Abstract

We introduce a new iterative scheme with Meir-Keeler contractions for an asymptotically nonexpansive mapping in -uniformly smooth and strictly convex Banach spaces. We also proved the strong convergence theorems of implicit and explicit schemes. The results obtained in this paper extend and improve many recent ones announced by many others.

1. Introduction

Let be a real Banach space. With , we denote the normalized duality mapping given by where denotes the generalized duality pairing and the dual space of . In the sequel we will donate single-valued duality mappings by . Given , by we will denote the generalized duality mapping given by

We recall that the following relation holds: for .

We recall that the modulus of smoothness of is the function defined by is said to be uniformly smooth if .

Let . is said to be -uniformly smooth if there exists a constant such that . Examples of such spaces are Hilbert spaces and (or ).

We note that a -uniformly smooth Banach space is uniformly smooth. This implies that its norm uniformly Fréchet differentiable (see [1]).

If is uniformly smooth, then the normalized duality map is single-valued and norm to norm uniformly continuous.

Let be a real Banach space and is a nonempty closed convex subset of . A mapping is said to be asymptotically nonexpansive if there exists a sequence with such that and denotes the set of fixed points of the mapping ; that is, . For asymptotically nonexpansive self-map , it is well known that is closed and convex (see e.g., [2]).

Theorem 1.1 (Banach [3]). Let be a complete metric space and let be a contraction on ; that is, there exists such that for all . Then has a unique fixed point.

Theorem 1.2 (Meir and Keeler [4]). Let be a complete metric space and let be a Meir-Keeler contraction (MKC) on , that is, for every , there exists such that implies for all . Then has a unique fixed point.

This theorem is one of generalizations of Theorem 1.1, because contractions are Meir-Keeler contractions.

We recall that, given a -uniformly smooth and strictly convex Banach space with a generalized duality map and a subset of , a mapping is called(1)-Lipschitzian, if there exists a constant such that  holds for every and ;(2)-strongly monotone, if there exists a constant such that  holds for every and .

In , Ali and Ugwunnadi [5] introduced and considered the following iterative scheme: where a family of asymptotically nonexpansive self-mappings of with sequences , such that as and are a contraction mapping with coefficient . Let be a strongly positive-bounded linear operator with coefficient , and . They proved the strong convergence of the implicit and explicit schemes for a common fixed point of the family , which solves the variational inequality .

Motivated and inspired by the results of Ali and Ugwunnadi [5], we introduced an iterative scheme as follows. for , where is an asymptotically nonexpansive self-mapping of with sequences , such that as and are a Meir-Keeler contraction (MKC, for short). Let is a -Lipschitzian and -strongly monotone operator with . We will prove the strong convergence of the implicit and explicit schemes for a fixed point of , which solves the variational inequality , for . Our results improve and extend the results of Ali and Ugwunnadi [5] for an asymptotically nonexpansive mapping in the following aspects:(i)Hilbert space is replaced by a -uniformly smooth and strictly convex Banach space;(ii)contractive mapping is replaced by a MKC;(iii)Theorems 3.1 and 4.1 extend the results of Ali and Ugwunnadi [5] from a strongly positive-bounded linear operator to a -Lipschitzian and -strongly monotone operator .

2. Preliminaries

In order to prove our main results, we need the following lemmas.

Lemma 2.1 (see [6]). Let and be a -uniformly smooth Banach space, then there exists a constant such that

Lemma 2.2 (see [7, Lemma 2.3]). Let be a MKC on a convex subset of a Banach space . Then for each , there exists such that

Lemma 2.3 (see [8]). Let and be bounded sequences in a Banach space and be a sequence in which satisfies the following condition: Suppose that , and . Then .

Lemma 2.4 (see [9, 10]). Let be a sequence of nonnegative real numbers satisfying where , and satisfy the following conditions: and , or . Then .

Lemma 2.5 (see [11]). Let be a nonempty closed convex subset of a uniformly convex Banach space and is an asymptotically nonexpansive mapping with . Then the mapping is demiclosed at zero, that is, and , then .

Lemma 2.6. Let be a -Lipschitzian and -strongly monotone operator on a -uniformly smooth Banach space with and . Then is a contraction with contractive coefficient and .

Proof. From (2.1), we have where , and Hence is a contraction with contractive coefficient .

Lemma 2.7 (see [5, Lemma 2.9]). Let be a uniformly Lipschitzian with a Lipschitzian constant , that is, there exists a constant such that

Lemma 2.8 (see, e.g., Mitrinović [12, page 63]). Let . Then the following inequality holds: for arbitrary positive real numbers .

3. Main Result

Theorem 3.1. Let be a -uniformly smooth and strictly convex Banach space, and a nonempty closed convex subset of such that and have a weakly sequentially continuous duality mapping from to . Let be an asymptotically nonexpansive mapping with sequences , such that as and . Let be a bounded subset of such that . Let be a -Lipschitzian and -strongly monotone operator on with , and be a MKC on with . Let be a sequence in (0,1) satisfying the following conditions:(A1);(A2). Let be defined by
Then, converges to a fixed point say in which solves the variational inequality

Proof. Let . Since and as , then as , so such that for all , and . Thus, for Therefore, Thus, is bounded and therefore and are also bounded. Also from (3.1), we have From (3.5) and , we obtain Thus, Since is bounded, now assume that is a weak limit point of and a subsequence of converges weakly to . Then, by Lemma 2.5 and (3.7), we have that is a fixed point of , hence .
Next we observe that the solution of the variational inequality (3.2) in is unique. Assume that are solutions of the inequality (3.2), without loss of generality, we may assume that there is a number such that . Then by Lemma 2.2, there is a number such that . From (3.2), we know Adding (3.8) and (3.9), we have Noticing that Therefore . That is, is the unique solution of (3.2).
Finally, we show that as . From (3.3), we get and in particular Since , from the above inequality and is a weakly sequentially continuous duality mapping, we have as . Next, we show that solves the variational inequality (3.2). Indeed, from the relation we get So, for any Now replacing in (3.16) with and letting , using for , and the fact that as , we obtain . This implies that is a solution of the variational inequality (3.2). Every weak limit of say belongs to . Furthermore, is a strong limit of that solves the variational inequality (3.2). As this solution is unique we get that as . This completes the proof.

Corollary 3.2. Let be a -uniformly smooth and strictly convex Banach space, and let be a nonempty closed convex subset of such that and have a weakly sequentially continuous duality mapping from to . Let be a nonexpansive mapping. Let be a sequence in (0, 1) satisfying . Let and be as in Theorem 3.1. For , let be defined by Then, converges to a fixed point say in which solves the variational inequality (3.2).

4. Explicit Algorithm

Theorem 4.1. Let be a -uniformly smooth and strictly convex Banach space, and let be a nonempty closed convex subset of such that and have a weakly sequentially continuous duality mapping from to . Let be an asymptotically nonexpansive mapping with sequences , such that as and . Let be a bounded subset of such that . Let be a -Lipschitzian and -strongly monotone operator on with , and be a MKC on with . Let , be sequences in (0,1) satisfying the following conditions:(B1);(B2);(B3);(B4).Then, defined by (1.9) converges strongly to a fixed point say in which solves the variational inequality (3.2).

Proof. Since and as , as . Thus, such that and , . For any point and , But Therefore, By induction, we have Next we show that From (1.9), Therefore, Hence, and by Lemma 2.3 Thus, from (1.9),
Next, we show that Since Thus, Hence, Since is Lipschitz with constant and for any positive number , we have Therefore, Next we show that where is the unique solution of inequality (3.2). Let be a subsequence of such that Since is bounded, we may also assume that there exists some such that . From (4.11) it follows that By Lemma 2.5, the weak limit of is a fixed point of the mapping , so this implies that . Hence by Theorem 3.1 and is a weakly sequentially continuous duality mapping, we have Finally, we show that . By contradiction, there is a number such that
Case 1. Fixed , if for some such that , and for the other such that .
Let From (4.20), we know . Hence, there is a number , when , we have . We extract a number satisfying , then we estimate But Therefore, Hence, we have In the same way, we can get It contradicts the .
Case 2. Fixed , if , for all , from Lemma 2.2, there is a number such that It follows (1.9) that Therefore, By Lemma 2.4, we have that as . It contradicts the . This completes the proof.