- Research Article
- Open Access

# Strong Convergence Theorems for a Generalized Equilibrium Problem with a Relaxed Monotone Mapping and a Countable Family of Nonexpansive Mappings in a Hilbert Space

- Shenghua Wang
^{1}Email author, - Giuseppe Marino
^{2}and - Fuhai Wang
^{1}

**2010**:230304

https://doi.org/10.1155/2010/230304

© Shenghua Wang et al. 2010

**Received:**15 March 2010**Accepted:**20 June 2010**Published:**8 July 2010

## Abstract

We introduce a new iterative method for finding a common element of the set of solutions of a generalized equilibrium problem with a relaxed monotone mapping and the set of common fixed points of a countable family of nonexpansive mappings in a Hilbert space and then prove that the sequence converges strongly to a common element of the two sets. Using this result, we prove several new strong convergence theorems in fixed point problems, variational inequalities, and equilibrium problems.

## Keywords

- Hilbert Space
- Variational Inequality
- Iterative Method
- Equilibrium Problem
- Monotone Mapping

## 1. Introduction

Some iterative methods have been proposed to find an element of ; see [1, 2].

then is called a solution of the variational inequality. The set of all solutions of the variational inequality is denoted by . It is known that is closed and convex. Recently Takahashi and Toyoda [5] introduced an iterative method for finding an element of ; see also [6]. On the other hand, Plubtieng and Punpaeng [7] introduced an iterative method for finding an element of ; see also [8].

where , , and are three control sequences. They proved that converges strongly to .

where is a constant; see [10]. In the case of for all , is said to be relaxed -monotone. In the case of for all and , where and , is said to be -monotone; see [11–13]. In fact, in this case, if , then is a -strongly monotone mapping. Moreover, every monotone mapping is relaxed - monotone with for all and .

In the case of , coincides with . In the case of and , coincides with .

where is a relaxed - monotone mapping, is a -inverse-strongly monotone mapping, and is a countable family of nonexpansive mappings such that , , and , , and are three control sequences. We prove that defined by (1.14) converges strongly to . Using the main result in this paper, we also prove several new strong convergence theorems for finding the elements of , , , and , respectively, where is a nonexpansive mapping.

## 2. Preliminaries

Hence, if , then is a nonexpansive mapping of into .

This inequality is a very useful characterization of . Observe what is more that it immediately yields that is a convex closed set.

Let be a bifunction of into satisfying the following conditions:

is monotone, that is, for all ;

for each , is convex and lower semicontinuous.

Definition 2.1 (see [10]).

Let be a Banach space with the dual space and let be a nonempty subset of . Let and be two mappings. The mapping is said to be -hemicontinuous if, for any fixed , the function defined by is continuous at .

Lemma 2.2.

Let be a Hilbert space and let be a nonempty closed convex subset of . Let be an -hemicontinuous and relaxed - monotone mapping. Let be a bifunction from to satisfying (A1) and (A4). Let and . Assume that

(ii)for any fixed , the mapping is convex.

Proof.

Thus is a solution of the problem (2.7).

for all . Therefore, is also a solution of the problem (2.6). This completes the proof.

Definition 2.3 (see [14]).

Let be a Banach space with the dual space and let be a nonempty subset of . A mapping is called a KKM mapping if, for any , , where denotes the family of all the nonempty subsets of .

Lemma 2.4 (see [14]).

Let be a nonempty subset of a Hausdorff topological vector space and let be a KKM mapping. If is closed in for all in and compact for some , then .

Next we use the concept of KKM mapping to prove two basic lemmas for our main result. The idea of the proof of the next lemma is contained in the paper of Fang and Huang [10].

Lemma 2.5.

Let be a real Hilbert space and be a nonempty bounded closed convex subset of . Let be an -hemicontinuous and relaxed - monotone mapping, and let be a bifunction from to satisfying (A1) and (A4). Let . Assume that

(ii)for any fixed , the mapping is convex and lower semicontinuous;

(iii) is weakly lower semicontinuous; that is, for any net , converges to in which implies that .

Then problem (2.6) is solvable.

Proof.

which is a contradiction. This implies that is a KKM mapping.

This completes the proof.

Lemma 2.6.

(ii)for any fixed , the mapping is convex and lower semicontinuous and the mapping is lower semicontinuous;

(iii) is weakly lower semicontinuous;

Then, the following holds:

Proof.

This shows that is firmly nonexpansive.

Finally, we prove that is closed and convex. Indeed, Since every firm nonexpansive mapping is nonexpansive, we see that is nonexpansive from (2). On the other hand, since the set of fixed points of every nonexpansive mapping is closed and convex, we have that is closed and convex from (2) and (3). This completes the proof.

## 3. Main Results

In this section, we prove a strong convergence theorem which is our main result.

Theorem 3.1.

Let be a nonempty bounded closed convex subset of a real Hilbert space and let be a bifunction satisfying (A1), (A2), (A3), and (A4). Let be an -hemicontinuous and relaxed - monotone mapping, let be a -inverse-strongly monotone mapping, and let be a countable family of nonexpansive mappings such that . Assume that the conditions (i)–(iv) of Lemma 2.6 are satisfied. Let and assume that is a strictly decreasing sequence. Assume that with some and with some . Then, for any , the sequence generated by (1.14) converges strongly to . In particular, if contains the origin 0, taking , then the sequence generated by (1.14) converges strongly to the minimum norm element in .

Proof.

We split the proof into following steps.

Step 1.

is closed and convex, the sequence generated by (1.14) is well defined, and for all .

This implies that . Noting that is a nonexpansive mapping for and the set of fixed points of a nonexpansive mapping is closed and convex, we have that is closed and convex.

So, for all . Hence , that is, for all . Since is closed, convex, and nonempty, the sequence is well defined.

Step 2.

and there exists such that as .

for all . This shows that is increasing. Since is bounded, is bounded. So, we have that exists.

Step 3.

Step 4.

Step 5.

First we prove . Indeed, since and , we have for each . Hence, .

This implies that . Hence, we get .

In view of (2.3), one sees that . This completes the proof.

Corollary 3.2.

Then the sequence converges strongly to . In particular, if contains the origin 0, taking , the sequence converges strongly to the minimum norm element in .

Proof.

So, taking with and choosing a sequence of real numbers with , we obtain the desired result by Theorem 3.1.

Corollary 3.3.

Then the sequence converges strongly to . In particular, if contains the origin 0, taking , the sequence converges strongly to the minimum norm element in .

Proof.

In Corollary 3.2, put and for all . Then is a monotone mapping and we obtain the desired result by Theorem 3.1.

Corollary 3.4.

Then the sequence converges strongly to . In particular, if contains the origin 0, taking , the sequence converges strongly to the minimum norm element in .

Proof.

In Theorem 3.1, put , , , and . We obtain the desired result by Theorem 3.1.

Corollary 3.5.

Proof.

In Corollary 3.4, by putting we obtain the desired result.

Corollary 3.6.

Proof.

Then, we obtain the desired result by Theorem 3.1.

Remark 3.7.

The novelty of this paper lies in the following aspects.

(i)A new general equilibrium problem with a relaxed monotone mapping is considered.

## Declarations

### Acknowledgment

This work was supported by the Natural Science Foundation of Hebei Province (A2010001482).

## Authors’ Affiliations

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