounded variation on 0, 1 since k1 

 

, 

1 

2M 

diverges. 

In order to show that f satisfies uniform Lipschitz condition of order , we consider 

three cases. 

(1) If x, y x i , x i1 ,andx, y x i , x ix i1 

2 

or x, y x ix i1 

2 

, x i1 , then 

|fx fy| |x y | |x y| . 

(2) If x, y x i , x i1 ,andx x i , x ix i1 

or y x ix i1 

, x 

2 2 i1 , then there is a 

z x i , x ix i1 

such that fy fz. So, 

2 

|fx fy| |fx fz| |x z | |x z| |x y| . 

(3) If x x i , x i1 and y x j , x j1 ,wherei j. 

If x x i , x ix i1 

, then there is a z x 

2 i , x ix i1 

such that fy fz. So, 

2 

|fx fy| |fx fz| |x z | |x z| |x y| . 

Similarly for x x ix i1 

, x 

2 i1 . 

Remark: Here is another example. Since it will use Fourier Theory, we do not give a 

proof. We just write it down as a reference. 

 

ft 

k1 

cos3k t 

3 k . 

(c) Give an example of a function f which is of bounded variation on a, b but which 

satisfies no uniform Lipschitz condition on a, b. 

Proof: Since a function satisfies uniform Lipschitz condition of order 0, it must be 

continuous. So, we consider 

x if x a, b 

fx 

b 1ifx b. 

Trivially, f is not continuous but increasing. So, the function is desired. 

Remark: Here is a good problem, we write it as follows. If f satisfies 

|fx fy| K|x y| 1/2 for x 0, 1, wheref0 0. 

define 

fx 

x 1/3 

if x 0, 1 

gx 

0ifx 0. 

Then g satisfies uniform Lipschitz condition of order 1/6. 

Proof: Note that if one of x, andy is zero, the result is trivial. So, we may consider 

0 y x 1 as follows. Consider 

|gx gy| 

fx 

x 1/3 

 

 

fx 

x 1/3 

fx 

x 1/3 

fy 

y 1/3 

fy 

x 1/3 

fy 

x 1/3 

fy fy 

x 1/3 x 1/3 

fy 

y 1/3 

1 

k 

fy 

y 1/3 . *

For the part 

fx 

x fy 1 |fx fy| 

1/3 x 1/3 x1/3 K 

x |x 1/3 y|1/2 by hypothesis 

K|x y| 1/2 |x y| 1/3 since x x y 0 

K|x y| 1/6 . A 

fy 

For another part fy , we consider two cases. 

x 1/3 y 1/3 

(1) x 2y which implies that x x y y 0, 

fy 

x 1/3 

fy 

y 1/3 

|fy| 

|fy| 

x 1/3 y 1/3 

xy 1/3 

x y 1/3 

xy 1/3 since |x 1/3 y 1/3 | |x y| 1/3 for all x, y 0 

|fy| x 1/3 

xy 1/3 since x y 1/3 x 1/3 

|fy| 1 

y 1/3 

K |y|1/2 

|y| 1/3 

K|y| 1/6 

by hypothesis 

K|x y| 1/6 since y x y. B 

(2) x 2y which implies that x y x y 0, 

fy 

x 1/3 

fy 

y 1/3 

|fy| 

|fy| 

|fy| 

x 1/3 y 1/3 

xy 1/3 

x y 1/3 

xy 1/3 since |x 1/3 y 1/3 | |x y| 1/3 for all x, y 0 

x y 1/3 

y 2/3 

K|y| 1/2 x y 1/3 

y 2/3 

K|y| 1/6 |x y| 1/3 

since x y 

by hypothesis 

K|x y| 1/6 |x y| 1/3 since y x y 

K|x y| 1/6 . C 

So, by (A)-(C), (*) tells that g satisfies uniform Lipschitz condition of order 1/6. 

Note: Hereisageneralresult.Let0 2. Iff satisfies 

|fx fy| K|x y| for x 0, 1, wheref0 0. 

define

fx 

x 

if x 0, 1 

gx 

0ifx 0. 

Then g satisfies uniform Lipschitz condition of order . The proof is similar, so we 

omit it. 

6.3 Show that a polynomial f is of bounded variation on every compact interval a, b. 

Describe a method for finding the total variation of f on a, b if the zeros of the derivative 

f are known. 

Proof: If f is a constant, then the total variation of f on a, b is zero. So, we may 

assume that f is a polynomial of degree n 1, and consider f x 0 by two cases as 

follows. 

(1) If there is no point such that f x 0, then by Intermediate Value Theorem of 

Differentiability, we know that f x 0ona, b, orf x 0ona, b. So, it implies 

that f is monotonic. Hence, the total variation of f on a, b is |fb fa|. 

(2) If there are m points such that f x 0, say 

a x 0 x 1 x 2 ... x m b x m1 , where 1 m n, then we know the monotone 

property of function f. So, the total variation of f on a, b is 

m1 

|fx i fx i1 |. 

i1 

Remark: Here is another proof. Let f be a polynomial on a, b, then we know that f is 

bounded on a, b since f is also polynomial which implies that it is continuous. Hence, we 

know that f is of bounded variation on a, b. 

6.4 A nonempty set S of real-valued functions defined on an interval a, b is called a 

linear space of functions if it has the following two properties: 

(a) If f S, then cf S for every real number c. 

(b) If f S and g S, then f g S. 

Theorem 6.9 shows that the set V of all functions of bounded variation on a, b is a 

linear space. If S is any linear space which contains all monotonic functions on a, b, 

prove that V S. This can be described by saying that the functions of bounded 

variation form the samllest linear space containing all monotonic functions. 

Proof: It is directlt from Theorem 6.9 and some facts in Linear Algebra. We omit the 

detail. 

6.5 Let f be a real-valued function defined on 0, 1 such that f0 0, fx x for all 

x, andfx fy whenever x y. LetA x : fx x. Prove that sup A A, andthat 

f1 1. 

Proof: Note that since f0 0, A is not empty. Suppose that sup A : a A, i.e., 

fa a since fx x for all x. So, given any n 0, then there is a b n A such that 

a n b n . * 

In addition, 

b n fb n since b n A. ** 

So,by(*)and(**),wehave(let n 0 ), 

a fa fa since f is monotonic increasing. 

which contradicts to fa a. Hence, we know that sup A A.

Claim that 1 sup A. Suppose NOT, thatis,a 1. Then we have 

a fa f1 1. 

Since a sup A, consider x a, fa, then 

fx x 


fa a 

which contradicts to a fa. So, we know that sup A 1. Hence, we have proved that 

f1 1. 

Remark: The reader should keep the method in mind if we ask how to show that 

f1 1 directly. The set A is helpful to do this. Or equivalently, let f be strictly increasing 

on 0, 1 with f0 0. If f1 1, then there exists a point x 0, 1 such that fx x. 

6.6 If f is defined everywhere in R 1 , then f is said to be of bounded variation on 

, if f is of bounded variation on every finite interval and if there exists a positive 

number M such that V f a, b M for all compact interval a, b. The total variation of f on 

, is then defined to be the sup of all numbers V f a, b, a b , and 

denoted by V f , . Similar definitions apply to half open infinite intervals a, and 

, b. 

(a) State and prove theorems for the inifnite interval , analogous to the 

Theorems 6.7, 6.9, 6.10, 6.11, and 6.12. 

(Theorem 6.7*) Letf : R R be of bounded variaton, then f is bounded on R. 

Proof: Given any x R, then x 0, a or x a,0. Ifx 0, a, then f is bounded 

on 0, a with 

|fx| |f0| V f 0, a |f0| V f , . 

Similarly for x a,0. 

(Theorem 6.9*) Assume that f, andg be of bounded variaton on R, thensoarethier 

sum, difference, and product. Also, we have 

V fg , V f , V g , 

and 

V fg , AV f , BV g , , 

where A sup xR |gx| and B sup xR |fx|. 

Proof: For sum and difference, given any compact interval a, b, wehave 

V fg a, b V f a, b V g a, b, 

V f , V g , 


V fg , V f , V g , . 

For product, given any compact interval a, b, wehave(letAa, b sup xa,b |gx|, 

and Ba, b sup xa,b |fx|), 

V fg a, b Aa, bV f a, b Ba, bV g a, b 

AV f , BV g , 

which implies that

V fg , AV f , BV g , . 

(Theorem 6.10*) Letf be of bounded variation on R, and assume that f is bounded 

away from zero; that is, suppose that there exists a positive number m such that 

0 m |fx| for all x R. Then g 1/f is also of bounded variation on R, and 

V g , V f, 

m 2 . 

Proof: Given any compacgt interval a, b, wehave 


V g a, b V fa, b 

m 2 

V f, 

m 2 

V g , V f, 

. 

m 2 

(Theorem 6.11*) Letf be of bounded variation on R, and assume that c R. Then f is 

of bounded variation on , c and on c, and we have 

V f , V f , c V f c, . 

Proof: Given any a compact interval a, b such that c a, b. Then we have 

V f a, b V f a, c V f c, b. 

Since 

V f a, b V f , 


V f a, c V f , and V f c, b V f , , 

we know that the existence of V f , c and V f c, . Thatis,f is of bounded variation on 

, c and on c, . 

Since 

V f a, c V f c, b V f a, b V f , 


V f , c V f c, V f , , * 

and 

V f a, b V f a, c V f c, b V f , c V f c, 


V f , V f , c V f c, , ** 

we know that 

V f , V f , c V f c, . 

(Theorem 6.12*) Letf be of bounded variation on R. LetVx be defined on , x as 

follows: 

Vx V f , x if x R, andV 0. 

Then (i) V is an increasing function on , and (ii) V f is an increasing function on 

, . 

Proof: (i)Letx y, then we have Vy Vx V f x, y 0. So, we know that V is 

an increasing function on , . 

(ii) Let x y, then we have V fy V fx V f x, y fy fx 0. So,

we know that V f is an increasing function on , . 

(b) Show that Theorem 6.5 is true for , if ”monotonic” is replaced by ”bounded 

and monotonic.” State and prove a similar modefication of Theorem 6.13. 

(Theorem 6.5*) Iff is bounded and monotonic on , , then f is of bounded 

variation on , . 

Proof: Given any compact interval a, b, then we have V f a, b exists, and we have 

V f a, b |fb fa|, sincef is monotonic. In addition, since f is bounded on R, say 

|fx| M for all x, we know that 2M is a upper bounded of V f a, b for all a, b. Hence, 

V f , exists. That is, f is of bounded variation on R. 

(Theorem 6.13*) Letf be defined on , , then f is of bounded variation on 

, if, and only if, f can be expressed as the difference of two increasing and 

bounded functions. 

Proof: Suppose that f is of bounded variation on , , then by Theorem 6.12*, we 

know that 

f V V f, 

where V and V f are increasing on , . In addition, since f is of bounded variation 

on R, we know that V and f is bounded on R which implies that V f is bounded on R. So, 

we have proved that if f is of bounded variation on , then f can be expressed as the 

difference of two increasing and bounded functions. 

Suppose that f can be expressed as the difference of two increasing and bounded 

functions, say f f 1 f 2 , Then by Theorem 6.9*, andTheorem 6.5*, we know that f is of 

bounded variaton on R. 

Remark: The representation of a function of bounded variation as a difference of two 

increasing and bounded functions is by no mean unique. It is clear that Theorem 6.13* 

also holds if ”increasing” is replaced by ”strictly increasing.” For example, 

f f 1 g f 2 g, whereg is any strictly increasing and bounded function on R. One 

of such g is arctan x. 

6.7 Assume that f is of bounded variation on a, b and let 

P x 0 , x 1 ,...,x n þa, b. 

As usual, write f k fx k fx k1 , k 1,2,...,n. Define 

AP k : f k 0, BP k : f k 0. 

The numbers 

and 

p f a, b sup 

f k : P þa, b 

kAP 

n f a, b sup 

|f k | : P þa, b 

kBP 

are called respectively, the positive and negative variations of f on a, b. For each x in 

a, b. LetVx V f a, x, px p f a, x, nx n f a, x, and let 

Va pa na 0. Show that we have: 

(a) Vx px nx.

Proof: Given a partition P on a, x, then we have 

n 

 

k1 

|f k | |f k | |f k | 

kAP 

 

kAP 

kBP 

f k |f k |, 

kBP 

which implies that (taking supermum) 

Vx px nx. * 

Remark: The existence of px and qx is clear, so we know that (*) holds by 

Theorem 1.15. 

(b) 0 px Vx and 0 nx Vx. 

Proof: Consider a, x, and since 

n 

Vx |f k | |f k |, 

k1 kAP 

we know that 0 px Vx. Similarly for 0 nx Vx. 

(c) p and n are increasing on a, b. 

Proof: Letx, y in a, b with x y, and consider py px as follows. Since 

py f k f k , 

kAP, a,y kAP, a,x 

we know that 

py px. 

That is, p is increasing on a, b. Similarly for n. 

(d) fx fa px nx. Part (d) gives an alternative proof of Theorem 6.13. 

Proof: Consider a, x, and since 


fx fa 

k1 

n 

f k f k f k 

kAP kBP 

fx fa |f k | f k 

kBP kAP 

which implies that fx fa px nx. 

(e) 2px Vx fx fa, 2nx Vx fx fa. 

Proof: By (d) and (a), the statement is obvious. 

(f) Every point of continuity of f is also a point of continuity of p and of n. 

Proof: By (e) and Theorem 6.14, the statement is obvious. 

Curves 

6.8 Let f and g be complex-valued functions defined as follows: 

ft e 2it if t 0, 1, gt e 2it if t 0, 2. 

(a) Prove that f and g have the same graph but are not equivalent according to defintion

in Section 6.12. 

Proof: Sinceft : t 0, 1 gt : t 0, 2 the circle of unit disk, we know 

that f and g have the same graph. 

If f and g are equivalent, then there is an 1-1 and onto function : 0, 2 0, 1 such 

that 

ft gt. 

That is, 

e 2it cos2t i sin 2t e 2it cos2t i sin 2t. 

In paticular, 1 : c 0, 1. However, 

fc cos2c i sin 2c g1 1 

which implies that c Z, a contradiction. 

(b) Prove that the length of g is twice that of f. 

Proof: Since 

and 

the length of g 

0 

2 

|g t|dt 4 

the length of f 

0 

1 

|f t|dt 2, 

we know that the length of g is twice that of f. 

6.9 Let f be rectifiable path of length L defined on a, b, and assume that f is not 

constant on any subinterval of a, b. Lets denote the arc length function given by 

sx f a, x if a x b, sa 0. 

(a) Prove that s 1 exists and is continuous on 0, L. 

Proof: ByTheorem 6.19, we know that sx is continuous and strictly increasing on 

0, L. So, the inverse function s 1 exists since s is an 1-1 and onto function, and by 

Theorem 4.29, we know that s 1 is continuous on 0, L. 

(b) Define gt fs 1 t if t 0, L and show that g is equivalent to f. Since 

ft gst, the function g is said to provide a representation of the graph of f with arc 

length as parameter. 

Proof: tisclearbyTheorem 6.20. 

6.10 Let f and g be two real-valued continuous functions of bounded variation defined 

on a, b, with 0 fx gx for each x in a, b, fa ga, fb gb. Leth be the 

complex-valued function defined on the interval a,2b a as follows: 

ht t ift, ifa t b 

2b t ig2b t, ifb t 2b a. 

(a) Show that h describes a rectifiable curve . 

Proof: It is clear that h is continuous on a,2b a. Note that t, f and g are of bounded 

variation on a, b, so h a,2b a exists. That is, h is rectifiable on a,2b a. 

(b) Explain, by means of a sketch, the geometric relationship between f, g, andh. 

Solution: The reader can give it a draw and see the graph lying on x y plane is a

closed region. 

(c) Show that the set of points 

S x, y : a x b, fx y gx 

in a region in R 2 whose boundary is the curve . 

Proof: It can be answered by (b), so we omit it. 

(d) Let H be the complex-valued function defined on a,2b a as follows: 

Ht t 1 igt ft, ifa t b 

2 

2b t 1 ig2b t f2b t, ifb t 2b a. 

2 

Show that H describes a rectifiable curve 0 which is the boundary of the region 

S 0 x, y : a x b, fx gx 2y gx fx. 

Proof: LetFt 1 gt ft and Gt 1 gt ft defined on a, b. Itis 

2 2 

clear that Ft and Gt are of bounded variation and continuous on a, b with 

0 Fx Gx for each x a, b, Fb Gb 0, and Fb Gb 0. In 

addition, we have 

Ht t iFt, ifa t b 

2b t iG2b t, ifb t 2b a. 

So, by preceding (a)-(c), we have prove it. 

(e) Show that, S 0 has the x axis as a line of symmetry. (The region S 0 is called the 

symmetrization of S with respect to x axis.) 

Proof: Itisclearsincex, y S 0 x, y S 0 by the fact 

fx gx 2y gx fx. 

(f) Show that the length of 0 does not exceed the length of . 

Proof: By (e), the symmetrization of S with respect to x axis tells that 

H a, b H b,2b a. So, it suffices to show that h a,2b a 2 H a, b. 

Choosing a partition P 1 x 0 a,...,x n b on a, b such that 

2 H a, b 2 H P 1 

n 

2 x i x i1 2 1 2 f gx i 1 2 f gx i1 2 1/2 

i1 

n 

 

i1 

4x i x i1 2 f gx i f gx i1 2 1/2 

and note that b a 2b a b, we use this P 1 to produce a partition 

P 2 P 1 x n b, x n1 b x n x n1 ,...,x 2n 2b a on a,2b a. Then we have 

*

2n 

h P 2 hx i hx i1 

i1 

n 

2n 

hx i hx i1 hx i hx i1 

i1 

in1 

n 

 

i1 

n 

 

i1 

x i x i1 2 fx i fx i1 2 

2n 

1/2 

 

in1 

x i x i1 2 gx i gx i1 2 1/2 

x i x i1 2 fx i fx i1 2 1/2 

xi x i1 2 gx i gx i1 2 1/2 

From (*) and (**), we know that 

2 H a, b 2 H P 1 h P 2 *** 


H a,2b a 2 H a, b h a,2b a. 

So, we know that the length of 0 does not exceed the length of . 

Remark: Definex i x i1 a i , fx i fx i1 b i ,andgx i gx i1 c i , then we 

have 

4a i2 b i c i 2 1/2 

ai2 b i2 1/2 a i2 c i2 1/2 . 

Hence we have the result (***). 

Proof: It suffices to square both side. We leave it to the reader. 

Absolutely continuous functions 

A real-valued function f defined on a, b is said to be absolutely continuous on a, b 

if for every 0, there is a 0 such that 

n 

 

k1 

|fb k fa k | 

for every n disjoint open subintervals a k , b k of a, b, n 1, 2, . . . , the sum of whose 

n 

lengths k1 

b k a k is less than . 

Absolutely continuous functions occur in the Lebesgue theory of integration and 

differentiation. The following exercises give some of their elementary properties. 

6.11 Prove that every absolutely continuous function on a, b is continuous and of 

bounded variation on a, b. 

Proof: Letf be absolutely continuous on a, b. Then 0, there is a 0 such that 

n 

 

k1 

|fb k fa k | 

for every n disjoint open subintervals a k , b k of a, b, n 1, 2, . . . , the sum of whose 

n 

lengths k1 

b k a k is less than . So,as|x y| , wherex, y a, b, wehave 

|fx fy| . 

That is, f is uniformly continuous on a, b. So,f is continuous on a, b. 

n 

In addition, given any 1, there exists a 0 such that as k1 

b k a k , 

where a k , b k s are disjoint open intervals in a, b, wehave 

**

n 

|fb k fa k | 1. 

k1 

For this , and let K be the smallest positive integer such that K/2 b a. So,we 

partition a, b into K closed subintervals, i.e., 

P y 0 a, y 1 a /2,....,y K1 a K 1/2, y K b. So, it is clear that f is 

of bounded variation y i , y i1 ,wherei 0, 1, . . . , K. It implies that f is of bounded 

variation on a, b. 

Note: There exists functions which are continuous and of bounded variation but 

not absolutely continuous. 

Remark: 1. The standard example is called Cantor-Lebesgue function. The reader 

can see this in the book, Measure and Integral, An Introduction to Real Analysis by 

Richard L. Wheeden and Antoni Zygmund, pp 35 and pp 115. 

2. If we wrtie ”absolutely continuous” by ABC, ”continuous” by C, and ”bounded 

variation” by B, then it is clear that by preceding result, ABC implies B and C, andB and 

C do NOT imply ABC. 

6.12 Prove that f is absolutely continuous if it satisfies a uniform Lipschitz condition 

of order 1 on a, b. (See Exercise 6.2) 

Proof: Letf satisfy a uniform Lipschitz condition of order 1 on a, b, i.e., 

|fx fy| M|x y| where x, y a, b. Thengiven 0, there is a /M such that 

n 

as k1 

b k a k , wherea k , b k s are disjoint open subintervals on a, b, k 1, . . , n, 

we have 

n 

 

k1 

|fb k fa k | M|b k a k | 

n 

k1 

n 

 

k1 

Mb k a k 

M 

. 

Hence, f is absolutely continuous on a, b. 

6.13 If f and g are absolutely continunous on a, b, prove that each of the following 

is also: |f|, cf (c constant), f g, f g; alsof/g if g is bounded away from zero. 

Proof: (1) (|f| is absolutely continuous on a, b): Given 0, we want to find a 

n 

0, such that as k1 

b k a k , wherea k , b k s are disjoint open intervals on 

a, b, wehave 

n 

||fb k | |fa k || . 1* 

k1 

Since f is absolutely continunous on a, b, for this , there is a 0 such that as 

n 

k1 

b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave 

n 

|fb k fa k | 

k1 

which implies that (1*) holds by the following

n 

 

k1 

n 

||fb k | |fa k || |fb k fa k | . 

So, we know that |f| is absolutely continuous on a, b. 

(2) (cf is absolutely continuous on a, b): If c 0, it is clear. So, we may assume that 

n 

c 0. Given 0, we want to find a 0, such that as k1 

b k a k , where 

a k , b k s are disjoint open intervals on a, b, wehave 

n 

 

k1 

k1 

|cfb k cfa k | . 2* 

Since f is absolutely continunous on a, b, for this , there is a 0 such that as 

n 

k1 


n 

|fb k fa k | /|c| 

k1 

which implies that (2*) holds by the following 

n 

 

k1 

|cfb k cfa k | |c| |fb k fa k | . 

So, we know that cf is absolutely continuous on a, b. 

(3) (f g is absolutely continuous on a, b): Given 0, we want to find a 0, 

n 

such that as k1 

b k a k , wherea k , b k s are disjoint open intervals on a, b, we 

have 

n 

|f gb k f ga k | . 3* 

k1 

Since f and g are absolutely continunous on a, b, for this , there is a 0 such that as 

n 


k1 

n 

 

k1 

|fb k fa k | /2 and |gb k ga k | /2 


n 

 

k1 

n 

n 

k1 

n 

k1 

|f gb k f ga k | 

|fb k fa k gb k ga k | 

k1 

n 

 

k1 

n 

|fb k fa k | 

k1 

|gb k ga k | 

. 

So, we know that f g is absolutely continuous on a, b. 

(4) (f g is absolutely continuous on a, b.): Let M f sup xa,b |fx| and 

M g sup xa,b |gx|. Given 0, we want to find a 0, such that as 

n 


k1

n 

|f gb k f ga k | . 4* 

k1 

Since f and g are absolutely continunous on a, b, for this , there is a 0 such that as 

n 


k1 

n 

n 

|fb k fa k | 

2M g 1 and |gb k ga k | 

k1 

k1 


n 

 

k1 

n 

 

k1 

M f 

k1 

 

. 

|f gb k f ga k | 

|fb k gb k ga k ga k fb k fa k | 

n 

n 

|gb k ga k | M g 

k1 

M f 

2M f 1 

M g 

2M g 1 

|fb k fa k | 

 

2M f 1 

Remark: The part shows that f n is absolutely continuous on a, b, wheren N, iff is 

absolutely continuous on a, b. 

(5) (f/g is absolutely continuous on a, b): By (4) it suffices to show that 1/g is 

absolutely continuous on a, b. Sinceg is bounded away from zero, say 0 m gx for 

n 

all x a, b. Given 0, we want to find a 0, such that as k1 

b k a k , 

where a k , b k s are disjoint open intervals on a, b, wehave 

n 

 

k1 

|1/gb k 1/ga k | . 5* 

Since g is absolutely continunous on a, b, for this , there is a 0 such that as 

n 

k1 


n 

|gb k ga k | m 2 

k1 


n 

 

k1 

n 

 

k1 

|1/gb k 1/ga k | 

gb k ga k 

gb k ga k 

n 

1 |gb m 2 k ga k | 

k1 

.

Supplement on lim sup and lim inf 

Introduction 

In order to make us understand the information more on approaches of a given real 

sequence a n 

n1 

, we give two definitions, thier names are upper limit and lower limit. It 

is fundamental but important tools in analysis. We do NOT give them proofs. The reader 

can see the book, Infinite Series by Chao Wen-Min, pp 84-103. (Chinese Version) 

Definition 

Definition of limit sup and limit inf 

and 

Example 

Example 

Example 

Given a real sequence a n 

n1 

,wedefine 

b n supa m : m ≥ n 

1 −1 n 

n1 

−1 n 

n n1 

 

−n n1 

c n infa m : m ≥ n. 

0,2,0,2,..., sowehave 

b n 2andc n 0 for all n. 

−1, 2, −3,4,..., sowehave 

b n and c n − for all n. 

−1, −2, −3, . . . , sowehave 

b n −n and c n − for all n. 

Proposition Given a real sequence a n 

n1 

, and thus define b n and c n as the same as 

before. 

1 b n ≠−,andc n ≠∀n ∈ N. 

2 If there is a positive integer p such that b p , then b n ∀n ∈ N. 

If there is a positive integer q such that c q −, then c n − ∀n ∈ N. 

3 b n is decreasing and c n is increasing. 

By property 3, we can give definitions on the upper limit and the lower limit of a given 

sequence as follows. 

Definition Given a real sequence a n and let b n and c n as the same as before. 

(1) If every b n ∈ R, then 

infb n : n ∈ N 

is called the upper limit of a n , denoted by 

lim n→ 

sup a n . 

That is, 

lim n→ 

sup a n inf b n. 

n 

If every b n , then we define 

lim n→ 

sup a n . 

(2) If every c n ∈ R, then 

supc n : n ∈ N 

is called the lower limit of a n , denoted by

lim n→ 

inf a n . 

That is, 

lim n→ 

inf a n sup c n . 

n 

If every c n −, then we define 

lim n→ 

inf a n −. 

Remark The concept of lower limit and upper limit first appear in the book (Analyse 

Alge’brique) written by Cauchy in 1821. But until 1882, Paul du Bois-Reymond 

gave explanations on them, it becomes well-known. 

Example 1 −1 n 

n1 

0,2,0,2,..., sowehave 

b n 2andc n 0 for all n 


lim sup a n 2 and lim inf a n 0. 

Example −1 n 

n n1 

−1, 2, −3,4,..., sowehave 

b n and c n − for all n 


lim sup a n and lim inf a n −. 

 

−n n1 

Example −1, −2, −3, . . . , sowehave 

b n −n and c n − for all n 


lim sup a n − and lim inf a n −. 

Relations with convergence and divergence for upper (lower) limit 

Theorem Let a n be a real sequence, then a n converges if, and only if, the upper 

limit and the lower limit are real with 

lim n→ 

sup a n lim n→ 

inf a n lim n→ 

a n . 

Theorem Let a n be a real sequence, then we have 

(1) lim n→ sup a n a n has no upper bound. 

(2) lim n→ sup a n − for any M 0, there is a positive integer n 0 such 

that as n ≥ n 0 ,wehave 

a n ≤−M. 

(3) lim n→ sup a n a if, and only if, (a) given any 0, there are infinite 

many numbers n such that 

a − a n 

and (b) given any 0, there is a positive integer n 0 such that as n ≥ n 0 ,wehave 

a n a . 


Theorem Let a n be a real sequence, then we have

(1) lim n→ inf a n − a n has no lower bound. 

(2) lim n→ inf a n for any M 0, there is a positive integer n 0 such 

that as n ≥ n 0 ,wehave 

a n ≥ M. 

(3) lim n→ inf a n a if, and only if, (a) given any 0, there are infinite 


a a n 

and (b) given any 0, there is a positive integer n 0 such that as n ≥ n 0 ,wehave 

a n a − . 

From Theorem 2 an Theorem 3, the sequence is divergent, we give the following 

definitios. 

Definition Let a n be a real sequence, then we have 

(1) If lim n→ sup a n −, then we call the sequence a n diverges to −, 

denoted by 

lim n→ 

a n −. 

Theorem 



(2) If lim n→ inf a n , then we call the sequence a n diverges to , 

denoted by 

lim n→ 

a n . 

Let a n be a real sequence. If a is a limit point of a n , then we have 

lim n→ 

inf a n ≤ a ≤ lim n→ 

sup a n . 

Some useful results 

Let a n be a real sequence, then 

(1) lim n→ inf a n ≤ lim n→ sup a n . 

(2) lim n→ inf−a n − lim n→ sup a n and lim n→ sup−a n − lim n→ inf a n 

(3) If every a n 0, and 0 lim n→ inf a n ≤ lim n→ sup a n , then we 

have 

lim n→ 

sup a 1 n 

1 

lim n→ inf a n 

Let a n and b n be two real sequences. 

(1) If there is a positive integer n 0 such that a n ≤ b n , then we have 

lim n→ 

inf a n ≤ lim n→ 

inf b n and lim n→ 

sup a n ≤ lim n→ 

sup b n . 

(2) Suppose that − lim n→ inf a n , lim n→ inf b n , lim n→ sup a n , 

lim n→ sup b n , then 

lim n→ 


inf b n 

and lim n→ 

inf 1 a n 

1 

lim n→ sup a n 

. 

≤ lim n→ 

infa n b n 

≤ lim n→ 


sup b n (or lim n→ 


inf b n ) 

≤ lim n→ 

supa n b n 

≤ lim n→ 


sup b n .

In particular, if a n converges, we have 

lim n→ 

supa n b n lim n→ 

a n lim n→ 

sup b n 

and 

lim n→ 

infa n b n lim n→ 

a n lim n→ 

inf b n . 

(3) Suppose that − lim n→ inf a n , lim n→ inf b n , lim n→ sup a n , 

lim n→ sup b n , anda n 0, b n 0 ∀n, then 

lim n→ 

inf a n 

lim n→ 

inf b n 

≤ lim n→ 

infa n b n 

≤ lim n→ 


sup b n (or lim n→ 

inf b n 

≤ lim n→ 

supa n b n 

≤ lim n→ 


sup b n . 


and 

lim n→ 

supa n b n 

lim n→ 

infa n b n 

lim n→ 

a n 

lim n→ 

a n 

lim n→ 

sup b n 

lim n→ 

inf b n . 

lim n→ 

sup a n ) 

Theorem Let a n be a positive real sequence, then 

lim n→ 

inf a n1 

a n 

≤ lim n→ 

infa n 1/n ≤ lim n→ 

supa n 1/n ≤ lim n→ 

sup a n1 

a n 

. 

Remark We can use the inequalities to show 

n! 1/n 

lim n→ n 1/e. 

Theorem Let a n be a real sequence, then 

lim n→ 


inf a 1 ...a n 

n 

≤ lim n→ 

sup a 1 ...a n 

n 

≤ lim n→ 

sup a n . 

Exercise Let f : a, d → R be a continuous function, and a n is a real sequence. If f is 

increasing and for every n, lim n→ inf a n , lim n→ sup a n ∈ a, d, then 

lim n→ 

sup fa n f lim n→ 

sup a n 

and lim n→ 

inf fa n f lim n→ 

inf a n . 

Remark: (1) The condition that f is increasing cannot be removed. For 

example, 

fx |x|, 

and 

1/k if k is even 

a k 

−1 − 1/k if k is odd. 

(2) The proof is easy if we list the definition of limit sup and limit inf. So, we 

omit it. 

Exercise Let a n be a real sequence satisfying a np ≤ a n a p for all n, p. Show that 

an 

n converges. 

Hint: Consider its limit inf.

Remark: The exercise is useful in the theory of Topological Entorpy. 

Infinite Series And Infinite Products 

Sequences 

8.1 (a) Given a real-valed sequence a n bounded above, let u n supa k : k ≥ n. 

Then u n ↘ and hence U lim n→ u n is either finite or −. Prove that 

U lim n→ 


supa k : k ≥ n. 

Proof: It is clear that u n ↘ and hence U lim n→ u n is either finite or −. 

If U −, then given any M 0, there exists a positive integer N such that as n ≥ N, 

we have 

u n ≤−M 

which implies that, as n ≥ N, a n ≤−M. So, lim n→ a n −. Thatis,a n is not bounded 

below. In addition, if a n has a finite limit supreior, say a. Thengiven 0, and given 

m 0, there exists an integer n m such that 

a n a − 

which contradicts to lim n→ a n −. From above results, we obtain 

U lim n→ 

sup a n 

in the case of U −. 

If U is finite, then given 0, there exists a positive integer N such that as n ≥ N, we 

have 

U ≤ u n U . 

So, as n ≥ N, u n U which implies that, as n ≥ N, a n U . In addition, given 

′ 0, and m 0, there exists an integer n m, 

U − ′ a n 

by U ≤ u n supa k : k ≥ n if n ≥ N. From above results, we obtain 

U lim n→ 

sup a n 

in the case of U is finite. 

(b)Similarly, if a n is bounded below, prove that 

V lim n→ 


infa k : k ≥ n. 

Proof: Since the proof is similar to (a), we omit it. 

If U and V are finite, show that: 

(c) There exists a subsequence of a n which converges to U and a subsequence which 

converges to V. 

Proof: SinceU lim sup n→ a n by (a), then 

(i) Given 0, there exists a positive integer N such that as n ≥ N, wehave 

a n U . 

(ii) Given 0, and m 0, there exists an integer Pm m, 

U − a Pm . 

Hence, a Pm is a convergent subsequence of a n with limit U. 

Similarly for the case of V.

(d) If U V, every subsequnce of a n converges to U. 

Proof: By(a)and(b),given 0, then there exists a positive integer N 1 such that as 

n ≥ N 1 ,wehave 

a n U 

and there exists a positive integer N 2 such that as n ≥ N 2 ,wehave 

U − a n . 

Hence, as n ≥ maxN 1 , N 2 ,wehave 

U − a n U . 

That is, a n is a convergent sequence with limit U. So, every subsequnce of a n 

converges to U. 

8.2 Given two real-valed sequence a n and b n bounded below. Prove hat 

(a) lim sup n→ a n b n ≤ lim sup n→ a n lim sup n→ b n . 

Proof: Note that a n and b n bounded below, we have lim sup n→ a n or is 

finite. And lim sup n→ b n or is finite. It is clear if one of these limit superior is , 

sowemayassumethatbotharefinite.Leta lim sup n→ a n and b lim sup n→ b n . Then 

given 0, there exists a positive integer N such that as n ≥ N, wehave 

a n b n a b /2. * 

In addition, let c lim sup n→ a n b n ,wherec by (*). So, for the same 0, and 

given m N there exists a positive integer K such that as K ≥ N, wehave 

c − /2 a K b K . ** 

By (*) and (**), we obtain that 

c − /2 a K b K a b /2 


c ≤ a b 

since is arbitrary. So, 

lim sup a n b n ≤ lim sup a n lim sup b n. 

n→ n→ n→ 

Remark: (1) The equality may NOT hold. For example, 

a n −1 n and b n −1 n1 . 

(2) The reader should noted that the finitely many terms does NOT change the relation 

of order. The fact is based on process of proof. 

(b) lim sup n→ a n b n ≤ lim sup n→ a n lim sup n→ b n if a n 0, b n 0 for all n, and 

if both lim sup n→ a n and lim sup n→ b n are finite or both are infinite. 

Proof: Let lim sup n→ a n a and lim sup n→ b n b. It is clear that we may assume 

that a and b are finite. Given 0, there exists a positive integer N such that as n ≥ N, 

we have 

a n b n a b ab a b . * 

In addition, let c lim sup n→ a n b n ,wherec by (*). So, for the same 0, and 

given m N there exists a positive integer K such that as K ≥ N, wehave 

c − a K b K . ** 

By (*) and (**), we obtain that 

c − a K b K a b a b


c ≤ a b 

since is arbitrary. So, 

lim sup a nb n ≤ lim sup a n lim sup b n . 

n→ n→ n→ 

Remark: (1) The equality may NOT hold. For example, 

a n 1/n if n is odd and a n 1ifn is even. 

and 

b n 1ifn is odd and b n 1/n if n is even. 

(2) The reader should noted that the finitely many terms does NOT change the relation 

of order. The fact is based on the process of the proof. 

(3) The reader should be noted that if letting A n log a n and B n log b n , thenby(a) 

and log x is an increasing function on 0, , we have proved (b). 

8.3 Prove that Theorem 8.3 and 8.4. 

(Theorem 8.3) Leta n be a sequence of real numbers. Then we have: 

(a) lim inf n→ a n ≤ lim sup n→ a n . 

Proof: If lim sup n→ a n , then it is clear. We may assume that 

lim sup n→ a n . Hence, a n is bounded above. We consider two cases: (i) 

lim sup n→ a n a, wherea is finite and (ii) lim sup n→ a n −. 

For case (i), if lim inf n→ a n −, then there is nothing to prove it. We may assume 

that lim inf n→ a n a ′ ,wherea ′ is finite. By definition of limit superior and limit inferior, 

given 0, there exists a positive integer N such that as n ≥ N, wehave 

a ′ − /2 a n a /2 

which implies that a ′ ≤ a since is arbitrary. 

For case (ii), since lim sup n→ a n −, wehavea n is not bounded below. If 

lim inf n→ a n −, then there is nothing to prove it. We may assume that 

lim inf n→ a n a ′ ,wherea ′ is finite. By definition of limit inferior, given 0, there 

exists a positive integer N such that as n ≥ N, wehave 

a ′ − /2 a n 

which contradicts that a n is not bounded below. 

So, from above results, we have proved it. 

(b) The sequence converges if and only if, lim sup n→ a n and lim inf n→ a n are both 

finite and equal, in which case lim n→ a n lim inf n→ a n lim sup n→ a n . 

Proof: ()Given a n a convergent sequence with limit a. So, given 0, there 

exists a positive integer N such that as n ≥ N, wehave 

a − a n a . 

By definition of limit superior and limit inferior, a lim inf n→ a n lim sup n→ a n . 

()By definition of limit superior, given 0, there exists a positive integer N 1 such 

that as n ≥ N 1 ,wehave 

a n a 

and by definition of limit superior, given 0, there exists a positive integer N 2 such that 

as n ≥ N 2 ,wehave

a − a n . 

So, as n ≥ maxN 1 , N 2 ,wehave 

a − a n a . 

That is, lim n→ a n a. 

(c) The sequence diverges to if and only if, lim inf n→ a n lim sup n→ a n . 

Proof: ()Given a sequence a n with lim n→ a n . So, given M 0, there is a 

positive integer N such that as n ≥ N, wehave 

M ≤ a n . * 

It implies that a n is not bounded above. So, lim sup n→ a n . In order to show that 

lim inf n→ a n . We first note that a n is bounded below. Hence, lim inf n→ a n ≠−. 

So, it suffices to consider that lim inf n→ a n is not finite. (So, we have 

lim inf n→ a n . ). Assume that lim inf n→ a n a, wherea is finite. Then given 1, 

and an integer m, there exists a positive Km m such that 

a Km a 1 

which contradicts to (*) if we choose M a 1. So, lim inf n→ a n is not finite. 

(d) The sequence diverges to − if and only if, lim inf n→ a n lim sup n→ a n −. 

Proof: Note that, lim sup n→ −a n − lim inf n→ a n . So, by (c), we have proved it. 

(Theorem 8.4)Assume that a n ≤ b n for each n 1,2,.... Then we have: 

lim n→ 


inf b n and lim n→ 

sup a n ≤ lim n→ 

sup b n . 

Proof: If lim inf n→ b n , there is nothing to prove it. So, we may assume that 

lim inf n→ b n . That is, lim inf n→ b n − or b, whereb is finite. 

For the case, lim inf n→ b n −, it means that the sequence a n is not bounded 

below. So, b n is also not bounded below. Hence, we also have lim inf n→ a n −. 

For the case, lim inf n→ b n b, whereb is finite. We consider three cases as follows. 

(i) if lim inf n→ a n −, then there is nothing to prove it. 

(ii) if lim inf n→ a n a, wherea is finite. Given 0, then there exists a positive 

integer N such that as n ≥ N 

a − /2 a n ≤ b n b /2 

which implies that a ≤ b since is arbitrary. 

(iii) if lim inf n→ a n , then by Theorem 8.3 (a) and (c), we know that 

lim n→ a n which implies that lim n→ b n . Also, by Theorem 8.3 (c), wehave 

lim inf n→ b n which is absurb. 

So, by above results, we have proved that lim inf n→ a n ≤ lim inf n→ b n . 

Similarly, we have lim sup n→ a n ≤ lim sup n→ b n . 

8.4 If each a n 0, prove that 

lim n→ 

inf a n1 

a n 

≤ lim n→ 

infa n 1/n ≤ lim n→ 


sup a n1 

a n 

. 

Proof: ByTheorem 8.3 (a), it suffices to show that 

lim n→ 

inf a n1 

a n 

We first prove 

≤ lim n→ 

infa n 1/n and lim n→ 


sup a n1 

a n 

. 

lim n→ 


sup a n1 

a n 

.

If lim sup n→ 

a n1 

a n 

, then it is clear. In addition, since a n1 

a n 

is positive, 

a 

≠−. So, we may assume that lim sup n1 

n→ a n 

a, wherea is finite. 

Given 0, then there exists a positive integer N such that as n ≥ N, wehave 

a n1 

a n 

a 

lim sup n→ 

a n1 

a n 


So, 


a Nk a N a k ,wherek 1,2,.... 

a Nk 1 

Nk 

a N 1 

Nk a 

k 

Nk 

lim supa Nk 1 

Nk 

k→ 

≤ lim supa N 1 

k 

Nk a 

Nk 

k→ 

a . 

So, 

lim supa Nk 1 

Nk 

≤ a 

k→ 

since is arbitrary. Note that the finitely many terms do NOT change the value of limit 

superiror of a given sequence. So, we finally have 

lim n→ 

supa n 1/n ≤ a lim n→ 

sup a n1 

a n 

. 

Similarly for 

lim n→ 

inf a n1 

a n 

≤ lim n→ 

infa n 1/n . 

Remark: These ineqaulities is much important; we suggest that the reader keep it mind. 

At the same time, these inequalities tells us that the root test is more powerful than the 

ratio test. We give an example to say this point. Given a series 

1 

2 1 3 1 2 1 2 3 ... 1 2 2 n 3 1 n ... 

where 

n n 

a 2n−1 1 ,anda2n 1 , n 1, 2, . . . 

2 3 

with 

and 

lim n→ 

inf a n1 

a n 

lim n→ 

supa n 1/n 1 

2 

1 

0, lim n→ 

sup a n1 

a n 

. 

8.5 Let a n n n /n!. Show that lim n→ a n1 /a n e and use Exercise 8.4 to deduce that 

lim n 

n→ 

e. 

n! 1/n 

Proof: Since 

a n1 

a n 

n 1n1 n! 

n 1!n n 1 1 n 

n → e, 

by Exercise 8.4, wehave 

lim n→ 

a n 1/n lim n 

n→ 

e. 

n! 1/n

Remark: There are many methods to show this. We do NOT give the detailed proof. 

But there are hints. 

(1) Taking log on n! 

n n 1/n , and thus consider 

1 

n log 1 n ... log n n → 

0 

1 

log xdx −1. 

(2) Stirling’s Formula: 

n! n n e −n 2n e 

12n ,where ∈ 0, 1. 

Note: In general, we have 

lim 

x→ 

Γx 1 

x x e −x 2x 1, 

where Γx is the Gamma Function. The reader can see the book, Principles of 

Mathematical Analysis by Walter Rudin, pp 192-195. 

(3) Note that 1 1 x x ↗ e and 1 1 x x1 ↘ e on 0, . So, 

1 1 n n1 

n e 1 1 n 


en n e −n n! en n1 e −n . 

(4) Using O-Stolz’s Theorem: Let lim n→ y n and y n ↗.If 

lim 

x n1 − x n 

n→ y n1 − y n 

a, wherea is finite or , 

then 

lim 

x n 

n→ y n 

a. 

Let x n log 1 n ... log n n and y n n. 

Note: For the proof of O-Stolz’s Theorem, the reader can see the book, An 

Introduction to Mathematical Analysis by Loo-Keng Hua, pp 195. (Chinese Version) 

(5) Note that, if a n is a positive sequence with lim n→ a n a, then 

a 1 a n 1/n → a as n → . 

Taking a n 1 1 n n , then 

a 1 a n 1/n n n 

1 

n! 

1 n → e. 

Note: For the proof, it is easy from the Exercise 8.6. We give it a proof as follows. Say 

lim n→ a n a. Ifa 0, then by A. P. ≥ G. P. , we have 

a 1 a n 1/n ≤ a 1 ...a n 

n → 0byExercise 8.6. 

So, we consider a ≠ 0 as follows. Note that log a n → log a. So, by Exercise 8.6, 

log a 1 ... log a n 

n → log a 

which implies that a 1 a n 1/n → a. 

8.6 Let a n be real-valued sequence and let n a 1 ...a n /n. Show that 

lim n→ 


inf n ≤ lim n→ 

sup n ≤ lim n→ 

sup a n . 

Proof: ByTheorem 8.3 (a), it suffices to show that 

lim n→ 


inf n and lim n→ 


sup a n . 

1/n

We first prove 

lim n→ 


sup a n . 

If lim sup n→ a n , there is nothing to prove it. We may assume that 

lim sup n→ a n − or a, wherea is finite. 

For the case, lim sup n→ a n −, then by Theorem 8.3 (d), wehave 

lim n→ 

a n −. 

So, given M 0, there exists a positive integer N such that as n ≥ N, wehave 

a n ≤−M. * 

Let n N, wehave 

n a 1 ...a N ..a n 

n 

a 1 ...a N 

n 

a N1 ...a n 

n 

≤ a 1 ...a N 

n n − N 

n −M 


lim n→ 

sup n ≤−M. 

Since M is arbitrary, we finally have 

lim n→ 

sup n −. 

For the case, lim sup n→ a n a, wherea is finite. Given 0, there exists a positive 

integer N such that as n ≥ N, wehave 

a n a . 

Let n N, wehave 

n a 1 ...a N ..a n 

n 

a 1 ...a N 

n 

a N1 ...a n 

n 

≤ a 1 ...a N 

n n − N 

n 

a 


lim n→ 

sup n ≤ a 


lim n→ 

sup n ≤ a 

since is arbitrary. 

Hence, from above results, we have proved that lim sup n→ n ≤ lim sup n→ a n . 

Similarly for lim inf n→ a n ≤ lim inf n→ n . 

Remark: We suggest that the reader keep it in mind since it is the fundamental and 

useful in the theory of Fourier Series. 

8.7 Find lim sup n→ a n and lim inf n→ a n if a n is given by 

(a) cosn 

Proof: Note that, a b : a, b ∈ Z is dense in R. Bycosn cosn 2k, we 

know that 

lim n→ 

sup cos n 1 and lim n→ 

inf cos n −1.

Remark: The reader may give it a try to show that 

lim n→ 

sup sin n 1 and lim n→ 

inf sin n −1. 

(b) 1 1 n cosn 

Proof: Note that 

So, it is clear that 

lim n→ 

sup 1 1 n 

1 1 n cosn 

1ifn 2k 

−1 ifn 2k − 1 . 

cosn 1 and lim n→ 

inf 1 1 n cosn −1. 

(c) n sin n 3 

Proof: Note that as n 1 6k, n sin n 1 6k sin ,andasn 4 6k, 

3 3 

n −4 6k sin . So, it is clear that 

3 

lim n→ 

sup n sin n and lim 

3 

inf n sin n n→ 3 

−. 

(d) sin n cos n 2 2 

Proof: Note that sin n cos n 1 sin n 0, we have 

2 2 2 

lim n→ 

sup sin n 2 cos n 2 

lim inf sin n n→ 2 cos n 2 0. 

(e) −1 n n/1 n n 


we know that 

(f) n 3 

− n 3 


So, it is clear that 

lim n→ 

−1 n n/1 n n 0, 

lim n→ 

sup−1 n n/1 n n lim n→ 

inf−1 n n/1 n n 0. 

n 

3 − n 3 

 

1 

3 

2 

3 

if n 3k 1 

if n 3k 2 

0ifn 3k 

,wherek 0, 1, 2, . . . . 

lim n→ 

sup n 3 − n 

3 

2 and lim 

3 inf n n→ 3 − n 3 

Note.In(f),x denoted the largest integer ≤ x. 

n 

8.8 Let a n 2 n − ∑ k1 

1/ k . Prove that the sequence a n converges to a limit p 

in the interval 1 p 2. 

n 

n 

Proof: Consider ∑ k1 

1/ k : S n and x 

−1/2 

dx : T n , then 

1 

lim n→ 

d n exists, where d n S n − T n 

by Integral Test. We denote the limit by d, then 

0.

0 ≤ d 1 * 

by Theorem 8.23 (i). Note that d n − fn is a positive increasing sequence, so we have 

d 0. ** 

Since 

T n 2 n − 2 


lim n→ 

n 

2 n − ∑ 1/ k lim n→ 

a n 2 − d p. 

k1 

By (*) and (**), we have proved that 1 p 1. 

Remark: (1) The use of Integral Test is very useful since we can know the behavior 

of a given series by integral. However, in many cases, the integrand may be so complicated 

that it is not easy to calculate. For example: Prove that the convergence of 

 

∑ 

n2 

1 

nlog n p ,wherep 1. 

Of course, it can be checked by Integral Test. But there is the Theorem called Cauchy 

Condensation Theorem much powerful than Integral Test in this sense. In addition, the 

reader can think it twice that in fact, Cauchy condensation Theorem is equivalent to 

Integral Test. 

(Cauchy Condensation Theorem)Let a n be a positive decreasing sequence. Then 

 

∑ 

n1 

a n converges if, and only if, ∑ 

k0 

 

2 k a 2 

k converges. 

Note: (1) The proof is not hard; the reader can see the book, Principles of 

Mathematical Analysis by Walter Rudin, pp 61-63. 

(2) There is an extension of Cauchy Condensation Theorem (Oskar Schlomilch): 

Suppose that a k be a positive and decreasing sequence and m k ⊆ N is a sequence. If 

there exists a c 0 such that 

0 m k2 − m k1 ≤ cm k1 − m k for all k, 

then 

 

∑ 

k1 

 

a k converges if, and only if, ∑m k1 − m k a mk . 

k0 

Note: The proof is similar with Cauchy Condensation Theorem, soweomitit. 

(2) There is a similar Theorem, we write it as a reference. If t ≥ a, ft is a 

non-negative increasing function, then as x ≥ a, wehave 

x 

∑ fn − ftdt 

a≤n≤x 

a 

≤ fx. 

Proof: The proof is easy by drawing a graph. So, we omit it. 

P.S.: The theorem is useful when we deal with some sums. For example, 

ft log t. 

Then

∑ log n − x log x x − 1 ≤ log x. 

1≤n≤x 

In particular, as x ∈ N, we thus have 

n log n − n 1 − log n ≤ log n! ≤ n log n − n 1 log n 


n n−1 e −n1 ≤ n! ≤ n n1 e −n1 . 

In each of Exercise 8.9. through 8.14, show that the real-valed sequence a n is 

convergent. The given conditions are assumed to hold for all n ≥ 1. In Exercise 8.10 

through 8.14, show that a n has the limit L indicated. 

8.9 |a n| 2, |a n2 − a n1 | ≤ 1 8 |a 2 

n1 − a n2 |. 

Proof: Since 

|a n2 − a n1 | ≤ 1 8 |a 2 

n1 − a n2 | 

1 8 |a n1 − a n||a n1 a n| 

we know that 

So, 

≤ 1 2 |a n1 − a n| since |a n| 2 

|a n1 − a n| ≤ 1 

2 

n−1 

|a2 − a 1 | ≤ 1 

2 

n−3 

. 

k 

|a nk − a n| ≤ ∑|a nj − a nj−1 | 

j1 

k 

≤ ∑ 

j1 

1 

2 

nj−4 

≤ 1 

2 

n−2 

→ as n → . 

Hence, a n is a Cauchy sequence. So, a n is a convergent sequence. 

Remark: (1) If |a n1 − a n| ≤ b n for all n ∈ N, and∑ b n converges, then ∑ a n 

converges. 

Proof: Since the proof is similar with the Exercise, we omit it. 

(2) In (1), the condition ∑ b n converges CANNOT omit. For example, 

n 1 

(i) Let a n sin ∑ k1 

Or 

k 

(ii) a n is defined as follows: 

a 1 1, a 2 1/2, a 3 0, a 4 1/4, a 5 1/2, a 6 3/4, a 7 1, and so on. 

8.10 a 1 ≥ 0, a 2 ≥ 0, a n2 a n a n1 1/2 , L a 1 a 22 1/3 . 

Proof: If one of a 1 or a 2 is 0, then a n 0 for all n ≥ 2. So, we may assume that 

a 1 ≠ 0anda 2 ≠ 0. So, we have a n ≠ 0 for all n. Letb n a n1 

a n 

, then 

b n1 1/ b n for all n 

which implies that

Consider 



b n1 b 1 −1 

2 

n 

→ 1asn → . 

n1 j2 b j n 

j1 b j −1/2 

a 1 1/2 a 2 

−2/3 

an1 1 

b n1 

2/3 

lim n→ 

a n1 a 1 a 22 1/3 . 

Remark: There is another proof. We write it as a reference. 

Proof: If one of a 1 or a 2 is 0, then a n 0 for all n ≥ 2. So, we may assume that 

a 1 ≠ 0anda 2 ≠ 0. So, we have a n ≠ 0 for all n. Leta 2 ≥ a 1 .Sincea n2 a n a n1 1/2 , 

then inductively, we have 

a 1 ≤ a 3 ≤...≤ a 2n−1 ≤...≤ a 2n ≤...≤ a 4 ≤ a 2 . 

So, both of a 2n and a 2n−1 converge. Say 

lim n→ 

a 2n x and lim n→ 

a 2n−1 y. 

Note that a 1 ≠ 0anda 2 ≠ 0, so x ≠ 0, and y ≠ 0. In addition, x y by 

a n2 a n a n1 1/2 . Hence, a n converges to x. 

By a n2 a n a n1 1/2 , and thus 

n 

j1 a2 j2 n 

j1 a j a j1 a 1 a 22 a n1 n−2 j1 a2 

j2 



a n1 a2 n2 a 1 a2 

2 

lim n→ 

a n x a 1 a 22 1/3 . 

8.11 a 1 2, a 2 8, a 2n1 1 2 a 2n a 2n−1 , a 2n2 a 2na 2n−1 

a 2n1 

, L 4. 

Proof: First, we note that 

a 2n1 a 2n a 2n−1 

2 

≥ a 2n a 2n−1 by A. P. ≥ G. P. * 

for n ∈ N. So, by a 2n2 a 2na 2n−1 

a 2n1 

and (*), 

a 2n2 a 2na 2n−1 

a 2n1 

≤ a 2n a 2n−1 ≤ a 2n1 for all n ∈ N. 

Hence, by Mathematical Induction, itiseasytoshowthat 

a 4 ≤ a 6 ≤...≤ a 2n2 ≤...≤ a 2n1 ≤...≤ a 5 ≤ a 3 

for all n ∈ N. It implies that both of a 2n and a 2n−1 converge, say 

lim n→ 

a 2n x and lim n→ 

a 2n−1 y. 

With help of a 2n1 1 a 2 2n a 2n−1 , we know that x y. In addition, by a 2n2 a 2na 2n−1 

a 2n1 

, 

a 1 2, and a 2 8, we know that x 4. 

8.12 a 1 −3 2 n1 2 a n3 , L 1. Modify a 1 to make L −2. 

Proof: ByMathematical Induction, itiseasytoshowthat 

− 2 ≤ a n ≤ 1 for all n. *

So, 

3a n1 − a n a n3 − 3a n 2 ≥ 0 

by (*) and fx x 3 − 3x 2 x − 1 2 x 2 ≥ 0on−2, 1. Hence, a n is an 

increasing sequence with a upper bound 1. So, a n is a convergent sequence with limit L. 

So,by3a n1 2 a n3 , 

L 3 − 3L 2 0 


L 1or − 2. 

So, L 1sinca n ↗ and a 1 −3/2. 

In order to make L −2, it suffices to let a 1 −2, then a n −2 for all n. 

8.13 a 1 3, a n1 31an 

3a n 

, L 3 . 

Proof: By Mathematical Induction, itiseasytoshowthat 

a n ≥ 3 for all n. * 

So, 

a n1 − a n 3 − a n 2 

≤ 0 

3 a n 

which implies that a n is a decreasing sequence. So, a n is a convergent sequence with 

limit L by (*). Hence, 

31 L 

L 

3 L 


L 3 . 

So, L 3 since a n ≥ 3 for all n. 

and 

8.14 a n b n1 

b n 

,whereb 1 b 2 1, b n2 b n b n1 , L 1 5 

Hint. Show that b n2 b n − b2 n1 −1 n1 and deduce that |a n − a n1 | n −2 ,ifn 4. 

Proof: ByMathematical Induction, itiseasytoshowthat 

Thus, (Note that b n ≠ 0 for all n) 

|a n1 − a n| b n2 

b n1 

b n2 b n − b2 n1 −1 n1 for all n 

b n ≥ n if n 4 

2 

. 

− b n1 

−1n1 ≤ 1 

b n b n b n1 nn 1 1 if n 4. 

n2 So, a n is a Cauchy sequence. In other words, a n is a convergent sequence, say 

lim n→ b n L. Then by b n2 b n b n1 ,wehave 

b n2 

b n 

1 

b n1 b n1 

which implies that (Note that 0 ≠L ≥ 1sincea n ≥ 1 for all n) 

L 

L 1 1 


L 1 5 

2 

.

So, L 1 5 since L ≥ 1. 

2 

Remark: (1) The sequence b n is the famous sequence named Fabonacci sequence. 

There are many researches around it. Also, it is related with so called Golden Section, 

5 −1 

0.618.... 

2 

(2) The reader can see the book, An Introduction To The Theory Of Numbers by G. 

H. Hardy and E. M. Wright, Chapter X. Then it is clear by continued fractions. 

(3) There is another proof. We write it as a reference. 

Proof: (STUDY) Sinceb n2 b n b n1 , we may think 

x n2 x n x n1 , 

and thus consider x 2 x 1. Say and are roots of x 2 x 1, with . Thenlet 

F n n − n 

− , 

we have 

F n b n . 

So,itiseasytoshowthatL 1 5 . We omit the details. 

2 

Note: The reader should be noted that there are many methods to find the formula of 

Fabonacci sequence like F n . For example, using the concept of Eigenvalues if we can 

find a suitable matrix. 

Series 

8.15 Test for convergence (p and q denote fixed rela numbers). 

(a) ∑ 

n1 

n 3 e −n 

Proof: ByRoot Test, wehave 

So, the series converges. 

(b) ∑ 

n2 

log n p 

lim n→ 

sup 

n3 1/n 

e 1/e 1. 

n 

Proof: We consider 2 cases: (i) p ≥ 0, and (ii) p 0. 

For case (i), the series diverges since log n p does not converge to zero. 

For case (ii), the series diverges by Cauchy Condensation Theorem (or Integral 

Test.) 

(c) ∑ n1 

p n n p (p 0) 

Proof: ByRoot Test, wehave 

lim n→ 

sup 

pn 1/n 

n p. 

p 

So, as p 1, the series diverges, and as p 1, the series converges. For p 1, it is clear 

that the series ∑ n diverges. Hence, 

and 

 

∑ 

n1 

p n n p converges if p ∈ 0, 1

∑ p n n p diverges if p ∈ 1, . 

n1 

1 

(d) ∑ n2 n p −n 

(0 q p) 

q 

1 


n p −n 

1 1 

q n 

. We consider 2 cases: (i) p 1 and (ii) p ≤ 1. 

p 1−n q−p 

For case (i), by Limit Comparison Test with 1 n 

, p 

1 

n 

lim 

p −n q 

n→ 1 

1, 

n p 

the series converges. 

For case (ii), by Limit Comparison Test with 1 n 

, p 

the series diverges. 

(e) ∑ 

n1 

n −1−1/n 

1 

n 

lim 

p −n q 

n→ 1 

n p 1, 

Proof: Sincen −1−1/n ≥ n −1 for all n, the series diverges. 

1 

(f) ∑ n1 p n −q 

(0 q p) 

n 

1 


p n −q 

1 1 

n p n 1− 

q n . We consider 2 cases: (i) p 1 and (ii) p ≤ 1. 

p 

For case (i), by Limit Comparison Test with 1 p 

, n 

1 

p 

lim 

n −q n 

n→ 

1, 

1 

p n 


For case (ii), by Limit Comparison Test with 1 p n , 

the series diverges. 

1 

(g) ∑ n1 nlog11/n 

1 

p 

lim 

n −q n 

n→ 

1, 

1 

p n 

Proof: Since 

lim 1 

n→ n log1 1/n 1, 

we know that the series diverges. 

1 

(h) ∑ n2 logn log n 

Proof: Since the identity a logb b loga ,wehave 

log n logn nlog logn 

So, the series converges. 

1 

(i) ∑ n3 nlognlog logn p 

≥ n 2 as n ≥ n 0 . 

Proof: We consider 3 cases: (i) p ≤ 0, (ii) 0 p 1 and (iii) p 1.

For case (i), since 

1 

n log nlog log n p ≥ 1 for n ≥ 3, 

n log n 

1 

we know that the series diverges by the divergence of ∑ . 

n3 n logn 

For case (ii), we consider (choose n 0 large enough) 

 

∑ 

jn 0 

2 j 

2 j log 2 j log log 2 j p 1 

log 2 ∑ jn 0 

 

≥ ∑ 

jn 0 

 

1 

jlog j p , 

1 

jlog j log 2 p 

then, by Cauchy Condensation Theorem, the series diverges since ∑ jn0 

by using Cauchy Condensation Theorem again. 

For case (iii), we consider (choose n 0 large enough) 

 

∑ 

jn 0 

2 j 

2 j log 2 j log log 2 j p 1 

log 2 ∑ jn 0 

 

≤ 2 ∑ 

jn 0 

 

≤ 4 ∑ 

jn 0 

 

1 


1 


1 

jlog j p , 

1 

jlogj p 

1 

jlogj p 

diverges 

then, by Cauchy Condensation Theorem, the series converges since ∑ jn0 

converges by using Cauchy Condensation Theorem again. 

Remark: There is another proof by Integral Test. We write it as a reference. 

1 

Proof: It is easy to check that fx is continous, positive, and 

xlogxlog logx p 

decreasing to zero on a, where a 0 for each fixed p. Consider 

 

 

 

dx 

dy 

a x log xlog log x p 

log loga y p 

which implies that the series converges if p 1 and diverges if p ≤ 1byIntegral Test. 

1 

(j) ∑ n3 log logn 

log logn 

log logn 

1 

Proof: Leta n 

log logn for n ≥ 3andbn 1/n, then 

log logn 

a n 

n 1 

b n log log n 

e −ylogy−ey 

→. 

So, by Limit Comparison Test, the series diverges. 

 

(k) ∑ n1 

1 n 2 − n 


1 n 2 − n 1 

1 n 2 n ≥ 1 for all n. 

1 2 n 

So, the series diverges.

(l) ∑ n2 

n p 1 

n−1 − 1 n 


n p 1 

n − 1 − 

1 n 

1 

n 3 2 −p 

n 

n − 1 

1 

1 

n−1 

n 

So, as p 1/2, the series converges and as p ≥ 1/2, the series diverges by Limit 

Comparison Test. 

 

(m) ∑ n1 

n 1/n − 1 n 

Proof: With help of Root Test, 

lim n→ 

sup n 1/n − 1 n 1/n 0 1, 


(n) ∑ n1 

n p n 1 − 2 n n − 1 


n p n 1 − 2 n n − 1 

1 

n 3 2 

n 3 2 −p n n 1 n n − 1 n − 1 n 1 

. 

So, as p 1/2, the series converges and as p ≥ 1/2, the series diverges by Limit 

Comparison Test. 

8.16 Let S n 1 , n 2 ,... denote the collection of those positive integers that do not 

involve the digit 0 is their decimal representation. (For example, 7 ∈ S but 101 ∉ S.) 

Show that ∑ k1 

1/n k converges and has a sum less than 90. 

Proof: DefineS j the j − digit number ⊆ S. Then#S j 9 j and S 

j1 S j . Note 

that 

∑ 1/n k 9 j 

10 . j−1 k∈S j 

So, 

 

∑ 

k1 

 

1/n k ≤ ∑ 

 

j1 

9 j 

10 j−1 90. 

In addition, it is easy to know that ∑ k1 

1/n k ≠ 90. Hence, we have proved that ∑ k1 

1/n k 

converges and has a sum less than 90. 

8.17 Given integers a 1 , a 2 ,...such that 1 ≤ a n ≤ n − 1, n 2, 3, . . . Show that the 

 

sum of the series ∑ n1 

a n /n! is rational if and only if there exists an integer N such that 

a n n − 1 for all n ≥ N. Hint: For sufficency, show that ∑ 

n2 

n − 1/n! is a telescoping 

series with sum 1. 

Proof: ()Assume that there exists an integer N such that a n n − 1 for all n ≥ N. 


.

()Assume that ∑ 

n1 

 

That is, p! ∑ np1 

a n 

p! ∑ 

np1 

 

∑ 

n1 

an 

n! 

N−1 

∑ 

n1 

N−1 

∑ 

n1 

N−1 

∑ 

n1 

 

an 

n! ∑ 

nN 

an 

n! 

 

an 

n! ∑ n − 1 

n! 

nN 

 

an 

n! ∑ 

nN 

1 

n − 1! − 1 n! 

N−1 

∑ 

an 

n! 1 

N − 1! ∈ Q. 

n1 

a n /n! is rational, say q p ,whereg. c. d. p, q 1. Then 

n! 

∈ Z. Note that 

 

a n 

n! 

 

≤ p! ∑ 

np1 

 

p! ∑ 

n1 

n − 1 

n! 

an 

n! 

∈ Z. 

p! 

p! 1since1 ≤ a n ≤ n − 1. 

So, a n n − 1 for all n ≥ p 1. That is, there exists an integer N such that a n n − 1for 

all n ≥ N. 

Remark: From this, we have proved that e is irrational. The reader should be noted that 

we can use Theorem 8.16 to show that e is irrational by considering e −1 . Since it is easy, 

we omit the proof. 

8.18 Let p and q be fixed integers, p ≥ q ≥ 1, and let 

pn 

x n ∑ 

kqn1 

n 

1 

k 

, s n ∑ 

k1 

−1 k1 

k 

(a) Use formula (8) to prove that lim n→ x n logp/q. 

Proof: Since 

we know that 

n 

∑ 

k1 

1 

k 

log n r O 1 n , 

pn 

x n ∑ 

k1 

qn 

1 − k 

∑ 

k1 

logp/q O 1 n 

which implies that lim n→ x n logp/q. 

(b) When q 1, p 2, show that s 2n x n and deduce that 

 

∑ 

n1 

1 

k 

−1 

n1 

n log 2. 

Proof: We prove it by Mathematical Induction as follows. As n 1, it holds 

trivially. Assume that n m holds, i.e., 

.

2m 

s 2m ∑ 

k1 

consider n m 1 as follows. 

x m1 

2m1 

∑ 

km11 

−1 

k1 

k 

1 

k 

2m 

∑ 

km1 

1 

k 

x m 

x m − 1 

m 1 1 

2m 1 1 

2m 2 

s 2m 1 

2m 1 − 1 

2m 2 

s 2m1 . 

So, by Mathematical Induction, we have proved that s 2n x n for all n. 

By s 2n x n for all n, wehave 

 

lim n→ 

s 2n ∑ 

k1 

−1 

k1 

k 

log 2 lim n→ 

x n . 

(c) rearrange the series in (b), writing alternately p positive terms followed by q 

negative terms and use (a) to show that this rearrangement has sum 

log 2 1 2 logp/q. 

∑ 

k1 


Proof: We prove it by using Theorem 8.13. So, we can consider the new series 

a k as follows: 

a k 1 

2k − 1p 1 ... 1 

2kp − 1 

n 

S n ∑ a k 

k1 

2np 

∑ 

k1 

np 

1k − ∑ 

k1 

nq 

1 

2k − ∑ 

k1 

1 

2k 

− 1 

2k − 1q ... 1 

2kq 

log 2np O 1 n − 1 2 log np − 2 O 1 n − 1 2 log nq − 2 O 1 n 

log 2np − log n pq O 

1 n 

log 2 p q O 1 n . 

So, 

lim n→ 

S n log 2 1 2 logp/q 

by Theorem 8.13. 

Remark: There is a reference around rearrangement of series. The reader can see the 

book, Infinite Series by Chao Wen-Min, pp 216-220. (Chinese Version) 

(d) Find the sum of ∑ 

n1 

−1 n1 1/3n − 2 − 1/3n − 1. 

Proof: Write

n 

S n ∑−1 k1 1 

3k − 2 − 1 

3k − 1 

k1 

n 

∑−1 k 1 

k1 

n 

−∑ 

k1 

− 

− 

− 

3n 

∑ 

k1 

n 

3k − 1 ∑ 

k1 

−1 3k−1 1 

−1 k1 1 

3k − 2 

n 

3k − 1 − ∑−1 3k−2 1 

3k − 2 

k1 

n 

n 

∑−1 3k−1 1 

3k − 1 ∑−1 3k−2 1 

3k − 2 

k1 

k1 

3n 

∑ 

k1 

3n 

∑ 

k1 

−1 k 

k 

−1 k 

k 

−1 k1 

k 

n 

− ∑ 

k1 

− 1 3 ∑ k1 

n 

−1 3k 

3k 

n 

−1 k 

k 

−1 k1 

k 

− 1 3 ∑ k1 

→ 2 log 2. 

3 

So, the series has the sum 2 log 2. 

3 

Remark: There is a refernece around rearrangement of series. The reader can see the 

book, An Introduction to Mathematical Analysis by Loo-Keng Hua, pp 323-325. 

(Chinese Version) 

8.19 Let c n a n ib n ,wherea n −1 n / n , b n 1/n 2 . Show that ∑ c n is 

conditioinally convergent. 

Proof: It is clear that ∑ c n converges. Consider 

∑|c n| ∑ 1 n 2 1 n 4 ∑ 1 n 1 1 n 2 ≥ ∑ 1 n 

Hence, ∑|c n| diverges. That is, ∑ c n is conditioinally convergent. 

Remark: Wesay∑ c n converges if, and only if, the real part ∑ a n converges and the 

imaginary part ∑ b n converges, where c n a n ib n . 

8.20 Use Theorem 8.23 to derive the following formulas: 

n 

(a) ∑ k1 

logk 

1 k 2 log2 n A O logn 

n 

(A is constant) 

Proof: Letfx logx 

x define on 3, , then f ′ x 1−logx 0on3, . So, it is 

x 2 

clear that fx is a positive and continuous function on 3, , with 

log x 

lim x→ 

fx lim x→ x 

So, by Theorem 8.23, wehave 

lim x→ 

1 x 0byL-Hospital Rule.

n 

∑ 

k3 

log k 

k 


 

3 n log x 

x dx C O log n 

n 

1 2 log2 n − 1 2 log2 3 C O 

n 

∑ 

k1 

log k 

k 

,whereC is a constant 

log n 

n 

1 2 log2 n A O 


log n 

n , 

where A C log2 − 1 2 2 log2 3 is a constant. 

n 1 

1 

(b) ∑ k2 

loglog n B O (B is constant) 

klogk n logn 

1 

Proof: Letfx defined on 2, , then 2 

xlogx f′ 1 

x − 

xlogx 1 log x 0on 

2, . So, it is clear that fx is a positive and continuous function on 3, , with 

lim x→ 

fx lim 1 

x→ x log x 0. 

So, by Theorem 8.23, wehave 

n 

∑ 

k2 

1 

k log k n 

2 

dx 

x log x C O 1 

n log n 

log log n B O 1 

n log n 

where B C − log log 2 is a constant. 

8.21 If 0 a ≤ 1, s 1, define s, a ∑ n0 

n a −s . 



(a) Show that this series converges absolutely for s 1 and prove that 

k 

∑ s, h k 

k 

s s if k 1, 2, . . . 

h1 

where s s,1 is the Riemann zeta function. 

Proof: First, it is clear that s, a converges absolutely for s 1. Consider 

k 

∑ s, h k 

h1 

k 

∑ 

h1 

k 

∑ 

h1 

 

∑ 

n0 

 

∑ 

n0 

k 

∑ ∑ 

n0 h1 

 

k s ∑ 

n0 

 

k s ∑ 

n0 

k s s. 

k 

∑ 

h1 

1 

n h k s 

k s 

kn h s 

k s 

kn h s 

1 

kn h s 

1 

n 1 s

(b) Prove that ∑ n1 

−1 n−1 /n s 1 − 2 1−s s if s 1. 

Proof: Let 

n 

S n ∑ j1 

−1 j−1 

j s 

, and thus consider its subsequence S 2n as follows: 

2n 

S 2n ∑ 

j1 

2n 

∑ 

j1 

1 

j s 

1 

j s 

n 

− 2 ∑ 

j1 

n 

− 2 1−s ∑ 

j1 

1 

2j s 


lim n→ 

S 2n 1 − 2 1−s s. 

Since S n converges, we know that S 2n also converges and has the same value. Hence, 

 

∑ 

n1 

1 

j s 

−1 n−1 /n s 1 − 2 1−s s. 

8.22 Given a convergent series ∑ a n , where each a n ≥ 0. Prove that ∑ a n n −p 

converges if p 1/2. Give a counterexample for p 1/2. 

Proof: Since 

a n n −2p 

2 

≥ a n n −2p a n n −p , 

we have ∑ a n n −p converges if p 1/2 since 

∑ a n converges and ∑ n −2p converges if p 1/2. 

and 

For p 1/2, we consider a n 

1 

nlogn 2 , then 

∑ a n converges by Cauchy Condensation Theorem 

∑ a n n −1/2 ∑ 1 

n log n 

diverges by Cauchy Condensation Theorem. 

8.23 Given that ∑ a n diverges. Prove that ∑ na n also diverges. 

n 

Proof: Assume ∑ na n converges, then its partial sum ∑ k1 

ka k is bounded. Then by 

Dirichlet Test, we would obtain 

∑ka k 1 

k 

∑ a k converges 

which contradicts to ∑ a n diverges. Hence, ∑ na n diverges. 

8.24 Given that ∑ a n converges, where each a n 0. Prove that 

∑a n a n1 1/2 

also converges. Show that the converse is also true if a n is monotonic. 

Proof: Since 

a n a n1 

≥ a 

2 

n a n1 1/2 , 

we know that 

∑a n a n1 1/2

converges by ∑ a n converges. 

Conversly, since a n is monotonic, it must be decreasing since ∑ a n converges. So, 

a n ≥ a n1 for all n. Hence, 

a n a n1 1/2 ≥ a n1 for all n. 

So, ∑ a n converges since ∑a n a n1 1/2 converges. 

8.25 Given that ∑ a n converges absolutely. Show that each of the following series 

also converges absolutely: 

(a) ∑ a n 

2 

Proof: Since∑ a n converges, then a n → 0asn → . So, given 1, there exists a 

positive integer N such that as n ≥ N, wehave 

|a n| 1 


a n2 |a n| for n ≥ N. 

So, ∑ a n2 converges if ∑|a n| converges. Of course, ∑ a n2 converges absolutely. 

(b) ∑ an 

1a n 

(if no a n −1) 

Proof: Since∑|a n| converges, we have lim n→ a n 0. So, there exists a positive 

integer N such that as n ≥ N, wehave 

1/2 |1 a n|. 

Hence, as n ≥ N, 

a n 

2|a 

1 a n| 

n 

which implies that ∑ 

(c) ∑ a n 2 

1a n 

2 

Proof: It is clear that 

an 

1a n 

By (a), we have proved that ∑ a n 2 

converges. So, ∑ an 

1a n 

1a n 

2 

a2 

n 

≤ a 

1 a2 n2 . 

n 

converges absolutely. 


8.26 Determine all real values of x for which the following series converges. 

 

∑ 

n1 

Proof: Consider its partial sum 

n 

∑ 

k1 

1 1 2 ... 1 n 

1 1 2 

... 1 k 

k 

sin nx 

n . 

sin kx 

as follows. 

As x 2m, the series converges to zero. So it remains to consider x ≠ 2m as 

follows. Define 

and 

a k 1 1 2 

... 1 k 

k

then 

b k sin kx, 

a k1 − a k 1 1 2 ... 1 k 1 

k1 

k 1 

− 1 1 2 

... 1 k 

k 

k1 1 ... 1 1 − k 11 1 2 k k1 2 

kk 1 

k 

k1 

 

− 1 1 ... 1 

2 k 

0 

kk 1 

and 

n 

∑ b k ≤ 1 

k1 

sin x . 

2 

So, by Dirichlet Test, we know that 

 

∑ 

k1 

a k b k ∑ 

k1 

1 1 ... 1 

2 k 

sin kx 

k 

... 1 k 

converges. 

From above results, we have shown that the series converges for all x ∈ R. 

8.27. Prove that following statements: 

(a) ∑ a n b n converges if ∑ a n converges and if ∑b n − b n1 converges absolutely. 

Proof: Consider summation by parts, i.e., Theorem 8.27, then 

n 

∑ 

k1 

a k b k A n b n1 − ∑ 

k1 

n 

A k b k1 − b k . 

Since ∑ a n converges, then |A n| ≤ M for all n. In addition, by Theorem 8.10, lim n→ b n 

exists. So, we obtain that 

(1). lim n→ 

A n b n1 exists 

and 

n 

(2). ∑ 

k1 

n 

|A k b k1 − b k | ≤ M ∑ 

k1 

 

|b k1 − b k | ≤ M ∑ 

k1 

|b k1 − b k |. 

(2) implies that 

n 

(3). ∑ A k b k1 − b k converges. 

k1 

n 

By (1) and (3), we have shown that ∑ k1 

a k b k converges. 

Remark: In 1871, Paul du Bois Reymond (1831-1889) gave the result. 

(b) ∑ a n b n converges if ∑ a n has bounded partial sums and if ∑b n − b n1 converges 

absolutely, provided that b n → 0asn → . 

Proof: Bysummation by parts, wehave 

n 

∑ 

k1 

a k b k A n b n1 − ∑ A k b k1 − b k . 

k1 

Since b n → 0asn → and ∑ a n has bounded partial sums, say |A n| ≤ M for all n. Then 

n

In addition, 

n 

(2). ∑ 

k1 

(1). lim n→ 

A n b n1 exists. 

n 

|A k b k1 − b k | ≤ M ∑ 

k1 

 

|b k1 − b k | ≤ M ∑ 

k1 

|b k1 − b k |. 

(2) implies that 

n 

(3). ∑ A k b k1 − b k converges. 

k1 

n 

By (1) and (3), we have shown that ∑ k1 

a k b k converges. 

Remark: (1) The result is first discovered by Richard Dedekind (1831-1916). 

(2) There is an exercise by (b), we write it as a reference. Show the convergence of the 

 

k 

−1 

series ∑ k1 

. 

k 

Proof: Leta k −1 k 

, then in order to show the convergence of 

1/3 

 

∑ k1 

−1 

k 

and b 

k 2/3 k 1 

k 

, it suffices to show that ∑ 

k k1 

a k : S n 

n ∈ N, there exists j ∈ N such that j 2 ≤ N j 1 2 . Consider 

S n a 1 a 2 a 3 a 4 ...a 8 a 9 ....a 15 ...a j 

2 ...a n 

n 

is bounded sequence. Given 

≤ 3a 3 5a 4 7a 15 9a 16 ..4k − 1a 2k 2 −1 4k 1a 2 

2k 

if j 2k, k ≥ 2 

3a 3 5a 4 7a 15 9a 16 ..4k − 3a 2k−2 

2 if j 2k − 1, k ≥ 3 

then as n large enough, 

S n ≤ −3a 4 5a 4 −7a 16 9a 16 ... −4k − 1a 2k 

2 4k 1a 2k 

2 

−3a 4 5a 4 −7a 16 9a 16 ... −4k − 5a 2k−2 

2 4k − 3a 2k−2 

2 

which implies that as n large enough, 

 

 

S n ≤ 2 ∑ a 2j 

2 2 ∑ 

1 

j2 

j2 2j 4/3 : M 1 * 


M 2 ≤ S n for all n ** 

By (*) and (**), we have shown that 

n 

a k : S n is bounded sequence. 

∑ k1 

Note: (1) By above method, it is easy to show that 

 

−1 

k 

∑ 

k1 

converges for p 1/2. For 0 p ≤ 1/2, the series diverges by 

1 

n 2 p ... 1 

n 2 2n p ≥ 2n 1 

n 2 n p ≥ 2n 1 

n 2 n p ≥ 2n 1 

n 1 2p ≥ 2n n 1 1 1. 

k p 

log k 

−1 

. 

k 

 

(2) There is a similar question, show the divergence of the series ∑ k1 

Proof: WeuseTheorem 8.13 to show it by inserting parentheses as follows. We insert 

k 

−1log 

parentheses such that the series ∑ forms ∑−1 k b k . If we can show ∑−1 k b k 

k

diverges, then ∑ 

where 

By (2) and (4), 

By (1) and (3), 

−1log 

k 

k 

also diverges. Consider 

b k 1 m ... 1 m r , * 

(1). log m N 

(2). logm − 1 N − 1 log em − 1 N 

(3). logm r N 

(4). logm r 1 N 1 log mr1 

e N 

m r 1 

e 

m − 1 r 1 ≥ m if m is large enough. 

2m ≥ r. 

So, as k large enough ( m is large enough), 

b k ≥ m r 1 r ≥ 3m m 1 by (*). 

3 

It implies that ∑−1 k b k diverges since b k does NOT tends to zero as k goes infinity.So, 

k 

−1log 

we have proved that the series ∑ diverges. 

k 

(3) There is a good exercise by summation by parts, we write it as a reference. 

Assume that ∑ 

k1 

a k b k converges and b n ↗ with lim n→ b n . Show that b n ∑ 

kn 

a k 

converges. 

Proof: First, we show that the convergence of ∑ 

k1 

a k by Dirichlet Test as follows. 

Since b n ↗ , there exists a positive integer n 0 such that as n n 0 ,wehaveb n 0. So, 

 

1 

we have 

b nn0 

is decreasing to zero. So 

n1 

 

∑ a kn0 

k1 

converges by Dirichlet Test. 

 

 

∑a kn0 b kn0 1 

b kn0 

k1 

For the convergence of b n ∑ kn 

a k ,letn n 0 , then 

 

b n ∑ a k ∑ a k b 

b n k 

b k 

kn kn 

and define c k a k b k and d k bn 

b k 

. Note that d k is decreasing to zero. Define 

k 

C k ∑ j1 

c j and thus we have 

m m 

b n ∑ a k ∑ a k b 

b n k 

b k 

kn kn 

So, 

m 

∑C k − C k−1 d k 

kn 

m−1 

∑ C k d k − d k1 C m d m − C n−1 d n . 

kn 

.

n ∑ 

kn 

 

a k ∑ 

kn 

 

a k b k 

b n 

b k 

∑ C k d k − d k1 C d − C n−1 d n 

kn 

 

∑ 

kn 

C k d k − d k1 − C n−1 d n 

by C lim k→ C k and lim k→ d k 0. In order to show the existence of lim n→ b n ∑ kn 

a k , 

it suffices to show the existence of lim n→ ∑ 

kn 

C k d k − d k1 . Since the series 

∑ 

kn 

C k d k − d k1 exists, lim n→ ∑ 

kn 

C k d k − d k1 0. From above results, we have 

proved the convergence of lim n→ b n ∑ kn 

a k . 

Note: We also show that lim n→ b n ∑ 

kn 

a k 0 by preceding sayings. 

Supplement on the convergence of series. 

A Show the divergence of ∑ 1/k. We will give some methods listed below. If the 

proof is easy, we will omit the details. 

(1) Use Cauchy Criterion for series. Since it is easy, we omit the proof. 

(2) Just consider 

1 1 2 1 3 1 4 ... 2 1 n ≥ 1 1 2 2 1 4 1 ...2n−1 2 n 

1 n 2 → . 

Remark: We can consider 

1 1 ... 1 1 ... 2 10 

1 ...≥ 1 11 100 

10 9 100 90 ... 

Note: The proof comes from Jing Yu. 

(3) Use Mathematical Induction to show that 

1 

k − 1 1 k 1 

k 1 ≥ 3 if k ≥ 3. 

k 


1 1 2 1 3 1 4 1 5 1 6 ....≥ 1 3 3 3 6 3 9 ... 

Remark: The proof comes from Bernoulli. 

(4) Use Integral Test. Since the proof is easy, we omit it. 

(5) Use Cauchy condensation Theorem. Since the proof is easy, we omit it. 

(6) Euler Summation Formula, the reader can give it a try. We omit the proof. 

(7) The reader can see the book, Princilpes of Mathematical Analysis by Walter 

Rudin, Exercise 11-(b) pp 79. 

Suppose a n 0, S n a 1 ...a n ,and∑ a n diverges. 

(a) Prove that ∑ an 

1a n 

diverges. 

Proof: Ifa n → 0asn → , then by Limit Comparison Theorem, we know that 

diverges. If a n does not tend to zero. Claim that does not tend to zero. 

∑ an 

1a n 

a n 

1a n

Suppose NOT, it means that lim n→ 

a n 

1a n 

0. That is, 

lim 1 

n→ 

0 lim 

1 a 1 n→ 

a n 0 

n 

which contradicts our assumption. So, ∑ an 

1a n 

(b) Prove that 

and deduce that ∑ an 

S n 


diverges. 

a N1 

S N1 

a N1 

S N1 

... a Nk 

S Nk 

diverges by claim. 

... a Nk 

S Nk 

≥ 1 − S N 

S Nk 

then ∑ an 

S n 

diverges by Cauchy Criterion with (*). 

Remark: Leta n 1, then ∑ an 

S n 

(c) Prove that 

and deduce that ∑ an 

S2 

n 


and 

So, ∑ an 

S n 

2 

converges. 

≥ a N1 ...a Nk 

S Nk 

1 − S N 

S Nk 

, * 

∑ 1/n diverges. 

a n 

S n 

2 

≤ 1 

S n−1 

1 

S n−1 

− 1 S n 

 

− 1 S n 

a n 

S n−1 S n 

≥ a n 

S n 

2 , 

∑ 

S 1 

n−1 

− 

S 1 n 

converges by telescoping series with 

S 1 n 

→ 0. 

converges. 

(d) What can be said about 

∑ 

Proof: For∑ 

the series ∑ 

For ∑ 

so ∑ 

an 

1n 2 a n 

a n 

1 na n 

and ∑ a n 

1 n 2 a n 

 

an 

1na n 

:asa n 1 for all n, theseries∑ an 

1na n 

∑ 1 

1n 

an 

1na n 

∑ 1 

an 

1n 2 a n 

1k 2 

: Consider 

converges. 

(8) Consider ∑ sin 1 n diverges. 

Proof: Since 

a n 

converges. 

a n 

0ifn ≠ k2 

, 

1ifn k 2 

1 ≤ 1 1 n 2 a 

1 n a n 

n 2 n , 2 

lim 

sin 1 n 

n→ 1 

n 

1, 

the series ∑ 1 n diverges by Limit Comparison Theorem. 

diverges. As

Remark: In order to show the series ∑ sin 1 n diverges, we consider Cauchy Criterion 

as follows. 

n sin 1 ≤ sin 1 ... sin 1 

2n n 1 

n n 

and given x ∈ R, forn 0,1,2,..., we have 

|sin nx| ≤ n|sin x|. 

So, 

sin 1 2 ≤ sin 1 ... sin 1 

n 1 

n n 

for all n. Hence, ∑ sin 1 n diverges. 

Note: There are many methods to show the divergence of the series ∑ sin 1 n . We can 

use Cauchy Condensation Theorem to prove it. Besides, by (11), it also owrks. 

(9) O-Stolz’s Theorem. 

n 1 

Proof: LetS n ∑ j1 j 

and X n log n. Then by O-Stolz’s Theorem, it is easy to see 

lim n→ 

S n . 

(10) Since n k1 1 1 k 

diverges, the series ∑ 1/k diverges by Theorem 8.52. 

(11) Lemma: If a n is a decreasing sequence and ∑ a n converges. Then 

lim n→ na n 0. 

Proof: Sincea n → 0anda n is a decreasing sequence, we conclude that a n ≥ 0. 

Since ∑ a n converges, given 0, there exists a positive integer N such that as n ≥ N, 

we have 

a n ..a nk /2 for all k ∈ N 


k 1a nk /2 since a n ↘. 

Let k n, then as n ≥ N, wehave 

n 1a 2n /2 

which implies that as n ≥ N 

2n 1a 2n 


lim n→ 

2na 2n 0 since lim n→ 

a n 0. * 

Similarly, we can show that 

lim n→ 

2n 1a 2n1 0. ** 

So,by(*)adn(**),wehaveprovedthatlim n→ na n 0. 

Remark: From this, it is clear that ∑ 1 n diverges. In addition, we have the convergence 

of ∑ na n − a n1 . We give it a proof as follows. 

Proof: Write

then 

since 

n 

lim n→ 

∑ 

k1 

n 

S n ∑ ka k − a k1 

k1 

n 

∑ 

k1 

a k − na n1 , 

lim n→ 

S n exists 

a k exists and lim n→ 

na n 0. 

B Prove that ∑ 1 p diverges, where p is a prime. 

Proof: GivenN, letp 1 ,...,p k be the primes that divide at least one integer≤ N. Then 

N 

∑ 

n1 

k 

1 

n ≤ 

j1 

k 

 

j1 

≤ exp 

1 1 p j 

1 p j 

2 ... 

1 

1 − p 1 j 

by 1 − x −1 ≤ e 2x if 0 ≤ x ≤ 1/2. Hence, ∑ 1 p diverges since ∑ 1 n diverges. 

Remark: There are many proofs about it. The reader can see the book, An 

Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 91-93. (Chinese 

Version) 

C Discuss some series related with ∑ sink . 

k 

STUDY: (1) We have shown that the series ∑ sin 1 diverges. 

k 

(2) The series ∑ sinna b diverges where a ≠ n for all n ∈ Z and b ∈ R. 

Proof: Suppose that ∑ sinna b converges, then lim n→ sinna b 0. Hence, 

lim n→|sinn 1a b − sinna b| 0. Consider 

|sinn 1a b − sinna b| 

k 

∑ 

j1 

2 

p j 

2cos na b a 2 

sin a 2 


2 cosna b cos a 

2 

− sinna b sin a 2 

sin a 2

lim n→ 

|sinn 1a b − sinna b| 

lim n→ 

sinn 1a b − sinna b 

lim n→ 

sup 2 cosna b cos a 

2 

− sinna b sin a 2 

sin a 2 

lim n→ 

sup 2 cosna b cos a sin a 2 2 

|sin a| ≠ 0 

which is impossible. So, ∑ sinna b diverges. 

Remark: (1) By the same method, we can show the divergence of ∑ cosna b if 

a ≠ n for all n ∈ Z and b ∈ R. 

(2) The reader may give it a try to show that, 

and 

p 

by considering ∑ n0 

(**). 

(3) The series ∑ sink 

k 

p 

∑ 

n0 

cosna b 

Proof: First, it is clear that ∑ sink 

k 

|∑ sin k| ≤ 

1 

sin 1 2 

partial sums as follows: Since 

sin 

p1 

2 b 

sin b 2 

sin a p 2 b * 

p 

p1 

sin 

∑ sinna b b 2 

cos a p sin b 2 b ** 

n0 

2 

e inab . However, it is not easy to show the divergence by (*) and 

converges conditionally. 

converges by Dirichlet’s Test since 

, we consider its 

. In order to show that the divergence of ∑ sink 

k 

3n3 

n 

∑ sin k ∑ 

sin 3k 1 sin 3k 2 sin 3k 3 

k 

3k 1 3k 2 3k 3 

k1 

k0 

and note that there is one value is bigger than 1/2 among three values |sin 3k 1|, 

|sin 3k 2|, and|sin 3k 3|. So, 

3n3 

∑ 

k1 

sin k 

k 

n 1 

2 

≥ ∑ 

3k 3 

k0 

which implies the divergence of ∑ sink . 

k 

 

Remark: The series is like Dirichlet Integral sinx 

0 

x 

Integral converges conditionally. 

(4) The series ∑ |sink|r diverges for any r ∈ R. 

k 

Proof: We prove it by three cases as follows. 

(a) As r ≤ 0, we have 

So, ∑ |sink|r diverges in this case. 

k 

(b) As 0 r ≤ 1, we have 

∑ 

|sin k|r 

k 

≥ ∑ 1 k . 

dx. Also, we know that Dirichlet

|sin k|r 

∑ 

k 

So, ∑ |sink|r diverges in this case by (3). 

k 

(c) As r 1, we have 

3n3 

∑ |sin k| 

r 

k 

k1 

n 

∑ 

k0 

n 

≥ ∑ 

k0 

|sin 3k 1| r 

3k 1 

1 2 r 

3k 3 . 

≥ ∑ 

 

|sin k| 

k 

. 

|sin 3k 2|r 

3k 2 

 

|sin 3k 3|r 

3k 3 

So, ∑ |sink|r diverges in this case. 

k 

(5) The series ∑ sin2p−1 k 

,wherep ∈ N, converges. 

k 

Proof: We will prove that there is a positive integer Mp such that 

n 

∑ 

k1 

sin 2p−1 k ≤ Mp for all n. * 

So, if we can show (*), then by Dirichlet’s Test, we have proved it. In order to show (*), 

we claim that sin 2p−1 k can be written as a linear combination of sink, sin3k,..., 

sin2p − 1k. So, 

n 

∑ 

k1 

n 

sin 2p−1 k ∑ 

k1 

n 

≤ |a 1 | ∑ 

k1 

|a 1 | 

a 1 sin k a 2 sin 3k ...a p sin2p − 1k 

n 

sin k ...|a p| ∑ sin2p − 1k 

k1 

|a 

≤ ... p| 

: Mp by Theorem 8.30. 

sin 1 2 sin 2p−1 

2 

We show the claim by Mathematical Induction as follows. As p 1, it trivially holds. 

Assume that as p s holds, i.e., 

then as p s 1, we have 

sin 2s−1 k ∑ 

j1 

s 

a j sin2j − 1k

sin 2s1 k sin 2 ksin k 2s−1 

sin 2 k 

s 

∑ a j sin2j − 1k 

j1 

s 

∑ a j sin 2 k sin2j − 1k 

j1 

s 

∑ a j 

j1 

1 2 

1 2 

s 

∑ 

j1 

1 − cos2k 

2 

sin2j − 1k 

by induction hypothesis 

s 

a j sin2j − 1k − ∑ a j cos2k sin2j − 1k 

j1 

s 

∑ a j sin2j − 1k − 1 2 ∑ s 

j1 

j1 

a j sin2j 1k sin2j − 3k 

which is a linear combination of sink,...,sin2s 1k. Hence, we have proved the claim 

by Mathematical Induction. 

Remark: By the same argument, the series 

n 

∑ 

k1 

cos 2p−1 k 

is also bounded, i.e., there exists a positive number Mp such that 

n 

(6) Define ∑ k1 

n 

∑ 

k1 

|cos 2p−1 k| ≤ Mp. 

sinkx 

: F 

k n x, then F n x is boundedly convergent on R. 

Proof: SinceF n x is a periodic function with period 2, andF n x is an odd function. 

So, it suffices to consider F n x is defined on 0, . In addition, F n 0 0 for all n. 

Hence, the domain I that we consider is 0, . Note that sinkx x 

cosktdt. So, 

0 

n 

F n x ∑ 

k1 

 

0 

x 

∑ 

k1 

 

0 

x 

 

0 

x 

sin kx 

k 

n 

cosktdt 

sinn 1 t − sin 1 t 

2 2 

2sin 1 t dt 

2 

sinn 1 2 t 

t 

dt 

0 

x 

1 

2sin t 2 

− 1 t 

k 

sin n 1 2 t dt − x 2 


 

0 

n 1 2 

x sin t 

t 

|F n x| ≤ 

0 

n 1 2 x sin t 

t 

dt 

0 

x t − 2sin t 2 

2t sin t 2 

dt 

0 


2t sin t 2 

sin n 1 2 t dt − x 2 

sin n 1 2 t dt 2 .

n 

For the part 1 2 

0 

that 

x sint 

 

t 

dt :Since sint 

0 t 

dt converges, there exists a positive M 1 such 

 

0 

n 1 2 

x sin t 

t 

dt ≤ M 1 for all x ∈ I and for all n. 

For the part 0 

x t−2sin t 2 

2t sin t 2 

sinn 1 2 tdt 

: Consider 

 

0 


2t sin t 2 

sin n 1 2 

t dt 

≤ 

0 


2t sin t 2 

dt since t − 2sin t 2 0onI 

≤ 

0 

t − 2sin t 2 

2t sin t 2 

dt : M 2 since lim 

t→0 

 

t − 2sin t 2 

2t sin t 2 

Hence, 

|F n x| ≤ M 1 M 2 for all x ∈ I and for all n. 

2 

So, F n x is uniformly bounded on I. It means that F n x is uniformly bounded on R. 

In addition, since 

F n x 

0 

n 1 2 

fixed x ∈ I, wehave 

x sin t 

t 

dt 

0 


2t sin t 2 

sin t 

0 t 

dt exists. 

and by Riemann-Lebesgue Lemma, in the text book, pp 313, 

0. 

sin n 1 2 t dt − x 2 , 

x t − 2sin 

lim n→ 

t 2 

0 2t sin t sin n 1 t dt 0. 

2 

2 

So, we have proved that 

lim n→ 

F n x sin t 

0 t 

dt − 

2 x where x ∈ 0, . 

Hence, F n x is pointwise convergent on I. It means that F n x is pointwise 

convergent on R. 

Remark: (1) For definition of being boundedly convergent on a set S, the reader can 

see the text book, pp 227. 

(2) In the proof, we also shown the value of Dirichlet Integral 

sin t 

0 t 

dt 2 

by letting x . 

(3) There is another proof on uniform bound. We write it as a reference. 

Proof: The domain that we consider is still 0, . Let 0, and consider two cases as 

follows. 

(a) x ≥ 0 : Using summation by parts,

n 

∑ 

k1 

sin kx 

k 

≤ 1 

n 1 ∑ n 

k1 

≤ 1 

n 1 

sin kx 

k 

n 

∑ 

k1 

1 

sin 2 1 

sin 2 1 − 1 

n 1 

1 

sin . 

2 

(b) 0 x ≤ :LetN 1 x , consider two cases as follows. 

As n N, then 

and as n ≥ N, then 

≤ 

n 

∑ 

k1 

N−1 

∑ 

k1 

sin kx 

k 

sin kx 

k 

n 

≤ 1 ∑ 

kN 

sin kx 

k 

≤ 1 1 

n 1 ∑ n 

k1 

n 

∑ 

kN 

by (*) 

sin kx 

k 

by summation by parts 

≤ 1 1 

n 1 sin x 2 

1 2 . 

1 x sin x 2 

2 

Note that lim x→0 

 

1 x sin x 2 

n 

∑ 

k1 

sin kx 

k 

 

1 

N sin x 2 

sin kx 

k 

N−1 

1 

N ∑ k1 

k 

∑ sin jx 1 

k 1 − 1 k 

j1 

≤ n|x| N|x| ≤ 1 * 

sin kx 

k 

1 

N 

− 1 

n 1 

n 

∑ 

kN 

1 

sin x 2 

4. So, we may choose a ′ such that 

k 

∑ sin jx 1 

k 1 − 1 k 

j1 

2 ≤ 5 for all x ∈ 0, 

1 x sin x ′ . 

2 

By preceding sayings, we have proved that F n x is uniformly bounded on I. It means 

that F n x is uniformly bounded on R. 

D In 1911, Otto Toeplitz proves the following. Let a n and x n be two sequences 

1 

such that a n 0 for all n with lim n→ a 1 ...a n 

0 and lim n→ x n x. Then 

lim 

a 1 x 1 ...a n x n 

n→ a 1 ...a n 

x. 

n 

Proof: LetS n ∑ k1 

n 

a k and T n ∑ k1 

a k x k , then 

lim 

T n1 − T n 

n→ S n1 − S n 

lim n→ 

a n1 x n1 

a n1 

lim n→ 

x n1 x. 

So, by O-Stolz’s Theorem, wehaveproveit. 

Remark: (1) Let a n 1, then it is an extension of Theorem 8.48. 

(2) Show that

lim n→ 

sin ... sin n 

1 ... 1 n 

. 

Proof: Write 

sin ... sin n 

1 1sin ... 1 1 n n sin n 

, 

1 ... 1 n 

1 ... 1 n 

the by Toeplitz’s Theorem, we have proved it. 

E Theorem 8.16 emphasizes the decrease of the sequence a n , we may ask if we 

remove the condition of decrease, is it true The answer is NOT necessary. For example, 

let 

a n 1 n −1n1 . 0 

2n 

F Some questions on series. 

(1) Show the convergence of the series ∑ n1 

log n sin 1 n . 

Proof: Since n sin 1 n 1 for all n, logn sin 1 n 0 for all n. Hence, we consider the 

new series 

 

∑ 

n1 

− log n sin 1 

n ∑ log 

n1 

sin 1/n 

1/n 

as follows. Let a n log sin1/n and b 

1/n n log 1 1 , then 

n 2 

lim 

a n 

n→ 

 

b 1 n 6 . 

In addition, 

∑ b n ≤ ∑ 1 n 2 

by e x ≥ 1 x for all x ∈ R. From the convergence of ∑ b n , we have proved that the 

convergence of ∑ a n by Limit Comparison Test. 

 

∑ n1 

(2) Suppose that a n ∈ R, and the series ∑ 

n1 


a n 

n 

Proof: By A. P. ≥ G. P. , we have 

a n2 1 n 2 

2 

 

which implies that ∑ n1 

a n 

n 

≥ 


a n2 converges. Prove that the series 

Remark: We metion that there is another proof by using Cauchy-Schwarz inequality. 

the difference of two proofs is that one considers a n , and another considers the partial 

sums S n . 

Proof: ByCauchy-Schwarz inequality, 

 

which implies that ∑ n1 

a n 

n 

n 

∑ 

k1 

|a n| 

k 

2 

≤ 

a n 

n 

∑ a2 

k 

k1 


Double sequences and series 

8.28 Investigate the existence of the two iterated limits and the double limit of the 

n 

∑ 

k1 

1 

k 2

double sequence f defined by the followings. Answer. Double limit exists in (a), (d), (e), 

(g). Both iterated limits exists in (a), (b), (h). Only one iterated limit exists in (c), (e). 

Neither iterated limit exists in (d), (f). 

(a) fp, q pq 

1 

Proof: It is easy to know that the double limit exists with lim p,q→ fp, q 0by 

definition. We omit it. In addition, lim p→ fp, q 0. So, lim q→ lim p→ fp, q 0. 

Similarly, lim p→ lim q→ fp, q 0. Hence, we also have the existence of two iterated 

limits. 

(b) fp, q 

p 

pq 

Proof: Letq np, then fp, q 1 . It implies that the double limit does not exist. 

n1 

However, lim p→ fp, q 1, and lim q→ fp, q 0. So, lim q→ lim p→ fp, q 1, and 

lim p→ lim q→ fp, q 0. 

(c) fp, q −1p p 

pq 

Proof: Letq np, then fp, q −1p 

n1 

. It implies that the double limit does not exist. 

In addition, lim q→ fp, q 0. So, lim p→ lim q→ fp, q 0. However, since 

lim p→ fp, q does not exist, lim q→ lim p→ fp, q does not exist. 

(d) fp, q −1 pq 1 p 1 q 

Proof: It is easy to know lim p,q→ fp, q 0. However, lim q→ fp, q and lim p→ fp, q 

do not exist. So, neither iterated limit exists. 

(e) fp, q −1p 

q 

Proof: It is easy to know lim p,q→ fp, q 0. In addition, lim q→ fp, q 0. So, 

lim p→ lim q→ fp, q 0. However, since lim p→ fp, q does not exist, 

lim q→ lim p→ fp, q does not exist. 

(f) fp, q −1 pq 

Proof: Let p nq, then fp, q −1 n1q . It means that the double limit does not 

exist. Also, since lim p→ fp, q and lim q→ fp, q do not exist, lim q→ lim p→ fp, q and 

lim p→ lim q→ fp, q do not exist. 

(g) fp, q cos p 

q 

Proof: Since|fp, q| ≤ 1 q , then lim p,q→ fp, q 0, and lim p→ lim q→ fp, q 0. 

However, since cosp : p ∈ N dense in −1, 1, we know that lim q→ lim p→ fp, q does 

not exist. 

(h) fp, q p q 

∑ 

q 2 n1 

sin n p 

Proof: Rewrite 

and thus let p nq, fp, q sin 1 2n sin 

fp, q p sin q 

2p 

q1 

2nq 

nqsin 1 

2nq 

sin 

q1 

2p 

q 2 sin 1 

2p 

. It means that the double limit does not exist. 

However, lim p→ fp, q q1 

2q 

since sin x~x as x → 0. So, lim q→ lim p→ fp, q 1 2 .

Also, lim q→ fp, q lim q→ p sin 1 2p 

lim p→ lim q→ fp, q 0. 

8.29 Prove the following statements: 

sin q 2p 

sin 

q1 

2p 

q 2 

0since|sin x| ≤ 1. So, 

(a) A double series of positive terms converges if, and only if, the set of partial sums is 

bounded. 

Proof: ()Suppose that ∑ m,n 

fm, n converges, say ∑ m,n 

fm, n A 1 , then it means 

that lim p,q→ sp, q A 1 . Hence, given 1, there exists a positive integer N such that as 

p, q ≥ N, wehave 

|sp, q| ≤ |A 1 | 1. 

So, let A 2 maxsp, q :1≤ p, q N, wehave|sp, q| ≤ maxA 1 , A 2 for all p, q. 

Hence, we have proved the set of partial sums is bounded. 

()Suppose that the set of partial sums is bounded by M, i.e., if 

S sp, q : p, q ∈ N, then sup S : A ≤ M. Hence, given 0, then there exists a 

sp 1 , q 1 ∈ S such that 

A − sp 1 , q 1 ≤ A. 

Choose N maxp 1 , q 1 , then 

A − sp, q ≤ A for all p, q ≥ N 

since every term is positive. Hence, we have proved lim p,q→ sp, q A. Thatis, 

∑ m,n 

fm, n converges. 

(b) A double series converges if it converges absolutely. 

p 

Proof: Lets 1 p, q ∑ m1 

q 

∑ n1 

p 

|fm, n| and s 2 p, q ∑ m1 

q 

∑ n1 

fm, n, wewant 

to show that the existence of lim p,q→ s 2 p, q by the existence of lim p,q→ s 1 p, q as 

follows. 

Since lim p,q→ s 1 p, q exists, say its limit a. Then lim p→ s 1 p, p a. It implies that 

lim p→ s 2 p, p converges, say its limit b. So, given 0, there exists a positive integer N 

such that as p, q ≥ N 

|s 1 p, p − s 1 q, q| /2 

and 

|s 2 N, N − b| /2. 

So, as p ≥ q ≥ N, 

|s 2 p, q − b| |s 2 N, N − b s 2 p, q − s 2 N, N| 

/2 |s 2 p, q − s 2 N, N| 

/2 s 1 p, p − s 1 N, N 

/2 /2 

. 

Similarly for q ≥ p ≥ N. Hence, we have shown that 

lim 

p,q→ s 2p, q b. 

That is, we have prove that a double series converges if it converges absolutely. 

(c) ∑ m,n 

e −m2 n 2 

converges. 

Proof: Letfm, n e −m2 n 2 

, then by Theorem 8.44, we have proved that

∑ m,n 

e −m2 n 2 

converges since ∑ m,n 

e −m2 n 2 

∑ m 

e −m2 ∑ n 

e −n2 . 

 

Remark: ∑ m,n1 

e −m2 n 2 

∑ 

m1 

e −m2 ∑ n1 

e −n2 e 

 

e 2 −1 

8.30 Asume that the double series ∑ m,n 

anx mn converges absolutely for |x| 1. Call 

its sum Sx. Show that each of the following series also converges absolutely for |x| 1 

and has sum Sx : 

 

∑ 

n1 

Proof: ByTheorem 8.42, 

So, ∑ 

n1 

∑ 

n1 

that 

∑ 

m,n 

 

an x n 

1 − x n , ∑ 

 

anx mn ∑ 

n1 

n1 

 

an ∑ 

m1 

Anx n ,whereAn ∑ ad. 

d|n 

 

x mn ∑ an 

n1 

2 

. 

x n 

1 − xn if |x| 1. 

an xn converges absolutely for 

1−x 

|x| 1 and has sum Sx. 

n 

Since every term in ∑ m,n 

anx mn , the term appears once and only once in 

Anx n . The converse also true. So, by Theorem 8.42 and Theorem 8.13, we know 

 

∑ 

n1 

Anx n ∑ anx mn Sx. 

m,n 

8.31 If is real, show that the double series ∑ m,n 

m in − converges absolutely if, 

p q 

and only if, 2. Hint. Let sp, q ∑ m1 

∑ n1 

|m in| − . The set 

m in : m 1,2,...p, n 1, 2, . . . , p 

consists of p 2 complex numbers of which one has absolute value 2 , three satisfy 

|1 2i| ≤ |m in| ≤ 2 2, five satisfy |1 3i| ≤ |m in| ≤ 3 2, etc. Verify this 

geometricall and deduce the inequlity 

p 

p 

2 −/2 ∑ 

2n − 1 

n ≤ sp, p ≤ ∑ 

2n − 1 

n1 

n1 n 2 1 . /2 

Proof: Since the hint is trivial, we omit the proof of hint. From the hint, we have 

p 

∑ 

n1 

2n − 1 

n 2 ≤ sp, p ∑ 

m1 

p 

p 

∑ 

n1 

p 

|m in| − ≤ ∑ 

n1 

2n − 1 

1 n 2 /2 . 

Thus, it is clear that the double series ∑ m,n 

m in − converges absolutely if, and only if, 

2. 

8.32 (a) Show that the Cauchy product of ∑ 

n0 

−1 n1 / n 1 with itself is a 

divergent series. 

Proof: Since

n 

c n ∑ a k b n−k 

k0 

n 

∑ 

k0 

−1 k 

k 1 

n 

−1 n ∑ 

k0 

−1 n−k 

n − k 1 

1 

k 1 n − k 1 

and let fk n − k 1k 1 −k − n 2 2 n2 

2 2 ≤ n2 for k 0, 1, . . . , n. 

2 

Hence, 

n 

|c n| ∑ 

k0 

1 

k 1 n − k 1 

2n 1 

≥ → 2asn → . 

n 2 

That is, the Cauchy product of ∑ 

n0 

−1 n1 / n 1 with itself is a divergent series. 

(b) Show that the Cauchy product of ∑ 

n0 

−1 n1 /n 1 with itself is the series 

 

2 ∑ −1 

n1 

1 

n 1 

1 2 ... 1 n . 

n1 

Proof: Since 

n 

c n ∑ a k b n−k 

we have 

 

∑ 

n0 

k0 

n 

∑ 

k0 

−1 n ∑ 

k0 

2−1n 

n 2 

−1 n 

n − k 1k 1 

 

c n ∑ 

n 

n0 

 

2 ∑ 

n0 

 

2 ∑ 

n1 

∑ k0 

1 

n 2 

n 

2−1 

n 

n 2 

1 

k 1 , 

∑ n 

k0 

1 

k 1 1 

n − k 1 

1 

k 1 

−1 

n 

n 2 1 1 2 ... 1 

n 1 

−1 

n1 

n 1 

(c) Does this converge Why 

Proof: Yes by the same argument in Exercise 8.26. 

1 1 2 ... 1 n . 

8.33 Given two absolutely convergent power series, say ∑ 

n0 

a n x n and ∑ n0 

b n x n , 

having sums Ax and Bx, respectively, show that ∑ n0 

c n x n AxBx where

n 

c n ∑ a k b n−k . 

k0 

Proof: ByTheorem 8.44 and Theorem 8.13, it is clear. 

Remark: We can use Mertens’ Theorem, thenitisclear. 

8.34 Aseriesoftheform∑ 

n1 

a n /n s is called a Dirichlet series. Given two absolutely 

convergent Dirichlet series, say ∑ 

n1 

a n /n s and ∑ 

n1 

b n /n s , having sums As and Bs, 

respectively, show that ∑ n1 

c n /n s AsBs, wherec n ∑ d|n 

a d b n/d . 

Proof: ByTheorem 8.44 and Theorem 8.13, wehave 

 

∑ 

n1 

a n /n s 

 

∑ 

n1 

 

b n /n s ∑ 

n1 

where 

C n ∑ a d d −s b n/d n/d −s 

d|n 

n −s ∑ a d b n/d 

d|n 

c n /n s . 

So,wehaveprovedit. 

8.35 s ∑ 

n1 

1/n s , s 1, show that 2 s ∑ 

n1 

dn/n s ,wheredn is the 

number of positive divisors of n (including 1 and n). 

Proof: ItisclearbyExercise 8.34. So, we omit the proof. 

Ces’aro summability 

8.36 Show that each of the following series has C,1 sum 0 : 

(a) 1 − 1 − 1 1 1 − 1 − 1 1 1 −−. 

Proof: It is clear that |s 1 ...s n| ≤ 1 for all n, wheres n means that the nth partial sum 

of given series. So, 

s 1 ...s n 

n ≤ 1 n 

which implies that the given series has C,1 sum 0. 

(b) 1 − 1 1 1 − 1 1 1 − 1 − . 

2 2 2 2 2 

Proof: It is clear that |s 1 ...s n| ≤ 1 for all n, wheres 2 

n means that the nth partial 

sum of given series. So, 

s 1 ...s n 

n ≤ 

2n 

1 


(c) cos x cos3x cos5x (x real, x ≠ m). 

Proof: Lets n cosx ... cos2n − 1x, then 

C n

So, 

n 

s n ∑ cos2k − 1x 

j1 

sin 2nx 

2sinx . 

n 

∑ j1 

s ∑ n 

j sin 2jx 

j1 

n 

2n sin x 

sin nx sinn 1x 

 

2n sin x sin x 

≤ 1 → 0 

2nsin x 2 


8.37 Given a series ∑ a n ,let 

Prove that: 

(a) t n n 1s n − n n 

n 

s n ∑ 

k1 

Proof: DefineS 0 0, and thus 

n 

t n ∑ ka k 

k1 

n 

n 

a k , t n ∑ 

k1 

∑ ks k − s k−1 

k1 

n 

∑ 

k1 

n 

n 

ks k − ∑ ks k−1 

k1 

∑ ks k − ∑k 1s k 

k1 

n 

∑ 

k1 

n−1 

k1 

n 

ks k − ∑ 

k1 

n 1s n − ∑ 

k1 

n 1s n − n n . 

ka k , n 1 n 

n ∑ s k . 

k1 

k 1s k n 1s n 

(b) If ∑ a n is C,1 summable, then ∑ a n converges if, and only if, t n on as 

n → . 

Proof: Assume that ∑ a n converges. Then lim n→ s n exists, say its limit a. By(a),we 

have 

t nn n 1 n s n − n . 

Then by Theorem 8.48, we also have lim n→ n a. Hence, 

n 

s k

lim 

t nn 

n→ 

lim n 1 

n→ n s n − n 

lim n 1 

n→ n lim n→ 

s n − lim n→ 

n 

1 a − a 

0 

which is t n on as n → . 

Conversely, assume that t n on as n → , thenby(a),wehave 

n t nn n 

n 1 n 1 n s n 

which implies that (note that lim n→ n exists by hypothesis) 

lim n→ 

s n lim n t nn 

n→ 

n 

n 1 n 1 n 

lim n 

n→ n 1 lim t nn lim n 

n→ n→ n 1 lim 

n→ 

n 

1 0 1 lim n→ 

n 

lim n→ 

n 

That is, ∑ a n converges. 

(c) ∑ a n is C,1 summable if, and only if, ∑ t n /nn 1 converges. 



t n 

nn 1 s n − 

n 

∑ 

k1 

n 

n 1 

n n − n − 1 n−1 

n − n 

n 1 

n 

n 1 n − n − n 

1 n−1 

t k 

kk 1 

n 

n 1 n. * 

()Suppose that ∑ a n is C,1 summable, i.e., lim n→ n exists. Then 

n t 

lim n→ ∑ k 

k1 

exists by (*). 

kk1 

n t 

()Suppose that lim n→ ∑ k 

k1 

exists. Then lim 

kk1 

n→ n exists by (*). Hence, ∑ a n 

is C,1 summable. 

8.38 Given a monotonic a n of positive terms, such that lim n→ a n 0. Let 

n 

s n ∑ 

k1 

Prove that: 

(a) v n 1 u 2 n −1 n s n /2. 

n 

a k , u n ∑ 

k1 

Proof: Defines 0 0, and thus consider 

−1 k a k , v n ∑ 

k1 

n 

−1 k s k .

n 

u n ∑−1 k a k 

k1 

n 

∑−1 k s k − s k−1 

k1 

n 

∑−1 k s k ∑−1 k1 s k−1 

k1 

n 

∑ 

k1 

n 

k1 

n 

−1 k s k ∑ 

k1 

−1 k s k −1 n1 s n 


2v n −1 n1 s n 

v n 1 2 u n −1 n s n /2. 

(b) ∑ 

n1 

−1 n s n is C,1 summable and has Ces’aro sum 1 ∑ −1 n a 

2 n1 n . 

Proof: First, lim n→ u n exists since it is an alternating series. In addition, since 

lim n→ a n 0, we know that lim n→ s n /n 0byTheorem 8.48. Hence, 

Consider by (a), 

v n 

n 

∑ k1 

v k 

n 

u n 

2n −1n s n 

2n 

 

1 n 

2 

∑ k1 

∑ n 

u 

k1 k 

2n 

→ 1 2 lim n→ u k 

 

→ 0asn → . 

u k 1 n 

2 

∑ k1 

−1 k s k 

n 

v n 

2n 

1 2 ∑ −1 n a n 

n1 

by Theorem 8.48. 

(c) ∑ n1 

−1 n 1 1 2 ... 1 n − log 2 C,1. 

 

Proof: By (b) and ∑ n1 

Infinite products 

−1 n 

n 

− log 2, it is clear. 

8.39 Determine whether or not the following infinite products converges. Find the 

value of each convergent product. 

 

(a) n2 

1 − 

2 

nn1 


we have 

1 − 2 

nn 1 

 

n − 1n 2 

nn 1 

,

n 

 

n2 


(b) 

n2 

1 − n −2 


we have 


 

(c) n2 

n 3 −1 

n 3 1 


1 − 2 

kk 1 

n 

 

k2 

 

 

n2 

n 

 

n2 

k − 1k 2 

kk 1 

1 2 4 

3 2 3 5 

4 3 4 6 

5 

n 3n 

2 

1 − n −2 

1 − 2 

nn 1 

1 − k −2 

we have (let fk k − 1 2 k − 1 1), 

n 

 

k2 

1 3 . 

n − 1n 1 

nn , 

n 

k2 

n 1 

2n 

k − 1k 1 

kk 

 

1 − n −2 1/2. 

n2 

n 3 − 1 

n 3 1 n − 1n2 n 1 

n 1n 2 − n 1 

n − 1n 

 

2 n 1 

n 1 n − 1 2 n − 1 1 

n 

k 3 − 1 

k 3 1 

k2 

n 

− 1n 2 

nn 1 

k − 1k 2 k 1 

k 1 k − 1 2 k − 1 1 

2 3 n2 n 1 

nn 1 


 

n3 − 1 

n 3 1 2 3 . 

n2 

(d) n0 

1 z 2n 

if |z| 1. 


n 

1 z 2k 

1 z1 z 2 1 z 2n 

 

k0


which implies that (if |z| 1) 

So, 

n 

 

k0 

n 

1 − z 1 z 2k 

1 − z 2n1 

k0 

1 z 2k 

1 − z2n1 

1 − z 

→ 1 

1 − z 

 

1 z 2n 

1 

1 − z . 

n0 

as n → . 

8.40 If each partial sum s n of the convergent series ∑ a n is not zero and if the sum 

itself is not zero, show that the infinite product a 1 n2 

1 − a n /s n−1 converges and has the 

value ∑ n1 

a n . 


n 

n 

s 

a 1 1 a k /s k−1 a 1 k−1 a k 

s k−1 

k2 

k2 

n 

 

a 1 

k2 

s k 

s k−1 

s n → ∑ a n ≠ 0. 

So, the infinite product a 1 

n2 

1 − a n /s n−1 converges and has the value ∑ n1 

a n . 

8.41 Find the values of the following products by establishing the following identities 

and summing the series: 

(a) n2 

1 − 1 2 ∑ 2 

2 n −2 −n . 

n1 


we have 

1 − 1 

2 n − 2 2n − 1 

2 n − 2 1 2 2n − 1 

2 n−1 − 1 , 

n 

 

k2 

1 − 1 

2 k − 2 

n 

 

k2 

1 

2 2k − 1 

2 k−1 − 1 

n 

2 −n−1 

k2 

2 k − 1 

2 k−1 − 1 

2 −n−1 2 n − 1 

2 −n−1 2 n−1 ...1 

1 ... 1 

n 

∑ 

k1 

n 

2 ∑ 

k1 

1 

2 k−1 

2 n−1 

1 

2 k .

So, 

 

(b) n2 

1 1 

n 2 −1 


we have 

So, 

 

 

n2 

1 − 1 

2 n − 2 

1 

2 ∑ . 

n1 nn1 

1 1 

n 2 − 1 n 2 

n 2 − 1 

n 

 

k2 

 

 

n2 

1 1 

k 2 − 1 

1 1 

n 2 − 1 

 

2 ∑ 2 −n 

n1 

2. 

nn 

n − 1n 1 , 

n 

 

kk 

k − 1k 1 

k2 

2 n 

n 1 

2 1 − 1 

n 

2 ∑ 

k1 

 

2 ∑ 

n1 

2. 

n 1 

1 

kk 1 . 

1 

nn 1 

8.42 Determine all real x for which the product 

n1 

cosx/2 n converges and find the 

value of the product when it does converge. 

Proof: Ifx ≠ m, wherem ∈ Z, thensin x 

n 

 

k1 

cosx/2 k 2n sin x 

2 n 

2 n sin x 

n 

 

2 n k1 

2 n ≠ 0 for all n ∈ N. Hence, 

cosx/2 k 

sin x 

2 n sin x → sin x x . 

2 n 

If x m, wherem ∈ Z. Then as m 0, it is clear that the product converges to 1. So, 

we consider m ≠ 0 as follows. Since x m, choosing n large enough, i.e., as n ≥ N so 

that sin x ≠ 0. Hence, 

2 n 

and note that 

Hence, 

n 

 

k1 

N−1 

cosx/2 k 

k1 

lim n→ 

n 

cosx/2 k cosx/2 k 

kN 

N−1 

cosx/2 k sinx/2 

 

N−1 

2 n−N1 sinx/2 n 

k1 

sinx/2 N−1 

2 n−N1 sinx/2 n sinx/2N−1 

x/2 N−1 . 

 

cosx/2 k sinx/2N−1 N−1 

 

cosx/2 

x/2 N−1 k . 

k1 

k1

So, by above sayings, we have prove that the convergence of the product for all x ∈ R. 

8.43 (a) Let a n −1 n / n for n 1,2,... Show that1 a n diverges but that 

∑ a n converges. 

Proof: Clearly, ∑ a n converges since it is alternating series. Consider 

2n 

 

k2 

2n 

1 a k 1 −1k 

k2 

k 

1 1 2 

≤ 1 1 2 

1 − 1 3 

1 − 1 4 

1 1 4 

1 1 4 

1 − 1 

2n − 1 

1 − 1 

2n 

1 1 

2n 

1 1 

2n 

1 1 2 

1 − 1 4 

1 − 

1 2n 

* 

and note that 

n 

1 − 

2k 1 : p n 

k2 

is decreasing. From the divergence of ∑ 1 , we know that p 2k n → 0. So, 

That is, 

k2 

1 a k diverges to zero. 

 

1 a k 0. 

k2 

(b) Let a 2n−1 −1/ n , a 2n 1/ n 1/n for n 1, 2, . . . Show that 1 a n 

converges but ∑ a n diverges. 

Proof: Clearly, ∑ a n diverges. Consider 

and 

2n 

1 a k 1 a 2 1 a 3 1 a 4 1 a 2n 

k2 

31 a 3 1 a 4 1 a 2n 

3 1 − 1 

2 2 

1 − 1 

n n 

2n1 

1 a k 1 a 2 1 a 3 1 a 4 1 a 2n 1 a 2n1 

k2 

3 1 − 1 1 − 1 

2 2 

n n 

By (*) and (**), we know that 

1 a n converges 

n 

since k2 

1 − 1 

k k 

converges. 

1 − 1 

n 1 

8.44 Assume that a n ≥ 0 for each n 1, 2, . . . Assume further that 

* 

**

Show that k1 

a 2n2 

1 a 2n2 

a 2n1 

a 2n 

1 a 2n 

for n 1, 2, . . . 

1 −1 k a k converges if, and only if, ∑ k1 

−1 k a k converges. 

a 

1c 

Proof: First, we note that if b, then 1 a1 − b 1, and if b 

1a 

c , then 

1 1 − b1 c. Hence, by hypothesis, we have 

1 1 a 2n 1 − a 2n1 * 

and 

1 1 a 2n2 1 − a 2n1 . ** 

()Suppose that ∑ 

k1 

−1 k a k converges, then lim k→ a k 0. Consider Cauchy 

Condition for product, 

1 −1 p1 a p1 1 −1 p2 a p2 1 −1 pq a pq − 1 for q 1, 2, 3, . . . . 

If p 1 2m, andq 2l, then 

1 −1 p1 a p1 1 −1 p2 a p2 1 −1 pq a pq − 1 

|1 a 2m 1 − a 2m1 1 a 2m2l − 1| 

≤ 1 a 2m − 1by(*)and(**) 

a 2m → 0. 

 

Similarly for other cases, so we have proved that k1 

1 −1 k a k converges by 

Cauchy Condition for product. 

()This is a counterexample as follows. Let a n −1 n 

n, thenitiseasytoshowthat 

a 2n2 

a 

1 a 2n1 

2n2 

In addition, 

n 

 

k1 

1 −1 k a k 

k1 

However, consider 

n 

∑ 

k1 

n 

∑ 

k1 

n 

∑ 

k1 

n 

n 

a 2k − a 2k−1 

exp 1 

2k 

exp −1k 

k 

− exp 

exp 

expb k 1 

2k 1 

2k − 1 

 

exp −1n 

n 

a 2n 

1 a 2n 

for n 1, 2, . . . 

−1 

2k − 1 

n 

∑ 

k1 

−1 k 

k 

,whereb k ∈ 

≥ ∑ exp−1 1 

k1 

2k 1 

2k − 1 

So, by Theorem 8.13, we proved the divergence of ∑ k1 

→ as n → . 

 

−1 k a k . 

− 1 

≥ 0 for all 

→ exp− log 2 as n → . 

−1 

2k − 1 , 1 

2k 

8.45 A complex-valued sequence fn is called multiplicative if f1 1andif 

fmn fmfn whenever m and n are relatively prime. (See Section 1.7) It is called 

completely multiplicative if 

f1 1andiffmn fmfn for all m and n.

(a) If fn is multiplicative and if the series ∑ fn converges absolutely, prove that 

 

∑ 

n1 

fn 1 fp k fp k2 ..., 

k1 

where p k denote the kth prime, the product being absolutely convergent. 

 

Proof: We consider the partial product P m m 

k1 

1 fp k fp k2 ... and show 

that P m → ∑ 

n1 

fn as m → . Writing each factor as a geometric series we have 

m 

P m 1 fp k fp k2 ..., 

k1 

a product of a finite number of absolutely convergent series. When we multiple these series 

together and rearrange the terms such that a typical term of the new absolutely convergent 

series is 

fn fp 1 

a 1 

fp m 

a m 

,wheren p 1 

a 1 

p m 

a m 

, 

and each a i ≥ 0. Therefore, we have 

P m ∑ fn, 

1 

where ∑ 1 

is summed over those n having all their prime factors ≤ p m .Bytheunique 

factorization theorem (Theorem 1.9), each such n occors once and only once in ∑ 1 

. 

Substracting P m from ∑ 

n1 

fn, weget 

 

 

∑ fn − P m ∑ fn − ∑ fn ∑ fn 

n1 

n1 

1 

2 

where ∑ 2 

is summed over those n having at least one prime factor p m . Since these n 

occors among the integers p m ,wehave 

 

∑ 

n1 

fn − P m 

 

≤ ∑ 

np m 

|fn|. 

As m → the last sum tends to 0 because ∑ n1 

fn converges, so P m → ∑ n1 

fn. 

To prove that the product converges absolutely we use Theorem 8.52. The product has 

the form 1 a k ,where 

a k fp k fp k2 .... 

 

The series ∑|a k | converges since it is dominated by ∑ n1|fn|. 

Thereofore, 1 ak 

also converges absolutely. 

Remark: The method comes from Euler. By the same method, it also shows that there 

are infinitely many primes. The reader can see the book, An Introduction To The Theory 

Of Numbers by Loo-Keng Hua, pp 91-93. (Chinese Version) 

(b) If, in addition, fn is completely multiplicative, prove that the formula in (a) 

becomes 

 

 

∑ fn 

1 

1 − fp 

n1 

k1 

k . 

Note that Euler’s product for s (Theorem 8.56) is the special case in which fn n −s . 

Proof: By(a),iffn is completely multiplicative, then rewrite

1 fp k fp k2 ... ∑fp k n 

n0 

1 

1 − fp k 

since |fp k | 1 for all p k . (Suppose NOT, then |fp k | ≥ 1 |fp kn | |fp k | n ≥ 1 

contradicts to lim n→ fn 0. ). 

Hence, 

 

∑ 

n1 

 

fn 

k1 

1 

1 − fp k . 

8.46 This exercise outlines a simple proof of the formula 2 2 /6. Start with the 

inequality sinx x tan x, valid for 0 x /2, taking recipocals, and square each 

member to obtain 

cot 2 x 1 1 cot 

x 2 x. 

2 

Now put x k/2m 1, wherek and m are integers, with 1 ≤ k ≤ m, and sum on k to 

obtain 

m 

∑ cot 2 k 

m 

m 

2m 12 

 

2m 1 

∑ 

1 m ∑ cot 

2 k 2 

2 

2m k 

1 . 

k1 

k1 

k1 

Use the formula of Exercise 1.49(c) to deduce the ineqaulity 

m2m − 1 2 

32m 1 2 

m 

∑ 

k1 

1 

k 2 

 

2mm 12 

32m 1 2 

Now let m → to obtain 2 2 /6. 

Proof: The proof is clear if we follow the hint and Exercise 1.49 (c), soweomitit. 

8.47 Use an argument similar to that outlined in Exercise 8.46 to prove that 

4 4 /90. 

Proof: The proof is clear if we follow the Exercise 8.46 and Exercise 1.49 (c), sowe 

omit it. 

Remark: (1) From this, it is easy to compute the value of 2s, where 

s ∈ n : n ∈ N. In addition, we will learn some new method such as Fourier series and 

so on, to find the value of Riemann zeta function. 

(2) Ther is an open problem that 2s − 1, wheres ∈ n ∈ N : n 1.

Sequences of Functions 

Uniform convergence 

9.1 Assume that f n → f uniformly on S and that each f n is bounded on 

S. Prove that {f n } is uniformly bounded on S. 

Proof: Since f n → f uniformly on S, then given ε = 1, there exists a 

positive integer n 0 such that as n ≥ n 0 , we have 

|f n (x) − f (x)| ≤ 1 for all x ∈ S. (*) 

Hence, f (x) is bounded on S by the following 

|f (x)| ≤ |f n0 (x)| + 1 ≤ M (n 0 ) + 1 for all x ∈ S. (**) 

where |f n0 (x)| ≤ M (n 0 ) for all x ∈ S. 

Let |f 1 (x)| ≤ M (1) , ..., |f n0 −1 (x)| ≤ M (n 0 − 1) for all x ∈ S, then by 

(*) and (**), 

So, 

|f n (x)| ≤ 1 + |f (x)| ≤ M (n 0 ) + 2 for all n ≥ n 0 . 

|f n (x)| ≤ M for all x ∈ S and for all n 

where M = max (M (1) , ..., M (n 0 − 1) , M (n 0 ) + 2) . 

Remark: (1) In the proof, we also shows that the limit function f is 

bounded on S. 

(2) There is another proof. We give it as a reference. 

Proof: Since Since f n → f uniformly on S, then given ε = 1, there exists 

a positive integer n 0 such that as n ≥ n 0 , we have 

|f n (x) − f n+k (x)| ≤ 1 for all x ∈ S and k = 1, 2, ... 

So, for all x ∈ S, and k = 1, 2, ... 

|f n0 +k (x)| ≤ 1 + |f n0 (x)| ≤ M (n 0 ) + 1 (*) 

where |f n0 (x)| ≤ M (n 0 ) for all x ∈ S. 

Let |f 1 (x)| ≤ M (1) , ..., |f n0 −1 (x)| ≤ M (n 0 − 1) for all x ∈ S, then by 

(*), 

|f n (x)| ≤ M for all x ∈ S and for all n 

1

where M = max (M (1) , ..., M (n 0 − 1) , M (n 0 ) + 1) . 

9.2 Define two sequences {f n } and {g n } as follows: 

( 

f n (x) = x 1 + 1 ) 

if x ∈ R, n = 1, 2, ..., 

n 

g n (x) = 

{ 1 

n 

Let h n (x) = f n (x) g n (x) . 

if x = 0 or if x is irrational, 

b + 1 n if x is rational, say x = a b , b > 0. 

(a) Prove that both {f n } and {g n } converges uniformly on every bounded 

interval. 

and 

Proof: Note that it is clear that 

lim f n (x) = f (x) = x, for all x ∈ R 

n→∞ 

{ 0 if x = 0 or if x is irrational, 

lim g n (x) = g (x) = 

n→∞ b if x is ratonal, say x = a, b > 0. 

b 

In addition, in order to show that {f n } and {g n } converges uniformly 

on every bounded interval, it suffices to consider the case of any compact 

interval [−M, M] , M > 0. 

Given ε > 0, there exists a positive integer N such that as n ≥ N, we 

have 

M 

n < ε and 1 n < ε. 

Hence, for this ε, we have as n ≥ N 

|f n (x) − f (x)| = ∣ x ∣ ≤ M < ε for all x ∈ [−M, M] 

n n 

and 

|g n (x) − g (x)| ≤ 1 n 

< ε for all x ∈ [−M, M] . 

That is, we have proved that {f n } and {g n } converges uniformly on every 

bounded interval. 

Remark: In the proof, we use the easy result directly from definition 

of uniform convergence as follows. If f n → f uniformly on S, then f n → f 

uniformly on T for every subset T of S. 

2

(b) Prove that h n (x) does not converges uniformly on any bounded interval. 

Proof: Write 

{ 

h n (x) = 

a + a n 

x 

n 

( 

1 + 

1 

( n) 

if x = 0 or x is irrational 

1 + 

1 

+ ) 1 

b bn if x is rational, say x = 

a 

b 


{ 0 if x = 0 or x is irrational 

lim h n (x) = h (x) = 

n→∞ a if x is rational, say x = a . 

b 

Hence, if h n (x) converges uniformly on any bounded interval I, then h n (x) 

converges uniformly on [c, d] ⊆ I. So, given ε = max (|c| , |d|) > 0, there is a 

positive integer N such that as n ≥ N, we have 

max (|c| , |d|) > |h n (x) − h (x)| 

{ ∣ ( )∣ x 

= n 1 + 

1 ∣ 

n = 

|x| ∣ 

n 1 + 

1 ∣ ( n if x ∈ Q c ∩ [c, d] or x = 0 

∣ a 

n 1 + 

1 

+ )∣ 1 ∣ 

b bn if x ∈ Q ∩ [c, d] , x = 

a 

b 

which implies that (x ∈ [c, d] ∩ Q c or x = 0) 

max (|c| , |d|) > |x| 

n ∣ 1 + 1 n∣ ≥ |x| 

n 

≥ 

max (|c| , |d|) 

n 

which is absurb. So, h n (x) does not converges uniformly on any bounded 

interval. 

9.3 Assume that f n → f uniformly on S, g n → f uniformly on S. 

(a) Prove that f n + g n → f + g uniformly on S. 

Proof: Since f n → f uniformly on S, and g n → f uniformly on S, then 

given ε > 0, there is a positive integer N such that as n ≥ N, we have 

. 

and 

|f n (x) − f (x)| < ε 2 

|g n (x) − g (x)| < ε 2 

for all x ∈ S 

for all x ∈ S. 

Hence, for this ε, we have as n ≥ N, 

|f n (x) + g n (x) − f (x) − g (x)| ≤ |f n (x) − f (x)| + |g n (x) − g (x)| 

< ε for all x ∈ S. 

3

That is, f n + g n → f + g uniformly on S. 

Remark: There is a similar result. We write it as follows. If f n → f 

uniformly on S, then cf n → cf uniformly on S for any real c. Since the proof 

is easy, we omit the proof. 

(b) Let h n (x) = f n (x) g n (x) , h (x) = f (x) g (x) , if x ∈ S. Exercise 9.2 

shows that the assertion h n → h uniformly on S is, in general, incorrect. 

Prove that it is correct if each f n and each g n is bounded on S. 

Proof: Since f n → f uniformly on S and each f n is bounded on S, then 

f is bounded on S by Remark (1) in the Exercise 9.1. In addition, since 

g n → g uniformly on S and each g n is bounded on S, then g n is uniformly 

bounded on S by Exercise 9.1. 

Say |f (x)| ≤ M 1 for all x ∈ S, and |g n (x)| ≤ M 2 for all x and all n. Then 

given ε > 0, there exists a positive integer N such that as n ≥ N, we have 

and 

|f n (x) − f (x)| < 

|g n (x) − g (x)| < 

which implies that as n ≥ N, we have 

ε 

2 (M 2 + 1) 

ε 

2 (M 1 + 1) 

|h n (x) − h (x)| = |f n (x) g n (x) − f (x) g (x)| 



= |[f n (x) − f (x)] [g n (x)] + [f (x)] [g n (x) − g (x)]| 

≤ |f n (x) − f (x)| |g n (x)| + |f (x)| |g n (x) − g (x)| 

ε 

< 

2 (M 2 + 1) M ε 

2 + M 1 

2 (M 1 + 1) 

< ε 2 + ε 2 

= ε 

for all x ∈ S. So, h n → h uniformly on S. 

9.4 Assume that f n → f uniformly on S and suppose there is a constant 

M > 0 such that |f n (x)| ≤ M for all x in S and all n. Let g be continuous 

on the closure of the disk B (0; M) and define h n (x) = g [f n (x)] , h (x) = 

g [f (x)] , if x ∈ S. Prove that h n → h uniformly on S. 

4

Proof: Since g is continuous on a compact disk B (0; M) , g is uniformly 

continuous on B (0; M) . Given ε > 0, there exists a δ > 0 such that as 

|x − y| < δ, where x, y ∈ S, we have 

|g (x) − g (y)| < ε. (*) 

For this δ > 0, since f n → f uniformly on S, then there exists a positive 

integer N such that as n ≥ N, we have 

|f n (x) − f (x)| < δ for all x ∈ S. (**) 

Hence, by (*) and (**), we conclude that given ε > 0, there exists a positive 

integer N such that as n ≥ N, we have 

Hence, h n → h uniformly on S. 

|g (f n (x)) − g (f (x))| < ε for all x ∈ S. 

9.5 (a) Let f n (x) = 1/ (nx + 1) if 0 < x < 1, n = 1, 2, ... Prove that {f n } 

converges pointwise but not uniformly on (0, 1) . 

Proof: First, it is clear that lim n→∞ f n (x) = 0 for all x ∈ (0, 1) . Supppos 

that {f n } converges uniformly on (0, 1) . Then given ε = 1/2, there exists a 


|f n (x) − f (x)| = 

1 

∣1 + nx∣ < 1/2 for all x ∈ (0, 1) . 

So, the inequality holds for all x ∈ (0, 1) . It leads us to get a contradiction 

since 

1 

1 + Nx < 1 2 

for all x ∈ (0, 1) ⇒ lim 

x→0 + 1 

1 + Nx = 1 < 1/2. 

That is, {f n } converges NOT uniformly on (0, 1) . 

(b) Let g n (x) = x/ (nx + 1) if 0 < x < 1, n = 1, 2, ... Prove that g n → 0 

uniformly on (0, 1) . 

Proof: First, it is clear that lim n→∞ g n (x) = 0 for all x ∈ (0, 1) . Given 

ε > 0, there exists a positive integer N such that as n ≥ N, we have 

1/n < ε 

5


∣ ∣ |g n (x) − g| = 

x 

∣∣∣ ∣1 + nx∣ = 1 ∣∣∣ 

1 

+ n < 1 n < ε. 

x 

So, g n → 0 uniformly on (0, 1) . 

9.6 Let f n (x) = x n . The sequence {f n (x)} converges pointwise but not 

uniformly on [0, 1] . Let g be continuous on [0, 1] with g (1) = 0. Prove that 

the sequence {g (x) x n } converges uniformly on [0, 1] . 

Proof: It is clear that f n (x) = x n converges NOT uniformly on [0, 1] 

since each term of {f n (x)} is continuous on [0, 1] and its limit function 

{ 0 if x ∈ [0, 1) 

f = 

1 if x = 1. 

is not a continuous function on [0, 1] by Theorem 9.2. 

In order to show {g (x) x n } converges uniformly on [0, 1] , it suffices to 

shows that {g (x) x n } converges uniformly on [0, 1). Note that 

lim g (x) 

n→∞ xn = 0 for all x ∈ [0, 1). 

We partition the interval [0, 1) into two subintervals: [0, 1 − δ] and (1 − δ, 1) . 

As x ∈ [0, 1 − δ] : Let M = max x∈[0,1] |g (x)| , then given ε > 0, there is a 


M (1 − δ) n < ε 

which implies that for all x ∈ [0, 1 − δ] , 

|g (x) x n − 0| ≤ M |x n | ≤ M (1 − δ) n < ε. 

Hence, {g (x) x n } converges uniformly on [0, 1 − δ] . 

As x ∈ (1 − δ, 1) : Since g is continuous at 1, given ε > 0, there exists a 

δ > 0 such that as |x − 1| < δ, where x ∈ [0, 1] , we have 

|g (x) − g (1)| = |g (x) − 0| = |g (x)| < ε 

which implies that for all x ∈ (1 − δ, 1) , 

|g (x) x n − 0| ≤ |g (x)| < ε. 

6

Hence, {g (x) x n } converges uniformly on (1 − δ, 1) . 

So, from above sayings, we have proved that the sequence of functions 

{g (x) x n } converges uniformly on [0, 1] . 

Remark: It is easy to show the followings by definition. So, we omit the 

proof. 

(1) Suppose that for all x ∈ S, the limit function f exists. If f n → f 

uniformly on S 1 (⊆ S) , then f n → f uniformly on S, where # (S − S 1 ) < 

+∞. 

(2) Suppose that f n → f uniformly on S and on T. Then f n → f uniformly 

on S ∪ T. 

9.7 Assume that f n → f uniformly on S and each f n is continuous on S. 

If x ∈ S, let {x n } be a sequence of points in S such that x n → x. Prove that 

f n (x n ) → f (x) . 

Proof: Since f n → f uniformly on S and each f n is continuous on S, by 

Theorem 9.2, the limit function f is also continuous on S. So, given ε > 0, 

there is a δ > 0 such that as |y − x| < δ, where y ∈ S, we have 

|f (y) − f (x)| < ε 2 . 

For this δ > 0, there exists a positive integer N 1 such that as n ≥ N 1 , we 

have 

|x n − x| < δ. 

Hence, as n ≥ N 1 , we have 

|f (x n ) − f (x)| < ε 2 . (*) 

In addition, since f n → f uniformly on S, given ε > 0, there exists a 

positive integer N ≥ N 1 such that as n ≥ N, we have 

|f n (x) − f (x)| < ε 2 



|f n (x n ) − f (x n )| < ε 2 . (**) 

7

By (*) and (**), we obtain that given ε > 0, there exists a positie integer N 

such that as n ≥ N, we have 

|f n (x n ) − f (x)| = |f n (x n ) − f (x n )| + |f (x n ) − f (x)| 

< ε 2 + ε 2 

= ε. 

That is, we have proved that f n (x n ) → f (x) . 

9.8 Let {f n } be a seuqnece of continuous functions defined on a compact 

set S and assume that {f n } converges pointwise on S to a limit function f. 

Prove that f n → f uniformly on S if, and only if, the following two conditions 

hold.: 

(i) The limit function f is continuous on S. 

(ii) For every ε > 0, there exists an m > 0 and a δ > 0, such that n > m 

and |f k (x) − f (x)| < δ implies |f k+n (x) − f (x)| < ε for all x in S and all 

k = 1, 2, ... 

Hint. To prove the sufficiency of (i) and (ii), show that for each x 0 in S 

there is a neighborhood of B (x 0 ) and an integer k (depending on x 0 ) such 

that 

|f k (x) − f (x)| < δ if x ∈ B (x 0 ) . 

By compactness, a finite set of integers, say A = {k 1 , ..., k r } , has the property 

that, for each x in S, some k in A satisfies |f k (x) − f (x)| < δ. Uniform 

convergence is an easy consequences of this fact. 

Proof: (⇒) Suppose that f n → f uniformly on S, then by Theorem 

9.2, the limit function f is continuous on S. In addition, given ε > 0, there 

exists a positive integer N such that as n ≥ N, we have 

|f n (x) − f (x)| < ε for all x ∈ S 

Let m = N, and δ = ε, then (ii) holds. 

(⇐) Suppose that (i) and (ii) holds. We prove f k → f uniformly on S as 

follows. By (ii), given ε > 0, there exists an m > 0 and a δ > 0, such that 

n > m and |f k (x) − f (x)| < δ implies |f k+n (x) − f (x)| < ε for all x in S 

and all k = 1, 2, ... 

Consider ∣ ∣f k(x0 ) (x 0 ) − f (x 0 ) ∣ ∣ < δ, then there exists a B (x 0 ) such that as 

x ∈ B (x 0 ) ∩ S, we have 

∣ fk(x0 ) (x) − f (x) ∣ ∣ < δ 

8

y continuity of f k(x0 ) (x) − f (x) . Hence, by (ii) as n > m 

∣ 

∣f k(x0 )+n (x) − f (x) ∣ ∣ < ε if x ∈ B (x 0 ) ∩ S. (*) 

Note that S is compact and S = ∪ x∈S (B (x) ∩ S) , then S = ∪ p k=1 (B (x k) ∩ S) . 

So, let N = max p i=1 (k (x p) + m) , as n > N, we have 

|f n (x) − f (x)| < ε for all x ∈ S 

with help of (*). That is, f n → f uniformly on S. 

9.9 (a) Use Exercise 9.8 to prove the following theorem of Dini: If 

{f n } is a sequence of real-valued continuous functions converginf 

pointwise to a continuous limit function f on a compact set S, and 

if f n (x) ≥ f n+1 (x) for each x in S and every n = 1, 2, ..., then f n → f 

uniformly on S. 

Proof: By Exercise 9.8, in order to show that f n → f uniformly on S, 

it suffices to show that (ii) holds. Since f n (x) → f (x) and f n+1 (x) ≤ f n (x) 

on S, then fixed x ∈ S, and given ε > 0, there exists a positive integer 

N (x) = N such that as n ≥ N, we have 

0 ≤ f n (x) − f (x) < ε. 

Choose m = 1 and δ = ε, then by f n+1 (x) ≤ f n (x) , then (ii) holds. We 

complete it. 

Remark: (1) Dini’s Theorem is important in Analysis; we suggest the 

reader to keep it in mind. 

(2) There is another proof by using Cantor Intersection Theorem. 

We give it as follows. 

Proof: Let g n = f n − f, then g n is continuous on S, g n → 0 pointwise on 

S, and g n (x) ≥ g n+1 (x) on S. If we can show g n → 0 uniformly on S, then 

we have proved that f n → f uniformly on S. 

Given ε > 0, and consider S n := {x : g n (x) ≥ ε} . Since each g n (x) is 

continuous on a compact set S, we obtain that S n is compact. In addition, 

S n+1 ⊆ S n since g n (x) ≥ g n+1 (x) on S. Then 

∩S n ≠ φ (*) 

9

if each S n is non-empty by Cantor Intersection Theorem. However (*) 

contradicts to g n → 0 pointwise on S. Hence, we know that there exists a 

positive integer N such that as n ≥ N, 

S n = φ. 

That is, given ε > 0, there exists a positive integer N such that as n ≥ N, 

we have 

|g n (x) − 0| < ε. 

So, g n → 0 uniformly on S. 

(b) Use he sequence in Exercise 9.5(a) to show that compactness of S is 

essential in Dini’s Theorem. 

Proof: Let f n (x) = 1 , where x ∈ (0, 1) . Then it is clear that each 

1+nx 

f n (x) is continuous on (0, 1) , the limit function f (x) = 0 is continuous on 

(0, 1) , and f n+1 (x) ≤ f n (x) for all x ∈ (0, 1) . However, f n → f not uniformly 

on (0, 1) by Exercise 9.5 (a). Hence, compactness of S is essential in Dini’s 

Theorem. 

9.10 Let f n (x) = n c x (1 − x 2 ) n for x real and n ≥ 1. Prove that {f n } 

converges pointwsie on [0, 1] for every real c. Determine those c for which the 

convergence is uniform on [0, 1] and those for which term-by-term integration 

on [0, 1] leads to a correct result. 

Proof: It is clear that f n (0) → 0 and f n (1) → 0. Consider x ∈ (0, 1) , 

then |1 − x 2 | := r < 1, then 

lim f n (x) = lim n c r n x = 0 for any real c. 

n→∞ n→∞ 

Hence, f n → 0 pointwise on [0, 1] . 

Consider 

f n ′ (x) = n ( c 1 − x 2) ( ) 

n−1 1 

(2n − 1) 

2n − 1 − x2 , 

then each f n has the absolute maximum at x n = 1 √ 2n−1 

. 

As c < 1/2, we obtain that 

|f n (x)| ≤ |f n (x n )| 

( 

n c 

= √ 1 − 1 ) n 

2n − 1 2n − 1 

[√ ( 

= n c− 1 n 

2 

1 − 1 ) n ] 

2n − 1 2n − 1 

10 

→ 0 as n → ∞. (*)

In addition, as c ≥ 1/2, if f n → 0 uniformly on [0, 1] , then given ε > 0, there 

exists a positive integer N such that as n ≥ N, we have 

which implies that as n ≥ N, 

which contradicts to 

|f n (x)| < ε for all x ∈ [0, 1] 

|f n (x n )| < ε 

{ 

lim f √ 1 

n (x n ) = 2e 

if c = 1/2 

n→∞ ∞ if c > 1/2 . (**) 

From (*) and (**), we conclude that only as c < 1/2, the seqences of 

functions converges uniformly on [0, 1] . 

In order to determine those c for which term-by-term integration on [0, 1] , 

we consider ∫ 1 

n c 

f n (x) dx = 

2 (n + 1) 

and ∫ 1 

f (x) dx = 

0 

0 

∫ 1 

0 

0dx = 0. 

Hence, only as c < 1, we can integrate it term-by-term. 

9.11 Prove that ∑ x n (1 − x) converges pointwise but not uniformly on 

[0, 1] , whereas ∑ (−1) n x n (1 − x) converges uniformly on [0, 1] . This illustrates 

that uniform convergence of ∑ f n (x) along with pointwise convergence 

of ∑ |f n (x)| does not necessarily imply uniform convergence 

of ∑ |f n (x)| . 

Proof: Let s n (x) = ∑ n 

k=0 xk (1 − x) = 1 − x n+1 , then 

{ 1 if x ∈ [0, 1) 

s n (x) → 

. 

0 if x = 1 

Hence, ∑ x n (1 − x) converges pointwise but not uniformly on [0, 1] by Theorem 

9.2 since each s n is continuous on [0, 1] . 

Let g n (x) = x n (1 − x) , then it is clear that g n (x) ≥ g n+1 (x) for all x ∈ 

[0, 1] , and g n (x) → 0 uniformly on [0, 1] by Exercise 9.6. Hence, by Dirichlet’s 

Test for uniform convergence, we have proved that ∑ (−1) n x n (1 − x) 

converges uniformly on [0, 1] . 

11

9.12 Assume that g n+1 (x) ≤ g n (x) for each x in T and each n = 1, 2, ..., 

and suppose that g n → 0 uniformly on T. Prove that ∑ (−1) n+1 g n (x) converges 

uniformly on T. 

Proof: It is clear by Dirichlet’s Test for uniform convergence. 

9.13 Prove Abel’s test for uniform convergence: Let {g n } be a sequence 

of real-valued functions such that g n+1 (x) ≤ g n (x) for each x in T and for 

every n = 1, 2, ... If {g n } is uniformly bounded on T and if ∑ f n (x) converges 

uniformly on T, then ∑ f n (x) g n (x) also converges uniformly on T. 

Proof: Let F n (x) = ∑ n 

k=1 f k (x) . Then 

s n (x) = 

n∑ 

f k (x) g k (x) = F n g 1 (x)+ 

k=1 

and hence if n > m, we can write 

n∑ 

(F n (x) − F k (x)) (g k+1 (x) − g k (x)) 

k=1 

s n (x)−s m (x) = (F n (x) − F m (x)) g m+1 (x)+ 

n∑ 

k=m+1 

Hence, if M is an uniform bound for {g n } , we have 

|s n (x) − s m (x)| ≤ M |F n (x) − F m (x)| + 2M 

(F n (x) − F k (x)) (g k+1 (x) − g k (x)) 

n∑ 

k=m+1 

|F n (x) − F k (x)| . (*) 

Since ∑ f n (x) converges uniformly on T, given ε > 0, there exists a positive 

integer N such that as n > m ≥ N, we have 

|F n (x) − F m (x)| < ε 

M + 1 

By (*) and (**), we have proved that as n > m ≥ N, 

|s n (x) − s m (x)| < ε for all x ∈ T. 

Hence, ∑ f n (x) g n (x) also converges uniformly on T. 

for all x ∈ T (**) 

Remark: In the proof, we establish the lemma as follows. We write it 

as a reference. 

12

(Lemma) If {a n } and {b n } are two sequences of complex numbers, define 

A n = 

n∑ 

a k . 

k=1 

Then we have the identity 

n∑ 

n∑ 

a k b k = A n b n+1 − A k (b k+1 − b k ) 

(i) 

k=1 

= A n b 1 + 

k=1 

n∑ 

(A n − A k ) (b k+1 − b k ) . (ii) 

k=1 

Proof: The identity (i) comes from Theorem 8.27. In order to show 

(ii), it suffices to consider 

b n+1 = b 1 + 

n∑ 

b k+1 − b k . 

k=1 

9.14 Let f n (x) = x/ (1 + nx 2 ) if x ∈ R, n = 1, 2, ... Find the limit function 

f of the sequence {f n } and the limit function g of the sequence {f ′ n} . 

(a) Prove that f ′ (x) exists for every x but that f ′ (0) ≠ g (0) . For what 

values of x is f ′ (x) = g (x) 

Proof: It is easy to show that the limit function f = 0, and by f ′ n (x) = 

1−nx 2 

(1+nx 2 ) 2 , we have 

{ 1 if x = 0 

lim f n ′ (x) = g (x) = 

n→∞ 0 if x ≠ 0 . 

Hence, f ′ (x) exists for every x and f ′ (0) = 0 ≠ g (0) = 1. In addition, it is 

clear that as x ≠ 0, we have f ′ (x) = g (x) . 

(b) In what subintervals of R does f n → f uniformly 


1 + nx 2 

2 

≥ √ n |x| 

13

y A.P. ≥ G.P. for all real x. Hence, 

∣ x ∣∣∣ 

∣ ≤ 1 

1 + nx 2 2 √ n 

which implies that f n → f uniformly on R. 

(c) In what subintervals of R does f ′ n → g uniformly 

Proof: Since each f n ′ = 

1−nx2 is continuous on R, and the limit function 

(1+nx 2 ) 2 

g is continuous on R − {0} , then by Theorem 9.2, the interval I that we 

consider does not contains 0. Claim that f n ′ → g uniformly on such interval 

I = [a, b] which does not contain 0 as follows. 

Consider ∣ ∣ ∣∣∣ 1 − nx 2 ∣∣∣ 1 

(1 + nx 2 ) 2 ≤ 

1 + nx ≤ 1 

2 na , 2 

so we know that f ′ n → g uniformly on such interval I = [a, b] which does not 

contain 0. 

9.15 Let f n (x) = (1/n) e −n2 x 2 if x ∈ R, n = 1, 2, ... Prove that f n → 0 

uniformly on R, that f ′ n → 0 pointwise on R, but that the convergence of 

{f ′ n} is not uniform on any interval containing the origin. 

Proof: It is clear that f n → 0 uniformly on R, that f n ′ → 0 pointwise 

on R. Assume that f n ′ → 0 uniformly on [a, b] that contains 0. We will prove 

that it is impossible as follows. 

We may assume that 0 ∈ (a, b) since other cases are similar. Given ε = 1, 

e 

then there exists a positive integer N ′ such that as n ≥ max ( ) 

N ′ , 1 b := N 

(⇒ 1 ≤ b), we have N 

|f n ′ (x) − 0| < 1 for all x ∈ [a, b] 

e 


∣ ∣∣∣ 

2 Nx 

e (Nx)2 ∣ ∣∣∣ 

< 1 e 

for all x ∈ [a, b] 

which implies that, let x = 1 N , 2 

e < 1 e 

which is absurb. So, the convergence of {f ′ n} is not uniform on any interval 

containing the origin. 

14

9.16 Let {f n } be a sequence of real-valued continuous functions defined 

on [0, 1] and assume that f n → f uniformly on [0, 1] . Prove or disprove 

∫ 1−1/n 

lim 

n→∞ 

0 

f n (x) dx = 

∫ 1 

0 

f (x) dx. 

Proof: By Theorem 9.8, we have 

∫ 1 

lim 

n→∞ 

0 

f n (x) dx = 

∫ 1 

0 

f (x) dx. (*) 

Note that {f n } is uniform bound, say |f n (x)| ≤ M for all x ∈ [0, 1] and all 

n by Exercise 9.1. Hence, 

∫ 1 

∣ f n (x) dx 

∣ ≤ M → 0. (**) 

n 

1−1/n 

Hence, by (*) and (**), we have 

∫ 1−1/n 

lim 

n→∞ 

0 

f n (x) dx = 

∫ 1 

0 

f (x) dx. 

9.17 Mathematicinas from Slobbovia decided that the Riemann integral 

was too complicated so that they replaced it by Slobbovian integral, defined 

as follows: If f is a function defined on the set Q of rational numbers 

in [0, 1] , the Slobbovian integral of f, denoted by S (f) , is defined to be the 

limit 

1 

S (f) = lim 

n→∞ n 

n∑ 

f 

k=1 

( k 

n) 

, 

whenever the limit exists. Let {f n } be a sequence of functions such that 

S (f n ) exists for each n and such that f n → f uniformly on Q. Prove that 

{S (f n )} converges, that S (f) exists, and S (f n ) → S (f) as n → ∞. 

Proof: f n → f uniformly on Q, then given ε > 0, there exists a positive 

integer N such that as n > m ≥ N, we have 

|f n (x) − f (x)| < ε/3 (1) 

15

and 

So, if n > m ≥ N, 

|S (f n ) − S (f m )| = 

∣ lim 1 

k∑ 

k→∞ k 

j=1 

1 

k∑ 

= lim 

k→∞ k ∣ 

|f n (x) − f m (x)| < ε/2. (2) 

1 

≤ lim 

k→∞ k 

= ε/2 

< ε 

j=1 

( ( ) ( )) ∣ j j ∣∣∣ 

f n − f m 

k k 

( ( ) ( )) ∣ j j ∣∣∣ 

f n − f m 

k k 

k∑ 

ε/2 by (2) 

j=1 

which implies that {S (f n )} converges since it is a Cauchy sequence. Say its 

limit S. 

Consider, by (1) as n ≥ N, 

1 

k 

k∑ 

j=1 

[ ( ) ] j 

f n − ε/3 ≤ 1 k 

k 

k∑ 

( ) j 

f ≤ 1 k k 

j=1 

k∑ 

j=1 

[ ( ) ] j 

f n + ε/3 

k 


[ 

1 

k∑ 

( ) ] j 

f n − ε/3 ≤ 1 k k 

k 

j=1 

k∑ 

( ) j 

f ≤ 

k 

j=1 

[ 

1 

k 

k∑ 

( ) ] j 

f n + ε/3 

k 

j=1 

which implies that, let k → ∞ 

S (f n ) − ε/3 ≤ lim 

k→∞ 

sup 1 k 

k∑ 

( ) j 

f ≤ S (f n ) + ε/3 (3) 

k 

j=1 

and 

S (f n ) − ε/3 ≤ lim 

k→∞ 

inf 1 k 

k∑ 

( ) j 

f ≤ S (f n ) + ε/3 (4) 

k 

j=1 

16


∣ lim sup 1 k∑ 

( ) j 

f − lim inf 1 k∑ 

( ) ∣ j ∣∣∣ 

f 

k→∞ k k k→∞ k k 

j=1 

j=1 

≤ 

∣ lim sup 1 k∑ 

( ) 

∣ 

j ∣∣∣∣ f − S (f n ) 

k→∞ k k ∣ + lim inf 1 

k→∞ k 

j=1 

k∑ 

( ) 

j f − S (f n ) 

k ∣ 

≤ 2ε by (3) and (4) 

3 

< ε. (5) 

Note that (3)-(5) imply that the existence of S (f) . Also, (3) or (4) implies 

that S (f) = S. So, we complete the proof. 

9.18 Let f n (x) = 1/ (1 + n 2 x 2 ) if 0 ≤ x ≤ 1, n = 1, 2, ... Prove that {f n } 

converges pointwise but not uniformly on [0, 1] . Is term-by term integration 

permissible 

Proof: It is clear that 

lim f n (x) = 0 

n→∞ 

for all x ∈ [0, 1] . If {f n } converges uniformly on [0, 1] , then given ε = 1/3, 

there exists a positive integer N such that as n ≥ N, we have 


|f n (x)| < 1/3 for all x ∈ [0, 1] 

∣ ∣∣∣ 

f N 

( 1 

N 

)∣ ∣∣∣ 

= 1 2 < 1 3 

j=1 

which is impossible. So, {f n } converges pointwise but not uniformly on [0, 1] . 

Since {f n (x)} is clearly uniformly bounded on [0, 1] , i.e., |f n (x)| ≤ 1 

for all x ∈ [0, 1] and n. Hence, by Arzela’s Theorem, we know that the 

sequence of functions can be integrated term by term. 

9.19 Prove that ∑ ∞ 

n=1 x/nα (1 + nx 2 ) converges uniformly on every finite 

interval in R if α > 1/2. Is the convergence uniform on R 

Proof: By A.P. ≥ G.P., we have 

x 

∣n α (1 + nx 2 ) ∣ ≤ 1 

17 

2n α+ 1 2 

for all x.

So, by Weierstrass M-test, we have proved that ∑ ∞ 

n=1 x/nα (1 + nx 2 ) converges 

uniformly on R if α > 1/2. Hence, ∑ ∞ 

n=1 x/nα (1 + nx 2 ) converges 

uniformly on every finite interval in R if α > 1/2. 

9.20 Prove that the series ∑ ∞ 

n=1 ((−1)n / √ n) sin (1 + (x/n)) converges 

uniformly on every compact subset of R. 

Proof: It suffices to show that the series ∑ ∞ 

n=1 ((−1)n / √ n) sin (1 + (x/n)) 

converges uniformly on [0, a] . Choose n large enough so that a/n ≤ 1/2, and 

therefore sin ( 1 + ( )) ( 

x 

n+1 ≤ sin 1 + 

x 

n) 

for all x ∈ [0, a] . So, if we let fn (x) = 

(−1) n / √ n and g n (x) = sin ( 1 + n) x , then by Abel’s test for uniform convergence, 

we have proved that the series ∑ ∞ 

n=1 ((−1)n / √ n) sin (1 + (x/n)) 

converges uniformly on [0, a] . 

Remark: In the proof, we metion something to make the reader get 

more. (1) since a compact set K is a bounded set, say K ⊆ [−a, a] , if we can 

show the series converges uniformly on [−a, a] , then we have proved it. (2) 

The interval that we consider is [0, a] since [−a, 0] is similar. (3) Abel’s test 

for uniform convergence holds for n ≥ N, where N is a fixed positive 

integer. 

9.21 Prove that the series ∑ ∞ 

n=0 (x2n+1 / (2n + 1) − x n+1 / (2n + 2)) converges 

pointwise but not uniformly on [0, 1] . 

Proof: We show that the series converges pointwise on [0, 1] by considering 

two cases: (1) x ∈ [0, 1) and (2) x = 1. Hence, it is trivial. Define 

f (x) = ∑ ∞ 

n=0 (x2n+1 / (2n + 1) − x n+1 / (2n + 2)) , if the series converges 

uniformly on [0, 1] , then by Theorem 9.2, f (x) is continuous on [0, 1] . 

However, 

f (x) = 

{ 1 

2 

log (1 + x) if x ∈ [0, 1) 

log 2 if x = 1 

Hence, the series converges not uniformly on [0, 1] . 

Remark: The function f (x) is found by the following. Given x ∈ [0, 1), 

then both 

∞∑ 

t 2n = 1 

1 − t and 1 ∞∑ 

t n 1 

= 

2 2 2 (1 − t) 

n=0 

converges uniformly on [0, x] by Theorem 9.14. So, by Theorem 9.8, we 

n=0 

. 

18

have 

∫ x 

0 

∞∑ 

t 2n − 1 2 

n=0 

∞∑ 

t n = 

n=0 

= 

∫ x 

0 

∫ x 

0 

1 

1 − t − 1 

2 2 (1 − t) dt 

( 

1 1 

2 1 − t + 1 ) 

1 + t 

= 1 log (1 + x) . 

2 

− 1 2 

( ) 1 

dt 

1 − t 

And as x = 1, 

∞∑ ( 

x 2n+1 / (2n + 1) − x n+1 / (2n + 2) ) 

n=0 

= 

= 

∞∑ 

n=0 

∞∑ 

n=0 

1 

2n + 1 − 1 

2n 

(−1) n+1 

n + 1 

by Theorem8.14. 

= log 2 by Abel’s Limit Theorem. 

9.22 Prove that ∑ a n sin nx and ∑ b n cos nx are uniformly convergent on 

R if ∑ |a n | converges. 

Proof: It is trivial by Weierstrass M-test. 

9.23 Let {a n } be a decreasing sequence of positive terms. Prove that 

the series ∑ a n sin nx converges uniformly on R if, and only if, na n → 0 as 

n → ∞. 

Proof: (⇒) Suppose that the series ∑ a n sin nx converges uniformly on 

R, then given ε > 0, there exists a positive integer N such that as n ≥ N, 

we have 

∣ ∣∣∣∣ 2n−1 

∑ 

a k sin kx 

< ε. (*) 

∣ 

Choose x = 1 

2n , then sin 1 2 

k=n 

≤ sin kx ≤ sin 1. Hence, as n ≥ N, we always 

19

have, by (*) 

2n−1 

∑ 

2n−1 

(ε >) 

a k sin kx 

∣ 

∣ = ∑ 

a k sin kx 

k=n 

≥ 

= 

k=n 

2n−1 

∑ 

a 2n sin 1 2 since a k > 0 and a k ↘ 

k=n 

( 1 

2 sin 1 2) 

(2na 2n ) . 

That is, we have proved that 2na 2n → 0 as n → ∞. Similarly, we also have 

(2n − 1) a 2n−1 → 0 as n → ∞. So, we have proved that na n → 0 as n → ∞. 

(⇐) Suppose that na n → 0 as n → ∞, then given ε > 0, there exists a 

positive integer n 0 such that as n ≥ n 0 , we have 

ε 

|na n | = na n < 

2 (π + 1) . (*) 

In order to show the uniform convergence of ∑ ∞ 

n=1 a n sin nx on R, it suffices 

to show the uniform convergence of ∑ ∞ 

n=1 a n sin nx on [0, π] . So, if we can 

show that as n ≥ n 0 

n+p 

∑ 

a k sin kx 

< ε for all x ∈ [0, π] , and all p ∈ N 

∣ 

∣ 

k=n+1 

then we complete [ ] it. We consider two cases as follows. (n ≥ n 0 ) 

π 

As x ∈ 0, , then 

n+p 

∣ 

∣ 

∣ 

n+p 

∑ 

k=n+1 

n+p 

a k sin kx 

∣ = ∑ 

≤ 

= 

k=n+1 

n+p 

∑ 

k=n+1 

n+p 

∑ 

k=n+1 

a k sin kx 

a k kx by sin kx ≤ kx if x ≥ 0 

(ka k ) x 

ε pπ 

≤ 

by (*) 

2 (π + 1) n + p 

< ε. 

20

[ ] 

π 

And as x ∈ , π , then 

n+p 

∣ 

n+p 

∑ 

k=n+1 

a k sin kx 

∣ ≤ 

≤ 

≤ 

≤ 

≤ 

m 

∑ 

k=n+1 

a k sin kx + 

∣ 

n+p 

∑ 

k=m+1 

[ π 

] 

a k sin kx 

∣ , where m = x 

m∑ 

a k kx + 2a m+1 

sin x by Summation by parts 

k=n+1 

2 

ε 

2 (π + 1) (m − n) x + 2a m+1 

sin x 2 

ε 

π 

2 (π + 1) mx + 2a m+1 

x by 2x [0, 

π ≤ sin x if x ∈ π 2 

ε 

2 (π + 1) π + 2a m+1 (m + 1) 

< ε 2 + 2 ε 

2 (π + 1) 

< ε. 

Hence, ∑ ∞ 

n=1 a n sin nx converges uniformly on R. 

Remark: (1) In the proof (⇐), if we can make sure that na n ↘ 0, then 

we can use the supplement on the convergnce of series in Ch8, (C)- 

(6) to show the uniform convergence of ∑ ∞ 

n=1 a n sin nx = ∑ ∞ 

n=1 (na n) ( ) 

sin nx 

n 

by Dirichlet’s test for uniform convergence. 

(2)There are similar results; we write it as references. 

(a) Suppose a n ↘ 0, then for each α ∈ ( 0, π 2 

) 

, 

∑ ∞ 

n=1 a n cos nx and 

∑ ∞ 

n=1 a n sin nx converges uniformly on [α, 2π − α] . 

Proof: The proof follows from (12) and (13) in Theorem 8.30 and 

Dirichlet’s test for uniform convergence. So, we omit it. The reader 

can see the textbook, example in pp 231. 

(b) Let {a n } be a decreasing sequence of positive terms. ∑ ∞ 

n=1 a n cos nx 

uniformly converges on R if and only if ∑ ∞ 

n=1 a n converges. 

Proof: (⇒) Suppose that ∑ ∞ 

n=1 a n cos nx uniformly converges on R, then 

let x = 0, then we have ∑ ∞ 

n=1 a n converges. 

(⇐) Suppose that ∑ ∞ 

n=1 a n converges, then by Weierstrass M-test, we 

have proved that ∑ ∞ 

n=1 a n cos nx uniformly converges on R. 

] 

21

9.24 Given a convergent series ∑ ∞ 

∑ n=1 a n. Prove that the Dirichlet series 

∞ 

n=1 a nn −s converges uniformly 

∑ 

on the half-infinite interval 0 ≤ s < +∞. 

Use this to prove that lim ∞ 

s→0 + 

n=1 a nn −s = ∑ ∞ 

n=1 a n. 

Proof: Let f n (s) = ∑ n 

k=1 a k and g n (s) = n −s , then by Abel’s test for 

uniform convergence, we have proved that the Dirichlet series ∑ ∞ 

n=1 a nn −s 

converges uniformly on the half-infinite 

∑ 

interval 0 ≤ s < +∞. Then by 

Theorem 9.2, we know that lim ∞ 

s→0 + 

n=1 a nn −s = ∑ ∞ 

n=1 a n. 

9.25 Prove that the series ζ (s) = ∑ ∞ 

n=1 n−s converges uniformly on every 

half-infinite interval 1 + h ≤ s < +∞, where h > 0. Show that the equation 

ζ ′ (s) = − 

∞∑ 

n=1 

log n 

n s 

is valid for each s > 1 and obtain a similar formula for the kth derivative 

ζ (k) (s) . 

Proof: Since n −s ≤ n −(1+h) for all s ∈ [1 + h, ∞), we know that ζ (s) = 

∑ ∞ 

n=1 n−s converges uniformly on every half-infinite interval 1+h ≤ s < +∞ 

by Weierstrass M-test. Define T n (s) = ∑ n 

k=1 k−s , then it is clear that 

And 

1. For each n, T n (s) is differentiable on [1 + h, ∞), 

3. T ′ n (s) = − 

n∑ 

k=1 

2. lim 

n→∞ 

T n (2) = π2 

6 . 

log k 

k s converges uniformly on [1 + h, ∞) 

by Weierstrass M-test. Hence, we have proved that 

ζ ′ (s) = − 

∞∑ 

n=1 

log n 

n s 

by Theorem 9.13. By Mathematical Induction, we know that 

ζ (k) (s) = (−1) k 

∞ 

∑ 

n=1 

(log n) k 

n s . 

22

0.1 Supplement on some results on Weierstrass M- 

test. 

1. In the textbook, pp 224-223, there is a surprising result called Spacefilling 

curve. In addition, note the proof is related with Cantor set in 

exercise 7. 32 in the textbook. 

2. There exists a continuous function defined on R which is nowhere 

differentiable. The reader can see the book, Principles of Mathematical 

Analysis by Walter Rudin, pp 154. 

Remark: The first example comes from Bolzano in 1834, however, he 

did NOT give a proof. In fact, he only found the function f : D → R that 

he constructed is not differentiable on D ′ (⊆ D) where D ′ is countable and 

dense in D. Although the function f is the example, but he did not find the 

fact. 

In 1861, Riemann gave 

g (x) = 

∞∑ 

n=1 

sin (n 2 πx) 

n 2 

as an example. However, Reimann did NOT give a proof in his life until 

1916, the proof is given by G. Hardy. 

In 1860, Weierstrass gave 

h (x) = 

∞∑ 

a n cos (b n πx) , b is odd, 0 < a < 1, and ab > 1 + 3π 2 , 

n=1 

until 1875, he gave the proof. The fact surprises the world of Math, and 

produces many examples. There are many researches related with it until 

now 2003. 

Mean Convergence 

9.26 Let f n (x) = n 3/2 xe −n2 x 2 . Prove that {f n } converges pointwise to 0 

on [−1, 1] but that l.i.m. n→∞ f n ≠ 0 on [−1, 1] . 

Proof: It is clear that {f n } converges pointwise to 0 on [−1, 1] , so it 

23

emains to show that l.i.m. n→∞ f n ≠ 0 on [−1, 1] . Consider 

∫ 1 

−1 

f 2 n (x) dx = 2 

∫ 1 

0 

n 3 x 2 e −2n2 x 2 dx since f 2 n (x) is an even function on [−1, 1] 

= √ 1 ∫ √ 2n 

y 2 e −y2 dy by Change of Variable, let y = √ 2nx 

2 

= 1 

−2 √ 2 

0 

∫ √ 2n 

0 

yd 

(e −y2) 

[ 

= 1 ∣√ ∣∣ 2n 

∫ √ ] 

2n 

−2 √ ye −y2 − e −y2 dy 

2 

0 0 

√ ∫ π 

∞ 

√ π 

→ 

4 √ 2 since e −x2 dx = by Exercise 7. 19. 

2 

0 

So, l.i.m. n→∞ f n ≠ 0 on [−1, 1] . 

9.27 Assume that {f n } converges pointwise to f on [a, b] and that 

l.i.m. n→∞ f n = g on [a, b] . Prove that f = g if both f and g are continuous 

on [a, b] . 

Proof: Since l.i.m. n→∞ f n = g on [a, b] , given ε k = 1 

2 k , there exists a n k 

such that 

Define 

then 

∫ b 

a 

|f nk (x) − g (x)| p dx ≤ 1 2 k , where p > 0 

h m (x) = 

m∑ 

k=1 

∫ x 

a. h m (x) ↗ as x ↗ 

b. h m (x) ≤ h m+1 (x) 

a 

|f nk (t) − g (t)| p dt, 

c. h m (x) ≤ 1 for all m and all x. 

So, we obtain h m (x) → h (x) as m → ∞, h (x) ↗ as x ↗, and 

h (x) − h m (x) = 

∞∑ 

k=m+1 

∫ x 

a 

|f nk (t) − g (t)| p dt ↗ as x ↗ 

24


h (x + t) − h (x) 

t 

≥ h m (x + t) − h m (x) 

t 

for all m. (*) 

Since h and h m are increasing, we have h ′ and h ′ m exists a.e. on [a, b] . Hence, 

by (*) 

m∑ 

h ′ m (x) = |f nk (t) − g (t)| p ≤ h ′ (x) a.e. on [a, b] 

k=1 


∞∑ 

|f nk (t) − g (t)| p exists a.e. on [a, b] . 

k=1 

So, f nk (t) → g (t) a.e. on [a, b] . In addition, f n → f on [a, b] . Then we 

conclude that f = g a.e. on [a, b] . Since f and g are continuous on [a, b] , we 

have 

∫ b 

a 

|f − g| dx = 0 

which implies that f = g on [a, b] . In particular, as p = 2, we have f = g. 

Remark: (1) A property is said to hold almost everywhere on a set 

S (written: a.e. on S) if it holds everywhere on S except for a set of measurer 

zero. Also, see the textbook, pp 254. 

(2) In this proof, we use the theorem which states: A monotonic function 

h defined on [a, b] , then h is differentiable a.e. on [a, b] . The reader can 

see the book, The reader can see the book, Measure and Integral (An 

Introduction to Real Analysis) written by Richard L. Wheeden and 

Antoni Zygmund, pp 113. 

(3) There is another proof by using Fatou’s lemma: Let {f k } be a 

measruable function defined on a measure set E. If f k ≥ φ a.e. on E and 

φ ∈ L (E) , then ∫ 

∫ 

lim k ≤ lim inf f k . 

E 

k→∞ k→∞ 

E 

Proof: It suffices to show that f nk (t) → g (t) a.e. on [a, b] . Since 

l.i.m. n→∞ f n = g on [a, b] , and given ε > 0, there exists a n k such that 

∫ b 

a 

|f nk − g| 2 dx < 1 2 k 

25


∫ b m∑ 

a 

k=1 

|f nk − g| 2 dx < 

which implies that, by Fatou’s lemma, 

m∑ 

k=1 

1 

2 k 

∫ b 

a 

lim inf ∑ m ∫ b m∑ 

|f nk − g| 2 dx ≤ lim inf |f nk − g| 2 dx 

m→∞ m→∞ 

k=1 

a 

k=1 

∞∑ 

∫ b 

= |f nk − g| 2 dx < 1. 

k=1 

a 

That is, 


∫ b ∞∑ 

a 

k=1 

|f nk − g| 2 dx < 1 

∞∑ 

|f nk − g| 2 < ∞ a.e. on [a, b] 

k=1 

which implies that f nk → g a.e. on [a, b] . 

Note: The reader can see the book, Measure and Integral (An Introduction 

to Real Analysis) written by Richard L. Wheeden and 


(4) There is another proof by using Egorov’s Theorem: Let {f k } be a 

measurable functions defined on a finite measurable set E with finite limit 

function f. Then given ε > 0, there exists a closed set F (⊆ E) , where 

|E − F | < ε such that 

f k → f uniformly on F. 

Proof: If f ≠ g on [a, b] , then h := |f − g| ≠ 0 on [a, b] . By continuity 

of h, there exists a compact subinterval [c, d] such that |f − g| ≠ 0. So, there 

exists m > 0 such that h = |f − g| ≥ m > 0 on [c, d] . Since 

∫ b 

a 

|f n − g| 2 dx → 0 as n → ∞, 

26

we have 

∫ d 

c 

|f n − g| 2 dx → 0 as n → ∞. 

then by Egorov’s Theorem, given ε > 0, there exists a closed susbet F of 

[c, d] , where |[c, d] − F | < ε such that 


f n → f uniformly on F 

0 = lim |f n − g| 

n→∞ 

∫F 

2 dx 

∫ 

= lim |f n − g| 2 dx 

F 

n→∞ 

∫ 

= |f − g| 2 dx ≥ m 2 |F | 

F 

which implies that |F | = 0. If we choose ε < d−c, then we get a contradiction. 

Therefore, f = g on [a, b] . 

Note: The reader can see the book, Measure and Integral (An Introduction 

to Real Analysis) written by Richard L. Wheeden and 


9.28 Let f n (x) = cos n x if 0 ≤ x ≤ π. 

(a) Prove that l.i.m. n→∞ f n 

converge. 

= 0 on [0, π] but that {f n (π)} does not 

Proof: It is clear that {f n (π)} does not converge since f n (π) = (−1) n . 

It remains to show that l.i.m. n→∞ f n = 0 on [0, π] . Consider cos 2n x := g n (x) 

on [0, π] , then it is clear that {g n (x)} is boundedly convergent with limit 

function 

{ 0 if x ∈ (0, π) 

g = 

1 if x = 0 or π . 

Hence, by Arzela’s Theorem, 

∫ π 

lim 

n→∞ 

0 

So, l.i.m. n→∞ f n = 0 on [0, π] . 

cos 2n xdx = 

27 

∫ π 

0 

g (x) dx = 0.

(b) Prove that {f n } converges pointwise but not uniformly on [0, π/2] . 

Proof: Note that each f n (x) is continuous on [0, π/2] , and the limit 

function 

{ 0 if x ∈ (0, π/2] 

f = 

. 

1 if x = 0 

Hence, by Theorem9.2, we know that {f n } converges pointwise but not 

uniformly on [0, π/2] . 

9.29 Let f n (x) = 0 if 0 ≤ x ≤ 1/n or 2/n ≤ x ≤ 1, and let f n (x) = n if 

1/n < x < 2/n. Prove that {f n } converges pointwise to 0 on [0, 1] but that 

l.i.m. n→∞ f n ≠ 0 on [0, 1] . 

Proof: It is clear that {f n } converges pointwise to 0 on [0, 1] . In order 

to show that l.i.m. n→∞ f n ≠ 0 on [0, 1] , it suffices to note that 

∫ 1 

Hence, l.i.m. n→∞ f n ≠ 0 on [0, 1] . 

Power series 

0 

f n (x) dx = 1 for all n. 

9.30 If r is the radius of convergence if ∑ a n (z − z 0 ) n , where each a n ≠ 0, 

show that 

∣ ∣ ∣ ∣∣∣ 

lim inf a n ∣∣∣ ≤ r ≤ lim sup 

a n ∣∣∣ 

n→∞ a n+1 n→∞ ∣ . 

a n+1 

Proof: By Exercise 8.4, we have 

1 

∣ lim n→∞ sup ∣ a n+1 ∣∣ 

≤ r = 

a n 

1 

lim n→∞ sup |a n | 1 n 

≤ 

1 

lim n→∞ inf 

∣ ∣ a n+1 ∣∣ 

. 

a n 

Since 

and 

1 

∣ 

lim n→∞ sup 

1 

lim n→∞ inf 

∣ 

∣ a n+1 

a n 

∣ a n+1 

a n 

∣ ∣∣ 

= lim 

n→∞ 

inf 

∣ ∣∣ 

= lim 

n→∞ 

sup 

∣ a n ∣∣∣ 

∣a n+1 

∣ a n ∣∣∣ 

∣ , 

a n+1 

28

we complete it. 

9.31 Given that two power series ∑ a n z n has radius of convergence 2. 

Find the radius convergence of each of the following series: In (a) and (b), k 

is a fixed positive integer. 

(a) ∑ ∞ 

n=0 ak nz n 

Proof: Since 

2 = 

we know that the radius of ∑ ∞ 

n=0 ak nz n is 

1 

, (*) 

1/n 

lim n→∞ sup |a n | 

1 

lim n→∞ sup |a k n| 1/n = 1 

(lim n→∞ sup |a n | 1/n) k = 2k . 

(b) ∑ ∞ 

n=0 a nz kn 



lim sup ∣ an z kn∣ ∣ 1/n = lim sup |a n | 1/n |z| k < 1 

n→∞ n→∞ 

( 

) 1/k 

1 

|z| < 

= 2 1/k by (*). 

lim n→∞ sup |a n | 1/n 

So, the radius of ∑ ∞ 

n=0 a nz kn is 2 1/k . 

(c) ∑ ∞ 

n=0 a nz n2 


∣ 

∣ ∣∣ 

1/n 

lim sup ∣a n z n2 = lim sup |a n | 1/n |z| n 

n→∞ 

and claim that the radius of ∑ ∞ 

n=0 a nz n2 is 1 as follows. 

If |z| < 1, it is clearly seen that the series converges. However, if |z| > 1, 

lim sup |a n| 1/n lim inf |z| n ≤ lim sup |a n | 1/n |z| n 

n→∞ n→∞ n→∞ 

29

which impliest that 

lim sup |a n| 1/n |z| n = +∞. 

n→∞ 

so, the series diverges. From above, we have proved the claim. 

9.32 Given a power series ∑ a n x n whose coefficents are related by an 

equation of the form 

a n + Aa n−1 + Ba n−2 = 0 (n = 2, 3, ...). 

Show that for any x for which the series converges, its sum is 

a 0 + (a 1 + Aa 0 ) x 

1 + Ax + Bx 2 . 



n=2 


n=0 

∞∑ 

(a n + Aa n−1 + Ba n−2 ) x n = 0 

n=2 

∞∑ 

∞∑ 

a n x n + Ax a n−1 x n−1 + Bx 2 

n=2 

∞∑ 

∞∑ 

a n x n + Ax a n x n + Bx 2 


n=0 

∞∑ 

n=0 

∞ 

∑ 

n=0 

∞ 

∑ 

n=2 

a n x n = a 0 + (a 1 + Aa 0 ) x 

1 + Ax + Bx 2 . 

a n−2 x n−2 = 0 

a n x n = a 0 + a 1 x + Aa 0 x 

Remark: We prove that for any x for which the series converges, then 

1 + Ax + Bx 2 ≠ 0 as follows. 


( ) ∑ 

∞ 

1 + Ax + Bx 

2 

a n x n = a 0 + (a 1 + Aa 0 ) x, 

n=0 

30

if x = λ (≠ 0) is a root of 1 + Ax + Bx 2 , and ∑ ∞ 

n=0 a nλ n exists, we have 

1 + Aλ + Bλ 2 = 0 and a 0 + (a 1 + Aa 0 ) λ = 0. 

Note that a 1 + Aa 0 ≠ 0, otherwise, a 0 = 0 (⇒ a 1 = 0) , and therefore, a n = 

0 for all n. Then there is nothing to prove it. So, put λ = −a 0 

a 1 +Aa 0 

into 

1 + Aλ + Bλ 2 = 0, we then have 

a 2 1 = a 0 a 2 . 

Note that a 0 ≠ 0, otherwise, a 1 = 0 and a 2 = 0. Similarly, a 1 ≠ 0, otherwise, 

we will obtain a trivial thing. Hence, we may assume that all a n ≠ 0 for all 

n. So, 

a 2 2 = a 1 a 3 . 

And it is easy to check that a n = a 0 

1 

λ n for all n ≥ N. Therefore, ∑ a n λ n = 

∑ 

a0 diverges. So, for any x for whcih the series converges, we have 1+Ax+ 

Bx 2 ≠ 0. 

9.33 Let f (x) = e −1/x2 if x ≠ 0, f (0) = 0. 

(a) Show that f (n) (0) exists for all n ≥ 1. 

Proof: By Exercise 5.4, we complete it. 

(b) Show that the Taylor’s series about 0 generated by f converges everywhere 

on R but that it represents f only at the origin. 

Proof: The Taylor’s series about 0 generated by f is 

∞∑ 

n=0 

f (n) (0) 

x n = 

n! 

∞∑ 

0x n = 0. 

So, it converges everywhere on R but that it represents f only at the origin. 

Remark: It is an important example to tell us that even for functions 

f ∈ C ∞ (R) , the Taylor’s series about c generated by f may NOT represent 

f on some open interval. Also see the textbook, pp 241. 

9.34 Show that the binomial series (1 + x) α = ∑ ∞ 

n=0 (α n) x n exhibits the 

following behavior at the points x = ±1. 

(a) If x = −1, the series converges for α ≥ 0 and diverges for α < 0. 

n=0 

31

Proof: If x = −1, we consider three cases: (i) α < 0, (ii) α = 0, and (iii) 

α > 0. 

(i) As α < 0, then 

∞∑ 

( α n) (−1) n = 

n=0 

∞∑ 

n=0 

(−1) n α (α − 1) · · · (α − n + 1) 

, 

n! 

say a n = (−1) n α(α−1)···(α−n+1) 

n! 

, then a n ≥ 0 for all n, and 

a n 

1/n 

= 

−α (−α + 1) · · · (−α + n − 1) 

(n − 1)! 

≥ −α > 0 for all n. 

Hence, ∑ ∞ 

n=0 (α n) (−1) n diverges. 

(ii) As α = 0, then the series is clearly convergent. 

(iii) As α > 0, define a n = n (−1) n ( α n) , then 

a n+1 

a n 

= n − α 

n 

≥ 1 if n ≥ [α] + 1. (*) 

It means that a n > 0 for all n ≥ [α] + 1 or a n < 0 for all n ≥ [α] + 1. Without 

loss of generality, we consider a n > 0 for all n ≥ [α] + 1 as follows. 

Note that (*) tells us that 

and 

So, 

m∑ 

n=[α]+1 

a n > a n+1 > 0 ⇒ lim 

n→∞ 

a n exists. 

a n − a n+1 = α (−1) n ( α n) . 

(−1) n ( α n) = 1 α 

m∑ 

n=[α]+1 

(a n − a n+1 ) . 

By Theorem 8.10, we have proved the convergence of the series ∑ ∞ 

n=0 (α n) (−1) n . 

(b) If x = 1, the series diverges for α ≤ −1, converges conditionally for α 

in the interval −1 < α < 0, and converges absolutely for α ≥ 0. 

Proof: If x = 1, we consider four cases as follows: (i) α ≤ −1, (ii) 

−1 < −α < 0, (iii) α = 0, and (iv) α > 0 : 

(i) As α ≤ −1, say a n = α(α−1)···(α−n+1) 

n! 

. Then 

|a n | = 

−α (−α + 1) · · · (−α + n − 1) 

n! 

32 

≥ 1 for all n.

So, the series diverges. 

(ii) As −1 < α < 0, say a n = α(α−1)···(α−n+1) 

n! 

. Then a n = (−1) n b n , where 

with 

b n = 

b n+1 

b n 

−α (−α + 1) · · · (−α + n − 1) 

n! 

= n − α 

n 

> 0. 

< 1 since − 1 < −α < 0 

which implies that {b n } is decreasing with limit L. So, if we can show L = 0, 

then ∑ a n converges by Theorem 8.16. 

Rewrite 

n∏ 

( 

b n = 1 − α + 1 ) 

k 

k=1 

and since ∑ α+1 

k 

diverges, then by Theroem 8.55, we have proved L = 0. 

In order to show the convergence is conditionally, it suffices to show the 

divergence of ∑ b n . The fact follows from 

b n 

1/n 

= 

−α (−α + 1) · · · (−α + n − 1) 

(n − 1)! 

≥ −α > 0. 

(iii) As α = 0, it is clearly that the series converges absolutely. 

(iv) As α > 0, we consider ∑ |( α n)| as follows. Define a n = |( α n)| , then 

a n+1 

a n 

= n − α 

n + 1 

< 1 if n ≥ [α] + 1. 

It implies that na n − (n + 1) a n = αa n and (n + 1) a n+1 < na n . So, by 

Theroem 8.10, 

∑ 

an = 1 ∑ 

nan − (n + 1) a n 

α 

converges since lim n→∞ na n exists. So, we have proved that the series converges 

absolutely. 

9.35 Show that ∑ a n x n converges uniformly on [0, 1] if ∑ a n converges. 

Use this fact to give another proof of Abel’s limit theorem. 

Proof: Define f n (x) = a n on [0, 1] , then it is clear that ∑ f n (x) converges 

uniformly on [0, 1] . In addition, let g n (x) = x n , then g n (x) is unifomrly 

bouned with g n+1 (x) ≤ g n (x) . So, by Abel’s test for uniform convergence, 

33

∑ 

an x n converges uniformly on [0, 1] . Now, we give another proof of Abel’s 

Limit Theorem as follows. Note that each term of ∑ a n x n is continuous 

on [0, 1] and the convergence is uniformly on [0, 1] , so by Theorem 9.2, the 

power series is continuous on [0, 1] . That is, we have proved Abel’s Limit 

Theorem: 

lim 

x→1 − ∑ 

an x n = ∑ a n . 

9.36 If each a n > 0 and ∑ a n diverges, show that ∑ a n x n → +∞ as 

x → 1 − . (Assume ∑ a n x n converges for |x| < 1.) 

Proof: Given M > 0, if we can find a y near 1 from the left such that 

∑ 

an y n ≥ M, then for y ≤ x < 1, we have 

M ≤ ∑ a n y n ≤ ∑ a n x n . 

That is, lim x→1 − 

∑ 

an x n = +∞. 

Since ∑ a n diverges, there is a positive integer p such that 

p∑ 

a k ≥ 2M > M. (*) 

k=1 

Define f n (x) = ∑ n 

k=1 a kx k , then by continuity of each f n , given 0 < ε (< M) , 

there exists a δ n > 0 such that as x ∈ [δ n , 1), we have 

n∑ 

a k − ε < 

k=1 

n∑ 

a k x k < 

k=1 

n∑ 

a k + ε (**) 

k=1 

By (*) and (**), we proved that as y = δ p 

M ≤ 

p∑ 

a k − ε < 

k=1 

p∑ 

a k y k . 

k=1 

Hence, we have proved it. 

9.37 If each a n > 0 and if lim x→1 − 

∑ 

an x n exists and equals A, prove 

that ∑ a n converges and has the sum A. (Compare with Theorem 9.33.) 

Proof: By Exercise 9.36, we have proved the part, ∑ a n converges. In 

order to show ∑ a n = A, we apply Abel’s Limit Theorem to complete it. 

34

9.38 For each real t, define f t (x) = xe xt / (e x − 1) if x ∈ R, x ≠ 0, 

f t (0) = 1. 

(a) Show that there is a disk B (0; δ) in which f t is represented by a power 

series in x. 

Proof: First, we note that ex −1 

= ∑ ∞ x n 

x n=0 

:= p (x) , then p (0) = 1 ≠ 

(n+1)! 

0. So, by Theorem 9. 26, there exists a disk B (0; δ) in which the reciprocal 

of p has a power series exapnsion of the form 

∞ 

1 

p (x) = ∑ 

q n x n . 

So, as x ∈ B (0; δ) by Theorem 9.24. 

n=0 

f t (x) = xe xt / (e x − 1) 

( ∞ 

) ( 

∑ (xt) n ∞ 

) 

∑ x n 

= 

n! (n + 1)! 

n=0 

n=0 

∞∑ 

= r n (t) x n . 

n=0 

(b) Define P 0 (t) , P 1 (t) , P 2 (t) , ..., by the equation 

and use the identity 

f t (x) = 

n=0 

∞∑ 

n=0 

P n (t) xn 

, if x ∈ B (0; δ) , 

n! 

∞∑ 

P n (t) xn 

n! = ∑ ∞ etx P n (0) xn 

n! 

to prove that P n (t) = ∑ n 

k=0 (n k ) P k (0) t n−k . 

Proof: Since 

f t (x) = 

∞∑ 

n=0 

n=0 

P n (t) xn 

n! = etx x 

e x − 1 , 

35

and 

f 0 (x) = 

n=0 

∞∑ 

n=0 

P n (0) xn 

n! = x 

e x − 1 . 

So, we have the identity 

∞∑ 

P n (t) xn 

n! = ∑ ∞ etx P n (0) xn 

n! . 

n=0 

Use the identity with e tx = ∑ ∞ t n 

n=0 n! xn , then we obtain 

n∑ 


P n (t) 

n! 

= 

k=0 

= 1 n! 

P n (t) = 

t n−k P k (0) 

(n − k)! k! 

n∑ 

( n k) P k (0) t n−k 

k=0 

n∑ 

( n k) P k (0) t n−k . 

k=0 

This shows that each function P n is a polynomial. There are the Bernoulli 

polynomials. The numbers B n = P n (0) (n = 0, 1, 2, ...) are called the 

Bernoulli numbers. Derive the following further properties: 

(c) B 0 = 1, B 1 = − 1 2 , ∑ n−1 

k=0 (n k ) B k = 0, if n = 2, 3, ... 

Proof: Since 1 = p(x) , where p (x) := ∑ ∞ x n 

p(x) n=0 

, and 1 := ∑ ∞ 

(n+1)! p(x) n=0 P n (0) xn . n! 

So, 

where 

1 

1 = p (x) 

p (x) 

∞∑ x n 

= 

(n + 1)! 

n=0 

∞∑ 

= C n x n 

C n = 

n=0 

1 

(n + 1)! 

n∑ 

k=0 

36 

∞∑ 

n=0 

P n (0) xn 

n! 

( n+1 

) 

k Pk (0) .

So, 

B 0 = P 0 (0) = C 0 = 1, 

B 1 = P 1 (0) = C 1 − P 0 (0) 

= − 1 2 2 , by C 1 = 0 

and note that C n = 0 for all n ≥ 1, we have 

0 = C n−1 

= 1 ∑n−1 

( n 

n! 

k) P k (0) 

k=0 

= 1 ∑n−1 

( n 

n! 

k) B k for all n ≥ 2. 

k=0 

(d) P ′ n (t) = nP n−1 (t) , if n = 1, 2, ... 

Proof: Since 

P ′ n (t) = 

n∑ 

( n k) P k (0) (n − k) t n−k−1 

k=0 

n−1 

∑ 

= ( n k) P k (0) (n − k) t n−k−1 

k=0 

n−1 

∑ 

= 

k=0 

n−1 

n! (n − k) 

k! (n − k)! P k (0) t (n−1)−k 

∑ (n − 1)! 

= n 

k! (n − 1 − k)! P k (0) t (n−1)−k 

k=0 

n−1 

∑ 

= n 

k=0 

( n−1 

) 

k Pk (0) t (n−1)−k 

= nP n−1 (t) if n = 1, 2, ... 

(e) P n (t + 1) − P n (t) = nt n−1 if n = 1, 2, ... 

37


f t+1 (x) − f t (x) = 

so as n = 1, 2, ..., we have 

∞∑ 

n=0 

[P n (t + 1) − P n (t)] xn 

n! 

= xe xt by f t (x) = xe xt / (e x − 1) 

∞∑ 

= (n + 1) t n xn+1 

(n + 1)! , 

n=0 

P n (t + 1) − P n (t) = nt n−1 . 

by (b) 

(f) P n (1 − t) = (−1) n P n (t) 


so we have 

n=0 

Hence, P n (1 − t) = (−1) n P n (t) . 

(g) B 2n+1 = 0 if n = 1, 2, ... 

f t (−x) = f 1−t (x) , 

∞∑ 

∞ 

(−1) n P n (t) xn 

n! = ∑ 

P n (1 − t) xn 

n! . 

Proof: With help of (e) and (f), let t = 0 and n = 2k + 1, then it is clear 

that B 2k+1 = 0 if k = 1, 2, ... 

n=0 

(h) 1 n + 2 n + ... + (k − 1) n = P n+1(k)−P n+1 (0) 

n+1 

(n = 2, 3, ...) 

Proof: With help of (e), we know that 

P n+1 (t + 1) − P n+1 (t) 

n + 1 

= t n 


1 n + 2 n + ... + (k − 1) n = P n+1 (k) − P n+1 (0) 

n + 1 

(n = 2, 3, ...) 

38

Remark: (1) The reader can see the book, Infinite Series by Chao 

Wen-Min, pp 355-366. (Chinese Version) 

(2) There are some special polynomials worth studying, such as Legengre 

Polynomials. The reader can see the book, Essentials of Ordinary 

Differential Equations by Ravi P. Agarwal and Ramesh C. Gupta. 

pp 305-312. 

(3) The part (h) tells us one formula to calculte the value of the finite 

series ∑ m 

k=1 kn . There is an interesting story from the mail that Fermat, 

pierre de (1601-1665) sent to Blaise Pascal (1623-1662). Fermat 

used the Mathematical Induction to show that 

n∑ 

k (k + 1) · · · (k + p) = 

k=1 

n (n + 1) · · · (n + p + 1) 

. (*) 

p + 2 

In terms of (*), we can obtain another formula on ∑ m 

k=1 kn . 

39

Limit sup and limit inf. 

Introduction 

In order to make us understand the information more on approaches of a given real 

sequence a n 

n1 

, we give two definitions, thier names are upper limit and lower limit. It 

is fundamental but important tools in analysis. 

Definition 

Definition of limit sup and limit inf 

and 

Example 

Example 

Example 

Given a real sequence a n 

n1 

,wedefine 

b n supa m : m n 

1 1 n 

n1 

1 n 

n n1 

 

n n1 

c n infa m : m n. 

0,2,0,2,..., sowehave 

b n 2andc n 0 for all n. 

1, 2, 3,4,..., sowehave 

b n and c n for all n. 

1, 2, 3, . . . , sowehave 

b n n and c n for all n. 

Proposition Given a real sequence a n 

n1 

, and thus define b n and c n as the same as 

before. 

1 b n ,andc n n N. 

2 If there is a positive integer p such that b p , then b n n N. 

If there is a positive integer q such that c q , then c n n N. 

3 b n is decreasing and c n is increasing. 

By property 3, we can give definitions on the upper limit and the lower limit of a given 

sequence as follows. 

Definition Given a real sequence a n and let b n and c n as the same as before. 

(1) If every b n R, then 

infb n : n N 

is called the upper limit of a n , denoted by 

lim n 

sup a n . 

That is, 

lim n 

sup a n inf b n. 

n 

If every b n , then we define 

lim n 

sup a n . 

(2) If every c n R, then 

supc n : n N 

is called the lower limit of a n , denoted by 

lim n 

inf a n .

That is, 

lim n 

inf a n sup c n . 

n 

If every c n , then we define 

lim n 

inf a n . 

Remark The concept of lower limit and upper limit first appear in the book (Analyse 

Alge’brique) written by Cauchy in 1821. But until 1882, Paul du Bois-Reymond 

gave explanations on them, it becomes well-known. 

Example 1 1 n 

n1 

0,2,0,2,..., sowehave 

b n 2andc n 0 for all n 


lim sup a n 2 and lim inf a n 0. 

Example 1 n 

n n1 

1, 2, 3,4,..., sowehave 

b n and c n for all n 


lim sup a n and lim inf a n . 

 

n n1 

Example 1, 2, 3, . . . , sowehave 

b n n and c n for all n 


lim sup a n and lim inf a n . 

Relations with convergence and divergence for upper (lower) limit 

Theorem Let a n be a real sequence, then a n converges if, and only if, the upper 

limit and the lower limit are real with 

lim n 

sup a n lim n 

inf a n lim n 

a n . 


(1) lim n sup a n a n has no upper bound. 

(2) lim n sup a n for any M 0, there is a positive integer n 0 such 

that as n n 0 ,wehave 

a n M. 

(3) lim n sup a n a if, and only if, (a) given any 0, there are infinite 


a a n 

and (b) given any 0, there is a positive integer n 0 such that as n n 0 ,wehave 

a n a . 



(1) lim n inf a n a n has no lower bound. 

(2) lim n inf a n for any M 0, there is a positive integer n 0 such

that as n n 0 ,wehave 

a n M. 

(3) lim n inf a n a if, and only if, (a) given any 0, there are infinite 


a a n 

and (b) given any 0, there is a positive integer n 0 such that as n n 0 ,wehave 

a n a . 

From Theorem 2 an Theorem 3, the sequence is divergent, we give the following 

definitios. 

Definition Let a n be a real sequence, then we have 

(1) If lim n sup a n , then we call the sequence a n diverges to , 

denoted by 

lim n 

a n . 




(2) If lim n inf a n , then we call the sequence a n diverges to , 

denoted by 

lim n 

a n . 

Let a n be a real sequence. If a is a limit point of a n , then we have 

lim n 

inf a n a lim n 

sup a n . 

Some useful results 

Let a n be a real sequence, then 

(1) lim n inf a n lim n sup a n . 

(2) lim n infa n lim n sup a n and lim n supa n lim n inf a n 

(3) If every a n 0, and 0 lim n inf a n lim n sup a n , then we 

have 

lim n 

sup a 1 n 

1 

lim n inf a n 

Let a n and b n be two real sequences. 

(1) If there is a positive integer n 0 such that a n b n , then we have 

lim n 

inf a n lim n 

inf b n and lim n 

sup a n lim n 

sup b n . 

(2) Suppose that lim n inf a n , lim n inf b n , lim n sup a n , 

lim n sup b n , then 

lim n 

inf a n lim n 

inf b n 

and lim n 

inf 1 a n 

1 

lim n sup a n 

. 

lim n 

infa n b n 

lim n 

inf a n lim n 

sup b n (or lim n 

sup a n lim n 

inf b n ) 

lim n 

supa n b n 

lim n 

sup a n lim n 

sup b n . 


lim n 

supa n b n lim n 

a n lim n 

sup b n

and 

lim n 

infa n b n lim n 

a n lim n 

inf b n . 

(3) Suppose that lim n inf a n , lim n inf b n , lim n sup a n , 

lim n sup b n , anda n 0, b n 0 n, then 

lim n 

inf a n 

lim n 

inf b n 

lim n 

infa n b n 

lim n 

inf a n lim n 

sup b n (or lim n 

inf b n 

lim n 

supa n b n 

lim n 

sup a n lim n 

sup b n . 


and 

lim n 

supa n b n 

lim n 

infa n b n 

lim n 

a n 

lim n 

a n 

lim n 

sup b n 

lim n 

inf b n . 

lim n 

sup a n ) 

Theorem Let a n be a positive real sequence, then 

lim n 

inf a n1 

a n 

lim n 

infa n 1/n lim n 

supa n 1/n lim n 

sup a n1 

a n 

. 

Remark We can use the inequalities to show 

n! 1/n 

lim n n 1/e. 

Theorem Let a n be a real sequence, then 

lim n 

inf a n lim n 

inf a 1 ...a n 

n 

lim n 

sup a 1 ...a n 

n 

lim n 

sup a n . 

Exercise Let f : a, d R be a continuous function, and a n is a real sequence. If f is 

increasing and for every n, lim n inf a n , lim n sup a n a, d, then 

lim n 

sup fa n f lim n 

sup a n 

and lim n 

inf fa n f lim n 

inf a n . 

Remark: (1) The condition that f is increasing cannot be removed. For 

example, 

fx |x|, 

and 

1/k if k is even 

a k 

1 1/k if k is odd. 

(2) The proof is easy if we list the definition of limit sup and limit inf. So, we 

omit it. 

Exercise Let a n be a real sequence satisfying a np a n a p for all n, p. Show that 

an 

n converges. 

Hint: Consider its limit inf.

Something around the number e 

1. Show that the sequence 1 1 n n converges, and denote the limit by e. 

Proof: Since 

1 1 n 

n 

n 

kn 1 n 

k 

k0 

1 n 1 n 

 

nn 1 

2! 

1 

n 

1 1 

2! 

1 1 1 n ... 

n! 

1 

1 1 1 2 1 2 .. 1 ... 

2 2 n1 

3, 

2 nn 1 1 

.. 1 

n! n 

1 1 n 1 n n 

1 

and by (1), we know that the sequence is increasing. Hence, the sequence is convergent. 

We denote its limit e. Thatis, 

lim n 

1 1 n 

n e. 

n 

1 

Remark: 1. The sequence and e first appear in the mail that Euler wrote to Goldbach. 

It is a beautiful formula involving 

e i 1 0. 

1 

2. Use the exercise, we can show that k0 

e as follows. 

k! 

Proof: Let x n 1 1 n n , and let k n, wehave 

1 1 

2! 

1 1 1 .. 

k n! 

1 1 1 1 

k 

n k 

1 x k 

which implies that ( let k ) 

On the other hand, 

So,by(2)and(3),wefinallyhave 

n 

y n : 

i0 

 

 

k0 

1 

i! e. 2 

x n y n 3 

1 

k! e. 4 

3. e is an irrational number. 

Proof: Assume that e is a rational number, say e p/q, where g.c.d. p, q 1. Note 

that q 1. Consider 

q!e q! 

q! 

 

 

k0 

q 

 

k0 

1 

k! 

1 

k! 

q! 

 

 

kq1 

and since q! k0 

q 1 

k! 

and q!e are integers, we have q! kq1 

1 

k! 

is also an 

integer. However, 

1 

k! 

,

q! 

 

 

kq1 

1 

k! 

 

 

kq1 

q! 

k! 

1 

q 1 1 

q 1q 2 ... 

1 

q 1 1 

q 1 

2 

... 

1 q 

1, 

a contradiction. So, we know that e is not a rational number. 

n 1 

4. Here is an estimate about e k0 k! 

, where 0 1. ( In fact, we know 

nn! 

that e 2. 71828 18284 59045 . . . . ) 

 

Proof: Since e k0 

1 

k! 

 

0 e x n 

kn1 

1 

n 1! 

1 

n 1! 

1 

k! ,wherex n 

n 

k0 

1 

k! 

1 1 

n 2 1 

n 2n 3 ... 

1 1 

n 2 1 

n 2 2 

1 

n 1! n n 2 

1 

1 

nn! since n 2 1 

n 1 2 n . 

So, we finally have 

n 

e 

1 

k! , where 0 1. 

nn! 

k0 

Note: We can use the estimate dorectly to show e is an irrational number. 

2. For continuous variables, we have the samae result as follows. That is, 

x 

e. 

lim 

x 

1 1 x 

Proof: (1) Since 1 1 n n e as n , we know that for any sequence a n N, 

with a n , wehave 

lim n 

1 a 1 a n 

n 

e. 5 

(2) Given a sequence x n with x n , and define a n x n , then 

a n x n a n 1, then we have 

a n x 

1 1 1 1 n 

a n 1 

xn 

1 a 1 a n1 

n 

. 

Since 

a n a 

1 1 e and 1 1 n1 

a n 1 

an e as x by (5) 

we know that 

...

lim 1 1 x n 

n x n 

e. 

Since x n is arbitrary chosen so that it goes infinity, we finally obtain that 

lim 1 1 x 

x x e. 6 

(3) In order to show 1 1 x x e as x , weletx y, then 

1 1 x y 

x 1 1 y 

 

y 

y 1 

1 1 1 1 

y 1 y 1 . 

Note that x y , by (6), we have shown that 

e y 

lim 1 1 

y 1 

lim x 

1 1 x 

x 

. 

y 

y1 

y1 

1 1 

y 1 

3. Prove that as x 0, we have1 1 x x is strictly increasing, and 1 1 x x1 is 

dstrictly ecreasing. 

Proof: Since, by Mean Value Theorem 

1 

x 1 log 1 1 x logx 1 logx 1 1 x for all x 0, 

we have 

x log 1 1 

x log 1 1 x 1 0 for all x 0 

x 1 

and 

 

x 1 log 1 1 x log 1 1 x 1 x 0 for all x 0. 

Hence, we know that 

x log 1 1 x is strictly increasing on 0, 

and 

x 1 log 1 1 x is strictly decreasing on 0, . 


1 1 x 

x 

is strictly increasing 0, , and 1 1 x 

x1 

is strictly decreasing on 0, . 

Remark: By exercise 2, we know that 

lim 

x 

1 1 x 

x 

e lim 

x 

1 1 x 

4. Follow the Exercise 3 to find the smallest a such that 1 1 x xa e and strictly 

decreasing for all x 0, . 

Proof: Let fx 1 1 x xa , and consider 

log fx x a log 1 1 x : gx, 

Let us consider 

x1 

.

g x log 1 1 x x a 

x 2 x 

log1 y y 1 ay 2 1 

1 y , where 0 y 1 

1 x 1 

 

 

y 

k 

k y 1 ay2 y k 

k1 

k0 

1 

2 

a y 2 1 

3 

a y 3 ... 1 n a yn ... 

It is clear that for a 1/2, we have g x 0 for all x 0, . Note that for a 1/2, 

if there exists such a so that f is strictly decreasing for all x 0, . Then g x 0for 

all x 0, . However, it is impossible since 

g x 1 a y 2 1 a y 3 ... 1 2 3 n a yn ... 

1 2 a 0asy 1 . 

So, we have proved that the smallest value of a is 1/2. 

Remark: There is another proof to show that 1 1 x x1/2 is strictly decreasing on 

0, . 

Proof: Consider ht 1/t, and two points 1, 1 and 1 1 1 

x , lying on the graph 

1 1 x 

From three areas, the idea is that 

The area of lower rectangle Theareaofthecurve Theareaoftrapezoid 

So, we have 

1 

1 x 1 x 

Consider 

1 1 x 

1 

1 1 x 

x1/2 

 

 

log 1 1 x 1 2x 

1 1 x 

x1/2 

1 1 

1 1 x 

log 1 1 x x 1 2 

x 1 2 

1 

xx 1 

1 

xx 1 

0by(7); 

hence, we know that 1 1 x x1/2 is strictly decreasing on 0, . 

Note: Use the method of remark, we know that 1 1 x x is strictly increasing on 

0, . 

. 7

The Real And Complex Number Systems

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