Abstract and Applied Analysis

Volume 2013 (2013), Article ID 218964, 7 pages

http://dx.doi.org/10.1155/2013/218964

Research Article

Algorithmic Approach to the Split Problems

Ming Ma

Tianjin and Education Ministry, Key Laboratory of Advanced Composite Materials, Tianjin 300387, China

Received 6 June 2013; Accepted 26 June 2013

Academic Editor: Abdellah Bnouhachem

Copyright © 2013 Ming Ma. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

This paper deals with design algorithms for the split variational inequality and equilibrium problems. Strong convergence theorems are demonstrated.

1. Introduction

Let be a real Hilbert space. Let and be two nonempty closed convex subsets of . Consider the following problem.

Problem 1. Find a point such that

This problem is called split feasibility problem when is a bounded linear operator. In this case, Problem 1 can be applied to many practical problems such as signal processing and image reconstruction. Specifically, we can find the prototype of Problem 1 in intensity-modulated radiation therapy; see, for example, [1–3]. Based on this relation, many mathematicians were devoted to study the split feasibility problem and develop its iterative algorithms. Related works can be found in [4–8] and the references therein.

Let be two mappings. Consider the variational inequality of finding such that

for all . We use to denote the set of solutions of (2). Variational inequality problems have important applications in many fields such as elasticity, optimization, economics, transportation, and structural analysis, and various numerical methods have been studied by many researchers; see, for instance, [9–17].

Let be an equilibrium bifunction; that is, for each . Consider the equilibrium problem which is to find such that

Denote the set of solutions of (3) by . The equilibrium problems include fixed point problems, optimization problems, and variational inequality problems as special cases. Some algorithms have been proposed to solve the equilibrium problems; see, for example, [18–22]. Thus it is an interesting topic associated with algorithmic approach to the variational inequality and equilibrium problems. In this paper, our main purpose is to study the following split problem involved in the variational inequality and equilibrium problems. Find a point such that

We are devoted to study (4) with operator being a nonlinear mapping. For this purpose, we develop an iterative algorithm for solving the split problem (4). We can compute iteratively by using our algorithm. Convergence analysis is given under some mild assumptions.

2. Basic Concepts

Let be a nonempty closed convex subset of a real Hilbert space . An operator is said to be

(i)monotone for all ;(ii)strongly monotone for some constant and for all ;(iii)inverse-strongly monotone for some and for all ; in this case, is called -inverse strongly monotone;(iv)-inverse strongly -monotone for all and for some , where is a mapping.

A mapping is said to be

(i)nonexpansive for all ;(ii)firmly nonexpansive for all ;(iii)-Lipschitz continuous for some constant and for all . In such a case, is said to be -Lipschitz continuous.

In the sequel, we use to denote the set of fixed points of .

Let be a multivalued mapping. The effective domain of is denoted by . is said to be

(i)monotone for all , , and ;(ii)maximal monotone is monotone and its graph is not strictly contained in the graph of any other monotone operator on .

A function is said to be convex if for any and for any , .

Let be the metric projection from onto . It is known that satisfies the following inequality:

for all and . From this characteristic inequality, we can deduce that is firmly nonexpansive.

3. Useful Lemmas

In this section, we present several lemmas which will be used in the next section.

Lemma 2 (see [19]). Let be a nonempty closed convex subset of a real Hilbert space . Let be a bifunction. Assume that satisfies the following conditions:

  for all ;   is monotone, that is, for all ;  for each , ;  for each , is convex and lower semicontinuous.

Let and . Then there exists such that

Set for all . Then one have the following:

(i) is single valued and is firmly nonexpansive,(ii) is closed and convex and .

Lemma 3 (see [23]). Let be a nonempty closed convex subset of a real Hilbert space . For , let the mapping be the same as in Lemma 2. Then for and , one has

Lemma 4 (see [24]). Let and be two bounded sequences in a Banach space , and let be a sequence in satisfying . Suppose for all and . Then, .

Lemma 5 (see [25]). Let be a nonempty closed convex subset of a real Hilbert space . Let be a nonexpansive mapping with . Then is demiclosed on .

Lemma 6 (see [26]). Let be a sequence. Assume that , where is a sequence in , and is a sequence satisfying and (or ). Then .

4. Main Results

In this section, we firstly present our problem and algorithm constructed. Consequently, we give the convergence analysis of the presented algorithm.

Problem 7. Let be a nonempty closed convex subset of a real Hilbert space . Assume that

(1) is a weakly continuous and -strongly monotone mapping such that ; (2) is an -inverse strongly -monotone mapping; (3) is a bifunction satisfying conditions in Lemma 2.

Our objective is to

We use to denote the set of solutions of (8). In the following, we assume that is nonempty. For solving Problem 7, we introduce the following algorithm.

Algorithm 8.   

Step  0 (initialization). Let

Step  1. For given , let the sequence be generated iteratively by

where is the metric projection and is a real number sequence.

Step  2. For given , find such that

where and are two real number sequences.

Step  3. For the previous sequences and , let the sequence be generated by

where is a real number sequence.

Theorem 9. Assume that the following conditions are satisfied:

  and ;  ;  , , and ;   and .

Then the sequence generated by Algorithm 8 converges strongly to .

Proof. Let . Hence and , noting that implies for all . Hence for all . Thus, from (10), we have

Condition and (13) imply that

From Lemma 2 and (11), we get for all . Since , from Lemma 2 we deduce that for all . So,

It follows that

By induction

Hence, is bounded. Since is -strongly monotone, we can get . So, . This implies that is bounded. Next, we show . From , we have

Using Lemma 3, we obtain

Then

By condition , we have . So,

From (10), we have

Therefore,

It follows that

Since , , and and the sequences , , , and are bounded, we deduce that

Applying Lemma 4, we obtain

Thus,

This together with the -strong monotonicity of implies that

From (13) and (16), we derive

Hence,

Since , , and , we obtain

Set for all . By using the firm nonexpansivity of projection, we get

It follows that

From (29) and (32), we have

Then, we obtain

Since , , and , we deduce that

Next, we prove , where satisfies (GVI): , for all (note that is -strongly monotone; we can easily deduce that the solution of (GVI) is unique). We take a subsequence of such that

By the boundedness of , we can choose a subsequence of such that weakly. For the convenience, we may assume that . This implies that due to the weak continuity of . Now, we show . We firstly show .

Note that . Then we choose a subsequence of such that . From (26) and (36), we deduce that . Thus, . From Lemma 2, we know that is nonexpansive. By demiclosed principle (Lemma 5), we get immediately that .

Next we prove . Set

By [27], we know that is maximal -monotone. Let . Since and , we have . Noting that , we get

It follows that

Then,

Since and , we deduce that by taking in (41). Thus, by the maximal -monotonicity of . Hence, . Therefore, . From (37), we obtain

From (12), we have

Using Lemma 6, we conclude that , and hence . This completes the proof.

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