Journal of Applied Mathematics

Volume 2014, Article ID 478172, 9 pages

http://dx.doi.org/10.1155/2014/478172

## Iterative Computation for Solving the Variational Inequality and the Generalized Equilibrium Problem

^{1}School of Science, Tianjin Polytechnic University, Tianjin 300387, China^{2}College of Management and Economics, Tianjin University, Tianjin 300072, China^{3}Department of Mathematics and the RINS, Gyeongsang National University, Jinju 660-701, Republic of Korea^{4}Department of Mathematics, Dong-A University, Pusan 614-714, Republic of Korea

Received 22 March 2014; Revised 7 May 2014; Accepted 13 May 2014; Published 27 May 2014

Academic Editor: Giuseppe Marino

Copyright © 2014 Xiujuan Pan 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

An iterative algorithm for solving the variational inequality and the generalized equilibrium problem has been introduced. Convergence result is given.

#### 1. Introduction

Let be a real Hilbert space with inner product and norm , respectively. Let be a nonempty closed convex subset of . Recall that a mapping is said to be(i)nonexpansive , for all (we use to denote the set of the fixed points of );(ii)firmly nonexpansive , for all ;(iii)-Lipschitz there exists a constant such that , for all ;(iv)monotone , for all ;(v)strongly monotone there exists a constant such that , for all ;(vi)inverse strongly monotone there exists such that , for all ;(vii)-inverse strongly -monotone , for all and for some , where is a nonlinear mapping.

If is a multivalued mapping of into , then is said to be a monotone operator on , for all , , and . A monotone operator on is said to be maximal if and only if its graph is not strictly contained in the graph of any other monotone operator on .

Let ,, and be three nonlinear mappings on . Let be a bifunction. Recall that the equilibrium problem is to find such that The solution set of (1) is denoted by . Now we know that the equilibrium theory provides us a natural, novel, and unified framework to study a wide class of problems arising in economics, finance, transportation, network and structural analysis, elasticity, and optimization. For related works, please refer to [1–3] and the references therein.

Recall also that the variational inequality problem is to find , such that The solution set of (2) is denoted by . It is well known that variational inequality theory has emerged as an important tool in studying a wide class of obstacle, unilateral, free, moving, and equilibrium problems arising in several branches of pure and applied sciences in a unified and general framework. Several numerical methods have been developed for solving variational inequalities and related optimization problems. For related works, please see [4–14]. Noor [15] introduced an iterative scheme and studied the approximate solutions of variational inclusion in Hilbert spaces. Glowinski and Le Tallec [16] used the iterative schemes to find the approximate solutions of the elastoviscoplasticity problem, liquid crystal theory, and eigenvalue computation. In 1998, Haubruge et al. [17] studied the convergence analysis of the iterative schemes of Glowinski and Le Tallec and applied these schemes to obtain new splitting-type algorithms for solving variational inequalities, separable convex programming, and minimization of a sum of convex functions.

Our main purpose in the present paper is to solve the following equilibrium problem and variational inequality problem: finding a point such that Our main motivations are inspired by the following two reasons.

Firstly, it is still an interesting topic for solving the variational inequality problem and the equilibrium problem based on their applications in science and engineering. Secondly, the split problem of finding a point such that has received much attention. For related works, please refer to [18–20]. However, we observe that the involved operator in (4) is a bounded liner operator. In this paper, we devote to study the problem (3), where the transformation is a nonlinear mapping. For this purpose, we introduce a new iterative algorithm. Consequently, strong convergence analysis is demonstrated.

#### 2. Preliminaries

In this section, we recall some useful lemmas.

Recall that the metric projection satisfies . The metric projection is a typical firmly nonexpansive mapping. The characteristic inequality of the projection is , for all , .

Assume that is a bifunction which satisfies the following conditions: (C1), for all ; (C2) is monotone; that is, , for all ; (C3)for each , ; (C4)for each , is convex and lower semicontinuous.

Lemma 1 (see [2]). *Let be a nonempty closed convex subset of a real Hilbert space . Let be a bifunction which satisfies conditions (C1)–(C4). Let and . Then, there exists such that
**
Further, if , for all , then the following hold:*(a)* is single-valued and is firmly nonexpansive;*(b)* is closed and convex and .*

Lemma 2 (see [21]). *Let and be bounded sequences in a Banach space and let be a sequence in with . Suppose that , for all , and . Then, .*

Lemma 3 (see [22]). *Assume that the sequence satisfies and , where is a sequence in and is a sequence such that and or . Then, .*

#### 3. Main Results

Let be a nonempty closed convex subset of a real Hilbert space . Let be an -inverse strongly monotone mapping. Let be a weakly continuous and -strongly monotone mapping such that . Let be a -inverse strongly -monotone mapping. Let be an L-Lipschitz continuous mapping. Let be a bifunction which satisfies (C1)–(C4) in the above section. Let , , , and be four real number sequences and let be a constant.

We use to denote the solution set of (3). In order to solve (3), we introduce the following three-step algorithm.

*Algorithm 4. *Let be an initial guess. Define the sequence as follows:

Theorem 5. *Suppose that . Assume that the following conditions are satisfied:* (r1)* and ;* (r2)* and ;* (r3)*;
* (r4)* and ;* (r5)*.**
Then, the sequence generated by (6) converges strongly to which solves the following variational inequality:
*

*Proof. *First of all, we prove that the solution of the variational inequality (7) is unique. In fact, if also solves (7), then we get
It follows that
So,
which implies that
Since is -strongly monotone, we have
Thus,
Since , we deduce the contradiction. Therefore, . So, the solution of variational inequality (7) is unique.

Let . Hence, and . Note that
Since , we have , for all . For , we have
Hence,
for all . Thus,
From (6), we have , for all . Noting that , we deduce , for all . It follows from (17) that
An induction implies that
Hence, is bounded. Since is -strongly monotone, we can deduce . So,
This implies that is bounded.

From (6), we have
So,
Similarly,
Hence,
Since is monotone, we have
So,
Thus,
It follows that
and hence
By (6) and (16), we have
Therefore,
It follows that
Since , , and the sequences ,, , , and are bounded, we have
From Lemma 2, we obtain
Note that
Hence,
This together with the -strong monotonicity of implies that
From (18), we have
By the convexity of the norm and (17), we have
So,
Since , , and , we obtain
Note that
It follows that
From (39) and (43), we have
Then, we obtain
Since , , and , we have
Next, we prove that , where is the unique solution of (7). Let be a subsequence of such that
By the boundedness of , there exists a subsequence of which converges weakly to some point . Without loss of generality, we may assume that . From the weak continuity of , we deduce . Next, we prove . We firstly show . Noting that , for any , we have
Since is monotone, we have
Hence,
Let , for all and . We have . So, from (50) we have
Note that . Further, from monotonicity of , we have . Letting in (51), we have . This together with (C1) and (C4) implies that
and hence . Letting , we have . This implies that . Next, we prove . Set
It is well known that is maximal -monotone. Let (the graph of ). Since and , we have . Noting that , we get
It follows that
Then,
Since and , we deduce that by taking in (56). Thus, by the maximal -monotonicity of . Hence, . Therefore, . From (47), we obtain
Set , for all . Then, we have
It follows that
Therefore,
where and . It is easily seen that . Since
and by , we get . We can therefore apply Lemma 3 to conclude that and . This completes the proof.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors are extremely grateful to the anonymous referees for their useful comments and suggestions. This study was supported by research funds from Dong-A University.

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