Spectral Theory of Multivalued Linear Operators 1771889667, 9781771889667

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Spectral Theory of Multivalued Linear Operators

Spectral Theory of Multivalued Linear Operators

Aymen Ammar, PhD Aref Jeribi, PhD

First edition published [2022] Apple Academic Press Inc. 1265 Goldenrod Circle, NE, Palm Bay, FL 32905 USA 4164 Lakeshore Road, Burlington, ON, L7L 1A4 Canada

CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 USA 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN UK

© 2022 Apple Academic Press, Inc. Apple Academic Press exclusively co-publishes with CRC Press, an imprint of Taylor & Francis Group, LLC Reasonable efforts have been made to publish reliable data and information, but the authors, editors, and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors, editors, 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, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. Library and Archives Canada Cataloguing in Publication Title: Spectral theory of multivalued linear operators / Aymen Ammar, PhD, Aref Jeribi, PhD. Names: Ammar, Aymen, author. | Jeribi, Aref, author. Description: Includes bibliographical references and index. Identifiers: Canadiana (print) 20210157038 | Canadiana (ebook) 20210157194 | ISBN 9781771889667 (hardcover) | ISBN 9781774639382 (softcover) | ISBN 9781003131120 (ebook) Subjects: LCSH: Linear operators. | LCSH: Spectral theory (Mathematics) Classification: LCC QA329.2 .A46 2022 | DDC 515/.7246—dc23 Library of Congress Cataloging‑in‑Publication Data Names: Ammar, Aymen, author. | Jeribi, Aref, author. Title: Spectral theory of multivalued linear operators / Aymen Ammar, Aref Jeribi. Description: Burlington ON ; Palm Bay, Florida : Apple Academic Press, [2022] | Includes bibliographical references and index. | Summary: “The concept of multivalued linear operators-or linear relations, the one of the most exciting and influential fields of research in modern mathematics. Applications of this theory can be found in economic theory, noncooperative games, artificial intelligence, medicine, and more. This new book, Spectral Theory of Multivalued Linear Operators, focuses on the theory of multivalued linear operators, responding to the lack of resources exclusively dealing with the spectral theory of multivalued linear operators. The subject of this book is the study of linear relations over real or complex Banach spaces. The main purposes are the definitions and characterization of different kinds of spectra and extending the notions of spectra that are considered for the usual one single-valued operator bounded or not bounded. The volume introduces the theory of pseudospectra of multivalued linear operators. The main topics include demicompact linear relations, essential spectra of linear relation, pseudospectra, and essential pseudospectra of linear relations. The volume will be very useful for researchers since it represents not only a collection of a previously heterogeneous material but is also an innovation through several extensions. Beginning graduate students who wish to enter the field of spectral theory of multivalued linear operators will benefit from the material covered, and an expert reader will also find some of the results interesting enough to be sources of inspiration. Prerequisites for the book are the basic courses in classical real and complex analysis and some knowledge of basic functional analysis. In fact, this theory constitutes a harmonious mixture of analysis (pure and applied), topology, and geometry”-- Provided by publisher. Identifiers: LCCN 2021010465 (print) | LCCN 2021010466 (ebook) | ISBN 9781771889667 (hardcover) | ISBN 9781774639382 (paperback) | ISBN 9781003131120 (ebook) Subjects: LCSH: Linear operators. | Spectral theory (Mathematics) Classification: LCC QA329.2 .A395 2022 (print) | LCC QA329.2 (ebook) | DDC 515/.7246--dc23 LC record available at https://lccn.loc.gov/2021010465 LC ebook record available at https://lccn.loc.gov/2021010466 ISBN: 978-1-77188-966-7 (hbk) ISBN: 978-1-77463-938-2 (pbk) ISBN: 978-1-00313-112-0 (ebk) DOI: 10.1201/9781003131120

Dedication

To the memory of my mother Ayada, the memory of my father Khalifa, my wife Nihel, and my children Nourcen and Ayen, my brother Houcem, my sisters Imen, Ines, Dorsaf, and Awatef, all members of my extended family . . . —Aymen Ammar, PhD

To my mother Sania, my father Ali, my wife Fadoua, my children Adam and Rahma, my brothers Sofien and Mohamed Amin, my sister Elhem, and all members of my extended family. . . —Aref Jeribi, PhD

Contents

Dedication About the Authors Preface Symbol Description 1 Introduction 2 Fundamentals 2.1 Banach Space . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Direct Sum . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Distance Function . . . . . . . . . . . . . . . . . . . . 2.1.3 Normed Vector Space . . . . . . . . . . . . . . . . . . 2.1.4 Banach Space . . . . . . . . . . . . . . . . . . . . . . 2.2 Relations on Sets . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Linear Relations . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 The Algebra of Linear Relations . . . . . . . . . . . . 2.3.2 Inverse of Linear Relation . . . . . . . . . . . . . . . 2.3.3 Sum and Product of Linear Relations . . . . . . . . . 2.3.4 Restrictions and Extensions of Linear Relations . . . 2.4 Index and Co-Index of Linear Relations . . . . . . . . . . . . 2.4.1 Properties of the Quotient Map QT . . . . . . . . . . 2.5 Generalized Kernel and Range of Linear Relations . . . . . . 2.5.1 Ascent and Descent and Singular Chain of Linear Relations . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Norm of a Linear Relation . . . . . . . . . . . . . . . 2.5.3 Continuity and Openness of a Linear Relation . . . . 2.5.4 Selections of Linear Relation . . . . . . . . . . . . . . 2.6 Relatively Boundedness of Linear Relations . . . . . . . . .

v xiii xv xvii 1 11 11 11 13 13 17 18 20 20 21 23 25 28 31 32 38 39 43 45 47 vii

viii

Contents 2.7

2.8 2.9 2.10

2.11 2.12 2.13 2.14

2.15

2.16

2.17 2.18

2.19

2.20

Closed and Closable Linear Relations . . . . . . . . . . . . . 2.7.1 Closed Linear Relations . . . . . . . . . . . . . . . . 2.7.2 Closable Linear Relations . . . . . . . . . . . . . . . Adjoint of Linear Relations . . . . . . . . . . . . . . . . . . . Minimum Modulus of Linear Relations . . . . . . . . . . . . Quantities for Linear Relations . . . . . . . . . . . . . . . . 2.10.1 A Formula for Gap Between Multivalued Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . 2.10.2 Measures of Noncompactness . . . . . . . . . . . . . Precompact and Compact Linear Relations . . . . . . . . . . Strictly Singular Linear Relations . . . . . . . . . . . . . . . Polynomial Multivalued Linear Operators . . . . . . . . . . Some Classes of Multivalued Linear Operators . . . . . . . . 2.14.1 Multivalued Fredholm and Semi-Fredholm Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . 2.14.2 Multivalued Fredholm and Semi-Fredholm of Closed Linear Operators in Banach Space . . . . . . . . . . 2.14.3 Atkinson Linear Relations . . . . . . . . . . . . . . . Quasi-Fredholm and Semi Regular Linear Relations . . . . . 2.15.1 Quasi-Fredholm Linear Relations . . . . . . . . . . . 2.15.2 Semi Regular Linear Relations . . . . . . . . . . . . . 2.15.3 Essentially Semi Regular Linear Relations . . . . . . Perturbation Results for Multivalued Linear Operators . . . 2.16.1 Small Perturbation Theorems of Multivalued Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . Fredholm Perturbation Classes of Linear Relations . . . . . Spectrum and Pseudospectra of Linear Relations . . . . . . 2.18.1 Resolvent Set and Spectrum of Linear Relations . . . 2.18.2 Subdivision of the Spectrum of Linear Relations . . . 2.18.3 Essential Spectra of Linear Relations . . . . . . . . . S-Spectra of Linear Relations in Normed Space . . . . . . . 2.19.1 Some Properties of S-Resolvent of Linear Relations in Normed Space . . . . . . . . . . . . . . . . . . . . . . 2.19.2 The Augmented S-Spectrum . . . . . . . . . . . . . . 2.19.3 S-Essential Resolvent Sets of Multivalued Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . Pseudospectra and Essential Pseudospectra of Linear Relations 2.20.1 Pseudospectra of Linear Relations . . . . . . . . . . .

50 50 54 55 57 62 62 63 67 68 70 76 76 77 80 84 84 84 86 87 87 91 92 92 94 96 98 99 109 111 111 111

Contents 2.20.2 Essential Pseudospectra of Linear Relations . . . . . 2.20.3 The Essential ε-Pseudospectra of Linear Relations . . 2.20.4 S-Pseudospectra and S-Essential Pseudospectra of Linear Relations . . . . . . . . . . . . . . . . . . . . .

ix 114 115 116

3 The Stability Theorems of Multivalued Linear Operators 117 3.1 Stability of Closeness of Multivalued Linear Operators . . . 117 3.1.1 Sum of Two Closed Linear Relations . . . . . . . . . 117 3.1.2 Sum of Three Closed Linear Relations . . . . . . . . 119 3.1.3 Sum of Two Closable Linear Relations . . . . . . . . 122 3.1.4 Product of Closable Linear Relations . . . . . . . . . 125 3.2 Sequence of Multivalued Linear Operators Converging in the Generalized Sense . . . . . . . . . . . . . . . . . . . . . . . . 129 3.2.1 The Gap Between Two Linear Relations . . . . . . . 129 3.2.2 Relationship Between the Gap of Linear Relation and their Selection . . . . . . . . . . . . . . . . . . . . . . 138 3.2.3 Generalized Convergence of Closed Linear Relations 140 3.3 Perturbation Classes of Multivalued Linear Operators . . . . 142 3.3.1 Fredholm and Semi-Fredholm Perturbation Classes of Multivalued Linear Operators . . . . . . . . . . . . . 142 3.4 Atkinson Linear Relations . . . . . . . . . . . . . . . . . . . 147 3.5 The α-and β-Atkinson Perturbation Classes . . . . . . . . . 152 3.6 Index of a Linear Relations . . . . . . . . . . . . . . . . . . . 158 3.6.1 Index of Upper Semi-Fredholm Relation Under Strictly Singular Perturbation . . . . . . . . . . . . . . . . . . 162 3.6.2 Index of an Lower Semi-Fredholm Relation Under Strictly Singular Perturbations . . . . . . . . . . . . 165 3.7 Demicompact Linear Relations . . . . . . . . . . . . . . . . . 167 3.7.1 Auxiliary Results on Demicompact Linear Relations 167 3.7.2 Demicompactness and Fredholm Linear Relations . . 169 3.8 Relatively Demicompact Linear Relations . . . . . . . . . . 172 3.8.1 Auxiliary Results on Relatively Demicompact Linear Relations . . . . . . . . . . . . . . . . . . . . . . . . . 172 3.8.2 Fredholm Theory by Means Relatively Demicompact Linear Relations . . . . . . . . . . . . . . . . . . . . . 179 3.9 Essentially Semi Regular Linear Relations . . . . . . . . . . 182 3.9.1 Some Proprieties of Essentially Semi Regular Linear Relations . . . . . . . . . . . . . . . . . . . . . . . . . 182

x

Contents 3.9.2

Perturbation Results of Essentially Semi Regular Linear Relations . . . . . . . . . . . . . . . . . . . . .

186

4 Essential Spectra and Essential Pseudospectra of a Linear Relation 191 4.1 Characterization of the Essential Spectrum of a Linear Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 4.1.1 α- and β-Essential Spectra of Linear Relations . . . . 199 4.1.2 Invariance of Essential Spectra of Linear Relations . 205 4.2 The Essential Spectrum of a Sequence of Linear Relations . 210 4.3 Spectral Mapping Theorem of Essential Spectra . . . . . . . 213 4.3.1 Essential Spectra of the Sum of Two Linear Relations 214 4.4 S-Essential Spectra of Linear Relations . . . . . . . . . . . . 219 4.4.1 Characterization of S-Essential Spectra of Linear Relations . . . . . . . . . . . . . . . . . . . . . . . . . 219 4.4.2 Characterization of σe5,S (·) . . . . . . . . . . . . . . 219 4.4.3 Relationship Between σe4,S (·) and σe5,S (·) . . . . . . 221 4.5 Racoˇ cevi´ c and Schmoeger S-Essential Spectra of a Linear Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 4.5.1 Stability of Racoˇ cevi´ c and Schmoeger S-Essential Spectra of a Linear Relations . . . . . . . . . . . . . 223 4.6 S-Essential Spectra of the Sum of Two Linear Relations . . 226 4.7 Pseudospectra and ε-Pseudospectra of Linear Relations . . . 231 4.7.1 Some Properties of Pseudospectra and ε-Pseudospectra of Linear Relations . . . . . . . . . . . . . . . . . . . 231 4.7.2 Characterization of Pseudospectra of Linear Relations 235 4.7.3 Stability of Pseudospectra of Linear Relations . . . . 237 4.8 Localization of Pseudospectra of Linear Relations . . . . . . 240 4.9 Characterization of ε-Pseudospectra of Linear Relations . . 242 4.9.1 The Pseudospectra and the ε-Pseudospectra of a Sequence of Closed Linear Relations . . . . . . . . . 248 4.10 Essential Pseudospectra of Linear Relations . . . . . . . . . 250 4.10.1 Stability of Essential Pseudospectra of Linear Relations 257 4.11 The Essential ε-Pseudospectra of Linear Relations . . . . . . 262 4.11.1 The Essential ε-Pseudospectra of a Sequence of Linear Relations . . . . . . . . . . . . . . . . . . . . . . . . . 265 4.12 S-Pseudospectra of Linear Relations . . . . . . . . . . . . . 269

Contents 4.12.1 S-Essential Pseudospectra of Linear Relations . . . .

xi 272

Bibliography

279

Index

287

About the Authors

Aymen Ammar, PhD Associate Professor, Department of Mathematics, University of Sfax, Tunisia Aymen Ammar, PhD, is currently working as an Associate Professor in the Department of Mathematics, Faculty of Sciences of Sfax at the University of Sfax, Tunisia. He has published many articles in international journals. His areas of interest include spectral theory, matrice operators, transport theory, and linear relations.

Aref Jeribi, PhD Professor, Department of Mathematics, University of Sfax, Tunisia Aref Jeribi, PhD, is a Professor in the Department of Mathematics at the University of Sfax, Tunisia. He is the author of the book Spectral Theory and Applications of Linear Operators and Block Operator Matrices (SpringerVerlag, New-York, 2015), co-author of the book Nonlinear Functional Analysis in Banach Spaces and Banach Algebras: Fixed Point Theory under Weak Topology for Nonlinear Operators and Block Operator Matrices with Applications (Taylor-Francis, 2015), the author of the book Denseness, Bases, and Frames in Banach Spaces and Applications (De Gruyter, Berlin, 2018), the author of the book Linear Operators and Their Essential Pseudospectra (Apple Academic Press, CRC Press, Oakville, Boca Raton, 2018), and the coauthor of the book Analyse Num´erique Matricielle, M´ethodes et Algorithmes, Exercices et Probl`emes Corrig´es (R´ef´erences Sciences. Paris: Ellipses, 2020). He has published many journal articles in international journals. His areas of interest include spectral theory, matrice operators, transport theory, Gribov operator, Bargmann space, fixed point theory, Riesz basis, and linear relations.

xiii

Preface

This book is focused on the theory of multivalued linear operators. To some extent, it is a sequel of the authors’ recent work on linear operators on linear relation as well as their spectral theory. There are not many books exclusively dealing with the spectral theory of multivalued linear operators, and the authors wish to add to the scarce literature on the subject. A minimum necessary background material has been gathered, which will allow relatively friendly access to the book. Beginning graduate students who wish to enter the field of spectral theory of multivalued linear operators should benefit from the material covered, but an expert reader might also find some of the results interesting enough to be sources of inspiration. Prerequisites for the book are the basic courses in classical real and complex analysis and basic functional analysis knowledge. In fact, this theory constitutes a harmonious mixture of analysis (pure and applied), topology, and geometry. The concept of multivalued linear operators or linear relations is one of the most exciting and influential fields of research in modern mathematics. Applications of this theory can be found in economic theory, non-cooperative games, artificial intelligence, medicine, and existence of solutions for differential inclusions, the theory of a multivalued operator, assigning to the points of some set X a subset of another set Y, has arisen naturally by refining the classical concept of a multivalued function. The concept of a multivalued linear operator appeared in the literature a few decades ago, not only through the need of considering adjoints (conjugates) of non-densely defined linear differential operators, but also through the necessity of considering the inverses of certain operators used; for example, in the study of some Cauchy problems associated with parabolic type equations in Banach spaces. Let us, for example, the abstract degenerate equation in a Banach space. It is generally true that the degenerate linear evolution equations with respect to the time derivative are rewritten into non-degenerate equations in such a way using multivalued linear operators. This fact then leads us naturally to consider a problem of generalizing the well-developed results concerning xv

xvi

Preface

the ordinary linear evolution equations with univalent coefficient operators to those with multivalued operators and to handling the degenerate equations using analogous techniques to non-degenerate ones. The subject of this book is the study of linear relations over real or complex Banach spaces. The primary purposes are the definitions and characterization of different kinds of spectra, someone extending the notions of spectra considered for the usual one single-valued operators bounded or not bounded. One of the objectives of this book is to introduce the theory of pseudospectra of multivalued linear operators. The main topics include demicompact linear relations, essential spectra of linear relation, pseudospectra, and essential pseudospectra of linear relations. We hope that this book will be handy for researchers, since it represents a collection of a previously heterogeneous material and innovation through several extensions; of course, a single book can’t cover such a huge research field. In making personal choices for material inclusion, we tried to give useful complementary references in this research area, hence probably neglecting some relevant works. We would be very grateful to receive any comments from readers and researchers, providing us with some information concerning some missing references. We would like to mention that the thesis work results by my students Bilel Boukettaya, Houcem Daoud, Bilel Saadaoui, Slim Fakhfakh, Nawrez Lazrag, and Ameni Bouchekoua, the obtained results have helped us in writing this book. Sfax, Tunisia April 2021

Aymen Ammar, PhD Aref Jeribi, PhD

Symbol Description

The most frequently used notations, symbols, and abbreviations are listed below: The set of natural numbers. The fields of real and complex numbers, respectively. Rn The n-dimensional real space. inf(A) The infimum of the set A. sup(A) The supremum of the set A. Te The completion of T. Br The ball with radius r. BX The unit ball in X. dist(x, y)The distance between x and y. ∗ 0 X , X The dual of X. dim(X) The dimension of the space X. kxk The norm of x. (X, k · k) The linear normed space X. (xn )n The sequence (xn )n . G(T ) The graph of T. C(Ω, R) The space of real continuous functions on Ω. Lp The Lp space. ρ(T ) The resolvent set of T. σ(T ) The spectrum of T. σε (T ) The pseudospectrum of T. Σε (T ) The ε-pseudospectrum of T. N R, C

T|M

The restriction of T to M. g

Tn −→ T The sequence (Tn )n converge in the generalized sense to T. asc(T )

The ascent of T.

des(T )

The descent of T.

Rc (T )

The singular chain manifold of T.

W(X, Y ) The set of weakly compact linear operators from X into Y. L(X, Y ) The space of linear operators from X into Y. R(U, V ) The collection of relation from U into V. LR(X, Y ) The collection of linear relations from X into Y . BR(X, Y ) The collection of bounded linear relations from X into Y. CR(X, Y ) The class of all closed linear relations from X into Y. BCR(X, Y )The class of closed everywhere defined linear relations from X into Y. xvii

xviii

Symbol Description

KR(X, Y ) The families of all compact linear relations from X into Y.

PR(Aβ (X, Y )) The set of β-Atkinson perturbations linear relations from X into Y. F(X, Y ) The set of Fredholm linear DC(X, Y ) The families of all relations from X into Y. demicompact linear F+ (X, Y ) The set of upper relations from X into Y. semi-Fredholm linear N (T ) The null space of T. relations from X into Y. R(T ) The range space of T. F− (X, Y ) The set of lower PRK(X, Y ) The families of all semi-Fredholm linear precompact linear relations relations from X into Y. from X into Y. Φ(X, Y ) The set of Fredholm closed SSR(X, Y ) The set of strictly linear relations from X into singular linear relations Y. from X into Y. Φ+ (X, Y ) The set of upper X injection operator from M JM semi-Fredholm closed linear into X. relations from X into Y. xn → x xn converges strongly to x. Φ− (X, Y ) The set of lower semi-Fredholm closed linear M The closure of the set M. relations from X into Y. co(·) The convex hull. Aα (X, Y ) The class of α-Atkinson co(·) The closed convex hull. multivalued linear operators PR(Φ+ (X, Y )) The set of upper from X into Y. semi-Fredholm Aβ (X, Y ) The class of β-Atkinson perturbations of linear multivalued linear operators relations from X into Y. from X into Y. l PR(Φ− (X, Y )) The set of lower W (X, Y ) The set of left Weyl linear semi-Fredholm relations from X into Y. perturbations of linear W r (X, Y ) The set of right Weyl relations from X into Y. linear relations from X into Y. PR(Aα (X, Y )) The set of α-Atkinson perturbations linear B(X, Y ) The set of Browder linear relations from X into Y. relations from X into Y.

Chapter 1 Introduction

The spectral theory of linear operators is an important part of functional analysis which has application in several areas of modern mathematical analysis and physics; for instance, in differential and integral equation as well as quantum theory. The spectral theory of linear operators in a Banach space is one of the major advances in mathematics of the 20th century. It was initiated by T. Kato in his famous book (see Ref. [67]) which contains the most basic ideas for the spectral theory of linear operators. Another important field was stimulated by problems from mathematics physics is the concept of multivalued linear operators. The concept of a linear relation, (or multivalued linear operator) in a linear space which generalizes the notion of a (single valued) linear operator to that of a multivalued operator. This notion was introduced by R. Arens who gave a scientific treatment in Ref. [42] and by E. A. Coddington (see Refs. [48, 49]). Since that time, it will be helpful in different areas and it has been studied in diverse specific contexts, (see Ref. [6]). At present, the study of linear relations in Banach or Hilbert spaces is of significance since it has applications in many problems in physics and other areas of applied mathematical. A. Favini and A. Yagi, in his paper Ref. [57], introduced the idea of multivalued linear operators as an approach toward the degenerate linear evolution equations with respect to the time derivative. In Ref. [56, Sect. 5] he handled a time homogeneous equation   dM v + Lv = f (t), 0 ≤ t ≤ T, dt (1.1)  lim M v(t) = u0 , t→0+

in X and indicated that the operator defined by Z 1 Z0 (t) = e−λt (λM − L)−1 dλ, t > 0, 2πi Γ plays a role like a fundamental solution under the assumption that the bounded inverse (λM − L)−1 , called M -modified resolvent of L, exists for any λ ∈ Σ := {λ ∈ C : |arg(λ)| ≥ w}, 0 < w < π2 , and (λM − L)−1 satisfies 1

2

Spectral Theory of Multivalued Linear Operators

kL(λM − L)−1 k ≤ Const, λ ∈ Σ. To understand this result intuitively, let us rewrite (1.1) into a non degenerate form by putting u(t) = M v(t) (such a change of unknown function was already seen in R. E. Showalter [87], in which M is a nonlinear operator). It then turns out that   du + Au 3 f (t), 0 ≤ t ≤ T, dt  u(0) = u0 , where A = LM −1 and M −1 is the inverse of M . Of course M −1 is no longer defined as a univalent operator, but conserves its linearity; such an operator is called a multivalued linear operator. This fact then leads us naturally to consider a problem of generalizing the well developed results concerning the ordinary linear evolution equations with univalent coefficient operators to those with multivalued operators and to in tend handling the degenerate equations by means of analogous techniques to non-degenerate ones. Consider an initial boundary value problem for an elliptic-parabolic equation of the type               

∂ (m(x)u(x, t)) ∂t

= ∇ • (a(x)∇u(x, t)) +

n X i=1

ai (x)

∂(m(x)u(x, t)) ∂xi

+c0 (x)u(x, t) + f (x, t) (x, t) ∈ Ω × (0, τ ], u(x, t) = 0 (x, t) ∈ ∂Ω × (0, τ ], x ∈ Ω, m(x)u(x, 0) = v0 (x),

(1.2) where Ω is a bounded domain in Rn of class C 2 , n = 2, 3; ∇ denotes the gradient vector with respect to x variable; m(x) ≥ 0 on Ω and m ∈ C 2 (Ω); a(x), ai (x), c0 (x) are real-valued smooth functions on Ω; a(x) ≥ δ > 0 and c0 (x) ≤ 0 for all x ∈ Ω. Take the space X := L2 (Ω), the operators M u := m(·)u with domain D(M ) := X, and Lu = ∇ • (a(·)∇u) + c0 (·) with domain D(L) := H 2 (Ω) × H01 (Ω). Introduce the new unknown function v := M u and, in X, consider the operators A := LM −1 with domain D(A) := M (D(L)), and n X ∂v Bv := ai (x) with domain D(B) = H01 (Ω). Note that A is a multivalued ∂x i i=1 linear operator. Then, problem (1.2) is reduced to the multivalued Cauchy problem (inclusion)

Introduction  

∂v ∂t  v(0)

∈

3

Av + Bv + f (t), t ∈ (0, τ ]

= v0 ,

where f (t) := f (t, .) and v0 := v0 (·). The class of univalent linear operators is unstable under the operations closure, inverse and adjoint. This is not the case if we consider the more general class of multivalued linear operators. So, it is interesting to extend some results of spectral theory to the multivalued linear case as well as obtaining perturbation theorems for multivalued linear operators. As in the case of linear operators, the spectral theory of linear relations, including the associated analytic functional calculus, is an important tool for studying various properties of these objects and for deriving some of their applications. One of the objectives of this book is the study of the perturbation theory of multivalued linear operators, in order to establish the existence results of the second kind multivalued operator equations, hence allowing to describe the spectrum, and essential spectra of a linear relation. Fredholm theory and perturbation results are also widely investigated. Recall that the perturbation problems are among the main topics in both pure and applied mathematics. The perturbation theory of operators, from which many valuable results have been obtained has been extensively studied. In 1907, F. Riesz introduced and studied the theory of compact operators. This theory forms a part of the classic core of functional analysis and operator theory. A fundamental result proved by F. Riesz is that, if T is a compact operator defined on a Banach space, then I −T is a Fredholm operator with index 0. This contribution came with some us stability results of the essential spectra. But this is wrong, if T is a compact multivalued linear operator. It is well-known that if T is a closed Fredholm linear operator between Banach spaces, then T + S has the same property, whenever S is a compact linear operator (respectively, bounded with sufficiently small norm). Among the works in this direction, we can state, for example, I. C. Gohberg, A. S. Markus, and I. A. Feldman [58], A. Lebow and M. Schechter [77], E. Albrecht and F.-H. Vasilescu [7], and C.-G. Ambrozie and F.-H. Vasilescu [15]. In 1998, R. W. Cross [50], has investigated the question of stability in the sets of upper semi-Fredholm and lower semi-Fredholm (see Theorems 2.16.1 and 2.16.3). ´ Considerable attention has also been devoted by T. Alvarez, R. W. Cross and D. Wilcox [13] introduced the class of α- and β-Atkinson linear relations

4

Spectral Theory of Multivalued Linear Operators

in normed spaces. Specifically these authors gave important characterization theorems of such multivalued linear operators in terms of the existence of ´ left and right generalized inverses. After that, T. Alvarez and D. Wilcox [14] continued their study of α-and β-Atkinson linear relations. They inspect essentially, these characterization theorems which will be used in conjunction with perturbation theorems for Fredholm linear relations in order to establish a perturbation theory for Atkinson type relations. To have further details we may refer to Ref. [14]. The concept of demicompactness was introduced into functional analysis by W. V. Petryshyn [83], in order to discuss fixed points and it has been studied in a large number of papers (see for example, Refs. [43, 82]). In 2012, W. Chaker, A. Jeribi and B. Krichen achieved some results on Fredholm and upper semi-Fredholm operators involving demicompact operators [47]. Since much attention was paid to this notation, several research papers, used it (see Refs. [65,71]). In 2018, A. Ammar, H. Daoud and A. Jeribi, extended the concept of demicompact and k-set-contraction of linear operators to multivalued linear operators and developed some properties (see Ref. [24]). In Refs. [39, 40], A. Ammar, A. Jeribi and B. Saadaoui have firstly obtained more results in the spirit of those obtained in Ref. [24]. Secondly, they applied these new results in order to establish some perturbation results of multivalued linear operator. Recently, A. Ammar, S. Fakhfakh and A. Jeribi introduced in Ref. [28] concept of relatively demicompactness with respect to multivalued linear operator as a generalization of the class of demicompact linear relation introduced by A. Ammar, H. Daoud and A. Jeribi [24] and established some new results in Fredholm theory. The notion of essentially semi regularity operators amongst the various concepts of regularity originated by the classical treatment of perturbation theory owed to T. Kato and was studied by many authors; for instance, we cite Refs. [4, 5, 18, 65, 80, 95]. We remark that all the above authors considered only the case of bounded linear operators. It is the purpose of this section to consider these class of essentially semi regularity in the more general setting of linear relations in Hilbert spaces. Many properties of essentially semi regularity for the case of linear operators remain valid in the context of linear relations, sometimes under supplementary conditions. In Ref. [21], A. Ammar, B. Boukettaya and A. Jeribi studied some properties of the semi regular and essentially semi regular, of a multivalued linear operator.

Introduction

5

There are many ways to define the essential spectrum of a closed densely defined linear operator on a Banach space. Indeed, several definitions may be found in the literature. Particularly, various notions of essential spectrum appear in applications of the spectral theory (see for example, Refs. [4, 5, 65]). Five of these definitions were studied by D. E. Edmunds and W. D. Evans [55]. In 1998, R. W. Cross [50] introduced a concept of the essential spectrum of a multivalued linear operator on a complex normed space in terms of the nullity and the deficiency of its complete closure, and he showed its stability under relative compact perturbation with certain additional ´ conditions. Subsequently, in 2012, T. Alvarez has introduced several essential spectra of a linear relation on a normed space. We investigate the closedness and the emptiness of such essential spectra. As an application we prove two results, the first of which characterizes the class of quotient indecomposable normed spaces in terms of upper semi-Fredholm, lower semi-Fredholm and strictly cosingular linear relations, and it gives conditions under which a linear relation on a complex quotient indecomposable normed space is a strictly cosingular perturbation of multiple identities. In 2014, D. Wilcox has came with five distinct essential spectra of linear relations on Banach spaces in terms of semi-Fredholm properties, and has showed their stability under relative compact perturbations with some additional conditions and ´ under compact perturbations. More recently, T. Alvarez, A. Ammar and A. Jeribi [10, 11] have made a detailed treatment of some subsets of the essential spectra of a matrix of multivalued linear operator. What is notable is that the investigation of the essential spectra of a linear relation present a guide to the study of operators on normed spaces since every continuous operator between normed spaces is the inverse of an injective upper semiFredholm relation. In addition, the class of all bounded Fredholm operators on Banach spaces coincides with the class of the inverses of closed Fredholm linear relations which are both surjective and injective. In Refs. [16, 17], A. Ammar characterized some essential spectra of a closed linear relation in terms of certain linear relations type α- and β-Atkinson. This study leads us to generalize some well-known results for operators and to extend some ´ results of small perturbations given by T. Alvarez and D. Wilcox [14]. The investigation of the notion of the essential spectra of linear relations has important applications to several problems of operator theory. For instance, we cite Refs. [44] and [85]. Many spectral and basic properties about essential spectra were given in Refs. [1, 11, 17, 26–28, 37, 61–66, 93]. Besides, in Ref. [36], A. Ammar, A. Jeribi and N. Lazrag found a relationship between the essential

6

Spectral Theory of Multivalued Linear Operators

spectra of (Tn )n and essential spectra of T , where (Tn )n is a sequence of closed linear relation converging in the generalized sense to T . In order to study this convergence, they seek a counterpart of the gap between two closed linear relations as defined by T. Kato [67]. Several mathematical and physical problems lead to operator pencils, (operator-valued functions of a complex argument). Recently, the spectral theory of operator pencils attracted the attention of many mathematicians (see Ref. [2]). In this book, we give a characterization of the essential spectrum of the closed pencil linear relation in order to extend many known results in the literature. In fact, we give the characterization of the S-essential spectra of the sum of two multivalued linear operators and we study the stability of the S-essential approximate point spectrum and the S-essential defect spectrum of closed and closable linear relations under relatively precompact perturbations on a Banach space. On another level, we study the stability of the S-essential approximate point spectrum but under assumptions different from those adopted above (see Refs. [10, 29, 41]). An important direction was established by J. M. Varah [90] which investigated the notion of pseudospectra in the domain of linear operators and has been employed by other authors, such as, H. J. Landau [76], L. N. Trefethen [88, 89], D. Hinrichsen and A. J. Pritchard [60] and E. B. Davies [53]. More precisely, L. N. Trefethen developed this notion for matrices and operators and applied it to several interesting problems, and to study interesting problems in mathematical physics. The concept of pseudospectra appeared as a result of the realization that various properties of highly non-self-adjoint operators are related. We refer the reader to L. N. Trefethen for the definitions of the pseudospectra and the ε-pseudospectra of the closed linear operator T , respectively, by σε (T ) Σε (T )

[n 1o λ ∈ C such that k(λ − T )−1 k > , ε [n 1o = σ(T ) λ ∈ C such that k(λ − T )−1 k ≥ , ε

= σ(T )

where σ(.) is the spectrum of T and by convention k(λ − T )−1 k = +∞ if and only if, λ ∈ σ(T ). Importance characterization of pseudospectra given by [ σ(T + S) = σε (T ). kS k 0 such kxn k ≤ M . It is obvious that kQD xn k = dist(xn , D) ≤ dist(xn , 0) = kxn k ≤ M. Then, (QD xn )n is bounded. (ii) Let (xn )n be a sequence converges to a limit x, by continuity of QD , (QD xn )n converges to QD x. Q.E.D. Theorem 2.1.3 Let D be a linear subspace of a normed space X such that dim D < ∞. If (xn )n is a bounded sequence in X such that (QD xn )n is a convergent sequence, then (xn )n has a convergent subsequence. ♦ Proof. Let D be a linear subspace of X such that dim D < ∞, let (xn )n in X be a bounded sequence such that (QD xn )n is a convergent sequence. Then, there exists x ∈ X such that (QD xn )n converges to QD x. It is clear, since dim D < ∞, that there exists d0 ∈ D such that kQD xk =

dist(x, D)

= kx − d0 k ≤ dist(x, 0) = kxk

(2.2)

16

Spectral Theory of Multivalued Linear Operators

and there exists dn ∈ D such that kQD xn − QD xk = kQD (xn − x)k =

dist(xn − x, D)

= kxn − x − dn k.

(2.3)

Since (xn )n is bounded, then there exists M > 0 such that kxn k ≤ M . Let ε > 0. Since (QD xn )n converge to QD x, then there exists N1 ∈ N such that for all n ≥ N1 , we have ε kQD xn − QD xk < . 2 Hence, in view of (2.3), we have kxn − x − dn k
0, there exists x ∈ X with kxk = 1 such that 1 − ε < dist(x, M ). We can even achieve dist(x, M ) = 1 if dim M < ∞. ♦ L Remark 2.1.1 If X is a Banach space, X = M N and the projection P is continuous, then M is said to be complemented and N is said to be a topological complement of M . Note that each complemented subspace is closed, but the converse is not true, for instance c0 , the Banach space of all sequences which converge to 0, is a not complemented closed subspace of l∞ , where l∞ denotes the Banach space of all bounded sequences, (see Ref. [81]). ♦ Definition 2.1.6 Let M be a vector subspace of a Banach space X. M is called paracomplet subspace of X, if M is a Banach space and the injection map of M into X is continuous. ♦ Proposition 2.1.1 [72, Proposition 2.1.1] Let M and N be two paracomplet subspaces of X such that M + N and M ∩ N are closed on X. Then, M and N are two closed subspaces of X. ♦

2.2

Relations on Sets

Let U and V be arbitrary non-empty sets. A relation T from U into V is a mapping defined on a non-empty subset D(T ) of U , called the domain of T , which takes on values in 2V \∅ (the collection of non-empty subset of Y ). If T maps the points of its domain to singletons, then T is said to be a single valued operator, or function. The collection of relations as defined above will be denoted by R(U, V ) and as useful we write R(U ) := R(U, U ). Examples of relations are functions, inverse of functions, adjoints of linear operators, partial order relations, equivalent relations and convex processes. For u ∈ U , u∈ / D(T ), we define T u = ∅. With this convention, we get 6 ∅} . D(T ) = {u ∈ U such that T u =

Fundamentals

19

Let T ∈ R(U, V ), M ⊂ U and ∅ 6= N ⊂ V . The graph of T is a subset of U × V defined by G(T ) = {(u, v) ∈ U × V such that u ∈ D(T ) and v ∈ T u} . A relation in R(U, V ) is uniquely determined by its graph and conversely any non-empty subset of U × V uniquely determines a relation. The identity relation defined on a nonempty subset E of U is denoted by IE (or simply I) when E is understood, it is the relation in R(U ) whose graph is G(IE ) = {(e, e) such that e ∈ E} . Definition 2.2.1 Let M be a subspace of a normed vector space X, the X operator JM is called injection operator if it denote the natural injection X X X map of M into X, i.e., JM ∈ R(M, X), D(JM ) = M and JM (m) = m for m ∈ M. ♦ e be the completion of the normed space X. The completion of T , denoted Let X Te is defined in terms of her corresponding graph by e × Ye . G(Te) = G(T )e ⊂ X

(2.4)

We write T (M ) =

o [n \ T m such that m ∈ M D(T ) ,

called the image of M , with R(T ) = T (D(T )) called the range of T . The subspace T −1 (0) is called the null space (or kernel) of T , and is denoted by N (T ). The inverse of T is the relation T −1 given by G(T −1 ) = {(v, u) ∈ V × U such that (u, v) ∈ G(T )} . We write n o \ T −1 (N ) := u ∈ D(T ) such that T u N 6= ∅ . In particular, for v ∈ R(T ), T −1 (v) = {u ∈ D(T ) such that v ∈ T u} . If R(T ) = Y , then T is called surjective. We shall be using both N (T ) and T −1 (0) throughout the sequel. If T −1 is single valued, then T is called injective. Proposition 2.2.1 Let T ∈ R(U ). If T is injective, then for all u1 , u2 ∈ D(T ) such that T (u1 ) = T (u2 ), we have u1 = u2 . ♦ Proof. Let v ∈ T (u1 ) = T (u2 ), then (u1 , v) ∈ G(T ) and (u2 , v) ∈ G(T ). Hence, (v, u1 ) ∈ G(T −1 ) and (v, u2 ) ∈ G(T −1 ). Since T −1 is single valued, we infer that u1 = u2 . Q.E.D.

20

2.3 2.3.1

Spectral Theory of Multivalued Linear Operators

Linear Relations The Algebra of Linear Relations

Definition 2.3.1 Let X and Y be two vector spaces over the field K (R or C). A multivalued linear operator (or a linear relation) is a mapping T ⊂ X × Y which goes from a subspace D(T ) ⊂ X called the domain of T, into the collection of nonempty subsets of Y such that T (α1 x1 + α2 x2 ) = α1 T (x1 ) + α2 T (x2 ) for all nonzero scalars α1 , α2 and x1 , x2 ∈ D(T ). For x ∈ X\D(T ), we define T x = ∅. With this convention, we get D(T ) = {x ∈ X such that T x 6= ∅} .

♦

The collection of linear relations as defined above will be denoted by LR(X, Y ) and as useful we write LR(X) := LR(X, X). Definition 2.3.2 Let X and Y be two normed spaces. If T ∈ LR(X, Y ) maps the points in its domain to singletons, then T is said to be a single valued (or simply an operator). ♦ Remark 2.3.1 (i) A linear relation T ∈ LR(X, Y ) is uniquely determined and identified with its graph, G(T ), which is defined by G(T ) = {(x, y) ∈ X × Y such that x ∈ D(T ), y ∈ T x} . (ii) T is a linear relation if and only if, G(T ) is a linear subspace of X × Y . (iii) T (0) = {y such that (0, y) ∈ G(T )}. (iv) The zero relation is denoted by OX , which is defined by G(OX ) = {(x, 0) such that x ∈ X}. (v)The identity relation on a subspace M of X is denoted by IM (or simply I) when M is understood, it is the linear relation whose graph is G(IM ) = {(m, m) such that m ∈ M } .

♦

Proposition 2.3.1 Let X and Y be two vector spaces over the field K (R or C), and T ∈ R(X, Y ). Then, T ∈ LR(X, Y ) if and only if, for all x1 , x2 ∈ D(T ) and for all α1 , α2 ∈ K, we have α1 T (x1 ) + α2 T (x2 ) ⊂ T (α1 x1 + α2 x2 ).

(2.5)

Fundamentals

21 ♦

Proof. Let T ∈ LR(X, Y ), x1 , x2 ∈ D(T ) and α1 , α2 ∈ K, then in view of Remark 2.3.1 (ii), we have (α1 x1 + α2 x2 , α1 y1 + α2 y2 ) ∈ G(T ) for all y1 ∈ T x1 and y2 ∈ T x2 . Hence, α1 y1 + α2 y2 ∈ T (α1 x1 + α2 x2 ). So, (2.5) holds. Conversely, let T satisfy (2.5) and let (x1 , y1 ), (x2 , y2 ) ∈ G(T ). Then, for α1 , α2 ∈ K, we have α1 y1 + α2 y2 ∈ α1 T (x1 ) + α2 T (x2 ) ⊂ T (α1 x1 + α2 x2 ). Then, α1 (x1 , y1 ) + α2 (x2 , y2 ) ∈ G(T ). Therefore, G(T ) is a linear subspace of X × Y . Consequently, T is a linear relation by Remark 2.3.1 (ii). Q.E.D.

2.3.2

Inverse of Linear Relation

Let T ∈ LR(X, Y ). The inverse of T is the linear relation T −1 defined by G(T −1 ) = {(y, x) ∈ Y × X such that (x, y) ∈ G(T )} . If M ⊂ Y , then the inverse image of M under T is defined by n o \ T −1 (M ) := x ∈ D(T ) such that T x M 6= ∅ . Hence, T −1 (y) = {x ∈ D(T ) such that y ∈ T x} . Example 2.3.1 Let X = lp (N) be the space of sequences x : N −→ C summable with a power p ∈ [1, ∞) with the standard norm. Consider T the left shift single valued operator defined by T (x(n)) = x(n + 1), n ≥ 1 and x ∈ X. Then, T −1 is the linear relation defined by G(T −1 ) = {(x, y) ∈ X × X such that x(n) = y(n − 1) , n ≥ 2} . Definition 2.3.3 Let T ∈ LR(X, Y ). Then, the range R(T ) of T is defined by [ R(T ) = T x. x∈D(T )

The subspace T by N (T ).

−1

(0) is called the null space (or kernel) of T , and is denoted ♦

22

Spectral Theory of Multivalued Linear Operators

If R(T ) = Y , then we say that T is surjective, and if M ⊂ X, then the image of M under T is defined to be the set [ T (M ) = T x. x∈M

T

D(T )

If T −1 is single valued, then T is called injective, i.e., T −1 (0) = {0}. Proposition 2.3.2 [50, Subsection 1.1.3 (9), p. 3] If T ∈ LR(X, Y ), then (i) T is injective if and only if, T −1 T = ID(T ) . (ii) T is single valued if and only if, T T −1 = IR(T ) .

♦

Remark 2.3.2 (i) R(T ) = D(T −1 ) and D(T ) = R(T −1 ). (ii) R(T ) = {y such that (x, y) ∈ G(T )}. (iii) N (T ) = {x ∈ D(T ) such that (x, 0) ∈ G(T )}. (iv) The assertion (i) of Theorem 2.3.1, is equivalent to say that N (T ) = {x ∈ D(T ) such that T x = T (0)} . (v) If T is a linear operator such that N (T ) = 6 {0}, then T −1 is a linear relation (T −1 (0) = N (T ) 6= {0}). ♦ Example 2.3.2 Let T be an ordinary differential operator defined by T : C n ([a, b]) −→ C([a, b]) of the kind (T x(t)) = x(n) (t) + a1 (t)x(n−1) (t) + · · · + an (t)x(t), where ak ∈ C([a, b]), k = 1, . . . , n that acts in the Banach space C([a, b]) of bounded continuous complex functions on [a, b] ⊂ R and has a finitedimensional kernel N (T ) of dimension n ≥ 1. Therefore, T −1 ∈ LR(C([a, b])) and if T x = x0 , then Z T −1 x = x(t)dt, x ∈ C([a, b]). ♦ Theorem 2.3.1 [50, Proposition 1.2.8 and Corollary 1.2.11] Let T ∈ LR(X, Y ). Then, (i) for x ∈ D(T ), we have the following equivalence y ∈ T x if and only if, T x = y + T (0).

Fundamentals

23

In particular, 0 ∈ T x if and only if, T x = T (0). (ii) For x1 , x2 ∈ D(T ), we have the equivalence \ T x1 T x2 6= ∅ if and only if, T x1 = T x2 . (iii) T −1 T x = x + T −1 (0) for all x ∈ D(T ). (iv) T T −1 y = y + T (0) for all y ∈ R(T ). (v) T (0) and T −1 (0) are linear subspaces.

♦

Remark 2.3.3 By using (iii) and (iv) of Theorem 2.3.1, we have T T −1 (0) = T (0) and T −1 T (0) = T −1 (0). ♦

2.3.3

Sum and Product of Linear Relations

Definition 2.3.4 Let T , S ∈ LR(X, Y ). Then, the linear relation S + T is defined by (S + T )x := Sx + T x, for x ∈ X. ♦ Remark 2.3.4 Let T , S ∈ LR(X, Y ). As an immediate consequence of the Definition 2.3.4, we have the following properties T (i) D(T + S) = D(T ) D(S). (ii) T + S = S + T. (iii) The algebraic sum T + S is also a linear relation defined by G(T + S) = {(x, y + z) such that (x, y) ∈ G(T ), (x, z) ∈ G(S)} = {(x, y) such that y = u + v with (x, u) ∈ G(T ), (x, v) ∈ G(S)} . In particular, for all λ ∈ K, we have G(λ − T ) = {(x, λx − y) such that

(x, y) ∈ G(T )} .

(iv) Let T , S, R ∈ LR(X, Y ). Then, T + (R + S) = (T + R) + S.

♦

Proposition 2.3.3 Let T , S ∈ LR(X, Y ). If G(T ) ⊂ G(S), then G(T ) ⊂ G(T + S). ♦ Proof. To prove this, let us observe that G(T )

= {(x, y) ∈ X × Y such that x ∈ D(T ) ⊂ D(S) and y ∈ T x ⊂ Sx} n o \ ⊂ (x, y) ∈ X × Y such that x ∈ D(T ) D(S) and y ∈ T x + Sx = {(x, y) ∈ X × Y such that x ∈ D(T + S) and y ∈ (T + S)x} = G(T + S),

24

Spectral Theory of Multivalued Linear Operators

which completes the proof.

Q.E.D.

Proposition 2.3.4 Let S, T ∈ LR(X, Y ). If S(0) ⊂ T (0) and D(T ) ⊂ D(S), then T − S + S = T. ♦ Proof. Let (x, y) ∈ G(T + S − S), then x ∈ D(T − S + S) = D(T ) and y ∈ (T − S + S)x. So, y ∈ T x + S(0). Using the fact S(0) ⊂ T (0), it follows that T x + S(0) ⊂ T x + T (0) which yields y ∈ T x and x ∈ D(T ), that is (x, y) ∈ G(T ). Therefore, G(T + S − S) ⊂ G(T ). Conversely, let (x, y) ∈ G(T ), then x ∈ D(T ) and T x = y + T (0). Since S(0) ⊂ T (0) and D(T ) ⊂ D(S), we have (T + S − S)(0) = T (0) and D(T − S + S) = D(T ). This implies that x ∈ D(T − S + S) and (T + S − S)x = y + (T + S − S)(0). Thus, in view of Theorem 2.3.1 (i), we have y ∈ (T + S − S)x. Hence (x, y) ∈ G(T + S − S). Consequently, G(T ) ⊂ G(T + S − S). This completes the proof. Q.E.D. Definition 2.3.5 Let T ∈ LR(X, Y ) and S ∈ LR(Y, Z). If R(T ) ∅, then their product ST is also a linear relation defined by

T

D(S) = 6

ST x = S(T x), x ∈ X. For α ∈ K, we define the relation αT by (αT )x = α(T x), x ∈ X.

♦

Remark 2.3.5 Let X, Y and Z be linear spaces and let T ∈ LR(X, Y ), T S ∈ LR(Y, Z). If R(T ) D(S) 6= ∅, then (i) G(ST ) = {(x, z) ∈ X × Z such that (x, y) ∈ G(T ) and (y, z) ∈ G(S) for some y ∈ Y }, and D(ST )

6 ∅} = {x ∈ X such that ST x = T = {x ∈ X such that T x D(S) 6= ∅} = T −1 (D(S)).

In particular, for all α ∈ K\{0}, we have D(αT ) = D(T ), and G(αT ) = {(αx, y) such that (x, y) ∈ G(T )} . S (ii) If X = Y , we also define the product ST x = {Sy : y ∈ D(S) ∩ T x} with D(ST ) = {x ∈ D(T ) : D(S) ∩ T x = ♦ 6 ∅}.

Fundamentals

25

Theorem 2.3.2 Let X , Y and Z be linear spaces and let T ∈ LR(X, Y ), T S ∈ LR(Y, Z). If R(T ) D(S) 6= ∅, then (i) N (T ) ⊂ N (ST ) ⊂ D(ST ) ⊂ D(T ), and (ii) S(0) ⊂ ST (0) ⊂ R(ST ) ⊂ R(S).

♦

Proof. (i) Let x ∈ N (T ), then (x, 0) ∈ G(T ). Since (0, 0) ∈ G(S), it follows that (x, 0) ∈ G(ST ). Hence, x ∈ N (ST ), which shows that N (T ) ⊂ N (ST ). Clearly, N (ST ) ⊂ D(ST ). Let x ∈ D(ST ), then (x, z) ∈ G(ST ) for some z ∈ Z. Hence, (x, y) ∈ G(T ) and (y, z) ∈ G(S) for some y ∈ Y , which shows that x ∈ D(T ). So, D(ST ) ⊂ D(T ). (ii) Let z ∈ S(0), then (0, z) ∈ G(S). Since (0, 0) ∈ G(T ), it follows that (0, z) ∈ G(ST ). Hence, z ∈ ST (0), which shows that S(0) ⊂ ST (0). Clearly, ST (0) ⊂ R(ST ). Let z ∈ R(ST ), then (x, z) ∈ G(ST ) for some x ∈ X. Then, (x, y) ∈ G(T ) and (y, z) ∈ G(S) for some y ∈ Y . Thus, z ∈ R(S). Hence, R(ST ) ⊂ R(S). Q.E.D. Lemma 2.3.1 [50, Lemma 1.6.8] Let T ∈ LR(X) and M , N ⊂ X such that T X = M + N , M N = {0}, D(T ) = X, and N (T ) ⊂ N . Then, R(T ) = T T (M ) + T (N ) and T (M ) T (N ) = T (0). ♦ Proposition 2.3.5 [50, Proposition 1.3.1] Let T ∈ LR(X), α ∈ K\{0}, and let M , N ⊂ X. Then, (i) T (αM ) = αT (M ), (ii) T (M + N ) ⊃ T (M ) + T (N ), (iii) if M ⊂ D(T ) or N ⊂ D(T ), then T (M + N ) = T (M ) + T (N ). Moreover, T T T if M N = {0}, then T (M N ) = T (0) = T (M ) T (N ), T (iv) T T −1 (M ) = M R(T ) + T (0), for all M ⊂ X, and T ♦ (v) T −1 T (M ) = M D(T ) + N (T ), for all M ⊂ X.

2.3.4

Restrictions and Extensions of Linear Relations

Definition 2.3.6 (i) Let T ∈ LR(X, Y ), and M be a subspace of X such that T M D(T ) 6= ∅. Then, the restriction of T to M , denoted by T|M is defined by    T|M ∈ LR(X, Y ) \ D(T|M ) = D(T ) M   (T|M )m = T m for all m ∈ M.

26

Spectral Theory of Multivalued Linear Operators

(ii) Let T , S ∈ LR(X, Y ). The relation S is said an extension of T if S|D(T ) = T. We write T ⊂ S. ♦ Remark 2.3.6 (i) Let T ∈ LR(X, Y ), and M be a subspace of X such that T M D(T ) 6= ∅. Then, \ (i1 ) G(T|M ) = {(x, y) ∈ G(T ) such that x ∈ M } = G(T ) (M × Y ). (i2 ) T|M = T|D(T ) T M . (ii) By using Definition 2.2.1, we deduce that X X −1 T|M = T JM (JM ) . X −1 X ) . (JM Hence, IM = JM (iii) If T is an extension of S −1 , then it is not implies T −1 is an extension of S. ♦

Example 2.3.3 Let X = R3 and consider the relation T defined by         

        x1 x1 0         T  x2  =  x2  +  0  such that x3 ∈ R ,     x3 0 x3 D(T ) = R3 . 



   x1 1 0 0     Let S be the single valued map  0 1 0  . Writing x =  x2  and 0 0 0 x3   y1   y =  y2  for arbitrary vectors in R3 . Then, the graph of S is defined by y3          x1 y1 x1 y1            G(S) =  x2  ,  y2  such that  x2  =  y2  .     x3 y3 0 y3 The relation S −1 is defined by     x1 x1     S −1  x2  =  x2  + S −1 (0), 0 0 and

Fundamentals

27

      0         −1  S (0) = x ∈ such that R is a linear subspace of R3 , 0    3          x3    x1         2 −1   such that (x , x ) ∈ R . D( S ) = x  2  1 2        0 Then, T is an extension of S −1 , and by using Remark 2.3.6 (ii), we have −1 −1 T IR(A) = T JR(A) JR . (A) = S

Hence, −1 −1 S = JR(A) JR . (A) T

6 S. However, T −1 = Proposition 2.3.6 (i) If G(S) ⊂ G(T ), then T is an extension of S if and only if, S(0) = T (0). (ii) If G(S) ⊂ G(T ), N (T ) ⊂ N (S), and R(T ) ⊂ R(S), then T = S. (iii) If D(T ) = D(S) and T (0) = S(0), then T = S or the graphs of T and S are incomparable. (iv) (R + S)T ⊂ RT + ST with equality if T is single valued. (v) T R + T S = T (R + S) if D(T ) is the whole space. (vi) T (R + S) is an extension of T R + T S. ♦ Remark 2.3.7 (i) If S is an extension of T , then G(T ) ⊂ G(S). (ii) Let R = −S, where S is single valued and nonzero, and let G(T ) = X × Y 6 {0}. Then, where Y = [ (R + S)T (0) = (S − S)(Y ) = (S − S)(y) = {0} y∈D(S)

and (RT +ST )(0) = (ST −ST )(0) = ST (0)−ST (0) = S(Y )−S(Y ) = S(Y ) 6= {0}. This shows that RT + ST is not in general an extension of (R + S)T . (iii) Let R, S, T ∈ LR(X). If RT = T R and ST = T S, then G((R + S)T ) ⊂ G(T (R + S)).

(2.6)

The equality fails in (2.6), even if R, S, T are all single valued, as seen by considering the following example: S = −R = I, and T is a linear operator with D(T ) 6= X. ♦

28

Spectral Theory of Multivalued Linear Operators

We construct an example making T (R + S) in Proposition 2.3.6 (vi) a proper extension of T R + T S. Example 2.3.4 Let R, S, T ∈ LR(R2 ) be defined by R=I 

x 0

G(S) =

! ,

w x 

! !

!

 f or x ∈ R .

 f or x, w ∈ R

0 y

T = I|M , where M =

!

 f or y ∈ R .

Then, D(T R) = D(T ) and 

x 0

D(T S) = Hence, D(T R + T S) = {0}. Now,  G(R + S) =

x 0

! ,

z x

! !

 f or x, z ∈ R .



x 0

Thus, −1

D(T (R + S)) = (R + S)

(D(T )) =

!

 f or x ∈ R .

6 ! T (R + S). In this example S is the inverse of the single Therefore, T R + T S = 0 1 valued map . ♦ 0 0

2.4

Index and Co-Index of Linear Relations

Definition 2.4.1 Let T ∈ LR(X, Y ), we define α(T ) = dim(N (T )),

β(T ) = dim(Y /R(T )), and β(T ) = dim(Y /R(T )).

Fundamentals

29

If α(T ) and β(T ) are finite, then the index of T is defined by i(T ) = α(T ) − β(T ). The index of T −1 is called co-index of T and is denoted by ci(T ). Then, ci(T ) = i(T −1 ) = α(T −1 ) − β(T −1 ) = dim(T (0)) − codim(D(T )). ♦

If T is a single valued everywhere defined, then ci(T ) = 0.

Remark 2.4.1 For a given closed linear subspace E of a normed vector space X, we have i(QX E ) = dim E. X X In fact, it is obvious that N (QX E ) = E and R(QE ) = X/E. Hence, α(QE ) = dim(E) and β(QX ♦ E ) = 0.

Lemma 2.4.1 Let A, B ∈ LR(X, Y ) such that G(A) ⊂ G(B). Then, (i) N (A) ⊂ N (B) and α(A) ≤ α(B). (ii) R(A) ⊂ R(B) and β(B) ≤ β(A). (iii) i(A) ≤ i(B).

♦

Proof. (i) To prove this, let us observe that N (A)

= {x ∈ D(A) such that (x, 0) ∈ G(A)} ⊂

{x ∈ D(A) such that (x, 0) ∈ G(B)}

= N (B|D(A) ) ⊂ N (B). So, α(A) ≤ α(B). (ii) Let y ∈ R(A). Then, there exists x ∈ D(A) such that y ∈ Ax. By using G(A) ⊂ G(B), we have y ∈ Bx. Hence, y ∈ R(B|D(A) ). So, y ∈ R(B). Thus, Y /R(B) ⊂ Y /R(A) and hence β(B) ≤ β(A). (iii) i(A) = α(A) − β(A) ≤ α(B) − β(B) = i(B).

Q.E.D.

Proposition 2.4.1 Let T, S ∈ LR(X, Y ). Then, (i) If N (S) = N (T ), R(S) = R(T ), and S is an extension of T, then S = T. (ii) If S be an extension of T , and i(T ) = i(S), then T = S. ♦

30

Spectral Theory of Multivalued Linear Operators

Proof. (i) Let (x, y) ∈ G(S). Then, y ∈ Sx ⊂ R(S) = R(T ). So, there exists x1 ∈ D(T ) such that y ∈ T x1 = Sx1 . It follows that Sx = Sx1 which means that x − x1 ∈ N (S) = N (T ) ⊂ D(T ). Therefore, x = (x − x1 ) + x1 ∈ D(T ). Thus, (x, y) ∈ G(T ). (ii) To show that T = S, it suffices to prove that S ⊂ T (equivalently G(S) ⊂ G(T )). In view of Lemma 2.4.1 (i) and (ii), it is easy to see that α(T ) ≤ α(S) and β(S) ≤ α(T ). Since i(T ) = i(S), then R(T ) = R(S) and N (T ) = N (S). Hence, by using (i), we have T = S. Q.E.D. Lemma 2.4.2 [50, Corollary 1.6.12] (Index of product) Let X, Y , Z be three vector spaces, T ∈ LR(X, Y ), S ∈ LR(Y, Z) and D(S) = Y . Suppose that T and S have finite indices. Then, ST has a finite index and, T i(ST ) = i(S) + i(T ) − dim (T (0) N (S))
T = i(S) + i(T ) − dim T (0) S −1 (0) . ♦ ! 0 0 Example 2.4.1 Let X = Y = Z = R2 and S = . Then, the graph 0 1 of S is defined by ( ! !! ! !) x1 y1 0 y1 G(S) = , . : = x2 y2 x2 y2 Hence, S −1 is the linear relation defined by  ! !  0 0  −1  = + S −1 (0),   S x2 x2 ( ! )  0  −1  such that x2 ∈ R ,   D(S ) = x 2

and

( S

−1

(0) =

x1 0

!

) such that x1 ∈ R

is a linear subspace of R2 . Let T = S −1 , then N (T ) = {0} × {0}, R(T ) = R × R, N (S) = R × {0}, N (ST ) = {0} × {0}, and R(ST ) = {0} × R. Hence, α(T ) = 0, β(T ) = 0, α(S) = 1, β(S) = 1, α(ST ) = 0, and β(ST ) = 1. So, i(T ) = α(T ) − β(T ) = 0, and i(S) = α(S) − β(S) = 0. Thus, i(ST ) = T T i(S) + i(T ) − dim (T (0) N (S)) = −1 and dim (T (0) N (S)) = 1. We would like to mention that, A. Sandovici and H. de Snoo [84] study some proprieties of index formula for the product of unbounded linear relations.

Fundamentals

2.4.1

31

Properties of the Quotient Map QT

We shall denote QYT (0) by QT , or simply Q when T is understood. Theorem 2.4.1 Let T , S ∈ LR(X, Y ) such that T (0) + S(0) is closed. Then, (i) QYT +S

Y /T (0)

= Q T (0)+S(0) QYT T (0)

Y /S(0)

= Q T (0)+S(0) QYS . S(0)

 (ii)

(iii)

dim

dim



  T T (0) + S(0) = dim S(0) − dim T (0) S(0) . T (0) !   \ T (0) + S(0) = dim T (0) − dim T (0) S(0) . S(0)

♦

Proof. (i) Since T (0) + S(0) is closed, then T (0) + S(0) ⊂ T (0) + S(0).

(2.7)

Moreover, since T (0) ⊂ T (0) + S(0) and S(0) ⊂ T (0) + S(0), then T (0) + S(0) ⊂ T (0) + S(0).

(2.8)

From (2.7) and (2.8), it follows that T (0) + S(0) = T (0) + S(0). Therefore, QYT +S = QYT (0)+S(0) = QYT (0)+S(0) . Since T (0) + S(0) is closed containing T (0), then by using Lemma 2.1.4, we obtain Y /T (0) QYT +S = QYT (0)+S(0) = Q T (0)+S(0) QTY (0) . T (0)

Hence, Y /T (0)

QYT +S = Q T (0)+S(0) QYT . T (0)

In the same way, since T (0) + S(0) is closed containing S(0), we obtain Y /S(0)

QYT +S = QYT (0)+S(0) = Q T (0)+S(0) QYS(0) . S(0)

So, Y /S(0)

QYT +S = Q T (0)+S(0) QYS . S(0)

32

Spectral Theory of Multivalued Linear Operators

(ii) By using Theorem 2.1.2, we have T (0) + S(0) ∼ S(0) . = T T (0) T (0) S(0)

(2.9)

It follows that dim

T (0) + S(0)

!

! S(0) = dim T T (0) S(0)  T = dim(S(0)) − dim T (0) S(0) .

T (0)

(iii) The proof may be achieved in a similar way as (ii). It is sufficient to replace (2.9) by T (0) + S(0) ∼ T (0) . = T S(0) T (0) S(0)

Q.E.D.

Corollary 2.4.1 Let T , S ∈ LR(X, Y ) such that dim S(0) < +∞. Then, QYT +S

Y /T (0)

= Q T (0)+S(0) QYT , T (0)

Y /S(0)

= Q T (0)+S(0) QYS , S(0)

and  dim

T (0) + S(0) T (0)



  T = dim S(0) − dim T (0) S(0) .

Proposition 2.4.2 Let T ∈ LR(X, Y ). Then, QT T is single valued.

♦ ♦

Proof. Let x ∈ D(T ) and z1 , z2 ∈ QT T x, then, z1 − z2 ∈ QT T x − QT T x = QT T (0) ⊂ QT T (0) = {0}. Hence z1 = z2 . Q.E.D.

2.5

Generalized Kernel and Range of Linear Relations

The integer powers of a multivalued linear operator T ∈ LR(X) are recursively defined as follows: T 0 = I, and if T n−1 is defined, then we set T n x := (T T n−1 )x :=

[

T y such that y ∈ D(T ) ∩ T n−1 x ,

Fundamentals

33

with D(T n ) := {x ∈ D(T n−1 ) such that D(T ) ∩ T n−1 x 6= ∅}. It is easy to see that (T −1 )n = (T n )−1 , for all n ∈ Z. Remark 2.5.1 Let T and S be two linear relations in a vector space such that T and S commute, that is, T S = ST in the sense of the product of linear relations, then (λ − S)T = T (λ − S), and (λ − T )S = S(λ − T ) holds for all λ ∈ C. Let µ ∈ K, by applying this property to µ − T instead and µ − S, we have (λ − S)(µ − T ) = (µ − T )(λ − S) and (λ − T )(µ − S) = (µ − S)(λ − T ). Furthermore, it is clear that T n S m = S m T n holds for all n, m ∈ N. By using this equality with λ − S, µ − T and λ − S instead of T and S, it follows that (λ − S)m (µ − T )n = (µ − T )n (λ − S)m and (λ − T )m (µ − S)n = (µ − S)n (λ − T )m .

♦

The following results can be found in Refs. [84, 85]. Lemma 2.5.1 Let T be a linear relation in a vector space X. Then, (i) For all α, β ∈ K and for all m, n ∈ N, we have (a) D((T − α)n (T − β)m ) = D(T n+m ). (b) (T − α)n (T − β)m (0) = T n+m (0). (c) (T − α)n (T − β)m = (T − β)m (T − α)n . (d) If α = β, then N (T − α)n ⊂ R(T − β)m . (ii) If there exists λ ∈ C such that (λ − T )−1 is bounded single valued, then T N (T n ) T m (0) = {0} for all m, n ∈ N. (iii) Let n, p ∈ N. If there is η ∈ C such that T − η is bijective, then D(T n ) + R(T p ) = X, and

34

Spectral Theory of Multivalued Linear Operators \ T n (0) N (T p ) = {0}

for all n, p ∈ N. Further, if T has a finite index, then i(T n ) = ni(T ) and i(T n+p ) = i(T n ) + i(T p ) = ni(T ) + pi(T ) = (n + p)i(T ). ♦ Theorem 2.5.1 Let T be a linear relation in a vector space X. Then, for all n, m ∈ N D(T n+m ) ⊂ D(T n ), R(T n+m ) ⊂ R(T n ), (2.10) N (T n+m ) ⊃ N (T n ), T n+m (0) ⊃ T n (0),

(2.11)

and for all p, k ∈ N N (T p ) ⊂ D(T k ), T p (0) ⊂ R(T k ).

(2.12) ♦

Proof. Assume that x ∈ D(T n+m ), so that (x, y) ∈ G(T n+m ) for some y ∈ X. Since T n+m = T m T n , it follows that (x, z) ∈ G(T n ) and (z, y) ∈ G(T m ) for some z ∈ X, which shows that x ∈ D(T n ). Therefore, D(T n+m ) ⊂ D(T n ), and the first inclusion in Eqn. (2.10) is proved. In order to prove the first inclusion in Eqn. (2.11) assume that x ∈ N (T n ), so that (x, 0) ∈ G(T n ). Since (0, 0) ∈ G(T m ) as G(T m ) is a subspace of X × X, it follows that (x, 0) ∈ G(T n+m ), which shows that x ∈ N (T n+m ). Therefore, N (T n+m ) ⊃ N (T n ). If p ≤ k, then it follows from Eqn. (2.11) that N (T p ) ⊂ N (T k ) and since N (T k ) ⊂ D(T k ), it follows that in this case, Eqn. (2.12) holds true. Assume now that p > k and let x ∈ N (T p ), so that (x, 0) ∈ G(T p ) = G(T p−k T k ). Thus, (x, y) ∈ G(T k ) and (y, 0) ∈ G(T p−k ), which shows that x ∈ D(T k ). Therefore, the first inclusion in (2.12) is proved. The remaining inclusions in Eqs. (2.10)–(2.12) follow from the duality of T and T −1 . This completes the proof. Q.E.D. Corollary 2.5.1 (i) Let T ∈ LR(X). Then, the kernels and the ranges of the iterates T n , n ∈ N form two increasing and decreasing chains, respectively, i.e., the chain of kernels {0} = N (T 0 ) ⊂ N (T ) ⊂ N (T 2 ) ⊂ · · · and the chain of ranges X = R(T 0 ) ⊃ R(T ) ⊃ R(T 2 ) ⊃ · · · . (ii) T n (0) ⊂ T n+1 (0), for all n ∈ N.

♦

Fundamentals

35

Definition 2.5.1 Let T be a linear relation in X and let o n \ \ ∆(T ) = n ∈ N such that R(T n ) N (T ) = R(T m ) N (T ) for all m ≥ n . The degree η(T ) of T , is defined as ( inf ∆(T ) if ∆(T ) 6= ∅ η(T ) = +∞ if ∆(T ) = ∅.

♦

Lemma 2.5.2 [74, Lemmas 2.5, 2.7 and Corollary 2.6] Let T be a linear relation in X. Then, (i) d ∈ ∆(T ) if and only if, N (T m ) ⊂ N (T d ) + R(T n ) if and only if, T T R(T d ) N (T n ) ⊂ R(T m ) N (T n ) for all m, n ∈ N. (ii) η(T ) = 0 if and only if, N (T m ) ⊂ R(T n ) for all m, n ∈ N if, and only if, N (T n ) ⊂ R(T ) for all n ∈ N if and only if, N (T ) ⊂ R(T m ) for all m ∈ N. ♦ Lemma 2.5.3 Let m ∈ N. Then, η(T m ) ≤ η(T ) ≤ mη(T m ).

♦

Proof. If η(T ) = ∞, the inequality η(T m ) ≤ η(T ) is obvious. Let d := η(T ) < ∞. Then, by virtue of Lemma 2.5.2 (i), we have \ \ R(T d ) N (T m ) ⊂ R(T mr ) N (T m ), whenever r ≥ d. So, for all r ≥ d, \ \ N (T m ) R(T md ) = N (T m ) R(T mr ). So, d ∈ ∆(T m ). Hence, η(T m ) ≤ η(T ). If η(T m ) = ∞, then the inequality η(T ) ≤ mη(T m ) is clear. Assume that r := η(T m ) < ∞ and let n ≥ r. Then, \ \ \ N (T ) R(T mr ) = N (T ) N (T m ) R(T mr ) \ \ = N (T ) N (T m ) R(T mn ) \ = N (T ) R(T mn ). Let p ∈ N such that p ≥ mr and we write p = mk + s for some nonnegative integers k and s such that k + 1 ≥ r and 0 ≤ s ≤ m − 1. Then, \ \ N (T ) R(T mr ) = N (T ) R(T m(k+1) ) \ ⊂ N (T ) R(T p ) \ ⊂ N (T ) R(T mr ).

36

Spectral Theory of Multivalued Linear Operators T T Therefore, N (T ) R(T mr ) = N (T ) R(T p ) for all p ≥ mr. So, mr ∈ ∆(T ). Hence, η(T ) ≤ mη(T m ). This completes the proof.

Q.E.D.

Lemma 2.5.4 Let T be a relation in a linear space X and let i, k ∈ N ∪ {0}. Then, N (T i+k ) N (T k ) ∩ R(T i ) ∼ , = (N (T i ) + R(T k )) ∩ N (T i+k ) N (T k ) ∩ R(T i+k ) and

T i+k (0) T k (0) ∩ D(T i ) ∼ . = (T i (0) + D(T k )) ∩ T i+k (0) T k (0) ∩ D(T i+k )

♦

Proposition 2.5.1 Let T be a relation in a linear space X. Then, (i) If N (T k ) = N (T k+1 ) for some k ∈ N, then N (T n ) = N (T k ) for all nonnegative integers n ≥ k. (ii) If T k (0) = T k+1 (0) for some k ∈ N, then T n (0) = T k (0) for all nonnegative integers n ≥ k. ♦ Proof. (i) Assume that N (T n+1 ) = N (T n ). It will be shown that N (T n+2 ) = N (T n+1 ), and then, the statement will follow by induction. Clearly, (2.11) shows that N (T n+1 ) ⊂ N (T n+2 ). So, only the converse inclusion remains to be proved. Let x ∈ N (T n+2 ), then (x, 0) ∈ G(T n+2 ) = G(T n+1 T ). Thus, (x, y) ∈ G(T ) and (y, 0) ∈ G(T n+1 ) for some y ∈ X. Now, y ∈ N (T n+1 ) = N (T n ) by the induction hypothesis, which shows that (y, 0) ∈ G(T n ). Therefore, (x, 0) ∈ G(T n+1 ). So, x ∈ N (T n+1 ), which implies (i). The statement (ii) follows from the statement in (i) due to the duality of T and T −1 . Q.E.D. Theorem 2.5.2 Let T be a linear relation in a vector space X. Then, (i) If D(T k ) = D(T k+1 ) for some k ∈ N, then D(T n ) = D(T k ) for all nonnegative integers n ≥ k. (ii) If R(T k ) = R(T k+1 ) for some k ∈ N, then R(T n ) = R(T k ) for all nonnegative integers n ≥ k. ♦ Proof. (i) Assume that D(T n ) = D(T n+1 ). It suffices to show that D(T n+1 ) = D(T n+2 ). Clearly, (2.10) shows that D(T n+2 ) ⊂ D(T n+1 ). So, only the converse inclusion remains to be proved. Assume that x ∈ D(T n+1 ),

Fundamentals

37

then (x, y) ∈ G(T n+1 ) = G(T n T ). Hence, (x, z) ∈ G(T ) and (z, y) ∈ G(T n ) for some z ∈ X. Now, z ∈ D(T n ) = D(T n+1 ) implies that (z, u) ∈ G(T n+1 ) for some u ∈ X. This leads to (x, u) ∈ G(T n+2 ). So, x ∈ D(T n+2 ). Hence, D(T n+1 ) ⊂ D(T n+2 ). The statement (ii) follows from the statement in (i) due to the duality of T and T −1 . Q.E.D. Corollary 2.5.2 Let T be a linear relation in a vector space X. Then, (i) D(T ) = X if and only if, D(T p ) = X for some (for all) p ∈ N, (ii) R(T ) = X if and only if, R(T p ) = X for some (for all) p ∈ N.

♦

Proof. (i) If D(T ) = X, then it follows from Theorem 2.5.2 that D(T p ) = X for all p ∈ N. Conversely, if D(T p ) = X for some p ∈ N, then due to (2.10), it is clear to see that D(T i ) = X for 1 ≤ i ≤ p. The proof of (ii) is analogous.

Q.E.D.

An important role is played by certain root manifolds of a relation T in a linear space X. The root manifold (the generalized kernel) R0 (T ) is defined by ∞ [ R0 (T ) := N ∞ (T ) = N (T i ). (2.13) i=1

Similarly, the root manifold ( the generalized range) R∞ (T ), is defined by R∞ (T ) :=

∞ \ i=1

R(T i ) =

∞ [

T i (0).

(2.14)

i=1

Clearly the root manifolds R0 (T ) and R∞ (T ) are subspaces of D(T ) ⊂ X and R(T ) ⊂ X, respectively. Lemma 2.5.5 [9, Lemma 20] Let T be a linear relation in a vector space X. Then, (i) If λ ∈ K\{0}, then N (λ − T ) ⊂ R∞ (T ). (ii) If λ, µ ∈ K are distinct, then N (λ − T )n ⊂ R∞ (µ − T ) for all n ∈ N. T T (iii) If there exists d ∈ N such that N (T ) R(T d ) = N (T ) R(T n+d ) for all T ∞ n ∈ N, then T (D(T ) R (T )) = R∞ (T ). T (iv) If N (T ) ⊂ R∞ (T ) or α(T ) < ∞, then T (D(T ) R∞ (T )) = R∞ (T ). ♦

38

Spectral Theory of Multivalued Linear Operators

2.5.1

Ascent and Descent and Singular Chain of Linear Relations

Likewise, the statements in Theorem 2.5.1 and Proposition 2.5.1 lead to the introduction of the ascent and descent of linear relation. Definition 2.5.2 Let T ∈ LR(X, Y ). The ascent and the descent of T are defined as follows  asc(T ) = inf p ∈ N such that N (T p ) = N (T p+1 ) and  des(T ) = inf p ∈ N such that R(T p ) = R(T p+1 ) , respectively, whenever these minima exist. If no such numbers exist the ascent and descent of T are defined to be ∞. ♦ Let T ∈ LR(X). The singular chain manifold Rc (T ) is defined as the intersection of the root manifolds R0 (T ) in (2.13) and R∞ (T ) in (2.14): ! ! ∞ ∞ \ [ \ [ ∞ n n Rc (T ) = R0 (T ) R (T ) = N (T ) T (0) . n=1

n=1

The linear space Rc (T ) is non-trivial if and only if, there exists a number s ∈ N and elements xi ∈ X, 1 ≤ i ≤ s, not all equal to zero, such that (0, x1 ), (x1 , x2 ), · · · , (xs−1 , xs ), (xs , 0) ∈ G(T ). We say that T has trivial singular chain if Rc (T ) = {0}. Example 2.5.1 Let X = span{e1 , e2 } with e1 and e2 are linearly independent and define the relation T by T = span {{0, e1 }, {e1 , 0}, {e2 , 0}} and S = span {{e1 , e2 }, {e2 , e1 }} . Then, S 2 = I and Rc (S) = {0}. The nullity and defect of T are given by α(T ) = 2 and β(T ) = 1. Moreover, T 2 = T and the ascent and descent of T are given by asc(T ) = 1 and des(T ) = 1. So, Rc (T ) 6= {0}. Proposition 2.5.2 Let T be a relation in a linear space X. If one of the following conditions \ \ D(T r ) T (0) = {0} or R(T r ) N (T ) = {0} is satisfied for some r ∈ N, then Rc (T ) = {0}.

♦

Fundamentals 39 T Proof. Assume that D(T r ) T (0) = {0}. If r = 0, then T (0) = {0} and hence Rc (T ) = {0}. Now, let r ∈ N∗ . If Rc (T ) 6= {0}, then T has a non-trivial singular chain of the form (0, x1 ), (x1 , x2 ), (x2 , x3 ), . . . , (xn−1 , xn ), (xn , 0) for some non-zero vectors xi ∈ X, 1 ≤ i ≤ n. Clearly, x1 ∈ T (0) and x1 ∈ N (T n ) ⊂ D(T r ) by (2.12). Therefore, \ x1 ∈ D(T r ) T (0) = {0}, which shows that x1 = 0. This contradiction implies that Rc (T ) = {0}. The Q.E.D. argument for the other case is completely similar. Lemma 2.5.6 (i) [85, Lemmas 4.4 and 5.1] Let T be a linear relation in a vector space X with Rc (T ) = {0} and let i, k ∈ N. Then, \ N (T i+k ) ∼ k N (T ) R(T i ). = N (T i ) (ii) Let T be a linear relation in a vector space X and let r ∈ N. Then,     N (T r ) + R(T ) N (T r ) T dim = dim r R(T )  N (Tr ) R(T )) D(T ) + R(T ) ≤ dim R(T )   D(T r ) T = dim . D(T r ) R(T )

2.5.2

♦

Norm of a Linear Relation

Let X and Y be two normed spaces. Definition 2.5.3 Let T ∈ LR(X, Y ). We define the norm of T x and T by kT xk = kQT T xk

(x ∈ D(T )),

and kT k = kQT T k.

♦

We note kT k is not a true “norm” function since kT k = 0 does not imply that T = 0. For example, the linear relation T ∈ LR(X, Y ) defined as G(T ) = X × Y has a zero norm.

40

Spectral Theory of Multivalued Linear Operators

Lemma 2.5.7 Let T ∈ LR(X, Y ). Then, (i) For x ∈ D(T ), kT xk = =

dist(y, T (0)) for all y ∈ T x inf kyk

y∈T x

= dist(y + T (0), 0) for all y ∈ T x = dist(T x, 0) = dist(T x, T (0)). (ii) kT k =

sup x∈BD(T )

kT xk, where BD(T ) = {x ∈ D(T ) such that kxk ≤ 1}.

♦

Proof. (i) The first equality follows from definition of kT xk and Theorem 2.3.1. The second equality follows from the definition and properties of the norm on Y /T (0). The rest are obvious. (ii) kT k = kQT T k =

sup x∈BD(T )

kQT T xk =

sup

kT xk.

Q.E.D.

x∈BD(T )

Proposition 2.5.3 Let X be a normed space and T ∈ LR(X). Then, kT k = 0 if and only if, T (BD(T ) ) ⊂ T (0). ♦ Proof. If kT k = 0, then by using Lemma 2.5.7, we have kT xk = 0, for all x ∈ BD(T ) . Hence, for all x ∈ BD(T ) and y ∈ T x, we have dist(y, T (0)) = 0. So, for all x ∈ BD(T ) and y ∈ T x, we have dist(y, T (0)) = 0. This implies that for all x ∈ BD(T ) , T x ⊂ T (0). Thus, T (BD(T ) ) ⊂ T (0). Conversely, by using both the same above reasoning and Lemma 2.5.7, we find the result. Q.E.D. Proposition 2.5.4 For T and S ∈ LR(X, Y ), we have (i) kT x + Sxk ≤ kT xk + kSxk, for x ∈ D(T ) ∩ D(S). (ii) kαT xk = |α|kT xk, for α ∈ K and x ∈ D(T ). (iii) If S(0) ⊂ T (0) and D(T ) ⊂ D(S), then QT S is a single valued linear operator and kQT Sk ≤ kSk. ♦ Proof. (i) Let x ∈ D(T ) ∩ D(S), s ∈ Sx and t ∈ T x, then s + t ∈ Sx + T x = (S + T )x. Hence, by using Lemma 2.5.7 (i), we have kT x + Sxk = dist (t + s, (T + S)(0)) ≤ dist (t, T (0) + S(0)) + dist (s, T (0) + S(0)) ≤ dist (t, T (0)) + dist (s, S(0)) = kT xk + kSxk.

Fundamentals

41

(ii) The case α = 0 is obviously. If α = 6 0, then for all x ∈ D(T ) kαT xk = kQT (αT )(x)k (since αT (0) = T (0) for α = 6 0) = kαQT T (x)k = |α|kQT T (x)k. (iii) Since S(0) ⊂ T (0), then QT (S(0)) ⊂ QT (T (0)) = {0}. Consequently, QT S is a single valued operator, and kQT Sxk = dist(Sx, T (0)) = dist(Sx, T (0)) ≤ dist(Sx, S(0)) = kSxk = kQS Sxk. This completes the proof. Q.E.D. Proposition 2.5.5 [50, Corollary 2.3.13] Let T ∈ LR(X, Y ) and S ∈ LR(Y, Z). Then, kST k ≤ kSk kID(S) T k (∞.0 excluded)

(i)

with kST k = 0 whenever kSk = 0 even if kID(S) T k = ∞. (ii) Moreover, if T (0) ⊂ D(S), then kST k ≤ kSk kT k.

♦

Proposition 2.5.6 Let T ∈ LR(X, Y ). Then, (i) kT k < ∞ if and only if, there exists λ > 0 such that T (BD(T ) ) ⊂ λBR(T ) + T (0).

(2.15)

(ii) If kT k < ∞, then  kT k = inf λ such that T (BD(T ) ) ⊂ λBR(T ) + T (0) . λ>0

(2.16) ♦

Proof. (i) Suppose kT k < ∞. By using Lemma 2.5.7 (ii), we have for x ∈ BD(T ) and y ∈ T x, there exists m ∈ T (0) such that for all ε > 0, ky − mk < kT k + ε, that is, y − m ∈ (kT k + ε)BR(T ) . So, y ∈ (kT k + ε)BR(T ) + T (0).

(2.17)

Conversely, suppose (2.15) holds. Let x ∈ BD(T ) and choose y ∈ T x. Then, y = λy1 + m, where ky1 k ≤ 1 and m ∈ T (0). Thus, ky − mk ≤ λ, in particular, dist(y, T (0)) ≤ λ. It follows from Lemma 2.5.7 (i) that kT k ≤ λ < ∞.

42

Spectral Theory of Multivalued Linear Operators

(ii) Suppose kT k < ∞. If kT k = 0, then by using Proposition 2.5.3, we have T (BD(T ) ) ⊂ T (0). So, (2.16) holds. Suppose kT k > 0. It follows from (2.17) that  kT k ≥ inf λ such that T (BD(T ) ) ⊂ λBR(T ) + T (0) . λ>0

Let α ∈]0, kT k[ and choose x ∈ BD(T ) , y ∈ T x such that α < dist(y, T (0)).

(2.18)

If T (BD(T ) ) ⊂ αBR(T ) + T (0), then there exist y1 ∈ BR(T ) and m ∈ T (0) such that y = αy1 + m. So, ky − mk ≤ α, which contradicts (2.18). Thus,  α < inf λ such that T (BD(T ) ) ⊂ λBR(T ) + T (0) . Q.E.D. λ>0

Definition 2.5.4 Let T ∈ LR(X, Y ). The T graph norm k · kT is defined on D(T ) by kxkT = kxk + kT xk, f or x ∈ D(T ). We denote by XT the vector space (D(T ), k · kT ) and GT ∈ LR(XT , X) the identity injection (or the graph operator) of XT into X, i.e., D(GT ) = XT , and GT x = x for all x ∈ XT . GT is called the graph operator of T . ♦ Remark 2.5.2 (i) If T = Te, then XT and T (0) are complete. (ii) Let X be a complete space. If we denote by (GT )−1 the inverse of GT , then (GT )−1 and GT are bijective everywhere defined. So, R(T GT ) = T GT (D(T GT )) = T GT (GT )−1 (D(T )) = R(T ) and α(T GT ) = α(T ). ♦ Theorem 2.5.3 Let S, T ∈ LR(X, Y ) such that D(S) = D(T ). Then, kSk kSGT k = 1+kT ♦ k. Proof. We have kSGT k =

sup x∈XT

= =

kSGT xk kxkT

kSxk x∈D(T )\{0} kxk + kT xk sup sup x∈D(T )\{0}

=



1+

kSxk  kT x k kxk

kxk

kSk (as D(S) = D(T )). 1 + kT k

This completes the proof. Corollary 2.5.3 Let T ∈ LR(X, Y ). Then, kT GT k =

Q.E.D. kT k 1+kT k .

♦

Fundamentals

2.5.3

43

Continuity and Openness of a Linear Relation

Definition 2.5.5 (i) Let T be an arbitrary linear relation from one topological space X to another Y . Then, T is said to be continuous if for each neighbourhood V in R(T ) the image T −1 (V) is a neighbourhood in D(T ). (ii) T is called open if whenever V is a neighbourhood in D(T ), the image T (V) is a neighbourhood of R(T ). (iii) If D(T ) = X and T is a continuous linear relation, then we shall say that T is bounded. ♦ The collection of bounded linear relations as defined above will be denoted by BR(X, Y ) and as useful we write BR(X) := BR(X, X) and the collection of bounded linear operators as defined above will be denoted by L(X, Y ) and as useful we write L(X) := L(X, X). Remark 2.5.3 (i) We note that T is continuous if and only if, T −1 is open. (ii) Let M be a subspace of a normed vector space X, then the injection X is continuous. ♦ operator JM Theorem 2.5.4 [50, Theorem 3.4.2] (Closed graph and open mapping theorem) Let T ∈ LR(X, Y ). Then, (i) Te is continuous if and only if, D(Te) is closed, where Te is given in (2.4). (ii) Te is open if and only if, R(Te) is closed. ♦ Theorem 2.5.5 Let T ∈ LR(X, Y ). Then, (i) T is continuous if and only if, kT k < ∞. (ii) If dim D(T ) < ∞, then T is continuous.

♦

Proof. (i) Suppose T is continuous. Since T (0) + BY is a neighbourhood of T (0), it follows that T −1 (T (0) + BY ) = T −1 (BR(T ) ) is a neighbourhood of 0. So, there is λ > 0 such that λBD(T ) ⊂ T −1 (BR(T ) ). Thus, λT (BD(T ) ) ⊂ BR(T ) + T (0) = T T −1 (BR(T ) ). Using Proposition 2.5.6 (i), we have kT k < ∞. Conversely, suppose kT k < ∞. Let x ∈ D(T ), y ∈ T x and let V be a nontrivial closed ball in R(T ) with center y. Then, V0 = V \{y} = αBR(T ) for some α > 0. Using Proposition 2.5.6 (i), there exists λ > 0 such that T (BD(T ) ) ⊂ λBR(T ) +T (0). It follows that BD(T ) +T −1 (0) ⊂ λT −1 (BR(T ) ) = α−1 λT −1 (V0 ) and λ−1 αBD(T ) + T −1 (0) ⊂ T −1 (V0 ) = T −1 (V − y). Hence, λ−1 αBD(T ) + T −1 y ⊂ T −1 (V ) − T −1 y + T −1 y = T −1 (V ). So, λ−1 αBD(T ) + T −1 y is a neighbourhood of x in D(T ). Let W be a neighbourhood of T x, let U ⊂ W

44

Spectral Theory of Multivalued Linear Operators

be an open set containing y ∈ T x and let V ⊂ U be a non-trivial closed ball with center y. Then, T −1 (W ) is neighbourhood of x. Hence, T is continuous. (ii) If dim D(T ) < ∞, then QT T is a continuous single valued linear operator, that is, kQT T k < ∞. On the other hand, kT k = kQT T k. Now, the result follows from (i). Q.E.D. Corollary 2.5.4 Let X and Y be normed spaces, and T ∈ LR(X, Y ). If T is continuous, then kT (x)k ≤ kT kkxk for all x ∈ D(T ). Proposition 2.5.7 Let X and Y be normed spaces, and T , S ∈ LR(X, Y ) such that D(T ) ⊂ D(S) and S(0) ⊂ T (0). Then, k(T − S)xk ≥ kT xk − kSxk, x ∈ D(T ).

(2.19)

Further, if T and S are bounded, then T − S is bounded, and kT − Sk ≥ kT k − kSk.

(2.20)

In addition, if either S is bounded and T is unbounded or T is bounded, S|D(T ) is unbounded, and S(0) = T (0), then T − S is unbounded. ♦ Proof. We can easily check that T (0) = (T − S)(0) since S(0) ⊂ T (0). So, by (i) of Lemma 2.5.7, we have for any x ∈ D(T ), and any given y1 ∈ T x and y2 ∈ Sx, k(T − S)xk

=

dist(y1 − y2 , (T − S)(0))

=

dist(y1 − y2 , T (0))

≥ dist(y1 , T (0)) − dist(y2 , T (0)) ≥ dist(y1 , T (0)) − dist(y2 , S(0)) = kT xk − kSxk,

(2.21)

which yields that (2.19) holds. Further, suppose that T and S are bounded. Then, by using Lemma 2.5.7 (ii) and Theorem 2.5.5, we have T −S is bounded. It follows from (2.21) that kT xk ≤ k(T − S)xk + kSxk, for all x ∈ D(T ) with kxk ≤ 1. Hence, by (ii) of Lemma 2.5.7, we have kT k ≤ k(T − S)k + kSk. Consequently, (2.20) holds. In addition, suppose that S is bounded and T is unbounded. It can be easily verified that T − S is unbounded by (2.21), Lemma 2.5.7 and Corollary 2.5.4. Finally, suppose that T is bounded, S|D(T ) is unbounded, and S(0) = T (0). It follows from (2.21) that k(T − S)xk

≥ dist(y2 , S(0)) − dist(y1 , T (0)) = kSxk − kT xk,

Fundamentals

45

which, together with Lemma 2.5.7, implies that T − S is unbounded. Q.E.D. Remark 2.5.4 (i) The conditions D(T ) ⊂ D(S) and S(0) ⊂ T (0), in Proposition 2.5.7, are necessary. In fact, let X = 6 {0}, T = IX , and G(S) = X × X. Then, for all x = 6 0, we have kT x − Sxk = 0 and kT xk − kSxk = kxk. So, kT xk − kSxk  kT x − Sxk. (ii) Let X and Y be normed spaces, and S, T ∈ LR(X, Y ) satisfy that 6 S(0), D(S) ⊂ D(T ). If S is bounded, T is unbounded, T (0) ⊂ S(0), and T (0) = then S − T may be bounded or unbounded. For example, let X = Y = l2 , and T (x) = (nxn )n≥1 , x = (xn )n≥1 ⊂ D(T ), where D(T ) = {x = (xn )n≥1 ⊂ l2 : (nxn )n≥1 ⊂ l2 }. Then, T is unbounded single valued. Let D(S1 ) = D(S2 ) = D(T ), and S1 (x) = l2 , S2 (x) = x + S2 (0), x ∈ D(T ), where S2 (0) = span{e1 } with e1 (1) = 1 and e1 (n) = 0 for n ≥ 2. It is evident that S1 (0) = l2 , S1 is bounded with bound kS1 k = 0, and S2 is bounded with 6 Si (0) for i = 1, 2. bound kS2 k = 1. In addition T (0) = {0} ⊂ Si (0) and T (0) = Further, we get that for any x ∈ D(T ), S1 (x) − T (x) = l2 , S2 (x) − T (x) = ((1 − n)xn )n≥1 + S2 (0), which implies that S1 − T is bounded with bound kS1 − T k = 0 and S2 − T is unbounded. ♦

2.5.4

Selections of Linear Relation

We define the linear relation TG associated with T by    TG ∈ LR(X, X × Y ), D(TG ) = D(T ), and   TG x = {(x, y) ∈ X × Y such that y ∈ T x} .

(2.22)

Remark 2.5.5 If T is continuous, then TG is continuous. In fact, kTG xk = inf{kxk + kyk such that y ∈ T x} = kxk + kT xk ≤ kxk + kT kkxk as Corollary 2.5.4 = (1 + kT k)kxk.

♦

46

Spectral Theory of Multivalued Linear Operators

Definition 2.5.6 Let T ∈ LR(X, Y ). A linear operator S is called a selection (or single valued part) of T if T = S + T − T and D(T ) = D(S). ♦ If S is a selection of T , then for all x ∈ D(T ), we have T x = Sx + T x − T x = Sx + T (0). So, G(T ) = G(S) + {0} × T (0). Example 2.5.2 Let X be denote the vector space C[a, b] of all continuous valued functions defined on some given compact interval [a, b] in R. We define T by T : X −→ ZX f −→

f (t)dt,

then T ∈ LR(X) and T (0) is the one dimensional subspace consisting of the constant functions on [a, b]. Clearly, for x ∈ [a, b], the function Z x Af = f (t)dt a

is a selection of T . Proposition 2.5.8 [50, Proposition 2.4.2] Let T ∈ LR(X, Y ). If T is continuous, and T (0) is the kernel of a continuous projection P defined in R(T ), then P T is a continuous selection of T . ♦ Theorem 2.5.6 Let T and S be two linear relations on a vector space X and let A and B be two selections of T and S, respectively. Then, (i) A + B is a selection of T + S. (ii) If R(T ) ⊂ D(S), then BA is a selection of ST . (iii) If T be an everywhere defined linear relation on X and A be a selection of T. Then, for all integer n ∈ N, An is a selection of T n . ♦ Proof. (i) Let A be a selection of T and B be a selection of S, then \ \ D(T + S) = D(T ) D(S) = D(A) D(B) = D(A + B). It is easy to prove that T + S = A + T − T + B + S − S = A + B + (T + S) − (T + S). (ii) Since R(T ) ⊂ D(S), then D(ST ) = D(T ). Since B is a selection of S, it follows that D(B) = D(S). Hence, R(A) ⊂ R(T ) ⊂ D(B) and then D(AB) = D(ST ). As B is a selection of S, it follows that BT x ⊂ ST x f or all x ∈ D(T ).

(2.23)

Fundamentals

47

As A is a selection of T , it follows that BAx ⊂ BT x f or all x ∈ D(T ).

(2.24)

Hence, the result follows from (2.23) in combination with (2.24). (iii) Since A is a selection of T , then A is single valued. Hence, An is an operator for all n ∈ N. It is enough to prove that, for all x ∈ X and n ∈ N, An x ⊂ T n x.

(2.25)

To do this, proceed by induction on n ∈ N. The case n = 1 is obvious (A is a selection of T ). Let n ≥ 2 and suppose that (2.25) holds true. A is a selection of T , then it is clear that AT n x ⊂ T n+1 x for all x ∈ X.

(2.26)

On the other hand, it follows from the induction hypothesis, (2.25), and (2.26), that An+1 x ⊂ AT n x. (2.27) Finally, combining (2.25) and (2.27), we get the result.

2.6

Q.E.D.

Relatively Boundedness of Linear Relations

Definition 2.6.1 Let X and Y be two normed spaces and S, T be linear relations from X into Y and from X into Z, respectively. The linear relation S is called relatively bounded with respect to T (or T -bounded) if D(S) ⊃ D(T ) and there exist constants a, b for which the inequality kSxk ≤ akxk + bkT xk

(2.28)

holds for all x ∈ D(T ). The infimum δ of all b such that the inequality (2.28) holds for some a ≥ 0 is called the relative bound of S with respect to T (or T -bound of S). ♦ Remark 2.6.1 Let S, T ∈ LR(X, Y ). Then, (i) If S is 0-bounded, then S is bounded.

48

Spectral Theory of Multivalued Linear Operators

(ii) The inequality (2.28) is equivalent to kSxk2 ≤ a21 kxk2 + b21 kT xk2 , x ∈ D(T ), 1

1

where a1 = (a2 + ab) 2 , and b1 = (b2 + ab) 2 . (iii) S is T -bounded if and only if, S is (λ − T )-bounded for some λ ∈ C. (iv) S is T -bounded if and only if, D(T ) ⊂ D(S), and SGT is bounded. (v) S is T -bounded with T -bound δ if and only if, QS S is QT T -bounded with QT T -bound δ. Indeed, kQT Sxk =

dist(T (0), Sx)

≤ dist(S(0), Sx) = kSxk ≤ akxk + bkT xk = akxk + bkQT T xk.

♦

Lemma 2.6.1 Let S, T ∈ LR(X, Y ) satisfy S(0) ⊂ T (0) and D(T ) ⊂ D(S). If S is T -compact, then S is T -bounded. ♦ Proof. Suppose that S is not T -bounded. Then, assume without loss of generality that for each positive integer n, there exists an xn ∈ D(T ) such that kxn k + kT xn k = 1, and kSxn k > n. This implies that (xn )n , and (T xn )n are bounded, and that S is T -compact. So, for all yn ∈ Sxn , we can extract a convergent subsequence of (yn )n , which is a contradiction. Q.E.D. Theorem 2.6.1 Let A, B, and S ∈ LR(X, Y ) verifying S(0) ⊂ B(0) ⊂ A(0) and λ ∈ C. If S is A-bounded with A-bound δ1 and B is A-bounded with Abound δ2 such that δ2 +|λ|δ1 < 1. Then, k·kA and k·kλS−(A+B) (see Definition 2.5.4) are equivalent. ♦ Proof. Since S is A-bounded with bound δ1 and B is A-bounded with bound δ2 , then there exist non-negative constants a, b, a1 and b1 such that, for x ∈ D(A), kSxk ≤ akxk + bkAxk and kBxk ≤ a1 kxk + b1 kAxk. So, −kBxk ≥ −a1 kxk − b1 kAxk. Thus, kAxk − kBxk ≥ −a1 kxk + (1 − b1 )kAxk.

(2.29)

By using Proposition 2.5.7 and (2.29), we get kAx + Bxk ≥ −a1 kxk + (1 − b1 )kAxk.

(2.30)

Fundamentals

49

On the other hand, kxkλS −(A+B)

= kxk + k(λS − (A + B))xk ≥

kxk + k(A + B)xk − |λ|kSxk, (since S(0) ⊂ (A + B)(0))

≥

kxk − a1 kxk + (1 − b1 )kAxk − |λ|kSxk, (as (2.30))

≥

kxk − a1 kxk + (1 − b1 )kAxk − |λ|akxk − b|λ|kAxk

≥ (1 − a1 − |λ|a)kxk + (1 − b1 − |λ|b)|kAxk ≥ min (1 − a1 − |λ|a, 1 − b1 − |λ|b) |(kxk + kAxk). Therefore, kxkλS −(A+B) ≥ min (1 − a1 − |λ|a, 1 − b1 − |λ|b) kxkA . Hence, we obtain kxkλS−(A+B)

= kxk + k(λS − (A + B))xk ≤

kxk + kAxk + kBxk + |λ|kSxk

≤

kxk + a1 kxk + b1 kAxk + kAxk + |λ|akxk + b|λ|kAxk

≤ (1 + a1 + |λ|a)kxk + (1 + b1 + |λ|b)|kAxk ≤ max(1 + a1 + |λ|a, 1 + b1 + |λ|b)|(kxk + kAxk). Therefore, kxkλS −(A+B) ≤ max(1 + a1 + |λ|a, 1 + b1 + |λ|b)kxkA , and we deduce that k · kA and k · kλS −(A+B) are equivalent.

Q.E.D.

Proposition 2.6.1 If S is T -bounded with T -bound δ < 1 and S(0) ⊂ T (0), δ then S is (T + S)-bounded with (T + S)-bound ≤ 1−δ . ♦ Proof. First of all, it should be mentioned that the linear relation T + S T is well-defined as D(T + S) = D(S) D(T ) = D(T ) with D(T + S) ⊂ D(S). Using the fact that S is T -bounded, it follows that there exist a > 0, δ ≤ b < 1, such that for all x ∈ D(T ), kSxk

≤ akxk + bkT xk = akxk + bkT x + Sx − Sxk (as Proposition 2.3.4) ≤ akxk + bkT x + Sxk + bkSxk (as Proposition 2.5.4).

Since b < 1, it follows that     a b kSxk ≤ kxk + k(T + S)xk, x ∈ D(T ). 1−b 1−b

(2.31)

50

Spectral Theory of Multivalued Linear Operators

This completes the proof.

Q.E.D.

Proposition 2.6.2 Let A, B, C and T ∈ LR(X, Y ). Then, (i) If A is B-bounded with B-bound δ1 , and B is C-bounded with C-bound δ2 , then A is C-bounded with C-bound δ1 δ2 . (ii) If B is T -bounded with T -bound δ1 , and C is T -bounded with T -bound δ2 , then A = B ± C is T -bounded with T -bound δ1 + δ2 . ♦ Proof. (i) Since A is B-bounded, and B is C-bounded, then there exist a, b, c, d ≥ 0, such that kAxk ≤ akxk + bkBxk for all x ∈ D(B), and kBxk ≤ ckxk + dkCxk for all x ∈ D(C). It follows that, for all x ∈ D(C), kAxk ≤ (a + bc)kxk + bdkCxk, and D(C) ⊂ D(A). (ii) Since B is T -bounded, and C is T -bounded, then there exist a, b, c, d ≥ 0, such that kBxk ≤ akxk + bkT xk for all x ∈ D(T ), and kCxk ≤ ckxk + dkT xk, for all x ∈ D(T ). It follows that, for all x ∈ D(T ), kAxk = k(B ± C)xk ≤ kBxk + kCxk ≤ (a + c)kxk + (b + d)kT xk, and D(T ) ⊂ D(A). This completes the proof. Q.E.D.

2.7

Closed and Closable Linear Relations

e Ye ), the question Since every linear relation T has a completion Te ∈ LR(X, arises as to why non closed linear relations in incomplete spaces need be considered at all. The answer is simple: The sum of two closed linear relations is, in general, not closed or closable, likewise for the composition. Furthermore, continuity for Te does not imply continuity for T, and the inverse of a closable linear relation (or operator) need not be closable.

2.7.1

Closed Linear Relations

Definition 2.7.1 Let T ∈ LR(X, Y ). The relation T is called closed if its graph G(T ) is closed. ♦ We denote the class of all closed linear relations from X into Y by CR(X, Y ) and as useful we write CR(X) := CR(X, X) and we denote by BCR(X, Y ) = {T ∈ CR(X, Y ) such that T is everywhere defined} and as useful we write BCR(X) := BCR(X, X).

Fundamentals

51

Proposition 2.7.1 [50, Proposition 2.5.3], [8, Lemma 5.3] (i) Let T ∈ LR(X, Y ). The following properties are equivalent (i1 ) T is closed, (i2 ) T −1 is closed, and (i3 ) QT T and T (0) are closed. In particular, N (T ) is closed, if T is closed. (ii) Assume that T ∈ CR(X, Y ). Then, R(T ) is closed if and only if, R(QT T ) is closed. ♦ Proposition 2.7.2 Let S, T ∈ LR(X, Y ) such that S(0) ⊂ T (0) and D(T ) ⊂ D(S). If T is closed and S is continuous, then T + S is closed. ♦ Proof. First case : We shall first assume that T and S are single valued. Let (xn )n ⊂ D(T + S) = D(T ) such that xn → x and (T + S)xn → y. Writing T xn = (T + S)xn − Sxn . Since S is continuous, then Sxn → Sx. Therefore, T xn → y − Sx. Thus, since T is closed, then T x = y − Sx and x ∈ D(T ). Hence, y = (S + T )x and x ∈ D(S + T ). Second case : If T and S are linear relations, then since S(0) ⊂ T (0), we have (T + S)(0) = T (0) + S(0) = T (0). Hence, (T + S)(0) is closed. On the other hand, QT +S (T + S) = QT (T + S) = QT T + QT S (2.32) and S(0) . ⊂ T (0) = T (0). So, by using Lemma 2.1.3, we infer that Y /S(0) R ≡ Y /T (0) and QT = QR QS , where R = T (0)/S(0). Thus, QT S = QR QS S. As S is continuous, then QT S is a continuous single valued. Using (2.32), we obtain QT +S (T + S) is closed. Again, applying Proposition 2.7.1, we infer that T + S is closed. Q.E.D. Remark 2.7.1 If the hypotheses of Proposition 2.7.2 is satisfied, then by using Proposition 2.5.4 (iii), we can prove that QT S is continuous. ♦ Theorem 2.7.1 Let T , S ∈ LR(X, Y ) such that T is closed and S is continuous with dim S(0) < +∞ and D(S) ⊃ D(T ). Then, T + S is closed. ♦ Proof. First, since D(QT +S ) = Y and by using Proposition 2.3.6 QT +S (T + S) = QT +S T + QT +S S. Moreover, since T is closed, then by using Proposition 2.7.1, T (0) is closed and QT T is closed. Since dim S(0) < ∞, then by using

52

Spectral Theory of Multivalued Linear Operators

Theorem 2.1.2, we have  dim

T (0) + S(0) T (0)

 < ∞,

and by using Theorem 2.4.1 (i), we obtain that Y /T (0)

QT +S T = Q T (0)+S(0) QYT T T (0)

is closed. On the other hand, QT +S S is a continuous single valued linear operator. In fact, since S(0) ⊂ (S + T )(0) and D(S + T ) ⊂ D(S) and by using Proposition 2.5.4 (iii), we have kQT +S Sk ≤ kSk. Finally, since D(QT +S S) = D(S) ⊃ D(QT +S T ) = D(T ), then by using Proposition 2.7.2, we obtain that QT +S (T + S) = QT +S T + QT +S S is closed. It is obvious that (T + S)(0) = T (0) + S(0) is closed. Therefore, by Proposition 2.7.1 (i), T + S is closed. This completes the proof. Q.E.D. Theorem 2.7.2 [50, Theorem 3.5.3 and Corollary 3.5.4] (i) If T ∈ CR(X, Y ) and D(T ) = X, then T is bounded. (ii) If T ∈ CR(X, Y ) and open, then R(T ) is closed. (iii) If T ∈ CR(X, Y ) and R(T ) is closed, then T is open. (iv) If T is continuous, D(T ) and T (0) are closed, then T is closed. ♦ Lemma 2.7.1 Let T ∈ L(X, Y ) and S ∈ CR(Y, Z). Then, ST ∈ CR(X, Z). ♦ Proof. Let (xn , yn ) ∈ G(ST ) such that (xn , yn ) converges to (x, y). Then, yn ∈ ST xn = S(T xn ). This implies that (T xn , yn ) ∈ G(S). Since T is an bounded operator, then (T xn , yn ) converges to (T x, y). The fact that S is closed, then (T x, y) ∈ G(S). So, y ∈ S(T x) = ST x. Thus, (x, y) ∈ G(ST ). This completes the proof. Q.E.D. Remark 2.7.2 If X and Y are two complete spaces and T ∈ CR(X, Y ), then the space XT is also a Banach space and T (0) is complete. Since GT is a bounded operator, then it follows from Lemma 2.3.4 that T GT ∈ CR(XT , Y ). ♦ Proposition 2.7.3 [50, Proposition 6.5.2] Let X be complete space, and let S, T ∈ C R(X) be bijective. Then, ST has the same properties. ♦

Fundamentals

53

Proposition 2.7.4 Let X and Y be two Banach spaces. Then, (i) If T ∈ LR(X, Y ) has a continuous selection, T (0) and D(T ) are closed, then T ∈ CR(X, Y ). (ii) T ∈ CR(X, Y ), then T has a closed selection. ♦ Proof. (i) Let S be a continuous selection of T . Then, kT k = kS + T − T k ≤ kSk + kT − T k = kSk < ∞. Hence, by using Theorem 2.5.5 (i) we have T is continuous. Since D(T ) and T (0) are closed, then by using Theorem 2.7.2 (iv), we have T ∈ CR(X, Y ). (ii) If S is selection of T , then G(T ) = G(S) + ({0} × T (0)) .

(2.33)

It follows from (2.33) that TG x = SG x + TG(0) = SG x + {0} × T (0) for all x ∈ D(T ), where TG (respectively, SG ) is the linear relation associated with T (respectively, S). Let P be any bounded projection defined on G(T ) with T kernel {0} × T (0) such that R(SG ) ({0} × T (0)) = {0}. Then, for all x ∈ D(T ), we have P TG x

= P (SG x + TG (0)) = P (SG x) + P (TG (0)) = P SG x,


since P TG (0) = P P −1 (0) = 0

= P SG x + SG x − SG x = SG x + (P − I)SG x. Since P (P − I)SG x = P SG x − P SG x = 0, then \ (P − I)SG x ∈ R(SG ) ({0} × T (0)) = {0}. Consequently, P TG x = SG x. By using Remark 2.7.2, we have (D(T ), k · kT ) is complete. Also, kxkS = kSG xk = kP TG xk ≤ kP kkTG xk = kP kkxkT .

(2.34)

We know that D(T ) = D(S), then (D(S), k · kT ) is complete. This, together with (2.34) that (D(S), k · kS ) is complete. Since S is single valued, then S is closed. This completes the proof. Q.E.D.

54

2.7.2

Spectral Theory of Multivalued Linear Operators

Closable Linear Relations

The closure of a linear relation T , denoted T , be defined in terms of their corresponding graphs in X × Y as follows G(T ) = G(T ) ⊂ X × Y. Definition 2.7.2 The linear relation T is said to be closable if T is an extension of T i.e., T x = T x for all x ∈ D(T ). ♦ Remark 2.7.3 It is easy to construct natural examples of non closable linear relations. Here is an example suggested by V. V. Shevchik (a discontinuous linear functional on a normed space). Let X be the space of continuous functions C[0, 1] with the norm 1

Z kxk =

|x(t)|dt. 0

Fix t ∈ [0, 1] and define ft (x) = x(t). Then, ft (·) is an unbounded linear functional on X. ♦ Proposition 2.7.5 [50, Proposition 2.5.1] Let T ∈ LR(X, Y ), then QT T = QT T . ♦ Lemma 2.7.2 [50, Exercise 5.19] Let T ∈ LR(X, Y ). Then, T (0) ⊂ T (0).♦ Lemma 2.7.3 [50, Definition 2.5.7] Let T ∈ LR(X, Y ). Then, T is closable if and only if, T (0) = T (0) if and only if, QT T is closable and T (0) is closed. In particular, if T is continuous and T (0) is closed, then T is closable. ♦ Remark 2.7.4 The condition T (0) closed, in Lemma 2.7.3, is necessary. Indeed, let X, Y be two Banach spaces and F be a finite dimensional vector subspace of X. Let T ∈ LR(X, Y ) such that T (0) = 6 T (0). Assume that T0 = T |F . On the one hand, since D(T0 ) = D(T ) ∩ F ⊂ F , then we have dim(D(T0 )) < +∞. Hence, by using Theorem 2.5.5 (ii) T0 is a continuous linear relation. On the other hand, since T x = T0 x for all x ∈ D(T ) ∩ F and 0 ∈ D(T )∩F , then we deduce that T0 (0) = T (0). Hence, T0 (0) = T (0) 6= T (0). This implies that T0 (0) 6= T0 (0). Therefore, T0 is not closable. ♦ Proposition 2.7.6 (T )−1 = T −1 .

[50, Definition 2.5.2(3)] Let T ∈ LR(X, Y ). Then, ♦

Fundamentals

55

Lemma 2.7.4 Let E be a subspace of X. Then, IE = IE .

♦

Proof. It suffices to show that G(IE ) = G(IE ). Let (x, y) ∈ G(IE ), then there exists a sequence (xn , xn ) ∈ G(IE ) such that (xn , xn ) → (x, y). Hence, x = y ∈ E. Therefore, (x, y) ∈ G(IE ). Conversely, if (x, x) ∈ G(IE ), i.e., x ∈ E, then there exists a sequence (xn )n ∈ E converging to x. Hence, the sequence (xn , xn ) ∈ G(IE ) converges to (x, x). Therefore, (x, x) ∈ G(IE ). Q.E.D. Lemma 2.7.5 [67, Theorem 4.1.1 p. 190] Let T and A be operators from X into Y , and let A be T -bounded with T -bound smaller than 1. Then, S = T +A is closable if and only if, T is closable. In this case, the closure of T and S have the same domain. In particular S is closed if and only if, T is also closed. ♦

2.8

Adjoint of Linear Relations

Let X be a normed vector space. We will denote by X ∗ the norm dual of X, i.e., the space of all continuous functionals x0 defined on X, with norm kx0 k = inf {λ ∈ K such that |x0 (x)| ≤ λkxk for all x ∈ X} . A consequence of Hann-Banach theorem is the following: Theorem 2.8.1 Let X be a normed vector space. Then, for all x ∈ X and 6 0, there exists x0 ∈ X ∗ (the adjoint of X) such that kx0 k = 1 and x = ♦ x0 (x) = kxk. If M and N are subspaces of X, then M ⊥ = {x0 ∈ X ∗ such that x0 (x) = 0 for all x ∈ M } and N > = {x ∈ X such that x0 (x) = 0 for all x0 ∈ N } , where X ∗ is the dual of X. The adjoint T ∗ of T is defined by G(T ∗ ) = G(−T −1 )⊥ ⊂ X ∗ × Y ∗ ,

56

Spectral Theory of Multivalued Linear Operators

where h(y, x), (y 0 , x0 )i = hx, x0 i + hy, y 0 i = x0 (x) + y 0 (y).

(2.35)

Thus, means that (y 0 , x0 ) ∈ G(T ∗ ) if and only if, y 0 (y) − x0 (x) = 0 for all (x, y) ∈ G(T ). By using (2.35), we have y 0 (y) = x0 (x) for all y ∈ T x, x ∈ D(T ). Hence, x0 ∈ T ∗ y 0 if and only if, y 0 (T x) = x0 (x) for all x ∈ D(T ). So, we can characterize the adjoint as follows G(T ∗ ) = {(y 0 , x0 ) ∈ Y ∗ × X ∗ such that x0 is an extension of y 0 T } . Remark 2.8.1 Note that this definition coincides with the classical one when T is a densely defined operator and it allows one to define the conjugate of a non-densely defined operator. ♦ In the following we list some basis properties of the adjoint of a linear relation. Proposition 2.8.1 [50, Chapter 3] Let T ∈ LR(X, Y ). Then, (i) T ∗ is a closed linear relation in LR(Y ∗ , X ∗ ). (ii) G(T ∗ + S ∗ ) ⊂ G((T + S)∗ ). (iii) If S is continuous with D(T ) ⊂ D(S), then (T + S)∗ = T ∗ + S ∗ . (iv) N (T ∗ ) = R(T )⊥ and T ∗ (0) = D(T )⊥ . (v) If T is closed, then R(T ) is closed if and only if, R(T ∗ ) is closed if and only if, T is open if and only if, T ∗ is open. (vi) If T is continuous, then kT k = kT ∗ k. (vii) D(T ∗ )>⊥ = D(T ∗ ). ♦ Furthermore, it follows from Proposition 2.8.1 (iv) that the adjoint of a linear operator is single valued if and only if, it is densely defined. Lemma 2.8.1 Let T , S ∈ LR(X, Y ), then ∗

Y (QT +S S)∗ = S ∗ J(T . (0)+S(0))⊥

♦

Proof. First, we have (QYT +S )∗

∗

=



QYT (0)+S(0)

=



 Y∗ J(T . ⊥ (0)+S(0))

Moreover, by using Proposition 2.8.1, we have D((QYT +S )∗ )

⊥

= T (0) + S(0) =

Y (T (0) + S(0))

!∗ .

Fundamentals

57

Since S ∈ LR(X, Y ) and QT +S ∈ LR(Y, Y /(T (0) + S(0))) with D(QT +S ) = Y , then ∗

Y (QT +S S)∗ = S ∗ J(T . (0)+S(0))⊥

Q.E.D.

Lemma 2.8.2 Let H be a Hilbert space with inner product h·, ·i and T ∈ LR(H). If T is bounded, then for all n ∈ N∗ , (T n )∗ = (T ∗ )n .

♦

Proof. Since D(T ) = H, then D(T 2 ) = H. Hence (T 2 )∗ is a single valued linear operator. Let y ∈ H, then h(T 2 )∗ x, yi = hx, ti for all t ∈ T 2 (y). Let t ∈ T 2 (y). Then, there exists z such that (y, z) ∈ G(T ) and (z, t) ∈ G(T ). Which implies that hx, ti = hT ∗ x, zi = hT ∗ (T ∗ x), yi. This follows that (T 2 )∗ = (T ∗ )2 . Proceeding by induction, we can prove the result for each n ≥ 3. Q.E.D. Lemma 2.8.3 Let T ∈ LR(X) and S be a continuous linear relation such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). Then, S ∗ is continuous, S ∗ (0) ⊂ T ∗ (0), and D(S ∗ ) ⊃ D(T ∗ ). ♦ Proof. Since S is continuous, then by using Proposition 2.8.1 (vi), we have kS ∗ k = kSk. Hence, by using Theorem 2.5.5 (i), we have S ∗ is continuous. Also, ⊥ D(S ∗ ) = S(0)⊥ ⊃ T (0) ⊃ T (0)⊥ = D(T ∗ )>⊥ = D(T ∗ ), and using Proposition 2.8.1 (iv), S ∗ (0) = D(S)⊥ ⊂ D(T )⊥ = T ∗ (0). Q.E.D.

2.9

Minimum Modulus of Linear Relations

Definition 2.9.1 Let T ∈ LR(X, Y ). The minimum modulus of a linear relation T is the quantity γ(T ) γ(T ) = sup {λ such that kT xk ≥ λ dist(x, N (T )) for x ∈ D(T )} .

♦

58

Spectral Theory of Multivalued Linear Operators

Proposition 2.9.1 [50, Proposition 2.2.2] Let T ∈ LR(X, Y ). The minimum modulus of T is the quantity   ∞ if D(T ) ⊂ N (T )  γ(T ) = kT xk  inf such that x ∈ D(T )\N (T ) otherwise. ♦ dist(x, N (T )) Proposition 2.9.2 [50, Corollary 2.3.9] If T is open and N (T ) is closed, then N (T ) = N (QT T ) and γ(T ) = γ(QT T ). ♦ Theorem 2.9.1 [50, Theorem 2.3.11] Let T LR(Y, Z). Then,

∈

LR(X, Y ) and S

∈

(i) γ(ST ) ≥ γ(S|R(T ) )γ(T ) (∞.0 excluded), with γ(ST ) = ∞ whenever γ(T ) = ∞ (even γ(S|R(T ) ) = 0). (ii) If S −1 (0) ⊂ R(T ), then γ(ST ) ≥ γ(S)γ(T ). ♦ Theorem 2.9.2 Let S, T ∈ LR(X, Y ) such that kT k = 6 0, D(S) = D(T ), and N (S) ⊂ N (T ). Then, γ(SGT ) ≤

γ(T )kS k . (1 + γ(T ))kT k

Moreover, if N (S) = N (T ), then γ(SGT ) =

γ(T )kSk . (1 + γ(T ))kT k

♦

Proof. We have N (SGT )

= {x ∈ XT such that SGT x = S(0)} = {x ∈ D(T ) such that Sx = S(0)} = {x ∈ D(S) such that Sx = S(0)} as D(S) = D(T ) = N (S).

(2.36)

Since dist(x, N (SGT ))

=

inf z∈N (S GT )

=

inf

kx − z kT

kx − zk + kT (x − z)k see (2.36)

z∈N (S)

=

inf

kx − zk + kT xk as N (S) ⊂ N (T ),

z∈N (S)

then dist(x, N (SGT )) = dist(x, N (S)) + kT xk.

Fundamentals

59

Hence, 

γ(SGT )

= = ≤ = =

 kSGT xk such that x ∈ XT \N (SGT )  dist(x, N (SGT ))  kSxk inf such that x ∈ D(T )\N (T )  dist(x, N (S)) + kT xk  kSxk inf such that x ∈ D(T )\N (T ) dist(x, N (T )) + kT xk −1   dist(x, N (T )) + kT xk sup such that x ∈ D(T )\N (T ) kS xk  −1   dist(x,N (T )) + 1  kT xk sup  . such that x ∈ D(T )\N (T ) kSxk   inf

kT xk

So, −1 (γ(T )−1 + 1)kT k γ(SGT ) ≤ kSk  −1 (γ(T ) + 1)kT k = γ (T )kSk γ(T )kS k = . (1 + γ(T ))kT k 

If N (T ) = N (S), we have the equalities. This completes the proof. Corollary 2.9.1 Let T ∈ LR(X, Y ), then γ(T GT ) =

Q.E.D.

γ(T ) . (1 + γ(T ))

♦

Proposition 2.9.3 [50, Proposition 2.3.2] Let T ∈ LR(X, Y ). Then, (i) T is open if and only if, γ(T ) > 0. (ii) T is open if and only if, R(T ∗ ) = N (T )⊥ . (iii) If T is open, then γ(T ) = γ(T ∗ ).

♦

Proposition 2.9.4 [50, Proposition 2.5.17] Let S ∈ CR(X, Y ) and T ∈ LR(Y, Z) have a closed range and satisfy α(T ) < ∞ and γ(T ) > 0, then T S ∈ CR(X, Z). ♦ Corollary 2.9.2 If S ∈ LR(X, Y ) and T ∈ LR(Y, Z) are closed with α(T ) < ∞ and R(T ) is closed, then T S ∈ CR(X, Z). ♦ Proof. By using Theorem 2.5.4 (ii) and Proposition 2.9.3 (i), we have T is open. Thus, the result follows immediately from Proposition 2.9.4. Q.E.D. Lemma 2.9.1 [50, Theorem 2.2.5] Let T kT −1 k−1 .

∈ LR(X, Y ), then γ(T ) = ♦

60

Spectral Theory of Multivalued Linear Operators

Theorem 2.9.3 [50, Theorem 3.7.4] Let γ(T ) > 0 and S satisfy S(0) ⊂ T (0), D(S) ⊃ D(T ) and kSk < γ(T ). Then, (i) α(T + S) ≤ α(T ). (ii) β(T + S) ≤ β(T ).

♦

Lemma 2.9.2 Let T , S ∈ LR(X) such that T is injective, open, S(0) ⊂ T (0) and D(T ) ⊂ D(S). Then, γ(T − S) ≥ γ(T ) − kSk. In addition, if kSk < γ(T ), then T − S is open and injective. ♦ Proof. Let x ∈ D(T − S), then by using Proposition 2.5.7, we have k(T − S)xk

≥ kT xk − kSxk ≥ γ(T )kxk − kSkkxk ≥ (γ(T ) − kSk)kxk.

Therefore, γ(T − S) ≥ γ(T ) − kSk. Since kSk < γ(T ), then γ(T − S) > 0. Hence, by Proposition 2.9.3 (i), we have T − S is open. In view of Theorem 2.9.3 (i), we have T − S is injective. Q.E.D. Lemma 2.9.3 [50, Corollary 3.7.7] Let T ∈ LR(X) be open and injective with dense range. Then, for any relation S such that S(0) ⊂ T (0), D(S) ⊃ D(T ) and kSk < γ(T ), we have T + S is open injective with dense range. ♦ Remark 2.9.1 Let T ∈ LR(X) be open and injective with dense range. Then, for any relation S such that S(0) ⊂ T (0), D(S) ⊃ D(T ) and kSk < γ(T ), and by using Lemmas 2.9.2 and 2.9.3, we have T ± S is open and injective . ♦ Proposition 2.9.5 [50, Proposition 2.6.3] Let X, Y be two normed vector spaces and let T ∈ LR(X, Y ) such that N (T ) is topologically complemented in D(T ) (or in X). Then, 0 ≤ kP k−1 γ(T ) ≤ γ(T|R(P ) ) ≤ γ(T ), where P is any continuous projection defined on D(T ) (or on X) with kernel N (T ). ♦ Remark 2.9.2 Let X, Y be two normed vector spaces and let T ∈ LR(X, Y ) such that N (T ) is topologically complemented in D(T ) (or in X). If T is open, then 0 < kP k−1 ≤ 1, where P is any continuous projection defined on D(T ) (or on X) with kernel N (T ). ♦

Fundamentals

61

Proposition 2.9.6 Let X be a Banach space and let T , S ∈ LR(X) such that γ(T ) > 0, S(0) ⊂ T (0) and α(T ) < ∞. Then, γ(T − S) ≥ kP k−1 γ(T ) − kSk, where P is any continuous projection defined on D(T ) (or on X) with kernel N (T ). ♦ Proof. Since α(T ) < ∞, then N (T ) is topologically complemented in D(T ). Using the fact that γ(T ) > 0 and Proposition 2.9.5, we obtain 0 < kP k−1 γ(T ) ≤ γ(T|R(P ) ) ≤ γ(T ).

(2.37)

Let Tb : R(P ) −→ R(T ) be the bijection associated with T . It follows from (2.37) that 0 < kP k−1 γ(T ) ≤ γ(Tb) ≤ γ(T ). (2.38) Consequently, γ(Tb) > 0 which implies that Tb is open. Using the fact that Tb is injective, Tb(0) = T (0) ⊃ S(0) and also Lemma 2.9.2, we infer that γ(Tb − S) ≥ γ(Tb) − kSk.

(2.39)

By virtue of (2.38) and (2.39), we deduce that γ(Tb − S) ≥ kP k−1 γ(T ) − kSk.

(2.40)

Now, we propose to show that γ(T − S) ≥ γ(Tb − S). Let x ∈ N (Tb − S). This implies that Tbx − Sx = Tb(0) − S(0) = T (0) − S(0). (2.41) T Since x ∈ N (Tb − S) ⊂ D(Tb − S ) = D(Tb) D(S) ⊂ R(P ), then Tbx = T x. It follows from (2.41) that T x − Sx = T (0) − S(0). Therefore, we deduce that N (Tb − S) ⊂ N (T − S). This leads to dist(x, N (T − S)) ≤ dist(x, N (Tb − S)). For x ∈ D(Tb − S), by applying (2.42), we have γ(Tb − S) dist(x, N (T − S)) ≤ γ(Tb − S) dist(x, N (Tb − S)) ≤ k(Tb − S)xk ≤

k(T − S)xk.

(2.42)

62

Spectral Theory of Multivalued Linear Operators

This leads to γ(T − S) ≥ γ(Tb − S).

(2.43)

Hence, by (2.40) and (2.43), we conclude that γ(T − S) ≥ kP k−1 γ(T ) − kSk.

2.10 2.10.1

Q.E.D.

Quantities for Linear Relations A Formula for Gap Between Multivalued Linear Operators

The gap between two linear subspaces M and N of a normed space X is defined by the following formula δ(M, N ) =

sup

dist(x, N ),

(2.44)

x∈M, kxk=1

in the case where M 6= {0}. Otherwise we define δ({0}, N ) = 0 for any subspace N . Moreover, δ(M, {0}) = 1 if M 6= {0}, as shown from (2.44). We can also define δb(M, N ) = max {δ(M, N ), δ(N, M )} . Sometimes, the latter is called the symmetric or maximal gap between M and N in order to distinguish it from the former. The gap δ(M, N ) can be characterized as the smallest number δ such that dist(x, N ) ≤ δkxk, for all x ∈ M. The notion of a gap between linear subspaces and linear operators was introduced by M. G. Krein and M. A. Krasnoselski in Ref. [69] in the 1940s. Among the works in this direction we can state, for example (see Refs. [52, 67]). Remark 2.10.1 Let us assume that M and N be two vectorial subspaces of a Banach space X. Then, (i) Keeping in mind that dist(cx, N ) = |c|dist(x, N ) for any non-zero element c in R. Then, we infer that sup dist(x, N ) = |c|δ(M, N ).

(2.45)

x∈M kxk≤|c|

b b , N ). N ) = δ(M (ii) δ(M, N ) = δ(M , N ) and δ(M, (iii) δ(M, N ) = 0 if and only if, M ⊂ N . (iv) δb(M, N ) = 0 if and only if, M = N . (v) 0 ≤ δ(M, N ) ≤ 1 and 0 ≤ δb(M, N ) ≤ 1.

♦

Fundamentals

63

Definition 2.10.1 Let T, S ∈ CR(X, Y ). The gap between T and S is defined by   sup dist(ϕ, G(S)) if G(T ) 6= {0}  ϕ∈G(T ) δ(T, S) = kϕk=1   0 otherwise, and δb(T, S) = max {δ(T, S), δ(S, T )} .

♦

Theorem 2.10.1 Let T , S ∈ CR(X, Y ), then δ(T, S) = δ(T −1 , S −1 ) and δb(T, S) = δb(T −1 , S −1 ). ♦ Proof. Fix any ϕ = (x, y) ∈ G(T ) with kϕk = 1, then ϕ−1 = (y, x) ∈ G(T −1 ) and kϕ−1 k = kϕk = 1. Hence, δ(T, S)

=

sup dist(ϕ, G(S)) ϕ∈G(T ) kϕk=1

=

sup

dist(ϕ, G(S)).

ϕ−1 ∈G(T −1 ) kϕ−1 k=1

Let ψ = (u, v) ∈ G(S), then ψ −1 = (v, u) ∈ G(S −1 ). Therefore, kϕ − ψk22

= k(x, y) − (u, v)k2 = kx − uk2 + ky − vk2 = k(y, x) − (v, u)k2 = kϕ−1 − ψ −1 k22 .

Which implies that dist(ϕ, G(S)) = dist(ϕ−1 , G(S −1 )). As a result, δ(T, S) = δ(T −1 , S −1 ). Similarly, we can prove δ(S, T ) = δ(S −1 , T −1 ).

Q.E.D.

Remark 2.10.2 In the previous theorem (Theorem 2.10.1), we notice that the gap is invariant with respect to inversion. ♦

2.10.2

Measures of Noncompactness

Definition 2.10.2 The diameter of a set B is the number ( sup {dist(x, y) such that x ∈ B and y ∈ B} diam (B) := 0

6 ∅ if B = otherwise. ♦

64

Spectral Theory of Multivalued Linear Operators

Remark 2.10.3 (i) (B) = 0 if and only if, B is an empty set or consists of exactly one point. (ii) The following properties are satisfied in complete space and are an immediate consequence of the Definition 2.10.2 (ii1 ) If B1 ⊂ B2 , then (B1 ) ≤ (B2 ). (ii2 ) (B) = (B), where B is the closure of B. (ii3 ) Cantor’s intersection theorem: If (Bn )n is a decreasing sequence of nonempty, closed and bounded subsets of X and limn→+∞ (Bn ) = 0, then the intersection B∞ of all Bn is nonempty and consists of exactly one point. Moreover, if X is a Banach space, then (iv) (λB) = |λ| (B) for any real number λ. (v) (x + B) = (B) for any x ∈ X. (vi) (B1 + B2 ) ≤ (B1 ) + (B2 ). (vii) (co (B)) = (B), where co (B) is the convex hull of B. ♦ Definition 2.10.3 Let D be a bounded subset of X. The Kuratowski measure of noncompactness is defined by δ(D) = inf{d > 0 : D can be covered by a finite number of sets of diameter ≤ d}. ♦ It is obvious that 0 ≤ δ(B) ≤ (B) < +∞ for every nonempty bounded subset B of X. In what follows, we use the following known properties of δ(·). Proposition 2.10.1 [83, Proposition 1] For arbitrary bounded sets D, Q, Dn ⊂ X and λ ∈ C, we have (i) δ(D) = 0 if and only if, D is compact, where D is the closure of D. (ii) δ(D) = δ(D), δ(λD) = |λ| δ(D) and if D ⊂ Q, then δ(D) ≤ δ(Q). (iii) If Dn = Dn , Dn+1 ⊂ Dn , and lim δ(Dn ) = 0, then n→+∞

D∞ :=

+∞ \

D n 6= ∅

n=1

and δ(D∞ ) = 0. S

(iv) δ(D Q) = max{δ(D), δ(Q)}, δ(D + Q) ≤ δ(D) + δ(Q). (v) δ(D) = δ(co (D)), where co (D) is the convex closure of D. (vi) δ(Nε (D)) ≤ δ(D) + 2ε, where Nε (D) = {x ∈ X : dist(x, D) < ε} .

♦

Definition 2.10.4 Let T ∈ LR(X) and δ1 , δ2 be two Kuratowski measure of noncompactness. Then,

Fundamentals

65

(i) T is said to be D-condensing if, for any bounded subset B of D(T ), QT T (B) is a bounded subset of Y /T (0) and δ2 (QT T B) < δ1 (QD B) whenever δ1 (QD B) > 0. (ii) If D = {0}, then T is said to be {0}-condensing linear relation or simply condensing. ♦ Definition 2.10.5 Let X and Y be two normed spaces, and I(X), C(X), and P(X) denote, respectively, the infinite dimensional, finite codimensional, and closed finite codimensional subspaces of a normed linear space X. We define the quantities (also called measures of compactness) Γ(T ), Γ0 (T ), Γ0 (T ), and e T ) as follows: ∆( First case : If dim(D(T )) < ∞ and dim(Y ) < ∞, then all quantities are zero. Thus, e ) = Γ0 (T ) = Γ0 (T ) = 0. Γ(T ) = ∆(T Second case : If dim(D(T )) = ∞ and dim(Y ) = ∞, then Γ(T )

=

Γ0 (T )

=

Γ0 (T )

=

e T) ∆(

=

inf

kT|M k

inf

kT|M k

inf

kT|M k

sup

Γ(T|M ).

M ∈I(D (T )) M ∈C(D(T ))

M ∈P(D(T ))

M ∈I(D(T ))

♦

The following inequalities hold e T ) ≤ Γ0 (T ) ≤ Γ0 (T ). Γ(T ) ≤ ∆(

(2.46)

Proposition 2.10.2 (i) [50, Exercise 4.1.5] If T is continuous, then Γ0 (T ) = Γ0 (T ). e , Γ, Γ0 }, we (ii) [50, Proposition 4.3.4] Let T ∈ LR(X, Y ). For each f ∈ {∆ have f (T ) , f (T GT ) = 1 + f (T ) where

∞ ∞

:= 1.

♦

66

Spectral Theory of Multivalued Linear Operators

Proposition 2.10.3 Let S, T ∈ LR(X, Y ) such that D(S) = D(T ). For each e , Γ, Γ0 }, we have f ∈ {∆ f (S) ≤ f (SGT ), 1 + f (T ) ∞ ∞

where

♦

:= 1.

Proof. We may suppose that dim D(T ) = ∞. We have Γ(SGT )

= =

inf

M ∈I(D(SGT ))

inf

M ∈I(XT )

kSGT |M k

kSGT |M k as D(T ) = D(S).

Hence, Γ(SGT )

kSGT mk kmkT kSGT mk = inf sup M ∈I(XT ) m∈M kmk + kT mk kSGT mk = inf sup M ∈I(XT ) m∈M kGT mk + kT GT mk  kG mk kT GT mk −1 T = inf sup + . kSGT mk M ∈I(XT ) m∈M kSGT mk =

inf

sup

M ∈I(XT ) m∈M

Since D(T ) = D(S), then GT m = GS m for all m ∈ M . This implies that kGT mk = kGS mk. Thus,   −1 kGT mk kT GT mk kGS mk Γ(SGT ) = inf sup + × kSGT mk kSGT mk kGT mk M ∈I(XT ) m∈M   −1 kGT mk kT GT mk kGS mk = inf sup + × kSGT mk kGT mk kSGT mk M ∈I(XT ) m∈M  −1  kGT mk kT GT mk kGS mk inf = inf + inf × m∈M kGT mk kSGT mk M ∈I(XT ) m∈M kSGT mk   −1 1 kT GT mk 1 = inf + inf × . m∈M kGT mk kS|GS M k M ∈I(XT ) kS|GT M k kT GT mk ≤ kT |GT M k, then kGT mk   −1 kS|GS M k 1 kT G T mk 1 + inf × ≥ . m∈M kGT mk kS|GT M k kS|GS M k 1 + kT|GT M k

Since for all M ∈ I (XT ), we have inf

m∈M

This implies that

Fundamentals

inf

M ∈I(XT )

67

kS|GS M k ≤ Γ(SGT ). 1 + kT|GT M k

The fact that D(T ) = D(S) implies that Γ(S) =

inf

M ∈I(XT )

kS|GS M k. This

implies that Γ(S) ≤ Γ(SGT ). 1 + Γ(T ) Similarly, for Γ0 (S) ≤ Γ0 (SGT ). 1 + Γ0 (T ) Next e S) ∆( e SGT ) = sup Γ(S|GT M ) . ≤ ∆( e M 1 + Γ(S|GT M ) 1 + ∆(T )

2.11

Q.E.D.

Precompact and Compact Linear Relations

Let X and Y be two normed spaces and T ∈ LR(X, Y ). We say that T is precompact (respectively compact), if the analogous property holds for QT T , i.e., T is precompact (respectively compact) if QT T (BX ) (respectively QT T (BX )) is totally bounded (respectively a compact subset) in Y . A relation S ∈ LR(X, Y ) is said to be T -compact if D(T ) ⊂ D(S), and SGT is compact. S is called T -precompact if D(T ) ⊂ D(S), and SGT is precompact. The families of all precompact and compact linear relations will be denoted by PRK(X, Y ) and KR(X, Y ), respectively. If X = Y , we write PRK(X) := PRK(X, X) and KR(X) := KR(X, X). We denote by K(X) the class of compact operators on X. Note that if T ∈ PRK(X, Y ) and Y is complete, then T ∈ KR(X, Y ). Proposition 2.11.1 [50, Theorem 5.2.2] T is precompact if and only if, Γ0 (T ) = 0. ♦ Proposition 2.11.2 [50, Proposition 5.5.3] Let T be a continuous linear relation. Then, T is precompact if and only if, T ∗ is compact. ♦ Proposition 2.11.3 Let A, B and S ∈ LR(X) and λ ∈ C. Then,

68

Spectral Theory of Multivalued Linear Operators

(i) R((λS − A)GB ) = R(λS − A). (ii) N ((λS − A)GB ) = N (λS − A). (iii) i((λS − A)GB ) = i(λS − A). (iv) If X is complete space and B is A-precompact, then i(λS − A) = i(λS − (A + B)).

♦

Proof. (i) Using the fact that GB x = (GB )−1 x = x, R(A) = A(D(A)) and D(AB) = B −1 (D(A)), we have R((λS − A)GB )

=

(λS − A)GB (D((λS − A)GB ))

=

(λS − A)(D((λS − A)GB ))

=

(λS − A)GB (D(λS − A))

=

(λS − A)(D(λS − A))

= R(λS − A). (ii) Since N ((λS − A)GB ) = {x ∈ D((λS − A)GB ) : (λS − A)GB (x) = (λS − A)GB (0)} = {x ∈ D(λS − A) : (λS − A)(x) = (λS − A)(0)} , then N ((λS − A)GB ) = N (λS − A). (iii) The assertion (iii) is immediately deduced from (i) and (ii). (iv) Since BGA is precompact, and X is complete, then BGA is compact. So, by using (iii), we have i(λS − A)

= i((λS − A)GA ), = i((λS − A)GA + BGA ) = i((λS − (A + B))GA ) = i(λS − (A + B)).

This completes the proof.

2.12

Q.E.D.

Strictly Singular Linear Relations

Definition 2.12.1 Let T ∈ LR(X, Y ). The linear relation T is called strictly singular if there is no infinite-dimensional subspace M of D(T ) for which T|M is injective and open. ♦

Fundamentals

69

We will denote by SSR(X, Y ) the set of strictly singular linear relations. If X = Y , then SSR(X) := SSR(X, X). The following proposition gives a characterization of the strictly singular multivalued linear operators by means of the compactness measure. Proposition 2.12.1 [50, Theoreom 5. 2.6] Let X and Y be two normed spaces and let T ∈ LR(X, Y ). Then, the linear relation T is strictly singular e T ) = 0. if and only if, ∆( ♦ Lemma 2.12.1 [50, Corollary 5.2.3 and 5.2.8] Let T ∈ LR(X, Y ) be precompact, then T is continuous strictly singular linear relation. ♦ Lemma 2.12.2 Let S, T ∈ LR(X, Y ). If S is T -precompact, then S is strictly singular. ♦ Proof. Since S is T -precompact, then SGT is precompact. By using Lemma 2.12.1, it follows that SGT is continuous. On the one hand, by using Proposition 2.11.1, we obtain Γ0 (SGT ) = Γ0 (SGT ) = 0. e SGT ) = 0. On the other hand, by using Then, by using (2.46), we have ∆( Proposition 2.10.3, we get e S) ∆( = 0. e T) 1 + ∆( As a matter of fact, e ∆(S) = 0. Therefore, the use of Proposition 2.12.1 shows that S is strictly singular. This completes the proof. Q.E.D. Note that the sum of two strictly singular linear relation is strictly singular, but there exist strictly singular linear relations S and T such that ST is not strictly singular. We shall use the following result which gives sufficient conditions for the composition of two linear relations to be strictly singular. Theorem 2.12.1 [50, Proposition 5.2.12] Let T ∈ L(X, Y ) and S ∈ LR(Y, Z) be strictly singular. Then, ST ∈ SSR(X, Z) . ♦ Theorem 2.12.2 [50, Proposition 5.2.10] Let T ∈ LR(X, Y ) be strictly singular (respectively, precompact) and let S ∈ LR(Y, Z) be continuous. If T (0) ⊂ D(S), then ST is strictly singular (respectively, precompact) linear relation. ♦

70

2.13

Spectral Theory of Multivalued Linear Operators

Polynomial Multivalued Linear Operators

Our aim in this section is to show some fundamental results for a polynomial linear relation. The notion of polynomial linear operator can be naturally generalized to linear relation as follows: Definition 2.13.1 Let T ∈ LR(X) and given a polynomial P (λ) = α0 + α1 λ + · · · + αn λn with coefficients in C, we define the polynomial in T by P (T ) = α0 T 0 + α1 T + · · · + αn T n , where T 0 is the identity operator defined on X.

♦

Remark 2.13.1 Let µ ∈ C, n and mi , 1 ≤ i ≤ n be some positive integers, and let λi ∈ K, 1 ≤ i ≤ n be some distinct constants. Fix λ ∈ C and let P (λ) − µ = c

n Y

(λ − λk )mk .

k=1

Then, by using Lemma 2.5.1, the polynomial P (T ) in T is given by P (T ) − µ = c

n Y

(T − λk )mk

(2.47)

k=1

♦

is a linear relation in X.

The following properties concerning the behaviour of the domain, the range, the null space and the multivalued part of P (T ), were proved in [84, Theorems 3.2–3.6] by A. Sandovici. Lemma 2.13.1 Let P (T ) as in Remark 2.13.1. Then, (i) D(P (T )) = D(T m1 +m2 +...+mn ) and P (T )(0) = T m1 +m2 +...+mn (0). n \ (ii) R(P (T )) = R((T − λi )mi ). i=1 n X

(iii) N (P (T )) =

N ((T − λi )mi ) and the sum is direct if D(T ) = X.

i=1

(iv) For d ∈ N, we have N (P (T ))

T

R(P (T )d ) =

n X \ (N (T − λi )mi R(T − λi )mi d ). i=1

♦

Fundamentals

71

Lemma 2.13.2 [42, Theorem 3.3] Let T ∈ LR(X) and let P and Q be two polynomials in C. Then, ♦

(QP )(T ) = Q(T )P (T ). Lemma 2.13.3 Let β ∈ C. Then, (i) T n =

n X

Cnk β k (T − β)n−k , n ∈ N.

k=0

(ii) Every polynomial in T , P (T ) of degree n, can be described as P (T ) = (T − β)Q(T ) + ζ, where Q(T ) is a polynomial in T of degree n − 1 and ζ is a scalar. ♦ Proof. (i) We proceed by induction. For n = 0 it is trivial. Since T = (T − β) + β, we have the property is true for n = 1. Assume (i) holds for n = m. m X k k Then, T m+1 = A(B + C), where A = Cm β (T − β)m−k , B = T − β and k=0

C = β = βI. Indeed, T

m+1

m

=T T =

m X

! k k Cm β (T

− β)

m−k

((T − β) + β) .

k=0

Moreover, A(B + C)

= AB + AC.

(2.48)

We first prove that D (A(B + C)) = D (AB + AC). In fact, we have n o \ D (A(B + C)) = x ∈ D(B + C) such that (B + C)x D(A) 6= ∅ n o \ = x ∈ D(T ) such that T x D(T m ) 6= ∅
= D T m+1 . T On the other hand, D(AB + AC) = D(AB) D(AC), where n o \ D(AB) = x ∈ D(B) such that Bx D(A) 6= ∅ o n \ = x ∈ D(T − β) such that (T − β )x D(T m ) 6= ∅ = D (T m (T − β))
= D T m+1 , and D(AC)

=

n o \ x ∈ D(C) = X such that βx D(A) 6= ∅

= D (T m ) .

72

Spectral Theory of Multivalued Linear Operators
T
Hence, D (AB + AC) = D T m+1 D (T m ) = D T m+1 . Therefore, A(B + C) and AB + AC have the same domain. Furthermore, we infer from Proposition 2.3.6 (iv) that A(B + C) is an extension of AB + AC. So, G(AB + AC) ⊂ G(A(B + C)), and A(B + C)(0) = AB(0) + AC(0). In this situation, the property (2.48) follows immediately from Proposition 2.3.6 (iii). Now, we prove that T m+1

m+1 X

=

k Cm+1 β k (T − β)m+1−k .

(2.49)

k=0

From (2.48), we have T m+1 = E + F, where E=

m X

k k Cm β (T − β)m−k (T − β),

k=0

and F =β

m X

k k Cm β (T − β)m−k .

k=0

If we take S = T − β, Q(S) =

m X

k k Cm β (T − β)m−k ,

k=0

and P (S) = T − β, then in view of Lemma 2.13.2, we obtain Q(P (S)) = Q(S)P (S), that is m X k k E= Cm β (T − β)m+1−k . k=0

Therefore, (2.49) holds. The proof of the statement (i) is completed. (ii) Let P (T ) = α0 + α1 T + · · · + αn T n . The use of (i) leads to P (T ) = a0 + a1 (T − β) + · · · + an (T − β)n , for some scalars a0 , a1 , . . . , an . On the other hand, if in Lemma 2.13.2, we take S = T − β, P (S) = a1 S + · · · + an S n−1 and Q(S) = S, we deduce that
P (T ) − a0 = (T − β) a1 + a2 (T − β) + · · · + an (T − β)n−1 = (T − β)Q(T ),

Fundamentals

73

where, Q(T ) = a1 + a2 (T − β) + · · · + an (T − β)n−1 . Again, applying the part (i), we have Q(T ) = b1 + b2 T + · · · + bn T n−1 , for some scalars b1 , b2 , . . . , bn . So, Q is a polynomial in T of degree n − 1, as desired. This completes the proof. Q.E.D. Lemma 2.13.4 Suppose that T is closed and there exists β ∈ C such that T − β is bijective. Let P (T ) = α0 + α1 T + · · · + αn T n be a polynomial in T of degree n. Then, P (T ) is a closed linear relation. ♦ Proof. We shall proceed by induction. For n = 1, it is clear since P (T ) = α0 + α1 T is closed by Proposition 2.7.2. Assume that the property is true for n = m. Then, if P (T ) is a polynomial in T of degree m + 1, we infer from the condition (ii) in Lemma 2.13.3 that P (T ) = (T − β)Q(T ) + δ, where Q(T ) is a polynomial in T of degree m and δ is scalar. Then, Q(T ) is closed by the induction hypothesis. By using Proposition 2.7.2, we have T − β is closed. Since T − β is bijective, we deduce from Proposition 2.7.3 that (T − β)Q(T ) is closed. So, in view of Proposition 2.7.2, we get P (T ) is closed, as required. This completes the proof. Q.E.D. Lemma 2.13.5 Assume that A is everywhere defined. Then, (i) For all nonzero scalar λ and for all n ∈ N, we have (A − λ)n =

n X

Cni (−1)i λi An−i .

i=0

(ii) Suppose that (A − α)−1 is a single valued linear operator everywhere defined. Then, (a) For all β ∈ K and for all m, n ∈ N (A − α)−n (A − β)m ⊂ (A − β)m (A − α)−n . (b) For all β ∈ K with α 6= β and for all n ∈ N, we have N ((A − β)n ) = N (((β − α)−1 − (A − α)−1 )n ) and R((A − β)n ) = R(((β − α)−1 − (A − α)−1 )n ).

♦

74

Spectral Theory of Multivalued Linear Operators

Proof. (i) We proceed by induction. The case n = 1 is clear. Suppose the property holds for some positive integer k. Then, one deduces from Lemma 2.5.1 (i) and Proposition 2.3.6 combined with the induction hypothesis that (A − λ)k+1

= A(A − λ)k − λ(A − λ)k =

k X

Cki (−1)i λi Ak+1−i −

i=0

k X

Cki (−1)i λi+1 Ak−i

i=0

= Ak+1 +

k X

i Ck+1 (−1)i λi Ak+1−i + (−1)k+1 λk+1

i=1

=

k+1 X

i Ck+1 (−1)i λi Ak+1−i .

i=0

(ii) (a) We first note that (A − α)n is bijective and everywhere defined and thus it is easy to see that (A − α)−n (A − α)n = IX ⊂ (A − α)n (A − α)−n . These facts together with the part (c) of Lemma 2.5.1 imply that (A − α)−n (A − β)m

⊂ (A − α)−n (A − β)m (A − α)n (A − α)−n =

(A − α)−n (A − α)n (A − β)m (A − α)−n

=

(A − β)m (A − α)−n .

(b) We first verify that A−β

=

(β − α)((β − α)−1 − (A − α)−1 )(A − α)

=

(β − α)(A − α)((β − α)−1 − (A − α)−1 ).

(2.50)

One deduces from Proposition 2.3.6 and the part (a) that A−β

=

(A − α) − (β − α)

=

(A − α) − (β − α)(A − α)−1 (A − α)

=

(I − (β − α)(A − α)−1 )(A − α)

=

(β − α)((β − α)−1 − (A − α)−1 )(A − α)

=

(β − α)(β − α)−1 (A − α) − (β − α)IX

⊂ (β − α)((β − α)−1 (A − α) − (A − α))(A − α)−1 ) =

(β − α)(A − α)((β − α)−1 − (A − α)−1 ).

On the other hand, since A − α is everywhere defined and bijective it is easy to see that D((β−α)−1 −(A−α)−1 )(A−α)) = D((A−α)((β−α)((β−α)−1 −(A−α)−1 )) = X

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and ((β − α)−1 − (A − α)−1 )(A − α)(0)

=

(A − α)((β − α)−1 − (A − α)−1 )(0)

= A(0). Now, the use of Proposition 2.3.6, makes us to conclude (2.50). (b) We shall prove the first equality in (b) proceeding by induction. Define B := (β − α)−1 − (A − α)−1 . Applying (2.50), we infer that N (A − β)

= B −1 (N (A − α)) = B −1 (0) (as N (A − α) = {0}) = N (B).

Assume the property holds for some positive integer k. Then, N (B k+1 )

= B −1 (N (B k )) = B −1 (N ((A − β)k )) = B −1 (N ((A − β)k (A − α))) (by Lemma 2.5.1) = B −1 (A − α)−1 (N (A − β)k ) =

((A − α)B)−1 (N (A − β)k )

=

(B(A − α))−1 (N (A − β)k ) (by (2.50))

= N ((A − β)k+1 ). The second identity in (b) is obtained by induction. The case n = 1 follows immediately from (2.50). Assume now that R((A − β)k ) = R(B k ) for some positive integer k. Then, R((A − β)k+1 )

=

(A − β)(R(A − β)k )

=

(A − β)(R(B k ))

=

(β − α)B(R(B k (A − α))).

Thus, by Lemma 2.5.1 (i) (c) and the surjectivity of A − α, we infer that R((A − β)k+1 ) = R(B k+1 ). This completes the induction argument. Q.E.D.

76

2.14 2.14.1

Spectral Theory of Multivalued Linear Operators

Some Classes of Multivalued Linear Operators Multivalued Fredholm and Semi-Fredholm Linear Operators

Definition 2.14.1 A linear relation T ∈ LR(X, Y ) is said semi-Fredholm, denoted by T ∈ F+ (X, Y ), if there exists a codimensional subspace M of X such that the restriction T|M valued continuous inverse. A linear relation T is said to be Fredholm, denoted by T ∈ F− (X, Y ), if its adjoint T ∗ is Fredholm.

to be closed has a lower upper

upper finite single semisemi♦

Lemma 2.14.1 [50, Proposition 5.5.11 and Corollary 5.7.7] Let T ∈ LR(X, Y ) and let M ⊂ Y such that dim(M ) < ∞. Then, (i) T ∈ F+ (X, Y ) if and only if, QM T ∈ F+ (X, Y /M ). (ii) T ∈ F− (X, Y ) if and only if, QM T ∈ F− (X, Y /M ). Proposition 2.14.1 equivalent

♦

[50, Proposition 5.5.2] The following properties are

(i) T ∈ F− (X, Y ). (ii) β(T ) < ∞ and γ(T ∗ ) > 0. e , Ye ). (iii) Te ∈ Φ− (X (iv) QT T ∈ F− (X, Y /T (0)).

♦

Proposition 2.14.2 [50, Proposition 5.5.13] Let T ∈ LR(X, Y ) and we consider the properties (i) T ∈ F+ (X, Y ) and β(T ) < ∞. (ii) T ∗ ∈ F− (Y ∗ , X ∗ ) and α(T ∗ ) < ∞. Then, (i) =⇒ (ii) and if X is complete and T ∈ CR(X, Y ), then (ii) =⇒ (i). ♦ Lemma 2.14.2 [50, Proposition 5.2.16] Let T ∈ LR(X, Y ) and S ∈ LR(Y, Z). Then, (i) If T is sa single valued linear operator, T ∈ F+ (X, Y ) and S ∈ F+ (Y, Z), then ST ∈ F+ (X, Z). (ii) If S ∈ BR(Y, Z) and ST ∈ F+ (X, Z), then T ∈ F+ (X, Y ). ♦

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Theorem 2.14.1 [50, Theorem 5.10.3] Let X be a Banach space, Y be a normed vector space and T ∈ LR(X, Y ). Then, the following properties are equivalent (i) T is upper semi-Fredholm. (ii) There exists a bounded linear operator A and a bounded finite rank projection operator P such that AT = ID(T ) − P. ♦ Proposition 2.14.3 (i) [50, Proposition 5.5.27] Let T ∈ LR(X, Y ). If T is closable, then T ∈ F− (X, Y ) if and only if, T GT ∈ F− (XT , Y ). (ii) [50, Corollary 5.2.5] T ∈ F+ (X, Y ) if and only if, T GT ∈ F+ (XT , Y ). ♦ Theorem 2.14.2 [50, Theorem 5.1.11] Let X and Y be two normed spaces. Then, the following properties are equivalent (i) T ∈ / F+ (X, Y ). (ii) There exists a non precompact bounded subset W of D(T ) such that QT W is precompact. (iii) T has a singular sequence, i.e., a sequence (xn )n of norm one elements of D(T ) such that (xn )n has no Cauchy subsequence and limn→∞ kT xn k = 0. ♦ Proposition 2.14.4 [13, Theorem 2.17] Let S ∈ LR(X, Y ), T ∈ F+ (X, Y ) with G(S) ⊂ G(T ), and dim D(S) = ∞, then S ∈ F+ (X, Y ). ♦ Proposition 2.14.5 [92, Proposition 5.9.2] Let T ∈ F+ (X, Y ). Then, any bounded sequence (xn )n in D(T ) such that (QT T xn )n is a Cauchy sequence has a Cauchy subsequence. ♦ Proposition 2.14.6 [50, Theorem 5.2.4] Let T ∈ LR(X, Y ). If dim D(T ) = ∞, then T ∈ F+ (X, Y ) if and only if, Γ(T ) > 0. ♦

2.14.2

Multivalued Fredholm and Semi-Fredholm of Closed Linear Operators in Banach Space

Let X and Y be two Banach spaces, we extend the classes of closed single valued Fredholm type operators to include closed multivalued operators, and note that the definitions of the classes F+ (X, Y ) and F− (X, Y ) are consist, respectively, with Φ+ (X, Y ) := {T ∈ CR(X, Y ) such that α(T ) < ∞ and R(T ) is closed in Y } ,

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Spectral Theory of Multivalued Linear Operators

and Φ− (X, Y ) := {T ∈ CR(X, Y ) such that β(T ) < ∞ and R(T ) is closed in Y } . S T Φ± (X, Y ) := Φ+ (X, Y ) Φ− (X, Y ) and Φ(X, Y ) := Φ+ (X, Y ) Φ− (X, Y ) denotes the set of Fredholm linear relations from X into Y . If X = Y, then Φ+ (X, Y ), Φ− (X, Y ), Φ± (X, Y ) and Φ(X, Y ) are replaced, respectively, by Φ+ (X), Φ− (X), Φ± (X) and Φ(X). For T ∈ CR(X), a number complex λ is in Φ+T , or ΦT if λ − T is in Φ+ (X), or Φ(X), respectively. For T ∈ CR(X, Y ) and S ∈ LR(X, Y ) such that λS − T is closed, a number complex λ is in Φ+T,S , or ΦT,S if λS − T is in Φ+ (X, Y ), or Φ(X, Y ), respectively. Definition 2.14.2 We say that a linear relation T from X into Y is Browder, if T is a Fredholm linear relation of index zero and has finite ascent and descent. ♦ We will denote by B(X, Y ) the set of Browder linear relations. If X = Y , then B(X) := B(X, X). Remark 2.14.1 Let T be a Fredholm single valued operator from X into Y of finite ascent and descent. Then, T has index zero. However, this property is not true in the context of linear relations. ♦ Theorem 2.14.3 [11, Theorem 6.1] Let X be a Banach space and let S, T ∈ CR(X). Then, (i) If S, T ∈ Φ+ (X), then ST ∈ Φ+ (X) and T S ∈ Φ+ (X). (ii) If S, T ∈ Φ− (X) with T S (respectively, ST ) is closed, then T S ∈ Φ− (X) (respectively, ST ∈ Φ− (X)). (iii) If S, T ∈ Φ(X), then T S ∈ Φ(X) and   \ X i(T S) = i(T ) + i(S) + dim − dim(S(0) N (T )). R(S) + D(T ) (iv) If S and T are everywhere defined and T S ∈ Φ+ (X), then S ∈ Φ+ (X). (v) If S and T are everywhere defined such that T S ∈ Φ(X) and ST ∈ Φ(X), then S ∈ Φ(X) and T ∈ Φ(X). ♦ Example 2.14.1 Let T : l2 (N) −→ l2 (N) be defined as ( e k if k is even, 2 T ek = 0 if k is odd.

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This single valued linear operator T is not Fredholm, since α(T ) = +∞. Now, we defined S : l2 (N) −→ l2 (N) as the continuous linear extension of Sek = e2k , (k ∈ N). Then, T S = I the identity on l2 (N), which is clearly Fredholm single valued linear operator, but T is not. Theorem 2.14.4 [9, Proposition 8 and Theorem 25] Let X and Y be two Banach spaces and let T ∈ CR(X, Y ). Then, (i) T ∈ Φ+ (X, Y ) if and only if, T ∗ ∈ Φ− (Y ∗ , X ∗ ). (ii) T ∈ Φ− (X, Y ) if and only if, T ∗ ∈ Φ+ (Y ∗ , X ∗ ). Moreover, if T is semi-Fredholm, then α(T ∗ ) = β(T ) and α(T ) = β(T ∗ ). (iii) Let T ∈ Φ+ (X). Then, there exists ε > 0 such that for every λ ∈ K with 0 < |λ| < ε, we have λ − T ∈ Φ+ (X), α(λ − T ) is constant, α(λ − T ) ≤ α(T ) and i(λ − T ) = i(T ). (iv) Let T ∈ Φ− (X). Then, there exists ε > 0 such that for every λ ∈ K with 0 < |λ| < ε, we have λ − T ∈ Φ− (X), β(λ − T ) is constant, β(λ − T ) ≤ β(T ) and i(λ − T ) = i(T ). ♦ Definition 2.14.3 (i) Let T ∈ Φ+ (X) and let ε > 0 as in Theorem 2.14.4 (iii). The jump of T is defined by j(T ) := α(T ) − α(λ − T ),

0 < |λ| < ε.

(ii) Let T ∈ Φ− (X) and let ε > 0 as in Theorem 2.14.4 (iv). The jump of T is defined by j(T ) := β(T ) − β(λ − T ),

0 < |λ| < ε.

♦

Remark 2.14.2 (i) Clearly, j(T ) = 0 and the continuity of the index (see Theorem 2.14.4) ensures that both definitions of j(T ) coincide whenever T ∈ T Φ+ (X) Φ− (X), so that j(T ) is unambiguously defined. (ii) An immediate consequence of Theorem 2.14.4 (i) is that if T is semiFredholm, then j(T ) = j(T ∗ ). ♦ Lemma 2.14.3 Let X and Y be two Banach spaces and let T ∈ CR(X, Y ). Then,

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Spectral Theory of Multivalued Linear Operators

(i) T ∈ Φ+ (X, Y ) if and only if, QT T ∈ Φ+ (X, Y /T (0)). In such case i(T ) = i(QT T ). (ii) T ∈ Φ− (X, Y ) if and only if, QT T ∈ Φ− (X, Y /T (0)). In such case i(T ) = i(QT T ). (iii) T ∈ Φ(X, Y ) if and only if, QT T ∈ Φ(X, Y /T (0)). In such case i(T ) = i(QT T ). ♦ Proof. First, in view of Proposition 2.7.1 (i), QT T is closed and T (0) is closed. Second, R(T ) is closed if and only if, R(QT T ) is closed. In such case β(T ) = β(QT T ). Indeed, T (0) is closed implies  .  R(QT T ) = R(T ) + T (0) T (0) = R(T ) T (0). (2.51) So, by using Lemma 2.1.3 (i), we have R(T ) is closed if and only if, R(QT T ) is closed. In such case by Lemma 2.1.3 (ii), we have β(T ) = β(QT T ). Finally, N (T )

= {x ∈ D(T ) such that T x = T (0)} = {x ∈ D(T ) such that T x = T (0)} (since T is closed) = {x ∈ D(T ) such that QT T x = 0} = N (QT T ).

This completes the proof.

2.14.3

Q.E.D.

Atkinson Linear Relations

We begin this subsection with the notion of a Atkinson linear relation in a Banach space. Lemma 2.14.4 Let S ∈ LR(Y, Z) and T ∈ Φ− (X, Y ) be bounded operators such that N (S) and N (T ) are topologically complemented in Y and X, respectively. Then, N (ST ) is topologically complemented in X. ♦ Proof. Let Y1 and X1 be closed subspaces of Y and X, respectively, L L such that Y = N (S) Y1 and X = N (T ) X1 . Since R(T ) is a closed finite codimensional subspace we infer from Lemma 2.1.1 that there exist a finite dimensional subspaces N ⊂ N (S) and Y2 ⊂ Y1 such that N (S) = T L T L (N (S) R(T )) N and Y1 = (Y1 R(T )) Y2 . Therefore, the subspace T L T (N (S) R(T )) (Y1 R(T )) is a finite codimensional subspace of R(T ). So, there is a finite dimensional subspace Y3 ⊂ R(T ) such that  M \  \ R(T ) = N (S) R(T ) Y1 R(S) + Y3

Fundamentals

81

and M \  \ R(T ) Y1 R(S) Y3 . T We define T1 := T|X1 , X3 := T1−1 (N (S) R(T )) and X4 := L T L L T1−1 (Y3 (R(T ) Y1 )) . Then, it is easy to see that X = N (T ) X3 X4 . T Hence, since T (N (ST )) = R(T ) N (S), we deduce that N (ST ) = L L N (T ) X3 . Thus, X = N (ST ) X4 . Q.E.D. {0} =



N (S)

\

Let X and Y be two normed spaces and T ∈ CR(X, Y ). We say that T is bounded below if it is injective and open, T is left invertible if T is injective and R(T ) is topologically complemented in Y , and T is right invertible if T is surjective and N (T ) is topologically complemented in X. The sets of all left invertible linear relations, right invertible relations and invertible linear relations on X are defined, respectively, by GRl (X, Y ) = {T ∈ CR(X, Y ) : T is a left invertible linear relation} , GRr (X, Y ) = {T ∈ CR(X, Y ) : T is a right invertible linear relation} , T GR(X, Y ) = GRl (X, Y ) GRr (X, Y ). The class of α-Atkinson multivalued linear operators from X into Y is defined by Aα (X, Y ) = {T ∈ Φ+ (X, Y ) : R(T ) is topologically complemented in Y } and the class of β-Atkinson multivalued linear operators from X into Y is defined by Aβ (X, Y ) = {T ∈ Φ− (X, Y ) : N (T ) is topologically complemented in X} . Based on the previous definitions, we defined the set of left Weyl linear relations by W l (X, Y ) := {T ∈ Aα (X, Y ) such that i(T ) ≤ 0} , and the set of right Weyl linear relations is defined by W r (X, Y ) := {T ∈ Aβ (X, Y ) such that i(T ) ≥ 0} . The set of Weyl linear relations is defined by W(X, Y ) := W l (X, Y ) Thus, we have the following inclusions: W(X, Y ) ⊂ Aα (X, Y ) ⊂ Φ+ (X, Y ), and W(X, Y ) ⊂ Aβ (X, Y ) ⊂ Φ− (X, Y ).

T

W r (X, Y ).

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Spectral Theory of Multivalued Linear Operators

If X = Y , then the sets GRl (X, Y ), GRr (X, Y ), GR(X, Y ), Aα (X, Y ), Aβ (X, Y ), W l (X, Y ), W r (X, Y ), and W(X, Y ) are remplaced, respectively, by GRl (X), GRr (X), GR(X), Aα (X), Aβ (X), W l (X), W r (X), and W(X). Lemma 2.14.5 [13, Proposition 3.28] Let X and Y be two complete spaces and let T ∈ CR(X, Y ). Then, (i) T ∈ Aα (X, Y ) if and only if, T ∗ ∈ Aβ (Y ∗ , X ∗ ). (ii) T ∈ Aβ (X, Y ) if and only if, T ∗ ∈ Aα (Y ∗ , X ∗ ).

♦

Remark 2.14.3 Let X and Y be two complete spaces and let T ∈ CR(X, Y ). (i) Using Lemma 2.14.5 and definitions of both W l (X, Y ) and W r (X, Y ), we have the following equivalences (i1 ) T ∈ W l (X, Y ) if and only if, T ∗ ∈ W r (Y ∗ , X ∗ ). (i2 ) T ∈ W r (X, Y ) if and only if, T ∗ ∈ W l (Y ∗ , X ∗ ). (ii) Let M ⊂ Y such that dim(M ) < ∞. Then, (ii2 ) If T (0) is topologically complemented in Y , then T ∈ Aα (X, Y ) if and only if, QM T ∈ Aα (X, Y /M ). (ii3 ) If T ∗ (0) is topologically complemented in Y ∗ , then T ∈ Aβ (X, Y ) if and only if, QM T ∈ Aβ (X, Y /M ).

♦

Theorem 2.14.5 [79, Theorem 1.1] Let X and Y be two complete spaces and let T ∈ CR(X, Y ). If T is a single valued linear operator, then we have the following equivalences (i) T ∈ Aα (X, Y ). (ii) There exist U ∈ L(Y, X) and E ∈ K(X) such that (ii1 ) R(U ) ⊂ D(T ). (ii2 ) R(E) ⊂ D(T ). (ii3 ) U T = I − K on D(T ).

♦

Lemma 2.14.6 [13, Corollary 3.25] Let T ∈ CR(X, Y ). The following properties are equivalent (i) T ∈ Aβ (X, Y ). (ii) There exist Tr ∈ L(Y, X) and a finite rank projection F ∈ L(Y ) such that R(Tr ) is topologically complemented in D(T ), Tr T and F T are continuous operators, I − F ∈ Φ(X, Y ) and I − F is a linear selection of T Tr . ♦

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Definition 2.14.4 An operator Tr ∈ L(Y, X) satisfying Lemma 2.14.6 is called a right regularizer (or a right generalized inverse) of T . ♦ Proposition 2.14.7 Let T ∈ CR(X, Y ) and let S ∈ LR(Y, Z). Then, T (i) If T ∈ Aβ (X, Y ), S ∈ Aβ (Y, Z) L(Y, Z) and ST ∈ CR(X, Z), then ST ∈ Aβ (X, Z). (ii) If D(S ∗ ) = Z ∗ , R(T ) ⊂ D(S), T ∈ Aα (X, Y ) and S ∈ T Aα (Y, Z) L(Y, Z), then ST ∈ Aα (X, Z). ♦ Proof. (i) Follows immediately from Lemma 3.1.5 and Theorem 2.14.4. T (ii) Let T ∈ Aα (X, Y ) and let S ∈ Aα (Y, Z) L(Y, Z). Then, by using Lemma T 2.14.5 (i), we have T ∗ ∈ Aβ (Y ∗ , X ∗ ) and S ∗ ∈ Aβ (Z ∗ , Y ∗ ) L(Z ∗ , Y ∗ ). Hence, by using (i), we have (ST )∗ = T ∗ S ∗ ∈ Aβ (Z ∗ , X ∗ ). By applying Lemma 2.14.5 (ii), we have ST ∈ Aβ (X, Z). Q.E.D. Proposition 2.14.8 Assume that T is a single valued linear operator. Then, the following properties are equivalent (i) T ∈ Aα (X, Y ). (ii) There exist A ∈ L(Y, X) and B a continuous linear operator in X such that D(T ) ⊂ D(B), I − B ∈ Φ(X), R(B) ⊂ D(T ) and AT = (I − B)|D(T ) . ♦ Proof. Let us consider two cases of T : First case : If D(T ) = X, then the result was proved through Theorem 2.14.5. Second case : If D(T ) 6= X, then for A and B satisfy the conditions in (ii), we have AT GT = (I − B)|XT . Hence, kT GT k =

sup x∈XT

= =

kT GT xk kxkT

kT xk x∈D(T )\{0} kxk + kT xk sup sup x∈D(T )\{0}



1+

kT xk  kT xk kxk

kxk

kT k 1 + kT k < 1.

=

Then, T GT ∈ L(XT , Y ) and XT is complete, then by applying the first case to the operator T GT , we have T GT ∈ Aα (XT , Y ). Therefore, T ∈ Aα (X, Y ), as desired. Q.E.D.

84

2.15 2.15.1

Spectral Theory of Multivalued Linear Operators

Quasi-Fredholm and Semi Regular Linear Relations Quasi-Fredholm Linear Relations

In this subsection, we introduce and study the class of quasi-Fredholm linear relations in Banach spaces. Let us first recall some facts which will help to understand Definition 2.15.1 below. It is proved in Theorem 2.7.2 (i) that an everywhere defined closed linear relation in a Banach space is bounded. These remarks suggest generalizing the notion of bounded quasi-Fredholm operator due to M. Mbekhta and V. Muller ¨ [78]) to the case of multivalued linear operators as follows: Definition 2.15.1 Let X be a Banach space and let T be an everywhere defined closed linear relation in X with there exists λ ∈ C such that (T − λ)−1 is a single valued linear operator everywhere defined. We say that T is quasiFredholm if there exists d ∈ N for which R(T d+1 ) is closed and R(T ) + N (T d ) = R(T ) + N ∞ (T ).

♦

In this case T is said to be a quasi-Fredholm linear relation of degree d and we write T ∈ qφd (X).

2.15.2

Semi Regular Linear Relations

We begin with the following lemma. Lemma 2.15.1 [73, Lemma 3.7] Let T ∈ BR(X). Then, the following statements are equivalent (i) N (T ) ⊂ R(T n ) for every n ∈ N, (ii) N (T m ) ⊂ R(T ) for every m ∈ N, and (iii) N (Am ) ⊂ R(T n ) for every n, m ∈ N.

♦

Definition 2.15.2 We say that a linear relation T ∈ LR(X) is semi regular if R(T ) is closed and T verifies one of the equivalent conditions of Lemma 2.15.1. ♦ Theorem 2.15.1 [9, Theorem 27] Let T ∈ Φ± (X). Then, T is semi regular if and only if, j(T ) = 0. ♦

Fundamentals

85

Theorem 2.15.2 [9, Theorem 23] Let X be a complex Banach space and let T be a semi regular linear relation in X. Then, there exists δ > 0 such that λ − T is semi regular if |λ| < δ. ♦ Trivial examples of regular linear relations are surjective multivalued operators as well as injective multivalued operators with closed range. For an essential version of semi regular linear relation, we use the following notations. Definition 2.15.3 Let T ∈ LR(X). Assume that X1 and X2 are two L subspaces of X such that X = X1 X2 . We say that T is completely reduced by the pair (X1 , X2 ) if L ♦ T|X2 . T = T|X1 In such case, we have   D(T ) =       N (T ) = R(T ) =    T (0) =     Tn =

L D(T|X1 ) D(TX2 ), L N (T|X1 ) N (T|X2 ), L R(T|X1 ) R(T|X2 ), L T|X1 (0) T|X2 (0), and L (T|X1 )n (T|X2 )n for all n ∈ N.

Definition 2.15.4 Let T ∈ LR(X). T is said to be a Kato linear relation of degree d, if there exists d ∈ N and a pair of closed subspaces (M, N ) of E L such that E = M N , T is completely reduced by the pair (M, N ) with T|M is a regular linear relation and T|N is a nilpotent bounded operator of degree d (that is T|dN = 0). T is said to be a Kato linear relation if it is a Kato linear relation of degree d, for some d ∈ N. The pair (M, N ) is called the Kato decomposition of the linear relation T . ♦ Proposition 2.15.1 [12, Proposition 2.5] Let T be a Kato linear relation of degree d and let (M, N ) be a decomposition of degree d associated with T . Then, (i) R∞ (T ) = T (R∞ (T )) = R(T|M ). Further, R∞ (T ) is closed. T (ii) For every nonnegative integer n ≥ d, we have N (T ) R(T n ) = T T N (T n ) M = N (T ) R(T d ). (iii) For every nonnegative integer n ≥ d, we have R(T ) + N (T n ) = L T (M ) N is closed. ♦ Theorem 2.15.3 [74, Theorem 2.5] Let T be a range space relation in a Hilbert space X (i.e., the graph of T is the range of a bounded linear operator

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Spectral Theory of Multivalued Linear Operators

from a Hilbert space X to X × X), which is quasi-Fredholm linear relation of degree d ∈ N. Then, T is closed and there exist two closed subspaces M and N of X such that T (i) X = M + N and M N = {0}. (ii) R(T d ) ⊂ M , (iii) N ⊂ N (T d ) and, if d ≥ 1, then N * N (Ad−1 ). ♦ Proposition 2.15.2 [9, Proposition 11] Let X be a complex Banach space. If T is a semi regular linear relation in X, then γ(T n ) = γ(T )n . ♦ Proposition 2.15.3 [9, Proposition 12] Let X be a complex Banach space and let T be a regular linear relation in X. Then, for each n ∈ N, we have (i) N (T n )⊥ = R((T ∗ )n ). (ii) N ((T ∗ )n )> = R(T n ).

♦

Proposition 2.15.4 [45, Proposition 3.2] Let T ∈ L(X). Then, T is a quasiFredholm operator if and only if there exists n ∈ N such that R(T n ) is closed and T n is semi regular. ♦

2.15.3

Essentially Semi Regular Linear Relations

Definition 2.15.5 Let T ∈ LR(X). T is said essentially semi regular if R(T ) is closed and N (T ) ⊂e R∞ (T ). ♦ Samuel multiplicity operators have been studied by several authors. Particularly, in Ref. [95], the authors studied the Samuel multiplicity of essentially semi regular operators. In the next we extend this study to the general case of multivalued linear operators. Definition 2.15.6 For any essentially semi regular linear relation A in Hilbert space X, define its shift (Samuel) multiplicity by   β(Ak ) s.mult(A) = lim . k→∞ k Similarly, define its backward shift (Samuel) multiplicity by   α(Ak ) b.s.mult(A) = lim . k→∞ k

♦

Fundamentals

2.16 2.16.1

87

Perturbation Results for Multivalued Linear Operators Small Perturbation Theorems of Multivalued Linear Operators

Theorem 2.16.1 [50, Theorem V.3.2] Let S, T ∈ LR(X, Y ) such that e S) < Γ(T ), then T + S ∈ F+ (X, Y ). S(0) ⊂ T (0). If ∆( ♦ The following example shows that the hypothesis of S(0) ⊂ T (0) in Theorem 2.16.1 is crucial. Example 2.16.1 Let T = IX and let S ∈ LR(X) be defined by G(S) = X × X. Then, S is a compact multivalued linear operator. Now, by using e S) < Γ(T ) = 1 but T + S 6∈ F+ (X). Proposition 2.12.1, we have 0 = ∆( T Theorem 2.16.2 Let T ∈ BR(X) KR(X) and λ ∈ K\{0}. If dim(T (0)) < ∞, then λ − T ∈ Φ(X) and i(λ − T ) = dim(T (0)) < ∞. ♦ Proof. In view of Theorem 2.7.2 (iv), we have T is closed. By using (2.51),  we have R(QT (λ)) = X T (0). In view of N (QT (λ)) = T (0), then QT (λ) is a Fredholm operator. Moreover, Qλ−T (λ − T ) = QT (λ − T ) = QT (λ) − QT T.

(2.52)

Since kQT T k = kT k < ∞ and T ∈ KR(X), then QT T is bounded compact operator. Hence, by using (2.52), we have Qλ−T (λ−T ) is a Fredholm operator. So, by using Lemma 2.14.3 (iii), we have λ − T is a Fredholm linear relation and i(λ − T ) = i(Qλ−T (λ − T )) = dim T (0). Q.E.D. The following example shows that if T is a bounded compact linear relation on a Banach space X and λ ∈ K \ {0}, then λ − T may not be Fredholm. Example 2.16.2 Let X be an infinite-dimensional Banach space and let T be a linear relation whose graph is X × X. Then, T is a bounded compact closed linear relation such that I + T is not Fredholm. Indeed, we infer immediately from the definition of the sum (Definition 2.3.4) of linear relations that I + T = T. Observing that for any linear relation U in a vector space E, we T have D(U ) × E = G(U ) + ({0} × E), {0} × U (0) = G(U ) ({0} × E) and T N (U ) × {0} = G(U ) (E × {0}). We deduce trivially that D(T ) = N (T ) =

88

Spectral Theory of Multivalued Linear Operators

T (0) = X. Therefore, T is not an operator (since T (0) = 6 {0}), QT T is the zero operator on X (since D(T ) = T (0) = X). So, T is a bounded compact linear relation. Finally, it follows from the equalities I +T = T and N (T ) = X that I + T is not a Fredholm linear relation. Remark 2.16.1 We remark that in Example 2.16.2, dim T (0) = ∞.

♦

Theorem 2.16.3 [50, Theorem V.5.12] Let X and Y be two normed spaces and let S, T ∈ LR(X, Y ) such that D(S) ⊃ D(T ). Then, (i) If S ∈ PRK(X, Y ) and T ∈ F− (X, Y ), then T + S ∈ F− (X, Y ). (ii) If kSk < γ(T ∗ ) and T ∈ F− (X, Y ), then T + S ∈ F− (X, Y ). (iii) If dim(R(S)) < ∞ and T ∈ F− (X, Y ), then T + S ∈ F− (X, Y ).

♦

By using Proposition 2.3.4, Theorems 2.16.1 and 2.16.3, Propositions 2.12.1 and 2.7.2, one can easily get to the following result. Corollary 2.16.1 Let X and Y be two complete spaces and let S, T ∈ LR(X, Y ) such that S(0) ⊂ T (0) and D(T ) ⊂ D(S). Then, (i) If S ∈ PRK(X, Y ) and T ∈ Φ− (X, Y ), then T ± S ∈ Φ− (X, Y ). (ii) If kSk < γ(T ∗ ) and T ∈ Φ− (X, Y ), then T ± S ∈ Φ− (X, Y ). (iii) If dim (R(S)) < ∞ and T ∈ Φ− (X, Y ), then T + S ∈ Φ− (X, Y ). (iv) If S ∈ SSR(X, Y ), and T ∈ Φ+ (X, Y ), then T ± S ∈ Φ+ (X, Y ). (v) If S ∈ SSR(X, Y ), and T + S ∈ Φ+ (X, Y ) or T − S ∈ Φ+ (X, Y ), then T ∈ Φ+ (X, Y ). (vi) If kSk < γ(T ) and T ∈ Φ+ (X, Y ), then T ± S ∈ Φ+ (X, Y ). (vii) If S ∈ SSR(X, Y ), then T + S ∈ Φ(X, Y ) and i(T + S) = i(T ). ♦ Corollary 2.16.2 Let X be a Banach space and T ∈ Φ(X). If 0 < |λ| < γ(T ), then λ − T ∈ Φ(X). ♦ Proof. To prove this corollary, it is sufficient to replace S by λIX in Corollary 2.16.1 (ii), (vi) and Proposition 2.9.3 (iii). Q.E.D. Lemma 2.16.1 Let A ∈ CR(X) and K ∈ LR(X) be continuous such that K(0) ⊂ A(0) and D(A) ⊂ D(K). Then, (i) A ∈ Φ+ (X) if and only if, A + K − K ∈ Φ+ (X). In such case i(A) = i(A + K − K). (ii) A ∈ Φ− (X) if and only if, A + K − K ∈ Φ− (X). In such case i(A) = i(A + K − K).

♦

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89

Proof. We first note that by using Proposition 2.7.2, A + K and A + K − K are closed. Furthermore, we have (a) (A + K − K)(0) = A(0) and QA+K−K = QA(0)/K(0) QK . Indeed, (A + K − K)(0) = A(0) + K(0) = A(0), then QA+K −K = QA and by Lemma 2.1.3, we infer that QA = QA(0)/K(0) QK . (b) QA+K −K (A + K − K) = QA (A). Indeed, QK (A + K − K) = QK (A) + QK (K) − QK (K) = QK (A), so, by (a) we obtain (b). Now, combining (b) and Lemma 2.14.3, we obtain the result. Q.E.D. Lemma 2.16.2 Let S ∈ LR(X, Y ) and T ∈ F+ (X, Y ) with dim D(T ) = e S) < Γ(T ). If, ∞. If S is precompact, then S is strictly singular, with ∆( additionally, S(0) ⊂ T (0), then T + S ∈ F+ (X, Y ). ♦ Proof. Since T ∈ F+ (X, Y ), and dim D(T ) = ∞, then by using Proposition 2.14.6, we obtain Γ(T ) > 0. Since S is precompact and using Proposition 2.11.1, we get Γ0 (S) = 0. e S) ≤ Γ0 (S) = 0. Hence, So, ∆( e S) = 0. ∆( Then, S is strictly singular. Thus, by Proposition 2.3.4, Theorems 2.16.1 and 2.16.3, it follows that T + S ∈ F+ (X, Y ). Q.E.D. Theorem 2.16.4 Let X be a Banach space and let T ∈ CR(X), S ∈ LR(X) such that S(0) ⊂ T (0), D(T ) ⊂ D(S), kSk < γ(T ) and β(T ) < ∞. Then, γ(T − S) ≥ kP k−1 γ(T ) − kSk, where P is any continuous projection defined on D(T ∗ ) with kernel N (T ∗ ).♦

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Spectral Theory of Multivalued Linear Operators

Proof. Combining both T is closed and γ(T ) > 0, and by using Propositions 2.8.1 (v) and 2.9.3 (i), we have R(T ) is closed. Accordingly, β(T ) < ∞ and T ∈ Φ− (X). By virtue of Theorem 2.9.3 (iii), we have 0 < γ(T ∗ ). Since kSk < γ(T ) = γ(T ∗ ) and according to S(0) ⊂ T (0), D(T ) ⊂ D(S), and Corollary 2.16.1 (ii), we obtain T − S ∈ Φ− (X). This implies from Theorem 2.14.4 (i) that (T − S)∗ ∈ Φ+ (X ∗ ). Using the fact that D(T ) ⊂ D(S), and in view of Proposition 2.8.1 (iii), we can write (T − S)∗ = T ∗ − S ∗ . (2.53) Now, we have to prove that S ∗ (0) ⊂ T ∗ (0). Indeed, since D(T ) ⊂ D(S), then by using Proposition 2.8.1 (iv), we have S ∗ (0) = D(S)⊥ ⊂ D(T )⊥ = T ∗ (0). By referring to Proposition 2.9.6 and (2.53), we obtain γ((T − S)∗ ) = γ(T ∗ − S ∗ ) ≥ kP k−1 γ(T ∗ ) − kS ∗ k, where P is any continuous projection defined on D(T ∗ ) with kernel N (T ∗ ). Finally, the use of Proposition 2.9.3 (iii) allows us to conclude that γ(T − S) ≥ kP k−1 γ(T ) − kSk.

Q.E.D.

Corollary 2.16.3 Let X be a Banach space and let T ∈ CR(X), S ∈ LR(X) such that S(0) ⊂ T (0), D(T ) ⊂ D(S), and kSk < γ(T ). If T ∈ Φ− (X), then γ(T − S) ≥ kP k−1 γ(T ) − kSk, where P is any continuous projection defined on D(T ∗ ) with kernel N (T ∗ ).♦ Proof. The proof is a direct consequence of Theorem 2.16.4 and Proposition 2.5.3. Q.E.D.

Fundamentals

2.17

91

Fredholm Perturbation Classes of Linear Relations

Definition 2.17.1 Let X and Y be two Banach spaces. Then, (i) The set of upper semi-Fredholm perturbations of linear relations is defined by PR (Φ+ (X, Y )) = {S ∈ LR(X, Y ) is continuous : T + S ∈ Φ+ (X, Y ), for all T ∈ Φ+ (X, Y ), D(T ) ⊂ D(S) and T (0) ⊃ S(0)}. (ii) The set of lower semi-Fredholm perturbations of linear relations is defined by PR(Φ− (X, Y )) = {S ∈ LR(X, Y ) is continuous : T + S ∈ Φ− (X, Y ), for all T ∈ Φ− (X, Y ), D(T ) ⊂ D(S) and T (0) ⊃ S(0)}. (iii) The set of α-Atkinson perturbations linear relations is defined by PR (Aα (X, Y )) = {S ∈ LR(X, Y ) : T +S ∈ Aα (X, Y ) for all T ∈ Aα (X, Y ) such that T (0) is topologically complemented in Y, D(T ) ⊂ D(S) and T (0) ⊃ S(0)}. (iv) The set of β-Atkinson perturbations linear relations is defined by PR (Aβ (X, Y )) = {S ∈ LR(X, Y ) such that T + S ∈ Aβ (X, Y ) for all T ∈ Aβ (X, Y ) such that T ∗ (0) is topologically complemented in Y ∗ , D(T ) ⊂ D(S) and S(0) ⊂ T (0)}. (v) The set of Atkinson perturbations is defined by T PR(A(X, Y )) := PR (Aα (X, Y )) PR (Aβ (X, Y )) .

♦

If X = Y, we denote by PR(Φ+ (X)) := PR(Φ+ (X, X)), PR(Φ− (X)) := PR(Φ− (X, X)), PR(Aα (X)) := PR(Aα (X, X)), PR(Aβ (X)) := PR(Aβ (X, X)) and PR(Φ(X)) := PR(Φ(X, X)). Proposition 2.17.1 Let X and Y be two Banach spaces, and let F1 , F2 ∈ LR(X, Y ). Then, (i) If F1 ∈ PR (Φ+ (X, Y )) and F2 ∈ PR (Φ+ (X, Y )), then F1 ± F2 ∈ PR (Φ+ (X, Y )) . (ii) If F1 ∈ PR (Φ− (X, Y )) and F2 ∈ PR (Φ− (X, Y )), then F1 ± F2 ∈ PR (Φ− (X, Y )) .

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Spectral Theory of Multivalued Linear Operators

(iii) If F1 ∈ PR (Aα (X, Y )) and F2 ∈ PR (Aα (X, Y )), then F1 ± F2 PR (Aα (X, Y )) . (iv) If F1 ∈ PR(Aβ (X, Y )) and F2 ∈ PR(Aβ (X, Y )), then F1 ± F2 PR (Aβ (X, Y )) . (v) If F1 ∈ PR (Φ(X, Y )) and F2 ∈ PR (Φ(X, Y )), then F1 ± F2 PR (Φ(X, Y )) .

∈ ∈ ∈ ♦

Proof. (i) Let A ∈ Φ+ (X, Y ) such that D(A) ⊂ D(F1 ± F2 ) = D(F1 )

\

D(F2 ),

(2.54)

and A(0) ⊃ (F1 ± F2 )(0) = F1 (0) + F2 (0).

(2.55)

By using the relation (2.54), we deduce that D(A) ⊂ D(F1 ) and, according to the relation (2.55), we infer that A(0) ⊃ F1 (0). However, if F1 ∈ PR (Φ+ (X, Y )), then A + F1 ∈ Φ+ (X, Y ), D(A + F1 ) = D(A) ⊂ D(F2 ),

(2.56)

(F1 + A)(0) ⊃ F2 (0).

(2.57)

and Now, by using the relations (2.56), (2.57) and the fact that F2 ∈ PR (Φ+ (X, Y )), we deduce that A + F1 ± F2 ∈ Φ+ (X, Y ). So, F1 ± F2 ∈ PR (Φ+ (X, Y )) . The proof of the other cases is analogous to the proof of (i).

2.18 2.18.1

Q.E.D.

Spectrum and Pseudospectra of Linear Relations Resolvent Set and Spectrum of Linear Relations

Definition 2.18.1 Let T ∈ LR(X) and λ ∈ C. The quantites R(λ, T ) = (λIX − T )−1 = (λ − T )−1 is called the resolvent of T (corresponding to λ) and Tλ := (λ − Te)−1 is called the complete resolvent of T , where Te is given in (2.4).

♦

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93

Definition 2.18.2 Let T ∈ LR(X). The resolvent set and the spectrum of T are, respectively, defined as ρ(T ) = {λ ∈ C such that Tλ is single valued everywhere defined} , σ(T ) = C\ρ(T ). The left spectrum is defined by σl (T ) = {λ ∈ C such that λ − T 6∈ GRl (X)} , and the right spectrum is defined by σr (T ) = {λ ∈ C such that λ − T 6∈ GRr (X)} . The approximate point spectrum of T is the set defined by σap (T ) := {λ ∈ C such that λ − T is not bounded below} . The defect spectrum of T is the set defined by σδ (T ) := {λ ∈ C such that λ − T is not surjective} .

♦

Remark 2.18.1 Let T ∈ LR(X). Then, (i) The resolvent set ρ(T ) is open, whereas the spectrum σ(T ) of a closed linear relation T is closed. (ii) We observe that σ(T ) = σ(Te) = σ(T ) and ρ(T ) = ρ(T ), where Te is given in (2.4) and T is the closure of T .

♦

Lemma 2.18.1 [50, Exercise 6.1.2] Let T ∈ LR(X). Then, ρ(T ) = {λ ∈ C such that λ − T is injective, open with dense range on X} .♦ Lemma 2.18.2 Let S be a closed linear relation in X. Then, (i) If S is bounded, then S n is bounded for all n ∈ N. (ii) If ρ(S) 6= ∅, then S n is closed for all n ∈ N. (iii) If S is everywhere defined and M is a closed subspace of X such that S(0) ⊂ M , then S −1 (M ) is a closed subspace of X. ♦ Remark 2.18.2 Let µ ∈ ρ(T ). Then, for all µ = 6 λ ∈ C and S = (µ − λ)((µ − λ)−1 − Tµ ), we have λ − T = S(µ − T ).

♦

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Proposition 2.18.1 [50, Proposition 6.1.3, and 6.1.11] Let T ∈ LR(X). Then, (i) ρ(T ) is an open set. (ii) [Resolvent Equation] Let λ, µ ∈ ρ(T ). Then, Tµ − Tλ = (λ − µ)Tµ Tλ . (iii) σ(T ) = σ(T ∗ ).

♦

Proposition 2.18.2 [50, Corollary 6.1.8] The family {Tλ such that λ ∈ ρ(T )} of resolvent operators is holomorphic. ♦ Let H(U, X) denote the space of all analytic functions from U into X, where U is open set of C and X is a complex Banach space. Let D(0, ε) denoted the open disc of C centred at 0 and with radius ε.

2.18.2

Subdivision of the Spectrum of Linear Relations

There are many different ways to subdivide the spectrum of linear relation. Some of them are motivated by applications to physics (in particular, quantum mechanics). Definition 2.18.3 Let T ∈ LR(X). The point spectrum, continuous, and the residual spectrum are defined, respectively, as := {λ n ∈ C such that N (Tλ ) 6= 0} , o λ ∈ C such that Tλ is injective and R(Tλ ) 6= X , Rσ(T ) := n o Cσ(T ) := λ ∈ C : Tλ is injective and R(Tλ ) = X but is not open . P σ(T )

♦ Example 2.18.1 Let X = lp (N) (1 ≤ p ≤ ∞), (an )n be a bounded sequence in K, and T ∈ L(lp (N)) be defined by T (x1 , x2 , x3 , . . . ) = (a1 x1 , a2 x2 , a3 x3 , . . . ) . Let first p < ∞. Denoting by Θ = {a1 , a2 , a3 , . . . } the set of all elements of the sequence (an )n , we get σ(T ) = Θ, P σ(T ) = Θ, Cσ(T ) = Θ\Θ and Rσ(T ) = ∅. In particular, if an → 0 (i.e., T is compact single valued), then Cσ(T ) = ∅. In case p = ∞ the continuous and residual spectrum change their role, i.e., we get σ(T ) = Θ, P σ(T ) = Θ, Cσ(T ) = Θ and Rσ(T ) = Θ\Θ.

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95

Remark 2.18.3 (i) We remark that in Example 2.16.2, ρ(T ) = ∅. (ii) It is well known that the spectrum of a bounded operator in a Banach space is a proper subset of C. But a non-single valued bounded linear relation may have a spectrum that coincides with the whole complex plane. But this property is not valid in the case of a linear relation. For example, let X = C2 and let T be defined in the terms of its graph by G(T ) = G(I) + {0} × E, where E may be one two dimensional. Then, T is bounded, T (0) = E and σ(T ) = C. ♦ Example 2.18.2 Let X = l2 and K be the linear single valued defined by   1 1 K(x) = 0, x1 , x2 , x3 , . . . . 2 3 Then, K is an injective compact single valued linear operator with σ(K) = {0}. Now, we defined T ∈ LR(X) by G(T ) = G(K) + {0} × E. Then, T is a bounded compact linear relation. Since N (T ) = K −1 (E), then T T is injective as E R(K) = {0}. Also, T is closed if and only if, E is closed. Now, suppose that E is closed and let λ ∈ C, λ = 6 0. Since λ − K is surjective, then for all 0 6= e ∈ E, there exists x = 6 0 such that (λ − K)x = e. Then, (λ − T )x = E = (λ − T )(0). Hence, λ is an eigenvalue of T . We have constructed a bounded compact closed injective linear relation T such that P σ(T ) = C\{0}, and σ(T ) = C. Moreover, in this example we can choose dim(T (0)) = dim(E) < ∞. Lemma 2.18.3 [91, Proposition 7.3.2] Let X, Y be two normed space and let T ∈ LR(X, Y ). If A is a linear selection of T , then P σ(T ) = P σ(A). ♦ Lemma 2.18.4 [50, Theorem 6.5.4] Let X be a normed space and T ∈ LR(X). Then, for any complex polynomial P , we have σ(P (Te)) = σ(P (T )), where Te is the completion of a linear relation T . ♦ Lemma 2.18.5 Let P (T ) =

n Y i=1

(T − λi )mi as in Definition 2.13.1 and let m

be a positive integer. Then, R(P (T )m ) is closed if and only if, R((T − λi )mi m ) is closed for all i, 1 ≤ i ≤ n. ♦

96

Spectral Theory of Multivalued Linear Operators m

Proof. Note that P (T ) is closed and σ(P (T ) ) = (σ(P (T ))m by virtue of Lemma 2.18.4. So, P (T )m is closed and bounded by Lemma 2.18.2. Assume that R(P (T )m ) is closed and we write U := (T − λ1 )m1 and V :=

n Y

(T − λi )mi .

i=2

Then, arguing as in Lemma 2.13.1 (iv) and using Lemma 2.18.2 together with the condition R(P (T )m ) is closed, we obtain that V −m R(P (T )m ) = R(U m ) + N (V m ) = R(U m ), that is, R((T − λ1 )mi m ) is closed. Now, applying the above reasoning, we obtain that R((T − λi )mi m ) is closed, 1 ≤ i ≤ n. The converse follows immediately from Lemmas 2.5.1 and 2.13.1.

2.18.3

Q.E.D.

Essential Spectra of Linear Relations

In this book, we are concerned with the following essential spectra (see Refs. [1, 17, 26, 65, 66, 93]): σe1 (T ) = {λ ∈ C such that λ − T 6∈ Φ+ (X)} , σe2 (T ) = {λ ∈ C such that λ − T 6∈ Φ− (X)}, σe3 (T ) = {λ ∈ C such that λ − T 6∈ Φ± (X)}, σe4 (T ) = {λ ∈ C such that λ − T 6∈ Φ(X)}, \ σe5 (T ) = σ(T + K), K∈KT (X)

σqφd (T ) = {λ ∈ C such that λ − T 6∈ qφd (X)}, σeb (T ) = {λ ∈ C such that λ − T 6∈ B(X)} \ σeap (T ) = σap (T + K), K∈KT (X)

σeδ (T ) =

\

σδ (T + K),

K∈KT (X)

σeα (T ) = {λ ∈ C such that λ − T 6∈ Aα (X)}, σeβ (T ) = {λ ∈ C such that λ − T 6∈ Aβ (X)}, \ σel (T ) = σl (T + K), K∈KT (X)

σer (T ) =

Fundamentals \ σr (T + K),

97

K∈KT (X)

σr (T ) :=

\

σ(T + K).

K∈ΥT (X)

σl (T ) =

\

σ(T + K),

K∈ΨT (X)

with KT (X) = {K ∈ KR(X) such that D(T ) ⊂ D(K), K(0) ⊂ T (0)} , Λ(X) ΥT (X)

= {T ∈ LR(X) : µT is demicompact for every µ ∈ [0, 1]} ,   D(T ) ⊂ D(K),  K ∈ LR(X) : = K(0) ⊂ T (0) and ∀µ ∈ ρ(T + K),   −(µ − T − K)−1 K ∈ Λ(X)

  

,

 

and

ΨT (X)

=

    

K ∈ LR(X) :

D(T ) ⊂ D(K), K(0) ⊂ T (0) and ∀µ ∈ ρ(T + K), −K(µ − T − K)−1 ∈ Λ(X)

  

.

 

In general those spectra satisfy the following inclusions \ σe1 (T ) σe2 (T ) = σe3 (T ) ⊂ σe4 (T ) ⊂ σe5 (T ) ⊂ σ(T ). σe1 (T ) ⊂ σeap (T ), σe2 (T ) ⊂ σeδ (T ), σe1 (T ) ⊂ σeα (T ), σe2 (T ) ⊂ σeβ (T ), and [ σe5 (T ) = σeap (T ) σeδ (T ). Remark 2.18.4 (i) The sets σei (T ) are closed, with i = 1, . . . , 5, ap, δ, α, β, l, r. (ii) It is clear that, for all K ∈ KT (X), we have σei (T + K) = σei (T ), with i = 5, ap, δ, l, r. (iii) For all K ∈ KT (X), σei ((T + K)∗ ) = σei (T ∗ + K ∗ ) = σei (T ∗ ), with i = 5, ap, δ, l, r. (iv) If K is a single valued compact operator, then σei (K) = {0}, with i = 1, . . . , 5, α, β, l, r. But this is not true if K is a compact linear relation. In fact, let K ∈ KR(X) be a bounded linear relation such that 0 6= dim K(0) < ∞. By using Theorem 2.16.2, we get λ − K ∈ Φ(X) and i(λ − K) = dim K(0). So, σe5 (K) = C. ♦

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2.19

S-Spectra of Linear Relations in Normed Space

Definition 2.19.1 Let X be a normed vector space, S, T ∈ LR(X), S be a continuous linear relation such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). Then, we define the S-resolvent set of T by ρS (T ) := {λ ∈ C such that λS − T is injective, open with dense range on X} . We denote the S-spectrum set of T by σS (T ) = C\ρS (T ).

♦

Remark 2.19.1 It is clear that if S, T ∈ CR(X) and X is complete, we will return to the S-spectrum definition in a Banach space with closed linear relation. In this case ρS (T )

= {λ ∈ C such that λS − T is bijective}  = λ ∈ C : (λS − T )−1 is a bounded linear operator on X . ♦

Definition 2.19.2 Let S, T ∈ LR(X) and λ ∈ C. We recall the S-resolvent of T at λ the operator defined by RS (λ, T ) := (λS − T )−1 .

♦

Definition 2.19.3 The augmented S-spectrum of T is the set ( σ S (T ) =

S σS (T ) {∞} if 0 ∈ σS −1 (T −1 ) σS (T ) otherwise.

♦

The Mobius transformation is defined by ηeµ (λ) = (µ − λ)−1 , where µ is a fixed point of C, and a topological homeomorphism from C∞ onto itself. Proposition 2.19.1 Let λ ∈ C, T ∈ CR(X) and S ∈ L(X). Let µ ∈ ρS (T ) such that λ 6= µ. Then, N (λS − T ) = N ((µ − λ)−1 − (µS − T )−1 S).

♦

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99

Proof. Let µ ∈ ρS (T ) and λ ∈ C such that µ = 6 λ. Let (x, y) ∈ G(T −T +(µ− λ)S), then x ∈ D(T ) and y ∈ (T − T + (µ − λ)S)x. So, y ∈ T (0) + (µ − λ)Sx. Since T (0) ⊂ (λS − T )x and S is a single valued linear operator, we obtain y ∈ (λS − T )x + (µ − λ)Sx = (µS − T )x. Hence, (x, y) ∈ G(µS − T ). We infer that G(T − T + (µ − λ)S) ⊂ G(µS − T ). Furthermore, (T − T + (µ − λ)S)(0) = T (0) = (µS − T )(0) and clearly T D(T − T + (µ − λ)S) = D(T ) D(S) = D(T ) = D(µS − T ). Hence, by using Proposition 2.3.6 (iii), we have T − T + (µ − λ)S = µS − T . So, T (0) + (µ − λ)Sx = (λS − T )x + (µ − λ)Sx = (µS − T )x.

(2.58)

Thus, x ∈ N (λS − T ) if and only if, x ∈ D(λS − T ) = D(T ) and (λS − T )x = (λS −T )(0) = T (0) if and only if, (µS−T )x = (λS −T )x−λSx+µSx = T (0)− λSx+µSx if and only if, (µS −T )−1 (µS −T )x = (µS −T )−1 ((µ−λ)Sx+T (0)) (by (2.58)) if and only if, x + (µS − T )−1 (0) = (µ − λ)(µS − T )−1 Sx + (µS − T )−1 (µS − T )(0) if and only if, 0 = (I − (µ − λ)(µS − T )−1 S)x (since (I −(µ−λ)(µS −T )−1 S)(0) = 0) if and only if, ((µ−λ)−1 −(µS −T )−1 S)x = 0 if and only if, x ∈ N ((µ − λ)−1 − (µS − T )−1 S). Q.E.D. Corollary 2.19.1 Let µ ∈ ρ(T ) and let µ 6= λ. Then,
N (λ − T ) = N (µ − λ)−1 − (µ − T )−1 . In particular, λ is an eigenvalue of T if and only if, (µ − λ)−1 is an eigenvalue of (µ − T )−1 . ♦

2.19.1

Some Properties of S-Resolvent of Linear Relations in Normed Space

Proposition 2.19.2 Let T ∈ LR(X) and S be a continuous linear relation such that S(0) ⊂ T (0), D(T ) ⊂ D(S). If kSk = 0, then ρS (T ) = ∅ or C. ♦ Proof. Let kSk = 0. We suppose that ρS (T ) 6= ∅ and ρS (T ) 6= C. Let λ0 ∈ ρS (T ). Then, λ0 S − T is injective, open with dense range. Hence, γ(T − λ0 S) > 0. It is clear to see kλ0 Sk = |λ0 |kSk = 0. So, kλ0 Sk < γ(T − λ0 S). Then, by using Lemma 2.9.3, we have λ0 S + T − λ0 S is injective, open with dense range. Moreover, λ0 S(0) ⊂ T (0) and D(T ) ⊂ D(S) = D(λ0 S). Thus, by using Proposition 2.3.4, λ0 S + T − λ0 S = T is injective open with dense range. Now, let λ ∈ C, then kλSk < γ(T ). Hence, λS − T is injective, open

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Spectral Theory of Multivalued Linear Operators

with dense range. Thus, ρS (T ) = C, which is a contradiction. This completes the proof. Q.E.D. Corollary 2.19.2 Let S, T ∈ LR(X) and S be a continuous linear relation such that S(0) ⊂ T (0) and D(T ) ⊂ D(S). If S(BD(S) ) ⊂ S(0), then ρS (T ) = ∅ or C. ♦ The following proposition show that the S-spectrum of linear relation may be a proper subset of C. Theorem 2.19.1 Let T ∈ CR(X) be injective with dense range and S be a continuous linear relation such that S(0) ⊂ T (0), D(S) ⊃ D(T ) and kSk 6= 0. Then,   γ(T ) σS (T ) ⊂ λ ∈ C such that |λ| ≥ . ♦ kSk n o ) Proof. It suffices to show that λ ∈ C such that |λ| < γ(T ⊂ ρS (T ). Let kSk λ ∈ C such that |λ|
0, then T is open. On the other

hand, λS(0) = S(0) ⊂ T (0) and D(λS) = D(S) ⊃ D(T ). Moreover, kλSk = |λ|kSk < γ(T ) and T is open, injective with dense range. So, by using Lemma 2.9.3, we obtain that λS − T is open, injective with dense range. Hence, λ ∈ ρS (T ). This completes the proof. Q.E.D. Theorem 2.19.2 Let T ∈ LR(X) and S be a continuous linear relation such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). Then, ρS (T ) is an open set. ♦ Proof. We will discuss two cases. First case : kSk = 0, by using Proposition 2.19.2, we have ρS (T ) = ∅ or ρS (T ) = C. Thus, ρS (T ) is open. Second case : kSk 6= 0, if λ ∈ ρS (T ), then γ(λS − T ) > 0. Let µ ∈ C such that −T ) |µ − λ| < γ(λS , then k(µ − λ)Sk = |µ − λ|kSk < γ(λS − T ). Furthermore, kSk (µ − λ)S(0) ⊂ T (0) = (λS − T )(0) and D((µ − λ)S) = D(S) ⊃ D(T ) = D(λS − T ). Thus, by using Lemma 2.9.3, µS − T is injective, open with dense range. So, µ ∈ ρS (T ). Therefore, ρS (T ) is open. Q.E.D. Theorem 2.19.3 Let X be a Banach space. Let T ∈ CR(X) and S ∈ BR(X) such that D(T ) = X, S(0) = T (0) and dim S(0) < ∞. If 0 ∈ ρ(S), then T ρS (T ) = ρ(S −1 T ) ρ(T S −1 ). ♦

Fundamentals

101

Proof. First of all, it should be mentioned that λ − S −1 T and λ − T S −1 are closed for all λ ∈ C. Indeed, since S is continuous and D(S) and S(0) are closed, then by using Theorem 2.7.2 (iv), we have S is closed. Hence, the fact that 0 ∈ ρ(S) implies that S −1 ∈ L(X). So, S −1 is closed . Further, α(S −1 ) = dim(N (S −1 )) = dim(S(0) < ∞. By using Lemma 2.9.1, we have γ(S −1 ) = kSk−1 > 0. Hence, R(S −1 ) is closed. So, we infer from Proposition 2.9.4 that S −1 T is closed. Now, applying Proposition 2.7.2, we get λ − S −1 T is closed. Moreover, λ − T S −1 is closed. Indeed, since T is closed and S −1 is bounded operator (as 0 ∈ ρ(S)), then by applying Lemma 2.7.1, we conclude that T S −1 is closed, which implies by using Proposition 2.7.2, that λ − T S −1 is closed. Now, let λ ∈ ρS (T ). We claim that λ − S −1 T and λ − T S −1 are bijective. Indeed, we first verify λ − S −1 T = S −1 (λS − T ),

(2.59)

λS − T = S(λ − S −1 T ).

(2.60)

and −1

Since S is injective, then S S = ID(S) = IX . Hence, by using the fact that D(S −1 ) = X and Proposition 2.3.6 (v), we have S −1 (λS − T )

= λS −1 S − S −1 T = λ − S −1 T.

Hence, (2.59) holds. Now, we prove (2.60). Let x ∈ X, then x ∈ D(S −1 ) = R(S) = X. So, there exists a ∈ X such that (a, x) ∈ G(S) and (x, a) ∈ G(S −1 ). Thus, (x, x) ∈ G(SS −1 ). This implies that G(IX ) ⊂ G(SS −1 ). Then, λS − T ⊂ SS −1 (λS − T ) = S(λ − S −1 T ). Consequently, λS − T ⊂ S(λ − S −1 T ).

(2.61)

On the other hand, D(S(λ − S −1 T ))

= = = = = =

n o \ x ∈ D(λ − S −1 T ) : (λ − S −1 T )x D(S) 6= ∅ n o \ x ∈ D(λ − S −1 T ) : (λ − S −1 T )x X 6= ∅ n o \ x ∈ D(S −1 T ) : S −1 T x X 6= ∅ n o \ x ∈ D(T ) : T x D(S −1 ) 6= ∅ n o \ x ∈ D(T ) : T x X 6= ∅ n o \ x ∈ D(λS − T ) : (λS − T )x X 6= ∅

= D(λS − T ).

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Spectral Theory of Multivalued Linear Operators

In addition, we have S(λ − S −1 T )(0)

= S((λ − S −1 T (0)) = SS −1 T (0) = SS −1 S(0) = SS −1 (0) = S(0) = T (0) =

(λS − T )(0).

Thus, these properties and (2.61) implies from Proposition 2.3.6 (iii) that (2.60) holds. Now, using (2.59), we have R(λ − S −1 T )

= D((S −1 (λS − T ))−1 ) = D((λS − T )−1 S) = S −1 (D((λS − T )−1 )) = S −1 (R(λS − T )).

Since λS − T is open with dense range, then R(λS − T ) = R(λS − T ) = X. This implies that R(λ − S −1 T )

= R(S −1 ) = D(S) = X.

On the other hand, N (λ − S −1 T )

= N (S −1 (λS − T )), =

(λS − T )−1 S(0),

=

(λS − T )−1 T (0)

=

(λS − T )−1 (λS − T )(0),

=

(λS − T )−1 (0)

= {0}. Then, λ − S −1 T is bijective. So, λ ∈ ρ(S −1 T ). In the same ways, we can prove that λ − T S −1 is bijective. Accordingly, λ − T S −1 and λ − S −1 T are bijective and closed, then \ ρS (T ) ⊂ ρ(T S −1 ) ρ(S −1 T ). (2.62)

Fundamentals 103 T Conversely, let λ ∈ ρ(S −1 T ) ρ(T S −1 ), then λ − S −1 T and λ − T S −1 are bijective. Since S is surjective, then by using (2.60), we have R(λS − T ) = SR(λ − S −1 T ) = R(S) = X and N (λS − T ) ⊂ N (S −1 (λS − T )) = N (λ − S −1 T ) = {0}. This implies that λS − T is bijective. Thus, \ (2.63) ρ(T S −1 ) ρ(S −1 T ) ⊂ ρS (T ). By using (2.62) and (2.63), we conclude that ρ(T S −1 ) This completes the proof.

T

ρ(S −1 T ) = ρS (T ). Q.E.D.

Theorem 2.19.4 [The S-resolvent identity] Let T ∈ CR(X) and let S ∈ BR(X) such that S(0) ⊂ T (0). Then, for all λ, µ ∈ ρS (T ), we have RS (µ, T ) − RS (λ, T ) = (λ − µ)RS (µ, T )SRS (λ, T ).

♦

Proof. Let λ, µ ∈ ρS (T ) and x ∈ X. On the one hand, we have (λ − µ)RS (µ, T )SRS (λ, T )x = RS (µ, T )(λS − µS)RS (λ, T )x. By using Remark 2.3.3, we have RS (µ, T )T (0) = (µS − T )−1 (µS − T )(0) = (µS − T )−1 (0) = 0. It follows that (λ − µ)RS (µ, T )SRS (λ, T )x = RS (µ, T )((λS − µS)RS (λ, T )x + T (0)). Also, the fact that (λS − T )RS (λ, T )x − (λS − T )RS (λ, T )x = T (0) leads from Proposition 2.3.6 (iv) to (λ − µ)RS (µ, T )SRS (λ, T )x =h i RS (µ, T ) (λS − µS) + (λS − T ) − (λS − T ) RS (λ, T )x. Since RS (λ, T ) is a single valued linear operator, then by using Proposition 2.3.6 (iv), we have (λ − µ)RS (µ, T )SRS (λ, T )x h i = RS (µ, T ) (λS − µS)RS (λ, T )x + (λS − T )(0) h i = RS (µ, T ) (λS − µS)RS (λ, T )x + T (0) h i = RS (µ, T ) (λS − µS − T + T )RS (λ, T )x hh i i = RS (µ, T ) (λS − T ) − (µS − T ) RS (λ, T )x h i = RS (µ, T ) (λS − T )RS (λ, T )x − (µS − T )RS (λ, T )x . Since D(RS (µ, T )) = X, then by using Proposition 2.3.6 (v), we have

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(λ − µ)RS (µ, T )SRS (λ, T )x = RS (µ, T )(λS − T )RS (λ, T )x − RS (µ, T )(µS − T )RS (λ, T )x. Now, the use of Theorem 2.3.1 realizes that RS (µ, T )(λS − T )RS (λ, T )x = RS (µ, T )(x + T (0)). The fact that µS − T is injective implies from Proposition 2.3.2 that RS (µ, T )(µS − T )RS (λ, T )x = RS (λ, T )x. Hence, by using Remark 2.3.3, we have RS (µ, T )T (0) = RS (µ, T )(µS − T )(0) = 0. Then, we conclude that (λ − µ)RS (µ, T )SRS (λ, T )x = RS (µ, T )x + RS (µ, T )T (0) − RS (λ, T )x = RS (µ, T )x − RS (λ, T )x.

(2.64)

On the other hand, D(RS (µ, T )SRS (λ, T ))

=

(SRS (λ, T ))−1 D(RS (µ, T ))

=

(RS (λ, T ))−1 S −1 X

=

(RS (λ, T ))−1 D(S)

=

(RS (λ, T ))−1 X

= D(RS (λ, T )) = D(RS (µ, T ) − RS (λ, T )). In addition, we have (λ − µ)RS (µ, T )SRS (λ, T )(0)

=

(λ − µ)RS (µ, T )S(0)

⊂ (λ − µ)RS (µ, T )T (0) = {0}. Hence, (λ − µ)RS (µ, T )SRS (λ, T )(0) = (RS (µ, T ) − RS (λ, T ))(0). Thus, from these properties, Proposition 2.3.6 (iii), and (2.64) implies that RS (µ, T ) − RS (λ, T ) = (λ − µ)RS (µ, T )SRS (λ, T ).

Q.E.D.

Theorem 2.19.5 Let T ∈ CR(X) and S ∈ L(X). Assume for all λ ∈ ρS (T ), RS (λ, T ) commutes with S. Then,
lim (µ − λ)−1 (RS (µ, T ) − RS (λ, T )) = −RS (λ, T )SRS (λ, T ). ♦ µ→λ

Fundamentals

105

Proof. Let λ ∈ ρS (T ) and µ 6= λ in such a neighborhood, then by using Theorem 2.19.2, we have µ ∈ ρS (T ). Since RS (λ, T ) is a single valued linear operator, then it follows from Theorem 2.19.4 that (µ − λ)−1 (RS (µ, T ) − RS (λ, T )) + RS (λ, T )SRS (λ, T ) = −RS (µ, T )SRS (λ, T ) + RS (λ, T )SRS (λ, T ) =

(RS (λ, T )S − RS (µ, T )S)RS (λ, T ).

The fact that S commutes with RS (λ, T ) for all λ ∈ ρS (T ), then (µ − λ)−1 (RS (µ, T ) − RS (λ, T )) + RS (λ, T )SRS (λ, T ) = (SRS (λ, T ) − SRS (µ, T ))RS (λ, T ). This implies from Proposition 2.3.6 (iv) that (µ − λ)−1 (RS (µ, T ) − RS (λ, T )) + RS (λ, T )SRS (λ, T ) = S(RS (λ, T ) − RS (µ, T ))RS (λ, T ). Hence, k(µ − λ)−1 (RS (µ, T ) − RS (λ, T ) + RS (λ, T )SRS (λ, T )k ≤ kSkkRS (λ, T ) − RS (µ, T )kkRS (λ, T )k. Now, suppose that |µ − λ|kSk < kRS (λ, T )k−1 . Since (λ − µ)SRS (λ, T ) is a single valued linear operator, then X |µ − λ|n kSkn kRS (λ, T )kn < ∞. n≥0

Hence, (I − (λ − µ)SRS (λ, T ))−1 =

X

(µ − λ)n S n RS (λ, T )n .

n≥0

Since RS (λ, T ) and S commute, then for x ∈ D(T ), we have (λ − µ)RS (λ, T )−1 SRS (λ, T )x = (λ − µ)RS (λ, T )−1 RS (λ, T )Sx. By using Theorem 2.3.1 (iii), we get (λ − µ)RS (λ, T )−1 SRS (λ, T )x = (λ − µ)Sx. Since D(RS (λ, T )−1 ) = X, then it follows from Proposition 2.3.6 (iv) that RS (λ, T )−1 (I − (λ − µ)SRS (λ, T ))x

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Spectral Theory of Multivalued Linear Operators

= RS (λ, T )−1 x + (µ − λ)RS (λ, T )−1 SRS (λ, T ))x = RS (λ, T )−1 x + (µ − λ)Sx = λSx − T x + µSx − λSx =

(µS − T )x.

Hence, RS (λ, T )−1 (I − (λ − µ)SRS (λ, T ))x = RS (µ, T )−1 x. Since D(RS (λ, T )−1 (I − (λ − µ)SRS (λ, T )) = D(RS (µ, T )−1 ) = D(T ), then we conclude that RS (µ, T ) = (I − (λ − µ)SRS (λ, T ))−1 RS (λ, T ). So, in the first place, we have RS (µ, T ) =

X

(λ − µ)n S n RS (λ, T )n+1 .

n≥0

In the second place, we have RS (µ, T ) − RS (λ, T )

=

X

(λ − µ)n S n RS (λ, T )n+1 − RS (λ, T )

n≥0

X

= RS (λ, T )

(λ − µ)n S n RS (λ, T )n .

n≥1

Hence, kRS (µ, T ) − RS (λ, T )k ≤ kRS (λ, T )k

X

|λ − µ|n kSkn kRS (λ, T )kn .

n≥1

Arguing as above we conclude that lim (µ − λ)−1 (RS (µ, T ) − RS (λ, T )) = −RS (λ, T )SRS (λ, T ).

µ→λ

Q.E.D.

Corollary 2.19.3 Let T ∈ CR(X) and S ∈ L(X). If for all λ ∈ ρS (T ), we have RS (λ, T ) commutes with S, then the function ϕ : λ −→ RS (λ, T ) is holomorphic for all λ ∈ ρS (T ). ♦ Proof. If λ, µ ∈ ρS (T ), then   ϕ(µ) − ϕ(λ) lim = −RS (λ, T )SRS (λ, T )∈ L(X). µ→λ µ−λ

Q.E.D.

Fundamentals

107

Corollary 2.19.4 Let T ∈ CR(X) and S ∈ L(X). Assume that for all λ ∈ ρS (T ), we have RS (λ, T ) commutes with S. If |λ − µ|kSk 0 and Proposition 2.8.1 (iv), R((λS − T )∗ ) = N (λS − T )⊥ = {0}⊥ = X ∗ . Therefore, (λS − T )∗ is injective, open with dense range. It remains to show that (λS − T )∗ = λS ∗ − T ∗ . Using Proposition 2.8.1 (vi), λS is continuous, and D(λS) = D(S) ⊃ D(T ). This allows us to conclude that (λS − T )∗ = (λS)∗ − T ∗ = λS ∗ − T ∗ .

(2.67)

So, by using (2.67), we have λ ∈ ρS ∗ (T ∗ ). Then, σS ∗ (T ∗ ) ⊂ σS (T ). Conversely, let λ ∈ ρS ∗ (T ∗ ), then λS ∗ − T ∗ = (λS − T )∗ is injective, open with dense range. In view of Proposition 2.8.1 (i), (λS − T )∗ is closed. Then, by applying (2.67), we infer that λS ∗ − T ∗ is a closed linear relation. Using the fact that (λS − T )∗ is open and Theorem 2.5.4 (ii), R((λS − T )∗ ) is closed. This implies that R((λS − T )∗ ) = X ∗ . Moreover, by using Proposition 2.8.1,

Fundamentals

109

>

R(λS − T ) = (R(λS − T )⊥ ) = N ((λS − T )∗ )> = {0}> = X. On the other hand, by referring to Proposition 2.8.1, N (λS − T ) = R((λS − T )∗ )> = X ∗> = {0}. Since λS − T is closed (see Proposition 2.7.2), then N (λS − T ) N (λS − T ) = {0}. So,

=

R((λS − T )∗ ) = R((λS − T )∗ ) = X ∗ . In addition, based on the assumption that λS − T is closed, we can conclude that (λS − T )∗∗ = λS − T . Therefore, λS − T is injective, open with dense range. Thus, λ ∈ ρS (T ). As a result σS (T ) ⊂ σS ∗ (T ∗ ). Q.E.D.

2.19.2

The Augmented S-Spectrum

Theorem 2.19.8 Let T ∈ CR(X) and S ∈ L(X). Let µ ∈ ρS (T ), then ηeµ (σS (T )) = σ(SRS (µ, T )).

♦

Proof. Without loss of generality, we assume that X is complete, T is closed and S is continuous. Let λ ∈ C, and µ ∈ ρS (T ) with λ 6= µ. Then, λS − T

=

(µS − T ) − (µ − λ)S since S is single valued

=

(I − (µ − λ)SRS (µ, T ))(µS − T ) since µ ∈ ρS (T )

=

(µ − λ)((µ − λ)−1 − S(µS − T )−1 )(µS − T )

= A(µS − T ),

(2.68)

where A := (µ−λ)((µ−λ)−1 −SRS (µ, T )). We shall verify that A is injective. Suppose that λ ∈ ρS (T ). Let x ∈ N (A), then Ax = 0 (since A(0) = 0). This implies (µ − λ)−1 x − SRS (µ, T )x = 0. Hence, (µ − λ)−1 x = SRS (µ, T )x. So, (µ − λ)−1 (µS − T )x = Sx + (µS − T )(0). Thus, (µS − T )x = (µ − λ)Sx + (µS − T )(0). This gives that µSx − T x = µSx − λSx + T (0). Therefore, (λS − T )x = (λS − T )(0). Then, x ∈ N (λS − T ). Thus, x = 0. So, A is injective. Next, the use of (2.68) allows us to conclude that X = R(λS − T ) = R(A(µS − T )) ⊂ R(A). Hence, A is surjective. As required, since A is both bijective and open, then it follows that (µ − λ)−1 ∈ ρ(SRS (µ, T )). Conversely, let (µ − λ)−1 ∈

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Spectral Theory of Multivalued Linear Operators

ρ(SRS (µ, T )). For x ∈ D(T ), by using (2.68), we have k(λS − T )xk = kA(µS − T )xk ≥ γ(A(µS − T ))dist(x, RS (µ, T )A−1 (0)). Since A is injective, then RS (µ, T )A−1 (0) = RS (µ, T )(0) = {0}. Therefore, k(λS − T )xk ≥ γ(A(µS − T ))kxk.

(2.69)

As A is injective, then A−1 (0) = {0} ⊂ R(µS − T ). In view of Theorem 2.9.1 (ii), we have γ(A(µS − T )) ≥ γ(A)γ(µS − T ). Hence, γ(A) > 0 from the hypothesis, and γ(µS − T ) > 0 since µ ∈ ρS (T ). Hence, it is easy to verify from (2.68) and (2.69) that λS −T is injective, surjective and open. We deduce that λ ∈ ρS (T ). Q.E.D. Proposition 2.19.4 Let S and T ∈ LR(X) be commute such that D(T ) = D(S). Then, for all λ and µ ∈ C, we have λS − T and µS − T commutes. ♦ Proof. On the one hand, we have D((µS − T )(λS − T ))

=

(λS − T )−1 D((µS − T ))

=

(λS − T )−1 D(T )

= D(T (λS − T )) = D((λT S − T 2 )) \ = D(T S) D(T 2 ). Since S commute with T , we infer that D(T S) = D(ST ) = T −1 D(S) = T −1 D(T ) = D(T 2 ). Hence, the domains of (µS − T )(λS − T ) and (λS − T )(µS − T ) are each equal to D(T 2 ). On the other hand, for x ∈ D(T 2 ), we have (λS − T )(µS − T )x = =

(λS − T )(µSx − T x) (λS − T ){µy1 − y2 : y1 ∈ Sx and y2 ∈ T x}

= {λµSy1 − λSy2 − µT y1 + T y2 : y1 ∈ Sx and y2 ∈ T x} = λµS 2 x − λST x − µT Sx + T 2 x. Since T and S commute, then (λS − T )(µS − T )x = λµS 2 x − λT Sx − µST x + T 2 x = This completes the proof.

(µS − T )(λS − T )x. Q.E.D.

Fundamentals

2.19.3

111

S-Essential Resolvent Sets of Multivalued Linear Operators

Definition 2.19.4 Let X be a normed vector space, T ∈ LR(X), S be a continuous linear relation such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). The Sessential resolvent sets of T , where we define by ρei,S (T ), i = 1, 2, 3, 4, 5, 6 as follows ρe1,S (T ) ρe2,S (T ) ρe3,S (T ) ρe4,S (T ) ρe5,S (T ) ρe6,S (T )

= = = = = =

{λ ∈ C {λ ∈ C {λ ∈ C {λ ∈ C {λ ∈ C {λ ∈ C

: : : : : :

λS − T ∈ Φ+ (X)} , λS − T ∈ Φ− (X)} , S λS − T ∈ Φ+ (X) Φ− (X)} , T λS − T ∈ Φ+ (X) Φ− (X)} , T λS − T ∈ Φ+ (X) Φ− (X) and i(λS − T ) = 0} , and λ ∈ ρe5,S (T ) such that all scalars near λ are in ρS (T )} .

The S-essential spectra σei,S (T ), i = 1, 2, 3, 4, 5, 6 are the complement of respective ρei,S (T ). ♦ Definition 2.19.5 Let T ∈ CR(X) and S ∈ BR(X) such that S(0) ⊂ T (0) and S = 6 0. We define \ σeap,S (T ) = σap,S (T + K) K∈KT (X)

and σeδ,S (T ) =

\

σδ (T + K)

K∈KT (X)

with σap,S (T ) = {λ ∈ C such that λS − T not bounded below} , where bounded below is injective and open and σδ,S (T ) := {λ ∈ C such that λS − T is not surjective} .

2.20 2.20.1

♦

Pseudospectra and Essential Pseudospectra of Linear Relations Pseudospectra of Linear Relations

Definition 2.20.1 Let T ∈ LR(X), where X is a normed vector space and ε > 0. We define the pseudospectra of T by

112

Spectral Theory of Multivalued Linear Operators  [ 1 σε (T ) = σ(T ) λ ∈ C such that kTλ k > . ε

We denote the pseudoresolvent set of T by   T 1 ρε (T ) = C\σε (T ) = ρ(T ) λ ∈ C such that kTλ k ≤ . ε

♦

Remark 2.20.1 Let X be a normed vector space, T ∈ LR(X), and ε > 0. Then, (i) If ε1 < ε2 , then σε1 (T ) ⊂ σε2 (T ). (ii) If X is complete and T ∈ CR(X), then   S 1 −1 σε (T ) = σ(T ) λ ∈ C such that k(λ − T ) k > . ε

♦

Definition 2.20.2 Let X be a normed vector space, T ∈ LR(X) and ε > 0. The ε-pseudospectrum of T is defined by   S 1 Σε (T ) := σ(T ) λ ∈ C such that k(λ − Te)−1 k ≥ . ♦ ε Remark 2.20.2 (i) Let X be a normed vector space, T ∈ LR(X) and ε > 0. Then,   1 −1 e (i1 ) Σε (T )\σε (T ) = λ ∈ C such that k(λ − T ) k = . ε (i2 ) Σε (T ) is a closed subset of the complex plane. (i3 ) If 0 < ε1 < ε2 , then σ(T ) ⊂ Σε1 (T ) ⊂ Σε2 (T ). (ii) Let X be a non zero normed space, ε > 0 and let T be a continuous and densely defined linear relation in LR(X). Then, σ(T ) 6= ∅. Hence, σε (T ) 6= ∅ and Σε (T ) 6= ∅. (iii) If X is a Banach space and T ∈ CR(X), then   S 1 −1 . Σε (T ) = σ(T ) λ ∈ C such that k(λ − T ) k ≥ ε

♦

Example 2.20.1 Let us notice that the boundary value problem for an elliptic-parabolic equation can be formulated, in a Hilbert space L2 (Ω) with Ω is a bounded domain in R2 of class C 2 , as the following multivalued Cauchy problem on L2 (Ω)  n X ∂v   ∂v ∈ LM −1 v + ai (x) + f (t), t ∈ (0, τ ] ∂t ∂ xi i=1   v(0) = v0 ,

Fundamentals

113

where f (t) := f (t, ·), v0 = v0 (·), M −1 is the inverse of the operator M defined by M u = m(·)u, for all u ∈ X, \ Lu = ∇ • (a(·)∇u) + c0 (·), for all u ∈ H 2 (Ω) H01 (Ω), ∇ denotes the gradient vector with respect to x variable, m(x) ≥ 0 on Ω and m ∈ C 2 (Ω), a(x), ai (x), c0 (x) are real-valued smooth functions on Ω. Putting Pn ∂v Sv = i=1 ai (x) , for all v ∈ H 1 (Ω) and T = LM −1 . Then, ∂ xi [ Σε (T ) ⊂ (] − ∞, −c1 [+iR) D(0, ε2 c22 ), (2.70)  where c1 , c2 > 0, i2 = −1, and D(0, ε2 c22 ) = λ ∈ C : |λ| ≤ ε2 c22 . Indeed, it follows from Ref. [51] that, for any given λ ∈ C satisfying −c1 (1 + |Imλ|) ≤ Reλ, for some c1 > 0, we have (λ − T )−1 is an operator. Hence, there exists c2 > 0 such that 1

k(λ − T )−1 k ≤ c2 (1 + |λ|)− 2−r , for some r ∈ (0, 1). Keeping in mind that −

(2.71)

1 ln(1 + |λ|) ≤ 0. Then, we infer from (2.71) that 2−r k(λ − T )−1 k ≤ c2 .

Therefore, {λ ∈ C : −c1 (1 + |Imλ|) ≤ Reλ} ⊂ ρ(T ). This implies that  [ 1 λ ∈ C : k(λ − T )−1 k ≥ . ε (2.72) We divide this part of the proof into two steps. Σε (T ) ⊂ {λ ∈ C : −c1 (1 + |Imλ|) > Reλ}

Step 1. We show that  [ 1 λ ∈ C : k(λ − T )−1 k ≥ ⊂ Γε , ε (2.73) S 2 2 where Γε = {λ ∈ C : −c1 (1 + |Imλ|) > Reλ} λ ∈ C : |λ| ≤ ε c − 1 . Let 2  S 1 λ ∈ {λ ∈ C : −c1 (1 + |Imλ|) > Reλ} λ ∈ C : k(λ − T )−1 k ≥ . This is ε 1 equivalent to saying that −c1 (1 + |I mλ|) > Reλ or k(λ − T )−1 k ≥ . If ε {λ ∈ C : −c1 (1 + |Imλ|) > Reλ}

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Spectral Theory of Multivalued Linear Operators

−c1 (1 + |Imλ|) > Reλ, then it is clear that λ ∈ Γε . At this level, let us 1 assume that −c1 (1 + |Imλ|) ≤ Reλ and k(λ − T )−1 k ≥ . Then, it follows ε from (2.71) that 1 1 ≤ c2 (1 + |λ|)− 2−r . ε This leads to 1 (1 + |λ|) 2−r ≤ εc2 . (2.74) 1

1

1

Since (1 + |λ|) 2 ≤ (1 + |λ|) 2−r , then by using (2.74), we have (1 + |λ|) 2 ≤ εc2 . Hence, |λ| ≤ ε2 c22 − 1. This implies that (2.73) holds. Step 2. We show that Γε ⊂ (] − ∞, −c1 [+iR)

[

D(0, ε2 c22 ),

where i2 = −1. Let us assume that λ ∈ Γε . We discuss two cases. First case : If Reλ < −c1 (1 + |Imλ|), then Reλ < −c1 . This implies that λ ∈ (] − ∞, −c1 [+iR) .
Second case : If |λ| ≤ ε2 c22 − 1, then |λ| ≤ ε2 c22 . Hence, λ ∈ D 0, ε2 c22 . Thus, the use of (2.72) and (2.73) makes us conclude that [ Σε (T ) ⊂ Γε ⊂ (] − ∞, −c1 [+iR) D(0, ε2 c22 ). Therefore, (2.70) holds.

2.20.2

Essential Pseudospectra of Linear Relations

Definition 2.20.3 Let ε > 0 and T ∈ CR(X). We define the essential pseudospectra of T by

σe1,ε (T )

= C\ {λ ∈ C such that λ − T + S ∈ Φ+ (X), for all S ∈ UT (X)} ,

σe2,ε (T )

= C\ {λ ∈ C such that λ − T + S ∈ Φ− (X), for all S ∈ UT (X)} ,

σe3,ε (T )

= C\ {λ ∈ C such that λ − T + S ∈ Φ± (X), for all S ∈ UT (X)} ,

σe4,ε (T )

= C\ {λ ∈ C such that λ − T + S ∈ Φ(X), for all S ∈ UT (X)} , \ = σε (T + K),

σe5,ε (T )

K∈KT (X)

where UT (X) = {S ∈ LR(X) is continuous such that kSk < ε, S(0) ⊂ T (0) and D(S) ⊃ D(T )}. ♦

Fundamentals

115

Remark 2.20.3 Let ε > 0 and T ∈ CR(X), then one may order pseudospectra as T σe3,ε (T ) = σe1,ε (T ) σe2,ε (T ) ⊂ σe4,ε (T ) ⊂ σe5,ε (T ). ♦

2.20.3

The Essential ε-Pseudospectra of Linear Relations

The aim of this subsection is to introduce and study the essential εpseudospectra of linear relations in a Banach space. Definition 2.20.4 Let X be a Banach space, ε > 0 and let T ∈ CR(X). The essential ε-pseudospectra sets of T are defined as follows   S 1 −1 λ ∈ C : k(λ − T ) k ≥ . (i) Σe1,ε (T ) = {λ ∈ C : α(λ − T ) = ∞} ε   S 1 λ ∈ C : k(λ − T )−1 k ≥ . (ii) Σe2,ε (T ) = {λ ∈ C : β(λ − T ) = ∞} ε T S (iii) Σe3,ε (T ) = ({λ ∈ C : α(λ − T ) = ∞} {λ ∈  C : β(λ − T ) = ∞})  1 λ ∈ C : k(λ − T )−1 k ≥ . ε S S (iv) Σe4,ε (T ) = {λ ∈ C : α(λ − T ) = ∞} {λ ∈  C : β(λ − T ) = ∞}  1 λ ∈ C : k(λ − T )−1 k ≥ . ε S S (v) Σe5,ε (T ) = {λ ∈ C : α(λ − T ) = ∞} {λ ∈ C : β(λ − T ) = ∞}  S 1 . ♦ {λ ∈ C : α(λ − T ) 6= β(λ − T )} λ ∈ C : k(λ − T )−1 k ≥ ε Remark 2.20.4 Let X be a Banach space and let T ∈ CR(X). Then, (i) If 0 < ε1 < ε2 , then Σei,ε1 (T ) ⊂ Σei,ε2 (T ) for i = 1, . . . , 5. (ii) If ε > 0, then Σei,ε (T ) ⊂ Σε (T ) for i = 1, . . . , 5. T (iii) If ε > 0, then Σe3,ε (T ) = Σe1,ε (T ) Σe2,ε (T ). S (iv) If ε > 0, then Σe4,ε (T ) = Σe1,ε (T ) Σe2,ε (T ). S (v) If ε > 0, then Σe5,ε (T ) = Σe4,ε (T ) {λ ∈ C : α(λ − T ) 6= β(λ − T )} .

♦

Example 2.20.2 Let T be the operator defined on L2 (R) by d2 + ix, dx2 n n o
2 o where i2 = −1. Then, λ ∈ C : max c1 ; log 1ε 3 c2 ≤ Reλ ⊂ Σei,ε (T ), for i = 1, . . . , 5, where c1 , c2 > 0. Indeed, the operator T is closed and non self-adjoint when considered on its maximal domain  D(T ) = ψ ∈ L2 (R) : −ψ 00 + iψ ∈ L2 (R) . T =

116

Spectral Theory of Multivalued Linear Operators

Moreover, ρ(T ) = C and ( ( 

1 λ ∈ C : max c1 ; log ε

where c1 , c2 > 0. Hence, (  

 23

1 λ ∈ C : max c1 ; log ε

2.20.4

) c2

 23

) ≤ Reλ

)

 c2

  1 −1 ⊂ λ ∈ C : k(λ − T ) k ≥ , ε

≤ Reλ

⊂ Σei,ε (T ), for i = 1, . . . , 5.

S-Pseudospectra and S-Essential Pseudospectra of Linear Relations

Definition 2.20.5 We define the S-pseudospectra of T by   S 1 σε,S (T ) = σS (T ) λ ∈ C such that k(λS − T )−1 k > . ε We denote the S-pseudoresolvent set of T by   T 1 −1 . ♦ ρε,S (T ) = C\σε,S (T ) = ρS (T ) λ ∈ C such that k(λS − T ) k ≤ ε Remark 2.20.5 (i) Note that if S = I, then we find the usual definition of pseudospectra of a linear relation. (ii) If 0 < ε1 < ε2 , then it is clear that σε1 ,S (T ) ⊂ σε2 ,S (T ). ♦ Definition 2.20.6 Let T be a linear relation in CR(X) and ε > 0. The Sessential pseudospectrum of T is the set \ σe5,ε,S (T ) = σε,S (T + K). K∈KT (X)

We define the S-essential pseudoresolvent set by ρe5,ε,S (T ) = C\σe5,ε,S (T ).

♦

Chapter 3 The Stability Theorems of Multivalued Linear Operators

This chapter studies the stability theorems of multivalued linear operators. In first time, we investigate the closedness of linear relations. Secondly, enlightened by the notion of gap between two linear subspaces found in Ref. [67], we generalize Kato’s results, and apply it to generalized convergence between closed linear relations, which represents the convergence between their graphs in a certain distance. Hence, we investigate the stability of the index, the nullity and the deficiency of a linear relation T in Banach spaces spaces under perturbations by strictly singular and T -strictly singular linear relations. Therefore, we extend the concept of demicompactness and k-setcontactive linear operators on multivalued linear operators and we develop some properties. Finally, we study the essentially semi regular linear relation operators everywhere defined in Hilbert space. The obtained results is then applied to study some properties of the Samuel-multiplicity.

3.1 3.1.1

Stability of Closeness of Multivalued Linear Operators Sum of Two Closed Linear Relations

The sum (respectively, the product) of two closed linear relations is not necessarily closed. We shall use the following result which gives sufficient conditions for the sum (respectively, the product) of two closed linear relations to be closed. Theorem 3.1.1 Let X and Y be two Banach spaces, S, T ∈ LR(X, Y ) such that D(S) ⊃ D(T ), S(0) ⊂ T (0) and 117

118

Spectral Theory of Multivalued Linear Operators

kSxk ≤ akxk + bkT xk, x ∈ D(T ),

(3.1)

for some constants a, b with b < 1. Then, T ∈ CR(X, Y ) if and only if, T + S ∈ CR(X, Y ). ♦ Proof. Suppose T ∈ CR(X, Y ). Let us consider two cases for S and T . First case : If S and T are single valued. Then, kT xk = k(T + S − S)xk ≤ k(T + S)xk + kSxk, x ∈ D(T ), which in turn yields by using (3.1) (1 − b)kT xk ≤ k(T + S)xk + akxk, x ∈ D(T ). Thus, if (xn )n is sequence in D(T + S) = D(T ) such that xn → x and (T + S)xn → ψ, then there exist N0 ∈ N such that for all n, m ≥ N0 , we have (1 − b)kT xn − T xm k ≤ akxn − xm k + k(T + S)xn − xn k. Therefore, (T xn )n is a Cauchy sequence in the Banach space Y . Thus, T xn → x0 . Since T is closed, then x ∈ D(T ) and T x = x0 . Moreover, Sxn = (T + S)xn − T xn → ψ − x0 as n → ∞. But kS(xn − x)k ≤ (akxn − xk + bkT xn − x0 k) → 0 as n → ∞, which shows that Sx = ψ − x0 . Hence, (T + S)x = ψ and x ∈ D(T + S). Second case : S and T are linear relations. Since T ∈ CR(X, Y ) and by using Proposition 2.7.1 (i), we have T (0) = T (0). In view of S(0) ⊂ T (0), then S(0) ⊂ T (0). So, QT = QT +S and QT +S (T + S) = QT T + QT S. By using Proposition 2.5.4 (iii), we deduce that QT S is single valued and kQT Sxk ≤ kQS Sxk = kSxk ≤ akxk + bkT xk, x ∈ D(T ). Hence, kQT Sxk ≤ akxk + bkQT T xk, x ∈ D(T ). On the other hand, since QT T is closed, then in view of the first case, QT T + QT S is closed single valued. This means, since (T + S)(0) = T (0) is closed, then T + S is closed. Conversely, assume that T + S is closed. It follows from (2.31) that     a b k − Sxk = kSxk ≤ kxk + k(T + S)xk, x ∈ D(T ). 1−b 1−b

The Stability Theorems of Multivalued Linear Operators

119

In view of the above, we have T + S − S is closed. Now, by Proposition 2.3.4, we have T is closed. Q.E.D. Theorem 3.1.2 Let X and Y be two Banach spaces, T ∈ LR(X, Y ) and S ∈ LR(X, Y ) such that S(0) ⊂ T (0). If S is T -bounded with T -bound δ, then (i) If T is closed and δ < 1, then T + S is closed. (ii) If T + S is closed, and δ < 12 , then T is closed.

♦

Proof. (i) Since T is closed, then by Proposition 2.7.1 (i), T (0) is closed, and QT T is closed. In view of S(0) ⊂ T (0), then (T + S)(0) = T (0). Hence, (T + S)(0) is closed. Since S is T -bounded with T -bound δ < 1, then by using Remark 2.6.1 (v), we obtain QT S is QT T -bounded with QT T -bound < 1. Therefore, since S(0) ⊂ T (0), then QT = QT +S . Thus, QS+T (T + S) = QT (T + S) = QT T + QT S is closed. Applying Theorem 3.1.1, we have QS+T (T + S) is closed. Therefore, by using Proposition 2.7.1 (i), it follows that T + S is closed. (ii) Since T + S is closed, S is T -bounded with T -bound δ < 12 , and by using Proposition 2.3.4, we get T = T + S − S. Then, by Proposition 2.6.1, we obtain S is (T + S)-bounded with (T + S)bound < 1. Applying (i), it follows that T is closed. Q.E.D.

3.1.2

Sum of Three Closed Linear Relations

The sum (respectively, the product) of three closed linear relations is not necessarily closed. We shall use the following result which gives sufficient conditions for the sum (respectively, the product) of three closed linear relations to be closed. Theorem 3.1.3 Let X and Y be two Banach spaces, T ∈ CR(X, Y ), and S, K ∈ LR(X, Y ) such that K(0) ⊂ S(0) ⊂ T (0). If S is T -bounded with δ1 δ2 T -bound δ1 < 1, and K is S-bounded with S-bound δ2 such that 1−δ < 1. 1 Then, T + S + K is a closed linear relation. ♦

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Spectral Theory of Multivalued Linear Operators

Proof. Since T is closed and S is T -bounded with T -bound δ1 < 1, then by using Theorem 3.1.1, we have T + S is closed. On the one hand, since S is T -bounded with T -bound δ1 < 1, then in view of Proposition 2.6.2, we obtain δ1 S is (T + S)-bounded with (T + S)-bound ≤ 1−δ . On the other hand, since 1 K is S-bounded with S-bound δ2 , then by using Proposition 2.6.2, we get K δ1 δ2 . Using the fact that K is is (T + S)-bounded with (T + S)-bound ≤ 1−δ 1 (T + S)-bounded with (T + S)-bound < 1, and T + S is closed, we infer from Theorem 3.1.1 that T + S + K is closed which completes the proof. Q.E.D. Lemma 3.1.1 Let X and Y be two Banach spaces, A, B and C ∈ LR(X, Y ) S satisfy B(0) C(0) ⊂ A(0). Suppose that B is A-bounded with A-bound δ1 , C is A-bounded with A-bound δ2 , and Y is complete. Then, (i) If δ1 + δ2 < 1 and A is closed, then A + B + C is closed. (ii) If δ1 + δ2 < 12 and A + B + C is closed, then A is closed.

♦

Proof. (i) Since B is A-bounded with A-bound δ1 , and C is A-bounded with A-bound δ2 , then by Proposition 2.6.2 (ii), B +C is A-bounded with A-bound δ1 + δ2 < 1. Applying Theorem 3.1.2 (i), we obtain A + B + C is closed. (ii) Since B is A-bounded with A-bound δ1 , and C is A-bounded with Abound δ2 , then for all x ∈ D(A), we have k(B + C)xk

≤

kBxk + kCxk

≤ (a1 + a2 )kxk + (b1 + b2 )kAxk ≤ (a1 + a2 )kxk + (b1 + b2 )k(A + B + C − B − C)xk ≤ (a1 + a2 )kxk + (b1 + b2 )k(A + B + C)xk + (b1 + b2 )k(B + C)xk (a1 + a2 ) (b1 + b2 ) ≤ kxk + k(A + B + C)xk. 1 − (b1 + b2 ) 1 − (b1 + b2 ) Therefore, B + C is (A + B + C)-bounded with (A + B + C)-bound < 1. On the other side, by Proposition 2.3.4, we get A = A + B + C − B − C. Thus, A+B +C is closed, and B +C is (A+B +C)-bounded with (A+B +C)-bound < 1. Finally applying (i), it follows that A is closed. Q.E.D. Theorem 3.1.4 Let X and Y be two Banach spaces, S, T , K ∈ LR(X, Y ) such that D(T ) ⊂ D(S) ⊂ D(K), K(0) ⊂ S(0) ⊂ T (0) and (i) there exist two constants a1 , b1 > 0 such that kSxk ≤ a1 kxk + b1 kT xk, x ∈ D(T ),

The Stability Theorems of Multivalued Linear Operators

121

(ii) there exist two constants a2 , b2 > 0 such that b1 (1 + b2 ) < 1 and kKxk ≤ a2 kxk + b2 kSxk, x ∈ D(S). Then, T ∈ CR(X, Y ) if and only if, T + S + K ∈ CR(X, Y ).

♦

Proof. Let T ∈ CR(X, Y ), we will show that T + S + K ∈ CR(X, Y ). To do this, we will consider two cases for S, T and K. First case : If S, T and K are operators, then for all x ∈ D(T ), we have k(T + S + K)xk

≤ kT xk + kSxk + kKxk ≤ kT xk + a1 kxk + b1 kT xk + a2 kxk + b2 kSxk = (1 + b1 )kT xk + (a1 + a2 )kxk + b2 (a1 kxk + b1 kT xk).

Hence, k(T + S + K)xk ≤ (a1 + a2 + a1 b2 )kxk + (1 + b1 + b1 b2 )kT xk.

(3.2)

Similarly, for all x ∈ D(T ), we have k(S + K)xk

≤ kSxk + kKxk ≤ a1 kxk + b1 kT xk + a2 kxk + b2 kSxk = b1 kT xk + (a1 + a2 )kxk + b2 (a1 kxk + b1 kT xk).

So, k(S + K)xk ≤ (a1 + a2 + a1 b2 )kxk + b1 (1 + b2 )kT xk.

(3.3)

It follows from (3.3) that for all x ∈ D(T ), we have k(T + S + K)xk

≥ kT xk − kSx + Kxk ≥ kT xk − (a1 + a2 + a1 b2 )kxk − b1 (1 + b2 )kT xk = −(a1 + b2 + a1 b2 )kxk + (1 − b1 (1 + a2 ))kT xk

which yields (1 − b1 (1 + a2 ))kT xk ≤ k(T + S + K)xk + (a1 + b2 + a1 b2 )kxk.

(3.4)

Setting Θ = 1 − b1 (1 + a2 ) (0 < Θ < 1) and Ψ = a1 + b2 + a1 b2 (Ψ > 0), then (3.4) gives kT xk ≤ Θ−1 (k(T + S + K)xk + Ψkxk) . (3.5) Let (xn )n be sequence in D(T + S + K) = D(T ) such that xn → x and (T + S + K)xn → ψ. Then, by (3.5), there exist N1 ∈ N such that for all n, m ≥ N1 , we have kT xn − T xm k ≤ Θ−1 (k(T + S + K)(xn − xm )k + Ψkxn − xm k) .

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Spectral Theory of Multivalued Linear Operators

So, (T xn )n is a Cauchy sequence in the Banach space Y . Therefore, there exist ψ1 ∈ Y such that T xn → ψ1 . Since T is closed, then x ∈ D(T ) and T x = ψ1 . Now, from (3.2), we have k(T +S+K)(xn −x)k ≤ ((a1 + a2 + a1 b2 )kxn − xk + (1 + b1 + b1 b2 )kT xn − xk) → 0, which, by letting n → +∞, implies that (T + S + K)xn → ψ = (T + S + K)x. Therefore, T + S + K is a closed linear operator. Second case : T , S and K are linear relations. Since K(0) ⊂ S(0) ⊂ T (0), it is clear that QT +S+K = QT . Then, QT +S+K (T + S + K) = QT (T + S + K) = QT T + QT S + QT K. The fact that QT T is a closed operator and from Proposition 2.5.4 (iii), lead to the following QT S and QT K are single valued. Hence, kQT Sk ≤ kQS Sk and kQT Kk ≤ kQK Kk. So, kQT Sk ≤ kQS Sk = kSxk ≤ a1 kxk + b1 kT xk, x ∈ D(T ). Thus, kQT Sk ≤ kQS Sxk ≤ a1 kxk + b1 kQT T xk, x ∈ D(T ). Similarly, we have kQT Kk ≤ a2 kxk+b2 kQS Sxk, x ∈ D(S), where b1 (1+a2 ) < 1. Consequently, QT T + QT S + QT K is a closed operator. Since T (0) + S(0) + K(0) = T (0) is closed, then by using the first case, T + S + K is a closed linear relation. Conversely, assume that T + K + S is closed. Then, k − Sxk ≤ (a1 + b1 Ψ) kxk + b1 Θ−1 k(T + K + S)xk, x ∈ D(T ), and k − Kxk ≤ a2 kxk + b2 k − Sxk, x ∈ D(S) and b1 Θ−1 (1 + a2 ) < 1. Now, applying from what precedes to the relations T + S + K, −K and −S, that T + S + K − K − S is a closed operator. Since (K + S)(0) ⊂ T (0), then T = T + (K + S) − (K + S). Therefore, by Proposition 2.3.4, T is closed. This completes the proof. Q.E.D.

3.1.3

Sum of Two Closable Linear Relations

Proposition 3.1.1 Let S, T ∈ LR(X, Y ) such that S(0) ⊂ T (0) and D(T ) ⊂ D(S). If T is closable and S is continuous, then (i) T + S is closed and T + S ⊂ T + S. (ii) T + S is closable and S + T = S + T .

♦

The Stability Theorems of Multivalued Linear Operators

123

Proof. (i) Since T is closable, then in view of Lemma 2.7.3, T (0) = T (0). Hence, by using Lemma 2.7.2, T (0) ⊂ T (0). So, T (0) = T (0) = T (0). Thus, S(0) ⊂ T (0). Moreover, since D(T ) × Y = G(T ) + ({0} × Y ), then D(T ) ⊂ D(T ). Thus, G(T ) + ({0} × Y ) ⊂ G(T ) + ({0} × Y ) = D(T ) × Y

(3.6)

G(T ) + ({0} × Y ) = G(T ) + ({0} × Y ) = D(T ) × Y.

(3.7)

and clearly,

Using (3.6), (3.7) and Proposition 2.7.2, we have T + S is closed, as desired. Now, we prove T + S ⊂ T +S. Clearly, T ⊂ T and consequently, T +S ⊂ T +S. Thus, since T + S is closed, then T + S ⊂ T + S ⊂ T + S.

(3.8)

(ii) Since (T + S)(0) = T (0) + S(0) = T (0) = T (0) = T (0) = (T + S)(0)

(3.9)

and, by using (3.8), we have (T + S)(0) ⊂ (T + S)(0) ⊂ T (0) + S(0) = T (0) + S(0) = (T + S)(0). (3.10) So, in view of (3.9) and (3.10), we have (T + S)(0) = (T + S)(0). Thus, it follows from Lemma 2.7.3 that T + S is closable. Now, we prove T + S = T + S. By (3.8), it only remains to see that T + S ⊂ T + S. Using Proposition 2.3.4, we have T = T + S − S ⊂ T + S − S. Moreover, by the hypotheses we can apply the part (i) to ensure that T + S − S is closed. So, the fact that T ⊂ T + S − S = T + S − S and by Proposition 2.3.4, lead to T + S ⊂ T + S. Q.E.D. Corollary 3.1.1 Let S, T ∈ LR(X, Y ) such that S(0) ⊂ T (0), D(T ) ⊂ D(S), and λ ∈ C. Then, (i) If T ∈ CR(X, Y ) and S is continuous, then λS − T ∈ CR(X, Y ). (ii) If T is closable and S is continuous, then λS −T is closable and λS − T = λS − T . ♦

124

Spectral Theory of Multivalued Linear Operators

Lemma 3.1.2 Let S, T , A and B ∈ LR(X, Y ). Then, (i) If S and T are closable, and D(T ) ⊂ D(S), then S is T -bounded with T -bound α if and only if, S is T -bounded with T -bound α. (ii) If S = A + B is T -bounded with T -bound β, B is A-bounded with A-bound β δ < 1, and B(0) ⊂ A(0), then A is T -bounded with T -bound γ ≤ 1−δ . ♦ Proof. (i) First case : If S and T are two operators, then since S is T bounded with T -bound α, we have for any b > α, kSxk ≤ akxk + bkT xk,

(3.11)

for some a ≥ 0 and all x ∈ D(T ) ⊂ D(S). From (3.11) it readily follows that D(T ) ⊂ D(S) and kSxk ≤ akxk + bkT xk, for x ∈ D(T ). Thus, S is T -bounded and, since G(S) ⊂ G(S), it follows that S has T -bound α. Conversely, if S has T -bound α, then S has T -bound ≤ α. The first part of the proof now implies that S has T -bound α. Second case : If S and T are two relations, then QS S is QT T -bounded with QT T -bound α. The latter is equivalent to QS S is QT T -bounded with QT T bound α. Since D(T ) = D(QT T ), and QT T = QT T , the latter holds if, and only if, S is T -bounded with T -bound α. (ii) Since B is A-bounded with A-bound δ < 1, and B(0) ⊂ A(0), then if we δ apply Proposition 2.6.1, we obtain B is S-bounded with S-bound α ≤ 1−δ . Also, S is T -bounded with T -bound β. Hence, by using Proposition 2.6.2 (i), δβ it follows that B is T -bounded with T -bound αβ ≤ 1−δ . So, B is T -bounded δβ with T -bound αβ ≤ 1−δ , and knowing that S is T -bounded with T -bound β and A = S − B, the just given argument implies by Proposition 2.6.2 (ii) that β A is T -bounded with T -bound γ ≤ 1−δ . Q.E.D. Lemma 3.1.3 Let S, T ∈ LR(X, Y ) such that S(0) ⊂ T (0). Suppose that S is T -bounded with T -bound δ. Then, (i) If T is closable, and δ < 1, then T + S is closable. (ii) If T + S is closable, and δ < 12 , then T is closable. If, in addition, S is closable, then S + T = S + T and D(T + S) = D(T ). ♦ Proof. (i) Since S is T -bounded with T -bound δ < 1, then QT S is QT T bounded with QT T -bound < 1. Moreover, T is closable, then by Lemma 2.7.3,

The Stability Theorems of Multivalued Linear Operators

125

QT T is closable. In view of (T +S)(0) = T (0) and QS+T (T +S) = QT (T +S) = QT T + QT S, we have QS+T (T + S) is closable and (T + S)(0) is closed. Now, pplying Lemma 2.7.3, it follows that T + S is closable. (ii) Since T + S is closable, S is T -bounded with T -bound δ < 12 , and by appliying Proposition 2.3.4, we get T = T + S − S. Also, S is (T + S)-bounded with (T +S)-bound < 1. So, using (i), it follows that T is closable. Since T and S are closable, and S is T -bounded with T -bound < 1, then by Lemma 3.1.2 (i), we obtain S is T -bounded with T -bound < 1. So, S + T is closed. Since T ⊂ T , and S + T ⊂ S + T , then T + S ⊂ T + S = T + S (S + T is closed). On the one hand, we obtain by Proposition 2.3.4 and T = T + S − S that T = T +S −S ⊂ T + S −S. On the other hand, S is T -bounded with T -bound δ < 12 , then S is (T +S)-bounded with (T +S)-bound < 1. Using Lemma 3.1.2 (i), we obtain S is (T + S)-bounded with (T + S)-bound < 1. So, T + S − S is closed. Then, T ⊂ T + S − S = T + S − S, and we obtain T + S ⊂ T + S. Wherefrom, T + S = T + S, and D(T + S) = D(T + S) = D(T ). Q.E.D.

3.1.4

Product of Closable Linear Relations

Theorem 3.1.5 Let S ∈ LR(Y, Z) be closable and T ∈ L(X, Y ). Then, (i) ST is closed. (i) ST is closable and ST ⊂ ST. (ii) If T is bijective, then ST = ST.

♦

Proof. (i) Since S is closed and T ∈ L(X, Y ), then by using Lemma 2.7.1, we have ST is closed. (ii) Since S ⊂ S, then ST ⊂ ST which implies that ST ⊂ ST = ST.

(3.12)

Now, we prove that ST is closable. Indeed, ST is closable if and only if, ST (0) = ST (0). But, using Lemma 2.7.2, we have ST (0) ⊂ ST (0).

(3.13)

We infer from (3.12) and (3.13) that ST (0) ⊂ ST (0).

(3.14)

126

Spectral Theory of Multivalued Linear Operators

On the other hand, ST (0)

= S(0) (as T is single valued) = S(0) (as S is closable) = ST (0).

(3.15)

Applying (3.14) and (3.15), we obtain that ST (0) = ST (0) = S(0) = ST (0). So, we conclude that ST is closable. (iii) By (3.12), it only remains to see that ST ⊂ ST . But T −1 (0) = N (T ) = {0}, then T −1 is single valued, R(T ) = Y and D(T −1 ) = R(T ). Thus, D(T −1 ) = Y. Since T is closed, then by using Lemma 2.7.3, we have T −1 is closed. Thus, by the closed graph theorem for linear relations (see Theorem 2.7.2 (i)), T −1 is bounded single valued. Since ST is closable by (i), we can apply (i) to ST instead S and T −1 instead of T to obtain ST T −1 is closable. Hence, ST T −1 ⊂ ST T −1 . So, T −1 T = ID(T ) and T T −1 = IR(T ) = IY . Thus, S ⊂ ST T −1 which implies that S ⊂ ST T −1 ⊂ ST T −1 . Therefore, ST ⊂ ST T −1 T = ST . Q.E.D. Corollary 3.1.2 Let T ∈ L(X, Y ) be bijective and let S ∈ LR(Z, Y ) be closable. Then, T −1 S is closable and T −1 S = T −1 S. ♦ Proof. Since T is bounded single valued, then T is closed. Hence, by using Proposition 2.7.1, we have T −1 is closed. So, T −1 is bounded single valued. Indeed, T −1 (0) = T (0) = {0} and R(T −1 ) = D(T ) = X. Then, by using Corollary 2.9.2, we have T −1 S is closed. Therefore, T −1 S(0)

= T −1 S(0) = T −1 S(0) (as S is closable).

(3.16)

Moreover, T −1 S ⊂ T −1 S and T −1 S ⊂ T −1 S = T −1 S. Hence, T −1 S ⊂ T −1 S. A combination of (3.16) and Lemma 2.7.2 leads to T −1 S(0) ⊂ T −1 S(0) ⊂ T −1 S(0) = T −1 S(0). Hence, T −1 S(0) ⊂ T −1 S(0) ⊂ T −1 S(0) ⊂ T −1 S(0), Consequently, T −1 S(0) = T −1 S(0) = T −1 S(0).

The Stability Theorems of Multivalued Linear Operators

127

Therefore, T −1 S is closable. The proof of T −1 S ⊂ T −1 S may be checked in the same way as the proof of Theorem 3.1.5 (ii). Q.E.D. Theorem 3.1.6 Let T ∈ BR(X, Y ) be a single valued linear operator bijective and assume that H ∈ BR(Z, W ) is bijective with H(0) closed. Then, (i) If S ∈ LR(Y, Z) is closable, then HST is closable and HST = RHT . (ii) If H is bounded single valued, then S ∈ LR(Y, Z) is closable if and only if, HST is closable and HST = HST . ♦ Proof. Let us consider two cases of H. First case : If H is single valued, then by virtue of Theorem 3.1.5 (ii), we have ST is closable and ST = ST . On the other hand, H −1 is bounded single valued bijective. Indeed, H −1 (0) = N (H) = {0} (as H is injective) implies H −1 is a closed single valued and D(H −1 ) = R(H) = W (as H is surjective). Then, H is open. Since H −1 is continuous, then H −1 is bounded single valued. Clearly, N (H −1 ) = H(0) = {0} (since H is single valued) and R(H −1 ) = D(H) = Z. So, H = (H −1 )−1 . This leads to HST = (H −1 )−1 ST is closable with HST = HST = HST. Second case : If H is linear relation, then (a) QH HST is closable and QH HST = QH HST. (b) HS is closed by virtue of Proposition 2.9.4. (c) HST is closed equivalently (HST )−1 is closed. Indeed, (HST )−1 = T −1 (HS)−1 and, we have HS is closed. Therefore, (HS)−1 is closed. Since T is bounded single valued bijective, we deduce from Corollary 3.1.2 that T −1 (HS)−1 is closable and T −1 (HS)−1 = T −1 (HS)−1 . But T −1 (HS)−1 = (HST )−1 and T −1 (HS)−1

= T −1 (HS)−1 = (HST )−1 ,

then by applying Proposition 2.7.5, we have (HS)−1 = (HS)−1 . So, we obtain (c). (d) HST (0) = HS(0) = HS(0) = HST (0) is closed. Indeed, HST (0)

= HS(0) (again T is single valued so that T (0) = {0}) = HS(0) (as S is closable) = HST (0) (again T is single valued).

128

Spectral Theory of Multivalued Linear Operators

Therefore, HST (0) is closed by virtue of (c) and Lemma 2.7.3. (e) QHST HST is a a closable operator. Indeed, using Lemma 2.1.3, we have . . W HS(0) ≡ (W/H(0)) (HS(0)/H(0)), and QHS = QA QH with A = HS(0)/H(0), HS(0) = HST (0) is closed (by (d)), and H(0) is closed by hypothesis. Moreover, by using (d), we have QHS = QHST = QHS = QHST . Clearly, D(QHST HST )

= D(QA QH HST ) = D(QH HST ) = D(HST ).

On the other hand, by (a), we have QH HST = QH HST|

D(QH HST )

= QH HST|

D(HST )

.

So, QHST HST = QHST HST is an extension of QHST HST . Since HST is closed (by (c)), then by applying Lemma 2.7.3, we have QHST HST is closed. Therefore, QHST HST is closed extension of QHST HST . Hence, (e) is true. (f ) HST is closable. Follows immediately form the properties (d), (e) and Lemma 2.7.3 (iii). It only remains to see that HST =HST. (g) HST ⊂ HST . Indeed, since HST ⊂ HST , then HST ⊂ RST = HST . (h) HST ⊂ HST . Indeed, the equality QH HST = QH HST implies QA QH HST = QA QH HST , where QA QH = QHST . But QA QH HST ⊂ QA QH HST implies that QHST HST ⊂ QHST HST with QHST HST = QHST HST . Hence, QHST HST ⊂ QHST HST . By using (g), we have the other inclusion. So, QHST HST = QHST HST . Thus, in particular D(HST )

= D(QHST HST ) = D(QHST HST ) = D(HST )

and D(HST ) = D(HST ). Since HST (0) = HST (0) by (d), HST (0) = HST (0) ⊂ HST (0), and HST (0) ⊂ HST (0) by (g), we deduce D(HST ) =

The Stability Theorems of Multivalued Linear Operators

129

D(HST ), HST (0) = HST (0) and HST ⊂ HST . Thus, by virtue of Proposition 2.3.6, we have HST = HST. Conversely, if H is single valued bijective and HST is closable, then from Corollary 3.1.2, we have H −1 HST = ST is closable and ST = H −1 HST = H −1 HST = H −1 HST = ST . By the same token S(0) = ST (0) = ST (0) = ST (0) = S(0), we conclude that S is closable. This completes the proof. Q.E.D.

3.2 3.2.1

Sequence of Multivalued Linear Operators Converging in the Generalized Sense The Gap Between Two Linear Relations

Theorem 3.2.1 Let T ∈ BCR(X, Y ). Then, δ(T, OX ) ≤

kT k 1

(1 + kT kp ) p

, p ≥ 1.

♦

Proof. Since T ∈ BCR(X, Y ), then we can conclude from Theorem 2.5.5 (i) that kQT T k < ∞, which yields that QT T ∈ L(X, Y /T (0)). Hence, δ(QT T, OX ) =

kQT T k 1

(1 + kQT T kp ) p

.

Let ϕ = (x, y) ∈ G(T ) such that kϕk = k(x, y)kpp = kxkp + kykp = 1.

(3.17)

It follows that δ(T, OX )

=

sup dist(ϕ, G(OX )) ϕ∈G(T ) kϕk=1

=

sup

h

z ∈X

x∈ X kxkp +kykp =1

=

sup x∈X kxkp +kykp =1

=

sup x∈X kxkp +ky kp =1

i inf k(x, y) − (z, 0)k

h

1

inf (kx − zkp + kykp ) p

i

z∈X

kyk.

(3.18)

130

Spectral Theory of Multivalued Linear Operators

By using Lemma 2.5.7 (i), we infer that kT xk = inf ktk. t∈T x

Since x ∈ D(T ) = X and y ∈ T x, then in view of definition of the lower bound, for any given ε > 0, we obtain that kyk < kT xk + ε.

(3.19)

This implies that kxkp + kykp

< kxkp + (kT xk + ε)p p−1 X

= kxkp + kT xkp +

Cpk kT xkk εp−k .

k=0

Since ε is arbitrary, then kxkp + kykp

kxkp + kT xkp

≤

= kxkp + kQT T xkp . Hence, according to the condition (3.17), we deduce that kxkp + kQT T xkp = 1. Putting x = cw, where w ∈ X and c =

1 1

(1 + kQT T wkp ) p

kQT T xk =

sup

(3.20)

x∈X kxkp +kQT T xkp =1

sup w∈X kwk=1

=

, we get

kQT T wk 1

(1 + kQT T wkp ) p

kQT T k 1

(1 + kQT T kp ) p

= δ(QT T, OX ). This implies from (3.18) and (3.19) that, for any given ε > 0, δ(T, OX )

=

sup

ky k

x∈X kxk +kykp =1 p

≤

sup

kQT T xk + ε see (3.20)

x∈ X kxk +kQT T xkp =1 p

≤ δ(QT T, OX ) + ε.

The Stability Theorems of Multivalued Linear Operators

131

By the arbitrariness of ε, we conclude that δ(T, OX ) ≤ δ(QT T, OX ) =

kQT T k (1 + kQT

=

1

T kp ) p

kT k 1

(1 + kT kp ) p

.

Q.E.D.

Remark 3.2.1 If T ∈ BCR(X, Y ), then δ(T, OX ) ≤ δ(QT T, OX ).

♦

Theorem 3.2.2 Let T ∈ BCR(X, Y ). Then, δb(T, OX ) ≤ p

kT k

♦

.

1 + kT k2

Proof. If we take p = 2 in Theorem 3.2.1, we have δ(T, OX ) ≤

kT k

.

1

(1 + kT k2 ) 2

Then, it is sufficient to prove that kT k

δ(OX , T ) ≤ p

1 + kT k2

.

Let us observe that δ(OX , T )

=

sup dist((x, 0), G(T )) x∈X kxk=1

=

sup

h

(y,z)∈G(T )

x∈X kxk=1

=

sup x∈X kxk=1

i k(x, 0) − (y, z)k

inf

h

inf

i 1 (kx − y k2 + kzk2 ) 2 .

(y,z)∈G(T )

Moreover, in view of Lemma 2.5.7 (i), we have kT yk = inf ktk. t∈T y

Since y ∈ D(T ) = X and z ∈ T y, then for any given ε > 0, we have kzk ≤ kT yk + ε. So, for any x, y ∈ X, we obtain that kx − yk2 + kzk2

≤

kx − yk2 + (kT yk + ε)2

= kx − yk2 + kT yk2 + 2εkT yk + ε2 . Since ε is arbitrary, we get 1

1

(kx − y k2 + kz k2 ) 2 ≤ (kx − yk2 + kT yk2 ) 2 .

(3.21)

132

Spectral Theory of Multivalued Linear Operators

Consider the function f (x, y) = kx−yk2 +kT yk2 . Let x ∈ X such that kxk = 1 x and y = . Then, 1 + kT xk2

2  

x x kT xk2

x −

+ f (x, y ) = f x, =

2 2 1 + kT xk 1 + kT xk (1 + kT xk2 )2 kxk2 kT xk4 + kT xk2 = (1 + kT xk2 )2 kT xk2 (kxk2 kT xk2 + 1) = (1 + kT xk2 )2 kT xk2 (kT xk2 + 1) = since kxk = 1 (1 + kT xk2 )2 kT xk2 = . 1 + kT xk2 Hence, by using (3.21), we deduce that 1

dist((x, 0), G(T )) ≤ (kx − yk2 + kzk2 ) 2 kT xk ≤ p . 1 + kT xk2 Since the function g(t) =

t 1+t

is increasing, it follows that

δ(OX , T )

=

sup dist((x, 0), G(T )) x∈X kxk=1

≤

kT xk sup p 1 + kT xk2 x∈X

kxk=1

≤

kT k p

1 + kT k2

.

Finally, we obtain the following result δb(T, OX )

= ≤

max {δ(T, OX ), δ(OX , T )} kT k p . 1 + kT k2

This completes the proof.

Q.E.D.

Now, we give a relationship between δ(S + A, T + A) and δ(S, T ). Theorem 3.2.3 Let T , S ∈ CR(X, Y ) and let A : X −→ Y be a continuous linear relation satisfies D(A) ⊃ D(T ) ∪ D(S) and A(0) ⊂ T (0) ∩ S(0). Then, δ(S + A, T + A) ≤ (2 + kAk2 )δ(S, T )

(3.22)

The Stability Theorems of Multivalued Linear Operators

133

and δb(S + A, T + A) ≤ (2 + kAk2 )δb(S, T ).

(3.23) ♦

Proof. Since D(T + A) = D(T ) and D(S + A) = D(S), then by referring to Proposition 2.7.2, we infer that T + A and S + A are closed. Let ϕ = (x, y) ∈ G(S + A) with kϕk = 1. Then, there exists (x, y1 ) ∈ G(S) and (x, y2 ) ∈ G(A) such that y = y1 + y2 . Hence, kϕk22 = kxk2 + ky1 + y2 k2 = 1.

(3.24)

Since y2 ∈ Ax, then by using Lemma 2.5.7 (i), we infer, for any given ε > 0, that ky2 k < kAxk + ε < kAkkxk + ε.

(3.25)

Hence, by using (3.25), we have ky1 k = ky1 + y2 − y2 k ≤

ky1 + y2 k + ky2 k

≤

ky1 + y2 k + kAkkxk + ε.

Thus, ky1 k2

2

≤ ((ky1 + y2 k + kAkkxk) + ε) 2

≤ (ky1 + y2 k + kAkkxk) + 2ε (ky1 + y2 k + kAkkxk) + ε2 . Therefore, for any given ε > 0 and by Cauchy-Schwarz inequality, we obtain

ky1 k2 ≤ 1 + kAk2 kxk2 + ky1 + y2 k2 + 2ε (ky1 + y2 k + kAkkxk) + ε2 . It follows from (3.24) that
ky1 k2 ≤ 1 + kAk2 + 2ε (ky1 + y2 k + kAkkxk) + ε2 , and by arbitrariness of ε, we deduce that
1
1 kxk2 + ky1 k2 2 ≤ 2 + kAk2 2 .

(3.26)

In addition, for any u ∈ D(T + A), there exists (u, v1 ) ∈ G(T ) and (u, v2 ) ∈ G(A) such that v = v1 + v2 ∈ (T + A)u. On the one hand, since y2 ∈ Ax and v2 ∈ Au, then y2 − v2 ∈ A(x − u). Hence, for any given ε0 > 0, we have ky2 − v2 k ≤ kAkkx − uk + ε0 .

(3.27)

134

Spectral Theory of Multivalued Linear Operators

On the other hand, by using (3.27), we get ky1 + y2 − v1 − v2 k

≤

ky1 − v1 k + ky2 − v2 k

≤

ky1 − v1 k + kAkkx − uk + ε0 .

Similarly, for all u ∈ D(T + A) = D(T ) and by arbitrariness of ε, we conclude that kx − uk2 + ky1 + y2 − v1 − v2 k2


12

≤ 2 + kAk2


21

kx − uk2 + ky1 − v1 k2


21

Noting that ψ = (u, v) ∈ G(T + A), ϕ1 = (x, y1 ) ∈ G(S) and ψ1 = (u, v1 ) ∈ G(T ), we get dist(ϕ, G(T + A)) ≤ 2 + kAk2


12

dist(ϕ1 , G(T )).

(3.28)

Therefore, it follows from (2.45), (3.26) and (3.28) that δ(S + A, T + A)

=

sup

dist (ϕ, G(T + A))

ϕ∈G(S+A) kxk2 +ky1 +y2 k2 =1

≤

2 + kAk2


12

sup

dist (ϕ1 , G(T ))

ϕ1 ∈G(S) 1

1

(kxk2 +ky1 k2 ) 2 ≤(2+kAk2 ) 2

≤


2 + kAk2 δ(S, T ).

Thus, (3.22) holds and it also implies (3.23) as a direct consequence. Q.E.D. The following result can be easily derived from Theorems 3.2.2 and 3.2.3. Theorem 3.2.4 Let T , S ∈ LR(X, Y ). If T and S are two closed linear relations everywhere defined satisfying T (0) = S(0), then kT − S k δb(T, S) ≤ (2 + min{kT k, kSk}2 ) p . 1 + kT − Sk2

♦

Proof. It follows from Proposition 2.7.2 that T − S and S − T are closed, and in view of T (0) = S(0), we have T = T − S + S and S = S − T + T . Then, T ∈ BR(X, Y ). Hence, by using Proposition 2.3.4 and Theorem 3.2.3, we deduce that δb(T, S) = δb((T − S) + S, OX + S) ≤ (2 + kSk2 )δb(T − S, OX ) and b S) = δ(O b X + T, (S − T ) + T ) ≤ (2 + kT k2 )δ(S b − T, OX ) δ(T,

.

The Stability Theorems of Multivalued Linear Operators

135

which, together with Theorem 3.2.2, yields that kT − S k δb(T, S) ≤ (2 + min{kT k, kSk}2 ) p . 1 + kT − Sk2

Q.E.D.

Theorem 3.2.5 Let S, T ∈ CR(X, Y ) satisfy that kT k < ∞, T (0) ⊂ S(0) 1 and D(T ) ⊃ D(S). If δb(S, T ) < (1 + kT k2 )− 2 , then kSk < ∞, D(S) = D(T ) and 1 + kT k2 p kS − T k ≤ δ(S, T ). ♦ 1 − 1 + kT k2 δ(S, T ) Proof. We divide this proof into two steps. Step 1. We show that S − T is continuous. For any given x ∈ D(S − T ) with x = 6 0, there exist y ∈ (S − T )x and z ∈ T x. Then, by using Lemma 2.5.7 (i), we infer, for any given ε > 0, that kyk < k(S − T )xk +

ε ε and kzk < kT kkxk + . 2 2

(3.29)

Based on the assumptions that T (0) ⊂ S(0) and D(S) ⊂ D(T ), and by using Proposition 2.3.4, we can write S = S−T +T . Then, we deduce that (x, y+z) ∈ G(S). Moreover, for any (u, v) ∈ G(T ), we get (x − u, z − v) ∈ G(T ) and for any ε0 > 0, there is w ∈ T (x − u) such that kwk < kT kkx − uk + ε0 .

(3.30)

Hence, we deduce that (0, z − v − w) ∈ G(T ). The fact that T (0) ⊂ (S − T )(0) implies that (0, z − v − w) ∈ G(S − T ). Thus, (x, y + z − v − w) ∈ G(S − T ). Consequently, for any (u, v) ∈ G(T ), we infer from (3.30) that ky + z − v − wk

≤

ky + z − vk + kwk

< ky + z − vk + kT kkx − uk + ε0 . By Cauchy-Schwarz inequality, we obtain ky + z − v − wk < 1 + kT k2


21

kx − uk2 + ky + z − vk2


12

+ ε0 .

Hence, ky + z − v − wk < 1 + kT k2


21

inf

kx − uk2 + ky + z − vk2

(u,v)∈G(T )


21

+ ε0 .

Thus, by arbitrariness of ε0 , we conclude that 1

k(S − T )xk ≤ δ(S, T )(1 + kT k2 ) 2 .

(3.31)

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Spectral Theory of Multivalued Linear Operators

Now, let us assume that kxk2 + ky + zk2 = 1.

(3.32)

Then, by using (3.29), we have 1 ≤

kxk2 + (kyk + kzk)2 2

< kxk2 + (k(S − T )xk + kT kkxk + ε)

< (1 + kT k2 )kxk2 + 2kT kkxkk(S − T )xk + k(S − T )xk2 + 2ε(k(S − T )xk + kT kkxk) + ε2 . Then, by arbitrariness of ε, we get
1 (1 + kT k2 )kxk2 + 2kT kkxkk(S − T )xk + k(S − T )xk2 2  12  1 ≤ (1 + kT k2 )kxk2 + 2(1 + kT k2 ) 2 kxkk(S − T )xk + k(S − T )xk2  2  12 2 12 ≤ (1 + kT k ) kxk + k(S − T )xk

1 ≤

1

≤ (1 + kT k2 ) 2 kxk + k(S − T )xk.

(3.33)

Therefore, it follows from (3.31) and (3.33) that 1

k(S − T )xk ≤ (1 + kT k2 )δ(S, T )kxk + (1 + kT k2 ) 2 δ(S, T )k(S − T )xk. Hence, for all x ∈ D(S − T ) satisfying (3.32), we conclude that k(S − T )xk ≤

1 + kT k2 1

1 − (1 + kT k2 ) 2 δ(S, T )

δ(S, T )kxk.

(3.34)

Since (3.34) is homogeneous in x, then it is true for any x ∈ D(S − T ) = D(S) without normalization (3.32). So, S − T is continuous. Step 2. We prove that D(S) is dense in X. Let u be any given non-zero vector. Then, for any ε1 > 0, there is v ∈ T u such that kvk < kT uk + ε1 < kT kkuk + ε1 .

(3.35)

Let us assume that u ∈ D(T ) so normalized that (u, v) is in unit sphere of G(T ); kuk2 + kvk2 = 1. (3.36)

The Stability Theorems of Multivalued Linear Operators

137

1

Let δ be such that δ(S, T ) < δ < (1 + kT k2 )− 2 . Then, there exists (x, y) ∈ G(S) satisfying kx − uk2 + ky − vk2 < δ 2 . This implies that kx − uk < δ for all x ∈ D(S).   Since dist (u, D(S)) = dist u, D(S) , then from (3.37), we have

(3.37)

dist(u, D(S)) < δ. In addition, it follows from (3.35) and (3.36) that 1 < kuk2 + (kT kkuk + ε1 )2 < (1 + kT k2 )kuk2 + 2ε1 kT kkuk + ε21 . 1

Then, by arbitrariness of ε1 , we obtain 1 ≤ (1 + kT k2 ) 2 kuk. Hence, 1

dist(u, D(S)) < (1 + kT k2 ) 2 δkuk.

(3.38)

Since (3.38) is homogeneous in u, then it is true for any u ∈ D(T ). The fact 1 that (1 + kT k2 ) 2 δ < 1 implies from Lemma 2.1.6 that D(S) = D(T ). Step 3. We prove that S is continuous and D(S) is closed. By referring to Proposition 2.5.4 (ii), we conclude that kSk = kS − T + T k ≤

kS − T k + kT k < ∞.

As a result, the use of Theorem 2.5.4 (i) makes us conclude that D(S) is closed. This implies that D(S) = D(T ), as desired. Q.E.D. The following result can be derived from Theorem 3.2.5. Theorem 3.2.6 Let T, S ∈ CR(X, Y ) and T −1 ∈ L(Y, X). If 1 δb(S, T ) < (1 + kT −1 k2 )− 2 ,

then S −1 ∈ L(Y, X).

♦

Proof. In view of Theorem 2.10.1, we have δ(S −1 , T −1 ) = δ(S, T ). Then, from Theorem 3.2.5, S −1 is continuous and D(S −1 ) = D(T −1 ). Thus, it remains to prove that S is injective, which yields that S −1 is an operator. 1 Let S be such that δ(S, T ) < (1 + kT −1 k2 )− 2 , and suppose, by contradiction,

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Spectral Theory of Multivalued Linear Operators

that (x, 0) ∈ G(T ) such that kxk = 1 can be found. Then, (x, 0) is in the unit sphere of G(S). Hence, there is (y, z) ∈ G(T ) such that kx − yk2 + kzk2 < δ 02 ,

(3.39) 1

for some number δ 0 satisfy δ(S, T ) < δ 0 < (1 + kT −1 k2 )− 2 . Hence, 1 = kxk2

≤ (kx − yk + kyk)2 ≤ (kx − yk + kT −1 kkzk)2 .

By Cauchy-Schwarz inequality and (3.39), we get 1 ≤ (1 + kT −1 k2 )(kx − yk2 + kzk2 )
0, then we can suppose that ϕ1 = γ −1 (x, T1 x) ∈ G(T1 ) with kϕ1 k = 1. Therefore, by (3.40), we get dist(ϕ1 , G(S1 )) < δ 0 . Hence, there is a z ∈ D(S1 ) such that kx − zk2 + kT1 x − S1 zk2 < γ 2 δ 02 . (3.42) Assume that ψ = (z, t) ∈ G(S) such that (z, t) = (z, S1 z) + (0, α2 ), where α2 ∈ S(0). On the one hand, by (3.42), we obtain that kϕ − ψk22

= k ((x, T1 x) + (0, α1 )) − ((z, S1 z) + (0, α2 )) k22 = k(x − z, T1 x − S1 z) + (0, α1 − α2 )k22 = kx − zk2 + k(T1 x − S1 z) + (α1 − α2 )k2 ≤

kx − zk2 + 2kT1 x − S1 zk2 + 2kα1 − α2 k2

< 2γ 2 δ 02 + 2kα1 − α2 k2 . On the other hand, from (3.41), we get γ2

= kxk2 + kT1 xk2 = kxk2 + kT1 x + α1 − α1 k2 ≤

kxk2 + 2kT1 x + α1 k2 + 2kα1 k2

≤ 2(1 + kα1 k2 ). Hence, kϕ−ψk2 ≤ 4(1+kα1 k2 )δ 02 +2kα1 −α2 k2 . Since α1 ∈ T (0), then for any given ε > 0, we infer that kα1 k < kT (0)k+ε. This yields from Proposition 2.5.3 that kα1 k < ε. Consequently, for any given ε0 > 0, we obtain kα1 − α2 k ≤ ε0 . Thus, kϕ − ψk22 ≤ 4(1 + ε2 )δ 02 + 2ε20 . Using the fact that ψ ∈ G(S) and the arbitrariness of ε and ε0 , we can get that dist(ϕ, G(S)) ≤ 2δ 0 . Hence, by the hypothesis, ϕ ∈ G(T ) and kϕk = 1, we deduce that δ(T, S) ≤ 2δ 0 . Since δ 0 may be arbitrarily close to δ(T1 , S1 ), we conclude that δ(T, S) ≤ 2 δ(T1 , S1 ). Q.E.D. The following result can be easily derived from Theorem 3.2.7.

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Spectral Theory of Multivalued Linear Operators

Corollary 3.2.2 Let T, S ∈ CR(X, Y ). If T1 and S1 are two closed selections of T and S, respectively, then δb(T, S) ≤ 2δb(T1 , S1 ).

3.2.3

♦

Generalized Convergence of Closed Linear Relations

Definition 3.2.1 Let (Tn )n be a sequence of closed linear relations from X into Y and T ∈ CR(X, Y ). The sequence (Tn )n is said to be converge in the g generalized sense to T , denoted by Tn −→ T , if δb(Tn , T ) converges to 0 when n → ∞. ♦ Theorem 3.2.8 Let (Tn )n be a sequence of closed linear relations from X into Y and T ∈ CR(X, Y ). Then, S (i) If S is a continuous relation from X into Y satisfy D(S) ⊃ D(Tn ) D(T ) T g and S(0) ⊂ Tn (0) T (0), for all n ≥ 1, then Tn −→ T if and only if, Tn + g S −→ T + S. (ii) If D(Tn ) ⊂ D(T ) and T (0) ⊂ Tn (0), for all n ≥ 1, then T ∈ BR(X, Y ) g and Tn −→ T if and only if, Tn ∈ BR(X, Y ) for sufficiently large n and Tn converges to T . g

(iii) Let Tn −→ T . Then, T −1 ∈ L(Y, X) if and only if, Tn−1 ∈ L(Y, X) for ♦ sufficiently large n and Tn−1 converges to T −1 . S T Proof. (i) Since D(S) ⊃ D(Tn ) D(T ) and S(0) ⊂ Tn (0) T (0), for all n ≥ 1, then from Proposition 2.7.2, we infer that Tn + S and T + S are closed. So, by applying Theorem 3.2.3, we conclude that b n , T ). δb(Tn + S, T + S) ≤ (2 + kSk2 )δ(T

(3.43)

g

b n , T ) → 0. So, by using (3.43), we Let us assume that Tn −→ T , then δ(T b conclude that, δ(Tn + S, T + S) → 0 when n → ∞. This is equivalent to say g g that Tn + S −→ T + S. Conversely, we suppose that Tn + S −→ T + S. By referring to Proposition 2.3.4, we have Tn = Tn − S + S and T = T − S + S. Hence, we can write δb(Tn , T ) = δb ((Tn + S) − S, (T + S) − S). Thus, again by Theorem 3.2.3, we deduce that δb ((Tn + S) − S, (T + S) − S) ≤ 2(1+kSk2 )δb((Tn +S)−S +S, (T +S)−S +S). This implies that δb ((Tn , T ) ≤ 2(1 + kSk2 )δb(Tn + S, T + S).

(3.44)

The Stability Theorems of Multivalued Linear Operators

141

According to (3.44), we infer that δb(Tn , T ) → 0 as n → ∞. Hence, Tn converges in the generalized sense to T. g

(ii) Assume that Tn −→ T and T ∈ BCR(X, Y ). Let n0 ∈ N such that 1 δb(Tn , T ) < (1 + kT k2 ) 2 holds for all n ≥ n0 . Then, by using Theorem 3.2.5, we deduce that kTn k < ∞, D(Tn ) = D(T ) = X, and ! 1 + kT k2 p kTn − T k ≤ δ(Tn , T ), ∀n ≥ n0 . (3.45) 1 − 1 + kT k2 δ(Tn , T ) Using the fact that δb(Tn , T ) → 0 as n → ∞, and from (3.45), we deduce that Tn converges to T . Conversely, we suppose that Tn ∈ BR(X, Y ) and Tn converges to T . By hypothesis, X = D(Tn ) ⊂ D(T ), then by using Theorem 2.7.2 (i), T is bounded. Now, from Proposition 2.3.4, we can write δb(Tn , T ) = δb((Tn − T ) + T, T + OX ). Then, by applying Theorems 3.2.2 and 3.2.3, we get the following inequality
kTn − T k δb(Tn , T ) ≤ 2 + kT k2 p . 1 + kT k2 g

As a result, Tn −→ T , as desired. g

(iii) Since Tn −→ T , then there exists N ∈ N such that 1 δb(Tn , T ) < (1 + kT −1 k2 )− 2 , for all n ≥ N.

By referring to Theorem 3.2.6, we obtain Tn−1 ∈ L(Y, X) for sufficiently large n. Since the gap is invariant under inversion, then by applying Theorem 3.2.5 to the pair Tn−1 , T −1 , we deduce that ! 1 + kT −1 k2 −1 −1 p kTn − T k ≤ δ(Tn−1 , T −1 ), ∀n ≥ N. 1 − 1 + kT −1 k2 δ(Tn−1 , T −1 ) (3.46) Hence, the use of Theorem 2.10.1, and also the estimation (3.46), makes us conclude that ! 1 + kT −1 k2 −1 −1 p kTn − T k ≤ δ(Tn , T ). (3.47) 1 − 1 + kT −1 k2 δ(Tn , T ) Consequently, Tn−1 converges to T −1 . Conversely, let us assume that 0 ∈ ρ(Tn ) for sufficiently large n and kTn−1 − T −1 k → 0 as n → ∞. Since kT −1 k = kT −1 + Tn−1 − Tn−1 k ≤ kT −1 − Tn−1 k + kTn−1 k,

142

Spectral Theory of Multivalued Linear Operators g

then T −1 ∈ L(Y, X). Now, we demonstrate that Tn −→ T. Keeping in mind b n , T ) can be expressed in the form that δb(Tn , T ) = δb(Tn−1 , T −1 ), then δ(T δb(Tn , T ) = δb((Tn−1 − T −1 ) + T −1 , 0 + T −1 ). Hence, the use of Theorems 3.2.2 and 3.2.3 allows us to conclude that
kT −1 − T −1 k δb(Tn , T ) ≤ 2 + kT −1 k2 p n . 1 + kT −1 k2 As a result, (Tn )n converges in the generalized sense to T , as desired. Q.E.D. Corollary 3.2.3 Let Tn , Sn ∈ CR(X, Y ) and T , S ∈ BCR(X, Y ) such that g g g T (0) ⊂ Tn (0) and S(0) ⊂ Sn (0). If Tn −→ T and Sn −→ S, then Tn + Sn −→ T + S. ♦ g

g

Proof. Let us assume that Tn −→ T and Sn −→ S. Then, by using (ii) of Theorem 3.2.8, we deduce that Tn , Sn ∈ BR(X, Y ), kTn − T k → 0 and kSn − Sk → 0 when n → ∞. Now, by applying Proposition 2.5.4 (ii), we infer that k(Tn + Sn ) − (T + S)k ≤ kTn − T k + kSn − Sk → 0 as n → ∞. Using the fact that (T + S)(0) = T (0) + S(0) ⊂ Tn (0) + Sn (0) = (Tn + Sn )(0), and D(Tn + Sn ) ⊂ X = D(T + S), we conclude from Theorem 3.2.8 (ii) that Tn + Sn converges in the generalized sense to T + S. Q.E.D.

3.3

Perturbation Classes of Multivalued Linear Operators

The main aim of this section, is to carry on the generalizations of some perturbation results of linear operators in Banach spaces to the multivalued case.

3.3.1

Fredholm and Semi-Fredholm Perturbation Classes of Multivalued Linear Operators

Theorem 3.3.1 Let X be a Banach space and let T ∈ CR(X) and S be continuous such that S(0) ⊂ T (0), D(T ) ⊂ D(S), and kSk 6= 0. If 0 < |λ| < γ(T ) kSk , then

The Stability Theorems of Multivalued Linear Operators

143

(i) α(T − λS) ≤ α(T ). (ii) β(T − λS) ≤ β(T ). (iii) If R(T ) is closed, then R(T − λS) is closed.

♦

Proof. Since T ∈ CR(X) and S is continuous with S(0) ⊂ T (0) and D(T ) ⊂ D(S), then by using Proposition 2.7.2, we have T − λS is closed. Hence, by Proposition 2.7.1, we infer that (T − λS)(0) = (T − λS)(0). Moreover, N (T − λS)

= {x ∈ D(T − λS) such that (T − λS)x = (T − λS)(0)} = {x ∈ D(T − λS) such that (T − λS)x = (T − λS)(0)} = N (QT −λS (T − λS)) = N (QT (T − λS)).

(3.48)

Let x ∈ N (T − λS) such that x 6= 0. Then, by using (3.48), we have γ(T )dist(x, N (T )) ≤

kT xk = kQT (T x)k

≤

kλQT (Sx)k (since QT (T x − λSx) = 0)

≤

|λ|kSkkxk

≤ γ(T )kxk. Thus, dist(x, N (T )) < kxk for all x ∈ N (T − λS). Hence, by using Lemma 2.1.2, we have α(T − λS) ≤ α(T ). (ii) By using Proposition 2.8.1 (iii) and (vi), we have kλSk = |λ|kS ∗ k < γ(T ) = γ(T ∗ ). Since λS is continuous and λ 6= 0, then D(λS) = D(S) ⊃ D(T ) and (λS − T )∗ = (λS)∗ − T ∗ = λS ∗ − T ∗ . Replacing X by D(S) if necessary, we may assume S to everywhere defined, thus making S ∗ single valued. Applying Proposition 2.8.1 (iv), we have β(T − λS) = α((T − λS)∗ ) = α(T ∗ − λS ∗ ) ≤ α(T ∗ ) = β(T ). (iii) Using Proposition 2.7.1 (ii), it suffices to show that R(QT −λS (T − λS)) is closed. Since QT −λS = QT , then R(QT −λS (T − λS)) = R(QT (T − λS)) = R(QT T − λQT S). So, R(QT T − λQT S) is closed. Thus, R(QT −λS (T − λS)) is closed. Q.E.D.

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Spectral Theory of Multivalued Linear Operators

Corollary 3.3.1 If the hypothesis of Theorem 3.3.1 is satisfied, then (i) If T ∈ Φ+ (X, Y ), then λS − T ∈ Φ+ (X, Y ) and i(λS − T ) = i(T ). (ii) If T ∈ Φ− (X, Y ), then λS − T ∈ Φ− (X, Y ) and i(λS − T ) = i(T ). (iii) If T ∈ Φ(X, Y ), then λS − T ∈ Φ(X, Y ) and i(λS − T ) = i(T ).

♦

Theorem 3.3.2 Let T ∈ CR(X, Y ) and S ∈ LR(X, Y ). Then, (i) If T ∈ Φ+ (X, Y ) and S ∈ PR (Φ+ (X, Y )), then T + S ∈ Φ+ (X, Y ) and i(T + S) = i(T ). (ii) If T ∈ Φ− (X, Y ) and S ∈ PR (Φ− (X, Y )), then T + S ∈ Φ− (X, Y ) and i(T + S) = i(T ). (iii) If T ∈ Φ(X, Y ) and S ∈ PR(Φ(X, Y )), then T + S ∈ Φ(X, Y ) and i(T + S) = i(T ). ♦ Proof. (i) Assume that T ∈ Φ+ (X, Y ) and S ∈ PR (Φ+ (X, Y )). Since T is closed, then by applying Proposition 2.7.1, we have T (0) is closed. Hence, (T + S)(0) = T (0) is closed. Moreover, QT +S = QT . Hence, QT +S (T +S) = QT (T + S) = QT T +QT S. Since S(0) ⊂ T (0), then S(0) . ⊂ T (0) = T (0). Furthermore, applying Lemma 2.1.3 (ii), we have Y /S(0) T (0)/S(0) ≡ Y /T (0), and QT = QR QS , where R = T (0)/S(0). This enables us to conclude that QT +S (T + S) = QT T + QR QS S.

(3.49)

By using Proposition 2.7.1, we have QT T is closed single valued. Clearly, QR QS S is single valued continuous, so applying Proposition 2.7.1, we infer that QT T + QR QS S is closed. In view of (3.49), we have QT +S (T + S) is a closed single valued. So, by using Proposition 2.7.1, we obtain that T + S is closed. Let us consider two cases for S. First case : If kSk < γ(T ), then in view of Lemma 2.14.3 (i), we have T ∈ Φ+ (X, Y ) if and only if, QT T is single valued upper semi-Fredholm. Since D(QT T ) = D(T ) ⊂ D(S) = D(QR QS S) and kQR QS Sk ≤ kQS Sk = kSk < γ(T ) = γ(QT T ), then QT T +QR QS S is upper semi-Fredholm and i(QT T +QR QS S) = i(QT T ). Now, applying (3.49), we obtain that QT +S (T + S) ∈ Φ+ (X, Y /(T + S)(0)) with i(QT T ) = i (QT +S (T + S)). By applying Lemma 2.14.3 (i), to conclude that T + S ∈ Φ+ (X, Y ) and i(T ) = i(T + S).

The Stability Theorems of Multivalued Linear Operators

145

Second case : If kSk ≥ γ(T ), then since S ∈ PR (Φ+ (X, Y )) we have λS ∈ PR (Φ+ (X, Y )). So, by using Definition 2.17.1, we have T + λS ∈ Φ+ (X, Y )

(3.50)

for all λ ∈ C. Then, T + S ∈ Φ+ (X, Y ). Now, we prove i(T + S) = i(T ). For this, we shall prove that the map [ ϕ : I := [0, 1] −→ Z {−∞} λ −→

ϕ(λ) = i(T + λS)

is continuous. Let λ0 ∈ I arbitrary but fixed. Then, for λ ∈ I such that +λ0 S) |λ − λ0 | < γ(TkSk , and by using (3.50) and Corollary 2.16.1 (iv), we have A := T + λ0 S − (λ0 S − λS) ∈ Φ+ (X, Y ) and i(A) = i(T + λ0 S).

(3.51)

Since S(0) ⊂ T (0), then QT = QT +λS = QT +λ0 S = QT +λS+λ0 S−λ0 S and QT = QR QS (as in (3.49)). So, QS (T + λS + λ0 S − λ0 S)

= QS (T + λS) + λ0 QS S − λ0 QS S = QS (T + λS)

and QT A = QA A = QR QS A = QR QS (T + λS) = QT (T + λS) = QT T + λS(T + λS). Then, we obtain that QA A = QT +λS (T + λS).

(3.52)

The use of Lemma 2.14.3 (i) and (3.51) and (3.52), implies i(QA A) = i(A) = i(QT +λS (T + λS)) = i(T + λS) = i(T + λ0 S). Thus, i(T + λ0 S) = i(T + λS). 0 S) = δ such that for all λ ∈ I with Now, let ε > 0, then there exists γ(Tk+λ Sk |λ − λ0 | < δ, we have 0 = |i(T + λS) − i(T + λ0 S)| < ε. So, ϕ is continuous. So, ϕ(I) is a connected set which therefore, consists of only one point. It follows that i(T ) = ϕ(0) = ϕ(1) = i(T + S). This completes the proof of (i). (ii) Let T ∈ Φ− (X, Y ) and S ∈ PR (Φ− (X, Y )), then for all λ ∈ C, T + λS ∈ Φ− (X, Y ) equivalently (T + λS)∗ ∈ Φ+ (Y ∗ , X ∗ ). Since (T + λS)∗ = T ∗ + (λS)∗ = T ∗ + λS ∗ (obvious that (λS)∗ = λS ∗ , λ 6= 0), then T ∗ + λS ∗ ∈ Φ+ (Y ∗ , X ∗ ). Furthermore, since S ∗ is continuous, then λS ∗ is continuous, λS ∗ (0) = S ∗ (0) ⊂ T ∗ (0) and D(T ∗ ) ⊂ D(S ∗ ) = D(λS ∗ ). So, we can apply (i) obtaining that i(T ∗ ) = i(T ∗ + S ∗ ) with T ∗ and T ∗ + S ∗ ∈ Φ+ (Y ∗ , X ∗ ).

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Spectral Theory of Multivalued Linear Operators

Thus, −i(T ) = i(T ∗ ) = i(T ∗ + S ∗ ) = i((T + S)∗ ) = −i(T + S). Therefore, i(T ) = i(T + S), as required. (iii) The proof of (iii) follows from the same reasoning as previously. Q.E.D. Theorem 3.3.3 Let S, T , K ∈ LR(X, Y ) such that D(T ) ⊂ D(S) ⊂ D(K), K(0) ⊂ S(0) ⊂ T (0) and let Tb be the bijection associated with T . Suppose (a) there exists a constant α1 such that kSxk ≤ α1 (kxk + kT xk) , x ∈ D(T ), (b) there exists a constant β1 such that α1 (1 + β1 ) < 1, (1 + β1 ) < γ(Tb) and kKxk ≤ β1 (kxk + kSxk) , x ∈ D(T ), If T ∈ Φ(X, Y ), then the sum T + S + K ∈ Φ(X, Y ) and satisfy the following properties (i) α(T + S + K) ≤ α(T ), and (ii) β(T + S + K) ≤ β(T ).

♦

Proof. From Theorem 3.1.4, it follows that T + S + K is a closed linear relation. Let T1 , S1 , and K1 be the restrictions of the relation T , S, and K to XT , respectively. Obviously, T is Fredholm linear relation and S1 , K1 is a bounded linear relation. Moreover, it is easy to prove that kS1 + K1 k ≤ (β1 + α1 (1 + β1 )) ≤ γ(Tb). Then, by using Corollary 2.16.1 (vii), we have S1 + K1 + T1 is a Fredholm linear relation. The properties (i) and (ii) are straightforward consequences of Theorem 2.9.3. Q.E.D. Corollary 3.3.2 Let S, T , and K be three operators such that D(T ) ⊂ D(S) ⊂ D(K) and let Tb be the bijection associated with T . Suppose, (a) there exists a constant α1 such that kSxk ≤ α1 (kxk + kT xk) , x ∈ D(T ), (b) there exists a constant β1 such that α1 (1 + β1 ) < 1, (1 + β1 ) < γ(Tb) and kKxk ≤ β1 (kxk + kSxk) , x ∈ D(T ),

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If T ∈ Φ(X, Y ), then T + S + K ∈ Φ(X, Y ) and also satisfies the following properties (i) α(T + S + K) ≤ α(T ), (ii) β(T + S + K) ≤ β(T ), and (ii) i(T + S + K) = i(T ).

3.4

♦

Atkinson Linear Relations

Theorem 3.4.1 Let X and Y be two Banach spaces and T ∈ CR(X, Y ). Then, (i) If T ∈ Aα (X, Y ), then QT T ∈ Aα (X, Y /T (0)). (ii) If QT T ∈ Aα (X, Y /T (0)) and T (0) is topologically complemented in Y , then T ∈ Aα (X, Y ). (iii) If T ∈ Aβ (X, Y ), then QT T ∈ Aβ (X, Y /T (0)). (iv) If QT T ∈ Aβ (X, Y /T (0)) and T ∗ (0) is topologically complemented in Y ∗ , then T ∈ Aβ (X, Y ). ♦ Proof. (i) Let T ∈ Aα (X, Y ). Then, by using Lemma 2.14.3 (i), we deduce that QT T ∈ Φ+ (X, Y /T (0)) with R(QT T ) = R(T )/T (0) = R(T )/T (0). Since L R(T ) N = Y for some closed subspace N of Y and N (QT ) = T (0) = L T (0) ⊂ R(T ), then R(QT T ) QT N = Y /T (0). So, R(QT T ) is topologically complemented in Y /T (0), as desired. (ii) Let QT T ∈ Aα (X, Y /T (0)). By virtue of Lemma 2.14.3 (i), we have T ∈ Φ+ (X, Y ). So, our purpose is to verify that R(T ) is topologically complemented in Y . In order to prove it, let us notice that Y /T (0) = L R(QT T ) QT M, for some closed subspace M of Y . The use of Lemma 2.3.1 T allows us to ensure that Y = R(T ) + M and T (0) = R(T ) M . Since, by hypothesis, T (0) is topologically complemented in Y , then there exists a closed L L T subspace N of Y such that T (0) N = Y . Hence, Y = R(T ) (N M ). So, R(T ) is topologically complemented in Y , as desired. The proof of (iii) (respectively, (iv)) follows immediately from (i) (respectively, (ii)) and Lemma 2.14.5. Q.E.D. As a direct consequence of Theorem 3.4.1, we have the following:

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Corollary 3.4.1 Let T ∈ CR(X, Y ). Then, (i) If T (0) is topologically complemented in Y , then T ∈ Aα (X, Y ) if and only if, QT T ∈ Aα (X, Y /T (0)). (ii) If T ∗ (0) is topologically complemented in Y ∗ , then T ∈ Aβ (X, Y ) if and only if, QT T ∈ Aβ (X, Y /T (0)).

♦

Proposition 3.4.1 Let T ∈ CR(X, Y ) such that T (0) is topologically complemented in Y. Then, the following properties are equivalent. (i) T ∈ Aα (X, Y ). (ii) There exist A ∈ L(X, Y ), a bounded finite rank projection F ∈ L(X) such that N (A) is topologically complemented in Y , R(A) is a closed subspace contained in D(T ), R(F ) ⊂ D(T ) and AT = (I − F )|D(T ) . (iii) There exist A ∈ L(Y, X) and a continuous operator B in X such that D(T ) ⊂ D(B), I − B ∈ Φ(X), R(B) ⊂ D(T ) and AT = (I − B)|D(T ) . ♦ Proof. (i) =⇒ (ii) Let us consider two cases for T : First case : T is an operator. Since T ∈ Φ+ (X, Y ), then there is a closed L finite codimensional subspace M of X for which N (T ) M = X. So, L T N (T ) (M D(T )) = D(T ). Since R(T ) is topologically complemented in L Y , then there exists a closed subspace N of Y such that R(T ) N = Y . Let
−1 P denote the bounded projection of Y onto R(T ) and let A := T|M P. Then,
 T D(A) = y ∈ D(P ) such that P y D (T|M )−1 6= ∅  = y ∈ Y such that P y ∈ R(T|M ) = R(P ) = Y.
−1 By virtue of Theorem 2.5.4, T|M is continuous. Hence, A is continuous. Moreover,
R(A) = (T|M )−1 P Y T = (T|M )−1 (T|M )(M D(T )) T = (M D(T )) and N (A)

= P −1 N (T|M )−1 = N (P ) = N.




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Let F be the bounded projection of X onto N (T ). Then, it is obvious that R(F ) ⊂ D(T ). For x ∈ D(T ), we can write x = x1 + x2 , where x1 ∈ N (T ) T and x2 ∈ M D(T ). Then, AT x = (T|M )−1 P (T|M )x2 = x2 = (I − F )x. Therefore, (i) =⇒ (ii) whenever T is an operator. Second case : T is a linear relation. Then, by using Lemma 2.14.3 (i), we have L QT T ∈ Φ+ (X, Y /T (0)) with R(QT T ) = R(T )/T (0). Since R(T ) N = Y for some closed subspace N of Y and N (QT ) = T (0) ⊂ R(T ), then L R(QT T ) QT N = Y /T (0). In this situation, it follows from the first case applied to QT T that there exist A ∈ L(Y /T (0), X) and a bounded finite rank projection F ∈ L(X) such that R(F ) ⊂ D(QT T ) = D(T ). N (A) is topologically complemented in Y /T (0), R(A) is a closed subspace contained in D(QT T ) and (AQT )T = (I − F )|D(T ) . Hence, it only remains to prove that N (AQT ) is topologically complemented in Y . To ensure this, we note that QT is a bounded lower semi-Fredholm operator from Y onto Y /T (0) and N (QT ) = T (0) is topologically complemented in Y by hypothesis, then from Theorem3.1.5 that N (AQT ) is topologically complemented in Y , as required. (ii) =⇒ (iii) It is clear. (iii) =⇒ (i) Let us consider various cases for T : First case : T is a densely defined operator. In such case the result was proved by V. Muller-Horrig ¨ in Theorem 2.14.5. Second case : T is an operator. Let A and B satisfy the conditions in (iii). Then, it is clear that AT GT = (I − B)|XT . Since T GT ∈ L(XT , Y ) and XT is complete, it follows from the first case applied to T GT that T GT ∈ Aα (XT , Y ). Hence, T ∈ Aα (X, Y ), as desired. Third case : T is a linear relation. Let A and B as in (iii). Recalling that a linear relation S is an operator if, and only if, S(0) = {0}. Hence, 1 −1 AQ− T (0) = AT (0) = (I − B)|D (T ) = {0}. So, AQT is an op erator. Clearly, it is everywhere defined and continuous. Thus, AQ−1 T is bounded operator. So, −1 AQT ∈ L(Y /T (0), X). On the other hand,   \ Q−1 Q T (D(T )) = T (D(T )) D(Q ) + N (QT ) T T T

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which implies that (AQ−1 T )QT T = AT . Thus, it follows from the second case that QT T ∈ Aα (X, Y /T (0)). Hence, QT T ∈ Φ+ (X, Y /T (0)) and R(QT T ) is topologically complemented in Y /T (0). By virtue of Lemma 2.14.3 (i), T ∈ Φ+ (X, Y /T (0)). So, it remains to see that R(T ) is topologically complemented L in Y. To prove this, we note that Y /T (0) = R(QT T ) QT U for some closed T subspace U of X containing T (0), Y = R(T ) + U and T (0) = R(T ) U . Since L by hypothesis T (0) V = Y for some closed subspace V of Y , then Y = L T R(T ) (V U ). So, R(T ) is topologically complemented in Y , as desired. This completes the proof. Q.E.D. Definition 3.4.1 An operator Tl ∈ L(Y, X) satisfying Proposition 3.4.1 (ii) is called a left regularizer (or a left generalized inverse) of T . ♦ Proposition 3.4.2 Let T ∈ Aα (X) such that T (0) is topologically complemented in X and let K ∈ KT (X). Then, T + K ∈ Aα (X) and i(T + K) = i(T ). ♦ Proof. Let A and B as in part (iii) of Proposition 3.4.1. Then, A(T + K) = (I − (B − AK))|D(T ) . Further, AK(0) ⊂ AT (0) = {0}, so by Theorem 2.12.2, AK is compact single valued. Then, applying the implication (iii) =⇒ (i) in Proposition 3.4.1, we get T + K ∈ Aα (X) and the equality i(T + K) = i(T ) follows immediately from Lemma 3.5.5. Q.E.D. Theorem 3.4.2 Let T ∈ BCR(X). Then, (i) If T ∈ Aα (X) and T (0) is topologically complemented in X, then Tl T ∈ Φ(X) and i(Tl T ) = 0. (ii) If T ∈ Aβ (X), dim (T (0)) < ∞ and T Tr − T Tr ∈ KR(X), then (ii1 ) T Tr ∈ Φ(X) and i(T Tr ) = dim (T (0)) . (ii2 ) Tr ∈ Φ+ (X) and if T , Tr have finite indices, then i(T ) + i(Tr ) − dim (T (0)) = 0. ♦ Proof. (i) Follows immediately from Proposition 3.4.1 and Theorem2.16.2. (ii) (ii1 ) By using Lemma 2.14.6, there exist Tr ∈ L(X) and F ∈ L(X), which is a finite rank projection such that T Tr

= I − F + T Tr − T Tr = I − K,

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151

where K = F − T Tr + T Tr . Hence, K(0) = T (0) and K ∈ KR(X). So, from Theorem 2.16.2, we infer that T Tr ∈ Φ(X) and i(T Tr ) = dim (T (0)) . (ii) (ii2 ) By using (ii1 ), we have T Tr ∈ Φ+ (X) and by applying Theorem 2.14.3, we infer that Tr ∈ Φ+ (X). Moreover, by using Lemma 2.4.2, we have i(T ) + i(Tr ) = dim (T (0)) . Therefore, (ii2 ) holds. Q.E.D. Theorem 3.4.3 Let T ∈ complemented in Y. Then,

CR(X, Y ) such that T (0) is topologically

(i) If A is the left generalized inverse of QT T , then AQT is the left generalized inverse of T . (ii) If A is the left generalized inverse of T , then AQ−1 T is the left generalized inverse of QT T . ♦ Proof. (i) Let QT T ∈ Aα (X, Y /T (0)). Then, by using Proposition 3.4.1, there exist A ∈ L(Y /T (0), X) and F ∈ L(X) which is, a bounded finite rank projection such that R(F ) ⊂ D(QT T ) = D(T ). Moreover, N (A) is topologically complemented in Y /T (0), R(A) is a closed subspace contained in D(QT T ), and AQT T = (AQT )T = (I − F )|D(T ) . Finally, it is sufficient to prove that N (AQT ) is topologically complemented in Y . To ensure this, we notice that QT is a bounded lower semi-Fredholm operator from Y onto Y /T (0) and N (QT ) = T (0) is, by hypothesis, topologically complemented in Y. Hence, according to Theorem 3.1.5, we deduce that N (AQT ) is topologically complemented in Y. So, Tl = AQT . (ii) Let Tl = A. Then, by using Proposition 3.4.1, there exists F ∈ L(X), which is a bounded finite rank projection such that N (A) is topologically complemented in Y , R(A) is a closed subspace contained in D(T ), R(F ) ⊂ D(T ), and AT = (I − F )|D(T ) . Since T (0) is closed, then (AQ−1 T )QT T = AT = (I − F )|D(T ) . Knowing that AQ−1 T (0) = AT (0) = (I − F )|D(T ) = {0}, this allows us to −1 deduce that AQT is an operator and clearly, it is everywhere defined and continuous. Thus, AQ−1 T ∈ L(Y /T (0), X). Now, by applying Proposition 3.4.1, we infer that (QT T )l = AQ−1 Q.E.D. T .

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3.5

Spectral Theory of Multivalued Linear Operators

The α-and β-Atkinson Perturbation Classes

The goal of this section is to study the class of α- and β-Atkinson perturbation of linear relations and to present the relationship between upper semi-Fredholm perturbations (respectively, lower semi-Fredholm perturbations) and the sets of α-Atkinson perturbations(respectively, βAtkinson perturbations). Theorem 3.5.1 Let X be a complex Banach space. Then, (i) KR(X) ⊂ SSR(X) ⊂ PR (Φ+ (X)) ⊂ PR (Aα (X)) . (ii) KR(X) ⊂ PR (Φ− (X)) ⊂ PR (Aβ (X)) .

♦

Proof. (i) First, let us notice that the inclusion KR(X) ⊂ SSR(X) is a simple consequence of Proposition 2.12.1. Second, we have to prove that SSR(X) ⊂ PR(Φ+ (X)). Let A ∈ SSR(X) and B ∈ Φ+ (X) such that D(B) ⊂ D(A) and A(0) ⊂ B(0). Since B is closed, then B(0) = B(0). By using Corollary 2.16.1 (iv), we deduce that A + B ∈ Φ+ (X). This shows that SSR(X) ⊂ PR (Φ+ (X)) . Finally, we have to prove that PR(Φ+ (X)) ⊂ PR (Aα (X)). For this purpose, let A ∈ PR (Φ+ (X)) and let B ∈ Aα (X) such that D(B) ⊂ D(A) and A(0) ⊂ B(0). By using Proposition 2.12.1, we conclude that A + B ∈ CR(X).

(3.53)

By applying Proposition 3.4.1 to B, there exist Bl ∈ L(X) and F , which is a finite rank linear operator, such that D(B) ⊂ D(F ), R(F ) ⊂ D(B), and Bl B = (I − F )|D(B) . We can easily deduce that Bl (A + B) = Bl B + Bl A = (I − F + Bl A)|D(B) .

(3.54)

Now, we propose to show that Bl A ∈ PR (Φ+ (X)) .

(3.55)

Let us consider two cases of Bl : First case : Bl is invertible single valued. Since Bl A(0) ⊂ Bl B(0) = {0}, then Bl A is an operator. For this, we take S ∈ Φ+ (X) which is single valued. Then,

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153

Bl−1 S + A ∈ Φ+ (X). Moreover, Bl ∈ Φ+ (X) and it is single valued. By using Theorem 2.14.3, we have
Φ+ (X) 3 Bl Bl−1 S + A = S + Bl A. This implies that Bl A ∈ PR(Φ+ (X)). Second case : Bl is not invertible single valued. Since Bl ∈ L(X), then ρ(Bl ) 6= ∅. Hence, there exists λ ∈ ρ(Bl ) such that Bl is the sum of two invertible operators Bl = λ + (Bl − λ) . By applying the previous case and Proposition 2.17.1 (i), we have Bl A ∈ PR(Φ+ (X)). Consequently, (3.55) holds. So, by using (3.53), (3.54) and (3.55), we deduce that Bl (A + B) ∈ Φ+ (X).

(3.56)

Moreover, R(Al ) ⊂ D(A + B) and R (F − Bl A) ⊂ D(A + B) and clearly, we have (A + B)(0) = B(0). Then, (A + B)(0) is complemented in X. By using (3.53), (3.54), (3.56), Proposition 3.4.1, we infer that A + B ∈ Aα (X, Y ). So, PR(Φ+ (X)) ⊂ PR (Aα (X)) . (ii) In order to prove the inclusion KR(X) ⊂ PR(Φ− (X)), let A ∈ KR(X). Hence, we take B ∈ Φ− (X) such that D(B) ⊂ D(A) and A(0) ⊂ B(0). Then, by using Proposition 2.8.1, we have ∗

(A + B) = A∗ + B ∗ .

(3.57)

We claim that A∗ (0) ⊂ B ∗ (0) and D(B ∗ ) ⊂ D(A∗ ). Indeed, since D(B) ⊂ D(A), then by using Proposition 2.8.1, we have A∗ (0) = D(A)⊥ ⊂ D(B)⊥ = B ∗ (0). Hence, A∗ (0) ⊂ B ∗ (0), and by applying D(B ∗ ) ⊂ B(0)⊥ ⊂ A(0)⊥ ⊂ D(A∗ ), we have D(B ∗ ) ⊂ D(A∗ ) which proves our claim. By using Proposition 2.11.2, we have A∗ ∈ KR(X ∗ ) and by using Theorem 2.14.4, we have B ∗ ∈ Φ+ (X ∗ ). Hence, we conclude that A∗ + B ∗ ∈ Φ+ (X ∗ ).

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Then, from (3.57) and Theorem 2.14.4, we have A + B ∈ Φ− (X). So, A ∈ PR(Φ− (X)). The proof of the the inclusion PR (Φ− (X)) ⊂ PR (Aβ (X)) follows from the same reasoning as previously. Q.E.D. Theorem 3.5.2 Let X and Y be two Banach spaces. Then, (i) If T ∈ Aα (X, Y ) and S ∈ PR (Aα (X, Y )), then T + S ∈ Aα (X, Y ) i(T + S) = i(T ). (ii) If T ∈ Aβ (X, Y ) and S ∈ PR (Aβ (X, Y )), then T + S ∈ Aβ (X, Y ) i(T + S) = i(T ). (iii) If T ∈ Φ+ (X, Y ) and S ∈ PR (Φ+ (X, Y )), then T + S ∈ Φ+ (X, Y ) i(T + S) = i(T ). (iv) If T ∈ Φ− (X, Y ) and S ∈ PR (Φ− (X, Y )), then T + S ∈ Φ− (X, Y ) i(T + S) = i(T ).

and and and and ♦

Proof. (i) Let S ∈ PR(Aα (X, Y )). Then, S(0) ⊂ T (0), D(T ) ⊂ D(S) and S is continuous. By using Proposition 2.7.2, we have T + S ∈ CR(X, Y ). Thus, in view of Definition 2.17.1, we infer that T + S ∈ Aα (X, Y ). Let λ ∈ K, then λS ∈ PR(Aα (X, Y )). So, λS + T ∈ Aα (X, Y ) for all λ ∈ K. It follows that ϕ(λ) = i(λS + T ) is continuous from [0, 1] into Z (Z := integers), where [0, 1] has the usual topology and Z is the topological space consisting of the integers, and {−∞} with the discrete topology. Hence, ϕ is a constant function. In particular, i(T ) = ϕ(0) = ϕ(1) = i(T + S). The proofs of (ii), (iii) and (iv) may be achieved by using the same reasoning as (i). Q.E.D. Corollary 3.5.1 Let S, T ∈ LR(X) such that S(0) ⊂ T (0) and D(T ) ⊂ D(S). (i) Assume that T (0) is topologically complemented in X. Then, (i1 ) If T ∈ Aα (X) and S ∈ SSR(X), then T +S ∈ Aα (X) and i(T +S) = i(T ). (i2 ) If T ∈ W l (X) and S ∈ SSR(X), then T + S ∈ W l (X). (ii) Assume that T ∗ (0) is topologically complemented in X ∗ . Then, (ii1 ) If T ∈ Aβ (X) and S ∈ KR(X), then T +S ∈ Aβ (X) and i(T +S) = i(T ).

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155

(ii2 ) If T ∈ W r (X) and S ∈ KR(X), then T + S ∈ W r (X). (iii) If T ∈ Φ+ (X) and S ∈ SSR(X), then T + S ∈ Φ+ (X) and i(T + S) = i(T ). (iv) If T ∈ Φ− (X) and S ∈ KR(X), then the linear relation T + S ∈ Φ− (X) and i(T + S) = i(T ). ♦ Theorem 3.5.3 Let T ∈ LR(X). Then, (i) If T ∈ Aα (X), T (0) is topologically complemented in X, and α(T ) ≤ β(T ), then there exists K ∈ L(X) with dim (R(K)) ≤ α(T ), such that the linear relation S = T − K satisfies S ∈ GRl (X). (ii) If T ∈ Aβ (X), T ∗ (0) is topologically complemented in X ∗ , and β(T ) ≤ α(T ), then there exists K ∈ L(X) with dim (R(K)) ≤ α(T ) such that the linear relation S = T − K satisfies S ∈ GRr (X). ♦ Proof. (i) Let T ∈ Aα (X) and i(T ) ≤ 0. If α(T ) = 0, then we can consider K = 0 and S = T . Suppose that 1 ≤ n := α(T ) and let {x1 , . . . , xn } be a basis of N (T ). Let us choose x01 , . . . , x0n ∈ X ∗ such that x0i (xj ) = δij , where δij = 0 if i 6= j and δij = 1 if i = j. Let us choose y1 , . . . , yn ∈ X such that [y1 ], . . . , [yn ] ∈ X/R(T ) are linearly independent. The single valued linear operator K is defined by K : X 3 x −→ Kx =

n X

x0i (x)yi ∈ X.

i=1

Hence, K is an everywhere defined linear operator, ! n X 0 kKxk ≤ kxi kkyi k kxk, i=1

with dim (R(K)) ≤ n and T − K is injective. By using Corollary 3.5.3 (i), we deduce that R(T − K) is a closed and complemented subspace of X. So, T − K ∈ GRl (X). The proof of (ii) may be achieved in the same way as in the proof of (i). This completes the proof. Q.E.D. As a direct consequence of Theorems 3.5.3 and 3.5.5, we infer the following result. Corollary 3.5.2 Let T ∈ LR(X).

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(i) Let us assume that T (0) is topologically complemented in X. Then, the following conditions are equivalent (i1 ) T ∈ W l (X). (i2 ) There exist K ∈ KT (X) and S ∈ GRl (X), such that T = S + K. (i3 ) There exist K ∈ PRα (A(X)) and S ∈ GRl (X) such that T = S + K. (ii) Let us assume that T ∗ (0) is topologically complemented in X ∗ . Then, the following conditions are equivalent (ii1 ) T ∈ W r (X). (ii2 ) There exist K ∈ KT (X) and S ∈ GRr (X) such that T = S + K. (ii3 ) There exist K ∈ PRβ (A(X)) and S ∈ GRr (X) such that T = S +K. ♦ Theorem 3.5.4 Let T ∈ CR(X, Y ) and S ∈ LR(X, Y ) be continuous such that S(0) ⊂ T (0) and S is T -bounded with T -bound δ < 1. Then, the following statements hold (i) If T ∈ Aα (X, Y ) and SGT is α-Atkinson perturbation, then T + S ∈ Aα (X, Y ) and i(T + S) = i(T ). (ii) If T ∈ Aβ (X, Y ) and SGT is β-Atkinson perturbation, then T + S ∈ Aβ (X, Y ) and i(T + S) = i(T ). (iii) If T ∈ Φ+ (X, Y ) and SGT is an upper semi-Fredholm perturbation, then T + S ∈ Φ+ (X, Y ) and i(T + S) = i(T ). (iv) If T ∈ Φ− (X, Y ) and SGT is an lower semi-Fredholm perturbation, then T + S ∈ Φ− (X, Y ) and i(T + S) = i(T ). ♦ Proof. (i) Let T ∈ Aα (X, Y ). Since S ∈ LR(X, Y ) is continuous such that S(0) ⊂ T (0) and S is T -bounded with T -bound δ < 1, then by using Proposition 2.7.2, we have (T + S)GT ∈ CR(XT , Y ). Observe that (T + S)GT = T GT + SGT with T GT ∈ Aα (XT , Y ) and SGT ∈ PR (Aα (XT , Y )). Hence, (T + S)GT ∈ Aα (XT , Y ). Thus, by applying Remark 2.7.2, we infer that i(T + S) = i((T + S)GT ). Since N (T GT ) = N (T ) and R(T ) = R(T GT ), it follows from Theorem 3.5.4 that i((T + S)GT ) = i(T GT ) = i(T ). The proofs of (ii), (iii) and (iv) may be achieved by the same reasoning as for (i). Q.E.D. Theorem 3.5.5 Let X and Y be two complete spaces. Then, the following statements hold

The Stability Theorems of Multivalued Linear Operators (i) SSR(X, Y ) ⊂ PR(Aα (X, Y )). (ii) KR(X, Y ) ⊂ PR(Aβ (X, Y )).

157

♦

Proof. (i) Let A ∈ SSR(X, Y ). We will prove that A ∈ PR(Aα (X, Y )). Let B ∈ Aα (X, Y ) such that D(B) ⊂ D(A) and A(0) ⊂ B(0). Since A is continuous, then by using Proposition 2.7.2, we conclude that A + B ∈ CR(X, Y ). Applying Proposition 3.4.1 to B, there exist Bl ∈ L(Y, X) and F , which is a finite rank linear operator, such that D(B) ⊂ D(F ), R(F ) ⊂ D(B), and Bl B = (I − F )|D(B) . We can easily deduce that Bl (A + B) = Bl B + Bl A = (I − F + Bl A)|D(B) .

(3.58)

Now, in order to show that Bl (A + B) ∈ Φ+ (X), we will prove that Bl A ∈ PR(Φ+ (X)). Since A ∈ SSR(X, Y ) and Bl is bounded single valued, then by using Proposition 2.3.4 (iii), we deduce that Bl A ∈ SSR(X). It follows from Theorem 3.5.5 that Bl A ∈ PR(Φ+ (X)). Now, by using (3.58) and Bl A ∈ PR(Φ+ (X)), we conclude that Bl (A + B) ∈ Φ+ (X). Moreover, R(Bl ) ⊂ D(A + B) and R (F − Bl A) ⊂ D(A + B) and clearly, we have (A + B)(0) = B(0). Then, (A + B)(0) is complemented in X. Now, the use of Proposition 3.4.1 allows us to conclude that A + B ∈ Aα (X, Y ). Hence, SSR(X, Y ) ⊂ PR(Aα (X, Y )). (ii) Let A ∈ KR(X, Y ). We are going to prove that A ∈ PR(Aβ (X, Y )). Indeed, we take B ∈ Aβ (X, Y ) such that D(B) ⊂ D(A) and A(0) ⊂ B(0). We prove that A + B ∈ Aβ (X, Y ). In view of Lemma 2.14.5 (ii), it is sufficient to prove that (A + B)∗ ∈ Aα (Y ∗ , X ∗ ). Since, D(B) ⊂ D(A) and A is continuous, then by using Proposition 2.8.1 (iii), we have A∗ +B ∗ = (A+B)∗ . By referring to Lemmas 2.3.4 (ii) and 2.14.5, we have B ∗ ∈ Aα (Y ∗ , X ∗ ), A∗ (0) ⊂ B ∗ (0), and D(B ∗ ) ⊂ D(A∗ ). Since A∗ ∈ PRK(Y ∗ , X ∗ ), then in view of Propositions 2.12.1 and 2.8.1, implies that A∗ is continuous strictly singular linear relation. Hence, A∗ + B ∗ ∈ Aα (Y ∗ , X ∗ ). So, A + B ∈ Aβ (X, Y ). Q.E.D. As an immediate consequence of Theorems 3.5.4 and 3.5.5: Corollary 3.5.3 Let T ∈ CR(X, Y ) and S ∈ LR(X, Y ) be continuous such that S(0) ⊂ T (0) and S is T -bounded with T -bound δ < 1. (i) Assume that T (0) is topologically complemented in X. Then, (i1 ) If T ∈ Aα (X, Y ) and SGT ∈ KR(XT , Y ) (respectively, SGT ∈ SSR(XT , Y )), then T + S ∈ Aα (X, Y ) and i(T + S) = i(T ). (i2 ) If T ∈ W l (X, Y ) and SGT ∈ KR(XT , Y ) (respectively, SGT ∈ SSR(XT , Y )), then T + S ∈ W l (X, Y ).

158

Spectral Theory of Multivalued Linear Operators

(ii) Assume that T ∗ (0) is topologically complemented in X ∗ . Then, (ii1 ) If T ∈ Aβ (X, Y ) and SGT ∈ KR(XT , Y ), then T + S ∈ Aβ (X, Y ) and i(T + S) = i(T ). (ii2 ) If T ∈ W r (X, Y ) and SGT ∈ KR(XT , Y ), then T + S ∈ W r (X, Y ). ♦

3.6

Index of a Linear Relations

Proposition 3.6.1 Let n and mi , 1 ≤ i ≤ n be some positive integers, and let λi ∈ K, 1 ≤ i ≤ n be some distinct constants. Let T ∈ LR(X) such that ρ(T ) 6= ∅. Let P (T ) be defined in (2.47). If each factor T − λk has finite index, then n X i(P (T ) − µ) = i ((T − λk )mk ) . ♦ k=1

Proof. We first show that each ρ(T − λk ) 6= ∅.

(3.59)

Since T − α = (T − λk ) + (λk − α) for all α ∈ C, we have η − λk ∈ ρ(T − λk ) whenever η ∈ ρ(T ). So, (3.59) holds. Now, combine (3.59) with Lemma 2.5.1, we have n Y P (T ) − µ = c (T − λk )mk . k=1

It only remains to see that i

n Y

! mk

(T − λk )

k=1

=

n X

i ((T − λk )mk ) .

k=1

We prove the result by induction. For n = 1, it is clear by Lemma 2.5.1 that i ((T − λk )mk ) = mk i(T − λk ). Assume that it is true for n = r, that is ! r r Y X i (T − λk )mk = i ((T − λk )mk ) . k=1

k=1

The Stability Theorems of Multivalued Linear Operators Let A =

r Y

(T − λk )mk and B = (T − λr+1 )

mr+1

159

. Then, by the induction

k=1

hypothesis, we infer that A and B have finite index. Thus, it follows from Theorem 2.14.3 that     \ X i(AB) = i(A) + i(B) + dim − dim N (A) B(0) . D(A) + R(B) By applying Lemma 2.13.1, we have D(A) = D(T m1 +m2 +···+mr ) = D ((T − λr+1 )m1 +m2 +···+mr ) , and R(B) = R ((T − λr+1 )mr+1 ) . Hence,
D(A) + R(B) = D (T − λr+1 )m1 +m2 +···+mr + R ((T − λr+1 )mr+1 ) = X. (3.60) It is enough to observe that by Lemma 2.5.6 (ii), we have α(A) ≤ α ((T − λ1 )m1 ) + α ((T − λ2 )m2 ) + · · · + α ((T − λr )mr ) ≤ m1 α(T − λ1 ) + m2 α(T − λ2 ) + · · · + mr α(T − λr ) < ∞. Hence,   \ 0 < dim N (A) B(0) = δ < ∞.

(3.61)

Now, it follows from (3.60) and (3.61) that i(AB) = i(A) + i(B) − δ. Hence, i(AB) ≤ i(A) + i(B). It only remains to verify that i(A) + i(B) ≤ i(AB). We first note that AB = BA and by virtue of the part (i) of Lemma 2.5.1 and, in view of Theorem 2.14.3, we get     \ X i(BA) = i(AB) = i(A)+i(B)+dim −dim N (B) A(0) . D(B) + R(A) The use of Theorem 2.14.3, make us conclude that A is a Fredholm relation, in particular   X dim 1 and for all x ∈ D(T ) kT x − T0 xk ≥ |a|kSx − T0 xk − |b|kSx − T xk, then T is an S-demicompact linear relation.

♦

176

Spectral Theory of Multivalued Linear Operators

Proof. Since kT x − T0 xk = kT x − T0 x + Sx − Sxk ≤

kSx − T0 xk + kSx − T xk,

then |a|kSx − T0 xk − |b|kSx − T0 xk ≤ kSx − T xk + kSx − T xk. Hence, (|a| − 1)kSx − T0 xk ≤ (1 + |b|)kSx − T xk. So,  kSx − T0 xk

≤

1 + |b| |a| − 1

 kSx − T xk.

(3.77)

Now, take (xn )n a bounded sequence of D(T ) such that QT (S − T )xn → y. Applying (3.77), we get kQT0 (S − T0 )(xn − y)k → 0. So, by using the fact that T0 is S-demicompact, we have QT0 (S − T0 )xn → y. Thus, we obtain (QT0 Sxn )n has a convergent subsequence. On the other side, we have kQT Sxn k =

dist(T (0), Sxn )

≤ dist(T0 (0), Sxn ) = kQT0 Sxn k. Finally, we get (QT Sxn )n has a convergent subsequence.

Q.E.D.

Proposition 3.8.6 Let α, µ, k ∈ C such that µ 6= 0. Let T : D(T ) ⊂ X −→ Y be a continuous linear relation and S : D(S) ⊂ X −→ Y . If T is a (|k|−T (0))set-contraction, then |α|T is µS-demicompact for each |αk| < 1. ♦ Proof. Let (xn )n be a bounded sequence of D(T ) such that Q|α|T µSxn − Q|α|T |α|T xn → y. Then, Q|α|T µS xn

= Q|α|T (µSxn − |α|T xn ) + Q|α|T |α|T xn .

(3.78)

If γ({Q|α|T µSxn }) 6= 0, then by using (3.78), we have γ({Q|α|T µSxn }) ≤ γ({Q|α|T (µSxn − |α|T xn )}) + γ({Q|α|T |α|T xn }), ≤

|αk|γ({Q|α|T xn }),


0 such that (i) α(λ − A) = α(A|X2 ) ≤ α(A) for all 0 < |λ| < δ. (ii) β(λ − A) = β(A|X2 ) ≤ β(A) for all 0 < |λ| < δ.

♦

Proof. (i) Since A is essentially semi regular, then there exists a L decomposition X = X1 X2 with the properties that dim X1 < ∞, A|X2 is a semi regular linear relation and A|X1 is a nilpotent bounded operator L of degree d (Ad = 0). Then, λ − A = (λ − A)|X1 (λ − A)|X2 for all λ ∈ C. Therefore, α(λ − A) = α((λ − A)|X1 ) + α((λ − A)|X2 ). Since A|X1 is a nilpotent bounded operator of degree d, then (λ − A)|X1 is invertible for all λ 6= 0 which implies that α((λ − A)|X1 ) = 0. Since A|X2 is a semi regular linear relation, it follows from Lemma 3.9.2 that there exists a positive constant δ > 0 such that (λ − A)|X2 is semi regular for all 0 < |λ| < δ. In view of Theorem 2.15.1, we have α((λ − A)|X2 ) = α(A|X2 ) for all 0 < |λ| < δ. Hence, α(λ − A) = α((λ − A)|X2 ) = α(A|X2 ) ≤ α(A) for all 0 < |λ| < δ. (ii) Let A be essentially semi regular. Then, there exists a decomposition L X = X1 X2 with the properties that A|X1 ⊂ X1 , A|X2 ⊂ X2 , dim X1 < ∞, A|X2 is a semi regular linear relation and A|X1 is a nilpotent bounded L operator of degree d. Hence, λ − A = (λ − A)|X1 (λ − A)|X2 . Therefore, β(λ − A) = β((λ − A)|X1 ) + β((λ − A)|X2 ). Now, a same reasoning as the proof of (i) leads to the result. Q.E.D. Theorem 3.9.4 Let X be a Hilbert space and A ∈ CR(X) be essentially semi regular linear relation. Then, (i) If α(A) < ∞ and Rc (A) = 0, then there exists a positive constant δ > 0 such that α(λ − A) = b.s.mult(A)

188

Spectral Theory of Multivalued Linear Operators

and if β(A) < ∞, then there exists a positive constant δ > 0 such that β(λ − A) = s.mult(A) for all 0 < |λ| < δ. In particular, the following two functions λ −→ s.mult(λ − A) and λ −→ b.s.mult(λ − A) are constant on a neighborhood of the origin D(0, δ) = {λ ∈ C such that |λ| < δ}. (ii) When k is large enough, if α(A) < ∞ and Rc (A) = 0, then !   X X s.mult(A) = dim = dim R(A) + N (Ak ) R(A) + N (A)∞ and if β(A) < ∞, then     \ \ b.s.mult(A) = dim N (A) R(Ak ) = dim N (A) R(A)∞ . (iii) A is upper semi-Fredholm (respectively, lower semi-Fredholm) if and only if, b.s.mult(A) < ∞ (respectively, s.mult(A) < ∞). Moreover, if b.s.mult(A) < ∞ and s.mult(A) < ∞, then A is semi-Fredholm and i(A) = b.s.mult(A) − s.mult(A).

♦

Proof. (i) Since A is essentially semi regular, then by applying Theorem L 3.9.2, there exists a decomposition X = X1 X2 with the properties that A|X1 ⊂ X1 , A|X2 ⊂ X2 , dim X1 < ∞, A|X2 is a semi regular linear relation and A|X1 is a nilpotent bounded operator of degree d. Let dim X1 = n0 , then by using Lemma 3.9.2, we have   α(Ak ) b.s.mult(A) = lim k→∞ k ! ! k α(A|X ) α(Ak|X1 ) 2 = lim + lim k→∞ k→∞ k k ! k   α(A|X2 ) n0 = lim + lim k→∞ k→∞ k k   kα(A|X2 ) = lim k→∞ k = α(A|X2 ),

The Stability Theorems of Multivalued Linear Operators and

 s.mult(A)

= = =

lim

k→∞

lim

k→∞

lim

 β(Ak ) k ! β(A|kX1 ) k ! β(Ak|X2 )

+ lim

k→∞

+ lim k→∞ k   kβ(A|X2 ) = lim k→∞ k = β(A|X2 ).

β(Ak|X2 )

189

!

k n 

k→∞

0

k

By using Lemma 3.9.3, there exists a positive constant δ > 0 such that α(λ − A) = α(A|X2 ) = b.s.mult(A) and β(λ − A) = β(A|X2 ) = s.mult(A) for all 0 < |λ| < δ. (ii) Since A is essentially semi regular, then by applying Theorem 3.9.2, there L exists a decomposition X = X1 X2 with the properties that A|X1 ⊂ X1 , A|X2 ⊂ X2 , dim X1 < ∞, A|X2 is a semi regular linear relation and A|X1 is a nilpotent bounded operator of degree d. By using Proposition 2.15.1 and in view of (i), we have     \ \ b.s.mult(A) = α(A|X2 ) = dim N (A) R(Ak ) = dim N (A) R(A)∞ . Let k ≥ dim X1 , then R(A) + N (A)∞

= = = = =

R(A|X1 ) + R(A|X2 ) + N ∞ (A|X1 ) + N ∞ (A|X2 ) R(A|X1 ) + R(A|X2 ) + N ∞ (A|X2 ) + X1 R(A|X1 ) + R(A|X2 ) + N ∞ (A|X2 ) + X1 R(A|X1 ) + R(A|X2 ) + X1 R(A|X1 ) + R(A|X2 ) + N (Ak|X1 ) + N (A|kX2 )

= R(A) + N (Ak ). Thus, X X = k R(A) + N (A ) R(A) + N (A)∞ for all k ≥ dim X1 . Consequently,

190

Spectral Theory of Multivalued Linear Operators s.mult(A)

= β(A|X 2)  X2 = dim )  A(X2L  X2 X1 L = dim  A(X2 ) X1  X = dim R(A) + N (Ak ) ! X = dim . R(A) + N (A)∞

(iii) If A ∈ Φ+ (X), then dim N (A) < ∞. This implies that T dim N (A) R(A) < ∞. By using (ii), we have b.s.mult(A) < ∞. If A ∈ Φ− (X), then β(A) < ∞. This implies that codim (N ∞ (A) + R(A)) < ∞. Hence, by using (ii), we have s.mult(A) < ∞. To prove the converse, let A be essentially semi regular, then R(A) is closed. By applying Theorem L 3.9.2, there exists a decomposition X = X1 X2 such that A|X2 is semi regular linear relation, A|X1 is a nilpotent bounded operator and dim X1 < ∞. Suppose that b.s.mult(A) < ∞, then α(A|X2 ) < ∞ and consequently, α(A) = α(A|X1 ) + α(A|X2 ) < ∞. Since R(A) is closed, then A ∈ Φ+ (X). Let s.mult(A) < ∞, then β(A|X2 ) < ∞. Hence, β(A) = β(A|X1 ) + β(A|X2 ) < ∞. Since R(A) is closed, then A ∈ Φ− (X). On the other hand, if α(A) and β(A) are finite and Rc (A) = 0, then by using Lemma 3.9.2, we have i(Ak ) = α(Ak ) − β(Ak ) = k i(A) and     α(Ak ) β(Ak ) i(A) = lim − lim k→∞ k→∞ k k = b.s.mult(A) − s.mult(A). This completes the proof.

Q.E.D.

Chapter 4 Essential Spectra and Essential Pseudospectra of a Linear Relations

In this chapter, we emphasize the strong connection between the spectral theory of closed linear relations and that of some closed linear operators. As a matter of fact, we develop a spectral theory for a certain class of linear operators, obtaining as consequences most of the main spectral properties of linear relations. Furthermore, we introduce and study the pseudospectra and the essential pseudospectra of linear relations. We start by giving the definition and we investigate the characterization and some properties of these pseudospectra. We end this chapter by gives some new results related to the pseudospectra and the essential pseudospectra of linear relations. We start by studying the stability of these pseudospectra and some characterization.

4.1

Characterization of the Essential Spectrum of a Linear Relations

Theorem 4.1.1 Let X be a Banach space and let T ∈ CR(X). Then, S (i) σe5 (T ) = σe4 (T ) {λ ∈ C such that i(T − λ) 6= 0} . S (ii) σeap (T ) = σe1 (T ) {λ ∈ C such that i(T − λ) > 0} . S (iii) σeδ (T ) = σe2 (T ) {λ ∈ C such that i(T − λ) < 0} .

♦

S Proof. (i) Let λ 6∈ σe4 (T ) {λ ∈ C such that i(T − λ) 6= 0}. Then, T − λ ∈ Φ(X) and i(T − λ) = 0. Hence, n = α(T − λ) = β(T − λ) < ∞. So, there exists an everywhere defined single valued linear operator K with dim R(K) ≤ α(T − λ) such that T − λ − K is bijective. Furthermore, see the

191

192

Spectral Theory of Multivalued Linear Operators

proof of Lemma 3.5.3 (iii), K is defined by Kx :=

n X

x0i (x)yi ,

x ∈ X,

i=1

where {x1 , . . . , xn } is a basis of N (T − λ) and x01 , . . . , x0n are linear functionals such that x0i (xj ) = δij . Choose y1 , . . . , yn ∈ X such that [y1 ], . . . , [yn ] ∈ X/R(λ − T ) are linearly independent (n = α(T − λ) = β(T − λ)). Hence, it is clear that K is a bounded finite rank operator. So, K ∈ KR(X) and {0} = K(0) ⊂ (T − λ)(0) = T (0) − λ(0) = T (0). Thus, it is clear that K ∈ KT (X) and (λ−T )− K = λ−(T +K) is bijective. Therefore, λ ∈ ρ(T +K). \ This shows that λ 6∈ σ(T + K) and K∈KT (X)

σe5 (T ) ⊂ σe4 (T )

[

{λ ∈ C such that i(λ − T ) 6= 0} .

(4.1)

To prove the inverse inclusion of (4.1). Suppose λ 6∈ σe5 (T ), then there exists K ∈ KT (X) such that λ ∈ ρ(T + K). Hence, λ − (T + K) ∈ Φ(X) and i(λ − (T + K)) = 0. Using Remark 2.3.4, we get λ − T ∈ Φ(X) and i(λ − T ) = S i(λ−(T +K)) = 0. We conclude λ 6∈ σe4 (T ) {λ ∈ C such that i(λ−T ) 6= 0}. So, S σe4 (T ) {λ ∈ C such that i(λ − T ) 6= 0} ⊂ σe5 (T ). Therefore, σe5 (T ) = σe4 (T )

S

{λ ∈ C such that i(λ − T ) 6= 0} .

The proof of other cases is analogous.

Q.E.D.

Corollary 4.1.1 Let X be a Banach space and T ∈ CR(X). Then, (i) λ 6∈ σe5 (T ) if and only if, λ − T ∈ Φ(X) and i(λ − T ) = 0. (ii) λ 6∈ σeap (T ) if and only if, λ − T ∈ Φ+ (X) and i(λ − T ) ≤ 0. (iii) λ 6∈ σeδ (T ) if and only if, λ − T ∈ Φ− (X) and i(λ − T ) ≥ 0.

♦

Theorem 4.1.2 Let X be Banach spaces and T ∈ CR(X). Then, \ (i) Cσ(T ) ⊂ σei (T ), where Λ := {1, 2, 3, 4, 5, ap, δ}, and i∈Λ

(ii) Rσ(T ) ⊂ σeδ (T ).

♦

Proof. (i) Let λ ∈ Cσ(T ), then R(λ − T ) is not closed (otherwise λ ∈ ρ(T )). Therefore, λ ∈ σei (T ), i = 1, . . . , 5, ap, δ. Consequently, \ Cσ(T ) ⊂ σei (T ), where Λ := {1, 2, 3, 4, 5, ap, δ}. i∈Λ

Essential Spectra and Essential Pseudospectra of a Linear Relations

193

(ii) Let λ ∈ Rσ(T ), then β(λ − T ) 6= 0. Hence, i(λ − T ) < 0 and λ − T is one to one. This implies, by the use of Corollary 4.1.1 (iii) that λ ∈ σeδ (T ). This completes the proof. Q.E.D. Theorem 4.1.3 Let T ∈ CR(X). If 0 ∈ ρ(T ), then for λ 6= 0, λ − T ∈ Φ(X) if and only if, λ−1 − T −1 ∈ Φ(X) and i(λ − T ) = i(λ−1 − T −1 ). ♦ Proof. Let λ ∈ C∗ . By using Remark 2.18.2 (with µ = 0), we have λ − T = (0 − λ)[(0 − λ)−1 − (0 − T )−1 ](0 − T ). 6 0, we have Then, for all λ = λ − T = −λ(λ−1 − T −1 )T.

(4.2)

R(λ − T ) = R(λ−1 − T −1 ).

(4.3)

Now, let us prove that

Indeed, by (4.2), we have
R(λ − T ) = R (λ−1 − T −1 )T = (λ−1 − T −1 )R(T ). The fact that R(T ) = X, we have R(λ − T ) = R(λ−1 − T −1 ).

(4.4)

Now, by using Corollary 2.19.1 (with µ = 0), we infer that N (λ − T ) = N (λ−1 − T −1 ).

(4.5)

Applying (4.4) and (4.5), we have λ − T ∈ Φ(X) if and only if, λ−1 − T −1 ∈ Φ(X). Further, in view of (4.4) and (4.5), we have i(λ − T )

This completes the proof.

= α(λ − T ) − β(λ − T ) = α(λ−1 − T −1 ) − β(λ−1 − T −1 ) = i(λ−1 − T −1 ). Q.E.D.

The following corollary follows immediately from Theorem 4.1.3 and Lemma 2.13.5 (ii) (b). Corollary 4.1.2 Let α ∈ ρ(T ) and λ ∈ C such that λ = 6 α. Then, λ ∈ σei (T ) −1 −1 if and only if, (α − λ) ∈ σei ((T − α) ), i ∈ {1, 2, 3, 4, 5, ap, δ, qφd }. ♦

194

Spectral Theory of Multivalued Linear Operators

Theorem 4.1.4 Let X be a Banach space and T ∈ LR(X) such that 0 ∈ ρ(T ). Then, for 0 6= λ ∈ C, we have (i) des(T − λ) = des(T −1 − λ−1 ), and (ii) asc(T − λ) = asc(T −1 − λ−1 ).

♦

Proof. (i) For all n ∈ N. We prove by induction the following (λ − T )n T = T (λ − T )n .

(4.6)

Indeed, (λ − T )T = −(λ − T )(0 − T ) = −(0 − T )(λ − T ). Hence, (λ − T )T = T (λ − T ). Let n ∈ N∗ , we suppose that (λ − T )n T = T (λ − T )n . Then, (λ − T )n+1 T = (λ − T )n (λ − T )T = (λ − T )n T (λ − T ) = T (λ − T )n (λ − T ).
So, (4.6) holds. Now, we prove that R ((λ − T )n ) = R (λ−1 − T −1 )n for all n ∈ N. Let λ ∈ C∗ . By using (4.4), we have R(λ − T ) = R(λ−1 − T −1 ). Let
n ∈ N∗ , we assume that R ((λ − T )n ) = R (λ−1 − T −1 )n . Then,
R (λ − T )n+1 = (λ − T )R ((λ − T )n ) = −λ(λ−1 − T −1 )T R ((λ − T )n ) = −λ(λ−1 − T −1 )R (T (λ − T )n )) = −λ(λ−1 − T −1 )R ((λ − T )n T ) (by (4.6)) = −λ(λ−1 − T −1 )R((λ − T )n ) = −λ(λ−1 − T −1 )R((λ−1 − T −1 )n )) = R((λ−1 − T −1 )n+1 ).

(4.7)

So, (4.7) gives (i). (ii) First, we show that for all λ ∈ C∗ , we have (λ−1 − T −1 )T = T (λ−1 − T −1 ).

(4.8)

Let λ ∈ C∗ and (x, x) ∈ G(ID(T ) ). Since G(ID(T ) ) ⊂ G(T T −1 ), then x ∈ T y for some y ∈ D(T ) (as T is surjective). Hence, (y, x) ∈ G(T ) and (x, y) ∈ G(T −1 ). So, (x, x) ∈ G(T T −1 ). On the other hand, λ−1 T − T T −1

= T λ−1 − T T −1 ⊂ T (λ−1 − T −1 ),

Essential Spectra and Essential Pseudospectra of a Linear Relations

195

and (λ−1 − T −1 )T

⊂ λ−1 T − T −1 T = λ−1 T − ID(T ) = λ−1 T − T T −1 .

Hence, (λ−1 − T −1 )T ⊂ T (λ−1 − T −1 ).

(4.9)

N (T (λ−1 − T −1 )) = N ((λ−1 − T −1 )T ).

(4.10)

Now, we prove that

Indeed, N (T (λ−1 − T −1 ))

=

(λ−1 − T −1 )−1 N (T )

=

(λ−1 − T −1 )−1 (0) (as N (T ) = {0})

= N (λ−1 − T −1 ) = N (λ − T ) (as Eq. (4.5)) = N (−λ(λ−1 − T −1 )T ) (as Eq. (4.2)) = N ((λ−1 − T −1 )T ). Hence, (4.10) holds. Now, we claim R(T (λ−1 − T −1 )) = R((λ−1 − T −1 )T ).

(4.11)

Indeed, R(T (λ−1 − T −1 ))

= T R(λ−1 − T −1 ) = T R(λ − T ) = R(T (λ − T )) = R((λ − T )T ) =

(λ − T )R(T )

= R(λ − T ) = R(λ−1 − T −1 ) = R((λ−1 − T −1 )T ) (since T is surjective) , which proves our claim. Now, (4.9), (4.10), and (4.11) ensures that (4.8) is true. Now, we prove by induction that N ((λ − T )n ) = N ((λ−1 − T −1 )n ) for

196

Spectral Theory of Multivalued Linear Operators

all n ∈ N and λ ∈ C∗ . By using (4.5), we have N (λ − T ) = N (λ−1 − T −1 ). Let n ∈ N∗ , we assume N ((λ − T )n ) = N ((λ−1 − T −1 )n ). Then, N ((λ−1 − T −1 )n+1 )

= N ((λ−1 − T −1 )n (λ−1 − T −1 )) =

(λ−1 − T −1 )N ((λ−1 − T −1 )n )

=

(λ−1 − T −1 )N ((λ − T )n )

=

(λ−1 − T −1 )N ((λ − T )n T ) (by (4.2) and T is injective)

=

(λ−1 − T −1 )T −1 N ((λ − T )n )

=

(T (λ−1 − T ))−1 N ((λ − T )n ) (by (4.3))

((λ−1 − T )T )−1 N ((λ − T )n )  −1 1 = − (−λ)(λ−1 − T )T N ((λ − T )n ) λ  −1 1 = − (λ − T ) N ((λ − T )n T ) λ =

= N ((λ − T )n+1 ) (by Eq. (4.2)). So, (4.12) gives (ii).

(4.12) Q.E.D.

A consequence of Theorem 4.1.4 is the following: Corollary 4.1.3 Let α ∈ ρ(T ) and λ ∈ C such that λ = 6 α. Then, λ ∈ σeb (T ) −1 −1 if and only if, (α − λ) ∈ σeb ((T − α) ). ♦ Proposition 4.1.1 Let X be a Banach space and let T, S ∈ LR(X). If T ST ⊂ T S and for some λ ∈ ρ(T ) ρ(S), we have (λ − T )−1 − (λ − S)−1 ∈ PR(Φ(X)), then σeb (T ) = σeb (S).

♦

Proof. We may assume λ = 0. Let J = T −1 − S −1 . Then, JS −1 = (T −1 − S −1 )S −1 ⊂ (T −1 S −1 − S −1 S −1 ) = (ST )−1 − S −1 S −1 .

Since (ST )−1 − S −1 S −1 ⊂ (T S)−1 − S −1 S −1 = (S −1 T −1 − S −1 S −1 ) and
(S −1 T −1 − S −1 S −1 ) ⊂ S −1 T −1 − S −1 , we have JS −1 ⊂ S −1 J. Now, using both Rc (S −1 ) = {0} and J ∈ PR(Φ(X)), we infer that σeb (T −1 ) = σeb (S −1 ). Finally, in view of Corollary 4.1.3, we get σeb (T ) = σeb (S). Q.E.D. Theorem 4.1.5 Let X be a Banach space and let T, S ∈ LR(X). Then,

Essential Spectra and Essential Pseudospectra of a Linear Relations 197 T (i) If for some λ ∈ ρ(T ) ρ(S), we have (λ − T )−1 − (λ − S)−1 ∈ PR(Φ(X)), then σei (T ) = σei (S), i = 4, 5. T (ii) If for some λ ∈ ρ(T ) ρ(S), we have (λ−T )−1 −(λ−S)−1 ∈ PR (Φ+ (X)), then σei (T ) = σei (S), i = 1, ap. T (iii) If for some λ ∈ ρ(T ) ρ(S), we have (λ − T )−1 − (λ − S)−1 ∈ PR (Φ− (X)), then σei (T ) = σei (S), i = 2, δ. T (iv) If for some λ ∈ ρ(T ) ρ(S), we have (λ − T )−1 − (λ − S)−1 ∈ T PR (Φ+ (X)) PR (Φ− (X)), then σe3 (T ) = σe3 (S).

♦

Proof. (i) We may assume λ = 0. From, the fact that T −1 − S −1 is a Fredholm perturbation, it follows that ΦT −1 = ΦS −1 and i(η − T −1 ) = i(η − S −1 ) for all η ∈ ΦT −1 . If we apply Corollary 4.1.2 to both T and S, we can see that ΦT = ΦS and i(λ − T ) = i(λ − S) for λ ∈ ΦT . The proof of (ii), (iii) and (iv) may be checked in the same way as the proof of (i). Q.E.D. Theorem 4.1.6 Let X be a Banach space and T ∈ CR(X). If dim T (0) < ∞, then σe5 (T ) = σl (T ). ♦ Proof. Let λ ∈ / σl (T ), then there exists K ∈ ΨT (X) such that λ ∈ ρ(T + K). This implies that λ − T − K is injective, open and R(λ − T − K) = X. So, (λ − T − K)−1 is a bounded linear operator on X. Since R(λ − T − K) = X, then D(K(λ − T − K)−1 ) = (λ − T − K)(D(K)) T ⊃ (λ − T − K)(D(K) D(T )) = (λ − T − K)(D(λ − T − K)) = R(λ − T − K) = X, and according to Theorem 3.7.4, we have I + K(λ − T − K)−1 ∈ Φ+ (X) and i(I + K(λ − T − K)−1 ) = dim K(λ − T − K)−1 (0) = dim K(0) < ∞. Since λ − T − K is injective and open, then λ − T − K is in Φ+ (X). In view of Proposition 2.14.2, we have (I + K(λ − T − K)−1 )(λ − T − K) ∈ Φ+ (X).

(4.13)

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Spectral Theory of Multivalued Linear Operators

The use of Remark 2.3.3, leads to 0 ∈ K −1 (0). So, (λ − T − K)(0) ⊂ (λ − T − K)K −1 (0) = N (K(λ − T − K)−1 ). Hence, (I + K(λ − T − K)−1 )(λ − T − K) = λ − T − K + K(λ − T − K)−1 (λ − T − K) = λ − T − K + K.

(4.14)

As (4.14) and Proposition 2.3.4, we have (I + K(λ − T − K)−1 )(λ − T − K) = λ − T.

(4.15)

By using (4.13) and (4.15), we have λ − T ∈ Φ+ (X). Therefore, i(λ − T )

= i((I + K(λ − T − K)−1 )(λ − T − K)) = i(I + K(λ − T − K)−1 ) + i(λ − T − K) +   X dim − R(λ − T − K) + D(I + K(λ − T − K)−1   \ dim (λ − T − K)(0) N (I + K(λ − T − K)−1 ) =

dim K(0) −   \ dim (λ − T − K)(0) N (I + K(λ − T − K)−1 ) .(4.16)

On the other hand, (I + K(λ − T − K)−1 )(λ − T − K)(0) (I + K(λ − T − K)−1 )(0) T D(I + K(λ − T − K)−1 ) (λ − T − K)(0) T = N (I + K(λ − T − K )−1T) (λ − T − K)(0) D(K (λ − T − K)−1 ) (λ − T − K)(0) T = N (I + K(λ − T − K)−1 T ) (λ − T − K)(0) (λ − T − K)D(K) (λ − T − K)(0) T = N (I + K(λ − T − K)−1 ) (λ − T − K)(0) (λ − T − K)(0) = . N (I + K(λ − T − K)−1 )(λ − T − K)(0) But

(λ − T − K)(0) (I + K(λ − T − K)−1 )(λ − T − K)(0) = . (I + K(λ − T − K)−1 )(0) K(0)

Since (λ − T − K)(0) = −T (0) − K(0) = T (0) + K(0) = T (0) has finite dimensional space, then \ dim K(0) = dim N (I + K(λ − T − K)−1 ) (λ − T − K)(0). (4.17) Hence, by (4.16) and (4.17), we have i(λ − T ) = 0. So, σe5 (T ) ⊂ σl (T ). To prove the opposite inclusion. Let K ∈ KT (X) and λ ∈ ρ(T + K), then

Essential Spectra and Essential Pseudospectra of a Linear Relations

199

(λ − T − K)−1 is single valued and continuous. Hence, −K(λ − T − K)−1 is precompact. Since X is a Banach space, −K(λ − T − K)−1 is compact, and by using Theorem 3.7.1, we have −K(λ − T − K)−1 ∈ ΛX . So, K ∈ ΨT (X). Thus, σl (T ) ⊂ σe5 (T ). Q.E.D. Corollary 4.1.4 Let X be a Banach space and T ∈ CR(X) such that dim T (0) < ∞. If Σ(X) is a subset of ΨT (X) containing KT (X), then \ ♦ σe5 (T ) = σ(T + K). K∈Σ(X)

Proof. Since Σ(X) ⊂ ΨT (X), we obtain \ \ σ(T + K) ⊂ K∈ΨT (X)

σ(T + K).

K∈Σ(X)

Applying Theorem 4.1.6, we get σe5 (T ) ⊂

\

σ(T + K).

K∈Σ(X)

On the other hand, since KT (X) ⊂ ΨT (X), then \ σ(T + K) ⊂ σe5 (T ).

Q.E.D.

K∈Σ(X)

Corollary 4.1.5 Let X be a Banach space, T ∈ CR(X) such that dim T (0) < ∞, and HT (X) be a subset of ΨT (X), containing KT (X). If for all K1 , K2 ∈ HT (X), K1 ± K2 ∈ HT (X), then for every K ∈ HT (X), σe5 (T ) = σe5 (T + K).

♦

Proof. We denote σ ∗ (T ) =

\

σ(T + K).

K∈HT (X)

From Corollary 4.1.5, we have σe5 (T ) = σ ∗ (T ). Furthermore, for each K ∈ HT (X), we have HT (X) + K = HT (X). Then, σ ∗ (T + K) = σ ∗ (T ). Hence, we get the desired result. Q.E.D.

4.1.1

α- and β-Essential Spectra of Linear Relations

Proposition 4.1.2 Let T ∈ CR(X). Then, σeα (T ) and σeβ (T ) are closed.♦

200

Spectral Theory of Multivalued Linear Operators

Proof. Using Lemma 2.14.5, we deduce that σeα (T ) = σeβ (T ∗ ). It is sufficient to show that σeα (T ) is closed. Let λ 6∈ σeα (T ). Then, λ−T ∈ Aα (X). By using Proposition 3.4.1, we deduce that Tl (λ − T ) = (I − F )|D(T ) , where Tl ∈ L(X), F is a bounded finite rank operator such that N (Tl ) is topologically complemented in X, R(Tl ) is a closed subspace contained in D(T ), and R(F ) ⊂ D(T ). Therefore,   0 < γ(I − F ) ≤ γ (I − F )| N (I−F )+D(T )
≤ γ (I − F )|R(F )+D(T )
≤ γ (I − F )|D(T ) (as R(F ) ⊂ D(T )). Thus, γ (Tl (λ − T )) > 0. Now, we take η ∈ C such that |η − λ|
0} . (ii) If T ∗ (0) is topologically complemented in X ∗ , then S σer (T ) = σeβ (T ) {λ ∈ C such that i(λ − T ) < 0} . ♦ S Proof. (i) Let λ 6∈ (σeα (T ) {λ ∈ C such that i(λ − T ) > 0}) . Then, λ − T ∈ Aα (X) and i(λ − T ) ≤ 0. Hence, λ − T ∈ Φ+ (X) with i(λ − T ) ≤ 0. Thus, by virtue of Theorem 3.5.3 (i), λ − T can be expressed in the form λ − T = U + K,

(4.20)

Essential Spectra and Essential Pseudospectra of a Linear Relations

201

where K ∈ Kλ−T (X) = KT (X) and U is injective linear operator. Moreover, λ − T ∈ Aα (X) and (λ − T )(0) = T (0) is topologically complemented in X. By using Proposition 3.4.1, we obtain Tl (λ − T ) = (I − F )|D(T ) , where Tl ∈ L(X), F is a bounded finite rank operator such that N (Tl ) is topologically complemented in X, R(Tl ) is a closed subspace contained in D(T ), and R(F ) ⊂ D(T ). Now, let L = F + Tl K. Then, \ R(L) ⊂ R(F ) + R(Tl ) ⊂ D(T ) = D(λ − T − K) = D(T ) D(K), and in view K(0) ⊂ T (0), we have (λ − T − K)(0) = T (0). Hence, (λ − T − K)(0) is topologically complemented in X. Moreover, Tl (λ − T − K) = (I − L)|D(T ) . So, λ − T − K ∈ Aα (X). Thus, R(λ − T − K) is closed and topologically complemented in X. By using (4.20), we have λ − T − K is injective and by applying Proposition 3.4.1, we infer that λ − T − K ∈ GRl (X). Hence, λ 6∈ σl (T + K) and we can deduce that \ λ 6∈ σl (T + K). K∈KT (X)

T Conversely, let us assume that λ 6∈ K ∈KT (X) σl (T + K). Then, there exists K ∈ KT (X) such that λ − T − K ∈ GRl (X). Since K(0) ⊂ T (0) = (λ − T )(0) and D(λ − T ) = D(T ) ⊂ D(K), and by using Proposition 2.3.4, we have λ − T = (λ − T − K) + K. Now, in view of Corollary 3.5.3 (i), we infer that λ − T ∈ Aα (X) and i(λ − T ) = i(λ − T − K) = −β(λ − T − K) ≤ 0. This is S equivalent to say that λ 6∈ (σeα (T ) {λ ∈ C : i(λ − T ) > 0}) . S (ii) Let λ 6∈ (σeβ (T ) {λ ∈ C : i(λ − T ) < 0}). Then, λ − T ∈ Aβ (X) and i(λ − T ) ≥ 0 which implies that λ − T ∈ Φ− (X) and i(λ − T ) ≥ 0. By using Lemma 3.5.3, λ − T can be expressed in the form λ − T = S + F,

(4.21)

where S is a surjective linear relation and F ∈ Kλ−T (X) = KT (X). Moreover, λ − T ∈ Aβ (X), and by applying Lemma 2.14.6, we deduce that (λ − T )Tr = I − K + (λ − T )Tr − (λ − T )Tr

(4.22)

202

Spectral Theory of Multivalued Linear Operators

for some bounded operators Tr and K such that R(Tr ) ⊂ D(T ), Tr (λ − T ) and K(λ − T ) are continuous operators and I − K ∈ Φ(X). Then, one finds out from (4.22) that (λ − T − F )Tr = I − K − F Tr + (λ − T − F )Tr − (λ − T − F )Tr . Clearly, K+F Tr is a bounded operator, I −K−F Tr ∈ Φ(X) and, Tr (λ−T −F ) and (K+F Tr )(λ−T −F ) are continuous operators. By applying Lemma 2.14.6, we deduce that N (λ − T − F ) is topologically complemented in X. By using (4.21), we infer that λ − T − F is surjective. So, λ − T − F ∈ GRr (X). Hence, λ 6∈ σr (T + F ). We can deduce that \ λ 6∈ σr (T + F ). F ∈KT (X)

T Conversely, let us assume that λ 6∈ F ∈KT (X) σr (T + F ). Then, there exists F ∈ KT (X) for which λ − T − F ∈ GRr (X). Since F (0) ⊂ T (0) = (λ − T )(0) and D(λ − T ) = D(T ) ⊂ D(F ), and by using Proposition 0 2.3.4, we infer that λ − T = (λ − T − F ) + F . Since ((λ − T − F )(0)) = ∗ T ∗ (0), then ((λ − T − F )(0)) is topologically complemented in X ∗ . By using Corollary 3.5.3 (ii), we infer that λ − T ∈ Aβ (X) and i(λ − T ) = i(λ − T − F ) = α(λ − T − F ) ≥ 0. This is equivalent to say that λ 6∈ S (σeβ (T ) {λ ∈ C such that i(λ − T ) < 0}). This completes the proof.Q.E.D.

Corollary 4.1.6 (i) Assume that T (0) is topologically complemented in X. Then,  (i1 ) σel (T ) = λ ∈ C such that λ − T ∈ / W l (X) , and (i2 ) λ 6∈ σel (T ) if and only if, λ − T ∈ Aα (X) and i(λ − T ) ≤ 0. (ii) Assume that T ∗ (0) is topologically complemented in X ∗ . Then, (ii1 ) σer (T ) = {λ ∈ C such that λ − T ∈ / W r (X)}, and (ii2 ) λ ∈ 6 σer (T ) if and only if, λ − T ∈ Aβ (X) and i(λ − T ) ≥ 0. (iii) Assume that the conditions of (i) and (ii) are satisfied. Then, (iii1 ) σel (T ) and σer (T ) are closed, (iii2 ) σel (T ) = σer (T ∗ ), and (iii3 ) σer (T ∗ ) = σel (T ).

♦

The aim of the following theorem is to give a refinement of the definitions of σel (·) and σer (·).

Essential Spectra and Essential Pseudospectra of a Linear Relations

203

Theorem 4.1.8 Let T ∈ LR(X). Then, (i) If T (0) is topologically complemented in X, then \ σel (T ) = σl (T + K). K∈PR(Aα (X))

(ii) If T ∗ (0) is topologically complemented in X ∗ , then \ σer (T ) = σr (T + K).

♦

K∈PR(Aβ (X))

Proof. (i) Let U :=

\

σl (T + K). Since KT (X) ⊂ PR(Aα (X)),

K∈PR(Aα (X))

we have U ⊂ σel (T ). Conversely, assume that λ 6∈ U. Then, there exists K ∈ PR(Aα (X)) such that λ 6∈ σl (T +K), that is λ−(T +K) = (λ−T )−K ∈ GRl (X). By applying Corollary 3.5.2 (i1 ), we have λ − T − K + K ∈ Aα (X) and i(λ − T − K + K) ≤ 0. By using Proposition 2.3.4, we have λ − T ∈ Aα (X) and i(λ − T ) ≤ 0. Now, the use of Corollary 4.1.6 (i) allows us to conclude that λ 6∈ σel (T ). The proof of (ii) is analogous to the previous one.

Q.E.D.

Corollary 4.1.7 Let X be a Banach space, and let N(X) and M(X) be any subsets of LR(X). Then, (i) If KR(X) ⊂ N(X) ⊂ PR(Aα (X)) and if T (0) is topologically complemented in X, then \ σel (T ) = σl (T + F ). F ∈N(X)

Moreover, if for all F1 , F2 ∈ N(X), we have F1 ± F2 ∈ N(X), then σel (T + F ) = σel (T ) for all F ∈ N(X). (ii) If KR(X) ⊂ M(X) ⊂ PR (Aβ (X)) and if T ∗ (0) is topologically complemented in X ∗ , then \ σer (T ) = σr (T + F ). F ∈M(X)

Moreover, if for all F1 , F2 ∈ M(X), we have F1 ± F2 ∈ M(X), then σel (T + F ) = σel (T ) for all F ∈ M(X).

♦

204

Spectral Theory of Multivalued Linear Operators

Remark 4.1.1 (i) If T (0) is topologically complemented in X, then from Corollary 4.1.8 (i) and Proposition 2.17.1 (iii), it follows that σel (T + F ) = σel (T ) for all F ∈ PR (Aα (X)) . (ii) If T ∗ (0) is topologically complemented in X ∗ , then from Corollary 4.1.8 (ii) and Proposition 2.17.1 (iv), it follows that σer (T + F ) = σer (T ) for all F ∈ PR (Aβ (X)) . (iii) By using both Proposition 2.3.4 and Definition 2.17.1, we have σeα (T + F ) = σeα (T ) for all F ∈ PR (Aα (X)) and σeβ (T + F ) = σeβ (T ) for all F ∈ PR (Aβ (X)) .

♦

Theorem 4.1.9 Let α ∈ ρ(T ) and λ ∈ C such that λ 6= α. Then, λ ∈ σei (T ) if and only if, (α − λ)−1 ∈ σei ((T − α)−1 ), i = α, β, l, r. ♦ Proof. The proof of Theorem 4.1.9 is inspired from the proof of Theorem 4.1.3. Q.E.D. Remark 4.1.2 Theorem 4.1.9 allows a very important relationship between the essential spectra of a linear relation and the essential spectra of a linear operator. Then, we can transfer some known results about the essential spectrum of a linear operator to the case of a linear relation. ♦ As an illustration, let us see the following theorem: Theorem 4.1.10 Let T , S ∈ LR(X). Then, T (i) If, for some λ ∈ ρ(T ) ρ(S), we have (λ−T )−1 −(λ−S)−1 ∈ PR(Aα (X)), then σei (T ) = σei (S), i = α, l. (ii) If, for some λ ∈ ρ(T ) PR (Aβ (X)), then

T

ρ(S), we have (λ − T )−1 − (λ − S)−1 ∈

σei (T ) = σei (S), i = β, r.

♦

Essential Spectra and Essential Pseudospectra of a Linear Relations

205

Proof. (i) We may assume that λ = 0. The fact that T −1 − S −1 ∈ PR(Aα (X)), allows us to deduce the existence of F ∈ PR(Aα (X)) such that T −1 = S −1 + F . Then, T −1 (0) = S −1 (0) = F (0) = {0}. Hence, by applying Remark 4.1.1, we infer that σei (T −1 ) = σei (S −1 ), i = α, l. Now, if we apply Theorem 4.1.9 to both T and S, we get σei (T −1 ) = σei (S −1 ), i = β, r. (ii) The proof of (ii) may be checked in the same way as in the proof of (i), which completes the proof. Q.E.D.

4.1.2

Invariance of Essential Spectra of Linear Relations

In this subsection we apply the reached findings to study the invariance and the characterization of the essential spectra of a closed multivalued linear operator. Theorem 4.1.11 Let X be complete space, T ∈ CR(X) with dim D(T ) = ∞, and S ∈ LR(X) such that D(T ) ⊂ D(S) and S(0) ⊂ T (0). If S is T precompact, then (i) σeap (T + S) = σeap (T ), and (ii) σeδ (T + S) = σeδ (T ).

♦

Proof. (i) Let S be T -precompact, then SGT is precompact. Since X and XT are complete, then SGT is compact. Suppose that λ ∈ / σeap (T ), then by using Corollary 4.1.1, we have λ − T ∈ Φ+ (X). Hence, (λ − T )Gλ−T ∈ Φ+ (XT , X), which gives (λ − T )GT ∈ Φ+ (XT , X). Since SGT is compact, then (λ − T + S)GT ∈ Φ+ (XT , X). So, (λ − (T + S))Gλ−(T +S) ∈ Φ+ (XT , X). Thus, λ − (T + S) ∈ Φ+ (X). Moreover, by using Proposition 2.11.3, it is clear that R ((λ − (T + S))GT ) = R (λ − (T + S)) and N ((λ − (T + S))GT ) = N (λ − (T + S)) . These relations imply that i (λ − (T + S)) = i ((λ − T )GT + SGT ) . Now, the use of Lemma 3.5.4 allows us to conclude that i ((λ − (T + S))) = i (λ − T ), which implies that λ ∈ / σeap (T + S). So, σeap (T + S) ⊂ σeap (T ). Conversely, let λ ∈ / σeap (T + S). Then, by using Corollary 4.1.1, we have λ − (T + S) ∈ Φ+ (X). Arguing as before, it follows that λ − (T + S − S) ∈ Φ+ (X). Then, λ − T ∈ Φ+ (X) and,

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Spectral Theory of Multivalued Linear Operators

we have i(λ − T ) = i(λ − (T + S)). Hence, λ ∈ / σeap (T ) see Corollary 4.1.1. Thus, σeap (T + S) = σeap (T ). (ii) Now, suppose that λ ∈ / σeδ (T ), then by Corollary 4.1.1, λ − T ∈ Φ− (X). Hence, (λ−T )Gλ−T ∈ Φ− (XT , X). This implies that (λ−T )GT ∈ Φ− (XT , X). Since SGT is precompact, then (λ − (T + S))G(λ−T ) ∈ Φ− (XT , X). So, (λ − (T + S))Gλ−(T +S) ∈ Φ− (XT , X). Therefore, λ − (T + S) ∈ Φ− (X) and, we have i(λ − T ) = i(λ − (T + S)). Hence, λ ∈ / σeδ (T + S). Thus, σeδ (T + S) ⊂ σeδ (T ). Conversely, let λ ∈ / σeδ (T + S), then λ − (T + S) ∈ Φ− (X). So, (λ − (T + S))Gλ−(T +S) ∈ Φ− (XT , X). Hence, (λ − (T + S))GT ∈ Φ− (XT , X). The latter holds if and only if, ((λ − (T + S))GT )∗ ∈ Φ+ (X ∗ , XT∗ ) and ((λ − T )GT )∗ + (SGT )∗ ∈ Φ+ (X ∗ , XT∗ ). Since SGT is precompact, then (SGT )∗ is compact. This implies that ((λ − T )GT )∗ ∈ Φ+ (X ∗ , XT∗ ). Hence, (λ − T )GT ∈ Φ− (XT , X). So, λ−T ∈ Φ− (X) and i(λ−T ) = i(λ−(T +S)). Thus, λ ∈ / σeδ (T ). We infer that σeδ (T + S) = σeδ (T ), which completes the proof. Q.E.D. Theorem 4.1.12 Let T ∈ CR(X), then \ (i) σap (T + K) ⊂ σeap (T ) = K∈PR(Aα (X))

(ii)

\

σap (T + K).

K∈SSR(X)

\

σδ (T + K) ⊂ σeδ (T ).

K∈PR(Aβ (X))

(iii)

\ K∈PR(Aα (X))

S

σ(T + K) ⊂ σe5 (T ). PR(Aβ (X))

(iv) If T (0) is topologically complemented in X, then \ \ σel (T ) = σl (T + K) = K∈PR(Φ+ (X))

σl (T + K).

K∈PR(Aα (X))

(v) If T ∗ (0) is topologically complemented in X ∗ , then \ \ σer (T ) = σr (T + K) = K∈PR(Φ− (X))

Proof. (i) Let U :=

\

σr (T + K).

♦

K ∈PR(Aβ (X))

σap (T + K). Since KT (X) ⊂ PR(Aα (X)),

K∈PR(Aα (X))

we have U ⊂ σeap (T ). Conversely, assume that λ 6∈ U. Then, there exists K ∈ SSR(X) such that λ 6∈ σap (T + K). Moreover, since λ − (T + K) is bounded below, then (λ − T ) − K ∈ Φ+ (X) and i((λ − T ) − K) ≤ 0. Also, since SSR(X) ⊂ PR(Φ+ (X)), then (λ − T ) − K + K ∈ Φ+ (X) and i((λ − T ) − K + K) ≤ 0. Again, by applying Proposition 2.3.4, we infer that

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λ − T ∈ Aα (X) and i(λ − T ) ≤ 0. Finally, the use of Corollary 4.1.6 gives the wanted inclusion and achieves the proof of (i). Statement (ii), (iii), (iv) and (v) can be checked in the same way as the assertion (i). Q.E.D. As a direct consequence of Theorem 4.1.12, we infer the following result. Corollary 4.1.8 Let U(X) and V(X) be any subsets of LR(X). Then, (i) If KR(X) ⊂ U(X) ⊂ PR(Φ+ (X)), then \ σeap (T ) = σap (T + K). K∈U(X)

If, for all F1 , F2 ∈ U(X), F1 ± F2 ∈ U(X), then for every F ∈ U(X), we have σeap (T + F ) = σeap (T ). (ii) If KR(X) ⊂ U(X) ⊂ Aα (X)) and T (0) is topologically complemented in X, then \ σel (T ) = σl (T + F ). F ∈U(X)

If, for all F1 , F2 ∈ U(X), F1 ± F2 ∈ U(X), then for every F ∈ U(X), we have σel (T + F ) = σel (T ). (iii) If KR(X) ⊂ U(X) ⊂ PR(Φ+ (X)), then \ σeδ (T ) = σδ (T + K). K∈V(X)

If, for all F1 , F2 ∈ U(X), F1 ± F2 ∈ U(X), then for every F ∈ U(X), we have σeδ (T + F ) = σeδ (T ). (iv) If KR(X) ⊂ V(X) ⊂ PR(Aβ (X)) and T ∗ (0) is topologically complemented in X ∗ , then \ σer (T ) = σr (T + F ). F ∈V(X)

If, for all F1 , F2 ∈ V(X), F1 ± F2 ∈ V(X), then for every F ∈ V(X), we have σer (T + F ) = σer (T ).

♦

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Remark 4.1.3 It follows immediately, from Corollary 4.1.8, that (i) σei (T + F ) = σei (T ) for all F ∈ PR(Φ+ (X)) and i ∈ {1, ap}. (ii) σei (T + F ) = σei (T ) for all F ∈ PR(Φ− (X)) and i ∈ {2, δ }. T (iii) σe3 (T + F ) = σe3 (T ) for all F ∈ PR(Φ+ (X)) PR(Φ− (X)). S (iv) σei (T + F ) = σei (T ) for all F ∈ PR(Φ+ (X)) PR(Φ− (X)) and i ∈ {4, 5}. (v) σeα (T + F ) = σeα (T ) for all F ∈ PR (Aα (X)) . (vi) σeβ (T + F ) = σeβ (T ) for all F ∈ PR (Aβ (X)) . (vii) If T ∗ (0) is topologically complemented in X ∗ , then σer (T + F ) = σer (T ) for all F ∈ PR (Aβ (X)) . (vii) If T (0) is topologically complemented in X, then σel (T + F ) = σel (T ) for all F ∈ PR (Aα (X)) . ♦ Proposition 4.1.3 Let X = lp (N) be the space of sequences x : N −→ C summable with a power p ∈ [1, ∞) and the standard norm. Consider the linear relation T defined by G(T ) = {(x, y) ∈ X × X such that x(n) = y(n − 1) , n ≥ 2} and let F = K + F − F , where K is a compact single valued linear operator given by   x(n − 1) , x ∈ X and n ≥ 2 K(x(n)) = n−1  0, n = 1, and F (0) = span{e1 } (the subspace generated by e1 ). Then, {λ ∈ C such that |λ| = 1} ⊂ σ(T ) ⊂ {λ ∈ C such that |λ| ≥ 1} , and σei (T + F ) = {λ ∈ C such that |λ| = 1} i = 1, . . . , 4.

♦

Proof. It is clear that T = L−1 , where L is the left shift single valued operator defined by L(x(n)) = x(n + 1), n ≥ 1 and x ∈ X. Then, T is closed and D(T ) = R(L) = X. Hence, T is everywhere defined closed linear relation. So,   S 1 σ(T ) {∞} = ∈ C such that λ ∈ σ(T ) λ  1 = ∈ C such that 0 ≤ |λ| ≤ 1 λ = {λ ∈ C such that |λ| ≥ 1} .

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On the other hand, we will check that σei (T ) = {λ ∈ C such that |λ| = 1} , i = 1, . . . , 4. Indeed, N (T ) = L(0) = {0} and R(T ) = X. Hence, 0 ∈ ρ(T ). By using Remark 2.18.3, we have σei (T ) = {λ ∈ C such that |λ| = 1} , i = 1, . . . , 4. So, {λ ∈ C such that |λ| = 1} ⊂ σ(T ) ⊂ {λ ∈ C such that |λ| ≥ 1} . Observe that kF k = kK + F − F k ≤ kKk + kF − F k = kKk < ∞. Hence, F is continuous, D(F ) = D(K) = X and F (0) = T (0). Therefore, by using Proposition 2.7.2, we have T + F ∈ CR(X). Since F is bounded compact linear relation, then by using Theorem 3.5.5, we have F ∈ PR(Aα (X)), and F ∈ PR(Aβ (X)). Hence, by virtue of Remark 4.1.3, we have σei (T + F ) = σei (T ), i = 1, . . . , 5, ap, δ, α, β. In particulary, σei (T + F ) = σei (T ) = {λ ∈ C such that |λ| = 1} i = 1, . . . , 4. This completes the proof. Q.E.D. We close this section by the following result: Theorem 4.1.13 Let T ∈ CR(X) and S ∈ LR(X) be continuous such that S(0) ⊂ T (0) and S is T -bounded with T -bound < 1. Then, (i) If SGT is α-Atkinson perturbation, then σeα (T + S) = σeα (T ). Moreover, if T (0) is topologically complemented in X, then σel (T + F ) = σel (T ). (ii) If SGT is β-Atkinson perturbation, then σeβ (T + S) = σeβ (T ). Moreover, if T ∗ (0) is topologically complemented in X ∗ , then σer (T + F ) = σer (T ).

♦

Proof. (i) Let λ ∈ C such that λ − T ∈ Aα (X). Then, (λ − T )Gλ−T ∈ Aα (XT , X). Hence, by the equivalence of the norms k·kT and k·kλ−T , we have (λ − T )GT ∈ Aα (XT , X). Moreover, SGT ∈ PR(Aα (XT , X)). We can deduce that (λ − (T + S))GT ∈ Aα (XT , X). Again, by the equivalence of the norms k · kT and k · kT +S (see Theorem 2.6.1), we have (λ − (T + S))GT +S ∈ Φ+ (X). Thus, by using the equivalence of the norms k · kT +S and k · kλ−(T +S) , we have λ − (T + S)Gλ−(T +S) ∈ Aα (XT , X). Hence, λ − (T + S) ∈ Aα (X).

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Observing that each step in the above argument is reversible, so we obtain the equivalence λ − T ∈ Aα (X) if and only if, λ − (T + S) ∈ Aα (X).

(4.23)

It follows from (4.23) that σeα (T + S) = σeα (T ). Moreover, by using Proposition 2.11.3, it is clear that R ((λ − (T + S))GT ) = R (λ − (T + S)) and N ((λ − (T + S))GT ) = N (λ − (T + S)) . These relations imply that i (λ − (T + S)) = i ((λ − T )GT + SGT ) . Now, the use of Lemma 3.5.4 allows us to conclude that i (λ − (T + S)) = i (λ − T ) . (4.24) Then, it follows from both (4.24) and Corollary 4.1.6 that σel (T +S) = σel (T ). (ii) The proofs of (ii) may be achieved by using the same reasoning as (i), which completes the proof. Q.E.D.

4.2

The Essential Spectrum of a Sequence of Linear Relations

The goal of this section is to examine the invariance of Weyl essential spectrum of closed linear relation T by sequence Tn , that converges in the generalized sense to T . Lemma 4.2.1 Let (Tn )n be a sequence of linear closed relations, T ∈ CR(X) g such that 0 ∈ ρ(T ) and Tn −→ T . If O ⊂ C is open and 0 ∈ O, then there exists n0 ∈ N such that for every n ≥ n0 , we have σe5 (Tn−1 ) ⊂ σe5 (T −1 ) + O.

(4.25) ♦

g

Proof. Let us assume that 0 ∈ ρ(T ) and Tn −→ T , then by using Theorem 3.2.8 (iii), we infer that Tn−1 ∈ L(X). Since Tn−1 is a bounded linear operator on X, then σe5 (Tn−1 ) is a compact subset of C (see [65]). Now, let us assume that (4.25) fails. Then, by passing to subsequence, we suppose that for each n ∈ N, there exists λn ∈ σe5 (Tn−1 ) such that λn 6∈ σe5 (T −1 ) + O. Assume that lim λn = λ, which implies that λ 6∈ σe5 (T −1 ) + O. Since 0 ∈ O, then n→+∞

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211

we obtain λ 6∈ σe5 (T −1 ). This equivalent to saying that λ − T −1 ∈ Φ(X) and i(λ − T −1 ) = 0. The fact that λn − Tn−1 ∈ L(X) for sufficiently large n and λn − Tn−1 → λ − T −1 as n → ∞ implies from (ii) of Theorem 3.2.8 that δb(λn − Tn−1 , λ − T −1 ) → 0 as n → ∞. Since R(λ − T −1 ) is closed, then by Proposition 2.9.3 (i) and Theorem 2.5.4 (ii), we deduce that γ(λ − T −1 ) > 0. Hence, there exists N ∈ N such that, for all n ≥ N , we have −1 b n − T −1 , λ − T −1 ) ≤ p γ(λ − T ) δ(λ . n 1 + [γ(λ − T −1 )]2

Thus, allows us to conclude that λn − Tn−1 ∈ Φ(X) and i(λn − Tn−1 ) = i(λ − T −1 ) = 0. This implies that λn 6∈ σe5 (Tn−1 ), which is a contradiction. Q.E.D. Theorem 4.2.1 Let (Tn )n be a sequence of linear closed relations, T ∈ g CR(X) such that 0 ∈ ρ(T ) and Tn −→ T . Then, (i) If O ⊂ C is open and 0 ∈ O, then there exists n0 ∈ N such that for every n ≥ n0 , we have σe5 (Tn ) ⊂ σe5 (T ) + O. (4.26) (ii) Let S ∈ BR(X) satisfy that D(S) ⊃ D(Tn ) and kSk < γ(T ), for all n ≥ 1. Then,

S

D(T ), S(0) ⊂ Tn (0)

σe5 (Tn + S) ⊂ σe5 (T ) + O, for all n ≥ n0 .

T

T (0)

♦

g

Proof. (i) Since Tn −→ T , then by using Lemma 4.2.1, there exists n0 ∈ N such that σe5 (Tn−1 ) ⊂ σe5 (T −1 ) + O for all n ≥ n0 . (4.27) Now, take γn ∈ σe5 (Tn ) such that γn 6∈ σe5 (T ) + O. From σe5 (.) is upper semi continuous at T −1 (since T −1 is a single valued), there exists k > 0 such that k −1 ≤ |γn − λ0 |−1 , so (γn ) is bounded. Therefore it can be assumed that γn → γ. Then, γ 6∈ σe5 (T ) + O and hence γ 6∈ σe5 (T ). This implies that (γ)−1 6∈ σe5 (T −1 ). We set λn = (γn )−1 , λ = (γ)−1 . Then by using (4.27), there exists n0 ∈ N such that for all n ≥ n0 , λn 6∈ σe5 ((Tn )−1 ), which implies γn 6∈ σe5 ((Tn )), is a contradiction. (ii) Let S ∈ BR(X) such that kSk < γ(T ). Let An = Tn + S and A = T + S, g then by using Theorem 3.2.8 (i), we deduce that Tn + S = An −→ A = T + S. The fact that 0 ∈ ρ(T ) implies that T is open and injective with dense range.

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Hence, it follows from Lemma 2.9.3 that T + S is open and injective with dense range. In view of the above, 0 ∈ ρ(T + S). Now, by applying (i) to An and A, we conclude that there exists n0 ∈ N such that σe5 (Tn + S) ⊂ σe5 (T + S) + O, for all n ≥ n0 . Let us assume that 0 < |λ| < γ(T ) − kSk. Then, for any λ ∈ C, we have kSk < γ(T ) − |λ|.

(4.28)

Suppose that λ 6∈ σe5 (T ), then λ − T ∈ Φ(X) and i(λ − T ) = 0. Since T is open and injective, then it follows from (4.28) and Lemma 2.9.2 that kSk < γ(T ) − |λ| ≤ γ(λ − T ), for any λ ∈ C. Thus, by applying Corollary 2.16.1 (vii), we conclude that λ − (T + S) ∈ Φ(X) and i(λ − (T + S)) = i(λ − T ) = 0. This is equivalent to saying that λ 6∈ σe5 (T + S). Therefore, σe5 (T + S) ⊂ σe5 (T ). Finally, we have proved the following result σe5 (Tn + S) ⊂ σe5 (T ) + O, for all n ≥ n0 .

Q.E.D.

Remark 4.2.1 The compactness of the Weyl spectrum is not valid for the case of bounded linear relations. For this reason we are not able to investigate directly σe5 (T ). Accordingly, we added the condition that 0 ∈ ρ(T ) to work on the Weyl spectrum of a bounded linear operator T −1 . ♦ Proposition 4.2.1 Let X = lp (N) be the space of sequences x : N −→ C summable with a power p ∈ [1, ∞) with the standard norm. Consider the linear relation T defined by G(T ) := {(x, y) ∈ X × X : x(n) = y(n − 1), n ≥ 2} . n For each positive integer element m such that n ≥ 2, let Tn = ( n−1 )T . Then,

(i) Tn converges in the generalized sense to T . (ii) σe4 (Tn ) 6= σe4 (T ).

♦

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Proof. (i) It is clear that T = L−1 , where L is the left shift single valued operator defined by L(x(n)) = x(n + 1), for all n ≥ 1 and x ∈ X. Then, T is closed and D(T ) = R(L) = X. Hence, T is an everywhere defined closed linear relation. So, T ∈ BR(X), and according to Proposition 2.5.4 (iii), we deduce that (Tn )n ∈ BR(X) for sufficiently large n. Now, we have to g prove that Tn −→ T . Indeed, since T is a linear relation, then for all n ≥ 2, we get     n n Tn (0) = T (0) = T × 0 = T (0), n−1 n−1 and kTn − T k → 0 when m → ∞. Then, by applying Theorem 3.2.8 (ii), we conclude that Tn converges in the generalized sense to T . (ii) Since N (T ) = L(0) = {0} and R(T ) = X, then 0 ∈ ρ(T ). It is clear  that σe4 (T ) = {λ ∈ C : |λ| = 1} and σe4 (Tn ) = λ ∈ C : |λ| = 1 − n1 , for all n ≥ 2. This implies that σe4 (Tn ) 6= σe4 (T ).

4.3

Q.E.D.

Spectral Mapping Theorem of Essential Spectra

Theorem 4.3.1 Assume that T ∈ CR(X) has a nonempty resolvent set n Y and P (T ) = (T − λi )mi as in Definition 2.13.1. Then, for any complex i=1

polynomial P , we have (i) σeap (P (T )) ⊂ P (σeap (T )), and (ii) σeδ (P (T )) ⊂ P (σeδ (T )) .

♦

Proof. (i) Let P (T ) defined in (2.47). Assume that µ ∈ σeap (P (T )). If λk ∈ / σeap (T ) for all k = 1, 2, . . . , n, then it follows from the characterization of σeap (T ), established in Theorem 2.14.3, that T −λk ∈ Φ+ (X) with i(T −λk ) ≤ 0, for all k ∈ {1, 2, . . . , n}. Thus, we deduce from Theorem 2.14.3 that P (T ) − µ is an upper semi-Fredholm relation. Let us consider two cases for the index:

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First case : i(T − λk ) ∈] − ∞, 0], for all k ∈ {1, 2, . . . , n}. Then, i(P (T ) − µ) ∈ ] − ∞; 0], by Proposition 3.6.1 (i). Hence, µ ∈ / σeap (P (T )) which contradicts the fact that µ ∈ σeap (P (T )). Second case : i(T − λk ) = −∞, for some k ∈ {1, 2, . . . , n}. Then, i(P (T )−µ) would be −∞, applying Proposition 3.6.2 (i). So, µ ∈ / σeap (P (T )). Consequently, there exists j ∈ {1, 2, . . . , n} for which λj ∈ σeap (T ). Since µ = P (λj ), then we conclude that σeap (P (T )) ⊂ P (σeap (T )). (ii) This assertion may be proved with a similar scheme by using Propositions 3.6.1 (ii) and 3.6.2 (ii). Q.E.D.

4.3.1

Essential Spectra of the Sum of Two Linear Relations

In this subsection we investigate the essential spectra of the sum of two closed linear relations defined on a Banach space by means of essential spectra of these two linear relations. Theorem 4.3.2 Let X be a Banach space and let A, B ∈ BCR(X) and AB ⊂ BA. Then, (i) If A is a single valued linear operator and AB ∈ PR(Φ(X)), then h i [ σe4 (A + B)\{0} ⊂ σe4 (A) σe4 (B) \{0}. If, further, BA ∈ PR(Φ(X)), then h i [ σe4 (A + B)\{0} = σe4 (A) σe4 (B) \{0}. (ii) If A is single valued and BA ∈ PR(Φ(X)), then h i [ σe5 (A + B)\{0} ⊂ σe5 (A) σe5 (B) \{0}. Moreover, if AB ∈ PR(Φ(X)) and ΦA is connected, then h i [ σe5 (A + B)\{0} = σe5 (A) σe5 (B) \{0}.

(4.29)

(iii) If AB ∈ PR (Φ+ (X)), then h i [ σe1 (A + B)\{0} ⊂ σe1 (A) σe1 (B) \{0}. If, further, BA ∈ PR (Φ+ (X)) then h i [ σe1 (A + B)\{0} = σe1 (A) σe1 (B) \{0}.

(4.30)

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(iv) If the hypothesis of (iii) is satisfied, then h i [ σeap (A + B)\{0} ⊂ σeap (A) σeap (B) \{0}. If, further, ΦA is connected, then h i [ σeap (A + B)\{0} = σeap (A) σeap (B) \{0}. (vi) If the hypothesis of (iv) is satisfied, then h i [ σeδ (A + B)\{0} ⊂ σeδ (A) σeδ (B) \{0}. If, further, ΦA is connected, then h i [ σeδ (A + B)\{0} = σeδ (A) σeδ (B) \{0}. (vii) If AB ∈ PR (Φ+ (X))

T

PR (Φ− (X)), then

σe3 (A + B)\{0} ⊂ S S T S T ([σe3 (A) σe3 (B)] [σe1 (A) σe2 (B)] [σe2 (A) σe1 (B)]) \{0}. T Moreover, if BA ∈ PR (Φ+ (X)) PR (Φ− (X)), then σe3 (A + B)\{0} = S S T S T ([σe3 (A) σe3 (B)] [σe1 (A) σe2 (B)] [σe2 (A) σe1 (B)]) \{0}.

♦

Proof. Let A, B ∈ BCR(X) and λ ∈ C. By using Proposition 2.3.6, it is easy to see that (λ − A)(λ − B) = AB + λ(λ − A − B)

(4.31)

and (λ − B)(λ − A) = BA + λ(λ − A − B). (4.32) S S (i) Let λ 6∈ σe4 (A) σe4 (B) {0}. Then, λ − A ∈ Φ(X) and λ − B ∈ Φ(X). By using Theorem 2.16.1, we have (λ − A)(λ − B) ∈ Φ(X). So, (4.31) gives AB + λ(λ − A − B) ∈ Φ(X). Since AB ∈ PR(Φ(X)), then AB − (λ − A)(λ − B) ∈ Φ(X). This implies that AB − AB − λ(λ − A − B) ∈ Φ(X).

(4.33)

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Since AB(0) ⊂ BA(0) = B(0) ⊂ λ(A+B −λ)(0) = (A+B)(0) = A(0)+B(0), then in view of (4.33), we have λ − A − B ∈ Φ(X). Hence, λ 6∈ σe4 (A + B). Therefore, h i [ σe4 (A + B)\{0} ⊂ σe4 (A) σe4 (B) \{0}. (4.34) S Prove the inverse inclusion of (4.34). Suppose λ 6∈ σe4 (A + B) {0}, then λ − A − B ∈ Φ(X). Since AB and BA are in PR(Φ(X)) and AB(0) ⊂ BA(0) ⊂ (A + B)(0) = A(0) + B(0), then by using both (4.31) and (4.32), we have (λ − A)(λ − B) ∈ Φ(X) and (λ − B)(λ − A) ∈ Φ(X). (4.35) Applying Theorem 2.14.3, it is clear that λ − A ∈ Φ(X) and λ − B ∈ Φ(X). S Therefore, λ 6∈ σe4 (A) σe4 (B). This proves that h i [ σe4 (A) σe4 (B) \{0} ⊂ σe4 (A + B)\{0}. S (ii) Let λ 6∈ [σe5 (A) σe5 (B)] \{0}. Then, A − λ ∈ Φ(X), i(A − λ) = 0, B − λ ∈ Φ(X) and i(B − λ) = 0. Using Theorem 2.14.3, we infer that (B − λ)(A − λ) ∈ Φ(X) and i((A − λ)(B − λ)) = 0. Since   X i ((B − λ)(A − λ)) = i(B − λ) + i(A − λ) + dim R(A − λ) + D(B − λ) T − dim (A(0) N (B − λ)) , T T R(A − λ) + D(T − λ) = X, and A(0) N (B − λ) = {0} N (B − λ) = {0}, then i ((B − λ)(A − λ)) = i(B − λ) + i(A − λ) = 0. (4.36) This implies that i (BA + λ(λ − A − B)) = 0. Moreover, BA ∈ PR(Φ(X)) and BA(0) = B(0) ⊂ λ(A + B − λ)(0) = (A + B)(0) = A(0) + B(0). Hence, by using Proposition 3.3.2, we infer that λ−A−B ∈ Φ(X) and i (λ − A − B) = 0. Therefore, λ 6∈ σe5 (A + B), whence h i [ σe5 (A + B)\{0} ⊂ σe5 (A) σe5 (B) \{0}. (4.37) To prove the inverse inclusion of (4.37). Let λ 6∈ σe5 (A + B)\{0}, then A + B − λ ∈ Φ(X) and i(A + B − λ) = 0. Since AB and BA are in PR(Φ(X)), then it is easy to show that A − λ ∈ Φ(X) and B − λ ∈ Φ(X). On the other hand, applying (4.31), (4.35), (4.36) and Proposition 3.3.2 (iii), we have i[(B − λ)(A − λ)]

= i(A − λ) + i(B − λ) = i(A + B − λ) = 0.

(4.38)

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Since A is bounded single valued, we get ρ(A) 6= ∅. Besides, ΦA is connected, this together with Proposition 3.6.2 (i), allow us to deduce that σe4 (A) = σe5 (A). Using the last equality and the fact that A − λ ∈ Φ(X), we deduce that i(A − λ) = 0. It follows from (4.38) that i(B − λ) = 0. We conclude that S λ 6∈ σe5 (A) σe5 (B). Hence, h i [ σe5 (A) σe5 (B) \{0} ⊂ σe5 (A + B)\{0}. So, we have proved (4.29). S S (v) Suppose that λ 6∈ σe1 (A) σe1 (B) {0}, then A−λ ∈ Φ+ (X) and B −λ ∈ Φ+ (X). Using Theorem 2.14.3, we have (A − λ)(B − λ) ∈ Φ+ (X). Since AB ∈ PR (Φ+ (X)) and AB(0) ⊂ BA(0) = B(0) ⊂ (A + B)(0), then by using (4.31) and Proposition 3.3.2 (i), we have A + B − λ ∈ Φ+ (X). So, λ 6∈ σe1 (A + B). Therefore, [ [ σe1 (A + B)\{0} ⊂ σe1 (A) σe1 (B) {0}. S Suppose λ 6∈ σe1 (A + B) {0}, then A + B − λ ∈ Φ+ (X). Since AB and BA are in PR (Φ+ (X)), and by applying Eqs. (4.31), (4.32), we have (A − λ)(B − λ) ∈ Φ+ (X), (B − λ)(A − λ) ∈ Φ+ (X). It is clear that A − λ ∈ Φ+ (X) and B − λ ∈ Φ+ (X). Hence, λ 6∈ S σe1 (A) σe1 (B). Therefore, h i [ σe1 (A) σe1 (B) \{0} ⊂ σe1 (A + B)\{0}. This proves (4.30). S S (vi) Now, suppose that λ 6∈ σeap (A) σeap (B) {0}, then by Corollary 4.1.1, we have A − λ ∈ Φ+ (X), i(A − λ) ≤ 0, B − λ ∈ Φ+ (X) and i(B − λ) ≤ 0. Using Theorem 2.14.3, we have (A − λ)(B − λ) ∈ Φ+ (X) and   X i ((A − λ)(B − λ)) = i(A − λ) + i(B − λ) + dim − R(B − λ) + D(A − λ) T dim[A(0) N (A − λ)]. Since R(B − λ) + D(A − λ) = X , then i ((A − λ)(B − λ)) = i(A − λ) + i(B − λ) − dim[A(0)

\

N (A − λ)] ≤ 0.

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On the other hand, since AB ∈ PR (Φ+ (X)), then A + B − λ ∈ Φ+ (X) and i ((A − λ)(B − λ)) = i(A + B − λ) ≤ 0. Again, applying Corollary 4.1.1, it is clear that λ 6∈ σeap (A + B). Hence, [ [ σeap (A + B)\{0} ⊂ σeap (A) σeap (B) {0}. The rest of the proof is analogous to the previous case, it use Lemma 2.13.2 and D(A) + R(B) = X. The proofs of (vi) and (vii) may be achieved by using the same reasoning as (iii). Q.E.D. Theorem 4.3.3 Let λ ∈ C, T , S ∈ BCR(X), and S(0) ⊂ T (0). If there exists Hλ such that Hλ (λ − T − S) = I − K and −λ−1 T SHλ is demicompact,   S then λ ∈ σe1 (T + S)\{0} implies that λ ∈ σe1 (T ) σe1 (S) \{0}. ♦ Proof. Since T ∈ BCR(X), then (λ − T )(λ − S) = λ(λ − T ) − (λ − T )S. So, λ(λ − T ) − (λ − T )S

= λ(λ − T − S) + T S.

Hence, T SK(0) = T S(0) ⊂ (λ(λ − T − S) + T S) (0) and D(λ(λ − T − S) + T S) = D(T SK) = X. Thus, (λ − T )(λ − S) = λ(λ − T − S) + T S + T SK − T SK. Clearly, D(T S) = X, then (λ − T )(λ − S)

= λ(λ − T − S) + T S(I − K) + T SK

= λ(λ − T − S) + T SHλ (λ − T − S) + T SK.
So, λ(λ − T − S) + T SHλ (λ − T − S) + T SK = λ I + λ−1 T SHλ (λ − T − S) + T SK. We conclude
(λ − T )(λ − S) = λ I + λ−1 T SHλ (λ − T − S) + T SK.

Since D(T SK) = D λ I + λ−1 T SHλ (λ − T − S) and T SK(0) ⊂ λ(λ −
T − S) + T S(I − K)(0) = λ I + λ−1 T SHλ (λ − T − S)(0), then
(λ − T )(λ − S) − T SK = λ I + λ−1 T SHλ (λ − T − S). (4.39)   S Let λ 6∈ σe1 (T ) σe1 (S) \{0}, then λ − T ∈ Φ+ (X) and λ − S ∈ Φ+ (X). Hence, T SK ∈ PR(Φ+ (X)). In view of both Lemma 2.14.3 (iii) and Proposition 2.7.2, we get (λ − T )(λ − S) − T SK ∈ Φ+ (X). Again, by using Lemma 2.14.3 (v) and (4.39), we show that λ−T −S ∈ Φ+ (X). Q.E.D. So, we deduce that λ 6∈ σe1 (T + S)\{0}.

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4.4 4.4.1

219

S-Essential Spectra of Linear Relations Characterization of S-Essential Spectra of Linear Relations

Theorem 4.4.1 Let T ∈ CR(X) and S ∈ BR(X) such that S(0) ⊂ T (0) and kSk = 6 0. Then, (i) σei,S (T ), i = 1, 2, 3, 4, 5 are closed. (ii) The index is constant in any component of ρei,S (·), i = 1, 2, 3, 4, 5.

♦

Proof. (i) For i = 5. Let λ0 ∈ ρe5,S (T ). Then, R(T − λ0 S) is closed. Hence, T −λ0 S is open. Then, according to Proposition 2.9.3 (i), we have γ(T −λ0 S) > −λ0 S) −λ0 S) 0. Thus, γ(TkSk > 0. Consider µ ∈ C such that |µ − λ0 | < γ(TkSk . Then, according to Corollary 3.3.1, λ0 S−µS−λ0 S+T ∈ Φ(X) and i(λ0 S−µS−λ0 S+ T ) = 0. Since λ0 S(0) ⊂ T (0) = (µS − T )(0) and D(µS − T ) = D(T ) ⊂ D(S), then the use Proposition 2.3.4 allows us to conclude that λ0 S −µS −λ0 S +T = µS − T. Hence, µS − T ∈ Φ(X) and i(µS − T ) = 0. Therefore, ρe5 (T ) is open. For the other cases, the same arguments have been used. (ii) Let λ1 and λ2 be any two points in ρei,S (T ), i = 1, 2, 3, 4, 5 which are connected by a smooth curve Γ whose points are all in ρei,S (T ), i = 1, 2, 3, 4, 5. Since ρei,S (T ), i = 1, 2, 3, 4, 5 are an open set, then for each λ ∈ Γ, then by Corollary 3.3.1, there exists an ε > 0 such that, for all µ ∈ C, 0 < |µ − λ| < ε, µ ∈ ρei,S (T ), i = 1, 2, 3, 4, 5 and i(µS − T ) = i(λS − T ). By using the HeineBorel theorem, there exists a finite number of such sets which cover Γ. Since each of these sets overlaps with, at least, another set and since i(µS − T ) is constant on each one, we see that i(λ1 S − T ) = i(λ2 S − T ). Q.E.D.

4.4.2

Characterization of σe5,S (·)

Theorem 4.4.2 Let T ∈ CR(X) and S ∈ BR(X) such that S(0) ⊂ T (0), S= 6 0, and S = 6 T . Then, \ σe5,S (T ) = σS (T + K). ♦ K∈KT (X)

Proof. Let λ 6∈ σe5,S (T ). Then, T − λS ∈ Φ(X) and i(T − λS) = 0. Hence, α(T − λS) = β(T − λS) = n < ∞. So, there exists an everywhere defined

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single valued K with dim R(K) ≤ α(T −λS) such that T −λS −K is bijective, where K is defined by Kx :=

n X

x0i (x)yi ,

x ∈ X,

i=1

{x1 , . . . , xn } is a basis of N (T −λS), and x01 , . . . , x0n are linear functionals such that x0i (xj ) = δij . Choose y1 , . . . , yn ∈ X such that [y1 ], . . . , [yn ] ∈ X/R(T − λS) are linearly independent (n = α(T − λS) = β(T − λS) < ∞). Hence, it is clear that K is a bounded finite rank operator. So, K ∈ KR(X) and K(0) = 0 ⊂ (T − λS)(0) = T (0) − S(0) = T (0). It is clear that K ∈ KT (X) and also (λS − T ) − K =\ λS − (T + K) is bijective. Therefore, λ ∈ ρS (T + K). This shows that λ 6∈ σS (T + K). So, K∈KT (X)

\

σS (T + K) ⊂ σe5,S (T ).

(4.40)

K∈KT (X)

To prove the inverse inclusion of (4.40). Suppose λ 6∈

\

σS (T +K), then

K∈KT (X)

there exists K ∈ KT (X) such that λ ∈ ρS (T + K). Hence, T − λS + K ∈ Φ(X) and i(T − λS + K) = 0. Since K ∈ KT (X), it follows from Lemma 2.16.1 that T − λS + K − K ∈ Φ(X) and i(T − λS + K − K) = i(T − λS + K) = 0. So, by using Lemma 2.16.1, we infer that T − λS \ ∈ Φ(X) and i(T − λS) = 0. We conclude λ 6∈ σe5,S (T ). Hence, σe5,S (T ) ⊂ σS (T + K). Q.E.D. K∈KT (X)

Corollary 4.4.1 Let T ∈ CR(X) and S ∈ BR(X) such that S(0) ⊂ T (0), S= 6 0, and S = 6 T . Then, \ σe5,S (T ) = σS (T + P ). ♦ P ∈PR(Φ(X))

T

Proof. Let O := P ∈PR(Φ(X)) σS (T + P ). Since KT (X) ⊂ PR(Φ(X)), we infer that O ⊂ σe5,S (T ). Conversely, let λ 6∈ O, then there exists P ∈ PR(Φ(X)) such that λ 6∈ σS (T + P ). Therefore, λ ∈ ρS (T + P ). Hence, λS − T + P ∈ Φ(X) and i(λS − T + P ) = 0. Now, the use of Proposition 3.3.2 makes us conclude that λS − T + P − P ∈ Φ(X) and i(λS − T + P − P ) = 0. Finally, Corollary 2.16.1, shows that λS − T ∈ Φ(X) and i(λS − T ) = 0. So, λ 6∈ σe5,S (T ). Q.E.D.

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4.4.3

221

Relationship Between σe4,S (·) and σe5,S (·)

Theorem 4.4.3 Let T ∈ CR(X) and S ∈ BR(X) such that S(0) ⊂ T (0), S 6= 0, and S 6= T . Then, S (i) σe5,S (T ) = σe4,S (T ) {λ ∈ C such that i(T − λS) 6= 0} . (ii) If ρe4,S (T ) is connected and ρS (T ) 6= ∅, then ♦

σe4,S (T ) = σe5,S (T ).

S Proof. (i) Let λ 6∈ σe4,S (T ) {λ ∈ C such that i(T − λS) 6= 0}. Then, T − λS ∈ Φ(X) and i(T − λS) = 0. Hence, λ 6∈ σe5,S (T ). So, [ (4.41) σe5,S (T ) ⊂ σe4,S (T ) {λ ∈ C such that i(T − λS) 6= 0} . To prove the inverse inclusion of (4.41). Suppose λ 6∈ σe5,S (T ), then by using Corollary 4.4.1, there exists K ∈ PR(Φ(X)) such that T −λS −K is bijective. Moreover, by virtue of Proposition 2.3.4, we infer that T −λS = T −λS+K−K. So, by using Proposition 3.3.2 (iii), T − λS ∈ Φ(X) and i(T − λS) = 0. Thus, S λ 6∈ σe4,S (T ) and λ 6∈ σe4,S (T ) {λ ∈ C such that i(T − λS) = 6 0}. Hence, S σe4,S (T ) {λ ∈ C such that i(T − λS) 6= 0} ⊂ σe5,S (T ). (ii) Clearly σe4,S (T ) ⊂ σe5,S (T ). The opposite inclusion can be obtained by showing that \ ρe4,S (T ) σe5,S (T ) = ∅. Assume that there exists λ0 ∈ ρe4,S (T )

\

σe5,S (T ).

(4.42)

T Let λ1 ∈ ρS (T ), then λ1 S − T ∈ Φ+ (X) Φ− (X) and i(λ1 S − T ) = 0. Since ρe4,S (T ) is connected, it follows from Theorem 4.4.1 (ii) that i(λS − T ) is constant on any component of ρe4,S (T ). Therefore, i(λ1 S − T ) = i(λ0 S − T ) = 0. Hence, λ0 ∈ / σe5,S (T ) which contradicts the (4.42). Q.E.D.

4.5

Racocˇevic´ and Schmoeger S-Essential Spectra of a Linear Relations

The main purpose of this section is to prove that the properties of the Racocˇevic´ and Schmoeger S-essential spectra of a linear relation for closed

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Spectral Theory of Multivalued Linear Operators

densely defined operators in Banach spaces obtained in Refs. [2, 65] are valid for closed linear relations. Theorem 4.5.1 Let T ∈ CR(X) and S ∈ BR(X) such that S(0) ⊂ T (0) and S 6= 0. Then, S (i) σeap,S (T ) = σe1,S (T ) {λ ∈ C such that i(T − λS) > 0}. S (ii) σeδ,S (T ) = σe2,S (T ) {λ ∈ C such that i(T − λS) < 0}. ♦ Proof. For all λ ∈ C, we have λS(0) ⊂ T (0) and D(T ) ⊂ D(λS) = X. So, by using Proposition 2.7.2, we infer that λS − T is closed. (i) Let λ ∈ / σeap,S (T ). Then, there exist K ∈ KT (X) such that λS − T − K

=

(λS − T ) − K

= A−K is bounded below. Since Aλ := λS − T is closed, K(0) ⊂ T (0) = Aλ (0) and D(Aλ ) = D(T ) ⊂ D(K), then by using Proposition 2.7.2, we have λS−(T +K) is closed. Since λS − (T + K) is injective with closed range and open, then λS − (T + K) ∈ Φ+ (X) and i((λS − T ) − K)

= α(λS − T − K) − β(λS − T − K) =

0 − β(λS − T − K) ≤ 0.

Therefore, λS − (T + K) ∈ Φ+ (X) and i(λS − (T + K)) ≤ 0. By Lemma 2.16.1 (i), we have λS−(T +K)+K = (λS−T )+K−K ∈ Φ+ (X) and i(λS−(T +K)+ K) = i(λS −(T +K)) ≤ 0. Since K(0) ⊂ T (0) = (λS −T )(0) and D(λS −T ) = D(T ) ⊂ D(K), and by using Proposition 2.3.4, we have λS − T ∈ Φ+ (X) and S i(λS − T ) ≤ 0. Therefore, λ 6∈ σe1,S (T ) {λ ∈ C such that i(λS − T ) > 0} . S Conversely, let λ 6∈ σe1,S (T ) {λ ∈ C such that i(λS − T ) > 0}, then λS − T ∈ Φ+ (X) and i(λS − T ) ≤ 0. We define the operator K by K : X −→ Kx :=

n X

x0i (x)yi ,

x ∈ X,

i=1

where {x1 , . . . , xn } is a basis of N (λS − T ) and x01 , . . . , x0n are a linear functionals such that x0i (xj ) = δij . Choose y1 , . . . , yn ∈ X such that [y1 ], . . . , [yn ] ∈ X/R(λS − T ) are linearly independent (such elements exist since n ≤ β(λS − T )). It is clear that K is continuous, and so that K is a bounded finite rank operator. So, K ∈ KT (X) and also (λS − T ) − K = λS−(T +K) is injective. Since λS−T ∈ Φ+ (X), we have λS−(T +K) ∈ Φ+ (X)

Essential Spectra and Essential Pseudospectra of a Linear Relations

223

with the same index that λS − T . In this situation, we have K ∈ KT (X), λS − (T + K) is injective, closed with closed range. Hence, λS − (T + K) is open. So, λ ∈ ρap,S (T + K). Therefore, λ ∈ / σeap,S (T ). The proofs of (ii) is similar to the previous one.

Q.E.D.

Corollary 4.5.1 Let T ∈ CR(X) and S ∈ BR(X) such that S(0) ⊂ T (0) and S 6= 0. Then, (i) λ 6∈ σeap,S (T ) if and only if, λS − T ∈ Φ+ (X) and i(T − λS) ≤ 0. (iii) λ 6∈ σeδ,S (T ) if and only if, λS − T ∈ Φ− (X) and i(T − λS) ≥ 0.

♦

Proposition 4.5.1 Let T ∈ CR(X) and S ∈ BR(X) such that S(0) ⊂ T (0), S 6= 0 and ρS (T ) 6= ∅. If ΦT,S is connected, then σe1,S (T ) = σeap,S (T ) and σe2,S (T ) = σeδ,S (T ).

♦

Proof. It is easy to check that σe1,S (T ) ⊂ σeap,S (T ). For the second inclusion, we take µ ∈ ρe1,S (T ), then [ µ ∈ Φ+T,S = ΦT,S (Φ+T,S \ ΦT,S ). Hence, we will discuss these two cases: First case : If µ ∈ ΦT,S , then i(T − µS) = 0. Indeed, let µ0 ∈ ρS (T ), then µ0 ∈ ΦT,S and i(T − µ0 S) = 0. It follows from Theorem 2.12.2 that i(T − µS) is constant on any component of ΦT,S . Therefore, ρS (T ) ⊂ ΦT,S , then i(T − µS) = 0 for all µ ∈ ΦT,S . This shows that µ ∈ ρeap,S (T ). Second case : If µ ∈ (Φ+T,S \ΦT,S ), then α(T −µS) < ∞ and β(T −µS) = +∞. So, i(T − µS) = −∞ < 0. Hence, we obtain the second inclusion from the above two cases. A same reasoning as before leads to the second equality, which completes the proof. Q.E.D.

4.5.1

Stability of Racocˇevic´ and Schmoeger S-Essential Spectra of a Linear Relations

We start our investigation with the following result. Theorem 4.5.2 Let X be complete, T ∈ CR(X), A, S ∈ LR(X) satisfy A(0) ⊂ S(0) ⊂ T (0) and dim D(A) = ∞. Let S be T -bounded with T -bound δ1

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Spectral Theory of Multivalued Linear Operators

and A is T -precompact with T -bound δ2 . Let λ ∈ C such that δ2 + |λ|δ1 < 1, then (i) λ ∈ σeap,S (T + A) if and only if, λ ∈ σeap,S (T ), (ii) λ ∈ σeδ,S (T + A) if and only if, λ ∈ σeδ,S (T ).

♦

Proof. Let A be T -precompact. Then, AGT is precompact, and X and XT are complete. By using Proposition 2.11.3, we get AGT is compact. As a matter of fact, A is T -bounded with T -bound δ2 . Using the fact that S is T -bounded with T -bound δ1 , δ2 + |λ|δ1 < 1 and, by applying Lemma 3.1.1 (i), we obtain λS − (T + A) is closed. (i) Suppose that λ ∈ / σeap,S (T ). Then, by using Corollary 4.5.1, we have λS − T ∈ Φ+ (X). In view of Proposition 2.14.3 (ii), we get (λS − T )GλS−T ∈ Φ+ (XT , X), which provides by referring to Theorem 2.6.1 that (λS − T )GT ∈ Φ+ (XT , X). Since AGT is compact and dim D(A) = dim D(AGT ) = ∞, then by using Corollary 2.16.1 (iv) and Lemma 2.16.2, it follows that (λS − (T + A))GT ∈ Φ+ (XT , X). The use of Theorem 2.6.1, we obtain (λS − (T + A))GλS −(T +A) ∈ Φ+ (XT , X). Thus, Proposition 2.11.3 yields λS − (T + A) ∈ Φ+ (X) and in view of Proposition 2.11.3, we have i(λS − T ) = i(λS − (T + A)). Using Corollary 4.5.1, we have λ ∈ / σeap,S (T + A) which yields σeap,S (T + A) ⊂ σeap,S (T ). Conversely, let λ ∈ / σeap,S (T + A). Then, by using Corollary 4.5.1, we have λS − (T + A) ∈ Φ+ (X). From Proposition 2.14.3 (ii), we get (λS − (T + A))GλS −(T +A) ∈ Φ+ (XT , X), which entails by referring to Theorem 2.6.1, (λS − (T + A))GT ∈ Φ+ (XT , X). Since AGT is compact, then by using both Corollary 2.16.1 (iv) and Lemma 2.16.2, it follows that (λS − T )GT ∈ Φ+ (XT , X). From Theorem 2.6.1, we obtain (λS − T )GλS −T ∈ Φ+ (XT , X). Using Proposition 2.14.3 (ii), we get λS−T ∈ Φ+ (X). From Proposition 2.11.3, we have i(λS−T ) = i(λS−(T +A)). So, by Corollary 4.5.1, λ ∈ / σeap,S (T ). Thus, σeap,S (T + A) = σeap,S (T ). (ii) Now, suppose that λ ∈ / σeδ,S (T ), then by Corollary 4.5.1, we have λS − T ∈ Φ− (X). Applying Proposition 2.11.3, we obtain (λS − T )GλS−T ∈ Φ− (XT , X). Using Theorem 2.6.1, we get (λS − T )GT ∈ Φ− (XT , X). Since AGT is precompact, then in view of Theorem 2.16.3 (i), we obtain (λS − (T + A))GT ∈ Φ− (XT , X). Resorting to Theorem 2.6.1, we get (λS − (T + A))GλS−(T +B) ∈ Φ− (XT , X). As a matter of fact, applying Proposition 2.11.3, we get λS −(T +A) ∈ Φ− (X). From Proposition 2.11.3, we have i(λS −T ) = i(λS −(T +A)). Hence, by Corollary 4.5.1, λ ∈ / σeδ,S (T +A). Then, σeδ,S (T + A) ⊂ σeδ,S (T ). Conversely, let λ ∈ / σeδ,S (T + A), then by

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225

Corollary 4.5.1, we obtain λS − (T + A) ∈ Φ− (X). Using Proposition 2.14.3 (i), we get (λS − (T + A))GλS −(T +A) ∈ Φ− (XT , X). Applying Theorem 2.6.1, we get (λS − (T + A))GT ∈ Φ− (XT , X). The latter holds if and only if, by Proposition 2.11.3, ((λS − (T + A))GT )∗ ∈ Φ+ (X ∗ , XT∗ ). Subsequently, using Propositions 2.8.1 (iii) and 2.14.3 (i), we get ((λS − T )GT )∗ + (AGT )∗ ∈ Φ+ (X ∗ , XT∗ ). Since AGT is precompact, then by Proposition 2.14.3 (ii), we have (AGT )∗ is precompact. Applying Theorem 2.16.3 (i), we get ((λS − T )GT )∗ ∈ Φ+ (XT , X ∗ ). Besides, using Theorem 2.14.4, we get (λS − T )GT ∈ Φ− (XT , X). Hence, by Proposition 2.14.3 (i), λS − T ∈ Φ− (X). Appliying Proposition 2.11.3, we have i(λS − T ) = i(λS − (T + A)). That is by Corollary Q.E.D. 4.5.1, λ ∈ / σeδ,S (T ). We conclude that σeδ,S (T + A) = σeδ,S (T ). Theorem 4.5.3 Let T ∈ CR(X), A, S ∈ LR(X). Suppose that S is T bounded with T -bound δ1 and A is T -bounded with T -bound δ2 . Let λ ∈ C such that δ2 + |λ|δ1 < 1. If G(A) ⊂ G(λS) ⊂ G(T ) and dim D(A) = ∞, then (i) λ ∈ σeap,S (T + A) implies λ ∈ σeap,S (T ). (ii) If dim A(0) < ∞, then λ ∈ σeap,S (T + A) if and only if, λ ∈ σeap,S (T ).♦ Proof. Since S is T -bounded with T -bound δ1 and A is T -bounded with T bound δ2 such that δ2 + |λ|δ1 < 1, then applying Lemma 3.1.1, we obtain λS − (T + A) is closed. (i) Suppose that λ ∈ / σeap,S (T ), then by Corollary 4.5.1, we get λS − T ∈ Φ+ (X) and i(λS − T ) ≤ 0. Since G(A) ⊂ G(λS) ⊂ G(T ), then applying Proposition 2.3.3, we get G(λS) ⊂ G(λS − T ), and G(A) ⊂ G(λS − T ). On the one hand, we have G(λS − (T + A))

= {(x, y) ∈ X × X such that (x, y1 ) ∈ G(λS − T ) and (x, y2 ) ∈ G(A) ⊂ G(λS − T ), where y = y1 + y2 } ⊂ G(λS − T ).

On the other hand, dim D(λS − (T + A)) = dim(D(λS − T )

\

D(A)) = dim D(A) = ∞.

Then, by Proposition 2.14.4, we obtain λS −(T +A) ∈ Φ+ (X). Hence, G(λS − (T + A)) ⊂ G(λS − T ). By using Lemma 2.4.1, we get i(λS − (T + A)) ≤ i(λS − T ) ≤ 0. So, by Corollary 4.5.1, we obtain λ ∈ / σeap,S (T + A). Thus, σeap,S (T + A) ⊂ σeap,S (T ).

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(ii) Since G(A) ⊂ G(λS) ⊂ G(T ), applying Proposition 2.3.3, we get G(λS) ⊂ G(λS − T ) and G(A) ⊂ G(λS − T ). By using Proposition 2.3.3, we obtain G(A) ⊂ G(λS − (T + A)). On the one hand, G(λS − T − A + A) = {(x, y + z) ∈ X × Y such that (x, y) ∈ G(λS − (T + A)) and (x, z) ∈ G(A)} ⊂ {(x, y + z) ∈ X × Y such that (x, y) ∈ G(λS − (T + A)) and (x, z) ∈ G(S) ⊂ G(λS − (T + A))} = G(λS − (T + A)). On the other hand, dim D(λS − T − A + A) = dim(D(λS)

\

D(T )

\

D(A)) = dim D(A) = ∞.

Let λ ∈ / σeap,S (T + A). Then, by Corollary 4.5.1, we have λS − (T + A) ∈ Φ+ (X). Since λS −T is closed and dim A(0) < ∞, then λS −T −A+A is closed and G(λS −T −A+A) ⊂ G(λS −(T +A)). Since dim D(λS −T −A+A) = ∞, then by Proposition 2.14.4, we get λS − T − A + A ∈ Φ+ (X). Thus, by Proposition 2.14.3 (iv), we obtain λS − T ∈ Φ+ (X). Using Lemma 2.4.1, we get i(λS − T ) ≤ i(λS − T − A + A) ≤ i(λS − (T + A)) ≤ 0. Hence, by Corollary 4.5.1, we obtain λ ∈ / σeap,S (T ). Therefore, σeap,S (T ) ⊂ σeap,S (T + A) which implies that σeap,S (T + A) = σeap,S (T ). Q.E.D.

4.6

S-Essential Spectra of the Sum of Two Linear Relations

In this section, we investigate the S-essential spectra of the sum of two closed linear relations defined on a Banach space by means of S-essential spectra of these two linear relations where their products are Fredholm or semi-Fredholm perturbations. Theorem 4.6.1 Let X be a Banach space and let A, S ∈ L(X) and B ∈ BCR(X) such that AB ⊂ BA ⊂ SB ⊂ BS. Then, (i) If AB ∈ PR(Φ(X)), then h i [ σe4,S 2 (AS + BS)\{0} ⊂ σe4,S (A) σe4,S (B) \{0}.

Essential Spectra and Essential Pseudospectra of a Linear Relations

227

If, further, BA ∈ PR(Φ(X)), SA = AS and BS = SB, then i h [ σe4,S 2 (AS + BS)\{0} = σe4,S (A) σe4,S (B) \{0}. (ii) If BA ∈ PR(Φ(X)), then h i [ σe5,S 2 (SA + BS) ⊂ σe5,S (A) σe5,S (B) \{0}. Moreover, if AB ∈ PR(Φ(X)), SA = AS, B S = SB and ΦA,S is connected, then h i [ (4.43) σe5,S (SA + BS)\{0} = σe5,S (A) σe5,S (B) \{0}. (iii) If AB ∈ PR (Φ+ (X)), then i h [ σe1,S 2 (SA + SB)\{0} ⊂ σe1,S (A) σe1,S (B) \{0}. If, further, BA ∈ PR (Φ+ (X)), SA = AS and BS = SB, then h i [ σe1,S 2 (AS + SB)\{0} = σe1,S (A) σe1,S (B) \{0}.

(4.44)

(iv) If the hypotheses of (iii) are satisfied, then h i [ σeap,S 2 (AS + SB)\{0} ⊂ σeap,S (A) σeap,S (B) \{0}. If, further, ΦS,A is connected, BA ∈ PR (Φ+ (X )), SA = AS and BS = SB, then h i [ σeap,S 2 (AS + SB)\{0} = σeap,S (A) σeap,S (B) \{0}. (v) If the hypotheses of (iv) are satisfied, then h i [ σeδ (A + B)\{0} ⊂ σeδ (A) σeδ (B) \{0}. If, further, ΦS,A is connected, BA ∈ PR(Φ− (X)), SA = AS and BS = SB, then h i [ σeδ,S 2 (AS + SB)\{0} = σeδ,S (A) σeδ,S (B) \{0}. T (vi) If AB ∈ PR (Φ+ (X)) PR (Φ− (X)), then σe3,S 2 (AS + SB)\{0} ⊂ S S T S T ([σe3,S (A) σe3,S (B)] [σe1,S (A) σe2,S (B)] [σe2,S (A) σe1,S (B)]) \{0}. T Moreover, if BA ∈ PR (Φ+ (X)) PR (Φ− (X)), SA = AS and BS = SB, then σe3 (SA + SB)\{0} = S S T S T ([σe3,S (A) σe3,S (B)] [σe1,S (A) σe2,S (B)] [σe2,S (A) σe1,S (B)]) \{0}.♦

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Spectral Theory of Multivalued Linear Operators

Proof. We first prove, for λ ∈ C (λS − A)(λS − B) = AB + λ(λS 2 − AS − SB)

(4.45)

and (λS − B)(λS − A) = BA + λ(λS 2 − SA − BS). (4.46) T Indeed, since D(λS − A) = D(λS) D(A) = X and, by using Proposition 2.3.6, we have (λS − A)(λS − B) = (λS − A)λS − (λS − A)B. Further, by using Proposition 2.3.6, we have (λS − A)B ⊂ λSB − AB. Then, (λSB − AB)(0) = λSB(0) − AB(0) ⊂ λSB(0) − BA(0) = B(0) ⊂ (λS − A)B(0). Hence, (λS − A)B(0) = (λSB − AB)(0). Since n o \ D((λS − A)B) = x ∈ D(B) = X : Bx D(λS − A) 6= ∅ n o \ = x ∈ D(B) = X : Bx X 6= ∅ = {x ∈ D(B) = X : Bx 6= ∅} = D(B) = X T and D(λSB − AB) = D(λSB) D(AB) = X, we infer from Proposition 2.3.6 that (λS − A)B = λSB − AB. Hence, (λS − A)(λS − B) = AB + λ(λS 2 − AS − SB). So, (4.45) holds. We note that (λS − B)(λS − A)

= λS(λS − A) − B(λS − A) = λ2 S 2 − λSA − λBS + BA.

Hence, (4.46) holds. S S (i) Let λ 6∈ σe4,S (A) σe4,S (B) {0}. Then, λS − A ∈ Φ(X) and λS − B ∈ Φ(X). So, Theorem 2.14.3 gives (λS − A)(λS − B) ∈ Φ(X). Hence, by using (4.45), we have AB + λ(λS 2 − AS − SB) ∈ Φ(X). Since AB ∈ PR(Φ(X)), then AB − (λS − A)(λS − B) ∈ Φ(X).

Essential Spectra and Essential Pseudospectra of a Linear Relations

229

This implies that AB − AB − λ(λS 2 − AS − SB) ∈ Φ(X). Since AB(0) ⊂ λ(λS 2 − AS − SB)(0) = (SB)(0), then by using Lemma 2.16.1, we infer that λS 2 − AS − SB ∈ Φ(X). Hence, λ 6∈ σe4,S 2 (AS + SB). Therefore, i h [ σe4,S 2 (AS + BS)\{0} ⊂ σe4,S (A) σe4,S (B) \{0}. (4.47) S To prove the inverse inclusion of (4.47). Suppose λ 6∈ σe4,S 2 (AS + SB) {0}, then λS 2 − AS − SB ∈ Φ(X). Since AB ∈ PR(Φ(X)), BA ∈ PR(Φ(X)), and AB(0) ⊂ (λS 2 − AS − SB)(0) = (SB)(0), then by using (4.45) and (4.46), we have (A − λS)(B − λS) ∈ Φ(X) and (B − λS)(A − λS) ∈ Φ(X).

(4.48)

Applying Theorem 2.14.3, it is clear that λS − A ∈ Φ(X) and λS − B ∈ Φ(X). S Therefore, λ 6∈ σe4,S (A) σe4,S (B). This proves that h i [ σe4,S (A) σe4,S (B) \{0} ⊂ σe4,S 2 (SA + BS)\{0}. (4.49) S (ii) Let λ 6∈ [σe5,S (A) σe5,S (B)] \{0}. Then, A − λS ∈ Φ(X), i(A − λS) = 0, B − λS ∈ Φ(X) and i(B − λS) = 0. Using Theorem 2.14.3, we infer that (B − λ)(A − λS) ∈ Φ(X). By using Theorem 2.14.3, we have i ((B − λS)(A − λS)) = i(B − λS) + i(A − λS)+  T dim X [R(A − λS) + D(B − λS)] − dim[A(0) N (B − λS)] and clearly, R(S−λS)+D(T −λS) = X and A(0)

\

N (B−λS) = {0}

\

N (B−λS) = {0}.

Then, i ((B − λS)(A − λS)) = i(B − λS) + i(A − λS) = 0 (4.50)
implies that i BA + λ(λS 2 − SA − BS) = 0. Moreover, BA ∈ PR(Φ(X)) and BA(0) = λ(λS 2 − SA − BS)(0) = BS(0) = B(0).

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Spectral Theory of Multivalued Linear Operators

Hence, by using Theorem 2.14.3, we infer that λS 2 − SA − BS ∈ Φ(X) and
i λS 2 − SA − BS = 0. Therefore, λ 6∈ σe5,S 2 (SA + BS), whence h i [ (4.51) σe5,S 2 (SA + BS) ⊂ σe5,S (A) σe5,S (B) \{0}. To prove the inverse inclusion of (4.51). Let λ 6∈ σe5,S 2 (SA + BS), then λS 2 − SA − BS ∈ Φ(X) and i(λS 2 − SA − BS) = 0. Since AB ∈ PR(Φ(X)), BA ∈ PR(Φ(X)) and S commute with A and B, it is easy to show that A − λS ∈ Φ(X) and B − λS ∈ Φ(X). On the other hand, applying (4.45), (4.48), (4.50) and Theorem 2.14.3, we have i[(B − λS)(A − λS)]

= i(A − λS) + i(B − λS) = i(λS 2 − SA − BS) = 0.

(4.52)

6 ∅. Besides, ΦS,A is Since A is bounded single valued, we get ρS (A) = connected, this together with Theorem 4.4.3 (ii), allow us to deduce that σe4,S (A) = σe5,S (A). Using the last equality and the fact that A − λS ∈ Φ(X), we deduce that i(A − λS) = 0. It follows from (4.52) that i(B − λS) = 0. We conclude λ 6∈ S σe5,S (A) σe5,S (B). Hence, i h [ σe5,S (A) σe5,S (B) \{0} ⊂ σe5,S 2 (SA + BS)\{0}. So, we prove (4.43). S S (iii) Suppose that λ 6∈ σe1,S (A) σe1,S (B) {0}, then A − λS ∈ Φ+ (X) and B −λS ∈ Φ+ (X). Using Theorem 2.14.3, we have (A−λS)(B −λS) ∈ Φ+ (X). Since AB ∈ PR (Φ+ (X)), and AB(0) ⊂ BA(0) = B(0) ⊂ λ(λS 2 − AS − SB)(0) = (SB)(0), we can apply (4.45) and Theorem 2.14.3, we have λS 2 − AS − SB ∈ Φ+ (X). So, λ 6∈ σe1,S 2 (AS + SB). Therefore, [ [ σe1,S 2 (AS + SB) ⊂ σe1,S (A) σe1 (B, S) {0}. S Suppose λ 6∈ σe1 (A + B) {0}, then A + B − λ ∈ Φ+ (X). Since AB ∈ PR (Φ+ (X)) and BA ∈ P R (Φ+ (X)), and applying Eqs. (4.45), (4.46), we have (A − λS)(B − λS) ∈ Φ+ (X), (B − λS)(A − λS) ∈ Φ+ (X).

Essential Spectra and Essential Pseudospectra of a Linear Relations

231

By using Theorem 2.14.3, it is clear that A − λS ∈ Φ+ (X) and B − λS ∈ S Φ+ (X). Hence, λ 6∈ σe1,S (A) σe1,S (B). Therefore, i h [ σe1,S (A) σe1,S (B) \{0} ⊂ σe1,S 2 (AS + SB)\{0}. This proves (4.44). The rest of proof is analogous to the previous case. This completes the proof. Q.E.D.

4.7 4.7.1

Pseudospectra and ε-Pseudospectra of Linear Relations Some Properties of Pseudospectra and ε-Pseudospectra of Linear Relations

In this subsection, we define the pseudospectra of linear relation and study some properties. Proposition 4.7.1 Let X be a normed vector space and T ∈ LR(X). Then, \ \ σε (T ) = σ(T ) = Σε (T ). ♦ ε>0

ε>0

Proof. It is clear that σ(T ) ⊂ σε (T ) ⊂ Σε (T ) for all ε > 0, then \ \ σ(T ) ⊂ σε (T ) ⊂ Σε (T ). ε>0

(4.53)

ε>0

Conversely, if λ ∈ / σ(T ), then λ ∈ ρ(T ). Hence, (λ − Te)−1 is a bounded linear operator, where Te is given in (2.4). So, there \ exists ε > 0 such that k(λ − Te)−1 k < 1ε . Thus, λ ∈ / Σε (T ). Hence, λ ∈ / Σε (T ). Then, ε>0

\

Σε (T ) ⊂ σ(T ).

(4.54)

ε>0

Thus, the use of (4.53) and (4.54) makes us conclude that \ \ σε (T ) = σ(T ) = Σε (T ). ε>0

Q.E.D.

ε>0

In the sequel of this section, X will denote Banach space over the complex field C.

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Spectral Theory of Multivalued Linear Operators

Proposition 4.7.2 Let ε > 0 and T ∈ CR(X) be injective with dense range. Then, σε (T ) ⊂ {λ ∈ C such that |λ| > γ(T ) − ε}.

♦

Proof. We will discuss three cases: First case : If γ(T ) < ε, then the result is obviously. Second case : If γ(T ) = ε. Then, T is open. Hence, 0 ∈ ρ(T ). Furthermore, kT −1 k = 1ε . Then, 0 ∈ / σε (T ). Therefore, σε (T ) ⊂ {λ ∈ C such that |λ| > 0}. Third case : If γ(T ) > ε, then T is open and injective with dense range. On the other hand, for λ ∈ C such that |λ| ≤ γ(T ) − ε, then |λ| < γ(T ). By using Lemma 2.9.3, the relation λ − T is open, injective with dense range, i.e., λ ∈ ρ(T ). For x ∈ D(T ), we have kT xk = kT x − λx + λxk ≤

kT x − λxk + kλxk.

Consequently, for x ∈ D(T ), we obtain k(T − λ)xk ≥ kT xk − |λ|kxk ≥ (γ(T ) − |λ|)kxk. Therefore, γ(λ − T ) ≥ γ(T ) − |λ| ≥ γ(T ) − γ(T ) + ε = ε. So, 1 . ε Then, λ ∈ ρε (T ). Hence, σε (T ) ⊂ {λ ∈ C such that |λ| > γ(T ) − ε}. Q.E.D. k(λ − T )−1 k ≤

Theorem 4.7.1 Let T ∈ CR(X) and ε > 0. Then, (i) σε (T ) = σε (T ∗ ). (ii) Σε (T ) = Σε (T ∗ ).

♦

Proof. (i) Let λ ∈ ρε (T ∗ ). Then, λ ∈ ρ(T ∗ ) and k(λ − T ∗ )−1 k ≤ 1ε . By using Remark 2.18.1, we have λ ∈ ρ(T ). So, (λ − T )−1 is continuous. From Proposition 2.8.1 (vi), we have k(λ − T )−1 k = k(λ − T ∗ )−1 k ≤

1 . ε

Essential Spectra and Essential Pseudospectra of a Linear Relations So, λ ∈ ρε (T ). Conversely, if λ ∈ ρε (T ), then k(λ − T )−1 k ≤ Proposition 2.8.1 (vi), we have k(λ − T )−1 k = k(λ − T ∗ )−1 k ≤

1 ε.

233

By using

1 . ε

Furthermore, λ ∈ ρ(T ∗ ) by Remark 2.18.1 (iii). Then, λ ∈ ρε (T ∗ ). (ii) Let us assume that λ ∈ Σε (T ). Then, using Remark 2.20.2 (i1 ), we obtain  [ 1 λ ∈ σε (T ) λ ∈ C : k(λ − T )−1 k = . ε There are two possible cases: First case : Let λ ∈ σε (T ). Then, by using (i) and Remark 2.20.2 (i), we deduce that λ ∈ Σε (T ∗ ). Second case : Let us assume that λ ∈ / σε (T ) and k(λ − T )−1 k = 1ε . Then, 1 −1 λ ∈ ρ(T ), k(λ − T ) k ≤ ε and k(λ − T )−1 k = 1ε . This implies that (λ − T )−1 is a bounded linear operator and k(λ − T )−1 k = 1ε . Hence, by referring to Proposition 2.8.1 (vi), we infer that k(λ − T ∗ )−1 k = k(λ − T )−1 k = 1ε . Consequently, λ ∈ Σε (T ∗ ). This shows that Σε (T ) ⊂ Σε (T ∗ ).

(4.55)

In order to prove the inverse inclusion, it is sufficient to proceed by the same reasoning as (4.55). Q.E.D. Lemma 4.7.1 Let T ∈ CR(X) and ε > 0. If λ ∈ / σ(T ), then λ ∈ σε (T ) if and only if there exists x ∈ D(T ) such that k(λ − T )xk < εkxk. ♦ Proof. If λ ∈ σε (T ) and λ ∈ / σ(T ), then k(λ − T )−1 k > 1ε . So, there exists a non-zero vector y ∈ X such that k(λ − T )−1 yk >

1 kyk. ε

(4.56)

Putting x := (λ − T )−1 y, then x ∈ D(T ) and (λ − T )x = (λ − T )(λ − T )−1 y = y + (λ − T )(0). By using Lemma 2.3.1, we have y ∈ (λ − T )x. On the other hand, (λ − T )(0) = T (0). By Lemma 2.3.1, 0 ∈ T (0) and, from Lemma 2.5.7, we

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Spectral Theory of Multivalued Linear Operators

have k(λ − T )xk = dist(y, (λ − T )(0)) = dist(y, T (0)) ≤ dist(y, 0) ≤ kyk.

(4.57)

So, from (4.56) and (4.57), we have kxk >

1 1 kyk ≥ k(λ − T )xk. ε ε

Hence, k(λ − T )xk < εkxk. Conversely, assume that there exists x ∈ D(T ) such that k(λ − T )xk < εkxk. Since λ ∈ ρ(T ), then λ − T is injective, open and, we have γ(λ − T )kxk ≤ k(λ − T )xk < εkxk. Hence, 0 < γ(λ − T ) < ε. Using Lemma 2.5.7, we have γ(λ − T ) = k(λ − Q.E.D. T )−1 k−1 . So, k(λ − T )−1 k > 1ε . Proposition 4.7.3 Let X be a Banach space, T ∈ CR(X) and ε > 0. Then, o [n Σε (T ) = σ(T ) λ ∈ C : ∃ (xn ) ⊂ D(T ), kxn k = 1 and lim k(λ − T )xn k ≤ ε . n→∞

n o Proof. Let λ ∈ λ ∈ C : ∃ (xn ) ⊂ D(T ), kxn k = 1 and lim k(λ − T )xn k ≤ ε n→∞

and λ ∈ / σ(T ). Since λ ∈ ρ(T ), then (λ − T )−1 is a bounded operator. This implies from Lemma 2.9.1 that γ(λ − T ) = k(λ − T )−1 k−1 < ∞. By referring to Lemma 2.5.7 (i), we have k(λ − T )xn k =

dist ((λ − T )xn , 0)


≥ γ(λ − T ) dist xn , (λ − T )−1 (0) . The fact that (λ − T )−1 is an operator implies that (λ − T )−1 (0) = 0. Hence, k(λ − T )xn k

≥ γ(λ − T ) dist(xn , 0) ≥ γ(λ − T ) kxn k ≥ γ(λ − T ) as kxn k = 1.

This implies from Lemma 2.9.1 that k(λ − T )xn k

≥

k(λ − T )−1 k−1 .

Therefore, k(λ − T )−1 k ≥ 1ε . As a result, λ ∈ Σε (T ), as desired. Conversely, let us assume that λ ∈ ρ(T ) and k(λ − T )−1 k ≥ 1ε . In view of Lemma 2.5.7

Essential Spectra and Essential Pseudospectra of a Linear Relations

235

(ii) implies that for every n ∈ N\{0}, there exists (yn )n such that kyn k ≤ 1 and 1 k(λ − T )−1 k − ≤ k(λ − T )−1 yn k ≤ k(λ − T )−1 k. n Consequently, lim k(λ − T )−1 yn k =

n→+∞

lim k(λ − T )−1 k

n→+∞

= k(λ − T )−1 k 1 ≥ . ε

(4.58)

Since (λ − T )−1 is an operator, then we can take xn = k(λ − T )−1 yn k−1 (λ − T )−1 yn . Hence, (xn )n ⊂ D(T ), kxn k = 1 and yn ∈ k(λ − T )−1 yn k (λ − T )xn . Therefore, kk(λ − T )−1 yn k (λ − T )xn k =


dist yn , k(λ − T )−1 yn k (λ − T )(0)

=

dist (yn , (λ − T )(0)) .

≤

k(λ − T )−1 yn k−1 kyn k

≤

k(λ − T )−1 yn k−1 .

Consequently, k(λ − T )xn k

Hence, by applying (4.58), we obtain lim k(λ − T )xn k ≤ ε. This enables us n→+∞

to conclude that o [n Σε (T ) = σ(T ) λ ∈ C : ∃ (xn ) ⊂ D(T ), kxn k = 1 and lim k(λ − T )xn k ≤ ε . n→∞

This completes the proof.

4.7.2

Q.E.D.

Characterization of Pseudospectra of Linear Relations

Theorem 4.7.2 Let T ∈ CR(X) and ε > 0. Then, the following three conditions are equivalent. (i) λ ∈ σε (T ). (ii) There exists a continuous linear relation S ∈ LR(X) satisfying D(S) ⊃ D(T ), S(0) ⊂ T (0), kSk < ε such that λ ∈ σ(T + S). 1 (iii) Either λ ∈ σ(T ) or k(λ − T )−1 k > . ♦ ε

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Spectral Theory of Multivalued Linear Operators

Proof. (i) =⇒ (ii) Assume that λ ∈ σε (T ). We will discuss these two cases: First case : If λ ∈ σ(T ), then we may put S = 0. Second case : if λ ∈ / σ(T ), then by using Lemma 4.7.1, there exists x0 ∈ D(T ), kx0 k = 1 such that k(λ − T )x0 k < ε. By using Theorem 2.8.1, there exists x0 ∈ X ∗ such that kx0 k = 1 and x0 (x0 ) = kx0 k. We define the relation S : X −→ X by S(x) := x0 (x)(λ − T )x0 . It is clear that D(S) = X and S(0) = 0 which implies that S is everywhere defined and single value. Moreover, for x ∈ X, we have kSxk

≤

kx0 (x)(λ − T )x0 )k

≤

kx0 kkxkk(λ − T )x0 k

≤ εkxk. This implies that S ∈ BR(X) and kSk ≤ ε. In addition, we have (λ − (T + S))x0

= = = = = = =

(λ − T − S)x0 (λ − T )x0 − Sx0 (λ − T )x0 − x0 (x0 )(λ − T )x0 (λ − T )x0 − (λ − T )x0 (λ − T )(0) (λ − T )(0) − S(0) (λ − (T + S))(0),

then 0 = 6 x0 ∈ N (λ − (T + S)). Hence, λ − (T + S) is not injective. So, λ ∈ σ(T + S). (ii) =⇒ (iii) We derive a contradiction from the assumption that λ ∈ / σ(T ) 1 −1 and k(λ−T ) k ≤ ε . By using Lemma 2.5.7, we have λ ∈ ρ(T ) and γ(λ −T ) ≥ ε. So, from Remark 2.18.1 (i), λ − T is injective, open with dense range. Furthermore, S(0) ⊂ T (0) = (λ − T )(0), D(S) ⊃ D(T ) = D(λ − T ) and kSk < ε ≤ γ(λ − T ). Then, by Lemma 2.9.3, λ − T − S is injective, open with dense range. So, λ ∈ ρ(T + S). This is a contradiction. (iii) =⇒ (i) Trivial. Q.E.D. Remark 4.7.1 It follows, immediately, from Theorem 4.7.2, that for T ∈ CR(X) and ε > 0, [ σε (T ) = σ(T + S). ♦ kSk 0. Then, for any α, β ∈ 6 0, we have C with β = ♦

σε (α + βT ) = α + β σε/|β | (T ). Proof. For α, λ ∈ C and β ∈ C\{0}, we have   λ−α − Te , λ − α − βTe = β β

(4.59)

where Te is given in (2.4). Let λ ∈ σε (α + βT ). There are two possible cases: First case : If λ ∈ σ(αI + βT ), then by using (4.59), we infer that σ(Te) = σ(T ) ⊂ σε/|β | (T ). This leads to λ ∈ α + β σε/|β | (T ).

λ−α β

∈

Second case : If λ ∈ σε (αI + βT )\σ(αI + βT ), then by applying (4.59), we obtain



 −1

λ−α

−1

−1

e e

−T

= β (λ − α − βT )

β

 −1

λ − α − βTe

= |β |

>

|β | . ε

Hence, we deduce that λ−α β ∈ σε/|β | (T ). Thus, λ ∈ α + βσε/|β| (T ). As a result, σε (αI + βT ) ⊂ α + β σε/|β | (T ), as desired. Conversely, a same reasoning as before leads to the result. Q.E.D. Proposition 4.7.5 Let X be a Banach space, T ∈ CR(X) and ε > 0. Then, for δ > 0, we have σε (T ) ⊂ D(0, δ) + σε (T ) ⊂ σε+δ (T ).

♦

Proof. Let λ ∈ D(0, δ) + σε (T ). Then, there exists λ1 ∈ D(0, δ) and λ2 ∈ σε (T ) such that λ = λ1 + λ2 . Assume that λ ∈ / σε+δ (T ), then λ1 + λ2 ∈ ρ(T ) 1 and k(λ1 + λ2 − T )−1 k ≤ ε+δ . This implies that λ1 + λ2 − T is injective, open with dense range and γ(λ1 + λ2 − T ) ≥ ε + δ. Since λ1 ∈ D(0, δ), then |λ1 | ≤ δ < ε + δ ≤ γ(λ1 + λ2 − T ). By using Lemma 2.9.3, we infer that λ1 + λ2 − T − λ1 = λ2 − T is injective and open with dense range. Therefore,

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Spectral Theory of Multivalued Linear Operators

λ2 ∈ ρ(T ). Now, by using Lemma 2.9.2, we have γ(λ2 − T )

= γ(λ2 − λ1 + λ1 − T ) ≥ γ(λ1 + λ2 − T ) − |λ1 | ≥ ε+δ−δ ≥ ε.

Hence, k(λ2 − T )−1 k ≤ contradiction.

1 ε.

We conclude that λ2 ∈ / σε (T ). This is a Q.E.D.

Proposition 4.7.6 Let X be a Banach space and T ∈ CR(X), then for any ε > 0 and S ∈ LR(X) such that S(0) ⊂ T (0), D(S) ⊃ D(T ) and kSk < ε, we have σε−kS k (T ) ⊂ σε (T + S) ⊂ σε+kS k (T ).

♦

1 Proof. Let λ ∈ / σε+kS k (T ). Then, λ ∈ ρ(T ) and k(λ −T )−1 k ≤ ε+kSk . Hence, λ − T is injective, open with dense range and γ(λ − T ) ≥ ε + kSk. The fact that kSk < ε + kSk ≤ γ(λ − T ) implies from Lemma 2.9.3 that λ − T − S is injective and open with dense range. Then, λ ∈ ρ(T + S). On the other hand, by Lemma 2.9.2, we have

γ(λ − T − S) ≥ γ(λ − T ) − kSk ≥ ε + kSk − kSk ≥ ε. Then, k(λ−T −S)−1 k ≤ 1ε . Hence, λ ∈ / σε (T +S). So, σε (T +S) ⊂ σε+kS k (T ). Let λ ∈ / σε (T + S). Then, λ ∈ ρ(T + S) and k(λ − T − S)−1 k ≤ 1ε . Hence, λ − T − S is injective, open with dense range and γ(λ − T − S) ≥ ε. Using the fact kSk < γ(λ−T −S), we infer from Lemma 2.9.3 that λ−T −S +S = λ−T is injective, open with dense range. Then, λ ∈ ρ(T ). On the other hand, by Lemma 2.9.2, we have γ(λ − T )

= γ(λ − T − S + S) ≥ γ(λ − T − S) − kSk ≥ ε − kSk.

1 Then, k(λ−T )−1 k ≤ ε−kSk . Hence, λ ∈ / σε−kS k (T ). So, σε+kSk (T ) ⊂ σε (T +S). This completes the proof. Q.E.D.

Essential Spectra and Essential Pseudospectra of a Linear Relations

239

Theorem 4.7.3 Let X be a Banach space and T ∈ CR(X). Assume that V ∈ L(X) such that 0 ∈ ρ(V ). Let k = kV kkV −1 k and S = V T V −1 . Then, σ(S) = σ(T ) and, for ε > 0, we have σε/k (T ) ⊂ σε (S) ⊂ σkε (T ).

♦

Proof. We first show that S is closed. Since 0 ∈ ρ(V ) and V is closed, then V has a closed range (R(V ) = X) and, we have α(V ) = 0 < ∞ and γ(V ) > 0 (as V is injective and open). By using Proposition 2.9.4, we have V T is closed. Moreover, since V −1 is single valued and bounded, then V T V −1 is closed. Hence, S is closed. Using the fact V and V −1 are bounded single valued, then λ−S

= λ − V T V −1 = V (λ V −1 − T V −1 ) = V (λ − T )V −1 .

(4.60)

Now, let λ ∈ ρ(T ). Then, (λ − T )−1 is a bounded single valued. This implies that λ − T is a bounded below and surjective. Since S is closed, then by using Proposition 2.9.4 and (4.60), we have λ − S is also closed, bounded below, surjective. Hence, λ ∈ ρ(S). Conversely, if λ ∈ ρ(S), then (λ − S)−1 is a bounded single valued. This leads to λ − S is a bounded below and surjective. Since V and V −1 are bounded single valued, then by (4.60), we have λ − T = V −1 (λ − S)V.

(4.61)

The fact that S is closed implies from Propositions 2.7.2 and 2.9.4, and (4.61) that λ − T is also closed, bounded below, surjective. Hence, λ ∈ ρ(T ), which implies the first result. Now, it follows from (4.60) and (4.61) that (λ − T )−1 = V −1 (λ − S)−1 V and (λ − S)−1 = V (λ − T )−1 V −1 . Thus, k(λ − S)−1 k = ≤ ≤ ≤

kV (λ − T )−1 V −1 k kV (λ − T )−1 k kV −1 k kV k k(λ − T )−1 k kV −1 k k k(λ − T )−1 k.

In the same way, k(λ − T )−1 k ≤ kk(λ − S)−1 k.

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For λ ∈ σε/k (T ), we have λ ∈ σ(T ) or k(λ − T )−1 k >

k . ε

Then, λ ∈ σ(S) or k(λ − S)−1 k ≥

1 1 k(λ − T )−1 k > . k ε

Hence, λ ∈ σε (S). Therefore, σε/k (T ) ⊂ σε (S). On the other hand, for λ ∈ σε (S), we have λ ∈ σ(S) or k(λ − S)−1 k >

1 . ε

Hence, λ ∈ σ(T ) or k(λ − T )−1 k ≥

1 1 k(λ − S)−1 k > . k kε

So, λ ∈ σkε (T ). Therefore, σε (S) ⊂ σkε (T ).

4.8

Q.E.D.

Localization of Pseudospectra of Linear Relations

We give some further results on the location of the pseudospectra. We start with the following general result. Although the result is well known, we include the proof. Theorem 4.8.1 Let X be a Banach space, T ∈ CR(X), and ε > 0. Then, (i) If λ ∈ / σ(T ), then k(λ − T )−1 k ≥

1 . dist(λ, σ(T ))

(ii) If λ ∈ / σε (T ), then k(λ − T )−1 k ≥

1 . dist(λ, σε (T )) + ε

♦

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241

Proof. (i) Let λ ∈ ρ(T ). Since dist(λ, σ(T )) = inf{|z − λ| such that z ∈ σ(T )}, then for all η > 0, there exists zη ∈ σ(T ) such that |λ − zη | < dist(λ, σ(T )) + η. Suppose |λ − zη | < γ(λ − T ). Since λ − T is injective open with dense range, (zη −λ)I(0) = 0 ⊂ (λ−T )(0), D((zη −λ)I) = D(I) = X ⊃ D((λ−T )) = D(T ) and |λ − zη | < γ(λ − T ), then by using Lemma 2.9.3, we have λ − T + zη − λ = zη − T is injective open with dense range. Hence, zη ∈ ρ(T ). This is a contradiction. Therefore, |λ − zη | ≥ γ(λ − T ) for all η > 0. Thus, γ(λ − T ) ≤ |λ − zη | < dist(λ, σ(T )) + η

for all η > 0.

So, γ(λ − T ) ≤ dist(λ, σ(T )). Hence, k(λ − T )−1 k ≥

1 . dist(λ, σ(T ))

(ii) Let λ ∈ ρε (T ). Assume that k(λ − T )−1 k < dist(λ,σ1ε (T ))+ε , i.e., γ(λ − T ) > dist(λ, σε (T )) + ε. Since dist(λ, σε (T )) = inf{|z − λ| such that z ∈ σε (T )}, for η, 0 < η ≤ ε, there exists zη ∈ σε (T ) such that |λ − zη | < dist(λ, σε (T )) + η.

(4.62)

Therefore, |λ − zη | < dist(λ, σε (T )) + ε < γ(λ − T ). But (zη − λ)I(0) = 0 ⊂ (λ − T )(0), D((zη − λ)I) = D(I) = X ⊃ D((λ − T )) = D(T ) and λ − T is injective open with dense range (as λ ∈ ρ(T )), then by using Lemma 2.9.3, we have λ − T + zη − λ = zη − T is injective open with 1 dense range. Therefore, zη ∈ ρ(T ). But zη ∈ σε (T ), then k(zη − T )−1 k > , ε i.e., γ(zη − T ) < ε. On the other hand, by Lemma 2.9.2, γ(zη − T )

= γ(λ − T + zη − λ) ≥ γ(λ − T ) − |zη − λ|.

Therefore, γ(λ − T ) ≤ γ(zη − T ) + |zη − λ|. This implies from (4.62) that for all 0 < η ≤ ε γ(λ − T ) < ε + dist(λ, σε (T )) + η.

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Thus, γ(λ − T ) ≤ ε + dist(λ, σε (T )). This is a contradiction. Hence, k(λ − T )−1 k ≥

1 . dist(λ, σε (T )) + ε

Q.E.D.

Corollary 4.8.1 Let X be a Banach space, T ∈ CR(X), and ε > 0. Then, {λ ∈ C such that dist(λ, σ(T )) < ε} ⊂ σε (T ).

♦

Proof. Let λ ∈ / σε (T ). Then, by using Theorem 4.8.1, we have 1 1 ≤ k(λ − T )−1 k ≤ . dist(λ, σ(T )) ε Therefore, dist(λ, σ(T )) ≥ ε.

4.9

Q.E.D.

Characterization of ε-Pseudospectra of Linear Relations

In this section, we investigate the characterization of ε-pseudospectra and discuss the pseudospectra of a sequence of closed linear relations in a Banach space. Theorem 4.9.1 Let X be a Banach space, ε > 0 and let T ∈ CR(X). Then, [ σ(T + S) ⊂ Σε (T ). ♦ kSk≤ε S(0)⊂T (0) D(S)⊃D(T )

Proof. We derive a contradiction from the assumption that λ ∈ / Σε (T ). Then, 1 λ ∈ ρ(T ) and k(λ − T )−1 k < . ε This leads to λ − T is injective, open with dense range and γ(λ − T ) > ε.

(4.63)

Let S ∈ LR(X) satisfy kSk ≤ ε, S(0) ⊂ T (0) and D(S) ⊃ D(T ). Then, by using (4.63), we infer that kSk < γ(λ − T ). Thus, by applying Lemma 2.9.3, we deduce that λ − T − S is injective and open with dense range. This is equivalent to saying that λ ∈ ρ(T + S). Q.E.D.

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Remark 4.9.1 (i) We should notice that if X is a Banach space and T ∈ CR(X), then [ σ(T + S) ⊂ Σε (T ). kSk≤ε S(0)⊂T (0) D(S)⊃D(T )

In fact, it suffices to consider the following example: let X = lp (N) be the space of sequences x : N −→ C summable with a power p ∈ [1, ∞) with the standard norm and ε > 0. Consider linear relation T defined by G(T ) = {(x, y) ∈ X × X : x(n) = ε y(n − 1), n ≥ 2} . It is clear that T = εL−1 , where L is the left shift single valued operator defined by L(x(n)) = x(n + 1), n ≥ 1 and x ∈ X.
We know that L is bounded operator with kLk = 1. Therefore, D ε−1 L = X and

−1

ε L = ε−1 kLk = ε−1 < ∞. This implies that ε−1 L is bounded operator. Hence, ε−1 L is closed which yields T is closed. Moreover, kT −1 k = 1ε . Thus, we conclude that 0 ∈ Σε (T ). Now, we show that 0∈ /

[

σ(T + S).

(4.64)

kSk≤ε S(0)⊂T (0) D(S)⊃D(T )

Since N (T ) = (ε−1 L)(0) = ε−1 L(0) = {0}, γ(T ) = kT −1 k−1 = ε > 0 and R(T ) = X, then T is injective and open with dense range. Let S ∈ LR(X) satisfy that kSk ≤ ε, S(0) ⊂ T (0) and D(S) ⊃ D(T ). In view of Lemma 2.9.3 implies that T + S is injective and open with dense range. This is equivalent to say that 0 ∈ ρ(T + S). As a result, (4.64) holds, as desired. (ii) Note that σε (T ) ⊂ Σε (T ).

(4.65)

Indeed, we have σε (T ) ⊂ Σε (T ). Using the fact that Σε (T ) is closed, then we obtain σε (T ) ⊂ Σε (T ). ♦

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Now, the goal is achieved the equality in (4.65). It is well known that if the resolvent norm of the closed linear operator T acting in a Banach space cannot be constant on an open set, then σε (T ) = Σε (T ) (see [46]). Note that, even if X is Banach space, T ∈ CR(X) and λ ∈ ρ(T ), we have (λ − T )−1 is bounded operator in a Banach space X. Theorem 4.9.2 Let X be a Banach space, ε > 0, and T ∈ CR(X) such that the resolvent norm of T cannot be constant on an open set. Then, Σε (T ) ⊂ σε (T ).

♦

Proof. It is clear that   1 Σε (T )\ λ ∈ C : k(λ − T )−1 k = ⊂ σε (T ). ε Then, it is sufficient to show that   1 λ ∈ C : k(λ − T )−1 k = ⊂ σε (T ). ε Let us assume that k(λ − T )−1 k = 1ε . Then, there exists µ belongs to the neighborhood of λ and µ ∈ ρ(T ) such that k(µ − T )−1 k > k(λ − T )−1 k = 1ε . This leads to µ ∈ σε (T ). This enables us to conclude that λ ∈ σε (T ). Q.E.D. As an immediate consequence of Remark 4.9.1 and Theorem 4.9.2, we have: Corollary 4.9.1 Let X be a Banach space, ε > 0, and T ∈ CR(X) such that the resolvent norm of T cannot be constant on an open set. Then, Σε (T ) = σε (T ).

♦

The following theorem is devoted to give some equivalent definitions of εpseudospectra. Theorem 4.9.3 Let X be a Banach space, ε > 0, and T ∈ CR(X) such that the resolvent norm of T cannot be constant on an open set. Then, the following propositions are all equivalent: [ (i) λ ∈ σ(T + S). kSk≤ε S(0)⊂T (0) D(S)⊃D(T )

  1 −1 (ii) λ ∈ σ(T ) λ ∈ C : k(λ − T ) k ≥ . ε S (iii) λ ∈ σ(T ) {λ ∈ C : ∃ x ∈ D(T ), kxk = 1 and k(λ − T )xk < ε}. S

♦

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245

Proof. (i) =⇒ (ii) Since Σε (T ) is closed, then the use of Theorem 4.9.1 gives the wanted inclusion and achieves the proof of (i) =⇒ (ii).   S 1 −1 (ii) =⇒ (iii) Let us assume that λ ∈ σ(T ) λ ∈ C : k(λ − T ) k ≥ . ε Then, we will discuss the following two cases: First case : If λ ∈ σ(T ), then the result is trivial. Second case : If λ ∈ Σε (T )\σ(T ), then λ ∈ ρ(T ) and k(λ−T )−1 k ≥ 1ε . We infer that there exists µ belongs to the neighborhood of λ such that k(µ − T )−1 k > k(λ − T )−1 k which yields k(µ − T )−1 k >

1 . ε

This implies that there exists y ∈ X such that kyk = 1 and k(µ − T )−1 yk > Since µ ∈ ρ(T ), then (µ − T )−1 is an operator. Putting

1 . ε

x = k(µ − T )−1 yk−1 (µ − T )−1 y, then x ∈ D(T ), kxk = 1 and y ∈ k(µ − T )−1 yk (µ − T )x. We infer that kk(µ − T )−1 yk (µ − T )xk = dist (y, (µ − T )(0)) . Consequently, k(µ − T )xk = k(µ − T )−1 yk−1 dist (y, (µ − T )(0)) ≤

k(µ − T )−1 yk−1 kyk


0, and T ∈ CR(X) such that the resolvent norm of T cannot be constant on an open set. Then, [ σ(T + S) = Σε (T ). ♦ kS k≤ε S(0)⊂T (0) D(S)⊃D(T )

The following result shows the relation between the pseudospectra of a closed linear relation on a Banach space and the ε-pseudospectra of all perturbed linear relations. Proposition 4.9.1 Let X be a Banach space, T ∈ CR(X) and δ, ε > 0. For any continuous S ∈ LR(X) such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). Then, σε (T ) ⊂ Σε+kS k (T + S) ⊂ σε+δ+2kSk (T ).

♦

Proof. It follows from Proposition 2.7.2 and Remark 2.20.2 (i) that σε (T ) ⊂ Σε+kS k (T + S). Now, we have to prove Σε+kS k (T + S) ⊂ σε+δ+2kS k (T ). Let λ ∈ / σε+δ+2kS k (T ). 1 −1 Then, λ ∈ ρ(T ) and k(λ − T ) k ≤ ε+δ+2kSk , which implies that λ − T is open, injective with dense range and γ(λ − T ) ≥ ε + δ + 2kSk > kSk. The fact that S(0) ⊂ T (0) = T (0) and D(S) ⊃ D(T ) ⊃ D(T ) implies from (ii) and (iii) of Lemma 2.9.3 that λ − T − S is open, injective with dense range and γ(λ − T − S) ≥ γ(λ − T ) − kSk ≥ ε + δ + 2kSk − kSk >

ε + kSk.

Essential Spectra and Essential Pseudospectra of a Linear Relations This leads to λ ∈ ρ(S + T ) and k(λ − T − S)−1 k
0. For any continuous S ∈ LR(X) such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). Then, Σε (T ) ⊂ Σε+kS k (T + S) ⊂ Σε+2kS k (T ).

♦

Proof. Since S(0) ⊂ T (0) and D(S) ⊃ D(T ), then by Lemma 2.9.3, we get T + S is closed. Let us assume that λ ∈ / Σε+2kS k (T ). Then, λ ∈ ρ(T ) and 1 −1 k(λ − T ) k < ε+2kSk . This leads to λ − T is open, injective with dense range and γ(λ − T ) > ε + 2kSk. (4.66) The fact that kSk < γ(λ − T ) implies from (4.66) that λ − T − S is open, injective with dense range. Moreover, it follows from Lemma 2.9.3 (iii) that γ(λ − T − S) >

γ(λ − T ) − kSk

>

ε + 2kSk − kSk

>

ε + kSk.

1 Hence, λ ∈ ρ(T + S) and k(λ − T − S)−1 k < ε+kSk . This is equivalent to saying that λ∈ / Σε+kS k (T + S).

Now, we prove Σε (T ) ⊂ Σε+kS k (T +S). Let us assume that λ ∈ / Σε+kS k (T +S), 1 −1 then λ ∈ ρ(T + S) and k(λ − T − S) k < ε+kSk . This implies that λ − T − S is open, injective with dense range and γ(λ − T − S) >

ε + kSk

> kSk. Combining the fact that S(0) ⊂ T (0) = T (0), D(S) ⊃ D(T ) and (ii) of Lemma 2.9.3, we obtain λ − T − S + S is open and injective with dense range. Using (iv) of Lemma 2.9.3, we can write λ − T = λ − T − S + S. Finally, the use of Lemma 2.9.3 (iii) gives γ(λ − T ) ≥ γ(λ − T − S) − kSk >

ε.

Consequently, λ ∈ / Σε (T ). This shows that Σε (T ) ⊂ Σε+kSk (T + S). Q.E.D.

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The Pseudospectra and the ε-Pseudospectra of a Sequence of Closed Linear Relations

The purpose of this subsection is to establish the relation between the pseudospectra (respectively, ε-pseudospectra) of the sequence Tn and its limit. Theorem 4.9.4 Let X be a Banach space, (Tn )n be a sequence of closed linear relations from X into X, T ∈ CR(X) such that kTn − T k → 0 when n → ∞, D(Tn ) = D(T ) and there exists n1 ∈ N such that T (0) = Tn (0), for all n ≥ n1 . Then, (i) For every triplet (ε1 , ε2 , ε3 ) of real numbers with 0 < ε1 < ε2 < ε3 , there exists a positive integers n2 ≥ n1 such that σε1 (T ) ⊂ σε2 (Tn ) ⊂ σε3 (T ), for all n ≥ n2 . (ii) For every real number ε > 0, there exists a positive integers n2 ≥ n1 such that Σε (T ) = Σε (Tn ), for all n ≥ n2 .

♦

Proof. (i) First, we prove that σε1 (T ) ⊂ σε2 (Tn ). Let us assume that λ ∈ / 1 −1 σε2 (Tn ), for all n ≥ 0. Then, λ ∈ ρ(Tn ) and k(λ − Tn ) k ≤ ε2 . This leads to λ − Tn is open, injective with dense range and γ(λ − Tn ) ≥ ε2 . Using the fact that 0 < ε1 < ε2 , we obtain γ(λ − Tn ) > ε1 . Since kTn − T k → 0 when n → ∞. Let n0 ∈ N such that kTn − T k < γ(λ − Tn ) − ε1 < γ(λ − Tn ), for all n ≥ n0 . T Using the fact D(λ − Tn ) = D(Tn ) = D(Tn ) D(T ) = D(Tn − T ) and (T − Tn )(0) = Tn (0) = (λ − Tn )(0) = (λ − Tn )(0), for all n ≥ n1 , then by Lemma 2.9.3 (ii), we deduce that λ − Tn + Tn − T is open and injective with dense range. Now, using Proposition 2.3.4, we can write λ − T = λ − T − Tn + Tn for all n ≥ n1 . Now, we propose to show that γ(λ − T ) ≥ ε1 . Let n2 = max{n0 , n1 }. The use of Lemma 2.9.2 makes us to conclude, for all n ≥ n2 , that γ(λ − T ) ≥ γ(λ − Tn ) − kTn − T k ≥ γ(λ − Tn ) − γ(λ − Tn ) + ε1 ≥ ε1 .

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Thus, λ ∈ ρ(T ) and γ(λ − T ) ≥ ε1 which yields λ ∈ / σε1 (T ). This shows that σε1 (T ) ⊂ σε2 (Tn ), for all n ≥ n2 . A similar reasoning allows us to reach σε2 (Tn ) ⊂ σε3 (T ), for all n ≥ n2 . (ii) Let λ ∈ / Σε (Tn ), for all n ≥ 0. Then, λ ∈ ρ(Tn ) and k(λ − Tn )−1 k < 1ε . This implies that λ−Tn is open, injective with dense range and γ(λ−Tn ) > ε. Since kTn − T k → 0 as n → 0, then there is n0 ∈ N such that kTn − T k < γ(λ − Tn ) − ε < γ(λ − Tn ), for all n ≥ n0 . Since D(λ − Tn ) = D(Tn − T ) and (T − Tn )(0) = (λ − Tn )(0), for all n ≥ n1 , then by using Proposition 2.3.4, we infer that λ − T = λ − Tn + Tn − T is open and injective with dense range. Let n2 = max{n0 , n1 }. Then, using Lemma 2.9.2, for all n ≥ n2 , we conclude that γ(λ − T ) ≥ γ(λ − Tn ) − kTn − T k ≥ γ(λ − Tn ) − γ(λ − Tn ) + ε >

ε.

Hence, we obtain λ ∈ / Σε (T ). This shows that Σε (T ) ⊂ Σε (Tn ), for all n ≥ n2 . The opposite inclusion is analogous. Q.E.D. The following result can be directly derived from Proposition 4.7.6, Corollary 4.9.3 and Theorem 4.9.4. Corollary 4.9.4 Let X be a Banach space, ε, δ1 , δ2 > 0, (Tn )n be a sequence of closed linear relations from X into X and T ∈ CR(X) such that kTn −T k → 0 when n → ∞. We suppose that there exists n1 ∈ N such that D(Tn ) = D(T ) and T (0) = Tn (0), for all n ≥ n1 . Then, for any continuous S ∈ LR(X) such that S(0) ⊂ T (0) and D(S) ⊃ D(T ), there exists a positive integers n2 ≥ n1 such that σε (T ) ⊂ σε+δ1 +kS k (Tn + S) ⊂ σε+δ1 +δ2 +2kS k (T ), for all n ≥ n2 , and Σε (T ) ⊂ Σε+kS k (Tn + S) ⊂ Σε+2kS k (T ), for all n ≥ n2 .

♦

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4.10

Essential Pseudospectra of Linear Relations

In this section, we study some properties of essential pseudospectra of a linear relation and establish some results of perturbation on the context of linear relations. Throughout this section, X, Y, . . . will denote Banach spaces over the complex field C. Theorem 4.10.1 Let T ∈ CR(X) and ε > 0. Then, σei,ε (T ) is a closed set with i ∈ {1, 2, 3, 4}. ♦ Proof. The fact that T ∈ CR(X) and S ∈ UT (X) implies that λ − T − S ∈ CR(X) for all λ ∈ C. For i = 1: let λ 6∈ σe1,ε (T ), then for all S ∈ UT (X), we have λ−T −S ∈ Φ+ (X). Since λ − T − S is closed and R(λ − T − S) is closed, then applying Proposition 2.9.3 (i), we obtain γ(λ − T − S) > 0. Let r > 0 such that r < γ(λ − T − S), and let µ ∈ D(λ, r), then |λ − µ| < r < γ(λ − T − S). Hence, µ − T − S = λ − T − S − (λ − µ) ∈ Φ+ (X). Consequently, µ 6∈ σe1,ε (T ) which implies that µ ∈ C\σe1,ε (T ). This leads to D(λ, r) ⊂ C\σe1,ε (T ). Therefore, σe1,ε (T ) is a closed. By using the same reasoning as above, we get σei,ε (T ) is a closed set with i ∈ {2, 3, 4}. Q.E.D. Proposition 4.10.1 Let T ∈ CR(X). Then, (i) If 0 < ε1 < ε2 , then σei (T ) ⊂ σei,ε1 (T ) ⊂ σei,ε2 (T ), with i ∈ {1, 2, 3, 4}. (ii) L\ et ε > 0, then σei,ε (T ) ⊂ σε (T ), with i ∈ {1, 2, 3, 4}. (iii) σei,ε (T ) = σei (T ), with i ∈ {1, 2, 3, 4}. ♦ ε>0

Proof. The fact that T ∈ CR(X) and S ∈ UT (X) implies that λ − T − S ∈ CR(X) for all λ ∈ C. (i) For i = 1: let λ 6∈ σe1,ε2 (T ), then for all S ∈ UT (X), we have λ − T − S ∈ Φ+ (X). Since ε1 < ε2 , then for all continuous S ∈ LR(X) such that S(0) ⊂ T (0), D(S) ⊃ D(T ) and kSk < ε1 , we infer that λ − T − S ∈ Φ+ (X). Hence, we conclude that λ 6∈ σe1,ε1 (T ). Therefore, σe1,ε1 (T ) ⊂ σe1,ε2 (T ). Now, we claim that σe1 (T ) ⊂ σe1,ε1 (T ). Suppose that λ 6∈ σe1,ε2 (T ) which is equivalent to λ − T − S ∈ Φ+ (X) for all S ∈ UT (X). In particular, for S = 0, then λ − T − S ∈ Φ+ (X). Hence, we infer that λ 6∈ σe1 (T ).

Essential Spectra and Essential Pseudospectra of a Linear Relations

251

For i = 2: let λ 6∈ σe2,ε2 (T ), then for all S ∈ UT (X), we have λ − T − S ∈ Φ− (X). This implies from Theorem 2.14.4 (ii) that (λ − T −S)∗ ∈ Φ+ (X ∗ ). Since S is continuous, then by Proposition 2.8.1, we get (λ − T − S)∗ = λ − T ∗ − S ∗ . It follows from Proposition 2.8.1 that kSk = kS ∗ k. Consequently, λ − T ∗ − S ∗ ∈ Φ+ (X ∗ ). Since D(T ) ⊂ D(S) and S(0) ⊂ T (0), then S ∗ (0) = D(S)⊥ ⊂ D(T )⊥ = T ∗ (0), and D(T ∗ ) = T (0)⊥ ⊂ S(0)⊥ = D(S ∗ ). Hence, λ − T ∗ − S ∗ ∈ Φ+ (X ∗ ), for all continuous linear relation S ∗ such that S ∗ (0) ⊂ T ∗ (0), D(S ∗ ) ⊃ D(T ∗ ) and kS ∗ k ≤ ε2 . This implies that λ 6∈ σe1,ε2 (T ∗ ). By referring to the case that i = 1, we conclude that λ 6∈ σe1,ε1 (T ∗ ). This lead to λ − T ∗ − S ∗ ∈ Φ+ (X ∗ ), for all continuous linear relation S ∗ such that kS ∗ k ≤ ε1 . Therefore, λ − T − S ∈ Φ− (X), for all S ∈ UT (X). Thus, λ 6∈ σe2,ε1 (T ). Now, let us assume that λ 6∈ σe2,ε1 (T ), then λ−T −S ∈ Φ− (X), for all S ∈ UT (X). The fact that 0 ∈ UT (X) implies that λ 6∈ σe2 (T ). T T For i = 3: since σe3 (T ) = σe1 (T ) σe2 (T ), σe3,ε1 (T ) = σe1,ε1 (T ) σe2,ε1 (T ) T and σe3,ε2 (T ) = σe1,ε2 (T ) σe2,ε2 (T ), then σe3 (T ) ⊂ σe3,ε1 (T ) ⊂ σe3,ε2 (T ). For i = 4: this assertion is immediately deduced from (i) and (ii). (ii) Let λ 6∈ σε (T ). Then, λ ∈ ρ(T ) and k(λ − T )−1 k ≤ 1ε . This implies that λ − T is injective, open with dense range and γ(λ − T ) ≥ ε. Now, if we take S ∈ UT (X), then kSk ≤ ε ≤ γ(λ − T ). Thus implies that λ − T − S is injective, open with dense range. This implies that α(λ − T − S) = 0, β(λ − T − S) = 0 and R(λ − T − S) is closed. Hence, λ − T − S ∈ Φ(X). Therefore, λ ∈ σei,ε (T ), for all i ∈ {1, 2, 3, 4}. (iii) For i = 1: \ from (i) and (ii), we have σe1 (T ) ⊂ σε (T ) for all ε > 0, then σe1 (T ) ⊂ σε (T ). Conversely, let λ 6∈ σe1 (T ), then λ − T ∈ Φ+ (X). ε>0

Therefore, R(λ − T ) is closed. Hence, γ(λ − T ) > 0. Let ε such that 0 < ε ≤ γ(λ − T ) and let S ∈ UT (X). Since γ(λ − T ) ≥ ε > kSk, then α(λ − T − S) ≤ α(λ − T ) < ∞. Hence, λ − T − S is op\ en, and R(λ − T − S) is closed. So, λ 6∈ σe1,ε (T ). Thus, we deduce that λ 6∈ σe1,ε (T ). ε>0

By using the same reasoning as above, we get the next inclusion.

Q.E.D.

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Spectral Theory of Multivalued Linear Operators

Theorem 4.10.2 Let T ∈ CR(X) and ε > 0. Then, the following properties are equivalent: (i) λ ∈ / σe5,ε (T ). (ii) For all continuous linear relations S ∈ LR(X) such that D(S) ⊃ D(T ), S(0) ⊂ T (0) and kSk < ε, we have T + S − λ ∈ Φ(X) and i(T + S − λ) = 0. (iii) For all D ∈ L(X) such that kDk < ε, we have T + D − λ ∈ Φ(X) and i(T + D − λ) = 0. ♦ Proof. (i) =⇒ (ii) Let λ ∈ / σe5,ε (T ). Then, there exists K ∈ KT (X) such that λ ∈ / σε (T + K). Using Theorem 4.7.2, for all continuous linear relations T S ∈ LR(X) such that D(S) ⊃ D(T + K) = D(T ) D(K) = D(T ) as D(K) ⊃ D(T ), S(0) ⊂ (T + K)(0) = T (0) as K(0) ⊂ T (0) and kSk < ε, we have λ ∈ ρ(T +S +K). Then, T +S +K −λ is open, injective with dense range. Since K is compact, then K is continuous. This implies that S +K −λ is continuous. T Furthermore, (S + K − λ)(0) ⊂ T (0) and D(S + K − λ) = D(S) D(K). Since T is closed, then by using Proposition 2.7.2, T + S + K − λ is closed. Hence, R(T + S + K − λ) is closed. So, R(T + S + K − λ) = X. Therefore, T + S + K − λ ∈ Φ(X) and i(T + S + K − λ) = 0, for all continuous linear relations S ∈ LR(X) such that D(S) ⊃ D(T ), S(0) ⊂ T (0) and kSk < ε. It comes from Proposition 2.3.4 for all continuous linear relations S ∈ LR(X) such that D(S) ⊃ D(T ), S(0) ⊂ T (0) and kSk < ε, we have T + S − λ ∈ Φ(X) and i(T + S − λ) = 0. (ii) =⇒ (iii) is trivial. (iii) =⇒ (i) We suppose that for all D ∈ L(X) such that kDk < ε, we have T +D−λ ∈ Φ(X) and i(T +D−λ) = 0. By Proposition 2.8.1, N ((T +D−λ)∗ ) = R(T + D − λ)⊥ . Let n = α(T + D − λ) = β(T + D − λ), {x1 , . . . , xn } be basis for N (T + D − λ) and {y 0 1 , . . . , y 0 n } be basis for the N ((T + D − λ)∗ ). Then, there are functionals x0 1 , . . . , x0 n ∈ X ∗ (the adjoint space of X) and elements y1 , . . . , yn such that x0j (xk ) = δjk

and

yj0 (yk ) = δjk ,

1 ≤ j, k ≤ n,

where δjk = 0 if j = 6 k and δjk = 1 if j = k. The single valued operator K is defined by n X Kx = x0 k (x)yk , x ∈ X. k=1

K is bounded, since D(K) = X and kKxk ≤ kxk

n X k=1

! kx0k kkyk k

.

Essential Spectra and Essential Pseudospectra of a Linear Relations

253

Moreover, the range of K is contained in a finite subspace of X. Then, K is a finite rank relation in X. Hence, K is compact. We prove that \ \ N (T + D − λ) N (K) = {0} and R(T + D − λ) R(K) = {0}, (4.67) for all D ∈ L(X) such that kDk < ε. Let x ∈ N (T + D − λ), then x=

n X

αk xk .

k=1

Therefore, x0j (x) = αj , 1 ≤ j ≤ n. On the other hand, if x ∈ N (K), then x0j (x) = 0, 1 ≤ j ≤ n. This proves the first relation in (4.67). The second inclusion is similar. In fact, if y ∈ R(K), then y=

n X

α k yk .

k=1

Hence, yj0 (y) = αj , 1 ≤ j ≤ n. But, if y ∈ R(T + D − λ), then yj0 (y) = 0, 1 ≤ j ≤ n. This gives the second relation in (4.67). On the other hand K ∈ KT (X) = KT +D−λ (X) since K(0) ⊂ T (0) = (T + D − λ)(0) and D(K) ⊃ D(T ) = D(T + D − λ). We deduce from Corollary 2.16.1 (vii) that T + D + K − λ ∈ Φ(X) and i(T + D + K − λ) = 0. If x ∈ N (T + D + K − λ), then T 0 ∈ T x+Dx+Kx−λx. Hence, −Kx ∈ (T +D −λ)x. So, Kx ∈ R(K) R(T + D − λ) = {0}. Therefore, Kx = 0 and 0 ∈ (T + D − λ)x. This implies that T x ∈ N (T + D − λ) N (K). Hence, x = 0. Thus, α(T + D + K − λ) = 0. In the some way, one proves that R(T + D + K − λ) = X. Since T + D + K − λ is closed and from Theorem 2.5.4 (ii), T + D + K − λ is open. Furthermore, it is injective with dense range. Hence, λ ∈ ρ(T + D + K) for all continuous single valued operators D ∈ LR(X) such that D(D) ⊃ D(T ) and kDk < ε. This implies that λ ∈ ρ(T + D + K) for all D ∈ L(X) such that kDk < ε. This is equivalent to saying that [ λ∈ / σε (T + K + D). kD k 0, we have [ [ σe5,ε (T ) = σe5 (T + D) = σe5 (T + S). ♦ kD k 0 and J (X) be a subset of LR(X). If KT (X) ⊂ J (X) ⊂ PR(Φ(X)), then \ σe5,ε (T ) = σε (T + J). ♦ J∈J (X)

Proof. It is clear to see that \ σe5,ε (T ) = σε (T + P ) ⊂ P ∈PR(Φ(X))

This completes the proof.

\ J∈J (X)

σε (T + J) ⊂

\

σε (T + K).

K∈KT (X)

Q.E.D.

Essential Spectra and Essential Pseudospectra of a Linear Relations

255

Remark 4.10.2 It follows from the definition of essential pseudospectra, and both Theorem 4.10.3 and Corollary 4.10.1 that for T ∈ CR(X) and ε > 0, we have (i) σe5,ε (T + K) = σe5,ε (T ) for all K ∈ KT (X). (ii) σe5,ε (T + P ) = σe5,ε (T ) for all P ∈ PR(Φ(X)). (iii) If KT (X) ⊂ J (X) ⊂ PR(Φ(X)) and if for all J1 , J2 ∈ J (X), J1 ± J2 ∈ J (X), then σe5,ε (T + J) = σe5,ε (T ) , for all J ∈ J (X). ♦ Proposition 4.10.2 Let T ∈ CR(X). Then, (i) If 0 < ε1 < ε2 , then σe5 (T ) ⊂ σe5,ε1 (T ) ⊂ σe5,ε2 (T ). (ii) F\ or ε > 0, σe5,ε (T ) ⊂ σε (T ). σe5,ε(T ) = σe5 (T ). (iii)

♦

ε>0

Proof. (i) If λ ∈ / σe5,ε2 (T ), then by using Theorem 4.10.2 and for all continuous linear relations S ∈ LR(X) such that D(S) ⊃ D(T ), S(0) ⊂ T (0) and kSk < ε2 , we have T + S − λ ∈ Φ(X) and i(T + S − λ) = 0. Hence, for all continuous linear relations S ∈ LR(X) such that D(S) ⊃ D(T ), S(0) ⊂ T (0) and kSk < ε1 , we have T + S − λ ∈ Φ(X) and i(T + S − λ) = 0. Then, / σe5,ε1 (T ), then for all continuous λ ∈ / σe5,ε1 (T ). On the other hand, if λ ∈ linear relations S ∈ LR(X) such that D(S) ⊃ D(T ), S(0) ⊂ T (0) and kSk < ε1 , we have T + S − λ ∈ Φ(X) and i(T + S − λ) = 0. In particular for S = 0, we have λ ∈ / σe5 (T ). T (ii) σe5,ε (T ) = K ∈KT (X) σε (T + K) ⊂ σε (T + K) for all K ∈ KT (X). In particular, K = 0. T (iii) From (i), σe5 (T ) ⊂ σe5,ε (T ) for all ε > 0. Then, σe5 (T ) ⊂ ε>0 σe5,ε (T ). Conversely, If λ ∈ / σe5 (T ), then T − λ ∈ Φ(X) and i(T − λ) = 0. Therefore, R(T − λ) is closed and by using Theorem 2.5.4 (ii), we have T − λ is open. So, γ(T − λ) > 0. Let ε such that 0 < ε ≤ γ(T − λ) and let D be a single valued linear relation in LR(X) such that D(D) ⊃ D(T ) and kDk < ε ≤ γ(T − λ). By using Proposition 2.9.3 (iii), γ(T − λ) = γ((T − λ)∗ ), then kDk < ε ≤ γ((T − λ)∗ ). So, from Corollary 2.16.1 (iii), T + D − λ ∈ Φ− (X). Now, by Theorem 2.9.3, α(T + D − λ) ≤ α(T − λ) < ∞. Hence, T + D − λ ∈ Φ+ (X). So, T + D − λ ∈ Φ(X). Furthermore, i(T + D − λ) = i(T − λ) = 0. Hence, T by using Theorem 4.10.2, λ ∈ / σe5,ε (T ). So, λ ∈ / ε>0 σe5,ε (T ). It follows that T Q.E.D. ε>0 σe5,ε (T ) ⊂ σe5 (T ). This completes the proof.

256

Spectral Theory of Multivalued Linear Operators

Theorem 4.10.4 Let T ∈ CR(X) and let ε > 0. Then, σe5,ε (T ) is a closed set. ♦ Proof. Let λ ∈ ρe5,ε (T ), then λ ∈ / σe5,ε (T ). Let D be a single valued continuous linear relation such that D(D) ⊃ D(T ) and kDk < ε. Hence, by using Theorem 4.10.2, T + D − λ ∈ Φ(X) and i(T + D − λ) = 0. So, R(T +D−λ) is closed. Hence, T +D−λ is open. Thus, γ(T +D−λ) > 0. Then, by using Proposition 2.8.1, γ(T + D − λ) = γ((T + D − λ)∗ ). Let r > 0 such that r < γ(T + D − λ), and let µ ∈ D(λ, r). Then, |λ − µ| 0 such that D(λ, r) ⊂ ρe5,ε (T ), Hence, ρe5,ε (T ) is an open set. Q.E.D. Theorem 4.10.5 Let T ∈ CR(X) and ε > 0. Then, σe5,ε (T ) = σe5,ε (T ∗ ). ♦ Proof. Let K ∈ KT (X), then K is compact. Therefore, it is continuous by Lemma 2.12.1 and, by using Proposition 2.11.2, we have K ∗ is compact. Furthermore, D(K) ⊃ D(T ). Hence, by using Proposition 2.8.1, K ∗ (0) = D(K)⊥ ⊂ T ∗ (0) = D(T )⊥ , and K(0) ⊂ T (0) since K is continuous. So, D(K ∗ ) = K(0)⊥ ⊃ T (0)⊥ = D(T ∗ ) ⊃ D(T ∗ ). Hence, KT (X) ⊂ {K ∈ LR(X) : K is continuous and K ∗ ∈ KT ∗ (X ∗ )}. Now, let K ∈ {K ∈ LR(X) such that K is continuous and K ∗ ∈ KT ∗ (X ∗ )}, then by using Proposition 2.11.2, we have K is compact. Since, K ∗ (0) = D(K)⊥ ⊂ T ∗ (0) = D(T )⊥ , then D(K) ⊃ D(T ), also K(0) ⊂ K(0) = D(K ∗ )> ⊂ D(T ∗ )> = T (0) as T is closed. Then, K ∈ KT (X). Hence, KT (X) = {K ∈ LR(X) : K is continuous and K ∗ ∈ KT ∗ (X ∗ )}.

(4.68)

On the other hand for K ∈ KT (X), T + K is closed. Using Theorem 4.7.1, we deduce that σε (T + K) = σε ((T + K)∗ ). But D(K) ⊃ D(T ) and K is continuous, then by using Proposition 2.8.1, we have (T + K)∗ = T ∗ + K ∗ . Hence, σε (T + K) = σε (T ∗ + K ∗ ) for all K ∈ KT (X). Therefore, using Eq (4.68), we have

Essential Spectra and Essential Pseudospectra of a Linear Relations

σe5,ε (T )

\

=

257

σε (T + K)

K∈KT (X)

\

=

K ∗ ∈K

T

σε (T ∗ + K ∗ )

∗ (X ∗ )

K∈LR(X) K continuous

\

⊃

=

σε (T ∗ + K ∗ )

K ∗ ∈KT ∗ (X ∗ ) K∈LR(X) σe5,ε (T ∗ ).

Let O :=

\

σε (T ∗ + K ∗ ),

K ∗ ∈KT ∗ (X ∗ ) K∈LR(X) K continuous

then O = σe5,ε (T ) ⊃ σe5,ε (T ∗ ).

(4.69)

On the other hand, let λ ∈ O, then for all K ∈ LR(X), K is continuous such that K ∗ ∈ KT ∗ (X ∗ ), we have λ ∈ σε (T ∗ + K ∗ ). Let K ∈ LR(X) such that e is compact, where K e is given in (2.4). Since G(K) e K ∗ ∈ KT ∗ (X), then K e e is the completion of G(K), then G(K ) = G(K) = G(K). Hence, K = K ∗ −1 is compact. Hence, K is continuous. Furthermore, G(K ) = G(−K )⊥ = ⊥ ∗ G(−K −1 ) = G(−K −1 )⊥ = G(K ∗ ). Hence, K = K ∗ . Thus, K is in LR(X), ∗ K is continuous and K = K ∗ ∈ KT ∗ (X ∗ ), since λ ∈ O, then λ ∈ σε (T ∗ + ∗ K ) = σε (T ∗ + K ∗ ). We conclude that if λ ∈ O, for all K ∈ LR(X) such that K ∗ ∈ KT ∗ (X ∗ ) and λ ∈ σε (T ∗ + K ∗ ), then \ λ∈ σε (T ∗ + K ∗ ) = σe5,ε (T ∗ ). K ∗ ∈KT ∗ (X ∗ ) K∈LR(X)

Hence, O ⊂ σe5,ε (T ∗ ). Using (4.69), we have O = σe5,ε (T ) = σe5,ε (T ∗ ). This completes the proof. Q.E.D.

4.10.1

Stability of Essential Pseudospectra of Linear Relations

Proposition 4.10.3 Let T ∈ LR(X) and ε > 0, then for any α, β ∈ C with β= 6 0, we have

258

Spectral Theory of Multivalued Linear Operators ♦

σe5,ε (α + βT ) = α + σw,ε/|β | (T )β. Proof. Let α, β ∈ C with β = 6 0. Then, \ σe5,ε (α + βT ) = σε (α + βT + K) K∈KT (X)

\

=

σε (β(β −1 α + T + β −1 K)).

K∈KT (X)

It is simply to verify that {β −1 K : K ∈ KT (X)} = KT (X). In fact, if K ∈ KT (X), then (β −1 K)(0) = K(0) ⊂ T (0) and D(β −1 K) = D(K) ⊃ D(T ). Moreover, since K is compact, then Qβ −1 K β −1 KBX = β −1 QK KBX = β −1 QK KBX is compact. Hence, β −1 K is compact. So, β −1 K ∈ KT (X). Therefore, {β −1 K : K ∈ KT (X)} ⊂ KT (X). Conversely, by the same way, if K ∈ KT (X), βK ∈ KT (X), then KT (X) ⊂ {β −1 K : K ∈ KT (X)}. Therefore, \ σe5,ε (αI + βT ) = σε (β (β −1 αI + T + K)) K∈KT (X)

\

=

σε (αI + β(T + K)).

K∈KT (X)

Using Proposition 4.7.4, we have σe5,ε (αI + βT )

=

\

(α + σε/|β | (T + K)β)

K∈K T (X)

= α+

 \

σε/|β | (T + K) β.

K∈KT (X) ε (T )β. = α + σw, |β|

This completes the proof.

Q.E.D.

Proposition 4.10.4 Let X be a Banach space, T ∈ CR(X), and ε > 0. Then, for δ > 0, we have σe5,ε (T ) ⊂ D(0, δ) + σe5,ε (T ) ⊂ σe5,ε+δ (T ).

♦

Proof. Let K ∈ KT (X), then K is compact. Hence, it is continuous (by Lemma 2.12.1). Using Proposition 2.7.2, T +K is closed. By using Proposition 4.7.5, D(0, δ) + σε (T + K) ⊂ σε+δ (T + K). Then, D(0, δ) +

\ K∈KT (X)

σε (T + K) ⊂

\ K∈KT (X)

σε+δ (T + K).

Essential Spectra and Essential Pseudospectra of a Linear Relations

259

Hence, D(0, δ) + σe5,ε (T ) ⊂ σe5,ε+δ (T ).

Q.E.D.

Proposition 4.10.5 Let X be a Banach space and T ∈ CR(X), then for any ε > 0 and S ∈ LR(X) such that S(0) ⊂ T (0), D(S) ⊃ D(T ) and kSk < ε, we have σe5,ε−kS k (T ) ⊂ σe5,ε (T + S) ⊂ σe5,ε+kS k (T ).

♦

Proof. Let K ∈ KT (X), then K is compact. Hence, it is continuous (by Lemma 2.12.1). Using Proposition 2.7.2, T + K is closed. Moreover, let S ∈ LR(X) such that S(0) ⊂ T (0) = (T + K)(0), D(S) ⊃ D(T ) = D(T + K) and kSk < ε. Then, from Proposition 4.7.6, σε−kS k (T + K) ⊂ σε (T + S + K) ⊂ σε+kS k (T + K). Hence, \

σε−kS k (T + K) ⊂

K∈KT (X)

\

\

σε (T + S + K ) ⊂

K∈KT (X)

σε+kS k (T + K).

K∈KT (X)

But, since T (0) = (T + S)(0) and D(T ) = D(T + S), then KT (X) = K(T +S) (X). Thus, \ \ \ σε−kS k (T +K) ⊂ σε (T +S+K) ⊂ σε+kSk (T +K). K∈KT (X)

K∈KT +S (X)

K∈KT (X)

Therefore, σe5,ε−kS k (T ) ⊂ σe5,ε (T + S) ⊂ σe5,ε+kS k (T ).

Q.E.D.

Theorem 4.10.6 Let X be a Banach space, ε > 0, T ∈ CR(X) and V ∈ L(X) such that 0 ∈ ρ(V ). Let k = kV kkV −1 k and S = V T V −1 . Then, σe5,ε/k (T ) ⊂ σe5,ε (S) ⊂ σe5,kε (T ). In particular σe5 (S) = σe5 (T ).

♦

Proof. Let K ∈ KT (X), then K is compact. Hence, it is continuous (by Lemma 2.12.1). Using Proposition 2.7.2, T + K is closed. Using the fact V and V −1 are bounded single valued, together with Proposition 2.3.6, we obtain S + V KV −1

= V T V −1 + V KV −1 = V (T V −1 + KV −1 ) = V (T + K)V −1 .

260

Spectral Theory of Multivalued Linear Operators

From Theorem 4.7.3, we have for any ε > 0, σε/k (T + K) ⊂ σε (S + V KV −1 ) ⊂ σkε (T + K). Hence, \

\

σε/k (T + K) ⊂

K∈KT (X)

\

σε (S + V KV −1 ) ⊂

K∈KT (X)

σkε (T + K).

K∈KT (X)

But {V KV −1 : K ∈ KT (X)} = KS (X). In fact, if K ∈ KT (X), then K(0) ⊂ T (0). Hence, V KV −1 (0) = V K(0) ⊂ V T (0) = V T V −1 (0) = S(0). Moreover, T since V is bounded, D(V K) = {x ∈ X : Kx D(V ) 6= ∅} = D(K) and T D(V T ) = {x ∈ X : T x D(V ) = 6 ∅} = D(T ), then D(V KV −1 ) = {x ∈ X : T T V −1 x D(V K) 6= ∅} ⊃ {x ∈ X : V −1 x D(V T ) 6= ∅} = D(V T V −1 ) = D(S). On the other hand, K is compact, V is continuous, and V (0) = {0} ⊂ D(K), then V K is precompact. Hence, Γ0 (V K) = 0. Furthermore, since V −1 is single valued (0 ∈ ρ(V )), then Γ0 (V KV −1 ) ≤ Γ0 (V K)Γ0 (V −1 ) = 0. Hence, V KV −1 is precompact and, we have X is complete then V KV −1 is compact. Therefore,  V KV −1 : K ∈ KT (X) ⊂ KS (X). In the similar way, if K ∈ KS (X), then K(0) ⊂ S(0). Hence, V −1 KV (0) = V −1 K(0) ⊂ V −1 S(0) = V −1 SV (0) = T (0). Moreover, since V −1 is bounded T as 0 ∈ ρ(V ), D(V −1 K) = {x ∈ X : Kx D(V −1 ) = 6 ∅} = D(K) and T D(V S) = {x ∈ X : Sx D(V −1 ) = 6 ∅} = D(S), then D(V −1 KV ) = {x ∈ T T X : V x D(V −1 K) 6= ∅} ⊃ {x ∈ X : V x D(V −1 S) 6= ∅} = D(V −1 SV ) = D(T ). Moreover, K is compact, V −1 is continuous and V −1 (0) = {0} ⊂ D(K) (as 0 ∈ ρ(V )), then by using Theorem 2.12.2, we obtain V −1 K is precompact. So, Γ0 (V −1 K) = 0. Furthermore, since V is single valued, then Γ0 (V −1 KV ) ≤ Γ0 (V −1 K)Γ0 (V ) = 0. Hence, V −1 KV is precompact and, we have X is complete. Thus, V −1 KV is compact. Therefore, V −1 KV ∈ KT (X). Hence, KS (X) ⊂ {V KV −1 : K ∈ KT (X)}. So, \

σε/k (T + K) ⊂

K∈KT (X)

\ K∈KS (X)

σε (S + K) ⊂

\ K∈KT (X)

Thus, for any ε > 0 σe5,ε/k (T ) ⊂ σe5,ε (S) ⊂ σe5,kε (T ).

σkε (T + K).

Essential Spectra and Essential Pseudospectra of a Linear Relations

261

Moreover, \

σe5,ε/k (T ) ⊂

ε>0

\

σe5,ε (S) ⊂

ε>0

\

σe5,kε (T ).

ε>0

From the proof of Theorem 4.7.3, S is closed. Then, σe5 (T ) ⊂ σe5 (S) ⊂ σw (T ). This completes the proof.

Q.E.D.

Theorem 4.10.7 Let X be a Banach space, T ∈ CR(X), and ε > 0. Then, (i) If λ ∈ / σe5 (T ), then k(λ − T )−1 k ≥

1 . dist(λ, σe5 (T ))

(ii) If λ ∈ / σe5,ε (T ), then k(λ − T )−1 k >

1 . dist(λ, σe5,ε (T )) + ε

♦

Proof. Let λ ∈ ρe5 (T ). Then, dist(λ, σe5 (T )) = inf{|z − λ| such that z ∈ σe5 (T )}. Hence, for all η > 0, there exists zη ∈ σe5 (T ) such that |λ − zη | < dist(λ, σe5 (T )) + η. Suppose |λ − zη | < γ(λ − T ). Since λ − T ∈ Φ(X) and i(λ − T ) = 0, (zη − λ)I(0) = 0 ⊂ (λ − T )(0), D((zη − λ)I) = D(I) = X ⊃ D((λ − T )) = D(T ) and |(zη − λ)I| = |λ − zη | < γ(λ − T ), then λ − T + zη − λ = zη − T ∈ Φ(X) and i(zη − T ) = i(λ − T + zη − λ) = i(λ − T ) = 0. Hence, zη ∈ ρe5 (T ). This is a contradiction. Therefore, |λ − zη | ≥ γ(λ − T ) for all η > 0. Thus, γ(λ − T ) ≤ |λ − zη | < dist(λ, σe5 (T )) + η

for all η > 0.

So, γ(λ − T ) ≤ dist(λ, σe5 (T )). Hence, k(λ − T )−1 k ≥

1 . dist(λ, σe5,ε (T ))

1 (ii) Let λ ∈ / σe5,ε (T ). Assume that k(λ − T )−1 k ≤ dist(λ,σe5,ε (T ))+ε , i.e., γ(λ − T ) ≥ dist(λ, σe5,ε (T )) + ε. Since dist(λ, σe5,ε (T )) = inf{|z − λ| such that z ∈

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σe5,ε (T )}, then for η, 0 < η ≤ ε, there exists zη ∈ σe5,ε (T ) such that |λ − zη | < dist(λ, σe5,ε (T )) + η. Therefore, |λ − zη | < dist(λ, σe5,ε (T )) + ε < γ(λ − T ). Let S be a linear relation such that S(0) ⊂ T (0), D(S) ⊃ D(T ) and kSk < ε. If λ ∈ / σe5,ε (T ), then by using Theorem 4.10.2, we have λ − T − S ∈ Φ(X) and i(λ − T − S) = 0. Moreover, T is closed, (λ − S)(0) = S(0) ⊂ T (0), D((λ−S) = D(S) ⊃ D(T ), and λ−S is continuous, then by using Proposition 2.7.2, we have λ − T − S is closed. Now, suppose |λ − zη | < γ(λ − T − S). Since (zη − λ)I(0) = 0 ⊂ (λ − T − S)(0), D((zη − λ)I) = D(I) = X ⊃ D((λ − T − S)) = D(T ) and |(zη − λ)I| = |λ − zη | < γ(λ − T − S), then λ−T −S+zη −λ = zη −T −S ∈ Φ(X) and i(zη −T −S) = i(λ−T −S+zη −λ) = i(λ − T − S) = 0. Hence, zη ∈ ρe5,ε (T ). This is a contradiction. Therefore, γ(λ − T − S) ≤ |λ − zη | for all linear relations S such that S(0) ⊂ T (0), D(S) ⊃ D(T ) and kSk < ε. In particular, we put S = 0. Then, γ(λ − T ) ≤ |λ − zη | < dist(λ, σe5,ε (T )) + η. Since η ≤ ε, we have γ(λ − T ) < dist(λ, σe5,ε (T )) + ε. This is a contradiction. Hence, k(λ − T )−1 k >

4.11

1 dist(λ, σe5,ε (T )) + ε

.

Q.E.D.

The Essential ε-Pseudospectra of Linear Relations

The aim of this section is to introduce and study the essential ε-pseudospectra of linear relations in a Banach space. Let T ∈ CR(X) and ε > 0. Let us consider JT,ε (X) the set JT,ε (X) := {S ∈ LR(X) : S(0) ⊂ T (0), D(S) ⊃ D(T ) and kSk ≤ ε}. Proposition 4.11.1 Let X be a Banach space, T ∈ CR(X) and ε > 0. Then,

Essential Spectra and Essential Pseudospectra of a Linear Relations

263

(i) If λ ∈ / Σe1,ε (T ), then λ − T − S ∈ Φ+ (X), for all S ∈ JT,ε (X), (ii) If λ ∈ / Σe2,ε (T ), then λ − T − S ∈ Φ− (X), for all S ∈ JT,ε (X), (iii) If λ ∈ / Σe3,ε (T ), then λ − T − S ∈ Φ± (X), for all S ∈ JT,ε (X), (iv) If λ ∈ / Σe4,ε (T ), then λ − T − S ∈ Φ(X), for all S ∈ JT,ε (X), (v) If λ ∈ / Σe5,ε (T ), then λ − T − S ∈ Φ(X) and i(λ − T − S) = 0, for all S ∈ JT,ε (X). ♦ Proof. The fact that T ∈ CR(X) implies from Proposition 2.7.2 that for all S ∈ JT,ε (X) and λ ∈ C, we have λ − T − S ∈ CR(X). (i) Let λ ∈ / Σe1,ε (T ). Then, α(λ − T ) < ∞ and k(λ − T )−1 k < 1ε . By virtue of Lemma 2.9.1, we have γ(λ − T ) > ε which implies that λ − T is open. Combining the fact that T ∈ CR(X) and Proposition 2.7.2 (i), we infer that λ − T is closed. Hence, using Theorem 2.7.2 (ii), we get R(λ − T ) is closed. Consequently, λ − T ∈ Φ+ (X). It is clear for any S ∈ JT,ε (X) that kSk < γ(λ − T ). Finally, the use of Corollary 2.16.1 (iv) allows us to conclude that λ − T − S ∈ Φ+ (X). (ii) Let λ ∈ / Σe2,ε (T ). Then, β(λ − T ) < ∞ and γ(λ − T ) > ε. In view of Proposition 2.9.3 (i) implies that λ − T is closed. Now, using the fact λ − T is open and Theorem 2.7.2 (ii), we infer that R(λ − T ) is closed. Hence, λ − T ∈ Φ− (X). Since kSk < γ(λ − T ) for any S ∈ JT,ε (X), then the use of Corollary 2.16.1 (ii) allows us to conclude that λ − T − S ∈ Φ− (X). (iii) Let us assume that λ ∈ / Σe3,ε (T ). Then, using (iii) of Remark 2.20.4, we T obtain λ ∈ / Σe2,ε (T ) Σe1,ε (T ). In view of (i) and (ii) implies that [ λ − T − S ∈ Φ− (X) Φ+ (X) = Φ± (X), for all S ∈ JT,ε (X). (iv) Let us assume that λ ∈ / Σe4,ε (T ). Then, using (iv) of Remark 2.20.4, we S infer that λ ∈ / Σe2,ε (T ) Σe1,ε (T ). The use of (i) and (ii) makes us conclude that \ λ − T − S ∈ Φ− (X) Φ+ (X) = Φ(X), for all S ∈ JT,ε (X). (v) Let us assume that λ ∈ / Σe5,ε (T ). Then, using (v) of Remark 2.20.4, we deduce that λ ∈ / Σe4,ε (T ) and i(λ − T ) = 0. Let S ∈ JT,ε (X). It follows from (iv) that λ − T − S ∈ Φ(X). The fact that kSk ≤ ε, then i(λ − T − S) = i(λ − T ) = 0.

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Spectral Theory of Multivalued Linear Operators

This completes the proof of (v).

Q.E.D.

The following result gives a characterization of the essential ε-pseudospectra of a closed linear relations by means of Fredholm relations and minimum modulus. Proposition 4.11.2 Let X be a Banach space, T ∈ CR(X) and ε > 0. Then, (i) λ ∈ / Σe1,ε (T ) if and only if, λ − T ∈ Φ+ (X) and γ(λ − T ) > ε. (ii) λ ∈ / Σe2,ε (T ) if and only if, λ − T ∈ Φ− (X) and γ(λ − T ) > ε.

♦

Proof. (i) Let λ ∈ / Σe1,ε (T ). Then, α(λ − T ) < ∞ and γ(λ − T ) > ε. Since γ(λ−T ) > 0, then λ−T is open. It follows from (i) of Lemma 2.9.3 and (ii) of Theorem 2.7.2 that R(λ−T ) is closed. Hence, λ−T ∈ Φ+ (X) and γ(λ−T ) > ε. Conversely, let λ − T ∈ Φ+ (X) and γ(λ − T ) > ε. Then, α(λ − T ) < ∞ and γ(λ − T ) > ε. This is equivalent to say that λ ∈ / Σe1,ε (T ). The proof of the assertion (ii) can be checked in the same way as (i). Q.E.D. From Proposition 4.11.2 and Remark 2.20.4, we can deduce the following corollary. Corollary 4.11.1 Let X be a Banach space, T ∈ CR(X) and ε > 0. Then, (i) λ ∈ / Σe3,ε (T ) if and only if, λ − T ∈ Φ± (X) and γ(λ − T ) > ε. (ii) λ ∈ / Σe4,ε (T ) if and only if, λ − T ∈ Φ(X) and γ(λ − T ) > ε. (iii) λ ∈ / Σe5,ε (T ) if and only if, λ − T ∈ Φ(X), i(λ − T ) = 0 and γ(λ − T ) > ε. ♦ Lemma 4.11.1 Let X be a Banach space, T ∈ CR(X) and let ε > 0. Then, (i) Σe1,ε (T ) = Σe2,ε (T ∗ ) (ii) Σe2,ε (T ) = Σe1,ε (T ∗ ). (iii) Σei,ε (T ) = Σei,ε (T ∗ ), i = 3, 4, 5.

♦

Proof. (i) Let us assume that λ ∈ / Σe1,ε (T ). Then, using (i) of Proposition 4.11.2, we have λ − T ∈ Φ+ (X) and γ(λ − T ) > ε. By virtue of Theorem 2.14.4 (ii) and Proposition 2.8.1 (iii), we get λ − T ∗ ∈ Φ− (X ∗ ). Combining the fact that λ − T is open and Propositions 2.9.3 (i), (iii), and 2.8.1 (iii), we infer that γ(λ − T ∗ ) > ε. Thus, λ ∈ / Σe2,ε (T ∗ ). This enables us to conclude that Σe1,ε (T ) ⊂ Σe2,ε (T ∗ ). A similar reasoning as above gives Σe2,ε (T ∗ ) ⊂ Σe1,ε (T ).

Essential Spectra and Essential Pseudospectra of a Linear Relations

265

(ii) Let us assume that λ ∈ / Σe1,ε (T ∗ ). Then, using (i) of Proposition 4.11.2, ∗ ∗ we have λ − T ∈ Φ+ (X ) and γ(λ − T ∗ ) > ε. In view of Theorem 2.14.4 (i) implies that λ − T ∈ Φ− (X). It follows from (iii) of Theorem 2.7.2 that λ − T is open. Then, according to Propositions 2.9.3 (i), (iii), we infer that γ(λ − T ) > ε. Hence, λ ∈ / Σe2,ε (T ). Consequently, Σe2,ε (T ) ⊂ Σe1,ε (T ∗ ). Conversely, a same reasoning as before leads to the result. (iii) For i = 3, the use of (i), (ii) and Remark 2.20.4 (iii) allows us to conclude that \ Σe3,ε (T ) = Σe1,ε (T ) Σe2,ε (T ) \ = Σe2,ε (T ∗ ) Σe1,ε (T ∗ ) =

Σe3,ε (T ∗ ).

In the same way, one checks easily Σe4,ε (T ) = Σe4,ε (T ∗ ).

(4.70)

For i = 5, let us assume that λ ∈ / Σe5,ε (T ). This implies from (v) of Remark 2.20.4 that λ ∈ / Σe4,ε (T ) and i(λ − T ) = 0. By virtue of (4.70), we can get that λ ∈ / Σe4,ε (T ∗ ). It remains to prove the following i(λ − T ∗ ) = −i(λ − T ) = 0. Using the fact that λ ∈ / Σe4,ε (T ), then by Corollary 4.11.1, we infer that λ − T ∈ Φ(X) and γ(λ − T ) > ε. This implies that R(λ − T ) is closed and λ − T is open. Hence, the use of Theorem 2.14.4 (ii) and Proposition 2.8.1 (iii) makes us conclude that α(λ − T ∗ ) = β(λ − T ) and α(λ − T ) = β(λ − T ∗ ). This leads to i(λ−T ∗ ) = −i(λ−T ) = 0. Again, by using (v) of Remark 2.20.4, we deduce that λ ∈ / Σe5,ε (T ∗ ). This is equivalent to say that Σe5,ε (T ∗ ) ⊂ Σe5,ε (T ). Conversely, a same reasoning as before leads to the result.

4.11.1

Q.E.D.

The Essential ε-Pseudospectra of a Sequence of Linear Relations

The aim of this subsection is to discuss the essential ε-pseudospectra of a sequence of linear relations in a Banach space.

266

Spectral Theory of Multivalued Linear Operators

Theorem 4.11.1 Let X be a Banach space, (Tn )n be a sequence of closed linear relations from X into X and T ∈ CR(X). We suppose that kTn − T k → 0 when n → ∞. If there exists n1 ∈ N such that T (0) = Tn (0) and D(T ) = D(Tn ), for all n ≥ n1 . Then, for every ε > 0, there exist ε1 , ε2 > 0 and a positive integers n2 such that Σe1,ε1 (T ) ⊂ Σe1,ε2 (Tn ) ⊂ Σe1,ε (T ), for all n ≥ n2 .

♦

Proof. Let us assume that λ ∈ / σe1,ε (T ). Then, α(λ − T ) < ∞ and k(λ − T )−1 k < 1ε . It follows from Lemma 2.9.1 that γ(λ−T ) > ε. Since kTn −T k → 0 when n → ∞, then there exists n0 ∈ N such that   kTn − T k < kP k−1 γ(λ − T ) − ε3 , for all n ≥ n0 , where P is any continuous projection defined on D(λ−T ) with kernel N (λ−T ). It follows from Remark 2.9.2, for all n ≥ n0 that kTn − T k <


kP k−1 γ(λ − T ) − kTn − T k ε kP k−1 .

Hence, we deduce that λ ∈ / σe1,ε2 (T ), where ε2 = ε kP k−1 . Let us show that σe1,ε1 (T ) ⊂ σe1,ε2 (Tn ), for all n ≥ n2 . Assume that λ ∈ / σe1,ε2 (Tn ). Then, α(λ − Tn ) < ∞ and k(λ − Tn )−1 k < ε12 . By referring to Lemma 2.9.1, we have γ(λ − Tn ) > ε2 . Since kTn − T k → 0 when n → ∞, then there exists n0 ∈ N such that   kTn − T k < kP1 k−1 γ(λ − Tn ) − ε2 , for all n ≥ n0 ,

Essential Spectra and Essential Pseudospectra of a Linear Relations

267

where P1 is any continuous projection defined on D(λ−Tn ) with kernel N (λ− Tn ). It follows from Remark 2.9.2, for all n ≥ n0 that kTn − T k <


kP1 k−1 γ(λ − Tn ) − kTn − T k ε2 kP1 k−1 .

Hence, λ ∈ / σe1,ε1 (T ), where ε1 = ε2 kP1 k−1 = ε kP k−1 kP1 k−1 .

Q.E.D.

Theorem 4.11.2 Let X be a Banach space, (Tn )n be a sequence of closed linear relations from X into X and T ∈ CR(X). We suppose that kTn −T k → 0 when n → ∞. Then, (i) If there exists n1 ∈ N such that D(T ) ⊂ D(Tn ) ⊂ D(T ) and T (0) = Tn (0), for all n ≥ n1 , then for every ε2 > 0, there exist a positive real number ε1 , and a positive integer n2 such that Σe2,ε1 (Tn ) ⊂ Σe2,ε2 (T ), for all n ≥ n2 . (ii) If there exists n1 ∈ N such that D(Tn ) ⊂ D(T ) ⊂ D(Tn ) and T (0) = Tn (0), for all n ≥ n1 , then for every ε2 > 0, there exist a positive real number ε1 , and a positive integer n2 such that Σe2,ε1 (T ) ⊂ Σe2,ε2 (Tn ), for all n ≥ n2 .

♦

Proof. (i) Let λ ∈ / Σe2,ε2 (T ). Then, by using (ii) of Proposition 4.11.2, we infer that λ − T ∈ Φ− (X) and γ(λ − T ) > ε2 . Since kTn − T k → 0 when n → ∞, then there exists n0 ∈ N such that kTn − T k < kP k−1 (γ(λ − T ) − ε2 ) , for all n ≥ n0 , where P is any continuous projection defined on D((λ − T )∗ ) with kernel N ((λ − T )∗ ). Using Theorem 2.14.4 (i) and the fact that λ − T ∈ Φ− (X),

268

Spectral Theory of Multivalued Linear Operators

we infer that (λ − T )∗ ∈ Φ+ (X ∗ ). This implies that (λ − T )∗ is open and α((λ − T )∗ ) < ∞. It follows from Remark 2.9.2, for all n ≥ n0 , that kTn − T k <


ε1 ,

where ε1 = kP k−1 ε2 . Finally, the use of Proposition 4.11.2 makes us conclude that λ ∈ / Σe2,ε1 (Tn ), for all n ≥ n2 . The proof of (ii) may be achieved by using the same reasoning as (i). Q.E.D. As an immediate consequence of Theorems 4.11.1 and 4.11.2. Corollary 4.11.2 Let X be a Banach space, ε > 0, (Tn )n be a sequence of closed and bounded linear relations from X into X and T ∈ BCR(X). We suppose that kTn − T k → 0 when n → ∞. If there exists n1 ∈ N such that T (0) = Tn (0), for all n ≥ n1 , then for every triplet (ε1 , ε2 , ε3 ) such that 0 < ε1 < ε2 < ε3 , there exists a positive integer n2 such that Σei,ε1 (T ) ⊂ Σei,ε2 (Tn ) ⊂ Σei,ε3 (T ), for all n ≥ n2 , for i = 1, . . . , 4.

♦

Essential Spectra and Essential Pseudospectra of a Linear Relations

4.12

269

S-Pseudospectra of Linear Relations

The purpose of this section is to define and characterize the S-pseudospectra of multivalued linear operator and study some properties. Lemma 4.12.1 Let X be a Banach space, ε > 0, T ∈ CR(X) and let S ∈ LR(X) such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). If λ ∈ / σS (T ), then λ ∈ σε,S (T ) if and only if there exists x ∈ X such that k(λS − T )xk < εkxk. ♦ Proof. Let λ ∈ σε,S (T )\σS (T ), then k(λS − T )−1 k > 1ε . Since (λS − T )−1 is bounded operator, there exists a non-zero vector y ∈ X such that k(λS − T )−1 yk >

1 kyk. ε

(4.74)

Putting x := (λS − T )−1 y, then y ∈ (λS − T )x. On the other hand, (λS − T )(0) = λS(0) − T (0) = T (0) (as S(0) ⊂ T (0)). By using Lemma 2.5.7, we have k(λS − T )xk = = ≤

dist(y, (λS − T )(0)) dist(y, T (0)) inf ky − zk z∈T (0)

= kyk.

(4.75)

From (4.74) and (4.75), we have kxk >

1 1 kyk ≥ k(λS − T )xk, ε ε

Hence, k(λS − T )xk < εkxk. Conversely, assume that there exists x ∈ X such that k(λS − T )xk < εkxk. Since λ ∈ ρS (T ), then λS − T is injective. This implies that γ(λS − T )kxk ≤ k(λS − T )xk < εkxk. So, 0 < γ(λS − T ) < ε. Using Lemma 2.5.7, we have γ(λS − T ) = k(λS − T )−1 k−1 . Therefore, λ ∈ σε,S (T ). Q.E.D.

270

Spectral Theory of Multivalued Linear Operators

Theorem 4.12.1 Let X be a Banach space, ε > 0 and let T ∈ CR(X). Then, λ ∈ σε,S (T ) if and only if there exists a continuous linear relation B satisfying D(B) ⊃ D(T ), B(0) ⊂ T (0), and kBk < ε such that λ ∈ σS (T + B).

♦

Proof. In the first sense, assume that λ ∈ σε,S (T ). We will discuss two cases. First case : If λ ∈ σS (T ), then we may put B = 0. Second case : If λ ∈ / σS (T ), then by Lemma 4.12.1 there exists x0 ∈ X, kx0 k = 1 such that k(λS − T )x0 k < ε. By Theorem 2.8.1, there exists x0 ∈ X ∗ such that kx0 k = 1 and x0 (x0 ) = kx0 k. We can define the relation B : X −→ X by B(x) := x0 (x)(λS − T )x0 . It is clear that B is everywhere defined and single valued. Hence, kBxk = kx0 (x)(λS − T )x0 )k ≤ kx0 kkxkk(λS − T )x0 k. k 6 0, we have kBx For x = kxk ≤ k(λS − T )x0 k. So, kBk < ε. On the other hand, (λS − (T + B))x0 = (λS − T )x0 − Bx0 = (λS − T )x0 − x0 (x0 )(λS − T )x0 = (λS − (T + B))(0). Therefore,

0 6= x0 ∈ N (λS − (T + B)). Hence, λS − (T + B) is not injective. So, λ ∈ σS (T + B). Conversely, we derive a contradiction from the assumption that λ ∈ / σε,S (T ), then λ ∈ ρS (T ) and γ(λS − T ) ≥ ε. Therefore, λS − T is injective, open with dense range. Furthermore, B(0) ⊂ T (0) = (λS − T )(0), D(B) ⊃ D(T ) = D(λS − T ) and kBk < ε ≤ γ(λS − T ). Using Lemma 2.9.3, λS − T − B is injective, open with dense range. So, λ ∈ ρS (T + B) and this is a contradiction. Q.E.D. Remark 4.12.1 It follows, immediately, from Theorem 4.12.1, that for T ∈ CR(X) and ε > 0, [ σε,S (T ) = σS (T + B). kBk 0, T ∈ CR(X) and let S ∈ LR(X) such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). Then, \ σε,S (T ) = σS (T ). ♦ ε>0

Proof. It is clear that σS (T ) ⊂ σε,S (T ) for all ε > 0. Then, \ σS (T ) ⊂ σε,S (T ). ε>0

Conversely, if λ ∈ / σS (T ), then λ ∈ ρS (T ). Hence, (λS − T )−1 is a bounded linear operator. So, there exists ε > 0 such that k(λS − T )−1 k ≤ 1ε . Thus, λ∈ / σε,S (T ) and \ λ∈ / σε,S (T ). Q.E.D. ε>0

Proposition 4.12.2 Let T ∈ CR(X) and ε > 0. If T is injective with dense 6 0, then range and kSk =   γ(T ) − ε σε,S (T ) ⊂ λ ∈ C such that |λ| > . ♦ kS k Proof. If γ(T ) < ε, then there is nothing to prove. If γ(T ) = ε > 0, then T = T −0S is open. Hence, 0 ∈ ρS (T ). Furthermore, kT −1 k = k(0S −T )−1 k = 1 / σε,S (T ). Therefore, ε . So, 0 ∈ σε,S (T ) ⊂ {λ ∈ C such that |λ| > 0}. Now, suppose that γ(T ) > ε. Then, T is open, injective and surjective. On the )−ε other hand, for λ ∈ C such that |λ| ≤ γ(T kSk , then kλSk = |λ|kSk ≤ γ(T )−ε. Thus, kλSk < γ(T ). Using Lemma 2.9.3, the relation λS − T is open, injective with dense range, i.e., λ ∈ ρS (T ). For x ∈ D(T ), since S(0) ⊂ T (0), we have k(λS − T )xk = k(T − λS)xk ≥ kT xk − kλSxk ≥ (γ(T ) − |λ|kSk)kxk. Therefore, γ(λS − T ) ≥ γ(T ) − |λ|kSk ≥ γ(T ) − γ(T ) + ε = ε. So, k(λS − T )−1 k ≤ Hence, λ ∈ ρε,S (T ).

1 . ε

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Spectral Theory of Multivalued Linear Operators

Thus,   γ(T ) − ε σε,S (T ) ⊂ λ ∈ C such that |λ| > . kSk

Q.E.D.

Theorem 4.12.2 Let ε > 0, T ∈ CR(X) and let S ∈ LR(X) such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). Then, σε,S (T ) = σε,S ∗ (T ∗ ). ♦ Proof. At first, it is clear from the proof of Theorem 2.19.7 that (λS −T )∗ = λS ∗ − T ∗ . Now, let λ ∈ ρε,S ∗ (T ∗ ). Then, λ ∈ ρS ∗ (T ∗ )

and

k((λS − T )∗ )−1 k ≤

1 . ε

By Theorem 2.19.7, λ ∈ ρS (T ). So, (λS − T )−1 is a continuous single valued. From Proposition 2.8.1, we have k(λS − T )−1 k = k((λS − T )−1 )∗ k ≤

1 . ε

Thus, λ ∈ ρε,S (T ). Conversely, if λ ∈ ρε,S (T ), then k(λS − T )−1 k ≤ Proposition 2.8.1, k(λS − T )−1 k = k(λS ∗ − T ∗ )−1 k ≤

1 ε.

By

1 . ε

Furthermore, λ ∈ ρS ∗ (T ∗ ) by Theorem 2.19.7. Then, λ ∈ ρε,S ∗ (T ∗ ). Q.E.D.

4.12.1

S-Essential Pseudospectra of Linear Relations

In this section, we define the S-essential pseudospectra of a closed linear relation, study some properties and establish some results of perturbation on the context of linear relations. Definition 4.12.1 Let T be a linear relation in CR(X) and ε > 0. The Sessential pseudospectra of T is the set \ σe5,ε,S (T ) = σε,S (T + K). K∈KT (X)

We define the S-essential pseudoresolvent set by ρe5,ε,S (T ) = C\σe5,ε,S (T ). Theorem 4.12.3 The following properties are equivalent: (i) λ ∈ / σe5,ε,S (T ).

♦

Essential Spectra and Essential Pseudospectra of a Linear Relations

273

(ii) For all continuous linear relations B ∈ LR(X) such that D(B) ⊃ D(T ), B(0) ⊂ T (0) and kBk < ε, we have T +B −λS ∈ Φ(X) and i(T +B −λS) = 0. (iii) For all continuous single valued operators D ∈ LR(X) such that D(D) ⊃ D(T ) and kDk < ε, we have T + D − λS ∈ Φ(X) and i(T + D − λS) = 0. ♦ Proof. (i) =⇒ (ii) Let λ ∈ / σe5,ε,S (T ). Then, there exists K ∈ KT (X) such that λ ∈ / σε,S (T + K). Using Theorem 4.12.1, for all continuous linear T relations B ∈ LR(X) such that D(B) ⊃ D(T + K) = D(T ) D(K) = D(T ) as D(K) ⊃ D(T ), B(0) ⊂ (T + K)(0) = T (0) as K(0) ⊂ T (0) and kBk < ε, we have λ ∈ ρS (T + B + K). Then, T + B + K − λS is open, injective with dense range. But T is closed, K is compact, then K is continuous. So, B + K − λS is continuous. Furthermore, (B + K − λS)(0) ⊂ T (0), then by Proposition 2.7.2, T + B + K − λS is closed. Hence, R(T + B + K − λS) is closed. So, R(T + B + K − λS) = X. Therefore, T + B + K − λS ∈ Φ(X) and i(T + B + K − λS) = 0, for all continuous linear relations B ∈ LR(X) such that D(B) ⊃ D(T ), B(0) ⊂ T (0) and kBk < ε. It comes from Proposition 2.3.4 and Theorem 3.5.5 for all continuous linear relations B ∈ LR(X) such that D(B) ⊃ D(T ), B(0) ⊂ T (0) and kBk < ε, we have T + B − λS ∈ Φ(X) and i(T + B − λS) = 0. (ii) =⇒ (iii) is trivial. (iii) =⇒ (i) Let D be a continuous single valued operators in LR(X) such that D(D) ⊃ D(T ) and kDk < ε, then T + D − λS ∈ Φ(X) and i(T + D − λS) = 0. By using Proposition 2.8.1, we have N ((T +D −λS)∗ ) = R(T +D −λS)⊥ . Let n = α(T + D − λS) = β(T + D − λS), {x1 , . . . , xn } be basis for N (T + D − λS) and {y10 , . . . , yn0 } be basis for N ((T + D − λS)∗ ). Hence, there are functionals x01 , . . . , x0n ∈ X ∗ and elements y1 , . . . , yn such that x0j (xk ) = δjk

and

yj0 (yk ) = δjk ,

1 ≤ j, k ≤ n,

6 k and δjk = 1 if j = k. The single valued operator K is where δjk = 0 if j = defined by n X Kx = x0k (x)yk , x ∈ X. k=1

K is bounded, since D(K) = X and kKxk ≤ kxk

n X k=1

! kx0k kkyk k .

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Spectral Theory of Multivalued Linear Operators

Moreover, the range of K is contained in a finite subspace of X. Then, K is a finite rank relation in X. Hence, K is compact. Let x ∈ N (T + D − λS), then x=

n X

αk xk .

k=1

Therefore, x0j (x) = αj , 1 ≤ j ≤ n. On the other hand, if x ∈ N (K), then T x0j (x) = 0, 1 ≤ j ≤ n. This proves that N (T + D − λS) N (K) = {0}. Now, if y ∈ R(K), then n X y= αk yk . k=1

Hence, yj0 (y) = αj , 1 ≤ j ≤ n. But, if y ∈ R(T + D − λS), then yj0 (y) = 0, 1 ≤ j ≤ n. T T So, N (T + D − λS) N (K) = {0} and R(T + D − λS) R(K) = {0}. On the other hand, K ∈ KT (X) = KT +D−λS (X) since K(0) ⊂ T (0) = (T +D −λS)(0) and D(K) ⊃ D(T ) = D(T + D − λS). So, we deduce that T + D + K − λS ∈ Φ(X) and i(T + D + K − λS) = 0. If x ∈ N (T + D + K − λS), then 0 ∈ T T x + Dx + Kx − λSx. Hence, −Kx ∈ (T + D − λS)x. So, Kx ∈ R(K) R(T + D − λS) = {0}. Therefore, Kx = 0 and 0 ∈ (T + D − λS)x. This implies that T x ∈ N (T + D − λS) N (K). Hence, x = 0. Thus, α(T + D + K − λS) = 0. In the some way, one proves that R(T + D + K − λS) = X. Since T + D + K − λS is closed by Proposition 2.7.2, then λ ∈ ρS (T + D + K) for all continuous single valued operators D ∈ LR(X) such that D(D) ⊃ D(T ) and kDk < ε. But from the proof of Theorem 4.12.1 ((i) =⇒ (ii)), if λ ∈ σε,S (T + K), then there exists a continuous single valued operator D ∈ LR(X) satisfying D(D) ⊃ D(T + K) = D(T ) and kDk < ε such that λ ∈ σS (T + K + D). Hence, λ∈ / σε,S (T + K). Since K ∈ KT (X), then \ λ∈ / σε,S (T + K). K∈KT (X)

So, λ ∈ / σe5,ε,S (T ).

Q.E.D.

Remark 4.12.2 It follow immediately, from Theorem 4.12.3 that for T ∈ CR(X) and ε > 0, [ [ σe5,ε,S (T ) = σe5,S (T + D) = σe5,S (T + B). kDk0

Proof. (i) If λ ∈ / σe5,ε2 ,S (T ), then by Theorem 4.12.3 for all continuous linear relations B ∈ LR(X) such that D(B) ⊃ D(T ), B(0) ⊂ T (0) and kBk < ε2 , we have T + B − λS ∈ Φ(X) and i(T + B − λS) = 0. Hence, for all continuous linear relations B ∈ LR(X) such that D(B) ⊃ D(T ), B(0) ⊂ T (0) and kBk < ε1 , we have T + B − λS ∈ Φ(X) and i(T + B − λS) = 0. Then, λ∈ / σε1 ,S (T ). On other hand, if λ ∈ / σe5,ε1 ,S (T ), then for all continuous linear relations B ∈ LR(X) such that D(B) ⊃ D(T ), B(0) ⊂ T (0) and kBk < ε1 , we have T + B − λS ∈ Φ(X) and i(T + B − λS) = 0. In particular B = 0, then λ ∈ / σe5,S (T ). T (ii) σe5,ε,S (T ) = K ∈KT (X) σε,S (T + K) ⊂ σε,S (T + K) for all K ∈ KT (X). In particular, K = 0. (iii) From (i), σe5,S (T ) ⊂ σe5,ε,S (T ) for all ε > 0, then \ σe5,S (T ) ⊂ σe5,ε,S (T ). ε>0

Conversely, if λ ∈ / σe5,S (T ), then T −λS ∈ Φ(X) and i(T −λS) = 0. Therefore, R(T −λS) is closed. Hence, T −λS is open. So, γ(T −λS) > 0. Let ε such that 0 < ε ≤ γ(T − λS) and let D be a single valued linear relation in LR(X) such that D(D) ⊃ D(T ) and kDk < ε ≤ γ(T − λS). By γ(T − λS) = γ((T − λS)∗ ), then kDk < ε ≤ γ((T − λS)∗ ). So, T + D − λS ∈ Φ− (X). Hence, α(T + D − λS) ≤ α(T − λS) < ∞. Thus, T + D − λS ∈ Φ+ (X) and T + D − λS ∈ Φ(X). Furthermore, we have i(T + D − λS) = i(T − λS ) = 0. Hence, by using T Theorem 4.12.3, we have λ ∈ / σe5,ε,S (T ). So, λ ∈ / ε>0 σe5,ε,S (T ). It follows \ σe5,ε,S (T ) ⊂ σe5,S (T ). Q.E.D. ε>0

Theorem 4.12.4 Let X be a Banach space, ε > 0, T ∈ CR(X) and let S ∈ LR(X) such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). Then, σe5,ε,S (T ) = σe5,ε,S ∗ (T ∗ ). ♦

276

Spectral Theory of Multivalued Linear Operators

Proof. Let K ∈ KT (X), then K is compact. Therefore, it is continuous by Lemma 2.12.1 and from Proposition 2.11.2, K ∗ is compact. Moreover, by Lemma 2.8.3, K ∗ (0) ⊂ T ∗ (0) and D(K ∗ ) ⊃ D(T ∗ ). Hence, KT (X) ⊂ {K ∈ LR(X) such that K is continuous and K ∗ ∈ KT ∗ (X ∗ )}. Now, let K ∈ {K ∈ LR(X) such that K is continuous and K ∗ ∈ KT ∗ (X ∗ )}, then K is compact and K ∗ (0) = D(K)⊥ ⊂ T ∗ (0) = D(T )⊥ . So, D(K) ⊃ D(T ), also K(0) ⊂ K(0) = D(K ∗ )> ⊂ D(T ∗ )> = T (0) as T is closed. Then, K ∈ KT (X). Hence, KT (X) = {K ∈ LR(X) such that K is continuous and K ∗ ∈ KT ∗ (X ∗ )}. (4.76) On the other hand for K ∈ KT (X), T + K is closed (by Proposition 2.7.2). By using Theorem 4.12.2, we have σε,S (T + K) = σε,S ∗ ((T + K)∗ ). But D(K) ⊃ D(T ) and K is continuous, then (T + K)∗ = T ∗ + K ∗ . Hence, σε,S (T + K) = σε,S ∗ (T ∗ + K ∗ ) for all K ∈ KT (X). Therefore, by using (4.76), we have \ σe5,ε,S (T ) = σε,S (T + K) K∈KT (X)

\

=

σε,S ∗ (T ∗ + K ∗ )

K ∗ ∈KT ∗ (X ∗ ) K∈LR(X) Kcontinuous

\

⊃

=

σε,S ∗ (T ∗ + K ∗ )

K ∗ ∈KT ∗ (X ∗ ) K∈LR(X) σe5,ε,S ∗ (T ∗ ).

Let O :=

\

σε,S ∗ (T ∗ + K ∗ ),

K ∗ ∈KT ∗ (X ∗ ) K∈LR(X) K continuous

then O = σe5,ε,S (T ) ⊃ σe5,ε,S ∗ (T ∗ ).

(4.77)

On the other hand, let λ ∈ O, then for all K ∈ LR(X), K is continuous such that K ∗ ∈ KT ∗ (X ∗ ), λ ∈ σε,S ∗ (T ∗ + K ∗ ). Let K ∈ LR(X) such e is compact, where Te is given in (2.4). that K ∗ ∈ KT ∗ (X ∗ ). Hence, K e ) is the completion of G(K), then G(K e ) = G(K) = G(K). Since G(K e = K is compact and hence, continuous. Furthermore, G(K ∗ ) = Hence, K

Essential Spectra and Essential Pseudospectra of a Linear Relations ⊥

277

∗

G(−K −1 )⊥ = G(−K −1 ) = G(−K −1 )⊥ = G(K ∗ ). Hence, K = K ∗ . Thus, K in LR(X), K is continuous and K ∗ = K ∗ ∈ KT ∗ (X ∗ ). Since λ ∈ O, ∗ then λ ∈ σε,S ∗ (T ∗ + K ) = σε,S ∗ (T ∗ + K ∗ ). We conclude that if λ ∈ O, for all K ∈ LR(X) such that K ∗ ∈ KT ∗ (X ∗ ), then λ ∈ σε,S ∗ (T ∗ + K ∗ ). So, T λ ∈ K ∗ ∈KT ∗ (X ∗ ) σε,S ∗ (T ∗ + K ∗ ) = σe5,ε,S ∗ (T ∗ ). Hence, O ⊂ σe5,ε,S ∗ (T ∗ ). By K ∈LR(X)

using (4.77), we have O = σe5,ε,S (T ) = σe5,ε,S ∗ (T ∗ ).

Q.E.D.

Theorem 4.12.5 Let X be a Banach space, ε > 0, T ∈ CR(X) and let S ∈ LR(X) such that S(0) ⊂ T (0) and D(S) ⊃ D(T ). Then, \ σe5,ε,S (T ) = σε,S (T + P ). ♦ P ∈PR(Φ(X))

T

Proof. Let O := P ∈PR(Φ(X)) σε,S (T + P ). Since KT (X) ⊂ PR(Φ(X)), / O, then there exist we infer that O ⊂ σe5,ε,S (T ). Conversely, let λ ∈ P ∈ PR(Φ(X)) such that λ ∈ / σε,S (T + P ). But P is continuous, then by Proposition 2.7.2, T + P is closed. Thus, by Theorem 4.12.1, we can see that λ ∈ ρS (T + B + P ) for all continuous linear relations B ∈ LR(X) such that T B(0) ⊂ (T + P )(0) = T (0), D(B) ⊃ D(T + P ) = D(T ) D(P ) = D(T ) and T kBk < ε. But (B +P −λS)(0) ⊂ T (0), D(B +P −λS) = D(B) D(P ) ⊃ D(T ) and T is closed, by Proposition 2.7.2, T + B + P − λS is closed. Hence, T + B + P − λS is injective and surjective. Then, T + B + P − λS ∈ Φ(X) and i(T + B + P − λS) = 0. Since −P ∈ PR(Φ(X)), −P (0) = P (0) ⊂ (T + B + P − λS)(0) and D(−P ) = T T T D(P ) ⊃ D(T + B + P − λS) = D(T ) D(B) D(P ) D(S), then −P is in PR(Φ(X)). Using Theorem 3.3.2, we conclude that for all continuous linear relations B ∈ LR(X) such that B(0) ⊂ T (0), D(B) ⊃ D(T ) and kBk < ε, T + B − λS ∈ Φ(X) and i(T + B − λS) = 0. Finally, Theorem 4.12.3 shows that λ ∈ / σe5,ε,S (T ).

Q.E.D.

Corollary 4.12.1 Let J (X) be a subset of LR(X). If KT (X) ⊂ J (X) ⊂ PR(Φ(X)), then \ σe5,ε,S (T ) = σε,S (T + J). ♦ J ∈J (X)

278

Spectral Theory of Multivalued Linear Operators

Proof. It is easy to see \ σe5,ε,S (T ) = σε,S (T +P ) ⊂ P ∈PR(Φ(X))

This completes the proof.

\ J∈J (X)

σε,S (T +J) ⊂

\

σε,S (T +K).

K∈KT (X)

Q.E.D.

Remark 4.12.3 It follows, from the definition of S-essential pseudospectra, Theorem 4.12.5 and Corollary 4.12.1 that (i) σe5,ε,S (T + K) = σe5,ε,S (T ) for all K ∈ KT (X). (ii) σe5,ε,S (T + P ) = σe5,ε,S (T ) for all P ∈ PR(Φ(X)). (iii) If for all J1 , J2 ∈ J (X), J1 ±J2 ∈ J (X), then σe5,ε,S (T +J) = σe5,ε,S (T ) for all J ∈ J (X) such that KT (X) ⊂ J (X) ⊂ PR(Φ(X)). ♦

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Index

X JM , 19 LR(X, Y ), 20 M -modified resolvent, 1 M + N , 11 L M N , 11 ⊥ M , 55 N (T ), 19, 21 N > , 55 OX , 20 P σ(T ), 94 Q, 31 QYT (0) , 31

Λ(X), 97 ΨT (X), 97 ρ(T ), 93 ρe1,S (T ), 111 ρe2,S (T ), 111 ρe3,S (T ), 111 ρe4,S (T ), 111 ρe5,S (T ), 111 ρe6,S (T ), 111 diam (B), 63 (M, N ), 85 2V \∅, 18 BCR(X, Y ), 50 BR(X), 43 BR(X, Y ), 43 BD(T ) , 40 C([a, b]), 22 CR(X, Y ), 50 Cσ(T ), 94 C k -norm, 17 D-condensing, 65 G(IE ), 19 G(IM ), 20 G(T ), 19, 20 GT , 42 GT x, 42 H(U, X), 94 IE , 19 IM , 20

QX E , 14 QT , 14, 31 QT T (BX ), 67 R(T ), 19, 21 R(λ, T ), 92 Rσ(T ), 94 R∞ (T ), 37 RS (λ, T ), 98 S ⊂ T , 30 S-essential resolvent sets, 111 spectra, 111 S-resolvent of T at λ, 98 S-spectrum set, 98 SGT is precompact, 67 ST , 24 T is a continuous, 43 287

288 T -bound, 47 T -bounded, 47 T -compact, 67 T -precompact, 67 T (0), 20 T ⊂ S, 26 T ∗ , 55, 76 T −1 , 19, 21 T −1 (0), 19, 21 T −1 (M ), 21 T −1 (N ), 19 TG , 45 g Tn −→ T , 140 Tλ , 92 T|M , 25 X/M , 12 X ∗ , 55 X1 ∼ = X2 , 12 XT , 42 [x], 12 ∆(T ), 35 Γ(T ), 65 Γ0 (T ), 65 Φ(X, Y ), 78 Φ± (X, Y ), 78 Φ+ (X, Y ), 77 Φ− (X, Y ), 78 Φ+T,S , 78 Φ+T , 78 ΦT,S , 78 ΦT , 78 Σe1,ε (T ), 115 Σe2,ε (T ), 115 Σe3,ε (T ), 115 Σe4,ε (T ), 115 Σe5,ε (T ), 115 ΥT (X), 97 α(T ), 28

Index αT , 24 α-Atkinson multivalued linear operators, 81 perturbations, 91 α-Atkinson linear relation, 4 β(T ), 28 β-Atkinson multivalued linear operators, 81 perturbations, 91 β-Atkinson linear relation, 4 δ(D), 64 δ(M, N ), 62 δ(T, S), 63 δij , 155 dim X, 12 X , 12 M η(T ), 35 γ(T ), 57 D(0, ε), 94 Aα (X, Y ), 81 Aβ (X, Y ), 81 B(X, Y ), 78 C(X), 65 DC(X), 167 DC(X, Y ), 167 DC S (X), 172 DC S (X, Y ), 172 DC µ (X), 167 DC µ (X, Y ), 167 D(T ), 18 F+ (X, Y ), 76 F− (X, Y ), 76 GRr (X, Y ), 81 GR(X, Y ), 81 GRl (X, Y ), 81

Index I(X), 65 KR(X), 67 KR(X, Y ), 67 K(X), 67 KT (X), 97 L(X), 43 L(X, Y ), 43 PRK(X, Y ), 67 PR(Φ− (X, Y )), 91 PR(A(X, Y )), 91 PR (Φ+ (X, Y )), 91 PR (Aα (X, Y )), 91 PR (Aβ (X, Y )), 91 P(X), 65 R(U, V ), 18 SSR(X, Y ), 69 UT (X), 114 W(X, Y ), 81 W l (X, Y ), 81 W r (X, Y ), 81 R0 (T ), 37 Rc (T ), 38 asc(T ), 38 des(T ), 38 T , 54 Γ0 (T ), 65 β(T ), 28 σ S (T ), 98 ρS (T ), 98 ρε (T ), 112 ρε,S (T ), 116 ρe1,S (T ), 111 ρe2,S (T ), 111 ρe3,S (T ), 111 ρe4,S (T ), 111 ρe5,S (T ), 111 ρe5,ε,S (T ), 116 ρe5,ε,S (T ), 272

289 ρe6,S (T ), 111 ρei,S (T ), 111 σ(T ), 93 σS (T ), 98 σε (T ), 112 σδ,S (T ), 111 σδ (T ), 93 σε,S (T ), 116 σap,S (T ), 111 σap (T ), 93 σe1,S (T ), 111 σe1,ε (T ), 114 σe1 (T ), 96, 97 σe2,S (T ), 111 σe2,ε (T ), 114 σe2 (T ), 96, 97 σe3,S (T ), 111 σe3,ε (T ), 114 σe3 (T ), 96, 97 σe4,S (T ), 111 σe4,ε (T ), 114 σe4 (T ), 96, 97 σe5,S (T ), 111 σe5,ε,S (T ), 272 σe5,ε,S (T ), 116 σe5,ε (T ), 114 σe5 (T ), 96, 97 σe6,S (T ), 111 σe6 (T ), 97 σeα (T ), 97 σeβ (T ), 96 σeδ,S (T ), 111 σeδ (T ), 96, 97 σeap,S (T ), 111 σeap (T ), 96, 97 σeb (T ), 96 σel (T ), 96 σer (T ), 97

290 σl (T ), 93, 97 σqφ (T ), 96 σr (T ), 93, 97 co (B), 64 dist(λ, Ω), 13 dist(U, V ), 14 dist(V, x), 14 dist(x, V ), 14 ε-pseudospectra, 6 δb(T, S), 63 Tb, 61 δb(M, N ), 62 e T ), 65 ∆( ηeµ (λ), 98 Te, 19 e , 19 X {0}-condensing, 65 b.s.mult(A), 86 c0 , 18 ci(T ), 29 i(T ), 29 j(T ), 79 k-set-contraction, 4 l∞ , 18 lp , 21 lp (N), 17 p-norm, 17 qφd (X), 84 s.mult(A), 86 span{e1 }, 208 x R y, 12 G(T −1 ), 19, 21 S-essential pseudoresolvent set, 116, 272 S-essential pseudospectra, 116, 272 S-pseudoresolvent, 116 S-pseudospectra of T , 116

Index S-resolvent, 98 adjoint, 55, 56 adjoints of linear operators, 18 algebraic complement, 11 approximate point spectrum, 93 ascent, 38 Atkinson α-and β-Atkinson perturbation classes, 152 perturbations, 91 linear relations, 80, 147 augmented S-spectrum, 98 backward Samuel multiplicity, 86 shift multiplicity, 86 Banach space, 3, 11, 17 direct sum, 11 distance function, 13 normed vector space, 13 bounded, 43 below, 81 linear operators, 43 bounded below, 111 Browder, 78 bundedness, 8 canonical isometry, 15 Cartesian product, 14 Cauchy sequence, 13 chain of kernels, 34 ranges, 34 characterization of ε-pseudospectra, 242 closability, 8 closable, 54 closed, 50

Index closed linear relations, 191 closure of a linear relation, 54 co-index, 29 commute, 33 compact, 67 interval [a, b], 46 compact linear relation, 8 complemented, 11, 18 complete, 13 resolvent, 92 completely reduced, 85 completion of a linear relation, 19 continuous, 43 spectrum, 94 valued functions, 46 converge in the generalized sense, 140 convergent sequence, 15

291

defect spectrum, 93 degree η(T ), 35 demicompact, 8 demicompact linear relations, 167 auxiliary results, 167 Fredholm linear relations, 169 relatively demicompact, 172 auxiliary results, 172 Fredholm theory, 179 demicompactness, 4 descent, 38 diameter of a set X, 63 dimension of X, 12 direct sum, 11 distance, 13 between U and V , 14 domain, 18, 20

endomorphisms, 11 equivalence relation, 12 essential ε-pseudospectra sets, 115 pseudospectra, 114 spectra, 96 essential ε-pseudospectra of linear relations, 262 sequence, 265 essential pseudospectra, 8 essential pseudospectra of linear relations, 250 stability, 257 essential spectra sum of two linear relations, 226 essential spectrum and sequence of linear relations, 210 essential spectrum of a multivalued linear operator, 5 essential spectrum of linear relations α and β, 199 characterization, 191 essentially semi regular linear relations, 86 essentially semi regularity operators, 182 bounded linear operators, 182 Hilbert spaces, 182 perturbation results, 186 perturbation theory, 182 proprieties, 182 extension, 27 of T , 26

elliptic-parabolic equation, 2

finite-dimensional subspace, 12

292 Fredholm linear relations, 4, 78 Fredholm operator, 3 Fredholm or semi-Fredholm perturbations, 226 Fredholm theory, 3, 4 function, 18 functional analysis, 3, 13 fundamentals, 11 gap between T and S, 63 two linear subspaces, 62 generalized kernel, 37 range, 37 geometric property, 14 graph, 19 operator, 42 graph norm T graph norm, 42 Hann-Banach theorem, 55 Hilbert spaces, 4 idempotent operator, 12 identity injection, 42 relation, 19, 20 image, 19 index of T , 29 infimum δ, 47 injection map of M , 19 injection operator, 19 injective, 19, 22 open, 111 inverse, 21 of T , 19 inverse of functions, 18 isomorphic, 12

Index isomorphic linear spaces, 12 jump, 79 Kato decomposition, 85 linear relation, 85 of degree d, 85 kernel, 19, 21 Kuratowski measure of noncompactness, 64 left generalized inverse, 150 invertible, 81 regularizer, 150 shift single valued operator, 21 spectrum, 93 linear mapping, 12 linear relations, 20 S-essential spectra, 219 σe4,S (·) and σe5,S (·), 221 characterization, 219 characterization of σe5,S (·), 219 Racocˇevic´ and Schmoeger S-essential spectra, 221, 223 S-pseudospectra and S-essential pseudospectra of, 116 S-essential pseudospectra of, 272 S-pseudospectra of, 269 adjoint of, 55 algebra of, 20 closable linear relations, 54

Index closed and closable linear relations, 50 closed linear relations, 50 continuity and openness, 43 essential ε-pseudospectra of, 115 essential pseudospectra of, 114 Fredholm perturbation classes of, 91 generalized kernel and range of, 32 ascent and descent and singular chain of, 38 index and co-index of, 28, 158 properties of the quotient map QT , 31 inverse of, 21 localization of pseudospectra, 240 minimum modulus of, 57 norm of, 39 precompact and compact, 67 pseudospectra and ε-pseudospectra, 231 characterization, 235 properties, 231 stability, 237 pseudospectra of, 111 quantities for linear relation, 62 formula for gap between multivalued linear operators, 62 measures of noncompactness, 63 relatively boundedness, 47

293 restrictions and extensions of, 25 selections, 45 spectrum and pseudospectra, 92 essential spectra of, 96 resolvent set and spectrum, 92 subdivision of spectrum, 94 sum and product of, 23 lower semi-Fredholm, 5, 76 perturbations, 91 maximal gap, 62 maximum norm, 17 measure of compactness, 65 noncompactness, 8 minimum modulus, 57 Mobius transformation, 98 multivalued Cauchy problem, 2 multivalued linear operators, 1–4, 6, 8, 20, 117 classes of, 76 Atkinson linear relations, 80 closed linear operators in Banach space, 77 multivalued Fredholm and semi-Fredholm, 76 perturbation, 142 part of P (T ), 70 perturbation results, 87 small perturbation theorems, 87 sequence of, 129 gap between two linear relations, 129

294

Index generalized convergence, 140 relationship between gap and their selection, 138 stability and closeness of, 117 product of closable linear relations, 125 sum of three closed linear relations, 119 sum of two closable linear relations, 122 sum of two closed linear relations, 117 stability theorems of, 117

natural quotient map, 14 non zero normed space, 112 nonlinear operator, 2 norm, 39 dual, 55 normed space, 7, 11 S-essential resolvent sets, 111 S-spectra of linear relations, 98 augmented S-spectrum, 109 properties of S-resolvent, 99 vector space, 14, 29 null space, 19, 21 open, 13, 43 injective with dense range, 60 operator, 20 operator theory, 3 ordinary differential operator, 22 pair of closed subspaces, 85 paracomplet subspace, 18 partial order relations, 18 perturbation theorems, 3

perturbation theory, 4 point spectrum, 94 polynomial linear relation, 70 polynomial multivalued linear operators, 70 precompact, 8, 67 projection, 11 pseudoresolvent, 112 pseudospectra, 8, 111 ε-pseudospectra of closed linear relations, 248 quantum theory, 1 quasi-Fredholm, 8 quasi-Fredholm linear relations, 84 quotient indecomposable normed spaces, 5 quotient space, 12, 14 range, 19, 21 space relation, 85 relation, 18 on sets, 18 relative bound of S with respect to T , 47 relative boundedness, 8 relatively bounded, 47 residual spectrum, 94 resolvent, 92 set, 93 restriction TM , 76 restriction of T , 25 right generalized inverse, 83 invertible, 81 regularizer, 83 spectrum, 93 root manifold, 37

Index Samuel multiplicity, 86 selection, 46 semi regular, 8 semi regular linear relations, 84 shift multiplicity, 86 single valued, 19, 22 operator, 18, 20 single valued linear operator, 8 singular chain manifold, 38 spectral mapping theorem of essential spectra, 213 sum of two linear relations, 214 spectral theory, 1, 191 spectrum, 93 strictly singular linear relations, 8, 68 sup-norm, 17 surjective, 19, 22 symmetric gap, 62

295 theory of linear relations, 11 time homogeneous equation, 1 topological complement, 18 triangle inequality, 13 trivial singular chain, 38 univalent coefficient operators, 2 linear operators, 3 operator, 2 upper semi-Fredholm, 5, 76 perturbations, 91 index of relation, 162 operators, 4 Weyl linear relations, 81 zero norm, 39 operator, 88 relation, 20