Applied Functional Analysis, Third Edition [3rd Ed] (Instructor Solution Manual, Solutions) [3, 3e ed.] 9781498761147, 1498761143

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solutions MANUAL FOR Applied Functional Analysis THIRD EDITION

by

J. Tinsley Oden Leszek F. Demkowicz

solutionS MANUAL FOR Applied Functional Analysis THIRD EDITION

by

J. Tinsley Oden Leszek F. Demkowicz

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2018 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20171011 International Standard Book Number-13: 978-1-498-76114-7 (Hardback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http:// www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

1 Preliminaries

Elementary Logic and Set Theory

1.1

Sets and Preliminary Notations, Number Sets

Exercises Exercise 1.1.1 If Z I = {. . . , −2, −1, 0, 1, 2, . . .} denotes the set of all integers and IN = {1, 2, 3, . . .} the set of all natural numbers, exhibit the following sets in the form A = {a, b, c, . . .}: (i) {x ∈ Z I : x2 − 2x + 1 = 0} (ii) {x ∈ Z I : 4 ≤ x ≤ 10} (iii) {x ∈ IN : x2 < 10} (i) {1} (ii) {4, 5, 6, 7, 8, 9, 10} (iii) {1, 2, 3}

1.2

Level One Logic

Exercises Exercise 1.2.1 Construct the truth table for De Morgan’s Law: ∼ (p ∧ q) ⇔ ((∼ p) ∨ (∼ q)) 1

2

APPLIED FUNCTIONAL ANALYSIS

∼ 1 1 1 0

(p ∧ 00 00 10 11

q) 0 1 0 1

⇔ ((∼ 1 1 1 1 1 0 1 0

SOLUTION MANUAL

p) 0 0 1 1

∨ 1 1 1 0

(∼ 1 0 1 0

q)) 0 1 0 1

Exercise 1.2.2 Construct truth tables to prove the following tautologies: (p ⇒ q) ⇔ (∼ q ⇒∼ p) ∼ (p ⇒ q) ⇔ p ∧ ∼ q

(p ⇒ 0 1 0 1 1 0 1 1 ∼ 0 0 1 0

q) 0 1 0 1

(p (0 (0 (1 (1

⇔ (∼ q ⇒ 1 10 1 1 01 1 1 10 0 1 01 1

⇒ q) 1 0) 1 1) 0 0) 1 1)

⇔ 1 1 1 1

p∧ 00 00 11 10

∼ p) 10 10 01 01 ∼ 1 0 1 0

q 0 1 0 1

Exercise 1.2.3 Construct truth tables to prove the associative laws in logic: p ∨ (q ∨ r) ⇔ (p ∨ q) ∨ r p ∧ (q ∧ r) ⇔ (p ∧ q) ∧ r p 0 0 0 0 1 1 1 1

∨ 0 1 1 1 1 1 1 1

(q 0 0 1 1 0 0 1 1

∨ 0 1 1 1 0 1 1 1

r) 0 1 0 1 0 1 0 1

⇔ 1 1 1 1 1 1 1 1

(p 0 0 0 0 1 1 1 1

∨ 0 0 1 1 1 1 1 1

q) 0 0 1 1 0 0 1 1

∨r 00 11 10 11 10 11 10 11

p 0 0 0 0 1 1 1 1

∧ 0 0 0 0 0 0 0 1

(q 0 0 1 1 0 0 1 1

∧ 0 0 0 1 0 0 0 1

r) 0 1 0 1 0 1 0 1

⇔ 1 1 1 1 1 1 1 1

(p 0 0 0 0 1 1 1 1

∧ 0 0 0 0 0 0 1 1

q) 0 0 1 1 0 0 1 1

∧r 00 01 00 01 00 01 00 11

Preliminaries

1.3

3

Algebra of Sets

Exercises Exercise 1.3.1 Of 100 students polled at a certain university, 40 were enrolled in an engineering course, 50 in a mathematics course, and 64 in a physics course. Of these, only 3 were enrolled in all three subjects, 10 were enrolled only in mathematics and engineering, 35 were enrolled only in physics and mathematics, and 18 were enrolled only in engineering and physics. (i) How many students were enrolled only in mathematics? (ii) How many of the students were not enrolled in any of these three subjects? Let A, B, C denote the subsets of students enrolled in mathematics, the engineering course and physics, repectively. Sets: A ∩ B ∩ C, A ∩ B − (A ∩ B ∩ C), A ∩ C − (A ∩ B ∩ C) and A − (B ∪ C) are

pairwise disjoint (no two sets have a nonempty common part) and their union equals set A, see Fig. 1.1. Consequently,

#(A − (B ∪ C)) = #A − #A ∩ B ∩ C − #(A ∩ B − (A ∩ B ∩ C)) − #(A ∩ C − (A ∩ B ∩ C)) = 50 − 3 − 10 − 35 = 2 In the same way we compute, #(B − (A ∪ C)) = 9

and

#(C − (A ∪ B)) = 8

Thus, the total number of students enrolled is #(A − (B ∪ C)) + #(B − (A ∪ C)) + #(C − (A ∪ B)) +#(A ∩ B − C) + #(A ∩ C − B) + #(B ∩ C − A) +#(A ∩ B ∩ C) = 2 + 9 + 8 + 10 + 35 + 18 + 3 = 85 Consequently, 15 students did not enroll in any of the three classes. Exercise 1.3.2 List all of the subsets of A = {1, 2, 3, 4}. Note: A and ∅ are considered to be subsets of A. ∅, {1}, {2}, {1, 2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}, {4}, {1, 4}, {2, 4}, {1, 2, 4}, {3, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}

4

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Figure 1.1 Illustration of Exercise 1.3.1. Exercise 1.3.3 Construct Venn diagrams to illustrate the idempotent, commutative, associative, distributive, and identity laws. Note: some of these are trivially illustrated. This is a very simple exercise. For example, Fig. 1.2 illustrates the associative law for the union of sets.

Figure 1.2 Venn diagrams illustrating the associative law for the union of sets.

Exercise 1.3.4 Construct Venn diagrams to illustrate De Morgan’s Laws. Follow Exercise 1.3.3. Exercise 1.3.5 Prove the distributive laws.

Preliminaries

5

x ∈ A ∩ (B ∪ C) � definition of intersection of sets x ∈ A and x ∈ (B ∪ C) � definition of union of sets x ∈ A and (x ∈ B or x ∈ C) � tautology:p ∧ (q ∨ r) ⇔ (p ∧ q) ∨ (p ∧ r) (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C) � definition of intersection of sets x ∈ (A ∩ B) or x ∈ (A ∩ C) � definition of union of sets x ∈ (A ∩ B) ∪ (A ∩ C) In the same way, x ∈ A ∪ (B ∩ C) � definition of union of sets x ∈ A or x ∈ (B ∩ C) � definition of intersection of sets x ∈ A or (x ∈ B and x ∈ C) � tautology:p ∨ (q ∧ r) ⇔ (p ∨ q) ∧ (p ∨ r) (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C) � definition of union of sets x ∈ (A ∪ B) and x ∈ (A ∪ C) � definition of intersection of sets x ∈ (A ∪ B) ∩ (A ∪ C) Exercise 1.3.6 Prove the identity laws. In each case, one first has to identify and prove the corresponding logical law. For instance, using the truth tables, we first verify that, if f denotes a false statement, then p∨f ⇔p for an arbitrary statement p. This tautology then provides the basis for the corresponding identity law in algebra of sets: A∪∅=A Indeed, x ∈ (A ∪ ∅) � definition of union of sets x ∈ A or x ∈ ∅ � tautology above x∈A The remaining three proofs are analogous. Exercise 1.3.7 Prove the second of De Morgan’s Laws.

6

APPLIED FUNCTIONAL ANALYSIS

x ∈ A − (B ∩ C) � x ∈ A and x ∈ / (B ∩ C) � x ∈ A and ∼ (x ∈ B ∩ C) � x ∈ A and ∼ (x ∈ B ∧ x ∈ C) � (x ∈ A and x ∈ / B) or (x ∈ A and x ∈ / C) � x ∈ (A − B) or x ∈ (A − C) � x ∈ (A − B) ∪ (A − C)

SOLUTION MANUAL

definition of difference of sets x∈ / D ⇔∼ (x ∈ D) definition of intersection tautology: p∧ ∼ (q ∧ r) ⇔ (p∧ ∼ q) ∨ (p∧ ∼ r) definition of difference of sets definition of union

Exercise 1.3.8 Prove that (A − B) ∩ B = ∅. Empty set is a subset of any set, so the inclusion ∅ ⊂ (A − B) ∩ B is obviously satisfied. To prove the

converse, notice that the statement x ∈ ∅ is equivalent to the statement that x does not exist. Suppose now, to the contrary, that there exists an x such that x ∈ (A − B) ∩ B. Then x ∈ (A − B) ∩ B ⇓ x ∈ (A − B) and x ∈ B ⇓ (x ∈ A and x ∈ / B) and x ∈ B ⇓ x ∈ A and (x ∈ / B and x ∈ B) ⇓ x ∈ A and x ∈ ∅ ⇓ x∈∅

definition of intersection definition of difference associative law for conjuction p∧ ∼ p is false identity law for conjuction

In fact, the statements above are equivalent. Exercise 1.3.9 Prove that B − A = B ∩ A� . x∈B−A � definition of difference x ∈ B and x ∈ /A � definition of complement x ∈ B and x ∈ A� � definition of intersection x ∈ B ∩ A�

Preliminaries

1.4

7

Level Two Logic

Exercises Exercise 1.4.1 Use Mathematical Induction to derive and prove a formula for the sum of squares of the first n positive integers:

n �

i2 = 1 + 22 + . . . + n2

i=1

This is an “inverse engineering” problem. Based on elementary integration formulas for polynomials, we expect the formula to take the form: n �

i2 =

i=1

αn3 + βn2 + γn + δ A

In the proof by induction, we will need to show that: n �

i2 + (n + 1)2 =

i=1

This leads to the identity:

n+1 �

i2

i=1

α(n + 1)3 + β(n + 1)2 + γ(n + 1) + δ αn3 + βn2 + γn + δ + (n + 1)3 = A A Comparing coefficients in front of n3 , n2 , n, 1 on both sides, we get relations: A = 3α,

2A = 3α + 2β,

A=α+β+γ

This leads to α = A/3, β = A/2, γ = A/6. Choosing A = 6, we get α = 2, β = 3, γ = 1. Validity of the formula for n = 1 implies that δ = 0. Exercise 1.4.2 Use mathematical induction to prove that the power set of a set U with n elements has 2n elements: #U = n

⇒

#P(U ) = 2n

The hash symbol # replaces the phrase “number of elements of.” • T (1). Let U = {a}. Then P(U ) = {∅, {a}}, so #P(U ) = 2. • T (n) ⇒ T (n + 1). Assume the statement has been proved for every set with n elements. Let #U = n + 1. Pick an arbitrary element a from set U . The power set of U can then be split into two families: subsets that do not contain element a and subsets that do contain a: P(U ) = A ∪ B where

A = {A ⊂ U : a ∈ / A},

B = {B ⊂ U : a ∈ B}

8

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

The two families are disjoint, so #P(U ) = #A + #B. But A = P(U − {a}) so, by the

assumption of mathematical induction, set A has 2n elements. On the other side: B ∈ B ⇔ B − {a} ∈ P(U − {a})

so family B has also exactly 2n elements. Consequently, power set P(U ) has 2n + 2n = 2n+1

elements, and the proof is finished.

Another way to see the result is to recall Newton’s formula: � � � � � � � � n n n n (a + b)n = a n b0 + an−1 b1 + . . . + a1 bn−1 + a 0 bn 0 1 n−1 n In the particular case of a = b = 1, Newton’s formula reduces to the identity: � � � � � � � � n n n n 2n = + + ... + + 0 1 n−1 n Recall that Newton’s symbol

� � n k

represents the number of k–combinations of n elements, i.e., the number of different subsets with k elements from a set with n elements. As all subsets of a set U with n elements can be partitioned into subfamiles of subsets with k elements, k = 0, 1, . . . , n, the right-hand side of the identity above clearly represents the number of all possible subsets of set U . Obviously, in order to prove the formula above, we may have to use mathematical induction as well.

1.5

Infinite Unions and Intersections

Exercises Exercise 1.5.1 Let B(a, r) denote an open ball centered at a with radius r: B(a, r) = {x : d(x, a) < r} Here a, x are points in the Euclidean space and d(x, a) denotes the (Euclidean) distance between the points. Similarly, let B(a, r) denote a closed ball centered at a with radius r: B(a, r) = {x : d(x, a) ≤ r} Notice that the open ball does not include the points on the sphere with radius r, whereas the closed ball does.

Preliminaries

9

Determine the following infinite unions and intersections: �

�

B(a, r),

r 0, we can find an infinite number of k’s such that d(xk , x) < � = 1/l. Proceed by induction. Pick any k1 such that d(xk1 , x) < � = 1. Given k1 , . . . , kl , pick kl+1 �= k1 , . . . , kl (you have an infinite number of indices to pick from ...) such that d(xkl+1 , x) < � = 1/(l + 1). By construction, subsequence xkl → x.

I 2 given by the formula Exercise 1.18.6 Let xk = (xk1 , xk2 ) be a sequence in R xki = (−1)k+i

k+1 , i = 1, 2, k ∈ IN k

Determine the cluster points of the sequence. For k even, the corresponding subsequence converges to (−1, 1). For k odd, the corresponding subsequence converges to (1, −1). These are the only two cluster points of the sequence. Exercise 1.18.7 Calculate lim inf and lim sup of the following sequence in R: I  for n = 3k  n/(n + 3) for n = 3k + 1 an = n2 /(n + 3)  2 n /(n + 3)2 for n = 3k + 2 where k ∈ IN .

Each of the three subsequences is convergent, a3k → 1, a3k+1 → ∞, and a3k+2 → 1. There are only

two cluster points of the sequence, 1 and ∞. Consequently, lim sup an = ∞ and lim inf an = 1. Exercise 1.18.8 Formulate and prove a theorem analogous to Proposition 1.17.2 for limit superior. Follow precisely the reasoning in the text. Exercise 1.18.9 Establish the convergence or divergence of the sequences {xn }, where n2 (b) xn = sin(n) 1 + n2 3n2 + 2 (−1)n n2 (d) x = (c) xn = n 1 + 3n2 1 + n2

(a) xn =

Sequences (a) and (c) converge, (b) and (d) diverge. I be > 1 and let x2 = 2 − 1/x1 , . . . , xn+1 = 2 − 1/xn . Show that this sequence Exercise 1.18.10 Let x1 ∈ R converges and determine its limit. For x1 > 1,

1 x1

< 1, so x2 = 2 −

1 x1

> 1. By the same argument, if xn > 1 then xn+1 > 1. By

induction, xn is bounded below by 1. Also, if x > 1 then 2−

1 ≤x x

44

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Indeed, multiply both sides of the inequality to obtain an equivalent inequality 2x − 1 ≤ x2 which in turn is implied by (x − 1)2 ≥ 0. The sequence is thus decreasing. By the Monotone Sequence

Lemma, it must have a limit x. The value of x may be computed by passing to the limit in the recursive relation. We get x=2−

1 x

which results in x = 1.

1.19

Limits and Continuity

Exercises Exercise 1.19.1 Prove Proposition 1.18.2. (i) ⇒ (ii). Let G be an open set. Let x ∈ f −1 (G). We need to show that x is an interior point of

f −1 (G). Since G is open and f (x) ∈ G, there exists an open ball B(f (x), �) ⊂ G. Continuity of f implies that there exists an open ball B(x, δ) such that

B(x, δ) ⊂ f −1 (B(f (x), �)) ⊂ f −1 (G) This proves that x is an interior point of f −1 (G). (ii) ⇒ (i). Take an open ball B(f (x), �) neighborhood of f (x). The inverse image f −1 (B(f (x), �)) being open implies that there exists a ball B(x, δ) such that

B(x, δ) ⊂ f −1 (B(f (x), �)) or, equivalently, f (B(x, δ)) ⊂ B(f (x), �) which proves the continuity of f at x. (ii) ⇔ (iii). This follows from the duality principle (complement of a set is open iff the set is closed) and the identity

Rm − G) = f −1 (I Rm ) − f −1 (G) = R I n − f −1 (G) f −1 (I I m and g : R Im →R I k . Prove Exercise 1.19.2 Let g ◦ f denote the composition of a function f : R In →R that if f is continuous at x0 and g is continuous at f (x0 ), then g ◦ f is continuous at x0 .

Preliminaries

45

Pick an open ball B(g(f (x0 )), �) neighborhood of g(f (x0 )). By continuity of function g, there exists an open ball B(f (x0 ), δ) neighborhood of f (x0 ) such that g(B(f (x0 ), δ)) ⊂ B(g(f (x0 )), �) In turn, by continuity of f , there exists an open ball B(x, α) neighborhood of x0 such that f (B(x, α)) ⊂ B(f (x0 ), δ) Consequently, (g ◦ f )(B(x, α)) ⊂ g(B(f (x0 ), δ)) ⊂ B(g(f (x0 )), �) which proves the continuity of composition g ◦ f . Exercise 1.19.3 Let f, g : R In →R I m be two continuous functions. Prove that the linear combination of f, g defined as

(αf + βg)(x) = αf (x) + βg(x) is also continuous. Let d denote the Euclidean distance in R I k (k = n, m), and � · � the corresponding Euclidean norm

(comp. Exercise 1.16.8). We have,

d(αf (x) + βg(x), αf (x0 ) + βg(x0 )) = �αf (x) + βg(x) − (αf (x0 ) + βg(x0 ))� = �α(f (x) − f (x0 )) + β(g(x) − g(x0 ))� ≤ |α|�f (x) − f (x0 )� + |β|�g(x) − g(x0 )� = |α|d(f (x), f (x0 )) + |β|d(g(x), g(x0 )) Pick an arbitrary � > 0. Continuity of f implies that there exists δ1 such that d(x, x0 ) < δ1

⇒

d(f (x), f (x0 ))
c. Case x < c is treated analogously. Define a polynomial φ(x) = f (c) +

1 � 1 1 f (c)(x − c) + f �� (c)(x − c)2 + · · · + f (n) (c)(x − c)n + A(x − c)n+1 1! 2! n!

48

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

and select the constant A in such a way that f (x) matches φ(x) at x = b. By Rolle’s Theorem, there exists an intermediate point ξ1 such that (f − φ)� (ξ1 ) = 0 But we have also (f − φ)� (c) = 0 so, by the Rolle’s Theorem again, there exists an intermediate point

ξ1 ∈ (c, ξ1 ) such that,

(f − φ)�� (ξ2 ) = 0

Continuing in this manner, we arrive at the existence of a point ξn ∈ (c, x) such that (f − φ)(n+1) (ξn ) = f (n+1) (ξn ) − A(n + 1)! = 0 Solving for A we get the final result. Exercise 1.20.3 Let f be differentiable on (a, b). Prove the following: (i) If f � (x) = 0 on (a, b), then f (x) = constant on (a, b). Pick arbitrary c, x ∈ (a, b), c �= x. By the Lagrange Mean-Value Theorem, ∃ξ ∈ (c, x) : f (x) − f (c) = f � (ξ)(x − c) = 0 i.e., f (x) = f (c). Since x and c were arbitrary points, the function must be constant. (ii) If f � (x) = g � (x) on (a, b), then f (x) − g(x) = constant. Apply (i) to f (x) − g(x).

(iii) If f � (x) < 0 ∀ x ∈ (a, b) and if x1 < x2 ∈ (a, b), then f (x1 ) > f (x2 ). Apply Lagrange Mean-Value Theorem,

f (x2 ) − f (x1 ) = f � (ξ) < 0 x2 − x1

⇒

f (x2 ) − f (x1 ) < 0

(iv) If |f � (x)| ≤ M < ∞ on (a, b), then |f (x1 ) − f (x2 )| ≤ M |x1 − x2 |

∀ x1 , x2 ∈ (a, b)

Again, by the Lagrange Mean-Value Theorem, f (x1 ) − f (x2 ) = f � (ξ)(x1 − x2 ) for some ξ ∈ (x1 , x2 ). Take absolute value on both sides to obtain |f (x1 ) − f (x2 )| = |f � (ξ)| |x1 − x2 | ≤ M |x1 − x2 | Exercise 1.20.4 Let f and g be continuous on [a, b] and differentiable on (a, b). Prove that there exists a point c ∈ (a, b) such that f � (c)(g(b) − g(a)) = g � (c)(f (b) − f (a)). This result is sometimes called the

Cauchy Mean-Value Theorem.

Preliminaries

49

Hint: Consider the function h(x) = (g(b) − g(a))(f (x) − f (a)) − (g(x) − g(a))(f (b) − f (a)). Repeat the reasoning from the proof of the Lagrange Mean-Value Theorem. We have: h(a) = h(b) = 0. By Rolle’s Theorem, there exists c ∈ (a, b) such that h� (c) = (g(b) − g(a))f � (c) − g � (c)(f (b) − f (a)) = 0 Exercise 1.20.5 Prove L’Hˆospital’s rule: If f (x) and g(x) are differentiable on (a, b), with g � (x) �= 0 ∀x ∈

(a, b), and if f (c) = g(c) = 0 and the limit K = limx→c f � (x)/g � (x) exists, then limx→c f (x)/g(x) =

K. Hint: Use the result of Exercise 1.20.4. According to the Cauchy Mean-value Theorem, there exists ξ ∈ (c, x) such that f � (ξ)(g(x) − g(c)) = g � (ξ)(f (x) − f (c)) or,

f (x) f � (ξ) = � g (ξ) g(x)

With x → c, the intermediate point ξ converges to c as well. As the limit on the left-hand side exists, the right-hand side has a limit as well, and the two limits are equal.

Exercise 1.20.6 Let f and g be Riemann integrable on I = [a, b]. Show that for any real numbers α and β, αf + βg is integrable, and �

b

(αf + βg) dx = α a

�

b

f dx + β a

�

b

g dx a

Let P be an arbitrary partition of I, a = x 0 ≤ x 1 ≤ x 2 ≤ · · · ≤ xn = b and ξk , k = 1, . . . , n arbitrary intermediate points. We have the following simple relation between the Riemann sums of functions f ,g and αf + βg, R(P, αf + βg) =

n �

(αf (ξk ) + βg(ξk ))(xk − xk−1 )

k=1 n �

=α

k=1

f (ξk )(xk − xk−1 ) + β

= αR(P, f ) + βR(P, g)

n �

k=1

g(ξk )(xk − xk−1 )

Thus, if the Riemann sums on the right-hand side converge, the sum on the left-hand side converges as well, and the two limits (integrals) are equal.

50

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Exercise 1.20.7 Let f and g be continuous on [a, b] and suppose that F and G are primitive functions of f and g, respectively, i.e., F � (x) = f (x) and G� (x) = g(x) ∀ x ∈ [a, b]. Prove the integration-by-parts formula:

�

b a

F (x)g(x) dx = F (b)G(b) − F (a)G(a) −

�

b

f (x)G(x) dx a

Integrate between a and b both sides of �

[F (x)G(x)] = f (x)G(x) + F (x)g(x) to obtain F (b)G(b) − F (a)G(a) =

�

b

f (x)G(x) dx + a

�

b

F (x)g(x) dx a

Exercise 1.20.8 Prove that if f is Riemann integrable on [a, c], [c, b], and [a, b], then �

b

f dx = a

�

c

f dx + a

�

b

f dx,

aV ∗ ×V Thus, ∗ δij =< e∗∗ i , ej >V ∗∗ ×V ∗

and

< ι(ei ), e∗j >V ∗∗ ×V ∗ =< e∗j , ei >V ∗ ×V = δji = δij

The relation follows then form the uniqueness of the (bi)dual basis.

68

2.11

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Transpose of a Linear Transformation

Exercises Exercise 2.11.1 The following is a “sanity check” of your understanding of concepts discussed in the last two sections. Consider R I 2. I 2. (a) Prove that a1 = (1, 0), a2 = (1, 1) is a basis in R It is sufficient to show linear independence. Any n linearly independent vectors in a n-dimensional vector space provide a basis for the space. The vectors are clearly not collinear, so they are linearly independent. Formally, α1 a1 + α2 a2 = (α1 + α2 , α2 ) = (0, 0) implies α1 = α2 = 0, so the vectors are linearly independent. I f (x1 , x2 ) = 2x1 + 3x2 . Prove that the functional is linear, (b) Consider a functional f : R I 2 → R, and determine its components in the dual basis a∗1 , a∗2 .

Linearity is trivial. Dual basis functionals return components with respect to the original basis, a∗j (ξ1 a1 + ξ2 a2 ) = ξj It is, therefore, sufficient to determine ξ1 , ξ2 . We have, ξ1 a1 + ξ2 a2 = ξ1 e1 + ξ2 (e1 + e2 ) = (ξ1 + ξ2 )e1 + ξ2 e2 so x1 = ξ1 + ξ2 and x2 = ξ2 . Inverting, we get, ξ1 = x1 − x2 , ξ2 = x2 . These are the dual basis functionals. Consequently,

f (x1 , x2 ) = 2x1 + 3x2 = 2(ξ1 + ξ2 ) + 3ξ2 = 2ξ1 + 5ξ2 = (2a∗1 + 5a∗2 )(x1 , x2 ) Using the argumentless notation, f = 2a∗1 + 5a∗2 If you are not interested in the form of the dual basis functionals, you get compute the components of f with respect to the dual basis faster. Assume α1 a∗1 + α2 a∗2 = f . Evaluating both sides at x = a1 we get, (α1 a∗1 + α2 a∗2 )(a1 ) = α1 = f (a1 ) = f (1, 0) = 2 Similarly, evaluating at x = a2 , we get α2 = 5. I 2 whose matrix representation in basis a1 , a2 is (c) Consider a linear map A : R I 2 →R � � 10 12

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69

Compute the matrix representation of the transpose operator with respect to the dual basis. Nothing to compute. Matrix representation of the transpose operator with respect to the dual basis is equal of the transpose of the original matrix, � � 11 02 Exercise 2.11.2 Prove Proposition 2.11.3. All five properties of the matrices are directly related to the properties of linear transformations discussed in Proposition 2.11.1 and Proposition 2.11.2. They can also be easily verified directly. (i) (αAij + βBij )T = αAji + βBji = α(Aij )T + β(βij )T (ii)

�

n � l=1

(iii) (δij )T = δji = δij .

Bil Alj

�T

=

n �

Bjl Ali =

l=1

n �

Ali Bjl =

l=1

n �

(Ail )T (Blj )T

l=1

(iv) Follow the reasoning for linear transformations: AA−1 = I

⇒

(A−1 )T AT = I T = I

A−1 A = I

⇒

AT (A−1 )T = I T = I

Consequently, matrix AT is invertible, and (AT )−1 = (A−1 )T . (v) Conclude this from Proposition 2.11.2. Given a matrix Aij , ij, = 1, . . . , n, we can interpret it as I n defined as: the matrix representation of map A : R I n →R where

y = Ax

yi =

n �

Aij xj

j=1

with respect to the canonical basis ei , i = 1, . . . , n. The transpose matrix AT can then be interpreted as the matrix of the transpose transformation: AT : (I Rn )∗ → (I R n )∗ The conclusion follows then from the facts that rank A = rank A, rank AT = rank AT , and Proposition 2.11.2. Exercise 2.11.3 Construct an example of square matrices A and B such that (a) AB �= BA �

10 A= 11 Then AB =

�

11 12

�

�

�

11 B= 01

and

BA =

�

�

21 11

�

70

APPLIED FUNCTIONAL ANALYSIS

(b) AB = 0, but neither A = 0 nor B = 0 � � 10 A= 00

SOLUTION MANUAL

�

00 B= 01

�

(c) AB = AC, but B �= C

Take A, B from (b) and C = 0.

Exercise 2.11.4 If A = [Aij ] is an m × n rectangular matrix and its transpose AT is the n × m matrix, ATn×m = [Aji ]. Prove that (i) (AT )T = A. �

(Aij )T

�T

= (Aji )T = Aij

(ii) (A + B)T = AT + B T . Particular case of Proposition 2.11.3(i). (iii) (ABC · · · XY Z)T = Z T Y T X T · · · C T B T AT . Use Proposition 2.11.3(ii) and recursion,

(ABC . . . XY Z)T = (BC . . . XY Z)T AT = (C . . . XY Z)T B T AT .. . = Z T Y T X T . . . C T B T AT (iv) (qA)T = qAT . Particular case of Proposition 2.11.3(i). Exercise 2.11.5 In this exercise, we develop a classical formula for the inverse of a square matrix. Let A = [aij ] be a matrix of order n. We define the cofactor Aij of the element aij of the i-th column of A as the determinant of the matrix obtained by deleting the i-th row and j-th column of A, multiplied by (−1)i+j :

Aij = cofactor aij

(a) Show that

� � a11 � � � � def i+j � ai−1,1 = (−1) � � ai+1,1 � � � an1 δij det A =

n �

k=1

a12 ai−1,2 ai+1,2 an2

· · · a1,j−1 ··· · · · ai−1,j−1 · · · ai+1,j−1 ··· · · · an,j−1

aik Ajk ,

a1,j+1 ··· ai−1,j+1 ai+1,j+1 ··· an,j+1

1 ≤ i, j ≤ n

� � � � � · · · ai−1,n �� · · · ai+1,n �� � � · · · ann � · · · a1n

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71

where δij is the Kronecker delta. Hint: Compare Exercise 2.13.4. For i = j, the formula reduces to the Laplace Expansion Formula for determinants discussed in Exercise 2.13.4. For i �= j, the right-hand side represents the Laplace expansion of the determi-

nant of an array where two rows are identical. Antilinearity of determinant (comp. Section 2.13) implies then that the value must be zero.

(b) Using the result in (a), conclude that A−1 =

1 [Aij ]T det A

Divide both sides by det A. (c) Use (b) to compute the inverse of



and verify your answer by showing that

 1 22 A =  1 −1 0  2 13

A−1 A = AA−1 = I

A

Exercise 2.11.6 Consider the matrices � � 1 041 A= , 2 −1 3 0 and

−1



1

4 3

 −1 4 B =  12 0  , 01

If possible, compute the following:



   1 1 −2    3 3 =    −1 −1 1 

� � 2 D= , 3

− 23



1  −1 E=  1 0

C = [1, −1, 4, −3] 0 2 4 0 0 2 1 −1

 3 1  4 2

(a) AAT + 4D T D + E T The expression is ill-defined, AAT ∈ Matr(2, 2) and E T ∈ Matr(4, 4), so the two matrices cannot be added to each other.

(b) C T C + E − E 2 

−1  −7  = −2 −1

 −4 3 −17 −12 −1 1  −8 16 −27  −2 −9 10

72

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

(c) B T D Ill-defined, mismatched dimensions. (d) B T BD − D =

�

276 36

�

(e) EC − AT A

EC is not computable.

(f) AT DC(E − 2I)



 32 −40 40 144  −12 15 −15 −54   =  68 −85 85 306  8 −10 10 36

Exercise 2.11.7 Do the following vectors provide a basis for R I 4? a = (1, 0, −1, 1),

b = (0, 1, 0, 22)

c = (3, 3, −3, 9),

d = (0, 0, 0, 1)

It is sufficient to check linear independence, αa + βb + γc + δd = 0

?

⇒

α=β=γ=δ=0

Computing αa + βb + γc + δd = (α + 3γ, β + 3γ, −α − 3γ, α + 22β + 9γ + δ) we arrive at the homogeneous system of equations   1 0 30  0 1 3 0    −1 0 −3 0  1 22 9 1

    α 0 β  0  =  γ  0 0 δ

The system has a nontrivial solution iff the matrix is singular, i.e., det A = 0. By inspection, the third

row equals minus the first one, so the determinant is zero. Vectors a, b, c, d are linearly dependent and, therefore, do not provide a basis for R I 4. Exercise 2.11.8 Evaluate the determinant of the matrix   1 −1 0 4 1 0 2 1  A=  4 7 1 −1  1 01 2

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73

Use e.g. the Laplace expansion with respect to the last row and Sarrus’ formulas, � � � � � � 1 −1 0 4 � � � � �1 0 4� � −1 −1 0 � � � � −1 0 4 � � � � � � � �1 0 2 1� � � � � � � � � � 4 7 1 −1 � = −(−1) � 0 2 1 � − 1 � 1 2 1 � + 2 � 1 0 2 � � � � 7 1 −1 � � 4 1 −1 � � 4 7 1� �1 0 1 2�

= 2 − 56 + 1 − (−2 + 4 − 32 − 1) + 2(−8 − 14 + 1) = −53 − (−31) − 42 = −64

Exercise 2.11.9 Invert the following matrices (see Exercise 2.11.5).  � � 42 1 −1 A= , B = 2 4 1 2 12

−1

A



 =

2 1 3 3

− 13



 1

3

B

−1



4 12

 1 2 2

2 − 12

0



  − 2 6  7 −   12 12 12 =    6 0 − 12 1

Exercise 2.11.10 Prove that if A is symmetric and nonsingular, so is A−1 . Use Proposition 2.11.3(iv). (A−1 )T = (AT )−1 = A−1 Exercise 2.11.11 Prove that if A, B, C, and D are nonsingular matrices of the same order then (ABCD)−1 = D −1 C −1 B −1 A−1 Use the fact that matrix product is associative, (ABCD)(D −1 C −1 B −1 A−1 ) = ABC(DD −1 )C −1 B −1 A−1 = ABC I C −1 B −1 A−1 = . . . = I In the same way, (D −1 C −1 B −1 A−1 )(ABCD) = I So, (ABCD)−1 = D −1 C −1 B −1 A−1 . Exercise 2.11.12 Consider the linear problem  (i) Determine the rank of T .

 0 1 3 −2 T =  2 1 −4 3  , 2 3 2 −1

  1 y = 5 7

74

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Multiplication of columns (rows) with non-zero factor, addition of columns (rows), and interchange of columns (rows), do not change the rank of a matrix. We may use those operations and mimic Gaussian elimination to compute the rank of matrices.   0 1 3 −2 rank  2 1 −4 3  2 3 2 −1   1 3 −2 0 = rank  1 −4 3 2  switch columns 1 and 4 3 2 −1 2   1 3 −2 0 = rank  1 −4 3 2  divide row 3 by 3 1 23 − 13 32   1 3 −2 0 subtract row 1 from rows 2 and 3 = rank  0 −7 5 2  0 − 73 35 32   1 0 0 0 = rank  0 −7 5 2  manipulate the same way columns to zero out the first row 0 − 73 35 32   10 0 0 = rank  0 1 − 57 − 27  00 0 0   1000 = rank  0 1 0 0  0000 =2

(ii) Determine the null space of T . Set x3 = α and x4 = β and solve for x1 , x2 to obtain 7 5 N (T ) = {( α − β, −3α + 2β, α, β)T : α, β ∈ R} I 2 2 (iii) Obtain a particular solution and the general solution. Check that the rank of the augmented matrix is also equal 2. Set x3 = x4 = 0 to obtain a particular solution x = (2, 1, 0, 0)T The general solution is then 7 5 x = (2 + α − β, 1 − 3α + 2β, α, β)T , α, β ∈ R I 2 2 (iv) Determine the range space of T . As rank T = 2, we know that the range of T is two-dimensional. It is sufficient thus to find two linearly independent vectors that are in the range, e.g. we can take T e1 , T e2 represented by the first two columns of the matrix, R(T ) = span{(0, 2, 2)T , (1, 1, 3)T }

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75

Exercise 2.11.13 Construct examples of linear systems of equations having (1) no solutions, (2) infinitely many solutions, (3) if possible, unique solutions for the following cases: (a) 3 equations, 4 unknowns (1)



 1000 T = 1 0 0 0 0111

(2)



 1111 T = 1 1 1 1 1111

(3) Unique solution is not possible.

  0 y = 1 0

  0 y = 0 0

(b) 3 equations, 3 unknowns (1)



 100 T = 1 0 0 011

(2)



 100 T = 2 0 0 111

(3)



 100 T = 0 1 0 001

Exercise 2.11.14 Determine the rank of the following matrices:    21 4 7 12134 0 1 2 1 , T2 = 2 0 3 2 1 T = 2 2 6 8 11121 4 4 14 10

  0 y = 1 0

  1 y = 2 1

  1 y = 1 1

   4 2 −1 1 5, T3 = 2 0 1 3 0 11

In all three cases, the rank is equal 3.

Exercise 2.11.15 Solve, if possible, the following systems: (a) + 3x3 − x4 + 2x5 = 2

4x1

x 1 − x2 + x3 − x4 + x5 = 1 x 1 + x2 + x3 − x4 + x5 = 1 x1 + 2x2 + x3 

 t+3   0    x =  −2t − 3  ,  −1  t

+ x5 = 0

t ∈R I

76

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

(b) − 4x1 − 8x2 + 5x3 = 1 − 2x2

2x1

+ x2

5x1

+ 3x3

= 2

+ 2x3

= 4

�

19 47 3 x= ,− , 120 120 10

�T

(c) 2x1 + 3x2 + 4x3 + 3x4 = 0 x1 + 2x2 + 3x3 + 2x4 = 0 x 1 + x2 + x3 + x4 = 0 

 α+β  −2α − β  , x=   α β

α, β ∈ R I

2.12

Tensor Products, Covariant and Contravariant Tensors

2.13

Elements of Multilinear Algebra

Exercises Exercise 2.13.1 Let X be a finite-dimensional space of dimension n. Prove that the dimension of the space s Mm (X) of all m-linear symmetric functionals defined on X is given by the formula � � (n + m − 1)! n(n + 1) . . . (n + m − 1) n+m−1 s dim Mm = = (X) = m 1 · 2 · ... · m m! (n − 1)!

Proceed along the following steps. (a) Let Pi,m denote the number of increasing sequences of m natural numbers ending with i, 1 ≤ a 1 ≤ a2 ≤ . . . ≤ a m = i Argue that s dim Mm (X) =

n � i=1

Pi,m

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77

Let a be a general m-linear functional defined on X. Let e1 , . . . , en be a basis for X, and let v j , j = 1, . . . , n, denote components of a vector v with respect to the basis. The multilinearity of a implies the representation formula, a(v1 , . . . , vm ) =

n � n �

...

j1 =1 j2 =1

n �

jm a(ej1 , ej2 , . . . , ejm ) v1j1 v2j2 . . . vm

jm =1

On the other side, if the form a is symmetric, we can interchange any two arguments in the coefficient a(ej1 , ej2 , . . . , ejm ) without changing the value. The form is thus determined by coefficients a(ej1 , ej2 , . . . , ejm ) where 1 ≤ j 1 ≤ . . . ≤ jm ≤ n s The number of such increasing sequences equals the dimension of space Mm (X). Obviously, we

can partition the set of such sequences into subsets that contain sequences ending at particular indices 1, 2, . . . , n, from which the identity above follows. (b) Argue that Pi,m+1 =

i �

Pj,m

j=1

The first m elements of an increasing sequence of m + 1 integers ending at i, form an increasing sequence of m integers ending at j ≤ i. (c) Use the identity above and mathematical induction to prove that Pi,m =

i(i + 1) . . . (i + m − 2) (m − 1)!

For m = 1, Pi,1 = 1. For m = 2, Pi,2 =

i �

1=i

j=1

For m = 3, Pi,3 =

i �

j=

j=1

i(i + 1) 2

Assume the formula is true for a particular m. Then Pi,m+1 =

i � j(j + 1) . . . (j + m − 2) j=1

(m − 1)!

We shall use induction in i to prove that i � j(j + 1) . . . (j + m − 2) j=1

(m − 1)!

=

i(i + 1) . . . (i + m − 1) m!

78

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

The case i = 1 is obvious. Suppose the formula is true for a particular value of i. Then, i+1 � j(j + 1) . . . (j + m − 2) j=1

(m − 1)!

i � j(j + 1) . . . (j + m − 2)

(i + 1)(i + 2) . . . (i + m − 1) (m − 1)! (m − 1)! j=1 i(i + 1) . . . (i + m − 1) m(i + 1)(i + 2) . . . (i + m − 1) + = m! m! (i + 1)(i + 2) . . . (i + m − 1)(i + m) = m! (i + 1)(i + 2) . . . (i + m − 1)((i + 1) + m − 1) = m! =

+

(d) Conclude the final formula. Just use the formula above. Exercise 2.13.2 Prove that any bilinear functional can be decomposed into a unique way into the sum of a symmetric functional and an antisymmetric functional. In other words, M2 (V ) = M2s (V ) ⊕ M2a (V ) Does this result hold for a general m-linear functional with m > 2 ? The result follows from the simple decomposition, a(u, v) =

1 1 (a(u, v) + a(v, u)) + (a(u, v) − a(v, u)) 2 2

Unfortunately, it does not generalize to m > 2. This can for instance be seen from the simple comparison of dimensions of the involved spaces in the finite-dimensional case, � � � � n+m−1 n m + n > m m for 2 < m ≤ n. Exercise 2.13.3 Antisymmetric linear functionals are a great tool to check for linear independence of vectors. Let a be an m-linear antisymmetric functional defined on a vector space V . Let v1 , . . . , vm be m vectors in space V such that a(v1 , . . . , vm ) �= 0. Prove that vectors v1 , . . . , vn are linearly independent.

Is the converse true? In other words, if vectors v1 , . . . , vn are linearly independent, and a is a nontrivial m-linear antisymmetric form, is a(v1 , . . . , vm ) �= 0? Assume in contrary that there exists an index i such that vi =

�

βj vj

j�=i

for some constants βj , j �= i. Substituting into the functional a, we get, a(v1 , . . . , vi , . . . , vm ) = a(v1 , . . . ,

� j�=i

βj vj , . . . , v m ) =

� j�=i

βj a(v1 , . . . , vj , . . . , vm ) = 0

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79

since in each of the terms a(v1 , . . . , vj , . . . , vm ), two arguments are the same. The converse is not true. Consider for instance a bilinear, antisymmetric form defined on a threedimensional space. Let e1 , e2 , e3 be a basis for the space. As discussed in the text, the form is uniquely determined by its values on pairs of basis vectors: a(e1 , e2 ), a(e1 , e3 ), a(e2 , e3 ). It is sufficient for one of these numbers to be non-zero in order to have a nontrivial form. Thus we may have a(e1 , e2 ) = 0 for the linearly independent vectors e1 , e2 , and a nontrivial form a. The discussed criterion is only a sufficient condition for the linear independence but not necessary. Exercise 2.13.4 Use the fact that the determinant of matrix A is a multilinear antisymmetric functional of matrix columns and rows to prove the Laplace Expansion Formula. Select a particular column of matrix Aij , say the j-th column. Let Aij denote the submatrix of A obtained by removing i-th row and j-th column (do not confuse it with a matrix representation). Prove that det A =

n �

(−1)i+j Aij det Aij

i=1

Formulate and prove an analogous expansion formula with respect to an i-th row. It follows from the linearity of the determinant with respect to the j-th column that,       ... 1 ... ... 0 ... . . . A1j . . .       det  ... ... ...  = A1j det  ... ... ...  + . . . + Anj det  ... ... ...  ... 0 ...

. . . Anj . . .

... 1 ...

On the other side, the determinant of matrix, 

 (j)  ... 0 ...    . .   (i) .. 1 ..  ... 0 ...

is a multilinear functional of the remaining columns (and rows) and, for Aij = I (The I denote here the identity matrix in R I n−1 ), its value reduces to (−1)i+j . Hence,   (j)  ... 0 ...   i+j i+j det  . .  = (−1) det A  (i) .. 1 ..  ... 0 ...

The reasoning follows identical lines for the expansion with respect to the i-th column.

Exercise 2.13.5 Prove the Kramer’s formulas for the solution of a nonsingular system of n equations with n unknowns,



a11  ..  .

... .. .

    x1 b1 a1n ..   ..  =  ..  .  .   . 

an1 . . . ann

xn

bn

80

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Hint: In order to develop the formula for the j-th unknown, rewrite the system in the form:



a11  ..  .  an1

 1 . . . x1 . . . a1n .. ..   ..  . .   . 0 . . . xn . . . . . . ann ... .. .

(j)





 a11 . . . b1 . . . a1n   ..  .. = .  .    1 an1 . . . bn . . . ann 0

(j)

Compute the determinant of both sides of the identity, and use Cauchy’s Theorem for Determinants for the left-hand side.

Exercise 2.13.6 Explain why the rank of a (not necessarily square) matrix is equal to the maximum size of a square submatrix with a non-zero determinant.

Consider an m × n matrix Aij . The matrix can be considered to be a representation of a linear map A from an n-dimensional space X with a basis ei , i = 1, . . . , n, into an m-dimensional space Y with a

basis g1 , . . . , gm . The transpose of the matrix represents the transpose operator AT mapping dual space ∗ and e∗1 , . . . , e∗n . The rank of the Y ∗ into the dual space X ∗ , with respect to the dual bases g1∗ , . . . , gm

matrix is equal to the dimension of the range space of operator A and operator AT . Let ej1 , . . . , ejk be such vectors that Aej1 , . . . , Aejk is the basis for the range of operator A. The corresponding submatrix represents a restriction B of operator A to a subspace X0 = span(ej1 , . . . , ejk ) and has the same rank as the original whole matrix. Its transpose has the same rank equal to k. By the same argument, there exist k vectors gi1 , . . . , gik such that AT gi∗1 , . . . , AT gi∗k are linearly independent. The corresponding k × k submatrix represents the restriction of the transpose operator to the k-dimensional subspace

Y0∗ = span(gi∗1 , . . . , gi∗k ), with values in the dual space X0∗ , and has the same rank equal k. Thus,

the final submatrix represents an isomorphism from a k-dimensional space into a k-dimensional space and, consequently, must have a non-zero determinant.

Conversely, let v 1 , . . . , v m be k column vectors in R I m . Consider a matrix composed of the columns. If there exists a square submatrix of the matrix with a non-zero determinant, the vectors must be linearly independent. Indeed, the determinant of any square submatrix of the matrix represents a klinear, antisymmetric functional of the column vectors, so, by Exercise 2.13.3, v 1 , . . . , v k are linearly independent vectors. The same argument applies to the rows of the matrix.

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Euclidean Spaces

2.14

Scalar (Inner) Product. Representation Theorem in Finite-Dimensional Spaces

2.15

Basis and Cobasis. Adjoint of a Transformation. Contra- and Covariant Components of Tensors

Exercises Exercise 2.15.1 Go back to Exercise 2.11.1 and consider the following product in R I 2, R I 2 ×R I 2 � (x, y) → (x, y)V = x1 y1 + 2x2 y2 Prove that (x, y)V satisfies the axioms for an inner product. Determine the adjoint of map A from Exercise 2.11.1 with respect to this inner product. The product is bilinear, symmetric and positive definite, since (x, x)V = x21 +2x22 ≥ 0, and x21 +2x22 =

0 implies x1 = x2 = 0. The easiest way to determine a matrix representation of A∗ is to determine the

cobasis of the (canonical) basis used to define the map A. Assume that a1 = (α, β). Then (a1 , a1 ) = α = 1 (a2 , a1 ) = α + 2β = 0

=⇒

β = − 12

so a1 = (1, − 12 ). Similarly, if a2 = (α, β) then, (a1 , a2 ) = α = 0 (a2 , a2 ) = α + 2β = 1

=⇒

β=

1 2

so a2 = (0, 12 ). The matrix representation of A∗ in the cobasis is simply the transpose of the original matrix,

�

11 02

�

In order to represent A∗ in the original, canonical basis, we need to switch in between the bases. a1 = e1 − 12 e2 a2 = 12 e2

=⇒

e 1 = a1 + a 2 e2 = 2a2

82

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Then, A∗ y = A∗ (y1 e1 + y2 e2 ) = A∗ (y1 (a1 + a2 ) + y2 2a2 ) = A∗ (y1 a1 + (y1 + 2y2 )a2 ) = y1 A∗ a1 + (y1 + 2y2 )A∗ a2 = y1 a1 + (y1 + 2y2 )(a1 + 2a2 ) = y1 (e1 − 12 e2 ) + (y1 + 2y2 )(e1 + 12 e2 ) = 2(y1 + y2 )e1 − 12 y1 e2 = (2(y1 + y2 ), y2 ) Now, let us check our calculations. First, let us compute the original map (that has been given to us in basis a1 , a2 ), in the canonical basis, A(x1 , x2 ) = A(x1 e1 + x2 e2 ) = A(x1 a1 + x2 (a2 − a1 )) = A((x1 − x2 )a1 + x2 a2 ) = (x1 − x2 )(a1 + a2 ) + x2 2a2 = (x1 − x2 )(2e1 + e2 ) + 2x2 (e1 + e2 ) = (2x1 , x1 + x2 ) If our calculations are correct then, (Ax, y)V = 2x1 y1 + 2(x1 + x2 )y2 must match (x, A∗ y)V = x1 2(y1 + y2 ) + 2x2 y2 which it does! Needless to say, you can solve this problem in many other ways.

3 Lebesgue Measure and Integration

Lebesgue Measure

3.1

Elementary Abstract Measure Theory

Exercises Exercise 3.1.1 Prove Proposition 3.1.2. Let Sι ⊂ P(X) be a family of σ-algebras. Prove that the common part

�

ι

Sι is a σ-algebra as well.

The result is a simple consequence of commutativity of universal quantifiers. For any open statement P (x, y), we have, P (x, y)

∀x ∀y ⇐⇒ P (x, y)

∀y ∀x

For finite sets of indices, the property follows by induction from level one logic axioms. The property serves then as a motivation for assuming the level two logic axiom above. The specific arguments are very simple and look as follows. � � 1. ∅, X ∈ Sι , ∀ι implies ∅, X ∈ ι Sι , so ι Sι is nonempty. � � 2. Let A ∈ ι Sι . Then A ∈ Sι , ∀ι and, consequently, A� ∈ Sι , ∀ι, which implies that A� ∈ ι Sι . � �∞ 3. Ai ∈ ι Sι , i = 1, 2, . . ., implies Ai ∈ Sι , ∀ι, i = 1, 2, . . ., which in turn implies i=1 Ai ∈ �∞ � Sι , ∀ι, and, consequently, i=1 Ai ∈ ι Sι .

Exercise 3.1.2 Let f : X → Y be a function. Prove the following properties of the inverse image: f −1 (Y − B) = X − f −1 (B) and f −1 (

∞ �

i=1

for arbitrary B, Bi ⊂ Y .

Bi ) =

∞ �

f −1 (Bi )

i=1

83

84

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Both proofs are very straightforward. x ∈ f −1 (Y − B)

⇔

f (x) ∈ /B

⇔

∼ (f (x) ∈ B)

⇔

x∈ / f −1 (B)

and x ∈ f −1 (

∞ �

i=1

Bi ) ⇔ f (x) ∈

∞ �

i=1

Bi ⇔ ∃i f (x) ∈ Bi ⇔ ∃i x ∈ f −1 (Bi ) ⇔ x ∈

∞ �

f −1 (Bi )

i=1

Exercise 3.1.3 Construct an example of f : X → Y and a σ-algebra in X such that f (S) is not a σ-algebra in Y .

Take any X, Y and any non-surjective function f : X → Y . Take the trivial σ-algebra in X, S = {∅, X} Then, in particular, f (S) = {∅, f (X)} does not contain Y = ∅� and, therefore, it violates the first axiom for σ-algebras. Exercise 3.1.4 Prove Corollary 3.1.1. Let f : X → Y be a bijection and S ⊂ P(X) a σ-algebra. Prove that: (i) f (S) := {f (A) : A ∈ S} is a σ-algebra in Y . (ii) If a set K generates S in X then f (K) generates f (S) in Y . Solutions: (i) Let g = f −1 be the inverse function of f . Then f (S) = g −1 (S) and the property follows from Proposition 3.1.3. Equivalently, f (S) = R defined in the same Proposition. (ii) Let f : X → Y be an arbitrary function. Let S = S(K) be the σ-algebra generated by a family K in Y . Then, by Proposition 3.1.3, f −1 (S) is a σ-algebra in X and it contains f −1 (S).

Consequently, it must contain the smallest σ-algebra containing f −1 (S), i.e. f −1 (S(K)) ⊃ S(f −1 (K)) Applying the result to the inverse function g = f −1 , we get f (S(K)) ⊃ S(f (K)) Applying the last result to g = f −1 and set L = f (K) ⊂ Y gives: S(g(L)) ⊂ g(S(L)) which, after applying f to both sides, leads to the inverse inclusion, f (S(g(L)) ⊂ S(���� L ) ���� K

f (K)

Lebesgue Measure and Integration

85

Exercise 3.1.5 Prove the details from the proof of Proposition 3.1.5. Let G ⊂ R I n be an open set. Prove that the family

{F ⊂ R I m : G × F ∈ B(I Rn × R I m )} is a σ-algebra in R I m. I n ×R I m . Thus, the family contains all For any open set F ⊂ R I m , Cartesian product G × F is open in R

open sets and, therefore, is not empty. The properties of σ-algebra follow then from simple identities, G × F � = G × (I Rm − F ) = G × R Im −G×F � � G× Fn = (G × Fn ) n

n

Exercise 3.1.6 Let X be a set, S ⊂ PX a σ-algebra of sets in X, and y a specific element of X. Prove that function

µ(A) := is a measure on S.

�

if y ∈ A otherwise

1 0

�∞ Obviously, µ ≡ / ∞. Let An ∈ S, i = 1, 2, . . ., be a family of pairwise disjoint sets. If y ∈ / n=1 An �∞ �∞ �∞ then y ∈ / An , ∀n, and, therefore, 0 = µ( n=1 An ) = n=1 0. If y ∈ n=1 An then there must be exactly one m such that y ∈ Am . Consequently, 1 = µ(

∞ �

An ) = µ(Am ) =

n=1

3.2

∞ �

µ(An )

n=1

Construction of Lebesgue Measure in R In

Exercises Exercise 3.2.1 Let F1 , F2 ∈ R I n be two disjoint closed sets, not necessarily bounded. Construct open sets G1 , G2 such that

Fi ⊂ Gi , i = 1, 2

and

G 1 ∩ G2 = ∅

Obviously, F1 ⊂ F2� . Since F2� is open, for every x ∈ F1 , there exists a ball B(x, �x ) ⊂ F2� . Define G1 =

�

x∈F1

B(x,

�x ) 2

Being a union of opens sets, G1 is open. Construct G2 in the same way. We claim that G1 , G2 are disjoint. Indeed, let y ∈ G1 ∩ G2 . It follows then from the construction that there exist x1 ∈ F1 and

86

APPLIED FUNCTIONAL ANALYSIS

x2 ∈ F2 such that y ∈ B(xi ,

�x i 2

SOLUTION MANUAL

), i = 1, 2. Without any loss of generality, assume that �x1 ≥ �x2 . Let

d be the Euclidean metric. Then,

d(x1 , x2 ) ≤ d(x1 , y) + d(x2 , y) ≤

�x � x1 + 2 ≤ �x 1 2 2

which implies that x2 ∈ B(x1 , �x1 ). Since x2 ∈ F2 , this contradicts the construction of G1 . Exercise 3.2.2 Complete proof of Proposition 3.2.4. Prove that the following families of sets coincide with Lebesgue measurable sets in R I n. (ii) {J ∪ Z : J is Fσ -type, m∗ (Z) = 0}, (iii) S(B(I Rn ) ∪ {Z : m∗ (Z) = 0}). Let E be a Lebesgue measurable set. According to Proposition 3.2.3, for every i, there exists a closed subset Fi of E such that m∗ (E − Fi ) ≤ Define H =

�∞

i=1

Fi ⊂ E. As the sets E −

�n

i=1

Fi form an increasing sequence, we have,

m∗ (E − H) = m(E − H) = lim m(E − n→∞

1 i

n �

i=1

Fi ) ≤ lim m(E − Fn ) = 0 n→∞

Consequently, Z := E − H is of measure zero, and E = H ∪ Z. Conversely, from the fact that Fi , Z �∞ Rn ) is a σ-algebra are Lebesgue measurable, it follows that i=1 Fi ∪Z must be measurable as well (L(I containing open sets).

Since L(I Rn ) is a σ-algebra that contains both Borel sets and sets of measure zero, it must also contain

the smallest σ-algebra containing the two families. Conversely, representation (ii) above proves that every Lebesgue measurable set belongs to S(B(I Rn ) ∪ {Z : m∗ (Z) = 0}).

3.3

The Fundamental Characterization of Lebesgue Measure

Exercises Exercise 3.3.1 Follow the outlined steps to prove that every linear isomorphism g : R In →R I n is a compoλ . sition of simple isomorphisms gH,c

Lebesgue Measure and Integration

87

/ H. Show Step 1: Let H be a hyperplane in R I n , and let a, b denote two vectors such that a, b, a − b ∈ λ that there exists a unique simple isomorphism gH,c such that λ (a) = b gH,c

Hint: Use c = b − a.

Indeed, consider the decompositions a = a0 + α(b − a),

a0 ∈ S, α ∈ R, I

b = b0 + β(b − a),

b0 ∈ S, β ∈ R, I

Subtracting the two representations from each other, we get b − a = b0 − a0 + (β − α)(b − a) which implies b0 − a0 + (β − α − 1)(b − a) = 0 It follows from the assumption b − a ∈ / H that the two terms are linearly independent. Conse/ S implies that α, β �= 0. Therefore, with λ = β/α, quently, a0 = b0 . Also, assumption a, b ∈

we have

λ λ gH,c (a) = gH,c (a0 + α(b − a)) = b0 + β(b − a) = b

Step 2: Let g be a linear isomorphism in R I n and consider the subspace Y = Y (g) such that g(x) = x on Y . Assume that Y �= R I n . Let H be any hyperplane containing Y . Show that there exist vectors a, b such that

g(a) ∈ /H and b∈ / H,

b − g(a) ∈ / H,

b−a∈ /H

Make use then of the Step 1 result and consider simple isomorphisms g1 and h1 invariant on H and mapping f (a) into b and b into a, respectively. Prove that dimY (h1 ◦ g1 ◦ g) > dimY (g) From the fact that g is an isomorphism follows that there must exist a vector a �= 0 for which g(a) ∈ / H. Denote c = g(a) ∈ / H and consider the corresponding direct sum decomposition R I n = H ⊕ Rc I Let b be an arbitrary vector. Consider decompositions of vectors a and b corresponding to the direct sum above a = a0 + αc,

b = b0 + βc,

a0 , b0 ∈ H, α, β ∈ R I

88

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Then b − g(a) = b − c = b0 + (β − 1)c,

b − a = b0 − a0 + (β − α)c

By choosing any β ∈ R I − {0, 1, α} we satisfy the required conditions. Let g1 , h1 be now the

simple isomorphisms invariant on H and mapping g(a) into b, and b into a, respectively. By construction, g(a) �= a (otherwise, a ∈ Y (g) ⊂ H), and (h1 ◦ g1 ◦ g)(a) = a, so a ∈

Y (h1 ◦ g1 ◦ g). As h1 , g1 do not alter H, Y (g) � Y (h1 ◦ g1 ◦ g). Step 3: Use induction to argue that after a finite number of steps m

dimY (hm ◦ gm ◦ . . . ◦ h1 ◦ g1 ◦ g) = n Consequently, hm ◦ gm ◦ . . . ◦ h1 ◦ g1 ◦ g = idR In Finish the proof by arguing that the inverse of a simple isomorphism is itself a simple isomorphism, too. The induction argument is obvious. Also, �

λ gH,c

�−1

1

λ = gH,c

Lebesgue Integration Theory

3.4

Measurable and Borel Functions

Exercises Exercise 3.4.1 Let ϕ : R In → R I¯ be a function such that dom ϕ is measurable (Borel). Prove that the following conditions are equivalent to each other: (i) ϕ is measurable (Borel). (ii) {(x, y) ∈ dom ϕ × R I : y ≤ ϕ(x)} is measurable (Borel). (iii) {(x, y) ∈ dom ϕ × R I : y > ϕ(x)} is measurable (Borel). (iv) {(x, y) ∈ dom ϕ × R I : y ≥ ϕ(x)} is measurable (Borel).

Lebesgue Measure and Integration

89

I n+1 into itself and it maps {y < g(x)} For λ �= 0, function gλ (x, y) = x+λy is an isomorphism from R

into {λy < g(x)}. Choose any 0 < λn � 1. Then {y ≤ g(x)} =

∞ �

n=1

{y < λ−1 n g(x)} =

∞ �

gλn ({y < g(x)})

n=1

Since a linear isomorphism maps measurable (Borel) sets into measurable (Borel) sets, each of the sets on the right-hand side is measurable (Borel) and, consequently, their common part must be measurable (Borel) as well. Use the identity {y < g(x)} =

∞ �

n=1

{y ≤ λn g(x)} =

∞ �

n=1

g λ1 ({y ≤ g(x)}) n

to demonstrate the converse statement. The last two results follow from simple representations {y < g(x)} = dom ϕ × R I − {y ≥ g(x)},

{y ≤ g(x)} = dom ϕ × R I − {y > g(x)}

Exercise 3.4.2 Prove Proposition 3.4.3. Let g : R In →R I n be an affine isomorphism. Then a function ϕ : R I n ⊃ dom ϕ → R I¯ is measurable (Borel) if and only if the composition ϕ ◦ g is measurable (Borel).

Obviously, g ⊗ idR I n ×R I � (x, y) → (g(x), y) ∈ R I n ×R I is an affine isomorphism, too. The I : R assertion follows then from the identity −1 {(x, y) ∈ g −1 (dom ϕ) × R I : y < (ϕ ◦ g)(x)} = (g ⊗ idR I : y < ϕ(z)} I ) {(z, y) ∈ dom ϕ) × R

Exercise 3.4.3 Prove Proposition 3.4.4. Let ϕi : E → R, I¯ i = 1, 2 and ϕ1 = ϕ2 a.e. in E. Then ϕ1 is measurable if an only if ϕ2 is measurable Let E0 = {x ∈ E : ϕ1 (x) = ϕ2 (x)}. Then, for i = 1, 2, {(x, y) ∈ E × R I : y < ϕi (x)} = {(x, y) ∈ E0 × R I : y < ϕi (x)} ∪Zi � �� � � �� � :=Si

:=S0

where Zi ⊂ (E − E0 ) × R I are of measure zero. Thus, if S1 is measurable then S0 = S1 − Z1 is measurable and, in turn, S2 = S0 + Z2 must be measurable as well.

90

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

3.5

Lebesgue Integral of Nonnegative Functions

3.6

Fubini’s Theorem for Nonnegative Functions

3.7

Lebesgue Integral of Arbitrary Functions

Exercises Exercise 3.7.1 Complete proof of Proposition 3.7.1. Let ai ∈ R, I¯ i ∈ IN . Suppose that ai = a1i − a2i , where a1i , a2i ≥ 0, and one of the series ∞ �

∞ �

a1i ,

1

is finite. Then the sum

�

IN ai exists and �

∞ �

ai =

1

IN

Case 3:

�∞

a2i

1

a1i −

∞ �

a2i

1

�∞ − 2 a2i = ∞. From a− i ≤ ai follows that 1 ai = ∞. By the same argument � + ai < ∞. Consequently, the sum IN ai exists but it is infinite. So is the sum of the sums of the two series.

1 �∞ 1

a1i < ∞,

�∞ 1

Exercise 3.7.2 Prove Corollary 3.7.1. � � I¯ The sum IN is finite if and only if IN |ai | is finite. In such a case Let ai ∈ R. �

ai =

∞ �

ai

1

IN

We have − |ai | = a+ i + ai

Consequently,

n � 1

|ai | =

n � 1

a+ i +

n � 1

a− i

Lebesgue Measure and Integration

91

If both sequences on the right-hand side have finite limits, so does the left-hand side, and the equality � �∞ holds in the limit. As the negative part of |ai | is simply zero, IN |ai | = 1 |ai |. �∞ Conversely, assume that 1 |ai | is finite. By the Monotone Sequence Lemma, both sequences �n + � n − 1 ai , 1 ai converge. The equality above implies that both limits must be finite as well.

Finally, the equality

n �

ai =

1

n � 1

− (a+ i − ai ) =

implies that the left-hand side converges and ∞ �

ai =

1

∞ � 1

a+ i −

∞ �

n � 1

a+ i −

n �

�

|ai |

a− i =:

1

IN

a− i

1

Exercise 3.7.3 Prove Corollary 3.7.2. Let f : E → R I¯ be measurable and f = f1 + f2 , where f1 , f2 ≥ 0 are measurable. Assume that at � � least one of the integrals f1 dm, f2 dm is finite. Then f is integrable and � � � f dm = f1 dm + f2 dm Repeat proof of Proposition 3.7.1 (compare also Exercise 3.7.1), replacing sums with integrals. Exercise 3.7.4 Prove Proposition 3.7.4. The following conditions are equivalent to each other: (i) f is summable. � � (ii) f + dm, f − dm < +∞. � (iii) |f | dm < +∞.

(i) ⇒ (ii) We have

|

�

f dm| = |

�

f + dm −

�

f − dm|

If the left-hand side is finite then so is the right-hand side. This implies that both � − f dm must be finite as well.

(ii)⇒(iii) follows from

(iii)⇒(i) follows from

�

|f | dm =

�

(f + + f − ) dm =

|

�

f dm| ≤

�

�

f + dm −

|f | dm

Exercise 3.7.5 Prove Proposition 3.7.5. All functions are measurable. The following properties hold:

�

f − dm

�

f + dm and

92

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

(i) f summable, E measurable ⇒ f |E summable. (ii) f, ϕ : E → R, I¯ |f | ≤ ϕ a.e. in E, ϕ summable ⇒ f summable. (iii) f1 , f2 : E → R I¯ summable ⇒ α1 F1 + α2 f2 summable for α1 , α2 ∈ R. I (i) follows from Proposition 3.7.4 and the monotonicity of measure which implies that � f dm for f ≥ 0.

�

E

f dm ≤

(ii) follows from Proposition 3.7.4 and monotonicity of measure. (iii) follows from (ii) and inequality

|α1 f1 + α2 f2 | ≤ |α1 | |f1 | + |α2 | |f2 | Exercise 3.7.6 Prove Theorem 3.7.2. Im →R I¯ be summable (and Borel). Then the following properties (Fubini’s Theorem): Let f : R I n ×R hold:

(i) y → f (x, y) is summable for almost all x (Borel for all x). � (ii) x → f (x, y) dm(y) is summable (and Borel). � � � �� (iii) f dm = f (x, y) dm(y) dm(x) The result is a direct consequence of the Fubini’s theorem for nonnegative functions, definition of the integral for arbitrary functions, Corollary 3.7.2 and Proposition 3.7.4. Summability of f implies � � that both integrals f + dm and f − dm are finite. Applying the Fubini’s theorem for nonnega-

tive functions, we conclude all three statements for both positive and negative parts of f . Then use Corollary 3.7.2 and Proposition 3.7.4.

3.8

Lebesgue Approximation Sums, Riemann Integrals

Exercises Exercise 3.8.1 Consider function f from Example 3.8.1. Construct explicitly Lebesgue and Riemann approximation sums and explain why the first sum converges while the other does not. Let . . . < y−1 < y0 < y1 < . . . < yk−1 ≤ 2 < yk < . . . < yn−1 ≤ 3 < yn < . . .

Lebesgue Measure and Integration

93

be any partition of the real axis. It follows from the definition of the function that Ek = f −1 ([yk−1 , yk )) = {irrational numbers} ∩ (0, 1) and that En = {rational numbers} ∩ (0, 1). Thus, m(Ek ) = 1 and

m(En ) = 0. All other sets Ei are empty. Consequently, the lower and upper Lebesgue sums reduce to s = yk−1 m((0, 1)) = yk−1

and

S = yk

If supi (yi − yi−1 ) → 0, then both yk−1 , yk must converge simply to 2, and both Lebesgue sums converge to the value of the Lebesgue integral equal to 2. On the other side, if 0 = x0 < . . . < xk−1 < xk < . . . < xn = 1 is an arbitrary partition of interval (0, 1), then for each subinterval [xk−1 , xk ) we can choose either a rational or irrational intermediate point ξk . If all intermediate points are irrational then f (ξk ) = 2 and the Riemann sum is equal to R=

n �

k=1

n �

f (ξk )(xk − xk−1 ) =

k=1

2(xk − xk−1 ) = 2

n �

(xk − xk−1 ) = 2

k=1

Similarly, if all intermediate points are rational, the corresponding value of the Riemann sum is equal 3. For a Riemann integrable function, the Riemann sums must converge to a common value which obviously cannot be the case. Exercise 3.8.2 Let f : R I n ⊃ D →R I¯ be a measurable (Borel) function. Prove that the inverse image f −1 ([c, d)) = {x ∈ D : c ≤ f (x) < d} is measurable (Borel), for any constants c, d. It is sufficient to prove that f −1 ((c, ∞)) = {x ∈ D : c < f (x)} is a measurable (Borel) set in R I n , for any constant c. Indeed, it follows from the identity, f −1 ([c, ∞)) = f −1 (

∞ �

n=1

(c −

∞ � 1 1 , ∞)) = f −1 ((c − , ∞)) n n n=1

that f −1 ([c, ∞)) is measurable. Similarly, the identity, f −1 ([c, d)) = f −1 ([c, ∞) − [d, ∞)) = f −1 ([c, ∞)) − f −1 ([d, ∞)) implies that f −1 ([c, d)) must be measurable as well. I n � x → (x, c) ∈ R I n+1 is obviously continuous, and Function ic : R f −1 (c, ∞) = i−1 c ({(x, t) : x ∈ D, t < f (x)}) This proves the assertion for Borel functions, as the inverse image of a Borel set through a continuous function is Borel. If f is only measurable then the Fubini’s Theorem (Generic Case) implies only that

94

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

set f −1 (c, ∞) is measurable a.e. in c. The critical algebraic property that distinguishes set {y < f (x)} from an arbitrary set, is that,

f −1 ((c, ∞)) = f −1 (

�

cn �c

(cn , ∞)) =

�

cn �c

f −1 ((cn , ∞))

for any sequence cn � c. As f −1 ((c, ∞)) are measurable a.e. in c, we can always find a sequence cn � c, for which sets f −1 ((cn , ∞)) are measurable. Consequently, f −1 ((c, ∞)) is measurable as

well.

Lp Spaces

3.9

H¨older and Minkowski Inequalities

Exercises Exercise 3.9.1 Prove the generalized H¨older inequality: �� � � � � uvw� ≤ �u�p �v�q �w�r � � where 1 ≤ p, q, r ≤ ∞, 1/p + 1/q + 1/r = 1. In view of estimate: |

�

uvw| ≤

�

|uvw| =

�

|u| |v| |w|

we can restrict ourselves to nonnegative, real-valued functions only. The inequality follows then from the original H¨older’s result, �

uvw ≤

��

u

p

�1/p ��

(v w)

p p−1

� p−1 p

If 1/q+1/r = 1−1/p = (p−1)/p, then for q � = q(p−1)/p, r� = r(p−1)/p, we have 1/q � +1/r� = 1. Consequently, �

v

p p−1

w

p p−1

≤

��

(v

p p−1

)

q�

� 1� �� q

(w

p p−1

)

r�

Combining the two inequalities, we get the final result.

� 1� r

=

��

v

q

p � q(p−1) ��

w

r

�

p r ( p−1)

Lebesgue Measure and Integration

95

Exercise 3.9.2 Prove that the H¨older inequality �

Ω

|f g| ≤

��

Ω

|f |

p

� p1 ��

Ω

|g|

q

� q1

1 1 + = 1, p q

,

p, q > 1

turns into an equality if and only if there exist constants α and β such that α|f |p + β|g|q = 0

a.e. in Ω

It is sufficient to prove the fact for real-valued and nonnegative functions f, g ≥ 0, with unit norms,

i.e.,

�

�

p

f = 1, Ω

gq = 1 Ω

The critical observation here is the fact that the inequality used in the proof of the H¨older inequality, 1

1 1 t+ v p q

1

tp vq ≤

turns into equality if and only if t = v. Consequently, for nonnegative functions u, v, � � � 1 1 1 1 p q u+ v−u v = 0 implies u = v a.e. in Ω . p q Ω Substituting u = f p , v = g q we get the desired result. Exercise 3.9.3

(i) Show that integral

�

is finite, but, for any � > 0, integral �

1 2

0

1 2

0

dx x ln2 x

dx [x ln2 x]1+�

is infinite. (ii) Use the property above to construct an example of a function f : (0, 1) → R I which belongs to space Lp (0, 1), 1 < p < ∞, but it does not belong to any Lq (0, 1), for q > p.

(i) Use substitution y = ln x to compute: � � 1 1 dx dy =− =− = y2 y ln x x ln2 x Hence,

Define:

�

1 2

�

1 1 dx + . =− ln 1/2 ln � x ln2 x

fn =

  x ln12 x 0

x ∈ [ n1 , 12 ) x ∈ (0, n1 )

96

APPLIED FUNCTIONAL ANALYSIS

Then fn � f =

1 x ln2 x

SOLUTION MANUAL

in (0, 12 ) and, therefore f is integrable with

�

fn �

�

f.

Raising function f to any power greater than 1 renders the resulting integral to be infinite. It is sufficient to construct a lower bound for f whose integral is infinite. It follows easily from applying the L’Hospital’s rule that lim x� lnr x = 0

x→0

for an arbitrarily small � > 0 and large r > 0. Consequently, for any �, r, there exists a δ > 0 such that: Consequently,

1 ln

for x < δ. Then,

with

�

1 x

x� lnr x ≤ 1

implies

x N . But this implies that f (xn ) ∈ G, for

n > N , i.e. f (xn ) → f (x).

Equivalence of second and third conditions follows immediately from the duality principle (Prop. 4.4.2). Exercise 4.4.5 Let X be a sequential Hausdorff topological space, and f : X → Y a sequentially continuous function into an arbitrary topological space Y . Prove that f is continuous.

Let G ⊂ Y be an arbitrary open and, therefore, sequentially open set in Y . By Exercise 4.4.4, f −1 (G)

is sequentially open in X. But X is a sequential space, so f −1 (G) must be open. This proves that f is (globally) continuous.

Exercise 4.4.6 Let X be a topological space such that, for every function f : X → Y (with arbitrary topo-

logical space Y ), sequential continuity of f implies its continuity. Prove that X must be a sequential topological space. 1. Let Y = X with a new topology introduced through open sets X1 where for the open sets X1 we take all sequentially open sets in X. Prove that such sets satisfy the axioms for open sets. • Empty set and the space X are open and, therefore, sequentially open as well.

• Union of an arbitrary family Gι , ι ∈ I of sequentially open sets is sequentially open. Indeed � let x ∈ ι∈I Gι . There exists then κ ∈ I such that x ∈ Gκ . Let xn → x. As Gκ is � sequentially open, there exists N such that for n > N , xn ∈ Gκ ⊂ ι∈I Gι . Done.

• Intersection of a finite number of sequentially open sets is sequentially open. Indeed, let xn → x and x ∈ G1 ∩G2 ∩. . .∩Gm . Let Ni , i = 1, . . . , m be such that xn ∈ Gi for n > Ni . Take N = max{N1 , . . . , Nm } to see that, for n > N , xn ∈ G1 ∩ . . . ∩ Gm .

2. Take Y = X with the new, stronger (explain, why?) topology induced by X1 and consider the

identity map idX mapping original topological space (X, X ) onto (X, X1 ). Prove that the map is sequentially continuous.

Let xn → x in the original topology. We need to prove that xn → x in the new topology as

well. Take any open neighborhood G ∈ X1 of x. By construction, G is sequentially open in the original topology so there exists N such that, for n > N , xn ∈ G. But, due to the arbitrariness

of G, this proves that x → x in the new topology.

112

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

3. The identity map is, therefore, continuous as well. But this means that G ∈ X1

⇒

id−1 X (G) = G ∈ X

i.e. every sequentially open set in X is automatically open.

4.5

Topological Equivalence. Homeomorphism

Theory of Metric Spaces

4.6

Metric and Normed Spaces, Examples

Exercises Exercise 4.6.1 Let (X, d) be a metric space. Show that ρ(x, y) = min{1, d(x, y)} is also a metric on X. Symmetry and positive definitness of d(x, y) imply symmetry and positive definitness of ρ(x, y). The easiest perhaps way to see that the triangle inequality is satisfied, is to consider the eight possible cases: 1: d(x, y) < 1, d(x, z) < 1, d(z, y) < 1. Then ρ(x, y) = d(x, y) ≤ d(x, z) + d(z, y) = ρ(x, z) + ρ(z, y). 2: d(x, y) < 1, d(x, z) < 1, d(z, y) ≥ 1. Then

ρ(x, y) = d(x, y) < 1 = ρ(z, y) ≤ ρ(x, z) + ρ(y, z).

3: d(x, y) < 1, d(x, z) ≥ 1, d(z, y) < 1. Then

ρ(x, y) = d(x, y) < 1 = ρ(x, z) ≤ ρ(x, z) + ρ(y, z).

4: d(x, y) < 1, d(x, z) ≥ 1, d(z, y) ≥ 1. Then

ρ(x, y) = d(x, y) < 1 < 2 = ρ(x, z) + ρ(y, z).

5: d(x, y) ≥ 1, d(x, z) < 1, d(z, y) < 1. Then

ρ(x, y) = 1 ≤ d(x, y) ≤ d(x, z) + d(z, y) = ρ(x, z) + ρ(z, y).

Topological and Metric Spaces

113

6: d(x, y) ≥ 1, d(x, z) < 1, d(z, y) ≥ 1. Then

ρ(x, y) = 1 = ρ(z, y) ≤ ρ(x, z) + ρ(y, z).

7: d(x, y) ≥ 1, d(x, z) ≥ 1, d(z, y) < 1. Then

ρ(x, y) = 1 = ρ(x, z) ≤ ρ(x, z) + ρ(y, z).

8: d(x, y) ≥ 1, d(x, z) ≥ 1, d(z, y) ≥ 1. Then ρ(x, y) = 1 < 2 = ρ(x, z) + ρ(y, z).

Exercise 4.6.2 Show that any two norms � · �p and � · �q in R I n , 1 ≤ p, q ≤ ∞, are equivalent, i.e., there exist constants C1 > 0, C2 > 0 such that

and

�x�p ≤ C1 �x�q

�x�q ≤ C2 �x�p

for any x ∈ R I n . Try to determine optimal (minimum) constants C1 and C2 . It is sufficient to show that any p-norm is equivalent to e.g. the ∞-norm. We have, � p1 � n � 1 p p p �x�∞ = max |xi | = |xj | (for a particular index j) = (|xj | ) ≤ |xi | = �x�p i=1,...,n

i=1

At the same time,

�x�p =

�

n � i=1

|xi |

p

� p1

≤

�

n � i=1

�x�p∞

� p1

= n �x�∞

Consider now 1 < p, q < ∞. Let Cpq be the smallest constant for which the following estimate holds. ∀x ∈ R In

�x�p ≤ Cpq �x�q

Constant Cpq can be determined by solving the constrained minimization problem, Cpq = max �x�p �x�q =1

This leads to the Lagrangian, L(x, λ) =

n � i=1

p

|xi | − λ

and the necessary conditions, � � xi  p|xi |p−1 − λq|xi |q−1 ∂L |xi | =  ∂xi 0 For xi �= 0, we get,

|xi | = λ

1 p−q

�

n � i=1

q

�

|xi | − 1

 if xi �= 0 

if xi = 0 1 � � p−q p q



Raising both sides to power q and summing up over i, we get q � � p−q n � q q q |xi | = mλ p−q 1= p i=1

= 0,

i = 1, . . . , n

114

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

where m is the number of coordinates for which xi �= 0. Notice that m must be positive to satisfy the constraint. This yields the value for λ,

λ=

p − p−q n q q

and the corresponding value for the p-norm, �

n � i=1

|xi |

p

� p1

1

1

= mp−q

Consequently, the maximum is attained at the point for which m = n, i.e., all coordinates are different from zero. This gives the value for the optimal constant, 1

1

Cpq = n p − q

Upon passing with p, q to 1 or ∞, we get the corresponding values of the constant for the limiting

cases.

Exercise 4.6.3 Consider R I N with the l1 -norm, x = (x1 , . . . , xN ),

�x�1 =

N � i=1

|xi |

Let �x� be now any other norm defined on R I n. (i) Show that there exists a constant C > 0 such that �x� ≤ C�x�1

∀x ∈ R IN

(ii) Use (i) to demonstrate that function R I N � x → �x� ∈ R I is continuous in l1 -norm. (iii) Use Weierstrass Theorem to conclude that there exists a constant D > 0 such that �x�1 ≤ D�x�

∀x ∈ R IN

Therefore, the l1 norm is equivalent to any other norm on R I N . Explain why the result implies that any two norms defined on an arbitrary finite-dimensional vector space must be equivalent. (i) Let ei denote the canonical basis in R I n . Then �x� = �

n � i=1

xi ei � ≤

n � i=1

|xi | �ei � ≤ C

where C = max{�e1 �, . . . , �en �}

n � i=1

|xi |

Topological and Metric Spaces

115

(ii) This follows immediately from the fact that |�x� − �y�| ≤ �x − y� and property (i). (iii) The l1 unit ball is compact. Consequently, norm � · � attains a minimum on the l1 unit ball, i.e., C≤�

x � �x�1

∀x

Positive definitness of the norm implies that C > 0. Multiplying by �x�1 /C, we get �x�1 ≤ C −1 �x� Take now two arbitrary norms. As each of them is equivalent to norm � · �1 , they must be equivalent with each other as well.

4.7

Topological Properties of Metric Spaces

Exercises Exercise 4.7.1 Prove that F : (X, D) → (Y, ρ) is continuous if and only if the inverse image of every (open) ball B(y, �) in Y is an open set in X.

It is sufficient to show that the inverse image of any open set G in Y is open in X. For any y ∈ G,

there exists �y > 0 such that B(y, �y ) ⊂ G. Set G thus can be represented as a union of open balls, � B(y, �y ) G= y∈G

Consequently, set



f −1 (G) = f −1 

�

y∈G

as a union of open sets must be open.



B(y, �y ) =

�

f −1 (B(y, �y ))

y∈G

Exercise 4.7.2 Let X = C ∞ (a, b) be the space of infinitely differentiable functions equipped with Chebyshev metric. Let F : X → X be the derivative operator, F f = df /dx. Is F a continuous map on X? No, it is not. Take, for instance, [a, b] = [0, π] and the sequence fn (x) = �fn � = max | x∈[0,π]

1 sin(nx) | ≤ → 0, n n

sin(nx) . n

Then,

116

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

but, at the same time, �fn� � = max |fn� (x)| = max | cos(nx)| = 1 � 0 x∈[0,π]

4.8

x∈[0,π]

Completeness and Completion of Metric Spaces

Exercises Exercise 4.8.1 Let Ω ⊂ R I n be an open set and let (C(Ω), � �p ) denote the (incomplete) metric space of continuous, real-valued functions on Ω with metric induced by the Lp norm. Construct arguments supporting the fact that the completion of this space is Lp (Ω). Hint: See Exercise 4.9.3 and use Theorem 4.8.2. Definition of the completion of a metric space and the density result imply that Lp (Ω) is a completion of (C(Ω), � �p ). By Theorem 4.8.2, completions are unique (up to an isometry) and, therefore, space

Lp (Ω) is the completion of (C(Ω), � �p ).

Exercise 4.8.2 Let xnk be a subsequence of a Cauchy sequence xn . Show that if xnk converges to x, so does the whole sequence xn . Recall that xn is Cauchy if ∀� > 0

∃N1 = N1 (�)

n, m ≥ N1 ⇒ d(xn , xm ) < �

Also xnk → x means that ∀� > 0

∃N2 = N2 (�)

k ≥ N2 ⇒ d(xnk , x) < �

Let � be now an arbitrary positive number. Choose N = max{N1 (�/2), N2 (�/2)} Then, for any n ≥ N , d(xn , x) ≤ d(xn , xnN2 (�/2) ) + d(xnN2 (�/2) , x)
0, there I such that exists a simple function φ� : Ω → R

�f − φ� �∞ ≤ � Hint: Use the Lebesgue approximation sums. I such that Indeed, let . . . < yi−1 < yi < yi+1 < . . . be a partition of R sup |yi − yi−1 | ≤ � i

Define Ei = f −1 ([yi−1 , yi )), select any ci ∈ [yi−1 , yi ] and set: φe =

�

c i χ Ei

i∈Z I

Exercise 4.9.3 Let Ω ⊂ R I n be an open set. Let f ∈ Lp (Ω), 1 ≤ p < ∞. Use results of Exercise 4.9.1 and

I converging to Exercise 4.9.2 to show that there exists a sequence of continuous functions φn : Ω → R function f in the Lp (Ω) norm.

Topological and Metric Spaces

119

1. Assume additionally that domain Ω is bounded, and 0 ≤ f ≤ M < ∞ a.e. in Ω. Pick an arbitrary � > 0.

• Select a partition 0 = y0 < y1 < . . . < yN = M such that max |yi − yi−1 | ≤

i=1,...,N

and define: φ=

N �

yi χ E i ,

1 � [m(Ω)]− p 2

Ei = f −1 ([yi−1 , yi ))

i=1

Then

1

�f − φ�p ≤ �f − φ�∞ [m(Ω)] p ≤

� 2

• Use Exercise 4.9.1 to select continuous functions φi such that �φEi − φi �p ≤ Then �

N � i=1

yi χ E i −

N �

By triangle inequality, �f −

i=1

y i φ i �p ≤

�N

i=1

N � i=1

� 1 1 2 2i yi N

yi �χEi − φi �p ≤

�� 1 � ≤ i 2 i=1 2 2

yi φi �p ≤ �.

2. Drop the assumption on f being bounded. Define fM (x) = min{f (x), M }. Then fM → f

pointwise and, by construction, fM are bounded by f . Thus, by Lebesgue Dominated Conver-

gence Theorem, |fM − f |p → 0 as M → ∞. Apply then the result of the previous step to functions fM .

3. Drop the assumption on f being nonnegative. Split function f into its positive and negative parts: f = f+ − f− and apply the previous step to each of the parts separately. 4. Drop the assumption on Ω being bounded. Let Bn = B(0, n). Consider restriction of f to Ω ∩ Bn . By the Lebesgue Dominated Convergence argument again, �fn − f �p → 0. Apply the

result of the previous step to fn then.

Exercise 4.9.4 Argue that, in the result of Exercise 4.9.3, one can assume additionally that functions fn have compact support. Let Ω ⊂ R I n be an open set. Define, Ωn := {x ∈ Ω : d(x, Ω� ) >

1 , x ∈ Bn } n

Continuity of the distance function d(x, A) implies that sets Ωn are open. By construction, they are ¯ n ⊂ Ωn+1 . Consider restriction fn of f to Ωn and apply the result of Exercise 4.9.3 also bounded and Ω

¯ n ⊂ Ωn+1 , one can assume that the opens sets G used in Exercise 4.9.1 are to function fn . Since Ω

contained in Ωn+1 . Consequently, the corresponding continuous approximations have their supports in Ωn+1 ⊂ Ω.

120

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Exercise 4.9.5 Let F be a uniformly bounded class of functions in Chebyshev space C[a, b], i.e., ∃M > 0 : |f (x)| ≤ M,

∀x ∈ [a, b], ∀f ∈ F

Let G be the corresponding class of primitive functions � x f (s) ds, F (x) =

f ∈F

a

Show that G is precompact in the Chebyshev space.

According to Arzel`a–Ascoli Theorem, it is sufficient to show that G is equicontinuous. We have, � x � x f (s) ds| ≤ |f (s)| ds ≤ M |x − y| |F (x) − F (y)| = | y

y

Thus functions F (x) are Lipschitz continuous with a uniform (with respect to F ) bound on the Lipschitz constant. This implies that functions F, f ∈ F are equicontinuous.

4.10

Contraction Mappings and Fixed Points

Exercises Exercise 4.10.1 Reformulate Example 4.10.6 concerning the Fredholm Integral Equation using the Lp spaces, 1 < p < ∞. What is the natural regularity assumption on kernel function K(x, y)? Does it have to be bounded? Let Af = φ + λ

�

b

K(·, y)f (y) dy a

We need to come up with sufficient conditions to have the estimate �Af − Ag�p = �A(f − g)�p ≤ k�f − g�p where k < 1. The inequality will also automatically imply that operator A is well defined. Let h = f − g. We have, � b � �λ K(·, y)h(y) dy�pp = a

b a

≤ |λ|p = |λ|p

|λ �

� b

a

�

b

K(x, y)h(y) dy|p dx a

� 

b a

�

��

a b

1

b

|K(x, y)|

p p−1

p

dy

|K(x, y)| p−1 dy

� p−1 �� p

�p−1

dx

b a

�

� p1 p |h(y)|p dy  dx

b a

|h(y)|p dy

Topological and Metric Spaces

121

where we have applied the H¨older inequality to the inner integral to obtain an estimate with the desired Lp norm of function h. A natural assumption is thus to request the inequality:   p1 �p−1 � � b � b  p |λ| |K(x, y)| p−1 dy dx 0 and it is not uniformly Lipschitz with respect to q in the domain of definition. We need to restrict ourselves to a smaller domain [0, T ] × [�, ∞) where � < 1 is fixed and T is to be determined. The Mean-Value Theorem implies then that ln q1 − ln q2 =

1 (q1 − q2 ) ξ

for some intermediate point ξ ∈ (q1 , q2 ). This implies that, in the smaller domain, function F (t, q) is

uniformly Lipschitz in q,

|t ln q1 − t ln q2 | ≤

T |q1 − q2 | �

∀t ∈ [0, T ], q ∈ [�, ∞)

The initial-value problem is equivalent to the integral equation:  q ∈ C([0, T ])    � t  q(t) = 1 + s ln q(s) ds   0

and function q is a solution of the integral equation if an only if q is a fixed point of the operator: � t s ln q(s) ds A : C([0, T ]) → C([0, T ]), (Aq)(t) = 1 + 0

We have,

|(Aq1 )(t) − (Aq2 )(t)| ≤

�

t 0

|s ln q1 (s) − s ln q2 (s)| ds ≤

Consequently, map A is a contraction if

�

t 0

T2 T |q1 (s) − q2 (s)| ds ≤ �q1 − q2 �∞ � �

T2 0

such that v ∈ C, |β − α| < δ

⇒

βv ∈ B

In particular, for β = α, v∈C

⇒

αv ∈ B

In other words, for every α, ∀B ∈ B0 ∃C ∈ B0 : αC ⊂ B i.e., αB0 � B0 . For α �= 0, this implies that αB0 ∼ B0 .

1 α B0

� B0 or, equivalently, B0 � αB0 . Consequently,

125

126

5.2

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Locally Convex Topological Vector Spaces

Exercises Exercise 5.2.1 Let A ∈ L(X, Y ) be a linear transformation from a vector space X into a normed space Y . Assume that A is non-injective, and define

p(x) := �Ax�Y Explain why p is a seminorm but not a norm. Funtion p(x) is homogeneous and it satisfies the triangle inequality. But it is not positive definite as p(x) vanishes on N (A). Exercise 5.2.2 Show that each of the seminorms inducing a locally convex topology is continuous with respect to this topology. Consider seminorm pκ , κ ∈ I, and u ∈ V . We need to show that ∀� > 0 ∃C ∈ Bx : y ∈ C ⇒ |pκ (y) − pκ (x)| ≤ � It is sufficient to take C = x + B({κ}, �). Indeed, by definition of B({κ}, �), y ∈ C ⇔ pκ (y − x) ≤ � and, consequently, pκ (y) = pκ (x + y − x) ≤ pκ (x) + pκ (y − x) ≤ pκ (x) + � At the same time, pκ (x) = pκ (y + x − y) ≤ pκ (y) + pκ (x − y) = pκ (y) + pκ (y − x) ≤ pκ (y) + � Exercise 5.2.3 Show that replacing the weak equality in the definition of set Mc with a strict one does not change the properties of Mc . Properties (i)-(iii) follow from the same arguments as for the case with weak inequality. (iv) Consider two cases: • If p(u) = 0 then p(αu) = αp(u) = 0, so αu ∈ Mc , for any α > 0. • If p(u) �= 0, take any α >

p(u) c .

Then

p(α−1 u) = α−1 p(u)
0 such that |f (v)| ≤ C max pι (v) ι∈I0

The following is a straightforward generalization of arguments used in Proposition 5.6.1. For linear functionals defined on a t.v.s., continuity at 0 implies continuity at any u. Indeed, let u be an arbitrary vector. We need to show that ∀� > 0 ∃B ∈ B0 : y ∈ u + B ⇒ |f (y) − f (u)| < � If f is continuous at 0 then ∀� > 0 ∃B ∈ B0 : z ∈ B ⇒ |f (z)| < � Consequently, for y ∈ u + B, i.e., y − u ∈ B, |f (y) − f (x)| = |f (y − x)| < � . Thus, it is sufficient to show that the condition above is equivalent to continuity of f at 0. Necessity: Let f be continuous at 0. For any � > 0 and, therefore, for � = 1 as well, there exists a neighborhood B ∈ B0 such that

x∈B

⇒

|f (x)| < 1

128

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Recalling the construction of the base of neighborhoods, we learn that there exists a finite subset I0 ⊂ I,

and a constant δ > 0 such that,

max pι (x) < δ ι∈I0

⇒

|f (x)| < 1

Let λ := maxι∈I0 pι (x). Then max pι ι∈I0

�

xδ λ

�

=

δ max pι (x) < δ λ ι∈I0

|f

�

xδ λ

�

and, therefore,

or, equivalently, |f (x)|
0 ∃N : n, m ≥ N ⇒ �un − um �U + �A(un − um )�V < � This implies that both un is Cauchy in U , and Aun is Cauchy in V . By completness of U and V , un → u, Aun → v, for some u ∈ U, v ∈ V . But operator A is closed, so (u, v) ∈ G(A) which in

turn implies that u ∈ D(A) and v = Au.

Banach Spaces

5.11

137

Example of a Closed Operator

Exercises Exercise 5.11.1 Let Ω ⊂ R I n be an open set. Prove that the following conditions are equivalent to each other. (i) For every point x ∈ Ω, there exists a ball B = B(x, �x ) ⊂ Ω such that u|B ∈ L1 (B). (ii) For every compact subset K ⊂ Ω, u|K ∈ L1 (K). � � (ii) ⇒ (i). Take a closed ball B ⊂ Ω and observe that B |u| = B |u|.

(ii) ⇒ (i). Let K be a compact subset of Ω. Use neighborhoods from (i) to form an open cover for K, � B(x, �x ) K⊂ x∈K

By compactness of K, there exists a finite number of points xi ∈ K, i = 1, . . . , N such that K⊂

But then

�

K

|u| ≤

N �

B(xi , �xi )

i=1

N � � i=1

B(xi ,�xi )

|u| < ∞

Exercise 5.11.2 Consider X = L2 (0, 1) and define a linear operator T u = u� , D(T ) = C ∞ ([0, 1]) ⊂ L2 (0, 1). Show that T is closable. Can you suggest what would be the closure of T ?

We shall use Proposition 5.10.3. Assume un → 0 and T un = u�n → v, where the convergence is

understood in the L2 -sense. We need to show that v = 0. Let φ ∈ D(0, 1). Then, � 1 � 1 � un φ = u�n φ − 0

0

L -convergence of un to zero implies that the left-hand side converges to zero. At the same time, � L2 -convergence of u�n to v implies that the right-hand side converges to vφ. Consequently, � 1 vφ = 0, ∀φ ∈ D(0, 1) 2

0

Thus, density of test functions in L2 (0, 1) implies that v must be zero and, by Proposition 5.10.3, the operator is closable. The closure of the operator is the distributional derivative defined on the Sobolev space, L2 (0, 1) ⊃ H 1 (0, 1) � u → u� ∈ L2 (0, 1)

138

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Exercise 5.11.3 Show that the Sobolev space W m,p (Ω) is a normed space. All three conditions for a norm are easily verified. Case: 1 ≤ p < ∞. Positive definitness. �u�W m,p (Ω) = 0 ⇒ �u�Lp (Ω) = 0 ⇒ u = 0

( in the Lp -sense, i.e., u = 0 a.e.)

Homogeneity. 

�λu�W m,p (Ω) = 

�

|α|≤m

Triangle inequality.



= |λ| 

�u + v�W m,p (Ω) 

=

�

�α�≤m



≤ ≤

�

�α�≤m



�Dα (λu)�pLp (Ω)  = 

�

|α|≤m

 p1

�Dα u�pLp (Ω) 

�

|α|≤m

 p1

|λ|p �Dα u�pLp (Ω) 

= |λ|�u�W m,p (Ω)

 p1

�Dα (u + v)�pLp (Ω) 

� �

�α�≤m



 p1

�p

 p1

�Dα u�Lp (Ω) + �Dα v�Lp (Ω)   p1



�Dα u�Lp (Ω)  + 

�

�α�≤m

( triangle inequality in Lp (Ω) )  p1

�Dα v�Lp (Ω)  ( triangle inequality for the p norm in R I N)

where, in the last step, N equals the number of partial derivatives, i.e., N = #{|α| ≤ m}. Case: p = ∞ is proved analogously.

Banach Spaces

139

Topological Duals, Weak Compactness

5.12

Examples of Dual Spaces, Representation Theorem for Topological Duals of Lp -Spaces

Exercises Exercise 5.12.1 Let Ω ⊂ R I N be a bounded set, and fix 1 ≤ p < ∞. Prove that, for every r such that p < r ≤ ∞, Lr (Ω) is dense in Lp (Ω).

Hint: For an arbitrary u ∈ Lp (Ω) define un (x) =

�

u(x)

if |u(x)| ≤ n

n

otherwise

Show that 1. un ∈ Lr (Ω) and 2. �un − u�p → 0. 1. By construction,

�

Ω

|un |p ≤ np meas(Ω) < ∞

2. From the definition of un , it follows that u − un → 0 pointwise and

Since

�

|u − un |p ≤ |u|p |u|p < ∞, the Lebesgue Dominated Convergence Theorem implies that � |u − un |p → 0 Ω

Exercise 5.12.2 Consider R I n equipped with the p-norm, � � p1 n  �   p  |xi | �x�p = i=1    max |xi |  1≤i≤n

Prove that

sup

n �

�y�p =1 i=1

1≤p 0. In order to determine the supremum, it is sufficient thus to compute λ. Solving for yi and requesting y to be of unit norm, we get n � i=1

p

p

|xi | p−1 = λ p−1

Consequently, λ=

�

n � i=1

|xi |

p p−1

n � i=1

p

|yi |p = λ p−1

� p−1 p

= �x�q

Case: p = 1. We have, |

n � i=1

x i yi | ≤

n � i=1

|xi yi | ≤ max |xi | 1≤i≤n

n � i=1

|yi | = max |xi | = �x�∞ 1≤i≤n

The bound is attained. Indeed, let i0 be an index such that |xi0 | = max |xi | 1≤i≤n

select yi = sgn xi0 δi,i0 . Then �y�1 = 1, and n � i=1

xi yi = |xi0 |

Case: p = ∞. We have, |

n � i=1

x i yi | ≤

n � i=1

|xi yi | ≤ max |yi | 1≤i≤n

n � i=1

|xi | =

n � i=1

|xi | = �x�1

Banach Spaces

141

The bound is attained. Indeed, select yi = sgn xi . Then �y�∞ = 1, and n �

x i yi =

i=1

n � i=1

|xi | = �x�1

Finally, recall that every linear functional defined on a finite-dimensional vector space equipped with a norm, is automatically continuous. The topological dual coincides thus with the algebraic dual. Given a linear functional f on R I n , and recalling the canonical basis ei , i = 1, . . . , n, we have the standard representation formula, f (y) = f

�

n �

yi e i

i=1

�

=

n � i=1

The map, Rn � y → R I n � f → {I

n � i=1

f (ei ) yi = � �� � fi

n �

f i yi

i=1

fi yi ∈ R} I ∈ (I Rn )∗ = (I R n )�

is a linear isomorphism and, by the first part of this exercise, it is an isometry if the space for f is equipped with the q-norm. Exercise 5.12.3 Prove Theorem 5.12.2. Let u = (u1 , u2 , . . .) ∈ �p , 1 ≤ p < ∞ and let ei = (0, . . . , 1(i) , . . .). Define uN =

N �

ui ei = (u1 , . . . , uN , . . .)

i=1

It follows from the definition of �p -spaces that the tail of the sequence converges to zero, �u − uN ��p =

�

∞ �

i=N +1

|ui |

p

� p1

→0

as N → ∞

Notice that the argument breaks down for p = ∞. Let f ∈ (�p )� . Set φi = f (ei ). Then ∞ �

def

φ i ui =

i=1

lim

N →∞

N �

φi ui = lim f (uN ) = f (u)

i=1

N →∞

H¨older inequality implies that |f (u)| ≤ �φ��q �u��p We will show that the bound equals the supremum. Case: p = 1. Let ij be a sequence such that |φij | → sup |φi | = �φ�∞ , as j → ∞ i

142

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Take v j = sgn φij eij . Then �v j ��1 = 1, and f (uj ) = |φij | → �φ�∞ Case: 1 < p < ∞. Use the choice derived in Exercise 5.12.2 for the finite-dimensional case, 1

|φi | p−1 sgn φi ui = �∞ p 1 ( i=1 |φi | p−1 ) p

Then �u��p = 1, and f (u) =

∞ �

φ i ui =

i=1

�∞

p p−1

i=1 |φi | p �n 1 ( i=1 |φi | p−1 ) p

= �φ��q

Exercise 5.12.4 Generalize the Representation Theorem for Lp Spaces (Theorem 5.12.1) to the case of an arbitrary (measurable) set Ω. Hint: Consider a sequence of truncated domains, Ωn = Ω ∩ B(0, n) , use Theorem 5.12.1 on Ωn to conclude existence φn ∈ Lq (Ωn ), and investigate convergence of φn . Let φn ∈ Lq (Ωn ) be such that,

�

∀v ∈ Lp (Ωn ) ,

φn v = f (˜ v) Ωn

where v˜ is the zero extension of v. For m > n and v ∈ Lp (Ωn ), � � φm v = f (˜ v) = φn v . Ωn

Ωn

Restriction φm |Ωn lives in Lq (Ωn ) so, by uniqueness of φ in the Representation Theorem, φm is an extension of φn . Let φ(x) be the (trivial) pointwise limit of φ˜n (x). For p > 1, Lemma 3.5.1 implies that

�

Ω

|φ˜n |q →

�

Ω

|φ|q .

At the same time, �φn �Lq (Ωn ) are uniformly bounded as, by the Representation Theorem, �φn �Lq (Ωn ) = �f |Lp (Ωn ) �(Lp (Ωn ))� ≤ �f �(Lp (Ω))� , which proves that �φ�Lq (Ω) is finite (case q = ∞ included). Let now v ∈ Lp (Ω), and let vn be its restriction to Ωn . By Lebesgue Dominated Convergence Theorem, v˜n → v in Lp (Ω). Consequently, the Lp functions with support in Ωn , n = 1, 2, . . ., are dense in Lp (Ω). Consequently,

�

φv = Ω

�

φn v = f (v) , Ωn

true for any v ∈ Lp (Ω) with support in Ωn , can be extended to arbitrary v ∈ Lp (Ω). I a measurable function defined on Ω. Prove that Exercise 5.12.5 Let Ω ⊂ R I n be an open set and f : Ω → R the following conditions are equivalent to each other:

Banach Spaces

143

1. For every x ∈ Ω there exists a neighborhood N (x) of x (e.g., a ball B(x, ε) with some ε = ε(x) > 0) such that

�

2. For every compact K ⊂ Ω

N (x)

�

K

|f | dx < +∞

|f | dx < +∞

Functions of this type are called locally integrable and form a vector space, denoted L1loc (Ω). Closed balls are compact, so the second condition trivially implies the first. To show the converse, consider a compact set K and the corresponding open cover consisting of balls present in the first condition, K⊂

�

B(x, εx )

x∈K

As the set is compact, there must exist a finite subcover. i.e., a collection of points x1 , . . . , xN ∈ K

such that

K ⊂ B(x1 , εx1 ) ∪ . . . ∪ B(xN , εxN ) Standard properties of the Lebesgue integral imply � � � |f | dx ≤ |f | dx + . . . + K

B(x1 ,εx1 )

B(xN ,εxN )

|f | dx < ∞

which proves the second assertion. Exercise 5.12.6 Consider the set B defined in (5.1). Prove that B is balanced, convex, absorbing, and Bi ⊂ B, for each i = 1, 2, . . ..

�

• B is balanced as, for every |α| ≤ 1, and α

�

i∈I

ϕi ∈ B,

ϕi =

i∈I

�

i∈I0

αϕi ∈ B

since all Bi ’s are balanced. • B is convex. Indeed, let α ∈ [0, 1], J0 , J2 = I2 − J0 . Then α

�

i∈I1

ϕi + (1 − α)

�

i∈I2

ψi =

�

�

j∈J0

i∈I1

ϕi ,

�

i∈I2

ψi ∈ B. Set J0 = I1 ∩ I2 , J1 = I1 −

αϕj + (1 − α)ψj +

�

j∈J1

αϕj +

�

j∈J2

(1 − α)ψj ∈ B

since Bi ’s are convex and balanced. • B is absorbing. Indeed, for ever ϕ ∈ D(Ω), there exists i such that ϕ ∈ D(Ki ) and Bi ⊂ D(Ki ) is absorbing.

• The last condition is satisfied by construction.

144

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Exercise 5.12.7 Let q be a linear functional on D(K). Prove that q is sequentially continuous iff there exist constants CK > 0 and k ≥ 0 such that

|q(φ)| ≤ Ck sup sup |Dα φ(x)| |α|≤k x∈K

∀φ ∈ D(K)

Proof is a direct consequence of the definition of topology in D(K) and Exercise 5.2.6. Exercise 5.12.8 Prove that the regular distributions and the Dirac delta functional defined in the text are continuous on D(Ω). It is sufficient to use the criterion discussed in the text. Let f ∈ L1loc (Ω), and K ⊂ Ω be an arbitrary compact set. Then, for ϕ ∈ D(K), �� � �� � � � � � � � f ϕ dx� = � f ϕ dx� ≤ |f | dx sup |ϕ(x)| � � � � where the integral

�

K

Ω

K

x∈K

K

|f | is finite.

Similarly, for the delta functional δx0 , and a compact set K, for every ϕ ∈ D(K), | < δx0 , ϕ > | = |ϕ(x0 )| ≤ sup |ϕ(x)| x∈K

Exercise 5.12.9 Consider function u : (0, 1) → R I of the form � u1 (x) 0 < x ≤ x0 u(x) = u2 (x) x0 < x ≤ 1

where x0 ∈ (0, 1)

Here u1 and u2 are C 1 functions (see Example 5.11.1), but the global function u is not necessarily continuous at x0 . Follow the lines of Example 5.11.1 to prove that the distributional derivative of the regular distribution qu corresponding to u is given by the formula (qu )� = qu� + [u(x0 )]δx0 where u� is the union of the two branches, derivatives u�1 and u�2 (see Example 5.11.1), δx0 is the Dirac delta functional at x0 , and [u(x0 )] denotes the jump of u at x0 , [u(x0 )] = u2 (x0 ) − u1 (x0 ) It is sufficient to interprete the result obtained in the text, def

< qu� , ϕ > = − < qu , ϕ� > � x0 � 1 � 1 uϕ� dx = u1 ϕ� dx + u2 ϕ� dx = 0 0 x0 � 1 � x0 � u1 ϕ dx + u�2 ϕ dx + [u2 (x0 ) − u1 (x0 )]ϕ(x0 ) = 0 x0 � 1 = u� ϕ dx + [u(x0 )] < δx0 , ϕ > 0

=< qu� + [u(x0 )]δx0 , ϕ >

Banach Spaces

5.13

145

Bidual, Reflexive Spaces

Exercises Exercise 5.13.1 Explain why every finite-dimensional space is reflexive. Recall the discussion from Chapter 2. As the evaluation map is injective and bidual space is of the same dimension as the orginal space, the evaluation map must be surjective. I n . The closure in W m,p (Ω) Exercise 5.13.2 Let W m,p (Ω) be a Sobolev space for Ω, a smooth domain in R of the test functions C0∞ (Ω) (with respect to the W m,p norm), denoted by W0m,p (Ω), W0m,p (Ω) = C0∞ (Ω) may be identified as a collection of all “functions” from W m,p (Ω) which “vanish” on the boundary together with their derivatives up to m − 1 order (this is a very nontrivial result based on Lions’ Trace Theorem; see [8, 10]). The duals of the spaces W0m,p (Ω) are the so-called negative Sobolev spaces def

W −m,p (Ω) = (W0m,p (Ω))

�

m>0

Explain why both W0m,p (Ω) and W −m,p (Ω), for 1 < p < ∞, are reflexive. This is a simple consequence of Proposition 5.13.1. Space W0m,p (Ω) as a closed subspace of a reflexive space must be reflexive, and W −m,p (Ω) as the dual of a reflexive space is reflexive as well.

5.14

Weak Topologies, Weak Sequential Compactness

Exercises Exercise 5.14.1 Prove Proposition 5.14.1. All properties are a direct consequence of the definitions. Exercise 5.14.2 Let U and V be two normed spaces. Prove that if a linear transformation T ∈ L(U, V ) is strongly continuous, then it is automatically weakly continuous, i.e., continuous with respect to weak topologies in U and V .

146

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Hint: Prove first the following: Lemma: Let X be an arbitrary topological vector space, and Y be a normed space. Let T ∈ L(X, Y ).

The following conditions are equivalent to each other.

(i) T : X → Y (with weak topology) is continuous (ii) f ◦ T : X → R(I I C ) is continuous, ∀ f ∈ Y � Follow then the discussion in the section about strongly and weakly continuous linear functionals. (i) ⇒ (ii). Any linear functional f ∈ Y � is also continuous on Y with weak topology. Composition of

two continuous functions is continuous.

(ii) ⇒ (i). Take an arbitrary B(I0 , �), where I0 is a finite subset of Y � . By (ii), ∀g ∈ I0 ∃Bg , a neighborhood of 0 in X : u ∈ Bg ⇒ |g(T (u))| < � It follows from the definition of filter of neighborhoods that B=

�

Bg

g∈I0

is also a neighborhood of 0. Consequently, u ∈ B ⇒ |g(T (u))| < � ⇒ T u ∈ B(I0 , �) To conclude the final result, it is sufficient now to show that, for any g ∈ Y � , g ◦ T : X (with weak topology) → R I is continuous. But g ◦ T , as a composition of continuous functions, is a strongly continuous linear

functional and, consequently, it is continuous in the weak topology as well (compare the discussion in the text). Exercise 5.14.3 Consider space c0 containing infinite sequences of real numbers converging to zero, equipped with �∞ -norm. def

c0 = {x = {xn } : xn → 0},

�x� = sup |xi | i

Show that (a) c�0 = �1 (b) c��0 = �∞ (c) If en = (0, . . . , 1(n) , . . .) then en → 0 weakly∗ but it does not converge to zero weakly.

Banach Spaces

147

(a) We follow the same reasoning as in the Exercise 5.12.3. Define N �

xN =

xi ei = (x1 , . . . , xN , . . .)

i=1

It follows from the definition of c0 -space that �x − xN � = sup |xi | → 0 i>N

Let f ∈ c�0 and set φi = f (ei ). Then ∞ �

def

φ i xi =

i=1

lim

N →∞

N �

φi xi = lim f (xN ) = f (x)

i=1

N →∞

Consequently, |f (x)| ≤ �φ��1 �x� In order to show that the bound equals the supremum, it is sufficient to take a sequence of vectors xN = (sgn φ1 , . . . , sgn φN , 0, . . .) ∈ c0 Then f (xN ) =

N � i=1

|φi | →

∞ � i=1

|φi |

(b) This follows from ��1 = �∞ . (c) We have < eN , x >�1 ×c0 = xN → 0,

∀x ∈ c0

but < φ, eN >�∞ ×�1 = 1 � 0 for φ = (1, 1, . . .) ∈ �∞ . Exercise 5.14.4 Let U and V be normed spaces, and let either U or V be reflexive. Prove that every operator A ∈ L(U, V ) has the property that A maps bounded sequences in U into sequences having weakly

convergent subsequences in V .

Case: V reflexive. Map A maps a bounded sequence in U into a bounded sequence in V . In turn, any bounded sequence in reflexive space V has a weakly convergent subsequence. Case: U reflexive. Any bounded sequence un in reflexive space U has a weakly convergent subsequence unk . As A is also weakly continuous ( Recall Exercise 5.14.2 ), it follows that Aunk is weakly convergent in V .

148

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Exercise 5.14.5 In numerical analysis, one is often faced with the problem of approximating an integral of a given continuous function f ∈ C[0, 1] by using some sort of numerical quadrature formula. For instance, we might introduce in [0, 1] a sequence of integration points 0 ≤ xn1 < xn2 < · · · < xnj < · · · < xnn ≤ 1, and set def

Qn (f ) =

n �

k=1

ank f (xnk ) ≈

�

n = 1, 2, . . .

1

f (x) dx 0

where the coefficients ank satisfy the condition n �

k=1

|ank | < M,

∀n≥1

Suppose that the quadrature rule Qn (f ) integrates polynomials p(x) of degree n − 1 exactly; i.e., Qn (p) = (a) Show that, for every f ∈ C[0, 1], lim

n→∞

�

�

1

p(x)dx 0

Qn (f ) −

�

1

f (x)dx 0

�

=0

(b) Characterize the type of convergence; this limit defines in terms of convergence in the space C[0, 1] (equipped with the Chebyshev norm). (a) We start with a simple abstract result. Lemma. Let U be a normed space, X a dense subspace of U , and fn ∈ U � a uniformly bounded

sequence of continuous linear functionals on U , i.e., �fn �U � ≤ M , for some M > 0. Assume

that the sequence converges to zero on X, fn (x) → 0, ∀x ∈ X. Then, the sequence converges to zero on the entire space,

fn (u) → 0, ∀u ∈ U Proof follows from the simple inequality, |fn (u)| = |fn (u − x) + fn (x)| ≤ �fn � �u − x� + |fn (x)| ≤ M �u − x� + |fn (x)| Given u ∈ U , and � > 0, select x ∈ X such that �u − x� < �/2M , and then N such that |fn (x)| < �/2 for n ≥ N . It follows from the inequality above that, for n ≥ N , |fn (x)| < �.

According to Weierstrass Theorem on polynomial approximation of continuous functions (see, e.g. []), polynomials are dense in space C([0, 1]). Functionals C([0, 1]) � f → Qn (f ) −

�

1 0

f (x)dx ∈ R I

Banach Spaces

149

converge to zero for all polynomials f (for n exceeding the order of polynomial f , the functionals are identically zero). At the same time, condition on the integration points implies that the functionals are uniformly bounded. Consequently, by the lemma above, sequence Qn (f ) −

�

1 0

f (x)dx → 0

for any f ∈ C([0, 1]). (b) The sequence converges to zero in weak∗ topology.

5.15

Compact (Completely Continuous) Operators

Exercises Exercise 5.15.1 Let T : U → V be a linear continuous operator from a normed space U into a reflexive

Banach space V . Show that T is weakly sequentially compact, i.e., it maps bounded sets in U into sets

whose closures are weakly sequentially compact in V . A is bounded in U ⇒ T (A) is weakly sequentially compact in V . This is a simple consequence of the fact that bounded sets in a reflexive Banach space are weakly sequentially compact. Exercise 5.15.2 Let U and V be normed spaces. Prove that a linear operator T : U → V is compact iff T (B) is precompact in V for B - the unit ball in U .

Assume T is linear and maps unit ball in U into a precompact set in V . Let C be an arbitrary bounded set in U , �u�U ≤ M,

∀u ∈ C

Set M −1 C is then a subset of unit ball B and, consequently, M −1 T (C) is a subset of T (B). Thus, M −1 T (C) ⊂ T (B) as a closed subset of a compact set, is compact as well. Finally, since multiplication by a non-zero constant is a homeomorphism, T (C) is compact as well. Exercise 5.15.3

Use the Frechet–Kolmogorov Theorem (Theorem 4.9.4) to prove that operator T from

R) Example 5.15.1 with an appropriate condition on kernel K(x, ξ) is a compact operator from Lp (I R), 1 ≤ p, r < ∞. into Lr (I

150

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

According to Exercise 5.15.2 above, we can restrict ourselves to a unit ball B ∈ Lp (I R) and seek

conditions that would guarantee that T (B) is precompact in Lr (I R). By the Frechet-Kolmogorov Theorem, we need to come up with sufficient conditions on kernel K(y, x) to satisfy the following three conditions. R). (i) T (B) is bounded in Lr (I (ii)

�

R I

(iii)

t→0

|T u(t + s) − T u(s)|r ds −→ 0, �

n→∞

|s|>n

|T u(s)|r ds −→ 0,

uniformly in u ∈ B

uniformly in u ∈ B

We shall restrict ourselves to a direct use of H¨older inequality only. We have the following estimates. (i)

�r � rq � �� � �� � � � � � K(y, x)u(y) dy � dx ≤ � |K(y, x)|q dy � � � � � R I R I R I R I

≤1 � rq � �� � � � |K(y, x)|q dy � dx =: A ≤ � � R I R I

(ii)

(iii)

� pr �� � � � |u(y)|p dy � dx � � I �R �� �

�r � �� � � � � K(y, t + s)u(y) dy − � ds K(y, s)u(y) dy � � R I R I R I � � � � � �r � [K(y, t + s) − K(y, s)]u(y) dy � ds = � � R I R I � rq �� � pr � �� � � � � � |K(y, t + s) − K(y, s)|q dy � � |u(y)|p dy � ds ≤ � � � � I R I R I �R �� � ≤1 � rq � �� � � � |K(y, t + s) − K(y, s)|q dy � ds =: B(t) ≤ � � R I R I

�� �r �� � rq � � � � � � K(y, s)u(y) dy � ds ≤ � |K(y, s)|q dy � � � � � I I |s|>n R |s|>n R

�

≤

�

|s|>n

|

�

R I

q

|K(y, s)| dy|

r q

�� � pr � � � |u(y)|p dy � ds � � I �R �� � ≤1

ds =: C(n)

where q is the conjugate index to p. Consequently, if A < ∞, limt→0 B(t) = 0, limn→∞ C(n) = 0,

R) into Lr (I R). In fact, the first condition implies the last then operator T is a compact map from Lp (I one and, if we assume that kernel K(y, x) is continuous a.e., we can use the Lebesgue Dominated Convergence Theorem to show that the second condition is verified as well. Indeed, the integrand

Banach Spaces

151

in B(t) converges pointwise a.e. to zero, as t → 0, and we can construct a dominating function by utilizing convexity of function xq for q ≥ 1, �q � 1 1 1 1 |K(y, t + s)| + |K(y, s)| ≤ |K(y, t + s)|q + |K(y, s)|q 2 2 2 2

which in turn implies

|K(y, t + s) − K(y, s)|q ≤ 2q−1 (|K(y, t + s)|q + |K(y, s)|q )

Closed Range Theorem, Solvability of Linear Equations

5.16

Topological Transpose Operators, Orthogonal Complements

Exercises Exercise 5.16.1 Prove Proposition 5.16.1(i)–(iv). The properties follow immediately from the properties of the algebraic transpose operator. Exercise 5.16.2 Let U, V be two Banach spaces, and let A ∈ L(U, V ) be compact. Show that A� is also compact. Hint: See Exercise 5.21.2 and recall Arzel`a–Ascoli Theorem. See proof of Lemma 5.21.5.

5.17

Solvability of Linear Equations in Banach Spaces, The Closed Range Theorem

Exercises Exercise 5.17.1 Let X be a Banach space, and P : X → X be a continuous linear projection, i.e., P 2 = P . Prove that the range of P is closed.

Let un ∈ R(P ), un → u. We need to show that u ∈ R(P ) as well. Let vn ∈ X be such that

un = P vn . Then P un = P 2 vn = P vn = un → P u. By the uniqueness of the limit, it must be u = P u. Consequently, u is the image of itself and must be in the range of the projection.

152

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Exercise 5.17.2 Let X be a Banach space, and M ⊂ X a closed subspace. Consider the map: ι : X � ⊃ M ⊥ � f → f˜ ∈ (X/M )�

f˜(x + M ) := f (x) .

Here M ⊥ is the orthogonal complement of M , M ⊥ = {f ∈ X � : f |M = 0} . Prove that map ι is an isometric isomorphism. Consequently, (X/M )� and M ⊥ can be identified with each other. The map is well-defined, i.e. f (x + m) = f (x), for every m ∈ M . Next, |f˜(x + M )| = |f (x + m)| ≤ �f �X � �x + m� . Taking infimum in m ∈ M on the right-hand side, we have, |f˜(x + M )| ≤ �f �X � �x + M �X/M , i.e. �f˜�(X/M )� ≤ �f �X � . On the other side, |f (x)| = |f˜(x + M )| ≤ �f˜�(X/M )� �x + M �X/M ≤ �f˜�(X/M )� �x� , so also, �f �X � ≤ �f˜�(X/M )� . Map ι is thus an isometry (and, therefore, injective). To show surjectivity, take g ∈ (X/M )� , and set f (x) = g(x + M ). Then, for any x ∈ M ,

f (x) = g(x + M ) = g(M ) = 0 so, f ∈ M ⊥ and, finally, f˜ = g.

( g is a linear map)

Banach Spaces

5.18

153

Generalization for Closed Operators

Exercises Exercise 5.18.1 Let X, Y be two normed spaces with Y being complete. Let X be a dense subspace of

X. Let A be a linear and continuous map from X into Y . Prove that operator A admits a unique continuous extension A˜ to the whole space X that preserves the norm of A. Hint: For x ∈ X, take

xn → x, xn ∈ X , and investigate sequence Axn .

Let x ∈ X and let xn → x, xn ∈ X . Sequence Axn is Cauchy in Y . Indeed, �Axn − Axm �Y ≤ �A�L(X ,Y ) �xn − xm � and xn , as a convergent sequence, is Cauchy in X. By completness of Y , sequence Axn has a unique ˜ := y = limn→∞ Axn . First of all, limit y that can be identified as the value of the extension, Ax extension A˜ is well defined, i.e. the limit y is independent of xn . Indeed, if we take another sequence converging to x, say z n → x, then �Axn − Az n �Y ≤ �A� �xn − z n �Y ≤ �A� (�xn − x�X + �x − z n �X ) → 0 Secondly, if xn → x and z n → z then αxn + βz n → αx + βz and, passing to the limit in: A(αxn + βz n ) = αAxn + βAz n we learn that A˜ is linear. Finally, by passing to the limit in �Axn � ≤ �A�L(X ,Y ) �xn � ˜ ≤ �A� and, therefore, �A� ˜ = �A�. we learn that A˜ is continuous and �A� Exercise 5.18.2 Discuss in your own words why the original definition of the transpose for a closed operator and the one discussed in Remark 5.18.1 are equivalent. The original definition requires the identity: �y � , Ax� = �x� , x�

∀x ∈ D(A)

for some x� ∈ X � and then sets A� y � := x� . Embedded in the condition is thus the requirement that

A� y � is continuous. Conversely, the more explicitly defined transpose satisfies the identity above with x� = A� y � . Exercise 5.18.3 Prove Proposition 5.18.1.

154

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

(i) By definition, y � ∈ D(A�1 ) and x�1 = A�1 y � , if < y � , A1 x > = < x�1 , x >

∀x ∈ D

Similarly, y � ∈ D(A�2 ) and x�2 = A�2 y � , if < y � , A2 x > = < x�2 , x >

∀x ∈ D

Let y � ∈ D(A�1 ) ∩ D(A�2 ). Taking a linear combination of the equalities above, we get < y � , (α1 A1 + α2 A2 )x > = < α1 x�1 + α2 x�2 , x >

∀x ∈ D

Consequently, y � ∈ D((α1 A1 + α2 A2 )� ) and (α1 A1 + α2 A2 )� y � = α1 x�1 + α2 x�2 = α1 A�1 y � + α2 A�2 y � (ii) Again, by definition, y � ∈ D(A� ) and x� = A� y � , if < y � , Ax > = < x� , x >

∀x ∈ D(A)

Similarly, z � ∈ D(B � ) and y � = B � z � , if < z � , By > = < y � , y >

∀y ∈ D(B)

Taking y = Ax in the second equality and making use of the first one, we get, < z � , BAx > = < y � , Ax > = < x� , x >

∀x ∈ D(A)

Consequently, z � ∈ D((BA)� ) and (BA)� z � = x� = B � y � = B � A� z � (iii) It is sufficient to notice that < y � , Ax > = < x� , x >

∀x ∈ D(A)

is equivalent to < y � , y > = < x� , A−1 y >

∀y ∈ D(A−1 )

Banach Spaces

5.19

155

Closed Range Theorem for Closed Operators

Exercises Exercise 5.19.1 Prove property (5.4). Let z � ∈ (M + N )⊥ . By definition, �z � , m + n� = 0

∀ m ∈ M, n ∈ N

Setting n = 0, we have, �z � , m� = 0 ∀ m ∈ M i.e. z � ∈ M ⊥ . By the same argument, z � ∈ N ⊥ . Similarly, if �z � , m� = 0 ∀ m ∈ M

and

�z � , n� = 0

∀n ∈ N

then, by linearity of z � , �z � , m + n� = �z � , m� + �z � , n� = 0 Exercise 5.19.2 Let Z be a Banach space and X ⊂ Z a closed subpace of Z. Let Z = X ⊕ Y for some Y , and let PX : Z → X be the corresponding projection, PX z = x where z = x + y is the unique

decomposition of z. Prove that projection PX is a closed operator. Assume z n = xn + y n Let

(z n , xn ) = (z n , PX z n ) → (z, x) Then y n = z n − xn → z − x =: y so, z = x + y. This proves that x = PX z, i.e. (z, x) ∈ graph of PX .

Exercise 5.19.3 Prove the algebraic identities (5.6). Elementary. Exercise 5.19.4 Let X, Y be two topological vector spaces. Prove that a set B ⊂ Y is closed in Y if and only if set X + B is closed in X × Y . Here X is identified with X × {0} ⊂ X × Y and B is identified with {0} × B ⊂ X × Y .

Elementary. The whole trouble lies in the identification. We have, X × {0} + {0} × B = X × B and the assertion follows from the construction of topology in Cartesian product.

156

5.20

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Examples

Exercises Exercise 5.20.1 Prove that the linear mapping (functional) H 1 (0, 1) � w → w(x0 ), where x0 ∈ [0, l] is continuous. Use the result to prove that space W in Example 5.20.1 is closed. Hint: Consider first smooth functions w ∈ C ∞ ([0, l]) and then use the density of C ∞ ([0, l]) in H 1 (0, l).

Take w ∈ C ∞ ([0, l]). Function w(x) ˆ := w(x) −

1 l

�

l

w(s) ds 0

is continuous and has a zero average. By the Mean-Value Theorem of Integral Calculus, there exists an intermediate point c ∈ [0, l] such that w(c) ˆ = 0. Integrating w ˆ � = w� from c to x ∈ [0, l], we obtain � x � x w� (s) ds = w ˆ � (s) ds = w(x) ˆ c

c

Consequently,

1 l

w(x) = This implies the estimate 1 |w(x)| ≤ l

��

0

� 12 ��

l

�

l

w(s) ds + 0

l 2

1

0

|w(s)| ds

≤ �w�L2 (0,l) + l�w� �L2 (0,l)

� 12

�

+

x

w� (s) ds

c

��

� 12 ��

x

x

1 c

c

|w� (s)|2 ds

� 12

≤ max{1, l}�w�H 1 (0,l) By the density argument, the estimate generalizes to w ∈ H 1 (0, l). Applying the result to higher derivatives, we conclude that operator I4 f : H 4 (0, l) � w → (w�� (0), w��� (0), w�� (l), w��� (l)) ∈ R is continuous. Consequently W = f −1 ({0}) as inverse image of a closed set, must be closed. Exercise 5.20.2 Let u, v ∈ H 1 (0, l). Prove the integration by parts formula � l � l uv � dx = − u� v dx + (uv)|l0 0

0

Banach Spaces

157

Hint: Make use of the density of C ∞ ([0, l]) in H 1 (0, l). Integrate (uv)� = u� v + uv � , to obtain the formula for smooth functions u, v ∈ C ∞ ([0, l]). Let

u, v ∈ H 1 (0, l). By the density argument, there exist seqeunces un , vn ∈ C ∞ ([0, l]) such that un →

u, vn → v in H 1 (0, l). For each n, we have � l � l � un vn dx = − u�n vn dx + (un vn )|l0 0

0

The point is that both sides of the equality represent continuous functional on space H 1 (0, l). Integrals represent L2 -products, and the boundary terms are continuous by the result of Exercise 5.20.1 above. Consequently, we can pass on both sides to the limit with n → ∞, to obtain the final result. Exercise 5.20.3 Work out all the details of Example 5.20.1 once again, with different boundary conditions: w(0) = w�� (0) = 0 and w�� (l) = w��� (l) = 0 (left end of the beam is supported by a pin support). We follow precisely the same lines to obtain W = {w ∈ H 4 (0, l) : w(0) = w�� (0) = w�� (l) = w��� (l) = 0} The transpose operator is defined on the whole L2 (0, l) by the same formula as before (but different W ). The corresponding null space is N (A� ) = {v ∈ L2 (0, l) : v ���� = 0 and v(0) = v �� (0) = v �� (l) = v ��� (l) = 0} I = {v ∈ L2 (0, l) : v(x) = αx, α ∈ R} The necessary and sufficient condition for the existence of a solution w ∈ W , q ∈ N (A� )⊥ = 0 is equivalent to

�

l

q(x)x dx = 0 0

(the moment of active load q with respect to the pin must vanish). The solution is determined up to a linearized rotation about x = 0 (the pin). Exercise 5.20.4 Prove that operator A from Remark 5.20.1 is closed. Consider a sequence of functions (un , qn ) ∈ L2 (0, l) × L2 (0, l) such that 2 �� �� ��� ���� un ∈ D(A), i.e., u���� n ∈ L (0, l), un (0) = un (0) = un (l) = un (l) = 0 and un = qn

Assume un → u, qn → q. The question is: Does u ∈ D(A), Au = q ? Recall the definition of distributional derivative, def

���� < u���� >= n , φ > = < un , φ

�

l 0

un φ���� =

�

l

qn φ 0

∀φ ∈ D(0, l)

158

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Keeping φ fixed, we pass to the limit with n → ∞ to learn that � l � l < u���� , φ >= uφ���� = qφ ∀φ ∈ D(0, l) 0

i.e., u

����

0

= q. The boundary conditions represent continuous functionals on H 4 (0, l) and, therefore,

they are satisfied in the limit as well. Exercise 5.20.5 (A finite-dimensional sanity check). Determine necessary conditions on data f for solutions to the linear systems of equations that follow to exist. Determine if the solutions are unique and, if not, describe the null space of the associated operator: Au = f Here

�

1 −1 0 A= −1 0 1

�



 1 2 −1 A = 4 0 2 3 −2 −3

 3 1 −1 2 A =  6 2 −2 4  9 3 −3 6 

We shall discuss the first case, the other two being fully analogous. Matrix operator A : R I3 →R I 2.

Identifying the dual of R I n with itself (through the canonical inner product), we have   1 −1 I 2 →R I 3 , A =  −1 0  A� : R 0 1

The null space of A� is trivial and, consequently, the linear problem has a solution for any right-hand side f . The solution is determined up to elements from the kernel of A, N (A) = {(t, t, t) : t ∈ R} I In elementary language, one can set x1 = t, and solve the remaining 2 × 2 system for x2 , x3 to obtain x2 = −f1 + t, x3 = f2 − t

5.21

Equations with Completely Continuous Kernels, Fredholm Alternative

Exercises Exercise 5.21.1 Complete the proof of Lemma 5.21.6. Remark: We use the same notation as in the text. A linear functional may be denoted with both standard and boldface symbols when we want to emphasize the vectorial character of the dual space.

Banach Spaces

159

Case: n ≥ m. Define operator def

P = T � + Q,

Qf =

m �

f (y k )f k

k=1

Operator I − P is injective. Indeed, assume f − P f = f − T � f − Qf = A� f − Qf = 0 Evaluating both sides at xi , we obtain, < A� f , xi > −

m �

f (y k ) < f k , xi >= 0

k=1

which implies < f , Axi > −f (y i ) = 0 But Axi = 0 so f (y i ) = 0, i = 1, . . . , m and, consequently, A� f = 0, i.e., f ∈ N (A� ). This implies

that f can be represented in terms of functionals gi , f=

m �

bi g i

i=1

Evaluating both sides at vectors y i , we conclude that bi = 0, i = 1, . . . , m and, therefore, f = 0. This proves that I − P is injective. By Corollary 5.21.2, I − P is surjective as well. There exists thus a solution f¯ to the problem m � f¯(y k )f k = f m+1 A� f¯ − k=1

Evaluating both sides at xn+1 we arrive at a contradiction, the left-hand side vanishes while the right-

hand one is equal to one. This proves that n = m. Exercise 5.21.2 Let X, d be a complete metric space and let A ⊂ X. Prove that the following conditions are equivalent to each other.

(i) A is precompact in X, i.e., A is compact in X. (ii) A is totally bounded. (iii) From every sequence in A one can extract a Cauchy subsequence. (i) ⇒ (ii). Since A is compact then, by Theorem 4.9.2, it is totally bounded, i.e., for every � > 0, there exists an �-net Y� ⊂ A, i.e.,

A⊂A⊂

�

B(y� , �)

y∈Y�

It remains to show that one can select �-nets from A itself. Take � > 0. Let Y 2� be the corresponding � 2 -net

in A. For each z ∈ Y 2� , there exists a corresponding yz ∈ A such that d(y, z) < �/2. It follows

from the triangle inequality that

{zy : y ∈ Y 2� } ⊂ A

160

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

is an �-net for set A. (ii) ⇒ (i). By Theorem 4.9.2 again, it is sufficient to show that A is totally bounded. Take � > 0 and

select an arbitrary �1 < �. Then �1 -net Y�1 for A is an �-net for A. Indeed, A⊂ implies A⊂

�

y∈Y�1

�

B(y, �1 )

y∈Y�1

B(y, �1 ) ⊂

�

B(y, �)

y∈Y�1

(i) ⇒ (iii). Let xn ∈ A. As A is sequentially compact, one can extract a convergent and, therefore, Cauchy subsequence xnk .

(iii) ⇒ (i). Let xn ∈ A. We need to demonstrate that there exists a subsequence xnk converging to x ∈ A. For each n, select yn ∈ A such that d(yn , xn )
0 (compare Exercise 3.9.2). Does the result extend to complex vector spaces?

164

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

The result is true for complex spaces. Let v = αu, α > 0. Then both sides are equal to (1 + α)�u�. Conversely, squaring the left-hand side of the equality above, �u + v�2 = (u + v, u + v) = �u�2 + 2Re (u, v) + �v�2 and comparing it with the square of the right-hand side, we learn that Re (u, v) = �u� �v� Consider now a function of a real argument α, �u − αv�2 = (u − αv, u − αv) = �u�2 − 2αRe (u, v) + α2 �v�2 = α2 �v�2 − 2α�u� �v� + �u�2 Quadratic function on the right-hand side has a minimum equal zero at α = �v�/�u� > 0. Conse-

quently, the left-hand side must vanish as well which implies that u − αv = 0.

Exercise 6.1.5 Let {un } be a sequence of elements in an inner product space V . Prove that if (un , u) −→ (u, u)

and

�un � −→ �u�

then un −→ u, i.e., �un − u� −→ 0. We have �un − u�2 = (un − u, un − u) = (un , un ) − (u, un ) − (un , u) + (u, u) = �un �2 − (u, un ) − (un , u) + �u�2 → 2�u�2 − 2(u, u) = 0 since (u, un ) = (un , u) → (u, u). Exercise 6.1.6 Show that the sequence of sequences u1 = (α1 , 0, 0, . . .) u2 = (0, α2 , 0, . . .) u3 = (0, 0, α3 , . . .) etc., where the αi are scalars, is an orthogonal sequence in �2 , i.e., (un , um ) = 0 for m �= n. Apply definition of the inner product in �2 . Exercise 6.1.7 Let A : U → V be a linear map from a Hilbert space U, (·, ·)U into a Hilbert space V, (·, ·)V . Prove that the following conditions are equivalent to each other, (i) A is unitary, i.e., it preserves the inner product structure, (Au, Av)V = (u, v)U

∀u, v ∈ U

(ii) A is an isometry, i.e., it preserves the norm, �Au�V = �u�U

∀u ∈ U

(i)⇒(ii). Substitute v = u. (ii)⇒(i). Use the parallelogram law (polarization formula) discussed in Exercise 6.1.2.

Hilbert Spaces

6.2

165

Orthogonality and Orthogonal Projections

Exercises Exercise 6.2.1 Let V be an inner product space and M, N denote vector subspaces of V . Prove the following algebraic properties of orthogonal complements: (i) M ⊂ N ⇒ N ⊥ ⊂ M ⊥ . (ii) M ⊂ N ⇒ (M ⊥ )⊥ ⊂ (N ⊥ )⊥ . (iii) M ∩ M ⊥ = {0}. (iv) If M is dense in V , (M = V ) then M ⊥ = {0}. (i) Let v ∈ N ⊥ . Then (n, v) = 0 ∀n ∈ N

⇒

(n, v) = 0 ∀n ∈ M

⇒

v ∈ M⊥

(ii) Apply (i) twice. (iii) Let v ∈ M ∩ M ⊥ . Then v must orthogonal to itself, i.e., (v, v) = 0 ⇒ v = 0. (iv) Let v ∈ M ⊥ and M � v n → v. Passing to the limit in (v n , v) = 0 we get (v, v) = 0 ⇒ v = 0. Exercise 6.2.2 Let M be a subspace of a Hilbert space V . Prove that M = (M ⊥ )⊥ By Corollary 6.2.1,

It is sufficient thus to show that

� ⊥ �⊥ M= M M⊥ = M

As M ⊂ M , Exercise 6.2.1(i) implies that M

M � mn → m ∈ M . Passing to the limit in

⊥

⊥

⊂ M ⊥ . Conversely, assume v ∈ M ⊥ , and let

(mn , v) = 0 we learn that v ∈ M

⊥

as well.

166

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Exercise 6.2.3 Two subspaces M and N of an inner product space V are said to be orthogonal, denoted M ⊥ N , if

∀ m ∈ M, n ∈ N

(m, n) = 0,

Let V now be a Hilbert space. Prove or disprove the following: (i) M ⊥ N =⇒ M ⊥ ⊥ N ⊥ . (ii) M ⊥ N =⇒ (M ⊥ )⊥ ⊥ (N ⊥ )⊥ . I The first assertion is false. Consider, e.g. R I 3 with the canonical inner product. Take M = R×{0}×{0} and N = {0} × R I × {0}. Then I ×R I M ⊥ = {0} × R

and

N⊥ =R I × {0} × R I

Obviously, spaces M ⊥ and N ⊥ are not orthogonal. To prove the second assertion, in view of Exercise 6.2.2, it is sufficient to show that M ⊥N ⇒ M ⊥N But this follows immediately from the continuity of the inner product. Take M � mn → m ∈ M and N � nn → n ∈ N , and pass to the limit in

(mn , nn ) = 0

Exercise 6.2.4 Let Ω be an open, bounded set in R I n and V = L2 (Ω) denote the space of square integrable functions on Ω. Find the orthogonal complement in V of the space of constant functions � � M = u ∈ L2 (Ω) : u = const a.e. in Ω

Let f ∈ L2 (Ω). Projection of f onto M is equivalent to the variational problem:  u ∈R I   f � �  u v = f v ∀v ∈ R I  f Ω

Ω

Selecting v = 1, we learn that uf is the average of f , u0 =

1 meas(Ω)

�

f Ω

Orthogonal complement M ⊥ contains functions f − uf , i.e., functions of zero average, � M ⊥ = {f ∈ L2 (Ω) : = 0} Ω

(compare Example 2.??)

Hilbert Spaces

167

I C ) a sequence of measurable functions. Exercise 6.2.5 Let Ω ⊂ R I N be a measurable set and fn : Ω → R(I

We say that sequence fn converges in measure to a measurable function f : Ω → R(I I C ) if, for every

ε > 0,

m({x ∈ Ω : |fn (x) − f (x)| ≥ ε}) → 0

as

n→0

Let now m(Ω) < ∞. Prove that Lp (Ω) convergence, for any 1 ≤ p ≤ ∞, implies convergence in measure. Hint:  �� � p1   1  |fn (x) − f (x)|p dx Ω m({x ∈ Ω : |fn (x) − f (x)| ≥ ε}) ≤ ε  1   ess supx∈Ω |fn (x) − f (x)| ε

1≤p 0, one can extract a subsequence fnk such that m ({x ∈ Ω : |fnk (x) − f (x)| ≥ ε}) ≤

1 2k+1

∀k ≥ 1

Step 2. Use the diagonal choice method to show that one can extract a subsequence fnk such that m({x ∈ Ω : |fnk (x) − f (x)| ≥

1 1 }) ≤ k+1 k 2

Consequently, m({x ∈ Ω : |fnk (x) − f (x)| ≥ ε}) ≤ for every ε > 0, and for k large enough.

∀k ≥ 1

1 2k+1

168

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Step 3. Let ϕk = fnk be the subsequence extracted in Step 2. Use the identities {x ∈ Ω : inf sup |ϕn (x) − f (x)| > 0} = ν≥0 n≥ν

{x ∈ Ω : inf sup |ϕn (x) − f (x)| ≥ ε} = ν≥0 n≥ν

to prove that

�

{x ∈ Ω : inf sup |ϕn (x) − f (x)| ≥

k �

ν≥0

ν≥0 n≥ν

1 } k

{x ∈ Ω : sup |ϕn (x) − f (x)| ≥ ε} n≥ν

m({x ∈ Ω : lim sup |ϕn (x) − f (x)| > 0}) n→∞

≤ Step 4. Use the identity

� k

lim m({x ∈ Ω : sup |ϕn (x) − f (x)| ≥

ν→∞

n≥ν

{x ∈ Ω : sup |ϕn (x) − f (x)| > n≥ν

1 }) k

� 1 1 }⊂ {x ∈ Ω : |ϕn (x) − f (x)| > } k k n≥ν

and the result of Step 2 to show that m({x ∈ Ω : sup |ϕn (x) − f (x)| ≥ ε}) ≤ n≥ν

1 2ν

for every ε > 0 and (ε-dependent!) ν large enough. Step 5. Use the results of Step 3 and Step 4 to conclude that m({x ∈ Ω : lim fnk (x) �= f (x)}) = 0 k→∞

Remark: The Lebesgue Dominated Convergence Theorem establishes conditions under which pointwise convergence of a sequence of functions fn to a limit function f implies the Lp -convergence. While the converse, in general, is not true, the results of the last two exercises at least show that the Lp convergence of a sequence fn implies the pointwise convergence (almost everywhere only, of course) of a subsequence fnk . Step 1. This follows directly from the definition of a convergent sequence. We have, ∀δ ∃N n ≥ N m({x ∈ Ω : |fn (x) − f (x)| ≥ ε}) ≤ δ

∀n

Select δ = 1 and an element fn1 that satisfies the condition for δ = 1/2. By induction, given n1 , . . . , nk−1 , select nk > n1 , . . . , nk−1 such that fnk satisfies the condition for δ = 1/2k+1 . Notice that avoding the duplication (enforcing injectivity of function k = nk ) is possible since we have an infinite number of elements of the sequence at our disposal. Step 2. Use Step 1 result for ε = 1. In particular, the subsequence converges alond with the orginal sequence. Take then ε = 1/2 and select a subsequence of the first subsequence fnk (denoted with the same symbol) to staisfy the same condition. Proceed then by induction. The diagonal subsequence satisfies the required condition.

Hilbert Spaces

169

Step 3. By Proposition 3.1.6v, m({x ∈ Ω : inf sup |ϕn (x) − f (x)| ≥ ν≥0 n≥ν

1 1 }) = lim m({x ∈ Ω : sup |ϕn (x) − f (x)| ≥ }) ν→∞ k k n≥ν

The final condition is then a consequence of the equality above, the first identity in Step 3, and subadditivity of the measure. Step 4. Given ε, choose k such that � > k1 . The identity and subadditivity of the measure imply that 1 }) k n≥ν � 1 }) ≤ m({x ∈ Ω : |ϕn (x) − f (x)| > k−1 n≥ν ∞ � 1 1 ≤ ≤ k+1 2 ν ν

m({x ∈ Ω : sup | ϕn (x) − f (x)| ≥ ε}) ≤ m({x ∈ Ω : sup | ϕn (x) − f (x)| > n≥ν

Step 5. Combining results of Step 3 and Step 4, we get

m({x ∈ Ω : lim sup |ϕn (x) − f (x)| > 0}) = 0 n→∞

which is equivalent to the final assertion.

6.3

Orthonormal Bases and Fourier Series

Exercises Exercise 6.3.1 Prove that every (not necessarily separable) nontrivial Hilbert space V possesses an orthonormal basis. Hint: Compare the proof of Theorem 2.4.3 and prove that any orthonormal set in V can be extended to an orthonormal basis. Let A0 be an orthonormal set. Let U be a class of orthonormal sets A (i.e., A contains unit vectors, and

every two vectors are orthogonal to each other) containing A0 . Obviously, U is nonempty. Family U is partially ordered by the inclusion. Let Aι , ι ∈ I, be a chain in U . Then � � A := Aι ∈ U and Aκ ⊂ Aι , ∀κ ∈ I ι∈I

ι∈I

Indeed, linear ordering of the chain implies that, for each two vectors in u, v ∈ A, there exists a common index ι ∈ I such that u, v ∈ Aι . Consequently, u and v are orthogonal. By the Kuratowski-

Zorn Lemma, U has a maximal element, i.e., an orthonormal basis for space U that contains A0 . To conclude the final result, pick an arbitrary vector u1 �= 0, and set A0 = {u/�u�}.

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APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Exercise 6.3.2 Let {en }∞ n=1 be an orthonormal family in a Hilbert space V . Prove that the following conditions are equivalent to each other.

(i) {en }∞ n=1 is an orthonormal basis, i.e., it is maximal. ∞ � (u, en ) en ∀u∈V. (ii) u = n=1

(iii) (u, v) = 2

(iv) �u� = (i)⇒(ii). Let

∞ �

n=1 ∞ �

n=1

(u, en ) (v, en ). 2

|(u, en )| .

uN :=

N �

uN → u

uj ej ,

j=1

Multiply both sides of the equality above by ei , and use orthonormality of ej to learn that ui = (uN , ei ) → (u, ei ) as N → ∞ (ii)⇒(iii). Use orthogonality of ei to learn that (uN , v N ) =

N �

ui v i =

i=1

N � i=1

(u, ei ) (v, ei ) →

∞ �

(u, ei ) (v, ei )

i=1

(iii)⇒(iv). Substitute v = u. (iv)⇒(i). Suppose, to the contrary, the {e1 , e2 , . . .} can be extended with a vector u �= 0 to a bigger

orthonormal family. Then u is orthogonal with each ei and, by property (iv), �u� = 0. So u = 0, a contradiction.

Exercise 6.3.3 Let {en }∞ n=1 be an orthonormal family (not necessarily maximal) in a Hilbert space V . Prove Bessel’s inequality

∞ � i=1

2

2

|(u, ei )| ≤ �u�

∀u∈V

Extend the family to an orthonormal basis (see Exercise 6.3.1), and use the property (iv) proved in Exercise 6.3.2. Exercise 6.3.4 Prove that every separable Hilbert space V is unitary equivalent with the space �2 . Hint: Establish a bijective correspondence between the canonical basis in �2 and an orthonormal basis in V and use it to define a unitary map mapping �2 onto V . Let e1 , e2 , . . . be an orthonormal basis in V . Define the map T : �2 � (x1 , x2 , . . .) →

∞ � i=1

xi ei =: x ∈ V

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171

Linearity is obvious. for N > M ,

�∞

i=1

|xi |2 < ∞ implies that sequence xN =

�uN − uM �2 =

N �

i=M +1

∞ �

|xi |2 ≤

i=M +1

�N

i=1

xi ei is Cauchy in V . Indeed,

|xi |2 → 0 as M → ∞

By completeness of V , the series converges, i.e., the map is well defined. By Exercise 6.3.2(iv), the map is a surjection. Orthonormality of ei implies that it is also an injection. Finally, it follows from the definition that the map is unitary. Exercise 6.3.5 Prove the Riesz–Fisher Theorem. Let V be a separable Hilbert space with an orthonormal basis {en }∞ n=1 . Then � �∞ ∞ � � 2 vn en : |vn | < ∞ V = n=1

n=1

In other words, elements of V can be characterized as infinite series

∞ �

vn en with �2 -summable

n=1

coefficients vn . See Exercise 6.3.4

Exercise 6.3.6 Let I = (−1, 1) and let V be the four-dimensional inner product space spanned by the monomials {1, x, x2 , x3 } with (f, g)V =

�

1

f g dx −1

(i) Use the Gram-Schmidt process to construct an orthonormal basis for V . (ii) Observing that V ⊂ L2 (I), compute the orthogonal projection Πu of the function u(x) = x4 onto V .

(iii) Show that (x4 − Πx4 , v)L2 (I) = 0

∀v ∈ V .

(iv) Show that if p(x) is any polynomial of degree ≤ 3, then Πp = p. (v) Sketch the function Πx4 and show graphically how it approximates x4 in V . (i) Taking monomials 1, x, x2 , x3 , we obtain e1 (x) =

1 2 �

3 x 2 � 90 2 1 e3 (x) = (x − ) 21 6 � 6 175 3 e4 (x) = (x − x) 8 10

e2 (x) =

(ii) We get (P u)(x) =

414 2 1 1 + (x − ) 10 441 6

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APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

(iii) Nothing to show. According to the Orthogonal Decomposition Theorem, u − P u is orthogonal to subspace V .

(iv) This is a direct consequence of the orthogonality condition. If u ∈ V then u − P u ∈ V as well and, in particular, u − P u must be orthogonal to itself,

(u − P u, u − P u) = 0 which implies u − P u = 0. (v) See Fig. 6.1.

Figure 6.1 Function x4 and its L2 -projection onto P 3 (−1, 1).

Exercise 6.3.7 Use the orthonormal basis from Example 6.3.4 to construct the (classical) Fourier series representation of the following functions in L2 (0, 1).

f (x) = x,

f (x) = x + 1

Evaluation of coefficients (f, ek ) leads to the formulas,

x=

∞ 1�1 1 + sin 2πkx, 2 π k k=1

x+1=

∞ 3 1�1 + sin 2πkx 2 π k k=1

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173

Duality in Hilbert Spaces

6.4

Riesz Representation Theorem

Exercises Exercise 6.4.1 Revisit Example 6.4.1 and derive the matrix representation of the Riesz map under the assumption that the dual space consists of antilinear functionals. Follow the lines in the text to obtain, fj = gjk xk The only difference between the formula above and the formula in the text, is the dissapearance of the conjugate over xk .

6.5

The Adjoint of a Linear Operator

6.6

Variational Boundary-Value Problems

Exercises Exercise 6.6.1 Let X be a Hilbert space and V a closed subspace. Prove that the quotient space X/V , which a priori is only a Banach space, is in fact a Hilbert space. Let V ⊥ be the orthogonal complement of V in X, X =V ⊕V⊥ As a closed subspace of a Hilbert space, V ⊥ is also a Hilbert space. Consider map T : V ⊥ � w → [w] = w + V ∈ X/V

174

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

Map T is an isometry from V ⊥ onto X/V . Indeed, T is linear and inf �w + v�2 = inf (�w�2 + �v�2 ) = �w�2

v∈V

v∈V

From the representation x = (x − P x) + P x where P is the orthogonal projection onto V , follows also that T is surjective. Map T transfers the inner product from V ⊥ into X/V , def

([w1 ], [w2 ])X/V = (T −1 [w1 ], T −1 [w2 ]) Exercise 6.6.2 Prove a simplified version of the Poincar´e inequality for the case of Γ1 = Γ. Let Ω be a bounded, open set in R I n . There exists a positive constant c > 0 such that � � 2 u dx ≤ c |∇u|2 dx ∀ u ∈ H01 (Ω) Ω

Ω

Hint: Follow the steps: Step 1. Assume that Ω is a cube in R I n , Ω = (−a, a)n and that u ∈ C0∞ (Ω). Since u vanishes on the

boundary of Ω, we have

u(x1 , . . . , xn ) = Use the Cauchy–Schwarz inequality to obtain 2

u (x1 , . . . , xn ) ≤

�

a −a

�

�

xn −a

∂u (x1 , ..., t)dt ∂xn

∂u (x1 , . . . , xn ) ∂xn

�2

dxn (xn + a)

and integrate over Ω to get the result. Step 2. Ω bounded. u ∈ C0∞ (Ω). Enclose Ω with a sufficiently large cube (−a, a)n and extend u by zero to the cube. Apply Step 1 result.

Step 3. Use density of test functions C0∞ (Ω) in H01 (Ω). Solution: Step 1. Applying Cauchy–Schwarz inequality to the identity above, we get, 2

u (x1 , . . . , xn ) ≤ ≤

�

�

xn −a a

−a

�

�

∂u (x1 , . . . , t) ∂xn

�2

∂u (x1 , . . . , xn ) ∂xn

dt (xn + a)

�2

dxn (xn + a)

Integrating over Ω on both sides, we get �

2

Ω

u dx ≤

� �

∂u ∂xn

�2

�

∂u ∂xn

�2

Ω

dx · 2a2

Step 2. Applying the Step 1 results, we get �

2

u dx = Ω

�

2

Ω1

u dx ≤ 2a

2

�

Ω1

2

dx = 2a

� � Ω

∂u ∂xn

�2

dx

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175

Step 3. Let u ∈ H01 (Ω) and um ∈ C0∞ (Ω) be a sequence converging to u in H 1 (Ω). Then �

Ω

u2m

dx ≤ 2a

� �

� �

�2

2

Ω

∂um ∂xn

�2

dx

Passing to the limit, we get �

2

Ω

u dx ≤ 2a

2

Ω

∂u ∂xn

dx ≤ 2a2

�

Ω

|∇u|2 dx

Exercise 6.6.3 Let Ω be a sufficiently regular domain in R I n , n ≥ 1, and let Γ denote its boundary. Con-

sider the diffusion-convection-reaction problem discussed in the text with slightly different boundary conditions,

 −(aij u,j ),i + bi u,i + cu = f in Ω    u = 0 on Γ1    aij u,j ni = 0 on Γ2

with commas denoting the differentiation, e.g., u,i =

∂u ∂xi ,

and the Einstein summation convention in

use. In this and the following exercises, we ask the reader to reproduce the arguments in the text for this slightly modified problem. Make the same assumptions on coefficients aij , bi , c as in the text. Step 1: Derive (formally) the classical variational formulation:  1   u ∈ HΓ1 (Ω) � �   {aij u,j v,i + bj u,j v + cuv} dx = f v dx Ω

Ω

∀v ∈ HΓ11 (Ω)

where

HΓ11 (Ω) := {u ∈ H 1 (Ω) : u = 0 on Γ1 } Step 2: Use Cauchy–Schwarz inequality, the assumptions on the coefficients aij , bj , c, and an appropriate assumption on source term f (x) to prove that the bilinear and linear forms are continuous on H 1 (Ω). Step 3: Use Poincar´e inequality and the assumptions on the coefficients aij , bj , c to prove that the bilinear form is coercive. Step 4: Use the Lax–Milgram Theorem to conclude that the variational problem is well posed. All reasoning is fully analogous to that in the text. Exercise 6.6.4 Reformulate the second order diffusion-convection-reaction problem considered in Exercise 6.6.3, as a first order problem �

σi = aij u,j −σi,i + bi u,i + cu = f

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APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

where the first equation may be considered to be a (new) definition of flux σi . Use the ellipticity condition to introduce inverse αij = (aij )−1 (the compliance matrix), and multiply the first equation with αij to arrive at the equivalent system, �

αij σj − u,i = gi

−σi,i + bi u,i + cu = f

(6.4)

with the additional source term gi = 0 vanishing for the original problem. We can cast the system into a general abstract problem Au = f where u, f are group variables, and A represents the first order system, �

u = (σ, u) ∈ (L2 (Ω))n × L2 (Ω) �

f = (g, f ) ∈ (L2 (Ω))n × L2 (Ω) Au := (αij σj − u,j , −σi,i + bi u,i + cu) Recall that the accent over the equality sign indicates a “metalanguage” and it is supposed to help you survive the notational conflicts; on the abstract level both u and f gain a new meaning. Definition of the domain of A incorporates the boundary condition, D(A) := { (σ, u) ∈ (L2 (Ω))n × L2 (Ω) : A(σ, u) ∈ (L2 (Ω))n × L2 (Ω) and u = 0 on Γ1 , σ · n = 0 on Γ2 } Step 1: Prove that the operator A is closed. Step 2: Prove that the operator A is bounded below, �Au� ≥ γ�u�,

u ∈ D(A)

Hint: Eliminate the flux and reduce the problem back to the second order problem with the righthand side equal to f − (aij gj ),i . Upgrade then slightly the arguments used in Exercise 6.6.3. Step 3: Identify the adjoint operator A∗ (along with its domain), (Au, v) = (u, A∗ v),

u ∈ D(A), v ∈ D(A∗ )

Step 4: Show that the adjoint operator is injective. Recall then the Closed Range Theorem for Closed Operators and conclude that the adjoint is bounded below with the same constant γ as well. Discuss then the well-posedness of the first order problem Au = f . Step 5: Discuss the well-posedness of the adjoint problem, v ∈ D(A∗ ), Again, the reasoning is identical to that in the text.

A∗ v = f

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177

Exercise 6.6.5 Consider the ultraweak variational formulation corresponding to the first order system studied in Exercise 6.6.4.

�

u ∈ L2 (Ω) (u, A∗ v) = (f, v)

or, in a more explicit form,  σ ∈ (L2 (Ω))n , u ∈ L2 (Ω)    � � where

   

σi (αji τj + v,i ) dx +

Ω

Ω

v ∈ D(A∗ )

u(τi,i − (bi v),i + cv) dx =

�

(gi τi + f v) dx Ω

τ ∈ HΓ2 (div, Ω), v ∈ HΓ11 (Ω)

HΓ2 (div, Ω) := {σ ∈ (L2 (Ω))n : div σ ∈ L2 (Ω), σ · n = 0 on Γ2 } Step 1: Double check that the energy spaces in the abstract and the concrete formulations are identical. Step 2: Identify the strong form of the adjoint operator discussed in Exercise 6.6.4 as the transpose operator corresponding to the bilinear form b(u, v) = (u, A∗ v), and conclude thus that the conjugate operator is bounded below. Step 3: Use the well-posedness of the (strong form) of the adjoint problem to conclude that the ultraweak operator, B : L2 → (D(A∗ ))� ,

�Bu, v� = b(u, v)

is injective. Step 4: Use the Closed Range Theorem for Continuous Operators to conclude that the ultraweak operator B satisfies the inf-sup condition with the same constant γ as for the adjoint operator and, therefore, the same constant γ as for the original strong form of operator A. Step 5: Conclude with a short discussion on the well-posedness of the ultraweak variational formulation. Just follow the text. Exercise 6.6.6 Suppose we begin with the first order system (6.4). We multiply the first equation with test function τi , the second with test function v, integrate over domain Ω, and sum up all the equations. If we leave them alone, we obtain a “trivial” variational formulation with solution (σ, u) in the graph norm energy space and L2 test functions (τ, v). If we integrate by parts (“relax”) both equations, we obtain the ultraweak variational formulation. The energy spaces have been switched. The solution lives now in the L2 space, and the test function comes from the graph norm energy space for the adjoint. We have two more obvious choices left. We can relax one of the equations and leave the other one in the strong form. The purpose of this exercise is to study the formulation where we relax the second equation (conservation law) only. Identify the energy spaces and show that the problem is equivalent to the classical variational formulation discussed in Exercise 6.6.3 with the right-hand side equal to

178

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

f − (αij gj ),i . Discuss the relation between the two operators. Could you modify the standard norm in the H 1 space in such a way that the inf-sup constants corresponding to the two formulations would

actually be identical ? Follow the text. Exercise 6.6.7 Analogously to Exercise 6.6.6, we relax now the first equation (constitutive law) only. We arrive at the so-called mixed formulation,  σ ∈ HΓ2 (div, Ω), u ∈ L2 (Ω)     � �  � αij σj τi dx + uτj,j dx = gi τi dx Ω Ω Ω  � �      (−σi,i + bi u,i + cu)v dx = f v dx Ω

Ω

τ ∈ HΓ2 (div, Ω) v ∈ L2 (Ω)

Identify the corresponding conjugate operator. Discuss boundedness below (inf-sup conditions) for both operators and the well-posedness of the formulation. Follow the text.

6.7

Generalized Green’s Formulae for Operators on Hilbert Spaces

Exercises Exercise 6.7.1 Consider the elastic beam equation (EIw�� )�� = q

0 n

�

1 −a+ 2n 1 −a− 2n

1 dξ > 4n2 → ∞ (a + ξ)2

Concluding, the operator has only a continuous spectrum that coincides with the whole real line.

Hilbert Spaces

6.9

187

Spectra of Continuous Operators. Fundamental Properties

Exercises Exercise 6.9.1 Let X be a real normed space and X × X its complex extension (comp. Section 6.1). Let A : X → X be a linear operator and let A˜ denote its extension to the complex space defined as ˜ A((u, v)) = (Au, Av) Suppose that λ ∈ IC is an eigenvalue of A˜ with a corresponding eigenvector w = (u, v). Show ¯ is an eigenvalue of A˜ as well with the corresponding eigenvector equal that the complex conjugate λ w ¯ = (u, −v). Exercise 6.9.2 Let U be a Banach space and let λ and µ be two different eigenvalues (λ �= µ) of an operator A ∈ L(U, U ) and its transpose A� ∈ L(U � , U � ) with corresponding eigenvectors x ∈ U and g ∈ U � . Show that

�g, x� = 0

6.10

Spectral Theory for Compact Operators

Exercises Exercise 6.10.1 Let T be a compact operator from a Hilbert space U into a Hilbert space V . Show that: (i) T ∗ T is a compact, self-adjoint, positive semi-definite operator from a space U into itself. (ii) All eigenvalues of a self-adjoint operator on a Hilbert space are real. Conclude that all eigenvalues of T ∗ T are real and nonnegative. By Proposition 5.15.3??(ii), T ∗ T as a composition of a compact and a continuous operator is compact. Since (T ∗ T )∗ = T ∗ T ∗∗ = T ∗ T T ∗ T is also self-adjoint. Finally, (T ∗ T u, u)U = (T u, T u)V = �T u�2V ≥ 0

188

APPLIED FUNCTIONAL ANALYSIS

SOLUTION MANUAL

i.e., the operator is positive semi-definite. Next, if A is a self-adjoint operator on a Hilbert space H, and (λ, e) is an eigenpair of A, we have λ�e�2 = (λe, e) = (Ae, e) = (Ae, e) = (e, Ae) = (e, λe) = λ�e�2 which implies that λ = λ. Additionally, if A is positive semi-definite then (Ae, e) = λ(e, e) ≥ 0 which implies λ ≥ 0.

6.11

Spectral Theory for Self-Adjoint Operators

Exercises Exercise 6.11.1 Determine the spectral properties of the integral operator � x� 1 u(η) dη dξ (Au)(x) = 0

ξ

defined on the space U = L2 (0, 1). The operator if self-adjoint. Indeed, integrating twice by parts, we get � 1� x� 1 � 1� 1 � 1 u(η) dη dξ v¯(x) dx = u(η) dη v¯(ξ) dξ dx 0 0 ξ �0 1 x � x � 1 x = u(x) v¯(η) dη dξ dx 0

0

ξ

Second step in the reasoning above shows also that the operator is positive semi-definite. The operator is also compact. Indeed, � x� 1 � x� x � x� 1 u(η) dη dξ = u(η) dη dξ + u(η) dη dξ 0 ξ � 01 x � x �0 x �ξ η dξu(η) dη + u(η) dη dξ = 0 � 1 x �0 x 0 ηu(η) dη + x u(η) dη = x �0 1 = K(x, η)u(η) dη 0

where K(x, η) =

�

η x

0