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Journal of Applied Mathematics

Volume 2012 (2012), Article ID 152023, 18 pages

http://dx.doi.org/10.1155/2012/152023

## A Viscosity Approximation Scheme for Finding Common Solutions of Mixed Equilibrium Problems, a Finite Family of Variational Inclusions, and Fixed Point Problems in Hilbert Spaces

^{1}School of Management, Tianjin University, Tianjin 300072, China^{2}School of Economics and Management, Shandong University of Science and Technology, Qingdao 266510, China

Received 16 February 2012; Accepted 10 April 2012

Academic Editor: Yeong-Cheng Liou

Copyright © 2012 Bin-Chao Deng et al. 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

We introduce an iterative method for finding a common element of set of fixed points of nonexpansive mappings, the set of solutions of a finite family of variational inclusion with set-valued maximal monotone mappings and inverse strongly monotone mappings, and the set of solutions of a mixed equilibrium problem in Hilbert spaces. Under suitable conditions, some strong convergence theorems for approximating this common elements are proved. The results presented in the paper improve and extend the main results of Plubtemg and Sripard and many others.

#### 1. Introduction

Let be a real Hilbert space whose inner product and norm are denoted by and , respectively. Let be a nonempty closed convex subset of , and let be a bifunction of into which is the set of real numbers. The equilibrium problem for is to find such that The set of solutions of (1.1) is denoted by . The mixed equilibrium problem for two bifunction of is to find such that In the sequel we will indicate by the set of solution of our mixed equilibrium problem. If we denote with .

In 2005, Combettes and Hirstoaga [1] introduced an iterative scheme of finding the best approximation to the initial data when is nonempty and proved a strong convergence theorem. Let be a nonlinear mapping. The classical variational inequality which is denoted by is to find such that The variational inequality has been extensively studied in the literature; see, for example, [2, 3] and the reference therein. Recall that mapping of into itself is called nonexpansive if A mapping is called contractive if there exists a constant such that We denote by the set of fixed points of .

Some methods have been proposed to solve the equilibrium problem and fixed point problem of nonexpansive mapping; see, for instance, [2, 4–6] and the references therein. In 2007, Plubtieng and Punpaeng [6] introduced the following iterative scheme. Let and let and be sequences generated by They proved that if the sequences and of parameters satisfy appropriate conditions, then the sequences and both converge strongly to the unique solution of the variational inequality which is the optimality condition for the minimization problem where is a potential function for .

Let be a single-valued nonlinear mapping, and let be a set-valued mapping. We consider the following variational inclusion, which is to find a point such that where is the zero vector in . The set of solutions of problem (1.9) is denoted by . Let be single-valued nonlinear mappings, and let , be set-valued mappings. If , then problem (1.9) becomes the variational inclusion problem introduced by Rockafellar [7]. If , where is a nonempty closed convex subset of and is the indicator function of , that is, then the variational inclusion problem (1.9) is equivalent to variational inequality problem (1.3). It is known that (1.9) provides a convenient framework for the unified study of optimal solutions in many optimization-related areas including mathematical programming, complementarity, variational inequalities, optimal control, mathematical economics, equilibria, and game theory. Also various types of variational inclusions problems have been extended and generalized (see [8] and the references therein). We introduce following a finite family of variational inclusions, which are to find a point such that where is the zero vector in . The set of solutions of problem (1.11) is denoted by . The formulation (1.11) extends this formalism to a finite family of variational inclusions covering, in particular, various forms of feasibility problems (see, e.g., [9]).

In 2009, Plubtemg and Sripard [10] introduced the following iterative scheme for finding a common element of set of solutions to the problem (1.9) with multivalued maximal monotone mapping and inverse-strongly monotone mapping, the set solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping in Hilbert spaces. Starting with an arbitrary , define sequence , , and by for all , where , , and ; is a strongly positive bounded linear operator on , and is a sequence of nonexpansive mappings on . They proved that under certain appropriate conditions imposed on and , the sequence , , and generated by (1.12) converge strongly to , where .

In 2011, Yao et al. [11] considered the following iterative method for finding a common element of set of solutions to the problem (1.9) with multi-valued maximal monotone mapping and inverse-strongly monotone mapping, the set solutions of a mixed equilibrium problem, and the set of fixed points of an infinite family of nonexpansive mappings in Hilbert spaces. Let be a bifunction, be a strongly positive bounded linear operator, and be inverse strongly monotone and let inverse strongly monotone, and be a lower semicontinuous and convex function. For any initial is selected in arbitrarily where and are two sequences in and is an infinite family of nonexpansive mappings. They study the strong convergence of the iterative algorithm (1.13).

Motivated and inspired by Saeidi [12], Aoyama et al. [13], Plubieng and Punpaeng [6], Plubtemg and Sripard [10], Peng et al. [14], and Yao et al. [11], we introduce an iterative scheme for finding a common element of the set of solutions of a finite family of variational inclusion problems (1.11) with multi-valued maximal monotone mappings and inverse-strongly monotone mappings, the set of solutions of a mixed equilibrium problem, and the set of fixed points of nonexpansive mappings in Hilbert space. Starting with an arbitrary , define sequence , and by for all , where , and , is a strongly positive bounded linear operator on , and is a sequence of nonexpansive mappings on . Under suitable conditions, some strong convergence theorems for approximating to this common elements are proved. Our results extend and improve some corresponding results in [10, 11, 14] and the references therein.

#### 2. Preliminaries

This section collects some lemmas which will be used in the proofs for the main results in next section.

Let be a real Hilbert space with inner product and norm , respectively. It is well known that for all and , there holds

Let be a nonempty closed convex subset of . Then, for any , there exists a unique nearest point of , denoted by . such that for all . Such a is called the metric projection from into . We know that is nonexpansive. It is also known that and It is easy to see that (2.2) is equivalent to

For solving the mixed equilibrium problem for a bifunction , let us assume that satisfies the following conditions:(A1) for all ;(A2) is monotone, that is, for all ;(A3) for each , (A4)for each is convex and lower semicontinuous.

Lemma 2.1 (see [15]). *Let be a convex closed subset of a Hilbert spaces .**Let be a bifunction such that*

(f1)* for all ;*(f2)* is monotone and upper hemicontinuous in the first variable;*(f3)* is lower semicontinuous and convex in the second variable. **Let be a bifunction such that*(h1)* for all ;*(h2)* is monotone and weakly upper semicontinuous in the first variable;*(h3)* is convex in the second variable.** Moreover let us suppose that*(H)*for fixed and there exists a bounded set and such that for all .** For and , let be a mapping defined by
**
Called resolvent of and . **Then,*(1)*;*(2)* is a single value;*(3)* is firmly nonexpansive;*(4)* and it is closed and convex.*

Recall that a mapping is called -inverse-strongly monotone, if there exists a positive number such that

Let be the identity mapping on . It is well known that if is -inverse-strongly monotone, then is -Lipschitz continuous and monotone mapping. In addition, if , then is a nonexpansive mapping.

A set-valued is called monotone, if for all , and imply . A monotone mapping is maximal if its graph of is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping is maximal if and only if for for every implies .

Let the set-valued be maximal monotone. we define the resolvent operator associated with and as follows: where is a positive number. It is worth mentioning that the resolvent operator is single-valued, nonexpansive, and 1-inverse-strongly monotone, see for example [16] and that a solution of problem (1.9) is a fixed point of the operator for all , see for instance, [17]. Furthermore, a solution of a finite family of variational inclusion problems (1.11) is a common fixed point of , , .

Lemma 2.2 (see [16]). *Let be a maximal monotone mapping and a Lipschitz-continuous mapping. Then the mapping is a maximal monotone mapping.*

Lemma 2.3 (see [18]). *Let be a Hilbert space, a nonempty closed subset of , a contraction with coefficient , and a strongly positive linear bounded operator with coefficient . Then, *(1)*if , then ;*(2)*if , then .*

Lemma 2.4. *For all , there holds the inequality*

Lemma 2.5 (the resolvent identity). *Let be a Banach space, for and ,
*

Lemma 2.6. *Let be a Hilbert space. Let be -inverse-strongly monotone mappings, maximal monotone mappings, and be a bounded sequence in . Assume , satisfy *(H1)*,*(H2)*.**Set for and for all . Then, for ,
*

*Proof. *From Lemma 2.5, we have, for all ,
Furthermore, from the definition of , it follows
Combining (2.11) and (2.12), we obtain
where . According to (H1) and (H2), then (2.11) holds.

Lemma 2.7 (see [19]). *Let and be bounded sequences in a Banach space , and let be a sequence in with and . Suppose
**
for all and
**
Then .*

Lemma 2.8 (see [20]). *Assume is a sequence of nonnegative real numbers such that
**
where is a sequence in and is a sequence in such that*(i)*,*(ii)* or .**Then .*

#### 3. Main Results

Theorem 3.1. *Let be a real Hilbert space, let and be bifunction satisfying (A1)–(A4), and let be a sequence of nonexpansive mappings on . Let , be -inverse-strongly monotone mappings, and let be maximal monotone mappings such that . Let be a contraction of into itself with a constant , and let be a strongly positive bounded linear operator on with coefficient and . Let , , and be sequences generated by and
**
for all , where , , , satisfy (H1)-(H2) and and satisfy*(C1)*;*(C2)*;*(C3)*;*(C4)*;*(C5)*;*(C6)*.** Suppose that for any bounded subset of . Let be a mapping of into itself defined by , for all and suppose that . Then, , , and converge strongly to , where is a unique solution of the variational inequality
*

*Proof. *Since , we will assume that and . Observe that, if , then
By Lemma 2.3 we have
Moreover, using the definition of in Lemma 2.6, we have . We divide the proof into several steps.*Step 1. *The sequence is bounded.

Since , we may assume that for all . Let . Using the fact that , , is nonexpansive and , we have
for all . Then, we have
It follow from (3.6) and induction that
Hence is bounded and therefore , , , and are also bounded.*Step 2. *We show that .

Define for each . From the definition of , we obtain
It follows that
By the suppose of , we obtain
From Lemma 2.6, we obtain
By and being nonexpansive, we have
Substituting (3.12) into (3.9), we get
By (3.10), (3.11), and the conditions (C1) and (C6), we imply that
Hence, by Lemma 2.7, we have . Consequently, it follows that
From (3.11), (3.12), and (3.15), we also imply that
*Step 3. *We now show that

Indeed, let . It follows from the firmly nonexpansiveness of that
for each . Thus we get
which implies that, for each ,
Set , and let be a constant such that
Using Lemma 2.2 and noting that is convex, we derive, from (3.20),
It follows, by Step 2 and condition (C1), that
*Step 4. *We will prove .

We note from (3.1)
Since , as , and , we get
Let . Since , it follows from Lemma 2.1 that
and hence . Therefore, using Lemma 2.6 and (3.22), we have
and hence
Since is bounded, , , and , it follows that
Next we will prove .
From (2.10), we obtain
In addition, according to , we have
It follows from (3.25), (3.32), and the inequality that . Since
for all , it follows that
*Step 5. *We show .

Since is bounded, there exists a subsequence of which converges weakly to . From (3.29), we obtain which converges weakly to . From (3.32), it follows that . We show . According to (3.1) and (A2), we obtain
and hence
Since and , from (A4), it follows that for all . For with and , let , then we obtain . So, from (A1) and (A4) we have
and hence . From (A3), we have for all . Therefore, .

Next, we show . Assume ; then we have . It follows by the Opial’s condition and (3.34) that
This is a contradiction. Hence .

We now show that . In fact, since is -inverse-strongly monotone, , , is a -Lipschitz continuous monotone mapping and , . It follows from Lemma 2.2 that , , is maximal monotone. Let , , that is, , . Since , we have , that is,
By the maximal monotonicity of , we have
which implies
for . From (3.17), it follows , especially, . Since , , are Lipschitz continuous operators, we have . So, from (3.41), we have
Since , is maximal monotone, this implies that , , that is, . So, we obtain result.*Step 6. *We show that
where is unique solution of the variational inequality (3.2).

To show this, we choose a subsequence of such that
By the proof of Step 5, we obtain that
*Step 7. *We prove that .

By using Lemmas 2.3 and 2.4, we have
It follows that
Now, from conditions (C1), (C2), and (C6), Step 6 and Lemma 2.8, we obtain . Namely, in norm.

Corollary 3.2. *Let be a real Hilbert space, let be a bifunction satisfying (A1)–(A4), and let be a sequence of nonexpansive mappings on. Let be - inverse-strongly monotone mappings and , maximal monotone mappings such that. Let be a contraction of into itself with a constant, and let be a strongly positive bounded linear operator on with coefficient and . Let , , and be sequences generated by and
**
for all , where , , satisfy (H1)-(H2) and , and satisfy*(C1)*;*(C2)*;*(C3)*;*(C4)*;*(C5)*;*(C6)*.** Suppose that for any bounded subset of . Let be a mapping of into itself defined by , for all , and suppose that . Then, , , and converge strongly to , where is a unique solution of the variational inequality
*

#### Acknowledgment

This work is supported in part by China Postdoctoral Science Foundation (Grant no. 20100470783).

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