Abstract

An iterative algorithm for finding a common element of the set of common fixed points of a finite family of asymptotically nonextensive nonself mappings and the set of solutions for equilibrium problems is discussed. A strong convergence theorem of common element is established in a uniformly smooth and uniformly convex Banach space.

1. Introduction

Let be a real Banach space with norm , let denote the dual of , and let denote the value of at . Suppose that is a nonempty, closed convex subset of . Let be a bifunction of into , where is the set of real numbers. The equilibrium problem for is to find such that The set of solutions of (1) is denoted by . Given a mapping , let for all . Then, if and only if for all ; that is, is a solution of the variational inequality. Numerous problems in physics, optimization, and economics reduce to find a solution of (1). Some methods have been proposed to solve the equilibrium problems; see [1–5].

Let be the normalized duality mapping from into given by for all . It is well known that if is uniformly smooth, then is uniformly norm-to-norm continuous on each bounded subset of . It is also well known that is uniformly smooth if and only if is uniformly convex.

Let be a nonempty closed convex subset of a Hilbert space and let be the metric projection of onto ; then is nonexpansive. This fact actually characterizes Hilbert spaces and consequently it is not available in more general Banach spaces. In this connection, Alber [6] recently introduced a generalized projection operator in a Banach space which is an analogue of the metric projection in Hilbert spaces. Consider the functional defined by Observe that, in a Hilbert space , (3) reduces to . The generalized projection is a map that assigns to an arbitrary point the minimum point of the functional ; that is, , where is the solution to the minimization problem existence and uniqueness of the operator follows from the properties of the functional and strict monotonicity of the mapping . In Hilbert spaces, . It is obvious from the definition of function that

Let be a nonempty subset of and let be a mapping. The set of fixed points of is denoted by . is called asymptotically nonextensive if and only if there exists a sequence with , such that Asymptotically nonextensive mappings coincide with asymptotically nonexpansive mappings in Hilbert spaces.

In [7], Chidume et al. studied the fixed point problem of an asymptotically nonextensive nonself mapping and obtained weak convergence theorem. Recently, in [8], liu introduced the following iterative scheme for approximating a common fixed point of two asymptotically nonextensive nonself mappings in a uniformly smooth and uniformly convex Banach space: Liu obtained strong convergence theorem.

Inspired and motivated by the facts above, the purpose of this paper is to prove a strong convergence theorem for finding a common element of the set of common fixed points of a finite family of asymptotically nonextensive nonself mappings and the set of solutions for equilibrium problems in a uniformly smooth and uniformly convex Banach space.

2. Preliminaries

Let be a real Banach space. When is a sequence in , we denote strong convergence of to by and weak convergence by . is said to have the Kadec-Klee property if and only if for a sequence of satisfying that and , then . It is known that if is uniformly convex, then has the Kadec-Klee property.

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

Lemma 1. Let be a uniformly smooth and strictly convex Banach space which enjoys the Kadec-Klee property, let be a nonempty, closed, and convex subset of , and let be an asymptotically nonextensive nonself mapping with a sequence such that is closed. Then is closed and convex.

Proof. Take ,  . Put . Using the same argument presented in the proof of [9, Theorem 2.1, page 854-855], we can obtain that . By the continuity of , we have Therefore, By (10), (11) and the closedness of , we have which implies that is convex.
Let and ; then, we have . It follows from the closedness of   that . This implies that is closed.

Lemma 2 (see [6]). Let be a reflexive, strictly convex, and smooth Banach space; let be a nonempty, closed, and convex subset of . Then the following conclusions hold:(1), and ;(2)if and , then if and only if ;(3)for if and only if .

Lemma 3 (see [10]). Let be a uniformly convex and smooth Banach space and let be two sequences of . If and either or is bounded, then .

Lemma 4 (see [11]). Let be a smooth and uniformly convex Banach space and let . Then there exists a strictly increasing, continuous, and convex function such that and for all and , where .

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 each ,  (A4) for each , is convex and lower semicontinuous.

Lemma 5 (see [12]). Let be a closed convex subset of a smooth, strictly convex, and reflexive Banach space , let be a bifunction from to satisfying (A1)–(A4), and for and , define a mapping as follows: Then the following conclusions hold:(1) is single-valued;(2) is firmly nonexpansive; that is, for any , (3);(4) is closed and convex;(5).

3. Main Results

Theorem 6. Let be a nonempty closed convex subset of a uniformly smooth and uniformly convex Banach space . Let be a bifunction from to satisfying (A1)–(A4), and let be some positive integer. Let be a closed asymptotically nonextensive nonself mapping with sequence such that for every . Suppose that is nonempty and bounded. Let be a sequence generated by the following manner: where , . is a real number sequence in for every , is a real number sequence in , where is some positive real number. Assume that and for every . Then converges strongly to .

Proof. First, we show that is closed and convex. From the definitions of , it is obvious is closed. Moreover, since is equivalent to , it follows that is convex. From Lemmas 1 and 5, we have that is closed and convex. Then is well defined.
Next, we prove for all . is obvious. Suppose that for some ; for each , from Lemma 5, we have This implies that , and so . From , one sees Since , we arrive at
Next we show that the sequence is bounded. From Lemma 2, we have for each and for all . Therefore, the sequence is bounded. It follows from (5) that the sequence is also bounded. By the assumption, we have On the other hand, noticing that and , one has for all . Therefore, is nondecreasing. It follows that the limit of exists. By the definition of , one has that and for any positive integer . It follows that Letting in (23), we have . It follows from Lemma 3 that as . Hence, is a Cauchy sequence. Since is a Banach space and is a closed and convex, one can assume that as .
Next we show that . By taking in (23), we have that From Lemma 3, we have Noticing that , we obtain It follows from (21) and (24) that From Lemma 3, we have Combining (25) with (28), we obtain that It follows from as that as. Since is uniformly norm-to-norm continuous on each bounded set, we have On the other hand, we have We obtain that Since is a uniformly smooth Banach space, we know that is a uniformly convex Banach space. From Lemma 4, we find that Therefore we have From and (21), (32), we have Therefore, from the property of we have Since is also uniformly norm-to-norm continuous on each bounded set, we have Using (7), (34), and (36), we have By (6), we obtain Since , from (38), we have . Since , then Applying (24), (38), (40), the definition of , and Lemma 3 to (39), we obtain that From Lemma 3, we obtain that In the same way, we can obtain From the closedness of , we have .
Next, we show . From Lemma 5, we have It follows from (21) and (32) that . From Lemma 3, we see that Since is uniformly norm-to-norm continuous on each bounded set, we have From , we have By , we have From (A2), we have Letting , we have from (A4), (47) and , as that. For and , let . Since and , we have and hence . So, from (A1) and (A4) we have Dividing by , we have Letting , from (A3), we have . Therefore,  .
Finally, we show . By taking limit in (19), we have At this point, in view of Lemma 2, we have that . This completes the proof.