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`Journal of Applied MathematicsVolume 2014, Article ID 478172, 9 pageshttp://dx.doi.org/10.1155/2014/478172`
Research Article

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

1School of Science, Tianjin Polytechnic University, Tianjin 300387, China

2College of Management and Economics, Tianjin University, Tianjin 300072, China

3Department of Mathematics and the RINS, Gyeongsang National University, Jinju 660-701, Republic of Korea

4Department 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

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 [13] 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 [414]. 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 [1820]. 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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