# On a Hybrid Method for Generalized Mixed Equilibrium Problem and Fixed Point Problem of a Family of Quasi- -Asymptotically Nonexpansive Mappings in Banach Spaces

- Min Liu
^{1}, - Shih-sen Chang
^{1}Email author and - Ping Zuo
^{1}

**2010**:157278

**DOI: **10.1155/2010/157278

© Min Liu et al. 2010

**Received: **30 November 2009

**Accepted: **19 January 2010

**Published: **14 February 2010

## Abstract

We prove a strong convergence theorem by using a hybrid method for finding a common element of the set of solutions for generalized mixed equilibrium problems, the set of fixed points of a family of quasi-
-asymptotically nonexpansive mappings in strictly convex reflexive Banach space with the Kadec-Klee property and, a *Fréchet* differentiable norm under weaker conditions. The method of the proof is different from, S. Takahashi and W. Takahashi that by (2008) and that by Takahashi and Zembayashi (2008) and see references. It also shows that the type of projection used in the iterative method is independent of the properties of the mappings. The results presented in the paper improve and extend some recent results.

## 1. Introduction

Let be a Banach space and let be a closed convex subsets of . Let be an equilibrium bifunction from into , let be a real-valued function, and let be a nonlinear mapping. The "so-called" generalized mixed equilibrium problem is to find such that

The set of solutions of (1.1) is denoted by , that is,

Sepecial Examples

which is called the mixed equilibrium problem; see [1]. The set of solutions of (1.3) is denoted by MEP.

which is called the mixed variational inequality of Browder type. The set of solutions of (1.4) is denoted by .

which is called the generalized equilibrium problem; see [2]. The set of solutions of (1.5) is denoted by EP.

which is called the equilibrium problem. The set of solutions of (1.6) is denoted by EP(F).

Recently, Tada and Takahashi [3] and S. Takahashi and W. Takahashi [4] considered iterative methods for finding an element of in Hilbert space. Very recently, S.Takahashi and W.Takahashi [2] introduced an iterative method for finding an element of , where is an inverse-strongly monotone mapping and is nonexpansive mapping and then proved a strong convergence theorem in Hilbert space. On the other hand, Takahashi and Zembayashi [5] prove a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a relatively nonexpansive mapping in a Banach space by using the shrinking Projection method. Very recently, Kimura and Takahashi [6] prove a strong convergence theorem for a family of relatively nonexpansive mapping in a Banach space by using a hybrid method.

In this paper, motivated by Kimura and Takahashi [6], we prove a strong convergence theorem for finding an element of in Banach space by using a hybrid method, where is a continuous and monotone operator and is a family of quasi- -asymptotically nonexpansive mapping. Moreover, the method of proof adopted in the paper is different from that of [2, 5].

## 2. Preliminaries

Throughout this paper, we assume that all the Banach spaces are real. We denote by and the sets of positive integers and real numbers, respectively. Let be a Banach space and let be the topological dual of . For all and , we denote the value of at by . The duality mapping is defined by

By Hahn-Banach theorem, is nonempty; see [7] for more details. We denote the weak convergence and the strong convergence of a sequence to in by and , respectively. A Banach space is said to be strictly convex if for and . It is also said to be uniformly convex if for each there exists such that for and . is said to have the Kadec-Klee property, that is, for any sequence , if and , then .

Define by

for
and
. A norm of
is said to be
differentiable if
has a limit for each
. In this case,
is said to be smooth. A norm of
is said to be *Fréchet* differentiable if
is attained uniformly for
for each
. It is known that
has a *Fréchet* differentiable norm if and only if
is strictly convex and reflexive, and has the Kadec-Klee property. We know that if
is smooth, strictly convex, and reflexive, then the duality mapping
is single valued, one to one, and onto. In this case, the inverse mapping
coincides with the duality mapping
on
. See [8] for more details.

Remark 2.1.

If
is a reflexive and strictly convex Banach space, then
is hemicontinuous, that is,
is *norm-weak continuous*.

Let be a smooth, strictly convex and reflexive Banach space and let be a closed convex subset of . Throughout this paper, we denote by the function defined by

Let be a sequence of nonempty closed convex subset of a reflexive Banach space . We define two subsets and as follows: if and only if there exists such that converges strongly to and that for all . Similarly, if and only if there exists a subsequence of and a sequence such that converges weakly to and for all . We define the Mosco convergence [9] of as follows: if satisfies that , then it is said that converges to in the sense of Mosco and we write . For more details, see [10].

Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space . Then, for arbitrarily fixed , a function has a unique minimizer . Using such a point, we define the metric projection by for every . In a similar fashion, we can see that a function has a unique minimizer . The generalized projection of onto is defined by for every ; see [11].

The generalized projection from onto is well defined and single valued and satisfies

If is a Hilbert space, then and is the metric projection of onto .

It is well-known that the following conclusions hold.

Lemma 2.3.

Let be a nonempty closed convex subsets of a smooth, strictly convex and reflexive Banach spaces , let and let . Then the following conclusions hold:

(a)

(b)For if and only if .

The following theorem proved by Tsukada [13] plays an important role in our results.

Theorem 2.4.

Let be a smooth, reflexive, and strictly convex Banach space having the Kadec-Klee property. Let be a sequence of nonempty closed convex subset of . If exists and is nonempty, then converges strongly to for each .

Theorem 2.4 is still valid if we replace the metric projections with the generalized projections as follows:

Theorem 2.5.

Let be a smooth, reflexive, and strictly convex Banach spaces having the Kadec-Klee property. Let be a sequence of nonempty closed convex subsets of . If exists and is nonempty, then converges strongly to for each .

Let be a nonempty closed convex subsets of , and let be a mapping from into itself. We denoted by the set of fixed points of . is said to be -asymptotically nonexpansive, if there exists some real sequence with and such that for all and . is said to be quasi- -asymptotically nonexpansive [14], if there exists some real sequence with and and such that for all , and . is said to be uniformly Lipschitzian continuous if there exists some such that for all and . A point is said to be an asymptotic fixed point of [15, 16] if there exists in which converges weakly to and . We denote the set of all asymptotic fixed point of by . Following Matsushita and Takahashi [17], a mapping is said to be relatively nonexpansive if the following conditions are satisfied:

(1) is nonempty,

(2) , for all ,

(3) .

A mapping is said to be quasi- -nonexpansive, if .

We remark that a quasi- -nonexpansive mapping with a nonempty fixed point set is a quasi- -asymptotically nonexpansive mapping, but the converse may be not true.

A mapping is said to be closed, if for any sequence with and , .

Lemma 2.6.

Let be a strictly convex reflexive Banach space having the Kadec-Klee property and a Fréchet differentiable norm, be a nonempty closed convex subset of , and let be a uniformly Lipschitzian continuous and quasi- -asymptotically nonexpansive mapping from into itself. Then is closed and convex.

Proof.

that is, . This implies that . Thus from (2.8) we have . Since and has the Kadec-Klee property, we have . Note that is hemicontinuous, it yields that . Again since , by using the Kadec-Klee property of , we have . Hence as . Since is uniformly Lipschitzian continuous, we have . This completes the proof.

For solving the equilibrium problem for bifunction , let us assume that satisfies the following conditions:

for all ,

is monotone, that is, for all ,

for each is a convex and lower semicontinuous.

If an equilibrium bifunction satisfies conditions , then we have the following two important results.

Lemma 2.7 (see[18]).

Lemma 2.8 (see[5]).

for all . Then, the following hold:

(1) is single-valued,

(3) ,

(4) is a closed and convex set.

Lemma 2.9 (see[5]).

## 3. The Main Results

Lemma 3.1.

Let be a strictly convex reflexive Banach space having a Fréchet differentiable norm, a nonempty closed convex subset of and a sequence of mappings of into itself. Let be a strongly convergent sequence in with a limit and a sequence in defined by for each , where is a convergent sequence in with a limit . Suppose that for all and that converges weakly to , where . Then converges strongly to 0. Moreover, if has the Kadec-Klee property, then converges strongly to .

Proof.

Therefore, we have that and hence . Thus we have that converges weakly to . It also holds that

*Fréchet*differentiable norm, it follows that has the Kadec-Klee property, and thus we have that converges strongly to . Then, we have that

*Fréchet*differentiable and, therefore, is norm-to-norm continuous. Hence we have that

which is the desired result.

Theorem 3.2.

where is the duality mapping on and for all , where . Let be a sequence in such that and for some , then converge strongly to , where is the metric projection of onto .

Proof.

Next, we prove that the bifunction satisfies conditions as follows

for all ,

since for all .

is monotone, that is, for all .

Since is a continuous and monotone operator, hence from the definition of we have

Since is continuous and is lower semicontinuous, we have

For each is a convex and lower semicontinuous.

So, is convex.

Similarly, we can prove that is lower semi-continuous.

Therefore, the generalized mixed equilibrium problem (1.1) is equivalent to the following equilibrium problem: find such that

Since the bifunction satisfies conditions , from Lemma 2.8, for given and , we can define a mapping as follows:

Moreover, satisfies the conclusions in Lemma 2.8.

Putting for all , we have from Lemma 2.8 and Lemma 2.9 that is relatively nonexpansive.

We divide the proof of Theorem 3.2 into five steps.

Step 1.

and thus is closed and convex for every .

Step 2.

Next we show that for each and .

for any , since is relatively nonexpansive and is quasi- -asymptotically nonexpansive, we have

Step 3.

Step 4.

Next we prove that .

(a) First, we prove that .

Since for every , it follows that for every . Fix arbitrarily. From the assumption that , we may take subsequences of and of such that with and converges weakly to a point . Then, by Lemma 3.1, we have that

Since is uniformly Lipschitzian continuous, from (3.24) and (3.27), we have , that is, .

(b) Next we prove that .

( ) In fact, since , we have

Since is norm-to-norm continuous, hence we have that

( ) Next we prove that

Since

hence it follows from (3.38) and (3.40) that

Since is uniformly norm-to-norm continuous on bounded sets, from (3.47), we have

This implies that . Hence from condition , we have for all , and hence .

Step 5.

Finally we prove that .

Since and is a nonempty closed convex subset of , we conclude that

This completes the proof of Theorem 3.2.

The proof of Theorem 3.2 shows that the properties of projections used in the iterative scheme do not interact with the properties of mappings . Therefore, we may prove similar results as follows by replacing Theorem 2.4 with Theorem 2.5 in the proof.

Theorem 3.3.

where is the duality mapping on and for all , where . Let be a sequence in such that and for some , then converge strongly to , where is the generalized projection of onto .

## Declarations

### Acknowledgments

The authors would like to express their thanks to the referees for their helpful suggestions and comments.

## Authors’ Affiliations

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