Turnpike Phenomenon and Symmetric Optimization Problems (Springer Optimization and Its Applications, 190) 303096972X, 9783030969721

Written by a leading expert in turnpike phenomenon, this book is devoted to the study of symmetric optimization, variati

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Turnpike Phenomenon and Symmetric Optimization Problems (Springer Optimization and Its Applications, 190)
 303096972X, 9783030969721

Table of contents :
Preface
Contents
1 Introduction
1.1 Generic Existence of Solutions of Minimization Problems
1.2 Optimization Problems Arising in Crystallography
1.3 Symmetric Optimization Problems
1.4 Turnpike Property for Variational Problems
1.5 Variational Problems with Symmetric Integrands
1.6 Notation
1.7 Exercises
2 Symmetric Optimization Problems
2.1 A Generic Approach in Optimization
2.2 A Generic Variational Principle
2.3 The First Generic Result
2.4 The Second and Third Generic Results
2.5 Proofs of Theorems 2.5 and 2.6
2.6 The Fourth Result
2.7 The Fifth Generic Result
2.8 Auxiliary Results for Theorems 2.14 and 2.15
2.9 Proof of Theorems 2.14 and 2.15
2.10 σ-Porous Sets in a Metric Space
2.11 A Variational Principle and Porosity
2.12 Well-Posedness and Porosity for Classes of Minimization Problems
2.13 An Auxiliary Result
2.14 A Well-Posedness Result
2.15 An Extension of Theorem 2.4
2.16 Extensions of Theorems 2.5 and 2.6
2.17 Extension of Theorem 2.10
2.18 An Extension of Theorem 2.14
3 Parametric Optimization
3.1 Generic Variational Principle
3.2 Concretization of the Hypothesis (H)
3.3 The First Generic Existence Results
3.4 The Second Generic Existence Result
4 Infinite Dimensional Control
4.1 Banach Space Valued Functions
4.2 Unbounded Operators
4.3 C0 Semigroup
4.4 Evolution Equations
4.5 C0 Groups
4.6 Admissible Control Operators
4.7 Examples
5 Symmetric Variational Problems
5.1 Preliminaries
5.2 The First Weak Turnpike Result
5.3 Auxiliary Results for the Turnpike Property
5.4 A Turnpike Result
5.5 The Second Weak Turnpike Result
5.6 The Space of Integrands
5.7 Auxiliary Results for Theorem 5.9
5.8 Proof of Theorem 5.9
5.9 Stability of the Weak Turnpike
5.10 Two Extensions of Theorem 5.9
5.11 The Turnpike Phenomenon in the Regions Close to the Right End Points
6 Infinite Dimensional Optimal Control
6.1 Preliminaries
6.2 The First Turnpike Result
6.3 The Space of Integrands
6.4 The Turnpike Property
6.5 The Second Weak Turnpike Result
6.6 A Well-Posedness Result
6.7 Stability of the Turnpike Phenomenon
6.8 An Auxiliary Result for Theorem 6.11
6.9 Proof of Theorem 6.11
6.10 Stability of the Weak Turnpike Phenomenon
6.11 The First Extension of Theorem 6.11
6.12 The Second Extension of Theorem 6.11
6.13 The Turnpike Property in the Regions Close to the Right end Points
7 Optimization Problems Arising in Crystallography
7.1 Preliminaries
7.2 Auxiliary Results
7.3 A Basic Lemma
7.4 Proof of Theorem 7.2
7.5 A Porosity Result
7.6 The Set MMr is Porous
7.7 Auxiliary Results
7.8 Proof of Theorem 7.11
8 Discrete Dispersive Dynamical Systems
8.1 Dynamical Systems with a Lyapunov Function
8.2 Proof of Theorem 8.1
8.3 Proofs of Propositions 8.2, 8.3, and 8.5
8.4 Proof of Theorem 8.6
8.5 Proof of Theorem 8.7
8.6 Proof of Theorem 8.9
8.7 The First Weak Turnpike Result
8.8 An Auxiliary Result
8.9 Proof of Theorem 8.10
8.10 The Second Weak Turnpike Result
8.11 An Auxiliary Result
8.12 Proof of Theorem 8.12
8.13 Stability Results
8.14 An Auxiliary Result
8.15 Proof of Theorem 8.14
8.16 Proof of Theorem 8.15
References
Index

Citation preview

Springer Optimization and Its Applications 190

Alexander J. Zaslavski

Turnpike Phenomenon and Symmetric Optimization Problems

Springer Optimization and Its Applications Volume 190 Series Editors Panos M. Pardalos , University of Florida My T. Thai , University of Florida Honorary Editor Ding-Zhu Du, University of Texas at Dallas Advisory Editors Roman V. Belavkin, Middlesex University John R. Birge, University of Chicago Sergiy Butenko, Texas A&M University Vipin Kumar, University of Minnesota Anna Nagurney, University of Massachusetts Amherst Jun Pei, Hefei University of Technology Oleg Prokopyev, University of Pittsburgh Steffen Rebennack, Karlsruhe Institute of Technology Mauricio Resende, Amazon Tamás Terlaky, Lehigh University Van Vu, Yale University Michael N. Vrahatis, University of Patras Guoliang Xue, Arizona State University Yinyu Ye, Stanford University

Aims and Scope Optimization has continued to expand in all directions at an astonishing rate. New algorithmic and theoretical techniques are continually developing and the diffusion into other disciplines is proceeding at a rapid pace, with a spot light on machine learning, artificial intelligence, and quantum computing. Our knowledge of all aspects of the field has grown even more profound. At the same time, one of the most striking trends in optimization is the constantly increasing emphasis on the interdisciplinary nature of the field. Optimization has been a basic tool in areas not limited to applied mathematics, engineering, medicine, economics, computer science, operations research, and other sciences. The series Springer Optimization and Its Applications (SOIA) aims to publish state-of-the-art expository works (monographs, contributed volumes, textbooks, handbooks) that focus on theory, methods, and applications of optimization. Topics covered include, but are not limited to, nonlinear optimization, combinatorial optimization, continuous optimization, stochastic optimization, Bayesian optimization, optimal control, discrete optimization, multi-objective optimization, and more. New to the series portfolio include Works at the intersection of optimization and machine learning, artificial intelligence, and quantum computing. Volumes from this series are indexed by Web of Science, zbMATH, Mathematical Reviews, and SCOPUS.

More information about this series at https://link.springer.com/bookseries/7393

Alexander J. Zaslavski

Turnpike Phenomenon and Symmetric Optimization Problems

Alexander J. Zaslavski Department of Mathematics Technion – Israel Institute of Technol Haifa, Israel

ISSN 1931-6828 ISSN 1931-6836 (electronic) Springer Optimization and Its Applications ISBN 978-3-030-96972-1 ISBN 978-3-030-96973-8 (eBook) https://doi.org/10.1007/978-3-030-96973-8 Mathematics Subject Classification: 49J20, 49K20, 49K27, 49K40, 90C26, 90C30, 90C31, 90C48 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

The book is devoted to the study of symmetric optimization, variational and optimal control problems in infinite dimensional spaces, and turnpike properties of their approximate solutions. It is a well-known fact that solutions exist for minimization problems on compact metric spaces with lower semicontinuous objective functions. Solutions also exist for minimization problems on metric spaces such that all their bounded closed subsets are compact if objective functions are lower semicontinuous and satisfy a coercivity growth condition. Since in the book we do not use compactness assumptions on spaces where optimization problems are considered, the situation becomes more difficult. In order to overcome this difficulty, we use the so-called generic approach which is applied fruitfully in many areas of analysis (see, for example, [11, 25, 37–40, 42, 62–64, 79, 97, 98, 100, 101, 104, 117, 119] and the references mentioned there). According to the generic approach, we say that a property holds for a generic (typical) element of a complete metric space (or the property holds generically) if the set of all elements of the metric space possessing this property contains a Gδ everywhere dense subset of the metric space, which is a countable intersection of open everywhere dense sets. In our book [143] we use this approach in order to establish a generic existence of solutions for various classes of minimization problems. In Chapters 4–10 of Zaslavski [143], it is shown that solutions of minimization problems exist generically for different classes of problems which are identified with corresponding spaces of functions with natural complete metrics. Many generic results of this type are collected in Zaslavski [143]. Among them, generic existence results for classes of constrained minimization problems with different type of constraints, for classes of parametric minimization problems, for classes of problems with increasing objective functions, and for classes of vector minimization problems and infinite horizon optimization problems. Any of these classes of problems is identified with a space of functions equipped with a natural complete metric, and it is shown that there exists a Gδ everywhere dense subset of the space of functions such that for any element of this subset the corresponding minimization problem possesses a unique solution and that any v

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Preface

minimizing sequence converges to this unique solution. These results are obtained as realizations of variational principles which are extensions or concretization of the variational principle established in Ioffe and Zaslavski [62]. Instead of considering the existence of a solution for a single minimization problem, we investigate it for a class (space) of problems and show that a unique solution exists for most of the problems in the class. It turns out that our results provide a proper explanation of what happens with individual problems in practice. The thing is that because of computational errors and errors in data which always present, instead of solving an individual problem with certain objective (constraint) function we actually solve a similar problem with an approximation of the objective (constraint) function. Probably this new problem possesses desirable properties which hold for most of the problems. This explains, for example, the fact that sometimes in practice we can get results which are better than their theoretical expectations. This happens when an algorithm works under some conditions which hold for “almost all” problems. Another type of generic existence and well-posedness results is obtained in Mizel and Zaslavski [89] where some class of minimization problems arising in crystallography is studied. Again, like in Zaslavski [143], this class of problems is identified with a space of functions equipped with a natural complete metric. It is shown that there exists a Gδ everywhere dense subset of the space of functions such that for any element of this subset the corresponding minimization problem possesses exactly two different solutions. This happens because the optimization problems in Mizel and Zaslavski [89] are symmetric. Namely, all the objective functions there are even. The result of Mizel and Zaslavski [89] and the importance of the optimization problems arising in crystallography give a strong motivation for our current research on symmetric optimization problems which is presented in this book. In Chapter 2 of this book, we consider several classes of symmetric optimization problems. Any of these classes of problems are identified with a space of functions equipped with a natural complete metric, and it is shown that there exists a Gδ everywhere dense subset of the space of functions such that for any element of this subset the corresponding minimization problem possesses exactly two different solutions and that any minimizing sequence converges to this solution set. These results are obtained as realizations of a variational principle. Some results of Chapter 2 were obtained recently in Zaslavski [171, 174] while some others are new. In Chapter 3, we discuss existence of solutions and well-posedness of for certain classes of parametric minimization problems. Its results are new. In Chapter 4, we present preliminaries which we need in order to study turnpike properties of infinite dimensional optimal control problems. We discuss Banach space valued functions, unbounded operators, C0 semigroups, evolution equations, and admissible control operators. In Chapters 5 and 6, we study the turnpike phenomenon for symmetric variational and optimal control problems in infinite dimensional spaces. To have the turnpike property means, roughly speaking, that the approximate solutions of the problems are determined mainly by the objective function and are essentially independent of the choice of interval and endpoint conditions, except in regions close to the endpoints.

Preface

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The turnpike property was discovered by P. Samuelson in 1948 when he showed that an efficient expanding economy would spend most of the time in the vicinity of a balanced equilibrium path (also called a von Neumann path). It is well known in the economic literature, where it was studied for various models of economic growth. Usually for these models a turnpike is a singleton. Now it is well known that the turnpike property is a general phenomenon which holds for large classes of variational and optimal control problems. In our research, using the Baire category (generic) approach, it was shown that the turnpike property holds for a generic (typical) variational problem [134] and for a generic optimal control problem [152]. In this book, we are interested in individual (non-generic) turnpike results for symmetric variational and optimal control problems and in the stability of the turnpike phenomenon under small perturbations of integrands. It turns out that for these problems the turnpike is a singleton x, ¯ such that (x, ¯ u) ¯ is a global minimizer of the integrand for some u. ¯ It is interesting to note that if u¯ is not zero, then the turnpike is not an admissible trajectory (function) for the corresponding problem. This is a new phenomenon which does not occur in our previous study of the turnpike properties for problems without symmetry. In Chapter 5, we study turnpike properties of symmetric variational problems, while optimal control problems of symmetric type are analyzed in Chapter 6. The results of Chapters 5 and 6 are new. In Chapter 7, we present generic results for symmetric optimization problems arising in crystallography obtained in Mizel and Zaslavski [89], Zaslavski [136]. Chapter 8 contains turnpike results for discrete dispersive dynamical systems obtained in Zaslavski [141, 172, 173, 175]. Haifa, Israel October 19, 2021

Alexander J. Zaslavski

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Generic Existence of Solutions of Minimization Problems . . . . . . . . . 1.2 Optimization Problems Arising in Crystallography. . . . . . . . . . . . . . . . . . 1.3 Symmetric Optimization Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Turnpike Property for Variational Problems. . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Variational Problems with Symmetric Integrands. . . . . . . . . . . . . . . . . . . . 1.6 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 7 9 11 16 20 21

2

25 25 27 32 34 36 40 43 47 47 52 58

Symmetric Optimization Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 A Generic Approach in Optimization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 A Generic Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The First Generic Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 The Second and Third Generic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Proofs of Theorems 2.5 and 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 The Fourth Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 The Fifth Generic Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Auxiliary Results for Theorems 2.14 and 2.15 . . . . . . . . . . . . . . . . . . . . . . . 2.9 Proof of Theorems 2.14 and 2.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.10 σ -Porous Sets in a Metric Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11 A Variational Principle and Porosity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12 Well-Posedness and Porosity for Classes of Minimization Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.13 An Auxiliary Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.14 A Well-Posedness Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.15 An Extension of Theorem 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.16 Extensions of Theorems 2.5 and 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17 Extension of Theorem 2.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.18 An Extension of Theorem 2.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65 68 71 73 76 81 85

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3

Parametric Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 3.1 Generic Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 3.2 Concretization of the Hypothesis (H) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 3.3 The First Generic Existence Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 3.4 The Second Generic Existence Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

4

Infinite Dimensional Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Banach Space Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Unbounded Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 C0 Semigroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Evolution Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 C0 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Admissible Control Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

109 109 111 113 114 115 116 124

5

Symmetric Variational Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 The First Weak Turnpike Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Auxiliary Results for the Turnpike Property . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 A Turnpike Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 The Second Weak Turnpike Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 The Space of Integrands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Auxiliary Results for Theorem 5.9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Proof of Theorem 5.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 Stability of the Weak Turnpike . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.10 Two Extensions of Theorem 5.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.11 The Turnpike Phenomenon in the Regions Close to the Right End Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

131 131 137 139 144 149 154 155 162 170 176

Infinite Dimensional Optimal Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 The First Turnpike Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 The Space of Integrands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 The Turnpike Property. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 The Second Weak Turnpike Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 A Well-Posedness Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Stability of the Turnpike Phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 An Auxiliary Result for Theorem 6.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 Proof of Theorem 6.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.10 Stability of the Weak Turnpike Phenomenon . . . . . . . . . . . . . . . . . . . . . . . . 6.11 The First Extension of Theorem 6.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.12 The Second Extension of Theorem 6.11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.13 The Turnpike Property in the Regions Close to the Right end Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

193 193 201 208 213 218 223 227 228 235 243 249 253

6

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7

Optimization Problems Arising in Crystallography . . . . . . . . . . . . . . . . . . . . . 7.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 A Basic Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Proof of Theorem 7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 A Porosity Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 The Set M \ Mr is Porous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8 Proof of Theorem 7.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

267 267 270 274 280 281 282 283 285

8

Discrete Dispersive Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Dynamical Systems with a Lyapunov Function . . . . . . . . . . . . . . . . . . . . . . 8.2 Proof of Theorem 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Proofs of Propositions 8.2, 8.3, and 8.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Proof of Theorem 8.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Proof of Theorem 8.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Proof of Theorem 8.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 The First Weak Turnpike Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8 An Auxiliary Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9 Proof of Theorem 8.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.10 The Second Weak Turnpike Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.11 An Auxiliary Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.12 Proof of Theorem 8.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.13 Stability Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.14 An Auxiliary Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.15 Proof of Theorem 8.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.16 Proof of Theorem 8.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

295 295 299 300 302 304 304 306 306 308 312 313 315 319 320 323 324

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335

Chapter 1

Introduction

In this chapter we consider optimization problems on complete metric spaces without compactness assumptions, optimization problems arising in crystallography and symmetric optimization problems in abstract spaces. We also discuss turnpike properties in the calculus of variations. To have the turnpike property means, roughly speaking, that the approximate solutions of the problems are determined mainly by the integrand and are essentially independent of the choice of interval and endpoint conditions, except in regions close to the endpoints.

1.1 Generic Existence of Solutions of Minimization Problems In our monograph [143] we use the generic approach in order to study the existence of solutions of various optimization problems in Banach spaces and complete metric spaces and show that for most of problems (in the sense of Baire category) all minimizing sequences converge to a unique solution. Here we demonstrate simple applications of this generic approach in optimization. Let (X, ρ) be a bounded complete metric space. Put diam (X) = sup{ρ(x, y) : x, y ∈ X} and denote by M the space of all bounded continuous functions f : X → R 1 equipped with a metric d(f1 , f2 ) = sup{|f1 (x) − f2 (x)| : x ∈ X}, f1 , f2 ∈ M.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 A. J. Zaslavski, Turnpike Phenomenon and Symmetric Optimization Problems, Springer Optimization and Its Applications 190, https://doi.org/10.1007/978-3-030-96973-8_1

1

2

1 Introduction

Clearly, (M, d) is a complete metric space. It is well-known that if the space X is compact, then for every f ∈ M, the problem minimize f (x) subject to x ∈ X possesses a solution. Since we do not assume the compactness of X the situation becomes more difficult and less understood. Nevertheless, there exists a subset F ⊂ M which is a countable intersection of open everywhere dense subsets of M such that for each element of F the corresponding minimization problem has a unique solution. Denote by L the set of all f ∈ M for which the problem minimize f (x) subject to x ∈ X possesses a solution. It turns out that L is an everywhere dense subset of M. Indeed, let f ∈ M and  be a positive number. There exists x ∈ X such that f (x ) ≤ f (x) +  for all x ∈ X. Put f (x) = max{f (x), f (x )}, x ∈ X. Clearly, f ∈ M, x is a point of minimum of the function f and therefore f ∈ L. It is not difficult to see that for all x ∈ X, f (x) ≤ f (x) ≤ f (x) +  and that d(f, f ) ≤ . Thus L is an everywhere dense subset of M. For any f ∈ L let xf ∈ X be a point of minimum of the function f such that f (xf ) ≤ f (x) for all x ∈ X. Let f ∈ L and n be a natural number. Define fn ∈ M by fn (x) = f (x) + (4n)−1 (diam(X) + 1)−1 ρ(x, xf ), x ∈ X. It is easy to see that d(f, fn ) ≤ (4n)−1 . Choose a number δ(f, n) ∈ (0, (12n2 )−1 (diam(X) + 1)−1 )

1.1 Generic Existence of Solutions of Minimization Problems

3

and set V (f, n) = {g ∈ M : d(fn , g) < δ(f, n)}. Assume that g ∈ V (f, n), x ∈ X and g(x) ≤ inf{g(z) : z ∈ X} + δ(f, n). It follows from the relations above, the definitions of fn , V (f, n), xf and δ(f, n) that f (x) + (4n)−1 (diam(X) + 1)−1 ρ(x, xf ) = fn (x) ≤ g(x) + δ(f, n) ≤ g(xf ) + 2δ(f, n) ≤ fn (xf ) + 3δ(f, n) = f (xf ) + 3δ(f, n) ≤ f (x) + 3δ(f, n) and that ρ(x, xf ) ≤ 3δ(f, n)4n(diam(X) + 1) < 1/n. Thus for each g ∈ V (f, n) the following property holds: if x ∈ X satisfies g(x) ≤ inf{g(z) : z ∈ X} + δ(f, n), then ρ(x, xf ) < 1/n. Set F = ∩∞ n=1 ∪ {V (f, k) : f ∈ L and an integer k ≥ n}. Clearly, F is a countable intersection of open everywhere dense subsets of M. Assume that g ∈ F and that n is a natural number. By definition of F, there exists a natural number k ≥ n and f ∈ L such that g ∈ V (f, k). Combined with the property above this implies that the following property holds: for each x ∈ X satisfying g(x) ≤ inf{g(z) : z ∈ X} + δ(f, k) the inequality ρ(x, xf ) < 1/n holds. Since n is an arbitrary natural number we conclude that any minimizing sequence for the problem minimize g(z) subject to z ∈ X converges to its unique solution, where g is an arbitrary element of F. There is also another way to prove the result obtained above. Let f ∈ M and n be a natural number. Choose a positive number δ(f, n) < (32n2 )−1 (diam(X) + 1)−1

4

1 Introduction

and choose xf ∈ X such that f (xf ) ≤ inf{f (z) : z ∈ X} + δ(f, n). Put fn (z) = f (z) + (4n)−1 (diam(X) + 1)−1 ρ(z, xf ), z ∈ X. Clearly fn ∈ M and d(f, fn ) ≤ (4n)−1 . Set V (f, n) = {g ∈ M : d(fn , g) < δ(f, n)}. Assume that g ∈ V (f, n), x ∈ X and g(x) ≤ inf{g(z) : z ∈ X} + δ(f, n). By the relations above, the definitions of fn , V (f, n), xf , and δ(f, n), f (x) + (4n)−1 (diam(X) + 1)−1 ρ(x, xf ) = fn (x) ≤ g(x) + δ(f, n) ≤ g(xf ) + 2δ(f, n) ≤ fn (xf ) + 3δ(f, n) = f (xf ) + 3δ(f, n) ≤ f (x) + 4δ(f, n) and ρ(x, xf ) ≤ 4δ(f, n)4n(diam(X) + 1) < 1/n. Thus for each g ∈ V (f, n) the following property holds: if x ∈ X satisfies g(x) ≤ inf{g(z) : z ∈ X} + δ(f, n), then ρ(x, xf ) < 1/n. Set F = ∩∞ k=1 ∪ {V (f, n) : f ∈ M and an integer n ≥ k}. Clearly, F is a countable intersection of open everywhere dense subsets of M. Assume that g ∈ F and that k is a natural number. By definition of F, there exists a natural number n ≥ k and f ∈ M such that g ∈ V (f, n). Combined with the property above this implies that the following property holds: for each x ∈ X satisfying g(x) ≤ inf{g(z) : z ∈ X} + δ(f, n) the inequality ρ(x, xf ) < 1/k holds.

1.1 Generic Existence of Solutions of Minimization Problems

5

Since k is an arbitrary natural number we conclude that any minimizing sequence for the problem minimize g(z) subject to z ∈ X converges to its unique solution, where g is an arbitrary element of F. We presented above two proofs of the same generic well-posedness result. In both proofs for a given f ∈ M and a positive number  we defined a function f¯ ∈ M, x¯ ∈ X and a positive constant δ such that if x ∈ X is a δ-approximate solution of the problem g(z) → min, z ∈ X, where g belongs to a δ-neighborhood of f¯, then x belongs to an -neighborhood of x. ¯ Many results of this type for various classes of minimization problems are collected in [143]. Here we briefly describe some of them. In Chapter 4 of [143] we consider problems minimize f (x) subject to x ∈ X where X is a complete metric space and f belongs to a space of lower semicontinuous functions on X satisfying a certain growth condition. The class of minimization problems is identified with this space of functions. We endow the space of functions with an appropriate metric and show that for most functions f in this metric space the corresponding minimization problem has a unique solution and is well-posed. In Chapter 4 of [143] we also consider the following class of minimization problems minimize f (x) subject to x ∈ A studied in [62, 130, 131], where A is a nonempty closed subset of a complete metric space X and f belongs to a space of lower semicontinuous functions on X. This class of problems is identified with a space of pairs (f, A) which is equipped with appropriate complete uniformities. It is shown that for a typical pair (f, A) the corresponding minimization problem has a unique solution and is well-posed. In Chapter 5 of [143] we continue to consider various classes of minimization problems showing that most problems in these classes are well-posed. In that chapter in order to meet this goal we use a porosity notion. As in Chapter 4 we identify a class of minimization problems with a certain complete metric space of functions, study the set of all functions for which the corresponding minimization problem is well-posed, and show that the complement of this set is not only of the first category but also a σ -porous set. We now recall the concept of porosity [104, 143]. Let (Y, d) be a complete metric space. We denote by Bd (y, r) the closed ball of center y ∈ Y and radius r > 0. A subset E ⊂ Y is called porous in (Y, d) if there exist α ∈ (0, 1] and r0 > 0 such that for each r ∈ (0, r0 ] and each y ∈ Y there

6

1 Introduction

exists z ∈ Y for which Bd (z, αr) ⊂ Bd (y, r) \ E. A subset of the space Y is called σ -porous in (Y, d) if it is a countable union of porous subsets in (Y, d). Since porous sets are nowhere dense, all σ -porous sets are of the first category. If Y is a finite dimensional Euclidean space, then σ -porous sets are of Lebesgue measure 0. In fact, the class of σ -porous sets in such a space is smaller than the class of sets which have measure 0 and are of the first category. To point out the difference between porous and nowhere dense sets note that if E ⊂ Y is nowhere dense, y ∈ Y and r > 0, then there is a point z ∈ Y and a number s > 0 such that Bd (z, s) ⊂ Bd (y, r) \ E. If, however, E is also porous, then for small enough r we can choose s = αr, where α ∈ (0, 1) is a constant which depends only on E. In Chapter 5 of [143] we employ the notion of porosity in order to study wellposedness of convex minimization problems and well-posedness for a class of equilibrium problems. The book [143] contains many other generic existence and well-posedness results established for various classes of optimization problems. In particular, in Chapter 6 of [143] we study a parametric family of the problems minimize f (b, x) subject to x ∈ X on a complete metric space X with a parameter b which belongs to a Hausdorff compact space B. Here f (·, ·) belongs to a space of functions on B × X endowed with an appropriate uniform structure. Using the generic approach and the notion of porosity we show that for a typical function f (·, ·) the minimization problem has a solution for all parameters b ∈ B. Chapter 7 of [143] is devoted to the study of problems minimize f (x) subject to x ∈ K where K is a closed subset of an ordered Banach space X and f belongs to a space of increasing lower semicontinuous functions on K. Using the generic approach and the notion of porosity we show that for most functions f in this space the corresponding minimization problem has a unique solution. In Chapter 8 of [143] we study minimization problems with mixed constraints minimize f (x) subject to G(x) = y, H (x) ≤ z, where f is a continuous (differentiable) finite valued function defined on a Banach space X, y is an element of a finite dimensional Banach space Y , z is an element of a Banach space Z ordered by a convex closed cone and G : X → Y and H : X → Z are continuous (differentiable) mappings. We consider two classes of these problems

1.2 Optimization Problems Arising in Crystallography

7

and show that most of the problems (in the Baire category sense) are well-posed. Our first class of problems is identified with the corresponding complete metric space of quintets (f, G, H, y, z). We show that for a generic quintet (f, G, H, y, z) the corresponding minimization problem has a unique solution and is well-posed. Our second class of problems is identified with the corresponding complete metric space of triples (f, G, H ) while y and z are fixed. We show that for a generic triple (f, G, H ) the corresponding minimization problem has a unique solution and is well-posed. In Chapter 9 of [143] we study classes of vector minimization problems on a complete metric space which are identified with the corresponding complete metric spaces of objective functions. We show that for most (in the sense of Baire category) functions the corresponding vector optimization problem has a solution. Using generic approach we also study other interesting properties of such classes of problems. In all the results discussed above, which are presented in [143], a class of minimization problems on a complete metric space is identified with the corresponding complete metric space of function and it is shown that a generic (typical) problem has a unique solution which is the limit of any minimizing sequence.

1.2 Optimization Problems Arising in Crystallography We discuss the structure of minimizers of variational problems considered in [56, 66, 89, 136] which describe step-terraces on surfaces of crystals. It is well-known in surface physics that when a crystalline substance is maintained at a temperature T above its roughening temperature TR then the surface stored energy integrand, usually referred to as surface tension, is a smooth function β of the azimuthal angle of orientation θ . Furthermore, β obeys the following: β(−θ ) = β(π − θ ) = β(θ ), 0 < β(π/2) ≤ β(θ ) ≤ β(0). The classical model is given by  J (y) =

S

β(θ )ds 0

where s is arclength and y is a function defined on a fixed interval [0, L] whose graph is the locus under consideration: y ∈ W 1,1 (0, L), θ = arctan y ∈ [−π/2, π/2], while β is a positive π -periodic function which belongs to a space of functions described below. Minimization of J subject to appropriate boundary data is a parametric variational problem. It is closely related to the variational problem

8

1 Introduction

defining the Wulff crystal shape as that shape for a domain of prescribed area such that the boundary integral with respect to arclength involving the integrand in J [referred to as the surface tension] attains its minimum value. For each function f : X → R 1 set inf(f ) = inf{f (x) : x ∈ X}. Denote by M the set of all functions β ∈ C 2 (R 1 ) which satisfy the following assumption: (A) β(t) ≥ 0 for every number t ∈ R 1 , β(π/2) ≤ β(t) ≤ β(0) for every number t ∈ R 1 , β(t) = β(−t) for every number t ∈ R 1 , β(t + π ) = β(t) for every number t ∈ R 1 , β(0) + β

(0) ≤ 0. For every pair of functions β1 , β2 ∈ M put (i)

(i)

ρ(β1 , β2 ) = sup{|β1 (t) − β2 (t)| : t ∈ R 1 , i = 0, 1, 2}. It is not difficult to show that the metric space (M, ρ) is complete. Denote by Mr the collection of all functions β ∈ M which satisfy β(t) > 0 for all t ∈ R 1 , β(0) + β

(0) < 0. It is clear that the set Mr is nonempty. The following result was obtained in [89]. Proposition 1.1 Mr is an open everywhere dense subset of the metric space (M, ρ). Let β ∈ Mr be given. Set Gβ (z) = β(arctan(z))(1 + z2 )1/2 , z ∈ R 1 . Evidently, Gβ is a continuous function and Gβ (z) → ∞ as z → ±∞, inf(Gβ < β(0). Note that the final inequality above was shown in [56]. We can rewrite the variational functional J in the form  J (y) = 0

L

Gβ (y )dx.

1.3 Symmetric Optimization Problems

9

It was shown in [56] that a function y ∈ W 1,1 (0, L) is a minimizer of the functional J if and only if |y | ∈ {z ∈ R 1 : Gβ (z) = inf(Gβ )} almost everywhere (a. e.). In [89] using the Baire category approach, it was shown that for a generic (typical) function β the set {z ∈ R 1 : Gβ (z) = inf(Gβ )} = {zβ , −zβ } where zβ is a unique positive number depending only on the function β. Denote by F the set of all β ∈ Mr satisfying the following condition: (C) There exists a number zβ ∈ R 1 such that Gβ (z) > Gβ (zβ ) for every number z ∈ R 1 \ {zβ , −zβ }. The following result was obtained in [89]. Theorem 1.2 F contains a countable intersection of open everywhere dense subsets of (M, ρ). Therefore, in contrast with the results discussed in the previous section, for a generic (typical) β ∈ Mr the minimization problem Gβ (z) → min, z ∈ R 1 has exactly two different minimizers. This happens because the optimization problem above is symmetric. Namely, Gβ (−z) = Gβ (z), z ∈ R 1 . The result stated above and the importance of the optimization problems arising in crystallography give a strong motivation for our current research on symmetric optimization problems which is presented in this book. In the next section we discuss some of these symmetric problems.

1.3 Symmetric Optimization Problems Assume that (X, ρ) is a complete metric space. Denote by Ml the set of all lower semicontinuous and bounded from below functions f : X → R 1 . We equip the set Ml with the uniformity determined by the following base E() = {(f, g) ∈ Ml × Ml : |f (x) − g(x)| ≤  for all x ∈ X},

10

1 Introduction

where  > 0. It is known that this uniformity is metrizable (by a metric d) and complete [143]. Denote by Mc the set of all continuous functions f ∈ Ml . It is not difficult to see that Mc is a closed subset of Ml . For each f ∈ Ml set inf(f ) = inf{f (x) : x ∈ X}. Consider a minimization problem f (x) → min, x ∈ X, where f ∈ Ml . Assume that a mapping T : X → X is continuous and the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X. Let AT be either the set of all f ∈ Ml such that f (T (x)) = f (x) for all x ∈ X or the set of all f ∈ Mc satisfying the equation above. Clearly, AT is a closed subset of Ml . It is equipped with the relative topology induced by the metric d. Given f ∈ AT we say that the problem of minimization of f on X is well-posed with respect to data in Ml if the following assertions hold: (1) there exists xf ∈ X such that {x ∈ X : f (x) = inf(f )} = {xf , T (xf )}. (2) For each  > 0 there is a neighborhood V of f in Ml and δ > 0 such that for each g ∈ V if z ∈ X satisfies g(z) ≤ inf(g) + δ, then |g(z) − f (xf )| ≤  and min{ρ(z, {xf , T (xf )}), ρ(T (z), {xf , T (xf )})} ≤ . In [171] it was shown that there exists a set F ⊂ AT which is a countable intersection of open and everywhere dense sets in AT such that for each f ∈ F, the minimization problem of f on X is well-posed with respect to Ml .

1.4 Turnpike Property for Variational Problems

11

This result was also obtained in [174] using a general variational principle for symmetric optimization problems. More precisely, the following result was obtained in [174]. Theorem 1.3 There exists an everywhere dense set B ⊂ AT which is a countable intersection of open subsets of AT such that for any f ∈ B the minimization problem of f on X is well-posed with respect to Ml . This result as well as many other generic well-posedness results for symmetric problems are collected in Chapter 2. Here we present one of them (see Theorem 2.31) related to the theorem above. Denote by Fix(T ) the set of all x ∈ X such that T (x) = x. Since the mapping T is continuous the set Fix(T ) is closed. Note that it can be empty. We assume that the set X \ Fix(T ) is everywhere dense. Denote by Mf the set of all f ∈ Ml which are continuous at any point of Fix(T ). If Fix(T ) = ∅, then Mf = Ml . Clearly, Mf is a closed subset of Ml . Now assume that AT is either the set of all f ∈ Mf such that f (T (x)) = f (x) for all x ∈ X or the set of all f ∈ Mc satisfying the equation above. Clearly, AT is a closed subset of Ml . It is equipped with the relative topology. The following result is proved in Chapter 2. Theorem 1.4 There exists an everywhere dense set B ⊂ AT which is a countable intersection of open subsets of AT such that for any f ∈ B the minimization problem of f on X is well-posed with respect to A and has exactly two different minimizers.

1.4 Turnpike Property for Variational Problems The study of the existence and the structure of solutions of optimal control problems and dynamic games defined on infinite intervals and on sufficiently large intervals has been a rapidly growing area of research [7–9, 20, 21, 29, 41, 44–46, 65, 69, 70, 83, 95, 134, 138, 142, 145, 151, 155, 160, 161, 164, 166, 169] which has various applications in engineering [5, 29, 76, 134], in models of economic growth [6, 10, 27–30, 43, 48, 60, 61, 68, 75, 82, 88, 96, 108, 110, 111, 118, 134, 149, 156, 165, 166], in infinite discrete models of solid-state physics related to dislocations in onedimensional crystals [12, 104, 122], in model predictive control [36, 51], and in the

12

1 Introduction

theory of thermodynamical equilibrium for materials [34, 77, 85–87]. Discrete-time problems optimal control problems were considered in [9, 13, 14, 24, 47, 57, 123, 124, 129, 133, 140, 144, 146, 149, 153, 154, 157, 162, 163, 165], finite dimensional continuous-time problems were analyzed in [20, 22, 23, 26, 31, 75, 78, 81, 84, 99, 125, 127, 132, 135, 152, 167, 168], infinite dimensional optimal control was studied in [29, 30, 52–54, 91, 93, 94, 102, 109, 113, 114, 126, 128, 143, 170] while solutions of dynamic games were discussed in [19, 49, 50, 55, 58, 72, 103, 147, 150, 158, 159]. Sufficient and necessary conditions for the turnpike phenomenon were obtained in [132, 133, 135, 153] for finite dimensional variational problems and for discretetime optimal control problems in compact metric space. In this section, which is based on [132], we discuss the structure of approximate solutions of variational problems with continuous integrands f : [0, ∞) × R n × R n → R 1 which belong to a complete metric space of functions. We do not impose any convexity assumption. The main result of this section, obtained in [132], deals with the turnpike property of variational problems. We consider the variational problems 

T2

f (t, z(t), z (t))dt → min, z(T1 ) = x, z(T2 ) = y,

(P )

T1

z : [T1 , T2 ] → R n is an absolutely continuous function (a. c.), where T1 ≥ 0, T2 > T1 , x, y ∈ R n and f : [0, ∞) × R n × R n → R 1 belongs to a space of integrands described below. It is well-known that the solutions of the problems (P) exist for integrands f which satisfy two fundamental hypotheses concerning the behavior of the integrand as a function of the last argument (derivative): one that the integrand should grow superlinearly at infinity and the other that it should be convex [33, 112]. Moreover, certain convexity assumptions are also necessary for properties of lower semicontinuity of integral functionals which are crucial in most of the existence proofs, although there are some interesting theorems without convexity [32, 90, 92]. For integrands f which do not satisfy the convexity assumption the existence of solutions of the problems (P) is not guaranteed and in this situation we consider δ-approximate solutions. Let T1 ≥ 0, T2 > T1 , x, y ∈ R n , f : [0, ∞) × R n × R n → R 1 be an integrand and let δ be a positive number. We say that an absolutely continuous (a.c.) function u : [T1 , T2 ] → R n satisfying u(T1 ) = x, u(T2 ) = y is a δ-approximate solution of the problem (P) if 

T2

T1



f (t, u(t), u (t))dt ≤



T2

f (t, z(t), z (t))dt + δ

T1

for each a.c. function z : [T1 , T2 ] → R n satisfying z(T1 ) = x, z(T2 ) = y.

1.4 Turnpike Property for Variational Problems

13

The main result of [132] deals with the turnpike property of the variational problems (P). As usual, to have this property means, roughly speaking, that the approximate solutions of the problems (P) are determined mainly by the integrand and are essentially independent of the choice of interval and endpoint conditions, except in regions close to the endpoints. In the classical turnpike theory, it was assumed that a cost function (integrand) is convex. The convexity of the cost function played a crucial role there. In [132] we get rid of convexity of integrands and establish necessary and sufficient conditions for the turnpike property for a space of nonconvex integrands M described below. Let us now define the space of integrands. Denote by | · | the Euclidean norm in R n . Let a be a positive constant and let ψ : [0, ∞) → [0, ∞) be an increasing function such that ψ(t) → +∞ as t → ∞. Denote by M the set of all continuous functions f : [0, ∞) × R n × R n → R 1 which satisfy the following assumptions: A(i) the function f is bounded on [0, ∞) × E for any bounded set E ⊂ R n × R n ; A(ii) f (t, x, u) ≥ max{ψ(|x|), ψ(|u|)|u|} − a for each (t, x, u) ∈ [0, ∞) × R n × Rn; A(iii) for each M,  > 0 there exist Γ, δ > 0 such that |f (t, x1 , u) − f (t, x2 , u)| ≤  max{f (t, x1 , u), f (t, x2 , u)} for each t ∈ [0, ∞) and each u, x1 , x2 ∈ R n which satisfy |xi | ≤ M, i = 1, 2, |u| ≥ Γ,

|x1 − x2 | ≤ δ;

A(iv) for each M,  > 0 there exists δ > 0 such that |f (t, x1 , u1 ) − f (t, x2 , u2 )| ≤  for each t ∈ [0, ∞) and each u1 , u2 , x1 , x2 ∈ R n which satisfy |xi |, |ui | ≤ M, i = 1, 2,

max{|x1 − x2 |, |u1 − u2 |} ≤ δ.

It is easy to show that an integrand f = f (t, x, u) ∈ C 1 ([0, ∞) × R n × R n ) belongs to M if f satisfies assumption A(ii), and if sup{|f (t, 0, 0)| : t ∈ [0, ∞)} < ∞ and also there exists an increasing function ψ0 : [0, ∞) → [0, ∞) such that sup{|∂f/∂x(t, x, u)|, |∂f/∂u(t, x, u)|} ≤ ψ0 (|x|)(1 + ψ(|u|)|u|) for each t ∈ [0, ∞) and each x, u ∈ R n . For the set M we consider the uniformity which is determined by the following base: E(N, , λ) = {(f, g) ∈ M × M : |f (t, x, u) − g(t, x, u)| ≤  for each t ∈ [0, ∞) and each x, u ∈ R n satisfying |x|, |u| ≤ N

14

1 Introduction

and (|f (t, x, u)| + 1)(|g(t, x, u)| + 1)−1 ∈ [λ−1 , λ] for each t ∈ [0, ∞) and each x, u ∈ R n satisfying |x| ≤ N }, where N > 0,  > 0, λ > 1. It is not difficult to show that the space M with this uniformity is metrizable (by a metric ρw ). It is known (see [132]) that the metric space (M, ρw ) is complete. The metric ρw induces in M a topology. We consider functionals of the form  T2 f f (t, x(t), x (t))dt I (T1 , T2 , x) = T1

where f ∈ M, 0 ≤ T1 < T2 < ∞ and x : [T1 , T2 ] → R n is an a.c. function. For f ∈ M, y, z ∈ R n and numbers T1 , T2 satisfying 0 ≤ T1 < T2 we set U f (T1 , T2 , y, z) = inf{I f (T1 , T2 , x) : x : [T1 , T2 ] → R n is an a.c. function satisfying x(T1 ) = y, x(T2 ) = z}. It is easy to see that −∞ < U f (T1 , T2 , y, z) < ∞ for each f ∈ M, each y, z ∈ R n and all numbers T1 , T2 satisfying 0 ≤ T1 < T2 . Let f ∈ M. A locally absolutely continuous (a.c.) function x : [0, ∞) → R n is called an (f )-good function [134] if for any a.c function y : [0, ∞) → R n there is a number My such that I f (0, T , y) ≥ My + I f (0, T , x) for each T ∈ (0, ∞). The following result was proved in [132]. Proposition 1.5 Let f ∈ M and let x : [0, ∞) → R n be a bounded a.c. function. Then the function x is (f )-good if and only if there is M > 0 such that I f (0, T , x) ≤ U f (0, T , x(0), x(T )) + M for any T > 0. Let us now give the precise definition of the turnpike property. Assume that f ∈ M. We say that f has the turnpike property, or briefly TP, if there exists a bounded continuous function Xf : [0, ∞) → R n which satisfies the following condition: For each K,  > 0 there exist constants δ, L > 0 such that for each x, y ∈ R n satisfying |x|, |y| ≤ K, each T1 ≥ 0, T2 ≥ T1 + 2L and each a.c. function v : [T1 , T2 ] → R n which satisfies v(T1 ) = x, v(T2 ) = y, I f (T1 , T2 , v) ≤ U f (T1 , T2 , x, y) + δ the inequality |v(t) − Xf (t)| ≤  holds for all t ∈ [T1 + L, T2 − L].

1.4 Turnpike Property for Variational Problems

15

The function Xf is called the turnpike of f . Assume that f ∈ M and X : [0, ∞) → R n is a bounded continuous function. How to verify if the integrand f has TP and X is its turnpike? In [132] we introduced two properties (P1) and (P2) and show that f has TP if and only if f possesses the properties (P1) and (P2). The property (P2) means that all (f )-good functions have the same asymptotic behavior while the property (P1) means that if an a.c. function v : [0, T ] → R n is an approximate solution and T is large enough, then there is τ ∈ [0, T ] such that v(τ ) is close to X(τ ). The next theorem is the main result [132]. Theorem 1.6 Let f ∈ M and Xf : [0, ∞) → R n be a bounded absolutely continuous function. Then f has the turnpike property with Xf being the turnpike if and only if the following two properties hold: (P1) For each K,  > 0 there exist γ , l > 0 such that for each T ≥ 0 and each a.c. function w : [T , T + l] → R n which satisfies |w(T )|, |w(T + l)| ≤ K, I f (T , T + l, w) ≤ U f (T , T + l, w(T ), w(T + l)) + γ there is τ ∈ [T , T + l] for which |Xf (τ ) − v(τ )| ≤ . (P2) For each (f )-good function v : [0, ∞) → R n , |v(t) − Xf (t)| → 0 as t → ∞. In [132] we proved the following theorem which is an extension of Theorem 1.6. Theorem 1.7 Let f ∈ M, Xf : [0, ∞) → R n be an (f )-good function. Assume that the properties (P1), (P2) hold. Then for each K,  > 0 there exist δ, L > 0 and a neighborhood U of f in M such that for each g ∈ U , each T1 ≥ 0, T2 ≥ T1 + 2L and each a.c. function v : [T1 , T2 ] → R n which satisfies |v(T1 )|, |v(T2 )| ≤ K, I g (T1 , T2 , v) ≤ U g (T1 , T2 , v(T1 ), v(T2 )) + δ the inequality |v(t) − Xf (t)| ≤  holds for all t ∈ [T1 + L, T2 − L]. In our recent [170] book we study sufficient and necessary conditions for the turnpike phenomenon, using the approach developed in [132, 133, 135, 153], for discrete-time optimal control problems in metric spaces, which are not necessarily compact, and for continuous-time infinite dimensional optimal control problems. Its main results have Theorem 1.6 as their prototype. In the present book we study the turnpike phenomenon for problems with symmetric integrands. It turns out that for these problems the turnpike is a singleton which is a global minimizer of the integrand. Since the discovery of the turnpike phenomenon by Paul Samuelson in 1948, different versions of the turnpike property were considered in the literature. In this book as well as in [132, 133, 135, 153, 170], we study the turnpike property introduced and used in our previous research [127, 134, 152, 153, 164]. This turnpike

16

1 Introduction

property differs from other versions and has important features. Our turnpike property is a property of approximate solutions. As it was shown in [127, 152], our turnpike property holds for most problems belonging to large classes of variational and optimal control problems.

1.5 Variational Problems with Symmetric Integrands Assume that f : R n × R n → R 1 is a bounded from below borelian function such that f (x, y) = f (x, −y) for all x, y ∈ R n , there exists (x, ¯ y) ¯ ∈ R n × R n such that f (x, ¯ y) ¯ = inf(f ) := inf{f (ξ, η) : ξ, η ∈ R n }, ¯ y), ¯ (x, ¯ −y)} ¯ {(x, y) ∈ R n × R n : f (x, y) = inf(f )} = {(x, and that the following assumptions hold: (A1) for each  > 0 there exists δ > 0 such that for each (x, y) ∈ R n × R n satisfying f (x, y) ≤ inf(f ) + δ the inequalities |x − x| ¯ ≤ and min{|y − y|, ¯ |y + y|} ¯ ≤ hold; (A2) for each  > 0 there exists δ > 0 such that for each (x, y) ∈ R n × R n satisfying |x − x| ¯ ≤ δ, |y − y| ¯ ≤δ the inequality f (x, y) ≤ f (x, ¯ y) ¯ + is true.

1.5 Variational Problems with Symmetric Integrands

17

Assumption (A2) means that the function f is continuous at the point (x, ¯ y) ¯ while assumption (A1) means that the minimization problem f (x, y) → min, x, y ∈ R n is well-posed. According to the results of Chapter 2 this well-posedness property holds for most symmetric objective functions. Let a > 0 and ψ : [0, ∞) → [0, ∞) be an increasing function satisfying lim ψ(t) = ∞.

t→∞

Assume that the following assumption holds: (A3) the function f is bounded on all bounded sets and for each (x, u) ∈ R × R n , f (x, u) ≥ ψ(|u|)|u| − a. For each pair of nonnegative numbers T1 < T2 and each y, z ∈ R n we denote by W 1,1 (T1 , T2 ) the set of all a. c. functions x : [T1 , T2 ] : R n , consider the problems 

T2

f (x(t), x (t))dt → min,

T1

(PT1 ,T2 )

x ∈ W 1,1 (T1 , T2 ), 

T2

f (x(t), x (t))dt → min,

T1

(PT1 ,T2 ,y )

x ∈ W 1,1 (T1 , T2 ), x(T1 ) = y, 

T2

f (x(t), x (t))dt → min,

T1

(PT1 ,T2 ,y,z )

x ∈ W 1,1 (T1 , T2 ), x(T1 ) = y, x(T2 ) = z and define  U (T1 , T2 ) = inf{

T2

f (x(t), x (t))dt : x ∈ W 1,1 (T1 , T2 )},

T1

 U (T1 , T2 , y) = inf{

T2

T1

f (x(t), x (t))dt : x ∈ W 1,1 (T1 , T2 ), x(T1 ) = y},

18

1 Introduction

 U (T1 , T2 , y, z) = inf{

T2

f (x(t), x (t))dt :

T1

x ∈ W 1,1 (T1 , T2 ), x(T1 ) = y, x(T2 ) = z}. There are two cases: y¯ = 0; y¯ = 0. If y¯ = 0, then for each T2 > T1 ≥ 0, the function x(t) = x, ¯ t ∈ [T1 , T2 ] is a solution of the problems (PT1 ,T2 ), (PT1 ,T2 ,x¯ ), (PT1 ,T2 ,x, ¯ x¯ ). For each pair of numbers T1 < T2 and each x ∈ W 1,1 (T1 , T2 ) set  I (T1 , T2 , x) =

T2

f (x(t), x (t))dt.

T1

In Chapter 5 we prove the following result. Theorem 1.8 Let T > 0. Then U (0, T ) = U (0, T , x) ¯ = U (0, T , x, ¯ x) ¯ = Tf (x, ¯ y). ¯ Moreover, for each  > 0 there exists x ∈ W 1,1 (0, T ) such that x(0) = x(T ) = x, ¯ I (0, T , x) ≤ Tf (x, ¯ y) ¯ + , |x(t) − x| ¯ ≤ , t ∈ [0, T ], ¯ −y}, ¯ t ∈ [0, T ] a. e.. x (t) ∈ {y, Note that if y¯ = 0, then for every positive T problems (P0,T ), (P0,T ,x¯ ) and (P0,T ,x, ¯ x¯ ) have no minimizers. The following useful result is also obtained in Chapter 5. Theorem 1.9 Let L0 , M0 > 0. Then there exist M1 > 0 such that for each T > L0 and each y, z ∈ R n satisfying |y|, |z| ≤ M0 the inequality U (0, T , y, z) ≤ Tf (x, ¯ y) ¯ + M1 holds. We denote by mes(Ω) the Lebesgue measure of a Lebesgue measurable set Ω ⊂ R1 The next theorem is the first turnpike result of Chapter 5. It shows that for approximate solutions x of our variational problems on intervals [0, T ], where T is sufficiently large and given values x(0), x(T ) at the end points belong to a given bounded set C, the Lebesgue measure of all points t ∈ [0, T ] such that (x(t), x (t))

1.5 Variational Problems with Symmetric Integrands

19

does not belong to an -neighborhood of the set {(x, ¯ y), ¯ (x, ¯ −y)} ¯ does not exceed a constant L which depends only on  and the set C and does not depend on T , x(0), x(T ). In the literature this property is known as the weak turnpike property. Theorem 1.10 Let  ∈ (0, 1) and L0 , M0 , M1 > 0. Then there exists L1 > L0 such that for each T > L1 and each x ∈ W 1,1 (0, T ) such that |x(0)| ≤ M0 and at least one of the following conditions holds: (a) |x(T )| ≤ M0 , I f (0, T , x) ≤ U (0, T , x(0), x(T )) + M1 ; (b) I f (0, T , x) ≤ U (0, T , x(0)) + M1 the inequality mes({t ∈ [0, T ] : max{|x(t) − x|, ¯ ¯ |x (t) + y|}} ¯ > }) ≤ L1 . min{|x (t) − y|, The following theorem is also proved in Chapter 5. It shows that for approximate solutions x of our variational problems on intervals [0, T ], where T is sufficiently large and given values x(0), x(T ) at the end points belong to a given bounded set C, the set of all points t ∈ [0, T ] such that x(t) does not belong to an -neighborhood of x¯ is contained in the union of two intervals, where the first interval contains 0, the second one contains T and their lengths do not exceed a constant L which depends only on  and the set C and does not depend on T , x(0), x(T ). In the literature this property is known as the turnpike property. Theorem 1.11 Let  ∈ (0, 1]. Then there exist L, δ > 0 such that for each T > 2L and each u ∈ W 1,1 (0, T ) such that |u(0)| ≤ M and at least one of the following conditions holds: |u(T )| ≤ M, I (0, T , u) ≤ U (0, T , u(0), u(T )) + δ; I (0, T , x) ≤ U (0, T , u(0)) + δ

20

1 Introduction

there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ] such that |u(t) − x| ¯ ≤ , t ∈ [τ1 , τ2 ]. ¯ ≤ δ, then τ2 = T . Moreover, if |u(0) − x| ¯ ≤ δ, then τ1 = 0 and if |u(T ) − x| Chapter 5 contains many other turnpike results, including the results of the stability of the turnpike phenomenon under small perturbations of integrands and on the turnpike phenomenon in the regions close to the right end points. The results of Chapter 5 are obtained for variational problems in infinite dimensional spaces.

1.6 Notation In this section we collect the notation which will be used in the book. Let (X, ρX ) be a metric space equipped with the metric ρX which induces the topology in X. For each x ∈ X and each r > 0 set BX (x, r) = {y ∈ X : ρ(x, y) ≤ r}. Let (Xi , ρXi ), i = 1, 2 be metric spaces. The set X1 × X2 is equipped with the metric ρX1 ×X2 ((x1 , x2 ), (y1 , y2 )) = ρX1 (x1 , y1 ) + ρX2 (x2 , y2 ), (x1 , x2 ), (y1 , y2 ) ∈ X1 × X2 . We denote by mes(Ω) the Lebesgue measure of a Lebesgue measurable set Ω ⊂ R 1 and define χΩ (x) = 1 for all x ∈ Ω, χΩ (x) = 0 for all x ∈ R 1 \ Ω. For each function f : X → [−∞, ∞], where X is nonempty, we set inf(f ) = inf{f (x) : x ∈ X}. For each s ∈ R 1 set s+ = max{s, 0}, s− = min{s, 0}, s = max{z : z is an integer and z ≤ s}. If (X,  · ) is a normed space, then X is equipped with the metric ρX (x, y) = x − y, x, y ∈ X. Let (X, ρX ) be a metric space and T be a Lebesgue measurable subset of R 1 . A function f : T → X is called Lebesgue measurable if for any Borel set D ⊂ X the set f −1 (D) is a Lebesgue measurable set.

1.7 Exercises

21

In the sequel we denote by Card(D) the cardinality of a set D, suppose that the sum over an empty set is zero and that the infimum over an empty set is ∞. Let (X, ·, ·X ) be a Hilbert space equipped with an inner product ·, ·X which induces the norm  · X . If the space X is understood, we use the notation ·, · = ·, ·X and  ·  =  · X . Let X be a Banach space equipped with the norm  · X and X∗ is its dual space with the norm  · X∗ . If the space X is understood, we use the notation  ·  =  · X . For x ∈ X and l ∈ X∗ we set l(x) = l, xX∗ ,X . The symbol ·, ·X∗ ,X is referred to as the duality pairing between X∗ and X. When the pair X, X∗ is understood we use the notation ·, · = ·, ·X∗ ,X .

1.7 Exercises Exercise 1.12 Let (X, ρ) be a metric space, T : X → X be a continuous mapping such that for all x ∈ X T 2 (x) = x and let f : X → R 1 be a continuous and bounded from below function such that f (T (x)) = f (x) for all x ∈ X. Then for every  > 0 there exists a continuous function f : X → R 1 such that for every x ∈ X, f (T (x)) = f (x), f (x) ≤ f (x) ≤ f (x) +  and the function f has a point of minimum. Exercise 1.13 Let (X, ρ) be a metric space, T : X → X be a continuous mapping such that for all x ∈ X T 2 (x) = x and let f : X → R 1 be a continuous function possessing a point of minimum which is a fixed point of the mapping T and such that f (T (x)) = f (x) for all x ∈ X. Then for every  > 0 there exists a continuous function f : X → R 1 such that for every x ∈ X, f (T (x)) = f (x), f (x) ≤ f (x) ≤ f (x) +  and the function f has a unique point of minimum. Exercise 1.14 Let (X, ρ) be a metric space, T : X → X be a continuous mapping such that for all x ∈ X T 2 (x) = x

22

1 Introduction

and let f : X → R 1 be a continuous function possessing a point of minimum which is not a fixed point of the mapping T and such that f (T (x)) = f (x) for all x ∈ X. Then for every  > 0 there exists a continuous function f : X → R 1 such that for every x ∈ X, f (T (x)) = f (x), f (x) ≤ f (x) ≤ f (x) +  and the function f has exactly two different points of minimum. Exercise 1.15 Let (X, ρ) be a compact metric space and let f : X → R 1 be a continuous function possessing a unique point of minimum xf which is an isolated point of X. Then there exists  > 0 such that for every continuous function g : X → R 1 satisfying |f (x) − g(x)| ≤  for every x ∈ X, xf is a unique point of minimum of g. Exercise 1.16 Let (X, ρ) be a compact metric space, let f : X → R 1 be a continuous function, and let Xmin be the set of all points of minimum of f . Then for each  > 0 there exists δ > 0 such that for each x ∈ X satisfying f (x) ≤ inf(f )+δ, we have inf{ρ(x, ξ ) : ξ ∈ Xmin } ≤ . Exercise 1.17 Let (X, ρ) be a compact metric space, let f : X → R 1 be a continuous function, and let Xmin be the set of all points of minimum of f . Then for each  > 0 there exists δ > 0 such that for each continuous function g : X → R 1 satisfying |f (z) − g(z)| ≤ δ for all z ∈ X and each x ∈ X satisfying g(x) ≤ inf(g) + δ, we have inf{ρ(x, ξ ) : ξ ∈ Xmin } ≤ . Exercise 1.18 Let (X, ρ) be a compact metric space, T : X → X be a continuous mapping such that for all x ∈ X T 2 (x) = x and let f : X → R 1 be a continuous function possessing exactly two points of minimum and such that f (T (x)) = f (x) for all x ∈ X. Then there exists  > 0 such that every continuous function g : X → R 1 satisfying |f (x) − g(x)| ≤  and g(T (x)) = g(x) for every x ∈ X has at least two different points of minimum. Exercise 1.19 Show that there exists M > 0 such that for each T ≥ 1 and each ξ1 , ξ2 ∈ [−1, 1],  inf{

T

(x(t)2 + x (t)2 )dt : x : [0, T ] → R 1 is an a. c. function and

0

x(0) = ξ1 , x(T ) = ξ2 } ≤ M.

1.7 Exercises

23

Exercise 1.20 Show that for each T > 0, 

T

inf{

(x(t)2 + (x (t)2 − 1)2 )dt : x : [0, T ] → R 1 is an a. c. function and

0

x(0) = 0, x(T ) = 0} = 0. Exercise 1.21 Show that there exists M > 0 such that for each T ≥ 1 and each ξ1 , ξ2 ∈ [−1, 1],  inf{

T

(x(t)2 + (x (t)2 − 1)2 )dt : x : [0, T ] → R 1 is an a. c. function and

0

x(0) = ξ1 , x(T ) = ξ2 } ≤ M.

Chapter 2

Symmetric Optimization Problems

In this chapter we study several classes of symmetric optimization problems which are identified with the corresponding spaces of objective functions, equipped with appropriate complete metrics. Using the Baire category approach, for any of these classes, we show the existence of subset of the space of functions, which is a countable intersection of open and everywhere dense sets, such that for every objective function from this intersection the corresponding symmetric optimization problem possesses a solution. These results are obtained as realizations of a general variational principle which is established in this chapter. We extend these results for certain classes of symmetric optimization problems using a porosity notion. We identify a class of symmetric minimization problems with a certain complete metric space of functions, study the set of all functions for which the corresponding minimization problem has a solution, and show that the complement of this set is not only of the first category but also a σ -porous set.

2.1 A Generic Approach in Optimization Assume that (X, ρ) is a complete metric space. Denote by Ml the set of all lower semicontinuous and bounded from below functions f : X → R 1 . We equip the set Ml with the uniformity determined by the following base: E() = {(f, g) ∈ Ml × Ml : |f (x) − g(x)| ≤  for all x ∈ X}, where  > 0. It is known that this uniformity is metrizable (by a metric d) and complete [143].

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 A. J. Zaslavski, Turnpike Phenomenon and Symmetric Optimization Problems, Springer Optimization and Its Applications 190, https://doi.org/10.1007/978-3-030-96973-8_2

25

26

2 Symmetric Optimization Problems

Consider a minimization problem f (x) → min, x ∈ X, where f ∈ Ml . This problem has a solution when X is compact or when f satisfies growth conditions and bounded subsets of X are compact. When X does not satisfy a compactness assumption, the existence problem becomes more difficult and less understood. It is possible to overcome this difficulty by using the so-called generic approach which is applied fruitfully in many areas of analysis (see, for example, [11, 25, 62, 86, 97, 100, 104, 123, 124, 134, 143], and the references mentioned there). According to the generic approach, we say that a property holds for a generic (typical) element of a complete metric space (or the property holds generically) if the set of all elements of the metric space possessing this property contains a Gδ everywhere dense subset of the metric space which is a countable intersection of open everywhere dense sets. In particular, it is known that the optimization problem stated above (see [62, 100, 101, 143], and the references mentioned therein) can be solved generically (for a generic objective function). Namely, there exists a set F ⊂ Ml which is a countable intersection of open and everywhere dense sets such that for each f ∈ F the minimization problem has a unique solution which is a limit of any minimizing sequence. This result and its numerous extensions are collected in [143]. It should be mentioned that generic existence results in optimal control and the calculus of variations are discussed in [151], while generic results in nonlinear analysis are presented in [79, 98, 104, 117, 119]. In particular, [151] contains generic results on the existence of solutions for large classes of optimal control problems without convexity assumptions, generic existence results for best approximation problems are presented in [25, 104], generic existence of fixed points for nonlinear operators is shown in [97, 98, 104], and the generic existence of a unique zero of maximally monotone operators is shown in [119]. In this chapter, our goal is to obtain a generic existence of minimization problems with symmetry. These results are important because they have applications in crystallography [89]. The first such result was obtained in [171]. In this chapter, we study several classes of symmetric optimization problems which are identified with the corresponding spaces of objective functions, equipped with appropriate complete metrics. Using the Baire category approach, for any of these classes, we show the existence of subset of the space of functions, which is a countable intersection of open and everywhere dense sets, such that for every objective function from this intersection the corresponding symmetric optimization problem possesses a solution. These results are obtained as realizations of a general variational principle which is established in this chapter. We extend these results for certain classes of symmetric optimization problems using a porosity notion. We identify a class of symmetric minimization problems with a certain complete metric space of functions, study the set of all functions for which the corresponding minimization problem has a solution, and show that the complement of this set is not only of the first category but also a σ -porous set.

2.2 A Generic Variational Principle

27

2.2 A Generic Variational Principle We will obtain our well-posedness results as a realization of a variational principle which is considered in this section. This variational principle, which was obtained in [174], is an extension of the variational principle of [62]. We consider a complete metric space (X, ρ) which is called the domain space and a complete metric space (A, d) which is called the data space. We always consider the set X with the topology generated by the metric ρ. For the space A, we consider the topology generated by the metric d. This topology will be called the strong topology. In addition to the strong topology, we also consider a weaker topology on A which is not necessarily Hausdorff. This topology will be called the weak topology. (Note that these topologies can coincide.) For each function h : Y → [−∞, ∞], where Y is nonempty, set inf(h) = inf{h(y) : y ∈ Y } and dom(h) = {y ∈ Y : h(y) < ∞}. For each x ∈ X and each nonempty set D ⊂ X, put ρ(x, D) = inf{ρ(x, y) : y ∈ D}. For each x ∈ X and each r > 0, set B(x, r) = {y ∈ X : ρ(x, y) ≤ r}. If Z in a topological space and Y ⊂ Z, then Y is equipped with a relative topology. We assume that with every a ∈ A, a lower semicontinuous function fa on X is associated with values in R¯ = [−∞, ∞]. Assume that a mapping T : X → X is continuous and the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X.

(2.1)

This implies that T (X) = X, if x1 , x2 ∈ X and T (x1 ) = T (x2 ), then x1 = x2 and that there exists T −1 = T . Assume that AT ⊂ A is a closed set in the strong topology such that fa ◦ T = fa for all a ∈ AT .

(2.2)

28

2 Symmetric Optimization Problems

The space AT ⊂ A is equipped with the relative weak and strong topologies. (Note that these spaces can coincide.) In our study, we use the following basic hypothesis about the functions. (H) For any a ∈ AT , any  > 0, and any γ > 0, there exist a nonempty open set W in A with the weak topology, x ∈ X, α ∈ R 1 , and η > 0 such that W ∩ {b ∈ AT : d(a, b) < } = ∅, and for any b ∈ W, (i) inf(fb ) is finite. (ii) If z ∈ X is such that fb (z) ≤ inf(fb ) + η, then ρ(x, {z, T (z)}) ≤ γ and |fb (z) − α| ≤ γ . We show (see Theorem 2.1 below which was obtained in [174]) that if (H) holds, then for a generic a ∈ AT the problem minimize fa (x) subject to x ∈ X has a solution. Given a ∈ AT , we say that the problem of minimization of fa on X is wellposed with respect to data in A (or just with respect to A) if the following assertions hold: 1. inf(fa ) is finite and there exists xa ∈ X such that {x ∈ X : fa (x) = inf(fa )} = {xa , T (xa )}. 2. For each  > 0, there are a neighborhood V of a in A with the weak topology and δ > 0 such that for each b ∈ V, inf(fb ) is finite and if z ∈ X satisfies fb (z) ≤ inf(fb ) + δ, then |fb (z) − fa (xa )| ≤  and min{ρ(z, {xa , T (xa )}), ρ(T (z), {xa , T (xa )})} ≤ . Theorem 2.1 Assume that (H) holds. Then there exists an everywhere dense (in the strong topology) set B ⊂ AT which is a countable intersection of open (in the weak topology) subsets of AT such that for any a ∈ B the minimization problem of fa on X is well-posed with respect to A. Moreover, if there exists an everywhere dense (in the strong topology) open (in the weak topology) set E ⊂ AT such that for each a ∈ E, inf(fa ) < inf{f (x) : x ∈ X and T (x) = x},

2.2 A Generic Variational Principle

29

then for any a ∈ B the minimization problem of fa on X has exactly two different minimizers. Following the tradition, we can summarize the theorem by saying that under the assumption (H) that the minimization problem for fa is well-posed with respect to A for a generic a ∈ AT . Proof Let a ∈ AT . By (H) for any natural n = 1, 2, . . . , there are a nonempty open set U (a, n) in A with the weak topology, x(a, n) ∈ X, α(a, n) ∈ R 1 , and η(a, n) > 0 such that U (a, n) ∩ {b ∈ AT : d(a, b) < 1/n} = ∅

(2.3)

and for any b ∈ U (a, n), inf(fb ) is finite and if z ∈ X satisfies fb (z) ≤ inf(fb ) + η(a, n), then |fb (z) − α(a, n)| ≤ 1/n and ρ(x(a, n), {z, T (z)}) ≤ 1/n. Define Bn = (∪{U (a, m) : a ∈ AT , m ≥ n}) ∩ AT

(2.4)

for n = 1, 2, . . . . Clearly, for each integer n ≥ 1, the set Bn ⊂ AT is open in the relative weak topology, and in view of (2.3), it is everywhere dense in the relative strong topology. Set B = ∩∞ n=1 Bn .

(2.5)

Since for each integer n ≥ 1, the set Bn is also open in the relative strong topology generated by the complete metric d, we conclude that B is everywhere dense in the relative strong topology. Let b ∈ B.

(2.6)

Evidently, inf(fb ) is finite. By (2.4)–(2.6), there are a sequence {an }∞ n=1 ⊂ AT and a strictly increasing sequence of natural numbers {kn }∞ n=1 such that b ∈ U (an , kn ), n = 1, 2, . . . .

(2.7)

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2 Symmetric Optimization Problems

Assume that {zn }∞ n=1 ⊂ X and lim fb (zn ) = inf(fb ).

n→∞

(2.8)

Let m ≥ 1 be an integer. By (2.8), for all large enough n, the inequality fb (zn ) < inf(fb ) + η(am , km )

(2.9)

is true and it follows from the definition of U (am , km ), (2.3), (2.7), and (2.9) that −1 |fb (zn ) − α(am , km )| ≤ km

(2.10)

−1 ρ(x(am , km ), {zn , T (zn )}) ≤ km

(2.11)

and

for all large enough n. Since m is an arbitrary natural number, we conclude that ∞ there exists a subsequence {zip }∞ p=1 such that at least one of the sequences {zip }p=1 ∞ 2 and {T (zip ) }p=1 converge. Since the mapping T is continuous and T is the identity mapping, we conclude that both these sequences converge. Denote xb = lim zip . p→∞

(2.12)

Then, T (xb ) = lim T (zip ). p→∞

Since fb is lower semicontinuous, it follows from (2.4)–(2.6) and (2.8) that fb (xb ) = fb (T (xb )) = inf(fb ).

(2.13)

By (2.10) and (2.11) with zn = xb , n = 1, 2, . . . , −1 |fb (xb ) − α(am , km )| ≤ km ,

(2.14)

−1 . ρ(x(am , km ), {xb , T (xb )}) ≤ km

(2.15)

Assume that ξ ∈ X satisfies fb (ξ ) = inf(fb ).

(2.16)

2.2 A Generic Variational Principle

31

By (2.10), (2.11), and (2.16) with zn = ξ , n = 1, 2, . . . , −1 . ρ(x(am , km ), {ξ, T (ξ )}) ≤ km

(2.17)

Together with (2.15) and (2.17), these relations imply that −1 . min{ρ(ξ, xb ), ρ(ξ, T (xb )), ρ(T (ξ ), xb ), ρ(T (ξ ), T (xb ))} ≤ 2km

Since m is any natural number, we obtain that min{ρ(ξ, xb ), ρ(ξ, T (xb )), ρ(T (ξ ), xb ), ρ(T (ξ ), T (xb ))} = 0, ξ ∈ {xb , T (xb )} and {x ∈ X : fb (x) = inf(fb )} = {xb , T (xb )}. Let  > 0. Choose a natural number m for which −1 < . 4km

(2.18)

a ∈ U (am , km ).

(2.19)

Let

Clearly, inf(fa ) is finite. Let z ∈ X and fa (z) ≤ inf(fa ) + η(am , km ).

(2.20)

By the definition of U (am , km ), (2.3), (2.19), and (2.20), −1 , |fa (z) − α(am , km )| ≤ km −1 . ρ(x(am , km ), {z, T (z)}) ≤ km

By (2.13)–(2.15), (2.18), and the inequalities above, −1 < , | inf(fb ) − fa (z)| ≤ 2km

min{ρ(z, xb ), ρ(z, T (xb )), ρ(T (z), xb ), ρ(T (z), T (xb ))} ≤ . Therefore, for every b ∈ B, the minimization problem of fa on X is well-posed with respect to A.

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2 Symmetric Optimization Problems

Assume now that there exists an everywhere dense (in the strong topology) open (in the weak topology) set E ⊂ AT such that for each a ∈ E, inf(fa ) < inf{f (x) : x ∈ X and T (x) = x}. Clearly, the assertion of the theorem holds for the set E ∩ B. Theorem 2.1 is proved.

2.3 The First Generic Result Assume that (X, ρ) is a complete metric space. Denote by Ml the set of all lower semicontinuous and bounded from below functions f : X → R 1 . We equip the set Ml with the uniformity determined by the following base: E() = {(f, g) ∈ Ml × Ml : |f (x) − g(x)| ≤  for all x ∈ X}, where  > 0. It is known that this uniformity is metrizable (by a metric d) and complete [143]. Denote by Mc the set of all continuous functions f ∈ Ml . It is not difficult to see that Mc is a closed subset of Ml . Consider a minimization problem f (x) → min, x ∈ X, where f ∈ Ml . Set A = Ml and fa = a for all a ∈ A. For the space A, the strong and weak topologies coincide. Assume that a mapping T : X → X is continuous and the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X. In this section, AT is either the set of all f ∈ Ml such that f (T (x)) = f (x) for all x ∈ X or the set of all f ∈ Mc satisfying the equation above.

2.3 The First Generic Result

33

Clearly, AT is a closed subset of A. It is equipped with the relative topology induced by the metric d. In [171], it was shown that there exists a set F ⊂ AT which is a countable intersection of open and everywhere dense sets in AT such that for each f ∈ F, the minimization problem of f on X is well-posed with respect to A. Here we deduce this result from Theorem 2.1. The following lemma was obtained in [171]. Lemma 2.2 Assume that f ∈ AT ,  ∈ (0, 1), γ > 0, δ ∈ (0, 8−1 γ ), x¯ ∈ X satisfies f (x) ¯ ≤ inf(f ) + δ, f¯(x) = f (x) + γ min{1, ρ(x, x), ¯ ρ(T (x), x)}, ¯ x∈X and that U = {g ∈ Ml : (f¯, g) ∈ E(δ)}. Then f¯ ∈ AT and for each g ∈ U and each z ∈ X satisfying g(z) ≤ inf(g) + δ the following inequality holds: min{ρ(z, x), ¯ ρ(T (z), x)} ¯ < . Proof It is not difficult to see that for every x ∈ X, ¯ f¯(T (x)) = f (T (x)) + γ min{1, ρ(T (x), x), ¯ ρ(T 2 (x), x)} = f (x) + γ min{1, ρ(x, x), ¯ ρ(T (x), x)} ¯ = f¯(x). Thus f¯ ∈ AT . Assume that g ∈ U and that z ∈ X satisfies g(z) ≤ inf(g) + δ.

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2 Symmetric Optimization Problems

In view of the relations above, f (z) + γ min{1, ρ(z, x), ¯ ρ(T (z), x)} ¯ = f¯(z) ≤ g(z) + δ ≤ inf(g) + 2δ ≤ g(x) ¯ + 2δ ≤ f¯(x) ¯ + 3δ = f (x) ¯ + 3δ ≤ inf(f ) + 4δ ≤ f (z) + 4δ and min{1, ρ(z, x), ¯ ρ(T (z), x)} ¯ ≤ 4δγ −1 . Together with our choice of δ, this implies that min{ρ(z, x), ¯ ρ(T (z), x)} ¯ ≤ 4δγ −1 < . Lemma 2.2 is proved. We can easily prove the following auxiliary result. Lemma 2.3 Let f ∈ Ml , and let  be a positive number. Then there exists a neighborhood U of f in Ml with the weak topology such that for each g ∈ U , | inf(g) − inf(f )| ≤  and if x ∈ X satisfies g(x) ≤ inf(g) + , then |g(x) − inf(f )| ≤ 2. The following result was obtained in [174]. Theorem 2.4 There exists an everywhere dense set B ⊂ AT which is a countable intersection of open subsets of AT such that for any f ∈ B the minimization problem of f on X is well-posed with respect to A. By Theorem 2.1, in order to prove this result, it is sufficient to show that (H) holds. The hypothesis (H) follows from Lemmas 2.2 and 2.3.

2.4 The Second and Third Generic Results Let (X, ρ) be a complete metric space. Fix θ ∈ X. Denote by M the set of all bounded from below lower semicontinuous functions f : X → R 1 ∪ {∞} such that dom(f ) = ∅ and f (x) → ∞ as ρ(x, θ ) → ∞. We equip the set M with strong and weak topologies.

(2.21)

2.4 The Second and Third Generic Results

35

For each function h : Y → R 1 ∪ {∞}, where Y is nonempty, set epi(h) = {(y, α) ∈ Y × R 1 : α ≥ h(y)}. Assume that a mapping T : X → X is continuous and the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X.

(2.22)

MT = {f ∈ M : f ◦ T = f }.

(2.23)

Set

For the set M, we consider the uniformity determined by the following base: Es (n) = {(f, g) ∈ M × M : f (x) ≤ g(x) + n−1 and g(x) ≤ f (x) + n−1 for all x ∈ X},

(2.24)

where n is a natural number. Clearly, this uniform space M is metrizable and complete. Denote by τs the topology in M induced by this uniformity. The topology τs is called the strong topology. Now we equip the set M with a weak topology. We consider the complete metric space X × R 1 with the metric Δ(·, ·) defined by Δ((x1 , α1 ), (x2 , α2 )) = ρ(x1 , x2 )+|α1 −α2 |, x1 , x2 ∈ X , α1 , α2 ∈ R 1 .

(2.25)

For each lower semicontinuous bounded from below function f : X → R 1 ∪ {∞} with a nonempty epigraph, define a function Δf : X × R 1 → R 1 by Δf (x, α) = inf{Δ((x, α), (y, β)) : (y, β) ∈ epi(f )}, (x, α) ∈ X × R 1 . (2.26) For each natural number n, denote by Ew (n) the set of all pairs (f, g) ∈ M × M which have the following property: C(i)

For each (x, α) ∈ X × R 1 satisfying ρ(x, θ ) + |α| ≤ n, |Δf (x, α) − Δg (x, α)| ≤ n−1 .

C(ii)

(2.27)

For each (x, α) ∈ X × R 1 satisfying α ≤ n, min{Δf (x, α), Δg (x, α)} ≤ n the inequality (2.27) is valid.

(2.28)

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2 Symmetric Optimization Problems

It was shown in Section 4.4 of [143] that for the set M, there exists the uniformity generated by the base Ew (n), n = 1, 2, . . . . This uniformity is metrizable (by w ) and it induces in M a topology τw which is weaker than τs . The a metric Δ topology τw is called the weak topology. For the set M, we consider the metrizable uniformity determined by the following base: E(n) = {(f, g) ∈ M × M : (2.27) is valid for all (x, α) ∈ X × R 1 satisfying ρ(x, θ ) + |α| ≤ n}

(2.29)

where n = 1, 2, . . . . Denote by τ∗ the topology induced by this uniformity. The topology τ∗ is called the epi-distance topology [143]. Clearly, the topology τ∗ is weaker than τw . Let φ ∈ M. Denote by M(φ) the set of all f ∈ M satisfying f (x) ≥ φ(x) for all x ∈ X. It is easy to verify that M(φ) is a closed subset of M with the topology τw . We consider the topological subspace M(φ) ⊂ M with the relative weak and strong topologies. Clearly, the topologies τw and τ∗ induce the same relative topology on M(φ). Set MT (φ) = MT ∩ M(φ).

(2.30)

Clearly, MT is a closed set in M with the relative strong topology and MT (φ) is a closed set in M(φ) with the relative strong topology. We prove the following two results obtained in [174]. Theorem 2.5 Let A = M, fa = a, a ∈ A, and AT = MT . Then there exists an everywhere dense (in the strong topology) set B ⊂ AT which is a countable intersection of open (in the weak topology) subsets of AT such that for any f ∈ B the minimization problem of f on X is well-posed with respect to A. Theorem 2.6 Let A = M(φ), AT = MT (φ), and fa = a, a ∈ M(φ). Then there exists an everywhere dense (in the strong topology) set B ⊂ MT (φ) which is a countable intersection of open (in the weak topology) subsets of MT (φ) such that for any f ∈ B the minimization problem of f on X is well-posed with respect to MT (φ).

2.5 Proofs of Theorems 2.5 and 2.6 The next result follows from Lemmas 4.9 and 4.10 of [143]. Lemma 2.7 Let f ∈ M, and let δ be a positive number. Then there exists a neighborhood U of f in M with the weak topology such that for each g ∈ U | inf{g(x) : x ∈ X} − inf{f (x) : x ∈ X}| ≤ δ.

2.5 Proofs of Theorems 2.5 and 2.6

37

Denote by ET the set of all f ∈ MT for which there exists xf ∈ X such that f (xf ) = inf{f (x) : x ∈ X}.

(2.31)

f ∈ MT .

(2.32)

Lemma 2.8 Let

Then there exists a sequence fn ∈ ET , n = 1, 2 . . . such that fn (x) ≥ f (x) for all x ∈ X and n = 1, 2, . . . and fn → f as n → ∞ in the strong topology. Proof For each natural number n, there exists xn ∈ X such that f (xn ) ≤ inf{f (x) : x ∈ X} + 1/n.

(2.33)

For n = 1, 2, . . . , define fn (x) = max{f (x), f (xn )} for all x ∈ X.

(2.34)

By (2.33) and (2.34), for n = 1, 2 . . . and all x ∈ X, fn (T (x)) = max{f (T (x)), f (xn )} = max{f (x), f (xn )} = fn (x).

(2.35)

By (2.31)–(2.35), fn ∈ ET , n = 1, 2, . . . . By (2.33) and (2.34), lim fn = f

n→∞

in the strong topology. Lemma 2.8 is proved. Lemma 2.9 Let f ∈ ET , xf ∈ X, f (xf ) = inf{f (x) : x ∈ X},  ∈ (0, 1), 1 ∈ (0, ), δ ∈ (0, 16−1 12 ),

(2.36)

ρ(T (x), xf ) ≤ /8 for all x ∈ B(T (xf ), 21 ), ρ(T (x), T (xf )) ≤ 1 for all x ∈ B(xf , 6δ −1 ).

(2.37)

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2 Symmetric Optimization Problems

Define a function f¯ by f¯(x) = f (x) + 2−1  min{ρ(T (x), xf ), ρ(x, xf ), 1} for all x ∈ X.

(2.38)

Then there exists a neighborhood U of f¯ in M with the weak topology such that for each g ∈ U and each x ∈ X satisfying g(x) ≤ inf(g) + δ, |g(x) − f (xf )| ≤ , min{ρ(x, xf ), ρ(T (x), xf )} ≤ . Proof By (2.38), f¯ ∈ ET . Lemma 2.7 implies that there exists a neighborhood U0 of f¯ in M with the weak topology such that | inf(g) − inf(f¯)| < δ/2 for all g ∈ U0 .

(2.39)

n0 > | inf(f )| + 4 + 4δ −1

(2.40)

U = U0 ∩ {g ∈ M : (f¯, g) ∈ Ew (n0 )}.

(2.41)

Fix an integer

and put

Let g ∈ U , and let x ∈ X satisfy g(x) ≤ inf(g) + δ.

(2.42)

In view of (2.39), (2.41), and (2.42), g(x) ≤ inf(f¯) + 2δ.

(2.43)

It follows from C(ii), (2.40), (2.41), and (2.43) that there exist (y, α) ∈ epi(f¯)

(2.44)

ρ(y, x) ≤ 2n−1 0

(2.45)

such that

2.5 Proofs of Theorems 2.5 and 2.6

39

and |α − (inf(f¯) + 2δ)| ≤ 2n−1 0 .

(2.46)

By (2.36), (2.38), (2.40), (2.44), (2.46), and the definition of f¯, f (y) + 2−1  min{ρ(y, xf ), ρ(T (y), xf ), 1} = f¯(y) ≤ α ¯ ≤ inf(f¯) + 2δ + 2n−1 0 ≤ inf(f ) + 3δ = f (xf ) + 3δ ≤ f (y) + δ. Together with (2.37), this implies that min{ρ(y, xf ), ρ(T (y), xf ), 1} ≤ 6δ −1 and min{ρ(y, xf ), ρ(T (y), xf )} ≤ 6δ −1 .

(2.47)

By (2.39) and (2.43), |g(x) − f (xf )| ≤ |g(x) − inf(g)| + | inf(g) − inf(f¯)| ≤ 2δ < . In view of (2.47), there are two cases: ρ(y, xf ) ≤ 6δ −1 ,

(2.48)

ρ(T (y), xf ) ≤ 6δ −1 .

(2.49)

If (2.48) is true, then together with (2.36), (2.40), and (2.45), this implies that −1 < . ρ(x, xf ) ≤ 2n−1 0 + 6δ

Assume that (2.49) holds. Then, (2.37), (2.49), and the equality T 2 (y) = y imply that ρ(y, T (xf )) ≤ 1 . It follows from (2.40), (2.45), and (2.50) that ρ(x, T (xf )) ≤ 2n−1 0 + 1 ≤ 21 .

(2.50)

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2 Symmetric Optimization Problems

Together with (2.38), this implies that ρ(T (x), xf ) ≤ /8. Thus, in both cases, min{ρ(x, xf ), ρ(T (x), xf )} ≤ . Lemma 2.9 is proved. By Theorem 2.1, in order to prove Theorem 2.5 (Theorem 2.6 respectively), it is sufficient to show that (H) holds. The hypothesis (H) follows from Lemmas 2.8 and 2.9.

2.6 The Fourth Result Denote by Mb the set of all bounded from below lower semicontinuous functions f : X → R 1 ∪ {∞} which are not identically infinity. We equip the set Mb with a weak and a strong topology. For each natural number n, denote by Gw (n) the set of all (f, g) ∈ Mb × Mb such that sup{Δf (x, α) : (x, α) ∈ epi(g)}, sup{Δg (x, α) : (x, α) ∈ epi(f )} ≤ n−1 . For the set Mb , we consider the complete metrizable uniformity determined by the base Gw (n), n = 1, 2, . . . . We equip the space Mb with a topology τw induced by this uniformity. The topology τw is called a weak topology. Also, for the set Mb , we consider the complete metrizable uniformity determined by the following base: U (n) = {(f, g) ∈ Mb × Mb : f (x) ≤ g(x) + n−1 and g(x) ≤ f (x) + n−1 , x ∈ X}. The space Mb is endowed with the topology τs induced by this uniformity. The topology τs is called a strong topology. Set A = Mb and fa = a for each a ∈ A. Assume that a mapping T : X → X is continuous and the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X.

2.6 The Fourth Result

41

Set AT = {f ∈ Mb : f ◦ T = f }. Clearly, AT is a closed set in A with the strong topology. We prove the following result obtained in [174]. Theorem 2.10 There exists an everywhere dense (in the strong topology) set B ⊂ AT which is a countable intersection of open (in the weak topology) subsets of AT such that for any f ∈ B the minimization problem of f on X is well-posed with respect to A. In order to prove Theorem 2.10, we need the following two lemmas. The first lemma is proved in a straightforward manner. Lemma 2.11 Assume that f ∈ Mb and that  is a positive number. Then there exists a neighborhood U of f in Mb with the weak topology such that | inf(g) − inf(f )| ≤  for all g ∈ U. Denote by E the set of all f ∈ AT such that there exists xf ∈ X satisfying f (xf ) = inf{f (x) : x ∈ X}. Analogously to Lemma 2.8, we can show that E is everywhere dense in AT with the strong topology. Lemma 2.12 Let f ∈ ET , xf ∈ X, f (xf ) = inf{f (x) : x ∈ X},

(2.51)

 ∈ (0, 1), 1 ∈ (0, ), δ ∈ (0, 16−1 12 ),

(2.52)

ρ(T (x), xf ) ≤ /8 for all x ∈ B(T (xf ), 21 ),

(2.53)

ρ(T (x), T (xf )) ≤ 1 for all x ∈ B(xf , 6δ −1 ).

(2.54)

Define a function f¯ by f¯(x) = f (x) + 2−1  min{ρ(T (x), xf ), ρ(x, xf ), 1} for all x ∈ X.

(2.55)

Then there exists a neighborhood U of f¯ in A with the weak topology such that for each g ∈ U and each x ∈ X satisfying g(x) ≤ inf(g) + δ, |g(x) − f (xf )| ≤  min{ρ(x, xf ), ρ(T (x), xf )} ≤ .

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2 Symmetric Optimization Problems

Proof Clearly, f¯ ∈ AT . Lemma 2.11 implies that there exists a neighborhood U of f¯ in Mb with the weak topology such that for each g ∈ U , Δg (x, α) ≤ 16−1 δ, (x, α) ∈ epi(f¯),

(2.56)

Δf¯ (x, α) ≤ 16−1 δ, (x, α) ∈ epi(g),

(2.57)

| inf(g) − inf(f¯)| ≤ δ/16.

(2.58)

Let g ∈ U , and let x ∈ X satisfy g(x) ≤ inf(g) + δ.

(2.59)

|g(x) − inf(f¯)| ≤ 2δ.

(2.60)

In view of (2.58) and (2.59),

It follows from (2.58) and (2.60) that there exist (y, α) ∈ epi(f¯)

(2.61)

ρ(y, x) ≤ 8−1 δ

(2.62)

|α − (inf(f¯) + 2δ)| ≤ 8−1 δ.

(2.63)

such that

and

By (2.51), (2.55), (2.61), and (2.63), f (y) + 2−1  min{ρ(y, xf ), ρ(T (y), xf ), 1} = f¯(y) ≤ α ≤ inf(f¯) + 2δ + 8−1 δ ≤ f (xf ) + 3δ ≤ f (y) + δ and min{ρ(y, xf ), ρ(T (y), xf )} ≤ 6δ −1 .

(2.64)

2.7 The Fifth Generic Result

43

By (2.51), (2.52), (2.55) and (2.60), |g(x) − f (xf )| ≤ |g(x) − inf(f¯)| ≤ 2δ < . In view of (2.64), there are two cases: ρ(y, xf ) ≤ 6δ −1 ,

(2.65)

ρ(T (y), xf ) ≤ 6δ −1 .

(2.66)

If (2.65) is true, then together with (2.52) and (2.62), this implies that ρ(x, xf ) ≤ 6δ −1 + 8−1 δ < . Assume that (2.66) holds. Then, by (2.54) and (2.66), ρ(y, T (xf )) = ρ(T 2 (y), T (xf )) ≤ 1 . Together with (2.52) and (2.62), this implies that ρ(x, T (xf )) ≤ 1 + 8−1 δ ≤ 21 . Together with (2.53), this implies that ρ(T (x), xf ) ≤ /8. Thus, in both cases, min{ρ(x, xf ), ρ(T (x), xf )} ≤ . Lemma 2.12 is proved. By Theorem 2.1, in order to prove Theorem 2.10, it is sufficient to show that (H) holds. The hypothesis (H) follows from Lemma 2.12.

2.7 The Fifth Generic Result We consider a complete metric space (X, ρ). Fix θ ∈ X. Denote by L the set of all lower semicontinuous bounded from below functions f : X → R 1 which satisfy the following assumptions: A(i)

inf{f (x) : x ∈ X \ B(θ, r)} → ∞ as r → ∞.

44

2 Symmetric Optimization Problems

A(ii)

For each x ∈ X, each neighborhood U of x in X, and each positive number δ, there exists an open nonempty set V ⊂ U such that f (y) ≤ f (x) + δ for all y ∈ V .

Denote by Lc the set of all finite-valued continuous functions f ∈ L. Remark 2.13 Let f : X → R 1 be a lower semicontinuous bounded from below function. Then, f satisfies assumption A(ii) if and only if epi(f ) is the closure of its interior. For x ∈ X and A ⊂ X, set ρ(x, A) = inf{ρ(x, y) : y ∈ A}. Denote by S(X) the collection of all closed nonempty sets in X. We equip the set S(X) with a weak and a strong topology. For the set S(X), we consider the uniformity determined by the following base: Gw (n) = {(A, B) ∈ S(X) × S(X) : |ρ(x, A) − ρ(x, B)| ≤ n−1 for all x ∈ B(θ, n)}, where n = 1, 2, . . . . It is well-known that the set S(X) with this uniformity is metrizable (by a metric Hw ) and complete [143]. The bounded Hausdorff (Attouch– Wets) topology induced by this uniformity is a weak topology in the space S(X). Also, for the set S(X), we consider the uniformity determined by the following base: Gs (n) = {(A, B) ∈ S(X) × S(X) : ρ(x, B) ≤ n−1 for all x ∈ A and ρ(y, A) ≤ n−1 for all y ∈ B}, n = 1, 2, . . . . It is well-known that the space S(X) with this uniformity is metrizable (by a metric Hs ) and complete. The Hausdorff topology induced by this uniformity is a strong topology in the space S(X). For the space L, we consider the uniformity determined by the following base: E0 (n) = {(f, g) ∈ L × L : |f (x) − g(x)| ≤ n−1 for all x ∈ B(θ, n)}, where n = 1, 2, . . . . Evidently, this uniform space is metrizable (by a metric h0 ) and Lc is a closed subset of L with the topology induced by the metric h0 . For each natural number n, denote by Ew (n), the set of all pairs (f, g) ∈ L × L which have the following property: B(i) B(ii)

For each x ∈ B(θ, n) the inequality |f (x) − g(x)| ≤ n−1 is valid. For each x ∈ X satisfying min{f (x), g(x)} ≤ n, the inequality |f (x) − g(x)| ≤ n−1 is true.

2.7 The Fifth Generic Result

45

It is not difficult to see (see Lemma 4.17 of [143]) that if n is a natural number and (f, g), (g, h) ∈ Ew (2n), then (f, h) ∈ Ew (n). This implies that the collection of the sets Ew (n), n = 1, 2, . . . is the base of a uniformity. For the set L, we consider the uniformity determined by the base Ew (n), n = 1, 2, . . . . Clearly, this uniformity is metrizable (by a metric hw ). We equip the set L with the topology induced by this uniformity. This topology is called the weak topology. Clearly, Lc is a closed subset of L with the weak topology. It was shown (see Proposition 4.18 in [143]) that the metric space (L, hw ) is complete. For the set L, we consider the uniformity determined by the following base: Es () = {(f, g) ∈ L × L : |f (x) − g(x)| ≤ , x ∈ X}, where  is a positive number. Evidently, this uniform space is metrizable (by a metric hs ) and complete. For the space L × S(X), we consider the product topology induced by the metric d((f, A), (g, B)) = hs (f, g) + Hs (A, B), (f, A), (g, B) ∈ L × S(X), which is called a strong topology, and consider the product topology induced by the metric dw ((f, A), (g, B)) = hw (f, g) + Hw (A, B), (f, A), (g, B) ∈ L × S(X) which is called a weak topology. Set A = L × S(X), and for each a = (h, A) ∈ L × S(X) define fa : X → R 1 ∪ {∞} by fa (x) = h(x) for all x ∈ A and fa (x) = ∞ for all x ∈ X \ A. Assume that T : X → X is a uniformly continuous mapping such that the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X.

(2.67)

Denote by L(T ) the set of all functions f ∈ L such that f ◦ T = f.

(2.68)

46

2 Symmetric Optimization Problems

It is easy to see that L(T ) is a closed subset of L with the weak topology. Set ) L(T c = {f ∈ Lc : f ◦ T = f }.

(2.69)

(T )

Clearly, Lc is a closed subset of L with the weak topology. Denote by S (T ) (X) the set of all A ∈ S(X) such that T (A) = A. Since the mapping T is uniformly continuous, it is not difficult to see that S (T ) (X) is a closed subset of S(X) with the strong topology. We equip the spaces L(T ) × (T ) S (T ) (X) and Lc × S (T ) (X) with the relative weak and strong topologies. Let a function φ : X → R 1 satisfy φ(x) → ∞ as ρ(x, θ ) → ∞.

(2.70)

Define L(T ) (φ) = {f ∈ L(T ) : f (x) ≥ φ(x) for all x ∈ X}.

(2.71)

Clearly, L(T ) (φ) is a closed subset of L with the weak topology. Set ) (T ) ) (φ) ∩ L(T L(T c (φ) = L c .

We consider the topological subspaces LTc (φ) × S (T ) (X) and L(T ) (φ) × S (T ) (X) with the relative weak and strong topologies. We prove the following results which are new. Theorem 2.14 Let AT be one of the spaces ) (T ) ) (T ) (X), L(T ) (φ) × S (T ) (X), L(T (X). L(T ) × S (T ) (X), L(T c ×S c (φ) × S

Assume that ρ(T (x), T (y)) = ρ(x, y), x, y ∈ X.

(2.72)

Then there exists an everywhere dense (in the strong topology) set B ⊂ AT which is a countable intersection of open (in the weak topology) subsets of AT such that for any f ∈ B the minimization problem of fa on X is well-posed with respect to A. Theorem 2.15 Let AT be one of the spaces ) (T ) ) (T ) (X), L(T ) (φ) × S (T ) (X), L(T (X). L(T ) × S (T ) (X), L(T c ×S c (φ) × S

2.9 Proof of Theorems 2.14 and 2.15

47

Then there exists an everywhere dense (in the strong topology) set B ⊂ AT which is a countable intersection of open (in the weak topology) subsets of AT such that for any f ∈ B the minimization problem of fa on X is well-posed with respect to AT .

2.8 Auxiliary Results for Theorems 2.14 and 2.15 f

Consider the complete metric space (L, hw ). Denote by τ0 the topology induced by f the metric h0 and by τw the topology induced by the metric hw . It is easy to see that f f the topology τw is stronger than the topology τ0 . We can easily prove the following auxiliary results (see Lemma 4.20 and Proposition 4.21 of [143]). Proposition 2.16 Let f ∈ L. Then the following assertions hold: f

1. There exist a neighborhood V of f in the space L with the topology τw and a real number a0 such that g(x) ≥ a0 for each g ∈ V and each x ∈ X. 2. For each positive number c, there exist a positive number r and a neighborhood f U of f in the space L with the topology τw such that g(x) ≥ c for each g ∈ U and each x ∈ X \ B(θ, r). f

Proposition 2.17 Let τ be a Hausdorff topology in L which is stronger than τ0 and which has the following property: For each f ∈ L and each positive number c, there exist a positive number r and a neighborhood U of f in the space L with the topology τ such that g(x) ≥ c for each g ∈ U and each x ∈ X \ B(θ, r). f Then the topology τ is stronger than the topology τw . Let φ : X → R 1 be a function such that φ(x) → ∞ as ρ(x, θ ) → ∞. Clearly, f f f L(φ) is a closed subset of L with the topology τ0 and the topologies τw and τ0 induce the same relative topology for the subspace L(φ) ⊂ L.

2.9 Proof of Theorems 2.14 and 2.15 By Theorem 2.1, it is sufficient to show that (H) holds. In both the cases, (H) follows from the following result. Proposition 2.18 Let f ∈ L(T ) , A ∈ S (T ) (X), and , γ ∈ (0, 1). Then, there exist ¯ A¯ ∈ S(X), f¯ ∈ L(T ) , x¯ ∈ X, a positive number η, and a neighborhood U of (f¯, A)

48

2 Symmetric Optimization Problems

in L × S(X) with the weak topology such that A¯ = A ∪ {x, ¯ T (x)}, ¯

(2.73)

ρ(x, ¯ A), ρ(T (x), ¯ A) < 0 ,

(2.74)

f¯(x) = f (x) + 0 min{1, ρ(x, x), ¯ ρ(T (x), x)}, ¯ x∈X

(2.75)

with 0 ∈ (0, ) and such that the following assertion holds. Assume that (g, D) ∈ U , z ∈ D satisfies g(z) ≤ inf{g(y) : y ∈ D} + η

(2.76)

and at least one of the following conditions holds: (a) ρ(T (ξ1 ), T (ξ2 )) = ρ(ξ1 , ξ2 ) for all ξ1 , ξ2 ∈ X. (b) D ∈ S (T ) (X), g ∈ L(T ) . Then |g(z) − f (x)| ¯ ≤γ and min{ρ(z, x), ¯ ρ(T (z), x)} ¯ < . Proof Fix positive numbers 0 < 2−1 min{, γ }, 1 < 4−1 0 , η < 16−1 1 0 .

(2.77)

Ω = {x ∈ X : ρ(x, A) < 0 , ρ(T (x), A) < 0 }.

(2.78)

Set

Clearly, Ω is an open set in the metric space (X, ρ) and A ⊂ Ω. Note that if (a) holds, then Ω = {x ∈ X : ρ(x, A) < 0 }.

(2.79)

Assumption A(ii) implies that there exists an open nonempty set E ⊂ X such that E ⊂ Ω, f (z) ≤ inf{f (u) : u ∈ Ω} + 16−1 η for all z ∈ E.

(2.80)

2.9 Proof of Theorems 2.14 and 2.15

49

Choose x¯ ∈ E,

(2.81)

A¯ = A ∪ {x, ¯ T (x)}, ¯

(2.82)

and define

¯ ρ(T (x), x)}, ¯ x ∈ X. f¯(x) = f (x) + 0 min{1, ρ(x, x),

(2.83)

Clearly, f¯ ∈ L(T ) , and A¯ ∈ S (T ) (X). There exists an integer ¯ + 8−1 γ . n1 > 4 + |f (x)|

(2.84)

Proposition 2.16 implies that there exists a neighborhood V1 of f¯ in the topology f τw and constants c0 ∈ R 1 and c1 > 1 such that inf{g(x) : x ∈ X} ≥ c0 for all g ∈ V1

(2.85)

and g(x) ≥ 2n1 + 4 for each g ∈ V1 and each x ∈ X \ B(θ, c1 ).

(2.86)

Fix an integer n2 > 2c1 + 2|c0 | + 2n1 + 16(0 − max{ρ(x, ¯ A), ρ(T (x), ¯ A)})−1 + 8ρ(x, ¯ θ ) + 8η−1

(2.87)

such that {z ∈ X : ρ(z, x) ¯ ≤ 8n−1 2 } ⊂ E.

(2.88)

Define ¯ B) ∈ Gw (n2 )}. U = {(g, B) ∈ V1 × S(X) : (f¯, g) ∈ Ew (n2 ), (A,

(2.89)

Assume that (g, D) ∈ U

(2.90)

and at least one of conditions (a) and (b) holds. We show that inf{g(u) : u ∈ D} = inf{g(u) : u ∈ D ∩ B(θ, n2 /2 − 2)}.

(2.91)

50

2 Symmetric Optimization Problems

In view of (2.82), (2.87), (2.89), (2.90), and the definition of Gw (·), there exists y ∈ X such that y ∈ D, ρ(y, x) ¯ ≤ 2n−1 2 .

(2.92)

By (2.80), (2.81), (2.88), and (2.92), y ∈ E ⊂ Ω,

(2.93)

|f (y) − f (x)| ¯ ≤ 8−1 η.

(2.94)

By (2.75), (2.92), and (2.94), |f¯(y) − f¯(x)| ¯ = |f¯(y) − f (x)| ¯ ¯ ρ(T (y), x)} ¯ − f (x)| ¯ = |f (y) + 0 min{1, ρ(y, x), ¯ ρ(T (y), x)} ¯ ≤ |f (y) − f (x)| ¯ + 0 min{1, ρ(y, x), ≤ 8−1 η + 20 n−1 2 .

(2.95)

By (2.84), (2.87), (2.89), (2.90), (2.92), (2.95), and the definition of Ew (·) |g(y) − f¯(y)| ≤ n−1 2

(2.96)

and inf{g(u) : u ∈ D} ≤ g(y) ≤ f¯(y) + n−1 2 −1 −1 −1 ≤ f¯(x) ¯ + n−1 2 + 8 η + 20 n2 + n2 < n1 .

(2.97)

Now (2.86), (2.88)–(2.90), and (2.97) imply that inf{g(u) : u ∈ D} = inf{g(u) : u ∈ D and g(u) ≤ n1 + 2} = inf{g(u) : u ∈ D ∩ B(θ, c1 )} = inf{g(u) : u ∈ D ∩ B(θ, n2 /2 − 2)}.

(2.98)

In view of (2.98), equation (2.91) is true. It follows from (2.84), (2.92), (2.95), and (2.96) that inf{g(u) : u ∈ D} ≤ g(y) ≤ f¯(y) + n−1 2 −1 ≤ f (x) ¯ + 8−1 η + 20 n−1 2 + n2 < n1 .

(2.99)

2.9 Proof of Theorems 2.14 and 2.15

51

Assume that z∈D

(2.100)

g(z) ≤ inf{g(y) : y ∈ D} + η.

(2.101)

and

By (2.86)–(2.90), (2.98), and (2.101), z ∈ B(θ, c1 ) ⊂ B(θ, n2 /2 − 1).

(2.102)

It follows from (2.89), (2.96), (2.100), and (2.102) that |g(z) − f¯(z)| ≤ n−1 2 ,

(2.103)

¯ ≤ n−1 . ρ(z, A) 2

(2.104)

If (a) holds, then in view of (2.82) and (2.104), ¯ = ρ(T (z), T (A)) ¯ ≤ ρ(z, A) ¯ ≤ n−1 . ρ(T (z), A) 2 Let (b) hold. Then, by (2.100) and (2.101), T (z) ∈ D, g(T (z)) = g(z) ≤ inf{g(y) : y ∈ D} + η.

(2.105)

By (2.86), (2.88), (2.89), (2.90), (2.98), and (2.105), T (z) ∈ B(θ, c1 ) ⊂ B(θ, 2−1 n2 − 1).

(2.106)

It follows from (2.89), (2.90), (2.106), and the inclusion T (z) ∈ D that ¯ ≤ n−1 . ρ(T (z), A) 2

(2.107)

Thus, (2.107) holds in both cases. Equations (2.78), (2.87), (2.104), and (2.107) imply that z ∈ Ω.

(2.108)

f (z) ≥ f (x) ¯ − 16−1 η.

(2.109)

By (2.80), (2.81), (2.86), and (2.108),

52

2 Symmetric Optimization Problems

It follows from (2.83), (2.87), (2.97), (2.101), (2.103), and (2.109) that ¯ ρ(T (z), x)} ¯ f (z) + 0 min{1, ρ(z, x), = f¯(z) ≤ g(z) + n−1 2 −1 ≤ f (x) ¯ + η + 8−1 η + n−1 2 + 20 n2

≤ f (z) + 16−1 η + η + 2−1 η + n−1 2 ≤ f (z) + 2η.

(2.110)

By (2.77) and (2.110), min{1, ρ(z, x), ¯ ρ(T (z), x)} ¯ ≤ 2η0−1 ≤ 8−1 1 < . Equations (2.77), (2.87), (2.99), (2.101), (2.103), and (2.109) imply that −1 ¯ ¯ − n−1 ¯ ≤ η + 2−1 η < γ . −γ < −n−1 2 − 16 η < f (z) − f (x) 2 ≤ g(z) − f (x)

This completes the proof of Proposition 2.18.

2.10 σ -Porous Sets in a Metric Space We recall the concept of porosity [104, 143, 151]. Let (Y, d) be a complete metric space. We denote by Bd (y, r) the closed ball of center y ∈ Y and radius r > 0. A subset E ⊂ Y is called porous with respect to d (or just porous if the metric is understood) if there exist α ∈ (0, 1] and r0 > 0 such that for each r ∈ (0, r0 ] and each y ∈ Y there exists z ∈ Y for which Bd (z, αr) ⊂ Bd (y, r) \ E. A subset of the space Y is called σ -porous with respect to d (or just σ -porous if the metric is understood) if it is a countable union of porous (with respect to d) subsets of Y . Since porous sets are nowhere dense, all σ -porous sets are of the first category. If Y is a finite dimensional Euclidean space, then σ -porous sets are of Lebesgue measure 0. In fact, the class of σ -porous sets in such a space is much smaller than the class of sets which have measure 0 and are of the first category. Let (X, ρ) be a metric space. For each x ∈ X and each r > 0, set Bρ (x, r) = {y ∈ X : ρ(x, y) ≤ r}.

2.10 σ -Porous Sets in a Metric Space

53

We begin with the following simple proposition. It shows that porosity is equivalent to another property which is easier to verify. Proposition 2.19 A set E ⊂ X is porous with respect to ρ if and only if the following property holds: (P1)

There exist α ∈ (0, 1] and r0 > 0 such that for every x ∈ X and every r ∈ (0, r0 ], there is y ∈ X which satisfies ρ(x, y) ≤ r and Bρ (y, αr) ∩ E = ∅.

Proof Assume that the property (P1) holds with α ∈ (0, 1] and r0 > 0. We need to show that E is porous. Choose a positive number γ < 4−1 . Let x ∈ X and r ∈ (0, r0 ]. By property (P1), there exists y ∈ X such that ρ(y, x) ≤ γ r and Bρ (y, αγ r) ∩ E = ∅. It is easy to see that Bρ (y, αγ r) ⊂ Bρ (x, αγ r + γ r) ⊂ Bρ (x, r). In view of the relations above, the set E is porous. Proposition 2.19 is proved. Assume now that X is a nonempty set and ρ1 , ρ2 : X × X → [0, ∞) are metrics which satisfy ρ1 (x, y) ≤ ρ2 (x, y) for all x, y ∈ X. The following definition was used in [104]. A subset E ⊂ X is called porous with respect to the pair (ρ1 , ρ2 ) (or just porous if the pair of metrics is understood) if there exist α ∈ (0, 1] and r0 > 0 such that for each r ∈ (0, r0 ] and each x ∈ X, there exists y ∈ X for which ρ2 (y, x) ≤ r and Bρ1 (y, αr) ∩ E = ∅.

(2.111)

A subset of the space X is called σ -porous with respect to (ρ1 , ρ2 ) (or just σ porous if the pair of metrics is understood) if it is a countable union of porous (with respect to (ρ1 , ρ2 )) subsets of X. We use this generalization of the porosity notion because in applications a space is usually endowed with a pair of metrics and one of them is weaker than the other. Note that porosity of a set with respect to one of these two metrics does not imply its porosity with respect to the other metric. However, the following proposition shows that if a subset E ⊂ X is porous with respect to (ρ1 , ρ2 ), then E is porous with respect to any metric which is weaker than ρ2 and stronger than ρ1 . Proposition 2.20 Let k1 and k2 be positive numbers and ρ : X × X → [0, ∞) be a metric such that k1 ρ(x, y) ≥ ρ1 (x, y) and k2 ρ(x, y) ≤ ρ2 (x, y) for all x, y ∈ X. Assume that a set E ⊂ X is porous with respect to (ρ1 , ρ2 ). Then E is porous with respect to ρ.

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2 Symmetric Optimization Problems

Proof We may assume without loss of generality that k1 > 1 and k2 < 1. Since E is porous with respect to (ρ1 , ρ2 ), there exist α ∈ (0, 1] and r0 > 0 such that for every x ∈ X and every r ∈ (0, r0 ] there is y ∈ X satisfying (2.111). Fix a positive number γ < k2 . Let x ∈ X and r ∈ (0, r0 ]. By our choice of α and r0 , there exists y ∈ X such that ρ2 (y, x) ≤ γ r and Bρ1 (y, γ αr) ∩ E = ∅. It is easy to see that ρ(y, x) ≤ k2−1 ρ2 (x, y) ≤ k2−1 γ r < r and Bρ (y, k1−1 γ αr) ⊂ Bρ1 (y, γ αr) ⊂ X \ E. These relations and Proposition 2.19 imply that E is porous with respect to ρ. Proposition 2.20 is proved. Note that in our definition of porosity with respect to (ρ1 , ρ2 ), we do not use completeness assumptions. However, in the chapter, the metric space (X, ρ2 ) will always be complete. We usually consider metric spaces, say X, with two metrics, say dw and ds , such that dw (x, y) ≤ ds (x, y) for all x, y ∈ X. (Note that they can coincide.) We refer to them as the weak and strong metrics, respectively. The strong metric induces the strong topology and the weak metric induces the weak topology. If the set X is equipped with one metric d, then we also consider X with weak and strong metrics which coincide. If (X, d) is a metric space with a metric d and Y ⊂ X, then usually Y is also endowed with the metric d (unless another metric is introduced in Y ). If X is endowed with weak and strong metrics, then usually Y is also endowed with these metrics. In the sequel, a set X equipped with two metrics d1 and d2 satisfying d1 (x, y) ≤ d2 (x, y) for all x, y ∈ X will be denoted by (X, d1 , d2 ). If (Xi , di ), i = 1, 2, are metric spaces with the metrics d1 and d2 , respectively, then the space X1 × X2 will be endowed with the metric d1 × d2 defined by (d1 × d2 )((x1 , x2 ), (y1 , y2 )) = d1 (x1 , y1 ) + d2 (x2 , y2 ), (x1 , x2 ), (y1 , y2 ) ∈ X1 × X2 .

(2.112)

Assume now that X1 and X2 are metric spaces and each of them is endowed with weak and strong metrics. Then, for the product X1 × X2 , we also introduce a pair of metrics: a weak metric which is defined by (2.112) using the weak metrics of X1 and X2 and a strong metric which is defined by (2.112) using the strong metrics of X1 and X2 .

2.10 σ -Porous Sets in a Metric Space

55

We use the convention that ∞ − ∞ = 0. For each function f : X → [−∞, ∞], where X is nonempty, we set inf(f ) = inf{f (x) : x ∈ X}. We consider a metric space (X, ρ) which is called the domain space and a topological space A with the topology τ which is called the data space. We always consider the set X with the topology generated by the metric ρ. We assume that with every a ∈ A, a lower semicontinuous function fa on X is associated with values in R¯ = [−∞, ∞]. Assume that a mapping T : X → X is continuous and the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X. This implies that T (X) = X, if x1 , x2 ∈ X and T (x1 ) = T (x2 ), then x1 = x2 and that there exists T −1 = T . Assume that AT is a closed set in (A, τ ) such that f ◦ T = f for all f ∈ AT . Let a ∈ AT . We say that the minimization problem for fa on (X, ρ) is well-posed if inf(fa ) is finite and there exists xa ∈ X such that {x ∈ X : f (x) = inf(fa )} = {xa , T (xa )}, and the following assertion holds: For each  > 0, there exists δ > 0 such that for each z ∈ X satisfying fa (z) ≤ inf(fa ) + δ the inequality min{ρ(z, {xa , T (xa )}), ρ(T (z), {xa , T (xa )})} ≤  holds. The following property was introduced in Section 2.1 for domains which are metric spaces. Let a ∈ AT . We say that the minimization problem for fa on (X, ρ) is well-posed with respect to (A, τ ) if inf(fa ) is finite and there exists xa ∈ X such that {x ∈ X : f (x) = inf(fa )} = {xa , T (xa )}, and the following assertion holds: For each  > 0, there exist a neighborhood V of a in A with the topology τ and δ > 0 such that for each b ∈ V, inf(fb ) is finite, and if z ∈ X satisfies fb (z) ≤ inf(fb ) + δ, then

56

2 Symmetric Optimization Problems

|fb (z) − fa (xa )| ≤  and min{ρ(z, {xa , T (xa )}), ρ(T (z), {xa , T (xa )})} ≤ . If (A, d) is a metric space and τ is a topology generated by the metric d, then “well-posedness with respect to (A, τ )” will be sometimes replaced by “wellposedness with respect to (A, d).” The following new proposition is the main result of this section. Proposition 2.21 Assume that a ∈ AT , the minimization problem for fa on (X, ρ) is well-posed, and that there exists D > inf(fa ) for which the following property holds: (P2)

For each  > 0, there exists a neighborhood U of a in A with the topology τ such that for each b ∈ U and each x ∈ X satisfying min{fa (x), fb (x)} ≤ D the relation |fa (x) − fb (x)| ≤  holds.

Then the minimization problem for fa on (X, ρ) is well-posed with respect to (A, τ ). Proof There exists xa ∈ X such that {x ∈ X : f (x) = inf(fa )} = {xa , T (xa )}. Let  > 0. There exists a number δ ∈ (0, ) such that the following property holds: (P3)

For each z ∈ X satisfying fa (z) ≤ inf(fa ) + δ, we have min{ρ(z, {xa , T (xa )}), ρ(T (z), {xa , T (xa )})} ≤ .

We may assume without loss of generality that inf(fa ) + 2δ < D.

(2.113)

By property (P2), there exists a neighborhood U of a in A with the topology τ such that for each b ∈ U and each x ∈ X satisfying min{fa (x), fb (x)} ≤ D

(2.114)

|fa (x) − fb (x)| ≤ δ/4

(2.115)

the inequality

holds. Clearly,

2.10 σ -Porous Sets in a Metric Space

57

inf(fa ) = inf{fa (x) : x ∈ X and fa (x) ≤ inf(fa ) + δ/4}.

(2.116)

Let b ∈ U . It follows from the definition of U and (2.113) and (2.115) that |fa (x) − fb (x)| ≤ δ/4

(2.117)

for each x ∈ X satisfying fa (x) ≤ inf(fa ) + δ/4 < D. Together with (2.117), this implies that inf(fb ) ≤ inf{fb (x) : x ∈ X and fa (x) ≤ inf(fa ) + δ/4} ≤ inf{fa (x) + δ/4 : x ∈ X and fa (x) ≤ inf(fa ) + δ/4} = inf(fa ) + δ/4.

(2.118)

It is easy now to see that inf(fb ) = inf{fb (x) : x ∈ X and fb (x) ≤ inf(fa ) + δ/2}.

(2.119)

It follows from the definition of U and (2.113)–(2.115) that if x ∈ X satisfies fb (x) ≤ inf(fa ) + δ/2 < D, then |fa (x) − fb (x)| ≤ δ/4. Together with (2.118) and (2.119), this implies that −∞ < inf(fa ) ≤ inf{fa (x) : x ∈ X and fb (x) ≤ inf(fa ) + δ/2} ≤ inf{fb (x) + δ/4 : x ∈ X and fb (x) ≤ inf(fa ) + δ/2} = inf(fb ) + δ/4.

(2.120)

| inf(fa ) − inf(fb )| ≤ δ/4.

(2.121)

By (2.118) and (2.120),

Assume now that z ∈ X and fb (z) ≤ inf(fb ) + δ/4.

(2.122)

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By (2.113), (2.121), and (2.122), fb (z) ≤ inf(fa ) + δ/2 < D.

(2.123)

It follows from the definition of U and (2.123) that |fb (z) − fa (z)| ≤ δ/4.

(2.124)

Equations (2.121) and (2.122) imply that |fb (z) − fa (xa )| = |fb (z) − inf(fa )| ≤ |fb (z) − inf(fb )| + | inf(fb ) − inf(fa )| ≤ δ/4 + δ/4 < .

(2.125)

By (2.124) and (2.125), fa (z) ≤ fb (z) + δ/4 ≤ inf(fa ) + δ/2 + δ/4. Combined with property (P3), this implies that min{ρ(z, {xa , T (xa )}), ρ(T (z), {xa , T (xa )})} ≤ .

(2.126)

Thus we have constructed the neighborhood U of a in A with the topology τ such that for each b ∈ U , inf(fb ) is finite (see (2.121)), and if z ∈ X satisfies (2.122), then inequalities (2.125) and (2.126) hold. Therefore the minimization problem for fa on (X, ρ) is well-posed with respect to (A, τ ) by definition. Proposition 2.21 is proved.

2.11 A Variational Principle and Porosity We consider a complete metric space (X, ρ) which is called the domain space and a set A which is called the data space. We always consider the set X with the topology generated by the metric ρ. We assume that with every a ∈ A, a lower semicontinuous function fa on X is associated with values in R¯ = [−∞, ∞]. Assume that d1 , d2 : A × A → [0, ∞) are metrics such that d1 (a, b) ≤ d2 (a, b) for all a, b ∈ A. Assume that T : X → X is a continuous mapping such that T 2 (x) = x for all x ∈ X. For each x ∈ X and each ρ > 0, set Bρ (x, r) = {y ∈ X : ρ(x, y) ≤ r}.

2.11 A Variational Principle and Porosity

59

Assume that AT is a closed set in the metric space (A, d2 ) such that f ◦ T = f for all f ∈ AT . We use the following basic hypotheses about the functions: (H1)

For each  > 0 and each integer m ≥ 1, there exist numbers δ > 0 and r0 > 0 such that the following property holds: For each a ∈ AT satisfying inf(fa ) ≤ m and each r ∈ (0, r0 ], there exist a¯ ∈ AT and x¯ ∈ X such that ¯ ≤ r, inf(fa¯ ) ≤ m + 1 d2 (a, a) and for each z ∈ X satisfying fa¯ (z) ≤ inf(fa¯ ) + δr, the inequality min{ρ(z, x), ¯ ρ(T (z), x)} ¯ ≤

(H3)

holds. For each integer n ≥ 1, there exist α ∈ (0, 1) and r0 > 0 such that for each r ∈ (0, r0 ], each a, b ∈ A satisfying d1 (a, b) ≤ αr and each x ∈ X satisfying min{fa (x), fb (x)} ≤ n, the relation |fa (x) − fb (x)| ≤ r is valid.

For each integer n ≥ 1, denote by AT ,n the set of all a ∈ AT which have the following property: (P4)

inf(fa ) is finite and there exist x¯ ∈ X and δ > 0 such that if z ∈ X satisfies fa (z) ≤ inf(fa ) + δ, then min{ρ(z, x), ¯ ρ(T (z), x)} ¯ ≤ 1/n.

Proposition 2.22 Assume that for each integer n ≥ 1, the set AT \AT ,n is σ -porous with respect to (d1 , d2 ). Then the set AT \ (∩∞ n=1 AT ,n ) is σ -porous with respect to (d1 , d2 ), and for each a ∈ ∩∞ A the minimization problem for fa on (X, ρ) is n=1 T ,n well-posed. Proof It is easy to see that the set ∞ AT \ (∩∞ n=1 AT ,n ) = ∪n=1 (AT \ AT ,n )

is σ -porous with respect to (d1 , d2 ). Let a ∈ ∩∞ n=1 AT ,n . It is sufficient to show that the minimization problem for fa on (X, ρ) is well-posed. Evidently, inf(fa ) is finite. By property (P4) for each integer n ≥ 1, there exist xn ∈ X and δn > 0 such that the following property holds: (P5)

If z ∈ X satisfies fa (z) ≤ inf(fa ) + δn , then min{ρ(z, xn ), ρ(T (z, xn )} ≤ 1/n.

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Assume that {zi }∞ i=1 ⊂ X and lim fa (zi ) = inf(fa ).

i→∞

If {ik }∞ k=1 is a strictly increasing sequence of natural numbers, then for every natural number n it has a subsequence {ikp }∞ p=1 such that at least one of the following inclusions holds: −1 {zikp }∞ p=1 ⊂ Bρ (xn , n ); −1 {T (zikp )}∞ p=1 ⊂ Bρ (xn , n ).

Using this property and the diagonalization process, we obtain a subsequence {zik }∞ k=1 such that at least one of the following inclusions holds: {zik }∞ k=1 is a Cauchy sequence; {T (zik )}∞ k=1 is a Cauchy sequence. Since the metric space X is complete and the continuous mapping T satisfies ∞ T 2 (x) = x for all x ∈ X, we obtain that both sequences {zik }∞ k=1 and {T (zik )}kp=1 converge in X. Set xa = lim zik . k→∞

Clearly, T (xa ) = lim T (zik ). k→∞

Since fa is a lower semicontinuous function, we obtain that fa (xa ) = f (T (xa )) = inf(fa ).

(2.127)

Property (P5) and (2.127) imply that for every integer n ≥ 1, min{ρ(xa , xn ), ρ(T (xa ), xn )} ≤ 1/n.

(2.128)

Let  > 0. Choose a natural number n > 8 −1 .

(2.129)

2.11 A Variational Principle and Porosity

61

Assume that z ∈ X satisfies fa (z) ≤ inf(fa ) + δn .

(2.130)

Property (P5) and (2.130) imply that min{ρ(z, xn ), ρ(T (z), xn )} ≤ 1/n.

(2.131)

It follows from (2.128), (2.129), and (2.131) that min{ρ(xa , z), ρ(T (xa ), z), ρ(xa , T (z)), ρ(T (xa ), T (z))} ≤ 2/n < . Thus the minimization problem for fa on (X, ρ) is well-posed. Proposition 2.22 is proved. Theorem 2.23 (Variational Principle) Assume that (H1) and (H2) hold and that inf(fa ) is finite for each a ∈ AT . Then there exists a set B ⊂ AT such that AT \ B is σ -porous with respect to (d1 , d2 ) and that for each a ∈ B the minimization problem for fa on (X, ρ) is well-posed. Proof We recall that AT ,n is the set of all a ∈ AT which have the property (P4) (n = 1, 2, . . . ). By Proposition 2.22 in order to prove the theorem, it is sufficient to show that for each integer n ≥ 1 the set AT \ AT ,n is σ -porous with respect to (d1 , d2 ). Let m, n ≥ 1 be integers. Set Ωmn = {a ∈ AT \ AT ,n : inf(fa ) ≤ m}.

(2.132)

To prove the theorem, it is sufficient to show that Ωmn is porous with respect to (d1 , d2 ). By (H2), there exist α1 ∈ (0, 1), r1 ∈ (0, 1/2)

(2.133)

such that for each r ∈ (0, r1 ], each b1 , b2 ∈ A satisfying d1 (b1 , b2 ) ≤ α1 r, and each x ∈ X satisfying min{fb1 (x), fb2 (x)} ≤ m + 4 the inequality |fb1 (x) − fb2 (x)| ≤ r holds. By (H1), there exist α2 ∈ (0, 1), r2 ∈ (0, r1 ]

(2.134)

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such that the following property holds: (P5)

For each a ∈ AT satisfying inf(fa ) ≤ m + 2 and each r ∈ (0, r2 ], there exist a¯ ∈ AT and x¯ ∈ X such that ¯ ≤ r, inf(fa¯ ) ≤ m + 3, d2 (a, a)

(2.135)

and for each z ∈ X satisfying fa¯ (z) ≤ inf(fa¯ ) + 4α2 r, the inequality min{ρ(z, x), ¯ ρ(T (z), x)} ¯ ≤ 1/n is valid. Define r¯ = α1 α2 r2 , α¯ = α1 α2 /4.

(2.136)

Let a ∈ AT and r ∈ (0, r¯ ]. Consider the set E0 = {b ∈ AT : d2 (b, a) ≤ r/4}.

(2.137)

E0 ∩ {b ∈ AT : inf(fb ) ≤ m + 2} = ∅,

(2.138)

E0 ∩ {b ∈ AT : inf(fb ) ≤ m + 2} = ∅.

(2.139)

There are two cases:

Assume that (2.138) holds. We will show that for each b ∈ AT satisfying d1 (a, b) ≤ r¯ , the inequality inf(fb ) > m is true. Assume the contrary. Then there exists b ∈ AT such that d1 (a, b) ≤ r¯ and inf(fb ) ≤ m.

(2.140)

fb (y) ≤ inf(fb ) + 1/2 ≤ m + 1/2.

(2.141)

Choose y ∈ X such that

It follows from the choice of α1 , r1 , (2.133), (2.136), (2.140), and (2.141)) that |fa (y) − fb (y)| ≤ r2 α2 . Equations (2.133), (2.141), and (2.142) imply that inf(fa ) ≤ fa (y) ≤ fb (y) + r1 ≤ m + 1,

(2.142)

2.11 A Variational Principle and Porosity

63

a contradiction (see (2.137) and (2.138)). Therefore if (2.138) holds, then {b ∈ AT : d1 (a, b) ≤ r¯ } ∩ Ωmn = ∅.

(2.143)

Assume now that (2.139) holds. There exists a1 ∈ AT such that d2 (a, a1 ) ≤ r/4, inf(fa1 ) ≤ m + 2.

(2.144)

By the choice of α2 and r2 (see (2.134), (2.135)), and property (P5)) and (2.136), (2.140), and (2.144), there exist a¯ ∈ AT and x¯ ∈ X such that ¯ a1 ) ≤ r/4, inf(fa¯ ) ≤ m + 3, d2 (a,

(2.145)

and the following property holds: (P6)

For each x ∈ X satisfying fa¯ (x) ≤ inf(fa¯ ) + α2 r, the inequality min{ρ(x, x), ¯ ρ(T (x), x)} ¯ ≤ 1/n holds.

Equations (2.144) and (2.145) imply that d2 (a, a) ¯ ≤ r/2.

(2.146)

¯ ≤ αr ¯ = α1 α2 r/4. d1 (b, a)

(2.147)

Assume that b ∈ AT satisfies

By (2.145), inf(fa¯ ) = inf{fa¯ (x) : x ∈ X and fa¯ (x) ≤ m + 7/2}.

(2.148)

Let x ∈ X and fa¯ (x) ≤ m + 7/2.

(2.149)

It follows from (2.147), (2.149), and the choice of α1 and r1 (see (2.133)) that |fa¯ (x) − fb (x)| ≤ α2 r/4. Together with (2.145) and (2.148), this implies that inf(fb ) ≤ inf{fb (x) : x ∈ X and fa¯ (x) ≤ m + 7/2} ≤ inf{fa¯ (x) + α2 r/4 : x ∈ X and fa¯ x) ≤ m + 7/2} = inf(fa¯ ) + α2 r/4.

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Thus inf(fb ) ≤ inf(fa¯ ) + α2 r/4.

(2.150)

x ∈ X and fb (x) ≤ inf(fb ) + α2 r/4.

(2.151)

Assume now that

Combined with (2.150), this implies that fb (x) ≤ inf(fa¯ ) + α2 r/2.

(2.152)

It follows from (2.134), (2.136), (2.145), and (2.152) that fb (x) ≤ m + 7/2. Then, by (2.147) and the choice of α1 and r1 (see (2.133)), |fa¯ (x) − fb (x)| ≤ α2 r/4. Combined with (2.152), this implies that fa¯ (x) ≤ inf(fa¯ ) + 3α2 r/4. It follows from this relation and the property (P6) that min{ρ(x, x), ¯ ρ(T (x), x)} ¯ ≤ 1/n. Since this inequality holds for any x ∈ X satisfying (2.151), we conclude that b ∈ ¯ ≤ αr. ¯ AT ,n . Thus we have shown that b ∈ AT ,n for each b ∈ AT satisfying d1 (b, a) Therefore ¯ ≤ αr} ¯ ∩ Ωmn = ∅. {b ∈ AT : d1 (b, a)

(2.153)

Combined with (2.143) and (2.146), this implies that in both cases (2.153) is true ¯ ≤ r/2. (Note that if (2.138) holds, then a¯ = a.) with a¯ ∈ AT satisfying d2 (a, a) Thus, Ωmn is porous. This completes the proof of Theorem 2.23. Since property (P2) for the space (A, d1 ) (see Proposition 2.21) follows from (H2), we obtain that Theorem 2.23 and Proposition 2.21 imply the following result. Theorem 2.24 Assume that (H1) and (H2) hold and that inf(fa ) is finite for each a ∈ AT . Then there exists a set B ⊂ AT such that AT \ B is σ -porous with respect to (d1 , d2 ) and that for each a ∈ B the minimization problem for fa on (X, ρ) is strongly well-posed with respect to (A, d1 ). The results of this section are new.

2.12 Well-Posedness and Porosity for Classes of Minimization Problems

65

2.12 Well-Posedness and Porosity for Classes of Minimization Problems In this section we apply our variational principle established in Section 2.11 to important classes of minimization problems. All its results are new. Let (X, ρ) be a complete metric space, and let Mb be the set of all bounded from below lower semicontinuous functions f : X → R 1 ∪{∞} which are not identically infinity. For each f, g ∈ Mb , set ˜ g) = sup{|f (x) − g(x)| : x ∈ X}, d(f, ˜ g)(1 + d(f, ˜ g))−1 . d(f, g) = d(f, (Here, we assume that ∞/∞ = 1.) Clearly, d : Mb × Mb → [0, ∞) is a metric and the metric space (Mb , d) is complete. Let A = Mb . For each a ∈ Mb , set fa = a. Assume that T : X → X is a continuous mapping such that T 2 (x) = x for all x ∈ X. Set AT = {f ∈ A : f ◦ T = f }. Theorem 2.25 There exists a set B ⊂ AT such that AT \B is σ -porous with respect to d, and for each f ∈ B the minimization problem for f on (X, ρ) is well-posed with respect to (Mb , d). By Theorem 2.24 in order to prove Theorem 2.25, it is sufficient to show that (H1) and (H2) hold. Evidently, (H2) is valid. The following lemma implies (H1). Lemma 2.26 Let  ∈ (0, 1), r ∈ (0, 1], and f ∈ AT . Then there exist f¯ ∈ AT and x¯ ∈ X such that f (x) ≤ f¯(x) ≤ f (x) + r/2 for all x ∈ X, and the following property holds: For each y ∈ X satisfying f¯(y) ≤ inf(f¯) + r/4, the inequality min{ρ(y, x), ¯ ρ(T (y), x)} ¯ ≤ is valid. Proof There exists x¯ ∈ X such that f (x) ¯ ≤ inf(f ) + r/4.

(2.154)

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Define f¯ ∈ AT by ¯ ρ(T (x), x)}, ¯ x ∈ X. f¯(x) = f (x) + 2−1 r min{1, ρ(x, x),

(2.155)

Let x ∈ X and f¯(x) ≤ inf(f¯) + r/4 < ∞.

(2.156)

Then in view of (2.154)–(2.156), we have ¯ ρ(T (x), x)} ¯ = f¯(x) ≤ inf(f¯) + r/4 f (x) + 2−1 r min{1, ρ(x, x), ≤ f¯(x) ¯ + r/4 = f (x) ¯ + r/4 ≤ f (x) + r/2. Then min{ρ(x, x), ¯ ρ(T (x), x)} ¯ ≤ . Lemma 2.26 is proved. We continue to consider the metric space (Mb , d). Set d2 = d. Fix θ ∈ X. We use the convention that ∞ − ∞ = 0, ∞/∞ = 1, λ/∞ = 0 for any λ ∈ R 1 and ln(∞) = ∞. Denote by Mg the set of all functions f ∈ Mb such that f (X) ⊂ [0, ∞) and such that f (x) ≥ c0 (f )ρ(x, θ ) − c1 (f ), x ∈ X,

(2.157)

where c0 (f ) > 0 and c1 (f ) > 0 depend only on f . Clearly, Mg is a closed set in the metric space (Mb , d2 ). For each f, g ∈ Mg , define d˜1 (f, g) = sup{|ln(f (x) + 1) − ln(g(x) + 1)| : x ∈ X}, d1 (f, g) = d˜1 (f, g)(d˜1 (f, g) + 1)−1 .

(2.158)

Clearly, the metric space (Mg , d1 ) is complete. Denote by Mc the set of all finitevalued continuous functions f ∈ Mg . It is easy to see that Mc is a closed subset of (Mg , d1 ). Let us compare the metrics d1 and d2 . Let f, g ∈ Mg . Clearly, di (f, g) ≤ 1, ˜ g) < ∞ and i = 1, 2. Assume that d2 (f, g) < 1. Then d(f, ˜ g), x ∈ X. |f (x) − g(x)| ≤ d(f,

2.12 Well-Posedness and Porosity for Classes of Minimization Problems

67

Let x ∈ X. We estimate |ln(f (x) + 1) − ln(g(x) + 1)|. We may assume without loss of generality that ln(f (x) + 1) ≥ ln(g(x) + 1). We have |ln(f (x) + 1) − ln(g(x) + 1)| = ln((f (x) + 1)(g(x) + 1)−1 ) ˜ g))(g(x) + 1)−1 )) ≤ ln(((g(x) + 1 + d(f, ˜ g)(g(x) + 1)−1 ) = ln(1 + d(f, ˜ g)) ≤ d(f, ˜ g). ≤ ln(1 + d(f, Thus ˜ g) d˜1 (f, g) ≤ d(f, and d1 (f, g) ≤ d2 (f, g), f, g ∈ Mg . Theorem 2.27 Let A = Mg , T : X → X be a continuous mapping satisfying T 2 (x) = x for all x ∈ X, AT = {f ∈ Mg : f ◦ T = f }, AT ,c = AT ∩ Mc . There exists a set F ⊂ Mg such that for each f ∈ F the minimization problem for f on X is well-posed with respect to (Mg , d1 , d2 ), the set Mg \ F is σ -porous in (Mg , d1 , d2 ), and the set Mc \ F is σ -porous in (Mc , d). By Theorem 2.24 we need to show that (H1) and (H2) hold. (H1) holds by Lemma 2.26, while (H2) follows from the following lemma. Lemma 2.28 Let M be a positive number,  ∈ (0, 1), and 0 < λ < (2e(M + 1))−1 .

(2.159)

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Then for each f, g ∈ Mg and each x ∈ X which satisfy d1 (f, g) ≤ λ, min{f (x), g(x)} ≤ M,

(2.160)

the inequality |f (x) − g(x)| ≤  holds. Proof Assume that f, g ∈ Mg , x ∈ X, and (2.160) holds. We may assume without loss of generality that g(x) ≥ f (x). By (2.158), (2.159), and (2.160), d˜1 (f, g) = d1 (f, g)(1 − d1 (f, g))−1 ≤ 2λ. Together with (2.157), this relation implies that for all z ∈ X f (z) + 1 ≤ g(z) + 1 ≤ e2λ (f (z) + 1), and in view of (2.159), (2.160), and the mean value theorem, 0 ≤ g(x) − f (x) ≤ e2λ (f (x) + 1) − (f (x) + 1) = = (e2λ − 1)(f (x) + 1) ≤ (e2λ − 1)(M + 1) ≤ 2λ(M + 1)et with t ∈ [0, 2λ] ⊂ [0, 1]. Hence, 0 ≤ g(x) − f (x) ≤ 2λ(M + 1)et ≤ 2λ(M + 1)e. Together with (2.159), these inequalities imply that |f (x) − g(x)| ≤ . This completes the proof of Lemma 2.28.

2.13 An Auxiliary Result Let (X, ρ) be a complete metric space. Fix θ ∈ X. Denote by M the set of all bounded from below functions f : X → R 1 . For each x ∈ X and each r > 0, set B(x, r) = {y ∈ X : ρ(x, y) ≤ r}. For each natural number n, denote by Ew (n) the set of all pairs (f, g) ∈ M × M which have the following property: C(i)

For each x ∈ B(θ, n), |f (x) − g(x)| ≤ n−1 .

2.13 An Auxiliary Result

C(ii)

69

For each x ∈ X satisfying min{f (x), g(x)} ≤ n we have |f (x) − g(x)| ≤ n−1 . It is easy to see that for each natural number n, if (f, g), (g, h) ∈ Ew (2n), then (f, h) ∈ Ew (n). Therefore there exists a uniformity generated by the base Ew (n), n = 1, 2, . . . . This uniformity is metrizable and it induces in M a topology τw .

Assume that a mapping T : X → X is continuous and the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X. Proposition 2.29 Assume that f ∈ M, x¯ ∈ X, {z ∈ X : f (z) = inf(f )} = {x, ¯ T (x)}, ¯

(2.161)

and the following property holds: (a) For each  > 0, there exists δ > 0 such that for each x ∈ X satisfying f (x) ≤ inf(f ) + δ, the inequality min{ρ(x, x), ¯ ρ(x, T (x))} ¯ 0 and a neighborhood U of f in M such that for each g ∈ U and each x ∈ X satisfying min{ρ(x, x), ¯ ρ(x, T (x))} ¯ > ,

(2.163)

the inequality g(x) ≥ f (x) ¯ + δ0 is valid. Proof Property (a) implies that there exists δ ∈ (0, 1) such that for each x ∈ X satisfying f (x) ≤ inf(f ) + δ

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inequality (2.162) is valid. Set δ0 = 2−1 δ.

(2.164)

Choose a natural number n such that ¯ + 4|, ρ(θ, x) ¯ + ρ(θ, T (x))}. ¯ n > max{2δ −1 , |f (x)

(2.165)

Set U = {g ∈ M : (f, g) ∈ Ew (n)}. Clearly, U is a neighborhood of f . Assume that g∈U

(2.166)

min{ρ(x, x), ¯ ρ(x, T (x))} ¯ > .

(2.167)

and x ∈ X satisfies

If g(x) ≥ n, then (2.165) implies (2.163). Assume that g(x) < n.

(2.168)

It follows from C(ii), the definition of U , (2.166), and (2.168) that |g(x) − f (x)| ≤ n−1 .

(2.169)

Property (a), the choice of δ, (2.161), and (2.167) imply that f (x) > inf(f ) + δ = f (x) ¯ + δ.

(2.170)

In view of (2.165) and (2.166), |g(x) ¯ − f (x)| ¯ ≤ n−1 . By (2.164), (2.165), (2.169), and (2.170), ¯ − n−1 + δ g(x) ≥ f (x) − n−1 ≥ f (x) ¯ + δ0 . ≥ f (x) ¯ + 2−1 δ = f (x) Proposition 2.19 is proved.

(2.171)

2.14 A Well-Posedness Result

71

2.14 A Well-Posedness Result We continue to use the notation, definitions, and assumptions introduced in Section 2.13. Theorem 2.30 Assume that f ∈ M, x¯ ∈ X, {z ∈ X : f (z) = inf(f )} = {x, ¯ T (x)} ¯ and the following property holds: For each  > 0, there exists δ > 0 such that for each x ∈ X satisfying f (x) ≤ inf(f ) + δ, the inequality min{ρ(x, x), ¯ ρ(x, T (x))} ¯ 0 and a neighborhood U of f in M such that for each g ∈ U and each x ∈ X satisfying g(x) ≤ inf(g) + δ, the inequalities |g(x) − f (x)| ¯ ≤ and min{ρ(x, x), ¯ ρ(x, T (x))} ¯ ≤ are valid. Proof Proposition 2.19 implies that there exist δ0 ∈ (0, ) and a neighborhood U0 of f in M such that the following property holds: (b) For each g ∈ U0 and each x ∈ X satisfying min{ρ(x, x), ¯ ρ(x, T (x))} ¯ > the inequality g(x) ≥ f (x) ¯ + δ0 is valid.

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Choose δ ∈ (0, δ0 /8) and a neighborhood U ⊂ U0 of f in M such that the following property holds: (c) For each g ∈ U and each x ∈ B(x, ¯ 1) ∪ B(T (x), ¯ 1) we have |f (x) − g(x)| ≤ δ0 /8. Property (c) implies that for each g ∈ U , |f (x) ¯ − g(x)| ¯ ≤ δ0 /8.

(2.172)

g∈U

(2.173)

g(x) ≤ inf(g) + δ.

(2.174)

min{ρ(x, x), ¯ ρ(x, T (x))} ¯ > .

(2.175)

Let

and x ∈ X satisfy

Assume that

Property (b), the inclusion U ⊂ U0 , (2.172), (2.173), and (2.175) imply that g(x) > f (x) ¯ + δ0 ≥ g(x) ¯ − δ0 /8 + δ0 ≥ inf(g) + δ0 /2. This contradicts (2.174). The contradiction we have reached proves that min{ρ(x, x), ¯ ρ(x, T (x))} ¯ ≤ .

(2.176)

Property (c) and (2.176) imply that |f (x) − g(x)| ≤ δ0 /8.

(2.177)

2.15 An Extension of Theorem 2.4

73

Since (2.177) holds for all x ∈ X satisfying (2.174), we obtain that ¯ − δ0 /8. inf(g) + δ ≥ inf(f ) − δ0 /8 = f (x)

(2.178)

In view of (2.172), inf(g) ≤ g(x) ¯ ≤ f (x) ¯ + δ0 /8.

(2.179)

By (2.178) and (2.179), ¯ − δ0 /8 f (x) ¯ + δ0 /4 ≥ inf(g) + δ0 /8 ≥ f (x) and | inf(g) − f (x)| ¯ ≤ δ0 /4. Together with (2.174), this implies that |g(x) − f (x)| ¯ ≤ 3δ0 /8 < . Theorem 2.20 is proved. The results of this section and Section 2.13 are new.

2.15 An Extension of Theorem 2.4 In this section and in Sections 2.16–2.18, we obtain improvement of our generic results. It is shown that for a generic problem there exist exactly two different solutions. The results of these sections are new. Assume that (X, ρ) is a complete metric space. Denote by Ml the set of all lower semicontinuous and bounded from below functions f : X → R 1 . We equip the set Ml with the uniformity determined by the following base: E() = {(f, g) ∈ Ml × Ml : |f (x) − g(x)| ≤  for all x ∈ X}, where  > 0. It is known that this uniformity is metrizable (by a metric d) and complete (see Section 2.3). Denote by Mc the set of all continuous functions f ∈ Ml . It is not difficult to see that Mc is a closed subset of Ml . Consider a minimization problem f (x) → min, x ∈ X,

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where f ∈ Ml . Set A = Ml and fa = a for all a ∈ A. For the space A, the strong and weak topologies coincide. Assume that a mapping T : X → X is continuous and the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X.

(2.180)

Denote by Fix(T ) the set of all x ∈ X such that T (x) = x.

(2.181)

Since the mapping T is continuous, the set Fix(T ) is closed. Note that it can be empty. We assume that the set X \ Fix(T ) is everywhere dense. Denote by Mf the set of all f ∈ Ml , which are continuous at any point of Fix(T ). If Fix(T ) = ∅, then Mf = Ml . Clearly, Mf is a closed subset of Ml . In this section, AT is either the set of all f ∈ Mf such that f (T (x)) = f (x) for all x ∈ X or the set of all f ∈ Mc satisfying the equation above. Clearly, AT is a closed subset of A. It is equipped with the relative topology induced by the metric d. We prove the following result. Theorem 2.31 There exists an everywhere dense set B ⊂ AT which is a countable intersection of open subsets of AT such that for any f ∈ B the minimization problem of f on X is well-posed with respect to A and has exactly two different minimizers. Proof It is easy to see that Lemmas 2.2 and 2.3 hold in our case. They imply (H). By Theorem 2.1, in order to complete our theorem, it is sufficient to show that there exists an everywhere dense open set E ⊂ AT such that for each f ∈ E, inf(f ) < inf{f (x) : x ∈ Fix(T )}. Let f ∈ AT and n ≥ 1 be an integer. Since f ∈ Mf and T is continuous, there exists x(f, n) ∈ X \ Fix(T )

(2.182)

2.15 An Extension of Theorem 2.4

75

such that f (x(f, n)) ≤ inf(f ) + (4n)−1 .

(2.183)

f˜n (z) = max{f (z), f (x(f, n))}, z ∈ X.

(2.184)

Define

Clearly, f˜n ∈ AT , and (f, f˜n ) ∈ E((4n)−1 ),

(2.185)

T (x(f, n)) ∈ X \ Fix(T ).

(2.186)

By (2.182) and (2.186), there exists δ ∈ (0, 1) such that Fix(T ) ∩ (B(x(f, n), δ) ∪ (B(T (x(f, n)), δ)) = ∅.

(2.187)

Define fn (x) = f˜n (x) + (4n)−1 min{1, ρ(x, x(f, n)), ρ(T (x), x(f, n))}, x ∈ X. (2.188) Clearly, fn ∈ AT . In view of (2.185) and (2.188), (f, fn ) ∈ E((2n)−1 ).

(2.189)

Equations (2.184) and (2.188) imply that for each x ∈ X, fn (x) ≥ fn (x(f, n)) + (4n)−1 min{1, ρ(x, x(f, n)), ρ(T (x), x(f, n))}. (2.190) By (2.187) and (2.190), for each x ∈ Fix(T ), fn (x) ≥ fn (x(f, n)) + (4n)−1 δ.

(2.191)

There exists an open neighborhood U (f, n) of fn in AT such that U (f, n) ⊂ {g ∈ AT : (g, fn ) ∈ E((4n)−1 δ)}. B (2.191) and (2.192), for each g ∈ U (f, n) and each x ∈ Fix(T )), g(x) ≥ fn (x) − (16n)−1 δ ≥ fn (x(f, n)) + (4n)−1 δ − (16n)−1 δ ≥ g(x(f, n)) + (4n)−1 δ − (8n)−1 δ

(2.192)

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2 Symmetric Optimization Problems

and inf{g(z) : z ∈ Fix(T )} ≥ inf(g) + (8n)−1 δ. Define E = ∪{U (f, n) : f ∈ AT , n ≥ 1 is an integer}. By (2.189), it is an open everywhere dense set in AT . This completes the proof of Theorem 2.31.

2.16 Extensions of Theorems 2.5 and 2.6 Let (X, ρ) be a complete metric space. Fix θ ∈ X. We consider the space M of all bounded from below lower semicontinuous functions f : X → R 1 ∪ {∞} such that dom(f ) = ∅ and f (x) → ∞ as ρ(x, θ ) → ∞ studied in Section 2.4 and equipped with the strong and weak topologies. Assume that a mapping T : X → X is continuous and the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X. Define Fix(T ) = {x ∈ X : T (x) = x}. We assume that the set X \ Fix(T ) is everywhere dense in X. Set MT = {f ∈ M : f ◦ T = f }. For the set M, we consider the uniformity determined by the base Es (n) (see (2.24)), where n is a natural number. This uniform space M is metrizable and complete. We denote by τs the topology in M induced by this uniformity. The topology τs is called the strong topology. The space M is also equipped with the weak topology τw introduced in Section 2.4, which is induced by the uniformity generated by the base Ew (n), n = 1, 2 . . . (see (2.27) and (2.28)). For the set M, we also consider the epi-distance topology τ∗ (see (2.20)), which is weaker than τw . Let φ ∈ M. Denote by M(φ) the set of all f ∈ M satisfying f (x) ≥ φ(x) for all x ∈ X. Note that M(φ) is a closed subset of M with the topology τw . We

2.16 Extensions of Theorems 2.5 and 2.6

77

consider the topological subspace M(φ) ⊂ M with the relative weak and strong topologies. Clearly, the topologies τw and τ∗ induce the same relative topology on M(φ). Set MT (φ) = MT ∩ M(φ). Note that MT is a closed set in M with the relative strong topology and MT (φ) is a closed set in M(φ) with the relative strong topology. Denote by Mf the set of all f ∈ M which are continuous at any point of Fix(T ). If Fix(T ) = ∅, then Mf = M. Denote by Mc the set of all continuous functions f ∈ M. Clearly, Mf and Mc are closed subsets of M in the strong topology. Define Mc,T = Mc ∩ MT , Mf,T (φ) = Mf ∩ MT (φ), Mc,T (φ) = Mc ∩ MT (φ). Note that all these sets are closed in M with the strong topology. We prove the following two results. Theorem 2.32 Let A = M, fa = a, a ∈ A and AT is either Mf,T or Mc,T . Then there exists an everywhere dense (in the strong topology) set B ⊂ AT which is a countable intersection of open (in the weak topology) subsets of AT such that for any f ∈ B the minimization problem of f on X is well-posed with respect to A and has exactly two different solutions. Theorem 2.33 Let A = M(φ), fa = a, a ∈ M(φ) and AT is either Mf,T (φ) or Mc,T (φ). Then there exists an everywhere dense (in the strong topology) set B ⊂ AT which is a countable intersection of open (in the weak topology) subsets of AT such that for any f ∈ B the minimization problem of f on X is well-posed with respect to MT (φ) and has exactly two different solutions. These two theorems are proved simultaneously. Note that Lemma 2.7 is valid in our case. Denote by ET the set of all f ∈ AT for which there exists xf ∈ X such that f (xf ) = inf{f (x) : x ∈ X}. Analogously to Lemma 2.8, we can prove the following result. Lemma 2.34 Let f ∈ AT . Then there exists a sequence fn ∈ ET , n = 1, 2 . . . such that fn (x) ≥ f (x) for all x ∈ X and n = 1, 2, . . . and fn → f as n → ∞ in the strong topology. It is not difficult to see that Lemma 2.9 is true in our case. Now (H) follows from Lemmas 2.9 and 2.34. In view of Theorem 2.1, in order to prove Theorems 2.32

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and 2.33, it is sufficient to show that there exists an everywhere dense (in the strong topology) open (in the weak topology) set E ⊂ AT such that for each f ∈ E, inf(f ) < inf{f (z) : z ∈ Fix(T )}. Completion of the Proof of Theorems 2.32 and 2.33 Let f ∈ AT and n ≥ 1 be an integer. Since f ∈ Mf and T is continuous, there exists x(f, n) ∈ X \ Fix(T )

(2.193)

f (x(f, n)) ≤ inf(f ) + (4n)−1 .

(2.194)

f˜n (x) = max{f (x), f (x(f, n))}, x ∈ X.

(2.195)

such that

Define

Clearly, f˜n ∈ AT , and (f, f˜n ) ∈ Es (4n).

(2.196)

There exists δ ∈ (0, 1) such that Fix(T ) ∩ (B(x(f, n), δ) ∪ (B(T (x(f, n)), δ)) = ∅.

(2.197)

Define fn (x) = f˜n (x) + (4n)−1 min{1, ρ(x, x(f, n)), ρ(T (x), x(f, n))}, x ∈ X. (2.198) Clearly, fn ∈ AT . In view of (2.196) and (2.198), (f, fn ) ∈ Es (2n).

(2.199)

Equations (2.195) and (2.198) imply that for each x ∈ X, fn (x) ≥ fn (x(f, n)) + (4n)−1 min{1, ρ(x, x(f, n)), ρ(T (x), x(f, n))}. (2.200) By (2.197) and (2.200), for each x ∈ Fix(T ), fn (x) ≥ fn (x(f, n)) + (4n)−1 δ.

2.16 Extensions of Theorems 2.5 and 2.6

79

Since the mapping T is continuous, there exists δ1 ∈ (0, δ/4) such that the following property holds: (a) If x ∈ X satisfies ρ(T (x), x(f, n)) ≤ δ1 , then ρ(x, T (x(f, n))) ≤ δ/4. Choose a natural number k > 48δ1−1 (ρ(x(f, n), θ ) + |fn (x(f, n))| + 4 + 4n).

(2.201)

There exists an open neighborhood U (f, n) of fn in AT with the weak topology such that U (f, n) ⊂ {g ∈ AT : (g, fn ) ∈ Ew (2k)}

(2.202)

(see (2.27) and (2.28)). Let g ∈ U (f, n).

(2.203)

In view of (2.27) and (2.201)–(2.203), ρ(x(f, n), θ ) + |fn (x(f, n))| < k, Δg (x(f, n), fn (x(f, n))) ≤ k −1 . Therefore there are z0 ∈ X, α0 ≥ g(z0 )

(2.204)

ρ(z0 , x(f, n)) ≤ k −1 ,

(2.205)

|α0 − fn (x(f, n))| ≤ k −1 .

(2.206)

g(z0 ) ≤ α0 ≤ fn (x(f, n)) + k −1 .

(2.207)

such that

By (2.204) and (2.206),

Assume that z ∈ X satisfies g(z) ≤ inf(g) + k −1 .

(2.208)

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2 Symmetric Optimization Problems

We have (z, g(z)) ∈ epi(g), Δg (z, g(z)) = 0. Equations (2.201) and (2.206)–(2.208) imply that g(z) ≤ g(z0 ) + 1/k ≤ fn (x(f, n)) + 2 < k.

(2.209)

By (2.27), (2.28), (2.202), (2.203), and (2.209), Δfn (z, g(z)) ≤ (2k)−1 . This implies that there exist y ∈ X, α ≥ fn (y)

(2.210)

ρ(y, z) ≤ 1/k, |α − g(z)| ≤ 1/k.

(2.211)

such that

In view of (2.207) and (2.208), g(z) ≤ g(z0 ) + 1/k ≤ fn (x(f, n)) + 2/k.

(2.212)

It follows from (2.195) and (2.210)–(2.212) that fn (y) ≤ g(z) + 1/k ≤ fn (x(f, n)) + 3/k = f˜n (x(f, n)) + 3/k.

(2.213)

By (2.195), (2.198), (2.201), and (2.213), (4n)−1 min{1, ρ(y, x(f, n)), ρ(T (y), x(f, n))} ≤ 3/k, min{ρ(y, x(f, n)), ρ(T (y), x(f, n))} ≤ 12nk −1 < 4−1 δ1 .

(2.214)

Property (a) implies that if ρ(T (y), x(f, n)) < δ1 /4, then ρ(y, T (x(f, n))) ≤ δ/4. Together with (2.214), this implies that min{ρ(y, x(f, n)), ρ(y, T (x(f, n)))} ≤ 4−1 δ.

(2.215)

2.17 Extension of Theorem 2.10

81

In view of (2.197) and (2.215), B(y, δ/2) ∩ Fix(T ) = ∅.

(2.216)

It follows from (2.211) and (2.216) that B(z, δ/4) ∩ Fix(T ) = ∅.

(2.217)

Thus, for each z ∈ X satisfying (2.208), equation (2.217) is true. This implies that inf{g(z) : z ∈ Fix(T )} > inf(g) + k −1 for every g ∈ U (f, n). Define E = ∪{U (f, n) : f ∈ AT , n ≥ 1 is an integer}. By (2.208), it is an open (in the weak topology) everywhere dense (in the strong topology) set in AT . Clearly, for each g ∈ E, inf{g(z) : z ∈ Fix(T )} > inf(g). This completes the proof of Theorems 2.32 and 2.33.

2.17 Extension of Theorem 2.10 Let (X, ρ) be a complete metric space. We consider the space Mb of all bounded from below lower semicontinuous functions f : X → R 1 ∪ {∞} which are not identically ∞. This space is equipped with the weak topology τw introduced in Section 2.6 which is induced by the uniformity generated by the base Gw (n), n = 1, 2 . . . , and by the strong topology τs introduced in Section 2.6 which is induced by the uniformity generated by the base U (n), n = 1, 2 . . . , Let A = Mb , fa = a, a ∈ A. Assume that a mapping T : X → X is continuous and the mapping T 2 = T ◦ T is an identity mapping in X: T 2 (x) = x for all x ∈ X. Define Fix(T ) = {x ∈ X : T (x) = x}. We assume that the set X \ Fix(T ) is everywhere dense in X. Denote by Mf the set of all f ∈ Mb which are continuous at any point of x ∈ Fix(T ) where f (x) < ∞. If Fix(T ) = ∅, then Mf = Mb . Denote by Mc the set

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of all finite-valued continuous functions f ∈ Mb . Clearly, Mf and Mc are closed subsets of Mb in the strong topology. Assume that AT is either {f ∈ Mf : f ◦ T = f } or {f ∈ Mc : f ◦ T = f }. Clearly, AT is a closed set in Mb with the strong topology. We prove the following result. Theorem 2.35 There exists an everywhere dense (in the strong topology) set B ⊂ AT which is a countable intersection of open (in the weak topology) subsets of AT such that for any f ∈ B the minimization problem of f on X is well-posed with respect to A and has exactly two different solutions. Denote by ET the set of all f ∈ AT for which there exists xf ∈ X such that f (xf ) = inf{f (x) : x ∈ X}. Analogously to Lemma 2.8, we can show that ET is everywhere dense in AT with the strong topology. It is not difficult to see that Lemmas 2.11 and 2.12 are true in our case. Now (H) follows from Lemmas 2.12. In view of Theorem 2.1, in order to prove Theorems 2.35, it is sufficient to show that there exists an everywhere dense (in the strong topology) open (in the weak topology) set E ⊂ AT such that for each f ∈ E, inf(f ) < inf{f (z) : z ∈ Fix(T )}. Completion of the Proof of Theorems 2.35 Let f ∈ AT and n ≥ 1 be an integer. Since f ∈ Mf and T is continuous, there exists x(f, n) ∈ X \ Fix(T ) such that f (x(f, n)) ≤ inf(f ) + (4n)−1 .

(2.218)

f˜n (x) = max{f (x), f (x(f, n))}, x ∈ X.

(2.219)

Define

Clearly, f˜n ∈ AT , and (f, f˜n ) ∈ U (4n).

(2.220)

2.17 Extension of Theorem 2.10

83

There exists δ ∈ (0, 1) such that Fix(T ) ∩ (B(x(f, n), δ) ∪ (B(T (x(f, n)), δ)) = ∅.

(2.221)

Define fn (x) = f˜n (x) + (4n)−1 min{1, ρ(x, x(f, n)), ρ(T (x), x(f, n))}, x ∈ X. (2.222) Clearly, fn ∈ AT . In view of (2.220) and (2.222), (f, fn ) ∈ U (2n).

(2.223)

Equations (2.199) and (2.222) imply that for each x ∈ X, fn (x) ≥ fn (x(f, n)) + (4n)−1 min{1, ρ(x, x(f, n)), ρ(T (x), x(f, n))}. (2.224) By (2.221) and (2.223), for each x ∈ Fix(T ), fn (x) ≥ fn (x(f, n)) + (4n)−1 δ.

(2.225)

Since the mapping T is continuous, there exists δ1 ∈ (0, δ/4) such that the following property holds: (a) If x ∈ X satisfies ρ(T (x), x(f, n)) ≤ δ1 , then ρ(x, T (x(f, n))) ≤ δ/4. Choose a natural number k > 64nδ1−1 .

(2.226)

There exists an open neighborhood U (f, n) of fn in AT with the weak topology such that U (f, n) ⊂ {g ∈ AT : (g, fn ) ∈ Gw (k)}.

(2.227)

g ∈ U (f, n).

(2.228)

Let

In view of (2.227) and (2.228), (x(f, n), fn (x(f, n))) ∈ epi(fn ),

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and there are z0 ∈ X, α0 ≥ g(z0 )

(2.229)

such that ρ(z0 , x(f, n)) ≤ k −1 , |α0 − fn (x(f, n))| ≤ k −1 .

(2.230)

By (2.229) and (2.230), g(z0 ) ≤ α0 ≤ fn (x(f, n)) + k −1 .

(2.231)

Assume that z ∈ X satisfies g(z) ≤ inf(g) + k −1 .

(2.232)

We have (z, g(z)) ∈ epi(g). Equations (2.227) and (2.228) imply that Δfn (z, g(z)) ≤ k −1 . This implies that there exist y ∈ X, α ≥ fn (y)

(2.233)

ρ(y, z) ≤ 2/k, |α − g(z)| ≤ 2/k.

(2.234)

such that

In view of (2.231) and (2.232), g(z) ≤ g(z0 ) + 1/k ≤ fn (x(f, n)) + 2/k.

(2.235)

It follows from (2.233)–(2.235) that fn (y) ≤ g(z) + 2/k ≤ fn (x(f, n)) + 4/k.

(2.236)

By (2.219), (2.222), (2.226), and (2.236), (4n)−1 min{1, ρ(y, x(f, n)), ρ(T (y), x(f, n))} ≤ 4/k, min{ρ(y, x(f, n)), ρ(T (y), x(f, n))} ≤ 16nk −1 < 4−1 δ1 .

(2.237)

2.18 An Extension of Theorem 2.14

85

Property (a) implies that if ρ(T (y), x(f, n)) < δ1 /4, then ρ(y, T (x(f, n))) ≤ δ/4. Together with (2.237), this implies that min{ρ(y, x(f, n)), ρ(y, T (x(f, n)))} ≤ 4−1 δ. Combined with (2.221), this implies that B(y, δ/2) ∩ Fix(T ) = ∅. It follows from (2.226), (2.234), and the equation above that B(z, δ/4) ∩ Fix(T ) = ∅. Thus for each z ∈ X satisfying (2.232), the equation above is true. This implies that inf{g(z) : z ∈ Fix(T )} ≥ inf(g) + k −1 . Define E = ∪{U (f, n) : f ∈ AT , n ≥ 1 is an integer}. By (2.223), it is an open (in the weak topology) everywhere dense (in the strong topology) set in AT . Clearly, for each g ∈ E, inf{g(z) : z ∈ Fix(T )} > inf(g). This completes the proof of Theorem 2.35.

2.18 An Extension of Theorem 2.14 We consider a complete metric space (X, ρ) and the class of problems studied in Section 2.7. Fix θ ∈ X. Assume that T : X → X is a uniformly continuous mapping such that the mapping T 2 = T ◦ T is the identity mapping in X: T 2 (x) = x for all x ∈ X. Define Fix(T ) = {x ∈ X : T (x) = x}.

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We assume that the set X \ Fix(T ) is everywhere dense in X and that the set Fix(T ) is compact. We continue to consider the spaces of functions L and Lc introduced in Section 2.7 which are equipped with the weak and strong topologies induced by the metrics hs and hw , respectively. We also consider the space S(X) of all nonempty closed sets in X which are equipped with the weak and strong topologies induced by the metrics Hw and Hs , respectively. We denote by Lc the set of all finite-valued continuous functions f ∈ L which is a closed subset of L with the weak topology. . The spaces L × S(X) and Lc × S(X) are equipped with the weak and strong topologies which are the product of the weak topologies and the product of the strong topologies, respectively, as in Section 2.7. Set A = L × S(X), and for each a = (h, A) ∈ L × S(X) define fa : X → R 1 ∪ {∞} by fa (x) = h(x) for all x ∈ A and fa (x) = ∞ for all x ∈ X \ A. Denote by L(T ) the set of all functions f ∈ L such that f ◦T =f and set ) L(T c = {f ∈ Lc : f ◦ T = f }. (T )

Note that L(T ) and Lc are closed subsets of L with the weak topology (see Section 2.7). Let a function φ : X → R 1 satisfy φ(x) → ∞ as ρ(x, θ ) → ∞, L(T ) (φ) = {f ∈ L(T ) : f (x) ≥ φ(x) for all x ∈ X}, ) (T ) ) (φ) ∩ L(T L(T c (φ) = L c . (T )

Note that L(T ) (φ) and Lc (φ) are closed subsets of L with the weak topology. Denote by S (T ) (X) the set of all A ∈ S(X) such that T (A) = A. Note that S (T ) (X) is a closed subset of S(X) with the strong topology. We prove the following result. Theorem 2.36 Let AT be one of the spaces ) (T ) ) (T ) (X), L(T ) (φ) × S (T ) (X), L(T (X). L(T ) × S (T ) (X), L(T c ×S c (φ) × S

2.18 An Extension of Theorem 2.14

87

Then there exists an everywhere dense (in the strong topology) set B ⊂ AT which is a countable intersection of open (in the weak topology) subsets of AT and such that for any f ∈ B the minimization problem of fa on X is well-posed with respect to AT and has exactly two different minimizers. If ρ(T (x), T (y)) = ρ(x, y), x, y ∈ X, then for any f ∈ B the minimization problem of fa on X is well-posed with respect to A. In Section 2.7, it was shown that Proposition 2.18 implies (H). In view of Theorem 2.1, in order to prove Theorems 2.36, it is sufficient to show that there exists an everywhere dense (in the strong topology) open (in the weak topology) set B0 ⊂ AT such that for each f ∈ B0 , inf(f ) < inf{f (z) : z ∈ Fix(T )}. (Here we suppose that the infimum of an empty set is ∞.) Namely, we prove the following result. Proposition 2.37 There exists an everywhere dense (in the strong topology) open (in the weak topology) set B0 ⊂ S (T ) (X) such that for each A ∈ B0 , A ∩ Fix(T ) = ∅. For its proof, we need the following two lemmas. The first one follows easily from the continuity of the mapping T . Lemma 2.38 For each  > 0, the set Fix (T ) := {x ∈ X : min{ρ(x, Fix(T )), ρ(T (x), Fix(T ))} < }

(2.238)

is open. Moreover, for each γ > 0, there exists  > 0 such that Fix (T ) ⊂ {x ∈ X : ρ(x, Fix(T )) < γ }. Since the set X \ Fix(T ) is everywhere dense, the uniform continuity of T and the compactness of Fix(T ) easily imply the next lemma. Lemma 2.39 Let  > 0. Then there exists a finite set C ⊂ X \ Fix(T ) such that T (C) = C, ρ(x, C) ≤ , x ∈ Fix(T ), ρ(y, Fix(T )) ≤ , y ∈ C.

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Completion of the Proof of Proposition 2.37 Let A ∈ S (T ) and n be a natural number. Lemma 2.38 implies that there exists  ∈ (0, 1) such that Fix (T ) ⊂ {x ∈ X : ρ(x, Fix(T )) < (8n)−1 }.

(2.239)

Lemma 2.39 implies that there exists a finite set C˜ ⊂ X \ Fix(T )

(2.240)

˜ = C, ˜ T (C)

(2.241)

˜ ≤ (8n)−1 , x ∈ Fix(T ), ρ(x, C)

(2.242)

˜ ρ(y, Fix(T )) ≤ (8n)−1 , y ∈ C.

(2.243)

˜ An = (A \ Fix (T )) ∪ C.

(2.244)

such that

Set

Clearly, the set An is closed and T (An ) = An . In view of (2.242)–(2.244) (see Section 2.7), (A, An ) ∈ Gs (8n).

(2.245)

By (2.238), (2.240), (2.244), and the compactness of Fix(T ), for each x ∈ Fix(T ), ˜ ρ(x, An ) > min{, ρ(x, C)} ˜ : y ∈ Fix(T )}} > 0. ≥ min{, min{ρ(y, C)

(2.246)

Choose a natural number k such that ˜ : y ∈ Fix(T )}} 4k −1 < min{, min{ρ(y, C)

(2.247)

and Fix(T ) ⊂ B(θ, k).

(2.248)

2.18 An Extension of Theorem 2.14

89

There exists an open neighborhood of An in S (T ) (X) such that U (A, n) ⊂ {B ∈ S (T ) (X) : (B, An ) ∈ Gw (k)}

(2.249)

(see Section 2.7). Let B ∈ U (A, n)

(2.250)

x ∈ Fix(T ).

(2.251)

|ρ(x, B) − ρ(x, An )| ≤ k −1 .

(2.252)

and

By (2.248)–(2.251),

It follows from (2.246), (2.247), (2.251), and (2.252) that ρ(x, B) ≥ ρ(x, An ) − γ −1 ˜ : y ∈ Fix(T )}} − k −1 ≥ k −1 . ≥ min{, min{ρ(y, C) Therefore, Fix(T ) ∩ B = ∅ for each B ∈ U (A, n). Define B0 = ∪{U (A, n) : A ∈ AT , n ≥ 1 is an integer}. In view of (2.245), B0 is everywhere dense in the strong topology. Clearly, for each A ∈ B0 , Fix(T ) ∩ B = ∅. Proposition 2.37 is proved.

Chapter 3

Parametric Optimization

In this chapter we study classes of symmetric parametric optimization problems which are identified with the corresponding spaces of functions, equipped with appropriate complete metrics. Using the Baire category approach, for any of these classes we show the existence of a subset of the space of functions, which is a countable intersection of open and everywhere dense sets, such that for every function from this intersection the corresponding symmetric parametric optimization problems possess solutions for any parameter. These results are obtained as realizations of a general variational principle which is established in this chapter. The results of this chapter are new.

3.1 Generic Variational Principle We consider a complete metric space (X, ρ). Let (A, dA ) be a complete metric space and B be a nonempty set. We always consider the set X with the topology generated by the metric ρ. For the space A we consider the topology generated by the metric dA . This topology will be called the strong topology. In addition to the strong topology we also consider a weaker topology on A which is not necessarily Hausdorff. This topology will be called the weak topology. (Note that these topologies can coincide.) We assume that with every (a, b) ∈ A × B a lower semicontinuous function fab on X is associated with values in R¯ = [−∞, ∞]. For each function g : X → R¯ we set inf(g) = inf{g(x) : x ∈ X}.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 A. J. Zaslavski, Turnpike Phenomenon and Symmetric Optimization Problems, Springer Optimization and Its Applications 190, https://doi.org/10.1007/978-3-030-96973-8_3

91

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For each x ∈ X and each F ⊂ X we set ρ(x, F ) = inf{ρ(x, y) : y ∈ F }. We use the convention that ∞ − ∞ = 0 and ∞/∞ = 1. Let (a, b) ∈ A × B. We say that the minimization problem for fab on (X, ρ) is well-posed in the generalized sense [143] if inf(fab ) is finite, the set {x ∈ X : fab (x) = inf(fab )} is nonempty and compact and each sequence {xi }∞ i=1 ⊂ X satisfying limi→∞ fab (xi ) = inf(fab ) has a convergent subsequence. Assume that T : X → X is a continuous mapping such that T 2 (x) = x for all x ∈ X

(3.1)

and that for each (a, b) ∈ A × B, fab ◦ T = fab .

(3.2)

In our study we use the following basic hypotheses about the functions. (H) For each a ∈ A and each positive number  there exist a¯ ∈ A, an open neighborhood V of a¯ in A with the weak topology, a finite set {x1 , . . . , xq } ⊂ X, where q ≥ 1 is an integer, and a positive number δ such that (i) dA (a, a) ¯ < ; (ii) for each ξ ∈ V and each b ∈ B, inf(fξ b ) is finite and if x ∈ X satisfies fξ b (x) ≤ inf(fξ b ) + δ, then min{min{ρ(x, xi ), ρ(T (x), xi )} : i = 1, . . . , q} ≤ . We prove the following result. Theorem 3.1 Assume that (H) holds. Then there exists a set F ⊂ A which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of A such that for each a ∈ F and each b ∈ B the minimization problem for fab on (X, ρ) is well-posed in the generalized sense. Proof It follows from (H) that for each a ∈ A and each integer n ≥ 1 there exist a(n) ∈ A, an open neighborhood V (a, n) of a(n) in A with the weak topology, a positive number δ(a, n) and a nonempty finite set Q(a, n) ⊂ X such that dA (a, a(n)) < 1/n

(3.3)

and the following property holds: (C1)

For each ξ ∈ V (a, n) and each b ∈ B, inf(fξ b ) is finite and if x ∈ X satisfies fξ b (x) ≤ inf(fξ b ) + δ(a, n),

3.1 Generic Variational Principle

93

then min{ρ(x, Q(a, n)), ρ(T (x), Q(a, n))} ≤ 1/n. Define F = ∩∞ n=1 [∪{V (a, i) : a ∈ A and i ≥ n}].

(3.4)

It is easy to see that F is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of A. Let ξ ∈ F and b ∈ B. It is easy to see that inf(fξ b ) is finite. It follows from (3.4) that for each natural number n there exist an ∈ A and an integer in ≥ n such that ξ ∈ V (an , in ).

(3.5)

Assume that a sequence {zi }∞ i=1 ⊂ X satisfies lim fξ b (zi ) = inf(fξ b ).

(3.6)

i→∞

In view of (3.5), (3.6) and the definition of V (an , in ) (see property (C1)), for each natural number n the inequality min{ρ(zj , Q(an , in )), ρ(T (zj ), Q(an , in ))} ≤ 1/n holds for all 1, 2, . . . are subsequence hold: For each (p) {zik }∞ k=1 .

(3.7)

sufficiently large natural numbers j . Since the sets Q(an , in ), n = finite we conclude that for each natural number p there exists a (p) ∞ {zik }∞ k=1 of the sequence {zi }i=1 such that the following properties (p+1) ∞ }k=1

natural number p the sequence {zik

is a subsequence of

For each natural number p at least one of the following relations holds: ρ(zij , zis ) ≤ 2p−1 for all integers j, s ≥ 1; (p)

(p)

ρ(T (zij ), T (zis )) ≤ 2p−1 for all integers j, s ≥ 1. (p)

(p)

∞ This implies that there exists a subsequence {zi∗k }∞ k=1 of the sequence {zi }i=1 such ∗ ∞ ∗ ∞ that at least one of the sequences {zik }k=1 , {T (zik )}k=1 is a Cauchy sequence. Since the mapping T is continuous and T −1 = T we obtain that in the both cases there exists

x∗ = lim zi∗k . k→∞

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Together with (3.6) and the lower semicontinuity of fξ b this implies that fξ b (x∗ ) = inf(fξ b ).

(3.8)

Therefore we have shown that for each sequence {zi }∞ i=1 ⊂ X satisfying (3.6) there exists a subsequence which converges to a point x∗ satisfying (3.8). This completes the proof of Theorem 3.1. Remark 3.2 Note that Theorem 3.1 is also valid if B = ∪∞ i=1 Ci and (H) holds with B = Ci for all integers i ≥ 1.

3.2 Concretization of the Hypothesis (H) We use the notations and definitions introduced in Section 3.1. The proofs of our generic existence results consist in verification in each case of the hypothesis (H). To simplify the verification of (H) in this section we introduce new assumptions and show that they imply (H). Thus to verify (H) we need to show that these new assumptions are valid. In fact this approach allows us to simplify the problem. In the sequel we assume that inf(fab ) is finite for all (a, b) ∈ A × B. We use the following assumptions. (A1) For each a ∈ A, supb∈B inf(fab ) < ∞. (A2) For each a ∈ A and each positive number  there exist a¯ ∈ A, a nonempty ¯ <  and the finite set Q ⊂ X and a positive number δ such that dA (a, a) following property holds: For each b ∈ B and each x ∈ X satisfying fab ¯ (x) ≤ inf(fab ¯ ) + δ, the inequality min{ρ(x, Q), ρ(T (x), Q)} ≤  holds. (A3) For each a ∈ A, each natural number n and each positive number  there exists a neighborhood V of a in A with the weak topology such that for each b ∈ B, each ξ ∈ V and each x ∈ X satisfying min{fab (x), fξ b (x)} ≤ n the inequality |fab (x) − fξ b (x)| ≤  holds. Proposition 3.3 Assume that (A1), (A2), and (A3) hold. Then (H) holds. Proof Let a ∈ A and  be a positive number. (A2) implies that there exist a¯ ∈ A, a nonempty finite set Q ⊂ X and a number δ ∈ (0, 1) such that dA (a, a) ¯ < /2

(3.9)

3.2 Concretization of the Hypothesis (H)

95

and that the following property holds: (C2) For each b ∈ B and each x ∈ X satisfying fab ¯ (x) ≤ inf(fab ¯ ) + δ the inequality min{ρ(x, Q), ρ(T (x), Q)} ≤ /2 holds. In view of (A1) there exists an integer n ≥ 1 such that n > | sup inf(fab ¯ )| + 4 + 4. b∈B

(3.10)

It follows from (A3) that there exists a neighborhood V of a¯ in A with the weak topology such that the following property holds: (C3)

For each b ∈ B, each ξ ∈ V and each x ∈ X satisfying min{fab ¯ (x), fξ b (x)} ≤ n

the inequality |fab ¯ (x) − fξ b (x)| ≤ δ/8

(3.11)

holds. Let ξ ∈ V and b ∈ B. We will show that | inf(fξ b ) − inf(fab ¯ )| ≤ δ/8.

(3.12)

fab ¯ (x) ≤ inf(fab ¯ ) + 1.

(3.13)

Let x ∈ X and

It follows from (3.10), (3.11), (3.13) and property (C3) that inf(fξ b ) ≤ inf{fξ b (x) : x ∈ X and fab ¯ (x) ≤ inf(fab ¯ ) + 1} ≤ inf{fab ¯ (x) + δ/8 : x ∈ X and fab ¯ (x) ≤ inf(fab ¯ ) + 1} = δ/8 + inf{fab ¯ (x) : x ∈ X and fab ¯ (x) ≤ inf(fab ¯ ) + 1} = δ/8 + inf(fab ¯ ). Hence inf(fξ b ) ≤ inf(fab ¯ ) + δ/8.

(3.14)

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Let x ∈ X and fξ b (x) ≤ inf(fξ b ) + 1.

(3.15)

In view of (3.10), (3.14), (3.15) and the property (C3), inequality (3.11) is true. Since (3.15) implies (3.11) we obtain that inf(fab ¯ ) ≤ inf{fab ¯ (x) : x ∈ X and fξ b (x) ≤ inf(fξ b ) + 1} ≤ inf{fξ b (x) + δ/8 : x ∈ X and fξ b (x) ≤ inf(fξ b ) + 1} = δ/8 + inf{fξ b (x) : x ∈ X and fξ b (x) ≤ inf(fξ b ) + 1} = δ/8 + inf(fξ b ). Hence inf(fab ¯ ) ≤ δ/8+inf(fξ b ). Together with (3.14) this inequality implies (3.12). Let x ∈ X and fξ b (x) ≤ inf(fξ b ) + δ/4.

(3.16)

fξ b (x) ≤ inf(fab ¯ ) + 3δ/8.

(3.17)

By (3.12) and (3.16),

It follows from (3.10), (3.17) and property (C3) that (3.11) is true. In view of (3.11) and (3.17), fab ¯ (x) ≤ inf(fab ¯ ) + δ/2. By this inequality and the property (C2), min{ρ(x, Q), ρ(T (x), Q)} ≤ /2. Proposition 3.3 is proved. Assume that B is a Hausdorff topological space. We use the following assumption. (A4) Let (a, b) ∈ A × B,  be a positive number and let n be a natural number. Then there exists a neighborhood U of b in B such that for each ξ ∈ U and each x ∈ X satisfying min{fab (x), faξ (x)} ≤ n

3.2 Concretization of the Hypothesis (H)

97

the inequality |fab (x) − faξ (x)| ≤  holds. Proposition 3.4 Assume that (A4) holds. Then the function b → inf(fab ), b ∈ B is continuous for each a ∈ A. Proof Let a ∈ A, b ∈ B and  ∈ (0, 1). Fix an integer n > | inf(fab )| + 4.

(3.18)

It follows from (A4) that there exists a neighborhood U of b in B such that for each ξ ∈ U and each x ∈ X satisfying min{fab (x), faξ (x)} ≤ n + 1

(3.19)

the following inequality holds: |fab (x) − faξ (x)| ≤ /2.

(3.20)

Assume that ξ ∈ U . Inequality (3.18) implies that inf(fab ) = inf{fab (x) : x ∈ X and fab (x) ≤ n − 3}.

(3.21)

Let x ∈ X satisfy fab (x) ≤ n − 3.

(3.22)

It follows from (3.22), the definition of U (see (3.19) and (3.20)) that faξ (x) ≤ fab (x) + /2.

(3.23)

Since (3.22) implies (3.23) equality (3.21) implies that inf(faξ ) ≤ inf{faξ (x) : x ∈ X and fab (x) ≤ n − 3} ≤ inf{fab (x) + /2 : x ∈ X and fab (x) ≤ n − 3} = /2 + inf(fab ). Hence inf(faξ ) ≤ inf(fab ) + /2.

(3.24)

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3 Parametric Optimization

Let x ∈ X satisfy faξ (x) ≤ inf(faξ ) + /2.

(3.25)

It follows from (3.18), (3.24) and (3.25) that faξ (x) ≤ n − 3. By this inequality and the definition of U , (3.20) is true. Since (3.25) implies (3.20) we obtain that inf(fab ) ≤ inf{fab (x) : x ∈ X and faξ (x) ≤ inf(faξ ) + /2} ≤ inf{faξ (x) + /2 : x ∈ X and faξ (x) ≤ inf(faξ ) + /2} = = /2 + inf(faξ ). Hence inf(fab ) ≤ /2 + inf(faξ ). Combined with (3.24) this implies that | inf(fab ) − inf(faξ )| ≤ /2 for all ξ ∈ U . This completes the proof of Proposition 3.4. Corollary 3.5 Assume that B is compact. Then (A4) implies (A1). We also use the following assumption. (A5) Let a ∈ A and  be a positive number. Then there exists a positive number δ such that for each nonempty finite set F ⊂ B there exist a¯ ∈ A and a nonempty compact subset A of the metric space (X, ρ) such that dA (a, a) ¯ < ,

(3.26)

sup inf(fab ¯ ) ≤ sup inf(fab ) + 1,

(3.27)

fab ¯ (x) ≥ fab (x) for all b ∈ B and all x ∈ X,

(3.28)

|fab ¯ 1 (x) − fab ¯ 2 (x)| ≤ |fab1 (x) − fab2 (x)|

(3.29)

b∈B

b∈B

3.2 Concretization of the Hypothesis (H)

99

for each b1 , b2 ∈ X and each x ∈ X, and the following property holds: For each b ∈ F and each x ∈ X satisfying fab ¯ (x) ≤ inf(fab ¯ ) + δ the inequality min{ρ(x, A), ρ(T (x), A)} ≤  holds. Proposition 3.6 Assume that B is compact and (A4) and (A5) hold. Then (A2) holds. Proof Corollary 3.5 implies that (A1) holds and sup inf(fab ) < ∞ for all a ∈ A.

b∈B

Let a ∈ A and 0 be a positive number. Let δ ∈ (0, 1) be as guaranteed by (A5) with  = 0 /2. Fix an integer m > | sup inf(fab )| + 4. b∈B

(3.30)

It follows from (A4) that for each b ∈ B there exists an open neighborhood Ub of b in B such that for each ξ ∈ Ub and each x ∈ X satisfying min{faξ (x), fab (x)} ≤ m + 4

(3.31)

|faξ (x) − fab (x)| ≤ δ/16

(3.32)

the inequality

holds. Since B is compact there exists a nonempty finite set F ⊂ B such that B = ∪{Ub : b ∈ F }.

(3.33)

In view of the definition of δ and (A5) with  = 0 /2, there exist a¯ ∈ A and a nonempty compact set A of the metric space (X, ρ) such that ¯ < 0 /2, dA (a, a)

(3.34)

equations (3.27)-(3.29) hold and the following property holds: (C4) For each b ∈ F and each x ∈ X satisfying fab ¯ (x) ≤ inf(fab ¯ ) + δ the inequality min{ρ(x, A), ρ(T (x), A)} ≤ 0 /2 is valid.

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Since A is compact there exists a nonempty finite set Q ⊂ A such that ρ(x, Q) ≤ 0 /4 for all x ∈ A.

(3.35)

faξ ¯ (x) ≤ inf(faξ ¯ ) + δ/16.

(3.36)

Let ξ ∈ B, x ∈ X and

In order to complete the proof of the proposition it is sufficient to show that min{ρ(x, Q), ρ(T (x), Q)} ≤ 0 . In view of (3.35) we need only to show that min{ρ(x, A), ρ(T (x), A)} ≤ 0 /2. It follows from (3.27) and (3.33) that there exists b ∈ F such that ξ ∈ Ub .

(3.37)

By (3.30), inf(fab ), inf(faξ ) < m − 4, inf(fab ¯ ), inf(faξ ¯ ) < m.

(3.38)

inf(fhg ) = inf{fhg (z) : z ∈ X and fhg (z) ≤ m + 1},

(3.39)

Hence

(h, g) ∈ {(a, b), (a, ξ ), (a, ¯ b), (a, ¯ ξ )}. We will show that | inf(fab ¯ ) − inf(faξ ¯ )| ≤ δ/16.

(3.40)

min{fab ¯ (z), faξ ¯ (z)} ≤ m + 2.

(3.41)

Let z ∈ X and

Relations (3.28) and (3.41) imply that min{fab (z), faξ (z)} ≤ m + 2. It follows from this inequality, (3.37) and the definition of Ub (see (3.31) and (3.32)) that |fab (z) − faξ (z)| ≤ δ/16.

3.3 The First Generic Existence Results

101

Together with (3.29) this inequality implies that |fab ¯ (z) − faξ ¯ (z)| ≤ |fab (z) − faξ (z)| ≤ δ/16. Thus for each z ∈ X satisfying (3.41), |fab ¯ (z) − faξ ¯ (z)| ≤ δ/16. Combining this inequality with (3.39) we see that (3.40) holds. By (3.28) and (3.36)–(3.38), faξ (x) ≤ faξ ¯ (x) ≤ m + 1. It follows from this inequality, (3.37), the definition of Ub (see (3.31) and (3.32)) and (3.29) that |faξ ¯ (x) − fab ¯ (x)| ≤ |faξ (x) − fab (x)| ≤ δ/16. Together with (3.26) and (3.40) this inequality implies that fab ¯ (x) ≤ faξ ¯ (x) + δ/16 ≤ δ/8 + inf(faξ ¯ ) ≤ inf(fab ¯ ) + δ/4. These inequalities and the property (C4) imply that min{ρ(x, A), ρ(T (x), A)} ≤ 0 /2. Proposition 3.6 is proved. Theorem 3.1, Propositions 3.3 and 3.6 and Corollary 3.5 imply the following result. Theorem 3.7 Assume that B is compact and (A3), (A4), and (A5) hold. Then there exists a set F which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of A such that for each a ∈ F and each b ∈ B the minimization problem for fab on (X, ρ) is well-posed in the generalized sense. Note that Theorem 3.7 is also valid if B = ∪∞ i=1 Ci and (A3), (A4) and (A5) hold with the compact set B = Ci for all natural numbers i.

3.3 The First Generic Existence Results Denote by Cl (X) the set of all lower semicontinuous bounded from below functions f : X → R 1 ∪ {∞} which are not identically ∞.

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Let T : X → X be a continuous mapping such that T 2 (x) = x, x ∈ X. In this section we assume that A, B ⊂ Cl (X), fab (x) = a(x) + b(x), x ∈ X, a ∈ A, b ∈ B, a ◦ T = a for all a ∈ A, b ◦ T = b for all b ∈ B, any function h ∈ A is finite-valued and any function h ∈ B is finite-valued. Fix θ ∈ X. For the set Cl (X) we consider the uniformity determined by the following base: U0 (n) = {(f, g) ∈ Cl (X) × Cl (X) : |f (x) − g(x)| ≤ n−1 for all x ∈ X such that ρ(x, θ ) ≤ n or min{f (x), g(x)} ≤ n}, where n = 1, 2, . . . Clearly the space Cl (X) with this uniformity is Hausdorff and has a countable base. Therefore this uniform space is metrizable (by a metric d0 (·, ·)). The set Cl (X) is also equipped with the uniformity determined by the following base: U1 (n) = {(f, g) ∈ Cl (X) × Cl (X) : f (x) ≤ g(x) + n−1 and g(x) ≤ f (x) + n−1 for each x ∈ X}, where n = 1, 2, . . . It is easy to see that the space Cl (X) with this uniformity is metrizable (by a metric d1 ) and complete. For the space A we consider the strong and weak topologies induced by the metrics d1 and d0 , respectively. The space B is equipped with the topology induced by the metric d0 . Define Cl,T = {f ∈ Cl (X) : f ◦ T = f }, Cc,T = {f ∈ Cl,T : f is finite-valued and continuous}. Assume that A is either Cl,T or Cc,T . Clearly, the metric space (A, d1 ) is complete. Theorem 3.8 Assume that B is a countable union of compact subsets of the metric space (Cl (X), d0 ) and that f ◦ T = f for all f ∈ B. Then there exists a set F ⊂ A which is a countable intersection of open (in the weak topology) everywhere dense

3.3 The First Generic Existence Results

103

(in the strong topology) subsets of A such that for each g ∈ F and each h ∈ B the problem minimize g(x) + h(x) subject to x ∈ X on (X, ρ) is well-posed in the generalized sense. Proof We may assume without loss of generality that B is a compact subset of (Cl (X), d0 ). In view of Theorem 3.7 we need to show that (A3), (A4), and (A5) hold. Evidently, (A4) holds. We will show that (A3) is true. Let a ∈ A,  ∈ (0, 1) and let n be a natural number. By (A4) and Proposition 3.4 the function b → inf(ξ + b), b ∈ B is continuous for all ξ ∈ A. Since B is compact there exists a number c0 > | inf(b)| + 1, b ∈ B. Fix an integer m > n + c0 +  −1 . Let b ∈ B, ξ ∈ A, (a, ξ ) ∈ U0 (m), x ∈ X and min{(a + b)(x), (ξ + b)(x)} ≤ n.

(3.42)

In view of (3.42) and the choice of c0 and m, min{a(x), ξ(x)} ≤ min{(a + b)(x), (ξ + b)(x)} − b(x) ≤ n + c0 < m. Since (a, ξ ) ∈ U0 (m), |a(x) − ξ(x)| ≤ m−1 < . Hence (A3) holds. We will show that (A5) holds. Let a ∈ A and let  ∈ (0, 1). Fix a natural number n0 such that the following condition holds: if ξ ∈ A and (a, ξ ) ∈ U1 (n0 ), then d1 (a, ξ ) < .

(3.43)

Choose numbers −1 0 ∈ (0, 8−1 min{, n−1 0 }), δ ∈ (0, 8 0 ).

(3.44)

Let {b1 , . . . , bq } ⊂ B where q ≥ 1 is an integer. For each i ∈ {1, . . . , q} choose xi ∈ X such that (a + bi )(xi ) ≤ inf(a + bi ) + 4−1 δ0 .

(3.45)

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3 Parametric Optimization

Define A = {x1 , . . . , xq }, a(y) ¯ = a(y) + 4−1 0 min{1, ρ(y, A), ρ(T (y), A)}, y ∈ X.

(3.46) (3.47)

It is easy to see that a¯ ∈ A. It follows from (3.43), (3.44) and (3.47) that (a, ¯ a) ∈ U1 (n0 ) and ¯ a) < . d1 (a, Evidently, (3.27)–(3.29) hold. Let i ∈ {1, . . . , q}, x ∈ X and (a¯ + bi )(x) ≤ inf(a¯ + bi ) + δ0 < ∞.

(3.48)

By (3.45), (3.47) and (3.48), (a¯ + bi )(x) ≤ inf(a¯ + bi ) + δ0 ≤ (a¯ + bi )(xi ) + δ0 = (a + bi )(xi ) + δ0 ≤ (a + bi )(x) + 2δ0 . Combined with (3.47) this implies that bi (x) + a(x) + 4−1 0 min{1, ρ(x, A), ρ(T (x), A)} = (a¯ + bi )(x) ≤ (a + bi )(x) + 2δ0 , min{1, ρ(x, A), ρ(T (x), A)} ≤ 8δ <  and min{ρ(x, A), ρ(T (x), A)} < . Thus (A5) holds and Theorem 3.8 is proved.

3.4 The Second Generic Existence Result We continue to use the assumptions, notation, and definitions from the previous section. In particular, we assume that T : X → X is a continuous mapping such that T 2 (x) = x, x ∈ X.

3.4 The Second Generic Existence Result

105

Let B be a Hausdorff topological space. We assume that B is a countable union of its compact subsets. Denote by M the set of all functions a : B × X → R 1 ∪ {∞} such that for each b ∈ B the following properties hold: The function x → a(b, x), x ∈ X belongs to Cl (X); for each natural number n and each positive number  there exists a neighborhood U of b in B such that for each ξ ∈ U and each x ∈ X satisfying min{a(ξ, x), a(b, x)} ≤ n the inequality |a(b, x) − a(ξ, x)| ≤  holds. For the set M we consider the uniformity determined by the following base: Us (n) = {(a1 , a2 ) ∈ M × M : |a1 (b, x) − a2 (b, x)| ≤ n−1 for all b ∈ B and all x ∈ X} where n ≥ 1 is an integer. The space M with this uniformity is metrizable (by a metric ds ) and complete. Fix θ ∈ X. For the set M we also consider the uniformity determined by the following base: Uw (n) = {(a1 , a2 ) ∈ M × M : |a1 (b, x) − a2 (b, x)| ≤ n−1 for each b ∈ B and each x ∈ X such that ρ(x, θ ) ≤ n or min{a1 (b, x), a2 (b, x)} ≤ n}, where n = 1, 2, . . . . It is easy to see that the space M with this uniformity is metrizable (by a metric dw ). For the space M we consider the strong and weak topologies induced by the metrics ds and dw , respectively. Denote by Mf the set of all finite-valued functions a ∈ M, by Ms the set of all functions a ∈ Mf such that the function x → a(b, x), x ∈ X is continuous for all b ∈ B, and by Mc the set of all continuous functions a ∈ Mf . Clearly Mf , Ms and Mc are closed subsets of the metric space (M, ds ). Define MT = {a ∈ M : a(b, T (x)) = a(b, x) for all x ∈ X}, Ms,T = MT ∩ Ms , Mc,T = MT ∩ Mc , Mf,T = MT ∩ Mf . Clearly, Mf,T , Ms,T , Mc,T are closed subsets in the metric space (M, ds ). We consider the subspaces Mf,T , Ms,T , Mc,T ⊂ M equipped with the relative weak and strong topologies.

106

3 Parametric Optimization

Theorem 3.9 Let A be either MT or Mf,T or Ms,T or Mc,T . Then there exists a subset F ⊂ A which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of A such that for each a ∈ F and each b ∈ B the problem minimize a(b, x) subject to x ∈ X on (X, ρ) is well-posed in the generalized sense. Proof We may assume without loss of generality that B is compact. For each (a, b) ∈ A × B set fab (x) = a(b, x), x ∈ X. Theorem 3.7 implies that we need to show that (A3), (A4), and (A5) hold. Clearly, (A3) and (A4) are valid. We will show that (A5) holds. Let a ∈ A and  ∈ (0, 1). There exists a natural number n0 such that if ξ ∈ A, (a, ξ ) ∈ Us (n0 ), then ds (a, ξ ) < .

(3.49)

−1 0 ∈ (0, 8−1 min{, n−1 0 }), δ ∈ (0, 8 0 ).

(3.50)

Fix

Let {b1 , . . . , bq } ⊂ B where q ≥ 1 is an integer. For each i ∈ {1, . . . , q} choose xi ∈ X for which a(bi , xi ) ≤ inf{a(bi , x) : x ∈ X} + 4−1 δ0 .

(3.51)

Define A = {x1 , . . . , xq } and a(b, ¯ y) = a(b, y) + 4−1 0 min{1, ρ(y, A), ρ(T (y), A)}, b ∈ B, y ∈ X. (3.52) It is easy to see that a¯ ∈ A. In view of (3.50) and (3.52), (a, a) ¯ ∈ Us (n0 ). ¯ < . Clearly (3.27)– It follows from (3.49) and the inclusion above that ds (a, a) (3.29) hold. Let i ∈ {1, . . . , q}, x ∈ X and a(b ¯ i , x) ≤ inf{a(b ¯ i , z) : z ∈ X} + 4−1 δ0 .

(3.53)

3.4 The Second Generic Existence Result

107

Relations (3.51)–(3.53) imply that a(bi , x) + 4−1 0 min{1, ρ(x, A), ρ(T (x), A)} ¯ i , xi ) + 4−1 δ0 = a(b ¯ i , x) ≤ a(b = a(bi , xi ) + 4−1 δ0 ≤ a(bi , x) + 2−1 δ0 and min{1, ρ(x, A), ρ(T (x), A)} ≤ 2δ < . Hence min{ρ(x, A), ρ(T (x), A)} < . Thus (A5) holds and Theorem 3.9 is proved.

Chapter 4

Infinite Dimensional Control

The study of infinite dimensional optimal control has been a rapidly growing area of research [1–4, 13–18, 35, 52–54, 67, 71, 73, 74, 80, 93, 94, 102, 113–116, 128]. In this chapter we present preliminaries which we need in order to study turnpike properties of infinite dimensional optimal control problems. We discuss Banach space valued functions, unbounded operators, C0 semigroups, evolution equations and admissible control operators.

4.1 Banach Space Valued Functions Let (X,  · ) be a Banach space and a < b be real numbers. For any set E ⊂ R 1 define χE (t) = 1 for all t ∈ E and χE (t) = 0 for all t ∈ R 1 \ E. If a set E ⊂ R 1 is Lebesgue measurable, then its Lebesgue measure is denoted by |E| or by mes(E). A function f : [a, b] → X is called a simple function if there exists a finite collection of Lebesgue measurable sets Ei ⊂ [a, b], i ∈ I , mutually disjoint, and xi ∈ X, i ∈ I such that f (t) =



χEi (t)xi , t ∈ [a, b].

i∈I

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 A. J. Zaslavski, Turnpike Phenomenon and Symmetric Optimization Problems, Springer Optimization and Its Applications 190, https://doi.org/10.1007/978-3-030-96973-8_4

109

110

4 Infinite Dimensional Control

A function f : [a, b] → X is strongly measurable if there exists a sequence of simple functions φk : [a, b] → X, k = 1, 2, . . . such that lim φk (t) − f (t) = 0, t ∈ [a, b] almost everywhere (a. e.).

k→∞

For every simple function f (·) = Bochner integral by 

b

 i∈I

(4.1)

χEi (·)xi , where the set I is finite, define its 

f (t)dt =

a

|Ei |xi .

i∈I

Let f : [a, b] → X be a strongly measurable function. We say that f is Bochner integrable if there exists a sequence of simple functions φk : [a, b] → X, k = b 1, 2, . . . such that (4.1) holds and the sequence { a φk (t)dt}∞ k=1 strongly converges in X. In this case we define the Bochner integral of the function f by 

b

 f (t)dt = lim

b

k→∞ a

a

φk (t)dt.

It is known that the integral defined above is independent of the choice of the sequence {φk }∞ k=1 [80]. Similar to the Lebesgue integral, for any measurable set E ⊂ [a, b], the Bochner integral of f over E is defined by 



b

f (t)dt = E

χE (t)f (t)dt. a

The following result is true (see Proposition 3.4, Chapter 2 of [80]). Proposition 4.1 Let f : [a, b] → X be a strongly measurable function. Then f is Bochner integrable if and only if the function f (·) is Lebesgue integrable. Moreover, in this case 

b





b

f (t)dt ≤

a

f (t)dt.

a

The Bochner integral possesses almost the same properties as the Lebesgue integral. If f : [a, b] → X is strongly measurable and f (·) ∈ Lp (a, b), for some p ∈ [1, ∞), then we say that f (·) is Lp Bochner integrable. For every p ≥ 1, the set of all Lp Bochner integrable functions is denoted by Lp (a, b; X) and for every f ∈ Lp (a, b; X), 

b

f Lp (a,b;X) = (

f (t)p dt)1/p .

a

Clearly, the set of all Bochner integrable functions on [a, b] is L1 (a, b; X).

4.2 Unbounded Operators

111

Let a < b be real numbers. A function x : [a, b] → X is absolutely continuous (a. c.) on [a, b] if for each  > 0 there exists δ > 0 such that for each pair of q q sequences {tn }n=1 , {sn }n=1 ⊂ [a, b] satisfying tn < sn , n = 1, . . . , q,

q 

(sn − tn ) ≤ δ,

n=1

(tn , sn ) ∩ (tm , sm ) = ∅ for all m, n ∈ {1, . . . , q} such that m = n we have q 

x(tn ) − x(sn ) ≤ .

n=1

The following result is true (see Theorem 1.124 of [18]). Proposition 4.2 Let X be a reflexive Banach space. Then every a. c. function x : [a, b] → X is a. e. differentiable on [a, b] and  x(t) = x(a) +

t

(dx/dt)(s)ds, t ∈ [a, b]

a

where dx/dt ∈ L1 (a, b; X) is the strong derivative of x. Let −∞ < τ1 < τ2 < ∞. Denote by W 1,1 (τ1 , τ2 ; X) (or W 1,1 (τ1 , τ2 ) if the space X is understood) the set of all functions x : [τ1 , τ2 ] → X for which there exists a Bochner integrable function u : [τ1 , τ2 ] → X such that for all t ∈ (τ1 , τ2 ],  x(t) = x(τ1 ) +

t

u(s)ds. τ1

4.2 Unbounded Operators Let (X,  · ) and (Y,  · ) be Banach spaces. Denote by L(X, Y ) the set of all linear continuous operators from X to Y . For every A ∈ L(X, Y ) define A = sup{Ax : x ∈ X, x ≤ 1}. Set L(X) = L(X, X), X∗ = L(X, R 1 ) and Y ∗ = L(Y, R 1 ). For all x ∈ X set I x = x. In the space X we always consider the norm convergence. Let D(A) be a linear subspace of X (not necessarily closed) and let A : D(A) → Y be a linear operator. We say that A is densely defined if D(A) is dense in X. We

112

4 Infinite Dimensional Control

say that A is closed if the graph G(A) = {(x, y) ∈ X × Y : x ∈ D(A), y = Ax} of A is closed in X × Y . We say that A is closable if there exists a closed operator ¯ → Y such that A¯ : D(A) ¯ ⊂ X, Ax ¯ = Ax, x ∈ D(A). D(A) ⊂ D(A) Let the symbol ·, · be referred to as the duality pairing between X∗ and X. Similar to a bounded operator, for any linear operator (not necessarily bounded) A : D(A) ⊂ X → X we still define the resolvent ρ(A) = {λ ∈ C 1 : (λI − A)−1 ∈ L(X)}, the spectrum σ (A) = C 1 \ ρ(A), the point spectrum (or the set of eigenvalues) σp (A) of A and set Ran(A) = {Ax : x ∈ D(A)}. Let A : D(A) ⊂ X → X be densely defined. Clearly, the map x → Ax, y, x ∈ D(A), y ∈ X∗ is well-defined. Suppose that y ∈ X∗ and |Ax, y| ≤ cy x for all x ∈ D(A). Then the functional fy (x) = Ax, y, x ∈ D(A) can be extended linearly and continuously to the whole X which is the closure of D(A). Such an extension (still denoted by itself) fy is in X∗ and unique. Hence we obtain Ax, y = fy (x) = x, fy  for all x ∈ D(A). Define D(A∗ ) = {y ∈ X∗ : ∃cy ≥ 0 such that |Ax, y| ≤ cy x ∀x ∈ D(A)}, A∗ y = fy , y ∈ D(A∗ ). Clearly, A∗ : D(A∗ ) ⊂ X∗ → X∗ is a linear operator satisfying Ax, y = x, A∗ y for all x ∈ D(A) and all y ∈ D(A∗ ). The mapping A∗ is called the adjoint operator of A.

4.3 C0 Semigroup

113

4.3 C0 Semigroup Let (X,  · ) be a Banach space and let {T (t) : t ∈ [0, ∞)} ⊂ L(X). We call T (·) a C0 semigroup (or a strongly continuous semigroup of operators) on X if the following properties hold: T (0) = I,

(4.2)

T (t + s) = T (s)T (t) for all s, t ≥ 0,

(4.3)

lim T (s)x − x = 0, x ∈ X.

(4.4)

s→0

In the case when T (t) is defined for all t ∈ R 1 and (4.3) holds for all t, s ∈ R 1 , we call T (·) as C0 group or a strongly continuous group of operators. Equations (4.3) and (4.4) are usually referred to as the semigroup property and the strong continuity, respectively. The following result is true (see Proposition 4.7, Chapter 2 of [80]). Proposition 4.3 Let T (·) be a C0 semigroup on X. Then there exist constants M ≥ 1 and ω ∈ R 1 such that T (t) ≤ Meωt , t ∈ [0, ∞). Let T (·) be a C0 semigroup on X and let D(A) = {x ∈ X : lim t −1 (T (t) − I )x exists}, t→0

Ax = lim t −1 (T (t) − I )x, x ∈ D(A). t→0

The operator A : D(A) → X is called the generator of the semigroup T (·). We also say that A generates the C0 semigroup T (·). In general, the operator A is not bounded. The following two results hold. Proposition 4.4 (Proposition 4.10, Chapter 2 of [80]) Let T (·) be a C0 semigroup on X. Then the generator A of T (·) is densely defined closed operator and n ∩∞ n=1 D(A ) is dense in X. Furthermore, S(·) is another C0 semigroup on X with the same generator A as T (·), then S(·) = T (·). Theorem 4.5 ([59, 120] and Theorem 4.11, Chapter 2 of [80]) Let A : D(A) ⊂ X → X be a linear operator. The following properties are equivalent. (i) A generates a C0 semigroup T (·) on X such that T (t) ≤ Meωt for all t ≥ 0 with some M ≥ 1 and ω ∈ R 1 .

114

4 Infinite Dimensional Control

(ii) A is a densely defined and closed operator such that for the above M ≥ 1 and ω ∈ R 1 , {λ ∈ C 1 : Re λ > ω} ⊂ ρ(A) and (λI − A)−n  ≤ M(Re λ − ω)−n for all integers n ≥ 0 and all λ ∈ C 1 with Re λ > ω. (iii) A is a densely defined and closed operator such that for the above M ≥ 1 and ω ∈ R 1 , there exists a sequence of positive numbers λk → ∞ as k → ∞ such that for all integers k ≥ 1, λk ∈ ρ(A) and (λk I − A)−n  ≤ M(λk − ω)−n for all integers n ≥ 0 and all integers k ≥ 1. Because the generator A determines the C0 semigroup T (·) uniquely and in the case A ∈ L(X) the C0 semigroup has an explicit expression eAt = ∞ −1 n n At the C 0 n=0 (n!) t A (Proposition 4.9, Chapter 2 of [80]), we denote by e semigroup generated by A.

4.4 Evolution Equations Let (X,  · ) be a Banach space, A : D(A) ⊂ X → X generate a C0 semigroup eAt on X, T > 0, f : [0, T ] → X be a Bochner integrable function and y0 ∈ X. We consider the following evolution equation y (t) = Ay(t) + f (t), t ∈ [0, T ],

(P0 )

y(0) = y0 . A continuous function y : [0, T ] → X is called a (mild) solution of (P0 ) if 

t

y(t) = eAt y0 +

eA(t−s) f (s)ds, t ∈ [0, T ].

0

A continuous function y : [0, T ] → X is called a weak solution of (P0 ) if for any x ∗ ∈ D(A∗ ), y(·), x ∗  is an absolutely continuous function on [0, T ] and that for all t ∈ [0, T ], ∗





y(t), x  = y0 , x  +

t

[y(s), A∗ x ∗  + f (s), x ∗ ]ds.

0

We have the following result. Proposition 4.6 ([15], Proposition 5.2, Chapter 2 of [80]) A continuous function y : [0, T ] → X is a solution of (P0 ) if and only if it is a weak solution of (P0 ).

4.5 C0 Groups

115

The following useful result is also valid (see Proposition 4.14, Chapter 2 of [80]). Proposition 4.7 For any x ∈ D(A), eAt x ∈ D(A) for all t ≥ 0 and (d/dt)(eAt x) = AeAt x = eAt Ax for all t ≥ 0, 

t

e x−e x = At

As

eAr Axdr for all t > s ≥ 0.

s

4.5 C0 Groups Let (H, ·, ·) be a Hilbert space equipped with an inner product ·, · which induces the norm  · . A one-parametric family S(t), t ∈ R 1 of continuous linear operators from H onto H is a strongly continuous group of continuous linear operators on H or C0 group on H if S(0)x = x for all x ∈ H, S(t1 + t2 ) = S(t1 )S(t2 ) for all t1 , t2 ∈ R 1 , lim S(t)x = x, x ∈ H.

t→0

Let S(t), t ∈ R 1 be a C0 group on H . Then the generator of S is the linear operator A : D(A) ⊂ H → H defined by D(A) = {x ∈ H : lim t −1 (S(t) − I )x exists}, t→0

Ax = lim t −1 (S(t) − I )x, x ∈ D(A). t→0

The following result is true. Theorem 4.8 ([35]) Let S(·) be a C0 group on H . Then the following assertions hold. (i) D(A) is dense in H and A is a closed linear operator. (ii) for every x 0 ∈ D(A) there exists a unique function x ∈ C 1 (R 1 ; H ) ∩ C 0 (R 1 ; D(A)) such that x(0) = x 0 , x(t) ∈ D(A) for all t ∈ R 1 and (dx/dt)(t) = Ax(t) for all t ∈ R 1 ; moreover, this solution satisfies x(t) = S(t)x 0 for all t ∈ R 1 . (iii) S(t)∗ , t ∈ R 1 is a C0 group on H and its generator is the adjoint operator A∗ of A.

116

4 Infinite Dimensional Control

It is clear that if S(·) is a C0 group with the generator A : D(A) → H , then S(−t), t ∈ R 1 is also a C0 group with the generator −A : D(A) → H . Let A : D(A) ⊂ H → H . It is known that A and −A are both generators of C0 semigroups if and only if A is a generator of a C0 group [116]. Note that a C0 group is determined by its generator A uniquely and is denoted by eAt , t ∈ R 1 . Clearly, for all t ∈ R 1 , e−At = eA(−t) .

4.6 Admissible Control Operators Let J be an open interval and U be a Hilbert space. Denote by H 1 (J ; U ) the Sobolev space of all locally absolutely continuous functions z : J → U for which dz/dt ∈ L2 (J ; U ). The space H 2 (J ; U ) is the set of all locally absolutely continuous functions z : J → U for which dz/dt ∈ H 1 (J ; U ). The space H01 (J ; U ) is the set of all functions in H 1 (J ; U ) which have limits equal to zero at the endpoints of J . Define H02 (J ; U ) = {h ∈ H 2 (J ; U ) ∩ H01 (J ; U ) : dh/dx ∈ H01 (J ; U )}. In this section we collect several useful results of [116]. Proposition 4.9 (Proposition 2.1.2 of [116]) Let eAt , t ≥ 0 be a C0 semigroup on a Hilbert space E. Then the mapping (t, z) → eAt z, t ∈ [0, ∞), z ∈ E is continuous on [0, ∞) × E in the product topology. Let X, Z be Hilbert spaces and D(A) be a linear subspace of X. A linear operator A : D(A) → Z is called closed if its graph G(A) = {(f, Af ) : f ∈ D(A)} is closed in X × Z. If A is closed, then D(A) is a Hilbert space with the graph norm  · gr : z2gr = z2X + Az2Z , z ∈ D(A). Proposition 4.10 (Proposition 2.10.1 of [116]) Let X be a Hilbert space and A : D(A) → X be a densely defined linear operator such that ρ(A) = ∅. Then for every β ∈ ρ(A) ∩ R 1 , the space D(A) with the norm z1 = (βI − A)(z), z ∈ D(A) is a Hilbert space denoted by X1 . The norms generated as above for different β ∈ ρ(A) ∩ R 1 are equivalent to the graph norm and the embedding X1 ⊂ X is continuous. If L ∈ L(X) satisfies LD(A) ⊂ D(A), then L ∈ L(X1 ). Note that the relation ρ(A) = ∅ implies that the operator A is closed. Let A be as in Proposition 4.10. It is clear that A∗ has the same property. We define a Hilbert space X1d = D(A∗ ) equipped with the norm zd1 = (βI − A∗ )z, z ∈ D(A∗ ), where β ∈ ρ(A) ∩ R 1 , or equivalently, β ∈ ρ(A∗ ) ∩ R 1 .

4.6 Admissible Control Operators

117

Proposition 4.11 (Proposition 2.10.2 of [116]) Let A be as in Proposition 4.10 and let β ∈ ρ(A) ∩ R 1 .We denote by X−1 the completion of X with respect to the norm z−1 = (βI − A)−1 z, z ∈ X.

(4.5)

The norms generated as above for different β ∈ ρ(A) ∩ R 1 are equivalent (in particular, X−1 is independent of the choice of β). Moreover, X−1 is the dual of X1d with respect to the pivot space X. If L ∈ L(X) satisfies L∗ D(A∗ ) ⊂ D∗ (A), then L has a unique extension to an operator L˜ ∈ L(X−1 ). Proposition 4.12 (Proposition 2.10.3 of [116]) Let A : D(A) → X be a densely defined linear operator such that ρ(A) = ∅, β ∈ ρ(A) ∩ R 1 , X1 be as in Proposition 4.10 and X−1 be as in Proposition 4.11. Then A ∈ L(X1 , X) and A has a unique extension A˜ ∈ L(X, X−1 ). Moreover, the operators (βI − A)−1 ∈ ˜ −1 ∈ L(X−1 , X) are unitary. L(X, X1 ) and (βI − A) Proposition 4.13 (Proposition 2.10.4 of [116]) We use the notation from Proposition 4.12 and assume that A generates a C0 semigroup S(·) on X. Then for every ˜ ˜ t ≥ 0, S(t) has a unique extension S(t) ∈ L(X−1 ), S(t), t ≥ 0 is a C0 semigroup ˜ on X−1 and A is its generator. Let U, X be Hilbert spaces, S(·) be a C0 semigroup on X, and A : D(A) → X be its generator. We use the notation above and denote by A and S(·) also the extension of the generator to X and the extension of the semigroup to X−1 , respectively. Consider the differential equation z (t) = Az(t) + f (t)

(4.6)

where f ∈ L1loc ([0, ∞); X−1 ). A solution of (4.6) in X−1 is a function z ∈ L1loc ([0, ∞); X) ∩ C([0, ∞); X−1 ) which satisfies the following equations in X−1 : 

t

z(t) − z(0) =

[Az(s) + f (s)]ds for all t ∈ [0, ∞).

(4.7)

0

It is also called a “strong” solution of (4.6) in X−1 . Equation (4.7) implies that z is an a. c. function with values in X−1 and (4.6) holds for almost every t ≥ 0, with the derivative computed with respect to the norm of X−1 . We can also define the concept of a “weak” solution of (4.6) in X−1 by regarding instead of (4.7) that for every φ ∈ X1d and every t ≥ 0,  z(t) − z(0), φX−1 ,Xd = 1

These two concepts are equivalent.

0

t

[z(s), A∗ φX + f (s), φX−1 ,Xd ]ds. 1

118

4 Infinite Dimensional Control

Proposition 4.14 (Proposition 4.1.4 of [116]) Suppose that z is a solution of (4.6) in X−1 and let z0 = z(0). Then z is given by 

t

z(t) = S(t)z0 +

S(t − σ )f (σ )dσ, t ≥ 0.

(4.8)

0

In particular, for every z0 ∈ X there exists at most one solution in X−1 of (4.6) which satisfies initial condition z(0) = z0 . Note that z satisfying (4.8) is called the mild solution of (4.6) with the initial state z0 ∈ X. It is not difficult to see that the following result is valid. Proposition 4.15 Let f ∈ L1loc ([0, ∞); X−1 ), T > 0 and z ∈ C 0 ([0, T ]; X−1 ). The z satisfies (4.8) for all t ∈ [0, T ] in X−1 if and only if for all ξ ∈ X1d = D(A∗ ) and all t ∈ [0, T ],  z(t), ξ X−1 ,Xd = S(t)z0 , ξ X−1 ,Xd + 1

1

 = S(t)z0 , ξ X−1 ,Xd + 1

t

0

0

t

[S(t − σ )f (σ ), ξ X−1 ,Xd ]dσ 1

[f (σ ), S ∗ (t − σ )ξ X−1 ,Xd ]dσ. 1

Proposition 4.16 (Proposition 12.1.2 of [116]) Assume that Z1 , Z2 , and Z3 are Hilbert spaces, F ∈ L(Z1 , Z3 ), G ∈ L(Z2 , Z3 ) and Ran(F ) = {F (z) : z ∈ Z1 } ⊂ Ran(G) = {G(z) : z ∈ Z2 }. Then there exists an operator L ∈ L(Z1 , Z2 ) such that F = GL. Corollary 4.17 Assume that Z1 , Z2 are Hilbert spaces and G ∈ L(Z1 , Z2 ) satisfies Ran(G) = Z2 . Then there exists an operator L ∈ L(Z2 , Z1 ) such that GL = I - the identity operator in Z2 . Let B ∈ L(U, X−1 ) and τ ≥ 0. Define Φτ ∈ L(L2 (0, ∞; U ), X−1 ) by  Φτ u =

τ

S(τ − σ )Bu(σ )dσ.

(4.9)

0

The operator B ∈ L(U, X−1 ) is called an admissible control operator for S(·) if for some τ > 0, Ran(Φτ ) = {Φτ u : u ∈ L2 (0, ∞; U )} ⊂ X.

4.6 Admissible Control Operators

119

Proposition 4.18 (Proposition 4.2.2 and (4.2.5) of [116]) Assume that B ∈ L(U, X−1 ) is an admissible control operator for S(·). Then for every t ≥ 0, Φt ∈ L(L2 (0, ∞; U ), X) and for all T > t > 0, Φt  ≤ ΦT . Proposition 4.19 (Proposition 4.2.5 of [116]) Assume that B ∈ L(U, X−1 ) is an admissible control operator for S(·). Then for every z0 ∈ X and every u ∈ L2loc (0, ∞; U ), the initial value problem z (t) = Az(t) + Bu(t), z(0) = z0 has a unique solution in X−1 . This solution is given by 

t

z(t) = S(t)z0 + Φt u = S(t)z0 +

S(t − σ )Bu(σ )dσ

0

and it satisfies 1 z ∈ C([0, ∞); X) ∩ Hloc ((0, ∞); X−1 ).

Proposition 4.20 (Proposition 4.2.6 of [116]) Assume that B ∈ L(U, X−1 ) is an admissible control operator for S(·). Then for every z0 ∈ X and every u ∈ L2loc (0, ∞; U ), there exists a unique function z ∈ C([0, ∞); X) such that for every t ≥ 0 and every ψ ∈ D(A∗ ), 

t

z(t) − z0 , ψX =

[z(σ ), A∗ ψX + u(σ ), B ∗ ψU ]dσ.

0

Propositions 4.18 and 4.19 and Theorem 4.5 imply the following result. Proposition 4.21 Assume that B ∈ L(U, X−1 ) is an admissible control operator for S(·) and T > 0. Then there exists a constant cT > 0 such that for every z0 ∈ X and every u ∈ L2 (0, T ; U ) the unique solution z of the initial value problem z (t) = Az(t) + Bu(t), t ∈ (0, T ) a. e., z(0) = z0 satisfies z ∈ C([0, T ]; X) and that for all t ∈ [0, T ], z(t) ≤ cT (z0  + u). Proposition 4.18 and Corollary 4.17 imply the following result.

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4 Infinite Dimensional Control

Proposition 4.22 Assume that B ∈ L(U, X−1 ) is an admissible control operator for S(·), T > 0 and Ran(ΦT ) = X. Then there exists L ∈ L(X, L2 (0, T ; U )) such that ΦT Lx = x for all x ∈ X. Theorem 4.23 Assume that B ∈ L(U, X−1 ) is an admissible control operator for a C0 semigroup S(·), T > 0, Ran(ΦT ) = X, x¯ ∈ X, u¯ ∈ U and that Ax¯ = 0, AB u¯ = 0.

(4.10)

Then there exists a constant c > 0 such that for each  ∈ (0, 1) and each pair of points z0 , z1 ∈ X there exist u ∈ L2 (0, T ; U ) and z ∈ C 0 ([0, T ]; X) which is a solution of the problem z (t) = Az(t) + Bu(t), t ∈ [0, T ] a. e., z(0) = z0

(4.11)

in X−1 and satisfy z(T ) = z1 , ¯ + z0 − x), ¯ t ∈ [0, T ] z(t) − x ¯ ≤  + c(z1 − x and for almost every (a. e.) t ∈ [0, T ], ¯ + z0 − x)) ¯ ∪ B(−u, ¯ c(z1 − x ¯ + z0 − x)). ¯ u(t) ∈ B(u, ¯ c(z1 − x Proof Let L ∈ L(X, L2 (0, T ; U )) be as guaranteed by Proposition 4.22. Therefore ΦT Lx = x for all x ∈ X.

(4.12)

Let cT ≥ 1 be as guaranteed by Proposition 4.21. Set c = cT + cT L + cT2 L.

(4.13)

Let z0 , z1 ∈ X and  ∈ (0, 1). By Proposition 4.3, S(t) ≤ M∗ ew∗ t , t ∈ [0, ∞),

(4.14)

where M∗ > 0 and w∗ ∈ R 1 . Choose a natural number n such that 2n−1 T < (M∗ + 1)−1 e−|w∗ | (B u ¯ + 1)−1 . Set τ = (2n)−1 T .

(4.15)

4.6 Admissible Control Operators

121

Define u(t) ˜ = u, ¯ t ∈ [0, τ ], u(t) ˜ = −u, ¯ t ∈ (τ, 2τ ].

(4.16)

By Proposition 4.21, there exists a unique solution x˜ ∈ C([0, 2τ ]; X) of the initial value problem ˜ t ∈ (0, 2τ ), x˜ = Ax˜ + B u, x(0) ˜ = x. ¯ Proposition 4.7, (4.10), and (4.1) imply that for every t ∈ [0, 2τ ], 

t

x(t) ˜ = S(t)x¯ +



0





τ

S(2τ − σ )B u(σ ˜ )dσ =

0

 S(2τ − σ )B udσ ¯ −

0

=

τ

S(t − σ )B u(σ ˜ )dσ,

(4.17)

0

 

t

S(t − σ )B u(σ ˜ )dσ = x¯ +



S(2τ − σ )B udσ ¯

τ

 S(τ )S(τ − σ )B udσ ¯ −

0

τ

S(τ − σ )B udσ ¯ =0

0

and x(2τ ˜ ) = x. ¯

(4.18)

By (4.15) for every t ∈ [0, 2τ ], ¯ = T n−1 M∗ e|w∗ | B u ¯ < . x(t) ˜ − x ¯ ≤ 2M∗ e|w∗ | τ B u

(4.19)

For every integer i ∈ {0, . . . , n − 1} define ˜ u∗ (iτ + t) = u(t), ˜ t ∈ [0, 2τ ]. x∗ (iτ + t) = x(t),

(4.20)

Clearly, x∗ satisfies the equation x∗ (t) = Ax∗ (t) + Bu∗ (t), t ∈ (0, 2τ n) = (0, T ) a. e..

(4.21)

By (4.18), (4.20) and (4.27), ¯ ≤ M∗ e|w∗ | B uT ¯ n−1 < , x∗ (t) − x x∗ (0) = x∗ (T ) = x, ¯ u∗ (t) ∈ {u, ¯ −u}, ¯ t ∈ [0, T ] a. e..

(4.22) (4.23)

122

4 Infinite Dimensional Control

Let z0 , z1 ∈ X. By Proposition 4.19, there exists z˜ ∈ C([0, T ]; X) which is a solution of the initial value problem z˜ = A˜z + Bu∗ , t ∈ (0, T ),

(4.24)

z˜ (0) = z0

(4.25)

in X−1 . Proposition 4.19 implies that there exists a function  z ∈ C([0, T ]; X) such that z + B(L(z1 − z˜ (T )), t ∈ (0, T ),  z = A

(4.26)

 z(0) = 0

(4.27)

in X−1 . In view of (4.12), (4.27) and Proposition 4.19,  z(T ) = S(T ) z(0)+ΦT (L(z1 −˜z(T ))) = ΦT (L(z1 −˜z(T ))) = z1 −˜z(T ).

(4.28)

Set z = z˜ + z.

(4.29)

z(0) = z˜ (0) + z(0) = z0 ,

(4.30)

z(T ) = z˜ (T ) + z(T ) = z1 .

(4.31)

By (4.24), (4.25), (4.27), and (4.28),

It follows from (4.24), (4.26), and (4.29) that z = A(˜z + z) + B(L(z1 − z˜ (T )) + u∗ ) z = z˜ + = Az + B(L(z1 − z˜ (T )) + u∗ ), t ∈ (0, T )

(4.32)

in X−1 . For all t ∈ [0, T ] set u(t) = u∗ (t) + L(z1 − z˜ (T )).

(4.33)

In view of (4.21) and (4.23)–(4.25), (˜z − x∗ ) = A(˜z − x∗ ), t ∈ (0, T )

(4.34)

¯ (˜z − x∗ )(0) = z0 − x.

(4.35)

in X−1 and

4.6 Admissible Control Operators

123

It follows from (4.34), (4.35), the choice of cT and Proposition 4.21 that for each t ∈ [0, T ], ¯ = cT z0 − x. ¯ ˜z(t) − x∗ (t) ≤ cT ˜z(0) − x

(4.36)

In view of (4.21) and (4.32), (z − x∗ ) = A(z − x∗ ) + B(L(z1 − z˜ (T ))), t ∈ (0, T )

(4.37)

in X−1 and in view of (4.23) and (4.30), ¯ (z − x∗ )(0) = z0 − x.

(4.38)

By (4.23) and (4.36), ¯ + x¯ − z˜ (T )) L(z1 − z˜ (T )) ≤ Lz1 − z˜ (T ) ≤ L(z1 − x ¯ + cT z0 − x). ¯ ≤ L(z1 − x

(4.39)

Proposition 4.21, the choice of cT , (4.13) and (4.37)–(4.39) imply that for all t ∈ [0, T ], ¯ + L(z1 − z˜ (T ))) z(t) − x∗ (t) ≤ cT (z0 − x ≤ cT (z0 − x ¯ + L(z1 − x ¯ + cT z0 − x)) ¯ ≤ z0 − x(c ¯ T + cT2 L) + cT Lz1 − x ¯ ≤ c(z0 − x ¯ + z1 − x). ¯

(4.40)

It follows from (4.13), (4.33), and (4.39) for a. e. t ∈ [0, T ], ¯ + z0 − x). ¯ u(t) − u∗ (t) ≤ c(z1 − x

(4.41)

In view of (4.41), for a. e. t ∈ [0, T ], ¯ + z0 − x)) ¯ ∪ B(−u, ¯ c(z1 − x ¯ + z0 − x)). ¯ u(t) ∈ B(u, ¯ c(z1 − x By (4.22) and (4.40), for all t ∈ [0, T ], ¯ + z0 − x). ¯ z(t) − x ¯ ≤  + c(z1 − x Theorem 4.23 is proved.

124

4 Infinite Dimensional Control

4.7 Examples Let U, X be Hilbert spaces, S(·) be a C0 semigroup on X, A : D(A) → X be its generator and B ∈ L(U, X−1 ) be an admissible control operator for S(·). We say that the pair (A, B) is exactly controllable in a time τ > 0 if Ran(Φτ ) = X [35, 116]. It is known that the exact controllability in time τ > 0 is equivalent to the following property: for each pair z0 , z1 ∈ X there exists u ∈ L2 (0, τ ; U ) such that the solution z of the initial value problem z = Az + Bu, z(0) = z0

(4.42)

satisfies z(τ ) = z1 . In this section we consider examples of the pairs (A, B) which are exactly controllable in some time τ > 0. All of these examples were discussed in [116] and [35]. It should be mentioned that in Chapter 6 our turnpike results will be established for large classes of infinite dimensional optimal control problems. One of this classes contains problems which are defined by an integrand and a pair of operators (A, B) introduced above, which is exactly controllable in some time τ > 0. Therefore, each example of pairs (A, B) considered below gives us a subclass of infinite dimensional optimal control problems, for which the results of Chapter 6 hold. Example 4.24 (Example 11.2.2 of [116]) We consider the problem of controlling the vibrations of an elastic membrane by a force field acting on a part of this membrane. More precisely, let n be a natural number and let Ω ⊂ R n be a bounded open set with ∂Ω of class C 2 or Ω be a rectangular domain. The physical problem described above can be modeled by the equations ∂ 2w − Δw = u in Ω × (0, ∞), ∂t 2 w = 0 on ∂Ω × (0, ∞), w(x, 0) = f (x),

∂w (x, 0) = g(x) for x ∈ Ω, ∂t

where f is the initial displacement and g is the initial velocity. Let O be a nonempty open subset of Ω and u ∈ L2 (0, ∞; L2 (O)) be the input function. For any such u we assume that u(x, t) = 0 for all x ∈ Ω \ O. The equation above can be written in the form (4.42) using the following spaces and operators: X = H01 (Ω) × L2 (Ω), D(A) = (H 2 (Ω) ∩ H01 (Ω)) × H01 (Ω), U = L2 (O) ⊂ L2 (Ω), A

        f g f 0 = for all ∈ D(A), Bu = for all u ∈ U. g Δf g u

4.7 Examples

125

Here B ∈ L(U, X). It was shown in Example 11.2.2 of [116] that B is an admissible control operator for the C0 semigroup eAt and that the pair (A, B) is exactly controllable in time τ > 0 if Γ and O satisfy the assumptions of Theorem 7.4.1 of [116] where Γ is a relatively open subset of ∂Ω. In particular, (A, B) is exactly controllable in time τ > 0 if there exist x0 ∈ R n and  > 0 such that N ({x ∈ ∂Ω : (x − x0 ) · ν(x) > 0}) ⊂ closO, τ > 2r(x0 ), where r(x0 ) = sup{x − x0  : x ∈ Ω}, | · | is the Euclidean norm in R n , ν is the unit outward normal vector field in ∂Ω and N (D) = {x ∈ Ω : d(x, D) < } for any D ⊂ Ω with d(x, D) = inf{|x − y| : y ∈ D}. Example 4.25 (Example 11.2.4 of [116]) Let n be a natural number, let Ω ⊂ R n be a bounded open set with ∂Ω of class C 2 or let Ω be a rectangular domain and let O be a nonempty open subset of Ω. We consider the problem of controlling the vibrations of an elastic plate occupying the domain Ω by a force field acting on O. More precisely, we consider the following initial and boundary initial value problem ∂ 2w + Δ2 w = u in Ω × (0, ∞), ∂t 2 w = Δw = 0 on ∂Ω × (0, ∞), w(x, 0) = 0,

∂w (x, 0) = 0 for x ∈ Ω, ∂t

where u ∈ L2 (0, ∞; L2 (O)) is the input function. As usual we assume that u(x, t) = 0 for all x ∈ Ω \ O. The equations above determines a system with the state space X = (H 2 (Ω) ∩ H01 (Ω)) × L2 (Ω) and the input space U = L2 (Ω) which is exactly controllable in any time τ > 0, as it was shown in Example 11.2.4 of [116], if the pair (Ω, O) satisfies one of the assumptions (A1) or (A2) in Example 11.2.3 of [116]. More precisely, let H = L2 (Ω), D(A0 ) = H1 be the Sobolev space H 2 (Ω) ∩ H01 (Ω), A0 : D(A0 ) → H be defined by A0 φ = −Δφ, φ ∈ D(A0 ) and let H2 = D(A20 ) be endowed with the graph norm. Let χ be the Hilbert space H1 × H , consider the dense subset of χ defined by D(A) = H2 × H1 and let the linear operator A : D(A) → χ be defined by  0 I . A= −A20 0 

Then the equations above can be written in the form z = Az + Bu, z(0) = 0

126

4 Infinite Dimensional Control

  0 where B ∈ L(U, χ ) is defined by Bu = for all u ∈ U . It was shown in u Example 11.2.4 of [116] that the pair (A, B) is exactly controllable in any time τ > 0 if the pair (Ω, O) satisfies one of the assumptions (A1) or (A2) in Example 11.2.3 of [116]. Example 4.26 (Example 11.2.6 of [116]) We consider the problem of controlling the vibrations of a string occupying the interval [0, π ] by means of a force u(t) acting at its left end. The equations describing this problem are formulated as a well-posed boundary control system in subsection 10.2.2 of [116]: ∂ 2w ∂ 2w (x, t) = (x, t), 0 < x < π, t ≥ 0, ∂t 2 ∂x 2 ∂w (0, t) = u(t), t ≥ 0, ∂x

w(π, t) = 0, w(x, 0) = f (x),

∂w (x, 0) = g(x), 0 < x < π. ∂t

For this problem X = HR1 (0, π ) × L2 (0, π ), A : D(A) → X is defined by D(A) = 1 {f ∈ H 2 (0, π ) ∩ HR1 (0, π ) : ∂f ∂x (0) = 0} × HR (0, π ),

    g f f A = d 2 f for all ∈ D(A). g g 2 dx

    f f = −g(0) for all ∈ D(A). It was g g shown in Example 10.2.6 of [116] that the pair (A, B) is exactly controllable in any time τ ≥ 2π.

The control operator B satisfies B ∗

Example 4.27 (Example 11.2.7 of [116]) We consider the boundary control of the non-homogeneous elastic string. The model is described by the equation ∂w ∂ ∂ 2w (a(x) (x, t)) − b(x)w(x, t), 0 < x < π, t > 0, (x, t) = 2 ∂x ∂x ∂t w(0, t) = u(t), w(π, t) = 0, w(·, 0) = f,

∂w (·, 0) = g. ∂t

Here a ∈ C 2 [0, π ], b ∈ L∞ (0, π ), a(x) ≥ m > 0, b(x) ≥ 0 for all x ∈ [0, π ]. These equations correspond to a well-posed boundary control system with state

4.7 Examples

127

space X = L2 (0, π ) × H −1 (0, π ). The generator A is defined by       f f g for all A ∈ D(A) = H01 (0, π ) × L2 (0, π ), = g g −A0 f where A0 ∈ L(H01 (0, π ), H −1 (0, π )) is defined by A0 f =

d df (a ) + bf for all f ∈ H01 (0, π ). dx dx

The control operator B of this system is determined by     d φ φ −1 B = a(0) (A0 ψ)|x=0 for all ∈ D(A∗ ) = D(A). ψ ψ dx ∗

It was shown in Example 11.2.7 of [116] that the pair (A, B) is exactly controllable π in any time τ ≥ 2 0 (a(x))−1/2 dx. Example 4.28 (Example 11.2.8 of [116]) We consider the problem of controlling the vibrations of a beam occupying the interval [0, π ] by means of a torque u(t) acting at its left end. The model is described by the initial and boundary value problem ∂ 2w ∂ 4w (x, t) = − (x, t), 0 < x < π, t > 0, ∂t 2 ∂x 4 w(0, t) = 0, w(π, t) = 0,

∂ 2w ∂ 2w (0, t) = u(t), (π, t) = 0, ∂x 2 ∂x 2

w(·, 0) = f,

∂w (·, 0) = g. ∂t

These equations correspond to a control system where H = L2 (0, π ), H1 = 2 H 2 (0, π ) ∩ H01 (0, π ), A0 : H1 → H is defined by A0 f = − ddxf2 for all f ∈ H1 , H1/2 = H01 (0, π ), H−1/2 = H0−1 (0, π ). The unique extensions of A0 to unitary operators from H1/2 onto H−1/2 and from H onto H−1 are still denoted by A0 . The space H3/2 = A−1 0 H1/2 is H3/2 = {g ∈ H 3 (0, π ) ∩ H01 (0, π ) :

d 2g d 2g (0) = (π ) = 0}. dx 2 d2

128

4 Infinite Dimensional Control

We set X = H1/2 × H−1/2 , D(A) = H3/2 × H1/2 ,       f f g A for all = ∈ D(A). g g −A20 f The control operator B of this system is determined by     d f f −1 = − (A0 g)|x=0 for all B ∈ D(A∗ ) = D(A). g g dx ∗

It was shown in Example 11.2.8 of [116] that the pair (A, B) is exactly controllable in any time τ > 0. Example 4.29 (Example 11.2.9 of [116]) We consider the problem of controlling the vibrations of a beam occupying the interval [0, 1] by means of an angular velocity u(t) applied at its left end. The equations describing this problem have been formulated as a well-posed boundary control system in Section 10.5 of [116] as follows: ∂ 2w ∂ 4w (x, t) = − (x, t), 0 < x < 1, t > 0, ∂t 2 ∂x 4 w(0, t) = 0, w(1, t) = 0,

∂w ∂w (0, t) = u(t), (1, t) = 0, ∂x ∂x

w(·, 0) = f,

∂f (·, 0) = g. ∂t

We denote X = V ×L2 (0, 1), where V = {h ∈ H 2 (0, 1) : h(0) = h(1) = 0}. The norm on X is defined by  z2 = z1 2V + z2 2L2 , where z1 2V =

1

0

|

dh dx (1)

=

d 2 z1 2 | dx. dx 2

Let Z ⊂ X be defined by Z = (V ∩ H 4 (0, 1)) × V , L : Z → X, G : Z → R 1 be defined by L

0 4

d − dx 4

  I dz2 z (0), , G 1 = z2 dx 0

KerG = {z ∈ Z : G(z) = 0} = (V ∩ H 4 (0, 1)) × H02 (0, 1), A = L|KerG

4.7 Examples

129

and a control operator B be defined by     d 2 ψ1 ψ1 ψ1 =− ∈ D(A∗ ) = D(A). (0), for all B ψ2 ψ2 dx 2 ∗

It was shown in Example 11.2.9 of [116] that the pair (A, B) is exactly controllable in any time τ > 0. The next two examples are discussed in [35]. Example 4.30 The transport equation. Let L > 0. We consider the linear control system yt + yx = 0, t ∈ (0, T ), x ∈ (0, L), y(t, 0) = u(t), t ∈ (0, T ), where u(t) ∈ R 1 , y(t, ·) : (0, L) → R 1 , with X = L2 (0, L), D(A) = {f ∈ H 1 (0, L) : f (0) = 0}, Af = −fx , f ∈ D(A), U = R 1 and B : R 1 → D(A∗ )

defined by (Bu)z = uz(0) for all u ∈ R 1 and all z ∈ D(A∗ ). It was shown in [35] that the pair (A, B) is controllable. Example 4.31 The Korteweg-de Vries equation. Let L > 0. We consider the linear control system yt + yx + yxxx = 0, t ∈ (0, T ), x ∈ (0, L), y(t, 0) = y(t, L) = 0, yx (t, L) = u(t), t ∈ (0, T ), where u(t) ∈ R 1 , y(t, ·) : (0, L) → R 1 , with X = L2 (0, L), D(A) = {f ∈ H 3 (0, L) : f (0) = f (L) = fx (L) = 0}, Af = −fx − fxxx , f ∈ D(A), U = R 1 and B : R 1 → D(A∗ ) is defined by (Bu)z = uzx (L) for all u ∈ R 1 and all z ∈ D(A∗ ). It was shown in [35] that the pair (A, B) is controllable.

Chapter 5

Symmetric Variational Problems

In this chapter we study the turnpike properties for symmetric variational problems in Banach spaces. To have the turnpike property means, roughly speaking, that the approximate solutions of the problems are determined mainly by the integrand and are essentially independent of the choice of interval and endpoint conditions, except in regions close to the endpoints. It is shown that for symmetric variational problems the turnpike is a minimizer of the integrand. The results of this chapter are new.

5.1 Preliminaries Assume that (X,  · ) is a Banach space. We denote by mes(Ω) the Lebesgue measure of a Lebesgue measurable set Ω ⊂ R 1 . For each x ∈ X and each r > 0 set B(x, r) = {y ∈ X : y − x ≤ r}. Suppose that the infimum over an empty set is ∞. Assume that f : X × X → R 1 is a bounded from below borelian function such that f (x, y) = f (x, −y) for all x, y ∈ X,

(5.1)

there exists (x, ¯ y) ¯ ∈ X × X such that f (x, ¯ y) ¯ = inf(f ) = inf{f (ξ, η) : ξ, η ∈ X}, {(x, y) ∈ X × X : f (x, y) = inf(f )} = {(x, ¯ y), ¯ (x, ¯ −y)} ¯

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 A. J. Zaslavski, Turnpike Phenomenon and Symmetric Optimization Problems, Springer Optimization and Its Applications 190, https://doi.org/10.1007/978-3-030-96973-8_5

(5.2)

131

132

5 Symmetric Variational Problems

and that the following assumptions hold: (A1)

for each  > 0 there exists δ > 0 such that for each (x, y) ∈ X×X satisfying f (x, y) ≤ inf(f ) + δ

the inequalities x − x ¯ ≤ and min{y − y, ¯ y + y} ¯ ≤ hold; (A2) for each  > 0 there exists δ > 0 such that for each (x, y) ∈ X×X satisfying x − x ¯ ≤ δ, y − y ¯ ≤δ the inequality f (x, y) ≤ f (x, ¯ y) ¯ + is true. Assumption (A2) means that the function f is continuous at the point (x, ¯ y) ¯ while assumption (A1) means that the minimization problem f (x, y) → min, x, y ∈ X is well-posed. According to the results of Chapter 2 this well-posedness property holds for most symmetric objective functions. Let a > 0 and ψ : [0, ∞) → [0, ∞) be an increasing function satisfying lim ψ(t) = ∞.

t→∞

(5.3)

Assume that the following assumption holds: (A3)

the function f is bounded on all bounded sets and for each (x, u) ∈ X × X, f (x, u) ≥ ψ(u)u − a.

5.1 Preliminaries

133

For each pair of nonnegative numbers T1 < T2 and each y, z ∈ X we consider the problems 

T2

f (x(t), x (t))dt → min,

T1

(PT1 ,T2 )

x ∈ W 1,1 (T1 , T2 ), 

T2

f (x(t), x (t))dt → min,

T1

(PT1 ,T2 ,y )

x ∈ W 1,1 (T1 , T2 ), x(T1 ) = y, 

T2

f (x(t), x (t))dt → min,

T1

(PT1 ,T2 ,y,z )

x ∈ W 1,1 (T1 , T2 ), x(T1 ) = y, x(T2 ) = z and define  U (T1 , T2 ) = inf{

T2

f (x(t), x (t))dt : x ∈ W 1,1 (T1 , T2 )},

T1

 U (T1 , T2 , y) = inf{

T2

f (x(t), x (t))dt : x ∈ W 1,1 (T1 , T2 ), x(T1 ) = y},

T1

 U (T1 , T2 , y, z) = inf{

T2

f (x(t), x (t))dt :

T1

x ∈ W 1,1 (T1 , T2 ), x(T1 ) = y, x(T2 ) = z}. There are two cases: y¯ = 0; y¯ = 0. If y¯ = 0, then for each T2 > T1 ≥ 0, the function x(t) = x, ¯ t ∈ [T1 , T2 ] is a solution of the problems (PT1 ,T2 ), (PT1 ,T2 ,x¯ ), (PT1 ,T2 ,x, ¯ x¯ ). For each pair of numbers T1 < T2 and each x ∈ W 1,1 (T1 , T2 ) set  I (T1 , T2 , x) =

T2

f (x(t), x (t))dt.

T1

Theorem 5.1 Let T > 0. Then U (0, T ) = U (0, T , x) ¯ = U (0, T , x, ¯ x) ¯ = Tf (x, ¯ y). ¯

134

5 Symmetric Variational Problems

Moreover, for each  > 0 there exists x ∈ W 1,1 (0, T ) such that x(0) = x(T ) = x, ¯ I (0, T , x) ≤ Tf (x, ¯ y) ¯ + , x(t) − x ¯ ≤ , t ∈ [0, T ], ¯ −y}, ¯ t ∈ [0, T ] a. e.. x (t) ∈ {y, Proof In the case y¯ = 0, the result is obvious. Assume that y¯ = 0.

(5.4)

Let  > 0. By (A2), there exists 1 ∈ (0, ) such that |f (x, ¯ y) ¯ − f (x, y)| ≤ /T

(5.5)

for each (x, y) ∈ X × X satisfying ¯ ≤ 1 . x − x ¯ ≤ 1 , y − y Choose a natural number n such that ¯ + 1) < 1 . T n−1 (y

(5.6)

Define a function ξ : [0, T n−1 ] → X by ξ(t) = x¯ + t y, ¯ t ∈ [0, (2n)−1 T ],

(5.7)

¯ t ∈ ((2n)−1 T , n−1 T ]. ξ(t) = x¯ + (T n−1 − t)y,

(5.8)

Clearly, ξ ∈ W 1,1 (0, T n−1 ), ξ(0) = x, ¯ ξ(T n−1 ) = x. ¯

(5.9)

For every i ∈ {0, . . . , n − 1} define x(iT n−1 + t) = ξ(t), t ∈ [0, T n−1 ].

(5.10)

By (5.7)–(5.10), x ∈ W 1,1 (0, T ), x(0) = x(T ) = x, ¯

(5.11)

I (0, T , x) = nI (0, T n−1 , ξ ).

(5.12)

5.1 Preliminaries

135

In view of (5.7), (5.8), and (5.10) for a. e. t ∈ [0, T n−1 ], ¯ −y}. ¯ ξ (t) ∈ {y,

(5.13)

Equations (5.6)–(5.8) imply that for every t ∈ [0, T n−1], ¯ < 1 . ξ(t) − x ¯ ≤ 2−1 T n−1 y

(5.14)

By (5.5), (5.13), and (5.14), for a. e. t ∈ [0, T n−1 ], ¯ y)| ¯ ≤ T −1 |f (ξ(t), ξ (t)) − f (x, and ¯ y)| ¯ ≤ n−1 . |I (0, T n−1 , ξ ) − T n−1 f (x, Together with (5.10) this implies that |I (0, T , f ) − Tf (x, ¯ y)| ¯ ≤ . Since  is any positive number we conclude that Theorem 5.1 is proved. Note that if y¯ = 0, then for every positive T problems (P0,T ) , (P0,T ,x) ¯ , and (P0,T ,x, ¯ x¯ ) have no minimizers. Theorem 5.2 Let L0 , M0 > 0. Then there exist M1 > 0 such that for each T > L0 and each y, z ∈ X satisfying y, z ≤ M0 the inequality U (0, T , y, z) ≤ Tf (x, ¯ y) ¯ + M1 holds. Proof By (A3) there exists M2 > 0 such that ¯ + M0 ) f (y, z) ≤ M2 for all y ∈ B(x ¯ and all z ∈ B(0, 2L−1 0 (M0 + x)).

(5.15)

¯ y)|). ¯ M1 = 1 + L0 (M2 + |f (x,

(5.16)

Set

Assume that T > L0 and y, z ∈ X satisfy y, z ≤ M0 .

(5.17)

136

5 Symmetric Variational Problems

Theorem 5.1 implies that there exists x0 ∈ W 1,1 (0, T − L0 ) such that ¯ x0 (0) = x0 (T − L0 ) = x,

(5.18)

¯ y) ¯ + 1. I (0, T − L0 , x0 ) ≤ (T − L0 )f (x,

(5.19)

Define a function x ∈ W 1,1 (0, T ) by x(t) = y + 2L−1 0 t (x¯ − y), t ∈ [0, L0 /2],

(5.20)

x(t) = x0 (t − 2−1 L0 ), ∈ (2−1 L0 , T − 2−1 L0 ],

(5.21)

¯ t ∈ (T − L0 /2, T ]. x(t) = x¯ + 2L−1 0 (t − (T − L0 /2))(z − x),

(5.22)

Equations (5.18)–(5.22) imply that x(0) = y, x(T ) = z, 

T

f (x(t), x (t))dt

0



2−1 L0

=

f (x(t), x (t))dt +



0



0

2−1 L0



T −L0

f (x(t), x (t))dt +

0



f (x0 (t), x0 (t))dt +

T

T −2−1 L0



T

T −2−1 L0

f (x(t), x (t))dt

f (x(t), x (t))dt + (T − L0 )f (x, ¯ y) ¯ + 1. (5.23)

In view of (5.17) and (5.20), for a. e. t ∈ [0, 2−1 L0 ], −1 ¯ + M0 ), x (t) = 2L−1 0 x¯ − y ≤ 2L0 (x −1 x(t) = 2tL−1 ¯ + M0 , 0 x¯ + (1 − 2L0 t)y ≤ x

for a. e. t ∈ [T − 2−1 L0 , T ], −1 ¯ + M0 ), x (t) = 2L−1 0 x¯ − z ≤ 2L0 (x −1 −1 x(t) = (t − 2(T − 2−1 L0 ))L−1 ¯ 0 z + (1 − (t − 2(T − 2 L0 ))L0 )x

≤ x ¯ + M0 .

(5.24)

5.2 The First Weak Turnpike Result

137

By (5.15), (5.24), and the equations above, for a. e. t ∈ [0, 2−1 L0 ]∪[T −2−1 L0 , T ], f (x(t), x (t)) ≤ M2 .

(5.25)

It follows from (5.16), (5.23), and (5.25) that 

T

f (x(t), x (t))dt ≤ L0 M2 + (T − L0 )f (x, ¯ y) ¯ +1

0

≤ Tf (x, ¯ y) ¯ + 1 + L0 (M2 + |f (x, ¯ y)|) ¯ ≤ Tf (x, ¯ y) ¯ + M1 . Theorem 5.2 is proved.

5.2 The First Weak Turnpike Result In this section we prove our first turnpike result. It shows that for approximate solutions x of our variational problems on intervals [0, T ], where T is sufficiently large and given values x(0), x(T ) at the end points belong to a given bounded set C, the Lebesgue measure of all points t ∈ [0, T ] such that (x(t), x (t)) does not belong to an -neighborhood of the set {(x, ¯ y), ¯ (x, ¯ −y)} ¯ does not exceed a constant L that depends only on  and the set C and does not depend on T , x(0), x(T ). In the literature this property is known as the weak turnpike property. Theorem 5.3 Let  ∈ (0, 1) and L0 , M0 , M1 > 0. Then there exists L1 > L0 such that for each T > L1 and each x ∈ W 1,1 (0, T ) such that x(0) ∈ B(0, M0 ) and at least one of the following conditions holds: (a) x(T ) ∈ B(0, M0 ), I f (0, T , x) ≤ U (0, T , x(0), x(T )) + M1 ; (b) I f (0, T , x) ≤ U (0, T , x(0)) + M1 the inequality mes({t ∈ [0, T ] : max{x(t) − x, ¯ ¯ x (t) + y}} ¯ > }) ≤ L1 . min{x (t) − y,

(5.26)

138

5 Symmetric Variational Problems

Proof Theorem 5.2 implies that there exists M2 > 0 such that for each T > 0 and each y, z ∈ B(0, M0 ), U (0, T , y, z) ≤ Tf (x, ¯ y) ¯ + M2 .

(5.27)

Assumption (A1) implies that there exists δ ∈ (0, ) such that for each (x, y) ∈ X × X satisfying max{x − x ¯ + min{y − y, ¯ y + y}} ¯ > .

(5.28)

f (x, y) > f (x, ¯ y)) ¯ +δ

(5.29)

L1 = max{L0 , δ −1 (M1 + M2 )}.

(5.30)

we have

Set

Assume that T > L1 , x ∈ W 1,1 (0, T ), (5.26) is true and at least one of conditions (a) and (b) holds. Conditions (a) and (b) and (5.27) imply that ¯ y) ¯ + M1 + M 2 . I f (0, T , x) ≤ Tf (x,

(5.31)

Set ¯ y) ¯ + δ}. E = {t ∈ [0, T ] : f (x(t), x (t)) > f (x, Equations (5.2), (5.31), and (5.32) imply that ¯ y) ¯ ≥ I (0, T , x) M1 + M2 + Tf (x,  =





f (x(t), x (t))dt + E

[0,T ]\E

f (x(t), x (t))dt

≥ Tf (x, ¯ y) ¯ + δmes(E) and in view of (5.30), mes(E) ≤ δ −1 (M1 + M2 ) ≤ L1 . Assume that t ∈ [T1 , T2 ] \ E.

(5.32)

5.3 Auxiliary Results for the Turnpike Property

139

By (5.32), ¯ y) ¯ + δ. f (x(t), x (t)) ≤ f (x, Combined with the choice of δ (see (5.28) and (5.29)) this implies that x(t) − x ¯ ≤ , ¯ x (t) + y} ¯ ≤ . min{x (t) + y Theorem 5.3 is proved.

5.3 Auxiliary Results for the Turnpike Property Proposition 5.4 Let  > 0. Then there exist γ ∈ (0, 2−1 ) and δ > 0 such that for each y, z ∈ B(x, ¯ δ) there exists ξ ∈ W 1,1 (0, γ ) such that ξ(0) = y, ξ(γ ) = z, I (0, γ , ξ ) ≤ γf (x, ¯ y) ¯ + , ξ(t) − x ¯ ≤ , t ∈ [0, γ ], ¯ −y} ¯ = ∅, t ∈ [0, γ ] a. e.. B(ξ (t), ) ∩ {y, Proof Assumption (A2) and (5.2) imply that there exists 1 ∈ (0, ) such that for each pair of points ¯ 1 ) u ∈ B(x, ¯ 1 ), v ∈ B(y, we have |f (u, v) − f (x, ¯ y)| ¯ ≤ 8−1 .

(5.33)

Choose a number γ ∈ (0, 2−1 ) that satisfies 4(1 + y)γ ¯ ≤ 1 .

(5.34)

140

5 Symmetric Variational Problems

Set δ = 6−1 1 γ .

(5.35)

y, z ∈ B(x, ¯ δ).

(5.36)

u = 2γ −1 (z − y) − y¯

(5.37)

ξ(t) = y + t y, ¯ t ∈ [0, 2−1 γ ],

(5.38)

Assume that

Define

and

ξ(t) = y + 2−1 γ y + (t − 2−1 γ )u, t ∈ (2−1 γ , γ ].

(5.39)

By (5.37)–(5.39), ξ ∈ W 1,1 (0, γ ), ¯ ξ(0) = y, ξ(2−1 γ ) = y + 2−1 γ y,

(5.40)

ξ(γ ) = y + 2−1 γ y¯ + 2−1 γ u = y + 2−1 γ (y¯ + u) = z.

(5.41)

In view of (5.38), for a. e. t ∈ [0, 2−1 γ ], ξ (t) = y. ¯

(5.42)

It follows from (5.35)–(5.37) and (5.39) that for a. e. t ∈ [2−1 γ , γ ], ¯ = u + y ¯ ξ (t) + y = 2γ −1 z − y ≤ 4δγ −1 ≤ 1 .

(5.43)

By (5.35), (5.36), and (5.38) for every t ∈ [0, 2−1 γ ], ξ(t) − x ¯ ≤ x¯ − y + ty ¯ ≤ δ + γ y ¯ ≤ 6−1 1 γ + 1 /4.

(5.44)

5.3 Auxiliary Results for the Turnpike Property

141

Equations (5.34)–(5.37), (5.39), and (5.44) imply that for all t ∈ [2−1 γ , γ ], x¯ − ξ(t) ≤ x¯ − ξ(2−1 γ ) + (t − γ /2)u ≤ δ + γ y ¯ + γ u ≤ δ + 2γ y ¯ + 2z − y ≤ 5δ + 2γ y ¯ ≤ 1 .

(5.45)

By (5.42)–(5.45) and the choice of 1 (see (5.33)) for a. e. t ∈ [0, γ ] ¯ y)| ¯ ≤ 8 −1 . |f (ξ(t), ξ (t)) − f (x, This implies that I (0, γ , ξ ) ≤ γf (x, ¯ y) ¯ + γ /8. Proposition 5.4 is proved. Proposition 5.5 Let  > 0. Then there exist γ ∈ (0, 1) and δ > 0 such that for each T > γ and each y, z ∈ B(x, ¯ δ)

(5.46)

there exists ξ ∈ W 1,1 (0, T ) such that ξ(0) = y, ξ(T ) = z, I (0, T , ξ ) ≤ Tf (x, ¯ y) ¯ + , ξ(t) − x ¯ ≤ , t ∈ [0, T ], ¯ −y} ¯ = ∅, t ∈ [0, T ] a. e.. B(ξ (t), ) ∩ {y, Proof By Proposition 5.4, there exist γ0 ∈ (0, 2−1 ) and δ > 0 such that the following property holds: (a) For each y, z ∈ B(x, ¯ δ) there exists ξ ∈ W 1,1 (0, γ0 ) such that ξ(0) = y, ξ(γ0 ) = z, ¯ y) ¯ + /4, I (0, γ0 , ξ ) ≤ γ0 f (x,

142

5 Symmetric Variational Problems

ξ(t) − x ¯ ≤ , t ∈ [0, γ0 ], ¯ −y} ¯ = ∅, t ∈ [0, γ0 ] a. e.. B(ξ (t), ) ∩ {y, Set γ = 2γ0 .

(5.47)

Assume that T > γ and (5.46) holds. Property (a) and (5.46) imply that there exist ξ1 ∈ W 1,1 (0, γ0 ) and ξ2 ∈ W 1,1 (T − γ0 , T ) such that ¯ ξ1 (0) = y, ξ1 (γ0 ) = x,

(5.48)

¯ y) ¯ + /4, I (0, γ0 , ξ1 ) ≤ γ0 f (x,

(5.49)

¯ −y} ¯ = ∅, t ∈ [0, γ0 ] a. e., B(ξ1 (t), ) ∩ {y,

(5.50)

¯ ≤ , t ∈ [0, γ0 ], ξ1 (t) − x

(5.51)

¯ ξ2 (T ) = z, ξ2 (T − γ0 ) = x,

(5.52)

¯ y) ¯ + /4, I (T − γ0 , T , ξ2 ) ≤ γ0 f (x,

(5.53)

ξ2 (t) − x ¯ ≤ , t ∈ [T − γ0 , T ],

(5.54)

¯ −y} ¯ = ∅, t ∈ [T − γ0 , T ] a. e.. B(ξ2 (t), ) ∩ {y, Theorem 5.1 implies that there exists ξ3 ∈ W 1,1 (γ0 , T − γ0 ) such that ¯ ξ3 (T − γ0 ) = ξ3 (γ0 ) = x, ¯ y) ¯ + /4, I (γ0 , T − γ0 , ξ3 ) ≤ (T − 2γ0 )f (x, ¯ −y}, ¯ t ∈ [γ0 , T − γ0 ] a. e. ξ (t) ∈ {y, and ¯ ≤ , t ∈ [γ0 , T − γ0 ]. ξ3 (t) − x Define ξ(t) = ξ1 (t), t ∈ [0, γ0 ], ξ(t) = ξ3 (t), t ∈ (γ0 , T − γ0 ], ξ(t) = ξ2 (t), t ∈ (T − γ0 , T ].

(5.55)

5.3 Auxiliary Results for the Turnpike Property

143

It follows from the relations above, (5.48)–(5.55) that ξ ∈ W 1,1 (0, T ), ξ(0) = y, ξ(T ) = z, I (0, T , ξ ) ≤ Tf (x, ¯ y) ¯ + , ξ(t) − x ¯ ≤ , t ∈ [0, T ], ¯ −y} ¯ = ∅, t ∈ [0, T ] a. e.. B(ξ (t), ) ∩ {y, Proposition 5.5 is proved. Denote by M the set of all Borelian functions g : X × X → R 1 such that g(x, u) ≥ ψ(u)u − a

(5.56)

(see (5.3 and (A3)) for each (x, u) ∈ X × X. Proposition 5.6 Let M1 ,  > 0 and 0 < τ0 < τ1 . Then there exists δ > 0 such that for each pair of numbers T1 , T2 satisfying 0 ≤ T1 , T2 ∈ [T1 + τ0 , T1 + τ1 ],

(5.57)

each g ∈ M, each x ∈ W 1,1 (T1 , T2 ) satisfying 

T2

g(x(t), x (t))dt ≤ M1

(5.58)

T1

and each t1 , t2 ∈ [T1 , T2 ] satisfying |t1 − t2 | ≤ δ, the inequality x(t1 ) − x(t2 ) ≤  holds. Proof By (5.56) and the properties of the function ψ, there exists a number c0 > 0 such that for each (x, u) ∈ X × X satisfying u ≥ c0 and each g ∈ M, g(x, u) ≥ 4 −1 (M1 + 2 + aτ1 )u.

(5.59)

δ ∈ (0, 8−1 (c0 + 1)−1 ).

(5.60)

Choose

144

5 Symmetric Variational Problems

Assume that numbers T1 , T2 satisfy (5.57), g ∈ M, x ∈ W 1,1 (T1 , T2 ) satisfies (5.58) and t1 , t2 ∈ [T1 , T2 ] satisfy |t1 − t2 | ≤ δ.

(5.61)

Set E1 = {t ∈ [t1 , t2 ] : x (t) ≥ c0 }, E2 = [t1 , t2 ] \ E1 .

(5.62)

By (5.14), (5.56), (5.58)–(5.62), and the choice of c0 , x(t1 ) − x(t2 )  x (t)dt + x (t)dt

 ≤ E1

E2



x (t)dt

≤ δc0 + E1

≤ δc0 + [4(M1 + 2 + aτ1 )]

−1



g(x(t), x (t))dt

 E1

≤ δc0 + [4(M1 + 2 + aτ1 )]−1 (



T2

g(x(t), x (t))dt + aτ1 )

T1

≤ δc0 + [4(M1 + 2 + aτ1 )]−1 (M1 + aτ1 ) ≤ δc0 + 4−1  < . Proposition 5.6 is proved.

5.4 A Turnpike Result In this section we prove a turnpike result. It shows that for approximate solutions x of our variational problems on intervals [0, T ], where T is sufficiently large and given values x(0), x(T ) at the end points belong to a given bounded set C, the set of all points t ∈ [0, T ] such that x(t) does not belong to an -neighborhood of x¯ is contained in the union of two intervals, where the first interval contains 0, the second one contains T , and their lengths do not exceed a constant L that depends

5.4 A Turnpike Result

145

only on  and the set C and does not depend on T , x(0), x(T ). In the literature this property is known as the turnpike property. Theorem 5.7 Let  ∈ (0, 1]. Then there exist L, δ > 0 such that for each T > 2L and each u ∈ W 1,1 (0, T ) such that u(0) ∈ B(0, M) and at least one of the following conditions holds: u(T ) ∈ B(0, M), I (0, T , u) ≤ U (0, T , u(0), u(T )) + δ; I (0, T , x) ≤ U (0, T , u(0)) + δ there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ] such that u(t) − x ¯ ≤ , t ∈ [τ1 , τ2 ]. Moreover, if u(0) − x ¯ ≤ δ, then τ1 = 0 and if u(T ) − x ¯ ≤ δ, then τ2 = T . Proof Proposition 5.6 implies that there exists 0 ∈ (0, min{/2, 1/4}) such that the following property holds: (a) For each S ≥ 0, each x ∈ W 1,1 (S, S + 1) satisfying I (S, S + 1, x) ≤ |f (x, ¯ y)| ¯ +2 and each t1 , t2 ∈ [S, S + 1] satisfying |t1 − t2 | ≤ 0 we have x(t1 ) − x(t2 ) ≤ /4. By (A1) there exists 1 ∈ (0, 1) such that f (y, z) ≥ f (x, ¯ y) ¯ + 1 for all (y, z) ∈ X × X satisfying y − x ¯ ≥ /4.

(5.63)

Choose a positive number δ0 < 2−1 1 0 .

(5.64)

146

5 Symmetric Variational Problems

Proposition 5.5 implies that there exist γ ∈ (0, 1) and δ1 ∈ (0, δ0 ) such that the following property holds: (b) For each T > γ and each y, z ∈ B(x, ¯ δ1 ) we have U (0, T , y, z) ≤ Tf (x, ¯ y) ¯ + δ0 . By Theorem 5.3, there exists L1 > 4 such that the following property holds: (c) For each T ≥ L1 and each x ∈ W 1,1 (0, T ) such that x(0) ≤ M and such that at least one of the following conditions holds: x(T ) ≤ M, I (0, T , x) ≤ U (0, T , x(0), x(T )) + 1; I (0, T , x) ≤ U (0, T , x(0)) + 1 the inequality mes({t ∈ [0, T ] : x(t) − x ¯ > δ1 }) < L1 is valid. Fix numbers δ = δ1 /2, L ≥ 4L1 + 4.

(5.65)

Assume that T > 2L and u ∈ W 1,1 (0, T ) satisfies u(0) ∈ B(0, M)

(5.66)

and at least one of the conditions below: u(T ) ∈ B(0, M),

(5.67)

I (0, T , u) ≤ U (0, T , u(0), u(T )) + δ;

(5.68)

I (0, T , u) ≤ U (0, T , u(0)) + δ.

(5.69)

By (5.65)–(5.69), the choice of L1 and property (c), mes({t ∈ [0, T ] : u(t) − x ¯ > δ1 }) < L1 .

(5.70)

5.4 A Turnpike Result

147

It follows from (5.70) that the following property holds: (d) For each number S satisfying [S, S + L1 ] ⊂ [0, T ] there exists τ ∈ [S, S + L1 ] such that u(τ ) − x ¯ ≤ δ1 . Property (d), (5.61), and (5.70) imply that there exist τ1 ∈ [0, L1 ], τ2 ∈ [T − L1 , T ]

(5.71)

¯ ≤ δ1 , i = 1, 2. u(τi ) − x

(5.72)

such that

¯ ≤ δ, then we set τ2 = T . If u(0) − x ¯ ≤ δ, then we set τ1 = 0 and if u(T ) − x We show that u(t) − x ¯ ≤ , t ∈ [τ1 , τ2 ]. Assume the contrary. Then there exists t∗ ∈ [τ1 , τ2 ]

(5.73)

¯ > . u(t∗ ) − x

(5.74)

such that

Property (d), (5.71), and (5.73) imply that there exists an interval [a, b] ⊂ [τ1 , τ2 ] such that t∗ ∈ [a, b],

(5.75)

¯ ≤ δ1 , u(a) − x ¯ ≤ δ1 , u(b) − x

(5.76)

2−1 L1 ≤ b − a ≤ 3L1 .

(5.77)

Property (b), (5.76), and (5.77) imply that U (a, b, u(a), u(b)) ≤ (b − a)f (x, ¯ y) ¯ + δ0 .

(5.78)

148

5 Symmetric Variational Problems

Equations (5.65), (5.68), (5.69), and (5.78) imply that I (a, b, u) ≤ U (a, b, u(a), u(b)) + δ ≤ (b − a)f (x, ¯ y) ¯ + 2δ0 .

(5.79)

In view of (5.2) and (5.79), for each c, d ∈ [a, b] satisfying c < d we have I (c, d, u) ≤ (c − d)f (x, ¯ y) ¯ + 2δ0 .

(5.80)

By (5.75) and (5.77), there exist c, d ∈ [a, b] such that d − c = 1, t∗ ∈ [c, d].

(5.81)

Property (a), (5.80), and (5.81) imply that for every t ∈ [c, d] ∩ [t∗ − 0 , t∗ + 0 ] we have u(t) − u(t∗ ) ≤ /4.

(5.82)

E = [c, d] ∩ [t∗ − 0 , t∗ + 0 ].

(5.83)

Set

Clearly, E is a closed interval. Since 0 < 1/4, it follows from (5.81) and (5.83) that mes(E) ≥ 0 .

(5.84)

By (5.74), (5.82), and (5.83), for all t ∈ E, u(t) − x ¯ ≥ /4.

(5.85)

By (5.63) and (5.85), for all t ∈ E, ¯ y) ¯ + 1 . f (u(t), u (t)) > f (x, Equations (5.80), (5.81), (5.84), and (5.86) imply that f (x, ¯ y) ¯ + 2δ0 ≥ I (c, d, u)  =



f (u(t), u (t))dt + E

 [c,d]\E

f (u(t), u (t))dt

(5.86)

5.5 The Second Weak Turnpike Result

149

≥ mes(E)(f (x, ¯ y) ¯ + 1 ) + mes([c, d] \ E)f (x, ¯ y) ¯ ≥ f (x, ¯ y) ¯ + 1  0 and δ0 ≥ 2−1 1 0 . This contradicts (5.64). The contradiction we have reached completes the proof of Theorem 5.7.

5.5 The Second Weak Turnpike Result In this section we prove our second weak turnpike result. It shows that for approximate solutions x of our variational problems on intervals [0, T ], where T is sufficiently large and given values x(0), x(T ) at the end points belong to a given bounded set C, the set of all points t ∈ [0, T ] such that x(t) does not belong to an -neighborhood of x¯ is contained in the union of a finite family of intervals, their lengths do not exceed a constant l, and their number does not exceed a constant Q that depend only on  and the set C and does not depend on T , x(0), x(T ). Theorem 5.8 Let M0 > x ¯ + y ¯ + 1, M1 > 0 and  ∈ (0, 1]. Then there exist l > 0 and a natural number Q such that for each T > lQ and each x ∈ W 1,1 (0, T ) such that x(0) ≤ M0 and at least of the conditions below holds: (a) x(T ) ≤ M0 , I (0, T , x) ≤ U (0, T , x(0), x(T )) + M1 ; (b) I (0, T , x) ≤ U (0, T , x(0)) + M1

(5.87)

150

5 Symmetric Variational Problems

there exists a sequence of closed intervals [ai , bi ] ⊂ [0, T ], i = 1, . . . , q such that q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, ai+1 ≥ bi for all i ∈ {1, . . . , q} \ {q} and q

{t ∈ [0, T ] : x(t) − x ¯ > } ⊂ ∪i=1 [ai , bi ]. Proof By Theorem 5.7, there exist l0 > 0, δ ∈ (0, 1] such that the following property holds: (c) For each T ≥ 2l0 and each x ∈ W 1,1 (0, T ) satisfying x(0) − x ¯ ≤ δ, x(T ) − x ¯ ≤ δ, I (0, T , x) ≤ U (0, T , x(0), x(T )) + δ we have x(t) − x ¯ ≤ , t ∈ [0, T ]. By Theorem 5.3, there exist l1 > 2l0 such that the following property holds: (d) For each T ≥ l1 and each x ∈ W 1,1 (0, T ) such that x(0) ≤ M0 and such that at least one of conditions (a), (b) holds we have mes({t ∈ [0, T ] : x(t) − x ¯ > δ} < l1 . Set l = 4l1 + 4.

(5.88)

Q > 4 + δ −1 M1 .

(5.89)

Choose an integer

5.5 The Second Weak Turnpike Result

151

Assume that T > lQ and that x ∈ W 1,1 (0, T ) satisfies at least one of conditions (a), (b). Together with property (d) this implies that there exists t0 ∈ [0, l1 ]

(5.90)

¯ ≤ δ. x(t0 ) − x

(5.91)

such that

By induction, using property (d), conditions (a) and (b), and equations (5.77), (5.90), and (5.91), we construct a finite sequence t0 < t 1 < · · · < t q belonging to the interval [0, T ] such that ¯ ≤ δ, i = 0, . . . , q, x(ti ) − x

(5.92)

ti+1 − ti ∈ [l1 , 2l1 ], i ∈ {0, . . . , q} \ {q}

(5.93)

tq > T − 2l1 .

(5.94)

Now we construct a strictly increasing sequence of numbers Si ∈ {t0 , . . . , tq }, i = 0, . . . , p. Set S0 = t0 .

(5.95)

Assume that k ≥ 0 is an integer and that we already defined Si ∈ {t0 , . . . , tq }, i = 0, . . . , k. If Sk = tq , then the construction is completed. Assume that Sk < tq . If I (Sk , tq , x) ≤ U (Sk , tq , x(Sk ), x(tq )) + δ, then we set Sk+1 = tq

152

5 Symmetric Variational Problems

and the construction is completed. Assume that I (Sk , tq , x) > U (Sk , tq , x(Sk ), x(tq )) + δ.

(5.96)

There exists j ∈ {0, . . . , q} such that Sk = tj .

(5.97)

If I (Sk , tj +1 , x) > U (Sk , tj +1 , x(Sk ), x(tj +1 )) + δ, then we set Sk+1 = tj +1 . Assume that I (Sk , tj +1 , x) ≤ U (Sk , tj +1 , x(Sk ), x(tj +1 )) + δ.

(5.98)

We define Sk+1 = min{ti : i ∈ {j + 1, . . . , q} : I (Sk , ti , x) > U (Sk , ti , x(Sk ), x(ti )) + δ}. Therefore by induction we defined a strictly increasing sequence of numbers Si ∈ {t0 , . . . , tq }, i = 0, . . . , p such that S0 = t0 , Sp = tq , for each j ∈ {0, . . . , p − 1} \ {p − 1}, I (Sj , Sj +1 , x) > U (Sj , Sj +1 , x(Sj ), x(Sj +1 )) + δ,

(5.99)

for each j ∈ {0, . . . , p − 1}, if i ∈ {0, . . . , q} and Sj < ti < Sj +1 , then I (Sj , ti , x) ≤ U (Sj , ti , x(Sj ), x(Sti )) + δ,

(5.100)

5.5 The Second Weak Turnpike Result

153

Conditions (a), (b), and (5.99) imply that M1 ≥ I (0, T , x) − U (0, T , x(0), x(T )) ≥



{I (Sj , Sj +1 , x)−U (Sj , Sj +1 , x(Sj ), x(Sj +1 )) : j ∈ {0, . . . , p−1}\{p−1}} ≥ (p − 1)δ

and p ≤ 1 + δ −1 M1 .

(5.101)

Assume that j ∈ {0, . . . , p − 1} satisfies Sj +1 − Sj ≥ l.

(5.102)

Property (d), conditions (a) and (b), and equations (5.87), (5.88), and (5.102) imply that there exists S˜j ∈ [Sj +1 − 3l1 , Sj +1 − 2l1 ]

(5.103)

¯ ≤ δ. x(S˜j ) − x

(5.104)

such that

It follows from (5.93), (5.100), (5.102), and (5.103) that I (Sj , S˜j , x) ≤ U (Sj , S˜j , x(Sj ), x(S˜j )) + δ.

(5.105)

Property (c) and equations (5.92), (5.104), and (5.105) imply that x(t) − x ¯ ≤ , t ∈ [Sj , S˜j ].

(5.106)

By the equation above, (5.102), (5.103), (5.106), {t ∈ [0, T ] : x(t) − x ¯ > } ⊂ [0, S0 ] ∪ [Sp , T ] ∪ {[Si , Si+1 ] : i ∈ {0, . . . , p − 1}, Si+1 − Si < l} ∪{[Si+1 − 3l1 , Si+1 ] : i ∈ {0, . . . , p − 1}, Si+1 − Si ≥ l}. It is easy to see that the right-hand side of the equation above is a union of a finite number of closed intervals, their length does not exceed l, and their number does

154

5 Symmetric Variational Problems

not exceed 2 + 2p ≤ 4 + δ −1 M1 ≤ Q. Theorem 5.8 is proved.

5.6 The Space of Integrands Denote by M (see Section 5.3) the set of all borelian functions g : X × X → R 1 such that g(x, u) ≥ ψ(u)u − a

(5.107)

(see (5.3 and (A3)) for each (x, u) ∈ X × X. For each pair of nonnegative numbers T1 < T2 each x ∈ W 1,1 (T1 , T2 ), each g ∈ M, and each y, z ∈ X define  T2 g g(x(t), x (t))dt, I (T1 , T2 , x) = T1

U g (T1 , T2 ) = inf{I g (T1 , T2 , x) : x ∈ W 1,1 (T1 , T2 )}, U g (T1 , T2 , y) = inf{I g (T1 , T2 , x) : x ∈ W 1,1 (T1 , T2 ), x(T1 ) = y}, U G (T1 , T2 , y, z) = inf{I g (T1 , T2 , x) : x ∈ W 1,1 (T1 , T2 ), x(T1 ) = y, x(T2 ) = z}. In the sequel we use the results of Section 2.13 applied to the metric space X×X. For each natural number n denote by Ew (n) the set of all pairs (g, h) ∈ M × M, which have the following properties: for each y, z ∈ B(0, n), |g(y, z) − h(y, z))| ≤ n−1 ; for each y, z ∈ X satisfying min{g(y, z), h(y, z)} ≤ n the inequality |g(y, z) − h(y, z)| ≤ n−1 holds.

5.7 Auxiliary Results for Theorem 5.9

155

For the space M there exists a uniformity generated by the base Ew (n), n = 1, 2, . . . , which induces in M a topology. We continue to consider the function f introduced in Section 5.1 satisfying all the assumptions introduced there (see (5.1), (5.2), (A1), (A2), (A3)). We will prove the following result that shows that the turnpike phenomenon is stable under small perturbations of the integrand f . Theorem 5.9 Let  > 0 and M > x ¯ + y + 1. Then there exist δ, L > 0 and a neighborhood U of f in M such that for each g ∈ U , each T > 2L, and each u ∈ W 1,1 (0, T ) at least one of the conditions below is satisfied: (a) u((0), u(T ) ∈ B(0, M0 ), I g (0, T , u) ≤ U g (0, T , u(0), u(T )) + δ; (b) u(0) ∈ B(0, M), I g (0, T , u) ≤ U g (0, T , u(0)) + δ there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ] such that u(t) − x ¯ ≤ , t ∈ [τ1 , τ2 ]. ¯ ≤ δ, then τ2 = T . Moreover, if u(0) − x ¯ ≤ δ, then τ1 = 0 and if u(T ) − x

5.7 Auxiliary Results for Theorem 5.9 Lemma 5.10 Let  ∈ (0, 1] and M > x ¯ + y + 1.

156

5 Symmetric Variational Problems

Then there exist L > 0 and a neighborhood U of f in M such that for each g ∈ U , each T > L, and each u ∈ W 1,1 (0, T ) for which at least one of the conditions below is satisfied: (a) u((0), u(T ) ∈ B(0, M), I g (0, T , u) ≤ U g (0, T , u(0), u(T )) + M (b) u(0) ∈ B(0, M), I g (0, T , u) ≤ U g (0, T , u(0)) + M the following property holds: (c) For each S ≥ 0 satisfying [S.S + L] ⊂ [0, T ], u(S) ≤ M there exists t0 ∈ [S + 1, S + L] such that ¯ ≤ . u(t0 ) − x Proof Proposition 2.19 implies that there exist δ0 ∈ (0, ) and a neighborhood U1 of f in M such that the following property holds: (d) For each g ∈ U1 and each x, y ∈ X satisfying x − x ¯ > /2 we have g(x, y) > f (x, ¯ y) ¯ + δ0 . Assumption (A3) implies that there exists M1 > 0 such that |f (y, z)| ≤ M1 for all y, z ∈ B(4M + 2).

(5.108)

Choose a number ¯ y)|δ ¯ 0−1 . L > 4 + 2(M + a + M1 + 1)δ0−1 + 2|f (x,

(5.109)

5.7 Auxiliary Results for Theorem 5.9

157

By assumption (A2), there exists δ1 ∈ (0, δ0 ) such that the following property holds: (e) For every (u, v) ∈ X × X satisfying u − x ¯ + v − y ¯ ≤ δ1 we have |f (u, v) − f (x, ¯ y)| ¯ ≤ δ0 /4. There exists a neighborhood U2 of f in M such that for each g ∈ U2 and each u, v ∈ B(0, 4M + 2) we have |f (u, v) − g(u, v)| ≤ δ0 /4.

(5.110)

U = U1 ∩ U2 .

(5.111)

g ∈ U,

(5.112)

Set

Assume that

T > L, u ∈ W 1,1 (0, T ), at least one of conditions (a), (b) holds and a nonnegative number S satisfies S ≤ T − L − 1, u(S) ≤ M.

(5.113)

In order to complete the proof of the theorem it is sufficient to show that there exists t0 ∈ [S + 1, S + L + 1] such that ¯ ≤ . u(t0 ) − x Assume the contrary. Then u(t) − x ¯ > , t ∈ [S + 1, S + L + 1].

(5.114)

Equations (5.111), (5.112), and (5.114) imply that ¯ y) ¯ + δ0 , t ∈ [S + 1, S + L + 1]. g(u(t), u (t)) ≥ f (x,

(5.115)

158

5 Symmetric Variational Problems

There are two cases: (1) There exists S0 ∈ [S + L + 1, T ] for which u(t) − x ¯ ≤ ; (2) u(t) − x ¯ >  for all t ∈ [S + L + 1, T ]. Set S1 = inf{t ∈ [S + L + 1, T ] : u(t) − x ¯ ≤ }.

(5.116)

(Here we suppose that the infimum over an empty set is T .) Clearly, S1 is welldefined. In the case (2) S1 = T . By (5.114) (5.116), in the case (2), u(t) − x ¯ ≥  for all t ∈ [S + 1, T ]

(5.117)

S + L + 1 ≤ S1 ≤ S0 ,

(5.118)

u(t) − x ¯ ≥ , t ∈ [S + 1, S1 ],

(5.119)

¯ = . u(S1 ) − x

(5.120)

and in the case (1),

We consider the following two cases separately: Property (a) holds or property (b) together with case (1) hold; property (b) together with case (2) hold. Consider the first case when property (a) holds or property (b) together with case (1) hold. Actually this case has two subcases. In these both subcases conditions (a), (b), (5.113), (5.116), (5.117), and (5.120) imply that u(S) ≤ M, u(S1 ) ≤ M + 1,

(5.121)

u(t) − x ¯ ≥ , t ∈ [S + 1, S1 ],

(5.122)

S1 − S − 1 ≥ L.

(5.123)

Property (d) and equations (5.112), (5.122) imply that in both subcases we have ¯ y) ¯ + δ0 , t ∈ [S + 1, S1 ]. g(u(t), u (t)) ≥ f (x,

(5.124)

5.7 Auxiliary Results for Theorem 5.9

159

It follows from Theorem 5.1 and equations (5.109) and (5.123) that there exists ξ ∈ W 1,1 (S, S1 ) such that ξ(S) = u(S), ξ(S1 ) = u(S1 ),

(5.125)

¯ ξ(S1 − 2−1 ) = x, ¯ ξ(S + 2−1 ) = x,

(5.126)

ξ(t) = u(S) + 2(t − S)(x¯ − u(S)), t ∈ [S, S + 2−1 ],

(5.127)

ξ(t) = x¯ + 2(t − (S1 − 2−1 ))(u(S1 ) − x), ¯ t ∈ [S1 − 2−1 , S1 ],

(5.128)

¯ y) ¯ + δ1 , I f (S + 2−1 , S1 − 2−1 , ξ ) ≤ (S1 − 2−1 − (S − 2−1 ))f (x,

(5.129)

¯ −y}, ¯ t ∈ [S + 2−1 , S1 − 2−1 ] a. e., ξ (t) ∈ {y,

(5.130)

ξ(t) − x ¯ ≤ δ1 , t ∈ [S + 2−1 , S1 − 2−1 ].

(5.131)

By (5.121) and (5.127), for every t ∈ [S, S + 2−1 ], ξ (t) = 2(x¯ − u(S)), ξ (t) ≤ 2(2M + 1) ≤ 4M + 2,

(5.132)

ξ(t) ≤ max{x, ¯ u(S)} ≤ M.

(5.133)

In view of (5.121) and (5.128), for a. e. t ∈ [S1 − 2−1 , S1 ], ¯ ξ (t) = 2(u(S1 ) − x), ¯ ≤ 4M + 2, ξ (t) ≤ 2u(S1 ) − x

(5.134)

ξ(t) ≤ max{u(S1 ), x} ¯ ≤ M + 1.

(5.135)

By (5.108) and (5.132)–(5.135), f (ξ(t), ξ (t)) ≤ M1 , t ∈ [S, S + 1/2] ∪ [S1 − 1/2, S1 ].

(5.136)

Property (c) and equations (5.130) and (5.131) imply that for a. e. t ∈ [S +1/2, S1 − 1/2], ¯ y)| ¯ ≤ δ0 /4. |f (ξ(t), ξ (t)) − f (x,

(5.137)

In view of (5.130)–(5.135), for t ∈ [S, S1 ] a. e., ξ(t), ξ (t) ∈ B(0, 4M + 2).

(5.138)

160

5 Symmetric Variational Problems

It follows from (5.110)–(5.112) and (5.138) that for t ∈ [S, S1 ] a. e. |f (ξ(t), ξ (t)) − g(ξ(t), ξ (t))| ≤ δ0 /4.

(5.139)

Equations (5.136) and (5.139) imply that for t ∈ [S, S + 1/2] ∪ [S1 − 1/2, S1 ] a. e., g(ξ(t), ξ (t)) ≤ M1 + 1.

(5.140)

By (5.137) and (5.139), for t ∈ [S + 1/2, S1 − 1/2] a. e., ¯ y) ¯ + δ0 /2. g(ξ(t), ξ (t)) ≤ f (ξ(t), ξ (t)) + δ0 /2 ≤ f (x,

(5.141)

In view of (5.140) and (5.141), ¯ y) ¯ + 2−1 δ0 (S1 − 1 − S). I g (S, S1 , ξ ) ≤ (M1 + 1) + (S1 − 1 − S)f (x,

(5.142)

By (5.107) and (5.124), I g (S, S1 , u) ≥ −a + I g (S + 1, S1 , u) ¯ y) ¯ + δ0 (S1 − S − 1). ≥ −a + (S1 − S − 1)f (x,

(5.143)

Conditions (a), (b), (5.123), (5.125), (5.142), and (5.143) imply that M ≥ I g (S, S1 , u) − I g (S, S1 , ξ ) ¯ y) ¯ + δ0 (S1 − S − 1) − M1 − 1 ≥ −a + (S1 − S − 1)f (x, −2−1 δ0 (S1 − 1 − S) − (S1 − 1 − S)f (x, ¯ y) ¯ ≥ 2−1 δ0 (S1 − S − 1) − a − M1 − 1 and 2(M + a + M1 + 1)δ0−1 ≥ S1 − S − 1 ≥ L. This contradicts (5.109). The contradiction we have reached proves that there exists t0 ∈ [S + 1, S + L + 1] such that ¯ ≤ u(t0 ) − x in our first case when property (a) holds or property (b) together with case (1) hold.

5.7 Auxiliary Results for Theorem 5.9

161

Consider now the second case when property (b) together with case (2) holds. Theorem 5.1 implies that there exists ξ ∈ W 1,1 (S, T ) such that ξ(t) = u(S) + 2(t − S)(x¯ − u(S)), t ∈ [S, S + 2−1 ],

(5.144)

¯ −y}, ¯ t ∈ [S + 2−1 , T ] a. e., ξ (t) ∈ {y,

(5.145)

ξ(t) − x ¯ ≤ δ1 , t ∈ [S + 2−1 , T ],

(5.146)

I g (S + 2−1 , T , ξ ) ≤ (T − (S − 2−1 ))f (x, ¯ y) ¯ + δ1 .

(5.147)

By (5.108), (5.113), (5.144), and (5.145), f (ξ(t), ξ (t)) ≤ M1 , t ∈ [S, S + 1/2].

(5.148)

Property (e) and equations (5.145) and (5.146) imply that for a. e. t ∈ [S + 1/2, T ], ¯ y)| ¯ ≤ δ0 /4. |f (ξ(t), ξ (t)) − f (x,

(5.149)

In view of (5.110), (5.113), (5.144), and (5.148), for t ∈ [S, S + 1/2] a. e., g(ξ(t), ξ (t)) ≤ M1 + 1.

(5.150)

By (5.110), (5.145), (5.146), and property (e), for t ∈ [S + 1/2, T ] a. e., ¯ y) ¯ + δ0 /2. g(ξ(t), ξ (t)) ≤ f (ξ(t), ξ (t)) + δ0 /4 ≤ f (x,

(5.151)

In view of (5.150) and (5.151), ¯ y) ¯ + 2−1 δ0 (T − S − 1/2). I g (S, T , ξ ) ≤ (M1 + 1) + (T − S − 1/2)f (x,

(5.152)

It follows from (5.114) and case (2) that u(t) − x ¯ ≥ , t ∈ [S + 1, T ]. Together with property (d), (5.111), and (5.112) this implies that ¯ y) ¯ + δ0 , t ∈ [S + 1, T ]. g(u(t), u (t)) ≥ f x,

(5.153)

Equations (5.107) and (5.153) imply that I g (S, T , u) ≥ −a + I g (S + 1, T , u) ≥ −a + (T − S − 1)f (x, ¯ y) ¯ + δ0 (T − S − 1).

(5.154)

162

5 Symmetric Variational Problems

Condition (b), (5.113), (5.144), (5.153), and (5.154) imply that M ≥ I g (S, T , u) − I g (S, T , ξ ) ≥ −a + (T − S − 1)f (x, ¯ y) ¯ + δ0 (T − S − 1) − M1 − 1 −2−1 δ0 (T − 1/2 − S) − (T − 1/2 − S)f (x, ¯ y) ¯ ¯ y)| ¯ ≥ 2−1 δ0 (T − S − 1) − a − M1 − 1 − |f (x, ¯ y)| ¯ ≥ 2−1 δ0 L − a − M1 − 1 − |f (x, and −1 ¯ y)|)δ ¯ L ≤ 2(M + a + M1 + 1 + |f (x, 0 .

This contradicts (5.109). The contradiction we have reached proves that there exists t0 ∈ [S + 1, S + L + 1] such that ¯ ≤ u(t0 ) − x in our second case when property (b) with case (2) holds. Lemma 5.10 is proved.

5.8 Proof of Theorem 5.9 We may assume that M > x ¯ + y ¯ + 1.

(5.155)

Proposition 2.19 implies that there exist 0 ∈ (0, ) and a neighborhood U0 of f in M such that the following property holds: (i) For each g ∈ U0 and each x, y ∈ X satisfying x − x ¯ > /4 we have g(x, y) > f (x, ¯ y) ¯ + 0 .

5.8 Proof of Theorem 5.9

163

By (5.3), there exists a natural number n such that ψ(n) ≥ a + f (x, ¯ y) ¯ + 1.

(5.156)

Proposition 5.6 implies that there exists γ0 ∈ (0, 4−1 ) such that the following property holds: (ii) For each S ≥ 0, each g ∈ M, each x ∈ W 1,1 (S, S + 1) satisfying 

S+1

g(x(t), x (t))dt ≤ f (x, ¯ y) ¯ +4

S

and each t1 , t2 ∈ [S, S + 1] satisfying |t1 − t2 | ≤ γ0 the inequality x(t1 ) − x(t2 ) ≤ /4 holds. Choose a positive number δ1 < 0 γ0 /16.

(5.157)

By Proposition 5.5, there exist δ > 0 and γ ∈ (0, 1) such that the following property holds: (iii) For each T > γ and each y, z ∈ B(x, ¯ δ) there exists ξ ∈ W 1,1 (0, T ) such that ξ(0) = y, ξ(T ) = z, ¯ y) ¯ + δ1 , I f (0, T , ξ ) ≤ Tf (x, ¯ −y} ¯ = ∅, t ∈ [0, T ] a. e., B(ξ (t), δ1 ) ∩ {y, ξ(t) − x ¯ ≤ δ1 , t ∈ [0, T ]. By Lemma 5.10, there exist L0 > 0 and a neighborhood U1 of f in M such that the following property holds: (iv) For each g ∈ U1 , each T > L0 , and each u ∈ W 1,1 (0, T ) that satisfies u(0) ∈ B(0, M)

164

5 Symmetric Variational Problems

and at least one of the conditions below: u(T ) ∈ B(0, M), I g (0, T , u) ≤ U g (0, T , u(0), u(T )) + M; I g (0, T , u) ≤ U g (0, T , u(0)) + M and each S ≥ 0 satisfying [S, S + L0 + 1] ⊂ [0, T ], u(S) ≤ M we have min{u(t) − x ¯ : t ∈ [S + 1, S + L0 + 1]} ≤ δ. Set L = 8L0 + 8.

(5.158)

There exists a neighborhood U ⊂ U1 ∩ U 0 of f in M such that the following property holds: (v) For each g ∈ U , |g(v1 , v2 ) − f (v1 , v2 )| ≤ δ1 L−1 for all v1 , v2 ∈ B(0, 2M + n + 2). Assume that g ∈ U,

(5.159)

u(0) ∈ B(0, M)

(5.160)

T > 2L and u ∈ W 1,1 (0, T ) satisfies

5.8 Proof of Theorem 5.9

165

and at least one of the conditions below: u(T ) ∈ B(0, M), I g (0, T , u) ≤ U g (0, T , u(0), u(T )) + δ;

(5.161)

I g (0, T , u) ≤ U g (0, T , u(0)) + δ.

(5.162)

Property (iv) and equations (5.159)–(5.162) imply that there exists t0 ∈ [0, L0 + 1] such that ¯ ≤ δ. u(t0 ) − x

(5.163)

t0 = 0.

(5.164)

If u(0) − x ¯ ≤ δ, then we set

Assume that p ≥ 0 is an integer and we already defined numbers t0 < · · · < tp from the interval [0, T ] such that for every i ∈ {0, . . . , p}, ¯ ≤ δ, u(ti ) − x

(5.165)

ti+1 − ti ∈ [1, L0 + 1].

(5.166)

for every i ∈ {0, . . . , p} \ {p},

If tp = T , then the construction is completed and x(tp ) ≤ x ¯ + 1 ≤ M. If T − tp ≥ L0 + 1, then by property (iv) and (5.159)–(5.162), there exists tp+1 ∈ [tp + 1, tp + L + 1] such that ¯ ≤δ u(tp+1 ) − x

(5.167)

166

5 Symmetric Variational Problems

and the assumption made for p also holds for p + 1. If T − tp < L0 + 1, then the construction is completed. Thus by induction we constructed a strictly increasing finite sequence of numbers ¯ ≤ δ, then t0 < · · · < tq from the interval [0, T ] such that if u(0) − x t0 = 0, for every i ∈ {0, . . . , q}, ¯ ≤ δ, u(ti ) − x for every i ∈ {0, . . . , q − 1}, ti+1 − ti ∈ [1, L0 + 1] and T − tq < L0 + 1. If u(T ) − x ¯ > δ, then we set t˜i = ti , i = 0, . . . , q. If u(T ) − x ¯ ≤ δ, then we set t˜i = ti , i = 0, . . . , q − 1, t˜q = T . It is not difficult to see that for all i = 0, . . . , q, ¯ ≤ δ, u(t˜i ) − x

(5.168)

t˜i+1 − t˜i ∈ [1, 2L0 + 2].

(5.169)

for every i ∈ {0, . . . , q − 1},

5.8 Proof of Theorem 5.9

167

In order to complete the proof of the theorem it is sufficient to show that u(t) − x ¯ ≤ , t ∈ [t˜0 , t˜q ]. Assume the contrary. Then there exists τ ∈ [t˜0 , t˜q ]

(5.170)

u(τ ) − x ¯ > .

(5.171)

such that

In view of (5.170), there exists j ∈ {0, . . . , q − 1} such that τ ∈ [t˜j , t˜j +1 ].

(5.172)

Property (iii) and equations (5.166), (5.169), and (5.172) imply that there exists ξ ∈ W 1,1 (t˜j , t˜j +1 ) such that ξ(t˜j ) = u(t˜j ), ξ(t˜j +1 ) = u(t˜j +1 ),

(5.173)

I f (t˜j , t˜j +1 , ξ ) ≤ (t˜j +1 − t˜j )f (x, ¯ y) ¯ + δ1 ,

(5.174)

¯ −y} ¯ = ∅, t ∈ [t˜j , t˜j +1 ] a. e., B(ξ (t), δ1 ) ∩ {y, ξ(t) − x ¯ ≤ δ1 , t ∈ [t˜j , t˜j +1 ].

(5.175) (5.176)

Property (v) and (5.159) imply that |g(ξ(t), ξ (t)) − f (ξ(t), ξ (t))| ≤ δ1 L−1 , t ∈ [t˜j , t˜j +1 ].

(5.177)

It follows from (5.158), (5.169), (5.174), and (5.177) that I g (t˜j , t˜j +1 , ξ ) ≤ I f (t˜j , t˜j +1 , ξ ) + δ1 L−1 (t˜j +1 − t˜j ) ¯ y) ¯ + 2δ1 . ≤ (t˜j +1 − t˜j )f (x,

(5.178)

Equations (5.161), (5.162), (5.173), and (5.178) imply that I g (t˜j , t˜j +1 , u) ≤ I g (t˜j , t˜j +1 , ξ ) + δ ¯ y) ¯ + 2δ1 + δ ≤ (t˜j +1 − t˜j )f (x, ¯ y) ¯ + 3δ1 . ≤ (t˜j +1 − t˜j )f (x,

(5.179)

168

5 Symmetric Variational Problems

Now we obtain another estimation of I g (t˜j , t˜j +1 , u). Let t ∈ [t˜j , t˜j +1 ]. Property (i) and (5.159) imply that ¯ y) ¯ + 0 . if u(t) − x ¯ > /4, then g(u(t), u (t)) > f (x,

(5.180)

Assume that u(t) − x ¯ ≤  ≤ 1. If u (t) ≥ n, then in view of (5.107) and (5.156), g(u(t), u (t)) ≥ ψ(u)u (t) − a ≥ ψ(n)n − a ≥ ψ(n) − a ≥ f (x, ¯ y) ¯ + 1. If u (t) ≤ n, then property (v) and (5.156) imply that |g(u(t), u (t)) − f (u(t), u (t))| ≤ δ1 L−1 and ¯ y) ¯ − δ1 L−1 . g(u(t), u (t)) ≥ f (u(t), u (t)) − δ1 L−1 ≥ f (x, Thus for a. e. t ∈ [t˜j , t˜j +1 ], ¯ y) ¯ − δ1 L−1 g(u(t), u (t)) ≥ f (x,

(5.181)

and equation (5.180) is true. By (5.169) and (5.170), there exists a number S such that [S, S + 1] ⊂ [t˜j , t˜j +1 ],

(5.182)

τ ∈ [S, S + 1].

(5.183)

It follows from (5.158), (5.169), (5.179), (5.181), and (5.182) that I g (S, S + 1, u) = I g (t˜j , t˜j +1 , u) − I g (t˜j , S, u) − I g (S + 1, t˜j +1 , u)

5.8 Proof of Theorem 5.9

169

≤ (t˜j +1 − t˜j )f (x, ¯ y) ¯ + 3δ1 − (S − t˜j )(f (x, ¯ y) ¯ − δ1 L−1 ) ¯ y) ¯ − δ1 L−1 ) ≤ f (x, ¯ y) ¯ + 4δ1 . − (t˜j +1 − S − 1)(f (x,

(5.184)

Property (ii), (5.171), and (5.182)–(5.184) imply that for all t ∈ [S, S + 1] ∩ [τ − γ0 , τ + γ0 ] we have u(t) − u(τ ) ≤ /4.

(5.185)

By (5.171) and (5.185), for all t ∈ [S, S + 1] ∩ [τ − γ0 , τ + γ0 ] we have u(t) − x ¯ > /2.

(5.186)

In view of (5.180) and (5.186), for all t ∈ [S, S + 1] ∩ [τ − γ0 , τ + γ0 ] we have ¯ y) ¯ + 0 . g(u(t), u (t)) > f (x,

(5.187)

E = [S, S + 1] ∩ [τ − γ0 , τ + γ0 ].

(5.188)

Set

By (5.188) and the inequality γ0 < 1/4, mes(E) ≥ γ0 . It follows from (5.181), (5.187), and (5.189) that   g

I (S, S + 1, u) ≥ g(u(t), u (t))dt + E

(5.189)

[S,S+1]\E

g(u(t), u (t))dt

¯ y) ¯ − δ1 L−1 ) ≥ mes(E)(f (x, ¯ y) ¯ + 0 ) + (1 − mes(E))(f (x, ¯ y) ¯ + 0 γ0 /2. ≥ f (x, ¯ y) ¯ + 0 γ0 − δ2 ≥ f (x,

170

5 Symmetric Variational Problems

Together with (5.184) this implies 0 γ0 /2 ≤ 4δ1 . This contradicts (5.157). The contradiction we have reached completes the proof of Theorem 5.9.

5.9 Stability of the Weak Turnpike Theorem 5.11 Let M > x ¯ + y ¯ +1

(5.190)

and  ∈ (0, 1]. Then there exist L > 0, a natural number Q, and a neighborhood U of f in M such that for each T > lQ, each g ∈ U , and each u ∈ W 1,1 (0, T ) such that u(0) ≤ M

(5.191)

and at least one of the conditions below holds: (a) u(T ) ≤ M, I g (0, T , u) ≤ U g (0, T , u(0), u(T )) + M; (b) I g (0, T , u) ≤ U g (0, T , u(0)) + M there exists a sequence of closed intervals [ai , bi ] ⊂ [0, T ], i = 0, . . . , q, where q ≥ 0 is an integer, such that 0 ≤ q ≤ Q, 0 ≤ bi − ai ≤ L, i = 0, . . . , q, ai+1 ≥ bi for all i ∈ {0, . . . , q} \ {q} and q

{t ∈ [0, T ] : u(t) − x ¯ > } ⊂ ∪i=0 [ai , bi ].

5.9 Stability of the Weak Turnpike

171

Proof By Theorem 5.9, there exist δ ∈ (0, 1), L0 > 2 and a neighborhood U0 of f in M such that the following property holds: (c) For each g ∈ U0 , each T > 2L0 , and each u ∈ W 1,1 (0, T ), which satisfies u(s) − x ¯ ≤ δ, s = 0, T , I g (0, T , u) ≤ U g (0, T , u(0), u(T )) + δ we have u(t) − x ¯ ≤ , t ∈ [0, T ]. By Lemma 5.10, there exist L1 > L0 and a neighborhood U ⊂ U0 of f in M such that for each g ∈ U , each T > L and each u ∈ W 1,1 (0, T ) such that u(0) ∈ B(0, M) and at least of the conditions (a), (b) is true the following property holds: (d) For each S ≥ 0 satisfying [S, S + L1 + 1] ⊂ [0, T ], u(S) ≤ M we have min{u(t) − x ¯ : t ∈ [S + 1, S + L1 + 1]} ≤ δ. Set L = 4L1 + 4.

(5.192)

Q > 4 + 2δ −1 M.

(5.193)

T > lQ, g ∈ U ,

(5.194)

Choose an integer

Assume that

172

5 Symmetric Variational Problems

u ∈ W 1,1 (0, T ) satisfies (5.191) and at least one of conditions (a), (b). By conditions (a), (b) and equations (5.191) and (5.194), property (d) holds. This implies that there exists t0 ∈ [0, L1 + 1]

(5.195)

¯ ≤ δ. u(t0 ) − x

(5.196)

such that

Assume that p ≥ 0 is an integer and we already defined numbers t0 < · · · < tp from the interval [0, T ] such that (5.195) and (5.196) are true, for every i ∈ {0, . . . , p}, ¯ ≤ δ, u(ti ) − x for every i ∈ {0, . . . , p} \ {p}, ti+1 − ti ∈ [1, L1 + 1]. (Note that for p = 0 our assumption is true.) In view of (5.190) and (5.197), u(tp )| ≤ M.

(5.197)

If T − tp < L1 + 1, then the construction is completed. If T − tp ≥ L1 + 1, then by property (d) and (5.197), there exists tp+1 ∈ [tp + 1, tp + L1 + 1] such that ¯ ≤δ u(tp+1 ) − x and the assumption made for p also holds for p + 1. Thus by induction we constructed a strictly increasing finite sequence of numbers q {ti }i=0 ⊂ [0, T ] such that t0 ∈ [0, L1 + 1], tq ≥ T − L1 − 1,

(5.198)

5.9 Stability of the Weak Turnpike

173

for every i ∈ {0, . . . , q}, ¯ ≤ δ, u(ti ) − x

(5.199)

ti+1 − ti ∈ [1, L1 + 1].

(5.200)

for every i ∈ {0, . . . , q − 1},

Now we construct a strictly increasing sequence of numbers Si ∈ {t0 , . . . , tq }, i = 0, . . . , p. Set S0 = t0 . Assume that k ≥ 0 is an integer and that we already defined Si ∈ {t0 , . . . , tq }, i = 0, . . . , k. If Sk = tq , then the construction is completed. Assume that Sk < tq . If I g (Sk , tq , u) ≤ U g (Sk , tq , u(Sk ), u(tq )) + δ, then we set Sk+1 = tq and the construction is completed. Assume that I g (Sk , tq , u) > U (Sk , tq , u(Sk ), u(tq )) + δ. There exists j ∈ {0, . . . , q} such that Sk = tj . If I g (Sk , tj +1 , u) > U g (Sk , tj +1 , u(Sk ), u(tj +1 )) + δ, then we set Sk+1 = tj +1 .

174

5 Symmetric Variational Problems

Assume that I g (Sk , tj +1 , u) ≤ U g (Sk , tj +1 , u(Sk ), u(tj +1 )) + δ. Define Sk+1 = min{ti : i ∈ {j + 1, . . . , q} : I g (Sk , ti , u) > U (Sk , ti , u(Sk ), u(ti )) + δ}. Therefore by induction we defined a strictly increasing sequence of numbers Si ∈ {t0 , . . . , tq }, i = 0, . . . , p such that S0 = t0 , Sp = tq ,

(5.201)

for each j ∈ {0, . . . , p − 1} \ {p − 1}, I g (Sj , Sj +1 , u) > U g (Sj , Sj +1 , u(Sj ), u(Sj +1 )) + δ,

(5.202)

for each j ∈ {0, . . . , p − 1}, if i ∈ {0, . . . , q} and Sj < ti < Sj +1 , then I g (Sj , ti , u) ≤ U g (Sj , ti , u(Sj ), u(ti )) + δ.

(5.203)

Conditions (a), (b), and (5.202) imply that M ≥ I g (0, T , u) − U g (0, T , u(0), u(T )) ≥



{I g (Sj , Sj +1 , u) − U (Sj , Sj +1 , u(Sj ), u(Sj +1 )) : j ∈ {0, . . . , p − 1} \ {p − 1}} ≥ (p − 1)δ

and p ≤ 1 + δ −1 M.

(5.204)

5.9 Stability of the Weak Turnpike

175

Assume that j ∈ {0, . . . , p − 1} satisfies Sj +1 − Sj ≥ L.

(5.205)

There exists i ∈ {0, . . . , q} such that Sj +1 = ti . Set S˜j = ti−1 . Equations (5.199), (5.200), and (5.203) imply that S˜j ∈ [Sj +1 − L1 − 1, Sj +1 − 1]

(5.206)

¯ ≤δ u(S˜j ) − x

(5.207)

such that

and I g (Sj , S˜j , u) ≤ U g (Sj , S˜j , u(Sj ), u(S˜j )) + δ.

(5.208)

Property (c) and equations (5.192) and (5.205)–(5.208) imply that u(t) − x ¯ ≤ , t ∈ [Sj , S˜j ].

(5.209)

By (5.201), (5.206), and (5.209), {t ∈ [0, T ] : u(t) − x ¯ > } ⊂ [0, S0 ] ∪ [Sp , T ] ∪ {[Si , Si+1 ] : i ∈ {0, . . . , p − 1}, Si+1 − Si < L} ∪{[Si+1 − 2L1 , Si+1 ] : i ∈ {0, . . . , p − 1}, Si+1 − Si ≥ L}. It is easy to see that the right-hand side of the equation above is a union of a finite number of closed intervals, their lengths do not exceed L (see (5.192) and (5.198)), and in view of (5.204), their number does not exceed 2 + 2p ≤ 4 + 2δ −1 M < Q. Theorem 5.11 is proved.

176

5 Symmetric Variational Problems

5.10 Two Extensions of Theorem 5.9 In this section we prove two extensions of Theorem 5.9. In the first result the turnpike property is established for functions u defined on intervals [0, T ], where T is sufficiently large, which are approximate solutions of the corresponding variational problems of subintervals of [0, T ], which have some fixed length L that does not depend on T . In the second result we analyze the behavior of the derivative u . Theorem 5.12 Let M > x ¯ + y ¯ +1 and  ∈ (0, 1). Then there exist L > 0, δ ∈ (0, ) and a neighborhood U of f in M such that for each T > 2L, each g ∈ U , and each u ∈ W 1,1 (0, T ) such that u(0) ≤ M and at least one of the conditions below holds: (a) u(T ) ≤ M, I g (0, T , u) ≤ U g (0, T , u(0), u(T )) + M; (b) I g (0, T , u) ≤ U g (0, T , u(0)) + M and such that for each S ∈ [0, T − L], I g (S, S + L, u) ≤ U g (S, S + L, u(S), u(S + L)) + δ there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ] such that u(t) − x ¯ ≤ , t ∈ [τ1 , τ2 ]. ¯ ≤ δ, then τ2 = T . Moreover, if u(0) − x ¯ ≤ δ, then τ1 = 0 and if u(T ) − x

5.10 Two Extensions of Theorem 5.9

177

Proof By Theorem 5.9, there exist δ ∈ (0, ), L0 > 2 and a neighborhood U0 of f in M such that the following property holds: (c) For each g ∈ U0 , each T ≥ L0 , and each u ∈ W 1,1 (0, T ) that satisfies u(s) − x ¯ ≤ δ, s = 0, T , I g (0, T , u) ≤ U g (0, T , u(0), u(T )) + δ we have u(t) − x ¯ ≤ , t ∈ [0, T ]. By Theorem 5.11, there exist l > 0, a natural number Q, and a neighborhood U ⊂ U0 of f in M such that the following property holds: (d) For each T > lQ, each g ∈ U , and each u ∈ W 1,1 (0, T ) such that u(0) ≤ M and at least one of conditions (a), (b) holds there exists a sequence of closed intervals [ai , bi ] ⊂ [0, T ], i = 0, . . . , q such that 0 ≤ q ≤ Q, 0 ≤ bi − ai ≤ l, i = 0, . . . , q,

(5.210)

ai+1 ≥ bi for all i ∈ {0, . . . , q} \ {q}

(5.211)

and q

{t ∈ [0, T ] : u(t) − x ¯ > } ⊂ ∪i=0 [ai , bi ].

(5.212)

Fix a number L > 4lQ + 4 + 4L0 .

(5.213)

T > 2L, g ∈ U

(5.214)

Assume that

178

5 Symmetric Variational Problems

and u ∈ W 1,1 (0, T ) satisfies u(0) ≤ M

(5.215)

and at least one of conditions (a), (b) and satisfies for each S ∈ [0, T − L], I g (S, S + L, u) ≤ U g (S, S + L, u(S), u(S + L)) + δ.

(5.216)

Property (d), conditions (a), (b), and equations (5.214), (5.215) imply that there exists a sequence of closed intervals [ai , bi ] ⊂ [0, T ], i = 0, . . . , q such that (5.210)–(5.212) are true. Thus the following property holds: (e) For each S ∈ [0, T − lQ − 1] there exists τS ∈ [S, S + lQ + 1] such that ¯ ≤ δ. u(τS ) − x Property (e) and (5.213), (5.214) imply that there exist τ1 ∈ [0, lQ + 1], τ2 ∈ [T − lQ − 1, T ]

(5.217)

¯ ≤ δ, i = 1, 2. u(τi ) − x

(5.218)

such that

If u(0) − x ¯ ≤ δ, then we set τ1 = 0 and if u(T ) − x ¯ ≤ δ, then we set τ2 = T . In view of (5.213), (5.214), and (5.217), τ2 − τ1 ≥ T − 2(lQ + 2) ≥ 2L − 2(lQ + 2) > 4L0 . Let t ∈ [τ1 , τ2 ].

5.10 Two Extensions of Theorem 5.9

179

Property (e) implies that there exist numbers a, b such that τ1 ≤ a ≤ t ≤ b ≤ τ2 , L0 + 1 ≤ b − a ≤ 2(lQ + 1) + 2L0 , u(a) − x ¯ ≤ δ, u(b) − x ¯ ≤ δ.

(5.219)

It follows from the relation above and (5.216) that I g (a, b, u) ≤ U g (a, b, u(a), u(b)) + δ. Together with property (a) and (5.219) this implies that u(t) − x ¯ ≤ . Theorem 5.12 is proved. Theorem 5.13 Let M > x ¯ + y ¯ + 1,

(5.220)

 ∈ (0, 1) and L1 > 1. Then there exist L > 0, δ ∈ (0, ) and a neighborhood U of f in M such that for each T > 2L, each g ∈ U , and each u ∈ W 1,1 (0, T ) such that u(0) ≤ M and at least one of the conditions below holds: (a) u(T ) ≤ M, I g (0, T , u) ≤ U g (0, T , u(0), u(T )) + M; (b) I g (0, T , u) ≤ U g (0, T , u(0)) + M and such that for each S ∈ [0, T − L], I g (S, S + L, u) ≤ U g (S, S + L, u(S), u(S + L)) + δ there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ]

(5.221)

180

5 Symmetric Variational Problems

such that for each τ ∈ [τ1 , τ2 ] satisfying [τ, τ + L1 ] ⊂ [τ1 , τ2 ] the inequality ¯ u (t) − y} ¯ ≥ }) ≤  mes({t ∈ [τ, τ + L1 ] : min{u (t) − y, ¯ ≤ δ, then holds. Moreover, if u(0) − x ¯ ≤ δ, then τ1 = 0 and if u(T ) − x τ2 = T . Proof By Proposition 2.29, there exist δ0 ∈ (0, ) and a neighborhood U0 of f in M such that the following property holds: (c) For each g ∈ U0 , each x ∈ X, and each y ∈ X satisfying min{y − y, ¯ y + y} ¯ > /2 the inequality g(x, y) ≥ f (x, ¯ y) ¯ + δ0 is valid. By Proposition 5.5, there exist δ1 ∈ (0, δ0 ) and γ > 0 such that the following property holds: (d) For each T > γ and each y, z ∈ B(x, ¯ δ1 ) there exists ξ ∈ W 1,1 (0, T ) such that ξ(0) = y, ξ(T ) = z, ¯ y) ¯ + 8−1 δ0 , I f (0, T , ξ ) ≤ Tf (x, ξ(t) − x ¯ ≤ δ0 , t ∈ [0, T ], ¯ −y} ¯ = ∅, t ∈ [0, T ] a. e. B(ξ (t), δ0 ) ∩ {y, There exists a neighborhood U1 of f in M such that for each g ∈ U1 and each y, z ∈ B(0, M),

5.10 Two Extensions of Theorem 5.9

181

we have |f (y, z) − g(y, z)| ≤ 8−1 (L1 + 1)−1 δ0 .

(5.222)

By Theorem 5.12, there exist L > 0, δ ∈ (0, 8−1 δ1 ) and a neighborhood U ⊂ U0 ∩ U1

(5.223)

of f in M such that the following property holds: (e) For each T > 2L, each g ∈ U , and each u ∈ W 1,1 (0, T ) satisfying (5.221) and at least one of conditions (a), (b) and such that for each S ∈ [0, T − L], I g (S, S + L, u) ≤ U g (S, S + L, u(S), u(S + L)) + δ there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ] such that u(t) − x ¯ ≤ δ1 , t ∈ [τ1 , τ2 ]; ¯ ≤ δ, then τ2 = T . moreover, if u(0) − x ¯ ≤ δ, then τ1 = 0 and if u(T ) − x Assume that T > 2L, g ∈ U , and u ∈ W 1,1 (0, T ) satisfy (5.221), at least one of conditions (a), (b) holds and that for each S ∈ [0, T − L], I g (S, S + L, u) ≤ U g (S, S + L, u(S), u(S + L)) + δ.

(5.224)

Property (e), conditions (a), (b), and equations (5.221) and (5.224) imply that there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ]

(5.225)

u(t) − x ¯ ≤ δ1 , t ∈ [τ1 , τ2 ],

(5.226)

such that

¯ ≤ δ, then τ2 = T . if u(0) − x ¯ ≤ δ, then τ1 = 0 and if u(T ) − x

182

5 Symmetric Variational Problems

Let τ ≥ 0 satisfy [τ, τ + L1 ] ⊂ [τ1 , τ2 ].

(5.227)

We show that ¯ u (t) + y} ¯ ≥ }) ≤ . mes({t ∈ [τ, τ + L1 ] : min{u (t) − y, Assume the contrary. Then ¯ u (t) + y} ¯ ≥ }) > . mes({t ∈ [τ, τ + L1 ] : min{u (t) − y,

(5.228)

Property (d) and equations (5.226) and (5.227) imply that there exists ξ ∈ W 1,1 (τ, τ + L1 ) such that ξ(τ ) = u(τ ), ξ(τ + L1 ) = u(τ + L1 ),

(5.229)

¯ y) ¯ + 8−1 δ0 , I f (τ, τ + L1 , ξ ) ≤ L1 f (x,

(5.230)

B(ξ (t), δ0 ) ∩ {y, ¯ −y} ¯ = ∅, t ∈ [τ, τ + L1 ], a. e.,

(5.231)

ξ(t) − x ¯ ≤ δ0 , t ∈ [τ, τ + L1 ].

(5.232)

It follows from (5.220), (5.222), (5.231), and (5.232) that |f (ξ(t), ξ (t)) − g(ξ(t), ξ (t))| ≤ 8−1 δ0 (L1 + 1)−1

(5.233)

for t ∈ [τ, τ + L1 ] a. e.. By (5.224), (5.230), and (5.233), I g (τ, τ + L1 , u) − δ ≤ I g (τ, τ + L1 , ξ ) ≤ I f (τ, τ + L1 , ξ ) + 8−1 δ0  ≤ L1 f (x, ¯ y) ¯ + 4−1 δ0 , ¯ y) ¯ + 3 · 8−1 δ0 . I g (τ, τ + L1 , u) ≤ L1 f (x,

(5.234)

¯ u (t) + y} ¯ ≥ }. E = {t ∈ [τ, τ + L1 ] : min{u (t) − y,

(5.235)

Set

Property (c) and (5.235) imply that for t ∈ E a. e., ¯ y) ¯ + δ0 . g(u(t), u (t)) ≥ f (x,

(5.236)

5.11 The Turnpike Phenomenon in the Regions Close to the Right End Points

183

Let t ∈ [τ, τ + L1 ] \ E. By (5.222), (5.226), (5.227), and (5.235), |g(u(t), u (t)) − f (u(t), u (t))| ≤ 8−1 δ0 (L1 + 1)−1 and g(u(t), u (t)) ≥ f (u(t), u (t)) − 8−1 δ0 (L1 + 1)−1 ≥ f (x, ¯ y) ¯ − 8−1 δ0 (L1 + 1)−1 . Combined with (5.228) and (5.236) this implies that 

g(u(t), u (t))dt +

I g (τ, τ + L1 , u) = E

 [τ,τ +L1 ]\E

g(u(t), u (t))dt

≥ (f (x, ¯ y) ¯ + δ0 )mes(E) +(f (x, ¯ y) ¯ − 8−1 δ0 (L1 + 1)−1 )(L1 − mes(E)) ¯ y) ¯ + mes(E)δ0 − 8−1 δ0  ≥ L1 f (x, ¯ y) ¯ + 2−1 δ0 . ≥ L1 f (x, This contradicts (5.234). The contradiction we have reached proves Theorem 5.13.

5.11 The Turnpike Phenomenon in the Regions Close to the Right End Points In this section we study the structure of approximate solutions of the problems 

T2 T1

f (x(t), x (t))dt → min,

(PT1 ,T2 ,y )

where T1 < T2 are nonnegative numbers and y, z ∈ X. We prove the following result that shows that the turnpike phenomenon also holds in the regions close to the right end points.

184

5 Symmetric Variational Problems

Theorem 5.14 Let M > x ¯ + y ¯ + 1,

(5.237)

 ∈ (0, 1) and L0 > 1. Then there exist L > L0 , δ ∈ (0, ) and a neighborhood U of f in M such that for each T > 2L, each g ∈ U , and each u ∈ W 1,1 (0, T ) such that u(0) ≤ M

(5.238)

I g (0, T , u) ≤ U g (0, T , u(0)) + M,

(5.239)

such that for each S ∈ [0, T − L], I g (S, S + L, u) ≤ U g (S, S + L, u(S), u(S + L)) + δ

(5.240)

I g (T − L, T , u) ≤ U g (T − L, T , u(T − L)) + δ

(5.241)

and

there exist τ0 ∈ [0, L] such that u(t) − x ¯ ≤ , t ∈ [τ0 , T ] and for each τ ∈ [τ0 , T − L0 ], the inequality ¯ u (t) − y} ¯ ≥ }) ≤  mes({t ∈ [τ, τ + L0 ] : min{u (t) − y, holds. Moreover, if u(0) − x ¯ ≤ δ, then τ0 = 0. Proof Proposition 5.6 implies that there exists 0 ∈ (0, 4−1 )

(5.242)

such that the following property holds: (i) For each S1 ≥ 0 and each S2 ∈ [S1 + 1, S1 + 2], each g ∈ M, each x ∈ W 1,1 (S1 , S2 ) satisfying ¯ y)| ¯ +4 I g (S1 , S2 , x) ≤ 2|f (x, and each t1 , t2 ∈ [S1 , S2 ] satisfying |t1 − t2 | ≤ 0 the inequality x(t1 ) − x(t2 ) ≤ /8 holds.

5.11 The Turnpike Phenomenon in the Regions Close to the Right End Points

185

Proposition 2.29 implies that there exist 1 ∈ (0, 4−1 0 )

(5.243)

and a neighborhood U1 of f in M such that the following property holds: (ii) For each g ∈ U1 and each (x, y) ∈ X × X satisfying max{x − x, ¯ min{y − y, ¯ y + y}} ¯ ≥ /4 the inequality g(x, y) ≥ f (x, ¯ y) ¯ + 21 is valid. By Theorems 5.12 and 5.13, there exist L1 > 2L0 + 4,

(5.244)

δ ∈ (0, 16−1 12 )

(5.245)

and a neighborhood U2 of f in M such that the following property holds: (iii) For each T > 2L1 , each g ∈ U2 , and each u ∈ W 1,1 (0, T ) satisfying (5.238) and at least one of the conditions below: (5.239) is true; u(T ) ∈ B(0, M), I g (0, T , u) ≤ U g (0, T , u(0), u(T )) + M and such that for each S ∈ [0, T − L1 ], I g (S, S + L1 , u) ≤ U g (S, S + L1 , u(S), u(S + L1 )) + δ there exist τ0 ∈ [0, L1 ], τ1 ∈ [T − L1 , T ] such that u(t) − x ¯ ≤ 1 /2, t ∈ [τ0 , τ1 ] and for each τ ∈ [τ0 , τ1 ] satisfying [τ, τ + L0 ] ⊂ [τ0 , τ1 ]

186

5 Symmetric Variational Problems

the inequality ¯ u (t) − y} ¯ ≥ 1 /2}) ≤ 1 /2 mes({t ∈ [τ, τ + L0 ] : min{u (t) − y, ¯ ≤ δ, then holds; moreover, if u(0) − x ¯ ≤ δ, then τ0 = 0 and if u(T ) − x τ1 = T . By Proposition 5.4, there exist γ ∈ (0, 2−1 ), δ˜ ∈ (0, δ)

(5.246)

such that the following property holds: (iv) For each ˜ y ∈ B(x, ¯ δ) there exists ξ ∈ W 1,1 (0, γ ) such that ξ(0) = y, ξ(γ ) = x, ¯ ¯ y) ¯ + 8−1 δ, I f (0, γ , ξ ) ≤ γf (x, ξ(t) − x ¯ ≤ δ, t ∈ [0, γ ], ¯ −y} ¯ = ∅, t ∈ [0, γ ] a. e. B(ξ (t), δ) ∩ {y, Theorem 5.11 implies that there exist l > 0, a natural number Q, and a neighborhood U3 of f in M such that the following property holds: (v) For each T > lQ, each g ∈ U3 , and each u ∈ W 1,1 (0, T ) satisfying (5.238) and (5.239) there exists a sequence of closed intervals [ai , bi ] ⊂ [0, T ], i = 1, . . . , q such that 1 ≤ q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, ai+1 ≥ bi for all i ∈ {1, . . . , q} \ {q} and ˜ ⊂ ∪q [ai , bi ]. {t ∈ [0, T ] : u(t) − x ¯ > δ} i=i Choose a number L > lQ + 4L1 + 4γ + 4.

(5.247)

5.11 The Turnpike Phenomenon in the Regions Close to the Right End Points

187

There exists a neighborhood U4 of f in M such that for each g ∈ U4 and each x, y ∈ X satisfying x, y ≤ M + 1 we have |g(x, y) − f (x, y)| ≤ 16−1 L−1 δ.

(5.248)

U = ∩4i=1 Ui .

(5.249)

Set

Assume that T > 2L, g ∈ U , u ∈ W 1,1 (0, T ) satisfies (5.238), (5.239), equation (5.240) holds for each S ∈ [0, T − L] and (5.241) is true. Property (v) and equations (5.238), (5.239), (5.248)– (5.250) imply that there exists a number S ∈ [T − L, T − γ − 1]

(5.250)

˜ u(S) − x ¯ ≤ δ.

(5.251)

such that

Property (iv) and equations (5.247), (5.250), and (5.251) imply that there exists ξ ∈ W 1,1 (S, S + γ ) such that ξ(S) = u(S), ξ(S + γ ) = x, ¯

(5.252)

¯ y) ¯ + 8−1 δ, I f (S, S + γ , ξ ) ≤ γf (x,

(5.253)

ξ(t) − x ¯ ≤ δ, t ∈ [S, S + γ ],

(5.254)

¯ −y} ¯ = ∅, t ∈ [S, S + γ ] a. e. B(ξ (t), δ) ∩ {y,

(5.255)

By Theorem 5.1, there exists ξ(t) ∈ X, t ∈ (S + γ , T ] such that ξ ∈ W 1,1 (S, T ), ¯ y) ¯ + 8−1 δ, I f (S + γ , T , ξ ) ≤ (T − S − γ )f (x,

(5.256)

ξ(t) − x ¯ ≤ δ, t ∈ [S + γ , T ],

(5.257)

¯ −y}, ¯ t ∈ (S + γ , T ] a. e., ξ (t) ∈ {y,

(5.258)

ξ(T ) = x. ¯

(5.259)

188

5 Symmetric Variational Problems

In view of (5.250), (5.253), and (5.256), ¯ y) ¯ + 4−1 δ. I f (S, T , ξ ) ≤ (T − S)f (x,

(5.260)

It follows from (5.250), (5.253), (5.255), (5.257), and (5.259) that for a. e. t ∈ [S, T ], ξ(t) − x ¯ ≤ δ,

(5.261)

¯ −y} ¯ = ∅. B(ξ (t), δ) ∩ {y,

(5.262)

Equations (5.237), (5.261), and (5.262) imply that ξ(t), ξ (t) ≤ M, t ∈ [S, T ] a. e.

(5.263)

By (5.247)–(5.250) and (5.263), for a. e. t ∈ [S, T ], |f (ξ(t), ξ (t)) − g(ξ(t), ξ (t))| ≤ (16L)−1 δ,

(5.264)

|I g (S, T , ξ ) − I f (S, T , ξ )| ≤ 8−1 δ.

(5.265)

In view of (5.260) and (5.265), ¯ y) ¯ + 4−1 δ + 8−1 δ. I g (S, T , ξ ) ≤ I f (S, T , ξ ) + 8−1 δ ≤ (T − S)f (x,

(5.266)

It follows from (5.248), (5.250), (5.252), and (5.266) that ¯ y) ¯ + δ + (3/8)δ. I g (S, T , u) ≤ I g (S, T , ξ ) + δ ≤ (T − S)f (x,

(5.267)

Set E = {t ∈ [S, T ] : ¯ u (t) + y}} ¯ ≥ /4}. max{u(t) − x, ¯ min{u (t) − y,

(5.268)

Property (ii) and equations (5.249), (5.250), and (5.268) imply that for each t ∈ E, ¯ y) ¯ + 21 . g(u(t), u (t)) ≥ f (x,

(5.269)

By (5.237), (5.247)–(5.250), and (5.268), for each t ∈ [S, T ] \ E, u(t), u (t) ≤ M,

(5.270)

|g(u(t), u (t)) − f (u(t), u (t))| ≤ (16L)−1 δ, g(u(t), u (t)) ≥ f (u(t), u (t)) − (16L)−1 δ.

(5.271)

5.11 The Turnpike Phenomenon in the Regions Close to the Right End Points

189

It follows from (5.245), (5.250), (5.267)–(5.269), and (5.271) that  (f (x, ¯ y) ¯ + 21 )mes(E) ≤ g(u(t), u (t))dt E

 ≤ I g (S, T , u) −

[S,T ]\E

g(u(t), u (t))dt

≤ (T − S)f (x, ¯ y) ¯ + 4−1 δ + 8−1 δ + δ −mes([S, T ] \ E)(f (x, ¯ y) ¯ − (16L)−1 δ) ≤ mes(E)f (x, ¯ y) ¯ + 4−1 δ + 8−1 δ + δ + 8−1 δ and 21 mes(E) ≤ 2−1 δ + δ, mes(E) ≤ δ1−1 < 1 /4.

(5.272)

In view of (5.242), (5.243), (5.268), and (5.272), mes({t ∈ [S, T ] : ¯ u (t) + y} ¯ ≥ /4}) < /4, min{u (t) − y,

(5.273)

mes({t ∈ [S, T ] : u(t) − x ¯ ≥ /4}) < 1 /4.

(5.274)

Assume that τ ∈ [S, T ],

(5.275)

u(τ ) − x ¯ > /2.

(5.276)

T − S > γ + 1.

(5.277)

In view of (5.250),

By (5.275) and (5.277), there exists a number a ≥ 0 such that τ ∈ [a, a + 1] ⊂ [S, T ].

(5.278)

190

5 Symmetric Variational Problems

Equations (5.250), (5.267), (5.269), (5.271), and (5.278) imply that  I g (a, a + 1, u) = I g (S, T , u) −

[S,T ]\[a,a+1]

g(u(t), u (t))dt

≤ (T − S)f (x, ¯ y) ¯ + δ + (3/8)δ − (S − T − 1)f (x, ¯ y) ¯ − 16−1 L−1 δ ≤ f (x, ¯ y) ¯ + δ + δ/2.

(5.279)

Property (i) and (5.279) imply that for all t ∈ [a, a + 1] ∩ [τ − 0 , τ + 0 ] we have u(t) − u(τ ) ≤ /8 and in view of (5.276), u(t) − x ¯ ≥ u(τ ) − x ¯ − u(τ ) − u(t) ≥ /2 − /8 > /4.

(5.280)

In view of (5.242), (5.278), and (5.279), mes([a, a + 1] ∩ [τ − 0 , τ + 0 ]) > 0 .

(5.281)

Equations (5.243), (5.278), (5.280), and (5.281) imply that mes({t ∈ [S, T ] : u(t) − x ¯ ≥ /4}) > 0 > 1 . This contradicts (5.274). The contradiction we have reached proves that u(τ ) − x ¯ ≤ /2 for all τ ∈ [S, T ].

(5.282)

Property (iii) and equations (5.238)–(5.240), (5.246), (5.248)–(5.251) imply that there exists τ0 ∈ [0, L1 ]

(5.283)

u(t) − x ¯ ≤ 1 /2, t ∈ [τ0 , S]

(5.284)

such that

5.11 The Turnpike Phenomenon in the Regions Close to the Right End Points

191

and for all τ ∈ [τ0 , S] satisfying [τ, τ + L0 ] ⊂ [τ0 , S] we have mes({t ∈ [τ, τ0 ] : ¯ u (t) + y} ¯ ≥ 1 /2}) < 1 /2 min{u (t) − y,

(5.285)

and if u(0) − x ¯ ≤ δ, then τ0 = 0. By (5.282)–(5.284), u(t) − x ¯ ≤ , t ∈ [τ0 , T ]. Let τ ∈ [τ0 , T − L0 ]. In view of (5.273) and (5.285), mes({t ∈ [τ, τ0 ] : ¯ u (t) + y} ¯ ≥ }) ≤ /4 + 1 /2 < . min{u (t) − y, This completes the proof of Theorem 5.14.

(5.286)

Chapter 6

Infinite Dimensional Optimal Control

In this chapter we study the turnpike phenomenon for continuous-time optimal control problems in infinite dimensional spaces that have a certain symmetric property. To have the turnpike property means, roughly speaking, that the approximate solutions of the problems are determined mainly by the integrand and are essentially independent of the choice of interval and endpoint conditions, except in regions close to the endpoints. It is shown that for our class of problems the turnpike is a minimizer of the integrand. We also study the structure of approximate solutions on large intervals in the regions close to the right end points. The results of this chapter are new. They will be obtained for two large classes of problems that will be treated simultaneously.

6.1 Preliminaries We begin with the description of the first class of problems. Let (E,  · ) be a Banach space and (F,  · ) be a normed space. We suppose that A is a nonempty subset of E and U : A → 2F is a point to set mapping with a graph M = {(x, u) : x ∈ A, u ∈ U (x)}.

(6.1)

We suppose that M is a Borel measurable subset of E × F , G : F → E is a Borelian function, and a linear operator A : D(A) → E generates a C0 semigroup eAt on E. Let f : M → R 1 be a bounded from below Borelian function. Let 0 ≤ T1 < T2 . We consider the following equation: x (t) = Ax(t) + G(u(t)), t ∈ [T1 , T2 ].

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 A. J. Zaslavski, Turnpike Phenomenon and Symmetric Optimization Problems, Springer Optimization and Its Applications 190, https://doi.org/10.1007/978-3-030-96973-8_6

(6.2)

193

194

6 Infinite Dimensional Optimal Control

A pair of functions x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is called a (mild) solution of (6.2) if x : [T1 , T2 ] → E is a continuous function and u : [T1 , T2 ] → F is a Lebesgue measurable function, x(t) ∈ A, t ∈ [T1 , T2 ], u(t) ∈ U (x(t)), t ∈ [T1 , T2 ] almost everywhere (a.e.),

(6.3) (6.4)

G(u(s)), s ∈ [T1 , T2 ] is Bochner integrable and  x(t) = eA(t−T1 ) x(T1 ) +

t

eA(t−s) G(u(s))ds, t ∈ [T1 , T2 ].

(6.5)

T1

The set of all pairs (x, u) that are solutions of (6.2) is denoted by X(T1 , T2 ). Let 0 ≤ T1 < T2 , x : [T1 , T2 ] → E, u : [T1 , T2 ] → F . Set y(t) = x(t + T1 ), v(t) = u(t + T1 ), t ∈ [0, T2 − T1 ]. It was shown in [170] that if (y, v) ∈ X(0, T2 − T1 ), then (x, u) ∈ X(T1 , T2 ). Also it was shown there that if (x, u) ∈ X(T1 , T2 ), then (x, u) ∈ X(τ, T2 ) for all τ ∈ (T1 , T2 ]. Let 0 ≤ T1 < T2 < T3 , (x1 , u1 ) ∈ X(T1 , T2 ), (x2 , u2 ) ∈ X(T2 , T3 ), x1 (T2 ) = x2 (T2 ). It was shown in [170] that (x, u) ∈ X(T1 , T3 ), where for each t ∈ [T1 , T2 ], x(t) = x1 (t), u(t) = u1 (t), and for each t ∈ (T2 , T3 ], x(t) = x2 (t), u(t) = u2 (t), t ∈ (T2 , T3 ]. Now we describe the second class of problems. Let (E, ·, ·)E be a Hilbert space equipped with an inner product ·, ·E that induces the norm ·E and let (F, ·, ·F ) be a Hilbert space equipped with an inner product ·, ·F that induces the norm ·F . For simplicity, we set ·, ·E = ·, ·,  · E =  · , ·, ·F = ·, ·,  · F =  · , if E, F are understood. We suppose that A is a nonempty subset of E and U : A → 2F is a point to set mapping with a graph M = {(x, u) : x ∈ A, u ∈ U (x)}.

6.1 Preliminaries

195

We suppose that M is a Borel measurable subset of E × F and that a linear operator A : D(A) → E generates a C0 semigroup S(t) = eAt , t ≥ 0 on E. As usual, we denote by S(t)∗ the adjoint operator of S(t). Then S(t)∗ , t ∈ [0, ∞) is C0 semigroup and its generator is the adjoint operator A∗ of A. The domain D(A∗ ) is a Hilbert space equipped with the graph norm  · D(A∗ ) of the operator A∗ : z2D(A∗ ) := z2E + A∗ z2E , z ∈ D(A∗ ). Let D(A∗ ) be the dual space of D(A∗ ) with the pivot space E. In particular, E1d := D(A∗ ) ⊂ E ⊂ D(A∗ ) = E−1 . (Here we use the notation of Section 1.7. of [170].) Let f : M → R 1 be Borelian bounded from below function, B ∈ L(F, E−1 ) is an admissible control operator for eAt , t ≥ 0, G = B. Let 0 ≤ T1 < T2 . We consider the following equation: x (t) = Ax(t) + Bu(t), t ∈ [T1 , T2 ] a. e.

(6.6)

A pair of functions x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is called a (mild) solution of (6.6) if x : [T1 , T2 ] → E is a continuous function, u : [T1 , T2 ] → F is a Lebesgue measurable function, u ∈ L2 ([T1 , T2 ]; F ), x(t) ∈ A, t ∈ [T1 , T2 ], u(t) ∈ U (x(t)), t ∈ [T1 , T2 ] a.e. and for each t ∈ [T1 , T2 ],  x(t) = e

A(t−T1 )

x(T1 ) +

t

eA(t−s) Bu(s)ds

(6.7)

T1

in E−1 . The set of all pairs (x, u) that are solutions of (6.6) is denoted by X(T1 , T2 ). From now these two classes of problems will be treated simultaneously. Let T1 ≥ 0. A pair of functions x : [T1 , ∞) → E, u : [T1 , ∞) → F is called a (mild) solution of the system x (t) = Ax(t) + G(u(t)), t ∈ [T1 , ∞)

196

6 Infinite Dimensional Optimal Control

if for every T2 > T1 , x : [T1 , T2 ] → E, u : [T1 , T2 ] → F is a solution of (6.2) for the first problem and is a solution of (6.6) for the second problem. The set of all such pairs (x, u) that are solutions of the equation above is denoted by X(T1 , ∞). A function x : I → E, where I is either [T1 , T2 ] or [T1 , ∞) (0 ≤ T1 < T2 ) is called a trajectory if there exists a Lebesgue measurable function u : I → F (referred to as a control) such that (x, u) ∈ X(T1 , T2 ) or (x, u) ∈ X(T1 , ∞) respectively. Let T2 > T1 ≥ 0, (x, u) ∈ X(T1 , T2 ). Define  I f (T1 , T2 , x, u) =

T2

f (x(t), u(t))dt T1

which is well-defined but can be ∞. Proposition 4.3 implies that there exist M∗ ≥ 1, ω∗ ∈ R 1 such that eAt  ≤ M∗ eω∗ t , t ∈ [0, ∞).

(6.8)

Let a0 , K0 > 0 and let ψ : [0, ∞) → [0, ∞) be an increasing function such that ψ(t) → ∞ as t → ∞.

(6.9)

We suppose that for the first class of problems the function f satisfies f (x, u) ≥ −a0 + ψ(G(u))(G(u))

(6.10)

for each (x, u) ∈ M and for the second class of problems f satisfies f (x, u) ≥ −a0 + K0 u2 , (x, u) ∈ M.

(6.11)

Let T2 > T1 ≥ 0 and y, z ∈ A. We consider the following problems: I f (T1 , T2 , x, u) → min, (x, u) ∈ X(T1 , T2 ), x(T1 ) = y, x(T2 ) = z,

(P1 )

I f (T1 , T2 , x, u) → min, (x, u) ∈ X(T1 , T2 ), x(T1 ) = y,

(P2 )

I f (T1 , T2 , x, u) → min, (x, u) ∈ X(T1 , T2 ).

(P3 )

We consider functionals of the form I f (T1 , T2 , x, u), where 0 ≤ T1 < T2 and (x, u) ∈ X(T1 , T2 ). For each pair of numbers T2 > T1 ≥ 0 and each pair of points y, z ∈ A we define U f (T1 , T2 , y, z) = inf{I f (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 ), x(T1 ) = y, x(T2 ) = z},

(6.12)

6.1 Preliminaries

197

σ f (T1 , T2 , y) = inf{U f (T1 , T2 , y, h) : h ∈ A},

(6.13)

σ f (T1 , T2 ) = inf{U f (T1 , T2 , h, y) : h, y ∈ A}.

(6.14)

(Recall that the infimum over an empty set is ∞.) For each x ∈ E, each u ∈ F , and each r > 0 set BE (x, r) = {y ∈ E : x − y ≤ r}, BF (u, r) = {v ∈ F : u − v ≤ r}. We suppose that x¯ ∈ E, u¯ ∈ F, (x, ¯ u), ¯ (x, ¯ −u) ¯ ∈ M, f (x, ¯ u) ¯ = f (x, ¯ −u) ¯ = inf{f (x, u) : (x, u) ∈ M}.

(6.15) (6.16)

Let L > 0. Denote by AL the set of all z ∈ A for which there exist τ ∈ (0, L] and (x, u) ∈ X(0, τ ) such that x(0) = z, x(τ ) = x, ¯ I f (0, τ, x, u) ≤ L. L the set of all z ∈ A for which there exist τ ∈ (0, L] and (x, u) ∈ Denote by A X(0, τ ) such that x(0) = x, ¯ x(τ ) = z, I f (0, τ, x, u) ≤ L. We suppose that the following assumptions hold. (A1)

There exists r∗ ∈ (0, 1] such that for each x ∈ E satisfying x − x ¯ ≤ r∗ we have x ∈ A and (x, u), ¯ (x, −u) ¯ ∈ M.

(A2)

For each  > 0 there exists 1 ∈ (0, ) such that for each (x, v) ∈ M satisfying ¯ 1 ) ∪ BF (−u, ¯ 1 ) x − x ¯ ≤ 1 , v ∈ BF (u, the inequality f (x, v) ≤ f (x, ¯ u) ¯ + holds.

198

(A3)

6 Infinite Dimensional Optimal Control

For each  > 0 there exists δ > 0 such that for each (y, v) ∈ M satisfying f (y, v) ≤ inf(f ) + δ the equations ¯ −u} ¯ = ∅ y − x ¯ ≤ , BF (v, ) ∩ {u, are true.

Assumption (A2) means that the function f is continuous at the point (x, ¯ u) ¯ while assumption (A3) means that the minimization problem f (x, y) → min, (x, y) ∈ M is well-posed. According to the results of Chapter 2 this well-posedness property holds for most symmetric objective functions. We also suppose that G(−u) ¯ = −G(u) ¯

(6.17)

(of course, for second class of problems (6.17) is always true), x¯ ∈ D(A), Ax¯ = 0, G(u) ¯ ∈ E, AG(u) ¯ = 0.

(6.18)

Proposition 4.7 implies that eAt x¯ ∈ D(A) for all t ≥ 0, ¯ = AeAt x¯ = eAt Ax¯ = 0 for all t ≥ 0, (d/dt)(eAt x)  e x¯ − e x¯ = At

As

t

eAr xdr ¯ = 0 for all t > s ≥ 0.

s

Then eAt x¯ = x¯ for all t ≥ 0.

(6.19)

Analogously we can show that ¯ = G(u) ¯ for all t ≥ 0. eAt G(u) in E. It is not difficult to see that the following proposition holds.

(6.20)

6.1 Preliminaries

199

Proposition 6.1 Let τ ∈ (0, 1] satisfy τ G(u) ¯ ≤ r∗ and u(t) = u, ¯ t ∈ [0, τ ], u(t) = −u, ¯ t ∈ (τ, 2τ ]. Then there exists a solution x(·) of the equation x (t) = Ax(t) + G(u(t)), t ∈ [0, 2τ ] a. e. , x(0) = x¯ and it satisfies x(t) = x¯ + tG(u), ¯ t ∈ [0, τ ], x(t) = x¯ + (2τ − t)G(u), ¯ t ∈ [τ, 2τ ] and for every t ∈ [0, 2τ ], x(t) − x ¯ ≤ τ G(u) ¯ ≤ r∗ . Theorem 6.2 Let T > 0. Then ¯ = U f (0, T , x, ¯ x) ¯ = Tf (x, ¯ u). ¯ σ f (0, T ) = σ f (0, T , x) Moreover, for each  > 0 there exists (x, u) ∈ X(0, T ) such that x(0) = x(T ) = x, ¯ u(t) ∈ {u, ¯ −u}, ¯ t ∈ [0, T ] a. e., x(t) − x ¯ ≤ , t ∈ [0, T ] and ¯ u) ¯ + . I f (0, T , x, u) ≤ Tf (x, Proof Let  ∈ (0, 1]. By (A2), there exists 1 ∈ (0, min{, r∗ })

(6.21)

200

6 Infinite Dimensional Optimal Control

such that for each ¯ 1 ) x ∈ A ∩ BE (x, we have f (x, u), ¯ f (x, −u) ¯ ≤ f (x, ¯ u) ¯ + /T

(6.22)

Choose a natural number n such that ¯ + 1)−1 1 . T n−1 < (G(u)

(6.23)

τ = T (2n)−1 .

(6.24)

Set

Proposition 6.1 and equations (6.23) and (6.24) imply that there exists (x, ˜ u) ˜ ∈ X(0, 2τ ) such that x(0) ˜ = x, ¯ x(2τ ˜ ) = x, ¯

(6.25)

u(t) ˜ = u, ¯ t ∈ [0, τ ], u(t) ˜ = −u, ¯ t ∈ (τ, 2τ ],

(6.26)

x(t) ˜ − x ¯ ≤ τ G(u) ¯ < 1 , t ∈ [0, 2τ ].

(6.27)

For every i ∈ {0, . . . , n − 1} define x(iτ + t) = x(t), ˜ u(iτ + t) = u(t), ˜ t ∈ [0, 2τ ].

(6.28)

By (6.24)–(6.28), (x, u) ∈ X(0, T ), x(0) = x(T ) = x, ¯

(6.29)

x(t) − x ¯ < 1 ,

(6.30)

u(t) ∈ {u, ¯ −u}. ¯

(6.31)

for all t ∈ [0, T ],

for a. e. t ∈ [0, T ],

In view of (6.21), (6.30), and (6.31), for a. e. t ∈ [0, T ], f (x(t), u(t)) ≤ f (x, ¯ u) ¯ + T −1

6.2 The First Turnpike Result

201

and ¯ u) ¯ + . I f (0, T , x, u) ≤ Tf (x, Since  is any positive number we conclude that Theorem 6.2 is true. We suppose that there exists a number bf > 0 and the following assumptions hold. (A4)

For each  > 0 there exists δ > 0 such that for each zi ∈ A, i = 1, 2 satisfying zi − x ¯ ≤ δ, i = 1, 2 there exist τ ∈ (0, bf ] and (x, u) ∈ X(0, τ ), which satisfies x(0) = z1 , x(τ ) = z2 , x(t) − x ¯ ≤ , t ∈ [0, τ ], for a. e. t ∈ (0, τ ], ¯ −u} ¯ = ∅ BF (u(t), ) ∩ {u, and ¯ u) ¯ + . I f (0, τ, x, u) ≤ τf (x,

6.2 The First Turnpike Result Theorem 6.3 Let L > 0. Then the following assertions hold. 1. For each z ∈ AL and each T ≥ L, ¯ u) ¯ + L + L|f (x, ¯ u)| ¯ + 1. σ f (0, T , z) ≤ Tf (x, L , and each T ≥ 2L, 2. For each z1 ∈ AL , each z2 ∈ A ¯ u) ¯ + 2L(1 + |f (x, ¯ u)|) ¯ + 1. U f (0, T , z1 , z2 ) ≤ Tf (x, Proof Let us prove Assertion 1. Let z ∈ AL , T ≥ L, and  ∈ (0, 1]. By the definition of AL , there exist τ1 ∈ (0, L] and (x1 , u1 ) ∈ X(0, τ1 ) such that ¯ I f (0, τ, x1 , u1 ) ≤ L. x1 (0) = z, x(τ1 ) = x,

(6.32)

Theorem 6.2 implies that there exists (x2 , u2 ) ∈ X(τ1 , T ) such that ¯ x2 (T ) = x, ¯ x2 (τ1 ) = x,

(6.33)

¯ u) ¯ + . I f (τ1 , T , x2 , u2 ) ≤ (T − τ1 )f (x,

(6.34)

202

6 Infinite Dimensional Optimal Control

Define (x, u) ∈ X(0, T ) as follows: x(t) = x1 (t), u(t) = u1 (t), t ∈ [0, τ1 ],

(6.35)

x(t) = x2 (t), u(t) = u2 (t), t ∈ (τ1 , T ].

(6.36)

By (6.32), (6.35), and (6.36), σ f (0, T , z) ≤ I f (0, τ1 , x, u) + I f (τ1 , T , x, u) ¯ u) ¯ +1 ≤ L + (T − τ1 )f (x, ≤ Tf (x, ¯ y) ¯ + L + L|f (x, ¯ y) ¯ + 1. Assertion 1 is proved. L , T ≥ 2L. By the definition of Let us prove Assertion 2. Let z1 ∈ AL , z2 ∈ A  AL and AL , there exist τ1 ∈ (0, L] and (x1 , u1 ) ∈ X(0, τ1 ) such that x1 (0) = z1 , x(τ1 ) = x, ¯

(6.37)

I f (0, τ1 , x1 , u1 ) ≤ L

(6.38)

and τ2 ∈ (0, L] and (x2 , u2 ) ∈ X(0, τ2 ) such that x2 (0) = x, ¯ x(τ2 ) = z2 ,

(6.39)

I f (0, τ2 , x2 , u2 ) ≤ L.

(6.40)

Theorem 6.2 implies that there exists (x3 , u3 ) ∈ X(0, T − τ1 − τ2 ) such that x3 (0) = x¯ = x3 (T − τ1 − τ2 ), ¯ u) ¯ + 1. I f (0, T − τ1 − τ2 , x3 , u3 ) ≤ (T − τ1 − τ2 )f (x,

(6.41) (6.42)

Define (x, u) ∈ X(0, T ) as follows: x(t) = x1 (t), u(t) = u1 (t), t ∈ [0, τ1 ],

(6.43)

x(t) = x3 (t − τ1 ), u(t) = u3 (t − τ1 ), t ∈ (τ1 , T − τ2 ],

(6.44)

x(t) = x2 (t − (T − τ2 )), u(t) = u2 (t − (T − τ2 )), t ∈ (T − τ2 , T ].

(6.45)

By (6.37), (6.39), (6.41) and (6.43)–(6.45), (x, u) ∈ X(0, T ), x(0) = z1 , x(T ) = z2 .

(6.46)

6.2 The First Turnpike Result

203

It follows from (6.38), (6.40), (6.42)–(6.46) that U f (0, T , z1 , z2 ) ≤ I f (0, T , x, u) ¯ u) ¯ +1 ≤ 2L + (T − τ1 − τ2 )f (x, ≤ Tf (x, ¯ u) ¯ + 2L + 2Lf (x, ¯ y) ¯ + 1. Assertion 2 is proved. This completes the Proof of Theorem 6.3. The next theorem is our first turnpike result. It shows that for approximate solutions (x, u) of our optimal control problems on intervals [0, T ], where T is sufficiently large, the Lebesgue measure of all points t ∈ [0, T ] such that (x(t), u(t)) does not belong to an -neighborhood of the set {(x, ¯ u), ¯ (x, ¯ −u)} ¯ does not exceed a constant L, which depends only on  and does not depend on T . In the literature this property is known as the weak turnpike property. Theorem 6.4 Let  > 0 and L0 , M0 > 0. Then there exists L1 > L0 such that the following assertions hold. L0 and each (x, u) ∈ X(0, T ) 1. For each T ≥ L1 , each y ∈ AL0 , each z ∈ A satisfying x(0) = y, x(T ) = z, I f (0, T , x, u) ≤ U f (0, T , y, z) + M0 the inequality mes({t ∈ [0, T ] : max{x(t) − x, ¯ min{u(t) − u, ¯ u(t) + u}} ¯ > }) ≤ L1 is true. 2. For each T ≥ L1 , each y ∈ AL0 , and each (x, u) ∈ X(0, T ) satisfying x(0) = y, I f (0, T , x, u) ≤ σ f (0, T , y) + M0 the inequality mes({t ∈ [0, T ] : max{x(t) − x, ¯ min{u(t) − u, ¯ u(t) + u}} ¯ > }) ≤ L1 is true.

204

6 Infinite Dimensional Optimal Control

Proof Assumption (A3) implies that there exists δ ∈ (0, ) such that the following property holds: for each (x, u) ∈ M satisfying f (x, u) ≤ f (x, ¯ u) ¯ +δ we have x − x ¯ ≤ and min{u − u, ¯ u + u} ¯ ≤ .

(6.47)

Choose L1 ≥ 2bf + δ −1 (M0 + 1 + 2L0 (1 + |f (x, ¯ u)|)) ¯ + L0 .

(6.48)

Assume that T ≥ L1 , y ∈ AL0 , (x, u) ∈ X(0, T ), x(0) = y, L0 z∈A and at least one of the following conditions holds: (i) x(T ) = z, I f (0, T , x, u) ≤ U f (0, T , y, z) + M0 ; (ii) I f (0, T , x, u) ≤ σ f (0, T , y) + M0 . In the case (ii) Theorem 6.3 implies that ¯ u) ¯ + L0 (1 + |f (x, ¯ u)|) ¯ + 1. I f (0, T , x, u) ≤ M0 + Tf (x, In the case (i) it follows from Theorem 6.3 that ¯ u) ¯ + 2L0 (1 + |f (x, ¯ u)|) ¯ + 1. I f (0, T , x, u) ≤ M0 + Tf (x,

(6.49)

In both cases (6.49) is true. Set E = {t ∈ [0, T ] : f (x(t), u(t)) > f (x, ¯ u) ¯ + δ}.

(6.50)

6.2 The First Turnpike Result

205

Equations (6.49) and (6.50) imply that ¯ u)|) ¯ + Tf (x, ¯ u) ¯ M0 + 1 + 2L0 (1 + |f (x,

 =

≥ I (0, T , x, u)  f (x(t), u(t))dt + f (x(t), u(t))dt [0,T ]\E

E

≥ Tf (x, ¯ u) ¯ + δmes(E). Combined with (6.48) this implies that ¯ u)|)) ¯ ≤ L1 . mes(E) ≤ δ −1 (M0 + 1 + 2L0 (1 + |f (x,

(6.51)

Assume that t ∈ [T1 , T2 ] \ E. By (6.47), (6.50), and the inclusion above, f (x(t), u(t)) ≤ f (x, ¯ u) ¯ + δ, x(t) − x ¯ ≤ , min{u(t) − u ¯ u(t) + u} ¯ ≤ . Combined with (6.51) this completes the Proof of Theorem 6.4. Proposition 6.5 Let  > 0. Then there exist δ > 0 such that for each T > 2bf and each ¯ δ) y, z ∈ BE (x, there exists (x, u) ∈ X(0, T ) such that x(0) = y, x(T ) = z, ¯ u) ¯ + , I f (0, T , x, u) ≤ Tf (x, B(u(t), ) ∩ {u, ¯ −u} ¯ = ∅, t ∈ [0, T ] a. e. x(t) − x ¯ ≤ , t ∈ [0, T ].

206

6 Infinite Dimensional Optimal Control

Proof Assumption (A2) implies that there exists 1 ∈ (0, /4)

(6.52)

such that the following property holds: (i) For each pair of points y ∈ BE (x, ¯ 1 ) ∩ A, v ∈ BF (u, ¯ 1 ) ∪ BF (−u, ¯ 1 ) we have f (y, v) ≤ f (x, ¯ u) ¯ + 8−1 (bf + 1)−1 . By Assumption (A4), there exists δ ∈ (0, min{1 , r∗ }) such that the following property holds: ¯ ≤ δ, i = 1, 2 there exist (ii) For each zi ∈ A, i = 1, 2 satisfying zi − x τ ∈ (0, bf ] and (x, u) ∈ X(0, τ ) that satisfies x(0) = z1 , x(τ ) = z2 , x(t) − x ¯ ≤ 1 , t ∈ [0, τ ], for a. e. t ∈ (0, τ ], ¯ −u} ¯ = ∅. BF (u(t), 1 ) ∩ {u, Assume that T > 2bf and ¯ δ). y, z ∈ BE (x,

(6.53)

y, z ∈ A,

(6.54)

(y, u), ¯ (y, −u), ¯ (z, u), ¯ (z, −u) ¯ ∈ M.

(6.55)

In view of (A1) and (6.53),

Property (ii) and equations (6.53) and (6.54) imply that there exist τ1 , τ2 ∈ (0, bf ], and (xi , ui ) ∈ X(0, τi ), i = 1, 2

6.2 The First Turnpike Result

207

which satisfies ¯ x1 (0) = y1 , x1 (τ1 ) = x,

(6.56)

¯ x2 (τ2 ) = z, x2 (0) = x,

(6.57)

xi (t) − x ¯ ≤ 1 , t ∈ [0, τi ], i = 1, 2,

(6.58)

¯ −u} ¯ = ∅, BF (ui (t), 1 ) ∩ {u,

(6.59)

for a. e. t ∈ [0, τi ],

i = 1, 2. Theorem 6.2 implies that there exists (x3 , u3 ) ∈ X(0, T − τ1 − τ2 ) such that x3 (0) = x¯ = x3 (T − τ1 − τ2 ),

(6.60)

for a. e. t ∈ [0, T − τ1 − τ2 ], u3 (t) ∈ {u, ¯ −u}, ¯

(6.61)

¯ ≤ 1 , t ∈ [0, T − τ1 − τ2 ], x3 (t) − x

(6.62)

¯ u) ¯ + 1 . I f (0, T − τ1 − τ2 , x3 , u3 ) ≤ (T − τ1 − τ2 )f (x,

(6.63)

Define (x, u) ∈ X(0, T ) as follows: x(t) = x1 (t), u(t) = u1 (t), t ∈ [0, τ1 ],

(6.64)

x(t) = x3 (t − τ1 ), u(t) = u3 (t − τ1 ), t ∈ (τ1 , T − τ2 ],

(6.65)

x(t) = x2 (t − (T − τ2 )), u(t) = u2 (t − (T − τ2 )), t ∈ (T − τ2 , T ].

(6.66)

By (6.58)–(6.66), (x, u) ∈ X(0, T ), x(0) = y, x(T ) = z, I f (0, T , x, u) = I f (0, τ1 , x1 , u1 ) + I f (0, τ2 , x2 , u2 ) + I f (0, T − τ1 − τ2 , x3 , u3 ) ¯ u) ¯ + 1 . ≤ I f (0, τ1 , x1 , u1 ) + I f (0, τ2 , x2 , u2 ) + (T − τ1 − τ2 )f (x,

(6.67)

208

6 Infinite Dimensional Optimal Control

In view of (6.58) and (6.59) and property (i), for all t ∈ [0, τi ] and i = 1, 2, ¯ u) ¯ + 8−1 (bf + 1)−1 . f (xi (t), ui (t)) ≤ f (x, This implies that ¯ u) ¯ + 8−1 , i = 1, 2. I f (0, τi , xi , ui ) ≤ τi f (x, Together with (6.67) this implies that ¯ u) ¯ + 4−1  + 1 ≤ Tf (x, ¯ u) ¯ + . I f (0, T , x, u) ≤ Tf (x, By (6.61), (6.62), (6.64)–(6.66), (6.68), and (6.69), for all t ∈ [0, T ], x(t) − x ¯ ≤ , t ∈ [0, T ] and for a. e. t ∈ [0, T ], ¯ −u} ¯ = ∅. BF (u(t), ) ∩ {u, Proposition 6.5 is proved.

6.3 The Space of Integrands Denote by M the set of all Borelian bounded from below functions g : [0, ∞) × M → R 1 such that for the first class of problems g(t, x, u) ≥ ψ(G(u))G(u) − a0

(6.68)

(see (6.9) and (6.10)) for each (t, x, u) ∈ [0, ∞) × M and for the second class of problems g(t, x, u) ≥ ψ(u)u2 − a0 for each (t, x, u) ∈ [0, ∞) × M. For each T2 > T1 ≥ 0, each g ∈ M, and each (x, u) ∈ X(T1 , T2 ) define  I g (T1 , T2 , x, u) =

T2

g(t, x(t), u(t))dt. T1

(6.69)

6.3 The Space of Integrands

209

For each g ∈ M, each pair of numbers T2 > T1 ≥ 0, and each pair of points y, z ∈ A we define U g (T1 , T2 , y, z) = inf{I g (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 ), x(T1 ) = y, x(T2 ) = z}, σ g (T1 , T2 , y) = inf{I g (T1 , T2 , x, u) : (x, u) ∈ X(T1 , T2 ), x(T1 ) = y}. For the second class of problems we assume that the operator B ∈ L(F, E−1 ) is an admissible control operator for the semigroup eAt , t ≥ 0. The following auxiliary result is an important ingredient of proofs of our turnpike results. Proposition 6.6 Let M1 ,  > 0 and 0 < τ0 < τ1 . Then there exists δ > 0 such that for each pair of numbers T1 , T2 satisfying 0 ≤ T1 , T2 ∈ [T1 + τ0 , T1 + τ1 ], each g ∈ M, each (x, u) ∈ X(T1 , T2 ) satisfying I g (T1 , T2 , x, u) ≤ M1 and each t1 , t2 ∈ [T1 , T2 ] satisfying ¯ ≤δ 0 < t2 − t1 ≤ δ, x(t1 ) − x the inequality ¯ ≤ x(t2 ) − x holds. Proof Let us consider the first class of integrands. Choose a constant Λ > 1 such that Λ−1 (M1 + a0 + 1)(M∗ + 1)e|w∗ | < /4

(6.70)

(see (6.8)). By (6.9), (6.68), and (6.69), there exists c0 > 0 such that for each g ∈ M and each (t, x, u) ∈ [0, ∞) × M satisfying G(u) ≥ c0

210

6 Infinite Dimensional Optimal Control

we have g(t, x, u) ≥ ΛG(u).

(6.71)

Choose a number δ ∈ (0, 1) such that δ(M∗ + 1)e|w∗ | (1 + c0 ) < /4.

(6.72)

Assume that g ∈ M, numbers T2 > T1 ≥ 0 satisfy T2 ∈ [T1 + τ0 , T1 + τ1 ],

(6.73)

I g (T1 , T2 , x, u) ≤ M1

(6.74)

and that t1 , t2 ∈ [T1 , T2 ] satisfy 0 < t2 − t1 ≤ δ,

(6.75)

¯ ≤ δ. x(t1 ) − x

(6.76)

We show that ¯ ≤ . x(t2 ) − x For every t ∈ [t1 , t2 ] we have  x(t) = eA(t−t1 ) x(t1 ) +

t

eA(t−s) G(u(s))ds.

(6.77)

t1

Let t ∈ [t1 , t2 ]. By (6.8), (6.19), and (6.74)–(6.77), x(t) − x ¯  ≤ eA(t−t1 ) x(t1 ) − eA(t−t1 ) x ¯ +

t

eA(t−s) G(u(s))ds

t1

≤ x(t1 ) − xM ¯ ∗e

|w∗ |δ

+ M∗ e

≤ δM∗ e|w∗ |δ + M∗ e|w∗ |δ ≤ δM∗ e

|w∗ |

+ M∗ e

|w∗ |

|w∗ |δ



t

G(u(s))ds

t1



t

G(u(s))ds

t1



t

t1

G(u(s))ds.

(6.78)

6.3 The Space of Integrands

211

Set Ω = {s ∈ [t1 , t] : G(u(s)) ≥ c0 }.

(6.79)

By (6.68), (6.71), (6.74), and (6.79), 

t

 G(u(s))ds =

 G(u(s))ds +

Ω

t1

 ≤ c0 (mes([t1 , t] \ Ω) +

[t1 ,t]\Ω

G(u(s))ds

Λ−1 g(s, x(s), u(s))ds

Ω

≤ c0 δ + Λ−1 (I g (T1 , T2 , x, u) + a0 (T2 − T1 )) ≤ c0 δ + Λ−1 (M1 + a0 ).

(6.80)

It follows from (6.70), (6.72), (6.78), and (6.80) that x(t) − x ¯ ≤ δM∗ e|w∗ | + c0 δM∗ e|w∗ | + Λ−1 (M1 + a0 )M∗ e|w∗ | < . Thus for the first class of problems Proposition 6.6 is proved. Let us consider the second class of problems. Recall that B ∈ L(F, E−1 ) is an admissible control operator for the semigroup S(t) = eAt , t ≥ 0, eAt  ≤ M∗ ew∗ t , t ∈ [0, ∞),

(6.81)

for every τ ≥ 0,  φτ (u) =

τ

eA(τ −s) Bu(s)ds

(6.82)

0

and φτ ∈ L(L2 (0, τ ; F ), E) (see (6.8) and (4.9), Proposition 4.18). Choose a number Λ > 1 satisfying Λ−1/2 (Φ1 + 1)(M1 + a0 τ1 + 1) < /8.

(6.83)

212

6 Infinite Dimensional Optimal Control

In view (6.9) and (6.69), there exists c0 > 0 such that the following property holds: (i) For each (x, u) ∈ M satisfying u > c0 , each g ∈ M and each t ≥ 0, g(t, x, u) ≥ Λu2 . Choose a number δ ∈ (0, 1) that satisfies δM∗ e|w∗ | < /8, φ1 c0 δ 1/2 < /8.

(6.84)

Assume that g ∈ M, T2 > T1 ≥ 0 satisfy T2 ∈ [T1 + τ0 , T1 + τ1 ], (x, u) ∈ X(T1 , T2 ), T1 ≤ t1 < t2 ≤ T2 and (6.74) and (6.75) hold. Clearly, (x, u) ∈ X(t1 , t2 ) and x(t2 ) = e

A(t2 −t1 )



t2

x(t1 ) +

eA(t2 −s) Bu(s)ds

(6.85)

t1

in E−1 = D(A∗ ) . Propositions 4.18 and 4.19 and equations (6.19), (6.75)–(6.82), and (6.85) imply that x(t2 ) − x ¯ ≤ e

A(t2 −t1 )

x(t1 ) − e

A(t2 −t1 )

 x ¯ +

t2

eA(t2 −s) Bu(s)ds

t1

≤ eA(t2 −t2 ) x(t1 ) − x ¯ +



t2 −t1

eA(t2 −t1 −s) Bu(t1 + s)ds

0

≤ δM∗ e|w∗ |δ + Φt2 −t1 u(t1 + ·)L2 (0,t2 −t1 ;F ) ≤ δM∗ e|w∗ | + Φ1 u(t1 + ·)L2 (0,t2 −t1 ;F ) ≤ δM∗ e|w∗ | + Φ1 (



t2

u(s)2 ds)1/2 .

(6.86)

t1

Set Ω = {t ∈ [t1 , t2 ] : u(t)) ≥ c0 }.

(6.87)

6.4 The Turnpike Property

213

Property (i), the choice of c0 , (6.69), (6.75), (6.83), and (6.87) imply that  Φ1 (

t2

 u(s)2 ds)1/2 = Φ1 (

t1

 u(s)2 ds +

Ω



[t1 ,t2 ]\Ω

u(s)2 ds)1/2



≤ Φ1 (

u(s)2 ds)1/2 + Φ1 ( Ω

 ≤ Φ1 (

[t1 ,t2 ]\Ω

u(s)2 ds)1/2

Λ−1 g(s, x(s), u(s))ds)1/2 + Φ1 c0 δ 1/2

Ω

≤ Φ1 c0 δ 1/2 + Φ1 (Λ−1 (I g (T1 , T2 , x, u) + a0 τ1 )1/2 ≤ Φ1 c0 δ 1/2 + Φ1 Λ−1/2 (M1 + a0 τ1 )1/2 .

(6.88)

By (6.83), (6.84), (6.86), and (6.88), ¯ ≤ δM∗ e|w∗ | + Φ1 c0 δ 1/2 x(t2 ) − x +Φ1 Λ−1/2 (M1 + a0 τ1 )1/2 < /2. Therefore Proposition 6.6 is proved for the second class of problems too.

6.4 The Turnpike Property From here we assume that for the second class of problems the following assumption holds. (A5)

B ∈ L(F, E−1 ) is an admissible control operator for the semigroup S(t) = eAt , t ≥ 0 and for each (x, u) ∈ M, f (x, u) ≥ ψ(u)u2 − a0 .

We prove the following turnpike result. It shows that for approximate solutions (x, u) of our optimal control problems on intervals [0, T ], where T is sufficiently large, the set of all points t ∈ [0, T ] such that x(t) does not belong to an neighborhood of x¯ is contained in the union of two intervals, where the first interval contains 0, the second one contains T, and their lengths do not exceed a constant L, which depends only on  and does not depend on T . In the literature this property is known as the turnpike property.

214

6 Infinite Dimensional Optimal Control

Theorem 6.7 Let  ∈ (0, 1] and L0 > 0. Then there exist L > L0 , δ > 0 such that for each T > 2L and each (x, u) ∈ X(0, T ) such that x(0) ∈ AL0 and at least one of the following conditions holds: (a) L0 , x(T ) ∈ A I f (0, T , x, u) ≤ U f (0, T , x(0), x(T )) + δ; (b) I f (0, T , x, u) ≤ σ f (0, T , x(0)) + δ there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ] such that x(t) − x ¯ ≤ , t ∈ [τ1 , τ2 ]. Moreover, if x(0) − x ¯ ≤ δ, then τ1 = 0 and if x(T ) − x ¯ ≤ δ, then τ2 = T . Proof By Proposition 6.6, there exists 0 ∈ (0, min{4−1 , /4}) such that the following property holds: (i) For each S ≥ 0, each (y, v) ∈ X(S, S + 1) satisfying I f (S, S + 1, y, v) ≤ |f (x, ¯ u)| ¯ +2 and each t1 , t2 ∈ [S, S + 1] satisfying 0 < t2 − t1 ≤ 0 , x(t1 ) − x ¯ ≤ 0 we have ¯ ≤ /4. x(t2 ) − x Assumption (A3) implies that there exists 1 ∈ (0, 0 ) such that the following property holds: (ii) f (y, z) ≥ f (x, ¯ u) ¯ + 1 for all (y, z) ∈ M satisfying y − x ¯ ≥ 0 /4.

6.4 The Turnpike Property

215

Choose a positive number δ0 < 4−1 1 0 .

(6.89)

Proposition 6.5 implies that there exist δ1 ∈ (0, δ0 ) such that the following property holds: (iii) For each T > 2bf and each y, z ∈ BE (x, ¯ δ1 ) we have ¯ u) ¯ + δ0 . U f (0, T , y, z) ≤ Tf (x, By Theorem 6.4, there exists L1 > L0 + 1 + 2bf such that the following property holds: L0 , and each (x, u) ∈ X(0, T ), (iv) For each T ≥ L1 , each y ∈ AL0 , each z ∈ A which satisfies at least one of the following conditions: (c) x(0) = y, x(T ) = z, I f (0, T , x, u) ≤ U f (0, T , x(0), x(T )) + 1; (d) x(0) = y, I f (0, T , x, u) ≤ σ f (0, T , y) + 1 the inequality mes({t ∈ [0, T ] : x(t) − x ¯ > δ1 }) < L1 is valid. Fix numbers δ ∈ (0, δ1 ), L ≥ L1 + 2bf + 2.

(6.90)

Assume that T > 2L and (x, u) ∈ X(0, T ) satisfies x(0) ∈ AL0 and at least one of the conditions (a), (b) holds.

(6.91)

216

6 Infinite Dimensional Optimal Control

By (6.90), property (iv), and conditions (a), (b), mes({t ∈ [0, T ] : x(t) − x ¯ > δ1 }) < L1 .

(6.92)

It follows from (6.92) that the following property holds: (v) For each number S satisfying [S, S + L1 ] ⊂ [0, T ] there exists τ ∈ [S, S + L1 ] such that x(τ ) − x ¯ ≤ δ1 . Property (v) and (6.90) imply that there exist τ1 ∈ [0, L1 ], τ2 ∈ [T − L1 , T ]

(6.93)

¯ ≤ δ1 , i = 1, 2. x(τi ) − x

(6.94)

such that

¯ ≤ δ, then we set τ2 = T . If x(0) − x ¯ ≤ δ, then we set τ1 = 0 and if x(T ) − x We show that x(t) − x ¯ ≤ , t ∈ [τ1 , τ2 ]. Assume the contrary. Then there exists t∗ ∈ [τ1 , τ2 ]

(6.95)

¯ > . x(t∗ ) − x

(6.96)

I f (τ1 , τ2 , x, u) ≤ U f (τ1 , τ2 , x(τ1 ), x(τ2 )) + δ.

(6.97)

such that

Conditions (a) and (b) imply that

In view of (6.90) and (6.93), τ2 − τ1 > T − L1 − L1 > 2L − 2L1 > 2(2bf + 2).

(6.98)

Property (iii), (6.94), and (6.98) imply that ¯ u) ¯ + δ0 . U f (τ1 , τ2 , x(τ1 ), x(τ2 )) ≤ (τ2 − τ1 )f (x,

(6.99)

6.4 The Turnpike Property

217

Equations (6.97) and (6.99) imply that ¯ u) ¯ + δ0 + δ. I f (τ1 , τ2 , x, u) ≤ (τ2 − τ1 )f (x,

(6.100)

Assume that τ ∈ [0, T ] satisfies τ1 ≤ τ < t∗ ≤ τ + 0 ≤ τ2 .

(6.101)

By (6.98), there exists b ≥ 0 such that [τ, τ + 0 ] ⊂ [b, b + 1] ⊂ [τ1 , τ2 ].

(6.102)

By (6.16), (6.89), (6.90), (6.100), and (6.102), ¯ u) ¯ + δ0 + δ ≤ f (x, ¯ u) ¯ + 1. I f (b, b + 1, x, u) ≤ f (x,

(6.103)

If x(τ ) − x ¯ ≤ 0 , then property (i) and equations (6.101)–(6.103) imply that ¯ ≤ /4. x(t∗ ) − x This contradicts (6.96). The contradiction we have reached proves that x(τ ) − x ¯ > 0 for every number τ ∈ [0, T ] satisfying (6.101). Therefore in view of (6.94) and (6.96), t∗ > τ1 + 0 and for each τ ∈ [t∗ − 0 , t∗ ] we have x(τ ) − x ¯ > 0 .

(6.104)

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Property (ii) and (6.104) imply that for all t ∈ [t∗ − 0 , t∗ ], f (x(t), u(t)) > f (x, ¯ u) ¯ + 1 .

(6.105)

It follows from (6.90), (6.100), and (6.105) that ¯ u) ¯ + δ0 + δ ≥ I f (τ1 , τ2 , x, u) (τ2 − τ1 )f (x,  = I f (t∗ − 0 , t∗ , x, u) +

[τ1 ,τ2 ]\[t∗ −0 ,t∗ ]

f (x(t), u(t))dt

¯ u) ¯ ≥ (f (x, ¯ u) ¯ + 1 )0 + (τ2 − τ1 − 0 )f (x, ¯ u) ¯ + 0  1 = (τ2 − τ1 )f (x, and 2δ0 ≥ 0 1 . This contradicts (6.89). The contradiction we have reached completes the Proof of Theorem 6.7.

6.5 The Second Weak Turnpike Result We continue to assume that (A5) holds for the second class of problems. In this section we prove our second weak turnpike result. It shows that for approximate solutions (x, u) of our optimal control problems on intervals [0, T ], where T is sufficiently large, the set of all points t ∈ [0, T ] such that x(t) does not belong to an -neighborhood of x¯ is contained in the union of a finite family of intervals, their lengths do not exceed a constant l, and their number does not exceed a constant Q that depend only on  and does not depend on T . Theorem 6.8 Let M0 , M1 > 0 and  > 0. Then there exist l > 0 and a natural number Q such that for each T > lQ and each (x, u) ∈ X(0, T ) such that x(0) ∈ AM0 and at least one of the following conditions holds: (a) I f (0, T , x, u) ≤ σ f (0, T , x(0)) + M1 ;

(6.106)

6.5 The Second Weak Turnpike Result

219

(b) M0 , x(T ) ∈ A I f (0, T , x, u) ≤ U f (0, T , x(0), x(T )) + M1 there exists a sequence of closed intervals [ai , bi ] ⊂ [0, T ], i = 1, . . . , q such that q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, ai+1 ≥ bi for all i ∈ {1, . . . , q} \ {q} and q

{t ∈ [0, T ] : x(t) − x ¯ > } ⊂ ∪i=1 [ai , bi ]. Proof By Theorem 5.7, there exist L1 > 2bf , δ ∈ (0, 1] such that the following property holds: (i) For each T ≥ 2L1 and each (x, u) ∈ X(0, T ) satisfying x(0) − x ¯ ≤ δ, x(T ) − x ¯ ≤ δ, I f (0, T , x) ≤ U (0, T , x(0), x(T )) + δ we have x(t) − x ¯ ≤ , t ∈ [0, T ]. By Theorem 6.4, there exist L2 > 2L1 such that the following property holds: (ii) For each T ≥ L2 and each (x, u) ∈ X(0, T ) such that (6.106) is true and such that at least one of conditions (a), (b) holds we have mes({t ∈ [0, T ] : x(t) − x ¯ > δ}) < L2 .

(6.107)

Set l = 4L1 + 4L2 + 4.

(6.108)

Q > 4 + δ −1 M1 .

(6.109)

Choose an integer

220

6 Infinite Dimensional Optimal Control

Assume that T > lQ and that (x, u) ∈ X(0, T ) satisfies (6.106) and at least one of conditions (a), (b). Property (ii) implies that (6.107) is valid. In view of (6.107), there exists t0 ∈ [0, L2 ]

(6.110)

¯ ≤ δ. x(t0 ) − x

(6.111)

such that

By induction, using equation (6.107), we construct a finite sequence t0 < t 1 < · · · < t q belonging to the interval [0, T ] such that (6.110) and (6.111) hold, x(ti ) − x ¯ ≤ δ, i = 0, . . . , q,

(6.112)

ti+1 − ti ∈ [L2 , 2L2 ],

(6.113)

tq > T − 2L2 .

(6.114)

i ∈ {0, . . . , q} \ {q},

Now we construct a strictly increasing sequence of numbers Si ∈ {t0 , . . . , tq }, i = 0, . . . , p. Set S0 = t0 .

(6.115)

Assume that k ≥ 0 is an integer and that we already defined an increasing sequence Si ∈ {t0 , . . . , tq }, i = 0, . . . , k. If Sk = tq , then the construction is completed. Assume that Sk < tq . If I f (Sk , tq , x, u) ≤ U f (Sk , tq , x(Sk ), x(tq )) + δ, then we set Sk+1 = tq

6.5 The Second Weak Turnpike Result

221

and the construction is completed. Assume that I f (Sk , tq , x) > U f (Sk , tq , x(Sk ), x(tq )) + δ. There exists j ∈ {0, . . . , q} such that Sk = tj . If I f (Sk , tj +1 , x, u) > U f (Sk , tj +1 , x(Sk ), x(tj +1 )) + δ, then we set Sk+1 = tj +1 . Assume that I f (Sk , tj +1 , x) ≤ U f (Sk , tj +1 , x(Sk ), x(tj +1 )) + δ. We define Sk+1 = min{ti : i ∈ {j + 1, . . . , q} : I f (Sk , ti , x, u) > U (Sk , ti , x(Sk ), x(ti )) + δ}. Therefore by induction we defined a strictly increasing sequence of numbers Si ∈ {t0 , . . . , tq }, i = 0, . . . , p such that S0 = t0 , Sp = tq , for each j ∈ {0, . . . , p − 1} \ {p − 1}, I f (Sj , Sj +1 , x, u) > U (Sj , Sj +1 , x(Sj ), x(Sj +1 )) + δ,

(6.116)

for each j ∈ {0, . . . , p − 1}, if i ∈ {0, . . . , q} and Sj < ti < Sj +1 , then I f (Sj , ti , x, u) ≤ U f (Sj , ti , x(Sj ), x(Sti )) + δ.

(6.117)

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6 Infinite Dimensional Optimal Control

Conditions (a) and (b) and the equations above imply that M1 ≥ I f (0, T , x, u) − U f (0, T , x(0), x(T )) ≥



{I f (Sj , Sj +1 , x, u) − U f (Sj , Sj +1 , x(Sj ), x(Sj +1 )) : j ∈ {0, . . . , p − 1} \ {p − 1}} ≥ (p − 1)δ

and p ≤ 1 + δ −1 M1 .

(6.118)

Assume that i ∈ {0, . . . , p − 1} satisfies Si+1 − Si ≥ l.

(6.119)

Equations (6.113), (6.117), and (6.119) imply that there exists S˜j ∈ [Sj +1 − 2L2 , Sj +1 − L2 ]

(6.120)

¯ ≤ δ, x(S˜j ) − x

(6.121)

such that

I f (Sj , S˜j , x, u) ≤ U f (Sj , S˜j , x(Sj ), x(S˜j )) + δ.

(6.122)

Property (i) and equations (6.120)–(6.122) imply that x(t) − x ¯ ≤ , t ∈ [Sj , S˜j ]. By the equation above, {t ∈ [0, T ] : x(t) − x ¯ > } ⊂ [0, S0 ] ∪ [Sp , T ] ∪ {[Si , Si+1 ] : i ∈ {0, . . . , p − 1}, Si+1 − Si < l} ∪{[Si+1 − 2L2 , Si+1 ] : i ∈ {0, . . . , p − 1}, Si+1 − Si ≥ l}. It is easy to see that the right-hand side of the equation above is a union of a finite number of closed intervals, their lengths do not exceed l, and in view of (6.109) and (6.118), their number does not exceed 2 + 2p ≤ 4 + δ −1 M1 < Q. Theorem 6.8 is proved.

6.6 A Well-Posedness Result

223

6.6 A Well-Posedness Result Let X = [0, ∞) × M. For each (α1 , x1 , y1 ), (α2 , x2 , y2 ) ∈ [0, ∞) × M define ρ((α1 , x1 , y1 ), (α2 , x2 , y2 )) = |α1 − α2 | + x1 − x2  + y1 − y2 . Clearly, (X, ρ) is a metric space. For each (α, x, y) ∈ [0, ∞) × M set (α, x, y) = |α| + x + y. We continue to consider the space M of all Borelian bounded from below functions g : X → R 1 that satisfy (6.68) for the first class of problems and that satisfy (6.69) for the second class of problems. For each natural number n denote by E(n) the set of all pairs (g1 , g2 ) ∈ M × M such that for every (t, ξ, η) ∈ [0, ∞) × M, |g1 (t, ξ, η) − g2 (t, ξ, η)| ≤ n−1 if at least one of the following conditions holds: ξ , η ≤ n; min{g1 (t, ξ, η), g2 (t, ξ, η)} ≤ n. Clearly, for each integer n ≥ 1 if (g1 , g2 ), (g2 , g3 ) ∈ E(2n), then (g1 , g3 ) ∈ E(n). For the space M there exists a uniformity generated by the base E(n), n = 1, 2, . . . . The uniform space M is metrizable by a metric dM . We continue to consider the same integrand f : M → R 1 satisfying the same assumptions as in the previous sections and define f˜(t, x, u) = f (x, u) for every (t, x, u) ∈ [0, ∞) × M. Proposition 6.9 Let  ∈ (0, 1]. Then there exist δ0 > 0 and a neighborhood U of f˜ in M such that for each g ∈ U and each (t, ξ, η) ∈ [0, ∞) × M satisfying max{ξ − x), ¯ min{η − u, ¯ η + u}} ¯ >

(6.123)

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6 Infinite Dimensional Optimal Control

the inequality g(t, ξ, η) ≥ f (x, ¯ u) ¯ + δ0

(6.124)

is valid. Proof Assumption (A3) implies that there exists δ ∈ (0, 1) such that the following property holds: (i) For each (ξ, η) ∈ M satisfying f (ξ, η) ≤ inf(f ) + δ we have ξ − x ¯ ≤ , min{η − u, ¯ η + u} ¯ ≤ . Set δ0 = 2−1 δ.

(6.125)

Choose a natural number ¯ u)| ¯ + 4, x + u}. ¯ n > max{2δ −1 , |f (x,

(6.126)

Set U = {g ∈ M : (f˜, g) ∈ E(n)}.

(6.127)

g∈U

(6.128)

Assume that

and (t, ξ, η) ∈ [0, ∞) × M satisfies (6.123). If g(t, ξ, η) ≥ n, then (6.124) holds. Assume that g(t, ξ, η) ≤ n.

(6.129)

It follows from (6.127)–(6.129) that |g(t, ξ, η) − f (ξ, η)| ≤ n−1 .

(6.130)

Property (i) and (6.123) imply that f (ξ, η) > inf(f ) + δ = f (x, ¯ u) ¯ + δ.

(6.131)

6.6 A Well-Posedness Result

225

In view of (6.126) and (6.127), |g(t, x, ¯ u) ¯ − f (x, ¯ u)| ¯ ≤ n−1 . By (6.125), (6.126), (6.130), and (6.131), ¯ u) ¯ − n−1 + δ g(t, ξ, η) ≥ f (ξ, η) − n−1 ≥ f (x, ¯ u) ¯ + δ0 . ≥ f (x, ¯ u) ¯ + 2−1 δ = f (x, Proposition 6.9 is proved. Theorem 6.10 Let  ∈ (0, 1]. Then there exist δ > 0 and a neighborhood U of f˜ in M such that for each g ∈ U and each (t, ξ, η) ∈ [0, ∞) × M satisfying g((t, ξ, η) ≤ inf(g(t, ·, ·)) + δ the inequalities |g(t, ξ, η) − f (x, ¯ u)| ¯ ≤ and ξ − x ¯ ≤ , min{η − u, ¯ η + u} ¯ ≤ are true. Proof Proposition 6.9 implies that there exist δ0 ∈ (0, ) and a neighborhood U0 of f˜ in M such that the following property holds: (i) For each g ∈ U0 and each (t, ξ, η) ∈ [0, ∞) × M satisfying max{ξ − x, ¯ min{η − u, ¯ η + u}} ¯ >

(6.132)

we have g(t, ξ, η) ≥ f (x, ¯ u) ¯ + δ0 . Choose δ ∈ (0, δ0 /4).

(6.133)

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6 Infinite Dimensional Optimal Control

There exists a neighborhood U ⊂ U0 of f˜ in M for each g ∈ U and each (t, ξ, η) ∈ [0, ∞) × B(x, ¯ 1) × ((B(u, ¯ 1) ∪ B(−u, ¯ 1)) ∩ M) we have |f (ξ, η) − g(t, ξ, η)| ≤ δ0 /8.

(6.134)

By (6.134), for each g ∈ U and each t ≥ 0, |f (x, ¯ u) ¯ − g(t, x, ¯ u)| ¯ ≤ δ0 /8.

(6.135)

g∈U

(6.136)

Assume that

and that (t, ξ, η) ∈ X = [0, ∞) × M satisfies g(t, ξ, η) ≤ inf(g(t, ·, ·)) + δ. In order to complete the proof of the theorem it is sufficient to show that |g(t, ξ, η) − f (x, ¯ u)| ¯ ≤ and ξ − x ¯ ≤ , min{η − u, ¯ η + u} ¯ ≤ . Assume that max{ξ − x, ¯ min{η − u, ¯ η + u}} ¯ > . Property (i), the relation above, (6.135) and (6.136) imply that g(t, ξ, η) > f (x, ¯ u) ¯ + δ0 ≥ g(t, x, ¯ u) ¯ + δ0 − δ0 /8 ≥ inf(g(t, ·, ·)) + δ0 /2. This contradicts (6.137). The contradiction we have reached proves that ξ − x ¯ ≤ , min{η − u, ¯ η + u} ¯ ≤ .

(6.137)

6.7 Stability of the Turnpike Phenomenon

227

By the relation above, (6.134) and (6.136), |f (ξ, η) − g(t, ξ, η)| ≤ δ0 /8. Since the relation above holds for all (t, ξ, η) ∈ X = [0, ∞)×M satisfying (6.137) we obtain that ¯ u) ¯ − δ0 /8. inf(g(t, ·, ·)) + δ ≥ g(t, ξ, η) ≥ inf(f ) − δ0 /8 = f (x, In view of (6.135), inf(g(t, ·, ·)) ≤ g(t, x, ¯ u) ¯ ≤ f (x, ¯ u) ¯ + δ. Thus by the equations above, f (x, ¯ u) ¯ + 2δ ≥ inf(g(t, ·, ·)) + δ ≥ g(t, ξ, η) ≥ f (x, ¯ u) ¯ − δ0 /8 and |g(t, ξ, η) − f (x, ¯ u)| ¯ ≤ . Theorem 6.10 is proved.

6.7 Stability of the Turnpike Phenomenon We continue to consider the space M and the integrand f . We assume that f is bounded on bounded sets. Let L > 0. Denote by BL the set of all z ∈ A for which there exist τ ∈ (0, L] and (x, u) ∈ X(0, τ ) such that x(0) = z, x(τ ) = x, ¯ I f (0, τ, x, u) ≤ L, x(t) ≤ L, u(t) ≤ L, t ∈ [0, τ ]. L the set of all z ∈ A for which there exist τ ∈ (0, L] and (x, u) ∈ Denote by B X(0, τ ) such that x(0) = x, ¯ x(τ ) = z, I f (0, τ, x, u) ≤ L, x(t) ≤ L, u(t) ≤ L, t ∈ [0, τ ]. We prove the following result that shows that the turnpike property is stable under small perturbations of the integrand f .

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Theorem 6.11 Let L0 ,  > 0. There exist δ > 0, L > L0 and a neighborhood U of f˜ in M such that for each g ∈ U , each T > 2L, and each (x, u) ∈ X(0, T ) such that x(0) ∈ BL0 and at least one of the conditions below holds: (a) L0 , x(T ) ∈ B I g (0, T , x, u) ≤ U g (0, T , x(0), x(T )) + δ; (b) I g (0, T , x, u) ≤ σ g (0, T , x(0)) + δ there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ] such that x(t) − x ¯ ≤ , t ∈ [τ1 , τ2 ]. ¯ ≤ δ, then τ2 = T . Moreover, if x(0) − x ¯ ≤ δ, then τ1 = 0 and if x(T ) − x

6.8 An Auxiliary Result for Theorem 6.11 Lemma 6.12 Let  ∈ (0, 1], M > x ¯ + u ¯ + bf + 1 and M . ¯ ) ⊂ BM ∩ B BE (x, Then there exist L > 0 and a neighborhood U of f˜ in M such that for each g ∈ U , each T > L and each (x, u) ∈ X(0, T ) such that at least one of the conditions below holds: (a) M , x(0) ∈ BM , x(T ) ∈ B I g (0, T , x, u) ≤ U g (0, T , x(0), x(T )) + M

6.8 An Auxiliary Result for Theorem 6.11

229

(b) x(0) ∈ BM , I g (0, T , x, u) ≤ σ g (0, T , x(0)) + M the following property holds: (c) For each S ≥ 0 satisfying [S, S + L + 1] ⊂ [0, T ], x(S) ∈ BM there exists t0 ∈ [S + 1, S + L + 1] such that ¯ ≤ . x(t0 ) − x Proof Proposition 6.9 implies that there exist δ0 ∈ (0, min{, (4M + 4)−1 }) and a neighborhood U1 of f˜ in M such that the following property holds: (d) For each g ∈ U1 and each (t, ξ, η) ∈ [0, ∞) × M satisfying ξ − x ¯ > /2 we have g(t, ξ, η) > f (x, ¯ u) ¯ + δ0 . There exists M1 > 0 such that |f (ξ, η)| ≤ M1 for all (ξ, η) ∈ BE (0, 4M + 2) × BF (0, 4M + 2).

(6.138)

Choose a number ¯ u)|δ ¯ 0−1 (2M + 1). L > 4 + 4(3M + a0 + M1 + 5)δ0−1 + 6|f (x,

(6.139)

By assumption (A2), there exists δ1 ∈ (0, δ0 ) such that the following property holds: (e) For every (u, v) ∈ M satisfying u − x ¯ ≤ δ1

230

6 Infinite Dimensional Optimal Control

and min{v − u, ¯ v + u} ¯ ≤ δ1 we have |f (u, v) − f (x, ¯ u)| ¯ ≤ δ0 /4. There exists a neighborhood U ⊂ U1 of f˜ in M such that for each g ∈ U and each (t, ξ, η) ∈ [0, ∞) × (M ∩ (BE (0, 4M + 2) × BF (0, 4M + 2))) we have |f (ξ, η) − g(t, ξ, η)| ≤ δ0 /4.

(6.140)

g ∈ U,

(6.141)

Assume that

T > L, (x, u) ∈ X(0, T ), at least one of conditions (a), (b) holds and a nonnegative number S satisfies S ≤ T − L − 1,

(6.142)

x(S) ∈ BM .

(6.143)

We show that there exists t0 ∈ [S + 1, S + L + 1] such that ¯ ≤ . x(t0 ) − x Assume the contrary. Then x(t) − x ¯ > , t ∈ [S + 1, S + L + 1].

(6.144)

Property (d) and equations (6.141) and (6.144) imply that g(t, x(t), u(t)) ≥ f (x, ¯ u) ¯ + δ0 , t ∈ [S + 1, S + L + 1].

(6.145)

6.8 An Auxiliary Result for Theorem 6.11

231

There are two cases: there exists S0 ∈ [S + L + 1, T ] for which ¯ ≤ ; x(S0 ) − x

(6.146)

x(t) − x ¯ >  for all t ∈ [S + L + 1, T ].

(6.147)

Set ¯ ≤ }. S1 = inf{t ∈ [S + L + 1, T ] : x(t) − x (Here we suppose that the infimum over an empty set is T .) Clearly, S1 is welldefined. Property (d) and (6.144) imply that if (6.147) holds, then S1 = T , x(t) − x ¯ ≥ , g(t, x(t), u(t)) > f (x, ¯ u) ¯ + δ0

(6.148)

for every t ∈ [S + 1, T ] and if (6.146) is valid, then S + L + 1 ≤ S1 ≤ S0 , x(t) − x ¯ ≥ , g(t, x(t), u(t)) > f (x, ¯ u) ¯ + δ0

(6.149)

for all t ∈ [S + 1, S1 ] and ¯ = . x(S1 ) − x

(6.150)

Assume that at least one of the following conditions holds: condition (a) holds; condition (b) holds and S1 < T . Property (a), equations (6.143) and (6.150), and the inclusion M ¯ ) ⊂ BM ∩ B BE (x, imply that M . x(S) ∈ BM , x(S1 ) ∈ B

(6.151)

It follows from Theorem 6.2 and equations (6.139) and (6.151) that there exist (ξ, η) ∈ X(S, S1 ) and τ1 , τ2 ∈ [0, M] such that ¯ ξ(S) = x(S), ξ(S + τ1 ) = x,

(6.152)

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6 Infinite Dimensional Optimal Control

I f (S, S + τ1 , ξ, η) ≤ M, ξ(t) ≤ M, η(t) ≤ M, t ∈ [S, S + τ1 ], ¯ ξ(S1 ) = x(S1 ), ξ(S1 − τ2 ) = x,

(6.153) (6.154) (6.155)

I f (S1 − τ1 , S1 , ξ, η) ≤ M, ξ(t) ≤ M, η(t) ≤ M, t ∈ [S1 − τ2 , S1 ], (6.156) ξ(t) − x ¯ ≤ δ1 , t ∈ [S + τ1 , S1 − τ2 ],

(6.157)

η(t) ∈ {u, ¯ −u}, ¯ t ∈ [S + τ1 , S1 − τ2 ] a. e.,

(6.158)

¯ u) ¯ + δ1 . I f (S + τ1 , S1 − τ2 , ξ, η) ≤ (S1 − S − τ1 − τ2 )f (x,

(6.159)

Property (e) and equations (6.157) and (6.158) imply that for a. e. t ∈ [S + τ1 , S1 − τ2 ], |f (ξ(t), η(t)) − f (x, ¯ u)| ¯ ≤ δ0 /4.

(6.160)

By (6.140), (6.154) and (6.156)–(6.158), for a. e. t ∈ [S, S1 ], |f (ξ(t), η(t)) − g(t, ξ(t), η(t))| ≤ δ0 /4.

(6.161)

In view of (6.153), (6.156), and (6.161), I g (S, S + τ1 , ξ, η) ≤ I f (S, S + τ1 , ξ, η) + 4−1 δ0 M ≤ M + 1,

(6.162)

I g (S1 − τ2 , S1 , ξ, η) ≤ I f (S1 − τ2 , S1 , ξ, η) + 4−1 δ0 M ≤ M + 1.

(6.163)

By (6.160) and (6.161), for a. e. t ∈ [S + τ1 , S1 − τ2 ], g(t, ξ(t), η(t)) ≤ f (x, ¯ u) ¯ + δ0 /2.

(6.164)

Equations (6.162)–(6.164) and the inequality S1 ≥ S + L + 1 imply that ¯ u) ¯ + 2−1 δ0 (S1 − S − τ1 − τ2 ). I g (S, S1 , ξ, η) ≤ 2M + 2 + (S1 − S − τ1 − τ2 )f (x, (6.165) By (6.68), (6.139), and (6.149), I g (S, S1 , x, u) ≥ −a0 + I g (S + 1, S1 , x, u) ¯ u) ¯ + δ0 ). ≥ −a0 + (S1 − S − 1)(f (x,

(6.166)

6.8 An Auxiliary Result for Theorem 6.11

233

Conditions (a), (b), (6.152), (6.155), and (6.166) imply that M ≥ I g (S, S1 , x, u) − I g (S, S1 , ξ, η) ¯ u) ¯ + δ0 (S1 − S − 1) − 2M − 2 ≥ −a0 + (S1 − S − 1)f (x, −(S1 − S − τ1 − τ2 )f (x, ¯ u) ¯ − 2−1 δ0 (S1 − S − τ1 − τ2 ) ¯ u)|(2M ¯ + 1) − 2M − 2 + 4−1 δ0 L ≥ −a0 − |f (x, and ¯ u)|). ¯ L ≤ 4δ0−1 (a0 + 3M + 2 + (2M + 1)|f (x, This contradicts (6.139). The contradiction we have reached proves that there exists t0 ∈ [S + 1, S + L + 1] such that ¯ ≤ x(t0 ) − x in our first case. Consider now the second case when condition (b) and equality S1 = T

(6.167)

hold. By (6.143) and Theorem 6.2, there exist (ξ, η) ∈ X(S, T ) and τ ∈ [0, M] such that ξ(S) = x(S), ξ(S + τ ) = x, ¯ I f (S, S + τ, ξ, η) ≤ M, ξ(t) ≤ M, η(t) ≤ M, t ∈ [S, S + τ ],

(6.168) (6.169)

ξ(t) − x ¯ ≤ δ1 , t ∈ [S + τ, T ],

(6.170)

η(t) ∈ {u, ¯ −u}, ¯ t ∈ [S + τ, T ] a. e.,

(6.171)

¯ u) ¯ + δ1 . I f (S + τ, T , ξ, η) ≤ (T − S − τ )f (x,

(6.172)

Property (e) and equations (6.170) and (6.171) imply that for a. e. t ∈ [S + τ, T ], |f (ξ(t), η(t)) − f (x, ¯ u)| ¯ ≤ δ0 /4.

(6.173)

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By (6.140), (6.169), and (6.170), for a. e. t ∈ [S, T ], |f (ξ(t), η(t)) − g(t, ξ(t), η(t))| ≤ δ0 /4.

(6.174)

In view of (6.169) and (6.174), I g (S, S + τ, ξ, η) ≤ I f (S, S + τ, ξ, η) + 4−1 δ0 M ≤ M + 1.

(6.175)

By (6.173) and (6.174), for a. e. t ∈ [S + τ, T ], g(t, ξ(t), η(t)) ≤ f (x, ¯ u) ¯ + δ0 /2.

(6.176)

Equations (6.175) and (6.176) imply that I g (S, T , ξ, η) ≤ M + 1 + (T − S − τ )(f (x, ¯ u) ¯ + 2−1 δ0 ).

(6.177)

By (6.68) and (6.148), I g (S, T , x, u) ≥ −a0 + I g (S + 1, T , x, u) ¯ u) ¯ + δ0 ). ≥ −a0 + (T − S − 1)(f (x,

(6.178)

Condition (b), (6.143), (6.168), (6.177), and (6.178) imply that M ≥ I g (S, T , x, u) − I g (S, T , ξ, η) ¯ u) ¯ + δ0 (T − S − 1) − M − 1 ≥ −a0 + (T − S − 1)f (x, −(T − S − τ ))f (x, ¯ u) ¯ − 2−1 δ0 (T − S − τ ) ¯ u)|(M ¯ + 1) − M − 2 + 2−1 δ0 (T − S − 1) ≥ −a0 − |f (x, ≥ −a0 − |f (x, ¯ u)|(M ¯ + 1) − M − 2 + 2−1 δ0 L and L ≤ 2δ0−1 (a0 + 2M + 2 + |f (x, ¯ u)|)(M ¯ + 1)). This contradicts (6.139). The contradiction we have reached proves that there exists t0 ∈ [S + 1, S + L + 1] such that ¯ ≤ x(t0 ) − x in the second case too. This completes the Proof of Lemma 6.12.

6.9 Proof of Theorem 6.11

235

6.9 Proof of Theorem 6.11 We may assume without loss of generality that M > x ¯ + u ¯ + 1 + 4bf

(6.179)

L0 . ¯ ) ⊂ BL0 ∩ B BE (x,

(6.180)

and that

Proposition 6.9 implies that there exist 0 ∈ (0, ) and a neighborhood U1 of f˜ in M such that the following property holds: (i) For each g ∈ U1 and each t ≥ 0 and each (x, u) ∈ M satisfying max{x − x, ¯ min{u − u, ¯ u + u}} ¯ > 4−1  we have g(x, y) > f (x, ¯ u) ¯ + 0 . Choose a natural number n such that ¯ u)| ¯ + 1. ψ(n) ≥ a0 + |f (x,

(6.181)

Proposition 6.6 implies that there exists γ0 ∈ (0, 4−1 ) such that the following property holds: (ii) For each S ≥ 0, each g ∈ M, each (ξ, η) ∈ X(0, 1) satisfying I g (S, S + 1, ξ, η) ≤ |f (x, ¯ u)| ¯ +4 and each t1 , t2 ∈ [S, S + 1] satisfying 0 < t2 − t1 ≤ γ0 and x(t1 ) − x ¯ ≤ γ0 the inequality ¯ ≤ /4 x(t2 ) − x holds. By Proposition 6.9, there exist 1 ∈ (0, 0 ) and a neighborhood U2 of f˜ in M such that the following property holds:

236

6 Infinite Dimensional Optimal Control

(iii) For each g ∈ U2 and each (t, ξ, η) ∈ [0, ∞) × M satisfying max{ξ − x, ¯ min{η − u, ¯ η + u}} ¯ > 4−1 γ0 the inequality g(t, ξ, η) ≥ f (x, ¯ u) ¯ + 1 is valid. Choose a positive number δ1 < 1 γ0 /16.

(6.182)

Theorem 6.2 and assumption (A4) imply that there exist δ ∈ (0, δ1 ) such that the following property holds: (iv) For each T > 2bf and each ¯ δ) y, z ∈ BE (x, there exists (ξ, η) ∈ X(0, T ) such that ξ(0) = y, ξ(T ) = z, ¯ u) ¯ + δ1 , I f (0, T , ξ, η) ≤ Tf (x, ξ(t) − x ¯ ≤ δ1 , t ∈ [0, T ], η(t) ∈ {u, ¯ −u}, ¯ t ∈ [bf , T − bf ] a. e., ¯ −u} ¯ = ∅, t ∈ [0, bf ] ∪ [T − bf , T ] a. e. B(η(t), δ1 ) ∩ {u, By Lemma 6.12, there exist L1 > 2bf and a neighborhood U3 of f in M such that the following property holds: (v) For each g ∈ U3 , each T > L1 , and each (ξ, η) ∈ X(0, T ) such that ξ(0) ∈ BL0 ,

6.9 Proof of Theorem 6.11

237

at least of the conditions below holds: L0 , ξ(T ) ∈ B I g (0, T , ξ, η) ≤ U g (0, T , ξ(0), ξ(T )) + L0 ; I g (0, T , ξ, η) ≤ σ g (0, T , ξ(0)) + L0 and that for each S ≥ 0 satisfying [S, S + L1 + 1] ⊂ [0, T ], ξ(S) ∈ BL0 there exists t0 ∈ [S + 1, S + L1 + 1] such that ¯ ≤ δ. ξ(t0 ) − x Set L = 8L0 + 8 + 8L1 .

(6.183)

There exists a neighborhood U4 of f˜ in M such that for each g ∈ U4 , |g(t, ξ, η) − f (ξ, η)| ≤ δ1 L−1

(6.184)

for all t ≥ 0 and all (ξ, η) ∈ (BE (0, 4L0 + 4) × BF (0, 4L0 + 4)) ∩ M. Set U = ∩4i=1 Ui .

(6.185)

g ∈ U , T > 2L,

(6.186)

x(0) ∈ BL0

(6.187)

Assume that

(x, u) ∈ X(0, T ) satisfies

238

6 Infinite Dimensional Optimal Control

and at least one of the conditions below holds: (a) L0 , x(T ) ∈ B I g (0, T , x, u) ≤ U g (0, T , x(0), x(T )) + δ; (b) I g (0, T , x, u) ≤ σ g (0, T , x(0)) + δ. Property (v), conditions (a), (b), and equations (6.183) and (6.185)–(6.187) imply that there exists t0 ∈ [0, L1 + 1]

(6.188)

¯ ≤ δ. x(t0 ) − x

(6.189)

such that

If x(0) − x ¯ ≤ δ, then we set t0 = 0. Assume that p ≥ 0 is an integer and we already defined numbers t0 < · · · < tp from the interval [0, T ] such that for every i ∈ {0, . . . , p}, ¯ ≤ δ, x(ti ) − x

(6.190)

ti+1 − ti ∈ [1, L1 + 1].

(6.191)

for every i ∈ {0, . . . , p} \ {p},

If tp > T − L1 − 1, then the construction is completed. In view of (6.180) and (6.190), L0 . x(tp ) ∈ BL0 ∩ B

(6.192)

If T − tp ≥ L1 + 1, then by property (v), conditions (a), (b), and equations (6.190), (6.185)–(6.187), and (6.192), there exists tp+1 ∈ [tp + 1, tp + L1 + 1]

6.9 Proof of Theorem 6.11

239

such that ¯ ≤δ x(tp+1 ) − x and the assumption made for p also holds for p + 1. Thus by induction we constructed a strictly increasing finite sequence of numbers ¯ ≤ δ, t0 < · · · < tq from the interval [0, T ] such that (6.188) holds, if x(0) − x then t0 = 0, for every i ∈ {0, . . . , q}, ¯ ≤ δ, x(ti ) − x for every i ∈ {0, . . . , q − 1}, ti+1 − ti ∈ [1, L1 + 1] and T − tq < L1 + 1. If x(T ) − x ¯ > δ, then we set t˜i = ti , i = 0, . . . , q. If x(T ) − x ¯ ≤ δ, then we set t˜i = ti , i = 0, . . . , q − 1, t˜q = T . It is not difficult to see that for all i = 0, . . . , q, ¯ ≤ δ, x(t˜i ) − x

(6.193)

t˜i+1 − t˜i ∈ [1, 2L1 + 2],

(6.194)

t˜0 ∈ [0, L1 + 1], T − t˜q < L1 + 1,

(6.195)

for every i ∈ {0, . . . , q − 1},

240

6 Infinite Dimensional Optimal Control

if x(0) − x ¯ ≤ δ, then t˜0 = 0 and if x(T ) − x ¯ ≤ δ, then t˜q = T . In order to complete the prove of the theorem it is sufficient to show that x(t) − x ¯ ≤ , t ∈ [t˜0 , t˜q ]. Assume the contrary. Then there exists τ ∈ [t˜0 , t˜q ]

(6.196)

x(τ ) − x ¯ > .

(6.197)

such that

In view of (6.183) and (6.195), t˜q − t˜0 ≥ T − 2L1 − 2 ≥ 2L − 2L1 − 2 > L. By (6.183), (6.195), (6.197), and the equation above, there exist j1 , j2 ∈ {0, . . . , q} such that j1 < j2 , τ ∈ [t˜j1 , t˜j2 ],

(6.198)

L1 ≤ t˜j2 − t˜j1 ≤ 2L1 + 1.

(6.199)

Property (iv) and equations (6.193) and (6.199) imply that there exists (ξ, η) ∈ X(t˜j1 , t˜j2 ) such that ξ(t˜j1 ) = x(t˜j1 ), ξ(t˜j2 ) = x(t˜j2 ),

(6.200)

I f (t˜j1 , t˜j2 , ξ, η) ≤ (t˜j2 − t˜j1 )f (x, ¯ y) ¯ + δ1 ,

(6.201)

ξ(t) − x ¯ ≤ δ1 , t ∈ [t˜j1 , t˜j2 ],

(6.202)

η(t) ∈ {u, ¯ −u}, ¯ t ∈ [t˜j1 + bf , t˜j2 − bf ] a. e.,

(6.203)

¯ −u} ¯ = ∅, t ∈ [t˜j1 , t˜j1 + bf ] ∪ [t˜j2 − bf , t˜j2 ] a. e., B(η(t), δ1 ) ∩ {u,

(6.204)

6.9 Proof of Theorem 6.11

241

It follows from (6.184), (6.186), (6.201)–(6.204) that for all t ∈ [t˜j1 , t˜j2 ] |g(t, ξ(t), η(t)) − f (ξ(t), η(t))| ≤ δ1 L−1 and in view of (6.183) and (6.199), I g (t˜j1 , t˜j2 , ξ, η) ≤ I f (t˜j1 , t˜j2 , ξ, η) + δ1 L−1 (t˜j2 − t˜j1 ) ¯ u) ¯ + 2δ1 . ≤ (t˜j2 − t˜j1 )f (x,

(6.205)

We will estimate I g (t˜j1 , t˜j2 , x, u). Conditions (a), (b) and (6.200) and (6.205) imply that I g (t˜j1 , t˜j2 , x, u) ≤ I f (t˜j1 , t˜j2 , ξ, η) + δ ¯ u) ¯ + 3δ1 . ≤ (t˜j2 − t˜j1 )f (x,

(6.206)

Let t ∈ [t˜j1 , t˜j2 ]. If max{x(t) − x, ¯ min{u(t) − u, ¯ u(t) + u}} ¯ > /4, then by property (i) and (6.186), g(t, x(t), u(t)) > f (x, ¯ y) ¯ + 0 .

(6.207)

If max{x(t) − x, ¯ min{u(t) − u, ¯ u(t) + u}} ¯ ≤ /4, then by (6.184) and (6.186), |g(t, x(t), u(t)) − f (x(t), u(t))| ≤ δ1 L−1 . In view of (6.207) and (6.208), for a. e. t ∈ [t˜j1 , t˜j2 ], g(t, x(t), u(t)) ≥ f (x, ¯ u) ¯ − δ1 L−1

(6.208)

242

6 Infinite Dimensional Optimal Control

and ¯ u) ¯ − δ1 L−1 (t˜j2 − t˜j1 ) I g (t˜j1 , t˜j2 , x, u) ≥ (t˜j2 − t˜j1 )f (x, ¯ u) ¯ − δ1 . ≥ (t˜j2 − t˜j1 )f (x,

(6.209)

By (6.206) and (6.209), for every measurable set Ω ⊂ [t˜j1 , t˜j2 ],  Ω

g(t, x(t), u(t))dt = I g (t˜j1 , t˜j2 , x, u) −

 [t˜j1 ,t˜j2 ]\Ω

g(t, x(t), u(t))dt

≤ (t˜j2 − t˜j1 )f (x, ¯ u) ¯ + 3δ1 − mes([t˜j1 , t˜j2 ] \ Ω)(f (x, ¯ u) ¯ − δ1 L−1 ) ≤ mes(Ω)f (x, ¯ u) ¯ + 4δ1 .

(6.210)

S = max{t˜j1 , τ − γ0 }.

(6.211)

Set

Clearly, [S, S + γ0 ] ⊂ [t˜j1 , t˜j2 ]. By (6.200), (6.211) and the inclusion above, there exists S1 ≥ 0 such that [S, S + γ0 ] ⊂ [S1 , S1 + 1] ⊂ [t˜j1 , t˜j2 ].

(6.212)

Property (ii), (6.182), (6.193), (6.197), and (6.210)–(6.212) imply that τ − γ0 > t˜j1 ,

(6.213)

x(t) − x ¯ > γ0 , t ∈ [τ − γ0 , τ ].

(6.214)

S = τ − γ0 > t˜j1 .

(6.215)

In view of (6.213),

Property (iii) and (6.214) imply that g(t, x(t), u(t)) ≥ f (x, ¯ u) ¯ + 1 , t ∈ [τ − γ0 , τ ].

(6.216)

6.10 Stability of the Weak Turnpike Phenomenon

243

It follows from (6.182), (6.183), (6.199), (6.209), (6.215), and (6.216) that  g ˜ g I (tj1 , t˜j2 , x, u) = I (τ − γ0 , τ, x, u) + g(t, x(t), u(t))dt [t˜j1 ,t˜j2 ]\[τ −γ0 ,τ ]

≥ γ0 (f (x, ¯ u) ¯ + 1 ) + (t˜j2 − t˜j1 − γ0 )(f (x, ¯ u) ¯ − δ1 L−1 ) ¯ u) ¯ + γ0 1 − δ1 ≥ (t˜j2 − t˜j1 )f (x, ≥ (t˜j2 − t˜j1 )f (x, ¯ u) ¯ + γ0 1 /2.

(6.217)

By (6.206) and (6.218), ¯ u) ¯ + γ0 1 ≤ I g (t˜j1 , t˜j2 , x, u) (t˜j2 − t˜j1 )f (x, ¯ u) ¯ + 3δ1 . ≤ (t˜j2 − t˜j1 )f (x,

(6.218)

In view of (6.218), γ0 1 ≤ 6δ1 . This contradicts (6.182). The contradiction we have reached completes the Proof of Theorem 6.11.

6.10 Stability of the Weak Turnpike Phenomenon Theorem 6.13 Let M0 , M1 ,  > 0. Then there exist l > 0, a natural number Q, and a neighborhood U of f˜ in M such that for each T > lQ, each g ∈ U and each (x, u) ∈ X(0, T ) such that x(0) ∈ BM0 and at least one of the conditions below holds: (a) M0 , x(T ) ∈ B I g (0, T , x, u) ≤ U g (0, T , x(0), x(T )) + M1 ; (b) I g (0, T , x, u) ≤ σ g (0, T , x(0)) + M1

(6.219)

244

6 Infinite Dimensional Optimal Control

there exists a sequence of closed intervals [ai , bi ] ⊂ [0, T ], i = 1, . . . , q such that 1 ≤ q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, ai+1 ≥ bi for all i ∈ {1, . . . , q} \ {q} and q

{t ∈ [0, T ] : x(t) − x ¯ > } ⊂ ∪i=1 [ai , bi ]. Proof We may assume without loss of generality that M0 . B(x, ¯ ) ⊂ BM0 ∩ B

(6.220)

By Theorem 6.11, there exist δ ∈ (0, min{1, }), L1 > 2bf + M0 and a neighborhood U1 of f˜ in M such that the following property holds: (i) For each g ∈ U1 , each T > 2L1 , and each (x, u) ∈ X(0, T ) that satisfies x(0) − x ¯ ≤ δ, x(T ) − x ¯ ≤ δ, I g (0, T , x, u) ≤ U g (0, T , x(0), x(T )) + δ we have x(t) − x ¯ ≤ , t ∈ [0, T ]. By Lemma 6.12, there exist L2 > L1 and a neighborhood U ⊂ U1 of f˜ in M such that the following property holds: (ii) For each g ∈ U , each T > L2 , and each (x, u) ∈ X(0, T ) such that (6.129) is true, at least one of the conditions (a), (b) holds and for each S ≥ 0 satisfying [S, S + L2 + 1] ⊂ [0, T ], x(S) ∈ BM0 ,

6.10 Stability of the Weak Turnpike Phenomenon

245

there exists t0 ∈ [S + 1, S + 1 + L2 ] such that ¯ ≤ δ. x(t0 ) − x Set l = 4L2 + 4.

(6.221)

Q > 8 + 2δ −1 M1 .

(6.222)

T > lQ, g ∈ U ,

(6.223)

Choose an integer

Assume that

(x, u) ∈ X(0, T ) satisfies x(S) ∈ BM0 , equation (6.129) and at least one of conditions (a), (b). Property (ii), conditions (a), (b), and equations (6.219), (6.221), and (6.223) imply that there exists t0 ∈ [0, L2 + 1]

(6.224)

¯ ≤ δ. x(t0 ) − x

(6.225)

such that

By induction, using (6.220), we construct a sequence of numbers t0 < t1 · · · < tq belonging to [0, T ] such that (6.224) holds, ¯ ≤ δ, i = 0, . . . , q x(ti ) − x ti+1 − ti ∈ [1, 1 + L2 ], i ∈ {0, . . . , q} \ {q}, tq > T − L2 − 1. Assume that p ≥ 0 is an integer and we already defined numbers t0 < · · · < tp from the interval [0, T ] such that (6.225) holds, for every i ∈ {0, . . . , p}, ¯ ≤ δ, x(ti ) − x

(6.226)

246

6 Infinite Dimensional Optimal Control

for every i ∈ {0, . . . , p} \ {p}, ti+1 − ti ∈ [1, L2 + 1]. (Note that in view of (6.225), for p = 0 our assumption is true.) If T − tp < L2 + 1, then the construction is completed. If T − tp ≥ L2 + 1, then property (ii), conditions (a), (b), (6.220) and (6.226) imply that there exists tp+1 ∈ [tp + 1, tp + L2 + 1] such that ¯ ≤δ x(tp ) − x and the assumption made for p also holds for p + 1. Thus by induction we constructed a strictly increasing finite sequence of numbers q {ti }i=0 ⊂ [0, T ] such that t0 ∈ [0, L2 + 1], tq ≥ T − L2 − 1,

(6.227)

for every i ∈ {0, . . . , q}, ¯ ≤ δ, x(ti ) − x

(6.228)

ti+1 − ti ∈ [1, L2 + 1].

(6.229)

for every i ∈ {0, . . . , q − 1},

Now we construct a strictly increasing sequence of numbers Si ∈ {t0 , . . . , tq }, i = 0, . . . , p. Set S0 = t0 . Assume that k ≥ 0 is an integer and that we already defined Si ∈ {t0 , . . . , tq }, i = 0, . . . , k.

(6.230)

6.10 Stability of the Weak Turnpike Phenomenon

247

If Sk = tq , then the construction is completed. Assume that Sk < tq . If I g (Sk , tk , x, u) ≤ U g (Sk , tq , x(Sk ), x(tq )) + δ, then we set Sk+1 = tq and the construction is completed. Assume that I g (Sk , tq , x, u) > U (Sk , tq , x(Sk ), x(tq )) + δ. There exists j ∈ {0, . . . , q} such that Sk = tj . If I g (Sk , tj +1 , x, u) > U g (Sk , tj +1 , x(Sk ), x(tj +1 )) + δ, then we set Sk+1 = tj +1 . Assume that I g (Sk , tj +1 , x, u) ≤ U g (Sk , tj +1 , x(Sk ), x(tj +1 )) + δ. Define Sk+1 = min{ti : i ∈ {j + 1, . . . , q} : I g (Sk , ti , x, u) > U (Sk , ti , x(Sk ), x(ti )) + δ}. Therefore by induction we defined a strictly increasing sequence of numbers Si ∈ {t0 , . . . , tq }, i = 0, . . . , p such that S0 = t0 , Sp = tq ,

(6.231)

248

6 Infinite Dimensional Optimal Control

for each j ∈ {0, . . . , p − 1} \ {p − 1}, I g (Sj , Sj +1 , x, u) > U g (Sj , Sj +1 , x(Sj ), x(Sj +1 )) + δ,

(6.232)

for each j ∈ {0, . . . , p − 1}, if i ∈ {0, . . . , q} and Sj < ti < Sj +1 , then I g (Sj , ti , x, u) ≤ U g (Sj , ti , x(Sj ), x(ti )) + δ.

(6.233)

Conditions (a), (b) and (6.226) imply that M ≥ I g (0, T , x, u) − U g (0, T , x(0), x(T )) ≥

 {I g (Sj , Sj +1 , x, u) − U g (Sj , Sj +1 , x(Sj ), x(Sj +1 )) : j ∈ {0, . . . , p − 1} \ {p − 1}} ≥ (p − 1)δ

and p ≤ 1 + δ −1 M1 .

(6.234)

Assume that j ∈ {0, . . . , p − 1} satisfies Sj +1 − Sj ≥ l.

(6.235)

Equations (6.221), (6.228), (6.229), (6.233), and (6.235) imply that there exists S˜j ∈ [Sj +1 − L2 − 1, Sj +1 − 1]

(6.236)

¯ ≤δ x(S˜j ) − x

(6.237)

such that

and I g (Sj , S˜j , x, u) ≤ U g (Sj , S˜j , x(Sj ), x(S˜j )) + δ.

(6.238)

Property (i) and equations (6.221), (6.223), (6.228), and (6.235)–(6.238) imply that x(t) − x ¯ ≤ , t ∈ [Sj , S˜j ].

(6.239)

6.11 The First Extension of Theorem 6.11

249

By (6.239), {t ∈ [0, T ] : x(t) − x ¯ > } ⊂ [0, S0 ] ∪ [Sp , T ] ∪ {[Si , Si+1 ] : i ∈ {0, . . . , p − 1}, Si+1 − Si ≤ l} ∪{[Si+1 − L2 − 1, Si+1 ] : i ∈ {0, . . . , p − 1}, Si+1 − Si ≥ l}. It is easy to see that the right-hand side of the equation above is a union of a finite number of closed intervals, their lengths do not exceed l, and in view of (6.222) and (6.234), their number does not exceed 2 + 2p ≤ 4 + δ −1 M1 ≤ Q. Theorem 6.13 is proved.

6.11 The First Extension of Theorem 6.11 In this section we prove an extension of Theorem 6.11. In this result the turnpike property is established for trajectory-control pairs (x, u) defined on intervals [0, T ], where T is sufficiently large, which are approximate solutions of the corresponding optimal control problems on subintervals of [0, T ] that have some fixed length L that does not depend on T . Theorem 6.14 Let M > 0 and L0 > x ¯ + u ¯ + 4,  ∈ (0, 1).

(6.240)

Then there exist L > L0 , δ ∈ (0, ) and a neighborhood U of f˜ in M such that for each T > 2L, each g ∈ U , and each (x, u) ∈ X(0, T ) such that x(0) ∈ BL0 , at least one of the conditions below holds: (a) L0 , x(T ) ∈ B I g (0, T , x, u) ≤ U g (0, T , x(0), x(T )) + M; (b) I g (0, T , x, u) ≤ σ g (0, T , x(0)) + M

(6.241)

250

6 Infinite Dimensional Optimal Control

and such that for each S ∈ [0, T − L], I g (S, S + L, x, u) ≤ U g (S, S + L, x(S), x(S + L)) + δ there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ] such that x(t) − x ¯ ≤ , t ∈ [τ1 , τ2 ]. ¯ ≤ δ, then τ2 = T . Moreover, x(0) − x ¯ ≤ δ, then τ1 = 0 and if x(T ) − x Proof We may assume without loss of generality that L0 ,  < 1. BE (x, ¯ ) ⊂ BL0 ∩ B

(6.242)

By Theorem 6.11, there exist δ ∈ (0, ), L1 > L0 + 2bf and a neighborhood U1 of f˜ in M such that the following property holds: (i) For each g ∈ U1 , each T ≥ 2L1 , and each (x, u) ∈ X(0, T ) that satisfies x(s) − x ¯ ≤ δ, s = 0, T , I g (0, T , x, u) ≤ U g (0, T , x(0), x(T )) + δ we have x(t) − x ¯ ≤ , t ∈ [0, T ]. By Theorem 6.13, there exist l > 0, a natural number Q, and a neighborhood U ⊂ U1 of f˜ in M such that the following property holds: (ii) For each T > lQ, each g ∈ U , and each (x, u) ∈ X(0, T ) such that (6.241) is true and at least one of conditions (a), (b) holds there exists a sequence of closed intervals [ai , bi ] ⊂ [0, T ], i = 0, . . . , q such that 0 ≤ q ≤ Q, 0 ≤ bi − ai ≤ L, i = 0, . . . , q, ai+1 ≥ bi for all i ∈ {0, . . . , q} \ {q} (6.243)

6.11 The First Extension of Theorem 6.11

251

and q

{t ∈ [0, T ] : x(t) − x ¯ > δ} ⊂ ∪i=0 [ai , bi ].

(6.244)

Fix a number L > 8lQ + 8 + 8L0 + 8L1 .

(6.245)

T > 2L, g ∈ U

(6.246)

Assume that

and (x, u) ∈ X(0, T ) satisfies (6.241), at least one of conditions (a), (b) and satisfies for each S ∈ [0, T − L], I g (S, S + L, x, u) ≤ U g (S, S + L, x(S), x(S + L)) + δ.

(6.247)

Property (ii), conditions (a), (b), and equations (6.241) and (6.246) imply that there exists a sequence of closed intervals [ai , bi ] ⊂ [0, T ], i = 0, . . . , q such that q ≤ Q and (6.243) and (6.244) are true. Thus the following property holds: (iii) For each S ∈ [0, T − lQ − 1] there exists τS ∈ [S, S + lQ + 1]

(6.248)

¯ ≤ δ. x(τS ) − x

(6.249)

such that

In particular, there exist τ1 ∈ [0, lQ + 1], τ2 ∈ [T − lQ − 1, T ]

(6.250)

¯ ≤ δ, i = 1, 2. x(τi ) − x

(6.251)

such that

If x(0) − x ¯ ≤ δ, then we set τ1 = 0 and if x(T ) − x ¯ ≤ δ,

252

6 Infinite Dimensional Optimal Control

then we set τ2 = T . In view of (6.245), (6.246), and (6.250), τ2 − τ1 ≥ T − 2(lQ + 1) > 2L − 2(lQ + 1).

(6.252)

Let τ ∈ [τ1 , τ2 ].

(6.253)

Define numbers S1 , S2 as follows. If τ ≤ lQ + 1 + 2L1 + τ1 , then set S1 = τ1 , otherwise S1 = τ − lQ − 1 − 2L1 .

(6.254)

If τ ≥ τ2 − lQ − 1 − 2L1 , then set S2 = τ2 , otherwise S2 = τ + lQ + 1 + 2L1 .

(6.255)

In view of (6.245), (6.252), (6.254), and (6.255), S2 − S1 ≥ lQ + 1 + 2L1 . If S1 = τ1 , then set S˜1 = τ1 .

(6.256)

If S1 > τ1 , then property (iii) implies that there exists S˜1 ∈ [S1 , S1 + lQ + 1] such that ¯ ≤ δ. x(S˜1 ) − x

(6.257)

S˜2 = τ2 ;

(6.258)

If S2 = τ2 , set

6.12 The Second Extension of Theorem 6.11

253

otherwise S2 < τ2 and property (iii) implies that there exists S˜2 ∈ [S2 − lQ − 1, S2 ] such that ¯ ≤ δ. x(S˜2 ) − x

(6.259)

It is not difficult to see that in all these cases S˜1 ≤ τ ≤ S˜2 . It follows from (6.245), (6.252), and the definitions above that in all these cases L ≥ S2 − S1 ≥ S˜2 − S˜1 ≥ 2L1 .

(6.260)

By (6.247), I g (S˜1 , S˜2 , x, u) ≤ U g (S˜1 , S˜2 , x(S˜1 ), x(S˜2 )) + δ.

(6.261)

Property (i) and equations (6.251), (6.256)–(6.261) imply that x(t) − x ¯ ≤ , t ∈ [S˜1 , S˜2 ] and in particular x(τ ) − x ¯ ≤ . Theorem 6.14 is proved.

6.12 The Second Extension of Theorem 6.11 In this section we prove the second extension of Theorem 6.11. In this result we analyze the behavior of the control function u(·). Theorem 6.15 Let L0 > x ¯ + y ¯ + 4 + 4bf ,  ∈ (0, 1) and L1 , M > 0. Then there exist L > L0 , δ ∈ (0, ) and a neighborhood U of f˜ in M such that for each T > 2L, each g ∈ U , and each (x, u) ∈ X(0, T ) such that x(0) ∈ BL0 ,

(6.262)

254

6 Infinite Dimensional Optimal Control

at least one of the conditions below holds: (a) L0 , x(T ) ∈ B I g (0, T , x, u) ≤ U g (0, T , x(0), x(T )) + M; (b) I g (0, T , x, u) ≤ σ g (0, T , x(0)) + M and such that the following condition holds: (c) For each S ∈ [0, T − L], I g (S, S + L, x, u) ≤ U g (S, S + L, x(S), x(S + L)) + δ there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ] such that for each τ ∈ [τ1 , τ2 − L0 ], the inequality mes({t ∈ [τ, τ + L0 ] : min{u(t) − u, ¯ u(t) + u} ¯ ≥ }) ≤  ¯ ≤ δ, then holds. Moreover, if x(0) − x ¯ ≤ δ, then τ1 = 0 and if x(T ) − x τ2 = T . Proof By Proposition 6.9, there exist δ0 ∈ (0, ) and a neighborhood U1 of f˜ in M such that the following property holds: (i) for each g ∈ U1 and each (t, ξ, η) ∈ [0, ∞) × M satisfying min{η − u, ¯ η + u} ¯ ≥ /2 the inequality g(t, ξ, η) ≥ f (x, ¯ u) ¯ + δ0 is valid. Theorem 6.2 and assumption (A4) imply that there exist δ1 ∈ (0, δ0 )

6.12 The Second Extension of Theorem 6.11

255

such that the following property holds: (ii) for each T > 2bf and each ¯ δ1 ) y, z ∈ A ∩ BE (x, there exists (ξ, η) ∈ X(0, T ) such that ξ(0) = y, ξ(T ) = z, ¯ u) ¯ + 8−1 δ0 , I f (0, T , ξ, η) ≤ Tf (x, ξ(t) − x ¯ ≤ δ0 t ∈ [0, T ], η(t) ∈ ∩{u, ¯ −u}, ¯ t ∈ [bf , T − bf ] a. e., ¯ −u} ¯ = ∅, t ∈ [0, bf ] ∪ [T − bf , T ] a. e. B(η(t), δ0 ) ∩ {u, There exists a neighborhood U2 of f˜ in M such that the following property holds: (iii) For each g ∈ U2 , |g(t, ξ, η) − f (ξ, η)| ≤ 8−1 δ1 (L0 + 1)−1 for all t ≥ 0 and all (ξ, η) ∈ (BE (0, L0 + M) × BF (0, L0 + M)) ∩ M. Theorem 6.14 implies that there exist L > L0 , δ ∈ (0, 8−1 δ0 ) and a neighborhood U3 of f˜ in M such that the following property holds: (iv) For each T > 2L, each g ∈ U3 , and each (x, u) ∈ X(0, T ) such that (6.262) holds, at least one of the conditions (a), (b) holds and such that for each S ∈ [0, T − L], I g (S, S + L, x, u) ≤ U g (S, S + L, x(S), x(S + L)) + δ there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ] such that x(t) − x ¯ ≤ δ1 , t ∈ [τ1 , τ2 ].

256

6 Infinite Dimensional Optimal Control

Moreover, if x(0) − x ¯ ≤ δ, then τ1 = 0 and if x(T ) − x ¯ ≤ δ, then τ2 = T . Set U = ∩3i=1 Ui .

(6.263)

T > 2L, g ∈ U

(6.264)

Assume that

and (x, u) ∈ X(0, T ) satisfies (6.262), at least one of conditions (a), (b) holds and property (c) holds. Properties (iv) and (c), conditions (a), (b), and equations (6.263) and (6.264) imply that there exist τ1 ∈ [0, L], τ2 ∈ [T − L, T ]

(6.265)

x(t) − x ¯ ≤ δ1 , t ∈ [τ1 , τ2 ];

(6.266)

such that

if x(0) − x ¯ ≤ δ, then τ1 = 0 and if x(T ) − x ¯ ≤ δ, then τ2 = T . Let τ ≥ 0 satisfy [τ, τ + L0 ] ⊂ [τ1 , τ2 ].

(6.267)

In order to complete the proof of the theorem it is sufficient to show that mes({t ∈ [τ, τ + L0 ] : min{u(t) − u, ¯ u(t) + u} ¯ ≥ }) ≤ . Assume the contrary. Then ¯ u(t) + u} ¯ ≥ }) > . mes({t ∈ [τ, τ + L0 ] : min{u(t) − u,

(6.268)

Property (ii) and equations (6.266) and (6.267) imply that there exists (ξ, η) ∈ X(τ, τ + L0 ) such that ξ(τ ) = x(τ ), ξ(τ + L0 ) = x(τ + L0 ),

(6.269)

¯ u) ¯ + 8−1 δ0 , I f (τ, τ + L0 , ξ, η) ≤ L0 f (x,

(6.270)

ξ(t) − x ¯ ≤ δ0 , t ∈ [τ, τ + L0 ],

(6.271)

¯ −u} ¯ = ∅, BF (η(t), δ0 ) ∩ {u, t ∈ [τ, τ + bf ] ∪ [τ + L0 − bf , τ + L0 ] a. e.,

(6.272)

η(t) ∈ {u, ¯ −u}, ¯ t ∈ [τ + bf , τ + L0 − bf ] a. e.

(6.273)

6.12 The Second Extension of Theorem 6.11

257

It follows from property (iii) and equations (6.271)–(6.273) that |f (ξ(t), ξ (t)) − g(t, ξ(t), ξ (t))| ≤ 8−1 δ1 (L0 + 1)−1

(6.274)

for t ∈ [τ, τ + L0 ] a. e. By (6.270) and (6.274), I g (τ, τ + L0 , ξ, η) ≤ I f (τ, τ + L0 , ξ, η) + 8−1 δ1  ¯ u) ¯ + 4−1 δ0 . ≤ L0 f (x,

(6.275)

Conditions (a), (b) and equations (6.269) and (6.275) imply that I g (τ, τ + L0 , x, u) ≤ I g (τ, τ + L0 , ξ, η) + δ ¯ u) ¯ + 3 · 8−1 δ0 . ≤ L0 f (x,

(6.276)

Set ¯ u(t) + u} ¯ ≥ }. E = {t ∈ [τ, τ + L0 ] : min{u(t) − u,

(6.277)

Property (i) and (6.277) imply that for t ∈ E a. e., g(t, x(t), u(t)) ≥ f (x, ¯ u) ¯ + δ0 .

(6.278)

By (6.274), for any t ∈ [τ, τ + L0 ] \ E we have |g(t, x(t), u(t)) − f (x(t), u(t))| ≤ 8−1 δ0 (L0 + 1)−1 and g(t, x(t), u(t)) ≥ f (x(t), u(t)) − 8−1 δ0 (L0 + 1)−1 ≥ f (x, ¯ u) ¯ − 8−1 δ0 (L0 + 1)−1 . It follows from (6.277)–(6.279) that   g I (τ, τ + L0 , x, u) = g(t, x(t), u(t))dt + E

[τ,τ +L0 ]\E

≥ (f (x, ¯ u) ¯ + δ0 )mes(E)

(6.279)

g(t, x(t), u(t))dt

258

6 Infinite Dimensional Optimal Control

+(f (x, ¯ u) ¯ − 8−1 δ0 (L0 + 1)−1 )(L0 − mes(E)) ¯ u) ¯ + mes(E)δ0 − 8−1 δ0  ≥ L0 f (x, ¯ u) ¯ + 2−1 δ0 . ≥ L0 f (x, This contradicts (6.275). The contradiction we have reached proves Theorem 6.15.

6.13 The Turnpike Property in the Regions Close to the Right end Points In this section we prove the following result that shows that the turnpike phenomenon also holds in the regions close to the right end points. Theorem 6.16 Let L0 , L0 > x ¯ + u ¯ + 4 + 2bf , BE (x, ¯ ) ∩ A ⊂ B

(6.280)

and  ∈ (0, 1). Then there exist L > L0 , δ ∈ (0, ) and a neighborhood U of f˜ in M such that for each T > 2L, each g ∈ U , and each (x, u) ∈ X(0, T ) such that x(0) ∈ BL0 ,

(6.281)

I g (0, T , x, u) ≤ σ g (0, T , x(0)) + L0 ,

(6.282)

and such that for each S ∈ [0, T − L], I g (S, S + L, x, u) ≤ U g (S, S + L, x(S), x(S + L)) + δ

(6.283)

I g (T − L, T , x, u) ≤ U g (T − L, T , x(T − L)) + δ

(6.284)

and

there exist τ0 ∈ [0, L] such that x(t) − x ¯ ≤ , t ∈ [τ0 , T ]

6.13 The Turnpike Property in the Regions Close to the Right end Points

259

and for each τ ∈ [τ0 , T − L0 ], the inequality ¯ u(t) + u} ¯ ≥ }) ≤  mes({t ∈ [τ, τ + L0 ] : min{u(t) − u, holds. Moreover, if x(0) − x ¯ ≤ δ, then τ0 = 0. Proof Proposition 6.6 implies that there exists 0 ∈ (0, 4−1 )

(6.285)

such that the following property holds: (i) For each S1 ≥ 0, each S2 ∈ [S1 + 1, S1 + 2], each g ∈ M, each (x, u) ∈ X(S1 , S2 ) satisfying ¯ u)| ¯ +4 I g (S1 , S2 , x, u) ≤ 2|f (x, and each t1 , t2 ∈ [S1 , S2 ] satisfying 0 < t2 − t1 ≤ 0 and ¯ ≤ 0 x(t1 ) − x the inequality ¯ ≤ /8 x(t2 ) − x holds. By Proposition 6.9, there exist 1 ∈ (0, 4−1 0 ) and a neighborhood U1 of f˜ in M such that the following property holds: (ii) For each g ∈ U1 and each (t, ξ, η) ∈ [0, ∞) × M satisfying max{ξ − x, ¯ min{η − u, ¯ η + u}} ¯ > 4−1 0 the inequality g(t, ξ, η) ≥ f (x, ¯ u) ¯ + 21 is valid. Theorems 6.14 and 6.15 imply that there exist L1 > 2L0 + 4, δ ∈ (0, 16−1 12 )

(6.286)

260

6 Infinite Dimensional Optimal Control

and a neighborhood U2 of f˜ in M such that the following property holds: (iii) For each T > 2L1 , each g ∈ U2 , and each (x, u) ∈ X(0, T ) satisfying (6.281), at least one of the following conditions: equation (6.282) is true; L0 , x(T ) ∈ B I g (0, T , x, u) ≤ U g (0, T , x(0), x(T )) + L0 and such that for each S ∈ [0, T − L1 ], I g (S, S + L1 , x, u) ≤ U g (S, S + L1 , x(S), x(S + L1 )) + δ there exist τ0 ∈ [0, L1 ], τ1 ∈ [T − L1 , T ] such that x(t) − x ¯ ≤ 1 /2, t ∈ [τ0 , τ1 ] and for each τ ∈ [τ0 , τ1 − L0 ], the inequality ¯ u(t) + u} ¯ ≥ 2−1 1 }) ≤ 1 /2 mes({t ∈ [τ, τ + L0 ] : min{u(t) − u, holds and moreover, if x(0) − x ¯ ≤ δ, then τ0 = 0 and if x(T ) − x ¯ ≤ δ, then τ1 = T . Proposition 6.5 implies that there exist δ˜ ∈ (0, δ) such that the following property holds: (iv) For each τ > 2bf and each ˜ y, z ∈ BE (x, ¯ δ) there exists (ξ, η) ∈ X(0, τ ) such that ξ(0) = y, ξ(τ ) = z, ¯ u) ¯ + 8−1 δ, I f (0, τ, ξ, η) ≤ τf (x, ξ(t) − x ¯ ≤ δ, t ∈ [0, τ ], B(η(t), δ) ∩ {u, ¯ −u} ¯ = ∅, t ∈ [0, τ ] a. e.

(6.287)

6.13 The Turnpike Property in the Regions Close to the Right end Points

261

By Theorem 6.13, there exist l > 0, a natural number Q, and a neighborhood U3 of f˜ in M such that the following property holds: (v) For each T > lQ, each g ∈ U3 , and each (x, u) ∈ X(0, T ) such that (6.281) and (6.282) are true there exists a sequence of closed intervals [ai , bi ] ⊂ [0, T ], i = 1, . . . , q such that 1 ≤ q ≤ Q, 0 ≤ bi − ai ≤ l, i = 1, . . . , q, ai+1 ≥ bi for all i ∈ {1, . . . , q} \ {q} and ˜ ⊂ ∪q [ai , bi ]. {t ∈ [0, T ] : x(t) − x ¯ > δ} i=1 Choose a number L > lQ + 4L1 + 4bf + 4.

(6.288)

There exists a neighborhood U4 of f˜ in M such that for each g ∈ U4 and each (t, ξ, η) ∈ [0, ∞) × M satisfying ξ , η ≤ L0 + 4 we have |g(t, ξ, η) − f (ξ, η)| ≤ 16−1 L−1 δ.

(6.289)

U = ∩4i=1 Ui .

(6.290)

T > 2L, g ∈ U ,

(6.291)

Set

Assume that

(x, u) ∈ X(0, T ) satisfies (6.281), (6.282), equation (6.283) holds for each S ∈ [0, T − L] and (6.284) is true. Property (v) and equations (6.288), (6.290) and (6.291) imply that there exists a number S ∈ [T − L, T − 2bf − 2]

(6.292)

262

6 Infinite Dimensional Optimal Control

such that ˜ x(S) − x ¯ ≤ δ.

(6.293)

Property (iv) and equations (6.288), (6.291)–(6.293) imply that there exists (ξ, η) ∈ X(S, T ) such that ξ(S) = x(S), ξ(T ) = x, ¯

(6.294)

¯ u) ¯ + 8−1 δ, I f (S, T , ξ, η) ≤ (T − S)f (x,

(6.295)

ξ(t) − x ¯ ≤ δ, t ∈ [S, T ],

(6.296)

B(η(t), δ) ∩ {u, ¯ −u} ¯ = ∅, t ∈ [S, T ] a. e.

(6.297)

By (6.279), (6.289)–(6.291), (6.294), (6.295), for a. e. t ∈ [S, T ], |f (ξ(t), η(t)) − g(t, ξ(t), η(t))| ≤ (16L)−1 δ.

(6.298)

In view of (6.292) and (6.298), |I g (S, T , ξ, η) − I f (S, T , ξ, η)| ≤ 8−1 δ.

(6.299)

Equations (6.295) and (6.299) imply that ¯ u) ¯ + 4−1 δ. I g (S, T , ξ, η) ≤ I f (S, T , ξ, η) + 8−1 δ ≤ (T − S)f (x,

(6.300)

It follows from (6.284), (6.292), (6.294), and (6.300) that ¯ u) ¯ + δ + 4−1 δ. I g (S, T , x, u) ≤ I g (S, T , ξ, η) + δ ≤ (T − S)f (x,

(6.301)

Set E = {t ∈ [S, T ] : max{x(t) − x, ¯ min{u(t) − u, ¯ u(t) + u}} ¯ ≥ /4}.

(6.302)

Property (ii) and equations (6.290), (6.291), and (6.302) imply that for each t ∈ E, g(t, x(t), u(t)) ≥ f (x, ¯ u) ¯ + 21 .

(6.303)

By (6.279), (6.289)–(6.291), and (6.302), for each t ∈ [S, T ] \ E, |g(t, x(t), u(t)) − f (x(t), u(t))| ≤ (16L)−1 δ.

(6.304)

6.13 The Turnpike Property in the Regions Close to the Right end Points

263

It follows from (6.292), (6.301), (6.303), and (6.304) that  (f (x, ¯ u) ¯ + 21 )mes(E) ≤ g(t, x(t), u (t))dt E

 ≤ I g (S, T , u) −

[S,T ]\E

g(t, x(t), u(t))dt

≤ (T − S)f (x, ¯ u) ¯ + 4−1 δ + δ −mes([S, T ] \ E)(f (x, ¯ u) ¯ − (16L)−1 δ) ≤ mes(E)f (x, ¯ u) ¯ + 4−1 δ + 8−1 δ + δ and in view of (6.286), mes(E) ≤ (21 )−1 2δ ≤ 1 /16.

(6.305)

In view of (6.302) and (6.305), mes({t ∈ [S, T ] : min{u(t) − u, ¯ u(t) + u} ¯ ≥ /4}) ≤ 1 /16,

(6.306)

mes({t ∈ [S, T ] : x(t) − x ¯ ≥ /4}) ≤ 1 /16.

(6.307)

Assume that τ ∈ [S, T ],

(6.308)

x(τ ) − x ¯ > /2.

(6.309)

T − S ≥ 2bf + 2.

(6.310)

a = max{S, τ − 1}.

(6.311)

In view of (6.292),

Set

By (6.310) and (6.311), τ ∈ [a, a + 1] ⊂ [S, T ].

264

6 Infinite Dimensional Optimal Control

Equations (6.292), (6.301), (6.303), (6.304), and the inclusion above imply that  I g (a, a + 1, x, u) = I g (S, T , x, u) − g(t, x(t), u(t))dt [S,T ]\[a,a+1]

¯ u) ¯ − 16−1 L−1 δ) ≤ (T − S)f (x, ¯ u) ¯ + δ + 4−1 δ − (S − T − 1)(f (x, ≤ f (x, ¯ u) ¯ + δ + δ/2.

(6.312)

Property (i) and equations (6.293), (6.309), (6.311), and (6.312) imply that τ − 0 > S

(6.313)

and that for all t ∈ [a, a + 1] ∩ [τ − 0 , τ ] we have x(t) − x ¯ > 0 .

(6.314)

In view of (6.311) and (6.313), for all τ ∈ [a, a + 1] ⊂ [S, T ] we have mes([a, a + 1] ∩ [τ − 0 , τ ]) ≥ 0 . Property (ii) and equations (6.290), (6.291), and (6.314) imply that for each t ∈ [a, a + 1] ∩ [τ − 0 , τ ] we have g(t, x(t), u(t)) ≥ f (x, ¯ u) ¯ + 2¯1 . By (6.303), (6.304), (6.315), and the relation above,  I g (a, a + 1, x, u) =

[a,a+1]∩[τ −0 ,τ ]

g(t, x(t), u(t))dt

 +

[a,a+1]\[τ −0 ,τ ]

g(t, x(t), u(t))dt

≤ (f (x, ¯ u) ¯ + 21 )mes([a, a + 1] ∩ [τ − 0 , τ ])

(6.315)

6.13 The Turnpike Property in the Regions Close to the Right end Points

265

+mes([a, a + 1] \ [τ − 0 , τ ])(f (x, ¯ u) ¯ − 16−1 L−1 δ) ≥ f (x, ¯ u) ¯ + 20 1 − 16−1 δ. Together with (6.312) this implies that f (x, ¯ u) ¯ + 20 1 − 16−1 δ ≤ I g (a, a + 1, x, u) ≤ f (x, ¯ u) ¯ + δ + 2−1 δ, 1 0 ≤ 12δ. This contradicts (6.286). The contradiction we have reached proves that x(τ ) − x ¯ ≤ /2 for all τ ∈ [S, T ].

(6.316)

Property (iii) applied to the restriction (x, u) ∈ X(0, S) and equations (6.279)– (6.282), (6.284), (6.287), (6.288), (6.291)–(6.293) imply that there exists τ0 ∈ [0, L1 ] such that x(t) − x ¯ ≤ 1 /2, t ∈ [τ0 , S]

(6.317)

and for all τ ∈ [τ0 , S − L0 ], we have mes({t ∈ [τ, τ + L0 ] : min{u(t) − u, ¯ u(t) + u} ¯ ≥ 1 /2}) ≤ 1 /2

(6.318)

and if x(0) − x ¯ ≤ δ, then τ0 = 0. By (6.315) and (6.317), x(t) − x ¯ ≤ , t ∈ [τ0 , T ]. Let τ ∈ [τ0 , T − L0 ]. In view of (6.306) and (6.318), mes({t ∈ [τ, τ + L0 ] : min{u(t) − u, ¯ u(t) + u} ¯ ≥ }) ≤ /16 + /2. This completes the Proof of Theorem 6.16.

(6.319)

Chapter 7

Optimization Problems Arising in Crystallography

In this chapter we study symmetric optimization problems arising in crystallography, which are identified with the corresponding space of objective functions, equipped with an appropriate complete metric. Using the Baire category approach and the porosity notion, we show that a typical (generic) problem has exactly two different solutions and is well-posed.

7.1 Preliminaries We study the structure of minimizers of variational problems considered in [56, 66, 89, 136], which describe step-terraces on surfaces of crystals. It is well-known in surface physics that when a crystalline substance is maintained at a temperature T above its roughening temperature TR then the surface stored energy integrand, usually referred to as surface tension, is a smooth function β of the azimuthal angle of orientation θ . Furthermore, β obeys the following: β(−θ ) = β(π − θ ) = β(θ ), 0 < β(π/2) ≤ β(θ ) ≤ β(0). The classical model is given by  J (y) =

S

β(θ )ds 0

where s is arclength and y is a function defined on a fixed interval [0, L] whose graph is the locus under consideration: y ∈ W 1,1 (0, L), θ = arctan y ∈ [−π/2, π/2],

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 A. J. Zaslavski, Turnpike Phenomenon and Symmetric Optimization Problems, Springer Optimization and Its Applications 190, https://doi.org/10.1007/978-3-030-96973-8_7

267

268

7 Optimization Problems Arising in Crystallography

while β is a positive π -periodic function that belongs to a space of functions described below. Minimization of J subject to appropriate boundary data is a parametric variational problem. It is closely related to the variational problem defining the Wulff crystal shape as that shape for a domain of prescribed area such that the boundary integral with respect to arclength involving the integrand in J [referred to as the surface tension] attains its minimum value. For each function f : X → R 1 set inf(f ) = inf{f (x) : x ∈ X}. Denote by M the set of all functions β ∈ C 2 (R 1 ) that satisfy the following assumption: (A) β(t) ≥ 0 for every number t ∈ R 1 , β(π/2) ≤ β(t) ≤ β(0) for every number t ∈ R 1 ,

(7.1) (7.2)

β(t) = β(−t) for every number t ∈ R 1 ,

(7.3)

β(t + π ) = β(t) for every number t ∈ R 1 ,

(7.4)

β(0) + β

(0) ≤ 0.

(7.5)

For every pair of functions β1 , β2 ∈ M put (i)

(i)

ρ(β1 , β2 ) = sup{|β1 (t) − β2 (t)| : t ∈ R 1 , i = 0, 1, 2}.

(7.6)

It is not difficult to show that the metric space (M, ρ) is complete. Denote by Mr the collection of all functions β ∈ M that satisfy β(t) > 0 for all t ∈ R 1 ,

(7.7)

β(0) + β

(0) < 0.

(7.8)

It is clear that the set Mr is nonempty. Proposition 7.1 Mr is an open everywhere dense subset of the metric space (M, ρ). Proof Clearly, Mr is an open subset of the metric space (M, ρ). We will prove that Mr is an everywhere dense subset of the metric space (M, ρ). Let a pair of functions β ∈ M and β˜ ∈ Mr be given. Then for every integer n ≥ 1 the function β + n−1 β˜ ∈ Mr , ˜ ˜ >0 ≥ n−1 β(t) (β + n−1 β)(t)

7.1 Preliminaries

269

for every number t ∈ R 1 and ˜ ˜

(0) (β + n−1 β)(0) + (β + n−1 β) ˜ + β˜

(0)]/n < 0. = β(0) + β

(0) + [β(0) Hence β + n−1 β˜ ∈ Mr for every integer n ≥ 1. Clearly, β + n−1 β˜ → β as n → ∞ in the metric space (M, ρ). Therefore Mr is an everywhere dense subset of (M, ρ). This completes the proof of Proposition 7.1. Let β ∈ Mr be given. Set Gβ (z) = β(arctan(z))(1 + z2 )1/2 , z ∈ R 1 .

(7.9)

Evidently, Gβ is a continuous function and Gβ (z) → ∞ as z → ±∞, inf(Gβ < β(0).

(7.10)

Note that the final inequality in (7.10) was shown in [56]. We can rewrite the variational functional J in the form  J (y) =

L

Gβ (y )dx.

0

It was shown in [56] that a function y ∈ W 1,1 (0, L) is a minimizer of the functional J if and only if |y | ∈ {z ∈ R 1 : Gβ (z) = inf(Gβ )} a.e. In this chapter using the Baire category approach, we show that for a generic (typical) function β the set {z ∈ R 1 : Gβ (z) = inf(Gβ )} = {zβ , −zβ } where zβ is a unique positive number depending only on the function β.

270

7 Optimization Problems Arising in Crystallography

Denote by F the set of all β ∈ Mr satisfying the following condition: (C) There exists a number zβ ∈ R 1 such that Gβ (z) > Gβ (zβ ) for every number z ∈ R 1 \ {zβ , −zβ }.

(7.11)

We will prove the following result obtained in [89]. Theorem 7.2 F contains a countable intersection of open everywhere dense subsets of (M, ρ).

7.2 Auxiliary Results Proposition 7.3 Assume that β ∈ Mr . Then there exist a positive number M0 and a neighborhood U of the function β in the metric space M such that U ⊂ Mr and the following assertion holds: if φ ∈ U , z ∈ R 1 and Gβ (z) ≤ inf(Gβ ) + 1, then |z| ≤ M0 . Proof There exists a positive number c0 for which β(t) ≥ c0 for all t ∈ R 1 . There exists an open neighborhood U of the function β in the metric space (M, ρ) such that U ⊂ Mr

(7.12)

and that for every function φ ∈ U , we have φ(t) ≥ c0 /2 for all t ∈ R 1 ,

(7.13)

φ(0) ≤ 2β(0).

(7.14)

Gφ (z) ≤ inf(Gφ ) + 1.

(7.15)

Assume that φ ∈ M, z ∈ R 1 ,

Then equations (7.13) and (7.14) are valid. In view of equations (7.9) and (7.13)– (7.15), we have (1 + z2 )1/2 c0 /2 ≤ φ(arctan(z))(1 + z2 )1/2 = Gφ (z) ≤ Gφ (0) + 1 = φ(0) + 1 ≤ 2β(0) + 1

7.2 Auxiliary Results

271

and |z| ≤ 2(2β(0) + 1)c0−1 . Thus the assertion of Proposition 7.3 holds with M0 = 2(2β(0) + 1)c0−1 . It is not difficult to see that the next auxiliary result holds. Proposition 7.4 Assume that β ∈ Mr and , M > 0. Then there exists an open neighborhood U of the function β in the metric space M such that U ⊂ Mr and the following assertion holds: if φ ∈ U and z ∈ R 1 satisfies |z| ≤ M, then |Gφ (z) − Gβ (z)| ≤ . Proposition 7.5 Assume that β ∈ Mr and  > 0. Then there exists an open neighborhood U of the function β in the metric space M such that U ⊂ Mr and the following assertion holds: if φ1 , φ2 ∈ U and z ∈ R 1 satisfies, Gφ1 (z) ≤ inf(Gφ1 ) + 1,

(7.16)

then |Gφ1 (z) − Gφ2 (z)| ≤ . Proof Proposition 7.3 implies that there exist a positive number M and an open neighborhood U1 of the function β in the metric space (M, ρ) such that U 1 ⊂ Mr and the following property holds: (P1)

if φ ∈ U1 and z ∈ R 1 satisfies Gφ (z) ≤ inf(Gφ ) + 1,

then |z| ≤ M.

(7.17)

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7 Optimization Problems Arising in Crystallography

Proposition 7.4 implies that there exists an open neighborhood U of the function β in the metric space (M), ρ) such that U ⊂ U1 and the following property holds: (P2) if φ1 , φ2 ∈ U and z ∈ R 1 satisfies |z| ≤ M, then |Gφ1 (z) − Gφ2 (z)| ≤ .

(7.18)

Now assume that φ1 , φ2 ∈ U , z ∈ R 1 and that equation (7.16) is true. In view of equation (7.16) and property (P1), we have |z| ≤ M.

(7.19)

Equation (7.19) and property (P2) imply inequality (7.18). This completes the proof of Proposition 7.5. Proposition 7.6 Assume that β ∈ Mr and  > 0. Then there exists an open neighborhood U of the function β in the metric space M such that U ⊂ Mr and for every function φ ∈ U , we have | inf(Gφ (z) − inf(Gβ )| ≤ . Proof Let an open neighborhood U of the function β in the metric space M be as guaranteed by Proposition 7.5. Let φ1 , φ2 ∈ U be given. It is sufficient to prove that inf(Gφ2 ) ≤ inf(Gφ1 ) + . In view of the choice of the neighborhood U and Proposition 7.5 equation (7.18) is true for every number z ∈ R 1 that satisfies equation (7.16). This implies that inf(Gφ2 ) ≤ inf{Gφ2 (z) : z ∈ R 1 and Gφ1 (z) ≤ inf(Gφ1 ) + 1} ≤ inf{Gφ1 (z) +  : z ∈ R 1 and Gφ1 (z) ≤ inf(Gφ1 ) + 1} =  + inf{Gφ1 (z) : z ∈ R 1 and Gφ1 (z) ≤ inf(Gφ1 ) + 1} =  + inf(Gφ1 )

7.2 Auxiliary Results

273

and inf(Gφ2 ) ≤ inf(Gφ1 ) + . Proposition 7.6 is proved. Proposition 7.7 Assume that β ∈ Mr and that z¯ ∈ R 1 satisfies Gβ (z) > Gβ (¯z) for every number z ∈ R 1 \ {¯z, −¯z}.

(7.20)

Let  > 0 be given. Then there exist an open neighborhood U of the function β in the metric space M and a positive number δ such that U ⊂ Mr and for every function φ ∈ U that satisfies Gφ (z) ≤ inf(Gβ ) + δ

(7.21)

the inequality min{|z − z¯ |, z + z¯ |} ≤  holds. Proof Let us assume the contrary. Then for every integer n ≥ 1 there exist a function φn ∈ Mr and a number zn ∈ R 1 for which ρ(β, φn ) ≤ 1/n,

(7.22)

Gφn (zn ) ≤ inf(Gφn ) + 1/n

(7.23)

min{|zn − z¯ |, zn + z¯ |} > .

(7.24)

and

Equations (7.22), (7.23), and Proposition 7.3 imply that the sequence {zn }∞ n=1 is bounded. Extracting a subsequence and re-indexing, if necessary, we may assume without loss of generality that there exists z∗ = lim zn . n→∞

(7.25)

In view (7.22) and Proposition 7.6, we have lim inf(Gφn ) = inf(Gβ ).

n→∞

(7.26)

274

7 Optimization Problems Arising in Crystallography

Since the sequence {zn }∞ n=1 is bounded equation (7.22) and Proposition 7.4 imply that lim [inf(Gφn ) − Gβ (zn )] = 0.

n→∞

(7.27)

By equations (7.23), (7.25)–(7.27), we have Gβ (z∗ ) = lim Gβ (zn ) = lim Gφn (zn ) n→∞

n→∞

= lim inf(Gφn ) = inf(Gβ ). n→∞

Hence Gβ (z∗ ) = inf(Gβ ).

(7.28)

It follows from (7.20) and (7.28) that either z∗ = z¯ or z∗ = −¯z.

(7.29)

By (7.25) and (7.29), we have either zn → z¯ or zn → −¯z as n → ∞.

(7.30)

Evidently, (7.30) contradicts (7.24). The contradiction we have reached completes the proof of Proposition 7.7.

7.3 A Basic Lemma For every function ψ ∈ C 2 (R 1 ) define ψC 2 = sup{|ψ (i) (t)| : t ∈ R 1 , i = 0, 1, 2}. Lemma 7.8 Assume that β ∈ Mr ,  > 0 and that β(t) > β(π/2) for all t ∈ [0, π/2). Then there exists a function φ ∈ Mr and a number z¯ ∈ R 1 such that ρ(φ, β) ≤  and Gβ (z) > Gβ (¯z) for all z ∈ R 1 \ {¯z, −¯z}.

(7.31)

7.3 A Basic Lemma

275

Proof There exists a number z¯ ∈ R 1 such that Gβ (¯z) = inf(Gβ ).

(7.32)

In view of equation (7.10), z¯ = 0. We may assume z¯ > 0.

(7.33)

θ¯ = arctan(¯z) ∈ (0, π/2).

(7.34)

Set

There exists a function ψ ∈ C ∞ (R 1 ) such that 0 ≤ ψ(t) ≤ 1 for all t ∈ R 1 , ψ(t) = 0 if |t| ≥ 1, ψ(t) = 1 if |t| ≤ 1/2.

(7.35)

ψ1 (t) = ψ(t)(1 − t 2 ), t ∈ R 1 .

(7.36)

Put

Evidently, ψ1 ∈ C ∞ (R 1 ), 0 ≤ ψ1 (t) ≤ 1 for every number t ∈ R 1 , ψ1 (t) = 0 if |t| ≥ 1, ψ(t) = 1 − t 2 if |t| ≤ 1/2, 1 = ψ1 (0) > ψ1 (t) for every number t ∈ R 1 \ {0}.

(7.37) (7.38)

It follows from (7.31), (7.34), and the relation β ∈ Mr that we can choose positive constants c0 , c1 for which c0 > 1, c1 < 1, [θ¯ − c0−1 , θ¯ + c0−1 ] ⊂ (0, π/2),

(7.39)

inf{β(t) : t ∈ [θ¯ − c0−1 , θ¯ + c0−1 ]} − β(π/2) > 4c1 , c1 < −[β(0) + β

(0)],

(7.40)

ψC 2 c1 c02 < .

(7.41)

276

7 Optimization Problems Arising in Crystallography

Consider the function ψ2 (t) = c1 − c2 ψ1 (c0 (t − θ¯ )), t ∈ R 1 .

(7.42)

Evidently, ψ2 ∈ C ∞ (R 1 ). In view of equations (7.37) and (7.42), we have 0 ≤ ψ2 (t) ≤ c1 , t ∈ R 1 , ¯ ≥ c−1 . ψ2 (t) = c2 for each t ∈ R 1 satisfying |t − θ| 0

(7.43) (7.44)

Equations (7.38) and (7.42) imply that ψ2 (θ¯ ) = 0,

(7.45)

¯ ψ2 (t) > 0 for every number t ∈ R 1 \ {θ}.

(7.46)

It is easy to see that there exists a function ψ3 : R 1 → R 1 such that ψ3 (t) = ψ2 (t), t ∈ [0, π/2],

(7.47)

ψ3 (−t) = ψ3 (t), t ∈ R 1 , ψ3 (t + π ) = ψ3 (t), t ∈ R 1 . It is not difficult to see that ψ3 ∈ C ∞ (R 1 ), 0 ≤ ψ3 (t) ≤ c1 , t ∈ R 1 ,

(7.48)

ψ3 (θ¯ ) = 0, ¯ ψ3 (t) > 0 for all t ∈ [0, π/2] \ {θ}, ψ3 (t) > 0 for all t ∈ [−π/2, 0] \ {−θ¯ }.

(7.49)

φ(t) = β(t) + ψ3 (t), t ∈ R 1 .

(7.50)

Set

It is clear that φ ∈ C 2 (R 1 ). In view of equations (7.1), (7.3), (7.4), (7.47), (7.48), and (7.50), φ(t) ≥ 0 for every number t ∈ R 1 and φ(t) = φ(−t) = φ(t + π ) for every number t ∈ R 1 .

(7.51)

7.3 A Basic Lemma

277

It follows from equations (7.48), (7.50), and the inclusion β ∈ Mr that φ(t) > 0 for every number t ∈ R 1 .

(7.52)

φ(0) + φ

(0) < 0.

(7.53)

We prove that

In view of (7.47) and (7.50), we have φ(0) + φ

(0) = β(0) + β

(0) + ψ3 (0) + ψ3

(0) = β(0) + β

(0) + ψ2 (0) + ψ2

(0).

(7.54)

By equations (7.39) and (7.44), we have ψ2 (0) = c1 , ψ2

(0) = 0.

(7.55)

Together with equations (7.40) and (7.54) this implies that φ(0) + φ

(0) = β(0) + β

(0) + c1 < 0. Hence (7.53) is valid. We prove that φ(π/2) ≤ φ(t) ≤ φ(0)

(7.56)

for every number t ∈ R 1 . Evidently, it is sufficient to show that this inequality is true for every number t ∈ [0, π/2]. Let t ∈ [0, π/2] be given. Then it follows from (7.47) and (7.50) that φ(t) = β(t) + ψ2 (t).

(7.57)

In view of equations (7.2), (7.43), (7.47), (7.50), (7.55), and (7.57), we have φ(t) ≤ β(t) + c1 = β(t) + ψ2 (0) ≤ β(0) + ψ2 (0) = β(0) + ψ3 (0) = φ(0). Therefore φ(t) ≤ φ(0).

(7.58)

It follows from (7.39), (7.44), (7.47), and (7.50) that φ(π/2) = β(π/2) + ψ2 (π/2) = β(π/2) + c1 .

(7.59)

278

7 Optimization Problems Arising in Crystallography

There are two cases: (a) |t − θ¯ | ≥ c0−1 ; (b) |t − θ¯ | < c0−1 . Consider the case (a). Then equations (7.44) and (7.47) imply that ψ3 (t) = ψ2 (t) = c1 . Together with equations (7.2), (7.50), and (7.59) this implies that φ(t) = β(t) + c1 ≥ β(π/2) + c1 = φ(π/2) and φ(t) ≥ φ(π/2).

(7.60)

¯ < c−1 . |t − θ| 0

(7.61)

Consider the case (b) with

In view of equations (7.40), (7.43), (7.47), (7.50), (7.59), and (7.61), we have φ(t) = β(t) + ψ2 (t) ≥ β(t) > β(π/2) + 4c1 > φ(π/2) and equation (7.60) holds. Therefore (7.60) is valid in both cases. Equations (7.60), (7.58) imply (7.56). We have shown that φ ∈ Mr . In view of (7.6), (7.41)–(7.43), (7.47), and (7.50), we have ρ(β, φ) = sup{|ψ3 (t)|, |ψ3 (t)|, |ψ3 (t)|

: t ∈ [−π/2, π/2]} = sup{|ψ2 (t)|, |ψ2 (t)|, |ψ2 (t)|

: t ∈ [0, π/2]} ≤ max{c1 , ψ1 C 2 c1 c0 , ψ1 C 2 c1 c02 } = ψ1 C 2 c1 c02 < . Hence ρ(β, φ) < .

(7.62)

φ(t) ≥ β(t) for all t ∈ R 1 .

(7.63)

Gφ (t) ≥ Gβ (t) for all t ∈ R 1 .

(7.64)

We have

This equation implies that

7.3 A Basic Lemma

279

It follows from equations (7.9), (7.32), (7.34), (7.45), (7.47), and (7.50) that ¯ + z¯ 2 )1/2 Gφ (t) = φ(arctan(¯z))(1 + z¯ 2 )1/2 = φ(θ)(1 = (β + ψ3 )(θ¯ )(1 + z¯ 2 )1/2 = β(θ¯ )(1 + z¯ 2 )1/2 = Gβ (¯z) = inf(Gβ ).

(7.65)

inf(Gφ ) = inf(Gβ ) = Gφ (¯z) = Gβ (¯z).

(7.66)

z ∈ R 1 \ {¯z, −¯z}

(7.67)

By (7.64), (7.65), we have

Let

be given. Equations (7.9) and (7.50) imply that Gφ (z) = φ(arctan(z))(1 + z2 )1/2 = β(arctan(z))(1 + z2 )1/2 + ψ3 (arctan(z))(1 + z2 )1/2 = Gβ (z) + ψ(arctan(z))(1 + z2 )1/2 .

(7.68)

It follows from (7.49), (7.67) that ψ3 (arctan(z)) > 0. Together with equations (7.66) and (7.68), this inequality imply that Gφ (z) > Gβ (z) ≥ Gφ (¯z) = Gφ (¯z). This completes the proof of Lemma 7.8. Lemma 7.9 Assume that β ∈ Mr and that  > 0. Then there exists a function ˜ <  and β(t) ˜ > β(π/2) ˜ β˜ ∈ Mr such that ρ(β, β) for every number t ∈ [0, π/2). Proof Consider the function β0 (t) = cos(2t) + 3/2, t ∈ R 1 . Evidently, β0 ∈ Mr . For every number t ∈ [0, π/2] we have β0 (t) = cos(2t) + 3/2 > cos(π ) + 3/2 = β0 (π/2).

280

7 Optimization Problems Arising in Crystallography

For every integer n ≥ 1 define βn (t) = β(t) + n−1 β0 , t ∈ R 1 . It is clear that for every integer n ≥ 1, βn ∈ Mr , βn (t) > βn (π/2) for every number t ∈ [0, π/2), βn → β as n → ∞ in (M, ρ). This completes the proof of Lemma 7.9. Lemmas 7.8 and 7.9 imply the following result. Lemma 7.10 (Basic Lemma) Assume that β ∈ Mr and  > 0. Then there exists a function φ ∈ Mr and a number z¯ ∈ R 1 such that ρ(β, φ) <  and Gφ (z) > Gφ (¯z) for every number z ∈ R 1 \ {¯z, −¯z}.

7.4 Proof of Theorem 7.2 Proposition 7.1 and Lemma 7.10 imply that F is an everywhere dense subset of the metric space (M, ρ). Let β ∈ F be given, a number zβ > 0 satisfy (7.11), and n ≥ 1 be an integer. In view of Proposition 7.7 there exists an open neighborhood U (β, n) of the function β in the metric space (M, ρ) and δ(β, n) > 0 such that U (β, n) ⊂ Mr and the following property holds: (P3)

if φ ∈ U (β, n) and z ∈ R 1 satisfies Gφ (z) ≤ inf(Gφ ) + δ(β, n),

then min{|z − zβ |, |z + zβ |} ≤ 1/n. Define F0 = ∩∞ n=1 ∪ {U (β, n) : β ∈ F, n is a natural number}.

(7.69)

7.5 A Porosity Result

281

Evidently, F ⊂ F0 and F0 is a countable intersection of open everywhere dense subsets of the metric space (M, ρ). Assume that φ ∈ F0 and that z1 , z2 ∈ [0, ∞) satisfy Gφ (z1 ) = Gφ (z2 ) = inf(Gφ ).

(7.70)

Let n ≥ 1 be an integer. In view of (7.69), there exists a function β ∈ F such that φ ∈ U (β, n). Property (P3), equation (7.70), and the inclusion above imply that |zi − zβ | ≤ 1/n, i = 1, 2, |z1 − z2 | ≤ 2/n. Since n is any natural number we conclude that z1 = z2 , φ ∈ F, and F0 = F. This completes the proof of Theorem 7.2.

7.5 A Porosity Result We continue to use the definitions, notation, and assumptions introduced in Section 7.1. In particular, we consider the complete metric space of functions (M, ρ) that satisfy assumption (A) (see (7.1)–(7.6)) and its open everywhere dense subset Mr (see (7.7), (7.8)). For every function β ∈ M we consider a continuous function Gβ defined by (7.9). Note that if β ∈ Mr , then (7.10) is true. In the previous sections we showed that for a generic (typical) function β the set {z ∈ R 1 : Gβ (z) = inf(Gβ )} = {zβ , −zβ } where zβ is a unique positive number depending only on the function β. More precisely, denote by F the collection of all functions β ∈ Mr that satisfy the following condition: (C) There exists a number zβ ∈ R 1 such that Gβ (z) > Gβ (zβ ) for all z ∈ R 1 \ {zβ , −zβ }. By Theorem 7.2, F is a countable intersection of open everywhere dense subsets of (M, ρ). Now we prove the following theorem obtained in [136]. Theorem 7.11 M \ F is a σ -porous subset of (M, ρ).

282

7 Optimization Problems Arising in Crystallography

Recall that for every function ψ ∈ C 2 (R), ψC 2 = sup{|ψ (i) (t)| : t ∈ R 1 , i = 0, 1, 2}.

7.6 The Set M \ Mr is Porous Proposition 7.12 M \ Mr is a porous subset of the metric space (M, ρ). Proof Consider the function ˜ = cos(2t) + 3/2, t ∈ R 1 . β(t) Evidently, β˜ ∈ Mr , ˜ ≥ 1/2 for all t ∈ R 1 , β(t)

(7.81)

˜ β(0) + β˜

(0) < −1,

(7.82)

˜ C 2 = 4. β

(7.83)

α = 1/32.

(7.84)

β ∈ M, r ∈ (0, 1].

(7.85)

˜ t ∈ R1. β1 (t) = β(t) + 8−1 r β(t),

(7.86)

Put

Assume that

Define

It is clear that β1 ∈ M. By equations (7.83) and (7.86), we have ˜ C 2 = 2−1 r. ρ(β, β1 ) = 8−1 rβ

(7.87)

Equations (7.1), (7.81), (7.85), and (7.86) imply that for every number t ∈ R 1 , ˜ ≥ 16−1 r. β1 (t) ≥ 8−1 r β(t)

(7.88)

7.7 Auxiliary Results

283

In view of (7.5), (7.82), (7.85), and (7.86), we have ˜ + β˜

(0)) β1 (0) + β1

(0) = β(0) + β

(0) + 8−1 r(β(0) ˜ + β˜

(0)) < −8−1 r. ≤ 8−1 r(β(0)

(7.89)

φ ∈ M, ρ(φ, β1 ) ≤ αr = r/32.

(7.90)

Assume that

By equations (7.87) and (7.90), we have ρ(φ, β) ≤ ρ(φ, β1 ) + ρ(β1 , β) ≤ r/32 + r/2 < r.

(7.91)

It follows from (7.90), (7.6), and (7.88) that for every number t ∈ R 1 φ(t) ≥ β1 (t) − ρ(φ, β1 ) ≥ β1 (t) − r/32 ≥ r/16 − 32−1 r = 32−1 r.

(7.92)

In view of (7.6), (7.89), and (7.90), we have φ(0) + φ

(0) ≤ β1 (0) + β1

(0) + 2ρ(φ, β1 ) ≤ −8−1 r + 2ρ(φ, β1 ) ≤ −8−1 r + r/16 = −r/16. By the equation above, (7.90) and (7.92), we have φ ∈ Mr . Together with (7.91) this equation implies that {φ ∈ M : ρ(φ, β1 ) ≤ αr} ⊂ {φ ∈ M : ρ(φ, β) ≤ r} ∩ Mr . Proposition 7.12 is proved.

7.7 Auxiliary Results Let n ≥ 1 be an integer. Define Ωn = {z ∈ R : 1/n ≤ |z| ≤ n}. Denote by Fn the collection of all function β ∈ M satisfying the following condition: (C1)

There exists a number zβn ∈ Ωn for which Gβ (z) > Gβ (zβn )

284

7 Optimization Problems Arising in Crystallography

for all z ∈ Ωn \ {zβn , −zβn }. Proposition 7.13 Mr ∩ (∩∞ n=1 Fn ) ⊂ F. Proof Assume that f ∈ Mr ∩ (∩∞ n=1 Fn ). By the inclusion β ∈ Mr we have lim Gβ (z) = ∞, inf(Gβ ) < β(0) = Gβ (0).

|z|→∞

(7.93)

(see (7.10)). In view of (7.93), there exist an integer k ≥ 1 and a positive number δ such that Gβ (z) ≥ Gβ (0) + 4 for every number z ∈ R 1 that satisfies |z| ≥ k,

(7.94)

Gβ (z) > inf(Gβ ) + δ for every number z ∈ [−1/k, 1/k]. Since β ∈ Fk condition (C1) implies that there exists a number zk ∈ Ωk for which Gβ (z) > Gβ (zk ) for all z ∈ Ωk \ {zk , −zk }.

(7.95)

Since the function Gβ is continuous equation (7.93) implies that the function G possesses a point of minimum. Let a number z ∈ R 1 satisfy Gβ (z) = inf(Gβ ).

(7.96)

By equation (7.94) and the definition of the set Ωk , we have 1/k ≤ |z| ≤ k and z ∈ Ωk . Together with equations (7.95) and (7.96) this implies that z ∈ {zk , −zk }. Thus β ∈ F. This completes the proof of Proposition 7.13. Let n, i ≥ 1 be integers. Denote by Fni the collection of all functions β ∈ M satisfying the following condition: (C2) There exist numbers δ > 0 and z∗ ∈ Ωn such that for every z ∈ Ωn that satisfies Gβ (z) ≤ inf{Gβ (x) : x ∈ Ωn } + δ

7.8 Proof of Theorem 7.11

285

the inequality min{|z − z∗ |, |z + z∗ |} ≤ 1/ i is true. Proposition 7.14 Let n ≥ 1 be an integer. Then ∩∞ i=1 Fni ⊂ Fn . Proof Assume that β ∈ ∩∞ i=1 Fni . Condition (C2) implies that for every integer i ≥ 1 there exist numbers zi ∈ Ωn , δi > 0

(7.97)

such that the following property holds: (C3)

If a numbers z ∈ Ωn satisfies Gβ (z) ≤ inf{Gβ (y) : y ∈ Ωn } + δi ,

then we have min{|z − zi |, |z + zi |} ≤ 1/ i. We may assume without loss of generality that zi ≥ 0 for every integer i ≥ 1.

(7.98)

z ∈ Ωn , Gβ (z) = inf{Gβ (x) : x ∈ Ωn }.

(7.99)

Let

In view of equation (7.99) and property (C3), for every natural number i we have min{|z − zi |, |z + zi |} ≤ 1/ i.

(7.100)

If z = 0, then equation (7.100) implies that limi→∞ zi = 0. If z > 0, then it follows from (7.98) and (7.100) that z = limi→∞ zi . If z < 0, then by (7.98) and (7.100) z = − limi→∞ zi . We conclude that there exists limi→∞ zi and if a number z satisfies (7.99), then z ∈ {limi→∞ zi , − limi→∞ zi }. This implies that β ∈ Fn . Proposition 7.14 is proved.

7.8 Proof of Theorem 7.11 We proceed the proof of Theorem 7.11 by the following auxiliary result.

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7 Optimization Problems Arising in Crystallography

Proposition 7.15 Let n, i be natural numbers. Then the set M \ Fni is a porous subset of the metric space (M, ρ). Proof There exists a function ψ ∈ C ∞ (R 1 ) such that 0 ≤ ψ(t) ≤ 1 for every number t ∈ R 1 , ψ(t) = 0 if |t| ≥ 1, ψ(t) = 1 if |t| ≤ 1/2.

(7.101)

ψ1 (t) = (1 − t 2 )ψ(t), t ∈ R 1 .

(7.102)

Define

It is clear that ψ1 ∈ C ∞ (R 1 ), 0 ≤ ψ1 (t) ≤ 1 for every number t ∈ R 1 , ψ1 (t) = 0 if |t| ≥ 1, ψ1 (t) = (1 − t 2 ) if |t| ≤ 1/2, ψ1 (t) < 1 for every number t ∈ R 1 \ {0}.

(7.103)

Fix c0 > 0 for which c0 > max{8n2 i, 2(arctan(1/n))−1 , 2(arctan(n + 1) − arctan(n))−1 }.

(7.104)

Put Δ = inf{cos(2t) − cos(π ) : t ∈ [0, arctan(n + 1)]}.

(7.105)

Clearly, Δ > 0.

(7.106)

α ≤ 8−1 min{(4c02 ψ1 C 2 )−1 , 64−1 Δ}n−1 .

(7.107)

β ∈ M, r ∈ (0, 1].

(7.108)

Choose α > 0 satisfying

Assume that

7.8 Proof of Theorem 7.11

287

Consider the function ˜ = cos(2t) + 3/2, t ∈ R 1 . β(t)

(7.109)

It is not difficult to see that β˜ ∈ Mr , ˜ ≥ 1/2 for all t ∈ R 1 , β(t)

(7.110)

˜ β(0) + β˜

(0) < −1.

(7.111)

˜ t ∈ R1. β1 (t) = β(t) + (32)−1 r β(t),

(7.112)

˜ C 2 = 8−1 r. ρ(β, β1 ) = (32−1 )rβ

(7.113)

Set

Evidently, β1 ∈ M,

By equations (7.1), (7.108), (7.110), and (7.112), for every number t ∈ R 1 , we have ˜ ≥ (64)−1 r. β1 (t) ≥ (32)−1 r β(t)

(7.114)

It follows from (7.5), (7.108), (7.111), and (7.112) that ˜ + β˜

(0)] β1 (0) + β1

(0) = β(0) + β

(0) + (32−1 )r[β(0) ˜ + β˜

(0)] = −32−1 r. ≤ (32)−1 r[β(0)

(7.115)

z¯ ∈ Ωn

(7.116)

Gβ1 (¯z) = inf{Gβ1 (z) : z ∈ Ωn }.

(7.117)

z¯ > 0.

(7.118)

θ¯ = arctan(¯z).

(7.119)

There exists a number

for which

We may assume that

Put

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7 Optimization Problems Arising in Crystallography

In view of equations (7.116), (7.118), (7.119), and the definition of Ωn , we have θ¯ ∈ [ arctan(1/n), arctan(n)].

(7.120)

cr = r min{(4c02 ψ1 C 2 )−1 , 64−1 Δ}

(7.121)

ψ2 (t) = cr (1 − ψ1 (c0 (t − θ¯ ))), t ∈ R 1 .

(7.122)

Put

and set

It is clear that ψ2 ∈ C ∞ (R 1 ). By (7.103) and (7.122), we have 0 ≤ ψ2 (t) ≤ cr for all t ∈ R 1 ,

(7.123)

ψ2 (t) = cr if |t − θ¯ | ≥ c0−1 ,

(7.124)

ψ2 (θ¯ ) = 0,

(7.125)

ψ2 (t) > 0 for each t ∈ R \ {θ¯ }.

(7.126)

By (7.104) and (7.120), we have {t ∈ [0, π/2] : |t − θ¯ | ≥ c0−1 } = [0, π/2] ∩ ((−∞, θ¯ − c0−1 ] ∪ [θ¯ + c0−1 , ∞)) ⊃ [0, π/2] ∩ ((−∞, arctan(1/n) − c0−1 ] ∪ [arctan(n) + c0−1 , ∞)) ⊃ [0, π/2] ∩ ((−∞, 2−1 arctan(1/n)] ∪ [arctan(n + 1), ∞)) = [0, 2−1 arctan(1/n)] ∪ [arctan(n + 1), π/2].

(7.127)

It follows from (7.124) and (7.127) that ψ2 (t) = cr for each t ∈ [0, 2−1 arctan(1/n)] ∪ [ arctan(n + 1), π/2].

(7.128)

In view of equation (7.128), there exists a function ψ3 : R 1 → R 1 such that ψ3 (t) = ψ2 (t), t ∈ [0, π/2], ψ3 (−t) = ψ3 (t + π ) = ψ3 (t) for all t ∈ R 1 .

(7.129)

7.8 Proof of Theorem 7.11

289

Evidently, ψ3 ∈ C ∞ (R 1 ). Equations (7.123), (7.125), (7.126), and (7.129) imply that 0 ≤ ψ3 (t) ≤ cr for all t ∈ R 1 ,

(7.130)

ψ3 (θ¯ ) = 0, ¯ ψ3 (t) > 0 for all t ∈ [0, π/2] \ {θ},

(7.131)

ψ3 (t) > 0 for all t ∈ [−π/2, 0] \ {−θ¯ }. Define φ(t) = β1 (t) + ψ3 (t), t ∈ R 1 .

(7.132)

Clearly, φ ∈ C 2 (R 1 ). The inclusion β1 ∈ M and equations (7.129), (7.130), and (7.132) imply that φ(t) ≥ 0 for all t ∈ R 1 , φ(t) = φ(−t) = φ(t + π ) for all t ∈ R 1 .

(7.133)

In view of (7.128) and (7.129) we have ψ3 (0) = ψ2 (0) = cr , ψ3

(0) = ψ2

(0) = 0,

(7.134)

ψ3 (π/2) = ψ2 (π/2) = cr . Equations (7.105), (7.115), (7.121), (7.132), and (7.134) imply that φ(0) + φ

(0) = β1 (0) + β1

(0) + ψ3 (0) + ψ3

(0) = β1 (0) + β1

(0) + cr ≤ −32−1 r + cr ≤ 64−1 r.

(7.135)

Now we show that for every number t ∈ R 1 we have φ(π/2) ≤ φ(t) ≤ φ(0). By equation (7.133), it is sufficient to show that this inequality is valid for every number t ∈ [0, π/2]. Let t ∈ [0, π/2] be given. In view of equations (7.2), (7.130), (7.132), and (7.134), we have φ(t) = β1 (t) + ψ3 (t) ≤ β1 (t) + cr ≤ β1 (0) + cr = β1 (0) + ψ3 (0) = φ(0). (7.136)

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7 Optimization Problems Arising in Crystallography

We will prove that φ(t) ≥ φ(π/2). There are two cases: |t − θ¯ | ≥ c0−1

(7.137)

¯ < c−1 . |t − θ| 0

(7.138)

and

Assume that equation (7.137) is true. In view of (7.124), (7.129), and (7.137), we have ψ3 (t) = ψ2 (t) = cr . Together with equations (7.2), (7.132), and (7.134) the equality above implies that φ(t) = β1 (t) + ψ3 (t) = β1 (t) + cr ≥ β1 (π/2) + cr = β1 (π/2) +ψ3 (π/2) = φ(π/2) and φ(t) ≥ φ(π/2). Assume that equation (7.138) is valid. By (7.104), (7.120), and (7.138), we have 2−1 arctan(1/n) ≤ arctan(1/n) − c0−1 ≤ θ¯ − c0−1 ≤ t ≤ θ¯ + c0−1 ≤ arctan(n) + c0−1 ≤ arctan(n + 1).

(7.139)

By (7.2), (7.105), (7.106), (7.109), (7.112), and (7.139), we have ˜ − β(π/2)) ˜ β1 (t) − β1 (π/2) = β(t) − β(π/2) + 32−1 r(β(t) ˜ − β(π/2)) ˜ = 32−1 r(cos(2t) − cos(π )) ≥ 32−1 rΔ. ≥ 32−1 r(β(t)

(7.140)

Equations (7.121), (7.130), (7.132), (7.134), and (7.140) imply that φ(t) = β1 (t) + ψ3 (t) ≥ β1 (t) ≥ β1 (π/2) + 32−1 rΔ ≥ β1 (π/2) + cr = β1 (π/2) + ψ3 (π/2) = φ(π/2). Thus in both cases φ(t) ≥ φ(π/2). Combined with equation (7.136) this implies that φ(π/2) ≤ φ(t) ≤ φ(0).

(7.141)

7.8 Proof of Theorem 7.11

291

We proved (7.141) for every number t ∈ [0, π/2]. By equation (7.133) inequality (7.141) holds for every number t ∈ R 1 . Together with equations (7.133) and (7.135) inequality (7.141) implies that φ ∈ M. In view of (7.6), (7.103), (7.104), (7.113), (7.122), (7.129), and (7.132), we have ρ(β, φ) ≤ ρ(β, β1 ) + ρ(β1 , φ) ≤ 8−1 r + ρ(β1 , φ) ≤ 8−1 r + ψ3 C 2 ≤ 8−1 r + ψ2 C 2 ≤ 8−1 r + cr max{1, c0 ψ1 C 2 , c02 ψ1 C 2 } ≤ 8−1 r + cr c02 ψ1 C 2 . Together with equation (7.121) this inequality implies that ρ(β, φ) ≤ r/8 + cr c02 ψ1 C 2 ≤ 8−1 r + r(4c02 ψ1 C 2 )−1 c02 ψ1 C 2 ≤ r/8 + r/4 ≤ r/2.

(7.142)

It follows from (7.9), (7.130), and (7.132) that φ(t) ≥ β1 (t) for all t ∈ R 1 and Gφ (t) ≥ Gβ1 (t) for all t ∈ R 1 .

(7.143)

h ∈ M, ρ(h, φ) ≤ αr,

(7.144)

Now assume that

z ∈ Ωn , Gh (z) ≤ inf{Gh (y) : y ∈ Ωn } + αrn.

(7.145)

In view of (7.107), (7.142), and (7.144), we have ρ(h, β) ≤ ρ(h, φ) + ρ(φ, β) ≤ αr + r/2 ≤ r/8 + r/2 ≤ 3r/4.

(7.146)

By equations (7.6), (7.9), the definition of the set Ωn and (7.144), for every number y ∈ Ωn , |Gφ (y) − Gh (y)| = |φ(arctan(y))(1 + y 2 )1/2 − h(arctan(y))(1 + y 2 )1/2 | ≤ (1 + y 2 )1/2 ρ(h, φ) ≤ (1 + n2 )1/2 ρ(h, φ) ≤ 2nαr.

(7.147)

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7 Optimization Problems Arising in Crystallography

In particular, |Gφ (z) − Gh (z)| ≤ 2nαr.

(7.148)

In view of (7.147), | inf{Gφ (y) : y ∈ Ωn } − inf{Gh (y) : y ∈ Ωn }| ≤ 2nαr.

(7.149)

It follows from (7.145), (7.148), and (7.149) that Gφ (z) ≤ Gh (z) + 2nαr ≤ inf{Gh (y) : y ∈ Ωn } + 3αrn ≤ inf{Gφ (y) : y ∈ Ωn } + 5αrn.

(7.150)

Equations (7.9), (7.119), (7.130), (7.132), and (7.150) imply that Gφ (z) ≤ 5αrn + Gφ (¯z) = 5αrn + φ(arctan(¯z))(1 + z¯ 2 )1/2 ¯ ¯ = 5αrn + φ(θ)(1 + z¯ 2 )1/2 = 5αrn + (1 + z¯ 2 )1/2 (β1 (θ¯ ) + ψ3 (θ)) = 5αrn + (1 + z¯ 2 )1/2 β1 (θ¯ ) = 5αrn + (1 + z¯ 2 )1/2 β1 (arctan(¯z)) = 5αrn + Gβ1 (¯z). By (7.9), (7.117), (7.129), (7.132), and (7.145), we have Gφ (z) = φ(arctan(z))(1 + z2 )1/2 = β1 (arctan(z))(1 + z2 )1/2 + ψ3 (arctan(z))(1 + z2 )1/2 = Gβ1 (z) + ψ2 (|arctan(z)|)(1 + z2 )1/2 ≥ Gβ1 (¯z) + ψ2 (|arctan(z)|)(1 + z2 )1/2 . Together with (7.151) the equality above implies that Gβ1 (¯z) + ψ2 (|arctan (z)|)(1 + z2 )1/2 ≤ Gφ (z) ≤ 5αrn + Gβ1 (¯z) and ψ2 (|arctan(z)|)(1 + z2 )1/2 ≤ 5αrn.

(7.151)

7.8 Proof of Theorem 7.11

293

This inequality implies that ψ2 (|arctan(z)|) ≤ 5αrn.

(7.152)

If ||arctan(z)| − θ¯ | ≥ c0−1 , then in view of equations (7.107), (7.121), and (7.124), we have ψ2 (|arctan(z)|) = cr ≥ 8αrn. The equation above contradicts (7.152). Thus |arctan(|z|) − θ¯ | < c0−1 .

(7.153)

By the mean value theorem, ||z| − z¯ | = |arctan(|z|) − θ¯ ||(tan) (x)| = |arctan(|z|) − θ¯ |(cos(x))−2 ,

(7.154)

where ¯ arctan(|z|)}, max{θ, ¯ arctan(|z|}]. x ∈ [min{θ,

(7.155)

It follows from (7.116), (7.118), (7.119), (7.145), (7.155), and the definition of Ωn that x ∈ [arctan(1/n), arctan(n)].

(7.156)

Since cos(x)−2 = (tan(x))2 + 1 ∈ [1 + 1/n2 , n2 + 1] it follows from (7.104), (7.153), and (7.154) that ¯ 2 + 1) ≤ c−1 (n2 + 1) ≤ 1/ i. ||z| − z¯ | ≤ |arctan(|z|) − θ|(n 0

(7.157)

We have shown that ||z| − z¯ | ≤ 1/ i. Therefore h ∈ Fni . Combined with equation (7.146) this inequality implies that {h ∈ M : ρ(h, φ) ≤ αr} ⊂ {h ∈ M : ρ(β, h) ≤ r} ∩ Fni

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7 Optimization Problems Arising in Crystallography

and M \ Fni is a porous subset of the metric space (M, ρ). This completes the proof of Proposition 7.15. Completion of the Proof of Theorem 7.11 Let n ≥ 1 be an integer. Proposition 7.15 implies that for every integer i ≥ 1, M \ Fni is a porous subset of the metric space (M, ρ). By Proposition 7.14, ∞ M \ Fn ⊂ M \ (∩∞ i=1 Fni ) = ∪i=1 (M \ Fni ).

Therefore M \ Fn is a σ -porous subset of the metric space (M, ρ) for every integer n ≥ 1. In view of Propositions 7.12 and 7.13, we have ∞ M \ F ⊂ M \ (Mr ∩ (∩∞ n=1 Fn )) = (M \ Mr ) ∪n=1 (M \ Fn )

and M \ F is a σ -porous subset of the metric space (M, ρ). This completes the proof of Theorem 7.11.

Chapter 8

Discrete Dispersive Dynamical Systems

In this chapter we study turnpike properties for a discrete dispersive dynamical system generated by set-valued mappings, which was introduced by A. M. Rubinov in 1980. This dispersive dynamical system has a prototype in mathematical economics. In particular, it is an abstract extension of the classical von Neumann– Gale model. Our dynamical system is described by a compact metric space of states and a transition operator, which is set-valued. Our goal is to study the asymptotic behavior of the trajectories of this dynamical system.

8.1 Dynamical Systems with a Lyapunov Function In [106, 107] A. M. Rubinov introduced a discrete dispersive dynamical system generated by a set-valued mapping acting on a compact metric space, which was studied in [106, 107, 121, 137, 139, 141, 148]. This dispersive dynamical system has a prototype in mathematical economics [82, 106, 134]. In particular, it is an abstract extension of the classical von Neumann–Gale model [82, 106, 134]. Our dynamical system is described by a compact metric space of states and a transition operator which is set-valued. Dynamical systems theory has been a rapidly growing area of research that has various applications to physics, engineering, biology, and economics. In this theory, one of the goals is to study the asymptotic behavior of the trajectories of a dynamical system. Usually in the dynamical systems theory a transition operator is single-valued. In [106, 107, 121, 137, 139, 141, 148] and in this chapter we study dynamical systems with a set-valued transition operator. Such dynamical systems correspond to certain models of economic dynamics that are prototypes of our dynamical system [82, 106, 107, 134]. In particular, von Neumann–Gale model generated by a monotone process of convex type that was studied in [105]. The results of this section were obtained in [141]. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 A. J. Zaslavski, Turnpike Phenomenon and Symmetric Optimization Problems, Springer Optimization and Its Applications 190, https://doi.org/10.1007/978-3-030-96973-8_8

295

296

8 Discrete Dispersive Dynamical Systems

Let (X, ρ) be a compact metric space and let a : X → 2X \ {∅} be a set-valued mapping whose graph graph(a) = {(x, y) ∈ X × X : y ∈ a(x)} is a closed subset of X × X. For each nonempty subset E ⊂ X set a(E) = ∪{a(x) : x ∈ E} and a 0 (E) = E. By induction we define a n (E) for any natural number n and any nonempty subset E ⊂ X as follows: a n (E) = a(a n−1 (E)). In this chapter we study convergence of trajectories of the dynamical system generated by the set-valued mapping a. Following [106, 107] this system is called a discrete disperse dynamical system. First we define a trajectory of this system. A sequence {xt }∞ t=0 ⊂ X is called a trajectory of a (or just a trajectory if the mapping a is understood) if xt+1 ∈ a(xt ) for all integers t ≥ 0. 2 Let T2 > T1 be integers. We say that {xt }Tt=T ⊂ X is a trajectory of a (or just a 1 trajectory if a is understood) if xt+1 ∈ a(xt ), t = T1 , . . . , T2 − 1. Put Ω(a) = {z ∈ X : for each  > 0 there is a trajectory {xt }∞ t=0 such that lim inf ρ(z, xt ) ≤ }. t→∞

Clearly, Ω(a) is closed subset of (X, ρ). In the dynamical system theory the set Ω(a) is called a global attractor of a. Note that in [106, 107] Ω(a) is called a turnpike set of a. This terminology was motivated by mathematical economics [82, 106, 134]. For each x ∈ X and each nonempty closed subset E ⊂ X put ρ(x, E) = inf{ρ(x, y) : y ∈ E}. It is clear that for each trajectory {xt }∞ t=0 we have lim ρ(xt , Ω(a)) = 0.

t→∞

8.1 Dynamical Systems with a Lyapunov Function

297

It is not difficult to see that if for a nonempty closed set B ⊂ X lim ρ(xt , B) = 0

t→∞

for each trajectory {xt }∞ t=0 , then Ω(a) ⊂ B. Let φ : X → R 1 be a continuous function such that φ(z) ≥ 0 for all z ∈ X,

(8.1)

φ(y) ≤ φ(x) for all x ∈ X and all y ∈ a(x).

(8.2)

It is clear that the function φ is a Lyapunov function for the dynamical system generated by the mapping a. It should be mentioned that in mathematical economics usually X is a subset of the finite-dimensional Euclidean space and φ is a linear functional on this space [82, 106, 134]. The following theorem will be proved in Section 8.2. Theorem 8.1 The following properties are equivalent: (1) If a sequence {xt }∞ t=−∞ ⊂ X satisfies xt+1 ∈ a(xt ) and φ(xt+1 ) = φ(xt ) for all integers t, then {xt }∞ t=−∞ ⊂ Ω(a). (2) For each  > 0 there exists a natural number T () such that for each trajectory {xt }∞ t=0 ⊂ X satisfying φ(xt ) = φ(xt+1 ) for all integers t ≥ 0 the inequality ρ(xt , Ω(a)) ≤  holds for all integers t ≥ T (). For each x ∈ X set π(x) = sup{ lim φ(xt ) : {xt }∞ t=0 is a trajectory and x0 = x}. t→∞

(8.3)

The following two results will be proved in Section 8.3. Proposition 8.2 Let x ∈ X. Then there is a trajectory {xt }∞ t=0 such that x0 = x and π(x) = limt→∞ φ(xt ). Proposition 8.3 The function π : X → R 1 is upper semicontinuous. It is clear that for each x ∈ X and each y ∈ a(x), π(y) ≤ π(x),

(8.4)

π(x) ≤ φ(x)

(8.5)

for each x ∈ X,

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8 Discrete Dispersive Dynamical Systems

and that for each x ∈ X and each natural number n, π(x) ≤ sup{φ(y) : y ∈ a n (x)}.

(8.6)

It is easy to see that the following proposition holds. Proposition 8.4 Let x ∈ X and {xt }∞ t=0 ⊂ X be a trajectory such that x0 = x. Then lim φ(xt ) = π(x)

t→∞

if and only if for each integer t ≥ 0, π(xt+1 ) = max{π(z) : z ∈ a(xt )}. The following result will also be proved in Section 8.3. Proposition 8.5 Let x ∈ X. Then π(x) = lim sup{φ(y) : ∈ a n (x)}. n→∞

The following theorem will be proved in Section 8.4. Theorem 8.6 Assume that the property (1) of Theorem 8.1 holds. Let  > 0 and x ∈ X. Then there exist δ > 0 and a natural number L such that for each integer T > 2L and each trajectory {xt }Tt=0 satisfying x0 = x and φ(xT ) ≥ π(x0 ) − δ the following inequality holds: ρ(xt , Ω(a)) ≤ , t = L, . . . , T − L. In this section we use the following property. (P)

If x1 , x2 ∈ Ω(a) and φ(x1 ) = φ(x2 ), then x1 = x2 .

Note that the property (P) holds for many models of economic dynamics for which Ω(a) is a subinterval of a line [82, 106, 134]. The following theorem will be proved in Section 8.5. Theorem 8.7 Assume that the property (P) holds. Then each trajectory of a converges to an element of Ω(a). It is not difficult to see that the following result holds.

8.2 Proof of Theorem 8.1

299

Proposition 8.8 Assume that the property (P) holds and that {xt }∞ t=0 is a trajectory of a such that limt→∞ φ(xt ) = π(x). Then by Theorem 8.7 there exists F (x) = lim xt , t→∞

the equality φ(F (x)) = lim φ(xt ) = π(x) t→∞

holds and moreover, F (x) is a unique element of Ω(a) belonging to φ −1 (π(x)). In the sequel if the property (P) holds, then for each x ∈ X we denote by F (x) the unique element of Ω(a) ∩ φ −1 (π(x)). The following turnpike result describes the structure of optimal (with respect to the functional φ) trajectories of a. It will be proved in Section 8.6. Theorem 8.9 Assume that the property (P) and the property (1) of Theorem 8.1 hold. Let  > 0 and x ∈ X. Then there exist δ > 0 and a natural number L such that for each integer T > 2L and each trajectory {xt }Tt=0 satisfying x0 = x and φ(xT ) ≥ π(x) − δ the following inequality holds: ρ(xt , F (x)) ≤ , t = L, . . . , T − L.

8.2 Proof of Theorem 8.1 It is clear that the property (2) implies the property (1). Let us show that the property (1) implies the property (2). Assume that the property (1) holds and assume that the property (2) does not hold. Then there is (n)  > 0 such that for each natural number n there exist a trajectory {xt }∞ t=0 and an integer τn ≥ n such that ρ(xτ(n) , Ω(a)) > , n φ(xt(n) ) = φ(x0(n) ), t = 0, 1, . . . .

(8.7)

Let n ≥ 1 be an integer. Define (n)

yt

(n)

= xt+τn for all integers t ≥ −τn .

(8.8)

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8 Discrete Dispersive Dynamical Systems

Clearly, (n)

, n = 1, 2, . . . . y0 = xτ(n) n

(8.9)

Extracting subsequences and using the diagonalization process we obtain that there exists a strictly increasing sequence of natural numbers {nj }∞ j =1 such that for each integer t there is (nj )

yt = lim yt j →∞

(8.10)

.

It is not difficult to see that yt+1 ∈ a(yt ) and φ(yt ) = φ(yt+1 ) for all integers t.

(8.11)

In view of (8.7), (8.9), and (8.10), ρ(y0 , Ω(a)) ≥ .

(8.12)

By (8.11) and property (1), {yt }∞ t=−∞ ⊂ Ω(a). This contradicts (8.12). The contradiction we have reached proves that (1) implies (2). Theorem 8.1 is proved.

8.3 Proofs of Propositions 8.2, 8.3, and 8.5 Proof of Proposition 8.2 It is clear that for each integer n ≥ 1 there is a trajectory {xt(n) }∞ t=0 such that (n)

(n)

x0 = x, lim φ(xt ) ≥ π(x) − 1/n.

(8.13)

t→∞

Extracting subsequences and using the diagonalization process we obtain that there exists a strictly increasing sequence of natural numbers {nj }∞ j =1 such that for each integer t ≥ 0 there is (nj )

yt = lim xt j →∞

(8.14)

.

Clearly, {yt }∞ t=0 is a trajectory and y0 = x. By (8.13) and (8.14) for each integer s≥0 (nj )

φ(ys ) = lim φ(xs j →∞

(nj )

) ≥ lim sup lim φ(xt j →∞ t→∞

≥ lim π(x) − n−1 j = π(x). j →∞

)

8.3 Proofs of Propositions 8.2, 8.3, and 8.5

301

Proposition 8.2 is proved. Proof of Proposition 8.3 Let (n) x ∈ X , {x (n) }∞ = x. n=1 ⊂ X, lim x n→∞

(8.15)

We show that π(x) ≥ lim sup π(x (n) ). n→∞

We may assume without loss of generality that there is limn→∞ π(x (n) ). (n) By Proposition 8.2, for each integer n ≥ 1 there is a trajectory {xt }∞ t=0 such that for each integer n ≥ 1 x0(n) = x (n) , lim φ(xt(n) ) = π(x0(n) ) = π(x (n) ). t→∞

(8.16)

Extracting subsequences and using the diagonalization process we obtain that there is a strictly increasing sequence of natural numbers {nj }∞ j =1 such that for each integer t ≥ 0 there is (nj )

xt = lim xt

.

(8.17)

= lim x (nj ) = x.

(8.18)

j →∞

Clearly, {xt )∞ t=0 is a trajectory and (nj )

x0 = lim x0 j →∞

j →∞

By (8.16) and (8.17), for each integer t ≥ 0 (nj )

φ(xt ) = lim φ(xt j →∞

(nj )

) ≥ lim sup lim φ(xs j →∞ s→∞

) = lim sup π(x (n) ). j →∞

Together with (8.3) and (8.18) this implies π(x) ≥ lim φ(xt ) ≥ lim π(x (n) ). t→∞

n→∞

Proposition 8.3 is proved. Proof of Proposition 8.5 Clearly, the sequence {sup{φ(z) : z ∈ a n (x)}}∞ n=1 is monotone decreasing, and its limit is larger or equal than π(x).

302

8 Discrete Dispersive Dynamical Systems

Assume that the proposition does not hold. Then l := lim sup{φ(z) : z ∈ a n (x)} > π(x). n→∞

(n)

For each natural number n there is a trajectory {xt

(8.19)

: t = 0, . . . , n} such that

(n)

x0 = x, φ(xn(n) ) ≥ l.

(8.20)

Extracting a subsequence and using the diagonalization process we obtain that there exists a strictly increasing sequence of natural numbers {nj }∞ j =1 such that for each integer t ≥ 0 there is (nj )

xt = lim xt j →∞

(8.21)

.

Clearly, {xt }∞ t=0 is a trajectory and x0 = x.

(8.22)

By (8.19)–(8.21) for each integer t ≥ 0, (nj )

φ(xt ) = lim φ(xt j →∞

)≥l

and lim φ(xt ) ≥ l > π(x).

t→∞

This contradicts the definition of π(x) (see (8.3)). The contradiction we have reached proves Proposition 8.5.

8.4 Proof of Theorem 8.6 Assume that the theorem does not hold. Then for each natural number n there exist (n) n an integer Tn > 4n, a trajectory {xt }Tt=0 satisfying (n)

(n)

φ(xTn ) ≥ π(x) − 1/n, x0 = x

(8.23)

τn ∈ [n, Tn − n]

(8.24)

and an integer

8.4 Proof of Theorem 8.6

303

such that , Ω(a)) > . ρ(xτ(n) n

(8.25)

By (8.2) and (8.23), for each integer n ≥ 1 and each integer t ∈ [0, Tn ], (n)

φ(x) ≥ φ(xt ) ≥ π(x) − 1/n.

(8.26)

Let n ≥ 1 be an integer. Set (n) for all integers t = −τn , . . . , Tn − τn . yt(n) = xt+τ n

(8.27)

In view of (8.23), (8.25), and (8.27), (n)

(8.28)

ρ(y0 , Ω(a)) > , (n)

φ(yTn −τn ) ≥ π(x) − 1/n.

(8.29)

By (8.26) and (8.27), for each integer n ≥ 1 and each integer t ∈ [−τn , Tn − τn ], (n)

φ(x) ≥ φ(yt ) ≥ π(x) − 1/n.

(8.30)

Extracting subsequences and using the diagonalization process we obtain that there exists a strictly increasing sequence of natural numbers {nj }∞ j =1 such that for each integer t there exists (nj )

yt = lim yt

.

(8.31)

ρ(y0 , Ω(a)) ≥ .

(8.32)

j →∞

By (8.28) and (8.31),

In view of (8.30) and (8.31), for each integer t ≥ 0, (nj )

φ(yt ) = lim φ(yt j →∞

) ≥ π(x).

(8.33)

Let t be an integer. By (8.23), (8.27), (8.31), (8.33), and Proposition 8.5, (nj )

φ(yt ) = lim φ(yt j →∞

≤ lim sup sup{φ(z) : z ∈ a j →∞

(n )

) = lim φ(xt+τj n ) j →∞

t+τnj

j

(x)} = π(x).

304

8 Discrete Dispersive Dynamical Systems

Together with (8.33) this implies that φ(yt ) = π(x) for all integers t.

(8.34)

By (8.27) and (8.31), yt+1 ∈ a(yt ) for all integers t. Combined with (8.34) and the property (1) this implies that {yt }∞ t=−∞ ⊂ Ω(a). This contradicts (8.32). The contradiction we have reached proves Theorem 8.6.

8.5 Proof of Theorem 8.7 Let {xt }∞ t=0 be a trajectory of a and y be its limit point. Then y ∈ Ω(a) and φ(y) = lim φ(xt ). t→∞

If z is also a limit point of {xt }∞ t=0 , then z ∈ Ω(a) and φ(z) = limt→∞ φ(xt ). By the property (P) y = z. This implies that y = lim xt . t→∞

Theorem 8.7 is proved.

8.6 Proof of Theorem 8.9 Recall that F (x) is as guaranteed by Proposition 8.8. Namely, {F (x)} = Ω(a) ∩ φ −1 (π(x)). By the property (P) there is δ1 > 0 such that the following property holds: (P1) If z1 , z2 ∈ Ω(a) satisfy |φ(z1 ) − φ(z2 )| ≤ 2δ1 , then ρ(z1 , z2 ) ≤ /4. Since φ is uniformly continuous on X there is 1 ∈ (0, /4) such that the following property holds: (P2) For each z1 , z2 ∈ X satisfying ρ(z1 , z2 ) ≤ 41 , |φ(z1 ) − φ(z2 )| ≤ δ1 /4.

8.6 Proof of Theorem 8.9

305

By Proposition 8.5, there is a natural number L0 such that | sup{φ(z) : z ∈ a L0 (x)} − π(x)| ≤ δ1 /2.

(8.35)

By Theorem 8.6, there exist δ ∈ (0, δ1 ) and a natural number L > 2L0 such that the following property holds: (P3) For each integer T > 2L and each trajectory {xt }Tt=0 satisfying φ(xT ) ≥ π(x0 ) − δ, x0 = x the inequality ρ(xt , Ω(a)) ≤ 1 , t = L, . . . , T − L holds. Assume that an integer T > 2L and that a trajectory {xt }Tt=0 satisfies x0 = x, φ(xT ) ≥ π(x) − δ.

(8.36)

ρ(xt , Ω(a)) ≤ 1 , t = L, . . . , T − L.

(8.37)

Then by (8.36) and (P3),

Assume that an integer t ∈ [L, T − L]. By (8.36) and (8.37), there is z such that z ∈ Ω(a), ρ(xt , z) ≤ 1 .

(8.38)

In view of (8.35), (8.36), and the relation L0 < L ≤ t ≤ T , φ(xt ) ≥ φ(xT ) ≥ π(x) − δ ≥ π(x) − δ1 , φ(xt ) ≤ φ(xL0 ) ≤ π(x) + δ1 /2 and |φ(xt ) − π(x)| ≤ δ1 .

(8.39)

|φ(z) − φ(xt )| ≤ δ1 /4.

(8.40)

By (8.38) and (P2),

It follows from Proposition 8.8, (8.39) and (8.40) that |φ(z) − φ(F (x))| = |φ(z) − π(x)| ≤ |φ(z) − φ(xt )| + |φ(xt ) − π(x)| ≤ (3/2)δ1 .

306

8 Discrete Dispersive Dynamical Systems

Together with the definition of F (x), (P1), and the inclusions z, F (x) ∈ Ω(a) this implies that ρ(F (x), z) ≤ /4. Together with (8.38) this implies that ρ(xt , F (x)) ≤ ρ(xt , z) + ρ(z, F (x)) ≤ 1 + /4 < /2. Theorem 8.9 is proved.

8.7 The First Weak Turnpike Result Set φ = sup{|φ(z)| : z ∈ X}. We denote by Card(A) the cardinality of a set A and suppose that the sum over empty set is zero. In this chapter we prove a weak version of the turnpike property obtained in [172], which hold for all trajectories of our dynamical system that are of a sufficient length. Namely, we prove the following result. Theorem 8.10 Assume that property (1) of Theorem 8.1 hold and that  > 0. Then there exists a natural number L such that for every integer T > L and every trajectory {xt }Tt=0 the inequality Card({t ∈ {0, . . . , T } : ρ(xt , Ω(a)) > }) ≤ L holds. This result is proved in Section 8.9. Its proof is based on an auxiliary result that is proved in Section 8.8.

8.8 An Auxiliary Result Lemma 8.11 Assume that property (1) of Theorem 8.1 holds and that  > 0. Then there exist δ > 0 and a natural number L such that for each integer T > 2L and each trajectory {xt }Tt=0 that satisfies φ(x0 ) ≤ φ(xT ) + δ

8.8 An Auxiliary Result

307

the inequality ρ(xt , Ω(a)) ≤ , t = L, . . . , T − L holds. Proof Assume that the lemma does not hold. Then for each natural number n there n exist an integer Tn > 2n and a trajectory {xt(n) }Tt=0 such that (n)

(n)

φ(x0 ) ≤ φ(xTn ) + 1/n, (n)

max{ρ(xt , Ω(a)) : t = n, . . . , Tn − n} > .

(8.41) (8.42)

In view of equation (8.42), for each natural number n there exists an integer Sn ∈ {n, . . . , Tn − n}

(8.43)

such that (n)

ρ(xSn , Ω(a)) > .

(8.44)

Assume that n ≥ 1 is an integer. Put (n)

yt

(n)

= xt+Sn , t = −Sn , . . . , Tn − Sn .

(8.45)

n −Sn By (8.45), {yt }Tt=−S is a trajectory. It follows from (8.41) and (8.45) that n

(n)

(n)

(n)

(n)

(n)

φ(yTn −Sn ) − φ(y−Sn ) = φ(xTn ) − φ(x0 ) ≥ −1/n.

(8.46)

By (8.2) and (8.46), for each integer t ∈ {−Sn , . . . , Tn − Sn − 1}, (n)

(n)

(n)

(n)

φ(yt+1 ) − φ(yt ) ≥ φ(yTn −Sn ) − φ(y−Sn ) ≥ −1/n.

(8.47)

It follows from (8.44) and (8.45) that (n)

(n)

ρ(y0 , Ω(a)) = ρ(xSn , Ω(a)) > .

(8.48)

It is not difficult to see that there exists a strictly increasing sequence of positive integers {nj }∞ j =1 such that for every integer t there exists (nj )

yt = lim yt j →∞

.

(8.49)

308

8 Discrete Dispersive Dynamical Systems

Equations (8.48) and (8.49) imply that ρ(y0 , Ω(a)) ≥ .

(8.50)

In view of equation (8.49) and the closedness of the graph of a, yt+1 ∈ a(yt ) for all integers t.

(8.51)

Equations (8.47) and (8.49) imply that for all integers t, we have (n )

(nj )

φ(yt+1 ) − φ(yt ) = lim φ(yt+1j ) − lim φ(yt j →∞

j →∞

) ≥ lim (−n−1 j ) = 0. j →∞

Together with equation (8.2) this implies that φ(yt+1 ) = φ(yt ) for all integers t. It follows from property (1) of Theorem 8.1, (8.51) and the equation above that the inclusion yt ∈ Ω(a) holds for each integer t. This inclusion contradicts equation (8.50). The contradiction we have reached completes the proof of Lemma 8.11.

8.9 Proof of Theorem 8.10 By Lemma 8.11, there exist δ ∈ (0, ) and a natural number L0 such that the following property holds: (a) For every integer T > 2L0 and every trajectory {xt }Tt=0 that satisfies φ(x0 ) ≤ φ(xT ) + δ we have ρ(xt , Ω(a)) ≤ , t = L0 , . . . , T − L0 . Choose an integer L > 2L0 + 2 + (4L0 + 7)(1 + 2δ −1 φ).

(8.52)

Assume that T > L is an integer and that a sequence {xt }Tt=0 is a trajectory. By induction we define a strictly increasing finite sequence ti ∈ {0, . . . , T }, i =

8.9 Proof of Theorem 8.10

309

0, . . . , q. Set t0 = 0.

(8.53)

If φ(xT ) ≥ φ(x0 ) − δ, then set t1 = T and complete to construct the sequence. Assume that φ(xT ) < φ(x0 ) − δ. Clearly, there exists an integer t1 ∈ (t0 , T ] such that φ(xt1 ) < φ(x0 ) − δ

(8.54)

and that if an integer S satisfies t0 < S < t1 , then φ(xS ) ≥ φ(x0 ) − δ.

(8.55)

If t1 = T , then we complete to construct the sequence. Assume that k ≥ 1 is an integer and that we defined a strictly increasing sequence t0 , . . . , tk ∈ {0, . . . } such that t0 = 0, tk ≤ T and that for every integer i ∈ {0, . . . , k − 1}, φ(xti+1 ) < φ(xti ) − δ and if an integer S satisfies ti < S < ti+1 , then φ(xS ) ≥ φ(xti ) − δ. (By equations (8.54) and (8.55), the assumption is true with k = 1).

310

8 Discrete Dispersive Dynamical Systems

If tk = T , then we complete to construct the sequence. Assume that tk < T . If φ(xT ) ≥ φ(xtk ) − δ, then we set tk+1 = T and complete to construct the sequence. Assume that φ(xT ) < φ(xtk ) − δ.

(8.56)

Clearly, there exists a natural number tk+1 ∈ (tk , T ] such that φ(xtk+1 ) < φ(xtk ) − δ and that if an integer S satisfies tk < S < tk+1 , then φ(xS ) ≥ φ(xtk ) − δ. It is clear that the assumption made for k is true for k+1 too. Therefore by induction, we constructed the strictly increasing finite sequence of integers ti ∈ [0, T ], i = 0, . . . , q such that t0 = 0, tq = T and that for every integer i satisfying 0 ≤ i < q − 1, φ(xti+1 ) < φ(xti ) − δ

(8.57)

and for every integer i ∈ {0, . . . , q − 1} and every integer S satisfies ti < S < ti+1 , φ(xS ) ≥ φ(xti ) − δ. It follows from (8.57) that 2φ ≥ φ(xt0 ) − φ(xtq−1 )  {φ(xti ) − φ(xti+1 ) : i is an integer, 0 ≤ i ≤ q − 2} ≥ δ(q − 1)

(8.58)

8.9 Proof of Theorem 8.10

311

and q ≤ 1 + 2δ −1 φ.

(8.59)

Define E = {i ∈ {0, . . . , q − 1} : ti+1 − ti ≥ 2L0 + 4}.

(8.60)

Let i ∈ E.

(8.61)

ti+1 − 1 − ti ≥ 2L0 + 3.

(8.62)

In view of (8.60) and (8.61), we have

By (8.58) and (8.62), we have φ(xti+1 −1 ) ≥ φ(xti ) − δ.

(8.63) t

−1

i+1 It follows from (8.62), (8.63), and property (a) applied to the program {xt }t=t i

ρ(xt , Ω(a)) ≤ , t = ti + L0 , . . . , ti+1 − 1 − L0 .

that

(8.64)

By (8.64), {t ∈ {0, . . . , T } : ρ(xt , Ω(a)) > } ⊂ ∪{{ti , . . . , ti+1 } : i ∈ {0, . . . , q − 1} \ E} ∪ {{ti , . . . , ti + L0 − 1} ∪ {ti+1 − L0 , . . . , ti+1 } : i ∈ E}. It follows from (8.59), (8.60) and (8.65) that Card({t ∈ {0, . . . , T } : ρ(xt , Ω(a)) > }) ≤ q(2L0 + 5) + (2L0 + 2)q = q(4L0 + 7) ≤ (4L0 + 7)(1 + 2δ −1 φ) ≤ L. Theorem 8.10 is proved.

(8.65)

312

8 Discrete Dispersive Dynamical Systems

8.10 The Second Weak Turnpike Result For each bounded function ψ : X → R 1 set ψ = sup{|ψ(z)| : z ∈ X}. Recall that we denote by Card(A) the cardinality of a set A and suppose that the sum over empty set is zero. For each (x1 , x1 ), (y1 , y2 ) ∈ X × X set ρ1 ((x1 , x2 ), (y1 , y2 )) = ρ(x1 , y1 ) + ρ(x2 , y2 ). For each (x1 , x2 ) ∈ X × X and each nonempty closed subset E ⊂ X × X put ρ1 ((x1 , x2 ), E) = inf{ρ1 ((x1 , x2 ), (y1 , y2 )) : (y1 , y2 ) ∈ E}. We show that the turnpike property established in Theorem 8.10 is stable under small perturbations. More precisely, the following result obtained in [175] is true. Theorem 8.12 Assume that property (1) of Theorem 8.1 holds and that  > 0. Then there exists a natural number Q and δ > 0 such that for each integer T > Q, each function ψ : X → R 1 satisfying |ψ(z) − φ(z)| ≤ δ, z ∈ X and each sequence {xt }Tt=0 such that for all integers t = 0, . . . , T − 1, ψ(xt+1 ) ≤ ψ(xt ) and ρ1 ((xt , xt+1 ), graph(a)) ≤ δ the following inequality holds: Card({t ∈ {0, . . . , T } : ρ(xt , Ω(a)) > }) ≤ Q. This result is proved in Section 8.12. Its proof is based on an auxiliary result that is proved in Section 8.11.

8.11 An Auxiliary Result

313

8.11 An Auxiliary Result Lemma 8.13 Assume that property (1) of Theorem 8.1 holds and that  > 0. Then there exist δ > 0 and a natural number L such that for each integer T > 2L and each sequence {xt }Tt=0 that satisfies for all integers t = 0, . . . , T − 1, |φ(xt+1 ) − φ(xt )| ≤ δ and ρ1 ((xt , xt+1 ), graph(a)) ≤ δ the following inequality holds: ρ(xt , Ω(a)) ≤ , t = L, . . . , T − L. Proof Assume the contrary. Then for each natural number n there exist an integer Tn > 2n

(8.66)

(n)

n and a sequence {xt }Tt=0 ⊂ X such that for all integers t = 0, . . . , Tn − 1,

(n) |φ(xt+1 ) − φ(xt(n) )| ≤ 1/n, (n)

(n)

ρ1 ((xt , xt+1 ), graph(a)) ≤ 1/n, (n)

max{ρ(xt , Ω(a)) : t = n, . . . , Tn − n} > .

(8.67) (8.68) (8.69)

In view of (8.69), for each integer n ≥ 1 there exists an integer Sn ∈ {n, . . . , Tn − n}

(8.70)

such that (n)

(8.71)

ρ(xSn , Ω(a)) > . Let n ≥ 1 be an integer. Define (n)

yt

(n)

= xt+Sn , t = −Sn , . . . , Tn − Sn .

(8.72)

By (8.67) and (8.72), for all integers t = −Sn , . . . , Tn − Sn , (n)

(n)

(n)

(n)

|φ(yt+1 ) − φ(yt )| = |φ(xt+1+Sn ) − φ(xt+Sn )| ≤ 1/n.

(8.73)

314

8 Discrete Dispersive Dynamical Systems

Equations (8.68) and (8.72) imply that for all integers t ∈ {−Sn , . . . , Tn − Sn − 1}, (n)

(n)

(n)

(n)

(8.74)

(n)

(8.75)

ρ1 ((yt , yt+1 ), graph(a)) = ρ1 ((xt+Sn , xt+Sn +1 ), graph(a)) ≤ 1/n. It follows from (8.71) and (8.72) that (n)

ρ(y0 , Ω(a)) = ρ(xSn , Ω(a)) > .

Extracting subsequences and using diagonalization process we obtain that there exists a strictly increasing sequence of natural numbers {nj }∞ j =1 such that for each integer t there exists (nj )

yt = lim yt

.

(8.76)

ρ(y0 , Ω(a)) ≥ .

(8.77)

j →∞

By (8.75) and (8.76),

It follows from (8.73), (8.76) and the continuity of the function φ that for every integer t, (n )

(nj )

φ(yt+1 ) = lim φ(yt+1j ) = lim φ(yt j →∞

j →∞

) = φ(yt ).

(8.78)

Let t be an integer. We show that yt+1 ∈ a(yt ). In view of (8.74), for every natural number j satisfying −Snj ≤ t < Tnj − Snj − 1 there exists (ξj , zj ) ∈ graph(a)

(8.79)

such that (nj )

ρ(yt

, ξj ) + ρ(yt+1j , z) ≤ n−1 j . (n )

Together with (8.76) this implies that lim zj = yt+1 , lim ξj = yt .

j →∞

j →∞

(8.80)

8.12 Proof of Theorem 8.12

315

Since the graph of a is closed it follows from (8.79) and (8.80) that (yt , yt+1 ) ∈ graph(a). Property (1) of Theorem 8.1, (8.78), and the inclusion above imply that yt ∈ Ω(a) for all integers t. This contradicts (8.77). The contradiction we have reached completes the proof of Lemma 8.13.

8.12 Proof of Theorem 8.12 Lemma 8.13 implies that there exist δ1 ∈ (0, min{, 1}) and a natural number L0 such that the following property holds: (a) For each integer T > 2L0 and each sequence {xt }Tt=0 ⊂ X that satisfies for all integers t = 0, . . . , T − 1, |φ(xt+1 ) − φ(xt )| ≤ δ1 and ρ1 ((xt , xt+1 ), graph(a)) ≤ δ1 we have ρ(xt , Ω(a)) ≤ , t = L0 , . . . , T − L0 . Choose an integer Q > 2L0 + 2 + (4L0 + 7)(1 + 2δ1−1 (φ + 1))

(8.81)

δ ∈ (0, 4−1 δ1 ).

(8.82)

and

Assume that T > Q is an integer, ψ : X → R 1 satisfies |ψ(z) − φ(z)| ≤ δ, z ∈ X

(8.83)

316

8 Discrete Dispersive Dynamical Systems

and that a sequence {xt }Tt=0 ⊂ X satisfies for all integers t = 0, . . . , T − 1, ψ(xt+1 ) ≤ ψ(xt )

(8.84)

ρ1 ((xt , xt+1 ), graph(a)) ≤ δ.

(8.85)

and

By induction we define a strictly increasing finite sequence of integers ti ∈ [0, T ], i = 0, . . . , q. Set t0 = 0.

(8.86)

If ψ(xT ) ≥ ψ(x0 ) − δ, then set t1 = T and the construction is completed. Assume that ψ(xT ) < ψ(x0 ) − δ.

(8.87)

Evidently, there exists an integer t1 ∈ (t0 , T ] such that ψ(xt1 ) < ψ(x0 ) − δ

(8.88)

and that if an integer S satisfies t0 < S < t1 , then ψ(xS ) ≥ ψ(x0 ) − δ.

(8.89)

If t1 = T , then the construction is completed. Assume that k is a natural number and that we defined a strictly increasing sequence of nonnegative integers t0 , . . . , tk ∈ [0, T ] such that t0 = 0, tk ≤ T

(8.90)

8.12 Proof of Theorem 8.12

317

and that for each i ∈ {0, . . . , k − 1}, ψ(xti+1 ) < ψ(xti ) − δ

(8.91)

and if an integer S satisfies ti < S < ti+1 , then ψ(xS ) ≥ ψ(xti ) − δ.

(8.92)

(In view of (8.88) and (8.89), our assumption holds for k = 1.) If tk = T , then our construction is completed. Assume that tk < T . If ψ(xT ) ≥ ψ(xtk ) − δ, then we set tk+1 = T and our construction is completed. Assume that ψ(xT ) < ψ(xtk ) − δ.

(8.93)

Clearly, there exists an integer tk+1 ∈ (tk , T ] such that ψ(xtk+1 ) < ψ(xtk ) − δ

(8.94)

and that if an integer S satisfies tk < S < tk+1 , then ψ(xS ) ≥ ψ(xtk ) − δ.

(8.95)

It is clear that the assumption made for k also holds for k+1. Therefore by induction, we constructed the strictly increasing finite sequence of integers ti ∈ [0, T ], i = 0, . . . , q, where q is a natural number such that t0 = 0, tq = T and that for each i satisfying 0 ≤ i < q − 1, ψ(xti+1 ) < ψ(xti ) − δ

(8.96)

318

8 Discrete Dispersive Dynamical Systems

and for each i ∈ {0, . . . , q − 1} and each integer S satisfies ti < S < ti+1 , we have ψ(xS ) ≥ ψ(xti ) − δ.

(8.97)

By (8.82), (8.83), and (8.96), 2φ + 2 ≥ φ(xt0 ) − φ(xtq−1 ) + 2 ≥ ψ(xt0 ) − ψ(xtq −1 ) =

 {ψ(xti ) − ψ(xti+1 ) : i is an integer, 0 ≤ i ≤ q − 2} ≥ δ(q − 1)

and q ≤ 1 + 2δ −1 (φ + 1).

(8.98)

Set E = {i ∈ {0, . . . , q − 1} : ti+1 − ti ≥ 2L0 + 4}.

(8.99)

Let i ∈ E.

(8.100)

ti+1 − 1 − ti ≥ 2L0 + 3.

(8.101)

By (8.99) and (8.100),

Equations (8.97) and (8.101) imply that ψ(xti+1 −1 ) ≥ ψ(xti ) − δ.

(8.102)

In view of (8.84) and (8.102), for each integer t ∈ {ti , . . . , ti+1 − 2}, |ψ(xt+1 ) − ψ(xt )| ≤ δ.

(8.103)

By (8.82), (8.83), and (8.103), for each integer t ∈ {ti , . . . , ti+1 − 2}, |φ(xt+1 ) − φ(xt )| ≤ |ψ(xt+1 ) − ψ(xt )| + 2δ ≤ 3δ < δ1 .

(8.104)

It follows from (8.85), (8.101), (8.104), and property (a) that ρ(xt , Ω(a)) ≤ , t = ti + L0 , . . . , ti+1 − 1 − L0 .

(8.105)

8.13 Stability Results

319

In view of (8.100) and (8.105), {t ∈ {0, . . . , T } : ρ(xt , Ω(a)) > } ⊂ ∪{{ti , . . . , ti+1 } : i ∈ {0, . . . , q − 1} \ E} ∪{{ti , . . . , ti + L0 − 1} ∪ {ti+1 − L0 , . . . , ti+1 } : i ∈ E}. By (8.98), (8.99) and the equation above, Card({t ∈ {0, . . . , T } : ρ(xt , Ω(a)) > }) ≤ q(2L0 + 5) + (2L0 + 2)q = q(4L0 + 7) (4L0 + 7)(1 + 2δ −1 (φ + 1)) ≤ Q. Theorem 8.12 is proved.

8.13 Stability Results We show that a weak version of the turnpike property established in Theorem 8.10 is stable under small perturbations. For every pair of nonempty sets A, B ⊂ X define H (A, B) = max{sup{ρ(x, B) : x ∈ A}, sup{ρ(y, A) : y ∈ B}}. We assume that the following assumption holds: (A) For every  > 0 there exists δ > 0 such that for every pair x, y ∈ X that satisfies ρ(x, y) ≤ δ, H (a(x), a(y)) ≤ . We also assume that property (1) of Theorem 8.1 holds true and prove the following two theorems that were obtained in [173]. Theorem 8.14 Let  > 0. Then there exists a natural number L0 such that for every integer L > L0 there exists δ > 0 such that for every sequence {xt }L t=0 ⊂ X that satisfies ρ1 ((xt , xt+1 ), graph(a)) ≤ δ, t = 0, . . . , T − 1

320

8 Discrete Dispersive Dynamical Systems

the inequality Card({t ∈ {0, . . . , L} : ρ(xt , Ω(a)) > }) ≤ L0 is true. Theorem 8.15 Let  > 0. Then there exists a natural number L1 and δ > 0 such that for every natural number T > L1 and every sequence {xt }Tt=0 ⊂ X that satisfies ρ1 ((xt , xt+1 ), graph(a)) ≤ δ, t = 0, . . . , T − 1 the inequality T −1 Card({t ∈ {0, . . . , T } : ρ(xt , Ω(a)) > }) ≤  is true.

8.14 An Auxiliary Result Lemma 8.16 Let  > 0 and L ≥ 1 be an integer. Then there exists δ > 0 such that for every sequence {xt }L t=0 ⊂ X that satisfies for all integers t = 0, . . . , L − 1, ρ1 ((xt , xt+1 ), graph(a)) ≤ δ there exists a trajectory {yt }L t=0 ⊂ X such that y0 = x0 , ρ(xt , yt ) ≤ , t = 0, . . . , L. Proof Let δL = /4.

(8.106)

By induction and assumption (A), we define numbers δi > 0, i = 0, . . . , L − 1 such that for every integer i ∈ {1, . . . , L}, we have δi−1 < δi /8

(8.107)

and for every pair x, y ∈ X satisfying ρ(x, y) ≤ δi−1 we have H (a(x), a(y)) < δi /8.

(8.108)

8.14 An Auxiliary Result

321

Put δ = δ0 .

(8.109)

Assume that {xt }L t=0 ⊂ X satisfies ρ1 ((xt , xt+1 ), graph(a)) ≤ δ, t = 0, . . . , L − 1.

(8.110)

Set x0 = y0 .

(8.111)

ρ1 ((y0 , x1 ), graph(a)) ≤ δ.

(8.112)

By (8.110) and (8.111), we have

In view of (8.112), there exists (z0 , z1 ) ∈ graph(a)

(8.113)

ρ(y0 , z0 ) ≤ δ, ρ(x1 , z1 ) ≤ δ.

(8.114)

such that

Equations (8.108), (8.109), and (8.114) imply that H (a(y0 ), a(z0 )) ≤ δ1 /8.

(8.115)

It follows from (8.113) and (8.115) that ρ(z1 , a(y0 )) ≤ δ1 /8.

(8.116)

y1 ∈ a(y0 )

(8.117)

ρ(y1 , z1 ) ≤ δ1 /4.

(8.118)

By (8.116), there exists

such that

Equations (8.107), (8.109), (8.114), and (8.118) imply that ρ(y1 , x1 ) ≤ δ + δ1 /4 < δ1 /2.

(8.119)

322

8 Discrete Dispersive Dynamical Systems

Assume that an integer k ∈ {1, . . . , L} \ {L} and that we have already defined a trajectory {yi }ki=1 such that y0 = x0 and that for integers i = 1, . . . , k, ρ(xi , yi ) < δi /2.

(8.120)

(In view of (8.111), (8.117), and (8.119), our assumption is valid for k = 1.) In view of (8.110), there exists (ξk , ξk+1 ) ∈ graph(a)

(8.121)

such that ρ1 ((xk , xk+1 ), (ξk , ξk+1 )) ≤ δ. The inequality above implies that ρ(xk , ξk ) ≤ δ, ρ(xk+1 , ξk+1 ) ≤ δ.

(8.122)

It follows from (8.107), (8.109), (8.120), and (8.122) that ρ(ξk , yk ) ≤ ρ(ξk , xk ) + ρ(xk , yk ) ≤ δ + δk /2 ≤ δk /8 + δk /2.

(8.123)

In view of (8.108) and (8.123), H (a(ξk ), a(yk )) ≤ δk+1 /8.

(8.124)

By (8.121) and (8.124), we have ρ(ξk+1 , a(yk )) ≤ δk+1 /8.

(8.125)

By (8.125), there exists a point yk+1 ∈ a(yk )

(8.126)

ρ(ξk+1 , yk+1 ) ≤ δk+1 /4.

(8.127)

such that

Equations (8.107), (8.109), (8.122), and (8.127) imply that ρ(yk+1 , xk+1 ) ≤ ρ(yk+1 , ξk+1 ) + ρ(ξk+1 , xk+1 ) ≤ δ + δk+1 /4 < δk+1 /2.

8.15 Proof of Theorem 8.14

323

Thus the assumption made for k also holds for k + 1. Therefore by induction we constructed the trajectory {yt }L t=0 ⊂ X such that y0 = x0 , ρ(xt , yt ) ≤ , t = 0, . . . , L. This completes the proof of Lemma 8.16.

8.15 Proof of Theorem 8.14 By Theorem 8.10, there exists a natural number L0 such that the following property holds: (i) For every natural number T > L0 and every trajectory {xt }Tt=0 , we have Card({t ∈ {0, . . . , T } : ρ(xt , Ω(a)) > /2}) ≤ L0 . Let L > L0 be an integer. Lemma 8.16 implies that there exists δ > 0 such that the following property holds: (ii) For every sequence {xt }L t=0 ⊂ X satisfying ρ1 ((xt , xt+1 ), graph(a)) ≤ δ, t = 0, . . . , L − 1

(8.128)

there exists a trajectory {yt }L t=0 ⊂ X such that y0 = x0 , ρ(xt , yt ) ≤ /4, t = 0, . . . , L.

(8.129)

Assume that {xt }L t=0 ⊂ X satisfies (8.128). It follows from property (ii) and equation (8.128) that there exists a trajectory {yt }L t=0 that satisfies (8.129). By property (i), Card({t ∈ {0, . . . , L} : ρ(yt , Ω(a)) > /2}) ≤ L0 . By (8.129), for all integers t = 0, . . . , L, we have ρ(xt , Ω(a)) ≤ ρ(xt , yt ) + ρ(yt , Ω(a)) ≤ /4 + ρ(yt , Ω(a)) and if ρ(xt , Ω(a)) > ,

(8.130)

324

8 Discrete Dispersive Dynamical Systems

then ρ(yt , Ω(a)) > /2. Combined with equation (8.130) this implies that Card({t ∈ {0, . . . , L} : ρ(xt , Ω(a)) > }) ≤ Card({t ∈ {0, . . . , L} : ρ(yt , Ω(a)) > /2}) ≤ L0 . Theorem 8.14 is proved.

8.16 Proof of Theorem 8.15 We may assume without loss of generality that  < 1. Theorem 8.14 implies that there exists a natural number L0 such that the following property holds: (i) For every integer L > L0 there exists δ > 0 such that for every sequence {xt }L t=0 that satisfies ρ1 ((xt , xt+1 ), graph(a)) ≤ δ, t = 0, . . . , T − 1 the inequality Card({t ∈ {0, . . . , L} : ρ(xt , Ω(a)) > }) ≤ L0 holds. Choose an integer L > 4L0  −1 .

(8.131)

Let δ > 0 be as guaranteed by property (i). Choose an integer k0 > 4 −1 .

(8.132)

L1 = k0 L.

(8.133)

Set

Assume that T > L1 is an integer and that a sequence {xt }Tt=0 satisfies ρ1 ((xt , xt+1 ), graph(a)) ≤ δ, t = 0, . . . , T − 1.

(8.134)

8.16 Proof of Theorem 8.15

325

There exists a natural number k1 such that k1 L ≤ T < (k1 + 1)L.

(8.135)

In view of (8.133) and (8.135), we have k1 > k0 .

(8.136)

It follows from property (i), (8.134), and (8.135) that for integers i = 0, . . . , k1 − 1, we have Card({t ∈ {iL, . . . , (i + 1)L} : ρ(xt , Ω(a)) > }) ≤ L0 . Together with (8.135) the equation above implies that Card({t ∈ {0, . . . , T } : ρ(xt , Ω(a)) > }) ≤ k1 L0 + L. Together with (8.131), (8.132), (8.135), and (8.136) the equation above implies that T −1 Card({t ∈ {0, . . . , T } : ρ(xt , Ω(a)) > }) ≤ k1 L0 T −1 + LT −1 ≤ L0 L−1 + k0−1 < . This completes the proof of Theorem 8.15.

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Index

A Absolutely continuous functions, 12, 114, 116 Adjoint operator, 112, 115, 195 Admissible control operators, 109, 116–124, 195, 209, 211, 213

B Baire category, vii, 1, 7, 9, 25, 26, 91, 267, 269 Banach spaces, vi, 1, 6, 21, 109–111, 113, 114, 131, 193 Bochner integrable functions, 110, 111, 114 Bochner integral, 110 Borelian functions, 16, 131, 143, 154, 193 Borel set, 20 Bounded operator, 112

C C0 groups, 113, 115–116 C0 semigroups, vi, 109, 113–114, 116, 117, 120, 124, 125, 193, 195 Cardinality, 21, 306, 312 Closable operator, 112 Closed operator, 112–114 Compact space, 6 Complete metric space, v, 1, 2, 5–7, 9, 12, 25–27, 32, 34, 43, 47, 52, 58, 68, 76, 81, 85, 91, 281 Continuous functions, 1, 5, 6, 8, 10, 12, 14, 15, 21, 22, 27, 32, 34, 40, 44, 55, 58, 60, 65, 66, 73, 76, 77, 82, 86, 105, 114, 116, 194, 195, 269, 281, 297

Convex function, 12, 13

D Data space, 27, 55, 58 Derivative, 12, 111, 117, 176 Differentiable function, 6 Differentiable mapping, 6 Domain space, 27, 55, 58 Dual space, 21, 195

E Epi-distance topology, 36, 76 Epigraph, 35 Euclidean norm, 13, 125 Evolution equations, vi, 109, 114–115 Exactly controllable pair, 124–129

G Generators, 113–117, 124, 127, 195 Good functions, 14, 15 Graph, 7, 112, 116, 125, 193–195, 267, 296, 308, 312–316, 319–324 Graph norm, 116, 125, 195

H Hausdorff space, 6, 27, 44, 91, 96, 102, 105 Hausdorff topology, 44, 47, 96, 105 Hilbert space, 21, 115–118, 124, 125, 194, 195

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 A. J. Zaslavski, Turnpike Phenomenon and Symmetric Optimization Problems, Springer Optimization and Its Applications 190, https://doi.org/10.1007/978-3-030-96973-8

335

336 I Identity mapping, 10, 27, 32, 35, 40, 45, 55, 69, 74, 76, 81, 85 Increasing function, 13, 17, 132, 196 Infinite dimensional optimal control, 12, 15, 109, 124, 193–265 Inner product, 21, 115, 194 K Korteweg-de vries equation, 129 L Lebesgue integral, 110 Lebesgue measurable function, 194, 196 Lebesgue measurable set, 18, 20, 109, 131 Lebesgue measure, 18, 20, 109, 137, 203 Linear continuous operator, 111 Linear operator, 111–113, 115–117, 125, 193, 195 Lower semicontinuity property, 94 Lower semicontinuous function, v, 5, 6, 27, 30, 34, 35, 40, 43, 44, 55, 60, 65, 76, 81, 94, 101 Lyapunov function, 295–299 M Metric, v, vi, 1, 2, 5–10, 12, 14, 20–22, 25–27, 29, 32–36, 43–45, 47, 48, 52–60, 65, 66, 68, 73, 74, 76, 81, 85, 91, 98, 99, 102, 105, 223, 267, 268, 270–273, 280, 281, 286, 294–296 Metric spaces, v, 1, 2, 5–9, 12, 14, 15, 20–22, 25–27, 32, 34, 43, 45, 47, 48, 52–60, 65, 66, 68, 73, 76, 81, 85, 91, 98, 99, 102, 105, 154, 223, 268–273, 280–282, 286, 294–296 Mild solution, 114, 118, 194, 195 N Norm, 13, 20, 21, 111, 115–117, 125, 128, 193–95 Normed space, 20, 193 O Operators, v, 26, 109, 111–129, 193, 195, 209, 211, 213, 295 P Point spectrum, 112 Product topology, 45, 116

Index R Reflexive banach space, 111 Relative topology, 10, 11, 33, 47, 74, 77 Resolvent, 112

S Semigroup property, 113 Set-valued mapping, 295, 296 Simple function, 109, 110 Sobolev space, 116, 125 Spectrum, 112 Strong continuity, 113 Strongly continuous group of operators, 113 Strongly continuous semigroup of operators, 113 Strongly measurable function, 110 Strong solution, 117

T Topologies, 10, 11, 14, 20, 27–29, 32–38, 40–42, 44–49, 54–56, 58, 69, 74, 76–79, 81–83, 85–87, 89, 91–96, 101–103, 105, 106, 116, 155 Trajectories, vii, 196, 249, 295–302, 304–308, 320, 322, 323 Transport equation, 129 Turnpike properties, v, vi, vii, 1, 11–16, 19, 131, 137, 139–145, 176, 193, 203, 213–218, 227, 258–265, 295, 306, 312, 319

U Uniformities, 5, 9, 10, 13, 14, 25, 32, 35, 36, 40, 44, 45, 69, 73, 76, 81, 102, 105, 155, 223 Uniform space, 35, 44, 76, 102, 223 Upper semicontinuous function, 297

V Variational principles, vi, 11, 25–32, 58–65, 91–94

W Weak solution, 114, 117 Well-posed problem, 5, 7, 10, 11, 28, 29, 31, 33, 34, 36, 41, 46, 47, 55, 56, 58, 59, 61, 64, 65, 67, 74, 77, 82, 87, 92, 101, 106, 126, 128, 198, 267