Nonlinear Ill-posed Problems of Monotone Type

NONLINEAR ILL-POSED PROBLEMS OF MONOTONE TYPE Nonlinear Ill-posed Problems of Monotone Type by YAKOV ALBER and IRIN...

Author: Yakov Alber | Irina Ryazantseva

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NONLINEAR ILL-POSED PROBLEMS OF MONOTONE TYPE

Nonlinear Ill-posed Problems of Monotone Type by

YAKOV ALBER and

IRINA RYAZANTSEVA

A C.I.P. Catalogue record for this book is available from the Library of Congress.

ISBN-10 ISBN-13 ISBN-10 ISBN-13

1-4020-4395-3 (HB) 978-1-4020-4395-6 (HB) 1-4020-4396-1 (e-book) 978-1-4020-4396-3 (e-book)

Published by Springer, P.O. Box 17, 3300 AA Dordrecht, The Netherlands. www.springer.com

Printed on acid-free paper

All Rights Reserved © 2006 Springer No part of this work may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, microfilming, recording or otherwise, without written permission from the Publisher, with the exception of any material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Printed in the Netherlands.

Contents vii

PREFACE

xiii

ACKNOWLEDGMENTS 1 INTRODUCTION INTO THE THEORY OF MONOTONE AND ACCRETIVE OPERATORS 1.1 Elements of Nonlinear Functional Analysis . . . . . . . . . . . . . . . . 1.2 Subdifferentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Monotone Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Maximal Monotone Operators . . . . . . . . . . . . . . . . . . . . . . . 1.5 Duality Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Banach Spaces Geometry and Related Duality Estimates . . . . . . . . 1.7 Equations with Maximal Monotone Operators . . . . . . . . . . . . . . 1.8 Summation of Maximal Monotone Operators . . . . . . . . . . . . . . 1.9 Equations with General Monotone Operators . . . . . . . . . . . . . . 1.10 Equations with Semimonotone Operators . . . . . . . . . . . . . . . . 1.11 Variational Inequalities with Monotone Operators . . . . . . . . . . . . 1.12 Variational Inequalities with Semimonotone Operators . . . . . . . . . 1.13 Variational Inequalities with Pseudomonotone Operators . . . . . . . . 1.14 Variational Inequalities with Quasipotential Operators . . . . . . . . . 1.15 Equations with Accretive Operators . . . . . . . . . . . . . . . . . . . 1.16 Equations with d-Accretive Operators . . . . . . . . . . . . . . . . . .

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1 1 16 19 29 33 40 52 59 66 70 72 79 84 90 95 108

2 REGULARIZATION OF OPERATOR EQUATIONS 2.1 Equations with Monotone Operators in Hilbert Spaces . 2.2 Equations with Monotone Operators in Banach Spaces . 2.3 Estimates of the Regularized Solutions . . . . . . . . . . 2.4 Equations with Domain Perturbations . . . . . . . . . . 2.5 Equations with Semimonotone Operators . . . . . . . . 2.6 Equations with Non-Monotone Perturbations . . . . . . 2.7 Equations with Accretive and d-Accretive Operators . .

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117 117 123 129 133 136 138 142

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PARAMETERIZATION OF REGULARIZATION METHODS 151 3.1 Residual Principle for Monotone Equations . . . . . . . . . . . . . . . . . . 152

Contents

vi 3.2 3.3 3.4 3.5 3.6

Residual Principle for Accretive Generalized Residual Principle Modified Residual Principle . . Minimal Residual Principle . . Smoothing Functional Principle

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163 167 179 181 183

4 REGULARIZATION OF VARIATIONAL INEQUALITIES 4.1 Variational Inequalities on Exactly Given Sets . . . . . . . . . . . . . 4.2 Variational Inequalities on Approximately Given Sets . . . . . . . . . 4.3 Variational Inequalities with Domain Perturbations . . . . . . . . . . 4.4 Examples of Variational Inequalities . . . . . . . . . . . . . . . . . . 4.5 Variational Inequalities with Unbounded Operators . . . . . . . . . . 4.6 Variational Inequalities with Non-Monotone Perturbations . . . . . . 4.7 Variational Inequalities with Mosco-Approximation of the Constraint 4.8 Variational Inequalities with Hypomonotone Approximations . . . . 4.9 Variational Inequalities with Pseudomonotone Operators . . . . . . . 4.10 Variational Inequalities of Mixed Type . . . . . . . . . . . . . . . . .

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191 191 204 215 219 227 233 236 245 249 254

5 APPLICATIONS OF THE REGULARIZATION METHODS 5.1 Computation of Unbounded Monotone Operators . . . . . . . . . 5.2 Computation of Unbounded Semimonotone Operators . . . . . . 5.3 Computation of Unbounded Accretive Operators . . . . . . . . . 5.4 Hammerstein Type Operator Equations . . . . . . . . . . . . . . 5.5 Pseudo-Solutions of Monotone Equations . . . . . . . . . . . . . 5.6 Minimization Problems . . . . . . . . . . . . . . . . . . . . . . . 5.7 Optimal Control Problems . . . . . . . . . . . . . . . . . . . . . . 5.8 Fixed Point Problems . . . . . . . . . . . . . . . . . . . . . . . .

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259 259 268 271 278 283 294 299 303

6 SPECIAL TOPICS ON REGULARIZATION METHODS 6.1 Quasi-Solution Method . . . . . . . . . . . . . . . . . . . . . . 6.2 Residual Method . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Penalty Method . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Proximal Point Method . . . . . . . . . . . . . . . . . . . . . 6.5 Iterative Regularization Method . . . . . . . . . . . . . . . . 6.6 Iterative-Projection Regularization Method . . . . . . . . . . 6.7 Continuous Regularization Method . . . . . . . . . . . . . . . 6.8 Newton−Kantorovich Regularization Method . . . . . . . . .

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311 311 315 322 328 340 352 363 376

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7 APPENDIX 385 7.1 Recurrent Numerical Inequalities . . . . . . . . . . . . . . . . . . . . . . . . 385 7.2 Differential Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 BIBLIOGRAPHY

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INDEX

407

PREFACE Many applied problems can be reduced to an operator equation of the first kind Ax = f, x ∈ X,

f ∈ Y,

(1)

where an operator A maps from a metric space X into a metric space Y. We are interested in those problems (1) which belong to the class of ill-posed problems. The concept of a well-posedness was introduced by J. Hadamard at the beginning of the 20th century [91]. Definition 1 Let X and Y be metric spaces. The problem (1) is called well-posed if the following conditions are satisfied: (i) the problem is solvable in X for all f ∈ Y ; (ii) its solution x ∈ X is unique; (iii) x ∈ X continuously depends on perturbations of the element f ∈ Y. Problems that do not satisfy even one of the above requirements (i)-(iii) collectively form the class of ill-posed problems. Emphasize that solutions of ill-posed problems (if they exist) are unstable to small changes in the initial data. In connection with this, a common belief of many mathematicians in the past was that well-posedness is a necessary condition for the problems (1) to be mathematically or physically meaningful. This raised a debate about whether or not there is any need for methods of solving ill-posed problems. The tremendous development of science and technology of the last decades led, more often than not, to practical problems which are ill-posed by their nature. Solving such problems became a necessity and, thus, inventing methods for that purpose became a field of research in the intersection of theoretical mathematics with the applied sciences. The fact is that, in practical computations, the date A and f of the problem (1), as a rule, are not precisely known. Therefore, it is important to study the continuous dependence of the approximate solutions on the intrinsic errors involved in the problem date. Without such knowledge, direct numerical resolution of (1) is impossible. At the same time, establishing the continuous dependence mentioned above is not an easy task. Attempts to avoid this difficulty led investigators to the new theory and conceptually new methods for stable solution of ill-posed problems. The expansion of computational technology contributed in large extent to the acceleration of this process. Building the so-called regularization methods for solving ill-posed problems was initiated by A.N. Tikhonov [216]. He proved there the following important result: Theorem 2 A continuous one-to-one operator A which maps a compact subset M of the metric space X into metric space Y has the continuous inverse operator A−1 on the set N = AM ⊂ Y. On the basis of this theorem, M.M. Lavrent’ev introduced the concept of conditionally well-posed problems [125].

viii

PREFACE

Definition 3 The problem (1) is said to be conditionally well-posed if there exists nonempty set X1 ⊆ X such that (i) the equation (1) has a solution in X1 ; (ii) this solution is unique for each f ∈ Y1 = AX1 ⊂ Y ; (iii) the operator A−1 is continuous on Y1 . The subset X1 is called the well-posedness set of problem (1). According to Theorem 2 and Definition 3, we should impose conditions on the operator A in equation (1) such that they will define the compact set X1 . Even if those conditions are given, there are still difficulties that arise when we try to establish solvability of (1) with f ∈ Y1 . These difficulties can be overcome if X1 is understood as the quasi-solution set of equation (1) with f ∈ Y1 [97]. Definition 4 An element x ∈ X1 is called a quasi-solution of equation (1) if it minimizes the residual ρY (Ax, f ) on the set X1 , where ρY is a metric in the space Y. Problems for which it is not possible to construct the compact set X1 of admissible solutions are said to be essentially ill-posed. In order to solve such problems, A.N. Tikhonov proposed the so-called regularization method [217]. To describe this method we establish the following definition assuming, for simplicity, that only the right-hand side f in equation (1) is given with some error δ. Definition 5 An operator R(α, f δ ) : Y → X is called a regularizing operator for equation (1) if it satisfies two requirements: (i) R(α, f δ ) is defined for all α > 0 and all f δ ∈ Y such that ρY (f, f δ ) ≤ δ; (ii) There exists a function α = α(δ) such that R(α(δ), f δ ) = xδα → x as δ → 0, where x is a solution of (1). Operators R(α, f ) lead to a variety of regularization methods. An element xδα is called the regularized solution and α is called the regularization parameter. Thus, by Definition 5, any regularization method has to solve two main problems: A) to show how to construct a regularizing operator R(α, f δ ), B) to show how the regularization parameter α = α(δ) should be chosen in order to ensure convergence of xδα to some x as δ → 0. The regularization method proposed in [217] is given in variational form (see also [43, 99, 130, 167, 218]). It produces a solution xδα as a minimum point of the following smoothing functional: Φαδ (x) = ρ2Y (Ax, f δ ) + αφ(x),

(2)

where φ(x) : X → R1 is a stabilizing functional which has the property that φ(z) ≥ 0 for all z ∈ D(φ) ⊂ X, where D(φ) is a domain of functional φ. Moreover, a solution x of equation (1) needs to be an inclusion x ∈ D(φ). Besides the regularization method described by the functional (2), two more variational methods for solving ill-posed problems are known [99]: the quasi-solution method given by

PREFACE

ix

the minimization problem Axδ − f δ = min{Ax − f δ | x ∈ X1 }, and the residual method in which approximate solutions xδ for equation (1) are found by solving the minimization problem: xδ = min {x | x ∈ Gδ }, where Gδ = {x ∈ D(A) | Ax − f δ ≤ σ(δ)}. Here D(A) denotes a domain of operator A and the positive scalar function σ(δ) has the property that σ(δ) → 0 as δ → 0. The increasing variety of variational regularization methods and growing interest in them are due to the following facts: (i) a priori information about (1) can be used in the process of their construction and (ii) there exists quite a wide choice of different means of solving optimization problems. If the element xδα is found by solving a parametric equation of the type Ax + αBx = f δ ,

(3)

where B is some stabilizing operator, then this procedure is an operator regularization method [54, 124, 125], and the equation (3) is a regularized operator equation. The methods presented above were intensively studied in the case of linear equations. This is connected with the fact that for linear operators there are very powerful investigation tools contained in the spectral theory and optimization theory for quadratic functionals. In this direction many rather delicate characteristics and properties of the mentioned methods have been already obtained [42, 99, 125, 153, 217]. In spite of the fact that some nonlinear problems were solved by regularization methods a long time ago, the progress in this sense was slow. This seems to be due to researchers’ wish to cover as wide classes of nonlinear problems as possible. It is clear that the spectral theory can not be applied directly to nonlinear operators. Moreover, if A is a nonlinear operator, the functionals ρY (Ax, f δ ) and Φαδ (x) are nonconvex, in general. In addition, the question of convexity for the set Gδ remains still open, that produces enormous difficulties in solving variational problems. At the same time, the separate researches into nonlinear problems of the monotone type, like operator equations and variational inequalities with monotone, maximal monotone, semimonotone, pseudomonotone and accretive operators, turned out to be much more successful. From the applied point of view these classes are extremely wide (it is sufficient to recall that gradients and subgradients of convex functionals are monotone and maximal monotone operators, respectively, and nonexpansive mappings are accretive.) A considerable part of this book is devoted to study of nonlinear ill-posed problems with operators in these classes. We use the fundamental results in the theory of monotone and accretive operators discovered during the latter half of the 20th century by R.I. Kachurovskii [102, 103], M.M. Vainberg [220, 221], F.E. Browder [52] - [58], G. Minty [149], J.-L. Lions [128], H. Brezis [49], R. Rockafellar [175] - [181], T. Kato [106, 109] and many others.

x

PREFACE

From the very beginning, the concept of ill-posedness was included in the construction and theoretical foundation of approximative methods for solving nonlinear problems with properly monotone and properly accretive operators, and according to this concept, it is necessary: 1) to assume that a solution of the original problem exists; 2) to construct a sequence of regularized solutions; 3) to determine whether this sequence converges to a solution of problem (1) and the convergence is stable when the initial data are perturbed. Let us briefly describe the contents of the book. It consists of seven chapters that comprise fifty-seven sections. The numeration of sections is in the form X.Y, where X indicates the number of a chapter and Y denotes the current number of the section within the given chapter. The numeration of definitions, theorems, lemmas, corollaries, remarks and propositions is in the form X.Y.Z, where X and Y are as above and Z indicates a number of the statement in Section X.Y. Formulas are numbered analogously. Chapter 1 contains the basic knowledge from convex and functional analysis, optimization, Banach space geometry and theory of monotone and accretive operators which allows the study of the proposed material without turning, generally, to other sources. Equations and variational inequalities with maximal monotone operators are main subjects of our investigations due to their most important applications. Recall that maximal monotone operators are set-valued, in general. In spite of this, the properties of maximal monotone operators, solvability criteria for equations and variational inequalities with such operators and also structures of their solution sets have been studied fully enough. However, it is not always easy to establish the maximal monotonicity of a map. Besides that, the numerical realization of maximal monotone operator means its change by single-valued sections when the property of maximal monotonicity is lost. Finally, the monotone operator of the original problem is not always maximal. Therefore, in this chapter we devote significant time to equations and variational inequalities with semimonotone, pseudomonotone and general monotone mappings. We describe in detail the known properties of the most important canonical monotone operators, so-called, normalized duality mapping J and duality mapping with gauge function J µ . We also present new properties of these mappings expressed by means of geometric characteristics of a Banach space, its modulus of convexity and modulus of smoothness. We introduce the new Lyapunov functionals and show that they have similar properties. In the theory of monotone operators this approach is not traditional and it is first stated in monographic literature. The two last sections of the chapter are dedicated to equations with multiple-valued accretive and d−accretive operators. Chapter 2 is devoted to the operator regularization method for monotone equations (that is, for equations with monotone operators). In order to preserve the monotonicity property in a regularized problem, we add to the original operator some one-parameter family of monotone operators. Thus, the resulting operators are monotone again. As a stabilizing operator, we use a duality mapping which has many remarkable properties. In the case of a Hilbert space, where the duality mapping is the identical operator, this regularization method for linear equations was first studied by M.M. Lavrent’ev [125]. In this chapter, we explore the operator regularization methods for monotone, semimonotone and accretive nonlinear equations in Hilbert and Banach spaces. The equations with single-valued and

PREFACE

xi

multiple-valued maximal monotone mappings are studied. Non-monotone approximations of operators and their domain perturbations is also discussed. Observe that the establishment of boundedness of a sequence of regularized solutions of operator equations is one of the central points of the proof of convergence of any regularization method. It allows us to use the weak compactness principle for bounded sets in reflexive Banach spaces and to construct a subsequence weakly convergent to a solution of a given problem. Further progress appears only when we are able to prove that any weak accumulation point is unique and the weak convergence is in reality a strong convergence (possibly, under some additional conditions). In general, we do not redenote the said subsequence unless not doing so would cause confusion. In Chapter 3 the problem of choosing regularization parameters is solved. We consider the residual principles − the ways of choosing the regularization parameter α by means of the equations that connect the residual of the equation (1) with error levels of the righthand side f and operator A. In the case of an arbitrary monotone operator, the residual is understood in the generalized sense. The smoothing functional principle deals with minimization problems like (2) involving potential monotone operators. The regularization parameter α is found by the equation which connects minimal values mδ (α) of the functional Φαδ (x) on X with the errors of the problem data. In this chapter we also investigate the minimal and modified residual principles. In Chapter 4, we present regularization methods for solving monotone variational inequalities (that is, for solving variational inequalities with monotone operators) on convex closed subsets of Hilbert and Banach spaces. Variational inequalities with bounded and unbounded operators are considered. In contrast to operator equations, the problems become more complicated due to the fact that additional errors of constraint sets are possible here. At that, the proximity between original and perturbed sets is described both by the Hausdorff and Mosco criteria. We also study variational inequalities with pseudomonotone operators. Convergence of the regularization methods for variational inequalities involving non-monotone and hypomonotone approximations of operators is each investigated separately. In the last section of the chapter we discuss the possibility of stable solutions to mixed variational inequalities. In Chapter 5 the operator regularization method is applied to solve the classical problem of calculating values of unbounded operators. Monotone, semimonotone and accretive mappings are considered. Observe that the problem of solving the equation (1) is equivalent to the problem of calculating the inverse operator on a given element f. Nevertheless, each such problem has its specific features, so that their separate investigation is expedient and useful. If the equation is unsolvable, then those elements which minimize its residual (they are called pseudo-solutions) acquire practical importance. We present in this chapter stable algorithms for finding pseudo-solutions of equation (1) with nonlinear monotone mappings based on the operator regularization method. As applications we also consider Hammerstein type equations, minimization problems, optimal control problems and fixed point problems. The first two sections of Chapter 6 are dedicated to the variational methods of quasisolutions and the residual for monotone problems. We establish the equivalence of these methods and also their connection with the operator regularization method. Note that the quasi-solution method and the residual method in this work are constructed in a non-

xii

PREFACE

traditional manner as compared with the linear case [99], though the same information about the initial problem is used. The monotonicity properties of mappings play an essential role in this approach. We proposed and studied the regularized penalty method for monotone variational inequalities. In this chapter, sufficient convergence conditions of the iterative methods have been obtained. We investigated the proximal point method, iterative regularization method, iterative-projection regularization method and Newton−Kantorovich regularization method in uniformly convex and uniformly smooth Banach spaces. Finally, we established the sufficient conditions for convergence of the continuous regularization method which is reduced to the Cauchy problem for a differential equation of the first order. All this variety of methods gives very wide possibilities to obtain strong approximations to a solution of nonlinear ill-posed problem. In Chapter 7 we consider recurrent numerical and differential inequalities.

ACKNOWLEDGMENTS We wish to express our deepest appreciation to Professors Dan Butnariu, Sylvie GuerreDelabriere, Michael Drakhlin, Rafael Esp´ınola, Alfredo Iusem, Athanassios Kartsatos, Zuhair Nashed, Mikhail Lavrent’ev, Simeon Reich, Terry Rockafellar and Jen-Chih Yao for repeatedly giving us the opportunity to discuss various aspects of this book and for their personal support. We would like to mention the following institutions whose support made this work possible: The Technion - Israel Institute of Technology, the University of Haifa (Israel), the Nizhnii Novgorod State Technical University (Russia), the Institute of Pure and Applied Mathematics, Rio de Janeiro (Brazil), the Abdus Salam International Centre for Theoretical Physics, Trieste (Italy), the University Paris VI (France), the University of Sevilia (Spain), and the National Sun Yat-Sen University of Kaohsiung (Taiwan). The authors are very grateful to Dr. Elisabeth Mol and Mrs. Marlies Vlot from Springer Verlag Publishers who helped us with thoughtful arrangements and patience.

Yakov Alber and Irina Ryazantseva

Chapter 1

INTRODUCTION INTO THE THEORY OF MONOTONE AND ACCRETIVE OPERATORS 1.1

Elements of Nonlinear Functional Analysis

Let X be a real linear normed space, x be a norm of an element x in X, θX be an origin of X. Strong convergence xn → x, n = 0, 1, ..., of the sequence {xn } ⊂ X to x ∈ X means that xn − x → 0 as n → ∞. In this case, x is a (strong) limit point of the sequence {xn }. If {xn } converges strongly to x ∈ X then 1) any subsequence {xnk } ⊂ {xn } also converges to the same point, 2) the sequence {xn − ξ} is bounded for any ξ ∈ X. A sequence {xn } ⊂ X is called a fundamental or Cauchy sequence, if for every > 0 there is n0 () such that xm − xn < for any m ≥ n0 () and n ≥ n0 (). If a sequence {xn } ⊂ X converges to a limit then it is fundamental. The inverse assertion is not always true: there are examples of normed spaces which have non-convergent Cauchy sequences. We say that X is complete if every Cauchy sequence of the normed space X has a limit x ∈ X. Let X and Y be arbitrary spaces. The expression A : X → Y means that operator A is single-valued and maps X into Y . If A is multiple-valued then we write A : X → 2Y . We further denote by D(A) = dom A = {x ∈ X | Ax = ∅} (1.1.1) the domain of the operator A, and by R(A) = {φ ∈ Y | φ ∈ Ax, x ∈ D(A)}

(1.1.2)

the range of A. For more details, an operator A : X → Y with D(A) ⊂ X and R(A) ⊂ Y is a one-to-one correspondence between sets of X and Y, which carries point x ∈ D(A) to the point Ax ∈ R(A). A map A : X → 2Y , which is not necessarily a one-to-one correspondence, carries some elements x ∈ D(A) to sets Ax ⊂ R(A). 1

2

1

THEORY OF MONOTONE AND ACCRETIVE OPERATORS

Let R1 be a set of all real numbers. An operator ϕ : X → R1 is called the functional on linear space X. By analogy with (1.1.1), a domain of the functional ϕ is defined as dom ϕ = {x ∈ X | ϕ(x) = ∅}. A functional ϕ is linear if 1) ϕ(x1 + x2 ) = ϕ(x1 ) + ϕ(x2 ) and 2) ϕ(xn ) → ϕ(x) as xn → x. It is bounded if there exists a constant M > 0 such that |ϕ(x)| ≤ M x. The smallest constant M satisfying this inequality is said to be the norm of the linear functional ϕ and it is denoted by |ϕ|. Many important facts of Functional Analysis are deduced from the following theorem and its corollary. Theorem 1.1.1 (Hahn−Banach) Any linear functional ϕ defined on a subspace M of the normed linear space X can be extended onto the whole space with preservation of the norm; that is, there exists a linear functional F, x ∈ X, such that the equalities F (x) = ϕ(x) for all x ∈ M and |ϕ|M = |F |X hold. Corollary 1.1.2 Let X be a normed linear space and x0 ∈ X (x0 = θX ) be an arbitrary fixed element. Then there exists a linear functional ϕ : X → R1 such that: |ϕ| = 1 and ϕ(x0 ) = x0 . In what follows, X will denote a Banach space, i.e., a complete linear normed space. The dual space X ∗ of X is the set of all linear continuous functionals on X. It is known that X ∗ is also a Banach space. We denote by φ∗ the norm of an element φ ∈ X ∗ and by φ, x the dual product (dual pairing) of elements φ ∈ X ∗ and x ∈ X, that is, φ, x is the value of the linear functional φ on the element x. One can show that φ∗ = sup{φ, x | x = 1}. The Cauchy−Schwarz inequality states that |φ, x | ≤ φ∗ x for all x ∈ X and for all φ ∈ X ∗. We say that sequence {xn } ⊂ X converges weakly to x ∈ X (and write xn x) if φ, xn → φ, x for every φ ∈ X ∗ . A weak limit point of any subsequence of the sequence {xn } is called a weak accumulation point. If all weak accumulation points of the sequence {xn } coincide, i.e., if all weakly convergent subsequences have the same limit x, then x is the weak limit of {xn }. If the sequence {xn } weakly converges to x ∈ X then {xn } is bounded. The weak convergence of the sequence {xn } to x always follows from its strong convergence to the same point x. The converse is not true in general. However, in finite-dimensional spaces strong convergence and weak convergence are equivalent. It is also possible to construct the dual space X ∗∗ for X ∗ . In general, X ⊂ X ∗∗ . If X ∗∗ = X then space X is said to be reflexive. Note that, in this case, X ∗ is also a reflexive space. In reflexive spaces, the weak convergence of φn ∈ X ∗ to φ ∈ X ∗ means that for every x ∈ X the limit φn , x → φ, x holds as n → ∞. We present the following important statements. Theorem 1.1.3 (Banach−Steinhaus) Let X be a Banach space, φn ∈ X ∗ , and suppose that the sequence {φn , x } is bounded for every x ∈ X. Then the sequence {φn } is bounded in X ∗ .

1.1 Elements of Nonlinear Functional Analysis

3

Theorem 1.1.4 Suppose that either {φn } ⊂ X ∗ strongly converges to φ ∈ X ∗ and {xn } ⊂ X weakly converges to x ∈ X or {φn } ⊂ X ∗ weakly converges to φ ∈ X ∗ and {xn } ⊂ X strongly converges to x ∈ X. Then limn→∞ φn , xn = φ, x . Definition 1.1.5 [161] A Banach space X satisfies Opial’s condition if for each sequence {xn } in X, the limit relation xn x implies lim inf xn − x < lim inf xn − y n→∞

n→∞

for all y ∈ X with x = y. A set B(a, r) ⊂ X such that x − a ≤ r is called the ball (or closed ball) with center a and radius r. A set B0 (a, r) ⊂ X such that x − a < r is called the open ball with center a and radius r. A set S(a, r) ⊂ X such that x − a = r is called the sphere with center a and radius r. The sphere and ball are called a unit if r = 1. We shall also use the denotations B ∗ (ψ, r), B0∗ (a, r) and S ∗ (a, r), respectively, for closed ball, open ball and sphere in a dual space X ∗ . A set Ω is bounded if it wholly lies inside of some ball. A set Ω is closed (weakly closed) if the conditions xn → x (xn x), where xn ∈ Ω for all n ≥ 0, imply the inclusion x ∈ Ω. For any ψ ∈ X ∗ , ψ = θX ∗ , and for any c ∈ R1 , the set Hc = {x ∈ X | ψ, x = c} is called a closed hyperplane. Definition 1.1.6 Let Ω ⊂ X be a bounded set. Then the value diam Ω = sup {x − y | x, y ∈ Ω} is a diameter of Ω. We denote by ∂Ω the boundary of the set Ω, by int Ω the totality of its interior points and by Ω the closure of the set Ω, that is, minimal closed set containing Ω. Definition 1.1.7 A set Ω ⊆ X is called 1) convex if together with the points x, y ∈ Ω, the whole segment [x, y] = λx + (1 − λ)y, 0 ≤ λ ≤ 1, also belongs to Ω; 2) compact if any infinite sequence of this set contains a convergent subsequence; 3) weakly compact if any infinite sequence of this set contains a weakly convergent subsequence; 4) dense in a set M ⊆ X if M ⊆ Ω; 5) everywhere dense in X if Ω = X.

A closed bounded set Ω of the reflexive space is weakly compact. Hence, from any bounded sequence belonging to Ω, one can choose a subsequence which weakly converges to some element of this space. Theorem 1.1.8 (Mazur) Any closed convex set of a Banach space is weakly closed.

4

1


Theorem 1.1.9 (Riesz) A Banach space is finite-dimensional if and only if each of its bounded closed subsets is compact. The next assertion follows from the Hahn−Banach theorem and is known as the strong separation theorem. Theorem 1.1.10 Let Ω1 ⊂ X and Ω2 ⊂ X be two convex sets, nonempty and disjoint. Suppose that Ω1 is closed and Ω2 is compact. Then there exists a closed hyperplane that strongly separates Ω1 and Ω2 . Definition 1.1.11 A functional ϕ is called convex in dom ϕ if the inequality ϕ(tx + (1 − t)y) ≤ tϕ(x) + (1 − t)ϕ(y)

(1.1.3)

is satisfied for all x, y ∈ dom ϕ and all t ∈ [0,1]. If the equality in (1.1.3) occurs only under the condition that x = y, then the functional ϕ is called strictly convex. If there exists a continuous increasing function γ : [0, ∞) → R1 , γ(0) = 0, such that ϕ(tx + (1 − t)y) ≤ tϕ(x) + (1 − t)ϕ(y) − t(1 − t)γ(x − y) for all x, y ∈ dom ϕ, then ϕ is called uniformly convex. The function γ(t) is called a modulus of convexity of ϕ. If γ(t) = ct2 , c > 0, then the functional ϕ is strongly convex. Definition 1.1.12 A functional ϕ is called lower semicontinuous at the point x0 ∈ dom ϕ if for any sequence xn ∈ dom ϕ such that xn → x0 there holds the inequality ϕ(x0 ) ≤ lim inf ϕ(xn ). n→∞

(1.1.4)

If the inequality (1.1.4) occurs with the condition that the convergence of {xn } to x0 is weak, then the functional ϕ is called weakly lower semicontinuous at x0 . Theorem 1.1.13 Let ϕ : X → R1 be a convex and lower semicontinuous functional. Then it is weakly lower semicontinuous. We present the generalized Weierstrass theorem. Theorem 1.1.14 Assume that a weakly lower semicontinuous functional ϕ is given on a bounded weakly closed set Ω of a reflexive Banach space X. Then it is bounded from below and reaches its greatest lower bound on Ω. In any normed linear space, the norm can be presented as an example of a weakly lower semicontinuous functional. Indeed, let xn x ¯, i.e., ψ, xn → ψ, x ¯ for any ψ ∈ X ∗ . ¯ ¯ ¯ ¯ = ¯ x. This element exists by Corollary 1.1.2. Choose ψ = ψ such that ψ∗ = 1 and ψ, x ¯ xn → ¯ x, that is, Then ψ, ¯ xn ≤ lim inf xn . ¯ x = lim ψ, n→∞

n→∞

Geometric characteristics of Banach spaces such that weak and strong differentiability of the norm, convexity and smoothness of spaces, duality mappings and projection operators will play very important roles in this book.


5

Definition 1.1.15 A functional ϕ : X → R1 is called 1) trivial if ϕ(x) = +∞ for all x ∈ X; 2) proper if M(ϕ) = {x | ϕ(x) = +∞} = ∅; 3) finite if |ϕ(x)| < ∞ for all x ∈ X. In this definition, M(ϕ) is called the effective set of the functional ϕ. If ϕ is convex then M(ϕ) is also convex. Theorem 1.1.16 Any finite convex functional ϕ : X → R1 given on an open set of X is weakly lower semicontinuous. Definition 1.1.17 Let a functional ϕ : X → R1 . We say that ϕ is directionally differentiable at a point x ∈ X if the limit V (x, h) = lim

t→0

ϕ(x + th) − ϕ(x) t

(1.1.5)

exists for all h ∈ X. Similarly, the directional differentiability can be defined when a functional ϕ is given on an open set U in X. In this case, in (1.1.5) x ∈ U and x + th ∈ U. Definition 1.1.18 If the limit in (1.1.5) is linear continuous (with respect to h) operator, i.e., V (x, h) = ϕ (x)h, then ϕ is called Gˆ ateaux differentiable (or weakly differentiable) at ateaux differential (or weak a point x ∈ X and V (x, h) and ϕ (x) are called, respectively, Gˆ differential) and Gˆ ateaux derivative (or weak derivative) of the functional ϕ at a point x. By Definition 1.1.18, there exists operator A : X → X ∗ such that V (x, h) = Ax, h for any x, h ∈ X. The operator A is called the gradient of a functional ϕ(x) and is denoted by grad ϕ or ϕ . Operator A is called a potential if there exists a functional ϕ such that A = grad ϕ. In this case, ϕ is the potential of A. Theorem 1.1.19 If ϕ(x) is a convex functional in dom ϕ, then its gradient ϕ (x) satisfies the inequality ϕ (x) − ϕ (y), x − y ≥ 0 (1.1.6) for all x, y ∈ dom ϕ. If ϕ(x) is uniformly convex then ϕ (x) − ϕ (y), x − y ≥ 2γ(x − y).

(1.1.7)

In particular, if ϕ(x) is strongly convex then ϕ (x) − ϕ (y), x − y ≥ 2cx − y2 .

(1.1.8)

6

1


Theorem 1.1.20 If ϕ(x) is a convex functional in dom ϕ, then its gradient ϕ (x) satisfies the inequality (1.1.9) ϕ (x), x − y ≥ ϕ(x) − ϕ(y) for all x, y ∈ dom ϕ. If ϕ(x) is uniformly convex then ϕ (x), x − y ≥ ϕ(x) − ϕ(y) + γ(x − y).

(1.1.10)

In particular, if ϕ(x) is strongly convex then ϕ (x), x − y ≥ ϕ(x) − ϕ(y) + cx − y2 .

(1.1.11)

Theorem 1.1.21 Let a functional ϕ : X → R1 be convex and differentiable at a point x0 . Then x0 is the extremum point of ϕ if and only if grad ϕ(x0 ) = 0. Remark 1.1.22 By tradition, we shall denote “0” as a null on the right-hand side of any operator equation. The sufficient existence conditions of the functional minimum point is given by the following theorem. Theorem 1.1.23 Let X be a reflexive Banach space, ϕ : X → R1 be a convex lower semicontinuous proper functional and ϕ(x) → ∞ for x → ∞. Then there exists a point x0 ∈ dom ϕ such that ϕ(x0 ) = min{ϕ(x) | x ∈ dom ϕ}. The minimum point x0 is unique if the functional ϕ is strictly convex. Theorem 1.1.24 Let X be a reflexive Banach space, ϕ : X → R1 be a uniformly convex lower semicontinuous functional on a convex closed set Ω ⊆ X. Then there exists a unique minimum point x∗ of ϕ at Ω and inequality γ(x − x∗ ) ≤ ϕ(x) − ϕ(x∗ ) ∀x ∈ Ω holds. Here γ(t) is the modulus of convexity of ϕ at Ω. Definition 1.1.25 A proper functional ϕ : X → R1 is said to be Fréchet differentiable (or strongly differentiable) at a point x ∈ D, where D is an open set in X, if there is a linear operator F : X → X ∗ such that, for any x + h ∈ D, ϕ(x + h) = ϕ(x) + F (x), h + ω(x, h) and lim

h→0

ω(x, h) = 0. h

The quantity F (x), h is called the Fréchet differential (or strong differential) and F (x) = ϕ (x) is called the Fréchet derivative (or strong derivative) of the functional ϕ at a point x.


7

In other word, if the limit in (1.1.5) exists uniformly for h on the unit sphere of X, then ϕ is Fréchet differentiable and ϕ (x) is the Fréchet derivative of ϕ at x. If a functional ϕ ateaux differentiable at that point. is Fréchet differentiable at a point x0 ∈ X, then it is Gˆ The inverse assertion is not true in general. However, if Gâteaux derivative ϕ is continuous at a neighborhood of a point x0 ∈ X, then it is, in fact, the Fréchet derivative at x0 . Definition 1.1.26 The space X is smooth if its norm is Gˆ ateaux differentiable. The space X is strongly smooth if its norm is Fréchet differentiable over θX . Next we introduce one of the most important canonical operators in a Banach space. ∗

Definition 1.1.27 Let X be an arbitrary Banach space. The operator J : X → 2X is called a normalized duality mapping in X if the following equalities are satisfied: ζ, x = ζ∗ x = x2 ∀ζ ∈ Jx, ∀x ∈ X. It immediately follows from Definition 1.1.27 that J is a homogeneous and odd operator. This means that, respectively, J(λx) = λJx for λ ≥ 0 and J(−x) = −Jx. In general, the normalized duality mapping is a multiple-valued operator. However, it is single-valued in a smooth Banach space and then Jx = 2−1 grad x2 .

(1.1.12)

Definition 1.1.28 A space X is called strictly convex if the unit sphere in X is strictly convex, that is, the inequality x + y < 2 holds for all x, y ∈ X such that x = y = 1, x = y. Definition 1.1.29 Fix a point x ∈ X and a number > 0 and then define the function

δX (x, ) = inf 1 −

x − y x = y = 1, x − y = . 2

It is said to be the modulus of the local convexity of space X at a point x. If for any x with x = 1, δX (x, ) > 0 for > 0, then the space X is called locally uniformly convex. Any locally uniformly convex space is strictly convex. In any reflexive Banach space, it is possible to introduce an equivalent norm such that the space will be locally uniformly convex with respect to this norm. If the function δX (x, ) does not depend on x such that in Definition 1.1.29 δX (x, ) = δX (), then δX () is the modulus of convexity of X. Definition 1.1.30 A Banach space X is called uniformly convex if for any given > 0 there exists δ > 0 such that for all x, y ∈ X with x ≤ 1, y ≤ 1, x − y = the inequality x + y ≤ 2(1 − δ) holds.

8

1


The function δX () is defined on the interval [0, 2], continuous and increasing on this interval, δX (0) = 0 and δX () ≤ 1. A Banach space X is uniformly convex if and only if δX () > 0 for > 0. Observe that the function gX () =

δ()

(1.1.13)

will play a very important role in our researches. It is known that gX () is a continuous and non-decreasing function on the interval [0, 2], and gX (0) = 0. Definition 1.1.31 The function ρX (τ ) defined by the formula ρX (τ ) = sup

x + y

2

+

x − y − 1 x = 1, y = τ 2

is the modulus of smoothness of the space X. Definition 1.1.32 A Banach space X is called uniformly smooth if for any given > 0 there exists δ > 0 such that for all x, y ∈ X with x = 1, y ≤ δ the inequality 2−1 (x + y + x − y) − 1 ≤ y holds. Note that ρX (τ ) is defined on the interval [0, ∞), convex, continuous and increasing on this interval, ρX (0) = 0. In addition, for any X the function ρX (τ ) ≤ τ for all τ ≥ 0. A Banach space X is uniformly smooth if and only if lim

τ →0

ρX (τ ) = 0. τ

Any uniformly convex and any uniformly smooth Banach space is reflexive. A space X is uniformly smooth if and only if X ∗ is uniformly convex. A space X is uniformly convex if and only if X ∗ is uniformly smooth. A reflexive Banach space X is smooth if and only if X ∗ is strictly convex. A reflexive Banach space X is strictly convex if and only if X ∗ is smooth. Definition 1.1.33 The norm in X is uniformly Fréchet differentiable if lim

t→0

x + th − x t

exixts uniformly for x and h in the unit sphere of X. ˘ Theorem 1.1.34 (Klee−Smulian) A Banach space X is uniformly smooth if and only if the norm in X is uniformly Fréchet differentiable.


9

Next we present several examples of uniformly convex and uniformly smooth Banach spaces. 1. A complete linear space X is called a real Hilbert space H if to every pair of elements x, y ∈ X there is associated a real number called their scalar (inner) product and denoted by (x, y), in such a way that the following rules are satisfied: 1) (x, y) = (y, x); 2) (x + z, y) = (x, y) + (z, y); 3) (λx, y) = λ(x, y) for all λ ∈ R1 ; 4) (x, x) > 0 for x = θH and (x, x) = 0 for x = θH . The norm of an element x ∈ H is then defined as x = (x, x). The main characterization of Hilbert spaces is the parallelogram equality 2x2 + 2y2 = x + y2 + x − y2 ,

(1.1.14)

which also enables us to define its modulus of convexity and modulus of smoothness. It can be proven that H is a uniformly convex and uniformly smooth space. Among all uniformly convex and uniformly smooth Banach spaces, a Hilbert space has the greatest modulus of convexity and smallest modulus of smoothness, that is, δH () ≥ δX () and ρH (τ ) ≤ ρX (τ ). Observe that the parallelogram equality (1.1.14) is the necessary and sufficient condition to be a Hilbert space. The formula (1.1.12) shows that the normalized duality mapping J in H is the identity operator I. 2. The Lebesgue space Lp (G), ∞ > p > 1, of the measurable functions f (x) such that

|f (x)|p dx < ∞, x ∈ G,

G

and G is a measurable set in with respect to the norm

Rn ,

is a uniformly convex and uniformly smooth Banach space

f Lp =

|f (x)|p dx

1/p

.

G

Dual to Lebesgue space Lp (G) with p > 1 is the Lebesgue space Lq (G) with q > 1 such that p−1 + q −1 = 1. Recall that in spaces Lp (G) the Hölder integral inequality

G

|f (x)g(x)|dx ≤

G

|f (x)|p dx

1/p

1/q

G

|g(x)|q dx

and the Minkovsky integral inequality

1/p

G

|f (x) + g(x)|p dx

≤

1/p

G

|f (x)|p dx

hold for all f (x) ∈ L (G) and for all g(x) ∈ L (G). p

q

1/p

+ G

|g(x)|p dx

10

1


3. The space lp , ∞ > p > 1, of number sequences x = {ξ1 , ξ2 , ..., ξj , ...}, such that ∞

|ξj |p < ∞

j=1

is a uniformly convex and uniformly smooth Banach space with respect to the norm xlp =

∞

|ξj |p

1/p

.

j=1

Dual to the space lp , ∞ > p > 1, is the space lq , ∞ > q > 1, such that p−1 + q −1 = 1. We note the following very useful properties of the functional φ(x) = xs on the spaces Lp and lp , ∞ > p > 1 (see [228]): φ(x) is uniformly convex on the whole Lp and on the whole lp if s ≥ p ≥ 2. It is not true if s ∈ (1, 2). In addition, the functional φ(x) = xsX with any ∞ > s > 1 is uniformly convex on each convex bounded set of the uniformly convex space X. p (G). For simplicity, let G be a bounded set on a 4. Introduce the Sobolev spaces Wm plane with sufficiently smooth boundary, C0∞ (G) be a set of functions continuous on G and equal to zero on ∂G, together with their derivatives of all orders. Consider functions v(x, y) of the space Lp (G), ∞ > p > 1. If there exists the function χ(x, y) ∈ Lp (G) such that for all u(x, y) ∈ C0∞ (G) the equality

∂lu v(x, y) dx dy = (−1)l ∂xl1 ∂y l2

G

u(x, y) χ(x, y) dx dy, G

l = l1 + l2 ,

holds, then the function χ(x, y) is said to be an l-order generalized derivative of the function v(x, y). We denote it as the usual derivative: χ(x, y) =

∂lv . ∂y l2

∂xl1

The functions v(x, y) such that G

|v(x, y)|p dx dy +

1≤l≤m

G

∂ l v p dx dy < ∞, l = l1 + l2 , ∂y l2

∂xl1

p form a Sobolev space Wm (G) with the norm

vWmp =

G

|v(x, y)|p dx dy +

1≤l≤m

G

1/p ∂ l v p . dx dy l 2 ∂y

∂xl1

◦

(1.1.15)

p p A closure of C0∞ (G) in the metric of Wm (G) is denoted as Wm (G). It is known that ◦

p p Wm (G) ⊂ Wm (G), and ◦ ∂lu p (G), l = l1 + l2 , 0 ≤ l ≤ m − 1. |∂G = 0 for u ∈Wm ∂ l1 x∂ l2 y

1.1 Elements of Nonlinear Functional Analysis The norm v

◦ p Wm

=

m=m1 +m2

is equivalent to (1.1.15).

G

11

p 1/p ∂mv dx dy ∂xm1 ∂y m2

◦

p p The Sobolev spaces Wm (G) and Wm (G) are also uniformly convex and uniformly smooth for ∞ > p > 1, and they can be defined for any bounded measurable domain q p G in Rn . The dual space of Wm (G) is the Banach space W−m (G) with ∞ > q > 1, such that p−1 + q −1 = 1. Next we introduce the Friedrichs’ inequality which is often used in applications. Let G be a bounded domain of points x = x(x1 , ..., xn ) of Rn , the boundary ∂G be Lipschitzcontinuous, f ∈ W12 . Then

G

f 2 dG ≤ k

n

∂f 2 G i=1

∂xi

dG +

f 2 d(∂G) ,

∂G

where k is a constant independent of n and is completely determined by the domain G. A similar inequality is valid in more general spaces. Consider the imbedding operator E which is defined for any function v(x, y) ∈ Wlp (G) p and carries v(x, y) into the same function considered now as an element of the space Wm (G). p (G) for any l < m is imbedded into the space Theorem 1.1.35 (Sobolev) The space Wm p Wl (G).

This theorem means that every function v(x, y) having all m-order generalized derivatives has also all l-order generalized derivatives for l < m. Moreover, there exists a constant C > 0 such that vW p (G) ≤ CvWmp (G) . l

It is clear that the imbedding operator E is bounded and linear, consequently, continuous. In addition, the following inclusions take place: r p Wm (G) ⊂ Wm (G) ⊂ Lp (G), 0 < p ≤ r < ∞.

More generally, let X, Y be Banach spaces and X ⊆ Y. We say that the operator E : X → Y with D(E) = X is an imbedding operator of X into Y if it carries each element x ∈ X into itself, that is, Ex = x. Definition 1.1.36 The space X has the Kadeˇc−Klee property if, for any sequence {xn }, the weak convergence xn x and convergence of the norms xn → x imply strong convergence xn → x. Definition 1.1.37 A reflexive Banach space X is said to be E-space if it is strictly convex and has the Kadeˇc−Klee property. Hilbert spaces as well as reflexive locally uniformly convex spaces are E-spaces. Therep fore, Lp , lp , Wm , ∞ > p > 1, are also E-spaces.

12

1


Theorem 1.1.38 X is an E-space if and only if X ∗ is strongly smooth. Definition 1.1.39 A space X is called separable if in this space there exists a countable everywhere dense set. In other words, a space X is separable if for any element x ∈ X, there exists a sequence {xn } ⊂ X such that there is a subsequence {xnk } of {xn } which converges to x. Next we define diverse properties of an operator. Definition 1.1.40 An operator A : X → 2Y is bounded if it carries any bounded set of X to a bounded set of Y . If R(A) ⊂ Y is a bounded set, then A is called a uniformly bounded operator. Definition 1.1.41 The operator A−1 : Y → 2X is called inverse to an operator A : X → 2Y if the set of values of A−1 at any point y ∈ R(A) is the set {x | y ∈ Ax}. ∗

Definition 1.1.42 An operator A : X → 2X is said to be coercive if there exists a function c(t) defined for t ≥ 0 such that c(t) → ∞ as t → ∞, and the inequality y, x ≥ c(x)x holds for all x ∈ D(A) and for all y ∈ Ax. ∗

Definition 1.1.43 An operator A : X → 2X is said to be coercive relative to a point x0 ∈ X if there exists a function c(t) defined for t ≥ 0 such that c(t) → ∞ as t → ∞, and the inequality y, x − x0 ≥ c(x)x holds for all x ∈ D(A) and for all y ∈ Ax. Theorem 1.1.44 If the functional ϕ(x) is uniformly convex (in particular, strongly convex) then Ax = grad ϕ(x) is coercive. ∗

Definition 1.1.45 An operator A : X → 2X is called weakly coercive if y → ∞ as x → ∞ for all y ∈ Ax. Emphasize that a coercive operator has a bounded inverse. This fact can be easily obtained by a contradiction. Definition 1.1.46 An operator A : X → Y is called compact on the set Ω if it carries any bounded subset of Ω into a compact set of Y. Theorem 1.1.47 (Schauder principle) Let X be a Banach space. Assume that Ω ⊂ X is a convex closed and bounded set. If the map A is compact on Ω and A(Ω) ⊆ Ω, then the equation Ax = x has at least one solution in Ω.


13

Definition 1.1.48 An operator A : X → Y is called 1) continuous at a point x0 ∈ D(A) if Axn → Ax0 as xn → x0 ; 2) hemicontinuous at a point x0 ∈ D(A) if A(x0 + tn x) Ax0 as tn → 0 for any vector x such that x0 + tn x ∈ D(A) and 0 ≤ tn ≤ t(x0 ); 3) demicontinuous at a point x0 ∈ D(A) if for any sequence {xn } ⊂ D(A) such that xn → x0 , the convergence Axn Ax0 holds (it is evident that hemicontinuity of A follows from its demicontinuity); 4) Lipschitz-continuous if there exists a constant l > 0 such that Ax1 −Ax2 Y ≤ lx1 −x2 X for all x1 , x2 ∈ X; 5) strongly continuous if xn x implies Axn → Ax; 6) completely continuous on the set Ω if it is continuous and compact on Ω; 7) weak-to-weak continuous (or sequentially weakly continuous) at a point x0 ∈ D(A) if for any sequence {xn } ⊂ D(A) such that xn x0 , the convergence Axn Ax0 holds. The corresponding properties are fulfilled in D(A) if they are valid at each point of D(A). Definition 1.1.49 Let an operator A : X → 2Y . We say that 1) A is closed at D(A) if the relations xn → x, yn → y, xn ∈ D(A), yn ∈ Axn for all n ≥ 0, imply the inclusion y ∈ Ax; 2) A is weakly closed at D(A) if the relations xn x, yn y, xn ∈ D(A), yn ∈ Axn for all n ≥ 0, imply the inclusion y ∈ Ax. Definition 1.1.50 An operator A1 : X → 2Y is said to be the extension of an operator A : X → 2Y if D(A) ⊆ D(A1 ) and Ax = A1 x for all x ∈ D(A). Definition 1.1.51 An operator A : X → Y is called linear if it is additive, i.e., A(x + y) = Ax + Ay ∀x, y ∈ X, and homogeneous, i.e., A(λx) = λAx ∀x ∈ X,

∀λ ∈ R1 .

Definition 1.1.51 also yields the definition of a linear functional ϕ in X when Y = R1 , that is, A = ϕ : X → R1 . Definition 1.1.52 Assume that X and Y are linear normed spaces and A : X → Y is a linear bounded operator. The smallest constant M satisfying the inequality AxY ≤ M xX is said to be the norm of the operator A, and it is denoted by |A|. Definition 1.1.53 Let X and Y be linear normed spaces, A : X → Y be a nonlinear operator, D(A) be an open set. A is said to be Fréchet differentiable (or strongly differentiable) at a point x ∈ D(A) if there exists a linear continuous operator A (x) : X → Y such that for all h ∈ X, A(x + h) − Ax = A (x)h + ω(x, h),

14

1


where x + h ∈ D(A) and ω(x, h) → 0 as h → 0. h Respectively, A (x)h and A (x) are called the Fréchet differential (or strong differential) and the Fréchet derivative (or strong derivative) of the operator A at a point x. Definition 1.1.54 Let X and Y be linear normed spaces, A : X → Y be a nonlinear operator, D(A) be an open set. If there exists a linear operator A (x) : X → Y such that lim

t→0

A(x + h) − Ax = A (x)h, t

then A is called Gˆ ateaux differentiable (or weakly differentiable). Respectively, A (x)h and ateaux differential (or weak differential) and the Gˆ ateaux derivative A (x) are called the Gˆ (or weak derivative) of the operator A at a point x. Theorem 1.1.55 Let X be a linear normed space, Y be a Banach space and A be a nonlinear operator acting from a linear dense set of X to Y which has a bounded Gˆ ateaux differential at each point of the set D = U ∩ X, where U is some neighborhood of x ∈ X. If the Gˆ ateaux derivative is continuous at a point x0 ∈ D, then it is, in fact, the Fréchet derivative there. Let X be a Banach space, X ∗ be its dual space, A : X → X ∗ be a nonlinear operator having the Fréchet derivatives on D(A) up to order n + 1, and [x, x + h] ⊂ D(A). Then

A(x + h), y = Ax + A (x)h + +

A (x)h2 A(n) (x)hn ,y + ◦ ◦ ◦ + n! 2!

A(n+1) (x + θh)hn+1

(n + 1)!

,y

(1.1.16)

∀y ∈ X,

where θ = θ(y) satisfies the inequality 0 < θ < 1. We call (1.1.16) the Taylor formula and its particular case A(x + h) − Ax, y = A (x + θh)h, y ∀y ∈ X ∗ the Lagrange formula. Let Ω be a convex closed subset of X. The operator PΩ is called a metric projection operator if it assigns to each x ∈ X its nearest point y ∈ Ω such that x − y = min{x − z | z ∈ Ω}.

(1.1.17)

It is known that the metric projection operator PΩ is continuous in a uniformly convex Banach space X and uniformly continuous on each bounded set of X if, in addition, X is uniformly smooth. An element y is called the metric projection of x onto Ω and denoted by PΩ x. It exists and is unique at any point of the reflexive strictly convex space. Observe that y is often called the best approximation of x because the quantity d(x, Ω) = x − y is the distance from x ∈ X to the subset Ω ⊂ X.


15

Definition 1.1.56 A linear bounded operator P is called a projector of the space X onto Y if R(P ) = Y and P 2 = P. Definition 1.1.57 A Banach space X possesses the approximation if there exists a directed family of finite-dimensional subspaces {Xn } ordered by inclusion, and a corresponding family of projectors Pn : X → Xn such that |Pn | = 1 for all n = 0, 1, , ... , and n Xn is dense in X. Let us consider two sets Ω1 and Ω2 in X. Let β(Ω1 , Ω2 ) = sup{d(x, Ω2 ) | x ∈ Ω1 } be a semideviation of the set Ω1 from Ω2 . Then the Hausdorff distance between the sets Ω1 and Ω2 is defined by the formula HX (Ω1 , Ω2 ) = max{β(Ω1 , Ω2 ), β(Ω2 , Ω1 )}. The distance HX ∗ (Ω1 , Ω2 ) between the sets Ω1 and Ω2 in a space X ∗ is introduced analogously. Definition 1.1.58 [154] The set sequence {Ωn }, Ωn ⊂ X, is Mosco-convergent to the set Ω (we write “M-convergent” in short), if (i) Ω = s − lim inf n→∞ Ωn , that is, for any element x ∈ Ω it is possible to construct the sequence {xn } such that xn ∈ Ωn and xn → x; (ii) Ω = w − lim supn→∞ Ωn , that is, if xn ∈ Ωn and xn x ∈ X then x ∈ Ω. Definition 1.1.59 Let Ω be a set of elements x, y, z, ... . Suppose there is a binary relation defined between certain pairs (x, y) of elements of Ω, expressed by x ≺ y, with the properties:   

x ≺ x; if x ≺ y and y ≺ x then x = y;   if x ≺ y and y ≺ z then x ≺ z. Then Ω is said to be semi-ordered by the relation ≺ . Definition 1.1.60 A semi-ordered set Ω is said to be linearly ordered if for every pair (x, y) in Ω, either x ≺ y or y ≺ x. Lemma 1.1.61 (Zorn) Let Ω be a nonempty semi-ordered set with the property that every linearly ordered subset of Ω has an upper bound in Ω. Then Ω contains at least one maximal element. n

Finally, we provide the following important theorem in a space Rn . Let A : Rn → 2R be a certain operator. Denote R(Ax) the closed convex hull of all limit points of sequence {Axk }, where xk → x, taking also into account infinite limit points in all directions. n

Theorem 1.1.62 Let A : Rn → 2R be an operator defined on a closed convex set Ω with the boundary ∂Ω, θRn ∈ int Ω and (y, x) ≥ 0 if x ∈ ∂Ω, y ∈ Ax. Then there exists a point x0 in Ω such that θRn ∈ R(Ax0 ).

16

1

1.2


Subdifferentials

Let X be a reflexive Banach space, ϕ : X → R1 be a proper functional with domain D(ϕ) = X. Definition 1.2.1 An element w ∈ X ∗ satisfying the inequality ϕ(y) ≥ ϕ(x) + w, y − x ∀y ∈ X,

(1.2.1)

is said to be a subgradient of the functional ϕ at a point x. The subgradient is often called a supporting functional. ∗

Definition 1.2.2 An operator ∂ϕ : X → 2X is called a subdifferential of the functional ϕ if and only if (1.2.1) is fulfilled for w ∈ ∂ϕ(x). Thus, the set {w | w ∈ ∂ϕ(x)} is the totality of all subgradients of the functional ϕ at a point x. Establish the connection between the subdifferential and the gradient of ϕ. Lemma 1.2.3 Let ϕ be a proper convex functional on X. If ϕ is Gˆ ateaux differentiable at a point x ∈ X, then there exists only one subgradient of the functional ϕ at this point and ∂ϕ(x) = ϕ (x). Proof. Since ϕ is convex, we can write

ϕ(x + t(y − x)) ≤ ϕ(x) + t ϕ(y) − ϕ(x) or

∀x, y ∈ X, 0 < t < 1,

ϕ(x + t(y − x)) − ϕ(x) ≤ ϕ(y) − ϕ(x). t

Letting t → 0 one gets ϕ (x), y − x ≤ ϕ(y) − ϕ(x) ∀y ∈ X. This means that ϕ (x) ∈ ∂ϕ(x). Suppose that w = ϕ (x) and w ∈ ∂ϕ(x). Then (1.2.1) with x + ty in place of y implies ϕ(x + ty) − ϕ(x) ≥ w, y ∀y ∈ X, t

and if t → 0 then we obtain ϕ (x) − w, y ≥ 0 ∀y ∈ X. This is possible only if ϕ (x) = w. Lemma 1.2.4 Assume that ∂ϕ is a single-valued hemicontinuous subdifferential of ϕ on X. Then ϕ has the Gˆ ateaux derivative ϕ and ∂ϕ(x) = ϕ (x) for all x ∈ X.

1.2

Subdifferentials

17

Proof. By (1.2.1), for all y ∈ X and t > 0, ϕ(x + ty) − ϕ(x) ≥ ∂ϕ(x), ty . Therefore,

ϕ(x + ty) − ϕ(x) ≥ ∂ϕ(x), y . t

lim inf t→0

Thus, it follows from (1.2.1) that ϕ(x) − ϕ(x + ty) ≥ −∂ϕ(x + ty), ty . By virtue of the hemicontinuity of ∂ϕ, lim sup t→0

ϕ(x + ty) − ϕ(x) ≤ ∂ϕ(x), y , t

i.e., ϕ (x) = ∂ϕ(x). Note that D(∂ϕ) ⊆ dom ϕ. In addition, it is known by (1.2.1) that the set of values of the operator ∂ϕ at each point x ∈ D(∂ϕ) is convex and weakly closed. The following lemmas emphasizes the exceptional importance of the subdifferential concept for applications. Lemma 1.2.5 A functional ϕ : X → R1 has the minimum at a point x ∈ D(∂ϕ) if and only if θX ∗ ∈ ∂ϕ(x). Proof. By (1.2.1), if θX ∗ ∈ ∂ϕ(x) then we have ϕ(y) ≥ ϕ(x), that is, ϕ(x) = min{ϕ(y) | y ∈ X}.

(1.2.2)

Let (1.2.2) hold. Then ϕ(y) − ϕ(x) ≥ 0 for all y ∈ X. By the definition of subdifferential, it follows from this that θX ∗ ∈ ∂ϕ(x). Lemma 1.2.6 If a functional ϕ on the open convex set M ⊂ dom ϕ has a subdifferential, then ϕ is convex and lower semicontinuous on this set. Proof. Since M is a convex set, then z = (1 − t)x + ty ∈ M for all x, y ∈ M and t ∈ [0, 1]. By (1.2.1), we conclude for w ∈ ∂ϕ(z) that ϕ(x) ≥ ϕ(z) + w, x − z and ϕ(y) ≥ ϕ(z) + w, y − z . Multiplying these inequalities by 1 − t and t, respectively, and adding them together, we obtain (1 − t)ϕ(x) + tϕ(y) ≥ ϕ((1 − t)x + ty). That means that the functional ϕ is convex, as claimed. Let x0 ∈ M, xn ∈ M for all n ≥ 0, xn → x0 , w ∈ ∂ϕ(x0 ). Then by the definition of ∂ϕ(x0 ) ∈ X ∗ , we have ϕ(xn ) ≥ ϕ(x0 ) + w, xn − x0 . Hence, lim inf ϕ(xn ) ≥ ϕ(x0 ), xn →x0

which means that ϕ is lower semicontinuous on M.

18

1


Lemma 1.2.7 If the operator A : X → X ∗ is hemicontinuous and ϕ (x) = Ax, then ϕ(x) − ϕ(θX ) =

1 0

A(tx), x dt ∀x ∈ X.

Proof. Consider the function ψ(t) = ϕ(tx). Then ψ (t) = lim

τ →0

ϕ(tx + τ x) − ϕ(tx) = A(tx), x . τ

Taking into account the hemicontinuity of A we have ϕ(x) − ϕ(θX ) = ψ(1) − ψ(0) =

1 0

ψ (t)dt =

1 0

A(tx), x dt.

Let us give a sufficient condition for the subdifferential existence. Theorem 1.2.8 Let ϕ : X → R1 be a proper convex lower semicontinuous functional. Then ϕ has a subdifferential at int dom ϕ. From the definition of ∂ϕ it follows that ∂(λϕ) = λ∂ϕ for all λ > 0. The additivity property of the subdifferential is established by the following theorem. Theorem 1.2.9 Let ϕ1 and ϕ2 be convex functionals on X and let there exist at least one point z ∈ dom ϕ1 ∩ dom ϕ2 such that one of these functionals is continuous at z. Then ∂(ϕ1 + ϕ2 ) = ∂ϕ1 + ∂ϕ2 . Definition 1.2.10 If a functional ϕ on X is non-trivial, then a functional ϕ∗ on X ∗ , defined by the formula ϕ∗ (x∗ ) = sup{x∗ , x − ϕ(x) | x ∈ X},

(1.2.3)

is called conjugate to ϕ. This definition implies the Young−Fenchel inequality: ϕ∗ (x∗ ) + ϕ(x) ≥ x∗ , x .

(1.2.4)

Theorem 1.2.11 Assume that ϕ is a weakly lower semicontinuous, convex and finite functional and ∂ϕ is its subdifferential. Then x∗ ∈ ∂ϕ(x) if and only if x∗ , x = ϕ(x) + ϕ∗ (x∗ ). Proof. Let x∗ ∈ ∂ϕ(x) be given. We have x∗ , y − ϕ(y) ≤ x∗ , x − ϕ(x) ∀y ∈ X. Hence, ϕ(x) + ϕ∗ (x∗ ) = ϕ(x) + sup {x∗ , y − ϕ(y) | y ∈ X} ≤ ϕ(x) + x∗ , x − ϕ(x) = x∗ , x .

(1.2.5)

1.3

Monotone Operators

19

Then the Young−Fenchel inequality implies (1.2.5). Let now (1.2.5) be valid. Then, by definition of ϕ∗ , one gets x∗ , x ≥ ϕ(x) + x∗ , y − ϕ(y) ∀y ∈ X, that is, ϕ(y) − ϕ(x) ≥ x∗ , y − x . Thus, x∗ ∈ ∂ϕ(x). Theorem 1.2.12 Suppose that ϕ : X → R1 is a proper convex lower semicontinuous functional. Then R(∂ϕ) = X ∗ if and only if for all w ∈ X ∗ , ϕ(x) − w, x → +∞ as x → ∞.

(1.2.6)

Proof. (i) Consider the functional Φ(x) = ϕ(x) − w, x for an arbitrary w ∈ X ∗ . It is proper convex lower semicontinuous and lim Φ(x) = +∞.

x→∞

Then, by Theorem 1.1.23, there exists x0 ∈ X such that Φ(x) reaches its minimum at this point. Therefore, ϕ(x) − w, x ≥ ϕ(x0 ) − w, x0 ∀x ∈ D(ϕ), i.e., w ∈ ∂ϕ(x0 ) and R(∂ϕ) = X ∗ . (ii) Let R(∂ϕ) = X ∗ be given. Prove (1.2.6) by the contradiction. Suppose that {xn } ⊂ X, xn → ∞ and the sequence {Φ(xn )} is bounded. Take an element g ∈ X ∗ such that g, xn → ∞ as xn → ∞. By the condition, there exists an element x ¯ ∈ D(∂ϕ) such that w + g ∈ ∂ϕ(¯ x). Then the inequality x) + w + g, x ¯ g, xn ≤ ϕ(xn ) − w, xn − ϕ(¯ follows from the subdifferential definition. Thus, {g, xn } is bounded which contradicts the choice of the element g.

1.3

Monotone Operators ∗

Let X be a reflexive Banach space, X ∗ its dual space and A : X → 2X . Definition 1.3.1 The set of pairs (x, f ) ∈ X × X ∗ such that f ∈ Ax is called the graph of an operator A and it is denoted by grA. Definition 1.3.2 A set G ⊆ X × X ∗ is called monotone if the inequality f − g, x − y ≥ 0 holds for all pairs (x, f ) and (y, g) from G.

20

1

THEORY OF MONOTONE AND ACCRETIVE OPERATORS ∗

Definition 1.3.3 An operator A : X → 2X is monotone if its graph is a monotone set, i.e., if for all x, y ∈ D(A), f − g, x − y ≥ 0 ∀f ∈ Ax, ∀g ∈ Ay.

(1.3.1)

It is obvious that if the operator A is monotone then the operators A(x + x0 ) and Ax + w0 are also monotone, where x0 ∈ X and w0 ∈ X ∗ are fixed elements. It is not difficult to verify that if A and B are monotone operators then the sum A + B, the product λA, λ > 0, and the inverse operator A−1 are monotone operators as well. For a linear ∗ operator A : X → 2X the monotonicity condition is equivalent to its non-negativity: g, x ≥ 0 ∀g ∈ Ax, ∀x ∈ D(A). ∗

Definition 1.3.4 An operator A : X → 2X is strictly monotone if the equality in (1.3.1) holds only for x = y. Proposition 1.3.5 If among two monotone operators A and B at least one is strictly monotone, then the sum A + B is strictly monotone. It is possible to give another definition of the monotone operator in a Hilbert space H. Definition 1.3.6 An operator A : H → 2H is said to be monotone if x − y ≤ x − y + λ(f − g)

(1.3.2)

for all x, y ∈ D(A), f ∈ Ax, g ∈ Ay and λ ≥ 0. Theorem 1.3.7 Definitions 1.3.3 and 1.3.6 are equivalent. Proof. Suppose that (1.3.1) is satisfied. Then (1.3.2) follows from the equality x − y + λ(f − g)2 = x − y2 + 2λ(f − g, x − y) + λ2 f − g2 .

(1.3.3)

If (1.3.2) holds then we deduce from (1.3.3) that 2(f − g, x − y) + λf − g2 ≥ 0. Setting λ → 0 we obtain (1.3.1). If operator A is weakly differentiable then the following definition of the monotonicity is given. Definition 1.3.8 A Gˆ ateaux differentiable operator A : X → X ∗ with D(A) = X is called monotone if A (x)h, h ≥ 0 ∀x, x + h ∈ X. This definition is motivated by the following theorem.

1.3

Monotone Operators

21

Theorem 1.3.9 Let an operator A be defined on a convex set Ω ∈ X. If the directional derivative d x2 − x1 , A x1 + t(x2 − x1 ) ∀x1 , x2 ∈ Ω t=0 dt exists and is non-negative, then A is monotone on Ω. Let us present some examples of monotone operators. 1

1. Let ϕ : R1 → R1 be a non-decreasing function. Then an operator A : R1 → 2R , defined by the equality Ax = [ϕ(x − 0), ϕ(x + 0)] ∀x ∈ dom ϕ, is monotone. 2. Assume that ϕ : X → R1 is a proper convex functional and there exists a subdiffer∗ ential ∂ϕ : X → 2X . Then the operator ∂ϕ is monotone. Indeed, by Definition 1.2.2, we can write for all x, y ∈ dom ϕ : ϕ(y) − ϕ(x) ≥ f, y − x , f ∈ ∂ϕ(x), ϕ(x) − ϕ(y) ≥ g, x − y , g ∈ ∂ϕ(y). Summing these inequalities one gets f − g, x − y ≥ 0 ∀x, y ∈ dom ϕ, f ∈ ∂ϕ(x),

g ∈ ∂ϕ(y).

Furthermore, it follows from Lemma 1.2.3 for the Gˆ ateaux differentiable functional that the gradient of a proper convex functional is the single-valued monotone operator. 3. Assume that H is a Hilbert space, A : H → H is a nonexpansive operator, i.e., Ax − Ay ≤ x − y ∀x, y ∈ D(A). Then the operator I − A, where I : H → H is an identity operator, is monotone. Indeed, the claim follows from the relations (x − Ax − y + Ay, x − y) = x − y2 − (Ax − Ay, x − y) ≥ x − y2 − Ax − Ayx − y ≥ x − y2 − x − y2 = 0. 4. Let Ω be a convex closed set in a Hilbert space H, x ∈ H and PΩ x be a projection of x on Ω defined by (1.1.17): x − PΩ x = min{x − z | z ∈ Ω}.

(1.3.4)

The element u = PΩ x is unique for every x ∈ H. Prove that the operator PΩ is monotone. First of all, we show that (1.3.4) is equivalent to the inequality (PΩ x − x, z − PΩ x) ≥ 0 ∀z ∈ Ω.

(1.3.5)

22

1


In fact, the expression x − z2 ≥ x − PΩ x2 + 2(x − PΩ x, PΩ x − z) ∀z ∈ Ω, holds (see Theorem 1.1.20 and Lemma 1.2.3). Then, obviously, (1.3.4) results from (1.3.5). Let now (1.3.4) be valid and t ∈ (0, 1) be given. Since Ω is a convex set, we have that the element (1 − t)PΩ x + tz ∈ Ω and then, by the inequality x − PΩ x2 − x − (1 − t)PΩ x − tz2 ≥ 2(x − PΩ x − t(z − PΩ x), t(z − PΩ x)), one gets (x − PΩ x − t(z − PΩ x), z − PΩ x) ≤ 0. That leads to (1.3.5) as t → 0. Similarly to (1.3.5) we can write (PΩ y − y, z − PΩ y) ≥ 0 ∀z ∈ Ω.

(1.3.6)

Presume z = PΩ y and z = PΩ x, respectively, in (1.3.5) and in (1.3.6). Summing thus obtained inequalities we have (PΩ x − PΩ y, x − y) − PΩ x − PΩ y2 ≥ 0. Hence, PΩ is monotone. Note that it also follows from the last inequality that the projection operator is nonexpansive in H. If Ω is a subspace of Hilbert space then PΩ is a linear and orthogonal operator. In a Banach space, a metric projection operator is not monotone and not nonexpansive, in general. However, there exists nonexpansive projections from a Banach space even into a nonconvex subset Ω [61]. 5. Let G ⊂ Rn be a bounded measurable domain. Define the operator A : Lp (G) → q L (G), p−1 + q −1 = 1, p > 1, by the formula Ay(x) = ϕ(x, |y(x)|p−1 )|y(x)|p−2 y(x), x ∈ G, where the function ϕ(x, s) is measurable as a function on x for every s ∈ [0, ∞) and continuous for almost all x ∈ G as a function on s, |ϕ(x, s)| ≤ M for all s ∈ [0, ∞) and for almost all x ∈ G. Note that the operator A really maps Lp (G) to Lq (G) because of the inequality |Ay| ≤ M |y|p−1 . Show that the operator A is monotone provided that function sϕ(x, s) is non-decreasing with respect to s. We can write the following estimates: Ay − Az, y − z = ≥ − =

G

G

G

ϕ(x, |y|p−1 )|y|p−2 (|y|2 − |y||z|)dx ϕ(x, |z|p−1 )|z|p−2 (|y||z| − |z|2 )dx

G

ϕ(x, |y|p−1 )|y|p−2 y − ϕ(x, |z|p−1 )|z|p−2 z (y − z)dx

ϕ(x, s1 )s1 − ϕ(x, s2 )s2 (|y| − |z|)dx ≥ 0,

1.3

Monotone Operators

23

where s1 = |y|p−1 , s2 = |z|p−1 . Hence, the property of monotonicity is proved. 6. Let the operator Au = −

n

∂ i=1

∂xi

∂u p−1 ∂u p−2 ∂u + a0 (x, |u|p−1 )|u|p−2 u, ai x, ∂x ∂xi ∂xi i

be given, where the functions ai (x, s), i = 0, 1, 2, ..., n, have the same properties as the function ϕ(x, s) in Example 5 and G is a bounded measurable set in Rn . Then the operator ◦

◦

A : W1p (G) →(W1p (G))∗ defined by the formula Au, v =

G

∂u p−1 ∂u p−2 ∂u ∂v dx ai x, ∂x ∂xi ∂xi ∂xi i G

i=1

+

n

◦

∀u, v ∈W1p (G)

a0 (x, |u|p−1 )|u|p−2 uvdx

is monotone. This fact is verified by the same arguments as in Example 5 (see [83]). 7. Next we give the example from quantum mechanics [230, 231]. Consider the operator Au = −a2 u + (g(x) + b)u(x) + u(x)

R3

u2 (y) dy, |x−y |

3

∂2

is the Laplace operator (Laplacian) in R3 , a and b are constants, g(x) = ∂x2i g0 (x) + g1 (x), g0 (x) ∈ L∞ (R3 ), g1 (x) ∈ L2 (R3 ). Represent A in the form A = L + B, where the operator L is the linear part of A (it is the Schr¨ odinger operator) and B is defined by the last term. It is known [107] that there exists b ≥ 0 such that L becomes positive in the domain

where ∆ =

i=1

D(L) = D(∆) = {u ∈ H | ∇u ∈ H × H × H, ∆u ∈ H}, H = L2 (R3 ),

∂u ∂u . , ∂x1 ∂x2 ∂x3 It is obvious that B is hemicontinuous. Furthermore, B is the gradient of the functional ∇u =

Φ(u) =

1 4

R3

R3

∂u

,

u2 (x)u2 (y) dxdy, u ∈ W12 (R3 ), |x − y|

which is proper convex lower semicontinuous. Therefore, B is a monotone operator from H to H. This implies that A : L2 (R3 ) → L2 (R3 ) is also a monotone operator. 1 It is known that g(x) = −2|x|−2 and a = correspond to the case of the Coulomb 2 potential in the quantum mechanics, and b ≥ 2 guarantees a positivity of the Schrödinger operator [169]. The operator A describes here the atom of helium in the situation when both electrons are in the same state. Observe that |u(y)|p A1 u(x) = |u(x)|p−2 u(x) dy, p > 2 R3 |x − y|

24

1


is the gradient of the functional

Lp

3

Φ1 (u) =

1 2p

3

−1

q

R3

R3

|u(x)|p |u(y)|p dxdy |x − y|

−1

and acts from (R ) to L (R ), p + q = 1. In addition, it is not difficult to make sure that the functional Φ1 (u) is convex, i.e., the operator A1 is monotone too. 8. In the filtration theory one has to solve the following equation [143]: −div (g(x, ∇2 u)∇u) = f (x), u|Γ = 0, where x ∈ Ω, Ω ⊂ Rn is a bounded measurable domain, Γ = ∂Ω, u(x) is a pressure, f (x) is a density of sources, ∇ is the Hamilton operator (Hamiltonian) defined for a scalar function u(x) = u(x1 , x2 , ..., xn ) as ∂u ∂u ∂u , ..., , ∇u = ∂xn ∂x1 ∂x2

and the symbol div w ¯ denotes divergence of the vector field w ¯ [82]. It is known that the function g(x, ξ 2 )ξ can be written in the form g(x, ξ 2 )ξ = g0 (x, ξ 2 )ξ + g1 (x, ξ 2 )ξ, where g0 (x, ξ 2 )ξ is non-negative and non-decreasing with respect to ξ, g0 (x, ξ 2 ) = 0 for ξ ≤ β, g0 (x, ξ 2 ) is measurable with respect to the first argument and absolutely continuous with respect to the second argument. Beside this, g0 (x, ξ 2 )ξ ≤ c1 |ξ − β|p−1 , p > 1, ξ ≤ β, β = β(x) ∈ Lp (Ω), c1 > 0, 2

g1 (x, ξ )ξ =

ω > 0, 0,

if ξ > β, if ξ ≤ β. ◦

Then the functions g0 and g1 define, respectively, the operators A0 and A1 from W1p (Ω) ◦

to (W1p )∗ (Ω) : Ai u, v =

Ω

gi (x, ∇2 u)(∇u, ∇v)dx, i = 0, 1,

◦

∀u, v ∈W1p (Ω),

where both operators are monotone, A0 is continuous, A1 is discontinuous. Here (∇u, ∇v) =

n

∂u ∂v i=1

∂xi ∂xi

.

Furthermore, the monotone operator A = A0 + A1 is potential and its potential is defined by the following expression: F (u) =

Ω

|∇u| 0

g(x, ξ 2 )ξdξ dx.

Introduce monotone operators with stronger properties of the monotonicity.

1.3

Monotone Operators

25 ∗

Definition 1.3.10 An operator A : X → 2X is called uniformly monotone if there exists a continuous increasing function γ(t) (t ≥ 0), γ(0) = 0, such that the inequality f − g, x − y ≥ γ(x − y) ∀f ∈ Ax, ∀g ∈ Ay

(1.3.7)

holds for all x, y ∈ D(A). If here γ(t) = ct2 , where c is a positive constant, then A is strongly monotone. ∗

Lemma 1.3.11 Let ∂ϕ : X → 2X be a subdifferential of the functional ϕ : X → R1 . If ϕ is uniformly convex on a convex closed set Ω with modulus of convexity γ(t), then f − g, x − y ≥ 2γ(x − y) ∀x, y ∈ Ω, f ∈ ∂ϕ(x),

g ∈ ∂ϕ(y),

that is, ∂ϕ is a uniformly monotone operator. ∗

Remark 1.3.12 We say that a monotone operator A : X → 2X is properly monotone if there is not any strengthening of (1.3.1) (for instance, up to the level of strong or uniform monotonicity). In Section 1.6 we shall give examples of monotone operators A that do not satisfy the inequality (1.3.7) on the whole domain D(A) but they are uniformly monotone on any bounded set of D(A). Therefore, it makes sense to give the following definition. ∗

Definition 1.3.13 An operator A : X → 2X is called locally uniformly monotone if there exists an increasing continuous function γR (t) (t ≥ 0, R > 0), γR (0) = 0 and f − g, x − y ≥ γR (x − y)

∀f ∈ Ax,

∀g ∈ Ay

for x and y from D(A), where x ≤ R, y ≤ R. If γR (t) = C(R)t2 , C(R) > 0, then A is said to be locally strongly monotone. The following lemma asserts that the class of locally uniformly monotone operators is not empty. Lemma 1.3.14 A continuous strictly monotone potential operator A : Rn → Rn is locally uniformly monotone. Proof. Define the function 0 (τ ) = inf {(Ax − Ay, x − y) | x − y = τ, x, y ∈ B(θRn , R)}. γR 0 It is obvious that γR (0) = 0. Since the operator A is strictly monotone and continuous, 0 (τ ) > 0 as τ > 0. Moreover, under our conditions, there can be found the function γR 0 x0 , y0 ∈ B(θRn , R) such that x0 − y0 = τ and (Ax0 − Ay0 , x0 − y0 ) = γR (τ ). We will show 0 that γR (τ ) is an increasing function. Let τ1 < τ2 be given. Then there exist x2 and y2 such that 0 x2 − y2 = τ2 , γR (τ2 ) = (Ax2 − Ay2 , x2 − y2 ).

26

1


By strong monotonicity of A, we have (Ax2 − A(y2 + t(x2 − y2 )), x2 − y2 ) > 0, 0 ≤ t < 1. Hence, (Ax2 − Ay2 , x2 − y2 ) > (A(y2 + t(x2 − y2 )) − Ay2 , x2 − y2 ). Let w = y2 +

(1.3.8)

τ1 (x2 − y2 ) . τ2

Then w − y2 = τ1 . Substitute t in (1.3.8) for t = τ1 /τ2 , 0 < t < 1, then 0 (τ2 ) > γR

τ2 0 (Aw − Ay2 , w − y2 ) ≥ (Aw − Ay2 , w − y2 ) ≥ γR (τ1 ). τ1

0 (τ ) is increasing, but it is not continuous in general. Therefore, let Hence, the function γR 0 us proceed with the constructions. It follows from the definition of the function γR (τ ) that

1 0 (tx − y). (A(y + t(x − y)) − Ay, x − y) ≥ γR t Then 0≤

(1.3.9)

0 x−y 1 γR (τ ) − Ay, x − y , A y+τ ≤ x − y x − y τ

0 from which we conclude that γR (τ )/τ → 0 as τ → 0 because of continuity of A. Let Ax = grad ϕ(x) be given. Taking into account Lemma 1.2.7 and integrating (1.3.9), we write the inequality

ϕ(x) − ϕ(y) − (Ay, x − y) ≥

1 0

0 γR (tx − y)

dt . t

(1.3.10)

Interchange x and y in (1.3.10) and add thus obtained inequality to (1.3.10). Then (Ax − Ay, x − y) ≥ 2

0

1

0 γR (tx − y)

dt =2 t

x−y 0

0 γR (τ ) dτ. τ

Hence, we have constructed the function

γR (t) = 2

t 0

0 γR (τ ) dτ τ

which is increasing, continuous, γR (0) = 0 and for all x ≤ R and for all y ≤ R (Ax − Ay, x − y) ≥ γR (x − y). The lemma is proven.

We present the property of the local boundedness of monotone mappings.

1.3

Monotone Operators

27 ∗

Definition 1.3.15 An operator A : X → 2X is said to be locally bounded at a point x ∈ X if there exists a neighborhood M = M (x) of this point such that the set A(M ) = {y | y ∈ Ax, x ∈ M ∩ D(A)} is bounded in X ∗ . ∗

Theorem 1.3.16 A monotone mapping A : X → 2X is locally bounded at each interior point of its domain. Proof. Prove this theorem by contradiction. Suppose that A is not locally bounded at a point x0 while x0 ∈ int D(A). Then there exists a sequence {xn }, xn → x0 , xn ∈ D(A), such that τn = yn ∗ → +∞, where yn ∈ Axn . Denote tn = max

1

τn

;

xn − x0 .

It is clear that tn > 0, tn τn ≥ 1, xn − x0 ≤ t2n and limn→∞ tn = 0. Construct an element zn = x0 + tn z, where any z ∈ X. Since x0 ∈ int D(A) and since tn → 0, then zn ∈ D(A) for sufficiently large n. Let un ∈ Azn be given and r be a positive number such that v = x0 + rz ∈ D(A). By the monotonicity of A, un − f, zn − v ≥ 0 ∀f ∈ Av. This implies (tn − r)un − f, z ≥ 0. For tn < r, we have un , z ≤ f, z . Hence, lim supn→∞ |un , z | < ∞. Then the Banach−Steinhaus theorem ensures boundedness of the sequence {un ∗ }, That is, there exists a constant c > 0 such that un ∗ ≤ c for all n ≥ 1. Using the monotonicity condition of A again we obtain yn − un , xn − zn = yn − un , xn − (x0 + tn z) ≥ 0. The last inequality yields yn , z ≤ ≤

1 xn − x0 − z yn , xn − x0 − un , tn tn 1 yn ∗ xn − x0 + c(tn + z) ≤ tn τn + c(tn + z). tn

Then lim sup n→∞

|yn , z | < ∞. tn τn

Applying the Banach−Steinhaus theorem again one gets the inequality (tn τn )−1 yn < ∞. At the same time, we have 1 → ∞. (tn τn )−1 yn = tn Thus, we have arrived at the contradiction. The theorem is proved.

28

1


Corollary 1.3.17 Suppose that X is a real Banach space and A : X → 2X is a monotone operator. Let Ω be an open subset of D(A). If A is locally bounded at some point of Ω, then it is locally bounded at every point of Ω. ∗

Corollary 1.3.18 If an operator A : X → 2X is monotone and x ∈ int D(A), then the set Ax is bounded in X ∗ . Observe that a monotone operator is unbounded, in general (see the example in [128]). However, the following assertion is true [1]. Theorem 1.3.19 If X is a finite-dimensional Banach space, then any monotone operator ∗ A : X → 2X with D(A) = X is bounded. Proof. Let xn → x ¯ and yn ∗ → ∞, where yn ∈ Axn . Since X is finite-dimensional, there exists a subsequence {xk } ⊆ {xn } such that yk yk −1 ∗ → z. It is obvious that z∗ = 1. Then by the monotonicity of A, we can write yk ∗ −1 yk − y, xk − x ≥ 0 ∀x ∈ X,

∀y ∈ Ax.

Now we turn k to ∞ and obtain z, x ¯ − x ≥ 0 ∀x ∈ X. Assuming x = x ¯ + z in the previous inequality we come to the equality z = θX ∗ , which contradicts the fact that z∗ = 1. It has been noted in Section 1.1 that hemicontinuity of an operator A follows from its demicontinuity. If A is monotone then the following converse assertion is also fulfilled.

Theorem 1.3.20 Any monotone hemicontinuous operator A : X → X ∗ is demicontinuous on int D(A). Proof. Let A be hemicontinuous on int D(A), {xn } ⊂ int D(A) be a sequence such that xn → x ∈ int D(A). Then by virtue of the local boundedness of A at x, the sequence {Axn } is bounded beginning with a large enough n. Therefore, we conclude that there exists some subsequence Axnk f ∈ X ∗ . Write the monotonicity condition of A: Axnk − Ay, xnk − y ≥ 0

∀y ∈ D(A).

Passing to the limit in the previous inequality we obtain f − Ay, x − y ≥ 0

∀y ∈ D(A).

(1.3.11)

Since int D(A) is an open set, then for all u ∈ X there exists t¯ such that elements yt = x + tu ∈ D(A) as 0 ≤ t ≤ t¯. If we replace y in (1.3.11) by yt then f − Ayt , u ≤ 0

∀u ∈ X.

Let t → 0 be given. Then the inequality f − Ax, u ≤ 0 holds for all u ∈ X because of the hemicontinuity of A. Hence, Ax = f. Thus, Axn Ax, i.e., A is demicontinuous at a point x.

1.4 Maximal Monotone Operators

29

Corollary 1.3.21 If A : X → X ∗ is a monotone hemicontinuous operator, D(A) = X, X is a finite-dimensional space, then A is continuous. Corollary 1.3.22 Every linear monotone operator A : X → X ∗ with D(A) = X is continuous. Proof. Since A is linear, we have A(x + ty) = Ax + tAy for all t ∈ (−∞, +∞). Hence, A is hemicontinuous. It follows from Theorem 1.3.20 that A is demicontinuous. Prove by contradiction that A is continuous. Let x, xn ∈ X, xn → x, and Axn − Ax∗ ≥ τ > 0 be given for all n ≥ 0. If tn = xn − x−1/2 and yn = x + tn (xn − x) we obtain yn → x and Ayn − Ax∗ = tn Axn − Ax∗ ≥ tn τ → +∞, which contradicts the demicontinuity of A.

1.4

Maximal Monotone Operators

Let X be a reflexive Banach space and X ∗ its dual space. Definition 1.4.1 A monotone set G ⊆ X × X ∗ is called maximal monotone if it is not a proper subset of any monotone set in X × X ∗ . ∗

Definition 1.4.2 An operator A : X → 2X with D(A) ⊆ X is called maximal monotone if its graph is a maximal monotone set of X × X ∗ . From this definition, immediately follows ∗

Proposition 1.4.3 A monotone operator A : X → 2X is maximal on D(A) if and only if the inequality g − f, y − x0 ≥ 0 ∀(y, g) ∈ grA, (1.4.1) implies the inclusions x0 ∈ D(A) and f ∈ Ax0 . Since the graphs of the operator A and its inverse A−1 coincide, then the maximal monotonicity of A−1 follows from the maximal monotonicity of A and conversely. Definition 1.4.4 A set G ⊆ X × X ∗ is called demiclosed if the conditions xn → x, yn f or xn x, yn → f, where (xn , fn ) ∈ G, imply that (x, f ) ∈ G. The following assertion is established by the definition of the maximal monotonicity of operators. ∗

Lemma 1.4.5 The graph of any maximal monotone operator A : X → 2X is demiclosed. Theorem 1.4.6 Any monotone hemicontinuous operator A : X → X ∗ with D(A) = X is maximal monotone.

30

1


Proof. It suffices to prove that the equality f = Ay follows from the inequality f − Ax, y − x ≥ 0

∀x ∈ X.

(1.4.2)

Since D(A) = X, it is possible to take in (1.4.2) x = xt = y + tz, where z ∈ X and t > 0. Then f − Axt , z ≤ 0 for all z ∈ X. Letting t → 0, by hemicontinuity of A on X, we obtain f − Ay, z ≤ 0

∀z ∈ X.

Hence f = Ay. ∗

Theorem 1.4.7 Let A : X → 2X be a demiclosed single-valued monotone operator with D(A) = X. Then A is maximal monotone. Proof. We shall prove that the inequality (1.4.1) with any x ∈ X and any f ∈ X ∗ implies (x, f ) ∈ grA. Indeed, fix an element z ∈ X and take zt = x + tz ∈ X for any t > 0. Then (1.4.1) with y = zt gives f − gt , x − zt ≥ 0 ∀gt ∈ Azt , that is, gt − f, z ≥ 0 ∀gt ∈ Azt , ∀z ∈ X.

(1.4.3)

If t → 0 then zt → x. By virtue of the local boundedness of A at a point x, it is possible to assert that gt g¯(z) ∈ X ∗ . Then we conclude for a demiclosed operator A that g¯(z) ∈ Ax for all z ∈ X. By (1.4.3), ¯ g (z) − f, z ≥ 0 ∀z ∈ X. Since A is single-valued, we have g¯(z) = Ax. Thus, Ax − f, z ≥ 0 ∀z ∈ X. The last inequality asserts that Ax = f. Then Proposition 1.4.3 proves the claim. ∗

Theorem 1.4.8 Let A : X → 2X be a monotone demiclosed operator such that D(A) = X and for each x ∈ X the image Ax is a nonempty convex subset of X ∗ . Then A is maximal monotone. Proof. As in the previous theorem, we prove that the inequality (1.4.1) with any x ∈ X and any f ∈ X ∗ implies (x, f ) ∈ grA. Suppose that, on the contrary, f ∈ / Ax. Since Ax is convex and A is demiclosed, Ax is weakly closed and, therefore, closed. According to the strong separation theorem, there exists an element z ∈ X such that f, z > sup {g, z | g ∈ Ax}.

(1.4.4)

Take zt = x + tz ∈ X and let gt ∈ Azt for any t > 0. Similarly to Theorem 1.4.7, as t → 0, one gets ¯ g (z) − f, z ≥ 0, g¯(z) ∈ Ax. This contradicts (1.4.4). Consequently, f ∈ Ax and the conclusion follows from Proposition 1.4.3 again.

1.4 Maximal Monotone Operators

31

∗

Theorem 1.4.9 If A : X → 2X is a maximal monotone operator, then the set {f | f ∈ Ax} for every x ∈ D(A) is convex and closed in X ∗ . Proof. Let f1 ∈ Ax and f2 ∈ Ax. It is clear that the monotonicity of A yields the inequalities f1 − g, x − y ≥ 0 (1.4.5) and f2 − g, x − y ≥ 0

(1.4.6)

for all (y, g) ∈ grA. Let f = tf1 + (1 − t)f2 , where t ∈ [0, 1]. Multiply (1.4.5) and (1.4.6) by t and 1 − t, respectively, and add them. Then we get the following inequality: f − g, x − y ≥ 0

∀(y, g) ∈ grA.

The maximal monotonicity of the operator A implies then the inclusion f ∈ Ax, i.e., the set {f | f ∈ Ax} is convex. Let fn → f¯, fn ∈ Ax. We have fn − g, x − y ≥ 0

∀(y, g) ∈ grA,

and the limit passing as n → ∞ gives f¯ − g, x − y ≥ 0 ∀(y, g) ∈ grA. Hence, f¯ ∈ Ax, and the theorem is proved. ∗

Corollary 1.4.10 If A : X → 2X is a maximal monotone operator, then the set {x | f ∈ Ax} for every f ∈ R(A) is convex and closed in X. Proof. The proof follows from Theorem 1.4.9 applied to the inverse mapping A−1 . ∗

Consider a linear, possibly multiple-valued, operator L : X → 2X , for which grL is a linear subspace of X × X ∗ . The monotonicity condition for it can be written in the following form: f, x ≥ 0 ∀(x, f ) ∈ grL. ∗

Let L∗ : X → 2X be the adjoint (conjugate) operator to L. It is defined as follows: g ∈ L∗ y

implies

g, x = f, y ∀(x, f ) ∈ gr L.

(1.4.7)

An operator L is self-adjoint if L∗ = L. Theorem 1.4.11 Assume that X is a reflexive strictly convex Banach space together with ∗ its dual space X ∗ . A linear monotone operator L : X → 2X is maximal monotone if and ∗ only if it is closed and L is a monotone mapping.

32

1


Proof. Let L be maximal monotone. Then, by Lemma 1.4.5, it is closed. We will prove that the conjugate operator L∗ is monotone. Indeed, in view of (1.4.7) and the monotonicity of L, we have for all (x, f ) ∈ grL and for all (y, g) ∈ grL∗ , f + g, x − y = f, x − f, y + g, x − g, y = f, x − g, y ≥ −g, y .

(1.4.8)

If g, y < 0 then f + g, x − y > 0 for all (x, f ) ∈ grL. Since L is maximal monotone, (y, −g) ∈ grL. Then assuming in (1.4.8) that x = y and f = −g we arrive at the contradiction. Thus, L∗ is monotone. The first part of the lemma is proved. Let L be closed and L∗ be a monotone operator. In order to prove the maximal monotonicity of L, it is sufficient to show that the inequality f − h, x − z ≥ 0 ∀(x, f ) ∈ grL, z ∈ X, h ∈ X ∗ ,

(1.4.9)

implies that (z, h) ∈ grL. Introduce on the set grL ⊆ X × X ∗ the norm by the formula (x, f ) = x + f ∗ , (x, f ) ∈ grL. Note that grL is a strictly convex space with respect to this norm. Construct in grL the functional ϕ(x, f ) = 2−1 f − h2∗ + 2−1 x − z2 + f − h, x − z ,

(x, f ) ∈ grL.

This functional is continuous, convex and due to (1.4.9), ϕ(x, f ) → +∞ as (x, f ) → ∞. Then, by Theorem 1.1.23, there exists a point (z0 , h0 ) ∈ grL in which ϕ(x, f ) reaches a minimum, that is, ϕ (z0 , h0 ) = (θX ∗ , θX ). Consequently, ϕ (z0 , h0 ), (x, f ) = 0 for all (x, f ) ∈ grL. It can be written in the expanded form: For all (x, f ) ∈ grL, f, J ∗ (h0 − h) + J(z0 − z), x + f, z0 − z + h0 − h, x = 0, that is,

f, J ∗ (h0 − h) + z0 − z = J(z − z0 ) + h − h0 , x .

Then by (1.4.7), we deduce that J(z − z0 ) + h − h0 ∈ L∗ (J ∗ (h0 − h) + z0 − z). Now the monotonicity of L∗ leads to the following inequality: J(z − z0 ) + h − h0 , J ∗ (h0 − h) + z0 − z ≥ 0. Here J : X → X ∗ and J ∗ : X ∗ → X are normalized duality mappings in X and X ∗ , respectively. Taking into account the definitions of duality mappings one gets h − h0 2∗ + z − z0 2 + h − h0 , z − z0 ≤ z − z0 h − h0 ∗ , where (z0 , h0 ) ∈ grL. In view of (1.4.9), the last expression implies h − h0 2∗ + z − z0 2 ≤ z − z0 h − h0 ∗ . It follows that h = h0 , z = z0 , hence, we have proved the inclusion (z, h) ∈ grL.

1.5 Duality Mappings

33

Theorem 1.4.12 If L : X → X ∗ is a linear single-valued maximal monotone operator, then D(L) is dense in X. Proof. Indeed, let g, x = 0

∀x ∈ D(L), g ∈ X ∗ .

Then Lx − g, x − θX = Lx, x ≥ 0 for every x ∈ D(L). Due to the maximal monotonicity of L, we conclude that g = L(θX ) = θX ∗ . The additional properties of maximal monotone operators are also given in Section 1.8.

1.5

Duality Mappings

In Section 1.1 we introduced the definition of the normalized duality mapping. The more general concept of a duality mapping is given by Definition 1.5.1 Let µ(t) be a continuous increasing function for t ≥ 0 such that µ(0) = ∗ 0, µ(t) → ∞ as t → ∞. An operator J µ : X → 2X is called a duality mapping with the gauge function µ(t) if the equalities y∗ = µ(x), y, x = µ(x)x are satisfied for all x ∈ X and for all y ∈ J µ x. If µ(t) = t then J µ coincides with the normalized duality mapping J. ∗

Lemma 1.5.2 A duality mapping J µ : X → 2X exists in any Banach space and its domain is all of X. Proof. Take an arbitrary x ∈ X. Due to Corollary 1.1.2 of the Hahn−Banach theorem, there exists at least one element φ ∈ X ∗ such that φ∗ = 1 and φ, x = x. Then it is not difficult to verify that ψ = µ(x)φ satisfies the inclusion ψ ∈ J µ x.

Lemma 1.5.2 implies ∗

Corollary 1.5.3 If J µ : X → 2X is a one-to-one operator at a point x, then J µ x = µ(x)e∗ , where e∗ ∈ X ∗ , e∗ , x = x, e∗ ∗ = 1. We present the other properties of the operator J µ .

34

1


Lemma 1.5.4 In any Banach space X, the duality mapping J µ is a bounded monotone and coercive operator and it satisfies the inequality

u − v, x − y ≥ µ(x) − µ(y) (x − y)

(1.5.1)

for all u ∈ J µ x, v ∈ J µ y and for all x, y ∈ X. If J µ is single-valued then

J µ y − J µ y, x − y ≥ µ(x) − µ(y) (x − y) ∀x, y ∈ X.

(1.5.2)

In this case, Lemma 1.5.4 gives the following important estimate for normalized duality mapping J : (1.5.3) Jx − Jy, x − y ≥ (x − y)2 ∀x, y ∈ X. Lemma 1.5.5 If X is a strictly convex space, then J µ is a strictly monotone operator. If X ∗ is strictly convex, then J µ is single-valued. If X is reflexive and dual space X ∗ is strictly convex, then J µ is demicontinuous. Thus, if X is a reflexive strictly convex Banach space with strictly convex dual space X ∗ then J µ is a single-valued demicontinuous (hence, hemicontinuous) and strictly monotone operator. Lemma 1.5.6 If the space X is reflexive and strongly smooth, then J µ is continuous. Introduce in X the functional

Φ(x) =

x

0

µ(t)dt,

(1.5.4)

and prove the following result: Lemma 1.5.7 The functional Φ defined by (1.5.4) in a strictly convex Banach space X with strictly convex X ∗ has the Gˆ ateaux derivative Φ and Φ (x) = J µ x for all x ∈ X. Proof. Indeed, (1.5.4) implies Φ(y) − Φ(x) =

y

x

µ(t) dt.

Since µ(t) is an increasing function, one gets

y

x

µ(t) dt ≥ µ(x)(y − x).

By Definition 1.5.1, we have µ(x)y − µ(x)x ≥ J µ x, y − J µ x, x = J µ x, y − x .


35

Thus, Φ(y) − Φ(x) ≥ J µ x, y − x . According to Definition 1.2.1, this means that J µ x ∈ ∂Φ(x), where ∂Φ(t) is a subgradient of Φ(t). Let now u ∈ ∂Φ(x). Then we have Φ(y) − Φ(x) ≥ u, y − x ∀y ∈ X.

(1.5.5)

Choose in (1.5.5) an element y ∈ X such that x = y. Then u, y − x ≤ 0, i.e., u, y ≤ u, x

∀y ∈ X.

(1.5.6)

It follows from strict convexity of the space X that sup {u, y | y = r} = u, z = u∗ z. Therefore, by (1.5.6), if r = x then z = x. Hence, u, x = u∗ x. Assume that in (1.5.5) x = tv, y = sv, v = 1, t, s ∈

(1.5.7) 1 R+ .

Then

s−t s−t u∗ x = (s − t)u∗ . u, x = Φ(s) − Φ(t) ≥ (s − t)u, v = t t Hence, the following limit-relation is valid:

lim

s→t

Φ(s) − Φ(t) = u∗ , s−t

which implies that µ(x) = u∗ . By (1.5.7) and by the definition of J µ , we conclude that u = J µ x. Consequently, Φ (x) = J µ x. The lemma is proved. Corollary 1.5.8 In a smooth Banach space, any normalized duality mapping is defined by the formula (1.1.12). Moreover, J µx =

µ(x) Jx. x

(1.5.8)

If X is a Hilbert space H then J is the identity operator I, which is linear in Hilbert spaces. Inversely, if the operator J is linear in X then X is a Hilbert space. Indeed, let J : X → X ∗ be a linear operator, f = Jx, g = Jy. Then f + g = J(x + y), f − g = J(x − y). We deduce

x + y2 = f + g, x + y = x2 + f, y + g, x + y2 , x − y2 = f − g, x − y = x2 − f, y − g, x + y2 .

Summing up those inequalities, we obtain the parallelogram equality x + y2 + x − y2 = 2x2 + 2y2 ∀x, y ∈ X, which is possible only in a Hilbert space. Thus, the following assertion has been proven:

36

1


Theorem 1.5.9 A normalized duality map J : X → X ∗ is linear if and only if X is a Hilbert space. The analytical representations of dual mappings are known in a number of Banach p (G), p ∈ (1, ∞), respectively, spaces. For instance, in the spaces lp , Lp (G) and Wm q p−2 Jx = x2−p x1 , |x2 |p−2 x2 , ...}, x = {x1 , x2 , ...}, lp y ∈ l , y = {|x1 | p−2 Jx = x2−p x(s) ∈ Lq (G), s ∈ G, Lp |x(s)|

and Jx = x2−p p W m

|α|≤m

q (−1)|α| Dα (|Dα x(s)|p−2 Dα x(s)) ∈ W−m (G), m > 0, s ∈ G,

where p−1 + q−1 = 1. If µ(t) = tp−1 then the duality mappings J µ = J p with a gauge function µ(t) have a simpler form. Namely, In lp : J p x = {|x1 |p−2 x1 , |x2 |p−2 x2 , ...}; In Lp (G) : J p x = |x(s)|p−2 x(s); p (G) : In Wm

J px =

(−1)|α| D α (|Dα x(s)|p−2 D α x(s)).

|α|≤m

If X is a reflexive strictly convex Banach space together with its dual space X ∗ , then the duality mapping J ∗ in X ∗ is an operator acting from X ∗ into X. It is obvious that J ∗ = J −1 , i.e., for all x ∈ X and for all φ ∈ X ∗ , the equalities JJ ∗ φ = φ and J ∗ Jx = x hold. At the same time, a more general assertion is true: Lemma 1.5.10 Let X be a reflexive strictly convex Banach space with strictly convex dual space X ∗ . If J µ : X → X ∗ and (J ν )∗ : X ∗ → X are duality mappings with gauge functions µ(t) and ν(s), respectively, and ν(s) = µ−1 (t), then (J ν )∗ = (J µ )−1 . Corollary 1.5.11 Let X be a reflexive strictly convex Banach space with strictly convex dual space X ∗ . If J p : X → X ∗ and (J q )∗ : X ∗ → X are the duality mapping with gauge functions µ(t) = tp−1 and ν(s) = sq−1 , p−1 + q −1 = 1, respectively, then (J q )∗ = (J p )−1 . For instance, if in the space Lp (G) the duality mapping is J p , then (J q )∗ is expressed in the explicit form as follows: (J q )∗ y = |y(s)|q−2 y(s) ∈ Lp (G), s ∈ G, (J p )−1 = (J q )∗ , [(J q )∗ ]−1 = J p . However, there are Banach spaces in which (J ν )∗ is not explicitly known but it can be found by means of solving some boundary value problem. Observe that such a situation arises in p Sobolev spaces Wm (G) [83].


37

Remark 1.5.12 In spaces lp , p > 1, the duality mapping J p is weak-to-weak continuous, while in spaces Lp (G), p > 1, p = 2, any J µ has not this property. In a Hilbert space Jx = x, therefore, the weak-to-weak continuity of J is obvious. Next we will present results which are of interest for applications and prove them by making use of duality mappings. Lemma 1.5.13 (Opial) If in a Banach space X having a weak-to-weak continuous duality mapping J µ the sequence {xn } is weakly convergent to x, then for any y ∈ X, lim inf xn − x ≤ lim inf xn − y. n→∞

n→∞

(1.5.9)

If, in addition, the space X is uniformly convex, then the equality in (1.5.9) occurs if and only if x = y. Proof. Since J µ is weak-to-weak continuous and {xn } weakly converges to x, lim J µ (xn − x), y − x = 0.

n→∞

Then lim J µ (xn − x), xn − x = lim J µ (xn − x), xn − y .

n→∞

n→∞

Therefore, by Definition 1.5.1, lim inf µ(xn − x)xn − x = lim inf |J µ (xn − x), xn − y | n→∞

n→∞

≤ lim inf J µ (xn − x)∗ xn − y n→∞

= lim inf µ(xn − x)xn − y. n→∞

This implies (1.5.9). Let now in (1.5.9) x = y and both limits be equal. Then for any point z ∈ x + t(y − x), 0 < t < 1, we would have lim inf xn − z < lim inf xn − y n→∞

n→∞

which is impossible in a uniformly convex Banach space X. ∗

Lemma 1.5.14 Let A : X → 2X be a monotone operator and some x0 ∈ int D(A). Then there exists a constant r0 > 0 and a ball B(x0 , r0 ) such that for any x ∈ D(A) and y ∈ Ax the following inequality holds: y, x − x0 ≥ y∗ r0 − c0 (x − x0 + r0 ), where c0 = sup {y∗ | y ∈ Ax, x ∈ B(x0 , r0 )} < ∞.

(1.5.10)

38

1


Proof. Since x0 ∈ int D(A), we conclude by Theorem 1.3.16 that there exists r0 > 0 such that c0 is defined. Choose an element x ∈ D(A) and construct z0 = x0 + r0 J ∗ h, where h = y−1 ∗ y and y ∈ Ax. The monotonicity condition of A yields the inequality y − y0 , x − z0 ≥ 0 ∀y ∈ Ax, y0 ∈ Az0 . Therefore,

c0 h, x − z0 1 . ≥− = y − y0 , x − z0 + y0 , x − z0 x − z0 y∗ y∗ x − z0 Hence, r0 h, x − z0 ≥ −

c0 r0 c0 r0 (x − x0 + r0 ). x − z0 ≥ − y∗ y∗

Taking into consideration that z0 = x0 + r0 J ∗ h, one gets J(z0 − x0 ) = r0 h, and thus the following calculations are valid: r0 h, x − x0 = r0 h, z0 − x0 + r0 h, x − z0 ≥ r02 − c0 r0 y−1 ∗ (x − x0 + r0 ).

(1.5.11)

Multiplying (1.5.11) by r0 −1 y∗ , we deduce (1.5.10). The lemma is completely proved. Lemma 1.5.15 Assume that X is a reflexive strictly convex Banach space and a sequence {xn } ⊂ X. If for some x ∈ X, Jxn − Jx, xn − x → 0 as n → ∞, then xn x. In addition, if X is E-space, then xn → x in X. Proof. Since Jxn − Jx, xn − x ≥ (xn − x)2 , it follows from the conditions of the lemma that xn → x. The following equality is obvious: Jxn − Jx, xn − x = (xn − x)2 + (xxn − Jx, xn ) + (xn x − Jxn , x ), from which we have that Jx, xn → Jx, x . It is well known that in this case the sequence {xn } is bounded. Then there exists a subsequence {xm } ⊆ {xn } such that xm x ¯ ∈ X. ¯ and Jx, x ¯ = Jx, x . Since the duality map J ∗ is single-valued Thus, Jx, xm → Jx, x in X ∗ , it follows from the latter equality that x ¯ = x, that is, xm x. This means that any subsequence weakly converges to x. Consequently, the whole sequence xn x. The last assertion of the lemma results now from the definition of the E-space. Corollary 1.5.16 If X ∗ is E-space and X and X is strictly convex, then the normalized duality mapping J is continuous.


39

Proof. Let xn → x. Then Jxn − Jx, xn − x → 0, hence, yn − y, J ∗ yn − J ∗ y → 0, where yn = Jxk , y = Jx. By Lemma 1.5.15, one gets that yn → y, i.e., Jxn → Jx. In Section 1.3 we proved that the metric projection operator PΩ onto closed convex subset Ω ⊂ H satisfies the condition x − PΩ x, PΩ x − y ≥ 0 ∀y ∈ Ω,

∀x ∈ H.

Show that PΩ has the similar property also in a Banach space. To this end, we now apply the normalized duality mapping J. As above, assume that X is reflexive and strictly convex together with dual space X ∗ . Under these conditions, the metric projection operator PΩ is well defined for all x ∈ X and single-valued. Lemma 1.5.17 For all x ∈ X, the element z = PΩ x if and only if J(x − z), z − y ≥ 0 ∀y ∈ Ω.

(1.5.12)

Proof. Since y, z ∈ Ω, we conclude that (1 − t)z − ty ∈ Ω for all t ∈ [0, 1]. By the definition of the metric projection z = PΩ x, one gets x − z ≤ x − (1 − t)z − ty

∀y ∈ Ω.

(1.5.13)

Then Corollary 1.5.8 implies 2 J(x − z − t(y − z)), t(y − z) ≤ x − z2 − x − (1 − t)z − ty2 . In view of (1.5.13), there holds the inequality J(x − z − t(y − z)), y − z ≤ 0. Letting here t → 0 and using Lemma 1.5.5 we come to (1.5.12). Suppose that (1.5.12) is now satisfied. Then x − y2 − x − z2 ≥ 2 J(x − z), z − y ≥ 0 ∀y ∈ Ω. Hence, x − z ≤ x − y for all y ∈ Ω, that is, z = PΩ x by definition of PΩ .

Observe that unlike a Hilbert space (see Example 4 in Section 1.3) the metric projection operator onto a convex closed subset of a Banach space X is not necessarily contractive. Besides, since this operator acts from X to X, thus, the monotonicity property for it does not make sense. However, the following assertion is still valid: Lemma 1.5.18 The operator A = J(I − PΩ ) : X → X ∗ is monotone, bounded and demicontinuous.

40

1


Proof. The boundedness and demicontinuity of A is easily proved by using the properties of the operators J and PΩ . Show that A is monotone. Take arbitrary elements x, y ∈ X. Then, by Lemma 1.5.17, we can write the inequality J(x − PΩ x), u − PΩ x ≤ 0 ∀u ∈ Ω and J(y − PΩ y), u − PΩ y ≤ 0 ∀u ∈ Ω. Substituting u = PΩ y for the first inequality and u = PΩ x for the second one, we sum them and thus obtain (1.5.14) J(x − PΩ x) − J(y − PΩ y), PΩ x − PΩ y ≥ 0. It is easy to verify that Ax − Ay, x − y = J(x − PΩ x) − J(y − PΩ y), x − PΩ x − y + PΩ y + J(x − PΩ x) − J(y − PΩ y), PΩ x − PΩ y . It implies the monotonicity of A because of (1.5.14) and monotonicity of J. Definition 1.5.19 Let Ω be a non-empty closed convex subset of X. A mapping QΩ : X → Ω is called (i) a retraction onto Ω if Q2Ω = QΩ ; (ii) a nonexpansive retraction if it also satisfies the inequality QΩ x − QΩ y ≤ x − y ∀x, y ∈ X; (iii) a sunny retraction if for all x ∈ X and for all 0 ≤ t < ∞, QΩ (QΩ x + t(x − QΩ x)) = QΩ x. Proposition 1.5.20 Let Ω be a non-empty closed convex subset of X, J µ be a duality mapping with gauge function µ(t). A mapping QΩ : X → Ω is a sunny nonexpansive retraction if and only if for all x ∈ X and for all ξ ∈ Ω, J µ (QΩ x − ξ), x − QΩ x ≥ 0.

1.6

Banach Spaces Geometry and Related Duality Estimates

The properties of duality mappings are defined by the properties of spaces X and X ∗ . In particular, it is well known that a duality mapping is uniformly continuous on every bounded set in a uniformly smooth Banach space X, that is, for every R > 0 and arbitrary x, y ∈ X with x ≤ R, y ≤ R, there exists a real non-negative and continuous function ωR : [0, ∞) → R1 such that ωR (t) > 0 if t > 0 and ωR (0) = 0, for which the inequality Jx − Jy∗ ≤ ωR (x − y)

(1.6.1)

1.6 Banach Spaces Geometry and Related Duality Estimates

41

holds. Furthermore, a duality mapping is uniformly monotone on every bounded set in a uniformly convex Banach space X. In other words, for every R > 0 and arbitrary x, y ∈ X with x ≤ R, y ≤ R, there exists a real non-negative and continuous function ψR : [0, ∞) → R1 such that ψR (t) > 0 for t > 0, ψ(0) = 0 and Jx − Jy, x − y ≥ ψR (x − y).

(1.6.2)

Our aim is to find in the analytical form the functions ωR (t) and ψR (t) and also the function ω ˜ R (t) evaluating the left-hand side of (1.6.2) from above such that Jx − Jy, x − y ≤ ω ˜ R (x − y).

(1.6.3)

Estimates (1.6.1) - (1.6.3) play a fundamental role in the convergence and stability analysis of approximation methods for nonlinear problems in Banach spaces. Recall that δX () and ρX (τ ) denote, respectively, the modulus of convexity and the modulus of smoothness of a Banach space X. It is known that in a uniformly smooth Banach space X the following inequality holds for any 0 < τ ≤ σ : τ 2 ρX (σ) ≤ Lσ 2 ρX (τ ),

(1.6.4)

where 1 < L < 1.7 is the Figiel constant. Theorem 1.6.1 In a uniformly smooth Banach space X, for every R > 0 and arbitrary x, y ∈ X such that x ≤ R, y ≤ R, the inequality Jx − Jy, x − y ≤ 8x − y2 + c1 ρX (x − y)

(1.6.5)

is satisfied with c1 = 8max{L, R}. Proof. Denote

D = 2−1 (x2 + y2 − 2−1 x + y2 ).

(1.6.6)

1) Let x + y ≤ x − y. Then x + y ≤ x + y + x − y ≤ 2x − y,

(1.6.7)

from which we easily obtain the inequality 2−1 x2 + 2−1 y2 + xy ≤ 2x − y2 .

(1.6.8)

Subtracting 2−1 (x + y)2 from both parts of (1.6.8), we deduce D ≤ 2x − y2 − (4−1 x + y2 + xy). If we assume that then we will have at once

2−1 (x + y)2 + xy ≥ x − y2 ,

(1.6.9)

D ≤ x − y2 .

(1.6.10)

42

1


Suppose that a contrary inequality to (1.6.9) holds. In this case, from the inequality (x − y)2 ≤ x + y2 , which can be re-written as x2 − 2xy + y2 ≤ x + y2 , it immediately follows that D = 2−1 x2 + 2−1 y2 − 2−1 (x + y)2 ≤ 2−1 (x + y)2 + xy ≤ x − y2 . Thus, (1.6.10) is satisfied again. 2) Assume now that x + y > x + y and show that x + y − x − y ≤ (x, y), where (x, y) = x + yρX

x − y

x + y

(1.6.11)

.

Indeed, making the substitutions x = 2−1 (u + v) and y = 2−1 (u − v) for left-hand side of (1.6.11) and after setting u0 = uu−1 and v 0 = vu−1 , we can write the following obvious estimates: x + y − x + y = 2−1 (u + v + u − v) − u = 2−1 u(u0 + v 0 + u0 − v 0 − 2) ≤ usup {2−1 (u0 + v 0 + u0 − v0 ) − 1 | u0 = 1, v 0 = τ } = uρX (v 0 ). Returning to the previous denotations we obtain (1.6.11) which implies x + y x + y − (x, y) . ≥

2

(1.6.12)

2

We assert that the right-hand side of (1.6.12) is non-negative. In fact, by the property ρX (τ ) ≤ τ (see Section 1.1), one establishes the inequality x + y − (x, y) ≥ x + y − x − y ≥ 0. Then

x + y 2 x + y 2 x + y − (x, y) . ≥

2

2

2

1.6


43

Since |x − y| ≤ x − y, we deduce D≤

x − y 2

2

+ (x, y)

x − y2 x + y x + y . ≤ + (x, y) 2 2 4

(1.6.13)

Suppose that x + y ≤ 1. Then x + y−1 x − y ≥ x − y. By (1.6.13) and (1.6.4), we have x − y LρX (x − y . ≤ ρX x + y2 x + y

This inequality yields the estimate D ≤ 4−1 x − y2 + 2−1 L(x + y)x + y−1 ρX (x − y). Since x + y > x − y by the hypothesis, one gets 2−1 x + y−1 (x + y) ≤ (2x + y)−1 (x + y + x − y) ≤ 1. Therefore, D≤

x − y2 + LρX (x − y). 4

(1.6.14)

Assume next that x + y ≥ 1. Taking into account (1.6.13) and the convexity of the function ρX (τ ), we deduce the additional estimate of (1.6.14): D ≤ 4−1 x − y2 + 2−1 (x + y)ρX (x − y). Finally, (1.6.10) implies 2x2 + 2y2 − x + y2 ≤ 4x − y2 + 2max{2L, x + y}ρX (x − y).

(1.6.15)

Denote by k(x − y) the right-hand side of the last inequality. Then D≤

k(x − y) . 4

(1.6.16)

Put into a correspondence to the convex function ϕ(x) = 2−1 x2 the concave function Φ(λ) = λϕ(x) + (1 − λ)ϕ(y) − ϕ(y + λ(x − y)), 0 ≤ λ ≤ 1. It is obvious that Φ(0) = 0. Furthermore, λ1−1 Φ(λ1 ) ≥ λ2−1 Φ(λ2 ) as λ1 ≤ λ2 , −1 ≤ 0. The last inequality implies Φ (λ) ≤ λ−1 Φ(λ). In particular, we have that (λ Φ(λ)) is, 1 ≤ 4Φ 1 . At the same time, Φ 4 4

1

Φ

4

=

1 3 3 1 ϕ(x) + ϕ(y) − ϕ x + y . 4 4 4 4

44

1


By (1.6.16), one has for any z1 , z2 ∈ X, ϕ

z + z 1 2

2

≥

ϕ(z1 ) ϕ(z2 ) k(z1 − z2 ) . + − 8 2 2

Assume that z1 = 2−1 (x + y) and z2 = y. Then, by the property k(t/2) ≤ k(t)/2, we obtain ϕ

Thus, Φ

11 1 1 1 x + y 1 3 + ϕ(y) x+ y + y ≥ ϕ = ϕ x+ y 2 2 2 2 2 2 2 4 4

1

1 4

−

1 x − y 1 1 1 1 k ≥ ϕ(x) + ϕ(y) − k(x − y) + ϕ(y) 8 2 4 4 16 2

−

1 3 1 x − y 1 k ≥ ϕ(x) + ϕ(y) − k(x − y). 8 4 4 8 2

≤ 8−1 k(x − y) and

Φ

1

4

= ϕ(x) − ϕ(y) − ϕ (y + 4−1 (x − y)), x − y ≤ 2−1 k(x − y).

It is clear that Φ

1

4

= ϕ(y) − ϕ(x) − ϕ (x + 4−1 (y − x)), y − x ≤ 2−1 k(x − y).

These inequalities together give ϕ (x − 4−1 (x − y)) − ϕ (y + 4−1 (x − y)), x − y ≤ k(x − y). Make a non-degenerate replacement of the variables z1 = 2x − 2−1 (x − y),

z2 = 2y + 2−1 (x − y)

and recall that Jx = ϕ (x) is a homogeneous operator. Then z1 − z2 = x − y and x + y ≤ z1 + z2 . Therefore, Jz1 − Jz2 , z1 − z2 ≤ 2k(z1 − z2 ). Thus, for z1 ≤ R, z2 ≤ R, we obtain (1.6.5). The proof is accomplished.

Remark 1.6.2 It follows from (1.6.15) that for arbitrary x, y ∈ X it is necessary to make use of the estimate (1.6.5) with c1 = c1 (x, y) = 4max{2L, x + y}. Corollary 1.6.3 Let X be a uniformly convex and smooth Banach space. Then for any x, y ∈ X such that x ≤ R, y ≤ R, the following inequality holds: Jx − Jy, x − y ≤ 8Jx − Jy2∗ + c1 ρX ∗ (Jx − Jy∗ ), where c1 = 8max{L, R}.

(1.6.17)


1.6

45

Proof. Since X ∗ is a uniformly smooth Banach space, by Theorem 1.6.1 for any φ, ψ ∈ X ∗ such that φ∗ ≤ R, ψ∗ ≤ R, we have φ − ψ, J ∗ φ − J ∗ ψ ≤ 8φ − ψ2∗ + c1 ρX ∗ (φ − ψ∗ ),

(1.6.18)

where c1 = 8max{L, R}. The space X is uniformly convex and smooth, consequently, it is reflexive and strictly convex together with its dual space X ∗ . Therefore, the normalized duality mapping J ∗ is single-valued. This means that for any φ ∈ X ∗ there exists a unique x ∈ X such that x = J ∗ φ. Besides this, JJ ∗ = IX ∗ and J ∗ J = IX . Now (1.6.17) follows from (1.6.18) by the substitution φ = Jx and ψ = Jy. Theorem 1.6.4 Let X be a uniformly convex Banach space. Then for any R > 0 and any x, y ∈ X such that x ≤ R, y ≤ R the following inequality holds: Jx − Jy, x − y ≥ (2L)−1 δX (c−1 2 x − y),

(1.6.19)

where c2 = 2max{1, R}. Proof. As it was shown in [127], for x, y ∈ X, the equality

implies

x2 + y2 = 2

(1.6.20)

2−1 (x + y)2 ≤ 1 − δX (2−1 x − y).

(1.6.21)

If X is uniformly convex then the function δX () is increasing, δX (0) = 0 and 0 ≤ δX () < 1. Denote R12 = 2−1 (x2 + y2 ) and introduce the new variables by the formulas x ˜=

Then

y x , y˜ = . R1 R1

y 2 = R1−2 (x2 + y2 ) = 2. ˜ x2 + ˜

Hence, the inequality ˜ x − y˜ ˜ + y˜ y 2 ˜ x2 + ˜ 2 x − ≥ δX 2 2 2

is satisfied by (1.6.21). If now we return to the old variables x and y, then we will obtain D ≥ R12 δX

x − y

2R1

,

where D is defined by (1.6.6). Consider two cases. 1. Let R1 ≥ 1. Then for x ≤ R and y ≤ R, one gets R1 ≤ R and D ≥ δX

x − y

2R1

≥ δX

x − y

2R

.

(1.6.22)

46

1


2. Let R1 < 1. It is known that in a uniformly convex Banach space X for any η ≥ > 0 2 δX (η) ≥ (4L)−1 η 2 δX (),

(1.6.23)

where 1 < L < 1.7 (cf. (1.6.4)). Hence, δX

x − y

2R1

≥ R1−2 (4L)−1 δX

x − y

2

≥ (4L)−1 δX

x − y

2

.

(1.6.24)

Combining (1.6.22) and (1.6.24) we finally deduce D ≥ (4L)−1 δX (c−1 2 x − y), c2 = 2max{1, R}. Denote ϕ(x) =

2−1 x2 .

(1.6.25)

Then

2−1 ϕ(x) + 2−1 ϕ(y) − ϕ(2−1 (x + y)) ≥ (8L)−1 δX (c2 −1 x − y). Passing to a duality mapping and using Lemma 1.3.11 we obtain (1.6.19). The theorem is proved. Remark 1.6.5 It follows from (1.6.22) and (1.6.24) that for arbitrary x, y ∈ X, 2x2 + 2y2 − x + y2 ≥ L−1 δX (c2 −1 x − y), where c2 = c2 (x, y) = 2max{1,

2−1 (x2 + y2 )}.

(1.6.26)

(1.6.27)

In addition, the estimate (1.6.19) is satisfied if c2 = 2max {1, R} is replaced by (1.6.27). Remark 1.6.6 It should be observed that only in a uniformly convex space is δX () a strictly increasing function and δX (0) = 0. In an arbitrary Banach space, estimate (1.6.19) guarantees, in general, the monotonicity property of normalized duality mapping J. The next assertion follows from Theorem 1.6.4 (cf. Corollary 1.6.3). Corollary 1.6.7 Let X be a uniformly smooth and strictly convex Banach space. Then for any x, y ∈ X such that x ≤ R, y ≤ R, the following inequality holds: Jx − Jy, x − y ≥ (2L)−1 δX ∗ (c−1 2 Jx − Jy∗ ),

(1.6.28)

where c2 = 2max{1, R}. Corollary 1.6.8 Let X be a uniformly smooth and strictly convex Banach space. Suppose that the function gX ∗ () = −1 δX ∗ () is increasing. Then for all x, y ∈ X, x ≤ R, y ≤ R the following estimate is valid: −1 Jx − Jy∗ ≤ c2 gX ∗ (2Lc2 x − y).

(1.6.29)

If X is a uniformly convex and smooth Banach space, gX () = −1 δX () is increasing, then −1 x − y ≤ c2 gX (2Lc2 Jx − Jy∗ ).

(1.6.30)


47

Proof. By the Cauchy−Schwarz inequality, (1.6.28) can be estimated as follows: (2L)−1 δX ∗ (c−1 2 Jx − Jy∗ ) ≤ Jx − Jy∗ x − y. Then gX ∗

Jx − Jy ∗

c2

=

c2 δX ∗ (c−1 2 Jx − Jy∗ ) ≤ 2Lc2 x − y, Jx − Jy∗

and (1.6.29) is obviously satisfied. Here it is necessary to recall that δX () ≥ 0 for all ≥ 0 and gX (0) = 0. The estimate (1.6.30) results from (1.6.19). Remark 1.6.9 Note that if gX (t) does not increase strictly for all t ∈ [0,2] but there exists a non-negative increasing continuous function g˜X (t) such that gX (t) ≥ g˜X (t), then (1.6.30) −1 −1 remains still valid if gX (·) is replaced by g˜X (·). The same can be said for (1.6.29) in X ∗ . Inequality (1.6.29) defines the modulus of uniform continuity of a normalized duality mapping J on set B(θX , R) in a uniformly smooth Banach space X, that is, it is the function −1 ωR (t) in (1.6.1). Namely, ωR (t) = c2 gX ∗ (2Lc2 t). Estimates (1.6.5), (1.6.17), (1.6.19) and (1.6.28) are reduced to calculation of the moduli of convexity and smoothness of the spaces X and X ∗ . However, in practice, one usually uses upper estimates for the modulus of smoothness and lower estimates for the modulus of convexity of the spaces. For the spaces X of type lp and Lp , 1 < p ≤ 2, the modulus of smoothness ρX (τ ) can be calculated by the Lindenstrauss formula ρX (τ ) = sup

τ

2

− δX ∗ (), 0 ≤ ≤ 2 .

(1.6.31)

For that, it is necessary to use Hanner’s equality

δX ∗ () = 1 − 1 − (2−1 )q

1 q

,

which is true if X ∗ is lq or Lq , p−1 + q −1 = 1. By (1.6.31), it is not difficult to obtain the equality 1 ρX (τ ) = (1 + τ p ) p − 1. By virtue of the inequality ar − br ≤ rbr−1 (a − b), 0 ≤ r ≤ 1, a, b > 0, we have

ρX (τ ) ≤ p−1 τ p .

It makes sense to use this estimate only if τ ≤ 1, because the relation ρX ∗ (τ ) ≤ τ is more precise by order as τ > 1. We address now an estimate of the modulus of convexity of the spaces lq , Lq , 1 < q ≤ 2. First of all, we observe that there holds the following functional identity [94]: Φ(, δX ∗ ()) = (1 − δX ∗ () − 2−1 )q + (1 − δX ∗ () + 2−1 )q − 2 ≡ 0.

48

1


From this, after some simple algebra, we have δX ∗ () = −

∂Φ ∂Φ −1

∂

∂δX ∗

≥

(q − 1) . 8

Integrating both parts of the previous relation, one gets δX ∗ () ≥

(q − 1)2 , 0 ≤ ≤ 2, 1 < q ≤ 2. 16

Note that it coincides (up to constants) with Hanner’s asymptotic result in [94]. Using (1.6.31) again we obtain ρX ∗ (τ ) ≤ (p − 1)τ 2 , p ≥ 2. (1.6.32) Thus, in the spaces X = lp and X = Lp , 1 < p < ∞, one has ρX (τ ) ≤ (p − 1)τ 2 , δX () ≥ p−1

and ρX (τ ) ≤

τp , p

δX () ≥

p

2

(p − 1)2 , 16

,

p ≥ 2,

1 < p ≤ 2.

(1.6.33)

(1.6.34)

p , 1 < p < ∞. The same upper and lower estimates are also valid in the Sobolev spaces Wm By (1.6.5), (1.6.19), (1.6.33) and (1.6.34), if x ≤ R, y ≤ R and if p ≥ 2, then

Jx − Jy, x − y ≤ 8 + (p − 1)c1 x − y2 , and Jx − Jy, x − y ≥

x − yp , 2p+1 Lpcp2

c1 = 8max{L, R}),

c2 = 2max{1, R}.

Corollary 1.6.8 implies the Lipschitz-continuity of the normalized duality mapping on each p bounded set in the spaces lp , Lp and Wm , when p ≥ 2, namely, Jx − Jy∗ ≤ C(R)x − y, If 1 < p ≤ 2 then

C(R) = 32Lc22 (q − 1)−1 .

Jx − Jy, x − y ≤ 8x − y2 + c1 p−1 x − yp

and Jx − Jy, x − y ≥

(1.6.35)

(p − 1)x − y2 . 32Lc22

The Hölder-continuity of the normalized duality mapping expressed by the inequality ¯ ¯ − yp−1 , C(R) = Jx − Jy∗ ≤ C(R)x

p p−1 p p−1 2p−1 c2 L 2 , p−1

follows again from Corollary 1.6.8. In a Hilbert space 1 ρH (τ ) = (1 + τ 2 ) 2 − 1 ≤ 2−1 τ 2

1.6 Banach Spaces Geometry and Related Duality Estimates and

2 1

49

2 . 8 2 It is known that a Hilbert space has the smallest modulus of smoothness among all uniformly smooth Banach spaces, and it has the biggest modulus of convexity among all uniformly convex Banach spaces, that is ρH (τ ) ≤ ρX (τ ), δX () ≥ δH (). Furthermore, there is a duality relationship: if δX () ≥ k1 γ1 then ρX ∗ (τ ) ≤ k2 τ γ2 , where k1 and k2 are positive constants and γ1−1 + γ2−1 = 1. These assertions follow from (1.6.31). δH () = 1 − 1 −

2

≥

Remark 1.6.10 The estimates like (1.6.5) and (1.6.19) can be also obtained for duality mappings J µ with the gauge function µ(t) = ts−1 , s > 1. Suppose that X and X ∗ are reflexive strictly convex spaces. Introduce the Lyapunov functional W (x, y) : X × X → R1 defined by the formula W (x, y) = 2−1 (x2 − 2Jx, y + y2 )

(1.6.36)

and study its properties. 1. Show that W (x, y) ≥ 0 for all x, y ∈ X. Indeed, W (x, y) ≥ 2−1 (Jx2∗ − 2Jx∗ y + y2 ) = 2−1 (x − y)2 ≥ 0.

(1.6.37)

2. By (1.6.37), it is easy to see that W (x, y) → ∞ as x → ∞ or/and y → ∞. On the other hand, W (x, y) ≤ 2−1 (x + y)2 . The latter inequality implies the following assertion: If W (x, y) → ∞ then x → ∞ or/and y → ∞. 3. One can verify by direct substitution that W (x, x) = 0. 4. Let y be a fixed element of X. Consider the general functional W1 (φ, y) : X ∗ ×X → R1 such that W1 (φ, y) = 2−1 (φ2∗ − 2φ, y + y2 ). Since x2 = Jx2∗ , (1.6.36) is presented in the equivalent form as W1 (Jx, y) = 2−1 (Jx2∗ − 2Jx, y + y2 ).

(1.6.38)

Since the space X ∗ is smooth, W1 (φ, y) has the Gâteaux derivative W1 (φ, y) with respect to φ = Jx. It is not difficult to be sure that grad W1 (φ, y) = J ∗ φ − y.

(1.6.39)

According to Lemma 1.2.6, W1 (φ, y) is convex and lower semicontinuous in the whole space X ∗ . Hence, grad W1 (φ, y) : X ∗ → X is a monotone and single-valued operator for all φ ∈ X ∗ . Definition 1.2.2 of the subdifferential gives the relation W1 (φ, y) − W1 (ψ, y) ≥ φ − ψ, J ∗ ψ − y .

(1.6.40)

50

1


Using the fact that there exist a unique x ∈ X and unique z ∈ X such that for every φ ∈ X ∗ and for every ψ ∈ X ∗ , respectively, J ∗ φ = x and J ∗ ψ = z, we must have that W (x, y) − W (z, y) ≥ Jx − Jz, z − y .

(1.6.41)

It is clear that (1.6.41) is valid for all x, y, z ∈ X. Analogously, considering the functional W (x, y) with respect to the variable y with a fixed element x, we conclude that W (x, y) is convex and lower semicontinuous, and its gradient is defined in the smooth space X as follows: grad W (x, y) = Jy − Jx.

(1.6.42)

W (x, y) − W (x, z) ≥ Jz − Jx, y − z ,

(1.6.43)

This yields the second relation

which is satisfied for all x, y, z ∈ X again. 5. By (1.6.41), one has W (y, y) − W (x, y) ≥ Jy − Jx, x − y

∀x, y ∈ X.

Now the property 3 leads to inequality W (x, y) ≤ Jx − Jy, x − y ∀x, y ∈ X.

(1.6.44)

Taking now into account Theorem 1.6.1, we obtain for all x, y ∈ B(θX , R) the estimate W (x, y) ≤ 8x − y2 + c1 ρX (x − y),

(1.6.45)

where c1 = 8max{L, R}. Let X be a uniformly convex Banach space. Rewrite the inequality (1.6.25) in the form 2−1 (x + z)2 ≤ 2−1 x2 + 2−1 z2 − (4L)−1 δX (c−1 2 x − z). By the definition of a subdifferential, we deduce 2−1 (x + z)2 ≥ x2 + Jx, z − x , and then

z2 ≥ x2 + 2Jx, z − x + (2L)−1 δX (c−1 2 x − z),

in view of (1.6.46). Replace z by 2−1 (x + y) in the last inequality to obtain 2−1 (x + y)2 ≥ x2 + Jx, y − x + (2L)−1 δX ((2c2 )−1 x − y) = Jx, y + (2L)−1 δX ((2c2 )−1 x − y). Thus,

Jx, y ≤ 2−1 (x + y)2 − (2L)−1 δX ((2c2 )−1 x − y).

(1.6.46)


51

It follows from (1.6.36) that W (x, y) ≥ 2−1 x − 2−1 (x + y)2 + 2−1 y2 + (2L)−1 δX ((2c2 )−1 x − y), and due to (1.6.46), we write down −1 −1 W (x, y) ≥ (4L)−1 δX (c−1 2 x − y) + (2L) δX ((2c2 ) x − y).

(1.6.47)

Since the function t−1 δX (t) is non-decreasing, the inequality δX (2−1 ) ≤ 2−1 δX () holds. Then (1.6.47) yields for all x, y ∈ B(θX , R) the estimate W (x, y) ≥ L−1 δX ((2c2 )−1 x − y), where c2 = 2max{1, R}. Combining (1.6.45) and the last inequality we obtain L−1 δX ((2c2 )−1 x − y) ≤ W (x, y) ≤ 8x − y2 + c1 ρX (x − y).

(1.6.48)

6. If X is a Hilbert space then W (x, y) = 2−1 x − y2 . We may consider that δ() ≥ cγ , γ ≥ 0, c > 0.

(1.6.49)

This assumption is not so limiting because, on the one hand, the direct calculations of δ() p in the spaces lp , Lp , Wm , 1 < p ≤ ∞, and in the Orlich spaces with Luxemburg norm [24, 119] show that (1.6.49) is true. On the other hand, it is asserted in [168] that the same relates (to within isomorphism) to super-reflexive Banach space. In these cases, there exists a constant c > 0 such that ρX ∗ (τ ) ≤ cτ γ/(γ−1) . Let us make a few remarks concerning the spaces lp and Lp . In [63] the following parallelogram inequalities are presented: 2x2 + 2y2 − x + y2 ≥ (p − 1)x − y2 , 2

2

2

1 < p ≤ 2;

(1.6.50)

p ≥ 2.

(1.6.51)

2

2x + 2y − x + y ≤ (p − 1)x − y , They imply the estimates

Jx − Jy, x − y ≥ (p − 1)x − y2 , 1 < p ≤ 2;

(1.6.52)

Jx − Jy∗ ≤ (p − 1)x − y, p ≥ 2.

(1.6.53)

The other well-known relations in spaces

lp

and L are the Clarkson inequalities [211]:

xp + yp x + y p − ≤

2

2

p

x − y p ,

2

1 < p ≤ 2;

(1.6.54)

x − y p xp + yp x + y p (1.6.55) − , p ≥ 2. ≥ 2 2 2 They yield, respectively, the following estimates of duality mapping J p with the gauge function µ(t) = tp−1 :

J p x − J p y, x − y ≤ 23−p p−1 x − yp ,

1 < p ≤ 2,

(1.6.56)

52 and

1


J p x − J p y, x − y ≥ 22−p p−1 x − yp ,

p ≥ 2.

(1.6.57)

Suppose that there exists a strictly increasing continuous function δ˜X (t), δ˜X (0) = 0, and a positive constant K such that δX (t) ≥ K δ˜X (t). We known the following estimates:

and

J p x − J p y, x − y ≥ C p δ˜X (C −1 x − y) ∀x, y ∈ X

(1.6.58)

J p x − J p y, x − y ≤ C p ρX (C −1 x − y) ∀x, y ∈ X,

(1.6.59)

where C(x, y) = max {x, y}. Unlike Theorems 1.6.1 and 1.6.4 (when they are applied to the spaces l p and Lp ), the parallelogram inequality (1.6.50), (1.6.51) and the Clarkson inequality (1.6.54), (1.6.55) admit only one-sided estimates for each p = 2. By analogy with (1.6.50) and (1.6.51), the inequalities (1.6.15) and (1.6.26) can be treated, respectively, as the upper parallelogram inequality in a uniformly smooth Banach space and lower parallelogram inequality in a uniformly convex Banach space.

1.7

Equations with Maximal Monotone Operators

Assume that X is a reflexive Banach space, X and X ∗ are strictly convex. We begin to study solvability of equations with maximal monotone operators. First of all, we prove the following fundamental auxiliary statement: Lemma 1.7.1 (Debrunner−Flor) Let Ω be a convex compact set in X, G be a monotone set in the product Ω × X ∗ , F : Ω → X ∗ be a continuous operator, h ∈ X ∗ . Then there exists an element u ∈ Ω such that the inequality f + F u − h, x − u ≥ 0

(1.7.1)

holds for all (x, f ) ∈ G. Proof. Since the set of pairs (x, f − h), where (x, f ) ∈ G, is also monotone, we may consider, without loss of generality, that h = θX ∗ . We want to prove the lemma by contradiction. Suppose that (1.7.1) is not true. Then for every u ∈ Ω, there is a pair (x, f ) ∈ G such that f + F u, x − u < 0. For each (x, f ) ∈ G, we are able to construct the set N (x, f ) = {y ∈ Ω | f + F y, x − y < 0}. The sets N (x, f ) with (x, f ) ∈ G form a family of open coverings of the compact set Ω. Therefore, there exists a finite subcovering, that is, finite family (xi , fi ) ∈ G, 1 ≤ i ≤ n, such that n Ω=

i=1

N (xi , fi ).

1.7 Equations with Maximal Monotone Operators

53

On the basis of that finite covering, we build a continuous partition of the unit on Ω. In other words, we build n continuous functions βi : N (xi , fi ) → [0, 1], i = 1, 2, ..., n, such that follows:

n

i=1 βi (y)

= 1 for all y ∈ Ω. Define operators T1 : Ω → X and T2 : Ω → X ∗ as T1 (y) =

n

βi (y)xi , T2 (y) =

i=1

n

βi (y)fi .

i=1

Since Ω is convex, the operator T1 acts from Ω to Ω. Furthermore, T1 is continuous. Then, by the Schauder principle, there exists y0 ∈ Ω such that T1 (y0 ) = y0 , i.e., y0 is a fixed point of T1 . Next let p(y) = T2 (y) + F y, T1 (y) − y = p1 (y) + p2 (y), where p1 (y) =

n

βi2 (y)fi + F y, xi − y

i=1

and p2 (y) =

βi (y)βj (y) fi + F y, xj − y + fj + F y, xi − y .

1≤i<j≤n

For every y ∈ Ω, there exists at least one number m (1 ≤ m ≤ n) such that βm (y) = 0, whereas y ∈ N (xm , fm ). Therefore, fm + F y, xm − y < 0. Thus, p1 (y) < 0 for all y ∈ Ω. The monotonicity property of G gives further the following relations for y ∈ N (xi , fi ) ∩ N (xj , fj ) : the products βi (y)βj (y) > 0 and fi + F y, xj − y + fj + F y, xi − y = fi + F y, xi − y + fj + F y, xj − y + fi − fj , xj − xi < 0. Hence, p(y) < 0 for all y ∈ Ω. On the other hand, p(y0 ) = T2 (y0 ) + F y0 , T1 (y0 ) − y0 = 0. This contradiction proves the lemma. We now introduce the solution concept in the sense of inclusion. Definition 1.7.2 An element x0 ∈ D(A) such that f ∈ Ax0 is called the solution (in the sense of inclusion) of the equation Ax = f with a maximal monotone operator A. Remark 1.7.3 If the operator A is single-valued in a solution x0 of the equation Ax = f, then f = Ax0 . In this case, x0 is called the classical solution. ∗

Theorem 1.7.4 Let A : X → 2X be a maximal monotone operator with D(A) and J : X → X ∗ be a normalized duality mapping. Then R(A + αJ) = X ∗ for all α > 0.

54

1


Proof. It suffices to consider the case α = 1 and show that θX ∗ ∈ R(A + J). Let Xn be an n-dimensional subspace of X, En : Xn → X be a natural imbedding operator. Denote by En∗ : X ∗ → Xn∗ an adjoint operator to En , F = En∗ JEn , and Ω = {x ∈ Xn | x ≤ r}. Put h = θX ∗ and define large enough number r satisfying condition B(θX , r) ∩ D(A) = ∅. Then, by the Debrunner−Flor Lemma, there exist xrn ∈ Ω and ynr = F xrn such that y + ynr , x − xrn ≥ 0 ∀(x, y) ∈ G,

(1.7.2)

where G = gr(En∗ AEn ) ∩ Ω × Xn∗ . Hence, xrn 2 ≤ y∗ x + xrn (x + y∗ )

∀(x, y) ∈ G.

Thus, the sequence {xrn } is bounded. We know that xrn ∈ Xn for all r, therefore, xrn → xn ∈ Xn as r → ∞. Since the operator F = En∗ JEn is continuous (see Section 1.5) and the subspace Xn∗ is closed, we conclude that ynr → yn ∈ Xn∗ as r → ∞ and yn = F xn . Hence, (1.7.2) implies (1.7.3) y + yn , x − xn ≥ 0 ∀(x, y) ∈ grA, x ∈ Xn . It results from the coerciveness of J and the latter inequality that the sequences {xn } and {yn } remain bounded in X and X ∗ , respectively, when Xn are running through the ordered increasing filter of finite-dimensional subspaces in X. Therefore, there exist x ¯ ∈ Ω and ¯ and yn y¯. y¯ ∈ X ∗ such that xn x Let ϕ(x) = 2−1 x2 , x ∈ X, and let ϕ∗ (x∗ ) be a conjugate to ϕ(x) functional. Since x is a weakly lower semicontinuous functional, by Definition 1.2.10 of the functional ϕ∗ (x∗ ) and by Theorem 1.2.11, we obtain y ) ≤ lim inf ϕ(xn ) + lim inf ϕ∗ (yn ) = lim inf yn , xn . ¯ y, x ¯ ≤ ϕ(¯ x) + ϕ∗ (¯ n→∞

n→∞

n→∞

(1.7.4)

Furthermore, by (1.7.3), one has yn , xn ≤ y, x + yn , x − y, xn . Hence, y , x − y, x ¯ ∀(x, y) ∈ grA, lim inf yn , xn ≤ y, x + ¯ n→∞

x ∈ Xn .

(1.7.5)

∞

Observe that n=1 Xn is dense in X. Therefore, (1.7.5) is valid for all (x, y) ∈ grA, x ∈ X. In view of (1.7.4), we thus obtain −¯ y − y, x ¯ − x ≥ 0 ∀(x, y) ∈ grA. Since the operator A is maximal monotone, this inequality gives the inclusion: −¯ y ∈ A¯ x. Assuming x = x ¯, y = y¯ in (1.7.5) we conclude that lim inf yn , xn ≤ ¯ y, x ¯ . n→∞


55

This fact and (1.7.4) lead to the following result: y ). ¯ y, x ¯ = ϕ(¯ x) + ϕ∗ (¯ Hence, y¯ = J x ¯ (see Theorem 1.2.11), i.e., θX ∗ ∈ (A + J)¯ x. The theorem is proved. By making use of this theorem we establish the following important result. ∗

Theorem 1.7.5 Let A : X → 2X be a maximal monotone and coercive operator. Then R(A) = X ∗ . Proof. Choose an arbitrary element f ∈ X ∗ . Owing to Theorem 1.7.4, for every α > 0, there exists xα ∈ D(A) such that yα + αJxα = f, yα ∈ Axα . Then

(1.7.6)

f ∗ xα ≥ f, xα = yα , xα + αxα 2 ≥ yα , xα .

Therefore,

yα , xα ≤ f ∗ , xα

yα ∈ Axα .

Since A is coercive, it follows from the last inequality that the sequence {xα } is bounded. ¯ ∈ X as α → 0. By (1.7.6), one gets yα = f − αJxα and then the monotonicity Then xα x property of A gives f − αJxα − y, xα − x ≥ 0 ∀(x, y) ∈ grA. Letting α → 0, we obtain f − y, x ¯ − x ≥ 0 ∀(x, y) ∈ grA. Since the operator A is maximal monotone, we deduce by Proposition 1.4.3 that f ∈ A¯ x. The theorem is proved. ∗

Corollary 1.7.6 Let A : X → 2X be a maximal monotone operator whose domain D(A) is bounded. Then R(A) = X ∗ . The next assertions follow from Theorems 1.4.6 and 1.7.5. Corollary 1.7.7 Let A : X → X ∗ be a monotone hemicontinuous and coercive operator with D(A) = X. Then R(A) = X ∗ . Note that this statement is known in the literature as the Minty−Browder theorem. Corollary 1.7.8 Let J : X → X ∗ and J ∗ : X ∗ → X be normalized duality mappings. Then R(J) = X ∗ and R(J ∗ ) = X.

56

1


Theorem 1.7.9 Suppose that A : X → 2X is a maximal monotone operator, f ∈ X ∗ and there exists a number r > 0 such that y − f, x ≥ 0 for all y ∈ Ax as x ≥ r. Then there exists an element x ¯ ∈ X such that f ∈ A¯ x and ¯ x ≤ r. Proof. Consider again the equality (1.7.6). It is obvious that yα − f, xα = −αxα 2 .

(1.7.7)

¯ = xα . If xα = θX for some α > 0 then (1.7.6) immediately gives yα = f, i.e., f ∈ Axα and x If xα = θX for all α > 0 then, by (1.7.7), we have yα −f, xα < 0. In this case, the conditions of the theorem imply the estimate xα < r for all α > 0. Then the inclusion f ∈ A¯ x is proved following the pattern of the proof given in Theorem 1.7.5, and the estimate ¯ x ≤ r is obtained by the weak lower semicontinuity of the norm in X. Remark 1.7.10 Under the conditions of Theorems 1.7.5, 1.7.9 and Corollary 1.7.7, if operator A is strictly monotone, then the equation Ax = f for all f ∈ X ∗ has a unique solution. ∗

Lemma 1.7.11 Let A : X → 2X be a maximal monotone operator. Then R(A) = X ∗ if and only if A−1 is locally bounded on R(A). Proof. Let R(A) = X ∗ be given, then D(A−1 ) = X ∗ . Since A−1 is monotone, it is locally bounded on R(A) = X ∗ . Let now A−1 be locally bounded on R(A). Prove that R(A) = X ∗ . For this aim, it is sufficient to show that the set R(A) is both open and closed in X ∗ at the same time. Let fn → f, fn ∈ Axn , i.e., fn ∈ R(A). Since A−1 is locally bounded on R(A), the sequence {xn } is bounded at least as n is sufficiently large. Then there exists some subsequence {xnk } ⊂ {xn } which weakly converges to x ∈ X, and (x, f ) ∈ grA because grA is demiclosed; see Lemma 1.4.5. Hence, f ∈ R(A), which means that R(A) is a closed set. Assume now that (x, f ) ∈ grA is given and r> 0 such that the operator A−1 is bounded r belongs to R(A). Consider the equality on B ∗ (f, r). Show that an element g ∈ B ∗ f, 2

gα + αJ(xα − x) = g, gα ∈ Axα , α > 0.

(1.7.8)

By the monotonicity of A, we have g − αJ(xα − x) − f, xα − x ≥ 0. r Hence, αxα − x ≤ g − f ∗ < . Then, in view of (1.7.8), one gets 2 r g − gα ∗ = αxα − x < . 2 Therefore, gα − f ∗ ≤ gα − g∗ + g − f ∗ < r, i.e., gα ∈ B ∗ (f, r). Using the local boundedness of A−1 on B ∗ (f, r), we conclude then that the sequence {xα } is bounded in X. Therefore, g − gα ∗ = αxα − x → 0 as α → 0. We established above a closedness of R(A). Thus, g ∈ R(A), which implies that R(A) is open. The proof is complete.


57

∗

Corollary 1.7.12 If A : X → 2X is a maximal monotone and weakly coercive operator, then R(A) = X ∗ . Proof. Indeed, it follows from the Definition 1.1.45 of the weak coerciveness of A that A−1 is bounded. Then the assertion is the consequence of Lemma 1.7.11. Next we present the following important result. ∗

Theorem 1.7.13 Let A : X → 2X be a monotone operator. Then A is a maximal monotone operator if and only if R(A + J) = X ∗ . ∗ Proof. By Zorn’s lemma, there exists a maximal monotone extension A¯ : X → 2X ¯ ¯ such that grA ⊆ grA. Applying Theorem 1.7.4 to the operator A we deduce for all f ∈ X ∗ ¯ such that y + Jx = f. Hence, that there exist the unique elements x ∈ X and y ∈ Ax grA = grA¯ if and only if every element f ∈ X ∗ can be presented in the form of f = y + Jx for some x ∈ X, y ∈ Ax, i.e., if and only if R(A + J) = X ∗ .

Corollary 1.7.14 Suppose that H is a Hilbert space, A : H → 2H , operator (I + αA)−1 is defined for all α > 0 on the whole space H and I − A is a nonexpansive mapping. Then A is maximal monotone. Proof. It follows from Section 1.3 (Example 3) that A is monotone. Further, by the hypothesis, R(A + I) = H. This is enough in order to apply the previous theorem and obtain the claim. ∗

Theorem 1.7.15 A subdifferential ∂ϕ : X → 2X of a proper convex lower semicontinuous functional ϕ : X → R1 is a maximal monotone operator. Proof. In Section 1.3 (Example 2), monotonicity of ∂ϕ has been established on D(∂ϕ). According to Theorem 1.7.13, to prove the maximal monotonicity of ∂ϕ, we have to show that R(∂ϕ + J) = X ∗ , that is, that for any f ∈ X ∗ there is x ∈ D(∂ϕ) satisfying the inclusion f ∈ (∂ϕ + J)x. Construct a proper convex lower semicontinuous functional Φ(y) = 2−1 y2 + ϕ(y) − f, y ∀y ∈ D(ϕ). The definition of ∂ϕ at a point x0 ∈ D(∂ϕ) gives ϕ(y) ≥ ϕ(x0 ) + ∂ϕ(x0 ), y − x0 ∀y ∈ D(ϕ). Then the following inequalities hold: Φ(y) ≥ 2−1 y2 + ϕ(x0 ) + ∂ϕ(x0 ), y − x0 − f, y ≥ ϕ(x0 ) − ∂ϕ(x0 ), x0 + 2−1 y(y − 2∂ϕ(x0 )∗ − 2f ∗ ). Hence, Φ(y) → +∞ as y → ∞. Thus, by Theorem 1.1.23 and Lemma 1.2.5, there exists a point x ∈ X at which Φ(y) reaches minimum. Then we can write down that ∗ θX ∗ ∈ ∂Φ(x) = (J + ∂ϕ)x − f or f ∈ (J + ∂ϕ)x. Actually, the operator ∂ϕ : X → 2X is maximal monotone.

58

1


Corollary 1.7.16 Any monotone potential operator A : X → 2X is maximal monotone. Proof. See Lemma 1.2.6 and Theorem 1.7.15. ∗

Theorem 1.7.17 Let A : X → 2X be a maximal monotone operator. Then the set D(A) is convex.

Proof. For any x0 ∈ X, consider the equation J(x − x0 ) + αAx = 0, α > 0.

(1.7.9)

By Theorem 1.7.13, it has a solution xα ∈ D(A) which is unique because A is monotone and J is strictly monotone. Then there exists yα ∈ Axα such that J(xα − x0 ) + αyα = θX ∗ . If (v, u) ∈ grA then the following equalities hold: xα − x0 2 = J(xα − x0 ), xα − x0 = J(xα − x0 ), xα − v + J(xα − x0 ), v − x0 = αu − yα , xα − v + αu, v − xα + J(xα − x0 ), v − x0 . Now the monotonicity of A yields the relation xα − x0 2 ≤ αu, v − xα + J(xα − x0 ), v − x0 .

(1.7.10)

By the definition of J, we obtain xα − x0 2 ≤ αuv − xα + xα − x0 v − x0 . Hence, the sequences {J(xα − x0 )} and {xα } are bounded for all α > 0. There exists a subsequence {αn }, αn → 0 as n → ∞, such that J(xαn − x0 ) y¯ ∈ X ∗ . Thus, (1.7.10) gives the estimate lim sup xαn − x0 2 ≤ ¯ y , v − x0 αn →0

∀v ∈ D(A).

(1.7.11)

Is is obvious that (1.7.11) holds for all v ∈ D(A). If x0 ∈ D(A) then we conclude from (1.7.11) that xα → x0 as α → 0. Choose now elements x01 , x02 ∈ D(A) and put xt = tx01 + (1 − t)x02 , t ∈ [0, 1]. Let xtα ∈ D(A) be a solution of the equation (1.7.9) with xt in place of x0 . Then, by (1.7.11), we deduce that xtα → xt , xtα ∈ D(A), i.e., xt ∈ D(A). Hence, the set D(A) is convex.

Corollary 1.7.18 If an operator A is maximal monotone, then the set R(A) is convex.

1.8 Summation of Maximal Monotone Operators

59

Proof. This claim can be proved by applying Theorem 1.7.17 to the inverse map A−1 . Consider in more detail the property of the local boundedness of a maximal monotone operator. We proved in Theorem 1.3.16 that an arbitrary monotone operator is locally bounded at any interior point of its domain. Hence, a maximal monotone operator has the same property. Furthermore, for a maximal monotone operator the statement about the local boundedness can be specified. Namely, the following theorem holds: ∗

Theorem 1.7.19 If int D(A) = ∅, then a maximal monotone operator A : X → 2X is unbounded at any boundary points of its domain. Moreover, the range of A has at least one semi-line at these points. Proof. Let x ∈ ∂D(A). Consider the set M = D(A), the closure of D(A). Recall that M is convex and closed, and int M = ∅. It is clear that x ∈ ∂M. Therefore, it is possible to construct a supporting hyperplane to the set M at the point x. In other words, we assert that there exists an element y ∈ X ∗ (y = θX ∗ ) such that

y, x − u ≥ 0 ∀u ∈ D(A).

(1.7.12)

Let z ∈ Ax and wλ = z + λy ∈ X ∗ , λ ≥ 0. Since A is monotone, we obtain by (1.7.12) v − wλ , u − x = v − z, u − x − λy, u − x ≥ 0

∀u ∈ D(A), ∀v ∈ Au.

(1.7.13)

Since an operator A is also maximal monotone, it follows from (1.7.13) that wλ ∈ Ax. Hence, the range of the maximal monotone operator A at a boundary point x of D(A) contains the semi-line {z + λy, | z ∈ Ax, λ ≥ 0}. ∗

Corollary 1.7.20 If A : X → 2X is a maximal monotone operator and R(A) is a bounded set in X ∗ , then D(A) = X. ∗

Corollary 1.7.21 If A : X → 2X is a monotone operator and there exist elements y ∈ X ∗ and yn ∈ Axn , n = 1, 2, ... such that lim xn = ∞,

n→∞

lim yn − y∗ = 0,

n→∞

then y is a boundary point of R(A).

1.8

Summation of Maximal Monotone Operators

Construct the indicator function Ir (x) of the ball B(θX , r) ⊂ X, r > 0 :

Ir (x) =

0, +∞,

x ∈ B(θX , r), x ∈ B(θX , r).

60

1


By Theorem 1.7.15, its subdifferential ∂Ir : X → 2X is a maximal monotone operator and D(∂Ir ) = B(θX , r). It is obvious that ∂Ir (x) = θX ∗ when x < r and ∂Ir (x) = ∅ for all x with x > r. The points x ∈ X such that x = r are the boundary for D(∂Ir ), therefore, by Theorem 1.7.19, the ranges of ∂Ir at these points form semi-lines in X ∗ . Since Jx, x − y ≥ 0 as x = r and y < r, a semi-line at the point x ∈ ∂D(∂Ir ) is as follows: {λJx | λ ≥ 0}. ∗ Hence, the operator ∂Ir : X → 2X has the following representation:    θX ∗ ,

∂Ir (x) =

∅,   λJx,

x < r; x > r; x = r, λ ≥ 0.

(1.8.1)

∗

Lemma 1.8.1 Let A : X → 2X be a monotone operator, θX ∈ D(A) and let there exist a number r0 > 0 such that the operator A + ∂Ir is maximal monotone for all r ≥ r0 . Then A is a maximal monotone operator too. Proof. Without loss of generality, we presume that θX ∗ ∈ A(θX ). If we shall prove that R(A + J) = X ∗ then the claim to be proved follows from Theorem 1.7.13. Take an arbitrary element f ∈ X ∗ and a number r ≥ r0 such that r ≥ f ∗ . By the hypothesis, an operator A + ∂Ir is maximal monotone as r ≥ r0 , hence, Theorem 1.7.4 guarantees existence of an x ∈ X such that f ∈ (A + ∂Ir + J)x = (A + J)x + ∂Ir (x). (1.8.2) Since D(A+∂Ir ) = D(A)∩D(∂Ir ) = D(A)∩B(θX , r), we conclude that x ≤ r. If x < r then the conclusion of the lemma follows from (1.8.1) and (1.8.2). Consider the case when x = r. Taking into account (1.8.1), we may rewrite (1.8.2) in the form: f ∈ (A+J)x+λJx where λ ≥ 0. If λ = 0 then the lemma is proved. Assume that λ > 0 and let y ∈ Ax and f = y + (1 + λ)Jx. Then y, x + (1 + λ)x2 = f, x . (1.8.3) Since the operator A is monotone and since θX ∗ ∈ A(θX ), we have the inequality y, x ≥ 0 and (1.8.3) implies (1 + λ)x2 ≤ f, x ≤ f ∗ x. It follows that

x ≤ (1 + λ)−1 f ∗ < r.

Thus, we come to the contradiction which establishes the result. ∗

Lemma 1.8.2 Let A : X → 2X be a maximal monotone operator. Then Aα = (A + αJ)−1 : X ∗ → X, α > 0, is a single-valued, monotone and demicontinuous mapping with D(Aα ) = X ∗ . Proof. By Theorem 1.7.4, R(A + αJ) = X ∗ for all α > 0 and thus D(Aα ) = X ∗ . Since the operator A + αJ is strictly monotone in view of Lemma 1.5.5 and Proposition 1.3.5, the equation Ax + αJx = f is uniquely solvable for any f ∈ X ∗ . This proves that the operator


61

Aα is single-valued. To show that Aα is demicontinuous, choose {xn } ⊂ X and let f ∈ X ∗ and fn ∈ X ∗ , n = 1, 2, ... , be such that f ∈ (A + αJ)x and fn ∈ (A + αJ)xn . Suppose that fn → f. It was proved in Theorem 1.7.5 that {xn } is bounded. Obviously, yn − y, xn − x + αJxn − Jx, xn − x = fn − f, xn − x ,

(1.8.4)

where yn ∈ Axn , y ∈ Ax, yn + αJxn = fn and y + αJx = f. Because A and J are monotone operators, we have from (1.8.4) the limit equality lim Jxn − Jx, xn − x = 0.

(1.8.5)

n→∞

Then Lemma 1.5.15 allows us to state that xn x. Furthermore, the mapping Aα is monotone as the inverse map to the monotone operator B = A + αJ. The proof is now complete. ∗

Theorem 1.8.3 Let A1 and A2 be maximal monotone operators from X to 2X and D(A1 ) ∩ int D(A2 ) = ∅.

(1.8.6)

Then their sum A1 + A2 is also a maximal monotone operator. Proof. Making shifts, if it is necessary, in the domains of A1 and A2 and in the range of A1 , we assume, without loss of generality, that θX ∗ ∈ A1 (θX ), θX ∈ int D(A2 ).

(1.8.7)

We deal first with the case where D(A2 ) is a bounded set. For this, it is sufficient to show that R(A1 + A2 + J) = X ∗ . Choose an arbitrary element f ∈ X ∗ and prove that f ∈ R(A1 + A2 + J). We may put f = θX ∗ , shifting the range of A2 , if it is necessary. Hence, we have to prove that there exists x ∈ X such that θX ∗ ∈ (A1 + A2 + J)x.

(1.8.8)

The inclusion (1.8.8) holds if and only if there exist x ∈ X and y ∈ X ∗ such that

1 1 −y ∈ (A1 + J)x, y ∈ (A2 + J)x. 2 2 Construct the maps T1 : X ∗ → X and T2 : X ∗ → X as follows: 1 T1 y = −(A1 + J)−1 (−y) 2 and

1 T2 y = (A2 + J)−1 y. 2 It is clear that (1.8.8) is equivalent to the inclusion

θX ∈ T1 y + T2 y.

(1.8.9)

62

1


This means that, instead of (1.8.8), it is sufficient to show that θX ∈ R(T1 + T2 ). By Lemma 1.8.2, operators T1 and T2 are monotone, demicontinuous and D(T1 ) = D(T2 ) = X ∗ . Hence, the sum T = T1 + T2 is also a monotone and demicontinuous operator with D(T ) = X ∗ . Then maximal monotonicity of T : X ∗ → X arises from Theorem 1.4.6. Since θX ∗ = J(θX ) and θX ∗ ∈ A1 (θX ), we conclude that 1 θX ∗ ∈ (A1 + J)(θX ), 2 that is, θX ∈ T1 (θX ∗ ). Therefore, y, T1 y ≥ 0 ∀y ∈ X ∗ .

(1.8.10)

To prove (1.8.9) we can use Theorem 1.7.9. Show that there exists a number r > 0 such that y, T1 y + T2 y ≥ 0 as y∗ ≥ r. In view of (1.8.10), it is necessary to find r > 0 satisfying the condition y, T2 y ≥ 0 as y∗ ≥ r.

(1.8.11)

By the definition of T2 ,

1 R(T2 ) = D(A2 + J) = D(A2 ). 2 Hence, the set R(T2 ) is bounded. Write the monotonicity condition of T2 :

y − z, T2 y − T2 z ≥ 0 ∀y, z ∈ X ∗ , from which we obtain the inequality y, T2 y ≥ y, T2 z + z, T2 y − T2 z .

(1.8.12)

Since R(T2 ) is bounded, there exists a constant c > 0 such that |z, T2 y − T2 z | ≤ cz∗ .

(1.8.13)

Furthermore, θX ∈ int R(T2 ) because of (1.8.7). Hence, by Theorem 1.3.16, the inverse to T2 operator is locally bounded at zero. This means that there exist numbers c1 > 0 and c2 > 0 such that {y ∈ X ∗ | T2 y ≤ c1 } ⊆ B ∗ (θX ∗ , c2 ). Then (1.8.12) and (1.8.13) imply y, T2 y ≥ y, T2 z − cc2 if T2 z ≤ c1 , from which we deduce y, T2 y ≥ sup {y, T2 z | T2 z ≤ c1 } − cc2 = c1 y∗ − cc2 .


63

−1 Therefore, y, T2 y ≥ 0 if y∗ ≥ c−1 1 cc2 , that is, in (1.8.11) we can put r = c1 cc2 . Thus, the theorem is proved provided that D(A2 ) is bounded. We now omit this assumption. Construct the maximal monotone operator ∂Ir for any r > 0. It is clear that D(∂Ir ) is a bounded set and

D(A2 ) ∩ int D(∂Ir ) = ∅. On the basis of the previous proof we conclude that the operator A2 + ∂Ir is maximal monotone. Further, D(A1 ) ∩ int (A2 + ∂Ir ) = ∅, and D(A2 + ∂Ir ) = {x ∈ D(A2 ) | x ≤ r} is a bounded set. Hence, A1 + A2 + ∂Ir is a maximal monotone operator. It arises from Lemma 1.8.1 that the map A1 + A2 is maximal monotone. The proof is accomplished. Remark 1.8.4 Theorem 1.8.3 is valid if one of the operators A1 and A2 is the subdifferential of a proper convex lower semicontinuous functional. ∗

Theorem 1.8.5 Let A : X → 2X be a maximal monotone operator, Ω ⊆ D(A) be a convex ¯ = Ω, closed set, int Ω = ∅. Then there exists a maximal monotone operator A¯ with D(A) where A = A¯ on int Ω. Proof. Let IΩ (x) be the indicator function associated with the set Ω, that is,

IΩ (x) =

0, +∞,

x ∈ Ω, x ∈ Ω. ∗

Using Theorem 1.7.15, we see that its subdifferential ∂IΩ : X → 2X , represented by the formula    θX ∗ ,

∂IΩ (x) =

∅,   λJx,

x ∈ int Ω; x∈ / Ω; x ∈ ∂Ω, λ ≥ 0,

(1.8.14)

is maximal monotone. Due to Theorem 1.8.3, the sum A + ∂IΩ is maximal monotone too, D(A + ∂IΩ ) = Ω and (A + ∂IΩ )x = Ax as x ∈ int Ω. Therefore, on the set Ω the maximal monotone extension A¯ = A + ∂IΩ . Observe that A¯ is obtained from the original operator A by joining additional values on the boundary of Ω.

Remark 1.8.6 Theorem 1.8.5 remains still valid if the condition int Ω = ∅ is replaced by int D(A) ∩ Ω = ∅.

64

1


It is easy to check that if Ω is a nonempty convex and closed subset of X then the subdifferential ∂IΩ is the normality operator NΩ given as follows:

NΩ (x) =

{ψ ∈ X ∗ | ψ, y − x ≥ 0 ∀y ∈ Ω} ∅

if x ∈ Ω; if x ∈ / Ω.

(1.8.15)

The following results arise from Theorems 1.8.3 and 1.7.5. ∗

∗

Theorem 1.8.7 Let A1 : X → 2X and A2 : X → 2X be maximal monotone operators satisfying the condition (1.8.6) and the sum A1 + A2 be coercive. Then R(A1 + A2 ) = X ∗ . ∗

Theorem 1.8.8 Let A1 : X → 2X be a maximal monotone operator, A2 : X → X ∗ be a monotone hemicontinuous and coercive operator. Then R(A1 + A2 ) = X ∗ . ∗

Corollary 1.8.9 If A : X → 2X is a maximal monotone operator, J µ : X → X ∗ is a duality mapping with gauge function µ(t), then the equation Ax + αJ µ x = 0 has a unique solution for all α > 0 and for all f ∈ X ∗ . As a consequence, Theorem 1.4.11 allows us to present the following assertion. Theorem 1.8.10 Let L : X → X ∗ be a linear single-valued monotone and closed operator, ∗ the adjoint operator L∗ : X → X ∗ be monotone, A : X → 2X be a maximal monotone and ∗ coercive operator with int D(A) = ∅. Then R(L + A) = X . Return to Example 7 of Section 1.3. There we considered the monotone operator A defined as follows: Au = −a2 ∆u + (g(x) + a)u(x) + u(x)

R3

u2 (y) dy. |x−y |

(1.8.16)

We wonder whether it is maximal monotone. As before, represent A in the form: A = L+B, where L is a linear part of (1.8.16) and B is defined by the last term. It is not difficult to see that D(L) = {u ∈ H | ∇u ∈ H × H × H, ∆u ∈ H}, H = L2 (R3 ), that is, D(L) = D(∆). We emphasize that the operations ∇u and u are regarded here in the generalized sense. Furthermore, it is obvious that L is a self-adjoint operator. Under these conditions, there exists a positive self-adjoint operator L1/2 defined by the following equality: (Lu, v) = (L1/2 u, L1/2 v). We study the domain of L1/2 . First of all, L1/2 can be represented by the sum L1/2 = 1/2 1/2 L1 + L2 , where L1 is defined by the first summand in the right-hand side of (1.8.16) and 1/2 L2 by the second one. It is clear that D(L2 ) ⊃ H. Further, (L1 u, v) = (∇u, ∇v), that is, 1/2 2 3 D(L ) ⊂ W1 (R ). The following inequality holds [44]: u(y) | x − y | ≤ 2∇u2 . 2

1.8 Summation of Maximal Monotone Operators Hence,

u2 (y) dy R3 | x − y |

65

≤ 2u2 ∇u2 . ∞

Since u ∈ W12 (R3 ), we have Bu2 ≤ 2u22 ∇u2 < ∞, where vp denotes the norm of an element v in Lp (R3 ), p > 1. Thus, the inclusion D(L1/2 ) ⊂ W12 (R3 ) ⊂ D(B) is established. Since the condition int D(B) ∩ D(L) = ∅ of Theorem 1.8.3 is difficult to verify, we use the technique of [231] to prove that A is maximal monotone. To this end, first of all note that the operator (I + α−1 L)1/2 with α > 0 is well defined on all of H, positive, self-adjoint and it has a bounded inverse. Moreover,

D (I + α−1 L)1/2 = D(L1/2 ). Indeed, we have L1/2 =

∞√ 0

(1.8.17)

λdEλ ,

where {Eλ } is called the identity decomposition generated by L. Then u ∈ D((I +α−1 L)1/2 ) if and only if ∞

0

The latter is fulfilled if

(1 + α−1 λ)(dEλ u, u) < ∞.

∞ 0

λ(dEλ u, u) < ∞.

Hence, (1.8.17) holds. Introduce the operator Q = (I + α−1 L)−1/2 B(I + α−1 L)−1/2 with D(Q) = H. In virtue of the properties of L and B, it is hemicontinuous and monotone. Then Theorem 1.4.6 immediately gives the maximal monotonicity of Q. In its turn, Theorem 1.7.4 guarantees that a solution w of the equation Qw + αw = (I + α−1 L)−1/2 v exists. Hence, the equation Lu + Bu + αu = v, α > 0, where u = (I + α−1 L)−1/2 w, is solvable for any v ∈ H. By Theorem 1.7.13, we conclude that the operator A = L + B is maximal monotone.

66

1

1.9


Equations with General Monotone Operators

We have seen in Section 1.7 that the maximal monotonicity of operators allow us to prove the existence theorems for equations with such operators, to study their domains and ranges and to describe the structure of their solution sets. Observe that, by Zorn’s lemma, an arbitrary monotone operator has at least one maximal monotone extension. However, there is no constructive way to build such extensions. Besides, the example given in the previous section shows that the establishment of maximal monotonicity of monotone operators is often a very complicated problem. There is another problem. The reader already knows that a maximal monotone operator is multiple-valued, in general. In practice calculations, we usually consider only certain sections of maximal monotone operators which guarantee the monotonicity of resulting discontinuous mappings. Hence, there is a necessity to analyze problems with arbitrary monotone operators that, generally speaking, do not satisfy the conditions of continuity or maximal monotonicity. Assume that X is a reflexive Banach space, X and X ∗ are strictly convex. The next result will be useful in the sequel. Theorem 1.9.1 Suppose that a monotone hemicontinuous operator A : X → X ∗ is given on open or linear dense set D(A) ⊆ X. Then the equation Ax = f has a solution x0 ∈ D(A) for any f ∈

X∗

(1.9.1)

if and only if

Ax − f, x − x0 ≥ 0 ∀x ∈ D(A).

(1.9.2)

Proof. Let Ax0 = f. Then, by the monotonicity of A, Ax − f, x − x0 = Ax − Ax0 , x − x0 + Ax0 − f, x − x0 ≥ 0, because Ax0 − f, x − x0 = 0. Let there exist x0 ∈ D(A) such that (1.9.2) is valid. Take any w ∈ D(A) and put xt = x0 + tw ∈ D(A) with t ≥ 0. Substitute xt for x in (1.9.2). Then Axt − f, xt − x0 = Axt − f, tw ≥ 0, that is, Axt − f, w ≥ 0 ∀w ∈ D(A). By the hemicontinuity of the operator A, we have in a limit as t → 0 : Ax0 − f, w ≥ 0 ∀w ∈ D(A). This means that Ax0 = f. Observe that the second part of this theorem does not need the monotonicity property of A. Therefore, it is useful to state the following result:

1.9 Equations with General Monotone Operators

67

Corollary 1.9.2 Suppose that hemicontinuous operator A : X → X ∗ is defined on open or linear dense set D(A) ⊆ X and let x0 ∈ D(A). Then the inequality (1.9.2) implies f = Ax0 . It has happened that a solution of the equation (1.9.1) with an arbitrary monotone ∗ operator A : X → 2X does not exist both in the classical sense and in the sense of inclusions. For this reason, using Theorem 1.9.1, we introduce Definition 1.9.3 An element x0 ∈ X is said to be a generalized solution of the equation (1.9.1) if for all x ∈ D(A) and for all y ∈ Ax it satisfies the inequality y − f, x − x0 ≥ 0.

(1.9.3)

It follows from (1.9.3) that if a generalized solution set of (1.9.1) is not empty then it is convex and closed. For equations with monotone hemicontinuous operators, generalized and classical solutions coincide. We are interested in the generalized solvability of the equation Ax + αJx = f, α > 0, x ∈ D(A),

(1.9.4)

with a monotone operator A. Denote by R(Ax − f ) a convex closed hull of the weak limits of all subsequences of the sequences {Axn − f } when xn → x, x ∈ X, xn ∈ D(A). Definition 1.9.4 A point x0 ∈ X is called an sw-generalized solution of the equation (1.9.1) if θX ∗ ∈ R(Ax0 − f ). Lemma 1.9.5 If int D(A) = ∅, then for any point x0 ∈ int D(A) the inclusion θX ∗ ∈ R(Ax0 − f ) ¯ 0 − f, where A¯ is a maximal monotone extension of A. is satisfied if and only if θX ∗ ∈ Ax Proof. Let θX ∗ ∈ R(Ax0 − f ), xn ∈ D(A), xn → x0 . By the monotonicity of A, (y − f ) − (yn − f ), x − xn ≥ 0 ∀x ∈ D(A), ∀y ∈ Ax, ∀yn ∈ Axn . According to the definition of R(Ax0 − f ), we obtain y − f, x − x0 ≥ 0 ∀x ∈ D(A), ∀y ∈ Ax. ¯ 0 . Hence, θX ∗ ∈ Ax ¯ 0 − f. From this inequality, it follows that f ∈ Ax ¯ 0 − f. Show by the contradiction that θX ∗ ∈ R(Ax0 − f ). Assume Let now θX ∗ ∈ Ax that θX ∗ ∈ R(Ax0 − f ). Then there is an element g ∈ X such that z, g < 0 for all z ∈ R(Ax0 − f ). Further, construct the sequence xn = x0 + tn g, tn > 0. If tn → 0 then xn → x0 . Since x0 ∈ int D(A) then A is locally bounded at this point. Therefore, there is some subsequence yk − f f1 ∈ X ∗ , yk ∈ Axk , that is, f1 ∈ R(Ax0 − f ) by the definition of the last set. Now the monotonicity of A¯ yields the inequality ¯ k. ¯ yk − f, xk − x0 ≥ 0 ∀¯ yk ∈ Ax

68

1


Consequently, yk − f, xk − x0 ≥ 0 ∀yk ∈ Axk , or yk − f, g ≥ 0 ∀yk ∈ Axk . Setting tk → 0 we obtain f1 , g ≥ 0, which contradicts the definition of g. Observe that if A is a maximal monotone operator then (1.9.3) determines the inclusion f ∈ Ax0 . This means that the solutions of (1.9.1) in the sense of Definitions 1.7.2 and 1.9.3 coincide. Corollary 1.9.6 A monotone hemicontinuous operator carries any bounded weakly closed set M ⊂ X to a closed set of X ∗ . Proof. Let xn ∈ M, xn x0 ∈ M and Axn → y0 ∈ X ∗ . Write the monotonicity condition of A, Ax − Axn , x − xn ≥ 0. Hence, Ax − y0 , x − x0 ≥ 0. By Theorem 1.9.1, we conclude then y0 = Ax0 , that is, y0 ∈ A(M ).

Corollary 1.9.7 If the set D(A) is closed and convex and int D(A) = ∅, then there exists ¯ = D(A). only one maximal monotone extension A¯ with the domain D(A) Proof. Indeed, it follows from Lemma 1.9.5 that at each point x0 ∈ int D(A) the set ¯ 0 } coincides with R(Ax0 ). In view of Theorem 1.7.19, at boundary points of {y0 | y0 ∈ Ax D(A) the operator A¯ is finished determining A by semi-lines, if it is required. ∗

Lemma 1.9.8 Let A : X → 2X be a monotone operator, D(A) be a convex closed set in X, int D(A) = ∅. Then each generalized solution of (1.9.4) is a solution of the equation ¯ + αJx = f, Ax

(1.9.5)

¯ = D(A). The converse implicawhere A¯ is a maximal monotone extension of A with D(A) tion is also true. Proof. By Theorem 1.7.4, equation (1.9.5) has the unique solution xα ∈ D(A) such that ¯ α + αJxα . f ∈ Ax Then the monotonicity property of A¯ + αJ allows us to write down the inequality y + αJx − f, x − xα ≥ 0 ∀x ∈ D(A),

¯ ∀y ∈ Ax.

(1.9.6)

Furthermore, (1.9.6) holds for all y ∈ Ax. In its turn, this means that xα is the generalized solution of (1.9.4) too.

1.9 Equations with General Monotone Operators

69

Now let xα be a generalized solution of (1.9.4). Then (1.9.6) is true with y ∈ Ax and ¯ α + αJxα . x ∈ D(A). Hence, by Lemma 1.9.5 and Theorem 1.8.3, f ∈ (A + αJ)xα = Ax Thus, xα is a solution of (1.9.5).

It is obvious that the following result holds. Theorem 1.9.9 Under the conditions of Lemma 1.9.8, the equation (1.9.4) has the unique generalized solution belonging to D(A). In fact, Lemma 1.9.8 proves solvability of the equation (1.9.4) with arbitrary domain of an operator A, however, its solution does not necessarily belong to D(A) and uniqueness cannot be guaranteed. Below we give one more sufficient uniqueness condition for solution of the equation (1.9.4). Introduce the set T (x0 , z) = {x | x = x0 + tz, t ∈ R1 , x0 ∈ X, z ∈ X, z = θX }. Definition 1.9.10 We say that the set G ⊂ X densely surrounds a point x0 if x0 is a bilateral boundary point of the set G ∩ T (x0 , z) for every T (x0 , z). Theorem 1.9.11 If D(A) densely surrounds X, then equation (1.9.4) has a unique generalized solution for all α > 0. Proof. Let x1 and x2 be two generalized solutions of (1.9.4) and x1 = x2 . Then it follows from (1.9.3) that there are numbers t1 and t2 such that 0 < t1 , t2 < 1, t1 + t2 − 1 > 0 and points x3 = t1 x1 + (1 − t1 )x2 , x4 = t2 x2 + (1 − t2 )x1 belong to D(A). The following equalities can be easily verified: x3 − x1 = (1 − t1 )(x2 − x1 ), x4 − x2 = (1 − t2 )(x1 − x2 ), x4 − x3 = (1 − t1 − t2 )(x1 − x2 ). The operator A + αJ is strictly monotone. Therefore, y3 − y4 , x3 − x4 > 0, y3 ∈ (A + αJ)x3 , y4 ∈ (A + αJ)x4 . However, y3 − y4 , x3 − x4 = (1 − t1 − t2 ) which contradicts (1.9.7).

y3 − f,

x4 − x2 x3 − x1 + y4 − f, ≤0 1 − t1 1 − t2

(1.9.7)

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1.10


Equations with Semimonotone Operators

Suppose that X is a reflexive strictly convex Banach space together with its dual space X ∗ . ∗

Definition 1.10.1 An operator A : X → 2X is said to be semimonotone, if there exists strongly continuous operator C : X → X ∗ such that T = A + C is a monotone map and D(A) ⊆ D(C). It is easy to obtain the following assertions using the properties of monotone operators: a) a semimonotone operator is locally bounded at any interior point of its domain; ∗ b) if T : X → 2X is a maximal monotone operator, then the set of values of a semimonotone operator at every point of the domain is a convex and closed set. We are interested in solvability of the equation (1.9.1) with a semimonotone operator ∗ A : X → 2X . First of all, observe that there is the following statement which is an analogy of the Minty−Browder theorem for semimonotone operators: Theorem 1.10.2 Let X be a reflexive Banach space, A : X → X ∗ be a semimonotone hemicontinuous and coercive operator with D(A) = X. Then R(A) = X ∗ . We study the solvability problem for arbitrary, possibly multiple-valued or discontinuous, semimonotone operators A with the domain D(A) = X. Construct a maximal monotone extension T¯ of a monotone operator T and let A¯ = T¯ − C be given. Definition 1.10.3 An element x0 ∈ X is said to be an s-generalized solution of the equation ¯ 0. (1.9.1) if f ∈ Ax Definition 1.10.4 We shall say that a space X has the M -property if there exists a sequence of projectors {Pn }, n = 1, 2, ... , such that Pn X = Xn and for all x ∈ X, Pn x − x → 0 as n → ∞. As in Section 1.9, we denote by R(Ax − f ) the convex closed hull of weak limits of subsequences of {Axn − f }, where xn → x as n → ∞, xn ∈ X, x ∈ X. ¯ − f } and R(Ax − f ) coincide. Lemma 1.10.5 For all x ∈ X, the sets {y | y ∈ Ax Proof. See the proof of Lemma 1.9.5. ∗

Theorem 1.10.6 Suppose that a space X possesses the M -property, A : X → 2X is a semimonotone operator, D(A) = X, and there exist a constant r > 0 and y ∈ Ax such that y − f, x ≥ 0

as

x = r.

Then the equation (1.9.1) has at least one s-generalized solution x0 with x0 ≤ r.

1.10 Equations with Semimonotone Operators

71

Proof. There exists a sequence of the adjoint projectors {Pn∗ }, n = 1, 2, ... , such that Pn∗ X ∗ = Xn∗ . Consider in Xn the equation Pn∗ (Ax − f ) = 0. If x ∈ Xn and x ≤ r then for some y ∈ Ax we have Pn∗ (y − f ), x = y − f, x ≥ 0. By Theorem 1.1.62, there is at least one element xn ∈ Xn such that θXn∗ ∈ R(Pn∗ (Axn − f )), ∗ where xn ≤ r. Show that the operator Pn∗ T¯ : Xn → 2Xn is a maximal monotone extension ∗ ∗ X of the operator Pn T : Xn → 2 n . Recalling that T = A + C and T¯ = A¯ + C, suppose this is not the case. Let

¯ − x ≥ 0 ∀x ∈ Xn , ∀˜ y ∈ T¯x, x ¯ ∈ Xn , y¯ ∈ Xn∗ , ¯ y − Pn∗ y˜, x

(1.10.1)

but y¯ ∈ Pn∗ T¯x ¯. By Theorem 1.4.9, it follows that a set {y | y ∈ T¯x} is convex and closed for every x ∈ X. Then there is an element u ∈ Xn such that y − y¯, u < 0 for all y ∈ T¯x ¯. ¯ + tk u, tk > 0, tk → 0 as k → ∞. Obviously, Introduce the sequence {xk }, where xk = x xk → x ¯ as k → ∞.

(1.10.2)

Due to the local boundedness of T¯, there exists a sequence {ykm }, km → ∞, and some g ∈ X ∗ such that (1.10.3) ykm ∈ T¯xkm , {xkm } ⊆ {xk }, ykm g. Then it results from (1.10.2), (1.10.3) and Lemma 1.4.5 that g ∈ T¯x ¯. Substitute x = xkm and y˜ = ykm ∈ T¯xkm for (1.10.1) to obtain 0 ≤ Pn∗ ykm − y¯, u = ykm − y¯, u → g − y¯, u = Pn∗ g − y¯, u < 0. Therefore, for some k > 0, the following inequality holds: ¯ − xk < 0, xk ∈ Xn , yk ∈ T¯xk , ¯ y − Pn∗ yk , x ¯ n − f ). that contradicts (1.10.1), hence, θXn∗ ∈ Pn∗ (Ax So, we have constructed the sequence {xn }, n = 1, 2, ... , such that xn ≤ r. Then there exists a subsequence {xnl } ⊆ {xn } which converges weakly to some element x ¯ ∈ X. ¯ n be a sequence We shall prove that x ¯ is a solution of the equation (1.9.1). Let ynl ∈ Ax l such that Pn∗ (y nl − f ) = θXn∗ . We assert that {y nl } is bounded. Indeed, let z nl ∈ T¯xnl . This means that z nl = y nl + Cxnl . Since T¯ is locally bounded at every point X, there exist two constants a1 > 0 and a2 > 0 such that the inequality y∗ ≤ a2 , y ∈ T¯x holds if x ≤ a1 . Consider in X the sequence J ∗ y nl unl = a1 n , y l ∗

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where J ∗ : X ∗ → X is a normalized duality mapping in X ∗ . Then unl = a1 and hence for all v nl ∈ T¯unl we obtain v nl ∗ ≤ a2 . Now the monotonicity property of T¯ gives z nl − vnl , xnl − unl ≥ 0. This implies a1 y nl ∗ ≤ v nl , unl − xnl − Cxnl , unl + z nl , xnl ≤ a2 (a1 + r) + a1 Cxnl ∗ + rCxnl + f ∗ ≤ a3 , ¯ so that y nl ∗ ≤ a3 a−1 1 . Again, the monotonicity of T yields the inequality ¯ x ∈ X, y + Cx − ynl − Cxnl , x − xnl ≥ 0 ∀y ∈ Ax, or y − f, x − xnl ≥ Cx − Cxnl , xnl − x + ynl − f, x . Since {y nl } is bounded as nl → ∞ and since the space X has the M -property, one gets ¯ y + Cx − f − C x ¯, x − x ¯ ≥ 0, ∀y ∈ Ax, ∀x ∈ X. Thus, z − f − C x ¯, x − x ¯ ≥ 0 ∀z ∈ T¯x, ∀x ∈ X. ¯x. The proof It follows from the maximal monotonicity of T¯ that C x ¯ + f ∈ T¯x ¯, i.e., f ∈ A¯ is now complete.

1.11

Variational Inequalities with Monotone Operators

Let X be a reflexive strictly convex Banach space together with its dual space X ∗ , A : X → ∗ 2X be a maximal monotone operator with domain D(A), J : X → X ∗ be a normalized duality mapping. In this section, we consider the following variational inequality problem: To find x ∈ Ω such that Ax − f, y − x ≥ 0 ∀y ∈ Ω, (1.11.1) where Ω ⊆ D(A) is a closed and convex set of X, f ∈ X ∗ . We present two definitions of its solutions. Definition 1.11.1 An element x0 ∈ Ω is called the solution of the variational inequality (1.11.1) if there is an element z 0 ∈ Ax0 such that z 0 − f, y − x0 ≥ 0 ∀y ∈ Ω.

(1.11.2)

A solution x0 satisfying (1.11.2) we shall also call the classical solution of the variational inequality (1.11.1). Definition 1.11.2 An element x0 ∈ Ω is called the solution of the variational inequality (1.11.1) if z − f, y − x0 ≥ 0 ∀y ∈ Ω, ∀z ∈ Ay. (1.11.3)

1.11 Variational Inequalities with Monotone Operators

73

Lemma 1.11.3 If x0 ∈ Ω is a solution of (1.11.1) defined by the inequality (1.11.2), then it satisfies also the inequality (1.11.3). Proof. Write the monotonicity condition for the operator A : z − z 0 , y − x0 ≥ 0 ∀y ∈ Ω,

∀z ∈ Ay,

where z 0 ∈ Ax0 and satisfies (1.11.2). Then z − f, y − x0 + f − z 0 , y − x0 ≥ 0 ∀y ∈ Ω,

∀z ∈ Ay.

Taking into account (1.11.2) we obtain (1.11.3). Now we present the Minty−Browder lemma for variational inequalities with maximal monotone operators. Lemma 1.11.4 If Ω ⊆ D(A) and if either int Ω = ∅ or int D(A) ∩ Ω = ∅, then Definitions 1.11.1 and 1.11.2 are equivalent. Proof. Let x0 ∈ Ω be a solution of (1.11.1) in the sense of Definition 1.11.2 and ∂IΩ be a subdifferential of the indicator function IΩ associated with Ω. Since θX ∗ ∈ ∂IΩ x for all x ∈ Ω and ∂IΩ is (maximal) monotone, we have η, y − x0 ≥ 0 ∀y ∈ Ω,

∀η ∈ ∂IΩ y.

(1.11.4)

Adding (1.11.3) and (1.11.4) one gets z + η − f, y − x0 ≥ 0 ∀z ∈ Ay,

∀y ∈ Ω,

∀η ∈ ∂IΩ y.

(1.11.5)

X∗

Therefore, z + η ∈ By, where the operator B = A + ∂IΩ : X → 2 is maximal monotone in view of Theorem 1.8.3, and D(B) = Ω by the condition Ω ⊆ D(A). Then (1.11.5) implies the inclusion f ∈ Bx0 . In other words, there exist elements z 0 ∈ Ax0 and η 0 ∈ ∂IΩ x0 such that f = z 0 + η 0 . Consequently, z 0 + η 0 − f, y − x0 = 0 ∀y ∈ Ω. It results from this that However,

(1.11.6)

z 0 − f, y − x0 = η 0 , x0 − y . η 0 , x0 − y ≥ 0 ∀y ∈ Ω,

because ∂IΩ is the normality operator defined by (1.8.15). Thus, z 0 − f, y − x0 ≥ 0 ∀y ∈ Ω, that is, (1.11.2) follows. Taking now into account the previous lemma we obtain the result claimed in the lemma.

We prove the following important assertions.

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Lemma 1.11.5 If A : X → 2X is a maximal monotone operator with D(A) = Ω, then the inequality (1.11.1) and the equation (1.9.1) are equivalent in the sense of Definition 1.11.1. Proof. Indeed, a solution x0 of the equation (1.9.1) with a maximal monotone operator A, defined by the inclusion f ∈ Ax0 , satisfies Definition 1.11.1. Therefore, it is a solution of the inequality (1.11.1). Let now x0 be a solution of (1.11.1) in the sense of Definition 1.11.1. Then, by Lemma 1.11.3, the inequality (1.11.3) holds. Since A is a maximal monotone operator and D(A) = Ω, (1.11.3) and Proposition 1.4.3 imply the inclusion f ∈ Ax0 . Thus, x0 is the solution of the equation Ax = f. In particular, the conclusion of this lemma holds if Ω = X or x0 is a interior point of Ω. ∗

Lemma 1.11.6 If A : X → 2X is a maximal monotone operator with D(A) ⊆ X, Ω ⊆ D(A) is convex closed set and x0 ∈ int Ω, then z − f, x − x0 ≥ 0 ∀x ∈ Ω,

∀z ∈ Ax,

(1.11.7)

implies the inclusion f ∈ Ax0 . The converse is also true. Proof. Take an arbitrary v ∈ X. Element xt = x0 + tv ∈ Ω for sufficiently small t > 0 because x0 ∈ int Ω. Consequently, xt → x0 as t → 0. Since operator A is local bounded at x0 , we have yt y, where yt ∈ Axt . We deduce from (1.11.7) that yt − f, xt − x0 ≥ 0 or yt − f, v ≥ 0 ∀v ∈ X. Letting t → 0 one gets

y − f, v ≥ 0 ∀v ∈ X.

This means that y = f. A maximal monotone operator A is demiclosed in view of Lemma 1.4.5, that leads to the inclusion y ∈ Ax0 . Hence, f ∈ Ax0 . The converse assertion immediately follows from the monotonicity of A. ∗

Lemma 1.11.7 Let A : X → 2X be a maximal monotone operator. Let Ω ⊂ D(A) be a convex and closed set. Let ∂IΩ be a subdifferential of the indicator function IΩ associated with Ω. If int Ω = ∅ or int D(A)∩Ω = ∅, then a solution x0 ∈ Ω of the variational inequality (1.11.1) is a solution in the sense of the following inclusion: f ∈ Ax0 + ∂IΩ (x0 ).

(1.11.8)

The inverse conclusion is also true. Proof. Let x0 be a solution of the variational inequality (1.11.1) in the sense of Definition 1.11.1. Then (1.11.2) and (1.11.3) are satisfied. Construct the indicator function IΩ (x) ∗ for the set Ω and find its subdifferential ∂IΩ : Ω ⊂ X → 2X . By definition of ∂IΩ , one gets u, y − x0 ≥ 0 ∀y ∈ Ω, ∀u ∈ ∂IΩ (y).


75

Consequently, (1.11.3) involves z + u − f, y − x0 ≥ 0 ∀y ∈ Ω,

∀z ∈ Ay,

∀u ∈ ∂IΩ (y).

By Theorem 1.8.3, using the conditions int Ω = ∅ or int D(A) ∩ Ω = ∅, we conclude that A + ∂IΩ is a maximal monotone operator with D(A + ∂IΩ ) = Ω, which implies f ∈ Ax0 + ∂IΩ (x0 ). Let now (1.11.8) be hold with x0 ∈ Ω. Then there is an element z 0 ∈ Ax0 such that f − z 0 ∈ ∂IΩ (x0 ). Hence, we can write down the inequality z 0 − f, y − x0 ≥ 0 ∀y ∈ Ω. Thus, x0 is a solution of the variational inequality (1.11.1). The proved lemmas imply Theorem 1.11.8 Under the conditions of Lemma 1.11.7, the set of solutions of the variational inequality (1.11.1), if it is nonempty, is convex and closed. Note also that if maximal monotone operator A is strictly monotone then the solution x0 of the variational inequality (1.11.1) is unique. Indeed, suppose, by contradiction, that x1 is another solution of (1.11.1). Then x0 and x1 satisfy (1.11.2), so that, for some z 0 ∈ Ax0 and for some z 1 ∈ Ax1 we have, respectively, z 0 − f, y − x0 ≥ 0 ∀y ∈ Ω, and

z 1 − f, y − x1 ≥ 0 ∀y ∈ Ω.

It is easy to see that and

z 0 − f, x1 − x0 ≥ 0 z 1 − f, x0 − x1 ≥ 0.

Summing up two last inequalities we deduce z 1 − z 0 , x0 − x1 ≥ 0. Since A is monotone, this gives z 0 − z 1 , x0 − x1 = 0 which contradicts our assumption that x0 = x1 . Analyze the solvability problem for the inequality (1.11.1). By Lemma 1.11.7, solvability of the variational inequality (1.11.1) with the maximal monotone operator A and solvability of the equation Ax + ∂IΩ (x) = f with the maximal monotone operator A + ∂IΩ are equivalent. Therefore, we can apply the existence theorems of Section 1.7 to obtain the following statements.

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Theorem 1.11.9 Suppose that A : X → 2X is a maximal monotone and coercive operator, Ω is a convex closed set in D(A), and either int Ω = ∅ or int D(A) ∩ Ω = ∅. Then inequality (1.11.1) has at least one solution for all f ∈ X ∗ . ∗

Theorem 1.11.10 Let A : X → 2X be a maximal monotone operator, a set Ω satisfy the conditions of Theorem 1.11.9, and there exist a number r > 0 such that for all y such that y ≥ r, y ∈ D(A) ∩ Ω, the following inequality holds: z − f, y ≥ 0 ∀z ∈ Ay, f ∈ X ∗ .

(1.11.9)

Then there exists at least one solution x of the variational inequality (1.11.1) with x ≤ r. ∗

Theorem 1.11.11 Assume that A : X → 2X is a maximal monotone operator, Ω satisfies the conditions of Theorem 1.11.9. Then the regularized variational inequality Ax + αJx − f, y − x ≥ 0 ∀y ∈ Ω, x ∈ Ω,

(1.11.10)

has a unique solution for all α > 0 and for all f ∈ X ∗ . A solution xα satisfying (1.11.10) we call the regularized solution of the variational inequality (1.11.1). Remark 1.11.12 Similarly to Corollary 1.7.6, if Ω in Theorems 1.11.9, 1.11.10 and 1.11.11 is bounded, then the coerciveness of A and the condition (1.11.9) are unnecessary there. ∗

Corollary 1.11.13 Let A : X → 2X be a maximal monotone operator which is coercive relative to a point x0 ∈ X, let Ω be a convex closed set in D(A) and either int Ω = ∅ or int D(A) ∩ Ω = ∅. Then inequality (1.11.1) has at least one solution for all f ∈ X ∗ . Proof. It is sufficient to consider the operator A1 (x) = A(x + x0 ). The constraint minimization problems for a convex functional lead to inequalities of the type (1.11.1). Let us describe such problems in detail. ∗ 1) Let ϕ : X → R1 be a proper convex lower semicontinuous functional, ∂ϕ : X → 2X be its subdifferential which is the maximal monotone operator in view of Theorem 1.7.15. Let Ω be a convex closed set, Ω ⊆ D(∂ϕ). The problem is to find min {ϕ(y) | y ∈ Ω}.

(1.11.11)

It is assumed that this problem is solvable. Theorem 1.11.14 If int Ω = ∅ or int D(∂ϕ) ∩ Ω = ∅, then problem (1.11.11) is equivalent to the variational inequality ∂ϕ(x), y − x ≥ 0 ∀y ∈ Ω, x ∈ Ω.

(1.11.12)


77

Proof. Let x0 be a solution of (1.11.12). Then, by the definition of subdifferential ∂ϕ at the point x0 , we have ϕ(y) − ϕ(x0 ) ≥ ∂ϕ(x0 ), y − x0 ≥ 0 ∀y ∈ Ω. This means that ϕ(y) ≥ ϕ(x0 ) for all y ∈ Ω. Hence, x0 is a solution of the problem (1.11.11). Let now x0 be a solution of the problem (1.11.11). Construct the functional Φ(y) = ϕ(y) + IΩ (y), where IΩ (y) is the indicator function associated with the set Ω. Hence, its subdifferential is ∂Φ = ∂ϕ + ∂IΩ and D(∂Φ) = Ω. It is clear that Φ reaches a minimum at the point x0 ∈ Ω. Then, by Lemma 1.2.5, θX ∗ ∈ ∂ϕ(x0 )+∂IΩ (x0 ). The converse implication arises from Lemma 1.11.7. 2) Let X1 and X2 be reflexive strictly convex Banach spaces together with their dual spaces, a function Φ(u, v) : X1 × X2 → R1 be proper convex lower semicontinuous with respect to u and concave upper semicontinuous with respect to v. Let G1 ⊂ X1 and G2 ⊂ X2 be convex and closed sets such that int G1 = ∅, int G2 = ∅, G = G1 × G2 and G ⊆ D(∂Φ). Definition 1.11.15 A point (u∗ , v ∗ ) ∈ G is said to be a saddle point of the functional Φ(u, v) on G if Φ(u∗ , v) ≤ Φ(u∗ , v ∗ ) ≤ Φ(u, v ∗ ) (1.11.13) for all (u∗ , v) ∈ G and (u, v ∗ ) ∈ G. Consider the functional Φ1 (u) = Φ(u, v ∗ ) − Φ(u∗ , v ∗ ), Φ1 : X1 → R1 . In view of (1.11.13), u∗ ∈ G1 is a minimum point of the convex functional Φ1 (u) on G1 . By Theorem 1.11.14, we have then ∂Φ1 (u∗ ), u∗ − u ≤ 0 ∀u ∈ G1 , that is,

∂u Φ(u∗ , v ∗ ), u∗ − u ≤ 0 ∀u ∈ G1 .

(1.11.14)

Applying the same arguments for the functional Φ2 (v) = Φ(u∗ , v ∗ ) − Φ(u∗ , v), Φ2 : X2 → R1 , we arrive at the inequality ∂v Φ(u∗ , v ∗ ), v − v ∗ ≤ 0 ∀v ∈ G2 .

(1.11.15)

Thus, the saddle point (u∗ , v ∗ ) is a solution of the variational inequality (1.11.12) presented by (1.11.14) and (1.11.15) together. 3) We study the variational inequality (1.11.1) with a maximal monotone operator ∗ A : X → 2X on Ω = Ω1 ∩ Ω2 , where Ω1 ⊂ X is a convex closed set and Ω2 ⊂ X is defined by the system of inequalities ϕi (x) ≤ 0, i = 1, 2, ..., n.

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Here ϕi : X → R1 are proper convex lower semicontinuous functionals on X having, due to n the non-negative octant in Rn . Define Theorem 1.2.8, subdifferentials on X. Denote by R+ the quantity [p, ϕ(x)] =

n

pi ϕi (x),

i=1 n. where p = {p1 , p2 , ..., pn } ∈ R+ n Definition 1.11.16 We say that the Slater condition is fulfilled if for every p ∈ R+ there exists an element x ¯ ∈ X such that [p, ϕ(¯ x)] < 0.

Let x∗ ∈ Ω be a solution of the variational inequality (1.11.1) in the sense of Definition 1.11.1, therefore, there exists z ∗ ∈ Ax∗ such that z ∗ − f, x∗ − y ≤ 0 ∀y ∈ Ω.

(1.11.16)

Introduce the Lagrange function: L(x, p) = z ∗ − f, x +

n

pi ϕi (x).

(1.11.17)

i=1

By (1.11.16), we have z ∗ − f, x∗ = min {z ∗ − f, y | y ∈ Ω}. Next we need the Karush−Kuhn−Tucker theorem. Theorem 1.11.17 Let g(x) and ϕi (x), i = 1, ..., n, be convex functionals on convex set Ω ⊆ X. Let the Slater condition be fulfilled. Then an element x∗ ∈ Ω is the solution of the minimization problem min {g(x) | x ∈ Ω, ϕ(x) ≤ 0}, ϕ(x) = {ϕi (x)}, if and only if there exists a vector p∗ such that the pair (x∗ , p∗ ) is the saddle point of the Lagrange function L(x, p) = g(x) + [p, ϕ(x)], that is,

L(x∗ , p) ≤ L(x∗ , p∗ ) ≤ L(x, p∗ ) ∀x ∈ Ω,

n ∀p ∈ R+ .

n By virtue of the Karush−Kuhn−Tucker theorem, there is a vector p∗ ∈ R+ such that is the saddle point of the Lagrange function (1.11.17). According to (1.11.14), (1.11.15), this point satisfies the inequalities

(x∗ , p∗ )

Ax∗ − f +

n

p∗i ∂ϕi (x∗ ), x∗ − x ≤ 0 ∀x ∈ Ω,

(1.11.18)

i=1 n . [q − p∗ , ϕ(x∗ )] ≤ 0 ∀q ∈ R+

(1.11.19)

1.12

Variational Inequalities with Semimonotone Operators

79

Let now (x∗ , p∗ ) be a solution of the system (1.11.18), (1.11.19). Rewrite (1.11.19) in the following form: n

qi ϕi (x∗ ) ≤

i=1

n

n p∗i ϕi (x∗ ) ∀q ∈ R+ .

i=1

By simple algebra, we come to the relations ϕi (x∗ ) ≤ 0, i = 1, 2, ..., n, [p∗ , ϕ(x∗ )] = 0, i.e., x∗ ∈ Ω2 . Since the functional [p∗ , ϕ(x)] =

n

p∗i ϕi (x)

i=1

is proper convex lower semicontinuous on X, it has a subdifferential. Using Definition 1.2.2, we conclude that [p∗ , ϕ(x)] − [p∗ , ϕ(x∗ )] ≥

n

p∗i ∂ϕi (x∗ ), x − x∗ ,

i=1

that with x ∈ Ω gives 0 ≥ [p∗ , ϕ(x)] ≥

n

p∗i ∂ϕi (x∗ ), x − x∗ .

i=1

Then (1.11.18) implies the fact that x∗ is the solution of (1.11.1). Thus, we have proved the following theorem. Theorem 1.11.18 If the Slater condition is satisfied, then any solution x∗ of the varian of the system of the intional inequality (1.11.1) determines a solution (x∗ , p∗ ), p∗ ∈ R+ equalities (1.11.18), (1.11.19), and, conversely, any solution (x∗ , p∗ ) of (1.11.18), (1.11.19) determines a solution x∗ of (1.11.1).

1.12

Variational Inequalities with Semimonotone Operators

In this section (and also in Sections 1.13 and 1.14) we study the solution existence of the ∗ variational inequality (1.11.1) with non-monotone maps A : X → 2X . We present the sufficient existence conditions for a variational inequality with the semimonotone (possibly, unbounded) operator. However, we are interested in the solutions which are points where the operator A is single-valued. Let X be a reflexive strictly convex Banach space together with its dual space X ∗ . Observe one important property of semimonotone maps.

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Lemma 1.12.1 Assume that A : X → 2X is a semimonotone operator, D(A) = X, ∗ C : X → X ∗ is strongly continuous and T = A + C : X → 2X is maximal monotone. If xn x as n → ∞, yn ∈ Axn and lim sup yn , xn − x ≤ 0,

(1.12.1)

n→∞

then for every y ∈ X there exists an element z(y) ∈ Ax such that z(y), x − y ≤ lim inf yn , xn − y . n→∞

Proof. Let xn x, yn ∈ Axn and (1.12.1) be fulfilled. Since C : X → X ∗ is strongly continuous, we have Cxn , xn − x → 0 as n → ∞. For zn = yn + Cxn ∈ T xn , one gets lim sup zn , xn − x ≤ 0.

(1.12.2)

n→∞

By virtue of the monotonicity of T, we write z, xn − x ≤ zn , xn − x ∀z ∈ T x. Using (1.12.2) and weak convergence of {xn } to x, this implies the limit equality lim zn , xn − x = 0.

n→∞

(1.12.3)

Choose an arbitrary point (u, v) ∈ grT. By the obvious equality zn , xn − u = zn , xn − x + zn , x − u , we obtain lim inf zn , xn − u = lim inf zn , x − u . n→∞

n→∞

(1.12.4)

Since v, xn − u ≤ zn , xn − u , where v ∈ T u, we deduce that the weak convergence of {xn } to x leads to the inequality v, x − u ≤ lim inf zn , xn − u . n→∞

(1.12.5)

Let y ∈ X, ut = (1 − t)x + ty, t > 0 and zt ∈ T ut . We replace (u, v) in (1.12.5) by (ut , zt ). Then according to (1.12.4), zt , x − y ≤ lim inf zn , x − y . n→∞

(1.12.6)

It is clear that ut → x as t → 0. By the local boundedness of the operator T on X, zt

z(y) ∈ X ∗ , where z(y) ∈ T x because T is a maximal monotone mapping. Hence, by (1.12.6), we derive the inequality z(y), x − y ≤ lim inf zn , x − xn + lim inf zn , xn − y . n→∞

n→∞

Taking into account (1.12.3), we obtain the conclusion of the lemma.

1.12 Variational Inequalities with Semimonotone Operators

81

Theorem 1.12.2 Let X be a uniformly convex Banach space, X ∗ be an E-space, Ω be a ∗ convex closed set in X, θX ∈ Ω, A : X → 2X be a semimonotone operator, D(A) = X. Let T = A + C be a maximal monotone operator, where C : X → X ∗ is strongly continuous, and the following inequality hold for x = r > 0 : y − f, x ≥ 0 ∀y ∈ Ax.

(1.12.7)

Assume that zn ∈ X, v n ∈ Az n , and at any point z 0 ∈ Ω, where the operator A is multiplevalued, there exists even if one element u ∈ Ω such that lim sup v n − f, z n − u > 0,

(1.12.8)

n→∞

provided that z n z 0 . Then there exists at least one solution x ¯ of the variational inequality (1.11.1) with ¯ x ≤ r. Moreover, the operator A is single-valued at x ¯. Proof. For simplicity of reasoning we consider that a space X is separable. Let {Xn } be a sequence of finite-dimensional subspaces of X, Qn : X → Xn be projection operators, Q∗n be their conjugate operators. Let Ωn = Ω ∩ Xn be a convex closed set in Xn , PΩn : X → Ωn , PΩ : X → Ω be projection operators on Ωn and Ω, respectively. Let Fn x = J(x − PΩn x) ∀x ∈ X, and F x = J(x − PΩ x) ∀x ∈ X, ∗

where J : X → X is the normalized duality mapping in X. It is obvious that θXn ∈ Ωn for all n > 0. An operator F : X → X ∗ is single-valued, bounded and monotone by Lemma 1.5.18, and F x = θX ∗ if and only if x ∈ Ω. The operators Fn have the similar properties. Observe that the sets Ωn Mosco-approximate Ω in the following sense: For any element g ∈ Ω there is a sequence {gn } ∈ Ωn such that gn → g, and if hn h, hn ∈ Ωn , then h ∈ Ω. Choose a sequence {n } such that n > 0, n → 0, and consider in Xn the equation Q∗n (Ax + n −1 Fn x − f ) = 0.

(1.12.9)

Since θXn ∈ Ωn , we have Fn x, x ≥ 0 for all x ∈ Xn . Therefore, for x ∈ Xn with x = r and for y ∈ Ax, the following relations hold: Q∗n (y + n −1 Fn x − f ), x = y − f, x + n −1 Fn x, x ≥ 0. Now the existence of the solution xn ∈ Xn of equation (1.12.9) with xn ≤ r follows ¯ ∈ X because the sequence {xn } is bounded. from Theorem 1.1.62. It is clear that xn x Besides, there is y n ∈ Axn such that Q∗n (y n + n −1 Fn xn − f ) = θX ∗ .

(1.12.10)

We prove that {y } is bounded. Let n

un = a1

J ∗yn , yn ∗

v n ∈ T un , v n ∗ ≤ a2 , a1 > 0,

a2 > 0,

(1.12.11)

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where J ∗ : X ∗ → X is the normalized duality mapping in X ∗ . The local boundedness of the maximal monotone operator T guarantees existence of the sequences {un } and {v n } satisfying (1.12.11). The monotonicity property of T gives y n + Cxn − v n , xn − un ≥ 0. This inequality and (1.12.11) imply the estimate a1 y n ∗ ≤ (Cxn ∗ + a2 )a1 + (f ∗ + Cxn ∗ + a2 )r + Q∗n (y n + n −1 Fn xn − f ), xn − n −1 Fn xn , xn . Now we make sure that the sequence {y n } is bounded which follows from (1.12.10) and from the properties of the operator Fn . Using (1.12.10) again, one gets n (y n − f ) + Fn xn , zn = 0 ∀zn ∈ Xn .

(1.12.12)

From (1.12.12) we deduce that Fn xn , zn → 0 for every bounded sequence {zn }. In particular, Fn xn , xn → 0. It is not difficult to obtain the following equalities: Fn xn , xn = J(xn − PΩn xn ), xn − PΩn xn + J(xn − PΩn xn ), PΩn xn = Fn xn 2∗ + J(xn − PΩn xn ), PΩn xn .

(1.12.13)

By Lemma 1.5.17 and by the definition of PΩn xn with xn ∈ Xn , we can write J(xn − PΩn xn ), PΩn xn − y ≥ 0 ∀y ∈ Ωn . Assuming y = θXn we obtain J(xn − PΩn xn ), PΩn xn ≥ 0. Then the estimate

Fn xn , xn ≥ Fn xn 2∗

appears from (1.12.12). Thus, Fn xn → θXn∗ as n → ∞. Show that Fn x → F x for any x ∈ X. Let PΩ x = z ∈ Ω, and PΩn x = zn ∈ Ωn , that is, x − z = min{x − y | y ∈ Ω}

(1.12.14)

x − zn = min{x − y | y ∈ Ωn }.

(1.12.15)

and Due to Theorem 1.11.14, (1.12.14) and (1.12.15) yield the inequalities J(x − z), z − y ≥ 0 ∀y ∈ Ω, z ∈ Ω,

(1.12.16)

1.12 Variational Inequalities with Semimonotone Operators

83

and J(x − zn ), zn − y ≥ 0 ∀y ∈ Ωn , zn ∈ Ωn .

(1.12.17)

Since sequence {zn } is bounded, zn z¯ ∈ Ω. Let u ¯ n ∈ Ωn , u ¯n → z ∈ Ω. Put y = z¯ in (1.12.16) and y = u ¯n in (1.12.17). Then summation of these inequalities gives J(x − zn ) − J(x − z), zn − z + J(x − zn ), z − u ¯n + J(x − z), zn − z¯ ≥ 0.

(1.12.18)

The following result was obtained in Theorem 1.6.4. Let x, y ∈ X. If x ≤ R and y ≤ R then there exist constants c1 , c2 > 0 such that Jx − Jy, x − y ≥ c1 δX (c−1 2 x − y), where δX () is the modulus of convexity of a space X. Therefore, using weak convergence of {zn } to z¯ and (1.12.18), we conclude that for any fixed x ∈ X, c1 δX (c−1 ¯n + J(x − z), zn − z¯ , 2 zn − z) ≤ x − zn z − u from which the convergence zn → z follows. By Corollary 1.5.16, J is continuous, which guarantees the strong convergence of Fn x to F x for all x ∈ X. Write the monotonicity condition of the operators Fn : Fn y − Fn xn , y − xn ≥ 0, y ∈ X. Applying the properties of the sequence {Fn } and setting n → ∞, we obtain F y, y − x ¯ ≥ 0 ∀y ∈ X,

x ¯ ∈ X.

(1.12.19)

Since the operator F : X → X ∗ is monotone and demicontinuous and D(F ) = X, we have by Theorem 1.4.6 that F is maximal monotone. Hence, the equality F x ¯ = θX ∗ is satisfied because of (1.12.19). Thus, we conclude that x ¯ ∈ Ω. Let wn ∈ Ωn and wn → x ¯ ∈ Ω. Then Fn wn = θX ∗ and y n − f, wn − xn = n −1 Fn wn − Fn xn , wn − xn ≥ 0.

(1.12.20)

Therefore, y n − f, xn − x ¯ ≤ y n − f, wn − x ¯ . Since the sequence {y n } is bounded, we have lim sup y n − f, xn − x ¯ ≤ 0.

(1.12.21)

y − f, x ¯ − v , lim inf y n − f, xn − v ≥ ¯

(1.12.22)

n→∞

By Lemma 1.12.1, n→∞

where y¯ = y¯(v) ∈ A¯ x, v ∈ X. Let now u ∈ Ω be an arbitrary element, wn ∈ Ω and wn → u. Then like (1.12.21), we obtain the inequality lim sup yn − f, xn − u ≤ 0 ∀u ∈ Ω. n→∞

(1.12.23)

84

1


By the condition (1.12.8), the operator A is single-valued at the point x ¯ ∈ Ω. Consequently, in (1.12.22) y¯ = A¯ x for all v ∈ X. Assuming v = u ∈ Ω in (1.12.22) and taking into account (1.12.23), we deduce for x ¯ ∈ Ω, A¯ x − f, x ¯ − u ≤ lim inf y n − f, xn − u n→∞

≤ lim sup y n − f, xn − u ≤ 0 ∀u ∈ Ω. n→∞

The proof of the theorem is accomplished. One can verify that Theorem 1.12.2 holds if we require coerciveness of A in place of (1.12.7). The condition (1.12.8) of the theorem plays the defining role in proving the facts that the operator A is single-valued on the solutions of (1.11.1) and that some subsequence of {xn } weakly converges to x ¯.

1.13

Variational Inequalities with Pseudomonotone Operators

The property of semimonotone mappings noted in Lemma 1.12.1 is the most essential in the definition of the wider class of the so-called pseudomonotone mappings, which, in general, are not necessarily monotone ones. Definition 1.13.1 Let X and Y be linear topological spaces. An operator A : X → 2Y is said to be upper semicontinuous if for each point x0 ∈ X and arbitrary neighborhood V of Ax0 in Y, there exists a neighborhood U of x0 such that for all x ∈ U one has the inclusion: Ax ⊂ V. Definition 1.13.2 Let Ω be a closed convex set of a reflexive Banach space X. An operator ∗ A : Ω → 2X is called pseudomonotone if the following conditions are satisfied: a) for any x ∈ Ω the set Ax is non-empty, bounded, convex and closed in dual space X ∗ ; b) A is upper semicontinuous from each finite-dimensional subspace F of X to the weak topology on X ∗ , that is, to a given element x0 ∈ F and a weak neighborhood V of Ax0 in X ∗ there exists neighborhood U of x0 in F such that Ax ⊂ V for all x ∈ U. c) if {xn } is a sequence of elements from Ω that converges weakly to x ∈ Ω, elements zn ∈ Axn such that lim sup zn , xn − x ≤ 0, n→∞

then for every element y ∈ Ω there exists z(y) ∈ Ax such that lim inf zn , xn − y ≥ z(y), x − y . n→∞

Assume that X is a reflexive Banach space. Lemma 1.13.3 Let Ω be a closed convex subset of X. Any hemicontinuous monotone operator A : Ω → X ∗ is pseudomonotone.

1.13 Variational Inequalities with Pseudomonotone Operators

85

Proof. Suppose that the sequence {xn } ⊂ Ω weakly converges to x ∈ X and lim sup Axn , xn − x ≤ 0. n→∞

Since A is monotone, we have lim Axn , xn − x = 0.

(1.13.1)

n→∞

Take an arbitrary y ∈ Ω and put zt = (1 − t)x + ty, t ∈ (0, 1]. It is clear that zt ∈ Ω. The monotonicity property of A gives Axn − Azt , xn − zt ≥ 0. Then Azt , xn − x + t(x − y) − Axn , xn − x ≤ tAxn , x − y . Using (1.13.1), one gets Azt , x − y ≤ lim inf Axn , x − y . n→∞

By the hemicontinuity of A, this produces in a limit as t → 0, Ax, x − y ≤ lim inf Axn , x − y . n→∞

In view of (1.13.1) again, it is not difficult to obtain from the latter inequality that Ax, x − y ≤ lim inf Axn , xn − y . n→∞

The lemma is proved. Consider pseudomonotone operators on the whole space when Ω = D(A) = X. ∗

Lemma 1.13.4 Any maximal monotone operator A : X → 2X with D(A) = X is pseudomonotone. Proof. The condition a) of Definition 1.13.2 follows from the maximal monotonicity of the operator A. Since any maximal monotone operator is semimonotone, we conclude, by Lemma 1.12.1, that the property c) is true. Let xn → x, xn ∈ F ∩ X, x ∈ F ∩ X and let F be a finite-dimensional subspace of X. In view of the local boundedness of a maximal monotone mapping at a point x, we have the weak convergence of yn ∈ Axn to some y. It is known that the graph of A is demiclosed. This guarantees the inclusion y ∈ Ax, thus, the property b) also holds.

It is not difficult to verify the following assertion. Lemma 1.13.5 Any completely continuous operator A : X → X ∗ with D(A) = X is pseudomonotone.

86

1


Next we give the criteria when a sum of two operators is pseudomonotone. ∗

Lemma 1.13.6 Let A : Ω → X ∗ be a hemicontinuous monotone operator, B : Ω → 2X be a pseudomonotone mapping. Then their sum C = A + B is pseudomonotone. Proof. The requirements a) and b) of Definition 1.13.2 for C are obvious. Prove the property c). Let xn ∈ Ω, xn x and lim sup yn , xn − x ≤ 0, yn ∈ Cxn . n→∞

Taking into account the monotonicity of A we have zn , xn − x ≤ yn , xn − x − Ax, xn − x , zn ∈ Bxn ,

yn = Axn + zn ,

from which we deduce the following inequality: lim sup zn , xn − x ≤ lim sup yn , xn − x ≤ 0. n→∞

(1.13.2)

n→∞

Pseudomonotonicity of B gives now the estimate z(y), x − y ≤ lim inf zn , xn − y ∀y ∈ Ω, n→∞

z(y) ∈ Bx.

(1.13.3)

By (1.13.3) with y = x, one gets lim inf zn , xn − x ≥ 0. n→∞

The latter relation and (1.13.2) give lim zn , xn − x = 0.

n→∞

Then (1.13.2), where yn = Axn + zn , implies lim sup Axn , xn − x ≤ 0. n→∞

Due to Lemma 1.13.3 the inequality Ax, x − y ≤ lim inf Axn , xn − y ∀y ∈ Ω, n→∞

(1.13.4)

holds. Summing (1.13.3) and (1.13.4), we obtain the conclusion of the lemma. ∗

Lemma 1.13.7 Let A1 and A2 be two pseudomonotone mappings from X into 2X . Then their sum is also pseudomonotone. Let us turn to the question on solvability of variational inequalities with pseudomonotone mappings. We give one of the main results of this direction. In order to prove it we need the following two theorems.


87

Theorem 1.13.8 Let Ω be a nonempty compact and convex subset of the locally convex topological vector space X, A : Ω → 2Ω be an upper semicontinuous mapping such that Ax is a nonempty closed and convex subset of X for all x ∈ Ω. Then there exists an element x0 ∈ Ω such that x0 ∈ Ax0 . Theorem 1.13.9 Let Ω be a nonempty compact and convex subset of the locally convex ∗ topological vector space X, A : Ω → 2X be an upper semicontinuous mapping such that Ax is a nonempty compact closed convex subset of X ∗ for each x ∈ Ω. Then there exists x0 ∈ Ω and y0 ∈ Ax0 such that y0 , y − x0 ≤ 0 ∀y ∈ Ω. In the sequel, a solution of the variational inequality (1.11.1) is understood in the sense of Definition 1.11.1. The main result of this section is the following Theorem 1.13.10 Suppose that X is a reflexive Banach space, Ω is a closed convex subset ∗ of X, θX ∈ Ω, A : Ω → 2X is a pseudomonotone and coercive operator on Ω. Then for ∗ any f ∈ X there exists a solution of the variational inequality (1.11.1). Proof. Without loss of generality, assume that f = θX ∗ , and prove that there exists a pair (x, z) ∈ grA such that z, x − y ≤ 0 for all y ∈ Ω. Let Λ be a family of all finitedimensional subspaces F ⊂ X, ordered by inclusion. For each R > 0, let ΩR F = ΩF ∩ B(θX , R), ΩF = Ω ∩ F. convex subset of X. Hence, we may apply Theorem 1.13.9 Then ΩR F is a nonempty compact X ∗ , which allows us to assert that there exist elements xR ∈ ΩR to the mapping −A : ΩR F →2 F F and zFR ∈ AxR F such that zFR , xR ∀y ∈ ΩR (1.13.5) F − y ≤ 0 F. In particular, we may take y = θF ∈ ΩR F for each R > 0 and obtain the inequality R R zFR , xR F ≤ 0, zF ∈ AxF .

Since the operator A is coercive, it follows that xR F ≤ λ, λ > 0, for all R > 0. R Consider R > λ. Then xR F ≤ λ < R. For any y ∈ ΩF , we put xt = (1 − t)xF + ty, t ∈ (0, 1). By the convexity of set ΩF , the element xt ∈ ΩF . If t is sufficiently small then xt ≤ R. This means that xt ∈ ΩR F , hence, by (1.13.5) we have zFR , xR F − xt ≤ 0 or zFR , xR F − y ≤ 0

∀y ∈ ΩF .

Thus, there exists a solution of the variational inequality Ax, x − y ≤ 0

∀y ∈ ΩF , x ∈ ΩF .

(1.13.6)

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1


Let xF be some solution of (1.13.6). Then for all F ∈ Λ there exists an element zF ∈ AxF such that zF , xF − y ≤ 0 ∀y ∈ ΩF , xF ∈ ΩF . Since θF ∈ ΩF for all F ∈ Λ and A is coercive, we have c(xF )xF ≤ zF , xF ≤ 0, where c(t) → +∞ as t → +∞. Hence, there exists M > 0 such that xF ≤ M for all F ∈ Λ. Construct the set VF = {xF }, F ∈ Λ. F ⊃F

Then the family {VF }F ∈Λ has the finite intersection property. Indeed, consider F1 ∈ Λ and F2 ∈ Λ. If F3 ∈ Λ is so that F1 ∪ F2 ⊆ F3 , then VF1 ∩ VF2 ⊇ VF3 . One can show that the weak closure of VF in Ω (for short, weakcl VF ) is weakly compact, and the set

G=

weakcl VF = ∅.

F ∈Λ

Let an element x0 ∈ Ω ∩ G be given. Choose some fixed element y ∈ Ω and a set F ∈ Λ such that it contains the elements x0 and y. Since x0 ∈ weakcl VF , there exists a sequence {xj } ∈ VF such that xj x0 as j → ∞. For every j, we denote by Fj just the set of Λ for which the inclusion xj ∈ ΩFj holds. By the definition of the element xj , we have zj , xj − u ≤ 0 ∀u ∈ ΩFj , zj ∈ Axj .

(1.13.7)

Since y ∈ Fj for all j we can write zj , xj − y ≤ 0, zj ∈ Axj . Then (1.13.7) for u = x0 gives

(1.13.8)

zj , xj − x0 ≤ 0

because x0 ∈ Fj for all j ≥ 1. Therefore, lim sup zj , xj − x0 ≤ 0. j→∞

By the pseudomonotonicity of A, we conclude now that there exists an element z(y) ∈ Ax0 such that lim inf zj , xj − y ≥ z(y), x0 − y , y ∈ Ω. j→∞

In view of (1.13.8), the latter inequality implies z(y), x0 − y ≤ 0 ∀y ∈ Ω, x0 ∈ Ω, z(y) ∈ Ax0 .

(1.13.9)

In order to finish the proof of the theorem it suffices to show that there exists a unique element z ∈ Ax0 such that (1.13.9) is satisfied and the inequality z, x0 − y ≤ 0 holds for all


89

y ∈ Ω. We shall prove this claim by contradiction. Assume that for every element z ∈ Ax0 there exists y(z) ∈ Ω such that (1.13.10) z, x0 − y(z) > 0. Construct the set Ny = {z | z ∈ Ax0 , z, x0 − y > 0} ∀y ∈ Ω. Since Ny ⊂ Ax0 for all y ∈ Ω, it follows from (1.13.10) and from Definition 1.13.2 that Ny is a nonempty open and bounded set. Besides, the family of sets {Ny | y ∈ Ω} is an open covering of the weakly compact set Ax0 . Hence, there exists a finite covering {Ny1 , Ny2 , ..., Nys } of the set Ax0 and a corresponding decomposition of the unit on this set. The latter is defined by the family of functions {α1 , α2 , ..., αs }, where αj : Ax0 → [0, 1] are strongly continuous, 0 ≤ αj (z) ≤ 1 ∀z ∈ Ax0 , αj (z) = 0

if

z ∈ Nyj , 1 ≤ j ≤ s,

αj (z) > 0

if

z ∈ Nyj , 1 ≤ j ≤ s,

s

αj (z) = 1 ∀z ∈ Ax0 .

j=1

Define the map B : Ax0 → Ω by the equality Bz =

s

αj (z)yj ∀z ∈ Ax0 .

j=1

Obviously, it is strongly continuous. Then, by the definitions of αj (z) and yj , we have z, x0 − Bz =

s

αj (z)z, x0 − yj > 0 ∀z ∈ Ax0 ,

(1.13.11)

j=1

because z belongs to Nyj and z, x0 − yj > 0 for any j satisfying the inequality αj (z) > 0. On the other hand, taking into account (1.13.9), it is possible to construct the following nonempty closed and convex set: Sy = {z | z ∈ Ax0 , z, x0 − y ≤ 0}, so that it is a closed and convex subset of Ax0 . Thus, we defined the mapping S : Ω → 2Ax0 which has the property of upper semicontinuity. Then we consider the operator C = SB : Ax0 → 2Ax0 with the following properties: (i) C is upper semicontinuous, (ii) Cz is a nonempty convex closed set for any z ∈ Ax0 . Hence, by Theorem 1.13.8, operator C has a fixed point z1 ∈ Ax0 , i.e., z1 ∈ Cz1 . Thus, there exists z1 ∈ Ax0 such that z1 , x0 − Bz1 ≤ 0, z1 ∈ Ax0 , which contradicts (1.13.11). The proof is accomplished. Remark 1.13.11 If Ω = X, then Theorem 1.13.10 gives the sufficient conditions for equations with pseudomonotone operators to be solvable.

90

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1.14


Variational Inequalities with Quasipotential Operators

In this section we present one more class of non-monotone mappings for which the variational inequality (1.11.1) is solvable. Let X be a reflexive strictly convex Banach space together with its dual space X ∗ . Definition 1.14.1 An operator A : X → X ∗ is called radially summable on Ω if for all x and y from Ω the function ψx,y (t) = A(y + t(x − y)), x − y is Lebesgue integrable on the interval [0, 1]. Definition 1.14.2 A radially summable on Ω operator A is called quasipotential on Ω if there exists a functional ϕ : Ω → R1 satisfying the equality ϕ(x) − ϕ(y) =

0

1

A(y + t(x − y)), x − y dt

(1.14.1)

for all x, y ∈ Ω. A functional ϕ is said to be the potential of the mapping A. Note that hemicontinuous quasipotential mappings are potential. Definition 1.14.3 An operator A : X → X ∗ is said to be upper h-semicontinuous on Ω if lim sup A(y + t(x − y)) − Ay, x − y ≤ 0 ∀x, y ∈ Ω. t→0+

Lemma 1.14.4 Let A : X → X ∗ denote a quasipotential upper h-semicontinuous on Ω operator, and let its potential ϕ(x) have the minimum point x ∈ Ω. Then x is a solution of the variational inequality Ax, x − y ≤ 0 ∀y ∈ Ω, x ∈ Ω.

(1.14.2)

Proof. Since x is the minimum point of the functional ϕ(x) on Ω, there is a number r r > 0 such that for all y ∈ Ω and for all 0 < τ < τ0 (y), where τ0 (y) = min{1, }, x − y the following inequality holds: 0 ≤ ϕ(x + τ (y − x)) − ϕ(x) = τ

= τ

1 0

1 0

A(x + τ t(y − x)), y − x dt

A(x + τ t(y − x)) − Ax, y − x dt + τ Ax, y − x .

(1.14.3)

Assume that x is not a solution of the variational inequality (1.14.2). Then for some y ∈ Ω, we have Ax, y − x = − < 0. By virtue of the upper h-semicontinuity of the operator A on Ω, there exists 0 < δ < 1 such that A(x + θ(y − x)) − Ax, y − x ≤ 2

1.14 Variational Inequalities with Quasipotential Operators

91

for all 0 < θ < δ. Hence, if 0 < τ < min{δ, τ0 (y)} then ϕ(x + τ (y − x)) − ϕ(x) ≤

τ τ − τ = − < 0, 2 2

which contradicts (1.14.3). Theorem 1.14.5 Let A : X → X ∗ denote a quasipotential upper h-semicontinuous on Ω mapping, and let its potential ϕ be weakly lower semicontinuous on Ω. Suppose that either Ω is bounded or it is unbounded and lim

x→+∞

ϕ(x) = +∞, x ∈ Ω.

(1.14.4)

Then the variational inequality (1.14.2) has at least one solution. Proof. By the Weierstrass generalized theorem and by the condition (1.14.4), it follows that there exists an element x ∈ Ω such that ϕ(x) = min {f (y) | y ∈ Ω}. Then Lemma 1.14.4 ensures a validity of the assertion being proved. Theorem 1.14.6 Let A = A1 + A2 , where operators Ai : X → X ∗ , i = 1, 2, are quasipotential on Ω. Let A1 be a monotone mapping and A2 be strongly continuous, A be upper h-semicontinuous on Ω, a set Ω be unbounded. If the condition (1.14.4) is satisfied, then the variational inequality (1.14.2) has a solution. Proof. If we establish the weak lower semicontinuity on Ω of the functional ϕ = ϕ1 +ϕ2 , where ϕ1 and ϕ2 are potentials of the operators A1 and A2 , respectively, then the assertion follows from Theorem 1.14.5. Let x ∈ Ω, xn ∈ Ω, xn x. Then the monotonicity property of the operator A1 on Ω gives the following relations: ϕ1 (xn ) − ϕ1 (x) = =

1 0

1 0

A1 (x + t(xn − x)), xn − x dt A1 (x + t(xn − x)) − A1 x, xn − x dt + A1 x, xn − x

≥ A1 x, xn − x . Therefore, lim inf ϕ1 (xn ) ≥ ϕ1 (x). n→∞

Since the mapping A2 is strongly continuous on Ω, we have for all t ∈ [0, 1], lim A2 (x + t(xn − x)), xn − x = 0.

n→∞

Furthermore, by the boundedness of A2 , there exists a constant M > 0 such that |A2 (x + t(xn − x)), xn − x | ≤ M

(1.14.5)

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1


for all t ∈ [0, 1] and for all n > 0. Then we may write

lim

n→∞

ϕ2 (xn ) − ϕ2 (x)

= =

1

lim

n→∞ 0 1

A2 (x + t(xn − x)), xn − x dt

lim A2 (x + t(xn − x)), xn − x dt = 0.

0 n→∞

In view of (1.14.5), this implies lim inf ϕ(xn ) = lim inf [ϕ1 (xn ) + ϕ2 (xn )] = lim inf ϕ1 (xn ) + ϕ2 (x) n→∞

n→∞

n→∞

≥ ϕ1 (x) + ϕ2 (x) = ϕ(x). The theorem is proved. Remark 1.14.7 If Ω = X, then Theorems 1.14.5, and 1.14.6 give the solvability conditions of equations with quasipotential mappings. Furthermore, if the potential of the operator A is non-negative, then the potential of A + αJ(α > 0) actually has the properties (1.14.4). Let us present an example. Consider the problem: To find a function x ∈ Ω satisfying for any y ∈ Ω the following inequality:

1≤|α|=|β|≤m

G

aαβ (s)Dα x(s)Dβ (y(s) − x(s))ds +

G

g(s, x(s))(y(s) − x(s))ds ≥ 0. (1.14.6)

Assume that G ⊂ Rn is a bounded measurable set with the piecewise smooth boundary ◦

2 (G), functions a (s) are continuous in G, where a (s) = a (s) on G, ∂G, Ω ⊂Wm αβ βα αβ 1 ≤ |α| = |β| ≤ m, and for arbitrary s ∈ G, ξ ∈ Rn the inequality

aαβ (s)ξ α+β ≥ κ|ξ|2m , κ > 0

|α|=|β|=m

holds. Let a function g : G × R1 → R1 be measurable superposition. Suppose that the following conditions are satisfied: 1) the function g(s, x) is continuous to the right for almost all s ∈ G and has discontinuities just of the first kind. Moreover, if x is a discontinuity point of g(s, x), then g(s, x − 0) > g(s, x), and for all x ∈ R1 the function g(s, x) is measurable on Ω; 2) g(s, x) has (p − 1)-order of growth with respect to x : |g(s, x)| ≤ a(s) + b|x|p−1 for all x ∈ R1 and for almost all s ∈ G, where a(s) ∈ Lq (G), b > 0, p−1 + q −1 = 1. Here 2n if n > 2m, and p > 1 if n < 2m; 1 0; for all x ∈Wm 4)

2

0

x

g(s, y)dy ≥ −kx2 − d(x)|x|λ − c(s)

for all x ∈ R1 and for almost all s ∈ G, 0 < λ < 2, d(x) ∈ Lγ (G), γ = M, where C is a constant in the inequality x2L2 ≤ C

|α|≤m

Dα x(s)2L2 (G)

2 , k > 0, kC < 2−λ

◦

2 ∀x(s) ∈Wm (G),

and c(s) is the Lebesgue absolutely integrable function; ◦

2 (G) is an unbounded convex closed set. 5) Ω ⊂Wm

◦

2 Show that the inequality (1.14.6) has a solution if all these conditions hold. Let X =Wm (G) with the norm

u =

1/2

|α|=m

| Dα u(x) |2 dx

G

.

Consider an operator A1 : X → X ∗ generated by the dual form

A1 x, y =

aαβ (s)D α x(s)D β y(s)ds,

1≤|α|=|β|≤m

G

where x and y are elements from X. By the properties of the functions aαβ (s), the operator A1 is bounded, self-adjoint and monotone. The condition 3) guarantees the estimate A1 x, x ≥ M x2 ∀x ∈ X.

(1.14.7)

Hence, A1 is a potential mapping and its potential ϕ1 (x) = 2−1 A1 x, x . By condition 1), if x(s) is measurable on G then superposition g(s, x(s)) is also measurable on G [213]. Define an operator A2 : X → X ∗ by the formula A2 x, y =

G

g(s, x(s))y(s)ds ∀x, y ∈ X. ◦

2 By the condition 2), the imbedding operator E of the space Wm (G) into Lp (G) is completely continuous and the operator

A3 x(s) = g(s, x(s)) ∀x(s) ∈ Lp (G) is bounded. It is clear that A3 : Lp (G) → Lq (G). Denote the adjoint operator to E by E ∗ . Now compactness of A2 follows from the equality A2 = E ∗ A3 E. Define functional ϕ2 on X as

ϕ2 (x) =

ds G

0

x(s)

g(s, y)dy.

94

1


Then for any x, h ∈ X we deduce

ϕ2 (x + h) − ϕ2 (x) =

x(s)+h(s)

ds

g(s, y)dy x(s)

G

ds

= G

=

d dt

0

x(s)+th(s)

g(s, y)dydt

0

1

dt

0

=

1

g(s, x(s) + th(s))h(s)ds G

1 0

A2 (x + th), h dt.

Hence, A2 is quasipotential on X and ϕ2 is its potential. If A = A1 + A2 then

A(x + th) − Ax, h = t

1≤|α|=|β|≤m

+ G

aαβ (s)Dα h(s)D β h(s)ds G

g(s, x(s) + th(s)) − g(s, x(s)) h(s)ds.

By the conditions 1) and 2), one gets lim A(x + th) − Ax, h ≤ 0.

t→0+

Thus, the operator A is upper h-semicontinuous on X. Now (1.14.7) and the condition 4) produce for the potential ϕ = ϕ1 + ϕ2 of operator A the following inequalities: 2ϕ(x) ≥ M x2 +

G

(−kx2 (s) − d(s) | x(s) |λ −c(s))ds

≥ (M − kC)x2 −

G

1/γ

| d(s) |γ ds

C λ/2 xλ −

G

| c(s) | ds.

Since M − kC > 0, lim

x→+∞

ϕ(x) = +∞.

We see that all the conditions of Theorem 1.14.6 are satisfied for an operator A. Therefore, there exists a function x(s) ∈ Ω such that Ax, y − x ≥ 0 ∀y ∈ Ω. The latter relation is identical to (1.14.6) in view of the definitions of A1 and A2 . Remark 1.14.8 Consider the equation

1≤|α|=|β|≤m

(−1)|β| D β (aαβ (s)Dα x(s)) + g(s, x(s)) = 0, s ∈ G,

(1.14.8)

1.15 Equations with Accretive Operators

95

with the boundary condition ∂νr (∂G)x(s) = 0 ∀x ∈ ∂G,

0 ≤ r ≤ m − 1, ◦

2 (G) where ∂νr (∂G) is a derivative in the interior normal direction to ∂G. A function x ∈Wm ◦

2 (G) the following equality holds: is said to be the solution of (1.14.8) if for all y ∈Wm

aαβ (s)Dα x(s)Dβ y(s)ds +

1≤|α|=|β|≤m

G

g(s, x(s))y(s)ds = 0. G

Solvability of the equation (1.14.8) is obtained by the same arguments as in the previous example.

1.15

Equations with Accretive Operators

In this section we study the class of nonlinear operators acting from a Banach space X to X. We assume that X is reflexive and strictly convex together with its dual space X ∗ . Definition 1.15.1 An operator A : X → 2X is called accretive if J(x1 − x2 ), y1 − y2 ≥ 0

(1.15.1)

for all x1 , x2 ∈ D(A), y1 ∈ Ax1 , y2 ∈ Ax2 . If an operator A is Gâteaux differentiable, then this definition is equivalent to the following one. Definition 1.15.2 A Gˆ ateaux differentiable operator A : X → X is called accretive if Jh, A (x)h ≥ 0 ∀x, h ∈ X. Remark 1.15.3 The properties of monotonicity and accretiveness of an operator coincide in a Hilbert space. We present one more definition of an accretive operator. Definition 1.15.4 An operator A : X → 2X is said to be accretive if x1 − x2 ≤ x1 − x2 + λ(y1 − y2 ), λ > 0, for all x1 , x2 ∈ D(A), y1 ∈ Ax1 , y2 ∈ Ax2 . Theorem 1.15.5 Definitions 1.15.1 and 1.15.4 are equivalent.

(1.15.2)

96

1


Proof. Indeed, let (1.15.1) hold. Then the inequality J(x1 − x2 ), x1 − x2 + λ(y1 − y2 ) ≥ x1 − x2 2 , λ > 0, is valid and (1.15.2) immediately follows. Further, we know that if X ∗ is strictly convex then X is smooth and Jx = 2−1 gradx2 . By convexity of the functional x2 , we may write the inequality x1 − x2 2 ≥ x1 − x2 + λ(y1 − y2 )2 − 2λJ(x1 − x2 + λ(y1 − y2 )), y1 − y2 . If (1.15.2) holds then J(x1 − x2 + λ(y1 − y2 )), y1 − y2 ≥ 0. Letting λ → 0 and using the hemicontinuity of J, we obtain (1.15.1). We present some properties of accretive mappings. Definition 1.15.6 Accretive operator A : X → 2X is said to be coercive if Jx, y ≥ c(x)x ∀y ∈ Ax, where c(t) → +∞ as t → +∞. Definition 1.15.7 An operator A : X → 2X is called locally bounded in a point x ∈ D(A) if there exists a neighborhood M of that point such that the set A(M ) = {y | y ∈ Ax, x ∈ M ∩ D(A)} is bounded in X. Theorem 1.15.8 Let A : X → 2X be an accretive operator, dual mappings J : X → X ∗ and J ∗ : X ∗ → X be continuous in X and X ∗ , respectively. Then A is locally bounded at every point x ∈ int D(A). Proof. Suppose that is not the case. Let x0 ∈ int D(A), xn ∈ D(A), n = 1, 2, ... , xn → x0 , and yn → ∞, where yn ∈ Axn . Let tn = xn − x0 1/2 . For any z ∈ X ∗ , we construct the sequence {zn } by the formula zn = J(xn − x0 − tn w) + tn z, where w = J ∗ z. The elements zn are well defined because of D(J) = X. In view of continuity of J and since tn → 0, it follows that zn → θX ∗ and the elements x0 + tn w ∈ D(A) for all w ∈ X provided that n > 0 is sufficiently large. If f ∈ Av, v = x0 + σw, un ∈ A(x0 + tn w), then the accretiveness of A gives the inequality (tn − σ)z, un − f ≥ 0,


97

from which we obtain z, un ≤ z∗ f for tn < σ. Since R(J) = X ∗ , one gets lim sup z, un < ∞ ∀z ∈ X ∗ . n→∞

Now the Banach−Steinhaus theorem implies the boundedness of the sequence {un }, say, un ≤ C for all n > 0. Further, since A is accretive, we write zn − tn z, yn − un ≥ 0. Therefore, 1 1 zn , yn − zn − tn z, un tn tn

z, yn ≤

z zn ∗ n ∗ + z∗ . yn + C tn tn

≤ Denoting

x − x zn ∗ n 0 − J ∗z + z , = J ∗ tn tn we deduce by the latter inequality that

τn (z) =

lim sup n→∞

z, yn < ∞. 1 + yn τn (z)

The functional

z, yn 1 + yn τn (z) is continuous with respect to z because J and J ∗ are continuous. Then, as in the proof of the Banach−Steinhaus theorem, we assert that there are constants C1 > 0 and r > 0 such that z, yn (1.15.3) ≤ C1 , n = 1, 2, ... , 1 + yn τn (z) ϕ(z) =

if z ∈ X ∗ with z∗ ≤ r. Take in (1.15.3) z = z¯n ∈ B ∗ (θX ∗ , r) satisfying the condition ¯ zn , yn = ryn . Then C1 yn . ≤ r 1 + yn τn (¯ zn ) Hence, −1 C1 C1 zn ) . τn (¯ 1− yn ≤ r r The continuity of J yields the relation x − x n 0

zn ) = J τn (¯

tn

Therefore, for > 0 such that 1 − C1 r inequality holds:

− J ∗ z¯n + JJ ∗ z¯n → 0, n → ∞. ∗

−1

> 0 and for sufficiently large n > 0, the following

C1 C1 −1 . 1− r r That contradicts the assumption that yn → ∞ as n → ∞. yn ≤

98

1


Corollary 1.15.9 If A : X → 2X is an accretive operator, J is continuous, D(A) = X and X is finite-dimensional, then A is bounded. Proof. See Theorem 1.3.19. Lemma 1.15.10 If an operator T : X → X is nonexpansive in D(A), then A = I − T is accretive. Proof. We have for all x, y ∈ D(A), J(x − y), Ax − Ay = −J(x − y), T x − T y + J(x − y), x − y ≥ x − y2 − T x − T yx − y ≥ x − y2 − x − y2 = 0.

Definition 1.15.11 An accretive operator A : X → 2X is called maximal accretive if its graph is not the right part of the graph of any other accretive operator B : X → 2X . The following lemma shows that the graph of any maximal accretive operator is demiclosed (cf. Lemma 1.4.5). Lemma 1.15.12 Let A : X → 2X be a maximal accretive operator. Let xn ∈ D(A), yn ∈ Axn . Suppose that either xn → x, yn y and the duality mapping J is continuous, or xn x, yn → y and J is weak-to-weak continuous. Then x ∈ D(A) and y ∈ Ax. Proof. The accretiveness A produces the inequality J(xn − u), yn − v ≥ 0 ∀u ∈ D(A),

∀v ∈ Au.

Let n → ∞. Under the hypotheses of the theorem, we have in a limit J(x − u), y − v ≥ 0 ∀u ∈ D(A),

∀v ∈ Au.

Then maximal accretiveness of A implies inclusions: x ∈ D(A) and y ∈ Ax. Lemma 1.15.13 The value set of a maximal accretive operator A : X → 2X at any point of its domain is convex and closed. Proof. This property of maximal accretive operators is easily obtained by Definition 1.15.1. Theorem 1.15.14 Let A : X → X be a hemicontinuous accretive operator with D(A) = X. Then A is maximal accretive. Proof. It is enough to show (see Theorem 1.4.6) that the inequality J(x − y), Ax − f ≥ 0 ∀x ∈ X,

(1.15.4)


99

implies f = Ay. Since D(A) = X, we may put in (1.15.4) x = xt = y + tz ∀z ∈ X, t > 0. Then Jz, Axt − f ≥ 0. Letting t → 0 and using the hemicontinuity property of A, we obtain Jz, Ay − f ≥ 0 ∀z ∈ X. Since R(J) = X ∗ , we conclude that f = Ay. Theorem 1.15.15 Let A : X → X be an accretive operator with D(A) = X, duality mappings J and J ∗ be continuous. Then the properties of hemicontinuity and demicontinuity of A on int D(A) are equivalent. Proof. See Theorem 1.3.20.

Definition 1.15.16 An operator A : X → 2X is called strictly accretive if the equality (1.15.1) holds only with x1 = x2 . Definition 1.15.17 An operator A : X → 2X is called uniformly accretive if there exists an increasing function γ(t), t ≥ 0, γ(0) = 0, such that J(x1 − x2 ), y1 − y2 ≥ γ(x1 − x2 ), where x1 , x2 ∈ D(A), y1 ∈ Ax1 , y2 ∈ Ax2 . An operator A is called strongly accretive if γ(t) = ct2 , where c > 0. Remark 1.15.18 We say that an operator A : X → 2X is properly accretive if (1.15.1) is fulfilled and there is not any strengthening of (1.15.1), for instance, up to the level of strong or uniform accretiveness. Definition 1.15.19 An accretive operator A : X → 2X is said to be m-accretive if R(A + αI) = X for all α > 0, where I is the identity operator in X. Lemma 1.15.20 If operator A is m-accretive, then it is maximal accretive. Proof. By Definition 1.15.19, R(A + I) = X. Since A + I is the strongly accretive operator, then there exists a unique pair (x, y) ∈ grA such that y + x = f for every f ∈ X. Let A¯ be a maximal accretive extension of the operator A (it exists by Zorn’s lemma). If we admit that there exists a pair (¯ x, y¯), which belongs to the graph of A¯ and at the same time does not belong to the graph of A, then we arrive at a contradiction with the solution ¯ + x = f. Thus, A¯ = A. uniqueness of the equation Ax The converse assertion to Lemma 1.15.20 is not true. However, the following statement holds.

100

1


Theorem 1.15.21 Assume that X ∗ is an uniformly convex Banach space, the duality mapping J ∗ is continuous, an operator A : X → 2X is accretive and D(A) is an open set. Then A is m-accretive if and only if A is maximal accretive. Next we give one important result concerning the sum of m-accretive operators. Theorem 1.15.22 Let X and X ∗ be uniformly convex Banach spaces, A : X → 2X and B : X → 2X be m-accretive operators in X, D(A) ∩ D(B) = ∅ and one of them is locally bounded. Then A + B is an m-accretive mapping. The solvability of equations with accretive operators essentially depends on the properties both of operator A and duality mapping J. Recall that the properties of duality mappings are defined by the geometric characteristics of spaces X and X ∗ (see Sections 1.5 and 1.6). Next we present the existence theorem for the equation Ax = f

(1.15.5)

with the accretive operator A. Theorem 1.15.23 Assume that X and X ∗ are strictly convex spaces, X possesses an approximation, operator A : X → X is accretive and demicontinuous with a domain D(A) = X, duality mapping J : X → X ∗ is continuous and weak-to-weak continuous and there exist r > 0 such that for all x with x = r, Jx, Ax − f ≥ 0. Then equation (1.15.5) has at least one classical solution x ¯ with ¯ x ≤ r. Consider the solvability problem for the equation (1.15.5) when A : X → 2X is an arbitrary accretive operator and D(A) = X. First prove the following auxiliary lemma. Lemma 1.15.24 Assume that Xn is an n-dimensional Banach space, Pn : X → Xn is a projection operator with the norm |Pn | = 1 and Pn∗ : X ∗ → Xn∗ is conjugate operator to Pn . Then the equality Pn∗ Jx = Jx holds for every x ∈ Xn . Proof. It is easy to see that for all x ∈ Xn , Pn∗ Jx, x = Jx, Pn x = Jx, x = Jx∗ x = x2 .

(1.15.6)

Since |Pn∗ | = |Pn | = 1, we have Pn∗ Jx∗ ≤ |Pn∗ |Jx∗ = Jx∗ = x. On the other hand, by (1.15.6), x2 ≤ Pn∗ Jx∗ x, that is, x ≤ Pn∗ Jx∗ . Thus, x = Pn∗ Jx∗ . Together with (1.15.6), this means that Pn∗ J is a duality mapping in X and Pn∗ Jx = Jx because J is a single-valued operator. As in Section 1.9, denote by R(Ax − f ) a convex closed hull of the weak limits of all subsequences of the sequences {Axn − f } when xn → x for xn , x ∈ X.


101

Definition 1.15.25 A point x0 ∈ X is called an sw-generalized solution of the equation (1.15.5) with an accretive operator if θX ∈ R(Ax0 − f ). Observe that this definition of solution coincides with the classical one if J is continuous and A is hemicontinuous at the point x0 (see Lemma 1.15.12). Show that sw-generalized solution of (1.15.5) exists in a finite-dimensional space Xn . Lemma 1.15.26 Let A : Xn → 2Xn be defined on B(θXn , r) and the inequality Jx, z − f ≥ 0 holds for all x ∈ S(θXn , r) and for some z ∈ Ax. Then there exists sw-generalized solution x0 ∈ B(θXn , r) of the equation (1.15.5). Proof. Consider B ∗ (θXn∗ , r) = JB(θXn , r), where J : Xn → Xn∗ is a duality mapping. For every x ∈ B(θXn , r), there is y ∈ B ∗ (θXn∗ , r) such that y = Jx. At the points y of the sphere S ∗ (θXn∗ , r), there exists z ∈ AJ ∗ y such that the inequality y, z − f ≥ 0 holds because, by Lemma 1.5.10 and Corollary 1.7.8, J −1 = J ∗ , (J ∗ )−1 = J, R(J) = X ∗ and R(J ∗ ) = X. Hence, in view of Theorem 1.1.62, we are able to find an element y0 ∈ B ∗ (θXn∗ , r) such that θXn ∈ R(AJ ∗ y0 − f ) (spaces Xn and Xn∗ can be identified). Since J ∗ : Xn∗ → Xn is a continuous operator, θXn ∈ R(Ax0 − f ), where x0 = J ∗ y0 . A solution of the equation (1.15.5) with an arbitrary accretive operator can be also defined in the following way: Definition 1.15.27 A point x0 ∈ X is said to be a generalized solution of the equation (1.15.5) with accretive operator A if the inequality J(x − x0 ), y − f ≥ 0 ∀y ∈ Ax is satisfied for every x ∈ D(A). Repeating, in fact, the proof of Lemma 1.9.5 we obtain Lemma 1.15.28 Suppose that the operator A : X → 2X with D(A) = X is accretive and duality mappings J and J ∗ are continuous. Then generalized and sw-generalized solutions of the equation (1.15.5) coincide. Under the conditions of this lemma, the operator A has a unique maximal accretive extension A¯ (cf. Corollary 1.9.7). Theorem 1.15.29 Assume that spaces X and X ∗ are uniformly convex, X possesses an approximation, duality mapping J is weak-to-weak continuous, operator A : X → 2X is accretive with a domain D(A) = X and there exists r > 0 such that for every x with x = r there is a y ∈ Ax such that Jx, y − f ≥ 0. Then equation (1.15.5) has at least one generalized solution x ¯ such that ¯ x ≤ r.

102

1


Proof. Let Xn be finite-dimensional subspaces defining the M -property of X. Fixed Xn and consider the equation Pn (Ax − f ) = 0, where Pn : X → Xn is the projector, |Pn | = 1. By Lemma 1.15.24, Jx, Pn (y − f ) = Jx, y − f ≥ 0,

y ∈ Ax,

for all x ∈ Xn with x = r. Then, in view of Lemma 1.15.26, there exists at least one element xn ∈ X such that xn ≤ r and θX ∈ R(Pn (Axn − f )). Repeating the arguments of Theorem 1.10.6 we may verify that the operator Pn A¯ : Xn → 2Xn is a maximal accretive ¯ n such that extension of Pn A : Xn → 2Xn . Hence, there exists an element y n ∈ Ax Pn (yn − f ) = θX . We are going to show that the sequence {y n } is bounded. Since an operator A¯ is locally ¯ n if bounded, there exist numbers a1 > 0 and a2 > 0 such that v n ≤ a2 for all vn ∈ Aq n q ≤ a1 . Suppose that the element q n is defined by the equation J(q n − xn ) = wn − Jxn , where wn = a3 y n −1 Jy n and a3 ≤ r. Since X ∗ is uniformly smooth, the duality mapping J ∗ : X ∗ → X is uniformly continuous on every bounded set (see Section 1.6). In other ∗ (t) for t ≥ 0 such that ω (0) = 0 words, there exists an increasing continuous function ωR R and if φ1 , φ2 ∈ X ∗ , φ1 ∗ ≤ R, φ2 ∗ ≤ R then ∗ (φ1 − φ2 ∗ ). J ∗ φ1 − J ∗ φ2 ≤ ωR

Put R = 2r. Since J ∗ = J −1 , we have ∗ ∗ ∗ q n = J ∗ (wn − Jxn ) + J ∗ Jxn ≤ ωR (wn − Jxn + Jxn ∗ ) = ωR (w n ∗ ) = ωR (a3 ). ∗ Choose a3 so small as ωR (a3 ) ≤ a1 . Clearly such a definition of the sequence {wn } gives the n ¯ n. estimate v ≤ a2 , where v n ∈ Aq By the accretiveness of the operator A¯ we can write

J(xn − q n ), y n − vn = Jxn − wn , y n − v n ≥ 0, from which it follows that wn , y n ≤ wn − Jxn , v n + Jxn , y n . Since Jxn , y n = Jxn , Pn y n = Jxn , f , one gets from (1.15.7) that a1 + r) = a4 . a3 y n ≤ rf + a2 (˜

(1.15.7)


103

Thus, y n ≤ a3 −1 a4 . Next, let z ∈ X and Pn z = zn . Then, on account of the M -property of X, the relation Pn z − z → 0 holds as n → ∞. It is obvious that J(z − xn ), y n − f = J(z − xn ) − J(zn − xn ), yn − f + J(zn − xn ), yn − f . In its turn, J(zn − xn ), yn − f = J(zn − xn ), Pn (y n − f ) = 0. Consequently, J(z − xn ), y n − f = J(z − xn ) − J(zn − xn ), y n − f .

(1.15.8)

Since X is uniformly smooth, the duality mapping J : X → X ∗ is uniformly continuous on every bounded set, that is, there exists an increasing continuous function ωR (t) for t ≥ 0 such that ωR (0) = 0 and if x1 , x2 ∈ X, x1 ≤ R, x2 ≤ R, then Jx1 − Jx2 ∗ ≤ ωR (x1 − x2 ). This implies J(z − xn ) − J(zn − xn )∗ ≤ ωR¯ (z − zn ), ¯ = r + z. Thus, where R lim J(z − xn ) − J(zn − xn )∗ = 0.

n→∞

We proved above that the sequence {y n } is bounded, therefore, by (1.15.8), we obtain lim J(z − xn ), y n − f = 0 ∀z ∈ X.

n→∞

(1.15.9)

Furthermore, let {xk } ⊆ {xn } and xk x ¯ ∈ X. Then the accretiveness of A¯ gives ¯ J(z − xk ), y − y k ≥ 0 ∀y ∈ Az. By (1.15.9) and by weak-to-weak continuity of the duality mapping J, we have in a limit as k → ∞, ¯ J(z − x ¯), y − f ≥ 0 ∀y ∈ Ax, ¯x. Hence, x that is, f ∈ A¯ ¯ is the solution of (1.15.5) in the sense of Definition 1.15.27. Since xn ∈ B(θX , r) for all n > 0 and xk x ¯, we conclude by the weak lower semicontinuity of norm in X that ¯ x ≤ r. The proof is accomplished. Remark 1.15.30 All the conditions of Theorem 1.15.29 are satisfied, for example, in Banach spaces X = lp , p > 1. Remark 1.15.31 If an operator A in Theorem 1.15.29 is strictly accretive, then the corresponding operator equation has a unique solution.

104

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Consider in X the equation Ax + αx = f, α > 0, f ∈ X with a maximal accretive operator A : X → 2X . The following relations are easily verified: Jx, y + αx − f ≥ αx2 − f x − A(θX )x = x(αx − f − A(θX )) ∀y ∈ Ax. Hence, there exists a number r > 0 such that Jx, y + αx − f ≥ 0 ∀y ∈ Ax as x = r. Therefore, if the spaces X and X ∗ and operator J satisfy the conditions of Theorem 1.15.29 and if D(A) = X, then R(A + αI) = X. Hence, a maximal accretive operator A is maccretive (see also Theorem 1.15.21). Theorem 1.15.32 Let A : X → 2X be a coercive and m-accretive operator. Then R(A) = X. Proof. By the definition of m-accretiveness of A, it results for y1 , y2 ∈ X that there exist x1 ∈ X and x2 ∈ X such that y1 ∈ (A + αI)x1 and y2 ∈ (A + αI)x2 . Applying Definition 1.15.4 to A we can write for any η > 0 x1 − x2 + η(y1 − y2 ) = (1 + αη)(x1 − x2 ) + η[(y1 − αx1 ) − (y2 − αx2 )] ≥ Hence, the mapping

(1 + αη)x1 − x2 .

C = I + η(A + αI)

−1

satisfies the Lipschitz condition with the constant L = (1 + αη)−1 < 1. Consequently, the operator C is strongly contractive. Therefore, it has a fixed point xα which is a solution of the equation

x = I + η(A + αI)

−1

x.

It follows from this that yα = −αxα ∈ Axα . Since A is accretive, we have for β > α, αxα − xβ 2 ≤ αJ(xα − xβ ), xα − xβ − J(xα − xβ ), αxα − βxβ = (β − α)J(xα − xβ ), xβ ≤ (β − α)xα − xβ xβ .


105

This yields the inequality αxα − xβ ≤ (β − α)xβ , which implies αxα ≤ βxβ . Thus, the sequence {αxα } is bounded as α → 0. Since Jxα , yα = −αxα 2 and A is coercive, the boundedness of {xα } follows then. Hence, αxα → θX as α → 0, i.e., θX ∈ R(A). Now we choose an arbitrary element f ∈ X and apply the proved assertion to the shift operator A − f. Finally, we obtain θX ∈ R(A) − f, thus, f ∈ R(A) for all f ∈ X. The theorem is proved.

Theorem 1.15.33 If an operator A : X → 2X is m-accretive, a duality mapping J is weak-to-weak continuous and A−1 is locally bounded, then R(A) = X. Proof. Prove that R(A) is a closed and open set at the same time and then the claim follows. Let fn ∈ R(A) and fn → f, n = 1, 2, ... . Then xn ∈ A−1 fn is bounded in X, that is, there exists c > 0 such that xn ≤ c for all n > 0. Therefore, we may consider that xn x ¯ ∈ X. By Lemma 1.15.20, operator A is maximal accretive, then f ∈ A¯ x because of Lemma 1.15.12. Thus, the set R(A) is closed. Next we shall establish that R(A) is open. Let (x, f ) ∈ grA be given. Since A−1 is locally bounded, there exists r > 0 such that the set {x | u ∈ Ax} is bounded in X if r u−f ≤ r. Take g ∈ B(f, ) and show that g ∈ R(A). Since A is m-accretive, the equation 2

Ay + α(y − x) = g has a solution xα , that is, there exists gα ∈ Axα such that gα + α(xα − x) = g.

(1.15.10)

By the accretiveness of A, J(xα − x), g − α(xα − x) − f ≥ 0, from which we have αxα − x ≤ g − f ≤

r . 2

Then (1.15.10) implies the estimate gα − g = αxα − x ≤

r , 2

(1.15.11)

from which one gets gα − f ≤ gα − g + g − f ≤ r. The boundedness of {xα } for sufficiently small α > 0 follows now from the local boundedness ¯ ∈ X as α → 0. Finally, by (1.15.11), we deduce that gα → g. Hence, of A−1 . Then xα x g ∈ R(A) in view of Lemma 1.15.12.

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Theorem 1.15.34 Under the conditions of Theorem 1.15.32, if duality mapping J is weakto-weak continuous, then R(A) = X. Proof. The coercive operator A has bounded inverse. Therefore, the assertion follows from Theorem 1.15.33. Theorem 1.15.35 If an operator A : X → 2X is m-accretive and duality mapping J satisfies the Lipschitz−H¨ older condition Jx − Jy∗ ≤ cx − yγ , c > 0, 0 < γ ≤ 1,

(1.15.12)

then R(A) is a convex set in X.

Proof. Let xα ∈ D(A) be a unique solution of the equation Ax + αx = x0 , x0 ∈ X,

α > 0.

Then there is a element yα ∈ Axα such that yα + αxα = x0 or

J(yα − x0 ) = −αJxα .

(1.15.13)

Let (u, v) ∈ grA. Using (1.15.13) we can write yα − x0 2 = J(yα − x0 ), yα − x0 = J(yα − x0 ), yα − v + J(yα − x0 ), v − x0 = αJxα , v − yα + J(yα − x0 ), v − x0 = αJ(xα − u), v − yα + αJxα − J(xα − u), v − yα + J(yα − x0 ), v − x0 . Taking into account (1.15.12) and accretiveness of A, the previous norm can be evaluated in the following way: yα − x0 2 ≤ cαuγ v − yα + J(yα − x0 ), v − x0

(1.15.14)

which reduce to yα − x0 2 ≤ cαuγ v − yα + yα − x0 v − x0 . We now deduce that the sequence {yα } is bounded, and as a consequence, {J(yα − x0 )} are also bounded. Assume that J(yα − x0 ) z ∈ X ∗ as α → 0 (as a matter of fact, this weak convergence takes place on some subsequence of {J(yα − x0 )} but we do not change its notation). Then by (1.15.14), lim sup yα − x0 2 ≤ z, v − x0 ∀v ∈ R(A). α→0

The rest of the proof follows the pattern of Theorem 1.7.17.


107

Corollary 1.15.36 If the inverse operator A−1 is m-accretive and duality mapping J satisfies the Lipschitz−H¨ older condition, then the set D(A) is convex. Introduce co-variational inequalities with accretive operators. Let A : X → 2X be an maximal accretive operator with a domain D(A) ⊆ X. Consider the inequality J(y − x), Ax − f ≥ 0 ∀y ∈ Ω, x ∈ Ω,

(1.15.15)

where Ω ⊆ D(A) is a closed convex set and f ∈ X. By analogy with monotone variational inequalities, we present the following two definitions: Definition 1.15.37 An element x ∈ Ω is called a solution of the co-variational inequality (1.15.15) if there exists z ∈ Ax such that J(y − x), z − f ≥ 0 ∀y ∈ Ω.

(1.15.16)

Definition 1.15.38 An element x ∈ Ω is called a solution of the co-variational inequality (1.15.15) if J(y − x), u − f ≥ 0 ∀y ∈ Ω, ∀u ∈ Ay. (1.15.17) Lemma 1.15.39 If an element x ∈ Ω is the solution of (1.15.15) defined by the inequality (1.15.16), then it satisfies also the inequality (1.15.17). Proof. Write down the property of accretivness of A, J(y − x), u − z ≥ 0 ∀x, y ∈ Ω, ∀u ∈ Ay,

∀z ∈ Ax.

Using (1.15.16) we obtain (1.15.17).

Lemma 1.15.40 If A : X → X ∗ is a hemicontinuous operator and Ω ⊂ int D(A), then Definitions 1.15.37 and 1.15.38 are equivalent. Proof. Let x be a solution of (1.15.15) in the sense of Definition 1.15.37. Since the set Ω is convex, the element yt = (1 − t)x + ty ∈ Ω for t ∈ [0, 1] and all y ∈ Ω. Then (1.15.17) with y = yt gives J(yt − x), Ayt − f ≥ 0 ∀y ∈ Ω. It results from this that J(y − x), Ayt − f ≥ 0 ∀y ∈ Ω.

(1.15.18)

If t → 0 then yt → x. Now the hemicontinuity of A at x implies: Ayt z¯ ∈ X. In view of (1.15.18), it follows that (1.15.16) holds, that is, x is a solution of (1.15.15) in the sense of Definition 1.15.37. Joining to this result the previous lemma we obtain the necessary assertion.

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1.16


Equations with d-Accretive Operators

Let X be a reflexive and strictly convex Banach space and X ∗ also be strictly convex. In this section we study a class of d-accretive operators. Definition 1.16.1 An operator A : X → 2X with D(A) ⊆ X is said to be d-accretive if Jx − Jy, u − v ≥ 0 ∀x, y ∈ D(A), ∀u ∈ Ax, ∀v ∈ Ay.

(1.16.1)

Present several examples of d-accretive operators. Example 1. If F is a monotone operator from X ∗ to X, then the operator A = F J with D(A) = {x ∈ X | Jx ∈ D(F )} is a d-accretive operator from X to X. Indeed, since F satisfies the condition ϕ1 − ϕ2 , ψ1 − ψ2 ≥ 0 ∀ϕ1 , ϕ2 ∈ D(F ) ⊂ X ∗ , ∀ψ1 ∈ F ϕ1 , ∀ψ2 ∈ F ϕ2 , we can also write JJ ∗ ϕ1 − JJ ∗ ϕ2 , ψ1 − ψ2 ≥ 0 ∀ψ1 ∈ F ϕ1 , ∀ψ2 ∈ F ϕ2 , because JJ ∗ = IX ∗ . Setting J ∗ ϕ1 = x and J ∗ ϕ2 = y, one has Jx − Jy, ψ1 − ψ2 ≥ 0 ∀ψ1 ∈ Ax, ∀ψ2 ∈ Ay. Thus, the claim holds. As a matter of fact, it is also true that if A : X → 2X is d-accretive, then AJ ∗ : X ∗ → 2X is monotone. Example 2. Suppose that the operator T : X → X satisfies the inequality T x − T y ≤

Jx − Jy, x − y Jx − Jy∗

∀x, y ∈ D(T ).

(1.16.2)

Then the operator A = I − T, where I is the identity mapping in X, is d-accretive. Indeed, Jx − Jy, Ax − Ay = Jx − Jy, (I − T )x − (I − T )y = Jx − Jy, x − y − Jx − Jy, T x − T y ≥ Jx − Jy, x − y − T x − T yJx − Jy∗ ≥ 0. Note that (1.16.2) implies T x − T y ≤ x − y ∀x, y ∈ D(T ), i.e., the operator T is nonexpansive. The inverse assertion is not held in general. Observe also that in a Hilbert space the right-hand side of (1.16.2) is x − y.

1.16 Equations with d-Accretive Operators

109

Example 3. Let Ω be a nonempty closed convex subset of X, and consider the functional 1 defined in Section 1.6 by W : X × X → R+ W (x, ξ) = 2−1 (x2 − 2Jx, ξ + ξ2 ).

(1.16.3)

By virtue of the properties of W (x, ξ), for each x ∈ X there is a unique x ˆ ∈ Ω which solves the minimization problem min {W (x, ξ) | ξ ∈ Ω}. Denoting x ˆ by ΠΩ x we define a generalized projection operator ΠΩ : X → Ω ⊆ X. An element x ˆ is called a generalized projection of x onto Ω. We claim that the operator ΠΩ is d-accretive in the space X. Indeed, with the denotations x ˆ1 = ΠΩ x1 and x ˆ2 = ΠΩ x2 , the inequalities ˆ1 ) and W (x2 , η) ≥ W (x2 , x ˆ2 ) W (x1 , ξ) ≥ W (x1 , x are satisfied for all ξ, η ∈ Ω and for all x1 , x2 ∈ X. Assume ξ = x ˆ2 and η = x ˆ1 . Then ˆ2 ) ≥ W (x1 , x ˆ1 ) and W (x2 , x ˆ1 ) ≥ W (x2 , x ˆ2 ). W (x1 , x We now deduce x1 2 − 2Jx1 , x ˆ2 + ˆ x2 2 + x2 2 − 2Jx2 , x ˆ1 + ˆ x1 2 ≥ x1 2 − 2Jx1 , x ˆ1 + ˆ x1 2 + x2 2 − 2Jx2 , x ˆ2 + ˆ x 2 2 . Hence, ˆ1 + Jx2 , x ˆ2 ≥ Jx1 , x ˆ2 + Jx2 , x ˆ1 Jx1 , x and Jx1 − Jx2 , ΠΩ x1 − ΠΩ x2 ≥ 0 ∀x1 , x2 ∈ X. This means that ΠΩ is d-accretive operator. Note that generalized projection operators have important applications in the theory of approximation and optimization. We say that an operator A : X → 2X with a domain D(A) is gauge d-accretive if in Definition 1.16.1 the normalized duality mapping is replaced by the duality mapping with a gauge function. For example, using the duality mapping J p with the gauge function µ(t) = tp−1 , p > 1, we obtain instead of (1.16.1) the inequality J p x − J p y, u − v ≥ 0 ∀x, y ∈ D(A), ∀u ∈ Ax, ∀v ∈ Ay.

(1.16.4)

Example 4. Consider the operator A : Lp (G) → Lp (G) defined by the following equality: Ay = ϕ(x, |y|)y ∀x ∈ G, ∀y ∈ Lp (G),

(1.16.5)

where G is a bounded measurable domain of Rn , p > 1, ϕ(x, s) is a non-negative measurable function with respect to x for all s ≥ 0 and continuous with respect to s for almost all x ∈ G. Suppose that there exists a constant C > 0 such that ϕ(x, s) ≤ C ∀x ∈ G, ∀s ∈ [0, +∞).

(1.16.6)

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1


Let J p : Lp (G) → Lq (G) be a duality mapping with the gauge function µ(t) = tp−1 , p−1 + q −1 = 1. It is known (see Section 1.5) that in this case J p y = |y|p−2 y. One can verify that J p y1 − J p y2 , Ay1 − Ay2 ≥

g

ϕ(x, |y1 |)|y1 | − ϕ(x, |y2 |)|y2 | (|y1 |p−1 − |y2 |p−1 ) dx.

(1.16.7)

Let the function ϕ(x, s)s be non-decreasing with respect to s ≥ 0 for each fixed x ∈ G. Then d-accretiveness of the operator A defined by (1.16.5) arises from (1.16.7). Emphasize that the inclusion Ay ∈ Lp (G) for y ∈ Lp (G) guarantees the assumption (1.16.6). Definition 1.16.2 Let W(x, ξ) be defined as W(x, ξ) = 2−1 (x2 − 2Jξ, x + ξ2 ). We say that an operator A : X → 2X with domain D(A) is d-accretive if W(x1 , x2 ) ≤ W(x1 + λ(y1 − y2 ), x2 )

(1.16.8)

for all x1 , x2 ∈ D(A), for all y1 ∈ Ax1 , y2 ∈ Ax2 and λ > 0. Theorem 1.16.3 Definitions 1.16.1 and 1.16.2 are equivalent. Proof. Let ξ be fixed. It is easy to see that grad W(x, ξ) = Jx − Jξ, therefore, it is a monotone operator. Consequently, W(x, ξ) is a convex functional for all x ∈ X. Then W(x1 , x2 ) ≥ W(x1 + λ(y1 − y2 ), x2 ) − λJ(x1 + λ(y1 − y2 )) − Jx2 , y1 − y2 ). This inequality and (1.16.8) imply J(x1 + λ(y1 − y2 )) − Jx2 , y1 − y2 ) ≥ 0. Setting λ → 0 and using the fact that J is hemicontinuous, we prove d-accretivness of A in the sense of Definition 1.16.1. Conversely, let (1.16.1) hold, x1 , x2 ∈ D(A) and λ > 0. Then (1.16.8) follows from the convexity inequality for W(x, ξ) again, namely, W(x1 + λ(y1 − y2 ), x2 ) ≥ W(x1 , x2 ) + λJx1 − Jx2 , y1 − y2 . The result holds in view of the monotonicity of J. Definition 1.16.4 A d-accretive operator A : X → 2X is said to be maximal d-accretive if its graph is not the right part of the graph of any other d-accretive operator B : X → 2X . Lemma 1.16.5 The value set of the maximal d-accretive operator at any point of its domain is convex and closed. Proof: It follows from Definitions 1.16.1 and 1.16.4.


111

Theorem 1.16.6 Let A : X → X be a demicontinuous d-accretive operator with domain D(A) = X and let the normalized duality mapping J ∗ be continuous. Then A is a maximal d-accretive operator. Proof. We shall show that the inequality Jx − Jy, Ax − f ≥ 0 ∀x ∈ X

(1.16.9)

implies Ay = f. Since D(A) = X, we may put in (1.16.9) x = xt = J ∗ (Jy + tJz) for any z ∈ X and t > 0. Then Jz, Axt − f ≥ 0. Let t → 0. Using the demicontinuity property of A and continuity of J ∗ we obtain in a limit Jz, Ay − f ≥ 0 ∀z ∈ X. Since R(J) = X ∗ , the lemma is proved. Theorem 1.16.7 Let A : X → 2X be a maximal d-accretive operator. Let xn ∈ D(A), yn ∈ Axn . Suppose that either xn → x, yn y and the duality mapping J is continuous, or xn x, yn → y and J is weak-to-weak continuous. Then x ∈ D(A) and y ∈ Ax. Definition 1.16.8 A d-accretive operator A : X → 2X is said to be m-d-accretive if R(A + αI) = X ∀α > 0, where I is the identity map in X. Lemma 1.16.9 If an operator A is m-d-accretive, then it is maximal d-accretive. Proof: It is produced similarly to Lemma 1.15.20.

Theorem 1.16.10 Let A : X → 2X be a d-accretive operator, the normalized duality mappings J : X → X ∗ and J ∗ : X ∗ → X be continuous. Then A is locally bounded at any point x0 ∈ int D(A). Proof. Assume the contradiction: x0 ∈ int D(A), xn ∈ D(A), n = 1, 2, ... , xn → x0 , 1/2 but yn → ∞, where yn ∈ Axn . Introduce zn = Jxn − Jx0 , tn = Jxn − Jx0 ∗ and ∗ construct the elements wn = J (Jx0 + tn Jw) with w ∈ X. It is clear that zn = t2n . We conclude from the continuity of J that tn → 0 and zn → θX ∗ as n → ∞. Since J ∗ J = IX and J ∗ is continuous, we have that wn → x0 . Consequently, wn ∈ D(A) for sufficiently small tn ≤ σ. Let v = J ∗ (Jx0 + σJw), un ∈ Awn and f ∈ Av. Then d-accretivness of A implies (tn − σ)Jw, un − f ≥ 0, and we obtain for tn ≤ σ the inequality z, un − f ≤ 0, z = Jw.

(1.16.10)

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By R(J) = X ∗ , (1.16.10) holds for all z ∈ X ∗ . Therefore, {un } is bounded according to the Banach−Steinhaus theorem. Let un ≤ C. The property of d-accretivness of A allows us to write zn − tn z, yn − un ≥ 0. Then we have

zn ∗ 1 1 zn ∗ + z∗ , yn + C z, yn ≤ zn , yn − zn − tn z, un ≤ tn tn tn tn from which the following inequality appears: lim sup n→∞

z, yn < ∞ ∀z ∈ X ∗ , 1 + yn τn

where τn = t−1 n zn ∗ . Due to the Banach−Steinhaus theorem again, there exists a constant K > 0 such that yn ≤ K(1 + yn τn ). Since τn → 0, the estimate Kτn ≤ 2−1 is satisfied for sufficiently large n. Then yn ≤ 2K. Thus, we have established the boundedness of {yn } which contradicts our assumption above. The proof is accomplished. Theorem 1.16.11 Let J : X → X ∗ be a continuous and weak-to-weak continuous mapping, J ∗ : X ∗ → X be continuous, A : X → X be a d-accretive demicontinuous operator with D(A) = X and f ∈ X. Assume that the space X possesses an approximation and there exists a constant r > 0 such that Jx, Ax − f ≥ 0 as x = r. Then the equation (1.15.5) has at least one classical solution x ¯, it being known that ¯ x ≤ r. Proof. As in the proof of Lemma 1.15.24, Jx, Pn (Ax − f ) = Jx, Ax − f ≥ 0 for all xn ∈ X with xn = r. From this, by Lemma 1.15.26 there exists an element xn ∈ X such that xn ≤ r and Pn (Axn − f ) = θX . Show that the sequence {Axn } is bounded. By virtue of Theorem 1.16.10, the operator A is locally bounded, therefore, there exist constants a > 0 and K > 0 such that if yn ∈ X, yn ≤ a then Ayn ≤ K. Take a ˜ ≤ min{a, r} and put yn = a Ãxn −1 Axn . It is clear that yn = a ˜ ≤ a and Ayn ≤ K for these yn . Since A is d-accretive, we write down Jxn − Jyn , Axn − Ayn ≥ 0 or in the equivalent form Jyn , Axn ≤ Jyn − Jxn , Ayn + Jxn , Axn . It is easy to see that Jxn , Axn = Jxn , Axn − f + Jxn , f

(1.16.11)


113

= Jxn , Pn (Axn − f ) + Jxn , f = Jxn , f . Ãxn . Then (1.16.11) gives Since the operator J is homogeneous, we have Jyn , Axn = a a Ãxn ≤ (yn + xn )Ayn + xn f ≤ (˜ a + r)K + rf . We conclude from this that the sequence {Axn } is bounded. Since {xn } is bounded, there ¯ ∈ X as exists a subsequence of {xn } (we do not change its notation) such that xn x n → ∞. By hypothesis, the space X possesses an approximation. This fact allows us to construct for each element z ∈ X the sequence {zn } such that Pn z = zn ∈ Xn and zn → z. Write down again the property of the d-accretivness of A, Jz − Jxn , Az − Axn ≥ 0. The latter relation is identical to Jz − Jzn , Az − Axn + Jzn − Jxn , Az − f + Jzn − Jxn , f − Axn ≥ 0.

(1.16.12)

The first term in the left part of (1.16.12) tends to zero as n → ∞ because {Axn } is bounded, while Jzn → Jz by the continuity of duality mapping J. The last term equals zero due to the equalities Jzn − Jxn , f − Axn = Jzn − Jxn , Pn (f − Axn ) and Pn (Axn − f ) = θX . Using weak-to-weak continuity of J and setting n → ∞, we have in a limit Jz − J x ¯, Az − f ≥ 0. ¯ + tJv), where v is an arbitrary fixed element of Put in the last inequality z = zt = J ∗ (J x X and t > 0. Then Jv, Azt − f ≥ 0. (1.16.13) We recall that the operator A is demicontinuous and J ∗ is continuous. Then (1.16.13) leads as t → 0 to the inequality Jv, A¯ x − f ≥ 0. (1.16.14) Since R(J) = X ∗ , (1.16.14) implies A¯ x = f. In addition, the estimate xn ≤ r and weak lower semicontinuity of the norm in a Banach space guarantee the inequality ¯ x ≤ r. This completes the proof. Remark 1.16.12 If we omit the demicontinuity property of A in the hypotheses of Theorem 1.16.11 and understand a solution of the equation (1.15.5) as the element x0 ∈ X such that Jx − Jx0 , y − f ≥ 0 ∀y ∈ Ax, ∀x ∈ X, then one can establish solvability of (1.15.5) similarly to Theorem 1.15.29.

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Theorem 1.16.13 Let A : X → 2X be a coercive and m-d-accretive operator. Then R(A) = X. Proof. The condition of m-d-accretiveness of the operator A guarantees unique solvability of the equation Ax+αx = 0 for all α > 0. Consequently, there exist elements xα ∈ X and yα ∈ Axα such that yα + αxα = θX . (1.16.15) We have

Jxα , yα = −αxα 2 ∀α > 0.

Since A is coercive, the last equality implies boundedness of {xα } as α → 0. Therefore, αxα → θX . By (1.16.15), yα → θX , that is, θX ∈ R(A). The rest of the proof follows the pattern of Theorem 1.15.32. Consider the co-variational inequality with the maximal d-accretive operator A :

Jy − Jx, Ax − f ≥ 0 ∀y ∈ Ω, x ∈ Ω,

(1.16.16)

where f ∈ X, Ω ⊆ D(A) and Ω∗ = JΩ is a closed and convex set in X ∗ . We also present two definitions of their solutions. Definition 1.16.14 An element x ∈ Ω is called the solution of the co-variational inequality (1.16.16) if there is z ∈ Ax such that Jy − Jx, z − f ≥ 0 ∀y ∈ Ω.

(1.16.17)

Definition 1.16.15 An element x ∈ Ω is called the solution of the co-variational inequality (1.16.16) if Jy − Jx, u − f ≥ 0 ∀y ∈ Ω, ∀u ∈ Ay. (1.16.18) Lemma 1.16.16 If x ∈ Ω is a solution of (1.16.16) defined by the inequality (1.16.17), then it also satisfies the inequality (1.16.18). Proof. Write the d-accretivness property of A, Jy − Jx, u − z ≥ 0 ∀x, y ∈ Ω, ∀u ∈ Ay, ∀z ∈ Ax. In view of (1.16.17), we obtain (1.16.18). Lemma 1.16.17 If A : X → X is demicontinuous, Ω ⊂ int D(A) and duality mapping J ∗ is continuous, then Definitions 1.16.14 and 1.16.15 are equivalent. Proof. Let x be a solution in the sense of (1.16.18) and choose any y ∈ Ω. Since the set JΩ is convex, the element (1 − t)Jx + tJy ∈ JΩ for all t ∈ [0, 1]. Then it is obvious that the element yt = J ∗ ((1 − t)Jx + tJy) ∈ Ω. By (1.16.18) with yt in place of y and Ayt in place of u, one gets Jyt − Jx, Ayt − f ≥ 0,


115

that is, Jy − Jx, Ayt − f ≥ 0 ∀y ∈ Ω.

(1.16.19)

Letting t → 0, we obtain yt → x. By virtue of the demicontinuity of A at a point x and continuity of J ∗ at a point Jx, it is possible to assert that Ayt Ax ∈ X. Then, by (1.16.19), it follows that (1.16.17) holds with z = Ax, that is, x is a solution of (1.16.16) in the sense of Definition 1.16.14. Taking into account the previous lemma, we conclude that the proof is complete.

Bibliographical Notes and Remarks The definitions and results of Section 1.1 are standard and can be found in most textbook on functional analysis. We recommend here Dunford and Schwartz [75], Hille and Fillips [95], Kantorovich and Akilov [104], Kolmogorov and Fomin [117], Liusternik and Sobolev [141], Nirenberg [159], Riesz and Sz.-Nagy [174], Rudin [182], Yosida [235]. The theory of linear topological spaces is well covered by Schaefer [205]. The methods of Hilbert, Banach and normed spaces are treated by Maurin [145], Edwards [77] and Day [69]. The necessary material on nonlinear and convex analysis and optimization can be found in Cea [64], Ekeland and Temam [79], Holmes [96], Vainberg [221]. Interesting facts and observations are also contained in [46, 83, 113, 162, 228]. Observe that the main results in the present book are stated in uniformly convex and/or uniformly smooth Banach spaces. Recall that spaces of number sequences lp , Lebesgue p (G) with 1 < p < ∞, m > 0 and most of Orlicz spaces with spaces Lp (G), Sobolev spaces Wm the Luxemburg norm are uniformly convex and uniformly smooth [119, 127, 141, 210, 237]. Presenting the geometric characteristics of Banach spaces and their properties we follow Diestel [71], Figiel [81] and Lindenstrauss and Tzafriri [127]. The estimates of modulus of convexity and smoothness of Banach spaces have been obtained by Hanner [94], Alber and Notik [24, 25, 26] (see also [16, 127, 233, 236]). The upper and lower bounds of the constant L in (1.6.4) and (1.6.23) are presented in [25, 26, 81, 207]. For the proofs of the Hahn−Banach Theorem 1.1.1 and Banach−Steinhaus Theorem 1.1.3 we refer to [117, 141]. The strong separation Theorem 1.1.10 and the generalized Weierstrass Theorem 1.1.14 are stated, for instance, in [96, 224, 225, 238]. Theorem 1.1.8 has been established in [146] and Theorem 1.1.9 in [173]. Theorems 1.1.21 and 1.1.23 are proved in [221]. Theorem 1.1.62 appeared in [1]. A reader can look through differentiability problems of functionals and operators in [71, 221]. In particular, Theorem 1.1.34 is proved in [45, 112, 209] and Theorem 1.3.9 in [148]. Zorn’s Lemma 1.1.61 can be found in [235]. As regards the imbedding theorem, we are mostly concerned with [104, 210]. The concept of a monotone operator has been introduced by Kachurovskii in [102]. It plays a very important role in the theory of elliptic and parabolic differential equations and in optimization theory [111, 128, 85]. The properties of monotone operators are well described in [57, 83, 103, 162, 166, 221, 237]. Lemma 1.3.14 is contained in [85]. Local boundedness of a monotone operator at interior points of its domain has been proved by Rockafellar in [177]. Kato has shown in [106] that at such points a monotone hemicontinuous

116

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mapping is demicontinuous. The examples of monotone mappings are given in [49, 57, 76, 83, 113, 120, 128, 134, 135, 142, 162, 221, 224]. Maximal monotone operators first appeared in [148]. Their properties are studied in [48, 49, 57, 83, 162]. Necessary and sufficient conditions for a linear operator to be maximal monotone are provided, for instance, in [162]. The conditions of maximal monotonicity of the sum of monotone operators were obtained in [179]. Theorems 1.7.13, 1.7.15 and 1.7.19 are due to Minty and Rockafellar [150, 175, 177, 178]. Theorem 1.7.17 was proved in [51]. A duality mapping is one of deepest subjects of Banach spaces and theory of monotone operators. Recall that the normalized duality mapping has been introduced and investigated by Vainberg in [219] and after that by many other authors (see, for instance, [38, 57, 83, 113, 128, 162, 236]). The idea of duality mapping with a gauge function belongs to Browder. The properties of duality mappings with a gauge function were established in [160, 128, 233]. In particular, Lemma 1.5.10 has been proved in [128] and [233]. The analytical representations of duality mappings in the Lebesgue, Sobolev and Orlicz spaces can be found in [24, 128, 221, 236]. Theorems 1.6.1 and 1.6.4 were proved in [13, 25, 28]. The properties of the Lyapunov functional W (x, y) are also studied in [13]. The important Remark 1.5.12 can be seen in [221]. Duality mappings in non-reflexive spaces are studied in [89]. Using the concept of a duality mapping Kato introduced in [109] accretive operators which play a significant role in the fixed point theory and theory of integral equations and evolution differential equations [57, 108]. Note that Definition 1.3.6 arises from Definition 1.15.4. The general theory and examples of accretive mappings are described in [57, 65, 67, 110, 221, 234, 165]. Theorem 1.15.29 was proved in [191]. The Debrunner−Flor lemma appeared in [70]. Theorem 1.4.6 was proved by many authors. Our proof follows [89]. For Lemma 1.7.11 and Corollary 1.7.12 we refer to [162]. Lemma 1.5.14 has been established in [12]. Its accretive version was earlier proved in [165]. The main results of Sections 1.9 and 1.10 belong to Ryazantseva [185, 201]. Definition 1.9.10 has been introduced in [149]. Browder presented in [52, 53] examples of semimonotone mappings. The sufficient solvability conditions of equations with a single-valued hemicontinuous operator are given in [52, 53, 221]. The existence theorems for variational inequalities can be found, for instance, in [54, 100, 113, 162, 128]. Equivalence of their solutions was studied by Minty [149]. Theorem 1.12.2 was proved in [203]. Solvability of variational inequalities with semimonotone bounded operators having regular discontinuous points has been established in [164]. Section 1.13 is devoted to variational inequalities with pseudomonotone operators [47] and contains the results of [58]. We emphasize that pseudomonotone operators are nonmonotone, in general. There are many different definitions of pseudomonotone operators. Following [58], we present one of them. The properties of pseudomonotone operators are described in [58, 128, 162]. Variational inequalities with quasipotential maps were investigated in [163]. In addition, one can point out the papers [2, 72, 86] in which variational inequations with non-monotone mappings are also regarded. The concept of the d-accretive operator is introduced in [29]. The results of Section 1.16 correspond to [30]. As regards the projection operator ΠΩ , we refer the reader to [13, 16].

Chapter 2

REGULARIZATION OF OPERATOR EQUATIONS 2.1

Equations with Monotone Operators in Hilbert Spaces

1. Let H be a Hilbert space, A : H → H be a hemicontinuous monotone operator, D(A) = H and f ∈ H. We study the operator equation Ax = f

(2.1.1)

assuming that it has a solution (in the classical sense). Denote by N its solution set, by x∗ any point of N and by x ¯∗ a point in N with the minimal norm. Note that due to Theorem 1.4.6, operator A is maximal monotone. Then N is a closed and convex set and x ¯∗ is unique in N. As it has been already mentioned, the problem (2.1.1) is ill-posed, in general. Therefore, strong convergence and stability of approximate solutions can be proved only by applying some regularization procedure. In any method of finding x∗ ∈ N, the main aim is to establish a continuous dependence of approximate solutions on data perturbations. In connection with this, we assume that right-hand side and operator in (2.1.1) are given approximately, namely, instead of f and A, we have sequences {f δ } and {Ah }, where hemicontinuous monotone operators Ah : H → H with D(Ah ) = D(A) for all h > 0, and f δ ∈ H. We define the proximity between operators A and Ah by means of the following inequality: Ax − Ah x ≤ hg(x),

(2.1.2)

where g(t) is a non-negative continuous function for all t ≥ 0. It is clear that if g(t) ≡ 1 then (2.1.2) characterizes the uniform proximity of operators Ah and A. As regards {f δ }, we always assume that f − f δ ≤ δ, δ > 0. (2.1.3) Thus, instead of (2.1.1), the following equation is considered: Ah x = f δ , 117

(2.1.4)

118

2

REGULARIZATION OF OPERATOR EQUATIONS

which does not necessarily have a solution. Our goal in this section is to construct, by use of approximate data {Ah , f δ }, a sequence {xγ }, γ = (δ, h), which strongly converges to x ¯∗ ∈ N as γ → 0. Consider the regularization algorithm for (2.1.4) as follows: Ah x + αx = f δ , α > 0.

(2.1.5)

Denote T = A + αI, where I : H → H is the identity operator. Obviously, T is monotone as a sum of two monotone operators. It is coercive because (T x, x) = (Ax − A(θH ), x − θH ) + α(x, x) + (A(θH ), x) ≥ αx2 − A(θH )x, and then lim

x→∞

(T x, x) ≥ lim αx − A(θH ) = ∞. x x→∞

By the Minty−Browder theorem, the equation (2.1.5) has a classical solution xγα , that is, Ah xγα + αxγα = f δ .

(2.1.6)

Since (T x − T y, x − y) = (Ax − Ay, x − y) + αx − y2 ≥ αx − y2 , T is strongly monotone for all α > 0. Therefore, xγα is the unique solution. According to Definition 5, the solution xγα satisfying (2.1.5) is called the regularized solution of the operator equation (2.1.4). Remark 2.1.1 Note that {xγα } is often called the solution net. However, we prefer in the sequel the more usual term “the solution sequence”. Theorem 2.1.2 Let (2.1.2) and (2.1.3) hold. A sequence {xγα } generated by (2.1.6) is uniformly bounded and strongly converges to the minimal norm solution x ¯∗ if δ+h → 0 as α → 0. α

(2.1.7)

Proof. It is clear that (2.1.7) implies δ → 0 and h → 0. Let x∗ ∈ N. Then it follows from (2.1.1) that Ax∗ = f and the equation Ah xγα − f + α(xγα − x∗ ) = f δ − f − αx∗

(2.1.8)

is equivalent to (2.1.6). The scalar product of (2.1.8) and the difference xγα − x∗ gives (Ah xγα − f, xγα − x∗ ) + αxγα − x∗ 2 = (f δ − f, xγα − x∗ ) + α(x∗ , x∗ − xγα ).

(2.1.9)

In view of the monotonicity of Ah , (Ah xγα − Ah x∗ , xγα − x∗ ) ≥ 0,

(2.1.10)

2.1 Equations with Monotone Operators in Hilbert Spaces and we have

αxγα − x∗ ≤ f δ − f + Ah x∗ − Ax∗ + αx∗ .

119

(2.1.11)

By the hypotheses (2.1.2) and (2.1.3), xγα − x∗ ≤

δ h + g(x∗ ) + x∗ . α α

Now we conclude from this inequality and from (2.1.7) that the sequence {xγα } is bounded. Hence, according to (2.1.6), Ah xγα → f as α → 0, and there exists a subsequence {xξβ }, ˜ ∈ H. With where β ⊆ α and ξ = (δ , h ) ⊆ γ, which weakly converges to some element x this, also δ + h → 0 as α → 0. β Show that x ˜=x ¯∗ . First of all, establish the inclusion x ˜ ∈ N. Write down the mono tonicity property of Ah for an arbitrary x ∈ H :

(Ah x − Ah xξβ , x − xξβ ) ≥ 0.

(2.1.12)

Setting α → 0, and as consequence β → 0 and ξ → 0, we obtain from (2.1.12) and (2.1.2) the limit inequality (Ax − f, x − x ˜) ≥ 0 ∀x ∈ H. Since A is a hemicontinuous monotone operator, this means that x ˜ ∈ N (see Theorem 1.9.1). By (2.1.9), we further have (x∗ , x∗ − xξβ ) +

If β → 0 then

h δ ξ ξ xβ − x∗ + g(x∗ )xβ − x∗ ≥ 0. β β

(x∗ , x∗ − x ˜) ≥ 0 ∀x∗ ∈ N,

(2.1.13)

h δ → 0 as β → 0. Since N is a convex → 0 and β β ∗ ∗ x + (1 − t)x ∈ N for all x ∈ N and t ∈ [0, 1]. Substitute xt into (2.1.13) in place set, xt = t˜ of x∗ and use the obvious fact that 1 − t ≥ 0. Then we obtain

because the sequence {xξβ } is bounded,

(xt , x∗ − x ˜) ≥ 0 ∀x∗ ∈ N. If t → 1, we have that is,

˜) ≥ 0 ∀x∗ ∈ N, (˜ x, x ∗ − x ˜ x ≤ x∗ ∀x∗ ∈ N.

The last is equivalent to the relation ˜ x = min {x∗ | x∗ ∈ N }.

120

2


¯∗ . Finally, Thus, x ˜=x ¯∗ . This means that the whole sequence {xγα } converges weakly to x (2.1.9) leads to the inequality xγα − x ¯ ∗ 2 ≤

h δ γ ¯∗ + g(¯ ¯∗ + (¯ x∗ , x ¯∗ − xγα ). x∗ )xγα − x x − x α α α

(2.1.14)

Now the conclusion of the theorem follows from (2.1.7) and the weak convergence of {xγα } to x ¯∗ . The theorem is proved. 2. In comparison with (2.1.5), we study the more general regularized equation Ah x + αSx = f δ , α > 0,

(2.1.15)

where S : H → H is a strongly monotone bounded hemicontinuous operator with domain D(S) = H. By the obvious inequality (Ah x + αSx, x) ≥ αcx2 − Ah (θH )x − αS(θH )x, where c is a constant of the strong monotonicity of S, it follows that Ah + αS is coercive. Therefore, by Theorem 1.7.5, the equation (2.1.15) is solvable. It is uniquely solvable because S is strongly monotone. We are able to prove that the solution sequence {xγα } generated by (2.1.15) converges in the norm of H to the element x ¯ ∈ N satisfying the inequality (Sx∗ , x∗ − x ¯) ≥ 0 ∀x∗ ∈ N. (2.1.16) Let S be a potential operator, i.e., there exists a functional ϕ(x) such that Sx = grad ϕ(x) for all x ∈ H. Then Theorem 1.11.14 and (2.1.16) imply the following result: ϕ(¯ x) = min {ϕ(x∗ ) | x∗ ∈ N }. The properties of H, S and N guarantee uniqueness of x ¯. 3. Assume now that A : H → 2H and Ah : H → 2H are maximal monotone (possibly, multiple-valued) operators and D(A) = D(Ah ) ⊆ H. We recall that if A and Ah are multiplevalued, then their value sets at any point x ∈ D(A), which we denote, respectively, by Ax and Ah x, are convex and closed (see Theorem 1.4.9). In this case, the proximity between the sets Ax and Ah x is defined by means of the Hausdorff distance as HH (Ax, Ah x) ≤ hg(x),

(2.1.17)

where g(t) is a continuous non-negative function for all t ≥ 0. Theorem 2.1.3 Let (2.1.3), (2.1.7) and (2.1.17) hold. Then the sequence {xγα } generated by (2.1.5) is uniformly bounded and strongly converges to the minimal norm solution x ¯∗ of the equation (2.1.1). Proof. Recall that solutions of (2.1.1) and (2.1.6) are understood now in the sense of inclusions. By Theorem 1.7.4, the equation (2.1.5) has a solution, i.e., there exists xγα such that f δ ∈ Ah xγα + αxγα .

2.1 Equations with Monotone Operators in Hilbert Spaces

121

Hence, there exists yαγ ∈ Ah xγα such that yαγ + αxγα = f δ .

(2.1.18)

Since Ah + αI is strictly monotone, the solution xγα is unique for all fixed α > 0 and γ > 0. Let x∗ ∈ N. Then f ∈ Ax∗ . It results from (2.1.17) that there exist y∗h ∈ Ah x∗ such that y∗h − f ≤ hg(x∗ ). Using (2.1.18) and the monotonicity condition of Ah we can estimate the following scalar product: α(xγα , xγα − x∗ ) = (f δ − yαγ , xγα − x∗ ) = (f δ − f, xγα − x∗ ) − (yαγ − f, xγα − x∗ ) = (f δ − f, xγα − x∗ ) − (yαγ − y∗h , xγα − x∗ ) + (y∗h − f, xγα − x∗ ) ≤ (f δ − f, xγα − x∗ ) − (y∗h − f, xγα − x∗ ), Therefore, (xγα , xγα − x∗ ) ≤

δ

+

α from which we have the quadratic inequality

xγα 2 − x∗ +

y∗h ∈ Ah x∗ .

h g(x∗ ) xγα − x∗ , α

δ h h δ + g(x∗ ) x∗ ≤ 0. + g(x∗ ) xγα − α α α α

(2.1.19)

(2.1.20)

Without lost of generality, one can assume by (2.1.7) that there exists a constant C > 0 such that h δ + g(x∗ ) ≤ C. α α Then (2.1.20) yields the estimate

xγα ≤ C1 ,

where C1 = x∗ + 2C.

(2.1.21)

This implies existence of a subsequence (which for simplicity is denoted as before by {xγα }) such that xγα x ˜ as α → 0. Show that x ˜ ∈ N. Let x ∈ D(A). Since the operators Ah are monotone, we have for arbitrary fixed x ∈ D(A), y h ∈ Ah x and yαγ ∈ Ah xγα : (y h − yαγ , x − xγα ) ≥ 0. Then, by (2.1.18), for all y ∈ Ax and f ∈ Ax∗ we deduce (yh − yαγ , x − xγα ) = (y h − f δ + αxγα , x − xγα ) = (y − f, x − xγα ) + (y h − y, x − xγα ) + (f − f δ , x − xγα ) + α(xγα , x − xγα ) ≥ 0. This leads to the inequality (y − f, x − xγα ) ≥ −(hg(x) + δ + αxγα )x − xγα .

122

2


Consequently, ˜) ≥ 0 ∀x ∈ D(A), lim (y − f, x − xγα ) = (y − f, x − x

α→0

∀y ∈ Ax.

According to Lemma 1.11.6, this imposes that x ˜ ∈ N. It further follows from (2.1.19) that (x∗ , x∗ − xγα ) ≥ −

δ

α

+

h g(x∗ ) xγα − x∗ . α

Setting α → 0, we come to the inequality (x∗ , x∗ − x ˜) ≥ 0 ∀x∗ ∈ N, ¯∗ is guaranteed by convexity which means that x ˜=x ¯∗ . At the same time, the uniqueness of x of N and properties of H. Thus, the whole sequence {xγα } weakly converges to x ¯∗ . Finally, in view of (2.1.19) and (2.1.7), we obtain (xγα , xγα − x∗ ) → 0 ∀x∗ ∈ N because the sequence {xγα } is bounded. Then due to the weak convergence of xγα to x ¯∗ ∈ N, one gets x∗ , x ¯∗ − xγα ) + (xγα , xγα − x ¯∗ ) → 0 ¯ x∗ − xγα 2 = (¯ as α → 0, that is, limα→0 xγα = x ¯∗ . The proof of Theorem 2.1.3 can be also obtained by the scheme which has been earlier applied in the linear case. Namely, introduce the regularized operator equation with unperturbed data Ax + αx = f, (2.1.22) and denote its solution by x0α . Similarly to Theorem 2.1.3, one can prove that the sequence ¯∗ as α → 0, and {x0α } converges strongly to x x0α ≤ ¯ x∗

(2.1.23)

because in (2.1.21) C = 0. Since Ah is a monotone operator, by virtue of the equality (yαγ − yα0 , xγα − x0α ) + αxγα − x0α 2 = (f δ − f, xγα − x0α ), where yαγ ∈ Ah xγα and yα0 ∈ Ax0α , we obtain xγα − x0α ≤

h δ + g(x0α ) → 0 as α → 0. α α

The conclusion of Theorem 2.1.3 follows now from the inequality ¯∗ ≤ xγα − x0α + x0α − x ¯∗ . xγα − x

(2.1.24)

2.2 Equations with Monotone Operators in Banach Spaces

123

Theorem 2.1.4 Let (2.1.17), (2.1.3) and (2.1.7) hold. If the sequence {xγα } generated by (2.1.5) converges (even weakly) to some element x∗ ∈ H, then x∗ is a solution of the equation (2.1.1). Proof. Since {xγα } converges to x∗ , it is bounded. Then, by (2.1.18), yαγ → f as α → 0, where yαγ ∈ Ah xγα . Write the monotonicity condition for Ah as (y h − yαγ , x − xγα ) ≥ 0 ∀x ∈ D(A), ∀y h ∈ Ah x. From that, after passing to the limit as α → 0, we obtain (y − f, x − x∗ ) ≥ 0 ∀x ∈ D(A), ∀y ∈ Ax. Since the operator A is maximal monotone, the latter inequality means that f ∈ Ax∗ . Combining Theorems 2.1.3 and 2.1.4 implies Theorem 2.1.5 Suppose that the conditions (2.1.17), (2.1.3) and (2.1.7) are satisfied. Then the sequence {xγα } strongly converges to some element x ¯ ∈ H if and only if there exists a solution of the equation (2.1.1). The next two statements immediately follow from Theorem 2.1.3. Theorem 2.1.6 Let equation (2.1.1) have a unique solution x0 ∈ H and let there exist a constant C > 0 such that δ+h ≤C α as α → 0. Then xγα x0 as α → 0.

Theorem 2.1.7 If h, δ, α → 0, then 1) yαγ − f δ → 0, yαγ ∈ Ah xγα and 2) ˜ yαγ − f → 0, y˜αγ ∈ Axγα , where yαγ satisfies (2.1.18) and y˜αγ such that yαγ − y˜αγ ≤ hg(xγα ).

2.2

Equations with Monotone Operators in Banach Spaces

1. Let X be an E-space with strictly convex dual space X ∗ . Consider the equation (2.1.1) ∗ with the maximal monotone operator A : X → 2X . As in the case of Hilbert spaces, assume that (2.1.1) has a nonempty solution set N and denote by x ¯∗ ∈ N the minimal norm solution. It is unique because the set N is convex and closed and the Banach space X is reflexive and strictly convex. Suppose that, instead of A and f, the sequences {f δ } and

124

2

REGULARIZATION OF OPERATOR EQUATIONS ∗

{Ah } are given, where maximal monotone operators Ah : X → 2X have D(Ah ) = D(A) for all h > 0, and f δ ∈ X ∗ for all δ > 0. Thus, in reality, we study the equation (2.1.4) with the following proximity conditions: HX ∗ (Ax, Ah x) ≤ hg(x),

(2.2.1)

where g(t) is a continuous non-negative function for all t ≥ 0, HX ∗ (G1 , G2 ) stands the Hausdorff distance between the sets G1 and G2 in X ∗ and f − f δ ∗ ≤ δ, δ > 0.

(2.2.2)

If A and Ah are single-valued then the condition (2.2.1) is Ax − Ah x∗ ≤ hg(x).

(2.2.3)

Under these circumstances, we solve the regularized operator equation Ah x + αJx = f δ ,

(2.2.4)

where J : X → X ∗ is the normalized duality mapping. In our conditions, J is a demicontinuous and single-valued operator, D(J) = X and R(J) = X ∗ . Then Theorem 1.7.4 guarantees solvability of the equation (2.2.4) in the sense of inclusion. Let xγα be a solution of (2.2.4). It is unique because the operator Ah + αJ is strictly monotone. It is clear that there exists an element yαγ ∈ Ah xγα such that yαγ + αJxγα = f δ . According to Definition 5, the solution xγα satisfying (2.2.4) is also called the regularized solution of the operator equation Ah x = f δ . Theorem 2.2.1 Assume that (2.2.1), (2.2.2) and (2.1.7) hold. Let A and Ah be maximal monotone operators, X be an E-space, X ∗ be a strictly convex space. Then the sequence ¯∗ ∈ N as {xγα } of solutions of the equation (2.2.4) converges strongly in X to the element x α → 0. Proof. Similarly to Theorem 2.1.3, we obtain for an arbitrary x∗ ∈ N and f ∈ Ax∗ the equality yαγ − f, xγα − x∗ + αJxγα − Jx∗ , xγα − x∗ = f δ − f, xγα − x∗ + αJx∗ , x∗ − xγα .

(2.2.5)

Using the monotonicity property of Ah , hypotheses (2.2.1) and (2.2.2) and definition of J, we deduce from (2.2.5) the relation xγα 2 ≤

δ h + g(x∗ ) xγα − x∗ + x∗ xγα . α α


125

Now by the triangle inequality, we have xγα 2 −

δ h + g(x∗ ) + x∗ xγα − α α

δ h + g(x∗ ) x∗ ≤ 0. α α

This quadratic inequality yields the estimate xγα ≤

h δ + g(x∗ ) + 2x∗ , α α

which implies the boundedness of {xγα }. Then, as in the proof of Theorem 2.1.3, we get that there exists a subsequence of the sequence {xγα } (we do not change its notation) which converges weakly to an element x ˜ ∈ N. Using (2.2.5) and taking into account the monotonicity of J, one gets Jx∗ , x∗ − xγα +

h δ + g(x∗ ) xγα − x∗ ≥ 0 ∀x∗ ∈ N. α α

Next we pass to the limit in this inequality as α → 0 and thus obtain ˜ ≥ 0 ∀x∗ ∈ N, x ˜ ∈ N. Jx∗ , x∗ − x Replacing x∗ ∈ N by xt = t˜ x + (1 − t)x∗ , t ∈ [0, 1], we see that xt ∈ N and ˜ ≥ 0. Jxt , x∗ − x Since J is demicontinuous, this implies as t → 1, ˜ ≥ 0 ∀x∗ ∈ N, J x ˜ , x∗ − x

x ˜ ∈ N.

According to Theorem 1.11.14, we have x ˜=x ¯∗ . Hence, the whole sequence xγα x ¯∗ . ¯∗ , we can write By (1.5.3) and (2.2.5) with x∗ = x x∗ )2 ≤ J x ¯∗ , x ¯∗ − xγα + (xγα − ¯

h δ ¯∗ . x∗ ) xγα − x + g(¯ α α

(2.2.6)

¯∗ , we deduce from (2.2.6) and Then in view of the proved weak convergence of {xγα } to x γ ∗ x as α → 0. The conclusion of the theorem follows now from the (2.1.7) that xα → ¯ definition of E-space. The proof is complete.

Remark 2.2.2 The convergence of the operator regularization method (2.2.4) is established in Theorem 2.2.1 without any restrictions on D(A). Recall that the domain of a maximal monotone operator does not necessarily coincide with the whole space X. In particular, it may be a linear everywhere dense set, a convex closed set and open set having the convex closure. Corollary 2.2.3 Theorem 2.2.1 remains valid for the operator regularization method Ah x + αJ(x − u) = f δ , where u ∈ X is some fixed element. In addition, the solution x ¯∗ ∈ N satisfies the equality ¯ x∗ − u = min{x∗ − u | x∗ ∈ N }.

126

2


Consider the equation (2.1.15) again. The requirement of strong monotonicity of the operator S : X → X ∗ can be replaced now by the condition Sx − Sy, x − y ≥ (µ(x) − µ(y))(x − y),

(2.2.7)

where a function µ(t) is continuous and increasing as t ≥ 0, µ(0) = 0, µ(t) → ∞ as t → ∞. In particular, the duality mapping J µ : X → X ∗ with the gauge function µ(t) satisfies (2.2.7) as we observed in Lemma 1.5.4. In this case, the limit element of the sequence {xγα }, where xγα solves the equation (2.1.15) with S = J µ and α → 0, is the minimal norm solution in N. In other words, replacing in (2.2.4) the normalized duality mapping J by J µ , we find the same solution x ¯∗ ∈ N. However, by the corresponding choice of the function µ(t), µ the operator J may have the stronger monotonicity property in comparison with (2.2.7) (see, for instance, (1.6.57)). This fact is of great importance in Sections 6.5, 6.6, where the convergence analysis of iterative processes are given for regularized equations. Prove that solutions xγα of the equation Ah x + αJ µ x = f δ are bounded. From (2.2.1), (2.2.2) and (1.5.2) we have the inequality

µ(xγα )xγα −

−

h δ + g(x∗ ) xγα − µ(xγα )x∗ α α

δ ∗ h x − g(x∗ )x∗ ≤ 0. α α

(2.2.8)

However, we are not able to evaluate xγα from above by this inequality with any function µ(t). We provide such an estimate for the most important case of power functions. Let µ(t) = ts , s ≥ 1. Then (2.2.8) takes the following form:

xγα s+1 −

h δ h δ + g(x∗ ) xγα − xγα s x∗ − x∗ − g(x∗ )x∗ ≤ 0. α α α α

(2.2.9)

Consider the function ϕ(t) = ts+1 − ats − bt − ab. It is not difficult to verify that if t¯ = τ a + b1/s , τ s ≥ 2, τ > 1, then ϕ(t) > 0 as t ≥ t¯. Hence, (2.2.9) yields the estimate xγα ≤ τ x∗ +

1/s

h δ + g(x∗ ) α α

∀x∗ ∈ N.

(2.2.10)

The boundedness of xγα is finally proved by (2.1.7). ∗ ∗ 2. Let A : X → 2X and Ah : X → 2X be monotone (possibly, multiple-valued) operators which do not satisfy the condition of the maximal monotonicity. As before, we consider


127

the equation (2.2.4) as regularized, and we understand solutions of all corresponding equations in the generalized sense (see Definition 1.9.3). We further assume that D(A) is the convex closed set, int D(A) = ∅. Then the equation (2.2.4) has a unique solution in D(A), and, according to Lemma 1.9.8, it is equivalent to the equation (2.2.11) A¯h x + αJx = f δ , ¯ = D(A). It is obvious where A¯h are maximal monotone extensions of Ah , D(A¯h ) = D(A) that a set N of the generalized solutions of the equation (2.1.1) coincides in D(A) with the ¯ = f. The condition (2.2.1) is replaced by the following set of solutions of the equation Ax inequality: ¯ ≤ g(x)h. (2.2.12) HX ∗ (A¯h x, Ax) Then Theorems 2.2.1, 2.1.4 and 2.1.6 can be formulated for equations with an arbitrary monotone operator. ∗

∗

Theorem 2.2.4 Let A : X → 2X and Ah : X → 2X be monotone operators, D(A) = D(Ah ) be a convex closed set, int D(A) = ∅. Let (2.2.2), (2.2.12) and (2.1.7) hold. Assume that xγα ∈ D(A) is a solution of the regularized equation (2.2.4). Then the sequence {xγα } strongly converges in X as α → 0 to the solution x ¯∗ of the equation (2.1.1) with the minimal norm. Theorem 2.2.5 If the conditions of Theorem 2.2.4 are satisfied, then convergence of the regularization method (2.2.4) is equivalent to solvability of the equation (2.1.1). Theorem 2.2.6 Assume that the conditions of Theorem 2.2.4 are fulfilled, δ = O(α), h = O(α) as α → 0, and the equation (2.1.1) has a unique solution x∗ . Then the sequence {xγα } of solutions of the regularized equation (2.2.4) weakly converges to x∗ as α → 0. 3. Let us present examples of operators satisfying the conditions (2.2.3) and (2.2.12). Example 2.2.7 Suppose that in Example 6 of Section 1.3 the functions ahi (x, s) are given instead of ai (x, s) with the same properties, and |ai (x, s) − ahi (x, s)| ≤ h ∀x ∈ G,

∀s ≥ 0. 0

Then the following calculations can be verified for any functions u, v ∈W1p (G) : Au − Ah u, v =

n G i=1

+ G

≤ h

a0 (x, |u|p−1 ) − a0h (x, |u|p−1 ) |u|p−2 uvdx

p−1

n ∂v ∂u + |u|p−1 uv dx ∂x ∂x G

i=1

i

i

n

∂u p−1 ∂v ≤ h ∂x p ∂x i=1

∂u p−1 ∂u p−2 ∂u ∂v ∂u p−1 dx ai x, − ahi x, ∂x ∂xi ∂xi ∂xi ∂xi i

i L

≤ hup−1 1,p v1,p ,

i Lp

+ up−1 Lp vLp

128

2

REGULARIZATION OF OPERATOR EQUATIONS 0

where u1,p is the norm of u ∈ W1p (G). By Corollary 1.1.2, from the Hahn−Banach 0

theorem, there exists v ∈ W1p (G) with v1,p = 1 giving the estimate −1 Au − Ah u−1,q ≤ hup−1 + q −1 = 1, 1,p , p

0

∗

where · −1,q is the norm in the space W1p (G) . Then we obtain that in (2.2.3) g(t) = tp−1 . Example 2.2.8 Suppose that in Example 8 of Section 1.3, in place of the functions g0 and g1 , their approximations g0h and g1h determining, respectively, monotone operators Ah0 and Ah1 are known. Moreover, |g0h (x, ξ 2 ) − g0 (x, ξ 2 )| ≤ c|ξ|p−2 h,

c > 0,

and

g1h (x, ξ 2 )ξ

=

ωh , 0,

if if

ξ > β, ξ ≤ β,

where ω h > 0 and |ω − ωh | ≤ h. Repeating almost word for word all arguments given in Example 2.2.7, we come to the estimate

¯ ≤ cup−1 + 1 h, HX ∗ (A¯h u, Au) 1,p

0

∗

where X ∗ = W1p (G) , A¯ = A0 + A¯1 , A¯h = Ah0 + A¯h1 . With this, the norm in the space 0

W1p (G) is defined as u1,p =

G

|∇u|p dx

1/p

.

Since the duality mapping J pu = −

n

∂ i=1

∂xi

|∇u|p−2

∂u = −div (|∇u|p−2 ∇u), ∂xi

the regularized problem has the following form: −div (g h (x, ∇2 u)∇u + α|∇u|p−2 ∇u) = f δ (x), where

u |∂Ω = 0,

g h (x, ξ 2 )ξ = g0h (x, ξ 2 )ξ + g1h (x, ξ 2 )ξ

and f δ (x) is a δ-approximation of f (x), that is, f (x) − f δ (x)−1,q ≤ δ, δ > 0, p−1 + q −1 = 1.

(2.2.13)

2.3 Estimates of the Regularized Solutions

2.3

129

Estimates of the Regularized Solutions

Let the conditions of Section 2.2 hold. For simplicity of calculation, assume first that an operator A in (2.1.1) is given exactly. Consider the regularized equation Ax + αJ µ x = f δ

(2.3.1)

with the duality mapping J µ : X → X ∗ , where µ(t) is some gauge function. Let xδα be a solution of (2.3.1) and xα be a solution of the regularized equation Ax + αJ µ x = f

(2.3.2)

with exact right-hand side f. We already know that the sequences {xδα } and {xα } are δ → 0. Let xα ≤ d and xδα ≤ d. Assume that the duality bounded as α → 0 and α µ operator J satisfies the following condition: J µ x − J µ y, x − y ≥ C(R)x − ys ∀x, y ∈ X,

(2.3.3)

where s ≥ 2 and C(R) is a non-negative and non-increasing function of the variable R = max{x, y}.

(2.3.4)

By (2.3.1) - (2.3.3) and by the monotonicity of the operator A, one can deduce the estimate

xδα − xα ≤

δ αC(d)

κ

, κ=

1 . s−1

(2.3.5)

¯∗ , where x ¯∗ ∈ N. As follows We are going now to appraise from above the norm xδα − x from (2.3.5), it is enough for this to evaluate the norm xα − x ¯∗ . Describe the conditions on the operator A and geometry of the spaces X and X ∗ , which allow us to solve this problem. Assume that i) A is Fréchet differentiable and Fréchet derivative A (x) satisfies the Lipschitz−Hölder condition      (2.3.6) A (x) − A (y)  ≤ L(R)x − yσ , 0 < σ ≤ 1, where L(R) is a non-negative and non-decreasing function for all R ≥ 0. ii) There exists an element v ∈ X such that ¯∗ = A (¯ x∗ )v, J µx

(2.3.7)

where x ¯∗ is the minimal norm solution of the equation (2.1.1). Construct the linear operator Aµα (x, y) : X → X ∗ by the equality Aµα (x, y) = A(x, y) + αJ µ (x, y),

(2.3.8)

where x, y ∈ X, A(x, y) and J µ (x, y) are linear symmetric operators from X to X ∗ defined as follows: A(x, y)(x − y) = Ax − Ay

130

2


and J µ (x, y)(x − y) = J µ x − J µ y. In other words, A(x, y) and J µ (x, y) are the first order divided differences of the operators A and J µ , respectively. We suppose that the inverse operator [Aµα (x, y)]−1 exists. It is obvious that Aµα (x, y)(x − y), x − y ≥ αC(R)x − ys . Hence,

 

 

 αC(R)x − ys−2 ≤  A µ α (x, y)

or

   −1    [Aµ α (x, y)]  ≤

1 x − y2−s . αC(R)

(2.3.9)

Using (2.1.1), (2.3.2) and (2.3.8), it is not difficult to verify that ¯∗ )(xα − x ¯∗ ) = −αJ µ x ¯∗ . Aµα (xα , x Now the latter equality implies xα − x ¯∗ = α[Aµα (xα , x ¯∗ )]−1 A (¯ x∗ )v in view of the condition (2.3.7). Let      J µ (x, y)  ≤ M (R)x − y−γ ,

γ > 0,

(2.3.10)

where M (R) is a non-negative and non-decreasing function for all R > 0. By (2.3.6), (2.3.9) and (2.3.10), we deduce xα − x ¯∗ = α[Aµα (xα , x ¯∗ )]−1 (Aµα (xα , x ¯∗ ) − A (¯ x∗ ) − Aµα (xα , x ¯∗ ))v ≤ αv +   

   xα − x ¯∗ 2−s     A (¯ x∗ + t(xα − x ¯∗ )) − A x ¯∗      C(r)   

µ + α ¯∗ )    J (xα , x  v

≤ αv +

L(r) αv ¯∗ 2−s−γ , M (r)xα − x ¯∗ 2−s+σ + vxα − x C(r) C(r)

where 0 ≤ t ≤ 1 is some number and r = max{d, ¯ x∗ }. Thus, we establish the estimate

xα − x ¯∗ ≤ v α +

αM (r) L(r) xα − x ¯∗ 2−s+σ + ¯∗ 2−s−γ . xα − x C(r) C(r)

Note that if X is a Hilbert space, then C(r) ≡ 1, s = 2, µ(t) ≡ t, σ = 1, γ = 0, M (r) ≡ 1,

(2.3.11)

2.3 Estimates of the Regularized Solutions

131

and if 1 − vL(r) > 0, then (2.3.11) gives the following inequality: ¯∗ ≤ xα − x

2vα . 1 − vL(r)

We emphasize that, in the general case, it is impossible to obtain effective estimates of xα − x ¯∗ if we use the relation (2.3.11). Therefore, we introduce some additional assumptions. Let σ = s − 1. It is clear that σ ∈ (0, 1] because s ∈ (1, 2]. Suppose first that s + γ > 2 and that L(r) v > 0. (2.3.12) a(r) = 1 − C(r)

Then (2.3.11) yields the inequality αvM (r) αv ≤ 0. ¯∗ s+γ−2 − xα − x a(r)C(r) a(r)

xα − x ¯∗ s+γ−1 −

(2.3.13)

Introduce the function ϕ(t) = tβ − a1 tβ−1 − a2 , where β > 1, t ≥ 0, a1 > 0 and a2 > 0. The following simple properties hold for this 1/β function: 1) ϕ(0) = −a2 < 0 and 2) if t = t¯ = a2 + a1 then 1/β 1/β 1/β ϕ(t¯) = t¯β−1 (t¯ − a1 ) − a2 = t¯β−1 a2 − a2 = (a2 + a1 )β−1 a2 − a2 .

It is not difficult to see that if a1 = 0 then ϕ(t¯) = 0, and if a1 > 0 then ϕ(t¯) > 0. Besides, a derivative of the function ϕ(t) vanishes only if t = t0 = a1 (β − 1)β −1 . Moreover, the function ϕ(t) increases for all t > t0 and t¯ > t0 . Hence, by (2.3.13), we have xα − x ¯∗ ≤

αvM (r) a(r)C(r)

where τ=

τ

+

αv , a(r)

1 . s+γ−1

Taking into account the estimate (2.3.5) we obtain ¯∗ ≤ xδα − x

δ αC(r)

κ

+

αvM (r) a(r)C(r)

τ

+

αv . a(r)

(2.3.14)

Let now s + γ < 2. The inequality (2.3.13) can be rewritten as xα − x ¯∗ −

Consider the function

αv αvM (r) ≤ 0. xα − x ¯∗ 1−(s+γ−1) − a(r) a(r)C(r) ψ(t) = t − a1 t1−β − a2

with 0 < β < 1, a1 > 0, a2 > 0 and ψ(0) = −a2 < 0. Denote t¯β = a1 + aβ2 and calculate ψ(t¯) = t¯1−β (t¯β − a1 ) − a2 = t¯1−β aβ2 − a2 = (a1 + aβ2 )(1−β)/β aβ2 − a2 .

132

2


Obviously, ψ(t¯) = 0 if a1 = 0, and ψ(t) increases with respect to parameter a1 . Moreover, ψ(t) achieves a minimum when t = t0 = [a1 (1 − β)]1/β . It is clear that t0 < t¯. Thus, ψ(t) ≤ 0 if t ≤ (a1 + aβ2 )1/β . Since s + γ < 2, (2.3.13) produces the estimate 

∗

xα − x ¯ ≤

Therefore, xδα

∗

−x ¯ ≤

δ αC(r)

αvM (r) + a(r)C(r)

κ

αv a(r)



αvM (r) + +

a(r)C(r)

 τ1 τ  .

αv a(r)

τ1

τ  .

(2.3.15)

Finally, if s + γ = 2 then we conclude from (2.3.13) that xδα − x ¯∗ ≤

δ αC(r)

κ

+

M (r) αv . 1+ C(r) a(r)

(2.3.16)

Thus, we have obtained the following results: Theorem 2.3.1 Assume that A : X → X ∗ is a maximal monotone and Fréchet differolder condition (2.3.6), entiable operator, Fréchet derivative A (x) satisfies the Lipschitz−H¨ duality mapping J µ : X → X ∗ with a gauge function µ(t) has the property (2.3.3), and there ¯∗ )]−1 , where Aµα (x, y) is defined by (2.3.8). Let x ¯∗ be exists the inverse operator [Aµα (xα , x δ the minimal norm solution of the equation (2.1.1), xα and xα be solutions of the equations (2.3.1) and (2.3.2), respectively, and (2.3.7) and (2.3.10) hold. If σ = s − 1 and a(r) is defined by (2.3.12) with r = max{d, x∗ } and if xδα ≤ d, xα ≤ d, then the estimate (2.3.14) holds for all s + γ > 2. If s + γ < 2 or s + γ = 2 then, respectively, (2.3.15) or (2.3.16) are fulfilled. Remark 2.3.2 The duality mapping J µ satisfying the hypotheses of Theorem 2.3.1 exists in the spaces Lp (G) (1 < p ≤ 2). Indeed, since s = 2, we have µ(t) = t, κ = 1 and Jx − Jy, x − y ≤ M (R)x − yp 1 . Therefore, (2.3.14) in Lp (G) is expressed (see (1.6.35)), that is, γ = 2 − p and τ = 3−p as δ 1 + O(α 3−p ) + O(α). ¯∗ ≤ O xδα − x α

Suppose that requirements of Theorem 2.3.1 are satisfied and, instead of the exact operator A, the sequence of monotone single-valued operators Ah are given such that D(Ah ) = D(A) and Ax − Ah x∗ ≤ g(x)h ∀x ∈ D(A), where g(t) is a non-negative and non-decreasing function for all t ≥ 0. Let xγα be a unique solution of the equation Ah x + αJ µ x = f δ with xγα ≤ d. Then it is not difficult to verify that the estimates of xγα − x ¯∗ are obtained from (2.3.14) - (2.3.16) if the perturbation δ in their right-hand sides is replaced by δ+hg(r).

2.4 Equations with Domain Perturbations

2.4

133

Equations with Domain Perturbations

As before, A and Ah denote, respectively, exact and perturbed maximal monotone (possibly, multiple-valued) operators, and the equations (2.1.1) are solved in an E-space X. Let X ∗ be a strictly convex space. It has been earlier everywhere assumed that D(A) = D(Ah ). Now we shall study D(A) and D(Ah ) to be not coinciding sets and define the proximity between A and Ah in the following way: For any element x ∈ D(A) and given h > 0, let there exist an element xh ∈ D(Ah ) such that x − xh ≤ a(x)h

(2.4.1)

d∗ (y, Ah xh ) ≤ g(y∗ )ξ(h) ∀y ∈ Ax,

(2.4.2)

and ∗

where d∗ (y, A x ) is distance in the space X between y ∈ Ax and the convex closed set Ah xh . We further assume that functions a(t) and g(t) are non-negative, ξ(h) → 0 as h → 0 and ξ(0) = 0. If xγα is a solution of the regularized equation h h

Ah x + αJ µ x = f δ , α > 0,

(2.4.3)

yαγ + αJ µ xγα = f δ ,

(2.4.4)

and if yαγ ∈ Ah xγα , then ∗

where f ∈ X satisfies (2.2.2). Let N = ∅, where N is a solution set of the equations (2.1.1), and x∗ ∈ N. Thus, f ∈ Ax∗ . By the conditions (2.4.1) and (2.4.2), one can find elements xh ∈ D(Ah ) and yh ∈ Ah xh such that for every h > 0, x∗ − xh ≤ a(x∗ )h

(2.4.5)

f − y h ∗ ≤ g(f ∗ )ξ(h).

(2.4.6)

and We subtract the element f from both parts of equation (2.4.4) and calculate their dual products with the difference xγα − xh . We have yαγ − f, xγα − xh + αJ µ xγα , xγα − xh = f δ − f, xγα − xh .

(2.4.7)

Rewrite (2.4.7) in the equivalent form: yαγ − y h , xγα − xh + y h − f, xγα − xh + αJ µ xγα , xγα − xh = f δ − f, xγα − xh . By virtue of the monotonicity of operators Ah , definition of the duality mapping J µ and by (2.4.5) and (2.4.6), one gets µ(xγα )xγα − µ(xγα ) [x∗ + a(x∗ )h] −

δ

α

+

ξ(h) g(f ∗ ) (xγα + x∗ + a(x∗ )h) ≤ 0. α

(2.4.8)

134

2


If

δ + ξ(h) =0 (2.4.9) α ¯ ∈ X. then it results from (2.4.8) that the sequence {xγα } is bounded as α → 0 and xγα x The equality (2.4.4) allows us to assert that yαγ → f as α → 0. Next, by (2.4.1) and (2.4.2), for every z ∈ D(A) and every y ∈ Az, we find z h ∈ D(Ah ) and yh ∈ Ah z h such that lim

α→0

z − z h ≤ a(z)h and y − y h ∗ ≤ g(y∗ )ξ(h). Thus, z h → z and y h → y as h → 0. The monotonicity property of Ah gives yαγ − y h , xγα − z h ≥ 0. If α → 0 then by (2.4.9), f − y, x ¯ − z ≥ 0 ∀z ∈ D(A), ∀y ∈ Az. Maximal monotonicity of the operator A ensures now the inclusion x ¯ ∈ N. Show that x ¯ is the minimal norm solution of (2.1.1), that is, ¯ x ≤ x∗ for any x∗ ∈ N. Let x ¯ = θX . Then, due to the weak convergence of the sequence {xγα } to x ¯, we obtain ¯ x ≤ lim inf xγα . α→0

(2.4.10)

Thus, µ(¯ x) ≤ µ(lim inf xγα ) = lim inf µ(xγα ) α→0

α→0

because µ(t) is increasing. Hence, µ(xγα ) ≥ c > 0 for sufficiently small α > 0. Next, (2.4.7) yields J µ xγα , xγα = µ(xγα )xγα ≤

δ

α

+

ξ(h) g(f ∗ ) xγα − xh + µ(xγα )xh . α

(2.4.11)

Since, in view of (2.4.5), xh → x∗ , we deduce from (2.4.11) the inequality lim sup xγα ≤ x∗ .

(2.4.12)

α→0

The estimates (2.4.10) and (2.4.12) imply the fact that ¯ x = min{x∗ | x∗ ∈ N }, that ∗ is, x ¯=x ¯ . The estimates (2.4.10) and (2.4.12) allow us to establish strong convergence of xγα to ¯ x∗ . ¯=x ¯∗ . Suppose that in (2.4.11) xh → θX . Let x ¯ = θX . It is obvious in this case that x Then the function µ(t) satisfies the limit relation lim µ(xγα )xγα = 0,

α→0

and we conclude that xγα → 0 as α → 0. Thus, we come to the following result:

2.4 Equations with Domain Perturbations

135

∗

∗

Theorem 2.4.1 Let A : X → 2X and Ah : X → 2X be maximal monotone operators, X be an E-space, X ∗ be a strictly convex space. Let the hypotheses (2.2.2), (2.4.1), (2.4.2) and (2.4.9) hold. Then the sequence of solutions of the regularized equation (2.4.3) strongly converges in X to the minimal norm solution of the equation (2.1.1). Remark 2.4.2 If D(A) = D(Ah ), then it is possible to assume x = xh in (2.4.1) and (2.4.2). In this case, (2.4.1) holds even if a(t) ≡ 0, and (2.4.2) can be given in the form d∗ (y, Ah x) ≤ g(y∗ )ξ(h) ∀y ∈ Ax. The latter is weaker than (2.2.1). Present the example realizing (2.4.1) and (2.4.2). Example 2.4.3 Let a map A : R2 → R2 be defined by the matrix

A=

1 2 2 4

.

It is not difficult to verify that A is positive. Consider it on the set Ω = {(x1 , x2 ) | x2 ≤ k1 x1 , x2 ≥ k2 x1 }, k1 > k2 , lying in the first quarter. Finish defining the operator A on a boundary ∂Ω by semilines (see the proof of Theorem 1.7.19) and thus obtain the maximal monotone operator ¯ = Ω. By analogy, we construct a perturbed maximal monotone operator A¯ with D(A) 2 h 2 ¯ A : R → 2R by means of the closed matrix

h

A =

(2 + h)2 (4 + h)−1 2+h

2+h 4+h

,

where h is a small enough number and domain D(A¯h ) = Ωh = {(x1 , x2 ) | x2 ≤ k1h x1 , x2 ≥ k2h x1 }, |kih − ki | ≤ h, kih > ki , i = 1, 2. We assume that Ωh also lies in the first quarter. Show that the maps A¯ and A¯h satisfy the conditions (2.4.1) and (2.4.2). Indeed, if x ∈ Ω ∩ Ωh ⊂ int Ω then xh = x, d(y, A¯h x) = y − Ah x ≤ hy, ¯ If still x ∈ Ω and at the same time x ∈ Ωh and it lies between the where y = Ax = Ax. ¯h = {xh1 , k2h xh1 }, where straight lines x2 = k2 x1 , x2 = k2h x1 , then one can take xh = x h h x = x. It is clear that there exists c > 0 such that x − x ≤ chx. Therefore, by the inequality Ax − Ah xh ≤ Ax − Ah x + Ah x − Ah xh , ¯ If x lies on the straight line x2 = k2 x1 we have that d(y, A¯h xh ) ≤ chAx with y = Ax = Ax. h h then, as before, we take x = x ¯ . In this case, x − xh ≤ chx, ¯ = {Ax + λ{k2 , −1} | λ ≥ 0}, Ax

136

2


A¯h xh = {Ah xh + λ{k2h , −1} | λ ≥ 0}. Consequently, for y = Ax + λ{k2 , −1} and y h = Ah xh + λ{k2h , −1} with a fixed λ ≥ 0, we have y − yh ≤ c1 hy with some c1 > 0. Finally, if x ∈ Ω and it lies on the line x2 = k1 x1 then we choose a point xh on the line x2 = k1h x1 such that xh = x. Thus, the conditions (2.4.1), (2.4.2) are satisfied with a(t) = c2 t, g(t) = c3 t with some c2 , c3 > 0 and ξ(h) = h. Remark 2.4.4 This example shows that in (2.4.2) it is impossible to replace g(y) by g(x) because in the general case (in particular, for unbounded A) d(y, Ah xh ) can be arbitrarily large for a fixed x (see the case where points x and xh lie on different straight lines defining the boundaries of sets Ω and Ωh ).

2.5

Equations with Semimonotone Operators ∗

Suppose that a Banach space X possesses the M -property, an operator A : X → 2X in equation (2.1.1) is semimonotone, D(A) = X, and there exist r > 0 and y ∈ Ax such that y − f, x ≥ 0 as x ≥ r.

(2.5.1)

Then, by virtue of Theorem 1.10.6, the equation (2.1.1) has in X a nonempty set N of s-generalized solutions (a solution x0 ∈ N is understood in the sense of Definition 1.10.3 as ¯ 0 ). f ∈ Ax 1. Consider the equation Ax + αJx = f δ (2.5.2) with α > 0 and δ satisfying (2.2.2). Let δ = O(α) as α → 0. Show that there exists a solution of equation (2.5.2). Indeed, for all x ∈ X, it is not difficult to verify that y + αJx − f δ , x = y − f, x + αx2 + f − f δ , x

≥ αx x − Hence, if

δ + y − f, x , α

y ∈ Ax.

(2.5.3)

δ ≤ K and if x ≥ r1 = max{r, K}, then the inequality α

y + αJx − f δ , x ≥ 0 is satisfied. Since the duality mapping J is monotone, we are able to apply Theorem 1.10.6 and conclude that there exists a solution of equation (2.5.2) (which is not unique in general). Furthermore, it follows from (2.5.3) that the operator A + αJ is coercive, therefore, it has a bounded inverse operator. Hence, (2.5.2) may be considered as a regularization of the equation (2.1.1). Let T = A + C be a monotone operator, where C : X → X ∗ is a strongly continuous mapping, and let T¯ be a maximal monotone extension of T. Then the operator F¯ = T¯ + αJ is also maximal monotone, that is, F¯ is a maximal monotone extension of the operator

2.5 Equations with Semimonotone Operators

137

¯ δα , where A¯ = T¯ − C, then F = T + αJ. Denote by xδα a solution of (2.5.2). If any yαδ ∈ Ax we have (2.5.4) yαδ + αJxδα = f δ . Moreover, xδα ≤ r1 in view of Theorem 1.10.6 again. Hence, xδα x ¯ ∈ X as α → 0. Using now (2.5.4) we can write z + αJx − Cxδα − f δ , x − xδα ≥ 0 ∀z ∈ T¯x. Setting α → 0 and taking into account the strong continuity of C we obtain in a limit the inequality z − C x ¯ − f, x − x ¯ ≥ 0 ∀z ∈ T¯x. (2.5.5) ¯x, i.e., x Consequently, f ∈ A¯ ¯ ∈ N. Thus, we have proved the following assertion: Theorem 2.5.1 Let X be a reflexive strictly convex Banach space together with its dual ∗ space X ∗ and have the M -property. Let A : X → 2X in (2.1.1) be a semimonotone operator with domain D(A) = X and the conditions (2.5.1) and (2.2.2) hold. If there is a constant δ ≤ C as α → 0, then there exists a subsequence of the sequence {xδα }, C > 0 such that α where xδα are solutions of the equation (2.5.2) with fixed α and δ, which weakly converges to some point x ¯ ∈ N. If a solution x ¯ is unique, then the whole sequence {xδα } converges weakly to x ¯. 2. Suppose that in the equation (2.5.2) not only the right-hand side f but also the operator A is given approximately, that is, instead of A, we have the operators Ah which are also semimonotone for all h > 0 and D(Ah ) = X. Since value sets of the operators A¯ = T¯ − C and A¯h = T¯h − C h are convex and closed, the proximity between A and Ah can be defined by the Hausdorff distance as ¯ A¯h x) ≤ g(x)h ∀x ∈ X, HX ∗ (Ax,

(2.5.6)

where g(t) is a continuous non-negative function for all t ≥ 0. In this case, we use the duality mapping J µ with a gauge function µ(t) satisfying the condition µ(t) > K(g(t) + 1) δ h ≤ K. Then the equation for t ≥ t0 > 0, where ≤ K, α α

Ah x + αJ µ x = f δ

(2.5.7)

is considered in place of (2.5.2). Obviously, there exist y ∈ Ax and y h ∈ A¯h x such that y h + αJ µ x − f δ , x ≥ y − f, x + αµ(x)x − δx − hg(x)x

= αx µ(x) −

δ h + y − f, x . g(x) − α α

Choosing the function µ(t) as was described above, we conclude that there is r > 0 such that the inequality y h + αJ µ x − f δ , x ≥ 0

138

2


holds for all x ∈ X with x ≥ r and for some y h ∈ A¯h x. Theorem 1.10.6 guarantees the existence of a solution xγα of the equation (2.5.7). Let yαγ ∈ A¯h xγα such that yαγ + αJ µ xγα = f δ , and let x be an arbitrary element of X. Then according to the condition (2.5.6), for any ¯ there exists yh ∈ A¯h x such that y ∈ Ax, y − yh ∗ ≤ g(x)h ∀x ∈ X.

(2.5.8)

The monotonicity of T¯ h + αJ µ implies the relation y h + Cx + αJ µ x − Cxγα − f δ , x − xγα ≥ 0. Applying the estimate (2.5.8), we come again to (2.5.5) as α → 0. Hence, the conclusion of Theorem 2.5.1 holds for a solution sequence {xγα } of the equation (2.5.7).

2.6

Equations with Non-Monotone Perturbations

The perturbed operators Ah are often approximations of an original operator A as a result of applications of numerical methods for solving the equations (2.1.1) in Hilbert and in Banach spaces. From this point of view approximations Ah should be done so that they retain basic properties of the operator A, because their violation influences qualitative characteristics of approximations xγα to the exact solutions of (2.1.1). Sometimes to do this does not work, in particular, this concerns the monotonicity property of Ah . Therefore, it is necessary to investigate the case when the approximations Ah of the monotone operator A are nonmonotone themselves. Observe that in this case regularized equations, generally speaking, may not have solutions. 1. We shall study convergence of the regularization method (2.2.4) for finding solutions ∗ of the operator equation (2.1.1) with a monotone operator A : X → 2X . The perturbed h equation (2.1.4) is given under the condition that (2.2.2) holds and {A } is the sequence of semimonotone operators and D(A) = D(Ah ) = X. Let a Banach space X have the M -property, the Hausdorff distance ¯ A¯h x) ≤ h ∀x ∈ X, HX ∗ (Ax,

(2.6.1)

¯ and r > 0 such that for x ≥ r, and there exist y ∈ Ax y − f, x ≥ 0 as x ≥ r. For some y h ∈ A¯h x, we can write the following relations: y h + αJx − f δ , x ≥ y − f, x + αx2 − hx − δx

= αx x −

δ + h + y − f, x . α

(2.6.2)

2.6 Equations with Non-Monotone Perturbations Suppose that

139

h δ ≤ K and ≤ K as α → 0. If x ≥ r1 = max{r, 2K} then the inequality α α

y h + αJx − f δ , x ≥ 0 holds. By Theorem 1.10.6, a solution xγα of the equation (2.2.4) satisfying the condition xγα ≤ r1 exists though it may be non-unique. Furthermore, one can find yαγ ∈ A¯h xγα such that yαγ + αJxγα = f δ . γ ¯ and A¯h x are convex and closed for every Therefore, yα → f as α → 0. Since the sets Ax ¯ γα such that x ∈ X, it follows from (2.6.1) that, for every yαγ , there is an element y˜αγ ∈ Ax yαγ − y˜αγ ∗ ≤ h. Therefore, y˜αγ → f as α → 0. By Lemma 1.4.5, the graph of a maximal monotone operator is demiclosed. Then xγα x∗ ∈ N. Taking into account the monotonicity of A¯ and J we obtain from (2.2.5) Jx∗ , x∗ − xγα +

δ+h γ xα − x∗ ≥ 0 ∀x∗ ∈ N. α

Repeating in fact the proof of Theorem 2.2.1 we come to the following result: Theorem 2.6.1 Suppose that X is an E-space and has the M -property, X ∗ is strictly ∗ ∗ convex, A : X → 2X is a monotone operator and Ah : X → 2X are semimonotone h operators for all h > 0, D(A) = D(A ) = X. Let the conditions (2.2.2), (2.6.1), (2.6.2) ¯∗ , where x ¯∗ and (2.1.7) hold. Then the equation (2.2.4) is solvable and its solutions xγα → x is the element of N with the minimal norm. Consider now convergence of the operator regularization method for semimonotone approximations Ah with the weaker condition of the proximity between A and Ah . Namely, let ¯ A¯h x) ≤ g(x)h ∀x ∈ X, HX ∗ (Ax, (2.6.3) where g(t) is a continuous nonnegative and increasing function for all t ≥ 0. Instead of (2.2.4), we shall investigate the regularized equation (2.5.7) with duality mapping J µ : X → X ∗ whose gauge function µ(t) is defined by the inequality µ(t) > K(1 + g(t)) ∀t ≥ t0 .

(2.6.4)

Here t0 is a fixed positive number and max

δ h

≤ K. , α α

Theorem 2.6.2 Assume that X is an E-space and has the M -property, X ∗ is strictly ∗ ∗ convex, A : X → 2X is a monotone operator and Ah : X → 2X are semimonotone h operators for all h > 0, D(A) = D(A ) = X. Let the conditions (2.1.7), (2.6.2) and (2.6.3) be fulfilled. Then the equation (2.5.7) is solvable and the sequence {xγα } of its solutions converges strongly in X to the minimal norm solution of (2.1.1).

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2


Proof. The operator J µ : X → X ∗ is monotone, therefore, Ah + αJ µ is semimonotone. Then the inequality

y h + αJ µ x − f δ , x ≥ αx µ(x) −

δ h − g(x) + y − f, x , α α

y h ∈ A¯h x,

implies the estimate y h + αJ µ x − f δ , x ≥ 0 for all x with x ≥ r1 = max{r, t0 }. Hence, there exists a solution xγα of the equation (2.5.7) satisfying the condition xγα ≤ r1 . Thus, xγα x ¯ ∈ X as α → 0. Let yαγ ∈ A¯h xγα and yαγ + αJ µ xγα = f δ .

(2.6.5)

¯ ∈ N. Using (2.1.1) and (2.6.5), As in the previous theorem, we conclude that yαγ → f and x we further deduce similarly to (2.2.5) the following equality: yαγ − f, xγα − x∗ + αJ µ xγα − J µ x∗ , xγα − x∗ = f δ − f, xγα − x∗ + αJ µ x∗ , x∗ − xγα ∀x∗ ∈ N. By the monotonicity of operators A¯ and δ

J µ x∗ , x∗ − xγα + Since

{xγα }

α

+

Jµ

(2.6.6)

and by the condition (2.6.3), one has

h g(xγα ) xγα − x∗ ≥ 0 ∀x∗ ∈ N. α

is bounded and (2.1.7) holds, the latter inequality gives in a limit J µ x∗ , x∗ − x ¯ ≥ 0 ∀x∗ ∈ N.

As we have shown in Theorem 2.2.1, this yields the relation ¯, x∗ − x ¯ ≥ 0 ∀x∗ ∈ N J µ x because the operator J µ is a demicontinuous. Lemma 1.5.7 asserts that J µ is a potential operator and J µ x = Φ (x), where Φ(x) is defined by (1.5.4). Therefore, it results from the previous inequality that x ¯=x ¯∗ . Then Lemma 1.5.4 and (2.6.6) imply

µ(xγα ) − µ(¯ x∗ ) (xγα − ¯ x∗ ) ≤

δ

α

+

h ¯∗ + J µ x ¯∗ , x ¯∗ − xγα . g(xγα ) xγα − x α

Since

x ¯ and (2.1.7) is satisfied, the convergence of xγα to ¯ x∗ follows. Then the proof of the theorem is accomplished because X is E-space. xγα

∗

2. Next we present one more method for solving (2.1.1) with non-monotone mappings Ah . Let A : X → X ∗ be a monotone hemicontinuous operator with domain D(A) = X, Ah : X → X ∗ be hemicontinuous operators with D(Ah ) = X. Suppose that Ax − Ah x∗ ≤ hg(x),

(2.6.7)

2.6 Equations with Non-Monotone Perturbations

141

where g(t) is a non-negative continuous function for all t ≥ 0. We define the regularized solution xγα as a solution of the so-called variational inequality with small offset: Ah z + αJ µ z − f δ , z − xγα ≥ −g(z)z − xγα ∀z ∈ X, α > 0,

(2.6.8)

where ≥ h, µ(t) ≥ g(t) as t ≥ t0 , t0 is a fixed positive number, f δ − f ∗ ≤ δ. Lemma 2.6.3 The inequality (2.6.8) has a nonempty solution set N for each α > 0 and each f δ ∈ X ∗ . Proof. By Corollary 1.8.9, we deduce that the equation Ax + αJ µ x = f δ

(2.6.9)

has a unique solution xδα . Then (2.6.9) is equivalent to the variational inequality Az + αJ µ z − f δ , z − xδα ≥ 0 ∀z ∈ X.

(2.6.10)

Making use of (2.6.7), from (2.6.10) one gets Ah z + αJ µ z − f δ , z − xδα ≥ −hg(z)z − xδα . Since ≥ h, we conclude that xδα is all the more the solution of (2.6.8). Lemma 2.6.4 If xγα is a solution of (2.6.8) then it is also a solution of the inequality Ah xγα + αJ µ xγα − f δ , z − xγα ≥ −g(xγα )z − xγα ∀z ∈ X.

(2.6.11)

Proof. Replace z in (2.6.8) by zt = txγα +(1−t)z with t ∈ [0, 1), and divide the obtained inequality by 1 − t > 0. Then Ah zt + αJ µ zt − f δ , z − xγα ≥ −g(zt )z − xδα . Setting t → 1 and taking into account the properties of operators Ah , J µ and function g(t), we obtain in a limit (2.6.11).

Theorem 2.6.5 If lim

α→0

δ+ = 0, α

then xγα → x ¯∗ , where x ¯∗ ∈ N is the minimal norm solution of (2.1.1). Proof. It follows from (2.1.1) that Ax∗ − f, z − x∗ = 0 ∀z ∈ X, ∀x∗ ∈ N.

(2.6.12)

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By virtue of Lemma 2.6.4, one can put z = x∗ in (2.6.11) and z = xγα in (2.6.12). After this, summation of the obtained inequalities gives Ah xγα − Ax∗ , x∗ − xγα + αJ µ xγα , x∗ − xγα + f − f δ , x∗ − xγα ≥ −g(xγα )xγα − x∗ .

(2.6.13)

Now the monotonicity of A, the definition of J µ and condition (2.6.7) imply from (2.6.13) the inequalities µ(xγα )xγα ≤

≤

+h δ g(xγα )xγα − x∗ + xγα − x∗ + µ(xγα )x∗ α α

δ 2 g(xγα )xγα − x∗ + xγα − x∗ + µ(xγα )x∗ . α α

(2.6.14)

Since α → 0, (δ + )/α → 0 and since µ(t) > g(t) beginning with a certain t = t0 , it results from (2.6.14) that the sequence {xγα } is bounded for sufficiently small regularization parameter α > 0. Hence, xγα x ¯ ∈ X. Finally, we obtain from (2.6.8) as α → 0, Az − f, z − x ¯ ≥ 0 ∀z ∈ X, that is, in view of Theorem 1.9.1, x ¯ ∈ N. Using (2.6.13), we complete the proof as in Theorem 2.2.1.

Remark 2.6.6 The convergence analysis of the regularization methods (2.2.4) and (2.3.2) can also be done when perturbed maps Ah are pseudomonotone or quasipotential.

2.7

Equations with Accretive and d-Accretive Operators

In what follows, a Banach space X possesses an approximation, A : X → X is an accretive operator with domain D(A), the normalized duality mapping J : X → X ∗ is continuous and at the same time weak-to-weak continuous in X. Consider the equation Ax = f

(2.7.1)

with f ∈ X. Let N be its nonempty solution set. Suppose that the operator and righthand side of (2.7.1) are given approximately. This means that in reality the equation Ah x = f δ is solved, where Ah : X → X are accretive perturbations of A for all h > 0 with D(Ah ) = D(A), and f δ ∈ X are perturbations of f for all δ > 0. We assume that f δ − f ≤ δ,

(2.7.2)

Ax − Ah x ≤ g(x)h ∀x ∈ X,

(2.7.3)

and

2.7


143

where g(t) is a continuous non-negative function for all t ≥ 0. The regularized equation with accretive operator Ah is written in the form: Ah x + αx = f δ .

(2.7.4)

1. First assume that operators A and Ah are hemicontinuous and D(A) = D(Ah ) = X. Using the accretiveness of Ah we deduce Jx, Ah x + αx ≥ αx2 − Ah (θX )x = x(x − Ah (θX )).

(2.7.5)

Therefore, operator T = Ah + αI is coercive. Then, by Theorem 1.15.23, the equation (2.7.4) has a solution xγα , γ = {δ, h}, in the classical sense for any α > 0. It is unique because T is strongly accretive (see Remark 1.15.31). Thus, Ah xγα + αxγα = f δ .

(2.7.6)

Theorem 2.7.1 If the condition (2.1.7) holds, then xγα → x ¯∗ ∈ N, where x ¯∗ is a unique solution of (2.7.1) satisfying the inequality J(¯ x∗ − x∗ ), x ¯∗ ≤ 0 ∀x∗ ∈ N.

(2.7.7)

Proof. First of all, show that the element x ¯∗ is unique. Suppose that there exists x ¯∗∗ ∈ N such that x ¯∗∗ = x ¯∗ and J(¯ x∗∗ − x∗ ), x ¯∗∗ ≤ 0 ∀x∗ ∈ N.

(2.7.8)

Put x∗ = x ¯∗∗ in (2.7.7) and x∗ = x ¯∗ in (2.7.8) and add the obtained inequalities. Then 0 ≥ J(¯ x∗ − x ¯∗∗ ), x ¯∗ − x ¯∗∗ = ¯ x∗ − x ¯∗∗ 2 . This implies that x ¯∗ = x ¯∗∗ . ∗ Take an arbitrary x ∈ N. By (2.7.1) and (2.7.6), it is not difficult to make sure that J(xγα − x∗ ), Ah xγα − Ax∗ + αJ(xγα − x∗ ), xγα − x∗ = J(xγα − x∗ ), f δ − f − αJ(xγα − x∗ ), x∗ . Since the operators

Ah

(2.7.9)

are accretive, we have by (2.7.2) and by (2.7.3) that

xγα − x∗ ≤

δ h + g(x∗ ) + x∗ ∀x∗ ∈ N, α α

or

h δ (2.7.10) + g(x∗ ) + 2x∗ ∀x∗ ∈ N. α α ¯, that Hence, the sequence {xγα } is bounded, therefore, it has a weak accumulation point x ¯ ∈ X as α → 0 (as before, we do not change its denotation). Now we use again is, xγα x the accretiveness of the operator Ah with any h > 0 and (2.7.6) to get xγα ≤

J(x − xγα ), Ah x − Ah xγα = J(x − xγα ), Ah x + αxγα − f δ ≥ 0 ∀x ∈ X.

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2


Since the operator J is weak-to-weak continuous, we have in a limit when α → 0 the inequality J(x − x ¯), Ax − f ≥ 0 ∀x ∈ X. (2.7.11) According to Theorem 1.15.14, the operator A is maximal accretive. Then (2.7.11) implies f = A¯ x. Thus, the inclusion x ¯ ∈ N is established. By (2.7.9) with x∗ = x ¯, it is not difficult to obtain the estimate ¯ 2 ≤ xγα − x

δ γ h ¯ − J(xγα − x ¯), x ¯ . x)xγα − x ¯ + g(¯ x − x α α α

Since the sequence {xγα } is bounded and weakly converges to x ¯ ∈ N, we deduce from the ¯ as α → 0. After this, by simple algebra, last inequality the strong convergence of {xγα } to x we obtain from (2.7.9), J(xγα − x∗ ), xγα ≤

δ

α

+

h g(x∗ ) xγα − x∗ ∀x∗ ∈ N. α

Going to the limit as α → 0 one has J(¯ x − x∗ ), x ¯ ≤ 0 ∀x∗ ∈ N, ¯∗ . i.e., x ¯=x ¯∗ . Thus, the whole sequence {xγα } strongly converges to x

Similarly to Theorems 2.2.5 and 2.2.6, the following assertions can be proved for accretive equations. Theorem 2.7.2 Under the conditions of Theorem 2.7.1, convergence of the regularization method (2.7.4) is equivalent to solvability of the equation (2.7.1). Theorem 2.7.3 Let equation (2.7.1) be uniquely solvable, N = {¯ x∗ }, and assume that δ+h ¯∗ . ≤ C as α → 0. Then xγα x there exists a constant C > 0 such that α 2. Suppose that A : X → 2X is the maximal accretive (multiple-valued, in general) operator in a domain D(A), Ah : X → 2X is m-accretive (i.e., maximal accretive) and D(Ah ) = D(A). According to Lemma 1.15.13, the value set of a maximal accretive operator at any point of its domain is convex and closed. Therefore, the proximity of operators A and Ah can be defined by the relation HX (Ax, Ah x) ≤ g(x)h ∀x ∈ D(A), where g(t) is a continuous non-negative function for all t ≥ 0 and HX (G1 , G2 ) is the Hausdorff distance between the sets G1 and G2 in X. Then the regularized equation (2.7.4) has a unique solution which we denote, as before, by xγα . This means that there exists yαγ ∈ Ah xγα such that yαγ + αxγα = f δ .

2.7


145

Since the operator J is weak-to-weak continuous, it is not difficult to verify validity of all the assertions above in this section. 3. Let A : X → 2X be an arbitrary accretive operator and D(A) = X. In this case, solutions of the equations (2.7.1) and (2.7.4) are understood in the sense of Definition 1.15.25. We assume that spaces X and X ∗ are uniformly convex, X has the M -property and duality mapping J : X → X ∗ is weak-to-weak continuous. Let A¯ and A¯h be maximal accretive extensions of accretive operators A and Ah , respectively, and ¯ A¯h x) ≤ g(x)h ∀x ∈ X, HX (Ax, where g(t) is a continuous non-negative function for all t ≥ 0. We want to show that the equation (2.7.4) is uniquely solvable. Indeed, by the accretiveness of A¯h , one has Jx, yh + αx − f δ ≥ x(αx − y0h − f δ ), where y0h ∈ A¯h (θX ), y h ∈ A¯h x. Hence, there exists r1 (δ, h, α) > 0 such that Jx, y h + αx − f δ ≥ 0 for all x ∈ X with x ≥ r1 . Due to Theorem 1.15.29, for every α > 0 there exists a unique solution of the equation (2.7.12) A¯h x + αx = f δ , which is denoted by xγα again. Hence, A¯h is m-accretive. Obviously, the identity operator I is also m-accretive. Then, By Theorem 1.15.22, summary operator A¯h + αI is m-accretive. Consequently, it is maximal accretive in view of Lemma 1.15.20. Thus, under conditions of this subsection, the equations (2.7.4) and (2.7.12) are equivalent, that is, xγα is a unique solution of the equation (2.7.4) as well. Since xγα is a solution of (2.7.12), there exists an element y¯αγ ∈ A¯h xγα such that y¯αγ + αxγα = f δ . Now the validity of Theorems 2.7.1, 2.7.2 and 2.7.3 is established by the same arguments as those above. Corollary 2.7.4 For the operator regularization method Ah x + α(x − u) = f δ , where u ∈ X is some fixed element, all the assertions of this section still remain valid. Moreover, the solution x ¯∗ ∈ N satisfies the inequality ¯∗ − u ≤ 0 ∀x∗ ∈ N. J(¯ x∗ − x∗ ), x

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2


4. Up to the present, the operators Ah were assumed to be accretive. This requirement is one of the determining factors to prove solvability of the regularized equation (2.7.4). Let A : X → X be a hemicontinuous accretive operator, while Ah : X → X are arbitrary hemicontinuous operators. The solvability problem for the equation (2.7.4) is open in this case. Therefore, in order to construct approximations to solutions of the equation (2.7.1), as in Section 2.6, we make use of the additional assumptions. Consider the variational inequality with small offset J(z − xγα ), Ah z + αz − f δ ≥ −g(z)z − xγα ∀z ∈ X,

(2.7.13)

where ≥ h. Lemma 2.7.5 The inequality (2.7.13) has a solution xγα for any α > 0 and for any f δ ∈ X. Proof. Indeed, the equation Ax + αx = f with accretive hemicontinuous operator A has a unique classical solution xδα , i.e., Axδα + αxδα = f δ . It follows from the accretiveness of the operator A + αI that J(z − xδα ), Az + αz − f δ ≥ 0 ∀z ∈ X. In view of the condition (2.7.3), we have J(z − xδα ), Ah z + αz − f δ ≥ −hg(z)z − xδα ∀z ∈ X. Obviously, xδα is a solution of (2.7.13) because ≥ h. Lemma 2.7.6 The following inequality holds for a solution xγα of the inequality (2.7.13): J(z − xγα ), Ah xγα + αxγα − f δ ≥ −g(xγα )z − xγα ∀z ∈ X.

(2.7.14)

Proof. Put in (2.7.13) zt = txγα + (1 − t)z, t ∈ [0, 1), in place of z ∈ X, and apply the property J(cx) = cJx of the normalized duality mapping J, where c is a constant. Then we come to the inequality J(z − xγα ), Ah zt + αzt − f δ ≥ −g(zt )z − xγα because c = 1 − t > 0. Since g(s) is continuous and Ah is hemicontinuous, we have (2.7.14) as t → 1.

Theorem 2.7.7 Let {xγα } be a solution sequence of the inequality (2.7.13). Let g(t) ≤ ¯∗ ∈ N, where x ¯∗ satisfies M t + Q, M > 0, Q > 0, (δ + )/α → 0 as α → 0. Then xγα → x (2.7.7).

2.7


147

Proof. Put in (2.7.14) z = x∗ ∈ N. Since Ax∗ = f, we have J(x∗ − xγα ), Ah xγα − Ax∗ + αxγα + f − f δ ≥ −g(xγα )x∗ − xγα ∀x∗ ∈ N. By the accretiveness of the operator A and by the condition of (2.7.3), we deduce from the previous inequality the following estimate: xγα − x∗ ≤

δ +h g(xγα ) + + x∗ . α α

Hence, δ +h (2.7.15) g(xγα ) + + 2x∗ . α α According to the hypothesis, we presume ( + h)/α ≤ K for sufficiently small α > 0 and g(xγα ) ≤ M xγα + Q. Then (2.7.15) implies the boundedness of {xγα }. Let xγα x ¯ ∈ X. Since the duality mapping J is weak-to-weak continuous, we obtain from (2.7.13) as α → 0 the inequality J(z − x ¯), Az − f ≥ 0 ∀z ∈ X. xγα ≤

From this it follows that A¯ x = f because A is a maximal accretive operator. Thus, x ¯ ∈ N. Then the proof is accomplished by the same arguments as in Theorem 2.7.1.

δ Remark 2.7.8 If ≤ K, ≤ K, K > 0, g(t) ≤ M t + Q, M > 0, Q > 0, 2KM < 1, and α α x0 is a unique solution of (2.7.1), then the sequence {xγα } weakly converges to x0 as α → 0. 5. Let X and X ∗ be uniformly convex spaces, X possess an approximation, J be a weak-to-weak continuous operator. Consider the equation (2.7.1) with hemicontinuous daccretive operator A : X → X. Suppose that its solution set N = ∅. As the regularized equation, we investigate (2.7.4) with demicontinuous d-accretive operator Ah : X → X such that D(Ah ) = D(A) = X for all h > 0. Similarly to (2.7.5), we can show that the operator T = Ah + αI is coercive. Then, by Theorem 1.16.11, the equation (2.7.4) has solutions xγα for all positive α, δ and h. This means that xγα satisfy (2.7.6). Let N = ∅ be a solution set of (2.7.1). ¯∗ ∈ N, where x ¯∗ is a unique Theorem 2.7.9 If the condition (2.1.7) holds, then xγα → x solution of (2.7.1) satisfying the inequality J x ¯∗ − Jx∗ , x ¯∗ ≤ 0 ∀x∗ ∈ N.

(2.7.16)

Proof. As in Theorem 2.7.1, we establish from (2.7.16) a uniqueness of x ¯∗ . Assume that there exists one more element x ¯∗∗ ∈ N, x ¯∗∗ = x ¯∗ , such that J x ¯∗∗ − Jx∗ , x ¯∗∗ ≤ 0 ∀x∗ ∈ N.

(2.7.17)

We can put x∗ = x ¯∗∗ in (2.7.16) and x∗ = x ¯∗ in (2.7.17) and add the obtained inequalities. Then due to Theorem 1.6.4, ¯∗∗ , x ¯∗ − x ¯∗∗ ≥ (2L)−1 δX (c−1 x∗ − x ¯∗∗ ), 0 ≥ J x ¯∗ − J x 2 ¯

148

2


where δX () is the modulus of convexity of X, 1 < L < 1.7 is the Figiel constant, c2 = x∗∗ }. Therefore, by the properties of the modulus of convexity δX (), we 2 max{1, ¯ x∗ , ¯ ¯∗∗ . conclude that x ¯∗ = x Take an arbitrary x∗ ∈ N. Using (2.7.1) and (2.7.6), it is not difficult to verify that Jxγα − Jx∗ , Ah xγα − Ax∗ + αJxγα − Jx∗ , xγα = Jxγα − Jx∗ , f δ − f .

(2.7.18)

Since the operator Ah is d-accretive, we have Jxγα − Jx∗ , Ah x∗ − Ax∗ + αxγα 2 − αJxγα , x∗ ≤ (xγα + x∗ )f δ − f . By (2.7.2) and (2.7.3), αxγα 2 − αxγα x∗ ≤ (xγα + x∗ )(δ + hg(x∗ )). Consequently, xγα 2 −

δ + hg(x∗ )

α

+ x∗ xγα − x∗

δ + hg(x∗ ) ≤ 0. α

From this quadratic inequality, we obtain the estimate xγα ≤

2δ 2h g(x∗ ) + x∗ ∀x∗ ∈ N, + α α

(2.7.19)

¯ ∈ X as α → 0. Next that is, the sequence {xγα } is bounded in X, say xγα ≤ K and xγα x we use again d-accretiveness of the operator Ah and (2.7.6) to get Jx − Jxγα , Ah x − Ah xγα = Jx − Jxγα , Ah x + αxγα − f δ ≥ 0 ∀x ∈ X. Since J is weak-to-weak continuous, we have in a limit as α → 0, Jx − J x ¯, Ax − f ≥ 0 ∀x ∈ X.

(2.7.20)

According to Theorem 1.16.6, the operator A is maximal d-accretive. Then (2.7.20) implies f = A¯ x, i.e., x ¯ ∈ N. Rewrite (2.7.18) in the following form: Jxγα − Jx∗ , Ah xγα − Ax∗ + αJxγα − Jx∗ , xγα − x∗ = Jxγα − Jx∗ , f δ − f − αJxγα − Jx∗ , x∗ . (2.7.21) Assume in (2.7.21) x∗ = x ¯. It is not difficult to deduce the estimate γ ¯) ≤ (2L)−1 δX (c−1 2 xα − x

h δ ¯∗ − Jxγα − J x ¯, x ¯ , x)Jxγα − J x ¯∗ + g(¯ Jxγα − J x α α

2.7


149

x}. By the boundedness of xγα , weak-to-weak continuity of J and where c2 = 2 max{1, K, ¯ ¯ and by (2.1.7), the right-hand side of the previous inequality weak convergence of xγα to x vanishes. The latter results from the limit relation γ δX (c−1 ¯) → 0 as α → 0. 2 xα − x

Then lim xγα − x ¯ = 0, and we obtain strong convergence of {xγα } to x ¯ as α → 0. α→0

Finally, after some algebraic transformations in (2.7.21), one gets Jxγα − Jx∗ , xγα ≤

δ

α

+

h g(x∗ ) Jxγα − Jx∗ ∗ ∀x∗ ∈ N. α

Passing here to the limit as α → 0 we come to the inequality J x ¯ − Jx∗ , x ¯ ≤ 0 ∀x∗ ∈ N. ¯∗ . The proof is This implies x ¯=x ¯∗ . Thus, the whole sequence {xγα } converges strongly to x complete. Remark 2.7.10 All the assertions of this section take place if the normalized duality mapping J is replaced by J µ .

Bibliographical Notes and Remarks The operator regularization methods were firstly studied for linear equations in a Hilbert space in [125]. Another scheme of proofs was used in [99]. The deepest results have been obtained by applying the spectral theory that is a rather powerful instrument of research into linear operators [39, 42, 99, 131]. In a Banach − but not in a Hilbert − space, duality mapping J is indeed not linear. This does not allow us to use spectral theory even for linear regularized equations of a type (2.2.4) in Banach spaces including problems with exact operators. The operator regularization methods for nonlinear equations in Banach spaces have required new approaches (see [5, 54, 232]). The results of Section 2.1 have been developed in [31, 32]. The unperturbed case, that is, convergence of x0α to x ¯∗ , was considered in [5, 68]. Convergence analysis of the operator regularization methods in Banach E-spaces was conducted in [5, 34, 201]. Recall that E-spaces include Hilbert spaces and all uniformly convex Banach spaces [16, 74, 127]. The estimates of convergence rate of regularization methods for linear equations have been established, for instance, in [37, 222, 223]. A similar estimate for nonlinear equations in Hilbert spaces was obtained in [42] and for optimization problems in [226]. The condition (2.3.7) introduced there is the analogy of a sourcewise representability of solutions defined in [37]. Another sort of requirements was found in [80, 121, 156] that imply the estimates mentioned above for the Tikhonov regularization method. Theorem 2.3.1 for nonlinear equations in Banach spaces has been proved in [202]. ¯∗ have their own importance. However, they are The estimates of the norm xγα − x also essential when the problems connected with the optimality of regularizing algorithms

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2


are solved (see, for instance, in [99, 126]). One should note the necessity to find solutions with the property described in Corollary 2.2.3 appears, for example, in planning problems [217]. The results of Section 2.4 were obtained in [18]. Convergence of the regularization method for an equation with semimonotone operators was proved in [185]. Theorem 2.6.1 has been established in [202]. Variational inequality with a small offset has been constructed and studied by Liskovets in [132]. For the class of nonlinear accretive mappings the operator regularization method was investigated in [6, 170] and in [30] for d-accretive operators. The convergence problem of the operator regularization method (2.7.4) for maximal accretive operators was also solved in [171]. The method with a small offset for accretive maps was proposed and developed in [118]. The necessary and sufficient condition for convergence of solutions of the regularized equation (2.1.5) (namely, the solvability of (2.1.1)) has been found by Maslov [144] for the case of linear operators A in Hilbert spaces. Nonlinear versions of this criterion for Banach spaces (see, for example, Theorem 2.2.5) were obtained in [5, 7, 34].

Chapter 3

PARAMETERIZATION OF REGULARIZATION METHODS It has been established in Chapter 2 that the condition (2.1.7) for positive perturbation parameters δ and h and regularization parameter α are sufficient for the operator regularization methods to be convergent to solutions of monotone and accretive operator equations. However, such a wide choice of parameters does not possess the regularizing properties in the sense of Definition 5 (see Preface). Our aim in this chapter is to indicate the ways to find the functions α = α(δ, h), which solve this problem when δ → 0 and h → 0. The following criteria of choosing the regularization parameters were widely studied for linear equations Ax = f with perturbed right-hand side f in a Hilbert space: (I) The residual principle: α = α ¯ (δ) is defined by the equation Axδα¯ − f δ = kδ, k > 1, where xδα¯ is a solution of the regularized equation Ax + α ¯x = f δ . (II) The smoothing functional principle : α = α ¯ (δ) is defined by the equation ¯ δα¯ 2 = φ(δ), Axδα¯ − f δ 2 + αx where φ(t) is a positive function of t > 0 and xδα¯ is a minimum point of the functional Φδα¯ (x) = Ax − f δ 2 + α ¯ x2 . (III) The minimal residual principle: α(δ) is defined by the equality α=α ¯ (δ) = inf {α0 | ϕ(Axδα0 − f δ ) = inf {ϕ(Axδα − f δ ) | 0 < α ≤ α1 }, xδα ∈ M }, where ϕ(t) is a positive function for all t > 0, M is an admissible class of solutions. (IV) The quasi-optimality principle: α = α ¯ (δ) is chosen as a value realizing dxδ α infα>0 α , dα

151

152

3

PARAMETERIZATION OF REGULARIZATION METHODS

where xδα is a regularized solution. If we consider equations Ax = f with arbitrary nonlinear operators A, then the criteria above are impossible [217]. However, it is possible to study this aspect for nonlinear problems with monotone and accretive operators. In this chapter we present the sufficient conditions for the parametric criteria (I) - (III) which guarantee the regularizing properties of the corresponding methods not only with perturbed right-hand side f but also with perturbed operator A. Furthermore, we answer the following significant question: whether or not the constructed function α(δ, ¯ h) satisfies the sufficient convergence conditions of the operator regularization method δ+h → 0 as δ, h → 0. α ¯ (δ, h) We establish this result for the criteria (I) - (III) to the regularization methods of Chapter 2. We also introduce and investigate the new so-called generalized residual principle for nonlinear equations with multiple-valued and discontinuous operators. As regards the quasioptimality principle (IV) for monotone equations, we refer the reader to [193] and [197].

3.1

Residual Principle for Monotone Equations ∗

1. Let X be an E-space with a strictly convex space X ∗ , A : X → 2X be a maximal monotone operator, f ∈ X ∗ . Let the equation Ax = f

(3.1.1)

have a solution set N = ∅. Consider the regularized equations Ah x + αJx = f δ , α > 0,

(3.1.2) ∗

where J : X → X ∗ is the normalized duality mapping, Ah : X → 2X are maximal monotone operators, D(Ah ) = D(A) for all h > 0 and f δ ∈ X ∗ for all δ > 0. Let xγα be a solution of (3.1.2) with γ = (δ, h). Then there exists an element yαγ ∈ Ah xγα such that yαγ + αJxγα = f δ .

(3.1.3)

Our aim is to study properties of the functions σ(α) = xγα and ρ(α) = αxγα with fixed δ and h. ∗

Lemma 3.1.1 Let X be an E-space with a strictly convex space X ∗ , A : X → 2X be a maximal monotone operator, f ∈ X ∗ . Then the function σ(α) is single-valued, continuous and non-increasing for α ≥ α0 > 0, and if θX ∈ D(A), then σ(α) → 0 as α → ∞. Proof. Single-valued solvability of the equation (3.1.2) implies the same continuity property of the function σ(α). We prove that σ(α) is continuous. Let xγβ be the solution of

3.1 Residual Principle for Monotone Equations

153

(3.1.2) with α = β in the sense of inclusion. Then there exists an element yβγ ∈ Ah xγβ such that (3.1.4) yβγ + βJxγβ = f δ . Write down the difference between (3.1.3) and (3.1.4) and then calculate the corresponding dual products on the element xγα − xγβ . We obtain yαγ − yβγ , xγα − xγβ + αJxγα − βJxγβ , xγα − xγβ = 0. Taking into account the monotonicity of Ah we deduce the inequality αJxγα − Jxγβ , xγα − xγβ + (α − β)Jxγβ , xγα − xγβ ≤ 0.

(3.1.5)

Then the property (1.5.3) of the J yields for α ≥ α0 > 0 the following relation: (xγα − xγβ )2 ≤

|α − β| γ xβ (xγα + xγβ ). α0

(3.1.6)

δ h + g(x∗ ) ∀x∗ ∈ N α α

(3.1.7)

Recall that the estimate xγα ≤ 2x∗ +

was obtained in Section 2.2. Therefore, xγα is bounded when α ≥ α0 > 0 and γ is fixed. Now the continuity of σ(α) follows from (3.1.6). Since the dual mapping J is monotone, we have from (3.1.5), (α − β)Jxγβ , xγα − xγβ ≤ 0.

(3.1.8)

Suppose that α < β. Then Jxγβ , xγβ − xγα ≤ 0. Using now the definition of J we conclude that xγβ ≤ xγα , that is, σ(β) ≤ σ(α) as β > α. Thus, the function σ(α) does not increase. We prove the last claim of the lemma. By (3.1.3), it is obvious that σ(α) =

yαγ − f δ ∗ . α

(3.1.9)

We assert that the sequence {yαγ } is bounded as α → ∞. Indeed, since θX ∈ D(A), it follows from the monotonicity condition of Ah that yαγ − y h , xγα ≥ 0,

(3.1.10)

where y h ∈ Ah (θX ). A duality mapping J is homogeneous, therefore, from (3.1.3) one has J(αxγα ) = f δ − yαγ . Under our conditions, the operator J ∗ : X ∗ → X is one-to-one and it is defined on the whole space X ∗ . According to Lemma 1.5.10, J ∗ J = IX . Hence, the latter equality implies αxγα = J ∗ (f δ − yαγ ).

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The inequality (3.1.10) can be now rewritten as follows: yαγ − y h , J ∗ (f δ − yαγ ) ≥ 0, because of α > 0. It is easy to see that yαγ − f δ 2∗ = f δ − yαγ , J ∗ (f δ − yαγ ) ≤ f δ − y h , J ∗ (f δ − yαγ ) , and we thus have the estimate yαγ − f δ ∗ ≤ yh − f δ ∗ .

(3.1.11)

Hence, the sequence {yαγ } is bounded in X ∗ . Then the equality (3.1.9) allows us to assert that σ(α) → 0 as α → ∞. The proof is accomplished. Remark 3.1.2 If θX ∈ D(A), then we choose any element u ∈ D(A) and consider the regularized equation in the form Ah x + αJ(x − u) = f δ .

(3.1.12)

In this case, all the conclusions of Lemma 3.1.1 are fulfilled for the function σ(α) = xγα − u, where xγα is a solution of the equation (3.1.12). Proposition 3.1.3 Under the conditions of the present subsection, if D(A) = D(Ah ) is a convex and closed subset in X, then xγα → x∗ as α → ∞, where xγα is a solution of (3.1.2) with fixed δ and h, and x∗ ∈ D(A) is the minimal norm vector, i.e., x∗ = min {x | x ∈ D(A)}. Proof. It is not difficult to check that the vector x∗ exists and it is defined uniquely. Choose any element x ∈ D(A) = D(Ah ). Let yh ∈ Ah x be given. Then, by (3.1.3), yαγ − y h , xγα − x + αJxγα , xγα − x = f δ − y h , xγα − x . Since Ah is monotone, we have

xγα 2 − xγα x +

f δ − y h ∗ f δ − y h ∗ x ≤ 0. − α α

From this quadratic inequality, the estimate xγα ≤ x + 2

f δ − y h ∗ α

(3.1.13)

follows for all x ∈ D(A). Hence, the sequence {xγα } is bounded when α → ∞ and γ is ¯ ∈ X as β → ∞. fixed. Then there exists a subsequence {xγβ } ⊆ {xγα } such that xγβ x


155

Furthermore, by the Mazur theorem, x ¯ ∈ D(A). Taking now into account the fact that the norm in X is weakly lower semicontinuous, we obtain from the inequality (3.1.13) ¯ x ≤ lim inf xγβ ≤ lim sup xγβ ≤ x ∀x ∈ D(A). β→∞

(3.1.14)

β→∞

Consequently, x ¯ = x∗ . Therefore, the whole sequence xγα x∗ as α → ∞. In addition, by γ (3.1.14), xα → x∗ as α → ∞. Since X is an E-space, we finally establish the strong convergence of {xγα } to x∗ as α → ∞. Lemma 3.1.4 Under the conditions of Lemma 3.1.1, the function ρ(α) = αxγα is singlevalued and continuous for α ≥ α0 > 0, and if θX ∈ D(A) = D(Ah ), then y h − f δ ∗ , lim ρ(α) = ¯

(3.1.15)

¯ y h − f δ ∗ = min {yh − f δ ∗ | y h ∈ Ah (θX )}.

(3.1.16)

α→∞

where y¯h ∈ Ah (θX ) is defined as

Proof. By Lemma 3.1.1, the function ρ(α) is single-valued and continuous. Since the value set of a maximal monotone operator at a point is convex and closed, the element y¯h is defined uniquely. It was also proved in Lemma 3.1.1 that the sequence {yαγ } satisfying (3.1.3) is bounded in X ∗ and {xγα } converges to θX as α → ∞. Therefore, there exists a subsequence of {yαγ } (we do not change its notation) such that yαγ g γ ∈ X ∗ as α → ∞. Then the inclusion gγ ∈ Ah (θX ) follows from Lemma 1.4.5. By the weak convergence of yαγ to g γ and by the inequality (3.1.11) for all y h ∈ Ah (θX ), we deduce gγ − f δ ∗ ≤ lim inf yαγ − f δ ∗ ≤ lim sup yαγ − f δ ∗ ≤ y h − f δ ∗ . α→∞

(3.1.17)

α→∞

Thus, g γ = y¯h , and the whole subsequence yαγ y¯h as α → ∞. Finally, by (3.1.17), we prove the last conclusion of the lemma.

Remark 3.1.5 If θX ∈ D(A), then we are able to consider again the equation (3.1.12) and define ρ(α) = αxγα − z 0 . In this case, all the conclusions of Lemma 3.1.4 take place. Moreover, ¯ y h − f δ ∗ = min {y h − f δ ∗ | y h ∈ Ah (z 0 )}. Remark 3.1.6 If X ∗ is an E-space, then yαγ → y¯h as α → ∞. This assertion follows from (3.1.17) and from the proven above weak convergence of {yαγ } to y¯h . We shall further study the behavior of the functions σ(α) = xγα and ρ(α) = αxγα as α → ∞, where xγα is a solution of (3.1.2) and θX ∈ D(A). Let D(A) be a closure of D(A). According to Theorem 1.7.17, D(A) is a convex set in X. Rewrite (3.1.3) in the equivalent form yαγ − y h + αJxγα − αJx + αJx = f δ − y h , (3.1.18)

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where y h ∈ Ah x, x ∈ D(A) = D(Ah ). Then (yαγ + αJxγα ) − (y h − αJx), xγα − x + αJx, xγα − x = f δ − y h , xγα − x . Owing to the monotonicity of Ah +αJ, it is not difficult to deduce from (3.1.18) the following estimate: f δ − y h ∗ γ (3.1.19) xα − x. Jx, xγα − x ≤ α By (3.1.7), the sequence {xγα } is bounded as α → ∞, therefore xγα x ¯ ∈ X. Then we obtain in a limit Jx, x ¯ − x ≤ 0 ∀x ∈ D(A). (3.1.20) Since, by Theorem 1.3.20, J is demicontinuous in X, the inequality (3.1.20) holds with all ¯ ∈ D(A). Then using Lemma x ∈ D(A). Moreover, it results from the Mazur theorem that x 1.11.4 and Theorem 1.11.14, we conclude that x ¯ is the element x∗ ∈ D(A) with minimal norm defined uniquely in X. If x∗ = θX and x∗ ∈ D(A) then similarly to the proof of Proposition 3.1.3, we find that xγα → x∗ as α → ∞. Hence, lim ρ(α) = +∞. (3.1.21) α→∞

Assume now that x∗ = θX and, consequently, x∗ ∈ D(A). Show by contradiction that in this case (3.1.21) also holds. Let the function ρ(α) = αxγα = yαγ − f δ ∗ be bounded as α → ∞. Then xγα → 0, and on account of xγα x∗ = θX , we conclude that xγα → θX in an E-space. Furthermore, boundedness of the sequence {yαγ } allows us to assert that yαγ y˜γ ∈ X ∗ . Then y˜γ ∈ Ah (θX ) because maximal monotone operators Ah are demiclosed. This contradicts the assumption θX ∈ D(A). Let now x∗ = θX and x∗ ∈ / D(A). Then due to the proved weak convergence of xγα to x ¯ ∈ D(A) and by the weak lower semicontinuity of the norm in any Banach space, we obtain

0 < ¯ x ≤ lim inf xγα . α→∞

Hence, (3.1.21) also holds in this case. Thus, the following assertion is established: / D(A), then property (3.1.21) Lemma 3.1.7 Under the conditions of this subsection, if θX ∈ of the function ρ(α) holds, and lim σ(α) = x∗ = min{x | x ∈ D(A)}.

α→∞

2. As it has been already mentioned, the regularized equation Ah x + αJ µ x = f δ ,

(3.1.22)

where J µ : X → X ∗ is the duality mapping with a gauge function µ(t), is of interest. We are going to study the properties of the functions σµ (α) = µ(xγα ) and ρµ (α) = αµ(xγα ), where xγα is a solution of (3.1.22).


157

Lemma 3.1.8 The function σµ (α) is single-valued, continuous and non-increasing when α ≥ α0 > 0. In addition, if θX ∈ D(A), then σµ (α) → 0 as α → ∞, and if θX ∈ D(A), then σµ (α) → µ(x∗ ) as α → ∞. Proof. The operator J µ is strictly monotone. Consequently, a solution xγα of the equation (3.1.22) for each α > 0 is unique. Then the function σµ (α) is single-valued. The inequalities (3.1.5) and (3.1.6) accept, respectively, the following forms: αJ µ xγα − J µ xγβ , xγα − xγβ + (α − β)J µ xγβ , xγα − xγβ ≤ 0 and

µ(xγα ) − µ(xγβ ) (xγα − xγβ ) ≤

(3.1.23)

|α − β| µ(xγβ )(xγα + xγβ ), α0

because of the property (1.5.1) of J µ . In Section 2.2 we proved from the inequality (2.2.8) that the sequence {xγα } is bounded as α ≥ α0 > 0. Then the continuity of the function σµ (α) as α ≥ α0 > 0 is guaranteed by the properties of the function µ(t). In view of the inequality (3.1.23), we establish, as in Lemma 3.1.1, that σµ (α) is a non-increasing function. Let yαγ ∈ Ah xγα satisfy the equality yαγ + αJ µ xγα = f δ .

(3.1.24)

Then

f δ − yαγ ∗ . (3.1.25) α ∗ Replacing in the proof of Lemma 3.1.1 normalized duality mappings J and J by J µ and (J ν )∗ , respectively (see Lemma 1.5.10), one can show that the sequence {yαγ } is bounded when α → ∞ and γ is fixed. Write down the monotonicity property of Ah for the points xγα ∈ D(A) and θX ∈ D(A) : σµ (α) =

yαγ − y h , xγα ≥ 0 ∀y h ∈ Ah (θX ). Then (3.1.24) yields the inequality (yαγ − y h ), (J ν )∗ (α−1 (f δ − yαγ )) ≥ 0, α > 0. By making use of the definition of (J ν )∗ , we obtain (3.1.11). Finally, (3.1.25) allows us to conclude that σµ (α) → 0 as α → ∞. Similarly to the proof of Lemma 3.1.7, one can be sure that if θX ∈ D(A) then σµ (α) → µ(x∗ ) as α → ∞. The lemma is proved. We now address the function ρµ (α). First of all, the last lemma enables us to state that ρµ (α) is single-valued and continuous. The limit relations (3.1.15) and (3.1.21) for ρµ (α) are verified in the same way as in Lemmas 3.1.4 and 3.1.7. We present the final result. Lemma 3.1.9 If θX ∈ D(A), then lim ρµ (α) = ¯ y h − f δ ∗ ,

α→∞

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where y¯h is the nearest point to f δ in the set Ah (θX ). If θX ∈ D(A), then lim ρµ (α) = +∞.

α→∞

Remark 3.1.10 By (1.5.8), we can rewrite (3.1.24) as follows: yαγ + α

µ(xγα ) γ Jxα = f δ , xγα

xγα = θX .

Hence, the solution of (3.1.22) coincides with the solution of the equation Ah x + α Jx = f δ , where α = α

µ(xγα ) . xγα

Still, as it has already been mentioned, the possibility of obtaining different estimates for regularized solutions essentially depends on a choice of the function µ(t). 3. Let A : X → X ∗ be a maximal monotone hemicontinuous operator and θX ∈ D(A). Suppose that the equation (3.1.1) is given with an exact operator A and perturbed righthand side f δ satisfying the condition f − f δ ∗ ≤ δ,

(3.1.26)

such that in reality we solve the equation Ax = f δ . Consider the regularized equation Ax + αJx = f δ .

(3.1.27)

Denote by xδα its classical solution. Then Axδα + αJxδα = f δ .

(3.1.28)

Definition 3.1.11 The value ρ(α) = Axδα − f δ ∗ is called the residual of the equation Ax = f δ on the solution xδα of the equation (3.1.27). It follows from (3.1.28) that ρ(α) = Axδα − f δ ∗ = αxδα .

(3.1.29)

By Lemma 3.1.4, the residual ρ(α) is single-valued, continuous and lim ρ(α) = A(θX ) − f δ ∗ .

α→∞

(3.1.30)

In view of (3.1.7), the estimate of solution xδα to the equation (3.1.27) has the form xδα ≤ 2x∗ +

δ α

∀x∗ ∈ N.


159

Then (3.1.29) yields the relation ρ(α) ≤ 2α¯ x∗ + δ, where x ¯∗ ∈ N is the minimal norm solution of the equation (3.1.1). Let α be such that 2α¯ x∗ < (k − 1)δ p , k > 1, p ∈ (0, 1]. We may consider, without loss of generality, that δ ≤ 1 because δ → 0. Hence, ρ(α) ≤ (k − 1)δ p + δ < kδ p .

(3.1.31)

Assume that δ satisfies the inequality A(θX ) − f δ ∗ > kδ p , p ∈ (0, 1], k > 1.

(3.1.32)

Since (3.1.30) holds and since ρ(α) is continuous, we get from (3.1.31) and (3.1.32) that there exists at least one α = α ¯ (δ) such that ρ(¯ α) = αx ¯ δα¯ = kδ p , and α ¯>

k > 1,

p ∈ (0, 1],

(k − 1)δ p . 2¯ x∗

(3.1.33)

(3.1.34)

For every δ > 0, we find α = α ¯ from the scalar equation (3.1.33) solving at the same time (3.1.27) with α = α ¯ . Thus, we construct the sequence {xδα¯ }. We study its behaviour as δ → 0 and at the same time the properties of the function α ¯ (δ). At the beginning, let p ∈ (0, 1). By (3.1.34), it is easy to deduce the estimate xδα¯ =

2k¯ x∗ ρ(¯ α) x∗ 2kδ p ¯ . = ≤ k−1 α ¯ (k − 1)δ p

¯ ∈ X as δ → 0. Furthermore, it Hence, the sequence {xδα¯ } is bounded, therefore, xδα¯ x follows from the rule of choosing α ¯ and from Definition 3.1.11 that Axδα¯ − f δ ∗ = kδ p → 0

as

δ → 0.

By Lemma 1.4.5, then x ¯ ∈ N. ¯∗ = θX . Now Taking into account (3.1.26) and (3.1.32), we obtain that θX ∈ N. Thus, x we find by (3.1.34) that x∗ 2δ1−p ¯ δ . ≤ k−1 α ¯ Consequently, δ → 0 as δ → 0. (3.1.35) α ¯ Show that x ¯=x ¯∗ . As in the proof of Theorem 2.2.1, one gets

Jx∗ , xδα¯ − x∗ ≤

δ δ x − x∗ α ¯ α¯

∀x∗ ∈ N.

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It leads to the inequality ¯ − x∗ ≤ 0 Jx∗ , x

∀x∗ ∈ N,

¯ ∈ N as δ → 0. In addition, we proved in Theorem 2.2.1 because the sequence {xδα¯ } x that x ¯=x ¯∗ and xδα¯ → ¯ x∗ . This implies strong convergence of the sequence {xδα¯ } to x ¯∗ as δ → 0 in the E-space X. Since ¯ x∗ = 0, there exists κ > 0 such that xδα¯ ≥ κ for sufficiently small δ. Then from the equality kδ p (3.1.36) α ¯= δ xα¯ it results that α ¯ ≤ κ−1 kδ p . It is easy to see that α ¯ → 0 as δ → 0. Let now in (3.1.33) p = 1, that is, ρ(α) = kδ, k > 1. In this case

2¯ x∗ δ ≤ . α ¯ k−1 Assume that set N is a singleton and N = {x0 }. Repeating the previous arguments we obtain that {xδα¯ } weakly converges to x0 as δ → 0. Then x0 ≤ lim inf xδα¯ , δ→0

and from (3.1.32) we conclude that x0 = θX . Hence, the estimate xδα¯ ≥ κ > 0 holds for all δ ∈ (0, 1], perhaps, excepting their finite number. Then (3.1.36) guarantees that α ¯ → 0 as δ → 0. Thus, the following theorem (the residual principle for equations with maximal monotone hemicontinuous operators) has been proved: Theorem 3.1.12 Assume that X is an E-space with strictly convex dual space X ∗ , A : X → X ∗ is a maximal monotone hemicontinuous operator, θX ∈ D(A), the equation (3.1.1) has a nonempty solution set N with unique minimal norm solution x ¯∗ . Consider the regularized equation (3.1.27) with the conditions (3.1.26) and (3.1.32), 0 < δ < 1. Then there is at least one α = α ¯ satisfying (3.1.33), where xδα¯ is the solution of (3.1.27) with α = α ¯ . In δ δ ∗ → 0; addition, let δ → 0. It results that: 1) α ¯ → 0; 2) if p ∈ (0, 1), then xα¯ → x ¯ and α ¯ 3) if p = 1, N = {x0 }, that is, the equation (3.1.1) has a unique solution, then xδα¯ x0 δ ≤ C. and there exists C > 0 such that α ¯

Remark 3.1.13 A similar theorem can be established for the regularized equation (3.1.2) with a perturbed operator Ah (see also Section 3.3). Remark 3.1.14 Due to Lemma 3.1.7, if θX ∈ D(A), then the assumption (3.1.32) in Theorem 3.1.12 is not necessary.


161

Theorem 3.1.12 presents the residual principle for nonlinear equations with monotone hemicontinuous operators. It there asserts that the choice of regularization parameter according to the residual principle gives the regularizing algorithm. Moreover, it satisfies the sufficient convergence conditions of the operator regularization methods investigated in Chapter 2. Let the sequence {xδα¯ } constructed by the residual principle converge to some element x ∈ X as δ → 0 either strongly or weakly. Then Axδα¯ → f as δ → 0 in view of (3.1.28) and (3.1.33). Therefore, f ∈ A¯ x. This means that the equation (3.1.1) has a solution. Consequently, solvability of (3.1.1) is the necessary and sufficient convergence condition of the sequence {xδα¯ }. Next we show that the question arises if Theorem 3.1.12 is also true with k = 1. This problem is not only of theoretical but also of practical value, for instance, in the cases when the measurement accuracy of the right-hand side f of (3.1.1) cannot be made less than some δ0 > 0. We look for the conditions under which the given problem is positively solved. Theorem 3.1.15 Let X be an E-space with a strictly convex dual space X ∗ , A : X → X ∗ be a maximal monotone hemicontinuous operator, θX ∈ D(A) and the inequality A(θX ) − f δ ∗ > δ

(3.1.37)

hold. Assume that there exists a number r > 0 such that for for all x ∈ D(A) with x ≥ r, one has Ax − f, x ≥ 0. (3.1.38) Then there is at least one α ¯ > 0 satisfying the equation ρ(¯ α) = αx ¯ δα¯ = δ.

(3.1.39)

Moreover, if the equation (3.1.1) has a unique solution x0 , i.e., N = {x0 } and in (3.1.38) δ ≤ C as the strict inequality appears, then xδα¯ x0 and there exists C > 0 such that α ¯ δ → 0. Proof. Note first that θX ∈ N because of (3.1.37). By (3.1.26), it is not difficult to be sure that δ + Ax − f, x . Ax + αJx − f δ , x ≥ αx x − α

Therefore, solutions xδα of the equation (3.1.27) are bounded. Due to Theorem 1.7.9, there exists γ¯ = max{δα−1 , r} such that xδα ≤ γ¯ , and the estimate γ ρ(α) = αxδα ≤ α¯ δ holds. If γ¯ = r then the inequality ρ(α) < δ is satisfied for sufficiently small α. If γ¯ = α then ρ(α) ≤ α¯ γ = δ.

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Since ρ(α) is continuous, (3.1.37) implies solvability of the equation (3.1.39). Now we may ¯ and study its construct the sequence {xδα¯ } of solutions to the equation (3.1.27) with α = α behavior as δ → 0. Show that {xδα¯ } is bounded. For this end, assume that xδα¯ → ∞ as δ → 0. If δ is sufficiently small then we have 0 = Axδα¯ + α ¯ Jxδα¯ − f δ , xδα¯ ¯ δα¯ 2 = Axδα¯ − f, xδα¯ + f − f δ , xδα¯ + αx > α ¯ xδα¯ 2 − δxδα¯ = 0, where α ¯ xδα¯ = δ. We thus come to a contradiction. At the same time, the last equality δ means that there exists C > 0 such that ≤ C as δ → 0. Then the weak convergence of α ¯ xδα¯ to x0 is established as in Theorem 3.1.12.

Observe that under the conditions of Theorem 3.1.15 it is impossible to prove that δ ¯ xδα¯ → 0 and xδα¯ → x0 when δ → 0. This happens by reason of δ = α simultaneously α ¯ and x0 = θX . Theorem 3.1.16 Assume that conditions of Theorem 3.1.15 hold. If 0 < δ ≤ 1 and, instead of (3.1.37), the inequality A(θX ) − f δ ∗ > δ p , p ∈ (0, 1), is given, then the residual principle for choosing regularization parameter α ¯ from the equaδ → 0 as δ → 0. α ¯

¯∗ and tion ρ(¯ α) = δ p takes place and it produces the convergence xδα¯ → x Consider again the regularized equation Ax + αJ µ x = f δ with duality mapping

Jµ

. If

xδα

(3.1.40)

is its solution then the residual

ρµ (α) = Axδα − f δ ∗ = αµ(xδα ). We are not able to study the residual principle for the operator regularization method (3.1.40) with an arbitrary gauge function µ(t). However, we can do it if µ(t) = ts , s ≥ 1. Denote by J s the duality mapping with this gauge function and by ρs (α) the residual ρµ (α), that is, ρs (α) = Axδα − f δ ∗ = αxδα s . Theorem 3.1.17 Suppose that the conditions of Theorem 3.1.12 are fulfilled. Consider the equation (3.1.40) with 0 < δ ≤ 1 as regularized to Ax = f δ . If xδα is its solution, then there exists at least one value α = α ¯ such that ρs (¯ α) = Axδα¯ − f δ ∗ = kδ p , k > 1,

p ∈ (0, 1].

(3.1.41)

3.2 Residual Principle for Accretive Equations

163

δ → 0; α ¯ 3) if p = 1 and the equation (3.1.1) is uniquely solvable, i.e., N = {x0 }, then xδα¯ x0 and δ ≤ C as δ → 0. there exists C > 0 such that α ¯ ¯∗ and Moreover, let δ → 0. It results that: 1) α ¯ → 0; 2) if p ∈ (0, 1), then xδα¯ → x

Proof. According to (2.2.10), for all x∗ ∈ N, xδα ≤ τ x∗ +

δ κ

α

, τ > 1, τ s ≥ 2, κ =

1 . s

Then x∗ + δ κ . ρκs (α) = ακ xδα ≤ τ ακ ¯ Hence, the relation τ ακ ¯ x∗ ≤ (kκ − 1)δκp , k > 1, p ∈ (0, 1], holds with sufficiently small α, and ρs (α) < kδ p . Then, by the condition (3.1.32) and by the continuity of ρs (α), there exists at least one α = α ¯ such that (3.1.41) is true (see Lemmas 3.1.8, 3.1.9). The final proof repeats the reasoning given in Theorem 3.1.12.

3.2

Residual Principle for Accretive Equations

In what follows, A : X → X is a hemicontinuous accretive operator, D(A) = X, X is a reflexive strictly convex space together with its dual space X ∗ , J : X → X ∗ is a continuous and, at the same time, weak-to-weak continuous duality mapping and Banach space X possesses an approximation. Study the residual principle for this case. Let N be a nonempty solution set of the equation (3.1.1) in the classical sense. Assume that perturbed operators Ah : X → X are accretive and hemicontinuous, D(Ah ) = D(A) for all h > 0, f δ ∈ X for all δ > 0 and f δ − f ≤ δ. Consider the regularized equation Ah x + αx = f δ .

(3.2.1)

Let xγα , γ = (δ, h), be its unique solution (see Sections 1.15 and 2.7), that is, Ah xγα + αxγα = f δ .

(3.2.2)

We study the functions σ(α) = xγα and ρ(α) = αxγα . Lemma 3.2.1 The function σ(α) is single-valued and continuous for α ≥ α0 > 0. Moreover, σ(α) → 0 as α → ∞.

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Proof. Obviously, σ(α) is single-valued because xγα is a unique solution of (3.2.1) for each α ≥ α0 > 0. Let xγβ be a solution of (3.2.1) with α = β. Then Ah xγβ + βxγβ = f δ .

(3.2.3)

By (3.2.2) and (3.2.3), we have J(xγα − xγβ ), Ah xγα − Ah xγβ + J(xγα − xγβ ), αxγα − βxγβ = 0. The accretiveness property of Ah implies J(xγα − xγβ ), Ah xγα − Ah xγβ ≥ 0. Therefore, αJ(xγα − xγβ ), xγα − xγβ + (α − β)J(xγα − xγβ ), xγβ ≤ 0. From this inequality, one gets (xγα − xγβ )2 ≤ xγα − xγβ 2 ≤

|α − β| γ xβ (xγα + xγβ ). α0

We proved in Section 2.7 (see (2.7.10)) that the sequence {xγα } is bounded when γ → 0 and α ≥ α0 > 0. It results from the last inequality that the function σ(α) is continuous. In its turn, the equality (3.2.2) yields σ(α) =

Ah xγα − f δ α

(3.2.4)

and αJxγα = J(f δ − Ah xγα ). Since D(A) = X and

Ah

(3.2.5)

are accretive operators, we have Jxγα , Ah xγα − Ah (θX ) ≥ 0.

(3.2.6)

Then J(f δ − Ah xγα ), Ah xγα − Ah (θX ) ≥ 0 in view of (3.2.5). It is easy to see that this inequality leads to the estimate Ah xγα − f δ ≤ Ah (θX ) − f δ ,

(3.2.7)

that is, the sequence {Ah xγα − f δ } is bounded. Obviously, the last assertion of the lemma arises from (3.2.4). Lemma 3.2.2 The function ρ(α) is single-valued and continuous for α ≥ α0 > 0. Moreover, lim ρ(α) = Ah (θX ) − f δ . (3.2.8) α→∞

3.2 Residual Principle for Accretive Equations

165

Proof. It follows from Lemma 3.2.1 that the function ρ(α) is single-valued and continuous. The limit (3.2.8) is established on the basis of (3.2.7), as in Lemma 3.1.4. In the case of linear accretive operators, it is possible to obtain some important additional properties of the functions σ(α) and ρ(α). Indeed, a solution of (3.2.1) can be represented as xγα = Th f δ , where Th = (Ah + αI)−1 . It is well known that the operator Th exists for all α > 0, continuous, bounded and |Th | ≤ α−1 . Then σ(α) ≤ |Th |f δ ≤ α−1 f δ → 0 as α → ∞. By (3.2.2) and (3.2.3), we have (Ah + βI)(xγα − xγβ ) = −(α − β)xγα , from which one gets

xγα − xγβ

α−β

= −(Ah + βI)−1 xγα .

(3.2.9)

If β → α, then the limit of the right-hand side of (3.2.9) exists, hence, there exists a limit of the left-hand side. Therefore, as β → α, (3.2.9) implies dxγα = −(Ah + αI)−1 xγα dα or

(3.2.10)

dxγ dxγα + α α = −xγα . dα dα

Ah

Then the equality

dxγ α

J

dα

, Ah

dxγ

dxγ dxγ dxγα + α J α , α = − J α , xγα dα dα dα dα

is satisfied. In view of the accretiveness of Ah , we deduce that dxγ α ≤ α−1 xγα

dα

and

dxγ α

J

By (3.2.10),

Then the estimate

dα

dxγ 2 α ≤ 0.

, xγα ≤ −α

dα

(3.2.11)

dσ 2 (α) = −2Jxγα , (Ah + αI)−1 xγα . dα dσ 2 (α) 2σ 2 (α) 2xγα 2 =− ≥− α α dα

(3.2.12)

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holds. Consequently, the function σ(α) is continuous and if α ≥ α0 > 0 then σ(α) ≥

α0 σ(α0 ) . α

Write now the obvious equalities dσ 2 (α) dσ 2 (α) dρ2 (α) . = 2ασ 2 (α) + α2 = α 2σ2 (α) + α dα dα dα

(3.2.13)

Since (3.2.12) holds, it is clear that dρ2 (α) ≥ 0. dα

We finally obtain that the function ρ(α) for linear accretive operators Ah does not decrease as α ≥ α0 > 0. Observe that in Hilbert space dxγ 2 dσ 2 (α) ≤ −2α α ≤ 0. dα dα

This follows from (3.2.11). Hence, the function σ(α) does not increase as α ≥ α0 > 0. Considering the regularized equation Ax + αx = f δ

(3.2.14)

with an accretive operator A and denoting by xδα its solution, one can give the same Definition 3.1.11 of the residual to the equation Ax = f δ on a solution xδα , namely, ρ(α) = Axδα − f δ . The following residual principle for equations with accretive operators is valid: Theorem 3.2.3 Let A : X → X be a hemicontinuous accretive operator with D(A) = X, X be a reflexive strictly convex space with strictly convex dual space X ∗ , J : X → X ∗ be a continuous and weak-to-weak continuous duality mapping in X. Assume that Banach space X possesses an approximation, δ ∈ (0, 1] and A(θX ) − f δ > kδ p ,

k > 1,

p ∈ (0, 1].

Then there exists at least one α = α ¯ satisfying the equation ρ(¯ α) = Axδα¯ − f δ = kδ p ,

(3.2.15)

where xδα¯ is the solution of the equation (3.2.14) with α = α ¯ . Furthermore, let δ → 0. One δ ¯∗ , where x ¯∗ ∈ N is the unique → 0 and xδα¯ → x has: 1) α ¯ → 0; 2) if p ∈ (0, 1), then α ¯ solution of the inequality J(¯ x∗ − x∗ ), x ¯∗ ≤ 0 ∀x∗ ∈ N ;

3) if p = 1 and N = {x0 }, then xδα¯ x0 and there exists C > 0 such that

δ ≤ C. α ¯

3.3 Generalized Residual Principle

167

Proof. By the equations (3.1.1) and (3.2.14), we have for all x∗ ∈ N, J(xδα − x∗ ), Axδα − Ax∗ + αJ(xδα − x∗ ), xδα − x∗ = J(xδα − x∗ ), f δ − f − αJ(xδα − x∗ ), x∗ . Taking into consideration the accretiveness of A, one gets xδα − x∗ 2 ≤

δ δ x − x∗ + x∗ xδα − x∗ α α

or xδα − x∗ ≤

δ + x∗ . α

Consequently, xδα ≤

δ + 2x∗ . α

Due to Lemmas 3.2.1 and 3.2.2, the theorem is proved by use of the same arguments as in Theorem 3.1.12 for the monotone case.

3.3

Generalized Residual Principle

1. Let X be an E-space with strictly convex dual space X ∗ . Assume that the equation ∗ (3.1.1) with maximal monotone operator A : X → 2X is solved, N is its nonempty solution set and the sequence {f δ } of elements f δ ∈ X ∗ , δ > 0, and the sequence {Ah } of maximal ∗ monotone, possible multiple-valued, operators Ah : X → 2X , h > 0, are given. Besides, h assume that D(A ) = D(A), (3.1.26) holds and HX ∗ (Ax, Ah x) ≤ hg(x)

∀x ∈ D(A),

(3.3.1)

where g(t) is a continuous non-negative and increasing function for all t ≥ 0. This means that in reality we solve the equation Ah x = f δ .

(3.3.2)

Study the regularized equation written as Ah x + αJx = f δ ,

(3.3.3)

where J is the normalized duality mapping. Let xγα be a unique solution of (3.3.3) with γ = (δ, h). Then Lemmas 3.1.1 and 3.1.4 enable us to propose the following definition: Definition 3.3.1 The value ρ(α) = αxγα is called the generalized residual of the equation (3.3.2) in the solution xγα of the equation (3.3.3).

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3


The generalized residual is a single-valued and continuous function of the parameter α, though operators Ah are not continuous, in general. Moreover, it follows from (3.3.3) that there exists an element yαγ ∈ Ah xγα such that ρ(α) = yαγ − f δ ∗ . To state the residual principle for such operators, it is necessary to evaluate the functional xγα from above. In Section 2.2, the following estimate has been obtained: xγα ≤ 2x∗ +

δ h + g(x∗ ) ∀x∗ ∈ N. α α

(3.3.4)

Consequently, ρ(α) = yαγ − f δ ∗ ≤ 2αx∗ + δ + hg(x∗ ). By the same way as in the proof of Theorem 3.1.12, we are able to determine α ¯ from the scalar equation

ρ(¯ α) = k + g(x∗ ) (δ + h)p , k > 1,

p ∈ (0, 1].

The quantity x∗ is not known. However, it may happen that an estimate x∗ ≤ c is known. In this case, using the properties of g(t), we find that ρ(α) ≤ 2αc + δ + hg(c). Then it is possible to define α = α ¯ as a solution of the equation

ρ(¯ α) = k + g(c) (δ + h)p , k > 1, p ∈ (0, 1]. If c is not known then we act in the following way. First of all, recall that the estimate (3.3.4) was obtained in Section 2.2 from the equality yαγ − f, xγα − x∗ + αJxγα , xγα − x∗ = f δ − f, xγα − x∗ ∀x∗ ∈ N, applying the monotonicity property of Ah . If we use in the same equality the monotonicity of A, then we can write

xγα 2 − xγα x∗ +

δ δ h h − g(xγα )x∗ − x∗ ≤ 0. g(xγα ) + α α α α

From this it follows that xγα ≤ 2x∗ +

h δ + g(xγα ) ∀x∗ ∈ N. α α

(3.3.5)

This inequality plays an important role to establish the next theorem (generalized residual principle).


169

∗

∗

Theorem 3.3.2 Let A : X → 2X and Ah : X → 2X be maximal monotone operators, h > 0, D(A) = D(Ah ), the conditions (3.1.26) and (3.3.1) hold and 0 < δ + h ≤ 1. If θX ∈ D(A), then additionally assume that the following inequality is fulfilled:

y∗h − f δ ∗ > k + g(0) (δ + h)p , k > 1,

p ∈ (0, 1],

(3.3.6)

where y∗h is the nearest to f δ element from Ah (θX ) and g(t) is a continuous non-negative and increasing function for t ≥ 0. Then there exists a unique solution α ¯ of the equation

ρ(¯ α) = k + g(xγα¯ ) (δ + h)p ,

(3.3.7)

where xγα¯ is the solution of (3.3.3) with α = α ¯ . Moreover, let γ → 0. It results: 1) α ¯ → 0; δ+h γ γ ∗ ¯ and → 0; 3) if p = 1 and N = {x0 }, then xα¯ x0 and 2) if p ∈ (0, 1), then xα¯ → x α ¯ δ+h ≤ C. there exists C > 0 such that α ¯

Proof. By (3.3.5), we have ρ(α) = yαγ − f δ ∗ = αxγα ≤ 2αx∗ + δ + hg(xγα ), where yαγ ∈ Ah xγα and

yαγ + αJxγα = f δ .

Take α so small that the inequality 2α¯ x∗ < (k − 1)(δ + h)p , k > 1, p ∈ (0, 1],

(3.3.8)

is satisfied. Then ρ(α) < (k − 1)(δ + h)p + δ + hg(xγα )

≤ (k − 1)(δ + h)p + 1 + g(xγα ) (δ + h)p

=

k + g(xγα ) (δ + h)p .

(3.3.9)

Construct the function

d(α) = ρ(α) − k + g(xγα ) (δ + h)p . Owing to the continuity of g(t) and Lemmas 3.1.1 and 3.1.2, d(α) is also continuous for α ≥ α0 > 0. By the same lemmas, if θX ∈ D(A) then

lim d(α) = y∗h − f δ ∗ − k + g(0) (δ + h)p .

α→∞

Applying the condition (3.3.6), we come to the conclusion that lim d(α) > 0.

α→∞

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3


At the same time, by (3.3.9), there exists α > 0 for which d(α) < 0. Since d(α) is continuous, there exists at least one α ¯ which satisfies (3.3.7). If θX ∈ D(A), then the residual property (3.1.21) appears and, by virtue of Lemma 3.1.7, lim xγα = x∗ = min{x | x ∈ D(A)}. α→∞

Hence, lim d(α) = +∞.

α→∞

Thus, in this case, as well, the existence problem of α ¯ satisfying the equation (3.3.7) is solved positively. Show by contradiction that α ¯ is unique. Suppose that for given γ there are α ¯ and β¯ such that ρ(¯ α) = k + g(xγα¯ ) (δ + h)p (3.3.10) and

¯ = k + g(xγ¯ ) (δ + h)p , ρ(β) β xγα¯

(3.3.11)

xγβ¯

¯ where and are solutions of the regularized equation (3.3.3) with α = α ¯ and α = β, respectively. This means that the following equalities hold: ¯ Jxγα¯ = f δ , yαγ¯ + α

yαγ¯ ∈ Ah xγα¯ ,

(3.3.12)

¯ γ¯ = f δ , yβγ¯ + βJx β

yβγ¯ ∈ Ah xγβ¯.

(3.3.13)

and xγα¯

xγβ¯

¯ Therefore, assume further that xγα¯ = xγ¯. If = then (3.3.10) and (3.3.11) imply α ¯ = β. β Then, according to Corollary 1.5.3, there are uniquely defined elements e∗α¯ and e∗β¯ such that e∗α¯ ∗ = e∗β¯∗ = 1, e∗α¯ , xγα¯ = xγα¯ ,

e∗β¯, xγβ¯ = xγβ¯.

(3.3.14)

Using (3.3.10) - (3.3.13) we calculate ¯ γ¯, xγα¯ − xγ¯ = ¯ ¯ γ¯e∗¯, xγα¯ − xγ¯ ¯ αJxγα¯ − βJx αxγα¯ e∗α¯ − βx β β β β β = k(δ + h)p e∗α¯ − e∗β¯ , xγα¯ − xγβ¯ + (δ + h)p g(xγα¯ )e∗α¯ − g(xγβ¯)e∗β¯, xγα¯ − xγβ¯ .

(3.3.15)

By (3.3.14), it is not difficult to verify that e∗α¯ − e∗β¯, xγα¯ − xγβ¯ ≥ 0. Since the function g(t) increases, we deduce by applying again (3.3.14) that

g(xγα¯ )e∗α¯ − g(xγβ¯)e∗β¯, xγα¯ − xγβ¯ ≥ g(xγα¯ ) − g(xγβ¯) (xγα¯ − xγβ¯) > 0.


171

Consequently, the expression in the right-hand side of (3.3.15) is strictly positive. Next, (3.3.12) and (3.3.13) imply ¯ γ¯, xγα¯ − xγ¯ = 0. αJxγα¯ − βJx yαγ¯ − yβγ¯ , xγα¯ − xγβ¯ + ¯ β β Hence, yαγ¯ − yβγ¯ , xγα¯ − xγβ¯ < 0, yαγ¯ ∈ Ah xγα¯ , xγβ¯ ∈ Ah xγβ¯, ¯ is thus proved. and this contradicts the monotonicity of Ah . The uniqueness of α Note that the uniqueness proof of α ¯ was done above, in fact, by the hypothesis that solutions xγα¯ and xγβ¯ of the regularized equation (3.3.3) are not θX . Let now, for example, xγα¯ = θX be given. Then, by (3.3.3), there exists y h ∈ Ah (θX ) such that y h − f δ ∗ = 0. But this contradicts the condition (3.3.6) of the theorem. Observe that (3.3.8) yields the estimate α ¯> therefore,

k−1 (δ + h)p , 2¯ x∗

δ+h 2¯ x∗ (δ + h)1−p . ≤ k−1 α ¯

Hence, if p ∈ (0, 1) then

δ+h = 0. α ¯ In its turn, if p = 1 then there exists a constant C > 0 such that lim

γ→0

lim

γ→0

δ+h ≤ C, α ¯

δ + h

is bounded. It follows from the proof of Theorem 2.2.1 that α ¯ {xγα¯ } is also bounded as γ → 0. Then there exists a subsequence (we do not change its denotation) which weakly converges to some x ¯ ∈ X. Since (3.3.12) holds, the equation (3.3.7) enables us to obtain the strong convergence of yαγ¯ to f. Write down the monotonicity condition of Ah :

that is, the sequence

y h − yαγ¯ , x − xγα¯ ≥ 0 ∀x ∈ D(A), ∀y h ∈ Ah x. By (3.3.1), after passing in the latter inequality to the limit as γ → 0, we come to the inequality y − f, x − x ¯ ≥ 0 ∀x ∈ D(A), ∀y ∈ Ax. This means that x ¯ ∈ N. Thus, the conclusion 3) of the theorem is completely proved. If p ∈ (0, 1) then the assertion 2) is guaranteed by the proof of Theorem 2.2.1. Show now that x ¯∗ = θX . Indeed, assume that is not the case. Then, by the approximate data, we obtain for some yh ∈ Ah (θX ) the following: y h − f δ ∗ ≤ y h − f ∗ + f − f δ ∗ ≤ hg(0) + δ,

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3


where f ∈ A(θX ). Thus, we arrive at a contradiction with the condition (3.3.6). Since x ¯∗ = θX , there exists c > 0 such that xγα¯ ≥ c for sufficiently small γ. Observe that here we made use again of the weak lower semicontinuity of the norm in a Banach space. Then α ¯=

k + g(xγα¯ ) k + g(xγα¯ ) (δ + h)p ≤ (δ + h)p . γ xα¯ c

Consequently, α ¯ → 0 as γ → 0 because {xγα¯ } is bounded and g(t) is continuous. The proof of the theorem is now accomplished. Remark 3.3.3 If operator A is given exactly, i.e., h = 0 and if g(t) is bounded, then the residual principle takes the form (3.1.33). If A is strictly monotone, then α ¯ in Theorems 3.1.12, 3.1.15 and 3.1.17 is uniquely defined. In this case, the residual ρ(α) = yαδ − f δ ∗ is the increasing function of α. 2. As a rule, the parameter α ¯ is defined solving approximately the scalar equation (3.3.7). In this way, the value of α ¯ can be found inexactly. Therefore, it is desirable to establish a continuous dependence of the regularized solution xγα on perturbation of α with fixed γ. Let α → β. Using the fact that the function σ(α) = xγα is continuous, we obtain the convergence of xγα to xγβ as α ≥ α0 > 0 and β ≥ α0 > 0. By Lemma 3.1.1, the ¯β as α → β. Since xγα satisfies the sequence {xγα } is bounded for α ≥ α0 > 0. Hence, xγα x equation (3.3.3), the monotonicity condition of the operator Ah + αJ gives the following inequality: yh + αJx − f δ , x − xγα ≥ 0 ∀x ∈ D(A), ∀y h ∈ Ah x. Going to the limit when α → β we deduce ¯β ≥ 0 ∀x ∈ D(A), y h + βJx − f δ , x − x

∀y h ∈ Ah x.

It results from this relation that x ¯β is a solution of the equation Ah x + βJx = f δ .

(3.3.16)

¯β = xγβ and xγα xγβ . We know that (3.3.16) has a unique solution xγβ . Consequently, x γ Finally, in the E-space X we have the strong convergence of xα to xγβ as α → β, provided that α ≥ α0 > 0 and β ≥ α0 > 0. Thus, solutions of the regularized equation (3.3.3) are stable with respect to errors of the regularization parameter α. 3. Consider now the equation Ah x + αJ s+1 x = f δ ,

(3.3.17)

where J s+1 : X → X ∗ is a duality mapping with the gauge function µ(t) = ts , s ≥ 1. Theorem 3.3.4 If the condition (3.3.6) of Theorem 3.3.2 is replaced by

s

y∗h − f δ ∗ > k¯ + (1 + g(0))κ (δ + h)p , k¯ > 0,

κ = 1/s,

p ∈ (0, 1],


173

then there exists a unique α ¯ satisfying the equation

s

α) = αx ¯ γα¯ s = k¯ + (1 + g(xγα¯ ))κ (δ + h)p , ρs (¯ where xγα¯ is the solution of (3.3.17) with α = α ¯ . Moreover, the conclusions 1) - 3) of Theorem 3.3.2 hold. Proof. By analogy with (2.2.10), the following estimate is valid: xγα ≤ τ x∗ +

where τ > 1, τ s ≥ 2, s ≥ 1. Hence,

δ

α

+

κ h ∀x∗ ∈ N, g(xγα ) α

s

ρs (α) = αxγα s ≤ ακ τ ¯ x∗ + (δ + hg(xγα ))κ . If we choose small enough α to satisfy the relation ¯ + h)pκ , k¯ > 0, p ∈ (0, 1], x∗ < k(δ ακ τ ¯

(3.3.18)

then ρs (α) < ≤

¯ + h)pκ + (δ + hg(xγ ))κ k(δ α

s

s

k¯ + (1 + g(xγα ))κ (δ + h)p .

Consequently, there exists α > 0 such that

s

ρs (α) < k¯ + (1 + g(xγα ))κ (δ + h)p . Now the proof does not differ greatly from that of the previous theorem. ∗

∗

4. Let A : X → 2X and Ah : X → 2X be arbitrary monotone operators and we assume that the conditions of Theorem 2.2.4 are satisfied excepting (2.1.7). We study the ∗ regularized equation (3.3.3). Let A¯h : X → 2X be maximal monotone extensions of Ah . Then by Lemma 1.9.8, (3.3.3) is equivalent to the equation A¯h x + αJx = f δ ,

(3.3.19)

if a solution of (3.3.3) is understood in the generalized sense. Definition 3.3.1 allows us to construct the generalized residual principle for general monotone, even discontinuous, operators, as well. Let xγα be a solution of (3.3.19) in the sense of inclusion. Then there exists an element y¯αγ ∈ A¯h xγα such that y¯αγ + αJxγα = f δ . In this case, the generalized residual for equations Ah x = f δ in the point xγα can be defined as ρ(α) = αxγα

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3


because yαγ − f δ ∗ . αxγα = ¯ If Ah is a maximal monotone operator then y¯αγ = yαγ ∈ Ah xγα and ρ(α) = yαγ −f δ ∗ . Though, in general, operator Ah is discontinuous, the residual ρ(α) is always continuous and has all the other properties established by Lemmas 3.1.4 and 3.1.7. Moreover, y∗h ∈ A¯h (θX ) and y∗h − f δ is the element with minimal norm of the set A¯h (θX ) − f δ . Then the generalized residual principle for equations with arbitrary monotone operators follows from the theorems obtained for maximal monotone operators. ∗

∗

Theorem 3.3.5 Let A : X → 2X and Ah : X → 2X , h > 0, be monotone operators, D(A) be a convex closed set, int D(A) = ∅, A¯ and A¯h be maximal monotone extensions ¯ = D(A¯h ) = D(A), the conditions (3.1.26) of A and Ah , respectively. Assume that D(A) and (2.6.3) are fulfilled for all x ∈ D(A) and 0 < δ + h ≤ 1. If θX ∈ D(A), then assume additionally that (3.3.6) is satisfied, where y∗h is the element nearest to f δ in the set A¯h (θX ). Then there exists α ¯ such that (3.3.7) holds, where xγα¯ is the generalized solution of (3.3.3) ¯∗ and with α = α. ¯ Moreover, let γ → 0. It results: 1) α ¯ → 0; 2) if p ∈ (0, 1), then xγα¯ → x δ+h → 0; 3) if p = 1 and N = {x0 }, then xγα¯ x0 and there exists a constant C > 0 such α ¯ δ+h that ≤ C. α ¯

¯ = Consider the regularized equation (3.1.12), where some u ∈ D(A) and D(A) = D(A) D(A¯h ). Suppose that the proximity of operators A¯ and A¯h is defined by the inequality ¯ A¯h x) ≤ hg(x − u) ∀x ∈ D(A). HX ∗ (Ax,

(3.3.20)

Let in the previous theorem, in place of (3.3.6) and (3.3.7), the following relations be satisfied: (3.3.21) y0h − f δ ∗ > k + g(u) (δ + h)p , where y0h − f δ is a vector with minimal norm in the set {y − f δ | y ∈ A¯h u}, and

ρ(¯ α) = k + g(xγα¯ − u) (δ + h)p . Then all the conclusions 1) - 3) of Theorem 3.3.5 hold, where the element x ¯∗ is defined as ¯ x∗ − u = min{x − u | x ∈ N }. 5. Realizing the operator regularization method numerically and choosing parameter α from the generalized residual principle, it is important to know an estimate of α ¯ from above. We study this problem for the equation (3.1.1) with a maximal monotone operator ∗ A : X → 2X and the regularized equation with duality mapping J s . Consider first the equation with exact operator A : Ax + αJ s+1 (x − u) = f δ , s ≥ 1,

u ∈ X.

(3.3.22)


175

By analogy with (3.1.41), we apply the generalized residual principle in the form α) = yαδ¯ − f δ ∗ = α ¯ xδα¯ − us = kδ p , ρs (¯

(3.3.23)

where xδα¯ is the solution of (3.3.22) with α = α ¯ , p ∈ (0, 1), k > 1 and yαδ¯ ∈ Axδα¯ such that yαδ¯ + αJ ¯ s+1 (xδα¯ − u) = f δ . We suppose that all the requirements of Theorem 3.1.17 are satisfied. Then the sequence {xδα¯ } strongly converges to x ¯∗ . It was shown in [220, 221] that the condition “A acts from X to X ∗ ” imposes a special restriction on the growth order of Ax∗ . For example, y∗ ≤ ζ(x − u) ∀x ∈ D(A), ∀y ∈ Ax,

(3.3.24)

where ζ(t) is a non-negative continuous and increasing function for t ≥ 0. ∗

Theorem 3.3.6 Suppose that A : X → 2X is a maximal monotone operator, the assumptions (3.1.26) and (3.3.24) are carried out, the regularization equation has the form (3.3.22) and regularization parameter α ¯ is defined by (3.3.23). Then the estimate α ¯≤

kδp

s

ζ −1 (|f δ ∗ − kδ p |)

(3.3.25)

holds, where ζ −1 (s) is the function inverse to ζ(t). Proof. By (3.3.23), we find that α ¯=

kδ p . − us

xδα¯

(3.3.26)

Evaluate xδα¯ − u from below. It is easy to see that the hypothesis (3.3.24) implies ζ(xδα¯ − u) ≥ yαδ¯ ∗ ≥ |f δ ∗ − yαδ¯ − f δ ∗ | = |f δ ∗ − kδ p |. Since ζ(t) increases, the latter inequality gives the estimate xδα¯ − u ≥ ζ −1 (|f δ ∗ − kδ p |). Then the result (3.3.25) follows from (3.3.26). Remark 3.3.7 If in the conditions of Theorem 3.3.6, X is a Hilbert space, s = 1 and u = θX , then we can write (3.3.22) for an arbitrary linear operator A in the following form: A∗ Ax + αx = A∗ f δ . Therefore,

α ¯ xδα¯ = A∗ (Axδα¯ − f δ ).

176

3


Since (3.3.23) is true, we have α ¯ ≤ |A∗ |

Axδα¯ − f δ kδ p = |A| δ . δ xα¯ xα¯

Hence, α ¯ ≤ |A|

kδ p ζ −1 (|f δ

∗

− kδ p |)

.

6. Suppose now that a maximal monotone operator A in (3.1.1) is also given with some error depending on the parameter h > 0, such that in reality the equation Ah x = f δ is solved, D(A) = D(Ah ), the conditions (3.1.26) and (3.3.1) hold, regularized solutions are found from the equation Ah x + αJ s+1 (x − u) = f δ , s ≥ 1,

(3.3.27)

and the regularization parameter α ¯ is defined as follows: ρs (¯ α) = αx ¯ γα¯ − us = yαγ¯ − f δ ∗ = d(xγα¯ − u)(δ + h)p . Here

(3.3.28)

s 1 d(t) = k¯ + (1 + g(t))κ , κ = , k¯ > 0, p ∈ (0, 1), γ = (δ, h), s

¯ , yαγ¯ ∈ Ah xγα¯ such that xγα¯ denotes a solution of (3.3.27) with α = α ¯ s+1 (xγα¯ − u) = f δ . yαγ¯ + αJ Assuming that y h ∗ ≤ ζ(x − u) ∀h > 0, ∀x ∈ D(A),

∀yh ∈ Ah x,

(3.3.29)

where ζ(t) is a non-negative continuous and increasing function for t ≥ 0, and going through the same arguments, as in the proof of Theorem 3.3.6, we come to the inequality d(xγα¯ − u)(δ + h)p s . α ¯≤ ζ −1 (|f δ ∗ − d(xγα¯ − u)(δ + h)p |)

(3.3.30)

Evaluate from above the quantity of xγα¯ − u. According to (3.3.18), α ¯>

s k¯ (δ + h)p , τ ¯ x∗ − u

where τ > 1 and τ s ≥ 2. Let ¯ x∗ − u = min {x − u | x ∈ N } > 0 and c¯ be a constant such that

c¯ ≥ ¯ x∗ − u.

(3.3.31)


177

Then α ¯≥

k ¯ s

τ c¯

(δ + h)p .

Since the function σ(α) = xγα − u is decreasing, one gets xγα¯ − u ≤ xγα∗ − u, where α∗ =

k ¯ s

c¯τ

(δ + h)p .

If d(x∗ − u)(δ + h)p > f δ ∗ then, by virtue of the equalities lim xγα − u = x∗ − u = min{x − u | x ∈ D(A)}

α→∞

(see Lemma 3.1.7) and the properties of σ(α) and g(t), the estimate (3.3.30) takes the following form: d(xγα∗ − u)(δ + h)p s . (3.3.32) α ¯≤ ζ −1 (|f δ ∗ − d(x∗ − u)(δ + h)p |) Next, if d(xγα∗ − u)(δ + h)p < f δ ∗ then it follows from (3.3.30) that α ¯≤

d(xγα∗ − u)(δ + h)p

s .

ζ −1 (|f δ ∗ − d(xγα∗ − u)(δ + h)p |)

(3.3.33)

Thus, we are able to state the following theorem. Theorem 3.3.8 Under the conditions and denotations of this subsection, suppose that ˜ y∗h − f δ ∗ > d(x∗ − u)(δ + h)p , where ˜ y∗h − f δ ∗ = min{y − f δ ∗ | y ∈ Ah x∗ }. If d(x∗ − u)(δ + h)p > f δ ∗ , then estimate (3.3.32) holds. If f δ ∗ > d(xγα∗ − u)(δ + h)p , then (3.3.33) is satisfied.

178

3


Observe that the estimates (3.3.32) and (3.3.33) become most effective when maximal monotone operators A and Ah with D(A) = D(Ah ) are close in the uniform metric, that is, HX ∗ (Ax, Ah x) ≤ h ∀x ∈ D(A).

(3.3.34)

Then there exists a constant k1 > 2 such that α ¯≤

k1 (δ + h)p ζ −1 (|f δ ∗ − k1 (δ + h)p |)

s .

7. Assume that the conditions of Theorem 3.3.5 are fulfilled, an operator A is strictly monotone and given exactly. As it was noted in Remark 3.3.3, in this case the function ρ(α) is single-valued, continuous and increasing, and the value of α ¯ is defined by the equality ρ(¯ α) = αx ¯ δα¯ = kδ p . We introduce the function ϕ(α) =

kδ p . xδα

Then α ¯ = ϕ(¯ α). Due to the properties of the function σ(α), we conclude that ϕ(α) is a continuous, single-valued and increasing function. Thus, the equation ϕ(α) = α has the unique root α = α ¯ . Furthermore, ϕ(α) < α as α > α ¯ and ϕ(α) > α as α < α ¯ . Consider the method of successive approximations αn = ϕ(αn−1 )

(3.3.35)

with an arbitrary initial approximation α0 . It is well-known that the inequality |ϕ (α)| < 1 is its sufficient convergence condition. Let σ(α) be a differentiable function. Then ρ (α) = σ(α) + ασ (α) > 0, from which we obtain that σ (α) ≥ − Then 0 < ϕ (α) = −

σ(α) . α

ϕ(α) kδ p kδ p σ (α) = . ≤ 2 ασ(α) α σ (α)

¯ . Hence, method (3.3.35) converges. It is clear that 0 < ϕ (α) < 1 as α > α Consider Example 2.2.8. Suppose that there is a function g(t) = ctp−1 + 1 with some constant c > 0. In order to find the regularization parameter α, ¯ we have the equation

γ p−1 κ ¯ α ¯ uγα¯ p−1 ¯ 1,p ) 1,p = k + (2 + cuα

p−1

(δ + h)σ ,

(3.3.36)

1 , σ ∈ (0, 1], p ≥ 2, and uγα¯ is the solution of the equation (2.2.13) p−1 as α = α ¯ . By the properties of the functions g0h and g1h , we establish the inequality where k¯ > 0, κ =

h Ah u−1,q ≤ c1 up−1 1,p + ω ,

3.4 Modified Residual Principle

179

where a constant c1 > 0 and p−1 + q −1 = 1. Thus,

d(t) = k¯ + (2 + ctp−1 )κ

p−1

, ζ −1 (s) = c1 −κ (s − ωh )κ ,

in (3.3.27) u = θX , and the estimates (3.3.32) and (3.3.33) follow. Assume that for all x ∈ Ω and for all ξ ≥ 0, |g0h (x, ξ 2 )ξ − g0 (x, ξ 2 )ξ| ≤ and |ω − ω h | ≤

In this case, g(t) ≡ 1. If then α ¯≤

h 2

h . 2

|f δ ∗ − k1 (δ + h)σ | − ω h > 0, k1 c1 (δ + h)σ , k1 > 2, |f δ ∗ − k1 (δ + h)σ | − ω h

σ ∈ (0, 1].

If in the equation (2.2.13), instead of J p u, we take normalized dual mapping p Ju = u2−p 1,p J u,

then (3.3.36) should be replaced by the following equation:

σ α ¯ uγα¯ = k + cuγα¯ p−1 1,p (δ + h) ,

where k > 2,

σ ∈ (0, 1],

c > 0, p > 1.

Remark 3.3.9 Using constructions of Section 3.2, we are able to obtain the generalized residual principle for multiple-valued and discontinuous accretive operators A and Ah .

3.4

Modified Residual Principle

Up to the present, we have chosen the regularization parameter α from the equations (3.1.33) and (3.3.7). However, it is possible to choose it from some inequalities. We will show how it may be done. Theorem 3.4.1 Assume that the conditions of Theorem 3.1.12 for the equation (3.1.1) and (3.1.27) are fulfilled and there is a number r > 0 such that for all x ∈ D(A) with x ≥ r, Ax − f, x ≥ 0.

(3.4.1)

Then there exists α ¯ satisfying the inequalities δp ≤α ¯ ≤ max{α | ρ(α) ≤ δ p }, k > 1, p ∈ (0, 1), 0 < δ < 1. kr ¯∗ , Moreover, if δ → 0, then xδα¯ → x

δ → 0 and α ¯ → 0. α ¯

(3.4.2)

180

3


Proof. It is easy to check that

Ax + αJx − f δ , x ≥ Ax − f, x + αx x −

δ . α

If x ≥ r1 = max{r, δ/α}, then Ax + αJx − f δ , x ≥ 0 ∀x ∈ D(A). A solution of regularized equation (3.1.27) satisfies the inequality xδα ≤ r1 in view of Theorem 1.7.9. Then there is α such that ρ(α) < δ p . Beside this, if r1 = r then this δ δp inequality holds for all α < , while if r1 = then ρ(α) < δ p for all α > 0. Consequently, α r we proved that the parameter α = α ¯ choice is possible by (3.4.2). Then the left inequality of (3.4.2) gives δ ≤ krδ 1−p . α ¯ δ Hence, → 0 as δ → 0. α ¯ Show that the residual ρ(α) = αxδα is non-decreasing. Let

ρ(α) = αxδα = γ1 , αJxδα = γ1 e∗α ,

ρ(β) = βxδβ = γ2 , βJxδβ = γ2 e∗β ,

e∗α ∗ = e∗β ∗ = 1, e∗α , xδα = xδα ,

e∗β , xδβ = xδβ .

It is not difficult to verify by Corollary 1.5.3 that e∗α − e∗β , xδα − xδβ ≥ 0.

(3.4.3)

Let β < α, but γ2 > γ1 . As in the proof of Theorem 3.3.2, since xδα and xδβ are solutions of the equation (3.1.27) with regularization parameters α and β, respectively, the following equality holds: Axδα − Axδβ , xδα − xδβ + γ1 e∗α − e∗β , xδα − xδβ + (γ1 − γ2 )e∗β , xδα − xδβ = 0. The monotonicity of A and (3.4.3) imply (γ1 − γ2 )e∗β , xδα − xδβ ≤ 0. Since γ2 > γ1 , one gets

e∗β , xδβ − xδα ≤ 0.

Hence, xδβ ≤ xδα . By Lemma 3.1.1, we have that β ≥ α. Thus, we come to the contradiction. Consequently, the claim is proved. Therefore, ρ(¯ α) ≤ δ p .

3.5 Minimal Residual Principle

181

¯∗ as δ → 0, where x ¯∗ ∈ N is a Next, due to Theorem 3.1.12, we deduce that xδα¯ → x minimal norm solution of (3.1.1). Moreover, since α ¯≤

δp and x ¯∗ = θX , xδα¯

we obtain convergence of α ¯ to 0 as δ → 0.

Remark 3.4.2 A non-decreasing property of the residual established in the last theorem holds, as well, in the case of approximately given operators A. We are able to study in Theorem 3.4.1 general maximal monotone possibly multiplevalued mappings. If, in place of A, a sequence {Ah }, h > 0, of maximal monotone operators is given, D(A) = D(Ah ) and (3.3.34) holds, then the solution sequence {xγα¯ } of the equation ¯ = f δ, Ah x + αJx where α ¯ is chosen from the inequalities (δ + h)p ≤α ¯ ≤ max{α | ρ(α) ≤ (δ + h)p }, k > 1, kr

p ∈ (0, 1),

strongly converges to x ¯∗ ∈ N. Moreover, if θX ∈ N and δ, h → 0 then α ¯ → 0 and

δ+h → 0. α ¯

If we omit the condition (3.4.1) in Theorem 3.4.1, then (3.4.2) is necessarily replaced as follows: (k1 − 1)δ p ≤α ¯ ≤ max{α | ρ(α) ≤ k1 δ p }, kc where k1 > 1, k > 1, p ∈ (0, 1), and a constant c such that ¯ x∗ ≤ c. If the proximity ¯ must between operators A and Ah is defined by (3.3.1), then the criterion determining α be taken as

(k1 − 1)(δ + h)p ≤α ¯ ≤ max{α | ρ(α) ≤ [k1 + g(xγα )] (δ + h)p }. kc

3.5

Minimal Residual Principle ∗

Let A : X → 2X be a maximal monotone operator, D(A) ⊆ X, the equation (3.1.1) have ¯∗ ∈ N a nonempty solution set N and θX ∈ N. We find approximations to an element x by the regularized equation (3.1.27) with a right-hand side f δ , δ > 0, and consider that δ is not known. Construct the single-valued continuous residual ρ(α) = αxδα , where xδα is a unique solution of the equation (3.1.27). Parameter α = α0 is defined by the following equality: α | ρ(¯ α) = inf {ρ(α) | α > 0, xδα ∈ M }}, (3.5.1) α0 = inf {¯

182

3


where M ⊆ D(A) is an admissible set of solutions. We assume that M is bounded and x ¯∗ ∈ int M. Suppose further that f δ ∈ R(A). Otherwise, if f δ ∈ R(A) then, due to Theorem 2.2.5, the sequence {xδα } is bounded in X as α → 0. Therefore, by (3.5.1), α0 = 0. But such a choice of α is not acceptable in (3.1.27). Observe that the set ¯ Λ = {α | xδ ∈ M, 0 < δ < δ} α

is not empty. This follows from Theorem 2.2.1, which implies an existence of the sequence ¯∗ . Thus, xδα ∈ M for a small enough α. {xδα } such that xδα → x According to Lemma 3.1.7, if θX ∈ D(A), then lim ρ(α) = ∞,

α→∞

and if θX ∈ D(A) then

(3.5.2)

lim ρ(α) = y∗ − f δ ∗ ,

α→∞

where y∗ ∈ A(θX ) and y∗ − f δ ∗ = min {y − f δ ∗ | y ∈ A(θX )}. Prove that ρ(α) = 0 as α = 0. Let this claim be not true. Then the equality ρ(α) = 0 appears only if xδα = θX . In this case, f δ ∈ A(θX ), which contradicts the assumption that f δ ∈ R(A). Then the inequality (3.1.11) implies for α > 0, ρ(α) = yαδ − f δ ∗ ≤ y∗ − f δ ∗ , where yαδ ∈ Axδα , and

yαδ + αJxδα = f δ .

(3.5.3)

Taking into account (3.5.2), we get that the parameters α ¯ and α0 cannot approach infinity. ¯ then there exists a unique Lemma 3.5.1 Under the conditions of this section, if 0 < δ < δ, α0 > 0 satisfying (3.5.1). Proof. The uniqueness of α0 is obvious. Let {αn } be a minimizing sequence for ρ(α), that is, ¯ , ρ(αn ) → inf {ρ(α) | α > 0, xδα ∈ M }. αn → α Show that α ¯ > 0. Indeed, if α ¯ = 0 then xδα → ∞ because f δ ∈ R(A). Hence, beginning with a certain number n, an element xδαn ∈ M. This contradicts (3.5.1). Since ρ(α) is continuous for α > 0, one gets that ρ(αn ) → ρ(¯ α). Let now α ¯ n → α0 and ρ(¯ αn ) = inf {ρ(α) | α > 0, xδα ∈ M }. Reasoning by contradiction, as above, we establish that α0 > 0.

3.6 Smoothing Functional Principle

183

We study the behavior of the sequences {α0 (δ)} and {xδα0 } as δ → 0, where α0 (δ) are defined by (3.5.1) and xδα0 are solutions of (3.1.27) with α = α0 (δ). It follows from (3.5.1) ¯ ∈ X as δ → 0. It is known that there exists that {xδα0 } is bounded, therefore, xδα0 x a sequence {α(δ)} → 0 such that ρ(α) = αxδα → 0 and xδα → x ¯∗ with α = α(δ). Since ∗ δ x ¯ ∈ int M, we have the inclusion: xα ∈ M if δ is small enough. Well then, all the more yαδ 0 − f δ ∗ → 0 as δ → 0, where yαδ 0 satisfies (3.5.3) with α = α0 . Thus, yαδ 0 → f. This limit relation together with ¯ and with demiclosedness property of A allow us to assert that x ¯ ∈ N. one xδα0 x Show that α0 (δ) → 0 as δ → 0. Let α0 (δ) → α∗ = 0. We already proved that ρ(α0 ) = α0 xδα0 → 0. ¯ = θX , which contradicts the fact that θX ∈ N. Thus, we have Then xδα0 → θX . Thus, x proved the following theorem. Theorem 3.5.2 Under the conditions of this section, any weak accumulation point of the sequence {xδα0 }, where α0 is defined by (3.5.1), is a solution of the equation (3.1.1). Moreover, α0 (δ) → 0 and ρ(α0 ) → 0 as δ → 0.

3.6

Smoothing Functional Principle

1. We solve the equation (3.1.1) with a monotone hemicontinuous and potential operator ¯∗ is its A : X → X ∗ . Assume that D(A) = X, a solution set N of (3.1.1) is nonempty, x minimal norm solution, and, as before, f δ is an approximation of f. Let ω(x) be a potential of A, i.e., A = grad ω. Then regularized equation (3.1.27) is equivalent to the minimization problem of the functional Φαδ (x) = ω(x) − f δ , x +

α x2 ∀x ∈ X, 2

α > 0.

(3.6.1)

In the terms of variational regularization methods, (3.6.1) is a smoothing functional for the equation Ax = f δ . Let xδα be a unique minimum point of (3.6.1), which coincides with solution of the equation (3.1.27). We introduce the following denotations: F δ (x) = ω(x) − f δ , x , F 0 (x) = ω(x) − f, x , m = min {F 0 (x) | x ∈ X}, mδ (α) = min {Φαδ (x) | x ∈ X}, hδ = inf {F δ (x) | x ∈ X}. Observe that

x∗ , hδ ≤ inf {F 0 (x∗ ) + δx∗ | x∗ ∈ N } ≤ m + δ¯

and assume, without loss of generality, that

Fδ

(x) ≥ 0 for all x ∈ X.

(3.6.2)

184

3


Lemma 3.6.1 The function mδ (α) is continuous and non-decreasing for all α ≥ 0, lim mδ (α) = hδ

α→0+

and lim mδ (α) = ω(θX ).

α→∞

Moreover, if hδ < ω(θX ), then this function is increasing on the interval (0, αδ∗ ), where αδ∗ = sup {α | mδ (α) < ω(θX )}. Next we take functions ψ(δ) and C(δ) with the following properties: m + δC(δ) ≤ ψ(δ) < ω(θX )

lim ψ(δ) = m,

δ→0

(3.6.3)

and C(δ) → ∞ as δ → 0.

(3.6.4)

ω(θX ) > m + δ¯ x∗

(3.6.5)

Theorem 3.6.2 Let

and the function ψ(δ) satisfy (3.6.3) and (3.6.4). Then there exists a unique α ¯ = α(δ) > 0 which is defined by the following equation of the smoothing functional principle: α) = ψ(δ). mδ (¯

(3.6.6)

¯ Furthermore, the sequence {xδα¯ } of minimal points of the functional Φαδ (x) with α = α converges strongly to x ¯∗ as δ → 0. Proof. By (3.6.5), (3.6.2) and by Lemma 3.6.1, we conclude that the function mδ (α) is continuous and increasing as α ∈ (0, αδ∗ ). It follows from (3.6.3) and (3.6.4) that there exists a unique positive root of equation (3.6.6). Prove that regularized solutions xδα¯ strongly converge to x ¯∗ . Since |F δ (x) − F 0 (x)| ≤ δx, we obtain by (3.6.6) the inequality F 0 (xδα¯ ) − δxδα¯ +

α ¯ δ 2 α ¯ x ≤ F δ (xδα¯ ) + xδα¯ 2 = ψ(δ). 2 α¯ 2

Owing to the inequality F 0 (xδα¯ ) ≥ m, one gets α ¯ δ 2 x ≤ δxδα¯ + ψ(δ) − m. 2 α¯

(3.6.7)

From this quadratic inequality, the estimate "

xδα¯

δ ≤ + α ¯

2

δ α ¯

+

2(ψ(δ) − m) α ¯

(3.6.8)


185

holds. By (3.6.3) and (3.6.6), we deduce α) ≤ Φαδ¯ (¯ x∗ ) = F δ (¯ x∗ ) + m + δC(δ) ≤ ψ(δ) = mδ (¯ ≤ F 0 (¯ x∗ ) + δ¯ x∗ +

α ¯ ∗ 2 ¯ x 2

α ¯ ∗ 2 α ¯ ∗ 2 x∗ + ¯ x . ¯ x = m + δ¯ 2 2

(3.6.9)

Without loss of generality, one can consider that C(δ) − ¯ x∗ > 0. Therefore, ¯ x∗ 2 δ . ≤ 2(C(δ) − ¯ x∗ ) α ¯

δ → 0 as δ → 0. Moreover, from (3.6.9) follows Then, since C(δ) → ∞ as δ → 0, we have α ¯ the inequality δ ∗ ¯ x ∗ 2 ψ(δ) − m x . + ¯ ≤ α ¯ 2 α ¯

By (3.6.8), we have a boundedness of the sequence {xδα¯ } as δ → 0. Show that α ¯ → 0 as δ → 0. For this end, make use of inequality (3.6.8). The right-hand side of it vanishes as δ → 0 if α ¯ → 0. We prove by contradiction that lim xδα¯ = 0.

δ→0

Let xδα¯ → θX . Since the potential ω of A is weakly lower semicontinuous, we can write F 0 (θX ) ≤ lim F 0 (xδα¯ ). δ→0

(3.6.10)

On the other hand, F 0 (xδα¯ ) ≤ F δ (xδα¯ ) + δxδα¯ ≤ ψ(δ) + δxδα¯ , therefore, lim sup F 0 (xδα¯ ) ≤ m. δ→0

Then, in view of (3.6.10), we obtain that F 0 (θX ) ≤ m. Thus, ω(θX ) ≤ m. Taking into account the condition (3.6.5), we arrive at the contradiction. Consequently, limδ→0 xδα¯ = 0. By reason of (3.6.7), this implies the convergence α ¯ → 0 as δ → 0. Then the last assertion of the theorem being proved follows from Theorem 2.2.1. Remark 3.6.3 Under the conditions of Theorem 3.6.2, if the equation (3.1.1) has a unique solution x0 and the relations in (3.6.3) is replaced by the inequality m + Cδ ≤ ψ(δ) < ω(θX ),

C > x0 ,

then the weak convergence of {xδα¯ } to x0 follows when δ → 0.

186

3


Remark 3.6.4 The assumptions of Theorem 3.6.2 such that A is hemicontinuous and D(A) = X may be omitted if we consider maximal monotone or arbitrary monotone operators with domains which not necessarily coincide with the whole of space X, understanding solutions in the sense of Definitions 1.7.2 and 1.9.3, respectively. Then, applying the results of Section 2.2, we may establish the smoothing functional principle for these cases, as well. 2. We study now the smoothing functional principle with approximately given operators A. Instead of ω, let a sequence of functionals {ω h } with D(ω) = D(ω h ) = X, h > 0, be known. Assume that ω h have the same properties as ω and the inequality |ω(x) − ω h (x)| ≤ η(x)h ∀x ∈ X

(3.6.11)

holds, where η(t) is non-negative, continuous and increasing for all t ≥ 0. Then the equation (3.1.2) with Ah x = grad ω h (x) is equivalent to the minimization problem of the smoothing functional α (3.6.12) Φαγ (x) = ωh (x) − f δ , x + x2 , γ = (δ, h), α > 0. 2 Denote a unique solution of the problems (3.1.2) and minimal point of (3.6.12) by xγα , and assume F γ (x) = ω h (x) − f δ , x , mγ (α) = min {Φαγ (x) | x ∈ X}, hγ = inf {F γ (x) | x ∈ X}. Similarly to (3.6.2), we have hγ ≤ inf {F 0 (x∗ ) + hη(x∗ ) + δx∗ | x∗ ∈ N } x∗ . ≤ m + hη(¯ x∗ ) + δ¯ The assertions like Lemma 3.6.1 and Theorem 3.6.2 are as follows: Lemma 3.6.5 A function mγ (α) is continuous and non-decreasing for α ≥ 0, and lim mγ (α) = hγ ,

α→0+

lim mγ (α) = ω h (θX ).

α→∞

Moreover, if hγ < ωh (θX ), then mγ (α) is increasing for α ∈ (0, αγ∗ ), where αγ∗ = sup {α | mγ (α) < ω h (θX )}. Theorem 3.6.6 Suppose that a function ψ(γ) satisfies the conditions: lim ψ(γ) = m,

γ→0

m + (δ + h)C(γ) ≤ ψ(γ) < ω h (θX ),

(3.6.13)

where C(γ) → ∞ as γ → 0. Furthermore, let the inequality (5.6.3) hold, and ω h (θX ) > m + δ¯ x∗ + hη(¯ x∗ ),

(3.6.14)


187

where the function η(t) possesses the property lim sup t→∞

η(t) = M < ∞. t2

(3.6.15)

Then there exists a unique α ¯ > 0 such that mγ (¯ α) = ψ(γ),

(3.6.16)

and the sequence {xγα¯ } strongly converges to x ¯∗ as γ → 0. Proof. By the condition (3.6.14) and Lemma 3.6.5, the function mγ (α) is continuous and increasing on the interval (0, αγ∗ ). Hence, (3.6.13) implies the first assertion of the theorem. Similarly to (3.6.7), we deduce the estimate xγα¯ 2 ≤ 2

ψ(γ) − m

α ¯

+

h δ γ x + η(xγα¯ ) . α ¯ α ¯ α¯

(3.6.17)

Applying (3.6.13) and (3.6.16), one gets the following relations: m + (δ + h)C(γ) ≤ ψ(γ) = mγ (¯ α) ≤ Φαγ¯ (¯ x∗ ) = F γ (¯ x∗ ) +

α ¯ ∗ 2 α ¯ ∗ 2 x x∗ ) + δ¯ x∗ + hη(¯ x∗ ) + ¯ ¯ x ≤ F 0 (¯ 2 2

= m + δ¯ x∗ + hη(¯ x∗ ) + ≤ m + (δ + h)c1 +

α ¯ ∗ 2 ¯ x 2

α ¯ ∗ 2 ¯ x , 2

(3.6.18)

where c1 = max{¯ x∗ , η(¯ x∗ )}. Since lim C(γ) = ∞, one can regard that C(γ) − c1 > 0. γ→0

Then the inequality

¯ x∗ 2 δ+h ≤ 2(C(γ) − c1 ) α ¯

is valid, and δ+h → 0 as α ¯ → 0. α ¯

Moreover, it results from (3.6.18) that ¯ x∗ 2 δ + h ψ(γ) − m + c1 . ≤ 2 α ¯ α ¯ Taking into account the condition (3.6.15) and estimate (3.6.17), we conclude that the sequence {xγα¯ } as γ → 0 is bounded. As in the proof of Theorem 3.6.2, we next deduce that α ¯ → 0 when γ → 0. The last assertion follows from the sufficient convergence condition for the operator regularization method in Theorem 2.2.1.

188

3


Remark 3.6.7 Under the conditions of Theorem 3.6.6, if the equation (3.1.1) has a unique solution x0 , C(γ) = C > c1 in (3.6.13) and M

x∗ 2 < 1, C − c1

then the weak convergence of {xγα¯ } to x0 follows as γ → 0. 3. Theorem 3.6.6 imposes a growth condition of the function η(t). It may be omitted if we consider the smoothing functional in the form Φαγ (x) = ω h (x) − f δ , x + αΦ(x), where Φ(x) is defined by the equality (1.5.4). In view of Lemma 1.5.7, Φ (x) = J µ (x). We suppose that the gauge function µ(t) of duality mapping J µ is such that Φ(t) ≥ η(t). Under these conditions, strong convergence is proved following the same scheme as in Theorem 3.6.6. Weak convergence holds if a constant C > c1 and Φ(x∗ ) < 1. C − c1

For instance, in Example 2.2.8, the smoothing functional principle can be written as follows: α ¯ uγα¯ p1,p + ω h (uγα¯ ) = m + c¯(δ + h)η , η ∈ (0, 1), where c¯ > 0 satisfies the inequality m + c¯(δ + h)η < ω h (θX ).

(3.6.19)

Thus, in (3.6.13), C(γ) = c¯(δ + h)η−1 . Observe in addition that all results of this section can be reformulated for D(A) ⊂ X provided that D(A) are convex sets. 4. We further present the equation with a potential operator describing the twisting of reinforced bars (see [101, 147]), the minimal norm solution of which can be found by the approach above. If G is a bounded convex two-dimensional domain with the boundary ∂G, then the equation and boundary condition defining the elasto-plastic twisting x(t, s) are written in the form Ax = −

∂x ∂ ∂x ∂ = f, − g(T 2 ) g(T 2 ) ∂s ∂t ∂s ∂t x|∂G = 0.

Here

"

T = grad u =

∂x 2

∂t

(3.6.20) (3.6.21)

+

∂x 2

∂s


189

is a maximal tangential stress, f is an angle of twist per unit length of the bar and g(T 2 ) is a function which is characterized by the material of the bar in the stress state such that the constraint equation Γ = g(T 2 )T is fulfilled for intensity Γ of the shear deformations. It is well known that g(T 2 ) ≥ C0 , where C0 depends on the shear modulus of the stressed material, and ∂Γ ≥ 0, ∂T that is, g(T 2 ) + 2T 2 g (T 2 ) ≥ 0. (3.6.22) If we replace (3.6.22) by the stronger inequality g(T 2 ) + 2T 2 g (T 2 ) ≥ C1 > 0

(3.6.23)

and denote by A (x) the Gâteaux derivative at a point x ∈ G of the operator A acting in L2 (G), then one can show that (A (x)h, h) ≥ C

∂x 2 G

∂t

+

∂x 2

∂s

dtds,

h|∂G = 0,

where C = min{C0 , C1 }. Making use of the Friedrichs’ inequality with a constant k we obtain (A (x)h, h) ≥ kCx2 . If

1

0 < C2 ≤ lim g(v)v1− 2 p ≤ C3 v→∞

and

1

lim |g (v)|v 2− 2 p ≤ C4 ,

v→∞

then the problem (3.6.20), (3.6.21) is equivalent to the minimization problem of the functional T 2 (x) g(v) xdtds. (3.6.24) dv − f F 0 (x) = dtds 2 G 0 G It is proved that the functional F 0 (x) is well defined, uniformly convex, coercive and weakly lower semicontinuous on the space W1p . If C1 = 0 then the problem (3.6.20), (3.6.21) is illposed because the convexity of F (x) ceases to be uniform. In this case, for given f δ such that δ ≥ f δ − f , the twisting x(t, s) satisfying (3.6.20), (3.6.21) can be defined by the regularization methods of this section. Bibliographical Notes and Remarks The residual properties on solutions of the regularized linear equations with perturbed data in Hilbert spaces were studied in [152, 153]. Analogous results for nonlinear equations with monotone and accretive operators in Banach spaces have been obtained in [5, 6, 7] and [32, 33, 34]. The residual principle and generalized residual principle were stated in [11] for accretive and in [8, 33, 34] for monotone operator equations. The concept of the generalized

190

3


residual given in Definition 3.3.1 is due to Alber. Theorem 3.1.15 was proved in [188]. A more general form of the residual was used in [187]. The modified residual principle was stated in [192]. The estimates (3.3.32) and (3.3.33) of α ¯ have been found in [202]. The minimal residual principle for nonlinear monotone equations has been provided in [198]. The linear case can be seen in [215]. The choice of the regularization parameter according to the smoothing functional principle for potential monotone equations was investigated in [22]. It was earlier studied in [129, 130, 224] for linear equations. The proof of Lemma 3.6.1 can be found in [130].

Chapter 4

REGULARIZATION OF VARIATIONAL INEQUALITIES 4.1

Variational Inequalities on Exactly Given Sets ∗

1. Let X be an E-space, X ∗ be a strictly convex space, A : X → 2X be a maximal monotone operator with domain D(A), Ω ⊂ D(A) be a convex closed subset in X. Let either int Ω = ∅ or int D(A) ∩ Ω = ∅. (4.1.1) Consider a variational inequality problem: To find x ∈ Ω such that Ax − f, z − x ≥ 0 ∀z ∈ Ω.

(4.1.2)

As usual, we assume that its solution set N is nonempty. Hence, by Definition 1.11.1, for every x∗ ∈ N there exists y ∈ Ax∗ such that y − f, z − x∗ ≥ 0 ∀z ∈ Ω.

(4.1.3)

Moreover, according to Lemma 1.11.4, x∗ satisfies the inequality y − f, z − x∗ ≥ 0 ∀z ∈ Ω, ∀y ∈ Az.

(4.1.4)

Suppose that f and A in (7.1.1) are given with perturbations, that is, in place of f and A, ∗ their δ-approximations f δ and h-approximations Ah : X → 2X are known such that Ah h are maximal monotone, D(A ) = D(A) and, respectively, f − f δ ∗ ≤ δ,

δ > 0,

(4.1.5)

and HX ∗ (Ah x, Ax) ≤ g(x)h ∀x ∈ Ω, 191

h > 0,

(4.1.6)

192

4

REGULARIZATION OF VARIATIONAL INEQUALITIES

where g(t) is a continuous non-negative function for t ≥ 0 and HX ∗ (G1 , G2 ) stands the Hausdorff distance between the sets G1 and G2 in X ∗ . Thus, in reality, we solve the following approximate variational inequality: Ah x − f δ , z − x ≥ 0 ∀z ∈ Ω, x ∈ Ω.

(4.1.7)

By Theorem 1.11.8, the set N is convex and closed, therefore, there exists a unique ¯∗ of vector x ¯∗ ∈ N with the minimal norm. Our aim is to prove the convergence to x approximations defined by the regularized variational inequality Ah x + αJx − f δ , z − x ≥ 0 ∀z ∈ Ω, x ∈ Ω,

α > 0.

(4.1.8)

By Theorem 1.11.11, the inequality (4.1.8) is uniquely solvable. Denote its solutions by xγα , where γ = (δ, h) ∈ 0 and 0 = (0, δ ∗ ] × (0, h∗ ] with positive δ ∗ and h∗ . Observe that xγα is the regularized solution of the variational inequality (7.1.1). Since Ah is a maximal monotone operator, there is a vector yαγ ∈ Ah xγα such that yαγ + αJxγα − f δ , z − xγα ≥ 0 ∀z ∈ Ω.

(4.1.9)

Theorem 4.1.1 Let all the conditions of this subsection hold and δ+h → 0 as α → 0. α

(4.1.10)

¯∗ . Then the sequence {xγα } converges in the norm of X to x Proof. Presuming z = xγα ∈ Ω and z = x∗ ∈ N in (4.1.3) and in (4.1.9), respectively, and summing obtained inequalities, we can write down the following result: yαγ − y, x∗ − xγα + αJxγα , x∗ − xγα − f δ − f, x∗ − xγα ≥ 0.

(4.1.11)

Since Ah are monotone operators, we have yαγ − y h , x∗ − xγα ≤ 0 ∀y h ∈ Ah x∗ ,

∀x∗ ∈ N.

The condition (4.1.6) enables us to choose y h ∈ Ah x∗ such that y h − y∗ ≤ g(x∗ )h. Then, as in the proof of Theorem 2.2.1, we deduce from (4.1.11) the inequality xγα 2 −

δ

α

+

h δ h g(x∗ ) + x∗ xγα − x∗ − g(x∗ )x∗ ≤ 0. α α α

¯ ∈ X as α → 0. It By (4.1.10), it implies the boundedness of {xγα } as α → 0. Let xγα x results from the Mazur theorem that Ω is weakly closed. Then we establish the inclusion x ¯ ∈ Ω because xγα ∈ Ω. By virtue of Lemma 1.11.4, the variational inequality (4.1.8) is equivalent to y h + αJz − f δ , z − xγα ≥ 0 ∀y h ∈ Ah z, ∀z ∈ Ω. (4.1.12)

4.1 Variational Inequalities on Exactly Given Sets

193

Passing in (4.1.12) to the limit as α → 0 and taking into account that f δ → f, one gets y − f, z − x ¯ ≥ 0 ∀y ∈ Az, ∀z ∈ Ω.

(4.1.13)

By Lemma 1.11.4 again, x ¯ ∈ N. Using further (4.1.11), by the same arguments as in the x∗ and xγα → x ¯∗ . The final proof of Theorem 2.2.1, we conclude that x ¯=x ¯∗ , xγα → ¯ result is established by the fact that X is the E-space. Observe, that the assumptions (4.1.1) are used in Theorem 1.11.11 and Lemma 1.11.4. Theorem 4.1.2 Under the conditions of Theorem 4.1.1, if N = {x0 } and there exists a δ+h ≤ C as α → 0, then xγα x0 . constant C > 0 such that α

Proof follows from the previous theorem. Theorem 4.1.3 Under the conditions of Theorems 4.1.1 and 4.1.2, convergence of the operator regularization method (4.1.8) is equivalent to solvability of the variational inequality (7.1.1). Proof. Let xγα x ¯ ∈ X as α → 0. Then the inclusion x ¯ ∈ N is proved as in Theorem 4.1.1. At the same time, the inverse assertion follows from Theorems 4.1.1 and 4.1.2. 2. It is not difficult to verify that Theorems 4.1.1 - 4.1.3 remain still valid for more general regularized inequality Ah x + αJ µ x − f δ , z − x ≥ 0 ∀z ∈ Ω, x ∈ Ω,

(4.1.14)

where J µ : X → X ∗ is the duality mapping with a gauge function µ(t). Existence and uniqueness of its solution xγα ∈ Ω can be proved as in Theorem 1.11.11. In other words, we assert that there is an element yαγ ∈ Ah xγα such that yαγ + αJ µ xγα − f δ , z − xγα ≥ 0 ∀z ∈ Ω.

(4.1.15)

¯∗ , we further assume that the function g(t) in (4.1.6) is increasing To prove that lim xγα = x γ→0

for t ≥ 0 and apply the generalized residual principle of Chapter 3 to the variational inequality (4.1.14), where the residual of (4.1.7) on solutions xγα is understood as follows (cf. Definition 3.3.1): ρµ (α) = αJ µ xγα ∗ = αµ(xγα ). (4.1.16) We establish some important properties of the functions ρµ (α) and σµ (α) = µ(xγα ) for variational inequalities. ∗

∗

Lemma 4.1.4 Let X be an E-space, X ∗ be strictly convex, A : X → 2X and Ah : X → 2X be maximal monotone operators with domains D(A) = D(Ah ), Ω ⊂ D(A) be a convex closed subset in X such that (4.1.1) holds. Then the function σµ (α) is single-valued, continuous and non-increasing for α ≥ α0 > 0.

194

4


Proof. Fix two values of the regularization parameter α1 and α2 and write down for them the inequality (4.1.15). We have yαγ 1 + α1 J µ xγα1 − f δ , z − xγα1 ≥ 0 ∀z ∈ Ω, xγα1 ∈ Ω,

yαγ 1 ∈ Ah xγα1 ,

yαγ 2 + α2 J µ xγα2 − f δ , z − xγα2 ≥ 0 ∀z ∈ Ω, xγα2 ∈ Ω,

yαγ 2 ∈ Ah xγα2 .

and Assuming z = xγα2 and z = xγα1 , respectively, in the first and in the second inequalities, and summing them, one gets yαγ 1 − yαγ 2 , xγα2 − xγα1 + α1 J µ xγα1 − α2 J µ xγα2 , xγα2 − xγα1 ≥ 0. Then the monotonicity property of Ah implies α1 J µ xγα1 − α2 J µ xγα2 , xγα2 − xγα1 ≥ 0.

(4.1.17)

The rest of the proof follows the pattern of Lemma 3.1.8. Lemma 4.1.5 Assume the conditions of Lemma 4.1.4. Then lim xγα = x∗ , where x∗ ∈ Ω α→∞ and x∗ = min {x | x ∈ Ω}. Proof. First of all, we note that, by virtue of the convexity and closedness of Ω in E-space X, an element x∗ exists and it is uniquely defined. Applying (4.1.3), (4.1.15) and the definition of duality mapping J µ , we deduce the inequality (cf. Theorem 4.1.1):

+

h g(x∗ ) α

+

h g(x∗ ) ≤ 0 ∀x∗ ∈ N, α

δ

µ(xγα )(xγα − x∗ ) − xγα

− x∗

α

δ

α

¯ ∈ X, and which implies the boundedness of the sequence {xγα } as α → ∞. Hence, xγα x then x ¯ ∈ Ω according to the Mazur theorem. Show that x ¯ = x∗ using the properties of J µ . By (4.1.15),

yαγ − y h , xγα − z + α µ(xγα ) − µ(z) (xγα − z) + αJ µ z, xγα − z ≤ y h − f δ , z − xγα ∀y h ∈ Ah z,

∀z ∈ Ω.

Now the monotonicity condition of Ah yields the inequality

µ(xγα ) − µ(z) (xγα − z) + J µ z, xγα − z ≤

y h − f δ ∗ z − xγα . α

(4.1.18)

Then the estimate J µ z, xγα − z ≤

y h − f δ ∗ z − xγα ∀z ∈ Ω α

(4.1.19)


195

holds. Passing in (4.1.19) to the limit as α → ∞ we obtain ¯ − z ≤ 0 ∀z ∈ Ω. J µ z, x

(4.1.20)

As it has been established more than once (see, e.g., Theorem 2.2.1), (4.1.20) ensures the equality x ¯ = x∗ . If in (4.1.18) z = x∗ then xγα → x∗ as α → ∞. Since X is E-space, the lemma is proved. 3. Consider the generalized residual (4.1.16). Obviously, ρµ (α) = ασµ (α). By Lemma 4.1.4, ρµ (α) is single-valued and continuous for α ≥ α0 > 0. We study its behaviour as α → ∞. It follows from Lemma 4.1.5 that if θX ∈ Ω, then lim ρµ (α) = ∞.

α→∞

Let θX ∈ Ω. Then xγα → θX as α → ∞. By Lemma 1.11.7, the variational inequality (4.1.14) is equivalent to the inclusion f δ ∈ B h xγα + αJ µ xγα with the maximal monotone operator B h such that ∗

B h = Ah + ∂IΩ : X → 2X ,

(4.1.21)

∗

where ∂IΩ : X → 2X is a subdifferential of the indicator function of the set Ω. It is clear that D(B h ) = Ω. Hence, there is an element ξαγ ∈ B h xγα such that ξαγ + αJ µ xγα = f δ .

(4.1.22)

Show that the sequence {ξαγ } is bounded as α → ∞. The monotonicity of B h yields the inequality ξαγ − ξ h , xγα ≥ 0 ∀ξ h ∈ B h (θX ). (4.1.23) Making use of the property (J ν )∗ = (J µ )−1 of duality mapping J µ in an E-space X, we obtain from (4.1.22) the following formula: xγα = (J ν )∗

f δ − ξγ α

α

,

where (J ν )∗ is a duality mapping in X ∗ with the gauge function ν(t) = µ−1 (t). Then (4.1.23) can be rewritten as

ξγ − ξh α

α

, (J ν )∗

f δ − ξ γ α ≥ 0 ∀ξ h ∈ B h (θX ).

α

It is not difficult to deduce from this the inequality ξαγ − f δ ∗ ≤ ξ h − f δ ∗

∀ξ h ∈ B h (θX ),

(4.1.24)

that guarantees the boundedness of {ξαγ }. Therefore, ξαγ ξ ∈ X ∗ as α → ∞.

(4.1.25)

196

4


Since maximal monotone operator B h is demiclosed, ξαγ ∈ B h (xγα ) and xγα → θX as α → ∞, we have the limit inclusion ξ ∈ B h (θX ). Then from the weak convergence (4.1.25) and from (4.1.24) follows the chain of inequalities: ξ − f δ ∗ ≤ lim inf ξαγ − f δ ∗ ≤ lim sup ξαγ − f δ ∗ ≤ ξ h − f δ ∗ ∀ξ h ∈ B h (θX ). (4.1.26) α→∞

α→∞

Hence, ξ − f δ ∗ = min {ζ − f δ ∗ | ζ ∈ B h (θX )}.

(4.1.27)

{ξαγ }

converges weakly to ξ because ξ is uniquely defined by Thus, the whole sequence (4.1.27). The representation of the operator B h by (4.1.21) means that if θX ∈ int Ω, then the value sets of Ah and B h coincide at θX . Therefore, in this case, ξ ∈ Ah (θX ) and ξ − f δ ∗ = min {ζ − f δ ∗ | ζ ∈ Ah (θX )}.

(4.1.28)

The result (4.1.26) also implies the convergence of norms: ξαγ − f δ ∗ → ξ − f δ ∗ as α → ∞. At the same time, we find from (4.1.22) that ρµ (α) = ξαγ − f δ ∗ . Consequently, lim ρµ (α) = ξ − f δ ∗ ,

α→∞

where ξ ∈ B h (θX ). Finally, one can prove, as before, that if X ∗ is an E-space, then ξαγ → ξ as α → ∞. We are able to state the following lemma. Lemma 4.1.6 Let B h be defined by (4.1.21). Under the conditions of Lemma 4.1.4: (i) if θX ∈ Ω then lim ρµ (α) = ∞, α→∞

(ii) if θX ∈ Ω then ξαγ ξ as α → ∞, where ξαγ ∈ B h xγα and ξ ∈ B h (θX ). In addition, lim ρµ (α) = ξ − f δ ∗ .

α→∞

Moreover, (iii) if θX ∈ ∂Ω then ξ satisfies (4.1.27), (iv) if θX ∈ int Ω then ξ satisfies (4.1.28), (v) if X ∗ is an E-space then ξαγ → ξ as α → ∞. Instead of (4.1.14), consider the following regularized variational inequality: Ah x + αJ µ (x − z 0 ) − f δ , z − x ≥ 0 where z 0 is a fixed element of X.

∀z ∈ Ω,

x ∈ Ω,

(4.1.29)


197

Lemma 4.1.7 Let xγα be a solution of (4.1.29) and B h be defined by (4.1.21). Under the conditions of Lemma 4.1.4, the function σ(α) = xγα − z 0 is single-valued, continuous and non-increasing. In addition, xγα → x∗ as α → ∞, where x∗ ∈ Ω is defined by the following minimum problem: x∗ − z 0 = min {x − z 0 | x ∈ Ω}. The function

ρµ (α) = αµ(xγα − z 0 )

is single-valued and continuous. Moreover, (i) if z 0 ∈ Ω then lim ρµ (α) = ∞, α→∞

(ii) if z 0 ∈ Ω then ξαγ ξ0 as α → ∞, where ξαγ ∈ B h xγα and ξ0 ∈ B h z 0 . In addition, lim ρµ (α) = ξ0 − f δ ∗ ,

α→∞

(iii) if z 0 ∈ ∂Ω then ξ0 ∈ B h z 0 and there holds the equality ξ0 − f δ ∗ = min{ζ − f δ ∗ | ζ ∈ B h z 0 }, (iv) if z 0 ∈ int Ω then ξ0 ∈ Ah z 0 and ξ0 − f δ ∗ = min {ζ − f δ ∗ | ζ ∈ Ah z 0 }. Proof is established by the same arguments as in the previous lemma. 4. By making use of Lemmas 4.1.4 - 4.1.6, state and prove the generalized residual principle for the variational inequality (4.1.14). In the beginning, we study the case of exactly given monotone operators. Theorem 4.1.8 Let X be an E-space with strictly convex X ∗ , A : X → X ∗ be a maximal monotone and hemicontinuous operator with domain D(A), Ω ⊂ D(A) be a convex closed set. Let the variational inequality (7.1.1) have a nonempty solution set N and x ¯∗ ∈ N be a minimal norm solution, f δ be a δ-approximation of f such that f − f δ ∗ ≤ δ ≤ 1 and ¯ > 0 such that θX ∈ Ω. Then there exists α ρ(¯ α) = αx ¯ δα¯ = kδ p , k > 1, p ∈ (0, 1],

(4.1.30)

where xδα¯ is a (classical) solution of the regularized variational inequality Ax + αJx − f δ , z − x ≥ 0 with α = α ¯ . Moreover, (i) if δ → 0 then α ¯ → 0,

∀z ∈ Ω,

x ∈ Ω,

(4.1.31)

198

4


δ → 0, α ¯ (iii) if δ → 0, p = 1 and N = {x0 } then xδα¯ x0 and there exists a constant C > 0 such δ that ≤ C. α ¯

¯∗ and (ii) if δ → 0 and p ∈ (0, 1) then xδα¯ → x

Proof. In view of Theorem 1.7.19, maximal monotone and hemicontinuous operator A is defined on an open set D(A). Therefore, int D(A) ∩ Ω = ∅. Hence, in our circumstances, it is possible to apply Lemma 1.11.4 and Theorem 1.11.11. As in the proof of Theorem 4.1.1, we deduce the quadratic inequality xδα 2 ≤

δ δ + x∗ xδα + x∗ α α

∀x∗ ∈ N.

(4.1.32)

Then there holds the estimate xδα ≤ 2x∗ +

δ α

∀x∗ ∈ N.

(4.1.33)

Fix some x∗ ∈ N and choose α such that 2αx∗ < (k − 1)δ p ,

k > 1,

p ∈ (0, 1].

(4.1.34)

This allows us to find the estimate for the residual, namely, ρ(α) = αxδα ≤ 2αx∗ + δ < (k − 1)δ p + δ ≤ kδ p .

(4.1.35)

Furthermore, by Lemma 4.1.6, lim ρ(α) = ∞.

α→∞

Then the existence of α ¯ follows from the continuity of ρ(α). Next, by (4.1.34), we have the inequality α ¯>

(k − 1)δ p . 2x∗

Consequently,

2x∗ δ 1−p δ . ≤ k−1 α ¯ δ δ ≤ C with C = 2x∗ (k − 1)−1 when → 0 when δ → 0 and p ∈ (0, 1), and Therefore, α ¯ α ¯ δ → 0 and p = 1. It is easy to see that (4.1.33) implies the boundedness of the sequence {xδα¯ }. Hence, we obtain that xδα¯ x ¯ ∈ X as δ → 0. Since Ω is weakly closed by the Mazur theorem, the inclusion x ¯ ∈ Ω holds. Show that x ¯ ∈ N. Construct the maximal monotone operator ∗

B = A + ∂IΩ : X → 2X . It is clear that D(B) = Ω. By Lemma 1.11.7, from variational inequality (4.1.31) with α=α ¯ , we have ¯ Jxδα¯ . f δ ∈ Bxδα¯ + α


199

This means that there exists ξαδ¯ ∈ Bxδα¯ such that ¯ δα¯ = f δ . ξαδ¯ + αJx In view of (4.1.30), one gets ρ(¯ α) = αx ¯ δα¯ = ξαδ¯ − f δ ∗ = kδ p . This implies the limit result: ξαδ¯ → f as δ → 0. Write down the monotonicity property of the operator B as ξαδ¯ − Az − y, xδα¯ − z ≥ 0

∀z ∈ Ω,

∀y ∈ ∂IΩ z.

(4.1.36)

Since θ ∈ ∂IΩ z for all z ∈ Ω (see (1.8.14)), we may put in (4.1.36) y = θ passing in (4.1.36) to the limit as δ → 0, we deduce X∗

X∗

. After that,

f − Az, x ¯ − z ≥ 0 ∀z ∈ Ω. The latter inequality is equivalent to (7.1.1) because of Lemma 1.11.4. This means that x ¯ ∈ N. Let p ∈ (0, 1). Along with (4.1.33), the quadratic inequality (4.1.32) gives also the estimate 2δ (4.1.37) + x∗ ∀x∗ ∈ N. xδα¯ ≤ α ¯ δ Together with the weak convergence of xα¯ to x ¯ ∈ N, (4.1.37) yields the following relations: ¯ x ≤ lim inf xδα¯ ≤ lim sup xδα¯ ≤ x∗ ∀x∗ ∈ N. δ→0

(4.1.38)

δ→0

¯∗ ∈ N is unique. Moreover, the Therefore, x ¯ = x ¯∗ because the minimal norm solution x δ ∗ convergence of xα¯ to ¯ x as δ → 0 follows from (4.1.38). Since X is E-space, the claim (ii) holds. By (4.1.30), kδ p (4.1.39) α ¯= δ . xα¯

Observe that xδα¯ > 0 for a small enough δ > 0 by reason of x ¯∗ = θX . The latter follows ¯ → 0 as δ → 0 and from the hypotheses that θX ∈ Ω. Then one gets from (4.1.39) that α p ∈ (0, 1). If p = 1 then xδα¯ x0 = θX . By the property of the norm in a Banach space, we have the relation x0 ≤ lim inf xδα¯ , δ→0

which enables us to conclude that α ¯ → 0 as δ → 0. This completes the proof. Theorem 4.1.9 Assume that the conditions of Theorem 4.1.8 are satisfied, θX ∈ Ω and ξ0δ − f δ ∗ > kδ p , k > 1, p ∈ (0, 1], ξ0δ

where is defined as follows: (i) if θX ∈ int Ω then ξ0δ = A(θX ), (ii) if θX ∈ ∂Ω then ξ0δ − f δ ∗ = min {ζ − f δ ∗ | ζ ∈ B(θX )}, where B = A + ∂IΩ . Then all the conclusions of Theorem 4.1.8 remain still valid.

(4.1.40)

200

4


Proof. Due to Lemma 4.1.6, lim ρ(α) = ξ0δ − f δ ∗ .

α→∞

As it was established in (4.1.35), there is α > 0 such that ρ(α) < kδp . Then the condition (4.1.40) guarantees that α ¯ satisfying (4.1.30) exists. Since the solvability of variational inequality (4.1.31) is equivalent to the inclusion f ∈ Bx, where D(B) = Ω and Bx = Ax for x ∈ int Ω, it results from (4.1.40) that θX ∈ N. The remaining assertions are proved as in Theorem 4.1.8. Along with (4.1.31), it is possible to research the regularized variational inequality Axδα + αJ(xδα − z 0 ) − f δ , z − xδα ≥ 0

∀z ∈ Ω,

where xδα ∈ Ω and z 0 is an arbitrary fixed point of X. Under the conditions of Theorems 4.1.8 and 4.1.9, it is not difficult to prove the following: (i) If z 0 ∈ Ω then Theorem 4.1.8 holds with ρ(α) = αxδα − z 0 . The vector x ¯∗ is defined there by the minimization problem: ¯ x∗ − z 0 = min {x∗ − z 0 | x∗ ∈ N }. (ii) If z 0 ∈ Ω then Theorem 4.1.9 holds, provided that: in the case of z 0 ∈ int Ω, (4.1.40) is replaced by the inequality Az 0 − f δ ∗ > kδ p . In the case of z 0 ∈ ∂Ω, the element ξ0δ is defined as ξ0δ − f δ ∗ = min {ζ − f δ ∗ | ζ ∈ Bz 0 }, where B = A + ∂IΩ . Next we omit the hemicontinuity property of A and present the following theorem. Theorem 4.1.10 Assume that A is a maximal monotone (possibly, set-valued) operator. Then all the conclusions of Theorems 4.1.8 and 4.1.9 remain still valid if (4.1.1) is satisfied. Moreover, if θX ∈ int Ω, then ξ0δ is defined by the minimization problems: ξ0δ − f δ ∗ = min{ζ − f δ ∗ | ζ ∈ A(θX )}.

(4.1.41)

By analogy, if z 0 ∈ int Ω, then ξδ0 is defined as ξδ0 − f δ ∗ = min{ζ − f δ ∗ | ζ ∈ Az 0 }.

(4.1.42)

If θX ∈ ∂Ω (respectively, z 0 ∈ ∂Ω), then ξδ0 is defined by (4.1.41) (respectively, (4.1.42)), where A is replaced by B = A + ∂IΩ . 5. We further discuss the variational inequalities with approximately given operators.


201 ∗

Theorem 4.1.11 Suppose that X is an E-space with strictly convex X ∗ , A : X → 2X is a maximal monotone operator with domain D(A), Ω ⊂ D(A) is a convex closed subset, condition (4.1.1) holds, variational inequality (7.1.1) has a nonempty solution set N and ∗ x ¯∗ ∈ N is the minimal norm solution. Let Ah : X → 2X with h > 0 be maximal monotone h operators, Ω ⊂ D(A ), and if int Ω = ∅, then int D(Ah ) ∩ Ω = ∅. Let conditions (4.1.5) and (4.1.6) hold, where g(t) is a non-negative, continuous and increasing function and 0 < δ + h ≤ 1. Furthermore, assume that in the case of θX ∈ Ω, the additional inequality

ξ0 − f δ ∗ > k + g(0) (δ + h)p , k > 1,

p ∈ (0, 1],

(4.1.43)

is satisfied, where ξ0 is defined by the following minimization problems: 1) if θX ∈ int Ω then ξ0 − f δ ∗ = min {ζ − f δ ∗ | ζ ∈ Ah (θX )}, 2) if θX ∈ ∂Ω then ξ0 − f δ ∗ = min {ζ − f δ ∗ | ζ ∈ B h (θX )}, where B h = Ah + ∂IΩ . Then there exists a unique α ¯ satisfying the equation

ρ(¯ α) = αx ¯ γα¯ = k + g(xγα¯ ) (δ + h)p , k > 1,

p ∈ (0, 1],

(4.1.44)

¯ and where xγα¯ is a solution of the regularized variational inequality (4.1.8) with α = α γ = (δ, h). Moreover, (i) if γ → 0 then α ¯ → 0, δ+h ¯∗ and → 0, (ii) if γ → 0 and p ∈ (0, 1) then xγα¯ → x α ¯ (iii)) if γ → 0, p = 1 and N = {x0 } then xγα¯ x0 and there exists a constant C > 0 such δ+h ≤ C. that α ¯

Proof. Using the monotonicity of A, inequality (4.1.11) and conditions (4.1.5) and (4.1.6) of the theorem, we calculate the estimate xγα 2 ≤

δ

α

+

δ h h + g(xγα ) x∗ g(xγα ) + x∗ xγα + α α α

It implies δ h + g(xγα ) α α Hence, for the residual function, one gets xγα ≤ 2x∗ +

∀x∗ ∈ N.

ρ(α) = αxγα ≤ 2αx∗ + δ + hg(xγα ). If, for some x∗ ∈ N, the parameter α is such that 2αx∗ < (k − 1)(δ + h)p ,

k > 1,

p ∈ (0, 1],

∀x∗ ∈ N.

(4.1.45)

202

4


that is, α< then

k−1 (δ + h)p , 2x∗

(4.1.46)

ρ(α) < k + g(xγα ) (δ + h)p .

(4.1.47)

Consider first the case when θX ∈ Ω. Due to Lemmas 4.1.5 and 4.1.6, we conclude that lim xγα = 0

α→∞

and lim ρ(α) = ξ − f δ ∗ , ξ ∈ B h (θX ).

α→∞

Then, it follows from the continuity of ρ(α) and g(t) and from (4.1.43) and (4.1.47) that there exists a solution α ¯ > 0 of the equation (4.1.44). Moreover, (4.1.46) implies α ¯>

k−1 (δ + h)p . 2x∗

(4.1.48)

As in the proof of Theorem 4.1.9, assertions (i) and (ii) for the sequence

δ + h

can be α ¯ deduced from (4.1.48), separately for p = 1 and p ∈ (0, 1). If the monotonicity condition of operators Ah is used in (4.1.11) with α = α ¯ then we come to the inequality similar to (4.1.45): xγα¯ 2 ≤

δ

α ¯

+

δ h h + g(x∗ ) x∗ g(x∗ ) + x∗ xγα¯ + ¯ α ¯ α α ¯

∀x∗ ∈ N.

As before, there holds the estimate xγα¯ ≤ x∗ + 2

h δ + 2 g(x∗ ). α ¯ α ¯

(4.1.49)

¯ ∈ Ω as γ → 0. Therefore, {xγα¯ } is bounded, hence, xγα¯ x Show that x ¯ ∈ N. Indeed, variational inequality (4.1.8) is reduced to the inclusion f δ ∈ B h xγα¯ + α ¯ Jxγα¯ , where B h = Ah + ∂IΩ . This means that there exists ξαγ¯ ∈ B h xγα¯ such that ¯ γα¯ = f δ . ξαγ¯ + αJx The latter equality and (4.1.44) induce the strong convergence of {ξαγ¯ } to f as γ → 0. By (4.1.6), for every x ∈ Ω and every y ∈ Ax, one can construct a sequence {y h }, y h ∈ Ah x, such that y h → y as h → 0. Then the monotonicity of operator B h yields the following inequality: ξαγ¯ − y h , xγα¯ − x ≥ 0 ∀x ∈ Ω.


203

Letting γ → 0 one has f − y, x ¯ − x ≥ 0 ∀x ∈ Ω, ∀y ∈ Ax. Now, by Lemma 1.11.4, we conclude that x ¯ ∈ N, and the assertions of the theorem for p = 1 are proved completely. ¯ ∈ N and inequality (4.1.49), we Let p ∈ (0, 1). Using the weak convergence of xγα¯ to x make sure, as in Theorem 4.1.8, that x ¯=x ¯∗ and xγα¯ → ¯ x∗ as γ → 0. Thus, the strong ¯∗ is established in an E-space X. convergence xγα¯ → x Prove that α ¯ → 0 as γ → 0. For this end, show that condition (4.1.43) ensures that θX ∈ N. If θX ∈ N then f ∈ B(θX ), where B = A + ∂IΩ . By virtue of the definitions of operators B and B h , there exist ξ h ∈ B h (θX ) such that ξ h − f ∗ ≤ hg(0). Then

ξ h − f ∗ ≤ ξ h − f δ ∗ + f − f δ ∗ ≤ hg(0) + δ ≤ k + g(0) (δ + h)p , which contradicts (4.1.43). Then the claim is established in the same way as in Theorem 3.3.2. Prove that α ¯ involving (4.1.44) is unique. Assume that (4.1.44) has two solutions α ¯ and β¯ such that

ρ(¯ α) = αx ¯ γα¯ = k + g(xγα¯ ) (δ + h)p and

(4.1.50)

¯ = βx ¯ γ¯ = k + g(xγ¯) (δ + h)p , ρ(β) β β

(4.1.51)

where xγα¯ and xγβ¯ satisfy, respectively, the following inequalities: ¯ γα¯ − f δ , z − xγα¯ ≥ 0 yαγ¯ + αJx

∀z ∈ Ω, yαγ¯ ∈ Ah xγα¯ ,

xγα¯ ∈ Ω,

(4.1.52)

¯ γ¯ − f δ , z − xγ¯ ≥ 0 yβγ¯ + βJx β β

∀z ∈ Ω, yβγ¯ ∈ Ah xγβ¯,

xγβ¯ ∈ Ω.

(4.1.53)

and

Put x = xγβ¯ in (4.1.52) and z = xγα¯ in (4.1.53), and add the obtained inequalities. Then we obtain ¯ γ¯, xγα¯ − xγ¯ ≤ 0. αJxγα¯ − βJx yαγ¯ − yβγ¯ , xγα¯ − xγβ¯ + ¯ β β Further, it is necessary to use (4.1.50) and (4.1.51) and repeat the reasoning from the proof of Theorem 3.3.2. If θX ∈ Ω then, by Lemma 4.1.6, lim ρ(α) = ∞. Therefore the requirement (4.1.43) in α→∞ this case can be omitted. The proof is accomplished. Remark 4.1.12 Along with the generalized residual principle, the choice of regularization parameter in the variational inequality (4.1.8) can be realized by the modified residual principle of Section 3.4 (see Theorem 3.4.1 and Remarks 3.4.2).

204

4


Remark 4.1.13 Some important applied problems are reduced to variational inequalities for which sets Ω are defined as follows [128]: Ω = {u | u = u(x) ∈ X, u(x) ≥ 0} and Ω = {u | u = u(x) ∈ X, u(x)|∂Ω = 0}, where x ∈ G, G is a bounded measurable subset in Rn and X is a Banach space. These sets in the operator regularization methods can be considered as unperturbed ones.

4.2

Variational Inequalities on Approximately Given Sets

In the previous section, we dealt with the regularizing processes for variational inequalities provided that their constraint sets Ω are given exactly. However, there are many practical variational problems with approximately given sets. In this section, we show that the operator regularization methods also solve these problems stably. Under the conditions of Theorem 4.1.1, let {Ah } be a sequence of maximal monotone operators, {Ωσ } be a sequence of convex closed sets such that Ωσ ⊆ D(Ah ) and either int Ωσ = ∅ or int D(Ah ) ∩ Ωσ = ∅.

(4.2.1)

In the sequel, for some positive δ ∗ , h∗ and σ∗ , we denote = (0, δ ∗ ] × (0, h∗ ] × (0, σ ∗ ]. Let the inequality (4.1.6) hold for all x ∈ Ω ∩ Ωσ and for every couple of Ah and Ωσ . Suppose that X is a Hilbert space H and operators Ah have the following growth order: There exists a constant M > 0 such that y h − f δ ≤ M (x + 1) ∀y h ∈ Ah x, ∀x ∈ Ωσ ,

(4.2.2)

where γ = (δ, h, σ) ∈ . We study the strong convergence of the operator regularization method for variational inequalities with different proximity conditions between Ωσ and Ω. Let Ωσ uniformly approximate Ω in the Hausdorff metric, that is, HH (Ω, Ωσ ) ≤ σ.

(4.2.3)

Consider approximations to a solution of (7.1.1) generated by the regularized variational inequality (4.2.4) Ah x + αx − f δ , z − x ≥ 0 ∀z ∈ Ωσ , x ∈ Ωσ . As it follows from Section 1.11, this inequality has unique regularized solution xγα . Therefore, there exists yαγ ∈ Ah xγα such that yαγ + αxγα − f δ , z − xγα ≥ 0

∀z ∈ Ωσ .

(4.2.5)

We emphasize that each operator Ah in the variational inequality (4.2.4) is not assumed to be defined on every subset Ωσ . However, we declare that always there is a possibility to

4.2 Variational Inequalities on Approximately Given Sets

205

approach the parameters h and σ to zero at the same time. This remark has to do with all the variational inequalities of the type (4.2.4). 1. In this subsection we presume that Ω ⊆ Ωσ . This is the so-called exterior approximations of Ω. In the next subsection we will study the interior approximations of Ω when Ωσ ⊆ Ω. Theorem 4.2.1 Assume that (i) A : H → 2H is a maximal monotone operator; (ii) Ω is a convex closed set in H; (iii) {Ah } is a sequence of the maximal monotone operators Ah : H → 2H ; (iv) {Ωσ } is a sequence of the convex closed sets such that Ωσ ⊆ D(Ah ); (v) Ω ⊆ Ωσ for all σ ∈ (0, σ ∗ ]; (vi) for all x ∈ Ω, the proximity between the operators A and Ah is given by HH (Ah x, Ax) ≤ g(x)h ∀x ∈ Ω, h ∈ (0, h∗ ],

(4.2.6)

where g(t) is a continuous and non-negative function for t ≥ 0; (vii) the conditions (4.1.1), (4.1.5), (4.2.1), (4.2.2) and (4.2.3) are fulfilled; (viii) variational inequality (7.1.1) has a nonempty solution set N. If δ+h+σ = 0, lim α→0 α

(4.2.7)

then the solution sequence {xγα } of the variational inequality (4.2.4) converges strongly in H to the minimal norm solution x ¯∗ ∈ N as α → 0. Proof. According to condition (4.2.3), for all xγα ∈ Ωσ and for all x∗ ∈ N ⊂ Ω, there exist respective elements uγα ∈ Ω and vαγ ∈ Ωσ such that xγα − uγα ≤ σ

(4.2.8)

x∗ − vαγ ≤ σ.

(4.2.9)

and Since x∗ ∈ N, there exists y ∈ Ax∗ such that (4.1.3) holds. Presuming z = uγα in (4.1.3) and z = vαγ in (4.2.5) and summing the obtained inequalities, one gets for yαγ ∈ Ah xγα , (yαγ + αxγα − f δ , vαγ − xγα ) + (y − f, uγα − x∗ ) ≥ 0 ∀x∗ ∈ N,

y ∈ Ax∗ .

Since Ah is monotone for each h ∈ (0, h∗ ] and Ω ⊆ Ωσ ⊆ D(Ah ), we have (yαγ − y h , xγα − x∗ ) ≥ 0

∀y h ∈ Ah x∗ .

Moreover, there exists y h ∈ Ah x∗ such that y h − y ≤ g(x∗ )h.

(4.2.10)

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4


On the basis of (4.2.10), we obtain α(xγα , xγα − vαγ ) ≤ (y − f, uγα − xγα ) + (yαγ − f δ , vαγ − xγα ) + (y − f, xγα − x∗ ) ≤ (y − f, uγα − xγα ) + (yαγ − f δ , vαγ − x∗ ) + (y − yαγ , xγα − x∗ ) + (f δ − f, xγα − x∗ ) ≤ (y − f, uγα − xγα ) + (yαγ − f δ , vαγ − x∗ ) + (y h − y, x∗ − xγα ) + (f − f δ , x∗ − xγα )

≤ (y − f, uγα − xγα ) + (yαγ − f δ , vαγ − x∗ ) + hg(x∗ ) + δ xγα − x∗ .

(4.2.11)

Then (4.2.2), (4.2.8) and (4.2.9) imply the quadratic inequality xγα 2 − xγα

−

h

α

δ

α

+M

g(x∗ ) +

h σ + g(x∗ ) + σ + x∗ α α

σ σ ∗ δ x − 2M ≤ 0 +M α α α

∀x∗ ∈ N.

(4.2.12)

It yields the following estimate: xγα ≤

h σ δ + M + g(x∗ ) + σ + 2(x∗ + 1) ∀x∗ ∈ N, α α α

(4.2.13)

which together with (4.2.7) proves that {xγα } is bounded. Since xγα − uγα ≤ σ, {uγα } is also bounded. Therefore, uγα x ¯ ∈ H. Moreover, x ¯ ∈ Ω because uγα ∈ Ω and Ω is weakly closed. γ ¯ ∈ Ω as α → 0. Thus, we have established that xα x By Lemma 1.11.7, variational inequality (4.2.4) is equivalent to the inclusion f δ ∈ B λ xγα + αxγα , where operator B λ = Ah + ∂IΩσ ,

λ = (h, σ),

D(B λ )

= Ωσ . Recall that we denoted by ∂IΩσ a subdifferential is maximal monotone with of the indicator function of the set Ωσ . Hence, there exists an element ξαγ ∈ B λ xγα such that ξαγ + αxγα = f δ .

(4.2.14)

ξαγ = y˜αγ + zαγ ,

(4.2.15)

Obviously, ξαγ can be represented as

where y˜αγ ∈ Ah xγα and zαγ ∈ ∂IΩσ xγα . It follows from (4.1.5), (4.2.14) and (4.2.15) that yαγ + zαγ ) = f lim (˜

(4.2.16)

α→0

because {xγα } is bounded. By inclusion Ω ⊆ Ωσ and by (4.2.6), for every x ∈ Ω and for every y ∈ Ax, we may define a sequence {y h } such that y h ∈ Ah x and y h → y as h → 0. Since B λ is (maximal) monotone and Ω ⊆ Ωσ = D(B λ ), one has (˜ yαγ + zαγ − y h − z σ , xγα − x) ≥ 0 ∀x ∈ Ω,

z σ ∈ ∂IΩσ x.


207

It is known that θH ∈ ∂IΩσ x for all x ∈ Ωσ . Therefore, we may put in the latter inequality z σ = θH . Then, after passing to limit as α → 0, one gets (f − y, x ¯ − x) ≥ 0

∀x ∈ Ω,

∀y ∈ Ax,

where x ¯ ∈ Ω. This means that x ¯ ∈ N in view of Lemma 1.11.4. After some simple transformations, (4.2.11) is reduced to the inequality xγα − x∗ 2 ≤ xγα − x∗

σ δ h + g(x∗ ) + M (x∗ + xγα + 2) α α α

+ σxγα + (x∗ , x∗ − xγα ) x∗

Put here =x ¯. Then the strong convergence of Letting α → 0 in (4.2.17), one gets

∀x∗ ∈ N. xγα

(x∗ , x∗ − x ¯) ≥ 0

(4.2.17)

to x ¯ follows because (4.2.7) is satisfied.

∀x∗ ∈ N,

which implies the equality x ¯ = x ¯∗ (see, for instance, Theorem 2.1.2). Consequently, the γ ¯∗ as α → 0, and the proof is complete. whole sequence xα strongly converges to x Corollary 4.2.2 If N = {x0 } and there exists C > 0 such that δ+h+σ ≤ C as α → 0, α

then Theorem 4.2.1 guarantees the weak convergence of xγα to x0 . 2. We study the interior approximations of Ω. This means that (4.2.6) holds for any x ∈ Ωσ . Furthermore, we assume that function g(t) in (4.2.6) is bounded on bounded sets. The regularized solutions are found from the inequality (Ah x + αU µ x − f δ , z − x) ≥ 0 where U µx =

∀z ∈ Ωσ ,

x ∈ Ωσ ,

(4.2.18)

µ(x)x if x = θH , x

and J µ (θH ) = θH , µ(t) has the same properties as a gauge function of duality mappings J µ . We additionally assume that there exists t0 > 0 such that µ(t) > g(t) for t ≥ t0 . Let xγα ∈ Ωσ with γ ∈ be a solution of (4.2.18) and yαγ ∈ Ah xγα such that (yαγ + αU µ xγα − f δ , z − xγα ) ≥ 0

∀z ∈ Ωσ .

Then (4.2.10) accepts the following form: (yαγ + αU µ xγα − f δ , vαγ − xγα ) + (y − f, uγα − x∗ ) ≥ 0 ∀x∗ ∈ N,

y ∈ Ax∗ .

(4.2.19)

Since Ωσ ⊆ Ω, the condition (4.2.6) asserts that there exists an element ηαγ ∈ Axγα such that yαγ − ηαγ ≤ hg(xγα ).

(4.2.20)

208

4


Taking into account the monotonicity of A and (4.2.19), one gets α(U µ xγα , xγα − vαγ ) ≤ α(U µ xγα , xγα − vαγ ) + (y − ηαγ , x∗ − xγα ) ≤ (ηαγ − yαγ , xγα − x∗ ) + (yαγ − f δ , vαγ − x∗ ) + (y − f, uγα − xγα ) + (f δ − f, xγα − x∗ )

∀x∗ ∈ N.

(4.2.21)

Similarly to (4.2.12), we obtain with the help of (4.1.5), (4.2.2), (4.2.3), (4.2.8), (4.2.9) and (4.2.20) the following inequality: µ(xγα )(xγα − σ − x∗ ) −

−

δ h + g(xγα ) xγα − x∗ α α

σ M (2 + x∗ + xγα ) ≤ 0 α

∀x∗ ∈ N.

The properties of µ(t) and condition (4.2.7) enable us to assert that the sequence {xγα } is bounded. Then xγα x ¯. Since Ωσ are convex closed sets and xγα ∈ Ωσ ⊆ Ω, the Mazur theorem guarantees the inclusion x ¯ ∈ Ω. Show that x ¯ ∈ N. Let ξαγ ∈ B λ xγα , where B λ = Ah + ∂IΩσ , λ = (h, σ), such that ξαγ + αU µ xγα = f δ .

(4.2.22)

It is clear that (4.2.15) is true. By making use of (4.2.6), which is valid for all x ∈ Ωσ , we conclude that for every element yαγ ∈ Ah xγα there exists wαγ ∈ Axγα such that ˜ yαγ − wαγ ≤ g(xγα )h

(4.2.23)

is satisfied. Since {xγα } is bounded, yαγ − wαγ → 0 as α → 0. Then (4.2.16) and (4.2.23) imply lim (wαγ + zαγ ) = f,

α→0

(4.2.24)

where zαγ ∈ ∂IΩσ xγα . Write down the monotonicity property of the operator B = A + ∂IΩ with D(B) = Ω : (wαγ + z˜αγ − ζ, xγα − x) ≥ 0 ∀x ∈ Ω, ∀ζ ∈ Bx, ∀˜ zαγ ∈ ∂IΩ xγα . Since θH ∈ ∂IΩ x for all x ∈ Ω, we may presume in the latter inequality z˜αγ = θH . According to (4.2.3), for each x ∈ Ω, there exists {xσ } ⊂ Ωσ such that xσ → x as σ → 0. By simple algebra, we now come to the relation (wαγ + zαγ − ζ, xγα − x) + (zαγ , xσ − xγα ) − (zαγ , xσ − x) ≥ 0.

(4.2.25)

Taking into account that θH ∈ ∂IΩσ z for all z ∈ Ωσ and that ∂IΩσ is monotone, it is easy to see that the second term in (4.2.25) is non-positive. Omitting it, we only strengthen this


209

inequality. Further, (4.2.2) allows us to establish the boundedness of {˜ yαγ } as α → 0. In its turn, the boundedness of {zαγ } results from (4.2.23) and (4.2.24). Finally, (4.2.25) yields the inequality (f − ζ, x ¯ − x) ≥ 0 ∀x ∈ Ω, ∀ζ ∈ Ax, x ¯ ∈ Ω. By Lemma 1.11.4, it implies the inclusion x ¯ ∈ N. Using (4.2.21) and following the scheme of the previous theorem, we obtain Theorem 4.2.3 Assume that the conditions of Theorem 4.2.1 are satisfied with the inverse inclusion Ωσ ⊆ Ω for all σ ∈ (0, σ ∗ ], the sequence {xγα } is generated by the variational inequality (4.2.18), the function g(t) in (4.2.6) is bounded on bounded sets and there exists t0 > 0 such that µ(t) > g(t) for t ≥ t0 . Then {xγα } strongly converges to the minimal norm solution x ¯∗ as α → 0. Remark 4.2.4 Theorem 4.2.1 remains still valid if the inclusion Ω ⊆ Ωσ is replaced by N ⊂ Ωσ for all σ ∈ (0, σ ∗ ]. Indeed, in this case the estimate (4.2.13) is satisfied and verification of the fact that x ¯ ∈ N can be done as in the proof of the previous theorem. 3. Let the condition (4.2.6) hold for all x ∈ Ω ∪ Ωσ with all σ ∈ (0, σ ∗ ]. In other words, we suppose that D(A) contains Ω and D(Ah ) with h ∈ (0, h∗ ] contain Ωσ . In particular, this is carried out if D(A) = D(Ah ) for all h ∈ (0, h∗ ]. In this case, the requirements Ω ⊆ Ωσ and Ωσ ⊆ Ω in Theorems 4.2.1 and 4.2.3 should be omitted. Then the proof of the fact that x ¯ ∈ N is simplified. Indeed, let x be any fixed element of Ω. Put in (4.2.5) z = xσ ∈ Ωσ , provided that x − xσ ≤ σ. We have for all γ ∈ , (yαγ + αxγα − f δ , xσ − xγα ) ≥ 0, where yαγ ∈ Ah xγα . Rewrite this inequality in the following form: (yαγ + αxγα − f δ , xσ − x) + (y − f, x − xγα ) + (yαγ − wαγ , x − xγα ) + (wαγ − y, x − xγα ) + (f − f δ , x − xγα ) + α(xγα , x − xγα ) ≥ 0 ∀y ∈ Ax, where wαγ ∈ Axγα such that

(4.2.26)

yαγ − wαγ ≤ g(xγα )h.

Assuming that g(t) is bounded on bounded sets and passing in (4.2.26) to the limit as α → 0, one gets (y − f, x − x ¯) ≥ 0 ∀x ∈ Ω, ∀y ∈ Ax, x ¯ ∈ Ω, which is equivalent to (4.1.3). We come to the following statement. Theorem 4.2.5 If in the condition (vi) of Theorem 4.2.1, inequality (4.2.6) holds for all h ∈ (0, h∗ ] and for all x ∈ Ω ∪ Ωσ and if function g(t) is bounded on bounded sets, then the solution sequences {xγα } generated by the variational inequalities (4.2.4) and (4.2.18) converge strongly in H to the minimal norm solution x ¯∗ of the variational inequality (7.1.1) as α → 0.

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4


4. Convergence of the operator regularization methods is established in Theorems 4.2.1, 4.2.3 and 4.2.5 under the condition (4.2.3). It is natural if Ω and Ωσ are bounded subsets in a Hilbert space. However, there are variational problems of the type (7.1.1) in which (4.2.3) may not be fulfilled. For instance, if Ω is an unbounded set in R2 and if boundary ∂Ω of Ω is given by linear functions y = kx + b with k, b ∈ R1 , then any arbitrarily small error of k implies the fact that HH (Ω, Ωσ ) = ∞. Assume that Ω is unbounded. Construct the sets ΩR = Ω ∩ B(θH , R) and

∗ ΩR σ = Ωσ ∩ B(θH , R) ∀ σ ∈ (0, σ ].

Choose R large enough that sets GR = N ∩ ΩR = ∅ and ΩR σ = ∅. Furthermore, we assume that either h R ∗ ∗ int ΩR σ = ∅ or int D(A ) ∩ Ωσ = ∅ ∀ h ∈ (0, h ], ∀ σ ∈ (0, σ ],

and HH (ΩR , ΩR σ ) ≤ σ. Restricting the given problem (7.1.1) on the subset ΩR with fixed R, we obtain in Theorems ¯∗R ∈ N R , where xγα are solutions of 4.2.1, 4.2.3 and 4.2.5 the strong convergence of xγα to x R corresponding regularized variational inequalities on the sets ΩR σ , and N is a solutions set of the variational inequality (Ax − f, z − x) ≥ 0 at that

∀z ∈ ΩR , x ∈ ΩR ,

(4.2.27)

¯ x∗R = min {x | x ∈ N R }.

The question arises: can we find in this way the solution x ¯∗ of the variational inequality x∗R ≤ ¯ x∗ (7.1.1)? The answer is affirmative. Indeed, first of all, we note that x ¯∗ ∈ N R , ¯ and GR ⊆ N R . In our assumptions, solutions of (4.2.27) coincide with solutions of the inequality (Az − f, z − x) ≥ 0 ∀z ∈ ΩR , x ∈ ΩR . Therefore, any element x ∈ N R \GR cannot be an interior point of ΩR , because the inclusion f ∈ Ax + ∂IΩR x is equivalent to (4.2.27). If x ∈ int ΩR , then it is transformed into the inclusion f ∈ Ax. Thus, it follows from the inclusion x ∈ N R \GR that x ∈ ∂ΩR . Further, the sets N R and GR are convex and closed. Therefore, if N R \GR = ∅ and int GR = ∅ then there exists x ¯ ∈ N R \GR such that x ¯ ∈ int ΩR . Hence, GR = N R in this case. R R Sets N and G also coincide if ∂ΩR (more precisely, ∂ΩR \S(θH , R)) has no convex subsets. Indeed, construct maximal monotone operators T = A + ∂IΩ and T R = A + ∂IΩR .


211

Then inequalities (7.1.1) and (4.2.27) are equivalent to the equations T x = f and T R x = f, respectively. Show that values of T and T R coincide on the set S. To this end, it is sufficient to establish that sets ∂IΩ x and ∂IΩR x coincide on S. Let η ∈ ∂IΩR x and x ∈ S. Then, by the definition of a subdifferential, (η, x − y) ≥ 0

∀y ∈ ΩR .

By the hypotheses, for every y1 ∈ Ω, there exists a constant λ > 0 and an element y ∈ ΩR such that x − y1 = λ(x − y). Hence, η also belongs to ∂IΩ x, that is, ∂IΩR x ⊆ ∂IΩ x for all x ∈ S. Taking into account the obvious contrary inclusion ∂IΩ x ⊆ ∂IΩR x for x ∈ S, we finally obtain the claim. Thus, the inequality (4.2.27) can be considered in place of (7.1.1). After that, we are able to state Theorems 4.2.1, 4.2.3 and 4.2.5 for corresponding operator regularization methods. Observe that Theorem 1.11.10 may be also useful for an evaluation of R. Let now a constant R, for which N R = GR , be unknown. In this case, define the proximity between Ω and Ωσ in a different way than in (4.2.3). Namely, let s(R, Ω, Ωσ ) = sup {inf {u − v | u ∈ Ωσ } | v ∈ ΩR }

∀R > 0.

(4.2.28)

If ΩR = ∅ then we presume that s(R, Ω, Ωσ ) = 0. At the same time, if ΩR = ∅ or ΩR σ = ∅, define τ (R, Ω, Ωσ ) = max {s(R, Ω, Ωσ ), s(R, Ωσ , Ω)}. Suppose that τ (R, Ω, Ωσ ) ≤ a(R)σ,

(4.2.29)

where a(R) is a non-negative continuous and increasing function for R ≥ 0, a(0) = 0 and a(R) → ∞ as R → ∞. In order to introduce an operator U µ in the regularized inequality (4.2.18), choose µ(t) such that µ(t) > max{a(t), g(t)}, t ≥ t0 > 0,

(4.2.30)

where g(t) satisfies (4.2.6). Theorem 4.2.6 If the requirement (4.2.3) in Theorems 4.2.1, 4.2.3 and 4.2.5 is replaced by (4.2.29), then the solution sequence {xγα } generated by (4.2.18) with µ(t) satisfying (4.2.30) converges strongly to the minimal norm solution x ¯∗ of the variational inequality (7.1.1) as α → 0. Proof. We shall not repeat the proofs of the mentioned theorems because their arguments remain as before. For instance, the inequality of the type (4.2.12) has the following form: σ µ(xγα ) xγα − a(x∗ )σ − x∗ − M a(xγα )(x∗ + 1) α − xγα

−

σ

α

M a(x∗ ) +

δ δ h + g(x∗ ) − x∗ α α α

h σ g(x∗ )x∗ − M a(x∗ ) ≤ 0 α α

∀x∗ ∈ N.

(4.2.31)

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4


We deduce from this that {xγα } is bounded as α → 0. Then the rest of the proof follows the pattern of Theorem 4.2.1. 5. Next we present the residual principle for regularized variational inequality on approximately given sets Ωσ . Observe that the following Lemmas 4.2.7 - 4.2.9 are similar to Lemmas 4.1.4 - 4.1.6 of Section 4.1 and can be proved in the same manner. Lemma 4.2.7 A function σµ (α) = µ(xγα ) is single-valued, continuous and non-increasing for α ≥ α0 > 0. Lemma 4.2.8 lim xγα = xσ∗ , where xσ∗ ∈ Ωσ and xσ∗ = min{x | x ∈ Ωσ }. α→∞

Lemma 4.2.9 Let B λ = Ah + ∂IΩσ with λ = (h, σ). One has: (i) If θH ∈ Ωσ then lim ρµ (α) = lim αµ(xγα ) = ∞, α→∞

(ii) if θH ∈ Ωσ then B λ (θH ). In addition,

ξαγ

α→∞

ξ as α → ∞, where ξαγ ∈ B λ xγα in view of (4.1.22), and ξ ∈ lim ρµ (α) = ξ − f δ .

α→∞

Moreover, (iii) if θH ∈ int Ωσ then ξ ∈ Ah (θH ) is defined as ξ − f δ ∗ = min {ζ − f δ ∗ | ζ ∈ Ah (θH )},

(4.2.32)

(iv) if θH ∈ ∂Ωσ then ξ ∈ B (θH ) satisfies the equality λ

ξ − f δ ∗ = min {ζ − f δ ∗ | ζ ∈ B λ (θH )}.

(4.2.33)

These lemmas enable us to find the residual principle for choosing the regularization parameter α in a Hilbert space. Theorem 4.2.10 Let A : H → 2H be a maximal monotone operator with domain D(A), Ω ⊆ D(A) be a convex closed set satisfying (4.1.1) and let the variational inequality (7.1.1) with f ∈ H have a nonempty solution set N with the unique minimal norm solution x ¯∗ . h h H h Assume that a sequence of operators {A }, where A : H → 2 has domain D(A ), sequence of convex closed sets {Ωσ } ⊆ D(Ah ) and the sequence {f δ } ∈ H are known in place of A, Ω and f, respectively, such that in reality the variational inequality Ah x − f δ , z − x ≥ 0 ∀z ∈ Ωσ is solved. Let γ = (h, δ, σ) ∈ such that 0 < δ + h + σ ≤ 1. Using the perturbed data, construct the regularized inequality (4.2.4) with the condition (4.2.1). Denote its (unique) solution by xγα . Suppose that (4.1.5), (4.2.2) and (4.2.3) are satisfied and (4.2.6) holds on the set Ω ∪ Ωσ , where the function g(t) is continuous and increasing. If θH ∈ Ωσ , then the following additional condition is given:

ξ − f δ > k + M + g(0) (δ + h + σ)p , k > 1,

p ∈ (0, 1],

(4.2.34)


213

provided that 1) if θH ∈ int Ωσ then ξ ∈ Ah (θH ) and satisfies (4.2.32); 2) if θH ∈ ∂Ωσ then ξ ∈ B λ (θH ) and satisfies (4.2.33); 3) if θH ∈ Ωσ , but θH ∈ Ω, then it is assumed that θH ∈ N. Then there exists a unique α ¯ satisfying the equality

ρ(¯ α) = αx ¯ γα¯ = k + M + g(xγα¯ ) (δ + h + σ)p ,

(4.2.35)

¯ . Moreover, where xγα¯ is a solution of (4.2.4) with α = α (i) if γ → 0 then α ¯ → 0, (ii) if γ → 0 and p ∈ (0, 1) then xγα¯ → x ¯∗ and lim

γ→0

δ+h+σ = 0, α ¯

(iii) if γ → 0, p = 1 and N = {x0 } then xγα¯ x0 and there exists a constant C > 0 such that δ+h+σ ≤ C. (4.2.36) lim sup α ¯ γ→0

Proof. The monotonicity of A in the inequality (4.2.10) and the condition (4.2.2) induce the following estimate: h δ σ + M + g(xγα ) + σ + 2(x∗ + 1) α α α Take a small enough α > 0 such that for any fixed x∗ ∈ N, xγα ≤

α σ + 2(x∗ + 1) < (k − 1)(δ + h + σ)p , Then one gets

k > 1,

∀x∗ ∈ N.

p ∈ (0, 1].

ρ(α) ≤ δ + M σ + hg(xγα ) + α σ + 2(x∗ + 1)

(4.2.37)

(4.2.38)

< δ + M σ + hg(xγα ) + (k − 1)(δ + h + σ)p ≤

k + M + g(xγα¯ ) (δ + h + σ)p .

(4.2.39)

Now (4.2.34) guarantees the solvability of equation (4.2.35) if θH ∈ Ωσ . At the same time, ¯ follows from (4.2.39) and Lemma 4.2.9. By (4.2.38), we if θH ∈ Ωσ then the existence of α find that k−1 (δ + h + σ)p . α ¯> 2(x∗ + 1) + σ Consequently, 2x∗ + 3 δ+h+σ (δ + h + σ)1−p . < k−1 α ¯ Since, γ ∈ , (4.2.36) holds in the cases of p ∈ (0, 1) and p = 1 at the same time. The boundedness of {xγα¯ } results from (4.2.13). In the standard way, we establish that xγα¯

x ¯ ∈ Ω as γ → 0 and ξαγ¯ → f, where ξαγ¯ ∈ B λ (xγα¯ ) satisfies the equality

¯ xγα¯ = f δ . ξαγ¯ + α

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Other properties of {xγα¯ } are proved similarly to Theorem 4.2.5. Observe that θH ∈ N. Indeed, the inclusion θH ∈ N is possible if and only if θH ∈ Ωσ for all σ ∈ (0, σ ∗ ]. Then it is not difficult to deduce from (4.2.3) that θH ∈ Ω. Finally, one can make certain, as in the proof of Theorem 4.1.11, that (4.2.34) implies the claim. Then the convergence of α ¯ to zero follows if γ → 0. ∗

6. Let further X be an E-space, X ∗ be strictly convex, A : X → 2X and Ah : X → be maximal monotone operators. Note that condition (4.2.2) on the growth order of 2 operators Ah is quite natural just in a Hilbert space. If we consider variational inequalities in Banach spaces, then we have to replace the linear function in the right-hand side of (4.2.2) by a nonlinear one. For a short view, we presume in Theorem 4.2.6 that X∗

y h − f δ ∗ ≤ κ(x)

∀y h ∈ Ah x,

∀x ∈ Ωσ ,

(4.2.40)

where γ = (σ, h, δ) ∈ , κ(t) is a non-negative, continuous and increasing function for t ≥ 0. Define the function µ(t) such that µ(t) > max{a(t), g(t), κ(t)}, t ≥ t0 > 0, and corresponding duality mapping J µ : X → X ∗ . To define approximations to the minimal norm solution of the variational inequality (7.1.1), we apply the regularization method Ah x + αJ µ x − f δ , z − x ≥ 0

∀z ∈ Ωσ ,

x ∈ Ωσ .

(4.2.41)

As usual, denote its solutions by xγα . Theorem 4.2.11 Under the hypothesis of the present subsection, the results of Theorems 4.1.3, 4.2.1, 4.2.3, 4.2.5, 4.2.6 (and also Corollary 4.2.2) remain still valid for a solution sequence {xγα } of the variational inequality (4.2.41). Proof. The inequality (4.2.31) for xγα takes the following form:

µ(xγα ) xγα − a(x∗ )σ − x∗ − a(xγα )κ(x∗ )

− κ(xγα )a(x∗ )

−

σ α

δ h σ + g(x∗ ) − xγα α α α

h δ ∗ x − g(x∗ )x∗ ≤ 0 α α

∀x∗ ∈ N.

(4.2.42)

Owing to the properties of µ(t), we can be sure that the sequence {xγα } is bounded as α → 0. The rest of the proof follows the same scheme as in Theorems 4.1.3, 4.2.1, 4.2.3 and 4.2.5.

In order to write down the residual principle in a Banach space X, we need to obtain any upper estimate of xγα in the explicit form. Emphasize that it may be done, for example, if g(t) and κ(t) are power functions of the following kind: g(t) = c1 (ts−1 + 1),

κ(t) = c2 (ts−1 + 1),

s > 2,

4.3 Variational Inequalities with Domain Perturbations

215

with some positive constant c1 and c2 , and if µ(t) = c3 ts−1 , c3 > max{c1 , c2 }. In this special case, Theorem 3.3.4 can be stated in X. Remark 4.2.12 The conditions (4.2.1), (4.2.2) and (4.2.40) do not need to hold for every pair (Ah , Ωσ ) of sequences {Ah } and {Ωσ }. However, they should be fulfilled for such pairs which guarantee realization of the sufficiency criterion (4.2.7) for the strong convergence of regularized solutions.

4.3

Variational Inequalities with Domain Perturbations ∗

Let X be an E-space, X ∗ be a strictly convex space, A : X → 2X be a maximal monotone operator, Ω ⊂ D(A) be a closed and convex set, the inequality (7.1.1) with f ∈ X ∗ have a nonempty solution set N and the condition (4.1.1) hold. Suppose that for non-negative t, h and σ, there exist non-negative functions a1 (t), a2 (t), g1 (t) and β(h, σ) with the additional hypotheses that a2 (t) is non-decreasing and lim

t→∞

a2 (t) = 0, tµ(t)

(4.3.1)

where µ(t) is a gauge function of the duality mapping J µ . For some positive δ ∗ , h∗ and σ∗ , denote = (0, δ ∗ ] × (0, h∗ ] × (0, σ ∗ ]. We assume that for each γ = (δ, h, σ) ∈ one can ∗ define maximal monotone operators Ah : X → 2X , elements f δ ∈ X ∗ and closed convex h δ h sets Ωσ ⊆ D(A ) which are (f , A , Ωσ )−approximations of (f, A, Ω), satisfying (4.1.5), (4.2.1) and the following conditions: (i) for each x ∈ Ω there exists zσ ∈ Ωσ such that x − zσ ≤ a1 (x)σ,

(4.3.2)

d∗ (ζ, Ah zσ ) = inf {ζ − u∗ | u ∈ Ah zσ } ≤ g1 (ζ∗ )β(h, σ);

(4.3.3)

and then for each ζ ∈ Ax,

(ii) for each w ∈ Ωσ there exists v ∈ Ω such that w − v ≤ a2 (w)σ.

(4.3.4)

Observe that hypotheses (i) and (ii) do not require that D(A) and D(Ah ) coincide. Moreover, Ωσ ⊆ D(A) in general. We study again the regularized inequality (4.2.41). Let xγα be its (unique) solution.

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4


Theorem 4.3.1 Under the assumptions above, let θX ∈ N and lim

α→0

δ + σ + β(h, σ) = 0. α

(4.3.5)

Then solution sequence {xγα } of the variational inequality (4.2.41) strongly converges, as α → 0, to the minimal norm solution x ¯∗ of (7.1.1). Proof. Take x∗ ∈ N ⊂ Ω and ζ ∗ ∈ Ax∗ such that ζ ∗ − f, z − x∗ ≥ 0 ∀z ∈ Ω.

(4.3.6)

By conditions (4.3.2) and (4.3.3) for x = x∗ , there are elements zσ ∈ Ωσ and ζ λ ∈ Ah zσ with λ = (h, σ) satisfying the inequalities

and

x∗ − zσ ≤ a1 (x∗ )σ

(4.3.7)

ζ ∗ − ζ λ ∗ ≤ g1 (ζ ∗ ∗ )β(h, σ).

(4.3.8)

Let ζαγ ∈ Ah xγα be such that ζαγ + αJ µ xγα − f δ , z − xγα ≥ 0

∀z ∈ Ωσ .

(4.3.9)

By (4.3.4), there exists uγα ∈ Ω such that xγα − uγα ≤ a2 (xγα )σ.

(4.3.10)

Put z = uγα in (4.3.6) and z = zσ in (4.3.9). Adding the obtained inequalities we have ζαγ − f δ , xγα − zσ + αJ µ xγα , xγα − zσ + ζ ∗ − f, x∗ − uγα ≤ 0 or αJ µ xγα , xγα − zσ + ζαγ − ζ λ , xγα − zσ + ζ λ − f δ , xγα − zσ + ζ ∗ − f, x∗ − uγα ≤ 0. Since Ah are monotone operators, the second term of the last inequality is non-negative. Hence, αJ µ xγα , xγα ≤ αJ µ xγα , zσ + ζ ∗ − ζ λ , xγα − x∗ + ζ ∗ − f, uγα − xγα + f δ − f, xγα − x∗ + ζ λ − f δ , zσ − x∗ . Further, from (4.1.5), (4.3.7), (4.3.8) and (4.3.10), one gets, respectively: f δ ∗ ≤ f ∗ + δ,

(4.3.11)

4.3 Variational Inequalities with Domain Perturbations

217

zσ ≤ x∗ + a1 (x∗ )σ, ζ λ ∗ ≤ ζ ∗ ∗ + g1 (ζ ∗ ∗ )β(h, σ), uγα ≤ xγα + a2 (xγα )σ. Now we can evaluate all the terms in the right-hand side of (4.3.11) in the following way:

αJ µ xγα , zσ ≤ αµ(xγα ) x∗ + a1 (x∗ )σ , ζ ∗ − ζ λ , xγα − x∗ ≤ (xγα + x∗ )g1 (ζ ∗ ∗ )β(h, σ), ζ ∗ − f, uγα − xγα ≤ ζ ∗ − f ∗ a2 (xγα )σ, f δ − f, xγα − x∗ ≤ (xγα + x∗ )δ,

ζ λ − f δ , zσ − x∗ ≤ ζ ∗ ∗ + f ∗ + δ + g1 (ζ ∗ ∗ )β(h, σ) σ. With due regard for these facts, (4.3.11) leads to the inequality

µ(xγα ) xγα − x∗ − a1 (x∗ )σ −

≤

δ

α

+

β(h, σ) g1 (ζ ∗ ∗ ) (xγα + x∗ ) α

σ ∗ ζ − f ∗ a2 (xγα ) + ζ ∗ ∗ + f ∗ + δ + g1 (ζ ∗ ∗ )β(h, σ) . α

(4.3.12)

¯ ∈ X as α → 0. Since Then (4.3.12) and (4.3.1) guarantee that {xγα } is bounded and xγα x the function a2 (t) is non-decreasing, the estimate (4.3.10) implies the weak convergence of ¯ ∈ X. We recall that uγα ∈ Ω for all α > 0 and γ ∈ . Therefore, by the Mazur {uγα } to x theorem, x ¯ ∈ Ω. We show that x ¯ ∈ N. Due to the condition (4.2.1) and Lemma 1.11.4, solutions xγα of (4.2.41) satisfy the following inequality: ζ h + αJ µ z − f δ , z − xγα ≥ 0 ∀z ∈ Ωσ , ∀ζ h ∈ Ah z.

(4.3.13)

Choose and fix u ∈ Ω and η ∈ Au. The conditions (4.3.2) and (4.3.3) enable us to construct elements vσ ∈ Ωσ and η λ ∈ Ah vσ such that u − vσ ≤ a1 (u)σ

(4.3.14)

η − η λ ∗ ≤ g1 (η∗ )β(h, σ).

(4.3.15)

and Put z = vσ and ζ h = ηλ in (4.3.13). Then we come to the relation η λ + αJ µ vσ − f δ , vσ − xγα ≥ 0.

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4


Passing to the limit as α → 0 and taking into account (4.1.5), (4.3.5), (4.3.14), (4.3.15) and ¯ ∈ Ω, one gets the weak convergence of {xγα } to x η − f, u − x ¯ ≥ 0 ∀u ∈ Ω, ∀η ∈ Au. By (4.1.1) and by Lemma 1.11.4, we conclude that x ¯ ∈ N. It is not difficult to see from (4.3.12) that there exists a constant M > 0 such that µ(xγα )xγα ≤ µ(xγα )x∗ + M

δ + σ + β(h, σ) . α

(4.3.16)

¯ as α → 0 yields By the hypothesis, x ¯ = θX . Therefore, the weak convergence of {xγα } to x 0 < ¯ x ≤ lim inf xγα .

(4.3.17)

α→0

Hence, if α is a sufficiently small positive parameter, then there exists τ > 0 such that τ ≤ xγα , and µ(xγα ) ≥ µ(τ ) > 0, provided that a gauge function µ(t) of J µ is increasing. Consequently, the estimate xγα ≤ x∗ +

M δ + σ + β(h, σ) µ(τ ) α

∀x∗ ∈ N

follows from (4.3.16). Owing to (4.3.5), we now deduce that lim sup xγα ≤ x∗ ∀x∗ ∈ N.

(4.3.18)

α→0

Combining (4.3.17) and (4.3.18) gives ¯ x ≤ lim inf xγα ≤ lim sup xγα ≤ x∗ ∀x∗ ∈ N. α→0

(4.3.19)

α→0

This shows that x ¯=x ¯∗ . If we put x∗ = x ¯∗ in (4.3.19), then xγα → ¯ x∗ . The theorem is proved in an E-space X. Remark 4.3.2 Let A be a bounded operator on Ω and ζ∗ ≤ ϕ(x) ∀x ∈ Ω,

∀ζ ∈ Ax,

where ϕ(t) is a non-negative and non-decreasing function. If g1 (t) is also non-decreasing, then the inequality (4.3.3) can be rewritten as d∗ (ζ, Ah zσ ) ≤ g2 (x)β(h, σ) ∀ζ ∈ Ax, ∀x ∈ Ω,

(4.3.20)

with g2 (t) = g1 (ϕ(t)). Theorem 4.3.1 remains still valid if (4.3.3) is replaced by (4.3.20).

4.4 Examples of Variational Inequalities

4.4

219

Examples of Variational Inequalities

We present examples of variational inequalities on exactly and approximately given sets. Example 4.4.1 Consider the contact problem of deformable bodies with an ideally smooth boundary. Let there be a beam of length l with firmly fixed edges. Direct the axis OX on the axis of the beam and OY up through the beam left end. Let E(x) be an elastic modulus, I(x) the moment of inertia of a section and k(x) = E(x)I(x), x ∈ [0, l]. The beam bends under the influence of the force generated by a fixed rigid body pressing doun it, with absolutely smooth contact surface defined by the equation y = χ(x); points lying beyond this surface satisfy the inequality y ≤ χ(x). The following differential equation defines the equilibrium state of the beam:

d2 w(x) d2 = −q(x), k(x) dx2 dx2

(4.4.1)

where w(x) is a beam deformation and q(x) is a rigid body reaction, with the boundary conditions w(0) = w(l) = w (0) = w (l) = 0. (4.4.2) Besides, we have: a) the one-sided restriction w(x) ≤ χ(x)

(4.4.3)

characterizing the non-penetration of beam points into the rigid body; b) the condition q(x) ≥ 0

(4.4.4)

determining the direction of the rigid body reaction; c) the equation w(x) − χ(x) q(x) = 0, x ∈ (0, l),

(4.4.5)

implying that either (4.4.3) or (4.4.4) turns into a strict equality at any point x ∈ (0, l). Introduce the space X = {v | v = v(x), x ∈ [0, l], v ∈ W22 [0, l], v(0) = v(l) = v (0) = v (l) = 0} and define v =

l 0

1/2

|v (x)|2 dx

.

Next, we assume that any solution w(x) of the given problem has derivatives up to the fourth order. By (4.4.1),

l 0

d2 w(x) d2 v(x)dx = − k(x) 2 dx2 dx

l 0

q(x)v(x)dx ∀v ∈ X,

w ∈ X.

(4.4.6)

220

4


That the obtained equality is equivalent to (4.4.1) results from the basic lemma of the calculus of variations ([84], Lemma 1, p.9). Apply twice the formula of integration by parts to the left-hand side of (4.4.6). Taking into account the boundary conditions (4.4.2) we deduce l

0

k(x)w (x)v (x)dx = −

l

0

q(x)v(x)dx ∀v ∈ X,

w ∈ X.

(4.4.7)

The function w(x) satisfying (4.4.7) is called a weak solution of the problem (4.4.1) - (4.4.5). Suppose that in (4.4.7) v(x) = u(x)−w(x), where u ∈ X, and analyze the sign of the function

ξ(x) = q(x) u(x) − w(x) . If there is no contact between the beam and rigid body at a point x ∈ (0, l), then q(x) = 0 and ξ(x) = 0. If there is contact at that point then we assert that u(x) ≤ w(x) and, hence, ξ(x) ≤ 0. Consequently, solution w(x) of the problem (4.4.1) - (4.4.5) satisfies the variational inequality

0

l

k(x)w (x) u (x) − w (x) dx ≥ 0 ∀u ∈ Ω,

w ∈ Ω,

(4.4.8)

where Ω = {v | v = v(x) ∈ X, v(x) ≤ χ(x) for almost all x ∈ [0, l]} is a convex closed subset in X. Let w(x) now be a solution of (4.4.8). If there is no contact between the beam and rigid body at a point x ∈ (0, l), then difference u(x) − w(x) may take both positive and negative values. Integrating the left-hand side of (4.4.8) by parts twice and taking into account again the boundary conditions (4.4.2), we come to the inequality

l 0

d2 w(x) d2 k(x) u(x) − w(x) dx ≥ 0 ∀u ∈ Ω, 2 2 dx dx

w ∈ Ω.

(4.4.9)

Then by repeating the reasoning of the proof of the basic lemma of the calculus of variations, we establish that d2 w(x) d2 = 0. (4.4.10) k(x) dx2 dx2 Let x ∈ (0, l) be a contact point of the beam and rigid body. Then u(x) − w(x) ≤ 0. Using (4.4.9) and the proof by contradiction, we deduce the inequality

d2 w(x) d2 = −q(x) ≤ 0. k(x) dx2 dx2

(4.4.11)

By virtue of (4.4.10) and (4.4.11), w(x) satisfies the equation (4.4.1) with condition (4.4.4). Thus, the equivalence of the problem (4.4.1) - (4.4.5) and variational inequality (4.4.8) has been established. We introduce the operator A : X → X ∗ by the equality Aw, v =

l 0

k(x)w (x)v (x)dx ∀w, v ∈ X.


221

One can verify that A is a linear, continuous, monotone and potential map [83, 120, 128]. Using the natural assumption: There exists a constant C > 0 such that |k(x)| ≤ C ∀x ∈ [0, l], we are able to prove the boundedness of A. By the definition of A, inequality (4.4.8) can be rewritten in the form Aw, u − w ≥ 0 ∀u ∈ Ω, w ∈ Ω. (4.4.12) We assume that, in place of k(x), a perturbed function kh (x) is given, which depends on the positive parameter h, such that |k(x) − kh (x)| ≤ h ∀x ∈ [0, l].

(4.4.13)

The function k(x) induces the perturbed operator Ah : X → X ∗ as follows: Ah w, v =

l

kh (x)w (x)v (x)dx ∀w, v ∈ X.

0

Then Av − Ah v ≤ hv ∀v ∈ X, that is, in the condition (4.2.29) g(t) ≡ t. Moreover, the operators Ah and A have the same properties. The duality mapping J in X can be defined as follows: Jw, v =

l 0

w (x)v (x)dx ∀w, v ∈ X.

Let wαh ∈ Ω be a unique solution of the regularized variational inequality l 0

k h (x)w (x) + αw (x)

u (x) − w (x) dx ≥ 0 ∀u ∈ Ω,

w ∈ Ω.

Then Ah wαh + αJwαh , u − wαh ≥ 0 ∀u ∈ Ω. Suppose that the variational inequality (4.4.12) is solvable and parameter h in (4.4.13) is such that h lim = 0. α→0 α Then Theorem 4.1.1 implies the strong convergence of {wαh } to w ¯ ∗ as α → 0, where w ¯ ∗ is the minimal norm solution of (4.4.12). Observe that the equation of the residual principle (4.1.44) accepts the form:

ρ(¯ α) = αw ¯ αh¯ = M + wαh¯ hs , M > 1,

s ∈ (0, 1].

222

4


Example 4.4.2 We investigate the membrane bend problem, whose deflection is restricted by the hard fixed obstacle given by function z = χ(x, y), where (x, y) ∈ D, D is a bounded domain in the plane XOY with boundary ∂D = Γ. Let the membrane be jammed at the contour Γ, charged with pressure P, perpendicular to a median plane. Let deflection w(x, y) of the membrane occur in the direction of the axis OZ and function Q(x, y) be the obstacle reaction. Considering the problem to be nonlinear we come to the following equation [120]: −

∂ ∂ a2 (x, y, wy ) = −P (x, y, w) − Q(x, y), a1 (x, y, wx ) − ∂y ∂x

(4.4.14)

with w(x, y) |Γ = 0, w(x, y) ≤ χ(x, y),

(4.4.15)

Q(x, y) ≥ 0, Q(x, y) w(x, y) − χ(x, y) = 0.

(4.4.16)

Assume that the nonlinear functions in (4.4.14) satisfy the following conditions: a) ai (x, y, ξ), i = 1, 2 , and P (x, y, ξ) are measurable on D for all ξ ∈ R1 , continuous and non-decreasing with respect to ξ for almost all x, y ∈ D; p b) there exist ci > 0 and ki (x, y) ∈ Lq (D), q = , p > 2, i = 1, 2, 3, such that p−1

|ai (x, y, ξ)| ≤ ci ki (x, y) + |ξ|p−1 , i = 1, 2, and

|P (x, y, ξ)| ≤ c3 k3 (x, y) + |ξ|p−1 . Introduce the space X = {w | w = w(x, y) ∈ W1p (D), (x, y) ∈ D, w(x, y)|Γ = 0} and define w =

D

|wx (x, y)|p + |wy (x, y)|p dxdy

1/p

.

Construct the operator A as follows: For all v, w ∈ X Aw, v =

D

a1 (x, y, wx )vx + a2 (x, y, wy )vy + P (x, y, w)v dxdy.

(4.4.17)

It follows from the conditions b) that the operator A acts from X in X ∗ [128, 220, 221], while the properties of the functions a1 (x, y, ξ), a2 (x, y, ξ) and P (x, y, ξ) enable us to assert that A is monotone (see Section 1.3, Examples 5, 6). Moreover, it is possible to prove that A is bounded and potential [83, 143]. Repeating the arguments of the previous example, we make sure that the problem (4.4.14) - (4.4.16) can be stated as variational inequality of the type (4.4.12) on the set Ω = {w | w = w(x, y) ∈ X, w(x, y) ≤ χ(x, y), for almost all (x, y) ∈ D}.


223

Suppose that an operator Ah : X → X ∗ is defined by the equality (4.4.17) with perturbed functions ah1 (x, y, ξ), ah2 (x, y, ξ) and P h (x, y, ξ), instead of a1 (x, y, ξ), a2 (x, y, ξ), P (x, y, ξ), which also satisfy a) and b). In addition, for any element u ∈ X, let the following inequalities hold: g (u), ah1 (x, y, ux ) − a1 (x, y, ux )Lq ≤ h¯ ah2 (x, y, uy ) − a2 (x, y, uy )Lq ≤ h¯ g (u), g (u), P h (x, y, u) − P (x, y, u)Lq ≤ h¯ where g¯(t) is a non-negative, continuous and increasing function for t ≥ 0. Then it is not difficult to be sure that there exists a constant C > 0 such that Au − Ah u∗ ≤ Ch¯ g (u) ∀u ∈ X.

(4.4.18)

The duality mapping J is defined by the following expression: Ju, v = u2−p

D

|ux |p−2 ux vx + |uy |p−2 uy vy dxdy

∀u, v ∈ X, u = θX , JθX = θX ∗ . Now the regularized variational inequality and the equation of the residual principle are written similarly to Example 4.4.1. Example 4.4.3 Consider again Example 4.4.1 but now on approximately given sets Ωσ , which is defined by the inexact function χσ (x) in place of χ(x). Denote Ωσ = {v | v = v(x) ∈ X, v(x) ≤ χσ (x) for almost all x ∈ [0, l]}, assuming that χσ (x) − χ(x) ≤ σ.

(4.4.19)

We need to get the estimate (4.3.2) for z − zσ , where z ∈ Ω, zσ ∈ Ωσ . However, (4.4.19) is not enough for this aim. It is possible to apply the inequality [83] k

∂ m v(x, y) p ∂ v(x, y) p dxdy ≤ C k m dxdy, k m G

∂x 1 ∂y

2

m

G

∂x

1

∂y

2

where p > 1, k = k1 + k2 , m = m1 + m2 , 0 ≤ k ≤ m, and positive C depends on p, k, m and G. As a result, the required estimate is obtained only in the space L2 [0, l]. Observe that the function w (x), the modulus of which coincides with a curvature of the curve y = w(x) in the framework of the analyzed model, is the main characteristic of the problem we are considering [90, 120]. This fact enables us to solve the problem changing its statement and preserving the basic requirements (4.4.1) - (4.4.5).

224

4


For variational inequality (4.4.12), we introduce the set Ω = {v | v = v(x) ∈ Ω, v (x) ≤ χ (x) for almost all x ∈ [0, l]}, and define perturbed sets as follows: Ωσ = {v | v = v(x) ∈ Ωσ , v (x) ≤ χσ (x) for almost all x ∈ [0, l]}. Moreover, we assume that χσ (x) − χ (x)2 ≤ σ ∀x ∈ [0, l].

(4.4.20)

It is then obvious that for every element z ∈ Ω (z ∈ Ωσ ) there exists an element zσ ∈ Ωσ (zσ ∈ Ω ) such that z − zσ ≤ σ, that is, conditions (4.3.2) and (4.3.4) of Theorem 4.3.1 are satisfied with a1 (t) = a2 (t) = 1 for t ≥ 0. Since Av − Ah v ≤ hv and Ah are bounded, we conclude that there exists a constant c > 0 such that Az − Ah zσ ≤ Az − Ah z + Ah z − Ah zσ ≤ hz + cσ ≤ (z + c)(h + σ). Hence, the requirement (4.3.3) of Theorem 4.3.1 holds. Observe that the replacement of Ω and Ωσ by Ω and Ωσ , respectively, does not disturb the structure and qualitative characteristics of a desired solution; however, it allows us to describe perturbations of the set Ω satisfying the conditions of Theorem 4.3.1. This replacement may be treated as a construction of the set on which the regularizing operator is formed or as accounting for a priori information about unknown solutions. Assume that the variational inequality Aw, v − w ≥ 0 ∀v ∈ Ω , w ∈ Ω ,

(4.4.21)

is solvable, and in (4.4.13) and (4.4.20) h+σ → 0 as α → 0. α Next we find a solution sequence {wαγ } of the regularized variational inequality: Ah wαγ + αJ µ wαγ , v − wαγ ≥ 0 ∀v ∈ Ωσ ,

wαγ ∈ Ωσ ,

where γ = (h, σ). Then the strong convergence of {wαγ } to the minimal norm solution of (4.4.21) as α → 0 is proved by Theorem 4.3.1. It is known [120] that every solution of (4.4.21) is a solution of problem (4.4.1) - (4.4.5). Thus, we have constructed the sequence {wαγ } which converges in the norm of the space X to some solution of the problem (4.4.1) - (4.4.5).


225

Remark 4.4.4 In the definitions of Ω and Ωσ , we can change χ (x) and χσ (x) by the functions ψ(x) and ψσ (x) such that χ (x) ≤ ψ(x),

χσ (x) ≤ ψσ (x)

and ψ(x) − ψσ (x)2 ≤ σ ∀x ∈ [0, l]. Example 4.4.5 Consider Example 4.4.2 again. By the same reasoning as in the previous example, define sets Ω = {v | v = v(x, y) ∈ Ω, vx (x, y) ≤ χx (x, y), vy (x, y) ≤ χy (x, y) for almost all (x, y) ∈ D} and Ωσ = {v | v = v(x, y) ∈ Ωσ , vx (x, y) ≤ (χσ )x (x, y), vy (x, y) ≤ (χσ )y (x, y) for almost all x, y ∈ D}. Let, for all x, y ∈ D and for τ =

1 , the following inequalities hold: p

χx (x, y) − (χσ )x (x, y)Lp ≤

and

σ 2τ

σ . 2τ Under these conditions, it is not difficult to verify that d(u, Ωσ ) ≤ σ for all u ∈ Ω and d(u, Ω ) ≤ σ for all u ∈ Ωσ . In order to evaluate from above the norm Aw − Ah wσ ∗ , where w ∈ Ω , wσ ∈ Ωσ and w − wσ ≤ σ, we assume that [220]: χy (x, y) − (χσ )y (x, y)Lp ≤

|(ahi )ξ (x, y, ξ)| ≤ ci di (x, y) + |ξ|p−2 , i = 1, 2,

(4.4.22)

and

|(P h )ξ (x, y, ξ)| ≤ c3 d3 (x, y) + |ξ|p−2 .

(4.4.23)

p . Applyp−2 ing the Lagrange formula, the H¨ older and Minkovsky integral inequalities and taking into account the conditions (4.4.22) and (4.4.23), write down the chain of relations

Here ci > 0, di (x, y) ∈ Lp (D) are non-negative functions for i=1,2,3, p =

D

= D

ah1 (x, y, wx ) − ah1 (x, y, ux ) vx dxdy

(ah1 )ξ x, y, wx + θ(ux − wx ) (wx − ux )vx dxdy

226

4 ≤ v


# $ p−1 p p p p−1 h |wx − ux | p−1 dxdy (a1 )ξ x, y, wx + θ(ux − wx ) D

#

≤ c1 vw − u

D

d1 (x, y) + |wx + θ(ux − wx )|p−2

≤ c1 vw − u

+ D

D

|wx |p dxdy +

D

p p−2

$ p−2

dxdy

p

p−2

p

(d1 (x, y)) p−2 dxdy

|ux |p dxdy

p

p−2 p

≤ c1 vw − u(d1 (x, y)Lp + 2Rp−2 ), where 0 < θ < 1, R = max {w, u}. Evaluating analogously two other terms of the dual product Ah w − Ah u, v , namely, D

and

ah2 (x, y, wy ) − ah2 (x, y, uy ) vy dxdy

D

P h (x, y, w) − P h (x, y, u) vdxdy,

we come to the following estimate: Ah w − Ah u∗ ≤ c (m + 2Rp−2 )w − u, where c =

3

ci and m =

i=1

3

(4.4.24)

di (x, y)q .

i=1

Consequently, we established that an operator Ah is Lipschitz-continuous on each bounded set. By (4.4.18) and (4.4.24), if w − wσ ≤ σ then Aw − Ah wσ ∗ ≤ Aw − Ah w∗ + Ah w − Ah wσ ∗ ≤

C g¯(w) + c m + 2(w + σ ¯ )p−2

(h + σ),

σ ∈ (0, σ ¯ ).

Thus, the conditions (4.3.2) - (4.3.4) of Theorem 4.3.1 are also satisfied for problem (4.4.14) - (4.4.16). Example 4.4.6

Suppose that in a bounded set G ⊂ R3 , the filtration equation

−div(g(x, |∇u|2 )∇u) = f (x),

x ∈ G, ∂G = Γ = Γ0 + Γ1 + Γ2 ,

(4.4.25)

is solved (see Example 8 in Section 1.3) with the following boundary conditions: u(x) |Γ0 = 0, v¯ · n ¯ |Γ1 = 0,

(4.4.26)

4.5 Variational Inequalities with Unbounded Operators

227

¯ |Γ2 ≤ 0 as u(x) = 0, v¯ · n ¯ |Γ2 = 0 as u(x) ≥ 0 and v¯ · n

(4.4.27)

g(x, |∇u|2 )∇u,

n ¯ is a unit vector of exterior normal, v¯ · n ¯ is the scalar product where v¯(x) = of the vectors v¯ and n ¯ . Assume that in place of G, its approximations Gσ ⊂ R3 with ∂Gσ = Γ0 + Γ1σ + Γ2σ are known. Let G σ be a symmetric difference of sets G and Gσ , that is, G σ = (G ∪ Gσ )\(G ∩ Gσ ),

and let measure µ(G σ ) ≤ σ. Introduce the set D = σ≥0 Gσ , where G0 = G, and define the space X = {u | u = u(x) ∈ W1p (D), u |Γ0 = 0} with the norm

u =

1/p

D

|∇u|p dx

, x ∈ R3 .

Construct sets Ωσ , σ ≥ 0, as follows: Ω0 = Ω and if σ = 0 then Ωσ = {u | u = u(x) ∈ X, u(x) ≥ 0 for almost all x ∈ Γ2σ , u(x) = 0 as x ∈ D\Gσ }. It is well known (see, for instance, [123]) that if Γ and w(x) are sufficiently smooth, then the problem (4.4.25) - (4.4.27) is equivalent to the variational inequality Aw, w − v =

G

g(x, |∇w|2 )∇w · ∇(w − v) − f (x)(w − v) dx ≤ 0 ∀v ∈ Ω,

w ∈ Ω,

with the monotone operator A : X → X ∗ . Let u(x) ∈ Ω. Consider the function uσ (x) ∈ Ωσ , which coincides with u(x) on Gσ ∩ G, and assume that u(x) = 0 at any point x ∈ Gσ \G. Then u − uσ ≤

1/p

|∇u| dx p

G\Gσ

.

Since the Lebesgue integral is absolutely continuous [117], we deduce the inequality u − uσ ≤ a(u, σ), where a(u, σ) → 0 as σ → 0 for fixed u(x). However, we cannot obtain the function a(u, σ) in the analytical form and prove the convergence theorems for the regularization methods.

4.5

Variational Inequalities with Unbounded Operators

As in Section 4.2, let = (0, δ ∗ ]×(0, h∗ ]×(0, σ ∗ ], let X be a reflexive strictly convex Banach ∗ ∗ space, X ∗ be also strictly convex, A : X → 2X and Ah : X → 2X be maximal monotone

228

4


operators for all h ∈ (0, h∗ ], D(A) = D(Ah ), sets Ω ⊆ int D(A) and Ωσ ⊆ int D(Ah ) as σ ∈ (0, σ ∗ ], Ωσ uniformly approximate Ω with the estimate HX (Ω, Ωσ ) ≤ σ.

(4.5.1)

We also assume that there is no inequality of the kind (4.2.2) which imposes the growth condition upon the operators Ah . At the same time, the conditions (4.1.1), (4.1.5), (4.1.6) and (4.2.1) are fulfilled for all x ∈ Ω ∪ Ωσ , where, in addition, a function g(t) is bounded on bounded sets. Let xγα ∈ Ωσ with γ = (δ, h, σ) be a solution of the variational inequality Ah x + αJx − f δ , z − x ≥ 0 ∀z ∈ Ωσ . Theorem 4.5.1 There exists a constant r0 > 0 such that (4.2.7) is satisfied.

xγα

→

x ¯∗

(4.5.2) if σ < min {r0

, σ∗}

and

Proof. First of all, prove that {xγα } is bounded as α → 0. To this end, consider the auxiliary variational inequality problem: To find x ∈ Ωσ such that Ax + αJx − f, z − x ≥ 0 ∀z ∈ Ωσ .

(4.5.3) {xσα }

Denote by its unique solution for fixed α > 0 and σ ∈ . Show that is bounded as α → 0. By Lemma 1.5.14, there exist constants r0 > 0 and c0 > 0 such that xσα

D1 = yασ − f, xσα − x ¯∗ ≥ r0 yασ − f ∗ − c0 (xσα − x ¯∗ + r0 ), where

(4.5.4)

x∗ , r0 )} < ∞. c0 = sup {y − f ∗ | y ∈ Ax, x ∈ B(¯

Take yασ ∈ Axσα satisfying the inequality yασ + αJxσα − f, z − xσα ≥ 0 ∀z ∈ Ωσ .

(4.5.5)

Evaluate D1 from above. Since (4.5.1) holds for σ ∈ , there exists zσ ∈ Ωσ such that ¯ x∗ − zσ ≤ σ and at that yασ + αJxσα − f, zσ − xσα ≥ 0. Therefore, by the relations Jxσα , xσα − x ¯∗ = xσα 2 − Jxσα , x ¯∗ ≥ −xσα ¯ x∗ , one has D1 = yασ + αJxσα − f, xσα − x ¯∗ − αJxσα , xσα − x ¯∗ = yασ + αJxσα − f, xσα − zσ + yασ + αJxσα − f, zσ − x ¯∗ ¯∗ ≤ σyασ + αJxσα − f ∗ + αxσα ¯ x∗ − αJxσα , xσα − x ≤ αxσα (¯ x∗ + σ) + σyασ − f ∗ .

(4.5.6)


229

Using (4.5.4) and (4.5.6) we obtain x∗ + σ) + σyασ − f ∗ ≥ r0 yασ − f ∗ − c0 (xσα + ¯ x∗ + r0 ). αxσα (¯ Hence,

x∗ + ασ + c0 )xσα + c0 (¯ x∗ + r0 ). (r0 − σ)yασ − f ∗ ≤ (α¯

(4.5.7)

Since σ < r0 , we come to the following estimate: yασ − f ∗ ≤ M1 xσα + M2 , where M1 ≥

(4.5.8)

α¯ x∗ + ασ + c0 r0 − σ

and M2 ≥

c0 (¯ x∗ + r0 ) . r0 − σ

Thus, A has not more than linear growth on solutions xσα of the variational inequality (4.5.3). We emphasize that this fact is proved only by the condition that the semi-deviation β(Ω, Ωσ ) ≤ σ. Let now xα ∈ Ω be a solution of the variational inequality Ax + αJx − f, z − x ≥ 0 ∀z ∈ Ω.

(4.5.9)

By virtue of the monotonicity of A and by the properties of J, we get D2 = yασ + αJxσα − yα − αJxα , xσα − xα ≥ α(xσα − xα )2 , where yα ∈ Axα satisfies the inequality yα + αJxα − f, z − xα ≥ 0 ∀z ∈ Ω. On the other hand, there exist elements zασ ∈ Ωσ and uσα ∈ Ω such that zασ − xα ≤ σ and uσα − xσα ≤ σ. Then firstly yασ + αJxσα − f, xσα − zασ ≤ 0, and secondly yα + αJxα − f, xα − uσα ≤ 0. We come to the upper estimate of D2 , namely, D2 ≤ σ(yασ + αJxσα − f ∗ + yα + αJxα − f ∗ ). ¯∗ as α → 0 follows from the proof of Theorem 4.1.1, Convergence of the sequence {xα } to x while (4.5.8) implies the boundedness of {yα } if we put there σ = 0. Hence, there exists a constant C1 > 0 such that yα + αJxα − f ∗ ≤ C1 .

230

4


Then D2 ≤ σ (yασ − f ∗ + αxσα + C1 ) ≤ σ [(M1 + α)xσα + M2 + C1 ] , and we obtain the following quadratic inequality with respect to xσα : σ σ σ + xα 2 − (C1 + M2 ) ≤ 0. M1 + xσα 2 − 2xσα xα + α 2α 2 It is easy to calculate that σ σ M1 + xσα ≤ xα + 2 2α

+

σ σ M1 + 2 2α

2

+ 2xα

σ σ M1 + 2 2α

+

σ (C1 + M2 ) α

(4.5.10)

$1/2

≤ C2 ,

σ ≤ where C2 > 0. Owing to (4.2.7) one can consider, without loss of generality, that α C, C > 0. Then the sequence {xσα } is uniformly bounded because σ < r0 . Now we are going to show that the solutions xγα are bounded. To this end, evaluate from above the following expression:

D3 = yαγ + αJxγα − f δ − yασ − αJxσα + f, xγα − xσα , where yαγ ∈ Ah xγα satisfies the inequality yαγ + αJxγα − f δ , z − xγα ≥ 0 ∀z ∈ Ωσ .

(4.5.11)

Since operators Ah are monotone, we have D3 ≥ α (xγα − xσα )2 − δxγα − xσα − hg(xσα )xγα − xσα . In view of (4.5.5) and (4.5.11), it is not difficult to see that D3 ≤ 0. Hence, αxγα 2 − 2αxγα xσα + αxσα 2 − δxγα − δxσα − hg(xσα )xγα − hg(xσα )xσα ≤ 0. Consequently, we obtain the quadratic inequality again but now with respect to xγα :

xγα 2 − 2xγα xσα + −

δ + hg(xσα ) + xσα 2 2α

h δ σ x − g(xσα )xσα ≤ 0, α α α

which yields the estimate xγα ≤ xσα + +

# δ

2α

+

δ h + g(xσα ) 2α 2α

δ + hg(xσα ) hg(xσα ) 2 + 2xσα α 2α

$1/2

.


231

δ + h

are bounded as α → 0, there exists a constant C3 > 0 α such that xγα ≤ C3 . Prove the boundedness of yαγ as α → 0. For this, establish first the boundedness of elements vαγ ∈ Axγα such that yαγ − vαγ ≤ g(xγα )h. (4.5.12) Since the sequences {xσα } and

Observe that {vαγ } exists according to (4.1.6). By Lemma 1.5.14, ¯∗ ≥ r0 vαγ − f ∗ − c0 (r0 + xγα − x ¯∗ ). D4 = vαγ − f, xγα − x

(4.5.13)

Evaluate D4 from above. We have ¯∗ D4 = vαγ − yαγ + yαγ − f δ + f δ − f, xγα − x ≤ hg(xγα )xγα − x ¯∗ + δxγα − x ¯∗ ¯∗ + yαγ + αJxγα − f δ , xγα − x ¯∗ . − αJxγα , xγα − x We present the last term in the equivalent form yαγ + αJxγα − f δ , xγα − x ¯∗ = yαγ + αJxγα − f δ , xγα − zσ + zσ − x ¯∗ , where zσ ∈ Ωσ such that

¯∗ ≤ σ. zσ − x

Taking into account the inequality yαγ + αJxγα − f δ , zσ − xγα ≥ 0, we deduce yαγ + αJxγα − f δ , xγα − x ¯∗ ≤ yαγ + αJxγα − f δ , zσ − x ¯∗ ≤ σyαγ + αJxγα − f δ ∗ . Then ¯∗ + δxγα − x ¯∗ D4 ≤ hg(xγα )xγα − x + αxγα xγα − x ¯∗ + σyαγ + αJxγα − f δ ∗ . By the fact that yαγ + αJxγα − f δ ∗ ≤ yαγ − vαγ ∗ + vαγ − f δ ∗ + αxγα , we obtain the final estimate D4 ≤ hg(xγα )xγα − x ¯∗ + δxγα − x ¯∗ + αxγα xγα − x ¯∗ + σhg(xγα ) + σvαγ − f δ ∗ + ασxγα .

(4.5.14)

232

4


It follows from (4.5.13) and (4.5.14) that if σ < r0 then vαγ − f ∗ ≤

1 (c0 + hg(xγα ) + δ + αxγα ) xγα − x ¯∗ r0 − σ

+ c0 r0 + σ (hg(xγα ) + αxγα ) , which implies the boundedness of {vαγ }. Owing to (4.5.12), this allows us to conclude that {yαγ } is also bounded. As usual, from the sequence {xγα }, one can choose a subsequence which weakly converges to some x ¯ ∈ X as α → 0. Then, similarly to Theorem 4.2.5, we establish the inclusion x ¯ ∈ N. ¯∗ as α → 0 is done by the previous The proof of the last assertions that x ¯=x ¯∗ and xγα → x scheme (see Sections 4.1 and 4.2).

Further we use the weaker approximation criterion of the exact set Ω by perturbed sets Ωσ . Assume that there exist bounded functionals ϕ1 and ϕ2 with dom ϕ1 and dom ϕ2 , respectively, such that Ω ⊂ dom ϕ1 , Ωσ ⊂ dom ϕ2 , d(x, Ωσ ) ≤ σ1 ϕ1 (x) ∀x ∈ Ω,

(4.5.15)

d(x, Ω) ≤ σ2 ϕ2 (x) ∀x ∈ Ωσ

(4.5.16)

and lim sup x→∞

ϕ (x) 2

x2

| x ∈ D(ϕ2 ) = 0.

(4.5.17)

Let σ = max{σ1 , σ2 }. We note the changes in the proof of Theorem 4.5.1 that are brought about by these conditions. Instead of (4.5.6), we have the following estimate:

D1 ≤ αxσα ¯ x∗ + σ1 ϕ1 (¯ x∗ ) + σ1 yασ − f ∗ ϕ1 (¯ x∗ ). Therefore, if σ
0 such that a gauge function µ(t) ≥ max {g(t), a(t), κ(t)} for all t ≥ t0 . Lemma 4.6.1 A variational inequality (4.6.2) has at least one solution. Proof. Using the condition (4.6.1), it is enough to make certain that a solution of (4.2.41) with exact operator A satisfies the variational inequality (4.6.2). Observe that the solvability of this inequality follows from Theorem 1.11.9 (see also Lemma 2.6.3).

Let γ = (δ, h, σ) ∈ and xγα be one of solutions of (4.6.2), that is, Ah xγα + αJ µ xγα − f δ , z − xγα ≥ −g(xγα )z − xγα ∀z ∈ Ωσ .

(4.6.3)

Definition 4.6.2 It is said that an operator A : X → X ∗ has S-property if the weak convergence xn x and convergence Axn − Ax, xn − x → 0 imply the strong convergence xn → x as n → ∞.

234

4


Theorem 4.6.3 Assume that the conditions of the present section are satisfied and, in addition, operator A has the S-property, (4.2.1) and (4.6.1) hold and a solution set N of (7.1.1) is not empty. Let δ++σ = 0. lim α→0 α Then {xγα } converges strongly to the minimal norm solution x ¯∗ ∈ N. Proof. As in the proof of Theorem 4.2.1, by (4.1.3) and (4.6.3), we obtain Ah xγα + αJ µ xγα − f δ , v σ − xγα + Ax∗ − f, uγα − x∗ ≥ −g(xγα )xγα − v σ ,

(4.6.4)

where x∗ ∈ N, v σ ∈ Ωσ , uγα ∈ Ω, x∗ − v σ ≤ a(x∗ )σ

(4.6.5)

uγα − xγα ≤ a(xγα )σ.

(4.6.6)

and

The monotonicity property of A and inequalities (4.1.5), (4.6.1) and (4.6.4) yield the relation αJ µ xγα , xγα − vσ ≤

(h + )g(xγα ) + δ xγα − x∗

+ Ax∗ − f, uγα − xγα + Ah xγα − f δ , v σ − x∗ + g(xγα )vσ − x∗

∀x∗ ∈ N.

(4.6.7)

By (4.6.5), (4.6.6) and (4.2.40), one has µ(xγα )xγα ≤

+

h +

α

g(xγα ) +

δ γ xα − x∗ + µ(xγα ) x∗ + a(x∗ )σ α

σ α

κ(xγα )a(x∗ ) + κ(x∗ )a(xγα ) + g(xγα )a(x∗ ) .

h → 0 ) and since a gauge function µ(t) ≥ → 0 as α → 0 (and consequently, Since α α max {g(t), a(t), κ(t)}, we conclude from the last inequality that solutions xγα are bounded. ¯ ∈ Ω is established as in Theorem 4.2.1. After this, the weak convergence xγα x We prove the strong convergence of {xγα } to x ¯. The monotonicity of A and J µ implies 0 ≤ Axγα − A¯ x, xγα − x ¯ ≤ Axγα + αJ µ xγα − A¯ x − αJ µ x ¯, xγα − x ¯ ¯ − A¯ x + αJ µ x ¯, xγα − x ¯ . = Axγα + αJ µ xγα , xγα − x

(4.6.8)

4.6 Variational Inequalities with Non-Monotone Perturbations

235

¯, we have In view of the weak convergence of {xγα } to x x + αJ µ x ¯, xγα − x ¯ = 0. lim A¯

α→0

By virtue of (4.6.1), the next to last term in (4.6.8) admits the following estimate: ¯ ≤ Ah xγα + αJ µ xγα , xγα − x ¯ + hg(xγα )xγα − x ¯. Axγα + αJ µ xγα , xγα − x

(4.6.9)

Using further (4.6.3) for all x ¯σ ∈ Ωσ such that ¯ xσ − x ¯ → 0 as σ → 0, we deduce ¯ = Ah xγα + αJ µ xγα − f δ , xγα − x ¯σ Ah xγα + αJ µ xγα , xγα − x ¯σ + Ah xγα + αJ µ xγα , x ¯σ − x ¯ + f δ , xγα − x ≤ g(xγα )xγα − x ¯σ + f δ , xγα − x ¯ + Ah xγα + αJ µ xγα , x ¯σ − x ¯ . {Axγα }

is bounded together with bounded In view of (4.2.40), it results that ¯ and xγα x ¯, we have from (4.6.10) the limit inequality x ¯σ → x

(4.6.10) {xγα }.

Since

¯ ≤ 0. lim sup Ah xγα + αJ µ xγα , xγα − x α→0

Now we can conclude, by (4.6.8) and (4.6.9), that lim Axγα − A¯ x, xγα − x ¯ = 0.

α→0

¯ ∈ Ω. Finally, the S-property of A implies the strong convergence of {xγα } to x Show that x ¯ ∈ N. If z ∈ Ω then there exists a sequence {zσ } ∈ Ωσ such that {zσ } → z as σ → 0. Put in (4.6.3) z = zσ and take into account (4.6.1). We obtain Axγα + αJ µ xγα − f δ , zσ − xγα ≥ −( + h)g(xγα )zσ − xγα . Passing to the limit as α → 0, provided that A is demicontinuous, one gets A¯ x − f, z − x ¯ ≥ 0 ∀z ∈ Ω, x ¯ ∈ Ω. This means that x ¯ ∈ N. Prove that x ¯=x ¯∗ . Applying the monotonicity property of J µ , we rewrite (4.6.7) as J µ x∗ , xγα − x∗ ≤

+

h +

α

σ α

g(xγα ) +

κ(xγα )a(x∗ ) + κ(x∗ )a(xγα ) + g(xγα )a(x∗ )

+ µ(xγα )a(x∗ )σ If α → 0 then

δ γ xα − x∗ α

∀x∗ ∈ N.

J µ x∗ , x ¯ − x∗ ≤ 0 ∀x∗ ∈ N.

This implies the equality x ¯=x ¯∗ .

236

4

REGULARIZATION OF VARIATIONAL INEQUALITIES ∗

∗

Remark 4.6.4 If operators A : X → 2X and Ah : X → 2X are not single-valued but at the same time A is maximal monotone, the value set of Ah at every point of D(Ah ) is convex and closed (it occurs, for instance, if Ah is either semimonotone or pseudomonotone), then the conclusion of Theorem 4.6.3 is valid provided that the inequality (4.6.1) is replaced by (4.1.6). Remark 4.6.5 If the sequence

δ + + σ

α is bounded as α → 0, say by C, and µ(t) ≥ Cmax {g(t), α(t), κ(t)} for t ≥ t0 , then Theorem 4.6.3 asserts that every strong limit of {xγα } belongs to N.

4.7

Variational Inequalities with Mosco-Approximation of the Constraint Sets

We study in this section the regularization method for variational inequalities with s-wdemiclosed operators and Mosco-approximation of the constraint sets. Let X be a reflexive strictly convex Banach space together with its dual space X ∗ . ∗

Definition 4.7.1 An operator A : X → 2X is called strongly-weakly demiclosed (s-wdemiclosed for short) on a set Ω ⊆ D(A) if for any sequences {zn } and {ξn } such that zn ∈ Ω, ξn ∈ Azn , zn → z, ξn ξ, it follows that ξ ∈ Az. ∗

Definition 4.7.2 An operator A : X → 2X is called weakly-strongly demiclosed (w-sdemiclosed for short) on a set Ω ⊆ D(A) if for any sequences {zn } and {ξn } such that zn ∈ Ω, ξn ∈ Azn , zn z, ξn → ξ, it follows that ξ ∈ Az. ∗

Definition 4.7.3 An operator A : X → 2X is called demiclosed on a set Ω ⊆ D(A) if it is s-w-demiclosed and w-s-demiclosed at the same time. By Lemma 1.4.5, any maximal monotone operator is demiclosed. The following proposition is proved similarly to Theorem 1.4.7. ∗

Proposition 4.7.4 Let A : X → 2X be a monotone s-w-demiclosed operator in D(A). Suppose that Ω ⊆ int D(A) and that image Az is a nonempty closed convex subset of X ∗ at each point z ∈ Ω. Let x0 ∈ Ω. If the inequality ξ − f, z − x0 ≥ 0 ∀z ∈ Ω, holds then f ∈ Ax0 .

∀ξ ∈ Az,

(4.7.1)

4.7 Variational Inequalities with Mosco-Approximation

237

Proof. Suppose, by contradiction, that f ∈ / Ax0 . Then according to the strong separation theorem, there exists y ∈ X such that f, y > sup {g, y | ∀g ∈ Ax0 }

(4.7.2)

because Ax0 is a convex and closed set. Since x0 ∈ int Ω, there exists t¯ > 0 such that yt = x0 + ty ∈ Ω ∀t ∈ [0, t¯]. Obviously, the sets Ayt are not empty for each t ∈ [0, t¯], therefore, there exist a vector gt ∈ Ayt . If t → 0 then yt → x0 . By virtue of the local boundedness of A at any interior point of D(A), we conclude that gt g¯ ∈ X ∗ , where g¯ depends on y. Since A is demiclosed, g¯ ∈ Ax0 . This fact and (4.7.2) imply g , y . f, y > sup {g, y | ∀g ∈ Ax0 } ≥ ¯

(4.7.3)

On the other hand, by (4.7.1), gt − f, y ≥ 0 ∀gt ∈ Ayt . Setting t → 0, one gets

¯ g − f, y ≥ 0,

that is, ¯ g , y ≥ f, y which contradicts (4.7.3). In the sequel, operator A will be called convex-valued on Ω if its image Az is a nonempty convex subset of X ∗ at each point z ∈ Ω. Assume that Ω ⊆ D(A) is a convex closed subset and f ∈ X ∗ . Consider the variational inequality problem with s-w-demiclosed operator A : To find x ∈ Ω such that Ax − f, z − x ≥ 0 ∀z ∈ Ω. (4.7.4) A solution x0 of (4.7.4) is understood in the sense of Definition 1.11.1, that is, there exists ζ 0 ∈ Ax0 such that ζ 0 − f, z − x0 ≥ 0 ∀z ∈ Ω. (4.7.5) Next we present the Minty−Browder type lemma for variational inequalities with s-wdemiclosed operators. ∗

Lemma 4.7.5 Assume that A : X → 2X is a monotone operator, Ω ⊆ D(A) is a nonempty convex and closed set and x0 ∈ Ω. (i) If x0 is a solution of the variational inequality (4.7.4) defined by (4.7.5), then the inequality ζ − f, z − x0 ≥ 0 ∀z ∈ Ω, ∀ζ ∈ Az, (4.7.6) holds. (ii) If Ω ⊆ int D(A) and if the operator A is s-w-demiclosed and convex valued, then the converse implication is also true.

238

4


Proof. The claim (i) has been established in Lemma 1.11.3. We prove (ii). First of all, ¯ = Ax for any x ∈ Ω. we show that for any maximal monotone extension A¯ of A, one has Ax To this end, let x ∈ Ω and observe that, since the operator A is monotone and x ∈ int D(A), A is locally bounded at x. Hence, if {xn } is a sequence in X such that limn→∞ xn = x, and if {ξ n } is a sequence such that ξ n ∈ Axn for all n ≥ 1, then {ξ n } is bounded. It has weak accumulation points because X ∗ is reflexive. From s-w-demiclosedness of A it follows that any such point belongs to Ax. Denote by Rx the closed convex hull of the set of weak accumulation points of all sequences {ξ n } as described above. The set Ax is convex, therefore, it is closed by the reason of the demiclosedness of A on Ω. It results from this that Rx ⊆ Ax. Obviously, ¯ Thus, Rx ⊆ Ax. ¯ Ax ⊆ Ax. ¯ holds too. Suppose, by contradiction, that this We claim that the inclusion Rx ⊇ Ax ¯ such that η ∈ is not the case. Then there exists η ∈ Ax / Rx. According to the strong separation theorem, there is a vector w ∈ X satisfying the inequality ξ − η, w < 0 ∀ξ ∈ Rx.

(4.7.7)

Similarly to the proof of Proposition 4.7.4, we can show that the condition x ∈ int D(A) and monotonicity property of A¯ imply existence of the weak accumulation point ξ¯ ∈ Ax of the sequence {ζn }, ζn ∈ Axn , xn → x, such that ξ¯ − η, w ≥ 0. ¯ Clearly, ξ¯ ∈ Rx and this contradicts (4.7.7). Hence, Ax = Rx = Ax. ¯ for any x ∈ Ω. Therefore, variational inequality (4.7.4) and variational So, Ax = Ax inequality ¯ − f, z − x ≥ 0 ∀z ∈ Ω, x ∈ Ω Ax (4.7.8) have the same solution set. The maximal monotone operator A¯ also satisfies the requirements of Lemma 1.11.4. Consequently, x0 is a solution of (4.7.8). Therefore, it is a solution of (4.7.4) too. The lemma is proved. Definition 4.7.6 Let in (4.7.4) A be a monotone operator and let A¯ be a maximal monotone extension of A. A solution of the variational inequality ¯ − f, z − x ≥ 0 ∀z ∈ Ω, x ∈ Ω Ax

(4.7.9)

(in the sense of Definition 1.11.1) is called a generalized solution of (4.7.4). ∗

Lemma 4.7.7 If the operator A : X → 2X is monotone, if Ω ⊆ int D(A) is a nonempty convex and closed set, and if A is s-w-demiclosed and convex-valued on Ω, then any generalized solution of (4.7.4) is its solution in the sense of (4.7.5), and vice versa. Proof. Suppose that x0 is a generalized solution of (4.7.4). Then for some maximal ¯ 0 we have monotone extension A¯ of A and for some ζ¯0 ∈ Ax ζ¯0 − f, z − x0 ≥ 0 ∀z ∈ Ω.


239

¯ it results that for each z ∈ Ω and for any ζ ∈ Az, ¯ According to Lemma 1.11.3 applied to A, ζ − f, z − x0 ≥ 0. ¯ Lemma In particular, the last holds for each z ∈ Ω and for any ζ ∈ Az because Az ⊆ Az. 4.7.5 asserts now that x0 is a solution of (4.7.4). The inverse proposition follows from Definition 1.11.1. Namely, if there exists ζ 0 ∈ Ax0 such that ζ 0 − f, z − x0 ≥ 0 ∀z ∈ Ω, ¯ 0. then the same inequality holds with ζ 0 = ζ¯0 ∈ Ax Corollary 4.7.8 Let A be a monotone s-w-demiclosed and convex-valued on Ω operator and Ω ⊆ int D(A). If the generalized solution set N of the variational inequality (4.7.4) is not empty, then it is convex and closed. Consider the regularized variational inequality Ax + αJx − f, z − x ≥ 0 ∀z ∈ Ω,

x ∈ Ω.

(4.7.10)

∗

Lemma 4.7.9 If A : X → 2X is a monotone s-w-demiclosed and convex-valued operator on convex closed set Ω ⊆ int D(A), then (4.7.10) has a unique solution for any α > 0. Proof. The operator A + αJ is s-w-demiclosed on Ω and monotone because A and J are so. Let A¯ be some maximal monotone extension of A. Then A¯ + αJ is a maximal monotone extension of A + αJ. Due to Lemma 4.7.7, the variational inequality (4.7.10) and the variational inequality ¯ + αJx − f, z − x ≥ 0 ∀z ∈ Ω, Ax

x∈Ω

(4.7.11)

have the same solution sets. But we know that (4.7.11) has a unique solution because of Theorem 1.11.11.

Definition 4.7.10 Let {αn } be a sequence of positive real numbers converging to zero. The sequence of sets {Ωn } is said to be fast Mosco-convergent to Ω (fast M-convergent for short) relative to the sequence {αn } if the following conditions are satisfied: (j) if y ∈ Ω then θX ∈ s − lim inf αi−1 (y − Ωn ); n n→∞ n (jj) if {z } is a weakly convergent sequence in X such that, for some subsequence {Ωin } of (z n − Ω). {Ωn }, we have z n ∈ Ωin for all n > 0, then θX ∈ w − lim sup αi−1 n n→∞

Remark 4.7.11 A reasoning similar to that involved in the proof of Lemma 1.9 in [154] shows that if {Ωn } is fast M-convergent to Ω relative to some sequence of positive real numbers {αn } with lim αn = 0, then the sequence {Ωn } is M-convergent to Ω. n→∞

240

4


Assume that in the variational inequality (4.7.4), instead of A, f and Ω, we have the sequences {An }, {f n } and {Ωn }, n = 1, 2, ..., satisfying the following conditions: ∗ 1) An : X → 2X are monotone s-w-demiclosed and convex-valued on Ωn operators, and HX ∗ (An x, Ax) ≤ hn g(x) ∀x ∈ Ωn , where hn ≥ 0, g(t) is the non-negative bounded function for all t ≥ 0; 2) f − f n ∗ ≤ δn , δn ≥ 0; 3) Ωn is a convex closed set, Ωn ⊆ int D(A) ∩ int D(An ), a sequence {Ωn } is M-convergent to Ω. We introduce the following regularized variational inequality: An x + αn Jx − f n , z − x ≥ 0 ∀z ∈ Ωn , x ∈ Ωn ,

αn > 0.

(4.7.12)

Under our assumptions, according to Lemma 4.7.9, the inequality (4.7.12) has a unique solution xn in the sense of (4.7.5). More precisely, there is some element un ∈ An xn such that un + αn Jxn − f n , z − xn ≥ 0 ∀z ∈ Ωn . (4.7.13) Show that the sequence {xn } is bounded. To this end, along with (4.7.12), consider the variational inequality Ax + αn Jx − f, z − x ≥ 0 ∀z ∈ Ωn ,

x ∈ Ωn ,

(4.7.14)

and denote its (unique) solution by y n . This means that there exists v n ∈ Ay n such that v n + αn Jy n − f, z − y n ≥ 0 ∀z ∈ Ωn .

(4.7.15)

X∗

be a monotone convex-valued and s-w-demiclosed operLemma 4.7.12 Let A : X → 2 ator on closed convex set Ω ⊆ int D(A). Assume that the conditions 1) - 3) are fulfilled, αn > 0, n = 1, 2, ..., lim αn = 0 and the sequence n→∞

δ + h n n

αn

(4.7.16)

is bounded as n → ∞. Then the sequence {xn } of solutions to the variational inequality (4.7.12) is bounded if one of the following requirements is satisfied: a) For each real number β > 0, the set Mβ = {x ∈ X | ξ∗ ≤ βx ∀ξ ∈ Ax} is bounded; b) There exists a bounded sequence {w n } such that wn ∈ Ωn and {αn−1 d∗ (f n , Awn )} is bounded. Here d∗ (u, G) is a distance between an element u ∈ X ∗ and set G ⊂ X ∗ ; % c) There exists an element x ¯∈ ∞ n=1 Ωn , at which the operator A is coercive with respect to x ¯, that is, for any {ξ n } and {z n }, where ξ n ∈ Az n and z n ∈ D(A), we have lim

z n →∞

ξ n , z n − x ¯ = ∞. z n


241

Proof. First of all, show that if {y n } is bounded then {xn } is also. Really, due to the condition 1), for any v n ∈ Ay n , it is possible to find vñ ∈ An y n such that v n − vñ ∗ ≤ hn g(y n ).

(4.7.17)

Presuming z = y n in (4.7.13) and z = xn in (4.7.15), and adding the obtained inequalities, we get un + αn Jxn − f n − v n − αn Jy n + f, xn − y n ≤ 0. Rewrite it in the equivalent form: αn Jxn − Jy n , xn − y n + f − f n , xn − y n + un − vñ , xn − y n + ˜ vn − v n , xn − y n ≤ 0.

(4.7.18)

Since un ∈ An xn , vñ ∈ An y n and An is monotone, the last term in the left-hand side of (4.7.18) is non-negative and, therefore, it can be omitted. Taking into account the assumption 2) and applying inequalities (1.5.3) and (4.7.17), we deduce from (4.7.18) the relation δn + hn g(y n ) (xn + y n ). (xn − y n )2 ≤ αn It is clear that if {y n } and

δ + h n n

are bounded, then {xn } is bounded too. αn Suppose that the condition a) holds. Prove that {y n } is bounded in this case. Indeed, take z ∈ Ω ⊆ int D(A). By Lemma 1.5.14, there exist r0 > 0 and c0 > 0 such that for any y ∈ int D(A) and for any v ∈ Ay, we have v − f, y − z ≥ r0 v − f ∗ − c0 (y − z + r0 ).

(4.7.19)

According to the condition 3), the set Ωn ⊆ int D(A). Therefore, in (4.7.19) we can put y = y n ∈ Ωn and v = v n ∈ Ayn . Then vn − f, y n − z ≥ r0 v n − f ∗ − c0 (y n − z + r0 ).

(4.7.20)

Fix any z¯ ∈ Ω. By virtue of M -convergence of {Ωn } to Ω, we are able to construct a sequence {z n } such that z n ∈ Ωn and z n → z¯. Rewrite (4.7.15) with z = z n as v n + αn Jy n − f, z n − y n ≥ 0. Then v n − f, y n − z¯ = v n + αn Jy n − f, y n − z n + v n + αn Jxn − f, z n − z¯ − αn Jy n , y n − z¯ ≤ v n + αn Jy n − f, z n − z¯ + αn yn ¯ z ≤ vn − f ∗ z n − z¯ + αn y n (z n − z¯ + ¯ z ).

(4.7.21)

242

4


Now (4.7.20) with z = z¯ and (4.7.21) yield the inequality

z ) + c0 + c0 (r0 + ¯ z ). (r0 − z n − z¯)v n − f ∗ ≤ y n αn (z n − z¯ + ¯

(4.7.22)

r0 holds for a sufficiently large n > 0. By 2 (4.7.22), we conclude that there are c1 > 0 and c2 > 0 such that

Since z n → z¯, the estimate r0 − z n − z¯ >

v n ∗ ≤ c1 yn + c2 .

(4.7.23)

Suppose that {y n } is unbounded. Then (4.7.23) implies that v n ∗ ≤ βy n for some β > 0. This means that yn belongs to Mβ which is bounded according to the condition a). This fact leads to a contradiction. Thus, {yn } is bounded together with {xn }. Assume that the condition b) holds. It is obvious that the sequence {wn } within satisfies the inclusion An wn ⊆ A¯n wn , where A¯n is a maximal monotone extension of An . We know that the set A¯n wn is convex and closed as well. Therefore, there exists ζ n ∈ A¯n wn such that ζ n − f n ∗ = d∗ (f n , A¯n wn ) ≤ d∗ (f n , An wn ). (4.7.24) If in (4.7.13) we put z = wn , then one gets un − ζ n , xn − wn + ζ n − f n , xn − wn + αn Jxn , xn − wn ≤ 0.

(4.7.25)

A¯n

Recall that is monotone. Hence, the first term in the left-hand side of (4.7.25) is nonnegative. Therefore, by (4.7.24) and (4.7.25), we come to the estimate xn 2 ≤ xn wn + αn−1 d∗ (f n , An wn ) (xn + wn ) , which gives the boundedness of {xn }. ¯ in (4.7.15), we Suppose that the condition c) holds. Since x ¯ ∈ ∩∞ n=1 Ωn , putting z = x have v n − f, y n − x ¯ + αn yn 2 ≤ αn y n ¯ x. Moreover, v n − f, y n − x ¯ ≤ αn y n ¯ x. From this, it is easy to see that

f ∗ ¯ ¯ x v n , yn − x x + ≤ f ∗ + αn ¯ . yn yn Furthermore, if we assume that {y n } is unbounded then v n , y n − x ¯ ≤ f ∗ n→∞ yn lim

which contradicts the condition c). Thus, {y n } is bounded together with {xn }. It accomplishes the proof of the lemma.


243

Theorem 4.7.13 Let X be an E-space, X ∗ be strictly convex. Assume that an operator ∗ A : X → 2X is monotone s-w-demiclosed and convex-valued on a closed convex set Ω ⊆ int D(A), the variational inequality (4.7.4) has a nonempty solution set N, {αn } is a sequence of positive numbers such that limn→∞ αn = 0. Let the conditions 1) - 3) be fulfilled, {Ωn } be fast M-convergent to Ω relative to the sequence {αn } and lim

n→∞

δn + hn = 0. αn

(4.7.26)

Suppose that one of the conditions a) - c) of Lemma 4.7.12 holds. Then a sequence {xn } generated by (4.7.12) strongly converges in X to the minimal norm solution x ¯∗ ∈ N as n → ∞. Proof. First of all, observe that due to Lemma 4.7.12 {xn } is bounded. Then there exists a weak accumulation point of {xn }. Let xn x∗ (in reality, some subsequence of {xn } weakly converges to x∗ , however, as before, we do not change its denotation). Since xn ∈ Ωn , the fast M-convergence of {Ωn } to Ω implies inclusion x∗ ∈ Ω (see Remark 4.7.11). Show that x∗ ∈ N. Take an arbitrary element z ∈ Ω and fix it. For any n > 0, construct z n ∈ Ωn such that z n → z as n → ∞. Presuming z n in (4.7.13), in place of z, we get for some un ∈ An xn un + αn Jxn − f n , xn − z n ≤ 0. (4.7.27) ñ ∈ Axn satisfying the According to the assumption 1), for un ∈ An xn one can find u estimate un − u ñ ∗ ≤ hn g(xn ). (4.7.28) Further, take any u ∈ Az. It follows from (4.7.27) that u − f, xn − z + un + αn Jxn − f n , z − z n + f − f n , xn − z + αn Jxn , xn − z + un − u ñ , xn − z + ˜ un − u, xn − z ≤ 0. By the monotonicity of A, the last term in the left-hand side of the previous inequality can be omitted. Then, owing to the assumption 2) and estimate (4.7.28), we come to the following relation: u − f, xn − z ≤ un + αn Jxn − f n ∗ z n − z

+ δn + αn xn + hn g(xn ) xn − z.

(4.7.29)

Using the boundedness of {x } and (4.7.19), prove that {u } is bounded. Indeed, we can write down for some constants r0 > 0 and c0 > 0 the following inequality: n

n

un − f ∗ − c0 (xn − z + r0 ), u ñ ∈ Axn , ˜ un − f, xn − z ≥ r0 ˜ because of z ∈ int D(A). It is clear that there exists c1 > 0 such that un − f ∗ − c1 . ˜ un − f, xn − z ≥ r0 ˜

(4.7.30)

244

4


At the same time, un − un , xn − z + un + αn Jxn − f n , xn − z n ˜ un − f, xn − z = ˜ + un + αn Jxn − f n , z n − z + f n − f, xn − z − αn Jxn , xn − z ≤

+

hn g(xn ) + δn + αn xn xn − z

˜ un − f ∗ + hn g(xn ) + δn z n − z.

Therefore, there exists a constant c2 > 0 such that un − f ∗ z n − z + c2 . ˜ un − f, xn − z ≤ ˜

(4.7.31)

Combining (4.7.30) with (4.7.31), one gets un − f ∗ ≤ ˜ un − f ∗ z n − z + c1 + c2 . r0 ˜ By virtue of the strong convergence of {z n } to z, we conclude that there exists n ¯ > 0 such that for any n ≥ n ¯, c1 + c2 . ˜ un ∗ ≤ f ∗ + r0 − z n − z This means that {˜ un } is bounded. Furthermore, in view of (4.7.28), {un } is also bounded. Then the boundedness of {xn }, its weak convergence to x∗ ∈ Ω and (4.7.29) imply the inequality u − f, x∗ − z ≤ 0 ∀z ∈ Ω, ∀u ∈ Az, because αn → 0, δn → 0, hn → 0, f n → f and z n → z as n → ∞. Since A is s-w-demiclosed and convex-valued, Lemma 4.7.5 allows us to establish that x∗ ∈ Ω is a solution of (4.7.4). Under the hypothesis of the theorem, solution set N is convex closed and nonempty and, therefore, it contains the minimal norm element x ¯∗ . We are going to show that the only n ∗ weak accumulation point of {x } is x ¯ . To this end, we apply the fast Mosco-convergence of {Ωn } to Ω relative to the sequence {αn }. Due to Definition 4.7.10, there exist {q n } and {˜ q n } such that q n ∈ Ωn , qñ ∈ Ω and if n → ∞ then ¯∗ ) → 0 αn−1 (q n − x and

(4.7.32)

αn−1 (xn − qñ ) 0.

(4.7.33)

Since x ¯∗ is the solution of (4.7.4), there is u∗ ∈ A¯ x∗ such that ¯∗ ≥ 0 ∀z ∈ Ω. u∗ − f, z − x

(4.7.34)

4.8 Variational Inequalities with Hypomonotone Approximations

245

Putting z = q n in (4.7.13) and z = qñ in (4.7.34) and adding the obtained inequalities, by simple algebra, we have xn 2 ≤ xn ¯ x∗ + αn−1 (¯ x∗ − q n un + αn Jxn − f n ∗ + u∗ − f, qñ − xn + ˜ un − un , xn − x ¯∗ ñ , xn − x ¯∗ + f n − f, xn − x ¯∗ ). + u∗ − u Further we use the condition 2), monotonicity property of A and (4.7.28) to get xn 2 ≤ xn ¯ x∗ + αn−1 ¯ x∗ − q n un + αn Jxn − f n ∗ + αn−1 u∗ − f, qñ − xn +

δn + hn g(xn ) n x − x ¯∗ . αn

(4.7.35)

If θX ∈ N then x ¯∗ = θX and it results from (4.7.35), (4.7.26), (4.7.32) and (4.7.33) that lim xn = 0. Then the theorem is proved. Let now θX ∈ N. Then the weak convergence n→∞ of {xn } to x∗ ∈ N implies 0 < x∗ ≤ lim inf xn . (4.7.36) n→∞

Therefore, one can consider that xn > µ > 0. Dividing (4.7.35) on xn and passing to the limit as n → ∞, we conclude, according to (4.7.26), (4.7.32) and (4.7.33), that x∗ . lim sup xn ≤ ¯

(4.7.37)

n→∞

Finally, by (4.7.36) and (4.7.37), we establish the equality x∗ = x ¯∗ and limit relation n ∗ n x . Hence, {x } has a unique weak accumulation point x ¯∗ . Since X is an Ex → ¯ n ∗ space, the whole sequence {x } strongly converges to x ¯ . The proof is completed.

4.8

Variational Inequalities with Hypomonotone Approximations

Let X be a reflexive strictly convex and smooth Banach space, Ω be a closed and convex ∗ subset of X, A : X → 2X be a monotone s-w-demiclosed and convex-valued operator on ∗ Ω, Ω ⊆ int D(A), f ∈ X . We study the variational inequality (4.7.4) and suppose that it has a nonempty solution set N. As in the previous section, a solution of (4.7.4) is understood in the sense of Definition 1.11.1. We have shown there that in these circumstances a solution of (4.7.4) satisfies the inequality (4.7.6) and vice versa. In addition, each solution of (4.7.4) coincides with a solution of the variational inequality (4.7.9) with maximal monotone extension A¯ of A. By Corollary 4.7.8, the set N is convex and closed. ∗

Definition 4.8.1 An operator A : X → 2X is called hypomonotone in D(A) if there exists a constant c > 0 such that u − v, x − y ≥ −cx − y2 ∀x, y ∈ D(A), ∀u ∈ Ax, ∀v ∈ Ay.

246

4

REGULARIZATION OF VARIATIONAL INEQUALITIES ∗

Definition 4.8.2 An operator A : X → 2X is called strongly hypomonotone in D(A) if there exists a constant c > 0 such that u − v, x − y ≥ −c(x − y)2 ∀x, y ∈ D(A),

∀u ∈ Ax, ∀v ∈ Ay.

It is clear that the strong hypomonotonicity implies hypomonotonicity. Introduce = (0, δ∗ ] × (0, h∗ ] × (0, σ ∗ ] for some positive δ ∗ , h∗ and σ ∗ . Assume that the perturbed date f δ , Ωσ and Ah for the variational inequality (4.7.4) satisfy the following conditions: 1) f − f δ ∗ ≤ δ ∀δ ∈ (0, δ ∗ ]; 2) Ωσ is a convex and closed set for any σ ∈ (0, σ ∗ ]; ∗ 3) Ah : X → 2X is a s-w-demiclosed operator for any h ∈ (0, h∗ ] and convex-valued for any x ∈ Ωσ ⊆ int D(Ah ), there exists η(h) > 0 such that η(h) → 0 as h → 0 and for any x, y ∈ Ωσ , uh − v h , x − y ≥ −η(h)(x| − y)2 ∀uh ∈ Ah x, ∀v h ∈ Ah y;

(4.8.1)

4) For any x ∈ Ω, there exist xσ ∈ Ωσ , a(t) and g(t) such that x − xσ ≤ a(x)σ, and d∗ (ζ, Ah xσ ) ≤ g(ζ∗ )ξ(h, σ) ∀ζ ∈ Ax; 5) For any zσ ∈ Ωσ , there exist z ∈ Ω and b(t) such that z − zσ ≤ b(zσ )σ. In 4) - 5), the functions a(t), b(t), g(t) are non-negative and bounded on bounded sets for all t ≥ 0, and ξ(h, σ) → 0 as h, σ → 0. Observe that s-w-demiclosedness of A on Ω and Ah on Ωσ imply the closedness property of Ax and Ah x at each point x ∈ Ω and at each point x ∈ Ωσ , respectively. Lemma 4.8.3 For all α > η(h), the operator B = Ah + αJ is strictly monotone on Ωσ and ¯ for all x ∈ Ωσ , where B ¯ is an coercive relative to any element of Ωσ . In addition, Bx = Bx arbitrary maximal monotone extension of B. Proof. Let arbitrary x, y ∈ Ωσ and uh ∈ Ah x, v h ∈ Ah y. Then by (4.8.1) and by the property (1.5.3) of normalized duality mapping J, uh + αJx − v h − αJy, x − y ≥ (α − η(h))(x| − y)2 ≥ 0. Thus, B is the monotone operator. Let x = y. By virtue of (4.8.1), if x = y then we have uh − vh , x − y ≥ 0.

(4.8.2)

4.8 Variational Inequalities with Hypomonotone Approximations

247

Consequently, uh + αJx − v h − αJy, x − y ≥ αJx − Jy, x − y > 0 because J is strictly monotone. Taking into account (4.8.2) we conclude that B is strictly monotone on Ωσ . Fix xσ ∈ Ωσ and uλ ∈ Ah xσ , where λ = (h, σ). Let v ∈ Ah x, where x ∈ Ωσ . Then (4.8.2) allows us to write down the inequality v + αJx, x − xσ = v + αJx − uλ − αJxσ , x − xσ + uλ + αJxσ , x − xσ ≥ (α − η(h))(x| − xσ )2 − uλ + αJxσ ∗ (x + xσ ). It results from this that B is coercive relative to xσ ∈ Ωσ . ¯ be a maximal monotone extension of B. Take x ∈ Ωσ . Since Ωσ ⊆ int D(Ah ), we Let B assert that the monotone operator B is locally bounded at the point x. Therefore, if xn → x then un u, where un ∈ Bxn . Further, B is s-w-demiclosed because Ah is s-w-demiclosed and J is demicontinuous. Consequently, u ∈ Bx. The rest of the proof follows the pattern of Lemma 4.7.5. Consider the problem: To find xγα ∈ Ωσ , where γ = (δ , h, σ) ∈ , satisfying the regularized variational inequality (4.5.2) under the conditions 1) - 5) of this section. Lemma 4.8.4 The variational inequality (4.5.2) has a unique solution xγα for any fixed γ ∈ . ¯ of the operator Proof. We have shown above that maximal monotone extension B h ˜ ∈ Ωσ . According to Corollary B = A + αJ is coercive on Ωσ relative to some point x 1.11.13, the variational inequality ¯ − f δ , z − x ≥ 0 ∀z ∈ Ωσ , x ∈ Ωσ , Bx ¯ = Bx has at least one solution. However, this inequality coincides with (4.5.2) because Bx for all x ∈ Ωσ . Uniqueness of the solution of (4.5.2) results from the strict monotonicity of B proved in Lemma 4.8.3. ∗

Theorem 4.8.5 Suppose that X is an E-space, X ∗ is strictly convex, A : X → 2X is a monotone s-w-demiclosed operator and convex-valued on Ω ⊆ int D(A), Ω is a convex closed set, variational inequality (4.7.4) has a nonempty solution set N, approximation data f δ , Ah and Ωσ satisfy the conditions 1) - 5) and, in addition, δ + η(h) + σ + ξ(h, σ) = 0, α b(t) b) lim sup 2 ≤ Q, 0 < Q < ∞. t t→∞ Then a solution sequence {xγα } of the variational inequality (4.5.2) strongly converges to the minimal norm solution x ¯∗ of (4.7.4) as α → 0. a) lim

α→0

248

4


Proof. If x∗ ∈ N then there is ζ ∗ ∈ Ax∗ such that ζ ∗ − f, z − x∗ ≥ 0 ∀z ∈ Ω.

(4.8.3)

Since xγα ∈ Ωσ is a solution of (4.5.2), there exists ζαγ ∈ Ah xγα satisfying the inequality ζαγ + αJxγα − f δ , z − xγα ≥ 0 ∀z ∈ Ωσ .

(4.8.4)

By virtue of the condition 4) for x = x∗ ∈ N, we are able to find xσ ∈ Ωσ and ζ λ ∈ Ah xσ such that for any λ = (h, σ), (4.8.5) x∗ − xσ ≤ a(x∗ )σ and ζ ∗ − ζ λ ∗ ≤ g(ζ ∗ ∗ )ξ(h, σ). In its turn, by virtue of the condition 5), for

xγα

∈ Ωσ there exists

(4.8.6) yαγ

∈ Ω such that

xγα − yαγ ≤ b(xγα )σ.

(4.8.7)

Add (4.8.3) and (4.8.4) putting there z = yαγ and in z = xσ , respectively. Then we obtain ζαγ + αJxγα − f δ , xγα − xσ + ζ ∗ − f, x∗ − yαγ ≤ 0. We rewrite this inequality in the following equivalent form: 0 ≥ αJxγα , xγα − xσ + ζαγ − ζ λ , xγα − xσ + ζ λ − ζ ∗ , xγα − xσ + ζ ∗ − f, x∗ − yαγ + xγα − xσ + f − f δ , xγα − xσ . Taking into account the hypomonotonicity property (4.8.1) and the estimates (4.8.5) (4.8.7), we deduce

xγα 2 ≤ xγα x∗ + a(x∗ )σ +

+

δ

α

+

η(h) ξ(h, σ) g(ζ ∗ ∗ ) xγα + x∗ + a(x∗ )σ + α α

σ ∗ ζ − f ∗ a(x∗ ) + b(xγα ) . α

The conditions a) and b) of the theorem imply the boundedness of {xγα } as α → 0. The rest of the proof follows the pattern of Theorem 4.3.1. If the function b(t) does not possess the condition b), then in place of normalized duality mapping J, it should be used in (4.5.2) duality mapping J µ with a suitable gauge function µ(t).

4.9 Variational Inequalities with Pseodomonotone Operators

4.9

249

Variational Inequalities with Pseudomonotone Operators

Let X be a reflexive strictly convex Banach space together with its dual space X ∗ , A : X → X ∗ be a pseudomonotone operator, Ω be a convex closed set in D(A). We study in X the variational inequality (4.7.4) with such operator A. If A is coercive and θX ∈ Ω then, in accordance with Theorem 1.13.10, (4.7.4) has a nonempty solution set N. Lemma 4.9.1 N is a weakly closed set. Proof. Let xn ∈ N and xn x. Due to the Mazur theorem, x ∈ Ω. Show that x ∈ N. Since xn is a solution of (4.7.4), we have Axn − f, xn − z ≤ 0 ∀z ∈ Ω.

(4.9.1)

Assuming in (4.9.1) z = x, one gets Axn − f, xn − x ≤ 0. Consequently, lim sup Axn − f, xn − x ≤ 0. n→∞

The last implies lim sup Axn , xn − x ≤ 0 n→∞

because f, xn − x → 0 as n → ∞. The pseudomonotonicity of A allows us to write down the following relation: lim inf Axn , xn − z ≥ Ax, x − z ∀z ∈ Ω. n→∞

By the equality lim f, xn − z = f, x − z ∀z ∈ Ω,

n→∞

we obtain lim inf Axn − f, z − xn ≤ Ax − f, z − x ∀z ∈ Ω. n→∞

(4.9.2)

In addition, (4.9.1) yields the limit estimate lim inf Axn − f, z − xn ≥ 0 ∀z ∈ Ω, n→∞

which together with (4.9.2) imply (4.7.4). Thus, we have proved that x ∈ N. Let f, Ω and A be given approximately, that is, in reality, perturbed data f δ , Ωσ and are known such that for any γ = (δ, h, σ) ∈ , where = (0, δ ∗ ] × (0, h∗ ] × (0, σ ∗ ] with some positive δ ∗ , h∗ and σ∗ , there hold: 1) f − f δ ∗ ≤ δ; 2) Ωσ ⊂ D(Ah ) are convex and closed sets and θX ∈ Ωσ ; Ah

250

4 REGULARIZATION OF VARIATIONAL INEQUALITIES

3) {Ah } is the sequence of pseudomonotone operators acting from X to X ∗ , single-valued on Ωσ and satisfying the condition Ah x − Ax∗ ≤ hg(x) ∀x ∈ Ωσ ,

(4.9.3)

where the function g(t) is non-negative, non-decreasing and bounded on each bounded interval of t ≥ 0; 4) {Ωσ } Mosco-approximates Ω. In other words, every point x ∈ Ω is a strong limit of some sequence {xσ }, xσ ∈ Ωσ , as σ → 0, and every weak limit point x ˜ of any subsequence {˜ uσ } ⊂ Ωσ , belongs to Ω. We will find approximation solutions xγα ∈ Ωσ from the regularized variational inequality (4.5.2) again. By Lemma 1.13.6, the mapping B = Ah + αJ is pseudomonotone as the sum of two pseudomonotone operators Ah and αJ. Observe further that, under our hypothesis, Theorem 1.13.10 guarantees solvability of (4.5.2) if B is coercive on Ωσ . The following lemma forms the sufficient condition for B to be coercive. Denote

G=

Ωσ ⊂ D(A).

(4.9.4)

σ∈(0,σ ∗ ]

Lemma 4.9.2 If operator A is coercive on each set Ωσ , and there exists a function c(t) → ∞ as t → ∞ such that Ax, x ≥ c(x) ∀x ∈ G, (4.9.5) x and if lim

t→∞

c(t) ≥ c0 > 0, g(t)

(4.9.6)

then for all α > 0 the operator B = Ah + αJ is coercive on Ωσ . Proof. Indeed, using (4.9.3) and (4.9.5), we deduce for any x ∈ Ωσ , Ah x, x lim inf x x→∞

Ah x − Ax, x Ax, x + = lim inf x x x→∞

≥ lim inf c(x) − hg(x)

x→∞

= lim inf g(x) x→∞

c(x)

g(x)

−h .

Consequently, in view of (4.9.6), there exists > 0 such that Ah x, x ≥ (c0 − h − ) lim g(t) ≥ ch , t→∞ x x→∞

lim inf

where ch < ∞ when h > 0 is small enough. In this case, for all α > 0,

Ah x, x Ah x + αJx, x + αx ≥ ch − + αx → ∞ = x x


251

as x → ∞. The lemma is proved. We present now the main result of this section. Theorem 4.9.3 Assume that the conditions 1) - 4) of this section are fulfilled. Let A : X → X ∗ be a pseudomonotone bounded and coercive operator on each set Ωσ and let (4.9.5) and (4.9.6) hold. If A has the S-property, lim sup t→∞

g(t) ≤Q 0. Prove that {xγα } is bounded as ∆ → 0. Since θX ∈ Ωσ for all σ ∈ (0.σ ∗ ], assume z = θX in (4.5.2). Then we have Ah xγα + αJxγα − f δ , xγα ≤ 0, which is equivalent to the inequality Ah xγα − Axγα , xγα + Axγα , xγα − f δ , xγα + αxγα 2 ≤ 0. By (4.9.3), we deduce

Axγα , xγα + αxγα 2 ≤ hg(xγα ) + f δ ∗ xγα .

(4.9.9)

Suppose that {xγα } is unbounded, i.e., xγα → ∞ as ∆ → 0. Then (4.9.5) implies the estimate Axγα , xγα ≥ c(xγα )xγα . Now (4.9.9) gives the following relation:

c(xγα )xγα + αxγα 2 ≤ hg(xγα ) + f δ ∗ xγα . If α is small enough then the last can be transformed as 1≤

h g(xγα ) h g(xγα ) f δ ∗ − c(xγα ) . ≤ + γ γ α xγα αxα α xα

Here we used the boundedness of f δ ∗ , the condition 1) and the fact that c(t) → ∞ as t → ∞. Passing to the limit in the latter inequality as ∆ → 0 and taking into account

4

252


(4.9.7) and (4.9.8), we come to a contradiction. Thus, the sequence {xγα } is bounded as ∆ → 0. ¯. Since xγα ∈ Ωσ , we have that x ¯ ∈ Ω by reason of the Mosco-approximation Let xγα x properties. Applying these properties once more for x ¯, one can construct a sequence {¯ xσ }, x ¯σ ∈ Ωσ , such that x ¯σ → x ¯ as σ → 0. Assuming in (4.5.2) z = x ¯σ we obtain Ah xγα + αJxγα − f δ , x ¯σ − xγα ≥ 0. xσ } yield the inequality Then (4.9.3) and the boundedness of {xγα } and {¯ ¯σ ≤ lim sup hg(xγα )xγα − x ¯σ = 0. lim sup Axγα + αJxγα − f δ , xγα − x ∆→0

∆→0

(4.9.10)

¯, we conclude that Since xγα x ¯σ → 0 αJxγα , xγα − x and f δ , xγα − x ¯σ → 0 as ∆ → 0. Then it results from (4.9.10) that ¯σ ≤ 0. lim sup Axγα , xγα − x ∆→0

(4.9.11)

The sequence {Axγα } is bounded because the operator A and {xγα } are bounded. Thus, ¯σ − x ¯ → 0 as ∆ → 0. Axγα , x Therefore, (4.9.11) implies lim sup Axγα , xγα − x ¯ ≤ 0. ∆→0

(4.9.12)

By the definition of pseudomonotone operators, one gets x, x ¯ − x ∀x ∈ Ω. lim inf Axγα , xγα − x ≥ A¯ ∆→0

(4.9.13)

In view of (4.9.3), we are able now to write down the following relations: Ah xγα + αJxγα − f δ , xγα − x = Ah xγα − Axγα + αJxγα − f δ , xγα − x + Axγα , xγα − x ≥ −hg(xγα )xγα − x + αJxγα , xγα − x − f δ , xγα − x + Axγα , xγα − x ∀x ∈ Ω. It is obvious that if ∆ → 0 then hg(xγα )xγα − x → 0,

(4.9.14)


253

αJxγα , xγα − x → 0 and ¯ − x . f δ , xγα − x → f, x Together with (4.9.13), this allows us to deduce from (4.9.14) the inequality x − f, x ¯ − x ∀x ∈ Ω, x ¯ ∈ Ω. lim inf Ah xγα + αJxγα − f δ , xγα − x ≥ A¯ ∆→0

(4.9.15)

Fix x ∈ Ω. According to the Mosco-approximation of Ω by Ωσ , construct the sequence {xσ }, xσ ∈ Ωσ , such that xσ → x as σ → 0. Let uγα = Ah xγα + αJxγα − f δ .

(4.9.16)

Then uγα , xγα − x = uγα , xγα − xσ + uγα , xσ − x . Since {xγα } and {Axγα } are bounded, (4.9.3) implies the boundedness of {Ah xγα }. Consequently, by (4.9.16), {uγα } is also bounded as ∆ → 0. In addition, we have αJxγα → θX ∗ and f δ → f. Owing to the strong convergence of {xσ } to x, we conclude that uγα , xσ − x → 0 as ∆ → 0. Put in (4.5.2) z = xσ . Then uγα , xγα − xσ ≤ 0. Now (4.9.15) gives A¯ x − f, x ¯ − x ≤ 0 ∀x ∈ Ω, x ¯ ∈ Ω. This means that x ¯ ∈ N. Further, asuming in (4.9.13) x = x ¯, we obtain ¯ ≥ 0. lim inf Axγα , xγα − x ∆→0

Then (4.9.12) gives the limit equality lim Axγα , xγα − x ¯ = 0.

∆→0

¯, the last implies Since xγα x lim Axγα − A¯ x, xγα − x ¯ = 0.

∆→0

¯ ∈ N0 ⊆ N as ∆ → 0. Finally, the S-property of A guarantees strong convergence {xγα } to x The proof is accomplished. Note that normalized duality mapping J in the inequality (4.5.2) can be replaced by the duality mapping J µ with a gauge function µ(t). Theorem 4.9.3 remains still valid if, instead of (4.9.7), it used the inequality lim

t→+∞

g(t) ≤ Q < ∞. µ(t)

254

4

4.10


Variational Inequalities of Mixed Type

Let X be an E-space, X ∗ be a strongly convex space, A : X → X ∗ be a monotone bounded hemicontinuous operator with D(A) = X and ϕ : X → R1 be a properly convex lower semicontinuous functional. Consider the problem of solving the mixed variational inequality: To find x ∈ X such that Ax − f, x − y + ϕ(x) − ϕ(y) ≤ 0 ∀y ∈ X.

(4.10.1)

As usual, we start to study the existence of a solution x for (4.10.1). To this end, first of all, we need the following theorem [128]: Theorem 4.10.1 If there exists x0 ∈ dom ϕ satisfying the limit relation lim

x→∞

Ax, x − x0 + ϕ(x) = ∞, x

(4.10.2)

then (4.10.1) has at least one solution. We first prove two auxiliary lemmas. Lemma 4.10.2 The mixed variational inequality (4.10.1) is equivalent to Ay − f, x − y + ϕ(x) − ϕ(y) ≤ 0 ∀y ∈ X,

x ∈ X.

(4.10.3)

Proof. Since A is monotone, we have from (4.10.1) that Ay − f, x − y ≤ Ax − f, x − y ≤ ϕ(y) − ϕ(x) ∀y ∈ X,

x ∈ X.

This means that (4.10.3) is valid. Let now (4.10.3) be assumed. If y = yt = tx + (1 − t)z, t ∈ [0, 1], where z is any fixed element of X, then (4.10.3) accepts the form: (1 − t)Ayt − f, x − z + ϕ(x) − ϕ(tx + (1 − t)z) ≤ 0 ∀z ∈ X. Taking into account the convexity of ϕ we obtain (1 − t)Ayt − f, x − z + ϕ(x) − tϕ(x) − (1 − t)ϕ(z) ≤ 0 ∀z ∈ X. Consequently, Ayt − f, x − z + ϕ(x) − ϕ(z) ≤ 0 ∀z ∈ X. Letting t → 1 one gets (4.10.1).

Lemma 4.10.3 The solution set N of (4.10.1) is closed and convex if it is not empty.

4.10 Variational Inequalities of Mixed Type

255

Proof. Let x1 and x2 be two different elements of N. Then, by Lemma 4.10.2, Ay − f, x1 − y + ϕ(x1 ) − ϕ(y) ≤ 0 ∀y ∈ X and Ay − f, x2 − y + ϕ(x2 ) − ϕ(y) ≤ 0 ∀y ∈ X. Multiplying these inequalities, respectively, on t and 1 − t and adding them, we get Ay − f, z − y + tϕ(x1 ) + (1 − t)ϕ(x2 ) − ϕ(y) ≤ 0 ∀y ∈ X,

(4.10.4)

where z = tx1 + (1 − t)x2 . Since ϕ is convex, ϕ(z) ≤ tϕ(x1 ) + (1 − t)ϕ(x2 ). Hence, Ay − f, z − y + ϕ(z) − ϕ(y) ≤ 0 ∀y ∈ X because of (4.10.4). Thus, z ∈ N and convexity of N is proved. Show that N is closed. Let xn → x, where xn ∈ N. Then Axn − f, xn − y + ϕ(xn ) − ϕ(y) ≤ 0 ∀y ∈ X. Since a hemicontinuous monotone operator on X is demicontinuous and since ϕ is a lower semicontinuous functional, the last inequality implies (4.10.1) as n → ∞. Let ∗ = (0, δ ∗ ]×(0, h∗ ]×(0, ∗ ] with some positive δ ∗ , h∗ and ∗ and let γ = (δ, h, ) ∈ ∗ . Assume that the solution set N of the inequality (4.10.1) is nonempty, and its data A, f, ϕ are given with approximations Ah , f δ , ϕ satisfying the conditions: 1) f − f δ ∗ ≤ δ; 2) Ah : X → X ∗ are monotone hemicontinuous operators, D(Ah ) = D(A) = X, and Ah x − Ax∗ ≤ hg(x)

∀x ∈ X,

(4.10.5)

where g(s) is a non-negative bounded function for s ≥ 0; 3) ϕ : X → R1 are properly convex lower semicontinuous functionals and there exist positive numbers c and R such that ϕ (x) ≥ −c x as x > R and |ϕ (x) − ϕ(x)| ≤ q(x)

∀x ∈ X,

(4.10.6)

where q(s) has the same properties as g(s). The regularization method for the mixed variational inequality (4.10.1) is written as Ah x + αJx − f δ , x − y + ϕ (x) − ϕ (y) ≤ 0 ∀y ∈ X,

x ∈ X.

(4.10.7)

256

4


Let x ∈ dom ϕ . The monotonicity of Ah and assumption 3) imply for x > R the following inequality: x Ah x + αJx, x − x + ϕ (x) − c . ≥ α(x − x ) − Ah x 1 + x x

Consequently, (4.10.2) is fulfilled for the operator Ah + αJ and for the functional ϕ . Thus, a solution of (4.10.7) exists. Lemma 4.10.4 The inequality (4.10.7) has a unique solution. Proof. Let x1 and x2 be two different solutions of (4.10.7). Then Ah x1 + αJx1 − f δ , x1 − y + ϕ (x1 ) − ϕ (y) ≤ 0 ∀y ∈ X,

(4.10.8)

Ah x2 + αJx2 − f δ , x2 − y + ϕ (x2 ) − ϕ (y) ≤ 0 ∀y ∈ X.

(4.10.9)

and Put y = x2 in (4.10.8) and y = x1 in (4.10.9) and add the obtained inequalities. We obtain Ah x1 − Ah x2 , x1 − x2 + αJx1 − Jx2 , x1 − x2 ≤ 0. Due to the monotonicity of Ah and strict monotonicity of J, the letter occurs only if x1 = x2 . Thus, the lemma holds. Denote the unique solution of (4.10.7) by xγα . Then Ah xγα + αJxγα − f δ , xγα − y + ϕ (xγα ) − ϕ (y) ≤ 0 ∀y ∈ X.

(4.10.10)

Theorem 4.10.5 Let X be an E-space, X ∗ be a strictly convex space, A : X → X ∗ be a monotone hemicontinuous and bounded operator, D(A) = X, ϕ : X → R1 be a properly convex lower semicontinuous functional, a solution set N of the mixed variational inequality (4.10.1) be nonempty. Assume that the conditions 1) - 3) are fulfilled and, in addition, lim sup s→∞

q(s) ≤ Q, 0 < Q < ∞, s2

(4.10.11)

and

δ+h+ = 0. (4.10.12) α ¯∗ of (4.10.1). Then the sequence {xγα } converges strongly to the minimal norm solution x lim

α→0

Proof. Let x∗ ∈ N. Put x = x∗ , y = xγα in (4.10.1) and y = x∗ in (4.10.10). Adding the obtained results one gets Ax∗ − Ah x∗ , x∗ − xγα + Ah x∗ − Ah xγα , x∗ − xγα + αJxγα , xγα − x∗ + f − f δ , xγα − x∗ + ϕ(x∗ ) − ϕ (x∗ ) + ϕ (xγα ) − ϕ(xγα ) ≤ 0.

(4.10.13)

4.10 Variational Inequalities of Mixed Type

257

Using the conditions 1) - 3) we deduce xγα 2 ≤ xγα x∗ +

+

α

δ

α

+

h g(x∗ ) xγα + x∗ α

q(x∗ ) + q(xγα )

∀x∗ ∈ N.

Now (4.10.11) and (4.10.12) guarantee that {xγα } is bounded as α → 0. Hence, xγα → x ¯ ∈ X. Due to Lemma 4.10.2, (4.10.10) yields the inequality Ah y + αJy − f δ , xγα − y + ϕ (xγα ) − ϕ (y) ≤ 0 ∀y ∈ X.

(4.10.14)

According to Theorem 1.1.13, the functional ϕ is weakly lower semicontinuous. Therefore, ϕ(¯ x) ≤ lim inf ϕ(xγα ).

(4.10.15)

α→0

Since {xγα } and q(s) are bounded, by (4.10.6), we have ϕ(xγα ) ≤ ϕ (xγα ) + c2 ,

(4.10.16)

where c2 > 0. Obviously, → 0 as α → 0 because of (4.10.12). Then (4.10.5), (4.10.6), (4.10.14), (4.10.15), (4.10.16) and the condition 1) imply Ay − f, x ¯ − y + ϕ(¯ x) − ϕ(y) ≤ 0 ∀y ∈ X. It is not difficult to be now sure that the inclusion x ¯ ∈ N follows from Lemma 4.10.2. Further, by (4.10.13), there exists a constant c3 > 0 such that for all x∗ ∈ N

xγα − x∗

2

≤ Jxγα − Jx∗ , xγα − x∗ ≤ Jx∗ , x∗ − xγα + c3

δ+h+ . α

According to Lemma 4.10.3, the set N is closed and convex. Therefore, there exists a unique ¯∗ (see, for example, Section x ¯∗ . Then we obtain that in E-space X the sequence xγα → x 2.2). Observe that if the function q(s) does not satisfy (4.10.11) then in (4.10.7) we may use the duality mapping J µ with a suitable gauge function µ(t) in place of J.

Bibliographical Notes and Remarks The results of Section 4.1 are due to [35, 189, 201]. The idea to apply the generalized residual in oder to state the residual principle for variational inequalities belongs to Ryazantseva. In connection with this, we need to emphasize that, generally speaking, the classical residual of the variational inequality (4.1.7) in the form ρ(α) = Ah xγα − f δ ∗ on solutions xγα of the regularized inequality (4.1.14) does not make sense even in the case of

258

4


a continuous operator Ah . At the same time, it turned out that the generalized residual principle is a very effective tool for choosing the regularization parameter α as it is done, for instance, in Theorems 4.1.8 - 4.1.11. The operator regularization method for solving variational inequalities on approximately given sets was studied by Ryazantseva in [190, 201]. Theorem 4.3.1 was proved by Alber, Butnariu and Ryazantseva and can be found in [18]. The examples of variational inequalities in Section 4.4 have been adopted from [120]. The convergence of the operator regularization method for variational inequalities with unbounded monotone operators on inexact sets was established by Alber and Notik in [27]. It was also shown in [27] that the assertion similar to Theorem 2.2.5 holds for variational inequalities in Banach spaces. The so-called variational inequalities with small offset were introduced and investigated by Liskovets in [133, 138]. The regularization method for solving variational inequalities with Mosco-approximation of constraint sets was studied in [19]. The similar version for monotone inclusions with Mosco-approximation of operators was considered in [17]. Mosco perturbations for variational inequalities with inverse-strongly-monotone operators in a Hilbert space has been developed in [140, 155], where the estimates of convergence rate are also presented. The statement and convergence analysis of regularization algorithms for solving variational inequalities with hypomonotone mappings follows the paper [20]. The results of Section 4.9 have appeared in [137]. Theorem 4.10.5 for the mixed variational inequalities with monotone hemicontinuous operators A and Ah was proved in [139].

Chapter 5

APPLICATIONS OF THE REGULARIZATION METHODS 5.1

Computation of Unbounded Monotone Operators

It has been already marked that monotone operators in a Banach space are not bounded in general. Hence, the problem of the value computation y0 ∈ Ax0

(5.1.1) ∗ 2X

of an unbounded monotone operator A : X → at a point x0 ∈ X belongs to the class of unstable problems [99, 130, 131]. For solving the problem (5.1.1), as before, the regularization methods in operator form are used in this chapter. In the sequel, we assume that y0 ∈ R(A) but, in place of x0 , a sequence {xδ } ⊂ X is given such that xδ − x0 ≤ δ, where δ ∈ (0, δ ∗ ]. Note that elements xδ do not necessarily belong to D(A). Using xδ , we should construct in X ∗ an approximation sequence which converges weakly or strongly as δ → 0 to a certain value of the operator A at a point x0 . Consider separately the cases of the Hilbert and Banach spaces. 1. Let X = H be a Hilbert space, A : H → H be a maximal monotone and hemicontinuous (i.e., single-valued) operator. Then an element y0 ∈ H satisfying (5.1.1) is unique. Put into correspondence with (5.1.1) the following regularized equation: xδα + αAxδα = xδ , α > 0,

(5.1.2)

= → y0 , and establish the connection α = α(δ), which guarantees the limit relation when α(δ) → 0 as δ → 0. By virtue of Theorem 1.7.4, equation (5.1.2) with any right-hand side xδ has a unique solution. Introduce the intermediate equation corresponding to the exact value x0 as x0α + αAx0α = x0 , α > 0. (5.1.3) Axδα

yαδ

By the monotonicity of A, (Ax − Ax0 , x − x0 ) ≥ 0 ∀x ∈ D(A). 259

(5.1.4)

260

5

APPLICATIONS OF THE REGULARIZATION METHODS

Assume that in (5.1.4) x = x0α . Then, by (5.1.3), (Ax0α − Ax0 , x0α − x0 ) = −α(Ax0α − Ax0 , Ax0α ) ≥ 0,

(5.1.5)

Ax0α ≤ y0

(5.1.6)

and the estimate

appears from (5.1.5). It allows us to prove by the standard arguments that Ax0α weakly converges to some element y¯ ∈ H as α → 0. On the other hand, (5.1.3) implies x0α − x0 = αAx0α → 0 as α → 0. Show that y¯ = Ax0 . Let x = x0 + tv, where v is an arbitrary element of H and t ≥ 0. Due to Theorem 1.7.19, D(A) is open because A is a maximal monotone and hemicontinuous operator. Therefore, x ∈ D(A) if t is sufficiently small. Since A is monotone, we have (Ax0α − A(x0 + tv), x0α − x0 − tv) ≥ 0. Let first α → 0. By the strong convergence of x0α to x0 and by weak convergence of Ax0α to y¯, we deduce the inequality −t(¯ y − A(x0 + tv), v) ≥ 0. Hence, (A(x0 + tv) − y¯, v) ≥ 0. Let now t → 0. In view of the hemicontinuity of A, (Ax0 − y¯, v) ≥ 0 ∀v ∈ H. This means that Ax0 = y¯. Since the operator A is single-valued, the equality y¯ = y0 holds. Observe that the same results can be established by Lemma 1.4.5. Thus, we have proved a weak convergence (5.1.7) Ax0α y0 as α → 0. It is easy to see that (5.1.5) involves the inequality (Ax0α , Ax0α − y0 ) ≤ 0. Then by the obvious equality (Ax0α , Ax0α − y0 ) − (y0 , Ax0α − y0 ) = Ax0α − y0 2 , we obtain

Ax0α − y0 2 ≤ −(y0 , Ax0α − y0 ).

Consequently, the convergence Ax0α → y0 results from (5.1.7). It is clear that Axδα − y0 ≤ Axδα − Ax0α + Ax0α − y0 . Then (5.1.2), (5.1.3) and the monotonicity A imply the following estimates: Axδα − Ax0α ≤ α−1 (xδ − x0 + xδα − x0α )

(5.1.8)


5.1 and

261

xδα − x0α ≤ xδ − x0 ≤ δ.

Thus, by (5.1.6), Axδα ≤

2δ + y0 . α

(5.1.9)

At the same time, the equality xδα − x0 + αAxδα − αAx0 + αAx0 = xδ − x0 gives xδα − x0 ≤ δ + αy0 .

(5.1.10)

Hence, → x0 , when α → 0 and δ → 0. The same condition ensures the weak convergence of Axδα to y0 . Finally, 2δ Axδα − Ax0α ≤ . α δ δ If → 0 then Axα → Ax0 . It follows from (5.1.8) that Axδα → Ax0α as α → 0. Thus, we α have proved xδα

Theorem 5.1.1 The sequence {Axδα }, where xδα are solutions of the equation (5.1.2), conδ → 0 (δ → 0) as α → 0, and the estimate (5.1.9) verges strongly (weakly) to y0 = Ax0 if α holds. Corollary 5.1.2 Let xδα be solutions of the equation (5.1.2) and Axδα → y0 when δ → 0, α → 0 and xδ − x0 ≤ δ. Then y0 = Ax0 , that is, y0 ∈ R(A). These assertions may be combined by means of the following theorem. Theorem 5.1.3 Let xδα be solutions of the equation (5.1.2). The sequence {Axδα } converges δ → 0 and xδ − x0 ≤ δ if and only if y0 = Ax0 . to an element y0 ∈ H as α → 0, α

2. Next we discuss behavior of the functions ρ(α) = xδα − xδ and σ(α) = Axδα on a semi-infinite interval [α0 , ∞) with fixed δ ∈ (0, δ ∗ ] and α0 > 0, where xδα is a solution of the equation (5.1.2). Denote yαδ = Axδα . The function ρ(α) is single-valued because the δ equation (5.1.2) is uniquely solvable. The function σ(α) is single-valued and bounded if α is bounded (it holds for α ≥ α0 > 0). A continuity of σ(α) is shown by analogy with the case of Section 3.1 (see Lemma 3.1.1). Without loss of generality, take α1 and α2 such that α2 > α1 . It is obvious that 0 ≤ (yαδ 1 − yαδ 2 , xδα1 − xδα2 ) = (yαδ 1 − yαδ 2 , α2 yαδ 2 − α1 yαδ 1 ). From this, we have α1 yαδ 1 − yαδ 2 2 ≤ (α2 − α1 )(yαδ 1 − yαδ 2 , yαδ 2 ).

(5.1.11)

262

5


Then we find that, firstly, α2 − α1 δ δ δ yα2 , ≤ yα1 − yα 2

α1

(5.1.12)

and, secondly, (yαδ 1 − yαδ 2 , yαδ 2 ) ≥ 0. We come to the relation yαδ 2 ≤ yαδ 1 as α2 > α1 .

(5.1.13)

yαδ

Since is bounded, (5.1.12) guarantees that σ(α) and, consequently, ρ(α) are continuous because ρ(α) = ασ(α). The result (5.1.13) shows that σ(α) does not increase. Hence, σ(α) approaches a finite limit σ ¯ ≥ 0 as α → ∞. We prove below that σ ¯ = 0. It is well known that if operator A is monotone then inverse map A−1 is also monotone. Suppose that A−1 is defined at some point u0 ∈ H, that is, u0 ∈ R(A). We have (Axδα − u0 , xδα − v0 ) ≥ 0 ∀v0 ∈ A−1 u0 . Let θH ∈ R(A) and x0 ∈ A−1 (θH ), i.e., Ax0 = θH . Assuming in the last inequality that u0 = θH and v0 = x0 we obtain (Axδα , xδα − x0 ) ≥ 0. Then it results from the equation (5.1.2) that (xδ − xδα , xδα − x0 ) ≥ 0. The simple calculations allow us to derive for x0 ∈ A−1 (θH ) the following estimate: xδα − xδ ≤ x0 − xδ ≤ x0 − x0 + δ.

(5.1.14)

Thus, if θH ∈ R(A) then the sequence {xδα } is bounded for all α > 0 and a bound does not ¯ ∈ H and depend on α and δ. Therefore, xδα x σ(α) = Axδα = α−1 xδ − xδα → 0 as α → ∞. In addition, Axδα → θH . Due to Lemma 1.4.5, the graph of A is demiclosed and this implies that A¯ x = θH . Consequently, x ¯ ∈ Q = {x | Ax = 0}. After that, taking into account the weak convergence of xδα to x ¯ and the inequality (5.1.14), we establish, as in Lemma 3.1.4, strong convergence of xδα to x0∗ , as α → ∞, where x0∗ ∈ Q and x0∗ − xδ = min{x0 − xδ | x0 ∈ A−1 (θH )}.

(5.1.15)

This leads to the limit equality lim ρ(α) = x0∗ − xδ .

α→∞

(5.1.16)

5.1


263

Let p ∈ (0, 1]. By reason of (5.1.10), we can write xδα − xδ ≤ 2δ + αy0 . It is clear that there exists α ≥ α0 > 0 such that αy0 < δ p . Then ρ(α) < 2δ + δ p ≤ δ p (1 + 2δ ∗1−p ) = kδ p , k = 1 + 2δ ∗1−p . Assume that δ satisfies the inequality kδ p < x0∗ − xδ ,

(5.1.17)

where x0∗ is defined by (5.1.15). Then kδ p < x0∗ − x0 + δ. If we suppose now that y0 = θH , that is, x0∗ = x0 , then we will come to a contradiction. Hence, (5.1.17) really implies that y0 = Ax0 = θH . Since ρ(α) is continuous and (5.1.16) is true, there exists at least one α ¯ such that ρ(¯ α) = xδα¯ − xδ = kδ p ,

(5.1.18) p

where xδα¯ is a solution of (5.1.2) with α = α ¯ . Based upon that, α ¯> ρ(¯ α) < kδ p which contradicts (5.1.18). Hence,

δ because otherwise y0

δ ≤ δ 1−p y0 α ¯

and yαδ¯ = Axδα¯ =

kδ p ≤ ky0 . α ¯

In other words, the sequence {yαδ¯ } is bounded if xδα¯ satisfy (5.1.18) for every δ ∈ (0, δ ∗ ]. Therefore, {yαδ¯ } is weakly compact in H. Then we assert that there exists y¯ ∈ H such that yαδ¯ y¯ as δ → 0. On the other hand, xδα¯ − x0 ≤ xδα¯ − xδ + xδ − x0 ≤ kδ p + δ → 0, δ → 0, i.e., xδα¯ → x0 . According to Lemma 1.4.5, the graph of operator A is demiclosed, therefore, Ax0 = y¯ = y0 . We obtain Axδα¯ − Ax0 2 = α ¯ −1 (¯ αAxδα¯ , Axδα¯ − Ax0 ) − (Ax0 , Axδα¯ − Ax0 )

= −¯ α−1 (xδα¯ − xδ , Axδα¯ − Ax0 ) − (Ax0 , Axδα¯ − Ax0 )

= −¯ α−1 (xδα¯ − x0 , Axδα¯ − Ax0 ) − (Ax0 , Axδα¯ − Ax0 ) + α ¯ −1 (xδ − x0 , Axδα¯ − Ax0 ) δ Axδα¯ − Ax0 − (Ax0 , Axδα¯ − Ax0 ). ≤ α ¯

264

5


δ δ → 0 as δ → 0. Hence, Axδα¯ → Ax0 . If p = 1 then ≤ y0 . In this case α ¯ α ¯ we may assert only the weak convergence of Axδα¯ to Ax0 . Finally, it is necessary to recall that Ax0 = θH as δ ∈ (0, δ ∗ ], therefore, there exists a constant µ > 0 such that Axδα¯ ≥ µ for sufficiently small δ > 0. Since If p ∈ (0, 1) then

δ p ρ(¯ α) = αAx ¯ α ¯ = kδ ,

we establish that α ¯ → 0 as δ → 0. Thus, on the basis of Theorem 5.1.1, the residual principle in the form (5.1.18) satisfies the general conditions for the operator regularization methods to be convergent. However, we do not state the obtained result as a separate theorem because we will give it below in the more general form. 3. Observe further in what directions these results can be generalized. a) We can assume that θH ∈ R(A), z 0 ∈ R(A), z 0 = y0 , and consider the equation xδα + α(Axδα − z 0 ) = xδ . The boundedness of implications:

{xδα }

(5.1.19)

can be proved making use of monotonicity of A and following

(Axδα − z 0 , xδα − u0 ) ≥ 0 ∀u0 ∈ A−1 z 0 =⇒ (xδ − xδα , xδα − u0 ) ≥ 0 =⇒ −xδα − xδ 2 + (xδ − xδα , xδ − u0 ) ≥ 0 =⇒ xδα − xδ ≤ xδ − u0 ≤ xδ − x0 + x0 − u0 =⇒

Axδα → z 0 as α → ∞.

By means of the scheme applying in Lemma 3.1.9, we can prove that lim ρ(α) = x∗ − xδ ,

α→∞

where x∗ ∈ A−1 z 0 satisfies the equality x∗ − xδ = min{x − xδ | x ∈ A−1 z 0 }.

(5.1.20)

Now in the proof of the residual principle, it suffices to replace x0∗ ∈ A−1 (θH ) by x∗ ∈ A−1 z 0 . b) The hemicontinuity property of A can be omitted. Recall that A is maximal monotone and, as well known, its inverse operator A−1 is maximal monotone too, so, they are multiplevalued, in general. By Theorem 1.4.9, Ax0 is a convex closed set for all x0 ∈ D(A). Since H is a reflexive strictly convex space, there exists a unique element y0∗ , satisfying the condition y0∗ − z 0 = min {y − z 0 | y ∈ Ax0 }.

(5.1.21)

5.1


265

Presuming z 0 ∈ Ax0 , we can prove again all the assertions of this section. 4. Next we deal with the more general case. Suppose that X is a strictly convex Banach ∗ space, X ∗ is an E-space, A : X → 2X is a maximal monotone operator, and, as before, ∗ ∗ J : X → X is a duality mapping in X ∗ . Consider the equation x + αJ ∗ (Ax − z 0 ) = xδ ,

(5.1.22)

where z 0 ∈ Ax0 . Let xδα be a solution of (5.1.22). Then there exists yαδ ∈ Axδα such that xδα + αJ ∗ (yαδ − z 0 ) = xδ .

(5.1.23)

Applying now Theorem 1.7.4 to the equation A−1 y + αJ ∗ (y − z 0 ) = xδ , α > 0,

(5.1.24)

with maximal monotone operator A−1 : X ∗ → 2X , we conclude that there exists a unique element y = yαδ ∈ Axδα satisfying the equation (5.1.24). Hence, xδα in (5.1.23) is also unique. By the monotonicity of A, yαδ − y0 , xδα − x0 = yαδ − y0 , xδα − xδ + yαδ − y0 , xδ − x0 ≥ 0.

(5.1.25)

It is easy to see that yαδ − y0 , xδα − xδ = −αyαδ − z 0 , J ∗ (yαδ − z 0 ) − αz 0 − y0 , J ∗ (yαδ − z 0 ) because of (5.1.23). Present the last term in the left-hand side of the inequality (5.1.25) as yαδ − y0 , xδ − x0 = yαδ − z 0 , xδ − x0 + z 0 − y0 , xδ − x0 . Since

yαδ − z 0 , J ∗ (yαδ − z 0 ) = yαδ − z 0 2∗

and

z 0 − y0 , J ∗ (yαδ − z 0 ) ≥ −z 0 − y0 ∗ yαδ − z 0 ∗ ,

we have αyαδ − z 0 2∗ − αz 0 − y0 ∗ yαδ − z 0 ∗ − (yαδ − z0 ∗ + z 0 − y0 ∗ )xδ − x0 ≤ 0. One can be sure that the estimate xδ − x0 ≤ δ leads to the quadratic inequality for norms of regularized solutions similar to those that we have obtained in Sections 2.1 and 2.2:

yαδ − z 0 2∗ − z 0 − y0 ∗ +

δ δ δ yα − z 0 ∗ − z 0 − y0 ∗ ≤ 0. α α

(5.1.26)

There are two possibilities to evaluate yαδ − z 0 ∗ from (5.1.26): yαδ − z 0 ∗ ≤ 2

δ + z 0 − y0 ∗ α

(5.1.27)

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5


and yαδ − z 0 ∗ ≤ They show that if

δ + 2z 0 − y0 ∗ . α

(5.1.28)

δ is bounded as α → 0 then yαδ − z 0 ∗ is bounded as well. Hence, α

αyαδ − z 0 ∗ = xδα − xδ → 0 as α → 0. By analogy with Theorem 5.1.1 we prove that yαδ y¯0 ∈ X ∗ as α → 0, and then y¯0 ∈ Ax0 . Return again to the equation (5.1.23) which yields the equality αyαδ − y¯0 , J ∗ (yαδ − z 0 ) − J ∗ (¯ y0 − z 0 ) y0 − z 0 ) + yαδ − y¯0 , xδα − x0 + αyαδ − y¯0 , J ∗ (¯ = yαδ − y¯0 , xδ − x0 .

(5.1.29)

The second term in the left-hand side of (5.1.29) is non-negative because A is monotone. Therefore, in view of the monotonicity of J ∗ , y0 − z 0 ) 0 ≤ yαδ − y¯0 , J ∗ (yαδ − z 0 ) − J ∗ (¯ ≤ −yαδ − y¯0 , J ∗ (¯ y0 − z 0 ) + α−1 δyαδ − y¯0 ∗ . Since yαδ weakly converges to y¯0 as α → 0 and yαδ − y¯0 ∗ is bounded, we have yαδ − y¯0 , J ∗ (yαδ − z 0 ) − J ∗ (¯ y0 − z 0 ) → 0, provided that

δ → 0. On the other hand, by (1.5.3), α

y0 − z 0 ) ≥ (yαδ − z 0 ∗ − ¯ y 0 − z 0 ∗ )2 . yαδ − y¯0 , J ∗ (yαδ − z 0 ) − J ∗ (¯ Consequently, δ → 0. α We know that the weak convergence and convergence of norms imply in the E-space X ∗ strong convergence. Thus, yαδ → y¯0 ∈ Ax0 . Choose and fix an element y ∈ Ax0 . Replace in (5.1.29) y¯0 by y. In view of the inequality y0 − z 0 ∗ as α → 0, yαδ − z 0 ∗ → ¯

αyαδ − y, J ∗ (yαδ − z 0 ) − J ∗ (y − z 0 ) + yαδ − y, xδα − x0 ≥ 0, we derive the estimate yαδ − y, J ∗ (y − z 0 ) ≤

δ δ y − y∗ . α α

Letting α → 0, one gets ¯ y0 − y, J ∗ (y − z 0 ) ≤ 0 ∀y ∈ Ax0 , y¯0 ∈ Ax0 ,

5.1 Computation of Unbounded Monotone Operators

267

which proves that y¯0 = y0∗ (see (5.1.21)). Since y0∗ is unique, the whole sequence {yαδ } converges to y0∗ . 5. Consider further the equation (5.1.23) separately with α = α1 and α = α2 and with fixed δ > 0. It is not difficult to be sure that α1 yαδ 1 − yαδ 2 , J ∗ (yαδ 1 − z 0 ) − J ∗ (yαδ 2 − z 0 ) ≤ (α2 − α1 )yαδ 1 − yαδ 2 , J ∗ (yαδ 2 − z 0 ) . (5.1.30) We find from (5.1.30) that (yαδ 1 − z 0 ∗ − yαδ 2 − z 0 ∗ )2 ≤

|α2 − α1 | δ yα2 − z 0 ∗ yαδ 1 − yαδ 2 ∗ . α1

This inequality guarantees the continuity of the functions σ(α) = yαδ − z 0 ∗ and ρ(α) = ασ(α) with respect to α ≥ α0 > 0. By (5.1.30), if α2 ≥ α1 then yαδ 1 − yαδ 2 , J ∗ (yαδ 2 − z 0 ) ≥ 0. This implies the following result: yαδ 2 − z 0 ∗ ≤ yαδ 1 − z 0 ∗ . As in the case of Hilbert space, one can show that xδα is bounded and lim yαδ − z 0 ∗ = 0.

α→∞

In this case, the equality yαδ − z 0 = α−1 (J ∗ )−1 (xδ − xδα ) = α−1 J(xδ − xδα ) is used. It is valid because J ∗ J = IX and JJ ∗ = IX ∗ in our assumption about X and X ∗ . Thus, yαδ → z 0 as α → ∞. Assuming z 0 ∈ R(A) we conclude that lim ρ(α) = x∗ − xδ ,

α→∞

where x∗ ∈ A−1 z 0 , that is, z 0 ∈ Ax∗ . Moreover, (5.1.20) holds. Then (5.1.27) leads to the estimate ρ(α) = αyαδ − z 0 ∗ ≤ 2δ + αy0 − z 0 ∗ . The rest of the proof (including the assertion that σ(α) and ρ(α) are single-valued and continuous functions) follows the pattern of the Hilbert space case. Let us state the final result. Theorem 5.1.4 (the generalized residual principle). Let X be a strictly convex Banach ∗ space, X ∗ be an E-space, A : X → 2X be an unbounded maximal monotone (possibly 0 multiple-valued) operator and z ∈ R(A). Assume that for any δ ∈ (0, δ ∗ ] there holds the condition x∗ − xδ > kδ p , 0 < p ≤ 1, k > 1 + 2δ ∗1−p , (5.1.31)

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5


¯ such that the where x∗ ∈ A−1 z 0 satisfies (5.1.20). Then there exists at least one α = α generalized residual ρ(¯ α) = αy ¯ αδ¯ − z 0 ∗ = kδ p , where yαδ¯ ∈ Axδα¯ and xδα¯ is a solution of the equation (5.1.22) with α = α ¯ , i.e., xδα¯ + α ¯ J ∗ (yαδ¯ − z 0 ) = xδ . δ → 0 and α ¯ → 0; (ii) if p = 1, α ¯ δ ≤ const and δ → 0 and operator A is single-valued at a point x0 , then yαδ¯ y0 = Ax0 , α ¯ α ¯ → 0. Furthermore, (i) if p ∈ (0, 1) and δ → 0, then yαδ¯ → y0∗ ,

Remark 5.1.5 Instead of (5.1.22), it is possible to make use of the equation αAx + J(x − xδ ) = 0. Remark 5.1.6 The problems of the value computation of approximately given monotone operators are considered in a similar way. Remark 5.1.7 If we choose z 0 ∈ R(A), then requirement (5.1.31) in the last theorem should be omitted (see Section 3.1).

5.2

Computation of Unbounded Semimonotone Operators

We study the operator regularization method for problems of the value computation of unbounded semimonotone maps. Let X be a strictly convex Banach space, X ∗ be an E∗ space, A : X → 2X be a semimonotone unbounded operator with D(A) = X. Assume that T is a monotone operator such that T = A + C, where an operator C : X → X ∗ is strongly continuous. If T¯ is a maximal monotone extension of T then A¯ = T¯ − C. Denote ¯ 0 ) = {y ∈ X ∗ | v − y, x − x0 ≥ 0 ∀v ∈ T¯x, ∀x ∈ X}. R(x ¯ 0 ) is convex and closed. According to Theorem 1.4.9, the set R(x Definition 5.2.1 A set ¯ 0 )} R(x0 ) = {v | v = y − Cx0 , y ∈ R(x is called the generalized value set of semimonotone operator A at a point x0 . The set R(x0 ) is also convex and closed in X ∗ . Therefore, under our conditions, there exists a unique element y ∗ satisfying the equality y ∗ ∗ = min {y∗ | y ∈ R(x0 )}.

5.2 Computation of Unbounded Semimonotone Operators

269

Let a sequence {xδ } be given in place of x0 such that x0 − xδ ≤ δ, δ ∈ (0, δ ∗ ]. Our aim is to construct a sequence {yδ } ⊂ X ∗ , which strongly converges to y ∗ ∈ R(x0 ) as δ → 0. In the beginning, we solve the value computation problem of T at a point x0 , when operator C is known. Due to the results of Section 5.1, we are able to define the sequence ¯ 0 ) as {yαδ }, yαδ ∈ X ∗ , which strongly converges in X ∗ to the minimal norm element y¯0 ∈ R(x δ → 0, α → 0. Then α uδα = yαδ − Cxδα → u ∈ R(x0 ), where xδα is a solution of the equation αT x + J(x − xδ ) = 0, α > 0, in the sense of Definition 1.9.3. Note that continuity of C is enough in these arguments. However, there are problems of this type (see, e.g., [78]) in which it is impossible to give explicit representation of A by the operator C. We present the method for solving such problems, as well. Suppose that the space X has the M -property and there exists r1 > 0 such that the inequality ¯ y, x − x0 ≥ 0 ∀y ∈ Ax, x ∈ X, (5.2.1) is satisfied provided that x − x0 ≥ r1 . Consider in X the equation αAx + J(x − xδ ) = 0. Let

(5.2.2)

δ ¯ x ∈ X and x − x0 ≥ r1 that → 0 as α → 0. Then we have for all y ∈ Ax, α

αy + J(x − xδ ), x − x0 = αy, x − x0 + x − xδ 2

+ J(x − xδ ), xδ − x0 ≥ x − xδ x − xδ − δ . It follows from this that there is a constant r2 ≥ r1 such that if x − x0 ≥ r2 then the inequality ¯ αy + J(x − xδ ), x − x0 ≥ 0 ∀y ∈ Ax holds. By Theorem 1.10.6, we conclude that there exists at least one solution xδα of the equation (5.2.2) in the sense of Definition 1.10.3 with the estimate xδα − x0 ≤ r2 . Define the sequence {yαδ } as yαδ = −α−1 J(xδα − xδ ).

(5.2.3)

Theorem 5.2.2 Let X be strictly convex and have the M -property, X ∗ be an E-space, δ ∗ → 0 as A : X → 2X be a semimonotone operator with D(A) = X, xδ − x0 ≤ δ, α α → 0, and the condition (5.2.1) hold. Then the sequence {yαδ } defined by (5.2.3) converges strongly in X ∗ to the minimal norm element y ∗ ∈ R(x0 ).

5

270


Proof. Rewrite (5.2.3) as αyαδ = −J(xδα − xδ ). ¯ δα such that Since J ∗ = J −1 , where J ∗ is a duality mapping in X ∗ , there exists yαδ ∈ Ax αJ ∗ yαδ + xδα − xδ = θX .

(5.2.4)

¯ from the inequality For all x ∈ X and y ∈ Ax, y, αJ ∗ y + x − xδ = αy2∗ + y, x − x0 + y, x0 − xδ

≥ αy∗ y∗ −

δ + y, x − x0 , α

it results that {yαδ } is bounded in X ∗ . Hence, yαδ y¯ ∈ X ∗ . Then, by (5.2.4), one gets xδα − x0 ≤ xδα − xδ + xδ − x0 ≤ αyαδ ∗ + δ. Consequently, xδα → x0 as α → 0. We show that y¯ ∈ R(x0 ). Indeed, the monotonicity of T¯ implies v − yαδ − Cxδα , x − xδα ≥ 0 ∀v ∈ T¯x, ∀x ∈ X.

(5.2.5)

Since the operator C : X → X ∗ is strongly continuous, we have that Cxδα → Cx0 as α → 0. Letting α → 0 in (5.2.5), we derive the inequality v − y¯ − Cx0 , x − x0 ≥ 0 ∀v ∈ T¯x, ∀x ∈ X. This means that y¯ + Cx0 ∈ T¯x0 or y¯ ∈ R(x0 ). Rewrite now (5.2.4) in the equivalent form: αJ ∗ yαδ − αJ ∗ u + xδα − x0 = xδ − x0 − αJ ∗ u ∀u ∈ R(x0 ). Then one has αzαδ − v, J ∗ yαδ − J ∗ u + zαδ − v, xδα − x0 = zαδ − v, xδ − x0 − αzαδ − v, J ∗ u ,

(5.2.6)

where u ∈ R(x0 ) and zαδ = yαδ + Cxδα , v = u + Cx0 . Using further the properties of J ∗ we conclude that zαδ − v, J ∗ yαδ − J ∗ u ≥ (yαδ ∗ − u∗ )2 + Cxδα − Cx0 , J ∗ yαδ − J ∗ u . Since zαδ ∈ T¯xδα and v ∈ T¯x0 , then the monotonicity of T¯ gives zαδ − v, xδα − x0 ≥ 0.

(5.2.7)

Computation of Unbounded Accretive Operators

5.3

271

Combining (5.2.6) and (5.2.7) with u = y¯, we come to the relation y ∗ )2 + Cxδα − Cx0 , J ∗ yαδ − J ∗ y¯ (yαδ ∗ − ¯ ≤

δ δ z − v¯∗ − zαδ − v¯, J ∗ y¯ , α α

(5.2.8)

δ y ∗ as α → 0 and → 0. Taking into account where v¯ = y¯ + Cx0 . Hence, yαδ ∗ → ¯ α the fact that X ∗ is a E-space, in which weak convergence and convergence of norms imply strong convergence, one gets that yαδ → y¯. Next we use the proof scheme of Section 2.2. Presuming y¯ = y and v¯ = y + Cx0 in (5.2.8), where y is an arbitrary element of R(x0 ), and passing to the limit as α → 0 we obtain that y¯ = y ∗ .

In place of the equation (5.2.2), we may consider the following more general regularized equation: α(Ax − z 0 ) + J(x − xδ ) = 0, (5.2.9) where z 0 is a fixed element of X ∗ . In this case, the sequence {yαδ } defined by the equality yαδ = z 0 − α−1 J(xδα − xδ ), where xδα is a solution of (5.2.9), converges by norm of X ∗ as y0∗ ∈ R(x0 ) satisfying the condition:

δ → 0, α → 0 to the element α

y0∗ − z 0 = min{y − y 0 | y ∈ R(x0 )}. Remark 5.2.3 It is also possible to prove the strong convergence of the regularization method in the case when, instead of the operator A, a sequence Ah of the semimonotone maps is given with D(Ah ) = X, HX ∗ (R(x), Rh (x)) ≤ g(x)h ∀x ∈ X. Here Rh (x) is the generalized value set of the operator Ah at a point x.

5.3

Computation of Unbounded Accretive Operators

Let X be a reflexive Banach space, X ∗ be strongly convex, A : X → X be an accretive operator with domain D(A), x0 ∈ D(A) and y0 = Ax0 .

(5.3.1)

Suppose that, instead of x0 , its perturbed values xδ are given such that xδ −x0 ≤ δ, where δ ∈ (0, δ ∗ ]. Using elements xδ we will construct approximations to y0 ∈ X. Below we present the computation value theorem for unbounded accretive operators in a more precise setting. 1. We sudy first the case of hemicontinuous operators.

272

5

APPLICATIONS OF REGULARIZATION METHODS

Theorem 5.3.1 Assume that X is an E-space and it possesses an approximation, X ∗ is strongly convex, A : X → X is an accretive hemicontinuous operator, D(A) = X, duality δ → 0 as α → 0. Then mapping J : X → X ∗ is continuous and weak-to-weak continuous, α δ δ the sequence {Axα }, where xα are solutions of the regularized equation x + αAx = xδ , α > 0,

(5.3.2)

converges by the norm of X to y0 as α → 0. Proof. First of all, observe that, due to Theorem 1.15.23, the equation (5.3.2) is uniquely solvable. Then (5.3.3) xδα + αAxδα = xδ . Hence, Axδα is well defined. Apply further the proof scheme of Theorem 5.1.1 considering the following auxiliary equation: x0α + αAx0α = x0 .

(5.3.4)

Since X is smooth, it is obvious that J(x0α − x0 ) = −αJ(Ax0α ), and, by the accretiveness condition of A, −αJ(Ax0α ), Ax0α − Ax0 = J(x0α − x0 ), Ax0α − Ax0 ≥ 0. This inequality yields the estimate Ax0α ≤ Ax0 = y0 ,

(5.3.5)

that is, {Ax0α } is bounded. Then by (5.3.4), we easily obtain the limit result: x0α → x0 as α → 0. From the boundedness of {Ax0α }, it also follows that Ax0α y¯ ∈ X as α → 0. Then, by Lemma 1.15.12 and by Theorem 1.15.14, we deduce that y¯ = Ax0 = y0 . Moreover, taking into account (5.3.5) and the weak convergence of {Ax0α } to y¯ in E-space, it is proved, as in Theorem 5.1.1, that Ax0α → Ax0 . Show now the strong convergence of {Axδα } to y¯. Firstly, the equations (5.3.3) and (5.3.4) together give αJ(xδα − x0α ), Axδα − Ax0α + xδα − x0α 2 = J(xδα − x0α ), xδ − x0 . Using again the accretiveness of A we have xδα − x0α 2 ≤ J(xδα − x0α ), xδ − x0 ≤ xδα − x0α xδ − x0 . Thus,

xδα − x0α ≤ xδ − x0 ≤ δ.

(5.3.6)

5.3 Computation of Accretive Operators

273

Secondly, the same equations (5.3.3) and (5.3.4) allow us to derive the following relations: Axδα − Ax0α =

1 1 xδ − x0 + xδα − x0α . xδ − xδα − x0 + x0α ≤ α α

We see that the estimate Axδα − Ax0α ≤

2δ α

(5.3.7)

(5.3.8)

arises from (5.3.7) and (5.3.6). Hence, the conditions of the theorem give Axδα − Ax0α → 0 as α → 0. Then the assertion of the theorem follows from the evident inequality Axδα − Ax0 ≤ Axδα − Ax0α + Ax0α − Ax0 . The proof is accomplished. Corollary 5.3.2 Under the condition of Theorem 5.3.1, if {Axδα } is bounded as

α → 0, then any weak accumulation point of {Axδα } belongs to R(A).

δ → 0, α

Proof. It results from (5.3.3) that xδα → x0 . On the other hand, the boundedness of {Axδα } guarantees that there exist some y¯ such that Axδα y¯ ∈ X. Due to Theorem 1.15.14, maximal accretive operators are weak closed. Then y¯ = Ax0 , that is, Ax0 = y0 ∈ R(A). 2. From here, we consider the case of an m-accretive (possibly, multiple-valued) operator A : X → 2X . Let Ax0 = ∅. As it already earlier observed (see Lemmas 1.15.13 and 1.15.20), Ax0 is a convex and closed set. We are going to find an element y0 ∈ Ax0 .

(5.3.9)

Theorem 5.3.3 Let X be an E-space, X ∗ be strictly convex, J be a continuous and weakδ →0 to-weak continuous duality mapping in X, A : X → 2X be an m-accretive operator, α as α → 0. Then sequence {yαδ }, where yαδ ∈ Axδα , xδα is a solution of the regularized equation (5.3.2) and (5.3.10) xδα + αyαδ = xδ , converges strongly to the minimal norm element y0∗ ∈ Ax0 as α → 0. Proof. Since B = αA + I is strictly accretive, the equation (5.3.2) has a unique solution xδα for all α > 0 and for all xδ ∈ X. Hence, there exists a unique element yαδ ∈ Axδα such

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5


that (5.3.10) holds. We study the properties of {yαδ }. Let x0α be a solution of (5.3.2) with the right-hand side x0 , that is, there exists yα0 ∈ Ax0α such that x0α + αyα0 = x0 .

(5.3.11)

As in the proof of the previous theorem, it is established that yα0 ≤ y ∀y ∈ Ax0 . Therefore, where

yα0 ≤ y0∗ ,

(5.3.12)

y0∗ = min{y | y ∈ Ax0 }.

(5.3.13)

y0∗

Note that satisfying (5.3.13) exists and it is unique in E-space X. Since the sequence {yα0 } is bounded, there is y¯ ∈ X such that yα0 y¯ as α → 0. Moreover, the equality (5.3.11) guarantees strong convergence of x0α to x0 . Recall that the operator A is maximal accretive (as m-accretive), and duality mapping J is continuous. Then we have, by Lemma 1.15.12, that y¯ ∈ Ax0 . Now the weak convergence of yα0 to y¯ and (5.3.12) allow us to conclude that y¯ = y0∗ and yα0 → y0∗ as α → 0. This means in E-space X the strong convergence of yα0 → y0∗ . Using (5.3.1) and (5.3.10), as in Theorem 5.3.1, one can prove that yαδ − yα0 → 0 because

δ → 0 when α → 0. Finally, from the inequality α yαδ − y0∗ ≤ yαδ − yα0 + yα0 − y0∗ ,

it follows that

yαδ → y0∗

as

α → 0.

(5.3.14)

The theorem is proved.

Remark 5.3.4 We are able to study the regularized equation (5.3.2) in a more general form x + α(Ax − z 0 ) = xδ , where z 0 ∈ X is a fixed element. The assertions of Theorems 5.3.1 and 5.3.3 remain still valid; further, the element y0∗ ∈ Ax0 is defined as y0∗ − z 0 = min{y − z 0 | y ∈ Ax0 }.


275

3. We next investigate a choice possibility of the regularization parameter α from the residual principle in the value computation problem of unbounded accretive operators. Suppose that the conditions of Theorem 5.3.3 are satisfied. As usual, we first study the properties of the functions σ(α) = yαδ and ρ(α) = xδα − x0 = αyαδ , where xδα and yαδ are from (5.3.10) and δ ∈ (0, δ ∗ ] is fixed. It is easily established that σ(α) and ρ(α) are single-valued because the equation (5.3.2) has a unique solution xδα for each α > 0. Write down (5.3.10) with α = β and yβδ ∈ Axδβ as xδβ + βyβδ = xδ .

(5.3.15)

J(xδα − xδβ ) = J(βyβδ − αyαδ ).

(5.3.16)

By (5.3.10) and (5.3.15), we find

Since A is accretive, one has J(xδα − xδβ ), yαδ − yβδ ≥ 0. Taking into consideration (5.3.16) we obtain J(βyβδ − αyαδ ), yαδ − yβδ ≥ 0. This is equivalent to the inequality J(αyαδ − βyβδ ), αyαδ − βyβδ + (β − α)yβδ ≤ 0. Then the properties of duality mapping J yield αyαδ − βyβδ ≤ |β − α| yβδ . Therefore,

δ − βyβδ ≤ αyαδ − βyβδ ≤ |β − α| yβδ . αyα

(5.3.17)

As in Theorem 5.3.1, the inequality yαδ − yα0 ≤

2δ α

together with (5.3.12) give the following estimate: yαδ ≤ y0∗ +

2δ . α

(5.3.18)

This shows that {yαδ } is bounded if α ≥ α0 > 0. Then the continuity of ρ(α) = αyαδ follows from (5.3.17). We study the behavior of ρ(α) as α → ∞. Let θX ∈ R(A), N0 = {x | θX ∈ Ax}. Then the accretiveness condition of A implies J(xδα − x), yαδ ≥ 0 ∀x ∈ N0 .

(5.3.19)

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5


The equation (5.3.10) yields the equality yαδ =

xδ − xδα , α

(5.3.20)

and (5.3.19) can be rewritten in the form J(xδα − x), xδ − xδα ≥ 0 ∀x ∈ N0 .

(5.3.21)

xδα − x2 + J(x − xδα ), xδ − x ≤ 0 ∀x ∈ N0 .

(5.3.22)

xδα − x ≤ xδ − x ∀x ∈ N0 ,

(5.3.23)

This leads to the estimate

Thus, that is, {xδα } is bounded and, by (5.3.20), yαδ → θX . In addition, the weak convergence of ¯ ∈ X as α → ∞ is established as before. Since J is weak-to-weak continuous, we xδα to x x which means that x ¯ ∈ N0 . deduce by help of Lemma 1.15.12 that θX ∈ A¯ Pay attention to the inequality (5.3.22). Assuming there x = x ¯ and passing to the limit as α → ∞, one gets ¯. lim xδα = x α→∞

At the same time, the inequality (5.3.21) gives, as α → ∞, J(¯ x − x), xδ − x ¯ ≥ 0 ∀x ∈ N0 .

(5.3.24)

Show that x ¯ defined by (5.3.24) is unique. Let α → ∞ and {xδα } have two accumulation ¯2 . Then for them (5.3.24) holds, that is, points denoted by x ¯1 and x ¯1 ≥ 0 ∀x ∈ N0 J(¯ x1 − x), xδ − x and J(¯ x2 − x), xδ − x ¯2 ≥ 0 ∀x ∈ N0 . Assuming in the first and in the second inequalities x = x ¯2 and x = x ¯1 , respectively, and summing them, we thus obtain ¯2 ), x ¯1 − x ¯2 ≤ 0, J(¯ x1 − x from which the equality x ¯1 = x ¯2 follows. Hence, the whole sequence {xδα } converges as α → ∞ to x ¯ ∈ N0 , that satisfies inequality (5.3.24). In that case x − xδ as α → ∞. ρ(α) = αyαδ = xδα − xδ → ¯ The properties proved above of the generalized residual enable us to establish a generalized residual principle for the value computation problem of accretive operators.


277

Theorem 5.3.5 Assume that X is an E-space, a duality mapping J in X is continuous and weak-to-weak continuous, A : X → 2X in the problem (5.3.9) is an unbounded m-accretive operator, N0 = {x | θX ∈ Ax} = ∅, x0 − xδ ≤ δ, δ ∈ (0, δ ∗ ], ¯ x − xδ > kδ p , k = 1 + 2δ ∗1−p , p ∈ (0, 1],

(5.3.25)

where x ¯ ∈ N0 and satisfies the inequality (5.3.24). Then there exists at least one α = α ¯ such that ¯ yαδ¯ = kδ p , (5.3.26) ρ(¯ α) = xδα¯ − xδ = α where xδα¯ is a solution of the equation (5.3.2) with α = α. ¯ Moreover, α ¯ → 0 as δ → 0, and δ → 0 as δ → 0; (ii) if (i) if p ∈ (0, 1) then yαδ¯ → y0∗ , where y0∗ is defined by (5.3.13), and α ¯ p = 1 and A is single-valued at x0 then yαδ¯ y0 = Ax0 and there exists a constant C > 0 δ such that ≤ C as δ → 0. α ¯ δp Proof. Choosing α < ∗ we deduce by (5.3.18) that y0

ρ(α) < kδ p . Since ρ(α) is continuous and x − xδ > kδ p , lim ρ(α) = ¯

α→∞

we have (5.3.26) with α ¯≥

δp . Then y0∗ yαδ¯ = δ

kδ p ≤ ky0∗ . α ¯

on p is proved as in Theorem 5.1.4. This allows us α ¯ to consider that yαδ¯ y¯ ∈ X as δ → 0. By (5.3.26), we conclude that xδα¯ → x0 as δ → 0. In its turn, demiclosedness of A yields the inclusion y¯ ∈ Ax0 . Obviously, (5.3.18) holds for yαδ = yαδ¯ and y0∗ ∈ Ax0 . Now we are able to show, similarly to Theorem 5.1.4, that y¯ = y0∗ and yαδ¯ → y0∗ . Since X is the E-space, the strong convergence of yαδ¯ to y0∗ is δ is thus established for p ∈ (0, 1). If p = 1 and A is a single-valued operator, then α ¯ δ bounded and the sequence {yα¯ } weakly converges to y0 = Ax0 as δ → 0. In order to be sure that limδ→0 α ¯ = 0, it is necessary to repeat the corresponding arguments of the proof of Theorem 5.1.4.

Observe that the dependence of

Remark 5.3.6 If θX ∈ R(A) then requirement (5.3.25) of Theorem 5.3.5 may be omitted. Remark 5.3.7 If θX ∈ D(A) then we choose z 0 ∈ R(A) and find regularized solutions from the following equation: xδα + α(yαδ − z 0 ) = xδ , yαδ ∈ Axδα . Taking ρ(α) = αyαδ − z 0 and N0 = {x | z 0 ∈ Ax}, one can get all the conclusions of Theorem 5.3.5. With this, y0∗ is defined as the minimization problem: y0∗ − z 0 = min {y − z 0 | y ∈ Ax0 }.

278

5

5.4


Hammerstein Type Operator Equations ∗

Let X be an E-space, X ∗ be strictly convex, A : X → 2X and B : X ∗ → 2X be maximal monotone operators with domains D(A) and D(B), respectively, f ∈ X. An equation x + BAx = f

(5.4.1)

is said to be the Hammerstein type equation. 1. First of all, study the solvability of the problem (5.4.1). Let C = A−1 : X ∗ → 2X and φ ∈ Ax. Then x ∈ A−1 φ = Cφ, and (5.4.1) may be rewritten as Cφ + Bφ = f.

(5.4.2)

Recall that C is a maximal monotone operator because A is so. In order to apply to (5.4.2) the known existence theorems for equations with maximal monotone operators, it is necessary to impose such conditions on A and B which guarantee that the sum C + B is also maximal monotone. By Theorem 1.8.3, these conditions are the following: int D(C) ∩ D(B) = ∅ or D(C) ∩ int D(B) = ∅. Turning to operators A and B of the equation (5.4.1) we see that the operator C + B is maximal monotone if int R(A) ∩ D(B) = ∅ or R(A) ∩ int D(B) = ∅.

(5.4.3)

Joining (5.4.3) to Theorem 1.7.5 we get ∗

Theorem 5.4.1 If A : X → 2X and B : X ∗ → 2X are maximal monotone operators, operator A−1 + B is coercive and the condition (5.4.3) holds, then equation (5.4.1) with any right-hand side f ∈ X has at least one solution. We present some assertions necessary to construct the operator regularization method for the equation (5.4.2). Consider the space product (see [206]) Z = X × X ∗ = {ζ = [x, φ] | x ∈ X, φ ∈ X ∗ } with the natural linear operation αζ1 + βζ2 = [αx1 + βx2 , αφ1 + βφ2 ], where α and β are real numbers, ζ1 = [x1 , φ1 ] and ζ2 = [x2 , φ2 ] are elements from Z. If the norm of an element ζ = [x, φ] is defined by the formula

ζZ = x2 + φ2∗

1/2

then Z is the Banach space and Z ∗ = X ∗ × X is its dual space. A pairing of the spaces Z and Z ∗ is given by the dual product of elements ζ = [x, φ] ∈ Z and η∗ = [ψ, y] ∈ Z ∗ with respect to the equality: η ∗ , ζ = ψ, x + φ, y .

5.4 Hammerstein Type Operator Equations

279

Lemma 5.4.2 Let {ζn }, n = 1, 2, ..., be a sequence of Z such that ζn = [xn , φn ], and let ζ0 = [x0 , φ0 ] ∈ Z. Then the limit relations ζn ζ0 , ζn Z → ζ0 Z

(5.4.4)

and xn x0 ,

φn φ0 , xn → x0 ,

φn ∗ → φ0 ∗

(5.4.5)

are equivalent as n → ∞. Proof. The implication (5.4.5) =⇒ (5.4.4) is obvious. Let (5.4.4) be valid and assume that η ∗ = [ψ, y] is an element of Z ∗ . Then the limit η ∗ , ζn → η ∗ , ζ0 as n → ∞ implies ψ, xn + φn , y → ψ, x0 + φ0 , y .

(5.4.6)

If we put y = θX in (5.4.6) then ψ, xn → ψ, x0

∀ψ ∈ X ∗ ,

that is, xn x0 as n → ∞. If ψ = θX ∗ in (5.4.6) then φn φ0 , that leads to the estimates x0 ≤ lim inf xn and φ0 ∗ ≤ lim inf φn ∗ . n→∞

n→∞

Since

ζn Z → ζ0 Z = x0 2 + φ0 2∗

1/2

(5.4.7)

,

one can consider that xnk → a and φnk ∗ → b, where {xnk } and {φnk } are subsequences of {xn } and {φn }, respectively. Further, a2 + b2 = x0 2 + φ0 2∗ . Now it follows from (5.4.7) that x0 = a and φ0 ∗ = b. Thus, xn → x0 and φn ∗ → φ0 ∗ . The lemma is proved. Lemma 5.4.3 If J : X → X ∗ and J ∗ : X ∗ → X are normalized duality mappings in X and X ∗ , respectively, then the operator JZ : Z → Z ∗ , defined by the formula JZ ζ = [Jx, J ∗ φ]

∀ζ = [x, φ] ∈ Z,

(5.4.8)

is a normalized duality mapping in Z. And conversely: every normalized duality mapping in Z has the form (5.4.8). Proof. We verify Definition 1.5.1 with µ(t) = t for JZ . Toward this end, choose an element ζ = [x, φ] ∈ Z and, using the properties of J and J ∗ , find JZ ζ, ζ = Jx, x + φ, J ∗ φ = x2 + φ2∗ = ζ2Z and

JZ ζZ ∗ = Jx2∗ + J ∗ φ2

1/2

= x2 + φ2∗

1/2

= ζZ .

280

5


Thus , JZ : Z → Z ∗ is a duality mapping in Z. Let then operator J¯Z : Z → Z ∗ be such that (5.4.9) J¯Z ζ, ζ = ζ2Z and J¯Z ζZ ∗ = ζZ . ∗ ∗ ¯ Assuming that JZ ζ = η = [ψ, y], ψ ∈ X , y ∈ X, write down the following equalities: J¯Z ζ, ζ = η ∗ , ζ = ψ, x + φ, y , which, in view of (5.4.9), gives x2 + φ2∗ = ψ, x + φ, y = ψ2∗ + y2 .

(5.4.10)

Show that ψ = Jx, y = J ∗ φ. For that, it is necessary to establish that equalities ψ, x = ψ∗ x and φ, y = φ∗ y are true. Indeed, if it is not the case then, by (5.4.10), we come to a contradiction because x2 + φ2∗ < ψ∗ x + yφ∗

≤ 2−1 ψ2∗ + x2 + 2−1 y2 + φ2∗

= x2 + φ2∗ . Hence, (5.4.10) may be rewritten in the form x2 + φ2∗ = ψ∗ x + yφ∗ = ψ2∗ + y2 . Then

(x − ψ∗ )2 + (φ∗ − y)2 = 0,

i.e., x = ψ∗ and φ∗ = y. Therefore, ψ = Jx and y = J ∗ φ. Lemma 5.4.4 A space Z is strictly convex if and only if X and X ∗ are strictly convex. Proof. A space Z is strictly convex if and only if JZ is strictly monotone. In its turn, in view of JZ ζ = [Jx, J ∗ φ] for all ζ = [x, φ] ∈ Z, operator JZ is strictly monotone if and only if J and J ∗ are strictly monotone too.

2. Pass from equation (5.4.1) to the system

Ax − φ = θX ∗ , x + Bφ = f.

It is equivalent to the following operator equation: Tζ = h ¯, where ζ = [x, φ] ∈ Z, ¯ h = [θX ∗ , f ] ∈ Z ∗ and ∗

T : Z → 2Z , T ζ = [Ax − φ, x + Bφ] ∈ Z ∗ .

(5.4.11)

5.4 Hammerstein Type Operator Equations

281

We study now properties of the operator T . First of all, we assert that T is monotone. Indeed, for ζ1 = [x1 , φ1 ] ∈ Z and ζ2 = [x2 , φ2 ] ∈ Z, the equality ϑ1 − ϑ2 , ζ1 − ζ2 = ξ1 − ξ2 , x1 − x2 + φ1 − φ2 , w1 − w2 ,

(5.4.12)

holds for all ξi ∈ Axi , for all wi ∈ Bφi and for ϑi = [ξi − φi , wi − xi ], i = 1, 2. We note that the inclusion φi ∈ Axi implies xi ∈ A−1 φi . Therefore, (5.4.12) may be rewritten as ϑ1 − ϑ2 , ζ1 − ζ2 = φ1 − φ2 , x1 − x2 + φ1 − φ2 , w1 − w2 , where φi ∈ D(B) and φi ∈ D(A−1 ), i = 1, 2. From this equality follows the fact that if condition (5.4.3) is satisfied then the operator T is maximal monotone in D(T ). Assume further that operators A and B are maximal monotone, in addition, A is hemicontinuous, D(A) = X, and one of the conditions (5.4.3) is satisfied. Lemma 5.4.5 If the solution set N of the equation (5.4.1) is nonempty, then it is convex and closed. Proof. Let M be a solution set of the equation (5.4.11). By Corollary 1.4.10, M is convex and closed in Z. Since M = {[x, φ] | x ∈ N, φ ∈ Ax}, we conclude, in view of (5.4.11), that M = N × A(N ). Then obviously N is convex. Show that N is closed. Let xn ∈ N and xn → x0 as n → ∞. Due to Theorem 1.3.20, under our conditions, operator A is demicontinuous. By the Mazur theorem, the set M is weakly closed, therefore, we get for ζn = [xn , Axn ] ∈ M the following limit relation: ζn [x0 , Ax0 ] ∈ M. Hence, x0 ∈ N. Construct the regularized operator equation in the form x + (B + αJ ∗ )(A + αJ)x = f δ ,

(5.4.13)

where α > 0, f δ is a δ-approximation of f such that f − f δ ≤ δ. Lemma 5.4.6 Equation (5.4.13) is uniquely solvable for any f δ ∈ X. Proof. Let B α = B + αJ ∗ and Aα = A + αJ. We introduce an operator T α ζ = [Aα x − φ, x + B α φ] = T ζ + αJZ ζ, where ζ = [x, φ] and JZ ζ = [Jx, J ∗ φ]. It is clear that solvability of (5.4.13) is equivalent to solvability of the equation T αζ = h ¯δ with the maximal monotone operator T α = T + αJZ and right-hand side h ¯ δ = [θX ∗ , f δ ]. Then the conclusion of the lemma follows from Theorem 1.7.4.

5

282


Theorem 5.4.7 Let a solution set N of the equation (5.4.1) be not empty, A : X → X ∗ be a hemicontinuous monotone operator with D(A) = X. Assume that B : X ∗ → 2X is maximal monotone, f δ ∈ X are such that f − f δ ≤ δ and one of the conditions (5.4.3) δ → 0 as α → 0, then the sequence {xδα } of solutions of the regularized equation holds. If α (5.4.13) converges strongly in X to the solution x ¯∗ ∈ N which is defined as ¯ x∗ 2 + A¯ x∗ 2 = min{x∗ 2 + Ax∗ 2 | x∗ ∈ N }. Proof. Validity of this theorem follows from the results of Section 2.2 and from Lemmas 5.4.5 and 5.4.6.

Under the conditions of Theorem 5.4.7 (not assuming solvability of (5.4.1)), if solutions of the regularized equation (5.4.13) converge strongly as α → 0 then the equation (5.4.1) is solvable. Indeed, since ¯hδ ∈ T α ζαδ , where ¯hδ = [θX ∗ , f δ ] ∈ Z ∗ and ζαδ = [xδα , Axδα ] ∈ Z, the monotonicity condition of T α can be written as Aα x − φ, x − xδα + φ − Aα xδα , x + wα − f δ ≥ 0, [x, φ] ∈ D(T α ), w α ∈ B α φ. ¯, Axδα A¯ x. Therefore, passing to the limit in the last By the assumptions, xδα → x inequality as α → 0, we come to the inequality Ax − φ, x − x ¯ + φ − A¯ x, x + w − f ≥ 0, w ∈ Bφ, which means that x ¯ is a solution of (5.4.1). In addition, if A is weak-to-weak continuous, then for solvability of (5.4.1) it is necessary to assume that the operator method of regularization (5.4.13) converges weakly. As in Chapters 2 and 3, we are able to consider for the equation (5.4.1) the case of approximately given operators and also state the corresponding residual principle. For instance, under the conditions of Theorem 5.4.7, the residual should be defined by the formula

ρ(α) = α xδα 2 + Axδα 2

1/2

.

(5.4.14)

In addition, in all assertions above, the hemicontinuity of A can be replaced by its maximal monotonicity; and then in the proofs the local boundedness of A at any point x ∈ X is used. At the same time, (5.4.14) is rewritten as

ρ(α) = α xδα 2 + yαδ 2

1/2

,

where yαδ ∈ Axδα and f δ − xδα ∈ (B + αJ ∗ )(yαδ + αxδα ).

5.5 Pseudo-Solutions of Monotone Equations

5.5

283

Pseudo-Solutions of Monotone Equations ∗

Let X be a Banach space, A : X → 2X be a maximal monotone operator, f ∈ X ∗ . Suppose that the range R(A) is a closed set. Due to Corollary 1.7.18, it is convex. Consider an equation Ax = f. (5.5.1) Assuming that f ∈ R(A) we do not require in the sequel solvability of (5.5.1) in the classical sense. Our aim in this section is to study pseudo-solutions of (5.5.1). Definition 5.5.1 An element z ∈ D(A) is called a pseudo-solution of the equation (5.5.1) if g − f ∗ = min{y − f ∗ | y ∈ Ax, x ∈ D(A)}, g ∈ Az. Under the conditions above, there exists a unique element g ∈ R(A) such that µ = min{y − f ∗ | y ∈ R(A)} = g − f ∗ > 0.

(5.5.2)

Since g ∈ R(A), nonempty pseudo-solution set N of (5.5.1) coincides with the classical solution set of the equation Au = g. (5.5.3) Hence, N is convex and closed. It is known that in this case g belongs to the boundary of R(A). If int R(A) = ∅ then N is unbounded. Indeed, by Theorem 1.7.19 applied to the operator A−1 , we conclude that N contains at least one semi-line. 1. Assume first that X is a Hilbert space H, A : H → 2H is a maximal monotone x∗ = operator and f ∈ H. Let x ¯∗ be a minimal norm pseudo-solution of (5.5.1), i.e., ¯ ∗ ∗ min{x | x ∈ N }. Assume that, instead of exact right-hand side f, a sequence {f δ } is given such that f − f δ ≤ δ, where 0δ ∈ (0, δ ∗ ]. Construct in H approximations strongly converging to x ¯∗ ∈ N. To this end, consider the regularized operator equation Ax + αx = f δ ,

α > 0.

(5.5.4)

Denote by xδα its unique solution and let µδ = min {y − f δ | y ∈ R(A)}.

(5.5.5)

It is clear that there exists g δ ∈ R(A) such that µδ = gδ − f δ .

(5.5.6)

Lemma 5.5.2 The following estimate holds: g − f δ ≤ µδ + δ.

(5.5.7)

5

284


Proof. By (1.3.5), the problems (5.5.2) and (5.5.5) are equivalent, respectively, to inequalities (g − f, g − y) ≤ 0 ∀y ∈ R(A), (5.5.8) and (g δ − f δ , gδ − y) ≤ 0 ∀y ∈ R(A).

(5.5.9)

gδ

Substituting y = for the first inequality and y = g for the second one, and summing the obtained inequalities we obtain after simple algebra the estimate g − gδ ≤ f − f δ ≤ δ.

(5.5.10)

Hence, g − f δ ≤ g δ − f δ + g δ − g ≤ µδ + δ. Since the equation (5.5.1) is unsolvable in the classical sense, we conclude by Theorem δ → 0. Thus, the strong convergence α δ ∗ of {xα } to x ¯ may be established only if the regularization parameter α approaches some α0 > 0, which is unknown as well. Hence, the main problem is to point the way of choosing α in (5.5.4) ensuring convergence of {xδα } to x ¯∗ as δ → 0. 2.1.4 that {xδα } becomes unbounded as α → 0 and

Theorem 5.5.3 Let A : H → 2H be a maximal monotone operator, R(A) be closed and for all x ∈ D(A) such that x ≥ r > 0, the inequality (y − g, x) ≥ 0 ∀y ∈ Ax

(5.5.11)

be fulfilled, where g solves the problem (5.5.2). If θH ∈ D(A), then assume in addition that y 0 − f δ > µδ + kδ, where

k > 1,

(5.5.12)

y 0 − f δ = min {y − f δ | y ∈ A(θH )}.

(5.5.13)

Under these conditions, there exists at least one α ¯ satisfying the equation ρ(¯ α) = µδ + kδ.

(5.5.14) {xδα¯ },

where If δ → 0, then every strong accumulation point of the sequence of (5.5.4) with α = α ¯ , belongs to the pseudo-solutions set N of (5.5.1).

xδα¯

is a solution

Proof. Choose any fixed x∗ ∈ N. There exists yαδ ∈ Axδα such that yαδ + αxδα = f δ .

(5.5.15)

Subtract g ∈ Ax∗ from both sides of the equality (5.5.15) and consider the scalar product of the obtained difference and the element xδα − x. Then, similarly to (3.3.5), one gets xδα ≤ 2x∗ +

µδ + δ . α

(5.5.16)


285

We have proved in Lemma 3.1.4 that the function ρ(α) = αxδα with fixed δ is continuous as α ≥ α0 > 0. If θH ∈ D(A) then lim ρ(α) = y 0 − f δ .

α→∞

Otherwise, lim ρ(α) = +∞.

α→∞

By (5.5.16), if α < C, where C=

(k − 1)δ , x∗ ∈ N, 2x∗

then the estimate ρ(α) < µδ + kδ is satisfied. Using (5.5.12) and continuity of ρ(α) we establish (5.5.14) with α ¯ > C. Calculate the limit of µδ as δ → 0. The obvious inequalities µδ = f δ − g δ ≤ f − g + f − f δ + g − g δ and µ = f − g ≤ f δ − g δ + f − f δ + g − g δ together with (5.5.10) imply µ − 2δ ≤ µδ ≤ µ + 2δ. Therefore, lim µδ = µ.

(5.5.17)

δ→0

We investigate the behavior of {xδα¯ } as δ → 0. By virtue of (5.5.17) and (5.5.14), if yαδ¯ ∈ Axδα¯ and ¯ δα¯ = f δ , (5.5.18) yαδ¯ + αx then yαδ¯ − f δ → g − f = µ. Hence, the sequence {yαδ¯ } is bounded and there exists u ∈ R(A) such that yαδ¯ u. Then (5.5.14) yields the following relations: u − f ≤ lim inf yαδ¯ − f δ ≤ lim sup yαδ¯ − f δ δ→0

δ→0

= lim (µ + kδ) = µ = g − f . δ

δ→0

Since g satisfying (5.5.2) is unique, we conclude that u = g. Now, due to the weak convergence of {yαδ¯ } to g, strong convergence of yαδ¯ to g follows as δ → 0. We show by a contradiction that {xδα¯ } is bounded as δ → 0. Assume that xδα¯ → ∞ as δ → 0. Then for a sufficiently small 0 < δ ≤ δ ∗ , making use of (5.5.11), (5.5.14) and (5.5.18), we have ¯ δα¯ − f δ , xδα¯ ) = (yαδ¯ − g, xδα¯ ) + (g − f δ , xδα¯ ) + αx ¯ δα¯ 2 0 = (yαδ¯ + αx ≥ xδα¯ (µδ + kδ − g − f δ ).

(5.5.19)

5

286


If θH ∈ D(A) and xδα¯ = θH then (5.5.18) gives the equality yαδ¯ = f δ , where yαδ¯ ∈ A(θH ). Then we obtain from (5.5.12) and (5.5.13) that 0 < µδ + kδ < y 0 − f δ ≤ yαδ¯ − f δ = 0. But it is impossible. Hence, xδα¯ = 0. In its turn, (5.5.7) and (5.5.19) imply 0 ≥ xδα¯ (k − 1)δ > 0, which contradicts the assumption that xδα¯ → ∞. Therefore {xδα¯ } is bounded and there ¯ ∈ H. Recall that the strong convergence exists a subsequence {xδβ¯ } ⊂ {xδα¯ } such that xδβ¯ x of yαδ¯ to g ∈ H has been already proved. Hence, x ¯ ∈ N. Replacing in (5.5.18) α ¯ by β¯ we find that ¯ δ¯2 = 0. (yβδ¯ − f δ , xδβ¯) + βx β Then by (5.5.14), (yβδ¯ − f δ , xδβ¯) + (µδ + kδ)xδβ¯ = 0. Consequently, xδβ¯ =

(f δ − yβδ¯, xδβ¯) µδ + kδ

.

This shows that xδβ¯ has a limit as δ → 0. Moreover, lim xδβ¯ ≤ ¯ x.

δ→0

Due to the weak convergence of xδβ¯ to x ¯, we get x. ¯ x ≤ lim xδβ¯ ≤ ¯ δ→0

xδβ¯

→ ¯ x, which implies in a Hilbert space the strong convergence of {xδβ¯} to x ¯ ∈ N as δ → 0. The proof is accomplished.

This means that

Remark 5.5.4 According to Theorem 1.7.9, inequality (5.5.11) is one of the sufficient conditions guaranteing the inclusion g ∈ R(A). Theorem 5.5.5 Let H be a Hilbert space, A : H → 2H be a maximal monotone operator, R(A) be a closed set in H, 0 < δ < 1. If θH ∈ D(A), then it is additionally assumed that y0 − f δ > µδ + kδ p ,

k > 1,

p ∈ (0, 1),

y0

is defined by (5.5.13). Then there exists at least one value α ¯ of the regularization where parameter α such that α ¯ xδα¯ = kδ p . δ ¯∗ , → 0 and α Moreover, xδα¯ → x ¯ → 0 as δ → 0, where xδα¯ is a solution of the equation α ¯ Ax + αx = g δ with α = α ¯.

(5.5.20)


287

Proof. By (5.5.6), the following inequalities obviously hold: y 0 − g δ ≥ y 0 − f δ − g δ − f δ > kδ p . Observe that the problem (5.5.2) is well-posed. Applying Theorem 3.1.12 and Remark 3.1.14 to equations (5.5.20) and (5.5.3), and taking into account the estimate (5.5.10), we obtain the required conclusion. 2. We study the more general situation in a Banach space X. Namely, suppose now ∗ that A : X → 2X , a sequence of maximal monotone operators {Ah } is given in place of A, h D(A) = D(A ), h ∈ (0, h∗ ], and HX ∗ (Ax, Ah x) ≤ η(x)h ∀x ∈ D(A),

(5.5.21)

where η(t) is a continuous non-negative increasing function for t ≥ 0. We do not assume that the sets R(A) and R(Ah ) are closed, however, presume that there exists g ∈ R(A) satisfying (5.5.2). The incompatibility measure of approximate equations Ah x = f δ is defined by the quantity (5.5.22) µγ = g γ − f δ ∗ = min{y − f δ ∗ | y ∈ R(Ah )}, γ = (δ, h).

Here

f δ − f ∗ ≤ δ, δ ∈ (0, δ ∗ ],

(5.5.23)

R(Ah ) means the closure of R(Ah ) and gγ ∈ R(Ah ). By Corollary 1.7.18, R(Ah ) is convex. First of all, note that under our conditions the operator regularization method does not always converge. Let us give an example. 1

Example 5.5.6 Let A : R1 → 2R be given by the formula   

Ax =

1, [1,3],   3,

x < 0, x = 0, x > 0;

f = f δ = 0, Ah x = Ax + hx, i.e., in (5.5.21) η(t) ≡ t, t ≥ 0. Then g = 1, x ¯∗ = 0, g γ = 0 for all h > 0, γ = (0, h), xγα = −(h + α)−1 . Here xδα is a solution of the equality Ah x + αx = g γ . Hence xγα → −∞ if α and h tend to zero in any way. This situation is explained by the fact that it is not always possible to get estimates ˜ of the type g − g γ ≤ δ(γ), which guarantee a realization of the sufficient convergence condition of the operator regularization method: ˜ h + δ(γ) → 0 as α → 0 α (see Theorem 2.2.1). Our investigations show, in order to obtain such estimates, it is necessary along with (5.5.21) to use additional proximity properties of R(A) and R(Ah ). Since R(A) and R(Ah ) are convex, we assume that

τ (r, R(A), R(Ah )) ≤ ha(r),

(5.5.24)

288

5


where a(r) is a non-negative non-decreasing and continuous function for r ≥ 0, a(r) → ∞ as r → ∞ (see the definition of τ (r, R(A), R(Ah )) in Section 4.2). Observe that the requirement (5.5.21) is not a consequence of (5.5.24) and vice versa. For the operator A, given in Example 5.5.6, the condition (5.5.21) is satisfied with g(t) = t for all t ≥ 0. The one gets s(r, R(A), R(Ah )) = 0

and s(r, R(Ah ), R(A)) = r + 1, r > 1. Thus, (5.5.24) does not hold. 1

Example 5.5.7 Consider the operator A : R1 → 2R given in Example 5.5.6, and let h

A x=

 x   √x2 +h2 + 2,

3 − h,   1 + h,

where x ˜=

√ (1 − h) h √ , 2−h

|x| ≤ x ˜, x > x ˜, x < −˜ x,

0 < h < 1.

It is obvious that τ (r, R(A), R(Ah )) = h for all r > 1, but HR1 (A(0), Ah (0)) = 1. Hence, the property (5.5.24) is fulfilled, however, (5.5.21) does not hold. Thus, (5.5.21) and (5.5.24) are independent, in general. The quantity g − g γ ∗ can be evaluated by making use of geometrical characteristics of the space X ∗ , in other words, by the properties of duality mapping (J s )∗ : X ∗ → X with the gauge function µ(t) = ts−1 , s ≥ 2. Assume that (J s )∗ z1 − (J s )∗ z2 ≤ d(R)z1 − z2 σ∗ ∀z1 , z2 ∈ X ∗ ,

(5.5.25)

where 0 < σ ≤ 1, R = max{z1 ∗ , z2 ∗ }, d(R) is a non-negative non-decreasing function for R ≥ 0. Besides, z1 − z2 , (J s )∗ z1 − (J s )∗ z2 ≥ mz1 − z2 s∗ ∀z1 , z2 ∈ X ∗ ,

m > 0.

(5.5.26)

The problems of finding elements g and gγ are equivalent to the following variational inequalities: (5.5.27) g − y, (J s )∗ (g − f ) ≤ 0 ∀y ∈ R(A), g ∈ R(A),

and g γ − y, (J s )∗ (g γ − f δ ) ≤ 0 ∀y ∈ R(Ah ),

g γ ∈ R(Ah ).

By virtue of condition (5.5.24) for g ∈ R(A), there exists y ∈ h

g − y h ∗ ≤ a(g∗ )h.

R(Ah )

(5.5.28)

such that (5.5.29)


289

For g γ ∈ R(Ah ), there is uγ ∈ R(A) satisfying the inequality

g γ − uγ ∗ ≤ a(gγ ∗ )h.

(5.5.30)

Presuming y = uγ in (5.5.27) and y = y h in (5.5.28), and then summing the corresponding inequalities side by side, we obtain the following result: g − g γ , (J s )∗ (g − f ) + g γ − uγ , (J s )∗ (g − f ) + g γ − g, (J s )∗ (gγ − f ) + g γ − g, (J s )∗ (g γ − f δ ) − (J s )∗ (g γ − f ) + g − y h , (J s )∗ (g γ − f δ ) ≤ 0.

(5.5.31)

It follows from (5.5.25) and (5.5.26) that, respectively, |g γ − g, (J s )∗ (g γ − f δ ) − (J s )∗ (g γ − f ) | ≤ d(Rγ )g γ − g∗ f − f δ σ∗ ≤ d(Rγ )δ σ g γ − g∗ , where Rγ = max{µγ , g γ − f ∗ }, and g − gγ , (J s )∗ (g − f ) − (J s )∗ (gγ − f ) ≥ mg γ − gs . Further, (5.5.2), (5.5.22), (5.5.29), (5.5.30) and the definition of (J s )∗ yield the estimates |g γ − uγ , (J s )∗ (g − f ) | ≤ µs−1 a(gγ ∗ )h and

|g − y h , (J s )∗ (gγ − f δ ) | ≤ (µγ )s−1 a(g∗ )h.

Thus, (5.5.31) leads to the following inequality: mg γ − gs∗ ≤ µs−1 a(g γ ∗ )h + (µγ )s−1 a(g∗ )h + d(Rγ )δ σ g γ − g∗ .

(5.5.32)

Now from the relations µγ = g γ − f δ ∗ ≤ y0 − f ∗ + δ, where y0 is a fixed element of R(Ah ), we conclude that there exist constants c1 > 0 and c2 > 0 such that µγ ≤ c1 and g γ ∗ ≤ c2 . Therefore, taking Rγ = max{c1 , c2 + f ∗ } = c3 , (5.5.32) can be rewritten as mg γ − gs∗ ≤

µs−1 a(c2 ) + cs−1 1 a(g∗ ) h

+ d(c3 )δ σ g γ − g∗ .

290

5


Thus, g γ − gs∗ ≤ β1 h + β2 δ σ g γ − g∗ ,

(5.5.33)

where

µs−1 a(c2 ) + cs−1 d(c3 ) 1 a(g∗ ) . and β2 ≥ m m Consider the function ϕ(t) = ts − a ¯t − ¯b, where a ¯ > 0, ¯b > 0, s ≥ 2, t ≥ 0. It is easy to see that ϕ(0) = −¯b < 0 and ϕ (t0 ) = 0 for β1 ≥

t0 =

a ¯ ν

s

≤a ¯ν , and ν =

1 . s−1

Thus, ϕ(t) has a minimum at the point t = t0 . Choose t = t¯ such that t¯s−1 = a ¯ + ¯br , where r = (s − 1)s−1 . Then ¯) − ¯b = ¯br (¯ a + ¯br )ν − ¯b = ¯b(1 + a ¯¯b−r )ν − ¯b > 0. ϕ(t¯) = t¯(t¯s−1 − a Moreover, t¯ > t0 . Hence, the function ϕ(t) ≤ 0 on some interval [0, t2 ] with t2 ≤ t¯. Therefore, by (5.5.33), we conclude that gγ − g∗ ≤ (β1r hr + β2 δ σ )ν .

(5.5.34)

Thus, g γ → g as γ → 0. This allows us to apply the operator regularization method for solving equation Ax = g in the form Ah x + αJ p x = g γ , α > 0,

(5.5.35)

where J p : X → X ∗ is a duality mapping with the gauge function µ(t) = tp−1 , p ≥ 2. Note that parameter p is chosen independently of s. Finally, using Theorems 2.2.1 and (2.2.10) we obtain the following result: ∗

Theorem 5.5.8 Let A : X → 2X be a maximal monotone operator, {Ah } be a sequence of maximal monotone operators acting from X to X ∗ , D(A) = D(Ah ) for all h ∈ (0, h∗ ], an element g satisfy (5.5.2) and g ∈ R(A). Assume that the conditions (5.5.21), (5.5.23), (5.5.24) are fulfilled and duality mapping (J µ )∗ in X ∗ with the gauge function µ(t) = ts−1 , s ≥ 2, has the properties (5.5.25) and (5.5.26). Let

h1/s δσ → 0 as α → 0. → 0 and s−1 α α Then the sequence {xγα }, where xγα is a solution of (5.5.35), strongly converges to the minimal norm pseudo-solution of the equation (5.5.1). p with p > 1, m > 0, the estimates (5.5.25) and Remark 5.5.9 In the spaces lp , Lp and Wm s ∗ (5.5.26) for the duality mapping (J ) are actually fulfilled with some s ≥ 2 (see (1.6.52), (1.6.53) and (1.6.56) - (1.6.59)).


291

Remark 5.5.10 If X is a Hilbert space then σ = 1, s = 2, and, by (5.5.34), there exist constants R > 0 and C(R) > 0 such that √ (5.5.36) max{g − g γ | f ≤ R} ≤ C(R) h for δ = 0. The estimate (5.5.36) can not be improved relative to the order of h (see the example in [73]). Remark 5.5.11 It is possible to construct the pseudo-residual principle for the regularization method (5.5.35) if β1 and β2 are known. For this aim, it is proposed to use the quantity = (β1r hr + β2 δ σ )ν in place of δ (see Section 3.1). Remark 5.5.12 If A : X → 2X is an m-accretive operator and the conditions of Theorem 1.15.35 hold, then the set R(A) is convex. Therefore, the pseudo-solution concept can be extended to equations with such operators and Theorem 5.5.8 is proved again.

Instead of (5.5.1), consider a variational inequality Ax − f, z − x ≥ 0

∀z ∈ Ω,

x ∈ Ω,

(5.5.37)

where Ω ⊂ D(A) is a convex closed set and either int Ω = ∅ or int D(A) ∩ Ω = ∅. By Lemma 1.11.7, inequality (5.5.37) is equivalent to the equation Ax + ∂IΩ x = f,

(5.5.38)

where ∂IΩ is a subdifferential of the indicator function IΩ of Ω. If we define a pseudo-solution of (5.5.37) as a pseudo-solution of (5.5.38), then all the results of this section can be stated for variational inequalities. 3. The procedure of finding gγ in the minimization problem (5.5.22) meets essential difficulties connected with the specific character of the minimization set R(Ah ). By reason of that, to construct the element g γ with a fixed γ, it is possible to use the following lemma: Lemma 5.5.13 Let a functional ψs∗ , where s ≥ 2, be uniformly convex on the space X ∗ with a modulus of convexity cts , c > 0, s ≥ 2, and let an element xλ be a solution of the equation Ah x + J p (λx) = f δ , λ > 0, p ≥ 2. (5.5.39) Then f δ − J p (λxλ ) → g γ as λ → 0, where g γ ∈ R(Ah ) and defined by (5.5.22). Proof. Take > 0 and g ∈ R(Ah ) such that g − g γ ∗ ≤ . Then there exists u ∈ D(Ah ), for which g ∈ Ah (u ). Since xλ is a solution of (5.5.39), there is an element yλ ∈ Ah xλ satisfying the equality yλ + J p (vλ ) = f δ , where vλ = λxλ . Then we have yλ − g , xλ − u + J p (vλ ), xλ − u = f δ − g , xλ − u .

(5.5.40)

292

5


Due to the monotonicity of Ah , the first term in the left-hand side of (5.5.40) is non-negative. Therefore, (5.5.41) J p (vλ ), vλ − λu ≤ f δ − g , vλ − λu . By the Cauchy−Schwartz inequality and definition of J p , (5.5.41) implies vλ p ≤ λu vλ p−1 + ( + wγ ∗ )(λu + vλ ), where wγ = g γ − f δ . In accordance with (2.2.10), there exists a constant K1 > 1 such that the estimate vλ ≤ ( + wγ ∗ )1/(p−1) + K1 λu (5.5.42) holds if additionally K1 (p − 1) ≥ 2. Taking into account the inequality ar − br ≤ rar−1 (a − b), a > 0, b > 0,

r ≥ 1,

(5.5.43)

from (5.5.42) conclude that J p (vλ )∗ − (wγ ∗ + ) = vλ p−1 − (wγ ∗ + )

≤ (p − 1)vλ p−2 vλ − (wγ ∗ + )1/(p−1)

≤ (p − 1)K1 vλ p−2 λu .

(5.5.44)

It follows from (5.5.42) that vλ is bounded for a sufficiently small λ. Therefore, in view of (5.5.44), we can write J p (vλ )∗ − wγ ∗ ≤ K2 λ + , K2 > 0. Applying the inequality (5.5.43) once more, we deduce that J p (vλ )s − wγ s ≤ K3 (λ + ), K3 > 0. Obviously, the inclusion −J p (vλ ) ∈ R(Ah ) − f δ holds and wγ ∗ = min{y∗ | y ∈ R(Ah ) − f δ }. It arises from Theorem 1.1.24 for a uniformly convex functional ϕ(u) with modulus of convexity χ(t) and convex closed set Ω ⊆ dom (ϕ) that χ(u − u∗ ) ≤ ϕ(u) − ϕ(u∗ ) ∀u ∈ Ω, ϕ(u∗ ) = min{ϕ(u) | u ∈ Ω}.

(5.5.45)

Then we derive the inequality J p (vλ ) + wγ s∗ ≤ c−1 (J p (vλ )s − wγ s ) ≤ c−1 K3 (λ + ), where wγ is a minimal point of ψs∗ on the set S = {y − f δ | y ∈ R(Ah )}.

Since is arbitrary, we come to the conclusion of the lemma.

(5.5.46)


293

Remark 5.5.14 If g γ ∈ R(Ah ) then in the proof of Lemma 5.5.13, one can put g = g γ , g γ ∈ Ah (uγ ), u = uγ and = 0. In this case, (5.5.46) gives J p (vλ ) + wγ s∗ ≤ c−1 K3 λ.

(5.5.47)

In a Hilbert space, s = p = 2, and then there exists K4 > 0 such that √ vλ + wγ ≤ K4 λ. In conclusion, we present the following example. Suppose that A : R2 → R2 is defined by matrix

A=

1 2 2 4

.

2 It is easy to see that A is positive and det A = 0. Analyze A on a set R+ = {(x, y) ∈ R2 | 2 , as in Theorem 1.7.19, then y ≥ 0}. Observe that if A is defined on the boundary of R+ 2 we obtain nonlinear maximal monotone operator A¯ : R2 → 2R . By analogy, consider the perturbed matrix

h

A =

(2 + h)2 (4 + h)−1 2+h

2+h 4+h

2 2 with h > 0 and construct maximal monotone operator A¯h : R2 → 2R with D(A¯h ) = R+ . Then Theorem 1.7.19 and the Kronecker−Capelli Theorem [122] allow us to get the following representations:

¯ = {(x, y) ∈ R2 | y ≤ 2x}, R(A)

R(A¯h ) = {(x, y) ∈ R2 | y ≤ (2 + h)−1 (4 + h)x}.

Find the estimates for A¯ and A¯h like (5.5.24). It is obvious that, in our circumstances, the ¯ R(A¯h )) and s(r, R(A¯h ), R(A)), ¯ defined by (4.2.28) coincide, and the quantities s(r, R(A), simple calculations give rh ¯ R(A¯h )) = . τ (r, R(A), 10(h2 + 6h + 10)

r Hence, we may assume in (5.5.24) a(r) = . 10 If det Ah = 0 then R(A¯h ) = R2 and like (5.5.24) is not true. Observe, for instance, that if f = (1; 5)T then the inequality g γ − g ≤ ch holds with some c > 0 and solutions of the regularized equation Ah x + αx = g γ strongly converge to x∗ =

11 22

; 25 25 as α → 0 and h → 0. For general monotone nonlinear equations, necessary and sufficient conditions for existence of solutions are not obtained. Therefore, in this case, it is not possible to describe R(A) and R(Ah ).

294

5

APPLICATIONS OF THE REGULARIZATIOM METHODS

5.6

Minimization Problems

In this section, we study the minimization problem of a proper lower semicontinuous nondifferentiable, in general, functional ϕ(x) : X → R1 : To find a point x∗ ∈ X such that ϕ(x∗ ) = min{ϕ(x) | x ∈ X}.

(5.6.1)

Note that, by these hypotheses, ϕ(x) has no conditions of strong or uniform convexity. Therefore, the problem (5.6.1) is ill-posed and, according to the ill-posedness concept, we assume that it has a solution. Denote by N a solution set of (5.6.1) and by ∂ϕ = A : X → ∗ 2X a subdifferential of ϕ(x) with D(A) = X. Suppose that X is an E-space with strictly convex dual space X ∗ . Then, in view of Lemma 1.2.5 and Theorem 1.7.15, the problem (5.6.1) is equivalent to the equation Ax = 0 (5.6.2) with the maximal monotone operator A. In the sequel, we assume that, instead of ϕ, there is a sequence {ϕh } of bounded from below proper convex lower semicontinuous functionals ϕh : X → R1 , depending on the parameter h > 0 and satisfying the inequality |ϕ(x) − ϕh (x)| ≤ g(x)h ∀x ∈ X,

(5.6.3)

where g(t) is a non-negative continuous and non-decreasing function for all t ≥ 0. Thus, in reality, the problems (5.6.1) and (5.6.2) are replaced, respectively, by ϕh (x) → min s.t. x ∈ X

(5.6.4)

Ah x = 0,

(5.6.5)

and X∗

where ∂ϕh = Ah : X → 2 is the maximal monotone operator again. According to Theorem 2.2.1, in order to obtain strong approximations to x∗ ∈ N by means of operator Ah in (5.6.5) it is possible to use the regularized equation Ah x + αsJ s x = 0

(5.6.6)

where J s : X → X ∗ is a duality mapping with the gauge function µ(t) = ts−1 , s ≥ 2 : 1 J s x = grad xs , s and apply the theory and methods of Chapters 2 and 3. At the same time, it is clear that (5.6.6) is equivalent to minimization problem Φαh (x) → min s.t. x ∈ X

(5.6.7)

Φαh (x) = ϕh (x) + αxs , α > 0, s ≥ 2,

(5.6.8)

for the regularized functional

5.6 Minimization Problems

295

which is strictly convex and also non-differentiable, in general. To solve this problem one can apply the well developed methods for approximate minimization of nonsmooth functionals. This approach is used, for instance, in [85]. We present further the functional criterion of choosing a regularization parameter for the problem (5.6.7), (5.6.8). Let xhα be its unique minimum point of (5.6.8), mh (α) = Φαh (xhα ), mh = inf {ϕh (x) | x ∈ X} and m = inf {ϕ(x) | x ∈ X}. Next we state a known proposition. Lemma 5.6.1 (cf. Lemma 3.6.1). Any function mh (α) is continuous and non-decreasing as α ≥ 0. Moreover, (5.6.9) lim mh (α) = mh α→0+

and lim mh (α) = ϕh (θX ).

α→∞

In addition, if mh < ϕ(θX , then mh (α) increases on the interval (0, αh∗ ), where αh∗ = sup {α | mh (α) < ϕ(θX )}. We introduce the following denotations: Γ(α) = ϕh (xhα ) = mh (α) − αxhα s ; ξ(α) = Γ(α) − hg(xhα ) − τ h ; τ h = max{m, mh }. We claim that the function Γ(α) is continuous. Indeed, by Theorem 1.2.8 and Lemma 1.2.5, the minimization problem of the functional (5.6.8) on X leads to the equation ∂ϕh (x) + αsJ s x = 0, where ∂ϕh (x) is a subdifferential of ϕh (x) at a point x. Continuity of the function σ(α) = xhα is established as in Lemma 3.1.1. Then, by making use of Lemma 5.6.1, we obtain the claim. It additionally follows from Lemma 3.1.1 that σ(α) is non-increasing for all α ≥ α0 > 0 and lim σ(α) = 0. α→∞

Taking into account these properties, it is not difficult to verify the following assertions: a) ξ(α) is continuous as α ≥ α0 > 0;

296

5

APPLICATIONS OF THE REGULARIZATIOM METHODS

b) limα→∞ ξ(α) = ϕh (θX ) − hg(0) − τ h ; c) the equality ξ(α) = mh (α) − τ h − αxhα s − hg(xhα ) implies the limit relation lim ξ(α) < 0.

α→0+

We state the obvious proposition. Lemma 5.6.2 If ϕh (θX ) − hg(0) − τ h > 0,

(5.6.10)

then there exists at least one solution α ¯=α ¯ (h) of the equation ξ(¯ α) = 0.

(5.6.11)

Remark 5.6.3 If θX ∈ N then the inequality ϕh (θX ) ≤ τ h + hg(0) is satisfied. Consequently, the condition (5.6.10) of Lemma 5.6.2 implies the fact that θX ∈ N. Theorem 5.6.4 Let {xhα¯ } be a minimal point sequence of the smoothing functional (5.6.8) with α = α ¯ defined by (5.6.11). If the condition (5.6.10) is filfilled, then ¯∗ , lim xhα¯ = x

h→0

where x ¯∗ ∈ N is a minimal norm solution of the problem (5.6.1). Proof. Put in (5.6.8) α = α ¯ . Construct the sequence {xhα¯ } and study its behaviour when h → 0. First of all, the inclusion x∗ ∈ N yields the inequality ¯ hα¯ s ≤ ϕh (x∗ ) + αx ¯ ∗ s ∀x∗ ∈ N. ϕh (xhα¯ ) + αx By (5.6.11), one gets hg(xhα¯ ) + αx ¯ hα¯ s = ϕh (xhα¯ ) + αx ¯ hα¯ s − τ h ≤ hg(x∗ ) + αx ¯ ∗ s + m − τ h . Since m − τ h ≤ 0, we have hg(xhα¯ ) + αx ¯ hα¯ s ≤ hg(x∗ ) + αx ¯ ∗ s ∀x∗ ∈ N. Recalling that the function g(t) is non-decreasing we obtain the estimate xhα¯ ≤ x∗ ∀x∗ ∈ N.

(5.6.12)

5.6 Minimization Problems

297

¯ ∈ X as h → 0. Further, taking into account (5.6.12) and weak lower Thus, xhα¯ x semicontinuity of the norm in X, we can write ¯ x ≤ lim inf xhα¯ ≤ lim sup xhα¯ ≤ x∗ . h→0

(5.6.13)

h→0

Show that x ¯ ∈ N. Indeed, using (5.6.11) and (5.6.12), it is not difficult to verify that the following inequalities are satisfied: 0 ≤ ϕ(xhα¯ ) − ϕ(x∗ ) = ϕ(xhα¯ ) − ϕh (xhα¯ ) + ϕh (xhα¯ ) − ϕ(x∗ ) ≤ 2hg(xhα¯ ) + τ h − m ≤ 2hg(x∗ ) + τ h − m.

(5.6.14)

Besides, mh = inf {ϕh (x) | x ∈ X} ≤ ϕh (x∗ ) ≤ hg(x∗ ) + m,

(5.6.15)

that is, lim sup mh ≤ m. h→0

Therefore, lim sup τ h ≤ m. h→0

By definition of τ h , lim inf τ h ≥ m. h→0

Hence, lim τ h = m.

(5.6.16)

lim ϕ(xhα¯ ) = ϕ(x∗ ).

(5.6.17)

h→0

Then from (5.6.14) we deduce that h→0

By (5.6.17) and by weak lower semicontinuity of the functional ϕ (see Theorem 1.1.13), we have ϕ(¯ x) ≤ lim ϕ(xhα¯ ) = ϕ(x∗ ) ∀x∗ ∈ N. h→0

Thus, x ¯ ∈ N. Finally, by (5.6.13), we conclude that x ¯=x ¯∗ and xhα¯ → ¯ x∗ . Thus, the theorem is proved. Theorem 5.6.5 Under the conditions of Lemma 5.6.2, if a functional ϕ is Gˆ ateauxdifferentiable at the point x ¯ (h) → 0 as h → 0. ¯∗ , then α Proof. Let α ¯=α ¯ (h) → α ˆ as h → 0. First of all, observe that α ˆ is finite. Indeed, α ¯ =

mh (¯ α) − ϕh (xhα¯ ) xhα¯ s

298

5

APPLICATIONS OF THE REGULARIZATIOM METHODS = ≤

≤

α) − hg(xhα¯ ) − τ h mh (¯ xhα¯ s h ϕ (θX ) − hg(xhα¯ ) − τ h xhα¯ s h ϕ (θX ) − hg(0) − τ h , c

where a constant c > 0 satisfies the inequality xhα¯ s ≥ c with sufficiently small h. The existence of c follows from Theorem 5.6.4 and from the condition θX ∈ N because of Remark 5.6.3. Prove now the theorem by contradiction. Let α ˆ > 0. Then the properties of the sequence {xhα¯ } imply x∗ ) = ϕ(¯ x∗ ) + α¯ ˆ x∗ s . (5.6.18) lim Φαh¯ (xhα¯ ) = Φα0ˆ (¯ h→0

Since

xhα¯

is a minimum point of the functional Φαh¯ , we have s ¯ ∀x ∈ X. Φαh¯ (xhα¯ ) ≤ ϕh (x) + αx

Passing to the limit as h → 0 and using (5.6.18), we deduce that x ¯∗ is a minimum point of ˆ xs at the same time. In view of Lemmas 1.2.4 the functionals ϕ(x) and ϕ+ (x) = ϕ(x) + α and 1.2.5, the latter means that x∗ ) = θX ∗ and ϕ+ (¯ x∗ ) = ϕ (¯ x∗ ) + α ˆ grad ¯ x∗ s = θ X ∗ , ϕ (¯ ateaux derivative of ϕ(x). Thus, where ϕ (x) is the Gˆ ˆ = 0. α ˆ grad ¯ x∗ s = θX ∗ , α ¯∗ = θX . This contradicts the assumption (5.6.10) (see Therefore, grad ¯ x∗ s = θX ∗ and x Remark 5.6.3). The next example shows that the conclusion of Theorem 5.6.5 may be wrong if the smoothness demand of the functional ϕ at the point x ¯∗ is disturbed. Example 5.6.6 Let ϕ : R1 → R1 and ϕh : R1 → R1 be expressed by the formulas ϕ(x) = |x + 1| + 1 and ϕh (x) = |x + 1| + 1 + h, h > 0. It is easy to see that N = {−1} and g(t) = 1 in (5.6.3). Take s = 2 in (5.6.8). The condition (5.6.10) of Lemma 5.6.2 are satisfied if h < 1. One can verify that xhα¯ = h − 1, 1 1 1 ˆ= ¯∗ and α ¯ → as h → 0. Note that if α = α . Hence, xhα¯ → −1 = x where α ¯= 2 2 2 − 2h then functionals ϕ(x) and ϕ(x) + α ˆ x2 have the same minimum point x ¯∗ = −1. Moreover, 1 ˆ 2 for all α ∈ (0, ). x ¯∗ = −1 is a unique minimum point of the functional ϕ(x) + αx 2

5.7 Optimal Control Problems

299

Remark 5.6.7 By (5.6.17), we have lim ϕ(xhα¯ ) = m,

h→0

that is, the sequence {xhα¯ } converges to x ¯∗ also with respect to functional ϕ. Remark 5.6.8 The relations (5.6.11) and (5.6.15) imply

ϕh (xhα¯ ) − m ≤ h g(xhα¯ ) + g(x∗ ) . In other words, there exists a constant C > 0 such that ϕh (xhα¯ ) − m ≤ C. h Remark 5.6.9 Instead of mh , the quantity mh + (h) can be used, where (h) → 0 as h → 0. In the filtration problem (see Sections 1.3 and 2.2), operators A and Ah are potential. Moreover, their potentials are defined by the following expressions:

ϕ(u) = and

Ω 0

ϕh (u) =

|u|

Ω 0

|u|

g(x, ξ 2 )ξdξdx

g h (x, ξ 2 )ξdξdx.

If ϕh (θX ) − h − τ h > 0 then (5.6.11) is written by the equation h ϕh (uhα¯ ) − h(cuhα¯ p−1 1,p + 1) − τ = 0.

5.7

Optimal Control Problems

In this section we are interested in the following optimal control problem: min {Φ(u) = ϕ(u) + ψ(z(u)) | u ∈ Ω},

(5.7.1)

where the system state z = z(u) ∈ D is connected with a control u ∈ Ω by the variational inequality (A(u, z) − f, v − z) ≥ 0 ∀v ∈ D. (5.7.2) Here Ω ⊂ H is a convex closed bounded set, D ⊂ H1 is a convex closed set, f ∈ H1 is fixed, H and H1 are Hilbert spaces, ϕ : Ω → R1 is a bounded from below and weakly lower semicontinuous functional with the H-property, which intends that the relations un u in H and ϕ(un ) → ϕ(u) imply strong convergence un → u. We assume in the sequel that ψ : H1 → R1 is a strongly convex and Fréchet differentiable functional.

300

5


Under the solution of the problem (5.7.1), (5.7.2), we understand an element u0 ∈ Ω such that for a certain solution z = z(u0 ) of the inequality (5.7.2) there hold the equalities Φ(u0 ) = ϕ(u0 ) + ψ(z(u0 )) = min {Φ(u)|u ∈ Ω}, where a minimum is taken for all u ∈ Ω and for all z(u) satisfying (5.7.2). Suppose that the operator A : Ω × H1 → H1 has the following properties: 1) A(u, z) is strongly continuous with respect to u and demicontinuous with respect to z; 2) A(u, z) is monotone with respect to z for all u ∈ Ω. Let the problem (5.7.1), (5.7.2) possess a nonempty solution set N. As usual, we assume further that in place of f its δ-approximations f δ are known such that f − f δ H1 ≤ δ, where δ ∈ (0, δ ∗ ]. The problem is posed: For given {f δ }, to construct a sequence {uδ } which converges strongly to some element of the set N. In our assumptions, the variational inequality (5.7.2) and the minimization problem (5.7.1), (5.7.2) are ill-posed. Indeed, for given control u, either solvability of (5.7.2) or uniqueness of its solutions and stability with respect to data perturbations can not be guaranteed. Furthermore, in general, the functional Φ is not uniformly convex in the control space. Therefore, the minimization problem (5.7.1) can be also unstable. By virtue of the mentioned aspects, strong approximations to solutions of (5.7.1) and (5.7.2) can be constructed only by making use of some regularization procedure. Denote by Z(u) a solution set of the variational inequality (5.7.2) with any fixed u ∈ Ω. It follows from the properties of A that if Z(u) is nonempty then it is convex and closed. List several properties of the functional ψ and operator B = grad ψ. Since ψ is strongly convex, there exists a constant c > 0 such that ψ(y + t(x − y)) − ψ(y) ≤ t[ψ(x) − ψ(y)] − t(1 − t)cx − y2H1 ∀x, y ∈ H1 , t ∈ (0, 1). (5.7.3) Dividing (5.7.3) by t and then passing to the limit as t → 0 one gets (By, x − y) ≤ ψ(x) − ψ(y) − cx − y2H1 .

(5.7.4)

Interchanging x and y in (5.7.4) and adding the obtained inequality to (5.7.4), we establish the property of strong monotonicity of the operator B : (Bx − By, x − y) ≥ 2cx − y2H1 ∀x, y ∈ H1 .

(5.7.5)

We emphasize that we proved above the inequality (1.1.8) and (1.1.11) for ψ and grad ψ, respectively. If y = θH1 then it is not difficult to see from (5.7.5) that (Bx, x) ≥ 2cx2H1 − B(θH1 )xH1 . This means that the operator B : H1 → H1 is coercive. Then, due to Corollary 1.7.7, the equation Bx = θH1 with strongly monotone operator has a unique solution. In other

5.7 Optimal Control Problems

301

words, the functional ψ(z) has on H1 a unique minimizer (see Theorem 1.1.21). Hence, ψ is bounded on H1 from below. Next, it follows from (5.7.4) that for y = θH1 , ψ(x) ≥ ψ(θH1 ) + cx2H1 − B(θH1 )xH1 . Then it is clear that lim ψ(x) = ∞.

x→∞

(5.7.6)

Let zn z in H1 as n → ∞ and ψ(zn ) → ψ(z). Presuming x = zn and y = z in (5.7.4) we obtain (Bz, zn − z) ≤ ψ(zn ) − ψ(z) − czn − z2H1 , from which we deduce the strong convergence of {zn } to z as n → ∞. Therefore, the functional ψ has H-property. Moreover, according to Theorem 1.1.16, ψ is weakly lower semicontinuous. So, in order to solve the variational inequality (5.7.2), we have to apply some regularization method. To this end, we introduce the family of operators {Rδ (u, ·)}, Rδ (u, ·) : H1 → H1 , δ ∈ (0, δ ∗ ], such that for every u ∈ Ω provided that Z(u) = ∅, there holds the convergence Rδ (u, f δ ) → z ∈ Z(u) as δ → 0. (5.7.7) We study the problem: To find an element uδ ∈ Ω such that Φδ∗ = inf {Φδ (u) | u ∈ Ω} ≤ Φδ (uδ ) ≤ Φδ∗ + (δ),

(5.7.8)

where (δ) > 0, lim (δ) = 0 and δ→0

Φδ (u) = ϕ(u) + ψ(zδ (u)), zδ (u) = Rδ (u, f δ ).

(5.7.9)

Since the functionals ϕ and ψ are bounded from below, respectively, on Ω and on H1 , there exists an element uδ satisfying (5.7.8) and (5.7.9). We accept it as an approximate solution of the problem (5.7.1), (5.7.2). Denote Φ0 = min {Φ(u) | u ∈ Ω} and Z0 = {z(u0 ) | ϕ(u0 ) + ψ(z(u0 )) = Φ0 , u0 ∈ N }. Theorem 5.7.1 Let the assumptions of this section be held and the regularizing algorithm Rδ satisfy the following conditions: (i) the element z ∈ Z(u) in (5.7.7) is a minimizer of ψ(v) on Z(u), that is, ψ(z) = min{ψ(v) | v ∈ Z(u)};

(5.7.10)

(ii) for any sequence {uβ }, the limit relations uβ u, Rδ (uβ , f δ ) z as β → 0 and δ → 0 imply z ∈ Z(u). Then sets of strong limit points of the sequences {uδ } and {zδ (uδ )} are nonempty and belong, respectively, to N and Z0 .

302

5


Proof. First of all, note that uniqueness of the minimizer z ∈ Z(u) satisfying condition (5.7.10) arises from Theorem 1.1.23 because Z(u) is convex and closed, ψ is strongly convex and (5.7.6) holds. For the problem (5.7.1), (5.7.2), take the minimizing control u0 ∈ N and corresponding state z0 = z(u0 ) ∈ Z(u0 ) such that Φ(u0 ) = Φ0 . In view of (5.7.1), only one z0 is defined by the equality ψ(z0 ) = min {ψ(v) | v ∈ Z(u0 )}.

(5.7.11)

Then (5.7.8) implies the inequality ϕ(uδ ) + ψ(zδ (uδ )) ≤ ϕ(u0 ) + ψ(zδ (u0 )) + (δ).

(5.7.12)

By making use of the condition (i), we conclude that zδ (u0 ) → z0 as δ → 0. Taking into account continuity of ψ on H1 and equality Φ0 = ϕ(u0 ) + ψ(z0 ), one has

lim sup ϕ(uδ ) + ψ(zδ (uδ )) ≤ lim sup ϕ(u0 ) + ψ(zδ (u0 )) + (δ) = Φ0 . δ→0

(5.7.13)

δ→0

Since the functional ϕ is bounded from below and the property (5.7.6) holds, we obtain the boundedness of the sequence {zδ (uδ )} as δ → 0. In its turn, the sequence {uδ } is also bounded because Ω is so. Thus, the limit results uδ u ¯ ∈ Ω and zδ (uδ ) z¯

(5.7.14)

are established. Now the condition (ii) yields the inclusion z¯ ∈ Z(¯ u). Therefore,

u) + ψ(¯ z ) ≤ lim inf ϕ(uδ ) + ψ(zδ (uδ )) , Φ0 ≤ ϕ(¯ δ→0

(5.7.15)

and combination of (5.7.13) with (5.7.15) forms the following relations:

u) + ψ(¯ z ) ≤ lim inf ϕ(uδ ) + ψ(zδ (uδ )) Φ0 ≤ ϕ(¯

δ→0

≤ lim sup ϕ(uδ ) + ψ(zδ (uδ )) ≤ Φ0 . δ→0

Consequently,

lim ϕ(uδ ) + ψ(zδ (uδ )) = ϕ(¯ u) + ψ(¯ z ) = Φ0 .

δ→0

Besides, u ¯ ∈ N and z¯ ∈ Z0 . Then weak lower semicontinuity of the functionals ϕ and ψ gives u) and lim ψ(zδ (uδ )) = ψ(¯ z ). (5.7.16) lim ϕ(uδ ) = ϕ(¯ δ→0

δ→0

In conclusion, it just remains to recall that ϕ and ψ have the H-property. Then the strong convergence result uδ → u ¯, zδ (uδ ) → z¯ follows from (5.7.14) and from (5.7.16) as δ → 0. The theorem is proved.

5.8 Fixed Point Problems

303

We present the regularization algorithm satisfying the conditions of Theorem 5.7.1. Fix δ ∈ (0, δ ∗ ] and u ∈ Ω. Consider a mapping Rδ (u, ·) : H1 → H1 which assigns to each element f δ ∈ H1 a solution zδ (u) ∈ D of the regularized variational inequality (A(u, z) + α(δ)Bz − f δ , v − z) ≥ 0 where α(δ) > 0,

α(δ) → 0,

∀v ∈ D,

(5.7.17)

δ → 0 as δ → 0. α(δ)

The condition (i) for the chosen family {Rδ } is established on the basis of the property (5.7.5) to the potential operator B (see the deduction of (2.2.10), the results of which are easily transferred on variational inequalities). Let uβ u and zδ (uβ ) z as β, δ → 0. Due to Lemma 1.11.4, we are able to proceed from (5.7.17) to the equivalent variational inequality (A(uβ , v) + α(δ)Bv − f δ , v − zδ (uβ )) ≥ 0 ∀v ∈ D. Passing there to the limit as β → 0 and δ → 0 and making use of condition 1), one gets (A(u, v) − f, v − z) ≥ 0

∀v ∈ D.

The same Lemma 1.11.4 allows us to assert that z ∈ Z(u). Thus, the property (ii) holds. Remark 5.7.2 Theorem 5.7.1 remains still valid if the set Ω is unbounded and the functional ϕ has the property lim ϕ(u) = ∞, u ∈ Ω. u→∞

If ψ(z) = 2−1 z2 then B = I and (5.7.17) takes the form of the regularization method studied in Chapter 4.

5.8

Fixed Point Problems

We are going to study fixed point problems with nonexpansive mapping T : Ω → Ω, where Ω ⊆ X is a convex closed set and X is a uniformly smooth Banach space. The problem is to find a fixed point x∗ of T, in other words, to find a solution x∗ of the equation x = T x.

(5.8.1)

It is clear that (5.8.1) is equivalent to the equation Ax = 0

(5.8.2)

with the accretive operator A = I − T : X → X (see Lemma 1.15.10), that is, J(x − y), Ax − Ay ≥ 0 ∀x, y ∈ Ω.

(5.8.3)

If x∗ is a solution of (5.8.1) then Ax∗ = θX . In the sequel, we assume that the fixed point set F (T ) of T is not empty. Then it is closed and convex.

5

304


Introduce parameter ω such that 0 < ω < 1 and ω → 1. Obviously, if x∗ is a solution of (5.8.2) then it is solution of the equation ωAx = 0

(5.8.4)

for any fixed ω > 0. Using Corollary 2.7.4 with α = 1 − ω → 0, consider for (5.8.4) the operator regularization method ωAx + (1 − ω)(x − z0 ) = 0,

(5.8.5)

where some z0 ∈ Ω. It is easy to see that (5.8.5) is equivalent to the equation x = (1 − ω)z0 + ωT x.

(5.8.6)

Denote Tω x = (1 − ω)z0 + ωT x. Since Ω is convex and closed, we have that Tω : Ω → Ω, and (5.8.6) can be rewritten as x = Tω x.

(5.8.7)

Theorem 5.8.1 Let a Banach space X possess an approximation, Ω be a closed convex subset of X, T : Ω → Ω be a nonexpansive mapping, z0 ∈ Ω. Then for each 0 < ω < 1, operator Tω is a strong contraction of Ω into Ω with the estimate Tω x − Tω y ≤ ωx − y. Hence, Tω has a unique fixed point xω ∈ Ω. Proof. Since T is a nonexpansive mapping, we have Tω x − Tω y ≤ ωT x − T y ≤ ωx − y. Then the assertion results from the Banach principle for strong contractive maps. Due to Theorem 5.8.1, the equation (5.8.6) has a unique solution xω and the successive approximation method xn+1 = (1 − ω)z0 + ωT xn converges strongly to xω . Let ωk → 1 as k → ∞. Consider the equation ωk Ax + (1 − ωk )(x − z0 ) = 0

(5.8.8)

with fixed k and denote by xk its unique solution. Let J be weak-to-weak continuous. Then ¯∗ ∈ F (T ) as k → ∞. Moreover, Corollary 2.7.4 implies that xk → x J(¯ x∗ − x∗ ), x ¯∗ − z0 ≥ 0; ∀x∗ ∈ F (T ).


305

The goal of this section is to prove strong convergence of the generalized successive approximation method for the regularized equation (5.8.6) in the following form: yn+1 = (1 − ωn )z0 + ωn T yn , n = 0, 1, 2, ..., where lim ωn = 1 and

n→∞

∞

(1 − ωn ) = ∞.

(5.8.9)

(5.8.10)

n=0

Theorem 5.8.2 Let Ω be a closed convex subset of a uniformly smooth Banach space X, T : Ω → Ω be a nonexpansive mapping with a fixed point set F (T ) = ∅, z0 ∈ Ω, and {ωn } be an increasing in (0,1) sequence satisfying (5.8.10). If X has weak-to-weak duality mapping then the sequence {yn } generated by (5.8.9) converges strongly to the fixed point x ¯∗ = QΩ z0 of T, where QΩ : Ω → F (T ) is a unique sunny nonexpansive retraction. Proof. Let x∗ ∈ F (T ). It follows from (5.8.9) that yn+1 − x∗ ≤ (1 − ωn )z0 − x∗ + ωn T yn − T x∗ ≤ (1 − ωn )z0 − x∗ + ωn yn − x∗ . Denoting λn = yn − x∗ we have λn+1 ≤ ωn λn + (1 − ωn )z0 − x∗ . Let qn = 1 − ωn . Then

∞

qn = ∞, and the previous inequality is rewritten as

n=1

λn+1 ≤ λn − qn λn + qn z0 − x∗ . According to Lemma 7.1.1, the sequence {λn } is bounded, namely, yn − x∗ ≤ max{2z0 − x∗ , y0 − x∗ } = M1 . Consequently, {yn } is also bounded. It is not difficult to calculate the following difference: yn+1 − yn = (qn − qn−1 )(z0 − x∗ ) + (1 − qn )(T yn − T yn−1 ) + (qn−1 − qn )(T yn−1 − T x∗ ). We have

(qn − qn−1 )(z0 − x∗ ) + (qn−1 − qn )(T yn−1 − T x∗ ) ≤ |qn − qn−1 |(z0 − x∗ + yn−1 − x∗ ).

Therefore, there exists a constant M2 > 0 such that yn+1 − yn ≤ (1 − qn )yn − yn−1 + M2 |qn − qn−1 |. Denoting λn = yn − yn−1 one gets λn+1 ≤ λn − qn λn + M2 |qn − qn−1 |.

5

306


Lemma 7.1.2 implies now that limn→∞ yn − yn−1 = 0 because of lim

n→∞

Since

|qn − qn−1 | = 0. qn

yn+1 − T yn+1 = qn (z0 − x∗ ) + (T yn − T yn+1 ) − qn (T yn − T x∗ ),

we deduce

yn+1 − T yn+1 ≤ qn (z0 − x∗ + M1 ) + yn − yn+1 .

Therefore, lim (yn − T yn ) = θX .

(5.8.11)

n→∞

Let J : X → X ∗ be a weak-to-weak continuous normalized duality mapping and Φ(t) = 2−1 t2 . Then by Lemmas 1.5.7 and 1.2.7, we have Φ (x) = Jx and Φ(x + y) − Φ(x) =

= Jx, y +

1 0

1 0

J(x + ty), y dt

J(x + ty) − Jx, y dt.

(5.8.12)

Evaluate the last term in the previous equalities. By virtue of (1.6.5), for all x, y ∈ X such that x ≤ M and y ≤ M we have 0

1

J(x + ty) − Jx, y dt ≤

1 0

t−1 8ty2 + c1 ρX (ty) dt,

where c1 = 8max {L, M }. Since ρX (τ ) is convex, ρX (ty) ≤ tρX (y). Therefore,

1 0

J(x + ty) − Jx, y dt ≤ 8

Thus,

1 0

y2 tdt + c1

1 0

ρX (y)dt = 4y2 + c1 ρX (y).

Φ(x + y) − Φ(x) ≤ Jx, y + 4y2 + c1 ρX (y).

It is easy to verify the equality yn+1 − QΩ z0 = (1 − qn )(T yn − QΩ z0 ) + qn (z0 − QΩ z0 ). It now follows that

Φ(yn+1 − QΩ z0 ) ≤ Φ (1 − qn )(T yn − QΩ z0 )

+ qn (1 − qn ) J(T yn − QΩ z0 ), z0 − QΩ z0

+ 4qn2 z0 − QΩ z0 2 + c¯1 ρX (qn z0 − QΩ z0 ), where c¯1 = 8max {L, M3 , z0 − QΩ z0 } and M3 satisfies the inequality T yn − QΩ z0 = T yn − T x∗ + T x∗ − QΩ z0

(5.8.13)


307

≤ yn − x∗ + x∗ − QΩ z0 ≤ M1 + x∗ − QΩ z0 = M3 . Denote

ζn = J(T yn − QΩ z0 ), z0 − QΩ z0 .

(5.8.14)

We want further to show that lim ζn ≤ 0. If this is not the case then there would exist a n→∞ subsequence {ynk } of {yn } and > 0 such that lim

n→∞

J(T ynk − QΩ z0 ), z0 − QΩ z0 > .

(5.8.15)

The space X is reflexive, therefore, by the Mazur theorem, Ω is weakly closed. Then we can assume that {ynk } converges weakly to a point y¯ ∈ Ω. Since, in view of Lemma 1.5.13, X also satisfies the Opial condition, (5.8.11) implies that this weak accumulation point y¯ belongs to F (T ). Actually, if it is not true then one has lim inf ynk − y¯ < lim inf ynk − T y¯ k→∞

k→∞

≤ lim inf (ynk − T ynk + ynk − y¯) = lim inf ynk − y¯. k→∞

k→∞

This is a contradiction and, therefore, y¯ ∈ F (T ). Then (5.8.15) yields J(¯ y − QΩ z0 ), z0 − QΩ z0 > which, in its turn, contradicts Proposition 1.5.20 with the corresponding result J(¯ y − QΩ x), x − QΩ x ≤ 0

∀x ∈ Ω,

∀¯ y ∈ F (T ).

Hence, (5.8.15) is not true and lim ζn ≤ 0. n→∞ We prove that in reality lim ζn = 0. Indeed, since Φ(x) is convex and increasing and n→∞ T is nonexpansive, we have

Φ (1 − qn )(T yn − QΩ z0 )

≤ (1 − qn )Φ(T yn − T QΩ z0 ) ≤ (1 − qn )Φ(yn − QΩ z0 ).

Consequently, Φ(yn+1 − QΩ z0 ) ≤ Φ(yn − QΩ z0 ) − qn Φ(yn − QΩ z0 ) + γn ,

(5.8.16)

where γn = qn (1 − qn )ζn + µn and

µn = 4qn2 z0 − QΩ z0 2 + c¯1 ρX (z0 − QΩ z0 qn ).

Recall that qn → 0 and

∞

qn = ∞. Since X is a uniformly smooth Banach space, we

n=0

deduce lim

n→∞

µn ρX (z0 − QΩ z0 qn ) = lim 4z0 − QΩ mz0 2 qn + c¯1 = 0. n→∞ qn qn

5

308 Then


µn γn = lim ζn . = lim (1 − qn )ζn + n→∞ n→∞ qn n→∞ qn

lim

Rewrite now (5.8.16) in the following form: λn+1 ≤ λn − qn λn + γn ,

(5.8.17)

where λn = Φ(yn − QΩ z0 ). There may be only one alternative for any n ≥ 0 : (H1 ) : λn ≤

1 n

+

γn , qn

+

γn . qn

qi

i=0

or (H2 ) : λn >

1 n

qi

i=0

If we assume that lim ζn < 0 then also n→∞

lim

n→∞

γn < 0. qn

We show that this is wrong. First of all, we claim that (H1 ) happens infinitely many times. If this is not the case, there exists n ¯ > 1 such that the hypotheses (H2 ) holds for all n ≥ n ¯. Then qn γn ≤ q n λn − n

qi i=0

and (5.8.17) yields λn+1 ≤ λn − qn λn + qn λn −

qn n

= λn −

qi

0

i=0

Hence λn+1 ≤ λn¯ −

n

qj

j=¯ n

j

, qi

i=0

which is a contradiction because λn ≥ 0 for any n ≥ 1 and n

qj

j=¯ n

j

i=0

→ ∞ as n → ∞. qi

qn n

qi

.


309

Consequently, our claim is true. But then we come to a contradiction with the assumption that lim ζn < 0 because from the unbounded hypothesis (H1 ) we obtain that there exists n→∞ γn = 0. n ˜>n ¯ such that λn˜ < 0, which is not possible. That means that lim n→∞ qn Due to Lemma 7.1.2 for recurrent inequality (5.8.17), Φ(yn − QΩ z0 ) → 0 as n → ∞. Hence, {yn } converges strongly to Qz0 . The proof is complete. 1 1 Assume that J µ : X → X ∗ is the duality mapping with a gauge function µ : R+ → R+ , and t

Φ(t) =

0

µ(τ )dτ.

Denote J µ by J p if µ(t) = tp−1 with 1 < p < ∞. Then Φ(t) = p−1 tp and Φ (x) = J p x. Similarly to (5.8.12) one gets Φ(x + y) − Φ(x) =

= J p x, y +

0

1

1 0

J p (x + ty), y dt

J p (x + ty) − J p x, y dt.

Using (1.6.59) and (1.6.4) we obtain that if x ≤ M and y ≤ M, then there exist constants K1 > 0 and K2 > 0 such that Φ(x + y) − Φ(x) ≤ J p x, y + K1 M p ρX (K2 M −1 y). As before, we prove that {yn } converges strongly to QΩ z0 if J p is a weak-to-weak continuous duality mapping.

Bibliographical Notes and Remarks The regularization methods (5.1.2) and (5.1.22) for the value computation of an unbounded monotone operator were studied by Alber in [8]. The results of Sections 5.2 and 5.3 are stated in [186] and [11], respectively. Observe that Theorems 5.1.1, 5.1.3, 5.1.4 can be proved with inessential changes for discontinuous monotone operators and Theorem 5.3.1 for discontinuous accretive operators. In these cases, Rx0 is understood as a generalized ¯ 0 }, where A¯ is a maximal monotone value set of A at a point x0 , namely, R(x0 ) = {y | y ∈ Ax and maximal accretive extension of A, respectively. The regularized Hammerstein equation was constructed and investigated in [157, 158]. Other results are discussed in [50, 66, 136]. The pseudo-solutions of monotone equations described in Section 5.5 are due to Ryazantseva [196]. Lemma 5.5.13 uses the approach of [114] and [115]. Section 5.6 follows the paper [199]. Note that in [208], the authors propose to examine a smoothing functional in the form φ(Ah x − f δ ) + αxs , where Ah is weakly

310

5


continuous and ω(x) = φ(Ah x − f δ ) is convex. However, no direct means for construction of φ are given. The ill-posed optimal control problem (5.7.1), (5.7.2) was investigated in [116]. The regularized method of the successive approximations for finding fixed points of nonexpansive mappings was studied by Browder [57], Halpern [92], Reich [172], Takahashi and Kim [214] and others. Section 5.8, mainly, follows the paper [172]. General fixed point theory is well described in [87, 88].

Chapter 6

SPECIAL TOPICS ON REGULARIZATION METHODS 6.1

Quasi-Solution Method

In this section, we study the quasi-solution method for monotone equations and establish its connection with the operator regularization. ∗ Let X be a reflexive strictly convex space together with its dual space X ∗ , A : X → 2X be a maximal strictly monotone operator, M ⊂ D(A) be a closed convex compact set and int M = ∅, f ∈ X ∗ . Definition 6.1.1 An element x0 ∈ M is said to be a v-quasi-solution on M of the equation Ax = f

(6.1.1)

if it satisfies the inequality Ax − f, x0 − x ≤ 0

∀x ∈ M.

(6.1.2)

Observe that a solution of (6.1.2) is understood in the sense of Definition 1.11.2. Definition 6.1.1 considerably differs from the known definition of the classical quasisolution which is given by the equality Ax0 − f 2∗ = min{Ax − f 2∗ | x ∈ M}. f 2∗

(6.1.3)

The fact is that, the functional Ax − is not necessarily convex on M in the case of nonlinear monotone operators A, therefore, there are no effective tools to investigate both theoretical and numerical aspects of the classical quasi-solutions. However, if A is a monotone and potential operator, i.e., there exists a convex function ϕ : X → R1 such that A = ∂ϕ, then by Lemma 1.11.4 and Theorem 1.11.14, we conclude that the variational inequality (6.1.2) is equivalent to the following minimization problem: ϕ(x0 ) − f, x0 = min{ϕ(x) − f, x | x ∈ M}. 311

312

6

SPECIAL TOPICS ON REGULARIZATION METHODS

Under these conditions the v-quasi-solution coincides with a quasi-solution defined in [130], Chapter 5. Lemma 6.1.2 A v-quasi-solution of the equation (6.1.1) on M exists, is unique and depends continuously on a right-hand side f. Proof. The compact set M is bounded in X. Hence, according to Theorem 1.11.9 and Remark 1.11.12, the inequality (6.1.2) is solvable on M for any f ∈ X ∗ . Moreover, its solution x0 is unique, because A is a strictly monotone operator. Show that the v-quasisolution of (6.1.1) continuously depends on f. Let fn , n = 1, 2, ..., be given and fn → f as n → ∞. Denote by xn a (unique) v-quasi-solution of the equation Ax = fn on M with fixed n. In other words, xn satisfies the inequality y − fn , xn − x ≤ 0

∀x ∈ M,

∀y ∈ Ax.

(6.1.4)

Since xn ∈ M and M is compact, there exists x ¯ ∈ M such that xn → x ¯ as n → ∞. Passing in (6.1.4) to a limit, we obtain y − f, x ¯ − x ≤ 0

∀x ∈ M,

∀y ∈ Ax.

¯ This means that x ¯ is a v-quasi-solution of (6.1.1). Now we conclude that x ¯ = x0 because x is unique. Hence, the whole sequence xn → x0 as n → ∞. Lemma 6.1.3 If the equation (6.1.1) is solvable (in the sense of Definition 1.7.2) and its solution belongs to M, then it coincides with v-quasi-solution x0 of (6.1.1) on M. Proof. Let x∗ be a solution of (6.1.1), that is, f ∈ Ax∗ . It is unique because A is a strictly monotone operator. Due to the fact that A is maximal monotone and int M = ∅, we conclude, by Lemma 1.11.4, that a solution x0 ∈ M of the inequality (6.1.2) satisfies also the inequality (1.11.1), i.e., there exists ξ ∈ Ax0 such that ξ − f, x0 − x ≤ 0

∀x ∈ M.

Since f ∈ Ax∗ , we conclude that x∗ is a v-quasi-solution of (6.1.1) on M. Finally, uniqueness of v-quasi-solution following from Lemma 6.1.2 guarantees the equality x0 = x∗ . ∗

Theorem 6.1.4 Assume that A : X → 2X is a maximal strictly monotone operator, f ∈ X ∗ , the equation (6.1.1) has a solution x0 belonging to M, f δ ∈ X ∗ approximate f such that (5.5.23) holds. Then the sequence {xδ } of v-quasi-solutions on M of the equations Ax = f δ strongly converges to x0 as δ → 0. Proof. Let xδ be a v-quasi-solution of equation Ax = f δ on M. Lemma 6.1.2 guarantees existence and uniqueness of {xδ } and also its convergence to v-quasi-solution x∗ of equation (6.1.1) on M as δ → 0. It remains to add that, due to Lemma 6.1.3, x∗ coincides with the solution x0 of (6.1.1) in the sense of Definition 1.7.2. Denote 1 = (0, δ∗ ] × (0, h∗ ] and 2 = (0, h∗ ] × (0, σ ∗ ]. We present further the stability theorem of v-quasi-solutions.

6.1 Quasi-Solution Method

313

Theorem 6.1.5 Suppose that all the conditions of Theorem 6.1.4 are fulfilled, and operator A is also known with perturbations, namely, instead of A, a sequence {Ah } of the maximal strictly monotone operators is given, such that the estimate HX ∗ (Ax, Ah x) ≤ g(x)h ∀x ∈ M

(6.1.5)

holds, where g(t) is a non-negative continuous function for t ≥ 0. Let M ⊂ D(Ah ) for all h > 0. Then a sequence of v-quasi-solutions {xγ }, γ = (δ, h) ∈ 1 , of the equation Ah x = f δ on M converges strongly to a v-quasi-solution x ¯ of the equation (6.1.1) on M as γ → 0. Proof. An element xγ ∈ M is defined by the inequality Ah x − f δ , xγ − x ≤ 0

∀x ∈ M.

(6.1.6)

The existence and uniqueness of xγ ∈ M result from Lemma 6.1.2. Then the relation ζ h − f δ , xγ − x ≤ 0

∀x ∈ M.

(6.1.7)

holds all ζ h ∈ Ah x. Let x ∈ M be fixed. Take an arbitrary ζ ∈ Ax. The condition (6.1.5) enables us to find ζ h ∈ Ah x satisfying the estimate ζ h − ζ∗ ≤ g(x)h. Thus, if h → 0 then we construct a sequence {ζ h } such that ζ h → ζ. Since M is a compact set, we conclude that xγ → x ¯ ∈ M. Passing in (6.1.7) to the limit as γ → 0, one gets ζ − f, x ¯ − x ≤ 0

∀ζ ∈ Ax, ∀x ∈ M.

Hence, x ¯ is a v-quasi-solution of (6.1.1) on M. The theorem is proved because x ¯ is unique. Remark 6.1.6 If operators Ah , h ∈ (0, h∗ ], are not strictly monotone, then a v-quasisolution set on M of each equation Ah x = f δ is convex and closed, and it is not singleton, in general. The assertion of Theorem 6.1.5 remains valid in this case, if xγ in (6.1.6) is chosen arbitrarily. Assume that under the conditions of Theorem 6.1.5, in place of compact set M, a sequence of convex and closed compact sets {Mσ } is known, where Mσ ⊂ D(Ah ), int Mσ = ∅ for all (h, σ) ∈ 2 . Suppose also that {Mσ } converges to the compact set M as σ → 0 with respect to the Hausdorff metric. Then convergence of the v-quasi-solution sequence of equations Ah x = f δ on Mσ , as δ, h, σ → 0 is established by similar arguments as in Sections 4.2 and 4.3. Note also that the method described above of constructing approximations to the solution of (6.1.1) does not require us to know the level of errors δ, h and σ in the initial data. In what follows, we consider finite-dimensional approximations of v-quasi-solutions. Definition 6.1.7 We say that a sequence of finite-dimensional spaces {Xn }, Xn ⊂ X, is extremely dense in X if Pn x → x as n → ∞ for all x ∈ X, where {Pn } is a sequence of projectors Pn : X → Xn .

314

6


Theorem 6.1.8 Assume that in addition to the conditions of Theorem 6.1.5, operators Ah : X → X ∗ are hemicontinuous, {Xn }, n = 1, 2, ..., are the ordered sequences of finitedimensional subspaces of X which is extremely dense in X, Pn : X → Xn are linear operators and Pn∗ are their conjugate, fnδ = Pn∗ f δ , Ahn = Pn∗ Ah , {Mn } is a sequence of convex closed compact sets, Mn ⊂ Xn , Mn ⊂ D(Ahn ), Mn = Pn M for all n ≥ 1. Let γ = (δ, h) ∈ 1 . Then a sequence {xγn } of v-quasi-solutions of equations Ahn x = fnδ on Mn strongly converges to a v-quasi-solution xγ of the equation Ah x = f δ on M as n → ∞. Proof. A solution xγn ∈ Mn is uniquely defined by the inequality Ahn xn − fnδ , xγn − xn ≤ 0

∀xn ∈ Mn .

(6.1.8)

Rewrite (6.1.8) in the following form Ah xn − f δ , xγn − xn ≤ 0

∀xn ∈ Mn ,

xγn ∈ Mn .

(6.1.9)

xγn ∈ γ

It is possible because xn ∈ Xn and Xn . Since Mn ⊂ M for all n ≥ 1, the sequence ¯ ∈ M such that xγn → x ¯γ as n → ∞. Let in (6.1.8) {xγn } is compact, hence, there exists x and (6.1.9) xn = Pn x, where x ∈ M. Then xn → x as n → ∞, because a sequence {Xn } is extremely dense in X. Letting in (6.1.9) n → ∞ and taking into account the demicontinuity property of Ah we have ¯γ − x ≤ 0 ∀x ∈ M, Ah x − f δ , x

x ¯γ ∈ M.

From this inequality and from uniqueness of v-quasi-solution xγ , the conclusion of the theorem follows. We establish a connection between the quasi-solution method and operator regularization method. Let M be a convex closed compact set defined by the formula M = {x ∈ X | ψ(x) ≤ 0}, where ψ : X → R1 is a convex and continuous functional. We assume further that there exists at least one point z0 ∈ M such that the Slater condition ψ(z0 ) < 0

(6.1.10)

Ax − f, x0 − x ≤ 0 ∀x ∈ M, x0 ∈ M,

(6.1.11)

holds. Then the variational inequalities and Ax + α∂ψ(x) − f, xα − x ≤ 0 ∀x ∈ M, xα ∈ M,

α > 0,

(6.1.12)

where ∂ψ is a subdifferential of ψ, define, respectively, the quasi-solution method and operator regularization method. Due to the condition (6.1.10) and Theorem 1.11.18, the problems (6.1.11) and (6.1.12) are reduced mutually each to the other. Remark 6.1.9 If the variational inequality Ax − f, x − z ≤ 0 ∀z ∈ Ω,

x ∈ Ω,

is being solved, where Ω is a convex closed set in D(A) and M ⊂ Ω, then the definition of a v-quasi-solution on M given above and all the results of the present section remain still true.

6.2 Residual Method

6.2

315

Residual Method ∗

Let X be an E-space, X ∗ be strictly convex, A : X → 2X be a maximal monotone operator. Consider in X the equation (6.1.1). Let N = ∅ be its solution set, f δ ∈ X ∗ be δ-approximations of f ∈ X ∗ , i.e., f − f δ ∗ < δ, δ ∈ (0, δ ∗ ] with some positive δ∗ . 1. The residual method for solving the problem (6.1.1) with linear operator A in a Hilbert space is reduced to the minimization problem ϕ(x) = x2 → min

(6.2.1)

for a strongly convex functional ϕ on a convex closed set Mδ = {x ∈ X | Ax − f δ ≤ δ}.

(6.2.2)

Observe that in the case of a nonlinear operator A, the set Mδ is not convex, in general. In this case, the problem (6.2.1), (6.2.2) in the solving process meets with considerable difficulties. Therefore, we propose below another approach in which the constraint set Mδ is defined by means of variational inequalities. Let G ⊆ D(A) be a convex closed bounded set in X. Then there exists a constant C > 0 such that diam G ≤ C. Assume that NG = G ∩ N, N ∩ int G = ∅ and θX ∈ NG .

(6.2.3)

We introduce the following set: Ωδ = {w ∈ G | ξ − f δ , w − v ≤ δC ∀v ∈ G,

∀ξ ∈ Av}.

(6.2.4)

It is nonempty because NG ⊂ Ωδ for all δ ∈ (0, δ ∗ ]. Indeed, if w0 ∈ NG then we have ζ − f δ , w0 − w = ζ − f, w0 − w + f − f δ , w0 − w ≤ δw − w0 ≤ δC ∀w ∈ G,

∀ζ ∈ Aw.

(6.2.5)

Here we have taken into consideration that ζ − f, w − w0 ≥ 0 ∀w ∈ G,

∀ζ ∈ Aw,

(6.2.6)

which follows from the monotonicity of A. Moreover, Ωδ is a convex and closed set. Our aim is to find an element xδ satisfying the condition xδ s = min{xs | x ∈ Ωδ },

s ≥ 2,

(6.2.7)

and prove that xδ is the approximation to a solution of (6.1.1). Note that a number s should be chosen such that a functional xs has the best uniform convexity. For instance, in the Lebesgue spaces Lp (G) and lp with p ≥ 2, the preferable choice is s = p and if p ∈ (1, 2] then s = 2. So, we assume further that the functional xs is either uniformly convex or

316

6


strongly convex on X. Hence, the problem (6.2.4), (6.2.7) is uniquely solvable. Comparing it with (6.2.1), (6.2.2) above, it is natural to call (6.2.4), (6.2.7) the residual method. ¯∗ be a minimal norm element We study the behavior of the sequence {xδ } as δ → 0. Let x δ ∗ ∗ of NG . Then x ≤ ¯ x for all δ ∈ (0, δ ] because NG ⊂ Ωδ . Hence, {xδ } is a bounded ¯. Since xδ ∈ G and G is convex and sequence and there exists x ¯ ∈ X such that xδ x closed, the inclusion x ¯ ∈ G takes place. Then the definition of Ωδ gives ζ − f δ , xδ − v ≤ δC As δ → 0, we get

ζ − f, x ¯ − v ≤ 0

∀v ∈ G, ∀v ∈ G,

∀ζ ∈ Av.

(6.2.8)

∀ζ ∈ Av.

(6.2.9)

Similarly to (6.2.4), introduce now a set Ω0 = {w ∈ G | ζ − f, w − v ≤ 0 ∀v ∈ G, Ω0 .

∀ζ ∈ Av}.

Ω0

Let w0 ∈ and at the same time w0 ∈ NG . If we assume As we noted above, NG ⊂ that w0 ∈ int G, then one can deduce from the inequality (6.2.6) and from Lemma 1.11.6 that w0 ∈ NG . Therefore, we conclude that w0 ∈ ∂G. More precisely, w0 ∈ ∂Ω0 . But this is impossible because both the sets Ω0 and NG are convex and closed, and their intersections with int G coincide. Hence, it results from (6.2.9) that x ¯ ∈ NG . Due to the weak convergence ¯ ∈ G, and since xδ ≤ x∗ for all x∗ ∈ NG , we are able to write down the chain of xδ to x of inequalities ¯ x ≤ lim inf xδ ≤ lim sup xδ ≤ x∗ ∀x∗ ∈ NG . (6.2.10) δ→0

δ→0

It results from this that x ¯=x ¯∗ . Assuming in (6.2.10) x ¯=x ¯∗ , we obtain that xδ → ¯ x∗ as δ → 0. Thus, the following theorem is proved: ∗

Theorem 6.2.1 Let X be an E-space, X ∗ be strictly convex, A : X → 2X be a maximal monotone operator, G be a bounded convex closed set in D(A), N be a nonempty solution set of (6.1.1) with the properties (6.2.3), a functional xs (s ≥ 2) be uniformly convex on X. Then the problem (6.2.7), (6.2.4) has a unique solution xδ and xδ → x ¯∗ as δ → 0, where x ¯∗ is the minimal norm vector of NG . Corollary 6.2.2 If in Theorem 6.2.1 an operator A is continuous, then Axδ − f ∗ → 0 as δ → 0. If it is maximal monotone and single-valued at the point x ¯∗ , then y δ f as δ → 0, where yδ is any element of Axδ . Proof. The first assertion of this theorem follows from continuity of A and from the strong convergence of {xδ } to x ¯∗ proved in Theorem 6.2.1. Let A be single-valued at ¯∗ ∈ int D(A). Consequently, A is lothe point x ¯∗ . Since A is maximal monotone, then x ∗ cally bounded at x ¯ . Therefore, there exists g ∈ X ∗ such that y δ g. Now the equality ∗ g = A¯ x = f is guaranteed by the fact that grA is demiclosed. We can establish the connection between sets Mδ and Ωδ . By the monotonicity of A, it has the inequality ζ − f δ , w − v ≤ ξ − f δ , w − v ∀ζ ∈ Av, ∀ξ ∈ Aw.

(6.2.11)

6.2 Residual Method

317

It is not difficult to see that Mδ ⊆ Ωδ . Indeed, let w0 ∈ Mδ , that is, there exists f˜ ∈ Aw0 such that f˜ − f δ ∗ ≤ δ (see (6.2.2)). The inequality (6.2.11) implies ζ − f δ , w0 − v ≤ δC. Thus, w0 ∈ Ωδ . Note that the inverse inclusion Ωδ ⊆ Mδ does not necessarily hold. Let us give a corresponding example. Example 6.2.3 Let A : R1 → R1 , Ax = x − 3, f = 0, f δ = δ, δ ∈ (0, δ ∗ ], G = [2, 4], C = 2. Then it is not difficult to make certain that Mδ = [2, 4] ∩ [3, 3 + 2δ] and √ √ Ωδ = [2, 4] ∩ [3 + δ − 2 2δ, 3 + δ + 2 2δ].

1 If δ = , then M δ = [3, 4] ⊂ Ωδ . 2 Remark 6.2.4 An element xδ can also be defined by means of the minimization problem xδ − z 0 s = min {x − z 0 s | x ∈ Ωδ } ¯∗ is the nearest element from with some fixed element z 0 ∈ X. Then in Theorem 6.2.1, x 0 s NG to z . In addition, if in (6.2.7), instead of x , uniformly convex functional ω(x) is ¯∗ as δ → 0. minimized, then a solution xδ is also unique and {xδ } strongly converges to x Moreover, the condition ω(¯ x∗ ) = min{ω(x∗ ) | x∗ ∈ NG } holds. Generally speaking, convergence of the method (6.2.4), (6.2.7) does not imply solvability of the equation (6.1.1). In order to confirm this fact, we present the example of monotone operator A : R1 → R1 such that the method (6.2.4), (6.2.7) converges on each bounded set G and, at the same time, N = ∅. Example 6.2.5 Let Ax = 1 for all x ∈ R1 , G = [0, a] with a > 0, f = 0, f δ = δ and δ > 0 be sufficiently small. Then Ωδ = [0, aδ(1 − δ)−1 ], xδ = 0 for all δ ∈ (0, δ ∗ ]. Thus, xδ → 0 as δ → 0, but the set N = ∅. However, we are able to be sure that solvability of the variational inequality Ax − f, z − x ≤ 0 ∀x ∈ G, z ∈ G,

(6.2.12)

¯ ∈ G as δ → 0, i.e., is equivalent to convergence of the residual method. In fact, if xδ x the method (6.2.4), (6.2.7) converges, then passing in (6.2.8) to the limit as δ → 0 we get (6.2.9). Therefore, x ¯ is a solution of the variational inequality (6.2.12). Assume now that the variational inequality (6.2.12) is solvable and w0 ∈ G is its solution. Then (6.2.13) ζ − f, w0 − x ≤ 0 ∀x ∈ G, ∀ζ ∈ Ax,

318

6


and (6.2.5) holds. Thus, we have established that the set Ωδ is nonempty. Further the ¯∗ as δ → 0, where x ¯∗ is the minimal norm solution of (6.2.12), is proved convergence xδ → x as in Theorem 6.2.1. 2. Next we study the convergence of projection methods for the problem (6.2.4), (6.2.7). Theorem 6.2.6 Suppose that the conditions of Theorem 6.2.1 hold, A is an operator continuous on G and {Xn }, n = 1, 2, ..., is an ordered sequence of finite-dimensional subspaces of X. Let Qn : X → Xn and Pn : Xn → X be linear operators, |Qn | ≤ 1, fnδ = Q∗n f δ , Gn = G ∩ Xn , An = Q∗n A, lim Qn x − x = 0 ∀x ∈ G, (6.2.14) n→∞

and lim sup (Pn xn − xn ) ≤ 0 n→∞

∀xn ∈ Ωδn ,

(6.2.15)

where a constant C in (6.2.4) is such that diam Gn ≤ C for all n > 0, and Ωδn = {w ∈ Gn | An x − fnδ , w − x ≤ δC ∀x ∈ Gn }. Let

xδn

(6.2.16)

be defined as a solution of the following minimization problem: xδn s = min{xs | x ∈ Ωδn }.

(6.2.17)

Then xδn → xδ in X as n → ∞. Proof. First of all, observe that the problem (6.2.16), (6.2.17) is a finite-dimensional approximation of the residual method. The monotonicity condition of the operators An : Xn → Xn∗ is simply verified. Since Gn is bounded for all n ≥ 0, the inequality An y − fnδ , y − x ≤ 0

∀x ∈ Gn ,

y ∈ Gn ,

has a solution. It is not difficult now to see that the problem (6.2.16), (6.2.17) is uniquely ¯ ∈ G follows as solvable. Then, similarly to Theorem 6.2.1, the weak convergence xδn x n → ∞. Since xδn ∈ Ωδn , the inequality An xn − fnδ , xδn − xn ≤ δC,

xn = Qn x ∈ Gn

holds for all x ∈ G. Therefore, Axn − f δ , xδn − xn ≤ δC. Letting n → ∞ we get Ax − f δ , x ¯ − x ≤ δC

∀x ∈ G,

x ¯ ∈ G.

This means that x ¯ ∈ . It follows from (6.2.14) and (6.2.15) that (6.2.16) and (6.2.17) approximate the problem (6.2.4), (6.2.7) with the result Ωδ

lim xδn = xδ .

n→∞

6.2 Residual Method

319

Taking into account the unique solvability of the problem (6.2.4), (6.2.7) and weak conver¯, we deduce that x ¯ = xδ . The proof is accomplished because X is E-space. gence of xδn to x 3. We discuss the connection between the residual method and regularization method. Let x∗ be a minimal norm element of G and x∗ = xδ for all δ ∈ (0, δ ∗ ]. Consider a functional ϕδ (u) = sup {ζ − f δ , u − y − δC | y ∈ G, ζ ∈ Ay}. It is obvious that the set Ωδ in (6.2.4) can be defined as Ωδ = {u | ϕδ (u) ≤ 0}.

(6.2.18)

Study the properties of ϕδ (u). To this end, write down the inequalities ζ − f δ , u − y ≤ ξ − f δ , u − y ≤ Cξ − f δ ∗ valid for all u, y ∈ G, ξ ∈ Au and ζ ∈ Ay. Moreover, ϕδ (u) ≥ −δC. Hence, the functional ϕδ (u) is proper and G ⊂ int dom ϕδ . Then, by the inequality ζ − f δ , tu1 + (1 − t)u2 − y = tζ − f δ , u1 − y + (1 − t)ζ − f δ , u2 − y ≤ tϕδ (u1 ) + (1 − t)ϕδ (u2 ) ∀ζ ∈ Ay,

∀u1 , u2 , y ∈ G,

we prove that ϕδ is convex on G. Show that it is lower semi-continuous. Indeed, let limn→∞ un = u, where un , u ∈ G. Passing to the limit in the inequality ζ − f δ , un − y − δC ≤ ϕδ (un ) ∀ζ ∈ Ay,

∀y ∈ G,

we obtain ϕδ (u) ≤ lim inf ϕδ (un ). n→∞

In addition, by Theorem 1.2.8, we can make sure that ϕδ has a subdifferential on G. Due to Theorem 1.11.14, the residual method (6.2.4), (6.2.7) can be reduced to the variational inequality J s xδ , xδ − x ≤ 0

∀x ∈ Ωδ ,

xδ ∈ Ωδ ,

(6.2.19)

where J s is a duality mapping with the gauge function µ(t) = ts−1 . On the basis of Theorem 1.11.18, the problem (6.2.18), (6.2.19) is equivalent to the system J s xδ + p∂ϕδ (xδ ), xδ − y ≤ 0 ϕδ (xδ )(q − p) ≤ 0

∀y ∈ G,

xδ ∈ G,

1 1 ∀q ∈ R+ , p ∈ R+ ,

(6.2.20) (6.2.21)

with a solution {p, x } provided that the following Slater condition is satisfied: for every 1 , there exists x ¯ ∈ Ωδ such that p ∈ R+ δ

pϕδ (¯ x) < 0.

(6.2.22)

320

6


Let us establish (6.2.22) for our problem. To this end, construct a maximal monotone ¯ = G. It is possible in view of Theorem 1.8.5. Recall that G is operator A¯ with D(A) bounded. Therefore, by Corollary 1.7.6, there exists at least one solution v δ ∈ G of the ¯ = f δ . Hence, for ξ δ = f δ ∈ Av ¯ δ , we have equation Av pξ δ − f δ , v δ − x = 0

∀p > 0, ∀x ∈ G.

¯ one gets From the monotonicity of A, pξ − f δ , v δ − x ≤ pξ δ − f δ , v δ − x = 0

¯ ∀ξ ∈ Ax, ∀p > 0,

∀x ∈ G.

This means that (6.2.22) holds with x ¯ = v δ . Since xδ = x∗ as δ ∈ (0, δ ∗ ], it results from (6.2.20) that p = 0. It is easily established by contradiction. Indeed, if p = 0 then (6.2.20) implies the inequality J s xδ , xδ − y ≤ 0

∀y ∈ G,

xδ ∈ G,

that is, xδ ≤ y for all y ∈ G. Hence, xδ = x∗ for δ ∈ (0, δ ∗ ], which contradicts our assumption that xδ = x∗ . Thus, the following theorem is proved. Theorem 6.2.7 Suppose that the conditions of Theorem 6.2.1 are satisfied. Let xδ be a solution of the problem (6.2.4), (6.2.7) and xδ = x∗ as δ ∈ (0, δ ∗ ], where x∗ is a minimal norm element of G. Then there exists α > 0 such that the pair {α, xδ } is a solution of the system of inequalities ∂ϕδ (xδ ) + αJ s xδ , xδ − x ≤ 0 ∀x ∈ G, xδ ∈ G,

(6.2.23)

1 ϕδ (xδ )(q − α−1 ) ≤ 0 ∀q ∈ R+ .

(6.2.24)

Conversely, the pair {α, xδ }, which is a solution of the system (6.2.23), (6.2.24), determines a solution xδ of the problem (6.2.4), (6.2.7). We show now that, under the conditions of Theorem 6.2.7, the method (6.2.4), (6.2.7) is equivalent to regularization method (6.2.23) constructed for the variational inequality ∂ϕ0 (v), v − x ≤ 0 ∀x ∈ G, v ∈ G.

(6.2.25)

According to the definition of subdifferential ∂ϕ0 at a point v, we have sup {ζ − f, x − y | y ∈ G, ζ ∈ Ay} − sup {ζ − f, v − y | y ∈ G, ζ ∈ Ay} ≥ ∂ϕ0 (v), x − v ∀x, v ∈ G.

(6.2.26)

It is obvious that ϕ0 (x) = sup {ζ − f, x − y | y ∈ G, ζ ∈ Ay} ≥ 0 ∀x ∈ G, because ζ − f, x − y = 0 when y = x.

(6.2.27)

6.2 Residual Method

321

Let x ¯ ∈ G be a solution of the variational inequality Ay − f, y − x ≤ 0 ∀x ∈ G,

(6.2.28)

that is, there exists η ∈ A¯ x such that η − f, x ¯ − x ≤ 0 ∀x ∈ G.

(6.2.29)

Then, by Lemma 1.11.4, x ¯ is a solution of the inequality η − f, x ¯ − x ≤ 0 ∀x ∈ G,

∀η ∈ Ax.

x) ≤ 0. Taking into account (6.2.27), one gets that ϕ0 (¯ x) = 0. Then This means that ϕ0 (¯ (6.2.26) for v = x ¯ gives ϕ0 (x) ≥ ∂ϕ0 (¯ x), x − x ¯ ∀x ∈ G. x) if ϕ0 (x) ≥ g, x − x ¯ . However, by the definition of ϕ0 , it follows that Hence, g ∈ ∂ϕ0 (¯ ϕ0 (x) ≥ ¯ η − f, x − x ¯ ∀x ∈ G,

∀¯ η ∈ A¯ x, x ¯ ∈ G,

x). Then due to (6.2.29), we obtain that x ¯ is a which implies the inclusion A¯ x − f ⊂ ∂ϕ0 (¯ solution of (6.2.25). Let now x ¯ be a solution of (6.2.25), that is, ϕ0 (¯ x) = min{ϕ0 (x) | x ∈ G}. Since ϕ0 (x) ≥ 0 for all x ∈ G and since ϕ0 (x) attains the null value at a solution x ¯ of x) = 0. Thus, the inequalities (6.2.25) and (6.2.28) are equivalent. (6.2.28), we have ϕ0 (¯ Therefore, solutions of the regularized inequality Ay + αJ s y − f δ , y − x ≤ 0

∀x ∈ G,

y ∈ G,

and regularized inequality (6.2.23) approximate one and the same solution of the equation (6.1.1). Using (6.2.24) and following the proof of Theorem 1.11.18 we come to the equality ϕδ (xδ ) = 0,

(6.2.30)

which allows us to find α in (6.2.23) and define a connection between α and δ. The quantity ϕδ (x) = ϕδ (x) + δC may be regarded as the quasi-residual of the inequality Ay − f δ , x0 − y ≤ 0 ∀y ∈ G, x0 ∈ G, on an element xδ . Therefore, the equality (6.2.30) may be considered as the residual principle for the problem actually being solved Ay − f, x ˆ − y ≤ 0 ∀y ∈ G, x ˆ ∈ G. Hence, the method (6.2.4), (6.2.7) is equivalent to the regularization method with the operator ∂ϕδ , where the regularization parameter α is defined by (6.2.30).

6

322

SPECIAL TOPICS OF REGULARIZATION METHODS

√ Consider again Example 6.2.3. If 1 + δ − 2 2δ > 0 and xδ = 2, then an element √ xδ = 3 + δ − 2 2δ is a solution of (6.2.23) with √ 2 2δ √ . α= 3 + δ − 2 2δ

Moreover, max {(x − 3 − δ)(xδ − x) | x ∈ [2, 4]} = 2δ. Remark 6.2.8 Let in place of operator A, its maximal monotone approximations Ah be known such that for all x ∈ D(A) = D(Ah ) and h ∈ (0, h∗ ], HX ∗ (Ax, Ah x) ≤ g(x)h,

(6.2.31)

where g(t) is a non-negative and non-decreasing function for t ≥ 0. Replace Ω by δ

Ωδ,h = {w ∈ G | ζ h − f δ , w − v ≤ C(δ + g(C1 )h) ∀x ∈ G,

∀ζ h ∈ Ah v},

where C1 > v for all v ∈ G. Then all the assertions of the present section can be obtained by the same arguments.

6.3

Penalty Method

In this section we study one more regularization method for solving variational inequalities, the so-called penalty method. Let X be a reflexive strictly convex space with strictly convex dual space X ∗ , A : X → ∗ 2X be a maximal monotone operator, Ω ⊂ int D(A) be a convex closed bounded set. Consider the variational inequality (7.1.1). Let it have a solution set N = ∅ and let x ¯∗ ∈ N be a minimal norm solution. Assume that operator A, element f and set Ω are given approximately, namely, their approximations Ah , f δ and Ωσ are known for all (δ, h, σ) ∈ , where = (0, δ ∗ ] × (0, h∗ ] × (0, σ ∗ ] with some positive δ ∗ , h∗ , σ ∗ , such that f δ ∈ X ∗ , f − f δ ∗ ≤ δ, HX (Ω, Ωσ ) ≤ σ (6.3.1) and HX ∗ (Ax, Ah x) ≤ hg(x) ∀x ∈ Ω ∪ Ωσ . X∗

(6.3.2)

In (6.3.1) and (6.3.2), :X →2 are maximal monotone operators, D(A ) = D(A), Ωσ ⊂ int D(Ah ) are convex closed and bounded sets and g(t) is a continuous, non-negative and non-decreasing function for all t ≥ 0. Given convex closed set Ω, the penalty operator S : X → X ∗ is defined by the formula Ah

h

Sx = J(x − PΩ x), where J is a normalized duality mapping in X, PΩ is a metric projection operator onto Ω. As it has been proved in Lemma 1.5.18, S is a monotone demicontinuous and bounded mapping. We establish some of its additional properties. Rewrite (1.5.14) as Sx − Sy, PΩ x − PΩ y ≥ 0 ∀x ∈ X,

∀y ∈ X.

6.3 Penalty Method

323

Then the definition of PΩ implies the estimate J(x − PΩ x), PΩ x − y ≥ 0

∀y ∈ Ω.

(6.3.3)

Lemma 6.3.1 An element x ¯ ∈ Ω is a metric projection of x onto Ω if and only if the following inequality is fulfilled: x − x ¯2 ≤ J(x − x ¯), x − y ∀y ∈ Ω.

(6.3.4)

Proof. Let (6.3.4) hold. By the Cauchy−Schwarz inequality, ¯), x − y ≤ x − y ∀y ∈ Ω, x − x ¯ ≤ x − x ¯−1 J(x − x ¯ = PΩ x. Then (6.3.3) implies that for all y ∈ Ω, i.e., x ¯ = PΩ x. Assume now that x ¯), x − y , 0 ≤ J(x − x ¯), x ¯ − y = −x − x ¯2 + J(x − x and (6.3.4) follows. Lemma 6.3.2 Let X be a uniformly convex Banach space, X ∗ be strictly convex, Ω1 and Ω2 be convex closed sets in X, HX (Ω1 , Ω2 ) ≤ σ, δX () be a modulus of convexity of X. Then the following estimate holds: −1 (2Lc1 σ), PΩ1 x − PΩ2 x ≤ c2 δX

(6.3.5)

¯1 , x − x ¯2 }, x ¯1 = PΩ1 x, x ¯2 = PΩ2 x, c2 = where 1 < L < 1.7, c1 = 2max{x − x 2max {1, x − x ¯1 , x − x ¯2 }. Proof. Due to Theorem 1.6.4, we have ¯2 ), x ¯2 − x ¯1 ≥ (2L)−1 δX (c−1 x1 − x ¯2 ). J(x − x ¯1 ) − J(x − x 2 ¯

(6.3.6)

¯2 ∈ Ω2 there exists a z1 ∈ Ω1 such that ¯ x2 − z1 ≤ Since HX (Ω1 , Ω2 ) ≤ σ, for any element x σ. Furthermore, (6.3.3) yields the inequality J(x − x ¯1 ), z1 − x ¯1 ≤ 0. Then ¯2 − x ¯1 = J(x − x ¯1 ), x ¯2 − z1 + J(x − x ¯1 ), z1 − x ¯1 ≤ σx − x ¯1 . J(x − x ¯1 ), x x1 − z2 ≤ σ and By analogy, we assert that there exists z2 ∈ Ω2 such that ¯ ¯1 − x ¯2 ≤ σx − x ¯2 . J(x − x ¯2 ), x Therefore, J(x − x ¯1 ) − J(x − x ¯2 ), x ¯2 − x ¯1 ≤ σ(x − x ¯1 + x − x ¯2 ) ≤ σc1 .

(6.3.7)

6

324


By (6.3.6) and (6.3.7), we deduce (6.3.5). Consider in X the equation Ah x + −1 Sσ x + αJx = f δ , α > 0, > 0,

(6.3.8)

where Sσ x = J(x − PΩσ x). Since, in view of Theorem 1.4.6, the operator Sσ is maximal monotone with D(Sσ ) = X, we conclude that the operator Ah + −1 Sσ is so (see Theorem 1.8.3). A single-valued solvability of the equation (6.3.8) is guaranteed by Theorem 1.7.4. Thus, in place of the variational inequality (4.5.2) on Ω, it is proposed to solve operator equation (6.3.8) on D(A). Therefore, (6.3.8) can be regarded as the regularized penalty method. Let xγα, with γ(δ, h, σ) be a (unique) solution of this equation. Theorem 6.3.3 Assume that X is a uniformly convex and uniformly smooth Banach space, X ∗ is a dual space, δX () and δX ∗ () are moduli of convexity of X and X ∗ , respectively. Denote gX ∗ () = −1 δX ∗ () and suppose that lim

α→0

δ+h+ = 0, α

lim

α→0

−1 −1 gX ∗ (δX (σ)) = 0. α

(6.3.9)

¯∗ as α → 0. Then xγα, → x γ Proof. Since xγα, is a solution of (6.3.8), there exists an element ζα, ∈ Ah xγα, such that γ ζα, + −1 Sσ xγα, + αJxγα, = f δ . (6.3.10)

Let x ¯γα, = PΩσ xγα, . Then, by Lemma 6.3.1, we have xγα, − x ¯γα, 2 ≤ Sσ xγα, , xγα, − u

∀u ∈ Ωσ .

Taking into account (6.3.10), we obtain the inequality γ xγα, − x ¯γα, 2 ≤ f δ − ζα, − αJxγα, , xγα, − u

∀u ∈ Ωσ .

(6.3.11)

Let z0 be some fixed point of Ω. Due to (6.3.1), for any σ ∈ (0, σ ∗ ], there exists zσ ∈ Ωσ such that z0 − zσ ≤ σ, and then zσ → z0 as σ → 0. Putting in (6.3.11) u = zσ , we get xγα, − x ¯γα, 2 ≤ f δ − ξσh − αJzσ , xγα, − zσ γ + αJxγα, − ξσh − αJzσ , xγα, − zσ − ζα,

(6.3.12)

γ for all ζα, ∈ Ah xγα, and for all ξσh ∈ Ah zσ . The monotonicity property of the operator h A + αJ yields the relation γ + αJxγα, − ξσh − αJzσ , xγα, − zσ ≥ 0. ζα,

6.3 Penalty Method

325

By (6.3.2), for ξσh ∈ Ah zσ , we find ησ ∈ Azσ such that ξσh − ησ ≤ hg(zσ ). Then the following estimates are obtained: ξσh ∗ ≤ ξσh − ησ ∗ + ησ ∗ ≤ hg(zσ ) + ησ ∗ .

(6.3.13)

After this, (6.3.12) and (6.3.13) imply the relation

xγα, − x ¯γα, 2 ≤ f ∗ + δ + hg(zσ ) + ησ ∗ + αzσ ×

xγα, − x ¯γα, + ¯ xγα, − zσ .

(6.3.14)

Since Ω ⊂ int D(A), we conclude that the operator A is locally bounded at every point of Ω, hence, at the chosen point z0 . Therefore, the sequence {ησ } is bounded as σ → 0. By the properties of the function g(t) and by the boundedness of the sets Ωσ , there exist constants c1 > 0 and c2 > 0 such that (6.3.14) is evaluated as

¯γα, 2 ≤ c1 xγα, − x ¯γα, + c2 . xγα, − x From this, one deduces that a sequence {xγα, } is bounded as α → 0, say, xγα, ≤ r, r > 0. Introduce in X the auxiliary equation Ax + −1 Sx + αJx = f,

(6.3.15)

x0α, .

Making use of the above arguments, it is not difficult and denote its unique solution by to verify that a sequence {x0α, } is also bounded as α → 0 and x0α, ≤ r. This allows us ¯ ∈ X. Furthermore, since x0α, is a solution of (6.3.15), there exists to assert that x0α, x 0 ηα, ∈ Ax0α, such that 0 + −1 Sx0α, + αJx0α, = f. (6.3.16) ηα, Then Lemma 6.3.1 with y = z0 ∈ Ω implies 0 ηα, + αJx0α, − f, z0 − x0α, = −1 Sx0α, , x0α, − z0 ≥ 0.

(6.3.17)

In view of (6.3.17), it is quite easy to derive the estimate 0 , x0α, − z0 ≤ (f ∗ + αr) x0α, − z0 . ηα,

(6.3.18)

Moreover, according to Lemma 1.5.14 for x0 ∈ Ω ⊂ int D(A), there exists constant r0 > 0 0 ∈ Ax0α, , the inequality and a ball B(z0 , r0 ) such that for every x0α, ∈ D(A) and ηα, 0 0 ηα, , x0α, − z0 ≥ r0 ηα, ∗ − c0 (x0α, − z0 + r0 )

holds, where c0 = sup {ξ∗ | ξ ∈ Ax, x ∈ B(z0 , r0 )} < ∞. Then, by (6.3.18) and (6.3.19), we have 0 ∗ ≤ r0−1 (c0 + f ∗ + αr) x0α, − z0 + c0 , ηα,

(6.3.19)

326

6


0 } is bounded for all α > 0. Now we find from from which it results that the sequence {ηα, (6.3.16) that 0 0 Sx0α, ∗ ≤ f − ηα, − αJx0α, ∗ ≤ (f ∗ + ηα, ∗ + αx0α, ).

Consequently, Sx0α, → θX ∗ because → 0 as α → 0 (see (6.3.9)). Since S is demiclosed, ¯ ∈ Ω. the equality S x ¯ = θX ∗ holds. This means that x By (6.3.16), we calculate 0 − ξ, x0α, − x + ξ − f, x0α, − x + αJx0α, , x0α, − x ηα,

+ −1 Sx0α, − Sx, x0α, − x = 0 ∀x ∈ Ω, ∀ξ ∈ Ax. Then the monotonicity property of A and S leads to the inequality ξ − f, x0α, − x + αJx0α, , x0α, − x ≤ 0 ∀x ∈ Ω,

∀ξ ∈ Ax.

(6.3.20)

Assuming that α → 0 in (6.3.20) and taking into account the weak convergence of x0α, to x ¯ ∈ Ω, one gets ξ − f, x − x ¯ ≥ 0 ∀x ∈ Ω, ∀ξ ∈ Ax. (6.3.21) The maximal monotonicity of A and Lemma 1.11.4 allow us to assert that there exists ξ¯ ∈ A¯ x such that ξ¯ − f, x − x ¯ ≥ 0 ∀x ∈ Ω. Hence, x ¯ ∈ N. ¯∗ as α → 0. Since x ¯0α, = PΩ x0α, ∈ Ω, one has Show that x0α, → x ξ − f, x ¯0α, − x ≥ 0 ∀x ∈ N,

∀ξ ∈ Ax.

(6.3.22)

Rewrite (6.3.20) in the following form: ξ − f, x0α, − x ¯0α, + ξ − f, x ¯0α, − x + αJx0α, , x0α, − x ≤ 0 ∀ξ ∈ Ax.

(6.3.23)

Then, by (6.3.22) and by the Cauchy−Schwartz inequality, we obtain Jx0α, , x0α, − x ≤ α−1 ξ − f ∗ x0α, − x ¯0α, ∀x ∈ N,

∀ξ ∈ Ax.

Observe that (6.3.11) with u = x ¯0α, gives for solution x0α, of (6.3.15) the estimate ¯0α, ≤ c3 , x0α, − x where c3 is a positive constant. That is to say, there exists c4 > 0 such that Jx0α, , x0α, − x ≤ c4 α−1 ∀x ∈ N. ¯ ∈ N, monotonicity of J and (6.3.9), one gets Due to the weak convergence of x0α, to x Jx, x ¯ − x ≤ 0 ∀x ∈ N.

6.3 Penalty Method

327

¯∗ as α → 0. Furthermore, similarly to Section 2.2, This means that x0α, x ¯∗ as α → 0. x0α, → x

(6.3.24)

Combining (6.3.10) and (6.3.16), we come to the following equality: γ 0 ζα, − ηα, , xγα, − x0α, + −1 Sσ xγα, − Sx0α, , xγα, − x0α,

+ αJxγα, − Jx0α, , xγα, − x0α, = f δ − f, xγα, − x0α, .

(6.3.25)

Using Theorem 1.6.4, one can write γ 0 Jxγα, − Jx0α, , xγα, − x0α, ≥ (2L)−1 δX (c−1 5 xα, − xα, ),

(6.3.26)

h where 1 < L < 1.7 and c5 = 2max{1, r}. Let ζα, ∈ Ah x0α, such that 0 h − ζα, ∗ ≤ hg(x0α, ). ηα,

This element exists because of the condition (6.3.2). Since operators Ah are monotone and function g(t) is non-decreasing, we obtain γ 0 γ h h 0 − ηα, , xγα, − x0α, = ζα, − ζα, , xγα, − x0α, + ζα, − ηα, , xγα, − x0α, ζα,

≥ −hg(x0α, )xγα, − x0α, ≥ −hg(r)xγα, − x0α, .

(6.3.27)

The monotonicity property of Sσ makes it possible to evaluate the second term in the left-hand side of (6.3.25): Sσ xγα, − Sx0α, , xγα, − x0α, = Sσ xγα, − Sσ x0α, , xγα, − x0α, + Sσ x0α, − Sx0α, , xγα, − x0α, ≥ Sσ x0α, − Sx0α, , xγα, − x0α, . By Corollary 1.6.8, Sσ x0α, − Sx0α, , xγα, − x0α, = J(x0α, − PΩσ x0α, ) − J(x0α, − PΩ x0α, ), xγα, − x0α, ≥ − J(x0α, − PΩσ x0α, ) − J(x0α, − PΩ x0α, )∗ xγα, − x0α,

−1 0 0 γ 0 ≥ − c6 g X ∗ 2c6 LPΩσ xα, − PΩ xα, xα, − xα, ,

(6.3.28)

6

328


where c6 = 2max{1, c3 ∗ , r + d + σ ∗ }, d = max{x | x ∈ Ω}. Now Lemma 6.3.2 enables us to specify the obtained inequality. Indeed, Sσ x0α, − Sx0α, , xγα, − x0α,

−1 2 −1 γ 0 ≥ −c6 gX ∗ 2Lc6 δX (2Lc7 σ) xα, − xα, , ∗

(6.3.29)

∗

where c7 = 2max {c3 , r + d + σ }. It results from (6.3.25) - (6.3.29) that

−1 2 −1 γ 0 (δ + hg(r))xγα, − x0α, + −1 c6 gX ∗ 2Lc6 δX (2Lc7 σ) xα, − xα, γ 0 ≥ α(2L)−1 δX (c−1 5 xα, − xα, ).

(6.3.30)

Finally, (6.3.30) yields the estimate #

−1 2Lc5 xγα, − x0α, ≤ c5 gX

δ + hg(r) c6 −1 2 −1 + gX ∗ (2Lc6 δX (2Lc7 σ)) α α

$

.

In view of (6.3.9), this result together with (6.3.24) completes the proof. Remark 6.3.4 If X is a Hilbert space, then the second condition in (6.3.9) is written as √ σ = 0. lim α→0 α

6.4

Proximal Point Method

In this section we study the proximal point algorithm for solving the equation (6.1.1). 1. Consider first the Hilbert space case. Let A : H → 2H be a maximal monotone operator, equation (6.1.1) has a nonempty solution set N in the sense of inclusion. We assume that, in place of f, a sequence {fn }, n = 1, 2, ..., is known such that f − fn ≤ δn .

(6.4.1)

The proximal point algorithm is described as follows: Take some sequence of positive numbers {cn }. For a constant c1 > 0 and x0 ∈ D(A), we find the unique solution x1 of the equation c1 (Ax − f1 ) + x = x0 (see Theorem 1.7.4). Proceeding with this process further, we obtain in D(A) a sequence {xn } which is defined recurrently, namely, xn is a solution of the equation cn (Ax − fn ) + x = xn−1 ,

cn > 0,

n = 1, 2, ... .

(6.4.2)

Theorem 6.4.1 Let A : H → 2H be a maximal monotone operator, a solution set of (6.1.1) be nonempty, 0 < c ≤ cn ≤ C and

∞

δn < ∞.

n=1

Then a sequence {xn } generated by (6.4.2) weakly converges to some element of N.

(6.4.3)

6.4 Proximal Point Method

329

Proof. Let x∗ ∈ N, yn ∈ Axn and cn (yn − fn ) + xn = xn−1 .

(6.4.4)

By (6.4.4), one gets cn (yn − f, xn − x∗ ) + cn (f − fn , xn − x∗ ) + xn − x∗ 2 = (xn−1 − x∗ , xn − x∗ ).

(6.4.5)

Due to the monotonicity of A, we have from (6.1.1), (6.4.3) and (6.4.5) xn − x∗ ≤ xn−1 − x∗ + cn δn ≤ xn−1 − x∗ + Cδn .

(6.4.6)

This implies the inequality xn − x∗ ≤ x0 − x∗ + C

n

δi .

i=1

In view of convergence of the series ∞ n=1 δn , the sequence {xn } is bounded. Hence, there exists a subsequence {xk } ⊂ {xn } which converges weakly to some x ¯ ∈ H. We show that x ¯ ∈ N. First of all, it is well known that if the sequence of positive numbers {an } satisfies the recurrent inequality an ≤ an−1 + βn ,

∞

βn < ∞,

n=1

then there exists lim an = a ≥ 0. For this reason, (6.4.6) and the condition n→∞ imply existence of a limit for xn − x∗ as n → 0. We can write down that lim xn − x∗ = µ(x∗ ) < ∞.

n→∞

∞

n=1 δn

0 satisfies the estimate xn − x∗ ≤ M. It follows from the convergence criterion of the series above that δn → 0 as n → ∞. Hence, by virtue of (6.4.7) and (6.4.9), lim cn yn − fn = 0. n→∞

Since cn ≥ c > 0, we conclude that yn → f as n → ∞. Then Lemma 1.4.5 implies that x ¯ ∈ N.

6

330


¯ such that x ¯1 ∈ Show that x ¯ is uniquely defined. Suppose that there exist x ¯1 = x ¯1 , {xm } ⊂ {xn }, N, xm x ¯1 = µ(¯ x1 ) < ∞ lim xn − x

n→∞

and lim xn − x ¯ = µ(¯ x) < ∞.

n→∞

Write the obvious equality ¯1 2 = xn − x ¯2 + 2(xn − x ¯, x ¯−x ¯1 ) + ¯ x1 − x ¯ 2 . xn − x

(6.4.10)

Assuming n = k and n = m in (6.4.10) and passing to the limit as k → ∞ and m → ∞ we obtain, respectively, x1 ) − µ2 (¯ x) = ¯ x−x ¯ 1 2 > 0 µ2 (¯ and

x1 ) − µ2 (¯ x) = −¯ x−x ¯1 2 < 0, µ2 (¯

because x ¯ and x ¯1 are the weak limit points of {xn }. Thus, we arrive at a contradiction. Consequently, x ¯ is unique. Theorem 6.4.2 Let A : H → 2H be a maximal monotone operator with D(A) = H and conditions (6.4.3) hold. Then the weak convergence of the proximal point algorithm (6.4.2) is equivalent to solvability of the equation (6.1.1). Proof. By Theorem 6.4.1, strong (consequently, weak) convergence of the algorithm (6.4.2) follows from solvability of the equation (6.1.1). We show the contrary statement. ¯ ∈ H. The weak Let the proximal algorithm (6.4.2) converge weakly, that is, xn x convergence of {xn } implies its boundedness. Let xn ≤ r for all n > 0. Construct a maximal monotone operator A¯ = A + ∂I2r , where ∂I2r is defined by (1.8.1). Note that ¯ for all x ∈ B0 (θX , 2r). Moreover, D(A) ¯ = B(θX , 2r) is bounded in H. Therefore, Ax = Ax ¯ ¯ by Corollary 1.7.6, the set N = {x | f ∈ Ax} = ∅. Since a solution xn of the equation (6.4.2) with maximal monotone operator belongs to B0 (θX , 2r) and it is uniquely defined, we conclude that a sequence {xn } coincides with the sequence constructed for the equation ¯ = f by the same proximal point algorithm (6.4.2). As in the proof of Theorem 6.4.1, Ax ¯ , where ¯ it is established that xn x ¯∈N x < 2r. In other words, x ¯ ∈ B0 (θX , 2r). Hence f ∈ A¯ x.

Theorem 6.4.3 Let A : H → 2H be a maximal monotone operator, a solution set N of (6.1.1) be nonempty and the conditions (6.4.3) hold. Assume that a sequence {An } of maximal monotone operators An : H → 2H is given in place of A, D(A) = D(An ), HH (Ax, An x) ≤ g(x)hn ∀x ∈ D(A), where g(t) is a bounded positive function for t ≥ 0 and sequence {xn } of the equations

∞

n=1

hn < ∞. Then a solution

cn (An x − fn ) + x = xn−1 , n = 1, 2, ... , converges weakly to some element of N.

(6.4.11)

(6.4.12)


331

Proof. Let yn ∈ An xn be such that (6.4.4) is satisfied. If x∗ ∈ N then, by virtue of (6.4.11), we can choose yn∗ ∈ An x∗ which gives the estimate yn∗ − f ≤ g(x∗ )hn

(6.4.13)

because f ∈ Ax∗ . Rewrite (6.4.5) in the equivalent form cn (yn − yn∗ , xn − x∗ ) + cn (yn∗ − f, xn − x∗ ) + cn (f − fn , xn − x∗ ) + xn − x∗ 2 = (xn−1 − x∗ , xn − x∗ ).

(6.4.14)

Since An is monotone, the first term in the left-hand side of (6.4.14) is non-negative. Therefore, using (6.4.1), (6.4.3) and (6.4.13) we deduce from (6.4.14) that there exists C1 > 0 such that xn − x∗ ≤ x0 − x∗ + C1

n

(δi + hi ),

x∗ ∈ N.

i=1

The rest of the proof follows the pattern of Theorem 6.4.1. Similarly to Theorem 6.4.2, it is possible to establish that the weak convergence of the proximal point algorithm (6.4.12) is equivalent to solvability of the equation (6.1.1). To this end, is is enough to construct the maximal monotone operator A¯n + ∂I2r and apply Theorem 6.4.3. ∗

2. Consider further the case of Banach spaces. Let in (6.1.1) A : X → 2X be a maximal monotone operator, f ∈ X ∗ , X be a reflexive strictly convex Banach space with strictly convex X ∗ . Introduce the condition HX ∗ (Ax, An x) ≤ g(x)hn ∀x ∈ D(A),

(6.4.15)

where g(t) is a non-negative and increasing function for all t ≥ 0. We study the proximal point algorithm defined by the following iterative scheme: cn (An x − fn ) + Jx = Jxn−1 , cn > 0 n = 1, 2, ... ,

(6.4.16)

where J : X → X ∗ is the normalized duality mapping in X. Denote by xn its (unique) solution when n is fixed (see Theorem 1.7.4). ∗

∗

Theorem 6.4.4 Assume that A : X → 2X and An : X → 2X are maximal monotone operators, D(A) = D(An ), elements f, fn ∈ X ∗ for all n > 0, f − fn ∗ ≤ δn

(6.4.17)

and (6.4.15) holds. If cn → ∞, δn → 0, hn → 0 as n → ∞, and if a sequence {xn } defined recurrently by (6.4.16) is bounded, then equation (6.1.1) is solvable and every weak accumulation point of {xn } is a solution of (6.1.1).

332

6.4 SPECIAL TOPICS OF REGULARIZATION METHODS Proof. Suppose that yn ∈ An xn satisfies the equality cn (yn − fn ) + Jxn = Jxn−1 .

(6.4.18)

By (6.4.15), take an element y¯n ∈ Axn such that ¯ yn − yn ∗ ≤ g(xn )hn . It is obvious that there exists a constant C2 > 0 such that ¯ yn − yn ∗ ≤ C2 hn .

(6.4.19)

Now (6.4.17) - (6.4.19) yield the relations ¯ yn − f ∗ ≤ ¯ yn − yn ∗ + yn − fn ∗ + fn − f ∗ ¯ ≤ (xn + xn−1 )c−1 n + δn + Chn . Due to the properties of {xn }, {cn }, {δn } and {hn }, we conclude that y¯n → f. Let ¯ ∈ X. Since A is demiclosed, we come to the required inclu{xk } ⊂ {xn } and xk x sion: f ∈ A¯ x. With some different assumptions, prove the following ∗

∗

Theorem 6.4.5 Let A : X → 2X , An : X → 2X be maximal monotone operators, D(A) = D(An ), (6.4.15) and (6.4.17) hold, and ∞

cn (δn + hn ) < ∞.

(6.4.20)

n=1

Suppose that cn → ∞ as n → ∞, a duality mapping J in X is weak-to-weak continuous, and a sequence {xn } defined by (6.4.16) is bounded. Then it weakly converges to some solution of (6.1.1). Proof. It follows from the conditions of the theorem that δn → 0 and hn → 0 as n → ∞. By Theorem 6.4.4, we have that a set N = {x | f ∈ Ax} = ∅. Let x∗ ∈ N. Consider the functional (1.6.36) with y = x∗ : W (x, x∗ ) = 2−1 (x2 − 2Jx, x∗ + x∗ 2 ) ∀x ∈ X. In view of (1.6.41), there holds the inequality W (x, x∗ ) − W (v, x∗ ) ≤ Jx − Jv, x − x∗ ∀x, v ∈ X. If we put x = xn and v = xn−1 then W (xn , x∗ ) − W (xn−1 , x∗ ) ≤ Jxn − Jxn−1 , xn − x∗ .

(6.4.21)


333

By (6.4.15), we find yn∗ ∈ An x∗ such that for f ∈ Ax∗ , yn∗ − f ∗ ≤ hn g(x∗ ). Taking into account (6.4.18), one gets W (xn , x∗ ) − W (xn−1 , x∗ ) ≤ cn yn − fn , x∗ − xn = cn yn − yn∗ , x∗ − xn + cn yn∗ − f, x∗ − xn + cn f − fn , x∗ − xn ≤ cn (δn + g(x∗ )hn )xn − x∗ , because

cn yn − yn∗ , x∗ − xn ≤ 0.

Recall that {xn } is bounded. Therefore, there exists M > 0 such that W (xn , x∗ ) ≤ W (xn−1 , x∗ ) + M cn (δn + hn ). Then we have W (xn , x∗ ) ≤ W (x0 , x∗ ) + M

n

ck (δk + hk ).

(6.4.22)

(6.4.23)

k=1 ∗

By (6.4.20), this means that the sequence {W (xn , x )} is also bounded. Moreover, (6.4.20) and (6.4.22) yield the limit relation lim W (xn , x∗ ) = µ(x∗ ),

n→∞

(6.4.24)

where 0 ≤ µ(x∗ ) < ∞. Theorem 6.4.4 asserts that the sequence {xn } has a weak accumulation point, that is, there exists a subsequence {xk } ⊂ {xn } such that xk x ¯ ∈ N. We will show that there ¯1 = x ¯, such exists only one weak accumulation point. Suppose there is another point x ¯1 , x that xm x ¯1 ∈ N, where {xm } ⊂ {xn }. By (6.4.24), ¯) = µ(¯ x) lim W (xn , x

n→∞

and lim W (xn , x ¯1 ) = µ(¯ x1 ).

n→∞

It is easy to see that

lim

n→∞

W (xn , x ¯) − W (xn , x ¯1 ) = µ(¯ x) − µ(¯ x1 )

(6.4.25)

= 2−1 (¯ x2 − ¯ x1 2 ) + lim Jxn , x ¯1 − x ¯ . n→∞

It is clear that Jxn , x ¯1 − x ¯ has a limit which we denote by l. Taking n = k in the previous ¯ . Repeating equality and using weak-to-weak continuity of J one gets that l = J x ¯, x ¯1 − x ¯1 − x ¯ . Consequently, the same arguments with n = m, we derive that l = J x ¯1 , x J x ¯1 − J x ¯, x ¯1 − x ¯ = 0. Due to the strict monotonicity of J, we conclude that x ¯=x ¯1 . This fact establishes uniqueness of the weak accumulation point x ¯ ∈ N Therefore, the whole sequence {xn } weakly converges to x ¯. The proof is accomplished.

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6


Remark 6.4.6 If N is the singleton, then the requirement of weak-to-weak continuity of J can be omitted. 3. We study now the convergence of (6.4.2) for the equation (6.1.1) with a maximal accretive operator A. Theorem 6.4.7 Assume that X is a uniformly convex Banach space, A : X → 2X is a maximal accretive operator, An : X → 2X are m-accretive operators for all n > 0, duality mapping J is weak-to-weak continuous in X, D(A) = D(An ), f ∈ X, fn ∈ X, such that (6.4.1) stays valid and HX (Ax, An x) ≤ g(x)hn ∀x ∈ D(A),

(6.4.26)

where g(t) is a non-negative and increasing function for all t ≥ 0. Let N = {x∗ | f ∈ Ax∗ } = ∅, cn → ∞ and (6.4.20) hold. Then a sequence {xn } generated by (6.4.12) weakly converges to some x∗ ∈ N. Proof. Under the conditions of the theorem, the equation (6.4.2) is uniquely solvable. By analogy with (6.4.14), we have cn J(xn − x∗ ), yn − yn∗ + cn J(xn − x∗ ), yn∗ − f + cn J(xn − x∗ ), f − fn + xn − x∗ 2 = J(xn − x∗ ), xn−1 − x∗ , where x∗ ∈ N and yn∗ ∈ An x∗ satisfy (6.4.13). It is not difficult to deduce from this the estimate xn − x∗ ≤ x0 − x∗ + M1

n

ck (δk + hk )

k=1

with some constant M1 > 0. It implies the boundedness of {xn }. Let {xk } ⊂ {xn } and ¯ ∈ X. If yn ∈ An xn satisfy (6.4.4) then one can show that yn → f. Let ξ n ∈ Axn xk x such that yn − ξ n ≤ g(xn )hn . Then ξ n → f. It follows from Lemma 1.15.12 that x ¯ is a solution of equation (6.1.1). In these circumstances, there exists a limit of xn − x∗ for any fixed x∗ ∈ N. Hence, lim xn − x∗ = µ(x∗ ),

n→∞

¯1 = x ¯. It is obvious that where 0 ≤ µ(x∗ ) < ∞. Let {xm } ⊂ {xn } and xm x ¯2 ≤ xk − x ¯xk − x ¯1 + J(xk − x ¯), x ¯1 − x ¯ . xk − x Indeed, xk − x ¯2 = J(xk − x ¯), xk − x ¯ = J(xk − x ¯), xn − x ¯1 + J(xk − x ¯), x ¯1 − x ¯ ¯xk − x ¯1 + J(xk − x ¯), x ¯1 − x ¯ . ≤ xk − x


335

If k → ∞ one gets in a limit

x) ≤ µ(¯ x)µ(¯ x1 ). µ2 (¯

Analogously, the relation xm − x ¯1 2 ≤ xm − x ¯1 xm − x ¯ + J(xm − x ¯1 ), x ¯−x ¯1 implies

x1 ) ≤ µ(¯ x1 )µ(¯ x) µ2 (¯

as m → ∞. Consequently, µ(¯ x) = µ(¯ x1 ). By virtue of Lemma 1.5.7, we conclude that ¯ is proved. x ¯=x ¯1 . Hence, uniqueness of the weak limit point x 4. To obtain the strongly convergent approximating sequence, construct regularized proximal point algorithms for monotone and accretive equations. Let H be a Hilbert space, A : H → H be single-valued and An : H → 2H be maximal monotone operators, D(A) = D(An ), n > 0. Assume that the conditions (6.4.1) and (6.4.11) are fulfilled and lim αn = 0,

n→∞

lim

n→∞

δn + hn = 0. αn

(6.4.27)

Then, due to Theorem 2.1.3, solution sequence {vn } of the equation An v + αn v = fn , αn > 0, n ≥ 1,

(6.4.28)

is bounded and it converges as n → ∞ to the minimal norm solution x ¯∗ ∈ N. Given some element x0 ∈ H, construct a sequence {xn }, where xn+1 is calculated from the following equation: (6.4.29) cn (An x + αn x − fn ) + x = xn , cn > 0. Rewrite the equation (6.4.29) in the equivalent form µn (An x − fn ) + x = βn xn , where µn =

cn 1 + cn αn

βn =

1 . 1 + cn αn

and

(6.4.30)

Then (6.4.28) can be represented as µn (An v − fn ) + v = βn v.

(6.4.31)

Let yi,j ∈ Ai xj and wi,j ∈ Ai vj . According to (6.4.30) and (6.4.31), there are elements y n,n+1 ∈ An xn+1 and w n,n ∈ An vn such that µn (y n,n+1 − fn ) + xn+1 = βn xn

(6.4.32)

6

336


and µn (wn,n − fn ) + vn = βn vn .

(6.4.33)

Subtracting (6.4.33) from (6.4.32) side by side and multiplying the obtained result by xn+1 − vn , one gets µn (y n,n+1 − wn,n , xn+1 − vn ) + (xn+1 − vn , xn+1 − vn ) = βn (xn − vn , xn+1 − vn ). Since An are monotone, we have xn+1 − vn 2 ≤ βn (xn − vn , xn+1 − vn ). Now the Cauchy−Schwartz inequality yields xn+1 − vn ≤ βn xn − vn .

(6.4.34)

It is clear from (6.4.28) that wn+1,n+1 + αn+1 vn+1 = fn+1 . Then (wn+1,n+1 − wn,n , vn+1 − vn ) + (αn+1 vn+1 − αn vn , vn+1 − vn ) = (fn+1 − fn , vn+1 − vn )

(6.4.35)

or (w n+1,n+1 − wn+1,n , vn+1 − vn ) + (w n+1,n − wn,n , vn+1 − vn ) + αn+1 vn+1 − vn 2 + (αn+1 − αn )(vn , vn+1 − vn ) = (fn+1 − fn , vn+1 − vn ).

(6.4.36)

We know that there exists d > 0 such that vn ≤ d for all n > 0. Since A is singlevalued, one has wn+1,n − wn,n ≤ wn+1,n − Avn + Avn − wn,n ≤ g(vn )(hn+1 + hn ) ≤ M (hn+1 + hn ),

(6.4.37)

where M = sup {g(t) | 0 ≤ t ≤ d}. Further, fn+1 − fn ≤ fn+1 − f + fn − f ≤ δn+1 + δn . By the monotonicity of A vn+1 − vn ≤

n+1

(6.4.38)

, we derive from (6.4.36) and (6.4.38)

hn+1 + hn d |αn+1 − αn | δn+1 + δn + +M = n . αn+1 αn+1 αn+1

(6.4.39)


337

Furthermore, (6.4.34) and (6.4.39) allow us to obtain xn+1 − vn+1 ≤ xn+1 − vn + vn − vn+1 ≤ βn xn − vn + n . Thus, xn+1 − vn+1 ≤ (1 − γn )xn − vn + n , where γn =

c n αn , cn > 0, αn > 0. 1 + cn αn

(6.4.40) (6.4.41)

Applying Lemma 7.1.2 to (6.4.40) and taking into account the inequality xn − x ¯∗ ≤ xn − vn + vn − x ¯∗ , we thus come to the following result: Theorem 6.4.8 Let the equation (6.1.1) be solvable in a Hilbert space H, A : H → H and An : H → 2H be maximal monotone operators for all n > 0, D(A) = D(An ), f ∈ H, fn ∈ H. Let the conditions (6.4.1), (6.4.11) and (6.4.27) be satisfied, where {δn } and {hn } are nonincreasing sequences of positive numbers. Assume also that cn > 0, ∞

γn = ∞,

(6.4.42)

n=0

and lim

n→∞

|αn+1 − αn | + δn + hn = 0, αn+1 γn

(6.4.43)

where γn is defined by (6.4.41). Then a sequence {xn } generated by the iterative process (6.4.29) converges strongly to the minimal norm solution x ¯∗ of the equation (6.1.1). 5. We proceed to the regularized proximal algorithm in a Banach space X. Assume that ∗ A : X → X ∗ and An : X → 2X are maximal monotone, f ∈ X ∗ , fn ∈ X ∗ such that (6.4.15) and (6.4.17) are still valid. Denote by vn , n = 1, 2, ..., solutions of the equation An v + αn Jv = fn , αn > 0,

(6.4.44)

where J : X → X ∗ is the normalized duality mapping. Consider the regularized proximal point algorithm defining an approximate sequence {xn } in the following form: cn−1 (An−1 x + αn−1 Jx − fn−1 ) + Jx = Jxn−1 , cn−1 > 0, n = 1, 2, ... .

(6.4.45)

It is clear that there exist y n,n+1 ∈ An xn+1 and wn,n ∈ An vn such that cn (y n,n+1 + αn Jxn+1 − fn ) + Jxn+1 = Jxn

(6.4.46)

w n,n + αn Jvn = fn .

(6.4.47)

and

6

338


In the sequel, we assume that X is a uniformly convex Banach space and δX () is its modulus of convexity. Recall that function δX () is continuous and increasing on the interval [0, 2] and δX (0) = 0. At the same time, the function gX () defined by (1.1.13) is continuous and non-decreasing on [0, 2] and gX (0) = 0. However, we assume for simplicity that gX () is the increasing function (see also Remarks 1.6.9 and 6.5.3). As before, suppose that solution set N of (6.1.1) is not empty, and let x ¯∗ be its minimal norm solution. According to Theorem 2.2.1, under the conditions (6.4.15), (6.4.17) and ¯∗ as n → ∞. (6.4.27), the sequence {vn } is bounded, say, vn ≤ d for all n > 0 and vn → x Along with (6.4.47), the equality wn+1,n+1 + αn+1 Jvn+1 = fn+1 ,

(6.4.48)

∈ holds, where n+1 . Subtracting the equality (6.4.47) from (6.4.48) and calculating the values of the obtained functionals on the element vn+1 − vn , we have wn+1,n+1

An+1 v

wn+1,n+1 − wn,n , vn+1 − vn + αn+1 Jvn+1 − Jvn , vn+1 − vn + (αn+1 − αn )Jvn , vn+1 − vn = fn+1 − fn , vn+1 − vn .

(6.4.49)

By the property (1.6.19), evaluate the second term in (6.4.49) as αn+1 Jvn+1 − Jvn , vn+1 − vn ≥ αn+1 (2L)−1 δX (c−1 0 vn+1 − vn ), where 1 < L < 1.7 and c0 = 2max{1, d}. The monotonicity of A

n+1

(6.4.50)

yields

wn+1,n+1 − wn,n , vn+1 − vn = wn+1,n+1 − wn+1,n , vn+1 − vn + wn+1,n − wn,n , vn+1 − vn ≥ wn+1,n − wn,n , vn+1 − vn .

(6.4.51)

Return again to (6.4.37) and (6.4.38) assuming that {δn } and {hn } are non-increasing sequences of positive numbers. We have w n+1,n − wn,n ∗ ≤ 2M hn and fn+1 − fn ∗ ≤ 2δn . Then the estimates (6.4.50) and (6.4.51) allow us to deduce from (6.4.49) the inequality

αn+1 (2L)−1 δX (c−1 0 vn+1 − vn )

≤ 2(δn + M hn ) + |αn+1 − αn | vn+1 − vn . Now some simple algebra leads to the estimate vn+1 − vn ≤ c0 n ,

(6.4.52)

where

−1 L1 n = gX

δn + hn + |αn+1 − αn | , L1 = 2Lc0 max{2, 2M, d}. αn+1

(6.4.53)


339

Theorem 6.4.9 Let X be a uniformly convex Banach space, A : X → X ∗ and An : X → ∗ 2X be maximal monotone operators for all n > 0, D(A) = D(An ), f ∈ X ∗ , fn ∈ X ∗ . Assume that a solution set N of the equation (6.1.1) is nonempty, the conditions (6.4.15), (6.4.17), (6.4.27) and (6.4.42) are satisfied, where {δn } and {hn } are non-increasing sequences of positive numbers. Let γn and n are defined by (6.4.41) and (6.4.53), respectively, and n lim = 0. (6.4.54) n→∞ γn ¯∗ , where x ¯∗ ∈ N is the If a sequence {xn } generated by (6.4.45) is bounded, then xn → x minimal norm solution of (6.1.1). Proof. Using the property (1.6.43) of the functional W (x, z) we have W (xn+1 , vn+1 ) ≤ W (xn+1 , vn ) + Jvn+1 − Jxn+1 , vn+1 − vn ≤ W (xn+1 , vn ) + Jvn+1 − Jxn+1 ∗ vn+1 − vn .

(6.4.55)

Recall that vn ≤ d and assume that xn ≤ d1 for all n > 0. Then (6.4.52) implies W (xn+1 , vn+1 ) ≤ W (xn+1 , vn ) + c0 (d1 + d)n .

(6.4.56)

Further, the property (1.6.41) of the functional W (x, z) gives W (xn+1 , vn ) ≤ W (xn , vn ) + Jxn+1 − Jxn , xn+1 − vn .

(6.4.57)

Evaluate the last term in the right-hand side of (6.4.57). By virtue of (6.4.46) and (6.4.47), one gets the following result: Jxn+1 − Jxn = −cn (y n,n+1 + αn Jxn+1 − fn ) = −cn (yn,n+1 − w n,n + αn Jxn+1 − αn Jvn ). Applying now the monotonicity of An we have Jxn+1 − Jxn , xn+1 − vn = −cn y n,n+1 − wn,n , xn+1 − vn − cn αn Jxn+1 − Jvn , xn+1 − vn ≤ −cn αn Jxn+1 − Jvn , xn+1 − vn . Taking into account (1.6.44), it is easy to see that Jxn+1 − Jxn , xn+1 − vn ≤ −cn αn W (xn+1 , vn ). Then (6.4.57) can be rewritten as W (xn+1 , vn ) ≤ W (xn , vn ) − cn αn W (xn+1 , vn ).

340

6


Thus, W (xn+1 , vn ) ≤ (1 − γn )W (xn , vn ). Substitute this estimate for (6.4.56). Then W (xn+1 , vn+1 ) ≤ W (xn , vn ) − γn W (xn , vn ) + c0 (d + d1 )n . By the conditions (6.4.42) and (6.4.54) and by Lemma 7.1.2, we obtain that lim W (xn , vn ) = 0.

n→∞

Finally, the left inequality of (1.6.48) and the properties of δX () allow us to conclude that xn − vn → 0 as n → ∞. Then the assertion to be proved follows from the inequality ¯∗ ≤ xn − vn + vn − x ¯∗ . xn − x In conclusion, we present a result like Theorem 6.4.8 for the equation (6.1.1) with accretive operator A. Theorem 6.4.10 Let in a Banach space X, the equation (6.1.1) be solvable, that is, N = {x∗ | f ∈ Ax∗ } = ∅, A : X → X be a maximal accretive operator, f ∈ X, duality mapping J : X → X ∗ be continuous and weak-to-weak continuous. Suppose that An : X → 2X are m-accretive operators, D(An ) = D(A), fn ∈ X, and the conditions (6.4.1), (6.4.26), (6.4.27), (6.4.42) and (6.4.43) hold and γn is defined by (6.4.41). Then a sequence {xn } generated by (6.4.29) converges strongly in X to the unique solution x ¯∗ ∈ N satisfying the inequality (2.7.7).

6.5

Iterative Regularization Method

In order to obtain approximations to the minimal norm solution x ¯∗ of the equation (6.1.1) by the operator regularization method, we needed to solve a sequence of regularized equations (2.2.4) with corresponding parameters αn → 0. The iterative regularization method which we study in this section does not solve regularized equations, while it gives a new approximation to x ¯∗ on every iteration step of the algorithm. It is sufficient for this to calculate values of the given operator on the current iteration. 1. Assume that X is a real uniformly convex and uniformly smooth Banach space. We ∗ solve the equation (6.1.1) with maximal monotone bounded operator A : X → 2X having χ-growth order, that is, ξ∗ ≤ χ(x) ∀ξ ∈ Ax, ∀x ∈ X,

(6.5.1)

where χ(t) is a continuous non-decreasing function for t ≥ 0. As usual, suppose that the equation (6.1.1) with f ∈ X ∗ has a nonempty solution set N, and that, in place of f and ∗ A, approximations f δn ∈ X ∗ and Ahn : X → 2X are given such that Ahn are maximal monotone operators, D(A) = D(Ahn ) = X, f δn − f ∗ ≤ δn

(6.5.2)


6.5

341

and HX ∗ (Ahn x, Ax) ≤ φ(x)hn ∀x ∈ X,

(6.5.3)

where a function φ(t) is continuous non-negative and non-decreasing for all t ≥ 0. Thus, in reality, instead of (6.1.1), the perturbed equation Ahn x = f δn

(6.5.4)

is solved. In general, it not necessarily has a solution. In the case when A and Ahn are hemicontinuous, the inequality (6.5.3) is replaced by Ahn x − Ax∗ ≤ φ(x)hn ∀x ∈ X.

(6.5.5)

Study the following iterative regularization algorithm: Jxn+1 = Jxn − m (ξnhm + αm Jxn − f δm ),

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

(6.5.6)

where ξnhm ∈ Ahm xn , m = n + n0 , n0 is defined below by (6.5.11) and positive parameters αm and m and non-negative parameters δm and hm satisfy the inequalities ¯ hm ≤ h. ¯ ¯ , m ≤ ¯, δm ≤ δ, αm ≤ α

(6.5.7)

As before, we denote by δX (t) the modulus of convexity of X and by ρX (τ ) its modulus of smoothness. Since the functions δX (t) and ρX (τ ) are increasing on the interval [0,2] and −1 (·) and ρ−1 [0,∞), respectively, and since δX (0) = ρX (0) = 0, the inverse functions δX X (·) are −1 −1 also increasing and δX (0) = ρX (0) = 0. All these function are continuous. We introduce gX (t) by the formula (1.1.13) and assume that it is increasing for all t ∈ [0, 2] (see Remarks 1.6.9 and 6.5.3). Consider the unperturbed regularized equation Az + αJz = f,

α > 0.

(6.5.8)

x∗ ≤ K0 and As it was proved in Section 2.2, its solutions zα exist for all α > 0, zα ≤ ¯ zα → x ¯∗ as α → 0, where x ¯∗ is the minimal norm element of N. Denote zm = zαm and assume that αm → 0 as m → ∞. It is obvious that xn − x ¯∗ ≤ zm − x ¯∗ + xn − zm .

(6.5.9)

Therefore, to establish strong convergence of the method (6.5.6) we need only to prove that lim xn − zm = 0.

n→∞

Let R0 be any non-negative number. Assume K1 =

2R0 + K0 , K2 = K1 + ¯c1 , c = 2c0 LK0 ,

¯ ¯ 1 + δ¯ + f ∗ , c0 = 2max{1, K0 }, c1 = χ(K1 ) + hφ(K 1 ) + αK c2 = 8max{L, K2 }, c3 = K0 + K2 , c4 = 2max{1, K1 }.

6

342


Here 1 < L < 1.7 is the Figiel constant. Construct the following sequence:

γk = k (K0 + K1 ) hk φ(K1 ) + δk + 8c21 2k + c2 ρX ∗ (c1 k ) −1 + c0 c3 gX

c|α − α k k+1 |

αk

, k = 0, 1, 2, ... .

(6.5.10)

Choose n0 according to the condition

−1 n0 = min k | c4 δX

2Lγ k

k αk

≤ −1 (R0 ) ,

(6.5.11)

where (τ ) = 8τ 2 + c2 ρX (τ ). It is clear that the functions (t) and its inverse function −1 (s) are positive for all t, s ∈ [0, ∞), increasing and (0) = −1 (0) = 0. Therefore, n0 is well defined if γ k → 0 as k → ∞. (6.5.12) k αk Introduce the functional defined by (1.6.36): W (x, z) = 2−1 (x2 − 2Jx, z + z2 ) ∀x, z ∈ X,

(6.5.13)

where J is the normalized duality mapping. Let zn0 be a solution of the equation (6.5.8) with α = αn0 and let an initial point x0 in (6.5.6) satisfy the inequality W (x0 , zn0 ) ≤ R0 . Such x0 exists, because W (zn0 , zn0 ) = 0. For example, we can put x0 = zn0 . We premise the main result of this section on the following lemma. Lemma 6.5.1 Let x ¯∗ be a minimal norm solution of the equation (2.1.1) and let zn and zn+1 be solutions of the equation (6.5.8) with α = αn and α = αn+1 , respectively. Then the following inequality holds: −1 zn − zn+1 ≤ c0 gX

c|α − α n n+1 |

αn

.

(6.5.14)

Proof. Since zn and zn+1 are solutions of (6.5.8), there exist ζn ∈ Azn and ζn+1 ∈ Azn+1 such that the equalities ζn + αn Jzn = f (6.5.15) and ζn+1 + αn+1 Jzn+1 = f are satisfied. Evaluate from below the expression D = ζn + αn Jzn − ζn+1 − αn Jzn+1 , zn − zn+1 .

(6.5.16)

6.5


343

Applying the estimate (1.6.19) and taking into account the fact that A is monotone, one gets D = ζn − ζn+1 , zn − zn+1 + αn Jzn − Jzn+1 , zn − zn+1 ≥ αn Jzn − Jzn+1 , zn − zn+1 ≥ αn (2L)−1 δX (c−1 0 zn − zn+1 ).

(6.5.17)

Next we evaluate D from above. Since zn and zn+1 satisfy (6.5.15) and (6.5.16), D = ζn + αn Jzn − f, zn − zn+1 − ζn+1 + αn Jzn+1 − f, zn − zn+1 = −ζn+1 + αn Jzn+1 − f, zn − zn+1 = ζn+1 + αn+1 Jzn+1 − f, zn − zn+1 − ζn+1 + αn Jzn+1 − f, zn − zn+1 = (αn+1 − αn )Jzn+1 , zn − zn+1 ≤ |αn+1 − αn |zn+1 zn − zn+1 ≤ K0 |αn+1 − αn |zn − zn+1 .

(6.5.18)

By (6.5.17) and (6.5.18), the estimate (6.5.14) follows. In a Hilbert space δX () ≥ 8−1 2 , gX () ≥ 8−1 . Therefore, (6.5.14) accepts the form zn − zn+1 ≤

8K0 |αn − αn+1 | . αn

However, if we prove Lemma 6.5.1 directly in a Hilbert space, then one gets zn − zn+1 ≤

K0 |αn − αn+1 | . αn

(6.5.19)

On the basis of (1.6.43), we write down the inequality W (xn+1 , zm+1 ) − W (xn+1 , zm ) ≤ Jzm+1 − Jxn+1 , zm+1 − zm ≤ Jzm+1 − Jxn+1 ∗ zm+1 − zm .

(6.5.20)

Furthermore, (1.6.41) yields W (xn+1 , zm ) − W (xn , zm ) ≤ Jxn+1 − Jxn , xn+1 − zm = Jxn+1 − Jxn , xn − zm + Jxn+1 − Jxn , xn+1 − xn . (6.5.21)

6

344


Combining (6.5.20) and (6.5.21) we have W (xn+1 , zm+1 ) − W (xn , zm ) ≤ Jxn+1 − Jxn , xn+1 − xn + Jxn+1 − Jzm+1 ∗ zm+1 − zm + Jxn+1 − Jxn , xn − zm .

(6.5.22)

Estimate every summand forming the right-hand side of (6.5.22). 1) By (6.5.1) and (6.5.3), there exists ξ n ∈ Axn such that ξnhm ∗ ≤ ξnhm − ξ n ∗ + ξ n ∗ ≤ hm φ(xn ) + χ(xn ), and f δm ∗ ≤ f ∗ + f δm − f ∗ ≤ f ∗ + δm . Returning to the regularization method (6.5.6), we calculate Jxn+1 − Jxn ∗ = m ξnhm + αm Jxn − f δm ∗ ≤ m c1 (n), where c1 (n) = χ(xn ) + hm φ(xn ) + αm xn + f ∗ + δm .

(6.5.23)

Now we apply (1.6.17) and get the following result: Jxn+1 − Jxn , xn+1 − xn ≤ 8Jxn+1 − Jxn 2∗ + c2 (n)ρX ∗ (Jxn+1 − Jxn ∗ ), where c2 (n) = 8max{L, xn+1 , xn }. Then

(6.5.24)

Jxn+1 − Jxn , xn+1 − xn ≤ 82m c21 (n) + c2 (n)ρX ∗ (m c1 (n)).

2) It is obvious that Jxn+1 − Jzm+1 ∗ ≤ xn+1 + zm+1 = c3 (n). By Lemma 6.5.1, −1 zm+1 − zm ≤ c0 gX

c|α

m

(6.5.25)

− αm+1 | . αm

Consequently, −1 Jxn+1 − Jzm+1 ∗ zm+1 − zm ≤ c0 c3 (n)gX

c|α

m

− αm+1 | . αm

3) Evaluate the last term in (6.5.22): Jxn+1 − Jxn , xn − zm = −m ξnhm + αm Jxn − f δm , xn − zm = −m ξnhm − ξ n , xn − zm − m f − f δm , xn − zm

6.5

345

Iterative Regularization Method − m ξ n − ζm , xn − zm − m ζm + αm Jzm − f, xn − zm − m αm Jxn − Jzm , xn − zm ,

where ξ n ∈ Axn , ζm ∈ Azm and ζm + αm Jzm = f . Recalling that the operator A is monotone, we deduce

Jxn+1 − Jxn , xn − zm ≤ m hm φ(xn ) + δm xn − zm − m αm (2L)−1 δX

xn − z m

c4 (n)

,

where c4 (n) = 2max{1, xn , zm }.

(6.5.26)

Rewrite (6.5.22) using the estimates obtained in 1) - 3). We have W (xn+1 , zm+1 ) ≤ W (xn , zm ) − m αm (2L)−1 δX

xn − z m

c4 (n)

+ m hm φ(xn ) + δm xn − zm + 82m c21 (n) −1 + c2 (n)ρX ∗ (m c1 (n)) + c0 c3 (n)gX

c|α

m

− αm+1 | . αm

(6.5.27)

By hypotheses, W (x0 , zn0 ) ≤ R0 . Consider arbitrary n ≥ 0 such that W (xn , zm ) ≤ R0 . Then (1.6.37) implies xn ≤ zm +

2R0 ≤ K0 +

2R0 = K1 .

In this case, the following estimates hold: ¯ c1 (n) ≤ χ(K1 ) + hφ(K ¯ 1 + f ∗ + δ¯ = c1 , 1 ) + αK xn+1 ≤ xn + Jxn+1 − Jxn ∗ ≤ K1 + m c1 (n) ≤ K1 + ¯c1 = K2 ,

(6.5.28)

c2 (n) ≤ 8max{L, K2 } = c2 , c3 (n) ≤ K2 + K0 = c3 , c4 (n) ≤ 2max{1, K1 } = c4 . After that, the inequality (6.5.27) is obtained in the final form: n W (xn+1 , zm+1 ) ≤ W (xn , zm ) − m αm (2L)−1 δX (c−1 4 x − zm ) + γm ,

(6.5.29)

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346


where γm coincides with (6.5.10) if k = m. It results from (6.5.29) that W (xn+1 , zm+1 ) ≤ W (xn , zm ) + γ¯ , with

−1 2 2 ¯ ¯ γ¯ = ¯(hφ(K ¯ + c2 ρX ∗ (c1 ¯) + c0 c3 gX (cd), 1 ) + δ)(K0 + K1 ) + 8c1

where we assume that

|αn − αn+1 | ≤d αn

(6.5.30)

with some constant d > 0. Thus, if W (xn , zn+n0 ) ≤ R0 for all n ≥ 0 then xn ≤ K1 . Otherwise, if W (xn , zm ) ≤ R0 for 0 ≤ n ≤ n∗ < ∞, where n∗ is an integer, then ∗ +1

R0 < W (xn

, zn∗ +1+n0 ) ≤ R0 + γ¯ = R1 .

(6.5.31)

We show that inequality W (xn , zm ) ≤ R1 holds for all n ≥ n∗ + 1. To this end, consider the following alternative: either (H1 ) : (2L)−1 δX

or (H2 ) : (2L)−1 δX

xn∗ +1 − z

n∗ +1+n0

c4

xn∗ +1 − z

n∗ +1+n0

c4

>

γn∗ +1+n0 αn∗ +1+n0 n∗ +1+n0

≤

γn∗ +1+n0 . αn∗ +1+n0 n∗ +1+n0

If (H1 ) is true then we deduce from (6.5.29) and (6.5.31) that ∗ +2

W (xn

∗ +1

, zn∗ +2+n0 ) < W (xn

, zn∗ +1+n0 ) ≤ R1 .

(6.5.32)

At the same time, the hypothesis (H2 ) can not be held. Indeed, assuming the contrary, we obtain ∗ 2Lγn∗ +1+n0 −1 . xn +1 − zn∗ +1+n0 ≤ c4 δX ∗ ∗ αn +1+n0 n +1+n0 If the sequence in (6.5.12) decreases, we come to the inequality ∗ +1

xn

− zn∗ +1+n0 ≤ −1 (R0 )

(6.5.33)

by reason of (6.5.11). We estimate W (xn , zm ) through xn − zm . Due to the inequality (1.6.48), one gets W (xn , zm ) ≤ 8xn − zm 2 + c6 (n)ρX (xn − zm ), where c6 (n) = 8max{L, xn , zm }. It is clear from (6.5.28) that c6 (n) ≤ 8max{L, K2 } = c2

6.5


347

and W (xn , zm ) ≤ (xn − zm ). Then

∗ +1

W (xn

∗ +1

, zn∗ +1+n0 ) ≤ (xn

− zn∗ +1+n0 ).

Taking into account (6.5.33) we deduce W (xn

∗ +1

, zn∗ +1+n0 ) ≤ R0 ,

(6.5.34)

which contradicts the assumption (6.5.31). Consequently, by induction, the estimate W (xn , zm ) ≤ R1 is satisfied for all n ≥ 0 and then xn ≤ zm +

2R1 ≤ K0 +

2R1 = K3 ,

(6.5.35)

that is, the sequence {xn } is bounded. We introduce the new denotation:

−1 n Ψ(W (xn , zm )) = (2L)−1 δX c−1 4 (W (x , zm )) .

It is easy to verify that the function Ψ(t) is positive for all t > 0, continuous, increasing and Ψ(0) = 0. If we repeat now deduction of (6.5.29) replacing K1 by K3 everywhere above, then we shall obtain the following recursive inequality: W (xn+1 , zm+1 ) ≤ W (xn , zm ) − m αm Ψ(W (Jxn , zm )) + γm , where γm is defined by (6.5.10). It can be written as λn+1 ≤ λn − ρn Ψ(λn ) + κn ,

(6.5.36)

where λn = W (xn , zn+n0 ),

ρn = n+n0 αn+n0 ,

κn = γn+n0 .

By the properties of the functional W (x, z), λn ≥ 0 for all n ≥ 0. Consequently, we can apply Lemma 7.1.3 which gives the sufficient conditions to assert that limn→∞ W (xn , zm ) = 0. The left inequality of (1.6.48) implies −1 xn − zm ≤ 2c4 δX (LW (xn , zm )).

Then lim xn − zm = 0.

n→∞

By (6.5.9), this enables us to state the following result: Theorem 6.5.2 Suppose that X is a uniformly convex Banach space and 1) A solution set N of the equation (6.1.1) is not empty and x ¯∗ is its minimal norm solution; ∗ X 2) A : X → 2 is a maximal monotone bounded operator with χ-growth (6.5.1); 3) Instead of the equation Ax = f, in fact, the perturbed equations (6.5.4) with maximal monotone operators Ahn are solved, the estimates (6.5.2) and (6.5.3) hold and D(A) =

348

6


D(Ahn ) = X for all n ≥ 0; 4) An initial approximation x0 in the iterative regularization method (6.5.6) satisfies the inequality W (x0 , zn0 ) ≤ R0 , where W (x, z) is defined by (6.5.13) and R0 is an arbitrary non-negative number; 5) zn0 is a solution of the operator equation (6.5.8) with α = αn0 , where n0 obeys the rule (6.5.11); 6) Positive parameters αn and n and non-negative parameters δn and hn approach zero as n → ∞, such that (6.5.7) and (6.5.30) are valid. Besides, let 7)

∞

αn n = ∞;

n=1

8) 9)

lim

δn + hn + n = 0; αn

lim

ρX ∗ (n ) = 0; n αn

n→∞

n→∞

&

10)

'

−1 gX αn−1 |αn − αn+1 | = 0. n→∞ αn n

lim

Then the sequence {xn } generated by (6.5.6) converges strongly to x ¯∗ as n → ∞. Remark 6.5.3 If gX (t) does not increase strictly for all t ∈ [0,2] but there exists a nonnegative increasing continuous function g˜X (t) such that gX (t) ≥ g˜X (t), then Theorem 6.5.2 remains still valid if in its conditions, proof and conclusions g(·) and g −1 (·) are replaced by g˜(·) and g˜−1 (·), respectively. As it was already mentioned in Section 1.6, in most cases of uniformly convex Banach spaces the modulus of convexity δX (t) ≥ d1 tγ , γ ≥ 2, d1 > 0. −1 κ −1 Consequently, g˜X (t) = d1 tγ−1 and g˜X (ξ) = (d−1 1 ξ) , where κ = (γ − 1) . γ

Recall that if δX (t) ≥ d1 tγ , γ ≥ 2, d1 > 0, then the modulus of smoothness ρX ∗ (τ ) ≤ d2 τ γ−1 , d2 > 0. In this situation, the requirements 8) - 10) are simplified, namely: δn + hn + κn =0 n→∞ αn lim

and

|αn − αn+1 |κ = 0. n→∞ αnγκ n lim

Let us give some examples. 1) X = Lp , p ≤ 2, X ∗ = Lq , p−1 + q−1 = 1, q ≥ 2. Here δX (t) ≥ d1 t2 , that is, −1 2 (ξ) = d−1 γ = 2, g˜X 1 ξ, ρX ∗ (τ ) ≤ d2 τ , d1 , d2 are positive constants. Hence, the convergence ∗ of (6.5.6) to x ¯ is guaranteed if ∞

n=1

αn n = ∞,

lim αn = 0,

n→∞

6.5


349

lim

δn + h n + n =0 αn

lim

|αn − αn+1 | = 0, αn2 n

n→∞

and n→∞

that includes the case of a Hilbert space. 2) X = Lp , p > 2, X ∗ = Lq , p−1 + q −1 = 1, q ≤ 2. In this case δX (t) ≥ d3 tp , γ = p, 1/(1−p) 1/(p−1) −1 g˜X (ξ) = d3 ξ , ρX ∗ (τ ) ≤ d4 τ q , d3 , d4 are positive constants. Then the convergence conditions are the following: ∞

αn n = ∞,

n=1

lim αn = 0,

n→∞

1/(p−1)

δn + hn + n n→∞ αn

=0

lim

and lim

n→∞

|αn − αn+1 |1/(p−1) p/(p−1)

αn

= 0.

n

2. Results like Theorem 6.5.2 can also be obtained for the equation (2.1.1) with an accretive operator A : X → X. For simplicity of the proof, we assume that the sequence {xn } generated by the iterative algorithm xn+1 = xn − n (Ahn xn + αn xn − f δn ), n = 0, 1, 2, ... ,

(6.5.37)

is bounded. Suppose that equation (2.1.1) is solvable, N is its solution set and the (unique) solution x ¯∗ ∈ N satisfies the inequality (2.7.7). Let A be a demicontinuous operator, D(A) = X, perturbed operators Ahn : X → X and right-hand sides f δn ∈ X satisfy the previous conditions (6.5.2) and (6.5.5) (with the norm of X), D(Ahn ) = X and (6.5.7) holds for all m = n ≥ 0. Introduce the intermediate equation Az + αz = f, α > 0.

(6.5.38)

It was proved in Section 2.7 that if X possesses an approximation and duality mapping J : X → X ∗ is continuous and weak-to-weak continuous, then solutions zα of (6.5.38) x∗ . By analogy with Lemma 6.5.1, it is possible to converge to x ¯∗ as α → 0 and zα ≤ 2¯ show that for solutions zn = zαn and zn+1 = zαn+1 the following estimate holds: zn+1 − zn ≤ 2¯ x∗

|αn+1 − αn | . αn

(6.5.39)

Convergence analysis of the method (6.5.37) is done by a scheme like the monotone case, but now, in place of (6.5.13), the functional V (x, z) = 2−1 x − z2 is studied.

6

350


Indeed, assuming that xn ≤ R < ∞ one gets xn+1 − zn+1 2 ≤ xn+1 − zn 2 + 2 J(zn+1 − xn+1 ), zn+1 − zn ≤ xn+1 − zn 2 + 2zn+1 − xn+1 zn+1 − zn .

(6.5.40)

Similarly to (6.5.21), we derive xn+1 − zn 2 ≤ xn − zn 2 + 2 J(xn+1 − zn ), xn+1 − xn = xn − zn 2 + 2 J(xn − zn ), xn+1 − xn + 2 J(xn+1 − zn ) − J(xn − zn ), xn+1 − xn .

(6.5.41)

Consequently, (6.5.40) and (6.5.41) imply xn+1 − zn+1 2 ≤ xn − zn 2 + 2zn+1 − xn+1 zn+1 − zn + 2 J(xn − zn ), xn+1 − xn + 2 J(xn+1 − zn ) − J(xn − zn ), xn+1 − xn . By (6.5.37), we calculate J(xn − zn ), xn+1 − xn = −n J(xn − zn ), Ahn xn + αn xn − f δn = −n αn xn − zn 2 − n J(xn − zn ), Ahn xn + αn zn − f δn . In their turn, (6.5.2) and (6.5.5) yield the inequality

J(xn − zn ), Ahn xn + αn zn − f δn ≥ − δn + hn φ(xn ) xn − zn + J(xn − zn ), Axn − Azn + J(xn − zn ), Azn + αn zn − f . By virtue of the equality Azn + αn zn = f and by the accretiveness of A, we have

J(xn − zn ), Ahn xn + αn zn − f δn ≥ − δn + hn φ(xn ) xn − zn . Thus,

J(xn − zn ), xn+1 − xn ≤ −n αn xn − zn 2 + n δn + hn φ(xn ) xn − zn .

(6.5.42)

6.5


351

Let the sequences {xn } and {zn } be bounded by constants R and K0 , respectively. Then we conclude that zn − xn ≤ K0 + R = d5 , n = 0, 1, 2, ... . There exist constants d6 > 0 and d7 > 0 such that J(xn+1 − zn ) − J(xn − zn ), xn+1 − xn ≤ 8xn+1 − xn 2 + d6 ρX (xn+1 − xn ) and ¯ ¯ n ≤ d7 n , xn+1 − xn ≤ (Axn + hφ(R) + αR ¯ + f + δ) provided that the operator A is bounded. Finally, we deduce from (6.5.42) the following inequality: xn+1 − zn+1 2 ≤ xn − zn 2 − 2n αn xn − zn 2

+ 2n d5 δn + hn φ(R) + 16d27 2n + 2d6 ρX (d7 n ) + 4d5 K0

|αn+1 − αn | . αn

(6.5.43)

Finally, we use Lemma 7.1.3 again to obtain the following theorem. Theorem 6.5.4 A bounded sequence {xn } generated by the algorithm of iterative regularization (6.5.37) in a uniformly smooth Banach space X, which possesses an approximation, converges strongly to the unique solution x ¯∗ defined by the inequality (2.7.7) if the equation Ax = f is solvable and 1) A and Ahn are bounded demicontinuous accretive operators, D(A) = D(Ahn ) = X, conditions (6.5.2) (6.5.5) and (6.5.7) are satisfied; 2) J is the continuous and weak-to-weak continuous mapping; 3)

4)

lim αn = 0;

n→∞ ∞

αn n = ∞;

n=1

5)

6) 7)

lim

δn + hn + n = 0; αn

lim

ρX (n ) = 0; n αn

lim

|αn − αn+1 | = 0. αn2 n

n→∞

n→∞

n→∞

Remark 6.5.5 If it is not known a priori that the sequence {xn } is bounded, then it is necessary to use the proof scheme of Theorem 6.5.2.

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352

6.6


Iterative-Projection Regularization Method ∗

Suppose that X is uniformly convex Banach space, A : X → 2X is a maximal monotone bounded operator with domain D(A) and χ-growth (6.5.1). In place of the equation (6.1.1), we study in this section the variational inequality problem: To find y ∈ Ω such that Ay − f, x − y ≥ 0 ∀x ∈ Ω,

(6.6.1)

where Ω ⊂ int D(A) is a convex closed set and f ∈ X ∗ . As in the previous section, we suppose that (6.1.1) has a nonempty solution set N and that A, f and Ω are given with perturbations which we denote, respectively, by Ahn , f δn and Ωn , n = 0, 1, 2, ..., such that D(Ahn ) = D(A), the inequalities (6.5.3) and (6.5.2) hold as x ∈ D(A), Ωn ⊂ int D(A) are convex closed sets and HX (Ωn , Ω) ≤ σn . (6.6.2) We assume that there exist a convex function δ˜X (t) such that δX (t) ≥ δ˜X (t) for all 0 ≤ t ≤ 2 and increasing function gX (t) = t−1 δX (t). 1. We construct the iterative regularization method in the form: Jxn+1 = Jxn − m (¯ ηnhm + αm J x ¯n − f δm + qm ), n = 0, 1, 2, ... ,

(6.6.3)

¯n , m = n + n0 , where η¯nhm ∈ Ahm x qm = (αm + αm ¯ xn + ¯ ηnhm − f δm ∗ )

J(xn − x ¯n ) , n x − x ¯n

(6.6.4)

and x ¯n = PΩm xn is the metric projection of xn ∈ X onto Ωm , n0 ≥ 0 is defined by (6.6.7) below. Suppose that step parameters n , regularization parameters αn and perturbation parameters δn , hm and σn , describing, respectively, (6.5.2), (6.5.3) and (6.6.2), are positive for all n ≥ 0 and vanish as n → ∞, αn+1 ≤ αn , n+1 ≤ n , and (6.5.7) holds together with σm ≤ σ ¯. We introduce the intermediate variational inequality problem: To find z ∈ Ω such that Az + αJz − f, x − z ≥ 0 ∀x ∈ Ω, z ∈ Ω,

α > 0.

(6.6.5)

In Section 4.1 we have shown that its solution zα exists and is unique for each α > 0, the ¯∗ , where x ¯∗ ∈ N is sequence {zα } is bounded for all α > 0, say zα ≤ K0 , and lim zα = x α→0 a minimal norm solution. Let R0 be any non-negative number. We introduce the following constants: K1 =

2R0 + K0 , K2 = K1 + ¯c1 , K3 = K0 + 2K1 ,

¯ ¯ 3 + δ¯ + f ∗ , c0 = 2max{1, K0 }, c1 = χ(K3 ) + hφ(K 3 ) + αK c2 = 8max{L, K2 },

1 < L < 1.7, c3 = 2K2 ,

c4 = 2Lmax{1, K0 , 4L(χ(K0 ) + αK ¯ 0 )}, c5 = 2max{1, K3 },

Iterative-Projection Regularization Method

6.6

c6 = 2c0 LK0 ,

353

C = min (2L)−1 , c5 gX c−1 5 (K1 + K3 )

−1

.

Construct the numerical sequence γk , k = 0, 1, 2, ... as follows:

γk = k hk φ(K3 ) + δk (K0 + K3 ) + 8c21 2k + c2 ρX ∗ (c1 n )

+

−1 c0 c3 c4 gX

αk − αk+1 c6 +c αk

Choose n0 according to the condition −1 n0 = min{k | 2c4 δ˜X

(

γk 2Cαk k

σk + αk

"

σk+1 αk+1

.

≤ −1 (R0 )},

(6.6.6)

(6.6.7)

where (τ ) = 8τ 2 + c2 ρX ∗ (τ ). Let zn0 be a solution of the variational inequality (6.6.5) with α = αn0 and let the initial point x0 in (6.6.3) satisfy the inequality W (x0 , zn0 ) ≤ R0 , where W (x, z) is defined by (6.5.13). We need a more general statement than Lemma 6.5.1. Lemma 6.6.1 Suppose that 1) X is a uniformly convex Banach space; 2) sequences {zα1 } and {zα2 } of solutions of variational inequalities T1 z + α1 Jz, x − z ≥ 0 ∀x ∈ Ω1 ,

z ∈ Ω1 ,

(6.6.8)

T2 z + α2 Jz, x − z ≥ 0 ∀x ∈ Ω2 ,

z ∈ Ω2 ,

(6.6.9)

and are bounded for all α1 > 0 and for all α2 > 0, respectively, that is, there exists K0 > 0 such that zα1 ≤ K0 , zα2 ≤ K0 ; ∗ 3) an operator T1 : X → 2X is maximal monotone on D(T1 ) ⊆ X and bounded on the sequences {zα1 } and {zα2 }, i.e., there exists a constant M > 0 such that ζ1 ∗ ≤ M for all ζ1 ∈ T1 zα1 and ξ1 ∗ ≤ M for all ξ1 ∈ T1 zα2 ; ∗ 4) an operator T2 : X → 2X is maximal monotone on D(T2 ) ⊆ X, D(T1 ) = D(T2 ) = Q and HX ∗ (T1 z, T2 z) ≤ ω for all z ∈ Ω2 ; 5) Ωi ⊂ int Q (i = 1, 2) are convex closed sets such that HX (Ω1 , Ω2 ) ≤ σ. Then the following estimate holds:

−1 zα1 − zα2 ≤ c0 gX c6

ω |α1 − α2 | + c8 + c7 α1 α1

(

σ , α1

(6.6.10)

where c6 = 2c0 LK0 , c7 = 2c0 L, c8 = max{1, 2Lc9 } and c9 = 2M + (α1 + α2 )K0 + ω. Proof. Since zα1 is a solution of (6.6.8) and zα2 is a solution of (6.6.9), there exist ζ1 ∈ T1 zα1 and ζ2 ∈ T2 zα2 such that ζ1 + α1 Jzα1 , x − zα1 ≥ 0 ∀x ∈ Ω1

(6.6.11)

6

354


and ζ2 + α2 Jzα2 , x − zα2 ≥ 0 ∀x ∈ Ω2 .

(6.6.12)

We compose the following expression: D = ζ1 + α1 Jzα1 − ζ2 − α2 Jzα2 , zα1 − zα2 . Regarding the condition 4), there exists ξ1 ∈ T1 zα2 such that ζ2 − ξ1 ∗ ≤ ω. Then, due to the monotonicity of T1 and properties of duality mapping J, we deduce D = ζ1 − ξ1 + α1 (Jzα1 − Jzα2 ) + ξ1 − ζ2 + (α1 − α2 )Jzα2 , zα1 − zα2 ≥ α1 (2L)−1 δX (c−1 0 zα1 − zα2 ) − ζ2 − ξ1 ∗ zα1 − zα2 − |α1 − α2 |zα2 zα1 − zα2

≥ −zα1 − zα2 (ω + K0 |α1 − α2 |) + α1 (2L)−1 δX c−1 0 zα1 − zα2 . (6.6.13) On the other hand, since HX (Ω1 , Ω2 ) ≤ σ, we assert that for zα2 ∈ Ω2 there exists z1 ∈ Ω1 such that zα2 − z1 ≤ σ and ζ1 + α1 Jzα1 , zα1 − z1 ≤ 0 because of (6.6.11). Therefore, ζ1 + α1 Jzα1 , zα1 − zα2 = ζ1 + α1 Jzα1 , zα1 − z1 + ζ1 + α1 Jzα1 , z1 − zα2 ≤ (ζ1 ∗ + α1 zα1 )σ ≤ (M + α1 K0 )σ. Analogously, by (6.6.12), the following estimate is obtained: ζ2 + α2 Jzα2 , zα2 − zα1 ≤ (ζ2 ∗ + α2 K0 )σ. It is obvious that ζ2 ∗ ≤ ξ1 ∗ + ζ2 − ξ1 ∗ ≤ M + ω. From this one gets that D ≤ (2M + (α1 + α2 )K0 + ω) σ = c9 σ. The last inequality and (6.6.13) yield

c9 σ + zα1 − zα2 (ω + K0 |α1 − α2 |) ≥ α1 (2L)−1 δX c−1 0 zα1 − zα2 .

(6.6.14)

Consider two possible cases:

−1 c6 (i) zα1 − zα2 ≤ c0 gX

|α1 − α2 | ω + c7 + α1 α1

(

σ , α1

(6.6.15)

6.6


−1 c6 (ii) zα1 − zα2 > c0 gX

355

|α1 − α2 | ω + c7 + α1 α1

(

σ . α1

Since δX (t) ≤ δH (t) ≤ t2 , where H is a Hilbert space, case (ii) implies zα1 − zα2 ≥ gH c0

zα1 − zα2 c0

≥ gX

zα1 − zα2 c0

(

>

σ . α1

Taking into account (6.6.14) we obtain c0 δX (c−1 0 zα1 − zα2 ) zα1 − zα2

≤ c6

σ 2Lc0 c9 ω |α1 − α2 | + + c7 z − z α1 α1 α1 α 2 α1

≤ c6

|α1 − α2 | ω + c7 + 2Lc9 α1 α1

(

σ . α1

It follows from this relation that

−1 c6 zα1 − zα2 ≤ c0 gX

ω |α1 − α2 | + c7 + 2Lc9 α1 α1

(

σ . α1

Comparing this estimate with (6.6.15) we see that the conclusion of the lemma is true.

Return now to the method (6.6.3), (6.6.4) and denote solutions of the variational inequalities (6.6.16) Az + αm Jz − f, x − z ≥ 0 ∀x ∈ Ωm , z ∈ Ωm , and Az + αm+1 Jz − f, x − z ≥ 0 ∀x ∈ Ωm+1 ,

z ∈ Ωm+1 ,

(6.6.17)

respectively, by zm and zm+1 . Suppose that X is a uniformly convex and uniformly smooth Banach space. Similarly to (6.5.22), one gets W (xn+1 , zm+1 ) − W (xn , zm ) ≤ Jxn+1 − Jxn , xn+1 − xn + Jxn+1 − Jzm+1 ∗ zm+1 − zm + Jxn+1 − Jxn , x ¯ n − zm + Jxn+1 − Jxn , xn − x ¯n . Evaluate each of the four terms in the right-hand side of (6.6.18). A. By (6.6.3) and (6.6.4), we have ηnhm − f δm + αm J x ¯n + q m ∗ Jxn+1 − Jxn ∗ = m ¯ ≤ 2m (¯ ηn ∗ + ¯ ηnhm − η¯n + f δm ∗ + αm ¯ xn + αm ) ≤ c1 (n)m ,

(6.6.18)

6

356


xn such that ¯ ηn − η¯nhm ∗ ≤ hm φ(¯ xn ) and where η¯n ∈ A¯ c1 (n) = 2(χ(¯ xn ) + hm φ(¯ xn ) + αm ¯ xn + f ∗ + αm + δm ). Due to (1.6.17), Jxn+1 − Jxn , xn+1 − xn ≤ 8Jxn+1 − Jxn 2∗ + c2 (n)ρX ∗ (Jxn+1 − Jxn ∗ ), where ρX ∗ (τ ) is the modulus of smoothness of X ∗ and c2 (n) = 8max{L, xn+1 , xn }. Therefore,

Jxn+1 − Jxn , xn+1 − xn ≤ 8c21 (n)2m + c2 (n)ρX ∗ (c1 (n)m ).

B. In its turn, by Lemma 6.6.1, Jxn+1 − Jzm+1 ∗ zm+1 − zm

≤

−1 c0 c3 (n)gX

αm − αm+1 c6 + c4 αm

(

σm + αm

"

σm+1 αm+1

,

where c3 (n) = 2max{xn+1 , zm+1 }. We assumed here that the sequence {αn } is non-increasing, also used the estimate (6.6.10) with ω = 0 and the following relations: HX (Ωn , Ωn+1 ) ≤ σn + σn+1 and

"

σn+1 + σn ≤ αn

(

σn + αn

"

σn+1 . αn+1

C. It is possible to verify that ¯n − zm = −m ¯ ηnhm − f δm + αm J x ¯n + qm , x ¯n − zm Jxn+1 − Jxn , x = −m ¯ ηnhm − η¯n , x ¯n − zm − m αm J x ¯n − Jzm , x ¯ n − zm ηn − ξm , x ¯n − zm − m ξm + αm Jzm − f, x ¯ n − zm − m ¯ − m f − f δm , x ¯n − zm − m qm , x ¯n − zm , where ξm ∈ Azm . Recall that ξm + αm Jzm − f, zm − z ≤ 0 ∀z ∈ Ωm .

(6.6.19)

6.6

357


Take into consideration the following inequalities: ¯n − zm ≥ 0, ¯ ηn − ξm , x ξm + αm Jzm − f, x ¯n − zm ≥ 0, J x ¯n − Jzm , x ¯n − zm ≥ (2L)−1 δX

¯ x n − zm

c5 (n)

,

where xn , zm }, c5 (n) = 2max{1, ¯ and qm , x ¯n − zm ≥ 0, valid due to the monotonicity of operator A, (6.6.19), (1.6.28), and (1.5.12), respectively. Then we come to the estimate

¯n − zm ≤ m hm φ(¯ xn ) + δm ¯ xn − zm Jxn+1 − Jxn , x − m αm (2L)−1 δX

¯ x n − zm

c5 (n)

.

D. Finally, the last term in (6.6.18) is evaluated as Jxn+1 − Jxn , xn − x ¯n = −αm m J x ¯n , xn − x ¯n ¯n − m ¯ ηnhm − f δm , xn − x ¯n − m qm , xn − x ≤ m αm ¯ xn ¯ xn − xn − m αm ¯ xn − xn − m αm ¯ xn xn − x ¯n − m ¯ ηnhm − f δm ∗ ¯ xn − xn ηnhm − f δm ∗ xn − x ¯n + m ¯ = −m αm ¯ xn − xn . Combination of A - D gives the following result: W (xn+1 , zm+1 ) − W (xn , zm ) #

≤ −m αm (2L)−1 δX

¯ xn − zm c5 (n)

$

+ xn − x ¯n + Υm ,

(6.6.20)

6

358


where xn )¯ xn − zm + m δm ¯ xn − zm + 82m c21 (n) + c2 (n)ρX ∗ (c1 (n)m ) Υm = m hm φ(¯ −1 c6 + c0 c3 (n)gX

αm − αm+1 + c4 αm

(

σm + αm

"

σm+1 αm+1

.

(6.6.21)

Since W (x0 , zn0 ) ≤ R0 , it is possible to consider n ≥ 0 such that W (xn , zn+n0 ) ≤ R0 . For these n we have xn ≤ K1 . Observe that x ¯n ∈ Ωm and zm ∈ Ωm . This enables us to derive the estimate xn − xn + xn ≤ zm − xn + xn ≤ 2xn + zm ≤ 2K1 + K0 = K3 . ¯ xn ≤ ¯ Therefore, if (6.5.7) is satisfied then ¯ c1 (n) ≤ 2(χ(K3 ) + hφ(K ¯ 3+α ¯ + δ¯ + f ∗ ) = c1 . 3 ) + αK Moreover, similarly to (6.5.28), xn+1 ≤ K1 + m c1 (n) ≤ K1 + ¯c1 = K2 . Now it can be verified that cs (n) ≤ cs , s = 2, 3, 5. Thus, it follows from (6.6.20) and (6.6.21)) that

W (xn+1 , zm+1 ) ≤ W (xn , zm ) − m αm (2L)−1 δX (c−1 xn − zm ) + xn − x ¯n + γm , 5 ¯ where γm is calculated by the formula (6.6.6) with k = m. It is obvious that n δX (c−1 ¯n ) xn − x ¯n 5 x − x . = −1 n c5 gX (c5 x − x ¯n ) At the same time, one gets n gX (c−1 ¯n ) ≤ gX (c−1 5 x − x 5 (K1 + K3 )) = c10

because xn − x ¯n ≤ K1 + K3 . Hence, n n ¯n ≥ c5 c10 −1 δX (c−1 ¯n ) ≥ c5 c10 −1 δ˜X (c−1 ¯n ). xn − x 5 x − x 5 x − x

The last inequality implies xn − zm ) + xn − x ¯n ≥ (2L)−1 δ˜X (c−1 xn − zm ) (2L)−1 δX (c−1 5 ¯ 4 ¯ ˜ −1 n ¯n ) + c5 c−1 10 δX (c4 x − x

n ≥ C δ˜X (c−1 xn − zm ) + δ˜X (c−1 ¯n ) . 4 ¯ 4 x − x

6.6 Iterative-Projection Regularization Method

359

Since the function δ˜X (t) is convex and increasing for all 0 ≤ t ≤ 2, we can write (2L)−1 δX

¯ xn − zm c5

+ xn − x ¯n ≥ 2C δ˜X

≥ 2C δ˜X

¯ xn − zm + xn − x ¯n 2c5

xn − zm . 2c5

Thus, we come to the following numerical inequality:

W (xn+1 , zm+1 ) ≤ W (xn , zm ) − 2Cm αm δ˜X

xn − zm 2c5

+ γm .

Assume that for all m ≥ n0 , αm − αm+1 ≤ d1 , αm

σm ≤ d2 . αm

(6.6.22)

Then W (xn+1 , zm+1 ) ≤ R0 + γ¯ , where −1 2 2 ¯ ¯ γ¯ = ¯(hφ(K ¯ + c2 ρX ∗ (c1 ¯) + c0 c3 gX (c6 d1 + 2c4 d2 ). 3 ) + δ)(K0 + K3 ) + 8c1

By analogy with the proof of (6.5.35), one can show that the sequence {xn } is bounded by a constant which does not depend on n. Further the proof can be done by the same scheme as in Theorem 6.5.2. Thus, the following statement holds: Theorem 6.6.2 Suppose that 1) X is a uniformly convex and uniformly smooth Banach space with the modulus of convexity δX () and there exists a convex increasing function δ˜X () such that δX () ≥ δ˜X () for all 0 ≤ t ≤ 2; 2) x ¯∗ is the minimal norm solution of variational inequality (6.6.1); ∗ 3) A : X → 2X is a maximal monotone bounded operator with χ-growth (6.5.1), f ∈ X ∗ ; ∗ 4) A and f are given with perturbations as Ahn : X → 2X and f δn ∈ X ∗ satisfying the hn conditions (6.5.2) and (6.5.3), where D(A ) = D(A); 5) Ω is also known approximately with the estimate HX (Ω, Ωn ) ≤ σn , where Ω and Ωn are convex closed sets in X, Ω ⊂ int D(A) and Ωn ⊂ int D(A); 6) in the method of iterative regularization (6.6.3), (6.6.4), the initial approximation x0 satisfies the inequality W (x0 , zn0 ) ≤ R0 , where zn0 is a solution of the variational inequality (6.6.16) with m = n0 and n0 obeys the rule (6.6.7); ¯ and 7) parameters αn , n , δn , hn , σn are such that (6.5.7) is valid with σn ≤ σ 8)

9)

lim αn = 0;

n→∞ ∞

n=1

αn n = ∞;

360

6


10) 11)

lim

ρX ∗ (n ) = 0; n αn

lim

hn + δn + σn + n = 0; αn

n→∞

n→∞

&

12)

'

−1 gX αn−1 |αn − αn+1 | = 0. n→∞ αn n

lim

Moreover, {αn } does not increase and (6.6.22) holds. Then the sequence {xn } generated by (6.6.3) and (6.6.4) strongly converges to x ¯∗ as n → ∞. Remark 6.6.3 The requirement δX () ≥ δ˜X () in 1) is not too restrictive. For instance, p , 1 < p < ∞, the moduli of convexity δ () ≥ c 2 if 1 < p ≤ 2 in the spaces lp , Lp and Wm 1 X and δX () ≥ c2 p if 2 ≤ p < ∞ with some constants c1 , c2 > 0, that is, δ˜X () = c1 2 and δ˜X () = c2 p , respectively. Next we omit the assumption 2) of the previous theorem and prove the existence of a solution to variational inequality (6.6.1) provided that the regularization process (6.6.3), (6.6.4) converges. Theorem 6.6.4 Assume that in a uniformly convex and uniformly smooth Banach space X, the iterative sequence {xn } generated by (6.6.3), (6.6.4), where non-negative parameters hn , δn , σn and positive parameters αn , n approach zero as n → ∞ and

∞

αn n = ∞,

n=0

strongly converges to an element x∗ ∈ Ω. Then x∗ is a solution of the variational inequality (6.6.1). Proof. Without loss of generality, we suppose that hn = δn = 0, the operator A is one-to-one and n0 = 0. Let x∗ be not a solution of (6.6.1). Then there exists x ∈ Ω and c > 0 such that Ax − f, x − x∗ = −c. (6.6.23) Represent (6.6.23) in the equivalent form −c = Ax − f + αn Jx, x − x∗ − αn Jx, x − x∗ . Since αn → 0, there exists a number k1 > 0 such that for every n ≥ k1 αn Jx, x − x∗ ≤

c . 2

In this case, for all n ≥ k1 we have c Ax + αn Jx − f, x − x∗ ≤ − . 2

(6.6.24)

6.6


361

Represent now (6.6.24) in the equivalent form c ¯n + Ax + αn Jx − f, x ¯n − x∗ ≤ − , Ax + αn Jx − f, x − x 2

(6.6.25)

where x ¯n = PΩn xn . By the conditions, xn → x∗ . Therefore, on account of the inclusion ∗ ¯n → x∗ holds too (see Lemma 6.3.1). Then there exists an x ∈ Ω, the limit relation x integer k2 > 0 such that for all n ≥ k2 , |Ax + αn Jx − f, x ¯n − x∗ | ≤ Ax + αn Jx − f ∗ ¯ xn − x∗ ≤

c 4

because Ax + αn Jx − f ∗ is bounded. Hence, by (6.6.25), for all n ≥ max{k1 , k2 }, c ¯n ≤ − . Ax + αn Jx − f, x − x 4 Finally, using the monotonicity of A and the property (1.5.3) of J, we obtain the inequality −

c ¯n − f, x − x ¯n + αn (x − ¯ xn )2 . ≥ A¯ xn + αn J x 4

Then there exists an integer k3 > 0 such that for all n ≥ k3 , −

c ¯n − f, x − x∗ + αn (x − x∗ )2 + βn , ≥ A¯ xn + αn J x 4

where βn = A¯ xn + αn J x ¯n − f, x∗ − x ¯n − 2αn (x − x∗ )(¯ xn − x∗ ) + αn (x∗ − ¯ xn )2 ≤

c . 8

Hence, for all n ≥ max{k1 , k2 , k3 }, c ¯n − f, x − x∗ + αn (x − x∗ )2 , xn + αn J x − ≥ A¯ 8 or

c ¯n − f + qn , x − x∗ + αn (x − x∗ )2 − qn , x − x∗ . ≥ A¯ xn + αn J x 8 Estimate the last term. By virtue of the fact that HX (Ω, Ωn ) ≤ σn , there exists vn ∈ Ωn such that vn − x ≤ σn , where x satisfies (6.6.23). Let −

xn + A¯ xn − f ∗ . cn = αn + αn ¯ Then qn , x − x∗ = cn + cn

J(xn − x ¯n )

¯ xn

−

xn

J(xn − x ¯n )

¯ xn

−

xn

, x − x ∗ = cn

J(xn − x ¯n )

¯n + cn , vn − x

¯ xn

−

xn

,x ¯ n − x∗

J(xn − x ¯n )

¯ xn

−

xn

, x − vn .

6

362


In view of Lemma 1.5.17, ¯n ), vn − x ¯n ≤ 0. J(xn − x Therefore,

qn , x − x∗ ≤ cn ¯ xn − x∗ + x − vn ≤ cn ¯ xn − x∗ + σn , and there exists k4 > 0 such that for all n ≥ k4 , qn , x − x∗ ≤

c . 16

Consequently, if n ≥ max{k1 , k2 , k3 , k4 } = k5 then 0≥−

c ¯n + qn − f, x − x∗ + αn (x − x∗ )2 . ≥ A¯ xn + αn J x 16

By (6.6.3), for n ≥ k5 , Jxn+1 − Jxn , x − x∗ ≥ αn n (x − x∗ )2 .

(6.6.26)

Due to the strong convergence of {xn } to x∗ and continuity of J, the limit relation Jxn → Jx∗ holds. Hence, the series standing in the left-hand side of the inequality ∞

Jxn+1 − Jxn , x − x∗ ≤ x − x∗

n=0

∞

Jxn+1 − Jxn ∗

n=0

converges. Then (6.6.26) implies that ∞

αn n < ∞,

n=0

which contradicts the assumption of the theorem. Hence, x∗ is a solution of the variational inequality (6.6.1). 3. Constructing the sequence {xn }, we make use of the projection operation onto a set Ω. If Ω is given in the form of functional inequalities, then the effective method of solving variational inequalities is the method of indefinite Lagrange multipliers. Consider the general situation when Ω = Ω1 ∩ Ω2 , where it is assumed that the projection operation onto the first set Ω1 is done quite easily, and the second set Ω2 is given by the following constraint system: Ω2 = {x ∈ X | ϕi (x) ≤ 0, i = 1, 2, ..., l}, where ϕi (x) are continuous convex functionals on X. It is possible that Ω1 = X. Then we define the iterative process of finding a solution x ¯∗ of the variational inequality (6.6.1) as Jxn+1 = Jxn − n (Φ(xn ) + αn Jxn ) and λn+1 = {λni + n (ϕi (x) − αn λni )}+ ,

(6.6.27)

6.7 Continuous Regularization Method

363

where Φ(x) = Ax − f +

l

λni ϕi (x)

i=1

and {a}+ = max{0, a}. If Ω1 = X then we replace (6.6.27) by the expression

Jxn+1 = Jxn − n Φ(¯ xn ) + αn J x ¯n + qn , where x ¯n = PΩ1 xn and xn + Φ(¯ xn )∗ ) qn = (αn + αn ¯

J(xn − x ¯n ) . n x − x ¯n

The convergence proof is similar to the proof of Theorem 6.5.2 including the situation when an operator A, element f and sets Ω1 and Ω2 are given with perturbations. Observe that the iterative process (6.6.27) with αn = 0 is not stable, in general, even if A is strongly monotone.

6.7

Continuous Regularization Method

Continuous regularization processes for solving ill-posed equation (6.1.1) deal with ordinary differential equations in which the role of “regularization” parameter is performed by a certain positive function α(t) with t ≥ t0 . The stated methods are reduced to the Cauchy problem for a differential equation of some order. The order of the differential equation is called the order of the continuous method. In the present section we study the first order continuous regularization methods. We emphasize that, in the sequel, existence of solutions to any differential equation of this section is assumed. We investigate the continuous regularization methods separately for equations with monotone and accretive operators in Hilbert and Banach spaces. 1. Let the equation (6.1.1) be given in a Hilbert space H, A : H → H be a monotone continuous operator, D(A) = H and f ∈ H. Let (6.1.1) have a nonempty solution set N and x ¯∗ be its minimal norm solution. Under these conditions, the set N is convex and closed and there exists a unique element x ¯∗ ∈ N. Assume that monotone continuous operator A(t) : H → H and right-hand side f (t) ∈ H are perturbations of A and f, respectively, such that A(t)x − Ax ≤ h(t)g(x) ∀x ∈ H, ∀t ≥ t0 , (6.7.1) and f (t) − f ≤ δ(t) ∀t ≥ t0 .

(6.7.2)

We suppose that α(t) → 0, δ(t) → 0, h(t) → 0 as t → ∞ and g(s) is a continuous, non-negative and non-decreasing function for all s ≥ 0. Consider the differential equation dy(t) + A(t)y(t) + α(t)y(t) = f (t), dt

t ≥ t0 , y(t0 ) = y0 .

(6.7.3)

6

364


Our aim is to prove the strong convergence of y(t) to x ¯∗ as t → ∞. The research scheme is the same as in the iterative regularization algorithms. We introduce the intermediate equation Aw(t) + α(t)w(t) = f. (6.7.4) It is known that w(t) ≤ ¯ x∗ and ¯∗ . lim w(t) = x

(6.7.5)

¯∗ , y(t) − x ¯∗ ≤ y(t) − w(t) + w(t) − x

(6.7.6)

t→∞

Obviously,

so that we need only to prove that y(t) − w(t) → 0 as t → ∞ and then

¯∗ = 0. lim y(t) − x

t→∞

Theorem 6.7.1 Suppose that (i) all the conditions of the present subsection are fulfilled; (ii) a solution y(t) of the differential equation (6.7.3) exists on the interval [t0 , +∞); (iii) a solution w(t) of the operator equation (6.7.4) (which necessarily exists) is differentiable on the interval [t0 , ∞); (iv) α(t) is a positive continuous and differentiable function satisfying the following limit relations: ∞ |α (t)| h(t) + δ(t) α(τ )dτ = ∞. (6.7.7) = 0, lim = 0, lim α(t) = 0, lim 2 t→∞ t→∞ t→∞ α (t) α(t) t0

¯∗ . Then lim y(t) = x t→∞

Proof. Denote

V (y(t), w(t)) = 2−1 y(t) − w(t)2 .

It is easy to see that ∂V (y(t), w(t)) = y(t) − w(t) ∂y and

∂V (y(t), w(t)) = w(t) − y(t). ∂w

Then dV (y(t), w(t)) dt

∂V (y(t), w(t)) dy(t)

∂V (y(t), w(t)) dw(t)

, dt ∂w dt ∂y = −(A(t)y(t) − A(t)w(t), y(t) − w(t)) − α(t)y(t) − w(t)2

=

,

+

− (Aw(t) + α(t)w(t) − f, y(t) − w(t)) + (f (t) − f, y(t) − w(t)) + (Aw(t) − A(t)w(t), y(t) − w(t)) + (w(t) − y(t), w (t)).


365

By the monotonicity of A(t), we have (A(t)y(t) − A(t)w(t), y(t) − w(t)) ≥ 0. Since w(t) satisfies (6.7.4), one gets (Aw(t) + α(t)w(t) − f, y(t) − w(t)) = 0. Using (6.7.1) and (6.7.2) we deduce (Aw(t) − A(t)w(t), y(t) − w(t)) ≤ h(t)g(¯ x∗ )y(t) − w(t) and (f (t) − f, y(t) − w(t)) ≤ δ(t)y(t) − w(t). Hence, dV (y(t), w(t)) ≤ −α(t)y(t) − w(t)2 + δ(t) + h(t)g(¯ x∗ ) + w (t) y(t) − w(t). (6.7.8) dt Evaluate further w (t) from above. By (6.5.19),

w(t1 ) − w(t2 ) ≤

¯ x∗ |α(t1 ) − α(t2 )| . α(t1 )

This yields the inequality w(t ) − w(t ) ¯ x∗ α(t1 ) − α(t2 ) 1 2 . ≤

t 1 − t2

t1 − t2

α(t1 )

Owing to the differentiability of w(t) and α(t), we obtain dw(t) |α (t)| . x∗ ≤ ¯

dt

α(t)

(6.7.9)

Then it follows from (6.7.8) that

|α (t)| dy(t) − w(t) . x∗ ≤ −α(t)y(t) − w(t) + δ(t) + h(t)g(¯ x∗ )) + ¯ α(t) dt Denoting here λ(t) = y(t) − w(t) we come to the differential inequality dλ(t) ≤ −α(t)λ(t) + γ(t), dt where

|α (t)| . α(t) In order to show that λ(t) → 0, it is necessary to be sure that the conditions of Lemma 7.2.2 hold. Indeed, we see that α(t) and γ(t) satisfy the lemma. Next, taking into account condition (iv) of the theorem, we obtain γ(t) = δ(t) + h(t)g(¯ x∗ ) + ¯ x∗

lim

t→∞

γ(t) = 0. α(t)

Thus, y(t) − w(t) → 0 as t → ∞. By (6.7.6), the theorem is proved.

(6.7.10)

6

366


Corollary 6.7.2 Under the conditions of the previous theorem, if the function α(t) is decreasing, then the last relation in (iv) can be omitted. Proof. Show that the limit equality |α (t)| =0 t→∞ α2 (t)

(6.7.11)

lim

implies

∞

α(τ )dτ = +∞.

(6.7.12)

t0

Indeed, since α(t) is decreasing, we conclude from (6.7.11) that there exists a constant c > 0 such that α (t) ≤ c. − 2 α (t)

Then

1 1 ≤ c(t − t0 ), − α(t) α(t0 )

that leads to the estimate α(t) ≥

α(t0 ) . 1 + cα(t0 )(t − t0 )

Thus, (6.7.12) results. The following functions α(t) and γ(t) satisfy (6.7.10) - (6.7.12): 1) α(t) = c1 t−r , γ(t) = c2 e−st , 0 < r < 1, s > 0, c1 > 0, c2 > 0; 2) α(t) = c1 t−r , γ(t) = c2 t−s , 0 < r < 1, s > r, c1 > 0, c2 > 0; 3) α(t) = c1 t−r , γ(t) ≡ 0, 0 < r < 1, c1 > 0. 2. Next we assume that α(t) is a convex function. In this case we do not need to require differentiability of a solution w(t) to the equation (6.7.4). So, let A : H → H be a monotone continuous operator, α(t) be a positive differentiable convex and decreasing function for all t ≥ t0 such that lim α(t) = 0 and (6.7.11) holds. Let t→∞

A(t) : H → H, t ≥ t0 , be a family of monotone continuous operators with D(A(t)) = H satisfying, as earlier, the conditions (6.7.1) and (6.7.2). Consider the Cauchy problem (6.7.3) again and suppose for simplicity that it has a unique solution y(t) defined for all t ≥ t0 . We rewrite the regularized operator equation (6.7.4) in the form: Aw(τ ) + α(τ )w(τ ) = f.

(6.7.13)

For every fixed τ ≥ t0 , we also construct the auxiliary Cauchy problem dz(t, τ ) + Az(t, τ ) + α(τ )z(t, τ ) = f, z(t0 , τ ) = y0 . dt Denote

r(t, τ ) = 2−1 z(t, τ ) − w(τ )2 .

(6.7.14)

6.7 Continuous Regularization Method It is clear that dr(t, τ ) = dt

367

dz(t, τ ) , z(t, τ ) − w(τ ) . dt

The equations (6.7.13) and (6.7.14) involve the scalar equality

dz(t, τ ) , z(t, τ ) − w(τ ) dt

+ (Az(t, τ ) − Aw(τ ), z(t, τ ) − w(τ )) + α(τ )z(t, τ ) − w(τ )2 = 0.

(6.7.15)

Since A is a monotone operator, the following differential inequality is obtained from (6.7.15): dr(t, τ ) ≤ −2α(τ )r(t, τ ) dt with r(t0 , τ ) = 2−1 y0 − w(τ )2 = r0 (τ ). Now we have from Lemma 7.2.2 the estimate

r(t, τ ) ≤ r0 (τ )exp − 2α(τ )(t − t0 ) .

(6.7.16)

By the hypotheses, w(τ ) is bounded by ¯ x∗ , therefore, there exists a constant c > 0 such that r0 (τ ) ≤ c for any τ ≥ t0 . Then, in view of (6.7.16), we conclude that the trajectories z(t, τ ) are bounded for all t ≥ t0 and all τ ≥ t0 . Consequently, there exists c1 > 0 such that z(t, τ ) ≤ c1 for all t ≥ t0 and for all τ ≥ t0 . Using now (6.7.16) with t = τ, one gets

r(τ, τ ) ≤ c exp − 2α(τ )(τ − t0 ) ,

(6.7.17)

where the argument of the exponential function becomes indefinite as τ → +∞. Taking into account (6.7.7) and applying L’Hopital’s rule, we obtain from (6.7.17) the limit relation r(τ, τ ) → 0 as τ → ∞.

(6.7.18)

Along with (6.7.3), consider the following problem with the exact data A and f : du(t) + Au(t) + α(t)u(t) = f, u(t0 ) = y0 . dt

Define the functions v(t, τ ) = u(t) − z(t, τ ) and v¯(t, τ ) = 2−1 v 2 (t, τ ). Then d¯ v (t, τ ) = dt

dv(t, τ ) du(t) dz(t, τ ) . , u(t) − z(t, τ ) = v(t, τ ) − dt dt dt

(6.7.19)

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368


Further, (6.7.14) and (6.7.19) yield d¯ v (t, τ ) dt

+ (Au(t) − Az(t, τ ), u(t) − z(t, τ )) + α(t)v 2 (t, τ ) + (α(t) − α(τ ))(z(t, τ ), u(t) − z(t, τ )) = 0.

(6.7.20)

By the monotonicity of A and boundedness of z(t, τ ), (6.7.20) is reduced to the inequality d¯ v (t, τ ) ≤ −α(t)v2 (t, τ ) + c1 |α(t) − α(τ )|v(t, τ ), dt

(6.7.21)

which implies a similar differential inequality for the function v(t, τ ), namely, dv(t, τ ) ≤ −α(t)v(t, τ ) + c1 |α(t) − α(τ )| ∀τ ≥ t0 dt

(6.7.22)

with the initial condition v(t0 , τ ) = 0. The following expression is obtained from convexity of α(t) : |α(t) − α(τ )| ≤ |α (t)|(τ − t), τ > t, and (6.7.22) can be rewritten as dv(t, τ ) ≤ −α(t)v(t, τ ) + c1 |α (t)|(τ − t). dt It is not difficult to see that Lemma 7.2.2 gives now the estimate v(τ, τ ) ≤

c1

)τ t0

|α (s)|(τ − s)exp

exp

If there exists a constant C > 0 such that

τ t0

|α (s)|(τ − s)exp

)

τ t0

s t0

)

s t0

α(θ)dθ

α(θ)dθ ds

.

(6.7.23)

α(θ)dθ ds ≤ C,

then the condition (6.7.12) guarantees that v(τ, τ ) → 0 as τ → ∞.

(6.7.24)

Otherwise, (6.7.24) can be satisfied if we twice apply L’Hopital’s rule to the right-hand side of the inequality (6.7.23) and use (6.7.11). Observe that boundedness of the trajectory u(t) for all t ≥ t0 also follows from (6.7.24). Let p(t) = y(t) − u(t) and p¯(t) = 2−1 p2 (t). By (6.7.3) and (6.7.19), we obtain d¯ p(t) dt

+ (A(t)y(t) − A(t)u(t), y(t) − u(t)) + (A(t)u(t) − Au(t), y(t) − u(t)) + α(t)p2 (t) = (f (t) − f, y(t) − u(t)).


369

Since A(t) is a monotone operator, one has d¯ p(t) ≤ (A(t)u(t) − Au(t), y(t) − u(t)) + α(t)p2 (t) + (f − f (t), y(t) − u(t)). dt Now (6.7.1) and (6.7.2) imply d¯ p(t) ≤ −2α(t)¯ p(t) + δ(t) + h(t)g(u(t)) p(t). dt

The trajectory u(t) is bounded, therefore, there exists a constant c2 > 0 such that the previous inequality is rewritten as follows:

or

d¯ p(t) ≤ −2α(t)¯ p(t) + c2 δ(t) + h(t) p(t) dt

(6.7.25)

dp(t) ≤ −α(t)p(t) + c2 δ(t) + h(t) . dt

(6.7.26)

Assume that lim

t→∞

δ(t) + h(t) = 0. α(t)

(6.7.27)

Exploiting Lemma 7.2.2 again, it is easy to verify that lim p(τ ) = 0.

t→∞

(6.7.28)

Now the relations (6.7.5), (6.7.18), (6.7.24), (6.7.28) and the obvious inequality y(τ ) − x ¯∗ ≤ y(τ ) − u(τ ) + u(τ ) − z(τ, τ ) + z(τ, τ ) − w(τ ) + w(τ ) − x ¯∗ allow us to formulate the following theorem: Theorem 6.7.3 Let A and A(t), t ≥ t0 , be monotone continuous operators defined on a Hilbert space H, f ∈ H and f (t) ∈ H for all t ≥ t0 . Let (6.7.1), (6.7.2) and (6.7.27) be satisfied, the operator equation (6.1.1) have a nonempty solution set N and x ¯∗ be its minimal norm solution. Suppose that α(t) is a positive convex differentiable and decreasing to zero function with the property (6.7.11) and that the Cauchy problems (6.7.3), (6.7.14) and (6.7.19) have unique solutions on the interval [t0 , ∞). Then the trajectory y(t) of the equation (6.7.3) converges strongly to x ¯∗ as t → ∞. 3. Assume that the non-monotone perturbations A(t) : H → H of the operator A satisfy the condition (A(t)x1 − A(t)x2 , x1 − x2 ) ≥ −h1 (t)Ψ(x1 , x2 ) ∀x1 , x2 ∈ H,

(6.7.29)

where a continuous function h1 (t) ≥ 0, h1 (t) → 0 as t → ∞, the function Ψ(x1 , x2 ) is non-negative continuous and bounded, i.e., it carries bounded sets to bounded sets. Then

6

370


if the rest of the conditions of Theorem 6.7.1 hold and if the trajectory y(t) is bounded, we obtain, in place of (6.7.25), the following inequality: d¯ p(t) ≤ −2α(t)¯ p(t) + c2 δ(t) + h(t) p(t) + c3 h1 (t) dt

(6.7.30)

with some constant c3 > 0. Since 2ab ≤ a2 + b2 , we have

c2 δ(t) + h(t) p(t) = c2 p(t) δ(t) + h(t) δ(t) + h(t)

≤ 2−1 c22 δ(t) + h(t) + δ(t) + h(t) p¯(t). Then it follows from (6.7.30) that δ(t) + h(t) d¯ p(t) p¯(t) + c4 δ(t) + h(t) + h1 (t) , ≤ −α(t) 2 − α(t) dt

where c4 = max {2−1 c22 , c3 }. By (6.7.27), there exists c5 > 0 such that d¯ p(t) p(t) + c4 (δ(t) + h(t) + h1 (t)) . ≤ −c5 α(t)¯ dt Consequently, Theorem 6.7.3 is still valid for non-monotone perturbations A(t) provided that (6.7.29) holds with the additional requirement lim

t→∞

h1 (t) = 0. α(t)

4. Assume that the operator A : H → H satisfies the Lipschitz condition and there exists r0 > 0 such that (Ax − f, x) ≥ 0 as x ≥ r0 . (6.7.31) Then the equation (6.7.19) has a unique solution u(t) on an interval [t0 , t¯), t¯ ≤ +∞. We show that in this case t¯ = +∞. Set the contrary assumption: t¯ < +∞. First of all, we prove the inclusion u(t) ∈ B0 (θH , r0 ). Let u(t) ∈ B0 (θH , r0 ) for t1 ≤ t ≤ t2 < t¯ and u(t1 ) = r0 . It is clear that u(t) ∈ H \ B0 (θH , r0 ). Calculate the scalar product of (6.7.19) and u(t). We have 1 du(t)2 + (Au(t) − f, u(t)) + α(t)u(t)2 = 0. dt 2 In view of (6.7.31), du(t) ≤ −α(t)u(t) ∀t ∈ [t1 , t2 ]. dt Now Lemma 7.2.2 yields the estimate

u(t) ≤ u(t1 )exp −

t

α(τ )dτ t1

< u(t1 ),

which contradicts the claim that u(t) ∈ / B0 (θH , r0 ) as t ∈ [t1 , t2 ].


371

Since t2 is arbitrary and t2 > t1 , we conclude that u(t) ∈ B0 (θH , r0 ) on the semiinterval [t1 , t¯). By (6.7.19), this fact implies the boundedness of u (t) on [t0 , t¯). Then for all t, t ∈ [t0 , t¯) one has u(t) − u(t ) ≤ L|t − t |, L = max{u (t) | t ∈ [t0 , t¯)}. Therefore, there exists u ˜ such that ˜ ∈ H. lim u(t) = u t→t¯

Now we can again apply the existence theorem to (6.7.19) with the initial condition u(t¯) = u ˜ in order to be sure that u(t) is defined for t ≥ t¯. The obtained contradiction proves the first claim: t¯ = +∞. In addition, these arguments imply a boundedness of u(t) when t ≥ t0 . Similar results can be established for the equation (6.7.14). It is possible also to prove the existence theorem of a unique bounded solution of the differential equation (6.7.3) if we make the corresponding assumptions for the perturbed operator A(t). Next we will study continuous regularization in Banach spaces. 5. Let X be a reflexive Banach space, X and X ∗ be strictly convex Banach spaces, X possesses an approximation, duality mapping J : X → X ∗ be continuous and weak-to-weak continuous. Consider the equation (6.1.1) with continuous accretive operator A : X → X and regularized differential equation (6.7.3) with continuous accretive operator A(t) : X → X, with this D(A) = D(A(t)) = X. As in Subsection 1 of this section, we assume that solution w(t) of the intermediate equation (6.7.4) is differentiable on the interval [t0 , ∞). Since (6.5.39) holds in the accretive case, an estimate like (6.7.9) is deduced in the following form: dw(t) |α (t)| x∗ . ≤ 2¯ dt α(t)

Let r(t) = y(t) − w(t). Since

dω(t) dω(t)2 , = 2 Jω(t), dt dt

we use (6.7.3) and (6.7.4) and come to the inequality 1 dr(t)2 dt 2

+ J(y(t) − w(t)), A(t)y(t) − A(t)w(t) + α(t)r(t)2 ≤

Therefore,

dw(t) r(t).

δ(t) + h(t)g(2¯ x∗ ) +

dt

|α (t)| dr(t) x∗ ≤ −α(t)r(t) + δ(t) + h(t)g(2¯ x∗ ) + 2¯ α(t) dt

because A(t) is accretive. Then from Lemma 7.2.1 the following assertion arises:

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372


Theorem 6.7.4 Under the assumptions of this subsection, if the equation (6.1.1) has a nonempty solution set N and the conditions (6.7.1), (6.7.2) and (6.7.7) hold, then a solution ¯∗ is a y(t) of the Cauchy problem (6.7.3) converges strongly to x ¯∗ ∈ N as t → ∞, where x unique element satisfying (2.7.7). 6. Suppose now that X is a uniformly convex and uniformly smooth Banach space, A : X → X ∗ is a monotone continuous operator, D(A) = X. Let x(t) and y(t) be functions defined on [t0 , +∞) with values in X. Introduce the Lyapunov functional W (x(t), y(t)) = 2−1 (x(t)2 − 2Jx(t), y(t) + y(t)2 ).

(6.7.32)

Lemma 6.7.5 Let a function x(t) be continuous, y(t) be differentiable and Jx(t) be Gˆ ateaux differentiable on [t0 , +∞). Then the functional (6.7.32) is differentiable and the equality dy(t) dW (x(t), y(t)) dJx(t) , x(t) − y(t) + Jy(t) − Jx(t), = dt dt dt

holds. Proof. Since spaces X and X ∗ are uniformly smooth and x(t) = Jx(t)∗ , we have

and

∂W (x(t), y(t)) = x(t) − y(t) ∂Jx

(6.7.33)

∂W (x(t), y(t)) = Jy(t) − Jx(t). ∂y

(6.7.34)

Convexity of W (x, y) with respect to Jx and y (see Section 1.6) implies the following inequalities for t0 < s < t :

W (x(t), y(t)) ≥ W (x(s), y(t)) + Jx(t) − Jx(s), and W (x(s), y(t)) ≥ W (x(s), y(s)) +

∂W (x(s), y(t)) ∂Jx

∂W (x(s), y(s))

∂y

, y(t) − y(s) .

According to (6.7.33) and (6.7.34), one gets W (x(t), y(t)) ≥ W (x(s), y(s)) + Jx(t) − Jx(s), x(s) − y(t) + Jy(s) − Jx(s), y(t) − y(s) and W (x(s), y(s)) ≥ W (x(t), y(t)) + Jx(s) − Jx(t), x(t) − y(s) + Jy(t) − Jx(t), y(s) − y(t) ,


373

from which it results that

Jx(t) − Jx(s)

t−s

, x(t) − y(s)

y(t) − y(s) t−s W (x(t), y(t)) − W (x(s), y(s)) ≥ t−s

Jx(t) − Jx(s) , x(s) − y(t) ≥ t−s

y(t) − y(s) . + Jy(s) − Jx(s), t−s

+

Jy(t) − Jx(t),

(6.7.35)

Duality mapping J is continuous in a uniformly smooth space X. Therefore, the conditions of the lemma allow us to pass in (6.7.35) to the limit when s → t. We obtain

dJx(t)

dt

, x(t) − y(t)

+

≥

Jy(t) − Jx(t),

dJx(t)

dt

dy(t) dW (x(t), y(t)) ≥ dt dt

, x(t) − y(t) + Jy(t) − Jx(t),

dy(t) . dt

The lemma is proved.

Suppose that the equation (6.1.1) is given with perturbed date f (t) and A(t). Moreover, D(A(t)) = D(A) = X for t ≥ t0 and, as before, A(t) is a family of monotone and continuous operators such that (6.7.36) Ax − A(t)x∗ ≤ g(x)h(t) ∀x ∈ X, and f − f (t)∗ ≤ δ(t),

(6.7.37)

where h(t), δ(t) and g(s) have the same properties as in (6.7.1) and (6.7.2). We study the Cauchy problem for the following differential equation: dJy(t) + A(t)y(t) + α(t)Jy(t) = f (t), dt

t ≥ t0 ,

y(t0 ) = y0 ∈ X.

(6.7.38)

We introduce the intermediate equation Aw(t) + α(t)Jw(t) = f.

(6.7.39)

Theorem 6.7.6 Suppose that the following conditions are satisfied: 1) the equation (6.1.1) has a solution set N and x ¯∗ ∈ N is its minimal norm solution; 2) the solution y(t) of the Cauchy problem (6.7.38) exists and is bounded for all t ≥ t0 ; 3) the inequalities (6.7.37) and (6.7.36) hold; 4) the properties (6.7.7) are fulfilled; 5) either δX () ≥ C1 2 , C1 > 0, and the solution w(t) of the equation (6.7.39) is differentiable on the interval [t0 , ∞) or

6

374


6) the equation (6.7.39) is differentiable by t in the strong sense, and there exists strictly increasing and continuous for all ξ ≥ 0 function ψ(ξ) such that ψ(0) = 0,

dJw(t) dw

,

dt

and ψ

dt

dw dw ψ

≥

|α (t)|

α(t)

dt

dt

/α(t) → 0.

Then y(t) → x ¯∗ as t → ∞. Proof. By the hypotheses, there exists a constant R1 > 0 such that y(t) ≤ R1 . In its turn, it is known from Section 2.2 that w(t) → x ¯∗ as t → ∞ and w(t) ≤ ¯ x∗ . Rewrite (6.7.33) and (6.7.34) as ∂W (y(t), w(t)) = y(t) − w(t) ∂Jy and ∂W (y(t), w(t)) = Jw(t) − Jy(t). ∂w Calculate the following derivative: dW (y(t), w(t)) dt

=

dJy(t) ∂W (y(t), w(t))

dt

,

∂Jy(t)

+

∂W (y(t), w(t)) dw(t)

∂w

,

dt

= −A(t)y(t) − A(t)w(t), y(t) − w(t) − α(t)Jy(t) − Jw(t), y(t) − w(t) − Aw(t) + α(t)Jw(t) − f, y(t) − w(t) + f (t) − f, y(t) − w(t)

+ Aw(t) − A(t)w(t), y(t) − w(t) + Jw(t) − Jy(t),

dw(t) . dt

By (1.6.19), we have Jx − Jy, x − y ≥ (2L)−1 δX (c−1 6 x − y),

(6.7.40)

where 1 < L < 1.7, c6 = 2max{1, R1 , ¯ x∗ }. Then (6.7.38), (6.7.39), (6.7.40), Lemma 6.7.5 and monotonicity of A(t) yield dW (y(t), w(t)) dt

≤ −α(t)(2L)−1 δX (c−1 6 y(t) − w(t))

+

dw(t) .

δ(t) + h(t)g(¯ x∗ ) y(t) − w(t) + Jw(t) − Jy(t)∗

dt


375

Estimate w (t) from above. With the help of Lemma 6.5.1, we can write −1 w(t1 ) − w(t2 ) ≤ c0 gX

C|α(t1 ) − α(t2 )| , C = 2Lc0 ¯ x∗ , α(t1 )

(6.7.41)

−1 x∗ }. If 5) holds then gX (ζ) ≤ C1−1 ζ. Hence, for the differentiable where c0 = 2max{1, ¯ function w(t), there exists a constant R2 > 0 such that

dw |α (t)| . ≤ R2

dt

(6.7.42)

α(t)

The rest of the proof follows the pattern of Theorem 6.7.1. x∗ }, and ∆−1 (·) is its inverse Let now ∆(τ ) = 8τ 2 + c7 ρX (τ ), where c7 = 8max{L, R1 , ¯ function. By inequality (1.6.45), we then conclude that

∆−1 W (x(t), y(t)) ≤ x(t) − y(t).

(6.7.43)

Hence, (6.7.42) and (6.7.43) lead to the final inequality

dW (x(t), y(t)) dt

−1 ≤ −α(t)(2L)−1 δX c−1 W (x(t), y(t) 6 ∆

+

δ(t) + h(t)g(¯ x∗ ) + R2

|α (t)| (R1 + ¯ x∗ ) α(t)

or

dW (x(t), y(t)) dt

≤ −α(t)(2L)−1 ϕ W (x(t), y(t)) +

δ(t) + h(t)g(¯ x∗ ) + R2

|α (t)| (R1 + ¯ x∗ ), α(t)

−1 where ϕ(ξ) = c−1 6 δX ∆ (ξ) is a positive continuous non-decreasing function for all ξ ≥ 0 and ϕ(0) = 0. Then the conditions of the theorem allow us to conclude on the basis of Lemma 7.2.1 that (6.7.44) lim W (x(t), y(t)) = 0. t→∞

Observe, that we are not able to estimate w (t) from (6.7.41) when δX () < C1 2 . Let 6) hold. Differentiating (6.7.39) by t, we obtain A (w)

dJw(t) dw(t) + α (t)Jw(t) = 0, + α(t) dt dt

where A is a Fréchet derivative of A. Then, by (6.7.45), the equality

A (w)

dJw(t) dw(t) ,z = 0 , z + α (t)Jw(t), z + α(t) dt dt

(6.7.45)

376

6


appears for every z ∈ X. Assuming z = w (t) in this equality and making use of Definition 1.3.8 we have dw(t)

dJw(t) dw(t) ≤ |α (t)|w(t) , α(t) , dt dt dt that is, dw(t) x∗ . α(t)ψ ≤ |α (t)|¯ dt Consequently, dw(t) |α (t)| , x∗ ≤ ψ −1 ¯ α(t) dt

and (6.7.44) arises from Lemma 7.2.1 again. Finally, (1.6.48) implies the inequality −1 x − y ≤ 2c6 δX (LW (Jx, y)) .

Thus, y(t) − w(t) → 0 as t → ∞. The proof is accomplished by (6.7.6). Remark 6.7.7 Along with (6.7.3) and (6.7.38), we are able to study differential equations dy(t) + A(t)y(t) + α(t)(y(t) − u0 ) = f (t), dt

and

dJy(t) + A(t)y(t) + α(t)J(y(t) − u0 ) = f (t), dt

t ≥ t0 , ,

t ≥ t0 ,

y(t0 ) = y0

y(t0 ) = y0 ,

were u0 is some fixed point in H and X, respectively. By simple additional algebra, one can establish convergence of y(t) to unique solution x ¯ ∈ N, such that ¯ x − u0 = min{x − u0 | x ∈ N }.

6.8

Newton−Kantorovich Regularization Method

The convergence of the Newton−Kantorovich classical approximations for the nonlinear equation (6.1.1) has been studied by many authors (see, for instance, [104]), mainly, in the case when an operator A is invertible. Some results deal with the investigations of influence of the monotonicity of A on the behavior of the Newton−Kantorovich method. For example, in [221], convergence of this method was established under the assumption that A is strongly monotone and potential. For operator A being arbitrarily monotone, the question of convergence and numerical realization of the Newton−Kantorovich algorithm was open a long time. Note that in this situation only regularizing processes constructed on the basis of discrete and continuous Newton−Kantorovich schemes enable us to prove strong convergence to a solution of the equation (6.1.1). In the present section we study iterative and continuous Newton−Kantorovich regularization methods. 1. Let X be a reflexive Banach space, X and X ∗ be strictly convex Banach spaces, A : X → X ∗ be a monotone twice Fréchet differentiable (hence, wittingly continuous) on

6.8 Newton−Kantorovich Regularization Method

377

X operator, N be a nonempty solution set of (6.1.1), x ¯∗ be a solution of (6.1.1) with a minimal norm. Let xαn be a unique solution of the equation Axαn + αn J s xαn = f,

(6.8.1)

where αn > 0, n = 0, 1, 2, ..., αn → 0 as n → ∞, J s : X → X ∗ be duality mapping with ¯∗ as the gauge function µ(t) = ts−1 , s ≥ 2. Then it is known (see Section 2.2) that xαn → x n → ∞. Assume that an operator J s possesses the property: there exists c > 0 such that J s x − J s y, x − y ≥ cx − ys ∀x, y ∈ X.

(6.8.2)

The Newton−Kantorovich method for equation (6.1.1) takes the following form: Azn + A (zn )(zn+1 − zn ) = f, n = 0, 1, 2, ...,

(6.8.3)

where A (zn ) is a non-negative operator by reason of Definition 1.3.8. Consequently, the equation (6.8.3) is linear with respect to zn+1 and it belongs to the class of ill-posed problems. Including into (6.8.3) the regularizing operator connected with duality mapping J s , we form the following equation: Axn + A (xn )(xn+1 − xn ) + αn J s xn+1 = f.

(6.8.4)

The latter equation may be considered as some generalization of the Newton−Kantorovich method for (6.8.1), however, in contrast to the classical Newton−Kantorovich method the linearization process of J s x in (6.8.4) has not been realized. This is accounted for by the fact that as s = 2 the operator (J s ) (xn ) does not have properties necessary for well-posedness of the obtained equation. We analyze the behavior of a sequence {xn } as n → ∞. Assume that A (x) ≤ ϕ(x) ∀x ∈ X, (6.8.5) where ϕ(t) is a non-negative and non-decreasing function for all t ≥ 0. Using the Taylor formula (1.1.16) we have from (6.8.1), Axn , xn+1 − xαn + A (xn )(xαn − xn ), xn+1 − xαn +

1 A (ξn )(xαn − xn )2 , xn+1 − xαn 2

+ αn J s xαn , xn+1 − xαn = f, xn+1 − xαn ,

(6.8.6)

where ξn = xαn + Θ(xn − xαn ), Θ = Θ(xn+1 − xαn ), 0 < Θ < 1 (see (1.1.16)). Calculating the values of the functionals, that are in both parts of equation (6.8.4), on the element xn+1 − xαn and subtracting the equality (6.8.6) from the obtained expression, one gets A (xn )(xn+1 − xαn ), xn+1 − xαn + αn J s xn+1 − J s xαn , xn+1 − xαn −

1 A (ξn )(xαn − xn )2 , xn+1 − xαn = 0. 2

6

378


Taking into account non-negativity of the first term and the conditions (6.8.2) and (6.8.5) we come to the estimate ϕ(r ) τ n λ2τ (6.8.7) xn+1 − xαn ≤ n , 2cαn where λn = xn − xαn , τ =

1 , rn ≥ max{xαn , λn }. Write down (6.8.1) for n = n + 1 : s−1

Axαn+1 + αn+1 J s xαn+1 = f.

(6.8.8)

From (6.8.1) and (6.8.8) follows the equality Axαn − Axαn+1 , xαn − xαn+1 + αn J s xαn − J s xαn+1 , xαn − xαn+1 + (αn − αn+1 )J s xαn+1 , xαn − xαn+1 = 0. The monotonicity of A and (6.8.2) yield now the estimate

xαn+1 − xαn ≤

|αn − αn+1 | cαn

τ

xαn+1 .

(6.8.9)

x∗ ≤ d for all n > 0 and rn ≥ max{d, xn }. It Let ¯ x∗ ≤ d. Then, by (2.2.9), xαn ≤ ¯ is not difficult to verify that (6.8.7) and (6.8.9) imply λn+1 = xn+1 − xαn+1 ≤ xn+1 − xαn + xαn − xαn+1 ≤ eτn λ2τ n +

d cτ

|αn − αn+1 | αn

τ

,

(6.8.10)

where en = ϕ(rn )(2cαn )−1 . Further we assume that 2 ≤ s < 3 and a) {αn } is a monotone decreasing sequence, moreover, there exists σ > 0 such that the ≥ σαn holds as n = 0, 1, ... ; inequality α τ1 n+1 ϕ(d + γ) s−1 1 c 1 λ0 τ , , c1 = , η = σκ, κ = ≤ q < 1, τ1 = b) 2c (3 − s)2 ηατ01 3−s

where γ > 0 is found from the estimate

η

c)

|αn − αn+1 | 1 α2τ n

τ

≤

α0 c1

τ1

≤ γ;

(6.8.11)

dcτ1 q s−1 − q 2 , c2 = τ τ11 +κ . c σ c2

Note that from the condition a) we have σ < 1, and, hence, η = σκ < 1. By (6.8.11) and by the property of {αn }, it follows that 1 τ1 ηc−τ 1 αn ≤ γ ∀n > 0.

(6.8.12)


379

Therefore, the condition b) and (6.8.11) imply λ0 ≤ γ. Hence, r0 = d + γ. Show that the inequality cτ11 λn ≤ q s−1 < 1 (6.8.13) ηαnτ1 results from the assumptions a) - c) with 2 ≤ s < 3. Indeed, if n = 0 then (6.8.13) is true because of b). Since τ + τ1 = 2τ τ1 , we deduce from (6.8.7) with n = 0 and from b) that cτ11 x1 − xα0 ≤ ηα0τ1

c1 α0

2τ τ1

λ0 η

2τ

≤ q 2 < 1,

that is, x1 −xα0 < γ and r1 = d+γ. Let then inequality (6.8.13) hold with n = k. Establish its validity for n = k + 1. On the basis of (6.8.12) and (6.8.13), it can be easily verified that λk ≤ γ, that is, rk = d + γ. Therefore (6.8.10) gives

λk+1 ≤

αk − αk+1 cτ1 2τ λ + c2 αk αkτ k

τ

.

Then, by making use of a) - c) and (6.8.13) with 2 ≤ s < 3, one gets cτ11 λk+1 ≤ 1 ηατk+1

cτ11 λk ηατk1

2τ

+ c2

αk − αk+1 1 α2τ k

τ

≤ q 2 + q s−1 − q 2 = q s−1 < 1,

because τ + τ1 = 2τ τ1 . Since {αn } → 0 as n → ∞, we have from (6.8.13) that λn = ¯∗ follows from the inequality xn − xαn → 0. Finally, the the strong convergence {xn } to x ¯∗ ≤ xn − xαn + xαn − x ¯∗ . xn − x Thus, we have proved the following theorem: Theorem 6.8.1 Let equation (6.1.1) have a nonempty solution set N, x ¯∗ be its solution with the minimal norm, ¯ x∗ ≤ d, A be a twice differentiable monotone operator, αn → 0 as n → ∞ and the conditions (6.8.2), (6.8.5) and a) - c) be satisfied as 2 ≤ s < 3. Then a ¯∗ ∈ N. sequence {xn } generated by iterative process (6.8.4) strongly converges as n → ∞ to x Remark 6.8.2 The class of Banach spaces, in which the property (6.8.2) takes place, is nonempty. Indeed, in the Lebesgue spaces with 1 < p ≤ 2 and p > 2 one can assume s = 2 and for s = p, respectively (see Section 1.6). Conditions a) - c) of Theorem 6.8.1 impose requirements on the choice of the initial approximation x0 in (6.8.4) and on the choice of the sequence {αn }. Show that it is possible to achieve the fulfillment of these conditions. Let αn =

1 1 , , 0 0 such that 1/τ1 γ c1 α0 = η

380

6


(see the condition (6.8.11)). If b) is satisfied then we do not change γ. Otherwise, we choose γ such that λ0 ≤ q 1/τ , γ and then b) holds. It is not difficult to be verified that αn − αn+1 → 0 as n → ∞, 1 α2τ n

if β is defined by (6.8.14). We are able to take m large enough, that c) holds for all n ≥ 0. Then for every initial approximation x0 the conditions of Theorem 6.8.1 are satisfied. 2. Suppose that, in place of f and A, the sequences of perturbed data {fn } and {An } are known such that for all n ≥ 0 elements fn ∈ X ∗ and operators An : X → X ∗ have the same properties as A, and then f − fn ∗ ≤ δn and An x − Ax∗ ≤ g(x)hn ∀x ∈ X, where g(s) is a non-negative continuous function for all s ≥ 0. Define approximations xn by the equation An xn + (An ) (xn )(xn+1 − xn ) + αn J s xn+1 = fn ,

(6.8.15)

and the intermediate regularized equation as follows: An xαn + αn J s xαn = fn . Under these conditions, the validity of Theorem 6.8.1 can be established if there holds the additional relation δn + hn lim = 0. n→∞ αn 3. If X is a Hilbert space H, then s = 2, τ = τ1 = κ = c = 1, η = σ, J s is the identity operator I, and the Newton−Kantorovich regularization method (6.8.4) has the form: Axn + A (xn )(xn+1 − xn ) + αn xn+1 = f.

(6.8.16)

Besides, another sort of the Newton−Kantorovich regularization methods can be constructed on the basis of the classical Newton−Kantorovich scheme applied to the following regularization problem: Ax + αn Sx = f, where S : H → H is a twice differentiable operator such that (Sx − Sy, x − y) ≥ x − y2 . In this case, approximations xn are found from the equation &

'

Axn + αn Sxn + A (xn ) + αn S (xn ) (xn+1 − xn ) = f.

(6.8.17)


381

If S is the identity operator I, then S = I and (6.8.17) coincides with (6.8.16). Let S (x) ≤ ϕ1 (x), the function ϕ1 (t) be of the same class as ϕ(t). Then sufficient conditions for the convergence of the iterative process (6.8.17) are given by Theorem 6.8.1 with s = 2. 4. Let A : H → H be a monotone Fréchet differentiable operator, the equation (6.1.1) have a nonempty solution set N, x ¯∗ ∈ N be the minimal norm solution. We present the Newton−Kantorovich iterative regularization method (6.8.16) with inexact right-hand side, namely, Axn + A (xn )(xn+1 − xn ) + αn+1 xn+1 = fn . It is obvious that its continuous analogue can be written as a differential equation

Ax(t) + A (x(t))xt (t) + α(t)x (t) + α (t) + α(t) x(t) = f (t),

t ≥ t0 ≥ 0,

(6.8.18)

with the initial condition x(t0 ) = x0 , x0 ∈ H.

(6.8.19)

We assume that a function α(t) is positive continuously differentiable for all t ≥ t0 , continuous function f (t) is a certain approximation of the right-hand side f in (6.1.1) such that (6.7.2) holds. In addition, a function δ(t) is continuous and non-negative and

∞ t0

δ(τ )dτ = 0.

(6.8.20)

It is easy to see that (6.8.18) can be rewritten in the equivalent form as

Ax(t) + α(t)x(t)

t

+ Ax(t) + α(t)x(t) = f (t).

Denote v(t) = Ax(t) + α(t)x(t). Then we obtain the following Cauchy problem: dv(t) + v(t) = f (t), v(t0 ) = Ax0 + α0 x0 , dt

α0 = α(t0 ),

(6.8.21)

which has a unique solution v(t) on any finite interval [t0 , T ], T > t0 . In this situation, the solution of problem (6.8.18), (6.8.19) can be uniquely defined by the formula

x(t) = A + α(t)I

−1

v(t) ∀t ∈ [t0 , T ].

Solving the linear equation (6.8.21) we have Ax(t) + α(t)x(t) = exp(−t)

t t0

f (τ )exp(τ )dτ + (Ax0 + α0 x0 )exp(t0 ) .

(6.8.22)

Further, for each t ≥ t0 we consider in H the operator equation Axα (t) + α(t)xα (t) = f (t).

(6.8.23)

6

382


It is known (see Section 2.1) that if lim α(t) = 0,

t→∞

lim

t→∞

δ(t) = 0, α(t)

(6.8.24)

then limt→∞ xα (t) − x ¯∗ = 0, where x ¯∗ is the solution of equation (6.1.1) with minimal norm. Let β0 = Ax0 + α0 x0 , σ(t) =

t

δ(τ )exp(τ )dτ. t0

From (6.8.22) and (6.8.23), it appears that (Ax(t) − Axα (t), x(t) − xα (t)) + α(t)x(t) − xα (t)2

= exp(−t)

t t0

f (τ )exp(τ )dτ + (Ax0 + α0 x0 )exp(t0 ) − f (t), x(t) − xα (t) .

The monotonicity property of A implies x(t) − xα (t) ≤ γ1 (t) + γ2 (t), where

γ1 (t) = β0 exp(t0 − t)α−1 (t)

and

γ2 (t) = exp(−t)

t t0

f (τ )exp(τ )dτ − f (t)α−1 (t).

Suppose that lim

t→∞

exp(−t)σ(t) = 0. α(t)

(6.8.25)

The following should be noted. Since σ(t) > 0 for sufficiently large t, it follows from (6.8.25) that exp(−t) = 0. (6.8.26) lim t→∞ α(t) Thus, γ1 (t) is infinitely small as t → ∞. Moreover, γ2 (t) = exp(−t) ≤

t t0

(f (τ ) − f )exp(τ )dτ − exp(t0 − t)f + f − f (t)α−1 (t)

exp(−t)σ(t) + exp(t0 − t)f + δ(t) α−1 (t).

It shows that γ2 (t) → 0 as t → ∞ provided that the conditions (6.8.24) - (6.8.26) are satisfied. This proves the following theorem: Theorem 6.8.3 Suppose A : H → H is a Fréchet differentiable monotone operator, equation (6.1.1) has a nonempty solution set in H, for t ≥ t0 the function α(t) is positive and continuously differentiable, δ(t) is non-negative and continuous, f (t) : [t0 , ∞) → H is continuous, and the conditions (6.7.2), (6.8.20), (6.8.24) and (6.8.25) are satisfied. Then x(t) → x ¯∗ as t → ∞, where x(t) is the unique solution of the Cauchy problem (6.8.18), (6.8.19).


383

Note that the class of functions satisfying (6.8.24) and (6.8.25) is not empty. For example, we could have α(t) = (t + λ)−γ , δ(t) = exp(−ct), γ > 0, c > 0, λ > 0 for t0 = 0 and λ ≥ 0 for t0 > 0. Let the operator A in (6.1.1) be given with an error, the approximations A(t, x) : [t0 , ∞) × H → H being monotone with respect to the second argument be differentiable in the strong sense with respect to each argument, and A(t, x) − Ax ≤ g(x)h(t) ∀t ≥ t0 , ∀x ∈ H, here h(t) and g(t) are non-negative continuous functions. In this situation, we replace the equation (6.8.18) by the following: At (t, x(t)) + Ax (t, x(t))xt (t) + (α (t) + α(t))x(t) + α(t)xt (t) + A(t, x(t)) = f (t) or

A(t, x(t)) + α(t)x(t)

t

+ A(t, x(t)) + α(t))x(t) = f (t).

(6.8.27)

The solution of Cauchy problem (6.8.27), (6.8.19) converges to the solution x ¯∗ of equation (6.1.1) as t → ∞, if to the conditions of Theorem 6.8.3 the following relation is added: lim

t→∞

h(t) = 0. α(t)

In conclusion, we provide discrete and continuous schemes for the regularized Gauss− Newton method which were studied in [41] and [3] for the operator equation Ax = 0 in a Hilbert space. 1. Discrete scheme:

xn+1 = xn − A∗ (xn )A (xn ) + αn I

−1

A∗ (xn )A(xn ) + αn (xn − z 0 ) ,

where A (xn ) = Ax (xn ), A∗ (xn ) = Ax ∗ (xn ) and z 0 is some element of H. 2. Continuous scheme: −1 dx ∗ A (x(t))A(x(t)) + α(t)(x(t) − z 0 ) . = − A∗ (x(t))A (x(t)) + α(t)I dt

Bibliographical Notes and Remarks The classical quasi-solution method and residual method for potential equations are due to Ivanov and Liskovets [97, 98, 99, 130]. The convergence conditions of the residual method for a monotone potential and weakly closed operator A with a non-convex, generally

384

6


speaking, set Mδ have been obtained in [183]. However, the requirement of the sequential weak closedness of a nonlinear mapping is quite strong. Therefore, in a number of works, the operator conditions have been essentially weakened. So, for a potential operator A the residual method has been studied rather fully in [130]. Modifications of these methods for monotone problems have been presented in [202]. Certainly, problem (6.1.1) can be solved by the operator regularization method described in Chapters 2 and 3 as well, but interest in the residual method is connected with extension of tools for numerical solutions of ill-posed problems, taking into account availability of different initial information about the problem. The connection of the quasi-solution and residual methods with the regularization methods was also established in [195]. Strong convergence and stability of the penalty method were proved in [15]. The proximal point method was studied in [181, 184, 204, 212, 227]. Another approach is developed in [105]. The results of Section 6.5 concerning iterative regularization of monotone and accretive equations in Banach spaces first appeared in [10]. The case of Hilbert spaces was earlier considered by Bakushinskii [40] and Bruck [62]. The proof of Lemma 6.6.1 and Theorems 6.6.2, 6.6.4 can be found in [11, 21]. The other algorithms of the iterative regularization are studied in [20, 23]. The special iterations for ill-posed problems are also described in [93]. The continuous regularization method of the first order in Hilbert and Banach spaces was investigated in [14, 31, 36]. The linear case was earlier studied in [4]. The high order methods are considered, for instance, in [59, 60, 124]. The convergence of the regularized Newton−Kantorovich in the iterative form was proved in [194] and in the differential form in [200]. Many results of Section 6.8 can be transferred to variational inequalities.

Chapter 7

APPENDIX 7.1

Recurrent Numerical Inequalities

Lemma 7.1.1 Let {λn } and {γn } be sequences of non-negative real numbers, {αn } be a sequence of positive real numbers such that λn+1 ≤ λn − αn λn + γn ∀ n ≥ 0, γn ≤ c1 and αn ≤ α. αn

(7.1.1)

Then λn ≤ max{λ0 , K∗ }, where K∗ = (1 + α)c1 . Proof. Similarly to the proof of Lemma 7.1.3, consider the following alternative for all n ≥ 0 : either γn H1 : λn ≤ αn or γn . H1 : λn > αn The hypothesis H1 gives the estimate λn ≤ c1 . In turn, the hypothesis H2 implies λn+1 < λn . At intermediate indexes we have

λn+1 ≤ λn + γn ≤ c1 + αn c1 ≤ (1 + α)c1 . From this the claim follows. Lemma 7.1.2 Let {λn } and {γn } be sequences of non-negative real numbers, {αn } be a sequence of positive real numbers satisfying the inequality λn+1 ≤ λn − αn λn + γn ∀n ≥ 0, αn ≤ 1, where αn → 0 as n → ∞,

∞

αn = ∞,

n=0

385

(7.1.2)

(7.1.3)

386

7 APPENDIX γn = 0. αn

(7.1.4)

lim λn = 0.

(7.1.5)

lim

n→∞

Then n→∞

Proof. From (7.1.2) it is obvious that λn+1 ≤

n *

n

(1 − αi )λ0 +

i=0

Since

n *

n *

γi

i=0

(1 − αk ).

(7.1.6)

k=i+1

(1 − αi ) ≤ exp (−

i=0

n

αi ),

i=0

then by virtue of the condition (7.1.3), the first term in the right-hand side of (7.1.6) tends to zero as n → ∞. Further, n

i=0

γi

n *

n

(1 − αk ) ≤

γi exp −

i=0

k=i+1

n

=

n

αk

k=i+1

exp ( ik=0 αk ) n . γi

exp (

i=0

k=0 αk )

(7.1.7)

Applying the Stolz theorem (see [82]) we obtain lim

n→∞

n

i=0

γi

exp ( ik=0 αk ) exp ( nk=0 αk )

= =

=

γn+1 exp ( n+1 k=0 αk ) n→∞ exp ( α ) − exp ( nk=0 αk ) k k=0 lim

n+1

lim

γn+1 1 − exp (−αn+1 )

lim

γn+1 . αn+1

n→∞

n→∞

Then (7.1.5) follows from (7.1.4), (7.1.6) and (7.1.7). We prove more a general statement. Lemma 7.1.3 If sequences of non-negative real numbers {λn } and {γn } and a bounded sequence {ρn } of positive numbers satisfy the inequality λn+1 ≤ λn − ρn Ψ(λn ) + γn , n ≥ 0, where Ψ(t) is a continuous increasing function, Ψ(0) = 0, ∞

ρn = ∞,

n=0

lim

n→∞

γn = 0, ρn

then lim λn = 0.

n→∞

(7.1.8)

7.1 Recurrent Numerical Inequalities

387

Proof. Consider the following alternative for all n ≥ 0 : either H1 : Ψ(λn ) ≤

n

−1

ρi

+

γn ρn

+

γn . ρn

i=0

or H2 : Ψ(λn ) >

n

−1

ρi

i=0

Introduce sets I1 and I2 as the totalities of numbers n ≥ 0 such that the hypotheses H1 and H2 hold for all n ∈ I1 and n ∈ I2 , respectively. It is clear that a union of these sets is a set of all positive integers. Show that I1 is infinite. Indeed, assuming the contrary, it is not difficult to be sure that there exists N0 ≥ 0 such that for all n ≥ N0 , ρn−1 λn ≤ λn−1 − n−1 i=0 ρi or

n−1

λn ≤ λN0 −

j=N0

where Aj =

j

ρj , Aj

ρi .

i=0

By the Abel−Dini test [82],

∞

ρj j=0

Aj

= ∞.

Therefore, beginning with some n all λn becomes negative which should not be because every λn ≥ 0. Assume I1 = {n1 , ..., nl , ...} and consider two following cases: 1) n1 = 0. It is obvious that on an arbitrary interval I˜l = [nl , nl+1 ], where nl+1 > nl + 1, λnl +1 ≤ λnl + γnl ≤ Ψ−1

γn 1 + l ρnl Anl

+ γnl .

At the same time, for all nl < n < nl+1 we have λn ≤ λnl +1 −

n−1

ρi < λnl +1 . A i=n +1 i l

By virtue of unboundedness of the set I1 , it results that limn→∞ λn = 0. 2) n1 > 1. It is clear that for n > n1 it is possible to use the previous reasoning, while {1, 2, ..., n1 − 1} ⊂ I2 . Hence, λn+1 ≤ λ0 −

n

ρj j=0

Aj

≤ λ1 ,

and the equality limn→∞ λn = 0 holds again.

1 ≤ n ≤ n1 − 1,

388

7 APPENDIX

7.2

Differential Inequality

Next we study the asymptotic behavior of solutions to the ordinary differential inequalities. Lemma 7.2.1 Let a non-negative function λ(t) satisfy the differential inequality dλ(t) ≤ −α(t)ψ(λ(t)) + γ(t), λ(t0 ) = λ0 , dt

t ≥ t0 ,

(7.2.1)

where a function α(t) is continuous positive for t ≥ t0 , γ(t) is continuous non-negative, ψ(λ) is positive continuous and non-decreasing for λ > 0, ψ(0) = 0. Moreover, let

∞

t0

α(τ )dτ = ∞

(7.2.2)

γ(t) = 0. α(t)

(7.2.3)

and lim

t→∞

Then λ(t) → 0 as t → ∞. Proof. Consider the following alternative: H1 : ψ(λ(t)) < q(t); H2 : ψ(λ(t)) ≥ q(t), where q(t) =

t

α(τ )dτ

−1

t0

+

γ(t) . α(t)

Define the sets 1 T1i = {t0 ≤ t ∈ (ti , t¯i ) ⊆ R+ | H1 is true},

T1 = ∪i T1i ,

(7.2.4)

1 T2j = {t0 ≤ t ∈ [tj , t¯j ] ⊆ R+ | H2 is true},

T2 = ∪j T2j .

(7.2.5)

It is easy to see that T = T1 ∪ T2 = [t0 , ∞). Prove that T1 is an unbounded set. For that assume the contrary. Then there exists t = τ1 such that for all t ≥ τ1 the hypothesis H2 holds, and (7.2.1) yields the inequality α(t) dλ(t) , ≤ −) t dt t0 α(τ )dτ

Hence, λ(t) ≤ λ(τ1 ) −

where

t τ1

t ≥ τ1 .

α(τ ) dτ, S(τ )

τ

S(τ ) =

α(s)ds > 0. t0

(7.2.6)

(7.2.7)

7.2

Differential Inequality

389

It is obvious that

t τ1

α(t) dt = ln S(t) − ln S(τ1 ) → ∞ S(t)

t → ∞.

as

Then we deduce from the inequality (7.2.7) that there exists a point t = τ2 , for which λ(τ2 ) < 0. This contradicts the conditions of the lemma. By (7.2.2) and (7.2.3), the positive function ψ(λ(t)) → 0 as t → ∞ and t ∈ T1 . Now the convergence of λ(t) to zero as t ∈ T1 and t → ∞ is guaranteed due to the properties of ψ(t). At the same time the function λ(t) decreases on sets T2j because of (7.2.6). Thus, the lemma is proved.

Lemma 7.2.2 Let a non-negative function λ(t) satisfy the differential inequality dλ(t) ≤ −α(t)λ(t) + γ(t), t ≥ t0 , dt

(7.2.8)

where α(t) is positive and γ(t) are non-negative continuous functions on the interval [t0 , ∞). Then the inequality

λ(t) ≤ λ(t0 )exp −

t

t

α(s)ds + t0

t0

γ(θ)exp −

t

α(s)ds dθ

(7.2.9)

θ

holds. If (7.2.2) and (7.2.3) are satisfied, then λ(t) → 0 as t → ∞. Proof. Multiplying both parts of (7.2.8) by z(t) = exp

)

t t0

α(s)ds we obtain

d λ(t)z(t) ≤ γ(t)z(t). dt

Then λ(t)z(t) ≤ λ(t0 ) +

t

γ(τ )z(τ )dτ, t0

that is equivalent to (7.2.9). The first term in the right-hand side of (7.2.9) tends to zero by the equality (7.2.2). We find the limit of the second term as t → ∞. Denote the antiderivative of α(t) by α(t). ¯ If the integral ∞

¯ γ(θ)eα(θ) dθ

(7.2.10)

t0

is divergent then applying L’Hopital’s rule and (7.2.3), one gets t

lim

t→∞ t0

γ(θ)e−

)t θ

)t α(s)ds

dθ =

=

lim

t→∞

lim

t→∞

t0

¯ dθ γ(θ)eα(θ) α(t) ¯ e

γ(t) = 0. α(t)

(7.2.11)

If the integral (7.2.10) is convergent, then (7.2.11) holds in view of the equality (7.2.2) again. The lemma is proved.

390

7 APPENDIX Bibliographical Notes and Remarks

Lemmas 7.1.2, 7.1.3 and 7.1.1 were proved in [9, 29]. Lemmas 7.2.1 and 7.2.2 can be found in [4, 11]. The considerable part of the book [229] deals with the linear recurrent numerical inequalities and their applications.

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INDEX Fréchet derivative of functional 6 Fréchet derivative of operator 14 Fréchet differential of functional 6 Fréchet differential of operator 14 functional 2 bounded 2 conjugate 18 convex 4 strictly 4 strongly 4 uniformly 4 differentiable directionally 5 Fréchet 6 Gˆ ateaux 5 strongly 6 weakly 5 finite 5 H−property 299 linear 2 lower semicontinuous 4 weakly 4 Lyapunov W (x, y) 49 proper 5 smoothing vii stabilizing vii supporting 16 trivial 5

approximation of sets exterior 204 interior 204 ball 3 closed 3 open 3 unit 3 best approximation 14 closure of set 3 contact problem 218 convergence strong 1 weak 2 diameter of set 3 distance between point and set 14 domain of functional 2 operator 1 dual pairing 2 dual product 2 duality mapping in lp 36 in Lp 36 p 36 in Wm normalized 7 with gauge function 33 extension of operator 13

Gˆ ateaux derivative of functional 5 Gˆ ateaux derivative of operator 14 Gˆ ateaux differential of functional 5 Gˆ ateaux differential of operator 14 gauge function 33 generalized derivative 10 generalized residual 167

Figiel constant 41 filtration equation 24, 226 formula Lagrange 14 Lindenstrauss 47 Taylor 14 407

408

INDEX generalized projection 109 gradient of functional 5 graph of operator 19 Hammerstein equation 278 Hanner’s equality 47 Hausdorff distance 15 hyperplane 3 supporting 59 identity decomposition 65 indicator function 59, 63 subdifferential 60, 63 inequality Cauchy-Schwarz 2 Clarkson 51 co-variational 107,114 Friedrichs 11 H¨ older 9 Minkovsky 9 parallelogram 51 variational 72 regularized 76, 192, 204 with small offset 141, 233 Young-Fenchel 18 Kadeˇc-Klee property 11 Lagrange function 78 lemma Debrunner-Flor 52 Minty-Browder 73, 237 Opial 37 Zorn 15 Lipschitz−H¨ older condition 129 membrane problem 221 metric projection 14 modulus of convexity of functional 4 convexity of space 7 local 12 in lp , Lp 48 smoothness of space 8 in lp , Lp 48

Mosco-approximate 81, 249 Mosco-convergent sequence of sets 15 norm of functional 2 norm of operator 13 operator additive 13 accretive 95 locally bounded 96 maximal 98 properly 99 strictly 99 strongly 99 uniformly 99 adjoint 31 bounded 12 locally 27 uniformly 12 closed 13 coercive 12, 96 relatively a point 12 weakly 12 compact 12 conjugate 31 continuous 13 completely 13 strongly 13 weakly 13 weak-to-weak 13 convex-valued 237 d−accretive 108, 110 demiclosed 236 gauge 109 maximal 110 demicontinuous 13 Fréchet differentiable 13 Gˆ ateaux differentiable 14 generalized value set 268 Hamilton 24 hemicontinuous 13 homogeneous 7, 13 hypomonotone 245 strongly 245 imbedding 11

409

INDEX inverse 12 Laplace 23 linear 13, 31 Lipschitz-continuous, 13 m-accretive 99 m-d-accretive 111 metric projection 14 monotone 20, 20, 20 locally bounded 27 locally strongly 25 locally uniformly 25, maximal 29 properly 25 strictly 20 strongly 25 uniformly 25 nonexpansive 21 normality 64 odd 7 penalty 322 potential 5 projection operator generalized 109 metric 14 pseudomonotone 84 quasipotential 90 radial summable 90 regularizing vii self-adjoint 31 semi-monotone 70 Schr¨ odinger 23 stabilizing viii strongly differentiable 13 s-w-demiclosed 236 upper h-semicontinuous 90 upper semicontinuous 84 weakly closed 13 weakly differentiable 14 w-s-demiclosed 236 operator equation vi, 117 regularized viii, 124, 122 Opial condition 3 parallelogram equality 9

potential of operator 5, 90 principle quasi-optimality 151 residual 151 for accretive equation 166 for monotone equation 160 generalized 168, 173 minimal 151 Schauder 12 smoothing functional 151, 183 problem ill-posed vi essentially vii monotone type viii well-posed vi conditionally vii projector 15 pseudosolution 283 quasi-solution 311 quasi-solution method vii range 1 regularization method vii Tikhonov vii operator viii regularization parameter vii relation binary 15 residual vii, 158 method viii, 316 retraction 40 nonexpansive 40 sunny 40 semi-deviation 15 sequence fundamental 1 extremely dense 313 set bounded 3 closed 3 weakly 3 compact 3 weakly 3 convex 3 demiclosed 29

INDEX

410 dense 3 effective 5 everywhere dense 3 linearly ordered 15 monotone 19 maximal 29 semi-ordered 15 Slater condition 78, 314 solution of co-variational inequality 107 solution of operator equation generalized 67, 101 in classical sense 53 in the sense of inclusion 53 regularized vii, 118, 124 s-generalized 70 sw-generalized 67, 101 weak 219 solution of variational inequality classical 72 generalized 238 regularized 192, 204 space Banach 2 complete 1 convex local uniformly 7 strictly 7 uniformly 7 dual 2 E-space 11 Hilbert 9 Lebesgue 9 M -property 70 possesses an approximation 15 reflexive 2 separable 12 smooth 7 strongly 7 uniformly 8 Sobolev 10 sphere 3 strictly convex 7 unit 3 S-property of operator 233

strong derivative of functional 6 strong derivative of operator 14 strong differential of functional 6 strong differential of operator 14 subdifferential 16 subgradient 16 symmetric difference of sets 226 theorem Banach-Steinhaus 2 Hahn-Banach 2 Karush-Kuhn-Tacker 78 Klee-Shmulyan 8 Mazur 3 Minty-Browder 55, 70 Riesz 4 Sobolev 11 strong separation 4 Tikhonov vi Weierstrass 4 weak accumulation point 2 weak derivative of functional 5 weak derivative of operator 14 weak differential of functional 5 weak differential of operator 14 weak limit 2 well-posedness set vii

Nonlinear Ill-posed Problems of Monotone Type

Nonlinear Ill-posed Problems of Monotone Type

Nonlinear Ill-posed Problems of Monotone Type