A New Strong Convergence Theorem for Equilibrium Problems and Fixed Point Problems in Banach Spaces

Abstract

We introduce a new iterative sequence for finding a common element of the set of fixed points of a relatively nonexpansive mapping and the set of solutions of an equilibrium problem in a Banach space. Then, we study the strong convergence of the sequences. With an appropriate setting, we obtain the corresponding results due to Takahashi-Takahashi and Takahashi-Zembayashi. Some of our results are established with weaker assumptions.

1. Introduction

Throughout this paper, we denote by and the sets of positive integers and real numbers, respectively. Let be a Banach space, the dual space of and a closed convex subsets of . Let be a bifunction. The equilibrium problem is to find such that

(1.1)

The set of solutions of (1.1) is denoted by . The equilibrium problems include fixed point problems, optimization problems, variational inequality problems, and Nash equilibrium problems as special cases.

Let be a smooth Banach space and the normalized duality mapping from to . Alber [1] considered the following functional defined by

(1.2)

Using this functional, Matsushita and Takahashi [2, 3] studied and investigated the following mappings in Banach spaces. A mapping is relatively nonexpansive if the following properties are satisfied:

(R1),

(R2) for all and ,

(R3),

where and denote the set of fixed points of and the set of asymptotic fixed points of , respectively. It is known that satisfies condition (R3) if and only if is demiclosed at zero, where is the identity mapping; that is, whenever a sequence in converges weakly to and converges strongly to 0, it follows that . In a Hilbert space , the duality mapping is an identity mapping and for all . Hence, if is nonexpansive (i.e., for all ), then it is relatively nonexpansive.

Recently, many authors studied the problems of finding a common element of the set of fixed points for a mapping and the set of solutions of equilibrium problem in the setting of Hilbert space and uniformly smooth and uniformly convex Banach space, respectively (see, e.g., [4–21] and the references therein). In a Hilbert space , S. Takahashi and W. Takahashi [17] introduced the iteration as follows: sequence generated by ,

(1.3)

for every , where is nonexpansive, and are appropriate sequences in , and is an appropriate positive real sequence. They proved that converges strongly to some element in . In 2009, Takahashi and Zembayashi [19] proposed the iteration in a uniformly smooth and uniformly convex Banach space as follows: a sequence generated by ,

(1.4)

for every , is relatively nonexpansive, is an appropriate sequence in , and is an appropriate positive real sequence. They proved that if is weakly sequentially continuous, then converges weakly to some element in .

Motivated by S. Takahashi and W. Takahashi [17] and Takahashi and Zembayashi [19], we prove a strong convergence theorem for finding a common element of the fixed points set of a relatively nonexpansive mapping and the set of solutions of an equilibrium problem in a uniformly smooth and uniformly convex Banach space.

2. Preliminaries

We collect together some definitions and preliminaries which are needed in this paper. We say that a Banach space is strictly convex if the following implication holds for :

(2.1)

It is also said to be uniformly convex if for any , there exists such that

(2.2)

It is known that if is a uniformly convex Banach space, then is reflexive and strictly convex. We say that is uniformly smooth if the dual space of is uniformly convex. A Banach space is smooth if the limit exists for all norm one elements and in . It is not hard to show that if is reflexive, then is smooth if and only if is strictly convex.

Let be a smooth Banach space. The function (see [1]) is defined by

(2.3)

where the duality mapping is given by

(2.4)

It is obvious from the definition of the function that

(2.5)

(2.6)

for all and . The following lemma is an analogue of Xu's inequality [22, Theorem 2] with respect to .

Lemma 2.1.

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

(2.7)

for all , , and .

It is also easy to see that if and are bounded sequences of a smooth Banach space , then implies that .

Lemma 2.2 (see [23, Proposition 2]).

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

Remark 2.3.

For any bounded sequences and in a uniformly convex and uniformly smooth Banach space , we have

(2.8)

Let be a nonempty closed convex subset of a reflexive, strictly convex, and smooth Banach space . It is known that [1, 23] for any , there exists a unique point such that

(2.9)

Following Alber [1], we denote such an element by . The mapping is called the generalized projection from onto . It is easy to see that in a Hilbert space, the mapping coincides with the metric projection . Concerning the generalized projection, the following are well known.

Lemma 2.4 (see [23, Propositions 4 and 5]).

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

(a) if and only if for all ,

(b) for all .

Remark 2.5.

The generalized projection mapping above is relatively nonexpansive and .

Let be a reflexive, strictly convex and smooth Banach space. The duality mapping from onto coincides with the inverse of the duality mapping from onto , that is, . We make use of the following mapping studied in Alber [1]

(2.10)

for all and . Obviously, for all and . We know the following lemma (see [1] and [24, Lemma 3.2]).

Lemma 2.6.

Let be a reflexive, strictly convex and smooth Banach space, and let be as in (2.10). Then,

(2.11)

for all and .

Lemma 2.7 (see [25, Lemma 2.1]).

Let be a sequence of nonnegative real numbers. Suppose that

(2.12)

for all , where the sequences in and in satisfy conditions: , , and . Then, .

Lemma 2.8 (see [26, Lemma 3.1]).

Let be a sequence of real numbers such that there exists a subsequence of such that for all . Then, there exists a nondecreasing sequence such that ,

(2.13)

for all . In fact, .

For solving the equilibrium problem, we usually 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 and lower semicontinuous.

The following lemma gives a characterization of a solution of an equilibrium problem.

Lemma 2.9 (see [19, Lemma 2.8 ]).

Let be a nonempty closed convex subset of a reflexive, strictly convex, and uniformly smooth Banach space . Let be a bifunction satisfying conditions (A1)–(A4). For , define a mapping so-called the resolvent of as follows:

(2.14)

for all . Then, the following hold:

(i) is single-valued,

(ii) is a firmly nonexpansive-type mapping [27], that is, for all

(2.15)

(iii),

(iv) is closed and convex,

Lemma 2.10 (see [4, Lemma 2.3]).

Let be a nonempty closed convex subset of a Banach space , a bifunction from satisfying conditions (A1)–(A4) and . Then, if and only if for all .

Remark 2.11 (see [27]).

Let be a nonempty subset of a smooth Banach space . If is a firmly nonexpansive-type mapping, then

(2.16)

for all and . In particular, satisfies condition (R2).

Lemma 2.12 (see [3, Proposition 2.4]).

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

3. Main Results

In this section, we prove a strong convergence theorem for finding a common element of the fixed points set of a relatively nonexpansive mapping and the set of solutions of an equilibrium problem in a uniformly convex and uniformly smooth Banach space.

Theorem 3.1.

Let be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space and a bifunction satisfying conditions (A1)–(A4) and a relatively nonexpansive mapping such that . Let and be sequences generated by , and

(3.1)

for all , where satisfying and , , and . Then, and converge strongly to .

Proof.

Note that can be rewritten as . Since is nonempty, closed, and convex, we put . Since , , and satisfy condition (R2), by (2.6), we get

(3.2)

and so

(3.3)

By induction, we have

(3.4)

for all . This implies that is bounded and so are , , and . Put

(3.5)

Then, . Using Lemma 2.6 gives

(3.6)

Let be a function satisfying the properties of Lemma 2.1, where . Then, by Remark 2.11 and (3.6), we get

(3.7)

(3.8)

where for all . Notice that satisfying and .

The rest of the proof will be divided into two parts.

Case 1.

Suppose that there exists such that is nonincreasing. In this situation, is then convergent. Then,

(3.9)

It follows from (3.7) and that

(3.10)

Since ,

(3.11)

Consequently, by Remark 2.3,

(3.12)

From (2.6) and , we obtain

(3.13)

This implies that

(3.14)

Therefore,

(3.15)

Since is bounded and is reflexive, we choose a subsequence of such that and

(3.16)

Then, . Since and , by Remark 2.3,

(3.17)

Notice that

(3.18)

Replacing by , we have from (A2) that

(3.19)

Letting , we have from (3.17) and (A4) that

(3.20)

From Lemma 2.10, we have . Since satisfies condition (R3) and (3.15), . It follows that . By Lemma 2.4(a), we immediately obtain that

(3.21)

Since ,

(3.22)

It follows from Lemma 2.7 and (3.8) that . Then, and so .

Case 2.

Suppose that there exists a subsequence of such that

(3.23)

for all . Then, by Lemma 2.8, there exists a nondecreasing sequence such that ,

(3.24)

for all . From (3.7) and , we have

(3.25)

Using the same proof of Case 1, we also obtain

(3.26)

From (3.8), we have

(3.27)

Since , we have

(3.28)

In particular, since , we get

(3.29)

It follows from (3.26) that . This together with (3.27) gives

(3.30)

But for all , we conclude that , and .

From two cases, we can conclude that and converge strongly to and the proof is finished.

Applying Theorem 3.1 and [28, Theorem 3.2], we have the following result.

Theorem 3.2.

Let be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space , a bifunction satisfying conditions (A1)–(A4), and a sequence of relatively nonexpansive mappings such that . Let and be sequences generated by (3.1), where is defined by

(3.31)

Then, and converge strongly to .

Setting and in Theorem 3.1, we have the following result.

Corollary 3.3.

Let be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space and a relatively nonexpansive mapping. Let and be sequences generated by , and

(3.32)

for all , where satisfying and , . Then, and converge strongly to .

Letting in Corollary 3.3, we have the following result.

Corollary 3.4.

Let be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space and a relatively nonexpansive mapping. Let be a sequence in defined by , and

(3.33)

for all , where satisfying and , . Then converges strongly to .

Let be the identity mapping in Theorem 3.1, we also have the following result.

Corollary 3.5.

Let be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space and a bifunction satisfying conditions (A1)–(A4) such that . Let and be sequences generated by and

(3.34)

for all , where satisfying and , , and . Then, and converge strongly to .

4. Deduced Theorems in Hilbert Spaces

In Hilbert spaces, every nonexpansive mappings are relatively nonexpansive, and is the identity operator. We obtain the following result.

Theorem 4.1.

Let be a nonempty closed convex subset of a Hilbert space , a bifunction satisfying conditions (A1)–(A4), and a nonexpansive mapping such that . Let be a sequence in defined by and

(4.1)

for all , where is the resolvent of , satisfying and , , and . Then, converges strongly to .

Remark 4.2.

In Theorem 4.1, we have the same conclusion if the mapping is only quasinonexpansive (i.e., and for all and ) such that is demiclosed at zero.

Letting in Theorem 4.1, we have the following result.

Corollary 4.3.

Let be a nonempty closed convex subset of a Hilbert space and a nonexpansive mapping such that . Let be a sequence in defined by and

(4.2)

for all , where satisfying , , and . Then, converges strongly to .

Let be the identity mapping in Theorem 4.1, we have the following result.

Corollary 4.4.

Let be a nonempty closed convex subset of a Hilbert space and a bifunction satisfying conditions (A1)–(A4). Let be a sequence in defined by and

(4.3)

for all , where is the resolvent of , satisfying , , and . Then converges strongly to .

Proof.

We may assume without loss of generality that for all . Setting and for all , we get

(4.4)

, and . Applying Theorem 4.1, converges strongly to .

Remark 4.5.

Corollary 4.4 improves and extends [29, Corollary 5.3]. More precisely, the conditions and are removed.

Applying Corollary 4.4 and [30, Theorem 8], we have the following result.

Corollary 4.6.

Let be a nonempty closed convex subset of a Hilbert space , a bifunction satisfying conditions (A1)–(A4), and a contraction of into itself. Let be a sequence in defined by and

(4.5)

for all , where is the resolvent of , satisfying and and . Then, converges strongly to .

Remark 4.7.

Corollary 4.6 improves and extends [16, Corollary 3.4]. More precisely, the conditions and are removed.