Shrinking Projection Method of Common Solutions for Generalized Equilibrium Quasi--Nonexpansive Mapping and Relatively Nonexpansive Mapping

Abstract

We prove a strong convergence theorem for finding a common element of the set of solutions for generalized equilibrium problems, the set of fixed points of a relatively nonexpansive mapping, and the set of fixed points of a quasi--nonexpansive mapping in a Banach space by using the shrinking Projection method. Our results improve the main results in S. Takahashi and W. Takahashi (2008) and Takahashi and Zembayashi (2008). Moreover, the method of proof adopted in the paper is different from that of S. Takahashi and W. Zembayashi (2008).

1. Introduction

Let be a Banach space and let be a closed convex subsets of . Let be an equilibrium bifunction from into and let be a nonlinear mapping. Then, we consider the following generalized equilibrium problem: find such that

(1.1)

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

(1.2)

In the case of , is denoted by . In the case of , is denoted by .

A mapping is said to be nonexpansive if

(1.3)

We denote the set of fixed points of by .

A mapping is said to be Quasi--nonexpansive if

(1.4)

where is defined by (2.3).

Recently, in Hilbert space, Tada and Takahashi [1], and S. Takahashi and W. Takahashi [2] considered iterative methods for finding an element of . Very recently, S. Takahashi and W. Takahashi [3] introduced an iterative method for finding an element of , where is an inverse-strongly monotone mapping and then proved a strong convergence theorem. On the other hand, Takahashi and Zembayashi [4] 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 which is different from S. Takahashi and W. Takahashi's hybrid method [3].

In this paper, motivated by Takahashi and Zembayashi [4], in Banach space, we prove a strong convergence theorem for finding an element of , where is a continuous and monotone operator, is a relatively nonexpansive mapping, and is quasi--nonexpansive mapping. Moreover, the method of proof adopted in the paper is different from that of [3].

2. Preliminaries

Throughout this paper, 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 . Then, the duality mapping on is defined by

(2.1)

for every . By the Hahn-Banach theorem, is nonempty; see [5] for more details. We denote the weak convergence and the strong convergence of a sequence to in by and , respectively. We also denote the convergence of a sequence to in by . A Banach space is said to be strictly convex if for with and . It is also said to be uniformly convex if for each , there exists such that for with and . A uniformly convex Banach space has the Kadec-Klee property, that is, and imply . The space is said to be smooth if the limit

(2.2)

exists for all . It is also said to be uniformly smooth if the limit exists uniformly in . We know that if is smooth, strictly convex, and reflexive, then the duality mapping is single valued, one to one, and onto; see [6] for more details.

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

(2.3)

Following Alber [7], the generalized projection from onto is defined by , where is the solution to the following minimization problem:

(2.4)

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

(2.5)

If is a Hilbert space, then and is the metric projection of onto . It is well know that the following conclusions for generalized projections hold.

Lemma 2.1 (Alber [7] and Kamimura and Takahashi [8]).

Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space Then

(2.6)

Lemma 2.2.

Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space , let , and let . Then

(2.7)

Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space , and let be a mapping from into itself. We denoted by the set of fixed points of . A point is said to be an asymptotic fixed point of [9, 10] if there exists in which converges weakly to and . We denote the set of all asymptotic fixed point of by . Following Matsushita and Takahashi [11], a mapping is said to be relatively nonexpansive if the following conditions are satisfied:

(1) is nonempty;

(2) for all ;

(3).

The following lemma is due to Matsushita and Takahashi [11].

Lemma 2.3 (Matsushita and Takahashi [11]).

Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space , and let be a relatively nonexpansive mapping from into itself. Then is closed and convex.

We also know the following lemmas.

Lemma 2.4 (see [8]).

Let be a smooth and uniformly convex Banach space and let and be sequences in such that either or is bounded. If , then .

Lemma 2.5 (see [12]).

Let be a uniformly convex Banach space and be a closed ball of . Then there exists a continuous, stricting increasing and convex function with such that

(2.8)

for all and .

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 ,

(2.9)

for each is a convex and lower semicontinuous.

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

Lemma 2.6 (see [13]).

Let be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space , let be an equilibrium bifunction satisfying conditions ()–(), and let for any given . Then, there exists such that

(2.10)

Lemma 2.7 (see [4]).

Let be a nonempty closed convex subset of a uniformly smooth, strictly convex and reflexive Banach space ; let be an equilibrium bifunction satisfying conditions ()–(). For and , define a mapping as follows:

(2.11)

for all . Then, the following holds:

(1) is single-valued;

(2) is a firmly nonexpansive-type mapping that is, for any ,

(2.12)

(3);

(4) is a closed and convex set.

Lemma 2.8 (see [4]).

Let be a nonempty closed convex subset of a uniformly smooth, strictly convex and reflexive Banach space let be an equilibrium bifunction satisfying conditions ()–(). For , and ,

(2.13)

3. The Main Results

In this section, we prove a strong convergence theorem which is the main result in the paper.

Theorem 3.1.

Let be a uniformly smooth and uniformly convex Banach space, and let be a nonempty closed convex subset of . Let be a continuous and monotone operator. Let be a bifunction from to which satisfies ()–(), let be a relatively nonexpansive mapping of into itself such that , and let be a closed nonexpansive mapping. Let be the sequence generated by and

(3.1)

for every , where is the duality mapping on , , and for some . If the following conditions are satisfied

,

,

then converges strongly to where is the generalized projection of onto .

Proof.

We define a bifunction by

(3.2)

Next, we prove that the bifunction satisfies conditions ()–().

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

(3.3)

For each ,

(3.4)

Since

(3.5)

For each is a convex and lower semicontinuous.

For each and , since satisfies , we have

(3.6)

So, is convex.

Similarly, we can prove that is lower semicontinuous.

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

(3.7)

and (3.1) can be written as

(3.8)

Since the bifunction satisfies conditions ()–(), from Lemma 2.7, for a given and , we can define a mapping as follows:

(3.9)

Moreover, satisfies the conclusions in Lemma 2.7.

Putting for all , we have from Lemmas 2.7 and 2.8 that are relatively nonexpansive.

We divide the proof of Theorem 3.1 into six steps.

Step 1.

We first show that is closed and convex. It is obvious that is closed. Since

(3.10)

is convex. So, is a closed convex subset of for all .

Step 2.

Next we show by induction that for all . From , we have

(3.11)

Suppose that for some . For any since and are relatively nonexpansive, is nonexpansive, we have

(3.12)

Hence, we have This implies that

(3.13)

So, is well defined.

Step 3.

Next we prove that the sequences are bounded. From the definition of , we have

(3.14)

for all . Then is bounded. Therefore, ,, and are bounded.

Step 4.

Next we prove that

(3.15)

From and , we have

(3.16)

Thus, is nondecreasing. So, the limit of exists. Since

(3.17)

for all , we have . From , we have

(3.18)

Therefore, we also have

(3.19)

Since and is uniformly convex and smooth, we have from Lemma 2.4 that

(3.20)

So, we have

(3.21)

Since is uniformly norm-to-norm continuous on bounded sets and , we have

(3.22)

For any , from Lemma 2.5 and (3.8), we have

(3.23)

Therefore, we have

(3.24)

Since

(3.25)

from (3.21) and (3.22), we have

(3.26)

Since , we have

(3.27)

Therefore, from the property of , we have

(3.28)

Since is uniformly norm-to-norm continuous on bounded sets, we have

(3.29)

Similarly, we have

(3.30)

Therefore, we have

(3.31)

From (3.26) and , we have

(3.32)

Therefore, from the property of , we have

(3.33)

Since is uniformly norm-to-norm continuous on bounded sets, we have

(3.34)

Step 5.

Next we prove that

(3.35)

where .

(a)We prove that

In fact, for any given , there exists a subsequence of such that . Since and is relatively nonexpansive, we have , that is, .

(b)We prove that .

In fact, from , (3.12) and Lemma 2.8, we have that

(3.36)

Hence it follows from (3.26) that

(3.37)

Since is uniformly convex and smooth and is bounded, we have from Lemma 2.4 that

(3.38)

For any given , there exists a subsequence such that . Since , we have .

Since is uniformly norm-to-norm continuous on bounded sets, from (3.38), we have

(3.39)

From , we have

(3.40)

By , we have

(3.41)

Replacing by , we have from that

(3.42)

Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. So, letting , we have from (3.42) and that

(3.43)

For any with and , let . Since and hence , from conditions and , we have

(3.44)

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

(c)Now we prove that .

In fact, for any given , there exists a subsequence such that . Since , we have Since is a closed convex subset of . we have , that is, . From (3.14) and (3.16), we have

(3.45)

Since the norm is weakly lower semicontinuous, we have

(3.46)

that is, , then, . Since is uniformly convex Banach space, has a Kadec-Klee property, we have . From (3.34) and being closed, we have , that is, .

Step 6.

Finally we prove that

(3.47)

where . From and , we have

(3.48)

From the definition of , we have . Hence, . Therefore, we have

(3.49)

Since has the Kadec-Klee property, we have that . Therefore, converges strongly to .

This completes the proof of Theorem 3.1.

Theorem 3.2.

Let be a uniformly smooth and uniformly convex Banach space, and let be a nonempty closed convex subset of . and be a continuous and monotone operator. Let be a bifunction from which satisfies ()–() and let be a relatively nonexpansive mapping of into itself such that . Let be the sequence generated by , and

(3.50)

for every , where is the duality mapping on , satisfies and for some . Then converges strongly to , where is the generalized projection of onto .

Proof.

In Theorem 3.1, take , we get Therefore, the conclusion of Theorem 3.2 can be obtained from Theorem 3.1.

Remark 3.3.

Theorem 3.1 in [3] and Theorem 3.1 in [4] are special cases of Theorem 3.2.