A Complete Solution Guide to Real and Complex Analysis I 9789887879787, 9789887879794, 9887879797

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A Complete Solution Guide to Real and Complex Analysis I

by Kit-Wing Yu, PhD [email protected]

c 2019 by Kit-Wing Yu. All rights reserved. No part of this publication may be Copyright reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the author. ISBN: 978-988-78797-8-7 (eBook) ISBN: 978-988-78797-9-4 (Paperback)

ii

About the author

Dr. Kit-Wing Yu received his B.Sc. (1st Hons), M.Phil. and Ph.D. degrees in Math. at the HKUST, PGDE (Mathematics) at the CUHK. After his graduation, he has joined United Christian College to serve as a mathematics teacher for at least seventeen years. He has also taken the responsibility of the mathematics panel since 2002. Furthermore, he was appointed as a part-time tutor (2002 – 2005) and then a part-time course coordinator (2006 – 2010) of the Department of Mathematics at the OUHK. Besides teaching, Dr. Yu has been appointed to be a marker of the HKAL Pure Mathematics and HKDSE Mathematics (Core Part) for over thirteen years. Between 2012 and 2014, Dr. Yu was invited to be a Judge Member by the World Olympic Mathematics Competition (China). In the research aspect, he has published over twelve research papers in international mathematical journals, including some well-known journals such as J. Reine Angew. Math., Proc. Roy. Soc. Edinburgh Sect. A and Kodai Math. J.. His research interests are inequalities, special functions and Nevanlinna’s value distribution theory.

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iv

Preface

Professor Walter Rudina is the author of the classical and famous textbooks: Principles of Mathematical Analysis, Real and Complex Analysis, and Functional Analysis. (People commonly call them “Baby Rudin”, “Papa Rudin” and “Grandpa Rudin” respectively.) Undoubtedly, they have produced important and extensive impacts to the study of mathematical analysis at university level since their publications. As far as you know, Real and Complex Analysis keeps the features of Principles of Mathematical Analysis which are well-organized and expositions of theorems are clear, precise and well-written. Therefore, Real and Complex Analysis is always one of the main textbooks or references of graduate real analysis course in many universities. Actually, some universities will request their Ph.D. students to study this book for their qualifying examinations. After the publication of the book A Complete Solution Guide to Principles of Mathematical Analysis, some purchasers have suggested me to write a solution book of Real and Complex Analysis. (To the best of the author’s knowledge, Papa Rudin has no solution manual.) I was afraid of doing so at the beginning because the exercises are at graduate level and they are much more difficult than those in Baby Rudin. Fortunately, I was used to keeping solutions of mathematics exercises done by me in my undergraduate and graduate study. In fact, I have kept at least 25% of the exercises in the first six chapters of Papa Rudin. Therefore, after thorough consideration, I decided to start the next project of a solution book to Rudin’s Real and Complex Analysis. Since Papa Rudin consists of two components, I plan to write the solutions for “Real Analysis” part first. In fact, the present book A Complete Solution Guide to Real and Complex Analysis I covers all the exercises of Chapters 1 to 9 and its primary aim is to help every mathematics student and instructor to understand the ideas and applications of the theorems in Rudin’s book. To accomplish this goal, I have adopted the way I wrote the book A Complete Solution Guide to Principles of Mathematical Analysis. In other words, I intend writing the solutions as comprehensive as I can so that you can understand every detailed part of a proof easily. Apart from this, I also keep reminding you what theorems or results I have applied by quoting them repeatedly in the proofs. By doing this, I believe that you will become fully aware of the meaning and applications of each theorem. Before you read this book, I have two gentle reminders for you. Firstly, as a mathematics instructor at a college, I understand that the growth of a mathematics student depends largely on how hard he/she does exercises. When your instructor asks you to do some exercises from Rudin, you are not suggested to read my solutions unless you have tried your best to prove them yourselves. Secondly, when I prepared this book, I found that some exercises require knowledge that Rudin did not cover in his book. To fill this gap, I refer to some other analysis or topology a

https://en.wikipedia.org/wiki/Walter_Rudin.

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vi books such as [9], [10], [22], [42] and [47]. Other useful references are [1], [12], [24], [28], [29], [59], [64], [66] and [67]. Of course, it is not a surprise that we will regard the exercises in Baby Rudin as some known facts and if you want to read proofs of them, you are strongly advised to read my book [63]. The features of this book are as follows: • It covers all the 176 exercises from Chapters 1 to 9 with detailed and complete solutions. As a matter of fact, my solutions show every detail, every step and every theorem that I applied. • There are 11 illustrations for explaining the mathematical concepts or ideas used behind the questions or theorems. • Sections in each chapter are added so as to increase the readability of the exercises. • Different colors are used frequently in order to highlight or explain problems, lemmas, remarks, main points/formulas involved, or show the steps of manipulation in some complicated proofs. (ebook only) • Necessary lemmas with proofs are provided because some questions require additional mathematical concepts which are not covered by Rudin. • Many useful or relevant references are provided to some questions for your future research. Since the solutions are written solely by me, you may find typos or mistakes. If you really find such a mistake, please send your valuable comments or opinions to [email protected]. Then I will post the updated errata on my website https://sites.google.com/view/yukitwing/ irregularly.

Kit Wing Yu April 2019

List of Figures

2.1

The graph of gn on [−1, 1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

2.2 2.3

The graphs of gn,k on [0, 1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The pictures of V, E ′ , E ′ ∩ V and (E ′ ∩ V )c . . . . . . . . . . . . . . . . . . . . . .

31 57

2.4

The set (E ′ ∩ V )c \ U . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

3.1

The distribution of x and ǫn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

85

3.2

The geometric interpretation of a special case. . . . . . . . . . . . . . . . . . . . .

89

5.1 5.2

The unit circle in different p-norm. . . . . . . . . . . . . . . . . . . . . . . . . . . 144 The square K. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

6.1

The graphs of gn,1 (x) and gn,2 (x).

7.1

The closed intervals En and En+1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

7.2

Construction of the sequence Ei (p). . . . . . . . . . . . . . . . . . . . . . . . . . 242

. . . . . . . . . . . . . . . . . . . . . . . . . . 194

vii

List of Figures

viii

Contents

Preface

v

List of Figures

vii

1 Abstract Integration 1.1 Problems on σ-algebras and Measurable Functions . . . . . . . . . . . . . . . . . 1.2

1 1

Problems related to the Lebesgue’s MCT/DCT . . . . . . . . . . . . . . . . . . .

7

2 Positive Borel Measures 2.1 Properties of Semicontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17 17

2.2

Problems on the Lebesgue Measure on R . . . . . . . . . . . . . . . . . . . . . . .

25

2.3 2.4

Integration of Sequences of Continuous Functions . . . . . . . . . . . . . . . . . . Problems on Borel Measures and Lebesgue Measures . . . . . . . . . . . . . . . .

30 36

2.5

Problems on Regularity of Borel Measures . . . . . . . . . . . . . . . . . . . . . .

43

2.6

Miscellaneous Problems on

L1

and Other Properties . . . . . . . . . . . . . . . .

59

3 Lp -Spaces 3.1 Properties of Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69 69

3.2

Relations among Lp -Spaces and some Consequences . . . . . . . . . . . . . . . .

71

3.3 3.4

Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12 . . . . . . . . . . . . . . . . . Hardy’s Inequality and Egoroff’s Theorem . . . . . . . . . . . . . . . . . . . . . .

87 91

3.5 3.6

Convergence in Measure and the Essential Range of f ∈ L∞ (µ) . . . . . . . . . . 106 A Converse of Jensen’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

3.7

The Completeness/Completion of a Metric Space . . . . . . . . . . . . . . . . . . 112

3.8

Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

4 Elementary Hilbert Space Theory 123 4.1 Basic Properties of Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 123 4.2

Application of Theorem 4.14

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

4.3

Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

5 Examples of Banach Space Techniques 143 5.1 The Unit Ball in a Normed Linear Space . . . . . . . . . . . . . . . . . . . . . . . 143 5.2 5.3

Failure of Theorem 4.10 and Norm-preserving Extensions . . . . . . . . . . . . . 146 The Dual Space of X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 ix

Contents

x

5.4

Applications of Baire’s and other Theorems . . . . . . . . . . . . . . . . . . . . . 157

5.5

Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

6 Complex Measures

183

6.1 6.2

Properties of Complex Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 Dual Spaces of Lp (µ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

6.3 6.4

Fourier Coefficients of Complex Borel Measures . . . . . . . . . . . . . . . . . . . 191 Problems on Uniformly Integrable Sets . . . . . . . . . . . . . . . . . . . . . . . . 196

6.5

Dual Spaces of Lp (µ) Revisit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

7 Differentiation

207

7.1 7.2

Lebesgue Points and Metric Densities . . . . . . . . . . . . . . . . . . . . . . . . 207 Periods of Functions and Lebesgue Measurable Groups . . . . . . . . . . . . . . . 210

7.3 7.4

The Cantor Function and the Non-measurability of f ◦ T . . . . . . . . . . . . . 216 Problems related to the AC of a Function . . . . . . . . . . . . . . . . . . . . . . 218

7.5

Miscellaneous Problems on Differentiation . . . . . . . . . . . . . . . . . . . . . . 232

8 Integration on Product Spaces

249

8.1 8.2

Monotone Classes and Ordinate Sets of Functions . . . . . . . . . . . . . . . . . . 249 Applications of the Fubini Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 252

8.3 8.4

The Product Measure Theorem and Sections of a Function . . . . . . . . . . . . 268 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273

9 Fourier Transforms 281 9.1 Properties of The Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . 281 9.2 9.3

The Poisson Summation Formula and its Applications . . . . . . . . . . . . . . . 297 Fourier Transforms on Rk and its Applications . . . . . . . . . . . . . . . . . . . 301

9.4

Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310

Index

315

Bibliography

317

CHAPTER

1

Abstract Integration

1.1

Problems on σ-algebras and Measurable Functions

Problem 1.1 Rudin Chapter 1 Exercise 1.

Proof. Let X be a set. Assume that M was an infinite σ-algebra in X which has only countably many members. Then we have M = {∅, X, A1 , A2 , . . .}. Let x ∈ X be fixed, we define the set \ Sx = A, (1.1) x∈A

where x ∈ A ∈ M. (By this definition (1.1), it is clear that x ∈ Sx .) Since M is countable, the intersection in (1.1) is actually at most countable. Thus it follows from Comment 1.6(c) that Sx ∈ M. Next we define the set S = {Sx | x ∈ X} so that S ⊆ M. We need a lemma about the set S: Lemma 1.1 If Sx , Sy ∈ S and Sx ∩ Sy 6= ∅, then Sx = Sy .

Proof of Lemma 1.1. Let z ∈ Sx ∩ Sy . We want to show that Sz = Sx . On the one hand, since z ∈ Sx , we know from the definition (1.1) that z ∈ A for all A containing x. Now if w ∈ Sx , then w ∈ A for all A containing x. By the previous observation, such A must contain z, so we get w ∈ Sz and then Sx ⊆ Sz .

(1.2)

On the other hand, if w ∈ Sz , then w ∈ A for all A containing z. Thus the hypothesis z ∈ Sx ∩ Sy implies that w ∈ Sx ∩ Sy , i.e., Sz ⊆ Sx ∩ Sy . By this result and the definition (1.1), we obtain Sz ⊆ Sx ∩ Sy ⊆ Sx . (1.3) Combining the observations (1.2) and (1.3), we have the desired result that Sz = Sx .  Similarly, we can show that Sz = Sy . Hence we conclude that Sx = Sy . 1

Chapter 1. Abstract Integration

2

Now we return to the proof of the problem. By Lemma 1.1, we may assume that all elements in S are distinct. If B ∈ M, then the definition (1.1) certainly gives y ∈ Sy for every y ∈ B so that [ B⊆ Sy . (1.4) y∈B

Therefore, if the cardinality of S is N , then the cardinality of its power set 2S is 2N and so there are at most 2N elements in M by the inclusion (1.4), a contradiction. Thus the set S must be infinite. Recall that S ⊆ M and M is countable, S is also countable by [49, Theorem 2.8, p. 26]. Since M is a σ-algebra, it must contain the power set of S. However, it is well-known that the power set of an infinite countable set is uncountable [47, Problem 22, p. 16]. We conclude from these facts that M is uncountable, a contradiction. This completes the proof of the problem.  Problem 1.2 Rudin Chapter 1 Exercise 2.

Proof. Put f (x) = (f1 (x), . . . , fn (x)). Since f1 , . . . , fn : X → R, f maps the measurable space X into Rn . Let Y be a topological space and Φ : Rn → Y be continuous. Define
h(x) = Φ (f1 (x), . . . , fn (x)) .

Since h = Φ ◦ f , Theorem 1.7(b) shows that it suffices to prove the measurability of f .

To this end, we first consider f −1 (R) for an open rectangle R in Rn . By the definition, R = I1 × · · · × In , where Ii is an open interval in R for 1 ≤ i ≤ n. Now we know from [42, Exercise 3, p. 21] that f −1 (R) = {x ∈ X | f (x) ∈ R}

= {x ∈ X | fi (x) ∈ Ii for 1 ≤ i ≤ n}

= {x ∈ X | f1 (x) ∈ I1 } ∩ · · · ∩ {x ∈ X | fn (x) ∈ In } = f1−1 (I1 ) ∩ · · · ∩ fn−1 (In ).

Since each fi is measurable and each Ii is open in R, the set fi−1 (Ii ) is measurable in X by Definition 1.3(c). Thus this implies that f −1 (R) is measurable in X by Comment 1.6(c). Our proof will be complete if we can show that every open set V in Rn can be written as a countable union of open rectangles Rj in Rn . We need some topology. By [42, Exercise 8(a), p. 83], the countable collection B = {I = (a, b) | a < b and a, b ∈ Q} is a basis that generates the standard topology on R. Then we follow from [42, Theorem 15.1, p. 86] that the collection C = {I1 × · · · × In | Ii = (ai , bi ), ai , bi ∈ Q, 1 ≤ i ≤ n}

(1.5)

is a basis for the (product) topology of Rn .a Since elements in the collection (1.5) are open rectangles and it is countable by [49, Theorem 2.13, p. 29], every open set V in Rn is, in fact, a countable union of open rectangles Rj , i.e, f

−1

(V ) = f

−1

∞ [

j=1

a

For details, please read [42, §13 and §15].



Rj =

∞ [

j=1

f −1 (Rj ).

3

1.1. Problems on σ-algebras and Measurable Functions

Hence f −1 (V ) is also a measurable set in X by Definition 1.3(a)(iii). This completes the proof of the problem.  Problem 1.3 Rudin Chapter 1 Exercise 3.

Proof. Let f : X → R ⊂ [−∞, ∞] and M be the σ-algebra of X. Let α be real. By [49, Theorem 1.20(b), p. 9], we can find a sequence {rn } of rational numbers such that α < rn for all n ∈ N, rn → α as n → ∞. In other words, we have (α, ∞] =

∞ [

[rn , ∞]

n=1

which implies that f

−1

((α, ∞]) = f

−1

∞ [

∞  [ f −1 ([rn , ∞]). [rn , ∞] = n=1

n=1

Recall that f −1 ([rn , ∞]) = {x | f (x) ≥ rn }

is assumed to be measurable for each n, so we have f −1 ([rn , ∞]) ∈ M and then we follow from Definition 1.3(a)(iii) that f −1 ((α, ∞]) ∈ M. Since α is arbitrary, we conclude from Theorem  1.12(c) that f is measurable. This finishes the proof of the problem. Problem 1.4 Rudin Chapter 1 Exercise 4.

Proof. (a) Let αk = sup{−ak , −ak+1 , . . .} for k = 1, 2, . . .. By the definition, we have αk ≥ −an for all n ≥ k and if α ≥ −an for all n ≥ k, then α ≥ αk . Note that this is equivalent to the fact that −αk ≤ an for all n ≥ k and if −α ≤ an for all n ≥ k, then −α ≤ −αk . In other words, we have −αk = inf{ak , ak+1 , . . . , } for k = 1, 2, . . .. Thus this implies that sup(−an ) = sup{−ak , −ak+1 , . . .} = − inf{ak , ak+1 , . . .} = − inf (an ). n≥k

n≥k

(1.6)

Similarly, we have inf (−cn ) = − sup(cn )

n≥k

(1.7)

n≥k

for a sequence {cn } in [−∞, ∞]. By applying the equality (1.6) and then the equality (1.7), we achieve that n o o o n n lim sup(−an ) = inf sup(−an ) = inf − inf (an ) = − sup inf (an ) = − lim inf (an ) n→∞ k≥1 k≥1 n≥k n≥k n→∞ k≥1 n≥k | {z } cn

which is our desired result.

Chapter 1. Abstract Integration

4

(b) This part is proven in [63, Problem 3.5, pp. 32, 33]. (c) Since an ≤ bn for all n = 1, 2, . . ., we must have αk = inf (an ) ≤ inf (bn ) = βk n≥k

n≥k

(1.8)

for all k = 1, 2, . . .. Thus {αn } and {βn } are two sequences in [−∞, ∞] such that αn ≤ βn for all n = 1, 2, . . .. By a similar argument, we also have sup αk ≤ sup βk

k≥m

(1.9)

k≥m

for all m = 1, 2, . . .. Combining the inequalities (1.8) and (1.9), we have n o n o sup inf (an ) ≤ sup inf (bn ) k≥m

n≥k

k≥m

n≥k

for all m = 1, 2, . . .. By Definition 1.13, we have the desired result. For a counterexample to part (b), we consider an = (−1)n and bn = (−1)n+1 for all n = 1, 2, . . .. On the one hand, we have an + bn = 0 for all n = 1, 2, . . . so that lim sup(an + bn ) = 0.

(1.10)

n→∞

On the other hand, we have lim sup an = lim a2k = 1 and n→∞

k→∞

lim sup bn = lim b2k+1 = 1 n→∞

k→∞

which imply that lim sup an + lim sup bn = 2. n→∞

(1.11)

n→∞

Hence we obtain from the results (1.10) and (1.11) that the strict inequality can hold in part  (b). This completes the proof of the problem. Problem 1.5 Rudin Chapter 1 Exercise 5.

Proof. (a) Let MX be a σ-algebra of X. Further, suppose that Sf (±∞) = {x ∈ X | f (x) = ±∞}, and Sg (±∞) = {x ∈ X | g(x) = ±∞}. We see that Sf (∞) =

∞ \

{x ∈ X | f (x) > n}.

n=1

Since f is measurable and (n, ∞] is openb in [−∞, ∞], f −1 ((n, ∞]) = {x | f (x) > n} ∈ MX by Theorem 1.12(b) for each n ∈ N. Thus Sf (∞) ∈ MX by Comment 1.6(c). Similarly, all the other sets Sf (−∞), Sg (∞) and Sg (−∞) belong to MX too. Next, we let X ′ = X\(Sf (∞)∪Sf (−∞)∪Sg (∞)∪Sg (−∞)). Since Sf (∞), Sf (−∞), Sg (∞) and Sg (−∞) are measurable, X ′ ∈ M by Definition 1.3(a)(ii) and Comment 1.6(b). By b

See the proof of Theorem 1.12(c) in [51, p. 13]

5

1.1. Problems on σ-algebras and Measurable Functions the comment following Proposition 1.24, we know that X ′ is itself a measure space and if we let MX ′ be a σ-algebra of X ′ , then MX ′ ⊆ MX .

(1.12)

Furthermore, the restricted mappings fX ′ : X ′ → R and gX ′ : X ′ → R are also measurable because of Definition 1.3(c) and the fact that any open set V in R is a countable union of segments of the type (α, β) so that V is also open in the extended number system [−∞, ∞].c In fact, we have −1 −1 fX (V ) ∈ MX ′ ′ (V ) = f

and

−1 −1 gX (V ) ∈ MX ′ . ′ (V ) = g

We need to prove one more thing: the mapping −g : X → [−∞, ∞] is measurable. Since g is measurable, we know from [49, Definition 11.13] that {x ∈ X | g(x) > −a} ∈ MX for every real a. It is obvious that {x ∈ X | − g(x) < a} = {x ∈ X | g(x) > −a} for every real a, so we deduce from [49, Theorem 11.15, p. 311] that −g is measurable. Now we are ready to prove the desired results. Notice that

{x ∈ X | f (x) = g(x)} = {x ∈ X | h(x) = 0} = h−1 (0), where h = f − g. Since f and −g are measurable, the new (real) function h = f − g is also measurable by Theorem 1.9(c). Since −1

h

(0) =

∞ \

n=1

h−1



−

1 1  , n n

and h−1 ((− n1 , n1 )) ∈ MX ′ for every n ∈ N, we yield from this and the relation (1.12) that h−1 (0) ∈ MX ′ ⊆ MX . This shows the second assertion. For the first assertion, we note that {x ∈ X | f (x) < g(x)} = {x ∈ X | h(x) = f (x) − g(x) < 0} 
= h−1 (−∞, 0) ∪ [Sf (−∞) \ Sg (−∞)] ∪ [Sg (∞) \ Sf (∞)]  \ [Sf (∞) ∩ Sg (∞)] ∪ [Sf (−∞) ∩ Sg (−∞)] .

Recall that Sf (∞), Sf (−∞), Sg (∞), Sg (−∞) ∈ MX , so we have

Sf (−∞) \ Sg (−∞), Sg (∞) \ Sf (∞), Sf (∞) ∩ Sg (∞), Sf (−∞) ∩ Sg (−∞) ∈ MX . (1.13) By the measurability of h and the relation (1.12), we have h−1 ((−∞, 0)) ∈ MX .

(1.14)

Hence the facts (1.13) and (1.14) show that {x | f (x) < g(x)} ∈ MX . (b) This part is proven in [63, Problem 11.3, p. 339]. This completes the proof of the problem.



Problem 1.6 Rudin Chapter 1 Exercise 6. c

This can be seen, again, from the proof of Theorem 1.12(c) or from the comment following Proposition 1.24.

Chapter 1. Abstract Integration

6

Proof. We prove the assertions one by one. • M is a σ-algebra in X. We check Definition 1.3(a). Since X c = ∅, it is at most countable. Thus we have X ∈ M. Let A ∈ M. If Ac is at most countable, then Ac ∈ M. Similarly, if A is at most countable, then since (Ac )c = A, we have Ac ∈ M. Suppose that An ∈ M for n = 1, 2, . . .. Then we have either An or Acn is at most countable for n = 1, 2, . . .. If all An are at most countable, then we deduce from the corollary following [49, Theorem 2.12, p. 29] that the set A=

∞ [

An

(1.15)

n=1

is also at most countable so that A ∈ M. Otherwise, without loss of generality, we suppose that A1 is uncountable but Ac1 is at most countable. Then we consider c

A =

∞ [

An

n=1

c

=

∞ \

n=1

Acn ⊆ Ac1

which means that Ac is at most countable too. Hence Ac ∈ M and then M is a σ-algebra in X. • µ is a measure on M. We check Definition 1.18(a). Since ∅ is at most countable, we have µ(∅) = 0 so that µ is not identically ∞. In fact, it is clear that we have µ : M → {0, 1} ⊂ [0, ∞]. Let {An } be a disjoint countable collection of members of M. Case (i): All An are at most countable. Recall that the set A given by (1.15) is at most countable. By the definition of µ, we have µ(A) = µ(An ) = 0 for all n = 1, 2, . . .. Thus we have ∞ X µ(An ) (1.16) µ(A) = n=1

in this case.

Case (ii): There is at least one Ak is uncountable. Since Ak ∈ M, Ack must be at most countable. Since An ∩ Ak = ∅ for all n 6= k, we have An ⊆ Ack for all n 6= k. In other words, the measurable sets An are at most countable for all n 6= k. Therefore, we have µ(Ak ) = 1 and µ(An ) = 0 for all n 6= k. Since c

A =

∞ \

n=1

Acn ⊆ Ack ,

Ac is at most countable and then µ(A) = 1. Hence the equality (1.16) also holds in this case. This completes the proof that µ is a measure on M. • The determination of measurable functions and their integrals. Let f : X → R be a measurable function. Since X is uncountable, we have µ(X) = 1 which is the only thing that we know and start with. For every n ∈ Z, we know that   [n, n + 1) = R \ (−∞, n) ∪ [n + 1, ∞) .

d

By the fact thatd f −1 (A \ B) = f −1 (A) \ f −1 (B), we obtain


 f −1 [n, n + 1) = f −1 (R) \ f −1 (−∞, n) ∪ f −1 [n + 1, ∞) .

See [42, Exercise 2(d), p. 20].

(1.17)

7

1.2. Problems related to the Lebesgue’s MCT/DCT By [49, Theorem 11.15, p. 311], each set on the right-hand side in (1.17) is measurable. This implies that
f −1 [n, n + 1) ∈ M.
Let En = f −1 [n, n + 1) , where n ∈ Z. By the definition of M and then the definition of µ, we have either µ(En ) = 0 or µ(En ) = 1. For every x ∈ X, we must have f (x) ∈ [n, n+1) for some n ∈ Z, i.e., x ∈ En for some n ∈ Z. Therefore, we have ∞ [

En = X.

n=−∞

It is clear that R =

∞ [

[n, n + 1), so {En } is a disjoint countable collection of members

n=−∞

of M. By Definition 1.18(a), we see that µ(X) = µ

∞  [

n=−∞

∞  X µ(En ). En =

(1.18)

n=−∞

The fact µ(X) = 1 and the equality (1.18) force that there exists an integer n0 such that µ(En0 ) = 1. Without loss of generality, we may assume that n0 = 0, i.e.,

µ(E0 ) = µ f −1 [0, 1) = 1.

(1.19)

1 1 Next if we write [0, 1) = [0,

2 ) ∪ [ 2 , 1), then the above

argument and the value (1.19) 1 1 −1 −1 imply that either µ f [0, 2 ) = 1 or µ f [ 2 , 1) = 1. This process can be done continuously so that a sequence of intervals {[an , bn )} is constructed such that



µ f −1 [an , bn ) = 1,

0 ≤ bn − a n ≤

1 2n

and

lim (bn − an ) = 0,

n→∞

where n = 1, 2, . . .. In other words, it means that
µ f −1 (a) = 1


and µ f −1 (b) = 0

for some a ∈ R and all other real numbers b 6= a, but this is equivalent to saying that f −1 (a) is uncountable and f −1 (b) is at most countable for all b 6= a. Now we have completely characterized every measurable function on X. Finally, it is clear that X \ f −1 (a) is a set of measure 0. Therefore, we have Z Z f dµ = a. f dµ = X

f −1 (a)

Hence, this completes the proof of the problem.

1.2

Problems related to the Lebesgue’s MCT/DCT

Problem 1.7 Rudin Chapter 1 Exercise 7.



Chapter 1. Abstract Integration

8


Proof. Let Ek = {x ∈ X | f1 (x) > k} = f1−1 (k, ∞] and E = {x ∈ X | f1 (x) = ∞}, where k = 1, 2, . . .. It is clear that each (k, ∞] is a Borel set in [0, ∞], so each Ek is measurable by Theorem 1.12(b). Since ∞ \ E= Ek , k=1

E is also measurable by Comment 1.6(c). If µ(E) > 0, then we know from the definition that Z f1 dµ = ∞. (1.20) E

By using the result (1.20), Proposition 1.24(b) and Theorem 1.33, we conclude that Z Z Z f1 dµ = ∞ f1 dµ ≥ |f1 | dµ ≥ E X X

which contradicts the hypothesis that f1 ∈ L1 (µ). In other words, we must have µ(E) = 0 and thus f1 ∈ L1 (µ) on X \ E.

Here we may assume that the form of Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) is also valid for measurable functions defined a.e. on X.e Now we see that the measurable function f1 in the problem plays the role of g in the theorem and hence our desired result follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) immediately. For a counterexample, we consider X = R, In = (n, ∞), µ = m the Lebesgue measure (see [49, Definition 11.5, pp. 302, 303]) and define for each n = 1, 2, . . .,

fn (x) = χIn (x) =

  1, if x ∈ In ; 

0, if x ∈ / In .

It is clear that f1 ≥ f2 ≥ · · · ≥ 0 on X and Z Z Z fn dm = |fn | dm = R

R

In

dm = ∞

(1.21)

for every n = 1, 2, . . .. In particular, the integrals (1.21) show that f1 ∈ / L1 (m). Furthermore, we have f (x) = lim fn (x) = lim χIn (x) = 0 (1.22) n→∞

n→∞

for every x ∈ R. Thus we deduce from the expression (1.22) and Proposition 1.24(d) that Z f dm = 0. (1.23) R

Hence the inconsistence of the integrals (1.21) and (1.23) show that the condition “f1 ∈ L1 (µ)”  cannot be omitted. This completes the proof of the problem. Problem 1.8 Rudin Chapter 1 Exercise 8.

e

Actually, Rudin [51, p. 29] assumed this fact in the proof of Theorem 1.38 or the reader may refer to the comment between Theorem 11.32 and its proof in [49, p. 321].

9

1.2. Problems related to the Lebesgue’s MCT/DCT

Proof. If x ∈ E, then for all k ∈ N, we have


lim f2k (x) = lim 1 − χE (x) = 0.

k→∞

k→∞

Similarly, if x ∈ / E, then for all k ∈ N, we have

lim f2k+1 (x) = lim χE (x) = 0.

k→∞

k→∞

Thus we have lim inf fn (x) = 0 n→∞

for all x ∈ X. However, we have Z

fn dµ =

X

=

 Z  χ dµ,    X E

if n is odd;

Z     (1 − χE ) dµ, if n is even. X  if n is odd;  µ(E), 

(1.24)

µ(X \ E), if n is even.

Therefore we obtain from the results (1.24) that Z fn dµ = min(µ(E), µ(X \ E)) 6= 0 lim inf n→∞

X

if we assume that µ(E) > 0 and µ(X \E) > 0. Hence the example here shows that the inequality in Fatou’s lemma can be strict, completing the proof of the problem.f  Problem 1.9 Rudin Chapter 1 Exercise 9.

Proof. If µ(X) = 0, then Proposition 1.24(e) implies that c =

Z

f dµ = 0, a contradiction. Let X

E = {x ∈ X | f (x) = ∞}. We claim that µ(E) = 0. Otherwise, Proposition 1.24(b) implies that Z Z f dµ = ∞ f dµ ≥ c= X

E

which is a contradiction. Therefore, in the following discussion, we may assume that x ∈ X \ E so that 0 ≤ f (x) < ∞. For each n = 1, 2, . . ., we defineg fn : X \ E → [0, ∞) by

h  f (x) α i . fn (x) = n log 1 + n

Since f is measurable on X \ E, g(x) = [1 + ( nx )α ] and h(x) = n log(1 + x) are continuous on [0, ∞), Theorem 1.7(b) implies that

f g

h  f (x) α i

fn (x) = n log 1 + = h g f (x) n

There is another example in [63, Problem 11.5, p. 340]. X \ E is itself a measure space by the remark following Proposition 1.24.

Chapter 1. Abstract Integration

10

is also measurable on X \ E.

Now we are going to show that when α ≥ 1, there exists a function g ∈ L1 (µ) such that |fn (x)| ≤ g(x)

holds for all n = 1, 2, . . . and all x ∈ X \ E. We first show the following lemma: Lemma 1.2 For each n ∈ N and α ≥ 1, we have h  x α i ≤ αx n log 1 + n

(1.25)

on [0, ∞).

Proof of Lemma 1.2. For x ∈ [0, ∞), we let

Then basic calculus gives

Since

h  x α i F (x) = αx − n log 1 + . n

 x α−1  α α−1 nxα−1  n  = α − αnx = α 1 − . F ′ (x) = α − x α n α + xα n α + xα 1+ n

(1.26)

nxα−1 xα ≤ α 1. In this case, we apply L’Hospital’s rule [49, Theorem 5.13, p. 109] to conclude that h  f α i αf α y α−1 0 log(1 + f α y α ) lim n log 1 + = = lim = lim =0 (1.28) α α + + n→∞ n y 1+0 y→0 1 + f y y→0 on X \ E. Thus it follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) that  Z  f dµ, if α = 1;   Z  X h  f α i n log 1 + dµ = lim Z n→∞ X  n    0 dµ, if α > 1, =

 X  c, if α = 1; 

0, if α > 1.

It remains the case that 0 < α < 1. In this case, Theorem 1.28 (Fatou’s Lemma) can be applied directly to get ) Z Z ( h  f α i n h  f α io n log 1 + dµ ≤ lim inf lim inf n log 1 + dµ. (1.29) n→∞ X n→∞ n n X Since 0 < α < 1, we have y α−1 =

1 y 1−α

so that the limit (1.28) becomes

h  f α i αf α = ∞. = lim 1−α lim n log 1 + n→∞ n (1 + f α y α ) y→0+ y

(1.30)

Since lim xn = ∞ if and only if lim sup xn = lim inf xn = ∞, the inequality (1.29) and the limit n→∞

n→∞

n→∞

(1.30) combine to imply that n h  f α io lim inf n log 1 + =∞ n→∞ n

and then it certainly gives

lim

n→∞

Z

h  f α i dµ = ∞. n log 1 + n X

This completes the proof of the problem. Problem 1.10 Rudin Chapter 1 Exercise 10.

h

We use the definition ex = lim

n→∞



1+

x n here. n



Chapter 1. Abstract Integration

12

Proof. Since fn → f uniformly on X, there exists a positive integer N such that n ≥ N implies |fn (x) − f (x)| ≤ 1 for all x ∈ X. By this, we have |fn (x)| ≤ |fn (x) − f (x)| + |f (x)| ≤ |f (x)| + 1

(1.31)

|f (x)| ≤ |f (x) − fN (x)| + |fN (x)| ≤ |fN (x)| + 1

(1.32)

and for all n ≥ N and x ∈ X. Combining the inequalities (1.31) and (1.32), we see that |fn (x)| ≤ |fN (x)| + 2

(1.33)

g(x) = max{|f1 (x)|, . . . , |fN −1 (x)|, |fN (x)| + 2}.

(1.34)

for all n ≥ N and x ∈ X. We define g : X → [0, ∞) by

Then it is easy to see from the inequality (1.33) and the definition (1.34) that |fn (x)| ≤ g(x) for x ∈ X and n = 1, 2, . . .. Next we want to show that g ∈ L1 (µ). By Theorem 1.9(b), |f1 (x)|, . . . , |fN −1 (x)|, |fN (x)| are measurable. Thus it follows from the corollaries following Theorem 1.14 that g is also measurable. Furthermore, we know from the hypothesis that |fn (x)| ≤ Mn on X for some constants Mn , where n = 1, 2, . . . , N . Therefore, this and the definition (1.34) certainly imply that g(x) = |g(x)| ≤ M on X for some constant M . Since it is obvious that g ≥ 0 on X, we get from Proposition 1.24(a) that Z Z M dµ = M µ(X) < ∞ |g| dµ ≤ 0≤ X

X

L1 (µ).

so that g ∈ In conclusion, our sequence of functions {fn } satisfies the hypotheses of Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) and hence the desired result follows immediately from this. For a counterexample, consider X = R and µ = m so that m(R) = ∞. For each n ∈ N, define fn : R → R by 1 fn (x) = . n Then it is easy to prove that fn → f ≡ 0 uniformly on R. Since Z Z m(R) =∞ fn dm = f dm = 0 and n R R for n = 1, 2, . . ., we conclude that lim

Z

n→∞ R

This finishes the proof of the problem.

fn dm 6=

Z

f dm.

R



13

1.2. Problems related to the Lebesgue’s MCT/DCT Problem 1.11

Rudin Chapter 1 Exercise 11.

Proof. For each n ∈ N, define Bn =

∞ [

Ek .

k=n

Recall that A is the set of all x ∈ X which lie in infinitely many Ek . On the one hand, if x ∈ A, ∞ \ Bn . In other words, we have then x ∈ Bn for all n ∈ N so that x ∈ n=1

A⊆

On the other hand, if x ∈

∞ \

n=1

∞ \

Bn .

(1.35)

n=1

Bn , then x ∈ Bn for each positive integer n and this is equivalent

to the condition that x belongs to infinitely many Ek , i.e., ∞ \

n=1

Bn ⊆ A.

(1.36)

Hence the set relations (1.35) and (1.36) imply the desired result that A=

∞ \

Bn =

∞ [ ∞ \

Ek .

n=1 k=n

n=1

We have to show that µ(A) = 0. By the definition of Bn , we have B1 ⊇ B2 ⊇ B3 ⊇ · · · . Furthermore, our hypothesis and the subadditive property of a measure (see , for example, [54, Corollary 4.6, p. 26]) show that µ(B1 ) = µ

∞ [

k=1



Ek ≤

∞ X k=1

µ(Ek ) < ∞.

Thus it follows from Theorem 1.19(e) and the subadditive property of µ again that 0 ≤ µ(A) = µ

∞ \

n=1

∞ ∞  [  X Bn = lim µ(Bn ) = lim µ Ek ≤ lim µ(Ek ) = 0. n→∞

n→∞

k=n

n→∞

Hence we must have µ(A) = 0, completing the proof of the problem. Problem 1.12 Rudin Chapter 1 Exercise 12.

k=n



Chapter 1. Abstract Integration

14

Proof. Let n be a positive integer. Define fn : X → [0, ∞] by
fn (x) = min |f (x)|, n

(1.37)

for every x ∈ X. This definition (1.37) clearly satisfies

0 ≤ fn (x) ≤ n

(1.38)

for every x ∈ X. Since f ∈ L1 (µ), it is obviously measurable. By the corollaries following Theorem 1.14, each fn is also measurable on X. Furthermore, if x0 ∈ X such that |f (x0 )| ≤ n, then the definition (1.37) implies that fn (x0 ) = |f (x0 )|

and fn+1 (x0 ) = |f (x0 )|;

(1.39)

if n < |f (x0 )| ≤ n + 1, then we know again from the definition (1.37) that fn (x0 ) = n

and fn+1 (x0 ) = |f (x0 )|;

(1.40)

and fn+1 (x0 ) = n + 1.

(1.41)

if n + 1 < |f (x0 )|, then we must have fn (x0 ) = n

Thus we can conclude from the computations (1.39), (1.40) and (1.41) that the inequality 0 ≤ fn (x) ≤ fn+1 (x)

(1.42)

holds for all n ∈ N and x ∈ X. By Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem), we have Z Z F dµ, fn dµ = lim n→∞ X

X

where F = lim fn . By the definition (1.37), we have F = |f | and so n→∞

lim

Z

n→∞ X

or equivalently, lim

fn dµ =

Z

n→∞ X

Z

X

|f | dµ

|fn − f | dµ = 0.

Thus for every ǫ > 0, there exists a positive integer N such that Z ǫ |fn − f | dµ < 2 X

(1.43)

for all n ≥ N . We fix this N . Since 0 ≤ |f | ≤ |f −fN |+|fN | by the triangle inequalityi , we apply Proposition 1.24(a) and then Theorem 1.27 and the property (1.38) to the inequality (1.43) to derive Z Z Z Z ǫ ǫ |fN | dµ < + |f − fN | dµ + N dµ = + N µ(E) |f | dµ ≤ (1.44) 2 2 E E E E ǫ for every E ∈ M. Therefore, if we take δ = 2N and µ(E) < δ, then our inequality (1.44) becomes Z |f | dµ < ǫ E

as desired. Hence we complete the proof of the problem. i

See [49, Definition 2.15, p. 30].



15

1.2. Problems related to the Lebesgue’s MCT/DCT Problem 1.13

Rudin Chapter 1 Exercise 13.

Proof. Let f : X → [0, ∞] ⊂ [−∞, ∞] be a measurable function. For each n = 1, 2, . . ., we define fn : X → [0, ∞] ⊂ [−∞, ∞] by fn (x) = nf (x). It is clear that n α  α io i
fn−1 (α, ∞] = {x ∈ X | fn (x) ∈ (α, ∞]} = x ∈ X f (x) ∈ , ∞ = f −1 ,∞ n n
for all real α. Since f is measurable, we have f −1 ( αn , ∞] ∈ M for every real α, where M is a σalgebra in X. Thus we obtain from Theorem 1.12(c) that each fn is measurable for n = 1, 2, . . .. Furthermore, we have 0 ≤ f1 (x) ≤ f2 (x) ≤ · · · ≤ ∞ on X and fn (x) → ∞ · f (x) as n → ∞. By Proposition 1.24(c), we have Z Z Z f dµ nf dµ = n fn dµ = X

(1.45)

X

X

for n = 1, 2, . . .. Hence we conclude from the equality (1.45) and Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem) that Z Z Z Z f dµ = ∞ · fn dµ = lim n f dµ. ∞ · f dµ = lim X

n→∞ X

n→∞

X

X

Thus Proposition 1.24(c) also holds when c = ∞ and so we complete the proof of the problem. 

Chapter 1. Abstract Integration

16

CHAPTER

2

Positive Borel Measures

2.1

Properties of Semicontinuity

Problem 2.1 Rudin Chapter 2 Exercise 1.

Proof. We have fn : R → [0, ∞) for all n ∈ N. In Proposition 2.5, we will prove a property which is equivalent to Definition 2.8 and our proof of Statement (a) below becomes simpler. However, we choose to apply Definition 2.8 to prove the statements here. • Statement (a): For any real α, β1 and β2 , let E(α) = {x ∈ R | f1 (x) + f2 (x) < α}

and

Fi (βi ) = {x ∈ R | fi (x) < βi },

(2.1)

where i = 1, 2. If α ≤ 0, then we see that E(α) = ∅ which is open in R. Similarly, Fi (βi ) = ∅ if βi ≤ 0. So in the following discussion, we assume that α > 0, βi > 0 and furthermore, E(α) 6= ∅ and Fi (βi ) 6= ∅. Now we claim that

[

E(α) =

β1 +β2 ≤α β1 ,β2 >0

To prove the claim, on the one hand, if [ x∈

[F1 (β1 ) ∩ F2 (β2 )].

(2.2)

[F1 (β1 ) ∩ F2 (β2 )],

β1 +β2 ≤α β1 ,β2 >0

then there exist real β1 and β2 with β1 + β2 ≤ α and β1 , β2 > 0 so that x ∈ F1 (β1 )∩ F2 (β2 ). By the definition (2.1), this x satisfies f1 (x) < β1 and f2 (x) < β2 and their sum implies f1 (x) + f2 (x) < β1 + β2 ≤ α. Thus we have x ∈ E(α), i.e.,

[

[F1 (β1 ) ∩ F2 (β2 )] ⊆ E(α).

β1 +β2 ≤α β1 ,β2 >0

17

Chapter 2. Positive Borel Measures

18

On the other hand, if x ∈ E(α), then we let η = f1 (x) + f2 (x). Define the two numbers β1 and β2 by η+α η+α β1 = − f2 (x) and β2 = − f1 (x). 2 2 By direct computation, we know that βi >

f1 (x) + f2 (x) + f1 (x) + f2 (x) − fi (x) = f1 (x) + f2 (x) − fi (x) ≥ 0, 2

where i = 1, 2. Since η = f1 (x) + f2 (x), it is easy to check that f1 (x) + f2 (x) < α if and only if [f1 (x) + f2 (x)] + [f1 (x) + f2 (x)] < α + η if and only if f1 (x) + f2 (x) < η+α 2 if and only if f1 (x) < β1 . (2.3) Similarly, the above argument can be used to show that f1 (x) + f2 (x) < α if and only if f2 (x) < β2 .

(2.4)

Now we deduce from the inequalities (2.3) and (2.4) that x ∈ F1 (β1 )∩F2 (β2 ). Furthermore, since β1 + β2 = η + α − f1 (x) − f2 (x) = α, we have [ [F1 (β1 ) ∩ F2 (β2 )], x∈ β1 +β2 ≤α β1 ,β2 >0

i.e., E(α) ⊆

[

β1 +β2 ≤α β1 ,β2 >0

[F1 (β1 ) ∩ F2 (β2 )].

Hence the claim (2.2) holds. Since F1 (β1 ) and F2 (β2 ) are open in R, F1 (β1 ) ∩ F2 (β2 ) is also open in R. Since the union of any collection of open sets is open ([49, Theorem 2.24, p. 34]), we follow from this and the equality (2.2) that E(α) is open in R. By Definition 2.8, f1 + f2 is upper semicontinuous. • Statement (b): By a similar argument as in part (a), we can show that the sum of two lower semicontinuous functions is lower semicontinuous. • Statement (c): This is not true in general. We use Proposition A.7 to give a counterex1 , n1 ]. Then each χFn is upper semicontinuous because ample. For each n ∈ N, let Fn = [ n+1 Fn is closed in R. Consider the set ∞ X 1o χFn (x) < E = x ∈ R . 2 n=1

n

If x ∈ Fn for some n ∈ N, then we have 1≤

∞ X

n=1

χFn (x) ≤ 2.

In other words, we have Fn * E for n = 1, 2, . . .. Since

(2.5) ∞ [

Fn = (0, 1],a we see that

n=1

(0, 1] * E. a

If x ∈ (0, 1], then there exists a positive integer k such that

1 k+1

(2.6) < x which implies that x ∈ F1 ∪ F2 ∪ · · · ∪ Fk .

19

2.1. Properties of Semicontinuity However, if x ≤ 0 or x > 1, then x ∈ / Fn for every n ∈ N which means that χFn (x) = 0, i.e., (−∞, 0] ∪ (1, ∞) ⊆ E. (2.7) Hence, by combining the set relations (2.6) and (2.7), we conclude that E = (−∞, 0] ∪ (1, ∞) which is not open in R. By Definition 2.8,

∞ X

χFn is not upper semicontinuous.

n=1

• Statement (d): The set

Fn (α) = {x ∈ R | fn (x) > α}

is open for every real α. By applying part (b) repeatedly, we know that

N X

fn is lower

n=1

semicontinuous for every positive integer N . For every x ∈ [0, ∞), let f (x) = lim

N →∞

N X

fn (x) =

∞ X

fn (x),

n=1

n=1

E(α) = {x ∈ R | f (x) > α}, N X o n fn (x) > α FN (α) = x ∈ R

(2.8)

n=1

for real α and N ∈ N. We claim that

∞ [

E(α) =

FN (α)

(2.9)

N =1

for every real α. Similar to the proof of part (a), we suppose that α > 0, E(α) 6= R and FN (α) 6= R. Suppose that x ∈ E(α), i.e., f (x) > α. Then there exists a ǫ > 0 such that f (x) > α + ǫ. Since each fn is nonnegative,

N nX

n=1

o fn is an increasing sequence. By this and the definition ′

of f in (2.8), there exists a positive integer

N′

such that

N X

fn (x) > α + ǫ which implies

n=1

that x ∈ FN ′ (α), i.e., E(α) ⊆ To prove the other side, if x ∈ ′

N ′ , i.e., that

N X

∞ [

N =1

∞ [

FN (α).

(2.10)

N =1

FN (α), then x ∈ FN ′ (α) for some positive integer

fn (x) > α. Again, the fact that

N nX

n=1

n=1

fn

o

N′

f (x) ≥

X

n=1

fn (x) > α,

is an increasing sequence implies

Chapter 2. Positive Borel Measures i.e., x ∈ E(α) and then

20

∞ [

N =1

FN (α) ⊆ E(α).

(2.11)

Hence the set relations (2.10) and (2.11) definitely imply the claim (2.9) is true and since each FN (α) is open in R by Definition 2.8, E(α) is also open in R. Since α is arbitrary, f is lower semicontinuous by Definition 2.8. In the proof of Statements (a) and (b) above, we don’t use the property that f1 and f2 are nonnegative. Therefore, they remain valid even if the word “nonnegative” is omitted. However, Statement (c) cannot hold anymore if the word “nonnegative” is omitted. In fact, we consider the sequence of real functions {fn } defined by f1 = χ[−1,1]

and

fn = −χ[ 1 ,

1 ] n n−1

(n = 2, 3, . . .).

1 Since [−1, 1 and [ n1 , n−1 ] are closed in R, we know that f1 , f2 , . . . are upper semicontinuous. It is clear that f1 (0) = −1, so f1 is not a nonnegative function on R. By definition, we have

f=

∞ X

n=1

fn = χ[−1,1] −

∞ X

n=2

χ[ 1 ,

1 ] n n−1

= χ[−1,0] + χ(0,1] −

∞ X

χ[

n=1

1 ,1] n+1 n

.

(2.12)

By the inequalities (2.5), we see that  −2, if x = 21 , 13 , . . .;     ∞  X χ[ 1 , 1 ] (x) = − −1, if x ∈ (0, 1] \ { 21 , 13 , . . .}; n+1 n   n=1    0, if x ≤ 0 or x > 1.

Thus we follow from the expressions (2.12) and (2.13) that  1 1   −1, if x = 2 , 3 , . . .;    f (x) = 0, if x ∈ (0, 1] \ { 12 , 13 , . . .} or x < −1 or x > 1;      1, if x ∈ [−1, 0].

(2.13)

(2.14)

Therefore, the expression (2.14) of f gives n 1o E = x ∈ R f (x) > = [−1, 0] 2

which is not open in R. By Definition 2.8, f is not upper semicontinuous. By Definition 2.8, a function f is lower semicontinuous if and only if −f is upper semicontinuous. This observation indicates that we can deduce a counterexample to Statement (d) from the counterexample (2.12).

Finally, the truths of the Statements (a), (b) and (d) depend only on the range of f and the fact that the union of any collection of open sets in a topological space X is open in X (see  the set equalities (2.2) and (2.9)). This completes the proof of the problem. Problem 2.2 Rudin Chapter 2 Exercise 2.

21

2.1. Properties of Semicontinuity

Proof. We formulate and prove the general statement: Let X be a topological space, U an open set in X containing the point x and f : X → C. Define ϕ(x, U ) = sup{|f (s) − f (t)| | s, t ∈ U } and ϕ(x) = inf{ϕ(x, U ) | U is open, x ∈ U }. (2.15) We claim that ϕ is upper semicontinuous, f is continuous at x ∈ X if and only if ϕ(x) = 0 and the set of points of continuity of an arbitrary complex function is a Gδ . • ϕ is upper semicontinuous. Let E = {x ∈ X | ϕ(x) < α}, where α ∈ R. We have to show that E is open in X. If E = ∅, then there is nothing to prove. Thus we suppose that E 6= ∅. In this case, we have p ∈ E so that ϕ(p) < α. This fact shows that there exists an open set U containing p andb ϕ(p, U ) < α. Pick q ∈ U \ {p}. Since U is open, there exists an open set V containing q such that q ∈ V ⊆ U . Since p, q ∈ U , we know from the definition (2.15) that ϕ(p, U ) = ϕ(q, U ).

(2.16)

Furthermore, we observe from the definition (2.15) that ϕ(x, U ′ ) ≤ ϕ(x, U ) for every open sets U, U ′ with U ′ ⊆ U .

(2.17)

Now these facts (2.16) and (2.17) imply that ϕ(q, V ) ≤ ϕ(q, U ) = ϕ(p, U ) < α and then ϕ(q) ≤ ϕ(q, V ) < α. In other words, q ∈ E. Since q ∈ U \ {p} is arbitrary, we have shown that p ∈ U ⊆ E, i.e., E is open for every α ∈ R and hence ϕ is upper semicontinuous by Definition 2.8. • f is continuous at x if and only if ϕ(x) = 0. Suppose that f is continuous at x. Recall from the definition of continuity ([42, Theorem 18.1, p. 104] or [51, p. 9]) that for every ǫ > 0, the neighborhood n ǫo (2.18) B(f (x), ǫ) = z ∈ C |f (x) − z| < 2

has a neighborhood Uǫ of x such that f (Uǫ ) ⊆ B(f (x), ǫ), i.e., f (y) ∈ B(f (x), ǫ) for all y ∈ Uǫ . Now we take this Uǫ in the definition (2.15): ϕ(x, Uǫ ) = sup{|f (s) − f (t)| | s, t ∈ Uǫ } and it follows from the definition (2.18) thatc ϕ(x, Uǫ ) ≤ ǫ.

(2.19)

By the definition (2.15), we have ϕ(x) ≤ ϕ(x, U ) for every open set U containing x. Therefore, we deduce from this and the inequality (2.19) that ϕ(x) ≤ ǫ. Since ǫ is arbitrary, we have the desired result that ϕ(x) = 0. b

Otherwise, we have ϕ(p, U ) ≥ α for every open set U containing p and this implies that α ≤ ϕ(p), a contradiction. c Note that |f (s) − f (t)| ≤ |f (s) − f (x)| + |f (x) − f (t)| < ǫ for every s, t ∈ Uǫ .

Chapter 2. Positive Borel Measures

22

Conversely, suppose that ϕ(x) = 0. Thus for every ǫ > 0, there exists a neighborhood Uǫ of x such that ϕ(x, Uǫ ) < ǫ. By the definition (2.15), this implies that |f (s) − f (t)| < ǫ

(2.20)

for every s, t ∈ Uǫ . In particular, if we take s = x to be fixed and t = y vary, then the inequality (2.20) can be rewritten as |f (x) − f (y)| < ǫ for every y ∈ Uǫ . By the definition ([42, Theorem 18.1, p. 104] or [51, p. 9]), f is continuous at x. • The set of points of continuity of an arbitrary complex function is a Gδ . By the previous analysis, we establish that G = {x ∈ X | f is continuous at x} = {x ∈ X | ϕ(x) = 0} =

∞ n \

n=1

1o x ∈ X ϕ(x) < . n

Since ϕ is upper semicontinuous, each set {x ∈ X | ϕ(x) < n1 } is open in X. By Definition 1.11, we see that G is actually a Gδ . This completes the proof of the problem.



Problem 2.3 Rudin Chapter 2 Exercise 3.

Proof. The first part is proven in [63, Problem 4.20, pp. 73, 74]. The function in the question can be used to prove Urysohn’s Lemma for any metric space X directly. Lemma 2.1 (Urysohn’s Lemma) Suppose that X is a metric space with metric ρ, A and B are disjoint nonempty closed subsets of X. Then there exists a continuous function f : X → [0, 1] such that 0 ≤ f (x) ≤ 1 for all x ∈ X, f (x) = 0 precisely on A and f (x) = 1 precisely on B.

Proof of Lemma 2.1. This lemma is proven in [63, Problem 4.22, pp. 75, 76]. Readers are recommended to read [42, Theorem 32.2, p. 202; Theorem 33.1, pp. 207 – 210].  This ends the proof of the problem.



Problem 2.4 Rudin Chapter 2 Exercise 4.

Proof. (a) Recall from [51, Eqn. (2), p. 42] that µ(E) = inf{µ(V ) | E ⊆ V and V is open}

(2.21)

which is defined for every subset E of X.d By the result [51, Eqn. (4), p. 42], we have µ(E1 ∪ E2 ) ≤ µ(E1 ) + µ(E2 ). d

Here we don’t require E to be a measurable set.

(2.22)

23

2.1. Properties of Semicontinuity We need a lemma: Lemma 2.2 For disjoint open sets V1 and V2 , we have µ(V1 ∪ V2 ) = µ(V1 ) + µ(V2 ).

Proof of Lemma 2.2. The proof of the inequality µ(V1 ∪ V2 ) ≤ µ(V1 ) + µ(V2 ) can be found in [51, STEP I, p. 42], so we only prove the other side. Suppose that g ∈ Cc (X) and g ≺ V1 . Similarly, suppose that h ∈ Cc (X) and h ≺ V2 . By Definition 2.9, since Cc (X) is a vector space, f = g + h ∈ Cc (X). Furthermore, we know from Definition 2.9(a) that supp (f ) = supp (g + h) ⊆ supp (g) + supp (h) ⊆ V1 ∪ V2 .

(2.23)

By assumption, we have V1 ∩V2 = ∅ which means that g(x) = 0 on X \V1 and h(x) = 0 on X \ V2 . Thus we deduce from these facts that  g(x), if x ∈ V1 ;      f (x) = h(x), if x ∈ V2 ;      0, if x ∈ X \ (V1 ∪ V2 ).

In other words, we have

0 ≤ f (x) ≤ 1

(2.24)

on X. Therefore, we can conclude from the set relation (2.23) and the inequalities (2.24) that f ≺ V1 ∪ V2 . (2.25) Now, by Definition 2.1, the relation (2.25) and then [51, Eqn. (1), p.41], we obtain that Λ(g) + Λ(h) = Λ(g + h) = Λ(f ) ≤ µ(V1 ∪ V2 ). (2.26) We first fix the h in the inequality (2.26) and since it holds for every g ≺ V1 , the definition of supremum gives µ(V1 ) + Λ(h) ≤ µ(V1 ∪ V2 ).

(2.27)

Next, the inequality (2.27) holds for every h ≺ V2 , we establish from the definition of supremum that µ(V1 ) + µ(V2 ) ≤ µ(V1 ∪ V2 ). Hence we have µ(V1 ) + µ(V2 ) = µ(V1 ∪ V2 ).



Let’s go back to the original proof. We want to compute µ(E1 ∪ E2 ). By the definition (2.21), it needs to consider all open sets containing E1 ∪ E2 . Since V1 ⊆ V2 certainly implies µ(V1 ) ≤ µ(V2 ), we can restrict our attention to any open set W such that E1 ∪ E2 ⊆ W ⊆ V1 ∪ V2 . Since V1 ∩ V2 = ∅, we have W = (W ∩ V1 ) ∪ (W ∩ V2 ), where (W ∩ V1 ) ∩ (W ∩ V2 ) = ∅.

Chapter 2. Positive Borel Measures

24

Furthermore, since W ∩ V1 and W ∩ V2 are open sets in X, it yields from Lemma 2.2 that µ(W ) = µ(W ∩ V1 ) + µ(W ∩ V2 ) ≥ µ(E1 ) + µ(E2 ).

(2.28)

Since W is arbitrary, µ(E1 ) + µ(E2 ) is a lower bound of the set in the definition (2.21). Hence we follow from this fact and the inequality (2.28) that µ(E1 ∪ E2 ) ≥ µ(E1 ) + µ(E2 ).

(2.29)

By the inequalities (2.22) and (2.29), we have the desired result that µ(E1 ∪ E2 ) = µ(E1 ) + µ(E2 ). (b) Define E1 = E. By [51, STEP V, p. 44], there exists a compact set K1 and an open set V1 such that 1 K1 ⊆ E1 ⊆ V1 and µ(V1 \ K1 ) < 2 . 2 By [51, STEP II, p. 43], we have K1 ∈ MF . Then we follow from [51, STEP VI, p. 44] that E1 \ K1 ∈ MF . Define E2 = E1 \ K1 so that E = E1 = E2 ∪ K1 , where µ(E2 ) = µ(E1 \ K1 ) ≤ µ(V1 \ K1 ) < 212 . By induction, we can show that there exists a sequence of compact sets {Kn } and a sequence of open sets {Vn } such that Kn ⊆ En ⊆ Vn , where En+1 = En \ Kn , En+1 ∈ MF and µ(En+1 ) < the following relations: E = En+1 ∪ K1 ∪ K2 ∪ · · · ∪ Kn

(2.30) 1 . 2n+1

Therefore, the set E satisfies

and µ(En+1 )
0. 2k+1 2

Chapter 2. Positive Borel Measures

26

For the second assertion, let v be lower semicontinuous and v ≤ χK . Assume that there was a p ∈ R such that v(p) > 0. Since v is lower semicontinuous, Definition 2.8 implies that the set A = {x ∈ R | v(x) > 0} is open in R. By the assumption, we have p ∈ A, so there exists a δ > 0 such that (p − δ, p + δ) ⊆ A. (2.34) Now we follow from the hypothesis v ≤ χK and the relation (2.34) that χK (x) > 0 on (p − δ, p + δ), therefore we have (p − δ, p + δ) ∈ K which contradicts to the fact that K contains no segment. Hence we have v ≤ 0 and this completes the proof of the problem.  Problem 2.7 Rudin Chapter 2 Exercise 7.

Proof. By the idea used in Problem 2.6, it is not hard to see that the construction of the compact ǫ sets Kn and K still work for removing the “middle 22n−1 th” segments. In this case, instead of the Lebesgue measure (2.33), we have m(Kn ) = 1 − and so

n X k=1

ǫ 22k−1

k−1

·2

n X 1 =1−ǫ 2k

(2.35)

k=1

m(K) = lim m(Kn ) = 1 − ǫ. n→∞

Since K is closed in [0, 1], the complement E = [0, 1] \ K

(2.36)

is definitely open in [0, 1]. Now it remains to show that E is dense in [0, 1]. We prove an equivalent definition of a dense set: Let A ⊆ B ⊆ R. We say A is dense in B means that for every x, y ∈ B with x < y, we can find z ∈ A such that x < z < y. In particular, suppose that x, y ∈ [0, 1] with x < y. If (x, y) ∩ E = ∅, then the definition (2.36) says that (x, y) ⊆ K which is impossible. Therefore, we have (x, y) ∩ E 6= ∅ and this means that there exists z ∈ E such that x < z < y. Hence E is dense in [0, 1],  completing the proof of the problem. Problem 2.8 Rudin Chapter 2 Exercise 8.

Proof. Let {rn } be the enumeration of all rationals in R. Let Fn be the segment given by  1 1  Fn = rn − 2n , rn + 2n , (2.37) 2 2 where n = 1, 2, . . .. Define the sets Gn = Fn \ (Fn+1 ∪ Fn+2 ∪ · · · ) = Fn \ We divide the proof into several steps.

∞ [

k=1

Fn+k .

(2.38)

2.2. Problems on the Lebesgue Measure on R

27

• Step 1: m(Gn ) > 0. By the definition (2.37), we know that m(Fn+1 ) = n = 1, 2, . . .. By the definition (2.38), for every n ∈ N, we have Fn = Gn ∪

∞ [

m(Fn ) 22

for every

Fn+k

k=1

so that the subadditive property of a measure (see the proof of Problem 1.11) implies that m(Fn ) ≤ m(Gn ) +

∞ X

m(Fn+k ) = m(Gn ) + m(Fn )

k=1

∞ X 1 m(Fn ) = m(Gn ) + . 22k 3 k=1

Therefore, we have m(Gn ) ≥

2m(Fn ) >0 3

(2.39)

for every positive integer n. n) • Step 2: Existence of a Borel subset An ⊂ Fn with m(An ) = m(F 2 . It is easy to 1 . By Definition 2.19, we know thate check from the definition (2.37) that m(Fn ) = 22n−1

Fn′ =

 1 h 1 i Fn − rn − 2n = (0, 1). m(Fn ) 2

Thus, by the proof of Problem 2.7f , there exists a (totally disconnected) compact set A′n ⊂ Fn′ such that m(A′n ) = 21 . Then Theorem 2.20(c) implies that the set An given by An =

m(Fn )A′n

 1  + rn − 2n 2

is a (totally disconnected) compact subset of Fn with measure  
1  m(Fn ) ′ m(An ) = m m(Fn )An + rn − 2n . (2.40) = m m(Fn )A′n = m(Fn )m(A′n ) = 2 2

Since An is compact, it is closed by the Heine–Borel theorem [49, Theorem 2.41, p. 40]. By Definition 1.11, it is also a Borel set, completing the proof of Step 2. Now we can construct a Borel set with the desired properties. The construction is as follows: For every positive integer n, we define En = Gn ∩ An

and E =

∞ [

En .

(2.41)

n=1

By the definition (2.38), since each Fn is Borel, each Gn is also Borel. Thus each En and their countable union E are also Borel sets. It remains to show that the E satisfies the requirements of the problem. • Step 3: 0 < m(E ∩ Fn ). We first show that m(En ∩ Fn ) > 0 for every positive integer n. To prove this claim, we note from the definitions (2.38) and (2.41) that En ⊆ Gn ⊆ Fn , so we have m(En ∩ Fn ) = m(En ) = m(Gn ∩ An ). Furthermore, since Gn ⊆ Fn and An ⊂ Fn (by Step 2), we have m(Gn ∪ An ) ≤ m(Fn ). e f

It is just a translation and an enlargement. Take ǫ = 12 in the measure (2.35).

(2.42)

Chapter 2. Positive Borel Measures

28

Combining the inequality (2.39) and the expression (2.40), we can reduce the inequality (2.42) to m(Gn ∪ An ) < m(Gn ) + m(An ). (2.43) Now we apply a property of additive function [49, Eqn. (7), p. 302] to the inequality (2.43) to obtain m(Gn ∪ An ) < m(Gn ∪ An ) + m(Gn ∩ An ) which means that m(Gn ∩ An ) > 0 for every positive integer n. Therefore, we follow from this and the fact that En ∩ Fn ⊆ E ∩ Fn 0 < m(Gn ∩ An ) = m(En ∩ Fn ) ≤ m(E ∩ Fn ) for every positive integer n. • Step 4: m(E ∩ Fn ) < m(Fn ). Fix an integer n. For m < n, we apply the identity A \ B = B c ∩ A to the definition (2.38) to get c Gcm = [Fm \ (Fm+1 ∪ Fm+2 ∪ · · · )]c = (Fm+1 ∪ Fm+2 ∪ · · · ) ∪ Fm

which implies that Fn ⊆ Gcm for m = 1, 2, . . . , n − 1. Recall from the definition (2.41) that Enc = Gcn ∪ Acn , so we must have c Fn ⊆ Em for m = 1, 2, . . . , n − 1. In other words, it says that Fn ∩ Em = ∅ for m = 1, 2, . . . , n − 1. Therefore, this fact implies m(E ∩ Fn ) = m

∞  [

m=1

∞ ∞  [   [

 Am ∩ Fn . Em ∩ Fn ≤ m Em ∩ Fn = m

(2.44)

m=n

m=n

By the subadditive property of a measure again, we further deduce the inequality (2.44) to ∞ ∞ X X m(Am ). (2.45) m(Am ∩ Fn ) ≤ m(E ∩ Fn ) ≤ m=n

Recall the result (2.40) and the fact m(Fn+1 ) = (2.45) that

m=n

m(Fn ) , 22

so we derive from the inequality

∞ ∞ X m(Fn ) X 1 2m(Fn ) m(Fn+m ) = < m(Fn ) = m(E ∩ Fn ) ≤ 2m 2 2 2 3 m=0 m=0

for every positive integer n. • Step 5: 0 < m(E ∩ I) < m(I) for every nonempty segment I. By Step 3 and Step 4, the inequalities 0 < m(E ∩ Fn ) < m(Fn ) (2.46) hold for every positive integer n, where Fn and E are given by (2.37) and (2.41) respectively. Lemma 2.3 Let I = (α, β) with α < β. Then there is a positive integer n0 such that Fn0 ⊆ I.

2.2. Problems on the Lebesgue Measure on R

29

γ+β Proof of Lemma 2.3. Let γ = α+β 2 and δ = 2 . By the density of Q in R, there exists a rational in (γ, δ). Let it be rn for some positive integer n. If Fn ⊆ I, then we are done. Otherwise, we may find another rational in (rn , δ), namely rn+1 . In fact, this process can be repeated m times, i.e., there is a rational rn+m in (rn+m−1 , δ). It is clear that 1 1 1 (2.47) rn+m + 2(n+m) ≤ δ + 2(n+m) < δ + 2m 2 2 2 and 1 1 1 rn+m − 2(n+m) > γ − 2(n+m) > γ − 2m . (2.48) 2 2 2 Now we may pick the m large enough such that

1

2 2 2 22(n+m) Hence the inequalities (2.49) and (2.50) together imply that Fn0 ⊆ I, where n0 = n+m, completing the proof of the lemma.  rn+m −

We return to the proof of the problem. By Lemma 2.3, we have E ∩ Fn0 ⊆ E ∩ I and the left-hand side of the inequalities (2.46) shows that 0 < m(E ∩ Fn0 ) ≤ m(E ∩ I).

(2.51)

Since I = Fn0 ∪ (I \ Fn0 ) and Fn0 ∩ (I \ Fn0 ) = ∅, we have E ∩ I = (E ∩ Fn0 ) ∪ [E ∩ (I \ Fn0 )]. Hence, by applying the right-hand side of the inequalities (2.46) and Theorem 1.19(b) twice, we obtain
m(E ∩ I) = m(E ∩ Fn0 ) + m E ∩ (I \ Fn0 ) < m(Fn0 ) + m(I \ Fn0 ) = m(I).

(2.52)

Now our desired result follows immediately if we combine the inequalities (2.51) and (2.52). • Step 6: m(E) < ∞. By the subadditive property of the measure m, the expression (2.40) and the definition (2.41), we get m(E) ≤

∞ X

n=1

m(En ) ≤

∞ X

m(An ) =

n=1

This completes the proof of the problem.g g

∞

∞

n=1

n=0

1X m(F1 ) X 1 2m(F1 ) m(Fn ) = = < ∞. 2n 2 2 2 3 

Instead of Borel sets of the real number line R, Rudin proved a similar result for measurable set A in [0, 1], see [50]. Furthermore, there are two interesting results ([14] and [35]) related to this problem and some of its applications can be found in [13], [18], [25], [39], [46] and [60].

Chapter 2. Positive Borel Measures

2.3

30

Integration of Sequences of Continuous Functions

Problem 2.9 Rudin Chapter 2 Exercise 9.

Proof. For every n ∈ N, we consider the function gn : [−1, 1] → [0, 1] defined by gn (x) =

  0, 

if x 6∈ [− n1 , n1 ];

(2.53)

1 − n|x|, if x ∈ [− n1 , n1 ].

It is clear that each gn is continuous on [−1, 1] and 0 ≤ gn ≤ 1 on [−1, 1]. The graph of gn is shown in Figure 2.1 below.

Figure 2.1: The graph of gn on [−1, 1]. Next, we define gn,k : [0, 1] → [0, 1] by  k  gn,k (x) = gn x − , 2n

(2.54)

k where x ∈ [0, 1] and k = 0, 1, . . . , 2n. It is easy to check that x − 2n ∈ [−1, 1] so that each gn,k is well-defined by (2.54). We claim that if we define the sequence {f1 , f2 , . . .} to be

{g1,0 , g1,1 , g1,2 , g2,0 , g2,1 , g2,2 , g2,3 , g2,4 , . . .}, then {fn } satisfies the hypotheses of the problem.

To this end, we first note that since each gn,k is continuous on [0, 1] and 0 ≤ gn,k ≤ 1 on [0, 1], each fn is continuous on [0, 1] and 0 ≤ fn ≤ 1. To see why lim

Z

1

n→∞ 0

we have to check the behaviour of

Z

0

fn (x) dx = 0,

1

gn,k (x) dx

(2.55)

31

2.3. Integration of Sequences of Continuous Functions

for every positive integer n and k = 0, 1, . . . , 2n. In fact, we know from the definitions (2.53) and (2.54) that  1 1  2 + nx, if x ∈ [0, 2n );    1    1 − n|x|, if x ∈ [0, n ); 3 1 3 (2.56) gn,0 (x) = and gn,1 (x) = 2 − nx, if x ∈ [ 2n , 2n ];    0, if x ∈ [ n1 , 1]    3 , 1]. 0, if x ∈ ( 2n

Similarly, we have

gn,2n−1 (x) =

and

   0,

  1 − n x −

gn,2n (x) =

  0, 



if x ∈ [0, 2n−3 2n );

2n−1 2n ,

(2.57)

if x ∈ [ 2n−3 2n , 1]

if x ∈ [0, n−1 n );

(2.58)

1 + n(x − 1), if x ∈ [ n−1 n , 1].

Finally, for k = 2, 3, . . . , 2n − 2, we have  0,       1 − n x − gn,k (x) =       0,



k 2n ,

if x ∈ [0, k−2 2n ); k+2 if x ∈ [ k−2 2n , 2n ];

(2.59)

if x ∈ ( k+2 2n , 1].

The graphs of these gn,k are shown as follows:

Figure 2.2: The graphs of gn,k on [0, 1]. In other words, it is clear from Figure 2.2 that {gn,0 , gn,1 . . . , gn,2n } is a family of tent functions 1 0 h , 2n , . . . , 2n centered at { 2n 2n }. Now we have the following cases. h

We use the fact that each gn,k is a part or the whole of an isosceles triangle with height 1 and base less than or equal to n2 .

Chapter 2. Positive Borel Measures • When k = 0, we have Z

1

gn,0 (x) dx = 0

• When k = 1, we have Z

1

gn,1 (x) dx =

Z

1 2n

0

0

• When 2 ≤ k ≤ 2n − 2, we have Z 1

Z

32

1

gn (x) dx = 0

1

2



+ nx dx +

gn,k (x) dx =

• When k = 2n − 1, we have Z Z 1 gn,2n−1 (x) dx = • When k = 2n, we have Z Z 1 gn,2n (x) dx = 0

1 2n−3 2n



Z

(1 − nx) dx =

3 2n 1 2n

1 . 2n

(2.60)

 7 − nx dx = . 2 8n

3

1 2 1 × ×1= . 2 n n

gn (x − 1) dx =

Z

1 1 1− n

(2.61)

(2.62)

2n − 1  7 . 1 − n x − dx = 2n 8n

1 1 1− n

1 n

0

0

0

Z

[1 + n(x − 1)] dx =

(2.63)

1 . 2n

(2.64)

Combining the expressions (2.60) to (2.64), we may conclude that the limit (2.55) holds. Now it remains to show that {fn (x)} converges for no x ∈ [0, 1]. If x = 0, then we obtain from the definitions (2.56) and (2.57) that gn,0 (0) = 1 and

gn,2n−1 (0) = 0.

In other words, we can find subsequences {fnk (0)} and {fnl (0)} such that fnk (0) → 1 and fnl (0) → 0 as k → ∞ and l → ∞ respectively. By Definition 1.13, they imply that lim sup fn (0) = 1 and n→∞

lim inf fn (0) = 0.

(2.65)

n→∞

Similarly, if x = 1, then the definitions (2.56) and (2.58) show that gn,0 (1) = 0 and gn,2n (1) = 1 which then imply the limits (2.65). If x ∈ (0, 1) ∩ Q, then we have x = Thus it follows from the definitions (2.56) and (2.59) that p p = 0 and gnq,2np =1 gn,0 q q

p q

=

2p 2q ,

where 0 < p < q.

(2.66)

for n ≥ q. Therefore, the limits (2.65) also hold for x ∈ (0, 1) ∩ Q. For irrational x in (0, 1), the Archimedean Property shows that there exists N ∈ N such that x > N1 . Thus it implies that gn,0 (x) = 0

(2.67)

for all n ≥ N . By the density of rationals, there exists a sequence { pqnn } of rationals in (0, 1) such that pn →x qn

33

2.3. Integration of Sequences of Continuous Functions

as n → ∞, where 0 < pn < qn for all n ∈ N. Since we always have pn h npn − 1 npn + 1 i , ∈ , qn nqn nqn we derive from the second expression (2.66) that p  n gnqn ,2npn =1 qn

(2.68)

for all n ∈ N. Combining the two results (2.67) and (2.68), we have shown that the limits (2.65) remain valid for irrationals x, completing the proof of the problem.  Problem 2.10 Rudin Chapter 2 Exercise 10.

Proof. Given that ǫ > 0. For each positive integer n, we define En = {x ∈ [0, 1] | fk (x) > ǫ for some k ≥ n}.

(2.69)

It is clear that each En is bounded and En ⊇ En+1 ⊇ · · · . If En0 = ∅ for some positive integer n0 , then we know from the definition (2.69) that fk (x) ≤ ǫ for every x ∈ [0, 1] and every positive integer k ≥ n0 . Therefore it implies that 0≤

Z

0

1

fk (x) dx ≤ ǫ

for every positive integer k ≥ n0 . Since ǫ is arbitrary, we obtain from this that Z 1 fn (x) dx = 0. lim n→∞ 0

Without loss of generality, we assume that En 6= ∅ for all n ∈ N. Let’s prove two properties of En first: • Property 1: En is open. Let p ∈ En . Then we have fk (p) > ǫ for some k ≥ n. Define ǫ′ = 12 (fk (p) − ǫ) > 0. Since fk is continuous on [0, 1], it is continuous at p. Thus for this particular ǫ′ , there is a δ > 0 such that |fk (x) − fk (p)| < ǫ′ for all points x ∈ [0, 1] with |x − p| < δ. Now the inequality (2.70) implies that fk (x) > fk (p) − ǫ′ =

fk (p) + ǫ > ǫ. 2

In other words, we have (p − δ, p + δ) ⊆ En as desired.

(2.70)

Chapter 2. Positive Borel Measures

• Property 2:

∞ \

n=1

34

En = ∅. Otherwise, there was a p ∈ [0, 1] such that p ∈ En for all

n ∈ N. Therefore, for each positive integer n, we have fk (p) > ǫ

(2.71)

for some k ≥ n. If n → ∞, then k → ∞ and the inequality (2.71) implies that 0 = lim fn (p) = lim fk (p) ≥ ǫ > 0, n→∞

k→∞

a contradiction. To finish our proof, we have to study the lengths of certain subsets of En . Let F be a finite union of bounded (open or closed) intervals of [0, 1]. Then we may write F =

m [

[ak , bk ] ∪

k=1

s [

(ar , br )

(2.72)

r=1

where 0 ≤ ak < bk ≤ 1 and 0 ≤ ar < br ≤ 1 for k = 1, 2, . . . , m and r = 1, 2, . . . , s. Define the length of F , denoted by ℓ(F ), to be ℓ(F ) =

m X k=1

m s s X
X
X (br − ar ). ℓ (ar , br ) = (bk − ak ) + ℓ [ak , bk ] + r=1

k=1

r=1

Given a nonempty finite union of bounded interval F expressed in the form (2.72) and assume ǫ that F has at least one open interval. In addition, we suppose that ℓ(F ) > ǫ. Let δ = 4s . Then the set m s [ [ [ar + δ, br − δ] G= [ak , bk ] ∪ r=1

k=1

is clearly a nonempty finite union of bounded and closed intervals of [0, 1]. Now it is easy to see that m s X X ǫ (br − ar − 2δ) = ℓ(F ) − 2sδ = ℓ(F ) − ℓ(G) = (bk − ak ) + 2 r=1 k=1

implying that ℓ(G) > ℓ(F ) − ǫ. Here we need a lemma about the sequence of bounded subsets (2.69): Lemma 2.4 Let Sn = {F | F is a finite union of bounded intervals of En } for every positive integer n and Ln = sup{ℓ(F ) | F ∈ Sn }. (2.73) Then we have lim Ln = 0.

n→∞

35

2.3. Integration of Sequences of Continuous Functions Proof of Lemma 2.4. By Property 1, we know that each Sn is nonempty. It is obvious that {Ln } is a bounded decreasing sequence. By the Monotone Convergence Theorem [49, Theorem 3.14, p. 55], {Ln } converges in R. Assume that this limit was nonzero. • Step 1: The construction of a compact set Kn . There exists a δ > 0 such that Ln ≥ δ for every positive integer n. By the definition (2.73), Ln − 2δn is not an upper bound of {ℓ(F ) | F ∈ Sn }. In other words, there exists a Fn ∈ Sn such that ℓ(Fn ) > Ln − δ · 2−n (2.74) for every positive integer n. By the observation preceding Lemma 2.4, we may n \ Fk ⊆ Fn . Since each Fk assume further that each Fn is closed. Let Kn = k=1

is closed in R, Kn is closed in R too. By the Heine-Borel Theorem, the set Kn must be compact. Furthermore, we have Kn ⊇ Kn+1 for each n = 1, 2, . . ..

• Step 2: Each Kn is nonempty. There exists a F ∈ Sn such that ℓ(F ) ≥ δ

(2.75)

for each n ∈ N. Otherwise, there is a N ∈ N such that ℓ(F ) < δ for all F ∈ SN which means that δ is an upper bound of LN , but this is a contradiction. Now we are going to show that if Kn = ∅, then it is impossible to have the inequality (2.75). This contradiction bases on the following two facts: – Fact 1: Suppose that G is a finite union of bounded intervals of En \ Fn for every n ∈ N, where Fn are those sets considered in Step 1. Then we have G ∩ Fn = ∅ and G ∪ Fn ∈ Sn so that ℓ(G) + ℓ(Fn ) = ℓ(G ∪ Fn ) ≤ Ln . This and the inequality (2.74) give, for every n ∈ N, ℓ(G) < δ · 2−n .

(2.76)

– Fact 2: Suppose that F is a finite union of bounded intervals of En \ Kn for every n ∈ N. By De Morgan’s law (see [42, p. 11]), we have (F \ F1 ) ∪ · · · ∪ (F \ Fn ) = F \ (F1 ∩ · · · ∩ Fn ) = F \ Kn = F.

(2.77)

Note that F \ Fk is also a finite union of bounded intervals of Ek (and hence of Ek \ Fk ) for k = 1, 2, . . . , n, so it follows from the inequality (2.76) that ℓ(F \ Fk ) < δ · 2−k

(2.78)

for k = 1, 2, . . . , n. Hence it follows from the expression (2.77) and the inequality (2.78) that ℓ(F ) < δ. (2.79) If Km = ∅ for some m ∈ N, then the F considered in Fact 2 is a subset of Em , but the inequality (2.79) definitely contradicts the inequality (2.75). • Step 3: A contradiction to Property 2. By Step 1 and Step 2, we ∞ \ Kn 6= ∅. Since Fk ⊆ Ek for deduce from [49, Theorem 2.36, p. 38] that n=1

k = 1, 2, . . . , n, we must have Kn ⊆ En . However, these two facts contradict Property 2.

Hence the limit must be 0 which completes the proof of the lemma.



Chapter 2. Positive Borel Measures

36

We return to the proof of the problem. By Lemma 2.4, for ǫ > 0, there is a positive integer N such that for n ≥ N , we have ℓ(F ) ≤ Ln < ǫ, (2.80) where F is any finite union of bounded intervals of En . Now one may think that we can derive Z 1 fn (x) dx < 2ǫ 0

for n ≥ N directly from the uniform boundedness of fn , the definition (2.69) and the inequality (2.80). However, it fails because we have no way to estimate the integral Z fn (x) dx. En \F

To overcome this problem, we play the trick that since fn is Riemann integrable on [0, 1], the integral must be equal to its lower Riemann integral, see [2, §1.17, p. 74]. Thus we have to find nZ 1 o sup sn (x) dx 0 ≤ sn (x) ≤ fn (x) and sn is a step function , 0

where the sup is taken over all step functions sn below fn on [0, 1]. For every n ≥ N , we define

En = {x ∈ [0, 1] | sn (x) > ǫ} and Fn = [0, 1] \ En . Since sn is a step function, it is clear that En and Fn are finite unions of bounded intervals. Furthermore, since 0 ≤ sn (x) ≤ fn (x), we have En ⊆ En and En ∈ Sn . Then we deduce from the inequality (2.80) that ℓ(En ) < ǫ for all n ≥ N . Hence we obtain from this and repeated uses of [49, Theorem 6.12(c), p. 128] that for all n ≥ N , we have Z Z Z Z Z 1 ǫ dx ≤ ℓ(En ) + ǫ < 2ǫ. dx + sn (x) dx ≤ sn (x) dx + sn (x) dx = 0≤ 0

En

Fn

En

Fn

Since ǫ is arbitrary, we have shown that lim

Z

n→∞ 0

1

fn (x) dx = 0.

This completes the proof of the problem.



Remark 2.1 There are many mathematicians who have provided different proofs of Problem 2.10. See, for examples, [37], [38] and [53].

2.4

Problems on Borel Measures and Lebesgue Measures

Problem 2.11 Rudin Chapter 2 Exercise 11.

37

2.4. Problems on Borel Measures and Lebesgue Measures

Proof. We follow the given hint. Suppose that Ω is the family of all compact subsets Kα of X with µ(Kα ) = 1. Since X is compact and µ(X) = 1, Ω is not empty. Now we define \ K= Kα . (2.81) Kα ∈Ω

Since X is Hausdorff, each Kα is closed in X by the corollaries following Theorem 2.5. Thus K is closed in X and then K is Borel (see Definition 1.11), i.e., K ∈ M. Since K ⊆ X, Theorem 2.4 says that K is also compact. Suppose that K ⊆ V and V is open in X. Then V c is closed in X and so it is compact in X by Theorem 2.4. As we have mentioned in the previous paragraph that each Kα is closed in X, so each Kαc is open in X. By the definition (2.81) and the fact that K ⊆ V , we have [ V c ⊆ Kc = Kαc . Kα ∈Ω

In other words, {Kαc } forms an open cover of the compact set V c . Thus we have V c ⊆ Kαc 1 ∪ Kαc 2 ∪ · · · ∪ Kαc n

(2.82)

for some positive integer n. Since µ(Kαm ) = µ(X) = 1 for m = 1, 2, . . . , n, we have µ(Kαc m ) = 0 for m = 1, 2, . . . , n. Therefore, it follows from these and the set relation (2.82) that c

µ(V ) ≤

n X

µ(Kαc m ) = 0,

m=1

i.e., µ(V c ) = 0 or equivalent µ(V ) = 1. Since µ is regular and K ∈ M, Definition 2.15 implies that µ(K) = inf{µ(V ) | K ⊆ V and V is open} = 1 which proves the first assertion. For the second assertion, let H be a proper compact subset of K, i.e., H ⊂ K. Assume that µ(H) = 1. Then it means that H ∈ Ω and the definition (2.81) shows that K ⊆ H, a  contradiction. Hence we must have µ(H) < 1. This completes the proof of the problem. Problem 2.12 Rudin Chapter 2 Exercise 12.

Proof. Let B be the σ-algebra of all Borel sets in R and K be a nonempty compact subset of R. We have to show that there exists a measure µ on B such that K is the smallest closed subset with the property µ(K c ) = 0. Since every compact set K has a countable base (see [49, Exercise 25, Chap. 2, p. 45]), there exists a countable sequence F = {x1 , x2 , . . .} ⊆ K such that K = F. Next we define µ : F → [0, ∞] by

µ(xn ) =

1 2n

and for any E ∈ B, we define µ : B → [0, ∞] by ∞ X 1 χ (E), µ(E) = 2n {xn } n=1

(2.83)

Chapter 2. Positive Borel Measures

38

where χ{xn } is the characteristic function of the set {xn }. Suppose that {Ei } is a disjoint countable collections of members of B and E=

∞ [

Ei .

i=1

If xn ∈ Ei for some i, then xn ∈ / Ej for all j 6= i. In other words, each element of F belongs to i at most one element of {Ei }. Thus it follows from the definition (2.83) that µ(E) =

X 1 χ (E), 2n {xn } n

(2.84)

where the summation runs through all n such that xn ∈ Ei for some i. Since the series ∞ X 1 2n n=1

converges absolutely, its rearrangement converges to the same limit (see [49, Theorem 3.55, p. 78]). Therefore, we can rewrite the expression (2.84) as ∞ X X 1 χ (Ei ), µ(E) = 2ik {xik }

(2.85)

i=1 ik

where the inner summation runs through all ik such that xik ∈ Ei . It may happen that the set F ∩ Ei is finite for some i = 1, 2, . . .. Since the inner summation in the expression (2.85) is exactly µ(Ei ), we deduce from the expression (2.85) that µ

∞ [

i=1



Ei = µ(E) =

∞ X

µ(Ei ).

i=1

By Definitions 1.18(a) and 2.15, µ is a Borel measure on R. By the definition (2.83), it is easy to see that for every E ∈ B, we have µ(E) > 0 if and only if

K ∩ E 6= ∅

(2.86)

so that µ(K c ) = 0. Assume that there was a proper closed subset H ⊂ K in X such that µ(H c ) = 0. Since H ⊂ K, we have K c ⊂ H c . If H c ∩ K = ∅, then H c ⊆ K c , a contradiction. Thus H c ∩ K 6= ∅. Since H c is open in X, we have H c ∈ B. Therefore, we can conclude from the condition (2.86) that µ(H c ) > 0 which contradicts to our hypothesis. In other words, K is the smallest closed subset in R such that µ(K c ) = 0, or equivalently, K = supp (µ). This completes the analysis of the problem. Problem 2.13 Rudin Chapter 2 Exercise 13. i

However, Ei may contain more than one element of F .



39

2.4. Problems on Borel Measures and Lebesgue Measures

Proof. It is clear that the point set {0} is a compact subset of R. Assume f : R → R was a continuous function such that {0} = supp (f ). Let V = (−∞, 0) ∪ (0, ∞). We know from Definition 2.9 that supp (f ) = f −1 (V ), so {0} = f −1 (V )

(2.87)

so that f −1 (V ) 6= ∅. Since V is open in R and f is continuous on R, f −1 (V ) is open in R by Definition 1.2(c). Thus, if p ∈ f −1 (V ), then there exists a δ > 0 such that (p − δ, p + δ) ⊆ f −1 (V ).

(2.88)

Thus we deduce from the set relations (2.87) and (2.88) that [p − δ, p + δ] ⊆ f −1 (V ) = {0}, a contradiction. Hence there is no continuous function f such that {0} = supp (f ).

For the second assertion, we suppose that K is a compact subset of R. We claim that K is the support of a continuous function if and only if K is the closure of an open set V in X. If K is the support of a continuous function f : R → R, then Definition 2.9 shows that


K = {x ∈ R | f (x) 6= 0} = f −1 (−∞, 0) ∪ (0, ∞) = f −1 (−∞, 0) ∪ f −1 (0, ∞) .



Since f is continuous on R, f −1 (−∞, 0) and f −1 (0, ∞) are open in R. Thus this proves one direction. Conversely, if we have K=V (2.89) for some open set V , then we consider F = V c which is closed in R. Define ρF : R → R by ρF (x) = inf{|x − y| | y ∈ F }. By Problem 2.3, we know that ρF is uniformly continuous on R and ρF (x) = 0 if and only if x ∈ F . Therefore, we must have ρF (x) > 0

if and only if

x ∈ V.

(2.90)

Applying the relation (2.90) to the expression (2.89), we have K = {x ∈ R | ρF (x) 6= 0}, i.e., K is the support of the continuous function ρF . We note that the third assertion is not valid in other topological spaces. We consider the three-element set X = {a, b, c}. By [42, Example 1, p. 76], we see that {∅, {b}, X} is a topology of X. Let K = {a, b}. Then it is easy to check from the definition that K is compact. Now if K = V for some open set V , then K must be closed in X (see [42, p. 95]). However, since X \ K = {c} which is not open in X, K is not closed in X. This gives a counter-example to explain that the description “K is the closure of an open set V in X” cannot guarantee that K is the support of a continuous function in an arbitrary topological space and hence we complete the proof of the problem.  Problem 2.14 Rudin Chapter 2 Exercise 14.

Chapter 2. Positive Borel Measures

40

Proof. Let f : Rk → R. We follow the proof of Theorem 2.24 (Lusin’s Theorem). Suppose that 0 ≤ f < 1. Attach a sequence {sn } of simple measurable functions to f , as in the proof of Theorem 1.17 (The Simple Function Approximation Theorem). Put t1 = s1 and tn = sn − sn−1 for n = 2, 3, . . .. Then 2n tn is the characteristic function of a measurable set Tn ⊆ Rk and f (x) =

∞ X

tn (x) =

∞ X

2−n χTn (x)

(2.91)

n=1

n=1

on Rk . By Theorem 2.20(b), there exist An , Bn ∈ B in Rk such that An ⊆ Tn ⊆ Bn

and

m(Bn \ An ) = 0

(2.92)

for every positive integer n.j We define g : Rk → R and h : Rk → R by g(x) =

∞ X

−n

2

χAn (x)

and h(x) =

∞ X

2−n χBn (x).

(2.93)

n=1

n=1

We claim that g and h are Borel measurable functions on Rk . It is easy to check that  ∅, if α ≤ 0;      Acn , if 0 < α ≤ 1; {x ∈ Rk | χAn (x) < α} =      R, if α > 1.

Thus each χAn is a Borel measurable function by Definition 1.11. Similarly, each χBn is also a Borel measurable function. Since the partial sums gn and hn of g and h are just simple functions and An , Bn ∈ B, they are Borel measurable functions by the comment following Definition 1.16. Furthermore, for each n = 1, 2, . . ., we have |2−n χAn (x)| ≤ 2−n

and

|2−n χBn (x)| ≤ 2−n

∞ X

2−n converges, it follows from Weierstrass M -test (see [49, Theorem 7.10, p. Since Xn=1 X 147]) that 2−n χAn and 2−n χBn converge uniformly on Rk to g and h respectively. By the Corollary (a) following Theorem 1.14, g and h are Borel measurable. This proves our claim. on

Rk .

By the measure (2.92), we know that g = h a.e. [m] on Rk . Obviously, if F ⊆ E, then we must have χF (x) ≤ χE (x). Therefore, we observe from the set relations (2.92) that χAn (x) ≤ χTn (x) ≤ χBn (x)

(2.94)

on Rk . Hence we apply the inequalities (2.94) to functions (2.91) and (2.93), we obtain g(x) ≤ f (x) ≤ h(x)

(2.95)

on Rk . Next, if 0 ≤ f < M for some positive constant M on Rk , then the above argument can f be applied to M so that the inequalities (2.95) hold with g and h replaced by M g and M h respectively. j

In fact, each An is an Fσ and each Bn is a Gδ , see Definition 1.11.

41

2.4. Problems on Borel Measures and Lebesgue Measures

In the general case, we consider the measurable sets EN = {x ∈ Rk | − N ≤ f (x) ≤ N }, where N ∈ N. Now we have χEN f → f as N → ∞ on Rk . In fact, the above argument shows that we can find Boreal functions gN and hN such that gN (x) = hN (x) a.e. [m] on Rk and gN (x) ≤ χEN f (x) ≤ hN (x) on Rk . By Theorem 1.14, we see that the functions g and h defined by g(x) = lim sup gN (x)

and

h(x) = lim sup hN (x)

N →∞

(2.96)

N →∞

are Borel measurable functions on Rk . Since gN (x) = hN (x) a.e. [m] on Rk , we have g(x) = h(x) a.e. [m] on Rk . Furthermore, we note that g(x) = lim sup gN (x) ≤ lim sup χEN f (x) ≤ lim sup hN (x) = h(x) N →∞

N →∞

(2.97)

N →∞

on Rk . Since lim χEN f (x) = lim sup χEN f (x) = f (x)

N →∞

N →∞

on Rk , the inequalities (2.95) follow immediately from the inequalities (2.96) and the fact (2.97). Hence our desired results also hold in this general case, completing the proof of the problem.  Problem 2.15 Rudin Chapter 2 Exercise 15.

Proof. For each positive integer n, we define fn : [0, ∞) → [0, ∞] by   x n x  e 2 , if x ∈ [0, n];  1− n fn (x) =   0, if x > n.

It is easy to see that each fn is continuous on [0, ∞) and so Definition 1.11 shows that it is Borel measurable. Fix a x ∈ R. Then the Archimedean Property ensures the existence of a positive integer N such that N ≥ x. By this fact and the fact that (1 + nx )n → ex as n → ∞, we have  x x x n x e 2 = e−x · e 2 = e− 2 . lim fn (x) = lim 1 − n→∞ n→∞ n x

Since e− 2 ∈ L1 (R) and

x

|fn (x)| ≤ e− 2

(2.98)

for all n = 1, 2, . . . and x ∈ [0, ∞), it follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) that Z ∞ Z ∞ x (2.99) lim fn dx = e− 2 dx = 2. n→∞ 0

0

By the inequality (2.98), we get Z n Z Z ∞ fn dx = fn dx − 0

0

∞

n

which implies that

lim

Z

n→∞ 0

∞

Z fn dx ≤

∞

n

fn dx = lim

Z

|fn | dx ≤

n→∞ 0

Z

∞

x

n

e− 2 dx = 2e− 2

n

n

fn dx.

(2.100)

Chapter 2. Positive Borel Measures

42

Combining the results (2.99) and (2.100), we obtain Z n fn dx = 2. lim n→∞ 0

For the second integral, we define gn : [0, ∞) → [0, ∞] by   x n −2x  e , if x ∈ [0, n]; 1 +   n gn (x) = x    2 , if x > n. e2x

Therefore each gn is continuous on [0, ∞). Furthermore, by using a similar argument as above, we have for x ∈ R,  x n −2x e = ex · e−2x = e−x . lim gn (x) = lim 1 + n→∞ n→∞ n

Since e−x ∈ L1 (R) and |gn (x)| ≤ e−x for all n = 1, 2, . . . and x ∈ [0, ∞), it follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) that Z ∞ Z ∞ e−x dx = 1. gn dx = lim n→∞ 0

0

Now we apply the same trick as obtaining the result (2.100), we are able to show that Z n gn dx = 1. lim n→∞ 0

We have completed the proof of the problem.



Problem 2.16 Rudin Chapter 2 Exercise 16.

Proof. Let T : Rk → Rk be linear, Y = T (Rk ) with dim Y = r < k. Let B = {b1 , . . . , br } be a basis of Y . Since B is a finite set of linearly independent vectors of Rk , it can be enlarged to a basis {b1 , . . . , br , br+1 , . . . , bk } for Rk .k Define the linear transformation T : Rk → Rk by T (α1 b1 + · · · + αk bk ) = (α1 , . . . , αk ). Then it is easy to see that T is one-to-one and so det T 6= 0. Furthermore, we have T (Y ) = {(α1 , . . . , αr , 0, . . . , 0) | α1 , . . . , αr ∈ R} so that T (Y ) =

∞ [

Wn ,

(2.101)

n=1

where Wn = {(ξ1 , . . . , ξr , 0, . . . , 0) | −n ≤ ξi ≤ n and i = 1, 2, . . . , r}. By the definition, we have Wn ⊆ Wn+1 k

See, for example, [23, Theorem 30.19, p. 279]. Sometimes, this result is called the Basis Extension Theorem.

43

2.5. Problems on Regularity of Borel Measures

for every n ∈ N. It is trivial to see from Theorem 2.20(a) that m(Wn ) = vol (Wn ) = 0

(2.102)

for each n = 1, 2, . . .. In addition, since Rk is a Borel set, Y is also a Borel set. Hence we follow from this fact, the expressions (2.101), (2.102) and Theorem 1.19(d) that
m T (Y ) = lim m(Wn ) = lim vol (Wn ) = 0. (2.103) n→∞

n→∞

When we read the proof of Theorem 2.20(e) and §2.23 carefully, we see that the validity of the formula
m T (E) = | det T |m(E) (2.104)

when T is an one-to-one map of Rk onto Rk is independent of whether T (Rk ) is a subspace of lower dimension or not. As a consequence, we
may apply the formula (2.104) here. Hence we can conclude from m(Y ) = | det T |−1 m T (Y ) and the result (2.103) that m(Y ) = 0, completing  the proof of the problem.

2.5

Problems on Regularity of Borel Measures

Problem 2.17 Rudin Chapter 2 Exercise 17.

Proof. Let d be the mentioned distance in the problem, p = (x1 , y1 ) and q = (x2 , y2 ). We are going to prove the assertions one by one. • X is a metric space with metric d. We check the definition of a metric space, see [49, Definition 2.15, p. 30]: – If p = q, then d(p, q) = |y1 − y2 | = 0. Otherwise, we have  |y1 − y2 |, if x1 = x2 and y1 6= y2 ; d(p, q) = 1 + |y1 − y2 |, if x1 6= x2 , 6= 0.

– It is clear that d(p, q) = d(q, p) holds because |y1 − y2 | = |y2 − y1 |.

– Let r = (x3 , y3 ). Suppose that r 6= p and r 6= q. Otherwise, the triangle inequality is trivial. There are two cases: ∗ Case (i): x1 = x2 . Then we have d(p, q) = |y1 − y2 | and  |y1 − y3 |, if x3 = x1 and y3 6= y1 ; d(p, r) = 1 + |y1 − y3 |, if x3 6= x1 .

(2.105)

Similarly, we have d(q, r) =



|y2 − y3 |, if x3 = x2 and y3 6= y2 ; 1 + |y2 − y3 |, if x3 6= x2 .

(2.106)

Since |y1 − y2 | ≤ |y1 − y3 | + |y3 − y2 |, any combination of the distances (2.105) and (2.106) imply that d(p, q) ≤ d(p, r) + d(r, q).

(2.107)

Chapter 2. Positive Borel Measures

44

∗ Case (ii): x1 6= x2 . In this case, we have d(p, q) = 1 + |y1 − y2 |. If x3 = x1 , then x3 6= x2 so that d(p, r) = |y1 − y3 | and d(q, r) = 1 + |y2 − y3 |. The situation for x3 = x1 is similar. If x3 6= x1 and x3 6= x2 , then we see from the distances (2.105) and (2.106) again that d(p, r) = 1 + |y1 − y3 | and d(q, r) = 1 + |y2 − y3 |. Hence the triangle inequality (2.107) remains true. By the above analysis, we conclude that d is a metric. • X is locally compact. Let X = (R2 , d) and p = (x, y), q = (u, v) ∈ R2 . Fix p first. By the definition, d(p, q) < 1 implies that x = u and |y − v| < 1. Thus the neighborhood B(p, 1) of p is given by B(p, 1) = {q ∈ X | d(p, q) < 1}

= {q ∈ X | x = u and |y − v| < 1}

= {x} × (y − 1, y + 1).

(2.108)

Geometrically, B(p, 1) is the vertical line segment with half length less than 1 and (x, y) as its midpoint. In addition, this line segment is open in R with the usual metric. Furthermore, it is clear from the definition (2.108) that B(p, 1) = {x} × [y − 1, y + 1].

(2.109)

We want to show that B(p, 1) is compact with respect to the metric d. To this end, let {Vα } be an open cover of B(p, 1), i.e., [ B(p, 1) ⊆ Vα . α

Let a ∈ B(p, 1), where a = (s, t) with s = x and t ∈ [y − 1, y + 1]. Then (x, t) ∈ Vα for some α. Since Vα is open in X, there exists a δt ∈ (0, 1) such that B(a, δt ) = {q = (u, v) | u = s = x and |v − t| < δt } = {x} × (t − δt , t + δt ) ⊆ Vα (2.110) so that B(p, 1) ⊆

[

a∈B(p,1) a=(s,t)

B(a, δt ) ⊆

[

Vα .

(2.111)

α

Applying the expressions (2.109) and (2.110) to the set relation (2.111), we see that [ [ B(p, 1) = {x} × [y − 1, y + 1] ⊆ {x} × (t − δt , t + δt ) ⊆ Vα (2.112) t∈[y−1,y+1]

α

which means that {Ut }, where Ut = (t − δt , t + δt ) with t ∈ [y − 1, y + 1], is an open cover of [y − 1, y + 1]. Since [y − 1, y + 1] is compact with respect to the usual metric | · | by the Heine-Borel Theorem, there are finitely many indices t1 , . . . , tk such that [y − 1, y + 1] ⊆ (t1 − δt1 , t1 + δt1 ) ∪ (t2 − δt2 , t2 + δt2 ) ∪ · · · ∪ (tk − δtk , tk + δtk ). Suppose that (x, ti ) ∈ Vαi for i = 1, 2, . . . , k. Hence we follow immediately from the set relation (2.112) that B(p, 1) ⊆ Vt1 ∪ Vt2 ∪ · · · ∪ Vtk . In other words, the closure B(p, 1) is compact with respect to the metric d and since p is an arbitrary point in X, Definition 2.3(f) shows that X is locally compact.l l

We remark that we cannot use the Heine-Borel Theorem directly to the closure (2.109) because we are working in the space X with metric d, not R2 with the usual metric.

45

2.5. Problems on Regularity of Borel Measures • Construction of a positive linear functional. Let f ∈ Cc (X), i.e., f : (R2 , d) → C is a continuous function and supp (f ) is compact. First of all, we notice that the set E = {x ∈ R | f (x, y) 6= 0 for some y} is finite. To see this, let K = supp (f ) and 0 < δ < 1. Since [ K⊆ B(p, δ), p∈K

where B(p, δ) = {x} × (y − δ, y + δ). Now the compactness of K ensures that there exists some positive integer n such that K⊆

n [

B(pi , δ) =

n [

{q ∈ X | u = xi and |yi − v| < δ}

i=1

i=1

which forces E = {x1 , x2 , . . . , xn }, as claimed.

Next, we claim that the mapping Λ : Cc (X) → R defined by Λ(f ) =

n Z X j=1

∞

f (xj , y) dy

−∞

is a positive linear functional. To this end, let f, g ∈ Cc (X). Since Cc (X) is a vector space, we have αf + βg ∈ Cc (X) for any α, β ∈ C. Furthermore, let f (xi , y) 6= 0 and g(xj , y) 6= 0 for at least one y, where 1 ≤ i ≤ n and n + 1 ≤ j ≤ m, and f (zk , y)g(zk , y) 6= 0 for at least one y, where m + 1 ≤ k ≤ r. By the definition, we have f (xi , y) = g(xj , y) = 0, where n + 1 ≤ i ≤ m, 1 ≤ j ≤ n and all y, so we get Λ(αf + βg) =

r Z X s=1

"

=α

∞

n Z X s=1

+β

"

[αf (xs , y) + βg(xs , y)] dy

−∞

∞

f (xs , y) dy +

−∞

m Z X

∞

s=n+1 −∞

= αΛ(f ) + βΛ(g).

Z r X

∞

f (zs , y) dy

s=m+1 −∞ Z ∞ r X

g(xs , y) dy +

s=m+1 −∞

#

g(zs , y) dy

#

By Definition 2.1, Λ is a linear functional on Cc (X). If f ≥ 0, then Λ(f ) ≥ 0. Thus Λ is positive. This proves the claim. • The measures of the x-axis and its compact subsets. In order to apply Theorem 2.14 (The Riesz Representation Theorem), we have to show that X is Hausdorff. Given p, q ∈ X and p 6= q. Let p = (x, y) and q = (u, v). There are two cases: – Case (i): x = u but y 6= v. Then we define δ = |y − x| and it obtains from the definition (2.108) that  δ h    δ δ δ δ i h δ i ∩ B q, = {x} × y − , y + ∩ {x} × v − , v + = ∅. B p, 2 2 2 2 2 2

Chapter 2. Positive Borel Measures

46

– Case (ii): x 6= u but y = v. Then the geometric property of the neighborhoods B(p, 1) and B(q, 1) ensures that their intersection must be empty. In conclusion, we have shown that X is Hausdorff. By Theorem 2.14 (The Riesz Representation Theorem), there exists a unique positive measure µ on M associated with Λ. Let E = {(x, 0) | x ∈ R}. Let δ ∈ (0, 1) and

Uδ =

[

[

B(p, δ) =

x∈R

p∈E

{x} × (−δ, δ).

Then Uδ is clearly an open set in X and E ⊆ Uδ . Take (x1 , 0), . . . , (xn , 0) ∈ E and construct


K = B (x1 , 0), δ ∪ B (x2 , 0), δ ∪ · · · ∪ B (xn , 0), δ


= {x1 } × [−δ, δ] ∪ {x2 } × [−δ, δ] ∪ · · · ∪ {xn } × [−δ, δ] .


Since each B (xi , 0), δ is compact, K is also compact and K ⊆ Uδ . By Theorem 2.12 (Urysohn’s Lemma), there exists an f ∈ Cc (X) such that K ≺ f ≺ Uδ . Since µ(Uδ ) = sup{Λ(f ) | f ≺ Uδ } (see [51, Eqn. (1), p. 41]), we have µ(Uδ ) ≥ Λ(f ) =

m Z X j=1

∞

−∞

f (x′j , y) dy,

(2.113)

where x′1 , . . . , x′m are those values of x for which f (x, y) 6= 0 for at least one y and for some positive integer m. Since K ≺ f , we have f (xi , y) = 1 for all i = 1, 2, . . . , n and y ∈ (−δ, δ). Thus we have {x1 , x2 , . . . , xn } ⊆ {x′1 , x′2 , . . . , x′m } and then the inequality (2.113) reduces to n Z δ X dy = nδ µ(Uδ ) ≥ i=1

−δ

so that µ(Uδ ) → ∞ as n → ∞. If V is an open set containing E, then V must contain the open set Uδ for some small δ and thus µ(V ) = ∞ for every open set V containing E. By Theorem 2.14(c), we have µ(E) = inf{µ(V ) | E ⊆ V , V is open} = ∞. Next, let K ⊆ E be compact. Choose δ ∈ (0, 1). Since [
K⊆W = B (a, 0), δ ,

(2.114)

(a,0)∈K

the compactness of K implies that K⊆

m [

i=1

B((ai , 0), δ) =

m [

({ai } × (−δ, δ))

i=1

for some positive integer m. In other words, we know from the definition (2.114) that K = {(a1 , 0), (a2 , 0), . . . , (am , 0)}.

47

2.5. Problems on Regularity of Borel Measures Besides, since W is open in X, Theorem 2.12 (Urysohn’s Lemma) again ensures there is a g ∈ Cc (X) such that K ≺ g ≺ W. Recall from [51, Eqn. (7), p. 43] that µ(K) = inf{Λ(f ) | K ≺ f } ≤ Λ(g) =

m Z X i=1

∞

g(ai , y) dy = 2mδ.

(2.115)

−∞

Since m is fixed and δ is arbitrary, it follows from the inequality (2.115) that µ(K) = 0. This completes the proof of the problem.



Problem 2.18 Rudin Chapter 2 Exercise 18.

Proof. Recall that X has an order relation ‘ α1 > α2 > · · · .

(2.117)

in X, which is a contradiction to the hint. Therefore, we may assume that α1 ∈ / V0 m

See [42, p. 63].

and

(α1 , α0 ] ⊆ V0 .

(2.118)

Chapter 2. Positive Borel Measures

48

Since V is a cover of [ω0 , ω1 ], there exists a V1 ∈ V containing α1 . By the above argument, we may find an α2 ∈ X such that α2 ∈ / V1

and

(α2 , α1 ] ⊆ V1 .

(2.119)

If α2 ∈ V0 , then since [α2 , α0 ] ⊆ V0 and α2 < α1 < α0 , we have α1 ∈ V0 which contradicts the result (2.118). Thus α2 ∈ / V0 and then we deduce from this fact, the results (2.118) and (2.119) that α2 ∈ / V0 ∪ V1 and (α2 , α0 ] ⊆ V0 ∪ V1 . Inductively, we can find αn+1 ∈ X such that αn+1 ∈ / V0 ∪ V1 ∪ · · · ∪ Vn

and

(αn+1 , α0 ] ⊆ V0 ∪ V1 ∪ · · · ∪ Vn .

If this process continues infinitely, then we can find an infinite decreasing sequence (2.117) again, a contradiction. Therefore, the sequence must stop eventually at the N th step and it must stop at αn+1 = ω0 for all n ≥ N , i.e., ω0 ∈ / V0 ∪ V1 ∪ · · · ∪ VN

and

(ω0 , α0 ] ⊆ V0 ∪ V1 ∪ · · · ∪ VN .

Since ω0 ∈ V for some V ∈ V, we conclude that [ω0 , ω1 ] = [ω0 , α0 ] ⊆ V0 ∪ V1 ∪ · · · ∪ VN ∪ V. In other words, X is compact. Next, we prove that X is Hausdorff and we need the help of the following lemma: Lemma 2.5 Define the mapping γ : [ω0 , ω1 ] → [ω0 , ω1 ] by   min(Sα ), if α ∈ [ω0 , ω1 ); γ(α) =  ω1 , if α = ω1 .

Then we have X \ Sα = Pγ(α) = [ω0 , α].

Proof of Lemma 2.5. For every α ∈ [ω0 , ω1 ), Sα 6= ∅ so that the mapping γ is welldefined. By the definition, we have Sα = [γ(α), ω1 ] with α < γ(α). Furthermore, there is no element between α and γ(α). Otherwise, γ(α) is not a smallest element of Sα anymore. Hence we deduce from this that X \ Sα = [ω0 , α] = [ω0 , γ(α)) = Pγ(α) . If α = ω1 , then Sω1 = ∅ and Pγ(ω1 ) = [ω0 , ω1 ] = X so that X \ Sω1 = Pγ(ω1 ) in this  case, completing the proof of the lemma. Let α, β ∈ X and we may assume that α < β. Then we see that α ∈ Pγ(α) = [ω0 , α] and β ∈ Sα . Thus Pγ(α) and Sα are neighborhoods of α and β respectively. By Lemma 2.5, they are disjoint and this shows that X is Hausdorff, as required.

49

2.5. Problems on Regularity of Borel Measures

(b) X \ {ω1 } is open but not σ-compact. Let X \ {ω1 } = [ω0 , ω1 ) and α ∈ [ω0 , ω1 ). As we have proven in part (a) that Pγ(α) = [ω0 , γ(α)) is a neighborhood of α. Hence X \ {ω1 } is open in X. Assume that X \ {ω1 } = [ω0 , ω1 ) was σ-compact. Let K be a compact subset of [ω0 , ω1 ). We know that the family {Px = [ω0 , x) | x < ω1 } is an open cover of K so that there are finitely many indices x1 , x2 , . . . , xk such that K ⊆ Px1 ∪ Px1 ∪ · · · ∪ Pxk . Without loss of generality, we may assume that x1 < x2 < · · · < xk . Thus we have K ⊆ [ω0 , xk ) ⊂ [ω0 , xk ]. Since xk < ω1 , our hypothesis shows that K is at most countable. Since X \ {ω1 } is σ-compact, it is a countable union of compact sets so that X \ {ω1 } is countable. However, we have
X = X \ {ω1 } ∪ {ω1 }

which means that X is countable. Evidently, this result contradicts our hypothesis. Hence X \ ω1 is not σ-compact. (c) Every f ∈ C(X) is constant on Sα for some α 6= ω1 . Suppose that x = f (ω1 ) and B(x, n1 ) = {z ∈ C | |z − x| < n1 }, where n on X
is a positive integer. Since f is continuous
1 1 −1 −1 B(x, n ) is an open subset of X and ω1 ∈ f B(x, n1 ) . and B(x, n ) is open in C, f By the topology (2.116), there exists an α1 ∈ X \ {ω1 } such that   1  . (2.120) (α1 , ω1 ] ⊆ f −1 B x, n By Lemma 2.5, we note that (α1 , ω1 ] is uncountable, so a sequence {αn } exists such that α1 < α2 < · · · < ω1 .

(2.121)

Now the results (2.120) and (2.121) together give   1  Sαn ⊆ Sαn+1 ⊆ f −1 B x, n

for all n ∈ N. Before we proceed further, we need the following result:n Lemma 2.6 Every well-ordered set A has the least upper bound property.

Proof of Lemma 2.6. Let B be a nonempty subset of A having an upper bound in A. Therefore, the set U of upper bounds of B is nonempty. Since U ⊆ A and A is well-ordered, U has a least element, completing the proof of Lemma 2.6.  By the relation (2.121), we see that the nonempty subset {αn } of X is bounded above by ω1 . Therefore, it follows from Lemma 2.6 that sup{αn } exists in X. Call it α. It is n∈N

clear that α ≤ ω1 and n

See [42, Exercise 1, p. 66]

  1  Sα ⊆ Sαn ⊆ f −1 B x, n

(2.122)

Chapter 2. Positive Borel Measures

50

for all n = 1, 2, . . .. In fact, it is impossible to have α = ω1 because the set Sω1 = {x ∈ Y | ω1 < x} is uncountable, where Y is the well-ordered set as described in the “Construction” in the question. As ω1 = α, we have Sω1 = Sα ⊆ Sαn so that Sαn is uncountable too, but it contradicts to the fact that αn is a predecessor of ω1 . Hence we obtain α < ω1 and we deduce from the set relations (2.122) that f (Sα ) ⊆ B(x, n1 ) for all n ∈ N and this means that ∞  1 \ B x, f (Sα ) ⊆ = {x}. n n=1

Now this is exactly our desired result.

(d) The intersection of {Kn } of uncountable compact subsets of X is also uncountable compact. We first prove a lemma which indicates a relationship between the cardinality of a compact set K and the topological property of ω1 in K. Lemma 2.7 Suppose that K is a nonempty compact subset of X. Then K is uncountable if and only if ω1 is a limit point of K.

Proof of Lemma 2.7. Let K be uncountable. By the proof of part (b), we know that K cannot lie inside [ω0 , α] for every predecessor α < ω1 . Otherwise, K will be at most countable, a contradiction. Hence this implies that K ∩ (α, ω1 ] 6= ∅ for every α < ω1 . By the definition (see [42, p. 97]), it follows that ω1 is a limit point of K. Conversely, suppose that ω1 is a limit point of K. By [42, Theorem 17.9, p. 99], K must be infinite. Assume that K was countable. Then K has the following representation K = {αn }, where α1 ≤ α2 ≤ · · · . Since K is compact and X is Hausdorff, K is closed in X by Corollary (a) following Theorem 2.5. Thus it must be ω1 ∈ K, i.e., every neighborhood Sα of ω1 satisfies
K ∩ Sα \ {ω1 } 6= ∅.
Let αN ∈ K ∩ Sα \ {ω1 } for some positive integer N , i.e., α < αN < ω1 . Recall that {αn } is increasing, so we must have αn ∈ Sα for all n ≥ N . By the definition (see [42, p. 98]), {αn } converges to ω1 . However, by an argument similar to the proof of part (c), this fact will imply the contradiction that Sαn is uncountable. Hence K is uncountable and we complete the proof of the  lemma. Let’s go back to the proof of part (d). Denote the intersection of {Kn } by K. Since X is Hausdorff, each Kn is closed in X. Since K ⊆ Kn , Theorem 2.4 implies that K is

51

2.5. Problems on Regularity of Borel Measures compact. For each n ∈ N, let Kn′ = K1 ∩ K2 ∩ · · · ∩ Kn . It is easy to check that n \

Ki′

=

n \

(K1 ∩ · · · ∩ Ki ) =

i=1

i=1

n \

Ki ,

i=1

so we may assume that K1 ⊇ K2 ⊇ · · · .

(2.123)

Therefore, any finite subcollection of {Kn } is nonempty and Theorem 2.6 shows that K 6= ∅. By Lemma 2.7, ω1 is a limit point of every Kn . Then, using [42, Theorem 17.9, p. 99] again, it means that every neighborhood (α, ω1 ] of ω1 contains infinitely many points of every Kn . Combining this fact and the sequence (2.123), every neighborhood (α, ω1 ] of ω1 must contain infinitely many points of the compact set K. Hence ω1 is a limit point of K and we finally conclude from Lemma 2.7 that K is uncountable. (e) M is a σ-algebra containing all Borel sets in X. Suppose that M = {E ⊆ X | either E ∪ {ω1 } or E c ∪ {ω1 } contains an uncountable compact set}. – M is a σ-algebra. We check Definition 1.3(a). In fact, it is obvious that X ∈ M because X is itself compact uncountable. Let E ∈ M. Then either E ∪ {ω1 } or E c ∪ {ω1 } contains an uncountable compact set. Since (E c )c = E, it must be true that E c ∈ M. ∞ [ En . If there exists an To verify Definition 1.3(a)(iii), let En ∈ M and E = n=1

n0 ∈ N such that En0 ∪ {ω1 } contains an uncountable compact set K, then K ⊆ En0 ∪ {ω1 } ⊆

∞ [

n=1



En ∪ {ω1 } = E ∪ {ω1 }

so that E ∈ M. Otherwise, all En ∪ {ω1 } contain no uncountable compact sets. This forces that every Enc ∪ {ω1 } contains an uncountable compact set Kn . We note that c

E ∪ {ω1 } =

∞ \

Enc

n=1



∪ {ω1 } =

∞ \

n=1


Enc ∪ {ω1 } ,

so it is true that K=

∞ \

n=1

Kn ⊆

∞ \

n=1


Enc ∪ {ω1 } = E c ∪ {ω1 }.

Now the result of part (d) illustrates that K is uncountable compact, thus E c ∈ M which implies that E ∈ M.

– M contains all Borel sets in X. If α < ω1 , then (α, ω1 ) 6= ∅. Let β ∈ (α, ω1 ). Since X = Pβ ∪ [β, ω1 ] and Pβ is countable, [β, ω1 ] must be uncountable. Since [β, ω1 ] = X \Pβ , [β, ω1 ] is closed in X and Theorem 2.4 verifies that [β, ω1 ] is compact. Furthermore, we note that [β, ω1 ] ⊆ Sα , (2.124) so these facts show that Sα ∈ M. Similarly, for every α > ω0 , we have X \ Pα = [α, ω1 ] and then X \ Pα ∈ M. Since M is a σ-algebra, M also contains Pα . Hence M contains the order topology τ of X which proves that B(X) ⊆ M as desired.

Chapter 2. Positive Borel Measures

52

(f) λ is a measure on M but not regular. Suppose that λ : M → [0, ∞] is defined by   1, if E ∪ {ω1 } contains an uncountable compact set; λ(E) = (2.125)  0, if E c ∪ {ω1 } contains an uncountable compact set.

If E ∈ M, then we know from the definition of M in part (e) that it is impossible for both E ∪ {ω1 } and E c ∪ {ω1 } containing uncountable compact sets, so this function is well-defined. – λ is a measure on M. Suppose that {Ei } is a disjoint countable collection of members of M. We claim that at most one Ei ∪ {ω1 } contains an uncountable compact set. To see this, assume that both E1 ∪ {ω1 } and E2 ∪ {ω1 } contained uncountable compact sets K1 and K2 respectively. Since E1 ∩ E2 = ∅, we have

K1 ∩ K2 ⊆ E1 ∪ {ω1 } ∩ E2 ∪ {ω1 }

= E1 ∩ {ω1 } ∪ E2 ∩ {ω1 } ∪ {ω1 } = {ω1 }.

(2.126)

However, part (d) tells us that K1 ∩ K2 is uncountable so that the result (2.126) is impossible. This proves the claim. By the previous analysis, if E1 ∪ {ω1 } is the only set containing an uncountable compact set K, then we get ∞ ∞  [
[ Ei ∪ {ω1 } = Ei ∪ {ω1 }, K⊆ i=1

i=1

so we follow from the definition (2.125) that ∞  [ λ Ei = 1. i=1

On the other hand, we know from the facts λ(E1 ) = 1 and λ(Ei ) = 0 for i = 2, 3, . . . that ∞ X λ(Ei ) = 1. i=1

Therefore, we have

∞ ∞  X [ λ(Ei ) λ Ei =

(2.127)

i=1

i=1

in this case. Similarly, if there is no Ei ∪ {ω1 } contains an uncountable compact set, then each Eic ∪ {ω1 } contains an uncountable compact set Ki so that λ(Ei ) = 0 for every i = 1, 2, . . .. Furthermore, we have ∞ ∞ ∞ ∞ c  [ \ \
\ Ei ∪ {ω1 } Eic ∪ {ω1 } = Eic ∪ {ω1 } = Ki ⊆ K= i=1

i=1

i=1

i=1

and part (d) ensures that K is uncountable compact. Therefore, we obtain ∞  [ Ei = 0 λ i=1

and then the formula (2.127) still holds in this case. Finally, since λ(X) = 1 < ∞. By Definition 1.18(a), λ is a positive measure.

53

2.5. Problems on Regularity of Borel Measures – λ is not regular. For every α < ω1 , we know from the set relation (2.124) that Sα ∪ {ω1 } must contain an uncountable compact set, so the definition (2.125) gives that λ(Sα ) = 1. If E = {ω1 }, then E c ∪ {ω1 } = {ω1 }c ∪ {ω1 } = X so that {ω1 } ∈ M. Therefore, we derive from the definition (2.125) that λ({ω1 }) = 0. However, these facts give λ({ω1 }) 6= inf{λ(Sα ) | {ω1 } ⊆ Sα }. By Definition 2.15, λ is not regular.

(g) The validity of the integral. Let f ∈ Cc (X). By part (c), there exists an α0 < ω1 such that f (x) = f (ω1 ) on Sα0 . Recall from part (e) that Sα0 ∈ M. By the set relation (2.124), every Sα ∪ {ω1 } contains an uncountable compact set and thus, in particular, λ(Sαc 0 ) = 0. By this, we gain Z

f dλ =

X

Z

f dλ + c Sα 0

Z

f dλ =

Sα0

Z

Sα0

f dλ = f (ω1 )λ(Sα0 ) = f (ω1 )

which is the required result. (h) The regular µ associates with the linear functional in part (f ). It is clear that Λ : Cc (X) → C defined by Λ(f ) = f (ω1 ) (2.128) is a linear functional on Cc (X). By Theorem 2.14 (The Riesz Representation Theorem) and Theorem 2.17(b), we have a unique regular positive Borel measure µ on X. By the result (2.128), we know that Z f dµ = f (ω1 ). X

By [51, Eqn. (1), p. 41], we have µ(V ) = sup{Λ(f ) | f ≺ V } = f (ω1 ) for every open set V in X. Thus this implies that µ(E) = inf{µ(V ) | E ⊆ V and V is open} = f (ω1 ) for every E ∈ M. We have completed the proof of the problem.



Remark 2.2 The measure considered in Problem 2.18 is called the Dieudonn´ e’s measure, see [10, Exampl 7.1.3, pp. 68, 69] and [17].

Problem 2.19 Rudin Chapter 2 Exercise 19.

Chapter 2. Positive Borel Measures

54

Proof. Suppose that X is a compact metric space with metric ρ and Λ is a positive linear functional on Cc (X), the space of all continuous complex functions on X with compact support. To begin with the construction of the class µ, we use Λ to define a set function µ∗ on every open set V in X by µ∗ (V ) = sup{Λ(f ) | f ≺ V }

= sup{Λ(f ) | f ∈ Cc (X), 0 ≤ f ≤ 1 on X and supp (f ) ⊆ V }

(2.129)

and for every subset E ⊆ X that µ∗ (E) = inf{µ∗ (V ) | V is open in X and E ⊆ V }.

(2.130)

Now we are going to present the proof by quoting several facts (some are with proofs and some are not). In fact, the idea of the following proof is stimulated by Feldman’s online article [20]. • Fact 1: The set function µ∗ is an outer measure. We check the definition [47, p. 346]. – Since supp (f ) ⊆ ∅ if and only if supp (f ) = ∅, we get from Definition 2.9 that f ≡ 0 on X which implies that µ∗ (∅) = sup{Λf | f ∈ Cc (X), 0 ≤ f ≤ 1 on X and supp (f ) ⊆ ∅} = Λ(0) = 0. – Suppose that E, F ⊆ X and E ⊆ F . Since any open set V containing F must also contain E, we obtain from the definition (2.130) that µ∗ (E) = inf{µ∗ (W ) | W is open in X and E ⊆ W } ≤ inf{µ∗ (V ) | V is open in X and F ⊆ V }

= µ∗ (F ). – The proof of the subadditivity µ∗

∞ [

i=1

∞  X µ∗ (Ei ) Ei ≤

(2.131)

i=1

follows exactly the same as the proof of Step I. Hence µ∗ is an outer measure on X. • Fact 2: µ∗ (K) < ∞ for every compact set K ⊆ X. Since X is open in X, we follow from the definition (2.129) that µ∗ (X) ≤ Λ(1). By Fact 1, we have µ∗ (K) ≤ µ∗ (X) ≤ Λ(1) for every compact set K ⊆ X. • Fact 3: A topological result in metric spaces. Now we need the following topological result about metric spaces which will be used in Fact 4: Lemma 2.8 Let X be a metric space with metric ρ and E a nonempty proper subset of X. Then the set V = {x ∈ X | ρE (x) < ǫ} is open in X, where ǫ > 0 and ρE (x) is defined in Problem 2.3.

55

2.5. Problems on Regularity of Borel Measures Proof of Lemma 2.8. If V = ∅, then there is nothing to prove. Thus we assume that V 6= ∅ and pick x0 ∈ V so that ρE (x0 ) < ǫ. Then there exists a δ > 0 such that ǫ − δ > 0 and ρE (x0 ) < ǫ − δ. We consider the ball B(x0 , δ) = {x ∈ X | ρ(x, x0 ) < δ} and we claim that B(x0 , δ) ⊆ V . To see this, let x ∈ B(x0 , δ). For every y ∈ E, we have ρ(x, y) ≤ ρ(x, x0 ) + ρ(x0 , y) < δ + ρ(x0 , y) and therefore we obtain ρE (x) = inf{ρ(x, y) | y ∈ E}

≤ inf{ρ(x, x0 ) + ρ(x0 , y) | y ∈ E}

< δ + inf{ρ(x0 , y) | y ∈ E}

= δ + ρE (x0 ) < ǫ.

In other words, x ∈ V . Since x is arbitrary, B(x0 , δ) ⊆ V and then V is open in X,  proving Lemma 2.8. • Fact 4: If V is open in X, then V is measurable with respect to µ∗ . We recall from [47, p. 347] that V ⊆ X is measurable with respect to the outer measure µ∗ if for every subset E of X, we have µ∗ (E) = µ∗ (E ∩ V ) + µ∗ (E ∩ V c ).

(2.132)

It suffices to show the inequality µ∗ (E) ≥ µ∗ (E ∩ V ) + µ∗ (E ∩ V c ) − ǫ

(2.133)

holds for every ǫ > 0 because the other side follows directly from the subadditivity (2.131). By the definition (2.130), there is an open set E ′ such that E ⊆ E ′ and ǫ µ∗ (E) ≥ µ∗ (E ′ ) − . 2

(2.134)

By the property in Fact 1, we have µ∗ (E ∩ V ) ≤ µ∗ (E ′ ∩ V ) and µ∗ (E ∩ V c ) ≤ µ∗ (E ′ ∩ V c ).

(2.135)

Thus it is easy to see from the inequalities (2.134) and (2.135) that one can obtain the inequality (2.133) if we can show the following result holds ǫ µ∗ (E ′ ) ≥ µ∗ (E ′ ∩ V ) + µ∗ (E ′ ∩ V c ) − . 2

(2.136)

To this end, we first find bounds of µ∗ (E ′ ∩ V ) and µ∗ (E ′ ∩ V c ). Since E ′ ∩ V is open in X, the definition (2.129) implies the existence of a continuous function f1 : X → [0, 1] such that supp (f1 ) ⊆ E ′ ∩ V and ǫ µ∗ (E ′ ∩ V ) ≤ Λ(f1 ) + . 8

(2.137)


Let ǫ ∈ 0, 8Λ(1) and δ > 0 be a constant such that δ ≤ ǫ[8Λ(1) − ǫ]−1 . Then we have ǫ δ Λ(1) ≤ 1+δ 8

and F1 =

f1 . 1+δ

(2.138)

Chapter 2. Positive Borel Measures

56

Recall the fact that the positivity of Λ implies the monotonicity of Λ. In particular, Λ(f ) ≤ Λ(1) for every f ≺ E ′ ∩ V . Thus we apply the results (2.138) and the monotonicity of Λ to the inequality (2.137) to get µ∗ (E ′ ∩ V ) ≤

1 δ ǫ δ ǫ ǫ Λ(f1 )+ Λ(f1 )+ ≤ Λ(F1 )+ Λ(1)+ = Λ(F1 )+ . (2.139) 1+δ 1+δ 8 1+δ 8 4

We replace E by the closed set (E ′ ∩V )c in Lemma 2.8 to get the open set U . Given δ > 0. Since f1 is uniformly continuous on X, there exists a η > 0 such that |f1 (x) − f1 (y)| ≤ δ for all x, y ∈ X with ρ(x, y) < η. Recall that f1 vanishes on the closed set (E ′ ∩ V )c , so if we take y ∈ (E ′ ∩ V )c , then we have f1 (x) ≤ δ for all x ∈ X with ρ(x, y) < η. Since ρ(E ′ ∩V )c (x) < η, Lemma 2.8 ensures that the set U = {x ∈ X | ρ(E ′ ∩V )c (x) < η} is open in X. In other words, we have established from Fact 3 that there exists an open set U such that (2.140) U ⊃ (E ′ ∩ V )c ⊇ V c and f1 (x) ≤ δ on U . Since E ′ ∩ U is open in X, the definition (2.129) again shows that we can find a continuous function f2 : X → [0, 1] such that supp (f2 ) ⊆ E ′ ∩ U and furthermore, by using similar argument as in the proof of the inequality (2.139), we obtain that µ∗ (E ′ ∩ V c ) ≤ µ∗ (E ′ ∩ U ) ≤ Λ(f2 ) + where F2 =

ǫ ǫ ≤ Λ(F2 ) + , 4 4

(2.141)

f2 1+δ .

Now we have found the bounds (2.139) and (2.141) of µ∗ (E ′ ∩ V ) and µ∗ (E ′ ∩ V c ) respectively. Next, we want to show that F3 ∈ Cc (X), where F3 = F1 + F2 . To this 1 end, we recall the facts that F1 , F2 : X → [0, 1+δ ] because 0 ≤ f1 ≤ 1 and 0 ≤ f2 ≤ 1. Furthermore, we note from the set relation (2.140) that V ∩ U 6= ∅. Therefore, we have the following facts: – Since supp (f1 ) ⊆ E ′ ∩ V , supp (F1 ) ⊆ E ′ ∩ V , i.e., F1 (x) = 0 on (E ′ ∩ V )c . Therefore, we conclude that 1 F3 = F2 ≤ ≤1 1+δ on (E ′ ∩ V )c . Similarly, since supp (F2 ) ⊆ E ′ ∩ U , we have F2 (x) = 0 on (E ′ ∩ U )c and then 1 ≤1 F3 = F1 ≤ 1+δ on (E ′ ∩ U )c .

– On (E ′ ∩ V ) ∩ U , we immediately follow from the inequality in (2.140) that f1 (x) ≤ δ on (E ′ ∩ V ) ∩ U ⊆ U . It is clear that f2 (x) ≤ 1 on (E ′ ∩ V ) ∩ U , so these facts imply that f2 1 δ f1 + ≤ + = 1. F3 = F1 + F2 = 1+δ 1+δ 1+δ 1+δ – On (E ′ ∩ V ) \ U , we have F1 ≤ 1 and F2 = 0 so that F3 ≤ 1.

57

2.5. Problems on Regularity of Borel Measures – On U \ (E ′ ∩ V ), we have F1 = 0 and F2 ≤ 1 so that F3 ≤ 1. To have a better understanding of the above facts, we can draw some pictures. For examples, Figure 2.3(a) shows the sets V, E ′ and E ′ ∩ V , the shaded blue part in Figure 2.3(b) indicates the closed set (E ′ ∩ V )c and the part inside the circle in Figure 2.4 is the set (E ′ ∩ V )c \ U .

(a) The sets V, E ′ and E ′ ∩ V .

(b) The set (E ′ ∩ V )c .

Figure 2.3: The pictures of V, E ′ , E ′ ∩ V and (E ′ ∩ V )c .

Figure 2.4: The set (E ′ ∩ V )c \ U . Thus we have shown that 0 ≤ F3 (x) ≤ 1 on X. Next, since F1 and F2 are continuous on X, F3 is also continuous on X. By the definition, we know that x0 ∈ supp (F3 ) if and only if x0 ∈ supp (F1 ) ⊆ E ′ ∩ V or x0 ∈ supp (F2 ) ⊆ E ′ ∩ U . In addition, we note that (E ′ ∩ V ) ∪ (E ′ ∩ U ) ⊆ E ′ . On the other hand, if y ∈ E ′ but y ∈ / E ′ ∩ V , then we have ′ c y ∈ (E ∩ V ) and we follow from the set relation (2.140) that y ∈ U . Thus this implies that y ∈ E ′ ∩ U and so E ′ = (E ′ ∩ V ) ∪ (E ′ ∩ U ). In other words, it gives x0 ∈ supp (F3 ) if and only if x0 ∈ supp (F1 ) ∪ supp (F2 ) ⊆ E ′ , i.e., supp (F3 ) ⊆ E ′ .

Finally, it yields from the inequalities (2.139) and (2.141) as well as the definition (2.129)

Chapter 2. Positive Borel Measures

58

that µ∗ (E ′ ∩ V ) + µ∗ (E ′ ∩ V c ) ≤ Λ(F1 ) + Λ(F2 ) +

ǫ ǫ ǫ = Λ(F3 ) + ≤ µ∗ (E ′ ) + 2 2 2

which is exactly the inequality (2.136). Hence we obtain the desired result that (2.132). • Fact 5: The set M∗ of all measurable sets (with respect to µ∗ ) is a σ-algebra and µ = µ∗ |B (X) is a (complete) measure. The first assertion follows directly from Carath´eodory’s Theorem [47, Theorem 8, p. 349]. By Fact 4, we have B(X) ⊆ M∗ which is exactly Step VII, where B(X) denotes the set of all Borel sets in X. By Carath´eodory’s Theorem again, it shows that µ is a (complete) measure on B(X) and this is the same as Step IX. Consequently, Steps III, IV, V, VI and VIII can all be skipped.o • Fact 6: Deduction of parts (b) and (d). Let K be a compact set of X. Since X is a metric space, K is also Hausdorff. By Corollary (a) following Theorem 2.5, we know that K is closed in X and so K ∈ B(X). Therefore, we deduce from Facts 2 and 5 that µ(K) = µ∗ (K) ≤ µ∗ (X) ≤ Λ(1) which is exactly part (b) of Theorem 2.14 (The Riesz Representation Theorem). If E is open in X, then E ∈ B(X). Thus we observe from the equation (2.132) and then the definition (2.130) that µ(E) = µ∗ (E) = µ∗ (X) − µ∗ (E c )

= µ∗ (X) − inf{µ∗ (V ) | V is open in X and E c ⊆ V }

= µ∗ (X) − inf{µ∗ (X) − µ∗ (V c ) | V c is closed in X and V c ⊆ E}

= sup{µ∗ (V c ) | V c is closed in X and V c ⊆ E} = sup{µ(V c ) | V c is closed in X and V c ⊆ E}.

(2.142)

Since X is compact, V c is compact by Theorem 2.4 so that the expression (2.142) can be expressed as µ(E) = sup{µ(K) | K compact and K ⊆ E } which is the same as part (d) of Theorem 2.14 (The Riesz Representation Theorem). • Fact 7: Deduction of part (a). As Rudin pointed out in Step X that it suffices to prove Z f dµ (2.143) Λ(f ) ≤ X

o

holds for f ∈ Cc (X) because the opposite inequality follows by changing the sign of f . In fact, the proof of our inequality (2.143) for compact metric X is essentially the same as that of Step X, except that Rudin applied Step II to show that X µ(K) ≤ Λ(hi ),

We cannot say at this stage that Step II can be omitted because the proof of Step X still needs it, but we will see very soon in Fact 7 that it is eventually allowable to do so.

2.6. Miscellaneous Problems on L1 and Other Properties

59

P where hi ≺ Vi and hi = 1 on K.p However, this kind of argument can be replaced completely by using Fact 6 that µ(K) ≤ Λ(1) holds for every compact set K ⊆ X. Therefore, we finally arrive at the expected inequality (2.143) and furthermore, Step II can also be skipped. Hence we conclude that Steps II to IX can be replaced by a simpler argument (outer measure in the sense of Carath´eodory). In other words, only Steps I and X should be kept. Of course, the proof of the uniqueness of µ cannot be omitted. This completes the proof of the problem.  Remark 2.3 (a) We remark that a Radon measure is a measure µ defined on the σ-algebra M containing all Borel sets B(X) of the Hausdforff space X and it satisfies µ(K) < ∞ for all compact subsets K, outer regular on M and inner regular on open sets in X, see [22, p. 212] or [47, p. 455]. Hence the unique positive measure µ in Theorem 2.14 (The Riesz Representation Theorem) is in fact a Radon measure when X is Hausdforff. (b) You are also advised to read the paper [52] for a similar proof of Problem 2.19.

2.6

Miscellaneous Problems on L1 and Other Properties

Problem 2.20 Rudin Chapter 2 Exercise 20.

Proof. Let f = sup fn : [0, 1] → [0, ∞]. By Definition 1.30, f ∈ / L1 on [0, 1] is equivalent to n

Z

0

1

|f (x)| dx = ∞.

In fact, the inspiration of the construction of {fn } comes from the idea of Problem 2.9. For each positive integer n, we consider the continuous functions gn : [−1, 1] → [0, ∞) given by  if x ∈ / [− n12 , n12 ];  0, gn (x) =  n − n3 |x|, if x ∈ [− n12 , n12 ]. Next, we define fn : [0, 1] → [0, ∞) by  0, if x ∈ / [ n1 −  1  fn (x) = gn x − =  n n − n2 |nx − 1|, if x ∈ [ n1 −

1 1 , n2 n 1 1 , n2 n

+

1 ]; n2

+

1 ]. n2

(2.144)

Recall that fn (x) is the tent function centered at n1 and the graph of it is an isosceles triangle with height n and base n22 so that each fn is continuous on [0, 1] and its area is n1 which implies that Z 1 1 fn (x) dx = → 0 n 0 p

See Step X for the meanings of the notations used here.

Chapter 2. Positive Borel Measures

60

as n → ∞. Furthermore, if x = 0, then x ∈ / [ n1 −

1 1 , n2 n

+

1 ] n2

for all n ∈ N. and thus

fn (0) = 0 for all n ∈ N by the definition (2.144). If x ∈ (0, 1], then there is a positive integer N such that 1 1 / [ n1 − n12 , n1 + n12 ] for all n ≥ N . Therefore, n + n2 < x for all n ≥ N . In other words, we have x ∈ the definition (2.144) certainly shows that fn (x) → 0 as n → ∞ for all x ∈ (0, 1]. Now it remains to show that f ∈ / L1 on [0, 1]. To this end, we study the behaviour of f as x → 0. For every n ∈ N, we consider every x in [ n1 − n12 , n1 + n12 ]. If n is sufficiently large enough, we have 1 1 , (2.145) x= +o n n where o(x) is the little-o notation.q Therefore, it follows from the definition (2.144) and the relation (2.145) that 1 fn (x) = n + o(1) = + o(1) (2.146) x for large enough n and small positive x. By the definition and the expression (5.102), we have f (x) = sup fn (x) ≥ fn (x) = n

1 + o(1) x

(2.147)

/ L1 on [0, 1], so we conclude from the estimate that (2.147) for small positive x. Obviously, x1 ∈ 1 that f ∈ / L on [0, 1]. This completes the proof of the problem.  Problem 2.21 Rudin Chapter 2 Exercise 21.

Proof. Let α ∈ f (X) ⊆ R and


Eα = {x ∈ X | f (x) ≥ α} = f −1 [α, ∞) .

By the definition, it is clear that Eα is nonempty. Furthermore, since
Eα = X \ f −1 (−∞, α) ⊆ X,

the set Eα is closed in X. Now we consider the collection {Eα } of subsets of X. Suppose that α1 , α2 , . . . , αn ∈ f (X) and we define α = max(α1 , α2 , . . . , αn ). Then we have n \

k=1

Eαk =

n \

{x ∈ X | f (x) ≥ αk } = Eα 6= ∅.

k=1

Therefore, {Eα } has the finite intersection property and then since X is compact, we conclude thatr \ Eα 6= ∅. Thus suppose that x0 ∈

\ α

α

Eα which means that f (x0 ) ≥ α for all α ∈ f (X), i.e., f (x0 ) ≥ f (x)

for all x ∈ X. Hence this shows that f attains its maximum at some point of X, completing the proof of the problem.  q r

Recall that g(n) = o(f (n)) means fg(n) → 0 as n → ∞. (n) See, for instances, [42, Theorem 26.9, p. 169] or [64, Problem 4.23, p. 43].

2.6. Miscellaneous Problems on L1 and Other Properties

61 Remark 2.4

Problem 2.21 can be treated as the Extreme Value Theorem for upper semicontinuous functions. Similarly, we can show that if f : X → R is lower semicontinuous and X is compact, then f attains its minimum at some point of X.

Problem 2.22 Rudin Chapter 2 Exercise 22.

Proof. By the definition and the triangle inequality, for all p ∈ X, we have gn (x) ≤ f (p) + nd(x, p) ≤ [f (p) + nd(y, p)] + nd(x, y). In other words, gn (x) − nd(x, y) is a lower bound of {f (p) + nd(y, p) | p ∈ X}. Thus, by the definition of infimum, we must have gn (x) − nd(x, y) ≤ gn (y) or equivalently, gn (x) − gn (y) ≤ nd(x, y).

(2.148)

gn (y) − gn (x) ≤ nd(x, y).

(2.149)

Likewise, we have Therefore, we conclude from the inequalities (2.148) and (2.149) that |gn (x) − gn (y)| ≤ nd(x, y). Since f (x) ≥ 0 on X and d(x, y) ≥ 0 for every x, y ∈ X, we have gn (x) ≥ 0 for all positive integers n. Fix x ∈ X, we observe that f (p) + nd(x, p) ≥ f (p) + (n − 1)d(x, p) ≥ gn−1 (x) for every p ∈ X and n = 2, 3, . . .. In other words, this implies that gn (x) ≥ gn−1 (x) for every x ∈ X and n = 2, 3, . . .. Besides, it is trivial that gn (x) ≤ f (x) holds for all n ∈ N if f (x) = ∞. Without loss of generality, we may assume that f (x) < ∞. In this case, since f (p) ≤ f (p) + nd(x, p) for all p ∈ X, if we take p = x in the definition of gn , then we have gn (x) ≤ f (x) for every n = 1, 2, . . .. This proves property (ii). To prove property (iii), we need the following lemma: Lemma 2.9 A function f : X → R is lower semicontinuous on X if and only if given x ∈ X, for every {xn } ⊆ X \ {x} converging to x and for every ǫ > 0, there exists a positive integer N such that n ≥ N implies f (x) < f (xn ) + ǫ.

(2.150)

Chapter 2. Positive Borel Measures

62

Proof of Lemma 2.9. For every α ∈ R, denote Eα = {x
∈ X | f (x) ≤ α}. Recall that
f −1 (α, ∞) is open in X if and only if X \ f −1 (α, ∞) = Eα is closed in X. In other words, f is lower semicontinuous on X if and only if Eα is closed in X. Let Eα be closed in X. Assume that the inequality (2.150) was invalid. Then it means that there exists a x0 ∈ X and a sequence {xn } in X converging to x0 such that for some ǫ > 0, the inequality f (x0 ) − ǫ ≥ f (xnk )

(2.151)

holds for infinitely many k. Take α ∈ (f (x0 ) − ǫ, f (x0 )). On the one hand, since f (x0 ) > α, x0 ∈ / Eα . On the other hand, we gain from the inequality (2.151) that f (xnk ) < α for infinitely many k and this implies that {xnk } ⊆ Eα . Since the set Eα is closed and xnk → x0 as k → ∞, we have x0 ∈ Eα , a contradiction. Hence the inequality (2.150) must hold for a lower semicontinuous function f . Conversely, we prove that the inequality (2.150) implies Eα is closed in X by showing that Eα contains all its limit points. To this end, pick α ∈ R such that Eα 6= ∅. Let {xn } ⊆ Eα \ {x} and xn → x as n → ∞. Given ǫ > 0. By the inequality (2.150), there exists a positive integer N such that f (x) − ǫ < f (xn )

(2.152)

for all n ≥ N . Since f (xn ) ≤ α, we deduce immediately from the inequality (2.152) that f (x) < α + ǫ. Since ǫ is arbitrary, we must have f (x) ≤ α or equivalently x ∈ Eα . This ends the  proof of the lemma. Remark 2.5 (a) Similarly, a function f : X → R is upper semicontinuous on X if and only if given x ∈ X, for every {xn } ⊆ X \ {x} converging to x and for every ǫ > 0, there exists a positive integer N such that n ≥ N implies f (x) > f (xn ) − ǫ. (b) Except the equivalent definition stated in Lemma 2.9, a lower semicontinuous function f on X can also be reformulated in terms of the lower limit of f : lim inf f (y) ≥ f (x) y→x

for every y → x in X \ {x}, see [61, pp. 69, 70] and [65, Definition 3.62, p. 64].

Now it is time to go back to the proof of the problem. On the one hand, since f is lower semicontinuous on X, Lemma 2.9 implies that for every ǫ > 0, there exists a δ > 0 such that f (y) > f (x) − ǫ

(2.153)

for all y ∈ B(x, δ). In this case, we always have f (y) + nd(x, y) > f (x) − ǫ

(2.154)

for every positive integer n. Since ǫ > 0 is arbitrary, the inequality (2.154) becomes f (y) + nd(x, y) ≥ f (x).

(2.155)

2.6. Miscellaneous Problems on L1 and Other Properties

63

On the other hand, we consider y ∈ / B(x, δ) so that d(x, y) ≥ δ. It is clear that nδ − ǫ > 0 holds for large enough n,s thus we get from the inequality (2.153) that f (y) + nd(x, y) ≥ f (y) + nδ > f (x) − ǫ + nδ > f (x) for large enough n. Therefore, we conclude that the inequality (2.155) always holds for all y ∈ X and for all large enough n. By the definition of limit inferior, we have lim inf gn (x) ≥ f (x).

(2.156)

lim sup gn (x) ≤ f (x)

(2.157)

n→∞

By property (ii), we see that n→∞

holds for every x ∈ X. Hence it deduces from the inequalities (2.156) and (2.157) that lim gn (x) = f (x)

n→∞

on X. By property (i), each gn is continuous on X. Next, property (ii) says that {gn } is increasing and bounded by f . Finally, property (iii) ensures that {gn } converges to f on X pointwisely.  This ends the proof of the problem. Remark 2.6 With the aid of Lemma 2.9, we can prove Problem 2.1(b) easily. In fact, given x ∈ X, for every {xn } ⊆ X \ {x} converging to x and for every ǫ > 0, there exists a positive integer N1 such that n ≥ N1 implies that ǫ (2.158) f1 (x) < f1 (xn ) + . 2 Similarly, there exists a positive integer N2 such that n ≥ N2 implies ǫ f2 (x) < f2 (xn ) + . 2

(2.159)

Take N = max(N1 , N2 ). If n ≥ N , then we derive from the inequalities (2.158) and (2.159) that f1 (x) + f2 (x) < f1 (xn ) + f2 (xn ) + ǫ. Hence it follows from Lemma 2.9 that f is lower semicontinuous on X.

Problem 2.23 Rudin Chapter 2 Exercise 23.

Proof. Let x ∈ Rk and f : Rk → R be defined by f (x) = µ(V + x). Since V + x is just a translation of V , it is also open in Rk , i.e., V + x ∈ B. s

Obviously, the n depends on δ and ǫ.

Chapter 2. Positive Borel Measures

64

We first construct a counter example which is not upper semicontinuous or continuous. By Example 1.20(b), we consider the unit mass concentrated at 0, i.e.,   1, if 0 ∈ E;

µ(E) =



0, if 0 ∈ / E.

If V = B(0, 1), then we have V + x = B(x, 1) and

µ(V + x) =

=

  1, if 0 ∈ B(x, 1);  0, if 0 ∈ / B(x, 1)   1, if |x| < 1; 

(2.160)

0, if |x| ≥ 1.

Thus it establishes from the result (2.160) that
f −1 (−∞, 1) = {x ∈ Rk | f (x) < 1} = {x ∈ Rk | µ(V + x) < 1} = Rk \ B(x, 1)

which is closed in Rk . By Definition 2.8, f is not upper semicontinuous or continuous.

However, we claim that f is always lower semicontinous. Let {xn } ⊆ Rk \ {x} be such that xn → x as n → ∞. Recall that both V + x and V + xn are open in Rk . We claim that y ∈ V + x implies that y ∈ V + xn for all but finitely many n. Let y = a + x for some a ∈ V . Since V is open in Rk , there exists a ǫ > 0 such that B(a, ǫ) ⊆ V. By the hypothesis, there is a positive integer N such that n ≥ N implies xn − x =

ǫ 2

or

ǫ xn − x = − . 2

In the first case, we have y =a+x=a−

ǫ ǫ ǫ + x + = a − + xn . 2 2 2

(2.161)

Since a − 2ǫ ∈ V , the expression (2.161) guarantees that y ∈ V + xn for all n ≥ N . The other case can be done similarly, so we omit the details here. This proves our claim. Now if χV +x (y) = 1, then we have χV +xn (y) = 1 for all but finitely many n. In other words, it means that lim inf χV +xn (y) ≥ χV +x (y) (2.162) n→∞

for every y ∈ Rk . By Proposition 1.9(d), each χV +xn : Rk → [0, ∞] is measurable, so we may apply Theorem 1.28 (Fatou’s Lemma) to conclude that  Z   Z (2.163) lim inf lim inf χV +xn dµ. χV +xn dµ ≥ n→∞

Rk

Rk

By Proposition 1.24(f), we have Z Z χV +xn dµ = Rk

V +xn

n→∞

dµ = µ(V + xn ).

(2.164)

2.6. Miscellaneous Problems on L1 and Other Properties

65

In addition, it follows from the inequality (2.162) and the application of Proposition 1.24(f) again that Z Z Z   dµ = µ(V + x). (2.165) χV +x (y) dµ = lim inf χV +xn dµ ≥ Rk

n→∞

Rk

V +x

Thus, by substituting the results (2.164) and (2.165) into the inequality (2.163), we achieve that lim inf µ(V + xn ) ≥ µ(V + x) n→∞

or equivalently, lim inf µ(V + y) ≥ µ(V + x) y→x

Rk \{x}.

for every y → x in Hence we conclude from Remark 2.5(b) that f is lower semicontinuos k on R , completing the proof of the problem.  Problem 2.24 Rudin Chapter 2 Exercise 24.

Proof. Suppose that S1 and S2 are the sets of simple functions and step functions on R respectively. We divide the proof into several steps. • Step 1: S1 is dense in L1 (R). Let f = f + − f 1 . By the Corollary (b) following Theorem 1.14, both f + and f − are measurable. Since f + : R → [0, ∞], we follow from Theorem 1.17 (The Simple Function Approximation Theorem) that there is a sequence {sn } of nonnegative increasing simple functions such that for every x ∈ R, sn (x) → f + (x) as n → ∞. Thus Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem) implies that Z ∞ Z ∞ f + dx (2.166) sn dx → −∞

−∞

as n → ∞. It is clear that |f + (x) − sn (x)| = f + (x) − sn (x) for every x ∈ R, so we havet Z ∞ Z ∞ + [f + (x) − sn (x)] dx |f (x) − sn (x)| dx = −∞ −∞ Z ∞ Z ∞ + sn (x) dx. (2.167) f (x) dx − = −∞

−∞

If we apply the limit (2.166) to the expression (2.167), then we gain Z ∞ |f + (x) − sn (x)| dx = 0. lim n→∞ −∞

Likewise, the same is true for f − , so we have Z ∞ |f − (x) − tn (x)| dx = 0 lim n→∞ −∞

where {tn } is a sequence of nonnegative increasing simple functions such that for every x ∈ R, tn (x) → f − (x) as n → ∞. Hence {gn = sn − tn } is a sequence of simple functions such that Z ∞ Z ∞ |f + − f − − sn + tn | dx |f − gn | dx = −∞

−∞

Since |f (x)| ≤ |f (x)| on R and f ∈ L (R), we have f + ∈ L1 (R). Thus this makes the second equality holds because the right-hand side is not in the form ∞ − ∞. t

+

1

Chapter 2. Positive Borel Measures

≤

66

Z

∞ −∞

+

|f − sn | dx +

→0

Z

∞ −∞

|f − − tn | dx

as n → ∞. • Step 2: S2 is dense in S1 . To this end, it suffices to prove that for every measurable set E ⊂ R with m(E) < ∞ and every ǫ > 0, there exists a step function g on R such that Z ∞ |χE − g| dx < ǫ. (2.168) −∞

By Theorem 2.20(b), there exists an open set V in R such that E ⊆ V and m(V \ E) < 2ǫ . By [49, Exercise 29, Chap. 2, p. 45], we have V =

∞ [

(an , bn ),

(2.169)

n=1

where (ai , bi ) ∩ (aj , bj ) = ∅ if i 6= j. Assume that there was an unbounded segment in the expression (2.169). Then we have m(V ) = ∞. However, since m(V ) = m(V \ E) + m(E), the facts m(V ) = ∞ and m(V \ E) < 2ǫ will force that m(E) = ∞, a contradiction. Furthermore, the expression (2.169) implies that m(V ) =

∞ X

n=1


m (an , bn ) ,

so there exists a positive integer N such that m(V ) −

N X
ǫ m (an , bn ) . < 2

(2.170)

n=1

Define g : R → R by g(x) =

N X

χ(an ,bn ) (x)

n=1

and VN = (a1 , b1 ) ∪ · · · ∪ (aN , bN ). Let x ∈ (E \ VN ) ∪ (VN \ E). If x ∈ E \ VN , then we have χE (x) = 1 and g(x) = 0. Similarly, if x ∈ VN \ E, then we have χE (x) = 0 and g(x) = 1. Therefore, we have |g − χE | = 1 (2.171) on (E \ VN ) ∪ (VN \ E). On the other hand, we let

x ∈ (E \ VN )c ∩ (VN \ E)c = (VN ∪ E c ) ∩ (E ∪ VNc ) = (VN ∩ E) ∪ (VNc ∩ E c ). If x ∈ VN ∩ E, then we have χE (x) = g(x) = 1. Similarly, x ∈ VNc ∩ E c implies that χE (x) = g(x) = 0. Both cases give |g − χE | = 0

(2.172)

on (E \ VN )c ∩ (VN \ E)c . Now the facts E, VN ⊆ V and (2.170) definitely imply that ǫ ǫ m(VN \ E) ≤ m(V \ E) < and m(E \ VN ) ≤ m(V \ VN ) < . (2.173) 2 2 Hence we deduce from the results (2.171), (2.172) and the estimates (2.173) that Z Z Z ∞ dx = m(VN \ E) + m(E \ VN ) < ǫ. dx + |g − χE | dx = −∞

E\VN

This is the desired result (2.168).

VN \E

2.6. Miscellaneous Problems on L1 and Other Properties

67

• Step 3: S2 is dense in L1 (R). We need the following lemma: Lemma 2.10 Suppose that A, B and C are subsets of a metric space X. If A ⊆ B ⊆ C, A is dense in B and B is dense in C, then A is dense in C.

Proof of Lemma 2.10. Recall the definition that A is dense in S if A ⊆ S ⊆ A, where A denotes the closure of A in X. Then we have A ⊆ B ⊆ A and B ⊆ C ⊆ B which show immediately that A ⊆ C. Let p ∈ B. We claim that p ∈ A. To this end, we know from the assumption that B(p, δ) ∩ (B \ {p}) 6= ∅ for every δ > 0. Let q ∈ B(p, δ) ∩ (B \ {p}). Since B ⊆ A, A also contains q. If q ∈ A, then we have B(p, δ) ∩ (A \ {p}) 6= ∅ (2.174) for every δ > 0. Therefore, the definition implies that p ∈ A. Next, suppose that q ∈ A′ . Since B(p, δ) is open in X, there exists a ǫ > 0 such that B(q, ǫ) ⊆ B(p, δ). Since q is a limit point of A, the definition shows that B(q, ǫ) ∩ A 6= ∅ and thus the set relation (2.174) still holds in this case. Therefore, we have proven our claim that p ∈ A and this means that B ⊆ A. Hence we establish the set relations A⊆C⊆A which implies that A is dense in C, completing the proof of Lemma 2.10.  Let’s return to the original proof. By applying Lemma 2.10 to the results in Step 1 and Step 2, we conclude easily that S2 is dense in L1 (R) which is our desired result. Hence we have completed the proof of the problem.



Remark 2.7 By Definition 1.16, we remark that a simple function s is a finite linear combination of characteristic functions of an arbitrary set Ai and when all Ai are intervals, then s becomes a step function. In fact, the class of step functions is one of the building blocks for the theory of Riemann integration, see [49, Chap. 6].

Problem 2.25 Rudin Chapter 2 Exercise 25.

Proof. (a) We note that et > 0 and log(1 + et ) > 0 for every t > 0. Furthermore, we know that the functions ex and log x are increasing in their corresponding domains respectively. Thus

Chapter 2. Positive Borel Measures

68

the inequality log(1 + et ) < c + t is equivalent to 1 + et < ec+t and finally, it is equivalent to 0 < 1 + e−t < ec . (2.175) Apply log to both sides of the inequality (2.175), we have log(1 + e−t ) < c

(2.176)

for every t > 0. Since the left-hand side of the inequality (2.176) is a decreasing function of t, the smallest value of c for the validity of the inequality (2.176) on (0, ∞) is found by lim log(1 + e−t ) = log 2.

t→0 t>0


(b) Let f ∈ L1 on [0, 1]. Define E = {x ∈ [0, 1] | f (x) > 0} = f −1 (0, ∞) . By Theorem 1.12(b), E is a measurable set in [0, 1]. Thus we follow from part (a) that, on E,

nf (x) = log enf (x) < log 1 + enf (x) < log 2 + nf (x) on (0, ∞) and this implies that Z Z Z log 2 1 f dµ. log(1 + enf ) dµ < + f dµ < n E n E E

(2.177)

Next, we have 1 ≤ 1 + enf (x) ≤ 2 on [0, 1] \ E. Thus we have 0 ≤ log(1 + enf ) ≤ log 2

(2.178)

on [0, 1]\E. Since [0, 1]\E is also a measurable set in [0, 1], we deduce from the inequalities (2.178) that Z log 2 1 log(1 + enf ) dµ ≤ . 0≤ n [0,1]\E n Now the sum of the inequalities (2.177) and (2.178) give Z

1 f dµ < n E

Z

1 0

2 log 2 + log(1 + e ) dµ < n nf

Z

f dµ.

(2.179)

E

Hence, by taking n → ∞ in the inequalities (2.179), we get that 1 lim n→∞ n

Z

1

nf

log(1 + e ) dµ =

0

This completes the proof of the problem.

Z

f dµ. E



CHAPTER

3

Lp-Spaces

3.1

Properties of Convex Functions

Problem 3.1 Rudin Chapter 3 Exercise 1.

Proof. Let {ϕα } be a collection of convex functions on (a, b). Define ϕ = sup{ϕα }

(3.1)

and assume that it is finite. Suppose that x, y ∈ (a, b) and 0 ≤ λ ≤ 1. By the definition (3.1) and the convexity of ϕα , we have
ϕα (1 − λ)x + λy ≤ (1 − λ)ϕα (x) + λϕα (y) ≤ (1 − λ)ϕ(x) + λϕ(y)

for all α. In other words, we have


ϕ (1 − λ)x + λy ≤ (1 − λ)ϕ(x) + λϕ(y).

By Definition 3.1, ϕ is convex on (a, b).

Suppose that {ϕn } is a sequence of convex functions on (a, b). For x ∈ (a, b), we define ϕ : (a, b) → R to be ϕ(x) = lim ϕn (x). (3.2) n→∞

Now for x, y ∈ (a, b) and 0 ≤ λ ≤ 1, we follow from the limit (3.2) that lim [(1 − λ)ϕn (x) + λϕn (y)] = (1 − λ)ϕ(x) + λϕ(y)

n→∞

and



lim ϕn (1 − λ)x + λy = ϕ (1 − λ)x + λy .

n→∞

(3.3)

(3.4)


Since ϕn (1 − λ)x + λy ≤ (1 − λ)ϕn (x) + λϕn (y) for every n = 1, 2, . . ., we take limits to both sides and apply the results (3.3) and (3.4) to conclude that
ϕ (1 − λ)x + λy ≤ (1 − λ)ϕ(x) + λϕ(y).

Hence ϕ is convex on (a, b) by Definition 3.1.

69

Chapter 3. Lp -Spaces

70

For each positive integer n, suppose that fn : (a, b) → R is convex. Define f : (a, b) → R by f (x) = lim sup fn (x), n→∞

where x ∈ (a, b). By Definition 1.13, we have f (x) = lim gk (x), k→∞

(3.5)

where gk = sup{fk , fk+1 , . . .} for k = 1, 2, . . .. Thus if each fn is convex on (a, b), then the first assertion shows that each gk is convex on (a, b). By the definition (3.5), since f is the pointwise limit of the sequence of convex functions {g1 , g2 , . . .} defined on (a, b), we deduce from the second assertion that f is also convex on (a, b). However, the lower limit of a sequence of convex functions may not be convex. For example, consider the functions fn : R → R defined by fn (x) = (−1)n x. It is clear that each fn is convex on R. However, for each x ∈ R, we have f (x) = lim inf fn (x) = −|x|. n→∞

This shows that f is not convex on R and we complete the proof of the problem.



Problem 3.2 Rudin Chapter 3 Exercise 2.

Proof. For every x, y ∈ (a, b) and 0 ≤ λ ≤ 1, since ϕ is convex on (a, b), we have
ϕ (1 − λ)x + λy ≤ (1 − λ)ϕ(x) + λϕ(y). (3.6)
Since ψ is convex and nondecreasing on ϕ (a, b) , we obtain from the inequality (3.6) that 



ψ ϕ (1 − λ)x + λy ≤ ψ (1 − λ)ϕ(x) + λϕ(y) ≤ (1 − λ)ψ ϕ(x) + λψ ϕ(y) . By Definition 3.1, ψ ◦ ϕ is convex on (a, b).

Let ϕ > 0 on (a, b). By [51, Theorem (c), p. 2], the function exp is a monotonically increasing positive function on R. Since (ex )′ = ex , it is convex on R by the paragraph following Definition 3.1. Hence, if log ϕ is convex on (a, b), then we conclude from the first assertion that ϕ = elog ϕ is also convex on (a, b). However, the converse of the second assertion is false. For instance, we know that ϕ(x) = x

is convex on (0, ∞), but log x is not convex on (0, ∞) because (log x)′′ = −x−2 < 0 on (0, ∞), i.e., (log x)′ is not a monotonically increasing function on (0, ∞). We have completed the proof  of the problem. Problem 3.3 Rudin Chapter 3 Exercise 3.

Proof. This problem is proven in [63, Problem 4.24, pp. 79 – 81].



3.2. Relations among Lp -Spaces and some Consequences

71

Relations among Lp -Spaces and some Consequences

3.2

Problem 3.4 Rudin Chapter 3 Exercise 4.

Proof. (a) Since 0 < r < p < s, we can find λ ∈ (0, 1) such that p = λr + (1 − λ)s. By Theorem 3.5 (H¨older’s Inequality), we have Z |f |p dµ ϕ(p) = ZX |f |λr × |f |(1−λ)s dµ = X nZ o1−λ oλ n Z
1 1 ≤ |f |λr λ dµ × [|f |(1−λ)s ] 1−λ dµ X

X

= [ϕ(r)]λ × [ϕ(s)]1−λ .

(3.7)

Since r, s ∈ E, ϕ(r) and ϕ(s) are finite. Hence the inequality (3.7) ensures that ϕ(p) is also finite, i.e., p ∈ E. (b) We prove the assertion one by one. – Case (i): ln ϕ is convex in E ◦ . If E ◦ = ∅, then there is nothing to prove. Assume that E ◦ 6= ∅. By [49, Theorem 2.47, p. 42] and part (a), the set E is connected. Since E ⊆ (0, ∞), E is either an interval in one of the forms [a, b], [a, b), (a, b] or (a, b) for some positive a and b with a < b. In each of the cases, we always have E ◦ = (a, b). Let x, y ∈ (a, b) and λ ∈ [0, 1]. Since (a, b) is a convex set, λx + (1 − λ)y ∈ (a, b). Thus it follows from the inequality (3.7) that
ϕ λx + (1 − λ)y ≤ [ϕ(x)]λ × [ϕ(y)]1−λ . (3.8) If ϕ(p) = 0 for some p ∈ (0, ∞), then we have Z |f |p dµ = 0 X

so that |f (x)| = 0 almost everywhere on X. By the remark following Definition 3.7, it implies that kf k∞ = 0, a contradiction. Hence we have ϕ(p) > 0 and we are allowed to take the logarithm to both sides of the inequality (3.8) to get
 ln ϕ λx + (1 − λ)y ≤ ln [ϕ(x)]λ × [ϕ(y)]1−λ ≤ λ ln ϕ(x) + (1 − λ) ln ϕ(y). By Definition 3.1, ln ϕ is convex in (a, b).

– Case (ii): ϕ is continuous on E. By Theorem 3.2, ln ϕ is continuous on (a, b). Thus ϕ is also continuous on (a, b), so it remains to show that ϕ is continuous at the endpoints. Let a ∈ E. Then E is either [a, b) or [a, b], but no matter which case E is, a is a limit point of E so that we can find a decreasing sequence {pn } ⊆ (a, b) such that pn → a as n → ∞. By this, there exists a ǫ > 0 and a positive integer N such that pn ∈ (a, a + ǫ) ⊂ E for all n ≥ N . Fix this ǫ and then n ≥ N implies that   |f (x)|a+ǫ , if |f (x)| ≥ 1; |f (x)|pn ≤  |f (x)|a , if |f (x)| < 1.

Chapter 3. Lp -Spaces

72

Hence we obtain that |f (x)|pn ≤ |f (x)|a+ǫ + |f (x)|a

on X. We recall that a, a + ǫ ∈ E, so the definition implies that ϕ(a) and ϕ(a + ǫ) are finite. Since f is measurable, |f |a and |f |a+ǫ are measurable by Proposition 1.9(b) and Theorem 1.7(b). Thus these two facts show that |f |a , |f |a+ǫ ∈ L1 (µ). By Theorem 1.32, we have |f |a+ǫ + |f |a ∈ L1 (µ). For each x ∈ X, |f (x)| ≥ 0. Since an exponential function with nonnegative base is continuous on its domain, we have lim |f (x)|pn = |f (x)|a

n→∞

on X. In conclusion, we have shown that the sequence {|f |pn } satisfies the hypotheses of Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) and then Z Z |f |a dµ = ϕ(a). |f |pn dµ = lim ϕ(pn ) = lim n→∞

n→∞ X

X

By definition, ϕ is continuous at a. Similarly, ϕ is continuous at b if b ∈ E. Consequently, ϕ is continuous on E.

(c) We claim that E can be any open or closed connected subset of (0, ∞). We consider X = (0, ∞), µ = m and 0 < a ≤ b. We need the following result (see [2, Examples 1 & 5, pp. 417, 419]): Lemma 3.1 Let x > 0 and b > 1. Then we have  1 − x1−α   Z 1  , if α 6= 1; 1−α t−α dt =  x   − ln x if α = 1 and

 1−β b −1   , if β = 6 1;  b 1−β t−β dt =  1   ln b, if β = 1.

Z

Furthermore, the improper integrals Z 1 x−α dx and 0

Z

∞

x−β dx

1

are convergent if and only if α < 1 and β > 1 respectively.

– Case (i): E = (a, b). For x ∈ (0, ∞), let  1   x− b , if x ∈ (0, 1); f (x) =   x− a1 , if x ∈ [1, ∞). We have to find p ∈ (0, ∞) such that Z ϕ(p) =

0

∞

|f (x)|p dx < ∞.

3.2. Relations among Lp -Spaces and some Consequences

73

We assume that the integral can be split as follows: Z 1 Z ∞ p p x− b dx + ϕ(p) = x− a dx. 0

(3.9)

1

Apply Lemma 3.1 to the two improper integrals in the expression (3.9), we know that ϕ(p) < ∞ if and only if pb < 1 and ap > 1 if and only if a < p < b. In other words, we have E = (a, b) and such split is allowable. – Case (ii): E = [a, b]. In this case, we need another lemma:a Lemma 3.2 The improper integral

Z

e−1

dx xα | ln x|β

0

converges if and only if (i) α < 1 or (ii) α = 1 and β > 1. The improper integral Z ∞ dx α x (ln x)β e converges if and only if (i) α > 1 or (ii) α = 1 and β > 1.

We consider  1  , if x ∈ (0, e−1 );  1  1+ 1b  b x | ln x|       1 1  1 e− a − e b f (x) = b (x − e−1 ), if x ∈ (e−1 , e); e + −1  e − e        1   if x ∈ [e, ∞).  1 1 , x a (ln x)1+ a

(3.10)

It is clear that f is continuous on (0, ∞) and so it is measurable. Similar to Case (i), we assume that the integral can be split as follows: Z ∞ |f (x)|p dx ϕ(p) = =

Z

0

e−1

dx

p

1

x b | ln x|p(1+ b ) Z ∞ dx

0

+

p

e

1

+

x a (ln x)p(1+ a )

Z

e

e−1

h

1 1 ip e− a − e b −1 (x − e ) dx e + e − e−1 1 b

.

(3.11)

By Lemma 3.2, the first integral in the expression (3.11) converges if and only if (i) p p 1 b < 1 or (ii) b = 1 and p(1 + b ) > 1. The first condition gives p < b. If p = b, then p(1 + 1b ) = 1 + b > 1 so that the second condition is actually p = b. In this case, we have p ≤ b. Similarly, we apply Lemma 3.2 to the third integral in the expression (3.11), so it converges if and only if (i) ap > 1 or (ii) ap = 1 and p(1 + a1 ) > 1. The first condition a

The integrals in Lemma 3.2 are called Bertrand’s integrals, see the following webpage https://fr.wikipedia.org/wiki/Int%C3%A9grale_impropre.

Chapter 3. Lp -Spaces

74

shows p > a. If p = a, then p(1 + a1 ) = 1 + a > 1 so that the second condition implies that p = a. In this case, we have p ≥ a. Since the second integral in the expression (3.11) is finite, we combine these observations to conclude that ϕ(p) < ∞ if and only if p ∈ [a, b].

– Case (iii): E = {a}. If we take b = a in the definition (3.10), then we obtain p ≥ a and p ≤ a. Hence we establish E = {a}.

– Case (iv): E = ∅. Consider f ≡ 1 on (0, ∞). Since the integral Z ∞ |1|p dx 0

is obviously divergent for every p ∈ (0, ∞), we conclude that E = ∅ in this case. (d) Let r < p < s. By the proof of part (a), we see that p = λr + (1 − λ)s for some λ ∈ (0, 1) so that the inequality (3.7) holds. In other words, we have
λ
1−λ kf kpp ≤ kf krr × kf kss . (3.12) It is clear that

kf krr


λ

× kf kss


1−λ

≤

=


λ
1−λ  , if kf kr ≤ kf ks ;  kf krs × kf kss

  kf kr
λ × kf ks
1−λ , if kf k ≤ kf k s r r r  p  kf ks , if kf kr ≤ kf ks ; 

kf kpr ,

(3.13)

if kf ks ≤ kf kr .

Hence, by combining the inequalities (3.12) and (3.13), we get kf kp ≤ max(kf kr , kf ks ).

(3.14)

Let f ∈ Lr (µ) ∩ Ls (µ). By the definition, we have kf kr < ∞

and

kf ks < ∞.

Hence we follow from these and the inequality (3.14) that kf kp < ∞, i.e., f ∈ Lp (µ) and then Lr (µ) ∩ Ls (µ) ⊆ Lp (µ). (e) Recall that kf k∞ > 0. – Case (i): kf k∞ < ∞. Let Eα = {x ∈ X | |f (x)| ≥ α}, where α ∈ (0, kf k∞ ). It is clear that Eα is measurable because |f | is measurable. By Definition 3.7, kf k∞ is the smallest number such that µ({x ∈ X | |f (x)| > kf k∞ }) = 0. Thus we have µ(Eα ) > 0 for every α ∈ (0, kf k∞ ). Assume that µ(Eα ) = ∞. Then we have Z Z Z r r r |f |r dµ ≥ αr µ(Eα ) = ∞ |f | dµ + |f | dµ = kf kr = X\Eα

Eα

X

which contradicts to our hypothesis. Thus we have 0 < µ(Eα ) < ∞ and then Proposition 1.24(b) implies that Z Z |f |p dµ ≥ αp µ(Eα ) |f |p dµ ≥ kf kpp = X

Eα

3.2. Relations among Lp -Spaces and some Consequences

75 which shows that

1

kf kp ≥ α[µ(Eα )] p .

(3.15)

1

Since lim [µ(Eα )] p = 1, we obtain from the inequality (3.15) that p→∞

lim inf kf kp ≥ α. p→∞

In particular, since α is arbitrary in (0, kf k∞ ), we have lim inf kf kp ≥ kf k∞ . p→∞

Next, if p > r, then we have Z Z p p |f | dµ = kf kp = |f |p−r × |f |r dµ.

(3.16)

(3.17)

X

X

By Definition 3.7, since |f (x)| ≤ kf k∞ for almost all x ∈ X, we get from the expression (3.17) and Proposition 1.24(b) that Z Z p p−r r p−r r |f |r dµ = kf kp−r kf kp ≤ kf k∞ × |f | dµ = kf k∞ × (3.18) ∞ × kf kr , X\E

X\E

where E = {x ∈ X | |f (x)| > kf k∞ } is of measure 0. Rewrite the inequality (3.18) in the form r 1− r kf kp ≤ kf k∞ p kf krp which, by using the fact that kf kr < ∞, implies lim sup kf kp ≤ kf k∞ .

(3.19)

p→∞

Hence our desired result follows immediately from the inequalities (3.16) and (3.19). – Case (ii): kf k∞ = ∞. Then we deduce immediately from the inequality (3.16) that lim kf kp = lim inf kf kp = ∞ = kf k∞ .

p→∞

p→∞

Hence we have completed the proof of the problem.



Problem 3.5 Rudin Chapter 3 Exercise 5.

Proof. (a) Suppose that s = ∞. Since |f (x)| ≤ kf k∞ holds for almost all x ∈ X, we have µ(E) = µ({x ∈ X | |f (x)| > kf k∞ }) = 0 so that kf krr

=

Z

X\E

r

|f | dµ ≤

Z

X\E

kf kr∞ dµ = kf kr∞ · µ(X \ E) = kf kr∞ .

Hence we have kf kr ≤ kf k∞ in this case.

(3.20)

Chapter 3. Lp -Spaces

76

Next, we suppose that s < ∞. Since f is measurable, |f |r is also measurable on X by Proposition 1.9(b) and Theorem 1.7(b) for every r > 0. We apply Theorem 3.5 (H¨older’s Inequality) to the measurable functions |f |r and 1 to obtain that Z

X

|f |r dµ ≤ = =

nZ Z

|f |r

X

s


s r

|f | dµ

X kf krs .

dµ

r

s

or

s

×

nZ

X

dµ

o1− r

s

(3.21)

Hence we conclude that kf kr ≤ kf ks . (b) Suppose that 0 < r < s < ∞. Since kf kr = kf ks , it means that the equality holds in the inequality (3.21). Recall that the inequality (3.21) is derived from Theorem 3.5 (H¨older’s Inequality), so there are constants α and β, not both 0, such that s

α(|f |r ) r = β

a.e. on X.

If α = 0, then it is trivial that β = 0, a contradiction. Thus α 6= 0 and then we have |f | =

β 1 s

α

a.e.

In other words, |f | is a constant a.e. on X. Next, suppose that s = ∞ and kf kr = kf k∞ < ∞, where 0 < r < ∞. Then the equality holds in the inequality (3.20) so that |f |r = kf kr∞ < ∞ a.e. on X, i.e., |f | = kf k∞

a.e. on X.

In other words, we achieve that |f | is a constant a.e. on X in this case. Consequently, we conclude that 0 < r < s ≤ ∞ and the conditions kf kr = kf ks < ∞ hold if and only if |f | is a constant a.e. on X (c) If f ∈ Ls (µ), then kf ks < ∞ and we know from part (a) that kf kr < ∞, i.e., f ∈ Lr (µ). Thus it is true that Ls (µ) ⊆ Lr (µ). Now the other inclusion Lr (µ) ⊆ Ls (µ) is a consequence of the following lemmab : Lemma 3.3 Suppose that (X, M, µ) is a measure space and M0 = {E ∈ M | µ(E) > 0}. Then the following conditions are equivalent: – Condition (1). inf{µ(E) | E ∈ M0 } > 0, – Condition (2). Lr (µ) ⊆ Ls (µ) for all 0 < r < s ≤ ∞.

b

This lemma is part of the statements in [62, Theorem 1] and our proof follows part of the argument there.

3.2. Relations among Lp -Spaces and some Consequences

77

Proof of Lemma 3.3. Suppose that Condition (1) is true. Suppose that f ∈ Lr (µ) and En = {x ∈ X | |f (x)| ≥ n} for every n ∈ N. We claim that there exists an N ∈ N such that µ(EN ) = 0. Otherwise, there exists ǫ > 0 such that µ(En ) ≥ ǫ for each ∞ \ En . By the construction of En , we have E1 ⊇ E2 ⊇ · · · . Since n ∈ N. Let E = n=1

µ(E1 ) ≤ µ(X) < ∞, it yields from Theorem 1.19(e) that

µ(E) = lim µ(En ) ≥ ǫ > 0.

(3.22)

n→∞

However, we deduce from the inequality (3.22) that 1 1  Z Z 1 r r r |f |r dµ = ∞ · µ(E) r = ∞, |f | dµ ≥ kf kr = E

X

a contradiction. Therefore, we have our claim that µ(EN ) = 0 for some N ∈ N. In other words, |f (x)| ≤ N holds for almost all x ∈ X and we obtain from Definition 3.7 that kf k∞ ≤ N , i.e., f ∈ L∞ (µ) and so f ∈ Ls (µ). Hence Condition (2) holds. Assume that Condition (1) was false. Thus there exists a sequence of {En } ⊆ M0 such that 1 (3.23) 0 < µ(En ) < n . 2 Put F1 = E1 and Fn = En \ (En−1 ∪ · · · ∪ E1 ), where n = 2, 3, . . .. Then it is easy to check that Fi ∩ Fj = ∅ for i 6= j and
1 0 < µ(Fn ) = µ En \ (En−1 ∪ · · · ∪ E1 ) ≤ µ(En ) < n 2

for every n = 1, 2, . . .. Therefore, we may assume that {En } is a sequence of disjoint measurable sets. ∞ [ En . Suppose first that Next, we define αn = µ(En ) for n = 1, 2, . . . and E = n=1

s < ∞. Then we define f : X → R by

∞ X

f (x) =

−1

αn s χEn (x).

(3.24)

n=1

Since r < s, we have 1 − rs < 1. Note that f (x) = 0 if x ∈ X \ E. Thus it reduces from the bounds (3.23) and Theorem 1.29 that kf krr

=

Z

r

X\E

|f | dµ +

Z

r

E

|f | dµ =

∞ Z X

−r αn s

dµ =

n=1 En

∞ X

1− r αn s

n=1

r

21− s < r < ∞ 1 − 21− s

which means f ∈ Lr (µ). Similarly, it is easy to derive that kf kss

=

Z

s

E

|f | dµ =

∞ Z X

n=1 En

α−1 n dµ =

∞ X

n=1

1=∞

which implies that f ∈ / Ls (µ). Hence we have shown that Lr (µ) * Ls (µ). −1

−

1

If s = ∞, then we replace the coefficients αn s by αn 2r in the expression (3.24). By −

1

n

similar argument, since αn 2r > 2 2r for all n ∈ N, we conclude kf k∞ = ∞ and then Lr (µ) * L∞ (µ). 

Chapter 3. Lp -Spaces

78

(d) If 0 < p < q, then part (a) says that kf kp ≤ kf kq < ∞. Let E0 = {x ∈ X | |f (x)| = 0}. By Problem 1.5, E0 is measurable. – Case (i): µ(E0 ) > 0. Now the set E∞ = {x ∈ X | |f (x)| = ∞} is measurable by Problem 1.5. Pick p ∈ (0, r), so part (a) indicates that kf kp ≤ kf kr < ∞. If µ(E∞ ) > 0, then we have nZ o1 nZ o1 1 p p p kf kp = |f | dµ ≥ |f |p dµ = [∞ · µ(E∞ )] p = ∞, X

E∞

a contradiction. Thus we have µ(E∞ ) = 0, i.e., |f (x)| < ∞ a.e. on X. Consequently, we have Z Z Z Z log |f | dµ (3.25) log |f | dµ + log |f | dµ = log |f | dµ = X\(E∞ ∪E0 )

X\E∞

X

E0

Since µ(X) = 1 and |f (x)| is nonzero finite on X \ (E∞ ∪ E0 ), the first integral on the right-hand side of the equation (3.25) is finite. However, the second integral on the right-hand side of the equation (3.25) is in the form −∞ so that the right-hand side of our desired result is in the form exp{−∞} which is defined to be 0. On the other hand, we notice that o1 n Z nZ o1 p p p |f | dµ kf kp = |f |p χX\E0 dµ . = (3.26) X

X

Apply Theorem 3.5 (H¨older’s Inequality) with u > 1 and v > 1 as fixed conjugate exponents to the right-hand side of the expression (3.26), we see that o1 o1 o 1 nZ nZ nZ p pu pv p u p (χX\E0 )v dµ (|f | ) dµ |f | χX\E0 dµ ≤ X

X

X

1 pv

= µ(X \ E0 ) kf kpu .

(3.27)

Combining the expression (3.26) and the inequality (3.27), we gain 1

kf kp ≤ µ(X \ E0 ) pv kf kpu . Since µ(X) = 1 and µ(E0 ) > 0, we have µ(X \ E0 ) < 1 and then 1

lim µ(X \ E0 ) pv = 0.

(3.28)

p→0

In addition, when p is chosen to be small enough such that pu < r, part (a) implies that kf kpu ≤ kf kr < ∞. Hence we observe from the limit (3.28) and this fact that 1

1

lim kf kp ≤ lim µ(X \ E0 ) pv kf kpu ≤ kf kr lim µ(X \ E0 ) pv = 0.

p→0

p→0

p→0

In other words, the desired result holds in this case. – Case (ii): µ(E0 ) = 0. Then |f (x)| > 0 a.e. on X. By the fact used in Case (i), we may assume that 0 < |f (x)| < ∞ a.e. on X. (3.29) This fact allows us to apply [51, Eqn. (7), p. 63] to the almost positive function |f |p to obtain o n1 Z o nZ o1 nZ p p p log |f | dµ ≥ exp log |f | dµ = exp |f | dµ kf kp = p X X X

3.2. Relations among Lp -Spaces and some Consequences

79

for every p ∈ (0, r). In particular, lim kf kp ≥ exp

p→0+

nZ

X

o log |f | dµ .

(3.30)

For the reverse direction, we consider the function φ : (0, ∞) → R defined by φ(x) = x − 1 − log x.

Since φ′ (x) = 1 − x1 = 0 and φ′′ (x) = x12 , φ has the absolute minimum at x = 1, i.e., φ(x) ≥ φ(1) on (0, ∞) or equivalently log x ≤ x − 1

(3.31)

on (0, ∞). Recall the fact (3.29), so we replace x by kf kpp in the inequality (3.31) and then using the fact that µ(X) = 1 to obtain Z Z  Z |f |p − 1 kf kpp − 1 1 p log kf kp ≤ dµ = |f | dµ − = dµ. (3.32) p p X p X X Let E = {x ∈ X | |f (x)| = 1} and F = {x ∈ X | |f (x)| = 6 1}. Then we get Z Z Z |f |p − 1 |f |p − 1 |f |p − 1 dµ = dµ + dµ. p p p E X F

(3.33)

It is clear from Proposition 1.24(d) that Z |f |p − 1 dµ = 0, p E so it suffices to evaluate

Z

|f |p − 1 dµ. p→0+ F p To this end, we need the following lemma: lim

Lemma 3.4 Let |f | > 0 and kf kr < ∞ for some r > 0. Define ψ : R → R by ψ(p) = |f |p . Then we have |f |p − 1 ∈ L1 (µ) p for all p ∈ (0, r).

Proof of Lemma 3.4. Since ψ ′′ = |f |p (log |f |)2 > 0 on R, ψ is convex on R. Thus we see from [49, Exercise 23, p. 101] that, for 0 < p < r, ψ(p) − ψ(0) ψ(r) − ψ(0) ≤ p−0 r−0 which reduces to

and so

|f |p − 1 |f |r − 1 ≤ p r Z

X

|f |p − 1 dµ ≤ p

In other words, we have lemma.

|f |p −1 p

Z

X

|f |r − 1 1 1 dµ = kf krr − < ∞. r r r

∈ L1 (µ) for all p ∈ (0, r), completing the proof of the 

Chapter 3. Lp -Spaces

80

Let’s return to the proof of the problem. By Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem), we obtain that Z Z  |f |p − 1  |f |p − 1 lim dµ. (3.34) lim dµ = p→0+ F p p F p→0+ By L’ Hˆospital Rule, we derive |f |p − 1 = lim |f |p log |f | = log |f | p→0+ p→0+ p lim

(3.35)

on X. Therefore, when we combine the results (3.32), (3.33), (3.34) and (3.35), we establish that Z Z Z  |f |p − 1  lim log |f | dµ. (3.36) lim log kf kp ≤ log |f | dµ = dµ = p→0+ p F p→0+ X F Since log x is continuous for x > 0, we get from the inequality (3.36) that nZ o lim kf kp ≤ exp log |f | dµ . p→0+

(3.37)

X

Hence the required result follows immediately from the inequalities (3.30) and (3.37). This ends the proof of the problem.



Problem 3.6 Rudin Chapter 3 Exercise 6.

Proof. Let x > 0 and 0 ≤ c ≤ 1. We claim that the function Φ : [0, ∞) → R satisfies the relation cΦ(x) + (1 − c)Φ(1) = Φ(xc ).

(3.38)

Obviously, the relation (3.38) holds for c = 0 and c = 1. Without loss of generality, we may assume that 0 < c < 1 in the following discussion. Since f is bounded measurable and positive on [0, 1], there exists a positive constant M such that Z 1 |f | dx ≤ M < ∞. kf k1 = 0

Thus we get from Problem 3.5(d) that lim kf kp = exp

p→0

nZ

1

0

o log |f (t)| dt .

(3.39)

By putting the expression (3.39) into the relation in question, we obtain  nZ Φ exp

0

1

log |f (t)| dt

We define f : [0, 1] → R by f (t) =

o

=

Z

0

1


Φ f (t) dt.

  x, if x ∈ [0, c]; 

1, if x ∈ (c, 1].

(3.40)

(3.41)

3.2. Relations among Lp -Spaces and some Consequences

81

Since x > 0, the f is bounded measurable and positive on [0, 1]. Therefore, we substitute the function (3.41) into the relation (3.40) to yield
Φ(xc ) = Φ exp(c log x)  Z c  = Φ exp log x dt 0 Z 1 Z c Φ(1) dt Φ(x) dt + = 0

c

= cΦ(x) + (1 − c)Φ(1).

This proves the claim (3.38). Next, we prove a lemma which is useful of determining the explicit form of the function Φ: Lemma 3.5 The relation (3.38) holds for all x > 0 and c ≥ 0.

Proof of Lemma 3.5. The case 0 ≤ c ≤ 1 is clear by the above analysis. Let c > 1 so that 0 < 1c < 1. Then the relation (3.38) becomes

1

 1
1 1 Φ x c = Φ(x) + 1 − Φ(1). c c

(3.42)

Take y = x c . Since 1c > 0, y will take all positive real numbers. Therefore, it follows from the expression (3.42) that  1 1 Φ(1) Φ(y) = Φ(y c ) + 1 − c c

and after simplification, we have

Φ(y c ) = cΦ(y) + (1 − c)Φ(1). This proves the lemma.



We return to the proof of the problem. Define Ψ : [0, ∞) → R by Ψ(x) = Φ(x) − Φ(1), then the relation (3.38) can be rewritten as Ψ(xc ) = cΨ(x),

(3.43)

where x > 0 and c ≥ 0. Take x = e and c = ln y, where y ≥ 1. Then the expression (3.43) becomes Ψ(y) = Ψ(e) ln y, (3.44) where y ≥ 1. For 0 < y < 1, we put x = e−1 and c = ln(y −1 ) > 0 into the expression (3.43) so that −1
Ψ(y) = Ψ e− ln(y ) = −Ψ(e−1 ) ln y. (3.45) Combining the results (3.44) and (3.45), we establish that  if x ≥ 1;  Ψ(e) ln x, Ψ(x) =  −Ψ(e−1 ) ln x, if 0 < x < 1

Chapter 3. Lp -Spaces

82

which imply that Φ(x) =

  Φ(1) + [Φ(e) − Φ(1)] ln x, 

if x ≥ 1;

Φ(1) − [Φ(e−1 ) − Φ(1)] ln x, if 0 < x < 1,

completing the proof of the problem.



Problem 3.7 Rudin Chapter 3 Exercise 7.

Proof. In this problem, we assume that µ is a positive measure. We give examples one by one: Example 3.1. Let X = N with the counting measure µ. By Definition 3.6, we have ∞ o 1 n X p 0, where M0 = {E ∈ M | µ(E) > 0}. • Ls (µ) ⊆ Lr (µ) for 0 < r < s < ∞. Similar to the proof of Lemma 3.3, the following result is part of the statements in [62, Theorem 2]: Lemma 3.6 Suppose that (X, M, µ) is a measure space and M∞ = {E ∈ M | µ(E) is finite}. Then the following conditions are equivalent: – Condition (1). sup{µ(E) | E ∈ M∞ } < ∞, – Condition (2). Ls (µ) ⊆ Lr (µ) for all 0 < r < s < ∞.

Proof of Lemma 3.6. Suppose that Condition (1) holds. To begin with, let f ∈ Ls (µ) 1 ≤ |f (x)| < n1 } for each n = 1, 2, . . .. Then it is clear that En ∈ M. and En = {x ∈ X | n+1 Now for each fixed n, we see that Z Z Z µ(En ) 1 s s dµ = |f | dµ ≥ |f | dµ ≥ s (n + 1) En (n + 1)s En X which implies that µ(En ) ≤ (n + 1)s

Z

X

|f |s dµ = (n + 1)s kf kss < ∞.

Therefore, we conclude that En ∈ M∞ . Note that Ei ∩ Ej = ∅ if i 6= j and µ is a positive measure, if Fn = E1 ∪ E2 ∪ · · · ∪ En , then we deduce from Theorem 1.19(b) that µ(Fn ) = µ

n [



Ek =

k=1

n X k=1

µ(Ek ) < ∞.

In other words, each Fn is an element of M∞ . Next, we know that F1 ⊆ F2 ⊆ · · · and n n [ [ Fk = Ek , so Theorem 1.19(d) implies that

k=1

k=1

lim µ(Fn ) = µ

n→∞

∞ [

n=1

∞ ∞  X  [ µ(En ). En = Fn = µ n=1

(3.47)

n=1

Given a positive integer n, it is clear that µ(Fn ) ≤ sup{µ(Fk ) | k ∈ N} ≤ sup{µ(E) | E ∈ M∞ }.

(3.48)

Furthermore, by combining Condition (1), the expression (3.47) and the inequality (3.48), it yields that ∞ X

n=1

µ(En ) = lim µ(Fn ) ≤ sup{µ(E) | E ∈ M∞ } < ∞. n→∞

Chapter 3. Lp -Spaces

On X \

∞ [

n=1

84

En , we claim that |f (x)| ≥ 1. Otherwise, there is a x0 ∈ X \

∞ [

En such that

n=1

|f (x0 )| < 1. By the definition, x0 ∈ EN for some N ∈ N, a contradiction. This observation implies that |f (x)|r ≤ |f (x)|s on X \ Z

∞ [

En . Finally, we have

n=1 r

X

|f | dµ =

Z

S X\ ∞ n=1 En

r

|f | dµ +

∞ Z X

r

n=1 En

|f | dµ ≤

Z

X

|f |s dµ +

∞ X µ(En )

n=1

nr

< ∞.

This proves that Condition (2) holds. Conversely, since Ls (µ) ⊆ Lr (µ) implies that Lαs (µ) ⊆ Lαr (µ) for every α ∈ (0, ∞), we may assume that r ≥ 1. It is known that if r, s ∈ [1, ∞] and Ls (µ) ⊆ Lr (µ), then the mapping T : Ls (µ) → Lr (µ) is continuous, see [62, Lemma 1]. This result guarantees the existence of a positive constant k such that kf kr ≤ kkf ks for all f ∈ Ls (µ), so rs

µ(E) ≤ k s−r for every E ∈ M∞ and this implies Condition (1).



We have completed the proof of the problem.



Problem 3.8 Rudin Chapter 3 Exercise 8.

Proof. For every n ∈ N, since g(x) → ∞ as x → 0, there exists a sequence {ǫn } of positive numbers such that g(x) ≥ n (3.49) for every x ∈ (0, ǫn ). Without loss of generality, we may assume that {ǫn } is decreasing and ǫn+1
Nx . Thus we have x ∈ (0, ǫn ) for all 1 ≤ n ≤ Nx and x ∈ / (0, ǫn ) for all n > Nx . See Figure 3.1 for an illustration of the distribution of x and ǫn .

3.2. Relations among Lp -Spaces and some Consequences

85

Figure 3.1: The distribution of x and ǫn .

These facts imply that

so that

  x  n 1 − , if 1 ≤ n ≤ Nx ;  ǫn hn (x) =   0, if n > Nx

(3.51)

 x h(x) = sup hn (x) = sup{h1 (x), h2 (x), . . .} = max n 1 − < ∞, 1≤n≤Nx ǫn n

i.e., h is finite. Since each hn is convex on (0, 1), Problem 3.1 shows that the h is also convex on (0, 1). Next, we want to compare the magnitudes of hn (x) and g(x) at the fixed point x ∈ (0, 1). On the one hand, for 1 ≤ n ≤ Nx , we have x ∈ (0, ǫn ), so n(1 − ǫxn ) ≤ n and we deduce from the inequality (3.49) and the definition (3.51) that hn (x) ≤ g(x)

(3.52)

for 1 ≤ n ≤ Nx . On the other hand, if n > Nx , then x ∈ / (0, ǫn ) and we have hn (x) = 0, but the inequality (3.49) still holds for n = Nx (i.e., x ∈ (0, ǫNx )) so that the inequality (3.52) remains valid in this case. Hence we have established that h(x) ≤ g(x)

(3.53)

for the fixed point x ∈ (0, 1). Since x is arbitrary, the inequality (3.53) holds on (0, 1). Take n = Nx − 1 in the definition (3.51) to get

 hNx −1 (x) = (Nx − 1) 1 −

x ǫNx −1



.

(3.54)

By the hypothesis (3.50), we have 2ǫNx < ǫNx −1 . Then since x < ǫNx , we have x 1 x < . < ǫNx −1 2ǫNx 2

(3.55)

By substituting the estimate (3.55) into the expression (3.54), we obtain hNx −1 (x) >

Nx − 1 . 2

Since x → 0 if and only if Nx → ∞, we conclude that hNx (x) → ∞ as x → 0 and the definition implies that h(x) → ∞ as x → 0 as required. 1

The second assertion is false. To see this, consider g(x) = x 3 which satisfies g(x) → ∞ as x → ∞. Assume that there was a convex function h : (0, ∞) → R such that 1

h(x) ≤ x 3

(3.56)

Chapter 3. Lp -Spaces

86

on (0, ∞) and h(x) → ∞ as x → ∞. We know from [66, Exercise 6(c), p. 262]c that if h : (0, ∞) → R is convex on (0, ∞), then the ratio h(x) x tends to a finite limit or to infinity as x → ∞. If this limit is finite, since h(x) → ∞ as x → ∞, this limit must be positive. Thus it must be true that h(x) ≥ kx

(3.57)

as x → ∞ for some positive constant k. However, the inequality (3.57) definitely contradicts the inequality (3.56). This completes the proof of the problem.  Problem 3.9 Rudin Chapter 3 Exercise 9.

Proof. Let Φ : (0, ∞) → (0, ∞) be such that Φ(p) → ∞ as p → ∞. Suppose that E1 = (0, 21 ) and h1 1 1 1 1 1 h 1 1 + 2 + · · · + n−1 , + 2 + · · · + n = 1 − n−1 , 1 − n , En = 2 2 2 2 2 2 2 2 where n ≥ 2. Obviously, we have En ⊂ (0, 1) for every n ∈ N and Ei ∩ Ej = ∅ for i 6= j. Furthermore, each En is measurable and m(En ) =

1 > 0. 2n

Suppose that x ∈ (0, 1). If x < 21 , then x ∈ E1 . Otherwise, 12 ≤ x < 1 and there exists a δ > 0 such that x < δ < 1. Take N to be least positive integer such that N >− Then we have x < 1 −

1 2N

log(1 − δ) > 0. log 2

< 1 so that x ∈ E2 ∪ E3 ∪ · · · ∪ EN . In other words, we get (0, 1) =

∞ [

En .

n=1

Finally, we define fn , f : (0, 1) → (0, ∞) by fn (x) =

n X

kχEk (x)

and f (x) = lim fn (x) =

k=1

n→∞

∞ X

nχEn (x)

(3.58)

n=1

respectively. Since each En is measurable, each χEn is measurable by Proposition 1.9(d). Thus each function fn is measurable and then Theorem 1.14 implies that f is also measurable. Now it remains to show that the function f satisfies the required properties of the problem. • Property 1: kf kp → ∞. Assume that f ∈ L∞ ((0, 1)). Then there exists a positive constant M such that kf k∞ < M . By Definition 3.7, we see that |f (x)| < M c

a.e. on (0, 1).

See also [31, Theorem 126, p. 99] and [51, Eqn. (2), p. 62].

(3.59)

87

3.3. Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12 Since Ei ∩ Ej = ∅ for i 6= j, we have |f (x)| = f (x) ≥ fn (x) ≥ n

(3.60)

for every x ∈ En . Since n → ∞, there exists a positive integer N such that N ≥ M . Recall that m(EN ) = 21N > 0, so we obtain from the inequality (3.60) that |f (x)| ≥ M on EN , but this contradicts our assumption (3.59). Therefore, we must have f ∈ / L∞ ((0, 1)) and in fact kf k∞ = ∞. Furthermore, it follows from the Ratio Test (see [49, Theorem 3.34]) that Z 1 ∞ ∞ X X n n × m(En ) = f (x) dx = kf k1 = < ∞, n 2 0 n=1 n=1 so we have kf k1 < ∞. Hence we deduce from Problem 3.4(e) that kf kp → kf k∞ = ∞ as p → ∞. • Property 2: kf kp ≤ Φ(p) for sufficiently large p. Fix a positive integer n, we consider h n ip Φ(p)

for p > 0. Since Φ(p) → ∞ as p → ∞, when p is sufficiently large, we have that h n ip 1 × m(En ) ≤ n . Φ(p) 2

n Φ(p)

≤ 1 so (3.61)

Therefore, we see that

kf kpp

=

Z

0

∞ 1hX

n=1

∞ ip X np × m(En ). nχEn (x) dx =

(3.62)

n=1

Consequently, we follow from the inequality (3.61) and the expression (3.62) that ∞ ∞ h ∞ X X X kf kpp n ip 1 1 p × m(En ) ≤ n × m(En ) = = =1 n [Φ(p)]p [Φ(p)]p n=1 Φ(p) 2 n=1 n=1

for sufficiently large p, or equivalently, kf kp ≤ Φ(p) for sufficiently large p. We end the proof of the problem.

3.3

Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12

Problem 3.10 Rudin Chapter 3 Exercise 10.



Chapter 3. Lp -Spaces

88

Proof. First of all, we have to show that f ∈ Lp (µ). Recall that Lp (µ) is a metric space with metric k · k. By the triangle inequality kfn − fm kp ≤ kfn − f kp + kf − fm kp and the hypothesis kfn − f kp → 0 as n → ∞ together imply that {fn } is a Cauchy sequence in Lp (µ). Now the completeness of Lp (µ) ensures that {fn } converges to an element of Lp (µ). By the uniqueness of limits, this element must be f so that f ∈ Lp (µ). By Theorem 3.12, {fn } has a subsequence {fnk } such that fnk → f a.e. on X. Since fn → g a.e. on X as n → ∞, every subsequence of {fn } converges to g a.e. on X. In particular, we take this subsequence {fnk } and the uniqueness of limits again imply that f =g a.e. on X. We have completed the proof of the problem.



Problem 3.11 Rudin Chapter 3 Exercise 11.

Proof. Since f, g : Ω → [0, ∞], the hypothesis f g ≥ 1 implies that (H¨older’s Inequality), we see that Z Z p 2 Z f g dµ ≤ g dµ. f dµ × Ω

By Proposition 1.24(a), we have Z p Ω

f g dµ ≥

(3.63)

Ω

Ω

Z

√ f g ≥ 1. By Theorem 3.5

dµ = µ(Ω) = 1.

(3.64)

Ω

Now our desired result follows immediately by combining the inequalities (3.63) and (3.64). This completes the analysis of the problem.  Problem 3.12 Rudin Chapter 3 Exercise 12. √ Proof. Suppose that A = ∞. Since h ≤ 1 + h2 on Ω, we have Z p 1 + h2 dµ = ∞, Ω

so the √ equalities √ definitely hold in this case. Next, suppose that A = 0. By the fact that 1 ≤ 1 + h2 ≤ 1 + 2h + h2 = 1 + h, we achieve that Z Z Z p Z 2 dµ. (3.65) 1 + h dµ ≤ (1 + h) dµ = dµ ≤ Ω

Ω

Ω

Since µ(Ω) = 1, we conclude from the inequalities (3.65) that Z p 1 + h2 dµ = 1. Ω

Thus the equalities also hold in this case.

Ω

89

3.3. Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12

Suppose that 0 < A < ∞ which means h(x) ∈ (0, ∞) a.e. on Ω. We consider the function ϕ : R → (0, ∞) defined by p ϕ(x) = 1 + x2 which implies that

ϕ′ (x) = √

x 1 + x2

and ϕ′′ (x) =

1 3

(1 + x2 ) 2

.

(3.66)

Since ϕ′′ (x) ≥ 0 on R, we know from [49, Exercise 14, p. 115] that ϕ is convex on R. In particular, ϕ is convex on (0, ∞). By Theorem 3.3 (Jensen’s Inequality), we obtain that p

1+

A2

=

s

1+

Z

Ω

Z p h dµ ≤ 1 + h2 dµ

(3.67)

Ω

which is exactly √ the left-hand side inequality. The right-hand side inequality is easy because we always have 1 + h2 ≤ 1 + h. Hence we have verified the validity of the inequalities.

For the second assertion, we suppose that µ = m on Ω = [0, 1], h = f ′ and h is continuous on Ω. Then we have Z 1 f ′ (x) dx = f (1) − f (0). A= 0

In addition, we notice that Z p Ω

1+

h2 dµ

=

Z

0

1p

1 + [f ′ (x)]2 dx

which is the arc length of the curve from (0, f (0)) to (1, f (1)).d Thus the inequalities give bounds of such arc length. Geometrically, the upper bound consists of the length from (0, f (0)) to (1, f (0)) (which is 1) and the length from (1, f (0)) to (1, f (1)) (which is A). The lower bound is just the hypotenuse of the right triangle with vertices (0, f (0)), (0, f (1)) and (1, f (1)), see Figure 3.2 below:

Figure 3.2: The geometric interpretation of a special case. d

See, for examples, [2, p. 535] and [49, Theorem 6.27, p. 137].

Chapter 3. Lp -Spaces

90

Our first inequality (3.67) comes from the application of Theorem 3.3 (Jensen’s Inequality), so the equality of the first inequality (3.67) holds if and only if the equality in [51, Eqn. (2), p. 62] holds if and only if
ϕ(A) = ϕ h(x) . (3.68)

We see from the first derivative in (3.66) that ϕ′ (x) > 0 in (0, ∞) so that it is injective in (0, ∞). Recall that A ∈ (0, ∞) and h(x) ∈ (0, ∞) a.e. on Ω, so we deduce that the equality (3.68) holds if and only if h = A a.e. on Ω. √ Next we recall the second inequality is derived from the fact that 1 + h2 ≤ 1 + h and some simple computations show that the equality holds if and only if h≡0

a.e. on Ω.

Hence we have ended the proof of the problem.



Problem 3.13 Rudin Chapter 3 Exercise 13.

Proof. There are two cases. • Case (i): 1 < p < ∞. Then the inequality in Theorem 3.8 is just H¨older’s inequality. If kf kp = 0 or kgkp = 0, then we have f = 0 a.e. on X or g = 0 a.e. on X by Theorem 1.39(a). In either case, the equality holds trivially. Therefore, we may assume that both kf kp > 0 and kgkp > 0. Furthermore, by the remark following the proof of Theorem 3.5 (H¨older’s Inequality), we may further assume that 0 < kf kp < ∞ and

0 < kgkp < ∞.

In this case, we see that the equality in Theorem 3.8 holds if and only if there are positive constants α and β such that αf p = βgq a.e. on X. In Theorem 3.9, the inequality is in fact Minkowski’s inequality. By simple observation, it is clear that the equality holds if f = 0 a.e. on X or g = 0 a.e. on X. Therefore, we may assume that f 6= 0 and g 6= 0 on measurable sets E and F with µ(E) > 0 and µ(F ) > 0 respectively. By examining the proof of [51, Eqn. (2), Theorem 3.5, pp. 64, 65], we know that it applies H¨older’s inequality to the functions f and (f + g)p−1 as well as the functions g and (f + g)p−1 . Thus we deduce from the previous paragraph that the equality in Z

X

f · (f + g)p−1 dµ ≤

nZ

f p dµ

X

o1

p

×

nZ

(f + g)(p−1)q dµ

X

o1 q

holds if and only if there are positive constants α and β such that αf p = β(f + g)q

a.e. on X.

Similarly, the equality in Z o1 o1 n Z nZ p q (p−1)q p p−1 (f + g) dµ × g dµ g · (f + g) dµ ≤ X

X

X

(3.69)

91

3.4. Hardy’s Inequality and Egoroff ’s Theorem holds if and only if there are positive constants λ and ν such that λg p = ν(f + g)q

a.e. on X.

(3.70)

By combining the equations (3.69) and (3.70), we conclude that fp =

β βλ p (f + g)q = g α αν

a.e.

which means that f = Kg a.e. for some positive constant K. • Case (ii): p = ∞. The inequality in Theorem 3.8 comes from the inequalitye |f (x)g(x)| ≤ kf k∞ · |g(x)| for almost all x. Therefore, it is easy to see that the equality in Theorem 3.8 holds if and only if |f (x)g(x)| = kf k∞ · |g(x)| for almost all x if and only if g(x) = 0 for almost x or f (x) = kf k∞ for almost x.f Now the triangle inequality

|f + g| ≤ |f | + |g| implies the inequality in Theorem 3.9. Thus the equality in Theorem 3.9 holds if and only if f and g are either nonnegative functions or nonpositive functions for almost all x. This completes the proof of the problem.

3.4



Hardy’s Inequality and Egoroff’s Theorem

Problem 3.14 Rudin Chapter 3 Exercise 14.

Proof. We follow the suggestions given by Rudin. (a) We divide the proof into two cases:
– Special Case: f ≥ 0 and f ∈ Cc (0, ∞) . By the First Fundamental Theorem of Calculus, the function F (x) is differentiable on (0, ∞) and xF ′ (x) = f (x) − F (x)

(3.71)

for every x ∈ (0, ∞). By Integration by Partsg and the expression (3.71), we have Z ∞ Z ∞ p p ∞ pxF p−1 (x)F ′ (x) dx. (3.72) F (x) dx = [xF (x)]0 − 0

0

Now we have to evaluate the limits: lim xF p (x) and

x→0 e

lim xF p (x).

x→∞

This is [51, Eqn. (2), Theorem 3.8, p. 66]. The latter case means that f is a nonnegative constant for almost all x. g Here we assume that the formula for integration by parts is also valid for improper integrals, see [3, p. 278]. f

Chapter 3. Lp -Spaces

92

Since supp (f ) = {x ∈ (0, ∞) | f (x) 6= 0} is compact, the Heine-Borel Theorem guarantees that supp (f ) = [a, b] ⊂ (0, ∞). By this and the continuity of f , we conclude that f is bounded on (0, ∞), i.e., |f (x)| ≤ M on (0, ∞) for some positive constant M . Thus we follow from this that Z 1 x |F (x)| ≤ |f (t)| dt ≤ M x 0 for every x ∈ (0, ∞), so we see that lim xF p (x) = 0.

(3.73)

x→0

Again, the boundedness of f implies that Z b Z ∞ Z x |f (t)| dt ≤ M (b − a) < ∞ |f (t)| dt = |f (t)| dt ≤ |xF (x)| ≤ 0

0

on (0, ∞). Since xF p = that

(xF )p xp−1

(3.74)

a

and p > 1, it reduces from these and the bound (3.74) lim xF p (x) = 0.

(3.75)

x→∞

Therefore, we deduce from the formula (3.72) and the two limits (3.73) and (3.75) that Z ∞ Z ∞ p F p−1 (x)[f (x) − F (x)] dx F (x) dx = −p 0 Z ∞ Z0 ∞ F p (x) dx F p−1 (x)f (x) dx + p = −p 0

0

so that (p − 1)

Z

∞

F p (x) dx = p

Z

∞

F p−1 (x)f (x) dx.

(3.76)

0

0

Note that 1 < p < ∞, let q be its conjugate exponent. Obviously, our hypotheses make sure that f and F p−1 have range [0, ∞], so we may apply Theorem 3.5 (H¨older’s Inequality) to them. In fact, it yields from the fact (p − 1)q = p and the expression (3.76) that Z ∞ p F p (x) dx (p − 1)kF kp = (p − 1) 0 o1 nZ ∞ o1n Z ∞ p q F (p−1)q (x) dx ≤p f p (x) dx 0

0

p q

= pkf kp · kF kp

which is exactly

p− pq

kF kp = kF kp

≤

p kf kp . p−1


– General Case: f ∈ Lp (0, ∞) . By Definition 3.6, we always have

kf kp = |f | p ,

(3.77)

so we may
assume that f ≥ 0 on
(0, ∞). Since 1 < p < ∞, Theorem 3.14 says that
Cc (0, ∞) is dense in Lp (0, ∞) . Thus there exists a sequence {fn } ⊆ Cc (0, ∞) such that kfn − f kp → 0 (3.78)

93

3.4. Hardy’s Inequality and Egoroff ’s Theorem as n → ∞. Notice that fn may not be nonnegative, but since |fn | − |f | ≤ |fn − f | and f ≥ 0, we have |fn (x)| − f (x) ≤ |fn (x) − f (x)| (3.79)

for every x ∈ (0, ∞). Certainly, the inequality (3.79) implies that

|fn | − f ≤ kfn − f kp p

(3.80)

for every n = 1, 2, . . .. Now the result (3.78) and the inequality (3.80) ensure that

|fn | − f → 0 p

as n → ∞. In other words,
we may also assume that fn ≥ 0 for each n = 1, 2, . . .. Recall that fn ∈ Cc (0, ∞) , so we derive from the Special Case that kFn kp ≤

where

p kfn kp , p−1

1 Fn (x) = x

Z

1 F (x) = x

Z

and n = 1, 2, . . .. Suppose that

(3.81)

x

fn (t) dt

0

x

f (t) dt.

0

Since f ≥ 0 and fn ≥ 0, it is easily seen from the definition that F ≥ 0 and Fn ≥ 0. We claim that {Fn } converges pointwise to F on (0, ∞). To see this, fix x ∈ (0, ∞), then their definitions give Z 1 x |fn (t) − f (t)| dt. (3.82) |Fn (x) − F (x)| ≤ x 0 By the result (3.78), we know that for every ǫ > 0, there exists a positive integer N such that n ≥ N implies kfn − f kp < ǫ.

Apply Theorem 3.5 (H¨older’s Inequality) with 1p + 1q = 1 to the inequality (3.82), we get immediately that Z o1 n Z x o1 1n x p q p |fn (t) − f (t)| dt × 1q dt |Fn (x) − F (x)| ≤ x 0 0 1 1 ≤ kfn − f kp × x q x 1

= x− p kfn − f kp 1

< ǫx− p

(3.83)

for all n ≥ N . Since x is fixed and independent of ǫ, the estimate (3.83) verifies the truth of the claim. Next, recall that each fn is continuous on [a, x] for every a > 0, so the First Fundamental Theorem of Calculus shows that each Fn is continuous on [a, x]. As a continuous function, every Fn is measurable on [a, x].h Hence, it follows from the h

See §1.11 or [49, Example 11.14].

Chapter 3. Lp -Spaces

94

pointwise convergence of {Fn }, Theorem 1.28 (Fatou’s Lemma), the inequality (3.81) and then the result (3.78) that Z ∞ p F p (x) dx kF kp = Z0 ∞ lim inf Fnp (x) dx = n→∞ 0 Z ∞ Fnp (x) dx ≤ lim inf n→∞ 0 Z ∞ Fnp (x) dx ≤ lim n→∞ 0

= lim kFn kpp n→∞  p p ≤ lim kfn kpp p − 1 n→∞  p p kf kpp ≤ p−1

which is equivalent to the desired result kF kp ≤

p kf kp . p−1



Hence the mapping T : Lp (0, ∞) → Lp (0, ∞) given by T (f ) = F is continuous.
(b) Suppose that f ∈ Lp (0, ∞) and kF kp =

p kf kp < ∞. p−1

(3.84)

Recall that kf kp = |f | p , so we may suppose further that f ≥ 0 on (0, ∞). By the
General Case of the proof in part (a), there is a nonnegative sequence {fn } ⊆ Cc (0, ∞) such that kfn −f kp → 0 as n → ∞. Now we replace F and f by Fn and fn in the expression (3.76) respectively, we gain Z ∞ Z ∞ p p p Fn (x) dx = kFn kp = Fnp−1 (x)fn (x) dx (3.85) p−1 0 0

for each n = 1, 2, . . .. Since fn ∈ Cc (0, ∞) , T (fn ) = Fn ∈ Lp (0, ∞) , where T is the continuous mapping considered in part
(a). By Theorem 3.5 (H¨older’s Inequality), it is easy to see that fn Fnp−1 ∈ Lp (0, ∞) for each n = 1, 2, . . .. Since the mapping T is continuous, it follows from the expression (3.85) that Z ∞ Z ∞ p p F p−1 (x)f (x) dx. (3.86) F (x) dx = p − 1 0 0
In other words, the formula (3.76) holds for f ∈ Lp (0, ∞) . We apply Theorem 3.5 (H¨older’s Inequality) to the right-hand side of the inequality (3.86) and then using our hypothesis (3.84) to conclude that Z ∞ p p kF kp = F p−1 (x)f (x) dx p−1 0 Z o1 n Z ∞ o1 p n ∞ p p q p f (x) dx × (3.87) F (x) dx ≤ p−1 0 0

95

3.4. Hardy’s Inequality and Egoroff ’s Theorem p p kf kp × kF kpq p−1 = kF kpp .

=

Hence the equality in the inequality (3.87) must hold. Consequently, there are constants α and β, not both 0, such that αf p = β(F p−1 )q = βF p a.e. on (0, ∞) and so 1

1

αpf = β pF

a.e. on (0, ∞).

(3.88)

Here we have several cases: – Case (i): β = 0. Then we have α 6= 0 and f = 0 a.e. on (0, ∞). Thus we are done.

– Case (ii): α = 0. Then we have β 6= 0 and F = 0 a.e. on (0, ∞). By the hypothesis (3.84), we see that kf kp = 0 which gives f = 0 a.e. on (0, ∞) and we are done again. – Case (iii): αβ 6= 0. By Theorem 3.8, we see that f ∈ L1 ([a, b]), where [a, b] is any bounded interval of (0, ∞). By the comment following the proof of Theorem 11.3 on [49, p. 324], we know that F is differentiable a.e. on [a, b] and
′ xF (x) = f (x) (3.89) a.e. on [a, b]. Since [a, b] is arbitrary, the formula (3.89) holds a.e. on (0, ∞). Combining the equations (3.88) and (3.89), we see immediately that xF ′ + xF = f which implies xf ′ = (c − 1)f a.e. on (0, ∞) (3.90) for some nonzero constant c. By solving the differential equation (3.90), we obtain that f (x) = γxc−1 a.e. on (0, ∞)

for some constant γ. Since γxc−1 ∈ Lp ((0, ∞)), it forces that γ = 0. Hence we have shown that the equality holds only if f = 0 a.e. on (0, ∞). 1

(c) Take f (x) = x− p χ[1,A] (x) for large A. Then we have kf kp =

nZ

0

∞

|x

− p1

p

χ[1,A] (x)| dx

o1

p

=

nZ

A

x 1

−1

dx

o1

p

1

= (log A) p .

Next, we know from the definition that Z 1 x − p1 t χ[1,A] (t) dt F (x) = x 0  0, if x ∈ (0, 1);       Z x   1  1 t− p dt, if x ∈ [1, A]; = x 1     Z   1 A − p1    t dt, if x ∈ (A, ∞); x 1  0, if x ∈ (0, 1);          p (x− p1 − x−1 ), if x ∈ [1, A]; p−1 =      1− 1  A p −1 p    · , if x ∈ (A, ∞). p−1 x

(3.91)

(3.92)

Chapter 3. Lp -Spaces

96

Note that A is large and 1 −

1 p

> 0, so we have 1

A1− p − 1 > 0.

(3.93)

Therefore, we deduce from the expressions (3.92) and the fact (3.93) that Z ∞ p |F (x)|p dx kF kp = 0 Z ∞ Z A |F (x)|p dx |F (x)|p dx + = 1

A

 p p Z ∞ A1− p1 − 1 p  p p Z A − p1 −1 p |x − x | dx + = dx p−1 p−1 x 1 A  p p Z A
p 1−p 1 pp 1− p1 (x− p − x−1 )p dx + A = A − 1 p−1 (p − 1)p+1 1  p p Z A 1 (x− p − x−1 )p dx. > p−1 1

Fix p first. Assume that the constant for some δ < 1, i.e.,

p p−1

kF kp ≤

(3.94)

could be replaced by a smaller number

δp kf kp . p−1

δp p−1

1

Take ǫ > 0 such that δ < ǫ < 1. Then it is clear that for large A, we have x p −1 < 1 − ǫ for x ≥ A2 or equivalently, x

for x ≥

A 2.

Z

− p1

− x−1 > ǫx

− p1

>0

(3.95)

By this estimate (3.95), we obtain

A

(x

− p1

1

−x

−1 p

) dx >

Z

A A 2

 A > ǫp log A. ǫp x−1 dx = ǫp log A − log 2

(3.96)

By substituting the inequality (3.96) into the inequality (3.94) and then using the fact (3.91), we can show that  ǫp p  δp p kF kpp > log A > kf kpp ≥ 0, p−1 p−1

but this implies

kF kp > a contradiction.

δp kf kp , p−1

(d) Since f (x) > 0 on (0, ∞), there exists a δ > 0 such that we can find an α ∈ (0, ∞) with Z α f (t) dt > δ > 0. 0

By the definition, we have F (x) > 0 on (0, ∞) so that Z ∞ Z ∞ Z ∞ Z ∞ Z x  δ 1 f (t) dt dx > dx = ∞. kF k1 = F (x) dx ≥ F (x) dx = 0 α x 0 α α x

Hence F ∈ / L1 . In other words, this shows that Hardy’s inequality does not hold when p = 1.

97

3.4. Hardy’s Inequality and Egoroff ’s Theorem

Hence we have completed the proof of the problem.



Problem 3.15 Rudin Chapter 3 Exercise 15.

Proof. We follow the hint. Suppose that {an } is decreasing. Let f : (0, ∞) → (0, ∞) be given by ∞ X an χ(n−1,n] (x). (3.97) f (x) = n=1

Firstly, we note that

Z

∞

p

f (x) dx =

0

which implies that f ∈

∞ Z X

n

n=1 n−1

apn dx

=

∞ X

apn

(3.98)

n=1

∞ X
apn < ∞. This shows that the desired (0, ∞) if and only if

Lp

n=1

∞ X
p apn < ∞ in inequality holds if f ∈ / L (0, ∞) . Without loss of generality, we assume that n=1

the following discussion. Secondly, let N be a positive integer and x ∈ (N − 1, N ]. Then we deduce from the definition (3.97) that Z 1 x F (x) = f (t) dt x 0 Z x N −1 Z i 1h X n f (t) dt f (t) dt + = x N −1 n=1 n−1 Z Z N −1 i x 1h X n = aN dt an dt + x n=1 n−1 N −1 1 [a1 + a2 + · · · + aN −1 + aN (x − N + 1)]. x

=

(3.99)

It is clear from the expression (3.99) and the fact {an } is decreasing that N −1 N  i 1X 1h X an + aN (x − N + 1) = an − N aN + xaN x x n=1

=

1 x

n=1 N X n=1

 an − N aN + aN

N  1 X ≥ an − N aN + aN N n=1 N 1 X an . = N

(3.100)

n=1

Thus we obtain from the expression (3.99) and the inequality (3.100) that kF kp =

nZ

∞ 0

F p (x) dx

o1

p

=

∞ Z nX

N

N =1 N −1

F p (x) dx

o1

p

≥

∞  N nX 1 X p o 1p an . N n=1 N =1

(3.101)

Chapter 3. Lp -Spaces

98

Furthermore, by Problem 3.14(a) and the expression (3.98), we derive that p p n kF kp ≤ kf kp = p−1 p−1

Z

∞

p

f (x) dx

0

o1

p

p n X p o p1 a = . p − 1 n=1 n ∞

(3.102)

Hence our desired inequality follows immediately from the inequalities (3.101) and (3.102). ∞ X For the general case, recall the assumption that an > 0 for all n ∈ N and apn < ∞, so the n=1

sequence {apn } converges absolutely and every rearrangement converges to the same sum (see [49, Theorem 3.55, p. 78]). On the other hand, if we fix a positive integer N , then N  X k X 1 k=1

k

n=1

an

p

=

 a p

 a + a p  a + a + a p 1 2 1 2 3 + 1 2 3  a + a + · · · + a p 1 2 N , + ··· + N 1

+

so it follows from this form that the sum attains its maximum if and only if a1 ≥ a2 ≥ · · · ≥ aN . Therefore, we have ∞  ∞  ∞ ∞ N N X X 1 X p  p p X p  p p X p 1 X p αn = an , an ≤ αn ≤ N N p−1 p−1 n=1 n=1 n=1 n=1 N =1 N =1 {z } |

By the special case.

where {αn } is the decreasing rearrangement of {an }. This completes the proof of the problem.  Problem 3.16 Rudin Chapter 3 Exercise 16.

Proof. Let’s prove the assertions one by one. • A proof of Egoroff ’s Theorem. Put \ n 1o . S(n, k) = x ∈ X |fi (x) − fj (x)| < k

(3.103)

i,j>n

For each k ∈ N, we follow from the definition (3.103) that \ n \ n 1o 1o ∩ x ∈ X |fi (x) − fj (x)| < S(n, k) = x ∈ X |fi (x) − fj (x)| < k k i,j>n+1

\ n 1o ∩ x ∈ X |fi (x) − fj (x)| < k

i=n+1 j>n

j=n+1 i>n

which implies that S(n, k) ⊆ S(n + 1, k).

Next, for each fixed k ∈ N, we claim that X=

∞ [

n=1

S(n, k).

(3.104)

99

3.4. Hardy’s Inequality and Egoroff ’s Theorem To this end, we see that the set inclusion ∞ [

S(n, k) ⊆ X

n=1

is evident. To prove the reverse direction, we note that for each k ∈ N and x ∈ X, we yield from our hypothesis fn (x) → f (x) that there exists a positive integer N such that i, j > N implies that 1 |fi (x) − fj (x)| < . k In other words, we must have x ∈ S(n, k), i.e., X⊆

∞ [

S(n, k).

n=1

Thus these prove the validity of our claim (3.104) and Theorem 1.19(d) gives
lim µ S(n, k) = µ(X) n→∞

for every k = 1, 2, . . ..

Given that ǫ > 0. For each k ∈ N, this fact allows us to choose nk such that
ǫ |µ S(nk , k − µ(X)| < k . 2

Define

E=

∞ \

S(nk , k).

(3.105)

k=1

Then it is clear that X \E = which implies that µ(X \ E) ≤ Pick a k with

1 k

∞ X k=1

∞ [

k=1


X \ S(nk , k)

∞ ∞
X
X ǫ µ(X) − µ S(nk , k) < µ X \ S(nk , k) ≤ = ǫ. 2k k=1

k=1

< ǫ. If x ∈ S(nk , k), then we have

|fi (x) − fj (x)| < ǫ

(3.106)

for all i, j > nk . By the definition (3.105), we know that E ⊆ S(nk , k) for every k ∈ N, so the inequality (3.106) holds on E. By the definition, fn → f uniformly on E, proving Egoroff’s Theorem. • A counterexample on a σ-finite space. Recall from Definition 2.16 that X is said to be a σ-finite space if X is a countable union of sets Ei with µ(Ei ) < ∞. It is clear that R=

∞ [

[n − 1, n) ∪

n=1

∞ [

(−n, 1 − n],

n=1

so R is σ-finite. Let fn (x) = nx on R. It is clear that the sequence converges pointwise to 0 at every point of R. Furthermore, we know that a measurable set E ⊆ R with m(R\E) < 1 must be unbounded. Otherwise, E ⊆ [−M, M ] for some M > 0 and this implies that
m(R \ E) ≥ m R \ [−M, M ] = ∞,

Chapter 3. Lp -Spaces

100

a contradiction. However, {fn } cannot converge uniformly to 0 on the unbounded set E. Thus this counterexample shows that Egoroff’s Theorem cannot be extended to σ-finite spaces. • An extension of Egoroff ’s Theorem. Suppose that {ft } is a family of complex measurable functions such that (i) lim ft (x) = f (x) and t→∞

(ii) Fix a x ∈ X. Then the function F : (0, ∞) → C defined by F (t) = ft (x) is continuous. For each n ∈ N, we consider the real function gn : X → R given by gn (x) = sup{|ft (x) − f (x)|}.

(3.107)

t≥n

Then, for every x ∈ X, gn (x) → 0 as n → ∞. Now it remains to show that each gn is measurable. To this end, let a ∈ R and Ea (n) = {x ∈ X | gn (x) < a}

= {x ∈ X | |ft (x) − f (x)| < a for all t ≥ n} \ = {x ∈ X | |fr (x) − f (x)| < a}.

(3.108)

r≥n r∈Q

Since fr is measurable, f is also measurable by Corollary (a) following Theorem 1.14. Thus |fr − f | is measurable by Proposition 1.9(b) and then the set {x ∈ X | |fr (x) − f (x)| < a} is measurable for every a ∈ R by Definition 1.3(c). By the expression (3.108), since there are countable such measurable sets, Comment 1.6(c) implies that Ea (n) is measurable for every real a. By Definition 1.3(c) again, each gn is measurable and thus Egoroff’s Theorem can be applied to conclude that for every ǫ > 0, there exists a measurable set E ⊆ X such that µ(X \ E) < ǫ and {gn } converges uniformly to 0 on E. By the definition (3.107), {ft } converges to f uniformly on E. This completes the proof of the problem.



Problem 3.17 Rudin Chapter 3 Exercise 17.

Proof. (a) Since the inequality is clearly holds if either α = 0 or β = 0, we assume, without loss of generality, that α and β cannot be both zero in the following discussion. Let 0 < p ≤ 1. We claim that the continuous function ϕ : [0, 1] → R defined by ϕ(x) = xp + (1 − x)p

101

3.4. Hardy’s Inequality and Egoroff ’s Theorem satisfies ϕ(x) ≥ 1 on [0, 1]. To this end, we note that ϕ′ (x) = pxp−1 − p(1 − x)p−1 = 0 if and only if x = 21 . Since p − 1 ≤ 0, we know that ϕ′ (x) ≥ 0 and

ϕ′ (x) ≤ 0

on (0, 12 ) and ( 21 , 1) respectively. In other words, ϕ is increasing and decreasing on (0, 21 ) and ( 21 , 1) respectively. By the continuity of ϕ, these are also valid on [0, 12 ] and [ 21 , 1] respectively. Thus we observe ϕ(x) ≥ ϕ(0) = 1 on [0, 21 ] and

ϕ(x) ≥ ϕ(1) = 1 on [ 12 , 1]

which mean that xp + (1 − x)p ≥ 1

(3.109)

on [0, 1]. For arbitrary nonzero complex numbers α and β, we put x = inequality (3.109) to get |α|p |β|p + ≥1 (|α| + |β|)p (|α| + |β|)p

|α| |α|+|β|

into the

which implies (|α| + |β|)p ≤ |α|p + |β|p .

(3.110)

By the triangle inequality, it is true that |α − β| ≤ |α| + |β|, so this and the inequality (3.110) give |α − β|p ≤ |α|p + |β|p . (3.111) Next, let 1 < p < ∞. The function ψ : [0, 1] → R given by ψ(x) = xp has second derivative p(p − 1)xp−2 > 0 on (0, 1). By Definition 3.1, it is convex on (0, 1), i.e., [(1 − λ)x + λy]p ≤ (1 − λ)xp + λy p , (3.112) where x, y ∈ (0, 1) and λ ∈ [0, 1]. Put λ = 12 , x = (3.112), we obtain

|α| |α|+|β|

and y =

|β| |α|+|β|

into the inequality

 1 |α| + |β| p 1 |α|p |β|p 1 ≤ · · · + 2 |α| + |β| 2 (|α| + |β|)p 2 (|α| + |β|)p which reduces to (|α| + |β|)p ≤ 2p−1 (|α|p + |β|p ). Again, the triangle inequality and this show that |α − β|p ≤ 2p−1 (|α|p + |β|p ).

(3.113)

Hence the desired inequality follows from combining the inequalities (3.111) and (3.113). (b)

(i) If µ(X) = 0, then Proposition 1.24(e) implies that kf kp = 0 for every f ∈ Lp (µ) and there is nothing to prove. Thus, without loss of generality, we may assume that µ(X) > 0. We divide the proof into several steps:

Chapter 3. Lp -Spaces

102

∗ Step 1: Lemma 3.7 and its application. Lemma 3.7 For every ǫ > 0, there exists a δ > 0 such that for every E ∈ M with µ(E) < δ, we have Z |f |p dµ ≤ ǫ. (3.114) E

Proof of Lemma 3.7. Note that |f | must be bounded a.e. on X. Otherwise, there exists an E ∈ M such that µ(E) > 0 and |f | = ∞ on E. Then Proposition 1.24(b) shows that Z Z |f |p dµ = ∞, |f |p dµ ≥ E

X

a contradiction. Thus there exists a positive constant M such that |f | ≤ M a.e. on X. Given ǫ > 0. Now for every E ∈ M with µ(E) < Mǫ p , we have Z |f |p dµ ≤ M p µ(E) ≤ ǫ a.e. on X E

 which is exactly what we want. By Lemma 3.7, there exists a δ > 0 such that the inequality (3.114) holds for every E ∈ M with µ(E) < δ. For this δ > 0, suppose that we can find a measurable set F such that fn → f a.e. on F and µ(F ) < ∞. (The existence of such a F will be clear at the end of Step 2 below.) Then Egoroff’s Theorem guarantees that there exists a measurable set B ⊆ F such that µ(F \ B) < δ and {fn } converges uniformly to f on B. Since F \ B ∈ M and µ(F \ B) < δ, we also have Z ǫ (3.115) |f |p dµ ≤ . 2 F \B Furthermore, since fn → f uniformly on B, there is a positive integer N such that n ≥ N implies that |fn (x) − f (x)| ≤

 ǫ 1 p µ(B)

for all x ∈ B and this means that Z

B

|fn − f |p dµ ≤ ǫ

(3.116)

for n ≥ N .

∗ Step 2: Lemma 3.8 and its application. Lemma 3.8 If f ∈ L1 (µ), then for every ǫ > 0, there exists an E ∈ M such that F = X \ E, µ(F ) < ∞ and Z |f | dµ ≤ ǫ. E

103

3.4. Hardy’s Inequality and Egoroff ’s Theorem Proof of Lemma 3.8. Since |f | is measurable, Theorem 1.17 (The Simple Function Approximation Theorem) ensures that there exists a sequence of simple measurable functions {sk } on X such that 0 ≤ s1 ≤ s2 ≤ · · · ≤ |f |

(3.117)

and sk converges to |f | pointwisely. By Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem), we have Z Z |f | dµ. sk dµ = lim k→∞ X

X

Hence, for every ǫ > 0, there exists a positive integer N such that Z Z (|f | − sk ) dµ = (|f | − sk ) dµ ≤ ǫ.

(3.118)

X

X

By Definition 1.16, we have

sk =

nk X

αi χFi ,

i=1

where each αi is positive, Fi is measurable and Fi ∩ Fj = ∅ for all i 6= j. Recall that |f | ∈ L1 (µ), so the hypothesis (3.117) implies that sk ∈ L1 (µ) for every n ∈ N. Since each αi is positive, µ is positive on X (i.e., µ(E) ≥ 0 for every E ∈ M) and Z nk X sk dµ < ∞, αi µ(Fi ) = X

i=1

they force that µ(Fi ) < ∞ for every i = 1, 2, . . . , nk . We suppose that E = {x ∈ X | sk (x) = 0}

and F =

nk [

Fi .

i=1

Then the previous analysis shows that µ(F ) < ∞. Furthermore, it is clear that sk (x) > 0 if and only if x ∈ Fi for some i, so this fact implies that X \ E = F. Thus we deduce from the estimate (3.118) and the definition of E that Z Z Z Z Z sk dµ ≤ ǫ. (|f | − sk ) dµ + sk dµ ≤ |f | dµ = (|f | − sk ) dµ + E

E

E

X

E

This proves Lemma 3.8.  p p 1 Since f ∈ L (µ), we have |f | ∈ L (µ). Then Lemma 3.8 implies that there exists an E ∈ M such that F = X \ E, µ(F ) < ∞ and Z ǫ (3.119) |f |p dµ ≤ . 2 E ∗ Step 3: Constructions of A and B satisfying the hypotheses. Let E and F be defined as in Lemma 3.8 so that the estimate (3.119) holds. We remark that X = E ∪ F = E ∪ (F \ B) ∪ B = A ∪ B,

Chapter 3. Lp -Spaces

104

where B is the measurable set guaranteed by Egoroff’s Theorem in Step 1 and A = E ∪ (F \ B). Recall that B ⊆ F , so it is easy to see that µ(B) ≤ µ(F ) < ∞ and since E ∩ (F \ B) = ∅, we obtain from Theorem 1.29 and the two estimates (3.119) and (3.115) that Z Z Z p p |f |p dµ ≤ ǫ. (3.120) |f | dµ + |f | dµ = F \B

E

A

Since fn → f a.e. on X and particularly on B, this and Theorem 1.28 (Fatou’s Lemma) together show that Z Z Z p p |fn |p dµ (3.121) lim inf |fn | dµ ≤ lim inf |f | dµ = B n→∞

B

n→∞

B

Next, we know from the definition that A = (E ∪ F ) \ B = X \ B so that A ∩ B = ∅. Since kfn kp → kf kp as n → ∞, it deduces from this, Problem 1.4 and the inequalities (3.120) and (3.121) that Z Z   Z p p |fn |p dµ lim sup |fn | dµ ≤ lim sup |fn | dµ + lim sup − n→∞ n→∞ n→∞ B A Z ZX p |fn |p dµ = lim sup |fn | dµ − lim inf n→∞ B Zn→∞ X Z p p |f | dµ |f | dµ − ≤ B ZX |f |p dµ = A

≤ ǫ.

(3.122)

∗ Step 4: The establishment of kfn − f kp → 0 as n → ∞. By part (a), the estimates (3.116) and (3.122), we see that Z Z Z lim sup |fn − f |p dµ ≤ lim sup |fn − f |p dµ + lim sup |fn − f |p dµ n→∞ n→∞ n→∞ X AZ B p p ≤ γp lim sup (|f | + |fn | ) dµ + ǫ n→∞

A

≤ 2γp ǫ + ǫ.

Since ǫ is arbitrary, we conclude that lim kfn − f kp = lim

n→∞

n→∞

nZ

X

|fn − f |p dµ

o1

p

= 0.

(ii) Put hn = γp (|f |p + |fn |p ) − |f − fn |p . By part (a), we have hn ≥ 0 for all n ∈ N. Since f and fn are measurable, each hn is also measurable. By Theorem 1.28 (Fatou’s Lemma), we get Z Z   hn dµ. (3.123) lim inf hn dµ ≤ lim inf X

n→∞

f |p

n→∞

X

Since fn → f a.e. on X, |fn − → 0 and |fn → |f |p a.e. on X. Thus we must have lim hn = 2γp |f |p a.e. on X. (3.124) n→∞

|p

105

3.4. Hardy’s Inequality and Egoroff ’s Theorem Putting the limit (3.124) into the inequality (3.123), we obtain from Problem 1.4 that Z

p

X

2γp |f | dµ =

Z 

ZX 

 lim hn dµ

n→∞

 lim inf hn dµ n→∞ X Z hn dµ = lim inf n→∞ X Z i hZ |fn − f |p dµ γp (|f |p + |fn |p ) dµ − ≤ lim inf n→∞ X X Z hZ i p p (|f | + |fn | ) dµ − lim sup |fn − f |p dµ ≤ γp lim inf n→∞ n→∞ X X Z
|fn − f |p dµ. (3.125) = γp lim inf kfn kpp + kf kpp − lim sup =

n→∞

n→∞

X

Since kfn kp → kf kp as n → ∞, we can further reduce the inequality (3.125) to 0 ≤ lim sup n→∞

Z

X

|fn − f |p dµ ≤ γp



 lim kfn kpp + kf kpp − 2γp kf kpp = 0.

n→∞

Hence this leads to the following lim kfn − f kp = lim sup kfn − f kp = 0

n→∞

n→∞

as desired. (c) Consider X = (0, 1) and µ = m the Lebesgue measure. For each n = 1, 2, . . ., we let En = (0, n1p ) and fn = nχEn : (0, 1) → R. If x ∈ (0, 1), then there exists a positive integer N such that x ∈ / En for all n ≥ N . In this case, we have fn (x) = 0 for all n ≥ N . In other words, we have fn (x) → f (x) ≡ 0
for every x ∈ (0, 1). It is clear that f ∈ Lp (0, 1) and kf kp = 0. Besides, we have kfn kp =

nZ

0

1

p

|fn (x)| dx

o1

p

=

nZ

En

np dx

o1

p

=1

for every n ∈ N. Thus we know that kfn kp 9 kf kp as n → ∞. Finally, since kfn − f kp = kfn kp = 1, we see that kfn − f kp 9 0 as n → ∞, i.e., the conclusion of part (b) is false. We have completed the proof of the problem. Remark 3.1 We remark that there is a short and elementary proof of Problem 3.17(b) in [45].



Chapter 3. Lp -Spaces

106

Convergence in Measure and the Essential Range of f ∈ L∞ (µ)

3.5

Problem 3.18 Rudin Chapter 3 Exercise 18.

Proof. (a) Given small ǫ > 0. For each positive integer k, we let Ek = {x ∈ X | |fn (x) − f (x)| ≤ ǫ for all n ≥ k}

and E =

∞ [

Ek .

k=1

The definitions of Ek and E guarantee that X \E =

∞ \

(X \ Ek ) =

k=1

∞ \

k=1

{x ∈ X | |fn (x) − f (x)| > ǫ for some n ≥ k}.

(3.126)

Since fn (x) → f (x) a.e. on X, we deduce from the expression (3.126) that µ(X \ E) = 0. Furthermore, we have E1 ⊆ E2 ⊆ · · · , so Theorem 1.19 implies that µ(Ek ) → µ(E) as k → ∞. Thus it follows from this fact and the result µ(X \ E) = 0 that lim µ(Ek ) = µ(E) = µ(X) < ∞.

k→∞

This result ensures that there exists a N ∈ N such that µ(X \ EN ) < ǫ. {x ∈ X | |fn (x) − f (x)| > ǫ for all n > N } ⊆ X \ EN . Hence we must have µ({x ∈ X | |fn (x) − f (x)| > ǫ for all n > N }) < ǫ, i.e., fn → f in measure. (b) Suppose that 1 ≤ p < ∞. Since kfn − f kp → 0 as n → ∞, we must have fn − f ∈ Lp (µ) for every n ∈ N. Since fn ∈ Lp (µ), we also have f = (f − fn ) + fn ∈ Lp (µ) by Theorem 3.9. Given ǫ > 0, there exists a positive integer N such that n ≥ N implies that kfn − f kp < ǫp+1 .

(3.127)

Let Fn = {x ∈ X | |fn (x) − f (x)| > ǫ} for each n = 1, 2, . . .. Now it is easy to check thati Z Z p ǫp dµ = ǫp µ(Fn ). (3.128) |fn (x) − f (x)| dµ ≥ Fn

i

Fn

The inequality (3.128) is a consequence of the so-called Chebyshev’s Inequality: If f is a nonnegative, extended real-valued measurable function on X with measure µ, 0 < p < ∞ and ǫ > 0, then Z 1 µ({x ∈ X | f (x) ≥ ǫ}) ≤ p f p dµ. ǫ X See [22, Theorem 6.17, p. 193].

107

3.5. Convergence in Measure and the Essential Range of f ∈ L∞ (µ) Hence we establish from the inequalities (3.127) and (3.128) that µ({x ∈ X | |fn (x) − f (x)| > ǫ}) = µ(Fn ) < ǫ for n ≥ N . By the definition, fn → f in measure.

Next, we suppose that p = ∞. Now kfn − f k∞ → 0 means the existence of a positive integer N such that n ≥ N implies |fn (x) − f (x)| < ǫ a.e. on X. By the definition, we have µ(Fn ) = 0 < ǫ for all n ≥ N , i.e., fn → f in measure in this case.

(c) Suppose that fn → f in measure, i.e., for every ǫ > 0, there exists a positive integer N such that n ≥ N implies that µ({x ∈ X | |fn (x) − f (x)| > ǫ}) < ǫ. In particular, for each positive integer k, there exists a positive integer Nk such that µ({x ∈ X | |fn (x) − f (x)| > 2−k }) < 2−k

(3.129)

for all n ≥ Nk . Now we may choose nk > Nk freely in the estimate (3.129) and consider the subsequence {fnk }. We define −k

Ek = {x ∈ X | |fnk (x) − f (x)| > 2

}

and E =

∞ [ ∞ \

Ek .

(3.130)

m=1 k≥m

Then it follows from the estimate (3.129) that µ(E) ≤ µ

∞  [

k≥m

∞ ∞  X X 1 1 Ek ≤ µ(Ek ) < = m−1 , k 2 2 k=m

k=m

where m = 1, 2, . . .. As a result, we must have µ(E) = 0. Take any p ∈ X \ E, the definitions (3.130) say that p∈

∞ \

(X \ Ek ) =

k≥m

for some m ∈ N and then

∞ \

{x ∈ X | |fnk (x) − f (x)| ≤ 2−k }

k≥m

|fnk (p) − f (p)| ≤ 2−k

for all k ≥ m. In other words, we have lim fnk (p) = f (p)

k→∞

which is our desired result. Here we propose two examples which say that the converses of parts (a) and (b) fail: • Failure of the converse of part (a). By Problem 2.9, there exists a sequence of continuous functions fn : [0, 1] → R such that 0 ≤ fn ≤ 1 and kfn − 0k1 → 0 as n → ∞, but there is no x ∈ [0, 1] such that {fn (x)} converges. In other words, fn (x) does not converge to 0 a.e. on
[0, 1]. However, since each fn is continuous [0, 1], it is R on [0, 1] so that fn ∈ L1 [0, 1] by [49, Theorem 11.33, p. 323]. Thus the sequence of functions {fn } satisfies the hypotheses of part (b) which shows that fn → 0 in measure. Hence this example implies the the converse of part (a) is false.

Chapter 3. Lp -Spaces

108

• Failure of the converse of part (b). Suppose that X = [0, 1], µ is the Lebesgue measure and fn = en χ[0, 1 ] n

for each n = 1, 2, . . .. Then fn → 0 a.e. on [0, 1] and part (a) implies that fn → 0 in measure. However, we note that n Z n1 o1 nZ o1 en p p p = = 1 →∞ enp dx kfn kp = |fn (x)| dµ 0 X np
as n → ∞, so the sequence {fn } fails to converge in Lp [0, 1] .

Finally, by examining the proofs of parts (b) and (c), we see that they remain to be true in the case µ(X) = ∞. However, part (a) does not hold in this case. For instance, let X = [0, ∞) and En = [0, n] for each n ∈ N. Consider the functions fn = χEn . Then for every x ∈ [0, ∞), there exists a positive integer N such that x ∈ En for all n ≥ N , so it is true that fn (x) = 1 for all n ≥ N , i.e., fn (x) → f (x) ≡ 1 on [0, ∞). However, for every ǫ ∈ (0, 1), since fn (x) = 0 if and only if x ∈ (n, ∞), we see that {x ∈ [0, ∞) | |fn (x) − 1| > ǫ} = (n, ∞) for every n ∈ N. Hence it leads that m({x ∈ [0, ∞) | |fn (x) − 1| > ǫ}) = ∞ for every n ∈ N, i.e., fn does not converge in measure to f . Hence we end the proof of the problem.



Problem 3.19 Rudin Chapter 3 Exercise 19.

Proof. For every ǫ > 0, define Eǫ (w) = {x ∈ X | |f (x) − w| < ǫ} and \

Rf = {w ∈ C | µ Eǫ (w) > 0 for every ǫ > 0} = {w ∈ C | µ Eǫ (w) > 0}.

(3.131)

ǫ>0

We are going to prove the assertions one by one: • Rf is compact. Given {wn } ⊆ Rf and w ∈ C such that wn → w as n → ∞. For every ǫ > 0, there exists a positive integer N such that |wn − w| < 2ǫ . By the triangle inequality, we have ǫ |f (x) − w| ≤ |f (x) − wn | + |wn − w| < |f (x) − wn | + 2 ǫ which means that E 2 (wn ) ⊆ Eǫ (w) or equivalently

(3.132) µ Eǫ (w) > µ E 2ǫ (wn ) > 0. Since ǫ is arbitrary, we know from the estimate (3.132) that w ∈ Rf . In other words, Rf is closed in C.

Recall that kf k∞ < ∞, so we pick a complex number w0 such that |w0 | > kf k∞ and then consider the positive number ǫ = |w0 | − kf k∞ . If x ∈ Eǫ (w0 ), then the triangle inequality shows that |w0 | − |f (x)| < |f (x) − w0 | < ǫ = |w0 | − kf k∞

109

3.5. Convergence in Measure and the Essential Range of f ∈ L∞ (µ) which implies that |f (x)| > kf k∞ . Thus we have Eǫ (w0 ) ⊆ {x
∈ X | |f (x)| > kf k∞ }. By Definition 3.7, µ({x ∈ X | |f (x)| > kf k∞ }) = 0, so µ Eǫ (w0 ) = 0 too. By the definition (3.131), we have w0 ∈ / Rf and hence the set Rf is bounded by kf k∞ . By the Heine–Borel Theorem, we conclude that Rf is compact. • A relation between Rf and kf k∞ . By the previous analysis, it is obvious that the relation Rf ⊆ {w ∈ C | |w| ≤ kf k∞ } (3.133) holds, i.e., Rf lies in B(0, kf k∞ ). We claim that kf k∞ = max{|z| | z ∈ Rf }.

(3.134)

Given w ∈ C and ǫ > 0. Let B(w, ǫ) = {z ∈ C | |z − w| < ǫ}. Then we have
f −1 B(w, ǫ) = {x ∈ X | f (x) ∈ B(w, ǫ)} = {x ∈ X | |f (x) − w| < ǫ} = Eǫ (w).

By the definition, we see that \

Rf = {w ∈ C | µ f −1 B(w, ǫ) > 0} = C \ V,

(3.135)

ǫ>0



where V = {w ∈ C | µ f −1 B(w, ǫ) = 0 for some ǫ > 0}. Now we have the following two facts: – Fact 1: It is easy to show that V is the largest open subset of C such that
µ f −1 (V ) = µ({x ∈ X | f (x) ∈ V }) = 0,

i.e., V is the largest open subset of C such that f (x) ∈ / V for almost all x ∈ X. By this fact and the relation (3.135), we conclude that Rf is the smallest closed subset of C such that f (x) ∈ Rf for almost all x ∈ X.

– Fact 2: Recall from Definition 3.7 that the number kf k∞ is the minimum of the set {α ≥ 0} such that µ({x ∈ X | |f (x)| > α}) = 0, so we see from the relation (3.133) that it is the minimum radius of the closed disc centered at 0 containing the set Rf .

Hence we follow immediately from these two observations that the equality actually holds in the set relation (3.133). Since they are equal, our claim (3.134) is established. • Relations between Af and Rf . By the definition, we have o n 1 Z f dµ E ∈ M and µ(E) > 0 . Af = µ(E) E

(3.136)

We claim that Rf ⊆ Af . Given ǫ > 0 and w ∈ Rf . If w ∈ Af , then there is nothing to prove. Therefore, we may assume that w ∈ / Af . Consider Eǫ = {x ∈ X, | |f (x) − w| < ǫ}. By the definition of Rf , we have µ(Eǫ ) > 0. If µ(Eǫ ) = ∞ for every ǫ, then f (x) = w a.e. on X. In this case, we have Z 1 f dµ = w ∈ Af , µ(Eǫ ) Eǫ a contradiction. Therefore, we may assume that µ(Eǫ ) < ∞ for infinitely many ǫ > 0. In fact, we may take ǫ = n1 and let E(n) = E 1 . Then, since f ∈ L∞ (µ), we have f − w ∈ L1 (E(n)), where

Z
n L E(n) = f : X → C

n

1

E(n)

o |f | dµ < ∞ .

Chapter 3. Lp -Spaces

110

Consequently, it follows from Theorem 1.33 and the definition of E that Z Z 1 1 1

f dµ − w ≤ |f − w| dµ < . n µ E(n) E(n) µ E(n) E(n)

In other words, w is a limit point of Af and this means that w ∈ Af . Hence this proves the claim that Rf ⊆ Af . • Af is not always closed. For example, we consider
X = [0, 1], f (x) = x with µ = m the Lebesque measure. By Definition 3.7, f ∈ L∞ [0, 1] . In addition, for a measurable set E in [0, 1] with m(E) > 0, we have Z 1 f (x) dx. (3.137) w(E) = m(E) E We claim that w(E) can take any value in (0, 1). Indeed, if a, b ∈ (0, 1) and E = (a, b), then we have m(E) = b − a > 0 and it follows from the definition (3.137) that Z Z b 1 1 b2 − a 2 a+b 1 w(E) = f dx = × = x dx = m(E) E b−a a b−a 2 2 which implies the claim. Next, we claim that w(E) 6= 0. Assume that w(E0 ) = 0 for a measurable set E0 ⊆ [0, 1] with m(E0 ) > 0. Then we have Z f (x) dx = 0, E0

but Theorem 1.39(a) implies that f (x) = 0 a.e. on E0 which contradicts the fact that f (x) = 0 only at x = 0. Hence 0 is not a limit point of Af and Af is not closed in R. • A measure µ such that Af is convex for every f ∈ L∞ (µ). Consider X = {0} and µ the counting measure (see Example 1.20(a)). Then we have M = {∅, X} and µ(X) = µ({0}) = 1 > 0. By the definition, kf k∞ = |f (0)| < ∞ for every f ∈ L∞ (µ) so that f (0) is either kf k∞ or −kf k∞ . Now for every f ∈ L∞ (µ), we have Z 1 w(X) = f dµ = f (0). µ(X) X Therefore, the definition (3.136) gives either Af = {kf k∞ } or Af = {−kf k∞ }, but both cases are also convex sets. • A measure µ such that Af is not convex for some f ∈ L∞ (µ). We consider the set X = {0, 1}, µ the counting measure and f (x) = x. Then we have M = {∅, {0}, {1}, X}, µ({0}) = µ({1}) = 1 > 0 and µ(X) = 2 > 0. Now we have Z Z 1 1 w({0}) = f dµ = f (0) = 0, w({1}) = f dµ = f (1) = 1, µ({0}) {0} µ({1}) {1} Z 1 1 1 1 w(X) = f dµ = [f (0)µ({0}) + f (1)µ({1})] = [f (0) + f (1)] = . µ(X) X 2 2 2 Thus we have Af = {0, 1, 21 } is not convex. • The situations when L∞ (µ) is replaced by L1 (µ). We consider f : (0, ∞) → C by  1    √ , if x ∈ (0, 1]; x f (x) =    0, if x > 1

111

3.6. A Converse of Jensen’s Inequality
and take µ = m the Lebesgue measure. By Lemma 3.1, we have f ∈ L1 (0, ∞) . For any large positive integer N , if x ∈ (0, N12 ), then we have f (x) > N and

 1  1 0, 2 = 2 > 0. N N
= ∞, i.e., f ∈ / L∞ (0, ∞) . Given that ǫ > 0 and

m({x ∈ (0, ∞) | f (x) > N }) = m By Definition 3.7, we have kf k∞ sufficiently large N ∈ N.

– Rf is unbounded and not compact. If x ∈ have |f (x) − N | < ǫ so that m({x ∈ (0, ∞) | |f (x) − N | < ǫ}) = m



1 , 1 (N +ǫ)2 (N −ǫ)2




⊆ (0, 1), then we

 1 1 , > 0. (N + ǫ)2 (N − ǫ)2

Thus N ∈ Rf and then N ⊆ Rf , i.e., Rf is unbounded and not compact.

– Af is unbounded. Next, if E = (0, N42 ), then m(E) = 1 w(E) = m(E)

Z

N2 f (x) dx = 4 E

Z

0

Therefore, N ⊆ Af , i.e., Af is unbounded.

4 N2

x

− 12

4 N2

> 0 and

√ N42 N2 = N. × 2 x dx = 4 0

– Af is not always closed. In fact, the
previous∞example
that X = [0, 1], f (x) = x 1 and µ = m satisfies both f ∈ L [0, 1] and f ∈ L [0, 1] . Thus the same conclusion that 0 is not a limit point of Af is achieved. – A measure µ such that Af is convex for every f ∈ L1 (µ). The example that X = {0} and µ the counting measure also works for this case because if f ∈ L1 (µ), then |f (0)| = kf k1 < ∞. Thus we have either Af = {kf k1 } or Af = {−kf k1 }.

– A measure µ such that Af is not convex for some f ∈ L1 (µ). The example we considered for the case L∞ (µ) also works in this case.

We complete the proof of the problem.

3.6



A Converse of Jensen’s Inequality

Problem 3.20 Rudin Chapter 3 Exercise 20.

Proof. We check Definition 3.1. Let p, q ∈ R and λ ∈ [0, 1]. By the trick of Proposition 1.24(f), we see that Z Z χ[λ,1] (x) dx χ[0,λ] (x) dx + q λp + (1 − λ)q = p =

Z

[0,1] 1

0

[0,1]

[pχ[0,λ] (x) + qχ[λ,1] (x)] dx.

(3.138)

Suppose that f (x) = pχ[0,λ] (x) + qχ[λ,1] (x)

(3.139)

which is clearly a real and bounded function. Besides, since [0, λ] and [λ, 1] are Borel sets in [0, 1], χ[0,λ] and χ[λ,1] are measurable by Proposition 1.9(d). By the last paragraph in [51, §1.22,

Chapter 3. Lp -Spaces

112

p. 19], the function f , as the sum of two measurable functions, is also measurable. Hence, by substituting the expressions (3.138) and (3.139) into the inequality in question, we obtain


Z

1


ϕ pχ[0,λ] (x) + qχ[λ,1] (x) dx 0 Z 1 Z λ ϕ(q) dx ϕ(p) dx + =

ϕ λp + (1 − λ)q ≤

0

λ

= λϕ(p) + (1 − λ)ϕ(q).

Hence ϕ is convex in R, completing the proof of the problem.



Remark 3.2 We note that Problem 3.20 is a converse of Theorem 3.3 (Jensen’s Inequality).

3.7

The Completeness/Completion of a Metric Space

Problem 3.21 Rudin Chapter 3 Exercise 21.

Proof. Let (X, d) and (X ∗ , ρ) be two metric spaces. An isometry of X into X ∗ is a mapping ϕ : X → Y such that
ρ ϕ(p), ϕ(q) = d(p, q) (3.140) for all p, q ∈ X. We notice immediately that an isometry ϕ is necessarily injective and continuous. If ϕ is surjective, then we call ϕ an isomorphism. Two metric spaces X and Y are called isomorphic if there is an isomorphism between them. Before stating the statement, we need: Lemma 3.9 Let (X, d) be a metric space with metric d and p ∈ X. Then the function f : X → R defined by f (x) = d(x, p) is continuous.

Proof of Lemma 3.9. Given ǫ > 0. Let x ∈ X. If y ∈ X satisfies d(x, y) < ǫ, then the triangle inequality implies that |f (x) − f (y)| = |d(x, p) − d(y, p)| ≤ d(x, y) < ǫ. Hence f is continuous, completing the proof of Lemma 3.9. Lemma 3.10 Suppose that (X, d) and (Y, ρ) are metric spaces and Y is complete. If E is dense in X and f : E → Y is an isometry, then there exists an isometry g : X → Y such that g|E = f .



113

3.7. The Completeness/Completion of a Metric Space Proof of Lemma 3.10. If x ∈ E, we define g(x) = f (x). Let x ∈ X \ E. Since E is dense in X, we can choose a sequence {xn } ⊆ E such that xn → x. By [49, Theorem 3.11(a)], {xn } is Cauchy in X. Thus, given ǫ > 0, there exists a positive integer N such that n, m ≥ N imply that d(xm , xn ) < ǫ. Since f is an isometry, it follows from the expression (3.140) that
ρ f (xm ), f (xn ) < ǫ

for all n, m ≥ N . In other words, {f (xn )} is Cauchy in Y . Since Y is complete, {f (xn )} converges to a limit y in Y and we define g(x) = y in this case. Now this y is uniquely determined by x. Indeed, let {x′n } ⊆ E be another sequence converging to x and y ′ be the limit of {f (x′n )} in Y . For each m ∈ N, it follows from Lemma 3.9 and the expression (3.140) that

ρ y, f (x′m ) = lim ρ f (xn ), f (x′m ) = lim d(xn , x′m ) = d(x, x′m ). n→∞

n→∞

Then this implies that


ρ(y, y ′ ) = lim ρ y, f (x′m ) = lim d(x, x′m ) = 0. m→∞

m→∞

Since ρ is a metric, we must have y = y ′ and so we obtain a mapping g : X → Y given by   if x ∈ E;  f (x), (3.141) g(x) =   lim f (xn ), if x ∈ X \ E, {xn } ⊆ X and xn → x. n→∞

Next, we show that g is an isometry and we consider the following situations:

• Case (i): x, y ∈ E. In this case, we know from the definition (3.141) that g(x) = f (x) and g(y) = f (y) and then the use of the expression (3.140) implies that

ρ g(x), g(y) = ρ f (x), f (y) = d(x, y). • Case (ii): x ∈ X \ E and y ∈ E. In this case, we have

ρ g(x), g(y) = ρ g(x), f (y) .

Choose a sequence {xn } in E such that xn → x. By Lemma 3.9 and the expression (3.140), we establish that

(3.142) ρ g(x), f (y) = lim ρ f (xn ), f (y) = lim d(xn , y) = d(x, y). n→∞

n→∞

• Case (iii): x, y ∈ X \ E. Let {xn }, {yn } ⊆ E be sequences converging to x and y respectively. Then we deduce from the definition (3.141) and the expression (3.142) that
ρ g(x), f (yn ) = d(x, yn ). (3.143) Apply Lemma 3.9 to (3.143), we see that

ρ g(x), g(y) = lim ρ g(x), f (yn ) = lim d(x, yn ) = d(x, y). n→∞

By the definition, g is an isometry and g|E = f .

n→∞



Chapter 3. Lp -Spaces

114

Lemma 3.11 The mapping g in Lemma 3.9 is unique up to isometry.

Proof of Lemma 3.11. Let g ′ : X → Y be an isometry such that g′ |E = f . If x ∈ E, then g(x) = f (x) = g ′ (x). If x ∈ X \ E, then there is a sequence {xn } ⊆ E such that xn → x as n → ∞. The triangle inequality, the expressions (3.140) and (3.142) indicate that



ρ g(x), g ′ (x) ≤ ρ g(x), g(xn ) + ρ g(xn ), g′ (xn ) + ρ g′ (xn ), g′ (x)


= ρ g(x), f (xn ) + ρ f (xn ), f ′ (xn ) + ρ f ′ (xn ), g′ (x)

= ρ g(x), f (xn ) + d(xn , xn ) + ρ f ′ (xn ), g′ (x) = d(x, xn ) + d(xn , x)

Hence when n → ∞, it is clear that ρ(g(x), g′ (x)) = 0, so g(x) = g′ (x) on X \ E and our result follows from this.  Let’s return to the proof of the problem. Let (Y1 , ρ1 ) and (Y2 , ρ2 ) be complete metric spaces, ϕ1 : X → Y1 and ϕ2 : X → Y2 be isometries such that ϕ1 (X) is dense in Y1 and ϕ2 (X) is dense in Y2 . We claim that (Y1 , ρ1 ) and (Y2 , ρ2 ) are isomorphic. Define f : ϕ1 (X) → Y2 by f = ϕ2 ◦ ϕ−1 1 .

(3.144)

If p, q ∈ X, x = ϕ1 (p) and y = ϕ1 (q), then we have



−1 ρ2 f (x), f (y) = ρ2 ϕ2 ϕ−1 1 (x) , ϕ2 ϕ1 (y)
= ρ2 ϕ2 (p), ϕ2 (q) = d(p, q)


= ρ1 ϕ1 (p), ϕ1 (q) = ρ1 (x, y).

Consequently, f is an isometry. Since ϕ1 (X) is dense in Y1 and Y2 is complete, Lemma 3.10 ensures that there exists an isometry g : Y1 → Y2 such that g|ϕ1 (X) = f. It remains to prove that g is surjective. To this end, let y ∈ Y2 . Since ϕ2 (X) = Y2 , we can select a sequence {yn } ⊆ ϕ2 (X) such that yn → y as n → ∞. Furthermore, there exists a sequence {xn } ⊆ X such that yn = ϕ2 (xn ) for all n ∈ N. Consider the corresponding sequence zn = ϕ1 (xn )

(3.145)

for n ∈ N. Since {yn } is Cauchy in ϕ2 (X) (and hence in Y2 ) and ϕ2 is an isometry, the expression (3.140) forces that {xn } is Cauchy in X. Then, since ϕ1 is an isometry, the expression (3.140) again forces that {zn } is Cauchy in ϕ1 (X) (and hence in Y1 ). Since Y1 is complete, we can find z ∈ Y1 such that zn → z as n → ∞. We claim that g(z) = y.

115

3.8. Miscellaneous Problems

Thus we deduce from Lemma 3.10, the definition (3.145) and the fact g|ϕ1 (X) = f that





(3.146) ρ2 g(z), y = lim ρ2 g(zn ), y = lim ρ2 g ϕ1 (xn ) , y = lim ρ2 f ϕ1 (xn ) , y . n→∞

n→∞

n→∞

Finally, the definition (3.144) reduces the limit (3.146) to



ρ2 g(z), y = lim ρ2 f ϕ1 (xn ) , y = lim ρ2 ϕ2 (xn ), y = lim ρ2 (yn , y) = ρ2 (y, y) = 0. n→∞

n→∞

n→∞

Therefore, g(z) = y and g is surjective. Hence g is an isomorphism and our claim follows, completing the proof of the problem.  Problem 3.22 Rudin Chapter 3 Exercise 22.

Proof. This problem is proven in [63, Problem 3.20, p. 50].

3.8



Miscellaneous Problems

Problem 3.23 Rudin Chapter 3 Exercise 23.

Proof. If αn0 = 0 for some n0 , then we must have f = 0 a.e. on X by Theorem 1.39(a). However, this implies that the measure of
f −1 (α, ∞] = {x ∈ X | f (x) > α}

is zero for every α > 0. By Definition 3.7, it means that kf k∞ = 0 which contradicts the hypothesis. Thus we have αn 6= 0 for all n ∈ N so that the limit in question is well-defined. By Definition 3.7, we see that |f (x)| ≤ kf k∞ holds for almost all x ∈ X. Therefore, we have αn+1 ≤ kf k∞ · αn

a.e. on X

(3.147)

for all n ∈ N. Since kf k∞ > 0, we can find a ǫ > 0 such that kf k∞ > ǫ > 0. By the definition, the measurable set F = {x ∈ X | |f (x)| ≥ kf k∞ − ǫ > 0} satisfies µ(F ) > 0. Apply Theorem 3.5 (H¨older’s Inequality) to the measurable functions f n and 1 with p = n+1 n , we obtain αn =

Z

n

X

|f | dµ ≤

nZ

X

n

(|f | )

n+1 n

o

n n+1

nZ

1n+1 dµ X

o

1 n+1

so that −1 1 αn+1 ≥ (αn+1 ) n+1 × µ(X) n+1 αn nZ o 1 −1 n+1 ≥ |f |n+1 dµ × µ(X) n+1

F

 1 −1 ≥ (kf k∞ − ǫ)n+1 µ(F ) n+1 × µ(X) n+1 

n

1

= (αn+1 ) n+1 × µ(X) n+1

Chapter 3. Lp -Spaces

116

=

 µ(F ) 

1 n+1

µ(X)

× (kf k∞ − ǫ).

(3.148)

Thus we follow from the inequalities (3.147) and (3.148) that  µ(F ) 

1 n+1

µ(X)

× (kf k∞ − ǫ) ≤

αn+1 ≤ kf k∞ αn

a.e. on X

(3.149)

1

µ(F ) n+1 ) tends to 1 as n → ∞. By for all n ∈ N. Since 0 < µ(F ) ≤ µ(X) < ∞, the number ( µ(X) taking limit to both sides of the inequality (3.149), we see immediately that

kf k∞ − ǫ ≤ lim inf n→∞

αn+1 αn+1 ≤ lim sup ≤ kf k∞ αn αn n→∞

a.e. on X.

Remember that ǫ is arbitrary, so this implies that lim

n→∞

αn+1 = kf k∞ αn

a.e. on X

and we complete the proof of the problem.



Problem 3.24 Rudin Chapter 3 Exercise 24.

Proof. Let x ≥ 0 and y ≥ 0. Put α − β = x and α = y into the equality (3.111), we gain |x|p − |y|p ≤ |x − y|p .

(3.150)

Next, we substitute α − β = y and α = x into the equality (3.111) to get −|x − y|p ≤ |x|p − |y|p . Obviously, the inequalities (3.150) and (3.151) together imply that |xp − y p | = |x|p − |y|p ≤ |x − y|p

(3.151)

(3.152)

if 0 < p < 1.

If x = y, then the inequality |xp − y p | ≤ p|x − y|(xp−1 + y p−1 ) certainly holds. Without loss of generality, we may assume that x < y. For p > 1, we consider the function ϕ(t) = tp on [x, y]. Since ϕ is differentiable in (x, y), the Mean Value Theorem implies that |y p − xp | ≤ |x − y||ϕ′ (ξ)| ≤ p|x − y|ξ p−1 (3.153)

for some ξ ∈ (x, y). Clearly, we have ξ p−1 ≤ xp−1 + y p−1 , so it reduces from the inequality (3.153) that |xp − y p | ≤ p|x − y|(xp−1 + y p−1 ). (3.154) Hence we establish from the inequalities (3.152) and (3.154) that  if 0 < p < 1;  |x − y|p , p p |x − y | ≤  p|x − y|(xp−1 + y p−1 ), if 1 ≤ p < ∞.

117

3.8. Miscellaneous Problems

(a) We prove the assertions one by one: – The truth of the inequality. Put x = |f | ≥ 0 and y = |g| ≥ 0 into the inequality (3.152) and then take integration, we get Z Z p |f | − |g|p dµ ≤ |f | − |g| p dµ. (3.155)

Since |a| − |b| ≤ |a − b| holds for every a, b ∈ R, we follow immediately from the inequality (3.155) that Z Z Z p |f | − |g|p dµ ≤ |f | − |g| p dµ ≤ |f − g|p dµ, (3.156)

where 0 < p < 1.

– ∆ is a metric. Define ∆ : Lp (µ) × Lp (µ) → C by Z ∆(f, g) = |f − g|p dµ. By Remark 3.10, Lp (µ) is a complex vector space, so f − g ∈ Lp (µ) if f, g ∈ Lp (µ). Thus we have Z 0 ≤ ∆(f, g) = |f − g|p dµ < ∞

for all f, g ∈ Lp (µ). Next, Theorem 1.39(a) ensures that ∆(f, g) = 0 if and only if |f − g|p = 0 a.e. on X, i.e., f = g a.e. on X. It is clear that Z Z ∆(f, g) = |f − g|p dµ = |g − f |p dµ = ∆(g, f ).

Finally, since |f |p − |g|p ≤ |f |p − |g|p , it is clear from the inequality (3.156) that Z Z Z (|f |p − |g|p ) dµ ≤ |f |p − |g|p dµ ≤ |f − g|p dµ. (3.157) By replacing f and g by f − g and h − g in the inequality (3.157), we deduce that Z ∆(f, g) − ∆(h, g) = (|f − g|p − |h − g|p ) dµ Z ≤ |(f − g) − (h − g)|p dµ Z = |f − h|p dµ = ∆(f, h).

After rearrangement, we have ∆(f, g) ≤ ∆(f, h) + ∆(h, g). Hence, by the definition, ∆ is in fact a metric. – (Lp (µ), ∆) is a complete metric space. Suppose that {fn } is a Cauchy sequence in Lp (µ) with respect to the metric ∆, i.e., for every ǫ > 0, there exists a positive integer N such that n, m ≥ N implies that ∆(fn , fm ) < ǫ.

(3.158)

Although ∆(f, g) = kf −gkpp here, we cannot apply Theorem 3.11 directly to conclude that Lp (µ) is complete with respect to ∆ because p ≥ 1 in Theorem 3.11.

Chapter 3. Lp -Spaces

118

We imitate the proof of Theorem 3.11. To start with, there is a subsequence {fni } such that ∆(fni+1 , fni ) < 2−i (3.159) for each i = 1, 2, . . .. We define gk : X → [0, ∞] and g : X → [0, ∞] by gk =

k X i=1

|fni+1 − fni | and g =

∞ X i=1

|fni+1 − fni |

respectively. It is clear that gk ≥ 0 and g ≥ 0 and furthermore, it follows from Theorem 3.5 (Minkowski’s Inequality) that ∆(gk , 0) = ∆

k X i=1

=

 |fni+1 − fni |, 0

Z X k i=1

≤ =

p |fni+1 − fni | dµ

k nZ hX

i=1 k hX

|fni+1 − fni |p dµ 1

∆(fni+1 , fni ) p

i=1

ip

o 1 ip p

(3.160)

for each k = 1, 2, . . .. To proceed further, we need the following result: Lemma 3.12 Let p ∈ (0, 1). For nonnegative real numbers a1 , a2 , . . . , ak , we have (a1 + a2 + · · · + ak )p ≤ ap1 + ap2 + · · · + apk .

Proof of Lemma 3.12. Replace β by −β in the inequality (3.111), we get (α + β)p ≤ |α|p + |β|p for p ∈ (0, 1). In particular, if α ≥ 0 and β ≥ 0, then we have (α + β)p ≤ αp + β p

(3.161)

for p ∈ (0, 1). By applying the inequality (3.161) repeatedly to a1 , a2 , . . . , ak , we  derive the desired inequality. This completes the proof of Lemma 3.12. Apply Lemma 3.9 to the inequality (3.160) and then using the inequality (3.159) to get k k X X 2−i ≤ 1 (3.162) ∆(fni+1 , fni ) < ∆(gk , 0) ≤ i=1

for each k = 1, 2, . . ..

i=1

119

3.8. Miscellaneous Problems Since gk (x) → g(x) as k → ∞ for every x ∈ X, we have gk (x)p → g(x)p as k → ∞ on X. We note that since each fni is measurable, each gkp is also measurable and we may apply Theorem 1.28 (Fatou’s Lemma) to {gkp } and the inequality (3.162) to get Z Z Z Z p p p g dµ = lim gk dµ = lim inf gk dµ ≤ lim inf gkp dµ = lim inf ∆(gk , 0) ≤ 1, k→∞

k→∞

k→∞

k→∞

implying that g ∈ Lp (µ). In particular, g < ∞ a.e. on X, so the series ∞ X [fni+1 (x) − fni (x)] fn1 (x) +

(3.163)

i=1

converges absolutely for almost every x ∈ X. Denote the sum (3.163) by f (x) for those x at which the sum (3.163) converges and put f (x) = 0 on the remaining set of measure zero. Since we have k−1 X [fni+1 (x) − fni (x)] = fnk (x), fn1 (x) + i=1

it implies that fni (x) → f (x) as i → ∞ a.e. on X. Now it remains to show that f ∈ Lp (µ) and ∆(fn , f ) → 0 as n → ∞. Given ǫ > 0. Since {fn } is Cauchy in Lp (µ) with respect to ∆, there exists a N ∈ N such that the estimate (3.158) holds for n, m ≥ N . Take m ≥ N , since fni → f as i → ∞ a.e. on X, Theorem 1.28 (Fatou’s Lemma) again implies that Z Z   lim inf |fni − fm |p dµ |f − fm |p dµ = i→∞ Z  ≤ lim inf |fni − fm |p dµ i→∞

= lim inf ∆(fni , fm ) < ǫ. i→∞

(3.164)

so we conclude that f − fm ∈ Lp (µ), hence that f ∈ Lp (µ). Finally, according to the estimate (3.164), it follows that ∆(f, fm ) < ǫ for all m ≥ N . Since ǫ is arbitrary, we conclude that lim ∆(f, fm ) = 0.

m→∞

This shows the completeness of the space Lp (µ) with respect to ∆. (b) Similar to part (a), a direct substitution of x = |f | and y = |g| into the inequality (3.154) and the use of the fact |x| − |y| ≤ |x − y| give Z Z p |f | − |g|p dµ ≤ p |f | − |g| × (|f |p−1 + |g|p−1 ) dµ Z (3.165) ≤ p |f − g|(|f |p−1 + |g|p−1 ) dµ. By Theorem 3.5 (H¨older’s Inequality) and the facts that kf kp ≤ R and kgkp ≤ R, we observe that Z Z Z p−1 p−1 p−1 dµ + |f − g| · |g|p−1 dµ |f − g|(|f | + |g| ) dµ ≤ |f − g| · |f | nZ o p−1 o1 n Z p p p p−1 p−1 p (|f | ) ≤ dµ |f − g| dµ

Chapter 3. Lp -Spaces

120

+

nZ

p

|f − g| dµ

o1 n Z p

p

(|g|p−1 ) p−1 dµ

= kf − gkp · (kf kp−1 + kgkp−1 p p )

o p−1

≤ 2Rp−1 kf − gkp .

p

(3.166)

Combining the inequalities (3.165) and (3.166), we discover that Z p |f | − |g|p dµ ≤ 2pRp−1 kf − gkp

which is our expected result.

We have completed the proof of the problem.



Problem 3.25 Rudin Chapter 3 Exercise 25.

Proof. Suppose that E ⊆ X and 0 < µ(E) < ∞. Let ME be a σ-algebra in E. Then (E, ME , µ) is a measure space.j Define Z 1 dµ (F ∈ ME ). ϕ(F ) = µ(E) F Then we follow from Theorem 1.29 that ϕ is a measure on ME and Z Z 1 f dµ f dϕ = µ(E) E E

(3.167)

for every measurable f on E with range in [0, ∞]. It is trivial that ϕ(E) = 1 and Z Z 1 1 f dϕ ≤ f dµ = 0 on (0, ∞), we apply Theorem 3.3 (Jensen’s Inequality) with ψ and measure ϕ to conclude that Z op nZ f p dϕ. (3.169) f dϕ ≤ − − E

E

Next, we use the fact (3.167) to reduce the inequality (3.169) to Z n 1 Z op n 1 Z op 1 1 f p dµ ≤ f dµ ≤ f dµ = µ(E) E µ(E) E µ(E) X µ(E)p so that

Z

E

completing the proof of the problem.

f p dµ ≤ µ(E)1−p , 

Problem 3.26 Rudin Chapter 3 Exercise 26.

Proof. Let M be a σ-algebra in [0, 1]. Suppose that there exists an E ∈ M such that f (x) = ∞ and m(E) > 0, where m is the Lebesgue measure. Then Z

1

f (x) log f (x) dx =

0

Z

f (x) log f (x) dx +

E

Z

f (x) log f (x) dx.

(3.170)

[0,1]\E

On the one hand, since f > 0 on [0, 1] \ E, we know from Proposition 1.24(a) that the second integral on the right-hand side of the expression (3.170) is nonnegative. Since f (x) log f (x) = ∞ on E and m(E) > 0, we get from the expression (3.170) that Z

0

1

f (x) log f (x) dx = ∞.

(3.171)

On the other hand, by using similar argument, we can show that Z 1 Z 1 log f (t) dt = ∞. f (s) ds = ∞ and

(3.172)

0

0

Therefore, we conclude from the results (3.171) and (3.172) that Z

1

f (x) log f (x) dx = 0

Z

1

f (s) ds 0

Z

1

log f (t) dt

0

in this case. Next, we suppose that 0 < f (x) < ∞ a.e. on [0, 1]. Define ϕ : (0, ∞) → R and ψ : (0, ∞) → R by ϕ(x) = x log x and ψ(x) = − log x

respectively. Since ϕ′′ (x) = x1 > 0 and ψ ′′ (x) = x−2 > 0 on (0, ∞), they are convex on (0, ∞). Since 0 < f (x) < ∞ a.e. on [0, 1] and m([0, 1]) = 1, we apply Theorem 3.3 (Jensen’s Inequality) with ϕ to f to obtain o Z 1 nZ 1 on  Z 1 f (x) log f (x) dx (3.173) ≤ f (x) dx f (x) dx log 0

0

0

Chapter 3. Lp -Spaces

122

Similarly, we apply Theorem 3.3 (Jensen’s Inequality) with ψ to f to obtain −log

Z

1

Z  f (x) dx ≤ −

1

log f (x) dx.

(3.174)

0

0

Combining the inequalities (3.173) and (3.174), we see that nZ

1

f (x) dx 0

on Z

0

1

o

log f (x) dx ≤

We have ended the proof of the problem.

Z

1

f (x) log f (x) dx. 0



CHAPTER

4

Elementary Hilbert Space Theory

In the following problems, we use hx, yi to denote the inner product of the complex vectors x and y and k · k the norm with respect to the Hilbert space H. Besides, we use span (S) to denote the span of a set S.

4.1

Basic Properties of Hilbert Spaces

Problem 4.1 Rudin Chapter 4 Exercise 1.

Proof. Let x ∈ M . Recall that M ⊥ = {y ∈ H | y ⊥ x for all x ∈ M }, so x ⊥ y for every y ∈ M ⊥ which means that x ∈ (M ⊥ )⊥ , i.e., M ⊆ (M ⊥ )⊥ . Conversely, suppose that x ∈ (M ⊥ )⊥ . Since M is a closed subspace of H, it follows from Theorem 4.11(a) that x = y + z, where y ∈ M and z ∈ M ⊥ . On the one hand, since x ∈ (M ⊥ )⊥ and z ∈ M ⊥ , we have hx, zi = 0. On the other hand, we deduce from Definition 4.1 that hx, zi = hy + z, zi = hy, zi + hz, zi = hz, zi. Since hz, zi = 0, it must be z = 0 and then x ∈ M , i.e, (M ⊥ )⊥ ⊆ M . In conclusion, we obtain M = (M ⊥ )⊥ . Suppose that M is a subspace of H which may not be closed. We claim that M = (M ⊥ )⊥ . To see this, we recall from §4.8 that M is a closed subspace of H, so the first assertion implies that ⊥ M = (M )⊥ . (4.1) ⊥

Let x ∈ M . By §4.9, we see that x ⊥ M so that x ⊥ M particularly. In other words, we have M

⊥

⊆ M ⊥.

(4.2) ⊥

For the other direction, suppose that x ∈ M ⊥ , we want to show that x ∈ M , i.e., x ⊥ M . To this end, given y ∈ M , there exist a sequence {yn } ⊆ M such that yn → y as n → ∞. Since 123

Chapter 4. Elementary Hilbert Space Theory

124

x ∈ M ⊥ and yn ∈ M , we must have hyn , xi = 0. By Theorem 4.6, the mapping f : H → C defined by f (y) = hy, xi is continuous on H so that hy, xi = f (y) = lim f (yn ) = lim hyn , xi = 0. n→∞

In other words, x ⊥ M , i.e.,

n→∞

⊥

M⊥ ⊆ M .

(4.3)

Now the claim is derived by combining the set relations (4.2) and (4.3) and then using the result (4.1), completing the proof of the problem.  Problem 4.2 Rudin Chapter 4 Exercise 2.

Proof. We note that this is actually the Gram-Schmidt Process. Suppose that n = 1. Then the set {u1 } is clearly orthonormal and span (u1 ) = span (x1 ). Thus the statement is true for n = 1. Assume that the statement is true for n = k for some k ∈ N, i.e., {u1 , u2 , . . . , uk } is an orthonormal set and span (u1 , u2 , . . . , uk ) = span (x1 , x2 , . . . , xk ). (4.4) Let n = k + 1. Note that xk+1 ∈ / span (x1 , x2 , . . . , xk ) because {x1 , x2 , . . .} is linearly independent. By the assumption (4.4), it implies that xk+1 ∈ / span (u1 , u2 , . . . , uk ) so that vk+1 6= 0. By this fact and Definition 4.1, for j = 1, 2, . . . , k, we derive that   vk+1 , uj huk+1 , uj i = kvk+1 k + * k X 1 = hxk+1 , ui i ui , uj xk+1 − kvk+1 k i=1

= =

1 kvk+1 k 1

kvk+1 k = 0.

hxk+1 , uj i − hxk+1 , uj i −

1 kvk+1 k 1

kvk+1 k

k X i=1

hxk+1 , ui i hui , uj i

hxk+1 , uj i huj , uj i

Therefore, {u1 , u2 , . . . , uk+1 } is orthonormal. Next, by the assumption (4.4), we know that x ∈ span (x1 , x2 , . . . , xk , xk+1 ) if and only if x ∈ span (u1 , u2 , . . . , uk , xk+1 ). Since vk+1 = xk+1 −

k X i=1

hxk+1 , ui i ui

and uk+1 =

(4.5)

vk+1 , kvk+1 k

the xk+1 in the result (4.5) can be replaced by vk+1 and ultimately by uk+1 . Thus we have span (u1 , u2 , . . . , uk+1 ) = span (x1 , x2 , . . . , xk+1 ).

125

4.1. Basic Properties of Hilbert Spaces

Hence the statement is true for n = k + 1 if it is true for n = k. By induction, the construction yields an orthonormal set {u1 , u2 , . . .} with span (u1 , u2 , . . . , un ) = span (x1 , x2 , . . . , xn ) for all n ∈ N. This completes the proof of the problem.



Problem 4.3 Rudin Chapter 4 Exercise 3.

Proof. Recall the definition that a space is separable if it contains a countable dense subset. Suppose that 1 ≤ p < ∞. Since T is compact, the second paragraph following Definition 3.16 says that Cc (T ) = C0 (T ) = C(T ). Thus we deduce from this and Theorem 3.14 that C(T ) is dense in Lp (T ) in the norm k · kp . Let P be the set of all trigonometric polynomials on T . By Theorem 4.25 (The Weierstrass Approximation Theorem) and Definitions 4.23 (or Remark 3.15), P is dense in C(T ) in the norm k · k∞ . By Definition 3.7, we know that |f (x)| ≤ kf k∞ a.e. on T and then o1 n 1 Z π o1 n 1 Z π p p |f (t)|p dt kf kp∞ dt ≤ = kf k∞ . (4.6) kf kp = 2π −π 2π −π

In other words, the inequality (4.6) implies that P is also dense in C(T ) in the norm k · kp . By Lemma 2.10, we conclude that P is dense in Lp (T ) in the norm k · kp . Next we suppose that PQ is the set of all trigonometric polynomials with rational coefficients. Note thata n n o X PQ = P (t) = (ak + ibk )eikt a−n , . . . , an , b−n , . . . , bn ∈ Q and n ∈ N k=−n

=

∼

∞ [

n=1 ∞ [

n

P (t) =

n X

o (ak + ibk )eikt a−n , . . . , an , b−n , . . . , bn ∈ Q

k=−n

Q2(2n+1) .

(4.7)

n=1

For each n ∈ N, since each Q2(2n+1) is countable, it follows from the set relation (4.7) that PQ is also countable. Given f ∈ P. Since the set {p + iq | p, q ∈ Q} is dense in C and f must be in the form n X ck eikt f (t) = k=−n

for some positive integer n, there exists a sequence {fnk } ⊆ PQ such that kfnk − f k∞ → 0 as nk → ∞. By the inequality (4.6), it is true that kfnk − f kp → 0

as nk → ∞. Consequently, PQ is dense in P in the norm k · kp and an application of Lemma 2.10 shows that PQ is dense in Lp (T ) in the norm k · kp . Since PQ is countable, Lp (T ) is separable. By Definition 4.23, every function defined on T
can be identified with a 2π-periodic function on R, so we may identity L∞ (T ) with L∞ [0, 2π] . Consider the set E = {fθ = χ[0,θ] | θ ∈ [0, 2π]}.

a

We write A ∼ B if there is a bijection between the two sets A and B, see [49, Definition 2.3, p. 25].

Chapter 4. Elementary Hilbert Space Theory

126



It is clear that χ[0,θ] ∈ L∞ [0, 2π] so that E ⊆ L∞ [0, 2π] . Since χ[0,θ1] 6= χ[0,θ2 ] if θ1 6= θ2 , E ∼ [0, 2π] and then it is uncountable. In fact, we have kχ[0,θ1 ] − χ[0,θ2 ] k∞ = 1

(4.8)

if θ1 < θ2 .
Assume
that L∞ [0, 2π] was separable. Let F = {fn } be a countable dense subset of L∞ [0, 2π] . Then we must have [ 
1 E ⊆ L∞ [0, 2π] ⊆ B fn , , 2

(4.9)

fn ∈F

where B(fn , 12 ) = {f ∈ L∞ ([0, 2π]) | kf − fn k∞ < 12 }. If χ[0,θ1 ] , χ[0,θ2 ] ∈ B(fn , 21 ) with θ1 6= θ2 , then it follows from the result (4.8) that 1 = kχ[0,θ1 ] − χ[0,θ2 ] k∞ ≤ kχ[0,θ1 ] − fn k∞ + kfn − χ[0,θ2 ] k∞ < 1, a contradiction. Thus each B(fn , 12 ) contains at most one element of E. Since F is countable but E is uncountable, we have [  1 E* B fn , 2 fn ∈F


which definitely contradicts the set relation (4.9). Hence L∞ [0, π] is not separable and we have completed the proof of the problem.  Problem 4.4 Rudin Chapter 4 Exercise 4.

Proof. Suppose that H is separable. By the definition, it has a countable dense subset {un }. By Problem 4.2, we may assume that {un } is also an orthonormal set. Let {unk } be a maximal linearly independent subset of {un }. Note that {unk } is at most countable. Since we have span ({unk }) = {un }, span ({unk }) is also dense in H. By Theorem 4.18, the set {unk } is in fact an at most countable maximal orthonormal set in H. Conversely, we suppose that E = {un } is an at most countable maximal orthonormal set in H. By Theorem 4.18 again, span (E) is dense in H. Let span Q (E) be the span of the set E with rational coefficients. It is easy to see that span Q (E) =

n nX

o (ak + ibk )uk a1 , . . . , an , b1 , . . . , bn ∈ Q and n ∈ N

n=1 ∞ [

k=1

k=1

= ∼

∞ nX n [

Q2n .

o (ak + ibk )uk a1 , . . . , an , b1 , . . . , bn ∈ Q

n=1

As a result, span Q (E) must be countable. Next, since Q is dense in R, we conclude that span Q (E) is dense in span (E). By Lemma 2.10, span Q (E) is also dense in H and hence it is  separable. This completes the proof of the problem.

127

4.1. Basic Properties of Hilbert Spaces

Problem 4.5 Rudin Chapter 4 Exercise 5.

Proof. Suppose that M 6= H. If L = 0, then M = H so that L 6= 0. By Theorem 4.12 (The Riesz Representation Theorem for Hilbert Spaces), there is a unique z ∈ H such that L(x) = hx, zi for all x ∈ H. This z must be nonzero, otherwise we have L = 0 which contradicts our assumption. It is clear that M = {x ∈ H | hx, zi = 0} = L−1 (0),

(4.10)

so M is closed in H. Apart from this, if x, y ∈ M , then since L is a linear functional, Definition 2.1 shows that L(x + y) = L(x) + L(y) = 0 and L(αx) = αL(x) = 0 for a scalar α. Thus M is a closed subspace of H. Next, we note that span (z) = {αz | α ∈ C} and so


⊥ span (z) = {x ∈ H | hx, αzi = 0 for every α ∈ C}.

(4.11)

Since hx, αzi = α hx, zi, the two expressions (4.10) and (4.11) give
⊥ M = span (z)

and then

M⊥ =


⊥
⊥ span (z) .

(4.12)

We claim that span (z) is a closed subspace of H. In fact, it is a subspace of H follows easily from the definition of span (z). Suppose that {xn } is a sequence in span (z). Then we know that each xn has the form xn = αn z, where αn ∈ C. Let α = lim αn and x = αz. We want to show n→∞

that xn → x in H with respect to the norm k · k. Indeed, we have

kxn − xk2 = h(αn − α)z, (αn − α)zi = |αn − α2 | × kzk so that kxn − xk → 0 as n → ∞, i.e., lim xn = x

n→∞

as desired. In conclusion, span (z) contains all its limit points and therefore it is closed in H.
⊥
⊥ By Problem 4.1, we have span (z) = span (z). Combining this fact and the expression (4.12), we have M ⊥ = span (z)
and since {z} is a basis of span (z), we have dim span (z) = 1 and our required result follows.  We end the proof of the problem. Problem 4.6 Rudin Chapter 4 Exercise 6.

Chapter 4. Elementary Hilbert Space Theory

128

Proof. We prove the assertions one by one: • E = {un } is closed and bounded, but not compact. Since E is orthonormal, kun k ≤ 1 for every n ∈ N. Thus E is bounded by 1. If n 6= m, then it follows from Definition 4.1 that kun − um k2 = hun − um , un − um i = hun , un i + hum , um i = 2. (4.13) Assume that u was a limit point√ of E. Then there exists a positive integer N such that n ≥ N implies that kun − uk < 22 . If n, m ≥ N , then we have kun − um k ≤ kun − uk + ku − xm k
0, there is a N ′ ∈ N such that X ǫ δn2 < . (4.16) 8 ′ n>N

Then we obtain from the definition (4.14), the fact |ckn | ≤ δn for all k and the estimate (4.16) that kx

km (m)

∞

2

X

km (m) [cn − cn ]un − xk = 2

n=1

=

∞ X

n=1

|cknm (m) − cn |2

′

=

N X

n=1

|cknm (m) − cn |2 +

N′

≤

X

n=1

|cknm (m) − cn |2 +

N′

≤

X

n=1 N′


k > 0. n∈Ek

Now for n ∈ Ek , we define

1 an bn = n X o 1 × . k 2 a2n n∈Ek

c

See [49, Theorem 3.28, p. 62].

P

a2n = ∞. Therefore, for every (4.20)

131

4.1. Basic Properties of Hilbert Spaces

On the one hand, we have ∞ X

b2n =

n=1

∞ X X

b2n =

∞ X k=1

k=1 n∈Ek

1 X

a2n

×

X

n∈Ek

n∈Ek

1 a2n × 2 k

!

=

∞ X 1 < ∞. k2 k=1

On the other hand, we obtain from the inequality (4.20) that ! ∞ n X ∞ X ∞ ∞ ∞ o1 X X X X X X 1 a2n 1 2 21 √ × = a an bn = a n bn = > n n X o1 k k 2 k n=1 n∈Ek k=1 n∈Ek k=1 n∈Ek k=1 k=1 a2n n∈Ek

which is divergent, a contradiction. Hence we must have of the problem.

P

a2n < ∞ and it completes the proof 

Problem 4.8 Rudin Chapter 4 Exercise 8.

Proof. By Problem 4.2, H contains an orthonormal set A and then Theorem 4.22 (The Hausdorff Maximality Theorem) implies that A is contained in a maximal orthonormal set in H. Suppose that {uα | α ∈ A} and {vβ | β ∈ B} are maximal orthonormal sets in H1 and H2 respectively. Without loss of generality, we may assume that |A| ≤ |B|. Then we take a subset B ′ of B which is of the same cardinality as A. Now we consider H = span ({vβ | β ∈ B ′ }) ⊆ H2 . By Definition 4.13, span ({vβ | β ∈ B ′ }) is a subspace of H2 . Thus we follow from the last paragraph in §4.7 that H is a closed subspace of H2 . By the given hint, H is also Hilbert. Finally, we know from §4.19 that H1 and H are Hilbert space isomorphic to ℓ2 (A) and ℓ2 (B ′ ) respectively. Since A and B have the same cardinality, ℓ2 (A) is Hilbert space isomorphic to ℓ2 (B ′ ). In other words, H1 is Hilbert space isomorphic to H ⊆ H2 , completing the proof of the problem.  Problem 4.9 Rudin Chapter 4 Exercise 9.

Proof. Since A ⊆ [0, 2π] and A is measurable, we have χA ∈ L2 (T ). By Theorem 3.11, L2 (T ) is a complete metric space, so it is a Hilbert space by Definition 4.4. Since {un | n ∈ Z} is a maximal orthonormal set in the Hilbert space L2 (T ), Theorem 4.18(iii) implies that Z X hun , χA i 2 = kχA k2 =

n∈Z

2π 0

|χA (x)|2 dx = m(A) < ∞.

By [49, Theorem 3.23, p. 60], we know that lim hun , χA i = lim hun , χA i = 0. n→∞

n→−∞

(4.21)

Chapter 4. Elementary Hilbert Space Theory

132

By Definition 4.1, we have 

 un + u−n hcos nx, χA i = , χA = 2   un − u−n hsin nx, χA i = , χA = 2i

1 (hun , χA i + hu−n , χA i), 2 1 (hun , χA i − hu−n , χA i). 2i

(4.22)

By applying the limits (4.21) to the inner products (4.22), we obtain that Z π Z (cos nt) · χA (t) dt = 2π lim hcos nx, χA i = 0 cos nt dt = lim lim n→∞ A

and lim

Z

n→∞ A

n→∞

n→∞ −π

sin nt dt = lim

Z

π

n→∞ −π

(sin nt) · χA (t) dt = 2π lim hsin nx, χA i = 0. n→∞

Thus we have completed the proof of the problem.



Problem 4.10 Rudin Chapter 4 Exercise 10.

Proof. This problem is proven in [63, Problem 11.16, p. 352].



Problem 4.11 Rudin Chapter 4 Exercise 11.

Proof. For each n = 1, 2, . . ., we define xn = of L2 (T ) because, by Definition 4.23,

n+1 n un

kxn k = kxn k2 =

and E = {xn } which is obviously a subset

n+1

= kxN +1 k, N N +1 this contradiction shows that E has no element of smallest norm. Hence we end the proof of the problem.  Remark 4.1 Problem 4.11 indicates that the condition “convexity” in Theorem 4.10 cannot be omitted.

Problem 4.12 Rudin Chapter 4 Exercise 12.

Proof. By the half-angle formula for cosine function, we have ck 1= π

Z

 1 + cos t k

π

2

0

Z

2ck dt = π

π 2

cos2k x dx.

(4.24)

0

A further substitution with z = sin x reduces the expression (4.24) to 2ck 1= π

Z

1 0

1

(1 − z 2 )k− 2 dz.

(4.25)

If we substitute y = z 2 and use the integral form of the beta function (see [49, Theorem 8.20, p. 193]), then the expression (4.25) can be written as ck 1= π

Z

1 0

1

1

(1 − y)k− 2 y − 2 dy =

ck Γ(k + 21 )Γ( 12 ) ck Γ(k + 21 ) · =√ · . π Γ(k + 1) kΓ(k) π

(4.26)

Finally, we recall from [49, Exercise 30, p. 203] that Γ(k + c) =1 k→∞ k c Γ(k) lim

(4.27)

for every real constant c. Take c = 12 in the limit (4.27) and then substitute the corresponding result into the expression (4.26), we get lim k

k→∞

− 12

ck = lim

This completes the proof of the problem. Problem 4.13 Rudin Chapter 4 Exercise 13.

k→∞

√

1

√ k 2 Γ(k) π· π. 1 = Γ(k + 2 ) 

Chapter 4. Elementary Hilbert Space Theory

134

Proof. We follow the hint. Consider the function f (t) = e2πikt , where k = 0, ±1, ±2, . . .. When k = 0, we have f ≡ 1 and then the formula holds trivially. Suppose that k 6= 0. Since α is irrational, kα ∈ / Z and then e2πikα 6= 1. By this, we have N N 1 X 2πiknα 1 X 2πikα n e2πikα e2πikN α − 1 e = (e ) = · 2πikα N n=1 N n=1 N e −1

so that

N 1 X 2 2πiknα e = 0. lim ≤ lim N →∞ N |e2πikα − 1| N →∞ N

(4.28)

n=1

On the other hand, we have

Z

1

2πikt

e

0

e2πikt 1 dt = = 0. 2πik 0

(4.29)

Hence we deduce from the inequality (4.28) and the integral (4.29) that Z 1 N 1 X 2πiknα lim e = e2πikt dt N →∞ N 0

(4.30)

n=1

for every k = 0, ±1, ±2, . . .. Therefore the formula (4.30) holds for every trigonometric polynomials in the form N X P (t) = ck e2πikt , (4.31) k=−N

where c−N , . . . , cN ∈ C.

Suppose that f : R → R is continuous and f (x + 1) = f (x) for every x ∈ R. By Theorem 4.25 (The Weierstrass Approximation Theorem), for every ǫ > 0, there exists a trigonometric polynomial P in the form (4.31) such that |f (t) − P (t)|
p, then we have nZ 1 o1 n Z t o1 1 2 2 2 kγ(t) − γ(p)k = |χ(p,t] (x)| dx = dx = (t − p) 2 . (4.54) 0

p

Similarly, if t < p, then we have kγ(t) − γ(p)k =

nZ

p

dx t

o1

2

1

= (p − t) 2 .

(4.55)

By combining the two expressions (4.54) and (4.55), we conclude that 1

kγ(t) − γ(p)k = |t − p| 2 and thus γ is continuous on [0, 1]. Let 0 ≤ a ≤ b ≤ c ≤ d ≤ 1. If a = b, then γ(a) − γ(b) = 0 and Z 1 0 × χ(c,d] (x) dx = 0. hγ(b) − γ(a), γ(d) − γ(c)i = h0, γ(d) − γ(c)i = 0

Similarly, if c = d, then hγ(b) − γ(a), γ(d) − γ(c)i = 0. Without loss of generality, we may assume that a < b and c < d. Then (a, b] ∩ (c, d] = ∅ and so we have Z 1

 hγ(b) − γ(a), γ(d) − γ(c)i = χ(a,b] , χ(c,d] = χ(a,b] (x)χ(c,d] (x) dx = 0. 0


Hence the desired result holds in the special case that H = L2 [0, 1] . To treat the general case, we first consider the function un (t) = e2πint for every n ∈ Z. It is clear that un ∈ H and  Z 1  1, if n = m; 2πi(n−m)t e dt = hun , um i =  0 0, if n 6= m.
Thus {un | n ∈ Z} is an orthonormal set in H. By Theorem 3.14, C [0, 1] is dense in H with respect to the norm induced by the inner product (4.53). By Theorem 4.25 (The Weierstrass Approximation Theorem) and Definition 4.23, we know that the set P of all trigonometric polynomials in the form N N X X cn e2πint cn un (t) = P (t) = n=−N

n=−N


is dense in C [0, 1] with respect to the norm k · k∞ . Note that P
= span ({un | n ∈ Z}). By Definition 3.7, it can be shown easily that P is also dense in C [0, 1] with respect to the norm induced by the inner product (4.53). Therefore, we derive from Lemma 2.10 that P is also dense in H in the norm induced by the inner product (4.53). Consequently, we conclude from Theorem 4.18(i) that {un | n ∈ Z} is in fact maximal and thene H∼ = ℓ2 (N).

(4.56)

Suppose that H ′ is a Hilbert space with an infinite maximal orthonormal set {vα | α ∈ A}, where A is countable or uncountable infinite. Then we have H′ ∼ = ℓ2 (A). e

We say A ∼ = B if A is Hilbert space isomorphic to B.

(4.57)

Chapter 4. Elementary Hilbert Space Theory

140

Applying Problem 4.8 to the Hilbert space isomorphisms (4.56) and (4.57), we conclude that H must be Hilbert space isomorphic to a subspace of H ′ and then the special case is applicable to this general case. Indeed, let Λ : H → K be a Hilbert space isomorphism, where K is a Hilbert subspace of H ′ . If we define λ : [0, 1] → H ′ by λ = Λ ◦ γ, then λ is injective and continuous. Furthermore, since hΛ(f ), Λ(g)i = hf, gi for all f, g ∈ H, we deduce from the special case above that





 hλ(b) − λ(a), λ(d) − λ(c)i = Λ γ(b) − Λ γ(a) , Λ γ(d) − Λ γ(c)







 = Λ γ(b) , Λ γ(d) − Λ γ(b) , Λ γ(c)







 − Λ γ(a) , Λ γ(d) + Λ γ(a) , Λ γ(c) = hγ(b), γ(d)i − hγ(b), γ(c)i − hγ(a), γ(d)i + hγ(a), γ(c)i

= hγ(b) − γ(a), γ(c) − γ(d)i

= 0.

This completes the proof of the problem.



Problem 4.18 Rudin Chapter 4 Exercise 18.

Proof. If r, s ∈ R and r 6= s, then we have Z A 2i sin[(r − s)A] 1 = 0. ei(r−s)t dt = lim hur , us i = lim A→∞ A→∞ 2A −A 2A(r − s) If r = s, then 1 A→∞ 2A

hus , us i = lim

Z

A

1 A→∞ 2A

eist e−ist dt = lim

−A

Z

(4.58)

A

dt = 1.

(4.59)

−A

Thus {us | s ∈ R} is an orthonormal set of H by Definition 4.13. Since f and g are finite combinations of us , we may suppose that f (t) =

n X

isp t

cp e

and g(t) =

p=1

m X

dq eirq t .

(4.60)

q=1

Therefore, we obtain Z A X Z AX m n n X m   X 1 1 lim dq e−irq t dt = lim cp eisp t cp dq ei(sp −rq )t dt A→∞ 2A −A A→∞ 2A −A q=1

p=1

p=1 q=1

which is undoubtedly finite by the values (4.58) and (4.59). As a consequence, we have shown that hf, gi is well-defined for all f, g ∈ X. Next we are going to show that h · i satisfies Definition 4.1. Firstly, it is clear that 1 hf, gi = lim A→∞ 2A

Z

A

1 f (t)g(t) dt = lim A→∞ 2A −A

Z

A −A

Secondly, for f, g, h ∈ X, we observe that Z A 1 [f (t) + g(t)]h(t) dt hf + g, hi = lim A→∞ 2A −A

g(t)f (t) dt = hg, f i.

141

4.3. Miscellaneous Problems 1 = lim A→∞ 2A

Z

A

1 f (t)h(t) dt + lim A→∞ 2A −A

= hf, hi + hg, hi .

Z

A

g(t)h(t) dt −A

Thirdly, for any scalar α, we have Z A Z A 1 1 αf (t)h(t) dt = α lim f (t)h(t) dt = α hf, gi . hαf , gi = lim A→∞ 2A −A A→∞ 2A −A Fourthly, if f takes the representation (4.60), then we establish from the results (4.58) and (4.59) that Z A Z A n X 1 1 2 |cp |2 ≥ 0. (4.61) f (t)f (t) dt = lim |f (t)| dt = hf, f i = lim A→∞ 2A −A A→∞ 2A −A p=1

Finally, if hf, f i = 0, then the result (4.61) definitely shows that c1 = c2 = · · · = cp = 0. In other words, we have f ≡ 0 in this case. Hence this inner product certainly makes X into a unitary space. Let H be the completion of X, i.e., X is dense in H and H = X. Since span ({us | s ∈ R}) = X, Theorem 4.18(i) says that the set {us | s ∈ R} is in fact a maximal orthonormal in H. By the discussion in §4.19, we have H∼ (4.62) = ℓ2 (R). Assume that H was separable. By Problem 4.4, H has an at most countable maximal orthonormal system {vn | n ∈ N}. By the discussion in §4.19 again, we know that H∼ = ℓ2 (N).

(4.63)

Now the two isomorphic relations (4.62) and (4.63) imply that ℓ2 (R) ∼ = ℓ2 (N), but then R and N have the same cardinal number, a contradiction. Hence we complete the proof of the problem.  Problem 4.19 Rudin Chapter 4 Exercise 19.

Proof. If N = 1, then ω = e2πi = 1 and k can only take the value 0. Thus the orthogonality relation holds in this special case. Suppose that N > 1. If k = 0, then ω 0 = 1 so that N 1 X 0 ω = 1. N n=1

Let k = 1, 2, . . . , N − 1. We remark that z N − 1 = (z − 1)(z N −1 + · · · + z + 1). Putting z = ω k into the expression (4.64), we obtain  
(ω k )N − 1 = (ω k − 1) (ω k )N −1 + · · · + ω k + 1 = (ω k − 1) ω k(N −1) + · · · + ω k + 1 .

(4.64)

(4.65)

It is clear that (ω k )N = (ω N )k = 1. Furthermore, since 1 ≤ k ≤ N − 1, k is evidently not divisible by N which implies that ω k 6= 1. Therefore, it deduces from the expression (4.65) that ω k(N −1) + · · · + ω k + 1 = 0

Chapter 4. Elementary Hilbert Space Theory

142

or equivalently ω kN + ω k(N −1) + · · · + ω k = 0. In conclusion, we have

 N 1, if k = 0; 1 X nk  ω =  N n=1 0, if 1 ≤ k ≤ N − 1.

Let N ≥ 3. It is clear from the orthogonality relations and the properties in Definition 4.1 that N N 1 X 1 X n 2 n kx + ω yk ω = hx + ω n y, x + ω n yi ω n N N n=1

n=1

N N N X X 1 hX hω n y, xi ω n hx, ω n yi ω n + hx, xi ω n + = N n=1

n=1

+

N X

n=1

hω n y, ω n yi ω n

n=1

i

X X X 1h = hy, xi ω n × ω n hx, yi ω −n × ω n + ωn + hx, xi N + hy, yi

N

n=1 N X i n

N

n=1

n=1

ω

n=1 N X

1h = hx, xi N

= hx, yi

N

n=1

n

ω + N hx, yi + hy, xi

N X

n=1

ω

2n

+ hy, yi

N X

ωn

n=1

i

(4.66)

as desired. Finally, Definition 4.1 gives D E kx + eiθ yk2 = x + eiθ y, x + eiθ y = hx, xi + e−iθ hx, yi + eiθ hy, xi + hy, yi so that

1 2π

Z

Z π   1 kx + e yk e dθ = kxk2 + kyk2 + e−iθ hx, yi + eiθ hy, xi eiθ dθ 2π −π −π Z π Z π 1 1 hx, yi dθ + e2iθ hy, xi dθ = 2π −π 2π −π π

iθ

2 iθ

= hx, yi .

This completes the proof of the problem.

(4.67) 

Remark 4.3 We note that the polarization identity (see [51, p. 86]) can be written in the form 3

hx, yi =

1 X −n i kx + in yk2 . 4 n=0

Thus the two identities (4.66) and (4.67) in Problem 4.19 are actually generalizations of this.

CHAPTER

5

Examples of Banach Space Techniques

5.1

The Unit Ball in a Normed Linear Space

Problem 5.1 Rudin Chapter 5 Exercise 1.

Proof. For 0 < p ≤ ∞, we suppose that B = {f ∈ Lp (µ) | kf kp ≤ 1}. There are two cases: • Case (i): 0 < p < ∞. In this case, we have o1 nZ p |f |p dµ kf kp = X


1 = |f (a)|p × µ({a}) + |f (b)|p × µ({b}) p 1 1 = 1 {|f (a)|p + |f (b)|p } p . 2p

Consequently, kf kp ≤ 1 if and only if |f (a)|p + |f (b)|p ≤ 2.

(5.1)

Therefore, it follows from the inequality (5.1) that the unit ball B is a circle if and only if p = 2. Besides, the unit ball B becomes a square if and only if p = 1. • Case (ii): p = ∞. By Definition 3.7, we know that kf k∞ = max(|f (a)|, |f (b)|)

(5.2)

max(|f (a)|, |f (b)|) ≤ 1

(5.3)

so that kf k∞ ≤ 1 if and only if

which is a square. See Figure 5.1 for an illustration. 143

Chapter 5. Examples of Banach Space Techniques

144

Figure 5.1: The unit circle in different p-norm. If µ({a}) 6= µ({b}), then the expression (5.1) is replaced by µ({a})|f (a)|p + µ({b})|f (b)|p ≤ 1 which cannot be a circle or a square for any p ∈ (0, ∞). However, we note that the expression (5.2) is still valid even in the case µ({a}) 6= µ({b}), so we have the inequality (5.3) and then B  is a square when p = ∞. This ends the analysis of the problem. Problem 5.2 Rudin Chapter 5 Exercise 2.

Proof. Let X be a normed linear space with norm k · k and B(0, 1) = {x ∈ X | kxk < 1} be the open unit ball. For every x, y ∈ B(0, 1) and t ∈ [0, 1], we follow from Definition 5.2 that k(1 − t)x + tyk ≤ (1 − t)kxk + tkyk < 1 so that (1 − t)x + ty ∈ B(0, 1). By §4.8, B(0, 1) is convex. Similarly, the convexity of the closed unit ball B(0, 1) = {x ∈ X | kxk ≤ 1} can be proven similarly. Hence we complete the proof of the problem.  Problem 5.3 Rudin Chapter 5 Exercise 3.

Proof. Suppose that kf kp = kgkp = 1 and f 6= g. By Theorem 3.9, we always have

1

1 1

khkp = (f + g) ≤ kf kp + kgkp = 1. 2 2 2 p

Assume that khkp = 1. Then it follows from Problem 3.13 that f = λg a.e. on X for some λ > 0. Since λkgkp = kf kp = kgkp , we have λ = 1 implying f = g a.e. on X, a contradiction. However, strictly convexity does not hold for L1 (µ), L∞ (µ) and C(X). Ignore trivialities, we suppose that X contains more than one point. In fact, we can pick E, F ∈ M \ {∅} such that E ∩ F = ∅ and µ(E), µ(F ) ∈ (0, ∞). Define f : X → C and g : X → C by f (x) =

1 χE (x) and µ(E)

g(x) =

1 χF (x) µ(F )

respectively. It is obvious that f 6= g. Furthermore, we have Z Z 1 |f (x)| dx = kf k1 = χE (x) dx = 1, µ(E) E X

(5.4)

145

5.1. The Unit Ball in a Normed Linear Space Z 1 χF (x) dx = 1, µ(F ) F X Z Z Z

f + g 1 1

f (x) + g(x)

f (x) dx + g(x) dx = 1.



= dx = 2 2 2 E 2 F 1 X kgk1 =

Z

|g(x)| dx =

These show that the invalidity of strictly convexity in L1 (µ). For L∞ (µ), we replace the functions f and g in the definition (5.4) by f (x) = χE (x)

and g(x) = χE∪F (x)

respectively. Then it is also true that f 6= g. By Definition 3.7, we have

f + g

kf k∞ = kgk∞ =

= 1. 2 ∞ ∞ Therefore, the strictly convexity does not hold in L (µ). Finally, we consider the space C(X), where X is Hausdorff. By Definition 3.16, we see that X is assumed to be compact and so the norm k · k∞ of C(X) is given by kf k∞ = sup |f (x)|.

(5.5)

x∈X

Take f ∈ C(X) and suppose that f is nonconstant. Thus one can find a x0 ∈ X such that |f (x0 )| < kf k∞ .

(5.6)

Otherwise, we have |f (x)| ≥ kf k∞ on X, but this and the definition (5.5) will imply that |f (x)| = kf k∞ on X, i.e., f is a constant which is a contradiction. Define 1 3 a = kf k∞ + |f (x0 )| and 4 4

Consider the sets o n
K = x ∈ X |f (x)| ≥ a = |f |−1 [a, ∞)

and

1 3 b = kf k∞ + |f (x0 )|. 4 4

o
E = x ∈ X |f (x)| ≤ b = |f |−1 (−∞, b] . n

It is clear that x0 ∈ E, i.e., E 6= ∅. Since |f | : X → R is continuous, the Extreme Value Theorem ([42, Theorem 27.4, p. 174]) implies that kf k∞ = max |f (x)| = |f (p)| for some p ∈ X. x∈X

Consequently, we have

p ∈ K.

(5.7)

Furthermore, both K and E are closed in X by [42, Theorem 18.1, p. 104]. If K ∩ E 6= ∅, then we have 1 1 3 3 kf k∞ + |f (x0 )| ≤ kf k∞ + |f (x0 )| 4 4 4 4 but these imply that kf k∞ ≤ |f (x0 )| which contradicts the hypothesis (5.6). Hence we have K ∩ E = ∅.

What we have done in the previous paragraph is that V = E c is open in X and K ⊆ V . Since X is compact, Theorem 2.4 ensures that K is also compact. By Theorem 2.12 (Urysohn’s Lemma), there exists a h ∈ Cc (X) such that K ≺ h ≺ V , i.e., 0 ≤ h(x) ≤ 1 on X, h(x) = 1 on K and supp (h) ⊆ V (or equivalently h(x) = 0 on E). If we define g = h × f : X → C, then we must have g 6≡ f and furthermore, the fact (5.7) shows that kgk∞ = sup |h(x)f (x)| = |h(p)| · |f (p)| = |f (p)| = kf k∞ x∈X

and

f + g 1 1 1



= kf + hf k∞ = sup [1 + h(x)]f (x) = |1 + h(p)| · |f (p)| = |f (p)| = kf k∞ . 2 2 2 x∈X 2 ∞ Hence strictly convexity does not hold for C(X), completing the proof of the problem.



Chapter 5. Examples of Banach Space Techniques

5.2

146

Failure of Theorem 4.10 and Norm-preserving Extensions

Problem 5.4 Rudin Chapter 5 Exercise 4.

Proof. Define f : [0, 1] → C by

  −8x + 4, if x ∈ [0, 12 ];

f (x) =



if x ∈ ( 21 , 1].

0,

Then it is easy to see that f ∈ M , i.e., M is nonempty. Suppose that f, g ∈ M , t ∈ [0, 1] and h = (1 − t)f + tg. Then we have Z

1 2

0

h(x) dx −

Z

1 1 2

h(x) dx =

Z

1 2

0

[(1 − t)f (x) + tg(x)] dx −

hZ = (1 − t) +t =1

hZ

f (x) dx −

0

1

g(x) dx −

0

Z

1

Z

1

1

1 1 2

[(1 − t)f (x) + tg(x)] dx

f (x) dx

1 2

g(x) dx

1 2

Z

i

i

so that h ∈ M , i.e., M is convex.

Next, we note that C and M are metric spaces. By [49, Theorem 7.15, p. 151], we see that C is a complete metric space. Since M ⊆ C, M is also a complete metric space. Suppose that {fn } ⊆ M and kfn − f k = sup |fn (x) − f (x)| → 0 x∈[0,1]

as n → ∞. Then it follows froma [49, Theorem 7.9, p. 148] that fn → f uniformly on [0, 1]. Since each fn is continuous on [0, 1], f is also continuous on [0, 1]. Furthermore, we have lim

n→∞

Z

1 2

0



hZ

0

1 2

fn (t) dt −

Z  lim fn (t) dt −

n→∞

Z

1 2

0

1 2

1

Z

1 1 2

i fn (t) dt = 1

 lim fn (t) dt = 1

n→∞

f (t) dt −

Z

1

1 2

f (t) dt = 1.

In other words, f ∈ M and M is closed in the space C.

Finally, we want to show that M contains no element of minimal norm. To this end, we notice that

a

Z 1=

1 2

0

f (t) dt −

Z

1 1 2

Z f (t) dt ≤

0

1 2

|f (t)| dt +

Or we can see the rephrased Theorem 7.9 on p. 151 in [49].

Z

1 1 2

|f (t)| dt = kf k1 ≤ kf k∞

(5.8)

147

5.2. Failure of Theorem 4.10 and Norm-preserving Extensions

for every f ∈ M . Assume that there was a f ∈ M such that kf k∞ = 1. Then we deduce from the inequality (5.8) that Z 1 Z 1 (1 − |f (t)|) dt = 0. (5.9) |f (t)| dt = 1 or 0

0

Since kf k∞ = 1, we have |f (x)| ∈ [0, 1] for all x ∈ [0, 1] so that the function g(x) = 1 − |f (x)| is continuous and nonnegative on [0, 1]. Applying Theorem 1.39(a) to the second integral in (5.9), we obtain g(x) = 0 a.e. on [0, 1]. Now the continuity of g forces that g(x) = 0 on [0, 1], i.e., |f (x)| = 1 on [0, 1]. In particular, we have
−1 ≤ Re f (x) ≤ 1 (5.10)

on [0, 1]. On the other hand, since f ∈ M , the definition gives 1 = Re

Z

1 2

f (t) dt −

0

or equivalently,

Z

1 2

0



Z

1 1 2

 Z f (t) dt =


 Re f (t) − 1 dt +

Z

1 2

0

1

1 2


Re f (t) dt −

Z

1 1 2


Re f (t) dt


 −1 − Re f (t) dt = 0.

(5.11)



Observe from the inequalities (5.10) that Re f (x) − 1 ≤ 0 and
−1 − Re f (x) ≤ 0 on
[0, 1]. Therefore, the equation (5.11) and the continuities of Re f (x) − 1 and −1 − Re f (x) imply that  if [0, 12 ];
 1, Re f (x) =  −1, if [ 21 , 1].
However, this says that Re f (x) is discontinuous at x = 21 , a contradiction. Hence no such f exists and this completes the proof of the problem.  Problem 5.5 Rudin Chapter 5 Exercise 5.
Proof. It is clear that 1 ∈ M , so M 6= ∅. Suppose that f, g ∈ M ⊆ L1 [0, 1] , t ∈ [0, 1] and h = (1 − t)f + tg. Then we have Z 1 Z 1 Z 1 |g(x)| dx < ∞ |f (x)| dx + t |h(x)| dx ≤ (1 − t) 0

0

0


which means that h ∈ L1 [0, 1] . Besides, we also have Z

1

h(x) dx =

Z

0

0

1

[(1 − t)f (x) + tg(x)] dx = (1 − t)

Z

0

1

f (x) dx + t

Z

1

g(x) dx = 1 0

so that h ∈ M , i.e., M is convex. Next, suppose that {fn } ⊆ M and
there exists a function f on 1 [0, 1] such that kfn − f k1 → 0 as n → ∞. Thus fn − f ∈ L [0, 1] and it yields from Theorem 1.33 that Z 1 Z 1 Z 1 |fn (t) − f (t)| dt = kfn − f k1 f (t) dt ≤ fn (t) dt − 0

0

0

Chapter 5. Examples of Banach Space Techniques which implies that lim

Z

n→∞ 0

1

fn (t) dt =

Z

148

1

f (t) dt.

0


In other words, we have f ∈ M and M is closed in L1 [0, 1] .
For every f ∈ M ⊆ L1 [0, 1] , we see from Theorem 1.33 that kf k1 ≥ 1. For each n = 1, 2, . . ., we define fn : [0, 1] → C by fn (x) = nxn−1 . By direct checking, it is clear that kfn k1 =

Z

1

ntn−1 dt = 1.

0

In addition, fn 6= fm if n 6= m. Hence M contains infinitely many elements of minimal norm, completing the proof of the problem.  Problem 5.6 Rudin Chapter 5 Exercise 6.

Proof. We note that M is not necessarily closed in H. Let f : M ⊆ H → C be a bounded linear functional. If we check the proof of Theorem 5.4 carefully, we see that the bounded linear transformation Λ : X → Y is in fact uniformly continuous on X. Thus f : M → C is uniformly continuous on M . Since H is a metric (in fact Hilbert) space, M is also a metric space. Since C is a complete metric space and M is dense in M , we follow from [49, Exercies 4 & 13, pp. 98, 99] that f can be uniquely extended to a continuous function fe : M → C.b Without loss of generality, we may assume that M is closed in H. As a closed subspace of a Hilbert space H, M is itself a Hilbert space (see the note in Problem 4.8). By Theorem 4.12 (The Riesz Representation Theorem for Hilbert Spaces), there corresponds a unique x0 ∈ M such that f (x) = hx, x0 i

(5.12)

for all x ∈ M . Recall from Definition 5.3 that kf k is the smallest number such that |f (x)| ≤ kf k · kxk for every x ∈ M . In particular, we must have kf (x0 )k ≤ kf k · kx0 k. On the other hand, the representation (5.12) gives |f (x0 )| = f (x0 ) = hx0 , x0 i = kx0 k2 = kx0 k · kx0 k, so we must have kx0 k = kf k.

(5.13)

By Theorem 4.11, every x ∈ H can be expressed uniquely as x = P x + Qx, where P x ∈ M and Qx ∈ M ⊥ . This means that H = M ⊕ M ⊥ .c Now if we define F : H → C by F (x) = hx, x0 i In fact, such fe is uniformly continuous on M . See also [42, Exercise 2, p. 270]. Let U and W be two vector subspaces of the vector space V . Then V is said to be the direct sum of U and W , denoted by V = U ⊕ W , if V = U + W and U ∩ W . See [4, pp. 95, 96]. b c

149

5.2. Failure of Theorem 4.10 and Norm-preserving Extensions

for every x ∈ H, then we obtain F (x) = hx, x0 i =

  f (x), if x ∈ M ; 

if x ∈ M ⊥ .

0,

Furthermore, by the same analysis in obtaining the representation (5.13), we get kF k = kx0 k and then using the representation (5.13) again, we conclude that kF k = kf k = kx0 k,

(5.14)

i.e., F is a norm-preserving extension on H of f which vanishes on M ⊥ . Suppose that F ′ : H → C is another norm-preserving extension of f that vanishes on M ⊥ . Then F ′ is bounded and thus continuous by Theorem 5.4. By Theorem 4.12 (The Riesz Representation Theorem for Hilbert Spaces), there corresponds a unique x1 ∈ H such that F ′ (x) = hx, x1 i for all x ∈ H and also Since F ′ (x) = f (x) on M , we have

kF ′ k = kf k = kx1 k.

(5.15)

hx, x1 i = hx, x0 i

on M . By Definition 4.1, hx, x1 − x0 i = 0 and then x1 − x0 ∈ M ⊥ . Since x1 = x0 + (x1 − x0 ), where x0 ∈ M and x1 − x0 ∈ M ⊥ , Theorem 4.11 shows that x1 and x1 − x0 are unique and kx0 k2 = kx1 k2 + kx0 − x1 k2 .

(5.16)

By putting the numbers (5.14) and (5.15) into the formula (5.16), we see that kf k2 = kf k2 + kx0 − x1 k2 and this reduces to kx0 − x1 k2 = 0. By Definition 4.1, we must have x0 = x1 , i.e., F ′ = F . This  proves the uniqueness part and so we have completed the proof of the problem. Problem 5.7 Rudin Chapter 5 Exercise 7.

Proof. Consider X = [−1, 1] with µ = m, the Lebesgue measure. Then the vector space
L1 [−1, 1] consists of all complex measurable function on [−1, 1] such that kf k1 = We define

Z

1

−1

|f (x)| dx < ∞.


M = {f ∈ L1 [−1, 1] | f is real-valued and f (x) = 0 on [−1, 0]}.

Now the function χ[0,1] is obviously an element
of M , so M 6= ∅ and also kχ[0,1] k1 = 1. It is also clear that M is a subspace of L1 [−1, 1] . Next, we define the functional Λ : M → C by Λ(f ) =

Z

1

−1

f (x) dx =

Z

1

f (x) dx 0

(5.17)

Chapter 5. Examples of Banach Space Techniques

150

which is linear. By Definition 5.3, the definition (5.17) and the fact that χ[0,1] ∈ M , we have kΛk = sup{|Λ(f )| | f ∈ M and kf k1 = 1} Z Z 1 nZ 1 = sup |f (x)| dx = f (x) dx f ∈ M and −1

0

= 1.

Thus Λ is bounded.
Let δ ∈ (0, 1). If we define Λδ : L1 [−1, 1] → C by Λδ (f ) =

Z

1

−1

Re [f (x)]χ(−δ,1) (x) dx =

Z

1 0

o f (x) dx = 1

1

Re [f (x)] dx, −δ


then it satisfies Definition 2.1 so that Λδ is linear on L1 [−1, 1] . It is also clear that Λδ |M = Λ. A direct computation also shows that Z 1 Z 1 Z 1 |f (x)| dx |Re [f (x)]| dx ≤ Re [f (x)] dx ≤ |Λδ (f )| = −δ

−δ

−1

which implies that


kΛδ k = sup{|Λδ (f )| | f ∈ L1 [−1, 1] and kf k1 = 1} = 1 = kΛk.
Thus Λδ is a norm-preserving extension on L1 [−1, 1] of Λ. Hence if α 6= β, then we have Λα 6= Λβ . This completes the proof of the problem. 

5.3

The Dual Space of X

Problem 5.8 Rudin Chapter 5 Exercise 8.

Proof. (a) We first show that X ∗ is a normed linear space by checking Definition 5.2. For all f, g ∈ X ∗ , since |f (x) + g(x)| ≤ |f (x)| + |g(x)| for all x ∈ X, it must be true that kf + gk ≤ kf k + kgk. Next, for a scalar α and f ∈ X ∗ , we have |αf (x)| = |α| · |f (x)| for all x ∈ X which implies that kαf k = |α| · kf k. Finally, suppose that f ∈ X ∗ is such that kf k = 0. By [51, Eqn. (3), p. 96], we have |f (x)| ≤ kf k · kxk = 0 holds for every x ∈ X. Thus f ≡ 0 on X. By Definition 5.2, X ∗ is a normed linear space.

It remains to show that X ∗ is complete. To this end, let {Λn } ⊆ X ∗ be Cauchy. Then given ǫ > 0, there corresponds a positive integer N such that n, m ≥ N imply that kΛn − Λm k < ǫ

151

5.3. The Dual Space of X and for arbitrary but fixed x ∈ X, this and [51, Eqn. (3), p. 96] together imply that |Λn (x) − Λm (x)| = |(Λn − Λm )(x)| ≤ kΛn − Λm k · kxk < ǫkxk.

(5.18)

Thus {Λn (x)} ⊆ C is Cauchy. Since C is complete, {Λn (x)} converges to a point in C and it is reasonable to define the pointwise limit Λ : X → C by Λ(x) = lim Λn (x). n→∞

Now our target is to show that Λ ∈ X ∗ and {Λn } converges to Λ in X ∗ , i.e., kΛn −Λk → 0 as n → ∞. Let α, β ∈ C and x, y ∈ X. Since X is a normed linear space, αx + βy ∈ X and thus Λ(αx + βy) = lim Λn (αx + βy) n→∞

= lim [αΛn (x) + βΛn (y)] n→∞

= α lim Λn (x) + β lim Λn (y) n→∞

n→∞

= αΛ(x) + βΛ(y). By Definition 2.1, Λ is linear. By the hypothesis, {Λn } is a Cauchy sequence in X ∗ so that it is bounded, i.e., there exists a positive constant M such that kΛn k ≤ M

(5.19)

for every n = 1, 2, . . .. Therefore, it follows from the triangle inequality, then [51, Eqn. (3), p. 96] and finally the inequalities (5.19) that |Λ(x)| ≤ |Λn (x)| + |Λ(x) − Λn (x)|

≤ kΛn k · kxk + |Λ(x) − Λn (x)|

≤ M kxk + |Λ(x) − Λn (x)|

(5.20)

for all n > N and for all x ∈ X. Recall that if we take m → ∞ in the inequality (5.18), then we have |Λn (x) − Λ(x)| ≤ ǫkxk (5.21) for all n > N . Thus, by putting the inequality (5.19) into the inequality (5.20), we get |Λ(x)| ≤ (M + ǫ)kxk for every x ∈ X. Hence we obtain kΛk < ∞, i.e., Λ ∈ X ∗ . Since the inequality (5.21) is true for all x ∈ X, it must be true for those x with kxk ≤ 1 and so n > N implies that kΛn − Λk = sup{|Λn (x) − Λ(x)| | x ∈ X and kxk ≤ 1} ≤ ǫ. Since ǫ is arbitrary, we have {Λn } converges to Λ in X ∗ . In other words, X ∗ is complete and we conclude that X ∗ is a Banach space. (b) Fix the x ∈ X, define Λ∗x : X ∗ → C by Λ∗x (f ) = f (x). If x = 0, then it is clear that Λ∗0 (f ) = f (0) = 0 on X ∗ . Therefore, Λ∗0 must be linear and of norm 0. Without loss of generality, we may assume that x 6= 0 in the following discussion. For any f, g ∈ X ∗ and α, β ∈ C, we have Λ∗x (αf + βg) = (αf + βg)(x) = αf (x) + βg(x) = αΛ∗x (f ) + βΛ∗x (g).

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152

By Definition 2.1, Λ∗x is a linear functional on X ∗ . On the one hand, by [51, Eqn. (3), p. 96], we know that |Λ∗x (f )| = |f (x)| ≤ kf k · kxk ≤ kxk (5.22) for all f ∈ X ∗ with kf k = 1. Thus it means that kΛ∗x k ≤ kxk. On the other hand, since X is a normed linear space, and x 6= 0, Theorem 5.20 ensures the existence of a g ∈ X ∗ such that kgk = 1 and g(x) = kxk, so we have kΛ∗x k ≥ |Λ∗x (g)| = |g(x)| ≥ kxk,

(5.23)

where kgk = 1. Combining the inequalities (5.22) and (5.23), we get the desired result that kΛ∗x k = kxk. (c) Suppose that {xn } ⊆ X and {f (xn )} is bounded for every f ∈ X ∗ . We consider the mapping Λ∗xn : X ∗ → C given by Λ∗xn (f ) = f (xn ). By part (a), X ∗ is a Banach space. By part (b), kΛ∗xn k = kxn k for every n ∈ N. By the hypothesis, for each f ∈ X ∗ , the set {Λ∗xn (f )} = {f (xn )} is bounded by a positive constant Mn , so you can’t find a f ∈ X ∗ such that sup |Λ∗xn (f )| = ∞. n∈N

Therefore, we deduce from Theorem 5.8 (The Banach-Steinhaus Theorem) that kΛ∗xn k ≤ M for all n ∈ N, where M is a positive constant. Hence the sequence {kxn k} is bounded.  This completes the proof of the problem. Problem 5.9 Rudin Chapter 5 Exercise 9.

Proof. (a) We are going to prove the assertions one by one. – kΛk = kyk1 . Fix y = {ηi } ∈ ℓ1 . Define Λ : c0 → C by Λ(x) =

∞ X

ξi ηi ,

(5.24)

i=1

for every x = {ξi } ∈ c0 ⊆ ℓ∞ . Since ξi → 0 as i → ∞, the sequence {ξi } is bounded by a positive constant M (see [49, Theorem 3.2(c), p. 48]). Since y ∈ ℓ1 , we have ∞ ∞ ∞ X X X |ηi | < ∞ |ξi ηi | ≤ M ξi ηi ≤ i=1

i=1

i=1

so that the series (5.24) converges (absolutely) and thus the map Λ is well-defined.

153

5.3. The Dual Space of X Let z = {ωi } ∈ c0 and α, β ∈ C. Since c0 is a vector space, we have αx + βz ∈ c0 . By the definition, we have ∞ X (αξi + βωi )ηi Λ(x + z) =

(5.25)

i=1

By the analysis in the previous paragraph, the series (5.25) is allowed to split into two series so that Λ(x + z) = α

∞ X

ξi ηi + β

∞ X

ωi ηi = αΛ(x) + βΛ(z),

i=1

i=1

i.e., Λ is linear. By the definition (5.24), it is easy to see that |Λ(x)| ≤ kxk∞ · kyk1 for every x ∈ c0 . Thus it is true that kΛk ≤ kyk1 . For the reverse direction, since y ∈ ℓ1 , we know that |ηi | → 0 as i → ∞. Next, for each n ∈ N, we consider the sequence xn = {ξn,i } defined by  ηi   , if i = 1, 2, . . . , n and ηi 6= 0;  |ηi | (5.26) ξn,i =    0, if i > n or η = 0. i

Since | |ηηii | | ≤ 1 for all i ∈ N, we have xn ∈ ℓ∞ for each n ∈ N. In fact, we have kxn k∞ ≤ 1. Furthermore, for every fixed n ∈ N, one can find i ∈ N such that i > n, thus we deduce easily from the definition (5.26) that ξn,i → 0

as i → ∞, i.e., xn ∈ c0 . Now we derive from Definition 5.2 that kΛk = sup{|Λ(x)| | x ∈ c0 and kxk∞ ≤ 1} ≥ |Λ(xn )| =

n X i=1

ξn,i ηi =

n X i=1

|ηi |

for every n ∈ N, i.e., kΛk ≥ kyk1 . In conclusion, we have kΛk = kyk1 < ∞ and thus Λ ∈ (c0 )∗ . – (c0 )∗ is isometric isomorphic to ℓ1 . In the first assertion, we have shown that given a y ∈ ℓ1 , if Λ is defined in the form (5.24), then Λ ∈ (c0 )∗ . More precisely, the mapping Φ : ℓ1 → (c0 )∗ given by Φ(y) = Λy is well-defined, where Λy is in the form (5.24). ∗ Step 1: Φ is linear. For every y = {ηi } ∈ ℓ1 , z = {θi } ∈ ℓ1 and α, β ∈ C, we have αy + βz = {αηi + βθi } ∈ ℓ1 . If x = {ξi } ∈ c0 , then we deduce from the definition (5.24) that Λαy+βz (x) =

∞ X i=1

ξi (αηi + βθi ) = α

∞ X

ξi ηi + β

i=1

Consequently, Φ(αy + βz) = αΦ(y) + βΦ(z).

∞ X i=1

ξi θi = αΛy (x) + βΛz (x).

Chapter 5. Examples of Banach Space Techniques

154

∗ Step 2: Φ is surjective. Given Λ ∈ (c0 )∗ . We want to show that there exists a y ∈ ℓ1 such that Φ(y) = Λ. It is well-known that the sequence {ei }∞ i=1 of standard unit vectors of c0 is a P d In fact, we have basis for c0 and that {ξi } = ξi ei whenever {ξi }∞ i=1 ∈ c0 . ∞ ei = {δi,k }k=1 , where δi,k is the Kronecker delta function. Then it is routine to check that ei ∈ c0 and kei k∞ = 1. Next, we define the sequence y = {ηi }∞ i=1 by ηi = Λ(ei )

(5.27)

for every i = 1, 2, . . .. Our aim is to show that y ∈ ℓ1 . If the sequence xn = {ξn,i }ni=1 is given by the form (5.26), then we immediately know that xn ∈ c0 and kxn k∞ = 1. Since n ∞ X X ξn,i ei , ξn,i ei = xn = i=1

i=1

the linearity of Λ shows that Λ(xn ) =

n X

ξn,i Λ(ei ) =

n X

ξn,i ηi =

i=1

i=1

n X i=1

|ηi |.

(5.28)

Therefore, we apply [51, Eqn. (3), p. 96] to the expression (5.28) to obtain n X i=1

|ηi | = |Λ(xn )| ≤ kΛk · kxn k∞ = kΛk

holds for every n = 1, 2, . . .. In particular, we have kyk1 =

∞ X i=1

|ηi | ≤ kΛk < ∞,

i.e, y ∈ ℓ1 . Finally, for every x = {ξi } ∈ c0 , we have x=

∞ X

ξi ei

i=1

and the application of the expressions (5.27) and the linearity of Λ indicate that ∞ ∞ ∞  X X X ξi ηi ξi Λ(ei ) = ξi ei = Λ(x) = Λ i=1

i=1

i=1

which is exactly in the form (5.24). Hence the map Φ is surjective. ∗ Step 3: Φ is injective. Let y = {ηi } ∈ ℓ1 , z = {θi } ∈ ℓ1 . Suppose further that Φ(y) = Φ(z). Since Φ is linear, we have Λy−z = Φ(y − z) = 0. By this, for every j = 1, 2, . . ., we get Λy−z (ej ) =

∞ X i=1

δj,i (ηi − θi ) = ηj − θj = 0

so that ηj = θj . In other words, we conclude that y = z, as desired. The sequence {ei } is a Schauder basis for c0 (and also for ℓ1 ). In fact, a sequence {xi }∞ i=1 in a Banach space X is called a Schauder basis if for each x ∈ X, there is a unique sequence {αi } of scalars such that ∞ n X X x= αi xi = lim αi xi . See, for instance, [40, Example 4.1.3, p. 351]. d

i=1

n→∞

i=1

155

5.3. The Dual Space of X ∗ Step 4: Φ is an isometry. This follows from the first assertion directly because kΦ(y)k = kΛy k = kyk1 . Hence we have (c0 )∗ = ℓ1 .

(b) For each y = {ηi } ∈ ℓ∞ , we define Θ : ℓ1 → C by Θ(x) =

∞ X

ξi ηi

(5.29)

i=1

for every x = {ξi } ∈ ℓ1 . Similar to the proof of the first assertion in part (a), we can show that the series (5.29) converges (absolutely) and Θ is linear. It is clear from the definition that |Θ(x)| ≤ kyk∞ · kxk1 , so we have kΘk ≤ kyk∞ < ∞.

(5.30)

This means that Θ ∈ (ℓ1 )∗ .

On the other hand, suppose that Θ ∈ (ℓ1 )∗ , i.e., kΘk < ∞. If we consider the Schader basis {ei } for ℓ1 and the relationships (5.27) with Λ replaced by Θ, then we have |ηi | = |Θ(ei )| ≤ kΘk · kei k1 = kΘk < ∞

(5.31)

for every i = 1, 2, . . .. Put z = {ηi }. Then the inequality (5.31) immediately implies that z ∈ ℓ∞ . In fact, it shows further that kzk∞ ≤ kΘk. Now for every x = {ξi } ∈ ℓ1 , we have x=

∞ X

ξi ei

i=1

and then Θ(x) =

∞ X

ξi ηi

i=1

which is exactly in the form (5.29), so the inequality (5.30) holds for z. Therefore, we have kΘk = kzk∞ , i.e., Θ ∈ ℓ∞ . If we define the mapping Ψ : ℓ∞ → (ℓ1 )∗ by

Ψ(y) = Θy , where Θy is given by the form (5.29), then the previous paragraph actually shows that Ψ is surjective. By using similar argument as in the proofs of Step 1, Step 3 and Step 4 in part (a), it is easy to conclude that Ψ is an isometric vector space isomorphism. (c) Let y = {ηi } ∈ ℓ1 . Define the functional Λ : ℓ∞ → C by Λ(x) =

∞ X

ξi ηi ,

i=1

where x = {ξi } ∈ ℓ∞ . Since this series converges (absolutely), it is a linear functional. Since |Λ(x)| ≤ kxk∞ · kyk1 , we know from [51, Eqn. (2), p. 96] that kΛk ≤ kyk1 , i.e., Λ ∈ (ℓ∞ )∗ .

Assume that ℓ1 was isometric isomorphic to (ℓ∞ )∗ . It is clear that x = {1, 1, . . .} ∈ ℓ∞ \ c0 . By [40, Example 1.2.13, p. 14], we know that c0 is a proper closed subspace of

Chapter 5. Examples of Banach Space Techniques

156

ℓ∞ so that ℓ∞ \ c0 6= ∅. Take x0 ∈ ℓ∞ \ c0 . Then Theorem 5.19 implies that there is a f ∈ (ℓ∞ )∗ such that kf k = 1, f (x) = 0 on c0 and f (x0 ) 6= 0.

(5.32)

If f ∈ ℓ1 , then we have f ∈ (c0 )∗ by part (a). Since f (x) = 0 on c0 , f is in fact the trivial bounded functional on c0 . By the assumption and part (a), (c0 )∗ is isometric isomorphic to (ℓ∞ )∗ so that the trivial bounded functional on c0 corresponds to the trivial bounded functional on ℓ∞ . In other words, f (x) = 0 on ℓ∞ which contradicts the result (5.32). Hence we conclude that ℓ1 is not isometric isomorphic to (ℓ∞ )∗ . (d) We prove the assertions one by one. – c0 is separable. For each k ∈ N, we define Sk = {{ξ1 , . . . , ξk , 0, 0, . . .} | ξ1 , . . . , ξk ∈ Q} and S =

∞ [

k=1

Sk .

Since each Sk is countable, S is also countable. We claim that S is dense in c0 . Pick y = {ηi } ∈ c0 . Given ǫ > 0. Since ηi → 0 as i → ∞, there exists a positive integer N such that i ≥ N implies |ηi | < ǫ. (5.33) By the density of Q in R, we can take x = {ξ1 , . . . , ξN , 0, 0, . . .} ∈ SN such that |ξi − ηi | < ǫ

(5.34)

for all i = 1, 2, . . . , N . Thus we obtain from the definition of k · k∞ , the inequalities (5.33) and (5.34) that kx − yk∞ = sup |ξi − ηi | ≤ ǫ. i∈N

Hence this proves the claim and then c0 is separable. –

ℓ1

is separable. Take y = {ηi } ∈

ℓ1 .

Given that ǫ > 0. Since

∞ X i=1

|ηi | < ∞, the

Cauchy Criterion (see [49, Theorem 3.22, p. 59]) shows that one can find a positive integer N ′ such that i ≥ N ′ implies ∞ X

i=N ′

|ηi |
0 be so small such that 4n (x + h) ∈ [0, 1]. Then we obtain ϕn (x + h) − ϕn (x) h
1 ϕ 4n (x + h) − ϕ(4n x) = n· 2 h  n (x + h) − 2 · 4n x 2 · 4 1   n· , if 4n x, 4n (x + h) ∈ [0, 12 ];   2 h      1 1 − 2 · 4n h − 1 = · , if 4n x = 12 and 4n (x + h) = n  2 h      n n    1 · −2 · 4 (x + h) + 2 · 4 x , 4n x, 4n (x + h) ∈ [ 1 , 1], 2 n h  2 n 1 n n 2·2 , if 4 x, 4 (x + h) ∈ [0, 2 ];      −2 · 2n , if 4n x = 12 and 4n (x + h) = 21 + 4n h; =      −2 · 2n , 4n x, 4n (x + h) ∈ [ 12 , 1]

1 2

+ 4n h;

which implies that |ϕ′n (x+)| ≥ 2n . Similarly, |ϕ′n (x−)| ≥ 2n , completing the proof of  Lemma 5.1. • Step 3: Construction of a e g ∈ C with large slopes and ke g − f k < 2ǫ. We define ge = g + ϕk ,

where the k will be determined later. We claim that gb satisfies the required properties. To this end, let Mk = max k|f (xi+1 ) − f (xi )|. 0≤i≤k−1

Now for every x ∈ [0, 1], if we take h > 0 small enough such that x + h ∈ [xi , xi+1 ], then we see from the definition (5.65) and the expression (5.66) that g(x + h) − g(x) |λ x+h − λx | · |f (xi+1 ) − f (xi )| = k|f (xi+1 ) − f (xi )| ≤ Mk = h h

(5.68)

which implies that |g′ (x+)| ≤ Mk on [0, 1]. Similarly, we have |g′ (x−)| ≤ Mk for every x ∈ [0, 1].f f

If x = 1, then g ′ (1+) does not exist. Similarly, if x = 0, then g ′ (0−) does not exist.

167

5.4. Applications of Baire’s and other Theorems Suppose that N ′ ∈ N. Since f is bounded on [0, 1], we must have Mk ≤ kM for some constant M > 0 and then k can be chosen so large that 2k ≥ kM ≥ Mk + N ′

and

ǫ 1 < . k 2 2

(5.69)

Since g and ϕk are continuous on [0, 1], ge is also continuous on [0, 1]. Next, for x ∈ [0, 1) and h > 0 so small such that x + h ∈ [0, 1], we derive from Lemma 5.1 and the inequality (5.68) that ge(x + h) − ge(x) g(x + h) − g(x) ϕ(x + h) − ϕ(x) + = h h h ϕ(x + h) − ϕ(x) g(x + h) − g(x) ≥ − h h ≥ 2k − Mk > N′

(5.70)

which yields |e g ′ (x+)| > N ′ for all x ∈ [0, 1). Similarly, we have |e g′ (x−)| > N ′ for all x(0, 1]. Since N ′ can be chosen arbitrary large, our e g is continuous function on [0, 1] with large slopes.

Finally, it remains to show that ke g − f k < ǫ. On [0, 1], we know from Steps 1 and 2 that |e g(x) − f (x)| ≤ |g(x) − f (x)| + |ϕk (x)| < ǫ + 2−k |ϕ(4k x)| < ǫ +

ǫ < 2ǫ. 2

Thus we conclude that ke g − f k < 2ǫ. By the definition of Xn and the lower bound (5.70), we note that ge ∈ / Xn if we pick N ′ ≥ n.

• Step 4: Xn is closed in C. Let {fk } ⊆ Xn and fk → f in C. Then one can find a sequence {tk } ⊆ [0, 1] such that |fk (s) − fk (tk )| ≤ n|s − tk |

(5.71)

for every s ∈ [0, 1] and k ∈ N. Since [0, 1] is compact, we may assume that tk → t ∈ [0, 1]. By Theorem 3.17, C is a metric space and then we recall from the rephrased Theorem 7.9 [49, p. 151] that fk → f uniformly on [0, 1]. Hence it follows from [49, Exercise 9, p. 166] that lim fk (tk ) = f (t) k→∞

and we deduce from the inequality (5.71) that |f (s) − f (t)| ≤ n|s − t|, for every s ∈ [0, 1]. In other words, we conclude that f ∈ Xn which means Xn is closed in C. • Step 5: Construction of B(e g , ǫ′ ) ⊆ C such that B(e g, ǫ′ ) ∩ Xn = ∅. By Step 3 and our choice of ǫ > 0, we know that e g ∈ B(f, 2ǫ) ⊆ V and ge ∈ / Xn . Since ge ∈ Xnc and Xnc is open in C by Step 4. There exists a δ > 0 such that B(e g, δ) ⊆ Xnc , i.e., B(e g, δ) ∩ Xn = ∅. 1 ′ g − f k, 2ǫ − ke g − f k) and h ∈ B(e g , ǫ′ ), then we have If we take ǫ = 2 min(ke kh − f k ≤ kh − gek + ke g − f k < ǫ′ + ke g − fk ≤

which means B(e g, ǫ′ ) ⊆ B(f, 2ǫ) ⊆ V .

1 (2ǫ − ke g − f k) + ke g − f k < 2ǫ 2

Chapter 5. Examples of Banach Space Techniques

168

Hence we have shown the first assertion. To prove the second assertion, we notice that Xn◦ = ∅; otherwise, since Xn◦ is open in C, the first assertion shows the existence of a non-empty open set G ⊆ Xn◦ ⊆ Xn such that G ∩ Xn = ∅, a contradiction. Hence each Xnc is open and dense in C. Since C is a complete metric space, we establish from Theorem 5.6 (Baire’s Theorem) that the set X=

∞ \

Xnc

n=1

is a dense Gδ in C. Suppose that f ∈ C and f is differentiable at p ∈ [0, 1). Then there exists a δ > 0 such that for all 0 < |h| < δ, we haveg f (p + h) − f (p) ′ − f (p) ≤1 h

which implies that

Obviously, if |h| ≥ δ, then

f (p + h) − f (p) ≤ 1 + |f ′ (p)|. h

f (p + h) − f (p) |f (p + h)| + |f (p) 2kf k ≤ . ≤ h h δ

(5.72)

(5.73)

Combining the inequalities (5.72) and (5.73), we obtain that |f (x) − f (p)| ≤ N |x − p|

if N = max(1 + |f ′ (p)|, 2δ−1 kf k). In the case that p = 1, we will replace f (p + h) by f (p − h) in the inequalities (5.72) and (5.73). c Consequently, f ∈ Xn for every n ≥ N . By this, we must have f ∈ / X. Otherwise, f ∈ XN or equivalently, f ∈ / XN which is a contradiction. Hence we have completed the proof of the problem. 

Problem 5.15 Rudin Chapter 5 Exercise 15.

Proof. Recall from Problem 5.9 that c0 is the subspace of ℓ∞ consisting of all x = {ξi } ∈ ℓ∞ for which ξi → 0 as i → ∞. By the definition, we have {σi } = A(x) whenever the series converges (but ξi not necessarily converging to 0). We are going to prove the results one by one: • Necessity part: Suppose that A transforms every convergent sequence {sj } to a sequence {σi } which converges to the same limit. To prove Condition (a), it suffices to prove the case when both {sj } and {σi } converge to 0 because we may replace sj and σi by sj − L and σi − L respectively, where L is the g

Of course, only one-sided derivative f ′ (0+) will be considered if p = 0.

169

5.4. Applications of Baire’s and other Theorems common limit of {sj } and {σi }. In other words, we work on the space c0 . In fact, for each fixed i ∈ N, we define Λi : c0 → C by Λi (x) = σi =

∞ X

aij sj

j=0

for all x ∈ c0 so that A(x) = {Λi (x)}. Since {σi } is convergent, it is bounded so that |Λi (x)| ≤ M

(5.74)

for all x ∈ c0 and i = 1, 2, . . . for some positive constant M . For every k = 0, 1, 2, . . ., we take sj = δjk (the Kronecker delta function). Then we have sj → 0 as j → ∞ and we obtain ∞ X 0 = lim σi = lim aij sj = lim aik i→∞

i→∞

i→∞

j=0

which is Condition (a). If sj = 1 for all j ∈ N, then sj → 1 as j → ∞ and thus 1 = lim σi = lim i→∞

i→∞

∞ X

aij

j=0

which is Condition (c). To prove Condition (b), we first show that

∞ X j=0

|aij | < ∞ for every i = 1, 2, . . .. Assume

that it was not the case. Then one is able to find a strictly increasing sequence of integers {nk } such that nk+1 X |aij | > k. j=nk +1

Let k ∈ N and define

sj =

 0,  

if j = 1, 2, . . . , nk ;

  sgn (aij ) , if j = n + 1, n + 2, . . . , n . k k k+1 k

(5.75)

Obviously, j → ∞ if and only if k → ∞. Then we have sj → 0 as j → ∞. Thus x = {sj } ∈ c0 and so lim σi = 0. However, we notice from the construction (5.75) that i→∞

∞ nX ∞ k+1 X X |aij | aij sj = |Λi (x)| = |σi | = =∞ k j=0

k=1 j=nk +1

which contradicts the inequality (5.74). Thus we must have ∞ X j=0

|aij | < ∞.

By the inequality (5.74) again, we have Λi ∈ c∗0 and since c0 is Banach, we may apply Theorem 5.8 (The Banach-Steinhaus Theorem) to conclude that kΛi k = sup{|Λi (x)| | x ∈ c0 ⊆ ℓ∞ and kxk∞ = 1} ≤ M

(5.76)

Chapter 5. Examples of Banach Space Techniques

for all i ∈ N. Recall that

∞ X j=0

170

|aij | < ∞ for every i = 1, 2, . . ., so yi = {aij } ∈ ℓ1 for every

i = 1, 2, . . . and Problem 5.9(a) implies that kΛi k = kyi k1 =

∞ X j=0

|aij |

(5.77)

for every i = 1, 2, . . .. Hence Condition (b) follows from the inequality (5.76) and the equality (5.77). • Sufficiency part: Let x = {sj } and sj → L as j → ∞. By Condition (b), the series ∞ X aij sj converges absolutely for every i ∈ N. We write σi = j=0

σi =

∞ X j=0

aij (sj − L) + L ·

By Condition (c), we have lim L ·

i→∞

∞ X

∞ X

aij .

(5.78)

j=0

aij = L.

(5.79)

j=0

Let the supremum in Condition (b) be M . Given ǫ > 0, we choose a N ∈ N such that |sj − L| < Mǫ for all n ≥ N . Thus we see that N ∞ N ∞ X X X X |aij | · |sj − L| + ǫ. |aij | · |sj − L| ≤ |aij | · |sj − L| + aij (sj − L) ≤ j=0

n=N +1

j=0

j=0

By Condition (a), we know that

lim

i→∞

so that

N X j=0

|aij | · |sj − L| = 0

∞ X aij (sj − L) ≤ ǫ. lim

i→∞

j=0

Since ǫ is arbitrary, we conclude from the representation (5.78) and the limit (5.79) that lim σi = L.

i→∞

• Two examples. It is easy to check that the examples satisfy the conditions of the problem. To show the last assertion, we consider the sequence {sj } defined by  √ if j = 2k;  k, sj =  √ − k, if j = 2k − 1.

Then it is clear that {sj } is unbounded and direct computation shows that, for each i = 1, 2, . . ., σi =

∞ X j=0

aij sj

171

5.5. Miscellaneous Problems

=

i X j=0

=

=

1 sj i+1

 2k  1 X    sj , if i = 2k;     2k + 1 j=0

  2k−1  1 X    sj , if i = 2k − 1;   2k j=0   0, if i = 2k;   1    − √ , if i = 2k − 1. 2 k

Since i → ∞ if and only if k → ∞, we have σi → 0 as i → ∞, i.e., {σi } is convergent.

For the other example, pick δ to be a number such that ri < δ < 1 for every i ∈ N. If sj = (−δ)−j , then {sj } is divergent and we have σi =

∞ X j=0

aij sj = (1 − ri )

∞  X j=0

ri j 1 − ri − = →0 δ 1 + δ−1 ri

as i → ∞. We complete the proof of the problem.



Remark 5.2 Classically, Problem 5.15 is called the Silverman-Toepliotz Theorem. For more information or other proofs about this theorem, please refer to [30, §3.2], [44, Chap. 4] and [68, §1.2].

5.5

Miscellaneous Problems

Problem 5.16 Rudin Chapter 5 Exercise 16.

Proof. We follow Rudin’s hint. Let X ⊕Y = {(x, y) | x ∈ X and y ∈ Y } with addition and scalar multiplication defined componentwise. We define the norm k · k on X ⊕ Y byh k(x, y)k = kxkX + kykY . We check Definition 5.2. For (x1 , y1 ), (x2 , y2 ) ∈ X ⊕ Y , we have k(x1 , y1 ) + (x2 , y2 )k = k(x1 + x2 , y1 + y2 )k

= kx1 + x2 kX + ky1 + y2 kY

≤ kx1 kX + kx2 kX + ky1 kY + ky2 kY = k(x1 , y1 )k + k(x2 , y2 )k.

h

Of course, k · kX and k · kY denote the norms in X and Y respectively.

(5.80)

Chapter 5. Examples of Banach Space Techniques

172

Next, if (x, y) ∈ X ⊕ Y and α is a scalar, then we have kα(x, y)k = k(αx, αy)k = kαxkX + kαykY = |α| · kxkX + |α| · kykY = |α| · k(x, y)k. Finally, if k(x, y)k = 0, then kxkX + kykY = 0. Since kxkX and kykY are nonnegative for every x ∈ X and y ∈ Y , kxkX + kykY = 0 implies that kxkX = kykY = 0 and thus (x, y) = (0, 0). Hence X ⊕ Y is a normed linear space.

Suppose that {(xn , yn )} ⊆ X ⊕ Y is Cauchy. Given ǫ > 0, Then there exists a positive integer N such that n, m ≥ N imply that k(xn , yn ) − (xm , ym )k < ǫ.

By the definition (5.80), we must have kxn − xm kX < ǫ and kyn − ym kY < ǫ for all n, m ≥ N . In other words, {xn } ⊆ X and {yn } ⊆ Y are also Cauchy. Since X and Y are Banach, there exist x ∈ X and y ∈ Y such that kxn − xkX → 0

and kyn − ykY → 0

as n → ∞. Now these limits and the definition (5.80) show that k(xn , yn ) − (x, y)k = k(xn − x, yn − y)k = kxn − xkX + kyn − ykY → 0 as n → ∞. By Definition 5.2, X ⊕ Y ia Banach.
Suppose that G = { x, Λ(x) | x ∈ X} ⊆ X ⊕ Y . Furthermore, we let α, β ∈ C and

x, Λ(x) , y, Λ(y) ∈ G. Then the linearity of Λ says that




α x, Λ(x) + β y, Λ(y) = αx + βy, αΛ(x) + βΛ(y) = αx + βy, Λ αx + βy ∈ G.

Thus G is a linear subspace of X ⊕ Y
and it is also a metric
space. We claim that G is closed in X ⊕ Y . To see this, let { xn , Λ(xn ) } ⊆ G and k xn , Λ(xn ) − (x, y)k → 0 as n → ∞. By the definition (5.80), we have kxn − xkX → 0 and kΛ(xn ) − ykY → 0 as n → ∞, i.e., x = lim xn n→∞

and y = lim Λ(xn ). n→∞


By the hypothesis, we gain that y = Λ(x) and so (x, y) = x, Λ(x) ∈ G by the definition. Therefore, we have shown the claim that G is closed in X ⊕ Y . Since X ⊕ Y is complete, we deduce from [49, Theorem 3.11, p. 53] that G is also complete and thus Banach. Consider the mapping Φ : G → X defined by
Φ x, Λ(x) = x. Then it is clear that Φ is linear and bijective. Furthermore, for every x ∈ X, we see that

kΦ x, Λ(x) kX = kxkX ≤ kxkX + kΛ(x)kY = k x, Λ(x) k which implies that




kΦk = sup{kΦ x, Λ(x) kX | x, Λ(x) ∈ G and k x, Λ(x) k = 1} ≤ 1,

i.e., Φ is bounded. By Theorem 5.10, Φ−1 is also a bounded linear transformation of X onto G and it follows from Theorem 5.10’s proof that there exists a positive constant δ such that 1 kΦ−1 k ≤ . δ

173

5.5. Miscellaneous Problems

Besides, Theorem 5.4 shows that Φ−1 is continuous. Combining these two facts and [51, Eqn. (3), p. 96], we obtain
1 kxkX + kΛ(x)kY = k x, Λ(x) k = kΦ−1 (x)k ≤ kΦ−1 k · kxkX ≤ kxkX < ∞ (5.81) δ for every x ∈ X. Since kxkX and kΛ(x)kY are nonnegative, we get from the inequality (5.81) that δ < 1 and 1  0 ≤ kΛ(x)kY ≤ − 1 · kxkX (5.82) δ on X. Hence we conclude from the inequality (5.82) that 1 < ∞ − 1. δ Consequently, Λ is bounded and it is continuous on X by Theorem 5.4, completing the proof of the problem.  kΛk = sup{kΛ(x)kY | x ∈ X and kxkX = 1} ≤

Problem 5.17 Rudin Chapter 5 Exercise 17.

Proof. We prove the assertions one by one. • kMf k ≤ kf k∞ . Given f ∈ L∞ (µ). Define Mf : L2 (µ) → L2 (µ) by Mf (g) = f g. By Definition 3.7, we have |f g| ≤ kf k∞ · |g| for all f ∈ L2 (µ) and then Remark 3.10 shows that o1 nZ o1 n Z 2 2 2 ≤ kf k2∞ · |g|2 dµ = kf k∞ · kgk2 . (5.83) kMf (g)k2 = kf gk2 = |f g| dµ X

X

By the inequality (5.83) and Definition 5.3, we obtain immediately that

kMf k = sup{kMf (g)k2 | g ∈ L2 (µ) and kgk2 = 1} ≤ kf k∞ .

(5.84)

• Measures µ with kMf k = kf k∞ for all f ∈ L∞ (µ). We call the measure µ semifinite if for each E ∈ M with µ(E) = ∞ one can find a F ∈ M with F ⊂ E such that 0 < µ(F ) < ∞, see [22, p. 25]. Now we claim that kMf k = kf k∞ for all f ∈ L∞ (µ) if and only if the measure µ is semifinite.

– Suppose that µ is semifinite. Since f ∈ L∞ (µ), we have kf k∞ < ∞. Let α = kf k∞ . If α = 0, then the inequality (5.84) forces that kMf k = kf k∞ = 0 and we are done. If α > 0, then we see from Problem 3.19 that α = max{|z| | z ∈ Rf }, where Rf denotes the essential range of f . Thus there exists a z0 ∈ C such that α = |z0 |. Without loss of generality, we may assume that z0 = α and so µ{x ∈ X | |f (x) − α| < ǫ} > 0

(5.85)

for every ǫ > 0. Combining Definition 3.7 and the result (5.85), we establish that the measure of the set E = {x ∈ R | |f (x)| > α − ǫ} is nonzero.

Chapter 5. Examples of Banach Space Techniques

174

∗ Case (i): µ(E) < ∞. Now it is trivial that the function g=p

1 µ(E)

χE

satisfies the conditions that g ∈ L2 (µ) and kgk2 = 1. Furthermore, we derive from the definition that o 1 n (α − ǫ)2 Z o1 n 1 Z 2 2 2 |f χE | dµ dµ > = α − ǫ. (5.86) kMf (g)k2 = µ(E) E µ(E) E Since ǫ is arbitrary, the estimate (5.86) implies that kMf k = sup{kMf (g)k2 | g ∈ L2 (µ) and kgk2 = 1} ≥ α and hence kMf k = kf k∞ by the inequality (5.84). ∗ Case (ii): µ(E) = ∞. Then the hypothesis ensures that there exists a F ∈ M with F ⊂ E and 0 < µ(F ) < ∞. Now the function 1 g = √ χF F also satisfies the estimate (5.86) and thus kMf k = kf k∞ holds in this case.

– Suppose that µ is not semifinite. Then there is a E ∈ M such that µ(E) = ∞ and every F ⊂ E satisfies either µ(F ) = 0 or µ(F ) = ∞. Take g ∈ L2 (µ) so that Z Z 2 |g(x)|2 dµ < ∞. |g(x)| dµ ≤ E

X

If there is a F ⊂ E such that |g(x)| > 0 on F and g(x) = 0 on E \ F , then µ(F ) 6= ∞ by Proposition 1.24(a). Recall that µ is not semifinite, we have µ(F ) = 0. In other words, it must be true that g = 0 a.e. on E. Let f = χE . Then we have f g = 0 a.e. on E for every g ∈ L2 (µ). By the definition, we obtain kMf k = 0. On the other hand, kf k∞ = 1 so that kMf k < kf k∞ . Hence we have proven the claim. • Functions f ∈ L∞ (µ) such that Mf is onto. Suppose that µ satisfies the condition that every measurable set E of positive measure contains a measurable subset F with 0 < µ(F ) < ∞.i We claim that the map Mf is onto if and only if f1 ∈ L∞ (µ). – If Mf is onto, then we let E = {x ∈ X | f (x) = 0}.

i

(5.87)

We claim that µ(E) = 0. Assume that µ(E) > 0. By the hypothesis, there exists a F ⊆ E with 0 < µ(F ) < ∞. Thus χF ∈ L2 (µ). Since f g = 0 on F for every g ∈ L2 (µ), it implies that
χF ∈ / Mf L2 (µ)

Some books take this as the definition of a semifinite measure µ. See, for example, [7, Exercise 25.9]

175

5.5. Miscellaneous Problems which contradicts the surjective property of Mf . Thus we conclude that µ(E) = 0. If g ∈ L2 (µ) is such that Mf (g) = f g = 0, then it follows from the definition (5.87) that g = 0 a.e. on E c . Since µ(E) = 0, we obtain fact that g = 0 a.e. on X and this means that Mf is one-to-one. Since Mf is assumed to be onto, it is bijective. On the one hand, recall that L2 (µ) is Banach and kMf k ≤ kf k∞ < ∞ by the first assertion, Theorem 5.10 ensures that there corresponds a δ > 0 such that kMf (g)k2 ≥ δkgk2

(5.88)

for every g ∈ L2 (µ). On the other hand, we consider F = {x ∈ X | |f (x)| < 2δ }. If µ(F ) 6= 0, then our hypothesis tells us that there is a G ⊂ F such that 0 < µ(G) < ∞ and thus 1 kχG k2 = [µ(G)] 2 which verifies that kMf (χG )k2 = kf χG k2 =

nZ

G

|f |2 dµ

o1

2


0 a.e. on X if and only if |f (x)|

– Suppose that f1 ∈ L∞ (µ). Then it is clear that M 1 is the inverse operator of Mf so f that Mf is bijective. In particular, Mf is surjective. Hence we have completed the proof of the problem.



Problem 5.18 Rudin Chapter 5 Exercise 18.

Proof. Let x ∈ X and ǫ > 0. Since E is dense in the normed linear space X, we can find y ∈ E such that ǫ . (5.89) kx − yk < 2M Since {Λn (y)} converges in the Banach space Y , there exists a positive integer N such that n, m ≥ N imply that ǫ kΛn (y) − Λm (y)k < . (5.90) 2 Therefore, if n, m ≥ N , then we deduce immediately from the inequalities (5.89) and (5.90) that kΛn (x) − Λm (x)k ≤ kΛn (x) − Λn (y)k + kΛn (y) − Λm (y)k + kΛm (y) − Λm (x)k ǫ < kΛn (x − y)k + + kΛm (y − x)k 2 ǫ ≤ kΛn k · kx − yk + kΛm k · kx − yk + 2 ǫ ǫ < + 2 2 = ǫ. This means that {Λn (x)} is Cauchy in Y . Since Y is Banach, {Λn (x)} converges in Y and we complete the analysis of the problem. 

Chapter 5. Examples of Banach Space Techniques

176

Problem 5.19 Rudin Chapter 5 Exercise 19.

Proof. Given f ∈ C(T ), i.e., f is a continuous complex function on T . Recall that 1 sn (f ; x) = 2π

Z

π

−π

f (t)Dn (x − t) dt

and Dn (t) =

n X

k=−n

eikt =

sin(n + 21 )t . sin 2t

(5.91)

By §5.11, C(T ) is Banach relative to the supremum norm kf k∞ .j Next, since f ∈ C(T ), we follow from the equations (5.91) that sn (f ; x) ∈ C(T ) for each n ∈ N. Let X = Y = C(T ) and for each n = 2, 3, . . ., we define Λn : C(T ) → C(T ) by Λn (f ) =

sn (f ; x) . log n

Since f ∈ C(T ), there exists a M > 0 such that |f (x)| ≤ M for all x ∈ T so that Z π Z 1 M π |sn (f ; x)| ≤ |f (t)| · |Dn (x − t)| dt ≤ |Dn (x − t)| dt. 2π −π 2π −π

(5.92)

Since Dn (x − t) = eikx Dn (t) and Dn (t) is an even function, the estimate (5.92) can be replaced by Z Z Z M π M π sin(n + 21 )t M π |Dn (t)| dt = |Dn (t)| dt = (5.93) |sn (f ; x)| ≤ dt. 2π −π π 0 π 0 sin 2t Next, we consider f (x) = sin x2 − x4 on [0, π]. Using differentiation, we can show that sin x2 ≥ x4 on [0, π]. Thus the estimate (5.93) can be further written as Z 4M π | sin(n + 12 )t| |sn (f ; x)| ≤ dt π 0 t 1 Z 4M (n+ 2 )π | sin t| dt = π 0 t Z n Z (k+1)π X 4M  π sin t | sin t|  (5.94) ≤ dt + dt . π t t 0 kπ k=1 1 t

On the interval [kπ, (k + 1)π], it is easy to see that n Z X k=1

(k+1)π

kπ

X 1  | sin t| dt ≤ t kπ n

k=1

Z

0

is bounded by Z j

π 0

π 2

sin t dt = t

π 2

1 kπ ,

(k+1)π kπ 2 π

Besides, we get from [49, Problem 8.6, p. 197] that Z

≤

where k = 1, 2, . . . , n, so we have

n  2X 1 | sin t| dt = . π k

(5.95)

k=1


0 such that kΛn k ≤ M ′ for all n = 2, 3, . . .. That is, {Λn } satisfies the first hypothesis of Problem 5.18.

By Theorem 4.25 (The Weierstrass Approximation Theorem), the set of all trigonometric polynomials, namely P, is dense in C(T ). Let Pm (t) = eimt for some m ∈ Z. If n ≥ m, then we follow from the result [51, Eqn. (8), p. 89] that 1 sn (Pm ; x) = 2π

Therefore, if P (t) =

N X

Z

π

−imt

e −π

n X

ikt

e

Z π n X 1 dt = ei(k−m)t dt = 1. 2π −π

(5.98)

k=−n

k=−n

cm Pm (t), then we get immediately from the result (5.98) that

m=−N

sn (P ; x) =

N X

cm sn (Pm ; x) =

N X

cm

m=−N

m=−N

for every n ≥ N and this implies that N X 1 sn (P ; x) cm → 0 = · Λn (P ) = log n log n m=−N

as n → ∞. Thus {Λn } satisfies the second hypothesis of Problem 5.18. Hence we establish from Problem 5.18 that {Λn (f )} converges to 0 for each f ∈ C(T ) and as a consequence, we have lim

n→∞

ksn k∞ = lim kΛn k = 0. n→∞ log n

For the second assertion, we note that if we define Λn f =

sn (f ; 0) , λn

then we may apply an argument similar to that used in §5.11 to show that kΛn k =

kDn k1 |λn |

holds. Furthermore, we follow from the hypothesis

λn log n

→ 0 as n → ∞ thatk

4 X1 4 4  log n γ  kDn k1 →∞ ≥ 2 ≥ 2 (log n + γ) = 2 + |λn | π |λn | k π |λn | π |λn | |λn | n

k=1

k

We have applied the estimate of kDn k1 used in [51, p. 102] in the first inequality in (5.99).

(5.99)

Chapter 5. Examples of Banach Space Techniques

178

as n → ∞, where γ is the famous Euler constant. Hence Theorem 5.8 (the Banach-Steinhaus Theorem) ensures that the sequence n s (f ; 0) o n λn

is unbounded for every f in some dense Gδ set in C(T ), completing the proof of the problem.  Problem 5.20 Rudin Chapter 5 Exercise 20.

Proof. (a) Assume that such a sequence of continuous positive functions {fn } existed. Let x ∈ R and n, k ∈ N. Furthermore, let U = {x ∈ R | {fn (x)} is unbounded} and We claim that U=

Uk = {x ∈ R | fn (x) > k for some n ∈ N}. ∞ \

Uk .

(5.100)

k=1

On the one hand, if x ∈ Uk for all k ∈ N, then for every k ∈ N, there exists a n ∈ N such that fn (x) > k. In other words, we have x ∈ U . On the other hand, if x ∈ U , then since {fn (x)} is unbounded, for every k ∈ N, there exists a n ∈ N such that fn (x) > k which is equivalent to saying that x ∈ Uk for all k ∈ N. Thus this proves the claim. Next, it is easy to see that ∞ [ {x ∈ R | fn (x) > k}. Uk = n=1

Since each fn is continuous, the set {x ∈ R | fn (x) > k} is open in R and therefore, each Uk is open in R. By the definition (5.100), U is a Gδ set. Thus we follow from the assumption that Q=U is also a Gδ set. We deduce from the definition (5.100) that R = Q ∪ Qc = Q ∪

∞ [

k=1

Ukc =

∞ [

{qk } ∪

k=1

∞ [

Ukc

(5.101)

k=1

Clearly, each qk is a nowhere dense subset of R. Furthermore, it is trivial that each Ukc is closed in R. Assume that Ukc0 was not a nowhere dense subset of R for some k0 . Let Vk0 be a nonempty open subset of Ukc0 . Then there exists a p ∈ Vk0 and a δ > 0 such that (p − δ, p + δ) ⊆ Vk0 ⊆ Ukc0 ⊆ Qc , but this means that Q ∩ Qc 6= ∅, a contradiction. Thus every Ukc is also a nowhere dense subset of R and the representation (5.101) shows that R is a set of the first category which contradicts Theorem 5.6 (Baire’s Theorem) that no complete metric space is of the first category. (b) Suppose that Q = {q1 , q2 , . . .} and for each n ∈ N, we define fn : R → R by fn (x) = min(n|x − q1 | + 1, n|x − q2 | + 2, . . . , n|x − qn | + n).

(5.102)

179

5.5. Miscellaneous Problems It is clear that fn is continuous on R and fn (x) ≥ 1 for all x ∈ R, i.e., {fn } is a sequence of continuous positive functions on R. Suppose that θ is irrational and N ∈ N. Consider the number α = min(|θ − q1 |, |θ − q2 |, . . . , |θ − qN |) > 0. Then the Archimedean Property ([49, Theorem 1.20(a)]) implies that there is a positive integer n such that nα > N and thus fn (θ) = min(n|θ − q1 | + 1, n|θ − q2 | + 2, . . . , n|θ − qn | + n) > N + 1. Since N is arbitrary, we have fn (θ) → ∞ as n → ∞ so that {fn (θ)} is unbounded.

To prove the other direction (i.e., {fn (x)} is unbounded implies x is irrational), we prove its contrapositive. Let qk be a rational number. If n ≥ k, then we have fn (qk ) = min(n|qk − q1 | + 1, . . . , n|qk − qk | + k, . . . , n|qk − qn | + n) ≤ k
so that fn (qk ) ≤ min f1 (qk ), . . . , fk−1 (qk ), k . In other words, the set {fn (qk )} is bounded.

(c) The sequence of the functions given by (5.102) shows that the assertion is true for irrationals. For the rational numbers, we first prove the case on [0, 1].l To begin with, suppose that Qn = {q1 , q2 , . . . , qn } ⊆ [0, 1] ∩ Q, where q1 = 0. Furthermore, we let δn =

1 2n+1

min{|qi − qj | | 1 ≤ i < j ≤ n} > 0.

Clearly, we have lim δn = 0 and

n→∞

(qi − δn , qi + δn ) ∩ (qj − δn , qj + δn ) = ∅

(5.103)

for all 1 ≤ i < j ≤ n. (If qi = 0 or qi = 1, then (qi − δn , qi + δn ) are replaced by [0, δn ) or (1 − δn , 1] respectively.) n [ (qi − δn , qi + δn ) and fn : [0, 1] → R is defined by Suppose that En = i=1

 n  (x − qi + δn ), if x ∈ (qi − δn , qi ] and 1 ≤ i ≤ n;   δn     n fn (x) = − (x − qi − δn ), if x ∈ [qi , qi + δn ) and 1 ≤ i ≤ n;  δn       0, if x ∈ [0, 1] \ En .

Thus fn is a continuous function on [0, 1], zig-zag on (qi − δn , qi + δn ) and fn (qi ) = n for each i = 1, 2, . . . , n.m Therefore, if x ∈ [0, 1] ∩ Q, then x = qk for some k ∈ N and thus we obtain lim fn (x) = ∞. n→∞

Next, suppose that x is irrational. Assume that there was an N ∈ N such that x ∈ En for all n ≥ N . By the definition, it means that x ∈ (qk − δN , qk + δN ) l m

The following argument is stimulated by the papers of Fabrykowski [19] and Myerson [43]. The graph of the fn looks like the graph shown in Figure 2.1.

(5.104)

Chapter 5. Examples of Banach Space Techniques

180

for some k ∈ {1, 2, . . . , N }. Since x ∈ En for all n ≥ N , the set relation (5.104) shows that x ∈ (qk − δn , qk + δn ) for all n ≥ N . By the limit (5.103), we know that x = qk ∈ Q, a contradiction. Hence we must have x ∈ / En for infinitely many n, i.e., fn (x) = 0 for infinitely many n so that lim fn (x) = ∞ is impossible. n→∞

In conclusion, we have constructed the sequence of continuous functions fn : [0, 1] → R which satisfies fn (x) → ∞ as n → ∞ if and only if x ∈ Q. If we suppose that fn is a function of period 1, then the domain of fn can be extended to R and we obtain the desired result. Hence we have completed the proof of the problem.



Problem 5.21 Rudin Chapter 5 Exercise 21.

Proof. Since Q is a countable union of closed √ sets of R, we have Q ∈ B. It is well-known that m(Q) = 0. Obviously, the translate Q + 2 does not intersect Q. This gives an affirmative answer to the first assertion. Assume that there was a homeomorphismn h : R → R such that e = h(E) ∩ E 6= ∅ E

for every measurable E ⊆ R with m(E) = 0. Given Q = {qk } and ǫ > 0. We first construct a particular measurable set E with m(E) = 0. To this end, for all qk ∈ Q, we consider the neighborhoods (qk − 2−k ǫ, qk + 2−k ǫ) and their union E(ǫ) =

∞ [

(qk − 2−k ǫ, qk + 2−k ǫ).

k=1

It is easy to see that every
E(ǫ) is nonempty open in R. In addition, since Q ⊆ E(ǫ), each E(ǫ) is dense in R with m E(ǫ) ≤ 2ǫ. Hence it follows from Theorem 5.6 (Baire’s Theorem) that the set ∞ 1 \ (5.105) E E= n n=1

is a dense Gδ set in R and m(E) = 0.

Suppose that E is the set (5.105). Since h is a homeomorphism, it is an open map which implies that ∞    \ 1 h E h(E) = n n=1


e is also a dense Gδ set is also a dense Gδ set in R with m h(E) = 0. Thus the intersection E
1 e ⊆E of measure zero. Since E n is countable for all n ∈ N, E is also countable and thus E is a countable dense Gδ in R of measure zero. However, R is a complete metric space which e certainly contradicts Theorem 5.13. Hence no such has no isolated points, the existence of E  homeomorphism h exists and this completes the proof of the problem. n

By the definition, h is a continuous bijection and h−1 is continuous.

181

5.5. Miscellaneous Problems

Problem 5.22 Rudin Chapter 5 Exercise 22.

Proof. Suppose that f ∈ C(T ), f ∈ Lip α and f (0) = 0. We need to show that lim sn (f ; x) = f (x).

n→∞

(5.106)

To achieve the goal, we first quote the following stronger form of the Riemann-Lebesgue Lemma whose proof can be found in [3, Theorem 11.16, p. 313]. Lemma 5.2 (Riemann-Lebesgue Lemma) For every f ∈ L1 (T ), we define 1 fb(n) = 2π

Z

π

f (t)e−i(n+β)t dt,

−π

where β ∈ R. Then we have fb(n) → 0 as |n| → ∞. To begin with, since f ∈ C(T ), we have f ∈ L1 (T ). Furthermore, the hypothesis f ∈ Lip α implies that |f (s) − f (t)| ≤ Mf |s − t|α for all s, t ∈ [−π, π], where Mf is finite and α ∈ (0, 1]. Particularly, take s = 0 so that |f (t)| ≤ Mf |t|α for all t ∈ [−π, π] which implies Z π Z π f (t) i 2M π α hZ 0 f tα−1 dt = < ∞. |t|α−1 dt = Mf (−1)α−1 tα−1 dt + dt ≤ Mf t α 0 −π −π −π

Z

π

In other words, we have

f (t) t

∈ L1 (T ). Since we have

 i 1 1 h i(n+ 1 )t 1 2 t= e sin n + − e−i(n+ 2 )t , 2 2i it yields Z π Z π 1 Z π sin(n + 12 )t 1 f (t) i(n+ 1 )t f (t) −i(n+ 1 )t 1 2 2 dt = e e f (t) · dt − dt π −π t 2πi −π t 2πi −π t 1 Z π f (t) 1 Z π f (t) 1 i(n+ 12 )t dt + ≤ e e−i(n+ 2 )t dt . 2π −π t 2π −π t

(5.107)

By Lemma 5.2, each of the integrals on the right-hand side of the inequality (5.107) tends to 0 as |n| → ∞. Thus it is true that 1 π

Z

π

−π

f (t) ·

sin(n + 12 )t dt → 0 = f (0) t

as |n| → ∞.o

Next, we claim that

o

Integrals of the form

Z

0

sn (f ; 0) → f (0) b

g(t)

(5.108)

sin αt dt are called Dirichlet integrals, where α > 0 and g is defined on [0, b]. t

Chapter 5. Examples of Banach Space Techniques

182

as n → ∞. To this end, we consider

Z Z π Z sin(n + 12 )t 1 sin(n + 12 )t 1 π 1 π dt dt f (t) · f (t)D (t) dt − f (t) (f ; 0) − = sn n π −π t 2π −π π −π t 1 Z π 1  1 1  = − f (t) sin n + t dt t 2π −π sin 2t 2 2 1 Z π  1  t dt , (5.109) F (t)f (t) sin n + = 2π −π 2

where F : [0, π] → R is defined by F (t) =

  

1 sin 2t

− 1t , if t ∈ [−π, π] \ {0};

  0,

2

if t = 0.

Clearly, F is continuous on [−π, π] and so F f ∈ L1 (T ). Using Lemma 5.2 to the right-hand side of the equation (5.109), we see that Z π  1 1 t dt = 0 F (t)f (t) sin n + lim n→∞ 2π −π 2 and this guarantees the validity of the claim (5.108). For the general case, we consider the function g : [−π, π] → R defined by g(t) = f (x + t) − f (x)

(5.110)

for every x ∈ R. Clearly, the real function g satisfies the conditions g ∈ C(T ), g ∈ Lip α and g(0) = 0. By the above argument, we obtain lim sn (g; 0) = 0.

n→∞

(5.111)

By the definition (5.110), we gain
sn (g; 0) = sn f (x + t) − f (x); 0 Z π Z π 1 1 = f (x + t)Dn (−t) dt − f (x)Dn (−t) dt 2π −π 2π −π Z π 1 f (x + t)Dn (−t) dt − f (x) = 2π −π Z π 1 = f (x − t)Dn (t) dt − f (x) 2π −π = sn (f ; x) − f (x).

(5.112)

By applying the limit (5.111) to the equation (5.112), we obtain our desired result (5.106),  completing the proof of the problem.

CHAPTER

6

Complex Measures

6.1

Properties of Complex Measures

Problem 6.1 Rudin Chapter 6 Exercise 1.

Proof. By the definition, we have λ(E) = sup

n nX k=1

k o [ Ej . |µ(Ei )| E1 , E2 , . . . , Ek are mutually disjoint and E = j=1

By Definition 6.1, we have |µ|(E) = sup

∞ nX k=1

∞ o [ Ej . |µ(Ei )| E1 , E2 , . . . are mutually disjoint and E = j=1

Obviously, we have λ(E) ≤ |µ|(E) for every E ∈ M.

For the other direction, we suppose that {Ei } is a partition of E ∈ M. Given ǫ > 0. Since µ is a complex measure, we get from Definition 6.1 that the series ∞ X

µ(Ei )

i=1

converges absolutely. Thus there exists a N ∈ N such that ∞ X

n=N

|µ(En )| < ǫ.

(6.1)

′ ′ Define E1′ = E1 , E2′ = E2 , . . . , EN −1 = EN −1 and EN = EN ∪ EN +1 ∪ · · · . By Definition 1.3(a), ′ we have EN ∈ M. Then we have ∞ [

i=1

′ ′ Ei = E1′ ∪ E2′ ∪ · · · ∪ EN −1 ∪ EN

which implies, with the aid of the estimate (6.1), that ∞ X i=1

|µ(Ei )| =

N −1 X i=1

|µ(Ei′ )| +

∞ X

i=N

|µ(Ei )|
n En Thus we must have λ(En ) ∈ [0, ∞) or equivalently, En is finite or En = ∅. Let E=

∞ [

n=1

En = {x ∈ (0, 1) | h(x) > 0}.

(6.5)

Therefore, the set E is countable. There are two cases for consideration: • Case (i): E 6= ∅. Let E = {x1 , x2 , . . .}. On the one hand, we know from [49, Remark 11.11(f), p. 309] that µ(F ) = 0 for every countable subset F ⊂ (0, 1), so we deduce from the representation (6.4) and the definition (6.5) that Z ∞ X h(xn ) 6= 0, h dλ = 0 = µ(E) = E

n=1

a contradiction. • Case (ii): E = ∅. In this case, we have h(x) = 0 a.e. [λ] on (0, 1) and the representation (6.4) again shows that Z
h dλ = 0, 1 = µ (0, 1) = (0,1)

a contradiction.

a

See Definition 1.35 for the meaning of the notation a.e. [λ].

185

6.1. Properties of Complex Measures

Hence no such h exists and this ends the proof of the problem.



Problem 6.3 Rudin Chapter 6 Exercise 3.

Proof. We first show that if µ and λ are complex regular Borel measures, then both µ + λ and αµ are complex regular Borel measures too. By §6.18, it is equivalent to show that both |µ + ν| and |αµ| are regular Borel measures. To this end, we follow from §6.18 again that both |µ| and |λ| are regular Borel measures. Let E ∈ B and ǫ > 0. By Definition 2.15, there exist open sets V1 , V2 ⊇ E such that |µ|(V1 ) < |µ|(E) +

ǫ 2

ǫ and |λ|(V2 ) < |λ|(E) + . 2

(6.6)

Similarly, there are compact sets K1 , K2 ⊆ E such that |µ|(E) < |µ|(K1 ) +

ǫ 2

ǫ and |λ|(E) < |λ|(K2 ) + . 2

(6.7)

The triangle inequality certainly implies |µ(E) + λ(E)| ≤ |µ(E)| + |λ(E)| and then Definition 6.1 gives ∞ ∞ X X   |µ(E)| + |λ(E)| , (6.8) |µ(E) + λ(E)| ≤ sup |µ + λ|(E) = sup i=1

i=1

where the supremum being taken over all partitions {Ei } of E. Since the series ∞ X i=1

∞ X i=1

|µ(E)| and

|λ(E)| converge, we deduce from the expression (6.8) that |µ + λ|(E) ≤ sup

∞ X i=1

|µ(E)| + sup

∞ X i=1

|λ(E)| = |µ|(E) + |λ|(E).

Thus we have |µ + λ| ≤ |µ| + |λ|.

(6.9)

Let V = V1 ∩ V2 and K = K1 ∪ K2 . Then K is compact and V is open in X. Now we follow from the estimates (6.6) and (6.7) and the inequality (6.9) that |µ + λ|(V ) = |µ + λ|(E) + |µ + λ|(V \ E)

≤ |µ + λ|(E) + |µ|(V \ E) + |λ|(V \ E)

≤ |µ + λ|(E) + |µ|(V1 \ E) + |λ|(V2 \ E)

< |µ + λ|(E) + ǫ.

Since ǫ and E are arbitrary, the measure |µ + λ|(E) is in fact outer regular. Similarly, we have |µ + λ|(K) = |µ + λ|(E) − |µ + λ|(E \ K)

≥ |µ + λ|(E) − |µ + λ|(E \ K1 ) − |µ + λ|(E \ K2 ) > |µ + λ|(E) − ǫ.

In other words, |µ + λ| is also inner regular. By Definition 2.15, µ + λ is a regular complex Borel measure, i.e., µ + λ ∈ M (X). Now the regularity of |αµ| is easy to prove, so we omit the details here.

Chapter 6. Complex Measures

186

Now it is time to prove the assertion in the question. Since X is a locally compact Hausdorff space, Theorem 6.19 (The Riesz Representation Theorem) ensures every Φ ∈ C0 (X)∗ is represented by a unique µΦ ∈ M (X) in the sense that Z f dµΦ (6.10) Φ(f ) = X

for every f ∈ C0 (X). By [51, Eqn. (3), p. 130], we see that Z Z Z Z f d(µΦ + µΨ ) f dµΨ = f dµΦ + f dµΦ+Ψ = (Φ + Ψ)(f ) = Φ(f ) + Ψ(f ) = X

X

X

and

Therefore, they imply that

Z

f dµαΦ = αΦ(f ) =

Z

X

f d(αµΦ ).

X

X

µΦ+Ψ = µΦ + µΨ

and

µαΦ = αµΦ .

(6.11)

By the previous analysis, we see that µΦ+Ψ , µαµ ∈ M (X), so we may define the mapping F : C0 (X)∗ → M (X) by F (Φ) = µΦ and it is easy to see from the results (6.11) that F (Φ + Ψ) = F (Φ) + F (Ψ) and

F (αΦ) = αF (Φ).

Furthermore, Theorem 6.19 (The Riesz Representation Theorem) also implies that F is a bijection and kΦk = |µΦ |(X) = kµΦ k = kF (Φ)k. Consequently, F is actually an isometric vector space isomorphism, i.e., C0 (X)∗ ∼ = M (X). Since C0 (X) is a Banach space with the supremum norm, Problem 5.8 guarantees that C0 (X)∗ is Banach. Hence M (X) is Banach and we have completed the proof of the problem. 

6.2

Dual Spaces of Lp (µ)

Problem 6.4 Rudin Chapter 6 Exercise 4.

Proof. Since µ is positive and σ-finite, we can write X=

∞ [

Xn ,

(6.12)

n=1

where {Xn } is an increasing sequence of measurable sets and µ(Xn ) < ∞ for all n ∈ N, see [54, Definition 4.22, p. 22]. Let A = {x ∈ X | |g(x)| = ∞} and An = {x ∈ Xn | |g(x)| = ∞}, where n ∈ N. Since g is measurable, every An and A are measurable by Problem 1.5. Besides, it is evident that A1 ⊆ A2 ⊆ · · · and if x ∈ A, then x ∈ XN0 for some N0 ∈ N so that x ∈ AN0 . As a result, we get ∞ [ An . A= n=1

Now we are going to divide the proof into several steps:

6.2. Dual Spaces of Lp (µ)

187

• Step 1: µ(A) = 0. Otherwise, it follows from the construction and Theorem 1.19(c) that lim µ(An ) = µ(A) > 0.

n→∞

This means that one can find a N1 ∈ N such that 0 < µ(AN1 ) ≤ µ(XN1 ) < ∞. We know that χAN1 ∈ Lp (µ), so our hypothesis implies that χAN1 g ∈ L1 (µ), but Z Z |g| dµ = µ(AN1 ) · ∞ = ∞ |χAN1 g| dµ = kχAN1 gk1 = AN1

X

which is a contradiction. Hence µ(A) = 0, i.e., g is finite a.e. on X. • Step 2: g ∈ Lq (µ) when 1 < p < ∞. In this case, we have q ∈ (1, ∞). Here we define En = {x ∈ Xn | |g(x)| ≤ n}. Obviously, {En } is also an increasing sequence of measurable sets. Furthermore, if x0 ∈ X \ A, then x0 ∈ XN2 for some N2 ∈ N by the definition (6.12) so that x0 ∈ Xn for all n ≥ N2 . Since g is finite a.e. on X, there exists a positive integer N3 such that |g(x)| ≤ N3 for almost all x ∈ X. Take N4 = max(N2 , N3 ), then x0 ∈ XN4 and |g(x0 )| ≤ N4 which imply that x0 ∈ EN4 . Consequently, we have shown that ∞ [

X=

En .

n=1

Let gn = χEn g : X → C, where n = 1, 2, . . .. Then each gn is measurable on X and Z Z q q |g|q dµ ≤ nq µ(En ) ≤ nq µ(Xn ) < ∞ (6.13) |gn | dµ = kgn kq = En

X

for each n = 1, 2, . . .. Thus we have gn ∈ Lq (µ) for all n ∈ N. Next, we define Λn : Lp (µ) → C by Z f gn dµ

Λn (f ) =

X

which is linear for n ∈ N. Since f gn = (f χEn )g and f χEn ∈ Lp (µ), we have f gn ∈ L1 (µ). By Theorem 1.33 and Theorem 3.8, we obtain Z Z |f gn | dµ = kf gn k1 ≤ kf kp × kgn kq , (6.14) f gn dµ ≤ X

X

where n = 1, 2, . . .. By Definition 5.3 and the inequality (6.14), we know that kΛn k = sup{|Λn (f ) | f ∈ Lp (µ) and kf kp = 1} ≤ kgn kq , where n = 1, 2, . . .. Recall that 1 < q < ∞, so we may consider −q

f0 = kgn kq p |gn |q−2 gn . (q−1)p so that Then we have |f0 |p = kgn k−q q |gn | Z Z Z (q−1)p −q |f0 |p dµ = kgn k−q |g | dµ = kg k |gn |q dµ = 1 kf0 kp = n n q q X

X

and Λn (f0 ) =

Z

X

X

−q

kgn kq p |gn |q−2 gn · gn dµ

(6.15)

Chapter 6. Complex Measures

188 − pq

= kgn kq =

−q kgn kq p

= kgn kq

Z

X

|gn |q dµ

× kgn kqq

(6.16)

for all n ∈ N. Combining the results (6.13), (6.15) and (6.16), we have established the fact that kΛn k = kgn kq < ∞, (6.17) where n ∈ N. Hence {Λn } is a family of bounded linear transformations of Lp (µ) into C. Since Lp (µ) is Banach (see Definition 5.2) and Z Z Z Z |f g| dµ = kf gk1 < ∞ |f g| dµ ≤ |f gn | dµ = f gn dµ ≤ |Λn (f )| = X

X

En

X

for every f ∈ Lp (µ) and n ∈ N, Theorem 5.8 (The Banach-Steinhaus Theorem) implies that there is a M > 0 such that kΛn k ≤ M (6.18) for all n ∈ N. Using the results (6.17) and (6.18), we conclude that kgn kq ≤ M

(6.19)

for all n ∈ N.

By Step 1 and the definition of gn , we have |gn (x)| ≤ |g(x)| < ∞ a.e. on X for all n ∈ N and |g1 (x)| ≤ |g2 (x)| ≤ · · · for every x ∈ X. Now the Monotone Convergence Theorem [49, Theorem 3.14, p. 55] implies that |gn (x)| → |g(x)| < ∞ a.e. on X as n → ∞. Hence we gain from Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem) that Z Z |g|q dµ = kgkqq . (6.20) |gn |q dµ = lim kgn kqq = lim n→∞

n→∞ X

X

Hence it follows from the inequality (6.19) and the limit (6.20) that kgkq ≤ M < ∞, i.e. g ∈ Lq (µ). • Step 3: g ∈ L∞ (µ) when p = 1. In this case, q = ∞. We recall from Step 1 that g is finite a.e. on X, so there exists a M > 0 such that |g(x)| ≤ M a.e. on X. By Definition 3.7, we assert that kgk∞ ≤ M , i.e., g ∈ L∞ (µ). • Step 4: g ∈ L1 (µ) when p = ∞. Define f1 : X → C by

 g(x)    , if g(x) 6= 0; |g(x)| f1 (x) =    0, otherwise.

Since |f1 (x)| ≤ 1 for all x ∈ X, we get from Definition 3.7 that kf1 k∞ ≤ 1, i.e., f1 ∈ L∞ (µ). By the hypothesis, we have f1 g ∈ L1 (µ). Since f1 g = |g|, we conclude that g ∈ L1 (µ).

6.2. Dual Spaces of Lp (µ)

189 Now we have completed the proof of the problem.



Problem 6.5 Rudin Chapter 6 Exercise 5.

Proof. The answer is negative. On the one hand, if f1 , f2 : X → C are defined by f1 (a) = 0,

f1 (b) = 1 and f2 (a) = 1,

f2 (b) = 0,

(6.21)

then we have f1 , f2 ∈ L∞ (µ) and f1 6≡ f2 . For every f ∈ L∞ (µ), we have f (a) = A + Bi and f (b) = C + Di for some A, B, C, D ∈ R. Obviously, we get from the definition (6.21) the representation f = (C + Di)f1 + (A + Bi)f2 . As a result, L∞ (µ) is a two-dimensional space spanned by f1 and f2 , i.e., dim L∞ (µ) = 2. On the other hand, if f ∈ L1 (µ), then kf k1 < ∞ and we follow from Definition 3.6 that f (b) = 0 and kf k1 = |f (a)| = |f (a)|f2 (a). Therefore, L1 (µ) is an one-dimensional space spanned by f2 . As a vector space (see Remark 3.10), we have dim L1 (µ)∗ = dim L1 (µ) = 1. Hence, L∞ (µ) 6= L1 (µ)∗ , completing the proof of the problem.



Problem 6.6 Rudin Chapter 6 Exercise 6.

Proof. We want to show that Lp (µ)∗ ∼ = Lq (µ) for 1 < p < ∞. Equivalently, we have to show that for each Λ ∈ Lp (µ)∗ , there exists a unique g ∈ Lq (µ) such that Z f g dµ (6.22) Λf = X

for all f ∈ Lp (µ). The following proof follows mainly Follan’s argument ([22, p. 190]): • Step 1: µ is finite. Let s be a simple function on X. Since µ is finite, we have µ({x ∈ X | s(x) 6= 0}) ≤ µ(X) < ∞ and then it follows from Theorem 3.13 that s ∈ Lp (µ). Let Λ ∈ Lp (µ)∗ , E be measurable and λ(E) = Λ(χE ). For every partition {Ei } of E, if we let Fn = E1 ∪ E2 ∪ · · · ∪ En , then we obtain Z Z Z p p χE\Fn dµ = µ(E \ Fn ) (6.23) |χE\Fn | dµ = |χE − χFn | dµ = X

X

X

for all n ∈ N. In fact, the expression (6.23) can be rewritten as 1

kχE − χFn kp = µ(E \ Fn ) p .

(6.24)

Since E \ F1 ⊇ E \ F2 ⊇ · · · and µ(E \ F1 ) is finite, it establishes from Theorem 1.19(e) that ∞ \  lim µ(E \ Fn ) = µ (E \ Fn ) = µ(∅) = 0. (6.25) n→∞

n=1

Chapter 6. Complex Measures

190

Combining the results (6.24) and (6.25) and using the fact p ∈ (1, ∞), we derive that lim kχFn − χE kp = 0.

n→∞

Since χE ∈ Lp (µ) and each χEi g is measurable, we deduce from the representation (6.22) and then Theorem 1.27 that Z X Z ∞ ∞ Z ∞ X X λ(χEi ). χEi g dµ = (χEi g) dµ = χE g dµ = λ(E) = Λ(χE ) = X i=1

X

i=1

X

i=1

By Definition 6.1, λ is a complex measure. If E ∈ M satisfies µ(E) = 0, then χE = 0 in Lp (µ) and so λ(E) = Λ(0) = 0. By Definition 6.7, we have λ ≪ µ. Thus we know from Theorem 6.10 (The Lebesgue-RadonNikodym Theorem) that there exists a unique h ∈ L1 (µ) such that Z Z χE g dµ g dµ = Λ(χE ) = λ(E) = E

X

for every E ∈ M. Recall that Λ ∈ Lp (µ)∗ , Λ is linear and bounded so that Z sg dµ Λ(s) = X

Lp (µ).

for every simple function s ∈ By [51, Eqn. (3), p. 96], we know that Z sg dµ = kΛ(s)k ≤ kΛk · kskp < ∞. X

By [22, Theorem 6.14, p. 189], we conclude that g ∈ Lq (µ). Given that f ∈ Lp (µ). By Theorem 3.8, we have f g ∈ L1 (µ). By Theorem 3.13, there exists a sequence of simple functions {sn } such that |sn | ≤ |f | for every n ∈ N and sn → f in Lp (µ) as n → ∞. Since |sn g| ≤ |f g| and f g ∈ L1 (µ), we obtain from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) and Theorem 5.4 that Z Z f g dµ. sn g dµ = Λ(f ) = lim Λ(sn ) = lim n→∞

n→∞ X

X

• Step 2: µ is σ-finite. This is exactly the hypothesis of Theorem 6.16, so we have Lp (µ)∗ ∼ = Lq (µ) in this case. • Step 3: µ is any measure and p > 1. In this case, q ∈ (1, ∞). As in Step 2, for each σ-finite subset E of X, there corresponds a unique gE ∈ Lq (E) such that Z f gE dµ Λ(f ) = X

for all f ∈ Lp (E), where Lp (E) = {f ∈ Lp (µ) | f (x) = 0 for all x ∈ / E} and Lq (E) is defined similarly. By Theorem 6.16, we know that

kgE kq = Λ|Lp (E) ≤ kΛk. (6.26) If F is a σ-finite subset of X containing E, then the uniqueness of gE implies that gF = gE a.e. on E so that kgE kq ≤ kgF kq . (6.27) Now we define M = sup{kgE kq | E a σ-finite subset of X}.

191

6.3. Fourier Coefficients of Complex Borel Measures Then it follows from the inequality (6.26) that M ≤ kΛk holds. By the definition, we may select a sequence {En } of σ-finite subsets of X such that ∞ [ En which is obviously σ-finite subset of X so kgEn kq → M as n → ∞. Define F = n=1

that kgF kq ≤ M . Furthermore, we deduce from the inequality (6.27) that kgEn kq ≤ kgF kq

for all n ∈ N. Therefore, the definition of M implies that M ≤ kgF kq and then M = kgF kq . Finally, if A is a σ-finite subset of X containing F , we get Z Z Z Z q q q q q |gF |q dµ |gA | dµ = kgA kq ≤ M = |gA\F | dµ = |gF | dµ + X

X

X

X

which means that gA\F = 0 a.e. on X or equivalently, gA = gF a.e. on X.

(6.28)

If f ∈ Lp (µ), then the set A = F ∪ {x ∈ X | f (x) 6= 0} is clearly σ-finite. Therefore, we observe from the result (6.28) that Z Z f gF dµ. f gA dµ = Λ(f ) = X

X

Hence we may pick g = gF and the above argument makes sure this g is unique. We complete the proof of the problem.

6.3



Fourier Coefficients of Complex Borel Measures

Problem 6.7 Rudin Chapter 6 Exercise 7.

Proof. If µ is a real measure, then we have Z Z int b(n) µ b(−n) = e dµ(t) = e−int dµ(t) = µ

for every n ∈ Z. Therefore, our result follows in this case. Since µ is a complex Borel measure, Theorem 6.12 implies the existence of a Borel measurable function h such that |h(t)| = 1 on [0, 2π) and dµ = h d|µ|. Thus we have d|µ| = 1 · d|µ| = h · h d|µ| = h dµ. Since |µ| is real, the previous paragraph and the expression (6.29) imply that Z Z Z −int int µ b(−n) = e h d|µ| = e h(t) d|µ|(t) = e−int [h(t)]2 dµ(t)

(6.29)

Chapter 6. Complex Measures

192

for every n ∈ Z. By Theorem 3.14, C(T ) is dense in L1 (|µ|). Using Theorem 4.25 (The Weierstrass Approximation Theorem) and then Lemma 2.10, we deduce that the set of all trigonometric polynomials P is dense in L1 (|µ|). We want to apply Problem 5.18. For any f ∈ L1 (|µ|), we define Λn : L1 (|µ|) → C by Z Λn (f ) = f (t)e−int dµ(t) for all n ∈ N. Then we have Z Z Z |Λn (f )| ≤ |f | · | dµ(t)| = |f | · |h| d|µ(t)| = |f | d|µ(t)| = kf k1 .

(6.30)

By Definition 5.3, we see from the inequality (6.30) that kΛn k = sup{|Λn (f )| | f ∈ L1 (|µ|) and kf k1 = 1} ≤ 1 for every n ∈ N. Next, if g(t) = eikt , then we have Z Z −int Λn (g) = g(t)e dµ(t) = e−i(n−k)t dµ(t). Now our hypothesis ensures that Λn (g) → 0 as n → ∞ so that Λn (f ) → 0

(6.31)

as n → ∞ for each f ∈ P. Therefore, Problem 5.18 says that the limit (6.31) also holds for 2 every f ∈ L1 (|µ|). Obviously, we have h ∈ L1 (|µ|) and so Z 2 µ b(−n) = e−int [h(t)]2 dµ(t) = Λn (h ). Hence we conclude from the result (6.31) that

2

lim µ b(−n) = lim Λn (h ) = 0.

n→∞

n→∞

This completes the proof of the problem.



Problem 6.8 Rudin Chapter 6 Exercise 8.

Proof. Suppose that X = [0, 2π), B is the collection of all Borel sets in X and µ b is periodic −ikt with period k ∈ Z. Denote dλ(t) = (e − 1) dµ(t). Clearly, we have Z Z b (6.32) µ b(n + k) − µ b(n) = e−int (e−ikt − 1) dµ(t) = e−int dλ(t) = λ(n)

for all n ∈ Z. Particularly, take n = 0 in the result (6.32) and use the periodicity of µ b and Theorem 6.12 to get Z Z h d|λ| = dλ = 0, (6.33)

where h is a measurable function such that |h(x)| = 1 on X. If |λ|(X) 6= 0, then Theorem 1.39(a) and the form (6.33) together imply that h = 0 a.e. on X, a contradiction. This means that |λ|(X) = 0 and we assert from the fact |λ(E)| ≤ |λ|(X) = 0 that λ(E) = 0

(6.34)

193

6.3. Fourier Coefficients of Complex Borel Measures

for every E ∈ B.

Using Theorem 6.12 to write dµ = H d|µ|, where H is a measurable function with |H(x)| = 1 on X. Next, we recall the meaning of the notation dλ(t) = (e−ikt − 1) dµ(t) and we observe from the result (6.34) that Z Z −ikt 0 = λ(E) = (e (6.35) − 1) dµ = (e−ikt − 1)H d|µ| E

E

for every E ∈ B. Let A1 = {t ∈ X | cos kt − 1 = 0}, A2 = {t ∈ X | sin kt = 0} and A = {t ∈ X | e−ikt − 1 = 0} = A1 ∩ A2 . Since f1 (t) = cos kt − 1, f2 (t) = sin kt and g(t) = 0 are continuous on X, they are Borel measurable.b By Problem 1.5(a), we have A1 , A2 ∈ B and also A ∈ B. We claim that µ is concentrated on A. Assume that it was not the case, i.e., there is E0 ∈ B such that a E0 ∩ A = ∅ but µ(E0 ) 6= 0. We recall from the result (6.35) that (e−ikt − 1)H = 0 a.e. on any E with |µ|(E) > 0. Since |H| = 1 on X, we have e−ikt = 1 a.e. on any E with |µ|(E) > 0. Particularly, since µ(E0 ) 6= 0, we have |µ|(E0 ) > 0 and then e−ikt = 1

(6.36)

a.e. on E0 . However, E0 ∩ A = 0 = ∅ means that e−ikt 6= 1 on E0 which contradicts the expression (6.36). Hence we have proven our claim that µ is concentrated on A. On the other hand, if µ is concentrated on A, then for every E ∈ B, we write E = (E \ A) ∪ A and then Z Z Z −ikt −ikt (6.37) (e − 1) dµ + (e−ikt − 1) dµ. (e − 1) dµ = E

E\A

A

Notice that e−ikt − 1 = 0 on A and (E \ A) ∩ A = ∅ so that µ(E \ A) = 0. Thus we deduce from Proposition 1.24(d) and (e) that all the integrals in the expression (6.37) are zero. This certainly implies that λ(E) = 0 for every E ∈ B and then Z b λ(n) = e−int dλ = 0 (6.38) for every n ∈ Z. Combining the expression (6.32) and the result (6.38), we may conclude that µ b(n + k) = µ b(n)

for all n ∈ Z.

Hence we have shown that µ is a complex Borel measure with periodic Fourier coefficient µ b with period k if and only if µ is concentrated on o n 2nπ A = {t ∈ X | e−ikt − 1 = 0} = t = n ∈ N ∪ {0} , k

completing the proof of the problem. Problem 6.9 Rudin Chapter 6 Exercise 9.

b

See §1.11.



Chapter 6. Complex Measures

194

Proof. The assertion is false. Take µ = m the Lebesgue measure on I. Since m ≪ m, if m ⊥ m, then Proposition 6.8(g) implies that m = 0, a contradiction. Now it remains to construct a sequence of {gn } satisfying the required properties. 2 . Define gn,k : [0, 1] → R and Suppose that n, k ∈ N and k = 1, 2, . . . , n. Let δn = n(2n+3) gn : [0, 1] → R by  h k δn k δn i   n + 1, if x ∈ ; − , +   n+1 2 n+1 2        k  δn δn k δn    ; − − , −  linear, if x ∈ n+1 2 2n + 2 n + 1 2 gn,k (x) =    k  δn k δn δn     ; + , + + linear, if x ∈   n+1 2 n+1 2 2n + 2       0, otherwise and

gn (x) =

n X

gn,k (x)

k=1

respectively. By direct computation, it is easy to check that 0
0. Thus A−1 exists. Let y = A−1 (x). Then we follow from Theorem 7.26 (The Change-of-variables Theorem) (with X = Y = V = Rk , T = A−1 which is one-to-one and differentiable on Rk , T (Rk ) = A−1 (Rk ) = Rk and y = A−1 (x)) that g (z) = f[ b ◦ A(z) Z
f A(y) exp(−iy · z) dmk (y) = k {z } |R Z This is

=

Z

Rk

f dm.

Y

f (x) exp[−iA−1 (x) · z] × | det A−1 | dmk (x) {z } |

(9.70)

This is (f ◦ T )(x) = (f ◦ A−1 )(x).

for every z ∈ Rk . We need a result from linear algebra: Lemma 9.7

 If A is a k × k matrix, then for all x, y ∈ Ck , we have hA(x), yi = x, AT (y) , where AT is the transpose of A. Proof of Lemma 9.7. Using [41, p. 109; Example 5.3.1, p. 286], it can be seen easily that

 hA(x), yi = [A(x)]T y = xT AT (y) = x, AT (y) ,

completing the proof of Lemma 9.7.

Thus we apply Lemma 9.7 to the expression (9.70) to deduce that Z
f (x) exp[−ix · A−T (z)] dmk (x) = | det A| · fb A−T (z) . g(z) = | det A| · b Rk



9.3. Fourier Transforms on Rk and its Applications

307 Consequently, we conclude that

f\ ◦ A = | det A| · (fb ◦ A−T ).

(9.71)

Particularly, let A be a rotation matrix of Rk . Then it preserves the distance of x from the origin, i.e., hA(x), A(x)i = hx, xi. By [4, Proposition 5.1.13, p. 134], A is orthogonal which means AT = A−1 . Furthermore, this implies that | det A| = 1, so the formula (9.71) can be simplified to f\ ◦ A = fb ◦ A.

(9.72)

In other words, if f is invariant under rotations, then f ◦A = f and the expression (9.72) implies fb ◦ A = fb,

i.e., fb is also invariant under rotations. This completes the proof of the problem.



Problem 9.16

Rudin Chapter 9 Exercise 16.

Proof. We make the following assumption: Suppose that f ∈ S(Rk ), where S(Rk ) is the class of all functions f : Rk → C such that f ∈ C ∞ and xβ ∂ α f is bounded for all multi-indices α ∂f and β.k Let 1 ≤ j ≤ k, x = (xj , xj ) and F (x) = , where x = (x1 , . . . , xj , . . . , xk ) and ∂xj xj = (x1 , . . . , xj−1 , xj+1 , . . . , xk ). We note that x · y = x1 y1 + x2 y2 + · · · + xk yk = xj · yj + xj yj . For every y ∈ Rk , we follow from this, the definition in Problem 9.14 and Lemma 9.4 that Z ∂f b F (y) = (x) exp(−ix · y) dmk (x) k ∂xj ZR i h Z ∂f (xj , xj ) exp(−ixj · yj ) dm(xj ) exp(−ixj · yj ) dmk−1 (xk ) = ∂xj Rk−1 | R {z } =

Z

Apply Lemma 9.4 to this.

Rk−1 Z

= iyj

i h Z f (x) exp(−ixj · yj ) dm(xj ) exp(−ixj · yj ) dmk−1 (xk ) iyj

Rk

R

f (x) exp(−ix · y) dmk (x)

= iyj fb(y)

which implies that

Therefore, we have

k

2f ∂d (y) = −yj2 fb(y). ∂x2j

g(x) = − b

k X j=1

x2j fb(x) = −|x|2 · fb(x)

In fact, S is the Schwartz space on Rk , see Remark 9.2.

(9.73)

Chapter 9. Fourier Transforms

308

for all x ∈ Rk .

Suppose that f has continuous second derivatives so that the Laplacian ∆f is well-defined. Suppose further that supp (f ) is compact. In this case, f ∈ L1 (Rk ) so that fb is well-defined on Rk . Furthermore, we suppose that A is a rotation of Rk about the origin 0 and gA = ∆(f ◦ A). Then we deduce from the formulas (9.72) and (9.73) that 2 \ 2 2 b b gc b(Ax) = g[ ◦ A(x). A (x) = −|x| · f ◦ A(x) = −|Ax| · (f ◦ A)(x) = −|Ax| · f (Ax) = g

\ \ By the definition, we have ∆(f ◦ A) = (∆f ) ◦ A. To go further, we need the analogue of Theorem 9.12 (The Uniqueness Theorem): Lemma 9.8 If f ∈ L1 (Rk ) and fb(t) = 0 for all t ∈ Rk , then f (x) = 0

(9.74)

a.e. on Rk . Furthermore, if f is continuous on Rk , then the expression (9.74) holds everywhere on Rk .

Proof of Lemma 9.8. Since fb = 0 on Rk , we have fb ∈ L1 (Rk ) and the result (9.74) follows from the Inversion Theorem for Rk (Problem 9.14). If f is continuous on Rk , the definition of continuity ensures that the relation (9.74) holds everywhere on Rk .  Now we let F = ∆(f ◦ A) − (∆f ) ◦ A. It is clear that f ◦ A ∈ S(Rk ) if f ∈ S(Rk ). Similar to the proof of Problem 9.7, we can show that ∂ α f ∈ S(Rk )

and ∂ α f ∈ L1 (Rk ).

Thus we have ∆(f ◦ A) ∈ L1 (Rk ) and (∆f ) ◦ A ∈ L1 (Rk ) so that F ∈ L1 (Rk ). Since Fb(t) = 0 for all t ∈ Rk and F is continuous on Rk , Lemma 9.8 implies that ∆(f ◦ A) = (∆f ) ◦ A

(9.75)

holds everywhere on Rk . It is well-known that any rotation about a point p is the composition of a rotation about the origin 0 and a translation. By this fact, the analysis preceding Lemma 9.8 and the hypothesis that ∆ commutes with translations, we conclude that the expression (9.75) is still valid for every rotation about every point p. It is time to consider the general situation. Fix a p ∈ Rk . There exists a r > 0 such that p ∈ Kr = B(0, r) ⊂ Vr = B(0, 2r). By [49, Exercise 6, p. 289], there exists functions ψ1 , . . . , ψs ∈ C ∞ (Rk ) such thatl • 0 ≤ ψi ≤ 1 for 1 ≤ i ≤ s; • supp (ψi ) ⊂ Vr ; • ψ1 (x) + ψ2 (x) + · · · + ψs (x) = 1 for every x ∈ Kr . If we define Ψ : Rk → R by l

Ψ = ψ1 + ψ2 + · · · + ψs ,

For a proof of this, please read [63, Problem 10.6, pp. 266, 267 ].

9.3. Fourier Transforms on Rk and its Applications

309

then the above conditions imply that Ψ ∈ Cc∞ (Rk ) and Ψ(x) = 1 on Kr . Thus we have f · Ψ = f on Kr . Particularly, we have ∂Ψ ∂2Ψ (p) = 0 (p) = ∂xj ∂x2j

(f · Ψ)(p) = f (p) and

for every 1 ≤ j ≤ k. As a consequence, these imply that ∂ 2 (f · Ψ) ∂Ψ ∂2Ψ ∂2f ∂f ∂2f (p) · (p) + f · (p) = Ψ(p) · (p) + 2 (p) = (p) ∂xj ∂xj ∂x2j ∂x2j ∂x2j ∂x2j

(9.76)

for 1 ≤ j ≤ k. By the definition of the Laplacian and the result (9.76), we yield k k X X
∂ 2 (f · Ψ) ∂2f ∆(f · Ψ) (p) = (p) = (p) = (∆f )(p). ∂x2j ∂x2j j=1 j=1

(9.77)

Note that f · Ψ is compactly supported in this case. If Ap = q, then q ∈ Kr . On the one hand, the expressions (9.75) (with f replaced by f · Ψ) and (9.77) give

 ∆ f · Ψ) ◦ A (p) = [(∆(f · Ψ)) ◦ A](p) = ∆(f · Ψ) (q) = (∆f )(q) = [(∆f ) ◦ A](p). (9.78) On the other hand, since

[(f · Ψ) ◦ A](p) = [(f ◦ A)(p)] · [(Ψ ◦ A)(p)] = (f · Ψ)(q) = f (q) = (f ◦ A)(p) on Kr , we have (f · Ψ) ◦ A = f ◦ A on Kr so that 
   ∆ (f · Ψ) ◦ A (p) = ∆(f ◦ A) (p)

(9.79)

Finally, by combining the expressions (9.78) and (9.79) and using the fact that p is arbitrary, we may conclude that our desired formula (9.75) also holds everywhere on Rk in this general  case. This completes the proof of the problem. Problem 9.17 Rudin Chapter 9 Exercise 17.

Proof. We prove the case for Rk directly. Suppose that ϕ : Rk → C is a Lebesgue measurable character. We define Φ : L1 (Rk ) → C by Z f (x)ϕ(x) dmk (x). (9.80) Φ(f ) = Rk

Now direct computation gives Z Z (f ∗ g)(x)ϕ(x) dmk (x) = Φ(f ∗ g) = Rk

Rk

Z

Rk

f (x − y)g(y)ϕ(x) dmk (y) dmk (x).

(9.81)

Since ϕ is a character, we have ϕ(x) = ϕ(x − y)ϕ(y). Substituting this into the expression (9.81), we see that Z Z f (x − y)g(y)ϕ(x − y)ϕ(y) dmk (y) dmk (x) Φ(f ∗ g) = k k R R Z hZ i f (x − y)ϕ(x − y) dmk (x) g(y)ϕ(y) dmk (y) = Rk

Rk

Chapter 9. Fourier Transforms

=

Z

Rk

310

Φ(f )g(y)ϕ(y) dmk (y)

= Φ(f ) · Φ(g). Therefore, Φ is a linear functional on L1 (Rk ). In other words, Φ is a complex homomorphism of the Banach algebra L1 (Rk ). Hence the analogue of Theorem 9.23 in Problem 9.14 shows that one may find a β ∈ L∞ (Rk ) and a unique y ∈ Rk such that Z f (x) exp(−iy · x) dmk (x) = fb(y). (9.82) β(x) = exp(−iy · x) and Φ(f ) = Rk

Next, we deduce from the formulas (9.80) and (9.82) that Z f (x)[ϕ(x) − exp(−iy · x)] dmk (x) = 0

(9.83)

Rk

holds for all f ∈ L1 (Rk ). Since f (x) = e−|x| is in L1 (Rk ) and f (x) 6= 0 on Rk , the result (9.83) implies that ϕ(x) = exp(−iy · x) (9.84)

a.e. on Rk . To proceed further, we need the following result which is a generalization of Problem 7.6 and its proof uses the Steinhaus Theorem in Rk (see Remark 7.2), but we don’t present it here. Lemma 9.9 Suppose that G is a subgroup of Rk (relative to addition), G 6= Rk , and G is Lebesgue measurable. Then mk (G) = 0.

To apply this lemma, we consider h(x) = ϕ(x) − exp(−iy · x) and G = {x ∈ Rk | h(x) = 0}. For a, b ∈ G, if a + b ∈ / G, then we have h(a + b) 6= 0 and ϕ(a)ϕ(b) = ϕ(a + b) 6= exp[−iy · (a + b)] = exp(−iy · a) exp(−iy · b) = ϕ(a)ϕ(b) which implies 1 6= 1, a contradiction. Thus a + b ∈ G which means G is a subgroup of Rk (relative to vector addition). Besides, since h is obviously Lebesgue measurable by Proposition 1.9(c) and G = h−1 (0), G is a Lebesgue measurable set. Since mk (G) > 0, Lemma 9.9 forces that G = Rk so that the equation (9.84) holds for all x ∈ Rk . Now the continuity of the exponential function implies that ϕ is continuous on Rk . This completes the proof of the problem. 

9.4

Miscellaneous Problems

Problem 9.18 Rudin Chapter 9 Exercise 18.

Proof. First of all, the equation f (x + y) = f (x) + f (y) for all x, y ∈ R is called the Cauchy functional equation.

(9.85)

311

9.4. Miscellaneous Problems • Existence of real discontinuous functions f satisfying the equation (9.85). It is well-known that the Hausdorff Maximality Theorem, the Axiom of Choice and Zorn’s Lemma are equivalent to each other. To prove this part, we would like to apply Zorn’s Lemma which says that Lemma 9.10 (Zorn’s Lemma) If X is a partially ordered set and every totally ordered subset of X has an upper bound, then X has a maximal element.

As a consequence, Zorn’s Lemma can be used to prove that every vector space has a basis. In fact, if A is a linearly independent subset of a vector space V over a field K, then there is a basis B of V that contains A.m By this result, there exists a basis of R over Q with the usual addition and scalar multiplication.n Let x ∈ R and H be a basis of R over Q. Thus x has a unique representation x = r 1 b1 + r 2 b2 + · · · + r n bn , where b1 , b2 , . . . , bn ∈ H and r1 , r2 , . . . , rn ∈ Q. Define f : R → R by f (x) = r1 + r2 + · · · + rn . It is obvious that f satisfies the equation (9.85). Assume that f was continuous on R. Then f (R) must be connected (see [49, Theorem 4.22, p. 93]), but f (R) = Q which is a contradiction. • If f is Lebesgue measurable and satisfies the equation (9.85), then f is continuous. We use the following special form of Theorem 2.24 (Lusin’s Theorem):o Lemma 9.11 Suppose that f : [a, b] → R is Lebesgue measurable. For every ǫ > 0, there is a compact set K ⊆ [a, b] such that m([a, b] \ K) < ǫ and f |K is continuous. By Lemma 9.11, there exists a compact set K ⊆ [0, 1] such that m(K) >

2 3

(9.86)

and f |K is continuous. Since K is compact, f is actually uniformly continuous on K. Given ǫ > 0. There is a δ > 0 such that |f (x) − f (y)| < ǫ for every x, y ∈ K with |x − y| < δ. Without loss of generality, we may assume that δ < 13 . Let η ∈ (0, δ) and K − η = {x − η | x ∈ K}. Then K − η ⊆ [−η, 1 − η]. Assume that K ∩ (K − η) = ∅. Since K, K − η ⊆ [−η, 1], we have
m(K) + m(K − η) = m K ∪ (K − η) ≤ m([−η, 1]) = 1 + η. (9.87) By Theorem 2.20(c), m(K − η) = m(K), so we deduce from the inequalities (9.86) and (9.87) that 1 + η ≥ 2m(K) > 43 and thus η > 13 , a contradiction. Hence we have K ∩ (K − η) 6= ∅. m

For a hint of this proof, please refer to [42, Exercise 8, p. 72] This kind of basis is called a Hamel basis. o Read [22, Exercise 44, p. 64] or [47, p. 66]

n

Chapter 9. Fourier Transforms

312

Let p ∈ K ∩ (K − η). Then p, p + η ∈ K and |p + η − p| = η < δ so that |f (η) − f (0)| = |f (p) + f (η) − f (p)| = |f (p + η) − f (p)| < ǫ.

(9.88)

In other words, f is continuous at 0. Now for any x ∈ [0, 1], the inequality (9.88) implies clearly that |f (x + η) − f (x)| = |f (x) + f (η) − f (x)| = |f (η) − f (0)| < ǫ. Hence f is continuous on [0, 1]. Next, if z ∈ [1, 2], then we have z = x+y for some x, y ∈ [0, 1]. Since f (z) = f (x)+f (y), we have |f (z + η) − f (z)| = |f (x + η) − f (x)| < ǫ and the above paragraph shows that f is also continuous on [1, 2]. Now this kind of argument can be applied repeatedly and we conclude that f is continuous on R. • If graph (f ) is not dense in R2 and satisfies the equation (9.85), then f is continuous. Suppose that f is discontinuous. Then f cannot be of the form f (x) = cx for any constant c. We have
graph (f ) = { x, f (x) | x ∈ R}.

Choose a nonzero real number x1 . Then there exists another nonzero real number x2 such that f (x2 ) f (x1 ) 6= x1 x2 or equivalently, x1 f (x1 ) x2 f (x2 ) 6= 0.

In other words, the vectors u = x1 , f (x1 ) and v = x2 , f (x2 ) are linearly independent and thus span the space R2 . Let p ∈ R2 . Given ǫ > 0. Since Q2 is dense in R2 , one can find q1 , q2 ∈ Q such that |p − q1 u − q2 v| < ǫ. Clearly, we have Therefore, the set


q1 u + q2 v = q1 x1 + q2 x2 , f (q1 x1 + q2 x2 ) .
G = { x, f (x) ∈ R2 | x = q1 x1 + q2 x2 and q1 , q2 ∈ Q}

is dense in R2 . Since G ⊆ graph (f ), graph (f ) is dense in R2 .

• The form of all continuous functions satisfying the equation (9.85). In fact, by a similar argument as in the proof used in [63, Problem 8.6, p. 178], it can be shown that the function f satisfies f (r) = rf (1) for every r ∈ Q. Now for every x ∈ R, since Q is dense in R, we can find a sequence {rn } ⊆ Q such that rn → x as n → ∞. Since f is continuous, we have f (x) = lim f (rn ) = lim rn f (1) = xf (1). n→∞

n→∞

Hence every continuous solution of the equation (9.85) must be in the form f (x) = f (1) · x.

313

9.4. Miscellaneous Problems

We have completed the proof of the problem.



Problem 9.19 Rudin Chapter 9 Exercise 19.

Proof. Let f = χA and g = χB . Since f, g ∈ L1 , their convolution h = f ∗ g is well-defined by Theorem 8.14. Suppose that p and q are conjugate exponents. Direct computation gives Z ∞ Z m(A) 1 dx = √