Abstract and Applied Analysis

Volume 2014 (2014), Article ID 468593, 8 pages

http://dx.doi.org/10.1155/2014/468593

## Algorithmic Approach to the Equilibrium Points and Fixed Points

^{1}School of Electrical Engineering and Automation, Tianjin Polytechnic University, Tianjin 300387, China^{2}Department of Mathematics and the RINS, Gyeongsang National University, Jinju 660-701, Republic of Korea^{3}Department of Mathematics, Dong-A University, Pusan 614-714, Republic of Korea

Received 10 March 2014; Accepted 25 March 2014; Published 22 April 2014

Academic Editor: Chong Li

Copyright © 2014 Lijin Guo 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

The equilibrium and fixed point problems are considered. An iterative algorithm is presented. Convergence analysis of the algorithm is provided.

#### 1. Introduction

Let be a real Hilbert space with inner product and norm , respectively. Let be a nonempty closed convex subset of . Let be a nonlinear operator and let be a bifunction. The equilibrium problem is formulated as finding such that The solution set of (1) is denoted by EP. The problem (1) is very general in the sense that it includes, as special cases, optimization problems, variational inequalities, minimax problems, Nash equilibrium problem in noncooperative games, and others. For related work, please see, for example, [1–17]. Next, we recall several interesting results where verifies the following usual conditions ()–() which will be used in the sequel: for all ; is monotone; that is, for all ;for each , ;for each , is convex and lower semicontinuous.

Theorem 1. *Let be a nonempty closed convex subset of . Let be a bifunction which satisfies conditions ( C1)–(C4). Let be a nonexpansive mapping. Let be a contraction. For arbitrarily, let the sequence be generated by
*

*where and satisfy , , , and . Then, the sequence converges strongly to provided .*

Chuang et al. [18] considered an iteration process of Halpern’s type for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points for a nonexpansive mapping with perturbation in a Hilbert space and they proved a strong convergence theorem for such iterations.

Theorem 2. *Let be a nonempty closed convex subset of . Let be a bifunction which satisfies conditions ( C1)–(C4). Let be a nonexpansive mapping. Let be a sequence. For arbitrarily, let the sequence be generated by
*

*where and satisfy , , and . Then, the sequence converges strongly to provided .*

S. Takahashi and W. Takahashi [17] introduced the following iterative algorithm for finding an element of EP:

And they proved that the sequence converges strongly to .

*Remark 3. *Algorithm (3) is involved in a variant anchor and the parameters are also relaxed. In [17], the authors considered a general equilibrium problem.

In this paper, our main purpose is to introduce a new iteration process for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points for a nonexpansive mapping in a Hilbert space and then we prove a strong convergence theorem for such iterations. Our iterations are very different from (2)–(4). As a special case, we can find the minimum norm solution of .

#### 2. Preliminaries

In the sequel, we assume is a real Hilbert space. Let be a nonempty closed convex set. Recall that a mapping is called *-inverse-strongly monotone* if there exists a positive real number such that
A mapping is said to be* nonexpansive* if
We use to denote the set of fixed points of .

The following lemmas are useful for the next section.

Lemma 4 (see [11]). *Let be a nonempty closed convex subset of . Let be a bifunction which satisfies conditions ( C1)–(C4). Let and . Then, there exists such that
*

*Set . Then the following hold:*(i)

*is single-valued and is firmly nonexpansive; that is for any , ;*(ii)

*is closed and convex and .*

*Lemma 5. Let , , , and be as in Lemma 4. Then the following holds:
for all and .*

*Lemma 6 (see [19]). Let be a nonempty closed convex subset of . Let the mapping be -inverse-strongly monotone and let be a constant. Then, we have
In particular, if , then is nonexpansive.*

*Lemma 7 (see [20]). Let be a closed convex subset of and let be a nonexpansive mapping with . Then, the mapping is demiclosed. That is, if is a sequence in such that weakly and strongly, then .*

*Lemma 8 (see [21]). Let and be two bounded sequences in . Let be a sequence in satisfying . Suppose that for all and . Then, .*

*Lemma 9 (see [22]). Assume that is a sequence of nonnegative real numbers such that
where is a sequence in and is a sequence such that(1);
(2) or . Then .*

*3. Main Results*

*In this section, we will prove our main results.*

*Theorem 10. Let be a nonempty closed convex subset of and let be a bifunction satisfying conditions (C1)–(C4). Let be an -inverse-strongly monotone mapping and let be a nonexpansive mapping. Suppose that . Let be a sequence in . For arbitrarily, let the sequence be generated by
where is defined as that in Lemma 4 and , , and satisfy(i) and ;(ii);
(iii), where and ;(iv). Then generated by (11) converges strongly to .*

*We divide our proofs into several conclusions.*

*Conclusion. The sequence is bounded.*

*Proof. *Let . We have for all . Set for all . By Lemma 4, we know that is nonexpansive. By the convexity of , we derive
Since is -inverse-strongly monotone, we know from Lemma 6 that
It follows that
So, we have that
Note that . Without loss of generality, we can assume that for some .

By induction, we have
Therefore, is bounded.

*Conclusion. and .*

*Proof. *Putting for all , we have
It follows that
From Lemma 6, we know that is nonexpansive for all . Thus, we have that is nonexpansive for all due to the fact that . Then, we get
By Lemma 5, we have
From (18)–(20), we obtain
Then,
Therefore,
Since , and , we obtain
By Lemma 8, we get
Consequently, we obtain
From (11) and (14), we have
Then, we obtain
Since , , and , we have
Next, we show . By using the firm nonexpansivity of , we have
We note that
Thus,
that is,
It follows that
Hence,
Since , , , and , we deduce
This implies that

*Conclusion. Consider
where .*

*Proof. *Since is bounded, there exists a subsequence of such that weakly and
By (25) and (37), we deduce
Hence,
This together with Lemma 7 implies that .

Next we show that . Since for any , we have
From (*C*2), we have
Put for all and . Then, we have . So, from (43), we have
Since is -inverse-strongly monotone, is -Lipschitzian. By (37), we derive that . Further, from monotonicity of , we have . Letting in (45) and noting , we have
By , , and (45), we deduce
and hence
Letting , we have, for each ,
This implies . Therefore, we have . So,
Setting for all and taking in (29) to get , so, . Since , we have

*Conclusion (). *

*Proof. *From (11), we have
It is clear that and . We can therefore apply Lemma 9 to conclude that . This completes the proof.

*Corollary 11. Let be a nonempty closed convex subset of and let be a bifunction satisfying conditions (C1)–(C4). Let be an -inverse-strongly monotone mapping and let be a nonexpansive mapping. Suppose that . For arbitrarily, let the sequence be generated by
where is defined as that in Lemma 4 and , , and satisfy(i) and ;(ii);
(iii), where and . Then generated by (52) converges strongly to which is the minimum norm element in .*

* Conflict of Interests*

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

*Acknowledgment*

*This study was supported by research funds from Dong-A University.*

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