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

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)

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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.

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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.

1.1 Elements of Nonlinear Functional Analysis

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)

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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 Fr´echet 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 Fr´echet differential (or strong differential) and F (x) = ϕ (x) is called the Fr´echet derivative (or strong derivative) of the functional ϕ at a point x.

1.1 Elements of Nonlinear Functional Analysis

7

In other word, if the limit in (1.1.5) exists uniformly for h on the unit sphere of X, then ϕ is Fr´echet differentiable and ϕ (x) is the Fr´echet derivative of ϕ at x. If a functional ϕ ateaux differentiable at that point. is Fr´echet differentiable at a point x0 ∈ X, then it is Gˆ The inverse assertion is not true in general. However, if Gˆateaux derivative ϕ is continuous at a neighborhood of a point x0 ∈ X, then it is, in fact, the Fr´echet 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 Fr´echet 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.

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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 Fr´echet 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 Fr´echet differentiable.

1.1 Elements of Nonlinear Functional Analysis

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 H¨older 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

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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.

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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.

1.1 Elements of Nonlinear Functional Analysis

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 Fr´echet 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),

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where x + h ∈ D(A) and ω(x, h) → 0 as h → 0. h Respectively, A (x)h and A (x) are called the Fr´echet differential (or strong differential) and the Fr´echet 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 Fr´echet derivative there. Let X be a Banach space, X ∗ be its dual space, A : X → X ∗ be a nonlinear operator having the Fr´echet 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.

1.1 Elements of Nonlinear Functional Analysis

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

THEORY OF MONOTONE AND ACCRETIVE OPERATORS

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.

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THEORY OF MONOTONE AND ACCRETIVE OPERATORS

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)

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THEORY OF MONOTONE AND ACCRETIVE OPERATORS

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 Schr¨odinger 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

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THEORY OF MONOTONE AND ACCRETIVE OPERATORS

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 ).

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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.

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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.

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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.

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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 .

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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 .

1.5 Duality Mappings

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:

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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].

1.5 Duality Mappings

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)

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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.

1.5 Duality Mappings

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.

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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)

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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

Banach Spaces Geometry and Related Duality Estimates

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

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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)

Banach Spaces Geometry and Related Duality Estimates

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)

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THEORY OF MONOTONE AND ACCRETIVE OPERATORS

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)

1.6 Banach Spaces Geometry and Related Duality Estimates

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.

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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 H¨older-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 Gˆateaux 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)

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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)

1.6 Banach Spaces Geometry and Related Duality Estimates

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

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THEORY OF MONOTONE AND ACCRETIVE OPERATORS

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.

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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→∞

1.7 Equations with Maximal Monotone Operators

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.

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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.

1.7 Equations with Maximal Monotone Operators

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.

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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).

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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).

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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

1.8 Summation of Maximal Monotone Operators

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)

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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 .

1.8 Summation of Maximal Monotone Operators

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 = ∅.

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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.

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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

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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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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).

1.11 Variational Inequalities with Monotone Operators

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)

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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)

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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)

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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.

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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.

1.13 Variational Inequalities with Pseudomonotone Operators

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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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

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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.

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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

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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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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
1≤|α|=|β|≤m

G

aαβ (s)D α x(s)D β x(s)ds ≥ M



|α|=m

Dα x(s)2L2 ,

1.14 Variational Inequalities with Quasipotential Operators

93

◦

2 (G), M > 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.

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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)

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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 Gˆateaux 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)

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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,

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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  ≤

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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)

1.15 Equations with Accretive Operators

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.

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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.

1.15 Equations with Accretive Operators

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.

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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)

1.15 Equations with Accretive Operators

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.

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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 .

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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.

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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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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.

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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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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.

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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 ˜Axn −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)

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= Jxn , Pn (Axn − f ) + Jxn , f = Jxn , f . ˜Axn . Then (1.16.11) gives Since the operator J is homogeneous, we have Jyn , Axn = a a ˜Axn  ≤ (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,

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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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THEORY OF MONOTONE AND ACCRETIVE OPERATORS

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

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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 }.

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REGULARIZATION OF OPERATOR EQUATIONS

¯∗ . 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

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REGULARIZATION OF OPERATOR EQUATIONS

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

2.2 Equations with Monotone Operators in Banach Spaces

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

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REGULARIZATION OF OPERATOR EQUATIONS

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

2.2 Equations with Monotone Operators in Banach Spaces

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 Fr´echet differentiable and Fr´echet derivative A (x) satisfies the Lipschitz−H¨older 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

REGULARIZATION OF OPERATOR EQUATIONS

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

REGULARIZATION OF OPERATOR EQUATIONS

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 Fr´echet differolder condition (2.3.6), entiable operator, Fr´echet 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

REGULARIZATION OF OPERATOR EQUATIONS

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

REGULARIZATION OF OPERATOR EQUATIONS

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

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REGULARIZATION OF OPERATOR EQUATIONS

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).

140

2

REGULARIZATION OF OPERATOR EQUATIONS

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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REGULARIZATION OF OPERATOR EQUATIONS

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

Equations with Accretive and d-Accretive Operators

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.

144

2

REGULARIZATION OF OPERATOR EQUATIONS

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

Equations with Accretive and d-Accretive Operators

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

146

2

REGULARIZATION OF OPERATOR EQUATIONS

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 conti