Strong Convergence Theorem by Monotone Hybrid Algorithm for Equilibrium Problems, Hemirelatively Nonexpansive Mappings, and Maximal Monotone Operators

Abstract

We introduce a new hybrid iterative algorithm for finding a common element of the set of fixed points of hemirelatively nonexpansive mappings and the set of solutions of an equilibrium problem and for finding a common element of the set of zero points of maximal monotone operators and the set of solutions of an equilibrium problem in a Banach space. Using this theorem, we obtain three new results for finding a solution of an equilibrium problem, a fixed point of a hemirelatively nonexpnasive mapping, and a zero point of maximal monotone operators in a Banach space.

1. Introduction

Let be a Banach space, let be a closed convex subset of , and let be a bifunction from to , where is the set of real numbers. The equilibrium problem is to find

(11)

The set of such solutions is denoted by .

In 2006, Martinez-Yanes and Xu [1] obtained strong convergence theorems for finding a fixed point of a nonexpansive mapping by a new hybrid method in a Hilbert space. In particular, Takahashi and Zembayashi [2] established 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 nonexpansive mapping in a uniformly convex and uniformly smooth Banach space. Very recently, Su et al. [3] proved the following theorem by a monotone hybrid method.

Theorem 1.1 (see Su et al. [3]).

Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , and let be a closed hemirelatively nonexpansive mapping such that . Assume that is a sequence in such that . Define a sequence in by the following:

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where is the duality mapping on . Then, converges strongly to , where is the generalized projection from onto .

In this paper, motivated by Su et al. [3], we 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 hemirelatively nonexpansive mapping and for finding a common element of the set of zero points of maximal monotone operators and the set of solutions of an equilibrium problem in a Banach space by using the monotone hybrid method. Using this theorem, we obtain three new strong convergence results for finding a solution of an equilibrium problem, a fixed point of a hemirelatively nonexpnasive mapping, and a zero point of maximal monotone operators in a Banach space.

2. Preliminaries

Let be a real Banach space with dual . We denote by the normalized duality mapping from to defined by

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where denotes the generalized duality pairing. It is well known that if is uniformly convex, then is uniformly continuous on bounded subsets of . In this case, is single valued and also one to one.

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

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Following Alber [4], the generalized projection from onto is defined by

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

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

If is a reflexive strict convex and smooth Banach space, then for if and only if . It is sufficient to show that if , then . From (2.4), we have . This implies . From the definition of , we have , that is, .

Let be a closed convex subset of and let be a mapping from into itself. We denote by the set of fixed points of . is called hemirelatively nonexpansive if for all and .

A point in is said to be an asymptotic fixed point of [5] if contains a sequence which converges weakly to such that the strong . The set of asymptotic fixed points of will be denoted by . A hemirelatively nonexpansive mapping from into itself is called relatively nonexpansive [1, 5, 6] if .

We need the following lemmas for the proof of our main results.

Lemma 2.1 (see Alber [4]).

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

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Lemma 2.2 (see Alber [4]).

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

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Lemma 2.3 (see Kamimura and Takahashi [7]).

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

Lemma 2.4 (see Xu [8]).

Let be a uniformly convex Banach space and let . Then, there exists a strictly increasing, continuous, and convex function such that and

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where .

Lemma 2.5 (see Kamimura and Takahashi [7]).

Let be a smooth and uniformly convex Banach space and let . Then, there exists a strictly increasing, continuous, and convex function such that and

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

• (A1) for all

• (A2) is monotone, that is, for all

• (A3) for all ;

• (A4) for all is convex.

Lemma 2.6 (see Blum and Oettli [9]).

Let be a closed convex subset of a smooth, strictly convex, and reflexive Banach space , let be a bifunction from to satisfying (A1)–(A4), let , and let . Then, there exists such that

(29)

Lemma 2.7 (see Takahashi and Zembayashi [10]).

Let be a closed convex subset of a uniformly smooth, strictly convex, and reflexive Banach space , let be a bifunction from to satisfying (A1)–(A4), and let , for . Define a mapping as follows:

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Then, the following holds:

• (1) is single valued;

• (2) is a firmly nonexpansive-type mapping [11], that is, for all ,

  

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• (3) ;

• (4) is closed and convex.

Lemma 2.8 (see Takahashi and Zembayashi [10]).

Let be a closed convex subset of a smooth, strictly convex, and reflexive Banach space and let be a bifunction from to satisfying (A1)–(A4). Then, for and , and ,

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Lemma 2.9 (see Su et al. [3]).

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

Recall that an operator in a Banach space is called closed, if , then .

3. Strong Convergence Theorem

Theorem 3.1.

Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , let be a bifunction from to satisfying (A1)–(A4), and let be a closed hemirelatively nonexpansive mapping such that . Define a sequence in by the following:

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for every , where is the duality mapping on are sequences in such that and for some . Then, converges strongly to , where is the generalized projection of onto .

Proof.

First, we can easily show that and are closed and convex for each .

Next, we show that for all . Let . Putting for all , from Lemma 2.8, we have relatively nonexpansive. Since are relatively nonexpansive and is hemirelatively nonexpansive, we have

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Hence, we have

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Next, we show that for all . We prove this by induction. For , we have

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Suppose that , by Lemma 2.2, we have

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As , by the induction assumptions, the last inequality holds, in particular, for all . This, together with the definition of , implies that . So, is well defined.

Since and for all , we have

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Therefore, is nondecreasing. In addition, from the definition of and Lemma 2.2, . Therefore, for each , we have

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Therefore, and are bounded. This, together with (3.6), implies that the limit of exists. From Lemma 2.1, we have, for any positive integer ,

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Therefore,

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From (3.9), we can prove that is a Cauchy sequence. Therefore, there exists a point such that converges strongly to .

Since , we have

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Therefore, we have

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From Lemma 2.3, we have

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So, we have

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

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Let Since is a uniformly smooth Banach space, we know that is a uniformly convex Banach space. Therefore, from Lemma 2.4, there exists a continuous, strictly increasing, and convex function with , such that

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for , and . So, we have that for ,

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Therefore, we have

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Since

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we have

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From , we have

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Therefore, from the property of , we have

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

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Since is a closed operator and , then is a fixed point of .

On the other hand,

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So, we have from (3.19) that

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From Lemma 2.3, we have that

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From and , we have .

From (3.25), we have

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From , we have

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By , we have

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From (A2), we have that

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From (3.27) and (A4), we have

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For with and , let . We have . So, from (A1), we have

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Dividing by , we have

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Letting , from (A3), we have

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Therefore, . Finally, we prove that . From Lemma 2.1, we have

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Since and , for all , we get from Lemma 2.1 that

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By the definition of , it follows that and , whence . Therefore, it follows from the uniqueness of that . This completes the proof.

Corollary 3.2.

Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , and let be a bifunction from to satisfying (A1)–(A4). Define a sequence in by the following:

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for every , where is the duality mapping on and for some . Then, converges strongly to .

Proof.

Putting in Theorem 3.1, we obtain Corollary 3.2.

Corollary 3.3.

Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , and let be a closed hemirelatively nonexpansive mapping. Define a sequence in by the following:

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for every , where is the duality mapping on are sequences in such that . Then, converges strongly to .

Proof.

Putting for all and for all in Theorem 3.1, we obtain Corollary 3.3.

Corollary 3.4.

Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , let be a bifunction from to satisfying (A1)–(A4), and let be a closed relatively nonexpansive mapping such that . Define a sequence in by the following:

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for every , where is the duality mapping on are sequences in such that and for some . Then, converges strongly to .

Proof.

Since every relatively nonexpansive mapping is a hemirelatively one, Corollary 3.4 is implied by Theorem 3.1.

Remark 3.5 (see Rockafellar [12]).

Let be a reflexive, strictly convex, and smooth Banach space and let be a monotone operator from to . Then, is maximal if and only if for all .

Let be a reflexive, strictly convex, and smooth Banach space and let be a maximal monotone operator from to . Using Remark 3.5 and strict convexity of , we obtain that for every and , there exists a unique such that If , then we can define a single-valued mapping by , and such a is called the resolvent of . We know that for all and is relatively nonexpansive mapping (see [2] for more details). Using Theorem 3.1, we can consider the problem of strong convergence concerning maximal monotone operators in a Banach space.

Theorem 3.6.

Let be a uniformly convex and uniformly smooth real Banach space, let be a nonempty closed convex subset of , let be a bifunction from to satisfying (A1)–(A4), and let be a resolvent of and a closed mapping such that , where . Define a sequence in by the following:

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for every , where is the duality mapping on , is a sequences in such that and for some , Then, converges strongly to .

Proof.

Since is a closed relatively nonexpansive mapping and , from Corollary 3.4, we obtain Theorem 3.6.