Journal of Applied MathematicsVolume 2012 (2012), Article ID 602513, 17 pagesdoi:10.1155/2012/602513

Research Article

General Iterative Methods for Equilibrium Problems and Infinitely Many Strict Pseudocontractions in Hilbert Spaces

Peichao Duan and Aihong Wang

College of Science, Civil Aviation University of China, Tianjin 300300, China

Received 11 January 2012; Revised 24 February 2012; Accepted 25 February 2012

Academic Editor: Rudong Chen

Copyright © 2012 Peichao Duan and Aihong Wang. 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 propose an implicit iterative scheme and an explicit iterative scheme for finding a common element of the set of fixed point of infinitely many strict pseudocontractive mappings and the set of solutions of an equilibrium problem by the general iterative methods. In the setting of real Hilbert spaces, strong convergence theorems are proved. Our results improve and extend the corresponding results reported by many others.

1. Introduction

Let be a real Hilbert space and let be a nonempty closed convex subset of . Let be a bifunction from to , where is the set of real numbers.

The equilibrium problem for is to find such that for all . The set of such solutions is denoted by .

A mapping of is said to be a -strict pseudocontraction if there exists a constant such that for all ; see [1]. We denote the set of fixed points of by (i.e., ).

Note that the class of strict pseudocontractions strictly includes the class of nonexpansive mappings which are mapping such that for all . That is, is nonexpansive if and only if is a 0-strict pseudocontraction.

Numerous problems in physics, optimization, and economics reduce to finding a solution of the equilibrium problem. Some methods have been proposed to solve the equilibrium problem (1.1); see, for instance, [2–4]. In particular, Combettes and Hirstoaga [5] proposed several methods for solving the equilibrium problem. On the other hand, Mann [6], Shimoji and Takahashi [7] considered iterative schemes for finding a fixed point of a nonexpansive mapping. Further, Acedo and Xu [8] projected new iterative methods for finding a fixed point of strict pseudocontractions.

In 2006, Marino and Xu [3] introduced the general iterative method and proved that the algorithm converged strongly. Recently, Liu [2] considered a general iterative method for equilibrium problems and strict pseudocontractions. Tian [9] proposed a new general iterative algorithm combining an -Lipschitzian and -strong monotone operator. Very recently, Wang [10] considered a general composite iterative method for infinite family strict pseudocontractions.

In this paper, motivated by the above facts, we introduce two iterative schemes and obtain strong convergence theorems for finding a common element of the set of fixed points of a infinite family of strict pseudocontractions and the set of solutions of the equilibrium problem (1.1).

2. Preliminaries

Throughout this paper, we always write for weak convergence and for strong convergence. We need some facts and tools in a real Hilbert space which are listed as below.

Lemma 2.1. Let be a real Hilbert space. There hold the following identities: (i), ;(ii).

Lemma 2.2 (see [11]). Assume that is a sequence of nonnegative real numbers such that where is a sequence in and is a sequence such that (i)(ii)Then, . Recall that given a nonempty closed convex subset of a real Hilbert space , for any , there exists a unique nearest point in , denoted by , such that for all . Such a is called the metric (or the nearest point) projection of onto . As known, if and only if there holds the relation:

Lemma 2.3 (see [10]). Let be a -Lipschitzian and -strongly monotone operator on a Hilbert space with , , , and . Then, is a contraction with contractive coefficient and .

Lemma 2.4 (see [1]). Let be a -strict pseudocontraction. Define by for each . Then, as , is a nonexpansive mapping such that .

Lemma 2.5 (see [9]). Let be a Hilbert space and be a contraction with coefficient , and an -Lipschitzian continuous operator and -strongly monotone with , . Then for : That is, is strongly monotone with coefficient . Let be a sequence of -strict pseudo-contractions. Define . Then, by Lemma 2.4, is nonexpansive. In this paper, consider the mapping defined by where are real numbers such that . Such a mapping is called a -mapping generated by and . It is easy to see is nonexpansive.

Lemma 2.6 (see [7]). Let be a nonempty closed convex subset of a strictly convex Banach space , let be nonexpansive mappings of into itself such that and let be real numbers such that , for every . Then, for any and , the limit exists. Using Lemma 2.6, one can define the mapping of into itself as follows:

Lemma 2.7 (see [7]). Let be a nonempty closed convex subset of a strictly convex Banach space . Let be nonexpansive mappings of into itself such that and let be real numbers such that for all . If is any bounded subset of , then

Lemma 2.8 (see [12]). Let be a nonempty closed convex subset of a Hilbert space be a family of infinite nonexpansive mappings with , let be real numbers such that , for every . Then . For solving the equilibrium problem, assume that the bifunction satisfies the following conditions: (A1) for all ;(A2) is monotone, that is, for any ;(A3) for each ;(A4) is convex and lower semicontinuous for each . Recall some lemmas which will be needed in the rest of this paper.

Lemma 2.9 (see [13]). Let be a nonempty closed convex subset of , let be bifunction from to satisfying (A1)–(A4), and let and . Then, there exists such that

Lemma 2.10 (see [5]). For , define a mapping as follows: for all . Then, the following statements hold: (i) is single-valued;(ii) is firmly nonexpansive, that is, for any , (iii); (iv) is closed and convex.

Lemma 2.11 (see [14]). Let and be bounded sequences in a Banach space and let be a sequence of real numbers such that for all . Suppose that for all and . Then .

Lemma 2.12 (see [4]). Let , and be as in Lemma 2.10. Then, the following holds: for all and .

Lemma 2.13 (see [10]). Let be a Hilbert space and let be a nonempty closed convex subset of , and a nonexpansive mapping with . If is a sequence in weakly converging to and if converges strongly to , then .

3. Main Result

Throughout the rest of this paper, we always assume that is a contraction of into itself with coefficient , and is a -Lipschitzian continuous operator and -strongly monotone on with , . Assume that and .

Define a mapping . Since both and are nonexpansive, it is easy to get is also nonexpansive. Consider the following mapping on defined by where . By Lemmas 2.3 and 2.10, we have Since , it follows that is a contraction. Therefore, by the Banach contraction principle, has a unique fixed pointed such that

For simplicity, we will write for provided no confusion occurs. Next we prove the sequences converges strongly to a which solves the variational inequality: Equivalently, .

Theorem 3.1. Let be a nonempty closed convex subset of a real Hilbert space and a bifunction from to satisfying (A1)–(A4). Let be a family -strict pseudocontractions for some . Assume the set . Let be a contraction of into itself with and let be a -Lipschitzian continuous operator and -strongly monotone with and . For every , let be the mapping generated by and as in (2.5). Let and be sequences generated by the following algorithm: If , , and satisfy the following conditions: (i), ;(ii); (iii), .Then, converges strongly to a point , which solves the variational inequality (3.4).

Proof. The proof is divided into several steps.Step 1. Show first that is bounded.Take any , by (3.5) and Lemma 2.3, we derive that It follows that .Hence, is bounded, so are and . It follows from the Lipschitz continuity of that and are also bounded. From the nonexpansivity of and , it follows that and are also bounded.Step 2. Show that Notice that By Lemma 2.10, we have It follows that Thus, from Lemma 2.1 and (3.10), we get It follows that Since , we have From (3.8), it is easy to get Step 3. Show that This implies that From condition (ii), (3.13), and (3.14), we have Notice that By Lemma 2.7 and (3.18), we get (3.15).Since is bounded, so there exists a subsequence which converges weakly to .Step 4. Show that .Since is closed and convex, is weakly closed. So, we have .From (3.15), we obtain . From Lemmas 2.8, 2.4, and 2.13, we have .By , for all , we have It follows from (A2) that Hence, we get It follows from condition (iii), (3.13), and (A4) that For with and , let . Since and , we obtain and hence . So, we have Dividing by , we get Letting and from (A3), we get for all .Step 5. Show that : Hence, we obtain It follows that This implies that In particular, Since , it follows from (3.31) that as . Next, we show that solves the variational inequality (3.4).By the iterative algorithm (3.5), we have Therefore, we have that is, Hence, for , Since is monotone (i.e., , for all ). This is due to the nonexpansivity of .Now replacing in (3.35) with and letting , we obtain That is, is a solution of (3.4). To show that the sequence converges strongly to , we assume that . By the same processing as the proof above, we derive . Moreover, it follows from the inequality (3.36) that Interchanging and , we get By Lemma 2.5, adding up (3.37) and (3.38) yields Hence and, therefore, as , This is equivalent to the fixed point equation:

Theorem 3.2. Let be a nonempty closed convex subset of a real Hilbert space and a bifunction from to satisfying (A1)–(A4). Let be a family -strict pseudocontractions for some . Assume the set . Let be a contraction of into itself with and let be a -Lipschitzian continuous operator and -strongly monotone with , ,, and . For every , let be the mapping generated by and . Given , let and be sequences generated by the following algorithm: If , and satisfy the following conditions: (i), and ;(ii); (iii), and .Then, converges strongly to , which solves the variational inequality (3.4).

Proof. The proof is divided into several steps.Step 1. Show first that is bounded.Taking any , we have By induction, we obtain Hence, is bounded, so are and . It follows from the Lipschitz continuity of that and are also bounded. From the nonexpansivity of and , it follows that and are also bounded.Step 2. Show that Observe that and from (2.5), we have where .Suppose , then .Hence, we have It follows from (3.45), (3.46), and the above result that where . Hence, we get From condition (i), (iii), , and Lemma 2.12, we obtain By Lemma 2.11,we have . Thus, By Lemma 2.12, (3.45) and (3.44), we obtain Step 3. Show that Observe that From condition (i) and (3.5),we can obtain By Lemma 2.10, we get This implies that By nonexpansivity of , we have It follows from (3.42) that This implies that From condition (i), (ii), and (3.44), we have Further we have . Thus we get On the other hand, we have Combining (3.62), the last inequality, and Lemma 2.7, we obtain (3.53).Step 4. Show that where is a unique solution of the variational inequality (3.4). Indeed, take a subsequence of such that Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we can assume . From (3.53), we obtain .By the same argument as in the proof of Theorem 3.1, we have . Since , it follows that Step 5. Show that Since It follows from (3.44) and (3.66) that This implies that where , . It is easily to see that . Hence, by Lemma 2.2, the sequence converges strongly to .

Remark 3.3. If , then Theorem 3.2 reduces to Theorem 3.1 of Wang [10].

Acknowledgments

The authors would like to thank the referee for valuable suggestions to improve the manuscript NSFC Tianyuan Youth Foundation of Mathematics of China (no. 11126136), and the Fundamental Research Funds for the Central Universities (GRANT: ZXH2011C002).

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