Preface......Page 5
Contents......Page 8
1.2 The Mirror Descent Method......Page 20
1.3 Gradient Algorithm with a Smooth Objective Function......Page 28
1.4 Examples......Page 32
2.1 Preliminaries......Page 35
2.2 A Convex Minimization Problem......Page 37
2.3 The Main Lemma......Page 41
2.4 Proof of Theorem 2.4......Page 43
2.5 Subgradient Algorithm on Unbounded Sets......Page 44
2.6 Proof of Theorem 2.8......Page 52
2.7 Proof of Theorem 2.9......Page 55
2.8 Proof of Theorem 2.12......Page 59
2.9 Zero-Sum Games with Two Players......Page 63
2.10 Proof of Proposition 2.13......Page 68
2.11 Zero-Sum Games on Bounded Sets......Page 76
2.12 Zero-Sum Games on Unbounded Sets......Page 81
2.13 Proof of Theorem 2.16......Page 87
2.14 An Example for Theorem 2.16......Page 90
3.1 Optimization on Bounded Sets......Page 92
3.2 The Main Lemma......Page 96
3.3 Proof of Theorem 3.1......Page 99
3.4 Optimization on Unbounded Sets......Page 100
3.5 Proof of Theorem 3.3......Page 105
3.6 Proof of Theorem 3.4......Page 109
3.7 Proof of Theorem 3.5......Page 114
3.8 Zero-Sum Games on Bounded Sets......Page 119
3.9 Zero-Sum Games on Unbounded Sets......Page 125
4.1 Optimization on Bounded Sets......Page 135
4.2 Auxiliary Results......Page 137
4.3 The Main Lemma......Page 138
4.4 Proof of Theorem 4.1......Page 144
4.5 Optimization on Unbounded Sets......Page 145
5.1 Preliminaries and the Main Result......Page 159
5.2 Auxiliary Results......Page 162
5.3 Proof of Theorem 5.1......Page 169
5.4 The First Extension of Theorem 5.1......Page 171
5.5 The Second Extension of Theorem 5.1......Page 175
6.1 Bochner Integrable Functions......Page 180
6.2 Convergence Analysis for Continuous Subgradient Method......Page 181
6.3 An Auxiliary Result......Page 184
6.4 Proof of Theorem 6.1......Page 185
6.5 Continuous Subgradient Method for Zero-Sum Games......Page 188
6.6 An Auxiliary Result......Page 191
6.7 Proof of Theorem 6.5......Page 195
6.8 Continuous Subgradient Projection Method......Page 200
6.9 An Auxiliary Result......Page 202
6.10 Proof of Theorem 6.7......Page 203
6.11 Continuous Subgradient Projection Methodon Unbounded Sets......Page 208
6.12 An Auxiliary Result......Page 209
6.13 The Convergence Result......Page 214
6.14 Subgradient Projection Algorithm for Zero-Sum Games......Page 222
6.15 An Auxiliary Result......Page 223
6.16 A Convergence Result for Games on Bounded Sets......Page 232
6.17 A Convergence Result for Games on Unbounded Sets......Page 241
7.1 Preliminaries......Page 249
7.2 The Algorithm and Main Results......Page 251
7.3 Auxiliary Results......Page 254
7.4 Proof of Theorem 7.4......Page 260
7.5 Proof of Theorem 7.5......Page 263
8.1 The Algorithm and the Main Result......Page 265
8.2 Auxiliary Results......Page 269
8.3 Proof of Theorem 8.1......Page 277
9.1 Preliminaries and the Main Result......Page 282
9.2 Auxiliary Results......Page 286
9.3 Proof of Theorem 9.2 and Examples......Page 290
10.1 Preliminaries......Page 292
10.2 An Auxiliary Result......Page 293
10.3 The Main Result......Page 295
11.1 Preliminaries......Page 299
11.2 The Subdifferential of Weakly Convex Functions......Page 301
11.3 An Auxiliary Result......Page 302
11.4 The First Main Result......Page 305
11.5 An Algorithm with Constant Step Sizes......Page 312
11.6 An Auxiliary Result......Page 313
11.7 The Second Main Result......Page 317
11.8 Convex Problems......Page 319
11.9 An Auxiliary Result......Page 320
11.10 Proof of Theorem 11.7......Page 322
12.1 Preliminaries and Main Results......Page 325
12.2 Auxiliary Results......Page 328
12.3 An Auxiliary Result with Assumption A2......Page 333
12.4 An Auxiliary Result with Assumption A3......Page 339
12.5 Proof of Theorem 12.1......Page 342
12.6 Proof of Theorem 12.2......Page 350
References......Page 359
Index......Page 363

##### Citation preview

Springer Optimization and Its Applications  155

Alexander J. Zaslavski

Convex Optimization with Computational Errors

Springer Optimization and Its Applications Volume 155 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 J. Birge, University of Chicago S. Butenko, Texas A&M F. Giannessi, University of Pisa S. Rebennack, Karlsruhe Institute of Technology T. Terlaky, Lehigh University Y. 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 http://www.springer.com/series/7393

Alexander J. Zaslavski

Convex Optimization with Computational Errors

Alexander J. Zaslavski Department of Mathematics Amado Building Israel Institute of Technology Haifa, Israel

ISSN 1931-6828 ISSN 1931-6836 (electronic) Springer Optimization and Its Applications ISBN 978-3-030-37821-9 ISBN 978-3-030-37822-6 (eBook) https://doi.org/10.1007/978-3-030-37822-6 Mathematics Subject Classification: 49M37, 65K05, 90C25, 90C26, 90C30 © Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved 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 approximate solutions of optimization problems in the presence of computational errors. It contains a number of results on the convergence behavior of algorithms in a Hilbert space, which are known as important tools for solving optimization problems. The research presented in the book is the continuation and further developments of our book Numerical Optimization with Computational Errors, Springer 2016 . In that book as well as in this new one, we study the algorithms taking into account computational errors which always present in practice. In this case the convergence to a solution does not take place. We show that our algorithms generate a good approximate solution, if computational errors are bounded from above by a small positive constant. Clearly, in practice it is sufficient to find a good approximate solution instead of constructing a minimizing sequence. On the other hand in practice computations induce numerical errors, and if one uses methods in order to solve minimization problems these methods usually provide only approximate solutions of the problems. Our main goal is, for a known computational error, to find out what approximate solution can be obtained and how many iterates one needs for this. The main difference between this new book and the previous one is that here we take into consideration the fact that for every algorithm its iteration consists of several steps and that computational errors for different steps are different, in general. This fact, which was not taken into account in our previous book, is indeed important in practice. For example, the subgradient projection algorithm consists of two steps. The first step is a calculation of a subgradient of the objective function while in the second one we calculate a projection on the feasible set. In each of these two steps there is a computational error, and these two computational errors are different in general. It may happen that the feasible set is simple and the objective function is complicated. As a result, the computational error, made when one calculates the projection, is essentially smaller than the computational error of the calculation of the subgradient. Clearly, an opposite case is possible too. Another feature of this book is that here we study a number of important algorithms which appeared recently in the literature and which are not discussed in the previous book. v

vi

Preface

The monograph contains 12 chapters. Chapter 1 is an introduction. In Chap. 2 we study the subgradient projection algorithm for minimization of convex and nonsmooth functions. We begin with minimization problems on bounded sets and generalize Theorem 2.4 of  proved in the case when the computational errors for the two steps of an iteration are the same. We also consider minimization problems on unbounded sets and generalize Theorem 2.8 of  proved in the case when the computational errors for the two steps of an iteration are the same and prove two results which have no prototype in . Finally, in Chap. 2 we study the subgradient projection algorithm for zero-sum games with two players. For this algorithm each iteration consists of four steps. In each of these steps there is a computational error. We suppose that these computational errors are different and prove two results. In the first theorem, which is a generalization of Theorem 2.11 of  obtained in the case when all the computational errors are the same, we study the games on bounded sets. In the second result, which has no prototype in , we deal with the games on unbounded sets. In Chap. 3 we analyze the mirror descent algorithm for minimization of convex and nonsmooth functions, under the presence of computational errors. The problem is described by an objective function and a set of feasible points. For this algorithm each iteration consists of two steps. The first step is a calculation of a subgradient of the objective function while in the second one we solve an auxiliary minimization problem on the set of feasible points. In each of these two steps there is a computational error. In general, these two computational errors are different. We begin with minimization problems on bounded sets and generalize Theorem 3.1 of  proved in the case when the computational errors for the two steps of an iteration are the same. We also consider minimization problems on unbounded sets and generalize Theorem 3.3 of  proved in the case when the computational errors for the two steps of an iteration are the same and prove two results which have no prototype in . Finally, in Chap. 3 we study the mirror descent algorithm for zero-sum games with two players. For this algorithm each iteration consists of four steps. In each of these steps there is a computational error. We suppose that these computational errors are different and prove two theorems. In the first result, which is a generalization of Theorem 3.4 of  obtained in the case when all the computational errors are the same, we study the games on bounded sets. In the second result, which has no prototype in , we deal with the games on unbounded sets. In Chap. 4 we analyze the projected gradient algorithm with a smooth objective function under the presence of computational errors. For this algorithm each iteration also consists of two steps. The first step is a calculation of a gradient of the objective function while in the second one we calculate a projection on the feasible set. In each of these two steps there is a computational error. Our first result in this chapter, for minimization problems on bounded sets, is a generalization of Theorem 4.2 of  proved in the case when the computational errors for the two steps of an iteration are the same. We also consider minimization problems on unbounded sets and generalize Theorem 4.5 of  proved in the case when the computational errors for the two steps of an iteration are the same and prove

Preface

vii

two results which have no prototype in . In Chap. 5 we consider an algorithm, which is an extension of the projection gradient algorithm used for solving linear inverse problems arising in signal/image processing. This algorithm is used for minimization of the sum of two given convex functions, and each of its iteration consists of two steps. In each of these two steps there is a computational error. These two computational errors are different. We generalize Theorem 5.1 of  obtained in the case when the computational errors for the two steps of an iteration are the same and prove two results which have no prototype in . In Chap. 6 we study continuous subgradient method and continuous subgradient projection algorithm for minimization of convex nonsmooth functions and for computing the saddle points of convex–concave functions, under the presence of computational errors. For these algorithms each iteration consists of a few calculations, and for each of these calculations there is a computational error produced by our computer system. In general, these computational errors are different. All the results of this chapter have no prototype in . In Chaps. 7–12 we analyze several algorithms under the presence of computational errors, which were not considered in . Again, each step of an iteration has a computational error, and we take into account that these errors are, in general, different. An optimization problem with a composite objective function is studied in Chap. 7. A zero-sum game with two players is considered in Chap. 8. A predicted decrease approximation-based method is used in Chap. 9 for constrained convex optimization. Chapter 10 is devoted to minimization of quasiconvex functions. Minimization of sharp weakly convex functions is discussed in Chap. 11. Chapter 12 is devoted to a generalized projected subgradient method for minimization of a convex function over a set which is not necessarily convex. The author believes that this book will be useful for researches interested in the optimization theory and its applications. Rishon LeZion, Israel May 19, 2019

Alexander J. Zaslavski

Contents

1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Subgradient Projection Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 The Mirror Descent Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Gradient Algorithm with a Smooth Objective Function . . . . . . . . . . 1.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 10 18 22

2

Subgradient Projection Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 A Convex Minimization Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The Main Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Proof of Theorem 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Subgradient Algorithm on Unbounded Sets . . . . . . . . . . . . . . . . . . . . . . . 2.6 Proof of Theorem 2.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Proof of Theorem 2.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Proof of Theorem 2.12. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Zero-Sum Games with Two Players . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.10 Proof of Proposition 2.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11 Zero-Sum Games on Bounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12 Zero-Sum Games on Unbounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.13 Proof of Theorem 2.16. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.14 An Example for Theorem 2.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25 25 27 31 33 34 42 45 49 53 58 66 71 77 80

3

The Mirror Descent Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Optimization on Bounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 The Main Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Proof of Theorem 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Optimization on Unbounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Proof of Theorem 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Proof of Theorem 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Proof of Theorem 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Zero-Sum Games on Bounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Zero-Sum Games on Unbounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83 83 87 90 91 96 100 105 110 116 ix

x

Contents

4

Gradient Algorithm with a Smooth Objective Function . . . . . . . . . . . . . . . 4.1 Optimization on Bounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 The Main Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Proof of Theorem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Optimization on Unbounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

127 127 129 130 136 137

5

An Extension of the Gradient Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Preliminaries and the Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Proof of Theorem 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 The First Extension of Theorem 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 The Second Extension of Theorem 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . .

151 151 154 161 163 167

6

Continuous Subgradient Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Bochner Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Convergence Analysis for Continuous Subgradient Method . . . . . 6.3 An Auxiliary Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Proof of Theorem 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Continuous Subgradient Method for Zero-Sum Games. . . . . . . . . . . 6.6 An Auxiliary Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Proof of Theorem 6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Continuous Subgradient Projection Method . . . . . . . . . . . . . . . . . . . . . . . 6.9 An Auxiliary Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.10 Proof of Theorem 6.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11 Continuous Subgradient Projection Method on Unbounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.12 An Auxiliary Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.13 The Convergence Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.14 Subgradient Projection Algorithm for Zero-Sum Games . . . . . . . . . 6.15 An Auxiliary Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.16 A Convergence Result for Games on Bounded Sets . . . . . . . . . . . . . . 6.17 A Convergence Result for Games on Unbounded Sets . . . . . . . . . . .

173 173 174 177 178 181 184 188 193 195 196

7

An Optimization Problems with a Composite Objective Function. . . . 7.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 The Algorithm and Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Proof of Theorem 7.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Proof of Theorem 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

243 243 245 248 254 257

8

A Zero-Sum Game with Two Players. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 The Algorithm and the Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Proof of Theorem 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

259 259 263 271

201 202 207 215 216 225 234

Contents

xi

9

PDA-Based Method for Convex Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Preliminaries and the Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Proof of Theorem 9.2 and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

277 277 281 285

10

Minimization of Quasiconvex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 An Auxiliary Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 The Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

287 287 288 290

11

Minimization of Sharp Weakly Convex Functions. . . . . . . . . . . . . . . . . . . . . . 11.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The Subdifferential of Weakly Convex Functions. . . . . . . . . . . . . . . . . 11.3 An Auxiliary Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 The First Main Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 An Algorithm with Constant Step Sizes . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 An Auxiliary Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7 The Second Main Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.8 Convex Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.9 An Auxiliary Result. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.10 Proof of Theorem 11.7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

295 295 297 298 301 308 309 313 315 316 318

12

A Projected Subgradient Method for Nonsmooth Problems . . . . . . . . . . 12.1 Preliminaries and Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 An Auxiliary Result with Assumption A2 . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 An Auxiliary Result with Assumption A3 . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 Proof of Theorem 12.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.6 Proof of Theorem 12.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

321 321 324 329 335 338 346

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359

Chapter 1

Introduction

In this book we study behavior of algorithms for constrained convex minimization problems in a Hilbert space. Our goal is to obtain a good approximate solution of the problem in the presence of computational errors. It is known that the algorithm generates a good approximate solution, if the sequence of computational errors is bounded from above by a small constant. In our study, presented in this book, we take into consideration the fact that for every algorithm its iteration consists of several steps and that computational errors for different steps are different, in general. In this chapter we discuss several algorithms which are studied in this book.

1.1 Subgradient Projection Method In Chap. 2 we study the subgradient projection algorithm for minimization of convex and nonsmooth functions and for computing the saddle points of convex– concave functions, under the presence of computational errors. It should be mentioned that the subgradient projection algorithm is one of the most important tools in the optimization theory, nonlinear analysis, and their applications. See, for example, [56, 58, 64, 65, 67–73, 76, 79, 80, 83, 85, 89–92, 94] and the references mentioned therein. The problem is described by an objective function and a set of feasible points. For this algorithm each iteration consists of two steps. The first step is the calculation of a subgradient of the objective function while in the second one we calculate the projection on the feasible set. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this. © Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_1

1

2

1 Introduction

We use the subgradient projection algorithm for constrained minimization problems in Hilbert spaces equipped with an inner product denoted by ·, · which induces a complete norm  · . In this book we use the following notation. For every z ∈ R 1 denote by z the largest integer which does not exceed z: z = max{i ∈ R 1 : i is an integer and i ≤ z}. For every nonempty set D, every function f : D → R 1 , and every nonempty set C ⊂ D we set inf(f, C) = inf{f (x) : x ∈ C} and argmin(f, C) = argmin{f (x) : x ∈ C} = {x ∈ C : f (x) = inf(f, C)}. Let X be a Hilbert space equipped with an inner product denoted by ·, · which induces a complete norm  · . For each x ∈ X and each r > 0 set BX (x, r) = {y ∈ X : x − y ≤ r} and for each x ∈ X and each nonempty set E ⊂ X set d(x, E) = inf{x − y : y ∈ E}. For each nonempty open convex set U ⊂ X and each convex function f : U → R 1 , for every x ∈ U set ∂f (x) = {l ∈ X : f (y) − f (x) ≥ l, y − x for all y ∈ U }, which is called the subdifferential of the function f at the point x [61, 62, 81]. Let C be a nonempty closed convex subset of X, U be an open convex subset of X such that C ⊂ U , and f : U → R 1 be a convex function. Suppose that there exist L > 0, M0 > 0 such that C ⊂ BX (0, M0 ), |f (x) − f (y)| ≤ Lx − y for all x, y ∈ U. It is not difficult to see that for each x ∈ U , ∅ = ∂f (x) ⊂ BX (0, L).

3

For every nonempty closed convex set D ⊂ X and every x ∈ X there is a unique point PD (x) ∈ D satisfying x − PD (x) = inf{x − y : y ∈ D}. We consider the minimization problem f (z) → min, z ∈ C. Suppose that {ak }∞ k=0 ⊂ (0, ∞). Let us describe our algorithm. Subgradient projection algorithm Initialization: select an arbitrary x0 ∈ U . Iterative Step: given a current iteration vector xt ∈ U calculate ξt ∈ ∂f (xt ) and the next iteration vector xt+1 = PC (xt − at ξt ). In  we study this algorithm under the presence of computational errors. Namely, in  we suppose that δ ∈ (0, 1] is a computational error produced by our computer system, and study the following algorithm. Subgradient projection algorithm with computational errors Initialization: select an arbitrary x0 ∈ U . Iterative Step: given a current iteration vector xt ∈ U calculate ξt ∈ ∂f (xt ) + BX (0, δ) and the next iteration vector xt+1 ∈ U such that xt+1 − PC (xt − at ξt ) ≤ δ. In Chap. 2 of this book we consider more complicated, but more realistic, version of this algorithm. Clearly, for the algorithm each iteration consists of two steps. The first step is a calculation of a subgradient of the objective function f while in the second one we calculate a projection on the set C. In each of these two steps there is a computational error produced by our computer system. In general, these two computational errors are different. This fact is taken into account in the following projection algorithm studied in Chap. 2 of this book. Suppose that {ak }∞ k=0 ⊂ (0, ∞) and δf , δC ∈ (0, 1]. Initialization: select an arbitrary x0 ∈ U . Iterative Step: given a current iteration vector xt ∈ U calculate ξt ∈ ∂f (xt ) + BX (0, δf )

4

1 Introduction

and the next iteration vector xt+1 ∈ U such that xt+1 − PC (xt − at ξt ) ≤ δC . Note that in practice for some problems the set C is simple but the function f is complicated. In this case δC is essentially smaller than δf . On the other hand, there are cases when f is simple but the set C is complicated and therefore δf is much smaller than δC . In Chap. 2 we prove the following result (see Theorem 2.4). Theorem 1.1 Let δf , δC ∈ (0, 1], {ak }∞ k=0 ⊂ (0, ∞) and let x∗ ∈ C satisfy f (x∗ ) ≤ f (x) for all x ∈ C. ∞ Assume that {xt }∞ t=0 ⊂ U , {ξt }t=0 ⊂ X,

x0  ≤ M0 + 1 and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ) and xt+1 − PC (xt − at ξt ) ≤ δC . Then for each natural number T , T 

at (f (xt ) − f (x∗ ))

t=0

≤ 2−1 x∗ − x0 2 + δC (T + 1)(4M0 + 1) +δf (2M0 + 1)

T  t=0

Moreover, for each natural number T ,

at + 2−1 (L + 1)2

T  t=0

at2 .

f ((

T 

at )−1

t=0

T 

5

at xt ) − f (x∗ ), min{f (xt ) : t = 0, . . . , T } − f (x∗ )

t=0

≤ 2−1 (

T 

at )−1 x∗ − x0 2 + (

t=0

T 

at )−1 δC (T + 1)(4M0 + 1)

t=0

+δf (2M0 + 1) + 2−1 (

T 

at )−1 (L + 1)2

T 

t=0

at2 .

t=0

Theorem 1.1 is proved in Sect. 2.4. It is a generalization of Theorem 2.4 of  proved in the case when δf = δC . We are interestedin an optimal choice of at , t = 0, 1, . . . . Let T be a natural number and AT = Tt=0 at be given. It is shown in Chap. 2 that the best choice is at = (T + 1)−1 AT , t = 0, . . . , T . Let T be a natural number and at = a > 0, t = 0, . . . , T . It is shown in Chap. 2 that the best choice of a is a = (2δC (4M0 + 1))1/2 (L + 1)−1 . Now we can think about the best choice of T . It is not difficult to see that it should be at the same order as δC−1 . In Chap. 2 we also study the subgradient algorithm for minimization problems on unbounded sets. Let D be a nonempty closed convex subset of X, V be an open convex subset of X such that D ⊂ V, and f : V → R 1 be a convex function which is Lipschitz on all bounded subsets of V . Set Dmin = {x ∈ D : f (x) ≤ f (y) for all y ∈ D}. We suppose that Dmin = ∅. In Chap. 2 we will prove the following result. Theorem 1.2 Let δf , δC ∈ (0, 1], M > 0 satisfy Dmin ∩ BX (0, M) = ∅, M0 ≥ 4M + 4,

6

1 Introduction

L > 0 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, M0 + 2), 0 < τ0 ≤ τ1 ≤ (L + 1)−1 , 0 = 2τ0−1 δC (4M0 + 1) + 2δf (2M0 + 1) + 2τ1 (L + 1)2 and let n0 = τ0−1 (2M + 2)2 0−1 . ∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

{at }∞ t=0 ⊂ [τ0 , τ1 ], x0  ≤ M and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ) and xt+1 − PD (xt − at ξt ) ≤ δC . Then there exists an integer q ∈ [1, n0 + 1] such that xi  ≤ 3M + 2, i = 0, . . . , q and f (xq ) ≤ f (x) + 0 for all x ∈ D. Theorem 1.2 is a generalization of Theorem 2.8 of  which was proved in the case when δC = δf . It is proved in Sect. 2.6. We are interested in the best choice of at , t = 0, 1, . . . . Assume for simplicity that τ1 = τ0 = τ . In order to meet our goal we need to minimize the function 2τ −1 δC (4M0 + 1) + 2(L + 1)2 τ, τ ∈ (0, ∞). This function has a minimizer τ = (δC (4M0 + 1))1/2 (L + 1)−1 ,

7

the minimal value of 0 is 2δf (2M0 + 1) + 4(δC (4M0 + 1))1/2 (L + 1), and n0 = Δ where −1/2

Δ = (2M + 2)2 (4M0 + 1)−1/2 (L + 1)δC

(2δf (2M0 + 1)

+4δC (4M0 + 1)1/2 (L + 1))−1 . 1/2

Note that in the theorem above δf , δC are the computational errors produced by our computer system. In view of the inequality above, in order to obtain a good approximate solution we need −1/2

c1 δC

max{δf , δC }−1  + 1 1/2

iterations, where c1 is a constant which depends only on M, M0 , L. As a result, we obtain a point ξ ∈ V such that BX (ξ, δC ) ∩ D = ∅ and 1/2

f (ξ ) ≤ inf{f (x) : x ∈ D} + 2δf (2M0 + 1) + (4M0 + 1)1/2 4(L + 1)δC . The next result, which is proved in Sect. 2.7, does not have a prototype in . Theorem 1.3 Let δf , δC ∈ (0, 1), M > 1, L > 0 and let {at }∞ t=0 ⊂ (0, ∞) be such that {x ∈ V : f (x) ≤ inf(f, D) + 3} ⊂ BX (0, M − 1), |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 8), δf ≤ (6M + 5)−1 and δC (12M + 9) ≤ at ≤ (L + 1)−2 for all integers t ≥ 0. ∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M, BX (x0 , δC ) ∩ D = ∅

8

1 Introduction

and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ) and xt+1 − PD (xt − at ξt ) ≤ δC . Then xt  ≤ 3M for all integers t ≥ 0 and for each natural number T , T 

at (f (xt ) − inf(f, D))

t=0

≤ 2M 2 + δC (T + 1)(12M + 9) +2−1 (L + 1)2

T 

at2 + δf (6M + 5)

t=0

T 

at .

t=0

Moreover, for each natural number T , f ((

T 

at )−1

t=0

T 

at xt ) − inf(f, D),

t=0

min{f (xt ) : t = 0, . . . , T } − inf(f, D) ≤ 2M 2 (

T 

T  at )−1 + δC (T + 1)(12M + 9)( at )−1

t=0

+2−1 (L + 1)2

t=0 T  t=0

at2 (

T 

at )−1 + δf (6M + 5).

t=0

We are interestedin an optimal choice of at , t = 0, 1, . . . . Let T be a natural number and AT = Tt=0 at be given. It is shown in Chap. 2 that the best choice is at = (T + 1)−1 A. t = 0, . . . , T . Let T be a natural number and at = a > 0, t = 0, . . . , T . It is shown in Chap. 2 that the best choice of a is

9

a = (2δC (12M + 9))1/2 (L + 1)−1 . Now we can think about the best choice of T . It is shown that it should be at the same order as δC−1  The next result, which is proved in Sect. 2.8, does not have a prototype in . Theorem 1.4 Let δf , δC ∈ (0, 16−1 ), M > 4, L > 0, Dmin ∩ BX (0, M) = ∅, |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 3M + 6), a = (L + 1)−1 δC

1/2

and −1/2 −1 δf } − 1.

T = 6−1 min{δC−1 , δC ∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M, BX (x0 , δC ) ∩ D = ∅ and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ) and xt+1 − PD (xt − at ξt ) ≤ δC . Then T  (T + 1)−1 f (xt ) − inf(f, D), t=0

f(

T  (T + 1)−1 xt ) − inf(f, D), min{f (xt ) : t = 0, . . . , T } − inf(f, D) t=0 1/2

≤ 12(2M + 2)2 (L + 1) max{δC , δf }.

10

1 Introduction

In Chap. 2 we also study the subgradient projection algorithm for zero-sum games with two players. For this algorithm each iteration consists of four steps. In each of these steps there is a computational error. We suppose that these computational errors are different and prove two results: Theorems 2.15 and 2.16. In Theorem 2.15, which is a generalization of Theorem 2.11 of  obtained in the case when all the computational errors are the same, we study the games on bounded sets. In Theorem 2.16, which has no prototype in , we deal with the games on unbounded sets.

1.2 The Mirror Descent Method In Chap. 3 we analyze the mirror descent algorithm for minimization of convex and nonsmooth functions and for computing the saddle points of convex–concave functions, under the presence of computational errors. The problem is described by an objective function and a set of feasible points. For this algorithm each iteration consists of two steps. The first step is a calculation of a subgradient of the objective function while in the second one we solve an auxiliary minimization problem on the set of feasible points. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this. Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · . Let C be a nonempty closed convex subset of X, U be an open convex subset of X such that C ⊂ U , and let f : U → R 1 be a convex function. Suppose that there exist L > 0, M0 > 0 such that C ⊂ BX (0, M0 ), |f (x) − f (y)| ≤ Lx − y for all x, y ∈ U. We study the convergence of the mirror descent algorithm under the presence of computational errors. This method was introduced by Nemirovsky and Yudin for solving convex optimization problems . Here we use a derivation of this algorithm proposed by Beck and Teboulle . Let δf , δC ∈ (0, 1] and {ak }∞ k=0 ⊂ (0, ∞). We describe the inexact version of the mirror descent algorithm. Mirror descent algorithm Initialization: select an arbitrary x0 ∈ U . Iterative Step: given a current iteration vector xt ∈ U calculate

1.2 The Mirror Descent Method

11

ξt ∈ ∂f (xt ) + BX (0, δf ), define gt (x) = ξt , x + (2at )−1 x − xt 2 , x ∈ X and calculate the next iteration vector xt+1 ∈ U such that BX (xt+1 , δC ) ∩ argmin{gt (y) : y ∈ C} = ∅. Here δf is a computational error produced by our computer system when we calculate a subgradient of f and δC is a computational error produced by our computer system when we calculate a minimizer of the function gt . Note that gt is a convex bounded from below function on X which possesses a minimizer on C. In Sect. 3.3 we prove the following result. Theorem 1.5 Let δf , δC ∈ (0, 1], {ak }∞ k=0 ⊂ (0, ∞) and let x∗ ∈ C satisfies f (x∗ ) ≤ f (x) for all x ∈ C. ∞ Assume that {xt }∞ t=0 ⊂ U , {ξt }t=0 ⊂ X,

x0  ≤ M0 + 1 and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ), gt (z) = ξt , z + (2at )−1 z − xt 2 , z ∈ X and BX (xt+1 , δC ) ∩ argmin(gt , C) = ∅. Then for each natural number T , T  t=0

at (f (xt ) − f (x∗ ))

12

1 Introduction

≤ 2−1 (2M0 + 1)2 + (δf (2M0 + 1) + δC (L + 1))

T 

at

t=0

+4δC (T + 1)(2M0 + 1) + 2−1 (L + 1)2

T 

at2 .

t=0

Moreover, for each natural number T , f ((

T 

at )−1

t=0

T 

at xt ) − f (x∗ ),

t=0

min{f (xt ) : t = 0, . . . , T } − f (x∗ ) −1

≤2

(2M0 + 1) ( 2

T 

at )−1 + δf (2M0 + 1) + δC (L + 1)

t=0

+4δC (T + 1)(2M0 + 1)(

T 

at )−1

t=0

+2−1 (L + 1)2

T 

at2 (

t=0

T 

at )−1 .

t=0

This theorem is a generalization of Theorem 3.1 of  proved in the case when δf = δC . We are interested  in an optimal choice of at , t = 0, 1, . . . . Let T be a natural T number and AT = t=0 at be given. It is shown that the best choice is at = (T + 1)−1 AT , t = 0, . . . , T . Let T be a natural number and at = a, t = 0, . . . , T . It is shown that the best choice of a > 0 is a = (8δC (2M0 + 1))1/2 (L + 1)−1 and that the best choice of T should be at the same order as δC−1 . As a result, we obtain a point ξ ∈ U such that BX (ξ, δC ) ∩ C = ∅ and 1/2

f (ξ ) ≤ f (x∗ ) + c1 δC + δf (2M0 + 1), where the constant c1 > 0 depends only on L and M0 .

1.2 The Mirror Descent Method

13

In Chap. 3 we also use the mirror descent algorithm for minimization problems of unbounded sets. Let D be a nonempty closed convex subset of X, V be an open convex subset of X such that D ⊂ V, and f : V → R 1 be a convex function which is Lipschitz on all bounded subsets of V . Set Dmin = {x ∈ D : f (x) ≤ f (y) for all y ∈ D}. We suppose that Dmin = ∅. In Sect. 3.5 we will prove the following result. Theorem 1.6 Let δf , δC ∈ (0, 1], M > 1 satisfy Dmin ∩ BX (0, M) = ∅, M0 > 80M + 6, L > 1 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, M0 + 2), 0 < τ0 ≤ τ1 ≤ (4L + 4)−1 , 0 = 8τ0−1 δC (M0 + 1) + 2δC (L + 1) +2δf (2M0 + 1) + τ1 (L + 1)2 and let n0 = τ0−1 (2M + 2)2 0−1  + 1. Assume that ∞ ∞ {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X, {at }t=0 ⊂ [τ0 , τ1 ],

x0  ≤ M

14

1 Introduction

and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ), gt (v) = ξt , v + (2at )−1 v − xt 2 , v ∈ X and BX (xt+1 , δC ) ∩ argmin(gt , D) = ∅. Then there exists an integer q ∈ [1, n0 ] such that f (xq ) ≤ inf(f ; D) + 0 , xt  ≤ 15M + 1, t = 0, . . . , q. This theorem is a generalization of Theorem 3.3 of  proved with δf = δC . We are interested in the best choice of at , t = 0, 1, . . . . Assume for simplicity that τ1 = τ0 . In order to meet our goal we need to minimize 0 which obtains its minimal value when τ0 = (8δC (M0 + 1))1/2 (L + 1)−1 and the minimal value of 0 is 2δC (L + 1) + 2δf (2M0 + 1) + 2(8δC (M0 + 1))1/2 (L + 1). 1/2

Thus 0 is at the same order as max{δC , δf }. It is not difficult to see that n0 is at −1/2 1/2 the same order as δC max{δC , δf }−1 . In Chap. 3 we will also prove the following two theorems which have no prototype in . Theorem 1.7 Let δf , δC ∈ (0, 1), M > 0 satisfy δf ≤ (24M + 48)−1 , δC ≤ 8−1 (L + 1)−2 (16M + 16)−1 , {x ∈ V : f (x) ≤ inf(f, D) + 4} ⊂ BX (0, M), M0 = 12M + 12, L > 1 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, M0 + 2)

1.2 The Mirror Descent Method

15

and let {at }∞ t=0 ⊂ (0, ∞) satisfy for all integers t ≥ 0, 4δC (M0 + 1) ≤ at ≤ (2L + 2)−1 (L + 1)−1 . Assume that ∞ {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ), gt (v) = ξt , v + (2at )−1 v − xt 2 , v ∈ X and BX (xt+1 , δC ) ∩ argmin(gt , D) = ∅. Then xt  ≤ 5M + 3 for all integers t ≥ 0 and for each natural number T , T 

at (f (xt ) − inf(f, D))

t=0

≤ 2M 2 + (δf (2M0 + 1) + δC (L + 1))

T 

at

t=0

−1

+4δC (T + 1)(M0 + 1) + 2

(L + 1)

2

T  t=0

Moreover, for each natural number T , f ((

T  t=0

at )−1

T 

at xt ) − inf(f, D),

t=0

min{f (xt ) : t = 0, . . . , T } − inf(f, D)

at2 .

16

1 Introduction

≤ 2M 2 (

T 

at )−1 + δf (2M0 + 1) + δC (L + 1)

t=0

+4δC (T + 1)(M0 + 1)(

T 

at )−1 + 2−1 (L + 1)2

t=0

T 

at2 (

t=0

T 

at )−1 .

t=0

We are interested  in the best choice of at , t = 0, 1, . . . . Let T be a natural T number and AT = t=0 at be given. It is shown that the best choice is at = −1 (T + 1) AT , i = 0, . . . , T . Let T be a natural number and at = a, t = 0, . . . , T . The best choice of a > 0 is a = (8δC (M0 + 1))1/2 (L + 1)−1 and the best choice of T is at the same order as δC−1 . As a result, we obtain a point ξ ∈ U such that BX (ξ, δC ) ∩ C = ∅ and 1/2

f (ξ ) ≤ inf(f, D) + c1 δC + δf (2M0 + 1), where the constant c1 > 0 depends only on L and M0 . Theorem 1.8 Let δf , δC ∈ (0, 1), M > 8 satisfy Dmin ∩ BX (0, M) = ∅, L > 0 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 8M + 8), a = (4L + 4)−1 δC , 1/2

−1/2

T = 8−1 δC

max{δf , δC }−1 . 1/2

1.2 The Mirror Descent Method

17

Assume that ∞ {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M, BX (x0 , δC ) ∩ D = ∅ and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ), gt (v) = ξt , v + (2at )−1 v − xt 2 , v ∈ X and BX (xt+1 , δC ) ∩ argmin(gt , D) = ∅. Then xt  ≤ 3M + 2 for all integers t ∈ [0, T ] and T  (T + 1)−1 f (xt ) − inf(f, D), t=0

f ((T + 1)−1

T 

xt ) − inf(f, D), min{f (xt ) : t = 0, . . . , T } − inf(f, D)

t=0 1/2

≤ 64M 2 (L + 1) max{δf , δC }. In Chap. 3 we also study the mirror descent algorithm for zero-sum games with two players. For this algorithm each iteration consists of four steps. In each of these steps there is a computational error. We suppose that these computational errors are different and prove two results: Theorems 3.6 and 3.7. In Theorem 3.6, which is a generalization of Theorem 3.4 of  obtained in the case when all the computational errors are the same, we study the games on bounded sets. In Theorem 3.7, which has no prototype in , we deal with the games on unbounded sets.

18

1 Introduction

1.3 Gradient Algorithm with a Smooth Objective Function In Chap. 4 we analyze the convergence of a projected gradient algorithm with a smooth objective function under the presence of computational errors. The problem is described by an objective function and a set of feasible points. For this algorithm each iteration consists of two steps. The first step is a calculation of a gradient of the objective function while in the second one we calculate a projection on the feasible set. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this. Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · . Let C be a nonempty closed convex subset of X, U be an open convex subset of X such that C ⊂ U , and let f : U → R 1 be a convex continuous function. We suppose that the function f is Fréchet differentiable at every point x ∈ U and for every x ∈ U we denote by f  (x) ∈ X the Fréchet derivative of f at x. It is clear that for any x ∈ U and any h ∈ X f  (x), h = lim t −1 (f (x + th) − f (x)). t→0

We suppose that the mapping f  : U → X is Lipschitz on all bounded subsets of U . It is well known (see Lemma 2.2) that for each nonempty closed convex set D ⊂ X and each x ∈ X there exists a unique point PD (x) ∈ D such that x − PD (x) = inf{x − y : y ∈ D}. Suppose that there exist L > 1, M0 > 0 such that C ⊂ BX (0, M0 ), |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ U, f  (v1 ) − f  (v2 ) ≤ Lv1 − v2  for all v1 , v2 ∈ U. Let δf , δC ∈ (0, 1]. We describe below our algorithm. Gradient algorithm Initialization: select an arbitrary x0 ∈ U ∩ BX (0, M0 ). Iterative Step: given a current iteration vector xt ∈ U calculate ξt ∈ f  (xt ) + BX (0, δf )

1.3 Gradient Algorithm with a Smooth Objective Function

19

and calculate the next iteration vector xt+1 ∈ U such that xt+1 − PC (xt − L−1 ξt ) ≤ δC . In Chap. 4 we prove the following result which is a generalization of Theorem 4.2 of  proved in the case when δf = δC . Theorem 1.9 Let δf , δC ∈ (0, 1] and let x0 ∈ U ∩ BX (0, M0 ). ∞ Assume that {xt }∞ t=1 ⊂ U , {ξt }t=0 ⊂ X and that for each integer t ≥ 0,

ξt − f  (xt ) ≤ δ and xt+1 − PC (xt − L−1 ξt ) ≤ δC . Then for each natural number T , min{f (xt ) : t = 2, . . . , T + 1} − inf(f, C), T +1

f(

T −1 xt ) − inf(f, C)

t=2

≤ (2T )−1 L(2M0 + 1)2 + LδC (6M0 + 7) + δf (4M0 + 4). We are interested in an optimal choice of T . If we choose T in order to minimize the right-hand side of the estimation in the theorem above, we obtain that T should be at the same order as max{δf , δC }−1 . In this case the right-hand side of the estimation is at the same order as max{δf , δC }. In Chap. 4 we also study optimization on unbounded sets. Let D be a nonempty closed convex subset of X, V be an open convex subset of X such that D ⊂ V, and f : V → R 1 be a convex Fréchet differentiable function which is Lipschitz on all bounded subsets of V . Set Dmin = {x ∈ D : f (x) ≤ f (y) for all y ∈ D}.

20

1 Introduction

We suppose that Dmin = ∅. We will prove the following result which is a generalization of Theorem 4.5 of  proved in the case when δf = δC . Theorem 1.10 Let δf , δC ∈ (0, 1], M > 0, Dmin ∩ BX (0, M) = ∅, M0 = 4M + 8, L ≥ 1 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, M0 + 2), f  (v1 ) − f  (v2 ) ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, M0 + 2), 0 = δf (4M0 + 6) + δC L(6M0 + 8) and let n0 = 2−1 L(2M + 1)2 (δf + LδC )−1  + 1. ∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M and that for each integer t ≥ 0, ξt − f  (xt ) ≤ δf and xt+1 − PD (xt − L−1 ξt ) ≤ δC . Then there exists an integer q ∈ [1, n0 + 1] such that f (xq ) ≤ inf(f, D) + 0 ,

1.3 Gradient Algorithm with a Smooth Objective Function

21

xi  ≤ 3M + 3, i = 0, . . . , q. The following two results have no prototype in . Theorem 1.11 Let δf , δC ∈ (0, 1], M > 2 satisfy {x ∈ V : f (x) ≤ inf(f, D) + 3} ⊂ BX (0, M − 2), L ≥ 1 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 8), f  (v1 ) − f  (v2 ) ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 8), δC ≤ (L(18M + 19))−1 , δf ≤ (12M + 3)−1 . ∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M and that for each integer t ≥ 0, ξt − f  (xt ) ≤ δf and xt+1 − PD (xt − L−1 ξt ) ≤ δC . Then xt  ≤ 3M for all integers t ≥ 0 and for each natural number T , T +1

min{f (xt ) : t = 1, . . . , T + 1} − inf(f, D), f (

(T + 1)−1 xt ) − inf(f, D)

t=1

≤ LδC (18M + 19) + δf (12M + 13) + 2(T + 1)−1 M 2 L. It is clear that a best choice of T should be at the same order as max{δC , δf }−1 . Theorem 1.12 Let δf , δC ∈ (0, 1), M > 1, δC ≤ (160M)−1 , δf ≤ 120−1 ,

22

1 Introduction

Dmin ∩ BX (0, M) = ∅, L ≥ 1 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 10), f  (v1 ) − f  (v2 ) ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 10), T = 36−1 min{δC−1 , Lδf−1 }. ∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M and that for each integer t ≥ 0, ξt − f  (xt ) ≤ δf and xt+1 − PD (xt − L−1 ξt ) ≤ δC . Then xt  ≤ 3M for all integers t = 0, . . . , T + 1 and T  (T + 1)−1 (f (xt+1 ) − inf(f, D)) t=0 T +1

min{f (xt ) : t = 1, . . . , T + 1} − inf(f, D), f (

(T + 1)−1 xt ) − inf(f, D)

t=1

≤ 72(M + 1)2 max{δC L, δf }.

1.4 Examples Example 1.13 Let X = R n , C = {x = (x1 , . . . , xn ) ∈ R n : xi ∈ [0, 1], i = 1, . . . , n}, U = {x ∈ R n : x < 2n1/2 }

1.4 Examples

23

and let for all x ∈ R n , f (x) = Ax, x + b, x, where A = (ai,j )ni,j =1 is a symmetric positive semidefinite matrix and b ∈ R n . Clearly, the function f is convex and ∇f (x) = b + 2Ax for all x ∈ R n . We consider the minimization problem f (x) → min, x ∈ C and use the subgradient projection method taking into account computational errors (see Sect. 1.2). It is clear that for all x = (x1 , . . . , xn ) ∈ R n , PC (x) = (min{max{x1 , 0}, 1}, . . . , min{max{xi , 0}, 1}, . . . , min{max{xn , 0}, 1}). Assume that our computer system has an accuracy δ∗ . Then it is not difficult to see that in our case δC = δ∗ n1/2 , M0 = 2n1/2 , δf = 2δ∗ n max{|ai,j | : i, j = 1, . . . , n}, L = b + 4(

n 

2 1//2 1/2 ai,j ) n .

i,j =1

We apply Theorem 1.1 and obtain that for all integers t ≥ 0, at = a = (2δC (4M0 + 1))1/2 (L + 1)−1 = (2δ∗ n

1/2

1/2

(8n

+ 1))

1/2

(b + 1 + 4(

n 

2 1/2 1/2 −1 ai,j ) n )

i,j =1

and T = δC−1  = δ∗−1 n−1/2 . Example 1.14 Consider the previous example and apply the gradient projection algorithm with the smooth objective function (see Theorem 1.9) and with the same C, Uf, A, b, δ∗ . It is not difficult to see that M0 = 2n1/2 , L = b + 4(

n  i,j =1

2 1//2 1/2 ai,j ) n ,

24

1 Introduction

δC = δ∗ n1/2 , δf = 2δ∗ n max{|ai,j | : i, j = 1, . . . , n}. According to Theorem 1.9, max{δC , δf }−1  iterates should be done. As a result we obtain a point z ∈ R n such that d(z, C) ≤ δC and f (z) ≤ inf(f, C) + c0 max{δf , δC }, where c0 is a positive constant. Its value can be easily calculated using Theorem 1.9. Example 1.15 Consider the minimization problem from the previous two examples applying Theorem 1.9. Assume that we do 103 iterates. By Theorem 1.9, we obtain z ∈ R n such that d(z, C) ≤ δC and f (z) ≤ inf(f, C) + 2−1 10−3 L(2M0 + 1)2 + LδC (6M0 + 7) + δf (4M0 + 4).

Chapter 2

In this chapter we study the subgradient projection algorithm for minimization of convex and nonsmooth functions and for computing the saddle points of convex– concave functions, under the presence of computational errors. The problem is described by an objective function and a set of feasible points. For this algorithm each iteration consists of two steps. The first step is a calculation of a subgradient of the objective function while in the second one we calculate a projection on the feasible set. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this.

2.1 Preliminaries The subgradient projection algorithm is one of the most important tools in the optimization theory, nonlinear analysis, and their applications. See, for example, [1–3, 11, 15, 17, 22, 27, 29, 30, 32, 35, 36, 38, 39, 43, 46, 53–55] and the references mentioned therein. In this chapter we use this method for constrained minimization problems in Hilbert spaces. Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · . Let C be a nonempty closed convex subset of X, U be an open convex subset of X such that C ⊂ U , and f : U → R 1 be a convex function. Suppose that there exist L > 0, M0 > 0 such that

© Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_2

25

26

C ⊂ BX (0, M0 ),

(2.1)

|f (x) − f (y)| ≤ Lx − y for all x, y ∈ U.

(2.2)

In view of (2.2), for each x ∈ U , ∅ = ∂f (x) ⊂ BX (0, L).

(2.3)

It is easy to see that the following result is true. Lemma 2.1 Let z, y0 , y1 ∈ X. Then z − y0 2 − z − y1 2 − y0 − y1 2 = 2z − y1 , y1 − y0 . The next result is well known in the literature [12, 13, 92, 93]. Lemma 2.2 Let D be a nonempty closed convex subset of X. Then for each x ∈ X there is a unique point PD (x) ∈ D satisfying x − PD (x) = inf{x − y : y ∈ D}. Moreover, PD (x) − PD (y)|| ≤ x − y for all x, y ∈ X and for each x ∈ X and each z ∈ D, z − PD (x), x − PD (x) ≤ 0, z − PD (x)2 + x − PD (x)2 ≤ z − x2 . For the proof of the next simple result see Lemma 2.3 of . Lemma 2.3 Let A > 0 and n ≥ 2 be an integer. Then the minimization problem n 

ai2 → min

i=1

a = (a1 , . . . , an ) ∈ R n and

n 

ai = A

i=1

has a unique solution a ∗ = (a1∗ , . . . , an∗ ) where ai∗ = n−1 A, i = 1, . . . , n.

2.2 A Convex Minimization Problem

27

2.2 A Convex Minimization Problem Suppose that δf , δC ∈ (0, 1] and {ak }∞ k=0 ⊂ (0, ∞). Let us describe our algorithm. Subgradient projection algorithm Initialization: select an arbitrary x0 ∈ U . Iterative Step: given a current iteration vector xt ∈ U calculate ξt ∈ ∂f (xt ) + BX (0, δf ) and the next iteration vector xt+1 ∈ U such that xt+1 − PC (xt − at ξt ) ≤ δC . Here δf is a computational error produced by our computer system when we calculate a subgradient of f and δC is a computational error produced by our computer system when we calculate a projection on the set C. In this chapter we prove the following result. Theorem 2.4 Let δf , δC ∈ (0, 1], {ak }∞ k=0 ⊂ (0, ∞) and let x∗ ∈ C

(2.4)

f (x∗ ) ≤ f (x) for all x ∈ C.

(2.5)

satisfy

∞ Assume that {xt }∞ t=0 ⊂ U , {ξt }t=0 ⊂ X,

x0  ≤ M0 + 1

(2.6)

ξt ∈ ∂f (xt ) + BX (0, δf )

(2.7)

xt+1 − PC (xt − at ξt ) ≤ δC .

(2.8)

and that for each integer t ≥ 0,

and

Then for each natural number T ,

28

2 Subgradient Projection Algorithm T 

at (f (xt ) − f (x∗ ))

t=0

≤ 2−1 x∗ − x0 2 + δC (T + 1)(4M0 + 1) + δf (2M0 + 1)

T 

at + 2−1 (L + 1)2

t=0

T 

at2 .

(2.9)

t=0

Moreover, for each natural number T , f ((

T 

at )−1

t=0

≤ 2−1 (

T 

at xt ) − f (x∗ ), min{f (xt ) : t = 0, . . . , T } − f (x∗ )

t=0 T 

at )−1 x∗ − x0 2 + (

t=0

T 

at )−1 δC (T + 1)(4M0 + 1)

t=0

+ δf (2M0 + 1) + 2−1 (

T  t=0

at )−1 (L + 1)2

T 

at2 .

(2.10)

t=0

Theorem 2.4 is proved in Sect. 2.4. It is a generalization of Theorem 2.4 of  proved in the case when δf = δC . We are interested in an optimal choice of at , t = 0, 1, . . . . Let T be a natural number and AT = Tt=0 at be given. By Theorem 2.4, in order to make the best choice of at , t = 0, . . . , T , we need to minimize the function −1 2 φ(a0 , . . . , aT ) : = 2−1 A−1 T x∗ − x0  + AT δC (T + 1)(4M0 + 1) 2 +δf (2M0 + 1) + 2−1 A−1 T (L + 1)

T 

at2

t=0

on the set {a = (a0 , . . . , aT ) ∈ R T +1 : ai ≥ 0, i = 0, . . . , T ,

T 

ai = AT }.

i=0

By Lemma 2.3, this function has a unique minimizer a ∗ = (a0∗ , . . . , aT∗ ) where ai∗ = (T + 1)−1 AT , i = 0, . . . , T . This is the best choice of at , t = 0, 1, . . . , T . Theorem 2.4 implies the following result.

2.2 A Convex Minimization Problem

29

Theorem 2.5 Let δf , δC ∈ (0, 1], a > 0 and let x∗ ∈ C satisfy f (x∗ ) ≤ f (x) for all x ∈ C. ∞ Assume that {xt }∞ t=0 ⊂ U , {ξt }t=0 ⊂ X,

x0  ≤ M0 + 1 and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ) and xt+1 − PC (xt − aξt ) ≤ δC . Then for each natural number T , f ((T + 1)−1

T 

xt ) − f (x∗ ), min{f (xt ) : t = 0, . . . , T } − f (x∗ )

t=0

≤ 2−1 (T + 1)−1 a −1 (2M0 + 1)2 + a −1 δC (4M0 + 1) +δf (2M0 + 1) + 2−1 (L + 1)2 a. Now we will find the best a > 0. Since T can be arbitrarily large, we need to find a minimizer of the function φ(a) := a −1 δC (4M0 + 1) + 2−1 (L + 1)2 a, a ∈ (0, ∞). Clearly, the minimizer a satisfies a −1 δC (4M0 + 1) = 2−1 (L + 1)2 a, a = (2δC (4M0 + 1))1/2 (L + 1)−1 and the minimal value of φ is (2δC (4M0 + 1))1/2 (L + 1). Theorem 2.5 implies the following result.

30

Theorem 2.6 Let δf , δC ∈ (0, 1], a = (2δC (4M0 + 1))1/2 (L + 1)−1 , x∗ ∈ C satisfy f (x∗ ) ≤ f (x) for all x ∈ C. ∞ Assume that {xt }∞ t=0 ⊂ U , {ξt }t=0 ⊂ X,

x0  ≤ M0 + 1 and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ) and xt+1 − PC (xt − aξt ) ≤ δC . Then for each natural number T , f ((T + 1)−1

T 

xt ) − f (x∗ ), min{f (xt ) : t = 0, . . . , T } − f (x∗ )

t=0

≤ 2−1 (T + 1)−1 (2M0 + 1)2 (L + 1)(2δC (4M0 + 1))−1/2 + δf (2M0 + 1) +2−1 (2δC (4M0 + 1))1/2 (L + 1) + δC (4M0 + 1)(L + 1)(2(4M0 + 1))−1/2 . 1/2

Now we can think about the best choice of T . It is clear that it should be at the same order as δC−1 . Putting T = δ −1 , we obtain that f ((T + 1)−1

T 

xt ) − f (x∗ ), min{f (xt ) : t = 0, . . . , T } − f (x∗ )

t=0

≤ 2−1 (2M0 + 1)2 (L + 1)(8M0 + 2)−1/2 δC + δf (2M0 + 1) 1/2

1/2

+(L + 1)δC (8M0 + 2)1/2 . Note that in the theorems above δf is the computational error, produced by our computer system when we calculate a subgradient of f and δC is the computational

2.3 The Main Lemma

31

error, produced by our computer system when we calculate a projection on C subgradient. In view of the inequality above, which has the right-hand side bounded by 1/2 c1 δC + δf (2M0 + 1) with a constant c1 > 0, we conclude that after T = δC−1  iterations we obtain a point ξ ∈ U such that BX (ξ, δC ) ∩ C = ∅ and 1/2

f (ξ ) ≤ f (x∗ ) + c1 δC + δf (2M0 + 1), where the constant c1 > 0 depends only on L and M0 . It is interesting that the number of iterations depends only on δC .

2.3 The Main Lemma Lemma 2.7 Let δf , δC ∈ (0, 1], a > 0 and let z ∈ C.

(2.11)

x ∈ U ∩ BX (0, M0 + 1),

(2.12)

ξ ∈ ∂f (x) + BX (0, δf )

(2.13)

u∈U

(2.14)

u − PC (x − aξ ) ≤ δC .

(2.15)

Assume that

and that

satisfies

Then a(f (x) − f (z)) ≤ 2−1 z − x2 − 2−1 z − u2 +δC (4M0 + 1) + aδf (2M0 + 1) + 2−1 a 2 (L + 1)2 .

32

Proof In view of (2.13), there exists l ∈ ∂f (x)

(2.16)

l − ξ  ≤ δf .

(2.17)

such that

By Lemmas 2.1 and 2.2 and (2.11), 0 ≤ z − PC (x − aξ ), PC (x − aξ ) − (x − aξ ) = z − PC (x − aξ ), PC (x − aξ ) − x +aξ, z − PC (x − aξ ) = 2−1 [z − x2 − z − PC (x − aξ )2 − x − PC (x − aξ )2 ] + aξ, z − x + aξ, x − PC (x − aξ ).

(2.18)

Clearly, |aξ, x − PC (x − aξ )| ≤ 2−1 (aξ 2 + x − PC (x − aξ )2 ).

(2.19)

It follows from (2.18) and (2.19) that 0 ≤ 2−1 [z − x2 − z − PC (x − aξ )2 − x − PC (x − aξ )2 ] +aξ, z − x + 2−1 a 2 ξ 2 + 2−1 x − PC (x − aξ )2 ≤ 2−1 z − x2 − 2−1 z − PC (x − aξ )2 + 2−1 a 2 ξ 2 + aξ, z − x.

(2.20)

Relations (2.1), (2.11) and (2.15) imply that |z − PC (x − aξ )2 − z − u2 | = |z − PC (x − aξ ) − z − u|(z − PC (x − aξ ) + z − u) ≤ u − PC (x − aξ )(4M0 + 1) ≤ (4M0 + 1)δC . By (2.1), (2.11), (2.12), and (2.17), aξ, z − x = al, z − x + a(ξ − l), z − x ≤ al, z − x + aξ − lz − x

(2.21)

2.4 Proof of Theorem 2.4

33

≤ al, z − x + aδf (2M0 + 1).

(2.22)

It follows from (2.3), (2.16), (2.17), (2.20), (2.21), and (2.22) that 0 ≤ 2−1 z − x2 − 2−1 z − PC (x − aξ )2 + 2−1 a 2 ξ 2 + aξ, z − x ≤ 2−1 z − x2 − 2−1 z − u2 + δC (4M0 + 1) + 2−1 a 2 (L + 1)2 + al, z − x + aδf (2M0 + 1).

(2.23)

By (2.16) and (2.23), a(f (z) − f (x)) ≥ al, z − x and a(f (x) − f (z)) ≤ al, x − z ≤ 2−1 z − x2 − 2−1 z − u2 + δC (4M0 + 1) + 2−1 a 2 (L + 1)2 +aδf (2M0 + 1). This completes the proof of Lemma 2.7.

 

2.4 Proof of Theorem 2.4 It is clear that xt  ≤ M0 + 1, t = 0, 1, . . . . Let t ≥ 0 be an integer. Applying Lemma 2.7 with z = x∗ , a = at , x = xt , ξ = ξt , u = xt+1 we obtain that at (f (xt ) − f (x∗ )) ≤ 2−1 x∗ − xt 2 − 2−1 x∗ − xt+1 2 + δC (4M0 + 1) + at δf (2M0 + 1) + 2−1 at2 (L + 1)2 . By (2.24), for each natural number T ,

(2.24)

34

2 Subgradient Projection Algorithm T 

at (f (xt ) − f (x∗ ))

t=0

T 

(2−1 x∗ − xt 2 − 2−1 x∗ − xt+1 2 )

t=0

+δC (4M0 + 1) + at δf (2M0 + 1) + 2−1 at2 (L + 1)2 ≤ 2−1 x∗ − x0 2 + δC (T + 1)(4M0 + 1) +δf (2M0 + 1)

T 

at + 2−1 (L + 1)2

t=0

T 

at2 .

t=0

Thus (2.9) is true. Evidently, (2.10) implies (2.9). Theorem 2.4 is proved.

 

2.5 Subgradient Algorithm on Unbounded Sets Let X be a Hilbert space with an inner product ·, · which induces a complete norm  · , D be a nonempty closed convex subset of X, V be an open convex subset of X such that D ⊂ V,

(2.25)

and f : V → R 1 be a convex function which is Lipschitz on all bounded subsets of V . Set Dmin = {x ∈ D : f (x) ≤ f (y) for all y ∈ D}.

(2.26)

Dmin = ∅.

(2.27)

We suppose that

In this chapter we will prove the following result. Theorem 2.8 Let δf , δC ∈ (0, 1], M > 0 satisfy

L > 0 satisfy

Dmin ∩ BX (0, M) = ∅,

(2.28)

M0 ≥ 4M + 4,

(2.29)

2.5 Subgradient Algorithm on Unbounded Sets

|f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, M0 + 2), 0 < τ0 ≤ τ1 ≤ (L + 1)−1 , 0 = 2τ0−1 δC (4M0 + 1) + 2δf (2M0 + 1) + 2τ1 (L + 1)2

35

(2.30) (2.31) (2.32)

and let n0 = τ0−1 (2M + 2)2 0−1 .

(2.33)

∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

{at }∞ t=0 ⊂ [τ0 , τ1 ],

(2.34)

x0  ≤ M

(2.35)

ξt ∈ ∂f (xt ) + BX (0, δf )

(2.36)

xt+1 − PD (xt − at ξt ) ≤ δC .

(2.37)

and that for each integer t ≥ 0,

and

Then there exists an integer q ∈ [1, n0 + 1] such that xi  ≤ 3M + 2, i = 0, . . . , q and f (xq ) ≤ f (x) + 0 for all x ∈ D. Theorem 2.8 is a generalization of Theorem 2.8 of  which was proved in the case when δC = δf . It is proved in Sect. 2.6. We are interested in the best choice of at , t = 0, 1, . . . . Assume for simplicity that τ1 = τ0 = τ . In order to meet our goal we need to minimize the function 2τ −1 δC (4M0 + 1) + 2(L + 1)2 τ, τ ∈ (0, ∞). This function has a minimizer τ = (δC (4M0 + 1))1/2 (L + 1)−1 , the minimal value of 0 is

36

2δf (2M0 + 1) + 4(δC (4M0 + 1))1/2 (L + 1) and n0 = Δ where −1/2

Δ = (2M + 2)2 (4M0 + 1)−1/2 (L + 1)δC

(2δf (2M0 + 1)

+4δC (4M0 + 1)1/2 (L + 1))−1 . 1/2

Note that in the theorem above δf , δC are the computational errors produced by our computer system. In view of the inequality above, in order to obtain a good approximate solution we need −1/2

c1 δC

max{δf , δC }−1  + 1 1/2

iterations, where c1 is a constant which depends only on M, M0 , L. As a result, we obtain a point ξ ∈ V such that BX (ξ, δC ) ∩ D = ∅ and 1/2

f (ξ ) ≤ inf{f (x) : x ∈ D} + 2δf (2M0 + 1) + (4M0 + 1)1/2 4(L + 1)δC . The next result, which is proved in Sect. 2.7, does not have a prototype in . Theorem 2.9 Let δf , δC ∈ (0, 1), M > 1, L > 0 and let {at }∞ t=0 ⊂ (0, ∞) be such that {x ∈ V : f (x) ≤ inf(f, D) + 3} ⊂ BX (0, M − 1),

(2.38)

|f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 8),

(2.39)

δf ≤ (6M + 5)−1 and δC (12M + 9) ≤ at ≤ (L + 1)−2 for all integers t ≥ 0.

(2.40)

∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M, BX (x0 , δC ) ∩ D = ∅ and that for each integer t ≥ 0,

(2.41)

2.5 Subgradient Algorithm on Unbounded Sets

37

ξt ∈ ∂f (xt ) + BX (0, δf )

(2.42)

xt+1 − PD (xt − at ξt ) ≤ δC .

(2.43)

and

Then xt  ≤ 3M for all integers t ≥ 0 and for each natural number T , T 

at (f (xt ) − inf(f, D))

t=0

≤ 2M 2 + δC (T + 1)(12M + 9) +2−1 (L + 1)2

T 

at2 + δf (6M + 5)

t=0

T 

at .

t=0

Moreover, for each natural number T , f ((

T 

at )−1

t=0

T 

at xt ) − inf(f, D),

t=0

min{f (xt ) : t = 0, . . . , T } − inf(f, D) ≤ 2M 2 (

T 

T  at )−1 + δC (T + 1)(12M + 9)( at )−1

t=0

+2−1 (L + 1)2

t=0 T  t=0

at2 (

T 

at )−1 + δf (6M + 5).

t=0

We are interested in an optimal choice of at , t = 0, 1, . . . . Let T be a natural number and AT = Tt=0 at be given. By Theorem 2.9, in order to make the best choice of at , t = 0, . . . , T , we need to minimize the function −1 2 φ(a0 , . . . , aT ) := 2−1 A−1 T M + AT δC (T + 1)(12M + 9)

38

2 +δf (6M + 5) + 2−1 A−1 T (L + 1)

T 

at2

t=0

on the set {a = (a0 , . . . , aT ) ∈ R T +1 : ai ≥ 0, i = 0, . . . , T ,

T 

ai = AT }.

i=0

Theorem 2.9 and Lemma 2.3 imply the following result. Theorem 2.10 Let δf , δC ∈ (0, 1), a > 0, M > 1 and L > 0 be such that {x ∈ V : f (x) ≤ inf(f, D) + 3} ⊂ BX (0, M − 1), |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 8), δf ≤ (6M + 5)−1 and δC (12M + 9) ≤ a ≤ (L + 1)−2 . ∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M, BX (x0 , δC ) ∩ D = ∅ and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ) and xt+1 − PD (xt − aξt ) ≤ δC . Then xt  ≤ 3M for all integers t ≥ 0 and for each natural number T , f ((T + 1)−1

T 

xt ) − inf(f, D),

t=0

min{f (xt ) : t = 0, . . . , T } − inf(f, D)

2.5 Subgradient Algorithm on Unbounded Sets

39

≤ 2M 2 (T + 1)−1 a −1 + δC (12M + 9)a −1 +2−1 (L + 1)2 a + δf (6M + 5). Now we will find the best a > 0. Since T can be arbitrarily large, we need to find a minimizer of the function φ(a) := a −1 δC (12M + 9) + 2−1 (L + 1)2 a, a ∈ (0, ∞). Clearly, the minimizer a satisfies a −1 δC (12M + 9) = 2−1 (L + 1)2 a, a = (2δC (12M + 9))1/2 (L + 1)−1 and the minimal value of φ is (2δC (12M + 9))1/2 (L + 1). Clearly, we need to check if the minimizer a defined above satisfies (2.40). It is not difficult to see that this takes place if δC ≤ 2−1 (12M + 9)−1 (L + 1)−2 . Theorem 2.10 implies the following result. Theorem 2.11 Let δf , δC ∈ (0, 1), M > 1 and L > 0 be such that {x ∈ V : f (x) ≤ inf(f, D) + 3} ⊂ BX (0, M − 1), |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 8), δC ≤ 2−1 (12M + 9)−1 (L + 1)−2 δf ≤ (6M + 5)−1 and a = (2δC (12M + 9))1/2 (L + 1)−1 . ∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M, BX (x0 , δC ) ∩ D = ∅

40

and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ) and xt+1 − PD (xt − at ξt ) ≤ δC . Then xt  ≤ 3M for all integers t ≥ 0 and for each natural number T , f ((T + 1)−1

T 

xt ) − inf(f, D),

t=0

min{f (xt ) : t = 0, . . . , T } − inf(f, D) ≤ 2M 2 (T + 1)−1 (L + 1)(2δC (12M + 9))−1/2 +(L + 1)(2δC (12M + 9))1/2 + δf (6M + 5). Now we can think about the best choice of T . It is clear that it should be at the same order as δC−1 . Putting T = δC−1 , we obtain that f ((T + 1)−1

T 

xt ) − inf(f ; D), min{f (xt ) : t = 0, . . . , T } − inf(f ; D)

t=0

≤ 2M 2 (L + 1)δC (24M + 18)−1/2 1/2

+δC (12M + 9)(L + 1)(24M + 18)−1/2 1/2

+2−1 (L + 1)δC (24M + 18)1/2 + δf (6M + 5). 1/2

In view of the inequality above, which has the right-hand side bounded by 1/2 c1 δC + δf (6M + 5) with a constant c1 > 0, we conclude that after T = δC−1  iterations we obtain a point ξ ∈ U such that BX (ξ, δC ) ∩ D = ∅

2.5 Subgradient Algorithm on Unbounded Sets

41

and 1/2

f (ξ ) ≤ inf(f, D) + c1 δC + δf (6M + 5), where the constant c1 > 0 depends only on L and M. The next result, which is proved in Sect. 2.8, does not have a prototype in . Theorem 2.12 Let δf , δC ∈ (0, 16−1 ), M > 4, L > 0, Dmin ∩ BX (0, M) = ∅, |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 3M + 6),

(2.44)

a = (L + 1)−1 δC

1/2

and −1/2 −1 δf } − 1.

T = 6−1 min{δC−1 , δC ∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M, BX (x0 , δC ) ∩ D = ∅

(2.45)

and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf )

(2.46)

xt+1 − PD (xt − at ξt ) ≤ δC .

(2.47)

and

Then T  (T + 1)−1 f (xt ) − inf(f, D), t=0

f(

T  (T + 1)−1 xt ) − inf(f, D), min{f (xt ) : t = 0, . . . , T } − inf(f, D) t=0 1/2

≤ 12(2M + 2)2 (L + 1) max{δC , δf }.

42

2.6 Proof of Theorem 2.8 By (2.28) there exists z ∈ Dmin ∩ BX (0, M).

(2.48)

Lemma 2.2, (2.18), (2.34), (2.35), and (2.37) imply that x1 − z ≤ x1 − PD (x0 − a0 ξ0 ) + PD (x0 − a0 ξ0 ) − z ≤ δC + x0 − z + a0 ξ0  ≤ 1 + 2M + τ1 ξ0 .

(2.49)

In view of (2.30), (2.35), and (2.36), ξ0 ∈ ∂f (x0 ) + BX (0, 1) ⊂ BX (0, L) + 1, ξ0  ≤ L + 1.

(2.50)

It follows from (2.31) and (2.48)–(2.50) that x1 − z ≤ 2M + 2, x1  ≤ 3M + 2.

(2.51)

Assume that T is a natural number and that f (xt ) − f (z) > 0 , t = 1, . . . , T .

(2.52)

U = V ∩ {v ∈ X : v < M0 + 2}

(2.53)

C = D ∩ BX (0, M0 ).

(2.54)

Set

and

By induction we show that for every integer t ∈ [1, T ], xt − z ≤ 2M + 2, f (xt ) − f (z)

(2.55)

2.6 Proof of Theorem 2.8

43

≤ (2τ0 )−1 (z − xt 2 − z − xt+1 2 ) + τ0−1 δC (4M0 + 1) + δf (2M0 + 1) + 2−1 τ1 (L + 1)2 .

(2.56)

In view of (2.51), (2.55) holds for t = 1. Assume that an integer t ∈ [1, T ] and that (2.55) holds. It follows from (2.29), (2.48), and (2.53)–(2.55) that z ∈ C ⊂ BX (0, M0 ),

(2.57)

xt ∈ U ∩ BX (0, M0 + 1).

(2.58)

Relation (2.37) implies that xt+1 ∈ V satisfies xt+1 − PD (xt − at ξt ) ≤ 1.

(2.59)

By (2.30), (2.36) and (2.58), ξt ∈ ∂f (xt ) + BX (0, 1) ⊂ BX (0, L + 1).

(2.60)

It follows from (2.31), (2.34), (2.48), (2.55), (2.60), and Lemma 2.2 that z − PD (xt − at ξt ) ≤ z − xt + at ξt  ≤ z − xt  + ξt at ≤ 2M + 3, PD (xt − at ξt ) ≤ 3M + 3.

(2.61)

In view of (2.29), (2.54), and (2.61), PD (xt − at ξt ) ∈ C,

(2.62)

PD (xt − at ξt ) = PC (xt − at ξt ).

(2.63)

and

Relations (2.29), (2.53), (2.59), and (2.61) imply that xt+1  ≤ 3M + 4, xt+1 ∈ U.

(2.64)

By (2.25), (2.30), (2.36), (2.37), (2.53), (2.54), (2.57), (2.58), (2.63), (2.64), and Lemma 2.7 which holds with

44

x = xt , a = at , ξ = ξt , u = xt+1 , we have at (f (xt ) − f (z)) ≤ 2−1 z − xt 2 − 2−1 z − xt+1 2 +δC (4M0 + 1) + at δf (2M0 + 1) + 2−1 at2 (L + 1)2 . It follows from the relation above, (2.31) and (2.34) that f (xt ) − f (z) ≤ (2τ0 )−1 z − xt 2 − (2τ0 )−1 z − xt+1 2 + τ0−1 δC (4M0 + 1) + (2M0 + 1)δf + 2−1 τ1 (L + 1)2 .

(2.65)

By (2.30), (2.37), (2.48), (2.62), and (2.63), f (xt ) ≥ f (PD (xt − at ξt )) − δC L ≥ f (z) − δC L. In view of the relation above, (2.32), (2.52), (2.55), (2.65), and the inclusion t ∈ [1, T ], z − xt 2 − z − xt+1 2 ≥ 0, z − xt+1  ≤ z − xt  ≤ 2M + 2.

(2.66)

Therefore we assumed that (2.55) is true and showed that (2.65) and (2.66) hold. Hence by induction we showed that (2.65) holds for all t = 1, . . . , T and (2.55) holds for all t = 1, . . . , T + 1. It follows from (2.65) which holds for all t = 1, . . . , T , (2.51) and (2.52) that T 0 < T (min{f (xt ) : t = 1, . . . , T } − f (z)) ≤

T 

(f (xt ) − f (z))

t=1

≤ (2τ0 )−1

T  (z − xt 2 − z − xt+1 2 ) t=1

+T τ0−1 δC (4M0 + 1) + T (2M0 + 1)δf + 2−1 T τ1 (L + 1)2

2.7 Proof of Theorem 2.9

45

≤ (2τ0 )−1 (2M + 2)2 + T τ0−1 δC (4M0 + 1) +T (2M0 + 1)δf + 2−1 T τ1 (L + 1)2 . Together with (2.32) and (2.33) this implies that 0 < (2τ0 T )−1 (2M + 2)2 + τ0−1 δC (4M0 + 1) +(2M0 + 1)δf + 2−1 τ1 (L + 1)2 , 2−1 0 < (2τ0 T )−1 (2M + 2)2 , T < τ0−1 (2M + 2)2 0−1 ≤ n0 + 1. Thus we have shown that if an integer T ≥ 1 satisfies (2.52), then T ≤ n0 and z − xt  ≤ 2M + 2, t = 1, . . . , T + 1, xt  ≤ 3M + 2, t = 0, . . . , T + 1. This implies that there exists an integer q ∈ [1, n0 + 1] such that xt  ≤ 3M + 2, t = 0, . . . , q and f (xq ) − f (z) ≤ 0 .  

Theorem 2.8 is proved.

2.7 Proof of Theorem 2.9 Fix z ∈ Dmin .

(2.67)

U = V ∩ {x ∈ X : x < 3M + 2}

(2.68)

Set

and

46

C = D ∩ BX (0, 3M + 1).

(2.69)

In view of (2.38) and (2.67), z ≤ M − 1.

(2.70)

x0 − z ≤ 2M.

(2.71)

By (2.41) and (2.70),

Assume that t ≥ 0 is an integer and that xt − z ≤ 2M.

(2.72)

It follows from (2.38), (2.67)–(2.69), and (2.72) that z ∈ C ⊂ BX (0, 3M + 1),

(2.73)

xt ∈ U ∩ BX (0, 3M).

(2.74)

Relation (2.43) implies that xt+1 ∈ V satisfies xt+1 − PD (xt − at ξt ) ≤ 1.

(2.75)

By (2.39), (2.42), and (2.74), ξt ∈ ∂f (xt ) + BX (0, 1) ⊂ BX (0, L + 1).

(2.76)

It follows from (2.38), (2.40), (2.67), (2.70), (2.72), (2.76), and Lemma 2.2 that z − PD (xt − at ξt ) ≤ z − xt + at ξt  ≤ z − xt  + ξt at ≤ 2M + (L + 1)at ≤ 2M + 1, PD (xt − at ξt ) ≤ 3M + 1.

(2.77)

In view of (2.69) and (2.77), PD (xt − at ξt ) ∈ C, and

(2.78)

2.7 Proof of Theorem 2.9

47

PD (xt − at ξt ) = PC (xt − at ξt ).

(2.79)

Relations (2.43), (2.68), and (2.77) imply that xt+1  ≤ PD (xt − at ξt ) + δC < 3M + 2,

(2.80)

xt+1 ∈ U.

(2.81)

By (2.43), (2.73), (2.74), (2.79), (2.81), and Lemma 2.7 which holds with x = xt , a = at , ξ = ξt , u = xt+1 , M0 = 3M + 2, we have at (f (xt ) − f (z)) ≤ 2−1 z − xt 2 − 2−1 z − xt+1 2 + δC (12M + 9) + at δf (6M + 5) + 2−1 at2 (L + 1)2 .

(2.82)

There are two cases: z − xt+1  ≤ z − xt ;

(2.83)

z − xt+1  > z − xt .

(2.84)

If (2.83) holds, then in view of (2.72), z − xt+1  ≤ 2M. Assume that (2.84) holds. It follows from (2.40), (2.82), and (2.84) that f (xt ) − f (z) < at−1 δC (12M + 9) + δf (6M + 5) + 2−1 at (L + 1)2 ≤ 3. By the inequality above, (2.38) and (2.67), xt  ≤ M − 1.

(2.85)

Relations (2.70) and (2.85) imply that xt − z ≤ 2M − 2.

(2.86)

48

Lemma 2.2, (2.40), (2.43), (2.67), and (2.76) imply that xt+1 − z ≤ xt+1 − PD (xt − at ξt ) + PD (xt − at ξt ) − z ≤ δC + xt − at ξt − z ≤ δC + xt − z + at ξt  ≤ 1 + xt − z + (L + 1)at ≤ xt − z + 2 ≤ 2M. Thus in both cases xt+1 − z ≤ 2M and (2.82) holds. Therefore by induction we showed that for all integers t ≥ 0, xt  ≤ 3M and that (2.82) holds. Let T be a natural number. By (2.67), (2.72), and (2.82), T 

at (f (xt ) − inf(f, D))

t=0

T  (2−1 z − xt 2 − 2−1 z − xt+1 2 ) t=0

+δC (12M + 9)(T + 1) +δf (6M + 5)

T 

at + 2−1 (L + 1)2

t=0

T 

at2

t=0

≤ 2M 2 + δC (12M + 9)(T + 1) +δf (6M + 5)

T 

at + 2−1 (L + 1)2

t=0

This completes the proof of Theorem 2.9.

T 

at2 .

t=0

 

2.8 Proof of Theorem 2.12

49

2.8 Proof of Theorem 2.12 Clearly, there exists z ∈ Dmin ∩ BX (0, M).

(2.87)

In view of the theorem assumptions and (2.87), x0 − z ≤ 2M.

(2.88)

U = V ∩ {v ∈ X : v < 3M + 6}

(2.89)

C = D ∩ BX (0, 3M + 4).

(2.90)

Set

and

Assume that t ≥ 0 is an integer and that xt − z ≤ 2M + 2.

(2.91)

It follows from (2.85) and (2.91) that xt  ≤ 3M + 2.

(2.92)

xt ∈ U ∩ BX (0, 3M + 2).

(2.93)

In view of (2.89) and (2.92),

By (2.44), (2.46), and (2.92), ξt  ≤ L + 1. Lemma 2.2, (2.45), (2.47), (2.85), (2.89), (2.91), and (2.94) imply that xt+1 − z ≤ xt+1 − PD (xt − aξt ) + PD (xt − aξt ) − z ≤ δC + xt − aξt − z ≤ δC + xt − z + aξt 

(2.94)

50

≤ δC + 2M + 2 + a(L + 1) ≤ 2M + 4, xt+1  ≤ 3M + 4

(2.95)

xt+1 ∈ U.

(2.96)

and

It follows from (2.45), (2.85), (2.90), (2.91), (2.94), and Lemma 2.2 that PD (xt − aξt ) ≤ z + PD (xt − aξt ) − z ≤ M + z − xt  + aξt  ≤ M + 2M + 2 + 1, PD (xt − aξt ) ∈ C,

(2.97)

PD (xt − aξt ) = PC (xt − aξt ).

(2.98)

and

By (2.46), (2.47), (2.85), (2.89), (2.90), (2.93), (2.96), (2.98), and Lemma 2.7 which holds with M0 = 3M + 4, x = xt , ξ = ξt , u = xt+1 , we have a(f (xt ) − f (z)) ≤ 2−1 z − xt 2 − 2−1 z − xt+1 2 + δC (12M + 17) + aδf (6M + 9) + 2−1 a 2 (L + 1)2 . In view of (2.44), (2.47), (2.85), (2.92), and the relation BX (x0 , δC ) ∩ D = ∅, we have f (xt ) ≥ f (z) − δC L.

(2.99)

2.8 Proof of Theorem 2.12

51

It follows from (2.85), (2.99), and the inequality above that z − xt+1 2 − z − xt 2 ≤ 2δC (12M + 18) + a 2 (L + 1)2 + 2aδf (6M + 9).

(2.100)

Thus we have shown that if for an integer t ≥ 0 relation (2.91) holds, then (2.99) and (2.100) are true. Let us show that for all integers t = 0, . . . , T , z − xt 2 ≤ z − x0 2 +t[2δC (12M + 18) + a 2 (L + 1)2 + 2aδf (6M + 9)] ≤ z − x0 2 +T [2δC (12M + 18) + a 2 (L + 1)2 + 2aδf (6M + 9)] ≤ 4M 2 + 8M + 4.

(2.101)

First, note that by (2.45), T [2δC (12M + 18) + a 2 (L + 1)2 + 2aδf (6M + 9)] = 2T δC (12M + 18) + 2T δC + 2T aδf (6M + 9) = 2T δC (12M + 19) + 2T δf (6M + 9)δC (L + 1)−1 1/2

−1/2 −1 δf }(2δC (12M

≤ 6−1 min{δC−1 , δC

+ 19)

+2δf (6M + 9)δC (L + 1)−1 ) 1/2

≤ 6−1 (24M + 38 + 12M + 18) ≤ 8M + 4.

(2.102)

It follows from (2.86) and (2.102) that z − x0 2 + T [2δC (12M + 18) +a 2 (L + 1)2 + 2aδf (6M + 9)] ≤ 4M 2 + 8M + 4 = (2M + 2)2 . Assume that t ∈ {0, . . . , T } \ {T } and that (2.101) holds. Then

(2.103)

52

xt − z ≤ 2M + 2 and (2.99) and (2.100) hold. By (2.100), (2.101), and (2.103), z − xt+1 2 ≤ z − xt 2 + 2δC (12M + 18) + a 2 (L + 1)2 + 2aδf (6M + 9) ≤ z − x0 2 +(t + 1)[2δC (12M + 18) + a 2 (L + 1)2 + 2aδf (6M + 9)] ≤ z − x0 2 +T [2δC (12M + 18) + a 2 (L + 1)2 + 2aδf (6M + 9)] ≤ 4M 2 + 8M + 4 and z − xt+1  ≤ 2M + 2. Thus we have shown by induction that (2.99) and (2.101) hold for all t = 0, . . . , T . It follows from (2.45), (2.85), (2.86), and (2.99) that T 

a(f (xt ) − inf(f, D))

t=0

≤ 2−1 z − x0 2 + (T + 1)(δC (12M + 17) +aδf (6M + 9) + 2−1 a 2 (L + 1)2 ) ≤ 2M 2 + δC (12M + 18)(6δC )−1 +δC δf (6M + 9)(6δf δC )−1 1/2

1/2

≤ 2M 2 + 2M + 3 + M + 2 = 2M 2 + 3M + 5. Together with (2.45) this implies that

2.9 Zero-Sum Games with Two Players

53

T  (T + 1)−1 f (xt ) − inf(f, D), t=0

f(

T  (T + 1)−1 xt ) − inf(f, D), t=0

min{f (xt ) : t = 0, . . . , T } − inf(f, D) ≤ (2M + 2)2 a −1 (T + 1)−1 −1/2

≤ 12(2M + 2)2 (L + 1)δC

1/2

max{δC , δC δf } 1/2

≤ 12(2M + 2)2 (L + 1) max{δC , δf }.  

Theorem 2.12 is proved.

2.9 Zero-Sum Games with Two Players Let (X, ·, ·), (Y, ·, ·) be Hilbert spaces equipped with the complete norms  ·  which are induced by their inner products. Let C be a nonempty closed convex subset of X, D be a nonempty closed convex subset of Y , U be an open convex subset of X, and V be an open convex subset of Y such that C ⊂ U, D ⊂ V

(2.104)

and let a function f : U × V → R 1 possess the following properties: (i) for each v ∈ V , the function f (·, v) : U → R 1 is convex; (ii) for each u ∈ U , the function f (u, ·) : V → R 1 is concave. Assume that a function φ : R 1 → [0, ∞) is bounded on all bounded sets and that positive numbers M1 , M2 , L1 , L2 satisfy C ⊂ BX (0, M1 ), D ⊂ BY (0, M2 ),

(2.105)

|f (u1 , v) − f (u2 , v)| ≤ L1 u1 − u2  for all v ∈ V and all u1 , u2 ∈ U, |f (u, v1 ) − f (u, v2 )| ≤ L2 v1 − v2 

(2.106)

54

for all u ∈ U and all v1 , v2 ∈ V .

(2.107)

x∗ ∈ C and y∗ ∈ D

(2.108)

f (x∗ , y) ≤ f (x∗ , y∗ ) ≤ f (x, y∗ )

(2.109)

Let

satisfy

for each x ∈ C and each y ∈ D. The following result is proved in Sect. 2.10. Proposition 2.13 Let T be a natural number, δC , δD ∈ (0, 1], {at }Tt=0 ⊂ (0, ∞) T +1 T +1 ⊂ U , {yt }t=0 ⊂ V , for and let {bt,1 }Tt=0 , {bt,2 }Tt=0 ⊂ (0, ∞). Assume that {xt }t=0 each t ∈ {0, . . . , T + 1}, BX (xt , δC ) ∩ C = ∅, BY (yt , δD ) ∩ D = ∅,

(2.110)

for each z ∈ C and each t ∈ {0, . . . , T }, at (f (xt , yt ) − f (z, yt )) ≤ φ(z − xt ) − φ(z − xt+1 ) + bt,1

(2.111)

and that for each v ∈ D and each t ∈ {0, . . . , T }, at (f (xt , v) − f (xt , yt )) ≤ φ(v − yt ) − φ(v − yt+1 ) + bt,2 .

(2.112)

Let  xT = (

T 

ai )

−1

i=0

 yT = (

T  i=0

T 

at xt ,

t=0

ai )−1

T 

at yt .

(2.113)

t=0

Then BX ( xT , δC ) ∩ C = ∅, BY ( yT , δD ) ∩ D = ∅,

(2.114)

2.9 Zero-Sum Games with Two Players

|(

T 

at )−1

t=0

55

T 

at f (xt , yt ) − f (x∗ , y∗ )|

t=0

≤(

T 

at )−1 max{

t=0

T 

bt,1 ,

t=0

T 

bt,2 }

t=0

+ max{L1 δC , L2 δD } T  +( at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]},

(2.115)

t=0

yT ) − ( |f ( xT , 

T 

at )−1

t=0

T 

at f (xt , yt )|

t=0

T  at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]} ≤( t=0 T T T    at )−1 max{ bt,1 , bt,2 } +( t=0

t=0

t=0

+ max{L1 δC , L2 δD }

(2.116)

and for each z ∈ C and each v ∈ D, xT ,  yT ) f (z,  yT ) ≥ f ( −2(

T 

at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]}

t=0 T T T    at )−1 max{ bt,1 , bt,2 } −2( t=0

t=0

t=0

− max{L1 δC , L2 δD }, xT ,  yT ) f ( xT , v) ≤ f ( T  +2( at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]} t=0

(2.117)

56

+2(

T 

at )−1 max{

t=0

T 

bt,1 ,

t=0

T 

bt,2 }

t=0

+ max{L1 δC , L2 δD }.

(2.118)

Corollary 2.14 Suppose that all the assumptions of Proposition 2.13 hold and that x˜ ∈ C, y˜ ∈ D satisfy ˜ ≤ δC ,  yT − y ˜ ≤ δD .  xT − x

(2.119)

yT )| ≤ L1 δC + L2 δD |f (x, ˜ y) ˜ − f ( xT , 

(2.120)

Then

and for each z ∈ C and each v ∈ D, f (z, y) ˜ ≥ f (x, ˜ y) ˜ −2(

T 

at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]}

t=0 T T T    −2( at )−1 max{ bt,1 , bt,2 } − 4 max{L1 δC , L2 δD } t=0

t=0

t=0

and f (x, ˜ v) ≤ f (x, ˜ y) ˜ T  +2( at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]} t=0 T T T    +2( at )−1 max{ bt,1 , bt,2 } + 4 max{L1 δC , L2 δD }. t=0

t=0

t=0

Proof In view of (2.106), (2.107), and (2.119), |f (x, ˜ y) ˜ − f ( xT ,  yT )|

2.9 Zero-Sum Games with Two Players

57

≤ |f (x, ˜ y) ˜ − f (x, ˜  yT )| + |f (x, ˜  yT ) − f ( xT ,  yT )| yT  + L1 x˜ −  xT  ≤ L1 δC + L2 δD ≤ L2 y˜ −  and (2.120) holds. Let z ∈ C and v ∈ D. Relations (2.106), (2.107), and (2.119) imply that |f (z, y) ˜ − f (z,  yT )| ≤ L2 δD , |f (x, ˜ v) − f ( xT , v)| ≤ L1 δC . By the relation above and (2.117)–(2.120), f (z, y) ˜ ≥ f (z,  yT ) − L2 δD ≥ f ( xT ,  yT ) − 2(

T 

at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]}

t=0 T T T    −1 at ) max{ bt,1 , bt,2 } −2( t=0

t=0

t=0

− max{L1 δC , L2 δD } − L2 δD T  ≥ f (x, ˜ y) ˜ − 2( at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]} t=0 T T T    −1 at ) max{ bt,1 , bt,2 } −2( t=0

t=0

t=0

−4 max{L1 δC , L2 δD } and f (x, ˜ v) ≤ f ( xT , v) + L1 δC ≤ f ( xT ,  yT ) + 2(

T 

at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]}

t=0 T T T    −1 at ) max{ bt,1 , bt,2 } +2( t=0

t=0

t=0

58

+ max{L1 δC , L2 δD } + L1 δC T  ≤ f (x, ˜ y) ˜ + 2( at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]} t=0 T T T    −1 +2( at ) max{ bt,1 , bt,2 } t=0

t=0

t=0

+4 max{L1 δC , L2 δD }. This completes the proof of Corollary 2.14.

 

2.10 Proof of Proposition 2.13 It is clear that (2.114) is true. In view of (2.110), for each t ∈ {0, . . . , T + 1}, there exist x˜t ∈ C, y˜t ∈ D

(2.121)

xt − x˜t  ≤ δC , yt − y˜t  ≤ δD .

(2.122)

such that

Let t ∈ {0, . . . , T }. By (2.106)–(2.109), (2.111), (2.112), and (2.122), at (f (xt , yt ) − f (x∗ , y∗ )) ≤ at (f (xt , yt ) − f (x∗ , y˜t )) ≤ at (f (xt , yt ) − f (x∗ , yt )) + at (f (x∗ , yt ) − f (x∗ , y˜t )) ≤ φ(x∗ − xt ) − φ(x∗ − xt+1 ) +bt,1 + at L2 yt − y˜t  ≤ φ(x∗ − xt ) − φ(x∗ − xt+1 ) + bt,1 + at L2 δD and

(2.123)

2.10 Proof of Proposition 2.13

59

at (f (x∗ , y∗ ) − f (xt , yt )) ≤ at (f (x˜t , y∗ ) − f (xt , yt )) = at (f (x˜t , y∗ ) − f (xt , y∗ )) + at (f (xt , y∗ ) − f (xt , yt )) ≤ at L1 x˜t − xt  +φ(y∗ − yt ) − φ(y∗ − yt+1 ) + bt,2 ≤ at L1 δC + φ(y∗ − yt ) − φ(y∗ − yt+1 ) + bt,2 .

(2.124)

In view of (2.123) and (2.124), T 

at f (xt , yt ) −

T 

t=0

at f (x∗ , y∗ )

t=0

T  (φ(x∗ − xt ) − φ(x∗ − xt+1 )) t=0

+

T 

bt,1 + δD L2

t=0

T 

at

t=0

≤ φ(x∗ − x0 ) +

T 

bt,1 + δD L2

t=0

T 

at

(2.125)

t=0

and T 

at f (x∗ , y∗ ) −

T 

t=0

at f (xt , yt )

t=0

T  (φ(y∗ − yt ) − φ(y∗ − yt+1 )) t=0

+

T 

bt,2 + L1 δC

t=0

≤ φ(y∗ − y0 ) +

T 

at

t=0 T  t=0

bt,2 + L1 δC

T  t=0

at .

(2.126)

60

Relations (2.105), (2.108), (2.110), (2.125), and (2.126) imply that T T   |( at )−1 at f (xt , yt ) − f (x∗ , y∗ )| t=0

t=0

T  at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]} ≤( t=0 T T T    +( at )−1 max{ bt,1 , bt,2 } t=0

t=0

t=0

+ max{L1 δC , L2 δD }.

(2.127)

zT ∈ C

(2.128)

xT  ≤ δC . zT − 

(2.129)

By (2.214), there exists

such that

In view of (2.128), we apply (2.111) with z = zT and obtain that for all t = 0, . . . , T , at (f (xt , yt ) − f (zT , yt )) ≤ φ(zT − xt ) − φ(zT − xt+1 ) + bt,1 .

(2.130)

It follows from (2.106) and (2.129) that for all t = 0, . . . , T , xT , yt )| ≤ L1 zT −  xT  ≤ L1 δC . |f (zT , yt ) − f (

(2.131)

By (2.130) and (2.131), for all t = 0, . . . , T , xT , yt )) at (f (xt , yt ) − f ( ≤ at (f (xt , yt ) − f (zT , yt )) + at L1 δC ≤ φ(zT − xt ) − φ(zT − xt+1 ) + bt,1 + at L1 δC . Combined with (2.105), (2.110), and (2.128) this implies that

(2.132)

2.10 Proof of Proposition 2.13

61

T 

T 

at f (xt , yt ) −

t=0

at f ( xT , yt )

t=0

T  (φ(zT − xt ) − φ(zT − xt+1 ) t=0

+

T 

bt,1 +

T 

t=0

at L1 δC

t=0

≤ φ(zT − x0 ) +

T 

bt,1 +

t=0

≤ sup{φ(s) : s ∈ [0, 2M1 + 1]} +

T 

at L1 δC

t=0

T 

bt,1 +

t=0

T 

at L1 δC .

(2.133)

t=0

Property (ii) and (2.113) imply that T 

at f ( xT , yt ) = (

t=0

T  i=0

ai )

T T   (at ( ai )−1 f ( xT , yt )) t=0

i=0

T  at )f ( xT ,  yT ). ≤(

(2.134)

t=0

By (2.133) and (2.134), T 

at f (xt , yt ) − (

t=0

T 

at )f ( xT ,  yT )

t=0

T 

at f (xt , yt ) −

t=0

T 

at f ( xT , yt )

t=0

≤ sup{φ(s) : s ∈ [0, 2M1 + 1]} +

T  t=0

By (2.114), there exists

bt,1 +

T  t=0

at L1 δC .

(2.135)

62

hT ∈ D

(2.136)

yT  ≤ δD . hT − 

(2.137)

such that

In view of (2.136), we apply (2.112) with v = hT and obtain that for all t = 0, . . . , T , at (f (xt , hT ) − f (xt , yt )) ≤ φ(hT − yt ) − φ(hT − yt+1 ) + bt,2 .

(2.138)

It follows from (2.107) and (2.137) that for all t = 0, . . . , T , yT )| ≤ L2 hT −  yT  ≤ L2 δD . |f (xt , hT ) − f (xt , 

(2.139)

By (2.138) and (2.139), for all t = 0, . . . , T , yT ) − f (xt , yt )) at (f (xt ,  yT ) − f (xt , hT )) + at (f (xt , hT ) − f (xt , yt )) ≤ at (f (xt ,  ≤ φ(hT − yt ) − φ(hT − yt+1 ) + bt,2 + at L2 δD .

(2.140)

In view of (2.140), T 

at f (xt ,  yT ) −

t=0

T 

at f (xt , yt )

t=0

T T T    (φ(hT − yt ) − φ(hT − yt+1 )) + bt,2 + at L2 δD . t=0

t=0

(2.141)

t=0

Property (i) and (2.113) imply that T 

at f (xt ,  yT ) = (

t=0

T 

ai )

i=0

T 

T T   (at ( ai )−1 f (xt ,  yT )) t=0

i=0

at f ( xT ,  yT ).

t=0

By (2.90), (2.110), (2.136), (2.141), and (2.142),

(2.142)

2.10 Proof of Proposition 2.13

63

T 

T 

at f ( xT ,  yT ) −

t=0

T 

at f (xt ,  yT ) −

T 

t=0

at f (xt , yt )

t=0

at f (xt , yt )

t=0

T T T    (φ(hT − yt ) − φ(hT − yt+1 )) + bt,2 + at L2 δD t=0

t=0

≤ sup{φ(s) : s ∈ [0, 2M2 + 1]} +

T 

bt,2 +

t=0

T 

t=0

at L2 δD .

(2.143)

t=0

It follows from (2.135) and (2.143) that |

T 

at f ( xT ,  yT ) −

t=0

T 

at f (xt , yt )|

t=0

≤ sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]} + max{

T 

bt,1 ,

t=0

T 

bt,2 } +

t=0

T 

at max{L1 δC , L2 δD }.

t=0

This relation implies (2.116). Let z ∈ C. By (2.111), T 

at (f (xt , yt ) − f (z, yt ))

t=0

T T   [φ(z − xt ) − φ(z − xt+1 )] + bt,1 . t=0

t=0

By property (ii) and (2.113), T  t=0

at f (z, yt ) = (

T  i=0

ai )

T T   (at ( ai )−1 f (z, yt )) t=0

i=0

(2.144)

64

≤(

T 

at )f (z,  yT ).

(2.145)

t=0

In view of (2.144) and (2.145), T 

at f (xt , yt ) −

t=0

T 

at f (z,  yT )

t=0 T 

at (f (xt , yt ) − f (z, yt ))

t=0

T T   [φ(z − xt ) − φ(z − xt+1 )] + bt,1 t=0

t=0

≤ φ(z − x0 ) +

T 

(2.146)

bt,1 .

t=0

It follows from (2.105) and (2.110) that f (z,  yT ) ≥ (

T 

ai )−1

i=0

−(

T 

T 

at f (xt , yt )

t=0

ai )−1 sup{φ(s) : s ∈ [0, 2M1 + 1]} − (

i=0

T  i=0

≥ f ( xT ,  yT ) − 2(

T 

ai )−1

T 

bt,1

t=0

at )−1 sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]}

t=0 T T T    at )−1 max{ bt,1 , bt,2 } − max{L1 δC , L2 δD } −2( t=0

t=0

t=0

and (2.117) holds. Let v ∈ D. By (2.112), T  t=0

at (f (xt , v) − f (xt , yt ))

2.10 Proof of Proposition 2.13

T 

65

[φ(v − yt ) − φ(v − yt+1 )] +

t=0

T 

(2.147)

bt,2 .

t=0

By property (i) and (2.113), T 

at f (xt , v) = (

T 

t=0

T T   ai ) (at ( ai )−1 f (xt , v))

i=0

≥(

T 

t=0

i=0

(2.148)

at )f ( xT , v).

t=0

In view of (2.147) and (2.148), T 

at f ( xT , v) −

t=0

T 

at f (xt , yt )

t=0

T 

at f (xt , v) −

t=0

T 

at f (xt , yt )

t=0

≤ φ(v − y0 ) +

T 

bt,2 .

t=0

Together with (2.105), (2.110), and (2.116) this implies that f ( xT , v) ≤ (

T 

at )−1

t=0

+(

T 

T 

at f (xt , yt )

t=0

at )−1 sup{φ(s) : s ∈ [0, 2M2 + 1]} + (

t=0

T  t=0

≤ f ( xT ,  yT ) + 2(

T 

at )−1

T 

bt,2

t=0

at )−1 sup{φ(s) : s ∈ [0, 2 max{M1 , M2 } + 1]}

t=0 T T T    at )−1 max{ bt,1 , bt,2 } + max{L1 δC , L2 δD }. +2( t=0

t=0

t=0

Therefore (2.118) holds. This completes the proof of Proposition 2.13.

 

66

2.11 Zero-Sum Games on Bounded Sets Let (X, ·, ·), (Y, ·, ·) be Hilbert spaces equipped with the complete norms  ·  which are induced by their inner products. Let C be a nonempty closed convex subset of X, D be a nonempty closed convex subset of Y , U be an open convex subset of X, and V be an open convex subset of Y such that C ⊂ U, D ⊂ V .

(2.149)

For each concave function g : V → R 1 and each x ∈ V set ∂g(x) = {l ∈ Y : l, y − x ≥ g(y) − g(x) for all y ∈ V }.

(2.150)

Clearly, for each x ∈ V , ∂g(x) = −(∂(−g)(x)).

(2.151)

Suppose that there exist L1 , L2 , M1 , M2 > 0 such that C ⊂ BX (0, M1 ), D ⊂ BY (0, M2 ),

(2.152)

a function f : U × V → R 1 possesses the following properties: (i) for each v ∈ V , the function f (·, v) : U → R 1 is convex; (ii) for each u ∈ U , the function f (u, ·) : V → R 1 is concave, for each v ∈ V , |f (u1 , v) − f (u2 , v)| ≤ L1 u1 − u2  for all u1 , u2 ∈ U

(2.153)

and that for each u ∈ U , |f (u, v1 ) − f (u, v2 )| ≤ L2 v1 − v2  for all v1 , v2 ∈ V .

(2.154)

For each (ξ, η) ∈ U × V , set ∂x f (ξ, η) = {l ∈ X : f (y, η) − f (ξ, η) ≥ l, y − ξ  for all y ∈ U }, ∂y f (ξ, η) = {l ∈ Y :

(2.155)

2.11 Zero-Sum Games on Bounded Sets

l, y − η ≥ f (ξ, y) − f (ξ, η) for all y ∈ V }.

67

(2.156)

In view of properties (i) and (ii), (2.153), and (2.154), for each ξ ∈ U and each η ∈ V, ∅ = ∂x f (ξ, η) ⊂ BX (0, L1 ),

(2.157)

∅ = ∂y f (ξ, η) ⊂ BY (0, L2 ).

(2.158)

x∗ ∈ C and y∗ ∈ D

(2.159)

f (x∗ , y) ≤ f (x∗ , y∗ ) ≤ f (x, y∗ )

(2.160)

Let

satisfy

for each x ∈ C and each y ∈ D. Let δf,1 , δf,2 , δC , δD ∈ (0, 1] and {ak }∞ k=0 ⊂ (0, ∞). Let us describe our algorithm. Subgradient projection algorithm for zero-sum games Initialization: select arbitrary x0 ∈ U and y0 ∈ V . Iterative Step: given current iteration vectors xt ∈ U and yt ∈ V calculate ξt ∈ ∂x f (xt , yt ) + BX (0, δf,1 ), ηt ∈ ∂y f (xt , yt ) + BY (0, δf,2 ) and the next pair of iteration vectors xt+1 ∈ U , yt+1 ∈ V such that xt+1 − PC (xt − at ξt ) ≤ δC , yt+1 − PD (yt + at ηt ) ≤ δD . In this chapter we prove the following result. Theorem 2.15 Let δf,1 , δf,2 , δC , δD ∈ (0, 1] and {ak }∞ k=0 ⊂ (0, ∞). Assume that ∞ ∞ ∞ {xt }∞ t=0 ⊂ U , {yt }t=0 ⊂ V , {ξt }t=0 ⊂ X, {ηt }t=0 ⊂ Y , BX (x0 , δC ) ∩ C = ∅, BY (y0 , δD ) ∩ D = ∅

(2.161)

and that for each integer t ≥ 0, ξt ∈ ∂x f (xt , yt ) + BX (0, δf,1 ),

(2.162)

68

ηt ∈ ∂y f (xt , yt ) + BY (0, δf,2 ),

(2.163)

xt+1 − PC (xt − at ξt ) ≤ δC

(2.164)

yt+1 − PD (yt + at ηt ) ≤ δD .

(2.165)

and

Let for each natural number T ,  xT = (

T 

at )−1

i=0

T 

at xt ,  yT = (

t=0

T 

at )−1

i=0

T 

at yt .

(2.166)

t=0

Then for each natural number T , xT , δC ) ∩ C = ∅, BY ( yT , δD ) ∩ D = ∅, BX ( |(

T  t=0

≤ (2

T 

at )−1

T 

at f (xt , yt ) − f (x∗ , y∗ )|

t=0

at )−1 (max{2M1 , 2M2 } + 1)2 + max{L1 δC , L2 δD }

t=0

+(

T 

at )−1 (T + 1) max{(4M1 + 1)δC , (4M2 + 1)δD }

t=0

+2−1 (max{L1 , L2 } + 1)2 (

T  t=0

at )−1

T 

at2

t=0

+ max{δf,1 , δf,2 }(2 max{M1 , M2 } + 1), |f ( xT ,  yT ) − (

T  t=0

≤ (2

T 

at )−1

T 

at f (xt , yt )|

t=0

at )−1 (max{2M1 , 2M2 } + 1)2 + max{L1 δC , L2 δD }

t=0

+(

T  t=0

at )−1 (T + 1) max{(4M1 + 1)δC , (4M2 + 1)δD }

(2.167)

2.11 Zero-Sum Games on Bounded Sets

69

+2−1 (max{L1 , L2 } + 1)2 (

T 

at )−1

t=0

T 

at2

t=0

+ max{δf,1 , δf,2 }(2 max{M1 , M2 } + 1)

(2.168)

and for each natural number T , each z ∈ C, and each u ∈ D, xT ,  yT ) f (z,  yT ) ≥ f ( −(

T 

at )−1 (max{2M1 , 2M2 } + 1)2 − max{L1 δC , L2 δD }

t=0 T  at )−1 (T + 1) max{(4M1 + 1)δC , (4M2 + 1)δD } −2( t=0

−(max{L1 , L2 } + 1)2 (

T 

at )−1

t=0

T 

at2

t=0

−2 max{δf,1 , δf,2 }(2 max{M1 , M2 } + 1), xT ,  yT ) f ( xT , u) ≤ f ( +(

T 

at )−1 (max{2M1 , 2M2 } + 1)2 + max{L1 δC , L2 δD }

t=0 T  at )−1 (T + 1) max{(4M1 + 1)δC , (4M2 + 1)δD } +2( t=0

+(max{L1 , L2 } + 1) ( 2

T 

at )

t=0

−1

T 

at2

t=0

+2 max{δf,1 , δf,2 }(2 max{M1 , M2 } + 1). Proof By (2.152), (2.164), and (2.165), for all integers t ≥ 0, xt  ≤ M1 + 1, yt  ≤ M2 + 1. Let t ≥ 0 be an integer. Applying Lemma 2.7 with a = at , x = xt , f = f (·, yt ), ξ = ξt , u = xt+1

(2.169)

70

we obtain that for each z ∈ C, at (f (xt , yt ) − f (z, yt )) ≤ 2−1 z − xt 2 − 2−1 z − xt+1 2 + δC (4M1 + 1) + at δf,1 (2M1 + 1) + 2−1 at2 (L1 + 1)2 .

(2.170)

Applying Lemma 2.7 with a = at , x = yt , f = −f (xt , ·), ξ = −ηt , u = yt+1 we obtain that for each v ∈ D, at (f (xt , v) − f (xt , yt )) ≤ 2−1 v − yt 2 − 2−1 v − yt+1 2 + δD (4M2 + 1) + at δf,2 (2M2 + 1) + 2−1 at2 (L2 + 1)2 .

(2.171)

For all integers t ≥ 0 set bt,1 = δC (4M1 + 1) + at δf,1 (2M1 + 1) + 2−1 at2 (L1 + 1)2 , bt,2 = δD (4M2 + 1) + at δf,2 (2M2 + 1) + 2−1 at2 (L2 + 1)2 , and define φ(s) = 2−1 s 2 , s ∈ R 1 . It is easy to see that all the assumptions of Proposition 2.13 hold and it implies Theorem 2.15.   Theorem 2.15 is a generalization of Theorem 2.11 of  obtained in the case when δf,1 = δf,2 = δC = δD . . . , T . For simplicity We are interested in the optimal choice of at , t = 0, 1, . assume that L1 = L2 . Let T be a natural number and AT = Tt=0 at be given. By Theorem 2.15, in orderto make the best choice of at , t = 0, . . . , T , we need to minimize the function Tt=0 at2 on the set {a = (a0 , . . . , aT ) ∈ R T +1 : ai ≥ 0, i = 0, . . . , T ,

T 

ai = AT }.

i=0

By Lemma 2.3, this function has a unique minimizer a ∗ = (a0∗ , . . . , aT∗ ) where ai∗ = (T + 1)−1 AT , i = 0, . . . , T which is the best choice of at , t = 0, 1, . . . , T .

2.12 Zero-Sum Games on Unbounded Sets

71

Let T be a natural number and at = a for all t = 0, . . . , T . Now we will find the best a > 0. In order to meet this goal we need to choose a which is a minimizer of the function ΨT (a) = ((T + 1)a)−1 (2(max{2M1 , 2M2 } + 1)2 ) +2 max{(4M1 + 1)δC , (4M2 + 1)δD } + (max{L1 , L2 } + 1)2 a. Since T can be arbitrarily large, we need to find a minimizer of the function φ(a) := 2a −1 max{(4M1 +1)δC , (4M2 +1)δD }+(max{L1 , L2 }+1)2 a, a ∈ (0, ∞). It is not difficult to see that the minimizer is a = (2 max{(4M1 + 1)δC , (4M2 + 1)δD })1/2 (max{L1 , L2 } + 1)−1 and the minimal value of φ is (8 max{(4M1 + 1)δC , (4M2 + 1)δD })1/2 (max{L1 , L2 } + 1). Now our goal is to find the best integer T > 0 which gives us an appropriate value of ΨT (a). Since in view of the inequalities above, this value is bounded from below by c0 max{δC , δD }1/2 , with the constant c0 depending on L1 , L2 , M1 , M2 , it is clear that in order to make the best choice of T , it should be at the same order as max{δC , δD }−1 . Let T = max{δC , δD }−1 . In this case we obtain a pair of points  x ∈ U,  y∈V such that x , δC ) ∩ C = ∅, BY ( y , δD ) ∩ D = ∅ BX ( and for each z ∈ C and each v ∈ D, f (z,  y ) ≥ f ( x,  y ) − c max{δC , δD }1/2 − 2 max{δf,1 , δf,2 }(2 max{M1 , M2 } + 1), f ( x , v) ≤ f ( x,  y ) + c max{δC , δD }1/2 + 2 max{δf,1 , δf,2 }(2 max{M1 , M2 } + 1), where the constant c > 0 depends only on L1 , L2 , M1 , M2 .

2.12 Zero-Sum Games on Unbounded Sets Let (X, ·, ·), (Y, ·, ·) be Hilbert spaces equipped with the complete norms  ·  which are induced by their inner products. Let A be a nonempty closed convex

72

subset of X, D be a nonempty closed convex subset of Y , W be an open convex subset of X, and V be an open convex subset of Y such that A ⊂ W, D ⊂ V

(2.172)

and let a function f : W × V → R 1 possess the following properties: (i) for each v ∈ V , the function f (·, v) : W → R 1 is convex and continuous; (ii) for each w ∈ W , the function f (w, ·) : V → R 1 is concave and continuous. Suppose that L1 > 1, L2 > 0, M0 , M1 , M2 > 0,

(2.173)

θ0 ∈ A ∩ BX (0, M0 ),

(2.174)

D ⊂ BY (0, M0 ),

(2.175)

sup{f (θ0 , y) : y ∈ V ∩ BY (0, M0 + 1)} < M1 ,

(2.176)

for each y ∈ V ∩ BY (0, M0 + 1), {x ∈ W : f (x, y) ≤ M1 + 4} ⊂ BX (0, M2 ),

(2.177)

for each v ∈ V ∩ BY (0, M0 + 1), |f (u1 , v) − f (u2 , v)| ≤ L1 u1 − u2  for all u1 , u2 ∈ W ∩ BX (0, M0 + 3M2 + 4)

(2.178)

and that for each u ∈ W ∩ BX (0, M0 + 3M2 + 4), |f (u, v1 ) − f (u, v2 )| ≤ L2 v1 − v2  for all v1 , v2 ∈ V ∩ BY (0, M0 + 2).

(2.179)

x∗ ∈ A and y∗ ∈ D

(2.180)

f (x∗ , y) ≤ f (x∗ , y∗ ) ≤ f (x, y∗ )

(2.181)

Let

satisfy

for each x ∈ A and each y ∈ D. In view of (2.172), (2.174)–(2.176), (2.180), and (2.181),

2.12 Zero-Sum Games on Unbounded Sets

73

f (x∗ , y∗ ) ≤ f (θ0 , y∗ ) < M1 ,

(2.182)

x∗  ≤ M2 , θ0  ≤ M2 .

(2.183)

In this chapter we prove the following result which does not have an analog in . (Note that in  we consider only games on bounded sets.) Theorem 2.16 Let δf,1 , δf,2 , δA , δD ∈ (0, 1/2) and {ak }∞ k=0 ⊂ (0, ∞) satisfy δA ≤ (12M2 + 9)−1 (L1 + 1)−2 , δf,1 ≤ (6M2 + 5)−1 ,

(2.184)

δA (12M2 + 9) ≤ at ≤ (L1 + 1)−2 for all integers t ≥ 0.

(2.185)

∞ ∞ ∞ Assume that {xt }∞ t=0 ⊂ W , {yt }t=0 ⊂ V , {ξt }t=0 ⊂ X, {ηt }t=0 ⊂ Y ,

BX (x0 , δA ) ∩ A = ∅, BY (y0 , δD ) ∩ D = ∅,

(2.186)

x0  ≤ M2 + 1

(2.187)

ξt ∈ ∂x f (xt , yt ) + BX (0, δf,1 ),

(2.188)

ηt ∈ ∂y f (xt , yt ) + BY (0, δf,2 ),

(2.189)

xt+1 − PA (xt − at ξt ) ≤ δA

(2.190)

yt+1 − PD (yt + at ηt ) ≤ δD .

(2.191)

and that for each integer t ≥ 0,

and

Let for each natural number T ,  xT = (

T 

at )−1

i=0

T 

at xt ,  yT = (

t=0

T 

at )−1

i=0

T 

at yt .

(2.192)

t=0

Then for all integers t ≥ 0, xt  ≤ 3M2 + 1

(2.193)

and for each natural number T , xT , δA ) ∩ A = ∅, BY ( yT , δD ) ∩ D = ∅, BX (

(2.194)

74

|(

T 

at )−1

t=0

T 

at f (xt , yt ) − f (x∗ , y∗ )|

t=0

≤ max{L1 δA , L2 δD } +(2

T 

at )−1 (max{2M0 , 6M2 + 4} + 1)2

t=0

+ max{δf,1 , δf,2 }(2 max{2M0 + 1, 6M2 + 5}) +2−1 (max{L1 , L2 } + 1)2 (

T 

at )−1

t=0

+(

T 

T 

at2

t=0

at )−1 (T + 1) max{(12M2 + 9)δA , (4M0 + 1)δD },

(2.195)

t=0

yT ) − ( |f ( xT , 

T  t=0

at )−1

T 

at f (xt , yt )|

t=0

≤ max{L1 δA , L2 δD } +(2

T 

at )−1 (max{2M0 , 6M2 + 4} + 1)2

t=0

+ max{δf,1 , δf,2 }(max{2M0 + 1, 6M2 + 5}) +2−1 (max{L1 , L2 } + 1)2 (

T 

at )−1

t=0

+(

T 

T 

at2

t=0

at )−1 (T + 1) max{(12M2 + 9)δA , (4M0 + 1)δD }

t=0

and for each natural number T , each z ∈ A, and each u ∈ D, xT ,  yT ) f (z,  yT ) ≥ f ( − max{L1 δA , L2 δD }

(2.196)

2.12 Zero-Sum Games on Unbounded Sets

−(

T 

75

at )−1 (max{2M0 , 6M2 + 4} + 1)2

t=0

−2 max{δf,1 , δf,2 }(max{2M0 + 1, 6M2 + 5}) −(max{L1 , L2 } + 1)2 (

T 

at )−1

t=0

− 2(

T 

T 

at2

t=0

at )−1 (T + 1) max{(12M2 + 9)δA , (4M0 + 1)δD },

(2.197)

t=0

f ( xT , u) ≤ f ( xT ,  yT ) + max{L1 δA , L2 δD } +(

T 

at )−1 (max{2M0 , 6M2 + 4} + 1)2

t=0

+ max{δf,1 , δf,2 }(2 max{2M0 + 1, 6M2 + 5}) +(max{L1 , L2 } + 1)2 (

T  t=0

+ 2(

T 

at )−1

T 

at2

t=0

at )−1 (T + 1) max{(12M2 + 9)δA , (4M0 + 1)δD }.

(2.198)

t=0

We are interested in the optimal choice of at , t = 0, 1, . . . , T . Let T be a natural number and AT = Tt=0 at be given. By Theorem 2.16, in order to make the best  choice of at , t = 0, . . . , T , we need to minimize the function Tt=0 at2 on the set {a = (a0 , . . . , aT ) ∈ R

T +1

: ai ≥ 0, i = 0, . . . , T ,

T 

ai = AT }.

i=0

By Lemma 2.3, this function has a unique minimizer a ∗ = (a0∗ , . . . , aT∗ ) where ai∗ = (T + 1)−1 AT , i = 0, . . . , T which is the best choice of at , t = 0, 1, . . . , T . Let T be a natural number and at = a for all t = 0, . . . , T . Now we will find the best a > 0. In order to meet this goal we need to choose a which is a minimizer of the function

76

ΨT (a) = ((T + 1)a)−1 (max{2M0 , 6M2 + 2} + 1)2 +a −1 max{(12M2 + 9)δA , (4M0 + 1)δD } + (max{L1 , L2 } + 1)2 a, a > 0. Since T can be arbitrarily large, we need to find a minimizer of the function φ(a) := a −1 max{(12M2 + 9)δA , (4M0 + 1)δD } +(max{L1 , L2 } + 1)2 a, a ∈ (0, ∞). It is not difficult to see that the minimizer is a = (max{(12M2 + 9)δA , (4M0 + 1)δD })1/2 (max{L1 , L2 } + 1)−1 and the minimal value of φ is (2 max{(12M2 + 9)δA , (4M0 + 1)δD })1/2 (max{L1 , L2 } + 1). Now our goal is to find the best integer T > 0 which gives us an appropriate value of ΨT (a). In view of the inequalities above, this value is bounded from below by c0 max{δC , δD }1/2 , with the constant c0 depending on L1 , L2 , M0 , M1 , M2 , and it is clear that in order to make the best choice of T , it should be at the same order as max{δC , δD }−1 . In this case we obtain a pair of points  x ∈ W ∩ BX (0, 3M2 + 4),  y ∈ V such that x , δA ) ∩ A ∩ BX (0, 3M2 + 4) = ∅, BX ( y , δD ) ∩ D = ∅ BY ( and for each z ∈ A and each v ∈ D, f (z,  y ) ≥ f ( x,  y) −c max{δA , δD }1/2 − 2 max{δf,1 , δf,2 } max{6M2 + 5, 2M0 + 1}, f ( x , v) ≤ f ( x,  y) +c max{δA , δD }1/2 + 2 max{δf,1 , δf,2 } max{6M2 + 5, 2M0 + 1}, where the constant c > 0 depends only on L1 , L2 , M0 , M1 , M2 .

2.13 Proof of Theorem 2.16

77

2.13 Proof of Theorem 2.16 By (2.175), (2.186), and (2.191), for all integers t ≥ 0, yt  ≤ M0 + 1.

(2.199)

C = A ∩ BX (0, 3M2 + 2)

(2.200)

U = W ∩ {x ∈ X : x < 3M2 + 4}.

(2.201)

Set

and

We show that xt − θ0  ≤ 2M2 + 1 for all integers t ≥ 0. In view of (2.183) and (2.187), x0 − θ0  ≤ x0  + θ0  ≤ M2 + 1 + M2 = 2M2 + 1.

(2.202)

Assume that t ≥ 0 is an integer and that xt − θ0  ≤ 2M2 + 1.

(2.203)

It follows from (2.183), (2.201), and (2.203) that xt ∈ U.

(2.204)

xt+1 − PA (xt − at ξt ) ≤ 1.

(2.205)

Relation (2.190) implies

By (2.178), (2.188), (2.199), (2.201), and (2.204), ξt ∈ ∂x f (xt , yt ) + BX (0, 1) ⊂ BX (0, L1 + 1).

(2.206)

It follows from (2.174), (2.185), (2.203), (2.206), and Lemma 2.2 that θ0 − PA (xt − at ξt ) ≤ θ0 − xt + at ξt  ≤ θ0 − xt  + ξt at ≤ 2M2 + 2.

(2.207)

78

In view of (2.183) and (2.207), PA (xt − at ξt ) ≤ θ0  + 2M2 + 2 ≤ 3M2 + 2.

(2.208)

By (2.205) and (2.208), xt+1  ≤ PA (xt − at ξt ) + 1 ≤ 3M2 + 3.

(2.209)

It follows from (2.201) and (2.209) that xt+1 ∈ U.

(2.210)

Relations (2.200) and (2.208) imply that PA (xt − at ξt ) ∈ C,

(2.211)

PA (xt − at ξt ) = PC (xt − at ξt ).

(2.212)

and

By (2.178), (2.183), (2.188), (2.190), (2.200), (2.201), (2.203), (2.204), and (2.210)– (2.212), we apply Lemma 2.7 with M0 = 3M2 + 2, a = at , x = xt , f = f (·, yt ), ξ = ξt , u = xt+1 and obtain that for each z ∈ C, at (f (xt , yt ) − f (z, yt )) ≤ 2−1 xt − z2 − 2−1 xt+1 − z2 + δA (12M2 + 9) + at δf,1 (6M2 + 5) + 2−1 at2 (L1 + 1)2 .

(2.213)

In view of (2.213) with z = θ0 , at (f (xt , yt ) − f (θ0 , yt )) ≤ 2−1 xt − θ0 2 − 2−1 xt+1 − θ0 2 + δA (12M2 + 9) + at δf,1 (6M2 + 5) + 2−1 at2 (L1 + 1)2 .

(2.214)

There are two cases: θ0 − xt+1  ≤ θ0 − xt ;

(2.215)

2.13 Proof of Theorem 2.16

79

θ0 − xt+1  > θ0 − xt .

(2.216)

Assume that (2.215) holds. By (2.203) and (2.215), θ0 − xt+1  ≤ 2M2 + 1.

(2.217)

Assume that (2.216) is true. It follows from (2.184), (2.185), (2.214), and (2.216) that f (xt , yt ) − f (θ0 , yt ) ≤ at−1 δA (12M2 + 9) + δf,1 (6M2 + 5) + 2−1 at (L1 + 1)2 ≤ 3.

(2.218)

By (2.176), (2.199), and (2.218), f (xt , yt ) ≤ M1 + 3.

(2.219)

In view of (2.177), (2.199), and (2.219), xt  ≤ M2 . Together with (2.183) this implies that xt − θ0  ≤ 2M2 . It follows from the inequality above, (2.174), (2.185), (2.190), (2.206), and Lemma 2.2 that θ0 − xt+1  ≤ xt+1 − PA (xt − at ξt ) + θ0 − PA (xt − at ξt ) ≤ δA + θ0 − PA (xt − at ξt ) δ + θ0 − xt + at ξt  ≤ δA + θ0 − xt  + ξt at ≤ δA + 2M2 + (L1 + 1)at ≤ 2M2 + 1 and (2.217) holds in both cases. Thus by induction we have showed that for all integers t ≥ 0, xt − θ0  ≤ 2M2 + 1

80

and that (2.210)–(2.212) hold. Now we can apply Theorem 2.15 with C, D, U and V ∩ {y ∈ Y : y < M0 + 2} and obtain that (2.195), (2.196) are true and that for each natural number T , each z ∈ C, and each u ∈ D, (2.197) and (2.198) hold. In order to complete the proof it is sufficient to show that (2.197) holds for all z ∈ A \ C. By (2.176), (2.177), (2.194), (2.197) (with z = θ0 ), and (2.200), yT ) f (z,  yT ) > M1 + 4 > 4 + f (θ0 ,  and (2.197) holds for all z ∈ A. Theorem 2.16 is proved.

 

2.14 An Example for Theorem 2.16 Let X, Y be Hilbert spaces, A be a nonempty closed convex subset of X, D be a nonempty closed convex subset of Y , W be an open convex subset of X, and V be an open convex subset of Y such that A ⊂ W, D ⊂ V , let a function g : W → R 1 be convex and continuous, a function h : V → R 1 be concave and continuous, and f (w, v) = g(w)h(v), w ∈ W, v ∈ V . Suppose that c0 , M0 , M1 , M2 , L0 , L1 > 0, g(w) ≥ c0 , w ∈ W, h(v) ≥ c0 , v ∈ V , D ⊂ BY (0, M0 ), θ0 ∈ A ∩ BX (0, M0 ), |h(v1 ) − h(v2 )| ≤ L0 v1 − v2  for all v1 , v2 ∈ V ∩ BY (0, M0 + 2), sup{h(y) : y ∈ V ∩ BY (0, M0 + 1)} < M1 , {x ∈ W : g(x) ≤ c0−1 (M1 (g(θ0 ) + 1) + 4)} ⊂ BX (0, M2 ),

2.14 An Example for Theorem 2.16

81

|g(u1 ) − g(u2 )| ≤ L1 u1 − u2  for all u1 , u2 ∈ W ∩ BX (0, 3M2 + M0 + 4) and that x∗ ∈ A and y∗ ∈ D satisfy f (x∗ , y) ≤ f (x∗ , y∗ ) ≤ f (x, y∗ ) for all x ∈ A and all y ∈ D. It is not difficult to see that all the assumptions made in Sect. 2.12 hold.

Chapter 3

The Mirror Descent Algorithm

In this chapter we analyze the mirror descent algorithm for minimization of convex and nonsmooth functions and for computing the saddle points of convex–concave functions, under the presence of computational errors. The problem is described by an objective function and a set of feasible points. For this algorithm each iteration consists of two steps. The first step is a calculation of a subgradient of the objective function while in the second one we solve an auxiliary minimization problem on the set of feasible points. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this.

3.1 Optimization on Bounded Sets Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · . Let C be a nonempty closed convex subset of X, U be an open convex subset of X such that C ⊂ U and let f : U → R 1 be a convex function. Suppose that there exist L > 0, M0 > 0 such that C ⊂ BX (0, M0 ),

(3.1)

|f (x) − f (y)| ≤ Lx − y for all x, y ∈ U.

(3.2)

In view of (3.2), for each x ∈ U , ∅ = ∂f (x) ⊂ BX (0, L). © Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_3

(3.3) 83

84

3 The Mirror Descent Algorithm

For each nonempty set D ⊂ X and each function h : D → R 1 put inf(h, D) = inf{h(y) : y ∈ D} and argmin(h, D) = argmin{h(y) : y ∈ D} = {y ∈ D : h(y) = inf(h, D)}. We study the convergence of the mirror descent algorithm under the presence of computational errors. This method was introduced by Nemirovsky and Yudin for solving convex optimization problems . Here we use a derivation of this algorithm proposed by Beck and Teboulle . Let δf , δC ∈ (0, 1] and {ak }∞ k=0 ⊂ (0, ∞). We describe the inexact version of the mirror descent algorithm. Mirror descent algorithm Initialization: select an arbitrary x0 ∈ U . Iterative Step: given a current iteration vector xt ∈ U calculate ξt ∈ ∂f (xt ) + BX (0, δf ), define gt (x) = ξt , x + (2at )−1 x − xt 2 , x ∈ X and calculate the next iteration vector xt+1 ∈ U such that BX (xt+1 , δC ) ∩ argmin{gt (y) : y ∈ C} = ∅. Here δf is a computational error produced by our computer system when we calculate a subgradient of f and δC is a computational error produced by our computer system when we calculate a minimizer of the function gt . Note that gt is a convex bounded from below function on X which possesses a minimizer on C. In this chapter we prove the following result. Theorem 3.1 Let δf , δC ∈ (0, 1], {ak }∞ k=0 ⊂ (0, ∞) and let x∗ ∈ C

(3.4)

f (x∗ ) ≤ f (x) for all x ∈ C.

(3.5)

satisfies

∞ Assume that {xt }∞ t=0 ⊂ U , {ξt }t=0 ⊂ X,

3.1 Optimization on Bounded Sets

85

x0  ≤ M0 + 1

(3.6)

ξt ∈ ∂f (xt ) + BX (0, δf ),

(3.7)

gt (z) = ξt , z + (2at )−1 z − xt 2 , z ∈ X

(3.8)

BX (xt+1 , δC ) ∩ argmin(gt , C) = ∅.

(3.9)

and that for each integer t ≥ 0,

and

Then for each natural number T , T 

at (f (xt ) − f (x∗ ))

t=0

≤ 2−1 (2M0 + 1)2 + (δf (2M0 + 1) + δC (L + 1))

T 

at

t=0

+ 4δC (T + 1)(2M0 + 1) + 2−1 (L + 1)2

T 

at2 .

(3.10)

t=0

Moreover, for each natural number T , f ((

T 

at )−1

t=0

T 

at xt ) − f (x∗ ),

t=0

min{f (xt ) : t = 0, . . . , T } − f (x∗ ) −1

≤2

(2M0 + 1) ( 2

T 

at )−1 + δf (2M0 + 1) + δC (L + 1)

t=0

+4δC (T + 1)(2M0 + 1)(

T 

at )−1

t=0

+ 2−1 (L + 1)2

T  t=0

at2 (

T  t=0

at )−1 .

(3.11)

86

3 The Mirror Descent Algorithm

Theorem 3.1 is proved in Sect. 3.3. It is a generalization of Theorem 3.1 of  proved in the case when δf = δC . We are interested in an optimal choice of at , t = 0, 1, . . . . Let T be a natural number and AT = Tt=0 at be given. By Theorem 3.1, in order to make the best  choice of at , t = 0, . . . , T , we need to minimize the function Tt=0 at2 on the set {a = (a0 , . . . , aT ) ∈ R T +1 : ai ≥ 0, i = 0, . . . , T ,

T 

ai = AT }.

i=0

By Lemma 2.3, this function has a unique minimizer a ∗ = (a0∗ , . . . , aT∗ ) where ai∗ = (T + 1)−1 AT , i = 0, . . . , T . This is the best choice of at , t = 0, 1, . . . , T . Let T be a natural number and at = a, t = 0, . . . , T . Now we will find the best a > 0. By Theorem 3.1, we need to choose a which is a minimizer of the function 2−1 ((T + 1)a)−1 (2M0 + 1)2 + δf (2M0 + 1) + δC (L + 1) +4a −1 δC (2M0 + 1) + 2−1 (L + 1)2 a. Since T can be arbitrarily large, we need to find a minimizer of the function φ(a) := 4a −1 δC (2M0 + 1) + 2−1 (L + 1)2 a, a ∈ (0, ∞). Clearly, the minimizer is a = (8δC (2M0 + 1))1/2 (L + 1)−1 and the minimal value of φ is (8δC (2M0 + 1))1/2 (L + 1). Now we can think about the best choice of T . It is clear that it should be at the same order as δC−1 . As a result, we obtain a point ξ ∈ U such that BX (ξ, δC ) ∩ C = ∅ and 1/2

f (ξ ) ≤ f (x∗ ) + c1 δC + δf (2M0 + 1), where the constant c1 > 0 depends only on L and M0 .

3.2 The Main Lemma

87

3.2 The Main Lemma Lemma 3.2 Let δf , δC ∈ (0, 1], a > 0 and let z ∈ C.

(3.12)

x ∈ U ∩ BX (0, M0 + 1),

(3.13)

ξ ∈ ∂f (x) + BX (0, δf ),

(3.14)

g(v) = ξ, v + (2a)−1 v − x2 , v ∈ X

(3.15)

u∈U

(3.16)

BX (u, δC ) ∩ {v ∈ C : g(v) = inf(g, C)} = ∅.

(3.17)

Assume that

and that

satisfies

Then a(f (x) − f (z)) ≤ δf a(2M0 + 1) + δC a(L + 1) + 4δC (M0 + 1) +2−1 a 2 (L + 1)2 + 2−1 z − x2 − 2−1 z − u2 . Proof In view of (3.14), there exists l ∈ ∂f (x)

(3.18)

l − ξ  ≤ δ.

(3.19)

such that

Clearly, the function g is Fréchet differentiable on X. We denote by g  (v) its Fréchet derivative at v ∈ X. It is easy to see that g  (v) = ξ + a −1 (v − x), v ∈ X.

(3.20)

88

3 The Mirror Descent Algorithm

By (3.17), there exists  u ∈ BX (u, δC ) ∩ C

(3.21)

g( u) = inf(g, C).

(3.22)

such that

It follows from (3.20)–(3.22) that for all v ∈ C, u), v −  u = ξ + a −1 ( u − x), v −  u. 0 ≤ g  (

(3.23)

By (3.1), (3.12), (3.13), (3.18), and (3.19), a(f (x) − f (z)) ≤ ax − z, l = ax − z, ξ  + ax − z, l − ξ  ≤ ax − z, ξ  + ax − zl − ξ  ≤ δf a(2M0 + 1) + ax − z, ξ  u + aξ,  u − z = δf a(2M0 + 1) + aξ, x −  u, aξ  ≤ δf a(2M0 + 1) + x −  + z −  u, x −  u − aξ  + z −  u,  u − x.

(3.24)

Relations (3.12) and (3.23) imply that z −  u, x −  u − aξ  ≤ 0.

(3.25)

In view of Lemma 2.1, u2 −  u − x2 ]. z −  u,  u − x = 2−1 [z − x2 − z − 

(3.26)

It follows from (3.3), (3.18), (3.19), and (3.21) that x −  u, aξ  = x − u, aξ  + u −  u, aξ  ≤ aδC ξ  + x − u, aξ  ≤ aδC (L + 1) + x − u, aξ  ≤ aδC (L + 1) + 2−1 x − u2 + 2−1 a 2 ξ 2 .

(3.27)

3.2 The Main Lemma

89

By (3.3), (3.18), (3.19), and (3.24)–(3.27), a(f (x) − f (z)) u, aξ  ≤ δf a(2M0 + 1) + x −  +z −  u, x −  u − aξ  + z −  u,  u − x ≤ δf a(2M0 + 1) + aδC (L + 1) +2−1 x − u2 + 2−1 a 2 ξ 2 +2−1 z − x2 − 2−1 z −  u2 − 2−1  u − x2 ≤ δf a(2M0 + 1) + δC a(L + 1) + 2−1 a 2 (L + 1)2 +2−1 z − x2 − 2−1 z −  u2 + 2−1 x − u2 − 2−1  u − x2 .

(3.28)

In view of (3.1), (3.13), (3.17), and (3.21), u − x2 | |x − u2 −  ≤ |x − u −  u − x|(x − u +  u − x) ≤ 4u −  u(M0 + 1) ≤ 4(M0 + 1)δC .

(3.29)

Relations (3.1), (3.12), (3.13), and (3.21) imply that |z −  u2 − z − u2 | ≤ |z −  u − z − u|(z −  u + z − u) ≤ u −  u(4M0 + 1) ≤ (4M0 + 1)δC .

(3.30)

By (3.28), (3.29), and (3.30), a(f (x) − f (z)) ≤ δf a(2M0 + 1) + δC (L + 1) + +2−1 a 2 (L + 1)2 +2−1 z − x2 − 2−1 z − u2 + 4(M0 + 1)δC . This completes the proof of Lemma 3.2.

 

90

3 The Mirror Descent Algorithm

3.3 Proof of Theorem 3.1 In view of (3.1), (3.6), and (3.9), xt  ≤ M0 + 1, t = 0, 1, . . . .

(3.31)

Let t ≥ 0 be an integer. Applying Lemma 3.2 with g = gt , z = x∗ , a = at , x = xt , ξ = ξt , u = xt+1 we obtain that at (f (xt ) − f (x∗ )) ≤ 2−1 x∗ − xt 2 − 2−1 x∗ − xt+1 2 +4δC (M0 + 1) + at δf (2M0 + 1) + at δC (L + 1) + 2−1 at2 (L + 1)2 .

(3.32)

By (3.1), (3.4), (3.31), and (3.32), for each natural number T , T 

at (f (xt ) − f (x∗ ))

t=0

T 

(2−1 x∗ − xt 2 − 2−1 x∗ − xt+1 2 )

t=0

+δf (2M0 + 1)

T  t=0

at + δC (L + 1)

T 

at

t=0

+4(T + 1)δC (M0 + 1) + 2−1 (L + 1)2

T 

at2

t=0

≤ 2−1 (2M0 + 1)2 + (δf (2M0 + 1) + δC (L + 1))

T  t=0

+4(T + 1)δC (M0 + 1) + 2−1 (L + 1)2

T  t=0

at2 .

at

3.4 Optimization on Unbounded Sets

91

Thus (3.10) is true. Evidently, (3.10) implies (3.11). This completes the proof of Theorem 3.1.  

3.4 Optimization on Unbounded Sets Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · . Let D be a nonempty closed convex subset of X, V be an open convex subset of X such that D ⊂ V,

(3.33)

and f : V → R 1 be a convex function which is Lipschitz on all bounded subsets of V . Set Dmin = {x ∈ D : f (x) ≤ f (y) for all y ∈ D}.

(3.34)

We suppose that Dmin = ∅. In Sect. 3.5 we will prove the following result. Theorem 3.3 Let δf , δC ∈ (0, 1], M > 1 satisfy Dmin ∩ BX (0, M) = ∅,

(3.35)

M0 > 80M + 6,

(3.36)

L > 1 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, M0 + 2),

(3.37)

0 < τ0 ≤ τ1 ≤ (4L + 4)−1 ,

(3.38)

0 = 8τ0−1 δC (M0 + 1) + 2δC (L + 1) + 2δf (2M0 + 1) + τ1 (L + 1)2

(3.39)

n0 = τ0−1 (2M + 2)2 0−1  + 1.

(3.40)

and let

92

3 The Mirror Descent Algorithm

Assume that ∞ ∞ {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X, {at }t=0 ⊂ [τ0 , τ1 ],

(3.41)

x0  ≤ M

(3.42)

ξt ∈ ∂f (xt ) + BX (0, δf ),

(3.43)

gt (v) = ξt , v + (2at )−1 v − xt 2 , v ∈ X

(3.44)

BX (xt+1 , δC ) ∩ argmin(gt , D) = ∅.

(3.45)

and that for each integer t ≥ 0,

and

Then there exists an integer q ∈ [1, n0 ] such that f (xq ) ≤ inf(f, D) + 0 , xt  ≤ 15M + 1, t = 0, . . . , q. We are interested in the best choice of at , t = 0, 1, . . . . Assume for simplicity that τ1 = τ0 . In order to meet our goal we need to minimize 0 which obtains its minimal value when τ0 = (8δC (M0 + 1))1/2 (L + 1)−1 and the minimal value of 0 is 2δC (L + 1) + 2δf (2M0 + 1) + 2(8δC (M0 + 1))1/2 (L + 1). 1/2

Thus 0 is at the same order as max{δC , δf }. By (3.40) and the inequalities above, −1/2 1/2 n0 is at the same order as δC max{δC , δf }−1 . Theorem 3.3 is a generalization of Theorem 3.3 of  proved with δf = δC . In this chapter we will prove the following two theorems which have no prototype in . Theorem 3.4 Let δf , δC ∈ (0, 1), M > 0 satisfy δf ≤ (24M + 48)−1 , δC ≤ 8−1 (L + 1)−2 (16M + 16)−1 ,

(3.46)

{x ∈ V : f (x) ≤ inf(f, D) + 4} ⊂ BX (0, M),

(3.47)

M0 = 12M + 12,

(3.48)

3.4 Optimization on Unbounded Sets

93

L > 1 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, M0 + 2)

(3.49)

and let {at }∞ t=0 ⊂ (0, ∞) satisfy for all integers t ≥ 0, 4δC (M0 + 1) ≤ at ≤ (2L + 2)−1 (L + 1)−1 .

(3.50)

∞ {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

(3.51)

x0  ≤ M

(3.52)

ξt ∈ ∂f (xt ) + BX (0, δf ),

(3.53)

gt (v) = ξt , v + (2at )−1 v − xt 2 , v ∈ X

(3.54)

BX (xt+1 , δC ) ∩ argmin(gt , D) = ∅.

(3.55)

Assume that

and that for each integer t ≥ 0,

and

Then xt  ≤ 5M + 3 for all integers t ≥ 0 and for each natural number T , T 

at (f (xt ) − inf(f, D))

t=0

≤ 2M 2 + (δf (2M0 + 1) + δC (L + 1))

T 

at

t=0

+ 4δC (T + 1)(M0 + 1) + 2−1 (L + 1)2

T  t=0

Moreover, for each natural number T ,

at2 .

(3.56)

94

3 The Mirror Descent Algorithm

f ((

T 

at )−1

t=0

T 

at xt ) − inf(f, D),

t=0

min{f (xt ) : t = 0, . . . , T } − inf(f, D) ≤ 2M ( 2

T 

at )−1 + δf (2M0 + 1) + δC (L + 1)

t=0

+ 4δC (T + 1)(M0 + 1)(

T  t=0

at )−1 + 2−1 (L + 1)2

T 

at2 (

t=0

T 

at )−1 .

(3.57)

t=0

We are interestedin the best choice of at , t = 0, 1, . . . . Let T be a natural number and AT = Tt=0 at be given. By Theorem 3.4, in order to make the best  choice of at , t = 0, . . . , T , we need to minimize the function Tt=0 at2 on the set {a = (a0 , . . . , aT ) ∈ R T +1 : ai ≥ 0, i = 0, . . . , T ,

T 

ai = AT }.

i=0

By Lemma 2.3, this function has a unique minimizer a ∗ = (a0∗ , . . . , aT∗ ) where ai∗ = (T + 1)−1 AT , i = 0, . . . , T . This is the best choice of at , t = 0, 1, . . . , T . Let T be a natural number and at = a, t = 0, . . . , T . Now we will find the best a > 0. By Theorem 3.4, we need to choose a which is a minimizer of the function 2((T + 1)a)−1 M 2 + δf (2M0 + 1) + δC (L + 1) +4a −1 δC (M0 + 1) + 2−1 (L + 1)2 a. Since T can be arbitrarily large, we need to find a minimizer of the function φ(a) := 4a −1 δC (M0 + 1) + 2−1 (L + 1)2 a, a ∈ (0, ∞). Clearly, the minimizer is a = (8δC (M0 + 1))1/2 (L + 1)−1 and the minimal value of φ is (8δC (M0 + 1))1/2 (L + 1). Now we can think about the best choice of T . It is clear that it should be at the same order as δC−1 . As a result, we obtain a point ξ ∈ U such that

3.4 Optimization on Unbounded Sets

95

BX (ξ, δC ) ∩ C = ∅ and 1/2

f (ξ ) ≤ inf(f, D) + c1 δC + δf (2M0 + 1), where the constant c1 > 0 depends only on L and M0 . Theorem 3.5 Let δf , δC ∈ (0, 1), M > 8

(3.58)

Dmin ∩ BX (0, M) = ∅,

(3.59)

satisfy

L > 0 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 8M + 8),

(3.60)

a = (4L + 4)−1 δC ,

(3.61)

1/2

−1/2

T = 8−1 δC

max{δf , δC }−1 . 1/2

(3.62)

Assume that ∞ {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M, BX (x0 , δC ) ∩ D = ∅

(3.63)

and that for each integer t ≥ 0, ξt ∈ ∂f (xt ) + BX (0, δf ),

(3.64)

gt (v) = ξt , v + (2at )−1 v − xt 2 , v ∈ X

(3.65)

BX (xt+1 , δC ) ∩ argmin(gt , D) = ∅.

(3.66)

xt  ≤ 3M + 2 for all integers t ∈ [0, T ]

(3.67)

and

Then

96

3 The Mirror Descent Algorithm

and T  (T + 1)−1 f (xt ) − inf(f, D), t=0

f ((T + 1)−1

T 

xt ) − inf(f, D), min{f (xt ) : t = 0, . . . , T } − inf(f, D)

t=0 1/2

≤ 64M 2 (L + 1) max{δf , δC }.

(3.68)

3.5 Proof of Theorem 3.3 By (3.35) there exists z ∈ Dmin ∩ BX (0, M).

(3.69)

Assume that T is a natural number and that f (xt ) − f (z) > 0 , t = 1, . . . , T .

(3.70)

In view of (3.45), there exists η ∈ BX (x1 , δC ) ∩ argmin(g0 , D).

(3.71)

Relations (3.44), (3.69), and (3.71) imply that ξ0 , η + (2a0 )−1 η − x0 2 ≤ ξ0 , z + (2a0 )−1 z − x0 2 .

(3.72)

It follows from (3.37), (3.42), and (3.43) that ξ0  ≤ L + 1.

(3.73)

a0−1 ≥ τ1−1 ≥ 4(L + 1).

(3.74)

In view of (3.8) and (3.41),

By (3.42), (3.69), (3.72), and (3.73), (L + 1)M + (2a0 )−1 (2M + 1)2

3.5 Proof of Theorem 3.3

97

≥ ξ0 , z + (2a0 )−1 z − x0 2 ≥ (2a0 )−1 η − x0 2 + ξ0 , η − x0  + ξ0 , x0  ≥ (2a0 )−1 η − x0 2 − (L + 1)η − x0  − (L + 1)M. Together with (3.38) and (3.41) this implies that M + (2M + 1)2 ≥ η − x0 2 − 2−1 η − x0 , (η − x0  − 4−1 )2 ≤ (4M + 1)2 , η − x0  ≤ 8M.

(3.75)

Together with (3.48), (3.69), and (3.71) this implies that η ≤ 9M, x1  ≤ 9M + 1, η − z ≤ 10M, x1 − z ≤ 10M + 1.

(3.76)

By induction we show that for every integer t ∈ [1, T ], xt − z ≤ 14M + 1,

(3.77)

f (xt ) − f (z) ≤ δf (2M0 + 1) + δC (L + 1) +4τ0−1 δC (M0 + 1) + 2−1 τ1 (L + 1)2 + (2τ0 )−1 (z − xt 2 − z − xt+1 2 ).

(3.78)

U = V ∩ {v ∈ X : v < M0 + 2}

(3.79)

C = D ∩ BX (0, M0 ).

(3.80)

Set

and

In view of (3.69) and (3.76), (3.77) holds for t = 1.

98

3 The Mirror Descent Algorithm

Assume that an integer t ∈ [1, T ] and that (3.77) holds. It follows from (3.36), (3.69), (3.79), and (3.80) that z ∈ C ⊂ BX (0, M0 ).

(3.81)

In view of (3.36), (3.69), (3.77), and (3.79), xt ∈ U ∩ BX (0, M0 + 1).

(3.82)

Relations (3.37), (3.43), and (3.82) imply that ξt ∈ ∂f (xt ) + BX (0, δf ) ⊂ BX (0, L + 1).

(3.83)

In view of (3.45), there exists h ∈ BX (xt+1 , δC ) ∩ argmin(gt , D).

(3.84)

By (3.44), (3.69), and (3.84), ξt , h + (2at )−1 h − xt 2 ≤ ξt , z + (2at )−1 z − xt 2 .

(3.85)

In view of (3.85), ξt , z + (2at )−1 z − xt 2 ≥ (2at )−1 h − xt 2 + ξt , h − xt  + ξt , xt . It follows from the inequality above, (3.36), (3.38), (3.41), (3.69), (3.77), and (3.83) that (L + 1)M + (2at )−1 (14M + 1)2 ≥ (2at )−1 h − xt 2 − (L + 1)h − xt  − (L + 1)(15M + 1) ≥ (2at )−1 (h − xt 2 − h − xt ) − (L + 1)(15M + 1) ≥ (2at )−1 (h − xt  − 1)2 − (2at )−1 − (L + 1)(15M + 1), 4(14M + 1)2 ≥ 2 + 16M + (14M + 1)2 ≥ (h − xt  − 1)2 , h − xt  ≤ 28M + 4, h ≤ 44M + 5 < M0 .

(3.86)

3.5 Proof of Theorem 3.3

99

By (3.80), (3.84), and (3.86), h ∈ C.

(3.87)

Relations (3.80), (3.84), and (3.87) imply that h ∈ argmin(gt , C) and h ∈ BX (xt+1 , δC ) ∩ argmin(gt , C).

(3.88)

It follows from (3.37), (3.38), (3.43), (3.44), (3.69), (3.79), (3.82), (3.88), and Lemma 3.2 which holds with x = xt , a = at , ξ = ξt , u = xt+1 that at (f (xt ) − f (z)) ≤ δf at (2M0 + 1) + δC at (L + 1) +4δC (M0 + 1) + 2−1 at2 (L + 1)2 +2−1 z − xt 2 − 2−1 z − xt+1 2 . Together with the inclusion at ∈ [τ0 , τ1 ] this implies that f (xt ) − f (z) ≤ 4τ0−1 δC (M0 + 1) +(2M0 + 1)δf + (L + 1)δC + 2−1 τ1 (L + 1)2 + (2τ0 )−1 z − xt 2 − (2τ0 )−1 z − xt+1 2

(3.89)

and (3.78) holds. In view of (3.39), (3.70), (3.77), and (3.89), z − xt 2 − z − xt+1 2 ≥ 0, z − xt+1  ≤ z − xt  ≤ 14M + 1. Hence by induction we showed that (3.78) holds for all t = 1, . . . , T and (3.77) holds for all t = 1, . . . , T + 1. It follows from (3.39), (3.40), (3.42), (3.69), (3.70), and (3.78) that

100

3 The Mirror Descent Algorithm

T 0 < T (min{f (xt ) : t = 1, . . . , T } − f (z)) ≤

T 

(f (xt ) − f (z))

t=1

≤ (2τ0 )

−1

T  (z − xt 2 − z − xt+1 2 ) t=1

+T (δf (2M0 + 1) + δC (L + 1) +4δC (M0 + 1)τ0−1 + 2−1 τ1 (L + 1)2 ) ≤ (2τ0 )−1 (2M + 1)2 + 4T τ0−1 δC (M0 + 1) +T (δf (2M0 + 1) + δC (L + 1) + 2−1 τ1 (L + 1)2 ), 2−1 0 < (2τ0 T )−1 (2M + 1)2 and T < τ0−1 (2M + 1)2 0−1 < n0 . Thus we have shown that if an integer T ≥ 1 satisfies f (xt ) − f (z) > 0 , t = 1, . . . , T , then T < n0 and (3.77) holds for all t = 1, . . . , T + 1. This implies that there exists an integer q ∈ {1, . . . , n0 } such that f (xq ) − f (z) ≤ 0 , xt  ≤ 15M + 1, t = 0, . . . , q.  

Theorem 3.3 is proved.

3.6 Proof of Theorem 3.4 Fix z ∈ Dmin .

(3.90)

3.6 Proof of Theorem 3.4

101

Set U = V ∩ {v ∈ X : v < M0 + 2}

(3.91)

C = D ∩ BX (0, M0 ),

(3.92)

M1 = 4M + 3.

(3.93)

z ≤ M.

(3.94)

x0 − z ≤ 2M.

(3.95)

and

In view of (3.47) and (3.90),

By (3.52) and (3.94),

Assume that t ≥ 0 is an integer and that xt − z ≤ M1 .

(3.96)

It follows from (3.48), (3.90), (3.91), (3.93), (3.94), and (3.96) that z ∈ C ∩ BX (0, M),

(3.97)

xt  ≤ M1 + M,

(3.98)

xt ∈ U ∩ BX (0, M1 + M).

(3.99)

Relations (3.48), (3.49), (3.53), (3.93), and (3.98) imply that ξt ∈ ∂f (xt ) + BX (0, δf ) ⊂ BX (0, L + 1).

(3.100)

In view of (3.55), there exists h ∈ BX (xt+1 , δC ) ∩ argmin(gt , D).

(3.101)

By (3.54), (3.90), and (3.101), ξt , h + (2at )−1 h − xt 2 ≤ ξt , z + (2at )−1 z − xt 2 .

(3.102)

102

3 The Mirror Descent Algorithm

In view of (3.102), ξt , z + (2at )−1 z − xt 2 ≥ (2at )−1 h − xt 2 + ξt , h − xt  + ξt , xt .

(3.103)

It follows from (3.50), (3.100), and (3.103) that (2at )−1 (h − xt  − 1)2 − (2at )−1 ≤ (2at )−1 (h − xt 2 − h − xt ) ≤ (2at )−1 h − xt 2 − (L + 1)h − xt  ≤ (2at )−1 h − xt 2 + ξt , h − xt  ≤ (2at )−1 h − xt 2 + ξt , h − xt  + ξt , xt  + ξt xt .

(3.104)

By (3.94), (3.96), (3.98), (3.100), and (3.104), (2at )−1 (h − xt  − 1)2 − (2at )−1 ≤ (2at )−1 h − xt 2 + ξt , h − xt  +ξt , xt  + (L + 1)(M1 + M) ≤ (2at )−1 z − xt 2 + ξt , z + (L + 1)(M1 + M) ≤ (L + 1)(M1 + 2M) + (2at )−1 M12 .

(3.105)

In view of (3.48), (3.50), (3.93), (3.98), and (3.105), (h − xt  − 1)2 ≤ 1 + 2M + M12 + M1 , h − xt  ≤ M1 + 2, h ≤ M1 + 2 + M1 + M ≤ 3M1 + 2 = 12M + 11 < M0 .

(3.106)

By (3.48), (3.92), (3.101), and (3.106), h ∈ C.

(3.107)

3.6 Proof of Theorem 3.4

103

Relations (3.93), (3.101), and (3.107) imply that h ∈ argmin(gt , C).

(3.108)

xt+1 ∈ U ∩ BX (0, M0 + 1).

(3.109)

By (3.101) and (3.106),

It follows from (3.48), (3.53), (3.54), (3.93), (3.97), (3.99), (3.101), (3.108), (3.109), and Lemma 3.2 which holds with x = xt , a = at , ξ = ξt , u = xt+1 that at (f (xt ) − f (z)) ≤ δf at (2M0 + 1) + δC at (L + 1) +4δC (M0 + 1) + 2−1 at2 (L + 1)2 + 2−1 z − xt 2 − 2−1 z − xt+1 2 .

(3.110)

z − xt+1  ≤ z − xt ;

(3.111)

z − xt+1  > z − xt .

(3.112)

There are two cases:

Assume that (3.111) is true. In view of (3.96) and (3.111), z − xt+1  ≤ z − xt  ≤ M1 . Assume that (3.112) holds. It follows from (3.46), (3.48), (3.50), (3.110), and (3.112) that f (xt ) − f (z) ≤ 4at−1 δC (M0 + 1) + δf (2M0 + 1) +δC (L + 1) + 2−1 at (L + 1)2 ≤ 4. By the inequality above, (3.47) and (3.90),

104

3 The Mirror Descent Algorithm

xt  ≤ M.

(3.113)

It follows from (3.97), (3.100), (3.103), and (3.113) that (2at )−1 (h − xt  − 1)2 − (2at )−1 ≤ (2at )−1 h − xt 2 + ξt , h − xt  + ξt , xt  + (L + 1)M ≤ (2at )−1 z − xt 2 + ξt , z + (L + 1)M ≤ 2(L + 1)M + 4(2at )−1 M 2 . By the relation above, (3.50) and (3.113), (h − xt  − 1)2 ≤ 1 + 4M 2 + 2M ≤ (2M + 1)2 , h − xt  ≤ 2M + 2, h ≤ 3M + 2.

(3.114)

xt+1  ≤ 3M + 3.

(3.115)

In view of (3.101) and (3.114),

By (3.97) and (3.115), xt+1 − z ≤ 4M + 3. Together with (3.93) this implies that xt+1 − z ≤ M1 . Hence by induction we showed that for all integers t ≥ 0, xt − z ≤ M1 and (3.110) is true. It follows from (3.93), (3.97), and (3.116) that xt  ≤ 5M + 3 for all integers t ≥ 0. By (3.90), (3.95), and (3.110), for each natural number T , T  t=0

at (f (xt ) − inf(f, D))

(3.116)

3.7 Proof of Theorem 3.5

105

≤ 2−1 z − x0 2 + (δf (2M0 + 1) + δC (L + 1))

T 

at

t=0 T 

+4δC (T + 1)(M0 + 1) + 2−1 (L + 1)2

at2

t=0

≤ 2M 2 + (δf (2M0 + 1) + δC (L + 1))

T 

at

t=0

+4δC (T + 1)(M0 + 1) + 2−1 (L + 1)2

T 

at2 .

t=0

 

Theorem 3.4 is proved.

3.7 Proof of Theorem 3.5 By (3.59), there exists z ∈ Dmin ∩ BX (0, M).

(3.117)

U = V ∩ {v ∈ X : v < 8M + 8}

(3.118)

C = D ∩ BX (0, 5M + 6).

(3.119)

Set

and

In view of (3.63) and (3.116), x0 − z ≤ 2M.

(3.120)

Assume that t ≥ 0 is an integer and that xt − z ≤ 2M + 2.

(3.121)

It follows from (3.116)–(3.119), (3.121) that z ∈ C ∩ BX (0, M),

(3.122)

106

3 The Mirror Descent Algorithm

xt  ≤ 3M + 2,

(3.123)

xt ∈ U ∩ BX (0, 3M + 2).

(3.124)

Relations (3.60) and (3.123) imply that ξt ∈ ∂f (xt ) + BX (0, δf ) ⊂ BX (0, L + 1).

(3.125)

In view of (3.66), there exists h ∈ BX (xt+1 , δC ) ∩ argmin(gt , D).

(3.126)

By (3.65), (3.116), and (3.126), ξt , h + (2a)−1 h − xt 2 ≤ ξt , z + (2a)−1 z − xt 2 .

(3.127)

In view of (3.127), ξt , z + (2a)−1 z − xt 2 ≥ (2a)−1 h − xt 2 + ξt , h − xt  + ξt , xt .

(3.128)

It follows from (3.58), (3.61), (3.125), and (3.128) that (2a)−1 (h − xt  − 1)2 − (2a)−1 ≤ (2a)−1 (h − xt 2 − h − xt ) ≤ (2a)−1 h − xt 2 − (L + 1)h − xt  ≤ (2a)−1 h − xt 2 + ξt , h − xt  ≤ (2a)−1 h − xt 2 + ξt , h − xt  + ξt , xt  + ξt xt .

(3.129)

By (3.116), (3.125), (3.128), and (3.129), (2a)−1 (h − xt  − 1)2 − (2a)−1 ≤ (2a)−1 h − xt 2 + ξt , h − xt  + ξt , xt  + (L + 1)xt  ≤ (2a)−1 z − xt 2 + ξt , z + (L + 1)xt  ≤ (L + 1)(M + xt ) + (2a)−1 z − xt 2 .

(3.130)

3.7 Proof of Theorem 3.5

107

In view of (3.61), (3.121), (3.123), and (3.130), (h − xt  − 1)2 ≤ 1 + z − xt 2 + M + xt  1 + 4M 2 + 8M + 4 + 4M + 2 = 4M 2 + 12M + 7 ≤ (2M + 3)2 and h − xt  − 1 ≤ 2M + 3, h ≤ 5M + 6.

(3.131)

By (3.119)) and (3.131), h ∈ C. Together with relations (3.119) and (3.126) this implies that h ∈ argmin(gt , C).

(3.132)

xt+1  ≤ 5M + 7.

(3.133)

By (3.126) and (3.131),

It follows from (3.64), (3.65), (3.116)–(3.119), (3.124), (3.126), (3.132), (3.133), and Lemma 3.2 applied with M0 = 5M + 6, x = xt , ξ = ξt , u = xt+1 , that a(f (xt ) − f (z)) ≤ 2−1 z − xt 2 − 2−1 z − xt+1 2 +aδf (10M + 13) + aδC (L + 1) + 4δC (5M + 7) + 2−1 a 2 (L + 1)2 . In view of (3.60), (3.63), (3.67), (3.116), and (3.123), f (xt ) − f (z) ≥ −LδC . It follows from (3.61), (3.116), (3.134), and the inequality above that

(3.134)

108

3 The Mirror Descent Algorithm

z − xt+1 2 − z − xt 2 ≤ 8δC (5M + 7) + a 2 (L + 1)2 +2aδf (10M + 3) + 2δC a(L + 1) + 2aLδC ≤ 8δC (5M + 8) + a 2 (L + 1)2 + 2aδf (10M + 3).

(3.135)

Thus we have shown that the following property holds: (a) if for an integer t ≥ 0 relation (3.121) holds, then (3.134) and (3.135) are true. Let us show that for all integers t = 0, . . . , T , z − xt 2 ≤ z − x0 2 + t[8δC (5M + 8) + a 2 (L + 1)2 + 2aδf (10M + 3)] ≤ z − x0 2 + T [8δC (5M + 8) + a 2 (L + 1)2 + 2aδf (10M + 3)] ≤ 4M 2 + 8M + 4.

(3.136)

First, note that by (3.61) and (3.62), T [8δC (5M + 8) + a 2 (L + 1)2 + 2aδf (10M + 3)] ≤ 8T δC (5M + 8) + T δC + 2T aδf (10M + 3) −1/2 −1 δf }(8δC (5M

≤ 8−1 min{δC−1 , δC

+ 9) + δf (10M + 3)δC (L + 1)−1 ) 1/2

≤ 5M + 9 + 2M + 1 = 7M + 10 ≤ 8M + 4.

(3.137)

It follows from (3.120) and (3.137) that z − x0 2 + T [8δC (5M + 8) + a 2 (L + 1)2 + 2aδf (10M + 3)] ≤ (2M + 2)2 . Assume that t ∈ {0, . . . , T } \ {T } and that (3.136) holds. Then xt − z ≤ 2M + 2 (see (3.121)) and by property (a), (3.134) and (3.135) hold. By (3.135), (3.136), and (3.138),

(3.138)

3.7 Proof of Theorem 3.5

109

z − xt+1 2 ≤ z − xt 2 + 8δC (5M + 8) + a 2 (L + 1)2 + 2aδf (10M + 3) ≤ z − x0 2 +(t + 1)[8δC (5M + 8) + a 2 (L + 1)2 + 2aδf (10M + 3)] ≤ z − x0 2 +T [8δC (5M + 8) + a 2 (L + 1)2 + 2aδf (10M + 3)] ≤ 4M 2 + 8M + 4 and z − xt+1  ≤ 2M + 2. Thus we have shown by induction that (3.136) holds for all t = 0, . . . , T . Together with property (a) this implies that (3.134) holds for all t = 0, . . . , T . By (3.116) and (3.136), xt  ≤ 3M + 2, t = 0, . . . , T . It follows from (3.134) that T 

a(f (xt ) − f (z))

t=0

≤ 2−1 z − x0 2 + (T + 1)(4δC (5M + 7) + aδf (10M + 13) +2−1 a 2 (L + 1)2 + aδC (L + 1)). By the relation above, (3.61), (3.62), (3.116), and (3.120), T  (T + 1)−1 f (xt ) − inf(f, D), t=0

≤ δf (10M + 13) + δC (L + 1) −1/2

+4δC (5M + 7)2(L + 1)δC

+ 2−2 (L + 1)δC

1/2

110

3 The Mirror Descent Algorithm 1/2

+4M 2 (L + 1)8 max{δC , δf } 1/2

1/2

≤ 32M 2 (L + 1) max{δC , δf } + 44M(L + 1) max{δC , δf } 1/2

≤ 64M 2 (L + 1) max{δC , δf }.  

This implies (3.68). Theorem 3.5 is proved.

3.8 Zero-Sum Games on Bounded Sets Let (X, ·, ·), (Y, ·, ·) be Hilbert spaces equipped with the complete norms  ·  which are induced by their inner products. Let C be a nonempty closed convex subset of X, D be a nonempty closed convex subset of Y , U be an open convex subset of X, and V be an open convex subset of Y such that C ⊂ U, D ⊂ V .

(3.139)

Suppose that there exist L1 , L2 > 0, M1 , M2 > 0 such that C ⊂ BX (0, M1 ), D ⊂ BY (0, M2 ),

(3.140)

a function f : U × V → R 1 possesses the following properties: (i) for each v ∈ V , the function f (·, v) : U → R 1 is convex; (ii) for each u ∈ U , the function f (u, ·) : V → R 1 is concave, for each u ∈ U , |f (u, v1 ) − f (u, v2 )| ≤ L2 v1 − v2  for all v1 , v2 ∈ V ,

(3.141)

and that for each v ∈ V , |f (u1 , v) − f (u2 , v)| ≤ L1 u1 − u2  for all u1 , u2 ∈ U.

(3.142)

Recall that for each (ξ, η) ∈ U × V , ∂x f (ξ, η) = {l ∈ X : f (y, η) − f (ξ, η) ≥ l, y − ξ  for all y ∈ U }, ∂y f (ξ, η) = {l ∈ Y : l, y − η ≥ f (ξ, y) − f (ξ, η) for all y ∈ V }.

(3.143)

3.8 Zero-Sum Games on Bounded Sets

111

In view of properties (i) and (ii) and (3.141)–(3.143), for each ξ ∈ U and each η ∈ V, ∅ = ∂x f (ξ, η) ⊂ BX (0, L1 ),

(3.144)

∅ = ∂y f (ξ, η) ⊂ BY (0, L2 ).

(3.145)

x∗ ∈ C and y∗ ∈ D

(3.146)

f (x∗ , y) ≤ f (x∗ , y∗ ) ≤ f (x, y∗ )

(3.147)

Assume that

satisfy

for each x ∈ C and each y ∈ D. Let δf,1 , δf,2 , δC , δD ∈ (0, 1], and {ak }∞ k=0 ⊂ (0, ∞). Let us describe our algorithm. Mirror descent algorithm for zero-sum games Initialization: select arbitrary x0 ∈ U and y0 ∈ V . Iterative Step: given current iteration vectors xt ∈ U and yt ∈ V calculate ξt ∈ ∂x f (xt , yt ) + BX (0, δf,1 ), ηt ∈ ∂y f (xt , yt ) + BY (0, δf,2 ) and the next pair of iteration vectors xt+1 ∈ U , yt+1 ∈ V such that BX (xt+1 , δC ) ∩ argmin{ξt , v + (2at )−1 v − xt 2 : v ∈ C} = ∅, BY (yt+1 , δD ) ∩ argmin{−ηt , u + (2at )−1 u − yt 2 : u ∈ D} = ∅. In this chapter we prove the following result. Theorem 3.6 Let δf,1 , δf,2 , δC , δD ∈ (0, 1], and {ak }∞ k=0 ⊂ (0, ∞). Assume that ∞ ∞ ∞ {xt }∞ t=0 ⊂ U , {yt }t=0 ⊂ V , {ξt }t=0 ⊂ X, {ηt }t=0 ⊂ Y , BX (x0 , δC ) ∩ C = ∅, BY (y0 , δD ) ∩ D = ∅ and that for each integer t ≥ 0, ξt ∈ ∂x f (xt , yt ) + BX (0, δf,1 ), ηt ∈ ∂y f (xt , yt ) + BY (0, δf,2 ),

(3.148)

112

3 The Mirror Descent Algorithm

BX (xt+1 , δC ) ∩ argmin{ξt , v + (2at )−1 v − xt 2 : v ∈ C} = ∅, BY (yt+1 , δD ) ∩ argmin{−ηt , u + (2at )−1 u − yt 2 : u ∈ D} = ∅. Let for each natural number T ,  xT = (

T 

at )−1

T 

i=0

 yT = (

T 

at xt ,

t=0

at )−1

T 

i=0

at yt .

t=0

Then for each natural number T , xT , δC ) ∩ C = ∅, BX ( yT , δD ) ∩ D = ∅, BY ( |(

T 

at )−1

t=0

T 

at f (xt , yt ) − f (x∗ , y∗ )|

t=0

≤ max{δf,1 (2M1 + 1), δf,2 (2M2 + 1)} +2 max{δC (L1 + 1), δD (L2 + 1)} T  +(T + 1)( at )−1 max{4δC (M1 + 1), 4δD (M2 + 1)} t=0

+2−1 (

T 

at )−1

t=0

+2−1 (

T 

at2 max{(L1 + 1)2 , (L2 + 1)2 }

t=0 T 

at )−1 (max{2M1 , 2M2 } + 1)2 ,

t=0

yT ) − ( |f ( xT , 

T  t=0

at )−1

T 

at f (xt , yt )|

t=0

≤ max{δf,1 (2M1 + 1), δf,2 (2M2 + 1)} +2 max{δC (L1 + 1), δD (L2 + 1)}

(3.149)

3.8 Zero-Sum Games on Bounded Sets

113

T  +(T + 1)( at )−1 max{4δC (M1 + 1), 4δD (M2 + 1)} t=0

+2−1 (

T 

at )−1

t=0

T 

at2 max{(L1 + 1)2 , (L2 + 1)2 }

t=0

+2−1 (

T 

at )−1 (max{2M1 , 2M2 } + 1)2 ,

t=0

and for each natural number T , each z ∈ C, and each u ∈ D, f (z,  yT ) ≥ f ( xT ,  yT ) −2 max{δf,1 (2M1 + 1), δf,2 (2M2 + 1)} −3 max{δC (L1 + 1), δD (L2 + 1)} T  at )−1 max{4δC (M1 + 1), 4δD (M2 + 1)} −2(T + 1)( t=0 T T   at )−1 at2 max{(L1 + 1)2 , (L2 + 1)2 } −2( t=0

t=0

−(

T 

at )−1 (max{2M1 , 2M2 } + 1)2 ,

t=0

f ( xT , u) ≤ f ( xT ,  yT ) +2 max{δf,1 (2M1 + 1), δf,2 (2M2 + 1)} +3 max{δC (L1 + 1), δD (L2 + 1)} T  +2(T + 1)( at )−1 max{4δC (M1 + 1), 4δD (M2 + 1)} t=0 T T   at )−1 at2 max{(L1 + 1)2 , (L2 + 1)2 } +2( t=0

t=0

T  at )−1 (max{2M1 , 2M2 } + 1)2 . +( t=0

114

3 The Mirror Descent Algorithm

Proof Evidently, (3.149) holds. It is not difficult to see that xt  ≤ M1 + 1, yt  ≤ M2 + 1, t = 0, 1, . . . . Let t ≥ 0 be an integer. Applying Lemma 3.2 with a = at , x = xt , f = f (·, yt ), ξ = ξt , u = xt+1 we obtain that for each z ∈ C, at (f (xt , yt ) − f (z, yt )) ≤ 2−1 z − xt 2 − 2−1 z − xt+1 2 +at δf,1 (2M1 + 1) + at δC (L1 + 1) +4δC (M1 + 1) + 2−1 at2 (L1 + 1)2 . Applying Lemma 3.2 with a = at , x = yt , f = −f (xt , ·), ξ = −ηt , u = yt+1 we obtain that for each v ∈ D, at (f (xt , v) − f (xt , yt )) ≤ 2−1 v − yt 2 − 2−1 v − yt+1 2 +at δf,2 (2M2 + 1) + at δD (L2 + 1) +4δD (M2 + 1) + 2−1 at2 (L2 + 1)2 . For all integers t ≥ 0 set bt,1 = at δf,1 (2M1 + 1) + at δC (L1 + 1) +4δC (M1 + 1) + 2−1 at2 (L1 + 1)2 , bt,2 = at δf,2 (2M2 + 1) + at δD (L2 + 1) +4δD (M2 + 1) + 2−1 at2 (L2 + 1)2 and define

3.8 Zero-Sum Games on Bounded Sets

115

φ(s) = 2−1 s 2 , s ∈ R 1 . It is easy to see that all the assumptions of Proposition 2.13 hold and it implies Theorem 3.6.   We are interested in the optimal choice of at , t = 0, 1, . . . . Let T be a natural number and AT = Tt=0 at be given. By Theorem 3.6, in order to make the best  choice of at , t = 0, . . . , T , we need to minimize the function Tt=0 at2 on the set {a = (a0 , . . . , aT ) ∈ R T +1 : ai ≥ 0, i = 0, . . . , T ,

T 

ai = AT }.

i=0

By Lemma 2.3, this function has a unique minimizer a ∗ = (a0∗ , . . . , aT∗ ) where ai∗ = (T + 1)−1 AT , i = 0, . . . , T which is the best choice of at , t = 0, 1, . . . , T . Let T be a natural number and at = a for all t = 0, . . . , T . Now we will find the best a > 0. Since T can be arbitrarily large we need to choice a which is a minimizer of the function Ψ (a) = a −1 max{4δC (M1 + 1), 4δD (M2 + 1)} + a max{(L1 + 1)2 , (L2 + 1)2 }. This function has a minimizer a = (max{4δC (M1 + 1), 4δD (M2 + 1)})1/2 max{(L1 + 1)2 , (L2 + 1)2 }−1 . Now our goal is to find the best T > 0. Since in view of the inequalities above, in order to make the best choice of T , it should be at the same order as max{δC , δD }−1 . For example, T = max{δC , δD }−1 . In this case, we obtain a pair of points  x ∈ U,  y ∈ V such that x , δC ) ∩ C = ∅, BX ( y , δC ) ∩ D = ∅ BY ( and for each z ∈ C and each v ∈ D, f (z,  y ) ≥ f ( x,  y) −c max{δC , δD }1/2 − 2 max{δf,1 , δf,2 }(2 max{M1 , M2 } + 1) and

116

3 The Mirror Descent Algorithm

f ( x , v) ≤ f ( x,  y) +c max{δC , δD }1/2 + 2 max{δf,1 , δf,2 }(2 max{M1 , M2 } + 1) where the constant c > 0 depends only on L1 , L2 and M1 , M2 . Theorem 3.6 is a generalization of Theorem 3.4 of  obtained in the case when δf,1 = δf,2 = δC = δD .

3.9 Zero-Sum Games on Unbounded Sets Let (X, ·, ·), (Y, ·, ·) be Hilbert spaces equipped with the complete norms  ·  which are induced by their inner products. In this section we prove an extension of Theorem 3.6 for games on unbounded sets. This result has no prototype in , where only games on bounded sets are studied. Let A be a nonempty closed convex subset of X, D be a nonempty closed convex subset of Y , W be an open convex subset of X, and V be an open convex subset of Y such that A ⊂ W, D ⊂ V and let a function f : W × V → R 1 possess the following properties: (i) for each v ∈ V , the function f (·, v) : W → R 1 is convex and continuous; (ii) for each w ∈ W , the function f (w, ·) : V → R 1 is concave and continuous. Suppose that M0 , M1 > 0, M2 ≥ M0

(3.150)

θ0 ∈ A ∩ BX (0, M0 ),

(3.151)

D ⊂ BY (0, M0 ),

(3.152)

L1 ≥ 1, L2 > 0

(3.153)

sup{f (θ0 , y) : y ∈ V ∩ BY (0, M0 + 1)} < M1 ,

(3.154)

for each y ∈ V ∩ BY (0, M0 + 1), {x ∈ W : f (x, y) ≤ M1 + 4} ⊂ BX (0, M2 ), for each v ∈ V ∩ BY (0, M0 + 4), |f (u1 , v) − f (u2 , v)| ≤ L1 u1 − u2 

(3.155)

3.9 Zero-Sum Games on Unbounded Sets

for all u1 , u2 ∈ W ∩ BX (0, 10M2 + 12),

117

(3.156)

and that for each u ∈ W ∩ BX (0, 10M2 + 12), |f (u, v1 ) − f (u, v2 )| ≤ L2 v1 − v2  for all v1 , v2 ∈ V ∩ BY (0, M0 + 2).

(3.157)

x∗ ∈ A and y∗ ∈ D

(3.158)

f (x∗ , y) ≤ f (x∗ , y∗ ) ≤ f (x, y∗ )

(3.159)

Let

satisfy

for each x ∈ A and each y ∈ D. In view of (3.151), (3.152), (3.154), (3.155), (3.158), and (3.159), f (x∗ , y∗ ) ≤ f (θ0 , y∗ ) < M1 ,

(3.160)

x∗  ≤ M2 , θ0  ≤ M2 .

(3.161)

In this section we prove the following result which does not have an analog in . Theorem 3.7 Let δf,1 , δf,2 , δA , δD ∈ (0, 1/2) and {ak }∞ k=0 ⊂ (0, ∞) satisfy δA ≤ (36M2 + 44)−1 (L1 + 1)−2 ,

(3.162)

δf,1 ≤ (18M2 + 21)−1 ,

(3.163)

δA (36M2 + 44) ≤ at ≤ (L1 + 1)−2 for all integers t ≥ 0.

(3.164)

∞ ∞ ∞ Assume that {xt }∞ t=0 ⊂ W , {yt }t=0 ⊂ V , {ξt }t=0 ⊂ X, {ηt }t=0 ⊂ Y ,

BX (x0 , δA ) ∩ A = ∅, BY (y0 , δD ) ∩ D = ∅,

(3.165)

x0  ≤ M2 + 1

(3.166)

ξt ∈ ∂x f (xt , yt ) + BX (0, δf,1 ),

(3.167)

ηt ∈ ∂y f (xt , yt ) + BY (0, δf,2 ),

(3.168)

and that for each integer t ≥ 0,

118

3 The Mirror Descent Algorithm

BX (xt+1 , δA ) ∩ argmin{ξt , v + (2at )−1 v − xt 2 : v ∈ A} = ∅,

(3.169)

BY (yt+1 , δD ) ∩ argmin{−ηt , u + (2at )−1 u − yt 2 : u ∈ D} = ∅.

(3.170)

Let for each natural number T ,  xT = (

T 

T 

at )−1

i=0

at xt ,  yT = (

t=0

T 

at )−1

i=0

T 

at yt .

(3.171)

t=0

Then for all integers t ≥ 0, xt  ≤ 5M2 + 4

(3.172)

and for each natural number T , xT , δA ) ∩ A = ∅, BY ( yT , δD ) ∩ D = ∅, BX ( |(

T 

T 

at )−1

t=0

(3.173)

at f (xt , yt ) − f (x∗ , y∗ )|

t=0

≤ max{δf,1 (18M2 + 21), δf,2 (2M0 + 1)} +2 max{δA (L1 + 1), δD (L2 + 1)} T  +(T + 1)( at )−1 max{4δA (9M2 + 11), 4δD (M0 + 18)} t=0

−1

+2

(

T  t=0

at )

−1

T 

at2 max{(L1 + 1)2 , (L2 + 1)2 }

t=0

+ 2−1 (

T 

at )−1 (10M2 + 10)2 ,

t=0

yT ) − ( |f ( xT , 

T  t=0

at )−1

T 

at f (xt , yt )|

t=0

≤ max{δf,1 (18M2 + 21), δf,2 (2M0 + 1)} +2 max{δA (L1 + 1), δD (L2 + 1)}

(3.174)

3.9 Zero-Sum Games on Unbounded Sets

119

T  +(T + 1)( at )−1 max{4δA (9M2 + 11), 4δD (M0 + 18)} t=0

+2−1 (

T 

at )−1

t=0

T 

at2 max{(L1 + 1)2 , (L2 + 1)2 }

t=0

+ 2−1 (

T 

at )−1 (10M2 + 10)2

(3.175)

t=0

and for each natural number T , each z ∈ A, and each u ∈ D, xT ,  yT ) f (z,  yT ) ≥ f ( −2 max{δf,1 (18M2 + 21), δf,2 (2M0 + 1)} −3 max{δA (L1 + 1), δD (L2 + 1)} T  −2(T + 1)( at )−1 max{4δA (9M2 + 11), 4δD (M0 + 18)} t=0

−(

T 

at )−1

t=0

T 

at2 max{(L1 + 1)2 , (L2 + 1)2 }

t=0 T  −( at )−1 (10M2 + 10)2 , t=0

f ( xT , u) ≤ f ( xT ,  yT ) +2 max{δf,1 (18M2 + 21), δf,2 (2M0 + 1)} +3 max{δA (L1 + 1), δD (L2 + 1)} T  at )−1 max{4δA (9M2 + 11), 4δD (M0 + 18)} +2(T + 1)( t=0

+(

T  t=0

at )−1

T  t=0

at2 max{(L1 + 1)2 , (L2 + 1)2 }

(3.176)

120

3 The Mirror Descent Algorithm

+(

T 

at )−1 (10M2 + 10)2 .

(3.177)

t=0

Proof By (3.152) and (3.165), for all integers t ≥ 0, yt  ≤ M0 + 1 ≤ M2 + 1.

(3.178)

C = A ∩ BX (0, 9M2 + 10)

(3.179)

U = W ∩ {x ∈ X : x < 10M2 + 12}.

(3.180)

Set

and

In view of (3.151) and (3.166), x0 − θ0  ≤ x0  + θ0  ≤ 2M2 + 1.

(3.181)

Assume that t ≥ 0 is an integer and that xt − θ0  ≤ 4M2 + 4.

(3.182)

It follows from (3.151), (3.180), and (3.182) that xt  ≤ 5M2 + 4,

(3.183)

xt ∈ U ∩ BX (0, 5M2 + 4).

(3.184)

By (3.156), (3.167), (3.178), and (3.183), ξt ∈ ∂f (xt , yt ) + BX (0, 1) ⊂ BX (0, L1 + 1).

(3.185)

In view of (3.169), there exists h ∈ BX (xt+1 , δA ) ∩ argmin{ξt , w + (2at )−1 w − xt 2 : w ∈ A}.

(3.186)

By (3.151) and (3.186), ξt , h + (2at )−1 h − xt 2 ≤ ξt , θ0  + (2at )−1 θ0 − xt 2 . In view of (3.187),

(3.187)

3.9 Zero-Sum Games on Unbounded Sets

121

ξt , θ0  + (2at )−1 θ0 − xt 2 ≥ (2at )−1 h − xt 2 + ξt , h − xt  + ξt , xt .

(3.188)

It follows from (3.162), (3.185), and (3.188) that (2at )−1 (h − xt  − 1)2 − (2at )−1 ≤ (2at )−1 (h − xt 2 − h − xt ) ≤ (2at )−1 h − xt 2 − (L1 + 1)h − xt  ≤ (2at )−1 h − xt 2 + ξt , h − xt  ≤ (2at )−1 h − xt 2 + ξt , h − xt  + ξt , xt  + ξt xt .

(3.189)

By (3.161), (3.182), (3.184), (3.185), (3.188), and (3.189), (2at )−1 (h − xt  − 1)2 − (2at )−1 ≤ (2at )−1 h − xt 2 + ξt , h − xt  + ξt , xt  + (L1 + 1)(5M2 + 4) ≤ (2at )−1 θ0 − xt 2 + ξt , θ0  + (L1 + 1)(5M2 + 4) ≤ (L1 + 1)M2 + (2at )−1 (4M2 + 4)2 + (L1 + 1)(5M2 + 4).

(3.190)

In view of (3.164) and (3.190), (h − xt  − 1)2 ≤ 1 + (2at )(L1 + 1)(6M2 + 4) + (4M2 + 4)2 ≤ (4M2 + 4)2 + 6M2 + 5 and h − xt  ≤ 4M2 + 6.

(3.191)

It follows from (3.183) and (3.191) that h ≤ 9M2 + 10. By (3.179), (3.186), and (3.192),

(3.192)

122

3 The Mirror Descent Algorithm

h ∈ C.

(3.193)

Relations (3.179), (3.186), and (3.193) imply that h ∈ argmin{ξt , w + (2at )−1 w − xt 2 : w ∈ C}.

(3.194)

By (3.180) and (3.192), xt+1 ∈ U.

(3.195)

It follows from (3.158), (3.167), (3.179), (3.180), (3.184), (3.186), (3.194), (3.195), and Lemma 3.2 applied with M0 = 9M2 + 10, f = f (·, yt ), a = at , x = xt , ξ = ξt , u = xt+1 that for each z ∈ C, at (f (xt , yt ) − f (z, yt )) ≤ 2−1 z − xt 2 − 2−1 z − xt+1 2 +at δf,1 (18M2 + 21) + at δA (L1 + 1) + 4δA (9M2 + 11) + 2−1 at2 (L1 + 1)2 .

(3.196)

By (3.157), (3.168), (3.170), (3.178)–(3.180), and Lemma 3.2 applied with a = at , x = yt , f = −f (xt , ·), ξ = −ηt , u = yt+1 that for each v ∈ D, at (f (xt , v) − f (xt , yt )) ≤ 2−1 v − yt 2 − 2−1 v − yt+1 2 +at δf,2 (2M0 + 1) + at δD (L2 + 1) + 4δD (M0 + 1) + 2−1 at2 (L2 + 1)2 .

(3.197)

θ0 − xt+1  ≤ θ0 − xt ;

(3.198)

θ0 − xt+1  > θ0 − xt .

(3.199)

There are two cases:

3.9 Zero-Sum Games on Unbounded Sets

123

If (3.198) holds, then in view of (3.182), θ0 − xt+1  ≤ 4M2 + 4.

(3.200)

Assume that (3.199) is true. It follows from (3.151), (3.154), (3.163)–(3.165), (3.178), (3.179), and (3.196) with z = θ0 that f (xt , yt ) ≤ f (θ0 , yt ) +δf,1 (18M2 + 21) + δA (L1 + 1) + 4at−1 δA (9M2 + 11) + 1 ≤ M1 + 4.

(3.201)

By (3.155), (3.178), and (3.201), xt  ≤ M2 .

(3.202)

It follows from (3.151), (3.185), (3.188), (3.189), and (3.202) that (2at )−1 (h − xt  − 1)2 − (2at )−1 ≤ (2at )−1 h − xt 2 + ξt , h − xt  + ξt , xt  + (L1 + 1)M2 ≤ (2at )−1 θ0 − xt 2 + ξt , θ0  + (L1 + 1)M2 ≤ (L1 + 1)M0 + (2at )−1 (2M2 + 1)2 + (L1 + 1)M2 .

(3.203)

In view of (3.164), (3.202), and (3.203), (h − xt  − 1)2 ≤ 1 + (M0 + M2 ) + (2M2 + 1)2 ≤ 4M22 + 6M2 + 2 ≤ 4(M2 + 1)2 , and h − xt  ≤ 1 + 2M2 + 2, h ≤ 3M2 + 3.

(3.204)

By (3.186) and (3.204), xt+1  ≤ 3M2 + 4.

(3.205)

xt+1 − θ0  ≤ 4M2 + 4.

(3.206)

In view of (3.151) and (3.205),

124

3 The Mirror Descent Algorithm

Thus we have shown by induction that for all integers t ≥ 0, (3.206) holds, (3.196) is true for all z ∈ C = A ∩ BX (0, 9M2 + 10), and (3.197) holds for all v ∈ D. It follows from (3.155) and (3.206) that (3.172) is true for all integers t ≥ 0. For all integers t ≥ 0 set bt,1 = at δf,1 (18M2 + 21) + at δA (L1 + 1) +4δA (9M2 + 11) + 2−1 at2 (L2 + 1)2 , bt,2 = at δf,2 (2M0 + 1) + at δD (L2 + 1) +4δD (M0 + 1) + 2−1 at2 (L2 + 1)2 and define φ(s) = 2−1 s 2 , s ∈ R 1 . It is easy to see that all the assumptions of Proposition 2.13 hold and it implies (3.173)–(3.175) for each natural number T and (3.176) and (3.177) for each natural number T , each u ∈ D, and each z ∈ A ∩ BX (9M2 + 10). By (3.154), (3.155), and (3.178), for each z ∈ A \ BX (9M2 + 10), f (z,  yT ) > M1 + 4 > 4 + f (θ0 ,  yT ) and (3.176) holds for all natural numbers T and all z ∈ A. Theorem 3.7 is proved.   We are interested in the optimal choice of at , t = 0, 1, . . . . Let T be a natural number and AT = Tt=0 at be given. By Theorem 3.7, in order to make the best  choice of at , t = 0, . . . , T , we need to minimize the function Tt=0 at2 on the set {a = (a0 , . . . , aT ) ∈ R T +1 : ai ≥ 0, i = 0, . . . , T ,

T 

ai = AT }.

i=0

By Lemma 2.3, this function has a unique minimizer a ∗ = (a0∗ , . . . , aT∗ ) where ai∗ = (T + 1)−1 AT , i = 0, . . . , T which is the best choice of at , t = 0, 1, . . . , T . Let T be a natural number and at = a for all t = 0, . . . , T . Now we will find the best a > 0. Since T can be arbitrarily large we need to choice a which is a minimizer of the function

3.9 Zero-Sum Games on Unbounded Sets

125

Ψ (a) = 2a −1 max{4δA (9M2 + 1), 4δD (M0 + 18)} +a max{(L1 + 1)2 , (L2 + 1)2 }, a > 0. This function has a minimizer a = (2 max{4δA (9M2 + 1), 4δD (M0 + 18)})1/2 max{(L1 + 1), (L2 + 1)}−1 and the minimal value of Ψ is (8 max{4δA (9M2 + 1), 4δD (M0 + 18)})1/2 max{(L1 + 1), (L2 + 1)}. Now our goal is to find the best T > 0. In view of the inequalities above, in order to make the best choice of T , it should be at the same order as max{δA , δD }−1 . For example, T = max{δA , δD }−1  + 1. In this case, we obtain a pair of points  x ∈ U,  y ∈ V such that x , δA ) ∩ A = ∅, BX ( y , δD ) ∩ D = ∅ BY ( and for each z ∈ A and each v ∈ D, f (z,  y ) ≥ f ( x,  y) −c max{δA , δD }1/2 − 2 max{δf,1 , δf,2 } max{18M2 + 21, 2M0 + 1} and f ( x , v) ≤ f ( x,  y) +c max{δA , δD }1/2 + 2 max{δf,1 , δf,2 } max{18M2 + 21, 2M0 + 1} where the constant c > 0 depends only on L1 , L2 and M0 , M1 , M2 .

Chapter 4

Gradient Algorithm with a Smooth Objective Function

In this chapter we analyze the convergence of a projected gradient algorithm with a smooth objective function under the presence of computational errors. The problem is described by an objective function and a set of feasible points. For this algorithm each iteration consists of two steps. The first step is a calculation of a gradient of the objective function while in the second one we calculate a projection on the feasible set. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this.

4.1 Optimization on Bounded Sets Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · . Let C be a nonempty closed convex subset of X, U be an open convex subset of X such that C ⊂ U , and f : U → R 1 be a convex continuous function. We suppose that the function f is Fréchet differentiable at every point x ∈ U and for every x ∈ U we denote by f  (x) ∈ X the Fréchet derivative of f at x. It is clear that for any x ∈ U and any h ∈ X f  (x), h = lim t −1 (f (x + th) − f (x)). t→0

(4.1)

Recall that for each nonempty set D and each function g : D → R 1 , inf(g, D) = inf{g(y) : y ∈ D}, © Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_4

(4.2) 127

128

4 Gradient Algorithm with a Smooth Objective Function

argmin(g, D) = argmin{g(z) : z ∈ D} = {z ∈ D : g(z) = inf(g, D)}. We suppose that the mapping f  : U → X is Lipschitz on all bounded subsets of U . It is well known (see Lemma 2.2) that for each nonempty closed convex set D ⊂ X and each x ∈ X there exists a unique point PD (x) ∈ D such that x − PD (x) = inf{x − y : y ∈ D}. In this chapter we study the behavior of a projected gradient algorithm with a smooth objective function which is used for solving convex constrained minimization problems [67, 68, 72]. Suppose that there exist L > 1, M0 > 0 such that C ⊂ BX (0, M0 ),

(4.2)

|f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ U,

(4.3)

f  (v1 ) − f  (v2 ) ≤ Lv1 − v2  for all v1 , v2 ∈ U.

(4.4)

Let δf , δC ∈ (0, 1]. We describe below our algorithm. Gradient algorithm Initialization: select an arbitrary x0 ∈ U ∩ BX (0, M0 ). Iterative Step: given a current iteration vector xt ∈ U calculate ξt ∈ f  (xt ) + BX (0, δf ) and calculate the next iteration vector xt+1 ∈ U such that xt+1 − PC (xt − L−1 ξt ) ≤ δC . In this chapter we prove the following result. Theorem 4.1 Let δf , δC ∈ (0, 1] and let x0 ∈ U ∩ BX (0, M0 ).

(4.5)

∞ Assume that {xt }∞ t=1 ⊂ U , {ξt }t=0 ⊂ X and that for each integer t ≥ 0,

ξt − f  (xt ) ≤ δ

(4.6)

xt+1 − PC (xt − L−1 ξt ) ≤ δC .

(4.7)

and

4.2 Auxiliary Results

129

Then for each natural number T , min{f (xt ) : t = 2, . . . , T + 1} − inf(f, C), T +1

f(

T −1 xt ) − inf(f, C)

t=2

≤ (2T )−1 L(2M0 + 1)2 + LδC (6M0 + 7) + δf (4M0 + 4).

(4.8)

Theorem 4.1 is a generalization of Theorem 4.2 of  proved in the case when δf = δC . We are interested in an optimal choice of T . If we choose T in order to minimize the right-hand side of (4.8) we obtain that T should be at the same order as max{δf , δC }−1 . In this case the right-hand side of (4.8) is at the same order as max{δf , δC }. For example, if T = max{δf , δC }−1  + 1, then the right-hand side of (4.8) does not exceed 2−1 (2M0 + 1)2 L max{δf , δC } + δC (6M0 + 7)L + (4M0 + 5)δf .

4.2 Auxiliary Results Proposition 4.2 Let D be a nonempty closed convex subset of X, x ∈ X, and y ∈ D. Assume that for each z ∈ D, z − y, x − y ≤ 0.

(4.9)

Then y = PD (x). Proof Let z ∈ D. By (4.9), z − x, z − x = z − y + (y − x), z − y + (y − x) = y − x, y − x + 2z − y, y − x + z − y, z − y ≥ y − x, y − x + z − y, z − y = y − x2 + z − y2 . Thus y = PD (x). Proposition 4.2 is proved.

 

Proposition 4.3 Assume that x, u ∈ U , L > 0 and that for each v1 , v2 ∈ {tx + (1 − t)u : t ∈ [0, 1]}, f  (v1 ) − f  (v2 ) ≤ Lv1 − v2 .

130

4 Gradient Algorithm with a Smooth Objective Function

Then f (u) ≤ f (x) + f  (x), u − x + 2−1 Lu − x2 . Proof For each t ∈ [0, 1] set φ(t) = f (x + t (u − x)).

(4.10)

Clearly, φ is a differentiable function and for each t ∈ [0, 1], φ  (t) = f  (x + t (u − x)), u − x.

(4.11)

By (4.10), (4.11) and the proposition assumptions, f (u) − f (x) = φ(1) − φ(0)  =

1

φ  (t)dt =

0



1

=



1

f  (x + t (u − x)), u − xdt

0

f  (x), u − xdt +

0



1

f  (x + t (u − x)) − f  (x), u − xdt

0

≤ f  (x), u − x +



1

Ltu − x2 dt

0





1

= f (x), u − x + Lu − x

2

tdt 0

= f  (x), u − x + Lu − x2 /2.  

Proposition 4.3 is proved.

4.3 The Main Lemma Lemma 4.4 Let δf , δC ∈ (0, 1], u ∈ BX (0, M0 + 1) ∩ U,

(4.12)

ξ − f  (u) ≤ δf

(4.13)

ξ ∈ X satisfy

4.3 The Main Lemma

131

and let v ∈ U satisfy v − PC (u − L−1 ξ ) ≤ δC .

(4.14)

Then for each x ∈ U satisfying B(x, δC ) ∩ C = ∅

(4.15)

the following inequalities hold: f (x) − f (v) ≥ 2−1 Lx − v2 − 2−1 Lx − u2 − δC L(6M0 + 7) − δf (4M0 + 5),

(4.16)

f (x) − f (v) ≥ 2−1 Lu − v2 + Lv − u, u − x − δC L(4M0 + 5) − δf (4M0 + 5).

(4.17)

Proof For each x ∈ U define g(x) = f (u) + f  (u), x − u + 2−1 Lx − u2 .

(4.18)

Clearly, g : U → R 1 is a convex Fréchet differentiable function, for each x ∈ U , g  (x) = f  (u) + L(x − u), lim g(x) = ∞

x→∞

(4.19) (4.20)

and there exists x0 ∈ C

(4.21)

g(x0 ) ≤ g(x) for all x ∈ C.

(4.22)

such that

By (4.21) and (4.22), for all z ∈ C, g  (x0 ), z − x0  ≥ 0.

(4.23)

132

4 Gradient Algorithm with a Smooth Objective Function

In view of (4.19) and (4.23), L−1 f  (u) + x0 − u, z − x0  ≥ 0 for all z ∈ C.

(4.24)

Proposition 4.2, (4.21), and (4.24) imply that x0 = PC (u − L−1 f  (u)).

(4.25)

It follows from (4.13), (4.14), (4.25), and Lemma 2.2 that v − x0  ≤ v − PC (u − L−1 ξ ) +PC (u − L−1 ξ ) − PC (u − L−1 f  (u)) ≤ δC + L−1 ξ − f  (u) ≤ δC + L−1 δf .

(4.26)

In view of (4.2) and (4.25), x0  ≤ M0 .

(4.27)

Relations (4.2) and (4.14) imply that v ≤ M0 + 1.

(4.28)

By (4.4), (4.12), (4.18), and Proposition 4.3, for all x ∈ U , f (x) ≤ f (u) + f  (u), x − u + 2−1 Lu − x2 = g(x).

(4.29)

Let x∈U

(4.30)

B(x, δC ) ∩ C = ∅.

(4.31)

satisfy

It follows from (4.3), (4.26), and (4.30) that |f (x0 ) − f (v)| ≤ Lv − x0  ≤ δC L + δf .

(4.32)

In view of (4.21) and (4.29), g(x0 ) ≥ f (x0 ).

(4.33)

4.3 The Main Lemma

133

By (4.18), (4.33), and convexity of f , f (x) − f (x0 ) ≥ f (x) − g(x0 ) = f (x) − f (u) − f  (u), x0 − u − 2−1 Lu − x0 2 ≥ f (u) + f  (u), x − u −f (u) − f  (u), x0 − u − 2−1 Lu − x0 2 = f  (u), x − x0  − 2−1 Lu − x0 2 .

(4.34)

Relation (4.31) implies that there exists x1 ∈ C

(4.35)

x1 − x ≤ δC .

(4.36)

such that

By (4.3), (4.30), (4.35), and (4.36), |f (x1 ) − f (x)| ≤ LδC .

(4.37)

It follows from (4.23) (with z = x1 ) and (4.35) that 0 ≤ g  (x0 ), x1 − x0  = g  (x0 ), x1 − x + g  (x0 ), x − x0 .

(4.38)

By (4.3), (4.12), (4.20), (4.27), (4.35), and (4.36), g  (x0 ), x1 − x = f  (u) + L(x0 − u), x1 − x = f  (u), x1 − x + Lx0 − u, x1 − x ≤ LδC + LδC (2M0 + 1).

(4.39)

g  (x0 ), x − x0  = f  (u) + L(x0 − u), x − x0 .

(4.40)

In view of (4.19) and (4.21),

134

4 Gradient Algorithm with a Smooth Objective Function

Relations (4.38)–(4.40) imply that f  (u), x − x0  = g  (x0 ), x − x0  − Lx0 − u, x − x0  ≥ −g  (x0 ), x1 − x − Lx0 − u, x − x0  ≥ −Lx0 − u, x − x0  − LδC (2M0 + 2).

(4.41)

It follows from (4.34) and (4.41) that f (x) − f (x0 ) ≥ f  (u), x − x0  − 2−1 Lx0 − u2 ≥ −LC δ(2M0 + 2) − Lx0 − u, x − x0  − 2−1 Lx0 − u2 .

(4.42)

In view of (4.42) and Lemma 2.1, f (x) − f (x0 ) ≥ −LδC (2M0 + 2) − 2−1 Lx0 − u2 −2−1 L[x − u2 − x − x0 2 − u − x0 2 ] = 2−1 Lx − x0 2 − 2−1 Lx − u2 − LδC (2M0 + 2).

(4.43)

By (4.32), f (x) − f (v) ≥ f (x) − f (x0 ) − δC L − δf .

(4.44)

It follows from (4.12), (4.14), (4.25), (4.26), (4.30), and (4.31) that |x − x0 2 − x − v2 | = |x − x0  − x − v|(x − x0  + x − v) ≤ (δC + L−1 δf )(4M0 + 4)

(4.45)

and |u − x0 2 − u − v2 | = |u − x0  − u − v|(u − x0  + u − v) ≤ (δC + L−1 δf )(4M0 + 4).

(4.46)

4.3 The Main Lemma

135

In view of (4.42), f (x) − f (x0 ) ≥ −LδC (2M0 + 2) + 2−1 Lx0 − u2 −Lx0 − u, x0 − u − Lx0 − u, x − x0  ≥ −LδC (2M0 + 2) + 2−1 Lx0 − u2 − Lx0 − u, x − u.

(4.47)

By (4.32), (4.43), and (4.45), f (x) − f (v) ≥ f (x) − f (x0 ) − δC L − δf ≥ 2−1 Lx − x0 2 − 2−1 Lx − u2 −LδC (2M0 + 2) − LδC − δf ≥ 2−1 Lx − v2 − 2−1 Lx − u2 −L(δC + L−1 δf )(4M0 + 4) −LδC (2M0 + 2) − LδC − δf and (4.16) holds. It follows from (4.12), (4.26), (4.31), (4.32), (4.46), and (4.47) that f (x) − f (v) ≥ f (x) − f (x0 ) − δC L − δf ≥ −LδC (2M0 + 2) + 2−1 Lx0 − u2 −Lx0 − u, x − u − LδC − δf ≥ −LδC − δf − L(δC + L−1 δf )(2M0 + 2) + 2−1 Lu − v2 −Lv − u, x − u − L(δC + L−1 δf )(2M0 + 2) and (4.17) holds. Lemma 4.4 is proved.

 

136

4 Gradient Algorithm with a Smooth Objective Function

4.4 Proof of Theorem 4.1 Clearly, the function f has a minimizer on the set C. Fix z∈C

(4.48)

f (z) = inf(f, C).

(4.49)

xt  ≤ M0 + 1, t = 0, 1, . . . .

(4.50)

such that

It is easy to see that

Let T be a natural number and t ≥ 0 be an integer. Applying Lemma 4.4 with u = xt , ξ = ξt , v = xt+1 , x = z we obtain that f (z) − f (xt+1 ) ≥ 2−1 Lz − xt+1 2 − 2−1 Lz − xt 2 −δC L(6M0 + 7) − δf (4M0 + 5). This implies that Tf (z) −

T 

f (xt+1 )

t=1

T  (2−1 Lz − xt+1 2 − 2−1 Lz − xt 2 ) t=1

−δC LT (6M0 + 7) − T δf (4M0 + 5) = 2−1 L(z − xT +1 2 − z − x1 2 ) − δC LT (6M0 + 7) − T δf (4M0 + 5). Relations (4.2), (4.48), (4.50), and (4.51) imply that T min{f (xt ) : t = 2, . . . , T + 1} − inf(f, C),

(4.51)

4.5 Optimization on Unbounded Sets

Tf (

137

T +1

T −1 xt ) − inf(f, C)

t=2

≤ 2−1 (2M0 + 1)2 L + δC LT (6M0 + 7) + T δf (4M0 + 5). Together with (4.49) this implies that min{f (xt ) : t = 2, . . . , T + 1} − inf(f, C), T +1

f(

T −1 xt ) − inf(f, C)

t=2

≤ 2−1 (2M0 + 1)2 T −1 L + δC L(6M0 + 7) + δf (4M0 + 5). This completes the proof of Theorem 4.1.

 

4.5 Optimization on Unbounded Sets Let X be a Hilbert space with an inner product ·, · which induces a complete norm  · . Let D be a nonempty closed convex subset of X, V be an open convex subset of X such that D ⊂ V, and f : V → R 1 be a convex Fréchet differentiable function which is Lipschitz on all bounded subsets of V . Set Dmin = {x ∈ D : f (x) ≤ f (y) for all y ∈ D}.

(4.52)

Dmin = ∅.

(4.53)

We suppose that

We will prove the following result.

138

4 Gradient Algorithm with a Smooth Objective Function

Theorem 4.5 Let δf , δC ∈ (0, 1], M > 0, Dmin ∩ BX (0, M) = ∅,

(4.54)

M0 = 4M + 8,

(4.55)

L ≥ 1 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, M0 + 2),

(4.56)

f  (v1 ) − f  (v2 ) ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, M0 + 2),

(4.57)

0 = δf (4M0 + 6) + δC L(6M0 + 8)

(4.58)

n0 = 2−1 L(2M + 1)2 (δf + LδC )−1  + 1.

(4.59)

and let

∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M

(4.60)

ξt − f  (xt ) ≤ δf

(4.61)

xt+1 − PD (xt − L−1 ξt ) ≤ δC .

(4.62)

and that for each integer t ≥ 0,

and

Then there exists an integer q ∈ [1, n0 + 1] such that f (xq ) ≤ inf(f, D) + 0 , xi  ≤ 3M + 3, i = 0, . . . , q. Proof By (4.54) there exists z ∈ Dmin ∩ BX (0, M). By (4.56), (4.60)–(4.63), and Lemma 2.2,

(4.63)

4.5 Optimization on Unbounded Sets

139

x1 − z ≤ x1 − PD (x0 − L−1 ξ0 ) + PD (x0 − L−1 ξ0 ) − z ≤ δC + x0 − z + L−1 ξ0  ≤ 1 + 2M + L−1 (L + 1) ≤ 2M + 3.

(4.64)

In view of (4.63) and (4.64), x1  ≤ 3M + 3.

(4.65)

Assume that an integer T ≥ 0 and that for all t = 1, . . . , T + 1, f (xt ) − f (z) > 0 .

(4.66)

U = V ∩ {v ∈ X : v < M0 + 2}

(4.67)

C = D ∩ BX (0, M0 ).

(4.68)

Set

and

Assume that an integer t ∈ [0, T ] and that xt − z ≤ 2M + 3.

(4.69)

(In view of (4.60) and (4.63), our assumption is true for t = 0.) By (4.55), (4.63), and (4.68), z ∈ C ⊂ BX (0, M0 ).

(4.70)

Relations (4.55), (4.63), (4.67), and (4.69) imply that xt ∈ U ∩ BX (0, M0 + 1).

(4.71)

It follows from (4.56), (4.61), and (4.71) that ξt ∈ f  (xt ) + BX (0, 1) ⊂ BX (0, L + 1). By (4.63), (4.69), (4.72), and Lemma 2.2, z − PD (xt − L−1 ξt ) ≤ z − xt + L−1 ξt 

(4.72)

140

4 Gradient Algorithm with a Smooth Objective Function

≤ z − xt  + L−1 ξt  ≤ 2M + 5.

(4.73)

In view of (4.63) and (4.73), PD (xt − L−1 ξt ) ≤ 3M + 5.

(4.74)

Relations (4.55), (4.68), and (4.74) imply that PD (xt − L−1 ξt ) ∈ C,

(4.75)

PD (xt − L−1 ξt ) = PC (xt − L−1 ξt ).

(4.76)

It follows from (4.62), (4.67), and (4.74) that xt+1  ≤ 3M + 6, xt+1 ∈ U.

(4.77)

By (4.56), (4.57), (4.61)–(4.63), (4.68), (4.71), (4.77), and Lemma 4.4 applied with u = xt , ξ = ξt , v = xt+1 , x = z we obtain that f (z) − f (xt+1 ) ≥ 2−1 Lz − xt+1 2 − 2−1 Lz − xt 2 − LδC (6M0 + 7) − δf (4M0 + 5).

(4.78)

By (4.58), (4.66), and (4.78), LδC (6M0 + 8) + δf (4M0 + 6) = 0 < f (xt+1 ) − f (z) ≤ 2−1 L(z − xt 2 − z − xt+1 2 ) + LδC (6M0 + 7) + δf (4M0 + 5). In view of (4.69) and (4.79), z − xt+1  ≤ z − xt  ≤ 2M + 3. Thus by induction we showed that for all t = 0, . . . , T + 1

(4.79)

4.5 Optimization on Unbounded Sets

141

z − xt  ≤ 2M + 3, xt  ≤ 3M + 3 and that (4.79) holds for all t = 0, . . . , T . It follows from (4.79) that (1 + T )(LδC (6M0 + 8) + δf (4M0 + 6)) ≤ (1 + T )(min{f (xt ) : t = 1, . . . , T + 1} − f (z)) ≤

T  (f (xt+1 ) − f (z)) t=0

−1

≤2

L

T 

(z − xt 2 − z − xt+1 2 )

t=0

+(T + 1)(LδC (6M0 + 7) + δf (4M0 + 5)). By the relation above, (4.61) and (4.63), (T + 1)(δf + LδC ) ≤ 2−1 Lz − x0 2 ≤ 2−1 L(2M + 1)2 and T < 2−1 L(2M + 1)2 (δf + LδC )−1 ≤ n0 . Thus we assumed that an integer T ≥ 0 satisfies f (xt ) − f (z) > 0 , t = 1, . . . , T + 1 and showed that T ≤ n0 − 1 and xt  ≤ 3M + 3, t = 0, . . . , T + 1. This implies that there exists a natural number q ≤ n0 + 1 such that f (xq ) − f (z) ≤ 0 , xt  ≤ 3M + 3, t = 0, . . . , q. Theorem 4.5 is proved. The next result has no prototype in .

 

142

4 Gradient Algorithm with a Smooth Objective Function

Theorem 4.6 Let δf , δC ∈ (0, 1], M > 2 satisfy {x ∈ V : f (x) ≤ inf(f, D) + 3} ⊂ BX (0, M − 2),

(4.80)

L ≥ 1 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 8),

(4.81)

f  (v1 ) − f  (v2 ) ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 8),

(4.82)

δC ≤ (L(18M + 19))−1 , δf ≤ (12M + 3)−1 .

(4.83)

∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M

(4.84)

ξt − f  (xt ) ≤ δf

(4.85)

xt+1 − PD (xt − L−1 ξt ) ≤ δC .

(4.86)

and that for each integer t ≥ 0,

and

Then xt  ≤ 3M for all integers t ≥ 0 and for each natural number T , min{f (xt ) : t = 1, . . . , T + 1} − inf(f, D), T +1

f(

(T + 1)−1 xt ) − inf(f, D)

t=1

≤ LδC (18M + 19) + δf (12M + 13) + 2LM 2 (T + 1)−1 . Proof Fix z ∈ Dmin .

(4.87)

4.5 Optimization on Unbounded Sets

143

Set U = V ∩ {v ∈ X : v < 3M + 4}

(4.88)

C = D ∩ BX (0, 3M + 2).

(4.89)

and

In view of (4.80) and (4.87), z ≤ M − 2.

(4.90)

x0 − z ≤ 2M.

(4.91)

By (4.84) and (4.90),

Assume that t ≥ 0 is an integer and that xt − z ≤ 2M.

(4.92)

z ∈ C ∩ BX (0, M − 1).

(4.93)

By (4.87), (4.89), and (4.90),

Relations (4.88), (4.90), and (4.92) imply that xt ∈ U ∩ BX (0, 3M).

(4.94)

xt+1 − PD (xt − L−1 ξt ) ≤ 1.

(4.95)

In view of (4.86),

It follows from (4.81), (4.85), and (4.94) that ξt ∈ f  (xt ) + BX (0, 1) ⊂ BX (0, L + 1).

(4.96)

By (4.88), (4.92), (4.96), and Lemma 2.2, z − PD (xt − L−1 ξt ) ≤ z − xt + L−1 ξt  ≤ z − xt  + L−1 ξt  ≤ 2M + L−1 (L + 1) ≤ 2M + 2.

(4.97)

144

4 Gradient Algorithm with a Smooth Objective Function

In view of (4.90) and (4.97), PD (xt − L−1 ξt ) ≤ 3M + 2.

(4.98)

Relations (4.89) and (4.98) imply that PD (xt − L−1 ξt ) ∈ C,

(4.99)

PD (xt − L−1 ξt ) = PC (xt − L−1 ξt ).

(4.100)

It follows from (4.83), (4.86), (4.88), and (4.98) that xt+1  ≤ PD (xt − L−1 ξt ) + δC < 3M + 3, xt+1 ∈ U.

(4.101) (4.102)

By (4.81), (4.82), (4.85), (4.86)–(4.89), (4.94), (4.101), (4.102), and Lemma 4.4 applied with M0 = 3M + 2, u = xt , ξ = ξt , v = xt+1 , x = z we obtain that f (z) − f (xt+1 ) ≥ 2−1 Lz − xt+1 2 − 2−1 Lz − xt 2 − LδC (18M + 19) − δf (12M + 13).

(4.103)

z − xt+1  ≤ z − xt ;

(4.104)

z − xt+1  > z − xt .

(4.105)

There are two cases:

If (4.104) holds, then in view of (4.92), z − xt+1  ≤ 2M. Assume that (4.105) holds. It follows from (4.83), (4.103), and (4.105) that f (xt+1 ) − f (z) ≤ LδC (18M + 19) + δf (12M + 13) ≤ 2.

(4.106)

4.5 Optimization on Unbounded Sets

145

By (4.80), (4.87), and (4.106), xt+1  ≤ M2 .

(4.107)

z − xt+1  ≤ 2M − 2.

(4.108)

In view of (4.93) and (4.107),

Thus in both cases z − xt+1  ≤ 2M and (4.103) holds. Therefore by induction we showed that for all integers t ≥ 0, xt  ≤ 3M and (4.103) holds. Let T be a natural number. By (4.91) and (4.103), T  (f (xt+1 ) − f (z)) t=0

≤ (T + 1)(LδC (18M + 19) + δf (12M + 13)) +2−1 L

T  (z − xt 2 − z − xt+1 2 ) t=0

≤ (T + 1)(LδC (18M + 19) + δf (12M + 13)) + 2LM 2 . This implies that min{f (xt ) : t = 1, . . . , T + 1} − f (z), T +1

f(

(T + 1)−1 xt ) − f (z)

t=1

≤ LδC (18M + 19) + δf (12M + 13) + 2(T + 1)−1 LM 2 . Theorem 4.6 is proved.

 

It is clear that a best choice of T should be at the same order as max{δC , δf }−1 . The next result has no prototype in .

146

4 Gradient Algorithm with a Smooth Objective Function

Theorem 4.7 Let δf , δC ∈ (0, 1), M > 1, δC ≤ (160M)−1 , δf ≤ 120−1 ,

(4.109)

Dmin ∩ BX (0, M) = ∅,

(4.110)

L ≥ 1 satisfy |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 10),

(4.111)

f  (v1 ) − f  (v2 ) ≤ Lv1 − v2  for all v1 , v2 ∈ V ∩ BX (0, 4M + 10),

(4.112)

T = 36−1 min{δC−1 , Lδf−1 }.

(4.113)

∞ Assume that {xt }∞ t=0 ⊂ V , {ξt }t=0 ⊂ X,

x0  ≤ M

(4.114)

ξt − f  (xt ) ≤ δf

(4.115)

xt+1 − PD (xt − L−1 ξt ) ≤ δC .

(4.116)

and that for each integer t ≥ 0,

and

Then xt  ≤ 3M + 3 for all integers t = 0, . . . , T + 1 and T  (T + 1)−1 (f (xt+1 ) − inf(f, D)), t=0

min{f (xt ) : t = 1, . . . , T + 1} − inf(f, D), T +1

(T + 1)−1 xt ) − inf(f, D)

f(

t=1

4.5 Optimization on Unbounded Sets

147

≤ 72(M + 1)2 max{δC L, δf }. Proof By (4.111) there exists z ∈ Dmin ∩ BX (0, M).

(4.117)

x0 − z ≤ 2M.

(4.118)

U = V ∩ {v ∈ X : v < 4M + 9}

(4.119)

C = D ∩ BX (0, 4M + 8).

(4.120)

By (4.114) and (4.117),

Set

and

Assume that t ≥ 0 is an integer and that xt − z ≤ 2M + 2.

(4.121)

xt  ≤ 3M + 2.

(4.122)

xt ∈ U ∩ BX (0, 3M + 2).

(4.123)

By (4.117) and (4.121),

In view of (4.119) and (4.122),

It follows from (4.111), (4.115), and (4.122) that ξt  ≤ L + 1.

(4.124)

By (4.117), (4.121), (4.124), and Lemma 2.2, z − PD (xt − L−1 ξt ) ≤ z − xt + L−1 ξt  ≤ z − xt  + L−1 ξt  ≤ 2M + 2 + L−1 (L + 1) ≤ 2M + 4.

(4.125)

148

4 Gradient Algorithm with a Smooth Objective Function

In view of (4.117) and (4.125), PD (xt − L−1 ξt ) ≤ 3M + 4.

(4.126)

Relations (4.120) and (4.126) imply that PD (xt − L−1 ξt ) ∈ C,

(4.127)

PD (xt − L−1 ξt ) = PC (xt − L−1 ξt ).

(4.128)

It follows from (4.109), (4.116), and (4.126) that xt+1  ≤ 3M + 5.

(4.129)

Relations (4.119) and (4.129) imply that xt+1 ∈ U ∩ BX (0, 3M + 5).

(4.130)

By (4.111), (4.112), (4.115)–(4.117), (4.123), and Lemma 4.4 applied with M0 = 4M + 8, u = xt , ξ = ξt , v = xt+1 , x = z we obtain that f (z) − f (xt+1 ) ≥ 2−1 Lz − xt+1 2 − 2−1 Lz − xt 2 − LδC (24M + 55) − δf (16M + 37).

(4.131)

It follows from (4.111), (4.116), (4.117), (4.126), and (4.129) that f (xt+1 ) ≥ f (PD (xt − L−1 ξt )) − LδC ≥ f (z) − LδC .

(4.132)

By (4.131) and (4.132), z − xt+1 2 − z − xt 2 ≤ 2δC + 2δC (24M + 55) + 2L−1 δf (16M + 37).

(4.133)

Thus we have shown that the following property holds: (a) if for an integer t ≥ 0 relation xt − z ≤ 2M + 2 holds, then (4.131) and (4.133) are true.

4.5 Optimization on Unbounded Sets

149

Let us show that for all integers t = 0, . . . , T , z − xt 2 ≤ z − x0 2 +t[2δC (24M + 55) + 2L−1 δf (16M + 37)] ≤ z − x0 2 + T [2δC (24M + 55) + 2L−1 δf (16M + 37)] ≤ 4M 2 + 8M + 4.

(4.134)

First, note that by (4.113), T [2δC (24M + 55) + 2L−1 δf (16M + 37)] = 2T δC (24M + 55) + 2T L−1 δf (16M + 37) ≤ 18−1 (24M + 55) + 18−1 (16M + 37) ≤ 5M + 4.

(4.135)

It follows from (4.121) and (4.135) that z − x0 2 + T [2δC (24M + 55) + 2L−1 δf (16M + 13)] ≤ 4M 2 + 8M + 4 = (2M + 2)2 .

(4.136)

Assume that t ∈ {0, . . . , T } \ {T } and that (4.134) holds. Then xt − z ≤ 2M + 2 and by property (a), relations (4.131) and (4.133) hold. By (4.133) and (4.134), z − xt+1 2 ≤ z − xt 2 +2δC (24M + 55) + 2L−1 δf (16M + 37) ≤ z − x0 2 + (t + 1)[2δC (24M + 55) + 2L−1 δf (16M + 37)] ≤ z − x0 2 + T [2δC (24M + 55) + 2L−1 δf (16M + 37)] ≤ 4M 2 + 8M + 4. Thus we have shown by induction that for all t = 0, . . . , T , (4.134) holds and xt − z ≤ 2M + 2.

(4.137)

150

4 Gradient Algorithm with a Smooth Objective Function

Property (a) implies that (4.131) and (4.137) hold for all t = 0, . . . , T . In view of (4.117) and (4.137), xt  ≤ 3M + 2, t = 0, . . . , T . It follows from (4.117), (4.118), and (4.131) that T  (T + 1)−1 f (xt+1 ) − inf(f, D) t=0

≤ 2−1 L(T + 1)−1 z − x0 2 +δC L(24M + 55) + δf (16M + 37) ≤ 72LM 2 max{δC , L−1 δf } + + δC L(24M + 55) + δf (16M + 37) ≤ max{δC , L−1 δf }(72LM 2 + L(24M + 55) + L(16M + 37) ≤ max{LδC , δf }(72M 2 + 40M + 92) ≤ 72 max{LδC , δf }(M + 1)2 . Property (a), (4.110), (4.133), and (4.137) with t = T imply that z − xT +1 2 − z − xT 2 ≤ 2δC (24M + 56) + 2L−1 δf (16M + 37) ≤ 4M 2 + 8M + 4 + 160MδC + 106L−1 δf ≤ 4M 2 + 8M + 6 ≤ (2M + 3)2 and xT +1 ≤ 3M + 3. Theorem 4.7 is proved.

 

Chapter 5

An Extension of the Gradient Algorithm

In this chapter we analyze the convergence of a gradient type algorithm, under the presence of computational errors, which was introduced by Beck and Teboulle  for solving linear inverse problems arising in signal/image processing. This algorithm is used for minimization of the sum of two given convex functions and each of its iteration consists of two steps. The first step is a calculation of a gradient of the first function while in the second one we solve an auxiliary minimization problem. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this.

5.1 Preliminaries and the Main Result Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · . Suppose that f : X → R 1 is a convex Fréchet differentiable function on X and for every x ∈ X denote by f  (x) ∈ X the Fréchet derivative of f at x. It is clear that for any x ∈ X and any h ∈ X f  (x), h = lim t −1 (f (x + th) − f (x)). t→0

Recall that for each function φ : X → R 1 , inf(φ) = inf{φ(y) : y ∈ X},

© Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_5

151

152

5 An Extension of the Gradient Algorithm

argmin(φ) = argmin{φ(z) : z ∈ X} = {z ∈ X : φ(z) = inf(φ)}. We suppose that the mapping f  : X → X is Lipschitz on all bounded subsets of X. Let g : X → R 1 be a convex continuous function which is Lipschitz on all bounded subsets of X. Define F (x) = f (x) + g(x), x ∈ X. We suppose that argmin(F ) = ∅

(5.1)

and that there exists c∗ ∈ R 1 such that g(x) ≥ c∗ for all x ∈ X.

(5.2)

For each u ∈ X, each ξ ∈ X, and each L > 0 define a convex function Gu,ξ (w) = f (u) + ξ, w − u + 2−1 Lw − u2 + g(w), w ∈ X (L)

(5.3)

which has a minimizer. In this chapter we analyze the gradient type algorithm, which was introduced by Beck and Teboulle in  for solving linear inverse problems, and prove the following result. Theorem 5.1 Let δf , δG ∈ (0, 1], M ≥ 1 satisfy argmin(F ) ∩ BX (0, M) = ∅,

(5.4)

L ≥ 1 satisfy |f (w1 ) − f (w2 )| ≤ Lw1 − w2  for all w1 , w2 ∈ BX (0, 3M + 2)

(5.5)

and f  (w1 ) − f  (w2 ) ≤ Lw1 − w2  for all w1 , w2 ∈ X,

(5.6)

|f (w)|, F (w) ≤ M1 for all w ∈ BX (0, 3M + 2),

(5.7)

M1 ≥ 3M satisfy

5.1 Preliminaries and the Main Result

153

M2 = 18(M1 + |c∗ | + 1 + (L + 1)2 )1/2 + 3M + 2,

(5.8)

L0 ≥ 1 satisfy |g(w1 ) − g(w2 )| ≤ L0 w1 − w2  for all w1 , w2 ∈ BX (0, M2 ),

(5.9)

δf ≤ 4−1 (M2 + 9M + 4)−1 , δG ≤ 4−1 (2 + L0 + L(M2 + 3M + 3))−1 , 0 = 2δf (M2 + 9M + 4) + 2δG (2 + L0 + L(M2 + 3M + 3))

(5.10) (5.11)

and let n0 = 4LM 2 0−1  + 2.

(5.12)

∞ Assume that {xt }∞ t=0 ⊂ X, {ξt }t=0 ⊂ X,

x0  ≤ M

(5.13)

ξt − f  (xt ) ≤ δf

(5.14)

and that for each integer t ≥ 0,

and (L)

BX (xt+1 , δG ) ∩ argmin(Gxt ,ξt ) = ∅.

(5.15)

Then there exists an integer q ∈ [0, n0 + 1] such that xi  ≤ M2 , i = 0, . . . , q and F (xq ) ≤ inf(F ) + 0 . Note that in the theorem above δf , δG are the computational errors produced by our computer system. The computational error δf is produced when we calculate a gradient of f while the computational error δG is produced when we solve the (L) auxiliary minimization problem with the objective functions Gxt ,ξt , t = 0, 1, . . . .. By Theorem 5.1, after n0 + 1 iterations we obtain an approximate solution x satisfying

154

5 An Extension of the Gradient Algorithm

F (x) ≤ inf(f ) + 0 , where 0 , n0 are defined by (5.11) and (5.12). Theorem 5.1 is a generalization of Theorem 5.1 of  proved in the case when δf = δG . In this chapter we also prove two extensions of Theorem 5.1 which do not have prototypes in .

5.2 Auxiliary Results (L)

Lemma 5.2 () Let u, ξ ∈ X and L > 0. Then the function Gu,ξ has a point of (L)

minimum and z ∈ X is a minimizer of Gu,ξ if and only if 0 ∈ ξ + L(z − u) + ∂g(z). Proof By (5.2) and (5.3), lim G(L) u,ξ (w) = ∞.

w→∞ (L)

This implies that the function Gu,ξ has a minimizer. Clearly, z is a minimizer of (L)

Gu,ξ if and only if 0 ∈ ∂G(L) u,ξ (z) = ξ + L(z − u) + ∂g(z).  

Lemma 5.2 is proved. Lemma 5.3 () Let M0 ≥ 1, L ≥ 1 satisfy |f (w1 ) − f (w2 )| ≤ Lw1 − w2  for all w1 , w2 ∈ BX (0, M0 + 2)

(5.16)

and M1 ≥ M0 satisfy |f (w)|, F (w) ≤ M1 for all w ∈ BX (0, M0 + 2).

(5.17)

Assume that u ∈ BX (0, M0 + 1),

(5.18)

ξ − f  (u) ≤ 1

(5.19)

ξ ∈ X satisfies

5.2 Auxiliary Results

155

and that v ∈ X satisfies (L)

(L)

BX (v, 1) ∩ {z ∈ X : Gu,ξ (z) ≤ inf(Gu,ξ ) + 1} = ∅.

(5.20)

v ≤ (8(M1 + |c∗ | + (L + 1)2 + 1)1/2 + M0 + 2.

(5.21)

Then

Proof In view of (5.20), there exists  v ∈ BX (v, 1)

(5.22)

(L)

(5.23)

such that (L)

v ) ≤ inf(Gu,ξ ) + 1. Gu,ξ ( By (5.1), (5.3), (5.17), (5.18), and (5.23), v − u2 + g( v) f (u) + ξ, v − u + 2−1 L = G(L) v ) ≤ G(L) u,ξ ( u,ξ (u) + 1 = F (u) + 1 ≤ M1 + 1.

(5.24)

It follows from (5.2), (5.17), (5.18), and (5.24) that v − u2 ≤ 2M1 + 1 + |c∗ |. ξ, v − u + 2−1 L

(5.25)

It is clear that v − u| ≤ L−1 (4−1  v − u2 + 4ξ 2 ). 2L−1 |ξ,

(5.26)

Since the function f is Lipschitz on BX (0, M0 + 2) relations (5.16), (5.18), and (5.19) imply that ξ  ≤ f  (u) + 1 ≤ L + 1. By (5.25)–(5.27), 2L−1 (2M1 + 1 + |c∗ |) v − u| ≥  v − u2 − 2L−1 |ξ, ≥  v − u2 − 4−1  v − u2 − 4ξ 2 ≥ 2−1  v − u2 − 4(L + 1)2 .

(5.27)

156

5 An Extension of the Gradient Algorithm

This implies that  v − u2 ≤ 4(M1 + 1 + |c∗ |) + 8(L + 1)2 and  v − u ≤ (4(M1 + 1 + |c∗ |) + 8(L + 1)2 )1/2 . Together with (5.18) and (5.22) this implies that v ≤  v  + 1 ≤  v − u + u + 1 ≤ (4(M1 + 1 + |c∗ |) + 8(L + 1)2 )1/2 + M0 + 2.  

Lemma 5.3 is proved. Lemma 5.4 Let δf , δG ∈ (0, 1], M0 ≥ 1, L ≥ 1 satisfy |f (w1 ) − f (w2 )| ≤ Lw1 − w2  for all w1 , w2 ∈ BX (0, M0 + 2), |f  (w1 ) − f  (w2 )| ≤ Lw1 − w2  for all w1 , w2 ∈ X,

(5.28) (5.29)

M1 ≥ M0 satisfy |f (w)|, F (w) ≤ M1 for all w ∈ BX (0, M0 + 2),

(5.30)

M2 = (8(M1 + |c∗ | + (L + 1)2 + 1)1/2 + M0 + 2

(5.31)

and let L0 ≥ 1 satisfy |g(w1 ) − g(w2 )| ≤ L0 w1 − w2  for all w1 , w2 ∈ BX (0, M2 ).

(5.32)

u ∈ BX (0, M0 + 1),

(5.33)

ξ − f  (u) ≤ δf

(5.34)

Assume that

ξ ∈ X satisfies

and that v ∈ X satisfies

5.2 Auxiliary Results

157 (L)

BX (v, δG ) ∩ argmin(Gu,ξ ) = ∅.

(5.35)

Then for each x ∈ BX (0, M0 + 1), F (x) − F (v) ≥ 2−1 Lv − x2 − 2−1 Lu − x2 −δf (M2 + 3M0 + 3) − δG (1 + L0 + L(M0 + M + 2 + 3)). Proof By (5.35) there exists  v ∈ X such that  v ∈ argmin(Gu,ξ )

(L)

(5.36)

v −  v  ≤ δG .

(5.37)

and

In view of the assumptions of the lemma, Lemma 5.3, (5.31), (5.35), and (5.36), v,  v  ≤ M2 .

(5.38)

x ∈ BX (0, M0 + 1).

(5.39)

F (x) = f (x) + g(x), F (v) = f (v) + g(v).

(5.40)

Let

Clearly,

Proposition 4.3 and (5.33) imply that g(v) + f (v) ≤ g(v) + f (u) + f  (u), v − u + 2−1 Lv − u2 . By (5.3), (5.33), (5.34), (5.38), (5.40), and (5.41), F (x) − F (v) ≥ [f (x) + g(x)] − [f (v) + g(v)] ≥ f (x) + g(x)

(5.41)

158

5 An Extension of the Gradient Algorithm

−[f (u) + f  (u), v − u + 2−1 Lv − u2 + g(v)] = [f (x) + g(x)] −[f (u) + ξ, v − u + 2−1 Lv − u2 + g(v)] +ξ − f  (u), v − u ≥ [f (x) + g(x)] − Gu,ξ (v) − ξ − f  (u)v − u (L)

≥ [f (x) + g(x)] − G(L) u,ξ (v) − δf (M2 + M0 + 1).

(5.42)

It follows from (5.3) that (L)

(L)

v) Gu,ξ (v) − Gu,ξ ( = ξ, v − u + 2−1 Lv − u2 + g(v) −[ξ, v − u + 2−1 L v − u2 + g( v )] v − u2 ] + g(v) − g( v ). = ξ, v −  v  + 2−1 L[v − u2 − 

(5.43)

Relations (5.28), (5.32), and (5.34) imply that ξ  ≤ f  (u) + 1 ≤ L + 1.

(5.44)

In view of (5.37) and (5.44), |ξ, v −  v | ≤ (L + 1)δG .

(5.45)

By (5.33), (5.37), and (5.38), v − u2 | |v − u2 −  ≤ |v − u −  v − u|(v − u +  v − u) ≤ v −  v (2M2 + 2M0 + 2) ≤ δG (2M2 + 2M0 + 2).

(5.46)

In view of (5.32), (5.37), and (5.38), v  ≤ L 0 δG . |g(v) − g( v )| ≤ L0 v − 

(5.47)

5.2 Auxiliary Results

159

It follows from (5.43) and (5.45–(5.47) that (L)

(L)

v )| |Gu,ξ (v) − Gu,ξ ( ≤ (L + 1)δG + 2−1 LδG (2M0 + 2M2 + 2) + L0 δG = δG (L + 1 + L0 + L(M0 + M2 + 1)).

(5.48)

Relations (5.42) and (5.48) imply that F (x) − F (v) ≥ f (x) + g(x) − G(L) u,ξ (v) − δf (M2 + M0 + 1) (L)

≥ f (x) + g(x) − Gu,ξ ( v) − δf (M2 + M0 + 1) − δG (L0 + L + 1 + L(M2 + M0 + 1)).

(5.49)

By the convexity of f , (5.33), (5.34), and (5.39), f (x) ≥ f (u) + f  (u), x − u ≥ f (u) + ξ, x − u − |f  (u) − ξ, x − u| ≥ f (u) + ξ, x − u − f  (u) − ξ x − u ≥ f (u) + ξ, x − u − δf (2M0 + 2).

(5.50)

Lemma 5.2 and (5.36) imply that there exists l ∈ ∂g( v)

(5.51)

ξ + L( v − u) + l = 0.

(5.52)

such that

In view of (5.51) and the convexity of g, g(x) ≥ g( v ) + l, x −  v . It follows from (5.50) and (5.53) that f (x) + g(x)

(5.53)

160

5 An Extension of the Gradient Algorithm

≥ f (u) + ξ, x − u − δf (2M0 + 2) + g( v ) + l, x −  v .

(5.54)

In view of (5.3), v ) = f (u) + ξ, v − u + 2−1 L v − u2 + g( v ). Gu,ξ ( (L)

(5.55)

By (5.52), (5.54), and (5.55), (L)

v) f (x) + g(x) − Gu,ξ ( ≥ ξ, x −  v  + l, x −  v  − 2−1 L v − u2 − δf (2M0 + 2) v  − 2−1 L v − u2 = −δf (2M0 + 2) + ξ + l, x −  = −2−1 L v − u2 − L v − u, x −  v  − δf (2M0 + 2) v − u2 + L v − u, u − x − δf (2M0 + 2). = 2−1 L

(5.56)

In view of (5.37)–(5.39), | v − x2 − v − x2 | ≤ | v − x − v − x|( v − x + v − x) ≤  v − v(2M2 + 2M0 + 2) ≤ δG (2M2 + 2M0 + 2).

(5.57)

Lemma 2.1 implies that v − x2 −  v − u2 − u − x2 ].  v − u, u − x = 2−1 [ By (5.56)–(5.58), (L)

v) f (x) + g(x) − Gu,ξ ( ≥ 2−1 L v − u2 + 2−1 L v − x2 −2−1 L v − u2 − 2−1 Lu − x2 − δf (2M0 + 2) v − x2 − 2−1 Lu − x2 − δf (2M0 + 2) ≥ 2−1 L ≥ 2−1 Lv − x2 − 2−1 Lu − x2

(5.58)

5.3 Proof of Theorem 5.1

161

− 2−1 LδG (2M2 + 2M0 + 2) − δf (2M0 + 2).

(5.59)

It follows from (5.49) and (5.59) that F (x) − F (v) ≥ 2−1 Lv − x2 − 2−1 Lu − x2 −δG L(M2 + M0 + 1) − δf (2M0 + 2) −δf (M2 + M0 + 1) − δG (1 + L0 + L(M2 + M0 + 2)).  

Lemma 5.4 is proved.

5.3 Proof of Theorem 5.1 By (5.4), there exists z ∈ argmin(F ) ∩ BX (0, M).

(5.60)

In view of (5.13) and (5.60), x0 − z ≤ 2M.

(5.61)

If F (x0 ) ≤ F (z) + 0 , then the assertion of the theorem holds. Let F (x0 ) > F (z) + 0 .

(5.62)

If F (x1 ) ≤ F (z) + 0 , then in view of Lemma 5.3, x1  ≤ M2 and the assertion of the theorem holds. Let F (x1 ) > F (z) + 0 . Assume that T ≥ 0 is an integer and that for all integers t = 0, . . . , T , F (xt+1 ) − F (z) > 0 .

(5.63)

We show that for all t ∈ {0, . . . , T }, xt − z ≤ 2M

(5.64)

162

5 An Extension of the Gradient Algorithm

and F (z) − F (xt+1 ) ≥ 2−1 Lz − xt+1 2 − 2−1 Lz − xt 2 − δf (M2 + 9M + 3) − δG (1 + L0 + L(3M + M2 + 3)).

(5.65)

In view of (5.61), (5.64) is true for t = 0. Assume that t ∈ {0, . . . , T } and (5.64) holds. Relations (5.60) and (5.64) imply that xt  ≤ 3M.

(5.66)

M0 = 3M.

(5.67)

Set

By (5.14), (5.15), (5.60), (5.66), (5.67), and Lemma 5.4 applied with x = z, u = xt , ξ = ξt , v = xt+1 , we have F (z) − F (xt+1 ) ≥ 2−1 Lz − xt+1 2 − 2−1 Lz − xt 2 − δf (M2 + 3M0 + 3) − δG (1 + L0 + L(M0 + M2 + 3)).

(5.68)

It follows from (5.63), (5.67), and (5.68) that 0 < F (xt+1 ) − F (z) ≤ 2−1 Lz − xt 2 − 2−1 Lz − xt+1 2 + δf (M2 + 9M + 3) − δG (1 + L0 + L(M2 + 3M + 3)).

(5.69)

In view of (5.11), (5.64), and (5.69), 2M ≥ z − xt  ≥ z − xt+1 . Thus we have shown by induction that (5.65) holds for all t = 0, . . . , T and that (5.64) holds for all t = 0, . . . , T + 1. By (5.63) and (5.65),

5.4 The First Extension of Theorem 5.1

(T + 1)0
z − xt .

(5.90)

Assume that (5.89) is true. Then in view of (5.86) and (5.89), z − xt+1  ≤ 2M. Assume that (5.90) holds. It follows from (5.76), (5.77), (5.83), (5.88), and (5.90) that F (xt+1 ) ≤ F (z) +δf (2M2 + 9M + 3) + δG (1 + L0 + L(3M + M2 + 3)) ≤ inf(F ) + 2. By the inequality above, (5.70) and (5.84), xt+1  ≤ M and xt+1 − z ≤ 2M. Therefore by induction we showed that for all integers t ≥ 0, (3.86)–(3.88) hold. Let T be a natural number. By (5.83), (5.84) and (5.88), T 

(T + 1)−1 F (xt+1 ) − inf(F )

t=0

≤ δf (2M2 + 9M + 3) + δG (1 + L0 + L(3M + M2 + 3)) +2−1 L(T + 1)−1

T 

(z − xt 2 − z − xt+1 2 )

t=0

≤ δf (2M2 + 9M + 3) +δG (1 + L0 + L(3M + M2 + 3)) + 2(T + 1)−1 LM 2 . This implies that F(

T +1 t=1

(T + 1)−1 xt ) − inf(F ),

5.5 The Second Extension of Theorem 5.1

167

min{F (xt ) : t = 1, . . . , T + 1} − inf(F ) ≤ δf (2M2 + 9M + 3) +δG (1 + L0 + L(3M + M2 + 3)) + 2(T + 1)−1 LM 2 .  

This completes the proof of Theorem 5.5. By Theorem 5.5, the best choice of T is as the same order as max{δf , δG }−1 .

5.5 The Second Extension of Theorem 5.1 Theorem 5.6 Let δf , δG ∈ (0, 1), M > 1 argmin(F ) ∩ BX (0, M) = ∅,

(5.91)

L ≥ 1 satisfy |f (w1 ) − f (w2 )| ≤ Lw1 − w2  for all w1 , w2 ∈ BX (0, 4M + 10)

(5.92)

and f  (w1 ) − f  (w2 ) ≤ Lw1 − w2  for all w1 , w2 ∈ X,

(5.93)

M1 ≥ 4M + 8 satisfy |f (w)|, F (w) ≤ M1 for all w ∈ BX (0, 4M + 10),

(5.94)

M2 = (8(M1 + |c∗ | + 1 + (L + 1)2 )1/2 + 4M + 10,

(5.95)

L0 ≥ 1 satisfy |g(w1 ) − g(w2 )| ≤ L0 w1 − w2  for all w1 , w2 ∈ BX (0, M2 ),

(5.96)

δf , δG < 4−1 ((M2 + 12M + 27)L + 1 + L0 )−1 ,

(5.97)

−1 T = 2−1 L min{δf−1 , δG }((M2 + 12M + 27)L + 1 + L0 )−1 .

(5.98)

168

5 An Extension of the Gradient Algorithm

∞ Assume that {xt }∞ t=0 ⊂ X, {ξt }t=0 ⊂ X,

x0  ≤ M

(5.99)

ξt − f  (xt ) ≤ δf

(5.100)

and that for each integer t ≥ 0,

and (L)

BX (xt+1 , δG ) ∩ argmin(Gxt ,ξt ) = ∅.

(5.101)

Then for each integer t ∈ [0, T ], xt  ≤ 3M + 2 and T −1

T −1 F (xt+1 ) − inf(F ),

t=0 T −1

F(

T −1 xt+1 ) − inf(F ),

t=0

min{F (xt ) : t = 1, . . . , T } − inf(F ) ≤ max{δG , δf }(8M 2 + 1)((M2 + 12M + 27)(L + 1) + L0 + 1). Proof In view of (5.91), there exists z ∈ argmin(F ) ∩ BX (0, M).

(5.102)

x0 − z ≤ 2M.

(5.103)

By (5.99) and (5.102),

Assume that t ≥ 0 is an integer and that xt − z ≤ 2M + 2

(5.104)

Relations (5.102) and (5.104) imply that xt  ≤ 3M + 2.

(5.105)

5.5 The Second Extension of Theorem 5.1

169

Set M0 = 4M + 8.

(5.106)

By (5.100), (5.102), (5.105), (5.106) and Lemma 5.4 applied with x = z, u = xt , ξ = ξt , v = xt+1 , we have F (z) − F (xt+1 ) ≥ 2−1 Lz − xt+1 2 − 2−1 Lz − xt 2 −δf (M2 + 3M0 + 3) − δG (1 + L0 + L(M0 + M2 + 3)) ≥ 2−1 Lz − xt+1 2 − 2−1 Lz − xt 2 − δf (M2 + 12M + 27) − δG (1 + L0 + L(4M + M2 + 11)).

(5.107)

By (5.102) and (5.107), z − xt+1 2 − z − xt 2 ≤ 2L−1 δf (M2 + 12M + 27) + 2L−1 δG (1 + L0 + L(4M + M2 + 11)).

(5.108)

Therefore we have shown that the following property holds: (a) for each integer t ≥ 0 satisfying (5.104), relations (5.107) and (5.108) hold. Let us show that for all integers t = 0, . . . , T , z − xt 2 ≤ z − x0 2 + t[2L−1 δf (M2 + 12M + 27) +2L−1 δG (1 + L0 + L(4M + M2 + 11))] ≤ z − x0 2 + T [2L−1 δf (M2 + 12M + 27) +2L−1 δG (1 + L0 + L(4M + M2 + 11))] ≤ 4M 2 + 8M + 4. First, note that in view of (5.98),

(5.109)

170

5 An Extension of the Gradient Algorithm

T [2L−1 δf (M2 + 12M + 27) +2L−1 δG (1 + L0 + L(4M + M2 + 11))] = 2L−1 T δf (M2 + 12M + 27) + 2L−1 T δG (1 + L0 + L(4M + M2 + 11))] ≤ 2.

(5.110)

By (5.103) and (5.110), z − x0 2 + T [2L−1 δf (M2 + 12M + 27) +2L−1 δG (1 + L0 + L(4M + M2 + 11))] ≤ 4M 2 + 2 ≤ (2M + 2)2 .

(5.111)

Assume that t ∈ {0, . . . , T } \ {T } and that (5.109) holds. Then xt − z ≤ 2M + 2 and by property (a), relations (5.107) and (5.108) hold. By (5.108), (5.109), and (5.111), z − xt+1 2 ≤ z − xt 2 +2L−1 δf (M2 + 12M + 27) +2L−1 δG (1 + L0 + L(4M + M2 + 11)) ≤ z − x0 2 + (t + 1)[2L−1 δf (M2 + 12M + 27) +2L−1 δG (1 + L0 + L(4M + M2 + 11))] ≤ z − x0 2 + T [2L−1 δf (M2 + 12M + 27) +2L−1 δG (1 + L0 + L(4M + M2 + 11))] ≤ (2M + 2)2 and z − xt+1  ≤ 2M + 2.

5.5 The Second Extension of Theorem 5.1

171

Thus we have shown by induction that for all t = 0, . . . , T . z − xt  ≤ 2M + 2. and xt  ≤ 3M + 2. Property (a) implies that for all t = 0, . . . , T , (5.107) holds. It follows from (5.97), (5.98), (5.102), (5.103), (5.106), and (5.107) which holds for all t = 0, . . . , T that T −1

T −1 F (xt+1 ) − inf(F )

t=0

≤ 2−1 T −1 Lz − x0 2 + δf (M2 + 12M + 27) +δG (1 + L0 + L(4M + M2 + 11)) ≤ 2LM 2 (T /2)−1 + δf (M2 + 12M + 27) +δG (1 + L0 + L(4M + M2 + 11)) ≤ max{δC , δf }(8((M2 + 12M + 27)L + 1 + L0 )M 2 +M2 + 12M + 27 + (1 + L0 + L(4M + M2 + 8))). Theorem 5.6 is proved.

 

Chapter 6

In this chapter we study the continuous subgradient algorithm for minimization of convex nonsmooth functions and for computing the saddle points of convex– concave functions, under the presence of computational errors. The problem is described by an objective function and a set of feasible points. For this algorithm we need a calculation of a subgradient of the objective function and a calculation of a projection on the feasible set. In each of these two calculations there is a computational error produced by our computer system. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two calculations of our algorithm, we find out what approximate solution can be obtained and how much time one needs for this.

6.1 Bochner Integrable Functions Let (Y,  · ) be a Banach space and −∞ < a < b < ∞. A function x : [a, b] → Y is strongly measurable on [a, b] if there exists a sequence of functions xn : [a, b] → Y, n = 1, 2, . . . such that for any integer n ≥ 1 the set xn ([a, b]) is countable and the set {t ∈ [a, b] : xn (t) = y} is Lebesgue measurable for any y ∈ Y , and

© Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_6

173

174

xn (t) → x(t) as n → ∞ in (Y,  · ) for almost every ( a. e. ) t ∈ [a, b]. Denote by mes(E) the Lebesgue measure of a Lebesgue measurable set E ⊂ R 1 . The function x : [a, b] → Y is Bochner integrable if it is strongly measurable b and there exists a finite a x(t)dt. If x : [a, b] → Y is a Bochner integrable function, then for almost every (a. e.) t ∈ [a, b], lim (Δt)

−1



Δt→0

t+Δt

x(τ ) − x(t)dτ = 0

t

and the function  y(t) =

t

x(s)ds, t ∈ [a, b]

a

is continuous and a. e. differentiable on [a, b]. Let −∞ < τ1 < τ2 < ∞. Denote by W 1,1 (τ1 , τ2 ; Y ) the set of all functions x : [τ1 , τ2 ] → Y for which there exists a Bochner integrable function u : [τ1 , τ2 ] → Y such that  t u(s)ds, t ∈ (τ1 , τ2 ] x(t) = x(τ1 ) + τ1

(see, e.g., [10, 21]). It is known that if x ∈ W 1,1 (τ1 , τ2 ; Y ), then this equation defines a unique Bochner integrable function u which is called the derivative of x and is denoted by x  .

6.2 Convergence Analysis for Continuous Subgradient Method The study of continuous subgradient algorithms is an important topic in optimization theory. See, for example, [6, 9, 20, 21] and the references mentioned therein. In this chapter we analyze its convergence under the presence of computational errors. We suppose that X is a Hilbert space equipped with an inner product denoted by ·, · which induces a complete norm  · . Suppose that f : X → R 1 ∪{∞} is a convex, lower semicontinuous and bounded from below function such that dom(f ) := {x ∈ X : f (x) < ∞} = ∅ and that f possesses a minimizer. Set

6.2 Convergence Analysis for Continuous Subgradient Method

175

inf(f ) = inf(f (x) : x ∈ X} and argmin(f ) = {x ∈ X : f (x) = inf(f )}. For each set D ⊂ X put inf(f, D) = inf{f (z) : z ∈ D}, sup(f, D) = sup{f (z) : z ∈ D}. In Sect. 6.4 we will prove the following result which has no prototype in . Theorem 6.1 Let δ ∈ (0, 1], 0 < μ1 < μ2 , M ≥ 1, z ∈ BX (0, M) ∩ argmin(f ),

(6.1)

and let μ : [0, ∞) → R 1 be a Lebesgue measurable function such that μ1 ≤ μ(t) ≤ μ2 for all t ∈ [0, ∞).

(6.2)

T0 ∈ [(2δ)−1 , 5(4δ)−1 ],

(6.3)

x(0) ∈ dom(f ) ∩ BX (0, M)

(6.4)

Assume that

x ∈ W 1,1 (0, T0 ; X),

and that for almost every t ∈ [0, T0 ], x(t) ∈ dom(f )

(6.5)

BX (x  (t), δ) ∩ (−μ(t)∂f (x(t))) = ∅.

(6.6)

and

Then the following two assertions hold. 1. T0−1



T0 0

f (x(t)dt − inf(f ),

176

f (T0−1



T0

x(t)dt) − inf(f ),

0

min{f (x(t)) : t ∈ [0, T0 ]}− ≤ inf(f ) ≤ (2M + 2)2 μ−1 1 δ, z − x(t) ≤ 2M + 2, t ∈ [0, T0 ]. and for almost every t ∈ [0, T0 ], 2x(t) − z, x  (t) ≤ 2μ1 (f (z) − f (x(t)) + 2δ(2M + 2). 2. Assume that M ≥ 4, δ ≤ min{μ1 (2M + 2)−1 , 16−1 }

(6.7)

and that {x ∈ X : f (x) ≤ inf(f ) + 4} ⊂ BX (0, 4−1 M). Then x(T0 ) ≤ M. Theorem 6.1 easily implies the following result. Theorem 6.2 Let δ ∈ (0, 16−1 ], 0 < μ1 < 1 ≤ μ2 , M ≥ 1, μ : [0, ∞) → R 1 be a Lebesgue measurable function such that μ1 ≤ μ(t) ≤ μ2 for all t ∈ [0, ∞), δ ≤ μ1 (2M + 2)−1 , {x ∈ X : f (x) ≤ inf(f ) + 4} ⊂ BX (0, 4−1 M), T ≥ 5(4δ)−1 , x ∈ W 1,1 (0, T ; X), x(0) ∈ dom(f ) ∩ BX (0, M) and that for almost every t ∈ [0, T ], x(t) ∈ dom(f )

6.3 An Auxiliary Result

177

and B(x  (t), δ) ∩ (−μ(t)∂f (x(t))) = ∅. Assume that k ≥ 1 is an integer, (2δ)−1 k < T ≤ (2δ)−1 (k + 1), Ti = i(2δ)−1 , i = 0, . . . , k − 1, Tk = T . Then x(t) ≤ 3M + 2 for all t ∈ [0, T ], x(Ti ) ≤ M, i = 0, . . . , k and for all integers i = 0, . . . , k − 1, f ((Ti+1 − Ti )−1



Ti+1

x(t)dt) − inf(f ),

Ti

min{f (x(t)) : t ∈ [Ti , Ti+1 ]} − inf(f ) ≤ (Ti+1 − Ti )

−1



Ti+1 Ti

f (x(t))dt − inf(f ) ≤ (2M + 2)2 δμ−1 1 .

6.3 An Auxiliary Result Let V ⊂ X be an open convex set and g : V → R 1 be a convex locally Lipschitzian function. Let T > 0, x0 ∈ X and let u : [0, T ] → X be a Bochner integrable function. Set 

t

x(t) = x0 +

u(s)ds, t ∈ [0, T ].

0

Then x : [0, T ] → X is differentiable and x  (t) = u(t) for almost every t ∈ [0, T ]. Assume that x(t) ∈ V for all t ∈ [0, T ]. We claim that the restriction of g to the set {x(t) : t ∈ [0, T ]}

178

is Lipschitzian. Indeed, since the set {x(t) : t ∈ [0, T ]} is compact, the closure of its convex hull C is both compact and convex, and so the restriction of g to C is Lipschitzian. Hence the function (g · x)(t) := g(x(t)), t ∈ [0, T ], is absolutely continuous. It follows that for almost every t ∈ [0, T ], both the derivatives x  (t) and (g · x) (t) exist: x  (t) = lim h−1 [x(t + h) − x(t)],

(6.8)

(g · x) (t) = lim h−1 [g(x(t + h)) − g(x(t))].

(6.9)

h→0

h→0

We continue with the following fact (see Proposition 14.2 of ). Proposition 6.3 Assume that t ∈ [0, T ] and that both the derivatives x  (t) and (g · x) (t) exist. Then (g · x) (t) = lim h−1 [g(x(t) + hx  (t)) − g(x(t))]. h→0

(6.10)

Corollary 6.4 Let z ∈ X and g(y) = z − y2 for all y ∈ X. Then for almost every t ∈ [0, T ], the derivative (g · x) (t) exists and (g · x) (t) = 2x  (t), x(t) − z.

6.4 Proof of Theorem 6.1 Set φ(t) = z − x(t)2 , t ∈ [0, T0 ].

(6.11)

In view of Corollary 6.4, for a. e. t ∈ [0, T0 ], there exist derivatives x  (t), φ  (t), and φ  (t) = 2x  (t), x(t) − z.

(6.12)

By (6.6), for a. e. t ∈ [0, T0 ], there exist ξ(t) ∈ ∂f (x(t))

(6.13)

6.4 Proof of Theorem 6.1

179

such that x  (t) + μ(t)ξ(t) ≤ δ.

(6.14)

It follows from (6.12) that for almost every t ∈ [0, T0 ], φ  (t) = 2x(t) − z, x  (t) = 2x(t) − z, −μ(t)ξ(t) + 2x(t) − z, x  (t) + μ(t)ξ(t).

(6.15)

In view of (6.14), for almost every t ∈ [0, T0 ], |z − x(t), x  (t) + μ(t)ξ(t)| ≤ δz − x(t).

(6.16)

By (6.1), (6.13) and (6.17), for almost every t ∈ [0, T0 ], z − x(t), ξ(t) ≤ −f (x(t)) + f (z).

(6.17)

It follows from (6.1), (6.2), (6.15), and (6.16) that for almost every t ∈ [0, T0 ], φ  (t) ≤ 2μ(t)(−f (x(t)) + f (z)) + 2δz − x(t) ≤ 2μ1 (−f (x(t)) + f (z)) + 2δz − x(t).

(6.18)

We show that for all t ∈ [0, T0 ], z − x(t) ≤ 2M + 2.

(6.19)

Assume the contrary. Then there exists τ ∈ (0, T0 ] such that z − x(t) < 2M + 2, t ∈ [0, τ ),

(6.20)

z − x(τ ) = 2M + 2.

(6.21)

By (6.1), (6.4), (6.11), (6.18), (6.20), and (6.21), 8M + 4 = (2M + 2)2 − 4M 2 ≤ z − x(τ )2 − z − x(0)2

180



τ

= φ(τ ) − φ(0) =

2δz − x(t)dt

0

≤ 2τ δ(2M + 2).

(6.22)

In view of (6.3) and (6.22), 5(4δ)−1 ≥ T0 ≥ τ ≥ (8M + 4)(4M + 4)−1 δ −1 ≥ 3(2δ)−1 . The contradiction we have reached proves that z − x(t) ≤ 2M + 2, t ∈ [0, T0 ].

(6.23)

By (6.18), for almost every t ∈ [0, T0 ] φ  (t) ≤ 2μ1 (f (z) − f (x(t)) + 2δ(2M + 2).

(6.24)

It follows from (6.1), (6.4), (6.11), (6.12), and (6.14) that 4M 2 ≥ φ(0) − φ(T0 ) 

T0

=−

φ  (t)dt ≥ 2μ1

0



T0

(f (x(t)) − f (z))dt − 2δT0 (2M + 2).

(6.25)

0

By (6.3) and (6.25), T0−1



T0

(f (x(t)) − f (z))dt

0

≤ (2μ1 )−1 [4M 2 T0−1 + 2δ(M + 2)] ≤ 2δ(2μ1 )−1 (4M 2 + 4M + 4) = (2M + 2)2 μ−1 1 δ. Therefore assertion 1 is proved. Let us complete the proof of assertion 2. By (6.7) and (6.16), T0−1



T0

(f (x(t)) − f (z))dt ≤ 1.

0

This implies that 4T0−1 mes{t ∈ [0, T0 ] : f (x(t)) > f (z) + 4} < 1, mes{t ∈ [0, T0 ] : f (x(t)) > f (z) + 4} < 4−1 T0 .

(6.26)

6.5 Continuous Subgradient Method for Zero-Sum Games

181

This implies that there exists t0 ∈ [3T0 /4, T0 ]

(6.27)

f (x(t0 )) ≤ f (z) + 4.

(6.28)

such that

By (6.1), (6.28), and the assumptions of the assertion, x(t0 )| ≤ M/4, z ≤ M/4.

(6.29)

It follows from (6.1), (6.3), (6.11), (6.12), (6.18), (6.23), and (6.27) that z − x(T0 )2 − z − x(t0 )2  =

T0

φ  (t)dt

t0

≤ 28(T0 − t0 )(2M + 2) = (3/2)T0 δ(2M + 2) ≤ 4M + 4. In view of the relation above, (6.1) and (6.29), z − x(T0 )2 ≤ M 2 /4 + 4M + 4 ≤ (M/2 + 4)2 and x(T0 ) ≤ z + M/2 + 4 ≤ 3M/4 + 4 ≤ M. Assertion 2 is proved. This completes the proof of Theorem 6.1.

 

6.5 Continuous Subgradient Method for Zero-Sum Games In this section we study continuous subgradient method for zero-sum games with two players. This topic was not considered in . Let (X, ·, ·), (Y, ·, ·) be Hilbert spaces equipped with the complete norms · which are induced by their inner products. Let a borelian function f : X × Y → R 1 ∪ {−∞} possesses the following property: {(x, y) ∈ X × Y : f (x, y) > −∞} = X × C, where C is a nonempty bounded convex subset of Y .

182

Recall that for for each concave function g : Y → R 1 ∪ {−∞} and each x ∈ Y satisfying g(x) > −∞, ∂g(x) = {l ∈ Y : l, y − x ≥ g(y) − g(x) for all y ∈ Y }. Clearly, for each x ∈ Y satisfying g(x) > −∞, ∂g(x) = −(∂(−g)(x)). Suppose that the following properties hold: (i) for each y ∈ C, the function f (·, y) : X → R 1 is convex and lower semicontinuous; (ii) for each x ∈ X, the function f (x, ·) : Y → R 1 ∪ {−∞} is concave and upper semicontinuous. Let x∗ ∈ X and y∗ ∈ C

(6.30)

f (x∗ , y) ≤ f (x∗ , y∗ ) ≤ f (x, y∗ )

(6.31)

satisfy

for each x ∈ X and each y ∈ C. Let M > 0, x∗  < M/8,

(6.32)

C ⊂ BY (0, M),

(6.33)

f is bounded from above on BX (0, M/3) × C and let for every y ∈ C, {x ∈ X : f (x, y) ≤ f (x∗ , y∗ ) + 4} ⊂ BX (0, M/3).

(6.34)

In Sect. 6.7 we prove the following result. Theorem 6.5 Let 0 < μ1 < μ2 ,

(6.35)

μ ∈ [μ1 , μ2 ],

(6.36)

6.5 Continuous Subgradient Method for Zero-Sum Games

183

δf,1 , δf,2 ∈ (0, 1), δf,1 < μ1 /M.

(6.37)

Assume that T > 0, x ∈ W 1,1 (0, T ; X), y ∈ W 1,1 (0, T ; Y ), y(0) ∈ C,

(6.38)

y(t) ∈ C,

(6.39)

BY (y  (t), δf,2 ) ∩ (μ∂y f (x(t), y(t))) = ∅,

(6.40)

BX (x  (t), δf,1 ) ∩ (−μ∂x f (x(t), y(t))) = ∅

(6.41)

x(0) ≤ M/8.

(6.42)

x(t) ≤ M for all t ∈ [0, T ],

(6.43)

for almost every t ∈ [0, T ],

and that

Then

|T −1



T 0

f (x(t), y(t))dt − f (x∗ , y∗ )|

−1 ≤ 2M 2 T −1 μ−1 1 + 2Mμ1 max{δf,1 , δf,2 },

|f (T −1



T



x(t)dt, T −1

0

T

y(t)dt) − T −1

0



(6.44)

T

f (x(t), y(t))dt| 0

−1 ≤ 2M 2 T −1 μ−1 1 + 2Mμ1 max{δf,1 , δf,2 },

(6.45)

for each v ∈ C, f (T

−1



T

x(t)dt, v) 0

≤ f (T −1

 0

T

x(t)dt, T −1



T

y(t)dt) 0

−1 + 4M 2 T −1 μ−1 1 + 4Mμ1 max{δf,1 , δf,2 },

for each z ∈ X,

(6.46)

184

f (z, T −1



T

y(t)dt) 0

≥ f (T −1



T

x(t)dt, T −1

0



T

y(t)dt) 0

−1 − 4M 2 T −1 μ−1 1 − 3Mμ1 max{δf,1 , δf,2 }.

(6.47)

Clearly, the best choice of T should be at the same order as max{δf,1 , δf,2 }−1 .

6.6 An Auxiliary Result Proposition 6.6 Let 0 < μ1 < μ2 ,

(6.48)

μ ∈ [μ1 , μ2 ],

(6.49)

δf,1 , δf,2 ∈ (0, 1), δf,1 < μ1 /M.

(6.50)

Assume that T > 0, x ∈ W 1,1 (0, T ; X), y ∈ W 1,1 (0, T ; Y ), y(0) ∈ C,

(6.51)

y(t) ∈ C,

(6.52)

BY (y  (t), δf,2 ) ∩ (μ∂y f (x(t), y(t))) = ∅,

(6.53)

BX (x  (t), δf,1 ) ∩ (−μ∂x f (x(t), y(t))) = ∅.

(6.54)

and that for almost every t ∈ [0, T ],

Then the following assertions hold. 1. Let z ∈ X and v ∈ C. Then for almost every t ∈ [0, T ], (d/dt)(x(t) − z2 ) ≤ 2δf,1 z − x(t) + 2μ(f (z, y(t)) − f (x(t), y(t))),

(6.55)

(d/dt)(y(t) − v2 ) ≤ 2δf,2 v − y(t) + 2μ(−f (x(t), v) + f (x(t), y(t))).

(6.56)

6.6 An Auxiliary Result

185

2. Let x(0) ≤ M/8.

(6.57)

x(t) ≤ M

(6.58)

Then for each t ∈ [0, T ],

and for each z ∈ X, each v ∈ C and almost every t ∈ [0, T ], (d/dt)(x(t) − z2 ) ≤ 2δf,1 (z + M) + 2μ(f (z, y(t)) − f (x(t), y(t))),

(6.59)

(d/dt)(y(t) − v2 ) ≤ 4δf,2 M + 2μ(−f (x(t), v) + f (x(t), y(t))).

(6.60)

Proof Let z ∈ X and v ∈ C. For all t ∈ [0, T ] set φ1 (t) = z − x(t)2 , φ2 (t) = v − y(t)2 .

(6.61)

By (6.53) and (6.54), for almost every t ∈ [0, T ], there exist ξ(t) ∈ ∂x f (x(t), y(t)),

(6.62)

η(t) ∈ ∂y f (x(t), y(t))

(6.63)

x  (t) + μξ(t) ≤ δf,1

(6.64)

y  (t) − μη(t) ≤ δf,2 .

(6.65)

and

such that

and

Corollary 6.4 and (6.61) imply that for almost every t ∈ [0, T ], φ1 (t) = 2x  (t), x(t) − z,

(6.66)

φ2 (t) = 2y  (t), y(t) − v.

(6.67)

186

In view of (6.66) and (6.67), for almost every t ∈ [0, T ], φ1 (t) = 2x  (t), x(t) − z = 2x(t) − z, −μξ(t) + 2x(t) − z, x  (t) + μξ(t)

(6.68)

and φ2 (t) = 2y(t) − v, y  (t) = 2y(t) − v, μη(t) + 2y(t) − v, y  (t) − μη(t).

(6.69)

By (6.64) and (6.65), for almost every t ∈ [0, T ], |x(t) − z, x  (t) + μξ(t)| ≤ δf,1 z − x(t)

(6.70)

|y(t) − v, y  (t) − μη(t)| ≤ δf,2 v − y(t).

(6.71)

and

It follows from (6.62) and (6.63) that for almost every t ∈ [0, T ], z − x(t), ξ(t) ≤ f (z, y(t)) − f (x(t), y(t))

(6.72)

v − y(t), η(t) ≥ f (x(t), v) − f (x(t), y(t)).

(6.73)

and

By (6.61) and (6.68)–(6.73), for almost every t ∈ [0, T ], (d/dt)(x(t) − z2 ) ≤ 2δf,1 z − x(t) + 2μ(f (z, y(t)) − f (x(t), y(t)))

(6.74)

and (d/dt)(y(t) − v2 ) ≤ 2δf,2 v − y(t) + 2μ(−f (x(t), v) + f (x(t), y(t))). Thus assertion 1 is proved. Let us complete the proof of assertion 2. We show that x(t) ≤ M for all t ∈ [0, T ].

(6.75)

6.6 An Auxiliary Result

187

In view of (6.32) and (6.57), x∗ − x(0) ≤ M/4.

(6.76)

By (6.31) and (6.55) with z = x∗ , for almost every t ∈ [0, T ], (d/dt)(x(t) − x∗ 2 ) ≤ 2δf,1 x∗ − x(t) + 2μ(f (x∗ , y(t)) − f (x(t), y(t))) ≤ 2δf,1 x∗ − x(t) + 2μ(f (x∗ , y∗ ) − f (x(t), y(t))).

(6.77)

Define E = {τ ∈ [0, T ] : x(t) − x∗  ≤ (3/4)M for all t ∈ 0, τ ]}.

(6.78)

In view of (6.76), E = ∅. Set τ0 = sup E.

(6.79)

Since the function x is continuous we have τ0 ∈ E.

(6.80)

In order to complete the proof of assertion 2 it is sufficient to show that τ0 = T0 . Assume the contrary. Then τ0 < T0 .

(6.81)

In view of (6.76), τ > 0. Relations (6.78)–(6.81) imply that x(τ0 ) − x∗  = (3/4)M.

(6.82)

188

There exists τ1 ∈ (0, τ0 ) such that for all t ∈ [τ1 , τ0 ] we have x(t) − x∗  ≥ (2/3)M. Together with (6.32) this implies that for all t ∈ [τ1 , τ0 ], x(t) ≥ (2/3)M − M/8 > M/3. By the relation above, (6.34) and (6.52), for almost every t ∈ [τ1 , τ0 ], f (x(t), y(t)) > f (x∗ , y∗ ) + 4.

(6.83)

It follows from (6.49) and (6.77)–(6.80) that 0 ≤ x(τ0 ) − x∗ 2 − x(τ1 ) − x∗ 2 ≤ 2(τ0 − τ1 )δf,1 M − 8μ1 (τ0 − τ1 ) and 4μ1 ≤ δf,1 M. This contradicts (6.50). The contradiction we have reached proves that τ0 = T0 and completes the proof of assertion 2 and Proposition 6.6.

 

6.7 Proof of Theorem 6.5 Proposition 6.6 and the assumptions of the theorem imply that for each t ∈ [0, T ], x(t) ≤ M

(6.84)

and for each z ∈ X, each v ∈ C and almost every t ∈ [0, T ], (d/dt)(x(t) − z2 ) ≤ 2δf,1 (z + M) + 2μ(f (z, y(t)) − f (x(t), y(t))),

(6.85)

6.7 Proof of Theorem 6.5

189

(d/dt)(y(t) − v2 ) ≤ 4δf,2 M + 2μ(−f (x(t), v) + f (x(t), y(t))).

(6.86)

By (6.85) and (6.86), for each z ∈ X, each v ∈ C, T −1 (x(T ) − z2 − x(0) − z2 ) +2μT −1



T

f (x(t), y(t))dt − 2μT −1



T

f (z, y(t))dt

0

0

≤ 2δf,1 (z + M)

(6.87)

and T −1 (y(T ) − v2 − y(0) − v2 ) −2μT

−1



T

f (x(t), y(t))dt + 2μT

−1

0



T

f (x(t), v)dt 0

≤ 4δf,2 M.

(6.88)

z ≤ M

(6.89)

Let z ∈ X satisfy

and v ∈ C. Relations (6.84), (6.85), and (6.89) imply that T

−1



T

f (x(t), y(t))dt − T

−1



T

f (z, y(t))dt

0

0

≤ (2μT )−1 4M 2 + 2Mδf,1 (2μ)−1 .

(6.90)

By (6.33) and (6.88), T −1



T

f (x(t), v)dt − T −1

0



T

f (x(t), y(t))dt 0

≤ (2μT )−1 4M 2 + 4Mδf,2 (2μ)−1 . It follows from (6.30), (6.31), (6.90) with v = y∗ and (6.91) with z = x∗ that T

−1



T 0

f (x(t), y(t))dt − f (x∗ , y∗ )

(6.91)

190

≥ T −1



T

f (x(t), y(t))dt − f (T −1



0

≥ T −1



T

x(t)dt, y∗ )

0 T



f (x(t), y(t))dt − T −1

0

T

f (x(t), y∗ )dt

0

≥ −(μT )−1 2M 2 − 2Mδf,2 μ−1

(6.92)

and T −1



T 0

≤T

−1



T 0

≤ T −1



T

f (x(t), y(t))dt − f (x∗ , y∗ ) −1

f (x(t), y(t))dt − f (x∗ , T 

f (x(t), y(t))dt − T −1

0



T

y(t)dt) 0

T

f (x∗ , y(t))dt

0

≤ (μT )−1 2M 2 + Mδf,1 μ−1 .

(6.93)

Set xT = T −1



T

x(t)dt, 0

yT = T

−1



T

(6.94)

y(t)dt. 0

Since the set C is convex and closed we have yT ∈ C.

(6.95)

Relations (6.33), (6.84), (6.94), and (6.95) imply that xT , yT  ≤ M.

(6.96)

By (6.84), (6.90) with z = xT and (6.96), T −1



T 0

f (x(t), y(t))dt − T −1



T

f (xT , y(t))dt 0

≤ 2M 2 (μT )−1 + 2δf,1 Mμ−1 .

(6.97)

6.7 Proof of Theorem 6.5

191

By (6.52), (6.91) with v = yT and (6.96), T

−1



T

f (x(t), y(t))dt − T

−1



T

f (x(t), yT )dt

0

0

≥ −2M 2 (μT )−1 − 2δf,2 Mμ−1 .

(6.98)

Since the function f (·, yT ) is convex and lower semicontinuous it follows from (6.94) and (6.98) that f (xT , yT ) − T −1



T

f (x(t), yT )dt 0

= f (T −1



T

x(t)dt, yT ) − T −1



f (x(t), yT )dt

0

≤T

−1



T

T

0

f (x(t), yT )dt − T

−1

0



T

f (x(t), y(t))dt 0

≤ 2M 2 (μT )−1 + 2δf,2 Mμ−1 .

(6.99)

Since the function f (xT , ·) is concave and upper semicontinuous it follows from (6.94) and (6.97) that f (xT , yT ) − T −1



T

f (x(t), y(t))dt 0

−1

= f (xT , T



T

y(t)dt) − T

−1



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

0

≥ T −1



T

T 0

f (xT , y(t))dt − T −1

0



T

f (x(t), y(t))dt 0

≥ −2M 2 (μT )−1 − 2δf,1 Mμ−1 .

(6.100)

By (6.99) and (6.100), |f (xT , yT ) − T

−1



T

f (x(t), y(t))dt| 0

≤ max{2M 2 (μT )−1 + 2δf,1 Mμ−1 , 2M 2 (μT )−1 + 2δf,2 Mμ−1 } = 2M 2 (μT )−1 + 2Mμ−1 max{δf,1 , δf,2 }.

(6.101)

192

Let v ∈ C, z ∈ BX (0, M).

(6.102)

Then (6.90) and (6.91) hold. It follows from (6.94) and properties (i) and (ii) that 

T −1

T

f (z, y(t)dt ≤ f (z, T −1



0

T

y(t)dt) = f (z, yT )

(6.103)

x(t)dt, v) = f (xT , v).

(6.104)

0

and T

−1



T

f (x(t), v)dt) ≥ f (T

−1

0



T 0

It follows from (6.90) and (6.103) that −1

T



T

f (x(t), y(t))dt − f (z, yT )

0



≤ T −1

T

f (x(t), y(t))dt − T −1



T

f (z, y(t))dt

0

0

≤ 2M 2 (μT )−1 + Mμ−1 δf,1 .

(6.105)

In view of (6.101) and (6.105), f (z, yT ) ≥T

−1



T

f (x(t), y(t))dt − 2M 2 (μT )−1 − Mμ−1 δf,1

0

≥ f (xT , yT ) − 2Mμ−1 max{δf,1 , δf,2 } − 4M 2 (μT )−1 − Mμ−1 δf,1 .

(6.106)

It follows from (6.91) and (6.104) that T −1



T

f (x(t), y(t))dt − f (xT , v)

0

≥T

−1



T 0

f (x(t), y(t))dt − T

−1



T

f (x(t), v)dt 0

≤ −2M 2 (μT )−1 − 2Mμ−1 δf,2 .

(6.107)

193

In view of (6.101) and (6.107), f (xT , v) ≤ T −1



T

f (x(t), y(t))dt + 2M 2 (μT )−1 + 2Mμ−1 δf,2

0

≤ f (xT , yT ) + 2Mμ−1 max{δf,1 , δf,2 } +4M 2 (μT )−1 + 2δf,2 Mμ−1 . If z > M, then it follows from (6.34), (6.93), (6.96), and (6.101) that f (z, yT ) ≥ f (x∗ , y∗ ) + 4 ≥ T −1



T

f (x(t), y(t))dt − 2M 2 (μT )−1 − Mμ−1 δf,1

0

≥ f (xT , yT ) − 2Mμ−1 max{δf,1 , δf,2 } − 4M 2 (μT )−1 − δf,1 Mμ−1 .  

Theorem 6.5 is proved.

6.8 Continuous Subgradient Projection Method Let X be a Hilbert space with an inner product ·, · which induces a complete norm  · . Let C be a nonempty, convex, and closed set in the Hilbert space X, U be an open and convex subset of X such that C ⊂ U, and f : U → R 1 be a convex locally Lipschitzian function. Let x ∈ U and ξ ∈ X. Set f 0 (x, ξ ) = lim t −1 [f (x + tξ ) − f (x)], t→0+

∂f (x; ξ ) = {l ∈ ∂f (x) : l, ξ  = f 0 (x, ξ )}. It is a well-known fact of the convex analysis that ∂f (x; ξ ) = ∅.

194

Let M > 1, L > 0 and assume that C ⊂ BX (0, M − 1), {y ∈ X : d(y, C) ≤ 1} ⊂ U, |f (v1 ) − f (v2 )| ≤ Lv1 − v2  for all v1 , v2 ∈ BX (0, M + 1) ∩ U.

(6.108)

We will prove the following result which also has no prototype in . Theorem 6.7 Let δf , δC ∈ (0, 1], 0 < μ1 < μ2 , μ1 ≤ 1,

(6.109)

μ ∈ [μ1 , μ2 ].

(6.110)

d(x(0), C) ≤ δC

(6.111)

T0 > 0,

Assume that x ∈ W 1,1 (0, T0 ; X),

and that for almost every t ∈ [0, T0 ], there exists ξ(t) ∈ X such that BX (ξ(t), δf ) ∩ ∂f (x(t); x  (t)) = ∅,

(6.112)

PC (x(t) − μξ(t)) ∈ BX (x(t) + x  (t), δC ).

(6.113)

Then f (T0−1



T

x(s)ds) − inf(f, C),

0

min{f (x(t)) : t ∈ [0, T0 ]} − inf(f, C) ≤ T0−1



T

f (x(s))ds − inf(f, C)

0

−1 ≤ (2M 2 μ−1 1 + 2(M + 1))T0 −1 +6δf M + μ−1 1 δC (4M + μ2 (L + 1)) + 2δC LT0 .

Clearly, the best choice of T should be at the same order as max{δf , δC }−1 .

6.9 An Auxiliary Result

195

6.9 An Auxiliary Result Lemma 6.8 Let D be a nonempty convex closed subset of X, T , δ > 0, x ∈ W 1,1 (0, T ; X), d(x(0), D) ≤ δ,

(6.114)

Dδ = {x ∈ X : d(x, D) ≤ δ},

(6.115)

BX (x(t) + x  (t), δ) ∩ D = ∅ for almost every t ∈ [0, T ].

(6.116)

Then x(t) ∈ Dδ for all t ∈ [0, T ]. Proof For almost every t ∈ [0, T ] set φ(t) = x(t) + x  (t).

(6.117)

It is clear that φ : [0, T ] → X is a Bochner integrable function. In view of (6.116) and (6.117), for almost every t ∈ [0, T ], B(φ(t), δ) ∩ D = ∅.

(6.118)

Clearly, Dδ is a closed convex set, for each x ∈ Dδ , B(x, δ) ∩ D = ∅

(6.119)

and in view of (6.115) and (6.118), φ(t) ∈ Dδ for almost every t ∈ [0, T ]. Evidently, the function es φ(s), s ∈ [0, T ] is Bochner integrable. We claim that for all t ∈ [0, T ],  t −t −t es φ(s)ds. x(t) = e x(0) + e 0

Clearly, (6.121) holds for t = 0. For every t ∈ (0, T ] we have 

t

 e φ(s)ds = s

0

t

es (x(s) + x  (s))ds

0

 = 0

t

(es x(s)) ds = et x(t) − x(0).

(6.120)

(6.121)

196

This implies (6.121) for all t ∈ [0, T ]. By (6.121), for all t ∈ [0, T ], x(t) = e−t x(0) + (1 − e−t )(1 − e−t )−1 e−t



t

es φ(s)ds

0

= e−t x(0) + (1 − e−t )



t

es (et − 1)−1 φ(s)ds.

(6.122)

0

In view of (6.120), for all t ∈ [0, T ], 

t

es (et − 1)−1 φ(s)ds ∈ Dδ .

(6.123)

0

Relations (6.114), (6.122), and (6.123) imply that x(t) ∈ Dδ for all t ∈ [0, T ].  

Lemma 6.8 is proved.

6.10 Proof of Theorem 6.7 Let z ∈ C.

(6.124)

CδC = {x ∈ X : d(x, C) ≤ δC }.

(6.125)

Set

For almost every t ∈ [0, T0 ] set φ(t) = x(t) + x  (t).

(6.126)

Lemma 6.8, (6.111), and (6.113) imply that x(t) ∈ CδC for all t ∈ [0, T0 ].

(6.127)

It follows from (6.125) and (6.127) that for every t ∈ [0, T0 ], there exists  x (t) ∈ C such that

(6.128)

6.10 Proof of Theorem 6.7

197

x(t) −  x (t) ≤ δC .

(6.129)

By (6.128) and Lemma 2.2, for almost every t ∈ [0, T0 ],  x (t) − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ 0.

(6.130)

Inequality (6.130) implies that for almost every t ∈ [0, T0 ], x(t) − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ x(t) −  x (t), x(t) − μξ(t) − PC (x(t) − μξ(t)).

(6.131)

It follows from (6.124) and Lemma 2.2 that z − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ 0.

(6.132)

In view of (6.42), for almost every t ∈ [0, T0 ] there exists  ξ (t) ∈ ∂f (x(t); x  (t))

(6.133)

ξ (t), x  (t), f 0 (x(t), x  (t)) = 

(6.134)

ξ(t) −  ξ (t) ≤ δf .

(6.135)

such that

In view of (6.133), for almost every t ∈ [0, T0 ], f (z) ≥ f (x(t)) +  ξ (t), z − x(t).

(6.136)

By (6.113), for almost every t ∈ [0, T0 ], (x(t) + x  (t)) − PC (x(t) − μξ(t)) ≤ δC . Relations (6.126) and (6.137) imply that for almost every t ∈ [0, T0 ], z − x(t) − x  (t), x(t) − μξ(t) − x(t) − x  (t) = z − φ(t), x(t) − μξ(t) − φ(t) = z − φ(t), x(t) − μξ(t) − PC (x(t) − μξ(t)) +z − φ(t), PC (x(t) − μξ(t)) − φ(t)

(6.137)

198

≤ z − φ(t), x(t) − μξ(t) − PC (x(t) − μξ(t)) + δC z − φ(t).

(6.138)

In view of (6.110), (6.125), and (6.127), for all t ∈ [0, T0 ], x(t) ≤ M.

(6.139)

It follows from (6.110), (6.113), (6.126), and (6.139) that for almost every t ∈ [0, T0 ], φ(t) = x(t) + x  (t) ≤ M,

(6.140)

x  (t) ≤ 2M.

(6.141)

By (6.110), (6.124), (6.126), (6.138), and (6.140), for almost every t ∈ [0, T0 ], z − x(t) − x  (t), −μξ(t) − x  (t) ≤ z − φ(t), x(t) − μξ(t) − PC (x(t) − μξ(t)) + 2MδC .

(6.142)

By (6.110)–(6.112), (6.126), (6.132), and (6.137), z − φ(t), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ z − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) +δC x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ δC (2M + μ2 (L + 1)).

(6.143)

In view of (6.142) and (6.143), for almost every t ∈ [0, T0 ], x(t) + x  (t) − z, μξ(t) + x  (t) ≤ δC (4M + μ2 (L + 1)).

(6.144)

It follows from (6.110), (6.124), (6.135), (6.136), and (6.141) that for almost every t ∈ [0, T0 ], f (x(t)) − f (z) ≤  ξ (t), x(t) − z = ξ(t), x(t) − z + x  (t) − ξ(t), x  (t) + ξ (t) − ξ(t), x(t) − z + x  (t)

6.10 Proof of Theorem 6.7

199

ξ(t) −  ξ (t), x  (t) ≤ ξ(t), x(t) − z + x  (t) − ξ(t), x  (t) + 4Mδf .

(6.145)

Relations (6.144) and (6.145) imply that for almost every t ∈ [0, T0 ], f (x(t)) − f (z) ≤ μ−1 −x  (t), x(t) + x  (t) − z + μ−1 (4M + μ2 (L + 1))δC − ξ(t), x  (t) + 4Mδf .

(6.146)

By (6.110) and (6.146), for almost every t ∈ [0, T0 ], f (x(t)) − f (z) ≤ −μ−1 x  (t)2 − μ−1 x  (t), x(t) − z − ξ(t), x  (t) + δC (4M + μ2 (L + 1))μ−1 1 + 4Mδf

(6.147)

and μ−1 x  (t)2 + μ−1 x  (t), x(t) − z +ξ(t), x  (t) + f (x(t)) − f (z) ≤ δC (4M + μ2 (L + 1))μ−1 1 + 4Mδf .

(6.148)

In view of (6.135), (6.141), and (6.148), μ−1 x  (t)2 + μ−1 x  (t), x(t) − z + ξ (t), x  (t) + f (x(t)) − f (z) ≤ δC (4M + μ2 (L + 1))μ−1 1 + 6Mδf .

(6.149)

It follows from (6.134), (6.149), and Corollary 6.4 that for almost every t ∈ [0, T0 ], (2μ)−1 (d/dt)(x(t) − z2 ) + f 0 (x(t), x  (t)) +f (x(t)) − f (z) + μ−1 x  (t)2 ≤ 6δf M + δC (4M + μ2 (L + 1))μ−1 1 . Using Proposition 6.3, the equality

(6.150)

200

f 0 (x(t), x  (t)) = (f ◦ x) (t), and integrating inequality (6.150) over the interval [0, t], we obtain that for all t ∈ [0, T0 ], (2μ)−1 x(t) − z2 − (2μ)−1 x(0) − z2 +f (x(t)) − f (x(0)) 

t

+

(f (x(s)) − f (z))ds

0

≤ 6δf Mt + μ−1 1 δC (4M + μ2 (L + 1)).

(6.151)

Relations (6.108), (6.110), (6.111), and (6.129) imply that for all t ∈ [0, T0 ], f (x(t)) ≥ inf(f, C) − δC L, f (x(t)) ≤ sup(f, C) + δC L.

(6.152)

By (6.110), (6.151), and (6.152), for all t ∈ [0, T0 ], (2μ2 )−1 x(t) − z2 − (2μ1 )−1 x(0) − z2 + inf(f, C) − f (x(0)) − δC L 

t

+

(f (x(s)) − f (z))ds

0

≤ 6δf Mt + δC t (4M + μ2 (L + 1))μ−1 1 . The relation above with t = T0 , (6.110), (6.111), (6.128), (6.129), and (6.152) imply that 

T0

(f (x(s)) − f (z))ds

0

≤ (2μ1 )−1 x(0) − z2 + sup(f, C) − inf(f, C) +2δC L + 6δf MT0 + δC T0 (4M + μ2 (L + 1))μ−1 1 ≤ 2M 2 μ−1 1 + 2LM + 2δC L + 6δf MT0 +δC T0 (4M + μ2 (L + 1))μ−1 1 .

6.11 Continuous Subgradient Projection Method on Unbounded Sets

201

This implies that 

f (T0−1

T0

x(s)ds) − inf(f, C)

0

min{f (x(t)) : t ∈ [0, T0 ]} − inf(f, C) ≤ T0−1



T0

f (x(s))ds − inf(f, C)

0

−1 ≤ (2M 2 μ−1 1 + 2LM)T0 −1 +6δf M + μ−1 1 δC (4M + μ2 (L + 1)) + 2δC LT0 .

This completes the proof of Theorem 6.7.

 

6.11 Continuous Subgradient Projection Method on Unbounded Sets In Sect. 6.13 of chapter we obtain an extension of Theorem 6.7 for minimization problems on unbounded sets. Note that in  we study continuous subgradient projection method only for problems on bounded sets. Let X be a Hilbert space with an inner product ·, · which induces a complete norm  ·  Let C be a nonempty, convex, and closed set in the Hilbert space X, U be an open and convex subset of X such that C ⊂ U, and f : U → R 1 be a convex locally Lipschitzian function. Let M > 4, L ≥ 1, BX (0, M − 1) ∩ argmin(f, C) = ∅,

(6.153)

{y ∈ X : d(y, C) ≤ 1} ⊂ U,

(6.154)

0 < μ1 < μ2 .

(6.155)

We suppose that {x ∈ U : f (x) ≤ inf(f, C) + 1} ⊂ BX (0, M/4),

(6.156)

202

c∗ < 0, f (v) ≥ c∗ for all v ∈ U,

(6.157)

 > 4M + 210μ2 μ−1 , M 1

(6.158)

2 (8μ2 )−1 M > 4M 2 μ−1 1 − 2c∗ + 2| sup{f (v) : v ∈ BX (0, M)}|,

(6.159)

|f (v1 ) − f (v2 )| ≤ Lv1 − v2   + 1) ∩ U, for all v1 , v2 ∈ BX (0, M

(6.160)

δf , δC ∈ (0, 1) satisfy 2 4 max{δf , δC }(16(L + 1)μ−1 1 M

+ sup{|f (v)| : v ∈ BX (0, M)} − c∗  + 6μ2 (L + 3) + 3 +14M  + μ−1 1 (5M + 2μ2 (L + 1) + L + 4)) ≤ 1.

(6.161)

6.12 An Auxiliary Result Proposition 6.9 Let z ∈ BX (0, M) ∩ argmin(f, C),

(6.162)

T0 > 0, x ∈ W 1,1 (0, T0 ; X), d(x(0), C) ≤ δC

(6.163)

and for almost every t ∈ [0, T0 ] there exist ξ(t) ∈ X such that BX (ξ(t), δf ) ∩ ∂f (x(t); x  (t)) = ∅,

(6.164)

PC (x(t) − μξ(t)) ∈ BX (x(t) + x  (t), δC ).

(6.165)

Then for almost every t ∈ [0, T0 ],

6.12 An Auxiliary Result

203

(2μ)−1 (d/dt)(x(t) − z2 ) +(f ◦ x) (t) + f (x(t)) − f (z) ≤ δf (3x  (t) + x(t) − z) −1 +δC (μ−1 1 x(t) − z + μ1 z − φ(t) + ξ(t))

and for all t ∈ [0, T0 ], d(x(t), C) ≤ δC . Proof In view of (6.156) and (6.162), z ≤ M − 1

(6.166)

f (z) = inf(f, C).

(6.167)

φ(t) = x(t) + x  (t).

(6.168)

and

For almost every t ∈ [0, T0 ] set

It is clear that φ : [0, T0 ] → X is a Bochner integrable function. In view of (6.105) and (6.168), for almost every t ∈ [0, T0 ], BX (φ(t), δC ) ∩ C = ∅.

(6.169)

CδC = {x ∈ X : d(x, C) ≤ δC }.

(6.170)

Define

Clearly, Cδ is a convex closed set, for each x ∈ Cδ , BX (x, δC ) ∩ C = ∅

(6.171)

φ(t) ∈ CδC for almost every t ∈ [0, T0 ].

(6.172)

and in view of (6.169),

Lemma 6.8, (6.163), and (6.172) imply that x(t) ∈ CδC for all t ∈ [0, T0 ].

(6.173)

204

It follows from (6.171) and (6.173) that for every t ∈ [0, T0 ] there exists  x (t) ∈ C

(6.174)

x(t) −  x (t) ≤ δC .

(6.175)

such that

By (6.174) and Lemma 2.2, for almost every t ∈ [0, T0 ],  x (t) − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ 0.

(6.176)

By (6.176), x(t) − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ x(t) −  x (t), x(t) − μξ(t) − PC (x(t) − μξ(t)).

(6.177)

It follows from (6.162) and Lemma 2.2 that z − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ 0.

(6.178)

In view of (6.164), for almost every t ∈ [0, T0 ] there exists  ξ (t) ∈ ∂f (x(t); x  (t))

(6.179)

ξ (t), x  (t), f 0 (x(t), x  (t)) = 

(6.180)

ξ(t) −  ξ (t) ≤ δf .

(6.181)

such that

In view of (6.109) and (6.179), for almost every t ∈ [0, T0 ], f (z) ≥ f (x(t)) +  ξ (t), z − x(t).

(6.182)

By (6.165), for almost every t ∈ [0, T0 ], (x(t) + x  (t)) − PC (x(t) − μξ(t)) ≤ δC . Relations (6.168) and (6.183) imply that for almost every t ∈ [0, T0 ], z − x(t) − x  (t), x(t) − μξ(t) − x(t) − x  (t) = z − φ(t), x(t) − μξ(t) − φ(t)

(6.183)

6.12 An Auxiliary Result

205

= z − φ(t), x(t) − μξ(t) − PC (x(t) − μξ(t)) +z − φ(t), PC (x(t) − μξ(t)) − φ(t) ≤ z − φ(t), x(t) − μξ(t) − PC (x(t) − μξ(t)) + δC z − φ(t).

(6.184)

By (6.168), (6.178), (6.183), and the definition of PC , for a. e. t ∈ [0, T0 ], z − φ(t), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ z − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) +δC x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ δC x(t) − μξ(t) − z.

(6.185)

In view of (6.184) and (6.185), for a. e. t ∈ [0, T0 ], x(t) + x  (t) − z, μξ(t) + x  (t) ≤ δC z − φ(t) + δC (x(t) − z − μξ(t)).

(6.186)

It follows from (6.181) and (6.182) that for almost every t ∈ [0, T0 ], f (x(t)) − f (z) ≤  ξ (t), x(t) − z = ξ(t), x(t) − z + x  (t) − ξ(t), x  (t) + ξ (t) − ξ(t), x(t) − z + x  (t) +ξ(t) −  ξ (t), x  (t) ≤ ξ(t), x(t) − z + x  (t) − ξ(t), x  (t) + δf (x(t) − z + 2x  (t)). In view of (6.184) and (6.185), for almost every t ∈ [0, T0 ], ξ(t), x(t) − z + x  (t) ≤ μ−1 −x  (t), x(t) − z + x  (t)

(6.187)

206

+ξ(t) + μ−1 x  (t), x(t) − z + x  (t) ≤ μ−1 −x  (t), x(t) − z + x  (t) +μ−1 δC z − φ(t) +z − φ(t), x(t) − μξ(t) − PC (x(t) − μξ(t))μ−1 ≤ μ−1 δC z − φ(t) −μ−1 x  (t), x(t) − z + x  (t) + μ−1 δC x(t) − μξ(t − z.

(6.188)

Relations (6.187) and (6.188) imply that for almost every t ∈ [0, T0 ], f (x(t)) − f (z) ≤ μ−1 −x  (t), x(t) + x  (t) − z +μ−1 δC (z − φ(t) + x(t) − z + μξ(t)) − ξ(t), x  (t) + δf (x(t) − z + 2x  (t)).

(6.189)

By (6.189), for almost every t ∈ [0, T0 ], f (x(t)) − f (z) ≤ −μ−1 x  (t)2 − μ−1 x  (t), x(t) − z −ξ(t), x  (t) +δC μ−1 (z − φ(t) + x(t) − z + μξ(t)) + δf (x(t) − z + 2x  (t)). In view of (6.155), (6.181), and (6.190), μ−1 x  (t)2 + μ−1 x  (t), x(t) − z + ξ (t), x  (t) + f (x(t)) − f (z) ≤ δf x  (t) + δf x(t) − z + 2x  (t)

(6.190)

6.13 The Convergence Result

207

−1 + δC (ξ(t) + μ−1 1 z − φ(t) + μ x(t) − z).

(6.191)

Corollary 6.4, (6.181), and (6.191) imply that for almost every t ∈ [0, T0 ], (2μ)−1 (d/dt)(x(t) − z2 ) + f 0 (x(t), x  (t)) +f (x(t)) − f (z) + μ−1 x  (t)2 ≤ δf (x(t) − z + 3x  (t)) −1 +δC (ξ(t) + μ−1 1 z − φ(t) + μ x(t) − z).

Together with Proposition 6.3 and the equality f 0 (x(t), x  (t)) = (f ◦ x) (t) for almost every t ∈ [0, T0 ], this implies that for almost every t ∈ [0, T0 ], (2μ)−1 (d/dt)(x(t) − z2 ) +(f ◦ x) (t) + f (x(t)) − f (z) ≤ δf (x(t) − z + 3x  (t)) −1 +δC (ξ(t) + μ−1 1 z − φ(t) + μ x(t) − z).

 

Proposition 6.9 is proved.

6.13 The Convergence Result We will prove the following result. Theorem 6.10 Let 0 < μ1 ≤ 1 < μ2 ,

(6.192)

0 = 4 max{δf , δC }[8(L + 1)(2M 2 μ−1 1 + sup{|f (v)| : v ∈ BX (0, M)} − c∗ )  + 3μ2 (L + 1) + 3 + μ−1 (5M  + 2μ2 (L + 1) + L + 4)]. + 14M 1 Then following two assertions hold.

(6.193)

208

1. Let 8−1 min{δf−1 , δC−1 }(L + 1)−1 ≤ T0 ≤ min{δf−1 , δC−1 }(L + 1)−1 ,

(6.194)

μ1 ≤ μ ≤ μ2 . Assume that x ∈ W 1,1 (0, T0 ; X), x(0) ∈ BX (0, M), d(x(0), C) ≤ δC

(6.195)

and that for almost every t ∈ [0, T0 ], there exists ξ(t) ∈ X such that BX (ξ(t), δf ) ∩ ∂f (x(t); x  (t)) = ∅, PC (x(t) − μξ(t)) ∈ B(x(t) + x  (t), δC ). Then  for all t ∈ [0, T0 ], x(t) ≤ M min{f (x(t)) : t ∈ [0, T0 ]} − inf(f, C), f (T0−1 ≤

T0−1



T0



T0

x(s)ds) − inf(f, C)

0

f (x(s))ds − inf(f, C) ≤ 0 /4,

0

mes{t ∈ [0, T0 ] : f (x(t)) > inf(f, C) + 0 } ≤ T0 /4, d(x(t), C) ≤ δC , t ∈ [0, T0 ] and there exists t ∈ [(3/4)T0 , T0 ] such that x(t) ≤ M. 2. Assume that T ≥ max{δf−1 , δC−1 }(L + 1)−1 , x ∈ W 1,1 (0, T ; X),

(6.196) (6.197)

6.13 The Convergence Result

209

x(0) ∈ BX (0, M), d(x(0), C) ≤ δC and that for almost every t ∈ [0, T ], there exists ξ(t) ∈ X such that (6.196) and (6.197) hold. Then  t ∈ [0, T ] x(t) ≤ M, and there exists a sequence of numbers {Ti }ki=0 , where k is a natural number, such that T0 = 0, Tk = T and that for all integers i = 0, . . . , k − 1, 4−1 min{δf−1 , δC−1 }(L + 1)−1 ≤ Ti+1 − Ti ≤ min{δf−1 , δC−1 }(L + 1)−1 , x(Ti ) ≤ M, i = 0, . . . , k − 1 and that for all i = 0, . . . , k − 1, min{f (x(t)) : t ∈ [Ti , Ti+1 ]} − inf(f, C), f ((Ti+1 − Ti )−1



Ti+1

x(s)ds) − inf(f, C)

Ti

≤ (Ti+1 − Ti )−1



Ti+1

f (x(s))ds − inf(f, C) ≤ 0 .

Ti

Proof In view of (6.156), there exists z ∈ BX (0, M − 1) ∩ argmin(f, C).

(6.198)

Proposition 6.9 implies that for almost every t ∈ [0, T0 ], (2μ)−1 (d/dt)(x(t) − z2 ) +(f ◦ x) (t) + f (x(t)) − f (z) ≤ δf (x(t) − z + 3x  (t)) + δC (ξ(t) + 1 + μ−1 z − φ(t) + μ−1 x(t) − z).

(6.199)

210

For almost every t ∈ [0, T0 ] set φ(t) = x(t) + x  (t).

(6.200)

Define  for all t ∈ [0, τ ]}. E = {τ ∈ (0, T0 ] : x(t) ≤ M

(6.201)

Since the function x is continuous it follows from (6.158) and (6.195) that E = ∅. Set τ = sup(E).

(6.202)

τ ∈E

(6.203)

 t ∈ [0, τ ]. x(t) ≤ M,

(6.204)

Clearly,

and

By (6.160), (6.196), and (6.204), for almost every t ∈ [0, τ ], ξ(t) ≤ L + 1.

(6.205)

Lemma 2.2, (6.198), (6.204), and (6.205) imply that for almost every t ∈ [0, τ ], PC (x(t) − μξ(t)) ≤ PC (x(t) − μξ(t)) − z + z ≤ x(t) − μξ(t) − z + z  + μ2 (L + 1) + 2M. ≤M

(6.206)

It follows from (6.197), (6.200), (6.204), and (6.206) that for almost every t ∈ [0, τ ], φ(t) ≤ PC (x(t) − μξ(t)) + 1  + μ2 (L + 1) + 2M + 1 ≤M and

(6.207)

6.13 The Convergence Result

211

x  (t) ≤ x(t) + φ(t)  + μ2 (L + 1) + 2M + 1. ≤ 2M

(6.208)

Thus for a. e. t ∈ [0, τ ], (6.207) and (6.208) are true. Combined with (6.198), (6.199), and (6.201)–(6.203), this implies that for almost every t ∈ [0, τ ], (2μ)−1 (d/dt)(x(t) − z2 ) +(f ◦ x) (t) + f (x(t)) − f (z)  + 3μ2 (L + 1) + 6M + 3 + M  + M) ≤ δf (6M −1   +δC (μ−1 1 M + μ1 (M + μ2 (L + 1) + 4M + 1) + L + 1).

Integrating the inequality above over the interval [t1 , t2 ] ⊂ [0, τ ], we obtain that for all t1 , t2 ∈ [0, τ ] satisfying t1 < t2 , (2μ)−1 x(t2 ) − z2 − (2μ)−1 x(t1 ) − z2 +f (x(t2 )) − f (x(t1 ))  +

t2

(f (x(s)) − f (z))ds

t1

 + 3μ2 (L + 1) + 3) ≤ (t2 − t1 )δf (14M  +(t2 − t1 )δC μ−1 1 (5M + μ2 (L + 1) + 4) + δC (L + 1)(t2 − t1 ).

(6.209)

Since the function x is continuous it follows from (6.201) to (6.203) that at least one of the following equalities holds: τ = T0 ;

(6.210)

 x(τ ) = M.

(6.211)

Assume that (6.211) holds. By (6.157), (6.192), (6.195), (6.198), (6.209) with t2 = τ , t1 = 0 and (6.211),  − M)2 − 2(μ1 )−1 4M 2 (2μ2 )−1 (M +c∗ − sup{f (v) : v ∈ BX (0, M)}

212

 + 3μ2 (L + 1) + 3) ≤ τ δf (14M  + τ δC μ−1 1 (5M + 6μ2 (L + 1) + 1) + τ δC (L + 1).

(6.212)

By (6.158) and (6.159),  − 2M)2 − 2(μ1 )−1 4M 2 (2μ2 )−1 (M +c∗ − sup{f (v) : v ∈ BX (0, M)} −1 2 2 ≥ 2−1 μ−1 2 M /4 − 2μ1 M

+c∗ − sup{f (v) : v ∈ BX (0, M)} 2 ≥ 16−1 μ−1 2 M .

(6.213)

It follows from (6.194), (6.212), (6.213), and the inequality τ ≤ T0 that 2 16−1 μ−1 2 M  + 3μ2 (L + 1) + 3) ≤ τ δf (14M  +τ δC μ−1 1 (5M + 6μ2 (L + 1) + 1) +τ δC (L + 1)  + 3μ2 (L + 1) + 3) ≤ (L + 1)−1 min{δf−1 , δC−1 }δf (14M  +(L + 1)−1 min{δf−1 , δC−1 }δC [μ−1 1 (5M +6μ2 (L + 1) + 1) + L + 1)]  + 3μ2 + 1 ≤ 7M −1 −1  + 5μ−1 1 M + 6μ2 μ1 + μ1 + 1.

In view of (6.158), (6.192), and (6.214),  2 ≤ 16 · 7μ2 M  + 48μ22 M  2 μ−1 +16μ2 + 90Mμ 1 −1 +96μ22 μ−1 1 + 16μ2 μ1 + 16μ2

(6.214)

6.13 The Convergence Result

213

 μ2 μ−1 1 (202M + 16 + 160μ2 + 32) −1   ≤ μ2 μ−1 1 (202M + 208μ2 ) ≤ 203μ2 μ1 M

and  ≤ 203μ2 μ−1 . M 1 This contradicts (6.158). The contradiction we have reached proves that  x(τ ) < M. Therefore τ = T0 and  t ∈ [0, T0 ]. x(t) ≤ M,

(6.215)

It follows from (6.157), (6.192), (6.195), (6.198), and (6.209) with t2 = T0 and t1 = 0 that 

T0

(f (x(s)) − f (z))ds

0

≤ (2μ1 )−1 x(0) − z2 + f (x(0)) − f (x(T0 ))  + 3μ2 (L + 1) + 3) +T0 δf (14M  +T0 δC μ−1 1 (5M + μ2 (L + 1) + 4) + δC T0 (L + 1) ≤ (2μ1 )−1 4M 2 − c∗ + sup{f (v) : v ∈ BX (0, M)}  + 3μ2 (L + 1) + 3) +T0 δf (14M −1  + T0 δC μ−1 1 (5M + μ2 (L + 1) + 4) + μ1 δC T0 (L + 1).

By (6.191), (6.192), (6.193), (6.198), and (6.216), T0−1



T0

(f (x(s)) − inf(f, C))ds

0

≤ ((2μ1 )−1 4M 2 + sup{f (v) : v ∈ BX (0, M)}

(6.216)

214

−c∗ )8 max{δf , δC }(L + 1)  + 3μ2 (L + 1) + 3) +δf (14M  + δC μ−1 1 (5M + μ2 (L + 1) + 4) ≤ 0 /4.

(6.217)

By (6.217), f (T0−1



T

x(s)ds) − inf(f, C)

0

min{f (x(t)) : t ∈ [0, T0 ]} − inf(f, C) ≤

T0−1



T

f (x(s))ds − inf(f, C) ≤ 0 /4.

(6.218)

0

Define E = {t ∈ [0, T0 ] : f (x(t)) > inf(f, C) + 0 }.

(6.219)

In view of (6.217) and (6.219), T0−1 mes(E)0 ≤ T0−1



T0

(f (x(s)) − inf(f, C))ds ≤ 0 /4

0

and mes(E) ≤ T0 /4.

(6.220)

d(x(t), C) ≤ δC , t ∈ [0, T0 ].

(6.221)

Proposition 6.9 implies that

Set t0 = max{t ∈ [0, T0 ] : f (x(t)) − inf(f, C) ≤ 0 }. It follows from (6.219) and (6.220) that t0 > 3T0 /4. By the choice of t0 , (6.156), (6.161), (6.193), and the inequalities 0 ≤ 1, f (x(t0 )) ≤ inf(f, C) + 1

(6.222)

6.14 Subgradient Projection Algorithm for Zero-Sum Games

215

we have x(t0 ) ≤ M.

(6.223)

Now Assertion 1 follows from (6.215) and (6.217)–(6.223). Assertion 2 follows from Assertion 1 applied by induction. Theorem 6.10 is proved.  

6.14 Subgradient Projection Algorithm for Zero-Sum Games Let (X, ·, ·), (Y, ·, ·) be Hilbert spaces equipped with the complete norms  ·  which are induced by their inner products. Let C be a nonempty closed convex subset of X, D be a nonempty closed convex subset of Y , U be an open convex subset of X, and V be an open convex subset of Y such that C ⊂ U, D ⊂ V ,

(6.224)

and let a borelian function f : U × V → R 1 possess the following properties: (i) for each v ∈ V , the function f (·, v) : U → R 1 is convex and locally Lipschitzian; (ii) for each u ∈ U , the function f (u, ·) : V → R 1 is concave and locally Lipschitzian. Let δC , δD , δf,1 , δf,2 ∈ (0, 1].

(6.225)

For each (ξ, η) ∈ U × V , set ∂x f (ξ, η) = {l ∈ X : f (y, η) − f (ξ, η) ≥ l, y − ξ  for all y ∈ U }, ∂y f (ξ, η) = {l ∈ Y : l, y − η ≥ f (ξ, y) − f (ξ, η) for all y ∈ V }.

(6.226)

Let x ∈ U , y ∈ V , ξ ∈ X, and η ∈ Y . Set f 0 (x, y, ξ ) = lim t −1 [f (x + tξ, y) − f (x, y)],

(6.227)

f 0 (x, y, η) = lim t −1 [f (x, y + tη) − f (x, y)],

(6.228)

∂x f (x, y; ξ ) = {l ∈ ∂x f (x, y) : l, ξ  = f 0 (x, y, ξ )}.

(6.229)

∂y f (x, y; η) = {l ∈ ∂y f (x, y) : l, η = f 0 (x, y, η)}.

(6.230)

t→0+

t→0+

216

We study subgradient projection algorithm for the zero-sum game defined by the triplet (f, U, V ) and prove in Sect. 6.16 a convergence result which is based on an auxiliary result of Sect. 5.15.

6.15 An Auxiliary Result Proposition 6.11 Let T > 0, x ∈ W 1,1 (0, T ; X), y ∈ W 1,1 (0, T ; Y ), μ > 0,

(6.231)

x(t) ∈ U, y(t) ∈ V , t ∈ [0, T ], d(x(0), C) ≤ δC , d(y(0), D) ≤ δD

(6.232)

and let for almost every t ∈ [0, T ] there exist ξ(t) ∈ X, η(t) ∈ Y such that BX (ξ(t), δf,1 ) ∩ ∂x f (x(t), y(t)) = ∅,

(6.233)

PC (x(t) − μξ(t)) ∈ BX (x(t) + x  (t), δC ),

(6.234)

BY (η(t), δf,2 ) ∩ ∂y f (x(t), y(t)) = ∅,

(6.235)

PD (y(t) + μη(t)) ∈ BY (y(t) + y  (t), δD ).

(6.236)

Then for almost every t ∈ [0, T ], every z ∈ C, and every v ∈ D, (2μ)−1 (d/dt)(x(t) − z2 ) +f (x(t), y(t)) − f (z, y(t)) ≤ 4−1 μξ(t)2 +(μ−1 δC + δf,1 )(2 + 3z + 3x(t) + 2μξ(t)), (2μ)−1 (d/dt)(y(t) − v2 ) +f (x(t), v) − f (x(t), y(t)) ≤ 4−1 μη(t)2 +(μ−1 δD + δf,2 )(2 + 3v + 3y(t) + 2μη(t)) and for all t ∈ [0, T ],

6.15 An Auxiliary Result

217

BX (x(t), δC ) ∩ C = ∅, BY (y(t), δD ) ∩ D = ∅. Proof For almost every t ∈ [0, T ] set φX (t) = x(t) + x  (t), φY (t) = y(t) + y  (t).

(6.237)

It is clear that φX : [0, T ] → X and φY : [0, T ] → Y are Bochner integrable functions. In view of (6.138), (6.234), and (6.236), for almost every t ∈ [0, T ], BX (φX (t), δC ) ∩ C = ∅,

(6.238)

BY (φY (t), δD ) ∩ D = ∅.

(6.239)

Cδ = {x ∈ X : d(x, C) ≤ δC },

(6.240)

Dδ = {y ∈ Y : d(y, D) ≤ δD }.

(6.241)

Define

Clearly, Cδ and Dδ are convex closed sets, for each x ∈ Cδ , B(x, δC ) ∩ C = ∅,

(6.242)

B(x, δD ) ∩ D = ∅,

(6.243)

for each x ∈ Dδ ,

In view of (6.238) and (6.239), φX (t) ∈ Cδ , φY (t) ∈ Dδ for almost every t ∈ [0, T ].

(6.244)

Lemma 6.8, (6.232), (6.237), (6.240), (6.241), and (6.244) imply that for every t ∈ [0, T ], x(t) ∈ Cδ , y(t) ∈ Dδ . It follows from the relation above that for every t ∈ [0, T ] there exists  x (t) ∈ C,  y (t) ∈ D

218

such that x(t) −  x (t) ≤ δC , y(t) −  y (t) ≤ δD .

(6.245)

By (6.245) and Lemma 2.2, for almost every t ∈ [0, T ],  x (t) − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ 0

(6.246)

and  y (t) − PD (y(t) + μη(t)), y(t) + μη(t) − PD (y(t) + μη(t)) ≤ 0.

(6.247)

Relations (6.246) and (6.247) imply that for almost every t ∈ [0, T ], x(t) − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ x(t) −  x (t), x(t) − μξ(t) − PC (x(t) − μξ(t))

(6.248)

and y(t) − PD (y(t) + μη(t)), y(t) + μη(t) − PD (y(t) + μη(t)) ≤ y(t) −  y (t), y(t) + μη(t) − PD (y(t) + μη(t)).

(6.249)

Let z ∈ C, v ∈ D.

(6.250)

It follows from (6.250) and Lemma 2.2 that z − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ 0

(6.251)

v − PD (y(t) + μη(t)), y(t) + μη(t) − PD (y(t) + μη(t)) ≤ 0.

(6.252)

and

In view of (6.229), (6.230), (6.233), and (6.235), for almost every t ∈ [0, T ] there exists  ξ (t) ∈ ∂x f (x(t), y(t))

(6.253)

ξ (t), x  (t), f 0 (x(t), y(t), x  (t) ≥ 

(6.254)

ξ(t) −  ξ (t) ≤ δf,1

(6.255)

such that

6.15 An Auxiliary Result

219

and  η(t) ∈ ∂y f (x(t), y(t))

(6.256)

η(t), y  (t), f 0 (x(t), y(t), y  (t)) ≤ 

(6.257)

η(t) −  η(t) ≤ δf,2 .

(6.258)

such that

In view of (6.226), (6.253), and (6.256), for almost every t ∈ [0, T ], f (z, y(t)) ≥ f (x(t), y(t)) +  ξ (t), z − x(t), f (x(t), v) − f (x(t), y(t)) ≤  η(t), v − y(t) and f (x(t), y(t)) − f (z, y(t)) ξ (t), x  (t) ≤  ξ (t), x(t) − z + x  (t) − 

(6.259)

f (x(t), y(t)) − f (x(t), v) η(t), y  (t). ≥  η(t), y(t) − v + y  (t) − 

(6.260)

By (6.234) and (6.236), for almost every t ∈ [0, T ], (x(t) + x  (t)) − PC (x(t) − μξ(t)) ≤ δC

(6.261)

and (y(t) + y  (t)) − PD (y(t) + μη(t)) ≤ δD .

(6.262)

Relations (6.234), (6.237), (6.245), (6.261), and (6.262) imply that for almost every t ∈ [0, T ], z − x(t) − x  (t), x(t) − μξ(t) − x(t) − x  (t) = z − φX (t), x(t) − μξ(t) − φX (t) = z − φX (t), x(t) − μξ(t) − PC (x(t) − μξ(t)) +z − φX (t), PC (x(t) − μξ(t)) − φX (t)

220

≤ z − φX (t), x(t) − μξ(t) − PC (x(t) − μξ(t)) + δC z − φX (t)

(6.263)

and v − y(t) − y  (t), y(t) + μη(t) − y(t) − y  (t) = v − φY (t), y(t) + μη(t) − φY (t) = v − φY (t), y(t) + μη(t) − PC (y(t) + μη(t)) +v − φY (t), PD (y(t) + μη(t)) − φY (t) ≤ v − φY (t), y(t) + μη(t) − PD (y(t) + μη(t)) + δD v − φY (t).

(6.264)

By (6.237), (6.250)–(6.252), (6.261), and (6.262), z − φX (t), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ z − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) +PC (x(t) − μξ(t)) − φX (t), x(t) − μξ(t) − PC (x(t) − μξ(t)) ≤ z − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) +δC x(t) − μξ(t) − z ≤ z − PC (x(t) − μξ(t)), x(t) − μξ(t) − PC (x(t) − μξ(t)) +δC (x(t) + μξ(t) + z) ≤ δC (x(t) + μξ(t) + z) and v − φY (t), y(t) + μη(t) − PD (y(t) + μη(t)) ≤ v − PD (y(t) + μη(t)), y(t) + μη(t) − PD (y(t) + μη(t)) +PD (y(t) + μη(t)) − φY (t), y(t) + μη(t) − PD (y(t) + μη(t)) ≤ v − PD (y(t) + μη(t)), y(t) + μη(t) − PD (y(t) + μη(t))

(6.265)

6.15 An Auxiliary Result

221

+δD y(t) + μη(t) − v ≤ v − PD (y(t) + μη(t)), y(t) + μη(t) − PD (y(t) + μη(t)) +δD (y(t) + μη(t) + v) ≤ δD (y(t) + μη(t) + v).

(6.266)

Lemma 2.2, (6.237), (6.250), and (6.261)–(6.266) imply that for almost every t ∈ [0, T ], x(t) + x  (t) − z, μξ(t) + x  (t) ≤ z − φX (t), x(t) − μξ(t) − PC (x(t) − μξ(t))) +δC z − φX (t) ≤ δC (z − x(t) − x  (t) + x(t) + μξ(t) + z) ≤ δC (1 + z + x(t) + μξ(t) + x(t) + μξ(t) + z)

(6.267)

and y(t) + y  (t) − v, −μη(t) + y  (t) ≤ v − φY (t), y(t) + μη(t) − PD (y(t) + μη(t))) +δC v − φY (t) ≤ δD (v − y(t) − y  (t) + y(t) + μη(t) + v) ≤ δD (1 + v + y(t) + μη(t) + y(t) + μη(t) + v).

(6.268)

It follows from (6.163), (6.250), (6.253), (6.254), (6.256), (6.258), (6.261), and Lemma 2.2 that for almost every t ∈ [0, T ], f (x(t), y(t)) − f (z, y(t)) ≤  ξ (t), x(t) − z = ξ(t), x(t) − z + x  (t) − ξ(t), x  (t) + ξ (t) − ξ(t), x(t) − z + x  (t) +ξ(t) −  ξ (t), x  (t)

222

≤ ξ(t), x(t) − z + x  (t) − ξ(t), x  (t) +δf,1 x(t) − z + 2x  (t) ≤ ξ(t), x(t) − z + x  (t) − ξ(t), x  (t) +δf,1 (2x(t) − z + x  (t) + z + x(t)) ≤ ξ(t), x(t) − z + x  (t) − ξ(t), x  (t) + δf,1 (z + x(t) + 2 + 2(z + x(t) + μξ(t)))

(6.269)

and f (x(t), y(t)) − f (x(t), v) ≥  η(t), y(t) − v = η(t), y(t) − v + y  (t) − η(t), y  (t) + η(t) − η(t), y(t) − v + y  (t) η(t) −  η(t), y  (t) ≥ η(t), y(t) − v + y  (t) − η(t), y  (t) −δf,2 y(t) − v + 2y  (t) ≥ η(t), y(t) − v + y  (t) − η(t), y  (t) −δf,2 (2y(t) − v + y  (t) + v + y(t)) ≥ η(t), y(t) − v + y  (t) − η(t), y  (t) − δf,2 (2 + 2(y(t) + μη(t) + v) + v + y(t)). In view of (6.267), for almost every t ∈ [0, T ], ξ(t), x(t) − z + x  (t) ≤ μ−1 −x  (t), x(t) − z + x  (t) +ξ(t) + μ−1 x  (t), x(t) − z + x  (t) ≤ μ−1 −x  (t), x(t) − z + x  (t)

(6.270)

6.15 An Auxiliary Result

223

+ μ−1 δC (1 + 2z + 2x(t) + 2μξ(t)).

(6.271)

In view of (6.268), for almost every t ∈ [0, T ], η(t), y(t) − v + y  (t) = μ−1 y  (t), y(t) − v + y  (t) +η(t) − μ−1 y  (t), y(t) − v + y  (t) ≥ μ−1 y  (t), y(t) − v + y  (t) − μ−1 δD (1 + 2v + 2y(t) + 2μη(t)).

(6.272)

Relations (6.162), (6.163), (6.269), and (6.271) imply that for almost every t ∈ [0, T ], f (x(t), y(t)) − f (z, y(t)) ≤ μ−1 −x  (t), x(t) + x  (t) − z −ξ(t), x  (t) +μ−1 δC (1 + 2z + 2x(t) + 2μξ(t)) + δf,1 (3z + x(t) + 2 + 2μξ(t)).

(6.273)

By (6.270) are (6.272), f (x(t), y(t)) − f (x(t), v) ≤ μ−1 y  (t), y(t) + y  (t) − v −η(t), y  (t) −δf,2 (2 + 3v + 3y(t) + 2μη(t)) − μ−1 δD (1 + 2v + 2y(t) + 2μη(t)). In view of (6.273) and (6.274), for almost every t ∈ [0, T ], μ−1 x  (t)2 + μ−1 x  (t), x(t) − z +ξ(t), x  (t) + f (x(t), y(t)) − f (z, y(t))

(6.274)

224

≤ (μ−1 δC + δf,1 )(2 + 3z + 3x(t) + 2μξ(t))

(6.275)

and μ−1 y  (t)2 + μ−1 y  (t), y(t) − v −η(t), y  (t) + f (x(t), v) − f (x(t), y(t)) ≤ (μ−1 δD + δf,2 )(2 + 3v + 3y(t) + 2μη(t)).

(6.276)

Corollary 6.4, (6.275), and (6.276) imply that for almost every t ∈ [0, T ], (2μ)−1 (d/dt)(x(t) − z2 ) +f (x(t), y(t)) − f (z, y(t)) ≤ μ−1 x  (t), x(t) − z +f (x(t), y(t)) − f (z, y(t)) +μ−1 x  (t) + 2−1 μξ(t)2 ≤ μ−1 x  (t), x(t) − z +f (x(t), y(t)) − f (z, y(t)) +μ−1 x  (t)2 + x  (t), ξ(t) + 4−1 μξ(t)2 ≤ 4−1 μξ(t)2 + (μ−1 δC + δf,1 )(2 + 3z + 3x(t) + 2μξ(t)) and (2μ)−1 (d/dt)(y(t) − v2 ) +f (x(t), v) − f (x(t), y(t)) ≤ μ−1 y  (t), y(t) − v +f (x(t), v)) − f (x(t), y(t)) +μ−1 y  (t) − 2−1 μη(t)2 = μ−1 y  (t), y(t) − v

6.16 A Convergence Result for Games on Bounded Sets

225

+f (x(t), v) − f (x(t), y(t)) +μ−1 y  (t)2 − y  (t), η(t) + 4−1 μη(t)2 ≤ 4−1 μη(t)2 + (μ−1 δD + δf,2 )(2 + 3v + 3y(t) + 2μη(t)).  

Proposition 6.11 is proved.

6.16 A Convergence Result for Games on Bounded Sets Suppose that L1 , L2 ≥ 1, M1 , M2 > 1, C ⊂ BX (0, M1 − 1), D ⊂ BY (0, M2 − 1),

(6.277)

{x ∈ X : d(x, C) ≤ 1} ⊂ U, {y ∈ Y : d(y, D) ≤ 1} ⊂ V } and that the function f : U × V → R 1 is continuous and has the following properties: (iii) for each v ∈ V ∩ BY (0, M2 + 1), |f (u1 , v) − f (u2 , v)| ≤ L1 u1 − u2  for all u1 , u2 ∈ U ∩ BX (0, M1 + 1); (iv) for each u ∈ U ∩ BX (0, M1 + 1), |f (u, v1 ) − f (u, v2 )| ≤ L2 v1 − v2  for all v1 , v2 ∈ V ∩ BY (0, M2 + 1). Let x∗ ∈ C and y∗ ∈ D

(6.278)

f (x∗ , y) ≤ f (x∗ , y∗ ) ≤ f (x, y∗ )

(6.279)

satisfy

for each x ∈ C and each y ∈ D. In this section we prove the following result.

226

Theorem 6.12 Let M, T > 0, x ∈ W 1,1 (0, T ; X), y ∈ W 1,1 (0, T ; Y ), x(t) ∈ U, y(t) ∈ V , t ∈ [0, T ], d(x(0), C) ≤ δC , d(y(0), D) ≤ δD

(6.280)

and for almost every t ∈ [0, T ] there exist ξ(t) ∈ X, η(t) ∈ Y such that BX (ξ(t), δf,1 ) ∩ ∂x f (x(t), y(t)) = ∅,

(6.281)

PC (x(t) − μξ(t)) ∈ BX (x(t) + x  (t), δC ),

(6.282)

BY (η(t), δf,2 ) ∩ ∂y f (x(t), y(t)) = ∅,

(6.283)

PD (y(t) + μη(t)) ∈ BX (yt) + y  (t), δD ).

(6.284)

Then for almost every t ∈ [0, T ], BX (x(t), δC ) ∩ C = ∅, BY (y(t), δD ) ∩ D = ∅. BX (T −1



T

x(t)dt, δC ) ∩ C = ∅,

0

BY (T

−1



T

y(t)dt, δD ) ∩ D = ∅,

0

|T −1



T 0

f (x(t), y(t))dt − f (x∗ , y∗ )|

≤ max{6M1 μ−1 δC + 2δC (L1 + 1) + 6δf,1 M1 + 2μδf,1 (L1 + 1) +4−1 μ(L1 + 1)2 + (2μT )−1 4M12 , 4(2T μ)−1 M22 + 6M2 μ−1 δD + 2δD (L2 + 1) +6δf,2 M2 + 2μδf,2 (L2 + 1) + 4−1 μ(L2 + 1)2 }, |f (T

−1



T

x(t)dt, T 0

−1

 0

T

y(t)dt) − T

−1



T

f (x(t), y(t))dt| 0

6.16 A Convergence Result for Games on Bounded Sets

227

≤ max{6M1 μ−1 δC + δC (3L1 + 1) + 6δf,1 M1 + 2μδf,1 (L1 + 1) +4−1 μ(L1 + 1)2 + (2μT )−1 4M12 , 4(2T μ)−1 M22 + 6M2 μ−1 δD + δD (3L2 + 1) +6δf,2 M2 + 2μδf,2 (L2 + 1) + 4−1 μ(L2 + 1)2 }, and for every z ∈ C and every v ∈ D, −1

f (z, T



T

y(t)dt) ≥ f (T

−1

0



T

x(t)dt, T

−1

0



T

y(t)dt) 0

−2 max{6M1 μ−1 δC + δC (3L1 + 1) + 6δf,1 M1 + 2μδf,1 (L1 + 1) +4−1 μ(L1 + 1)2 + (2μT )−1 4M12 , (2T μ)−1 4M22 + 6M2 μ−1 δD + δD (3L2 + 1) +6δf,2 M2 + 2μδf,2 (L2 + 1) + 4−1 μ(L2 + 1)2 }, f (T −1



T 0

x(t)dt, v) ≤ f (T −1



T

x(t)dt, T −1

0



T

y(t)dt) 0

+2 max{6M1 μ−1 δC + δC (3L1 + 2) + 6δf,1 M1 +4−1 μ(L1 + 1)2 + (2μT )−1 4M12 , 4(2T μ)−1 M22 + 6M2 μ−1 δD + δD (3L2 + 1) +6δf,2 M2 + 2μδf,2 (L2 + 1) + 4−1 μ(L2 + 1)2 }. Proof Let z ∈ C, v ∈ D. Proposition 6.11 implies that for almost every t ∈ [0, T ], (2μ)−1 (d/dt)(x(t) − z2 ) +f (x(t), y(t)) − f (z, y(t)) ≤ 4−1 μξ(t)2 + (μ−1 δC + δf,1 )(2 + 3z

(6.285)

228

+ 3x(t) + 2μξ(t)),

(6.286)

(2μ)−1 (d/dt)(y(t) − v2 ) +f (x(t), v) − f (x(t), y(t)) ≤ 4−1 μη(t)2 + (μ−1 δD + δf,2 )(2 + 3v + 3y(t) + 2μη(t))

(6.287)

and that for all t ∈ [0, T ], BX (x(t), δC ) ∩ C = ∅, BY (y(t), δD ) ∩ D = ∅.

(6.288)

In view of (6.225), (6.277), and (6.288), for all t ∈ [0, T ], x(t) ≤ M1 , y(t) ≤ M2 .

(6.289)

Properties (iii) and (iv), (6.281), (6.283), and (6.289) imply that for almost every t ∈ [0, T ], ξ(t) ≤ L1 + 1, η(t) ≤ L2 + 1.

(6.290)

By (6.277), (6.285)–(6.287), (6.289), and (6.290), for almost every t ∈ [0, T ], (2μ)−1 (d/dt)(x(t) − z2 ) +f (x(t), y(t)) − f (z, y(t)) ≤ 4−1 μ(L1 + 1)2 +(μ−1 δC + δf,1 )(2 + 3(M1 − 1) + 3M1 + 2μ(L1 + 1)) ≤ 4−1 μ(L1 + 1)2 + (μ−1 δC + δf,1 )(6M1 + 2μ(L1 + 1)) = 6M1 μ−1 δC + 2δC (L1 + 1) + 6M1 δf,1 + 2μδf,1 (L1 + 1) + 4−1 μ(L1 + 1)2 and

(6.291)

6.16 A Convergence Result for Games on Bounded Sets

229

(2μ)−1 (d/dt)(y(t) − v2 ) +f (x(t), v) − f (x(t), y(t)) ≤ 4−1 μ(L2 + 1)2 +(μ−1 δD + δf,2 )(2 + 3(M2 − 1) + 3M2 + 2μ(L2 + 1)) ≤ 4−1 μ(L2 + 1)2 + (μ−1 δD + δf,2 )(6M2 + 2μ(L2 + 1)) = 6M2 μ−1 δD + 2δD (L2 + 1) + 6M2 δf,2 + 2μδf,2 (L2 + 1) + 4−1 μ(L2 + 1)2 .

(6.292)

By integrating (6.291) and (6.292) we obtain that T −1 ((2μ)−1 x(T ) − z2 − (2μ)−1 x(0) − z2 ) +T

−1



T

f (x(t), y(t))dt − T

−1



T

f (z, y(t))dt

0

0

≤ 6M1 μ−1 δC + 2δC (L1 + 1) + 6M1 δf,1 + 2μδf,1 (L1 + 1) + 4−1 μ(L1 + 1)2

(6.293)

and T −1 ((2μ)−1 y(T ) − v2 − (2μ)−1 y(0) − z2 ) +T −1



T

0

f (x(t), v)dt − T −1



T

f (x(t), y(t))dt 0

≤ 6M2 μ−1 δD + 2δD (L2 + 1) + 6M2 δf,2 + 2μδf,2 (L2 + 1) + 4−1 μ(L2 + 1)2 .

(6.294)

It follows from (6.277), (6.280), (6.285), (6.293), and (6.294) that T −1

 0

T

f (x(t), y(t))dt − T −1



T

f (z, y(t))dt 0

≤ (2μT )−1 4M12 + 6M1 μ−1 δC + 2δC (L1 + 1) + 6M1 δf,1 + 2μδf,1 (L1 + 1) + 4−1 μ(L1 + 1)2

(6.295)

230

and T −1



T



f (x(t), v)dt − T −1

T

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

0

0

≤ (2μT )−1 4M22 + 6M2 μ−1 δD + 2δD (L2 + 1) + 6M2 δf,2 + 2μδf,2 (L2 + 1) + 4−1 μ(L2 + 1)2 .

(6.296)

Set Δ1 = (2μT )−1 4M12 + 6M1 μ−1 δC + 2δC (L1 + 1) + 6M1 δf,1 + 2μδf,1 (L1 + 1) + 4−1 μ(L1 + 1)2

(6.297)

and Δ2 = (2μT )−1 4M22 + 6M2 μ−1 δD + 2δD (L2 + 1) + 6M2 δf,2 + 2μδf,2 (L2 + 1) + 4−1 μ(L2 + 1)2 .

(6.298)

By (6.278), (6.279), and (6.295)–(6.298), T −1



T 0

≥T

−1



T

f (x(t), y(t))dt − f (x∗ , y∗ )

f (x(t), y(t))dt − T

−1



0

T 0

f (x(t), y∗ )dt ≥ −Δ2

and T −1



T 0

≤ T −1



T

f (x(t), y(t))dt − f (x∗ , y∗ )

f (x(t), y(t))dt − T −1

0

 0

T

f (x∗ , y(t))dt ≤ Δ1 .

The relations above imply that |T −1



T 0

f (x(t), y(t))dt − f (x∗ , y∗ )| ≤ max{Δ1 , Δ2 }.

Clearly, the functions PC (x(t)), t ∈ [0, T ] and PD (y(t)), t ∈ [0, T ] are Bochner integrable. Set

6.16 A Convergence Result for Games on Bounded Sets

 x = T −1



T

PC (x(t))dt,  y = T −1



0

xT = T −1

231 T

PC (y(t))dt,

(6.299)

0



T

x(s)ds, yT = T −1



0

T

y(s)ds.

(6.300)

0

In view of (6.299),  x ∈ C,  y ∈ D.

(6.301)

By (6.288) and (6.299), 

 x − T −1

T

x(s)ds ≤ δC ,

0

 y−T



−1

T

y(s)ds ≤ δD .

(6.302)

0

It follows from (6.295) to (6.298) and (6.301) that T −1



T

f (x(t), y(t))dt − T −1



0

T

f ( x , y(t))dt ≤ Δ1

(6.303)

f (x(t),  y )dt ≥ −Δ2 .

(6.304)

0

and T

−1



T

−1

f (x(t), y(t))dt − T



0

T

0

Properties (iii) and (iv), (6.299), and (6.302) imply that for all t ∈ [0, T ], |f ( x , y(t)) − f (T

−1



T

x(s)ds, y(t))| ≤ L1 δC ,

(6.305)

0

|f (x(t),  y ) − f (x(t), T −1



T

y(s)ds)| ≤ L2 δD .

(6.306)

f (xT , y(t))dt| ≤ L1 δC ,

(6.307)

f (x(t), yT )dt| ≤ L2 δD .

(6.308)

0

In view of (6.300), (6.305), and (6.306), T −1 |



T

 f ( x , y(t))dt −

0

T −1 |

 0

T

0 T

 f (x(t),  y )dt − 0

T

232

By (6.303), (6.304), (6.307), and (6.308), T

−1



T

f (x(t), y(t))dt ≤ Δ1 + T

−1



T

f ( x , y(t))dt

0

0

≤ Δ1 + T −1



T

f (xT , y(t))dt + L1 δC

(6.309)

0

and T

−1



T

f (x(t), y(t))dt ≥ −Δ2 + T

−1

0



T

f (x(t),  y )dt

0

≥ −Δ2 − L2 δD + T −1



T

f (x(t), yT )dt.

(6.310)

0

Properties (i) and (ii), (6.300), (6.309), and (6.310) imply that  T −1 f (xT , yT ) − T f (x(t), y(t))dt 0

= f (T −1



T

x(t)dt, yT ) − T −1



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

0

≤T

−1



T

T

0

f (x(t), yT )dt − T

−1



T

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

0

0

≤ Δ 2 + L 2 δD , f (xT , yT ) − T −1



T

f (x(t), y(t))dt 0

= f (xT , T

−1



T

y(t)dt) − T

−1



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

0

≥ T −1



T

T 0

f (xT , y(t))dt − T −1



0

T

f (x(t), y(t))dt 0

≥ −Δ1 − L1 δC and |f (xT , yT ) − T

−1



T

f (x(t), y(t))dt| 0

≤ max{Δ1 + L1 δC , Δ2 + L2 δD }.

(6.311)

6.16 A Convergence Result for Games on Bounded Sets

233

Let z ∈ C, v ∈ D.

(6.312)

By (6.295)–(6.298), 

T −1

T

f (x(t), y(t))dt − T −1



0

T

f (z, y(t))dt ≤ Δ1

(6.313)

f (x(t), y(t))dt ≤ Δ2 .

(6.314)

0

and T

−1



T

f (x(t), v)dt − T

−1



0

T 0

Properties (i) and (ii) and (6.300) imply that T

−1



T

f (z, y(t))dt ≤ f (z, T

−1



0

T

y(t)dt) = f (z, yT )

(6.315)

x(t)dt, v) = f (xT , v).

(6.316)

0

and T −1



T

f (x(t), v)dt ≥ f (T −1



0

T

0

By (6.313) and (6.315), T −1



T

f (x(t), y(t))dt − f (z, yT )

0

≤ T −1



T

f (x(t), y(t))dt − T −1

0



T

f (z, y(t))dt ≤ Δ1 .

0

Together with (6.311) this implies that f (z, yT ) ≥ T −1



T

f (x(t), y(t))dt − Δ1

0

≥ f (xT , yT ) − Δ1 − max{Δ1 + L1 δC , Δ2 + L2 δD }. It follows from (6.314) and (6.316), T −1



T 0

f (x(t), y(t))dt − f (xT , v)

234

≥ T −1



T

f (x(t), y(t))dt − T −1

0



T

f (x(t), v)dt ≥ −Δ2 .

0

Together with (6.311) this implies that f (xT , v) ≤ T

−1



T

f (x(t), y(t))dt + Δ2

0

≤ f (xT , yT ) + Δ2 + max{Δ1 + L1 δC , Δ2 + L2 δD }.  

This completes the proof of Theorem 6.12.

We are interested to make the best choice of the parameter μ. In view of Theorem 6.12, we need to minimize the function max{6M1 μ−1 δC + 2μ(L1 + 1)2 , 6M2 μ−1 δD + 2μ(L2 + 1)2 }. It is not difficult to see that the best choice of μ is c1 max{δC , δD }1/2 where the constant c1 depends on L1 , M1 , L2 , M2 . Then the minimal value of the function above is c2 max{δC , δD }1/2 , where the constant c2 depends on L1 , M1 , L2 , M2 . The best choice of T is c3 max{δC , δD }1/2 , where the constant c3 also depends on L1 , M 1 , L 2 , M 2 .

6.17 A Convergence Result for Games on Unbounded Sets We continue to use the notation and definitions of Section 14. Let x∗ ∈ C, y∗ ∈ D

(6.317)

f (x∗ , v) ≤ f (x∗ , y∗ ) ≤ f (x, y∗ )

(6.318)

and

for all x ∈ C and all v ∈ D. Let M0 ≥ 1, D ⊂ BY (0, M0 − 1),

(6.319)

x∗  ≤ M/4,

(6.320)

M > 8,

6.17 A Convergence Result for Games on Unbounded Sets

235

 ≥ 4M, for each y ∈ V ∩ BY (0, M0 ), M {x ∈ U : f (x, y)  ≤ sup{f (x∗ , v) : v ∈ V ∩ BY (0, M0 + 1)} + 4} ⊂ BX (0, M/4), L1 , L 2 ≥ 1

(6.321) (6.322)

and the continuous function f : U × V → R 1 has the following properties: (v) for each v ∈ V ∩ BY (0, M0 + 1), |f (u1 , v) − f (u2 , v)| ≤ L1 u1 − u2   + 1); for all u1 , u2 ∈ U ∩ BX (0, M  + 1, (vi) for each u ∈ BX (0, M) |f (u, v1 ) − f (u, v2 )| ≤ L2 v1 − v2  for all v1 , v2 ∈ BY (0, M0 + 1); (vii) {x ∈ X : d(x, C) ≤ 1} ⊂ U, {y ∈ Y : d(y, D) ≤ 1} ⊂ V . In this section we prove the following result. Theorem 6.13 Let μ ∈ (0, 1] satisfy μ(L1 + 1)2 ≤ 1,  + 2μ(L1 + 1)) ≤ 1, (μ−1 δC + δf,1 ))(4M

(6.323)

T > 0, x ∈ W 1,1 (0, T ; X), y ∈ W 1,1 (0, T ; Y ), x(t) ∈ U, y(t) ∈ V , t ∈ [0, T ], d(x(0), C) ≤ δC , d(y(0), D) ≤ δD ,

(6.324)

x(0) ≤ M,

(6.325)

and let for almost every t ∈ [0, T ] there exist ξ(t) ∈ X, η(t) ∈ Y such that

236

BX (ξ(t), δf,1 ) ∩ ∂x f (x(t), y(t)) = ∅,

(6.326)

PC (x(t) − μξ(t)) ∈ BX (x(t) + x  (t), δC ),

(6.327)

BY (η(t), δf,2 ) ∩ ∂y f (x(t), y(t)) = ∅,

(6.328)

PD (y(t) + μη(t)) ∈ BY (yt) + y  (t), δD ).

(6.329)

Let  −1 δC + 4δC (L1 + 1) + 12δf,1 M  Δ1 = 12Mμ  2 + 4μδf,1 (L1 + 1), + 2−1 μ(L1 + 1)2 + (2μT )−1 8M

(6.330)

Δ2 = 12M0 μ−1 δD + 4δD (L2 + 1) + 12δf,2 M0 + 2−1 μ(L2 + 1)2 + (2μT )−1 8M02 + 2μδf,2 (L2 + 1), xT = T

−1



T

x(t)dt, yT = T

−1



0

T

y(t)dt. 0

Then for almost every t ∈ [0, T ], BX (x(t), δC ) ∩ C = ∅, BY (y(t), δD ) ∩ D = ∅. BX (xT , δC ) ∩ C = ∅, BY (yT , δD ) ∩ D = ∅, xT ∈ U, yT ∈ V , |T −1



T 0

f (x(t), y(t))dt − f (x∗ , y∗ )| ≤ max{Δ1 , Δ2 },

|f (xT , yT ) − T

−1



T

f (x(t), y(t))dt| ≤ max{Δ1 , Δ2 },

0

and for every z ∈ C and every v ∈ D, f (z, yT ) ≥ f (xT , yT )− ≤ max{Δ1 , Δ2 }, f (xT , v) ≤ f (xT , yT ) + max{Δ1 , Δ2 }.

(6.331)

6.17 A Convergence Result for Games on Unbounded Sets

237

Proof Proposition 6.11 implies that for every z ∈ C, every v ∈ D, and almost every t ∈ [0, T ], (2μ)−1 (d/dt)(x(t) − z2 ) +f (x(t), y(t)) − f (z, y(t)) ≤ 4−1 μξ(t)2 + (μ−1 δC + δf,1 )(2 + 3z + 3x(t) + 2μξ(t))

(6.332)

and (2μ)−1 (d/dt)(y(t) − v2 ) +f (x(t), v) − f (x(t), y(t)) ≤ 4−1 μη(t)2 + (μ−1 δD + δf,2 )(2 + 3v + 3y(t) + 2μη(t)).

(6.333)

Set  − M for all t ∈ [0, τ ]}. E = {τ ∈ [0, T ] : x(t) − x∗  ≤ M

(6.334)

In view of (6.322), (6.325), and (6.334), E = ∅. Set τ0 = sup(E). Clearly, τ0 ∈ E.

(6.335)

Proposition 6.11 implies that for every t ∈ [0, T ], BX (x(t), δC ) ∩ C = ∅, BY (y(t), δD ) ∩ D = ∅. In view of (6.319) and (6.334)–(6.336), for all t ∈ [0, τ0 ],

(6.336)

238

y(t) ≤ M0 ,  x(t) ≤ M.

(6.337)

Properties (v) and (vi), (6.326), (6.328), (6.336), and (6.337) imply that for almost every t ∈ [0, τ0 ], ξ(t) ≤ L1 + 1, η(t) ≤ L2 + 1.

(6.338)

We claim that τ0 = T . Assume the contrary. Then in view of (6.322) and (6.334)–(6.336), τ0 < T ,

(6.339)

 − M, x(τ0 ) − x∗  = M

(6.340)

 x(τ0 ) > 3M/4.

(6.341)

By (6.317), (6.320), (6.332), (6.337), and (6.338), for almost every t ∈ [0, τ0 ], (2μ)−1 (d/dt)(x(t) − x∗ 2 ) +f (x(t), y(t)) − f (x∗ , y(t)) ≤ 4−1 μ(L1 + 1)2  + 2μ(L1 + 1)). + (μ−1 δC + δf,1 )(2 + M + 3M

(6.342)

By (6.321) and (6.341), f (x(τ0 ), y(τ0 )) > sup{f (x∗ , v) : v ∈ V ∩ BY (0, M0 + 1)} + 4.

(6.343)

Since the function f is continuous it follows from (6.343) that there exists τ1 ∈ (0, τ0 ) such that for each t ∈ [τ1 , τ0 ], f (x(t), y(t)) > sup{f (x∗ , v) : v ∈ V ∩ BY (0, M0 + 1)} + 4.

(6.344)

6.17 A Convergence Result for Games on Unbounded Sets

239

By integrating (6.342) on [τ1 , τ0 ] we obtain that (2μ)−1 x(τ0 ) − x∗ 2 − (2μ)−1 x(τ1 ) − x∗ 2  +

τ0



τ0

f (x(t), y(t))dt −

τ1

τ1

f (x∗ , y(t))dt

≤ (τ0 − τ1 )(4−1 μ(L1 + 1)2  + 2μ(L1 + 1)). + (μ−1 δC + δf,1 )(2 + M + 3M

(6.345)

It follows from (6.320), (6.322), (6.323), (6.325), (6.337), (6.344), and (6.345) that  4(τ0 − τ1 ) ≤

τ0

f (x(t), y(t))dt τ1

−(τ0 − τ1 ) sup{f (x∗ , v) : v ∈ V ∩ BY (0, M0 + 1)} ≤ 2(τ0 − τ1 ), a contradiction. The contradiction we have reached proves that τ0 = T .

(6.346)

By (6.320), (6.322), (6.334)–(6.336), and (6.346),  − M for all t ∈ [0, T ], x(t) − x∗  ≤ M  − (3/4)M for all t ∈ [0, T ]. x(t) ≤ M

(6.347)

Set D˜ = D, V˜ = V ,  − 1), U˜ = U ∩ {x ∈ X : x < M}.  C˜ = C ∩ BX (0, M

(6.348)

In view of (6.324), there exists  x (0) ∈ C such that x(0) −  x (0) ≤ δC .

(6.349)

It follows from (6.320), (6.347), and (6.349) that  − (3/4)M + 1 ≤ M  − 1.  x (0) ≤ x(0) + 1 ≤ M By (6.348), (6.350), and the choice of  x (0), ˜  x (0) ∈ C,

(6.350)

240

˜ ≤ x(0) −  d(x(0), C) x (0) ≤ δC .

(6.351)

Lemma 2.2, (6.317), and (6.347) imply that for all t ∈ [0, T ],  − M. x∗ − PC (x(t)) ≤ x∗ − x(t) ≤ M

(6.352)

It follows from (6.319), (6.320), (6.323), (6.338), and (6.352) that for all t ∈ [0, T ]  − M + 1, x∗ − PC (x(t) + μξ(t)) ≤ M  − 3M/4 + 1 ≤ M  − 1. PC (x(t) + μξ(t)) ≤ M

(6.353)

In view of (6.348) and (6.353), ˜ PC (x(t) + μξ(t)) ∈ C, PC (x(t) + μξ(t)) = PC˜ (x(t) + μξ(t)).

(6.354)

By (6.347), (6.348), (6.351), and (6.354) all the assumptions of Theorem 6.12 hold with ˜ U = U˜ , D = D, ˜ V = V˜ . C = C,

(6.355)

Together with (6.330) and (6.331) and property (vii) this implies that for all t ∈ [0, T ], BX (x(t), δC ) ∩ C˜ = ∅, BY (y(t), δD ) ∩ D˜ = ∅. BX (xT , δC ) ∩ C˜ = ∅,

|T

−1



T 0

BY (yT , δD ) ∩ D = ∅,

(6.356)

xT ∈ U, yT ∈ V ,

(6.357)

f (x(t), y(t))dt − f (x∗ , y∗ )| ≤ max{Δ1 , Δ2 },

|f (xT , yT ) − T −1

 0

for every

T

f (x(t), y(t))dt| ≤ max{Δ1 , Δ2 },

6.17 A Convergence Result for Games on Unbounded Sets

241

 − 1) z ∈ C˜ = C ∩ BX (0, M and every v ∈ D, f (z, yT ) ≥ f (xT , yT ) − max{Δ1 , Δ2 }, f (xT , v) ≤ f (xT , yT ) + max{Δ1 , Δ2 }.

(6.358)

It follows from (6.319) to (6.321), (6.348), and (6.356) to (6.358) that for each ˜ z ∈ C \ C, f (z, yT ) ≥ sup{f (x∗ , v) : v ∈ V ∩ BX (0, M0 + 1)} + 4 ≥ f (x∗ , yT ) + 4 ≥ f (xT , yT ) − max{Δ1 , Δ2 } + 4. Theorem 6.13 is proved.

 

As in the case of Theorem 6.12 we can show that the best choice of μ is c1 max{δC , δD }1/2 where the constant c1 depends of L1 , M1 , L2 , M2 and the best choice of T is c2 max{δC , δD }1/2 , where the constant c2 depends of L1 , M1 , L2 , M2 .

Chapter 7

An Optimization Problems with a Composite Objective Function

In this chapter we study an algorithm for minimization of the sum of two functions, the first one being smooth and convex and the second being convex. For this algorithm each iteration consists of two steps. The first step is a calculation of a subgradient of the first function while the second one is a proximal gradient step for the second function. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this.

7.1 Preliminaries Let H be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · , Id : X → X be the identity operator such that Id(x) = x, x ∈ X and let a set D ⊂ H be nonempty. An operator T : D → H is called firmly nonexpansive if T x − T y2 + (Id − T )x − (Id − T )y2 ≤ x − y2 for all x, y ∈ D. Let f, h : H → (−∞, ∞]. The infimal convolution of f and g is the function f g : H → [−∞, ∞] © Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_7

243

244

7 An Optimization Problems with a Composite Objective Function

defined by f g(x) = inf{f (y) + g(x − y) : y ∈ H } . Let f : H → (−∞, ∞] and γ > 0. The Moreau envelope of f of parameter γ  is the function γ

f := f ((2γ )−1  · 2 ).

Clearly, for all x ∈ H , γ

f (x) = inf{f (y) + (2γ )−1 x − y2 : y ∈ H }.

(7.1)

Let f : H → (−∞, ∞] be a convex and lower semicontinuous function which is not identically infinity. Then Proxf x is the unique point in H which satisfies 1

f (x) = inf{f (y) + 2−1 x − y2 : y ∈ H } = f (Proxf x) + 2−1 x − Proxf x2 .

(7.2)

Let h : H → (−∞, ∞] be a convex and lower semicontinuous function which is not identically infinity and let x ∈ H . In view of (7.2), Proxth (x) = argmin{th(y) + 2−1 x − y2 : y ∈ H } = argmin{h(y) + (2t)−1 x − y2 : y ∈ H }.

(7.3)

In view of (7.3), 0 ∈ ∂h(Proxth (x)) + t −1 (Proxth (x) − x) and t −1 (x − Proxth (x)) ∈ ∂h(Proxth (x)).

(7.4)

In the sequel we use the following results. Proposition 7.1 () Let f : H → (−∞, ∞] be a convex and lower semicontinuous function which is not identically infinity. Then Proxf and Id − Proxf are firmly nonexpansive mappings. Proposition 7.2 () Let f : H → (−∞, ∞] be a convex and lower semicontinuous function which is not identically infinity and x, p ∈ H . Then

7.2 The Algorithm and Main Results

245

p = Proxf (x) if and only if y − p, x − p + f (p) ≤ f (y) for all y ∈ H . Proposition 7.3 Let h : H → (−∞, ∞] be a convex and lower semicontinuous function which is not identically infinity, u, w ∈ H , t > 0 and v = Proxth (u).

(7.5)

Then h(w) ≥ h(v) + (2t)−1 (u − v2 + w − v2 − u − w2 ). Proof Lemma 2.1, Proposition 7.2, and (7.5) imply that th(w) ≥ th(v) + w − v, u − v = th(v) + 2−1 [u − v2 + w − v2 − u − w2 ].  

Proposition 7.3 is proved.

7.2 The Algorithm and Main Results Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · , f : X → R 1 be a convex Fréchet differentiable function, g : X → (−∞, ∞] be a convex and lower semicontinuous function which is not identically infinity and F (x) = f (x) + g(x), x ∈ X.

(7.6)

We consider the minimization problem F (x) → min, x ∈ X studied in  for in the finite-dimensional case. We denote by f  (x) the Fréchet derivative of the function f at the point x ∈ X. Let g be bounded from below, θ ∈ X, M > 1, θ  < M − 1, g(θ ) < ∞,

(7.7)

246

7 An Optimization Problems with a Composite Objective Function

c∗ > 0, g(x) ≥ −c∗ for all x ∈ X,

(7.8)

argmin(F ) ∩ BX (0, M) = ∅,

(7.9)

L1 ≥ 1 be such that |f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, 3M + 2).

(7.10)

M1 = 6M + 2L1 + 4 + 2c∗ + 2|g(θ )| + M,

(7.11)

Let

L ≥ 1 be such that f  (z1 ) − f  (z2 ) ≤ Lz1 − z2  for all z1 , z2 ∈ BX (0, M1 + 2).

(7.12)

0 < S− < S+ ≤ L−1 ,

(7.13)

S− ≤ St ≤ S+ t = 0, 1, . . . .

(7.14)

Let δf , δG ∈ (0, 1],

Let us describe our algorithm. Initialization: select an arbitrary x0 ∈ X. Iterative Step: given a current iteration vector xk ∈ X calculate ξk ∈ X such that ξk − f  (xk ) ≤ δf and xk+1 ∈ X such that g(xk+1 ) + (2Sk )−1 xk+1 − (xk − Sk ξk )2 ≤ g(x) + (2Sk )−1 x − (xk − Sk ξk )2 + δG for all x ∈ X. Our iterative step consists of two calculations. The second one is a proximal method applied to the function g. It should be mentioned that proximal methods are important tools in analysis and optimization [4, 5, 7, 8, 14, 19, 23–26, 28, 31, 33, 34, 40–45, 47, 48]. In this chapter we prove the following two results.

7.2 The Algorithm and Main Results

247

Theorem 7.4 Let δG ≤ 4−1 , ∞ {xt }∞ t=0 ⊂ X, {ξt }t=0 ⊂ X,

x0  ≤ M,

(7.15)

ξt − f  (xt ) ≤ δf

(7.16)

for each integer t ≥ 0,

and g(xt+1 ) + (2St )−1 xt+1 − (xt − St ξt )2 ≤ g(w) + (2St )−1 w − (xt − St ξt )2 + δG for all w ∈ X.

(7.17)

Let −1 δG (4M1 + 1), 0 = 2δG + 4δf M1 + S− 1/2

−1 2 n0 = 40−1 S− M + 1.

Then there exists an integer q ∈ [0, n0 + 1] such that F (xq ) ≤ inf(F ) + 0 , xt  ≤ 3M, t = 0, . . . , q. It is easy to see that 1/2

0 = c1 max{δG , δf } and n0 = c2 max{δG , δf }−1 , 1/2

where c1 , c2 are positive constants which depend on L, M1 and S− . Theorem 7.5 Let δf < M1 /8, δG < min{4−1 , ((16M1 + 4)−1 S− )2 },

(7.18)

248

7 An Optimization Problems with a Composite Objective Function

{x ∈ X : F (x) ≤ inf(F ) + 4} ⊂ BX (0, M),

(7.19)

∞ {xt }∞ t=0 ⊂ X, {ξt }t=0 ⊂ X,

x0  ≤ M, for each integer t ≥ 0, ξt − f  (xt ) ≤ δf

(7.20)

and g(xt+1 ) + (2St )−1 xt+1 − (xt − St ξt )2 ≤ g(w) + (2St )−1 w − (xt − St ξt )2 + δG for all w ∈ X.

(7.21)

Then for all integers t ≥ 0 xt  ≤ 3M and for all pairs of natural numbers m, T , min{F (xt ) : t = m + 1, . . . , m + T } − inf(F ), f(

m+T 

T −1 xt ) − inf(F )

t=m+1

≤ 4T −1 M 2 + (2S− )−1 (4M1 + 1)δG + δG + 2δf M1 . 1/2

It is easy to see that T = c1 max{δG , δf }−1 , 1/2

where c1 is a positive constant which depends on M, M1 and S− , is the best choice of T .

7.3 Auxiliary Results Lemma 7.6 Let S > 0, u0 , u1 , ξ0 ,  u1 ∈ X, g(u1 ) + (2S)−1 u1 − (u0 − Sξ0 )2

7.3 Auxiliary Results

249

≤ g(x) + (2S)−1 x − (u0 − Sξ0 )2 + δG for all x ∈ X and u1 − (u0 − Sξ0 )2 g( u1 ) + (2S)−1  ≤ g(x) + (2S)−1 x − (u0 − Sξ0 )2 for all x ∈ X. Then u1  ≤ 2(SδG )1/2 . u1 −  Proof Set h1 (w) = g(w) + (2S)−1 w − (u0 − Sξ0 )2 , w ∈ X. Clearly, h1 is a convex function. Since the function g is convex it is not difficult to see that u1 ) ≤ h1 (2−1 u1 + 2−1 u1 ) h1 ( u1 ) + (2S)−1 2−1 u1 + 2−1 u1 − u0 + Sξ0 2 = g(2−1 u1 + 2−1 ≤ 2−1 g(u1 ) + 2−1 g( u1 ) +(2S)−1 2−1 (u1 − u0 + Sξ0 ) + 2−1 ( u1 − u0 + Sξ0 )2 = 2−1 g(u1 ) + 2−1 g( u1 ) u1 − u0 + Sξ0 )2 +(2S)−1 [22−1 (u1 − u0 + Sξ0 )2 + 22−1 ( −2−1 (u1 −  u1 )2 ] = 2−1 (g(u1 ) + (2S)−1 u1 − u0 + Sξ0 )2 ) u1 ) + (2S)−1  u1 − u0 + Sξ0 )2 ) +2−1 (g( u1 )2 −(2S)−1 2−1 (u1 −  = 2−1 h1 (u1 ) + 2−1 h1 ( u1 ) − (2S)−1 2−1 (u1 −  u1 )2 ≤ h1 ( u1 ) + δG /2 − (2S)−1 2−1 (u1 −  u1 )2 and

250

7 An Optimization Problems with a Composite Objective Function

u1 −  u1  ≤ 2(SδG )1/2 .  

Lemma 7.6 is proved. Lemma 7.7 Let S ∈ (0, 1], x, y, ξ, p ∈ X, δG ≤ 4−1 , ξ − f  (y) ≤ δf

(7.22)

x, y ≤ 3M,

(7.23)

g(p) + (2S)−1 p − (x − Sξ )2 ≤ g(z) + (2S)−1 z − (x − Sξ )2 + δG

(7.24)

for all z ∈ X. Then p ≤ M1 and for every z ∈ X satisfying z ≤ 3M, F (z) − F (p) ≥ ((2S)−1 − 2−1 L)p − z2 +(2S)−1 (x − p2 − x − z2 ) +p − z, ξ − f  (z) − δG − (2S)1/2 δG (4M1 + 1), 1/2

F (z) − F (p) ≥ (2S)−1 (x − p2 + z − p2 − x − z2 ) +f (z) − f (y) − 2−1 Lp − y2 +f  (y), y − z − 2M1 δf − δG − (2S)1/2 δG (4M1 + 1), 1/2

F (z) − F (p) ≥ (2S)−1 (x − p2 + z − p2 − x − z2 ) −2−1 Lp − y2 − 2M1 δf − δG − (2S)1/2 δG (4M1 + 1). 1/2

Proof Set p0 = ProxSg (x − Sξ ).

(7.25)

7.3 Auxiliary Results

251

In view of (7.25), g(p0 ) + (2S)−1 p0 − (x − Sξ )2 ≤ g(z) + (2S)−1 z − (x − Sξ )2

(7.26)

for all z ∈ X. Let z ∈ BX (0, 3M). Applying Proposition 7.3 with v = p0 , u = x − Sξ , z = w, t = S we obtain that g(z) − g(p0 ) ≥ (2S)−1 (x − Sξ − p0 2 + z − p0 2 − x − Sξ − z2 ).

(7.27)

In view of (7.27), g(z) + (2S)−1 x − Sξ − z2 −(g(p0 ) + (2S)−1 x − Sξ − p0 2 ) ≥ (2S)−1 z − p0 2 .

(7.28)

Lemma 7.6, (7.24), and (7.26) imply that p0 − p ≤ 2(SδG )1/2 .

(7.29)

ξ  ≤ f  (y) + 1 ≤ L1 + 1.

(7.30)

By (7.10), (7.22), and (7.23),

It follows from (7.24), (7.28), and (7.29) that g(z) + (2S)−1 x − Sξ − z2 −(g(p) + (2S)−1 x − Sξ − p2 ) ≥ g(z) + (2S)−1 x − Sξ − z2 −(g(p0 ) + (2S)−1 x − Sξ − p0 2 ) − δG ≥ (2S)−1 z − p0 2 − δG

252

7 An Optimization Problems with a Composite Objective Function

≥ (2S)−1 z − p2 − δG − p − p0 (z − p0  + z + p) ≥ (2S)−1 z − p2 − δG − 2(SδG )1/2 (2z + 2p0  + 1).

(7.31)

In view of (7.8) and (7.24), (2S)−1 p − (x − Sξ )2 ≤ c∗ + g(θ ) + (2S)−1 θ − x + Sξ )2 + 1.

(7.32)

In view of (7.32), p − (x − Sξ )2 ≤ 2(c∗ + |g(θ )| + 1) + θ − x + Sξ )2 .

(7.33)

It follows from (7.7), (7.11), (7.23), and (7.30) that p ≤ x + ξ  +2(c∗ + |g(θ )| + 1) + θ  + x + ξ  ≤ 3 + M + L1 + 1 + 2(c∗ + |g(θ )| + 1) + θ  + 3M + L1 + 1 = 6M + 2L1 + 4 + 2(c∗ + |g(θ )|) + θ  ≤ M1 .

(7.34)

By (7.11), (7.23), (7.31), and (7.34), g(z) + (2S)−1 x − Sξ − z2 −(g(p) + (2S)−1 x − Sξ − p2 ) ≥ (2S)−1 z − p2 − δG − 2(SδG )1/2 (4M1 + 1).

(7.35)

In view of (7.35), g(z) − g(p) ≥ (2S)−1 (x − Sξ − p2 − x − Sξ − z2 + z − p2 ) −δG − (2SδG )1/2 (4M1 + 1) = (2S)−1 (x − p2 − x − z2 + z − p2 ) + p − z, ξ  − δG − (2SδG )1/2 (4M1 + 1).

(7.36)

7.3 Auxiliary Results

253

Proposition 4.3 and (7.12) imply that f (z) − f (p) ≥ −2−1 Lp − z2 − f  (z), p − z.

(7.37)

It follows from (7.6), (7.36), and (7.37) that F (z) − F (p) ≥ ((2S)−1 − 2−1 L)p − z2 + (2S)−1 (x − p2 − x − z2 ) + ξ − f  (z), p − z − δG − (2SδG )1/2 (4M1 + 1).

(7.38)

In view of (7.38), the first inequality in the statement of the lemma is proved. Let us prove the second inequality in the statement of the lemma. Proposition 4.3 and (7.12) imply that f  (y), p − y ≥ f (p) − f (y) − 2−1 Lp − y2 .

(7.39)

Combining (7.22), (7.34), (7.36), and (7.39) we obtain that g(z) − g(p) ≥ (2S)−1 (x − p2 − x − z2 + z − p2 ) +f (p) − f (y) − 2−1 Lp − y2 +f  (y), y − z + ξ − f  (y), p − z −δG − (2SδG )1/2 (4M1 + 1) ≥ (2S)−1 (x − p2 − x − z2 + z − p2 ) +f (p) − f (y) − 2−1 Lp − y2 + f  (y), y − z − 2M1 δf − δG − (2SδG )1/2 (4M1 + 1).

(7.40)

We add f (z) − f (p) to (7.40) and obtain the second inequality in the statement of the lemma. Together with the convexity of f this implies the third inequality in the statement of the lemma. Lemma 7.7 is proved.   Lemma 7.7 implies the following result. Lemma 7.8 Let S ∈ (0, 1], x, ξ, p ∈ X, δG ≤ 4−1 , S ≤ L−1 ,

254

7 An Optimization Problems with a Composite Objective Function

ξ − f  (x) ≤ δf , x ≤ M + 3, g(p) + (2S)−1 p − (x − Sξ )2 ≤ g(z) + (2S)−1 z − (x − Sξ )2 + δG for all z ∈ X. Then p ≤ M1 and for every z ∈ X satisfying z ≤ 3M, F (z) − F (p) (2S)−1 (z − p2 − x − z2 ) +((2S)−1 − 2−1 L)x − p2 −2M1 δf − δG − (2S)−1 δG (4M1 + 1) 1/2

≥ (2S)−1 (z − p2 − x − z2 ) −2M1 δf − δG − (2S)−1 δG (4M1 + 1). 1/2

7.4 Proof of Theorem 7.4 In view of (7.9), there exists z ∈ argmin(F ) ∩ BX (0, M).

(7.41)

x0 − z ≤ 2M.

(7.42)

By (7.15) and (7.41),

If F (x0 ) ≤ inf(F ) + 0 , then the assertion of the theorem holds. Assume that F (x0 ) > inf(F ) + 0 .

(7.43)

7.4 Proof of Theorem 7.4

255

If F (x1 ) ≤ inf(F ) + 0 , then by Lemma 7.8, x1  ≤ M1 and the assertion of the theorem holds. Let F (x1 ) > inf(F ) + 0 .

(7.44)

Assume that T ≥ 0 is an integer and that for all integers t = 0, . . . , T , F (xt+1 ) − F (z) > 0 .

(7.45)

(Note that in view of (7.44), out assumption holds for T = 0.) We show that for all t = 0, . . . , T , xt − z ≤ 2M

(7.46)

and F (z) − F (xt+1 ) ≥ (2St )−1 (z − xt+1 2 − z − xt 2 ) − 2M1 δf − δG − (2St )−1 δG (4M1 + 1). 1/2

(7.47)

Assume that t ∈ {0, . . . , T } and (7.46) holds. By (7.41) and Lemma 7.8 applied with p = xt+1 , x = xt , ξ = ξt , (7.47) is true. It follows from (7.13), (7.45), and (7.47) that 0 < (2St )−1 (z − xt 2 − z − xt+1 2 ) +(2St )−1 δG (4M1 + 1) + 2M1 δf + δG 1/2

≤ (2St )−1 (z − xt 2 − z − xt+1 2 ) + (2S− )−1 δG (4M1 + 1) + 2M1 δf + δG . 1/2

(7.48)

256

7 An Optimization Problems with a Composite Objective Function

Relations (7.13), (7.18), and (7.48) imply that z − xt+1  ≤ z − xt . Thus by induction we have shown that (7.46) holds for all t = 0, . . . , T + 1, (7.47) holds for all t = 0, . . . , T and that z − xt+1  ≤ z − xt 

(7.49)

is true for all t = 0, . . . , T . By (7.42) and (7.45)–(7.49), (T + 1)0 < (T + 1)(min{F (xt ) : t = 1, . . . , T + 1} − F (z)) ≤

T 

(F (xt+1 − F (z))

t=0

T  (2S− )−1 (z − xt 2 − z − xt+1 2 ) t=0

+(T + 1)[(2S− )−1 δG (4M1 + 1) + 2M1 δf + δG ] 1/2

≤ (2S− )−1 4M 2 + (T + 1)[(2S− )−1 δG (4M1 + 1) + 2M1 δf + δG ]. 1/2

Together with the choice of 0 , n0 this implies that (T + 1)0 /2 ≤ (2S− )−1 4M 2 and −1 2 −1 M 0 < n0 + 1. T ≤ 4S−

Thus we have shown that if T is an integer and (7.45) holds for all integers t = 0, . . . , T , then T < n0 and in view of (7.41) and (7.46) xt  ≤ 3M, t = 0, . . . , T + 1. This implies that there exists an integer q ∈ [0, n0 + 1] such that F (xq ) ≤ inf(F ) + 0 , xt  ≤ 3M, t = 0, . . . , q. Theorem 7.4 is proved.

 

7.5 Proof of Theorem 7.5

257

7.5 Proof of Theorem 7.5 In view of (7.9) and (7.19), there exists z ∈ argmin(F ) ⊂ BX (0, M).

(7.50)

It follows from (7.50) and the assumptions of the theorem that x0 − z ≤ 2M.

(7.51)

Assume that t ≥ 0 is an integer and that xt − z ≤ 2M.

(7.52)

By (7.20), (7.50), (7.52), and Lemma 7.8 applied with p = xt+1 , x = xt , ξ = ξt , we have F (z) − F (xt+1 ) ≥ (2St )−1 (z − xt+1 2 − z − xt 2 ) − 2M1 δf − δG − (2St )−1 δG (4M1 + 1).

(7.53)

z − xt+1  ≤ z − xt ;

(7.54)

z − xt+1  > z − xt .

(7.55)

1/2

There are two cases:

Assume that (7.54) holds. Then by (7.52) and (7.54), z − xt+1  ≤ 2M.

(7.56)

Assume that (7.55) holds. It follows from (7.18), (7.50), (7.53), and (7.55) that F (xt+1 ) ≤ inf(F ) +2M1 δf + δG + (2St )−1 δG (4M1 + 1) 1/2

≤ inf(F ) + 1.

(7.57)

258

7 An Optimization Problems with a Composite Objective Function

Together with (7.19) and (7.57) this implies that xt+1  ≤ M, xt+1 − z ≤ 2M and that (7.56) holds in the both cases. Thus (7.52) and (7.53) hold for all integers t ≥ 0. In view of (7.50) and (7.52), xt  ≤ 3M for all integers t ≥ 0. Let m ≥ 1 and T > 0 be integers. It follows from (7.50) and (7.53) that m+T 

(F (xt − inf(F ))

t=m+1

m+T −1

(z − xt 2 − z − xt+1 2 )

t=m

+T [(2S− )−1 δG (4M1 + 1) + 2M1 δf + δG ]. 1/2

Together with (7.51) this implies that min{F (xt ) : t = m + 1, . . . , m + T } − inf(F ), f(

m+T 

T −1 xt ) − inf(F )

t=m+1

≤ 4T −1 M 2 + (2S− )−1 (4M1 + 1)δG + δG + 2δf M1 . 1/2

Theorem 7.5 is proved.

 

Chapter 8

A Zero-Sum Game with Two Players

In this chapter we study an algorithm for finding a saddle point of a zero-sum game with two players. For this algorithm each iteration consists of two steps. The first step is a calculation of a subgradient while the second one is a proximal gradient step. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this.

8.1 The Algorithm and the Main Result Let (X, ·, ·), (Y, ·, ·) be Hilbert spaces equipped with the complete norms  ·  which are induced by their inner products. Denote by Id the identity operator in X. Suppose that f : X → R 1 is a convex Fréchet differentiable function such that for each r > 0 its Fréchet derivative f  (·) is Lipschitz on BX (0, r), g : Y → R 1 ∪ {∞} is convex lower semicontinuous function which is not identically ∞, A : Y → X is a linear continuous operator, A∗ : X → Y its dual and that A = sup{Ay : y ∈ Y, y ≤ 1}.

(8.1)

For each x ∈ X and each y ∈ Y define F (x, y) = f (x) + x, Ay − g(y).

© Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_8

(8.2)

259

260

8 A Zero-Sum Game with Two Players

Suppose that there exists (u∗ , v ∗ ) ∈ X × Y such that F (u∗ , v) ≤ F (u∗ , v ∗ ) ≤ F (u, v ∗ )

(8.3)

for all (u, v) ∈ X × Y . Fix a constant τ > 0 such that τ A ≤ 1.

(8.4)

Let us describe our algorithm. Initialization: select arbitrary (u0 , v0 ) ∈ X × Y . Iterative Step: for t = 1, 2, . . . , given current iteration vectors ut−1 ∈ X and vt−1 ∈ Y calculate pt = ut−1 − τ (Avt−1 + f  (ut−1 )),

(8.5)

vt = Proxg (vt−1 + A∗ pt ) = argmin{g(z) + 2−1 z − vt−1 − A∗ pt 2 : z ∈ Y }

(8.6)

(see (7.2)), ut = ut−1 − τ (Avt + f  (ut−1 )). This algorithm was considered in . One of its steps (see (8.6)) is a proximal method applied to the function g. It should be mentioned that proximal methods is an important tool in analysis and optimization [49–52, 57, 59, 60, 63, 74, 75, 77, 78, 82, 84, 86–88]. Set G = Id − τ A∗ A.

(8.7)

Let δf , δG ∈ (0, 1]. In this chapter we study the algorithm described above, taking into account the computational errors. Initialization: select arbitrary (u0 , v0 ) ∈ X × Y . Iterative Step: for t = 1, 2, . . . , given current iteration vectors ut−1 ∈ X and vt−1 ∈ Y calculate ξt−1 ∈ X such that ξt−1 − f  (ut−1 ) ≤ δf ,

(8.8)

pt = ut−1 − τ (Avt−1 + ξt−1 ),

(8.9)

8.1 The Algorithm and the Main Result

261

vt ∈ Y such that g(vt ) + 2−1 vt − vt−1 − A∗ pt 2 ≤ g(z) + 2−1 z − vt−1 − A∗ pt 2 + δG for all z ∈ Y,

(8.10)

and ut = ut−1 − τ (Avt + ξt−1 ).

(8.11)

dom(g) = {y ∈ Y : g(y) < ∞}.

(8.12)

Define

Assume that the set dom(g) is bounded and let M0 > 0 be such that dom(g) ⊂ BY (0, M0 ).

(8.13)

Let M1 > 0 be such that {x ∈ X : f (x) ≤ |f (0)| + 4 + xAM0 } ⊂ BX (0, M1 )

(8.14)

and let L ≥ 1 be such that |f (z1 ) − f (z2 )| ≤ Lz1 − z2  for all z1 , z2 ∈ BX (0, M1 + 1)

(8.15)

and |f  (z1 ) − f  (z2 )| ≤ Lz1 − z2  for all z1 , z2 ∈ X.

(8.16)

In view of (8.3) and (8.13), v ∗ ∈ dom(g) ⊂ BY (0, M0 ), f (u∗ ) − g(v ∗ ) + u∗ , Av ∗  = F (u∗ , v ∗ ) ≤ F (0, v ∗ ) = f (0) − g(v ∗ ).

(8.17)

262

8 A Zero-Sum Game with Two Players

By (8.2) and (8.3), f (0) ≥ f (u∗ ) + u∗ , A∗ v ∗  ≥ f (u∗ ) − u∗ AM0 , f (u∗ ) ≤ f (0) + M0 Au∗ .

(8.18)

It follows from (8.14) and (8.18) that u∗  ≤ M1 .

(8.19)

In this chapter we prove the following result. Theorem 8.1 Let δf,1 ≤ min{τ M1−1 , (AM0 + L + 1)−1 },

(8.20)

τ ≤ L−1 ,

(8.21)

∞ ∞ ∞ Assume that {ut }∞ t=0 ⊂ X, {vt }t=0 ⊂ Y , {ξt }t=0 ⊂ X, {pt }t=0 ⊂ X satisfy (8.8)– (8.11) for all natural numbers t and that

g(v0 ) < ∞, u0  ≤ M1 .

(8.22)

Let Δ = 2M1 δf τ −1 + δG (12M0 + 3 + 4A(M1 + M0 + 2)) 1/2

(8.23)

and for each natural number T let  uT = T −1

T 

ut ,  vT = T −1

t=1

T  t=1

Then for all integers t ≥ 0, vt  ≤ M0 , ut  ≤ M1 , for all integers t ≥ 1, pt  ≤ M1 + M0 + 2 and for all integers T ≥ 1,

vt .

(8.24)

8.2 Auxiliary Results

263

 uT  ≤ M1 , vT  ≤ M0 ,  vT ∈ dom(g),  and for each u ∈ X and each v ∈ dom(g), uT , vT ) + Δ + (2τ T )−1 4M12 + (2T )−1 4M02 G F ( uT , v) ≤ F ( and vT ) ≤ F (u, vT ) + Δ + (2τ T )−1 4M12 + (2T )−1 4M02 G. F ( uT , It is clear that the best choice of T is at the same order as (max{δf , δG }1/2 )−1 . 1/2

8.2 Auxiliary Results Lemma 8.2 Let a linear continuous operator G0 : Y → Y satisfy G∗0 = G0 and z, G0 z ≥ 0 for all z ∈ Y. Then for each u, v, w ∈ Y , 2w − v, G0 (u − v) = w − v, G0 (w − v) − w − u, G0 (w − u) + u − v, G0 (u − v). Proof We have w − v, G0 (w − v) − w − u, G0 (w − u) + u − v, G0 (u − v) = w, G0 w + v, G0 v − 2w, G0 v −(w, G0 w + u, G0 u − 2w, G0 u) +(v, G0 v + u, G0 u − 2v, G0 u) 2[−w, G0 v + v, G0 v + w, G0 u − v, G0 u] = 2[G0 u, w − v + G0 v, v − w] = 2w − v, G0 (u − v). Lemma 8.2 is proved.

 

264

8 A Zero-Sum Game with Two Players

∞ ∞ ∞ Lemma 8.3 Assume that {ut }∞ t=0 ⊂ X, {vt }t=0 ⊂ Y , {ξt }t=0 ⊂ X, {pt }t=0 ⊂ X satisfy (8.8)–(8.11) for all natural numbers t. Let t ≥ 0 be an integer and  vt ∈ Y satisfy

vt − vt−1 − A∗ pt 2 g( vt ) + 2−1  ≤ g(z) + 2−1 z − vt−1 − A∗ pt 2

(8.25)

for all z ∈ Y . Then 1/2

vt  ≤ 2δG . vt −  Proof For all y ∈ Y set h(y) = g(y) + 2−1 y − vt−1 − A∗ pt 2 .

(8.26)

In view of (8.25) and (8.26),  vt ∈ argmin(h). By (8.12), (8.25), and (8.26), vt ) + δG . h(vt ) ≤ h( Since the function h is convex it follows from (8.25) to (8.27) that vt ) h( vt ) ≤ h(2−1 vt + 2−1 = g(2−1 vt + 2−1 vt ) vt − vt−1 − A∗ pt 2 +2−1 2−1 vt + 2−1 ≤ g(2−1 vt + 2−1 vt ) vt − vt−1 − A∗ pt )2 +2−1 2−1 (vt − vt−1 − A∗ pt ) + 2−1 ( ≤ 2−1 g(vt ) + 2−1 g( vt ) +2−1 [22−1 (vt − vt−1 − A∗ pt )2 +22−1 ( vt − vt−1 − A∗ pt )2 − 2−1 (vt −  vt )2 ] vt ) − 2−1 2−1 (vt −  vt )2 = 2−1 h(vt ) + 2−1 h( ≤ 2−1 h( vt ) + 2−1 δG + 2−1 h( vt ) − 2−1 2−1 (vt −  vt )2 ,

(8.27)

8.2 Auxiliary Results

265

vt −  vt 2 ≤ 4δG and 1/2

v −  vt  ≤ 2δG .  

Lemma 8.3 is proved.

∞ ∞ ∞ Lemma 8.4 Assume that {ut }∞ t=0 ⊂ X, {vt }t=0 ⊂ Y , {ξt }t=0 ⊂ X, {pt }t=0 ⊂ X satisfy (8.8)–(8.11) for all natural numbers t,

u0  ≤ M1 , v0  ≤ M0 , g(v0 ) < ∞. Then the following assertions hold. 1. vt  ≤ M0 for all integers t ≥ 0 and vt ∈ dom(g) for all integers t ≥ 0. 2. Let t ≥ 0 be an integer. Then for all u ∈ X, F (ut , vt ) − F (u, vt ) ≤ (2τ )−1 (u − ut−1 2 − u − ut 2 ) −2−1 (τ −1 − L)ut − ut−1 2 + τ −1 δf ut − u. 3. Let t ≥ 0 be an integer. Then for all v ∈ dom(g), F (ut , v) − F (ut , vt ) 1/2

≤ δG (12M0 + 3 + 4Apt ) +2−1 [v − vt−1 , G(v − vt−1 ) − v − vt , G(v − vt ) −vt−1 − vt , G(vt−1 − vt )].

(8.28)

266

8 A Zero-Sum Game with Two Players

Proof In view of (8.10), (8.13), and (8.28), g(vt ) < ∞, t = 0, 1, . . . . Clearly, vt  ≤ M0 for all integers t ≥ 0. Let us prove assertion 2. For every u ∈ X set h(u) = F (u, vt ) = f (u) + u, Avt  − g(vt ).

(8.29)

By (8.29), the function h is Fréchet differentiable, for all u ∈ X, h (u) = f  (u) + Avt

(8.30)

h (u1 ) − h (u2 ) = f  (u1 ) − f  (u2 ).

(8.31)

and for all u1 , u2 ∈ X,

Proposition 4.3, (8.16), and (8.31) imply that h(ut ) ≤ h(ut−1 ) + h (ut−1 ), ut − ut−1  + 2−1 Lut − ut−1 2 .

(8.32)

Let u ∈ X. Since the function h is convex we have h(u) ≥ h(ut−1 ) + h (ut−1 ), u − ut−1 .

(8.33)

It follows from (8.32) and (8.33) that h(ut ) ≤ h(u) − h (ut−1 ), u − ut−1  +h (ut−1 ), ut − ut−1  + 2−1 Lut − ut−1 2 = h(u) + h (ut−1 ), ut − u + 2−1 Lut − ut−1 2 .

(8.34)

By (8.29), (8.30), and (8.34), F (ut , vt ) − F (u, vt ) ≤ f  (ut−1 ) + Avt , ut − u + 2−1 Lut − ut−1 2 .

(8.35)

In view of (8.11), τ −1 (ut−1 − ut ) − (f  (ut−1 ) + Avt ) = τ −1 (ξt−1 − f  (ut−1 )) ≤ τ −1 δf .

(8.36)

8.2 Auxiliary Results

267

It follows from (8.35) and (8.36) that F (ut , vt ) − F (u, vt ) ≤ τ −1 (ut−1 − ut ), ut − u + τ −1 δf ut − u +2−1 Lut − ut−1 2 . Lemma 2.1 and the relation above imply that F (ut , vt ) − F (u, vt ) (2τ )−1 (u − ut−1 2 − u − ut 2 − ut − ut−1 2 ) +τ −1 δf ut − u + 2−1 Lut − ut−1 2 . Thus assertion 2 holds. Let us prove assertion 3. In view of (8.10), g(vt ) + 2−1 vt − vt−1 − A∗ pt 2 ≤ g(z) + 2−1 z − vt−1 − A∗ pt 2 + δG

(8.37)

for all z ∈ Y . Let  v = argmin{g(z) + 2−1 z − vt−1 − A∗ pt 2 : z ∈ Y }.

(8.38)

Lemma 8.3, (8.37), and (8.38) imply that 1/2

vt −  v  ≤ 2δG .

(8.39)

In view of (8.3), for every v ∈ Y , − F (pt , v) = −f (pt ) − pt , Av + g(v). By (8.40), for every v ∈ Y , g(v) + 2−1 v − vt−1 − A∗ pt 2 = g(v) + 2−1 v − vt−1 2 − v − vt−1 , A∗ pt  + A∗ pt 2 = g(v) − pt , Av + Avt−1 , pt 

(8.40)

268

8 A Zero-Sum Game with Two Players

+2−1 v − vt−1 2 + A∗ pt 2 = −F (pt , v) + 2−1 v − vt−1 2 + f (pt ) + Avt−1 , pt  + A∗ pt 2 .

(8.41)

It follows from (8.38) and (8.41) that  vt = argmin{−F (pt , v) + 2−1 v − vt−1 2 : v ∈ Y }.

(8.42)

Set h(v) = −F (pt , v), v ∈ Y.

(8.43)

Proposition 7.2 and (8.43) imply that for every y ∈ Y , v ) − F (pt , v) = h(v) − h( vt ) F (pt , v . ≥ v −  v , vt−1 − 

(8.44)

Let v ∈ dom(g). In view of (8.2), F (ut , v) − F (ut , vt ) + F (pt , vt ) − F (pt , v) = f (ut ) + ut , Av − g(v) −(f (ut ) + ut , Avt  − g(vt )) +f (pt ) + pt , Avt  − g(vt ) −(f (pt ) + pt , Av − g(v)) = ut − pt , Av − Avt . By the relation above, (8.9)–(8.11) and (8.44), F (ut , v) − F (ut , vt ) = F (pt , v) − F (pt , vt ) + ut − pt , A(v − vt ) = F (pt , v) − F (pt , vt ) + τ A(vt−1 − vt ), A(v − vt ) = F (pt , v) − F (pt , vt ) + τ vt−1 − vt , A∗ A(v − vt ).

(8.45)

8.2 Auxiliary Results

269

In view of (8.2) and (8.38), v ) − F (pt , vt ) F (pt , v  − g( v) = f (pt ) + pt , A −(f (pt ) + pt , Avt  − g(vt )) v − vt ) + g(vt ) − g( v ). = pt , A(

(8.46)

It is clear that v) g(vt ) − g( = g(vt ) + 2−1 vt − vt−1 − A∗ pt 2 −(g( v ) + 2−1  v − vt−1 − A∗ pt 2 ) v − vt−1 − A∗ pt 2 ). − (2−1 vt − vt−1 − A∗ pt 2 − 2−1 

(8.47)

It follows from (8.37), (8.39), and (8.47) that v )| |g(vt ) − g( ≤ |g(vt ) + 2−1 vt − vt−1 − A∗ pt 2 −(g( v ) + 2−1  v − vt−1 − A∗ pt 2 | +2−1 |vt − vt−1 − A∗ pt 2 −  v − vt−1 − A∗ pt 2 | v (vt  + 2vt−1  + 2Apt  +  v ) ≤ δG + 2−1 vt −  1/2

≤ δG + δG (2 + 2vt  + 2vt−1  + 2Apt ).

(8.48)

Relations (8.39), (8.46), and (8.48) imply that v ) − F (pt , vt )| |F (pt , 1/2

≤ 2pt AδG + δG 1/2

+ δG (2 + 2vt  + 2vt−1  + 2Apt ). By (8.44) and (8.45),

(8.49)

270

8 A Zero-Sum Game with Two Players

F (ut , v) − F (ut , vt ) = F (pt , v) − F (pt , vt ) + τ vt−1 − vt , A∗ A(v − vt ) v ) + F (pt , v ) − F (pt , vt ) = F (pt , v) − F (pt , +τ vt−1 − vt , A∗ A(v − vt ) v ≤  v − vt−1 , v −  1/2

+δG (2 + 2vt  + 2vt−1  + 2Apt ) 1/2

+δG + 2δG pt A +τ vt−1 − vt , A∗ A(v − vt ) 1/2

= δG + 2δG pt A 1/2

+δG (2 + 2vt  + 2vt−1  + 2Apt ) +τ vt−1 − vt , A∗ A(v − vt ) +vt − vt−1 , v − vt  v − vt−1 , v −  v ]. + [vt−1 − vt , v − vt  + 

(8.50)

By (8.13), (8.39), assertion 1, and the inclusion v ∈ dom(g), v  −  vt − vt−1 , v − vt | | v − vt−1 , v −  v  −  v − vt−1 , v − vt | ≤ | v − vt−1 , v −  +| v − vt−1 , v − vt  − vt − vt−1 , v − vt | v ≤  v − vt−1 v −  v − vt  +v − vt  1/2

≤ 2δG (2M0 + 2M0 + v).

(8.51)

Assertion 1, (8.50), and (8.51) imply that F (ut , v) − F (ut , vt ) 1/2

1/2

≤ δG (4M0 + 3 + 4Apt ) + 8M0 δG

8.3 Proof of Theorem 8.1

271

+τ vt−1 − vt , A∗ A(v − vt ) +vt − vt−1 , v − vt  1/2

= δG (12M0 + 3 + 4Apt ) + vt − vt−1 , (Id − τ A∗ A)(v − vt ).

(8.52)

G = Id − τ A∗ A.

(8.53)

Recall (see (8.7)) that

In view of (8.4) and (8.53), G∗ = G and for each z ∈ Y , z, Gz = z, z − τ A∗ Az z2 − τ Az2 ≥ z2 (1 − τ A) ≥ 0.

(8.54)

By (8.52)–(8.54) and Lemma 8.2, F (ut , v) − F (ut , vt ) 1/2

≤ δG (12M0 + 3 + 4Apt ) +2−1 (v − vt−1 , G(v − vt−1  −v − vt , G(v − vt ) −vt−1 − vt , G(vt−1 − vt )). Thus assertion 3 holds. This completes the proof of Lemma 8.4.

 

8.3 Proof of Theorem 8.1 Assertion 1 of Lemma 8.4 implies that for all integers t ≥ 0, vt ∈ dom(g), vt  ≤ M0 .

(8.55)

Assertion 2 of Lemma 8.4 and (8.21) imply that for all integers t ≥ 0 and all u ∈ X,

272

8 A Zero-Sum Game with Two Players

F (ut , vt ) − F (u, vt ) ≤ (2τ )−1 (u − ut−1 2 − u − ut 2 ) + L−1 δf ut − u.

(8.56)

By Assertion 3 of Lemma 8.4, for all integers t ≥ 0 and all v ∈ dom(g), F (ut , v) − F (ut , vt ) 1/2

≤ δG (12M0 + 3 + 4Apt ) + 2−1 [v − vt−1 , G(v − vt−1 ) − v − vt , G(v − vt )].

(8.57)

We show that for all integers t ≥ 0, ut  ≤ M1 , pt+1  ≤ M1 + M0 + 2.

(8.58)

Assume that t ≥ 1 is an integer and that ut−1  ≤ M1 .

(8.59)

(Note that in view of (8.22), inequality (8.59) holds for t = 1.) By (8.56), F (ut , vt ) ≤ F (0, vt ) + (2τ )−1 (ut−1 2 − ut 2 ) + τ −1 δf ut .

(8.60)

There are two cases: ut  ≤ ut−1 ;

(8.61)

ut  > ut−1 .

(8.62)

Assume that (8.61) holds. In view of (8.59) and (8.61), ut  ≤ ut−1  ≤ M1 . Assume that (8.62) holds. By (8.2), (8.60), and (8.62), f (ut ) + ut , Avt  ≤ f (0) + τ −1 δf ut .

(8.63)

Relations (8.55) and (8.63) imply that f (ut ) ≤ f (0) + AM0 ut  + δf τ −1 ut .

(8.64)

8.3 Proof of Theorem 8.1

273

It follows from (8.8), (8.11), (8.15), (8.55), and (8.59) that ut  ≤ ut−1  + τ (Avt  + ξt−1 ) ≤ M1 + τ (AM0 + L + 1).

(8.65)

By (8.20) and (8.65), δf τ −1 ut  ≤ δf τ −1 M1 + δf (AM0 + L + 1) ≤ 2.

(8.66)

It follows from (8.14), (8.64), and (8.66) that f (ut ) ≤ f (0) + AM0 ut  + 2.

(8.67)

In view of (8.67), ut  ≤ M1 . Thus by induction we showed that ut  ≤ M1 for all integers t ≥ 0.

(8.68)

Relations (8.8), (8.15), and (8.68) imply that for all integers t ≥ 0, ξt  ≤ L + 1.

(8.69)

By (8.4), (8.9), (8.21), (8.55), (8.68), and (8.69), pt  = ut−1 − τ (Avt−1 + ξt−1 ) ≤ ut−1  + τ (Avt−1  + ξt−1 ) ≤ M1 + τ (AM0 + L + 1) ≤ M1 + M0 + 2. Thus (8.58) holds. In view of (8.56), (8.58), and (8.68), for all integers t ≥ 0, all u ∈ BX (0, M1 ), and all v ∈ dom(g), F (ut , vt ) − F (u, vt ) ≤ (2τ )−1 (u − ut−1 2 − u − ut 2 ) + 2τ −1 δf M1 and F (ut , v) − F (ut , vt )

(8.70)

274

8 A Zero-Sum Game with Two Players

≤ 2−1 [v − vt−1 , G(v − vt−1 ) − v − vt , G(v − vt )] 1/2

+ δG (12M0 + 3 + 4A(M1 + M0 + 2)).

(8.71)

It follows from (8.7) and (8.70) that for all integers t ≥ 0, all u ∈ BX (0, M1 ), and all v ∈ dom(g), F (ut , v) − F (u, vt ) ≤ (2τ )−1 (u − ut−1 2 − u − ut 2 ) + 2τ −1 δf M1 +2−1 [v − vt−1 , G(v − vt−1 ) − v − vt , G(v − vt )] 1/2

+ 4δG (12M0 + 3 + 4A(M1 + M0 + 2)).

(8.72)

Let T be a natural number. By (8.24), (8.62), and (8.72), for all u ∈ BX (0, M1 ) and all v ∈ dom(g), vT ) F ( uT , v) − F (u, = F(

T 

T −1 ut , v) − F (u,

t=1

≤ T −1

T 

T −1 vt )

t=1

T 

F (ut , v) − T −1

t=1

T 

F (u, vt )

t=1

≤ 2M1 τ −1 δf + δG (12M0 + 3 + 4A(M1 + M0 + 2)) 1/2

+(2T τ )−1

T  (u − ut−1 2 − u − ut 2 ) t=1

−1

+(2T )

T  (v − vt−1 , G(v − vt−1 ) − v − vt , G(v − vt )) t=1

≤ 2M1 τ −1 δf + δG (12M0 + 3 + 4A(M1 + M0 + 2)) 1/2

+(2T τ )−1 4M12 + (2T )−1 4M02 G = Δ + (2T τ )−1 4M12 + (2T )−1 4M02 G.

(8.73)

ΔT = Δ + (2T τ )−1 4M12 + (2T τ )−1 4M02 G.

(8.74)

Set

8.3 Proof of Theorem 8.1

275

In view of (8.73) and (8.74), with v = v ∗ and u = u∗ , vT ) ≤ ΔT . F ( uT , v ∗ ) − F (u∗ ,

(8.75)

Relations (8.3) and (8.75) imply that 0 ≤ F ( uT , v ∗ ) − F (u∗ , v ∗ ) ≤ ΔT ,

(8.76)

0 ≤ F (u∗ , v ∗ ) − F (u∗ , vT ) ≤ ΔT .

(8.77)

By (8.73) and (8.74), for each u ∈ BX (0, M1 ) and each v ∈ dom(g), vT ) − F (u, vT ) ≤ ΔT , F ( uT ,

(8.78)

F ( uT , v) − F ( uT , vT ) ≤ ΔT .

(8.79)

u ∈ X \ BX (0, M1 ).

(8.80)

Assume that

In view of (8.2) and (8.78), vT ) ≤ ΔT + F (0, vT ) = ΔT + f (0) + g( vT ). F ( uT ,

(8.81)

It follows from (8.2), (8.14), (8.24), (8.55), and (8.80) that vT ) F (u, vT ) − F (0, = f (u) + u, A vT  − f (0) ≥ f (u) − uAM0 − |f (0)| ≥ 4.

(8.82)

By (8.81) and (8.82), vT ) ≤ ΔT + F (0, vT ) F ( uT , vT ) − 4. ≤ ΔT + F (u,

(8.83)

In view of (8.83) and (8.88), vT ) ≤ ΔT + F (u, vT ) F ( uT , for all u ∈ X. Theorem 8.1 is proved.

 

Chapter 9

PDA-Based Method for Convex Optimization

In this chapter we use predicted decrease approximation (PDA) for constrained convex optimization. For PDA-based method each iteration consists of two steps. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this.

9.1 Preliminaries and the Main Result Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · , F : X → R 1 be a convex Fréchet differentiable function, G : X → R 1 ∪ {∞} be a convex lower semicontinuous function which is not identically infinity, dom(G) = {x ∈ X : G(x) < ∞}, H (x) = F (x) + G(x), x ∈ X.

(9.1)

We denote by F  (x) the Fréchet derivative of F at x ∈ X. We suppose that dom(G) is a bounded set and that inf(H ) = inf{H (x) : x ∈ X} is a finite number. Let

© Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_9

277

278

9 PDA-Based Method for Convex Optimization

D := diam(dom(G)) = sup{x − y : x, y ∈ dom(G)},

(9.2)

H (x∗ ) = inf(H ).

(9.3)

x∗ ∈ X satisfy

Given a function f : X → R 1 ∪ {∞}, its convex conjugate is the function f ∗ (y) = sup{x, y − f (x) : x ∈ X}, y ∈ X.

(9.4)

Clearly, for all x, y ∈ X, f (x) + f ∗ (u) ≥ x, u

(9.5)

(Fenchel inequality). For every y ∈ X define S(y) = sup{F  (y), y − p + G(y) − G(p) : p ∈ X}.

(9.6)

By (9.4) and (9.6), for y ∈ X, S(y) = G(y) + F  (y), y + sup{−F  (y), p − G(p) : p ∈ X} G(y) + F  (y), y + G∗ (−F  (y)).

(9.7)

It follows from Fenchel inequality (9.5) and (9.7) that for every y ∈ dom(G), S(y) ≥ 0.

(9.8)

Let y ∈ dom(G). In view (9.6) and (9.8), the following four properties are equivalent: S(y) = 0;

(9.9)

G(p) ≥ G(y) + −F  (y), p − y for all p ∈ X;

(9.10)

− F  (y) ∈ ∂G(y);

(9.11)

y is a minimizer of H.

(9.12)

9.1 Preliminaries and the Main Result

279

For each x, y ∈ X set [x, y] = {αx + (1 − α)y : α ∈ [0, 1]}. Lemma 9.1 () For every y ∈ X, H (y) − inf(H ) ≤ S(y). Proof Let y ∈ Y . If S(y) = ∞, then the lemma holds. Assume that S(y) < ∞.

(9.13)

sup{−F  (y), p − G(p) : p ∈ X} < ∞

(9.14)

G(y) < ∞.

(9.15)

In view of (9.7) and (9.13),

and

By (9.14), there exists p ∈ X such that −F  (y), p  − G(p ) ≥ − − F  (y), p − G(p) for all p ∈ X.

(9.16)

It follows from (9.6) and (9.16) that S(y) ≥ F  (y), y − p  + G(y) − G(p ) = F  (y), y + G(y) − [F  (y), p  + G(p )] ≥ F  (y), y + G(y) − F  (y), x∗  + G(x∗ ) −  = F  (y), y − x∗  + G(y) − G(x∗ ) − . Since  is any positive number in view of (9.3), S(y) ≥ F  (y), y − x∗  + G(y) − G(x∗ ) ≥ F (y) − F (x∗ ) + G(y) − G(x∗ ) = H (y) − inf(H ). Lemma 9.1 is proved.

 

Suppose that L > 0 and F  (x) − F  (y) ≤ Lx − y for all x, y ∈ X.

(9.17)

280

9 PDA-Based Method for Convex Optimization

For every γ ≥ 1 and every y¯ ∈ dom(G), we say that u(y) ¯ ∈ dom(G) is a γ −1 -predicted decrease approximation (PDA) vector of H at y¯  if γ −1 S(y) ¯ ≤ F  (y), ¯ y¯ − u(y) ¯ + G(y) ¯ − G(u(y)). ¯

(9.18)

Note that for any γ˜ ≥ γ ≥ 1 any γ −1 -PDA vector is also a γ˜ −1 -PDA vector. Let γ ≥ 1, y¯ ∈ dom(G),  > 0. We say that u(y) ¯ ∈ dom(G) is a (γ −1 , )-predicted decrease approximation (PDA) vector of H at y¯ if ¯ ≤ F  (y), ¯ y¯ − u(y) ¯ + G(y) ¯ − G(u(y)) ¯ + . γ −1 S(y)

(9.19)

For x, y ∈ X define Q(y, x) = F (x) + F  (x), y − x + 2−1 Lx − y2 .

(9.20)

Let γ ≥ 1, δ1 , δ2 ∈ (0, 1]. Let us describe our algorithm. Initialization: select an arbitrary y0 ∈ dom(G). Iterative Step: given a current iteration vector yk ∈ dom(G) do the following: (i) calculate u(yk ) which is (γ −1 , δ1 )-predicted decrease approximation (PDA) vector of H at yk such that u(yk ) ∈ dom(G), γ −1 S(yk ) − δ1 ≤ F  (yk ), yk − u(yk ) + G(yk ) − G(u(yk ));

(9.21) (9.22)

(ii) choose a bounded set Xk such that [yk , u(yk )] ⊂ Xk

(9.23)

yk+1 ∈ Xk

(9.24)

and calculate

for which at least one of the following inequalities holds: Q(yk+1 , yk ) + G(yk+1 ) ≤ Q(y, yk ) + G(y) + δ2 for all y ∈ Xk

(9.25)

9.2 Auxiliary Results

281

(local modal update); H (yk+1 ) ≤ H (y) + δ2 for all y ∈ Xk

(9.26)

(global exact model update). Note that an exact version of this algorithm (without computational errors δ1 , δ2 ) was studied in . ∞ Theorem 9.2 Assume that {yk }∞ k=0 , {u(yk )}k=0 ⊂ X, Xk ⊂ X, k = 0, 1, . . . , (9.21)–(9.24) hold for all k = 0, 1, . . . and that for all nonnegative integers k ≥ 0, at least one of the relations (9.25) and (9.26) hold. Then for every integer k ≥ 0,

H (yk+1 ) − inf(H ) ≤ 2γ (k + 2γ )−1 ((2γ − 2)(k + 1)−1 (H (y0 ) − inf(H )) + LD 2 γ ) +(δ1 + δ2 )(2γ + k − 1). Theorem 9.2 is proved in Sect. 9.3. Its prototype without computational errors was obtained in .

9.2 Auxiliary Results ∞ Lemma 9.3 Assume that {yk }∞ k=0 , {u(yk )}k=0 ⊂ X, Xk ⊂ X, k = 0, 1, . . . , (9.21)– (9.24) hold for all k = 0, 1, . . . and that for all nonnegative integers k, at least one of the relations (9.25) and (9.26) hold. Let {tk }∞ k=0 ⊂ (0, 1]. Then for every integer k ≥ 0,

H (yk+1 ) ≤ H (yk ) + 2−1 tk2 LD 2 − γ −1 tk S(yk ) + δ1 + δ2 . Proof Let k ≥ 0 be an integer and set uk = u(yk ).

(9.27)

[yk , uk ] ⊂ Xk ,

(9.28)

By (9.21)–(9.23) and (9.27),

γ −1 S(yk ) − δ1 ≤ F  (yk ), yk − uk  + G(yk ) − G(uk ).

(9.29)

Since the function G is convex it follows from (9.2), (9.20), (9.23), (9.28), and (9.29) that

282

9 PDA-Based Method for Convex Optimization

inf{Q(y, yk ) + G(y) : y ∈ Xk } ≤ inf{Q(y, yk ) + G(y) : y ∈ [yk , uk ]} ≤ inf{Q(tuk + (1 − t)yk , yk ) + G(tuk + (1 − t)yk ) : t ∈ [0, 1]} ≤ Q(tk uk + (1 − tk )yk , yk ) + G(tk uk + (1 − tk )yk ) = F (yk ) + tk F  (yk ), uk − yk  + 2−1 Ltk2 yk − uk 2 +G(tk uk + (1 − t)yk ) ≤ F (yk ) + G(yk ) +tk (F  (yk ), uk − yk  + G(uk ) − G(yk )) +2−1 tk2 Lyk − uk 2 ≤ F (yk ) + G(yk ) − tk γ −1 S(yk ) + δ1 + 2−1 Ltk2 D 2 .

(9.30)

Proposition 4.3, (9.17), and (9.20) imply that F (y) ≤ Q(y, yk ) for all y ∈ Xk

(9.31)

F (y) + G(y) ≤ Q(y, yk ) + G(y), y ∈ Xk .

(9.32)

and

In the case of the local update it follows from (9.25) and (9.32) that F (yk+1 ) + G(yk+1 ) ≤ Q(yk+1 , yk ) + G(yk+1 ) ≤ inf{Q(y, yk ) + G(y) : y ∈ Xk } + δ2 . In the case of the exact model update it follows from (9.26) and (9.32) that F (yk+1 ) + G(yk+1 ) ≤ inf{F (y) + G(y) : y ∈ Xk } + δ2 ≤ inf{Q(y, yk ) + G(y) : y ∈ Xk } + δ2 . Therefore in both cases in view of (9.30),

9.2 Auxiliary Results

283

F (yk+1 ) + G(yk+1 ) ≤ inf{Q(y, yk ) + G(y) : y ∈ Xk } + δ2 . ≤ F (yk ) + G(yk ) + 2−1 tk2 LD 2 − γ −1 tk S(yk ) + δ1 + δ2 .  

Lemma 9.3 is proved. Lemma 9.4 Let c > 0, γ ≥ 1, δ ∈ (0, 2],

{ak }∞ k=0 ,

{bk }∞ k=0

R1,

0 ≤ ak ≤ bk , k = 0, 1, . . . , tk = 2γ (2γ + k)−1 , k = 0, 1, . . . .

(9.33)

Assume that for all integers k ≥ 0, ak+1 ≤ ak − tk bk γ −1 + 2−1 ctk2 + δ. Then for all integers k ≥ 0, ak+1 + (

k k   (bi − ai )(i + 2γ − 1))( (i + 2γ − 1))−1 i=0

i=0

≤ 2γ (k + 2γ )−1 ((2γ − 2)(k + 1)−1 a0 + cγ ) + δ(2γ + k − 1). Proof In view of (9.33) and (9.34), for any integer i ≥ 0, bi ≤ γ ti−1 (ai − ai+1 + 2−1 ti2 c + δ)2 and bi − ai ≤ (γ ti−1 − 1)ai −γ ti−1 ai+1 + 2−1 cγ ti + δγ ti−1 = 2−1 ai (2γ + i − 2) − 2−1 ai+1 (2γ + i) +(2γ + i)−1 cγ 2 + 2−1 δ(2γ + i). Multiplying the relation above by i + 2γ − 1 ≥ 0 we obtain that (bi − ai )(i + 2γ − 1) ≤ 2−1 (i + 2γ − 2)(i + 2γ − 1)ai − 2−1 (i + 2γ )(i + 2γ − 1)ai+1

(9.34)

284

9 PDA-Based Method for Convex Optimization

+ cγ 2 (i + 2γ − 1)(i + 2γ )−1 + 2−1 δ(2γ + i)(i + 2γ − 1).

(9.35)

Let k ≥ 0 be integer. Summing up (9.35) for i = 0, . . . , k we obtain that k 

(bi − ai )(i + 2γ − 1)

i=0

≤ 2−1 (2γ − 2)(2γ − 1)a0 − 2−1 (k + 2γ )(k + 2γ − 1)ak+1 + c(k + 1)γ 2 + 2−1 δ(k + 1)(2γ + k)(k + 2γ − 1).

(9.36)

Dividing both sides of (9.36) by 2−1 (k + 1)(k + 2γ ) we obtain that for any integer k ≥ 0, k  ( (bi − ai )(i + 2γ − 1))(2−1 (k + 1)(k + 2γ ))−1 i=0

+(k + 2γ − 1)(k + 1)−1 ak+1 ≤ 2γ (k + 2γ )−1 ((2γ − 2)(2γ − 1)(2γ )−1 (k + 1)−1 a0 + cγ ) + δ(2γ + k − 1).

(9.37)

Clearly, k  (i + 2γ − 1) = 2−1 (k + 1)(k + 4γ − 2) i=0

≥ 2−1 (k + 1)(k + 2γ ). By (9.37) and (9.38), k k   (bi − ai )(i + 2γ − 1)( (i + 2γ − 1))−1 + ak+1 i=0

i=0

k  (bi − ai )(i + 2γ − 1)(2−1 (k + 1)(k + 2γ ))−1 i=0

(9.38)

9.3 Proof of Theorem 9.2 and Examples

285

+(k + 2γ − 1)(k + 1)−1 ak+1 ≤ 2γ (k + 2γ )−1 ((2γ − 2)(k + 1)−1 a0 + cγ ) + δ(2γ + k − 1).  

Lemma 9.4 is proved.

9.3 Proof of Theorem 9.2 and Examples Theorem 9.2 follows from Lemmas 9.1, 9.3, and 9.4 applied with δ = δ1 + δ2 , ak = H (yk ) − inf(H ), bk = S(yk ), k = 0, 1, . . . , tk = 2γ (2γ + k)−1 , k = 0, 1, . . . , c = LD 2 . Clearly, the best choice of k is at the same order as (δ1 + δ2 )−1/2 and in this case the right-hand side of the final equation of Theorem 9.2 is c1 (δ1 + δ2 )1/2 , where c1 is a positive constant. We consider two special cases of the PDA-based method. Let γ = 1 and for a given integer k ≥ 0 define yk ∈ X, uk ∈ dom(G) such that F  (yk ), uk  + G(uk ) ≤ F  (yk ), u + G(u) + δ1 for all u ∈ X. By (9.6), for every integer k ≥ 0, S(yk ) = sup{F  (yk ), yk − u + G(yk ) − G(u) : u ∈ X} = F  (yk ), yk  + G(yk ) + sup{−F  (yk ), u − G(u) : u ∈ X} ≤ F  (yk ), yk  + G(yk ) −F  (yk ), uk  − G(uk ) + δ1 = F  (yk ), yk − uk  + G(yk ) − G(uk ) + δ1

(9.39)

286

9 PDA-Based Method for Convex Optimization

and (9.22) holds with u(yk ) = uk . Then we chose tk ∈ [0, 1] such that H (yk + tk (uk − yk )) ≤ H (yk + t (uk − yk )) + δ2 for all t ∈ [0, 1] and set yk+1 = yk + tk (uk − yk ). Clearly, in this case we have a generalized conditional algorithm. Consider now another special case of the PDA-based method with Xk = X, k = 0, 1, . . . For any integer k ≥ 0 find yk+1 ∈ X such that F (yk ) + F  (yk ), yk+1 − yk  + 2−1 Lyk − yk+1 2 + G(yk+1 ) ≤ F (yk ) + F  (yk ), y − yk  + 2−1 Lyk − y2 + G(y) + δ2 for all y ∈ X which is equivalent to the relation L−1 G(yk+1 ) + 2−1 yk+1 − (yk − L−1 F  (yk ))2 ≤ L−1 G(y) + 2−1 y − (yk − L−1 F  (yk ))2 + L−1 δ2 for all y ∈ X.

Chapter 10

Minimization of Quasiconvex Functions

In this chapter we study minimization of a quasiconvex function. Our algorithm has two steps. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this.

10.1 Preliminaries Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  ·  and let f : X → R 1 . We consider the problem f (x) → min, x ∈ X using the algorithm considered in . Recall that inf(f ) = inf{f (x) : x ∈ X}, argmin(f ) = {x ∈ X : f (x) = inf(f )}. Let F ⊂ X. Denote by int(F ), cl(F ), and bd(F ) its interior, closure, and boundary respectively. For every x ∈ X, N(F, x) = {q ∈ X : q, y − x ≤ 0 for all y ∈ F }.

© Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_10

(10.1)

287

288

10 Minimization of Quasiconvex Functions

Set SX (0, 1) = {z ∈ X : z = 1}.

(10.2)

Suppose that for each α ∈ R 1 the set {x ∈ X : f (x) ≤ α} is convex. In other words, the function f is quasiconvex. Assume that inf(f ) > −∞.

(10.3)

For every  > 0 and every x ∈ X set G (x) = {z ∈ X : f (z) < f (x) − }.

(10.4)

G(x) = G0 (x).

(10.5)

Let

For every x ∈ X set N (G(x), x) = {q ∈ X : q, y − x ≤ 0 for all y ∈ X satisfying f (y) < f (x)}.

(10.6)

β > 0, L ≥ 1, x∗ ∈ argmin(f ).

(10.7)

Let

Assume that for every z ∈ Y , |f (z) − f (x∗ )| ≤ Lz − x∗ β .

(10.8)

10.2 An Auxiliary Result Proposition 10.1 Let  ≥ 0, x ∈ X, and f (x) − inf(f ) > .

(10.9)

10.2 An Auxiliary Result

289

Then f (x) − inf(f ) ≤ Lq, x − x∗ β +  for all q ∈ SX (0, 1) ∩ N (G (x), x). Proof Since the function f is continuous and quasiconvex the set G (x) is nonempty, open, and convex. Set r = inf{z − x∗  : z ∈ bd(G (x))}.

(10.10)

{zk }∞ k=1 ⊂ bd(G (x))

(10.11)

There exist sequences

and {δk }∞ k=1 ⊂ (0, ∞) such that lim δk = 0,

k→∞

zk − x∗  ≤ r + δk , k = 1, 2, . . . .

(10.12)

By (10.8), (10.11), and (10.12), for all integers k ≥ 0, f (x) − inf(f ) −  ≤ f (zk ) − inf(f ) ≤ Lzk − x∗ β ≤ L(r + δk )β .

(10.13)

It follows from (10.11) and (10.13) that f (x) − inf(f ) ≤ Lr β + .

(10.14)

By (10.7), (10.9), and (10.10), x∗ + rz ∈ cl(G (x)) for all z ∈ SX (0, 1). In view of (10.1), (10.4), and (10.15), for every q ∈ SX (0, 1) ∩ N (G (x), x) we have q, x∗ + rz − x ≤ 0.

(10.15)

290

10 Minimization of Quasiconvex Functions

Applying the inequality above with z = q we obtain that 0 ≥ q, x∗ + rq − x and q, x − x∗  ≥ r for all q ∈ SX (0, 1) ∩ N (G (x), x).

(10.16)

Let q ∈ SX (0, 1) ∩ N (G (x), x). By (10.14) and (10.16), f (x) − inf(f ) ≤ Lr β +  ≤ Lq, x − x∗ β + .  

Proposition 10.1 is proved.

10.3 The Main Result Let δ1 , δ2 > 0, γ ∈ (0, 2). Let us describe our algorithm. Initialization: select an arbitrary x0 ∈ X. Iterative Step: Let xk ∈ X be given a current iteration vector. If Gδ1 (xk ) = ∅, then inf(f ) ≥ f (xk ) − δ1 and xk is considered as an approximate solution of our problem. If Gδ1 (xk ) = ∅, then find qk ∈ X such that BX (qk , δ2 ) ∩ SX (0, 1) ∩ N (Gδ1 (xk ), xk ) = ∅ and set λk = γ [(f (xk ) − inf(f ) − δ1 )L−1 ]1/β , xk+1 = xk − λk qk .

10.3 The Main Result

291

Let M > 0 be such that x∗  ≤ M.

(10.17)

In this chapter we prove the following result. Theorem 10.2 Let T = (4γ )−1 (2 − γ )δ2−2  + 1,

(10.18)

T −1 {xt }Tt=0 , {qt }t=0 ⊂ X, {λt }Tt=0 ⊂ (0, ∞),

x0  ≤ M,

(10.19)

Gδ1 (xt ) = ∅, t = 0, . . . , T − 1

(10.20)

and for t = 0, . . . , T − 1, BX (qt , δ2 ) ∩ SX (0, 1) ∩ N (Gδ1 (xt ), xt ) = ∅,

(10.21)

λt = γ [(f (xt ) − inf(f ) − δ1 )L−1 ]1/β ,

(10.22)

xt+1 = xt − λt qt .

(10.23)

Then there exists an integer t ∈ [0, T ] such that f (xt ) ≤ (8δ2 M)β (2 − γ )−β + inf(f ) + δ1 . Proof Assume that the theorem is not true. Then for all t = 0, . . . , T , f (xt ) > (8δ2 M)β (2 − γ )−β + inf(f ) + δ1 .

(10.24)

In view of (10.24) for all t = 0, . . . , T , ((2 − γ )(f (xt ) − inf(f ) − δ1 ))1/β > 8δ2 M.

(10.25)

By (10.17) and (10.19), x0 − x∗  ≤ 2M.

(10.26)

Assume that S ∈ {0, . . . , T }, S < T and that xt − x∗  ≤ 2M, t = 0, . . . , S.

(10.27)

292

10 Minimization of Quasiconvex Functions

(Note that in view of (10.26), (10.27) holds for S = 0.) Let t ∈ [0, S] be an integer. By (10.21) and (10.23), xt+1 − x∗ 2 = xt − λt qt − x∗ 2 = xt − x∗ 2 − 2λt qt , xt − x∗  + λ2t .

(10.28)

In view of (10.21), there exists  qt ∈ BX (qt , δ2 ) ∩ SX (0, 1) ∩ N (Gδ1 (xt ), xt ).

(10.29)

Proposition 10.1, (10.4), (10.20), (10.22), (10.25), (10.28), (10.29), and the inequality f (xt ) − inf(f ) > δ1 imply that xt+1 − x∗ 2 ≤ xt − x∗ 2 − 2λt  qt , xt − x∗  + λ2t +2 qt − qt λt xt − x∗  qt , xt − x∗  + λ2t ≤ xt − x∗ 2 − 2λt  +2δ2 λt xt − x ∗  ≤ xt − x∗ 2 − 2γ ((f (xt ) − inf(f ) − δ1 )L−1 )1/β ×(f (xt ) − inf(f ) − δ1 )1/β +γ 2 ((f (xt ) − inf(f ) − δ1 )L−1 )2/β + 2δ2 γ ((f (xt ) − inf(f ) − δ1 )L−1 )1/β xt − x ∗ . By (10.24), (10.27), and (10.30), xt+1 − x∗ 2 ≤ xt − x∗ 2 − γ (f (xt ) − inf(f ) − δ1 )2/β (2L−1/β − γ L−2/β ) +2δ2 γ ((f (xt ) − inf(f ) − δ1 )L−1 )1/β xt − x ∗  ≤ xt − x∗ 2

(10.30)

10.3 The Main Result

293

−γ (f (xt ) − inf(f ) − δ1 )1/β L−1/β ×(2 − γ L−1/β )((f (xt ) − inf(f ) − δ1 )1/β − 4δ2 M) ≤ xt − x∗ 2 − (f (xt ) − inf(f ) − δ1 )1/β γ (4δ2 M) ≤ xt − x∗ 2 − 4δ2 Mγ (2 − γ )−1 8δ2 M.

(10.31)

In view of (10.27) and (10.31), xt+1 − x∗  ≤ xt − x∗  ≤ 2M. By induction we have shown that xt − x∗  ≤ 2M, t = 0, . . . , T and that for all t = 0, . . . , T − 1, xt+1 − x∗ 2 ≤ xt − x∗ 2 − (4δ2 M)2 γ (2 − γ )−1 .

(10.32)

It follows from (10.26) and (10.32) that 4M 2 ≥ x0 − x∗ 2 − xT − x∗ 2 =

T −1

(xt − x∗ 2 − xt+1 − x∗ 2 )

t=0

≥ T (δ2 M)2 γ (2 − γ )−1 and T ≤ 4γ −1 (2 − γ )4−2 δ −2 . This contradicts (10.18). The contradiction we have reached completes the proof of Theorem 10.2.  

Chapter 11

Minimization of Sharp Weakly Convex Functions

In this chapter we study the subgradient projection algorithm for minimization of sharp weakly convex functions, under the presence of computational errors. The problem is described by an objective function and a set of feasible points. For this algorithm each iteration consists of two steps. The first step is a calculation of a subgradient of the objective function while in the second one we calculate a projection on the feasible set. In each of these two steps there is a computational error. In general, these two computational errors are different. We show that our algorithm generates a good approximate solution, if all the computational errors are bounded from above by a small positive constant. Moreover, if we know the computational errors for the two steps of our algorithm, we find out what approximate solution can be obtained and how many iterates one needs for this.

11.1 Preliminaries The problem and the algorithm studied in this chapter were considered in  in a finite-dimensional setting and without computational error. Let X be a Hilbert space equipped with an inner product ·, · which induces a complete norm  · . Let C ⊂ X be a nonempty closed convex set, ρ ≥ 0, μ > 0 g : X → R1 be a locally Lipschitz function such that the function g(x) + 2−1 ρx2 , x ∈ X

© Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_11

(11.1)

295

296

11 Minimization of Sharp Weakly Convex Functions

is convex. Following  the function g is called ρ-weakly convex. Set Cmin = argmin(g, C) = {x ∈ C : g(x) = inf(g, C)}.

(11.2)

Cmin = ∅.

(11.3)

Suppose that

Let M1 > 2 + 2μρ −1 , Cmin ⊂ BX (0, M1 − 2 − μρ −1 ).

(11.4)

Let L ≥ 2 be such that |g(z1 ) − g(z2 )| ≤ Lz1 − z2  for all z1 , z2 ∈ BX (0, M1 + 4).

(11.5)

For every x ∈ X, denote by ∂g(x) the set of all v ∈ X such that g(y) ≥ g(x) + v, y − x + o(y − x) as y → 0.

(11.6)

We can show (see Proposition 11.1) that for every x ∈ X, ∂(g + 2−1 ρ · 2 )(x) = ∂g(x) + ρx,

(11.7)

where the left-hand side of (11.7) is the subdifferential of the convex function. Suppose that the following sharpness property holds : for all z ∈ C g(z) − inf(g, C) ≥ μd(z, Cmin ).

(11.8)

A point x¯ ∈ C is called stationary if g(x) − g(x) ¯ ≥ o(x − x) ¯ as x → x¯ in C.

(11.9)

11.2 The Subdifferential of Weakly Convex Functions

297

11.2 The Subdifferential of Weakly Convex Functions Proposition 11.1 Let x ∈ X. Then v ∈ ∂g(x) if and only if v + ρx ∈ ∂(g + 2−1 ρ · 2 )(x). Moreover, if v ∈ ∂g(x), then for every y ∈ X, g(y) ≥ g(x) + v, y − x − 2−1 ρy − x2 . Proof Let v ∈ ∂g(x)

(11.10)

and h ∈ X \ {0}. In view of (11.6), for all t > 0, g(x + th) ≥ g(x) + v, th + o(th) as t → 0+ . This implies that t −1 (g(x + th) − g(x)) ≥ v, h + o(th)t −1 and lim inf t −1 (g(x + th) − g(x)) ≥ v, h. t→0+

(11.11)

Thus we have shown that the following property holds: (i) if v ∈ ∂g(x), then (11.11) holds for every h ∈ X. Note that the function g(x) + 2−1 ρx2 , x ∈ X is convex. For every t > 0 and every h ∈ X, we have t −1 (g(x + th) + 2−1 ρx + th2 − g(x) − 2−1 ρx2 ) = t −1 (g(x + th) − g(x)) + t −1 (2−1 ρ2th, x + 2−1 ρt 2 h2 ). By (11.12), lim t −1 (g(x + th) + 2−1 ρx + th2 − g(x) − 2−1 ρx2 )

t→0+

(11.12)

298

11 Minimization of Sharp Weakly Convex Functions

= lim (t −1 (g(x + th) − g(x)) + ρh, x). t→0+

(11.13)

It follows from (11.13) that lim (t −1 (g(x + th) − g(x))

t→0+

exists and lim (t −1 (g(x + th) − g(x))

t→0+

= lim t −1 ((g + 2−1 ρ · 2 )(x + th) − (g + 2−1 ρ · 2 )(x)) − ρh, x. t→0+

(11.14)

Property (i) implies that if v ∈ ∂g(x), then (11.11) holds and in view of (11.14), v + ρx ∈ ∂(g + 2−1 ρ · 2 )(x).

(11.15)

Let v ∈ X and (11.15) hold. By (11.15), for every y ∈ X, g(y) + 2−1 ρy2 ≥ g(x) + 2−1 ρx2 + v + ρx, y − x and g(y) ≥ g(x) + v, y − x +2−1 ρx2 − 2−1 y2 + ρx, y − x = g(x) + v, y − x − 2−1 ρy2 − 2−1 x2 + ρx, y = g(x) + v, y − x − 2−1 ρy − x2 .

(11.16)

Relation (11.16) implies that v ∈ ∂g(x). Thus (11.15) implies the inclusion above. Proposition 11.1 is proved.  

11.3 An Auxiliary Result Lemma 11.2 Let δ ∈ (0, 1], δ0 ∈ (0, 1], δ1 = μ/16,

(11.17)

11.3 An Auxiliary Result

299

x, ξ ∈ X, x ≤ M1 + 3,

(11.18)

B(x, δ) ∩ C = ∅,

(11.19)

ξ ∈ ∂f (x),

(11.20)

ξ  ≤ δ1 .

(11.21)

Then at least one of the following inequalities holds: d(x, Cmin ) ≤ 16δ;

(11.22)

d(x, Cmin ) ≤ 16Lδμ−1 ;

(11.23)

d(x, Cmin ) ≥ 2−1 3μρ −1 . Proof There exists {xi }∞ i=1 ⊂ Cmin

(11.24)

lim xi − x = d(x, Cmin ).

(11.25)

such that i→∞

In view of (11.19), there exists  x ∈ C such that x −  x  ≤ δ.

(11.26)

By (11.5), (11.8), (11.17), (11.18), (11.24), and (11.26), for i = 1, 2, . . . Lδ + g(x) − g(xi ) ≥ g( x ) − g(x) + g(x) − g(xi ) g( x ) − g(xi ) ≥ μd( x , Cmin ).

(11.27)

Proposition 11.1 and (11.20) imply that ξ + ρx ∈ ∂(g + 2−1 ρ · 2 )(x). In view of (11.21) and (11.28), for every y ∈ X,

(11.28)

300

11 Minimization of Sharp Weakly Convex Functions

g(y) − g(x) ≥ ξ, y − x − 2−1 ρy2 + 2−1 ρx2 − ρx2 + ρx, y ≥ −δ1 y − x − 2−1 ρx − y2 .

(11.29)

By (11.29), for i = 1, 2, . . . , g(x) − g(xi ) ≤ δ1 xi − x + 2−1 ρxi − x2 .

(11.30)

It follows from (11.27) and (11.30) that for all integers i ≥ 1, μd( x , Cmin ) ≤ Lδ + g(x) − g(xi ) ≤ Lδ + δ1 xi − x + 2−1 ρxi − x2 .

(11.31)

By (11.24)–(11.26) and (11.31), x , Cmin ) −μδ + μd(x, Cmin ) ≤ μd( ≤ Lδ + δ1 d(x, Cmin ) + 2−1 ρd(x, Cmin )2 .

(11.32)

In view of (11.32), μd(x, Cmin ) ≤ μδ + Lδ + δ1 d(x, Cmin ) + 2−1 ρd(x, Cmin )2 .

(11.33)

Assume that (11.22) and (11.23) do not hold. Together with (11.17) and (11.33) this implies that μd(x, Cmin ) ≤ (μ/16)d(x, Cmin ) +(μ/16)d(x, Cmin ) + (μ/16)d(x, Cmin ) + 2−1 ρd(x, Cmin )2 ≤ (μ/4)d(x, Cmin ) + 2−1 ρd(x, Cmin )2 .

(11.34)

By (11.34), μ ≤ μ/4 + 2−1 ρd(x, Cmin ) and μ ≤ (2/3)ρd(x, Cmin ). Lemma 11.2 is proved.

 

11.4 The First Main Result

301

11.4 The First Main Result Let γ ∈ (0, 1), δ, δ0 ∈ (0, 1], δ0 ≤ μ/16. Let us describe our algorithm. Initialization: select an arbitrary x0 ∈ X such that d(x0 , Cmin ) < γ μρ −1 , d(x0 , C) ≤ δ. Iterative Step: given a current iteration vector xt ∈ X, if g(xt ) ≤ inf(g, C) + max{(16δ(L + 1)3 γ (1 − γ )−1 ρ −1 )1/2 , 16δL, 16L2 δμ−1 , 4δL(μ + L)μ−1 (1 − γ )−1 }, then xt is considered as an approximate solution; otherwise calculate ξt ∈ X such that B(ξt , δ0 ) ∩ ∂f (xt ) = ∅ (note that by Lemma 11.2, ξt = 0) and calculate xt+1 ∈ X such that xt+1 − PC (xt − ξt −2 (g(xt ) − inf(g, C))ξt ) ≤ δ. Proposition 11.3 Let γ ∈ (0, 1), δ, δ0 ∈ (0, 1], δ0 < min{μ(1 − γ )/4, μ/32},

(11.35)

T −1 ⊂ X, T > 0 be an integer, {xt }Tt=0 ⊂ X, {ξt }t=0

d(x0 , C) ≤ δ,

(11.36)

d(x0 , Cmin ) < γ μρ −1 ,

(11.37)

for all t = 0, . . . , T − 1, g(xt ) > inf(g, C) + (16δ(L + 1)3 γ (1 − γ )−1 ρ −1 )1/2 ,

(11.38)

302

11 Minimization of Sharp Weakly Convex Functions

g(xt ) > inf(g, C) + max{16δL, 16L2 δμ−1 , δ(μ + L)L, 4δL(μ + L)μ−1 (1 − γ )−1 }, BX (ξt , δ0 ) ∩ ∂f (xt ) = ∅

(11.39) (11.40)

and xt+1 − PC (xt − ξt −2 (g(xt ) − inf(g, C))ξt ) ≤ δ.

(11.41)

Then ξt  ≥ μ/32, t = 0, . . . , T , T ≤ 2M1 δ −1 . Proof By (11.4) and (11.37), x0  ≤ M1 .

(11.42)

Assume that t ∈ [0, T ) is an integer, xt is well defined, and d(xt , Cmin ) < γ μρ −1 .

(11.43)

(Note that in view of (11.37), inequality (11.43) holds for t = 0.) By (11.4) and (11.43), xt  ≤ M1 .

(11.44)

It follows from (11.4), (11.5), (11.43), and (11.44), |g(xt ) − inf(g, C)| < Ld(xt , Cmin ).

(11.45)

g(xt ) < inf(g, C) + Ld(xt , Cmin ).

(11.46)

In view of (11.45),

Relations (11.39) and (11.46) imply that d(xt , Cmin ) > 16δ, 16δLμ−1 , 4δ(μ + L)μ−1 (1 − γ )−1 .

(11.47)

By (11.40), there exists ηt ∈ ∂f (xt )

(11.48)

11.4 The First Main Result

303

such that ηt − ξt  ≤ δ0 .

(11.49)

Lemma 11.2, (11.17), (11.36), (11.41), (11.43), (11.44), (11.47), and (11.48) imply that ηt  > μ/16.

(11.50)

It follows from (11.35), (11.49), and (11.50) that ξt  ≥ μ/32.

(11.51)

Clearly, xt+1 is well defined by (11.41). (Thus we showed by induction that xi , i = 0, . . . , T , ξi , i = 0, . . . , T − 1 are well defined.) There exists a sequence {zi }∞ i=1 ⊂ Cmin

(11.52)

lim xt − zi  = d(xt , Cmin ).

(11.53)

such that i→∞

Lemma 2.2, (11.41), and (11.52) imply that for all natural numbers i, xt+1 − zi  ≤ δ + PC (xt − ξt −2 (g(xt ) − inf(g, C))ξt ) − zi  ≤ δ + xt − ξt −2 (g(xt ) − inf(g, C))ξt − zi . Proposition 11.1, (11.6), (11.48), (11.49), (11.51), and (11.52) imply that xt − zi − ξt −2 (g(xt ) − inf(g, C))ξt )2 = xt − zi 2 +2ξt −2 (g(xt ) − inf(g, C))ξt , zi − xt  +ξt −2 (g(xt ) − inf(g, C))2 ≤ xt − zi 2 +(g(xt ) − inf(g, C))2 ξt −2

(11.54)

304

11 Minimization of Sharp Weakly Convex Functions

+2(g(xt ) − inf(g, C))ξt −2 ξt − ηt , zi − xt  +2(g(xt ) − inf(g, C))ξt −2 ηt , zi − xt  ≤ xt − zi 2 +(g(xt ) − inf(g, C))2 ξt −2 +2(g(xt ) − inf(g, C))ξt −2 δ0 zi − xt  +2(g(xt ) − inf(g, C) − δ)ξt −2 (g(zi ) − g(xt ) + 2−1 ρzi − xt 2 ) = xt − zi 2 +(g(xt ) − g(zi ))ξt −2 (ρxt − zi 2 − (g(xt ) − g(zi )) +2δ0 zi − xt ξt −2 (g(xt ) − inf(g, C)) = xt − zi 2 +(g(xt ) − g(zi ))ξt −2 (ρxt − zi 2 − (g(xt ) − g(zi )) + 2δ0 zi − xt ).

(11.55)

By (11.41) and (11.46), there exists  x∈C

(11.56)

 x − xt  ≤ δ.

(11.57)

such that

By (11.5), (11.44), and (11.57), for all integers i = 1, 2, . . . , x ) − g(zi )) ≤ Lδ. |(g(xt ) − g(zi )) − (g(

(11.58)

Let i ≥ 0 be an integer. It follows from (11.8), (11.52), (11.56), and (11.57) that x , Cmin ) ≥ μd(xt , Cmin ) − δ. g( x ) − g(zi ) ≥ μd(

(11.59)

In view of (11.58) and (11.59), g(xt ) − g(zi ) ≥ −Lδ − μδ + μd(xt , Cmin ).

(11.60)

11.4 The First Main Result

305

By (11.55) and (11.60), xt − zi − ξt −2 (g(xt ) − inf(g, C))ξt )2 ≤ xt − zi 2 +ξt −2 (g(xt ) − g(zi ))(ρxt − zi 2 − μd(xt , Cmin ) + Lδ + μδ + 2δ0 zi − xt ).

(11.61)

It follows from (11.47) and (11.62) that for all sufficiently large natural numbers i, 16δ ≤ d(xt , Cmin ) ≤ zi − xt  ≤ 2d(xt , Cmin ). By (11.35), (11.43), (11.52), (11.53), and (11.61), d(xt − ξt −2 (g(xt ) − inf(g, C))ξt , Cmin )2 ≤ lim sup xt − zi − ξt −2 (g(xt ) − g(zi ))ξt 2 i→∞

≤ d(xt , Cmin )2 + ξt −2 (g(xt ) − inf(g, C)) ×(ρd(xt , Cmin )2 − μd(xt , Cmin ) +Lδ + μδ + 2δ0 d(xt , Cmin )) = d(xt , Cmin )2 + ξt −2 (g(xt ) − inf(g, C)) ×(ρd(xt , Cmin )2 − (μ − 2δ0 )d(xt , Cmin ) + δ(μ + L)) = d(xt , Cmin )2 +ξt −2 ρ(g(xt ) − inf(g, C))d(xt , Cmin )(d(xt , Cmin ) −(μ − 2δ0 )ρ −1 + d(xt , Cmin )−1 δ(μ + L))ρ −1 ≤ d(xt , Cmin )2 +ξt −2 ρ(g(xt ) − inf(g, C))d(xt , Cmin )(γ μρ −1 −μρ −1 + 2δ0 ρ −1 + δ(μ + L)d(xt , Cmin )−1 )ρ −1 ) ≤ d(xt , Cmin )2

(11.62)

306

11 Minimization of Sharp Weakly Convex Functions

−ξt −2 ρ(g(xt ) − inf(g, C))(d(xt , Cmin )((1 − γ )μρ −1 −2δ0 ρ −1 ) − δ(μ + L)ρ −1 ) ≤ d(xt , Cmin )2 − ξt −2 ρ(g(xt ) − inf(g, C)) × (d(xt , Cmin )(1 − γ )μρ −1 /2 − δ(μ + L)ρ −1 ).

(11.63)

By (11.47) and (11.63), d(xt − ξt −2 (g(xt ) − inf(g, C))ξt , Cmin )2 ≤ d(xt , Cmin )2 − ξt −2 ρ(g(xt ) − inf(g, C))d(xt , Cmin )(1 − γ )μρ −1 /4.

(11.64)

In view of (11.39) and (11.64), d(xt − ξt −2 (g(xt ) − inf(g, C))ξt , Cmin )2 ≤ d(xt , Cmin )2 .

(11.65)

It follows from (11.43), (11.64), and (11.65) that d(xt , Cmin ) − d(xt − ξt −2 (g(xt ) − inf(g, C))ξt , Cmin ) = (d(xt , Cmin )2 − d(xt − ξt −2 (g(xt ) − inf(g, C))ξt , Cmin )2 ) ×(d(xt , Cmin ) + d(xt − ξt −2 (g(xt ) − inf(g, C))ξt , Cmin ))−1 ≥ d(xt , Cmin )2 − d(xt − ξt −2 (g(xt ) − inf(g, C))ξt , Cmin )2 (2γ μ/ρ)−1 ≥ d(xt , Cmin )ξt −2 (g(xt ) − inf(g, C))8−1 γ −1 (1 − γ )ρ.

(11.66)

Lemma 2.2, (11.41), and (11.66) imply that d(xt+1 , Cmin ) ≤ δ + d(PC (xt − ξt −2 (g(xt ) − inf(g, C))ξt , Cmin ) ≤ δ + d(xt − ξt −2 (g(xt ) − inf(g, C))ξt , Cmin ) ≤ δ + d(xt , Cmin ) − d(xt , Cmin )ξt −2 (g(xt ) − inf(g, C))8−1 γ −1 (1 − γ )ρ.

(11.67)

11.4 The First Main Result

307

By (11.5), (11.44), (11.48), and (11.49), ξt  ≤ L + 1.

(11.68)

In view of (11.4), (11.5), and (11.44), g(xt ) − inf(g, C) ≤ Ld(xt , Cmin ).

(11.69)

It follows from (11.38), (11.68), and (11.69) that d(xt , Cmin )ξt −2 (g(xt ) − inf(g, C))8−1 γ −1 (1 − γ )ρ ≥ (g(xt ) − inf(g, C))L−1 8−1 γ −1 (1 − γ )ρ(L + 1)−2 ≥ 2δ.

(11.70)

By (11.67) and (11.70), d(xt+1 , Cmin ) ≤ d(xt , Cmin ) − δ.

(11.71)

We assume that (11.43) holds and obtained (11.71). This implies that for all integers t = 0, . . . , T − 1, d(xt+1 , Cmin ) ≤ d(xt , Cmin ) − δ and d(xt , Cmin ) ≤ γ μρ −1 for all integers t = 0, . . . , T . Together with (11.4) and (11.42) this implies that 2M1 ≥ d(x0 , Cmin ) ≥ d(x0 , Cmin ) − d(xT , Cmin ) =

T −1

(d(xt , Cmin ) − d(xt+1 , Cmin )) ≥ T δ

t=0

and T ≤ 2M1 δ −1 . This completes the proof of Proposition 11.3. Proposition 11.3 implies the following result.

 

308

11 Minimization of Sharp Weakly Convex Functions

Theorem 11.4 Let γ ∈ (0, 1), δ, δ0 ∈ (0, 1], δ0 < min{μ(1 − γ )/4, μ/32}, T = 2M1 δ −1 , T −1 ⊂ X, {xt }Tt=0 ⊂ X, {ξt }t=0

d(x0 , C) ≤ δ, d(x0 , Cmin ) < γ μρ −1 , for all t = 0, . . . , T − 1, B(ξt , δ0 ) ∩ ∂f (xt ) = ∅, if ξt = 0, then xt+1 − PC (xt − ξt −2 (g(xt ) − inf(g, C))ξt ) ≤ δ, otherwise xt+1 = xt . Then there exists t ∈ {0, . . . , T }, such that g(xt ) ≤ inf(g, C) + max{(16δ(L + 1)3 γ (1 − γ )−1 ρ −1 )1/2 , 16δL, 16L2 δμ−1 , δ(μ + L)L, 4δL(μ + L)μ−1 (1 − γ )−1 }.

11.5 An Algorithm with Constant Step Sizes Let γ ∈ (0, 1), δ, δ0 ∈ (0, 1], α > 0. Let us describe our algorithm.

11.6 An Auxiliary Result

309

Initialization: select an arbitrary x0 ∈ X such that d(x0 , Cmin ) < γ μρ −1 , d(x0 , C) ≤ δ. Iterative Step: given a current iteration vector xt ∈ X, calculate ξt ∈ X such that B(ξt , δ0 ) ∩ ∂f (xt ) = ∅, if ξt = 0, then calculate xt+1 ∈ X such that xt+1 − PC (xt − αξt −1 ξt ) ≤ δ; otherwise set xk+1 = xk .

11.6 An Auxiliary Result Lemma 11.5 Let α > 0, δ, δ0 ∈ (0, 1], γ ∈ (0, 1) satisfy α ≤ 2−1 μ(L + 1)−1 , δ0 < μ/32, δ0 ≤ 64−1 αρ, δ < μρ −1 , δ ≤ min{64−2 αμ(L + μ)−1 , 2−1 α 2 L−2 μ−1 ρ},

(11.72)

x ∈ X satisfy d(x, C) < δ, d(x, Cmin ) < γ μρ −1 ,

(11.73)

d(x, Cmin ) > max{16δ, 16Lδμ−1 , 6αμ−1 L}.

(11.74)

Then for every η ∈ ∂f (x), η ≥ μ/16 and the following assertion holds.

(11.75)

310

11 Minimization of Sharp Weakly Convex Functions

Let ξ ∈ X, x + ∈ X, BX (ξ, δ0 ) ∩ ∂f (x) = ∅,

(11.76)

x + − PC (x − αξ −1 ξ ) ≤ δ.

(11.77)

Then ξ  ≥ μ/32, d(x + , Cmin )2 ≤ d(x, Cmin )2 − 2−1 α 2 . Proof By (11.4) and (11.73), x ≤ M1 .

(11.78)

Lemma 11.2, (11.73), (11.74), and (11.78) imply that for every η ∈ ∂f (x) (11.76) holds. Let us prove the assertion. In view of (11.76), there exists η ∈ ∂f (x)

(11.79)

ξ − η ≤ δ0 .

(11.80)

such that

It follows from (11.5), (11.72), (11.75), (11.76), and (11.78)–(11.80) that ξ  ≥ μ/32,

(11.81)

ξ  ≤ L + 1. Let {xi }∞ i=1 ⊂ Cmin

(11.82)

x − xi  → d(x, Cmin ) as i → ∞.

(11.83)

and

In view of (11.73), we may assume that for all integers i = 1, 2, . . . , xi − x < γ μρ −1 . Let i ≥ 1 be an integer. Lemma 2.2, (11.77), and (11.82) imply that

(11.84)

11.6 An Auxiliary Result

311

x + − xi  ≤ δ + PC (x − αξ −1 ξ ) − xi  ≤ δ + x − xi − αξ −1 ξ .

(11.85)

Proposition 11.1, (11.6), (11.72), (11.79)–(11.81), and (11.84) imply that x − xi − αξ −1 ξ 2 = x − xi 2 + 2αξ −1 ξ, xi − x + α 2 ≤ x − xi 2 + 2αξ −1 η − ξ xi − x + 2αξ −1 η, xi − x + α 2 ≤ x − xi 2 + 64αμ−1 δ0 μρ −1 + α 2 + 2αξ −1 (g(xi ) − g(x) + 2−1 ρx − xi 2 ).

(11.86)

In view of (11.73), there exists  x ∈ C ∩ BX (x, δ).

(11.87)

It follows from (11.5), (11.78), and (11.87) that g(x) − g( x )| ≤ Lδ. By (11.5), (11.8), (11.78)–(11.81), and (11.86)–(11.88), x − xi − αξ −1 ξ 2 x − xi 2 (1 + αρξ −1 ) + α 2 + 64αδ0 ρ −1 +2αξ −1 (g(xi ) − g( x ) + g( x ) − g(x)) ≤ x − xi 2 (1 + αρξ −1 ) x , Cmin )) +α 2 + 64αδ0 ρ −1 + 2αξ −1 (Lδ − μd( ≤ x − xi 2 (1 + αρξ −1 ) +α 2 + 64αδ0 ρ −1 + 2αξ −1 (Lδ + δμ − μd(x, Cmin )) ≤ x − xi 2 (1 + αρξ −1 ) −2αμd(x, Cmin )ξ −1 +α 2 + 64αδ0 ρ −1 + 2αξ −1 (L + μ)δ.

(11.88)

312

11 Minimization of Sharp Weakly Convex Functions

It follows from the relation above, (11.72), (11.73), and (11.81)–(11.83) that d(x − αξ −1 ξ, Cmin )2 ≤ d(x, Cmin )2 (1 + αρξ −1 ) −2αμd(x, Cmin )ξ −1 + α 2 + 64αδ0 ρ −1 + 2αξ −1 (L + μ)δ.

(11.89)

By (11.89), d(x, Cmin )2 − d(x − αξ −1 ξ, Cmin )2 ≥ d(x, Cmin )(2αμξ −1 − αρξ −1 d(x, Cmin )) −α 2 − 64αδ0 ρ −1 − 64αμ−1 (L + μ)δ ≥ d(x, Cmin )αμξ −1 − α 2 − 64αδ0 ρ −1 − 64αμ−1 (L + μ) ≥ d(x, Cmin )αμ(L + 1)−1 − 3α 2 ≥ d(x, Cmin )αμ(2L + 2)−1 .

(11.90)

d(x − αξ −1 ξ, Cmin ) ≤ d(x, Cmin ).

(11.91)

By (11.90),

Lemma 2.2 and (11.77) imply that for every z ∈ Cmin , d(x + , Cmin ) ≤ x + − z ≤ δ + PC (x − αξ −1 ξ ) − z ≤ δ + x − αξ −1 ξ − z. This implies that d(x + , Cmin ) ≤ δ + d(x − αξ −1 ξ, Cmin ).

(11.92)

By (11.72), (11.73), (11.91), and (11.92), d(x + , Cmin ) ≤ δ + d(x, Cmin ) ≤ δ + γ μρ −1 ≤ 2μρ −1 .

(11.93)

11.7 The Second Main Result

313

It follows from (11.73) and (11.91)–(11.93) that d(x + , Cmin )2 − d(x − αξ −1 ξ, Cmin )2 = (d(x + , Cmin ) − d(x − αξ −1 ξ, Cmin )) ×(d(x + , Cmin ) + d(x − αξ −1 ξ, Cmin )) ≤ 3δμρ −1 .

(11.94)

Lemma 2.2, (11.72), (11.74), (11.90), and (11.94) imply that d(x + , Cmin )2 ≤ d(x − αξ −1 ξ, Cmin )2 + 3δμρ −1 ≤ d(x, Cmin )2 − d(x, Cmin )αμ(2L + 2)−1 + 3δμρ −1 ≤ d(x, Cmin )2 − 6α 2 μ−1 Lμ(2L + 2)−1 + 3μρ −1 (2−1 α 2 L−2 μ−1 ρ) ≤ d(x, Cmin )2 − α 2 + 2α 2 L−2 ≤ d(x, Cmin )2 − α 2 /2.  

Lemma 11.5 is proved.

11.7 The Second Main Result Theorem 11.6 Let γ ∈ (0, 1), δ, δ0 ∈ (0, 1], α > 0, α ≤ 2−1 μ(L + 1)−1 , δ0 < μ/32, δ0 ≤ 64−1 ρα, δ < μ/ρ, δ ≤ min{2−1 α 2 L−2 μ−1 ρ, 64−1 μ(L + μ)−1 α}, x0 ∈ X satisfy d(x0 , C) ≤ δ, d(x0 , Cmin ) < γ μρ −1 , T = 1 + 2(μ/ρ)2 α −2 , T −1 ⊂ X, for all t = 0, . . . , T − 1, {xt }Tt=0 ⊂ X, {ξt }t=0

(11.95)

314

11 Minimization of Sharp Weakly Convex Functions

B(ξt , δ0 ) ∩ ∂f (xt ) = ∅, if ξt = 0, then xt+1 − PC (xt − ξt −1 αξt ) ≤ δ, otherwise xt+1 = xt . Then there exists t ∈ {0, . . . , T }, such that ξs  = 0 for all nonnegative integers s < t, g(xt ) ≤ inf(g, C) + max{16δ, 16Lμ−1 δ, 6αμ−1 L}. Proof Assume that the theorem does not hold. Then for all integers t = 0, . . . , T , d(xt , Cmin ) > max{16δ, 16δμ−1 L, 6αμ−1 L} + inf(g, C). Applying by induction Lemma 11.5 we obtain that for all integers t = 0, . . . , T − 1 ξt  ≥ μ/32, d(xt+1 , Cmin )2 ≤ d(xt , Cmin )2 − 2−1 α 2 . By the relation above and (11.95), (μ/ρ)2 ≥ d(x0 , Cmin )2 ≥ d(x0 , Cmin )2 − d(xT , Cmin )2 T −1

(d(xt , Cmin )2 − d(xt+1 , Cmin )2 )

t=0

≥ 2−1 T α 2 and

11.8 Convex Problems

315

T ≤ 2(μ/ρ)2 α −2 . This contradicts the choice of T . The contradiction we have reached completes the proof of Theorem 11.6.  

11.8 Convex Problems Let g : X → R 1 be a continuous convex function, μ > 0, M > 4, γ ∈ (0, 1) for every z ∈ C, g(z) − inf(g, C) ≥ μd(z, argmin(g, C))

(11.96)

argmin(g, C) ⊂ BX (0, M − 4).

(11.97)

and

Fix α > 0. Let L ≥ 1 be such that |f (z1 ) − f (z2 )| ≤ Lz1 − z2  for all z1 , z2 ∈ BX (0, M + 1)

(11.98)

and δ, δ0 ∈ (0, 1]. We prove the following result. Theorem 11.7 Let α ≤ 2−1 (L + 1)−1 , α ≤ 4−1 μ(L + 1)−2 , δ0 ≤ 4−1 M −1 α 2 (L + 1)2 , δ ≤ (2L)−1 α(L + 1)2 ,

(11.99)

x0  ≤ M, d(x0 , C) ≤ δ,

(11.100)

∞ {xt }∞ t=0 ⊂ X, {ξt }t=0 ⊂ X,

and for all integer t ≥ 0,

316

11 Minimization of Sharp Weakly Convex Functions

B(ξt , δ0 ) ∩ ∂f (xt ) = ∅, xt+1 − PC (xt − αξt ) ≤ δ.

(11.101)

Then there exists a nonnegative integer τ ≤ 2M 2 α −2 (L + 1)−2  such that for all integers t ∈ [0, τ ] \ {τ }, d(xt , Cmin ) ≤ d(xt , Cmin ) and for all integers t ≥ τ , d(xt , Cmin ) ≤ α(L + 1)((4L + 4)μ−1 + 3).

11.9 An Auxiliary Result Lemma 11.8 Let α < (2L + 2)−1 , 4δ0 M < α 2 (L + 1)2 , δ ≤ 2−1 α(L + 1),

(11.102)

x ≤ M,

(11.103)

d(x, C) ≤ δ,

(11.104)

BX (ξ, δ0 ) ∩ ∂f (x) = ∅,

(11.105)

y − PC (x − αξ ) ≤ δ.

(11.106)

x ∈ X,

ξ ∈ X,

y ∈ X,

Then d(y, Cmin )2 ≤ d(x, Cmin )2 + 6α 2 (L + 1)2 − 2αμd(x, Cmin ).

11.9 An Auxiliary Result

317

Proof Clearly, Cmin is a closed convex set. In view of (11.105), there exists η ∈ ∂g(x)

(11.107)

ξ − η ≤ δ0 .

(11.108)

such that

By (11.98), (11.103), and (11.107), η ≤ L.

(11.109)

It follows from (11.108) and (11.109) that ξ  ≤ L + 1.

(11.110)

 x∈C

(11.111)

x −  x  ≤ δ.

(11.112)

In view of (11.104), there exists

such that

Lemma 2.2 and (11.106) imply that y − PCmin (x) ≤ y − PC (x − αξ ) + PC (x − αξ ) − PCmin (x) ≤ δ + x − αξ − PC m in (x).

(11.113)

By (11.96), (11.98), (11.103), (11.104), (11.107), (11.108), and (11.110)–(11.112), x − αξ − PC m in (x)2 = x − PCmin (x)2 + α 2 ξ 2 + 2αξ, PCmin (x) − x ≤ x − PCmin (x)2 + α 2 ξ 2 + 2αη, PCmin (x) − x +2αη − ξ PCmin (x) − x ≤ x − PCmin (x)2 + α 2 (L + 1)2 + 4αδ0 M + 2αη, PCmin (x) − x

318

11 Minimization of Sharp Weakly Convex Functions

≤ d(x, Cmin )2 + α 2 (L + 1)2 + 4αδ0 M +2α(g(PCmin (x)) − g(x)) ≤ d(x, Cmin )2 + α 2 (L + 1)2 + 4αδ0 M +2α(inf(g, C) − g( x )) + 2α(g( x ) − g(x)) ≤ d(x, Cmin )2 + α 2 (L + 1)2 + 4αδ0 M + 2αδL −2αμd( x , Cmin ) ≤ d(x, Cmin )2 + α 2 (L + 1)2 + 4αδ0 M + 2αδL −2αμd(x, Cmin ) + 2αδμ ≤ d(x, Cmin )2 + 4α 2 (L + 1)2 − 2αμd(x, Cmin ).

(11.114)

It follows from (11.97), (11.102), (11.103), (11.110), (11.113), and (11.114) that y − PCmin (x)2 ≤ x − αξ − PCmin (x)2 + δ 2 + 2δx − αξ − PCmin (x) ≤ d(x, Cmin )2 + 4α 2 (L + 1)2 − 2αd(x, Cmin )μ + δ 2 + 2δ(2M + 2) ≤ d(x, Cmin )2 + 6α 2 (L + 1)2 − 2αd(x, Cmin )μ.  

Lemma 11.8 is proved.

11.10 Proof of Theorem 11.7 Assume that t ≥ 0 is an integer, xt  ≤ M. Applying Lemma 11.8 with x = xt , y = xt+1 , ξ = ξt we obtain that

(11.115)

11.10 Proof of Theorem 11.7

319

d(xt+1 , Cmin )2 ≤ d(xt , Cmin )2 + 6α 2 (L + 1)2 − 2αd(x, Cmin )μ.

(11.116)

Assume in addition that d(xt , Cmin ) ≥ 4α(L + 1)2 μ−1 .

(11.117)

By (11.116) and (11.117), d(xt+1 , Cmin )2 ≤ d(xt , Cmin )2 − 2α 2 (L + 1)2 .

(11.118)

Thus we have shown that the following property holds: (a) if an integer t ≥ 0 and (11.115) holds, then (11.116) holds and if (11.115) and (11.117) are valid, then (11.118) is true. Assume that τ ≥ 1 is an integer and that for all integers t = 0, . . . , τ − 1, (11.117) holds. By induction, we apply property (a) and (11.110) and show that (11.118) holds for all t = 0, . . . , τ − 1 and that (11.115) holds for all t = 0, . . . , τ . It follows from (11.97), (11.100), and (11.118) holding for all t = 0, . . . , τ − 1 that 4M 2 ≥ d(x0 , Cmin )2 ≥ d(x0 , Cmin )2 − d(xτ , Cmin )2 =

τ −1  (d(xt , Cmin )2 − d(xt+1 , Cmin )2 ) t=0

≥ 2α 2 (L + 1)2 τ and τ ≤ 2M 2 α −2 (L + 1)−2 . This implies that there exists an integer τ ≤ 2M 2 α −2 (L + 1)2

(11.119)

d(xτ , Cmin ) < 4α(L + 1)2 μ−1 ,

(11.120)

such that

if an integer t ∈ [0, τ ] \ {τ }, then (11.117) holds and d(xt+1 , Cmin ) ≤ d(xt , Cmin ).

320

11 Minimization of Sharp Weakly Convex Functions

We show that for all integers t ≥ τ , d(xt , Cmin ) ≤ 4α(L + 1)2 μ−1 + 3α(L + 1). Assume the contrary. Then in view of (11.120) there exists an integer s > τ such that d(xs , Cmin ) > 4α(L + 1)2 μ−1 + 3α(L + 1)

(11.121)

(11.121) does not hold for all t = τ, . . . , s − 1 and in particular, d(xs−1 , Cmin ) ≤ 4α(L + 1)2 μ−1 + 3α(L + 1).

(11.122)

By (11.97), (11.99), and (11.122), xs−1  ≤ M.

(11.123)

d(xs−1 , Cmin ) > 4α(L + 1)2 μ−1 ;

(11.124)

d(xs−1 , Cmin ) ≤ 4α(L + 1)2 μ−1 .

(11.125)

There are two cases:

Assume that (11.124) holds. Together with property (a) and (11.123) this implies that (11.118) holds with t = s − 1 and that d(xs , Cmin ) ≤ d(xs−1 , Cmin ) ≤ 4α(L + 1)2 μ−1 + 3α(L + 1). Assume that (11.125) holds. Property (a), (11.116) with t = s − 1, (11.213) and (11.125) imply that d(xs , Cmin ) ≤ d(xs−1 , Cmin ) + 3α(L + 1) ≤ 4α(L + 1)2 μ−1 + 3α(L + 1). Thus in both cases d(xs , Cmin ) ≤ 4α(L + 1)2 μ−1 + 3α(L + 1). This contradicts (11.121). The contradiction we have reached proves that d(xt , Cmin ) ≤ 4α(L + 1)2 μ−1 + 3α(L + 1) = α(L + 1)((4L + 4)μ−1 + 3) for all integers t ≥ τ . This completes the proof of Theorem 11.7.

 

Chapter 12

A Projected Subgradient Method for Nonsmooth Problems

In this chapter we study the convergence of the projected subgradient method for a class of constrained optimization problems in a Hilbert space. For this class of problems, an objective function is assumed to be convex but a set of admissible points is not necessarily convex. Our goal is to obtain an -approximate solution in the presence of computational errors, where  is a given positive number.

12.1 Preliminaries and Main Results Let (X, ·, ·) be a Hilbert space with an inner product ·, · which induces a complete norm  · . For each x ∈ X and each nonempty set A ⊂ X put d(x, A) = inf{x − y : y ∈ A}. Assume that f : X → R 1 is a convex continuous function which is Lipschitz on all bounded subsets of X. For each point x ∈ X and each positive number  let ∂f (x) = {l ∈ X : f (y) − f (x) ≥ l, y − x for all y ∈ X} be the subdifferential of f at x and let ∂ f (x) = {l ∈ X : f (y) − f (x) ≥ l, y − x −  for all y ∈ X} be the -subdifferential of f at x. Let C be a closed nonempty subset of the space X. Assume that © Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6_12

321

322

12 A Projected Subgradient Method for Nonsmooth Problems

lim f (x) = ∞.

x→∞

(12.1)

It means that for each M0 > 0 there exists M1 > 0 such that if a point x ∈ X satisfies the inequality x ≥ M1 , then f (x) > M0 . Define inf(f, C) = inf{f (z) : z ∈ C}. Since the function f is Lipschitz on all bounded subsets of the space X it follows from (12.1) that inf(f, C) is finite. Set Cmin = {x ∈ C : f (x) = inf(f, C)}. It is well known that if the set C is convex, then the set Cmin is nonempty. Clearly, the set Cmin = ∅ if the space X is finite-dimensional. In this chapter we assume that Cmin = ∅.

(12.2)

It is clear that Cmin is a closed subset of X. We suppose that the following assumption holds. (A1) For every positive number  there exists δ > 0 such that if a point x ∈ C satisfies the inequality f (x) ≤ inf(f, C) + δ, then d(x, Cmin ) ≤ . (It is clear that (A1) holds if the space X is finite-dimensional.) We also suppose PC : X → X satisfies for all x ∈ C and all y ∈ X, x − PC (y) ≤ x − y

(12.3)

PC (x) = x for all x ∈ C.

(12.4)

and that

(Note that in  we assumed that PC (X) = X.) In this chapter we also use the following assumption. (A2) For each M > 0 and each r > 0 there exists δ > 0 such that for each x ∈ BX (0, M) satisfying d(x, C) ≥ r and each z ∈ BX (0, M) ∩ C, we have PC (x) − z ≤ x − z − δ. In Section 3.13, Chapter 3 of  mappings satisfying (A2), (12.3) and (12.4) are called (C)-quasi-contractive and it is shown that in appropriate spaces of mappings satisfying (12.3) and (12.4) most of the mappings are (C)-quasi-contractive. In the chapter we also use the following assumption.

12.1 Preliminaries and Main Results

323

(A3) The mapping PC is uniformly continuous on all bounded subsets of X and for every x ∈ X, PCn (x) converges in the norm topology to a point of C uniformly on every bounded subset of X. Clearly, if (A3) holds, then limn→∞ PCn (x) is a fixed point of PC for every x ∈ X. Set P 0 x = x, x ∈ X. For every number  ∈ (0, ∞) set φ() = sup{δ ∈ (0, 1] : if x ∈ C satisfies f (x) ≤ inf(f, C) + δ, then d(x, Cmin ) ≤ min{1, }}.

(12.5)

In view of (A1), φ() is well defined for every positive number . In this chapter we will prove the following two results. Theorem 12.1 Suppose that at least one of the assumptions (A2) and (A3) holds. Let {αi }∞ i=0 ⊂ (0, 1] satisfy lim αi = 0,

i→∞

∞ 

αi = ∞

(12.6)

i=1

and let M,  > 0. Then there exist a natural number n0 and δ > 0 such that the following assertion holds. Assume that an integer n ≥ n0 , {xk }nk=0 ⊂ X, x0  ≤ M, vk ∈ ∂δ f (xk ) \ {0}, k = 0, 1, . . . , n − 1, n−1 {ηk }n−1 k=0 , {ξk }k=0 ⊂ BX (0, δ),

and that for k = 0, . . . , n − 1, xk+1 = PC (xk − αk vk −1 vk − αk ξk ) − αk ηk . Then the inequality d(xk , Cmin ) ≤  hods for all integers k satisfying n0 ≤ k ≤ n. Theorem 12.2 Suppose that at least one of the assumptions (A2) and (A3) holds. Let M,  > 0. Then there exists β0 ∈ (0, 1) such that for each β1 ∈ (0, β0 ) there exist a natural number n0 and δ > 0 such that the following assertion holds. Assume that an integer n ≥ n0 , {xk }nk=0 ⊂ X, x0  ≤ M,

324

12 A Projected Subgradient Method for Nonsmooth Problems

vk ∈ ∂δ f (xk ) \ {0}, k = 0, 1, . . . , n − 1, {αk }n−1 k=0 ⊂ [β1 , β0 ], n−1 {ηk }n−1 k=0 , {ξk }k=0 ⊂ BX (0, δ)

and that for k = 0, . . . , n − 1, xk+1 = PC (xk − αk vk −1 vk − αk ξk ) − ηk . Then the inequality d(xk , Cmin ) ≤  holds for all integers k satisfying n0 ≤ k ≤ n. In this chapter we use the following definitions and notation. Define X0 = {x ∈ X : f (x) ≤ inf(f, C) + 4}.

(12.7)

In view of (12.1), there exists a number K¯ > 0 such that ¯ X0 ⊂ BX (0, K).

(12.8)

Since the function f is Lipschtiz on all bounded subsets of the space X there exists a number L¯ > 1 such that ¯ 1 − z2  |f (z1 ) − f (z2 )| ≤ Lz for all z1 , z2 ∈ BX (0, K¯ + 4).

(12.9)

12.2 Auxiliary Results Proposition 12.3 Let  ∈ (0, 1]. Then for each x ∈ X satisfying d(x, C) < min{L¯ −1 2−1 φ(/2), /2},

(12.10)

f (x) ≤ inf(f, C) + min{2−1 φ(/2), /2},

(12.11)

the inequality d(x, Cmin ) ≤  holds. Proof In view of the definition of φ, φ(/2) ∈ (0, 1] and the following property holds: (i) if x ∈ C satisfies f (x) < inf(f, C) + φ(/2),

12.2 Auxiliary Results

325

then d(x, Cmin ) ≤ min{1, /2}.

(12.12)

Assume that a point x ∈ X satisfies (12.10) and (12.11). In view of (12.10), there exists a point y ∈ C which satisfies x − y < 2−1 L¯ −1 φ(/2) and x − y < /2.

(12.13)

Relations (12.5), (12.7), (12.8), (12.11), and (12.13) imply that ¯ y ∈ BX (0, K¯ + 1). x ∈ BX (0, K),

(12.14)

By (12.13), (12.14), and the definition of L¯ (see (12.9)), ¯ − y < φ(/2)2−1 . |f (x) − f (y)| ≤ Lx

(12.15)

It follows from the choice of the point y, (12.11), and (12.15) that y∈C and f (y) < f (x) + φ(/2)2−1 ≤ inf(f, C) + φ(/2). Combined with property (i) this implies that d(y, Cmin ) ≤ /2. Together with (12.13) this implies that d(x, Cmin ) ≤ x − y + d(y, Cmin ) ≤ . This completes the proof of Proposition 12.3.

 

Lemma 12.4 Assume that  > 0, x ∈ X, y ∈ X, f (x) > inf(f, C) + , f (y) ≤ inf(f, C) + /4, v ∈ ∂/4 f (x).

(12.16) (12.17)

Then v, y − x ≤ −/2. Proof In view of (12.17), f (u) − f (x) ≥ v, u − x − /4 for all u ∈ X.

(12.18)

326

12 A Projected Subgradient Method for Nonsmooth Problems

By (12.16) and (12.18), −(3/4) ≥ f (y) − f (x) ≥ v, y − x − /4.  

This completes the proof of Lemma 12.4. Lemma 12.5 Let x¯ ∈ Cmin ,

(12.19)

¯ α ∈ (0, 1], let a number δ ∈ (0, 1] satisfy K0 > 0,  ∈ (0, 16L], ¯ −1 , δ(K0 + K¯ + 1) ≤ (8L)

(12.20)

x ≤ K0 ,

(12.21)

f (x) > inf(f, C) + ,

(12.22)

η, ξ ∈ BX (0, δ),

(12.23)

v ∈ ∂/4 f (x) \ {0},

(12.24)

y = PC (x − αv−1 v − αξ ) − η.

(12.25)

let a point x ∈ X satisfy

and let

Then ¯ −1  + 2α 2 + η2 + 2η(K0 + K¯ + 2). ¯ 2 − α(4L) y − x ¯ 2 ≤ x − x Proof In view of (12.7)–(12.9) and (12.19), for every point ¯ 4−1  L¯ −1 ), z ∈ BX (x, we have ¯ − x f (z) ≤ f (x) ¯ + Lz ¯ ≤ f (x) ¯ + /4 = inf(f, C) + /4. Lemma 12.4, (12.22), (12.24), and (12.26) imply that for every point ¯ 4−1  L¯ −1 ), z ∈ BX (x,

(12.26)

12.2 Auxiliary Results

327

we have v, z − x ≤ −/2. Combined with (8.22) the inequality above implies that ¯ −1 ). ¯ (4L) v−1 v, z − x < 0 for all z ∈ B(x,

(12.27)

Set z˜ = x¯ + 4−1 L¯ −1 v−1 v.

(12.28)

z˜ ∈ B(x, ¯ 4−1 L¯ −1 ).

(12.29)

It is easy to see that

Relations (12.27)–(12.29) imply that 0 > v−1 v, z˜ − x = v−1 v, x¯ + 4−1 L¯ −1 v−1 v − x.

(12.30)

By (12.30), v−1 v, x¯ − x < −4−1 L¯ −1 .

(12.31)

y0 = x − αv−1 v − αξ.

(12.32)

Set

It follows from (12.7), (12.8), (12.19)–(12.21), (12.23), (12.31), and (12.32) that ¯ 2 = x − αv−1 v − αξ − x ¯ 2 y0 − x = x − αv−1 v − x ¯ 2 + α 2 ξ 2 −2αξ, x − αv−1 v − x ¯ ¯ 2 ≤ x − αv−1 v − x +α 2 δ 2 + 2αδ(K0 + K¯ + 1) ¯ αv−1 v ≤ x − x ¯ 2 − 2x − x, +α 2 + α 2 δ 2 + 2αδ(K0 + K¯ + 1) < x − x ¯ 2 − 2α(4−1 L¯ −1 )

328

12 A Projected Subgradient Method for Nonsmooth Problems

+α 2 (1 + δ 2 ) + 2αδ(K0 + K¯ + 1) ¯ −1  + 2α 2 . ≤ x − x ¯ 2 − α(4L)

(12.33)

In view of (12.7), (12.8), (12.19), (12.21), and (12.33), ¯ 2+2 ¯ 2 ≤ (K0 + K) y0 − x and ¯ ≤ K0 + K¯ + 2. y0 − x

(12.34)

By (12.3), (12.19), (12.25), (A2), (12.32), (12.33), and (12.34), ¯ 2 y − x ¯ 2 = PC (y0 ) − η − x ≤ PC (y0 ) − x ¯ 2 + η2 + 2ηPC (y0 ) − x ¯ ¯ 2 + η2 + 2ηy0 − x ¯ ≤ y0 − x ¯ −1  ≤ x − x ¯ 2 − α(4L) +2α 2 + η2 + 2η(K0 + K¯ + 2). This completes the proof of Lemma 12.5. Lemma 12.5 implies the following result. ¯ α ∈ (0, 1], a number δ ∈ (0, 1] satisfy Lemma 12.6 Let K0 > 0,  ∈ (0, 16L], ¯ −1 , δ(K0 + K¯ + 1) ≤ (8L) let x ∈ X satisfy x ≤ K0 , f (x) > inf(f, C) + , let η, ξ ∈ BX (0, δ), v ∈ ∂/4 f (x) \ {0}, and let y = PC (x − αv−1 v − αξ ) − η. Then

 

12.3 An Auxiliary Result with Assumption A2

329

¯ −1  d(y, Cmin )2 ≤ d(x, Cmin )2 − α(4L) +2α 2 + η2 + 2η(K0 + K¯ + 2). Lemma 12.6 applied with  = 16L¯ implies the next result. Lemma 12.7 Let K0 > 0, α ∈ (0, 1], a number δ ∈ (0, 1] satisfy δ(K0 + K¯ + 1) ≤ 2, let x ∈ X satisfy ¯ x ≤ K0 , f (x) > inf(f, C) + 16L, let η, ξ ∈ BX (0, δ), v ∈ ∂4 f (x) \ {0}, and let y = PC (x − αv−1 v − αξ ) − η. Then d(y, Cmin )2 ≤ d(x, Cmin )2 − 2α + 2η(K0 + K¯ + 3).

12.3 An Auxiliary Result with Assumption A2 Lemma 12.8 Assume that (A2) holds, K0 ,  > 0 and x¯ ∈ Cmin .

(12.35)

Then there exist a natural number m0 and δ0 ∈ (0, 1) such that for each integer n ≥ m0 and each finite sequence {xi }ni=0 ⊂ X satisfying xi  ≤ K0 , i = 0, . . . , n

(12.36)

and BX (xi+1 , δ0 ) ∩ PC (BX (xi , δ0 )) = ∅, i = 0, . . . , n − 1 the inequality

(12.37)

330

12 A Projected Subgradient Method for Nonsmooth Problems

d(xi , C) ≤  holds for all integers i ∈ [m0 , n]. Proof Set K¯ 1 = 2K0 + K¯ + 5.

(12.38)

Assumption (A2) implies that there exists γ0 ∈ (0, 1) such that the following property holds: (ii) for each x ∈ X satisfying x ≤ K0 + 1 and d(x, C) ≥ /8 and each z ∈ BX (0, K¯ 1 ) ∩ C, we have PC (x) − z ≤ x − z − γ0 . Choose a natural number m0 > 2(K0 + K¯ + 1)γ0−1 + 2

(12.39)

δ0 < min{/8, 1, γ0 /4}.

(12.40)

and a positive number

Assume that an integer n ≥ m0 and that a finite sequence {xi }ni=0 ⊂ X satisfies (12.36) and (12.37) for all integers i = 0, . . . , n − 1. We show that there exists j ∈ {0, . . . , m0 } such that d(xj , C) ≤ /4. We may assume without loss of generality that d(x0 , C) > /4.

(12.41)

y0 ∈ BX (x0 , δ0 )

(12.42)

By (12.37), there exists

such that

12.3 An Auxiliary Result with Assumption A2

x1 − PC (y0 ) ≤ δ0 .

331

(12.43)

By (12.40)–(12.42), d(y0 , C) ≥ d(x0 , C) − δ0 > /4 − δ0 > /8.

(12.44)

In view of (12.36), (12.40), and (12.42), y0  ≤ x0  + δ0 ≤ K0 + 1.

(12.45)

Property (ii), (12.7), (12.8), (12.35), (12.38), (12.42), (12.44), and (12.45) imply that ¯ ≤ y0 − x ¯ − γ0 PC (y0 ) − x ≤ x0 − x ¯ + x0 − y0  − γ0 ≤ x0 − x ¯ + δ0 − γ0 .

(12.46)

By (12.40), (12.43), and (12.46), ¯ ≤ x1 − PC (y0 ) + PC (y0 ) − x ¯ x1 − x ¯ − γ0 + δ0 ≤ δ0 + x0 − x = x0 − x ¯ − γ0 + 2δ0 ≤ x0 − x ¯ − γ0 /2 and ¯ ≤ x0 − x ¯ − γ0 /2. x1 − x

(12.47)

We may assume without loss of generality that d(x1 , C) > /4.

(12.48)

Assume that p is a natural number such that for all i = 0, . . . , p − 1, ¯ ≤ xi − x ¯ − γ0 /2. xi+1 − x

(12.49)

(In view of (12.47), this assumption holds for p = 1.) It follows from (12.7), (12.8), (12.35), (12.36), and (12.49) that ¯ K0 + K¯ ≥ x0 − x ¯ − xp − x ¯ ≥ x0 − x

332

12 A Projected Subgradient Method for Nonsmooth Problems

=

p−1 

(xi − x ¯ − xi+1 − x) ¯ ≥ (γ0 /2)p

i=0

and ¯ −1 . p ≤ 2(K0 + K)γ 0 Together with (12.39) this implies that p < m0 .

(12.50)

In view of (12.50), we may assume without loss of generality that p is the largest natural number such that (12.49) holds for i = 0, . . . , p − 1. Thus ¯ > xp − x ¯ − γ0 /2. xp+1 − x

(12.51)

d(xp , C) > /4.

(12.52)

yp ∈ BX (xp , δ0 )

(12.53)

xp+1 − PC (yp ) ≤ δ0 .

(12.54)

Assume that

By (12.37), there exists

such that

Relations (12.36), (12.40), and (12.53) imply that yp  ≤ K0 + 1.

(12.55)

By (12.40), (12.52), and (12.53), d(yp , C) ≥ d(xp , C) − xp − yp  > /4 − δ0 ≥ /8.

(12.56)

It follows from (12.7), (12.8), (12.35), (12.53), (12.55), and (12.56) that ¯ ≤ yp − x ¯ − γ0 PC (yp ) − x ≤ xp − x ¯ + yp − xp  − γ0 ≤ xp − x ¯ − γ0 + δ0 .

(12.57)

12.3 An Auxiliary Result with Assumption A2

333

By (12.40), (12.54), and (12.57), ¯ ≤ xp+1 − PC (yp ) + PC (yp ) − x ¯ xp+1 − x ¯ − γ0 + 2δ0 ≤ xp − x ≤ xp − x ¯ − γ0 /2. This contradicts (12.51). The contradiction we have reached proves that d(xp , C) ≤ /4. Thus we proved the existence of an integer p ≥ 0 such that p < m0 ,

(12.58)

d(xp , C) ≤ /4.

(12.59)

We show that d(xj , C) ≤  for all integers j ∈ [p, n]. Assume that j ≥ p is an integer, j < n and that d(xj , C) ≤ .

(12.60)

d(xj , C) ≤ /4;

(12.61)

d(xj , C) > /4.

(12.62)

y ∈ BX (xj , δ0 )

(12.63)

xj +1 − PC (y) ≤ δ0 .

(12.64)

There are two cases:

By (12.37), there exists

such that

Assume that (12.61) holds. It follows from (12.3), (12.40), (12.61), (12.63), and (12.64) that d(xj +1 , C) ≤ xj +1 − PC (y) + d(PC (y), C)

334

12 A Projected Subgradient Method for Nonsmooth Problems

≤ δ0 + d(y, C) ≤ δ0 + d(xj , C) + y − xj  ≤ 2δ0 + d(xj , C) ≤ /4 + 2δ0 ≤ .

(12.65)

Assume that (12.62) holds. By (12.7), (12.8), (12.35), (12.36), (12.40), and (12.63), d(y, C) ≤ y − x ¯ ≤ K0 + K¯ + 1.

(12.66)

In view of (12.40) and (12.66), there exists z ∈ C such that z − y ≤ d(y, C) + δ0 ≤ K0 + K¯ + 2.

(12.67)

By (12.36), (12.40), (12.63), and (12.67), z ≤ K0 + K¯ + 2 + y ≤ K0 + K¯ + 2 + xj  + y − xj  ≤ 2K0 + K¯ + 3.

(12.68)

It follows from (12.36), (12.40), and (12.63) that y ≤ xj  + 1 ≤ K0 + 1.

(12.69)

By (12.40), (12.62), and (12.63), d(y, C) ≥ d(xj , C) − xj − y ≥ /4 − δ0 > /8.

(12.70)

Property (ii), the inclusion z ∈ C, (12.38), and (12.68)–(12.70) imply that PC (y) − z ≤ y − z − γ0 .

(12.71)

The inclusion z ∈ C, (12.40), (12.63)–(12.65), (12.67), and (12.71) imply that d(xj +1 , C) ≤ xj +1 − z ≤ PC (y) − z + xj +1 − PC (y) ≤ y − z − γ0 + δ0 ≤ d(y, C) + δ0 − γ0 + δ0 ≤ d(xj , C) + y − xj  + δ0 − γ0 + δ0 ≤ d(xj , C) + 2δ0 − γ0 + δ0 ≤ d(xj , C) − γ0 /4 ≤ .

12.4 An Auxiliary Result with Assumption A3

335

Thus in both cases d(xj +1 , C) ≤ . Therefore d(xi , C) ≤  for all i = j, . . . , n. Lemma 12.8 is proved.

 

12.4 An Auxiliary Result with Assumption A3 Lemma 12.9 Assume that (A3) holds, K0 > 0,  ∈ (0, 1), and x¯ ∈ Cmin .

(12.72)

Then there exist a natural number m0 and δ0 ∈ (0, 1) such that for each integer n ≥ m0 and each finite sequence {xi }ni=0 ⊂ X satisfying xi  ≤ K0 , i = 0, . . . , n

(12.73)

and BX (xi+1 , δ0 ) ∩ PC (BX (xi , δ0 )) = ∅, i = 0, . . . , n − 1

(12.74)

the inequality d(xi , C) ≤  holds for all integers i ∈ [m0 , n]. Proof Assumption (A3) implies that for every x ∈ X there exists lim PCi (x) ∈ C.

(12.75)

Q(x) = lim PCi (x), x ∈ X.

(12.76)

i→∞

Set i→∞

(A3) implies that there exists a natural number m0 such that P m0 (y) − Q(y) ≤ /4 for all x ∈ BX (0, K0 + K¯ + 4).

(12.77)

Set δm0 = /4.

(12.78)

336

12 A Projected Subgradient Method for Nonsmooth Problems

In view of (A3), there exists δm0 −1 ∈ (0, δm0 /4) such that PC (z1 ) − PC (z2 ) ≤ δm0 /4 for all z1 , z2 ∈ BX (0, K0 + 1) satisfying z1 − z2  ≤ 4δm0 −1 . 0 By induction we construct a finite sequence {δi }m i=0 ⊂ (0, ∞) such that for each integer i ∈ [0, m0 − 1] we have

δi < δi+1 /4

(12.79)

and for each z1 , z2 ∈ BX (0, K0 + 1) satisfying z1 − z2  ≤ 4δi

(12.80)

PC (z1 ) − PC (z2 ) ≤ δi+1 /4.

(12.81)

we have

Assume that n ≥ m0 is an integer, {xi }ni=0 ⊂ BX (0, K0 )

(12.82)

and that (12.74) holds for each integer i ∈ {0, . . . , n − 1}. Let k ∈ [m0 , n] be an integer. In order to complete the proof it is sufficient to show that d(xk , C) ≤ . By (12.75)–(12.77) and (12.82), P m0 (xk−m0 ) − Q(xk−m0 ) ≤ /4 and d(P m0 (xk−m0 ), C) ≤ /4. In view of (12.74), there exists

(12.83)

12.4 An Auxiliary Result with Assumption A3

337

yk−m0 ∈ BX (xk−m0 , δ0 )

(12.84)

PC (yk−m0 ) − xk−m0 +1  ≤ δ0 .

(12.85)

such that

By (12.80)–(12.82) and (12.84), PC (yk−m0 ) − PC (xk−m0 ) ≤ δ1 /4.

(12.86)

It follows from (12.79), (12.85), and (12.86) that xk−m0 +1 − PC (xk−m0 ) ≤ δ1 /2.

(12.87)

We show that for all i = 1, . . . , m0 , xi+k−m0 − PCi (xk−m0 ) ≤ δi /2.

(12.88)

(Note that in view of (12.87), inequality (12.88) holds for i = 1.) Assume that i ∈ {1, . . . , m0 } \ {m0 } and that (12.88) holds. By (12.74), there exists yi+k−m0 ∈ BX (xi+k−m0 , δ0 )

(12.89)

PC (yi+k−m0 ) − xi+k−m0 +1  ≤ δ0 .

(12.90)

such that

In view of (12.88) and (12.89), yi+k−m0 − PCi (xk−m0 ) ≤ δi .

(12.91)

It follows from (12.80)–(12.82), (12.88), (12.89), and (12.91) that PC (yi+k−m0 ) − PCi+1 (xk−m0 ) ≤ δi+1 /4.

(12.92)

It follows from (12.79), (12.90), and (12.92) that xi+k−m0 +1 − PCi+1 (xk−m0 ) ≤ δi+1 /2.

(12.93)

Thus our assumption holds for i + 1 too. Therefore we showed by induction that xi+k−m0 − PCi (xk−m0 ) ≤ δi /2 for all i = 1, . . . , m0 and in particular (see (12.78))

338

12 A Projected Subgradient Method for Nonsmooth Problems

xk − PCm0 (xk−m0 ) ≤ δm0 = /4. Together with (12.83) this implies that d(xk , C) ≤ /2.  

Lemma 12.9 is proved.

12.5 Proof of Theorem 12.1 We may assume without loss of generality that  < 1. In view of Proposition 12.3, there exists a number ¯ ∈ (0, min{/8, 1/8})

(12.94)

such that the following property holds: (iii) if x ∈ X, d(x, C) ≤ 2¯ and f (x) ≤ inf(f, C) + 2¯ , then d(x, Cmin ) ≤ . In view of (12.1), we may assume without loss of generality that M > 4K¯ + 8, ¯ ⊂ BX (0, 2−1 M − 1). {x ∈ X : f (x) ≤ inf(f, C) + 16L}

(12.95) (12.96)

Fix ¯ −1 ). ¯ L) ¯1 ∈ (0, (64

(12.97)

Lemmas 12.8 and 12.9 imply that there exist δ1 ∈ (0, 1) and a natural number m0 such that the following property holds: (iv) for each integer n ≥ m0 and each finite sequence {yi }ni=0 ⊂ BX (0, 3M) satisfying BX (yi+1 , δ1 ) ∩ PC (BX (yi , δ1 )) = ∅, i = 0, . . . , n − 1 the inequality d(yi , C) ≤ ¯1

12.5 Proof of Theorem 12.1

339

holds for all integers i ∈ [m0 , n]. Since limi→∞ αi = 0 (see (12.6)) there is an integer p0 > 0 such that for all integers i ≥ p0 , we have αi ≤ δ1 /2.

(12.98)

x¯ ∈ Cmin .

(12.99)

Fix

Since limi→∞ αi = 0 (see (12.6)) there is an integer p 1 > p0

(12.100)

such that for all integers i ≥ p1 , we have ¯ −1 16−1 ¯1 . αi < (32L) Since

∞

i=0 αi

(12.101)

= ∞ (see (12.6)) there exists a natural number n0 > p0 + p1 + 4 + m0

(12.102)

such that n 0 −1

¯ 2 128¯ −1 L¯ + 1. αi > (3M + K)

(12.103)

i=p0 +p1 +m0

Fix a positive number δ such that ¯ −1 ¯1 . δ(3M + K¯ + 3) < 8−1 (64L)

(12.104)

Assume that an integer n ≥ n0 and that {xk }nk=0 ⊂ X, x0  ≤ M,

(12.105)

n−1 {ηk }n−1 k=0 , {ξk }k=0 ⊂ BX (0, δ),

(12.106)

vk ∈ ∂δ f (xk ) \ {0}, k = 0, 1, . . . , n − 1

(12.107)

and that for all integers k = 0, . . . , n − 1, we have xk+1 = PC (xk − αk vk −1 vk − αk ξk ) − αk ηk . In order to prove the theorem it is sufficient to show that

(12.108)

340

12 A Projected Subgradient Method for Nonsmooth Problems

d(xk , Cmin ) ≤  for all integers k satisfying n0 ≤ k ≤ n. First we show that for all integers i = 0, . . . , n, d(xi , Cmin ) ≤ 2M.

(12.109)

In view of (12.105), inequality (12.109) holds for i = 0. Assume that i ∈ {0, . . . , n} \ {n} and that (12.109) is true. There are two cases: ¯ f (xi ) ≤ inf(f, C) + 16L;

(12.110)

¯ f (xi ) > inf(f, C) + 16L.

(12.111)

Assume that (12.110) holds. In view of (12.96) and (12.110), xi  ≤ M/2 − 1.

(12.112)

¯ x ¯ ≤ K.

(12.113)

By (12.7), (12.8), and (12.99),

It follows from (12.112) and (12.113) that ¯ ≤ K¯ + M/2. xi − x

(12.114)

By (12.3), (12.95), (12.99), (12.104), (12.106), (12.108), and (12.114), ¯ xi+1 − x ≤ αi ηi  + x¯ − PC (xi − αi vi −1 vi − αi ξi ) ≤ αi δ + x¯ − (xi − αi vi −1 vi − αi ξi ) ≤ αi δ + x¯ − xi  + αi + αi δ ≤ x¯ − xi  + 3 ≤ K¯ + M/2 + 3 < 2M and d(xi+1 , Cmin ) ≤ 2M.

(12.115)

12.5 Proof of Theorem 12.1

341

Assume that (12.111) holds. It follows from (12.104), (12.106), (12.107), (12.113), Lemma 12.7 applied with K0 = 2M, α = αi , x = xi , ξ = ξi , v = vi , y = xi+1 , η = αi ηi , (12.95), and (12.109) that d(xi+1 , Cmin )2 ≤ d(xi , Cmin )2 − 2αi + αi δ(4M) ≤ d(xi , Cmin )2 ≤ 4M 2 . Thus in both cases d(xi+1 , Cmin ) ≤ 2M. Thus we have shown by induction that for all integers i = 0, . . . , n, d(xi , Cmin ) ≤ 2M. Together with (12.7), (12.8), and (12.95) this implies that xi  ≤ 3M, i = 0, . . . , n.

(12.116)

yi = xi+p0 , i = 0, . . . , n − p0 .

(12.117)

Set

By (12.116)–(12.118), n−p

{yi }i=0 0 ⊂ BX (0, 3M).

(12.118)

n − p0 > m 0 .

(12.119)

In view of (12.102),

It follows from (12.98), (12.104), (12.106), (12.108), and (12.117) that for all i = 0, . . . , n − p0 − 1, yi − (yi − αi+p0 vi+p0 −1 vi+p0 − αi+p0 ξi+p0 ) = xi+p0 − (xi+p0 − αi+p0 vi+p0 −1 vi+p0 − αi+p0 ξi+p0 ) ≤ 2αi+p0 ≤ δ1 ,

342

12 A Projected Subgradient Method for Nonsmooth Problems

yi+1 − PC (yi − αi+p0 vi+p0 −1 vi+p0 − αi+p0 ξi+p0 ) ≤ αi+p0 ≤ δ1 and BX (yi+1 , δ1 ) ∩ PC (BX (yi , δ1 )) = ∅.

(12.120)

It follows from (12.118) to (12.120) that d(yi , C) ≤ ¯1 , i = m0 , . . . , n − p0 . Together with (12.96) and (12.117) this implies that d(xi , C) ≤ ¯1 < ¯ , i = m0 + p0 , . . . , n.

(12.121)

Assume that an integer k ∈ [p0 + p1 + m0 , n − 1],

(12.122)

f (xk ) > inf(f, C) + ¯ /8.

(12.123)

It follows from (12.94), (12.97), (12.99), (12.106), (12.114), (12.116), Lemma 12.5 applied with K0 = 3M,  = ¯ /8, α = αk , x = xk , ξ = ξk , v = vk , y = xk+1 , η = αk ηk , (12.96), (12.101), (12.104), and (12.122) that ¯ −1 ¯ /8 ¯ 2 ≤ xk − x ¯ 2 − αk (4L) xk+1 − x +2αk2 + αk2 ηk 2 + 2ηk αk (3M + K¯ + 2) ¯ −1 ¯ ≤ xk − x ¯ 2 − αk (32L) +2αk2 + αk2 δ 2 + 2δαk (3M + K¯ + 2) ¯ −1 ¯ + 2δαk (3M + K¯ + 3) ≤ xk − x ¯ 2 − αk (64L) ¯ −1 ¯ . ¯ 2 − αk (128L) ≤ xk − x Thus we have shown that the following property holds:

12.5 Proof of Theorem 12.1

343

(v) if an integer k satisfies (12.122) and (12.123), then we have ¯ −1 αk ¯ . ¯ 2 ≤ xk − x ¯ 2 − (128L) xk+1 − x We claim that there exists an integer j ∈ {p0 + p1 + m0 , . . . , n0 } such that f (xj ) ≤ inf(f, C) + ¯ /8. Assume the contrary. Then f (xj ) > inf(f, C) + ¯ /8, i = p0 + p1 + m0 , . . . , n0 .

(12.124)

Property (v) and (12.124) imply that ¯ −1 αi ¯ , ¯ 2 ≤ xi − x ¯ 2 − (128L) xi+1 − x i = p0 + p1 + m0 , . . . , n0 − 1.

(12.125)

Relations (12.7), (12.8), (12.99), (12.116), and (12.125) imply that ¯ 2 ≥ xp0 +p1 +m0 − x ¯ 2 (3M + K) ≥ xp0 +p1 +m0 − x ¯ 2 − xn0 − x ¯ 2 =

n 0 −1

[xi − x ¯ 2 − xi+1 − x ¯ 2]

i=p0 +p1 +m0

¯ −1 ¯ ≥ (128L)

n 0 −1

αi .

(12.126)

i=p0 +p1 +m0

In view of (12.126), n 0 −1

¯ 2 L¯ ¯  −1 . αi ≤ 128(3M + K)

i=p0 +p1 +m0

This contradicts (12.103). The contradiction we have reached proves that there exists an integer j ∈ {p0 + p1 + m0 , . . . , n0 } such that

(12.127)

344

12 A Projected Subgradient Method for Nonsmooth Problems

f (xj ) ≤ inf(f, C) + ¯ /8.

(12.128)

By (12.121), (12.127), and (12.128), we have d(xj , Cmin ) ≤ .

(12.129)

We claim that for all integers i satisfying j ≤ i ≤ n, d(xi , Cmin ) ≤ . Assume the contrary. Then there exists an integer k ∈ [j, n] for which d(xk , Cmin ) > .

(12.130)

By (12.127), (12.129), and (12.130), we have k > j ≥ p0 + p1 + m0 .

(12.131)

We may assume without loss of generality that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i < k. Thus d(xk−1 , Cmin ) ≤ .

(12.132)

f (xk−1 ) ≤ inf(f, C) + ¯ /8;

(12.133)

f (xk−1 ) > inf(f, C) + ¯ /8.

(12.134)

There are two cases:

Assume that (12.133) is valid. It follows from (12.121) and (12.131) that d(xk−1 , C) ≤ ¯1 .

(12.135)

z∈C

(12.136)

xk−1 − z < 2¯1 .

(12.137)

By (12.135), there exists a point

such that

12.5 Proof of Theorem 12.1

345

By (12.3), (12.101), (12.104), (12.106), (12.108), (12.131), (12.136), and (12.137), xk − z ≤ αk−1 δ +z − PC (xk−1 − αk−1 vk−1 −1 vk−1 − αk−1 ξk−1 ) ≤ δ + z − xk−1  + αk−1 + δ ≤ 2¯1 + 2δ + αk−1 < 3¯1 .

(12.138)

In view of (12.137) and (12.138), xk − xk−1  ≤ xk − z + z − xk−1  < 5¯1 .

(12.139)

It follows from (12.7), (12.8), (12.94), (12.132), and (12.139) that xk−1  ≤ K¯ + 2, xk  ≤ xk−1  + 5¯1 ≤ K¯ + 3 and xk−1 , xk  ≤ K¯ + 4. Combined with (12.9) and (12.139) the relation above implies that ¯ k−1 − xk  ≤ 5L¯ ¯1 . |f (xk−1 ) − f (xk )| ≤ Lx Together with (12.97) and (12.133) this implies that ¯ 1 f (xk ) ≤ f (xk−1 ) + 5L¯ ≤ inf(f, C) + /8 ¯ + 5L¯ ¯1 ≤ inf(f, C) + /4. ¯

(12.140)

Property (iii), (12.96), (12.136), (12.138), and (12.140) imply that d(xk , Cmin ) ≤ . This inequality contradicts (12.130). The contradiction we have reached proves (12.134). It follows from (12.96), (12.104), (12.106), (12.116), and (12.134) that Lemma 12.6 holds with K0 = 3M,  = ¯ /8, x = xk−1 , y = xk , ξ = ξk−1 , v = vk−1 , α = αk−1 , η = αk−1 ηk−1 .

346

12 A Projected Subgradient Method for Nonsmooth Problems

Combined with (12.101), (12.104), (12.106), (12.131), and (12.132) this implies that ¯ −1 ¯ /8 d(xk , Cmin )2 ≤ d(xk−1 , Cmin )2 − αk−1 (4L) 2 2 + δ 2 αk−1 + 2αk−1 δ(3M + K¯ + 2) +2αk−1 2 ¯ −1 αk−1 ¯ + 4αk−1 ≤ d(xk−1 , Cmin )2 − (32L) + 2αk−1 δ(3M + K¯ + 2)

¯ −1 αk−1 ¯ + 2δαk−1 (3M + K¯ + 2) ≤ d(xk−1 , Cmin )2 − (64L) ¯ −1 αk−1 ¯ ≤ d(xk−1 , Cmin )2 − (128L) and d(xk , Cmin ) ≤ d(xk−1 , Cmin ) ≤ . This contradicts (12.130). The contradiction we have reached proves that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i ≤ n. This completes the proof of Theorem 12.1.  

12.6 Proof of Theorem 12.2 In view of (12.1), we may assume without loss of generality that  < 1, M > 8K¯ + 8

(12.141)

and that ¯ ⊂ BX (0, 2−1 M − 1). {x ∈ X : f (x) ≤ inf(f, C) + 16L}

(12.142)

Proposition 12.3 implies that there exists a number ¯ ∈ (0, /8) such that the following property holds: (vi) if x ∈ X, d(x, C) ≤ 2¯ and f (x) ≤ inf(f, C) + 2¯ , then d(x, Cmin ) ≤ /4.

(12.143)

12.6 Proof of Theorem 12.2

347

Fix x¯ ∈ Cmin

(12.144)

¯ −1 ). ¯ L) ¯1 ∈ (0, (64

(12.145)

and

Lemmas 2.8 and 2.9 imply that there exist δ1 ∈ (0, 1) and a natural number m0 such that the following property holds: (vii) for each integer n ≥ m0 and each finite sequence {yi }ni=0 ⊂ BX (0, 3M) satisfying BX (yi+1 , δ1 ) ∩ PC (BX (yi , δ1 )) = ∅, i = 0, . . . , n − 1 the inequality d(yi , C) ≤ ¯1 holds for all integers i ∈ [m0 , n]. Choose a positive number β0 such that ¯ −1 ¯1 . β0 ≤ δ1 /2, 2β0 < (34L)

(12.146)

β1 ∈ (0, β0 ).

(12.147)

¯  −1 β −1 . n0 > m0 + 322 M 2 L¯ 1

(12.148)

Let

Fix a natural number n0 such that

Fix positive number δ such that δ(3M + K¯ + 3) ≤ (128)−1 ¯1 β1 .

(12.149)

Assume that an integer n ≥ n0 , {xk }nk=0 ⊂ X, x0  ≤ M, vk ∈ ∂δ f (xk ) \ {0}, k = 0, 1, . . . , n − 1 {αk }n−1 k=0 ⊂ [β1 , β0 ],

(12.150) (12.151) (12.152)

348

12 A Projected Subgradient Method for Nonsmooth Problems n−1 {ηk }n−1 k=0 , {ξk }k=0 ⊂ BX (0, δ)

(12.153)

and that for all integers k = 0, . . . , n − 1, xk+1 = PC (xk − αk vk −1 vk − αk ξk ) − ηk .

(12.154)

In order to complete the proof it is sufficient to show that d(xk , Cmin ) ≤  for all integers k satisfying n0 ≤ k ≤ n. First we show that for all integers i = 0, . . . , n, d(xi , Cmin ) ≤ 2M.

(12.155)

In view of (12.7), (12.8), (12.141), and (12.150), inequality (12.155) holds for i = 0. Assume that i ∈ {0, . . . , n} \ {n} and that (12.155) is true. There are two cases: ¯ f (xi ) ≤ inf(f, C) + 16L;

(12.156)

¯ f (xi ) > inf(f, C) + 16L.

(12.157)

Assume that (12.156) holds. In view of (12.142) and (12.156), xi  ≤ M/2 − 1.

(12.158)

¯ x ¯ ≤ K.

(12.159)

By (12.7), (12.8), and (12.144),

It follows from (12.158) and (12.159) that ¯ ≤ K¯ + M/2. xi − x

(12.160)

By (12.3), (12.141), (12.144), (12.146), (12.149), (12.152)–(12.154), and (12.160), ¯ d(xi+1 , Cmin ) ≤ xi+1 − x ≤ ηi  + x¯ − PC (xi − αi vi −1 vi − αi ξi )

12.6 Proof of Theorem 12.2

349

≤ δ + x¯ − xi  + αi + αi δ ≤ 2β0 + δ + K¯ + M/2 ≤ K¯ + M/2 + 3 ≤ M.

(12.161)

Assume that (12.157) holds. In view of (12.141), (12.144), and (12.155), xi  ≤ 3M. It follows from (12.141), (12.146), (12.147), (12.149), (12.151), (12.152), (12.157), and Lemma 12.7 applied with K0 = 3M, α = αi , x = xi , ξ = ξi , v = vi , y = xi+1 , η = ηi , that d(xi+1 , Cmin )2 ≤ d(xi , Cmin )2 − 2αi + 2ηi (3M + K¯ + 3) ≤ d(xi , Cmin )2 − 2β1 + 8δM ≤ d(xi , Cmin )2 − β1 and in view of (12.155), d(xi+1 , Cmin ) ≤ d(xi , Cmin ) ≤ 2M. Thus in both cases d(xi+1 , Cmin ) ≤ 2M. Thus we have shown by induction that for all integers i = 0, . . . , n, d(xi , Cmin ) ≤ 2M.

(12.162)

By (12.7), (12.8), (12.141), and (12.162), xi  ≤ 3M, i = 0, . . . , n. It follows from (12.149), (12.152), and (12.153), for all i = 0, . . . , n − 1, xi − (xi − αi vi −1 vi − αi ξi ) ≤ αi + αi δ ≤ 2αi ≤ 2β0 ≤ δ1 , xi+1 − PC (xi − αi vi −1 vi − αi ξi )

(12.163)

350

12 A Projected Subgradient Method for Nonsmooth Problems

≤ ηi  ≤ δ ≤ β0 < δ1 and BX (xi+1 , δ1 ) ∩ PC (BX (xi , δ1 )) = ∅.

(12.164)

Property (vii), (12.148), (12.155), (12.163), and (12.164) imply that ¯ i = m0 , . . . , n. d(xi , C) ≤ ¯1 < ,

(12.165)

Assume that an integer k ∈ [m0 , n − 1], f (xk ) > inf(f, C) + ¯ /8.

(12.166)

It follows from (12.43)–(12.146), (12.149), (12.151), (12.154), (12.163), and Lemma 12.5 applied with K0 = 3M,  = ¯ /4, α = αk , x = xk , ξ = ξk , v = vk , y = xk+1 , η = ηk that ¯ 2 xk+1 − x ¯ −1 ¯ /4 + 2αk2 ≤ xk − x ¯ 2 − αk (4L) +ηk 2 + 2ηk (3M + K¯ + 2) ¯ −1 ¯ ¯ 2 − αk (16L) ≤ xk − x +2αk2 + δ 2 + 2δ(3M + K¯ + 2) ¯ −1 ¯ + 2δ(3M + K¯ + 3) ≤ xk − x ¯ 2 − αk (32L) ¯ −1 ¯ + 2δ(3M + K¯ + 3) ¯ 2 − β1 (32L) ≤ xk − x ¯ −1 ¯ . ¯ 2 − β1 (64L) ≤ xk − x Thus we have shown that the following property holds: (viii) if an integer k ∈ [m0 , n − 1] satisfies f (xk ) > inf(f, C) + ¯ /4,

12.6 Proof of Theorem 12.2

351

then ¯ −1 ¯ . ¯ 2 ≤ xk − x ¯ 2 − β1 (64L) xk+1 − x We claim that there exists an integer j ∈ {m0 , . . . , n0 } such that f (xj ) ≤ inf(f, C) + ¯ /4. Assume the contrary. Then f (xj ) > inf(f, C) + ¯ /4, j = m0 , . . . , n0 .

(12.167)

Property (viii) and (12.167) imply that for all k ∈ {m0 , . . . , n0 − 1}, ¯ −1 ¯ . ¯ 2 ≤ xk − x ¯ 2 − β1 (64L) xk+1 − x

(12.168)

Relations (12.7), (12.8), (12.141), (12.144), (12.163), and (12.168) imply that ¯ 2 (4M)2 ≥ xm0 − x ≥ xm0 − x ¯ 2 − xn0 − x ¯ 2 =

n 0 −1

[xi − x ¯ 2 − xi+1 − x ¯ 2]

i=m0

¯ −1 ¯ ≥ (n0 − m0 )β1 (64L) and ¯  −1 β −1 . n0 − m0 ≤ 322 M 2 L¯ 1 This contradicts (12.148). The contradiction we have reached proves that there exists an integer j ∈ {m0 , . . . , n0 }

(12.169)

f (xj ) ≤ inf(f, C) + ¯ /4.

(12.170)

such that

Property (vi), (12.145), and (12.170) imply that d(xj , Cmin ) ≤ /4.

(12.171)

352

12 A Projected Subgradient Method for Nonsmooth Problems

We claim that for all integers i satisfying j ≤ i ≤ n, d(xi , Cmin ) ≤ . Assume the contrary. Then there exists an integer k ∈ [j, n] for which d(xk , Cmin ) > .

(12.172)

By (12.171) and (12.172), we have k > j. We may assume without loss of generality that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i < k. Thus d(xk−1 , Cmin ) ≤ .

(12.173)

f (xk−1 ) ≤ inf(f, C) + ¯ /8;

(12.174)

f (xk−1 ) > inf(f, C) + ¯ /8.

(12.175)

There are two cases:

Assume that (12.174) is valid. It follows from (12.165) and (12.169) that d(xk−1 , C) ≤ ¯1 .

(12.176)

z∈C

(12.177)

xk−1 − z < 2¯1 .

(12.178)

By (12.176), there exists a point

such that

By (12.3), (12.146), (12.149), and (12.152)–(12.154), xk − z ≤ δ +z − PC (xk−1 − αk−1 vk−1 −1 vk−1 − αk−1 ξk−1 ) ≤ δ + z − xk−1  + 2αk−1

12.6 Proof of Theorem 12.2

353

≤ 2¯1 + δ + 2β0 ≤ 3¯1 .

(12.179)

In view of (12.179), xk − xk−1  ≤ xk − z + z − xk−1  < 5¯1 .

(12.180)

It follows from (12.7), (12.8), and (12.173) that xk−1  ≤ K¯ + 1.

(12.181)

By (12.144), (12.180), and (12.181), xk  ≤ xk−1  + 5¯1 ≤ K¯ + 4.

(12.182)

Relations (12.9) and (12.80)–(12.82) imply that ¯ k−1 − xk  ≤ 5L¯ ¯1 . |f (xk−1 ) − f (xk )| ≤ Lx Together with (12.145) and (12.174) this implies that ¯ 1 f (xk ) ≤ f (xk−1 ) + 5L¯ ≤ inf(f, C) + /8 ¯ + 5L¯ ¯1 ≤ inf(f, C) + /4. ¯

(12.183)

By (12.145), (12.176), and (12.180), d(xk , C) ≤ 6¯1 < ¯ .

(12.184)

Property (vi), (12.183), and (12.184) imply that d(xk , Cmin ) ≤ . This inequality contradicts (12.172). The contradiction we have reached proves (12.175). It follows from (12.145), (12.149), (12.151)–(12.154), (12.163), (12.165), (12.169), and (12.175) that Lemma 12.6 holds with K0 = 3M,  = ¯ /8, x = xk−1 , y = xk , ξ = ξk−1 , v = vk−1 , α = αk−1 , η = ηk−1 . Combined with (12.145), (12.146), and (12.152) this implies that ¯ −1 ¯ /8 d(xk , Cmin )2 ≤ d(xk−1 , Cmin )2 − αk−1 (4L)

354

12 A Projected Subgradient Method for Nonsmooth Problems 2 +2αk−1 + η2 + 2ηk−1 (3M + K¯ + 2)

¯ −1 ¯ − 2β0 ) + δ 2 + 2δ(3M + K¯ + 2) ≤ d(xk−1 , Cmin )2 − αk−1 ((32L) ¯ −1 ¯ + 2δ(3M + K¯ + 3) ≤ d(xk−1 , Cmin )2 − αk−1 (64L) ¯ −1 ¯ + 2δ(3M + K¯ + 3) ≤ d(xk−1 , Cmin )2 − β1 (64L) ¯ −1 ¯ . ≤ d(xk−1 , Cmin )2 − β1 (128L) In view of (12.173), d(xk , Cmin ) ≤ d(xk−1 , Cmin ) ≤ . This contradicts (12.172). The contradiction we have reached proves that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i ≤ n. This completes the proof of Theorem 12.2.  

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Index

A Absolutely continuous function, 178 Algorithm, 1 Approximate solution, 1 B Banach space, 173 Bochner integrable function, 173–174, 195 Borelian function, 181 Boundary, 287 C Closure, 287 Compact set, 178 Concave function, 66, 182 Continuous function, 71, 80, 315 Continuous subgradient algorithm, 173, 181–184 Continuous subgradient projection algorithm, 193–194 Convex-concave function, 1, 25, 83 Convex conjugate the function, 278 Convex function, 1, 5, 297–298, 315 Convex hull, 178 Convex minimization problem, 1, 84 Convex set, 2, 177

Firmly nonexpansive operator, 243 Fréchet derivative, 18, 127 Fréchet differentiable function, 18, 127 G Game, 53–58 Gradient-type method, 151 H Hilbert space, 1, 2, 66 I Infimal convolution, 243 Inner product, 2, 66 Interior, 287 Iteration, 1 L Lebesgue measurable function, 173, 175 Lebesgue measure, 174 Locally Lipschitzian function, 193, 201 Lower semicontinuous function, 174, 191, 245

D Derivative, 178

M Minimizer, 76, 84 Mirror descent method, 10–17, 83–125 Moreau envelope, 244

F Feasible point, 1 Fenchel inequality, 278

N Nonexpansive mapping, 244 Norm, 2

© Springer Nature Switzerland AG 2020 A. J. Zaslavski, Convex Optimization with Computational Errors, Springer Optimization and Its Applications 155, https://doi.org/10.1007/978-3-030-37822-6

359

360 O Objective function, 1

P Predicted decrease approximation (PDA), 277, 280 Projected gradient algorithm, 1, 18, 127 Projected subgradient method, 25–81 Proximal method, 246, 260

Q Quasiconvex function, 287–293

Index S Saddle point, 1, 25, 83 Sharp weakly convex function, 295–320 Strongly measurable function, 173 Subdifferential, 2, 296 Subgradient, 1–10 Subgradient projection algorithm, 2, 67

U Upper semicontinuous function, 182, 191

Z Zero-sum game, 53–58, 67, 181–184