The Projected Subgradient Algorithm in Convex Optimization [1st ed.] 9783030602994, 9783030603007

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SPRINGER BRIEFS IN OPTIMIZATION

Alexander J. Zaslavski

The Projected Subgradient Algorithm in Convex Optimization 123

SpringerBriefs in Optimization Series Editors Sergiy Butenko, Department of Industrial and Systems Engineering, Texas A&M University, College Station, TX, USA Mirjam Dür, Department of Mathematics, University of Trier, Trier, Germany Panos M. Pardalos, ISE Department, University of Florida, Gainesville, FL, USA János D. Pintér, Lehigh University, Bethlehem, PA, USA Stephen M. Robinson, University of Wisconsin-Madison, Madison, WI, USA Tamás Terlaky, Lehigh University, Bethlehem, PA, USA My T. Thai , CISE Department, University of Florida, Gainesville, FL, USA

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Alexander J. Zaslavski

The Projected Subgradient Algorithm in Convex Optimization

123

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

ISSN 2190-8354 ISSN 2191-575X (electronic) SpringerBriefs in Optimization ISBN 978-3-030-60299-4 ISBN 978-3-030-60300-7 (eBook) https://doi.org/10.1007/978-3-030-60300-7 Mathematics Subject Classification: 49M37, 65K05, 90C25, 90C26, 90C30 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Subgradient Projection Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1

2

Nonsmooth Convex Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Approximate Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Convergence to the Solution Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Superiorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Auxiliary Results for Theorems 2.1–2.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Proof of Theorem 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Proof of Theorem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Proof of Theorem 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Proof of Theorem 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.10 Proof of Theorem 2.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11 Proof of Theorem 2.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12 An Auxiliary Result for Theorems 2.11–2.15 . . . . . . . . . . . . . . . . . . . . . . . . 2.13 Proof of Theorem 2.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.14 Proof of Theorem 2.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.15 Proof of Theorem 2.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.16 Proof of Theorem 2.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17 Proof of Theorem 2.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.18 Proof of Theorem 2.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.19 Proof of Theorem 2.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.20 Proof of Theorem 2.18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5 5 7 18 22 24 35 36 39 41 43 45 48 49 54 59 65 72 78 80 82

3

Extensions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Optimization Problems on Bounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 An Auxiliary Result for Theorem 3.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 An Auxiliary Result for Theorem 3.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Proof of Theorem 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Proof of Theorem 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Optimization on Unbounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

85 85 89 91 94 95 96 v

vi

Contents

3.7 3.8 3.9

Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Proof of *Theorem 3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 Proof of Theorem 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

4

Zero-Sum Games with Two Players. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 4.1 Preliminaries and an Auxiliary Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 4.2 Zero-Sum Games on Bounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

5

Quasiconvex Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 The Main Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Optimization on Bounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Optimization on Unbounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

129 129 133 134 137

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

Chapter 1

Introduction

In this book we study the behavior of subgradient algorithms for constrained 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 section we discuss several algorithms which are studied in the book.

1.1 Subgradient Projection Method In this book we study the subgradient projection algorithm for minimization of convex and nonsmooth functions and for computing the saddle points of convexconcave 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, [1–3, 7, 10–12, 16, 17, 26, 28–30, 33–35, 37, 43–47, 50–56, 58–60, 64– 68, 71–75, 77, 78]. 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. In our recent research [77], 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 computational errors for the two steps of our algorithm, we find out what an approximate solution can be obtained © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 A. J. Zaslavski, The Projected Subgradient Algorithm in Convex Optimization, SpringerBriefs in Optimization, https://doi.org/10.1007/978-3-030-60300-7_1

1

2

1 Introduction

and how many iterates one needs for this. In this book we generalize all these results for an extension of the projected subgradient method, when instead of the projection on the feasible set it is used a quasi-nonexpansive retraction on this set. We study the subgradient algorithm for constrained minimization problems in Hilbert spaces equipped with an inner product denoted by ·, · which induces a complete norm  ·  and 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}. 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 [48, 49, 61]. Let C be a nonempty closed convex subset of X and let f : X → R 1 be a convex function. Suppose that there exist L > 0, M0 > 0 such that C ⊂ BX (0, M0 ), |f (x) − f (y)| ≤ Lx − y for all x, y ∈ BX (0, M0 + 2). It is not difficult to see that for each x ∈ BX (0, M0 + 1), ∅ = ∂f (x) ⊂ BX (0, L). It is well-known that for every nonempty closed convex, set D ⊂ X and every x ∈ X, there is a unique point PD (x) ∈ D satisfying

1.1 Subgradient Projection Method

3

x − PD (x) = inf{x − y : y ∈ D}. We consider the minimization problem f (z) → min, z ∈ C. Suppose that {αk }∞ k=0 ⊂ (0, ∞). Let us describe our algorithm. Subgradient Projection Algorithm Initialization: select an arbitrary x0 ∈ BX (0, M0 + 1). Iterative step: given a current iteration vector xt ∈ U , calculate ξt ∈ ∂f (xt ) and the next iteration vector xt+1 = PC (xt − αt ξt ). In [75] we study this algorithm under the presence of computational errors. Namely, in [75], 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 ∈ BX (0, M0 + 1). Iterative step: given a current iteration vector xt ∈ BX (0, M0 + 1), 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 [77] 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 Chapter 2 of [77]. Suppose that {αk }∞ k=0 ⊂ (0, ∞) and δf , δC ∈ (0, 1]. Initialization: select an arbitrary x0 ∈ BX (0, M0 + 1). Iterative step: given a current iteration vector xt ∈ BX (0, M0 + 1), calculate ξt ∈ ∂f (xt ) + BX (0, δf ) and the next iteration vector xt+1 ∈ U such that xt+1 − PC (xt − αt 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 our analysis of the behavior of the algorithm in [75, 77], properties of the projection operator PC play an important role. In the present book, we obtain generalizations of the results obtained in [75, 77] for the subgradient methods in the case when the set C is not necessarily convex and the projection operator PC is replaced by a mapping P : X → C which satisfies P x = x for all x ∈ C,

(1.1)

P x − z ≤ x − z for all z ∈ C and all x ∈ X.

(1.2)

4

1 Introduction

In other words, P is a quasi-nonexpansive retraction on C. Note that there are many mappings P : X → C satisfying (1.1) and (1.2). Indeed, in [57] we consider a space of mappings P : X → X satisfying (1.1) and (1.2), which is equipped with a natural complete metric, and show that for a generic (typical) mapping from the space, its powers converge to a mapping which also satisfies (1.1) and (1.2) and such that its image is C. Note that the generalizations considered in this book have, besides their obvious mathematical interest, also a significant practical meaning. Usually, the projection operator PC : X → C can be calculated when C is a simple set like a linear subspace, a half-space, or a simplex. In practice, C is an intersection of simple sets Ci , i = 1, . . . , q, where q is a large natural number. The calculation of PC is not possible in principle. Instead, it is possible to calculate the product PCq · · · PC1 and its powers (PCq · · · PC1 )m , m = 1, 2, . . . . It is well-known [76] that under certain regularity conditions on Ci , i = 1, . . . , q the powers (PCq · · · PC1 )m converge as m → ∞ to a mapping P : X → C which satisfies (1.1) and (1.2). Thus in practice we cannot calculate the projection operator PC but only a mapping P : X → C satisfying (1.1) and (1.2) [4, 5, 8, 9, 21, 24, 27, 31, 32, 36, 52, 62, 69] or, more exactly, its approximations. This shows that the results of this book are indeed important from the point of view of practice. In Chapter 2, we study the subgradient projection algorithm presented above for convex minimization problems with objective functions defined on the whole Hilbert space. In Chapter 3, we generalize some results of Chapter 2 for the case of problems with objective functions defined on subsets of the Hilbert space. In Chapter 4, we study the subgradient projection algorithm for zero-sum games with two players. In Chapter 5, we study the projected subgradient method for quasiconvex optimization problems.

Chapter 2

Nonsmooth Convex Optimization

In this chapter, we study an extension of the projected subgradient method 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 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. In our recent research [77], 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 computational errors for the two steps of our algorithm, we find out what an approximate solution can be obtained and how many iterates one needs for this. In this chapter, we generalize all these results for an extension of the projected subgradient method, when instead of the projection on the feasible set it is used a quasi-nonexpansive retraction on this set.

2.1 Preliminaries 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, set d(x, A) = inf{x − y : y ∈ A}. For each x ∈ X and each r > 0, set BX (x, r) = {y ∈ X : x − y ≤ r}.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 A. J. Zaslavski, The Projected Subgradient Algorithm in Convex Optimization, SpringerBriefs in Optimization, https://doi.org/10.1007/978-3-030-60300-7_2

5

6

2 Nonsmooth Convex Optimization

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}

(2.1)

be the subdifferential of f at x [48, 49], and let ∂ f (x) = {l ∈ X : f (y) − f (x) ≥ l, y − x −  for all y ∈ X}

(2.2)

be the -subdifferential of f at x. Let C be a closed nonempty subset of the space X. Assume that lim f (x) = ∞.

x→∞

(2.3)

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 . In this chapter, we consider the optimization problem f (x) → min, x ∈ C. Define inf(f, C) = inf{f (z) : z ∈ C}.

(2.4)

Since the function f is Lipschitz on all bounded subsets of the space X, it follows from (2.3) that inf(f, C) is finite. Set Cmin = {x ∈ C : f (x) = inf(f, C)}.

(2.5)

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. We assume that Cmin = ∅.

(2.6)

It is clear that Cmin is a closed subset of X. Define X0 = {x ∈ X : f (x) ≤ inf(f, C) + 4}.

(2.7)

In view of (2.3), there exist a number K¯ > 1 such that ¯ X0 ⊂ BX (0, K).

(2.8)

2.2 Approximate Solutions

7

Since the function f is Lipschitz on all bounded subsets of the space X, there exist a number L¯ > 1 such that ¯ 1 − z2  for all z1 , z2 ∈ BX (0, K¯ + 4). |f (z1 ) − f (z2 )| ≤ Lz

(2.9)

Denote by M the set of all mappings P : X → X such that for all x ∈ C and all y ∈ X, x − P y ≤ x − y,

(2.10)

P x = x for all x ∈ C.

(2.11)

For every P ∈ M, set P 0 x = x, x ∈ X.

2.2 Approximate Solutions In the following results, our goal is to obtain an approximate solution x which is close to the set C such that f (x) is closed to inf(f, C). In the first theorem, the set C is bounded, the computational errors δf , δC are given, and the step-size α depends on δf , δC . It is proved in Section 2.6. ¯ δf , δC ∈ (0, 1], Theorem 2.1 Assume that K1 ≥ K¯ + 1, L1 ≥ L, C ⊂ BX (0, K1 ),

(2.12)

|f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, K1 + 2),

(2.13)

δf (K¯ + K1 + 2 + 5L1 + 5L¯ 1 ) ≤ 1, δC (K¯ + K1 + 2 + 5L1 + 5L¯ 1 ) ≤ 1,

(2.14)

¯ 40(L1 + L)(δ ¯ C (K¯ +K1 +2+5L1 +5L)) ¯ 1/2 }  = max{4δf (K¯ +K1 +2+5L1 +5L), (2.15) and that ¯ 2 (L1 + L) ¯ 2  −2  + 2. n = 800(1 + K1 + K)

(2.16)

Let {Pi }n−1 i=0 ⊂ M satisfy Pi (X) = C, i = 0, . . . , n − 1,

(2.17)

¯ −2 , α = 10−2 (L1 + L)

(2.18)

{xi }ni=0 ⊂ X, {ξi }n−1 i=1 ⊂ X,

8

2 Nonsmooth Convex Optimization

x0  ≤ K1 , x1 − P0 x0  ≤ δC

(2.19)

and that for i = 1, . . . , n − 1, BX (ξi , δf ) ∩ ∂/4 f (xi ) = ∅,

(2.20)

xi+1 − Pi (xi − αξi ) ≤ δC .

(2.21)

Then there exists j ∈ {1, . . . , n} such that f (xj ) ≤ inf(f, C) + . In the second theorem, the set C is not necessarily bounded, the computational errors δf , δC are given, and the step-size α depends on δf , δC . It is proved in Section 2.7. ¯ δf , δC ∈ (0, 1], Theorem 2.2 Assume that K1 ≥ K¯ + 1, L1 ≥ L, |f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, 3K1 + 2), ¯ ≤ 1, δf (K¯ + 3K1 + 2 + 5L1 + 5L) ¯ ≤ (10(L¯ + L1 ))−2 , δC (K¯ + 3K1 + 2 + 5L1 + 5L)

(2.22) (2.23) (2.24)

¯  = max{4δf (K¯ + 3K1 + 2 + 5L1 + 5L), ¯ C (K¯ + 3K1 + 2 + 5L1 + 5L)) ¯ 1/2 } 40(L1 + L)(δ

(2.25)

and that ¯ 2 (L1 + L) ¯ 2  −2  + 2. n = 800(1 + K1 + K)

(2.26)

Let {Pi }n−1 i=0 ⊂ M satisfy Pi (X) = C, i = 0, . . . , n − 1,

(2.27)

¯ −2 , α = 10−2 (L1 + L)

(2.28)

x0  ≤ K1 ,

(2.29)

x1 − P0 x0  ≤ δC

(2.30)

{xi }ni=0 ⊂ X, {ξi }n−1 i=1 ⊂ X,

2.2 Approximate Solutions

9

and that for all i = 1, . . . , n − 1, BX (ξi , δf ) ∩ ∂/4 f (xi ) = ∅,

(2.31)

xi+1 − Pi (xi − αξi ) ≤ δC .

(2.32)

Then there exists j ∈ {1, . . . , n} such that f (xj ) ≤ inf(f, C) + . In the third theorem, the set C is bounded, the computational errors δf , δC and  are given, and the step-size αi , i = 1, . . . , n − 1 are given too. It is proved in Section 2.8. ¯ δf , δC ∈ (0, 1], Theorem 2.3 Assume that K1 ≥ K¯ + 1, L1 ≥ L, C ⊂ BX (0, K1 ), |f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, K1 + 2),

(2.33) (2.34)

0 <  ≤ 16(L¯ + L1 + 1)

(2.35)

δf (K¯ + K1 + 2 + 5L1 + 5L¯ 1 ) ≤ /8.

(2.36)

and that

Let n ≥ 2 be an integer, {Pi }n−1 i=0 ⊂ M satisfy Pi (X) = C, i = 0, . . . , n − 1,

(2.37)

αi ∈ (0, 1], i = 1, . . . , n − 1, {xi }ni=0 ⊂ X, {ξi }n−1 i=0 ⊂ X, x0  ≤ K1 ,

(2.38)

x1 − P0 x0  ≤ δC

(2.39)

and that for all i = 1, . . . , n − 1,

Then

BX (ξi , δf ) ∩ ∂/4 f (xi ) = ∅,

(2.40)

xi+1 − Pi (xi − αi ξi ) ≤ δC .

(2.41)

10

2 Nonsmooth Convex Optimization

min{f (xi ) : i = 1, . . . , n} − inf(f, C), f

  n

αj

−1  n

j =1

≤

 n

−1  n

αj

j =1

≤

 n

 αi xi − inf(f, C)

i=1

αi (f (xi ) − inf(f, C))

i=1

−1 αi

¯ 2 (2K1 + 1)2 + 3/4 + 25(L1 + L)

 n

i=1

αi2

  n

i=1

−1 αi

i=1

−1  n +2 αi δC (K¯ + K1 + 5L1 + 5L¯ + 2)n. i=1

Let n ≥ 2 be an integer and A > 0 be given. We  are interested in an optimal choice of the step-sizes αi , i = 1, . . . , n satisfying ni=1 αi = A which minimizes the right-hand side of the final equation in the statement of Theorem 2.3. In order to meet this goal, we need to minimize the function φ(α1 , . . . , αn ) =

n 

αi2

i=1

n on the set {(α1 , . . . , αn ) ∈ R n : αi ≥ 0, i = 1, . . . , n, i=1 αi = A}. By Lemma 2.3 of [75], the minimizer of φ is αi = n−1 A, i = 1, . . . , n. Theorem 2.3 implies the following result. ¯ δf , δC ∈ (0, 1], Theorem 2.4 Assume that K1 ≥ K¯ + 1, L1 ≥ L, C ⊂ BX (0, K1 ), |f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, K1 + 2), 0 <  ≤ 16(L¯ + L1 + 1) and that ¯ ≤ /8. δf (K¯ + K1 + 2 + 5L1 + 5L) Let n ≥ 2 be an integer, {Pi }n−1 i=0 ⊂ M satisfy Pi (X) = C, i = 0, . . . , n − 1, α ∈ (0, 1], {xi }ni=0 ⊂ X, {ξi }n−1 i=0 ⊂ X, x0  ≤ K1 , x1 − P0 x0  ≤ δC and that for all i = 1, . . . , n − 1,

2.2 Approximate Solutions

11

BX (ξi , δf ) ∩ ∂/4 f (xi ) = ∅, xi+1 − Pi (xi − αξi ) ≤ δC . Then   n  min{f (xi ) : i = 1, . . . , n} − inf(f, C), f n−1 xi − inf(f, C) i=1

≤ n−1

n 

f (xi ) − inf(f, C)

i=1

¯ 2α ≤ (nα)−1 (2K1 + 1)2 + 3/4 + 25(L1 + L) +2α −1 δC (K¯ + K1 + 5L1 + 5L¯ + 2). Now we can make the best choice of the step-size α in Theorem 2.4. Since n can be arbitrary large in view of Theorem 2.4, we need to minimize the function ¯ 2 α + 2α −1 δC (K¯ + K1 + 5L1 + 5L¯ + 2), α > 0 25(L1 + L) which has a minimizer ¯ −1 (2(K¯ + K1 + 5L1 + 5L¯ + 2))1/2 δ 1/2 . α = 5−1 (L1 + L) C With this choice of α, the right-hand side of the last equation in the statement of Theorem 2.4 is −1/2 ¯ K¯ + K1 + 5L1 + 5L¯ + 2))−1/2 δC n−1 (2K1 + 1)2 5(L1 + L)(2(

¯ +3/4 + 10(L1 + L)(2( K¯ + K1 + 5L1 + 5L¯ + 2))1/2 δC . 1/2

Now we should make the best choice of n. It is clear that n should be at the same order as δC−1 . In this case, the right-hand side of the last equation in Theorem 2.4 1/2 does not exceed c1 δC +3/4. Clearly,  depends on δf . In particular, we can choose  = 8δf (K¯ + K1 + 5L1 + 5L¯ + 2). In the next theorem, the set C is not necessarily bounded, and the computational errors δf , δC and the step-size α are given. It is proved in Section 2.9. ¯ δf , δC ∈ (0, 1],  ∈ (0, 4], Theorem 2.5 Assume that K1 ≥ K¯ + 2, L1 ≥ L, |f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, 3K1 + 1),

(2.42)

12

2 Nonsmooth Convex Optimization

¯ −2 ] α ∈ (0, 25−1 (L1 + L)

(2.43)

¯ ≤ , 8δf (3K¯ + K1 + 2 + 5L1 + 5L)

(2.44)

¯ ≤ α. δC (3K¯ + K1 + 2 + 5L1 + 5L)

(2.45)

and that

Let n ≥ 2 be an integer, {Pi }n−1 i=0 ⊂ M satisfy Pi (X) = C, i = 0, . . . , n − 1,

(2.46)

{xi }ni=0 ⊂ X, {ξi }n−1 i=0 ⊂ X, x0  ≤ K1 ,

(2.47)

x1 − P0 x0  ≤ δC

(2.48)

and that for all i = 1, . . . , n − 1, BX (ξi , δf ) ∩ ∂/4 f (xi ) = ∅,

(2.49)

xi+1 − Pi (xi − αξi ) ≤ δC .

(2.50)

Then xi  ≤ 2K¯ + K1 + 1, i = 1, . . . , n and   n  −1 min{f (xi ) : t = 1, . . . , n} − inf(f, C), f n xi − inf(f, C) i=1

≤ n−1

n 

f (xi ) − inf(f, C)

i=1

¯ 2 + /2 + 15(L1 + L) ¯ 2α ≤ (2nα)−1 (K1 + 1 + K) +α −1 δC (3K¯ + K1 + 5L1 + 5L¯ + 2). Now we can make the best choice of the step-size α in Theorem 2.5. Since n can be arbitrary large in view of Theorem 2.5, we need to minimize the function ¯ 2 α + α −1 δC (3K¯ + K1 + 5L1 + 5L¯ + 2), α > 0 15(L1 + L)

2.2 Approximate Solutions

13

which has a minimizer ¯ −1 (15−1 (3K¯ + K1 + 5L1 + 5L¯ + 2))1/2 δ 1/2 . α = (L1 + L) C Since α should satisfy (2.45), we obtain an additional condition on δC : ¯ −2 . (3K¯ + K1 + 5L1 + 5L¯ + 2)δC ≤ 15 · 25−2 (L1 + L) Together with the relations above Theorem 2.5 implies the following result. ¯ δf , δC ∈ (0, 1],  ∈ (0, 4], Theorem 2.6 Assume that K1 ≥ K¯ + 2, L1 ≥ L, |f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, 3K1 + 1), ¯ δf ≤ 8−1 (3K¯ + K1 + 2 + 5L1 + 5L), ¯ −2 (3K¯ + K1 + 2 + 5L1 + 5L) ¯ −1 δC ≤ 25−2 15(L1 + L) and that ¯ −1 (15−1 (3K¯ + K1 + 5L1 + 5L¯ + 2))1/2 δ 1/2 . α = (L1 + L) C Let n ≥ 2 be an integer, {Pi }n−1 i=0 ⊂ M satisfy Pi (X) = C, i = 0, . . . , n − 1, {xi }ni=0 ⊂ X, {ξi }n−1 i=0 ⊂ X, x0  ≤ K1 , x1 − P0 x0  ≤ δC and that for all i = 1, . . . , n − 1, BX (ξi , δf ) ∩ ∂/4 f (xi ) = ∅, xi+1 − Pi (xi − αξi ) ≤ δC . Then xi  ≤ 2K¯ + K1 + 1, i = 1, . . . , n and

14

2 Nonsmooth Convex Optimization

 min{f (xi ) : t = 1, . . . , n} − inf(f, C), f

−1

n

n 

 xi

− inf(f, C)

i=1

≤ n−1

n 

f (xi ) − inf(f, C)

i=1 1/2 ¯ 2 (L1 + L)15 ¯ (δC (3K¯ + K1 + 5L1 + 5L¯ + 2))−1/2 ≤ /2 + 2−1 n−1 (K1 + 1 + K) −1 ¯ +30(L1 + L)(15 (3K¯ + K1 + 5L1 + 5L¯ + 2)δC )1/2 .

Now we should make the best choice of n in Theorem 2.6. It is clear that n should be at the same order as δC−1 . Theorem 2.6 implies the following result. ¯ δf , δC ∈ (0, 1], Theorem 2.7 Assume that K1 ≥ K¯ + 2, L1 ≥ L, |f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, 3K1 + 1), ¯ −1 , δf ≤ 2−1 (3K¯ + K1 + 2 + 5L1 + 5L) ¯ −2 (3K¯ + K1 + 2 + 5L1 + 5L) ¯ −1 , δC ≤ 25−2 15(L1 + L) ¯  = 8δf (3K¯ + K1 + 2 + 5L1 + 5L) and that ¯ −1 (15−1 (3K¯ + K1 + 5L1 + 5L¯ + 2))1/2 δ 1/2 . α = (L1 + L) C Let n ≥ 2 be an integer, {Pi }n−1 i=0 ⊂ M satisfy Pi (X) = C, i = 0, . . . , n − 1, {xi }ni=0 ⊂ X, {ξi }n−1 i=0 ⊂ X, x0  ≤ K1 , x1 − P0 x0  ≤ δC and that for all i = 1, . . . , n − 1, BX (ξi , δf ) ∩ ∂/4 f (xi ) = ∅, xi+1 − Pi (xi − αξi ) ≤ δC . Then

2.2 Approximate Solutions

15

xi  ≤ 2K¯ + K1 + 1, i = 1, . . . , n and   n  −1 min{f (xi ) : i = 1, . . . , n} − inf(f, C), f n xi − inf(f, C) i=1

≤ n−1

n 

f (xi ) − inf(f, C)

i=1

¯ ≤ 4δf (3K¯ + K1 + 2 + 5L1 + 5L) 1/2 ¯ 2 (L1 + L)15 ¯ +2−1 n−1 (K1 + 1 + K) (δC (3K¯ + K1 + 5L1 + 5L¯ + 2))−1/2 −1 ¯ +30(L1 + L)(15 (3K¯ + K1 + 5L1 + 5L¯ + 2)δC )1/2 .

Now we should make the best choice of n in Theorem 2.7. It is clear that n should be at the same order as δC−1 . In this case, the right-hand side of the last equation in 1/2 Theorem 2.7 does not exceed c1 δf + c2 δC where c1 , c2 > 0 are constants. In the previous theorems, we deal with the projected subgradient method: at every iterative step i of the algorithm, ξi is an approximation of a subgradient v. In the following theorems, we study the projected normalized subgradient method: at every iterative step i of the algorithm, ξi is an approximation of v−1 v, where v is a subgradient. In the next result, the set C is bounded, and the computational errors δf , δC and  are given. It is proved in Section 2.10. Theorem 2.8 Assume that K1 ≥ K¯ + 1, C ⊂ BX (0, K1 ),

(2.51)

¯ 0 <  ≤ 16L,

(2.52)

¯ f (K¯ + K1 + 1) ≤ , 8Lδ

(2.53)

¯ 2 δC (K¯ + K1 + 3) ≤  2 (32L)

(2.54)

¯ 2  −2  + 3. ¯ 2 (1 + K1 + K) n = 2(16L)

(2.55)

δf , δC ∈ (0, 1),

and that

16

2 Nonsmooth Convex Optimization

Let {Pi }n−1 i=0 ⊂ M satisfy Pi (X) = C, i = 0, . . . , n − 1,

(2.56)

{xi }ni=0 ⊂ X, {ξi }n−1 i=1 ⊂ X, ¯ −1 ¯ −1 {αi }n−1 i=1 ⊂ [(32L) , (16L) ],

(2.57)

x0  ≤ K1 ,

(2.58)

x1 − P0 x0  ≤ δC

(2.59)

and that for all i = 1, . . . , n − 1, BX (ξi , δf ) ∩ {v−1 v : v ∈ ∂/4 f (xi ) \ {0}} = ∅, xi+1 − Pi (xi − αi ξi ) ≤ δC . Then there exists j ∈ {1, . . . , n} such that f (xj ) ≤ inf(f, C) +  and if i ∈ {1, . . . , n} \ {j }, then f (xi ) > inf(f, C) +  and ¯ −1 }. ∂/4 f (xi ) ⊂ {v ∈ X : v ≥ (3/4)(K1 + K) Theorem 2.8 implies the following result. Theorem 2.9 Assume that K1 ≥ K¯ + 1, C ⊂ BX (0, K1 ), δf , δC ∈ (0, 1), δf (K¯ + K1 + 1) ≤ 2, δC (K¯ + K1 + 3) ≤ 4−1 , ¯ f (K¯ + K1 + 1), 32L(δ ¯ C (K¯ + K1 + 3))1/2 }  = max{8Lδ and that

(2.60) (2.61)

2.2 Approximate Solutions

17

¯ 2 (K1 + K) ¯ 2  −2  + 2. n = 2(16L) Let {Pi }n−1 i=0 ⊂ M satisfy Pi (X) = C, i = 0, . . . , n − 1, {xi }ni=0 ⊂ X, {ξi }n−1 i=1 ⊂ X, ¯ −1 ¯ −1 {αi }n−1 i=1 ⊂ [(32L) , (16L) ], x0  ≤ K1 , x1 − P0 x0  ≤ δC and that for i = 1, . . . , n − 1, BX (ξi , δf ) ∩ {v−1 v : v ∈ ∂/4 f (xi ) \ {0}} = ∅, xi+1 − Pi (xi − αi ξi ) ≤ δC . Then there exists j ∈ {1, . . . , n} such that f (xj ) ≤ inf(f, C) +  and if i ∈ {1, . . . , n} \ {j }, then f (xi ) > inf(f, C) +  and ¯ −1 }. ∂/4 f (xi ) ⊂ {v ∈ X : v ≥ (3/4)(K1 + K) In the next theorem, the set C is not necessarily bounded. It is proved in Section 2.11. Theorem 2.10 Assume that K1 ≥ K¯ + 2, δf , δC ∈ (0, 1], ¯ f (3K¯ + K1 + 2) ≤ 1, 4Lδ

(2.62)

¯ 2 δC (3K¯ + K1 + 4) ≤ 1, (8L)

(2.63)

¯ f (3K¯ + K1 + 2), 32L(δ ¯ C (3K¯ + K1 + 4))1/2 }  = max{8Lδ and that

(2.64)

18

2 Nonsmooth Convex Optimization

¯ 2 (1 + K1 + K) ¯ 2  −2  + 2. n = 2(16L)

(2.65)

Let {Pi }n−1 i=0 ⊂ M satisfy Pi (X) = C, i = 0, . . . , n − 1,

(2.66)

{xi }ni=0 ⊂ X, {ξi }n−1 i=1 ⊂ X, ¯ −1 ¯ −1 {αi }n−1 i=1 ⊂ [(32L) , (16L) ] ⊂ (0, 1],

(2.67)

x0  ≤ K1 ,

(2.68)

x1 − P0 x0  ≤ δC

(2.69)

and that for i = 1, . . . , n − 1, BX (ξi , δf ) ∩ {v−1 v : v ∈ ∂/4 f (xi ) \ {0}} = ∅, xi+1 − Pi (xi − αi ξi ) ≤ δC .

(2.70) (2.71)

Then there exists j ∈ {1, . . . , n} such that f (xj ) ≤ inf(f, C) +  and if i ∈ {1, . . . , n} \ {j }, then f (xi ) > inf(f, C) +  and ∂/4 f (xi ) ⊂ {v ∈ X : v ≥ (3/4)(K1 + 3K¯ + 1)−1 }. All the results of this section are new.

2.3 Convergence to the Solution Set We use the notation and definitions introduced in Section 2.1 and suppose that all the assumptions made there hold. We continue to study the minimization problem f (x) → min, x ∈ C. In Section 2.2, we obtain an approximate solution x which is close to the set C such that f (x) is closed to inf(f, C). In this section, we obtain an approximate solution x which is close to the set Cmin .

2.3 Convergence to the Solution Set

19

We also 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.) For every number  ∈ (0, ∞), let φ() = sup{δ ∈ (0, 1] : if x ∈ C satisfies f (x) ≤ inf(f, C) + δ, then d(x, Cmin ) ≤ min{1, }}.

(2.72)

In view of (A1), φ() is well-defined for every positive number . In the following result, the step-sizes converge to zero, and their sum is infinity. It is proved in Section 2.13. Theorem 2.11 Let {αi }∞ i=0 ⊂ (0, 1] satisfy lim αi = 0,

i→∞

∞ 

αi = ∞

(2.73)

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 , {Pi }n−1 i=0 ⊂ M, Pi (X) = C, i = 0, . . . , n − 1,

(2.74)

{xi }ni=0 ⊂ X, x0  ≤ M,

(2.75)

vi ∈ ∂δ f (xi ) \ {0}, i = 0, 1, . . . , n − 1,

(2.76)

n−1 {ηi }n−1 i=0 , {ξi }i=0 ⊂ BX (0, δ),

and that for i = 0, . . . , n − 1 , xi+1 = Pi (xi − αi vi −1 vi − αi ξi ) − αi ηi . Then the inequality d(xi , Cmin ) ≤  holds for all integers i satisfying n0 ≤ i ≤ n. In the following theorem, the step-sizes are bounded from below by a sufficiently small positive constant. It is proved in Section 2.14. Theorem 2.12 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 , {Pi }n−1 i=0 ⊂ M,

20

2 Nonsmooth Convex Optimization

Pi (X) = C, i = 0, . . . , n − 1, {xi }ni=0 ⊂ X, x0  ≤ M,

(2.77)

vi ∈ ∂δ f (xi ) \ {0}, i = 0, 1, . . . , n − 1,

(2.78)

{αi }n−1 i=0 ⊂ [β1 , β0 ],

(2.79)

n−1 {ηi }n−1 i=0 , {ξi }i=0 ⊂ BX (0, δ)

(2.80)

and that for i = 0, . . . , n − 1, xi+1 = Pi (xi − αi vi −1 vi − αi ξi ) − ηi .

(2.81)

Then the inequality d(xi , Cmin ) ≤  holds for all integers i satisfying n0 ≤ i ≤ n. In the previous two theorems, we deal with the projected normalized subgradient method: at every iterative step i of the algorithm, ξi is an approximation of v−1 v, where v is a subgradient. In the following three theorems, we study the projected subgradient method: at every iterative step i of the algorithm, ξi is an approximation of a subgradient v. In the next theorem, the set C is bounded, and the step-sizes converge to zero, and their sum is infinity. It is proved in Section 2.15. Theorem 2.13 Let {αi }∞ i=0 ⊂ (0, 1] satisfy lim αi = 0,

i→∞

∞ 

αi = ∞,

(2.82)

i=1

C ⊂ BX (0, M)

(2.83)

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 , {Pi }n−1 i=0 ⊂ M, Pi (X) = C, i = 0, . . . , n − 1, {xi }n=0 ⊂ X, x0  ≤ M,

(2.84)

{ξi }n−1 i=0 ⊂ X, for all i = 0, . . . , n − 1, BX (ξi , δ) ∩ ∂δ f (xi ) = ∅,

(2.85)

xi+1 − Pi (xi − αi ξi ) ≤ δi .

(2.86)

Then the inequality d(xi , Cmin ) ≤  holds for all integers i satisfying n0 ≤ i ≤ n.

2.3 Convergence to the Solution Set

21

In the next theorem, the set C is not necessarily bounded, but the step-sizes converge to zero, and their sum is infinity. It is proved in Section 2.16. Theorem 2.14 Let ¯ M > 2K¯ + 1,  ∈ (0, 1), L0 > L, |f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, 2M + 4).

(2.87)

and let {αi }∞ i=0 ⊂ (0, 1] satisfy lim αi = 0,

i→∞

∞ 

αi = ∞,

i=1

αt ≤ 25−2 (L¯ + L0 + 2)−2 , t = 0, 1, . . . .

(2.88)

Then there exist a natural number n0 and δ > 0 such that the following assertion holds. Assume that an integer n ≥ n0 , {Pi }n−1 i=0 ⊂ M, Pi (X) = C, i = 0, . . . , n − 1, {xi }n=0 ⊂ X, x0  ≤ M, {ξi }n−1 i=0 ⊂ X, for all i = 0, . . . , n − 1, BX (ξi , δ) ∩ ∂δ f (xi ) = ∅, xi+1 − Pi (xi − αi ξi ) ≤ αi δ.

(2.89)

Then the inequality d(xi , Cmin ) ≤  holds for all integers i satisfying n0 ≤ i ≤ n. In the next theorem, the set C is not necessarily bounded, while the step-sizes are bounded from below by a positive constant. It is proved in Section 2.17. Theorem 2.15 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 , {Pi }n−1 i=0 ⊂ M, Pi (X) = C, i = 0, . . . , n − 1, {xi }ni=0 ⊂ X, x0  ≤ M,

(2.90)

{αi }n−1 i=0 ⊂ [β1 , β0 ],

(2.91)

22

2 Nonsmooth Convex Optimization

{ξi }n−1 i=0 ⊂ X, for all i = 0, . . . , n − 1, BX (ξi , δ) ∩ ∂δ f (xi ) = ∅,

(2.92)

xi+1 − Pi (xi − αi ξi ) ≤ δ.

(2.93)

Then the inequality d(xi , Cmin ) ≤  holds for all integers i satisfying n0 ≤ i ≤ n. Theorems 2.13–2.15 are new. In the case with Pi = P0 , i = 0, . . . , n − 1, Theorems 2.11 and 2.12 were obtained in [71].

2.4 Superiorization In this section, we present three results on the projected subgradient in the case when step-sizes are summable. Such results are of interest in the study of superiorization and perturbation resilience of algorithms. See [9, 13, 25, 29] and the references mentioned therein. In the first two theorems at the iterative step i of the algorithm ξi is an approximation of a subgradient v, while in the third theorem ξi is an approximation of v−1 v, where v is a subgradient. In the first theorem, the set C is bounded. It is proved in Section 2.18. ¯ {αi }∞ ⊂ (0, 1] satisfy Theorem 2.16 Let K1 ≥ K¯ + 1, L1 ≥ L, i=0 ∞ 

αi < ∞,

i=1

C ⊂ BX (0, K1 )

(2.94)

and let |f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, K1 + 2).

(2.95)

Assume that {Pi }∞ i=0 ⊂ M, Pi (X) = C, i = 0, . . . ,

(2.96)

x0  ≤ K1 ,

(2.97)

∞ {xi }∞ =0 ⊂ X, {ξi }i=0 ⊂ X,

for all integers i ≥ 0,

2.4 Superiorization

23

ξi ∈ ∂f (xi ),

(2.98)

xi+1 = Pi (xi − αi ξi ).

(2.99)

Then there exists x∗ = limi→∞ xi , and at least one of the following properties holds: x∗ ∈ Cmin ; there exists an integer n0 ≥ 1 and 0 > 0 such that for each integer t ≥ n0 and each z ∈ Cmin , xt+1 − z2 ≤ xt − z2 − αt . In the following two theorems, the set C is not necessarily bounded. The first of them is proved in Section 2.19. ¯ {αi }∞ ⊂ (0, 1], Theorem 2.17 Let K1 ≥ K¯ + 1, L1 ≥ L, i=0 |f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, 3K1 + 2).

(2.100)

Assume that {Pi }∞ i=0 ⊂ M, Pi (X) = C, i = 0, 1, . . . ,

(2.101)

∞ {xi }∞ i=0 ⊂ X, {ξi }i=0 ⊂ X,

x0  ≤ K1 , for all integers i ≥ 0, ¯ −2 ], αi ∈ (0, 100(L1 + L)

(2.102)

ξi ∈ ∂f (xi ),

(2.103)

xi+1 = Pi (xi − αi ξi ).

(2.104)

Then there exists x∗ = limi→∞ xi , and at least one of the following properties holds: x∗ ∈ Cmin ; there exists an integer n0 ≥ 1 and 0 > 0 such that for each integer t ≥ n0 and each z ∈ Cmin ,

24

2 Nonsmooth Convex Optimization

xt+1 − z2 ≤ xt − z2 − αt 0 . The next result is proved in Section 2.20. Theorem 2.18 Let K1 ≥ K¯ + 1, {αi }∞ i=0 ⊂ (0, 1] satisfy ∞ 

αi < ∞,

i=1

Assume that {Pi }∞ i=0 ⊂ M, Pi (X) = C, i = 0, 1, . . . ,

(2.105)

∞ {xi }∞ i=0 ⊂ X, {ξi }i=0 ⊂ X \ {0},

x0  ≤ K1 ,

(2.106)

ξi ∈ ∂f (xi ),

(2.107)

xi+1 = Pi (xi − αi ξi −1 ξi ).

(2.108)

for all integers i ≥ 0,

Then there exists x∗ = limi→∞ xi , and at least one of the following properties holds: x∗ ∈ Cmin ; there exists an integer n0 ≥ 1 and 0 > 0 such that for each integer t ≥ n0 and each z ∈ Cmin , xt+1 − z2 ≤ xt − z2 − αt .

2.5 Auxiliary Results for Theorems 2.1–2.10 Let PC ∈ M be an arbitrary element of the space M. Lemma 2.19 Let K0 , L0 , r > 0,

(2.109)

2.5 Auxiliary Results for Theorems 2.1–2.10

|f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, K0 + 1),

25

(2.110)

x ∈ BX (0, K0 ),

(2.111)

v ∈ ∂fr (x).

(2.112)

f (u) − f (x) ≥ v, u − x − r.

(2.113)

Then v ≤ L0 + r. Proof By (2.112), for all u ∈ X,

In view of (2.113), for all ξ ∈ BX (0, 1), r + f (x + ξ ) − f (x) ≥ v, ξ . Together with (2.110) and (2.111), this implies that v, ξ  ≤ r + L0 ξ  ≤ r + L0 . This implies that v ≤ L0 + 2. Lemma 2.19 is proved. Lemma 2.20 Assume that  > 0, x ∈ X, y ∈ X, f (x) > inf(f, C) + ,

(2.114)

f (y) ≤ inf(f, C) + /4,

(2.115)

v ∈ ∂/4 f (x).

(2.116)

Then v, y − x ≤ −/2. Proof In view of (2.116), for all u ∈ X, f (u) − f (x) ≥ v, u − x − /4. By (2.115), −(3/4) ≥ f (y) − f (x) ≥ v, y − x − /4. The inequality above implies that v, y − x ≤ −/2. This completes the proof of Lemma 2.20.

(2.117)

26

2 Nonsmooth Convex Optimization

Lemma 2.21 Let x¯ ∈ Cmin ,

(2.118)

K0 > 0, L0 > 0, |f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, K0 + 1), ¯  ∈ (0, 16(L0 + L)],

(2.119) (2.120)

α ∈ (0, 1], δf , δC ∈ (0, 1] satisfy δf (K0 + K¯ + 5L0 + 5L¯ + 1) ≤ /4,

(2.121)

let a point x ∈ X satisfy x ≤ K0 , f (x) > inf(f, C) + ,

(2.122)

v ∈ ∂/4 f (x),

(2.123)

ξ ∈ BX (v, δf )

(2.124)

y ∈ BX (PC (x − αξ ), δC ).

(2.125)

and let

Then ¯ 2 − 2−1 α + δC2 y − x ¯ 2 ≤ x − x ¯ + 25(L0 + L) ¯ 2α2 +2δC (K0 + K¯ + 5L0 + 5L)

Proof Lemma 2.19, (2.119)–(2.121), and (2.123) imply that ¯ v ≤ 5L0 + 4L. Lemma 2.20, (2.118), (2.122), and (2.123) imply that v, x¯ − x ≤ −/2.

(2.126)

y0 = x − αξ.

(2.127)

Set

2.5 Auxiliary Results for Theorems 2.1–2.10

27

It follows from (2.7), (2.8), (2.118), (2.120), (2.121), (2.124), and (2.126) that ¯ 2 = x − αξ − x ¯ 2 y0 − x = x − αv + (αv − αξ ) − x ¯ 2 = x − αv − x ¯ 2 + α 2 v − ξ 2 + 2αv − ξ, x − αv − x ¯ ≤ x − αv − x ¯ 2 + α 2 δf2 + 2αδf x − αv − x ¯ ¯ ≤ x − αv − x ¯ 2 + α 2 δf2 + 2αδf (K0 + K¯ + α(5L0 + 4L)) ¯ 2 ≤ x − x ¯ 2 − 2αx − x, ¯ v + α 2 (5L0 + 4L) ¯ +α 2 δf2 + 2αδf (K0 + K¯ + α(5L0 + 4L)) ¯ 2 ≤ x − x ¯ 2 − 2α(/2) + α 2 (5L0 + 4L) ¯ +α 2 δf2 + 2αδf (K0 + K¯ + α(5L0 + 5L)) ¯ 2 ≤ x − x ¯ 2 − α + 25α 2 (L0 + L) ¯ + 1) +2αδf (K0 + K¯ + 5α(L0 + L) ¯ 2 + α/2 ≤ x − x ¯ 2 − α + 25α 2 (L0 + L) ¯ 2. ≤ x − x ¯ 2 − α/2 + 25α 2 (L0 + L)

(2.128)

In view of (2.128), ¯ 2. ¯ 2 ≤ x − x ¯ 2 + 25(L0 + L) y0 − x

(2.129)

Relations (2.5), (2.7), (2.8), (2.118), (2.121), and (2.129) imply that ¯ ¯ ≤ K0 + K¯ + 5(L0 + L). y0 − x

(2.130)

By (2.109), (2.110), (2.118), (2.125), (2.127), (2.128), and (2.130), ¯ 2 y − x ¯ 2 = y − PC (x − αξ ) + PC (x − αξ ) − x ≤ y − PC (x − αξ )2 + 2y − PC (x − αξ )PC (x − αξ ) − x ¯ +PC (x − αξ ) − x ¯ 2 ¯ + y0 − x ≤ δC2 + 2δC (K0 + K¯ + 5L0 + 5L) ¯ 2 ¯ + x − x ¯ 2. ≤ δC2 + 2δC (K0 + K¯ + 5L0 + 5L) ¯ 2 − α/2 + 25α 2 (L0 + L) Lemma 2.21 is proved. Lemma 2.21 implies the following result. Lemma 2.22 Let K0 > 0, L0 > 0,

28

2 Nonsmooth Convex Optimization

|f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, K0 + 1), ¯  ∈ (0, 16(L0 + L)], α ∈ (0, 1], δf , δC ∈ (0, 1] satisfy δf (K0 + K¯ + 5L0 + 5L¯ + 1) ≤ /4, let a point x ∈ X satisfy x ≤ K0 , f (x) > inf(f, C) + , v ∈ ∂/4 f (x), ξ ∈ BX (v, δf ), and let y ∈ BX (PC (x − αξ ), δC ). Then d(y, Cmin )2 ≤ d(x, Cmin )2 − 2−1 α + δC2 ¯ + 25(L0 + L) ¯ 2α2. +2δC (K0 + K¯ + 5L0 + 5L)

Applying Lemma 2.22 with  = 4, we obtain the following result. Lemma 2.23 Let K0 > 0, L0 > 0, |f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, K0 + 1), α ∈ (0, 1], δf , δC ∈ (0, 1] satisfy δf (K0 + K¯ + 5L0 + 5L¯ + 1) ≤ 1, let a point x ∈ X satisfy x ≤ K0 , f (x) > inf(f, C) + 4, v ∈ ∂1 f (x), ξ ∈ BX (v, δf ),

2.5 Auxiliary Results for Theorems 2.1–2.10

29

and let y ∈ BX (PC (x − αξ ), δC ). Then ¯ 2α2. d(y, Cmin )2 ≤ d(x, Cmin )2 − 2α + 2δC (K0 + K¯ + 5L0 + 5L¯ + 1) + 25(L0 + L) Recall that PC ∈ M

(2.131)

be an arbitrary element of the space M. Lemma 2.24 Let x¯ ∈ Cmin ,

(2.132)

¯  ∈ (0, 16L],

(2.133)

¯ −1 , δf (K0 + K¯ + 1) ≤ (8L)

(2.134)

x ≤ K0 ,

(2.135)

f (x) > inf(f, C) + ,

(2.136)

v ∈ ∂/4 f (x).

(2.137)

¯ −1  v ≥ 3(4K0 + 4K)

(2.138)

ξ ∈ BX (v−1 v, δf ),

(2.139)

y ∈ BX (PC (x − αξ ), δC )

(2.140)

K0 > 0,

α ∈ (0, 1], δf , δC ∈ (0, 1] satisfy

let a point x ∈ X satisfy

Then

and for each

and each

30

2 Nonsmooth Convex Optimization

the following inequality holds: ¯ −1 α + 2α 2 + δC2 + 2δC (K0 + K¯ + 2). ¯ 2 − (4L) y − x ¯ 2 ≤ x − x Proof In view of (2.5), (2.7)–(2.9), (2.132), and (2.133) for every point ¯ 4−1  L¯ −1 ), z ∈ BX (x,

(2.141)

we have ¯ − x f (z) ≤ f (x) ¯ + Lz ¯ ≤ f (x) ¯ + /4 = inf(f, C) + /4.

(2.142)

By (2.132), (2.136), and (2.137), − > f (x) ¯ − f (x) ≥ v, x¯ − x > −/4, , and combined with (2.7), (2.8), (2.132), and (2.135), this implies that v, x¯ − x ≤ −(3/4), ¯ (3/4) ≤ v, x − x ¯ ≤ vx − x ¯ ≤ v(K0 + K). Therefore (2.138) is true. Let ξ, y ∈ X satisfy (2.139) and (2.140). Lemma 2.20, (2.136), (2.137), (2.141), and (2.142) imply that for every point ¯ 4−1  L¯ −1 ), z ∈ BX (x, we have v, z − x ≤ −/2. Combined with (2.138), the inequality above implies that ¯ −1 ). ¯ (4L) v−1 v, z − x < 0 for all z ∈ BX (x,

(2.143)

Set z˜ = x¯ + 4−1 L¯ −1 v−1 v.

(2.144)

¯ 4−1 L¯ −1 ). z˜ ∈ BX (x,

(2.145)

It is easy to see that

2.5 Auxiliary Results for Theorems 2.1–2.10

31

Relations (2.143)–(2.145) imply that 0 > v−1 v, z˜ − x = v−1 v, x¯ + 4−1 L¯ −1 v−1 v − x.

(2.146)

By (2.146), v−1 v, x¯ − x < −4−1 L¯ −1 .

(2.147)

y0 = x − αξ.

(2.148)

Set

It follows from (2.8), (2.131), (2.132), (2.134), (2.135), (2.139), and (2.148) that ¯ 2 = x − αξ − x ¯ 2 y0 − x = x − αv−1 v + α(v−1 v − ξ ) − x ¯ 2 x − αv−1 v − x ¯ 2 + α 2 v−1 v − ξ 2 +2αv−1 v − ξ, x − αv−1 v − x ¯ ≤ x − αv−1 v − x ¯ 2 +α 2 δf2 + 2αδf (K0 + K¯ + 1) ≤ x − x ¯ 2 − 2x − x, ¯ αv−1 v +α 2 (1 + δf )2 + 2αδf (K0 + K¯ + 1) < x − x ¯ 2 − 2α(4−1 L¯ −1 ) +α 2 (1 + δf2 ) + 2αδf (K0 + K¯ + 1) ¯ −1  + 2α 2 . ≤ x − x ¯ 2 − α(4L)

(2.149)

In view of (2.7), (2.8), (2.15), (2.132), and (2.149), ¯ 2+2 ¯ 2 ≤ (K0 + K) y0 − x and ¯ ≤ K0 + K¯ + 2. y0 − x By (2.10), (2.131), (2.132), (2.140), and (2.148)–(2.150), ¯ 2 y − x ¯ 2 = y − PC (x − αξ ) + PC (x − αξ ) − x

(2.150)

32

2 Nonsmooth Convex Optimization

y − PC (x − αξ )2 + PC (x − αξ ) − x ¯ 2 + 2y − PC (x − αξ )PC (x − αξ ) − x ¯ ¯ 2 + δC2 + 2δC y0 − x ¯ ≤ y0 − x ¯ −1  ≤ x − x ¯ 2 − α(4L) +2α 2 + δC2 + 2δC (K0 + K¯ + 2).

This completes the proof of Lemma 2.24. ¯ we obtain the following result. Applying Lemma 2.24 with  = 16L, Lemma 2.25 Let x¯ ∈ Cmin , K0 > 0, α ∈ (0, 1], δf , δC ∈ (0, 1] satisfy δf (K0 + K¯ + 1) ≤ 2 ¯ and v ∈ ∂ ¯ f (x). ; let a point x ∈ X satisfy x ≤ K0 , f (x) > inf(f, C) + 16L, 4L Then ¯ −1 ¯ 0 + K) v ≥ 12L(K and for each ξ ∈ BX (v−1 v, δf ), and each y ∈ BX (PC (x − αξ ), δC ) the following inequality holds: ¯ 2 − 2α + 2δC (K0 + K¯ + 3). y − x ¯ 2 ≤ x − x Lemma 2.24 implies the following result. ¯ α ∈ (0, 1], δf , δC ∈ (0, 1] satisfy Lemma 2.26 Let K0 > 0,  ∈ (0, 16L], ¯ −1 , δf (K0 + K¯ + 1) ≤ (8L) let a point x ∈ X satisfy x ≤ K0 , f (x) > inf(f, C) +  and let v ∈ ∂/4 f (x). Then ¯ −1  v ≥ 3(4K0 + 4K)

2.5 Auxiliary Results for Theorems 2.1–2.10

33

and for each ξ ∈ BX (v−1 v, δf ), and each y ∈ BX (PC (x − αξ ), δC ) the following inequality holds: ¯ −1 α + 2α 2 + δC2 + 2δC (K0 + K¯ + 2). d(y, Cmin )2 ≤ d(x, Cmin )2 − (4L) Lemma 2.27 Let x¯ ∈ Cmin ,

(2.151)

¯ L0 > 0, K0 ≥ K, |f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, K0 + 1), ¯  ∈ (0, 16(L0 + L)],

(2.152) (2.153)

α ∈ (0, 1], δf , δC ∈ (0, 1] satisfy δf (K0 + K¯ + 5L0 + 5L¯ + 1) ≤ /8,

(2.154)

let a point x ∈ X satisfy x ≤ K0 ,

(2.155)

v ∈ ∂/4 f (x),

(2.156)

ξ ∈ BX (v, δf ),

(2.157)

y ∈ BX (PC (x − αξ ), δC ).

(2.158)

and let

Then ¯ 2 + 2α(f (x) ¯ − f (x)) + 3α/4 y − x ¯ 2 ≤ x − x ¯ 2α2. +2δC (K0 + K¯ + 5L0 + 5L¯ + 1) + 25(L0 + L)

34

2 Nonsmooth Convex Optimization

Proof Set y0 = x − αξ.

(2.159)

Lemma 2.19, (2.152), (2.153), and (2.156) imply that ¯ v ≤ 5L0 + 4L.

(2.160)

It follows from (2.8), (2.151), (2.155), (2.157), (2.159), and (2.160) that ¯ 2 = x − αξ − x ¯ 2 y0 − x = x − αv + (αv − αξ ) − x ¯ 2 ≤ x − αv − x ¯ 2 + α 2 v − ξ 2 + 2αv − ξ, x − αv − x ¯ ≤ x − αv − x ¯ 2 + α 2 δf2 + 2αδf x − αv − x ¯ ¯ ≤ x − αv − x ¯ 2 + α 2 δf2 + 2αδf (K0 + K¯ + α(5L0 + 4L)) ¯ 2 ≤ x − x ¯ 2 − 2αx − x, ¯ v + α 2 (5L0 + 4L) ¯ +α 2 δf2 + 2αδf (K0 + K¯ + α(5L0 + 4L)).

(2.161)

In view of (2.256), v, x¯ − x ≤ f (x) ¯ − f (x) + /4.

(2.162)

By (2.154), (2.161), and (2.62), ¯ 2 ≤ x − x ¯ 2 + 2α(f (x) ¯ − f (x)) + α/2 y0 − x ¯ 2 + α 2 δf2 + 2αδf (K0 + K¯ + α(5L0 + 4L)) ¯ +α 2 (5L0 + 4L) ¯ 2 ¯ − f (x)) + α/2 + 25α 2 (L0 + L) ≤ x − x ¯ 2 + 2α(f (x) ¯ +2αδf (K0 + K¯ + 5(L0 + L)) ¯ 2 + α/4. ≤ x − x ¯ 2 + 2α(f (x) ¯ − f (x)) + α/2 + 25α 2 (L0 + L) (2.163) By (2.7), (2.8), (2.151), (2.153), and (2.163), ¯ 2 α 2 + 2αL0 x¯ − x y0 − x ¯ 2 ≤ x − x ¯ 2 + α + 25(L0 + L) ¯ 2 + 16(L0 + L) ¯ + (5(L0 + L)) ¯ 2 ≤ (K0 + K) ¯ ≤ (K0 + K¯ + 5L0 + 5L) ¯ 2, +2L0 (K0 + K)

2.6 Proof of Theorem 2.1

35

¯ y0 − x ¯ ≤ K0 + K¯ + 5(L0 + L).

(2.164)

It follows from (2.10), (2.131), (2.151), (2.158), (2.159), and (2.163) that ¯ 2 y − x ¯ 2 = y − PC (x − αξ ) + PC (x − αξ ) − x ≤ y − PC (x − αξ )2 + 2y − PC (x − αξ )PC (x − αξ ) − x ¯ +PC (x − αξ ) − x ¯ 2 ¯ + y0 − x ≤ δC2 + 2δC (K0 + K¯ + 5L0 + 5L) ¯ 2 ¯ + x − x ≤ δC2 + 2δC (K0 + K¯ + 5L0 + 5L) ¯ 2 ¯ 2 +2α(f (x) ¯ − f (x)) + 3α/4 + 25α 2 (L0 + L) ≤ x − x ¯ 2 + 2α(f (x) ¯ − f (x)) + 3α/4 ¯ 2 + 2δC (K0 + K¯ + 5L0 + 5L¯ + 1). +25α 2 (L0 + L) Lemma 2.27 is proved.

2.6 Proof of Theorem 2.1 By (2.17), (2.19), and (2.21), BX (xi , δC ) ∩ C = ∅, i = 1, . . . , n.

(2.165)

In view if (2.12), (2.19), and (2.165), xi  ≤ K1 + 1, i = 0, 1, . . . , n.

(2.166)

x¯ ∈ Cmin .

(2.167)

Fix

It follows from (2.6), (2.7), and (2.167) that x ¯ ≤ K¯ ≤ K1 .

(2.168)

Assume that the assertion of the theorem does not hold. Then f (xi ) > inf(f, C) + , i = 1, . . . , n.

(2.169)

Let i ∈ {1, . . . , n − 1}. In view of (2.12)–(2.15), (2.18), (2.20), (2.21), and (2.167), we apply Lemma 2.21 with

36

2 Nonsmooth Convex Optimization

K0 = K1 + 1, L0 = L1 , ξ = ξi , x = xi , y = xi+1 and obtain that ¯ 2 ≤ xi − x ¯ 2 − 2−1 α + δC2 xi+1 − x ¯ 2α2 +2δC (K0 + K¯ + 5L1 + 5L¯ + 1) + 25(L1 + L) = xi − x ¯ 2 − 4−1 α + 2δC (K1 + K¯ + 5L1 + 5L¯ + 2) ≤ xi − x ¯ 2 − 8−1 α.

(2.170)

Relations (2.166) and (2.168) imply that ¯ 2 ≥ x1 − x ¯ 2 ≥ x1 − x ¯ 2 − xn − x ¯ 2 (1 + K1 + K) =

n−1  (xi − x ¯ 2 − xi+1 − x ¯ 2 ) ≥ 8−1 α(n − 1). i=1

Together with (2.18), this implies that ¯ 2 α −1  −1 n ≤ 1 + 8(1 + K1 + K) ¯ 2 (L1 + L) ¯ 2  −2 . = 1 + 800(1 + K1 + K) This contradicts (2.16). The contradiction we have reached completes the proof of Theorem 2.1.

2.7 Proof of Theorem 2.2 By (2.27), (2.30), and (2.32), BX (xi , δC ) ∩ C = ∅, i = 1, . . . , n.

(2.171)

x¯ ∈ Cmin .

(2.172)

Fix

It follows from (2.7), (2.8), and (2.172) that ¯ x ¯ ≤ K. It follows from (2.10), (2.29), (2.30) and (2.172) that

(2.173)

2.7 Proof of Theorem 2.2

37

x1 − x ¯ ≤ x1 − P0 (x0 ) + P0 (x0 ) − x ¯ ¯ ¯ ≤ 1 + K1 + K. ≤ δC + x0 − x

(2.174)

In view of (2.172)–(2.174), x1  ≤ 1 + K1 + 2K¯ ≤ 1 + 3K1 ,

(2.175)

¯ d(x1 , Cmin ) ≤ 1 + K1 + K.

(2.176)

Assume that the assertion of the theorem does not hold. Then f (xi ) > inf(f, C) + , i = 1, . . . , n.

(2.177)

We show by induction that for i = 1, . . . , n, ¯ d(xi , Cmin ) ≤ 1 + K1 + K.

(2.178)

Assume that an integer i satisfies 1 ≤ i < n and that (2.178) is true. (In view of (2.176), our assumption holds with i = 1.) By (2.7), (2.8), and (2.178), xi  ≤ 1 + K1 + 2K¯ ≤ 1 + 3K1 .

(2.179)

f (xi ) ≤ inf(f, C) + 4;

(2.180)

f (xi ) > inf(f, C) + 4.

(2.181)

There are two cases:

Assume that (2.180) holds. In view of (2.7), (2.8), and (2.180), ¯ xi  ≤ K.

(2.182)

It follows from (2.10), (2.32), (2.172), (2.173), and (2.182) that ¯ ≤ xi+1 − Pi (xi − αξi ) + Pi (xi − αξi ) − x ¯ xi+1 − x ≤ δC + xi − αξi − x ¯ ≤ 1 + xi − x ¯ + αξi  ≤ 1 + 2K¯ + αξi .

(2.183)

Lemma 2.19, (2.8), (2.9), and (2.182) imply that ∂/4 f (xi ) ⊂ BX (0, L¯ + /4).

(2.184)

38

2 Nonsmooth Convex Optimization

By (2.23)–(2.25), (2.31), (2.98), and (2.184), αξi  ≤ α(L¯ + /4 + δf ) ≤ α(L¯ + 2) ≤ 1.

(2.185)

In view of (2.183) and (2.185), ¯ ≤ 2 + 2K¯ ≤ 1 + K1 + K¯ xi+1 − x and ¯ d(xi+1 , Cmin ) ≤ 1 + K1 + K. Assume that (2.181) holds. In view of (2.23)–(2.25), (2.31), (2.32), (2.48), (2.178), (2.179), and (2.181), we apply Lemma 2.23 with PC = Pi , K0 = 3K1 + 1, L0 = L1 , ξ = ξi , x = xi , y = xi+1 and obtain that d(xi+1 , Cmin )2 ≤ d(xi , Cmin )2 − 2α ¯ + 25(L1 + L) ¯ 2α2 +2δC (3K1 + K¯ + 2 + 5L1 + 5L) ¯ ≤ d(xi , Cmin )2 − α + 2δC (3K1 + K¯ + 2 + 5L1 + 5L) ≤ d(xi , Cmin )2 − α/2 ≤ d(xi , Cmin )2 and ¯ d(xi+1 , Cmin ) ≤ d(xi , Cmin ) ≤ 1 + K1 + K. Thus in the both cases d(xi+1 , Cmin ) ≤ K1 + K¯ + 1. Thus we showed by induction that for all integers i = 1, . . . , n, (2.178) holds and that xi  ≤ 1 + K1 + 2K¯ ≤ 1 + 3K1 .

(2.186)

In view of (2.23)–(2.25), (2.28), (2.31), (2.32), (2.171), (2.172), (2.177), and (2.186), we apply Lemma 2.21 with PC = Pi , K0 = 3K1 + 1, L0 = L1 , ξ = ξi , x = xi , y = xi+1 and obtain that

2.8 Proof of Theorem 2.3

39

xi+1 − x ¯ 2 ≤ xi − x ¯ 2 − 2−1 α ¯ + 25(L1 + L) ¯ 2α2 +2δC (3K1 + K¯ + 2 + 5L1 + 5L) ¯ ≤ xi − x ¯ 2 − 4−1 α + 2δC (3K1 + K¯ + 2 + 5L1 + 5L) ≤ xi − x ¯ 2 − 8−1 α.

(2.187)

By (2.28), (2.175), and (2.187), ¯ 2 ≥ x1 − x ¯ 2 (1 + K1 + K) ≥ x1 − x ¯ 2 − xn − x ¯ 2 =

n−1  (xi − x ¯ 2 − xi+1 − x ¯ 2 ) ≥ 8−1 α(n − 1) i=1

and ¯ 2 (α)−1 n ≤ 1 + 8(1 + K1 + K) ¯ 2  −2 (L1 + L) ¯ 2. ≤ 1 + 800(1 + K1 + K) This contradicts (2.26). The contradiction we have reached completes the proof of Theorem 2.2.

2.8 Proof of Theorem 2.3 Set Pn = Pn−1 , ξn ∈ ∂f (xn ), αn = 1,

(2.188)

xn+1 = Pn (xn − ξn ).

(2.189)

By (2.37), (2.39), (2.41), (2.188), and (2.189), BX (xi , δC ) ∩ C = ∅, i = 1, . . . , n + 1.

(2.190)

In view if (2.33), (2.38), and (2.190), xi  ≤ K1 + 1, i = 0, 1, . . . , n + 1.

(2.191)

x¯ ∈ Cmin .

(2.192)

Fix

40

2 Nonsmooth Convex Optimization

It follows from (2.7), (2.8), and (2.192) that x ¯ ≤ K¯ ≤ K1 .

(2.193)

Let i ∈ {1, . . . , n}. In view of (2.34), (2.36), (2.40), (2.41), (2.188), (2.189), (2.191), and (2.192), we apply Lemma 2.27 with K0 = K1 + 1, L0 = L1 , α = αi , ξ = ξi , x = xi , y = xi+1 and obtain that ¯ 2 ≤ xi − x ¯ 2 + 2αi (f (x) ¯ − f (xi )) + 3αi /4 xi+1 − x ¯ 2 αi2 + 2δC (K1 + K¯ + 5L1 + 5L¯ + 2). +25(L1 + L)

(2.194)

Inequality (2.194) implies that ¯ ≤ 2−1 xi − x ¯ 2 − 2−1 xi+1 − x ¯ 2 + 3αi /4 αi (f (xi ) − f (x)) ¯ 2 αi2 + 2δC (K1 + K¯ + 5L1 + 5L¯ + 2). +25(L1 + L)

(2.195)

In view of (2.195), n 

αi (f (xi ) − f (x)) ¯ ≤ 2−1 x1 − x ¯ 2 − 2−1 xn+1 − x ¯ 2

i=1

  n n  ¯ 2 +3 αi /4 + 25(L1 + L) αi2 + 2nδC (K1 + K¯ + 5L1 + 5L¯ + 2). i=1

i=1

(2.196) It follows from (2.191), (2.193), and (2.196) the convexity of f that ⎞ ⎛⎛ ⎞−1 n n   ⎟ ⎜ min{f (xi ) : t = 1, . . . , n} − inf(f, C), f ⎝⎝ αj ⎠ αi xi ⎠ − inf(f, C) j =1

≤

 n

αj

−1  n

j =1

≤

 n i=1

i=1

αi (f (xi ) − inf(f, C))

i=1

−1 αi

¯ 2 (2K1 + 1)2 + 3/4 + 25(L1 + L)

 n i=1

αi2

  n i=1

−1 αi

2.9 Proof of Theorem 2.5

41

 −1 n +2n αi δC (K¯ + K1 + 5L1 + 5L¯ + 2). i=1

Theorem 2.3 is proved.

2.9 Proof of Theorem 2.5 Set Pn = Pn−1 , ξn ∈ ∂f (xn ),

(2.197)

xn+1 = Pn (xn − αξn ).

(2.198)

x¯ ∈ Cmin .

(2.199)

Fix

It follows from (2.7), (2.8), and (2.199) that ¯ x ¯ ≤ K.

(2.200)

It follows from (2.10), (2.47), (2.48), and (2.200) that ¯ ≤ x1 − P0 (x0 ) + P0 (x0 ) − x ¯ x1 − x ¯ ≤ δC + x0 − x ¯ ≤ 1 + K1 + K.

(2.201)

We show that for all i = 1, . . . , n + 1, ¯ ¯ ≤ 1 + K1 + K. xi − x

(2.202)

Assume that an integer i satisfies 1 ≤ i < n + 1 and that (2.202) is true. (In view of (2.201), our assumption holds with i = 1.) There are two cases: f (xi ) ≤ inf(f, C) + 4;

(2.203)

f (xi ) > inf(f, C) + 4.

(2.204)

Assume that (2.203) holds. In view of (2.7), (2.8), and (2.203), ¯ xi  ≤ K.

(2.205)

42

2 Nonsmooth Convex Optimization

Lemma 2.19, (2.7)–(2.9), (2.42), and (2.205) imply that ∂/4 f (xi ) ⊂ BX (0, L¯ + /4) ⊂ BX (0, L¯ + 1).

(2.206)

By (2.49), (2.197), and (2.206), ξi  ≤ L¯ + 2.

(2.207)

It follows from (2.10), (2.43), (2.197), (2.198), (2.200), (2.205), and (2.207) that ¯ ≤ xi+1 − Pi (xi − αξi ) + Pi (xi − αξi ) − x ¯ xi+1 − x ≤ δC + xi − αξi − x ¯ ≤ 1 + xi − x ¯ + αξi  ≤ 1 + 2K¯ + α(L¯ + 2) ≤ 1 + 2K¯ + 2 ≤ 1 + K¯ + K1 . Assume that (2.204) holds. In view of (2.42)–(2.45), (2.50), (2.147), (2.148), (2.199), (2.202), and (2.204), we apply Lemma 2.27 with K0 = 1 + 2K¯ + K1 , L0 = L1 , ξ = ξi , x = xi , y = xi+1 ,  = 4 and obtain that ¯ 2 ≤ xi − x ¯ 2 + 2α(f (x) ¯ − f (xi )) + 3α/4 xi+1 − x ¯ 2 α 2 + 2δC (3K¯ + K1 + 2 + 5L1 + 5L) ¯ +25(L1 + L) ¯ 2 α 2 + 2δC (3K¯ + K1 + 2 + 5L1 + 5L) ¯ ≤ xi − x ¯ 2 − 8α + 3α + 25(L1 + L) ≤ xi − x ¯ 2 − 2α and ¯ ≤ xi − x ¯ ≤ 1 + K¯ + K1 . xi+1 − x

(2.208)

Thus in both cases, (2.208) holds. Therefore by induction we showed that ¯ i = 0, . . . , n + 1, ¯ ≤ 1 + K1 + K, xi − x ¯ i = 0, . . . , n + 1. xi  ≤ 1 + K1 + 2K,

(2.209) (2.210)

Let i ∈ {1, . . . , n}. In view of (2.42), (2.45), (2.49), (2.50), (2.147), (2.148), and (2.210), we apply Lemma 2.27 with K0 = 2K¯ + K1 + 1, L0 = L1 , ξ = ξi , x = xi , y = xi+1 and obtain that

2.10 Proof of Theorem 2.8

43

xi+1 − x ¯ 2 ≤ xi − x ¯ 2 + 2α(f (x) ¯ − f (xi )) + 3α/4 ¯ 2α2. +2δC (K1 + 3K¯ + 5L1 + 5L¯ + 2) + 25(L1 + L) This inequality implies that ¯ ≤ xi − x ¯ 2 − xi+1 − x ¯ 2 + 3α/4 2α(f (xi ) − f (x)) ¯ 2 α 2 + 2δC (K1 + 3K¯ + 5L1 + 5L¯ + 2). +25(L1 + L)

(2.211)

By (2.211), 2α

n n   (f (xi ) − f (x)) ¯ ≤ (xi − x ¯ 2 − xi+1 − x ¯ 2) i=1

i=1

¯ 2 α 2 + 2δC (K1 + 3K¯ + 5L1 + 5L¯ + 2)] +n[3α/4 + 25(L1 + L) ¯ 2 α 2 + 2δC (K1 + 3K¯ + 5L1 + 5L¯ + 2)]. ≤ x1 − x ¯ 2 + n[3α/4 + 25(L1 + L) (2.212) It follows from (2.209) and (2.212) that n−1

n−1 

¯ 2 + 3/8 f (xi ) − inf(f, C) ≤ (2α)−1 (1 + K1 + K)

i=1

¯ 2 α + α −1 δC (K1 + 3K¯ + 5L1 + 5L¯ + 2). +2−1 25(L1 + L) Theorem 2.5 is proved.

2.10 Proof of Theorem 2.8 Fix x¯ ∈ Cmin .

(2.213)

By (2.51), (2.56), (2.58), (2.59), and (2.61), for every i ∈ {1, . . . , n}, xi  ≤ K1 + 1.

(2.214)

Assume that an integer i ∈ {1, . . . , n} \ {n} satisfies f (xi ) > inf(f, C) + .

(2.215)

44

2 Nonsmooth Convex Optimization

We apply Lemma 2.14 with PC = Pi , K0 = K1 , ξ = ξi , x = xi , y = xi+1 , α = αi and in view of (2.52), (2.53), (2.54), (2.57), (2.58), (2.60), (2.61), and (2.215), obtain that ¯ −1 }, ∂/4 f (xi ) ⊂ {v ∈ X : v ≥ (3/4)(K1 + K)

(2.216)

¯ −1 − δf , ξi  ≥ (3/4)(K1 + K) ¯ −1 αi  + 2αi2 + δC2 + 2δC (K1 + K¯ + 2) xi+1 − x ¯ 2 ≤ xi − x ¯ 2 − (4L) ¯ −1 αi  + 2δC (K1 + K¯ + 3) ¯ 2 − (8L) ≤ xi − x ¯ −1 αi . ≤ xi − x ¯ 2 − (16L) Thus we have shown that the following property holds: (a) if an integer i ∈ {1, . . . , n} \ {n} and f (xi ) > inf(f, C) +  then (2.216) holds and ¯ −1 αi . ¯ 2 ≤ xi − x ¯ 2 − (16L) xi+1 − x

(2.217)

Assume that an integer j ∈ {1, . . . , n} \ {n} and that f (xi ) > inf(f, C) + , i = 1, . . . , j.

(2.218)

Property (a) implies that for all i = 1, . . . , j (2.216) and (2.217) hold. Combined with (2.7), (2.8), (2.57), (2.213), and (2.214), this implies that ¯ 2 ≥ x1 − x ¯ 2 − xj +1 − x ¯ 2 (K1 + K¯ + 1)2 ≥ x1 − x =

j  (xi − x ¯ 2 − xi+1 − x ¯ 2) i=1

¯ −2  2 (j + 1) ¯ −1 (j + 1) min{αi : i = 1, . . . , j } ≥ 2−1 (16L) ≥ (16L) and ¯ 2  −2 (16L) ¯ 2 ≤ n − 2. j + 1 ≤ 2(1 + K1 + K)

2.11 Proof of Theorem 2.10

45

This implies that there exists j ∈ {1, . . . , n} such that f (xj ) ≤ inf(f, C) +  and if i ∈ {1, . . . , n} \ {j }, then f (xi ) > inf(f, C) +  and ¯ −1 }. ∂/4 f (xi ) ⊂ {v ∈ X : v ≥ (3/4)(K1 + K) Theorem 2.8 is proved.

2.11 Proof of Theorem 2.10 In view of (2.66), (2.69), and (2.71), BX (xi , δC ) ∩ C = ∅, i = 1, . . . , n.

(2.219)

x¯ ∈ Cmin .

(2.220)

Fix

It follows from (2.7), (2.8), and (2.220) that ¯ x ¯ ≤ K.

(2.221)

It follows from (2.10), (2.68), (2.69), (2.220), and (2.221) that ¯ ≤ x1 − P0 (x0 ) + P0 (x0 ) − x ¯ x1 − x ¯ ≤ 1 + x0 − x ¯ ≤ 1 + K1 + K.

(2.222)

We show that for all i = 1, . . . , n, ¯ ¯ ≤ 1 + K1 + K. xi − x

(2.223)

Assume that an integer i satisfies 1 ≤ i < n and that (2.223) is true. (In view of (2.222), our assumption holds with i = 1.) We show that our assumption holds for i + 1 too. There are two cases: f (xi ) ≤ inf(f, C) + 4;

(2.224)

46

2 Nonsmooth Convex Optimization

f (xi ) > inf(f, C) + 4.

(2.225)

Assume that (2.224) holds. In view of (2.7), (2.8), and (2.224), ¯ xi  ≤ K.

(2.226)

It follows from (2.10), (2.62), (2.64), (2.67), (2.70), (2.71), (2.216), and (2.220) that ¯ ≤ xi+1 − Pi (xi − αi ξi ) + Pi (xi − αi ξi ) − x ¯ xi+1 − x ≤ δC + xi − αi ξi − x ¯ ≤ δC + xi − x ¯ + ξi  ≤ δC + 1 + 2K¯ + δf ≤ 3 + 2K¯ ≤ 1 + K¯ + K1 . Assume that (2.225) holds. In view of (2.62)–(2.64), (2.67), (2.70), (2.71), (2.220), (2.221), (2.223), and (2.225), we apply Lemma 2.24 with PC = Pi , K0 = 1 + 2K¯ + K1 , α = αi , ξ = ξi , x = xi , y = xi+1 ,  = 4 and obtain that ¯ 2 ≤ xi − x ¯ 2 − αi L¯ −1 + 2αi2 + δC2 + 2δC (3K¯ + K1 + 3) xi+1 − x ¯ −1 αi + 2δC (3K¯ + K1 + 4) ¯ 2 − (2L) ≤ xi − x ¯ −1 αi . ≤ xi − x ¯ 2 − (4L) By the relation above, ¯ ≤ xi − x ¯ ≤ 1 + K¯ + K1 . xi+1 − x Thus in the both cases ¯ ¯ ≤ 1 + K1 + K. xi+1 − x Therefore by induction we showed that (2.223) holds for all i = 1, . . . , n. This implies that for all i = 1, . . . , n, ¯ xi  ≤ 1 + K1 + 2K. Assume that i ∈ {1, . . . , n} \ {n} and that f (xi ) > inf(f, C) + . In view of the inequality above, (2.62)–(2.64), (2.67), (2.70), (2.71), (2.220), and (2.223), we apply Lemma 2.24 with

2.11 Proof of Theorem 2.10

47

PC = Pi , K0 = 2K¯ + K1 + 1, α = αi , ξ = ξi , x = xi , y = xi+1 and obtain that ∂/4 f (xi ) ⊂ {v ∈ X : v ≥ (3/4)(K1 + 3K¯ + 1)−1 },

(2.227)

¯ −1 αi  + 2αi2 + δC2 + 2δC (K1 + 3K¯ + 3) xi+1 − x ¯ 2 ≤ xi − x ¯ 2 − (4L) ¯ −1 αi  + 2δC (K1 + 3K¯ + 4) ¯ 2 − (8L) ≤ xi − x ¯ −1 αi . ≤ xi − x ¯ 2 − (16L) Thus we have shown that the following property holds: (a) if an integer i ∈ {1, . . . , n} \ {n} and f (xi ) > inf(f, C) +  then (2.227) holds and ¯ −1 αi . ¯ 2 ≤ xi − x ¯ 2 − (16L) xi+1 − x Assume that an integer j ∈ {1, . . . , n} \ {n} and that f (xi ) > inf(f, C) + , i = 1, . . . , j. Property (a) and the inequality above imply that for all i = 1, . . . , j (2.227) holds and ¯ −2  2 xi+1 − x ¯ 2 ≤ xi − x ¯ 2 − 2−1 (16L) is true. By (2.65), (2.222), and (2.228), ¯ 2 ≥ x1 − x ¯ 2 − xj +1 − x ¯ 2 (K1 + K¯ + 1)2 ≥ x1 − x j  = (xi − x ¯ 2 − xi+1 − x ¯ 2) i=1

¯ −2  2 ≥ 2−1 j (16L) and ¯ 2  −2 (16L) ¯ 2  ≤ n − 2. j ≤ 2(1 + K1 + K) This implies that there exists j ∈ {1, . . . , n} such that

(2.228)

48

2 Nonsmooth Convex Optimization

f (xj ) ≤ inf(f, C) +  and if i ∈ {1, . . . , j } \ {j }, then f (xi ) > inf(f, C) + } and ∂/4 f (xi ) ⊂ {v ∈ X : v ≥ (3/4)(K1 + 3K¯ + 1)−1 }. Theorem 2.10 is proved.

2.12 An Auxiliary Result for Theorems 2.11–2.15 Assume that all the assumptions made in Sections 2.1–2.3 hold. Proposition 2.28 Let  ∈ (0, 1]. Then for each x ∈ X satisfying d(x, C) < min{2−1 L¯ −1 φ(/2), /2},

(2.229)

f (x) ≤ inf(f, C) + min{2−1 φ(/2), /2},

(2.230)

the inequality d(x, Cmin ) ≤  holds. Proof In view of the definition of φ, φ(/2) ∈ (0, 1] and if x ∈ C satisfies f (x) < inf(f, C) + φ(/2), then d(x, Cmin ) ≤ min{1, /2}.

(2.231)

Assume that a point x ∈ X satisfies (2.229) and (2.230). There exists a point y ∈ C which satisfies x − y < 2−1 L¯ −1 φ(/2) and x − y < /2.

(2.232)

Relations (2.7), (2.8), (2.230), and (2.232) imply that ¯ y ∈ BX (0, K¯ + 1). x ∈ BX (0, K),

(2.233)

By (2.223), (2.232), and the definition of L¯ (see (2.9)), ¯ − y < φ(/2)2−1 . |f (x) − f (y)| ≤ Lx

(2.234)

It follows from the choice of the point y, (2.230), and (2.234) that y ∈ C and

2.13 Proof of Theorem 2.11

49

f (y) < f (x) + φ(/2)2−1 ≤ inf(f, C) + φ(/2). Combined with (2.231), this implies that d(y, Cmin ) ≤ /2. Together with (2.232), this implies that d(x, Cmin ) ≤ x − y + d(y, Cmin ) ≤ . This completes the proof of Proposition 2.28.

2.13 Proof of Theorem 2.11 We may assume without loss of generality that  < 1. In view of Proposition 2.28, there exist a number ¯ ∈ (0, /8)

(2.235)

such that if x ∈ X, d(x, C) ≤ 2¯ ,andf (x) ≤ inf(f, C) + 2¯ , then d(x, Cmin ) ≤ .

(2.236)

x¯ ∈ Cmin

(2.237)

¯ 0 ∈ (0, 4−1 ).

(2.238)

Fix

and

Since limi→∞ αi = 0 (see (2.73)), there is an integer p0 > 0 such that K¯ + 4 < p0

(2.239)

and that for all integers p ≥ p0 − 1, we have ¯ −1 0 . αp < (32L) Since that

∞

i=0 αi

(2.240)

= ∞ (see (2.73)), there exist a natural number n0 > p0 + 4 such

50

2 Nonsmooth Convex Optimization n 0 −1

¯ αi > (4p0 + M + x) ¯ 2 0−1 16L.

(2.241)

i=p0

Fix ¯ K∗ > K¯ + 4 + M + 4n0 + 4x

(2.242)

and a positive number δ such that ¯ −1 0 . 6δ(K∗ + 1) < (16L)

(2.243)

Assume that an integer n ≥ n0 and that n {Pk }n−1 k=0 ⊂ M, {xk }k=0 ⊂ X,

x0  ≤ M, Pi (X) = C, i = 0, . . . , n − 1,

(2.244)

vk ∈ ∂δ f (xk ) \ {0}, k = 0, , . . . , n − 1,

(2.245)

n−1 {ηk }n−1 k=0 , {ξk }k=0 ⊂ BX (0, δ),

(2.246)

and that for all integers k = 0, . . . , n − 1, we have xk+1 = Pk (xk − αk vk −1 vk − αk ξk ) − αk ηk .

(2.247)

In order to prove the theorem, it is sufficient to show that d(xk , Cmin ) ≤  for all integers k satisfying n0 ≤ k ≤ n. Assume that an integer k ∈ [p0 , n − 1],

(2.248)

xk  ≤ K∗ ,

(2.249)

f (xk ) > inf(f, C) + 0 .

(2.250)

In view of (2.237), (2.243), (2.245), (2.246), (2.247), (2.249), and (2.250), the conditions of Lemma 2.24 hold with PC = Pk , K0 = K∗ ,  = 0 , δf = δ, δC = δαk , α = αk , x = xk , v = vk , ξ = ξk + vk −1 vk , and y = xk+1 , and combined with (2.240), (2.243), and (2.248), this Lemma implies that ¯ −1 0 ¯ 2 ≤ xk − x ¯ 2 − αk (4L) xk+1 − x

2.13 Proof of Theorem 2.11

51

+2αk2 + αk2 δ 2 + 2δαk (K∗ + K¯ + 2) ¯ −1 0 + 2δαk (K∗ + K¯ + 3) ¯ 2 − αk (8L) ≤ xk − x ¯ −1 0 . ≤ xk − x ¯ 2 − αk (16L) Thus we have shown that the following property holds: (P1)

If an integer k ∈ [p0 , n − 1] and (2.249) and (2.250) are valid, then we have ¯ −1 αk 0 . ¯ 2 ≤ xk − x ¯ 2 − (16L) xk+1 − x

We claim that there exists an integer j ∈ {p0 , . . . , n0 } such that f (xj ) ≤ inf(f, C) + 0 . Assume the contrary. Then f (xi ) > inf(f, C) + 0 , i = p0 , . . . , n0 .

(2.251)

It follows from (2.10), (2.243), and (2.245)–(2.247)) that for all integers i = 0, . . . , n − 1, we have ¯ ≤ 1 + Pi (xi − αi vi −1 vi − αi ξi ) − x ¯ xi+1 − x ≤ 1 + xi − αi vi −1 vi − αi ξi − x ¯ ≤ 1 + xi − x ¯ + 2 = xi − x ¯ + 3.

(2.252)

By (2.242), (2.244), and (2.252), for all integers i = 0, . . . , n0 , ¯ +x ¯ ≤ M +3i +2x ¯ ≤ M +3n0 +2x ¯ < K∗ . xi  ≤ x0 − x+3i

(2.253)

Let i ∈ {p0 , . . . , n0 − 1}.

(2.254)

It follows from (2.251), (2.253), (2.254), and property (P1) that ¯ −1 αi 0 . ¯ 2 ≤ xi − x ¯ 2 − (16L) xi+1 − x

(2.255)

Relations (2.253) and (2.255) imply that ¯ 2 ≥ xp0 − x ¯ 2 − xn0 − x ¯ 2 (M + 3p0 + x) =

n 0 −1

n 0 −1

i=p0

i=p0

¯ −1 0 [xi − x ¯ 2 − xi+1 − x ¯ 2 ] ≥ (16L)

αi

52

2 Nonsmooth Convex Optimization

and n 0 −1

¯ −1 (M + 3p0 + x) αi ≤ 16L ¯ 2. 0

i=p0

This contradicts (2.241). The contradiction we have reached proves that there exists an integer j ∈ {p0 , . . . , n0 }

(2.256)

f (xj ) ≤ inf(f, C) + 0 .

(2.257)

such that

By (2.243), (2.244), (2.246), and (2.247), we have d(xj , C) ≤ αj −1 δ < ¯ .

(2.258)

In view of (2.236), (2.238), (2.257), and (2.258), d(xj , Cmin ) ≤ .

(2.259)

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 ) > .

(2.260)

By (2.256), (2.259), and (2.260), we have k > j ≥ p0 .

(2.261)

By (2.259) we may assume without loss of generality that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i < k.

(2.262)

Thus in view of (2.261) and (2.262) d(xk−1 , Cmin ) ≤ . There are two cases:

(2.263)

2.13 Proof of Theorem 2.11

53

f (xk−1 ) ≤ inf(f, C) + 0 ;

(2.264)

f (xk−1 ) > inf(f, C) + 0 .

(2.265)

Assume that (2.264) is valid. It follows from (2.7), (2.8), and (2.264) that ¯ xk−1 ∈ X0 ⊂ BX (0, K).

(2.266)

By (2.244), (2.246), and (2.247), there exists a point z ∈ C such that xk−1 − z ≤ δ.

(2.267)

By (2.10), (2.246), (2.247), and (2.267), xk − z ≤ αk−1 δ + z − Pk−1 (xk−1 − αk−1 vk−1 −1 vk−1 − αk−1 ξk−1 ) ≤ δ + z − xk−1  + αk−1 + δ = 3δ + αk−1 .

(2.268)

It follows from (2.240), (2.243), (2.261), and (2.268) that d(xk , C) ≤ 3δ + αk−1 < 0 .

(2.269)

In view of (2.267) and (2.268), xk − xk−1  ≤ xk − z + z − xk−1  ≤ 4δ + αk−1 .

(2.270)

It follows from (2.240), (2.243), (2.261), (2.263), and (2.270) that d(xk , Cmin ) ≤ 2.

(2.271)

Relations (2.7), (2.8), (2.263), and (2.271) imply that xk−1 , xk ∈ BX (0, K¯ + 2). Together with (2.9) and (2.270), the inclusion above implies that ¯ k−1 − xk  ≤ L(4δ ¯ + αk−1 ). |f (xk−1 ) − f (xk )| ≤ Lx

(2.272)

In view of (2.240), (2.243), (2.264), and (2.272), we have ¯ + αk−1 ) f (xk ) ≤ f (xk−1 ) + L(4δ ¯ ≤ inf(f, C) + 0 + L(4δ + αk−1 ) ≤ inf(f, C) + 20 . It follows from (2.236), (2.238), (2.269), and (2.273) that

(2.273)

54

2 Nonsmooth Convex Optimization

d(xk , Cmin ) ≤ . This inequality contradicts (2.260). The contradiction we have reached proves (2.265). By (2.7), (2.8), and (2.263), we have xk−1  ≤ K¯ + 1.

(2.274)

It follows from (2.240), (2.242), (2.243), (2.245), (2.247), (2.263), (2.265), and (2.274) that Lemma 2.26 holds with x = xk−1 , y = xk , ξ = ξk−1 , v = vk−1 , α = αk−1 , K0 = K¯ + 1,  = 0 , δf = δ, δC = αk−1 δ and this implies that d(xk , Cmin )2 2 2 ¯ −1 0 + 2αk−1 ≤ d(xk−1 , Cmin )2 − αk−1 (4L) + αk−1 δ 2 + 2αk−1 δ(2K¯ + 3)

¯ −1 αk−1 0 + 2αk−1 δ(2K¯ + 4) ≤ d(xk−1 , Cmin )2 − (8L) ¯ −1 αk−1 0 ≤ d(xk−1 , Cmin )2 ≤  2 . ≤ d(xk−1 , Cmin )2 − (16L)

This contradicts (2.260). The contradiction we have reached proves that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i ≤ n. Since j ≤ n0 , this completes the proof of Theorem 2.11.

2.14 Proof of Theorem 2.12 We may assume that without loss of generality  < 1, M > K¯ + 4.

(2.275)

Proposition 2.28 implies that there exists ¯ ∈ (0, /8) such that if x ∈ X, d(x, C) ≤ 2¯ , andf (x) ≤ inf(f, C) + 2¯ ,

(2.276)

2.14 Proof of Theorem 2.12

55

then d(x, Cmin ) ≤ /4.

(2.277)

¯ −1 ¯ . β0 = (64L)

(2.278)

β1 ∈ (0, β0 ).

(2.279)

Set

Let

There exists an integer n0 ≥ 4 such that ¯ β1 n0 > 162 (3 + 2M)2 ¯ −1 L.

(2.280)

K∗ > 2M + 4 + 4n0 + 2K¯ + 2M

(2.281)

Fix

and a positive number δ such that ¯ −1 ¯ β1 . 6δK∗ < (64L)

(2.282)

x¯ ∈ Cmin .

(2.283)

Fix a point

Assume that an integer n ≥ n0 , {Pi }n−1 i=0 ⊂ M, Pi (X) = C, i = 0, . . . , n − 1,

(2.284)

{xi }ni=0 ⊂ X, x0  ≤ M,

(2.285)

vi ∈ ∂δ f (xi ) \ {0}, i = 0, 1, . . . , n − 1

(2.286)

n−1 n−1 {αi }n−1 i=0 ⊂ [β1 , β0 ], {ηi }i=0 , {ξi }i=0 ⊂ BX (0, δ)

(2.287)

and that for all integers i = 0, . . . , n − 1, xi+1 = Pi (xi − αi vi −1 vi − αi ξi ) − ηi . We claim that d(xk , Cmin ) ≤  for all integers k satisfying n0 ≤ k ≤ n. Assume that an integer k ∈ [0, n − 1],

(2.288)

56

2 Nonsmooth Convex Optimization

xk  ≤ K∗ f (xk ) > inf(f, C) + ¯ /4.

(2.289)

It follows from (2.282), (2.283), and (2.286)–(2.289) that Lemma 2.24 holds with δf = δ, δC = δ,  = ¯ /4, K0 = K∗ , α = αk , x = xk , v = vk , ξ = ξk + vk 1 vk , and y = xk+1 and this implies that ¯ −1 ¯ ¯ 2 ≤ xk − x ¯ 2 − αk (16L) xk+1 − x +2αk2 + δ 2 + 2δ(K∗ + K¯ + 2) ¯ −1 ¯ + 2αk2 + 2δ(K∗ + K¯ + 3). ¯ 2 − αk (16L) ≤ xk − x Together with (2.278), (2.282), and (2.287), this implies that ¯ −1 ¯ + 2δ(K¯ + 3 + K∗ ) ¯ 2 ≤ xk − x ¯ 2 − αk (32L) xk+1 − x ¯ −1 ¯ β1 + 2δ(K¯ + 3 + K∗ ) ≤ xk − x ¯ 2 − (32L) ¯ −1 ¯ . ≤ xk − x ¯ 2 − β1 (64L) Thus we have shown that the following property holds: (P2)

if an integer k ∈ [0, n − 1] and (2.289) is valid, then we have ¯ −1 β1 ¯ . xk+1 − x ¯ 2 ≤ xk − x ¯ 2 − (64L)

We claim that there exists an integer j ∈ {1, . . . , n0 } for which f (xj ) ≤ inf(f, C) + ¯ /4. Assume the contrary. Then we have ¯ j = 1, . . . , n0 . f (xj ) > inf(f, C) + /4,

(2.290)

It follows from (2.10), (2.283), (2.287), and (2.288) that for all integers i = 0, . . . , n − 1, we have ¯ ≤ 1 + xi − αi vi −1 vi − αi ξi − x ¯ xi+1 − x ≤ xi − x ¯ + 3.

(2.291)

By (2.281), (2.283), (2.285), and (2.291) for i = 0, . . . , n0 ,

Let

¯ ≤ x0 − x ¯ + 3i, xi − x

(2.292)

xi  ≤ 2x ¯ + M + 3n0 < K∗ .

(2.293)

2.14 Proof of Theorem 2.12

57

k ∈ {1, . . . , n0 − 1}.

(2.294)

It follows from (2.290), (2.293), (2.294), and property (P2) that ¯ −1 β1 ¯ . ¯ 2 ≤ xk − x ¯ 2 − (64L) xk+1 − x

(2.295)

Relations (2.295), (2.275), (2.283), (2.285), and (2.292) imply that ¯ 2 − xn0 − x ¯ 2 (M + x ¯ + 3)2 ≥ x1 − x =

n 0 −1

[xi − x ¯ 2 − xi+1 − x ¯ 2]

i=1

¯ −1 β ¯ −1 ¯ , ¯ 1 ≥ β1 n0 (128L) ≥ (n0 − 1)(64L) ¯ −1 ¯ β1 ≤ (2M + 3)2 . (n0 /2)(64L)

This contradicts (2.280). The contradiction we have reached proves that there exists an integer j ∈ {1, . . . , n0 }

(2.296)

f (xj ) ≤ inf(f, C) + ¯ /4.

(2.297)

for which

By (2.284), (2.287), and (2.288), we have d(xj , C) ≤ δ.

(2.298)

Relations (2.277), (2.282), (2.297) and (2.298) imply that d(xj , Cmin ) ≤ .

(2.299)

We claim that for all integers i satisfying j ≤ i ≤ n, we have d(xi , Cmin ) ≤ . Assume the contrary. Then there exists an integer k ∈ [j, n] for which

(2.300)

58

2 Nonsmooth Convex Optimization

d(xk , Cmin ) > .

(2.301)

k > j.

(2.302)

By (2.296) and (2.299)–(2.301),

We may assume without loss of generality that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i < k.

(2.303)

Then d(xk−1 , Cmin ) ≤ .

(2.304)

f (xk−1 ) ≤ inf(f, C) + ¯ /4;

(2.305)

f (xk−1 ) > inf(f, C) + ¯ /4.

(2.306)

There are two cases:

Assume that (2.305) is valid. In view of (2.7), (2.8), (2.275), (2.276), and (2.305), ¯ xk−1 ∈ X0 ⊂ BX (0, K).

(2.307)

By (2.284), (2.287), and (2.288), there exists a point z ∈ C such that xk−1 − z ≤ δ.

(2.308)

It follows from (2.10), (2.287), (2.288), and (2.308) that xk − z ≤ δ + z − Pk−1 (xk−1 − αk−1 vk−1 −1 vk−1 − αk−1 ξk−1 ) ≤ δ + z − xk−1  + αk−1 + δ < 3δ + αk−1 .

(2.309)

Relations (2.277), (2.282), (2.305), and (2.308) imply that d(xk−1 , Cmin ) ≤ /4.

(2.310)

By (2.276), (2.278), (2.282), (2.287), (2.308), and (2.309), xk − xk−1  ≤ xk − z + z − xk+1  ≤ 4δ + αk−1 < ¯ < /8.

(2.311)

2.15 Proof of Theorem 2.13

59

In view of (2.310) and (2.311), d(xk , Cmin ) ≤ . This inequality contradicts (2.301). The contradiction we have reached proves (2.306). In view of (2.7), (2.8), and (2.304), xk−1  ≤ K¯ + 1.

(2.312)

It follows from (2.286)–(2.288), (2.306), and (2.312) that Lemma 2.26 holds with PC = Pi , K0 = K¯ + 1, x = xk−1 , y = xk , v = vk−1 , ξ = ξk−1 + vk−1 −1 v, α = αk−1 ,  = 4−1 ¯ , δf = δ, δC = δ , and combining with (2.278), (2.282), (2.287), and (2.304), this implies that d(xk , Cmin )2 2 ¯ −1 ¯ + 2αk−1 ≤ d(xk−1 , Cmin )2 − αk−1 (16L) + δ 2 + 2δ(2K¯ + 4)

¯ −1 αk−1 ¯ + 2δ(2K¯ + 5) ≤ d(xk−1 , Cmin )2 − (32L) ¯ −1 β1 ¯ + 2δ(2K¯ + 5) ≤ d(xk−1 , Cmin )2 − (32L) ≤ d(xk−1 , Cmin )2 ≤  2 . This contradicts (2.301). The contradiction we have reached proves that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i ≤ n. In view of inequality n0 ≥ j , Theorem 2.12 is proved.

2.15 Proof of Theorem 2.13 We may assume without loss of generality that M > K¯ + 4,  < 1.

(2.313)

. There exists L0 > L¯ such that |f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, M + 4).

(2.314)

60

2 Nonsmooth Convex Optimization

In view of Proposition 2.28, there exist a number ¯ ∈ (0, /8)

(2.315)

such that if x ∈ X, d(x, C) ≤ 2¯ , and f (x) ≤ inf(f, C) + 2¯ , then d(x, Cmin ) ≤ .

(2.316)

x¯ ∈ Cmin

(2.317)

¯ 0 ∈ (0, 4−1 ).

(2.318)

Fix

and

Since limi→∞ αi = 0 (see (2.83)), there is an integer p0 > 0 such that M + 4 < p0

(2.319)

and that for all integers p ≥ p0 − 1, we have αp < (20(L¯ + L0 + 2))−2 0 . Since

∞

i=0 αi

(2.320)

= ∞ (see (2.83)), there exist a natural number n0 > p0 + 4

(2.321)

such that n 0 −1

¯ αi > (4p0 + M + x) ¯ 2 0−1 16L.

(2.322)

i=p0

Fix ¯ + 5L0 + 5L¯ K∗ > K¯ + 4 + M + 4n0 + 4x

(2.323)

and a positive number δ such that 6δ(K∗ + 1) < 16−1 0 . Assume that an integer n ≥ n0 and that

(2.324)

2.15 Proof of Theorem 2.13

61 n {Pi }n−1 i=0 ⊂ M, {xi }i=0 ⊂ X,

Pi (X) = C, i = 0, . . . , n − 1,

(2.325)

x0  ≤ M,

(2.326)

{ξi }n−1 i=0 ⊂ X and that for all i = 0, . . . , n − 1, BX (ξi , δ) ∩ ∂δ f (xi ) = ∅,

(2.327)

xi+1 − Pi (xi − αi ξi ) ≤ αi δ.

(2.328)

In order to prove the theorem, it is sufficient to show that d(xk , Cmin ) ≤  for all integers k satisfying n0 ≤ k ≤ n. By (2.83), (2.324)–(2.326), and (2.328), xi  ≤ M + 1 for all integers i = 0, . . . , n.

(2.329)

Assume that an integer k ∈ [p0 , n − 1],

(2.330)

f (xk ) > inf(f, C) + 0 .

(2.331)

In view of (2.314), (2.317), (2.323), (2.324), (2.326)–(2.328), and (2.331), the conditions of Lemma 2.21 hold with K0 = M, δf = δ, δC = αk δ,  = 0 , α = αk , x = xk , ξ = ξk , and y = xk+1 , and combined with (2.320), (2.323), (2.324), and (2.330), this Lemma implies that ¯ 2 ≤ xk − x ¯ 2 − 2−1 αk 0 xk+1 − x ¯ + 25αk2 (L0 + L) ¯ 2 +αk2 δ 2 + 2δαk (M + K¯ + 5L0 + 5L) ≤ xk − x ¯ 2 − 4−1 αk 0 ¯ + 25αk2 (L0 + L) ¯ 2 +2δαk (M + K¯ + 5L0 + 5L) ¯ ≤ xk − x ¯ 2 − 8−1 αk 0 + 2δαk (M + K¯ + 5L0 + 5L) ≤ xk − x ¯ 2 − 16−1 αk 0 . Thus we have shown that the following property holds: (P3)

If an integer k ∈ [p0 , n − 1] and (2.331) is valid, then we have

62

2 Nonsmooth Convex Optimization

xk+1 − x ¯ 2 ≤ xk − x ¯ 2 − 16−1 αk 0 . We claim that there exists an integer j ∈ {p0 , . . . , n0 } such that f (xj ) ≤ inf(f, C) + 0 . Assume the contrary. Then f (xi ) > inf(f, C) + 0 , i = p0 , . . . , n0 .

(2.332)

It follows from (2.332) and property (P3) that ¯ 2 ≤ xi − x ¯ 2 − 16−1 αi 0 . xi+1 − x

(2.333)

Relations (2.83), (2.313), (2.317), (2.325), (2.328), and (2.333) imply that ¯ 2 − xn0 − x ¯ 2 (2M + 1)2 ≥ xp0 − x =

n 0 −1

−1

[xi − x ¯ − xi+1 − x ¯ ] ≥ 16 2

2

i=p0

0

n 0 −1

αi

i=p0

and n 0 −1

αi ≤ 160−1 (2M + 1)2 .

i=p0

This contradicts (2.322). The contradiction we have reached proves that there exists an integer j ∈ {p0 , . . . , n0 }

(2.334)

f (xj ) ≤ inf(f, C) + 0 .

(2.335)

such that

By (2.318), (2.324), and (2.328), we have d(xj , C) ≤ αj −1 δ < ¯ .

(2.336)

In view of (2.316), (2.335), and (2.336), d(xj , Cmin ) ≤ . We claim that for all integers i satisfying j ≤ i ≤ n,

(2.337)

2.15 Proof of Theorem 2.13

63

d(xi , Cmin ) ≤ . Assume the contrary. Then there exists an integer k ∈ [j, n]

(2.338)

d(xk , Cmin ) > .

(2.339)

k > j ≥ p0 .

(2.340)

for which

By (2.337)–(2.339), we have

By (2.337)–(2.340), we may assume without loss of generality that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i < k.

(2.341)

Thus in view of (2.341), d(xk−1 , Cmin ) ≤ .

(2.342)

f (xk−1 ) ≤ inf(f, C) + 0 ;

(2.343)

f (xk−1 ) > inf(f, C) + 0 .

(2.344)

There are two cases:

Assume that (2.343) is valid. By (2.325) and (2.328), there exists a point z∈C

(2.345)

xk−1 − z ≤ δαk−2 .

(2.346)

such that

By (2.10), (2.328), (2.345), and (2.346), xk − z ≤ αk−1 δ + z − Pk−1 (xk−1 − αk−1 ξk−1 ) ≤ δαk−1 + z − xk−1  + αk−1 ξk−1  ≤ αk−1 ξk−1  + δ(αk−1 + αk−2 ). Lemma 2.19, (2.83), (2.313), (2.314), (2.345), and (2.346) imply that

(2.347)

64

2 Nonsmooth Convex Optimization

∂δ f (xk−1 ) ⊂ BX (0, L0 + δ) ⊂ BX (0, L0 + 1).

(2.348)

In view of (2.327) and (2.348), ξk−1  ≤ L0 + 2.

(2.349)

αk−1 ξk−1  ≤ αk−1 (L0 + 2).

(2.350)

Equation (2.349) implies that

By (2.347) and (2.350), xk − z ≤ δ(αk−1 + αk−2 ) + αk−1 (L0 + 2).

(2.351)

It follows from (2.320), (2.324), (2.334), (2.338), (2.345), and (2.351) that d(xk , C) ≤ δ(αk−1 + αk−2 ) + αk−1 (L0 + 2) ≤ 0 .

(2.352)

In view of (2.346) and (2.351), xk − xk−1  ≤ xk − z + z − xk−1  ≤ δαk−2 + δ(αk−1 + αk−2 ) + αk−1 (L0 + 2).

(2.353)

It follows from (2.83), (2.314), (2.346), (2.352), and (2.353) that |f (xk−1 ) − f (xk )| ≤ L0 xk−1 − xk  ≤ 2L0 δ(αk−1 + αk−2 ) + αk−1 L0 (L0 + 2). (2.354) In view of (2.320), (2.324), (2.334), (2.340), (2.343), and (2.354), we have f (xk ) ≤ f (xk−1 ) + 2L0 δ(αk−1 + αk−2 ) + αk−1 L0 (L0 + 2) ≤ inf(f, C) + 0 + 8−1 0 + 20−1 0 ≤ inf(f, C) + 20 .

(2.355)

It follows from (2.316), (2.318), (2.352), and (2.355) that d(xk , Cmin ) ≤ . This inequality contradicts (2.339). The contradiction we have reached proves (2.344). It follows from (2.83), (2.314), (2.318), (2.323)–(2.325), (2.328), and (2.344) that Lemma 2.22 holds with x = xk−1 , y = xk , ξ = ξk−1 , α = αk−1 , K0 = M + 1,  = 0 , δf = δ, δC = αk−1 δ

2.16 Proof of Theorem 2.14

65

, and combined with (2.320), (2.323), (2.324), (2.334), (2.340), and (2.342), this implies that 2 ¯ 2 (L0 + L) d(xk , Cmin )2 ≤ d(xk−1 , Cmin )2 − 2−1 αk−1 0 + 25αk−1 2 ¯ +αk−1 δ 2 + 2αk−1 δ(M + 1 + K¯ + 5L0 + 5L)

¯ ≤ d(xk−1 , Cmin )2 − 4−1 αk−1 0 + 2αk−1 δ(M + 2 + 2K¯ + 5L0 + 5L) ≤ d(xk−1 , Cmin )2 − 8−1 αk−1 0 < d(xk−1 , Cmin )2 ≤  2 and d(xk , Cmin ) ≤ . This contradicts (2.339). The contradiction we have reached proves that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i ≤ n. This completes the proof of Theorem 2.13.

2.16 Proof of Theorem 2.14 Fix x¯ ∈ Cmin

(2.356)

In view of Proposition 2.28, there exist a number ¯ ∈ (0, /8)

(2.357)

such that if x ∈ X, d(x, C) ≤ 2¯ and f (x) ≤ inf(f, C) + 2¯ , then d(x, Cmin ) ≤ .

(2.358)

¯ 0 ∈ (0, 4−1 ).

(2.359)

Fix

Since limi→∞ αi = 0 (see (2.88)), there is an integer p0 > 0 such that M + 4 < p0 and that for all integers p ≥ p0 − 1, we have

(2.360)

66

2 Nonsmooth Convex Optimization

αp < (20(L¯ + L0 + 2))−2 0 . Since

∞

i=0 αi

(2.361)

= ∞ (see (2.88)), there exist a natural number n0 > p0 + 4

(2.362)

such that n 0 −1

αi > 16(4p0 + 2M + x ¯ + 2)2 0−1 .

(2.363)

i=p0

Fix ¯ + 5L0 + 5L¯ K∗ > K¯ + 4 + 3M + 4n0 + 4x

(2.364)

and a positive number δ such that 6δ(K∗ + 1) < 16−1 0 .

(2.365)

Assume that an integer n ≥ n0 and that n {Pi }n−1 i=0 ⊂ M, {xi }i=0 ⊂ X,

Pi (X) = C, i = 0, . . . , n − 1,

(2.366)

x0  ≤ M,

(2.367)

{ξi }n−1 i=0 ⊂ X and that for all i = 0, . . . , n − 1, BX (ξi , δ) ∩ ∂δ f (xi ) = ∅,

(2.368)

xi+1 − Pi (xi − αi ξi ) ≤ αi δ

(2.369)

and (2.88) is true. In order to prove the theorem, it is sufficient to show that d(xk , Cmin ) ≤  for all integers k satisfying n0 ≤ k ≤ n. We show that for all t = 0, , . . . , n, ¯ ¯ ≤ 2 + M + K. xt − x

(2.370)

(In view of (2.356) and (2.367), the inequality above holds with t = 0.) Assume that an integer t satisfies 0 ≤ t < n and that (2.370) is true. We show that our assumption holds for t + 1 too. There are two cases:

2.16 Proof of Theorem 2.14

67

f (xt ) ≤ inf(f, C) + 4;

(2.371)

f (xt ) > inf(f, C) + 4.

(2.372)

Assume that (2.371) holds. In view of (2.7), (2.8), and (2.371), ¯ xt  ≤ K.

(2.373)

Lemma 2.19, (2.9), and (2.373) imply that ∂δ f (xt ) ⊂ BX (0, L¯ + 1).

(2.374)

By (2.365), (2.368), and (2.374), ξt  ≤ L¯ + 2.

(2.375)

It follows from (2.7), (2.8), (2.10), (2.88), (2.356), (2.365), (2.366), (2.369), (2.373), and (2.375) that ¯ ≤ xt+1 − Pt (xt − αt ξt ) + Pt (xt − αt ξt ) − x ¯ xt+1 − x ¯ ≤ αt δ + xt − x ¯ + αt ξt  ≤ αt δ + xt − αt ξt − x ¯ ≤ 1 + 2K¯ + α ξt  ≤ 3 + 2K¯ ≤ M + 2 + K. Thus ¯ ¯ ≤ M + 2 + K. xt+1 − x

(2.376)

Assume that (2.372) holds. In view of (2.356), (2.364), (2.365), (2.368), (2.369), and (2.372), we apply Lemma 2.21 with PC = Pt , δf = δ, δC = αt δ, ¯ α = αt , ξ = ξt , x = xt , y = xt+1 ,  = 4 K0 = M + 4 + K, , and this Lemma together with (2.88), (2.364), (2.365), and (2.370) implies that ¯ 2 ≤ xt − x ¯ 2 − 2αt + αt2 δ 2 xt+1 − x ¯ + 25αt2 (L0 + L) ¯ 2 +2αt δ(2K¯ + M + 2 + 5L0 + 5L) ¯ ≤ xt − x ¯ 2 − αt + 2αt δ(2K¯ + M + 3 + 5L0 + 5L) ≤ xt − x ¯ 2

68

2 Nonsmooth Convex Optimization

and ¯ ≤ xt − x ¯ ≤ K¯ + M + 2. xt+1 − x Thus (2.376) holds in the both cases. Therefore by induction we showed that (2.370) holds for all t = 0, . . . , n. Assume that an integer k ∈ [p0 , n − 1],

(2.377)

f (xk ) > inf(f, C) + 0 .

(2.378)

In view of (2.356), (2.359), (2.364), (2.365), (2.368)–(2.370), (2.377), and (2.378), the conditions of Lemma 2.21 hold with K0 = 2M + 4, δf = δ, δC = αk δ,  = 0 , α = αk , x = xk , ξ = ξk , and y = xk+1 , and combined with (2.361), (2.365), and (2.377), this Lemma implies that ¯ 2 ≤ xk − x ¯ 2 − 2−1 αk 0 xk+1 − x ¯ 2 +αk2 δ 2 + 2δαk (2M + 2K¯ + 5L0 + 5L¯ + 4) + 25αk2 (L0 + L) ≤ xk − x ¯ 2 − 4−1 αk 0 +2δαk (2M + 2K¯ + 5L0 + 5L¯ + 4) ≤ xk − x ¯ 2 − 8−1 αk 0 . Thus we have shown that the following property holds: (P4)

If an integer k ∈ [p0 , n − 1] and (2.378) is valid, then we have ¯ 2 ≤ xk − x ¯ 2 − 8−1 αk 0 . xk+1 − x

We claim that there exists an integer j ∈ {p0 , . . . , n0 } such that f (xj ) ≤ inf(f, C) + 0 . Assume the contrary. Then f (xi ) > inf(f, C) + 0 , i = p0 , . . . , n0 .

(2.379)

It follows from (2.379) and property (P4) that for all i = p0 , . . . , n0 − 1, xi+1 − x ¯ 2 ≤ xi − x ¯ 2 − 8−1 αi 0 . Relations (2.370) and (2.380) imply that

(2.380)

2.16 Proof of Theorem 2.14

69

(M + K¯ + 2)2 ≥ xp0 − x ¯ 2 − xn0 − x ¯ 2 =

n 0 −1

−1

[xi − x ¯ − xi+1 − x ¯ ]≥8 2

2

i=p0

0

n 0 −1

αi

i=p0

and n 0 −1

αi ≤ 80−1 (M + K¯ + 2)2 .

i=p0

This contradicts (2.363). The contradiction we have reached proves that there exists an integer j ∈ {p0 , . . . , n0 }

(2.381)

f (xj ) ≤ inf(f, C) + 0 .

(2.382)

such that

By (2.365), (2.366), and (2.369), we have d(xj , C) ≤ αj −1 δ < ¯ .

(2.383)

In view of (2.358), (2.359), (2.382), and (2.383), d(xj , Cmin ) ≤ .

(2.384)

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]

(2.385)

d(xk , Cmin ) > .

(2.386)

k > j ≥ p0 .

(2.387)

for which

By (2.384)–(2.386), we have

70

2 Nonsmooth Convex Optimization

We may assume without loss of generality that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i < k.

(2.388)

Thus in view of (2.388), d(xk−1 , Cmin ) ≤ .

(2.389)

f (xk−1 ) ≤ inf(f, C) + 0 ;

(2.390)

f (xk−1 ) > inf(f, C) + 0 .

(2.391)

There are two cases:

Assume that (2.390) is valid. By (2.366) and (2.369), there exists a point z ∈ C such that xk−1 − z ≤ δαk−2 .

(2.392)

By (2.10), (2.366), (2.369), and (2.392), xk − z ≤ αk−1 δ + z − Pk−1 (xk−1 − αk−1 ξk−1 ) ≤ δαk−1 + z − xk−1  + αk−1 ξk−1  ≤ αk−1 ξk−1  + δ(αk−1 + αk−2 ).

(2.393)

Lemma 2.19, (2.87), and (2.370) imply that ∂δ f (xk−1 ) ⊂ BX (0, L0 + δ) ⊂ BX (0, L0 + 1).

(2.394)

In view of (2.368) and (2.394), ξk−1  ≤ L0 + 2. By the relation above, αk−1 ξk−1  ≤ αk−1 (L0 + 2). Together with (2.393), this implies that xk − z ≤ δ(αk−1 + αk−2 ) + αk−1 (L0 + 2).

(2.395)

It follows from the inclusion z ∈ C, (2.361), (2.365), and (2.387) that d(xk , C) ≤ δ(αk−1 + αk−2 ) + αk−1 (L0 + 2) ≤ 0 .

(2.396)

2.16 Proof of Theorem 2.14

71

In view of (2.392) and (2.395), xk − xk−1  ≤ xk − z + z − xk−1  ≤ δαk−2 + δ(αk−1 + αk−2 ) + αk−1 (L0 + 2).

(2.397)

It follows from (2.87), (2.370), and (2.397) that |f (xk−1 ) − f (xk )| ≤ L0 xk−1 − xk  ≤ 2L0 δ(αk−1 + αk−2 ) + αk−1 L0 (L0 + 2).

(2.398)

In view of (2.361), (2.364), (2.365), (2.387), (2.390), and (2.398), we have f (xk ) ≤ f (xk−1 ) + 2L0 δ(αk−1 + αk−2 ) + αk−1 L0 (L0 + 2) ≤ inf(f, C) + 0 + 8−1 0 + 20−1 0 ≤ inf(f, C) + 20 .

(2.399)

It follows from (2.358), (2.359), (2.396), and (2.399) that d(xk , Cmin ) ≤ . This inequality contradicts (2.386). The contradiction we have reached proves (2.391). It follows from (2.87), (2.356), (2.365), (2.368)–(2.370), and (2.391) that Lemma 2.22 holds with x = xk−1 , y = xk , ξ = ξk−1 , α = αk−1 , K0 = M + 2K¯ + 2,  = 0 , δf = δ, δC = αk−1 δ , and combined with (2.361), (2.364), (2.365), (2.387), and (2.389) this implies that 2 ¯ 2 (L0 + L) d(xk , Cmin )2 ≤ d(xk−1 , Cmin )2 − 2−1 αk−1 0 + 25αk−1 2 ¯ +αk−1 δ 2 + 2αk−1 δ(M + 2 + 2K¯ + 5L0 + 5L)

¯ ≤ d(xk−1 , Cmin )2 − 4−1 αk−1 0 + 2αk−1 δ(M + 3 + 2K¯ + 5L0 + 5L) ≤ d(xk−1 , Cmin )2 − 8−1 αk−1 0 and d(xk , Cmin ) ≤ d(xk−1 , Cmin ) ≤ .

72

2 Nonsmooth Convex Optimization

This contradicts (2.386). The contradiction we have reached proves that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i ≤ n. This completes the proof of Theorem 2.14.

2.17 Proof of Theorem 2.15 We may assume that without loss of generality  < 1, M > K¯ + 4.

(2.400)

There exists L0 > L¯ such that |f (z1 ) − f (x2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, 3M + 4).

(2.401)

Proposition 2.28 implies that there exists ¯ ∈ (0, /8)

(2.402)

such that if x ∈ X, d(x, C) ≤ 2¯ ,andf (x) ≤ inf(f, C) + 2¯ , then d(x, Cmin ) ≤ .

(2.403)

¯ −2 ¯ . β0 = (64(L0 + L))

(2.404)

β1 ∈ (0, β0 ).

(2.405)

Set

Let

There exists an integer n0 ≥ 4 such that β1 n0 > 162 (3 + 2M)2 ¯ −1 L0 .

(2.406)

K∗ > 6M + 4 + 4n0 + 5L0 + 5L¯

(2.407)

Fix

and a positive number δ such that 6δK∗ < (64L0 )−1 ¯ β1 .

(2.408)

2.17 Proof of Theorem 2.15

73

Assume that an integer n ≥ n0 , {Pi }n−1 i=0 ⊂ M, Pi (X) = C, i = 0, . . . , n − 1,

(2.409)

{xi }ni=0 ⊂ X, x0  ≤ M,

(2.410)

{αi }n−1 i=0 ⊂ [β1 , β0 ]

(2.411)

{ξi }n−1 i=0 ⊂ X,

and that for all integers i = 0, . . . , n − 1, BX (ξi , δ) ∩ ∂δ f (xi ) = ∅,

(2.412)

xi+1 − Pi (xi − αi ξi ) ≤ δ.

(2.413)

In order to prove the theorem, it is sufficient to show that d(xk , Cmin ) ≤  for all integers k satisfying n0 ≤ k ≤ n. Fix a point x¯ ∈ Cmin .

(2.414)

¯ ¯ ≤ 2 + M + K. xt − x

(2.415)

We show that for all t = 0, , . . . , n,

In view of (2.7), (2.8), (2.410), and (2.414), (2.415) holds with t = 0. Assume that an integer t satisfies 0 ≤ t < n and that (2.415) is true. There are two cases: f (xt ) ≤ inf(f, C) + 4;

(2.416)

f (xt ) > inf(f, C) + 4.

(2.417)

Assume that (2.416) holds. In view of (2.7), (2.8), and (2.416), ¯ xt  ≤ K.

(2.418)

Lemma 2.19, (2.401) and (2.418) imply that ∂δ f (xt ) ⊂ BX (0, L¯ + 1). By (2.412) and (2.419),

(2.419)

74

2 Nonsmooth Convex Optimization

ξt  ≤ L¯ + 2.

(2.420)

It follows from (2.7), (2.8), (2.10), (2.404), (2.411), (2.413), (2.414), (2.418), and (2.420) that ¯ ≤ xt+1 − Pt (xt − αt ξt ) + Pt (xt − αt ξt ) − x ¯ xt+1 − x ≤ δ + xt − αt ξt − x ¯ ≤ δ + xt − x ¯ + αt ξt  ≤ 1 + 2K¯ + αt ξt  ≤ 1 + 2K¯ + β0 ξt  ≤ 3 + 2K¯ ≤ M + K¯ + 2. Thus ¯ ¯ ≤ M + 2 + K. xt+1 − x Assume that (2.417) holds. In view of (2.7), (2.8), (2.407), (2.408), (2.412)– (2.415), and (2.417), Lemma 2.21 holds PC = Pt , δf = δ, δC = δ, K0 = M + 2K¯ + 2, α = αt , ξ = ξt , x = xt , y = xt+1 ,  = 4 , and this Lemma together with (2.48), (2.404), (2.407), (2.408), and (2.411) implies that ¯ 2 ≤ xt − x ¯ 2 − 2αt + δ 2 xt+1 − x ¯ + 25αt2 (L0 + L) ¯ 2 +2δ(3K¯ + M + 2 + 5L0 + 5L) ¯ ≤ xt − x ¯ 2 − αt + δ(3K¯ + M + 3 + 5L0 + 5L) ¯ ≤ xt − x ¯ 2 − β1 + δ(2K¯ + 2M + 3 + 5L0 + 5L) ≤ xt − x ¯ 2 and ¯ ≤ K¯ + M + 2 xt+1 − x in the both cases. Therefore by induction we showed that (2.415) holds for all t = 0, . . . , n. Assume that an integer k ∈ [0, n − 1],

(2.421)

f (xk ) > inf(f, C) + ¯ .

(2.422)

2.17 Proof of Theorem 2.15

75

It follows from (2.401), (2.404), (2.407), (2.408), (2.412)–(2.415), and (2.422) that Lemma 2.21 holds with δf = δ, δC = δ,  = ¯ , K0 = K¯ + 2M + 2, x = xk , ξ = ξk , and y = xk+1 , and this Lemma together with (2.411) implies that ¯ 2 ≤ xk − x ¯ 2 − 2−1 αk ¯ xk+1 − x ¯ + 25αk2 (L0 + L) ¯ 2 +δ 2 + 2δ(2K¯ + 2M + 2 + 5L0 + L) ¯ ≤ xk − x ¯ 2 − 4−1 αk ¯ + 2δ(2K¯ + 2M + 3 + 5L0 + 5L) ¯ ≤ xk − x ¯ 2 − 4−1 β1 ¯ + 2δ(2K¯ + 2M + 3 + 5L0 + 5L) ≤ xk − x ¯ 2 − 8−1 β1 . ¯ Thus we have shown that the following property holds: (P5) if an integer k ∈ [0, n − 1] and (2.422) is valid, then we have xk+1 − x ¯ 2 ≤ xk − x ¯ 2 − 8−1 β1 ¯ . We claim that there exists an integer j ∈ {1, . . . , n0 } for which f (xj ) ≤ inf(f, C) + ¯ . Assume the contrary. Then we have f (xj ) > inf(f, C) + ¯ , j = 1, . . . , n0 .

(2.423)

It follows from (2.423) and property (P5) that for all k = 0, . . . , n0 − 1, ¯ 2 ≤ xk − x ¯ 2 − 8−1 β1 ¯ . xk+1 − x

(2.424)

Relations (2.7), (2.8), (2.410), (2.414), and (2.424) imply that ¯ 2 ≥ x0 − x ¯ 2 − xn0 − x ¯ 2 (M + K) =

n 0 −1

[xi − x ¯ 2 − xi+1 − x ¯ 2 ] ≥ 8−1 n0 ¯ β1 ,

i=0

¯ 2 ¯ −1 β −1 . n0 ≤ 8(M + K) 1 This contradicts (2.406). The contradiction we have reached proves that there exists an integer j ∈ {p0 , . . . , n0 } for which

76

2 Nonsmooth Convex Optimization

f (xj ) ≤ inf(f, C) + ¯ .

(2.425)

By (2.408), (2.409), and (2.413), we have d(xj , C) ≤ δ < ¯ .

(2.426)

Relations (2.403), (2.425), and (2.426) imply that d(xj , Cmin ) ≤ .

(2.427)

We claim that for all integers i satisfying j ≤ i ≤ n, we have d(xi , Cmin ) ≤ . Assume the contrary. Then there exists an integer k ∈ [j, n]

(2.428)

d(xk , Cmin ) > .

(2.429)

k > j.

(2.430)

for which

By (2.426), (2.428), and(2.429),

In view of (2.426) and (2.429), we may assume without loss of generality that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i < k.

(2.431)

In particular d(xk−1 , Cmin ) ≤ .

(2.432)

f (xk−1 ) ≤ inf(f, C) + ¯ ;

(2.433)

f (xk−1 ) > inf(f, C) + . ¯

(2.434)

There are two cases:

Assume that (2.433) is valid. In view of (2.409) and (2.413), there exists a point z∈C such that

(2.435)

2.17 Proof of Theorem 2.15

77

xk−1 − z ≤ δ.

(2.436)

It follows from (2.10), (2.411), (2.413), (2.435), and (2.436) that xk − z ≤ δ + z − Pk−1 (xk−1 − αk−1 ξk−1 ) ≤ δ + z − xk−1  + αk−1 ξk−1  ≤ 2δ + β0 ξk−1 .

(2.437)

Lemma 2.19, (2.401), (2.414), and (2.415) imply that ∂δ f (xk−1 ) ⊂ BX (0, L0 + δ) ⊂ BX (0, L0 + 1).

(2.438)

In view of (2.412) and (2.438), ξk−1  ≤ L0 + 2.

(2.439)

By (2.404), (2.408), (2.435), (2.437), and (2.439), xk − z ≤ 2δ + β0 (L0 + 2),

(2.440)

d(xk , C) ≤ 2δ + β0 (L0 + 2) < ¯ .

(2.441)

By (2.436) and (2.440), xk − xk−1  ≤ xk − z + z − xk+1  ≤ 3δ + β0 (L0 + 2).

(2.442)

In view of (2.401), (2.414), (2.415), and (2.442), |f (xk−1 − f (xk )| ≤ L0 xk − xk−1  ≤ 3L0 δ + β0 L0 (L0 + 2).

(2.443)

In view of (2.404), (2.408), and (2.443), f (xk ) ≤ f (xk−1 ) + 3L0 δ + β0 L0 (L0 + 2) ≤ inf(f, C) + ¯ + 8−1 ¯ + 8−1 ¯ < inf(f, C) + 2¯ . It follows from (2.403), (2.441), and (2.444) that d(xk , Cmin ) ≤ .

(2.444)

78

2 Nonsmooth Convex Optimization

This inequality contradicts (2.429). The contradiction we have reached proves (2.434). It follows from (2.401), (2.407), (2.408), (2.412)–(2.415), (2.432), and (2.434) that Lemma 2.22 holds with PC = Pk , K0 = M + 2K¯ + 2, x = xk−1 , y = xk , ξ = ξk−1 , α = αk−1 ,  = ¯ , δf = δ, δC = δ , and combining with (2.404) and (2.411), this implies that d(xk , Cmin )2 2 ¯ 2 + δ2 ≤ d(xk−1 , Cmin )2 − 2−1 αk−1 ¯ + 25αk−1 (L0 + L)

¯ +2δ(M + 3K¯ + 2 + 5L0 + 5L) ¯ ≤ d(xk−1 , Cmin )2 − 4−1 αk−1 ¯ + 2δ(M + 3K¯ + 3 + 5L0 + 5L) ¯ ≤ d(xk−1 , Cmin )2 − 4−1 β1 ¯ + 2δ(M + 3K¯ + 3 + 5L0 + 5L) ≤ d(xk−1 , Cmin )2 − 8−1 β1 ¯ and d(xk , Cmin ) ≤ d(xk−1 , Cmin ) ≤ . This contradicts (2.429). The contradiction we have reached proves that d(xi , Cmin ) ≤  for all integers i satisfying j ≤ i ≤ n. Theorem 2.15 is proved.

2.18 Proof of Theorem 2.16 By (2.94), (2.95), (2.97), and (2.99), for each integer i ≥ 0, xi  ≤ K1 ,

(2.445)

ξi  ≤ L1 .

(2.446)

In view of (2.10), (2.96), (2.99), and (2.446), for each integer i ≥ 1, xi ∈ C,

(2.447)

2.18 Proof of Theorem 2.16

79

xi − xi+1  = xi − Pi (xi − αi ξi ) ≤ xi − (xi − αi ξi ) ≤ αi ξi  ≤ αi L1 .

(2.448)

By (2.448), ∞ 

xi − xi+1  ≤ L1

∞ 

i=1

αi < ∞.

i=1

Thus {xi }∞ i=1 is a Cauchy sequence, and there exists x∗ = lim xi .

(2.449)

x∗ ∈ Cmin .

(2.450)

i→∞

Assume that

It follows from (2.447) and (2.449) that x∗ ∈ C.

(2.451)

In view of (2.450) and (2.451), there exists  ∈ (0, 1) such that f (x∗ ) > inf(f, C) + .

(2.452)

In view of (2.449), there exist a natural number n0 such that for all integers n ≥ n0 , f (xn ) > inf(f, C) + ,

(2.453)

¯ −2 . αn ≤ 100−1 (L1 + L)

(2.454)

z ∈ Cmin .

(2.455)

Fix

Applying Lemma 2.21 with K0 = K1 , L0 = L1 , x = xt , y = xt+1 , and v = ξt and with arbitrary sufficiently small δf , δC > 0, we obtain that for all integers t ≥ n0 , ¯ 2 xt+1 − z2 ≤ xt − z2 − 2−1 αt  + 25αt2 (L1 + L) ≤ xt − z2 − 4−1 αt .

Theorem 2.16 is proved.

80

2 Nonsmooth Convex Optimization

2.19 Proof of Theorem 2.17 Fix x¯ ∈ Cmin .

(2.456)

We show that for all i = 1, 2, . . . ,, ¯ d(xi , Cmin ) ≤ 1 + K1 + K. We have x0  ≤ K1 .

(2.457)

By (2.7), (2.8), (2.10), (2.101), (2.104), (2.456), and (2.457), ¯ = x¯ − P1 (x0 − α0 ξ0 ) ≤ x¯0 − x0  + α0 ξ0 . x1 − x

(2.458)

In view of (2.100), (2.103), and (2.457), ξ0  ≤ L1 + 1.

(2.459)

It follows from (2.7), (2.8), (2.102), and (2.457)–(2.459) that ¯ ≤ K1 + K¯ + 1. x1 − x

(2.460)

Assume that t ≥ 1 is an integer and d(xt , Cmin ) ≤ K1 + K¯ + 1.

(2.461)

In view of (2.456) and (2.461), xt  ≤ K1 + 2K¯ + 1.

(2.462)

f (xt ) ≤ inf(f, C) + 4;

(2.463)

f (xt ) > inf(f, C) + 4.

(2.464)

There are two cases:

Assume that (2.463) holds. In view of (2.7), (2.8), and (2.463), ¯ xt  ≤ K.

(2.465)

2.19 Proof of Theorem 2.17

81

In view of (2.465), ¯ x¯ − xt  ≤ 2K.

(2.466)

By (2.10), (2.101), (2.104), and (2.466), ¯ = Pt (xt − αt ξt ) − x ¯ xt+1 − x ≤ xt − x ¯ + αt ξt .

(2.467)

It follows from (2.100), (2.103), and (2.465) that ξt  ≤ L1 .

(2.468)

It follows from (2.102) and (2.466)–(2.468) that ¯ ≤ 2K¯ + αt L1 ≤ 2K¯ + 1. xt+1 − x

(2.469)

Assume that (2.464) holds. In view of (2.7), (2.8), (2.102), (2.461), and (2.464), we apply Lemma 2.23 with PC = Pt , K0 = 3K1 + 1, L0 = L1 , α = αt , ξ = ξt , x = xt , y = xt+1 and with arbitrary sufficiently small positive δf , δC and obtain that ¯ 2 d(xt+1 , Cmin )2 ≤ d(xt , Cmin )2 − 2αt + 25αt2 (L1 + L) ≤ d(xt , Cmin )2 − αt ≤ d(xt , Cmin )2 and d(xt+1 , Cmin ) ≤ K1 + K¯ + 1. Together with (2.456) and (2.459), this implies that the inequality above holds in the both cases. Therefore by induction we showed that d(xt , Cmin ) ≤ K1 + K¯ + 1 for all integers t ≥ 0.

(2.470)

In view of (2.7), (2.8), (2.456), and (2.470), xt  ≤ K1 + 2K¯ + 1 for all integers t ≥ 0.

(2.471)

It follows from (2.100), (2.103), and (2.471) that ξt  ≤ L1 for all integers t ≥ 0.

(2.472)

82

2 Nonsmooth Convex Optimization

By (2.10), (2.101), (2.104), and(2.472), xt+1 − xt  = Pt (xt − αt ξt ) − xt  ≤ αt ξt  ≤ αt L1 . Therefore ∞ 

xt+1 − xt  ≤ L1

t=0

∞ 

αt < ∞.

(2.473)

t=0

This implies that there exists x∗ = lim xt ∈ C. t→∞

(2.474)

Assume that x∗ ∈ Cmin . By the relation above, there exists  ∈ (0, 1) such that f (x∗ ) > inf(f, C) + .

(2.475)

In view of (2.474) and (2.475), there exist a natural number n0 such that for all integers n ≥ n0 , f (xn ) > inf(f, C) + ,

(2.476)

¯ −2 . αn ≤ 100−1 (L1 + L)

(2.477)

Let z ∈ Cmin . In view of (2.100), (2.103), (2.104), (2.427), (2.456), and (2.471), we apply Lemma 2.21 with K0 = 3K1 , L0 = L1 , x = xt , y = xt+1 , and ξ = ξt and with arbitrary sufficiently small δf , δC > 0 and obtain that for all integers t ≥ n0 , ¯ 2 xt+1 − z2 ≤ xt − z2 − 2−1 αt  + 25αt2 (L1 + L) ≤ xt − z2 − 4−1 αt .

Theorem 2.17 is proved.

2.20 Proof of Theorem 2.18 By (2.10), (2.105), (2.106), and (2.108), for each integer t ≥ 0,

2.20 Proof of Theorem 2.18

83

xt − xt+1  = xt − Pt (xt − αt ξt −1 ξt ) ≤ αt , ∞ 

xt − xt+1  ≤

t=0

∞ 

(2.478)

αt < ∞.

t=0

Thus {xt }∞ t=0 is a Cauchy sequence, and there exists x∗ ∈ lim xt ∈ C.

(2.479)

t→∞

By (2.106) and (2.478), for all integers t ≥ 0, xt  ≤ x0  +

t 

αi ≤ K1 +

i=0

∞ 

αi .

(2.480)

i=0

Assume that x∗ ∈ Cmin .

(2.481)

In view of (2.479) and (2.481), there exists  > 0 such that f (x∗ ) > inf(f, C) + .

(2.482)

In view of (2.479) and (2.482), there exist a natural number n0 such that for all integers n ≥ n0 , f (xn ) > inf(f, C) + ,

(2.483)

αn ≤ 16−1 L¯ −1 .

(2.484)

z ∈ Cmin

(2.485)

Fix

(2.480), and (2.483)–(2.485), and t ≥ n0 be an integer. In view of (2.107), (2.108),  we apply Lemma 2.24 with x¯ = z, K0 = K1 + ∞ α i=0 i , α = αt , x = xt , y = xt+1 , and ξ = ξt and with arbitrary sufficiently small δf , δC > 0 and obtain that ¯ −1 αt  + 2αt2 xt+1 − z2 ≤ xt − z2 − (4L) ¯ −1 αt . ≤ xt − z2 − (4L) Theorem 2.18 is proved.

Chapter 3

Extensions

In this chapter we study the projected subgradient method for nonsmooth convex constrained optimization problems in a Hilbert space. For these problems, an objective function is defined on an open convex set and a set of admissible points is not necessarily convex. We generalize some results of Chapter 2 obtained in the case when an objective function is defined on the whole Hilbert space.

3.1 Optimization Problems on Bounded Sets Let (X, ·, ·) be a Hilbert space with an inner product ·, · which induces a complete norm  · . We use the notation and definitions introduced in Chapter 2. Let C be a closed nonempty subset of the space X and U be an open convex subset of X such that C ⊂ U.

(3.1)

C ⊂ BX (0, M),

(3.2)

Suppose that L, M > 0,

and that a convex function f : U → R 1 satisfies |f (u1 ) − f (u2 )| ≤ Lu1 − u2  for all u1 , u2 ∈ U.

(3.3)

For each point x ∈ U and each positive number , let ∂f (x) = {l ∈ X : f (y) − f (x) ≥ l, y − x for all y ∈ U }

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 A. J. Zaslavski, The Projected Subgradient Algorithm in Convex Optimization, SpringerBriefs in Optimization, https://doi.org/10.1007/978-3-030-60300-7_3

(3.4)

85

86

3 Extensions

and let ∂ f (x) = {l ∈ X : f (y) − f (x) ≥ l, y − x −  for all y ∈ U }.

(3.5)

Denote by M the set of all mappings P : X → C such that P z = z for all z ∈ C, P x − z ≤ x − z for all x ∈ X and all z ∈ C.

(3.6)

inf(f, C) = inf{f (z) : z ∈ C}.

(3.7)

Define

It is clear that inf(f, C) is finite. Set Cmin = {x ∈ C : f (x) = inf(f, C)}.

(3.8)

For all P ∈ M set P 0 x = x, x ∈ X. We assume that Cmin = ∅. In view of (3.3), for each x ∈ U , ∂f (x) ⊂ BX (0, L).

(3.9)

Proposition 3.1 Assume that , r > 0 and that x ∈ U , BX (x, r) ⊂ U. Then ∂ f (x) ⊂ BX (0, L + r −1 ). Proof Let ξ ∈ ∂ f (x). By (3.5) and (3.10), for every h ∈ BX (0, 1), Lr ≥ Lrh ≥ f (x + rh) − f (x) ≥ ξ, rh − , ξ, h ≤ L + /r. This implies that ξ  ≤ L + /r. Proposition 3.1 is proved. In this chapter we prove the following two results.

(3.10)

3.1 Optimization Problems on Bounded Sets

87

T −1 Theorem 3.2 Assume that δf , δC ∈ (0, 1], T ≥ 1 is an integer, {αt }t=0 ⊂ (0, 1], T −1 ⊂ M, {Pi }i=0

(3.11)

T −1 {xi }Ti=0 ⊂ U, {ξi }i=0 ⊂ X,

x0  ≤ M + 1,

(3.12)

BX (ξi , δf ) ∩ ∂f (xi ) = ∅,

(3.13)

xi+1 − Pi (xi − αi ξi ) ≤ δC .

(3.14)

and that for i = 0, . . . , T − 1,

Then ⎛⎛

T −1

⎜ min{f (xt ) : t = 0, . . . , T − 1} − inf(f, C), f ⎝⎝

⎞−1 αi ⎠

i=0

⎛ ≤⎝

T −1

⎞−1 αj ⎠

j =0

T −1

T −1

⎞ ⎟ αt xt ⎠ − inf(f, C)

t=0

αt (f (xt ) − inf(f, C))

t=0

⎛

T −1

≤ 2−1 (2M + 1)2 ⎝

⎞−1

T −1

+ T δC ⎝

⎞⎛

+ 2−1 L2 ⎝

αt2 ⎠ ⎝

T −1

αt ⎠

t=0

⎛

⎛

t=0

T −1

⎞−1 αt ⎠

t=0

⎞−1 αt ⎠

(2M + L + 3) + δf (2M + L + 2).

(3.15)

t=0

Theorem 3.3 Assume that r > 0, BX (z, 2r) ⊂ U for all z ∈ C,

(3.16)

T −1 Δ > 0, δf , δC ∈ (0, 1], δC ≤ r, T ≥ 1 is an integer, {αt }t=0 ⊂ (0, 1], T −1 ⊂ M, {Pi }i=0

(3.17)

T −1 ⊂ X, {xi }Ti=0 ⊂ U, {ξi }i=0

x0  ≤ M + 1,

(3.18)

BX (x0 , r) ⊂ U,

(3.19)

88

3 Extensions

and that for i = 0, . . . , T − 1, BX (ξi , δf ) ∩ ∂Δ f (xi ) = ∅,

(3.20)

xi+1 − Pi (xi − αi ξi ) ≤ δC .

(3.21)

Then ⎛ ⎞ −1 T −1 T −1  αi αt xt ⎠ − inf(f, C) min{f (xt ) : t = 0, . . . , T − 1} − inf(f, C), f ⎝ i=0

⎞−1

⎛

T −1

≤⎝

T −1

αj ⎠

j =0 −1

≤2

t=0

αt (f (xt ) − inf(f, C))

t=0

(2M + 1)

2

T −1 

−1 αt

−1

+Δ+2

(L + Δr

−1 2

)

t=0

+ T δC

T −1 

T −1 

αt2

 T −1 

t=0

−1 αt

t=0

−1 αt

(2M + L + 3 + Δr −1 ) + δf (2M + L + 2 + Δr −1 ).

t=0

(3.22) Note that (3.15) is a particular case of (3.22) with Δ = 0. Theorems 3.2 and 3.3 are new. They are proved in Sections 3.4 and 3.5, respectively. Let T ≥ 1 be an integer and A > 0 be We are interested in an optimal  given. −1 choice of αt , t = 0, . . . , T − 1 satisfying Tt=0 αt = A which minimizes the righthand side of (3.22). By Lemma 2.3 of [75], αt = α = T −1 A, t = 0, . . . , T − 1. In this case the right-hand side of (3.22) is 2−1 (2M + 1)2 T −1 α −1 + Δ + 2−1 (L + Δr −1 )2 α + δC α −1 (2M + L + 3 + Δr −1 ) + δf (2M + L + 2 + Δr −1 ). Now we can make the best choice of the step-size α > 0. Since T can be arbitrarily large, we need to minimize the function δC α −1 (2M + L + 3 + Δr −1 ) + 2−1 (L + Δr −1 )2 α, α > 0 which has a minimizer α = (L + Δr −1 )−1 (2δC (2M + L + 3 + Δr −1 ))1/2 .

3.2 An Auxiliary Result for Theorem 3.2

89

With this choice of α the right-hand side of (3.22) is 2−1 (2M + 1)2 T −1 (L + Δr −1 )(2δC (2M + L + 3 + Δr −1 ))1/2 + Δ + (L + Δr −1 )(2−1 δC (2M + L + 3 + Δr −1 ))1/2 + δf (2M + L + 2 + Δr −1 ) + 2−1 (L + Δr −1 )(2δC (2M + L + 3 + Δr −1 ))1/2 . Now we should make the best choice of T . It is clear that T should be at the same 1/2 order as δC−1 . In this case the right-hand side of (3.22) does not exceed c1 δC + Δ + δf (2M + L + 2 + Δr −1 ), where c1 > 0 is a constant.

3.2 An Auxiliary Result for Theorem 3.2 Lemma 3.4 Let P ∈ M,

(3.23)

x ∈ U ∩ BX (0, M + 1),

(3.24)

BX (ξ, δf ) ∩ ∂f (x) = ∅,

(3.25)

y ∈ BX (P (x − αξ ), δC ).

(3.26)

α ∈ (0, 1], δf , δC ∈ (0, 1]

let ξ ∈ X satisfy

and let

Then for each z ∈ C, 2α(f (x) − f (z)) ≤ x − z2 − y − z2 + α 2 L2 + 2δC (2M + L + 3) + 2αδf (2M + L + 2). Proof Let z ∈ C. By (3.25), there exists

(3.27)

90

3 Extensions

v ∈ ∂f (x)

(3.28)

ξ − v ≤ δf .

(3.29)

v ≤ L.

(3.30)

ξ  ≤ L + 1.

(3.31)

such that

In view of (3.9) and (3.28),

By (3.29) and (3.30),

It follows from (3.2), (3.24), and (3.27)–(3.30) that x − αξ − z2 = x − αv + (αv − αξ ) − z2 ≤ x − αv − z2 + α 2 v − ξ 2 + 2αv − ξ, x − αv − z ≤ x − αv − z2 + α 2 δf2 + 2αδf x − αv − z ≤ x − αv − z2 + α 2 δf2 + 2αδf (2M + L + 1) ≤ x − z2 − 2αx − z, v + α 2 v2 + α 2 δf2 + 2αδf (2M + 1 + L) ≤ x − z2 − 2αx − z, v + α 2 L2 + α 2 δf2 + 2αδf (2M + 1 + L).

(3.32)

In view of (3.28), v, z − x ≤ f (z) − f (x).

(3.33)

By (3.32) and (3.33), x − αξ − z2 ≤ x − z2 + 2α(f (z) − f (x)) + α 2 L2 + 2αδf (2M + 2 + L).

(3.34)

It follows from (3.2), (3.24), (3.27), and (3.31) that x − αξ − z ≤ 2M + 2 + L.

(3.35)

3.3 An Auxiliary Result for Theorem 3.3

91

By (3.26), (3.27), (3.34), and (3.35), y − z2 = y − P (x − αξ ) + P (x − αξ ) − z2 ≤ y − P (x − αξ )2 + 2y − P (x − αξ )P (x − αξ ) − z + P (x − αξ ) − z2 ≤ δC2 + 2δC x − αξ − z + x − αξ − z2 ≤ δC2 + 2δC (2M + L + 2) + x − z2 + 2α(f (z) − f (x)) + α 2 L2 + 2αδf (2M + 2 + L). This implies that 2α(f (x) − f (z)) ≤ x − z2 − y − z2 + α 2 L2 + 2δC (2M + L + 3) + 2αδf (2M + L + 2). Lemma 3.4 is proved.

3.3 An Auxiliary Result for Theorem 3.3 Lemma 3.5 Let P ∈ M, α ∈ (0, 1], δf , δC ∈ (0, 1], r, Δ > 0, x ∈ U ∩ BX (0, M + 1),

(3.36)

BX (x, r) ⊂ U,

(3.37)

BX (ξ, δf ) ∩ ∂Δ f (x) = ∅,

(3.38)

y ∈ BX (P (x − αξ ), δC ).

(3.39)

ξ ∈ X satisfy

and let

Then for each z ∈ C,

92

3 Extensions

2α(f (x) − f (z)) ≤ x − z2 − y − z2 + 2αΔ + δC2 + +2δC (2M + L + 2 + Δr −1 ) + α 2 (L + Δr −1 )2 + 2αδf (2M + L + 2 + Δr −1 ). Proof Let z ∈ C.

(3.40)

v ∈ ∂Δ f (x)

(3.41)

ξ − v ≤ δf .

(3.42)

By (3.38), there exists

such that

Proposition 3.1, (3.36), and (3.37) imply that ∂Δ f (x) ⊂ BX (0, L + Δr −1 ).

(3.43)

In view of (3.41) and (3.43), v ≤ L + Δr −1 .

(3.44)

ξ  ≤ L + Δr −1 + 1.

(3.45)

By (3.42) and (3.44),

It follows from (3.2), (3.36), (3.40), (3.42), and (3.44) that x − αξ − z2 = x − αv + (αv − αξ ) − z2 ≤ x − αv − z2 + α 2 v − ξ 2 + 2αv − ξ, x − αv − z ≤ x − αv − z2 + α 2 δf2 + 2αδf x − αv − z ≤ x − αv − z2 + α 2 δf2 + 2αδf (2M + L + 1 + Δr −1 ) ≤ x − z2 − 2αx − z, v + α 2 v2 + α 2 δf2 + 2αδf (2M + 1 + L + Δr −1 )

3.3 An Auxiliary Result for Theorem 3.3

93

≤ x − z2 − 2αx − z, v + α 2 (L + Δr −1 )2 + α 2 δf2 + 2αδf (2M + 1 + L + Δr −1 ).

(3.46)

In view of (3.41), v, z − x ≤ f (z) − f (x) + Δ.

(3.47)

By (3.46) and (3.47), x − αξ − z2 ≤ x − z2 + 2α(f (z) − f (x)) + 2αΔ + α 2 (L + Δr −1 )2 + 2αδf (2M + 2 + L + Δr −1 ).

(3.48)

It follows from (3.27), (3.36), (3.40), and (3.45) that x − αξ − z ≤ 2M + 2 + L + Δr −1 .

(3.49)

By (3.6), (3.39), (3.40), (3.48), and (3.49), y − z2 = y − P (x − αξ ) + P (x − αξ ) − z2 ≤ y − P (x − αξ )2 + 2y − P (x − αξ )P (x − αξ ) − z + P (x − αξ ) − z2 ≤ δC2 + 2δC x − αξ − z + x − αξ − z2 ≤ δC2 + 2δC (2M + L + 2 + Δr −1 ) + x − z2 + 2α(f (z) − f (x)) + 2αΔ + α 2 (L + Δr −1 )2 + 2αδf (2M + 2 + L + Δr −1 ). This implies that 2α(f (x) − f (z)) ≤ x − z2 − y − z2 + 2αΔ + δC2 + α 2 (L + Δr −1 )2 + 2δC (2M + L + 2 + Δr −1 ) + 2αδf (2M + L + 2 + Δr −1 ). Lemma 3.5 is proved.

94

3 Extensions

3.4 Proof of Theorem 3.2 Fix x¯ ∈ Cmin . For every t = 0, . . . , T − 1 we apply Lemma 3.4 with P = Pt , x = xt , y = xt+1 , ξ = ξt , α = αt and obtain that ¯ ≤ xt − x ¯ 2 − xt+1 − x ¯ 2 2αt (f (xt ) − f (x)) + αt2 L2 + 2δC (2M + L + 3) + 2αt δf (2M + L + 2). Together with (3.2) and (3.12), this implies that T −1

αt (f (xt ) − inf(f, C)) ≤ 2−1

t=0

T −1

(xt − x ¯ 2 − xt+1 − x ¯ 2)

t=0

+ 2−1

T −1

αt2 L2 + T δC (2M + L + 3) + δf (2M + L + 2)

T −1

t=0

αt

t=0

≤ 2−1 (2M + 1)2 + 2−1

T −1

αt2 L2 + T δC (2M + L + 3) + δf (2M + L + 2).

t=0

This implies that ⎞ ⎛ −1 T −1 T −1  min{f (xt ) : t = 0, . . . , T − 1} − inf(f, C), f ⎝ αi αt xt ⎠ − inf(f, C) i=0

⎛

T −1

≤⎝

⎞−1 αj ⎠

j =0

T −1

αt (f (xt ) − inf(f, C))

t=0

≤ 2−1 (2M + 1)2

T −1  t=0

+ T δC

t=0

T −1 

−1 αt

+ 2−1 L2

T −1  t=0

αt2

 n 

−1 αt

t=0

−1 αt

t=0

Theorem 3.2 is proved.

(2M + L + 3) + δf (2M + L + 2).

3.5 Proof of Theorem 3.3

95

3.5 Proof of Theorem 3.3 Fix x¯ ∈ Cmin . By (3.2), (3.17), and (3.21), for all t = 0, . . . , T , xt  ≤ M + 1. In view of (3.19), BX (x0 , r) ⊂ U. It follows from (3.16), (3.17), and (3.21) that for all integers t = 0, . . . , T − 1, BX (xt+1 , r) ⊂ BX (Pt (xt − αt ξt ), r + δC ) ⊂ BX (Pt (xt − αt ξt ), 2r) ⊂ U.

For every t = 0, . . . , T − 1 in view of the relation above, we apply Lemma 3.5 with P = Pt , x = xt , y = xt+1 , α = αt , ξ = ξt and obtain that ¯ 2 − xt+1 − x ¯ 2 2α(f (xt ) − inf(f, C)) ≤ xt − x + 2αt Δ + δC2 + αt2 (L + Δr −1 )2 + 2δC (2M + L + 2 + Δr −1 ) + 2αt δf (2M + L + 2 + Δr −1 ). The relation above implies that T −1

αt (f (xt ) − inf(f, C)) ≤ 2−1

t=0

T −1

(xt − x ¯ 2 − xt+1 − x ¯ 2)

t=0

+

T −1

αt Δ + 2−1

t=0

T −1

αt2 (L + Δr −1 )2 + T δC (2M + L + 3 + Δr −1 )

t=0

+ δf (2M + L + 2 + Δr −1 )

T −1

αt

t=0

≤ 2−1 x0 − x ¯ 2+

T −1 t=0

αt Δ + 2−1

T −1 t=0

αt2 (L + Δr −1 )2

96

3 Extensions

+ T δC (2M + L + 3 + Δr −1 ) + δf (2M + L + 2 + Δr −1 )

T −1

αt .

t=0

Together with (3.2) and (3.18) this implies that ⎛⎛ ⎜ min{f (xt ) : t = 0, . . . , T − 1} − inf(f, C), f ⎝⎝

T −1

⎞−1 αi ⎠

i=0

⎛ ≤⎝

T −1

⎞−1 αj ⎠

j =0

T −1

T −1

⎞ ⎟ αt xt ⎠ − inf(f, C)

t=0

αt (f (xt ) − inf(f, C))

t=0

⎛

T −1

≤ 2−1 (2M + 1)2 ⎝

⎞−1 αt ⎠

⎛

T −1

+ T δC ⎝

⎞⎛ ⎞−1 n  αt2 ⎠ ⎝ αt ⎠

t=0

t=0

+ Δ + 2−1 (L + Δr −1 )2 ⎝

t=0

⎛

T −1

⎞−1 αt ⎠

(2M + L + 3 + Δr −1 ) + δf (2M + L + 2 + Δr −1 ).

t=0

Theorem 3.3 is proved.

3.6 Optimization on Unbounded Sets Let (X, ·, ·) be a Hilbert space with an inner product ·, · which induces a complete norm  · . Let C be a closed nonempty subset of the space X, U be an open convex subset of X such that C ⊂ U, and f : U → R 1 be a convex function which is Lipschitz on all bounded subsets of U. For each point x ∈ U and each positive number , let ∂f (x) = {l ∈ X : f (y) − f (x) ≥ l, y − x for all y ∈ U } and let ∂ f (x) = {l ∈ X : f (y) − f (x) ≥ l, y − x −  for all y ∈ U }.

3.6 Optimization on Unbounded Sets

97

Assume that lim

x∈U,x→∞

f (x) = ∞.

(3.50)

It means that for each M0 > 0 there exists M1 > 0 such that if a point x ∈ U 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 (3.50) that inf(f, C) is finite. Set Cmin = {x ∈ C : f (x) = inf(f, C)}.

(3.51)

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. We assume that Cmin = ∅. It is clear that Cmin is a closed subset of C. Fix θ0 ∈ C.

(3.52)

U0 = {x ∈ U : f (x) ≤ f (θ0 ) + 4}.

(3.53)

Set

In view of (3.50) there exists a number K¯ > 1 such that ¯ U0 ⊂ BX (0, K).

(3.54)

Since the function f is Lipschitz on all bounded subsets of U , there exists a number L¯ > 1 such that ¯ 1 − z2  for all z1 , z2 ∈ U ∩ BX (0, K¯ + 4). |f (z1 ) − f (z2 )| ≤ Lz

(3.55)

Denote by MC the set of all mappings P : X → C such that P z = z for all z ∈ C, P z − x ≤ z − x for all x ∈ C and all z ∈ X.

(3.56) (3.57)

98

3 Extensions

We prove the following two theorems. Theorem 3.6 Assume that ¯ K1 ≥ K¯ + 4, L1 ≥ L,

(3.58)

δf , δC ∈ (0, 1], |f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, 3K1 + 2) ∩ U, ¯ −2 ) α ∈ (0, (1 + L)

(3.59) (3.60)

and that δf (K¯ + 3K1 + 2 + L1 ) ≤ α, δC (K¯ + 3K1 + L1 + 3) ≤ α.

(3.61)

T −1 ⊂ MC , {Pt }t=0

(3.62)

T −1 ⊂ X, {xt }Tt=0 ⊂ U, {ξt }t=0

(3.63)

x0  ≤ K1 ,

(3.64)

BX (x0 , δC ) ∩ C = ∅

(3.65)

BX (ξt , δf ) ∩ ∂f (xt ) = ∅,

(3.66)

xt+1 − Pt (xt − αξt ) ≤ δC .

(3.67)

Let T ≥ 2 be an integer

and that for t = 0, . . . , T − 1,

Then xt  ≤ 2K¯ + K1 , t = 0, . . . , T and  min{f (xt ) : t = 0, . . . , T − 1} − inf(f, C), f

T

−1

T −1 i=0

 xi

− inf(f, C)

3.6 Optimization on Unbounded Sets

≤ T −1

T −1

99

f (xi ) − inf(f, C)

i=0

¯ 2 + L21 α + α −1 δC (K¯ + 3K1 + L1 + 3) ≤ (2T α)−1 (K1 + K) + δf (3K1 + K¯ + L1 + 2). Theorem 3.7 Assume that ¯ r0 ∈ (0, 1], K1 ≥ K¯ + 4, L1 ≥ L, BX (z, r0 ) ⊂ U, z ∈ C,

(3.68)

|f (z1 ) − f (z2 )| ≤ L1 z1 − z2  for all z1 , z2 ∈ BX (0, 3K1 + 1) ∩ U,

(3.69)

Δ ∈ (0, r0 ], δf , δC ∈ (0, 2−1 r0 ], α ∈ (0, (L¯ + 3)−2 ],

(3.70)

and that δf (3K¯ + K1 + 4 + L1 ) ≤ α,

(3.71)

δC (3K¯ + K1 + L1 + 2) ≤ α.

(3.72)

T −1 ⊂ MC , {Pt }t=0

(3.73)

T −1 ⊂ X, {xt }Tt=0 ⊂ U, {ξt }t=0

(3.74)

x0  ≤ K1 ,

(3.75)

BX (x0 , δC ) ∩ C = ∅

(3.76)

BX (ξt , δf ) ∩ ∂Δ f (xt ) = ∅,

(3.77)

xt+1 − Pt (xt − αξt ) ≤ δC .

(3.78)

Let T ≥ 2 be an integer

and that for i = 1, . . . , T − 1,

Then xt  ≤ 2K¯ + K1 , t = 0, . . . , T and

100

3 Extensions

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

T

−1

T −1

 xi

− inf(f, C)

i=0

≤T

−1

T −1

f (xi ) − inf(f, C)

i=0

¯ 2 + (L1 + 2)2 α ≤ (2T α)−1 (K1 + K) + α −1 δC (3K¯ + K1 + L1 + 2) + Δ + δf (K1 + 3K¯ + L1 + 4).

3.7 Auxiliary Results Lemma 3.8 Let K0 , L0 > 0, |f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, K0 + 1) ∩ U, x ∈ BX (0, K0 ) ∩ U, v ∈ ∂f (x).

(3.79) (3.80)

Then v ≤ L0 . Proof In view of (3.80), for all u ∈ U , f (u) − f (x) ≥ v, u − x.

(3.81)

There exists r ∈ (0, 1) such that BX (x, r) ⊂ U.

(3.82)

BX (x, r) ⊂ BX (0, K0 + 1).

(3.83)

By (3.80) and (3.82),

It follows from (3.70), (3.80), (3.82), and (3.83) that for all h ∈ BX (0, 1), x + rh ∈ U ∩ BX (0, K0 + 1), v, rh ≤ f (x + rh) − f (x) ≤ L0 rh ≤ L0 r, v, h ≤ L0 .

3.7 Auxiliary Results

101

Therefore v ≤ L0 . Lemma 3.8 is proved. Lemma 3.9 Let K0 , L0 > 0, r ∈ (0, 1], Δ > 0, |f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, K0 + 1) ∩ U,

(3.84)

x ∈ BX (0, K0 ) ∩ U,

(3.85)

BX (x, r) ⊂ U.

(3.86)

Then ∂Δ f (x) ⊂ BX (0, L0 + Δr −1 ). Proof Let ξ ∈ ∂Δ f (x).

(3.87)

BX (x, r) ⊂ U ∩ BX (0, K0 + 1).

(3.88)

By (3.85) and (3.86),

In view of (3.88), for each h ∈ BX (0, 1), x + rh ∈ U ∩ BX (0, K0 + 1). Together with (3.84) and (3.87) this implies that L0 r ≥ L0 rh ≥ f (x + rh) − f (x) ≥ ξ, rh − Δ, L0 + Δr −1 ≥ ξ, h. This implies that ξ  ≤ L0 + Δr −1 . Lemma 3.9 is proved. ¯ L0 > 0, Lemma 3.10 Let P ∈ Mc , K0 ≥ K, |f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, K0 + 2) ∩ U,

(3.89)

α ∈ (0, 1], δf , δC ∈ (0, 1], x ∈ U ∩ BX (0, K0 + 1),

(3.90)

102

3 Extensions

ξ ∈ X satisfy BX (ξ, δf ) ∩ ∂f (x) = ∅,

(3.91)

y ∈ BX (P (x − αξ ), δC ) ∩ U.

(3.92)

and let

Then for each z ∈ C satisfying f (z) ≤ f (θ0 ) + 4, 2α(f (x) − f (z)) ≤ x − z2 − y − z2 + 2δC (K0 + K¯ + L0 + 3) + α 2 L20 + 2αδf (K0 + K¯ + L0 + 2). Proof Let z∈C

(3.93)

f (z) ≤ f (θ0 ) + 4.

(3.94)

satisfy

In view of (3.52), (3.53), and (3.94), ¯ z ≤ K.

(3.95)

v ∈ ∂f (x)

(3.96)

ξ − v ≤ δf .

(3.97)

By (3.91), there exists

such that

In view of (3.89), (3.90), and (3.96), v ≤ L0 .

(3.98)

ξ  ≤ L0 .

(3.99)

By (3.97) and (3.98),

It follows from (3.90), (3.95), (3.97), and (3.98) that

3.7 Auxiliary Results

103

x − αξ − z2 = x − αv + (αv − αξ ) − z2 ≤ x − αv − z2 + α 2 v − ξ 2 + 2αv − ξ, x − αv − z ≤ x − αv − z2 + α 2 δf2 + 2αδf x − αv − z ≤ x − αv − z2 + α 2 δf2 + 2αδf (K0 + K¯ + L0 + 1) ≤ x − z2 − 2αx − z, v + α 2 v2 + α 2 δf2 + 2αδf (K0 + K¯ + L0 + 1) ≤ x − z2 − 2αx − z, v + α 2 L20 + α 2 δf2 + 2αδf (K0 + K¯ + L0 + 1).

(3.100)

In view of (3.46), v, z − x ≤ f (z) − f (x).

(3.101)

By (3.100) and (3.101), x − αξ − z2 ≤ x − z2 + 2α(f (z) − f (x)) + α 2 L20 + α 2 δf2 + 2αδf (K0 + K¯ + L0 + 1).

(3.102)

It follows from (3.90), (3.95), and (3.99) that x − αξ − z ≤ K0 + 2 + K¯ + L0 .

(3.103)

By (3.56), (3.57), (3.92), and (3.103), y − z2 = y − P (x − αξ ) + P (x − αξ ) − z2 ≤ y − P (x − αξ )2 + 2y − P (x − αξ )P (x − αξ ) − z + P (x − αξ ) − z2 ≤ δC2 + 2δC (K0 + K¯ + L0 + 2) + x − αξ − z2 ≤ δC2 + 2δC (K0 + K¯ + L0 + 2) + x − z2 + 2α(f (z) − f (x)) + α 2 L20 + α 2 δf2 + 2αδf (K0 + K¯ + L0 + 1). This implies that

104

3 Extensions

2α(f (x) − f (z)) ≤ x − z2 − y − z2 + 2δC (K0 + K¯ + L0 + 3) + α 2 L20 + 2αδf (K0 + K¯ + L0 + 2). Lemma 3.10 is proved. ¯ L0 > 0, Lemma 3.11 Let P ∈ Mc , K0 ≥ K, |f (z1 ) − f (z2 )| ≤ L0 z1 − z2  for all z1 , z2 ∈ BX (0, K0 + 2) ∩ U,

(3.104)

α ∈ (0, 1], δf , δC ∈ (0, 1], Δ > 0, r ∈ (0, 1], x ∈ U ∩ BX (0, K0 + 1),

(3.105)

BX (x, r) ⊂ U,

(3.106)

BX (ξ, δf ) ∩ ∂Δ f (x) = ∅,

(3.107)

y ∈ BX (P (x − αξ ), δC ) ∩ U.

(3.108)

ξ ∈ X satisfy

and let

Then for each z ∈ C satisfying f (z) ≤ f (θ0 ) + 4, 2α(f (x) − f (z)) ≤ x − z2 − y − z2 + 2δC (K0 + K¯ + L0 + 3 + Δr −1 ) + 2αΔ + α 2 (L0 + Δr −1 )2 + 2αδf (K0 + K¯ + L0 + 2 + Δr −1 ). Proof Let z∈C

(3.109)

f (z) ≤ f (θ0 ) + 4.

(3.110)

satisfy

In view of (3.53), (3.54), (3.109), and (3.110), ¯ z ≤ K. By (3.107), there exists

(3.111)

3.7 Auxiliary Results

105

v ∈ ∂Δ f (x)

(3.112)

ξ − v ≤ δf .

(3.113)

such that

Lemma 3.9 and (3.104)–(3.106) imply that ∂Δ f (x) ⊂ BX (0, L0 + Δr −1 ).

(3.114)

In view of (3.112) and (3.114), v ≤ L0 + Δr −1 .

(3.115)

ξ  ≤ L0 + Δr −1 + 1.

(3.116)

By (3.113) and (3.115),

It follows from (3.105), (3.111), (3.113), and (3.115) that x − αξ − z2 = x − αv + (αv − αξ ) − z2 ≤ x − αv − z2 + α 2 v − ξ 2 + 2αv − ξ, x − αv − z ≤ x − αv − z2 + α 2 δf2 + 2αδf x − αv − z ≤ x − αv − z2 + α 2 δf2 + 2αδf (K0 + K¯ + L0 + 1 + Δr −1 ) ≤ x − z2 − 2αx − z, v + α 2 v2 + α 2 δf2 + 2αδf (K0 + K¯ + L0 + 1 + Δr −1 ) ≤ x − z2 − 2αx − z, v + α 2 (L0 + Δr −1 )2 + α 2 δf2 + 2αδf (K0 + K¯ + L0 + 1 + Δr −1 ).

(3.117)

In view of (3.112), v, z − x ≤ f (z) − f (x) + Δ.

(3.118)

By (3.105), (3.111), and (3.116), x − αξ − z ≤ K0 + K¯ + 2 + L0 + Δr −1 . It follows from (3.117) that

(3.119)

106

3 Extensions

x − αξ − z2 ≤ x − z2 + 2α(f (z) − f (x)) + 2αΔ + α 2 (L0 + Δr −1 )2 + 2αδf (K0 + K¯ + L0 + 2 + Δr −1 ).

(3.120)

It follows from (3.56), (3.57), (3.109), (3.119), and (3.120) that y − z2 = y − P (x − αξ ) + P (x − αξ ) − z2 ≤ y − P (x − αξ )2 + 2y − P (x − αξ )P (x − αξ ) − z + P (x − αξ ) − z2 ≤ δC2 + 2δC (K0 + 2 + K¯ + L0 + Δr −1 ) + P (x − αξ ) − z2 ≤ 2δC (K0 + 3 + K¯ + L0 + Δr −1 ) + x − z2 + 2α(f (z) − f (x)) + 2αΔ + α 2 (L0 + Δr −1 )2 + 2αδf (K0 + K¯ + L0 + 2 + Δr −1 ). This implies that 2α(f (x) − f (z)) ≤ x − z2 − y − z2 + 2δC (K0 + 3 + K¯ + L0 + Δr −1 ) + 2αΔ + α 2 (L0 + Δr −1 )2 + 2αδf (K0 + K¯ + L0 + 2 + Δr −1 ). Lemma 3.11 is proved.

3.8 Proof of *Theorem 3.6 Fix x¯ ∈ Cmin .

(3.121)

¯ x ¯ ≤ K.

(3.122)

¯ ≤ K¯ + K1 . x0 − x

(3.123)

By (3.52), (3.54), and (3.121),

In view of (3.64) and (3.122),

3.8 Proof of *Theorem 3.6

107

By induction we show that for all i = 0, . . . , T , ¯ ≤ K¯ + K1 . xi − x

(3.124)

In view of (3.123), inequality (3.124) holds for i = 0. Assume that i ∈ {0, . . . , T − 1} and that (3.124) is valid. There are two cases: f (xi ) ≤ inf(f, C) + 4;

(3.125)

f (xi ) > inf(f, C) + 4.

(3.126)

Assume that (3.125) holds. In view of (3.52)–(3.54) and (3.125), ¯ xi  ≤ K.

(3.127)

Lemma 3.8, (3.55), and (3.127) imply that ¯ ∂f (xi ) ⊂ BX (0, L).

(3.128)

ξi  ≤ L¯ + 1.

(3.129)

By (3.66) and (3.128),

It follows from (3.56), (3.57), (3.63), (3.67), (3.121), (3.122), (3.127), and (3.129) that ¯ ≤ xi+1 − Pi (xi − αξi ) + Pi (xi − αξi ) − x ¯ xi+1 − x ≤ δC + xi − αξi − x ¯ ¯ ≤ δC + 2K¯ + α(L¯ + 1) ≤ 2K¯ + 2 ≤ K1 + K. Assume that (3.126) holds. In view of (3.66), (3.121), (3.122), and (3.124), we apply Lemma 3.10 with P = Pi , K0 = 3K1 , L0 = L1 , ξ = ξi , x = xi , y = xi+1 , z = x¯ and obtain that 2α(f (xi ) − f (x)) ¯ ≤ xi − x ¯ 2 − xi+1 − x ¯ 2 + 2δC (K¯ + 3K1 + L1 + 3) + L21 α 2 + 2αδf (3K1 + K¯ + L1 + 2).

Together with (3.60), (3.61), and (3.126), this implies that ¯ 2 ≤ xi − x ¯ 2 − 8α + 2δC (K¯ + 3K1 + L1 + 3) xi+1 − x

108

3 Extensions

+ L21 α 2 + 2αδf (3K1 + K¯ + L1 + 2) ≤ xi − x ¯ 2 − 7α + 4α, xi+1 − x ¯ ≤ xi − x ¯ ≤ K¯ + K1 .

Thus in both cases xi+1 − x ¯ ≤ K¯ + K1 and the assumption made for i also holds for i + 1 too. Therefore (3.124) holds for all i = 0, . . . , T . In view of (3.122) and (3.124), for all t = 0, . . . , T , xi  ≤ 2K¯ + K1 .

(3.130)

Let i ∈ {0, . . . , T − 1}. In view of (3.58), (3.59), (3.66), and (3.130), we apply Lemma 3.10 with P = Pi , K0 = 3K1 , L0 = L1 , x = xi , y = xi+1 , z = x, ¯ ξ = ξi and obtain that ¯ ≤ xi − x ¯ 2 − xi+1 − x ¯ 2 2α(f (xi ) − f (x)) + 2δC (K¯ + 3K1 + L1 + 3) + α 2 L21 + 2αδf (3K1 + K¯ + L1 + 2).

(3.131)

By (3.131), T −1

α(f (xi ) − f (x)) ¯ ≤ 2−1

i=0

T −1

(xi − x ¯ 2 − xi+1 − x ¯ 2)

i=0

+ T δC (K¯ + 3K1 + L1 + 3) + T α 2 L21 + T αδf (3K1 + K¯ + L1 + 2). Together with (3.121) and (3.123), the relation above implies that min{f (xt ) : t = 0, . . . , T − 1} − inf(f, C), f (T −1

T −1 i=0

≤T

−1

T −1 i=0

f (xi ) − inf(f, C)

xi ) − inf(f, C)

3.9 Proof of Theorem 3.7

109

¯ 2 + L21 α + α −1 δC (K¯ + 3K1 + L1 + 3) ≤ (2T α)−1 (K1 + K) + δf (3K1 + K¯ + L1 + 2). Theorem 3.6 is proved.

3.9 Proof of Theorem 3.7 Fix x¯ ∈ Cmin .

(3.132)

¯ x ¯ ≤ K.

(3.133)

¯ ≤ K¯ + K1 . x0 − x

(3.134)

By (3.52)–(3.54) and (3.132),

In view of (3.75) and (3.133),

By induction we show that for all i = 0, . . . , T , ¯ ≤ K¯ + K1 . xi − x

(3.135)

In view of (3.134), inequality (3.135) holds for i = 0. Assume that i ∈ {0, . . . , T − 1} and that (3.135) is valid. There are two cases: f (xi ) ≤ inf(f, C) + 4;

(3.136)

f (xi ) > inf(f, C) + 4.

(3.137)

Assume that (3.136) holds. In view of (3.52)–(3.54) and (3.136), ¯ xi  ≤ K.

(3.138)

It follows from (3.68), (3.76), and (3.78) that BX (xi , r0 /2) ⊂ U.

(3.139)

Lemma 3.9, (3.55), (3.138), and (3.139) imply that ∂Δ f (xi ) ⊂ BX (0, L¯ + 2Δr0−1 ).

(3.140)

110

3 Extensions

By (3.77) and (3.140), ξi  ≤ L¯ + 1 + 2Δr0−1 .

(3.141)

It follows from (3.56), (3.57), (3.70), (3.73), (3.78), (3.133), (3.138), and (3.141) that ¯ ≤ xi+1 − Pi (xi − αξi ) + Pi (xi − αξi ) − x ¯ xi+1 − x ≤ δC + xi − αξi − x ¯ ≤ δC + 2K¯ + α(L¯ + 1 + 2Δr0−1 ) ≤ 2K¯ + 1 + α(L¯ + 3) ≤ 2K + 2 ≤ K¯ + K1 . Assume that (3.137) holds. In view of (3.133) and (3.135), xi  ≤ 2K¯ + K1 .

(3.142)

By (3.68), (3.70), (3.76), and (3.78), equation (3.139) is true. In view of (3.69), (3.70), (3.73), (3.74), (3.77), (3.78), and (3.142), we apply Lemma 3.11 with P = Pi , K0 = 2K¯ + K1 , L0 = L1 , r = r0 /2, ξ = ξi , x = xi , y = xi+1 , z = x¯ and obtain that ¯ ≤ xi − x ¯ 2 − xi+1 − x ¯ 2 2α(f (xi ) − f (x)) +2δC (3K¯ + K1 + L1 + 3 + 2Δr0−1 ) + 2αΔ + (L1 + 2Δr0−1 )2 α 2 + 2αδf (K1 + 3K¯ + L1 + 2 + 2Δr0−1 ).

(3.143)

Together with (3.70)–(3.72), (3.135), and (3.127), this implies that ¯ 2 ≤ xi − x ¯ 2 − 8α + 2δC (3K¯ + K1 + L1 + 3 + 2Δr0−1 ) xi+1 − x + 2αΔ + (L1 + 2Δr0−1 )2 α 2 + 2αδf (K1 + 3K¯ + L1 + 2 + 2Δr0−1 ) ≤ xi − x ¯ 2 − 8α + 7α < xi − x ¯ 2, xi+1 − x ¯ ≤ xi − x ¯ ≤ K¯ + K1 . Thus in both cases ¯ ≤ K¯ + K1 xi+1 − x

3.9 Proof of Theorem 3.7

111

and the assumption made for i also holds for i + 1 too. Therefore (3.135) holds for all i = 0, . . . , T . In view of (3.133) and (3.135), for all t = 0, . . . , T , xi  ≤ 2K¯ + K1 .

(3.144)

Let i ∈ {0, . . . , T −1}. By (3.68), (3.70), (3.76), and (3.78), equation (3.139) is true. In view of (3.69), (3.74), (3.77), (3.78), (3.139), and (3.143), we apply Lemma 3.11 ¯ L0 = L1 , r = r0 /2, x = xi , y = xi+1 , z = x, ¯ ξ = ξi with P = Pi , K0 = K1 + 2K, and obtain that (3.143) is true. By (3.70) and (3.143), T −1

α(f (xi ) − f (x)) ¯ ≤ 2−1

i=0

T −1

(xi − x ¯ 2 − xi+1 − x ¯ 2)

i=0

+ T δC (3K¯ + K1 + L1 + 2) + αΔT + T α 2 (L1 + 2)2 + T αδf (K1 + 3K¯ + L1 + 4). Together with (3.135) the relation above implies that  min{f (xt ) : t = 0, . . . , T − 1} − inf(f, C), f

T

−1

T −1 i=0

≤ T −1

T −1

f (xi ) − inf(f, C)

i=0

¯ 2 + α −1 δC (3K¯ + K1 + L1 + 2) ≤ (2T α)−1 (K1 + K) + Δ + α(L1 + 2)2 + δf (K1 + 3K¯ + L1 + 2). Theorem 3.7 is proved.

 xi

− inf(f, C)

Chapter 4

Zero-Sum Games with Two Players

In this chapter we study an extension of the projected subgradient method for zerosum games with two players under the presence of computational errors. In our recent research [77], 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 computational errors for our algorithm, we find out what an approximate solution can be obtained and how many iterates one needs for this. In this chapter we generalize these results for an extension of the projected subgradient method, when instead of the projection on a feasible set it is used a quasi-nonexpansive retraction on this set.

4.1 Preliminaries and an Auxiliary Result 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

(4.1)

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 ), © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 A. J. Zaslavski, The Projected Subgradient Algorithm in Convex Optimization, SpringerBriefs in Optimization, https://doi.org/10.1007/978-3-030-60300-7_4

113

114

4 Zero-Sum Games with Two Players

D ⊂ BY (0, M2 ),

(4.2)

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

(4.3)

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

(4.4)

x∗ ∈ C and y∗ ∈ D

(4.5)

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

(4.6)

Let

satisfy

for each x ∈ C and each y ∈ D. The following result was obtained in [77]. Proposition 4.1 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 = ∅, 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

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 .

Let

4.1 Preliminaries and an Auxiliary Result

115

−1   T T  xT = ai at xt , i=0

 yT =

t=0

−1   T T ai at yt . i=0

t=0

Then BX ( xT , δC ) ∩ C = ∅, BY ( yT , δD ) ∩ D = ∅,   −1 T       T  at at f (xt , yt ) − f (x∗ , y∗ )    t=0 t=0  T  T −1  T    ≤ at max bt,1 , bt,2 t=0

t=0

t=0

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

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

 +

T 



−1 at

max

T 

t=0

t=0

bt,1 ,

T 

 bt,2

t=0

+ max{L1 δC , L2 δD }

and for each z ∈ C and each v ∈ D, f (z,  yT ) ≥ f ( xT ,  yT )  T −1  −2 at sup{φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]} t=0

116

4 Zero-Sum Games with Two Players

 −2

T 



−1 max

at

t=0

T 

bt,1 ,

t=0

T 

 bt,2

t=0

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

 +2

T 



−1 max

at

t=0

T 

bt,1 ,

t=0

T 

 bt,2

t=0

+ max{L1 δC , L2 δD }.

The following corollary was obtained in [77]. Corollary 4.2 Suppose that all the assumptions of Proposition 4.1 hold and that x˜ ∈ C, y˜ ∈ D satisfy  xT − x ˜ ≤ δC ,  yT − y ˜ ≤ δD . Then yT )| ≤ L1 δC + L2 δD |f (x, ˜ y) ˜ − f ( xT ,  and for each z ∈ C and each v ∈ D, f (z, y) ˜ ≥ f (x, ˜ y) ˜  T −1  −2 at sup {φ(s) : s ∈ [0, max{2M1 , 2M2 } + 1]} t=0

 −2

T 



−1 at

t=0

and f (x, ˜ v) ≤ f (x, ˜ y) ˜

max

T  t=0

bt,1 ,

T  t=0

 bt,2 − 4 max{L1 δC , L2 δD }

4.2 Zero-Sum Games on Bounded Sets

 +2

T 

117

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

at

t=0

 +2

T  t=0



−1 at

max

T 

bt,1 ,

t=0

T 

 bt,2 + 4 max{L1 δC , L2 δD }.

t=0

4.2 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 . For each concave function g : V → R 1 , each x ∈ V , and each  > 0, set ∂g(x) = {l ∈ Y : l, y − x ≥ g(y) − g(x) for all y ∈ V },

(4.7)

∂ g(x) = {l ∈ Y : l, y − x +  ≥ g(y) − g(x) for all y ∈ V }.

(4.8)

Clearly, for each x ∈ V and each  > 0, ∂g(x) = −(∂(−g)(x)),

(4.9)

∂ g(x) = −(∂ (−g)(x)).

(4.10)

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

(4.11)

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

(4.12)

118

4 Zero-Sum Games with Two Players

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

(4.13)

For each (ξ, η) ∈ U × V and each  > 0, set ∂x f (ξ, η) = {l ∈ X : f (y, η) − f (ξ, η) ≥ l, y − ξ  for all y ∈ U },

(4.14)

∂y f (ξ, η) = {l ∈ Y : l, y − η ≥ f (ξ, y) − f (ξ, η) for all y ∈ V },

(4.15)

∂x, f (ξ, η) = {l ∈ X : f (y, η) − f (ξ, η) +  ≥ l, y − ξ  for all y ∈ U },

(4.16)

∂y, f (ξ, η) = {l ∈ Y : l, y − η +  ≥ f (ξ, y) − f (ξ, η) for all y ∈ V }.

(4.17)

In view of properties (i) and (ii), (4.12), and (4.13), for each ξ ∈ U and each η ∈ V, ∅ = ∂x f (ξ, η) ⊂ BX (0, L1 ),

(4.18)

∅ = ∂y f (ξ, η) ⊂ BY (0, L2 ).

(4.19)

x∗ ∈ C and y∗ ∈ D

(4.20)

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

(4.21)

Let

satisfy

for each x ∈ C and each y ∈ D. Denote by MU the set of all mappings P : X → X such that P x = x, x ∈ C,

4.2 Zero-Sum Games on Bounded Sets

119

P x − z ≤ x − z for all x ∈ X and all z ∈ C and by MV the set of all mappings P : Y → Y such that P y = y, y ∈ D, P y − z ≤ y − z for all y ∈ Y and all z ∈ C. Let δf,1 , δf,2 , δC , δD ∈ (0, 1] and {αk }∞ 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 the 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 − Pt (xt − αt ξt ) ≤ δC , yt+1 − Qt (yt + αt ηt ) ≤ δD , where Pt ∈ MU , Qt ∈ MV . In this chapter we prove the following result. Theorem 4.3 Let δf,1 , δf,2 , δC , δD ∈ (0, 1], {αk }∞ k=0 ⊂ (0, ∞), {Pt }∞ t=0 ⊂ MU , Pt (X) = C, t = 0, 1, . . . ,

(4.22)

{Qt }∞ t=0 ⊂ MV , Qt (Y ) = D, t = 0, 1, . . . .

(4.23)

∞ ∞ ∞ 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 = ∅

(4.24)

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

(4.25)

ηt ∈ ∂y f (xt , yt ) + BY (0, δf,2 ),

(4.26)

xt+1 − Pt (xt − αt ξt ) ≤ δC

(4.27)

120

4 Zero-Sum Games with Two Players

and yt+1 − Qt (yt + αt ηt ) ≤ δD .

(4.28)

For each integer t ≥ 0 set bt,1 = αt2 L21 + δC (2M1 + L1 + 3) + αt δf,1 (2M1 + L1 + 2),

(4.29)

bt,2 = αt2 L22 + δD (2M2 + L2 + 3) + αt δf,2 (2M2 + L2 + 2).

(4.30)

Let for each natural number T ,   xT =

T 

−1 αt

i=0

  yT =

T 

T 

αt xt ,

(4.31)

αt yt .

(4.32)

t=0

−1 αt

i=0

T  t=0

Then for each natural number T , xT , δC ) ∩ C = ∅, BY ( yT , δD ) ∩ D = ∅, BX (    −1 T   T      αt αt f (xt , yt ) − f (x∗ , y∗ )    t=0 t=0  ≤

T 



−1 max

αt

T 

t=0

 + 2

bt,1 ,

t=0 T 

T 

 bt,2 + max{L1 δC , L2 δD }

t=0

−1 max{(2M1 , 2M2 } + 1)2 ,

αt

t=0

   T −1 T        f ( x ,  y ) − α α f (x , y ) T T t t t t     t=0 t=0  ≤ 2

T 

−1 αt

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

t=0

 +

T  t=0

−1 αt

max

 T  t=0

bt,1 ,

T  t=0

 bt,2

4.2 Zero-Sum Games on Bounded Sets

121

and for each z ∈ C and each v ∈ D, xT ,  yT ) f (z,  yT ) ≥ f (  T −1  − αt (max{2M1 , 2M2 } + 1)2 − max{L1 δC , L2 δD } t=0

 −2

T 



−1 max

αt

t=0

T 

bt,1 ,

t=0

T 

 bt,2

t=0

f ( xT , v) ≤ f ( xT ,  yT )  T −1  + αt (max{2M1 , 2M2 } + 1)2 + max{L1 δC , L2 δD } t=0

 +2

T 



−1 max

αt

t=0

T  t=0

bt,1 ,

T 

 bt,2 .

t=0

Proof By (4.11), (4.22)–(4.24), (4.27), and (4.28), for all integers t ≥ 0, xt  ≤ M1 + 1, yt  ≤ M2 + 1.

(4.33)

Let t ≥ 0 be an integer. Applying Lemma 3.4 with P = Pt , δf = δf,1 , α = αt , x = xt , f = f (·, yt ), ξ = ξt , y = xt+1 , we obtain that for each z ∈ C, αt (f (xt , yt ) − f (z, yt )) ≤ 2−1 z − xt 2 − 2−1 z − xt+1 2 +αt2 L21 + δC (2M1 + L1 + 3) + αt δf,1 (2M1 + L1 + 2) ≤ 2−1 z − xt 2 − 2−1 z − xt+1 2 + bt,1 . Applying Lemma 3.4 with P = Qt , α = αt , x = yt , f = −f (xt , ·), ξ = −ηt , y = yt+1 , δf = δf,2 we obtain that for each v ∈ D, αt (f (xt , v) − f (xt , yt )) ≤ 2−1 v − yt 2 − 2−1 v − yt+1 2 +δD (2M2 + L2 + 3) + αt δf,2 (2M2 + L2 + 2) + αt2 L22

122

4 Zero-Sum Games with Two Players

≤ 2−1 v − yt 2 − 2−1 v − yt+1 2 + bt,2 . Define φ(s) = 2−1 s 2 , s ∈ R 1 . It is easy to see that all the assumptions of Proposition 4.1 hold and it implies Theorem 4.3. Theorem 4.4 Let r1 , r2 > 0, BX (z, 2r1 ) ⊂ U for all z ∈ C,

(4.34)

BY (u, 2r2 ) ⊂ V for all u ∈ D,

(4.35)

Δ1 , Δ2 > 0, δf,1 , δf,2 , δC , δD ∈ (0, 1], δC ≤ r1 , δD ≤ r2 ,

(4.36)

∞ {Pt }∞ t=0 ⊂ MU , {Qt }t=0 ⊂ MV ,

(4.37)

and {αt }∞ k=0 ⊂ (0, 1],

Pt (X) = C, t = 0, 1, . . . , Qt (Y ) = D, t = 0, 1, . . . .

(4.38)

∞ ∞ ∞ 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 = ∅

(4.39)

and that for each integer t ≥ 0, BX (ξt , δf,1 ) ∩ ∂x,Δ1 f (xt , yt ) = ∅, BY (ηt , δf,2 ) ∩ ∂y,Δ2 f (xt , yt ) = ∅,

(4.40)

xt+1 − Pt (xt − at ξt ) ≤ δC and yt+1 − Qt (yt + at ηt ) ≤ δD . For each integer t ≥ 0 set bt,1 = αt Δ1 + 2−1 αt2 (L1 + Δ1 r1−1 )

(4.41)

4.2 Zero-Sum Games on Bounded Sets

123

+δC (2M1 + L1 + 3 + Δ1 r1−1 ) + αt δf,1 (2M1 + L1 + 2 + Δ1 r1−1 ), bt,2 = αt Δ2 + 2−1 αt2 (L2 + Δ2 r2−1 ) +δD (2M2 + L2 + 3 + Δ2 r2−1 ) + αt δf,2 (2M2 + L2 + 2 + Δ2 r2−1 ). Let for each natural number T   xT =

T 

−1 αt

i=0

  yT =

T 

T 

αt xt ,

t=0

−1 αt

i=0

T 

αt yt .

t=0

Then for each natural number T , xT , δC ) ∩ C = ∅, BY ( yT , δD ) ∩ D = ∅, BX (   −1 T   T      αt αt f (xt , yt ) − f (x∗ , y∗ )    t=0 t=0  ≤

T 



−1 max

αt

T 

t=0

 + 2

T 

bt,1 ,

t=0

T 

 bt,2 + max{L1 δC , L2 δD }

t=0

−1 max{(2M1 , 2M2 } + 1)2 ,

αt

t=0

   T −1 T        f ( x ,  y ) − α α f (x , y ) T T t t t t     t=0 t=0  ≤ 2

T 

−1 αt

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

t=0

 +

T 

−1 αt

max

t=0

 T  t=0

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

bt,1 ,

T  t=0

 bt,2

124

4 Zero-Sum Games with Two Players

 −

T 

−1 αt

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

t=0

 −2

T 



−1 max

αt

t=0

T 

bt,1 ,

t=0

T 

 bt,2

t=0

f ( xT , v) ≤ f ( xT ,  yT )  T −1  + αt (max{2M1 , 2M2 } + 1)2 + max{L1 δC , L2 δD } t=0

 +2

T 



−1 αt

max

t=0

T  t=0

bt,1 ,

T 

 bt,2 .

t=0

Proof By (4.11) and (4.39)–(4.41), for all integers t ≥ 0, xt  ≤ M1 + 1, yt  ≤ M2 + 1, BX (xt , δC ) ∩ C = ∅, BY (yt , δD ) ∩ D = ∅. In view (4.34)–(4.36) and (4.39), BX (x0 , r1 ) ⊂ U, BY (y0 , r1 ) ⊂ V . It follows from (4.34)–(4.36), (4.38), (4.40), and (4.41) that BX (xt , r1 ) ⊂ BX (Pt−1 (xt−1 − αt−1 ξt−1 ), δC + r1 ) ⊂ BX (Pt−1 (xt−1 − αt−1 ξt−1 ), 2r1 ) ⊂ U, BY (yt , r2 ) ⊂ BY (Qt−1 (yt−1 + αt−1 ηt−1 ), δD + r2 ) ⊂ BY (Qt−1 (yt−1 + αt−1 ηt−1 ), 2r2 ) ⊂ V . Let t ≥ 0 be an integer. Applying Lemma 3.5 with r = r1 , Δ = Δ1 , P = Pt , α = at , δf = δf,1 , x = xt , f = f (·, yt ), ξ = ξt , y = xt+1 , we obtain that for each z ∈ C,

4.2 Zero-Sum Games on Bounded Sets

125

2αt (f (xt , yt ) − f (z, yt )) ≤ z − xt 2 − z − xt+1 2 + 2αt Δ1 +δC2 + αt2 (L1 + Δ1 r −1 ) + 2δC (2M1 + L1 + 2 + Δ1 r −1 ) +2αt δf,1 (2M1 + L1 + 2 + Δ1 r −1 ) ≤ z − xt 2 − z − xt+1 2 + 2bt,1 . Applying Lemma 3.5 with r = r2 , Δ = Δ2 , P = Qt , α = at , δf = δf,2 , x = yt , f = −f (xt , ·), ξ = −ηt , y = yt+1 we obtain that for each v ∈ D, 2αt (f (xt , v) − f (xt , yt )) ≤ v − yt 2 − v − yt+1 2 + 2αt Δ2 2 +δD + 2δD (2M2 + L2 + 2 + Δr2−1 )

+2αt δf,2 (2M2 + L2 + 2 + Δr2−1 ) + αt2 (L2 + Δr2−1 ) ≤ v − yt 2 − v − yt+1 2 + 2bt,2 . Define φ(s) = 2−1 s 2 , s ∈ R 1 . It is easy to see that all the assumptions of Proposition 4.1 hold and it implies Theorem 4.4. Theorems 4.3 and 4.4 are new. We are interested in the optimal choice of αt , t = 0, 1, . . . , T . Let T be a natural number and AT = Tt=0 αt be given. In order to make the best choice of αt , t =  0, . . . , T , we need to minimize the function Tt=0 αt2 on the set  α = (α0 , . . . , αT ) ∈ R

T +1

: αi ≥ 0, i = 0, . . . , T ,

T 

 αi = AT

.

i=0

By Lemma 2.3 of [75], this function has a unique minimizer αi = (T + 1)−1 AT , i = 0, . . . , T . Let T be a natural number and αt = α for all t = 0, . . . , T . Now we will find the best α > 0. In order to meet this goal we need to choose a which is a minimizer of the function

126

4 Zero-Sum Games with Two Players

((T + 1)α)−1 (max{2M1 , 2M2 } + 1)2  T  T   −1 −1 +2α (T + 1) max bt,1 , bt,2 t=0

t=0

= ((T + 1)α)−1 (max{2M1 , 2M2 } + 1)2 +2α −1 (T + 1)−1 max{(T + 1)(αΔ1 + δC (2M1 + 3 + L1 + Δ1 r1−1 )) +2−1 α 2 (L1 + Δ1 r1−1 ) + αδf,1 (2M1 + 2 + L1 + Δ1 r1−1 ), (T + 1)(αΔ2 + δD (2M2 + 3 + L2 + Δ2 r2−1 )) +2−1 α 2 (L2 + Δ2 r2−1 ) + αδf,2 (2M2 + 2 + L2 + Δ2 r2−1 )} = ((T + 1)α)−1 (max{2M1 , 2M2 } + 1)2 +2 max{Δ1 + α −1 δC (2M1 + 3 + L1 + Δ1 r1−1 ) +2−1 α(L1 + Δ1 r1−1 ) + δf,1 (2M1 + 2 + L1 + Δ1 r1−1 ), Δ2 + α −1 δD (2M2 + 3 + L2 + Δ2 r2−1 ) +2−1 α(L2 + Δ2 r2−1 ) + δf,2 (2M2 + 2 + L2 + Δ2 r2−1 )} ≤ ((T + 1)α)−1 (max{2M1 , 2M2 } + 1)2 +2 max{Δ1 , Δ2 } +2 max{δf,1 (2M1 + 2 + L1 + Δ1 r1−1 ), δf,2 (2M2 + 2 + L2 + Δ2 r2−1 )} +2α −1 max{δC (2M1 + 3 + L1 + Δ1 r1−1 ), δD (2M2 + 3 + L2 + Δ2 r2−1 )} +α max{L1 + Δ1 r1−1 , L2 + Δ2 r2−1 }.

Since T can be arbitrarily large, we need to find a minimizer of the function φ(α) := 2α −1 max{δC (2M1 + 3 + L1 + Δ1 r1−1 ), δD (2M2 + 3 + L2 + Δ2 r2−1 )} +α max{L1 + Δ1 r1−1 , L2 + Δ2 r2−1 }, α > 0. This function has a minimizer α∗ = 21/2 max{δC (2M1 + 3 + L1 + Δ1 r1−1 ), δD (2M2 + 3 + L2 + Δ2 r2−1 )}1/2 × max{L1 + Δ1 r1−1 , L2 + Δ2 r2−1 }−1/2

4.2 Zero-Sum Games on Bounded Sets

127

and ψ(α∗ ) = 23/2 max{δC (2M1 + 3 + L1 + Δ1 r1−1 ), δD (2M2 + 3 + L2 + Δ2 r2−1 )}1/2 × max{L1 + Δ1 r1−1 , L2 + Δ2 r2−1 }1/2 .

For the appropriate choice of T , it should be at the same order as max{δC , δD }−1 .

Chapter 5

Quasiconvex Optimization

In this chapter we study an extension of the projected subgradient method for minimization of quasiconvex and nonsmooth functions, under the presence of computational errors. The problem is described by an objective function and a set of feasible points. We extend some of the results of Chapter 2 and 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 computational errors for the two steps of our algorithm, we find out what an approximate solution can be obtained and how many iterates one needs for this.

5.1 Preliminaries Let (X, ·, ·) be a Hilbert space with an inner product ·, · which induces a complete norm  · . We use the notation and definitions introduced in Chapter 2. For each x ∈ X and each nonempty set A ⊂ X, set d(x, A) = inf{x − y : y ∈ A}. For each x ∈ X and each r > 0, set BX (x, r) = {y ∈ X : x − y ≤ r}. The boundary of a set E ⊂ X is denoted by bd(E) and the closure of E is denoted by cl(E). Let C be a closed nonempty subset of the space X and U be an open convex subset of X such that C ⊂ U. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 A. J. Zaslavski, The Projected Subgradient Algorithm in Convex Optimization, SpringerBriefs in Optimization, https://doi.org/10.1007/978-3-030-60300-7_5

(5.1) 129

130

5 Quasiconvex Optimization

Suppose that f : U → R 1 is a continuous function such that f (λx + (1 − λ)y) ≤ max{f (x), f (y)}

(5.2)

for all x, y ∈ U and all λ ∈ [0, 1]. In other words, the function f is quasiconvex. We consider the minimization problem f (x) → min, x ∈ C. It should be mentioned that quasiconvex optimization problems are studied in [19, 38–43, 77]. We suppose that inf(f, C) := inf{f (x) : x ∈ C} is finite and that Cmin = {x ∈ C : f (x) = inf(f, C)} = ∅.

(5.3)

x ∗ ∈ Cmin .

(5.4)

S = {x ∈ X : x = 1}.

(5.5)

Fix

Set

For each x ∈ U satisfying f (x) > inf(f (z) : z ∈ U }, define ∂ ∗ f (x) = {l ∈ X : l, y − x ≤ 0 for every y ∈ U satisfying f (y) < f (x)}.

(5.6)

This set is a closed convex cone and it is called the subdifferential of f at the point x. Its elements are called subgradients. Let  ≥ 0. For each x ∈ U satisfying f (x) −  > inf(f (z) : z ∈ U }, define ∂∗ f (x) = {l ∈ X : l, y − x ≤ 0 for every y ∈ U satisfying f (y) < f (x) − }.

(5.7)

This set is a closed convex cone and it is called the -subdifferential of f at the point x. Its elements are called -subgradients.

5.1 Preliminaries

131

Proposition 5.1 Let x ∈ U , L > 0, β > 0,  ≥ 0, M0 > 0, x, x ∗ ∈ BX (0, M0 ),

(5.8)

f (x ∗ ) < f (x) − 

(5.9)

and let |f (z) − f (x ∗ )| ≤ Lz − x ∗ β for all z ∈ BX (0, 3M0 + 1) ∩ U.

(5.10)

Then for all g ∈ ∂∗ f (x) ∩ S, f (x) − f (x ∗ ) ≤ Lg, x − x ∗ β + . Proof Let g ∈ ∂∗ f (x) ∩ S.

(5.11)

In view of the continuity and quasiconvexity of the function f , the set E := {z ∈ U : f (z) < f (x) − }

(5.12)

is open and convex. By (5.9) and (5.12), x ∗ ∈ E.

(5.13)

For the continuity of f , (5.12) and (5.13) imply that bd(E) = ∅ and that {λx + (1 − λ)x ∗ : λ ∈ [0, 1]} ∩ bd(E) = ∅.

(5.14)

r = inf{x ∗ − u : u ∈ bd(E) ∩ U }.

(5.15)

Set

It follows from (5.8), (5.14), and (5.15) that r ≤ x ∗ − x ≤ 2M0 .

(5.16)

There exists a sequence {uk }∞ k=1 ⊂ bd(E) ∩ U such that x ∗ − uk  ≤ r + k −1 , k = 1, 2, . . . .

(5.17)

The openness of E implies that f (x) ≤ f (u) +  for all u ∈ bd(E) ∩ U.

(5.18)

132

5 Quasiconvex Optimization

In view of (5.18), for every integer k ≥ 1, f (x) − f (x ∗ ) ≤ f (uk ) − f (x ∗ ) + .

(5.19)

Let k ≥ 1 be an integer. By (5.8), (5.16), and (5.17), uk  ≤ r + k −1 + x ∗  ≤ 3M0 + 1.

(5.20)

It follows from (5.8), (5.10), (5.17), (5.19), and (5.20) that f (x) − f (x ∗ ) ≤ f (uk ) − f (x ∗ ) +  ≤ Luk − x ∗ β +  ≤ L(r + k −1 )β + 

and f (x) − f (x ∗ ) ≤ Lr β + .

(5.21)

{z ∈ U : z − x ∗  < r} ⊂ E.

(5.22)

In view of (5.15),

Let k be a natural number. By (5.11) and (5.22), x ∗ + (1 − k −1 )rg ∈ E.

(5.23)

It follows from (5.11), (5.12), and (5.23) that (1 − k −1 )r − g, x − x ∗  = (1 − k −1 )rg2 − g, x − x ∗  = g, x ∗ + (1 − k −1 )rg − x ≤ 0 and r ≤ g, x − x ∗ . Together with (5.21) this implies that f (x) − f (x ∗ ) ≤ Lg, x − x ∗ β + . Proposition 5.1 is proved. This chapter contains two main results, Theorems 5.3 and 5.4, which are proved in Sections 5.3 and 5.4, respectively. These results are new and their proofs are based on our main Lemma 5.2 proved in Section 5.2.

5.2 The Main Lemma

133

5.2 The Main Lemma Lemma 5.2 Let M0 > 0, L > 0, β > 0, δf , δC ∈ (0, 1], Δ ≥ 0, α ∈ (0, 1], x ∈ U and let a mapping P : X → X satisfy P y = y, y ∈ C, P y − z ≤ y − z for all y ∈ X and all z ∈ C.

(5.24) (5.25)

Assume that x ∗  ≤ M0 , x ≤ M0 , |f (z) − f (x ∗ )| ≤ Lz − x ∗ β , z ∈ BX (0, 3M0 + 1) ∩ U,

(5.26) (5.27)

f (x) > inf(f, C) + Δ,

(5.28)

∗ f (x) ∩ S = ∅ BX (ξ, δf ) ∩ ∂Δ

(5.29)

y − P (x − αξ ) ≤ δC .

(5.30)

ξ ∈ X satisfies

and that y ∈ U satisfies

Then y − x ∗ 2 ≤ x − x ∗ 2 + α 2 (1 + δf2 ) + 2αδf (2M0 + 1) +δC2 + 2δC (2M0 + 2) − 2αL−1/β (f (x) − inf(f, C) − Δ)1/β .

Proof In view of (5.29), there exists ∗ g ∈ ∂Δ f (x) ∩ S

(5.31)

g − ξ  ≤ δf .

(5.32)

such that

Proposition 5.1 and (5.31) imply that f (x) − inf(f, C) ≤ Lg, x − x ∗ β + Δ.

(5.33)

134

5 Quasiconvex Optimization

By (5.3), (5.4), (5.25), (5.26), and (5.30)–(5.32), y − x ∗ 2 ≤ y − P (x − αξ ) + P (x − αξ ) − x ∗ 2 ≤ y − P (x − αξ )2 + P (x − αξ ) − x ∗ 2 +2y − P (x − αξ )P (x − αξ ) − x ∗  ≤ δC2 + P (x − αξ ) − x ∗ 2 + 2δC P (x − αξ ) − x ∗  ≤ δC2 + x − αξ − x ∗ 2 + 2δC x − αξ − x ∗  ≤ δC2 + x − αξ − x ∗ 2 + 2δC (2M0 + 2).

(5.34)

It follows from (5.26), (5.28), and (5.31)–(5.33) that x − αξ − x ∗ 2 = x − αg + (αg − αξ ) − x ∗ 2 ≤ x − αξ − x ∗ 2 + α 2 g − ξ 2 + 2αg − ξ x − αξ − x ∗  ≤ x − αg − x ∗ 2 + α 2 δf2 + 2αδf (2M0 + 1) ≤ α 2 δf2 + 2αδf (2M0 + 1) + x − x ∗ 2 + α 2 − 2αg, x − x ∗  ≤ α 2 δf2 +2αδf (2M0 +1)+x−x ∗ 2 +α 2 −2αL−1/β (f (x)− inf(f, C)−Δ)1/β ).

Together with (5.34) this implies that y − x ∗ 2 ≤ x − x ∗ 2 + α 2 (1 + δf2 ) + 2αδf (2M0 + 1) −2αL−1/β (f (x) − inf(f, C) − Δ)1/β ) + δC2 + 2δC (2M0 + 2). Lemma 5.2 is proved.

5.3 Optimization on Bounded Sets Denote by M the set of all mappings P : X → C such that P z = z, z ∈ C, P x − z ≤ x − z for all x ∈ X and all z ∈ C.

(5.35) (5.36)

Theorem 5.3 Let M > 1, L > 0, β > 0, δf , δC ∈ (0, 1], Δ ≥ 0, T be a natural T −1 T −1 number, {Pt }t=0 ⊂ M, {αt }t=0 ⊂ (0, 1],

5.3 Optimization on Bounded Sets

135

C ⊂ BX (0, M − 1),

(5.37)

|f (z) − f (x ∗ )| ≤ Lz − x ∗ β , z ∈ BX (0, 3M + 1) ∩ U

(5.38)

and let T −1 −1    = Δ + 2−1 L1/β (2M + 3)2 αt t=0

+L1/β

T −1  t=0

αt

2 T −1 

−1 αt

+ L1/β δf (2M + 1) + L1/β δC (2M + 3)

T −1 

t=0

−1 β αt

.

t=0

(5.39) T −1 ⊂ X, Assume that {xt }Tt=0 ⊂ U , {ξt }t=0

B(x0 , δC ) ∩ C = ∅

(5.40)

and that for all t = 0, . . . , T − 1, ∗ f (xt ) ∩ S = ∅, BX (ξt , δf ) ∩ ∂Δ

(5.41)

xt+1 − Pt (xt − αt ξt ) ≤ δC .

(5.42)

Then min{f (xt ) : t = 0, . . . , T − 1} ≤ inf(f, C) + .

(5.43)

Proof Clearly, ∂fΔ∗ (z) is well-defined if f (z) − Δ > inf{f (x) : x ∈ U }. We may assume without loss of generality that f (xt ) − Δ > inf{f (x) : x ∈ U }, t = 0, . . . , T − 1. We show that (5.43) holds. Assume the contrary. Then f (xt ) >  + inf(f, C) =  + f (x ∗ ), t = 0, . . . , T − 1.

(5.44)

Let t ∈ {0, . . . , T − 1}. By (5.35)–(5.42), (5.44), and Lemma 5.2 applied with α = αt , P = Pt , x = xt , ξ = ξt , y = xt+1 , and M0 = M, xt+1 − x ∗ 2 ≤ xt − x ∗ 2 + 2αt2 + 2αt δf (2M + 1)

136

5 Quasiconvex Optimization

−2αt L−1/β ( − Δ)1/β + 2δC (2M + 3) and xt − x ∗ 2 − xt+1 − x ∗ 2 ≥ 2αt L−1/β ( − Δ)1/β −2αt2 − 2αt δf (2M + 1) − 2δC (2M + 3).

(5.45)

It follows from (5.37), (5.40), and (5.45) that (2M + 2)2 > x0 − x ∗ 2 ≥ x0 − x ∗ 2 − xT − x ∗ 2 =

T −1

(xt − x ∗ 2 − xt+1 − x ∗ 2 )

t=0

≥ 2L−1/β ( − Δ)1/β

T −1

αt − 2

t=0

−2δf (2M + 1)

T −1

T −1

αt2

t=0

αt − 2δC (2M + 3)T

t=0

and ( − Δ)1/β < 2−1 L1/β (2M + 3)2

T −1 

−1 αt

t=0

+L

1/β

δf (2M + 1) + L

+ L1/β

T −1 

αt2

 T −1 

t=0 1/β

δC (2M + 3)T

−1 αt

t=0

T −1 

−1 αt

.

t=0

This contradicts (5.39). The contradiction we have reached completes the proof of Theorem 5.3. T −1 Let T be a natural number and i=0 αt = A is Theorem 5.3 are given. We need to choose αt ≥ 0, t =0, . . . , T − 1 in order to minimize . Clearly, we need to T −1 2 αt over the set minimize the function i=0  (α0 , . . . , αT −1 ) ∈ R : αi ≥ 0, i = 0, . . . , T − 1, T

T −1 i=0

 αt = A .

5.4 Optimization on Unbounded Sets

137

By Lemma 2.3 of [75], we should choose αt = AT −1 , t = 0, . . . , T − 1. Assume that αt = α > 0, t = 0, . . . , T − 1. In this case in view of (5.39),  = Δ + (2−1 L1/β (2M + 3)2 (αT )−1 + L1/β α + L1/β δf (2M + 1) +L1/β δC (2M + 3)α −1 )β .

We need to choose α > 0 in order to minimize . Since a natural number T can be arbitrarily large, we need to minimize the function α + δC (2M + 3)α −1 , α > 0. Clearly, its minimizer is α = (δC (2M + 3))1/2 . In this case  = Δ + (2−1 L1/β (2M + 3)2 (δC (2M + 3))−1/2 T −1 +L1/β (δC (2M + 3)) + L1/β δf (2M + 1) + L1/β (δC (2M + 3))1/2 )β .

Evidently, T should be at the same order as δC−1 .

5.4 Optimization on Unbounded Sets We continue to use the definitions and notation introduced in Section 5.1. Recall that M is the set of all mappings P : X → C which satisfy (5.35) and (5.36). Assume that lim

x∈U, x→∞

f (x) = ∞.

(5.46)

It means that for each M0 > 0, there exists M1 > 0 such that if a point x ∈ U satisfies x ≥ M1 , then f (x) > M0 . Fix θ0 ∈ C. Set

(5.47)

138

5 Quasiconvex Optimization

U0 = {x ∈ U : f (x) ≤ f (θ0 ) + 4}.

(5.48)

By (5.46) and (5.48), there exists M¯ > 1 such that ¯ U0 ⊂ BX (0, M).

(5.49)

Assume that there exists a number L¯ > 1 such that ¯ − x ∗ β , z ∈ BX (0, M¯ + 4) ∩ U. |f (z) − f (x ∗ )| ≤ Lz

(5.50)

Theorem 5.4 Let M ≥ M¯ + 4, L ≥ L¯ + 1, β > 0, δf , δC ∈ (0, 1], Δ ∈ (0, 1], 0 < α ≤ 4−1 L−1/β , δC ≤ (6M + 3)−1 4−1 L−1/β α, δf ≤ (6M + 3)−1 4−1 L−1/β (5.51) and |f (z) − f (x ∗ )| ≤ Lz − x ∗ β , z ∈ BX (0, 9M + 4) ∩ U.

(5.52)

Assume that T is a natural number, T −1 ⊂ M, {Pt }t=0

(5.53)

x0  ≤ M,

(5.54)

B(x0 , δC ) ∩ C = ∅

(5.55)

T −1 ⊂ X, {xt }Tt=0 ⊂ U , {ξt }t=0

and that for all t = 0, . . . , T − 1, ∗ f (xt ) ∩ S = ∅, BX (ξt , δf ) ∩ ∂Δ

(5.56)

xt+1 − Pt (xt − αξt ) ≤ δC .

(5.57)

Then min{f (xt ) : t = 0, . . . , T − 1} ≤ inf(f, C) + Δ +L(2M 2 (αT )−1 + α + δC α −1 (6M + 2) + δf (6M + 2))β . Proof Clearly, ∂fΔ∗ (z) is well-defined if f (z) − Δ > inf{f (x) : x ∈ U }. We may assume without loss of generality that

(5.58)

5.4 Optimization on Unbounded Sets

139

f (xt ) − Δ > inf{f (x) : x ∈ U }, t = 0, . . . , T − 1. We show that (5.58) holds. Assume the contrary. Then for all t = 0, . . . , T − 1, f (xt ) − f (x ∗ ) > Δ + L(2M 2 (αT )−1 + α + δC α −1 (6M + 2) + δf (6M + 2))β . (5.59) In view of (5.3), (5.4), (5.49), and (5.54), ¯ x0 − x ∗  ≤ M + M. By induction we show that for all t = 0, . . . , T , ¯ xt − x ∗  ≤ M + M.

(5.60)

Clearly, (5.60) is true for t = 0. Assume that t ∈ {0, . . . , T − 1} and (5.60) is true. There are two cases: f (xt ) ≤ inf(f, C) + 4;

(5.61)

f (xt ) > inf(f, C) + 4.

(5.62)

Assume that (5.61) holds. By (5.47)–(5.49) and (5.61), ¯ xt  ≤ M.

(5.63)

It follows from (5.3), (5.4), (5.35), (5.36), (5.47)–(5.49), (5.56), (5.57), and (5.63) that xt+1 − x ∗  ≤ xt+1 − Pt (xt − αξt ) + Pt (xt − αξt ) − x ∗  ¯ ≤ δC + xt − αξt − x ∗  ≤ 2M¯ + 3 ≤ M + M.

Assume that (5.62) holds. By (5.60) which implies xt  ≤ 3M, (5.51)–(5.53), (5.56), (5.57), (5.62), and Lemma 5.2 applied with P = Pt , x = xt , ξ = ξt , y = xt+1 , and M0 = 3M, xt+1 − x ∗ 2 ≤ xt − x ∗ 2 + 2α 2 + 2αδf (6M + 1) −2αL−1/β 31/β + 2δC (6M + 3)

and ¯ xt+1 − x ∗  ≤ xt − x ∗  ≤ M + M.

140

5 Quasiconvex Optimization

Thus in both cases ¯ xt+1 − x ∗  ≤ M + M. Therefore (5.60) holds for all t = 0, . . . , T . In view of (5.3), (5.4), (5.47)–(5.49), and (5.60), xt  ≤ 3M, t = 0, . . . , T .

(5.64)

Let t ∈ {0, . . . , T − 1}. By (5.35), (5.36), (5.52), (5.56), (5.57), (5.59), (5.64), and Lemma 5.2 applied with P = Pt , x = xt , ξ = ξt , y = xt+1 , and M0 = 3M, xt+1 − x ∗ 2 ≤ xt − x ∗ 2 + 2α 2 + 2αδf (6M + 1) +δC2 + 2δC (6M + 2) − 2αL−1/β (f (xt ) − f (x ∗ ) − Δ)1/β ≤ xt − x ∗ 2 + 2α 2 + 2αδf (6M + 1) +2δC (6M + 3) − 2αL−1/β (f (xt ) − f (x ∗ ) − Δ)1/β .

(5.65)

It follows from (5.60) and (5.65) that (2M)2 ≥ x0 − x ∗ 2 ≥ x0 − x ∗ 2 − xT − x ∗ 2 =

T −1

(xt − x ∗ 2 − xt+1 − x ∗ 2 )

t=0

≥ 2α

T −1

(L−1/β (f (xt ) − f (x ∗ ) − Δ)1/β

t=0

−α − α −1 δC (6M + 2) − δf (6M + 1)) ≥ 2αT ((min{f (xt ) : t = 0, . . . , T − 1} − f (x ∗ ) − Δ)1/β L−1/β −α − α −1 δC (6M + 2) − δf (6M + 1)) and (min{f (xt ) : t = 0, . . . , T − 1} − f (x ∗ ) − Δ)1/β < L1/β (2M 2 (T α)−1 + α + α −1 δC (6M + 2) + δf (6M + 1)). This implies that min{f (xt ) : t = 0, . . . , T − 1} < inf(f, C) + Δ +L(2M 2 (T α)−1 + α + α −1 δC (6M + 2) + δf (6M + 1))β . This contradicts (5.59). The contradiction we have reached proves Theorem 5.4.

5.4 Optimization on Unbounded Sets

141

In order to minimize the right-hand side of (5.58), we should choose α = (δC (6M + 2))1/2 . In this case T should be at the same order as δC−1 . Of course, we need that (5.51) is true. This leads us to the following condition on δC : δC ≤ (6M + 3)−1 4−2 L−2β . Together with our choice of α and the condition above on δC , Theorem 5.4 implies the following result. Theorem 5.5 Let M ≥ M¯ + 4, L ≥ L¯ + 1, β > 0, δf , δC ∈ (0, 1], Δ ∈ (0, 1], |f (z) − f (x ∗ )| ≤ Lz − x ∗ β , z ∈ BX (0, 9M + 4) ∩ U, δC ≤ (6M + 3)−1 4−2 L−2/β , δf ≤ (6M + 3)−1 4−1 L−1/β and α = (δC (6M + 2))1/2 . Assume that T is a natural number, T −1 ⊂ M, {Pt }t=0 T −1 ⊂ X, {xt }Tt=0 ⊂ U , {ξt }t=0

x0  ≤ M, B(x0 , δC ) ∩ C = ∅ and that for all t = 0, . . . , T − 1, ∗ f (xt ) ∩ S = ∅, BX (ξt , δf ) ∩ ∂Δ

xt+1 − Pt (xt − αt ξt ) ≤ δC . Then min{f (xt ) : t = 0, . . . , T − 1} ≤ inf(f, C) + Δ +L(2M 2 (T (δC (6M + 2))1/2 )−1 ) + (δC (6M + 2))1/2 +(δC (6M + 3))1/2 + δf (6M + 2))β .

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