## New Contribution to the Advancement of Fixed Point Theory, Equilibrium Problems, and Optimization Problems 2014

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# Hybrid Iterations for the Fixed Point Problem and Variational Inequalities

**Academic Editor:**Erdal Karapinar

#### Abstract

A hybrid iterative algorithm with Meir-Keeler contraction is presented for solving the fixed point problem of the pseudocontractive mappings and the variational inequalities. Strong convergence analysis is given as .

#### 1. Introduction

Throughout, we assume that is a real Hilbert space with the inner and the norm and is a nonempty closed convex set.

*Definition 1. *A mapping is said to be* pseudocontractive* if
for all .

We use to denote the set of fixed points of .

*Remark 2. *It is easily seen that (1) is equivalent to the following:
for all .

*Definition 3. *A mapping is said to be *-Lipschitzian* if
for all , where is a constant.

If , is said to be* nonexpansive*.

One of our purposes of this paper is to find the fixed points of the pseudocontractive mappings in Hilbert spaces. In the literature, there are a large number references associated with the fixed point algorithms for the pseudocontractive mappings. See, for instance, [1–9]. The first interesting algorithm for finding the fixed points of the Lipschitz pseudocontractive mappings in Hilbert spaces was presented by Ishikawa [4] in 1974.

*Ishikawa’s Algorithm.* For any , define the sequence iteratively by
for all , where and satisfy the following conditions:(a);(b).

Ishikawa proved that the sequence generated by (4) converges strongly to a fixed point of provided is a compact set.

Recently, Zhou [9] suggested the following algorithm.

*Zhou’s Algorithm.* For any , define the sequence iteratively by
where and are two real sequences in satisfying the following conditions:(a) for all ;(b).

Zhou proved that the sequence generated by (5) converges strongly to without the compactness assumption.

*Definition 4. *A mapping is said to be* inverse strongly monotone* if there exists such that
for all .

The variational inequality problem is to find such that The set of solutions of the variational inequality problem is denoted by . It is well known that variational inequality theory has emerged as an important tool in studying a wide class of obstacles, unilateral and equilibrium problems, which arise in several branches of pure and applied sciences in a unified and general framework. For related work, please refer to [10–18] and the references therein.

Motivated and inspired by the related work on the fixed point problem and the variational inequality problem in the literature, the purpose of this paper is continuous to study algorithmic approach to the fixed point problem of the pseudocontractive mappings and the variational inequality problem in Hilbert spaces. We suggest a hybrid algorithm with Meir-Keeler contraction and consequently we prove the strong convergence of the presented algorithm.

#### 2. Preliminaries

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

It is well known that, in a real Hilbert space , the following equality holds: for all and .

Lemma 5 (see [9]). *Let be a real Hilbert space and let be a closed convex subset of . Let be a continuous pseudocontractive mapping. Then,*(i)* is a closed convex set;*(ii)* is demiclosed at zero.*

Let be a sequence of nonempty closed convex sets. We define the symbols - and - as follows.(1)- there exists such that strongly.(2)- there exist a subsequence of and a sequence in such that weakly.

If satisfies the following: then we say that converges to in the sense of Mosco [19] and we write . It is easy to show that if is nonincreasing with respect to inclusion, then converges to in the sense of Mosco.

Tsukada [20] proved the following theorem for the metric projection.

Lemma 6 (see [20]). *Let be a Hilbert space. Let be a sequence of nonempty closed convex subsets of . If exists and is nonempty, then, for each , converges strongly to , where and are the metric projections of onto and , respectively.*

Let be a complete metric space. A mapping is called a* Meir-Keeler contraction* [21] if, for any , there exists such that
for all . It is well known that the Meir-Keeler contraction is a generalization of the contraction.

Lemma 7 (see [21]). *A Meir-Keeler contraction defined on a complete metric space has a unique fixed point.*

Lemma 8 (see [22]). *Let be a Meir-Keeler contraction on a convex subset of a Banach space . Then, for any , there exists such that
**
for all .*

Lemma 9 (see [22]). *Let be a convex subset of a Banach space . Let be a nonexpansive mapping on and let be a Meir-Keeler contraction on . Then the following holds.*(i)* is a Meir-Keeler contraction on .*(ii)*For each , is a Meir-Keeler contraction on .*

#### 3. Main Results

In this section, we firstly introduce a hybrid iterative algorithm for finding the common element of the fixed point problem and the variational inequality problem.

*Algorithm 10. *Let be a real Hilbert space and a nonempty closed convex set. Let be a Meir-Keeler contractive mapping. Let be a inverse strongly monotone mapping. Let be a -Lipschitz pseudocontractive mapping with . For arbitrarily, define a sequence iteratively by
where is a constant and and are two real number sequences in satisfying .

Next, we show the strong convergence of (14).

Theorem 11. *Suppose that . Then the sequence defined by (14) converges strongly to .*

*Remark 12. *Note that is a closed convex subset of . Thus is well defined. Since is a Meir-Keeler contraction of , it follows that is a Meir-Keeler contraction of by Lemma 9. According to Lemma 7, there exists a unique fixed point such that .

*Proof. *The outline of our proof is as follows.*Step **1. * for all ;*Step **2. * is closed and convex for all ;*Step **3. * where ;*Step **4. *;*Step **5. *;*Step **6. *.*Proof of Step **1.* We prove this step by induction. (i) is obvious. (ii) Suppose that for some . Pick up . Then, we have
By (2), we have
From (10), we obtain
Since is -Lipschitzian and , by (18), we get
By (10) and (16), we have
From (17), (19), and (20), we deduce
Since , we have
for all . This together with (21) implies that
By (10), (15), and (23) and noting that , we have
and hence . This indicates that for all .*Proof of Step **2.* In fact, it is obvious from the assumption that is closed convex. Suppose that is closed and convex for some . For any , we know that is equivalent to
So is closed and convex. By induction, we deduce that is closed and convex for all .*Proof of Step **3.* Firstly, from Step 2, we note that is well defined. Since is closed convex, we also have that is well defined and so is a Meir-Keeler contraction on . By Lemma 7, there exists a unique fixed point of . Since is a nonincreasing sequence of nonempty closed convex subsets of with respect to inclusion, it follows that
Setting and applying Lemma 6, we can conclude that

Now, we show that . Assume that . Then, for any with , we can choose such that
Since is a Meir-Keeler contraction, for the positive , there exists another such that
for all .

In fact, we can choose a common such that (28) and (29) hold. If , then
If , then, from (29), it follows that
for all . Thus, we have
for all . Since , there exists such that
for all .

Now, we consider two possible cases.*Case **1.* There exists such that
By (33) and (34), we get
By induction, we can obtain that
for all , which implies that
which contradicts (32). Therefore, we conclude that as .*Case **2* (* for all *). Now, we prove that Case 2 is impossible. Suppose that Case 2 is true. By Lemma 8, there exists such that
for all . Thus we have
for all . It follows that
which gives a contradiction. Hence we obtain
*Proof of Step **4.* By Step 3, we deduce immediately that is bounded. Observe that
Therefore, we have
Since , we have
This together with (44) implies that
From (15) and (24), we have
Then we have
By (46) and (48), we obtain
Since is firmly nonexpansive, we have
It follows that
From (24) and (51), we get
and so
This together with (46) and (49) implies that
Note that
It follows that
Since , we have by (54). So, from (56) and Lemma 5, we deduce that .*Proof of Step **5.* Define a mapping by
Then is maximal monotone (see [15]). Let . Since and , we have . On the other hand, from , we have
that is,
Therefore, we have
Noting that and is Lipschitz continuous, we obtain . Since is maximal monotone, we have and hence .*Proof of Step **6.* Since , we have
for all . Since , we get
for all . Noting that , we deduce
for all . Thus . 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 would like to thank the referees for useful comments and suggestions. Li-Jun Zhu was supported in part by NSFC of China (61362033) and NZ13087.

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#### Copyright

Copyright © 2014 Li-Jun Zhu 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.