Solutions Ch1-5

Plover’s Solution SeriesVolume 1:

Rudin, Principles of Mathematical Analysis

SYLeeMeng-Gen Tsai

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Preface

This solution is written for one special book named Principles of Mathemat-ical Analysis. It is written by Rudin. The purpose I wrote this solution isjust for fun; surely, you will find a lot of errors containing in my solution :ppFinally, I dedicate this solution to SYLee, wwli, ljl, and many other friends.

Meng-Gen, Tsai.

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Contents

Preface i

1 The Real and Complex Number Systems 3

2 Basic Topology 9

3 Numerical Sequences and Series 31

4 Continuity 35

5 Differentiation 41

6 Sequences and Series of Functions 67

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2 CONTENTS

Chapter 1

The Real and ComplexNumber Systems

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4 CHAPTER 1. THE REAL AND COMPLEX NUMBER SYSTEMS

Written by Men-Gen Tsaiemail: [email protected]

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6. Fix b > 1.(a) If m,n, p, q are integers, n > 0, q > 0, and r = m/n = p/q, provethat

(bm)1/n = (bp)1/q.

Hence it makes sense to define br = (bm)1/n.

(b) Prove that br+s = brbs if r and s are rational.

(c) If x is real, define B(x) to be the set of all numbers bt, where t isrational and t ≤ x. Prove that

br = sup B(r)

where r is rational. Hence it makes sense to define

bx = sup B(x)

for every real x.

(d) Prove that bx+y = bxby for all real x and y.

Proof: For (a): mq = np since m/n = p/q. Thus bmq = bnp.By Theorem 1.21 we know that (bmq)1/(mn) = (bnp)1/(mn), that is,(bm)1/n = (bp)1/q, that is, br is well-defined.

For (b): Let r = m/n and s = p/q where m, n, p, q are integers, andn > 0, q > 0. Hence (br+s)nq = (bm/n+p/q)nq = (b(mq+np)/(nq))nq =bmq+np = bmqbnp = (bm/n)nq(bp/q)nq = (bm/nbp/q)nq. By Theorem 1.21

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we know that ((br+s)nq)1/(nq) = ((bm/nbp/q)nq)1/(nq), that is br+s =bm/nbp/q = brbs.

For (c): Note that br ∈ B(r). For all bt ∈ B(r) where t is rational andt ≤ r. Hence, br = btbr−t ≥ bt1r−t since b > 1 and r − t ≥ 0. Hence br

is an upper bound of B(r). Hence br = sup B(r).

For (d): bxby = sup B(x) sup B(y) ≥ btxbty = btx+ty for all rationaltx ≤ x and ty ≤ y. Note that tx + ty ≤ x + y and tx + ty is rational.Therefore, sup B(x) sup B(y) is a upper bound of B(x + y), that is,bxby ≥ sup B(x + y) = b(x + y).

Conversely, we claim that bxbr = bx+r if x ∈ R1 and r ∈ Q. Thefollowing is my proof.

bx+r = sup B(x + r) = sup{bs : s ≤ x + r, s ∈ Q}= sup{bs−rbr : s− r ≤ x, s− r ∈ Q}= br sup{bs−r : s− r ≤ x, s− r ∈ Q}= br sup B(x)

= brbx.

And we also claim that bx+y ≥ bx if y ≥ 0. The following is my proof:(r ∈ Q)

B(x) = {br : r ≤ x} ⊂ {br : r ≤ x + y} = B(x + y),

Therefore, sup B(x + y) ≥ sup B(x), that is, bx+y ≥ bx.

Hence,

bx+y = sup B(x + y)

= sup{br : r ≤ x + y, r ∈ Q}= sup{bsbr−s : r ≤ x + y, s ≤ x, r ∈ Q, s ∈ Q}≥ sup{sup B(x)br−s : r ≤ x + y, s ≤ x, r ∈ Q, s ∈ Q}= sup B(x) sup{br−s : r ≤ x + y, s ≤ x, r ∈ Q, s ∈ Q}= sup B(x) sup{br−s : r − s ≤ x + y − s, s ≤ x, r − s ∈ Q}


= sup B(x) sup B(x + y − s)

≥ sup B(x) sup B(y)

= bxby

Therefore, bx+y = bxby.

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Chapter 2

Basic Topology

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10 CHAPTER 2. BASIC TOPOLOGY


1. Prove that the empty set is a subset of every set.

Proof: For any element x of the empty set, x is also an element ofevery set since x does not exist. Hence, the empty set is a subset ofevery set.

2. A complex number z is said to be algebraic if there are integers a0, ..., an,not all zero, such that

a0zn + a1z

n−1 + ... + an−1z + an = 0.

Prove that the set of all algebraic numbers is countable. Hint: Forevery positive integer N there are only finitely many equations with

n + |a0|+ |a1|+ ... + |an| = N.

Proof: For every positive integer N there are only finitely many equa-tions with

n + |a0|+ |a1|+ ... + |an| = N.

(since 1 ≤ n ≤ N and 0 ≤ |a0| ≤ N). We collect those equations asCN . Hence

⋃CN is countable. For each algebraic number, we can form

an equation and this equation lies in CM for some M and thus the setof all algebraic numbers is countable.

3. Prove that there exist real numbers which are not algebraic.

Proof: If not, R1 = { all algebraic numbers } is countable, a contra-diction.

4. Is the set of all irrational real numbers countable?

Solution: If R−Q is countable, then R1 = (R−Q)⋃

Q is countable,a contradiction. Thus R−Q is uncountable.

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5. Construct a bounded set of real numbers with exactly three limit points.

Solution: Put

A = {1/n : n ∈ N}⋃{1 + 1/n : n ∈ N}

⋃{2 + 1/n : n ∈ N}.

A is bounded by 3, and A contains three limit points - 0, 1, 2.

6. Let E ′ be the set of all limit points of a set E. Prove that S ′ isclosed. Prove that E and E have the same limit points. (Recall thatE = E

⋃E ′.) Do E and E ′ always have the same limit points?

Proof: For any point p of X − E ′, that is, p is not a limit point E,there exists a neighborhood of p such that q is not in E with q 6= p forevery q in that neighborhood.

Hence, p is an interior point of X − E ′, that is, X − E ′ is open, thatis, E ′ is closed.

Next, if p is a limit point of E, then p is also a limit point of E sinceE = E

⋃E ′. If p is a limit point of E, then every neighborhood Nr(p)

of p contains a point q 6= p such that q ∈ E. If q ∈ E, we completedthe proof. So we suppose that q ∈ E − E = E ′ − E. Then q is a limitpoint of E. Hence,

Nr′(q)

where r′ = 12min(r−d(p, q), d(p, q)) is a neighborhood of q and contains

a point x 6= q such that x ∈ E. Note that Nr′(q) contains in Nr(p)−{p}.That is, x 6= p and x is in Nr(p). Hence, q also a limit point of E. Hence,E and E have the same limit points.

Last, the answer of the final sub-problem is no. Put

E = {1/n : n ∈ N},

and E ′ = {0} and (E ′)′ = φ.

7. Let A1, A2, A3, ... be subsets of a metric space. (a) If Bn =⋃n

i=1 Ai,prove that Bn =

⋃ni=1 Ai, for n = 1, 2, 3, ... (b) If B =

⋃∞i=1, prove that

B ⊃ ⋃∞i=1 Ai. Show, by an example, that this inclusion can be proper.

Proof of (a): (Method 1) Bn is the smallest closed subset of X thatcontains Bn. Note that

⋃Ai is a closed subset of X that contains Bn,

thus

Bn ⊃n⋃

i=1

Ai.


If p ∈ Bn − Bn, then every neighborhood of p contained a point q 6= psuch that q ∈ Bn. If p is not in Ai for all i, then there exists someneighborhood Nri

(p) of p such that (Nri(p)−p)

⋂Ai = φ for all i. Take

r = min{r1, r2, ..., rn}, and we have Nr(p)⋂

Bn = φ, a contradiction.Thus p ∈ Ai for some i. Hence

Bn ⊂n⋃

i=1

Ai.

that is,

Bn =n⋃

i=1

Ai.

(Method 2) Since⋃n

i=1 Ai is closed and Bn =⋃n

i=1 Ai ⊂⋃n

i=1 Ai,Bn ⊂

⋃ni=1 Ai.

Proof of (b): Since B is closed and B ⊃ B ⊃ Ai, B ⊃ Ai for all i.Hence B ⊃ ⋃

Ai.

Note: My example is Ai = (1/i,∞) for all i. Thus, Ai = [1/i,∞), andB = (0,∞), B = [0,∞). Note that 0 is not in Ai for all i. Thus thisinclusion can be proper.

8. Is every point of every open set E ⊂ R2 a limit point of E? Answerthe same question for closed sets in R2.

Solution: For the first part of this problem, the answer is yes.

(Reason): For every point p of E, p is an interior point of E. Thatis, there is a neighborhood Nr(p) of p such that Nr(p) is a subset ofE. Then for every real r′, we can choose a point q such that d(p, q) =1/2 min(r, r′). Note that q 6= p, q ∈ Nr′(p), and q ∈ Nr(p). Hence, everyneighborhood Nr′(p) contains a point q 6= p such that q ∈ Nr(p) ⊂ E,that is, p is a limit points of E.

For the last part of this problem, the answer is no. Consider A ={(0, 0)}. A′ = φ and thus (0, 0) is not a limit point of E.

9. Let Eo denote the set of all interior points of a set E.

(a) Prove that Eo is always open.(b) Prove that E is open if and only if Eo = E.

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(c) If G is contained in E and G is open, prove that G is contained inEo.(d) Prove that the complement of Eo is the closure of the complementof E.(e) Do E and E always have the same interiors?(f) Do E and Eo always have the same closures?

Proof of (a): If E is non-empty, take p ∈ Eo. We need to show thatp ∈ (Eo)o. Since p ∈ Eo, there is a neighborhood Nr of p such thatNr is contained in E. For each q ∈ Nr, note that Ns(q) is contained inNr(p), where s = min{d(p, q), r − d(p, q)}. Hence q is also an interiorpoint of E, that is, Nr is contained in Eo. Hence Eo is always open.

Proof of (b): (⇒) It is clear that Eo is contained in E. Since E isopen, every point of E is an interior point of E, that is, E is containedin Eo. Therefore Eo = E.

(⇐) Since every point of E is an interior point of E (Eo(E) = E), Eis open.

Proof of (c): If p ∈ G, p is an interior point of G since G is open. Notethat E contains G, and thus p is also an interior point of E. Hencep ∈ Eo. Therefore G is contained in Eo. (Thus Eo is the biggest openset contained in E. Similarly, E is the smallest closed set containingE.)

Proof of (d): Suppose p ∈ X − Eo. If p ∈ X − E, then p ∈ X − Eclearly. If p ∈ E, then N is not contained in E for any neighborhoodN of p. Thus N contains an point q ∈ X − E. Note that q 6= p, thatis, p is a limit point of X − E. Hence X − Eo is contained in X − E.

Next, suppose p ∈ X − E. If p ∈ X − E, then p ∈ X − Eo clearly. Ifp ∈ E, then every neighborhood of p contains a point q 6= p such thatq ∈ X − E. Hence p is not an interior point of E. Hence X − E iscontained in X − Eo. Therefore X − Eo = X − E.

Solution of (e): No. Take X = R1 and E = Q. Thus Eo = φ andE

o= (R1)o = R1 6= φ.

Solution of (f): No. Take X = R1 and E = Q. Thus E = R1, andEo = φ = φ.


10. Let X be an infinite set. For p ∈ X and q ∈ X, define

d(p, q) =

{1 (if p 6= q)0 (if p = q)

Prove that this is a metric. Which subsets of the resulting metric spaceare open? Which are closed? Which are compact?

Proof: (a) d(p, q) = 1 > 0 if p 6= q; d(p, p) = 0. (b) d(p, q) = d(q, p)since p = q implies q = p and p 6= q implies q 6= p. (c) d(p, q) ≤d(p, r) + d(r, q) for any r ∈ X if p = q. If p 6= q, then either r = p orr = q, that is, r 6= p or r 6= q. Thus, d(p, q) = 1 ≤ d(p, r) + d(r, q). By(a)-(c) we know that d is a metric.

Every subset of X is open and closed. We claim that for any p ∈ X,p is not a limit point. Since d(p, q) = 1 > 1/2 if q 6= p, there exists anneighborhood N1/2(p) of p contains no points of q 6= p such that q ∈ X.Hence every subset of X contains no limit points and thus it is closed.Since X − S is closed for every subset S of X, S = X − (X − S) isopen. Hence every subset of X is open.

Every finite subset of X is compact. Let S = {p1, ..., pn} be finite.Consider an open cover {Gα} of S. Since S is covered by Gα, pi iscovered by Gαi

, thus {Gα1 , ..., Gαn} is finite subcover of S. Hence S iscompact. Next, suppose S is infinite. Consider an open cover {Gp} ofS, where

Gp = N 12(p)

for every p ∈ S. Note that q is not in Gp if q 6= p. If S is compact,then S can be covered by finite subcover, say

Gp1 , ..., Gpn .

Then there exists q such that q 6= pi for all i since S is infinite, a con-tradiction. Hence only every finite subset of X is compact.

11. For x ∈ R1 and y ∈ R1, define

d1(x, y) = (x, y)2,

d2(x, y) =√|x− y|,

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d3(x, y) = |x2 − y2|,d4(x, y) = |x− 2y|,

d5(x, y) =|x− y|

1 + |x− y|.

Determine, for each of these, whether it is a metric or not.

Solution: (1) d1(x, y) is not a metric. Since d1(0, 2) = 4, d1(0, 1) = 1,and d1(1, 2) = 1, d1(0, 2) > d1(0, 1) + d1(1, 2). Thus d1(x, y) is not ametric.(2) d2(x, y) is a metric. (a) d(p, q) > 0 if p 6= q; d(p, p) = 0. (b)

d(p, q) =√|p− q| =

√|q − p| = d(q, p). (c) |p − q| ≤ |p − r| + |r − q|,√

|p− q| ≤√|p− r|+ |r − q| ≤

√|p− r|+

√|r − q|. That is, d(p, q) ≤

d(p, r) + d(r, q). (3) d3(x, y) is not a metric since d3(1,−1) = 0.(4) d4(x, y) is not a metric since d4(1, 1) = 1 6= 0.(5) d5(x, y) is a metric since |x− y| is a metric.

Claim: d(x, y) is a metric, then

d′(x, y) =d(x, y)

1 + d(x, y)

is also a metric.

Proof of Claim: (a) d′(p, q) > 0 if p 6= q; d(p, p) = 0. (b) d′(p, q) =d′(q, p). (c) Let x = d(p, q), y = d(p, r), and z = d(r, q). Then x ≤ y+z.

d′(p, q) ≤ d′(p, r) + d′(r, q)

⇔ x

1 + x≤ y

1 + y+

z

1 + z

⇔ x(1 + y)(1 + z) ≤ y(1 + z)(1 + x) + z(1 + x)(1 + y)

⇔ x + xy + xz + xyz ≤ (y + xy + yz + xyz) + (z + xz + yz + xyz)

⇔ x ≤ y + z + 2yz + xyz

⇐ x ≤ y + z

Thus, d′ is also a metric.


12. Let K ⊂ R1 consist of 0 and the numbers 1/n, for n = 1, 2, 3, ....Prove that K is compact directly from the definition (without usingthe Heine-Borel theorem).

Proof: Suppose that {Oα} is an arbitrary open covering of K. LetE ∈ {Oα} consists 0. Since E is open and 0 ∈ E, 0 is an interior pointof E. Thus there is a neighborhood N = Nr(0) of 0 such that N ⊂ E.Thus N contains

1

[1/r] + 1,

1

[1/r] + 2, ...

Next, we take finitely many open sets En ∈ {Oα} such that 1/n ∈ En

for n = 1, 2, ..., [1/r]. Hence {E, E1, ..., E[1/r] is a finite subcover of K.Therefore, K is compact.

Note: The unique limit point of K is 0. Suppose p 6= 0 is a limit pointof K. Clearly, 0 < p < 1. (p cannot be 1). Thus there exists n ∈ Z+

such that1

n + 1< p <

1

n.

Hence Nr(p) where r = min{ 1n− p, p − 1

n+1} contains no points of K,

a contradiction.

13. Construct a compact set of real numbers whose limit points form acountable set.

Solution: Let K be consist of 0 and the numbers 1/n for n = 1, 2, 3, ...Let xK = {xk : k ∈ K} and x + K = {x + k : k ∈ K} for x ∈ R1. Itake

Sn = (1− 1

2n) +

K

2n+1,

S =∞⋃

n=1

Sn

⋃{1}.

Claim: S is compact and the set of all limit points of S is K⋃{1}.

Clearly, S lies in [0, 1], that is, S is bounded in R1. Note that Sn ⊂[1 − 1

2n , 1 − 12n+1 ]. By Exercise 12 and its note, I have that all limit

points of S⋂

[0, 1) is

0,1

2, ...,

1

2n, ...

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Clearly, 1 is also a limit point of S. Therefore, the set of all limit pointsof S is K

⋃{1}. Note that K⋃{1} ⊂ S, that is, K is compact. I com-

pleted the proof of my claim.

14. Give an example of an open cover of the segment (0, 1) which has nofinite subcover.

Solution: Take {On} = {(1/n, 1)} for n = 1, 2, 3, .... The following ismy proof. For every x ∈ (0, 1),

x ∈ (1

[1/x] + 1, 0) ∈ {On}

Hence {On} is an open covering of (0, 1). Suppose there exists a finitesubcovering

(1

n1

, 1), ..., (1

nk

, 1)

where n1 < n2 < ... < nk, respectively. Clearly 12np

∈ (0, 1) is not in

any elements of that subcover, a contradiction.

Note: By the above we know that (0, 1) is not compact.

15. Show that Theorem 2.36 and its Corollary become false (in R1, for ex-ample) if the word ”compact” is replaced by ”closed” or by ”bounded.”

Theorem 2.36: If {Kα} is a collection of compact subsets of a metricspace X such that the intersection of every finite subcollection of {Kα}is nonempty, then

⋂Kα is nonempty.

Corollary: If {Kn} is a sequence of nonempty compact sets such thatKn contains Kn+1 (n = 1, 2, 3, ...), then

⋂Kn is not empty.

Solution: For closed: [n,∞). For bounded: (−1/n, 1/n)− {0}.

16. Regard Q, the set of all rational numbers, as a metric space, withd(p, q) = |p − q|. Let E be the set of all p ∈ Q such that 2 < p2 < 3.Show that E is closed and bounded in Q, but that E is not compact.Is E open in Q?


Proof: Let S = (√

2,√

3)⋃

(−√

3,−√

2). Then E = {p ∈ Q : p ∈ S}.Clearly, E is bounded in Q. Since Q is dense in R, every limit point ofQ is in Q. (I regard Q as a metric space). Hence, E is closed in Q.

To prove that E is not compact, we form a open covering of E asfollows:

{Gα} = {Nr(p) : p ∈ E and (p− r, p + r) ⊂ S}

Surely, {Gα} is a open covering of E. If E is compact, then there arefinitely many indices α1, ..., αn such that

E ⊂ Gα1

⋃...

⋃Gαn .

For every Gαi= Nri

(pi), take p = max1≤i≤n pi. Thus, p is the nearestpoint to

√3. But Nr(p) lies in E, thus [p + r,

√3) cannot be covered

since Q is dense in R, a contradiction. Hence E is not compact.

Finally, the answer is yes. Take any p ∈ Q, then there exists aneighborhood N(p) of p contained in E. (Take r small enough whereNr(p) = N(p), and Q is dense in R.) Thus every point in N(p) is alsoin Q. Hence E is also open.

17. Let E be the set of all x ∈ [0, 1] whose decimal expansion contains onlythe digits 4 and 7. Is E countable? Is E dense in [0, 1]? Is E compact?Is E perfect?

Solution:

E ={ ∞∑

n=1

an

10n: an = 4 or an = 7

}Claim: E is uncountable.Proof of Claim: If not, we list E as follows:

x1 = 0.a11a12...a1n...

x2 = 0.a21a22...a2n...

... ...

xk = 0.ak1ak2...akn...

... ...

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(Prevent ending with all digits 9) Let x = 0.x1x2...xn... where

xn =

{4 if ann = 77 if ann = 4

By my construction, x /∈ E, a contradiction. Thus E is uncountable.

Claim: E is not dense in [0, 1].Proof of Claim: Note that E

⋂(0.47, 0.74) = φ. Hence E is not dense

in [0, 1].

Claim: E is compact.Proof of Claim: Clearly, E is bounded. For every limit point p of E,I show that p ∈ E. If not, write the decimal expansion of p as follows

p = 0.p1p2...pn...

Since p /∈ E, there exists the smallest k such that pk 6= 4 and pk 6= 7.When pk = 0, 1, 2, 3, select the smallest l such that pl = 7 if possible.(If l does not exist, then p < 0.4. Thus there is a neighborhood of psuch that contains no points of E, a contradiction.) Thus

0.p1...pl−14pl+1...pk−17 < p < 0.p1...pk−14.

Thus there is a neighborhood of p such that contains no points of E, acontradiction.

When pk = 5, 6,

0.p1...pk−147 < p < 0.p1...pk−174.

Thus there is a neighborhood of p such that contains no points of E, acontradiction.

When pk = 8, 9, it is similar. Hence E is closed. Therefore E iscompact.

Claim: E is perfect.Proof of Claim: Take any p ∈ E, and I claim that p is a limit pointof E. Write p = 0.p1p2...pn... Let

xk = 0.y1y2...yn...


where

yn =

pk if k 6= n4 if pn = 77 if pn = 4

Thus, |xk−p| → 0 as k →∞. Also, xk 6= p for all k. Hence p is a limitpoint of E. Therefore E is perfect.

18. Is there a nonempty perfect set in R1 which contains no rational num-ber?

Solution: Yes. The following claim will show the reason.

Claim: Given a measure zero set S, we have a perfect set P containsno elements in S.

Proof of Claim: (due to SYLee). Since S has measure zero, thereexists a collection of open intervals {In} such that

S ⊂⋃

In and∑|In| < 1.

Consider E = R1 − ⋃In. E is nonempty since E has positive mea-

sure. Thus E is uncountable and E is closed. Therefore there existsa nonempty perfect set P contained in E by Exercise 28. P

⋂S = φ.

Thus P is our required perfect set.

19. (a) If A and B are disjoint closed sets in some metric space X, provethat they are separated.

(b) Prove the same for disjoint open sets.

(c) Fix p ∈ X, δ > 0, define A to be the set of all q ∈ X for whichd(p, q) < δ, define B similarly, with > in place of <. Prove that A andB are separated.

(d) Prove that every connected metric space with at least two pointsis uncountable. Hint: Use (c).

Proof of (a): Recall the definition of separated: A and B are sep-arated if A

⋂B and A

⋂B are empty. Since A and B are closed sets,

A = A and B = B. Hence A⋂

B = A⋂

B = A⋂

B = φ. Hence A andB are separated.

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Proof of (b): Suppose A⋂

B is not empty. Thus there exists p suchthat p ∈ A and p ∈ B. For p ∈ A, there exists a neighborhood Nr(p) ofp contained in A since A is open. For p ∈ B = B

⋃B′, if p ∈ B, then

p ∈ A⋂

B. Note that A and B are disjoint, and it’s a contradiction.If p ∈ B′, then p is a limit point of B. Thus every neighborhood of pcontains a point q 6= p such that q ∈ B. Take an neighborhood Nr(p)ofp containing a point q 6= p such that q ∈ B. Note that Nr(p) ⊂ A,thus q ∈ A. With A and B are disjoint, we get a contradiction. HenceA

⋂(B) is empty.

Similarly, A⋂

B is also empty. Thus A and B are separated.

Proof of (c): Suppose A⋂

B is not empty. Thus there exists x suchthat x ∈ A and x ∈ B. Since x ∈ A, d(p, x) < δ. x ∈ B = B

⋃B′, thus

if x ∈ B, then d(p, x) > δ, a contradiction. The only possible is x is alimit point of B. Hence we take a neighborhood Nr(x) of x contains y

with y ∈ B where r = δ−d(x,p)2

. Clearly, d(y, p) > δ. But,

d(y, p) ≤ d(y, x) + d(x, p)

< r + d(x, p)

=δ − d(x, p)

2+ d(x, p)

=δ + d(x, p)

2

<δ + δ

2= δ.

A contradiction. Hence A⋂

B is empty. Similarly, A⋂

B is also empty.Thus A and B are separated.

Note: Take care of δ > 0. Think a while and you can prove the nextsub-exercise.

Proof of (d): Let X be a connected metric space. Take p ∈ X, q ∈ Xwith p 6= q, thus d(p, q) > 0 is fixed. Let

A = {x ∈ X : d(x, p) < δ}; B = {x ∈ X : d(x, p) > δ}.

Take δ = δt = td(p, q) where t ∈ (0, 1). Thus 0 < δ < d(p, q). p ∈ Asince d(p, p) = 0 < δ, and q ∈ B since d(p, q) > δ. Thus A and B arenon-empty.


By (c), A and B are separated. If X = A⋃

B, then X is not connected,a contradiction. Thus there exists yt ∈ X such that y /∈ A

⋃B. Let

E = Et = {x ∈ X : d(x, p) = δt} 3 yt.

For any real t ∈ (0, 1), Et is non-empty. Next, Et and Es are disjointif t 6= s (since a metric is well-defined). Thus X contains a uncount-able set {yt : t ∈ (0, 1)} since (0, 1) is uncountable. Therefore, X isuncountable.

Note: It is a good exercise. If that metric space contains only onepoint, then it must be separated.

Similar Exercise Given by SYLee: (a) Let A = {x : d(p, x) < r}and B = {x : d(p, x) > r} for some p in a metric space X. Show thatA, B are separated.(b) Show that a connected metric space with at least two points mustbe uncountable. [Hint: Use (a)]

Proof of (a): By definition of separated sets, we want to show A⋂

B =φ, and B

⋂A = φ. In order to do these, it is sufficient to show A

⋂B =

φ. Let x ∈ A⋂

B = φ, then we have:

(1) x ∈ A ⇒ d(x, p) ≤ r(2) x ∈ B ⇒ d(x, p) > r

It is impossible. So, A⋂

B = φ.

Proof of (b): Suppose that C is countable, say C = a, b, x3, .... Wewant to show C is disconnected. So, if C is a connected metric spacewith at least two points, it must be uncountable. Consider the setS = {d(a, xi) : xi ∈ C}, and thus let r ∈ R− S and inf S < r < sup S.And construct A and B as in (a), we have C = A

⋃B, where A and B

are separated. That is C is disconnected.

Another Proof of (b): Let a ∈ C, b ∈ C, consider the continuousfunction f from C into R defined by f(x) = d(x, a). So, f(C) is con-nected and f(a) = 0, f(b) > 0. That is, f(C) is an interval. Therefore,C is uncountable.

20. Are closures and interiors of connected sets always connected? (Lookat subsets of R2.)

23

Solution: Closures of connected sets is always connected, but interiorsof those is not. The counterexample is

S = N1(2, 0)⋃

N1(−2, 0)⋃{x− axis} ⊂ R2.

Since S is path-connected, S is connect. But So = N1(2)⋃

N1(−2) isdisconnected clearly.

Claim: If S is a connected subset of a metric space, then S is con-nected.

Pf of Claim: If not, then S is a union of two nonempty separated setA and B. Thus A

⋂B = A

⋂B = φ. Note that

S = S − T

= A⋃

B − T

= (A⋃

B)⋂

T c

= (A⋂

T c)⋃

(B⋂

T c)

where T = S − S. Thus

(A⋂

T c)⋂

B⋂

T c ⊂ (A⋂

T c)⋂

B⋂

T c

⊂ A⋂

B

= φ.

Hence (A⋂

T c)⋂

B⋂

T c = φ. Similarly, A⋂

T c⋂

(B⋂

T c) = φ.

Now we claim that both A⋂

T c and B⋂

T c are nonempty. Supposethat B

⋂T c = φ. Thus

A⋂

T c = S ⇔ A⋂

(S − S)c = S

⇔ A⋂

(A⋃

B − S)c = S

⇔ A⋂

((A⋃

B)⋂

Sc)c = S

⇔ A⋂

((Ac⋂

Bc)⋃

S) = S

⇔ (A⋂

S)⋃

(A⋂

Ac⋂

Bc) = S

⇔ A⋂

S = S.


Thus B is empty, a contradiction. Thus B⋂

T c is nonempty. Similarly,A

⋂T c nonempty. Therefore S is a union of two nonempty separated

sets, a contradiction. Hence S is connected.

21. Let A and B be separated subsets of some Rk, suppose a ∈ A, b ∈ B,and define

p(t) = (1− t)a + tb

for t ∈ R1. Put A0 = p−1(A), B0 = p−1(B). [Thus t ∈ A0 if and onlyif p(t) ∈ A.]

(a) Prove that A0 and B0 are separated subsets of R1.(b) Prove that there exists t0 ∈ (0, 1) such that p(t0) /∈ A

⋃B.

(c) Prove that every convex subset of Rk is connected.

Proof of (a): I claim that A0⋂

B0 is empty. (B0⋂

A0 is similar). Ifnot, take x ∈ A0

⋂B0. x ∈ A0 and x ∈ B0. x ∈ B0 or x is a limit

point of B0. x ∈ B0 will make x ∈ A0⋂

B0, that is, p(x) ∈ A⋂

B, acontradiction since A and B are separated.

Claim: x is a limit point of B0 ⇒ p(x) is a limit point of B. Take anyneighborhood Nr of p(x), and p(t) lies in B for small enough t. Moreprecisely,

x− r

|b− a|< t < x +

r

|b− a|.

Since x is a limit point of B0, and (x − r/|b − a|, x + r/|b − a|) is aneighborhood N of x, thus N contains a point y 6= x such that y ∈ B0,that is, p(y) ∈ B. Also, p(y) ∈ Nr. Therefore, p(x) is a limit point ofB. Hence p(x) ∈ A

⋂B, a contradiction since A and B are separated.

Hence A0⋂

B0 is empty, that is, A0 and B0 are separated subsets ofR1.

Proof of (b): Suppose not. For every t0 ∈ (0, 1), neither p(t0) ∈ Anor p(t0) ∈ B (since A and B are separated). Also, p(t0) ∈ A

⋃B for

all t0 ∈ (0, 1). Hence (0, 1) = A0⋃

B0, a contradiction since (0, 1) isconnected. I completed the proof.

Proof of (c): Let S be a convex subset of Rk. If S is not connected,then S is a union of two nonempty separated sets A and B. By (b),there exists t0 ∈ (0, 1) such that p(t0) /∈ A

⋃B. But S is convex, p(t0)

25

must lie in A⋃

B, a contradiction. Hence S is connected.

22. A metric space is called separable if it contains a countable dense sub-set. Show that Rk is separable. Hint: Consider the set of points whichhave only rational coordinates.

Proof: Consider S = the set of points which have only rational coor-dinates. For any point x = (x1, x2, ..., xk) ∈ Rk, we can find a rationalsequence {rij} → xj for j = 1, ..., k since Q is dense in R1. Thus,

ri = (ri1 , ri2 , ..., rik) → x

and ri ∈ S for all i. Hence S is dense in Rk. Also, S is countable, thatis, S is a countable dense subset in Rk, Rk is separable.

23. A collection {Vα} of open subsets of X is said to be a base for X if thefollowing is true: For every x ∈ X and every open set G ⊂ X such thatx ∈ G, we have x ∈ Vα ⊂ G for some α. In other words, every open setin X is the union of a subcollection of {Vα}.Prove that every separable metric space has a countable base. Hint:Take all neighborhoods with rational radius and center in some count-able dense subset of X.

Proof: Let X be a separable metric space, and S be a countable densesubset of X. Let a collection {Vα} = { all neighborhoods with rationalradius and center in S }. We claim that {Vα} is a base for X.

For every x ∈ X and every open set G ⊂ X such that x ∈ G, thereexists a neighborhood Nr(p) of p such that Nr(p) ⊂ G since x is aninterior point of G. Since S is dense in X, there exists {sn} → x. Takea rational number rn such that rn < r

2, and {Vα} 3 Nrn(sn) ⊂ Nr(p)

for enough large n. Hence we have x ∈ Vα ⊂ G for some α. Hence{Vα} is a base for X.

24. Let X be a metric space in which every infinite subset has a limitpoint. Prove that X is separable. Hint: Fix δ > 0, and pick x1 ∈X. Having chosen x1, ..., xj ∈ X, choose xj+1, if possible, so thatd(xi, xj+1) ≥ δ for i = 1, ..., j. Show that this process must stop after


finite number of steps, and that X can therefore be covered by finitemany neighborhoods of radius δ. Take δ = 1/n(n = 1, 2, 3, ...), andconsider the centers of the corresponding neighborhoods.

Proof: Fix δ > 0, and pick x1 ∈ X. Having chosen x1, ..., xj ∈ X,choose xj+1, if possible, so that d(xi, xj+1) ≥ δ for i = 1, ..., j. Ifthis process cannot stop, then consider the set A = {x1, x2, ..., xk}. Ifp is a limit point of A, then a neighborhood Nδ/3(p) of p contains apoint q 6= p such that q ∈ A. q = xk for only one k ∈ N . If not,d(xi, xj) ≤ d(xi, p) + d(xj, p) ≤ δ/3 + δ/3 < δ, and it contradicts thefact that d(xi, xj) ≥ δ for i 6= j. Hence, this process must stop afterfinite number of steps.

Suppose this process stop after k steps, and X is covered by Nδ(x1),Nδ(x2), ..., Nδ(xk), that is, X can therefore be covered by finite manyneighborhoods of radius δ.

Take δ = 1/n(n = 1, 2, 3, ...), and consider the set A of the centers ofthe corresponding neighborhoods.

Fix p ∈ X. Suppose that p is not in A, and every neighborhoodNr(p). Note that Nr/2(p) can be covered by finite many neighborhoodsNs(x1), ..., Ns(xk) of radius s = 1/n where n = [2/r] + 1 and xi ∈ Afor i = 1, ..., k. Hence, d(x1, p) ≤ d(x1, q) + d(q, p) ≤ r/2 + s < r whereq ∈ Nr/2(p)

⋂Ns(x1). Therefore, x1 ∈ Nr(p) and x1 6= p since p is not

in A. Hence, p is a limit point of A if p is not in A, that is, A is acountable dense subset, that is, X is separable.

25. Prove that every compact metric space K has a countable base, andthat K is therefore separable. Hint: For every positive integer n, thereare finitely many neighborhood of radius 1/n whose union covers K.

Proof: For every positive integer n, there are finitely many neighbor-hood of radius 1/n whose union covers K (since K is compact). Collectall of them, say {Vα}, and it forms a countable collection. We claim{Vα} is a base.

For every x ∈ X and every open set G ⊂ X, there exists Nr(x) such thatNr(x) ⊂ G since x is an interior point of G. Hence x ∈ Nm(p) ∈ {Vα}for some p where m = [2/r] + 1. For every y ∈ Nm(p), we have

d(y, x) ≤ d(y, p) + d(p, x) < m + m = 2m < r.

27

Hence Nm(p) ⊂ G, that is, Vα ⊂ G for some α, and therefore {Vα} is acountable base of K. Next, collect all of the center of Vα, say D, andwe claim D is dense in K (D is countable since Vα is countable). For allp ∈ K and any ε > 0 we can find Nn(xn) ∈ {Vα} where n = [1/ε] + 1.Note that xn ∈ D for all n and d(p, xn) → 0 as n → ∞. Hence D isdense in K.

26. Let X be a metric space in which every infinite subsets has a limitpoint. Prove that X is compact. Hint: By Exercises 23 and 24, X hasa countable base. It follows that every open cover of X has a countablesubcover {Gn}, n = 1, 2, 3, .... If no finite subcollection of {Gn} coversX, then the complement Fn of G1

⋃...

⋃Gn is nonempty for each n,

but⋂

Fn is empty. If E is a set contains a point from each Fn, considera limit point of E, and obtain a contradiction.

Proof: By Exercises 23 and 24, X has a countable base. It follows thatevery open cover of X has a countable subcover {Gn}, n = 1, 2, 3, ....If no finite subcollection of {Gn} covers X, then the complement Fn ofG1

⋃...

⋃Gn is nonempty for each n, but

⋂Fn is empty. If E is a set

contains a point from each Fn, consider a limit point of E.

Note that Fk ⊃ Fk+1 ⊃ ... and Fn is closed for all n, thus p lies in Fk

for all k. Hence p lies in⋂

Fn, but⋂

Fn is empty, a contradiction.

27. Define a point p in a metric space X to be a condensation point of a setE ⊂ X if every neighborhood of p contains uncountably many pointsof E.

Suppose E ⊂ Rk, E is uncountable, and let P be the set of all conden-sation points of E. Prove that P is perfect and that at most countablymany points of E are not in P . In other words, show that P c ⋂

E is atmost countable. Hint: Let {Vn} be a countable base of Rk, let W bethe union of those Vn for which E

⋂Vn is at most countable, and show

that P = W c.

Proof: Let {Vn} be a countable base of Rk, let W be the union ofthose Vn for which E

⋂Vn is at most countable, and we will show that

P = W c. Suppose x ∈ P . (x is a condensation point of E). Ifx ∈ Vn for some n, then E

⋂Vn is uncountable since Vn is open. Thus


x ∈ W c. (If x ∈ W , then there exists Vn such that x ∈ Vn and E⋂

Vn

is uncountable, a contradiction). Therefore P ⊂ W c.

Conversely, suppose x ∈ W c. x /∈ Vn for any n such that E⋂

Vn iscountable. Take any neighborhood N(x) of x. Take x ∈ Vn ⊂ N(x),and E

⋂Vn is uncountable. Thus E

⋂N(x) is also uncountable, x is a

condensation point of E. Thus W c ⊂ P . Therefore P = W c. Note thatW is countable, and thus W ⊂ W

⋂E = P c ⋂

E is at most countable.

To show that P is perfect, it is enough to show that P contains noisolated point. (since P is closed). If p is an isolated point of P , thenthere exists a neighborhood N of p such that N

⋂E = φ. p is not a

condensation point of E, a contradiction. Therefore P is perfect.

28. Prove that every closed set in a separable metric space is the unionof a (possible empty) perfect set and a set which is at most countable.(Corollary: Every countable closed set in Rk has isolated points.) Hint:Use Exercise 27.

Proof: Let X be a separable metric space, let E be a closed set onX. Suppose E is uncountable. (If E is countable, there is nothingto prove.) Let P be the set of all condensation points of E. Since Xhas a countable base, P is perfect, and P c ⋂

E is at most countable byExercise 27. Since E is closed, P ⊂ E. Also, P c ⋂

E = E − P . HenceE = P

⋃(E − P ).

For corollary: if there is no isolated point in E, then E is perfect. ThusE is uncountable, a contradiction.

Note: It’s also called Cauchy-Bendixon Theorem.

29. Prove that every open set in R1 is the union of an at most countablecollection of disjoint segments. Hint: Use Exercise 22.

Proof: (due to H.L.Royden, Real Analysis) Since O is open, for each xin O, there is a y > x such that (x, y) ⊂ O. Let b = sup{y : (x, y) ⊂ O}.Let a = inf{z : (z, x) ⊂ O}. Then a < x < b, and Ix = (a, b) is an openinterval containing x.

Now Ix ⊂ O, for if w ∈ Ix, say x < w < b, we have by the definition ofb a number y > w such that (x, y) ⊂ O, and so w ∈ O).

29

Moreover, b /∈ O, for if b ∈ O, then for some ε > 0 we have (b−ε, b+ε) ⊂O, whence (x, b + ε) ⊂ O, contradicting the definition of b. Similarly,a /∈ O.

Consider the collection of open intervals {Ix}, x ∈ O. Since each x ∈ Ois contained in Ix, and each Ix ⊂ O, we have O =

⋃Ix.

Let (a, b) and (c, d) be two intervals in this collection with a point incommon. Then we must have c < b and a < d. Since c /∈ O, it does notbelong to (a, b) and we have c ≤ a. Since a /∈ O and hence not to (c, d),we have a ≤ c. Thus a = c. Similarly, b = d, and (a, b) = (c, d). Thustwo different intervals in the collection {Ix} must be disjoint. ThusO is the union of the disjoint collection {Ix} of open intervals, and itremains only to show that this collection is countable. But each openinterval contains a rational number since Q is dense in R. Since wehave a collection of disjoint open intervals, each open interval containsa different rational number, and the collection can be put in one-to-onecorrespondence with a subset of the rationals. Thus it is a countablecollection.

30. Imitate the proof of Theorem 2.43 to obtain the following result:

If Rk =⋃∞

1 Fn, where each Fn is a closed subset of Rk, thenat least one Fn has a nonempty interior.

Equivalent statement: If Gn is a dense open subset of Rk, forn = 1, 2, 3, ..., then

⋂∞1 Gn is not empty (in fact, it is dense

in Rk).

(This is a special case of Baire’s theorem; see Exercise 22, Chap. 3, forthe general case.)

Proof: I prove Baire’s theorem directly. Let Gn be a dense opensubset of Rk for n = 1, 2, 3, .... I need to prove that

⋂∞1 Gn intersects

any nonempty open subset of Rk is not empty.

Let G0 is a nonempty open subset of Rk. Since G1 is dense and G0

is nonempty, G0⋂

G1 6= φ. Suppose x1 ∈ G0⋂

G1. Since G0 and G1

are open, G0⋂

G1 is also open, that is, there exist a neighborhood V1

such that V1 ⊂ G0⋂

G1. Next, since G2 is a dense open set and V1

is a nonempty open set, V1⋂

G2 6= φ. Thus, I can find a nonempty


open set V2 such that V2 ⊂ V1⋂

G2. Suppose I have get n nonemptyopen sets V1, V2, ..., Vn such that V1 ⊂ G0

⋂G1 and Vi+1 ⊂ Vi

⋂Gn+1

for all i = 1, 2, ..., n − 1. Since Gn+1 is a dense open set and Vn is anonempty open set, Vn

⋂Gn+1 is a nonempty open set. Thus I can find

a nonempty open set Vn+1 such that Vn+1 ⊂ Vn⋂

Gn+1. By induction, Ican form a sequence of open sets {Vn : n ∈ Z+} such that V1 ⊂ G0

⋂G1

and Vi+1 ⊂ Vi⋂

Gi+1 for all n ∈ Z+. Since V1 is bounded and V1 ⊃V2 ⊃ ... ⊃ Vn ⊃ ..., by Theorem 2.39 I know that

∞⋂n=1

Vn 6= φ.

Since V1 ⊂ G0⋂

G1 and Vn+1 ⊂ Gn+1, G0⋂

(⋂∞

n=1 Gn) 6= φ. Proved.

Note: By Baire’s theorem, I’ve proved the equivalent statement. Next,Fn has a empty interior if and only if Gn = Rk − Fn is dense in Rk.Hence we completed all proof.

Chapter 3

Numerical Sequences andSeries

31

32 CHAPTER 3. NUMERICAL SEQUENCES AND SERIES


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13. Prove that the Cauchy product of two absolutely convergent series con-verges absolutely.

Note: Given∑

an and∑

bn, we put cn =∑n

k=0 akbn−k and call∑

cn

the Cauchy product of the two given series.

Proof: Put An =∑n

k=0 |ak|, Bn =∑n

k=0 |bk|, Cn =∑n

k=0 |ck|. Then

Cn = |a0b0|+ |a0b1 + a1b0|+ ... + |a0bn + a1bn−1 + ... + anb0|≤ |a0||b0|+ (|a0||b1|+ |a1||b0|) + ...

+(|a0||bn|+ |a1||bn−1|+ ... + |an||b0|)= |a0|Bn + |a1|Bn−1 + ... + |an|B0

33

≤ |a0|Bn + |a1|Bn + ... + |an|Bn

= (|a0|+ |a1|+ ... + |an|)Bn

= AnBn

≤ AB

where A = lim An and B = lim Bn. Hence {Cn} is bounded. Notethat {Cn} is increasing, and thus Cn is a convergent sequence, thatis, the Cauchy product of two absolutely convergent series convergesabsolutely.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23. Suppose {pn} and {qn} are Cauchy sequences in a metric space X.Show that the sequence {d(pn, qn)} converges. Hint: For any m, n,

d(pn, qn) ≤ d(pn, pm) + d(pm, qm) + d(qm, qn);

it follows that|d(pn, qn)− d(pm, qm)|

is small if m and n are large.

Proof: For any ε > 0, there exists N such that d(pn, pm) < ε andd(qm, qn) < ε whenever m,n ≥ N . Note that

d(pn, qn) ≤ d(pn, pm) + d(pm, qm) + d(qm, qn).

34 CHAPTER 3. NUMERICAL SEQUENCES AND SERIES

It follows that

|d(pn, qn)− d(pm, qm)| ≤ d(pn, pm) + d(qm, qn) < 2ε.

Thus {d(pn, qn)} is a Cauchy sequence in X. Hence {d(pn, qn)} con-verges.

24.

25.

Chapter 4

Continuity

35

36 CHAPTER 4. CONTINUITY


1. Suppose f is a real function define on R1 which satisfies

limh→0

[f(x + h)− f(x− h)] = 0

for every x ∈ R1. Does this imply that f is continuous?

Solution: No. Take f(x) = 1, if x ∈ Z; f(x) = 0. otherwise.

2.

3.

4.

5. If f is a real continuous function defined on a closed set E ⊂ R1, provethat there exist continuous real function g on R1 such that g(x) = f(x)for all x ∈ E. (Such functions g are called continuous extensions of ffrom E to R1.) Show that the result becomes false if the word ”closed”is omitted. Extend the result to vector valued functions. Hint: Let thegraph of g be a straight line on each of the segments which constitutethe complement of E (compare Exercise 29, Chap. 2). The result re-mains true if R1 is replaced by any metric space, but the proof is notso simple.

Proof: Note that the following fact:

Every open set of real numbers is the union of a countablecollection of disjoint open intervals.

Thus, consider Ec =⋃

(ai, bi), where i ∈ Z, and ai < bi < ai+1 < bi+1.We extend g on (ai, bi) as following:

g(x) = f(ai) + (x− ai)f(bi)− f(ai)

bi − ai

(g(x) = f(x) for x ∈ E). Thus g is well-defined on R1, and g is contin-uous on R1 clearly.

37

Next, consider f(x) = 1/x on a open set E = R−0. f is continuous onE, but we cannot redefine f(0) = any real number to make new f(x)continue at x = 0.

Next, consider a vector valued function

f(x) = (f1(x), ..., fn(x)),

where fi(x) is a real valued function. Since f is continuous on E, eachcomponent of f , fi, is also continuous on E, thus we can extend fi, saygi, for each i = 1, ..., n. Thus,

g(x) = (g1(x), ..., gn(x))

is a extension of f(x) since each component of g, gi, is continuous onR1 implies g is continuous on Rn.

Note: The above fact only holds in R1. If we change R1 into anymetric spaces, we have no the previous fact.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.


17.

18.

19. Suppose f is a real function with domain R1 which has the interme-diate value property: If f(a) < c < f(b), then f(x) = c for some xbetween a and b.

Suppose also, for every rational r, that the set of all x with f(x) = ris closed. Prove that f is continuous.

Hint: If xn → x0, but f(xn) > r > f(x0) for some r and all n, thenf(tn) = r for some tn between x0 and xn; thus tn → x0. Find a contra-diction. (N. J. Fine, Amer. Math. Monthly, vol. 73, 1966, p. 782.)

Proof: Let S = {x : f(x) = r}. If xn → x0, but f(xn) > r > f(x0)for some r and all n since Q is dense in R1, then f(tn) = r for some tnbetween x0 and xn; thus tn → x0. Hence x0 is a limit point of S. SinceS is closed, f(x0) = r, a contradiction. Hence, lim sup f(xn) ≤ f(x0).Similarly, lim inf f(xn) ≥ f(x0). Hence, lim f(xn) = f(x0), and f iscontinuous at x0.

Note: Original problem is stated as follows:

Let f be a function from the reals to the reals, differentiableat every point. Suppose that, for every r, the set of pointsx, where f ′(x) = r, is closed. Prove that f ′ is continuous.

If we replace Q into any dense subsets of R1, the conclusion also holds.

20.

21.

22. Let A and B be disjoint nonempty closed sets in a metric space X, anddefine

f(p) =ρA(p)

ρA(p) + ρB(p), (p ∈ X).

39

Show that f is a continuous function on X whose range lies in [0, 1],that f(p) = 0 precisely on A and f(p) = 1 precisely on B. Thisestablishes a converse of Exercise 3: Every closed set A contained in Xis Z(f) for some continuous real f on X. Setting

V = f−1([0, 1/2)), W = f−1((1/2, 1]),

show that V and W are open and disjoint, and that A is contained inV , B is contained in W . (Thus pairs of disjoint closed set in a metricspace can be covered by pairs of disjoint open sets. This property ofmetric spaces is called normality.)

Proof: Note that ρA(p) and ρB(p) are (uniformly) continuous on X,and ρA(p) + ρB(p) > 0. (Clearly, ρA(p) + ρB(p) ≥ 0 by the definition.If ρA(p) + ρB(p) = 0, then p ∈ A

⋂B by Exercise 20, a contradiction).

Thus f(p) = ρA(p)/(ρA(p)+ρB(p)) is continuous on X. Next, f(p) ≥ 0,and f(p) ≤ 1 since ρA(p) ≤ ρA(p) + ρB(p). Thus f(X) lies in [0, 1].

Next, f(p) = 0 ⇔ ρA(p) = 0 ⇔ p ∈ A precisely, and f(p) = 1 ⇔ρB(p) = 0 ⇔ p ∈ B precisely by Exercise 20.

Now we prove a converse of Exercise 3: Every closed set A ⊂ X is Z(f)for some continuous real f on X. If Z(f) = φ, then f(x) = 1 for allx ∈ X satisfies our requirement. If Z(f) 6= φ, we consider two possiblecases: (1) Z(f) = X; (2) Z(f) 6= X. If Z(f) = X, then f(x) = 0 forall x ∈ X. If Z(f) 6= X, we can choose p ∈ X such that f(p) 6= 0.Note that Z(f) and {p} are one pair of disjoint closed sets. Hence welet

f(x) =ρZ(f)(x)

ρZ(f)(x) + ρ{p}(x).

By the previous result, we know that f(x) satisfies our requirement.Hence we complete the whole proof.

Note that [0, 1/2) and (1/2, 1] are two open sets of f(X). Since f iscontinuous, V = f−1([0, 1/2)) and W = f−1((1/2, 1]) are two open sets.f−1({0}) ⊂ f−1([0, 1/2)), and f−1({1}) ⊂ f−1((1/2, 1]). Thus, A ⊂ Vand B ⊂ W . Thus a metric space X is normal.


23.

24.

25.

26.

Chapter 5

Differentiation

41

42 CHAPTER 5. DIFFERENTIATION


1. Let f be defined for all real x, and suppose that

|f(x)− f(y)| ≤ (x− y)2

for all real x and y. Prove that f is constant.

Proof: |f(x)−f(y)| ≤ (x−y)2 for all real x and y. Fix y, |f(x)−f(y)x−y

| ≤|x− y|. Let x → y, therefore,

0 ≤ limx→y

f(x)− f(y)

x− y≤ lim

x→y|x− y| = 0

It implies that (f(x)− f(y))/(x− y) → 0 as x → y. Hence f ′(y) = 0,f = const.

2. Suppose f ′(x) > 0 in (a, b). Prove that f is strictly increasing in (a, b),and let g be its inverse function. Prove that g is differentible, and that

g′(f(x)) =1

f ′(x)(a < x < b).

Proof: For every pair x > y in (a, b), f(x)− f(y) = f ′(c)(x− y) wherey < c < x by Mean-Value Theorem. Note that c ∈ (a, b) and f ′(x) > 0in (a, b), hence f ′(c) > 0. f(x) − f(y) > 0, f(x) > f(y) if x > y, f isstrictly increasing in (a, b).

Let ∆g = g(x0 + h)− g(x0). Note that x0 = f(g(x0)), and thus,

(x0 + h)− x0 = f(g(x0 + h))− f(g(x0)),

h = f(g(x0) + ∆g)− f(g(x0)) = f(g + ∆g)− f(g).

Thus we apply the fundamental lemma of differentiation,

h = [f ′(g) + η(∆g)]∆g,

1

f ′(g) + η(∆g)=

∆g

h

43

Note that f ′(g(x)) > 0 for all x ∈ (a, b) and η(∆g) → 0 as h → 0, thus,

limh→0

∆g/h = limh→0

1

f ′(g) + η(∆g)=

1

f ′(g(x)).

Thus g′(x) = 1f ′(g(x))

, g′(f(x)) = 1f ′(x)

.

3. Suppose g is a real function on R1, with bounded derivative (say|g′| ≤ M). Fix ε > 0, and define f(x) = x + εg(x). Prove that fis one-to-one if ε is small enough. (A set of admissible values of ε canbe determined which depends only on M .)

Proof: For every x < y, and x, y ∈ R, we will show that f(x) 6= f(y).By using Mean-Value Theorem:

g(x)− g(y) = g′(c)(x− y) where x < c < y,

(x− y) + ε((x)− g(y)) = (εg′(c) + 1)(x− y),

that is,f(x)− f(y) = (εg′(c) + 1)(x− y). (∗)

Since |g′(x)| ≤ M , −M ≤ g′(x) ≤ M for all x ∈ R. Thus 1 − εM ≤εg′(c) + 1 ≤ 1 + εM , where x < c < y. Take c = 1

2M, and εg′(c) + 1 > 0

where x < c < y for all x, y. Take into equation (*), and f(x)−f(y) < 0since x− y < 0, that is, f(x) 6= f(y), that is, f is one-to-one (injective).

4. If

C0 +C1

2+ ... +

Cn−1

n+

Cn

n + 1= 0,

where C0, ..., Cn are real constants, prove that the equation

C0 + C1x + ... + Cn−1xn−1 + Cnx

n = 0

has at least one real root between 0 and 1.

Proof: Let f(x) = C0x + ... + Cn

n+1xn+1. f is differentiable in R1

and f(0) = f(1) = 0. Thus, f(1) − f(0) = f ′(c) where c ∈ (0, 1) byMean-Value Theorem. Note that

f ′(x) = C0 + C1x + ... + Cn−1xn−1 + Cnx

n.


Thus, c ∈ (0, 1) is one real root between 0 and 1 of that equation.

5. Suppose f is defined and differentiable for every x > 0, and f ′(x) → 0as x → +∞. Put g(x) = f(x + 1) − f(x). Prove that g(x) → 0 asx → +∞.

Proof: f(x + 1) − f(x) = f ′(c)(x + 1 − x) where x < c < x + 1by Mean-Value Theorem. Thus, g(x) = f ′(c) where x < c < x + 1,that is,

limx→+∞

g(x) = limx→+∞

f ′(c) = limc→+∞

f ′(c) = 0.

6. Suppose(a) f is continuous for x ≥ 0,(b) f ′(x) exists for x > 0,(c) f(0) = 0,(d) f ′ is monotonically increasing.Put

g(x) =f(x)

x(x > 0)

and prove that g is monotonically increasing.

Proof: Our goal is to show g′(x) > 0 for all x > 0

⇔ g′(x) = xf ′(x)−f(x)x2 > 0 ⇔ f ′(x) > f(x)

x.

Since f ′(x) exists, f(x) − f(0) = f ′(c)(x − 0) where 0 < c < x by

Mean-Value Theorem. ⇒ f ′(c) = f(x)x

where 0 < c < x. Since f ′ is

monotonically increasing, f ′(x) > f ′(c), that is, f ′(x) > f(x)x

for allx > 0.

7. Suppose f ′(x), g′(x) exist, g′(x) 6= 0, and f(x) = g(x) = 0. Prove that

limt→x

f(t)

g(t)=

f ′(x)

g′(x).

45

(This holds also for complex functions.)

Proof:

f ′(t)

g′(t)=

limt→xf(t)−f(x)

t−x

limt→xg(t)−g(x)

t−x

=limt→x f(t)

limt→x g(t)= lim

t→x

f(t)

g(t)

Surely, this holds also for complex functions.

8. Suppose f ′(x) is continuous on [a, b] and ε > 0. Prove that there existsδ > 0 such that

|f(t)− f(x)

t− x− f ′(x)| < ε

whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. (This could beexpressed by saying f is uniformly differentiable on [a, b] if f ′ is contin-uous on [a, b].) Does this hold for vector-valued functions too?

Proof: Since f ′(x) is continuous on a compact set [a, b], f ′(x) is uni-formly continuous on [a, b]. Hence, for any ε > 0 there exists δ > 0such that

|f ′(t)− f ′(x)| < ε

whenever 0 < |t − x| < δ, a ≤ x ≤ b, a ≤ t ≤ b. Thus, f ′(c) =f(t)−f(x)

t−xwhere c between t and x by Mean-Value Theorem. Note that

0 < |c− x| < δ and thus |f ′(c)− f ′(x)| < ε, thus,

|f(t)− f(x)

t− x− f ′(x)| < ε

whenever 0 < |t− x| < δ, a ≤ x ≤ b, a ≤ t ≤ b.Note: It does not hold for vector-valued functions. If not, take

f(x) = (cos x, sin x),

[a, b] = [0, 2π], and x = 0. Hence f ′(x) = (− sin x, cos x). Take any1 > ε > 0, there exists δ > 0 such that

|f(t)− f(0)

t− 0− f ′(0)| < ε


whenever 0 < |t| < δ by our hypothesis. With calculating,

|(cos t− 1

t,sin t

t)− (0, 1)| < ε

|(cos t− 1

t,sin t

t− 1)| < ε

(cos t− 1

t)2 + (

sin t

t− 1)2 < ε2 < ε

2

t2+ 1− 2(cos t + sin t)

t< ε

since 1 > ε > 0. Note that

2

t2+ 1− 4

t<

2

t2+ 1− 2(cos t + sin t)

t

But 2t2

+ 1− 4t→ +∞ as t → 0. It contradicts.

9. Let f be a continuous real function on R1, of which it is known thatf ′(x) exists for all x 6= 0 and that f ′(x) → 0 as x → 0. Dose it followthat f ′(0) exists?

Note: We prove a more general exercise as following.Suppose that f is continuous on an open interval I containing x0, sup-pose that f ′ is defined on I except possibly at x0, and suppose thatf ′(x) → L as x → x0. Prove that f ′(x0) = L.

Proof of the Note: Using L’Hospital’s rule:

limh→0

f(x0 + h)− f(x0)

h= lim

h→0f ′(x0 + h)

By our hypothesis: f ′(x) → L as x → x0. Thus,

limh→0

f(x0 + h)− f(x0)

h= L,

Thus f ′(x0) exists andf ′(x0) = L.

47

10. Suppose f and g are complex differentiable functions on (0, 1), f(x) → 0,g(x) → 0, f ′(x) → A, g′(x) → B as x → 0, where A and B are complexnumbers, B 6= 0. Prove that

limx→0

f(x)

g(x)=

A

B.

Compare with Example 5.18. Hint:

f(x)

g(x)= (

f(x)

x− A)

x

g(x)+ A

x

g(x).

Apply Theorem 5.13 to the real and imaginary parts of f(x)x

and g(x)x

.

Proof: Write f(x) = f1(x)+ if2(x), where f1(x), f2(x) are real-valuedfunctions. Thus,

df(x)

dx=

df1(x)

dx+ i

df2(x)

dx,

Apply L’Hospital’s rule to f1(x)x

and f2(x)x

, we have

limx→0

f1(x)

x= lim

x→0f ′1(x)

limx→0

f2(x)

x= lim

x→0f ′2(x)

Combine f1(x) and f2(x), we have

limx→0

f1(x)

x+ i lim

x→0

f2(x)

x= lim

x→0

f1(x)

x+ i

f2(x)

x= lim

x→0

f(x)

x

or

limx→0

f1(x)

x+ i lim

x→0

f2(x)

x= lim

x→0f ′1(x) + i lim

x→0f ′2(x) = lim

x→0f ′(x)

Thus, limx→0f(x)

x= limx→0 f ′(x). Similarly, limx→0

g(x)x

= limx→0 g′(x).Note that B 6= 0. Thus,

limx→0

f(x)

g(x)= lim

x→0(f(x)

x− A)

x

g(x)+ A

x

g(x)

= (A− A)1

B+

A

B=

A

B.


Note: In Theorem 5.13, we know g(x) → +∞ as x → 0. (f(x) = x,

and g(x) = x + x2ei

x2 ).

11. Suppose f is defined in a neighborhood of x, and suppose f”(x) exists.Show that

limh→0

f(x + h) + f(x− h)− 2f(x)

h2= f ′′(x).

Show by an example that the limit may exist even if f”(x) dose not.Hint: Use Theorem 5.13.

Proof: By using L’Hospital’s rule: (respect to h.)

limh→0

f(x + h) + f(x− h)− 2f(x)

h2= lim

h→0

f ′(x + h)− f ′(x− h)

2h

Note that

f ′′(x) =1

2(f ′′(x) + f ′′(x))

=1

2(limh→0

f ′(x + h)− f ′(x)

h+ lim

h→0

f ′(x− h)− f ′(x)

−h)

=1

2limh→0

f ′(x + h)− f ′(x− h)

h

= limh→0

f ′(x + h)− f ′(x− h)

2h

Thus,f(x + h) + f(x− h)− 2f(x)

h2→ f ′′(x)

as h → 0. Counter-example: f(x) = x|x| for all real x.

12. If f(x) = |x|3, compute f ′(x), f ′′(x) for all real x, and show that f (3)(0)does not exist.

Proof: f ′(x) = 3|x|2 if x 6= 0. Consider

f(h)− f(0)

h=|h|3

h

49

Note that |h|/h is bounded and |h|2 → 0 as h → 0. Thus,

f ′(0) = limh→0

f(h)− f(0)

h= 0.

Hence, f ′(x) = 3|x|2 for all x. Similarly,

f ′′(x) = 6|x|.

Thus,f ′′(h)− f(0)

h= 6

|h|h

Since |h|h

= 1 if h > 0 and = −1 if h < 0, f ′′′(0) does not exist.

13. Suppose a and c are real numbers, c > 0, and f is defined on [−1, 1] by

f(x) =

{xa sin (x−c) (if x 6= 0),0 (if x = 0).

Prove the following statements:

(a) f is continuous if and only if a > 0.(b) f ′(0) exists if and only if a > 1.(c) f ′ is bounded if and only if a ≥ 1 + c.(d) f ′ is continuous if and only if a > 1 + c.(e) f ′′(0) exists if and only if a > 2 + c.(f) f ′′ is bounded if and only if a ≥ 2 + 2c.(g) f ′′ is continuous if and only if a > 2 + 2c.

Proof: For (a): (⇒) f is continuous iff for any sequence {xn} → 0with xn 6= 0, xa

n sin x−cn → 0 as n →∞. In particular, take

xn = (1

2nπ + π/2)

1c > 0

and thus xan → 0 as n →∞. Hence a > 0. (If not, then a = 0 or a < 0.

When a = 0, xan = 1. When a < 0, xa

n = 1/x−an → ∞ as n → ∞. It

contradicts.)

(⇐) f is continuous on [−1, 1]− {0} clearly. Note that

−|xa| ≤ xa sin (x−c) ≤ |xa|,


and |xa| → 0 as x → 0 since a > 0. Thus f is continuous at x = 0.Hence f is continuous.

For (b): f ′(0) exists iff xa−1 sin (x−c) → 0 as x → 0. In the previousproof we know that f ′(0) exists if and only if a−1 > 0. Also, f ′(0) = 0.

14. Let f be a differentiable real function defined in (a, b). Prove that f isconvex if and only if f ′ is monotonically increasing. Assume next thatf ′′(x) exist for every x ∈ (a, b), and prove that f is convex if and onlyif f ′′(x) ≥ 0 for all x ∈ (a, b).

15. Suppose a ∈ R1, f is a twice-differentiable real function on (a,∞),and M0, M1, M2 are the least upper bounds of |f(x)|, |f ′(x)|, |f ′′(x)|,respectively, on (a,∞). Prove that

M21 ≤ 4M0M2.

Hint: If h > 0, Taylor’s theorem shows that

f ′(x) =1

2h[f(x + 2h)− f(x)]− hf ′′(ξ)

for some ξ ∈ (x, x + 2h). Hence

|f ′(x)| ≤ hM2 +M0

h.

To show that M21 = 4M0M2 can actually happen, take a = −1, define

f(x) =

{2x2 − 1 (−1 < x < 0),x2−1x2+1

(0 ≤ x < ∞),

and show that M0 = 1, M1 = 4, M2 = 4. Does M21 ≤ 4M0M2 hold for

vector-valued functions too?

Proof: Suppose h > 0. By using Taylor’s theorem:

f(x + h) = f(x) + hf ′(x) +h2

2f ′′(ξ)

for some x < ξ < x + 2h. Thus

h|f ′(x)| ≤ |f(x + h)|+ |f(x)|+ h2

2|f ′′(ξ)|

51

h|f ′(x)| ≤ 2M0 +h2

2M2.

h2M2 − 2h|f ′(x)|+ 4M0 ≥ 0 (∗)

Since equation (*) holds for all h > 0, its determinant must be non-positive.

4|f ′(x)|2 − 4M2(4M0) ≤ 0

|f ′(x)|2 ≤ 4M0M2

(M1)2 ≤ 2M0M2

Note: There is a similar exercise:Suppose f(x)(−∞ < x < +∞) is a twice-differentiable real function,and

Mk = sup−∞<x<+∞

|f (k)(x)| < +∞ (k = 0, 1, 2).

Prove that M21 ≤ 2M0M2.

Proof of Note:

f(x + h) = f(x) + f ′(x)h +f ′′(ξ1)

2h2

(x < ξ1 < x + h or x > ξ1 > x + h) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (*)

f(x− h) = f(x)− f ′(x)h +f ′′(ξ2)

2h2

(x− h < ξ2 < x or x− h > ξ2 > x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (**)(*) minus (**):

f(x + h)− f(x− h) = 2f ′(x)h +h2

2(f ′′(ξ1)− f ′′(ξ2)).

2h|f ′(x)| ≤ |2hf ′(x)|

2h|f ′(x)| ≤ |f(x + h)|+ |f(x− h)|+ h2

2(|f ′′(ξ1)|+ |f ′′(ξ2)|)

2h|f ′(x)| ≤ 2M0 + h2M2

M2h2 − 2|f ′(x)|h + 2M0 ≥ 0


Since this equation holds for all h, its determinant must be non-positive:

4|f ′(x)|2 − 4M2(2M0) ≤ 0,

|f ′(x)|2 ≤ 2M0M2

ThusM2

1 ≤ 2M0M2

16. Suppose f is twice-differentiable on (0,∞), f ′′ is bounded on (0,∞),and f(x) → 0 as x → ∞. Prove that f ′(x) → 0 as x → ∞. Hint: Leta →∞ in Exercise 15.

Proof: Suppose a ∈ (0,∞), and M0, M1, M2 are the least upperbounds of |f(x)|, |f ′(x)|, |f ′′(x)| on (a,∞). Hence, M2

1 ≤ 4M0M2. Leta →∞, M0 = sup |f(x)| → 0. Since M2 is bounded, therefore M2

1 → 0as a →∞. It implies that sup |f ′(x)| → 0 as x →∞.

17. Suppose f is a real, three times differentiable function on [−1, 1], suchthat

f(−1) = 0, f(0) = 0, f(1) = 1, f ′(0) = 0.

Prove that f (3)(x) ≥ 3 for some x ∈ (−1, 1).Note that equality holds for 1

2(x3 + x2).

Hint: Use Theorem 5.15, with α = 0 and β = 1,−1, to show thatthere exist s ∈ (0, 1) and t ∈ (−1, 0) such that

f (3)(s) + f (3)(t) = 6.

Proof: By Theorem 5.15, we take α = 0, β = 1,

f(1) = f(0) + f ′(0) +f ′′(0)

2+

f (3)(s)

6

where s ∈ (0, 1). Take α = 0, andβ = −1,

f(−1) = f(0)− f ′(0) +f ′′(0)

2− f (3)(t)

6

53

where t ∈ (−1, 0). Thus

1 =f ′′(0)

2+

f (3)(s)

6, s ∈ (0, 1) (∗)

0 =f ′′(0)

2− f (3)(s)

6, s ∈ (−1, 0) (∗∗)

Equation (*) - equation (**):

f (3)(s)

6+

f (3)(t)

6, s ∈ (0, 1), t ∈ (−1, 0).

f (3)(s) + f (3)(t) = 6, s, t ∈ (−1, 1).

f (3)(x) ≥ 3 for some x ∈ (−1, 1).

Theorem 5.15: Suppose f is a real function on [a, b], n is a positiveinteger, f (n−1) is continuous on [a, b], f (n)(t) exists for every t ∈ (a, b).Let α, β be distinct points of [a, b], and define

P (t) =n−1∑k=0

f (k)(α)

k!(t− α)k.

Then there exists a point x between α and β such that

f(β) = P (β) +f (n)(x)

n!(β − α)n.

18. Suppose f is a real function on [a, b], n is a positive integer, and f (n−1)

exists for every t ∈ [a, b]. Let α, β, and P be as in Taylor’s theorem(5.15). Define

Q(t) =f(t)− f(β)

t− β

for t ∈ [a, b], t 6= β, differentiate

f(t)− f(β) = (t− β)Q(t)


n − 1 times at t = α, and derive the following version of Taylor’stheorem:

f(β) = P (β) +Q(n−1)(α)

(n− 1)!(β − α)n.

19. Suppose f is defined in (−1, 1) and f ′(0) exists. Suppose −1 < αn <βn < 1, αn → 0, and βn → 0 as n →∞. Define the difference quotients

Dn =f(βn)− f(αn)

βn − αn

Prove the following statements:

(a) If αn < 0 < βn, then lim Dn = f ′(0).(b) If 0 < αn < βn and {βn/(βn−αn)} is bounded, then lim Dn = f ′(0).(c) If f ′ is continuous in (−1, 1), then lim Dn = f ′(0).

Give an example in which f is differentiable in (−1, 1) (but f ′ is notcontinuous at 0) and in which αn, βn tend to 0 in such a way thatlim Dn exists but is different from f ′(0).

Proof: For (a):

Dn =f(βn)− f(0)

βn

βn

βn − αn

+f(αn)− f(0)

αn

−αn

βn − αn

Note that

f ′(0) = limn→∞

f(αn)− f(0)

αn

= limn→∞

f(βn)− f(0)

βn

Thus for any ε > 0, there exists N such that

L− ε <f(αn)− f(0)

αn

< L + ε,

L− ε <f(βn)− f(0)

βn

< L + ε,

55

whenever n > N where L = f ′(0) respectively. Note that βn/(βn−αn)and −αn/(βn − αn) are positive. Hence,

βn

βn − αn

(L− ε) <f(βn)− f(0)

βn

βn

βn − αn

<βn

βn − αn

(L + ε)

−αn

βn − αn

(L− ε) <f(αn)− f(0)

αn

−αn

βn − αn

<−αn

βn − αn

(L + ε)

Combine two inequations,

L− ε < Dn < L + ε

Hence, lim Dn = L = f ′(0).

For (b): We process as above prove, but note that −αn/(βn−αn) < 0.Thus we only have the following inequations:

βn

βn − αn

(L− ε) <f(βn)− f(0)

βn

βn

βn − αn

<βn

βn − αn

(L + ε)

−αn

βn − αn

(L + ε) <f(αn)− f(0)

αn

−αn

βn − αn

<−αn

βn − αn

(L− ε)

Combine them:

L− βn + αn

βn − αn

ε < Dn < L +βn + αn

βn − αn

ε

Note that {βn/(βn − αn)} is bounded, ie,

| βn

βn − αn

| ≤ M

for some constant M . Thus

|βn + αn

βn − αn

| = | 2βn

βn − αn

− 1| ≤ 2M + 1

Hence,

L− (2M + 1)ε < Dn < L + (2M + 1)ε

Hence, lim Dn = L = f ′(0).


For (c): By using Mean-Value Theorem,

Dn = f ′(tn)

where tn is between αn and βn. Note that

min{αn, βn} < tn < max{αn, βn}

and

max{αn, βn} =1

2(αn + βn + |αn − βn|)

min{αn, βn} =1

2(αn + βn − |αn − βn|)

Thus, max{αn, βn} → 0 and min{αn, βn} → 0 as αn → 0 and βn → 0.By squeezing principle for limits, tn → 0. With the continuity of f ′,we have

lim Dn = lim f ′(tn) = f ′(lim tn) = f ′(0).

Example: Let f be defined by

f(x) =

{x2 sin(1/x) (x 6= 0),0 (x = 0).

Thus f ′(x) is not continuous at x = 0, and f ′(0) = 0. Take αn =1

π/2+2nπand βn = 1

2nπ. It is clear that αn → 0, and βn → 0 as n →∞.

Also,

Dn =−4nπ

π(π/2 + 2nπ)→ − 2

π

as n →∞. Thus, lim Dn = −2/π exists and is different from 0 = f ′(0).

20.

21.

22. Suppose f is a real function on (−∞,∞). Call x a fixed point of f iff(x) = x.

(a) If f is differentiable and f ′(t) 6= 1 for every real t, prove that f hasat most one fixed point.

57

(b) Show that the function f defined by

f(t) = t + (1 + et)−1

has no fixed point, although 0 < f ′(t) < 1 for all real t.

(c) However, if there is a constant A < 1 such that |f ′(t)| ≤ A for allreal t, prove that a fixed point x of f exists, and that x = lim xn, wherex1 is an arbitrary real number and

xn+1 = f(xn)

for n = 1, 2, 3, ...

(d) Show that the process describe in (c) can be visualized by the zig-zag path

(x1, x2) → (x2, x2) → (x2, x3) → (x3, x3) → (x3, x4) → ....

Proof: For (a): If not, then there exists two distinct fixed points, sayx and y, of f . Thus f(x) = x and f(y) = y. Since f is differeniable,by applying Mean-Value Theorem we know that

f(x)− f(y) = f ′(t)(x− y)

where t is between x and y. Since x 6= y, f ′(t) = 1. It contradicts.

For (b): We show that 0 < f ′(t) < 1 for all real t first:

f ′(t) = 1 + (−1)(1 + et)−2et = 1− et

(1 + et)2

Sinceet > 0

(1 + et)2 = (1 + et)(1 + et) > 1(1 + et) = 1 + et > et > 0

for all real t, thus(1 + et)−2et > 0

(1 + et)−2et < 1

for all real t. Hence 0 < f ′(t) < 1 for all real t.


Next, since f(t) − t = (1 − et)−1 > 0 for all real t, f(t) has no fixedpoint.

For (c): Suppose xn+1 6= xn for all n. (If xn+1 = xn, then xn = xn+1 =... and xn is a fixed point of f).

By Mean-Value Theorem,

f(xn+1)− f(xn) = f ′(tn)(xn+1 − xn)

where tn is betweem xn and xn+1. Thus,

|f(xn+1)− f(xn)| = |f ′(tn)||(xn+1 − xn)|

Note that |f ′(tn)| is bounded by A < 1, f(xn) = xn+1, and f(xn+1) =xn+2. Thus

|xn+2 − xn+1| ≤ A|xn+1 − xn||xn+1 − xn| ≤ CAn−1

where C = |x2 − x1|. For two positive integers p > q,

|xp − xq| ≤ |xp − xp−1|+ ... + |xq+1 − xq|= C(Aq−1 + Aq + ... + Ap−2)

≤ CAq−1

1− A.

Hence

|xp − xq| ≤CAq−1

1− A.

Hence, for any ε > 0, there exists N = [logAε(1−A)

C] + 2 such that

|xp − xq| < ε whenever p > q ≥ N . By Cauchy criterion we know that{xn} converges to x. Thus,

limn→∞

xn+1 = f( limn→∞

xn)

since f is continuous. Thus,

x = f(x).

x is a fixed point of f .

For (d): Since xn+1 = f(xn), it is trivial.

59

23.

24.

25. Suppose f is twice differentiable on [a, b], f(a) < 0, f(b) > 0, f ′(x) ≥δ > 0, and 0 ≤ f ′′(x) ≤ M for all x ∈ [a, b]. Let ξ be the unique pointin (a, b) at which f(ξ) = 0.

Complete the details in the following outline of Newton’s method forcomputing ξ.

(a) Choose x1 ∈ (ξ, b), and define {xn} by

xn+1 = xn −f(xn)

f ′(xn).

Interpret this geometrically, in terms of a tangent to the graph of f .

(b) Prove that xn+1 < xn and that

limn−>oo

xn = ξ.

(c) Use Taylor’s theorem to show that

xn+1 − ξ =f ′′(tn)

2f ′(xn)(xn − ξ)2

for some tn ∈ (ξ, xn).

(d) If A = M2δ

, deduce that

0 ≤ xn+1 − ξ ≤ 1

A[A(x1 − ξ)]2

n

.

(Compare with Exercise 16 and 18, Chap. 3)

(e) Show that Newton’s method amounts to finding a fixed point of thefunction g defined by

g(x) = x− f(x)

f ′(x).

How does g′(x) behave for x near ξ?

(f) Put f(x) = x1/3 on (−∞,∞) and try Newton’s method. Whathappens?


Proof: For (a): You can see the picture in the following URL:http://archives.math.utk.edu/visual.calculus/3/newton.5/1.html.

For (b): We show that xn ≥ xn+1 ≥ ξ. (induction). By Mean-ValueTheorem, f(xn) − f(ξ) = f ′(cn)(xn − ξ) where cn ∈ (ξ, xn). Sincef ′′ ≥ 0, f ′ is increasing and thus

f(xn)

xn − ξ= f ′(cn) ≤ f ′(xn) =

f(xn)

xn − xn+1

f(xn)(xn − ξ) ≤ f(xn)(xn − xn+1)

Note that f(xn) > f(ξ) = 0 since f ′ ≥ δ > 0 and f is strictly increasing.Thus,

xn − ξ ≤ xn − xn+1

ξ ≤ xn+1

Note that f(xn) > 0 and f ′(xn) > 0. Thus xn+1 < xn. Hence,

xn > xn+1 ≥ ξ.

Thus, {xn} converges to a real number ζ. Suppose ζ 6= ξ, then

xn+1 = xn −f(xn)

f ′(xn)

Note that f(xn)f ′(xn)

> f(ζ)δ

. Let α = f(ζ)δ

> 0, be a constant. Thus,

xn+1 < xn − α

for all n. Thus, xn < x1 − (n − 1)α, that is, xn → −∞ as n → ∞. Itcontradicts. Thus, {xn} converges to ξ.

For (c): By using Taylor’s theorem,

f(ξ) = f(xn) + f ′(xn)(ξ − xn) +f ′′(tn)

2(xn − ξ)2

0 = f(xn) + f ′(xn)(ξ − xn) +f ′′(tn)

2(xn − ξ)2

0 =f(xn)

f ′(xn)− xn + ξ +

f ′′(tn)

2f ′(xn)(xn − ξ)2

61

xn+1 − ξ =f ′′(tn)

2f ′(xn)(xn − ξ)2

where tn ∈ (ξ, xn).

For (d): By (b) we know that 0 ≤ xn+1 − ξ for all n. Next by (c) weknow that

xn+1 − ξ =f ′′(tn)

2f ′(xn)(xn − ξ)2

Note that f ′′ ≤ M and f ′ ≥ δ > 0. Thus

xn+1 − ξ ≤ A(xn − ξ)2 ≤ 1

A(A(x1 − ξ))2n

by the induction. Thus,

0 ≤ xn+1 − ξ ≤ 1

A[A(x1 − ξ)]2

n

.

For (e): If x0 is a fixed point of g(x), then g(x0) = x0, that is,

x0 −f(x0)

f ′(x0)= x0

f(x0) = 0.

It implies that x0 = ξ and x0 is unique since f is strictly increasing.Thus, we choose x1 ∈ (ξ, b) and apply Newton’s method, we can findout ξ. Hence we can find out x0.

Next, by calculating

g′(x) =f(x)f ′′(x)

f ′(x)2

0 ≤ g′(x) ≤ f(x)M

δ2.

As x near ξ from right hand side, g′(x) near f(ξ) = 0.

For (f): xn+1 = xn − f(xn)f ′(xn)

= −2xn by calculating. Thus,

xn = (−2)n−1x1

for all n, thus {xn} does not converges for any choice of x1, and wecannot find ξ such that f(ξ) = 0 in this case.


26. Suppose f is differentiable on [a, b], f(a) = 0, and there is a real numberA such that |f ′(x)| ≤ A|f(x)| on [a, b]. Prove that f(x) = 0 for allx ∈ [a, b]. Hint: Fix x0 ∈ [a, b], let

M0 = sup |f(x)|, M1 = sup |f ′(x)|

for a ≤ x ≤ x0. For any such x,

|f(x)| ≤ M1(x0 − a) ≤ A(x0 − a)M0.

Hence M0 = 0 if A(x0 − a) < 1. That is, f = 0 on [a, x0]. Proceed.

Proof: Suppose A > 0. (If not, then f = 0 on [a, b] clearly.) Fixx0 ∈ [a, b], let

M0 = sup |f(x)|, M1 = sup |f ′(x)|for a ≤ x ≤ x0. For any such x,

f(x)− f(a) = f ′(c)(x− a)

where c is between x and a by using Mean-Value Theorem. Thus

|f(x)| ≤ M1(x− a) ≤ M1(x0 − a) ≤ A(x0 − a)M0

Hence M0 = 0 if A(x0 − a) < 1. That is, f = 0 on [a, x0] by takingx0 = a + 1

2A. Repeat the above argument by replacing a with x0, and

note that 12A

is a constant. Hence, f = 0 on [a, b].

27. Let φ be a real function defined on a rectangle R in the plane, givenby a ≤ x ≤ b, α ≤ y ≤ β. A solution of the initial-value problem

y′ = φ(x, y), y(a) = c (α ≤ c ≤ β)

is, by definition, a differentiable function f on [a, b] such that f(a) = c,α ≤ f(x) ≤ β, and

f ′(x) = φ(x, f(x)) (a ≤ x ≤ b)

Prove that such a problem has at most one solution if there is a constantA such that

|φ(x, y2)− φ(x, y1)| ≤ A|y2 − y1|

63

whenever (x, y1) ∈ R and (x, y2) ∈ R.

Hint: Apply Exercise 26 to the difference of two solutions. Note thatthis uniqueness theorem does not hold for the initial-value problem

y′ = y1/2, y(0) = 0,

which has two solutions: f(x) = 0 and f(x) = x2/4. Find all othersolutions.

Proof: Suppose y1 and y2 are solutions of that problem. Since

|φ(x, y2)− φ(x, y1)| ≤ A|y2 − y1|,

y(a) = c, y′1 = φ(x, y1), and y′2 = φ(x, y2), by Exercise 26 we know thaty1 − y2 = 0, y1 = y2. Hence, such a problem has at most one solution.

Note: Suppose there is initial-value problem

y′ = y1/2, y(0) = 0.

If y1/2 6= 0, then y1/2dy = dx. By integrating each side and noting thaty(0) = 0, we know that f(x) = x2/4. With y1/2 = 0, or y = 0. Allsolutions of that problem are

f(x) = 0 and f(x) = x2/4

Why the uniqueness theorem does not hold for this problem? Onereason is that there does not exist a constant A satisfying

|y′1 − y′2| ≤ A|y1 − y2|

if y1 and y2 are solutions of that problem. (since 2/x → ∞ as x → 0and thus A does not exist).

28. Formulate and prove an analogous uniqueness theorem for systems ofdifferential equations of the form

y′j = φj(x, y1, ..., yk), yj(a) = cj (j = 1, ..., k)

Note that this can be rewritten in the form

y′ = φ(x, y), y(a) = c


where y = (y1, ..., yk) ranges over a k-cell, φ is the mapping of a (k+1)-cell into the Euclidean k-space whose components are the functionφ1, ..., φk, and c is the vector (c1, ..., ck). Use Exercise 26, for vector-valued functions.

Theorem: Let φj(j = 1, ..., k) be real functions defined on a rectangleRj in the plane given by a ≤ x ≤ b, αj ≤ yj ≤ βj.

A solution of the initial-value problem

y′j = φ(x, yj), yj(a) = cj (αj ≤ cj ≤ βj)

is, by definition, a differentiable function fj on [a, b] such that fj(a) =cj, αj ≤ fj(x) ≤ βj, and

f ′j(x) = φj(x, fj(x)) (a ≤ x ≤ b)

Then this problem has at most one solution if there is a constant Asuch that

|φj(x, yj2)− φj(x, yj1)| ≤ A|yj2 − yj1|

whenever (x, yj1) ∈ Rj and (x, yj2) ∈ Rj.

Proof: Suppose y1 and y2 are solutions of that problem. For eachcomponents of y1 and y2, say y1j and y2j respectively, y1j = y2j byusing Exercise 26. Thus, y1 = y2

29. Specialize Exercise 28 by considering the system

y′j = yj+1 (j = 1, ..., k − 1),

y′k = f(x)−k∑

j=1

gj(x)yj

where f, g1, ..., gk are continuous real functions on [a, b], and derive auniqueness theorem for solutions of the equation

y(k) + gk(x)y(k−1) + ... + g2(x)y′ + g1(x)y = f(x),

65

subject to initial conditions

y(a) = c1, y′(a) = c1, ..., y

(k−1)(a) = ck.

Theorem: Let Rj be a rectangle in the plain, given by a ≤ x ≤ b,min yj ≤ yj ≤ max yj. (since yj is continuous on the compact set, say[a, b], we know that yj attains minimal and maximal.) If there is aconstant A such that{

|yj+1,1 − yj+1,2| ≤ A|yj,1 − yj,2| (j < k)|∑k

j=1 gj(x)(yj,1 − yj,2)| ≤ A|yk,1 − yk,2|

whenever (x, yj,1) ∈ Rj and (x, yj,2) ∈ Rj.

Proof: Since the system y′1, ..., y′k with initial conditions satisfies a fact

that there is a constant A such that |y′1−y′2| ≤ A|y1−y2|, that systemhas at most one solution. Hence,

y(k) + gk(x)y(k−1) + ... + g2(x)y′ + g1(x)y = f(x),

with initial conditions has at most one solution.


Chapter 6

Sequences and Series ofFunctions

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68 CHAPTER 6. SEQUENCES AND SERIES OF FUNCTIONS

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Solutions Ch1-5