Classification of Lipschitz Mappings

307

Mathematics

“Deep understanding of the properties of Lipschitzian mappings is important for all levels of study in many branches of mathematics. This book by Łukasz Piasecki is a good choice for achieving such an understanding in the framework of mappings in general metric spaces, in particular, Banach spaces. Moreover, it gives new insight into the theory of Lipschitzian mappings via a study of the mean Lipschitz condition. … The book is written in a very clear and reader-friendly way. The author gives many examples illustrating various aspects of presented results.” —Stanislaw Prus, Marie Curie-Sklodowska University “… a self-contained, readable, and precise course on the subject. … Besides the presentation of the theory, the true value of the book lies in a collection of cleverly chosen interesting examples.” —Kazimierz Goebel, Maria Curie-Sklodowska University “I strongly recommend this book for advanced undergraduate and graduate students … The reader will find a new classification of this kind of mapping as well as many examples and illustrations designed to help the reader understand the definitions, properties, and results. … I also recommend this book for analysts or mathematicians who are looking for new topics to research.” —Victor Perez-Garcia, University of Veracruz

Classification of Lipschitz Mappings

Łukasz Piasecki

Piasecki

Classification of Lipschitz Mappings presents a systematic, self-contained treatment of a new, more precise classification of Lipschitz mappings and its application in many topics of metric fixed point theory. The mean Lipschitz condition introduced by Goebel, Japón Pineda, and Sims is relatively easy to check and turns out to satisfy several principles: regulating the possible growth of the sequence of Lipschitz constants k(Tn), ensuring good estimates for k0(T) and k∞(T), and providing some new results in metric fixed point theory.

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Classification of Lipschitz Mappings

Classification of Lipschitz Mappings

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Classification of Lipschitz Mappings

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Classification of Lipschitz Mappings

Łukasz Piasecki

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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2014 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 Version Date: 20131105 International Standard Book Number-13: 978-1-4665-9522-4 (eBook - PDF) 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

To My Parents

Contents

Introduction

1

1 The Lipschitz condition 1.1 Nonlinear spectral radius . . . . . . . . . . . . . . . . . . . . 1.2 Uniformly lipschitzian mappings . . . . . . . . . . . . . . . .

5 5 8

2 Basic facts on Banach spaces 2.1 Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 The operator norm . . . . . . . . . . . . . . . . . . . . . . . 2.3 Dual spaces, reflexivity, the weak, and weak* topologies . . .

11 12 17 17

3 Mean Lipschitz condition 3.1 Nonexpansive and mean nonexpansive mappings in Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 General case . . . . . . . . . . . . . . . . . . . . . . . . . . .

25 25 38

4 On the Lipschitz constants for iterates of mean lipschitzian mappings 4.1 A bound for Lipschitz constants of iterates . . . . . . . . . . 4.2 A bound for the constant k∞ (T ) . . . . . . . . . . . . . . . . 4.3 Moving averages in Banach spaces . . . . . . . . . . . . . . . 4.4 A bound for the constant k0 (T ) . . . . . . . . . . . . . . . . 4.5 More about k(T n ), k0 (T ), and k∞ (T ) . . . . . . . . . . . . .

41 41 62 69 72 77

5 Subclasses determined by p-averages 5.1 Basic definitions and observations . . . . . . . . . . . . . . . 5.2 A bound for k(T n ), k∞ (T ), and k0 (T ) . . . . . . . . . . . . . 5.3 On the moving p-averages . . . . . . . . . . . . . . . . . . . .

85 85 89 94

6 Mean contractions 6.1 Classical Banach’s contractions . . . . . . . . . . . . . . . . . 6.2 On characterizations of contractions . . . . . . . . . . . . . . 6.3 On the rate of convergence of iterates . . . . . . . . . . . . .

95 95 98 100

ix

x

Contents

7 Nonexpansive mappings in Banach space 7.1 The asymptotic center technique . . . . . . . . . . . . . 7.2 Minimal invariant sets and normal structure . . . . . . 7.3 Uniformly nonsquare, uniformly noncreasy, and reflexive Banach spaces . . . . . . . . . . . . . . . . . . . . . . . 7.4 Remarks on the stability of f.p.p. . . . . . . . . . . . . 7.5 The case of `1 . . . . . . . . . . . . . . . . . . . . . . .

. . . . . .

103 103 111

. . . . . . . . .

123 124 130

8 Mean nonexpansive mappings 8.1 Some new results of stability type . . . . . . . 8.2 Sequential approximation of fixed points . . . 8.3 The case of n = 3 . . . . . . . . . . . . . . . . 8.4 On the structure of the fixed points set . . . .

. . . .

. . . .

. . . .

. . . .

139 139 150 153 159

9 Mean lipschitzian mappings with k > 1 9.1 Losing compactness in Brouwer’s Fixed Point Theorem 9.2 Retracting onto balls in Banach spaces . . . . . . . . . 9.3 Minimal displacement . . . . . . . . . . . . . . . . . . . 9.4 Optimal retractions . . . . . . . . . . . . . . . . . . . . 9.5 Generalized characteristics of minimal displacement . .

. . . . .

. . . . .

. . . . .

163 163 169 180 197 205

. . . .

. . . .

. . . .

. . . .

Bibliography

217

Index

223

Introduction

The Lipschitz condition is of great importance in many branches of mathematics. The standard situation is the following: let (M, ρ) be a metric space; we say that a mapping T : M → M is lipschitzian if there exists a constant k such that, for all x, y ∈ M , we have ρ(T x, T y) ≤ kρ(x, y). The class of all mappings satisfying the Lipschitz condition with a constant k is denoted by L(k). The smallest constant k for which the above inequality holds is called the Lipschitz constant for T and is denoted by k(T ). In some approaches, we also investigate a behavior of Lipschitz constants k(T n ) for iterates T n of T . To describe such behavior, we can find in the literature two special constants: k∞ (T ) = lim sup k(T n ) n→∞

and k0 (T ) = lim

n→∞

p n k(T n ).

The constant k∞ (T ) plays a special role in the fixed point theory for uniformly lipschitzian mappings; we say that a mapping T : M → M is uniformly lipschitzian if there exists a constant k ≥ 0 such that, for every n = 1, 2, . . . and all x, y ∈ M , ρ(T n x, T n y) ≤ kρ(x, y). The constant k0 (T ), which in case of linear mapping T defined on a Banach space is just the spectral radius of T , has the following interpretation in a general case of lipschitzian mappings: k0 (T ) = inf kr (T ), where kr (T ) denotes the Lipschitz constant for T with respect to the metric r, and the infimum is taken over all metrics r that are equivalent to ρ; we say that a metric r is equivalent to ρ if there exist constants a, b > 0 such that, for all x, y ∈ M , aρ(x, y) ≤ r(x, y) ≤ bρ(x, y). In general, it is hard to determine or give nice evaluations for the constants k0 (T ) and k∞ (T ) because it demands nice estimates of k(T n ). The possible growth of the sequence k(T n ) can be regulated by the following standard inequality: k(T n ) ≤ k(T )n for n = 1, 2, . . . , 1

2

Introduction

which, in case of any class L(k), is sharp. However, in many cases, this estimate doesn’t give good approximation of k(T n ). The sequence {k(T n )} may behave very unpredictably even in simple situations of functions defined on the interval [0, 1], as seen below. Let us consider two “similar” graphs. The first graph describes the function f:

1 f

1 2

0

3 4

1

It is easy to check that k(f ) = 2 and k(f n ) = 2n for n = 1, 2, . . . . However, if we slightly change the graph of function f , then the situation will totally change:

1 g

1 2

0

1 4

1

In this case, we can divide the interval [0, 1] into two regions: [0, 1/4] and (1/4, 1]. It is easy to verify that the function g extends a distance between any two points in [0, 1/4], |g(s) − g(t)| = 2 |s − t|, sending them to the interval

Introduction

3  (1/4, 1] on which the second iterate contracts it back, sending them to 21 . Consequently, k(g) = 2, but k(g n ) = 0 for n ≥ 2. In the above samples, we have k(f ) = k(g) = 2. Hence, f, g ∈ L(2). Nevertheless, remaining iterates behave in various ways: k∞ (f ) = ∞, k0 (f ) = 2, and k∞ (g) = 0, k0 (g) = 0. In this book we deal with a problem of more precise classification of lipschitzian mappings. It seems natural that a condition that describes such classes should satisfy several principles. It should regulate the possible growth of the sequence of Lipschitz constants k(T n ), as well as ensure nice estimates for k0 (T ) and k∞ (T ). We also expect that such condition will provide some new results in the metric fixed point theory. The last and the most important request is that it has to be relatively easy to check. In this monograph, we widely study the mean Lipschitz condition, which was introduced by Goebel, Jap´on Pineda, and Sims: suppose Pn that α = (α1 , . . . , αn ), where n ≥ 1, αi ≥ 0, α1 > 0, αn > 0, and i=1 αi = 1; we say that T : M → M is α-lipschitzian for the constant k ≥ 0 if, for all x, y ∈ M , n X αi ρ(T i x, T i y) ≤ kρ(x, y). i=1

The class of all mappings that satisfies the above inequality with α = (α1 , . . . , αn ) and a constant k ≥ 0 is denoted by L(α, k). We also focus on mean Lipschitz condition described by averages of order p ≥ 1: let α = (α1 , . . . , αn ) be as above and p ≥ 1; we say that a mapping T : M → M is (α, p)-lipschitzian for the constant k ≥ 0 if, for all x, y ∈ M , n X

!1/p i

i

p

αi ρ(T x, T y)

≤ kρ(x, y).

i=1

The class of all mappings that satisfies the above condition with α = (α1 , . . . , αn ), p ≥ 1 and k ≥ 0 is denoted by L(α, p, k). Let us note that, for p = 1, we get a definition of α-lipschitzian mapping for the constant k. We usually write L(α, k) instead of L(α, 1, k). It is clear that, for n = 1, we get nothing more than the classical definition of k-lipschitzian mapping. Hence, in this special case, the mean Lipschitz condition determines the class L(k). The presented monograph arose as a result of discussions that took place during a seminar under the supervision of Professor Kazimierz Goebel at the University of Maria Curie-Sklodowska in Lublin, Poland. The included results have been presented many times during international conferences as well as at seminars devoted to the metric fixed point theory. Many of them have been already published. A part of the presented material appeared only in a PhD thesis of the present author [70] written under the guidance of Professor Kazimierz Goebel. The book is addressed to advanced undergraduate and graduate students as well as to professionals looking for new topics in the metric fixed point

4

Introduction

theory. Nevertheless, we deeply believe that this monograph will be of interest among readers working on other areas (for example, differential equations and dynamical systems). The proposed text is self-contained, and only a basic knowledge of functional analysis and topology is required. The monograph covers approximately 220 pages of a systematic course, with updated facts concerning discussed topics, with accompanying illustrations, a rich collection of examples, and open problems. The book does not include typical exercises. Instead, we can find many sentences like “It is easy to see...”, “We leave to the reader...”,“Observe that..” and so forth. This book does not aim to compete with any book devoted to the metric fixed point theory. Our goal is to provide a systematic, self-contained course of a new classification of Lipschitz mappings, pointing out its application in many topics of the metric fixed point theory. Moreover, we encourage new adept to acquaint themselves with some existing books, in particular [49], [36] and [31], which, with no doubt, could be useful for further development of this new field. I am deeply grateful to Professor Kazimierz Goebel for creating the atmosphere conducive to writing this book. His constant interest in progress, support, and discussions while writing were valuable to me and undoubtedly enriched the content of this monograph considerably. It is my kind duty to acknowledge Professors Tom´as Dom´ınguez Benavides, Emanuele Casini, and Stanislaw Prus for pointing out errors and omissions as well as for valuable remarks that improved this book. I would like to express my deep sense of gratitude to my colleague Dr. Piotr Oleszczuk for his kind support in proofreading and verifying the substantial content of subsequent parts of the text. I would like to express special thanks to my friend and scientific partner, Dr. V´ıctor P´erez Garc´ıa. During his postdoctoral position at Maria CurieSklodowska University, we had a number of opportunities to hold discussions as well as intensive and fruitful cooperation. Many results obtained then were included in this monograph. I am thankful to Sunil Nair for excellent cooperation on publishing this book.

Chapter 1 The Lipschitz condition

Let (M, ρ) be a metric space. By BM (x, r), we denote the closed ball centered at x ∈ M with radius r > 0: BM (x, r) = {y ∈ M : ρ(x, y) ≤ r} . By SM (x, r), we denote the sphere centered at x ∈ M with radius r > 0: SM (x, r) = {y ∈ M : ρ(x, y) = r} . As usual, we shall drop the subscript when M is clear from the context. For a subset A of (M, ρ), diam(A) denotes the diameter of A; that is, diam(A) = sup {ρ(x, y) : x, y ∈ A} . A subset A ⊂ M is called bounded if diam(A) < ∞. We say that z ∈ M is a fixed point of mapping T : M → M if z = T z. The set of all fixed points of T is denoted by Fix(T ). A mapping T : M → M is said to be lipschitzian if there exists k ≥ 0 such that, for all x, y ∈ M ρ(T x, T y) ≤ kρ(x, y). (1.1) If we want to indicate k ≥ 0, then we refer to such a mapping as k-lipschitzian. By L(k), we denote the class of all lipschitzian mappings that satisfies (1.1) with a constant k. It is clear that lipschitzian mappings are uniformly continuous. For a given mapping T , the smallest k for which (1.1) holds (such k always exists) is called the Lipschitz constant for T and will be denoted by kρ (T ) or simply k(T ) when the underlying metric is clear from the context. A mapping T : M → M is said to be a contraction if k(T ) < 1. If k(T ) ≤ 1, then T is called nonexpansive. Thus, the class of nonexpansive mappings includes all contractions and isometries, in particular, the identity.

1.1

Nonlinear spectral radius

For two lipschitzian mappings T, S : M → M and all x, y ∈ M , we have ρ(T Sx, T Sy) ≤ k(T )ρ(Sx, Sy) ≤ k(T )k(S)ρ(x, y) 5

6

Classification of Lipschitz mappings

and so k(T ◦ S) ≤ k(T )k(S).

(1.2)

In particular, (1.2) regulates the possible growth of the sequence of Lipschitz constants for iterates T n of T , k(T n+m ) = k(T n T m ) ≤ k(T n ) · k(T m ) for m, n = 1, 2, . . . and consequently k(T n ) ≤ k(T )n for n = 1, 2, . . . .

(1.3)

Thus, if T is nonexpansive, then all its powers T n are also nonexpansive. If T is a contraction, then limn→∞ k (T n ) = 0. To recall another basic observation about the sequence of Lipschitz constants {k (T n )} of a given mapping T , we shall need the following well-known fact: ∞

Lemma 1.1 Let {an }n=1 be a sequence of nonnegative numbers satisfying, for all m, n = 1, 2, . . . , the following condition: am+n ≤ am · an .  √ √ Then, lim n an = inf n an : n = 1, 2, . . . .

(1.4)

n→∞

Proof. Put

√ a = inf { n an : n = 1, 2, . . . } . √ Fix  > 0. There is k ∈ N such that a ≤ k ak ≤ a + . For any positive integer n, we can write n = qk + r, where q and r are nonnegative integers with 0 ≤ r ≤ k − 1. Then, using (1.4), we get an ≤ aqk · ar ≤ aqk · ar . Thus,

√ q/n n an ≤ ak · a1/n ≤ (a + )kq/n · a1/n r r .

Letting n tend to infinity and for fixed k we obtain  kq r = lim 1 − = 1, n→∞ n n→∞ n lim

and consequently, a ≤ lim sup

√ n an ≤ a + .

n→∞

√ Since  > 0 is arbitrary, this implies lim sup n an = a. On the other hand, n→∞ √ √ √ n a n a n a n ≥ a for each n and so lim inf n ≥ a. Thus, lim n = a. n→∞

n→∞



The Lipschitz condition

7

We have already seen that the sequence of Lipschitz constants {k (T n )} satisfies (1.4). Thus, the constant p k0 (T ) = lim n k(T n ) n→∞

is well defined and k0 (T ) = lim

p n

n→∞

k(T n ) = inf

np o n k(T n ) : n = 1, 2, . . . .

(1.5)

Using (1.3), we obtain the following evaluation for k0 (T ): p k0 (T ) = lim n k(T n ) ≤ k(T ). n→∞

We say that two metrics d and ρ are equivalent if there exist two constants a > 0 and b > 0 such that, for all x, y ∈ M ad (x, y) ≤ ρ (x, y) ≤ bd (x, y) . It is easy to check that any mapping T , which is lipschitzian with respect to a given metric ρ is also lipschitzian with respect to any equivalent metric d. The respective Lipschitz constants kρ (T ) and kd (T ) may differ and the difference is regulated by the relation d (T x, T y) ≤

1 1 b ρ (T x, T y) ≤ kρ (T )ρ (x, y) ≤ kρ (T )d (x, y) . a a a

This implies that b kρ (T ). a Changing roles of metrics d and ρ, we obtain kd (T ) ≤

kρ (T ) ≤

b kd (T ). a

Obviously, the same holds for all iterates T n of mapping T ; hence, a b kρ (T n ) ≤ kd (T n ) ≤ kρ (T n ) . b a

(1.6)

In particular, (1.6) implies that the constant k0 (T ) is independent of the selection of a metric d, which is equivalent to ρ. Thus, using (1.5), for each equivalent metric d, we have k0 (T ) ≤ kd (T ). As a consequence, k0 (T ) ≤ inf kd (T ), where infimum is taken over all metrics d, which are equivalent to ρ. Actually, we have k0 (T ) = inf kd (T ). Indeed, for arbitrary  > 0 set λ = 1/ (k0 (T ) + ). Let us define a metric dλ by putting for each x, y ∈ M dλ (x, y) =

∞ X i=0

ρ(T i x, T i y)λi = ρ(x, y) +

∞ X i=1

ρ(T i x, T i y)λi .

(1.7)

8

Classification of Lipschitz mappings

It is easy to observe that the metric dλ is equivalent to ρ and ! ∞ X i i ρ(x, y) ≤ dλ (x, y) ≤ kρ (T )λ ρ(x, y).

(1.8)

i=0

Now, for all x, y ∈ M , we have dλ (T x, T y)

= =

∞ X

ρ(T i+1 x, T i+1 y)λi

i=0 ∞ X

1 λ

i=0 ∞ X

ρ(T i+1 x, T i+1 y)λi+1

≤

1 λ

=

1 dλ (x, y). λ

ρ(T i x, T i y)λi

i=0

Thus, kdλ (T ) ≤ 1/λ = k0 (T ) + .

(1.9)

Since  > 0 is arbitrarily chosen, we finally obtain the following characterization of k0 (T ): k0 (T ) = inf {kd (T ) : d is equivalent to ρ} .

1.2

(1.10)

Uniformly lipschitzian mappings

Of particular interest in the metric fixed point theory is a class of uniformly lipschitzian mappings. We say that a mapping T : M → M is uniformly lipschitzian, if there exists a constant k ≥ 0 such that for all x, y ∈ M and n = 1, 2, . . . , we have ρ(T n x, T n y) ≤ kρ(x, y). If we want to indicate k ≥ 0, then we refer to such mapping as a uniformly klipschitzian. Equivalently, a lipschitzian mapping T is uniformly lipschitzian, if and only if sup {k(T n ) : n = 1, 2, . . .} < ∞. (1.11) We shall denote the class of all uniformly k-lipschitzian mappings by Lu (k). It is well known that there exists another characterization of such mappings. It is clear that, if for some equivalent metric d we have kd (T ) ≤ 1, then kρ (T n ) ≤ b/a for n = 1, 2, . . . (a and b are taken from the definition of equivalent metrics). Hence, condition (1.11) is satisfied. On the other hand, if T is

The Lipschitz condition

9

uniformly lipschitzian, then it is nonexpansive with respect to the equivalent metric defined by d(x, y) = sup {ρ (T n x, T n y) : n = 0, 1, 2, . . .} .

(1.12)

Indeed, for all x, y ∈ M , we have ρ(x, y) ≤ d(x, y) ≤ sup {k(T n ) : n = 0, 1, 2, ...} ρ(x, y) and d(T x, T y)


= sup ρ T n+1 x, T n+1 y : n = 0, 1, 2, . . . ≤ sup {ρ (T n x, T n y) : n = 0, 1, 2, . . .} = d(x, y).

Finally, we conclude that a mapping T is uniformly lipschitzian if and only if there exists an equivalent metric with respect to which T is nonexpansive. Consequently, for any uniformly lipschitzian mapping T , we have k0 (T ) ≤ 1. Also, there is another useful constant that appears during studies devoted to uniformly lipschitzian mappings. For any lipschitzian mapping T : M → M , we define the constant k∞ (T ) as k∞ (T ) = lim sup k(T n ). n→∞

Obviously, a mapping T is uniformly lipschitzian if and only if k∞ (T ) < ∞. Usefulness of the constant k∞ (T ) follows from the fact that usually k∞ (T ) < sup {k(T n ) : n = 1, 2, . . .} . We only mention that all the above definitions carry over to a Banach space setting (X, k·k) by taking ρ(x, y) = kx − yk. It is easy to verify that, for any nonempty subset C of a Banach space X and two lipschitzian mappings T, S : C → X, we additionally have the following elementary inequalities: k(T + S) ≤ k(T ) + k(S) and k(λT ) = λk(T ) for λ ≥ 0. In this text, unless otherwise stated, we always assume that all considered sets are nonempty.

Chapter 2 Basic facts on Banach spaces

Let (X, k·k) be a real Banach space. In this case, the closed unit ball BX (0, 1) = {x ∈ X : kxk ≤ 1} is denoted by BX and the unit sphere SX (0, 1) = {x ∈ X : kxk = 1} by SX . If X is clear from this context, then we write B and S instead of BX and SX , respectively. Let us introduce some notations of classical Banach spaces. • For each 1 ≤ p < ∞, `p denotes the space that consists of all p-summable sequences x = (x1 , x2 , . . . ) of real numbers furnished with the standard norm ! p1 ∞ X p . kxk = |xi | i=1

• `∞ is the Banach space that consists of all bounded sequences x = (x1 , x2 , . . . ) of real numbers, c denotes its subspace of all converging sequences, and c0 the subspace of c of all null sequences. All of them are furnished with the standard sup norm, kxk = sup {|xi | : i = 1, 2, . . . } . In the case of c0 space, the above formula can be written as kxk = max {|xi | : i = 1, 2, . . . } . • C[0, 1] denotes the space that consists of all continuous functions f : [0, 1] → R with the sup norm, kf k = sup {|f (t)| : t ∈ [0, 1]} = max {|f (t)| : t ∈ [0, 1]} . • For each 1 ≤ p < ∞ by Lp [0, 1], we denote the R space ofp all Lesbesgue measurable functions f : [0, 1] → R such that [0,1] |f (t)| dt < ∞, identifying those functions that differ only on a null set, and the norm is defined as ! p1 Z p |f (t)| dt . kf k = [0,1]

11

12

Classification of Lipschitz mappings • L∞ [0, 1] is the space of all Lesbesgue measurable functions f : [0, 1] → R for which ! kf k = sup ess {|f (t)| : t ∈ [0, 1]} = inf

A∈N

sup

|f (t)| ,

t∈[0,1]\A

where N = {A ⊂ [0, 1] : A is a null set}. Here, as in the previous case of Lp [0, 1] spaces, we identify functions that differ only on a set of measure zero.

2.1

Convexity

Let X be a Banach space with the closed unit ball BX and the unit sphere SX . A subset K of X is said to be convex if, for any x, y ∈ K and any λ ∈ [0, 1], the point (1 − λ)x + λy belongs to K. The unit ball BX and, consequently, any ball in X is convex. Indeed, if x, y ∈ BX , then for any λ ∈ [0, 1], k(1 − λ)x + λyk ≤ (1 − λ) kxk + λ kyk ≤ 1. It is easy to see that, for any z ∈ X and for any r > 0, we have BX (z, r) = z + rBX , where rBX = {rx : x ∈ BX }. For any nonempty set A ⊂ X by conv(A), we denote the convex hull of A, that is, the smallest convex set containing A. Hence, \ conv(A) = {K ⊂ X : K ⊃ A, K is convex set} . Pn It means that x ∈ conv(A) if and only P if x = i=1 λi xi for some x1 , . . . , xn ∈ n A and some scalars λi ≥ 0 for which i=1 λi = 1. Obviously, A is convex if and only if A = conv(A). If A is convex, then its closure A is also convex. Clearly, BX = conv(SX ). The set \ conv(A) = {K ⊂ X : K ⊃ A, K is convex and closed in X} is said to be the convex closure of A. Equivalently, conv(A) is the smallest convex and closed set containing A. In other words, conv(A) = conv(A). It is clear that A ⊂ conv(A) ⊂ conv(A), and it is easy to prove that, for any bounded set A in X, diam(A) = diam(conv(A)) = diam (conv(A)) . The following very important result states that compactness is invariant under taking a convex closure.

Basic facts on Banach spaces

13

Theorem 2.1 (Mazur) If A is compact, then conv(A) is also compact. We say that a Banach space X is strictly convex if, for all x, y ∈ X, the following implication holds: 

kxk ≤ 1 

x + y

< 1. kyk ≤ 1 ⇒  2 kx − yk > 0 The above condition means that a sphere does not contain a segment that is not a singleton. In such space, any three points x, z, y such that kx − zk + kz − yk = kx − yk must lie on a line. More precisely, z=

kx − zk ky − zk x+ y. kx − yk kx − yk

Equivalently, X is strictly convex if and only if the following implication holds: for each x, y ∈ X (x 6= 0 ∧ y 6= 0 ∧ y 6= tx for each t > 0) =⇒ (kx + yk < kxk + kyk) . We say that a Banach space (X, k·k) is uniformly convex if, for each  ∈ (0, 2], there exists δ > 0 such that the following implication holds: for each x, y ∈ X 

kxk ≤ 1 

x + y

kyk ≤ 1 ⇒ (2.1)

2 ≤ 1 − δ.  kx − yk ≥  One can check that X is uniformly convex if and only if for any two sequences {xn } and {yn } in X the following implication holds:   lim kxn k = lim kyn k = 1 ∧ lim kxn + yn k = 2 ⇒ lim kxn − yn k = 0. n→∞

n→∞

n→∞

n→∞

The condition of uniformly convexity was introduced by Clarkson [20] in 1936, and his classical result states that all Hilbert spaces, all Lp [0, 1], and `p spaces with 1 < p < ∞ are uniformly convex. The modulus of convexity of a Banach space X is the function δX : [0, 2] → [0, 1] defined by ( ) kx + yk δX () = inf 1 − : kxk ≤ 1, kyk ≤ 1, kx − yk ≥  . 2 Hence, for any  > 0, the number δX () is the largest number for which the implication (2.1) holds. It is easy to note that X is uniformly convex if and

14

Classification of Lipschitz mappings

only if δX () > 0 for every  > 0. Furthermore, for all x, y ∈ X and any r > 0, the following implication is satisfied:

   

x + y kxk ≤ r

≤ 1 − δX kx − yk ⇒ r. (2.2)

2 kyk ≤ r r The characteristic of convexity of a Banach space X is the number 0 (X) = sup { ∈ [0, 2] : δX () = 0} . Hence, the number 0 (X) bounds the length of segments lying on the unit sphere S or arbitrarily close to it. Obviously, 0 (X) = 0 if and only if X is uniformly convex. Below we list several very important properties of the modulus of convexity δX and the characteristic of convexity 0 (X). The proofs can be found in [36]. • δX (0) = 0 and δX is nondecreasing on [0, 2]. • δX (2) = 1 if and only if X is strictly convex. • δX is continuous on [0, 2). However, it may happen that δX is not continuous at  = 2. Such a situation takes place in example 2.2. In spite of this, we always have • lim− δ() = 1 − →2

0 (X) 2 .

In addition, • δX is strictly increasing on [0 (X), 2]. It is easy to check that a Hilbert space H is uniformly convex as a consequence of the Parallelogram Law   2 2 2 2 kx + yk + kx − yk = 2 kxk + kyk , which also implies that r δH () = 1 −

1−

2 . 4

In the case of `p and Lp , 2 < p < ∞, we have    p  p1 δp () = 1 − 1 − . 2 If 1 < p < 2, then we have the following implicit formula obtained by Hanner in 1956:  p  p 1 − δp () + + 1 − δp () − = 2. 2 2

Basic facts on Banach spaces

15

(2)

furnished with the stan-

Example 2.1 Consider two-dimensional space `1 dard norm, k(x1 , x2 )k = |x1 | + |x2 | .

If e1 = (1, 0) and e2 = (0, 1), then the whole segment [e1 , e2 ] = {(1 − λ)e1 + λe2 : λ ∈ [0, 1]} lies on the unit sphere S. Consequently, δ`(2) () = 0 for any  ∈ [0, 2]. 1

Furthermore, since the spaces `1 , `∞ , c0 , c, C[0, 1], L1 [0, 1], L∞ [0, 1] contain (2) an isometric copy of `1 , we conclude that δX () = 0 for any  ∈ [0, 2] if X is one of them. It is clear that each uniformly convex space is also strictly convex. In finite dimensional Banach spaces, both conditions are equivalent. It is an easy consequence of the fact that a closed ball is compact. In infinite dimensional spaces, such equivalence does not hold. Example 2.2 Consider the space c0 furnished with the standard norm given for any x = (x1 , x2 , . . . ) by kxk = sup {|xi | : i = 1, 2, . . . } . Define the new norm k|x|k =

2

kxk +

∞  X x i 2 i=1

i

! 12 .

It is easy to observe that both norms are equivalent and kxk ≤ k|x|k ≤

 1 π2 2 1+ kxk . 6

The space (c0 , k| · |k) is strictly convex. To see it, consider any two points x = (x1 , x2 , . . . ) and y = (y1 , y2 , . . . ) in c0 such that x 6= 0, y 6= 0, and y 6= tx for each t > 0. If we now denote   x1 x2 u = kxk , , , . . . , 1 2   y1 y2 v = kyk , , , . . . , 1 2 then also u 6= 0, v 6= 0, and v 6= tu for each t > 0. Consequently, by the triangular inequality, both for the supremum norm and for the Euclidean norm, we obtain

16

Classification of Lipschitz mappings

k|x + y|k =

2

kx + yk +

2 ∞  X xi + yi i

i=1

≤

2

(kxk + kyk) +

∞  X xi

yi 2 + i i

i=1

<

2

kxk +

∞  X i=1

! 12

x i 2 i

! 12

! 12 +

2

kyk +

∞  X y i 2 i=1

! 12

i

= k|x|k + k|y|k , as we desired. Let un = (0, . . . , 0, 1, 1, 0, . . . , 0, . . . ) | {z } n−1

and vn = (0, . . . , 0, −1, 1, 0, . . . , 0, . . . ). | {z } n−1

Then,  k|un |k = k|vn |k =

1 1 1+ 2 + n (n + 1)2

k|un − vn |k = k|(0, . . . , 0, 2, 0, . . . , 0, . . . )|k = | {z }

 12 ,

 2 ! 12 2 4+ n

n−1

and  k|un + vn |k = k|(0, . . . , 0, 2, 0, . . . , 0, . . . )|k = | {z }

4 4+ (n + 1)2

n

We have lim k|un |k = lim k|vn |k = 1, lim k|un + vn |k = 2

n→∞

n→∞

n→∞

and lim k|un − vn |k = 2;

n→∞

hence, (c0 , k| · |k) is not uniformly convex. Moreover, ( 0 if  ∈ [0, 2), δ(c0 ,k|·|k) () = 1 if  = 2.

 12 .

Basic facts on Banach spaces

2.2

17

The operator norm

For any two Banach spaces (X, k·k) and (Y, k·k) by L(X, Y ), we denote the space of all norm-to-norm continuous linear operators T : X → Y . If X = Y , then we write L(X) instead of L(X, X). For any T ∈ L(X, Y ), we define the norm k·k of T as kT k = sup {kT xk : x ∈ BX } . It is well known that the norm of T can be equivalently given by kT k =

sup {kT xk : x ∈ SX }   kT xk : x ∈ X, x 6= 0 = sup kxk = inf {M > 0 : kT xk ≤ M kxk for all x ∈ X} .

One can observe that the last equation means that actually the norm of the operator T is just its Lipschitz constant. Hence, using our notations, we can write kT k = k(T ). If T ∈ L(X), then the estimate (1.3) can be written as n

kT n k ≤ kT k , and the relation (1.5), which now takes the form np o p k0 (T ) = lim n kT n k = inf n kT n k : n = 1, 2, . . . , n→∞

is just the spectral radius of T , which we denote by spr(T ). Consequently, p k0 (T ) = spr(T ) = lim n kT n k ≤ kT k . n→∞

It is easy to verify that, if X is a normed space and Y is a Banach space, then (L(X, Y ), k·k) is a Banach space. In the latter, we consider the special case when (Y, k·k) = (R, |·|), where |·| denotes the absolute value.

2.3

Dual spaces, reflexivity, the weak, and weak* topologies

Let (X, k·k) be a real Banach space with the closed unit ball BX and the unit sphere SX . Recall that a dual space or conjugate space X ∗ of X is the

18

Classification of Lipschitz mappings

Banach space that consists of all continuous linear functionals x∗ : X → R furnished with a dual norm k·k∗ given by kx∗ k∗ = sup {x∗ (x) : x ∈ BX } = sup {x∗ (x) : x ∈ SX } . ∗

The second conjugate space X ∗∗ of X is defined as X ∗∗ = (X ∗ ) . The canonical embedding of X into X ∗∗ is the mapping j : X → X ∗∗ defined for any x ∈ X and x∗ ∈ X ∗ by j(x)(x∗ ) = x∗ (x). We leave for the reader to check that j is a linear mapping and j(x) ∈ X ∗∗ for any x ∈ X. Moreover, j is an isometry; that is, for any x ∈ X, kj(x)k∗∗ = kxk. This is one of the many useful consequences of the fundamental Hahn-Banach Theorem. Theorem 2.2 (Hahn-Banach) Let Y be a linear subspace of a Banach space X. Then for every continuous linear functional y ∗ on Y there exists a continuous linear functional x∗ ∈ X ∗ such that x∗ (y) = y ∗ (y) for every y ∈ Y and kx∗ k∗ = ky ∗ k∗ := sup {y ∗ (y) : y ∈ BY } . To prove that the canonical embedding j is an isometry, it is enough to observe that, in view of the Hahn-Banach Theorem, for any x ∈ X, x 6= 0, there exists x∗ ∈ X ∗ such that x∗ (x) = kxk and kx∗ k∗ = 1. If j(X) = X ∗∗ , then X is said to be reflexive, and we usually write X = X ∗∗ . If j is not surjective, then X is called nonreflexive. If k·k1 and k·k2 are two equivalent norms on X, then (X, k·k1 ) is reflexive if and only if (X, k·k2 ) is reflexive; recall that two norms k·k1 and k·k2 on X are said to be equivalent if there exist constants a, b > 0 such that for each x ∈ X a kxk2 ≤ kxk1 ≤ b kxk2 . Each x∗ ∈ X ∗ generates a seminorm px∗ on X defined for any x ∈ X by px∗ (x) = |x∗ (x)| . The weak topology T (X, X ∗ ) on X is the topology generated by the family of seminorms {px∗ : x∗ ∈ X ∗ } described above. It means that T (X, X ∗ ) is the weakest topology on X with respect to which all functionals x∗ ∈ X ∗ are continuous. Hence, if T (X) denotes the topology generated by the norm k·k on X, we have T (X, X ∗ ) ⊂ T (X). The weak topology coincides with the norm topology on X if and only if X is a finite dimensional Banach space. Furthermore, if X is infinite dimensional Banach space, then every nonempty weakly open set is unbounded; by a weakly open set, we understand any element of the weak topology T (X, X ∗ ). We say that a subset K of X is weakly closed if its complement in X is weakly open set.

Basic facts on Banach spaces

19

Similarly, each x ∈ X generates a seminorm px on X ∗ defined for any x ∈ X ∗ as px (x∗ ) = |x∗ (x)| = |j(x)(x∗ )| . ∗

The weak* topology T (X ∗ , X) on X ∗ is the topology generated by the family of seminorms {px : x ∈ X}. In other words, T (X ∗ , X) is the weakest topology on X ∗ with respect to which all functionals x∗∗ ∈ j(X) are continuous. Hence, T (X ∗ , X) ⊂ T (X ∗ , X ∗∗ ) ⊂ T (X ∗ ), where T (X ∗ , X ∗∗ ) denotes the weak topology on X ∗ and T (X ∗ ) denotes the topology generated by a norm k·k∗ in X ∗ . Moreover, T (X ∗ , X) = T (X ∗ , X ∗∗ ) if and only if X is reflexive. We say that K ⊂ X ∗ is weakly* closed if its complement in X ∗ belongs to T (X ∗ , X). A sequence {xn } ⊂ X converges to a point x ∈ X in the weak topology if and only if for every x∗ ∈ X ∗ lim x∗ (xn ) = x∗ (x).

n→∞

Then we say that {xn } is weakly convergent or converges weakly to x, and we write w − lim xn = x. n→∞

⊂ X ∗ converges to a point x∗ ∈ X ∗ in the Similarly, a sequence weak* topology if and only if for every x ∈ X {x∗n }

lim x∗n (x) = x∗ (x);

n→∞

we say that {x∗n } is weakly* convergent to x∗ , and we write w∗ − lim x∗n = x∗ . n→∞

A set A ⊂ X is said to be linearly dense in X, if the set ( n ) X span(A) := λi xi : x1 , . . . , xn ∈ A, λ1 , . . . , λn ∈ R, n ∈ N i=1

is dense in X; that is, span(A) = X. We shall list some well-known facts concerning weak and weak* topologies. • Let X be a Banach space and A ⊂ X ∗ be a linearly dense set in X ∗ . A bounded sequence {xn } ⊂ X converges weakly to x in X if and only if x∗ (x) = lim x∗ (xn ) for each x∗ ∈ A. n→∞

20

Classification of Lipschitz mappings • Let A ⊂ X be a linearly dense in X. A bounded sequence {x∗n } ⊂ X ∗ converges weakly* to x∗ ∈ X ∗ if and only if x∗ (x) = lim x∗n (x) for each n→∞ x ∈ A. • A convex set K ⊂ X is weakly closed if and only if it is closed.

The last property follows immediately from the Separation Theorem, which is actually a consequence of the Hahn-Banach Theorem. Theorem 2.3 If K and C are disjoint convex subsets of X such that K is closed and C is compact, then there exists x∗ ∈ X ∗ such that max {x∗ (x) : x ∈ C} < inf {x∗ (x) : x ∈ K} . • If K is a weakly compact subset of X, then convK is also weakly compact. The following property holds for both topologies. • If K is a weakly compact subset of X, then it is bounded. If K is a weakly* compact subset of a dual space X ∗ , then K is bounded. The above is a simple consequence of the Banach Steinhaus Theorem, which in the case of continuous linear functionals takes the form: Theorem 2.4 (Banach-Steinhaus Theorem) Let X be a Banach space and A be a subset of X ∗ . The following statements are equivalent: (i) A is bounded, that is, ∃M >0 ∀x∗ ∈A kx∗ k∗ ≤ M. (ii) For each x ∈ X, the set {x∗ (x) : x∗ ∈ A} is bounded. If X is an infinitely dimensional Banach space, then the closed unit ball BX is never compact in norm topology. However, we have a very important fact concerning weak* topology. • (Alaoglu’s Theorem). The closed unit ball B ∗ in X ∗ is always compact in the weak* topology. Consequently, • If X is reflexive, then the closed unit ball B in X is compact in the weak topology. The following, very deep result states that the weak compactness is equivalent to the sequential weak compactness. • (Eberlein-Smulian Theorem). Let A be the subset of a Banach space X. The following statements are equivalent. (i) A is weakly compact.

Basic facts on Banach spaces

21

(ii) Each sequence {xn } ⊂ A has a subsequence that converges weakly to a point of A. This property does not hold in general for the weak* topologies. However, if the Banach space X is separable, then the relative weak* topology on the ∞ unit ball B ∗ in X ∗ is metrizable. Indeed, suppose A = {xi }i=1 is a countable, dense subset of B ⊂ X. Then, we define a metric d on B ∗ by putting for each x∗ , y ∗ ∈ B ∗ ∞ X 1 ∗ |x (xi ) − y ∗ (xi )| . d(x∗ , y ∗ ) = i 2 i=1 One can notice that the topology on B ∗ generated by the metric d coincides with the relative weak* topology on B ∗ . Then, for any sequence {x∗n } in B ∗ , we have w∗ − lim x∗n = x∗ ⇐⇒ lim d(x∗n , x∗ ) = 0. n→∞

n→∞

It is easy to note that the same is true for any bounded set in X ∗ . Consequently, we have the following properties: • Let X be a separable Banach space, and let A be a subset of X ∗ . The following statements are equivalent: (i) A is weakly* compact. (ii) Each sequence {x∗n } ⊂ A has a subsequence that converges weakly* to a point of A. Below, we shall collect several known characterizations of reflexive Banach spaces. Theorem 2.5 Let X be a Banach space. The following statements are equivalent. (a) X is reflexive. (b) X ∗ is reflexive. (c) (Alaoglu’s Theorem). BX is weakly compact. (d) (Eberlein-Smulian Theorem). Any bounded sequence in X has a weakly convergent subsequence. (e) (James’s Theorem). For any x∗ ∈ X ∗ , there exists x ∈ BX such that x∗ (x) = kx∗ k∗ . (f ) (James’s Theorem). For any bounded, closed and convex set C ⊂ X and any x∗ ∈ X ∗ , there exists x ∈ C such that x∗ (x) = sup {x∗ (y) : y ∈ C} .

22

Classification of Lipschitz mappings

(g) (Smulian’s Theorem). Any descending sequence {Cn } of nonempty, bounded, T∞ closed and convex subsets of X has nonempty intersection; that is, n=1 Cn 6= ∅. From the above, we immediately obtain that every finite dimensional Banach space is reflexive. If Y is a closed subspace of reflexive Banach space X, then Y is also reflexive. Using Smulian’s or James’s characterization, one can show that every uniformly convex space is reflexive. We shall prove it by showing that each functional x∗ ∈ X ∗ attains its norm on the unit sphere SX . Indeed, suppose X is uniformly convex with the modulus of convexity δX . Take any functional x∗ ∈ X ∗ . Without loss of generality, we can assume that x∗ ∈ SX ∗ . For any n ∈ N, consider the slice       1 1 ∗ ∗ = x ∈ BX : x (x) ≥ 1 − δX . S x , δX n n

Then, for any two points x, y ∈ S x∗ , δX n1 , we have     x+y kx + yk kx + yk 1 ∗ ≤x ≤ kx∗ k = . 1 − δX n 2 2 2 Since X is uniformly convex, we conclude that kx − yk ≤


diam S x∗ , δX n1 ≤ n1 . Then, the intersection   \  1 ∗ S x , δX n

1 n,

and, consequently,

n∈N

consists of exactly one point z, as it is a descending family of nonempty, closed sets with diameters converging to zero, and it must be the case that x∗ (z) = 1. Consequently, for each 1 < p < ∞, the spaces Lp [0, 1] and `p are reflexive. Moreover, Lp [0, 1]∗ = Lq [0, 1] and `∗p = `q , where p1 + 1q = 1. The space `1 is nonreflexive. To see it, consider the linear functional x∗ ∈ `∗1 defined for any x = (x1 , x2 , . . . ) ∈ `1 by ∗

x (x) =

∞  X n=1

1 1− n

 xn .

It is easy to show that kx∗ k = 1. However, x∗ does not attain its norm on the unit ball B`1 . Also, the functional x∗ generates the family of descending ∞ sequence {Cn }n=1 of nonempty, bounded, closed and convex sets with empty intersection. Indeed, it is enough to consider slices defined as   1 ∗ Cn = x ∈ B`1 : x (x) ≥ 1 − . n Since c∗0 = `1 and `∗1 = `∞ , we immediately get that c0 and `∞ are nonreflexive. The space c is nonreflexive because it is isomorphic to c0 . The space

Basic facts on Banach spaces

23

L1 [0, 1] contains an isometric copy of `1 ; hence, L1 [0, 1] is also nonreflexive. Moreover, L1 [0, 1]∗ = L∞ [0, 1], so L∞ [0, 1] is nonreflexive. Also, the space C[0, 1] is nonreflexive. We say that a Banach space Y is finitely representable in a Banach space X if for each finite dimensional subspace Y1 of Y and for each µ > 0, there exist a subspace X1 of X and an isomorphism T : Y1 → X1 , satisfying for any y ∈ Y1 the following condition: (1 − µ) kyk ≤ kT yk ≤ (1 + µ) kyk . A Banach space X is said to be super-reflexive if any Banach space Y that is finitely representable in X is reflexive. We say that a Banach space X is uniformly nonsquare or has a uniformly nonsquare norm if 0 (X) < 2. The concept of uniformly nonsquare space is due to James. Theorem 2.6 Let X be a Banach space. The following statements are equivalent: (i) X is super-reflexive. (ii) (James, 1964). X admits an equivalent uniformly nonsquare norm. (iii) (Enflo, 1972). X admits an equivalent uniformly convex norm. Obviously, every super-reflexive space X is also reflexive because X is finitely representable in X. However, there are examples of reflexive spaces that fail to be super-reflexive. Now we can generalize an example 2.2 as follows. Let (X, k·k) be a Banach space that is not super-reflexive. Suppose that k·k◦ is a strictly convex norm on X such that, for any x ∈ X, kxk◦ ≤ kxk . Then, for any µ > 0, the formula  1 2 2 2 k|x|kµ = kxk + µ kxk◦ defines the norm in X, which is strictly convex and equivalent to the norm k·k, 1 kxk ≤ k|x|kµ ≤ (1 + µ) 2 kxk . However, (X, k·k) does not admit any equivalent uniformly nonsquare norm. Consequently, ( 0 if  ∈ [0, 2), δ(X,k|·|k ) () = µ 1 if  = 2.

24

Classification of Lipschitz mappings

Recall that a sequence {xn } in a Banach space X is said to be a Schauder basis or a basis for X if, for every x ∈ X, there is a unique sequence {an } of real numbers such that x=

∞ X

an xn = lim

k→∞

n=1

k X

an xn .

n=1

∞

For instance, the sequence {en }n=1 , where en = (0, . . . , 0, 1, 0, . . . , 0, . . . ), | {z } n−1

is the standard basis for c0 and `p , 1 ≤ p < ∞. However, if we want to obtain a basis for the space c, we have to add to the above sequence one more element e = (1, 1, . . . , 1, . . . ). The space `∞ does not have Schauder basis because it is not separable. The Schauder basis forms a linearly dense set; hence, we immediately obtain the following characterizations of weakly and weakly* convergent sequences. • Let {xn } = {(xn1 , xn2 , . . . )} be a bounded sequence in c0 or `p , 1 < p < ∞. Then, {xn } converges weakly to x = (x1 , x2 , . . . ) if and only if, for each i = 1, 2, . . . , lim xni = xi . n→∞

• If {xn } = {(xn1 , xn2 , . . . )} is a bounded sequence in `1 = c∗0 , then {xn } converges weakly* to x = (x1 , x2 , . . . ) ∈ `1 if and only if, for each i = 1, 2, . . . , lim xni = xi . n→∞

In other words, in the above-mentioned cases, weak convergence in c0 and `p with 1 < p < ∞ (or weak* convergence in `1 = c∗0 ) is equivalent to boundness and coordinate-wise convergence. In the case of `1 space, we have the following important property. • (Schur’s Lemma) Let {xn } be a bounded sequence in `1 . Then, {xn } is weakly convergent to x ∈ `1 if and only if lim kx − xn k = 0. Hence, n→∞ the weak convergence is equivalent to the norm convergence. It may happen that two different Banach spaces have the same dual space X ∗ . For instance, we have such a situation for the space `1 , which is a dual space for both c0 and c. Consequently, we have two different weak* topologies on `1 , which, in fact, are not comparable. Later, we shall discuss this fact more precisely.

Chapter 3 Mean Lipschitz condition

In 2007, Goebel and Jap´ on Pineda [33] introduced a class of mean nonexpansive mappings to obtain some new fixed point theorems of stability type. Their result initiated a new classification of Lipschitz mappings. Now, we shall present the origin of the condition discussed.

3.1

Nonexpansive and mean nonexpansive mappings in Banach spaces

Standard situation in the study of fixed point theory for nonexpansive mapping is the following: suppose C is a bounded closed and convex subset of a Banach space (X, k·k) and S : C → C is a nonexpansive mapping; that is, for all x, y ∈ C, kSx − Syk ≤ kx − yk . The above setting is natural. If we omit one of the above-mentioned assumptions on C, then, even in simple situations, nonexpansiveness does not guarantee that S has a fixed point in C. The reader can find appropriate examples on subsets of real line. Let us recall a basic and very important property of nonexpansive mappings. In the proof, we shall use the Banach’s Contraction Mapping Principle, which is discussed in detail in chapter 6. Lemma 3.1 Let C be a bounded closed and convex subset of a Banach space (X, k·k). If S : C → C is nonexpansive, then d(S) := inf {kx − Sxk : x ∈ C} = 0.

Proof. Let us fix z ∈ C and  ∈ (0, 1). Let S : C → C be defined by S x = z + (1 − )Sx. Let us notice that S is a contraction because kS x − S yk = (1 − ) kSx − Syk ≤ (1 − ) kx − yk 25

26

Classification of Lipschitz mappings

for x, y ∈ C. By the Banach’s Contraction Mapping Principle, there exists a unique point x ∈ C such that S x = x . Thus, kx − Sx k = kz + (1 − )Sx − Sx k =  kz − Sx k ≤  diam(C). Letting  → 0 we get the conclusion. 

As a consequence of the above lemma, we get the existence of an approximate fixed point sequence for S (a.f.p.s. for short), that is, a sequence {xn } in C such that lim kxn − Sxn k = 0. n→∞

The question about the existence of a fixed point for S is equivalent to the question about whether the continuous function ϕ : C → R given by ϕ(x) = kx − Sxk attains its infimum. Obviously, it is true when C is compact. However, in view of the celebrated Schauder’s Fixed Point Theorem, which guarantees that each continuous self-mapping F defined on a nonempty compact and convex set K has a fixed point, such a result is trivial. That is why we usually think about a noncompact set. Despite the fact that d (S) = 0, Fix(S) may be empty as seen in the following: Example 3.1 Let c0 be the space of all sequences x = (x1 , x2 , . . . ) converging to zero with a standard sup norm, kxk = k(x1 , x2 , . . . )k = sup |xi | . i=1,2,...

Let S : B → B be a mapping defined by Sx = S(x1 , x2 , . . . ) = (1, x1 , x2 , . . . ). It is easy to verify that S is an isometry, and so it is nonexpansive. However, the mapping S is fixed point free. Indeed, Sx = x implies that xi = 1 for i = 1, 2, . . . . But x = (1, 1, . . . , 1, . . . ) ∈ / c0 . Thus, Fix(S) = ∅. We say that C has the fixed point property for nonexpansive mappings (the f.p.p. for short) if each nonexpansive mapping S : C → C has at least one fixed point. If for a given space X, all bounded, closed and convex sets C in X have the f.p.p., then we say that X has the fixed point property for nonexpansive mappings or X has the f.p.p. for short. It is known that the f.p.p. for the set C depends deeply on the geometrical properties of the space X or on the set C itself. We shall discuss this in the chapter devoted to nonexpansive mappings. Now, we only mention that, among all such sets, we find, for example, all bounded closed and convex sets in a uniformly convex space X ([17] and

Mean Lipschitz condition

27

[43]), all weakly compact, convex sets having normal structure [48], all weak* compact, convex sets in `1 considered as a dual of c0 [47], and all bounded closed and convex sets in uniformly nonsquare Banach spaces [28]. An excellent overview of the fixed point theory for nonexpansive mappings can be found in the monograph by Goebel and Kirk [36] or Handbook of Metric Fixed Point Theory [49]. The mean Lipschitz condition has its roots in considerations devoted to the possibility of the extension of the following Generalized Banach’s Contraction Mapping Principle to the class of nonexpansive mappings: Theorem 3.1 Let (M, ρ) be a complete metric space. Suppose T : M → M satisfies the following condition: there exists an integer p and a number k ∈ [0, 1) such that for all x, y ∈ M we have 
min ρ T i x, T i y : i = 1, . . . , p ≤ kρ (x, y) . Then, T has exactly one fixed point. It is easy to observe that for every such mapping d(T ) = 0. If p = 1, then T is a contraction and consequently has a unique fixed point. Nevertheless, if p ≥ 2, then the above condition does not imply the continuity of T , which makes it hard to prove that inf {ρ (x, T x) : x ∈ M } is achieved. The proofs can be found in [3] and [62]. What if we use the counterpart of the above theorem in the case of nonexpansive mappings settings? Let us check it! Suppose C is a bounded, closed and convex subset of a Banach space (X, k·k) and that T : C → C satisfies for any x, y ∈ C the above condition with k = 1 and p = 2; that is,

 min kT x − T yk , T 2 x − T 2 y ≤ kx − yk . Does inf {kx − T xk} = 0? The above condition is satisfied by all nonexpansive mappings. Nevertheless, it is also justified by all mappings that have nonexpansive square, and, consequently, the answer is NO! To illustrate this, let us consider the following example presented in [33]: Example 3.2 Let (X, k·k) = (R, |·|) and C = [0, 1] ⊂ R. Let ϕ : [1/4, 1/2) → [3/4, 1] be a one to one function with ϕ ([1/4, 1/2)) = [3/4, 1]. Let us define a mapping T : [0, 1] → [0, 1] by  x + 1/2 if x ∈ [0, 1/4),    ϕ(x) if x ∈ [1/4, 1/2), Tx = x − 1/2 if x ∈ [1/2, 3/4),    −1 ϕ (x) if x ∈ [3/4, 1]. It is easy to notice that the function T is not continuous. Moreover, it does not have an approximate fixed point sequence because |x − T x| ≥ 1/4 for each x ∈ [0, 1]. However, T 2 x = x for each x ∈ [0, 1], and, consequently, T 2 is nonexpansive.

28

Classification of Lipschitz mappings

Immediately, we can ask what will happen if instead of min we shall consider max. Nothing interesting because the condition

 max kT x − T yk , T 2 x − T 2 y ≤ kx − yk is equivalent to the nonexpansiveness of T . However, between min and max, we have a large collection of averages. The simplest one is

kT x − T yk + T 2 x − T 2 y ≤ kx − yk . (3.1) 2 The above condition is satisfied by all nonexpansive mappings, but it does not exhaust all cases. To see it, consider the following example (see [42]): Example 3.3 ([42]) As a subset C, we take a closed unit ball B in the `1 space of all absolutely summable sequences, P∞x = (x1 , x2 , . . . ), with a metric inherited from the standard norm kxk = i=1 |xi |. Let τ : [−1, 1] → [−1, 1] be the function given by  if t ∈ [−1, −1/2] ,  2t + 1, 0, if t ∈ [−1/2, 1/2] , τ (t) =  2t − 1, if t ∈ [1/2, 1]. It is easy to observe that, for all s, t ∈ [−1, 1], |τ (s) − τ (t)| ≤ 2 |s − t| , and |τ (t)| ≤ |t| . Let us define a mapping T : B → B by   2 T x = T (x1 , x2 , . . . ) = τ (x2 ) , x3 , x4 , x5 , . . . . 3 Then, for i ≥ 2, we have     2 2 i T x= τ xi+1 , xi+2 , xi+3 , xi+4 , . . . . 3 3 For each x = (x1 , x2 , . . . ) and y = (y1 , y2 , . . . ) in B we have ∞

kT x − T yk = |τ (x2 ) − τ (y2 )| +

X 2 |x3 − y3 | + |xk − yk | 3 k=4

≤ 2 |x2 − y2 | + ≤ 2 kx − yk

2 |x3 − y3 | + 3

∞ X k=4

|xk − yk |

Mean Lipschitz condition

29

and for i ≥ 2    

i

T x − T i y = τ 2 xi+1 − τ 2 yi+1 + 2 |xi+2 − yi+2 | 3 3 3 ∞ X + |xk − yk | k=i+3

≤

∞ X 4 2 |xi+1 − yi+1 | + |xi+2 − yi+2 | + |xk − yk | 3 3 k=i+3

4 kx − yk . ≤ 3 Let ei denote i-th vector of standard Schauder basis in `1 . Then,



 



T e2 − T 3 e2 = τ (1) − τ 3 = 1 − 1 = 1 = 2 e2 − 3 e2 .

4 4 2 2 4 For i ≥ 2, we have

   

i



T ei+1 − T i 9 ei+1 = τ 2 − τ 3 = 1 − 1 = 2

10 3 5 3 5 15

4 9 = ei+1 − ei+1

. 3 10 Thus, k(T ) = 2 and k(T i ) = 4/3 for i ≥ 2. Despite of this, for all x, y ∈ B, we have

1 1 5 kT x − T yk + T 2 x − T 2 y ≤ |x2 − y2 | + |x3 − y3 | + |x4 − y4 | 2 2 6 ∞ X + |xk − yk | ≤ kx − yk . k=5

Hence, in the case of a closed unit ball in `1 space, the class of mappings that satisfies (3.1) is a proper extension of the class of nonexpansive ones. Immediately, we can ask about a more general form of (3.1),

α1 kT x − T yk + α2 T 2 x − T 2 y ≤ kx − yk , where α1 > 0, α2 > 0 and α1 + α2 = 1. Why not consider more than two iterates of T ? Let us introduce the following definitions. In this text, we shall say that Pn α = (α1 , . . . , αn ) is a multi-index , if n ≥ 1, α1 > 0, αn > 0, αi ≥ 0, and i=1 αi = 1. The number n will be called the length of the multi-index α = (α1 , . . . , αn ). Definition 3.1 (Goebel and Jap´ on Pineda, 2007) Fix a multi-index α = (α1 , . . . , αn ). We say that a mapping T : C → C is α-nonexpansive if, for all x, y ∈ C, we have n X

αi T i x − T i y ≤ kx − yk . i=1

30

Classification of Lipschitz mappings

For n = 1, we obtain the classical definition of a nonexpansive mapping. If α is not specified, then we say that T is mean nonexpansive. The importance of this theory is building up the intuition of the above class mappings. It could be observed in example 3.3 that such mappings may expand the distance between some points while sending them to such subset on which the second iterate contracts it back. There is a nice characterization of mean nonexpansive mappings. Suppose that T : C → C is α-nonexpansive with α = (α1 , . . . , αn ). Then, for all x, y ∈ C, we have n X

αi T i x − T i y ≤ kx − yk .

i=1

Adding to both sides of the above inequality the term   n n X X

 αj  T i−1 x − T i−1 y i=2

j=i

we get n X i=1

    n n n X X X

i

  αj  T i−1 x − T i−1 y . αj  T x − T i y ≤ j=i

i=1

j=i

From the last inequality, we conclude that T is nonexpansive with respect to an equivalent metric d given for x, y ∈ C by

d (x, y) = kx − yk + (α2 + · · · + αn ) kT x − T yk + · · · + αn T n−1 x − T n−1 y . Reassuming, a mapping T : C → C is α-nonexpansive with α = (α1 , . . . , αn ) if and only if it is nonexpansive with respect to the equivalent metric d defined as above. In particular, every mean nonexpansive mapping is uniformly lipschitzian; however, the implication in opposite direction does not hold, as seen in the following: Example 3.4 Again, consider the space `1 with a standard norm and the mapping T : B → B given by T x = T (x1 , x2 , . . . ) = (τ (x2 ) , x3 , x4 , . . . ) , where
τ is defined as in example 3.3. Then, T is uniformly lipschitzian with k T i = 2 for any i = 1, 2, . . . . In spite of this, T is not mean nonexpansive, for any α. It is natural to ask about the behavior of consecutive Lipschitz constants for iterates T m of T .

Mean Lipschitz condition

31

Suppose that T : C → C is α-nonexpansive with α = (α1 , α2 ); that is,

α1 kT x − T yk + α2 T 2 x − T 2 y ≤ kx − yk . Standard reasoning leads us to the estimates 1 k (T ) ≤ α1

and

k T

2




 ≤ min

1 1 , α12 α2

 .

On the other hand, the mapping T is nonexpansive with respect to the metric d defined as d (x, y) = kx − yk + α2 kT x − T yk . The metric d is equivalent to the metric inherited by the norm k·k, and for all x, y ∈ C, we have   1 α2 kx − yk . kx − yk = kx − yk ≤ d (x, y) ≤ 1 + α1 α1 Consequently, for all m = 1, 2, . . . , kT m x − T m yk ≤ d (T m x, T m y) ≤ d (x, y) ≤ Thus, k (T m ) ≤

1 kx − yk . α1

1 α1 .

In particular,    
1 1 1 1 1 2 , , k T ≤ min = min , . α12 α2 α1 α1 α2

(3.2)

Now we find an interesting part of the story. It occurs that the above estimate is not sharp! To see it, first observe that, for all x, y ∈ C, we have



α1 T 2 x − T 2 y + α2 T 3 x − T 3 y ≤ kT x − T yk . Multiplying both sides of the above inequality by α1 and adding

α2 T 2 x − T 2 y to both sides, we obtain


α12 + α2 T 2 x − T 2 y + α1 α2 T 3 x − T 3 y

≤ α1 kT x − T yk + α2 T 2 x − T 2 y ≤ kx − yk . Consequently,

2

T x − T 2 y ≤

1 kx − yk . α12 + α2

Since α1 + α2 = 1, we get α12 + α2 = α1 (1 − α2 ) + α2 = 1 − α1 α2 . Thus, k(T 2 ) ≤

1 , 1 − α1 α2

(3.3)

32

Classification of Lipschitz mappings

which is better than the estimate
given by (3.2)! For instance, if α = (1/2, 1/2), then from (3.2), we get k T 2 ≤ 2, whereas using (3.3) we obtain k(T 2 ) ≤ 4/3. Moreover, the construction presented in example 3.3 shows that this estimate is sharp! The above observations come from [33]. Also, the authors posed the following problems (see [33]): For any α, can we fully characterize the best Lipschitz constants of iterates of T ? Is the class of mean nonexpansive mappings a proper extension of nonexpansive ones? Is this true regardless of the selection of a convex set C? Later, we shall present answers to the above questions. Let us come back to the general case of mean nonexpansive mappings. For each α-nonexpansive mapping T : C → C, n X

αi T i x − T i y ≤ kx − yk ,

(3.4)

i=1

we define the mapping Tα : C → C by putting for each x ∈ C Tα x =

n X

αi T i x.

(3.5)

i=1

It is easy to observe that Tα is nonexpansive. Indeed, for all x, y ∈ C, we have

n n

X

X

i i kTα x − Tα yk = αi T x − αi T y

i=1

i=1 n

X


αi T i x − T i y =

i=1

≤

n X

αi T i x − T i y

i=1

≤

kx − yk .

Thus, in view of lemma 3.1, d(Tα ) = inf {kx − Tα xk : x ∈ C} = 0. The above observations suggest that Tα may be treated as a “connection” between the class of mean nonexpansive mappings and the class of nonexpansive ones. Not surprisingly, this mapping plays a fundamental role in the metric fixed point theory for mean nonexpansive mappings. However, the nonexpansiveness of Tα is much weaker than (3.4). The following example, presented in [42], shows that it does not imply the continuity of T . Example 3.5 Let T : [0, 1] → [0, 1] be defined as  1 if x ∈ [0, 1/2], Tx = 0 if x ∈ (1/2, 1].

Mean Lipschitz condition

33

Then, T 2x =



if x ∈ [0, 1/2], if x ∈ (1/2, 1].

0 1

Thus, T and T 2 are discontinuous and d(T ) = inf {|x − T x| : x ∈ [0, 1]} = 1/2 > 0. However, for α = (1/2, 1/2) and for each x ∈ [0, 1], we have Tα x =

1 1 1 T x + T 2x = . 2 2 2

Hence, Tα is a constant and so nonexpansive mapping. Moreover, even if Tα is nonexpansive and T is continuous, or even uniformly lipschitzian, it does not imply that (3.4) is satisfied and that d(T ) = 0 as seen below. Example 3.6 (see [42]) Let (X, k·k) be an infinite dimensional Banach space. Since the work of Benyamini and Sternfeld [10], it is known that the unit sphere S is the lipschitzian retract of the closed unit ball B. It means that there exists a lipschitzian mapping R : B → S such that Rx = x for all x ∈ S. Such retraction must have a sufficiently large Lipschitz constant. Standard evaluation is k(R) ≥ 3. Define T : B → S as T = −R. Then, T 2 = R, and all the iterates of T satisfy T n = (−1)n R. Hence, all the powers of T have the same Lipschitz constant as R, k(T n ) = k(R) ≥ 3. Since for any α = (α1 , . . . , αn ) n X

αi T i x − T i y = kRx − Ryk , (3.6) i=1

T does not satisfy (3.4). However, for α = (1/2, 1/2) Tα x =

1 1 T x + T 2x = 0 2 2

for all x ∈ B. In spite of this, T does not have an approximate fixed point sequence. Indeed, for all x ∈ B, we have

2 = T x − T 2 x ≤ k(T ) kx − T xk , which implies kx − T xk ≥

2 > 0. k(T )

Thus, d(T ) > 0. The condition of mean nonexpansiveness can be treated as a perturbation of nonexpansiveness in the following sense: if α1 is close to 1, then the perturbation is small, and for α1 = 1, we obtain nonexpansiveness of T , whereas

34

Classification of Lipschitz mappings

if α1 is close to 0, then the perturbation of nonexpansiveness is large, and for α1 = 0, we can even lose the continuity of T . By analogy to the classical case of nonexpansive mappings, we say that C has the fixed point property for α-nonexpansive mappings if each αnonexpansive mapping T : C → C has at least one fixed point. Below we present some basic fixed point theorems for mean nonexpansive mappings, which can be treated as stability results of f.p.p. for nonexpansive mappings. Theorem 3.2 (Goebel and Jap´ on Pineda [33]) If T (α1 , α2 )-nonexpansive with α1 ≥ 21 , then

: C

→ C is

d(T ) = inf {kx − T xk : x ∈ C} = 0.

Proof. Since the mapping Tα is nonexpansive this implies that for any  > 0 there exists a point x ∈ C such that

kx − Tα x k = x − α1 T x − α2 T 2 x ≤ α2 . By (α1 , α2 )-nonexpansiveness of T , we have



α1 T x − T 2 x + α2 T 2 x − T 3 x ≤ kx − T x k ≤ kx − Tα x k + kTα x − T x k

≤ α2  + (1 − α1 ) T x − T 2 x . Putting the last term on the right hand side to the left we obtain



(2α1 − 1) T x − T 2 x + α2 T 2 x − T 3 x ≤ α2 . Since α1 ≥

1 2

the above inequality implies

2

T x − T 3 x ≤ .

This means that d(T ) = 0. 

If C has the f.p.p. for nonexpansive mappings, then we can repeat the above proof with  = 0 to obtain the following Theorem 3.3 (Goebel and Jap´ on Pineda [33]) If C has the f.p.p. for nonexpansive mappings, then, for n = 2, C has the f.p.p. for α-nonexpansive mappings with all α = (α1 , α2 ) such that α1 ≥ 21 . Proof. Let z ∈ C be fixed under Tα , that is, Tα z = z. Then,



α1 T z − T 2 z + α2 T 2 z − T 3 z ≤ kz − T zk

Mean Lipschitz condition

35

≤ kTα z − T zk

= (1 − α1 ) T 2 z − T z . Thus,



(2α1 − 1) T z − T 2 z + α2 T 2 z − T 3 z ≤ 0. If 2α1 − 1 > 0, then T z = T 2 z. If 2α1 − 1 = 0, then T 2 z is fixed under T . 

In a general case of multi-index α = (α1 , . . . , αn ), we have the following Theorem 3.4 (Goebel and Jap´ on Pineda, [33]) If T : C → C is α1 nonexpansive for α = (α1 , . . . , αn ) with n ≥ 2 and α1 ≥ 2 1−n , then d(T ) = 0. Proof. Fix  > 0 and let x ∈ C be such that kx − Tα xk < . Then, n X

αi T i x − T i+1 x ≤

kx − T xk

i=1

≤

kx − Tα xk + kTα x − T xk

n ! n

X X

αi T i x − αi T x +  ≤

i=1

i=1

n

X


= αi T i x − T x + 

i=2

≤

n X

αi T x − T i x + .

i=2

For any i = 2, . . . , n, we have

 

X i i X






T j−1 x − T j x  .  T j−1 x − T j x αi T x − T i x = αi

≤ αi

j=2

j=2 Thus, n X

   n i X X

i



T j−1 x − T j x  +  αi  αi T x − T i+1 x ≤

i=1

i=2

j=2

  n i X X

j−1

 = αi T x − T j x  +  i=2

=

n X

j=2



n X

 j=2

i=j





j−1

αi  T x − T j x  + .

36

Classification of Lipschitz mappings

This implies that    n n X X

n

 αn T x − T n+1 x ≤ αi  − αj−1  T j−1 x − T j x + . j=2

(3.7)

i=j

The first term on the right-hand side equals " n ! # X



αi − α1 T x − T 2 x = (1 − 2α1 ) T x − T 2 x . i=2

Further, for any j = 3, . . . , n we have   n X  αi  − αj−1 = 1 − 2αj−1 − i=j

j−2 X

! αi

≤ 1 − α1 .

i=1

Using the above and the fact that k(T j ) ≤ 1/α1j , we obtain the evaluation



αn T n x − T n+1 x ≤ (1 − 2α1 ) T x − T 2 x n X

(1 − α1 ) T j−1 x − T j x +  + j=3



 n X ≤ 1 − 2α1 + (1 − α1 )  j=3

Pn

" = =

1 − 2α1 + (1 − α1 )



1 

T x − T 2 x + 

α1j−2

j−3 j=3 α1 α1n−2

#

T x − T 2 x + 

 

1 − α1n−2

T x − T 2 x + . 1 − 2α1 + n−2 α1

It is easy to see that 1 − 2α1 + holds if

1 − α1n−2 ≤0 α1n−2 1

α1 ≥ 2 1−n . 

If the mapping Tα has a fixed point x = Tα x, then we can repeat the above proof with  = 0. As a consequence, we get that z = T n x is fixed point of T . We shall summarize the last observation in the following theorem. Theorem 3.5 (Goebel and Jap´ on Pineda, [33]) If C has the f.p.p. for nonexpansive mappings, then for any multi-index α = (α1 , . . . , αn ) with n ≥ 2 1 and α1 ≥ 2 1−n , C has the f.p.p. for α-nonexpansive mappings.

Mean Lipschitz condition

37

In other words, if C has the f.p.p. for nonexpansive mappings, then C has the f.p.p. for each mapping T : C → C, which is nonexpansive with respect to the metric d defined for all x, y ∈ C as

d (x, y) = kx − yk + (α2 + · · · + αn ) kT x − T yk + · · · + αn T n−1 x − T n−1 y 1

for some multi-index α = (α1 , . . . , αn ) with α1 ≥ 2 1−n . The authors also noticed that this evaluation, which is based only on the value of initial index α1 , is not exact. In fact, if α = (α1 , α2 , α3 ) with α1 ≥ α2 ≥ α3 , then formula (3.7) takes the form



α3 T 3 x − T 4 x ≤ (1 − 2α1 ) T x − T 2 x + (α3 − α2 ) T 2 x − T 3 x + 

≤ (1 − 2α1 ) T x − T 2 x + . If α1 ≥ 1/2, then T has the minimal displacement zero. Consequently, we get the following: Theorem 3.6 (Goebel and Jap´ on Pineda) If T : C → C is αnonexpansive with a multi-index α = (α1 , α2 , α3 ) such that α1 ≥ 21 and α1 ≥ α2 ≥ α3 , then d(T ) = inf {kx − T xk : x ∈ C} = 0.

Now, we can connect theorems 3.6 and 3.4 (for n = 3) to obtain Theorem 3.7 (Goebel and Jap´ on Pineda) If T : C nonexpansive with a multi-index α = (α1 , α2 , α3 ) such that " √ ! 1 2 1 1 α1 ∈ , and α2 ≥ − α1 2 2 2 2 or

→

C is α-

√ 2 α1 ≥ , 2

then d(T ) = inf {kx − T xk : x ∈ C} = 0.

Nevertheless, the following problem, posed by Goebel and Jap´on Pineda, is still open: For n = 2, 3, . . . determine the set of all multi-indexes α of length n such that each α-nonexpansive mapping T : C → C has d(T ) = 0.

38

3.2

Classification of Lipschitz mappings

General case

It seems natural to extend the idea presented by Goebel and Jap´on Pineda to the whole class of lipschitzian self-mappings defined on a metric space (M, ρ) (see [42]). Definition 3.2 (Goebel, Jap´ on Pineda, Sims) Fix k ≥ 0 and a multiindex α = (α1 , . . . , αn ). A mapping T : M → M is said to be α-lipschitzian for the constant k if, for every x, y ∈ M , we have n X


αi ρ T i x, T i y ≤ kρ (x, y) .

(3.8)

i=1

We referred to (3.8) as the mean Lipschitz condition. The smallest constant k for which (3.8) holds is called the mean Lipschitz constant for T , and it is denoted by k (α, T ). Observe that any α-lipschitzian mapping is also lipschitzian and k(T ) ≤

k (α, T ) . α1

Moreover, for i = 1, . . . , n, provided αi > 0, we have # "  i  k(α, T ) k(α, T ) i i k(α, T ) ≤ min . , k(T ) ≤ min k(T ) , αi α1 αi

(3.9)

On the other hand, if T is lipschitzian, then for any multi-index α the mapping T is α-lipschitzian with k (α, T ) ≤

n X

αi k(T i ).

i=1

The class of all mappings that satisfies the mean Lipschitz condition with α = (α1 , . . . , αn ) and k ≥ 0 is denoted by L (α, k). Obviously, for n = 1, we get nothing more than the classical definition of k-lipschitzian mapping. Hence, in this special case, the mean Lipschitz condition determines the class L (k). If k (α, T ) ≤ 1, then we say that T is α-nonexpansive, and if k (α, T ) < 1, then T is called α-contraction. We say that T is mean nonexpansive (mean contraction) if there exists a multi-index α such that T is α-nonexpansive (αcontraction). When the multi-index α and the constant k are not explicitly specified, we refer to such mapping as a mean lipschitzian.

Mean Lipschitz condition

39

For any multi-index α = (α1 , . . . , αn ) and k ≥ 0, each class L (α, k) contains all the lipschitzian mappings T such that n X

αi k(T i ) ≤ k.

i=1

Thus, using estimate k(T i ) ≤ k(T )i , we conclude that each class L (α, k) contains all the lipschitzian mappings T such that n X

αi k(T )i ≤ k.

(3.10)

i=1

In particular, if k ≥ 1, then each class L (α, k) contains all nonexpansive mappings; that is, L (1) ⊂ L (α, k).  If T is uniformly lipschitzian with sup k(T i ) : i = 1, 2, . . . ≤ k, then for any α, T is α-lipschitzian with mean Lipschitz constant k(α, T ) ≤ k. Hence, for any α, Lu (k) ⊂ L (α, k). Suppose T ∈ L (α, k), where α = (α1 , . . . , αn ) and k > 0. Consider a metric d defined for all x, y ∈ M as   n n X X  d(x, y) = αi  ρ(T j−1 x, T j−1 y). (3.11) j=1

i=j

The metric d is equivalent to ρ and ρ (x, y) ≤ d (x, y) ≤ bρ (x, y) with b=1+

n X

! αi

kρ (T ) +

i=2

n X

! αi



kρ T 2 + . . . + αn kρ T n−1 .

i=3

Then, d (T x, T y)

=

n X

  n X
 αi  ρ T j x, T j y

j=1

=

n X

i=j


αi ρ T i x, T i y + d (x, y) − ρ (x, y)

i=1

≤ kρ (x, y) + d (x, y) − ρ (x, y) = d (x, y) − (1 − k) ρ (x, y) . If k < 1, then d(T x, T y) ≤ d(x, y) −

b−1+k 1−k d(x, y) = d(x, y). b b

40

Classification of Lipschitz mappings

Thus, kd (T ) ≤ (b − 1 + k) /b < 1. We shall summarize the above observation in the following theorem: Theorem 3.8 (Goebel and Sims, [42]) Let (M, ρ) be a metric space and T : M → M be an α-contraction. Then, there exists a metric d equivalent to ρ such that T is a contraction with respect to d. If (M, ρ) is a complete metric space, then (M, d) is also a complete metric space and, by the Banach’s Contraction Mapping Principle, the mapping T has a unique fixed point. Thus, we get the following: Theorem 3.9 Let (M, ρ) be a complete metric space. If T : M → M is an α-contraction, then T has the unique fixed point in M . If a mapping T is α-nonexpansive (k = 1) with respect to ρ, then T is nonexpansive with respect to d. Consequently, T is uniformly lipschitzian and the inequality ρ(T n x, T n y) ≤ d(T n x, T n y) ≤ d(x, y) ≤ bρ(x, y) holds for all x, y ∈ M and n = 1, 2, . . . . If k(α, T ) > 1, then, using estimate ρ(x, y) ≤ d(x, y), we get d(T x, T y) ≤ d(x, y) − (1 − k) ρ(x, y) ≤ d(x, y) − (1 − k) d(x, y) = kd(x, y). In view of the above remarks, we obtain the following: Theorem 3.10 Any α-lipschitzian mapping T : M → M with k(α, T ) ≥ 1 is lipschitzian with respect to the equivalent metric d defined by (3.11) with kd (T ) ≤ k(α, T ).

Chapter 4 On the Lipschitz constants for iterates of mean lipschitzian mappings

The mean Lipschitz condition involves only a finite numbers of iterates. In spite of this, as we shall see later, this condition has a serious influence not only on behavior of the sequence of Lipschitz constants for consecutive iterates but also on its asymptotic behavior expressed in terms of constants k0 (T ) and k∞ (T ). We start our investigation with a simple case of mean lipschitzian mappings with α = (α1 , α2 ). We will obtain some results that will be useful to predict some analogous in general case of multi-index α of length n.

4.1

A bound for Lipschitz constants of iterates

We have already seen that, for any mapping T ∈ L((1/2, 1/2), 1), we have k(T ) ≤ 2 and k(T 2 ) ≤ 4/3 and that this estimate is optimal. In the following theorem, we present sharp evaluation of Lipschitz constants for all iterates of (α1 , α2 )-nonexpansive mappings. Theorem 4.1 Let T : M → M be (α1 , α2 )-nonexpansive. Then, for n ≥ 0, we have 1 + α2 k(T n ) ≤ . 1 + (−1)n α2n+1 Proof. It is easy to verify that the above bound is satisfied for n = 0 and n = 1. For n = 2, we have α2 ρ(T 2 x, T 2 y) ≤ ρ(x, y) − α1 ρ(T x, T y) ≤ ρ(x, y) − α12 ρ(T 2 x, T 2 y). Thus, ρ(T 2 x, T 2 y) ≤ =

1 ρ(x, y) α12 + α2 1 + α2 ρ(x, y). 1 + α23 41

42

Classification of Lipschitz mappings

For n ≥ 3, we have 1 α1 ρ(T n−2 x, T n−2 y) − ρ(T n−1 x, T n−1 y) α2 α2  1 − (−α2 )n−2  α1 ρ(T n−2 x, T n−2 y) + α2 ρ(T n−1 x, T n−1 y) = α2 + (−α2 )n α2 + (−α2 )n−2 ρ(T n−2 x, T n−2 y) + α2 + (−α2 )n −1 + (−α2 )n + ρ(T n−1 x, T n−1 y) α2 + (−α2 )n 1 − (−α2 )n−2 ≤ ρ(T n−3 x, T n−3 y) α2 + (−α2 )n α2 + (−α2 )n−2 + ρ(T n−2 x, T n−2 y) α2 + (−α2 )n −1 + (−α2 )n + α1 ρ(T n x, T n y). α2 + (−α2 )n

ρ(T n x, T n y) ≤

By putting the last term on the right-hand side to the left and multiplying both sides by α2 + (−α2 )n , we obtain     1 − (−α2 )n+1 ρ(T n x, T n y) ≤ 1 − (−α2 )n−2 ρ(T n−3 x, T n−3 y)   +α2 1 − (−α2 )n−3 ρ(T n−2 x, T n−2 y). If n = 3, then ρ(T 3 x, T 3 y) ≤

1 + α2 ρ(x, y). 1 − α24

For n ≥ 4, we have   1 − (−α2 )n+1 ρ(T n x, T n y)     ≤ 1 − (−α2 )n−2 ρ(T n−3 x, T n−3 y) + α2 1 − (−α2 )n−3 ρ(T n−2 x, T n−2 y)  
= 1 − (−α2 )n−3 α1 ρ(T n−3 x, T n−3 y) + α2 ρ(T n−2 x, T n−2 y)   + α2 + (−α2 )n−3 ρ(T n−3 x, T n−3 y)     ≤ 1 − (−α2 )n−3 ρ(T n−4 x, T n−4 y) + α2 1 − (−α2 )n−4 ρ(T n−3 x, T n−3 y)  
= 1 − (−α2 )n−4 α1 ρ(T n−4 x, T n−4 y) + α2 ρ(T n−3 x, T n−3 y)   + α2 + (−α2 )n−4 ρ(T n−4 x, T n−4 y)     ≤ 1 − (−α2 )n−4 ρ(T n−5 x, T n−5 y) + α2 1 − (−α2 )n−5 ρ(T n−4 x, T n−4 y) ≤ ··· ≤ (1 + α2 ) ρ(x, y) + (α2 − α2 ) ρ(T x, T y) = (1 + α2 ) ρ(x, y).

On the Lipschitz constants for iterates of mean lipschitzian mappings

43

Thus, ρ(T n x, T n y) ≤

1 + α2 ρ(x, y). 1 − (−α2 )n+1 

The following example shows that the estimate presented in the above theorem is optimal. Moreover, for arbitrary α = (α1 , α2 ), we construct a mapping T , which is (α1 , α2 )-nonexpansive, and all its iterates T n have the biggest possible Lipschitz constant; that is, k(T n ) =

1 + α2 1 − (−α2 )n+1

for n = 1, 2, . . . . Example 4.1 As a metric space M , we consider the space `1P with a metric ∞ inherited from the standard norm, kxk = k(x1 , x2 , . . . )k = i=1 |xi | . Let α = (α1 , α2 ) be as above and T : `1 → `1 be a mapping defined for every x = (x1 , x2 , . . . ) ∈ `1 by   1 − (−α2 )1 1 − (−α2 )2 1 − (−α2 )j Tx = x2 , x3 , . . . , xj+1 , . . . . 1 − (−α2 )2 1 − (−α2 )3 1 − (−α2 )j+1 Then, T 2x =



1 − (−α2 )1 1 − (−α2 )2 1 − (−α2 )3 x3 , x4 , x5 , . . . 3 4 1 − (−α2 ) 1 − (−α2 ) 1 − (−α2 )5

 .

In general, for n = 2, 3, . . . , we have   1 − (−α2 )1 1 − (−α2 )2 1 − (−α2 )3 n xn+1 , xn+2 , xn+3 , . . . . T x= 1 − (−α2 )n+1 1 − (−α2 )n+2 1 − (−α2 )n+3 Let us observe that, for any x = (x1 , x2 , . . . ) ∈ `1 , we get

α1 kT xk + α2 T 2 x ∞ ∞ X X 1 − (−α2 )i 1 − (−α2 )i = α1 |x | + α |xi+2 | i+1 2 1 − (−α2 )i+1 1 − (−α2 )i+2 i=1 i=1 = |x2 | + α1

∞ X 1 − (−α2 )i+1 i=1

= |x2 | +

∞  X i=1

= |x2 | +

∞ X i=1

≤ kxk .

1 − (−α2 )i+2

α1

1 − (−α2 )i+1 1 − (−α2 )i+2

|xi+2 |

∞ X 1 − (−α2 )i |xi+2 | 1 − (−α2 )i+2 i=1  1 − (−α2 )i + α2 |xi+2 | 1 − (−α2 )i+2

|xi+2 | + α2

44

Classification of Lipschitz mappings

Since T is linear, the above means that it is α-nonexpansive. In particular, this implies that T is well defined; that is, for any x ∈ `1 , T x ∈ `1 . In view of the previous theorem, 1 + α2 k(T n ) ≤ 1 + (−1)n α2n+1 for every n = 1, 2, . . . . If en denotes the n-th vector of standard basis in `1 , then

 

1 − (−α2 )1 1 + α2 n

= , 0, 0, . . . kT en+1 k = .

n+1 1 − (−α2 ) 1 + (−1)n αn+1 2

Consequently, k(T n ) =

1 + α2 , 1 + (−1)n α2n+1

as we desired. For the convenience of the reader, we present several samples of graph of the sequence 1 + α2 bn = (4.1) 1 + (−1)n α2n+1 as well as vectors of approximate values of its initial terms. For clarity, we connect points of graph with lines. 1,3

1,2

1,1

1,0

0

2

4

6

8

FIGURE 4.1: A graph of {bn } with α = ( 54 , 15 ).

10

On the Lipschitz constants for iterates of mean lipschitzian mappings

45

[b0 , . . . , b15 ] ≈[1., 1.250000000, 1.190476190, 1.201923077, 1.199616123, 1.200076805, 1.199984640, 1.200003072, 1.199999386, 1.200000123, 1.199999975, 1.200000005, 1.199999999, 1.200000000, 1.200000000, 1.200000000]

5

4

3

2

1

0

5

10

15

20

25

FIGURE 4.2: A graph of {bn } with α = ( 51 , 45 ). [b0 , . . . , b100 ] ≈[1., 5., 1.190476190, 3.048780488, 1.355748373, 2.439500390, 1.487953528, 2.162869245, 1.586996884, 2.016522449, 1.657612197, 1.932822555, 1.706200628, 1.882806715, 1.738820686, 1.852132907, 1.760360212, 1.833020767, 1.774427800, 1.820994639, 1.783549658, 1.813380386, 1.789437028, 1.808540591, 1.793225386, 1.805456658, 1.795658364, 1.803488454, 1.797218938, 1.802231054, 1.798219130, 1.801427238, 1.798859837, 1.800913171, 1.799270129, 1.800584323, 1.799532815, 1.800373923, 1.799700973, 1.800239293, 1.799808612, 1.800153140, 1.799877507, 1.800098007, 1.799921602, 1.800062723, 1.799949825, 1.800040142, 1.799967887, 1.800025691, 1.799979448, 1.800016442, 1.799986847, 1.800010523, 1.799991582, 1.800006735, 1.799994612, 1.800004310, 1.799996552, 1.800002758, 1.799997793, 1.800001765, 1.799998588, 1.800001130, 1.799999096, 1.800000723, 1.799999422, 1.800000463, 1.799999630, 1.800000296, 1.799999763, 1.800000190, 1.799999848, 1.800000121, 1.799999903, 1.800000078, 1.799999938, 1.800000050, 1.799999960, 1.800000032, 1.799999975, 1.800000020, 1.799999984, 1.800000013, 1.799999990, 1.800000008, 1.799999993, 1.800000005, 1.799999996, 1.800000003, 1.799999997, 1.800000002, 1.799999998, 1.800000001, 1.799999999, 1.800000001, 1.799999999, 1.800000001, 1.800000000, 1.800000000, 1.800000000]

46

Classification of Lipschitz mappings

2,0

1,8

1,6

1,4

1,2

1,0 0

5

10

15

FIGURE 4.3: A graph of {bn } with α = ( 21 , 12 ).

[b0 , . . . , b40 ] ≈ [1., 2., 1.333333333, 1.600000000, 1.454545455, 1.523809524, 1.488372093, 1.505882353, 1.497076023, 1.501466276, 1.499267936, 1.500366300, 1.499816917, 1.500091558, 1.499954225, 1.500022889, 1.499988556, 1.500005722, 1.499997139, 1.500001431, 1.499999285, 1.500000358, 1.499999821, 1.500000089, 1.499999955, 1.500000022, 1.499999989, 1.500000006, 1.499999997, 1.500000001, 1.499999999, 1.500000000, 1.500000000, 1.500000000, 1.500000000, 1.500000000, 1.500000000, 1.500000000, 1.500000000, 1.500000000, 1.500000000] ∞

It is not easy to guess the shape of the sequence {bn }n=0 in general case of k > 0. Indeed, as we shall see later, for any T ∈ L((α1 , α2 ), k), we have k(T n ) ≤ bn , where

√ 2n+1 ∆ bn = k  √ n+1  √ n+1 , α1 + ∆ − α1 − ∆ n

On the Lipschitz constants for iterates of mean lipschitzian mappings

47

with ∆ = α12 + 4α2 k. Despite being complicated and sophisticated, the above-mentioned boundary is sharp! Nevertheless, this is still only the case of multi-index α of length n = 2. It may suggest that, in general, we should find a recurrence formula ∞ for the sequence {bn }n=0 , provided it exists! ∞ Let us start with some particular cases of {bn }n=0 given by (4.1).
For α = 21 , 12 , we get 4 8 16 32 64 128 , b3 = , b4 = , b5 = , b6 = , b7 = , 3 5 11 21 43 85 256 512 1024 2048 4096 8192 b8 = , b9 = , b10 = , b11 = , b12 = , b13 = , 171 341 683 1365 2731 5461 32768 65536 131072 262144 16384 , b15 = , b16 = , b17 = , b18 = , b14 = 10923 21845 43691 87381 174763 524288 1048576 b19 = , b20 = ,.... 349525 699051 b0 = 1, b1 = 2, b2 =

It is clear that the numerator of bn equals 2n . However, it is also easy to observe that the numerator (resp. denominator) of bn equals 2 times numerator (denominator) of bn−2 plus numerator (denominator) of bn−1 ; that is, b0 = 1 =

1 2 , b1 = 2 = , 1 1

and 2·1+2 4 2·2+4 8 2·4+8 16 = , b3 = = , b4 = = , 2·1+1 3 2·1+3 5 2·3+5 11 2 · 8 + 16 32 2 · 16 + 32 64 b5 = = , b6 = = ,.... 2 · 5 + 11 21 2 · 11 + 21 43

b2 =

For α = ( 51 , 45 ), we have 125 625 3125 15625 25 , b3 = , b4 = , b5 = , b6 = , 21 41 461 1281 10501 78125 390625 1953125 9765625 b7 = , b8 = , b9 = , b10 = ,.... 36121 246141 968561 5891381 b0 = 1, b1 = 5, b2 =

Obviously, the numerator of bn equals 5n or equivalently 5 times the numerator of bn−2 plus 4 times the numerator of bn−1 , but the denominator of bn equals 20 times the denominator of bn−2 plus the denominator of bn−1 , that is, 1 5 b0 = 1 = , b1 = 5 = 1 1 and 25 5 · 5 + 4 · 25 125 5·1+4·5 = , b3 = = , b2 = 20 · 1 + 1 21 20 · 1 + 21 41 5 · 25 + 4 · 125 625 5 · 125 + 4 · 625 3125 b4 = = , b5 = = ,.... 20 · 21 + 41 461 20 · 41 + 461 1281

48

Classification of Lipschitz mappings

Following the method described above, the reader can find nice formulas for some other multi-indexes, for example,         4 1 1 2 2 1 1 3 α= , , α= , , α= , , and α = , . 5 5 3 3 3 3 4 4 However, the above procedure does not lead us to the recurrence formula in the general case of α = (α1 , α2 ), especially if α has irrational coefficients. Nevertheless, the general recurrence formula exists and the reader can verify ∞ that the sequence {bn }n=0 defined by (4.1) satisfies the following relation: 1 b0 = 1, b1 = α1 , and 1 bn+2 = . −1 α1 bn+1 + α2 b−1 n ∞

It turns out that, in opposite to an explicit formula of {bn }n=0 , its recurrence formula has a very easy and natural extension for k > 0, as can be observed in the following theorem. Theorem 4.2 Let T : M → M be (α1 , α2 )-lipschitzian mapping for the constant k > 0. Then, for n ≥ 0, we have k(T n ) ≤ bn , where b0 = 1, b1 =

k k and bn+2 = −1 . α1 α1 b−1 n+1 + α2 bn

Proof. It is clear that our formula is correct for n = 0 and n = 1. For n = 2, by definition of T , we have α2 ρ(T 2 x, T 2 y) ≤ kρ(x, y) − α1 ρ(T x, T y) 1 ≤ kρ(x, y) − α1 ρ(T 2 x, T 2 y). b1 Thus, k ρ(x, y) α1 b−1 + α2 b−1 1 0 = b2 ρ(x, y).

ρ(T 2 x, T 2 y) ≤

For n ≥ 3, by definition of T , we can write α2 ρ(T n x, T n y) ≤ kρ(T n−2 x, T n−2 y) − α1 ρ(T n−1 x, T n−1 y) bn−2 = kρ(T n−2 x, T n−2 y) + α2 ρ(T n−1 x, T n−1 y) bn−3   bn−2 α2 ρ(T n−1 x, T n−1 y) + −α1 − bn−3

On the Lipschitz constants for iterates of mean lipschitzian mappings

49

bn−2 ≤ kρ(T n−2 x, T n−2 y) + α2 ρ(T n−1 x, T n−1 y) bn−3   α1 bn−2 + −α1 − α2 ρ(T n x, T n y). k bn−3 Moving the last term from the right-hand side to the left and dividing both sides by bn−2 , we obtain α1 (α1 bn−3 + α2 bn−2 ) + α2 kbn−3 ρ(T n x, T n y) kbn−3 bn−2 ≤k

1 1 ρ(T n−2 x, T n−2 y) + α2 ρ(T n−1 x, T n−1 y). bn−2 bn−3

It is easy to verify that the term on the left-hand side equals α1

1 k 1 + α2 = . bn−1 bn−2 bn

Hence, we can write k 1 1 ρ(T n x, T n y) ≤ k ρ(T n−2 x, T n−2 y) + α2 ρ(T n−1 x, T n−1 y). (4.2) bn bn−2 bn−3 For n = 3, we have k 1 1 ρ(T 3 x, T 3 y) ≤ k ρ(T x, T y) + α2 ρ(T 2 x, T 2 y) b3 b1 b0
1 α1 ρ(T x, T y) + α2 ρ(T 2 x, T 2 y) = b0   1 1 + k − α1 ρ(T x, T y) b1 b0 ≤ kρ(x, y). Thus, ρ(T 3 x, T 3 y) ≤ b3 ρ(x, y). For n ≥ 4, starting from the right-hand side of inequality (4.2), we repeat the following procedure from i = 1 to n − 3, k

1 1 ρ(T n−i−1 x, T n−i−1 y) + α2 ρ(T n−i x, T n−i y) bn−i−1 bn−i−2

=

≤k


1 α1 ρ(T n−i−1 x, T n−i−1 y) + α2 ρ(T n−i x, T n−i y) bn−i−2   1 1 + k − α1 ρ(T n−i−1 x, T n−i−1 y) bn−i−1 bn−i−2

1 1 ρ(T n−i−2 x, T n−i−2 y) + α2 ρ(T n−i−1 x, T n−i−1 y) bn−i−2 bn−i−3

50

Classification of Lipschitz mappings

and after (n − 3) steps (or 1 step, if n = 4), we obtain k 1 1 ρ(T n x, T n y) ≤ k ρ(T x, T y) + α2 ρ(T 2 x, T 2 y) bn b1 b0
1 = α1 ρ(T x, T y) + α2 ρ(T 2 x, T 2 y) b0   1 1 + k − α1 ρ(T x, T y) b1 b0 ≤ kρ(x, y). Thus, ρ(T n x, T n y) ≤ bn ρ(x, y). 

To see that the above estimation is optimal, we consider the following: Example 4.2 (Left shift with special coefficients) Let {bn }n≥0 be as in theorem 4.2. Let T : `1 → `1 be a linear mapping given for every x = (x1 , x2 , . . . ) ∈ `1 by   b2 bj b1 x2 , x3 , . . . , xj+1 , . . . . Tx = b0 b1 bj−1 Then, 2



T x=

b2 b3 b4 x3 , x4 , x5 , . . . b0 b1 b2



and, in general, n

T x=



bn bn+1 bn+2 xn+1 , xn+2 , xn+3 , . . . b0 b1 b2



for n ≥ 2. We claim that mapping T is (α1 , α2 )-lipschitzian for the constant k > 0. Indeed, for each x ∈ `1 , we have ∞ ∞ X X

bi+1 bi |xi+1 | + α2 |xi+2 | α1 kT xk + α2 T 2 x = α1 b b i=1 i−1 i=1 i−1

= k |x2 | + α1

∞ X bi+1 i=1

= k |x2 | +

∞  X i=1 ∞ X

= k |x2 | + k

i=1

≤ k kxk .

α1

bi

|xi+2 | + α2

bi+1 bi+1 + α2 bi bi−1

|xi+2 |

∞ X bi+1 i=1

bi−1

 |xi+2 |

|xi+2 |

On the Lipschitz constants for iterates of mean lipschitzian mappings

51

If en denotes the n-th vector of the standard Schauder basis in `1 , then kT n en+1 k = bn , and by theorem (4.2), we get k(T n ) = bn . Let us observe that, for k = 1, the above mapping coincide with those described in example 4.1. Now, using standard methods, we give an explicit formula for the sequence ∞ {bn }n=0 : ∞

Lemma 4.1 Let {bn }n=0 be as in theorem 4.2. Then, for n ≥ 0, we have √ 2n+1 ∆ bn = k n  √ n+1  √ n+1 α1 + ∆ − α1 − ∆ with ∆ = α12 + 4α2 k. Proof. If we define {dn }n≥0 by dn = relation: d0 = 1, d1 =

1 bn ,

then we obtain the following

α1 α2 α1 , and dn+2 = dn+1 + dn , n ≥ 0. k k k

The solutions of the characteristic equation λ2 − are

α1 α2 λ− =0 k k

√ √ α1 + ∆ α1 − ∆ λ1 = and λ2 = , 2k 2k

with ∆ := α12 + 4α2 k. Hence, the general solution has the form √ !n α1 + ∆ dn = C1 + C2 2k

√ !n α1 − ∆ , n ≥ 0. 2k

Using initial condition, we solve a linear system  C1 + C√2 = 1 √ ∆ ∆ C1 α1 + + C2 α1 − = 2k 2k and get (

C1 = C2 =

√ α1 + √ ∆ 2 ∆√ √ ∆. − α12− ∆

α1 k

52

Classification of Lipschitz mappings

Thus, 

α1 +

dn =

√ n+1  √ n+1 ∆ − α1 − ∆ √ , 2n+1 k n ∆

and, finally, for n ≥ 0, bn = k n 

√ 2n+1 ∆ √ n+1  √ n+1 . α1 + ∆ − α1 − ∆ 

2,0

1,5

1,0

0,5

0,0 0

10

20

FIGURE 4.4: A graph of {bn }, with k =

30

5 6

and α = ( 12 , 21 ).

[b0 , . . . , b50 ] ≈ [1.000 ∗ 100 , 1.667 ∗ 100 , 1.042 ∗ 100 , 1.068 ∗ 100 , 8.790 ∗ 10 , 8.037 ∗ 10−1 , 6.998 ∗ 10−1 , 6.235 ∗ 10−1 , 5.495 ∗ 10−1 , 4.868 ∗ 10−1 , 4.302 ∗ 10−1 , 3.806 ∗ 10−1 , 3.366 ∗ 10−1 , 2.977 ∗ 10−1 , 2.633 ∗ 10−1 , 2.329 ∗ 10−1 , 2.060 ∗ 10−1 , 1.822 ∗ 10−1 , 1.611 ∗ 10−1 , 1.425 ∗ 10−1 , 1.260 ∗ 10−1 , 1.115 ∗ 10−1 , 9.858 ∗ 10−2 , 8.719 ∗ 10−2 , 7.711 ∗ 10−2 , 6.820 ∗ 10−2 , 6.032 ∗ 10−2 , 5.335 ∗ 10−2 , 4.718 ∗ 10−2 , 4.173 ∗ 10−2 , 3.691 ∗ 10−2 , 3.264 ∗ 10−2 , 2.887 ∗ 10−2 , 2.553 ∗ 10−2 , 2.258 ∗ 10−2 , 1.997 ∗ 10−2 , 1.767 ∗ 10−2 , 1.562 ∗ 10−2 , 1.382 ∗ 10−2 , 1.222 ∗ 10−2 , 1.081 ∗ 10−2 , 9.560 ∗ 10−3 , 8.455 ∗ 10−3 , 7.478 ∗ 10−3 , 6.614 ∗ 10−3 , 5.850 ∗ 10−3 , 5.174 ∗ 10−3 , 4.576 ∗ 10−3 , 4.047 ∗ 10−3 , 3.579 ∗ 10−3 , 3.166 ∗ 10−3 ] −1

On the Lipschitz constants for iterates of mean lipschitzian mappings

53

50.0

40.0

30.0

20.0

10.0

0.0 0.0

10.0

20.0

FIGURE 4.5: A graph of {bn }, with k =

30.0

6 5

and α = ( 12 , 21 ).

[b0 , . . . , b30 ] ≈ [1., 2.4, 1.694117647, 2.383448276, 2.376618911, 2.856034433, 3.113237586, 3.574913812, 3.993769417, 4.527302101, 5.092599155, 5.752009524, 6.482707796, 7.314646544, 8.248314569, 9.304168566, 10.49338151, 11.83566936, 13.34901637, 15.05624926, 16.98159380, 19.15328259, 21.60261536, 24.36521932, 27.48108358, 30.99542723, 34.95918267, 39.42983712, 44.47220460, 50.15940250, 56.57388973] Corollary 4.1 Let T : M → M be (α1 , α2 )-lipschitzian mapping for the constant k > 0. Then, for n ≥ 0, we have √ 2n+1 ∆ n n (4.3) k(T ) ≤ k  √ n+1  √ n+1 , α1 + ∆ − α1 − ∆ with ∆ = α12 + 4α2 k; see figures 4.4 and 4.5. Let us observe that, for k = 1, we get 2

∆ = α12 + 4α2 = (1 − α2 ) + 4α2 = 1 + 2α2 + α22 = (1 + α2 )2 .

54

Classification of Lipschitz mappings

Consequently, using evaluation (4.3), for any (α1 , α2 )-nonexpansive mapping T : M → M , we get k(T n ) ≤ = = =

2n+1 (1 + α2 ) (α1 + α2 + 1)n+1 − (α1 − 1 − α2 )n+1 2n+1 (1 + α2 ) 2n+1 − (−2α2 )n+1 2n+1 (1 + α2 ) 2n+1 + (−1)n 2n+1 α2n+1 1 + α2 1 + (−1)n α2n+1

as in theorem 4.1. Let us pass now to the general case of multi-index α of length n. The recurrence formula presented in theorem 4.2 may suggest the following conjecture in the general case: let T : M → M be (α1 , . . . , αn )-lipschitzian for the constant k, let {bm } be defined as   1 for m = 0,    k  P for m = 1, . . . , n, m −1 bm = j=1 αj bm−j    n k for m = n + 1, n + 2, . . . ;    P αi b−1 m−i i=1

under the above settings we have k(T m ) ≤ bm . Actually, the above conjecture holds, and it was proved by P´erez Garc´ıa and the present author in [64]. To prove this result, we shall need the following lemma: Lemma 4.2 Let T : M → M be an (α1 , . . . , αn )-lipschitzian mapping for the constant k > 0. Let b0 = 1, and for m = 1, . . . , n, we set bm = P m j=1

k

.

αj b−1 m−j

Then, k(T m ) ≤ bm for m = 1, . . . , n. Proof. Obviously, k(T ) ≤ k/α1 = b1 . Suppose that, for some m, with 1 ≤ m ≤ n, we have already proved that k(T ) ≤ b1 , . . . , k(T m−1 ) ≤ bm−1 . Since ρ(T m x, T m y) ≤ k(T m−1 )ρ(T x, T y) ≤ bm−1 ρ(T x, T y), this implies that m m α1 b−1 m−1 ρ(T x, T y) ≤ α1 ρ(T x, T y).

On the Lipschitz constants for iterates of mean lipschitzian mappings

55

Further, ρ(T m x, T m y) ≤ k(T m−2 )ρ(T 2 x, T 2 y) ≤ bm−2 ρ(T 2 x, T 2 y), which implies that m m 2 2 α2 b−1 m−2 ρ(T x, T y) ≤ α2 ρ(T x, T y). Continuing this process, we obtain m m j j αj b−1 m−j ρ(T x, T y) ≤ αj ρ(T x, T y)

for j = 1, . . . , m. Using the above and our assumption that T is (α1 , . . . , αn )-lipschitzian for k > 0, we get m X
−1 m m α1 b−1 + · · · + α b ρ(T x, T y) ≤ αj ρ(T j x, T j y) ≤ kρ(x, y). m 0 m−1 j=1

Thus, k(T m ) ≤ bm . 

Let us note that the last lemma is valid even for the case αn = 0. Actually, we only used the fact that α1 > 0. We apply this observation in the proof of the following theorem: Theorem 4.3 (P´ erez Garc´ıa and Piasecki, [64]) Let T : M → M be an (α1 , . . . , αn )-lipschitzian mapping for the constant k > 0. Then for m ≥ 0 we have k(T m ) ≤ bm , where bm is defined as follows:   1    k  P m −1 bm = j=1 αj bm−j    n k    P αi b−1 m−i

for for

m = 0, m = 1, . . . , n,

for

m = n + 1, n + 2, . . . .

i=1

Proof. We have already proved that for m = 1, . . . , n, k(T m ) ≤ bm . Now consider m = n + j, with j ≥ 1. Since T : M → M is (α1 , . . . , αn )-lipschitzian, it is also (α1 , . . . , αn , αn+1 , . . . , αm )-lipschitzian, provided αn+1 = αn+2 = · · · = αm = 0. Applying lemma 4.2 to this mapping, we obtain bounds for k(T ), . . . , k(T n ), k(T n+1 ), . . . , k(T m ), which are precisely our desired bounds. 

To prove that the above bound is sharp, it is enough to consider a “copy” of example 4.2.

56

Classification of Lipschitz mappings

Example 4.3 Fix α = (α1 , . . . , αn ) and k > 0. Let {bm }m≥0 be the sequence defined as in theorem 4.3. Let us consider the space M = `1 and the linear mapping T : `1 → `1 given for every x = (x1 , x2 , . . . ) ∈ `1 by   b1 b2 bj Tx = x2 , x3 , . . . , xj+1 , . . . . b0 b1 bj−1 Then, 

 b1 b2 b3 x2 , x3 , x4 , . . . , b0 b1 b2   b3 b4 b2 α2 T 2 x =α2 x3 , x4 , x5 , . . . , b0 b1 b2 .. .. . .   bn bn+1 bn+2 n αn T x =αn xn+1 , xn+2 , xn+3 , . . . . b0 b1 b2 α1 T x =α1

Reordering the elements of

n P j=1

αj kT j xk and using the definition of {bm }m≥0 ,

we have that the term that contains |x2 | equals   α1 b−1 −1 0 (k|x2 |) = b1 (b−1 b1 (α1 b0 )|x2 | = b1 1 )k|x2 | = k|x2 |. k Similarly, the term that contains |x3 | is  −1  α1 b−1 −1 −1 1 + α2 b0 b2 (α1 b1 + α2 b0 )|x3 | = b2 (k|x3 |) = b2 (b−1 2 )k|x3 | = k|x3 |. k After doing this n times, we get that the term that contains |xn+1 | equals k|xn+1 |. Now, the term that contains |xn+1+j | with j ≥ 1 is ! n X −1 bn+j αi bn+j−i |xn+1+j | i=1

Pn = bn+j

i=1

αi b−1 n+j−i k

! (k|xn+1+j |)

= bn+j (b−1 n+j )k|xn+1+j | = k|xn+1+j |. Thus, n X j=1

αj kT j xk = k

∞ X j=2

|xj | ≤ kkxk,

(4.4)

On the Lipschitz constants for iterates of mean lipschitzian mappings

57

and, consequently, T is an (α1 , . . . , αn )-lipschitzian mapping for the constant k. This in particular implies that T is well defined; that is, for any x ∈ `1 , ∞ T x ∈ `1 . In view of theorem 4.3, we have k(T m ) ≤ bm . If {ej }j=1 is the standard basis of `1 , then we have T m (em+1 ) = bm b−1 0 e1 = bm e1 , which proves that the bounds bm are sharp. ∞

To show a behavior of the sequence {bm }m=0 , with k = 1 and n ≥ 3, we present some samples of its graph as well as vectors of approximate values of its initial terms:

3,0

2,5

2,0

1,5

1,0 0

5

10

15

FIGURE 4.6: A graph of {bm }, with k = 1 and α = ( 13 , 31 , 13 ). [b0 , . . . , b50 ] ≈[1., 3., 2.250000000, 1.687500000, 2.189189189, 2.008264463, 1.938829787, 2.040111940, 1.994831256, 1.990393366, 2.008196164, 1.997778329, 1.998762623, 2.001568038, 1.999368374, 1.999898952, 2.000278016, 1.999848378, 2.000008430, 2.000044925, 1.999967241, 2.000006865, 2.000006343, 1.999993483, 2.000002230, 2.000000685, 1.999998800, 2.000000572, 2.000000019, 1.999999797, 2.000000129, 1.999999982, 1.999999969, 2.000000027, 1.999999992, 1.999999996, 2.000000005, 1.999999998, 2.000000000, 2.000000001, 2.000000000, 2.000000000, 2.000000000, 2.000000000, 2.000000000, 2.000000000, 2.000000000, 2.000000000, 2.000000000, 2.000000000, 2.000000000]

58

Classification of Lipschitz mappings 10

8

6

4

2

0

0

5

10

15

20

25

FIGURE 4.7: A graph of {bm }, with k = 1 and α = ( 31 , 0, 23 ). [b1 , . . . , b100 ] ≈[1., 3., 9., 1.421052632, 2.189189189, 4.418181818, 1.836272040, 2.057384760, 3.195811008, 2.139689097, 2.084106872, 2.713357943, 2.301915805, 2.151980866, 2.496292593, 2.363255025, 2.218079368, 2.396111347, 2.374106553, 2.267758689, 2.351742660, 2.366604812, 2.299776996, 2.334161787, 2.355690725, 2.318117632, 2.328789115, 2.346654729, 2.327552566, 2.328376786, 2.340530279, 2.331862450, 2.329537516, 2.336854515, 2.333524100, 2.330864863, 2.334854547, 2.333967414, 2.331898129, 2.333868242, 2.333934356, 2.332576477, 2.333437495, 2.333768712, 2.332973753, 2.333282894, 2.333606750, 2.333184714, 2.333250166, 2.333487876, 2.333285759, 2.333262030, 2.333412590, 2.333328034, 2.333284031, 2.333369735, 2.333341935, 2.333303332, 2.333347600, 2.333343823, 2.333316829, 2.333337343, 2.333341663, 2.333325107, 2.333333264, 2.333338864, 2.333329692, 2.333332074, 2.333336600, 2.333331995, 2.333332048, 2.333335083, 2.333333024, 2.333332373, 2.333334180, 2.333333409, 2.333332719, 2.333333693, 2.333333504, 2.333332980, 2.333333455, 2.333333488, 2.333333149, 2.333333353, 2.333333443, 2.333333247, 2.333333318, 2.333333401, 2.333333298, 2.333333311, 2.333333371, 2.333333323, 2.333333315, 2.333333353, 2.333333333, 2.333333321, 2.333333342, 2.333333336, 2.333333326, 2.333333337]

On the Lipschitz constants for iterates of mean lipschitzian mappings

4.0

3.0

2.0

1.0 0.0

5.0

10.0

15.0

FIGURE 4.8: A graph of {bm }, with k = 1 and α =

20.0

1 1 1 1 4, 4, 4, 4

[b0 , . . . , b53 ] ≈ [1., 4., 3.200000000, 2.560000000, 2.048000000, 2.775067751, 2.577721838, 2.458214554, 2.434020427, 2.554387333, 2.504594409, 2.486956285, 2.494247041, 2.509772290, 2.498860985, 2.497431729, 2.500064393, 2.501522973, 2.499469111, 2.499621188, 2.500169153, 2.500195344, 2.499863657, 2.499962313, 2.500047609, 2.500017225, 2.499972699, 2.499999961, 2.500009373, 2.499999815, 2.499995462, 2.500001153, 2.500001451, 2.499999470, 2.499999384, 2.500000364, 2.500000167, 2.499999846, 2.499999940, 2.500000080, 2.500000008, 2.499999969, 2.499999999, 2.500000014, 2.499999998, 2.499999995, 2.500000001, 2.500000002, 2.499999999, 2.499999999, 2.500000000, 2.500000000, 2.500000000, 2.500000000]


.

59

60

Classification of Lipschitz mappings 12.0

10.0

8.0

6.0

4.0

2.0

0.0

10.0

20.0

30.0

40.0

50.0

FIGURE 4.9: A graph of {bm }, with k = 1 and α =

60.0

1 3 4 , 0, 0, 4


.

[b0 , . . . , b200 ] ≈[1., 4., 16., 64., 1.326424870, 2.659740260, 7.098786828, 21.30559168, 1.732610707, 2.345912569, 4.712086964, 11.33053647, 2.198106354, 2.307128288, 3.737968954, 7.514613324, 2.670431675, 2.388360419, 3.275273222, 5.677461873, 3.077991090, 2.530077792, 3.050643232, 4.671779165, 3.364984081, 2.697394453, 2.953932044, 4.078779738, 3.518939677, 2.864588948, 2.931077843, 3.715105219, 3.566012964, 3.012738075, 2.951075057, 3.489263571, 3.546510815, 3.130529285, 2.993981754, 3.350690894, 3.495440976, 3.214422884, 3.046208045, 3.269002834, 3.435940569, 3.267080674, 3.098578280, 3.224663003, 3.380567490, 3.294731996, 3.145393978, 3.204473522, 3.334754185, 3.304647211, 3.183750766, 3.199267583, 3.299817976, 3.303438577, 3.212852204, 3.202652957, 3.274978154, 3.296277198, 3.233310046, 3.210262597, 3.258555879, 3.286765230, 3.246510167, 3.219248395, 3.248639291, 3.277150095, 3.254116310, 3.227895133, 3.243428296, 3.268654077, 3.257738615, 3.235304625, 3.241393556, 3.261796050, 3.258752027, 3.241134786, 3.241328860, 3.256655058, 3.258227532, 3.245391137, 3.242343475, 3.253065331, 3.256935446, 3.248269535, 3.243822962, 3.250749805, 3.255386830, 3.250045940, 3.245376471, 3.249404804, 3.253889260, 3.251005918, 3.246782005, 3.248748707, 3.252602597, 3.251404941, 3.247936506, 3.248545619, 3.251587403, 3.251450554, 3.248814306, 3.248612786, 3.250843238, 3.251298704, 3.249435049, 3.248818313, 3.250336770, 3.251058167, 3.249840677, 3.249073844, 3.250020946,

On the Lipschitz constants for iterates of mean lipschitzian mappings

61

3.250798800, 3.250080155, 3.249325363, 3.249847023, 3.250560803, 3.250200304, 3.249544054, 3.249771275, 3.250363385, 3.250241072, 3.249718281, 3.249758026, 3.250212024, 3.250233810, 3.249847148, 3.249780306, 3.250104084, 3.250201378, 3.249935698, 3.249819153, 3.250032847, 3.250159243, 3.249991581, 3.249862258, 3.249990198, 3.250116980, 3.250022930, 3.249902425, 3.249968254, 3.250079798, 3.250037147, 3.249936104, 3.249960217, 3.250049902, 3.250040336, 3.249962161, 3.249960703, 3.250027601, 3.250037152, 3.249980909, 3.249965754, 3.250012139, 3.250030899, 3.249993406, 3.249972667, 3.250002271, 3.250023742, 3.250000990, 3.249979748, 3.249996640, 3.250016966, 3.250004984, 3.249986057, 3.249993994, 3.250011223, 3.250006544, 3.249991179, 3.249993291, 3.250006740, 3.250006593, 3.249995032, 3.249993726, 3.250003487, 3.250005816, 3.249997728, 3.249994727, 3.250001297, 3.250004686, 3.249999468, 3.249995912, 3.249999950, 3.250003502, 3.250000476, 3.249997053, 3.249999226, 3.250002433, 3.250000966, 3.249998031, 3.249998927, 3.250001557, 3.250001113, 3.249998802, 3.249998896, 3.250000892, 3.250001058, 3.249999366] 8.0

7.0

6.0

5.0

4.0

3.0

2.0

1.0 0.0

100.0

200.0

FIGURE 4.10: A graph of {bm }, with k = 1 and α =

300.0

1 9 10 , 0, 0, 0, 10


.

62

Classification of Lipschitz mappings

48

40

32

24

16

8

0

500

1000

1500

2000

2500

3000

3500

4000

FIGURE 4.11: A graph of {bm }, with α = (0.05, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.95) and k = 1.

4.2

A bound for the constant k∞ (T )

We have already seen that, for any α-contraction T , we have k0 (T ) < 1, and, consequently, k∞ (T ) = 0. In particular, it is satisfied for the mapping T ∞ as in example 4.3, with k < 1. It means that, for {bm }m=0 as in theorem 4.3, with k < 1, we have lim bm = 0. m→∞

Using (3.10), it is also easy to observe that, for any k > 1, we have lim bm = ∞.

m→∞

In case of α-nonexpansive mapping T , with α = (α1 , α2 ), we proved that k(T n ) ≤

1 + α2 . 1 + (−1)n α2n+1

Consequently, we immediately obtain the following: Corollary 4.2 If T : M → M is α-nonexpansive, with α = (α1 , α2 ), then k∞ (T ) ≤ 1 + α2 .

(4.5)

On the Lipschitz constants for iterates of mean lipschitzian mappings

63

Proof. Since α2 ∈ (0, 1), we get lim (−1)n α2n+1 = 0.

n→∞

Hence, k∞ (T ) = lim sup k(T n ) ≤ lim

n→∞

n→∞

1 + α2 = 1 + α2 . 1 + (−1)n α2n+1 

Example 4.4 Let T : `1 → `1 be as in example 4.1; that is,   1 − (−α2 )1 1 − (−α2 )2 1 − (−α2 )j Tx = x , x , . . . , x , . . . . 2 3 j+1 1 − (−α2 )2 1 − (−α2 )3 1 − (−α2 )j+1 We proved that T is (α1 , α2 )-nonexpansive and k(T n ) = bn =

1 + α2 , 1 + (−1)n α2n+1

where {bn }n≥0 is defined as in theorem 4.2, with k = 1. Hence, k∞ (T ) = lim bn = 1 + α2 . n→∞

This shows that the evaluation for k∞ (T ) is sharp. It can be also observed that the sequence of Lipschitz constants k(T n ) = bn oscillates around the limit k∞ (T ) = 1 + α2 in the following way: k(T 2n ) = b2n =

1 + α2 < 1 + α2 1 + α22n+1

and k(T 2n+1 ) = b2n+1 =

1 + α2 > 1 + α2 ; 1 − α22n+2

see also figures 4.1, 4.2, and 4.3. Let us also observe that, for any α-nonexpansive mapping T , with α = (α1 , α2 ), we have 1 ρ(x, y). ρ(T n x, T n y) ≤ α1 Hence, for α1 close to 1, the right-hand side of the above inequality is close to 1, whereas for α1 close to 0 it tends to infinity. Nevertheless, for any α, k∞ (T ) is always less than 2. In general case of multi-index α = (α1 , . . . , αn ) of length n, we have the following bound, which follows from the theorem 4.3:

64

Classification of Lipschitz mappings

Corollary 4.3 Let (M, ρ) be a metric space. Let T : M → M be an αnonexpansive mapping with α = (α1 , . . . , αn ). Then, k(T m ) ≤ bm , where the ∞ sequence {bm }m=0 is defined as follows:   for m = 0,  1  1  P for m = 1, . . . , n, m −1 bm = j=1 αj bm−j   1  for m = n + 1, n + 2, . . . . P   n αi b−1 m−i i=1

∞

As observed in figures 4.6–4.11, the behavior of the sequence {bm }m=0 is much more complicated for the multi-index α of length n ≥ 3 than that for n = 2, especially when α has many zeros inside. Moreover, even if we assume ∞ that the sequence {bm }m=0 converges, the above relation does not give us any ∞ information about its limit. Indeed, if g denotes the limit of {bm }m=0 , then we only get 1 . g= P n αi g −1 i=1

In spite of this, the above-mentioned figures as well as vectors may suggest ∞ that the sequence {bm }m=0 converges, and it can be also guessed that the candidate for the limit is the number g = 1 + (α2 + · · · + αn ) + (α3 + · · · + αn ) + · · · + αn . Indeed, • for α = ( 31 , 13 , 13 ), see figure 4.6, we get   1 1 1 + 1+ + = 2; 3 3 3 • for α = ( 13 , 0, 23 ), see figure 4.7, we get   2 2 7 1+ 0+ + = = 2.(3); 3 3 3 • for α =

1 1 1 1 4, 4, 4, 4

• for α =

1 3 4 , 0, 0, 4


, see figure 4.8, we get     1 1 1 1 1 1 1+ + + + + + = 2.5; 4 4 4 4 4 4


, see figure 4.9, we get     3 3 3 1+ 0+0+ + 0+ + = 3.25. 4 4 4

On the Lipschitz constants for iterates of mean lipschitzian mappings

65

Hence, all the ”limit numbers” fit to our samples! Next observation follows from the examination of our mapping T defined as in example 4.3 (see [65]): Lemma 4.3 Let X = `1 and T as in example 4.3, with k = 1. Then, T is an isometry on the subspace S := span {ei : i ≥ 2}, with respect to the new norm k·kT defined for any x ∈ `1 by kxkT = kxk + (α2 + · · · + αn )kT xk + · · · + αn kT n−1 xk; that is, for any x ∈ S, we have kT xkT = kxkT . Proof. Since T is linear and bounded, we conclude that k · kT is a norm that is equivalent to the standard norm k·k in `1 . By the equality (4.4), with k = 1, n P for every x ∈ S, αi kT i xk = kxk, and i=1

  n n X X  kT xkT = αi  kT j xk = j=1

i=j

n X

! αi kT i xk

i=1

+

n X j=2

  n X  αi  kT j−1 xk i=j

  n n X X  =kxk + αi  kT j−1 xk = kxkT . j=2

i=j 

In view of lemma 4.3, for any j ≥ 0, we have ken+j+1 kT

= kT en+j+1 kT = · · · = kT

n+j

 

b

n+j , 0, 0, . . . en+j+1 kT =

b0

T

= bn+j . However, by definition of k·kT , we get ken+j+1 kT = 1+(α2 +· · ·+αn )

bn+j bn+j−1

+(α3 +· · ·+αn )

bn+j bn+j−2

+· · ·+αn

bn+j . bj+1

Hence, 1 + (α2 + · · · + αn )

bn+j bn+j bn+j + (α3 + · · · + αn ) + · · · + αn = bn+j . bn+j−1 bn+j−2 bj+1

Dividing both sides of the above equation by bn+j and defining the sequence dm =

1 , bm

66

Classification of Lipschitz mappings

we conclude that, for each j ≥ 0, dn+j + (α2 + · · · + αn )dn+j−1 + (α3 + · · · + αn )dn+j−2 + · · · + αn dj+1 = 1. Let us observe that, for d=

1 =P n g

j=1

1 P

n i=j

αi

,

we also have d + (α2 + · · · + αn )d + (α3 + · · · + αn )d + · · · + αn d = 1. Substracting both expressions and defining cm = dm − d, we get cn+j + (α2 + · · · + αn )cn+j−1 + (α3 + · · · + αn )cn+j−2 + · · · + αn cj+1 = 0, or, equivalently, cn+j = −(α2 + · · · + αn )cn+j−1 − (α3 + · · · + αn )cn+j−2 − · · · − αn cj+1 . Now, our claim is to prove that lim cm = 0. It is obvious for n = 1. m→∞ Suppose n ≥ 2. Once we have c1 , c2 , . . . , cn−1 , we can define cm for m ≥ n via the relation     cm cm−1  cm−2   cm−1         A  cm−3  =  cm−2  ,  ...   ...  cm−n+2 cm−n+1 where   −(α2 + · · · + αn ) −(α3 + · · · + αn ) −(α4 + · · · + αn ) . . . −αn  1 0 0 ... 0     0 1 0 ... 0  A= . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 ... 1 0 To show that {cm } tends to zero, it is enough to prove that an equivalent metric for which A is a contraction exists. Then, in view of the classical Banach’s Contraction Mapping Principle, the claim will be proved. For this, we need to show that 1 spr (A) = lim kAm k m < 1, m→∞

what exactly means that all eigenvalues of A have modules strictly less than 1. The characteristic polynomial of A is r(t) = (−1)n−1 (tn−1 + (α2 + · · · + αn )tn−2 + · · · + αn ).

On the Lipschitz constants for iterates of mean lipschitzian mappings

67

Hence, the eigenvalues of A are exactly the roots of the polynomial p(t) = tn−1 + (α2 + · · · + αn )tn−2 + · · · + αn .

(4.6)

We shall need several technical lemmas concerning geometry of zeros of polynomials. Lemma 4.4 Let g(t) = bn tn + · · · + b1 t + b0 . (1) If 0 < bn ≤ bn−1 ≤ · · · ≤ b1 ≤ b0 , then every root w of g satisfies |w| ≥ 1. (2) If bn ≥ bn−1 ≥ · · · ≥ b1 ≥ b0 > 0, then every root w of g satisfies |w| ≤ 1. Proof. (See [5, pp. 180–181]). To prove (1), suppose |w| < 1. Then, using triangle inequality, we get |(1 − w)g(w)| ≥b0 − |(b1 − b0 )w| − · · · − |(bn − bn−1 )wn | − |bn wn+1 | >b0 − |b1 − b0 | − · · · − |bn − bn−1 | − |bn | =b0 + (b1 − b0 ) + · · · + (bn − bn−1 ) − bn = 0. Consequently, every root w of g satisfies |w| ≥ 1. To prove (2), observe that w is a root of g if and only if

1 w

is a zero of

b0 tn + b1 tn−1 + · · · + bn−1 t + bn . 

Lemma 4.5 Let f (t) = an tn + an−1 tn−1 + · · · + a1 t + a0 , where an ≥ an−1 ≥ · · · ≥ a1 ≥ a0 > 0. Define the number   an−1 an−2 a1 a0 v := max , ,..., , . an an−1 a2 a1 Then, each root w of polynomial f satisfies |w| ≤ v. Proof. (See [5, p. 326]). Let g(t) = an v n tn + · · · + a1 vt + a0 . If w is a zero of f , then it is also a root of polynomial g( vt ); that is, g( wv ) = 0. Also, observe that by definition of v, an v n ≥ an−1 v n−1 ≥ · · · ≥ a1 v ≥ a0 > 0. Thus, in view of previous lemma, |w| v ≤ 1. 

Actually, in [5, p. 326], the conclusion of lemma 4.5 states that |w| < v. However, this is in general false. Consider the polynomial f (t) = t + 1 or f (t) = t3 + t2 + t + 1. Nevertheless, from lemma 4.5, we obtain the following Corollary 4.4 Let f (t) = an tn +an−1 tn−1 +· · ·+a1 t+a0 , where an > an−1 > · · · > a1 > a0 > 0. Then each root w of polynomial f satisfies |w| < 1.

68

Classification of Lipschitz mappings

Observe that if all αi are strictly positive, then coefficients of our polynomial p given by formula (4.6) satisfy the assumption of corollary 4.4; that is, 1 = α1 + α2 + · · · + αn > α2 + · · · + αn > · · · > αn > 0. Thus, all roots of p are strictly inside the unit complex disk and this implies ∞ that our “limit” g is actually the limit of the sequence {bm }m=0 . Nevertheless, it is only a partial solution of our problem. What if αi ≥ 0 for i = 2, . . . , n−1, ∞ n ≥ 3? Does {bm }m=0 converge for any multi-index α = (α1 , . . . , αn )? ∞ If the answer is affirmative, is g the limit of {bm }m=0 ? Our samples strongly suggest that the answer is YES! To prove it, we shall need one more result concerning roots of polynomial. Maybe the following lemma is known, but we were not able to find any references about it. Lemma 4.6 (P´ erez Garc´ıa and Piasecki [65]) Let n ≥ 2 and 1 = an > an−1 ≥ an−2 ≥ · · · ≥ a1 ≥ a0 > 0. If z is any root of the polynomial f (t) = an tn + an−1 tn−1 + · · · + a1 t + a0 ,

(4.7)

then |z| < 1. Proof. Let us define  v = max

a1 a0 an−1 an−2 , ,..., , an an−1 a2 a1

 .

To prove our claim, we will construct a polynomial q(t) = r(t)f (t) for which assumption of corollary 4.4 holds. Then, all roots of q, which contain roots of f , will be contained strictly inside the unit complex disk. Suppose that in our polynomial (4.7) we have aj = aj+1 for some j. We can select c such that 0 < c < 1 − an−1 < 1 and observe that the polynomial q1 (t) = (t + c)f (t) = an tn+1 + (an−1 + can )tn + · · · + (a0 + ca1 )t + ca0 has the following properties: 1. The degree of q1 equals the degree of f plus 1. 2. All coefficients of q1 are positive and nonincreasing. Indeed, 1 = an > an−1 + can = an−1 + c, and for j = 1, . . . , n − 1, we have aj + caj+1 ≥ aj−1 + caj . Also, a0 + ca1 > ca0 > 0. 3. Suppose that for some k we have aj−1 < aj = aj+1 = · · · = aj+k−1 = aj+k < aj+k+1 in our polynomial f . Then, we have the following inequalities for the coefficients of q1 : aj−1 + caj

< aj + caj+1 = aj+1 + caj+2 = · · · = aj+k−1 + caj+k < aj+k + caj+k+1 .

On the Lipschitz constants for iterates of mean lipschitzian mappings

69

If the polynomial q1 still has some equal coefficients, then we apply the same procedure to q1 to generate a polynomial q2 . After at most n−1 steps, we will obtain the desired polynomial q(t) = r(t)f (t) such that all its coefficients are strictly decreasing. 

The coefficients of our polynomial p are positive and nonincreasing. Moreover, 1 = α1 + α2 + · · · + αn > α2 + · · · + αn because α1 > 0. Finally, applying lemma 4.6 to the polynomial p, we finish the proof. Reassuming, all the above lead us to the following Theorem 4.4 (P´ erez Garc´ıa and Piasecki [65]) Let α = (α1 , . . . , αn ) be ∞ a multi-index. If {bm }m=0 is defined as   1 for m = 0,    1  P for m = 1, . . . , n, m −1 bm = j=1 αj bm−j   1  for m = n + 1, n + 2, . . . ,  n  P αi b−1 m−i i=1

then lim bm =

m→∞

n X j=1

  n X  αi  =

1 + α2 + 2α3 + 3α4 + · · · + (n − 1)αn

i=j

= α1 + 2α2 + 3α3 + · · · + nαn . Corollary 4.5 If T : M → M is an (α1 , . . . , αn )-nonexpansive mapping, then k∞ (T ) ≤ 1 + α2 + 2α3 + 3α4 + · · · + (n − 1)αn = α1 + 2α2 + 3α3 + · · · + nαn . Proof. It is a consequence of corollary 4.3 and theorem 4.4. 

To see that the estimate for k∞ (T ) is sharp, it is enough to consider the mapping T from example 4.3 with k = 1. Also, observe that for any n ≥ 2 and any (α1 , . . . , αn )-nonexpansive mapping T , k∞ (T ) is always less than the length of multi-index α; that is, k∞ (T ) < n.

4.3

Moving averages in Banach spaces

The study of mean nonexpansive mappings leads us to the moving average problem.

70

Classification of Lipschitz mappings

Let us recall that for any α-nonexpansive mapping T : M → M with ∞ α = (α1 , . . . , αn ) we proved that k(T m ) ≤ bm , where the sequence {bm }m=0 is defined as follows:   1 for m = 0,    1  P for m = 1, . . . , n, m −1 bm = j=1 αj bm−j    n 1 for m = n + 1, n + 2, . . . .    P αi b−1 m−i i=1

We also showed that this bound is sharp. We used the mapping T : `1 → `1 defined for x = (x1 , x2 , . . . ) ∈ `1 by   b1 b2 bj Tx = x2 , x3 , . . . , xj+1 , . . . . b0 b1 bj−1 Later, we studied the asymptotic behavior of the sequence of Lipschitz constants for mean nonexpansive mappings in term of the constant k∞ (T ). We defined dm = 1/bm to obtain a relation:  1 for m = 0,    m P  α d for m = 1, . . . , n, dm = j=1 j m−j  n P    αi dm−i for m = n + 1, n + 2, . . . . i=1

We proved one result concerning localization of roots of polynomials (lemma 4.6). We used this to show that the eigenvalues of a special matrix lie strictly inside the complex unit disk. Using also a special property of the mapping T , we finally obtained 1 P , lim dm = P n n m→∞ α i j=1 i=j and, consequently, lim bm =

m→∞

n X j=1

  n X  αi  . i=j

If we follow the tricks used in the proof of theorem 4.4, this result can be easily extended to the moving average problem: find the limit of the sequence ∞ {dm }m=0 , where d0 , . . . , dn−1 are arbitrary numbers and for m ≥ n dm = α1 dm−1 + · · · + αn dm−n . Theorem 4.5 (P´ erez Garc´ıa and Piasecki) Let α = (α1 , . . . , αn ) be a multi-index. Let d0 , . . . , dn−1 be arbitrary numbers and dm be defined for m ≥ n by n X dm = αi dm−i . i=1

On the Lipschitz constants for iterates of mean lipschitzian mappings

71

Then, lim dm = Pn

m→∞

i=1

n X

1 Pn

αj

j=i

  n X  αj  dn−i .

i=1

j=i

Proof. It is enough to note that the following relation holds under our ∞ assumption of the sequence {dm }m=0 : (α1 + · · · + αn )dn+j + (α2 + · · · + αn )dn+j−1 + · · · + αn dj+1 =(α1 + · · · + αn )d + (α2 + · · · + αn )d + · · · + αn d, where d = Pn

i=1

n X

1 Pn

j=i



n X



αj

i=1

 αj  dn−i .

j=i

Consequently, (dn+j − d) + (α2 + · · · + αn )(dn+j−1 − d) + · · · + αn (dj+1 − d) = 0. If we set cm = dm − d, we get that limm→∞ cm = 0 as in the proof of the theorem 4.4. 

Let us pass now to the general case of Banach spaces. Suppose X is a Banach space. Fix a multi-index α = (α1 , . . . , αn ). Let x0 , . . . , xn−1 be arbitrary elements of X and {xm } be defined for m ≥ n as xm = α1 xm−1 + α2 xm−2 + · · · + αn xm−n . Now, if we put ym = xm − x, with x = Pn

i=1

1 Pn

n X

j=i

αj



n X

 i=1

 αj  xn−i ,

j=i

∞

∞

we see that {ym }m=0 converges weakly to 0. Since {ym }m=0 ⊂ span {x0 , . . . , xn−1 }, we have that the convergence is strong. Summarizing the above observations, we get the following: Theorem 4.6 (P´ erez Garc´ıa and Piasecki) Let α = (α1 , . . . , αn ) be as above. Let x0 , . . . , xn−1 be arbitrary elements of a Banach space X and xm be defined for m ≥ n by n X xm = αi xm−i . i=1

Then, lim xm = Pn

m→∞

i=1

1 Pn

j=i

n X

αj

i=1

  n X  αj  xn−i . j=i

72

4.4

Classification of Lipschitz mappings

A bound for the constant k0 (T )

Let us recall that, for any α-lipschitzian mapping T : M → M with α = (α1 , α2 ) and k > 0, we have the following bound for the sequence of consecutive Lipschitz constants of the iterates of T : k(T n ) ≤ bn , where bn =

   1

for n = 0, for n = 1, for n = 2, 3, . . . .

k α1

  

k −1 α1 b−1 n−1 +α2 bn−2

∞

We have also proved that the following relation for {bn }n=0 holds: √ 2n+1 ∆ bn = k n  √ n+1  √ n+1 , α1 + ∆ − α1 − ∆ where ∆ = α12 + 4α2 k. Now we use the above facts to prove the following: Theorem 4.7 Let T : M → M be (α1 , α2 )-lipschitzian mapping for the constant k > 0. Then, 2 √ , k0 (T ) ≤ k α1 + ∆ with ∆ as above. Proof. One can notice that, for each n ≥ 1, 1

an ≤ (bn ) n ≤ cn , ∞

where {bn }n=0 is given as above,   n1 √ n+1 2 ∆ 2   an = k   √ n+1  = k · α + √∆ · 1 2 α1 + ∆

√

∆ √ α1 + ∆

! n1

and

cn

   n1 √   2n+1 ∆  √ n+1 for n = 2l, l = 0, 1, . . .  k· (α1 + ∆) = 1   √ n  n+1   for n = 2l + 1, l = 0, 1, . . .  k · 2α 2 α +√∆∆ n ) 1( 1   √  n1  2√ 2 √ ∆  k· · for n = 2l, l = 0, 1, . . . α1 + ∆ α + ∆ =  √1  n1  2√  k· · α∆ for n = 2l + 1, l = 0, 1, . . . . 1 α + ∆ 1

On the Lipschitz constants for iterates of mean lipschitzian mappings

73

Obviously, lim an = lim cn = k

n→∞

n→∞

2 √ . α1 + ∆

Hence, 1

lim (bn ) n = k

n→∞

2 √ . α1 + ∆ 

Example 4.5 Let T : `1 → `1 be defined as in example 4.2; that is,   b1 b2 bj T (x1 , x2 , . . . ) = x2 , x3 , . . . , xj+1 , . . . . b0 b1 bj−1 We proved that T is (α1 , α2 )-lipschitzian for the constant k > 0, and for n = 1, 2, . . . , we have k(T n ) = bn . In particular, the above implies that k0 (T ) = k

2 √ . α1 + ∆

(4.8)

Let us observe that, in this special case, we can give an explicit formula for an equivalent norm in `1 , denoted by k·kT , with respect to which T satisfies the √  Lipschitz condition with the constant 2k/ α1 + ∆ ; for x = (x1 , x2 , . . .) ∈ `1 , we put

kxkT =

∞ X i=1

√ i−1 √  2 ∆ α1 + ∆  √ i |xi | . √ i  α1 + ∆ − α1 − ∆

It is easy to observe that, for i ≥ 1, √  √ i−1 √ 2 ∆ α1 + ∆ 1 2 ∆ √ = ·   √  √ i  √ i α1 + ∆ 1 − α1 −√∆ i α1 + ∆ − α1 − ∆ α1 + ∆ √ 1 2 ∆ √ √ · ≥ = 1, α1 + ∆ 1 − α1 −√∆ α + ∆ 1

and

√  √ i−1 √ 2 ∆ α1 + ∆ 1 2 ∆ √ = ·   √  √ i  √ i α1 + ∆ 1 − α1 −√∆ i α1 + ∆ − α1 − ∆ α + ∆ 1

74

Classification of Lipschitz mappings √ 1 2 ∆ √ · ≤  √  α1 + ∆ 1 − α1 −√∆ 2 α1 + ∆ √ α1 + ∆ = . 2α1

Hence, k·kT is equivalent to k·k, and for any x ∈ `1 , √ α1 + ∆ kxk ≤ kxkT ≤ kxk . 2α1 Further, for every x ∈ `1 , we have

kT xkT

=

=

≤

=

√  √ i−1 2 ∆ α1 + ∆ 2k  √ i+1 |xi+1 | √ i+1  i=1 − α1 − ∆ α1 + ∆ √  √ i ∞ 2 ∆ α + ∆ X 1 2k √  i+1  √ √ i+1 |xi+1 | α1 + ∆ i=1 α + ∆ − α − ∆ 1 1 √  √ i−1 ∞ 2 ∆ α1 + ∆ X 2k √    |xi |  α1 + ∆ i=1 α + √∆ i − α − √∆ i 1 1

∞ X

2k √ kxkT , α1 + ∆

and by (1.10), kk·kT (T ) = k

2 √ . α1 + ∆

Now let α = (α1 , . . . , αn ) be a multi-index, k > 0 and {bm }∞ m=0 as in theorem 4.3; that is,   1 for m = 0,    k  P for m = 1, . . . , n, m −1 bm = j=1 αj bm−j   k  for m = n + 1, n + 2, . . . .  n  P αi b−1 m−i i=1

Suppose that M = `1 is the space of absolutely summable sequences with the standard norm k · k and T : `1 → `1 is a linear mapping defined for x = (x1 , x2 , . . . ) ∈ `1 by   b1 b2 bj Tx = x2 , x3 , . . . , xj+1 , . . . . b0 b1 bj−1

On the Lipschitz constants for iterates of mean lipschitzian mappings

75

In example 4.3, it was shown that T is α-lipschitzian with a constant k and k(T m ) = bm for every m ≥ 0; in other words, if we define kT m k = sup {kT m xk/kxk : x 6= 0} , 1/m

then kT m k = bm . In particular, this implies that limm→∞ (bm ) exists. In the same example, it was also proved that, for every x = (x1 , x2 , . . . ) ∈ `1 , n X

j

αj kT xk = k

j=1

∞ X

|xj |.

j=2

If we define Em = span {ei : i ≥ m + 1}, the last fact means that, for every x ∈ E1 , n X

αj kT j xk = kkxk.

(4.9)

j=1

Lemma 4.7 (P´ erez Garc´ıa and Piasecki, [66]) Let α = (α1 , . . . , αn ) be a multi-index, k > 0, {bm }∞ m=0 and T : `1 → `1 as above. Let k · kT be a norm on `1 defined for every x ∈ `1 by
kxkT = α1 + α2 g + · · · + αn g n−1 kxk
+ α2 + α3 g + · · · + αn g n−2 kT xk
+ α3 + α4 g + · · · + αn g n−3 kT 2 xk + . . . + αn kT n−1 xk, where g is the unique positive solution of the equation α1 g + α2 g 2 + · · · + αn g n = k. Then, k · kT is equivalent to k · k, and for every m = 1, 2, . . . , we have kk·kT (T m ) = g m . Proof. Since T is a bounded linear operator we conclude that k · kT is a norm equivalent to k · k. For any x ∈ `1 , we have
kT xkT = α1 + α2 g + · · · + αn g n−1 kT xk
+ α2 + α3 g + · · · + αn g n−2 kT 2 xk + . . . + αn kT n xk =α1 kT xk + α2 kT 2 xk + · · · + αn kT n xk + (α2 g + · · · + αn g n−1 )kT xk + (α3 g + · · · + αn g n−2 )kT 2 xk + . . . + αn gkT n−1 xk

76

Classification of Lipschitz mappings ≤kkxk + (α2 g + · · · + αn g n−1 )kT xk + (α3 g + · · · + αn g n−2 )kT 2 xk + . . . + αn gkT n−1 xk =(α1 g + α2 g 2 + · · · + αn g n )kxk + (α2 g + · · · + αn g n−1 )kT xk + (α3 g + · · · + αn g n−2 )kT 2 xk + . . . + αn gkT n−1 xk =gkxkT .

In view of (4.9), for every x ∈ E1 , kT xkT = gkxkT ; in general, for every x ∈ Em , kT m xkT = g m kxkT , hence for every m ≥ 1, kT m kT = g m , where kT m kT = sup {kT m xkT /kxkT : x 6= 0}. Consequently, kk·kT (T m ) = g m , as we desired. 

Theorem 4.8 (P´ erez Garc´ıa and Piasecki, [66]) For the sequence {bm } as above we have lim (bm )1/m = g, m→∞

where g is the unique positive solution of the equation α1 g + α2 g 2 + · · · + αn g n = k. Proof. Since k · k and k · kT are equivalent norms, we obtain 1/m

lim (bm )1/m = lim kT m k1/m = lim kT m kT

m→∞

m→∞

m→∞

= g. 

Finally, in view of theorem 4.3, we get following Corollary 4.6 Let (M, ρ) be a metric space and T : M → M an αlipschitzian mapping with α = (α1 , . . . , αn ) and k > 0. Then, k0 (T ) ≤ g, where g is given as above.

On the Lipschitz constants for iterates of mean lipschitzian mappings

77

In fact, as we can see in the proof of lemma 4.7, this estimation is sharp. Moreover, by following its proof, it can be easily observed that for any αlipschitzian mapping T : M → M with α = (α1 , . . . , αn ) and k > 0, we can define a metric dT by
dT (x, y) = α1 + α2 g + · · · + αn g n−1 ρ(x, y)
+ α2 + α3 g + · · · + αn g n−2 ρ(T x, T y)
+ α3 + α4 g + · · · + αn g n−3 ρ(T 2 x, T 2 y) + . . . + αn ρ(T n−1 x, T n−1 y), which is equivalent to the metric ρ and such that, for all x, y ∈ M , we have dT (T x, T y) ≤ gdT (x, y). Consequently, kdT (T ) ≤ g. Let us observe that only in the case of α-nonexpansive mapping, the above metric coincides with a metric given by formula (3.11).

4.5

More about k(T n ), k0 (T ), and k∞ (T )

Consider first the situation on the half-line [0, ∞): Lemma 4.8 Let α = (α1 , α2 ), k ∈ (0, 1], that is,    1 k bn = α1    −1 k −1 α1 bn−1 +α2 bn−2

∞

and {bn }n=0 be as in theorem 4.2; for for for

n = 0, n = 1, n = 2, 3, . . . .

∞

Let {xn }n=0 be defined as follows, ( xn =

1 b0

xn−1 +

1 bn

for for

n = 0, n = 1, 2, . . . .

Then, the function f : [0, ∞) → [0, ∞) given by  f (t) =

bn+1 bn

0, (t − xn+1 ) + xn ,

if t ∈ [0, x0 ] , if t ∈ [xn , xn+1 ] ,

n≥0

is (α1 , α2 )-lipschitzian for the constant k and k(f n ) = bn for n ≥ 1 (see figure 4.12).

78

Classification of Lipschitz mappings

f (t)

x2 1 b2

x1 1 b1

x0 1 b0

0

x0

x1

x2

x3

t

FIGURE 4.12: A graph of function f . Proof. It is clear that, for n ≥ 0, we have n X 1 xn = . b i=0 i

If k < 1, then lim

n→∞

√ n bn < 1. This implies that lim xn = ∞. If k = 1, n→∞

then also lim xn = ∞ because lim 1/bn = 1/ (1 + α2 ) > 0. Hence, for n→∞

n→∞

k ∈ (0, √1], the function f is defined on [0, ∞). It is not true for k > 1 because lim n bn > 1, and this implies that lim xn < ∞. n→∞

n→∞

The function f is well defined because, for each i ≥ 0 (we set x−1 = 0), we have   bi+1 bi+1 1 1 f (xi ) = (xi − xi+1 ) + xi = · − + xi = − + xi = xi−1 , bi bi bi+1 bi and f (xi+1 ) =

bi+1 (xi+1 − xi+1 ) + xi = xi . bi

On the Lipschitz constants for iterates of mean lipschitzian mappings

79

This in particular implies that f ([xi , xi+1 ]) = [xi−1 , xi ] for each i ≥ 0. We shall prove that function f is (α1 , α2 )-lipschitzian for the constant k. Let t ∈ [xn , xn+1 ] and s ∈ [xn+j , xn+j+1 ], s ≥ t, j ≥ 0, and n ≥ 1. Hence, there exist λ1 , λ2 ∈ [0, 1] such that t = xn +λ1 /bn+1 and s = xn+j +λ2 /bn+j+1 . Then,   1 1 bn+1 xn + λ1 − xn+1 + xn = (λ1 − 1) + xn , f (t) = bn bn+1 bn and, by analogy, f (s) = (λ2 − 1)

1 + xn+j . bn+j

Further, f 2 (t) =

bn bn−1

(f (t) − xn ) + xn−1 = (λ1 − 1)

and f 2 (s) = (λ2 − 1)

1 bn+j−1

1 bn−1

+ xn−1 ,

+ xn+j−1 .

Thus,
α1 (f (s) − f (t)) + α2 f 2 (s) − f 2 (t)     1 1 1 1 + (1 − λ1 ) α1 + α2 = (λ2 − 1) α1 + α2 bn+j bn+j−1 bn bn−1 +α1 xn+j + α2 xn+j−1 − α1 xn − α2 xn−1 . By definitions of {xn }n≥0 and {bn }n≥0 , we have α1 xn+j + α2 xn+j−1 = α1

n+j X i=0

= α1 +

1 bi

! + α2

n+j−1 X i=0

n+j X i=1

1 bi

1 1 α1 + α2 bi bi−1

= α1 +

n+j X i=1

= α1 + kxn+j + k

k

!



1 bi+1

1 1 1 −k −k bn+j+1 b0 b1

and −α1 xn − α2 xn−1 = −α1 − kxn − k

1 1 1 +k +k . bn+1 b0 b1

80

Classification of Lipschitz mappings

Consequently,
α1 (f (s) − f (t)) + α2 f 2 (s) − f 2 (t)   1 1 = k xn+j − xn + λ2 − λ1 bn+j+1 bn+1 = k (s − t) . If s, t ∈ [x0 , x1 ], s ≥ t, then f (s) = and f 2 (s) = f 2 (t) = 0. Hence,

b1 b0 (s − x1 ) + x0 ,

f (t) =

b1 b0 (t − x1 ) + x0 ,


b1 α1 (f (s) − f (t)) + α2 f 2 (s) − f 2 (t) = α1 (s − t) = k(s − t). b0 If s, t ∈ [0, x0 ], then f (s) = f (t) = f 2 (s) = f 2 (t) = 0. If t ∈ [0, x0 ] and s ∈ [x0 , x1 ], then f (t) = f 2 (t) = f 2 (s) = 0, and we obtain
α1 (f (s) − f (t)) + α2 f 2 (s) − f 2 (t)   b1 (s − x1 ) + x0 = α1 b0    α1  k s − x0 − +1 = α1 α1 k = k (s − x0 ) ≤ k(s − t). If t ∈ [0, x0 ] and s ∈ [xn , xn+1 ], n ≥ 1, then
α1 (f (s) − f (t)) + α2 f 2 (s) − f 2 (t)     bn+1 bn+1 = α1 (s − xn+1 ) + xn + α2 (s − xn+1 ) + xn−1 bn bn−1   bn+1 bn+1 = (s − xn+1 ) α1 + α2 + α1 xn + α2 xn−1 bn bn−1 = k (s − xn+1 ) + α1

= k (s − xn+1 ) +

n  X i=1

= ks − k

n+1 X i=0

n−1 n X X 1 1 + α2 b b i=0 i i=0 i

1 1 α1 + α2 bi bi−1

 + α1

n X 1 1 +k + α1 bi b i+1 i=1

= ks − k

1 1 − k + α1 b0 b1

1 b0

On the Lipschitz constants for iterates of mean lipschitzian mappings

81

= k(s − 1) ≤ k(s − t). If t ∈ [x0 , x1 ], s ∈ [xn , xn+1 ], and n ≥ 1, then f 2 (t) = 0, and
α1 (f (s) − f (t)) + α2 f 2 (s) − f 2 (t)   b1 bn+1 (s − xn+1 ) + xn − (t − x1 ) − x0 = α1 bn b0   bn+1 +α2 (s − xn+1 ) + xn−1 bn−1   bn+1 bn+1 = (s − xn+1 ) α1 + α2 + α1 xn + α2 xn−1 − kt + k bn bn−1 = k (s − xn+1 ) + α1

= ks − k

n+1 X i=0

n−1 n X 1 X 1 + α2 − kt + k b b i=0 i i=0 i

 n  X 1 1 1 + α1 + α1 + α2 − kt + k bi bi bi−1 i=1

= ks − k

1 1 − k − kt + k + α1 b0 b1 = k(s − t).

Thus, f is (α1 , α2 )-lipschitzian for the constant k. For each n ≥ 0, we have f (xn ) = xn−1 , and f n (xn ) − f n (xn−1 ) = x0 − x−1 = 1 = bn ·

1 = bn (xn − xn−1 ) . bn

This implies that k(f n ) ≥ bn . Finally, by theorem 4.2, k(f n ) = bn . 

Since the half-line [0, ∞) can be isometrically embedded into every normed space X 6= {0}, we get the following theorem: Theorem 4.9 Let C be a subset of a normed space (X, k·k). If C contains a half-line, then for every multi-index α = (α1 , α2 ) and k ∈ (0, 1], there exists a mapping T : C → C, which is (α1 , α2 )-lipschitzian for the constant k and k(T n ) = bn for n = 1, 2, . . . . Proof. Let u ∈ C and v ∈ SX be such that {u + tv : t ≥ 0} ⊂ C. For a multi-index α = (α1 , α2 ) and k ∈ (0, 1], we define a mapping T : C → C by putting for each x ∈ C T x = u + f (kx − uk)v,

82

Classification of Lipschitz mappings

where f is the function as in lemma 4.8. We claim that the mapping T is (α1 , α2 )-lipschitzian for the constant k and k(T n ) = bn for n = 1, 2, . . . . Indeed, since f is (α1 , α2 )-lipschitzian for the constant k, this implies that

α1 kT x − T yk + α2 T 2 x − T 2 y = α1 |f (kx − uk) − f (ky − uk)| + α2 |f (kf (kx − uk)vk) − f (kf (ky − uk)vk)| = α1 |f (kx − uk) − f (ky − uk)| + α2 f 2 (kx − uk) − f 2 (ky − uk) ≤ k |kx − uk − ky − uk| ≤ k kx − yk for all x, y ∈ C. Now, it is enough to notice that, for x = u + xn v and y = u + xn−1 v, n ≥ 1, with {xn }n≥0 as in lemma 4.8, we obtain kT n x − T n yk = f n (xn ) − f n (xn−1 ) = bn (xn − xn−1 ) = bn kx − yk . Thus, k(T n ) ≥ bn for n = 1, 2, . . . . Finally, by theorem 4.2, we get the conclusion. 

Corollary 4.7 Let C be a subset of a normed space (X, k·k). If C contains a half-line, then for each multi-index α = (α1 , α2 ) and k ∈ (0, 1], there exists a mapping T : C → √ C, which is (α1 , α2 )-lipschitzian for the constant k and k0 (T ) = 2k/(α1 + ∆), where ∆ is as in the theorem 4.7. Corollary 4.8 Let C be a subset of a normed space (X, k·k). If C contains a half-line, then for each multi-index α = (α1 , α2 ), there exists a mapping T : C → C, which is (α1 , α2 )-nonexpansive with k∞ (T ) = 1 + α2 . Consider now any subset C of X that contains a nontrivial segment; that is, there exist two points u, v ∈ C such that [u, v] := {(1 − t)u + tv : t ∈ [0, 1]} ⊂ C. Let f be the function as in lemma 4.8 with k = 1. Define the function g : [0, ∞) → [0, ∞) by  t if t ∈ [0, 1] , g(t) = f (t − 1) + 1, if t ≥ 1. In view of lemma 4.8, we conclude that g is (α1 , α2 )-nonexpansive and k(g n ) =

1 + α2 for n ≥ 1. 1 + (−1)n α2n+1

On the Lipschitz constants for iterates of mean lipschitzian mappings

83

In the next step, we use the function g to construct a family of functions gN : [0, ∞) → [0, ∞), N ≥ 1, as follows (see figure 4.13):  g(t) if t ∈ [0, xN + 1], gN (t) = xN −1 + 1 if t ∈ [xN + 1, ∞). It is easy to check that, for every N ≥ 1, gN is (α1 , α2 )-nonexpansive. Indeed, if s, t ∈ [0, xN + 1], then it follows from the fact that function g is (α1 , α2 )-nonexpansive. If s ≥ xN + 1 and t ≥ xN + 1, then it is enough to notice that gN (s) = gN (t) = xN −1 + 1. If s ≥ xN + 1 and t ∈ [0, xN + 1], then
2 2 α1 (gN (s) − gN (t)) + α2 gN (s) − gN (t)
= α1 (g(xN + 1) − g(t)) + α2 g 2 (xN + 1) − g 2 (t) ≤ xN + 1 − t ≤ s − t. Moreover, n k(gN ) = k(g n ) =

1 + α2 1 + (−1)n α2n+1

for n = 1, . . . , N and n k(gN )=1

for n ≥ N + 1. Now we are ready to prove the following: Theorem 4.10 Let C be a subset of a normed space (X, k · k). If C contains a nontrivial segment, then for each multi-index α = (α1 , α2 ) and N ≥ 1, there exists an (α1 , α2 )-nonexpansive mapping T : C → C, with k(T n ) =

1 + α2 1 + (−1)n α2n+1

for n = 1, . . . , N and k(T n ) = 1 for n ≥ N + 1. Proof. Let u, v ∈ C be such that [u, v] ⊂ C. Let α = (α1 , α2 ), N ≥ 1 and gN be as above. The reader can easily verify that the mapping T : C → C defined by   kx − uk 1 gN (1 + xN ) · (v − u) Tx = u + 1 + xN kv − uk satisfies our claim. 

84

Classification of Lipschitz mappings

g3 (t)

x2 + 1 1 b2

x1 + 1 1 b1

x0 + 1 1 b0

1

0

1

x0 + 1 x1 + 1

x2 + 1

x3 + 1

t

FIGURE 4.13: A graph of function g3 . In the case of multi-index α = (α1 , . . . , αn ), it is enough to use the sequence {bm } defined as in theorem 4.3. We leave to the reader verification of the following theorem. Theorem 4.11 Let C be a subset of a normed space (X, k · k). If C contains a nontrivial segment, then for each multi-index α = (α1 , . . . , αn ) and N ≥ 1, there exists an α-nonexpansive mapping T : C → C, with k(T m ) = bm for m = 1, . . . , N and k(T m ) = 1 for m ≥ N + 1. Consequently, if C is any bounded, closed and convex subset of a Banach space (X, k · k), with diam(C) > 0, then, for any multi-index α = (α1 , . . . , αn ), the class of α-nonexpansive mappings is strictly wider than the class of nonexpansive mappings!

Chapter 5 Subclasses determined by p-averages

Now, we deal with a condition that instead of arithmetical mean in the definition of α-lipschitzian mapping, uses average of order p ≥ 1.

5.1

Basic definitions and observations

Let (M, ρ) be a metric space, α = (α1 , . . . , αn ) be a multi-index and p ≥ 1. Definition 5.1 A mapping T : M → M is said to be (α, p)-lipschitzian for the constant k ≥ 0 (α and p as above) if, for every x, y ∈ M , we have ! p1 n X i i p αi ρ(T x, T y) ≤ kρ(x, y). (5.1) i=1

In fact, the above condition was suggested by Goebel and Jap´on Pineda in [33]. If (5.1) is satisfied with p = 1, then we get a definition of α-lipschitzian mappings. For a given α, p, and k, we will denote the class of all (α, p)lipschitzian mappings T : M → M for the constant k by L(α, p, k) and if p = 1, then we usually write L(α, k) instead of L(α, 1, k). The smallest constant k for which (5.1) holds is denoted by k(α, p, T ). If (5.1) is satisfied with k = 1, then we say that T is (α, p)-nonexpansive, and if (5.1) is satisfied with k < 1, then T is called (α, p)-contraction. Let us list some basic properties of (α, p)-lipschitzian mappings. • Each (α, p)-lipschitzian mapping T is also lipschitzian and k(T ) ≤

k(α, p, T ) 1/p

.

α1 • For i = 1, . . . , n, we have "

k(T i ) ≤ min k(T )i ,

k(α, p, T ) 1/p

# ,

αi provided αi > 0.

85

86

Classification of Lipschitz mappings • If T is lipschitzian, then, for any multi-index α = (α1 , . . . , αn ) and p ≥ 1, the mapping T is (α, p)-lipschitzian with mean Lipschitz constant k(α, p, T ) ≤

n X

! p1 αi k(T i )p

.

i=1

• Each class L(α, p, k) contains all the lipschitzian mappings T such that n X

! p1 i p

≤ k.

αi k(T )

i=1

In particular, using the fact that k(T i ) ≤ k(T )i , each class L(α, p, k) contains all lipschitzian mappings T such that n X

! p1 ip

αi k(T )

≤ k.

(5.2)

i=1

 • If T is uniformly lipschitzian with sup k(T i ) : i = 1, 2, . . . ≤ k, then for any α and p ≥ 1, T is (α, p)-lipschitzian with k(α, p, T ) ≤ k. • Let Lu (k) denote the class of all uniformly k-lipschitzian mappings. By I, we denote the set of all multi-indexes α; that is, ) ( n X αi = 1, n ≥ 1 . I = (α1 , . . . , αn ) : α1 > 0, αn > 0, αi ≥ 0, i=1

One can notice that, for any p ≥ 1, we have \ Lu (k) = L(α, p, k).

(5.3)

α∈I

T Indeed, let us assume that T ∈ α∈I L(α, p, k) and T ∈ / Lu (k). Let m be the smallest natural number for which there exist points u, v ∈ M , u 6= v, such that ρ(T m u, T m v) > kρ(u, v) or equivalently ρ(T m u, T m v)p > k p ρ(u, v)p

(5.4)

for every p ≥ 1. If m = 1, then for the multi-index α = (1) of length n = 1, T is not (α, p)-lipschitzian for the constant k. If m = 2, then we set α = (α1 , α2 ) with α1 =

ρ(T 2 u, T 2 v)p − k p ρ(u, v)p 2ρ(T 2 u, T 2 v)p

Subclasses determined by p-averages and α2 =

87

ρ(T 2 u, T 2 v)p + k p ρ(u, v)p . 2ρ(T 2 u, T 2 v)p

Obviously, α is a multi-index of length n = 2, and in view of (5.4), we get α1 ρ(T u, T v)p +α2 ρ(T 2 u, T 2 v)p >

ρ(T 2 u, T 2 v)p + k p ρ(u, v)p > k p ρ(u, v)p . 2

In general, if m ≥ 3, then we define a multi-index α = (α1 , . . . , αm ) of length n = m by  ρ(T m u,T m v)p −kp ρ(u,v)p  for i = 1,  2ρ(T m u,T m v)p 0 for i = 2, . . . , m − 1, αi = p   ρ(T m u,T m v)p +kp ρ(u,v) for i = m. 2ρ(T m u,T m v)p Further, using (5.4), we get m X

αi ρ(T i u, T i v)p

=

i=1

ρ(T m u, T m v)p − k p ρ(u, v)p ρ(T u, T v)p 2ρ(T m u, T m v)p ρ(T m u, T m v)p + k p ρ(u, v)p ρ(T m u, T m v)p 2ρ(T m u, T m v)p ρ(T m u, T m v)p + k p ρ(u, v)p 2 k p ρ(u, v)p .

+ > >

All the above cases contradict with the assumption that, for every α, T is (α, p)-lipschitzian for the constant k. Thus, T is uniformly k-lipschitzian. Inclusion in opposite direction follows from previous items. • It is clear that for a given p and q, 1 ≤ q < p, every (α, p)-lipschitzian mapping is (α, q)-lipschitzian for the same constant k. Indeed, if T ∈ L(α, p, k), then, using the H¨older inequality, we obtain n X

! q1 αi ρ(T i x, T i y)q

n X

=

i=1

p−q p

αi

! q1

q p

αi ρ(T i x, T i y)q

i=1

 ≤ 

n X

! p−q p αi

n X

i=1

=

n X

! pq  q1  αi ρ(T i x, T i y)p

i=1

! p1 i

i

p

αi ρ(T x, T y)

i=1

≤ kρ(x, y).

88

Classification of Lipschitz mappings This means that L(α, p, k) ⊂ L(α, q, k) for p > q. In particular, for every p > 1, L(α, p, k) ⊂ L(α, k). Inclusion in opposite direction does not hold. To illustrate this, let us consider the following example. As a metric space (M, ρ), we take `1 with a metric inherited from the standard norm. Let T : `1 → `1 be a linear mapping given by   2 T x = T (x1 , x2 , . . . ) = 2x2 , x3 , x4 , x5 , . . . . 3 Then, 2



2

T x = T (x1 , x2 , . . . ) =

2 4 x3 , x4 , x5 , x6 , . . . 3 3



and ∞ X

1 1 5 |xi | ≤ kxk . kT xk + T 2 x = |x2 | + |x3 | + |x4 | + 2 2 6 i=5

Thus, T is (1/2, 1/2)-nonexpansive; that is, T ∈ L ((1/2, 1/2) , 1). On the other hand, for every p > 1, T is not ((1/2, 1/2) , p)-nonexpansive. Indeed, for x = e3 = (0, 0, 1, 0, . . . , 0, . . . ) ∈ `1 , we get   2 T e3 = 0, , 0, 0, . . . 3 and 2



T e3 =

4 , 0, 0, . . . 3

 .

Then,

p 1 1 1 p kT e3 k + T 2 e3 = · 2 2 2

 p  p 2 4 1 p + · > 1 = ke3 k . 3 2 3

• For each multi-index α = (α1 , . . . , αn ), we have \ L(1) = L(α, p, 1). p≥1

The case with n = 1 is trivial. Let n ≥ 2. Obviously, each class L(α, p, 1) contains all nonexpansive mappings. On the other hand, let us assume that a mapping T : M → M is in L(α, p, 1) for each p ≥ 1, and there exist points u, v ∈ M , u 6= v, such that ρ(T u, T v) > ρ(u, v).

Subclasses determined by p-averages

89

For such u, v ∈ M , we take p such that   ln(α1 ) p > max 1, . ln(ρ(u, v)) − ln(ρ(T u, T v)) Then, n X

! p1 i

i

p

αi ρ(T u, T v)

1/p

≥ α1 ρ(T u, T v)

i=1 ln(ρ(u,v)/ρ(T u,T v)) ln(α1 ) > α1 = ρ(u, v).

ρ(T u, T v)

By contradiction, we get that T must be nonexpansive.

5.2

A bound for k(T n ), k∞ (T ), and k0 (T )

The condition (5.1) determines subclasses L(α, p, k) of class L(α, k). It is natural to expect that, for every such subclass, we can obtain better bounds for k(T n ), k0 (T ), and k∞ (T ). Indeed, all results from the previous chapter can be extended easily for subclasses L(α, p, k), and proofs carry on with only minor technical changes. We begin by listing results devoted to the simple case of multi-index α of length n = 2: Theorem 5.1 Let T : M → M be ((α1 , α2 ), p)-lipschitzian mapping for the constant k > 0. Then, for n ≥ 0, we have k(T n ) ≤ b1/p n , where b0 = 1, b1 =

kp kp , and bn+2 = . −1 α1 α1 bn+1 + α2 b−1 n

Proof. It is enough to repeat the proof of theorem 4.2 by putting ρ(x, y)p , ρ(T i x, T i y)p , and k p instead of ρ(x, y), ρ(T i x, T i y), and k, respectively.  ∞ {bn }n=0

be as above. Then, for n ≥ 0, we have √ 2n+1 ∆ pn bn = k  √ n+1  √ n+1 , α1 + ∆ − α1 − ∆

Lemma 5.1 Let

with ∆ = α12 + 4α2 k p .

90

Classification of Lipschitz mappings

Proof. It is enough to put k p instead of k in the proof of lemma 4.1. 

Theorem 5.2 Let T : M → M be ((α1 , α2 ), p)-lipschitzian mapping for the constant k > 0. Then, for n ≥ 0, we have  n+1

 k(T n ) ≤ k n  

1/p

√

∆ 2  √ n+1  √ n+1  α1 + ∆ − α1 − ∆

,

where ∆ = α12 + 4α2 k p . Proof. It is a consequence of theorem 5.1 and lemma 5.1. 

Corollary 5.1 Let T : M → M be ((α1 , α2 ), p)-lipschitzian mapping for the constant k > 0. Then,  k0 (T ) ≤ k

2 √ α1 + ∆

1/p ,

with ∆ = α12 + 4α2 k p . Corollary 5.2 Let T : M → M be ((α1 , α2 ), p)-nonexpansive. Then, for n ≥ 0, we have  1/p 1 + α2 n k(T ) ≤ (5.5) 1 + (−1)n α2n+1 and 1/p

k∞ (T ) ≤ (1 + α2 )

.

(5.6)

Let us pass now to the general case when the length of multi-index α is equal to n. Then, evaluations presented in theorems 4.3, 4.4, and 4.8, and in corollary 4.6 can be easily extended to the case of (α, p)-lipschitzian mappings. Indeed, it is enough to observe that the condition (5.1) is equivalent to the following: n X

αi ρ(T i x, T i y)p ≤ k p ρ(x, y)p ,

i=1

and then repeat the proofs of the above-mentioned theorems by putting ρ(x, y)p , ρ(T i x, T i y)p , and k p instead of ρ(x, y), ρ(T i x, T i y), and k, respectively. Consequently, we get:

Subclasses determined by p-averages

91

Theorem 5.3 Let T : M → M be ((α1 , . . . , αn ), p)-lipschitzian for the constant k > 0. Then, for n ≥ 0, we have k(T m ) ≤ b1/p m , where {bm }m≥0 is defined as follows:   1 for    kp  P for m −1 bm = j=1 αj bm−j   kp  for  n  P αi b−1 m−i

m = 0, m = 1, . . . , n, m = n + 1, n + 2, . . . .

i=1

Theorem 5.4 If T : M → M is an ((α1 , . . . , αn ), p)-nonexpansive and {bm }m≥0 is as in theorem 5.3 with k = 1, then   k∞ (T ) = lim sup k(T m ) ≤ lim b1/p m = m→∞

m→∞

n X

 1/p n X  αi 

j=1

i=j 1/p

= (1 + α2 + 2α3 + 3α4 + · · · + (n − 1)αn ) 1/p

= (α1 + 2α2 + 3α3 + · · · + nαn )

.

Let us observe that the term on the right-hand side of the above inequality is always less than n1/p . Theorem 5.5 If T : M → M is an ((α1 , . . . , αn ), p)-lipschitzian for the con∞ stant k and {bm }m=0 is as in theorem 5.3, then  1/m  1/p 1/m k0 (T ) = lim [k(T m )] ≤ lim b1/p = lim b1/m = g 1/p , m m m→∞

m→∞

m→∞

where g is the unique positive solution of the equation α1 g + α2 g 2 + · · · + αn g n = k p . Moreover, an equivalent metric dT for which any (α, p)-lipschitzian mapping T : M → M with α = (α1 , . . . , αn ), p ≥ 1, and k > 0 satisfies the Lipschitz condition with a constant g 1/p can be given by dT (x, y) =


α1 + α2 g + · · · + αn g n−1 ρ(x, y)p
+ α2 + α3 g + · · · + αn g n−2 ρ(T x, T y)p
+ α3 + α4 g + · · · + αn g n−3 ρ(T 2 x, T 2 y)p + . . . 1/p + αn ρ(T n−1 x, T n−1 y)p .



To show that all the above estimates are sharp, we present the following example.

92

Classification of Lipschitz mappings

Example 5.1 For each p ≥ 1, let us consider the space M = `p with a metric P∞ p 1/p inherited from the standard norm, kxk = k(x1 , x2 , . . . )k = ( i=1 |xi | ) . For arbitrary multi-index α = (α1 , . . . , αn ) and k > 0, let {bm }m≥0 be as in theorem 5.3. Let T : `p → `p be a linear operator defined for each x = (x1 , x2 , . . . ) ∈ `p by ! 1/p 1/p 1/p bj b1 b2 Tx = x , x , . . . , 1/p xj+1 , . . . . 1/p 2 1/p 3 b0 b1 bj−1 The reader can verify that T is ((α1 , . . . , αn ) , p)-lipschitzian with the mean 1/p Lipschitz constant k (α, p, T ) = k. Moreover, k (T m ) = bm for each m ≥ 0 and k0 (T ) = g 1/p , where g is as in theorem 5.5. If k = 1, then T is ((α1 , . . . , αn ), p)-nonexpansive with k0 (T ) = 1 and   k∞ (T ) = lim sup k(T m ) = lim b1/p m =

n X

m→∞

m→∞

j=1

 1/p n X  αi  i=j 1/p

= (1 + α2 + 2α3 + 3α4 + · · · + (n − 1)αn ) 1/p

= (α1 + 2α2 + 3α3 + ... + nαn )

.

The above mapping can be modified to the ((α1 , . . . , αn ), p)-nonexpansive mapping T : B`p → B`p , with k∞ (T ) = (1 + α2 + 2α3 + · · · + (n − 1)αn )

1/p

.

For clarity and simplicity of arguments, we present it in the case of multi-index α of length n = 2. Example 5.2 For our construction, we shall need a cut function τ : [0, 1] → [0, 1] given by (see figure 5.1).

τ (t) =

αn 1 2 α1 t − α1 (2−α2 )n−1 αn − α11 t + α (2−α2 )n−1  1 2 

  

0

if

αn−1 αn 2 ≤ t ≤ (2−α2 )n−1 , n = 1, 2, . . . , (2−α2 )n−1 2 αn αn 2 2 , n = 1, 2, . . . , n ≤ t ≤ n−1 (2−α2 ) (2−α2 )

if if t = 0.

Obviously, for all s, t ∈ [0, 1], we have |τ (s) − τ (t)| ≤

1 |s − t| α1

and |τ (t)| = τ (t) ≤ |t| = t.

Subclasses determined by p-averages

93

Fix p ≥ 1. Then, for any α = (α1 , α2 ), we define a mapping T : B`p → B`p by !! 1/p  1 − (−α2 )i |xi+1 | . T (xi )i≥1 = τ α1 1 − (−α2 )i+1 i≥1

The reader can verify that k(α, p, T ) = 1 and k(T n ) =



1 + α2 1 + (−1)n α2n+1

1/p

for each n ≥ 1. Consequently, 1/p

k∞ (T ) = (1 + α2 )

.

τ (t)

1

1 3

1 9

0

1 1 1 18 9 6

1 3

1 2

t

1

FIGURE 5.1: A graph of function τ for α =

1 1 2, 2


.

94

5.3

Classification of Lipschitz mappings

On the moving p-averages

The result from theorem 4.5 can be easily extended to the moving paverages. Fix a multi-index α = (α1 , . . . , αn ) and p > 0. Let d0 , . . . , dn−1 be arbitrary nonnegative numbers and dm be defined for m ≥ n by dm =

n X

!1/p αi dpm−i

i=1

or, equivalently, dpm =

n X

αi dpm−i .

i=1

Then,  lim dm =  Pn

m→∞

i=1

1 Pn

j=i

n X

αj



n X

 i=1



1/p

αj  dpn−i 

.

j=i

To see this, it is enough to put bpm instead of bm and dpm instead of dm in the proof of the classical moving average problem presented in the previous chapter.

Chapter 6 Mean contractions

Let us pass now to the class of mean contractions. We have already seen that every such mapping is a classical contraction with respect to some equivalent metric. Actually, as we shall see later, every mean contraction is characterized by the constant k0 (T ) as well as k∞ (T ). Moreover, the mean Lipschitz condition provides nice estimates on the rate of convergence of iterates.

6.1

Classical Banach’s contractions

To present the just mentioned results, we begin by recalling the Banach’s Contraction Mapping Principle and some of its useful application variants. Let (M, ρ) be a metric space. Let us recall that a mapping T : M → M is said to be a contraction if there exists a constant k < 1 such that, for all x, y ∈ M , we have ρ(T x, T y) ≤ kρ(x, y). Under the above setting the Banach’s Contraction Mapping Principle states: Theorem 6.1 Let (M, ρ) be a complete metric space and let T : M → M be a k-contraction. Then, T has a unique fixed point z. Moreover, for each x ∈ M, lim T n x = z n→∞

and, in terms of error estimates, • ρ(T n x, z) ≤ • ρ(T n x, z) ≤

1 n n+1 x) 1−k ρ(T x, T n

k 1−k ρ(x, T x)

for each x ∈ M and n ≥ 1;

for each x ∈ M and n ≥ 1.

Proof. Since T is a contraction, we conclude that, for each x ∈ M ,
−1 ρ(x, T x) ≤ (1 − k) ρ(x, T x) − ρ(T x, T 2 x) .

95

96

Classification of Lipschitz mappings

Therefore, for any m, n ∈ N with n < m, we have ρ(T n x, T m+1 x) ≤

m X

ρ(T i x, T i+1 x)

i=n −1

≤ (1 − k)

m X


ρ(T i x, T i+1 x) − ρ(T i+1 x, T i+2 x)

i=n −1

= (1 − k)


ρ(T n x, T n+1 x) − ρ(T m+1 x, T m+2 x) .

In particular, for n = 1 and letting m → ∞, we obtain ∞ X

ρ(T i x, T i+1 x) ≤ (1 − k)

−1

ρ(T x, T 2 x) < ∞.

i=1 ∞

This implies that {T n x}n=1 is a Cauchy sequence. Since M is complete, there exists a point z ∈ M such that lim T n x = z.

n→∞

By continuity of T , we get z = lim T n x = lim T n+1 x = lim T (T n x) = T z. n→∞

n→∞

n→∞

Thus, z ∈ Fix(T ). Letting m → ∞ in the above proved inequality, −1

ρ(T n x, T m+1 x) ≤ (1 − k)


ρ(T n x, T n+1 x) − ρ(T m+1 x, T m+2 x) ,

we obtain

−1

ρ(T n x, z) ≤ (1 − k)

ρ(T n x, T n+1 x),

and, consequently,

kn ρ(x, T x). 1−k To show that z is the unique fixed point, let us assume that T y = y for some y ∈ M . Then, ρ(T n x, z) ≤

ρ(y, z) = ρ(T y, T z) ≤ kρ(y, z). Since k < 1, we conclude that y = z. 

The simple example T x = kx for x ∈ R shows that the above estimates are sharp. The Banach’s theorem fails to hold when k(T ) = 1. To show this, it is enough to consider a mapping T : R → R defined by T x = x + a, a 6= 0. It is known that the assumption k < 1 can be replaced by (see [36], pp. 10–11) ∞ X k(T i ) < ∞. (6.1) i=1

Mean contractions

97

Indeed, for any x ∈ M and n, p ∈ N, we have n

ρ(T x, T

n+p

x) ≤

p−1 X

ρ(T

n+i

x, T

n+i+1

x) ≤

i=0

n+p−1 X

! i

k(T ) ρ(x, T x).

i=n

Using (6.1) we conclude that {T n x} is again a Cauchy sequence. Since M is complete and T is lipschitzian, {T n x} converges to the fixed point z of T . Thus, letting p to go to infinity, we obtain for any fixed n ! ∞ X ρ(T n x, z) ≤ k(T i ) ρ(x, T x). i=n

In view of (6.1), there is a number m such that k(T m ) < 1. Thus, by the Banach’s theorem, T m : M → M has exactly one fixed point. This implies that z is the unique fixed point of T . We can characterize lipschitzian mappings that satisfy (6.1). Let us recall that, for the constant k0 (T ), we have (see (1.5) and (1.10)): np o p k0 (T ) = lim n k(T n ) = inf n k(T n ) : n = 1, 2, . . . = inf kr (T ), (6.2) n→∞

where infimum is taken over all metrics r equivalent to ρ. Thus, the series (6.1) converges if and only if k0 (T ) < 1. We summarize the above observation in the following: Theorem 6.2 Let (M, ρ) be a complete metric space. Suppose that T : M → M is lipschitzian with k0 (T ) < 1. Then, T has a unique fixed point z, and for each x ∈ M , lim T n x = z. n→∞

Even if kρ (T ) < 1, it may happen that k0 (T ) < kρ (T ). Then, we obtain better estimate on the rate of convergence as seen in the following: Theorem 6.3 Let (M, ρ) be a complete metric space and T : M → M be a lipschitzian mapping with k0 (T ) < 1. Then, T has a unique fixed point z and for each x ∈ M and  > 0, ρ(T n x, z) = o((k0 (T ) + )n ). Proof. Let {dλ : λ ∈ (0, 1/k0 (T ))} be a family of equivalent metrics defined by (1.7). For fixed  > 0, we take δ > 0 such that δ < min {, 1 − k0 (T )}. Then, using (1.9), for λ = 1/ (k0 (T ) + δ), we have kdλ (T ) ≤ k0 (T ) + δ < 1 . Thus, by the Banach’s Contraction Mapping Principle, there exists the unique point z such that z = T z, and for each x ∈ M , we have lim dλ (T n x, z) = 0.

n→∞

Since metric dλ is equivalent to ρ, we obtain lim ρ(T n x, z) = 0.

n→∞

98

Classification of Lipschitz mappings

For fixed x ∈ M and n, p ∈ N, using (1.8) and (1.9), we have ρ(T n x, T n+p x) ≤ dλ (T n x, T n+p x) ≤

n+p−1 X

dλ (T i x, T i+1 x)

i=n

≤

"n+p−1 X

# i

(k0 (T ) + δ)

dλ (x, T x)

i=n n

(k0 (T ) + δ) dλ (x, T x) 1 − k0 (T ) − δ "∞  i # n X (k0 (T ) + δ) 1 ρ(x, T x). ≤ kρ (T i ) 1 − k0 (T ) − δ i=0 k0 (T ) + δ

≤

Letting p → ∞, we obtain "∞  i # n X (k0 (T ) + δ) 1 i ρ(T x, z) ≤ kρ (T ) ρ(x, T x) 1 − k0 (T ) − δ i=0 k0 (T ) + δ n

= o((k0 (T ) + )n ). 

An application of the above presented variants of the Banach’s Contraction Mapping Principle in the theory of differential equations can be found in [36].

6.2

On characterizations of contractions

Let us recall that T : M → M is called mean contraction if there exist a multi-index α = (α1 , . . . , αn ) and a constant k < 1 such that, for all x, y ∈ M , we have n X αi ρ(T i x, T i y) ≤ kρ(x, y). i=1

The following theorem characterizes the class of all mean contractions: Theorem 6.4 Let (M, ρ) be a metric space. If T : M → M is lipschitzian, then the following conditions are equivalent: (i) T is mean contraction; (ii) k0 (T ) < 1; (iii) k∞ (T ) = 0;

Mean contractions

99

(iv) k∞ (T ) < 1; (v) There exists m such that k(T m ) < 1; P∞ i (vi) i=1 k(T ) < ∞. Proof. Equivalence of (ii), (iii), (iv), (v), and (vi) follows directly from (6.2). Thus, it is enough to prove that T is mean contraction if and only if k0 (T ) < 1. If T is mean contraction, then, in view of theorem 3.8, there exists an equivalent metric d such that kd (T ) < 1. Thus, k0 (T ) < 1. If k0 (T ) < 1, then there exists an equivalent metric d such that kd (T ) < 1. Then, lim kd (T m ) = 0, m→∞

and, consequently, lim kρ (T m ) = 0.

m→∞

Let n be the smallest natural number such that kρ (T n ) < 1. If n = 1, that is, kρ (T ) < 1, then for any multi-index α, T is an α-contraction. If n = 2, then for a multi-index α = (α1 , α2 ) given by  1−kρ (T 2 )  for i = 1, ρ (T ) αi = kρ (T 22k )+2kρ (T )−1  for i = 2, 2kρ (T ) we have α1 ρ(T x, T y) + α2 ρ(T 2 x, T 2 y)
≤ α1 kρ (T ) + α2 kρ (T 2 ) ρ(x, y)
≤ α1 kρ (T ) + kρ (T 2 ) ρ(x, y)   1 − kρ (T 2 ) + kρ (T 2 ) ρ(x, y) = 2 =

1 + kρ (T 2 ) ρ(x, y). 2

This implies that k(α, T ) ≤ (1+kρ (T 2 ))/2 < 1. If n ≥ 3, then for a multi-index α = (α1 , . . . , αn ) given by  1−kρ (T n )  for i = 1,  2kρ (T ) 0 for i = 2, . . . , n − 1, αi =   kρ (T n )+2kρ (T )−1 for i = n, 2kρ (T ) the mapping T is an α-contraction. 

100

Classification of Lipschitz mappings

A similar result for mean nonexpansive mappings fails to hold, that is, k0 (T ) = 1 does not imply that T is α-nonexpansive for some multi-index α (see example 3.6). It is easy to observe that, for any lipschitzian mapping T , we have k0 (T ) = 1 ⇐⇒ k∞ (T ) ∈ [1, ∞) and k0 (T ) > 1 ⇐⇒ k∞ (T ) = ∞. In view of theorem 6.4, we conclude that a mapping T is a contraction with respect to some equivalent metric if and only if T is mean contraction. In the proof of this theorem, we described how to determine α and k. However, it is clear that the choice of a multi-index α and a constant k < 1 is not unique. Even if k(T ) > 1 and k(T 2 ) > 1, it may happen that for some multi-index α = (α1 , α2 ) of length n = 2 and k < 1, T is in L(α, k): Example 6.1 As a metric space (M, ρ), we take `1 with a metric inherited from the standard norm. Let T : `1 → `1 be as in example 4.2. Then, for α = (1/2, 1/2) and k = 5/6, we have k(T ) = 5/3 > 1 and k(T 2 ) = 25/24 > 1 (see figure 4.4). Hence, despite having the first and the second iterate strictly expansive for some pairs of points in `1 , T can be classified into the class L((1/2, 1/2), 5/6).

6.3

On the rate of convergence of iterates

The estimate presented in the theorem 6.3 requires knowledge of the behavior of Lipschitz constants for all iterates of the mapping T . In practice, it is hard to determine or to give nice evaluations for them. We may try to remedy this problem using the condition of mean contraction as seen below. If, for some α = (α1 , α2 ) and k < 1, T can be classified into the class L(α, k), then it is enough to use one of the following theorems to get a “nice” rate of convergence of iterates: Theorem 6.5 Let (M, d) be a complete metric space, and let T : M → M be an (α1 , α2 )-contraction for the constant k < 1. Then, T has a unique fixed point z. Moreover, for each x ∈ M , lim T n x = z,

n→∞

and in terms of error estimates for each x ∈ M and n ≥ 1, • ρ(T n x, z) ≤

1 1−k

• ρ(T n x, z) ≤

1 1−k


ρ(T n x, T n+1 x) + α2 ρ(T n+1 x, T n+2 x) ;
kρ(T n−1 x, T n x) + α2 ρ(T n x, T n+1 x) .

Mean contractions

101

∞

Proof. Convergence of the sequence {T n x}n=1 to the unique fixed point z follows from theorem 6.4 and the Banach’s Contraction Mapping Principle. Let n ≥ 1 and j ≥ 4. By definition of T , we can write a system of inequalities for i = 1, . . . , j − 2, α1 ρ(T n+i x, T n+i+1 x) + α2 ρ(T n+i+1 x, T n+i+2 x) ≤ kρ(T n+i−1 x, T n+i x) and adding both sides, we obtain α1 ρ(T n+1 x, T n+2 x) +

n+j−2 X

ρ(T i x, T i+1 x)

i=n+2

+α2 ρ(T n+j−1 x, T n+j x) ≤ k

n+j−3 X

ρ(T i x, T i+1 x).

i=n

Adding to both sides the number ρ(T n x, T n+1 x) + α2 ρ(T n+1 x, T n+2 x) + kρ(T n+j−2 x, T n+j−1 x) + (α1 + k) ρ(T n+j−1 x, T n+j x), we get n+j−1 X

ρ(T i x, T i+1 x) + kρ(T n+j−1 x, T n+j x)

i=n

+kρ(T n+j−2 x, T n+j−1 x) ≤ k

n+j−1 X

ρ(T i x, T i+1 x)

i=n n

+ρ(T x, T n+1 x) +α2 ρ(T n+1 x, T n+2 x) +α1 ρ(T n+j−1 x, T n+j x). Moving the first term from the right-hand side to the left and the second with the third term from the left-hand side to the right, we obtain (1 − k)

n+j−1 X

ρ(T i x, T i+1 x) ≤ ρ(T n x, T n+1 x) + α2 ρ(T n+1 x, T n+2 x)

i=n

−kρ(T n+j−2 x, T n+j−1 x) + (α1 − k) ρ(T n+j−1 x, T n+j x). Dividing both sides by (1 − k) > 0, we get n+j−1 X i=n

ρ(T i x, T i+1 x) ≤


1 ρ(T n x, T n+1 x) + α2 ρ(T n+1 x, T n+2 x) 1−k k ρ(T n+j−2 x, T n+j−1 x) 1−k α1 − k + ρ(T n+j−1 x, T n+j x). 1−k

−

102

Classification of Lipschitz mappings

Using the triangle inequality and the above estimate, we can write ρ(T n x, T n+j x) ≤

n+j−1 X

ρ(T i x, T i+1 x)

i=n

≤


1 ρ(T n x, T n+1 x) + α2 ρ(T n+1 x, T n+2 x) 1−k k α1 − k − ρ(T n+j−2 x, T n+j−1 x) + ρ(T n+j−1 x, T n+j x). 1−k 1−k

Letting j → ∞, we obtain ρ(T n x, z) ≤


1 ρ(T n x, T n+1 x) + α2 ρ(T n+1 x, T n+2 x) . 1−k

The right-hand side of the above inequality can be rewritten as follows:
1 α1 ρ(T n x, T n+1 x) + α2 ρ(T n+1 x, T n+2 x) + α2 ρ(T n x, T n+1 x) , 1−k and by definition of T , we get ρ(T n x, z) ≤


1 kρ(T n−1 x, T n x) + α2 ρ(T n x, T n+1 x) . 1−k 

The reader can verify that the above estimates are sharp. Hint to the proof: use the function f defined as in lemma 4.8. It is easy to generalize the above theorem to the case of (α1 , . . . , αn )-contractions, and the proof is similar to the one just presented. We leave it to the reader as an exercise. Theorem 6.6 If T : M → M is α-contraction for the constant k with α = (α1 , . . . , αn ) and z ∈ M is a fixed point of T , then for each  > 0 and x ∈ M n

ρ(T n x, z) = o ((g + ) ) , where g is the unique positive solution of the equation α1 g + α2 g 2 + · · · + αn g n = k. Proof. It follows from theorem 6.3 and corollary 4.6. 

Corollary 6.1 If T : M → M is (α1 , α2 )-contraction for the constant k and z ∈ M is a fixed point of T , then for each  > 0 and x ∈ M ,  n  2k √ + ρ(T n x, z) = o , α1 + ∆ with ∆ = α12 + 4α2 k.

Chapter 7 Nonexpansive mappings in Banach space

The metric fixed point theory for nonexpansive mappings has its foundations in works of Browder [17], G¨ohde [43], and Kirk [48], published in 1965. Currently, it is known that f.p.p. depends deeply on the norm geometry. In the literature, the reader may find a huge collection of conditions put on the space X or the set C that guarantee possessing the fixed point property. It is not our aim to establish all of them. We rather stress those classical as well as very recent results that are sufficient to understand the topic under discussion. An excellent overview of almost all known facts can be found in the handbook [49].

7.1

The asymptotic center technique

One of main tools in the fixed point theory is the asymptotic center introduced by Edelstein [26] in 1972. Let C be a nonempty, closed and convex subset of X. For x ∈ C and a bounded sequence {xn } ⊂ X, we define the asymptotic radius of {xn } at x as the number r(x, {xn }) = lim sup kx − xn k . (7.1) n→∞

If {xn } is fixed, then the formula (7.1) defines a function on C, which has the following properties: • For all x ∈ C, r(x, {xn }) = 0 ⇔ lim xn = x. n→∞

• For all x, y ∈ C, |r(x, {xn }) − r(y, {xn })| ≤ kx − yk . • For all x, y ∈ C, and α, β ≥ 0 with α + β = 1, r(αx + βy, {xn }) ≤ αr(x, {xn }) + βr(y, {xn }). Therefore, r(x, {xn }) is a nonnegative, continuous, and convex function of x. The greatest bound from below, r (C, {xn }) = inf {r(y, {xn }) : y ∈ C} , 103

104

Classification of Lipschitz mappings

is called the asymptotic radius of {xn } in C. The set A (C, {xn }) = {y ∈ C : r(y, {xn }) = r (C, {xn })} is called the asymptotic center of {xn } in C. The asymptotic center can be defined as an intersection of a descending family of nonempty, bounded, closed and convex subsets of C: \ A (C, {xn }) = {y ∈ C : r(y, {xn }) ≤ r (C, {xn }) + } . (7.2) >0

Hence, the asymptotic center is always closed and convex. It may be empty or nonempty (in particular if C is weakly compact) and consists of one or more points. Structure of such sets depends deeply on the norm geometry of X. Example 7.1 Consider the space X = R2 furnished with the norm k(x1 , x2 )k = max {|x1 | , |x2 |}. As a subset C, take the closed unit ball B = n {(x1 , x2 ) : |x1 | ≤ 1, |x2 | ≤ 1}. Consider the sequence zn = ((−1) , 0). Then, A (B, {zn }) = {(0, x2 ) : x2 ∈ [−1, 1]} . A connection between the asymptotic center of the sequence and the characteristic of convexity 0 (X) is presented in the following lemma (see [36], pp. 91) Lemma 7.1 If C is a nonempty, closed, and convex subset of a Banach space X, then, for each bounded sequence {xn } in X, we have diam (A (C, {xn })) ≤ 0 (X) r (C, {xn }) . Proof. If A (C, {xn }) is empty or consists of exactly one point, then there is nothing to prove. Suppose that d := diam (A (C, {xn })) > 0. For each  ∈ (0, d), there exist points y, z ∈ A (C, {xn }) such that ky − zk ≥ d − . Since A (C, {xn }) is convex, y+z 2 ∈ A (C, {xn }). Denote r := r (C, {xn }). Then r > 0 and  lim sup kxn − yk = r 

n→∞ =⇒ lim sup kxn − zk = r  n→∞

  

y + z d−

=⇒ r = lim sup x − ≤ 1 − δ r. X

n 2 r n→∞

Consequently, d ≤ 0 (X) . r 

Nonexpansive mappings in Banach space

105

Observe that in example 7.1 we have r (B, {zn }) = 1, 0 (X) = 2, and diam (A (B, {zn })) = 2. If X is uniformly convex, then the asymptotic center A (C, {xn }) is nonempty as an intersection of weakly compact sets (uniform convexity implies reflexivity). By lemma 7.1, it consists of exactly one point (0 (X) = 0). The preceding observation may be used in the proof of the well-known theorem of Browder [17] and G¨ohde [43]. Theorem 7.1 (Browder and G¨ ohde, 1965) Bounded, closed and convex subsets of uniformly convex Banach X space have the fixed point property for nonexpansive mappings. Proof. Let C be a bounded, closed and convex subset of X. Suppose {xn } is an approximate fixed point sequence for nonexpansive mapping T : C → C; that is, lim kxn − T xn k = 0. n→∞

Let r = r (C, {xn }) and {z} = A(C, {xn }). Observe that kT z − xn k ≤ kT z − T xn k + kT xn − xn k ≤ kz − xn k + kT xn − xn k . Hence, putting limsup to both sides of the above inequality, we obtain r(T z, {xn }) ≤ r(z, {xn }) = r. This implies that T z ∈ A (C, {xn }). Thus, T z = z. 

Similarly, we can prove the above theorem using the sequence of iterates of a given point x0 ∈ C, {T n (x0 )}. If C is compact and {xn } is an approximate fixed point sequence for T : C → C, then {xn } may fail to converge. Nevertheless, by compactness of C, we can always choose a convergent subsequence {xnk } and then lim xnk = k→∞

z = T z. Consider, for instance, the mapping T : [−1, 1] → [−1, 1] given by n T x = x and the sequence xn = (−1) . In the general case of bounded, closed and convex subset C it does not hold. Example 7.2 Fix p ∈ [1, ∞). Let X = `p with the standard norm, kxk = k(x1 , x2 , . . . )k =

∞ X

! p1 p

|xi |

.

i=1 ∞

Let {ξi }i=1 be a sequence of real numbers in (0, 1) such that lim ξi = 1. Define i→∞

the mapping T : B`p → B`p by T x = T (x1 , x2 , x3 , . . . ) = (ξ1 x1 , ξ2 x2 , ξ3 x3 , . . . ) .

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Classification of Lipschitz mappings

It is easy to see that Fix(T ) = {(0, 0, . . . , 0, . . . )}. If en denotes n-th vector of standard basis, then T en = (0, . . . , 0, ξn , 0, . . . , 0, . . . ) | {z } n−1

and lim kT en − en k = lim k(0, . . . , 0, ξn − 1, 0, . . . , 0, . . . )k = lim |ξn − 1| = 0.

n→∞

n→∞

n→∞

Hence, {en } is an approximate sequence for T . However, {en } does not contain any convergent subsequence. Observe that, for p ∈ (1, ∞), the approximate sequence {en } of mapping T from the above example is weakly convergent, and its weak limit is a fixed point of T , w − lim en = (0, 0, . . . , 0, . . . ) = T (0, 0, . . . , 0, . . . ). n→∞

If p = 1, then {en } is weakly* convergent to the origin in `1 = c∗0 . It is not a coincidence. Recall that a bounded sequence in `p with p ∈ (1, ∞) (or in `1 = c∗0 ) is weak convergent (weak* convergent, respectively) if and only if it converges coordinate-wise. Consider first the case of space `1 = c∗0 . ∞

Lemma 7.2 If {xn }n=1 is a bounded sequence in `1 , which converges coordinate-wise to x, then for any y ∈ `1 , r (y, {xn }) = r (x, {xn }) + kx − yk . Proof. Since kxn − yk ≤ kxn − xk + kx − yk , we immediately obtain r (y, {xn }) ≤ r (x, {xn }) + kx − yk . Consider now the family of natural projections Pi of `1 onto the subspace span {e1 , . . . , ei }, that is, for i = 1, 2, . . . and x = (x1 , x2 , . . . ) ∈ `1 , we put Pi (x1 , x2 , . . . ) = (x1 , x2 , . . . , xi , 0, . . . , 0, . . . ) . Let Qi = I − Pi , where I denotes identity. Then Qi (x1 , x2 , . . . ) = (0, . . . , 0, xi+1 , xi+2 , . . . ). | {z } i

For any i = 1, 2, . . . , r (x, {xn }) = lim sup kxn − xk = lim sup kQi (xn − x)k . n→∞

n→∞

Nonexpansive mappings in Banach space

107

Moreover, kx − yk = lim

i→∞



 lim kPi (xn − y)k

n→∞

and lim kQi (x − y)k = 0.

i→∞

Observe that kPi (xn − y)k + kQi (xn − x)k ≤ kxn − yk + kQi (x − y)k . Letting k → ∞ and then i → ∞, we finally obtain r (y, {xn }) ≥ r (x, {xn }) + kx − yk .  ∞ {xn }n=1

Similarly, if is a bounded sequence in `p with p ∈ (1, ∞), which converges coordinate-wise to x, then for any y ∈ `p , p

p

p

(r (y, {xn })) = (r (x, {xn })) + kx − yk .

(7.3)

Let C be a bounded closed and convex subset of `p with 1 < p < ∞ and let T : C → C be a nonexpansive mapping. Consider any approximate fixed point sequence {xn } ⊂ C of T . Since `p is reflexive, the set C is weakly compact. Hence, in view of Eberlein-Smulian Theorem, there exists a subsequence {xnk } of {xn }, which converges weakly to a point x ∈ C. Using (7.3), for any y ∈ C, y 6= x, we have r (y, {xnk }) > r (x, {xnk }) and, consequently, A ({xnk } , C) = {x}. Following the proof of theorem 7.1, we also conclude that T x ∈ A ({xnk } , C). Hence, it must be the case T x = x. Let us return now to the case of `1 space. Suppose C is a weakly* compact and convex subset of `1 = c∗0 and T : C → C is a nonexpansive mapping. Then there exists an approximate fixed point sequence {xn } of T . Recall that the relative weak* topology on bounded sets in `1 = c∗0 is metrizable because c0 is separable. Hence, we can choose subsequence {xnk }, which converges weakly* to a point x ∈ C (observe that similar reasoning could be applied in the previous case of `p spaces with 1 < p < ∞). In view of lemma 7.2, r (y, {xnk }) > r (x, {xnk }) for any y ∈ C such that y 6= x. Thus, A ({xnk } , C) = {x}. Obviously, x is a fixed point of T . We shall summarize the above observation in the following theorem [47]: Theorem 7.2 (Karlovitz, 1976) Weakly* compact and convex subsets of `1 considered as a dual of c0 have the f.p.p. for nonexpansive mappings. In view of Alaoglu’s Theorem and the above result, we conclude that all closed balls in `1 have the f.p.p. for nonexpansive mappings. Also, weakly compact and convex subsets of `1 have the fixed point property. It is a consequence of Schur’s Lemma, which states that a bounded sequence in `1 is weakly convergent if and only if it is norm convergent. Hence,

108

Classification of Lipschitz mappings

in view of Eberlein-Smulian Theorem, a subset C is weakly compact if and only if it is compact in the norm topology. Finally, by compactness of C, for any nonexpansive mapping T : C → C, there exists a point z ∈ C such that kz − T zk = inf {kx − T xk : x ∈ C} = 0. Nevertheless, in view of the celebrated Schauder’s Fixed Point Theorem, which states that each continuous self-mapping T defined on compact and convex set has a fixed point, the above result is very weak. A situation is different in the case of weakly* compact and convex subsets of `1 considered as a dual of c. To see it, we shall need some basic facts. Recall that classical Schauder basis of c consists of vectors e = (1, 1, . . . , 1, . . . ) and ei = (0, . . . , 0, 1, 0, . . . , 0, . . . ) for i = 1, 2, . . . . | {z } i−1

Then, for each x = (x1 , x2 , . . . ) ∈ c, we have   x = lim xn · e + x1 − lim xn , x2 − lim xn , . . . n→∞

n→∞

= lim xn · e + n→∞

∞  X

n→∞



xi − lim xn ei . n→∞

i=1

Consequently, if x∗ ∈ c∗ and x = (x1 , x2 , . . . ) ∈ c, then ∗

∗

x (x) = lim xn · x (e) + n→∞

x (e) −

n→∞


xi − lim xn x∗ ei

i=1 ∞ X

∗

= lim xn ·

∞  X

n→∞

x

∗

i=1

e

i

!


+

∞ X


xi x∗ e i ;

i=1

observe that x∗ e , x∗ e , . . . ∈ `1 because x∗ ∈ c∗ implies that the restriction of x∗ to c0 is in c∗0 . Define the mapping ϕ : c∗ → `1 by ! ∞ X
∗ 1
∗ 2
∗ ∗ ∗ i ϕ (x ) = x (e) − x e ,x e ,x e ,... .
1


2




i=1 ∗ Note that ϕ (c∗ ) = `1 . Indeed, 1 , y
2 , . . . ) ∈ `1 , then ϕ (x ) = y for P∞ if y = (y ∗ ∗ ∗ ∗ i x ∈ c such that x (e) = i=1 yi and x e = yi+1 for i = 1, 2, . . . . We shall prove that ϕ is an isometry of c∗ onto `1 . For x∗ ∈ c∗ and x = (x1 , x2 , . . . ) ∈ c, we have ∞ ∞ X
X
∗ ∗ ∗ i |x (x)| ≤ lim xn · x (e) − x e + |xi | · x∗ ei n→∞ i=1 i=1 ! ∞ ∞ X X

∗ ∗ i ∗ i x e kxk . x e + ≤ x (e) − i=1

i=1

Nonexpansive mappings in Banach space

109

∗ Hence, kx∗ k∗ ≤ kϕ
(x )k. On the other hand, if for each k = 1, 2, . . . we define k k k x = x1 , x2 , . . . ∈ c as

xki =



∗ sgn xP ei ,

∞ sgn x∗ (e) − i=1 x∗ ei ,



for i = 1, . . . , k for i = k + 1, k + 2, . . . ,

then xk = 1, provided x∗ 6= 0, and

x

∗

x

k




∞ k ∞ X X ∗ i

X
∗ ∗ i x e ± = x (e) − x e + x∗ e i . i=1

i=1

i=k+1

Thus, lim x

k→∞

∗

x

k




∞ ∞ X ∗ i

X ∗ ∗ i x e , = x (e) − x e + i=1

∗

i=1 ∗ kx k∗

∗

and, consequently, kx k∗ ≥ kϕ (x )k. Finally, summarize the above remarks in the following

= kϕ (x∗ )k. We shall

Lemma 7.3 For each x∗ ∈ c∗ , there exists exactly one vector y = (y1 , y2 , . . . ) =

∗

x (e) −

∞ X

! x

∗

i




∗

e ,x

e

1




∗

,x

e

2




,...

∈ `1

i=1

such that for any x = (x1 , x2 , . . . ) ∈ c ∗

x (x) = y1 lim xn + n→∞

∞ X

xn yn+1

and

∗

kx k∗ = kyk =

n=1

∞ X

|yn | .

n=1

The mapping ϕ : x∗ 7→ y is a linear isometry of c∗ onto `1 . Hence, up to isometry, c∗ = `1 . Immediately, we can ask about  the ∞ characterization of weakly* convergent sequences in `1 = c∗ . Suppose y k k=1 is a bounded sequence in `1 and let
∞ {x∗k }k=1 be a sequence in c∗ such that ϕ (x∗k ) = y k = y1k , y2k , . . . . Since vectors e = (1, 1, . . . , 1, . . . ) and ei = (0, . . . , 0, 1, 0, . . . , 0, . . . ) for i = 1, 2, . . . | {z } i−1

 ∞ form a linearly dense set in c, the weak* convergence of the sequence y k k=1 to y = (y1 , y2 , . . . ) = ϕ (x∗ ) is equivalent to the condition lim x∗k (e) = x∗ (e) k→∞

and lim x∗k ei = x∗ ei for each i = 1, 2, . . . . Since k→∞


x∗ ei = yi+1 ,

x∗ (e) =

∞ X

yn ,

x∗k (e) =

n=1

for i, k = 1, 2, . . . , we obtain the following:

∞ X n=1

ynk ,


k and x∗k ei = yi+1

110

Classification of Lipschitz mappings  ∞
Lemma 7.4 Let y k k=1 , y k = y1k , y2k , . . . , be a bounded sequence in `1 .  k ∞ Then y k=1 is weakly* convergent to y = (y1 , y2 , . . . ) in `1 = c∗ if and only if for each i = 1, 2, . . . k lim yi+1 = yi+1 k→∞

and lim

k→∞

∞ X

! ynk

n=1

=

∞ X

yn .

n=1

Example 7.3 Consider the sequence {en } of basis vectors in `1 . Then w∗ − lim en = (0, 0, . . . , 0, . . . ) n→∞

in `1 =

c∗0 .

However, if we consider `1 as a dual of c, then w∗ − lim en = e1 = (1, 0, . . . , 0, . . . ) . n→∞

This shows that both weak* topologies are not comparable. In opposite to `1 = c∗0 , the space `1 = c∗ does not have the weak*-f.p.p. for nonexpansive mappings. To see it, consider the following: Example 7.4 Let j : c → c∗∗ be the canonical embedding; that is, j (x) (x∗ ) = x∗ (x) for x ∈ c and x∗ ∈ c∗ . Let e = (1, 1, . . . , 1, . . . ) ∈ c, and consider the set −1

(j (e))

{1} = {x∗ ∈ c∗ : (j (e)) (x∗ ) = 1} = {x∗ ∈ c∗ : x∗ (e) = 1} .

If ϕ denotes an isometry from lemma 7.3, then ) ( ∞   X −1 xi . ϕ (j (e)) {1} = (x1 , x2 , . . . ) ∈ `1 : x1 = 1 − i=2 ∗

∗

If B denotes the closed unit ball in c , then the set   −1 ϕ (j (e)) {1} ∩ ϕ (B ∗ ) is weakly* compact as the intersection of weakly* closed and weakly* compact sets. It is easy to see that the above set is a “positive face” S + of the closed unit ball B in `1 ; that is, ( ) ∞ X + S = x = (x1 , x2 , . . . ) ∈ `1 : xi ≥ 0 for i = 1, 2, . . . , and xi = 1 . i=1 +

To prove that S fails to have the fixed point property, consider the “right shift” mapping T : S + → S + defined for any x = (x1 , x2 , . . . ) ∈ S + by T (x1 , x2 , . . . ) = (0, x1 , x2 , . . . ).

Nonexpansive mappings in Banach space

111

It is easy to verify that T is an isometry; hence, it is nonexpansive. Further, T x = x implies x = (0, 0, . . . , 0, . . . ), but (0, 0, . . . , 0, . . . ) ∈ / S + . Hence, T is fixed point free. Obviously, S + is not weakly* compact in the weak* topology generated by c0 .

7.2

Minimal invariant sets and normal structure

Another approach was presented by Kirk in 1965. The first main ingredient in his proof is a geometric property called normal structure, which was introduced by Brodskii and Milman [15] in 1948. Let C be a bounded, closed and convex subset of a Banach space (X, k·k). For any x ∈ C, we define the radius of C with respect to x as the number rx (C) = sup {kx − yk : y ∈ C} . We say that a point x ∈ C is a diametral point if rx (C) = diam(C) and a non-diametral point if rx (C) < diam(C). If each point of C is its diametral point, then we say that C is a diametral set. The Chebyshev radius of C (related to C) is the number r(C) = inf {rx (C) : x ∈ C} , and the Chebyshev center of C (related to C) is given by K(C) = {x ∈ C : rx (C) = r(C)} . The reader can easily verify that K(C) can be given via relation \ K(C) = K (C), >0

where K (C) = {y ∈ C : ry (C) ≤ r(C) + } =

\

B(x, r(C) + ) ∩ C.

x∈C

It may happen that K(C) = ∅. However, if C is weakly compact, then all of its closed and convex subsets are weakly compact. Hence, K(C) 6= ∅. If C is

112

Classification of Lipschitz mappings

weakly* compact subset of X ∗ , then by the Alaoglu Theorem, all closed balls and, consequently, all sets of the form B(x, r(C)+)∩C are weakly* compact; hence, K(C) 6= ∅. Clearly, the simplest diametral sets are singletons. However, there exist bounded, closed and convex sets with a positive diameter, which are diametral! Example 7.5 Consider the space c0 with its usual sup norm, k(x1 , x2 , . . . )k = max {|xi | : i = 1, 2, . . . } . Let B

+

denote the set ( +

B =

{xi } : xi ≥ 0 and

∞ X

) xi ≤ 1 .

i=1

The reader can easily verify that B + is a bounded, closed and convex subset of c0 with diam(B + ) = 1. We claim that B + is diametral. Indeed, if en denotes n-th vector of the standard basis, then, for each x ∈ B + , we have lim kx − en k = 1,

n→∞

and, consequently, r(B + ) = diam(B + ) = 1. Example 7.6 In the space `1 furnished with the standard norm, k(x1 , x2 , . . . )k =

∞ X

|xi | ,

i=1

consider the “positive face” S + of the unit sphere S, ( ) ∞ X + S = {xi } : xi ≥ 0 and xi = 1 . i=1 +

We leave to the reader to verify that S is a bounded, closed and convex subset of `1 , having diameter equal to 2. Observe that, for each x ∈ S + , lim kx − en k = 2,

n→∞

where en denotes n-th vector of the standard basis. Hence, S + is a diametral set. One can observe that, in examples 7.5 and 7.6, we have B + = conv {en : n = 1, 2, . . . } and S + = conv {en : n = 1, 2, . . . } . Moreover, in both cases, we have lim dist (en+1 , conv {e1 , e2 , . . . , en }) = diam ({e1 , e2 , . . . }) .

n→∞

On the other hand, we have the following:

Nonexpansive mappings in Banach space

113

Example 7.7 Consider the space l2 with the standard norm, kxk = k(x1 , x2 , . . . )k =

∞ X

! 21 x2i

.

i=1

Let

( +

S =

{xi } : xi ≥ 0,

∞ X

) xi ≤ 1 .

i=1 + One can notice closed and convex in `2 . Moreover, √ that S is bounded, + diam (S ) = 2, whereas r (S + ) = 1. Hence, S + is not diametral. Also,

S + = conv {en : n = 1, 2, . . . } , where en denotes n-th vector of the standard Schauder basis in `2 . Nevertheless, √ lim dist (en+1 , conv {e1 , e2 , . . . , en }) = r(S + ) = 1 < diam ({e1 , e2 , . . . }) = 2. n→∞

We shall see below that it is not a coincidence. To proceed, we shall need some definitions introduced by Brodskii and Milman [15]. We say that a closed and convex set D ⊂ X (or the space X itself) has normal structure if any bounded, closed and convex subset C of D with diam(C) > 0 contains a non-diametral point. A bounded sequence {xn } in a Banach space X is said to be a diametral sequence if it is not constant and lim dist (xn+1 , conv {x1 , x2 , . . . , xn }) = diam ({x1 , x2 , . . . }) .

n→∞

Lemma 7.5 (Brodskii and Milman, 1948) Let C be a bounded, closed and convex subset of a Banach space X. Then the following are equivalent. (i) C has normal structure. (ii) C does not contain a diametral sequence. Proof. First, we shall prove that not (ii) implies not (i). Indeed, if C contains a diametral sequence {xn }, then the set D = conv {x1 , x2 , . . . } is a diametral subset of C. Hence, C does not have normal structure. To prove that not (i) implies not (ii), suppose that D is a diametral subset of C. Let d = diam(D) > 0 and  ∈ (0, d). Let y1 := x1 ∈ D. Since D is diametral, there is a point x2 such that kx2 − y1 k > d − . Define y2 as y2 =

x1 + x2 . 2

Further, since D is diametral, there exists a point x3 ∈ D such that

114

Classification of Lipschitz mappings

kx3 − y2 k > d − 4 . Inductively, we construct a sequence {xn } of elements of D satisfying  kxn+1 − yn k > d − 2 , n where x1 + x2 + · · · + xn yn = . n We shall prove Pn that {xn } is diametral. Take any xP∈n conv {x1 , . . . , xn }. Then x = i=1 αi xi for some αi ≥ 0 such that i=1 αi = 1. Define α b := max {α1 , . . . , αn }. It is clear that α b > 0. Now, we can express yn as  n  X αi 1 1 x+ − xi yn = nb α n nb α i=1 and observe that 1 αi − ≥0 n nb α

n

and

X 1 + nb α i=1



1 αi − n nb α

 = 1.

We have d−

 < kyn − xn+1 k n2  n  X 1 1 αi ≤ kx − xn+1 k + − kxi − xn+1 k nb α n nb α i=1   1 1 kx − xn+1 k + 1 − d. ≤ nb α nb α

Thus, 

d  − nb α n2 b α = d− n  ≥ d− . n

kx − xn+1 k ≥

 nb α

Since x ∈ conv {x1 , . . . , xn } and  ∈ (0, d) has been arbitrarily chosen, we conclude that for any n ∈ N, d ≥ dist (xn+1 , conv {x1 , x2 , . . . , xn }) ≥ d −

 . n

Therefore, {xn } is a diametral sequence. 

Nonexpansive mappings in Banach space

115

One can observe that a diametral sequence does not contain a convergent subsequence. This implies that all compact and convex subsets of X have normal structure. In particular, every finite dimensional Banach space X has normal structure. Furthermore, we have the following: Lemma 7.6 If X is a Banach space with a characteristic of convexity 0 (X) < 1, then X has normal structure. In particular, all uniformly convex Banach spaces have normal structure. Proof. Let C ⊂ X be a bounded, closed and convex set with a positive diameter d = diam(C). Fix  ∈ (0, d (1 − 0 (X))) and two points u and v in C such that ku − vk > d − . The following implication follows from (2.2) and the fact that the modulus of convexity δX is nondecreasing on [0, 2]: for any x ∈ C,



  

u + v kx − uk ≤ d

≤ 1 − δX d −  d. (7.4) x − =⇒

kx − vk ≤ d 2 d
d− > 0; hence, u+v Since d− d > 0 (X) we get δX d 2 is a non-diametral point of C. 

The second main ingredient in Kirk’s proof is Zorn’s Lemma, which ensures the existence of the so-called minimal invariant set. Let C be a weakly compact and convex subset of a Banach space X and T : C → C be a mapping. A nonempty, weakly closed and convex subset C ◦ of C is called minimal invariant for T if T (C ◦ ) ⊂ C ◦ and C ◦ does not contain a nonempty, weakly closed and convex proper subset D, which is T -invariant; that is, T (D) ⊂ D. Recall that by a relation  on a set A we mean a subset of A × A. For a, b ∈ A, we write a  b if and only if (a, b) ∈ . We say that (A, ) is a partially ordered set if for each a, b, c ∈ A: (i) a  a, (ii) (a  b and b  a) ⇒ a = b, (iii) (a  b and b  c) ⇒ a  c. Any subset B ⊂ A, which is linearly ordered by  is called a chain; recall that (B, ) is linearly ordered if it is partially ordered, and, in addition, for each a, b ∈ B: (iv) a  b or b  a. Let (A, ) be a partially ordered set. We say that a ∈ A is maximal if ∀b∈A (a  b ⇒ a = b) . Let B ⊂ A. An element a ∈ A is called an upper bound for B if ∀b∈B (b  a) .

116

Classification of Lipschitz mappings

Lemma 7.7 (Zorn’s Lemma) Let (A, ) be a partially ordered set. If every chain has an upper bound, then A has a maximal element. Now we are ready to present Kirk’s Theorem. Theorem 7.3 (Kirk’s Theorem, 1965) Let C be a nonempty, weakly compact and convex subset of a Banach space X and suppose that C has normal structure. Then each nonexpansive mapping T : C → C has a fixed point. Proof. Let F denote the family of all nonempty, weakly closed and convex subsets D ⊂ C such that T (D) ⊂ D. It is clear that the relation on F defined as D1  D2 ⇔ D1 ⊃ D2 generates a partial order. Consider any chain D ⊂ F and the set \ D◦ = D. D∈D

Since C is weakly compact and the family D has the finite intersection property, we conclude that D◦ 6= ∅. Moreover, D◦ is weakly closed, convex and T (D◦ ) ⊂ D◦ . Hence, D◦ is an upper bound for the chain D. By Zorn’s Lemma, there is a maximal element C ◦ ∈ F. Put C ◦◦ = convT (C ◦ ) ⊂ C ◦ . Then, T (C ◦◦ ) ⊂ T (C ◦ ) ⊂ convT (C ◦ ) = C ◦◦ . Consequently, C ◦◦ ∈ F. By maximality of C ◦ , it must be the case C ◦ = C ◦◦ . Thus, convT (C ◦ ) = C ◦ . Since C ◦ is weakly compact, we conclude that K(C ◦ ) = {z ∈ C ◦ : rz (C ◦ ) = r(C ◦ )} 6= ∅; hence, there exists a point z ∈ C ◦ such that rz (C ◦ ) = r(C ◦ ). Observe that, for each y ∈ C ◦ , we have kT z − T yk ≤ kz − yk ≤ rz (C ◦ ) = r(C ◦ ) for all y ∈ C ◦ . Hence, T (C ◦ ) ⊂ B(T (z), r(C ◦ )), and, consequently, C ◦ = convT (C ◦ ) ⊂ B(T (z), r(C ◦ )).

Nonexpansive mappings in Banach space

117

Thus, rT z (C ◦ ) = rz (C ◦ ) = r(C ◦ ). This implies that K(C ◦ ) is T -invariant. The set C ◦ is minimal invariant; hence, K(C ◦ ) = C ◦ . Since C ◦ has normal structure, we conclude that diam(C ◦ ) = 0. Thus, C ◦ = {z} , and z is a fixed point of T . 

If X is reflexive Banach space, then any bounded, closed and convex subset C of X is weakly compact. Hence, we obtain the following: Corollary 7.1 Let X be reflexive Banach space. If X has normal structure, then X has the f.p.p. for nonexpansive mappings. Example 7.8 Consider the space c0 renormed as in example 2.2; that is, !1 ∞  2 2 X x i 2 k|x|k = kxk + . i i=1 Then, (c0 , k| · |k) has normal structure. To see it, consider any bounded, closed and convex set C ⊂ c0 with diam(C) > 0. Fix any two points u = (u1 , u2 , . . . ) and v = (v1 , v2 , . . . ) in C such that u 6= v. Let z = u+v 2 . We shall prove that z is a non-diametral point of C. For any x ∈ C, we have 2 ∞  X xi − zi 2 2 . k|x − z|k = kx − zk + i i=1 Observe that 2

1 2 k(x − u) + (x − v)k 4 1 2 ≤ (kx − uk + kx − vk) 4  1 1 2 2 2 = kx − uk + kx − vk − (kx − uk − kx − vk) 2 4  1 2 2 ≤ kx − uk + kx − vk 2

kx − zk =

and for each i = 1, 2, . . . ,  2  2 xi − zi 1 (xi − ui ) + (xi − vi ) = i 4 i " 2  2 #  2 1 xi − ui xi − vi 1 ui − vi = + − . 2 i i 4 i

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Classification of Lipschitz mappings

Hence, ∞

2

 1X 1 2 2 k|x − u|k + k|x − v|k − 2 4 i=1   ∞ 2 1 X ui − vi . ≤ d2 − 4 i=1 i



k|x − z|k ≤

ui − vi i

2

Since u 6= v are fixed and x ∈ C has been arbitrary chosen, we conclude that r(C) ≤ rz (C) < diam(C). In particular, the set B + = {x = {xi } : kxk ≤ 1, xi ≥ 0 for i = 1, 2, . . . } has normal structure with respect to the new norm k| · |k, and it is easy to observe that the mapping T : B + → B + defined as T (x1 , x2 , . . . ) = (1, x1 , x2 , . . . ) is nonexpansive with respect to k| · |k, and it is fixed point free. Hence, T does not have a minimal invariant set. Consequently, normal structure of the set or the space without some additional assumption does not imply the fixed point property. Also, the above example shows that normal structure of X does not imply reflexivity. Moreover, Zizler [77] proved that any separable Banach space can be renormed to have normal structure. Actually, these two properties, normal structure and reflexivity, are independent. Example 7.9 (see [31]) In the space `2 with the classical norm ! 21 ∞ X 2 , kxk = k(x1 , x2 , . . . )k = xi i=1

consider the positive part of the unit ball, B + = {x ∈ `2 : kxk ≤ 1, xi ≥ 0} . Define a new norm by  k|x|k = max

 1 √ kxk , max |xi | . i∈N 2

Observe that k| · |k is equivalent to k·k and for any x ∈ `2 1 √ kxk ≤ k|x|k ≤ kxk . 2 Hence, (`2 , k| · |k) is reflexive. However, it does not have normal structure. We leave to the reader to verify that the set B + is diametral with respect to the new norm k| · |k and r(B + ) = diam(B + ) = 1.

Nonexpansive mappings in Banach space

119

In fact, the estimate (7.4) does not depend on the choice of a set C. This leads us to the following definition introduced in 1980 by Bynum [18]. We say that a closed, convex set D ⊂ X (or the whole space X) has uniform normal structure if there is a constant a ∈ (0, 1) such that, for any bounded, closed and convex subset C of D, we have r(C) ≤ a · diam(C). The number  N (X) := inf

diam(C) : C ⊂ X is bounded, closed and convex r(C) with diam(C) > 0}

is called the normal structure coefficient of the space X. Equivalently, a Banach space (X, k·k) has uniform normal structure if and only if N (X) > 1. From (7.4), we conclude that, if 0 (X) < 1, then X has uniform normal structure with a = 1 − δX (1), and, consequently, N (X) ≥

1 . 1 − δX (1)

√ Moreover, Bynum [18] proved that for a Hilbert space H we have N (H) = 2. Dom´ınguez Benavides [7] and Prus [71] gave exact values of the constant N (X) in the case of `p and Lp [0, 1] spaces: n 1 1o N (`p ) = N (Lp [0, 1]) = min 2 p , 2 q , where p > 1 and p1 + 1q = 1. It is clear that uniform normal structure implies normal structure. In fact, it is a much stronger property. In 1984, Maluta [58] proved that every Banach space X having uniform normal structure must be reflexive! Corollary 7.2 If X has uniform normal structure, then X has the f.p.p. for nonexpansive mappings. Let us pass now to the case of weakly* compact sets in dual space X ∗ . Let D be a convex subset of a dual Banach space X ∗ . We say that D (or the space X ∗ itself) has weak* normal structure if each bounded, weakly* closed and convex set E ⊂ D with diam(E) > 0 contains a non-diametral point. Theorem 7.4 (Kirk’s Theorem) Let X be a Banach space. Let C be a weakly* compact convex subset of a dual Banach space X ∗ and suppose C has weak* normal structure. Then, C has f.p.p. for nonexpansive mappings. Proof. Repeat steps from the proof of theorem 7.3. 

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Classification of Lipschitz mappings

Now it is enough to apply the following lemma to obtain another proof of Theorem 7.2: Lemma 7.8 If C is a weakly* compact and convex subset of l1 = c∗0 , then C has weak* normal structure. Proof. Obviously, it is enough to prove that every weakly* compact and convex set C in l1 = c∗0 contains a non-diametral point. Suppose diam(C) > 0. If C is compact, then it has weak* normal structure. Hence, suppose C is not n compact.

i Then,

there exists a number a > 0 and a sequence {x } in C such j

that x − x ≥ a for all i, j = 1, 2, . . . with i 6= j. Since the weak* topology on C is metrizable, we conclude that {xn } contains a weakly* convergent subsequence to x ∈ C. We can assume that {xn } has been already chosen to satisfy this condition. It is clear that {xn } does not converge in norm to x. This implies that r (x, {xn }) = lim sup kxn − xk > 0. n−→∞

By lemma 7.2, for each y ∈ C, we have r (y, {xn }) = r (x, {xn }) + kx − yk , and, consequently, kx − yk = r (y, {xn }) − r (x, {xn }) ≤ diam(C) − r (x, {xn }) . Hence, the point x is a non-diametral point of C. 

We have already seen in the proof of Kirk’s Theorem that any minimal invariant set C must be diametral. However, every such set possesses a much stronger property, which has proved to be very useful in the study of fixed point property for nonexpansive mappings. This fact has been discovered independently by Goebel [30] and Karlovitz [46]: Lemma 7.9 (Goebel-Karlovitz Lemma) Let C ⊂ X be a minimal with respect to being nonempty, weakly compact, convex, and T -invariant for some nonexpansive mapping T . If {xn } ⊂ C is an approximate fixed point sequence for T , then for each x ∈ C, r (x, {xn }) = lim sup kx − xn k = lim kx − xn k = diam(C). n→∞

n→∞

Proof. It is enough to consider the case when diam(C) > 0. The set C must be diametral. Suppose there is a point z ∈ C such that lim supn−→∞ kz − xn k = r < diam(C). Then, the set Ar ({xn } , C) := {y ∈ C : r (y, {xn }) ≤ r}

Nonexpansive mappings in Banach space

121

is nonempty, closed and convex. Since Ar ({xn } , C) is a weakly closed subset of a weakly compact set C, we conclude that Ar ({xn } , C) is itself weakly compact. Also, the set Ar ({xn } , C) is T -invariant. Hence, by minimality of C, we have Ar ({xn } , C) = C. Further, we can select µ > 0 such that r + µ < diam(C). Choose any finite collection of points {y1 , y2 , . . . , ym } ⊂ C. Then, there exists an integer k such that for each n > k and for each i = 1, 2, . . . , m, we have xn ∈ B (yi , r + µ) ∩ C. Thus, the family {B(x, r + µ) ∩ C : x ∈ C} has the finite intersection property. Since all sets of the form B(x, r + µ) ∩ C are weakly compact, we conclude that \ \ (B(x, r + µ) ∩ C) = C ∩ B(x, r + µ) 6= ∅. x∈C

x∈C

T Then, for any z ∈ C ∩ x∈C B(x, r + µ), we have r(z, C) ≤ r + µ < diam(C). However, this contradicts the fact that C is diametral. Hence, for each x ∈ C, we have lim sup kx − xn k = diam(C). n→∞

Since each subsequence of {xn } is still an approximate fixed point sequence for T , we can replace lim sup by lim. 

For several years, it had been unknown whether there exists a weakly compact and convex set C with a positive diameter that is minimal invariant for a certain nonexpansive mapping T : C → C. The answer came in 1981 when Alspach [1] published an example of fixed point free self-mapping defined on a weakly compact and convex set. Example 7.10 (Alspach’s example) Let X = L1 [0, 1]. Recall that for any two functions g, h ∈ L1 [0, 1] satisfying g ≤ h up to the set of measure zero, the order segment, defined as {f ∈ L1 [0, 1] : g(t) ≤ f (t) ≤ h(t) almost everywhere in [0, 1]} , is weakly compact in L1 [0, 1]. Let C be the set  Z C = f ∈ L1 [0, 1] : 0 ≤ f ≤ 1 and 0

1

f (t) dt =

1 2

 .

The set C is convex and weakly compact as the intersection of the order segment with a hyperplane. Consider the following mapping, usually called the “baker transformation”, ( min {2f (2t) , 1} if 0 ≤ t ≤ 21 (T f ) (t) = max {2f (2t − 1) − 1, 0} if 12 < t ≤ 1.

122

Classification of Lipschitz mappings

We leave to the reader the verification that T is an isometry on C. We shall prove that T is fixed point free. Suppose g ∈ C is a fixed point of T , g = T g. Then, A := {t : g(t) = 1} = {t : (T g) (t) = 1} , and, by definition of T , we get     1 1 1 A = t : 0 ≤ t ≤ ∧ ≤ g(2t) ≤ 1 ∪ t : < t ≤ 1 ∧ g(2t − 1) = 1 2 2 2     1 1 1 1+t = t : ≤ g(t) < 1 ∪ {t : g(t) = 1} ∪ : g(t) = 1 . 2 2 2 2 Observe that the above three sets are mutually disjoint. Moreover, the measure of each of the last two sets is equal to the measure of A. Consequently, the set   1 t : ≤ g(t) < 1 2 has measure zero. Further,       t 1 1 1 1 : ≤ g(t) < ⊂ t : ≤ T g(t) < 1 = t : ≤ g(t) < 1 . 2 4 2 2 2 Hence, the measure of   1 1 t : ≤ g(t) < 4 2 is zero. Continuing the process, it follows that the measure of each set of the form   1 1 t : n ≤ g(t) < n−1 2 2 equals zero for n = 1, 2, . . . , and, consequently, the same holds for the set  ∞  [ 1 1 {t : 0 < g(t) < 1} = t : n ≤ g(t) < n−1 . 2 2 n=1 Hence, ( 1 g(t) = χA (t) = 0 Since Z

if t ∈ A if t ∈ [0, 1] \ A.

1

χA (t)dt = 0

1 , 2

the measure of A is 12 . We have T (χA ) = χ 12 A + χ( 1 + 1 A) = χA . 2

2

Nonexpansive mappings in Banach space 123     Hence, the intersections of A with each of the intervals 0, 21 and 12 , 1 have measure equal to 21 . Further, T 2 (χA ) = χ 14 A + χ( 1 + 1 A) + χ( 1 + 1 A) + χ( 3 + 1 A) = χA . 4 4 2 4 4 4     implies that the intersections of A with each of the intervals 0, 14 , 14 , 21 , This    1 3 3 2 , 4 , 4 , 1 have measure equal to one half that of the interval. Continuing   in this way we have that the same is true for any dyadic interval 2in , i+1 2n , n = 1, 2, . . . , i = 0, 1, 2, . . . , 2n − 1, and hence for any interval. Thus, for any t ∈ (0, 1), the density of A at t defined as lim

h→0+

µ (A ∩ [t − h, t + h]) 2h

is equal to 21 . However, it is impossible because, in view of the Banach’s Density Theorem, any measurable set has density 0 or 1 almost everywhere. One can notice that the Chebyshev radius r(C) = 12 , whereas diam(C) = 1. Consequently, C is not diametral; hence, it is not a minimal invariant set for T. Consequently, weak compactness itself does not imply the f.p.p.

7.3

Uniformly nonsquare, uniformly noncreasy, and reflexive Banach spaces

We have already seen that all Banach spaces with the characteristic of convexity 0 (X) < 1 share the fixed point property for nonexpansive mappings. In 2005, this result was extended for all uniformly nonsquare Banach spaces. In [28], Garc´ıa-Falset, Llorens-Fuster, and Mazcu˜ n´an-Navarro proved the following: Theorem 7.5 Let (X, k·k) be a Banach space with 0 (X) < 2. Then, any bounded, closed and convex subset C has the f.p.p. for nonexpansive mappings. Another approach has been presented by Prus [72]. He introduced a class of uniformly noncreasy Banach spaces. For any x∗ , y ∗ ∈ SX ∗ and δ ∈ [0, 1], S(x∗ , y ∗ , δ) denotes the set S(x∗ , δ) ∩ S(y ∗ , δ) = {x ∈ BX : x∗ (x) ≥ 1 − δ} ∩ {x ∈ BX : y ∗ (x) ≥ 1 − δ} . A closed unit ball BX (or SX ) is said to have a crease if, for some x∗ , y ∗ ∈ SX ∗ , we have diam(S(x∗ , y ∗ , 0)) > 0 provided x∗ 6= y ∗ . A Banach space is named

124

Classification of Lipschitz mappings

uniformly noncreasy if the following implication holds: for any  > 0, there is δ > 0 such that for any x∗ , y ∗ ∈ X ∗  kx∗ k∗ = 1  ky ∗ k∗ = 1 =⇒ diam(S(x∗ , y ∗ , δ)) ≤ .  ∗ ∗ kx − y k∗ >  Prus proved the following: Theorem 7.6 If X is uniformly noncreasy Banach space, then X has the f.p.p. for nonexpansive mappings. It is known that uniformly nonsquare as well as uniformly noncreasy Banach spaces are super-reflexive, but it is still unknown whether superreflexivity implies the fixed point property for nonexpansive mappings. However, in 2009, Dom´ınguez Benavides [9] obtained the following intriguing theorem: Theorem 7.7 Let (X, k·k) be a reflexive Banach space. Then, there exists an equivalent norm k|·|k on X such that (X, k|·|k) has the fixed point property for k|·|k-nonexpansive mappings. Proofs of the above-mentioned results go beyond frames of presented monograph, and we do not establish them here.

7.4

Remarks on the stability of f.p.p.

Actually, bounded, closed and convex subsets of a Banach space X with 0 (X) < 1 or N (X) > 1 have the f.p.p. not only for nonexpansive mappings but also for uniformly k-lipschitzian mappings, with a constant k slightly greater than 1. The first result of this type was discovered by Goebel and Kirk [35] in 1973: Theorem 7.8 (Goebel and Kirk) Let X be a uniformly convex Banach space with the modulus of convexity δX , and let C be a bounded, closed and convex subset of X. Suppose that γ satisfies    1 = 1. (7.5) γ 1 − δX γ If T : C → C is uniformly lipshitzian relative to a constant k < γ, then T has a fixed point in C.

Nonexpansive mappings in Banach space Recall that, for a Hilbert space H, we have r δH () = 1 −

1−

125

2 . 4

Solving the equation (7.5) we get √ γ=

5 . 2

Later, in 1975, the above theorem was improved and generalized by Lifshitz [54]. Let (M, ρ) be a complete metric space. We say that balls in M are cregular if the following holds: ∀k∈(0,c) ∃α,µ∈(0,1) ∀x,y∈M ∀r>0 ρ(x, y) ≥ (1 − µ) r ⇒ ∃z∈M B (x, (1 + µ) r) ∩ B (y, k (1 + µ) r) ⊂ B (z, αr) . The largest number c for which balls in M are c-regular is called the Lifshitz character of M and is denoted by κ(M ); that is, κ(M ) = sup {c > 0 : the balls in M are c-regular} . It is easy to check that κ(M ) ≥ 1 for any M . Under the above settings, the Lifshitz Theorem states: Theorem 7.9 (Lifshitz, 1975) Let (M, ρ) be a bounded and complete metric space. Suppose T : M → M is uniformly k-lipschitzian mapping with k < κ(M ). Then, T has a fixed point in M . Looking up the proof of this theorem, it can be stated: Theorem 7.10 (Lifshitz, strong version) Let (M, ρ) be a bounded and complete metric space. Suppose T : M → M is uniformly lipschitzian with k∞ (T ) < κ(M ). Then, T has a fixed point in M . Proof. If κ(M ) = 1, then, in view of theorem 6.4, k0 (T ) < 1. Thus, using theorem 6.2, T has a unique fixed point. Now, suppose κ(M ) > 1. For fixed x ∈ M , set r(x) = inf {r > 0 : for some y ∈ M , ρ(x, T n y) ≤ r for n = 1, 2, . . . } . Fix k ∈ (k∞ (T ), κ(M )) and let n0 ∈ N be such that, for each n > n0 , we have k(T n ) ≤ k. Let µ and α ∈ (0, 1) be associated with k as in the definition of κ(M )-regular balls. Then, by definition of r(x), there exists a natural number m > n0 that satisfies ρ(x, T m x) ≥ (1 − µ)r(x), and there is y ∈ M such that, for n = 1, 2, . . . , ρ(x, T n y) ≤ (1 + µ)r(x).

126

Classification of Lipschitz mappings

Thus, by definition of κ(M )-regular balls, there exists z ∈ M such that B (x, (1 + µ) r(x)) ∩ B (T m x, k (1 + µ) r(x)) ⊂ B (z, αr(x)) . Since for n > m, ρ(T m x, T n y) ≤ kρ(x, T n−m y) ≤ k(1 + µ)r(x), the orbit {T n y : n > m} is contained in B (x, (1 + µ) r(x)) ∩ B (T m x, k (1 + µ) r(x)) , and, consequently, in B (z, αr(x)). Thus, r(z) ≤ αr(x). Moreover, for any u ∈ B (x, (1 + µ) r(x)) ∩ B (T m x, k (1 + µ) r(x)), ρ(z, x) ≤ ρ(z, u) + ρ(u, x) ≤ αr(x) + (1 + µ)r(x) ≤ 3r(x). ∞

By putting x0 = x, we can construct a sequence {xn }n=0 , with xn+1 = z(xn ), where z(xn ) is obtain via the above procedure. Thus, r(xn ) ≤ αn r(x0 ) and ρ(xn+1 , xn ) ≤ 3r(xn ) ≤ 3αn r(x0 ). ∞

∞

This implies that {xn }n=0 is a Cauchy sequence. Since M is complete, {xn }n=0 converges. It is clear that its limit is a fixed point of T . 

Let us return to our standard situation. Let C be a bounded, closed and convex subset of Banach space X. It is easy to verify that, under this setting, the Lifshitz character κ(X) can be defined as κ(X) = sup {c > 0 : r(B(0, 1) ∩ B(x, c)) < 1 for all x with kxk = 1} . By definition, we conclude that 1 ≤ κ(X) ≤ 2. Now the Lifshitz Theorem takes the form: Theorem 7.11 (Lifshitz, strong version) If T : C → C is uniformly lipschitzian with k∞ (T ) < κ(X), then T has a fixed point in C. Proof. It is enough to repeat steps from the proof of theorem 7.10 with the point z, which lies in the segment [x, T m x]. Since C is convex, z ∈ C. 

Nonexpansive mappings in Banach space

127

It is known that, in the case of Hilbert space, we have √ √ 5 κ(H) = 2 > . 2 In [25], Downing and Turett gave a characterization of spaces with κ(X) > 1, using the coefficient 0 (X): Theorem 7.12 (Downing, Turett) In any Banach space X, κ(X) > 1 ⇐⇒ 0 (X) < 1.

In principle, κ(X) > 1 for every uniformly convex space X (0 (X) = 0). It is natural to define the following constant: for a Banach space X, γ0 (X) := sup { k : any bounded, closed and convex set C ⊂ X has the f.p.p. for uniformly k-lipschitzian mappings} . Hence, for any Banach space√X with 0 (X) < 1, we have γ0 (X) > 1, and for a Hilbert space H, γ0 (H) ≥ 2. Recall that every uniformly k-lipschitzian mapping T : C → C generates the metric d on C defined for any x, y ∈ C by

 d(x, y) = sup T i x − T i y : i = 0, 1, 2, . . . , which is equivalent to the metric on C inherited from the norm k·k, kx − yk ≤ d(x, y) ≤ k kx − yk

for all x, y ∈ C.

Moreover, T is d-nonexpansive; that is, d(T x, T y) ≤ d(x, y)

for any x, y ∈ C.

On the other hand, suppose a metric ρ on C is equivalent to k·k; that is, there exist constants a, b > 0 such that, for all x, y ∈ C, a kx − yk ≤ ρ(x, y) ≤ b kx − yk .

(7.6)

Then, every ρ-nonexpansive mapping T : C → C satisfies

i

T x − T i y ≤ b kx − yk a for all x, y ∈ C and i = 1, 2, . . . . The above observation leads us to an equivalent formulation of the constant γ0 (X):

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Classification of Lipschitz mappings

γ0 (X) = sup { k : any bounded, closed and convex set C ⊂ X furnished b with a metric ρ satisfying (7.6) with < k has a the f.p.p. for ρ-nonexpansive mappings} . We have already seen in example 3.6 that, in every infinite dimensional Banach space (X, k·k), there is uniformly lipschitzian mapping T : BX → BX , with a positive minimal displacement; that is, d(T ) = inf {kx − T xk : x ∈ BX } > 0. On the other hand, in the construction of our mapping T , we used a lipschitzian retraction R : BX → SX of the whole unit ball BX onto its boundary SX . Since every such retraction must have sufficiently large Lipschitz constant, this method does not lead us to nice upper bound for γ0 (X). An elegant evaluation from above for γ0 (H) in the case of Hilbert space H is due to Baillon [4]. Example 7.11 (Baillon, 1978-79) Consider the space `2 with the standard norm, ! 21 ∞ X 2 kxk = k(x1 , x2 , . . . )k = xi . i=1

Let T : `2 → `2 be the “right shift” operator; that is, T (x1 , x2 , . . . ) = (0, x1 , x2 , . . . ) . Consider the positive part of the unit ball B + = {x ∈ `2 : kxk ≤ 1 and xi ≥ 0 for i = 1, 2 . . . } . Define the mapping F : B + → B + as (    πkxk 1 cos πkxk e + sin Tx 1 2 kxk 2 Fx = e1

if x 6= 0 if x = 0,

where e1 = (1, 0, . . . , 0, . . . ). The verification that F is fixed point free as well as uniformly π2 -lipschitzian is left for the reader as an exercise. Consequently, for a Hilbert space H, we have √ π 2 ≤ γ0 (H) ≤ . 2 The exact value of γ0 (H) is still unknown! Another approach, using the uniform normal structure coefficient, was presented by Casini and Maluta [19] in 1985.

Nonexpansive mappings in Banach space

129

Theorem 7.13 (Casini and Maluta) Let C be a bounded, closed and convex subset of a Banach space X with N (X) > 1. If T : C → C is uniformly k-lipschitzian with p k < N (X), then T has a fixed point. One can observe that, for any Banach space X, we have κ(X) ≤ N (X). In the case of Hilbert space, the Lifshitz√Theorem provides better lower bound for γ0 (H) because κ(H) = N (H) = 2. On the other hand, there exist Banach spaces for which κ(X) = 1, whereas N (X) > 1. Hence, Casini-Maluta’s Theorem and Lifshitz Theorem are not comparable. In [8], Dom´ınguez Benavides improved evaluation for γ0 (X) obtained by Lifshitz. He introduced a new constant that involves both the normal structure coefficient and the Lifshitz constant. Theorem 7.14 (Dom´ınguez Benavides) Let C be a bounded, closed and convex subset of a Banach space X. Let T : C → C be uniformly k-lipschitzian mapping. If p 1 + 1 + 4N (X)(κ(X) − 1) , k< 2 then T has a fixed point. Since κ(X) ≤ N (X), we immediately conclude that p 1 + 1 + 4N (X)(κ(X) − 1) κ(X) ≤ ≤ N (X). 2 It is easy to verify that in the above inequality p 1 + 1 + 4N (X)(κ(X) − 1) κ(X) = 2 if and only if κ(X) = 1 or κ(X) = N (X). In particular, in the case of Hilbert space H, the Dom´ınguez Benavides Theorem √ and the Lifshitz Theorem give the same lower bound for γ0 (H), γ0 (H) ≥ 2. However, in [8], we can find a class of Banach spaces with 1 < κ(X) < N (X). Obviously, for such spaces, the theorem of Dom´ınguez Benavides is a strict improvement of Lifshitz’s result. Another approach to stability of f.p.p. uses only those equivalent metrics that are generated by equivalent norms. Recall that two norms k·k1 and k·k2 on a Banach space X are equivalent if there exist constants a, b > 0 such that inequality a kxk2 ≤ kxk1 ≤ b kxk2 holds for all x ∈ X. Then, every k·k2 -nonexpansive mapping T : C → C is uniformly k-lipschitzian with respect to the norm k·k1 with k ≤ ab . Under this setting, we define the stability constant γN (X) of (X, k·k1 ) as b < γ has the f.p.p. a for k·k2 -nonexpansive mappings} .

γN (X) = sup { γ : any Banach space (X, k·k2 ) with

130

Classification of Lipschitz mappings

It is clear that γN (X) ≥ γ0 (X). In [55], Lin proved that, for a Hilbert space H, s √ 5 + 13 π γN (H) ≥ = 2.07 · · · > ≥ γ0 (H). 2 2 Later, this result was slightly improved by Mazcu˜ n´an-Navarro [61], with a conclusion s √ 5 + 17 γN (H) ≥ = 2.135 . . . . 2 In particular, the above result implies that the space `2 renormed as in example 7.9; that is,   1 k|x|k = max √ kxk , max |xi | , i∈N 2 √ has the f.p.p. for k|·|k-nonexpansive mappings because in this case ab = 2.

7.5

The case of `1

The first result concerning lack of stability of f.p.p. for the space `1 was obtained by Goebel and Kuczumow [37] in 1978. Let {ai } be a bounded sequence of nonnegative real numbers. If {ei } is a standard basis in `1 , then ) ( ∞ ∞ X X + xi = 1 . S = conv {ei : i ∈ N} = x = xi ei : xi ≥ 0 and i=1

i=1

We shall modify the set S + by moving its vertexes ei as follows. Let f i = (1 + ai ) ei for i ∈ N. Put a = inf ai and N0 = {i : ai = a}. Let ( ) ∞ ∞ X X  i i C = conv f : i ∈ N = x = λi f : λi ≥ 0 and λi = 1 . i=1

i=1

It is easy to note that C is bounded, closed and convex. However, C is not weakly* compact in `1 considered as a dual of c0 because the origin is w∗ the weak* limit of the sequence of vertexes f i . The weak* closure C of conv (C ∪ {0}) has the form ( ) ∞ ∞ X X w∗ i C = x= µi f : µi ≥ 0 and µi ≤ 1 . i=1

i=1

Nonexpansive mappings in Banach space

131

Observe that a representation of x as a combination of f i is unique. Hence, w∗ for each x ∈ C , we can define a number δx as δx = 1 −

∞ X

µi .

i=1

Then, for each x ∈ C

w∗

, dist (x, C) = (1 + a) δx .

Indeed, since x + δx f i ∈ C, we have
 dist (x, C) ≤ inf x − x + δx f i : i = 1, 2, . . .  = inf δx f i : i = 1, 2, . . . = inf {(1 + ai ) δx : i = 1, 2, . . . } = (1 + a) δx . P∞ On the other hand, for any y = i=1 λi f i ∈ C, kx − yk =

∞ X

|µi − λi | (1 + ai )

i=1 ∞ X

λi − µi (1 + a) i=1 i=1 ! ∞ X = 1− µi (1 + a) ∞ X

≥

i=1

= (1 + a)δx . Moreover, the above calculations imply that kx − yk > (1 + a)δx , provided y ∈ C \ conv x + δx f i : i ∈ N0 . Consequently,  PC (x) := {z ∈ C : kz − xk = dist(x, C)} = conv x + δx f i : i ∈ N0 . Case 1. N0 is nonempty but finite. Let {xn } be an approximate fixed point sequence for a nonexpansive mapping T : C → C. Without loss w∗ of generality, we can assume that {xn } is weak* convergent to x ∈ C ⊂ ∗ `1 = c0 . Then,  A ({xn } , C) = conv x + δx f i : i ∈ N0 is compact, convex, and T -invariant; hence, T has a fixed point in A ({xn } , C) ⊂ C. Consequently, if N0 is nonempty but finite, then C has f.p.p. for nonexpansive mappings.

132

Classification of Lipschitz mappings ∞

Case 2. N0 is empty. Then, there exists a subsequence {aik }k=1 of {ai } such that ∀k∈N ∀i∈N (i < ik ⇒ ai > aik ) . ∞

It is clear that {aik }k=1 is strictly decreasing and lim aik = a. Put k→∞

N1 = {i : i ≤ i1 } and Nk = {i : ik−1 < i ≤ ik } Let T :x=

∞ X

λi f i →

i=1

for k = 2, 3, . . . . ∞ X

βk f ik+1 ,

k=1

where X

βk =

λi .

i∈Nk

Since

∞ X

βk =

∞ X X

λi =

k=1 i∈Nk

k=1

∞ X

λi = 1,

i=1

we conclude that T : C → C. The mapping T is not only nonexpansive P∞ P∞ but even contractive. Indeed, for any x = i=1 λi f i and y = i=1 µi f i , x 6= y, we have

∞ (" # )

X X

ik+1 (λi − µi ) f kT x − T yk =

≤

<

=

i∈Nk k=1 ∞ XX

|λi − µi | 1 + aik+1

k=1 i∈Nk ∞ X X




|λi − µi | (1 + ai )

k=1 i∈Nk ∞ X

|λi − µi | (1 + ai )

i=1

= kx − yk . P∞ P Suppose that, for some x = i=1 λi f i , T x = x. Then i∈Nk λi = λik+1 ∞ for each k ∈ N. This implies that λik ≤ λik+1 , and because {λik }k=1 is ∞ subsequence of {λi }i=1 ⊂ `1 , we conclude that λik = 0 for each k ∈ N. However, this contradicts the fact that 1=

∞ X X k=1 i∈Nk

λi =

∞ X k=1

λik+1 .

Nonexpansive mappings in Banach space

133

Thus, if N0 is empty, then C fails the f.p.p. for nonexpansive (and even contractive) mappings! Case 3. N0 is infinite. Let N0 = {i1 , i2 , . . . }. Let N1 = {i : i ≤ i1 } and Nk = {i : ik−1 < i ≤ ik }

for k = 2, 3, . . . .

We leave to the reader the verification of the fact that the mapping T : C → C defined as in the previous case is nonexpansive and fixed point free. Hence, if N0 is infinite, then C fails the f.p.p. for nonexpansive mappings! We can summarize the above cases as follows: Theorem 7.15 (Goebel and Kuczumow, 1978) Let C and N0 be as above. Then, the set C has the f.p.p. if and only if N0 is nonempty but finite. Subsequently, the stability of the fixed point property for nonexpansive self-mappings defined on bounded, closed and convex subsets of `1 has been widely studied by Dowling, Lennard, and Turett. The first step in their approach was the following: Theorem 7.16 (Strong Version of James’s Distortion Theorem, [22]) A Banach space X contains an isomorphic copy of `1 if and only if, for each ∞ sequence {n }n=1 in (0, 1) such that lim n = 0, there exists a sequence n→∞

∞

{xn }n=1 in X such that (1 − k )

∞ X n=k

∞ ∞

X

X

|tn | ≤ tn xn ≤ |tn |

n=k

(7.7)

n=k

∞

holds for all {tn }n=1 ∈ `1 and for all k = 1, 2, . . . . Using the above, Dowling, Lennard, and Turett [24] proved the following Theorem 7.17 Let X be a Banach space containing an isomorphic copy of `1 . Then, for each  > 0, there exists a bounded, closed and convex subset C of X and a uniformly lipschitzian mapping T : C → C that is fixed point free and ku (T ) < 1 + . ∞

Proof. Fix  > 0. Let {n }n=1 in (0, 1) be a null sequence such that 1 < 1 + . 1 − 1

134

Classification of Lipschitz mappings

By the Strong Version of James’s Distortion Theorem, there exists a sequence ∞ {xn }n=1 in X satisfying (7.7). Consider the set C ⊂ `1 defined as (∞ ) ∞ X X C= tn xn : tn ≥ 0 and tn = 1 . n=1

n=1

The set C is bounded, closed and convex subset of `1 . Define the mapping T : C → C by ! ∞ ∞ X X T tn xn = tn xn+1 . n=1

n=1

P∞

P∞

z = n=1 sn xn in C and for any k ∈ N,

∞

X

k



T y − T k z = (tn − sn ) xn+k

Then, for each y =

n=1 tn xn ,

n=1

≤

∞ X

|tn − sn |

n=1

1 ≤ 1 − 1

∞

X

(tn − sn ) xn

n=1

1 ≤ ky − zk . 1 − 1 Consequently, ku (T ) ≤

1 < 1 + . 1 − 1

Obviously, T is fixed point free. 

It means that `1 cannot be renormed to have the fixed point property for uniformly lipschitzian mappings having uniform Lipschitz constant slightly greater than 1. Hence, γN (`1 ) = γ0 (`1 ) = 1. Immediately, one can ask about renorming of `1 having the f.p.p. for nonexpansive mappings. To proceed, we shall need the concept of asymptotically isometric copy of `1 , which was initiated and developed by Dowling, Lennard, and Turett (see [49]). A Banach space X is said to contain an asymptotically isometric copy of ∞ ∞ `1 if, for every null sequence {n }n=1 in (0, 1), there exists a sequence {xn }n=1 in X such that

∞

∞ ∞

X

X X

(1 − n ) |tn | ≤ tn xn ≤ |tn | ,

n=1

∞

for all {tn }n=1 ∈ `1 .

n=1

n=1

Nonexpansive mappings in Banach space

135

Theorem 7.18 (see [23] or [49]) If a Banach space X contains an asymptotically isometric copy of `1 , then X fails the f.p.p. for nonexpansive mappings. ∞

Proof. Let {λi }i=1 be a strictly decreasing sequence in (1, ∞) converging to ∞ 1. Then, there exists a sequence {i }i=1 in (0, 1) such that lim i = 0 and i→∞

λi+1 < (1 − i )λi ∞

for i = 1, 2, . . . . By assumption, there exists a sequence {xi }i=1 in X such that

∞

∞ ∞

X

X X

(1 − i ) |ti | ≤ t i xi ≤ |ti |

i=1

i=1

i=1

∞

for every {ti }i=1 ∈ `1 . Put yi = λi xi for i = 1, 2, . . . . Define the set C ⊂ X as (∞ ) ∞ X X ∞ C= ti yi : {ti }i=1 ∈ `1 , ti ≥ 0 and ti = 1 . i=1

i=1

We leave to the reader the verification that C is a bounded, closed and convex subset of X. Now, consider the mapping T : C → C defined by ! ∞ ∞ X X T ti yi = ti yi+1 . i=1

i=1

It is easy to check that is fixed point free. PT P∞ We shall prove that T is nonex∞ pansive. For any u = i=1 ti yi and v = i=1 si yi in C, we have

∞

X

(ti − si ) yi+1 kT u − T vk =

i=1

≤ ≤ ≤

∞ X i=1 ∞ X i=1 ∞ X

(ti − si ) kyi+1 k (ti − si ) λi+1 (ti − si ) λi (1 − i )

i=1

∞

X

≤

i=1



(ti − si )λi xi

= ku − vk . 

136

Classification of Lipschitz mappings

Theorem 7.19 (see [49]) Let Y be a closed infinite-dimensional subspace P∞ of `1 furnished with the standard norm; that is, kxk = i=1 |xi | for x = (x1 , x2 , . . . ) ∈ `1 . Then, Y contains an asymptotically isometric copy of `1 . Theorem 7.20 (see [23] or [49]) If X is a nonreflexive subspace of L1 [0, 1], then X contains an asymptotically isometric copy of `1 . Hence, each nonreflexive subspace of L1 [0, 1] fails the f.p.p. for nonexpansive mappings. On the other hand, Maurey [60] proved that each reflexive subspace of L1 [0, 1] has the fixed point property. Consequently, we obtain the following Theorem 7.21 Let X be a subspace of L1 [0, 1]. The following conditions are equivalent: (i) X is reflexive. (ii) X has the fixed point property for nonexpansive mappings. There exist renormings of `1 that fail to contain asymptotically isometric copies of `1 . Example 7.12 [49] Consider the space `1 with a new norm defined by ( ∞ ) X kxkDLT = k(x1 , x2 , . . . )kDLT = max γi |xk | : i = 1, 2, . . . , k=i

where {γi } is an arbitrary sequence in (0, 1) strictly increasing to 1. It is easy to note that, for any x ∈ `1 γ1 kxk`1 ≤ kxkDLT ≤ kxk`1 . Dowling, Lennard, and Turett proved that (`1 , k·kDLT ) does not contain an asymptotically isometric copy of `1 . Nevertheless, it was not clear whether there is k·kDLT -nonexpansive mapping that is fixed point free. To see it, consider, for instance, the well-known set ( ) ∞ X + S = {xi } : xi ≥ 0 and xi = 1 . i=1

Obviously S + is convex, and, since k·kDLT is equivalent to the classical norm, it is also bounded and closed. Let T be the classical “right shift” self-mapping defined on S + , T (x1 , x2 , . . . ) = (0, x1 , x2 , . . . ). The mapping T is fixed point free. However, T is not nonexpansive with respect 8i to the new norm k·kDLT with γi = 1+8 i for i = 1, 2, . . . . We shall find the n o consecutive Lipschitz constants for iterates T n of T . Since the sequence γi+1 γi

Nonexpansive mappings in Banach space

137

is strictly decreasing, for every x = (x1 , x2 , . . . ), y = (y1 , y2 , . . . ) in S + , and for all n ∈ N, we have

n

n

kT x − T ykDLT = max γn+i i∈N

∞ X

|xk − yk |

k=i ∞

= max i∈N

X γn+i−1 γi+1 γn+i · · ... · · γi · |xk − yk | γn+i−1 γn+i−2 γi k=i

≤ max i∈N

γn γ2 γn+1 · · ... · · γi · γn γn−1 γ1

∞ X

|xk − yk |

k=i

∞ X γn+1 = · max γi |xk − yk | i∈N γ1 k=i γn+1 = kx − ykDLT . γ1

Further, kT n e1 − T n e2 kDLT = 2γn+1 and ke1 − e2 kDLT = 2γ1 , hence, kT n e1 − T n e2 kDLT = Consequently, kk·kDLT (T n ) =

γn+1 γ1

−→

γn+1 ke1 − e2 kDLT . γ1 1 γ1

=

9 8

as n −→ ∞.

In 2008, Lin [56] proved that `1 furnished with the above-mentioned norm has the fixed point property for nonexpansive mappings! Theorem 7.22 (Lin, 2008) The space `1 furnished with an equivalent norm k·kDLT defined as ( ∞ ) X kxkDLT = k(x1 , x2 , . . . )kDLT = max γi |xk | : i = 1, 2, . . . , k=i

with γi =

8i 1+8i

has the fixed point property for k·kDLT -nonexpansive mappings.

It was also the first example of nonreflexive Banach space with the fixed point property! Reassuming bounded, closed and convex subsets of (`1 , k·kDLT ) with γi = 8i 1+8i have the fixed point property for nonexpansive mappings but for any  > 0, there exists a bounded, closed and convex subset C of (`1 , k·kDLT ) and (1 + )-uniformly lipschitzian mapping T : C → C, which is fixed point free.

Chapter 8 Mean nonexpansive mappings

Recall that, for any convex subset C of a Banach space X with diam(C) > 0 and for any multi-index α = (α1 , . . . , αn ), there is an α-nonexpansive mapping T : C → C that is not nonexpansive. We also know that a mapping T is αnonexpansive if and only if it is nonexpansive with respect to the metric d defined by   n n X X

 d(x, y) = αi  T j−1 x − T j−1 y . j=1

i=j

Moreover, every time we prove the fixed point theorem for nonexpansive mappings, we get a fixed point theorem for some class of α-nonexpansive mappings. Consequently, the fixed point property for nonexpansive mappings is always stable with respect to special changes of metrics. In this chapter, we shall develop the idea concerning this new kind of stability.

8.1

Some new results of stability type

As usual, C denotes a bounded closed and convex subset of a Banach space (X, k·k) and α = (α1 , . . . , αn ) a multi-index. Fix p ≥ 1. Under the above settings, we say that a mapping T : C → C is (α, p)nonexpansive if for all x, y ∈ C n X

p p αi T i x − T i y ≤ kx − yk .

(8.1)

i=1

Equivalently, T : C → C is (α, p)-nonexpansive if and only if it is nonexpansive with respect to the metric d defined for all x, y ∈ C as     p1 n n X X

j−1

p  d(x, y) =  αi  T x − T j−1 y  . j=1

(8.2)

i=j

It is easy to see that d is equivalent to the metric ρ on C generated by the

139

140

Classification of Lipschitz mappings

norm, ρ(x, y) = kx − yk for all x, y ∈ C. Further, for each (α, p)-nonexpansive mapping T : C → C we define the mapping Tα : C → C by putting Tα x =

n X

αi T i x

for any x ∈ C.

i=1

Since every (α, p)-nonexpansive mapping T is (α, 1)-nonexpansive, we conclude that Tα is nonexpansive. Thus, d(Tα ) = 0. Theorem 8.1 Let C be a bounded, closed and convex subset of a Banach space X. If T : C → C is (α, p)-nonexpansive with α = (α1 , α2 ) and p ≥ 1 such that α2p + α2 ≤ 1, then d(T ) = 0. Proof. If p = 1, then the conclusion follows from theorem 3.2. Suppose p > 1. Fix  > 0. Since Tα is nonexpansive, there exists a point x ∈ C, x = x(), such that

kx − Tα xk = x − α1 T x − α2 T 2 x < . By definition of T , we have

p

p α1 T x − T 2 x + α2 T 2 x − T 3 x ≤ ≤ ≤ = Let q > 1 be such that

1 p

+

1 q

p

kx − T xk p (kx − Tα xk + kTα x − T xk) p (kTα x − T xk + )

2


p α2 T x − T x +  .

= 1. Using H¨older inequality, we obtain

 p


p 1 1 α2 T 2 x − T x +  = α2 T 2 x − T x +  q ·  p    p1 p

p 1 p 2 q

≤ (1 + ) · α2 T x − T x +   

p = (1 + )p−1 α2p T 2 x − T x +  i

p h

p p−1 = α2p T 2 x − T x + (1 + ) − 1 α2p T 2 x − T x +(1 + )p−1 i

p h p−1 ≤ α2p T 2 x − T x + (1 + ) − 1 (diam(C))p +(1 + )p−1 . Thus, h i

p

p p−1 (α1 − α2p ) T x − T 2 x + α2 T 2 x − T 3 x ≤ (1 + ) − 1 (diam(C))p +(1 + )p−1 .

Mean nonexpansive mappings

141

The right-hand side of the above inequality can be made arbitrary close to zero for small . By assumption, α1 − α2p = 1 − α2 − α2p ≥ 0. Thus, d(T ) = 0. 

If C has the f.p.p. for nonexpansive mappings, then we can repeat arguments from the previous theorem with  = 0 to get the following: Theorem 8.2 Let C be a bounded, closed and convex subset of a Banach space X. If C has the f.p.p. for nonexpansive mappings, then C has the f.p.p. for (α, p)-nonexpansive mappings with α = (α1 , α2 ) and p ≥ 1 such that α2p + α2 ≤ 1. In other words, if for some α = (α1 , α2 ) and p ≥ 1 such that α2p + α2 ≤ 1, a mapping T : C → C is nonexpansive with respect to the metric d defined for all x, y ∈ C by p

p

1

d(x, y) = (kx − yk + α2 kT x − T yk ) p , then T has a fixed point. Moreover, if α2p + α2 < 1, then Fix(T ) = Fix(Tα ). Proof. Since Tα is nonexpansive, there is z ∈ C such that Tα z = α1 T z + α2 T 2 z = z.

(8.3)

By definition of T , we have

p

p

p p α1 T z − T 2 z + α2 T 2 z − T 3 z ≤ kz − T zk = kTα z − T zk

p = αp T 2 z − T z . 2

Thus,

p

p (α1 − α2p ) T z − T 2 z + α2 T 2 z − T 3 z ≤ 0. If α1 − α2p = 1 − α2 − α2p > 0, then T z = T 2 z and from (8.3) we get T z = z. Thus, Fix(Tα ) ⊂ Fix(T ). Inclusion in opposite direction is obvious. If α1 − α2p = 0, then T 2 z is a fixed point of T . 

For the convenience of the reader, we illustrate the set of all pairs (p, α2 ) for which the above theorems hold using figure 8.1. Let us observe that, for any α2 ∈ (0, 1), there exists p ≥ 1 such that C has the f.p.p. for all mappings in L((α1 , α2 ), p). Nevertheless, the above theorem

142

Classification of Lipschitz mappings

does not guarantee that there exists p ≥ 1 such that for every α2 ∈ (0, 1), C has the f.p.p. for all mappings in L((α1 , α2 ), p). α2 1 p 1 α 2 + α2 =

1 2

0

1

2

3

4

p

FIGURE 8.1 Similar situation is observed in the general case of ((α1 , . . . , αn ), p)nonexpansive mappings as seen in the following theorems: Theorem 8.3 If T : C → C is (α, p)-nonexpansive with n ≥ 2, α = (α1 , . . . , αn ) and p ≥ 1 such that n−1

n−1

1

(1 − α1 )(1 − α1 p ) ≤ α1 p (1 − α1p ), then d(T ) = 0. Proof. If p = 1, then the conclusion follows from theorem 3.4. Suppose p > 1. For every  > 0, there exists x ∈ C such that kx − Tα xk ≤ . Then, !1/p n X

i

i+1 p

αi T x − T x ≤ kx − T xk ≤ kx − Tα xk + kTα x − T xk i=1

n n

X

X

≤ αi T i x − αi T x + 

i=1 i=1

n

X


αi T i x − T x +  =

i=2

≤

n X

αi T i x − T x + .

i=2

Since k(T j ) ≤



1 1/p α1

j , we get



X

i

i


j−1 j

αi T x − T x = αi T x−T x

j=2

Mean nonexpansive mappings ≤ αi

143

i X

j−1

T x − T j x j=2

 ≤ αi

!j−2 

1  T x − T 2 x .

i X

1/p

j=2

α1

Thus, n X



 n X i X

i αi T x − T x ≤ 

αi

(j−2)/p i=2 j=2 α1

i=2

 T x − T 2 x

    n n X  X

1  αi  (j−2)/p  T x − T 2 x . =   α j=2

Since

i=j

n X

1

αi ≤ 1 − α1

i=j

for j = 2, . . . , n, we obtain n X

 n X

i αi T x − T x ≤ (1 − α1 ) 

 1

(j−2)/p j=2 α1

i=2

= (1 − α1 ) ·

= (1 − α1 ) ·

1 (n−2)/p

α1

 T x − T 2 x

  n X

(j−2)/p 

T x − T 2 x α1 · j=2 (n−1)/p

1 (n−2)/p α1

·

1 − α1 1−



T x − T 2 x .

1/p α1

If we define A as A = (1 − α1 ) ·

1 (n−2)/p

α1

(n−1)/p

·

1 − α1

1/p

1 − α1

,

then we can write n X

p αi T i x − T i+1 x

!1/p

≤ A T x − T 2 x + .

i=1

Thus, n X i=1

p


p αi T i x − T i+1 x ≤ A T x − T 2 x +  .

(8.4)

144

Classification of Lipschitz mappings

We can select q > 1 such that


p A T x − T 2 x +  = ≤ = =

1 p

1 q

+

= 1. Using H¨older inequality, we obtain

 p

1 1 A T x − T 2 x +  p ·  q  p  p1

1 2 p p q A T x − T x +  · (1 + )  

p p−1 Ap T x − T 2 x +  (1 + ) i

p h

p p−1 Ap T x − T 2 x + (1 + ) − 1 Ap T x − T 2 x p−1

+ (1 + ) i

p h p−1 p − 1 Ap (diam(C)) ≤ Ap T x − T 2 x + (1 + ) p−1

+ (1 + )

.

Thus,

p

(α1 − Ap ) T x − T 2 x n h i X

p p−1 p + αi T i x − T i+1 x ≤ (1 + ) − 1 Ap (diam(C)) i=2 p−1

+ (1 + )

.

By assumption α1 − Ap ≥ 0, this implies that i

p h p−1 p p−1 αn T n x − T n+1 x ≤ (1 + ) − 1 Ap (diam(C)) +  (1 + ) . Since αn > 0 and the right-hand side of the above inequality converges to 0 if  → 0, we finally obtain d(T ) = 0. 

If C has the f.p.p. for nonexpansive mappings, then we can repeat the proof with  = 0 to obtain the following: Theorem 8.4 If C has the f.p.p. for nonexpansive mappings, then C has the f.p.p. for (α, p)-nonexpansive mappings with n ≥ 2, α = (α1 , . . . , αn ) and p ≥ 1 such that n−1

n−1

1

(1 − α1 )(1 − α1 p ) ≤ α1 p (1 − α1p ). Equivalently, if for some α and p satisfying the above condition, a mapping T : C → C is nonexpansive with respect to the metric d given by formula (8.2), then T has a fixed point. Moreover, if n−1

n−1

1

(1 − α1 )(1 − α1 p ) < α1 p (1 − α1p ), then Fix(T ) = Fix(Tα ).

Mean nonexpansive mappings

145

Proof. If p = 1, then the first conclusion follows from theorem 3.5. Suppose p > 1. Since Tα : C → C is nonexpansive, there exists x ∈ C such that Tα x = α1 T x + α2 T 2 x + · · · + αn T n x = x.

(8.5)

Now inequality (8.4) takes the form n X

p

p αi T i x − T i+1 x ≤ Ap T x − T 2 x .

i=1

Observe that the above inequality also holds for p = 1. Thus, n

p X

p (α1 − Ap ) T x − T 2 x + αi T i x − T i+1 x ≤ 0. i=2

Since α1 − Ap ≥ 0 and αn > 0, this implies that T n x is fixed under T . If α1 − Ap > 0, then T x = T 2 x = · · · = T n x. From (8.5), we get that x = T x. Thus, Fix(Tα ) ⊂ Fix(T ). Inclusion in opposite direction is obvious. 

The reader can easily verify that for p = 1 the above theorem coincides with the theorem 3.5. In view of theorems 8.4, 7.5, and 7.6, we obtain the following: Theorem 8.5 Let (X, k·k) be a Banach space that is uniformly nonsquare or uniformly noncreasy. Then, any bounded, closed and convex subset C has the f.p.p. for (α, p)-nonexpansive mappings with n ≥ 2, α = (α1 , . . . , αn ), and p ≥ 1 such that n−1

n−1

1

(1 − α1 )(1 − α1 p ) ≤ α1 p (1 − α1p ). Moreover, if n−1

n−1

1

(1 − α1 )(1 − α1 p ) < α1 p (1 − α1p ), then Fix(T ) = Fix(Tα ) 6= ∅. In particular, for n = 2, the above theorem states: Theorem 8.6 Let (X, k·k) be a Banach space that is uniformly nonsquare or uniformly noncreasy. Then, any bounded, closed and convex subset C has the f.p.p. for (α, p)-nonexpansive mappings with α = (α1 , α2 ) and p ≥ 1 such that α2p + α2 ≤ 1. Moreover, if α2p + α2 < 1, then Fix(T ) = Fix(Tα ) 6= ∅. For the space `1 , we have 0 (`1 ) = 2. Nevertheless, weak* compact and convex sets in `1 = c∗0 have the f.p.p. for nonexpansive mappings.

146

Classification of Lipschitz mappings

Theorem 8.7 Weak* compact and convex sets in `1 (considered as a dual of c0 ) have the f.p.p. for (α, p)-nonexpansive mappings with n ≥ 2, α = (α1 , . . . , αn ), and p ≥ 1 such that n−1

n−1

1

(1 − α1 )(1 − α1 p ) ≤ α1 p (1 − α1p ). Moreover, if n−1

n−1

1

(1 − α1 )(1 − α1 p ) < α1 p (1 − α1p ), then Fix(T ) = Fix(Tα ) 6= ∅. For n = 2, the above theorem states: Theorem 8.8 Weak* compact and convex sets in `1 = c∗0 have the f.p.p. for (α, p)-nonexpansive mappings with α = (α1 , α2 ) and p ≥ 1 such that α2p + α2 ≤ 1. Furthermore, if α2p + α2 < 1, then Fix(T ) = Fix(Tα ) 6= ∅. In view of Lin’s Theorem, we obtain the following result: Theorem 8.9 Consider the space `1 furnished with the norm k·kDLT defined as ( ) ∞ X kxkDLT = k(x1 , x2 , . . . )kDLT = max γn |xk | : n = 1, 2, . . . , k=n n

8 where γn = 1+8 n for n = 1, 2, . . . . Let C be a bounded, closed and convex subset of `1 and T : C → C be a mapping. If for some α = (α1 , . . . , αn ) and p ≥ 1 satisfying n−1

n−1

1

(1 − α1 )(1 − α1 p ) ≤ α1 p (1 − α1p ), the mapping T is nonexpansive with respect to the metric d given by  p1    n n X X

j−1 p  d(x, y) =  αi  T x − T j−1 y DLT  , j=1

i=j

then T has a fixed point. Another approach uses the constant k∞ (T ). As was observed in corollary 5.2 and theorem 5.4, for every (α, p)-nonexpansive mapping T with α = (α1 , α2 ) and p ≥ 1, 1/p

k∞ (T ) ≤ (1 + α2 )

,

and, in a general case of ((α1 , . . . , αn ), p)-nonexpansive mappings,   1/p n n X X 1/p  k∞ (T ) ≤  αi  = (α1 + 2α2 + 3α3 + · · · + nαn ) j=1

i=j

= (1 + α2 + 2α3 + · · · + (n − 1)αn ) Using the above inequality and the Lifshitz Theorem, we conclude

1/p

.

Mean nonexpansive mappings

147

Theorem 8.10 (P´ erez Garc´ıa and Piasecki, [65]) Let (M, ρ) be a bounded and complete metric space with κ(M ) > 1. If T : M → M is (α, p)nonexpansive with p ≥ 1 and α = (α1 , . . . , αn ) such that (1 + α2 + 2α3 + · · · + (n − 1)αn )

1/p

< κ(M ),

then T has a fixed point in M . Now we are ready to prove the following: Theorem 8.11 (P´ erez Garc´ıa and Piasecki, [65]) Let X be a Banach space with 0 (X) < 1. If T : C → C is (α, p)-nonexpansive with n ≥ 2, p ≥ 1 and α = (α1 , . . . , αn ) such that (1 + α2 + 2α3 + · · · + (n − 1)αn )

1/p

< κ(X),

then T has a fixed point in C. Proof. It is a consequence of theorems 7.12, 5.4, and 7.11. 

Hence, for n = 2, we have Theorem 8.12 Let X be a Banach space with 0 (X) < 1. If T : C → C is (α, p)-nonexpansive with α = (α1 , α2 ) and p ≥ 1 such that 1/p

(1 + α2 )

< κ(X),

then T has a fixed point in C. In particular, since κ(H) =

√

2 for a Hilbert space H, theorem 8.11 states:

Theorem 8.13 Let H be a Hilbert space. If T : C → C is (α, p)nonexpansive, with n ≥ 2, p ≥ 1, and α = (α1 , . . . , αn ) such that 1 + α2 + 2α3 + · · · + (n − 1)αn < 2p/2 , then T has a fixed point in C. For n = 2, we get Theorem 8.14 Let H be a Hilbert space. If T : C → C is (α, p)nonexpansive, with p ≥ 1 and α = (α1 , α2 ) such that α2 < 2p/2 − 1, then T has a fixed point in C. For mappings in the class L(α, 2, 1), the theorem 8.13 takes the form

148

Classification of Lipschitz mappings

Theorem 8.15 Let H be a Hilbert space. If T : C → C is (α, 2)nonexpansive, with n ≥ 2 and α = (α1 , . . . , αn ) such that α2 + 2α3 + 3α4 + · · · + (n − 1)αn < 1, then T has a fixed point in C. Let us note that, for n = 2, the last theorem implies: Theorem 8.16 Let C be a bounded, closed and convex subset of a Hilbert space H. Then, for any multi-index α = (α1 , α2 ) of length n = 2, C has the f.p.p. for all (α, 2)-nonexpansive mappings. In other words, if for some α = (α1 , α2 ), a mapping T : C → C is nonexpansive with respect to the metric d given by 1  2 2 2 d(x, y) = kx − yk + α2 kT x − T yk , then T has a fixed point. In general, if 0 (X) < 1, then for any n ≥ 2, there exists sufficiently large p such that for any multi-index α of length n, C has the f.p.p. for all mappings in the class L(α, p, 1). This fact is a consequence of the following theorem: Theorem 8.17 (P´ erez Garc´ıa and Piasecki, [65]) Let C be a bounded, closed and convex subset of a Banach space X with 0 (X) < 1. Then, for any ln(n) multi-index α = (α1 , . . . , αn ) of length n ≥ 2 and p = ln(κ(X)) , C has the f.p.p. for the class of (α, p)-nonexpansive mappings. In other words, if for some α = (α1 , . . . , αn ), a mapping T : C → C is nonexpansive with respect to the metric d given by   p1   n n X X

j−1 p  d(x, y) =  αi  T x − T j−1 y  , j=1

with p =

ln(n) ln(κ(X)) ,

i=j

then T has a fixed point.

Proof. It is enough to observe that, for any α = (α1 , . . . , αn ), we have n X

αi = 1

and

i=1

n X

αi < 1

for j = 2, . . . , n.

i=j

Thus, 

n X

 j=1

 1/p n X  αi  < n1/p = κ(X). i=j 

Mean nonexpansive mappings

149

Consequently, in case of Hilbert space, we have Theorem 8.18 If C is a bounded, closed and convex subset of a Hilbert space ln(n) H, then for any multi-index α = (α1 , . . . , αn ) of length n ≥ 2 and p = 2ln(2) , C has the f.p.p. for the class of (α, p)-nonexpansive mappings. In other words, if T : C → C is a mapping such that for some α = (α1 , . . . , αn ) it is nonexpansive with respect to the metric d defined as  2ln(2)    ln(n) n n 2 ln(n) X X

j−1

j−1 ln(2) 

   d(x, y) = , αi T x−T y j=1

i=j

then T has a fixed point. Now, we shall compare the approach that uses the constants k∞ (T ) and κ(X) with the previous one, which involved the mapping Tα . For technical reasons, we consider the case of (α, p)-nonexpansive mappings with a multiindex α of length n = 2. Hence, we shall compare theorem 8.12 with theorem 8.6. If 0 (X) ≥ 1, then the situation is clear. Only theorem 8.6 provides some new fixed point results for the class of mean nonexpansive mappings. However, if 0 (X) < 1, then the situation is much more complicated. For clarity of arguments, we consider the case of Hilbert space H. Then, theorem 8.12 takes the form of theorem 8.14. We have already noticed that for p = 2 theorem 8.14 ensures the f.p.p. for the whole class of (α, 2)-nonexpansive mappings, whereas, using theorem 8.6, √ we obtain the f.p.p. only for (α, 2)-nonexpansive mappings with α2 ≤ 5−1 2 . On the other hand, for p = 1, theorem 8.14 ensures the f.p.p. for all α√ nonexpansive mappings with α2 < 2 − 1, whereas using theorem 8.6, we get the f.p.p. for all α-nonexpansive mappings with α2 ≤ 12 , which is a better estimate. To display more precisely the above remark, we use figure 8.2. By E = (p0 , α20 ) = (1.240..., 0.537...), we denote the point at which graphs of functions α2p + α2 = 1 (see theorem 8.6) and α2 = 2p/2 − 1 (see theorem 8.14) intersect. It is easy to observe that, for p ≤ p0 , theorem 8.6 gives better estimate for α2 than theorem 8.14, whereas for p > p0 the situation is contrary. We summarize the above observation in the following: Theorem 8.19 Let C be a bounded, closed and convex subset of a Hilbert space H. If T : C → C is (α, p)-nonexpansive, with α = (α1 , α2 ) and p ≥ 1 such that p ∈ [1, p0 ] and α2p + α2 ≤ 1 or

n p o p > p0 and α2 < min 2 2 − 1, 1 ,

then T has a fixed point in C.

150

Classification of Lipschitz mappings

α2

1

p

α2 1 2

0

=

22

−

1

E

α20 √ 2−1

1

p α 2 + α2 = 1

p0

p

2

FIGURE 8.2

8.2

Sequential approximation of fixed points

Let C be a bounded, closed and convex subset of uniformly convex Banach space X. Let α = (α1 , . . . , αn ) and p be such that (see theorem 8.4): n−1

n−1

1

(1 − α1 )(1 − α1 p ) < α1 p (1 − α1p ). Then, in view of theorem 8.4 and the proof of theorem 7.1, we can give a procedure to determine the fixed point for (α, p)-nonexpansive mapping T : C → C (α and p as above). 1. Construct an approximate fixed point sequence {xm } for mapping Tα 1 for m = 1, 2, . . . . as in the proof of lemma 3.1 with m = m 2. Determine the asymptotic center of the sequence {xm }, which consists of exactly one point z. Then, z is a fixed point of T . Actually, under the above assumption on α and p, the sets of approximate fixed point sequences for mappings T and Tα coincide, as seen in the following theorems:

Mean nonexpansive mappings

151

Theorem 8.20 Let C be a nonempty and convex subset of a Banach space X. If T : C → C is (α, p)-nonexpansive with α = (α1 , α2 ) and p ≥ 1 such that α2p + α2 < 1, then for each sequence {xn } ⊂ C, lim kxn − T xn k = 0 ⇐⇒ lim kxn − Tα xn k = 0.

n→∞

n→∞

Proof. Suppose lim kxn − T xn k = 0. Then, n→∞

kTα xn − xn k ≤ α1 kT xn − xn k + α2 T 2 xn − xn

≤ α1 kT xn − xn k + α2 T 2 xn − T xn + α2 kT xn − xn k

= kT xn − xn k + α2 T 2 xn − T xn α2 ≤ kT xn − xn k + 1/p kT xn − xn k α1 ! α2 = 1 + 1/p kT xn − xn k . α1 Thus, lim kTα xn − xn k = 0. n→∞

Conversely, let lim kTα xn − xn k = 0. Then, n→∞

kxn − T xn k ≤ kxn − Tα xn k + kTα xn − T xn k

= kxn − Tα xn k + α2 T 2 xn − T xn α2 ≤ kxn − Tα xn k + 1/p kT xn − xn k . α1 1/p

Shifting the last term to the left-hand side and multiplying both sides by α1 we obtain   1/p 1/p α1 − α2 kT xn − xn k ≤ α1 kxn − Tα xn k . 1/p

It is clear that our assumption α2p + α2 < 1 is equivalent to α1 Hence, we conclude that lim kT xn − xn k = 0.

− α2 > 0.

n→∞



In the general case of ((α1 , . . . , αn ), p)-nonexpansive mappings, we have the following: Theorem 8.21 Let C be a nonempty and convex subset of a Banach space X. If T : C → C is (α, p)-nonexpansive with n ≥ 2, α = (α1 , . . . , αn ), and p ≥ 1 such that n−1

n−1

1

(1 − α1 )(1 − α1 p ) < α1 p (1 − α1p ), then for any sequence {xm } in C, lim kxm − T xm k = 0 ⇐⇒ lim kxm − Tα xm k = 0.

m→∞

m→∞

152

Classification of Lipschitz mappings

Proof. Suppose lim kxm − T xm k = 0. Then, m→∞

n

n

X




X i kxm − Tα xm k = αi xm − T xm ≤ αi xm − T i xm .

i=1

i=1

j/p

Since k(T j ) ≤ 1/α1 , for i = 1, . . . , n, we obtain

X

X i−1

i−1 j

xm − T i xm = (T j xm − T j+1 xm ) ≤

T xm − T j+1 xm

j=0

j=0   i−1 X j/p  ≤ 1/α1  kxm − T xm k . j=0

Thus,    n i−1 X X α i  kxm − T xm k .  kxm − Tα xm k ≤  j/p α i=1 j=0 1 This implies that lim kxm − Tα xm k = 0. m→∞

Inversely, suppose lim kxm − Tα xm k = 0. Then, m→∞

kxm − T xm k ≤ kxm − Tα xm k + kTα xm − T xm k

n

X


i = kxm − Tα xm k + αi T xm − T xm

i=2

≤ kxm − Tα xm k +

n X

αi T i xm − T xm .

i=2 j/p

Since k(T j ) ≤ 1/α1 , for i = 2, . . . , n, we have

i−1

i−1

i

X j

X j

T xm − T xm = (T xm − T j+1 xm ) ≤

T xm − T j+1 xm

j=1

j=1   i−1 X 1  ≤ kxm − T xm k . j/p j=1 α1 Thus,    n i−1 X X

i

αi   αi T xm − T xm ≤  kxm − T xm k . j/p i=2 i=2 j=1 α1

n X

Mean nonexpansive mappings

153

Adding terms with the same denominator, we get     n n−1 i−1 n X X X X α α i i = .   j/p j/p α α i=2 j=1 j=1 1 i=j+1 1 Since

n X

αi ≤ 1 − α1 ,

i=j+1

we obtain    n−1 n−1 j−1 X X α 1 − α (1 − α ) i  1 1  ≤ = (n−1)/p  α1 p  j/p j/p α α α j=1 i=j+1 1 j=1 j=1 1 1     −1 n−1 n−1 1 = (1 − α1 ) 1 − α1 p α1 p 1 − α1p .

n−1 X



n X

Thus, kxm − T xm k ≤ kxm − Tα xm k    n−1  −1 n−1 1 + (1 − α1 ) 1 − α1 p α1 p 1 − α1p kxm − T xm k . By assumption, 

n−1 p

(1 − α1 ) 1 − α1

  n−1  −1 1 p p α1 1 − α1 < 1.

Hence, lim kxm − T xm k = 0. m→∞



Actually, evaluations based only on α1 are not exact. In the next section, we shall illustrate this fact for a multi-index α of length n = 3 and p = 1.

8.3

The case of n = 3

Let C be a bounded, closed and convex subset of a Banach space (X, k·k). Let T : C → C be α-nonexpansive with α = (α1 , α2 , α3 ); that is, the inequality



α1 kT x − T yk + α2 T 2 x − T 2 y + α3 T 3 x − T 3 y ≤ kx − yk (8.6) holds for all x, y ∈ C. Then, the mapping Tα : C → C has the form Tα x = α1 T x + α2 T 2 x + α3 T 3 x.

154

Classification of Lipschitz mappings α2

1

1 2

1 4

0

√ 2 2

1 2

1

α1

FIGURE 8.3 In [33], Goebel and Jap´on Pineda proved the following: Theorem 8.22 Let C be a bounded, closed and convex subset of a Banach space X having the f.p.p. for nonexpansive mappings. Suppose that a mapping T : C → C is α-nonexpansive with a multi-index α = (α1 , α2 , α3 ) such that " √ ! 2 1 1 1 α1 ∈ , and α2 ≥ − α1 2 2 2 2 or

√ 2 . 2 Then, T has a fixed point in C (look at the figure 8.3). α1 ≥

Proof. It is enough to repeat the proof of the theorem 3.7, with  = 0. 

Mean nonexpansive mappings

155

Looking up the proof of this theorem, we conclude that the evaluation for coefficients of α = (α1 , α2 , α3 ), which guarantees the f.p.p. can be improved as seen below. Theorem 8.23 Let C be a bounded, closed and convex subset of a Banach space X. If T : C → C is α-nonexpansive with α = (α1 , α2 , α3 ) such that α1 ≥ 21 and α2 ≥ 12 − α12 , then d(T ) = 0. Proof. Fix  > 0. Since Tα is nonexpansive, there is a point x ∈ C such that kx − Tα xk < α3 . Thus, using definition of T , we can write 3 X

αi T i x − T i+1 x ≤ kx − T xk ≤ kx − Tα xk + kTα x − T xk

i=1

≤ =

kTα x − T xk + α3 





α2 T 2 x − T x + α3 T 3 x − T x + α3 



≤ α2 T 2 x − T x + α3 T 3 x − T x + α3 



≤ α2 T 2 x − T x + α3 T 3 x − T 2 x

+α3 T 2 x − T x + α3 



= (α2 + α3 ) T 2 x − T x + α3 T 3 x − T 2 x + α3 . Leaving only the last term on the left hand-side and shifting all others to the right we obtain



α3 T 3 x − T 4 x ≤ (α2 + α3 − α1 ) T 2 x − T x

+ (α3 − α2 ) T 3 x − T 2 x + α3 



≤ (1 − 2α1 ) T 2 x − T x + (α3 − α2 ) T 3 x − T 2 x + α3 . If α1 ≥ 1/2 and α2 ≥ α3 , then

3

T x − T 4 x ≤ . Suppose α2 < α3 . Since k(T ) ≤

1 α1 ,

we get



α3 − α2 2

T x − T x + α3  α3 T 3 x − T 4 x ≤ (1 − 2α1 ) T 2 x − T x + α1 2

1 − 2α2 − 2α1 2 = T x − T x + α3 . α1

If 1 − 2α2 − 2α12 ≤ 0, then T 3 x − T 4 x ≤ . Thus, d(T ) = 0. 

156

Classification of Lipschitz mappings

To compare the theorem 8.22 with the theorem 8.23 we present figure 8.4.

α2

1

α2 =

1 2

1 2

− α12

1 4

0

1 2

√ 2 2

1

α1

FIGURE 8.4 If the mapping Tα has a fixed point, x = Tα x, then we can repeat the above proof with  = 0. Thus, we have the following: Theorem 8.24 Let C be a bounded, closed and convex subset of a Banach space X. If C has the f.p.p. for nonexpansive mappings, then C has the f.p.p. for α-nonexpansive mappings with α = (α1 , α2 , α3 ) such that α1 ≥ 21 and α2 ≥ 12 − α12 . In other words, if for some multi-index α = (α1 , α2 , α3 ) satisfying the above conditions, a mapping T : C → C is nonexpansive with respect to the metric d defined by

d(x, y) = kx − yk + (α2 + α3 ) kT x − T yk + α3 T 2 x − T 2 y , then T has a fixed point in C.

Mean nonexpansive mappings

157

To present some further results related to asymptotic center methods, we recall some basic evaluations for Lipschitz constants of iterates of T . It is clear that, for all x, y ∈ C, we have kT x − T yk ≤

1 kx − yk . α1

(8.7)

Further, using (8.6), we can write

α2 T 2 x − T 2 y ≤ kx − yk − α1 kT x − T yk

≤ kx − yk − α12 T 2 x − T 2 y . Putting the last term on the left-hand side, we obtain


α12 + α2 T 2 x − T 2 y ≤ kx − yk . Thus,

2

T x − T 2 y ≤

α12

1 kx − yk . + α2

(8.8)

Now we are ready to prove the following lemmas: Lemma 8.1 Let C be a nonempty and convex subset of a Banach space X. If T : C → C is α-nonexpansive with α = (α1 , α2 , α3 ) such that α1 > 21 and α2 > 12 − α12 , then Fix(T ) = Fix(Tα ). Proof. Obviously, Fix(T ) ⊂ Fix(Tα ). Clearly, we can assume that Fix(Tα ) 6= ∅. Let x ∈ Fix(Tα ). Then, kx − T xk = =

kT x − T xk

α 2



α2 T x − T x + α3 T 3 x − T x



≤ α2 T 2 x − T x + α3 T 3 x − T 2 x + α3 T 2 x − T x



≤ (α2 + α3 ) T 2 x − T x + α3 T 3 x − T 2 x .

If α2 > 0, then combining (8.7), (8.8), and α-nonexpansivness of T , we can write    1 − α1 s α1 1 kx − T xk ≤ max (α2 + α3 ) s + α3 :s∈ , kx − T xk α2 α12 + α2 α1  2   α1 + α2 − α1 α3 α1 1 = max s+ :s∈ , kx − T xk . α2 α2 α12 + α2 α1 If α12 + α2 − α1 ≥ 0, then the maximum is achieved for s = 1/α1 . Thus, kx − T xk ≤

1 − α1 kx − T xk , α1

158

Classification of Lipschitz mappings

and for α1 > 12 , we get x ∈ Fix(T ). If α12 + α2 − α1 < 0, then the maximum is achieved for s = α1 /(α12 + α2 ). Thus,   1 1 − α12 − α2 kx − T xk = − 1 kx − T xk , kx − T xk ≤ α12 + α2 α12 + α2 and for α2 > 21 − α12 , we have x ∈ Fix(T ). For α2 = 0, we can use (8.7) and (8.8) to obtain kx − T xk ≤ α3 =

2






T x − T x + T 3 x − T 2 x ≤



α3 α3 + 2 α1 α1

 kx − T xk

1 − α12 kx − T xk . α12

√

If α1 > 22 , then x ∈ Fix(T ). All the above means that, for α1 > Fix(Tα ).

1 2

and α2 >

1 2

− α12 , we have Fix(T ) =



Lemma 8.2 Let C be a nonempty and convex subset of a Banach space X. If T : C → C is α-nonexpansive with α = (α1 , α2 , α3 ) such that α1 > 12 and α2 > 12 − α12 , then for each sequence {xn } ⊂ C, lim kxn − T xn k = 0 ⇐⇒ lim kxn − Tα xn k = 0.

n→∞

n→∞

Proof. Suppose lim kxn − T xn k = 0. Then, as in the proof of theorem 8.21, n→∞

lim kTα xn − xn k = 0.

n→∞

Conversely, let lim kTα xn − xn k = 0. Then, n→∞

kxn − T xn k ≤ kxn − Tα xn k + kTα xn − T xn k



≤ kxn − Tα xn k + α2 T 2 xn − T xn + α3 T 3 xn − T xn

≤ kxn − Tα xn k + (α2 + α3 ) T 2 xn − T xn

+α3 T 3 xn − T 2 xn . Now, it is enough to repeat arguments from the previous lemma to get the conclusion. 

Finally, we get the following theorem, which gives us sequential methods of approximation of fixed points:

Mean nonexpansive mappings

159

Theorem 8.25 Let C be a bounded closed and convex subset of a uniformly convex Banach space X. If T : C → C is α-nonexpansive with α = (α1 , α2 , α3 ) such that α1 > 21 and α2 > 12 − α12 , then the set of all a.f.p.s. for T coincides with the set of all a.f.p.s. for Tα and for each such sequence {xn }, A (C, {xn }) consists of exactly one point, which is fixed under T . Proof. It is a consequence of lemmas 8.1, 8.2, and the proof of theorem 7.1. 

All presented results can be generalized for n ≥ 4, and difficulties are based only on some technicalities.

8.4

On the structure of the fixed points set

It is well known that, in the case of strictly convex space, the fixed points set of nonexpansive mapping is convex (it may be empty): Theorem 8.26 Let C be a nonempty, closed and convex subset of a strictly convex Banach space X. If T : C → C is nonexpansive, then Fix(T ) is closed and convex. Thus, in view of theorem 8.4, we have the following: Conclusion 8.1 Let C be a bounded, closed and convex subset of a strictly convex Banach space X. If T : C → C is (α, p)-nonexpansive with α = (α1 , . . . , αn ) such that n−1

n−1

1

(1 − α1 )(1 − α1 p ) < α1 p (1 − α1p ), then Fix(T ) is closed and convex (it may be empty). Actually, the result for nonexpansive mappings can be generalized to the whole class of α-nonexpansive ones. Theorem 8.27 (P´ erez Garc´ıa and Piasecki, [65]) Let C be a nonempty, closed and convex subset of a strictly convex Banach space X. For any (α, p)nonexpansive mapping T : C → C with α = (α1 , . . . , αn ) and p ≥ 1, Fix(T ) is closed and convex (it may be empty). Proof. Since C is closed and T is continuous, Fix(T ) is closed in X. Obviously, it is enough to consider the case of α-nonexpansive mapping. Suppose that Fix(T ) consists of more than one point. Let x, y ∈ Fix(T), x 6= y. Then, T ([x, y]) ⊂ [x, y], where [x, y] = {(1 − t)x + ty : t ∈ [0, 1]} .

160

Classification of Lipschitz mappings

Indeed, suppose there exists z ∈ [x, y] such that T z ∈ / [x, y]. Since X is strictly convex and T is α-nonexpansive, we obtain kx − yk =

n X i=1

≤

n X i=1

n




X



αi T i x − T i y < αi T i x − T i z + T i z − T i y i=1 n

X

αi T i x − T i z + αi T i z − T i y i=1

≤ kx − zk + kz − yk = kx − yk , which is a contradiction. Since T ([x, y]) ⊂ [x, y], T x = x, T y = y, and T is continuous, we conclude that T ([x, y]) = [x, y]. Since Fix(T ) is closed, it is enough to show that Fix(T ) is dense in [x, y]. Suppose that Fix(T ) is not dense in [x, y]. Without loss of generality, we can assume that there is not any fixed point in (x, y). Let ϕ : [0, 1] → [x, y] be a parametrization of [x, y]; that is, ϕ(t) = (1 − t)x + ty. It is clear that ϕ is bijection, ϕ and ϕ−1 are continuous. Let f : [0, 1] → [0, 1] be defined by f = ϕ−1 T ϕ. Since f is continuous and Fix(f ) = {0, 1}, this implies that either f (t) > t for every t ∈ (0, 1) or f (t) < t for every t ∈ (0, 1). Suppose that

f (t) > t for any t ∈ (0, 1). Then, for every z ∈ (x, y), we have x − T i z > kx − zk, i = 1, . . . , n. Thus, n X i=1

n

i

X i

αi x − T i z > kx − zk , αi T x − T z = i=1

which contradicts α-nonexpansiveness of T . Thus, Fix(T ) is dense in [x, y]. Since Fix(T ) is closed, we conclude that [x, y] ⊂ Fix(T ). 

If the space is not strictly convex, then the thesis of theorem 8.27 does not hold. To prove it, we use an example of nonexpansive mapping presented in [36], pp. 35, ex. 3.7: Example 8.1 In the space c0 of sequences of real numbers with the limit 0 and standard sup norm let us consider a closed unit ball B and a mapping T : B → B defined by T x = T (x1 , x2 , . . . ) = (x1 , 1 − kxk , x2 , x3 , . . . ). It is easy to check that the mapping T is an isometry; that is, for each x, y ∈ B kT x − T yk = kx − yk .

Mean nonexpansive mappings

161

This implies that, for any α and p ≥ 1, T is (α, p)-nonexpansive. It is easy to verify that the set of fixed points of T consists of two points e1 = (1, 0, . . . , 0, . . . ) and −e1 = (−1, 0, . . . , 0, . . . ), hence, is not convex. Even if X is strictly convex, then the fixed points set for uniformly lipschitzian mapping may be not convex: Example 8.2 Let X be an infinite dimensional Banach space. Then there exists a retraction R : BX → SX of the closed unit ball BX onto its boundary SX satisfying the Lipschitz condition with a certain Lipschitz constant k(R) = k ≥ 3. Since Rn = R for n ≥ 1, we conclude that R is uniformly lipschitzian with k(Rn ) = k. In spite of this, Fix(R) = SX ; hence, is not convex. This in particular implies that R is not mean nonexpansive at least in case of strictly convex space. Actually, it is easy to verify that a retraction R is lipschitzian with k(R) = k if and only if for any α, it is α-lipschitzian with k(α, R) = k.

Chapter 9 Mean lipschitzian mappings with k > 1

The minimal displacement and optimal retraction problems have their roots in efforts to extend the celebrated Brouwer’s Fixed Point Theorem to noncompact settings. Below, we briefly outline the evolution of this problem. We will consider stages that presumably have exerted the greatest impact on the development of this issue as well as some of the intriguing questions that were arising as the problem was confronted with more and more answers. Then, we will widely study lipschitzian retractions onto a ball, lipschitzian self-mappings T defined on a bounded, closed, convex but noncompact subset C of infinite dimensional Banach space X with a positive minimal displacement, that is, inf {kx − T xk : x ∈ C} > 0, as well as lipschitzian retractions of the whole unit ball onto its boundary with relatively small Lipschitz constants. This topic, which among specialists is called the minimal displacement and optimal retraction problem, has been widely studied in books [36], [31], and [49]. In spite of this, within the last 10 years, we have had a big progress. In this chapter, we shall give an updated presentation of this topic. Then, we shall extend obtained results to classes L (α, p, k) with k > 1.

9.1

Losing compactness in Brouwer’s Fixed Point Theorem

The most famous and the most frequently quoted theorems in the topological fixed point theory were formulated by Brouwer in 1912 and Schauder in 1930. Before those fundamental findings are cited, let us start by presenting some classical definitions and facts: A topological space X is said to have the topological fixed point property (generally abbreviated as t.f.p.p.), if each continuous mapping f : X → X has a fixed point. This property is topologically invariant. Indeed, suppose Y is a topological space, and h : X → Y is a homeomorphism such that h (X) = Y . Consider any continuous function f : Y → Y . Then, the mapping g = h−1 ◦ f ◦ h : X
→ X is continuous. Thus, there exists a point x ∈ X such that h−1 ◦ f ◦ h x = x, and, consequently, f (h (x)) = h (x). This means

163

164

Classification of Lipschitz mappings

that h (x) is a fixed point of f . Thus, all sets homeomorphic to X share the topological fixed point property. Let us recall that a subset D ⊂ X is called a retract if there exists a continuous mapping (a retraction) r : X → D such that Fixr = D. Again, consider any continuous function f : D → D, and extend it to the continuous mapping g = f ◦ r : X → D ⊂ X. It is easy to observe that Fix(g) = Fix(f ). Consequently, if X has the t.f.p.p., then this property is also inherited by all retracts of X. In the sequel, we shall assume that Rn is furnished withP the inner product n defined for x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) as (x, y) = i=1 xi yi and the
Pn 2 1/2 . Moreover, we use B n and S n−1 to denote the unit norm kxk = i=1 xi ball and the unit sphere in Rn . Within our terminology, Brouwer’s Theorem might be formulated as follows [16]: Theorem 9.1 (Brouwer’s Fixed Point Theorem) The unit ball B n ⊂ Rn has the topological fixed point property. Obviously, every closed ball in Rn has the t.f.p.p., as it is homeomorphic to the unit ball B n . Suppose now that C ⊂ Rn is bounded, closed and convex. Then, in view of lemma 9.2, C is a nonexpansive retract of the whole space Rn via the nearest point projection PC : Rn → C, which maps each point x ∈ Rn to the closest point PC x in C; that is, kPC x − xk = inf {kx − yk : y ∈ C}. Since C is bounded, it is contained in some closed ball in Rn , and hence, it is a retract of that ball. This implies that all such sets have the topological fixed point property. Finally, let us point out that every Banach space X of finite dimension is isomorphic to Rn for n = dimX. In the light of the foregoing, Brouwer’s theorem may be formulated in an equivalent yet structurally more general form: Theorem 9.2 (Brouwer’s Fixed Point Theorem) Any bounded, closed and convex subset of a finite dimensional Banach space has the topological fixed point property. Presently, Brouwer’s theorem may take many equivalent forms. Many information on various types of equivalents can be found in [34] as well as in [31] and [36]. We shall elaborate on two of them. Theorem 9.3 (No retraction theorem) S n−1 is not the retract of B n . The equivalence of these two facts follows from the following observations. First, suppose that R : B n → S n−1 would be a retraction of the unit ball B n onto its boundary S n−1 . Then, the mapping T = −R : B n → B n is continuous and fixed point free. On the other hand, suppose that there exists a mapping

Mean lipschitzian mappings with k > 1

165

T = B n → B n without fixed points. Then, we can extend this mapping to the ball 2B n by putting ( Tx   if kxk ≤ 1, Fx = x (2 − kxk) T kxk if kxk ∈ [1, 2] and observe that F : 2B n → B n , F is continuous, fixed point free, and maps the doubled unit sphere 2S n−1 into the origin. By putting Fex = 12 F (2x) n e we obtain → B n such that
a continuous, fixed point free mapping F : B n−1 n Fe S = {0}, and, consequently, the retraction R : B → S n−1 , x − Fex

. Rx =

x − Fex Our next equivalent variant of Brouwer’s Theorem requires the notion of contractibility. We say that a topological space X is contractible to a point z ∈ X if there exists a continuous mapping (a homotopy) H : X × [0, 1] → X satisfying H (x, 0) = x and H (x, 1) = z for all x ∈ X. Obviously, if a topological space X is contractible to some point, then it is contractible to any point in X. Theorem 9.4 (No contractibility theorem) The sphere S n−1 is not contractible to a point. To see this, suppose that there is a homotopy H : S n−1 × [0, 1] → S n−1 such that for all x ∈ S n−1 we have H (x, 0) = x and H (x, 1) = z, where z ∈ S n−1 is arbitrarily fixed. Then, for any 0 ≤ r < 1, the mapping defined as (  z  if kxk ≤ r, Rx = 1−kxk x H kxk , 1−r if kxk ∈ (r, 1] , would be a retraction of B n onto S n−1 , which contradicts “No retraction theorem”. To see that this theorem is equivalent to the Brouwer’s Theorem, suppose that there is a retraction R : B n → S n−1 . Then, S n−1 would be contractible by a homotopy H : S n−1 × [0, 1] → S n−1 defined as H (x, t) = R ((1 − t) x) . Notice that actually neither of the above constructions takes advantage of the fact that we work with a finite dimensional Banach space Rn and that in fact they allow the formulation of the following result, among specialists, referred to as a “trivial theorem”. Theorem 9.5 (trivial theorem) For any Banach space X, the following statements are equivalent: a) The unit ball B has the topological fixed point property,

166

Classification of Lipschitz mappings

b) The unit sphere S is not the retract of B, c) S is not contractible. Classical Brouwer’s Theorem shows that the above statements are true if X = Rn , and, consequently, so are they for all finite-dimensional Banach spaces. The remaining part of the chapter focuses on results showing how strongly three statements mentioned above are false in the case of infinite-dimensional Banach spaces. A crucial reason for this involves the fact that a bounded and closed set does not have to be compact in this case. In particular, this is applicable to every closed ball. The most popular and useful result extending Brouwer’s theorem to infinite-dimensional spaces is the following theorem formulated in 1930 by Schauder [74]: Theorem 9.6 (Schauder’s Fixed Point Theorem) Any convex and compact subset C of a Banach space X has topological fixed point property. Already at this very stage, a natural, general question arises. What happens if C is noncompact? Around 1935, Ulam posed the following question in the Scottish Book [59]: “Can one transform continuously the solid sphere of a Hilbert space into its boundary such that the transformation should be identity on the boundary of the ball?” A note accompanying Ulam’s question reads: “There exists a transformation with the required property given by Tychonoff ”. Unfortunately, until now all attempts to find the Tychonoff’s construction have failed. An answer to Ulam’s question came eight years later and was formulated by Kakutani [45]. In his publication, Kakutani provided some examples of uniformly continuous and lipschitzian mappings without fixed points. Each of them could be used to construct the retraction required. Example 9.1 (Kakutani’s Construction) Consider the space `2 , the standard model of a Hilbert space. The mapping T : B → B defined for any x = (x1 , x2 , . . . ) by  p 1 − kxk, x1 , x2 , . . . Tx = is p fixed point free and uniformly continuous. Indeed, T x = x implies xi = 1 − kxk = 0 for all i ∈ N and kxk = 1, which is impossible. Its continuity and even uniform continuity follows from the easily justified inequality p kT x − T yk ≤ 2 kx − yk + kx − yk . For the mapping T : B → B,  ∈ (0, 1], defined by T x = ( (1 − kxk) , x1 , x2 , . . . )

Mean lipschitzian mappings with k > 1

167

we get the same √ conclusion, but this time the mapping T is even lipschitzian with k(T ) = 1 + 2 . To construct a retraction R : B → S, apply the following Kakutani’s receipt. For any x ∈ B, follow the straight half line from T x through x until you reach the sphere S. Then, define the value of the retraction Rx as the unique point of intersection of this half line with the unit sphere S. The situation is illustrated in figure 9.1. The corresponding analytic formula is Rx = x + λ (x) u (x) , where u (x) =

x − Tx kx − T xk

and λ (x) ≥ 0 is selected to satisfy kRxk = 1. Standard calculations, which we leave to the reader, show that q 2 2 λ (x) = − (x, u (x)) + 1 − kxk + (x, u (x)) .

Ty

Rx

Ry = y

x

0 Tx

FIGURE 9.1 The next important step involves Klee’s constructions formulated in 1953 and 1955 in [51] and [52]. They offered a full answer to Ulam’s question not only in the case of a Hilbert space but also for any infinitely dimensional Banach space. Theorem 9.7 (Klee’s Theorem) For any infinitely dimensional Banach space X, there exists a homeomorphism h : X → X \ {0} such that hx = x for any x ∈ X with kxk ≥ 1.

168

Classification of Lipschitz mappings

Observe that in virtue of the above result we can immediately construct the retraction R : B → S by putting Rx =

hx . khxk

Theorem 9.8 (Klee’s Theorem) For any noncompact, closed, and convex subset C of a Banach space X, there exists a continuous, fixed point free mapping T : C → C. In view of Schauder’s and Klee’s theorems, we conclude that a closed and convex subset of a Banach space has t.f.p.p. iff it is compact. The mappings defined by Klee are continuous but not uniformly continuous. On the other hand, there are many lipschitzian mappings known, even nonexpansive mappings without fixed points. Such observations justify the following questions. Do all bounded, closed, convex but noncompact sets fail to have the t.f.p.p. in some stronger sense? How “regular” such mappings can be? How far they can move all the points? Do spaces or sets differ with respect to the “intensity” of this failure? Such an approach to the problem was noticed and specified in 1973 by Goebel in [29]. It includes examples of lipschitzian mappings T : C → C for which the minimal displacement d (T ) = infx∈C kx − T xk > 0. This work has exerted a major impact on the further evolution of this problem. We leave to the reader the verification of the fact that in the Kakutani’s constructions we have d (T ) = 0. Indeed, research in subsequent years led to the formulation of theorems much stronger than Klee’s construction. In 1979, Nowak [63] proved that for a certain class of Banach spaces the sphere is a lipschitzian retract of the ball. Four years later Benyamini and Sternfeld [10] proved that this is true for any Banach space. Theorem 9.9 (Benyamini and Sternfeld) There exists a universal constant k0 such that any infinitely dimensional Banach space X admits a retraction R : BX → SX of class L (k0 ). Once we have a k-lipschitzian retraction R : BX → SX , we can define a mapping T = −R : BX → SX ⊂ BX , which is also of class L (k) and such that, for any x ∈ BX kx − T xk ≥

1

T x − T 2 x = 2 kRxk = 2 . k k k

This implies that d (T ) ≥ k2 > 0. Further, for any α ∈ (0, 1], the mapping Tα = (1 − α) I + αT ∈ L (1 − α + αk), Tα : BX → BX and satisfies d (Tα ) = αd (T ) = 2α k > 0. Theorem 9.9 can be also formulated as

Mean lipschitzian mappings with k > 1

169

Theorem 9.10 (Benyamini and Sternfeld) For any infinitely dimensional Banach space X, the following three statements are true and equivalent: a) For any k > 1, there exists a mapping T : BX → BX of class L (k) such that d (T ) > 0, b) The unit sphere SX is a lipschitzian retract of BX , c) SX is Lipschitz contractible. The last statement means that there exist constants M, N ≥ 0, and a homotopy H : SX ×[0, 1] → SX such that H (x, 0) = x for all x ∈ S, H (x, 1) = z for arbitrarily fixed z ∈ SX and kH (x, t) − H (y, s)k ≤ M |t − s| + N kx − yk for all x, y ∈ SX and for all s, t ∈ [0, 1]. It is clear that, if the retraction R : BX → SX is of class L (k), then the homotopy obtained by standard tricks, H (x, t) = R ((1 − t) x), satisfies kH (x, t) − H (y, s)k ≤ k |t − s| + k kx − yk . The strongest result in this matter, concerning not only ball but also all bounded, closed and convex subsets was obtained in 1985 by Lin and Sternfeld [57]: Theorem 9.11 (Lin and Sternfeld) Let C be a bounded, closed, convex, but noncompact subset of a Banach space X. Then, for any k > 1, there exists a mapping T : C → C of class L (k) with d (T ) > 0. In spite of the fact that theorems 9.9 and 9.11 provide qualitative answers to the theory, there are still some quantitative aspects, which were initiated by Goebel in 1973. The two basic are the minimal displacement problem and optimal retraction problem. This topic has been widely studied in books [36], [49], and [31] as well as in the survey articles [32] and [41]. In spite of this, within the past 10 years we have had big progress. The aim of this chapter is to give the updated presentation of the subject and possibly attract newcomers to the field. To proceed, we shall need some basic facts concerning retractions of the whole space onto balls. Exhaustive information as well as proofs of almost all above-mentioned results can be found in books [31] and [36].

9.2

Retracting onto balls in Banach spaces

The most popular retraction of the whole Banach space X onto the unit ball BX is the radial projection P : X → BX defined as

170

Classification of Lipschitz mappings  Px =

x if kxk ≤ 1, x kxk if kxk > 1.

It is easy to observe that kx − P xk = dist (x, BX ) = max {0, kxk − 1} . Thus, P is the nearest point projection. In general, P is of class L (2) because of the following: Lemma 9.1 For any x, y ∈ X, x, y 6= 0, we have

x y 2

kxk − kyk ≤ max {kxk , kyk} kx − yk . Proof. Observe that, for any x, y 6= 0,





x

x

x y x y





kxk − kyk ≤ kxk − kyk + kyk − kyk 1 1 1 = kxk − + kx − yk kxk kyk kyk 1 1 |kxk − kyk| + kx − yk kyk kyk 2 ≤ kx − yk . kyk =

Exchanging roles of x and y in the above evaluations gives us the thesis. 


(2) Example 9.2 As a Banach space (X, k·k) consider `∞ = R2 , k·k furnished with the norm kxk = k(x1 , x2 )k = max {|x1 | , |x2 |}. Then, for x = (1, 1) and y = (1 + , 1 − ), we get (see figure 9.2)

   

1−

= 0, 2 kP x − P y k = (1, 1) − 1,

1+ 1+ =

2 kx − y k . 1+

Since  > 0 can be chosen arbitrarily close to 0, we get k (P ) = 2. (2)

Spaces c0 , c, `∞ , C [0, 1], `1 , L1 (0, 1) contain an isometric copy of l∞ . Thus, k (P ) = 2 for each of them. Moreover, it is known that k (P ) < 2 if and only if 0 (X) < 2; see [75]. Results concerning the Lipschitz constants of radial projections in spaces `p and Lp can be found in papers [27] and [44].

Mean lipschitzian mappings with k > 1

171

x = (1, 1) = P x y = (1 + , 1 − )

1

  P y = 1, 1− 1+ 0

1

(2)

FIGURE 9.2: The radial projection in `∞ . Let us pass to the case of Hilbert space H. Suppose that C is a closed and convex subset of H. Since every Hilbert space is uniformly convex, for any point x ∈ H, there is exactly one closest point PC x in C, kx − PC xk = inf {kx − zk : z ∈ C}. Thus, the mapping PC : H → C described above is well defined, and it is the nearest point projection. We shall prove that PC is nonexpansive. Lemma 9.2 The nearest point projection PC : H → C is of class L(1). Proof. For any z ∈ C, consider the function φ : [0, 1] → R defined as 2

φ (t) = kx − (1 − t) PC x − tzk 2

2

= kx − PC xk + 2t (PC x − x, z − PC x) + t2 kPC x − zk . Obviously, φ is convex and φ(0) ≤ φ(t) for any t ∈ [0, 1]. This implies that φ0 (0) = 2 (PC x − x, z − PC x) ≥ 0 or equivalently (PC x − x, z − PC x) ≥ 0 for all z ∈ C. Thus, for any two points x, y ∈ H, we have (PC x − x, PC y − PC x) ≥ 0

172

Classification of Lipschitz mappings

and (PC y − y, PC x − PC y) ≥ 0. This implies ((x − y) − (PC x − PC y) , PC x − PC y) ≥ 0, and, consequently, 2

kPC x − PC yk ≤ (PC x − PC y, x − y) ≤ kPC x − PC yk kx − yk . This shows that PC is nonexpansive. 

Thus, for a Hilbert space, the radial projection P not only is a unique nearest point projection but also is nonexpansive. The complete characterization of spaces, for which the radial projection is nonexpansive, has been given in [21]: Theorem 9.12 For the radial projection P : X → BX in (X, k·k), the following statements hold. • If dim X ≥ 3, then P is nonexpansive if and only if X is a Hilbert space. ∗

• If dim X = 2, then P is nonexpansive if and only if the norm k·k dual to k·k satisfies ∗ k(x, y)k = k(−y, x)k for all (x, y) ∈ R2 . It means that the unit ball B ∗ coincides with the unit ball B rotated by π2 . Example 9.3 Mixed `p , `q norms. Take p, q > 1 such that Consider the space R2 furnished with the norm (see figure 9.3) ( p p 1 (|x1 | + |x2 | ) p if x1 · x2 ≥ 0, k(x1 , x2 )kp,q = q q 1 (|x1 | + |x2 | ) q if x1 · x2 ≤ 0.

1 p

+

1 q

= 1.

∗

It is easy to verify that the norm k·kp,q , dual to k·kp,q is given via relation n o ∗ k(y1 , y2 )kp,q = max x1 y1 + x2 y2 : k(x1 , x2 )kp,q ≤ 1 ( p p 1 (|y1 | + |y2 | ) p if y1 · y2 ≤ 0, = q q 1 (|y1 | + |y2 | ) q if y1 · y2 ≥ 0. ∗

Thus, k(x1 , x2 )kp,q = k(−x2 , x1 )kp,q for any (x1 , x2 ) ∈ R2 . Hence, in spite   of the fact that R2 , k·kp,q is not a Hilbert space, the radial projection is nonexpansive.

Mean lipschitzian mappings with k > 1

173

1

-1 1

-1

FIGURE 9.3: The unit ball in (R2 , k·kp,q ) with p = 4 and q = 43 . Nevertheless, for some spaces with uniform norm, there are other than radial nonexpansive projections. (2)

Example 9.4 Consider again the space `∞ and τ : R → [−1, 1] defined as (see figure 9.4)   −1 τ (t) = max {−1, min {1, t}} = t   1

the truncation function

if t < −1, if − 1 ≤ t ≤ 1, if t > 1.

τ

1 −1 0

1

−1

FIGURE 9.4: A graph of function τ .

174

Classification of Lipschitz mappings (2)

Define Qτ : `∞ → B as (see figure 9.5) Qτ (x1 , x2 ) = (τ (x1 ) , τ (x2 )) .

0

(2)

FIGURE 9.5: Retraction Qτ : `∞ → B. It is clear that Qτ is a nonexpansive retraction. Moreover, it is the nearest point projection onto B. Example 9.5 Consider the space BC (R) of all bounded continuous functions f : R → R furnished with the standard sup norm, kf k = sup {|f (t)| : t ∈ R}. The truncation retraction Qτ : BC (R) → B is given by (see figures 9.6 and 9.7) Qτ f (t) = τ (f (t)) . Again, Qτ is nonexpansive, and it is the nearest point projection.

f 1 0 −1

FIGURE 9.6: A graph of function f .

Mean lipschitzian mappings with k > 1

175

1 Qτ f 0 −1

FIGURE 9.7: A graph of Qτ f.

Now, let us consider a more general situation. Suppose K is an arbitrary nonempty set and B (K) is a Banach space consisting of all real bounded functions on K and furnished with the uniform norm, kf k = sup {|f (x)| : x ∈ K}. A subspace X of B (K) is called truncation invariant if for each f ∈ X the function Qτ f = τ ◦ f ∈ X, where τ is the truncation. In particular, B (K) is a truncation invariant subspace of B (K). It is clear that the truncation Qτ : X → BX is a nonexpansive retraction onto the unit ball BX . Moreover, it is the nearest point projection. The class of truncation invariant subspaces was introduced in [39]. Let us list some particular cases: (n)

• If K = {1, . . . , n}, then B (K) = `∞ = (Rn , k·k), where k(x1 , . . . , xn )k = max {|x1 | , . . . , |xn |} . (n)

• If K = N, then c, c0 , and `∞ are truncation invariant subspaces of B (K) = `∞ . • If K = R and as a norm in R we take an absolute value |·|, then BC (R) is a truncation invariant subspace of B (R). Also, the space BC0 (R) of all bounded continuous functions f : R → R vanishing at 0, f (0) = 0, is a truncation invariant subspace of B (R). • If (K, k·k) = ([0, 1] , |·|), then C [0, 1] and C0 [0, 1] are truncation invariant subspaces of B ([0, 1]). • Suppose (M, d) is a metric space. By BC (M ) we denote the space of all bounded continuous functions f : M → M and by BCz (M ) its subspace, consisting of of all functions vanishing at z, f (z) = 0. It is clear that BC (M ) and BCz (M ) are truncation invariant subspaces of B (M ).

176

Classification of Lipschitz mappings

For each r ≥ 0, the mapping Qτ generates the truncation retraction Qτr : X → rBX of the whole space X onto a ball centered at origin and radius r, (
rQτ 1r f if r > 0, τ Qr f = 0 if r = 0. Obviously, for every r ≥ 0, Qτr is the nearest point projection onto rBX . In particular, for every f ∈ / rBX , we have kQτr f k = r. A very useful property of this family is given in the following Lemma 9.3 For all f, g ∈ X, and r1 , r2 ≥ 0, we have

τ

Qr f − Qτr g ≤ max {kf − gk , |r1 − r2 |} . 1 2 Proof. We shall prove that the inequality τ Qr f (x) − Qτr g(x) ≤ max {|f (x) − g(x)| , |r1 − r2 |} 1 2 holds for every x ∈ K. It is clear for r1 = r2 = 0. If r2 = 0 and r1 > 0, then, for any x ∈ K, we have τ Qr f (x) − Qτ0 g(x) ≤ |r1 | = |r1 − r2 | . 1 Suppose that 0 < r2 ≤ r1 . Let x ∈ K be arbitrarily chosen. Then, we have the following cases: 1. If |f (x)| ≤ r1 and |g(x)| ≤ r2 , then     τ Qr f (x) − Qτr g(x) = r1 τ 1 f (x) − r2 τ 1 g(x) 1 2 r1 r2 = |f (x) − g(x)| . 2. If f (x) ≥ r1 and g(x) ≥ r2 or f (x) ≤ −r1 and g(x) ≤ −r2 , then τ Qr f (x) − Qτr g(x) = |r1 − r2 | . 1 2 3. If f (x) ≥ r1 and |g(x)| ≤ r2 , then

τ Qr f (x) − Qτr g(x) = |r1 − g(x)| = r1 − g(x) ≤ f (x) − g(x) 1 2 = |f (x) − g(x)| . A similar situation applies to f (x) ≤ −r1 and |g(x)| ≤ r2 ; that is, τ Qr f (x) − Qτr g(x) ≤ |f (x) − g(x)| . 1 2

Mean lipschitzian mappings with k > 1

177

4. If f (x) ≥ r1 and g(x) ≤ −r2 , then

τ Qr f (x) − Qτr g(x) = |r1 + r2 | = r1 + r2 ≤ f (x) − g(x) 1 2 = |f (x) − g(x)| . Obviously, for f (x) ≤ −r1 and g(x) ≥ r2 we also have τ Qr f (x) − Qτr g(x) ≤ |f (x) − g(x)| . 1 2 5. If r2 ≤ f (x) ≤ r1 and g(x) ≥ r2 , then

τ Qr f (x) − Qτr g(x) = |f (x) − r2 | = f (x) − r2 ≤ r1 − r2 1 2 = |r1 − r2 | . By analogy, if −r1 ≤ f (x) ≤ −r2 and g(x) ≤ −r2 , then

τ Qr f (x) − Qτr g(x) ≤ |r1 − r2 | . 1 2 6. If −r1 ≤ f (x) ≤ r2 and g(x) ≥ r2 , then

τ Qr f (x) − Qτr g(x) = |f (x) − r2 | = −f (x) + r2 ≤ −f (x) + g(x) 1 2 = |f (x) − g(x)| . If −r2 ≤ f (x) ≤ r1 and g(x) ≤ −r2 , then we also have

τ Qr f (x) − Qτr g(x) ≤ |f (x) − g(x)| . 1 2 If r1 ≤ r2 , then it is enough to repeat the above cases exchanging roles between f and g. Hence, for all r1 , r2 ≥ 0, f, g ∈ X and every x ∈ K, we obtain τ Qr f (x) − Qτr g(x) ≤ max {|f (x) − g(x)| , |r1 − r2 |} , 1 2 as we desired. Consequently, the inequality τ Qr f (x) − Qτr g(x) ≤ max {kf − gk , |r1 − r2 |} 1 2 holds for every x ∈ K. By taking supremum over all x ∈ K on the left-hand side of the above inequality, we finally get the thesis. 

178

Classification of Lipschitz mappings

For spaces like B (M ), BC (M ), and BCz (M ), there are other nonexpansive retractions onto the unit ball. To see it, consider any function β : R → [−1, 1] satisfying β (t) = t for t ∈ [−1, 1] and |β (s) − β (t)| ≤ |s − t| for all s, t ∈ R. Every such function generates a nonexpansive retraction Qβ f = β ◦ f onto the unit ball. Consider, for example, the retraction Qβ corresponding to the function β given in figure 9.8.

1 β −2

−1 0

1

r1

r2

−1

FIGURE 9.8: A graph of function β. Such retraction maps the ball centered at z ≡ − 23 and radius 12 onto the 1 1 ball centered every function
at α ◦ z ≡ − 2 and radius 2 . More precisely, 1 f ∈ B z, 2 is reflected with respect to the level −1, Qβ f (t) = −f (t) − 2 for 2 1 t ∈ M . Observe also that the ball centered at z ≡ r1 +r and radius r2 −r is 2 2 mapped to zero. Of particular interest for us will be the retraction QΛ f = Λ◦f corresponding to the function Λ : R → [−1, 1] defined as (see figure 9.9)   0 if t < −2,      if − 2 ≤ t < −1, −t − 2 Λ (t) = t if − 1 ≤ t ≤ 1,    −t + 2 if 1 < t ≤ 2,    0 if t > 2. To see how QΛ works, consider, for instance, the space BC (R). Take any function f ∈ BC (R) and divide its graph into five groups of pieces in the following way: the pieces that lie above the level 2 (resp. below the level −2), between levels 1 and 2 (resp. between levels −2 and −1) and between levels −1 and 1. Obviously, the part of graph that lies between levels −1 and 1 remains unchanged. The part which lies between levels 1 and 2 (resp. between levels −2 and −1) is reflected with respect to the level 1 (resp. level −1). The pieces lying above the level 2 as well as below the level −2 are mapped to zero. The situation is illustrated in figures 9.10 and 9.11.

Mean lipschitzian mappings with k > 1

179

As opposed to Qτ , the retraction QΛ does not send all the elements to the closest point in B. Even more, for any f ∈ B (M ) with kf k > 1, we have

kf − Qτ f k = dist (f, B) = kf k − 1 < f − QΛ f = min {2 kf k − 2, kf k} . The same is true for subspaces BC (M ) and BCz (M ). We shall use the retractions Qτ and QΛ discussing the so-called problems of minimal displacement and optimal retractions.

1 Λ −2

−1 0

1

2

−1

FIGURE 9.9: A graph of function Λ.

f 2 1 0 −1 −2

FIGURE 9.10: A graph of function f .

180

Classification of Lipschitz mappings

f 2 1 QΛ f 0 −1 −2

FIGURE 9.11: A graph of function QΛ f .

9.3

Minimal displacement

Suppose that C is a bounded, closed, convex, and noncompact subset of a Banach space X. Recall that the minimal displacement of a mapping T : C → C is the number d (T ) = inf {kx − T xk : x ∈ C} . For any set C, the characteristic of minimal displacement is the function ϕC : [1, +∞) → [0, diam (C)] defined as ϕC (k) = sup {d (T ) : T : C → C, T ∈ L (k)} . In this setting, theorem 9.11 states that ϕC (k) > 0 for k > 1. Recall that the Chebyshev radius of C relative to C is the number r (C) = inf sup {kz − yk : y ∈ C} . z∈C

Observe that for any two sets K and C such that K = aC + b, a 6= 0,

Mean lipschitzian mappings with k > 1

181

b ∈ X, we have r (K) = |a| r (C). Further, for every mapping T : C → C of class L (k), we can define the mapping T1 : K → K as   x−b T1 x = aT + b. a Then, k (T1 ) = k (T ) and d (T1 ) = |a| d (T ). Consequently, ϕK (k) = ϕaC+b (k) = |a| ϕC (k) . Thus, without loss of generality, we can always assume that r (C) = 1. In the special case, when C = BX , we write ψX instead of ϕBX . For the whole Banach space X, we define the characteristic of the whole space ϕX : [1, +∞) → [0, diam (C)] as ϕX (k) = sup {ϕC (k) : C ⊂ X, r (C) = 1} . As usual, we shall drop the subscripts when either C or X is clear from the context. Let us list some general properties of functions ϕC , ϕX , and ψX (see [36], pp. 215–216 and 222). Lemma 9.4 If r (C) = 1 and ϕ denotes any of the functions ϕX , ϕC , ψX , then: (1) ϕ (k) increases with k; (2) ϕ (1 − α + αk) ≥ αϕ(k) for all α ∈ [0, 1] and k > 1; (3) the ratio

ϕ(k) k−1

(4) the ratio

kϕ(k) k−1

decreases with k; increases with k;

(5) the derivative ϕ0 (1) := lim

ϕ(k) k−1

(6) ϕ0 (1) 1 −

1 k

k→1+

1 k

(7) ϕ (k) = 1 −

always exists and ϕ0 (1) > 0;




≤ ϕ (k) ≤ 1 −

1 k

if and only if ϕ0 (1) = 1;

for all k > 1;

(8) lim ψX (k) = 1 for any space X. k→∞

Proof. (1) is obvious. (2) follows from the observation that for any T : C → C, T ∈ L (k) and any α ∈ [0, 1] the mapping Tα : C → C defined as Tα x = (1 − α) x + αT x is of class L (1 − α + αk) and such that kx − Tα xk = α kx − T xk for any x ∈ C. (3) follows from (2).

182

Classification of Lipschitz mappings

To prove (4), for a fixed A > k and x ∈ C consider the equation   1 1 y = 1− x + T y. A A Then, the right-hand side defines a contraction and, consequently, the above equation has a unique solution which we shall denote by F x,   1 1 Fx = 1 − x + T F x. A A Then, we have 

 1 kF x − F yk ≤ 1 − kx − yk + A   1 kx − yk + ≤ 1− A

1 kT F x − T F yk A k kF x − F yk . A

Thus,  F ∈L and



A−1 A−k



A−1 TF ∈ L k A−k

 .

Observe that for any x ∈ C kx − T F xk =

A Ad (T ) kF x − T F xk ≥ . A−1 A−1

Using the introduced notation, we can write    A−1 A ϕ k ≥ ϕ(k). A−k A−1 Finally, if we put l =

k(A−1) A−k

> k, then the above formula takes the form ϕ (l) l ϕ (k) k ≥ . l−1 k−1

(5) follows from (3) and (4).
To prove (6) observe that the first inequality ϕ0 (1) 1 − k1 ≤ ϕ (k) follows from (4) and (5). The second estimate ϕ (k) ≤ 1 − k1 follows from the subsequent observation. Suppose T : C → C is of class L (k). Fix  > 0. Then, there exists z ∈ C such that C ⊂ B(z, 1 + ). Consider the equation   1 1 x= 1− z+ T x. k+ k+

Mean lipschitzian mappings with k > 1

183

It is clear that the right-hand side defines a contraction. Thus, there is a unique point x ∈ C satisfying the above equation. Then, we have     1 1 kx − T x k = 1 − kz − T x k ≤ 1 − (1 + ) . k+ k+ Letting  → 0 we get d (T ) ≤ 1 − k1 . (7) follows from (6). Let us pass now to (8). In view of theorem 9.9 by Benyamini and Sternfeld, there exists the retraction R : BX → SX of class L (k0 ). Fix  ∈ (0, 1). Define the mapping T : BX → BX as (
if kxk ≤ , −R x T x = x if kxk ∈ [, 1] . − kxk
Then, T ∈ L k0 . Indeed, if x, y ∈ BX , then

x  x  k

0 kT x − T yk = R −R kx − yk .

≤    In view of lemma 9.1, for any x, y ∈ BX \BX , we have

x 2 y

≤ kT x − T yk = − kx − yk kxk kyk max {kxk , kyk} 2 k0 ≤ kx − yk ≤ kx − yk .   If kxk <  and kyk > , then there is a unique number λ ∈ (0, 1) such that k(1 − λ) x + λyk = . Then, kT x − T yk ≤ kT x − T ((1 − λ) x + λy)k + kT ((1 − λ) x + λy) − T yk k0 k0 kx − yk ≤ λ kx − yk + (1 − λ)   k0 = kx − yk ,  as we desired. For any x ∈ BX , we have kx − T xk ≥ 1 − . Indeed, if kxk ≤ , then

 x   x 



kx − T xk = x + R

≥ R

− kxk ≥ 1 − ,   and if kxk ≥ , then

x

= 1 + kxk ≥ 1 + . kx − T xk = x +

kxk

184

Classification of Lipschitz mappings

By putting k =

k0 

we finally get ψX (k) ≥ 1 −

k0 k

for any k > k0 . Now, it is enough to use (6) to finish the proof. 

Observe that (1), (2), and (5) from the above lemma imply the continuity of ϕC , ϕX and ψX on [1, ∞). A set C is called extremal if ϕC (k) = 1 − k1 for all k > 1. A space X for which the unit ball BX is extremal, ψX (k) = 1 − k1 for all k > 1, is referred to as an extremal space. The characteristic of minimal displacement can be considered not only for the whole class of lipschitzian mappings but also for various subclasses. The most interesting are ψB→S (k) = sup {d (T ) : T : B → S, T ∈ L (k)} and ψS→{0} (k) = sup {d (T ) : T : B → B, T ∈ L (k) and T (S) = {0}} . The minimal displacement problem involves finding or evaluating the functions ϕ and ψ for concrete sets or spaces. Let us begin by listing some examples of extremal spaces. Example 9.6 ([36]) Consider the space C [0, 1] with the standard sup norm. Take any strictly increasing function g ∈ C [0, 1] satisfying g (0) = −2 and g (1) = 2. Then, there is a unique point s ∈ (0, 1) such that g (s) = 0. Fix k > 1 and consider the mapping Tk : B → B defined as Tk f = Qτ (k (f + g)) . It is clear that Tk is of class L (k). We shall prove that kf − Tk f k > 1 − k1 for any f ∈ C [0, 1]. First, assume that f (s) ≤ 0. Since f (1) + g (1) = f (1) + 2 ≥ 1, there exists a point t1 ∈ (s, 1) such that f (t1 ) + g (t1 ) = k1 . Thus, 1 kTk f − f k ≥ |Tk f (t1 ) − f (t1 )| = |1 − f (t1 )| = 1 − + g (t1 ) k 1 1 = 1 − + g (t1 ) > 1 − . k k Now, assume that f (s) ≥ 0. Since f (0) + g (0) = f (0) − 2 ≤ −1, there is a point t2 ∈ (0, s) satisfying 1 f (t2 ) + g (t2 ) = − . k

Mean lipschitzian mappings with k > 1

185

Then, 1 kTk f − f k ≥ |Tk f (t2 ) − f (t2 )| = |−1 − f (t2 )| = −1 + + g (t2 ) k 1 1 = 1 − − g (t2 ) > 1 − k k as we desired. Consequently, d (Tk ) = 1 − k1 . However, Tk maps B into the unit sphere S. Thus, we can write ϕC[0,1] (k) = ψC[0,1] (k) = ψB→S (k) = 1 −

1 . k

Example 9.7 Now, consider the subspace C0 [0, 1] of the space C [0, 1]. Take any strictly increasing function g ∈ C0 [0, 1] satisfying g (1) = 1. For fixed k > 1 define the mapping Tk : B → B as Tk f = Qτ (k (|f | + g)) . Obviously, Tk is of class L (k). Further, observe that for any f ∈ B we have |f (0)| + g (0) = 0 and |f (1)| + g (1) ≥ 1. Thus, for any f ∈ B, there exists s ∈ (0, 1) satisfying |f (s)| + g (s) = k1 and in a similar way as in example 9.6 we get kf − Tk f k > 1 − k1 . Observe also that Tk maps the whole unit ball B into the unit sphere S and more precisely into its positive part S + = {f ∈ S : f (t) ≥ 0 for all t ∈ [0, 1]}. Thus, ϕC0 [0,1] (k) = ψC0 [0,1] (k) = ψB→S (k) = ψB→S + (k) = 1 −

1 . k

Example 9.8 Consider the space BC (R) furnished with the sup norm. Take any strictly increasing function g ∈ BC (R) satisfying lim g (t) = ±3 and t→±∞

g (0) = 0. For fixed k > 1 define mapping Tk : B → S by putting for each f ∈B Tk f = Qτ (k (f + g)) . Again, Tk is of class L (k) and kf − Tk f k > 1 − get it in a way known from example 9.6. Thus,

1 k

for any f ∈ BC (R). We

ϕBC(R) (k) = ψBC(R) (k) = ψB→S (k) = 1 −

1 . k

Observe also that the subspace BC0 (R) is invariant under Tk . This implies that for any k > 1 ϕBC0 (R) (k) = ψBC0 (R) (k) = ψB→S (k) = 1 −

1 . k

Example 9.9 ([69]) Consider the space c furnished with the sup norm.

186

Classification of Lipschitz mappings

For fixed k > 1 define the mapping Tk : B → S by putting for each x = (x1 , x2 , . . .) ∈ B Tk x = Qτ (1, −1, kx1 , kx2 , . . .) = (1, −1, τ (kx1 ) , τ (kx2 ) , . . .) . It is easy to check that Tk is of class L (k). We shall prove that for any x ∈ B we have kTk x − xk > 1 − k1 . Indeed, let us assume that there exists x ∈ B such that kTk x − xk ≤ 1 − k1 . However, this implies that x2i−1 ≥ k1 and x2i ≤ − k1 for i = 1, 2, . . . , which is impossible because x ∈ c. Thus, d (Tk ) = 1 − k1 and ϕc (k) = ψc (k) = ψB→S (k) = 1 −

1 . k

One can observe that the subspace c0 of c is invariant under Tk . Consequently, for any k > 1 we get ϕc0 (k) = ψc0 (k) = ψB→S (k) = 1 −

1 . k

In the above examples, we have ϕ(k) = ψ(k) = 1 − k1 for k > 1. The same is true for all subspaces of C [0, 1] of finite codimension (see [12]). It is still unknown if the space `∞ furnished with the sup norm has extremal balls. Very recently Bolibok [14] has proved that ( √ √ (3 − 2 2)(k − 1) for 1 ≤ k ≤ 2 + 2, √ ψl∞ (k) ≥ for k > 2 + 2. 1 − k2 The same estimate holds for L1 (0, 1) equipped with the standard norm (see [11]) as well as for few other spaces (see [39]). It is known that for any space X with the characteristic of convexity 0 (X) < 1 we have ϕ0X (1) < 1 and ϕX (k) < 1− k1 for all k > 1 (for the proof see [36], pp. 217–218). In particular, it applies to all uniformly convex spaces. An interesting situation is observed for the space `1 . It is known that this space is not extremal, and we have (for the proof see [36], pp. 212, ex. 20.3 or [31], pp. 145, ex. 12.5) ( √ √
2+ 3 1 − k1 for 1 ≤ k ≤ 3√ + 2 3, 4 ψ`1 (k) ≤ (9.1) k+1 for k > 3 + 2 3. k+3 Nevertheless, `1 contains an extremal subset as seen in the following: Example 9.10 ([31]) Consider the “positive face” S + of the unit sphere in the space `1 : ( ) ∞ X ∞ S + = {xi }i=1 : xi ≥ 0, xi = 1 . i=1

Obviously, r (S + ) = diam (S + ) = 2. For fixed k > 1, define the map i : S + → N by

Mean lipschitzian mappings with k > 1

187

  ∞  X 1 i(x) = max i ∈ N : xi > .  k j=i Let µ(x) ∈ [0, 1) be given via relation ∞ X

µ(x)xi(x) +

xj =

j=i(x)+1

1 . k

Then the mapping T : S + → S + defined by
T x = T (x1 , x2 , . . . ) = k 0, 0, . . . , 0, µ(x)xi(x) , xi(x)+1 , xi(x)+2 , . . . , where 0 in the last sequence appears i(x) times, is of class L(k) and   1 d(T ) = 2 1 − . k Indeed, let x, y ∈ S + . We shall consider the case in which i(x) 6= i(y). The case i(x) = i(y) is similar. Without loss of generality, we can suppose that i(x) < i(y) and then kT x − T yk is equal to   i(y)−1 ∞ X X |xj − yj | k µ(x)xi(x) + xj + xi(y) − µ(y)yi(y) + j=i(y)+1

j=i(x)+1

 ≤ k

1 − k

X

xj +

j=i(x)+1

xj + xi(y) − yi(y) + (1 − µ(y)) yi(y)

j=i(x)+1



∞ X

+

i(y)−1

∞ X

|xj − yj |

j=i(y)+1

 ≤ k

1 − k

∞ X

xj −

j=i(y)



i(y)−1

i(y)−1

X

X

= k

xj −

j=1

≤k

1 + k

∞ X

j=1

∞ X

yj +

j=i(y)

yj +

∞ X



∞ X

|xj − yj |

j=i(y)

 |xj − yj |

j=i(y)

|xj − yj |

j=1

= k kx − yk . For every x = (x1 , x2 , . . . ) ∈ S + , we have kx − T xk =

i(x) X j=1

xj + xi(x)+1 − kµ(x)xi(x) +

∞ X j=i(x)+1

|xj+1 − kxj |

188

Classification of Lipschitz mappings ≥ 1−

1 + µ(x)xi(x) − xi(x)+1 + kµ(x)xi(x) + k k ∞ X

−

∞ X

xj

j=i(x)+1

xj+1

j=i(x)+1 ∞ X 1 + 1 + µ(x)xi(x) − xj k j=i(x)+1   1 , ≥ 2 1− k
as we desired. Hence, ϕS + (k) = 1 − k1 r (S + ) for all k > 1 and

= 1−

ψ`1 (k) < ϕ`1 (k) = 1 −

1 . k

A similar construction can be done for L1 (0, 1). Example 9.11 ([31]) Let S + be the “positive face” of the unit ball in L1 (0, 1); that is, S+ =

 Z f ∈ L1 (0, 1) : f ≥ 0 and

1

 f (t)dt = 1 .

0

Then, r (S + ) = diam (S + ) = 2. Fix k > 1. For any function f ∈ S + , we define the number tf via relation  Z t  1 tf = sup t : . f (s)ds = 1 − k 0 The reader can check that the mapping T : S + → S + given by ( 0 if t ≤ tf , T f (t) = kf (t) if t > tf , is k-lipschitzian and for every f ∈ S + ,    
1 1 kf − T f k = 2 1 − = 1− r S+ . k k

Consequently, d(T ) = 2 1 − k1 and ϕS + (k) = 2 1 − k1 . This implies that, for every k > 1, ϕL1 (0,1) (k) = 1 − k1 . We have already seen that, for the space `1 , we have ψ`1 (k) < ϕ 12 S + (k) = 1 −

1 . k

Mean lipschitzian mappings with k > 1

189

In the case of Hilbert space H, the situation is different, and for any set C ⊂ H with a Chebyshev radius equal to 1, we have ϕC (k) = ψH (k)

for all k > 1.

Indeed, suppose that T : C → C is of class L (k). Then, an extension Te : H → C of T given by Tex = T P x, where P denotes the nearest point projection onto C (see lemma 9.2), is also k-lipschitzian. Moreover, by nonexpansiveness of P , we get



x − Tex = kx − T P xk ≥ kP x − P T P xk = kP x − T P xk .   Consequently, d Te = d (T ). Now, it is enough to note that there is a unique point z ∈ H such that C ⊂ B (z, 1) and consider the restriction of Te to this ball. For a Hilbert space H, we have (for the proof see [29] or [31], pp. 167, or [36], pp. 214, ex. 20.4):    12 1 k 1 0 ϕH (k) = ψH (k) ≤ 1 − and 0 < ψH (k) ≤ √ . k k+1 2 It is worth stressing that the above estimate was obtained 40 years ago, and so far it is not known whether it is accurate. As opposed to the spaces C [0, 1], BC (R), C0 [0, 1], BC0 (R), c and c0 , in the case of Hilbert space both functions, ψH and ψB→S , differ at least in the vicinity of 1. To see this, consider the following example. Example 9.12 ([31]) Consider the Hilbert space H and let T : B → S be 1 a mapping of class L (k).
Then, the mapping k T is nonexpansive and maps 1 B into the sphere S k ⊂ B. Since every Hilbert space has the fixed point property for nonexpansive mappings, there exists a point x ∈ B satisfying x = k1 T x. Obviously, kxk = k1 . For the same reason, there is a point y ∈ B, satisfying   1 1 y = 1− x + T y. k k Then, 2

2

2

k 2 kx − yk ≥ kT x − T yk = kkx − ky + (k − 1) xk 2

2

2

= k 2 kx − yk + 2k(k − 1) (x − y, x) + (k − 1) kxk and, consequently, 2

2k (k − 1) (y − x, x) ≥

(k − 1) . k2

190

Classification of Lipschitz mappings

Further, 2

2

2

2

1 = kT yk = k(T y − x) + xk = kT y − xk + 2(T y − x, x) + kxk  2 k 1 2 = kT y − yk + 2k (y − x, x) + 2 k−1 k  2 k 1 k−1 2 ≥ + 2 d (T ) + k−1 k2 k  2 k 1 2 = d (T ) + . k−1 k
q Thus, d (T ) ≤ 1 − k1 1 − k1 , and, consequently,  ψB→S (k) ≤

1 1− k

 32 .

(9.2)

0 Thus, ψB→S (1) = 0, whereas ψ 0 (1) > 0. It is not clear whether ψB→S (k) < ψ (k) for all k > 1.

An interesting situation is observed for the function ψS→{0} . Let us begin with some general estimates, true for any space (see [31]). Let X be an arbitrary Banach space. Consider any mapping T : B → B of class L (k) sending all the points on the unit sphere to the origin, T (S) = {0}. Extend T to the mapping Te defined on the whole space X by putting Tex = 0 if kxk ≥ 1. Observe that for any x 6= 0, x ∈ B, we have



Tex − 1 Te0 ≤ 1 kT xk + 1 kT x − T 0k

2 2 2

 

1 x

+ k kxk ≤ T x − T 2 kxk 2 k k k ≤ (1 − kxk) + kxk = . 2 2 2 Also if kxk ≥ 1, or x = 0, then



Tex − 1 Te0 = 1 kT 0k ≤ 1 .

2 2 2 Te0 2

= T20 of radius   k d (T ) = d Te ≤ ψX (k) 2

Thus, Te maps the ball centered at

k 2

into itself. Hence,

and     k k−1 k−1 ψS→{0} (k) ≤ min ψ (k) , ψ (k) ≤ min , 2 2 k  k−1 for 1 ≤ k ≤ 2, 2 = 1 − k1 for k > 2.

(9.3)

Mean lipschitzian mappings with k > 1

191

Consequently, 0 ψS→{0} (1) = lim sup k→1+

ψS→{0} (k) 1 1 ≤ ψ 0 (1) ≤ . k−1 2 2

We leave to the reader the verification of the fact that for any Banach space we 0 have limk→∞ ψS→{0} (k) = 1. There exist spaces, for which ψS→{0} (1) = 12 . Example 9.13 ([39]) Let K be an infinite set, and suppose that X is extremal and truncation invariant subspace of B (K). Fix k > 1 and take
 ∈ 0, 1 − k1 . There exists a mapping T : BX → BX of class L (k) with d (T ) ≥ 1 − k1 − . Extend T to the mapping F : 2BX → BX by putting   T f F f = T Qτ f  Qτ T Qτ f k (2−kf k)

kf k ≤ 1,   kf k ∈ 1, 2 − k1 −  ,   kf k ∈ 2 − k1 − , 2 .

if if if

1+k

First, observe that on the boundary of each region, appropriate formulas coincide and F (2SX ) = {0}. Again, F ∈ L (k). Indeed, • if f, g ∈ BX , then kF f − F gk = kT f − T gk ≤ k kf − gk;   • if kf k , kgk ∈ 1, 2 − k1 −  , then having used the fact that Qτ is nonexpansive, we have kF f − F gk = kT Qτ f − T Qτ gk ≤ k kQτ f − Qτ gk ≤ k kf − gk ;   • if kf k , kgk ∈ 2 − k1 − , 2 , then according to lemma 9.3, we get the estimate

kF f − F gk = Qτ

k 1+k (2−kf k)

T Qτ f − Qτ

k 1+k (2−kgk)

 ≤ max kT Qτ f − T Qτ gk ,

T Qτ g

 k |kgk − kf k| 1 + k ≤ max {k kf − gk , k kf − gk} = k kf − gk .

The above implies that for all f, g ∈ 2BX , kF f − F gk ≤ k kf − gk as we desired. Observe also that, for any f ∈ 2BX , we have kf − F f k ≥ 1 − Indeed,

1 − . k

192

Classification of Lipschitz mappings

• for f ∈ BX it is obvious;   • if kf k ∈ 1, 2 − k1 −  , then kf − F f k = kf − T Qτ f k ≥ kQτ f − Qτ T Qτ f k = kQτ f − T Qτ f k 1 ≥ 1 − − ; k  • if kf k ∈ 2 −

1 k

 − , 2 , then

kf − F f k = f − Qτ

k 1+k (2−kf k)

≥ 2−



T Qτ f ≥ kf k − Qτ

k 1+k (2−kf k)

T Qτ f

1 1 −  − 1 = 1 − − . k k

Now, putting Ff =

1 F (2f ) , 2

we obtain a mapping F : BX → BX of class L (k) such that F (SX ) = {0} and for any f ∈ BX  

f − F f = 1 k2f − F (2f )k ≥ 1 1 − 1 −  . 2 2 k Finally, by taking T of class L (k) with d (T ) close to ψX (k) = 1 − k1 , we end up with the estimate   1 1 1 1− . ψSX →{0} (k) ≥ ψX (k) = 2 2 k Consequently, having considered 9.3, we obtain ψS0 X →{0} (1) = 12 . The above estimate is highly inaccurate given the fact that, for any Banach space, we have limk→∞ ψS→{0} (k) = 1. We have already seen that, for any space X, ψS→{0} (k) < ψX (k) for all k ∈ (1, 2). In the light of foregoing, the following questions arise. Are there spaces for which ψS→{0} (k) = ψX (k) for sufficiently large k? Moreover, is there a space X, for which ψS→{0} (k) = 1 − k1 on a certain subset of [2, ∞)? The construction presented in the following example shows that answers to all the above questions are affirmative. Example 9.14 ([41]) Let us consider the space C0 [0, 1] with the standard sup norm. In [38], it was proved that ψS→{0} (3) = 23 . Below we present a general situation, for arbitrary k > 1. Let us consider two cases, for k ∈ (1, 3) and for k ≥ 3.

Mean lipschitzian mappings with k > 1

193

If k ≥ 3, then we define the mapping Fk : B → B by ( QΛ (k (|f | + g)) if 0 ≤ kf k ≤ 1 − k1 , Fk f = τ Λ Qk(1−kf k) Q (k (|f | + g)) if 1 − k1 ≤ kf k ≤ 1. Obviously, Fk (S) = {0} and in view of lemma 9.3 we conclude that Fk is of class L (k). We shall prove that, for any f ∈ B, kFk f − f k ≥ 1 − k1 . If kf k ≤ 1 − k1 , then by applying a similar reasoning to this presented in example 9.7 we can select a point s
satisfying |f (s)| + g (s) = k1 to obtain kf − Tk f k > 1 − k1 . If kf k ∈ 1 − k1 , 1 , then by taking a point t0 ∈ (0, 1] such that |f (t0 )| = kf k and using the fact that k2 ≤ 1 − k1 if and only if k ≥ 3 we get the following estimate   Λ (k (|f (t0 )| + g (t0 ))) kf − Fk f k ≥ f (t0 ) − k (1 − kf k) τ k (1 − kf k)   0 = f (t0 ) − k (1 − kf k) τ k (1 − kf k) 1 = |f (t0 )| ≥ 1 − . k If kf k = 1, then kf − Fk f k = kf − 0k = 1 > 1 − k1 . This implies that ψS→{0} (k) = ψC0 [0,1] (k) = 1 −

1 k

for every k ≥ 3. Consider now the case k ∈ (1, 3). Define the mapping Fk : as

Fk f =

 QΛ (k (|f | + g))

4 k+1 B

→

4 k+1 B

3k−1 k(k+1) , 4 kf k ≤ k+1 .

if 0 ≤ kf k ≤

3k−1 ≤ QΛ (k (|f | + g)) if k(k+1) 4 −kf k) ( k+1   4 It is clear that Fk k+1 S = {0}. Again, using lemma 9.3, we conclude that Fk is of class L (k). 1 We shall prove that, for all f ∈ k+1 4 B, we have kf − Fk f k ≥ 1 − k . For 4 3k−1 f ∈ k+1 S, it is clear. For f ∈ k(k+1) B, we get it in a way known from example h i 3k−1 4 9.7. For f with kf k ∈ k(k+1) , k+1 , we take t0 ∈ (0, 1] such that |f (t0 )| = kf k and considerh two cases. i 3k−1 2 If kf k ∈ k(k+1) , k , then we obtain

Qτk

    Λ (k (|f (t0 )| + g(t0 )))  4   kf − Fk f k ≥ |f (t0 )| − k − |f (t0 )| τ  4 k+1 k k+1 − |f (t0 )|

194

Classification of Lipschitz mappings     4 Λ (k |f (t0 )|)   ≥ |f (t0 )| − k − |f (t0 )| τ   k+1 k 4 − |f (t )| 0

k+1

 = |f (t0 )| − k

4 k+1

 − |f (t0 )|

2 − k |f (t0 )|   4 k k+1 − |f (t0 )|

= |f (t0 )| − 2 + k |f (t0 )| = (k + 1) |f (t0 )| − 2 3k − 1 1 ≥ (k + 1) −2=1− . k(k + 1) k  h 4 2 If kf k ∈ k , k+1 , then we get    4 Λ (k |f (t )|) 0  kf − Fk f k ≥ |f (t0 )| − k − |f (t0 )| τ   4 k+1 k k+1 − |f (t0 )|     4 0  = |f (t0 )| − k − |f (t0 )| τ   4 k+1 k − |f (t )| 

k+1

= |f (t0 )| ≥

0

1 2 ≥1− . k k

Now, it is enough to define the mapping Fk : B → B by putting for every f ∈B   k+1 4 Fk f , Fk f = 4 k+1 which is of class L (k), Fk (S) = {0}, and

 



4

f − Fk f = k + 1 4 f − Fk f

4 k+1 k+1   k+1 1 ≥ 1− . 4 k Consequently, for every k ∈ (1, 3), we have k+1 ψS→{0} (k) ≥ 4

  1 1− . k

Taking into account (9.3), we end up with         1 k+1 1 k 1− min , 1 ≤ ψS→{0} (k) ≤ 1 − min ,1 k 4 k 2
k+1 1 for all k ≥ 1. It is not known whether ψS→{0} (k) = 4 1 − k for 1 < k < 3. However, we have an exact value of 0 ψS→{0} (1) = lim+ k→1

ψS→{0} (k) 1 1 = ψ 0 (1) = . k−1 2 2

Mean lipschitzian mappings with k > 1

195

It is still open problem whether, for each Banach space X, ψS→{0} (k) = ψX (k) for some k > 2. Let us pass now to the case of Hilbert space. A convenient tool in the construction presented below will be the following well-known theorem first formulated by Kirzbraun [50] for Euclidean space and then extended to any Hilbert space by Valentine [76]. Theorem 9.13 (Kirzbraun-Valentine Theorem) Let A be an arbitrary subset of a Hilbert space H. Suppose that T : A → H is k-lipschitzian. Then, there exists a k-lipschitzian extension Te : H → Conv (T (A)) of T . Example 9.15 ([31]) Let H be a Hilbert space and T0 : B → B be a √ mapping of class L (k) with d (T0 ) = d > 0. Then, for any x ∈ B with kxk ≥ 1 − d2 , we have 2

2

2

0 ≤ kxk + d2 − 1 ≤ kxk + kT0 x − xk − 1 2

= (x, x) + kT0 xk − (x, T0 x) − (T0 x − x, x) − 1 2

= −2 (T0 x − x, x) + kT0 xk − 1 ≤ −2 (T0 x − x, x) . Consequently, (T0 x − x, x) ≤ 0.

(9.4)

Let us modify T0 defining T1 : B → B as √  2 T0 x if kxk √ ≤ 1−d , T1 x = 2 √ T0 P 1−d2 x if 1 − d ≤ kxk ≤ 1, √ where P√1−d2 is the radial projection onto the ball 1 − d2 B. Since P√1−d2 is nonexpansive, T1 is also of class L (k). We shall prove that, for any x ∈ S and for any µ ∈ [0, 1] and λ ≥ 1, kµT1 x − λxk ≥ d. Indeed, using (9.4), we get

2 p 

2 kµT0 P x − λxk = µT0 1 − d2 x − λx

  p  2 p p

1 − d2 x − µ 1 − d2 x + µ 1 − d2 − λ x = µT0

p

2  p

= µ2 T0 1 − d2 x − 1 − d2 x √
 p  p µ µ 1 − d 2 − λ  p √ +2 T0 1 − d2 x − 1 − d2 x, 1 − d2 x 1 − d2  p 2 2 + µ 1 − d2 − λ kxk

(9.5)

196

Classification of Lipschitz mappings  p 2 ≥ µ2 d2 + µ 1 − d2 − λ ≥ d2 .

 calculations also imply that, for any x ∈ B with kxk ∈ √ The above 1 − d2 , 1 , we have kT1 x − xk ≥ d and for x ∈ S, r   p kx − T1 xk ≥ 2 1 − 1 − d2 > d. In particular, the above implies that d (T1 ) ≥ d (T0 ) = d > 0. Now, let us define the domain   1 kT1 P xk D = x ∈ H : kxk ≤ 1 + − k k and extend T1 to the mapping T2 : D → B putting  T1 x for x ∈ B, T2 x = T1 P x for x ∈ D \ B, where P denotes the radial projection of H onto B. Obviously, T2 ∈ L (k) and, in view of
(9.5), d (T2 ) ≥ d. Again, let us extend T2 defining the mapping T3 : D ∪ 1 + k1 S → B as  T2 x if x ∈ D,
T3 x = 0 if x ∈ 1 + k1 S. It is clear that T3 is still of class L (k) and has d (T3 ) = d (T2 ) ≥ d > 0. In view of the Kirzbraun-Valentine Theorem, T3 can be once more extended to
the mapping T4 : 1 + k1 B → B of class L (k). Observe that  for every x ∈ S

such an extension must map the segment joining points 1 + k1 − kT1kP xk x
and 1 + k1 x, of length kT1kP xk , onto an arc joining T1 P x and 0 of length not less than kT1 P xk. Thus, this arc must be the segment. Consequently, an extension can be done in exactly one way:  T2 x if x ∈ D,

T4 x = x k 1 + k1 − kxk kTT11 P if x ∈ 1 + k1 B \ D. P xk According to (9.5), we have d (T4 ) ≥ d. Now, we can define T : B → B of class L (k) with T (S) = {0} as   k k+1 Tx = T4 x . k+1 k k Obviously, d (T ) ≥ k+1 d, and, consequently, since d can be chosen arbitrarily close to ψH (k), we get

ψS→{0} (k) ≥

k ψH (k) . k+1

It is not known whether the above estimate is accurate.

(9.6)

Mean lipschitzian mappings with k > 1

9.4

197

Optimal retractions

The functions ψX , ψB→S and ψS→{0} are closely related to another highly nontrivial problem posed by Goebel in 1973, which, roughly speaking, consists in constructing a retraction of the closed unit ball onto the unit sphere with the smallest possible Lipschitz constant. More precisely, for a given infinitely dimensional Banach space X, we define the optimal retraction constant: k0 (X) = inf {k : there exists a retraction R : B → S of class L (k)} . The optimal retraction problem involves finding or evaluating the constant k0 (X). At present, the exact value of k0 (X) is not known for any single Banach space. It is a general feeling that a value of the constant k0 (X) should somehow depend on the geometry of the space X, and its value should increase as the geometry gets a more “regular” structure. There were several approaches to give a reasonable universal estimate from above. All of them ended on the level of the high thousands. Much better estimates exist for particular spaces or some classes of spaces. All of them have been obtained by a range of individual methods, constructions, and tricks developed by a number of authors. Below, we present the best of currently known estimates. Let us begin with some basic estimates from below. Suppose R : B → S is a retraction of class L (k). Consider the mapping n T = −R and observe that T : B → S, T n = (−1) R, and this implies that T is uniformly lipschitzian, T ∈ Lu (k). Moreover, T is fixed point free. Take any point x ∈ B and consider the segment [x, T x]. The mapping T maps this segment onto a rectifiable curve γ contained in S and joining two antipodal points T x = −Rx and T 2 x = Rx via relation γ(s) = T ((1 − s) x + sT x) , where s ∈ [0, 1]. The length of γ, defined as l (γ) = sup

n−1 X

kγ (si ) − γ (si+1 )k ,

i=0

where the supremum is taken over all finite partitions 0 = s0 < s1 < s2 < · · · < sn = 1 of [0, 1], exceeds the girth g(X) of the sphere defined as the infimum of lengths of all curves joining two antipodal points. On the other hand, since T ∈ L(k) we get

l(γ) ≤ sup

n−1 X i=0

kT ((1 − si ) x + si T x) − T ((1 − si+1 ) x + si+1 T x)k

198

Classification of Lipschitz mappings ≤ sup

n−1 X

k (si+1 − si ) kx − T xk = k kx − T xk .

i=0

The above implies that l(γ) satisfies g(X) ≤ l(γ) ≤ k kx − T xk , and, consequently, the minimal displacement of T is positive, d(T ) ≥

g(X) > 0. k

Further, since d(T ) ≤ ψB→S (k), we get the inequality, which can be used to obtain some basic estimates for k0 (X): kψB→S (k) ≥ g(X).

(9.7)

Obviously, g(X) ≥ 2 for all spaces X, and there are some flat spaces for which g(X) = 2. Since ψB→S (k) ≤ ψ(k) ≤ 1− k1 we get the first basic estimate for the optimal retraction constant, k0 (X) ≥ 3 for any space X. However, for some spaces there are better estimates. For the space `1 , we have g (`1 ) = 2 and by using the estimate (9.1) to solve the inequality (9.7), we get √ k0 (`1 ) ≥ 17 − 8 3 = 3.143 · · · . A better result has been obtained in [12], and it states that k0 (`1 ) ≥ 4. It is known that for all uniformly convex spaces g(X) > 2, and, consequently, k0 (X) > 3 for every such space. In the particular case of Hilbert space H, g(H) = π. Having used this fact and the evaluation (9.2) for ψB→S presented in example 9.12, we conclude that k0 (H) exceeds the solution of the equation  3 1 2 k 1− = π. k Thus, k0 (H) > 4.5 · · · . For details concerning the girth of the sphere, see [73]. Considerably more published results concern estimates from above. All the results of this type have been obtained via concrete examples of mappings. Let us begin with a construction in the space `1 presented by Annoni and Casini in [2]. Example 9.16 Consider the space `1 furnished with the standard norm. Let i : B\ 12 B → N be the function defined for x = (x1 , x2 , . . . ) ∈ `1 with kxk ∈ 1 2 , 1 as   ∞   X i (x) = min j ∈ N : |xk | < 1 − kxk   k=j+1

Mean lipschitzian mappings with k > 1

199

and let the map µ : B \ 21 B → (0, 1] be given via relation ∞ X

µ (x) xi(x) +

|xk | = 1 − kxk .

k=i(x)+1 ∞

Now, if {ek }k=1 denotes the standard Schauder basis in `1 , we define Q : P∞ B \ 21 B → 21 B by putting for every x = k=1 xk ek ( P∞ µ (x) xi(x) ei(x) + k=i(x)+1 xk ek Qx = 0

if if

1 2

≤ kxk < 1, kxk = 1.

Justification that Q is 3-lipschitzian and I − Q is 2-lipschitzian, where I denotes the identity map, is left for the reader as an exercise. We shall also need the “right shift” mapping A : B → B, ! ∞ ∞ X X Ax = A xk e k = xk ek+1 . k=1

k=1

Obviously, A as an isometry is of class L (1). Finally, we can define R : B → S as ( (1 − 2 kxk) e1 + 2Ax if Rx = (I − Q)x + 2AQx if

0 ≤ kxk ≤ 12 , 1 2 ≤ kxk ≤ 1.

The reader can easily verify that R is a retraction of B onto S and that R ∈ L (8). Consequently, k0 (`1 ) ≤ 8, and it is the best known upper bound for this space. Next, with the use of the same tricks, the above result was extended for L1 (0, 1) and for few other spaces (see [39]). Below, we present yet another simple construction in the space L1 (0, 1) given by the present author (unpublished), which is also based on an idea originated by Annoni and Casini. Example 9.17 Consider the space L1 (0, 1). For every function f ∈ B, we define the number tf as   Z 1 tf = inf t ∈ (0, 1) : |f (t)| dt ≤ 1 − kf k . t

If kf k ≤

1 2,

then tf = 0. If kf k ∈ Z

1

2, 1

 , then

1

|f (t)| dt = 1 − kf k . tf

Let us define a mapping T : B \ 12 B → S as

200

Classification of Lipschitz mappings ( f (t) (T f ) (t) = 2 |f (t)|

if t ∈ (0, tf ] , if t ∈ (tf , 1) .

We shall prove that T is 8-lipschitzian. Let f, g ∈ B \ 21 B, and without loss of generality, we can assume that tf ≤ tg . Then, tf

Z kT f − T gk =

Z

tg

|f (t) − g(t)| dt + 0

|2 |f (t)| − g(t)| dt tf

Z

1

|2 |f (t)| − 2 |g(t)|| dt

+ tg

Z

tf

≤

Z

tg

|f (t) − g(t)| dt + 0

|2 |f (t)| − 2 |g(t)|| dt tf

Z

tg

Z

1

|g(t)| dt +

+3

|2 |f (t)| − 2 |g(t)|| dt

tf

tg

Z

1

≤ 2 kf − gk + 3

Z |g(t)| dt −

tf

Z

!

1

|g(t)| dt tg

!

1

= 2 kf − gk + 3

|g(t)| dt − 1 + kgk tf

Z

1

= 2 kf − gk + 3

Z |g(t)| dt − kf k −

tf

Z

1

≤ 2 kf − gk + 3

!

1

|f (t)| dt + kgk tf

! |f (t) − g(t)| dt + kf − gk

tf

≤ 8 kf − gk . Now we can define the mapping R : B → S as ( 2 |f (t)| − 2 kf k + 1 if kf k ≤ 21 ,   (Rf )(t) = (T f )(t) if kf k ∈ 12 , 1 . and the reader can easily verify that R is an 8-lipschitzian retraction of the solid unit ball B onto its boundary S. Hence, k0 (L1 (0, 1)) ≤ 8, and it is the best known estimate for this space. Let us pass now to the spaces furnished with the sup norm. For many years, √
2 the best known estimate for the space C [0, 1] was k0 (C [0, 1]) ≤ 4 1 + 2 = 23.31 · · · (see [31]). This estimate was obtained by the following procedure, which works in case of any Banach space X. Once we have the mapping T : BX → BX of class T ∈ L (k) such that d (T ) > 0 and T (SX ) = {0}, we

Mean lipschitzian mappings with k > 1

201

can construct the retraction R : BX → SX by putting   x − Tx x − Tx Rx = =P , kx − T xk d (T ) where P denotes the radial projection. Then, k (R) ≤ k (P )

k+1 . d (T )

Since the mapping T can be chosen having d (T ) arbitrarily close to ψSX →{0} (k), we conclude that k0 (X) ≤ k (P ) inf

k>1

k+1 ψSX →{0} (k)

.

(9.8)

We have already seen in example 9.13 that for X = C [0, 1], and in general, for all extremal and truncation invariant subspaces X of B (K),   1 1 1 ψSX →{0} (k) ≥ 1− = ψX (k) . 2 k 2 Thus, for any such space (see [39]) 3 ≤ k0 (X) ≤ 4 min k>1

 √ 2 (k + 1) k = 4 1 + 2 = 23.31 · · · . k−1

As we have already mentioned in the previous section, it was proved in [38] that for the space C0 [0, 1] we have ψS→{0} (3) = 32 and, consequently, k0 (C0 [0, 1]) ≤ 12. Later, a better estimate was obtained in [40], k0 (C0 [0, 1]) ≤ 7. A disadvantage of the above method is the fact that for spaces c, c0 , BC (R), BC0 (R), C [0, 1], C0 [0, 1] and many others we have k (P ) = 2; hence, the biggest possible Lipschitz constant for the radial projection. However, it was observed by the present author in [67] that, in the case of the above-mentioned spaces, we can omit the radial projection P by applying the following receipt. Again, suppose that X is an infinite-dimensional and truncation invariant subspace of B (K). Consider a mapping T : BX → BX of class L (k) such that d (T ) = d > 0 and T (SX ) = {0}. Denote q 2 k + 1 + (k + 1) − 4kd r= 2k and define the mapping F : rBX → X as ( f − Tf Ff = (k + 1 − k kf k) f

if if

kf k ≤ 1, kf k ∈ [1, r] .

202

Classification of Lipschitz mappings

It is easy to verify that, for any f ∈ rBX , we have kF f k ≥ d. Moreover, the sphere with a radius r is mapped radially onto the sphere with a radius d; that is, for every f ∈ rSX , we have F f = dr f . For a convenience of the reader we shall prove that F ∈ L (k + 1). It is clear that F is (k + 1)-lipschitzian on BX . To see that it also holds on the subset rBX \ BX , first assume that kgk ≤ kf k. Then, kF f − F gk = k(k + 1 − k kf k) f − (k + 1 − k kgk) gk ≤ k(k + 1 − k kf k) (f − g)k + k(k + 1 − k kf k) g − (k + 1 − k kgk) gk ≤ (k + 1 − k kf k) kf − gk + k kgk (kf k − kgk) ≤ (k + 1 − k kf k + k kgk) kf − gk ≤ (k + 1) kf − gk . If kf k ≤ kgk, then it is enough to exchange roles between f and g in the above estimate. Finally, for all f, g ∈ rBX , we have kF f − F gk ≤ (k + 1) kf − gk . Now define the mapping F : BX → X as Ff =

1 F (rf ) . r

We leave to the reader checking that • F ∈ L (k + 1);

• for any f ∈ BX we have F f ≥ dr ; • for every f ∈ SX we have F f = dr f . Finally, we can define the retraction R : BX → SX by putting  r Ff Rf = Qτ d and observe that

r   r  r r



kRf − Rgk = Qτ F f − Qτ F g ≤ F f − F g d d d d r ≤ (k + 1) kf − gk . d Consequently, since T can be selected with d (T ) arbitrarily close to ψSX →{0} (k), we obtain the following Theorem 9.14 ([67]) If X is an infinite-dimensional and truncation invariant subspace of B (K), then q 2 k + 1 + (k + 1) − 4kψSX →{0} (k) k0 (X) ≤ inf (k + 1) . k>1 2kψSX →{0} (k)

Mean lipschitzian mappings with k > 1

203

Example 9.18 (see [69] or [67]) Consider an extremal and truncation invariant subspace X of B (K). We have already seen in example 9.13 that   1 1 ψSX →{0} (k) ≥ 1− . 2 k Thus, in virtue of theorem 9.14, we get the estimate

k0 (X) ≤ min (k + 1)

k+1+

q

k>1

√

= min (k + 1) k>1

2

(k + 1) − 2k(1 − k1 ) k(1 − k1 )

k + 1 + k2 + 3 . k−1

We leave to the reader checking that the minimum is achieved for k = 3. Consequently, √ k0 (X) ≤ 4(2 + 3) = 14.92 · · · . We have already seen that among such spaces there are, for instance, c, c0 , C [0, 1], and BC (R). Example 9.19 (see [67] or [68]) Consider the space C0 [0, 1]. A construction presented in example 9.14 shows that   k+1 1 ψS→{0} (k) ≥ 1− for k ∈ [1, 3] 4 k and

1 for k ≥ 3. k Consequently, the expression in theorem 9.14 takes its minimal value for k = 3. Thus, we have √ 3 ≤ k0 (C0 [0, 1]) ≤ 2(2 + 2) = 6.828 · · · . ψS→{0} (k) = 1 −

At present the above estimate is the minimum of upper bounds over all the Banach spaces for which the upper bound is known! The above construction can be extended to a much wider class of spaces. Example 9.20 (see [67] or [68]) Suppose that (M, d) is a connected metric space consisting of more than one point and z ∈ M is a given point. Consider the space BCz (M ) consisting of all bounded, continuous functions f : M → R vanishing at z, f (z) = 0, and furnished with the standard sup norm. First, assume that (M, d) is unbounded and define T as    d(x, z) . T f (x) = Λ 3 |f (x)| + 1 + d(x, z)

204

Classification of Lipschitz mappings

Then, T : B → B and T ∈ L (3). Since M is a connected metric space, for every f ∈ B there exists a point x1 ∈ M such that |f (x1 )| +

d(x1 , z) 1 = , 1 + d(x1 , z) 3

and, consequently, kf − T f k ≥ 23 .

 √  In the next step, define a mapping F : 2+3 2 B → BCz (M ) as   if kf k ≤ 23 ,  f − T f   τ if kf k ∈ 32 , 1 , F f = f − Q3(1−kf k) T f h √ i   (4 − 3 kf k) f if kf k ∈ 1, 2+3 2 .

We leave  √  to the reader to check that F ∈ L (4) and kF f k ≥ 2+ 2 B. 3 Now, having denoted √ ! 3 2+ 2 ˜ √ F Ff = f , 3 2+ 2

2 3

for any f ∈

˜ we obtain a mapping

F : B → BCz (M ) of class L (4) such that, for every

˜ f ∈ B, we have F f ≥ 2+2√2 and for every f ∈ S, F˜ f = 2+2√2 f . Finally, we can define the retraction R : B → S as ! √ 2+ 2 ˜ τ Ff Rf = Q 2 and observe that for all f, g ∈ B we have √

 √  2+ 2

˜

kRf − Rgk ≤

F f − F˜ g ≤ 2 2 + 2 kf − gk . 2 If (M, d) is bounded, then it is enough to put m = sup {d (x, z) : x ∈ M } and modify T by putting    d (x, z) T f (x) = Λ 3 |f (x)| + . m The proof carries on with only minor technical changes. All the above implies that  √  3 ≤ k0 (BCz (M )) ≤ 2 2 + 2 = 6.828 · · · .

Mean lipschitzian mappings with k > 1

205

It is worth noting that for the whole class of extremal spaces the best known estimate from above states k0 (X) ≤ 30.84 (see [6]). The case of Hilbert space H is very challenging. Because of its geometric regularity, it opposes the examination of minimal displacement and optimal retraction. The first upper bound k0 (H) ≤ 64.25 was obtained by Komorowski and Wo´sko in [53]. Subsequently, there were several improvements, for instance, k0 (H) ≤ 32.26 in [13] and k0 (H) ≤ 28.99 from [6]. The last estimate is the best known published estimate for the Hilbert space. There are some relations that tie the unknown value of k0 (H) and the unknown function ψH . Since, in this case k (P ) = 1, then in view of (9.8) and (9.6), we have 2

(k + 1) k+1 ≤ inf . k>1 kψH (k) k>1 ψS→{0} (k)
Since for any space ψ (k) ≥ ψ 0 (1) 1 − k1 , we get k0 (H) ≤ inf

2

(k + 1) = 8. k>1 k − 1

0 k0 (H) ψH (1) ≤ inf

It is not known if the above estimates are sharp. In spite of a forty-year efforts, the quantitative problems of minimal displacement and optimal retraction are still open for consideration. We hold a deep-seated belief that the area needs new ideas and approaches.

9.5

Generalized characteristics of minimal displacement

By analogy to the classical case, we define minimal displacement functions for classes L(α, p, k). As usual, C denotes a bounded, closed, convex, and noncompact subset of an infinitely dimensional Banach space X with r(C) = 1. By ϕC , we denote a function defined for every k ≥ 0 and α = (α1 , . . . , αn ) by ϕC (α, k) = sup {d(T ) : T : C → C, T ∈ L(α, k)} = sup {inf {kx − T xk : x ∈ C} : T : C → C, T ∈ L(α, k)} and by ϕX a function given for every k ≥ 0 and α = (α1 , . . . , αn ) by ϕX (α, k) = sup {ϕC (α, k) : C ⊂ X, r(C) = 1} . If C = BX , then we write ψX instead of ϕBX , and we refer to ψX as the minimal displacement characteristic of X for mappings in the class L(α, k).

206

Classification of Lipschitz mappings

Thus, ψX (α, k) = sup {d(T ) : T : BX → BX , T ∈ L(α, k)} . Similarly, for classes L(α, p, k), we define functions ϕC , ϕX and ψX as ϕC (α, p, k) = sup {d(T ) : T : C → C, T ∈ L(α, p, k)} , ϕX (α, p, k) = sup {ϕC (α, p, k) : C ⊂ X, r(C) = 1} and ψX (α, p, k) = sup {d(T ) : T : BX → BX , T ∈ L(α, p, k)} . If p = 1, then we identify ϕC (α, 1, k) with ϕC (α, k). The same holds for ψX (α, 1, k) and ϕX (α, 1, k). It is easy to see that for a multi-index α = (1) of length n = 1 the above definitions coincide with the classical definitions of characteristics of minimal displacement; that is, ϕC (α, p, k) = ϕC (k) = sup {d(T ) : T : C → C, T ∈ L(k)} , ϕX (α, p, k) = ϕX (k) = sup {ϕC (k) : C ⊂ X, r(C) = 1} and ψX (α, p, k) = ψX (k) = sup {d(T ) : T : BX → BX , T ∈ L(k)} . Let us list some immediate consequences of the above definitions: • Since L(α, p, k) ⊂ L(α, p, l) for k < l, we conclude that for fixed α and p, functions ϕC , ϕX and ψX are nondecreasing. • Since L(α, p, k) ⊂ L(α, q, k) for p > q, we conclude that for fixed α and k functions ϕC , ϕX and ψX are nonincreasing. • For any α, p, and k, we have ϕC (α, p, k) ≤ ϕX (α, p, k). In particular, ψX (α, p, k) ≤ ϕX (α, p, k). • In view of theorem 3.9, for any α, p, and k < 1, we have ϕC (α, p, k) = 0. Consequently, for any α, p, and k < 1, we have ϕX (α, p, k) = ψX (α, p, k) = 0. • Using (5.2), for any α = (α1 , . . . , αn ), p and k, the class L(α, p, k) contains the class L(l) with l satisfying the following equation: n X

αi lip = k p .

i=1

It is clear that l > 1 provided k > 1. Thus, in view of theorem 9.11, for any k > 1, we have ϕC (α, p, k) ≥ ϕC (l) > 0.

Mean lipschitzian mappings with k > 1

207

The first evaluation of the minimal displacement characteristic of c0 for mappings in the class L(α, k) was presented in [42]: Example 9.21 Let c0 denotes the space of all sequences converging to zero with the standard sup norm, kxk = k(x1 , x2 , . . . )k = supi=1,2,... |xi |. Let τ : R → [−1, 1] be the truncation function as in example 9.4. For arbitrary k > 1, let us define the mapping T : Bc0 → Bc0 by T x = T (x1 , x2 , . . . ) = (1, τ (kx1 ), τ (kx2 ), . . . ). We leave to the reader the verification that T is lipschitzian with k(T ) = k and for all x ∈ Bc0 we have kx − T xk > 1 − 1/k. The space c0 is isometric to the product c0 × c0 with the maximum norm. The unit ball in this setting is the product of two unit balls Bc0 . Let us define the mapping F : Bc0 × Bc0 → Bc0 × Bc0 by F (x, y) = (y, T x). Then,

F 2 (x, y) = (T x, T y) , F 3 (x, y) = T y, T 2 x , F 4 (x, y) = T 2 x, T 2 y ,

F 5 (x, y) = T 2 y, T 3 x , F 6 (x, y) = T 3 x, T 3 y , . . . . Thus, k(F ) = k(F 2 ) = k, k(F 3 ) = k(F 4 ) = k 2 , k(F 5 ) = k(F 6 ) = k 3 , . . . . Consequently, for any multi-index α of length 2, F ∈ L(α, k). Further, for all (x, y) ∈ Bc0 × Bc0 , we have kF (x, y) − (x, y)k = = ≥ ≥ ≥

k(y, T x) − (x, y)k max {ky − xk , kT x − yk} max {ky − xk , kT x − xk − ky − xk} max {ky − xk , 1 − 1/k − ky − xk} 1/2 (1 − 1/k) .

Thus, for all α of length 2, 1 1 ψc0 (α, k) ≥ ψc0 (k) = 2 2

  1 1− . k

Fix a multi-index α = (α1 , . . . , αn ) and k > 1. Let w : R → R be defined by w(t) = α1 t + α2 t2 + · · · + αn tn and l (more precisely l = l(α, k)) denotes the unique solution of the equation w(t) = k for t > 1.

208

Classification of Lipschitz mappings

Hence, l > 1 and w(l) = α1 l + α2 l2 + · · · + αn ln = k. Under the above notations, we formulate the following: Theorem 9.15 For every α = (α1 , . . . , αn ) and k > 1, we have   1 α1 0 ϕC (1) 1 − ≤ ϕC (α, k) ≤ ϕC (k/α1 ) ≤ 1 − , l k   α1 1 0 ψX ≤ ψX (α, k) ≤ ψX (k/α1 ) ≤ 1 − (1) 1 − l k and

(9.9)

  1 α1 . ϕ0X (1) 1 − ≤ ϕX (α, k) ≤ ϕX (k/α1 ) ≤ 1 − l k

Proof. We shall prove inequality (9.9). Using the fact that L(α, k) ⊂ L(k/α1 ) and lemma 9.4, we get the right-hand side of (9.9). To prove the lefthand side of (9.9), we observe that, in view of (3.10), L(l) ⊂ L(α, k). Thus, ϕC (α, k) ≥ ϕC (l), and from lemma 9.4, we finally obtain   1 ϕC (α, k) ≥ ϕC (l) ≥ ϕ0C (1) 1 − . l 

A disadvantage of theorem 9.15 is that we have to determine the root l of polynomial w(t) − k. We can remedy this problem by making the above inequality weaker but much easier to determine, as seen below: Theorem 9.16 For any α = (α1 , . . . , αn ) and k > 1, we have   ϕ0 (1) 1 ϕC (α, k) ≥ C0 1− , w (k) k   1 ψ 0 (1) 1− ψX (α, k) ≥ X0 w (k) k and

ϕ0 (1) ϕX (α, k) ≥ X0 w (k)



1 1− k

 .

Proof. Similarly to the previous case, we shall prove only the first inequality. Using (3.10), we get L(l) ⊂ L(α, k). Thus, in view of lemma 9.4, for s < l, we can write ϕC (α, k) ≥ ϕC (l) ≥

sϕC (s) l−1 k w(l) − 1 · · · . s − 1 w(l) − 1 l k

Mean lipschitzian mappings with k > 1

209

Using lemma 9.4 and the fact that k ≥ l, we obtain   1 l−1 ϕC (α, k) ≥ ϕ0C (1) · · 1− . w(l) − 1 k In view of Lagrange’s Mean Value Theorem, we have w(l) − 1 = w0 (θ) for some θ ∈ (1, l). l−1 Since w0 (θ) ≤ w0 (l) ≤ w0 (k), we finally get ϕC (α, k) ≥

ϕ0C (1) w0 (k)

 1−

1 k

 . 

In case of multi-index α of length n = 2, we can give better evaluations: Theorem 9.17 For any α = (α1 , α2 ) and k > 1, we have   ϕ0C (1) 1 ϕC (α, k) ≥ 0 1− , w (1) k   ψ 0 (1) 1 ψX (α, k) ≥ X0 1− , w (1) k and

ϕ0 (1) ϕX (α, k) ≥ X0 w (1)



1 1− k

(9.10)

 .

Proof. We start as in the proof of the previous theorem, ϕC (α, k) ≥ ϕ0C (1) ·

l−1 w(l) w(l) − 1 · · . w(l) − 1 l w(l)

To get the thesis, it is enough to show that, for any s > 1, we have s−1 w(s) 1 · ≥ 0 , w(s) − 1 s w (1)

(9.11)

or equivalently, (w0 (1) − 1) w(s) − w0 (1)

w(s) + 1 ≥ 0. s

Indeed, the left-hand side of the last inequality is equal to
(α1 + 2α2 − 1) α1 s + α2 s2 − (α1 + 2α2 ) (α1 + α2 s) + 1 and since α1 + 2α2 − 1 = α2 and α1 + 2α2 = 1 + α2 , the above is equal to

210

Classification of Lipschitz mappings
α2 α1 s + α2 s2 − (1 + α2 ) (α1 + α2 s) + 1 = α1 α2 s + α22 s2 − α1 − α1 α2 − α2 s − α22 s + 1
= α22 s2 + α1 α2 − α2 − α22 s + 1 − α1 − α1 α2   = α22 s2 + α2 (α1 − 1) − α22 s + α2 (1 − α1 ) = α22 s2 − 2α22 s + α22 2

= α22 (s − 1) ≥ 0 as desired. 

In general, the evaluation (9.11) does not hold; that is, it is not true that, for any multi-index α = (α1 , . . . , αn ) and s > 1, we have s−1 w(s) 1 · ≥ 0 . w(s) − 1 s w (1) To see this, consider, for example,     98 1 1 3 1 1 1 α= , , or α = , , , . 100 100 100 4 12 12 12 Nevertheless, it is easy to verify that, for any multi-index α = (α1 , α2 ) of length n = 2 and k > 1, we obtain l=

2k p . α1 + α12 + 4α2 k

Hence, in view of lemma 9.4, ϕC (α, k) ≥ ϕ0C (1) 1 −

α1 +

! p α12 + 4α2 k , 2k

which is better bound than (9.10). Moreover, in the case of extremal space X, 0 we have ψX (1) = 1, and, consequently,   p α1 + α12 + 4α2 k α1 1− ≤ ψX (α, k) ≤ 1 − . (9.12) 2k k The following theorem was proved in [42]. Theorem 9.18 For any α = (α1 , . . . , αn ) we have lim ψX (α, k) = 1. k→∞

Mean lipschitzian mappings with k > 1

211

Proof. It is clear that ψX (l) ≤ ψX (α, k) ≤ ψX (k/α1 ). If k → ∞ (for a fixed α), then l → ∞. Hence, in view of lemma 9.4, we get lim ψX (l) = lim ψX (k/α1 ) = 1.

k→∞

k→∞

Thus, lim ψX (α, k) = 1. k→∞



From the above theorem, we conclude that the evaluation presented in example 9.21 is not sharp. The above results can be extended easily for classes L(α, p, k). Theorem 9.19 For any α = (α1 , . . . , αn ), p ≥ 1, and k > 1, we have   1/p α 1 1/p 0 ≤ ϕC (α, p, k) ≤ ϕC (k/α1 ) ≤ 1 − 1 , (9.13) ϕC (1) 1 − l k   1/p 1 α 1/p 0 ψX (1) 1 − ≤ ψX (α, p, k) ≤ ψX (k/α1 ) ≤ 1 − 1 , l k and   1/p 1 α 1/p ϕ0X (1) 1 − ≤ ϕX (α, p, k) ≤ ϕX (k/α1 ) ≤ 1 − 1 , l k where l = l(α, p, k) > 1 satisfies the condition w(lp ) = α1 lp + α2 l2p + · · · + αn lnp = k p .

(9.14)

1/p

Proof. Using the fact that L(α, p, k) ⊂ L(k/α1 ) and lemma 9.4, we get the right-hand side of inequality (9.13). To prove the left-hand side of (9.13), observe that, in view of (5.2), L(l) ⊂ L(α, p, k). Thus, ϕC (α, p, k) ≥ ϕC (l) and from lemma 9.4, we finally obtain   1 0 ϕC (α, p, k) ≥ ϕC (l) ≥ ϕC (1) 1 − . l 

Corollary 9.1 For any p ≥ 1, k > 1, and α = (α1 , α2 ), we have  1/p   p 2 + 4α k p 1/p α α + 2 1 1 α1   0 ψX (1) 1 − ≤ ψ (α, p, k) ≤ 1 − ,  X k 21/p k and in the case of extremal space, we obtain   p1 p 1/p α1 + α12 + 4α2 k p α1 1− ≤ ψ (α, p, k) ≤ 1 − . X k 21/p k

(9.15)

212

Classification of Lipschitz mappings

For classes L(α, p, k), the theorem 9.16 takes the form: Theorem 9.20 For any α = (α1 , . . . , αn ), p ≥ 1 and k > 1, we have   ϕ0C (1) 1 ϕC (α, p, k) ≥ 1− p , pw0 (k p ) k   ψ 0 (1) 1 ψX (α, p, k) ≥ X0 p 1 − p , pw (k ) k and

ϕ0 (1) ϕX (α, p, k) ≥ X0 p pw (k )



1 1− p k

 .

Proof. Using (5.2), we get L(l) ⊂ L(α, p, k), where l is the solution of the equation (9.14). Thus, in view of lemma 9.4, for every s < l, we have ϕC (α, p, k) ≥ ϕC (l) ≥

sϕC (s) l−1 k p w(lp ) − 1 · · · . p s − 1 w(l ) − 1 l kp

Hence, using lemma 9.4, we obtain ϕC (α, p, k) ≥

ϕ0C (1)

  l−1 kp 1 · · · 1− p . w(lp ) − 1 l k

In view of Lagrange’s Mean Value Theorem, there exists θ ∈ (1, l) such that w(lp ) − 1 = pθp−1 w0 (θp ) ≤ pk p−1 w0 (k p ). l−1 Since k > l, we finally obtain     ϕ0C (1) kp 1 1 ϕ0C (1) ϕC (α, p, k) ≥ p−1 0 p · · 1− p ≥ 1− p . pk w (k ) l k pw0 (k p ) k 

The following theorem is a generalization of theorem 9.18. Theorem 9.21 For any p ≥ 1 and α = (α1 , . . . , αn ), we have lim ψX (α, p, k) = 1.

k→∞

Proof. It is enough to observe that for any α = (α1 , . . . , αn ), p ≥ 1, and k > 1 we have ! k ψX (l) ≤ ψX (α, p, k) ≤ ψX 1/p α1 with l > 1 satisfying the equation (9.14). 

Mean lipschitzian mappings with k > 1

213

The estimates presented in the theorem 9.19 are not sharp. Indeed, if the length n of multi-index tends to infinity, then it may happen that solutions l = ln of the equation (9.14) decreases to 1 and this implies that the left-hand side  (n)of (9.13) tends to 0. For example, if we define the sequence of multi-indexes by α n≥2   1 1 α(n) =  , 0, . . . , 0, 1 −  , n | {z } n n−2

then for any n ≥ 2 and for arbitrarily fixed k > 1, the equation (9.14) takes the form   1 p 1 np ln + 1 − l = kp . n n n Since

    1 p 1 np 1 np k = ln + 1 − l > 1− l , n n n n n p

we conclude that 1 < ln <

k 1/n
1/np . 1 − n1

Letting n tend to infinity we obtain lim ln = 1. n→∞ On the other hand, we have already seen in example 3.6 that, using a lipschitzian retraction R : B → S of the closed unit ball B onto its boundary S, we can construct a mapping T : B → S by putting T = −R, which is uniformly lipschitzian with k(T n ) = k(R) for n ≥ 1 and consequently for any p ≥ 1 we have (see (5.3)) \ T ∈ L(α, p, k(R)). α∈I

Since d(T ) ≥ 2/k(R) and ψ(α, p, ·) is nondecreasing with respect to k, we conclude that, for each α = (α1 , . . . , αn ), p ≥ 1 and k ≥ k(R), ψ(α, p, k) ≥

2 > 0. k(R)

(9.16)

Consequently, for sufficiently large k the function ψ(·, ·, k) is separated from zero! It is not clear whether the function ψ(α, p, ·) is continuous with respect to k. Hence, if we want to use the constant k0 (X), then the estimate (9.16) takes the form: for any p ≥ 1, α = (α1 , . . . , αn ), and k > k0 (X), ψ(α, p, k) ≥

2 > 0. k0 (X)

Below we list some estimates obtained via method described above.

214

Classification of Lipschitz mappings

• Since for extremal √ and
cut invariant subspace X of B (K) we have k0 (X) ≤ 4 2 + 3 , for any p ≥ 1, α = (α1 , . . . , αn ) and k > √
4 2+ 3 , 1 √ . ψX (α, p, k) ≥ 2(2 + 3) For the spaces H, BC(R), C[0, 1], c, c0 , `1 , L1 (0, 1), and BCz (M ), the upper bound for k0 (X) has been obtained by constructing a retraction satisfying the Lipschitz condition with the Lipschitz constant, which is equal to the presented upper bound. Hence, for any p ≥ 1 and α = (α1 , . . . , αn ), we can write: • ψH (α, p, 28.99) ≥ 0.068989; √ √

• ψBC(R) α, p, 4 2 + 3 ≥ 1 − 23 ; √ √

• ψC[0,1] α, p, 4 2 + 3 ≥ 1 − 23 ; √ √

• ψc α, p, 4 2 + 3 ≥ 1 − 23 ; √ √

• ψc0 α, p, 4 2 + 3 ≥ 1 − 23 ; • ψ`1 (α, p, 8) ≥ 14 ; • ψL1 (0,1) (α, p, 8) ≥ 14 ; • ψBCz (M ) α, p, 2 2 +

√

2 ≥1−

√ 2 2 .

Note that we do not claim that for any multi-index α and p ≥ 1 we have ϕX (α, p, 1) = 0. However, some particular cases has been solved in previous chapter. Below, we list known facts in the language of function ϕX : Conclusion 9.1 In view of theorem 8.1, for p ≥ 1 and a multi-index α = (α1 , α2 ) with α2p + α2 ≤ 1, we have ϕX (α, p, 1) = 0. Conclusion 9.2 In view of theorem 8.3, for any p ≥ 1 and α = (α1 , . . . , αn ) such that n−1 n−1 1 (1 − α1 )(1 − α1 p ) ≤ α1 p (1 − α1p ), we have ϕX (α, p, 1) = 0. Conclusion 9.3 In view of theorem 8.23, for any α = (α1 , α2 , α3 ) with α1 ≥ 1 1 2 2 and α2 ≥ 2 − α1 , we have ϕX (α, 1, 1) = 0. Conclusion 9.4 In view of theorem 8.11, if X is a Banach space with 0 (X) < 1, then for p ≥ 1 and multi-index α = (α1 , . . . , αn ) such that   1/p n n X X   αi  < κ(X), j=1

we have ϕX (α, p, 1) = 0.

i=j

Mean lipschitzian mappings with k > 1

215

Conclusion 9.5 In view of theorem 8.15, for every Hilbert space H and multi-index α = (α1 , . . . , αn ) such that    n n X X   αi  < 2, j=1

i=j

we have ϕH (α, 2, 1) = 0. Conclusion 9.6 In view of theorem 8.17, if X is a Banach space with 0 (X) < 1, then for each multi-index α = (α1 , . . . , αn ) of length n ≥ 2 and ln(n) , we have ϕX (α, p, 1) = 0. p = ln(κ(X)) Conclusion 9.7 In view of theorem 8.18, for every Hilbert space H and for ln(n) every multi-index α = (α1 , . . . , αn ) of length n ≥ 2 and p = 2ln(2) , we have ϕH (α, p, 1) = 0. Conclusion 9.8 In view of theorem 8.16, for every Hilbert space H and for every multi-index α = (α1 , α2 ), we have ϕH (α, 2, 1) = 0.

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Mathematics

“Deep understanding of the properties of Lipschitzian mappings is important for all levels of study in many branches of mathematics. This book by Łukasz Piasecki is a good choice for achieving such an understanding in the framework of mappings in general metric spaces, in particular, Banach spaces. Moreover, it gives new insight into the theory of Lipschitzian mappings via a study of the mean Lipschitz condition. … The book is written in a very clear and reader-friendly way. The author gives many examples illustrating various aspects of presented results.” —Stanislaw Prus, Marie Curie-Sklodowska University “… a self-contained, readable, and precise course on the subject. … Besides the presentation of the theory, the true value of the book lies in a collection of cleverly chosen interesting examples.” —Kazimierz Goebel, Maria Curie-Sklodowska University “I strongly recommend this book for advanced undergraduate and graduate students … The reader will find a new classification of this kind of mapping as well as many examples and illustrations designed to help the reader understand the definitions, properties, and results. … I also recommend this book for analysts or mathematicians who are looking for new topics to research.” —Victor Perez-Garcia, University of Veracruz

Classification of Lipschitz Mappings

Łukasz Piasecki

Piasecki

Classification of Lipschitz Mappings presents a systematic, self-contained treatment of a new, more precise classification of Lipschitz mappings and its application in many topics of metric fixed point theory. The mean Lipschitz condition introduced by Goebel, Japón Pineda, and Sims is relatively easy to check and turns out to satisfy several principles: regulating the possible growth of the sequence of Lipschitz constants k(Tn), ensuring good estimates for k0(T) and k∞(T), and providing some new results in metric fixed point theory.

A SERIES OF MONOGRAPHS AND TEXTBOOKS

Classification of Lipschitz Mappings

Classification of Lipschitz Mappings

PURE AND APPLIED MATHEMATICS

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