About this Journal Submit a Manuscript Table of Contents
Abstract and Applied Analysis
Volume 2013 (2013), Article ID 540108, 10 pages
http://dx.doi.org/10.1155/2013/540108
Research Article

Strong Convergence of Iterative Algorithm for a New System of Generalized Cocoercive Operator Inclusions in Banach Spaces

1Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
2Department of Mathematics, Yasouj University, Yasouj 75918, Iran

Received 10 September 2013; Revised 6 November 2013; Accepted 20 November 2013

Academic Editor: Mohammad Mursaleen

Copyright © 2013 Saud M. Alsulami et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We introduce and study a new system of generalized cocoercive operator inclusions in Banach spaces. Using the resolvent operator technique associated with cocoercive operators, we suggest and analyze a new generalized algorithm of nonlinear set-valued variational inclusions and establish strong convergence of iterative sequences produced by the method. We highlight the applicability of our results by examples in function spaces.

1. Introduction

The resolvent operator technique is a powerful tool to study the approximation solvability of nonlinear variational inequalities and variational inclusions, which have been applied widely to optimization and control, mechanics and physics, economics and transportation equilibrium, and engineering sciences, see, for example, [14] and the references therein.

In a series of papers [58], the authors investigated -accretive and -accretive operators for solving variational inclusions in Banach spaces. Convergence and stability of iterative algorithms for the systems of -accretive operators have been studied in [9, 10]. The notion of -monotone operators has been introduced and investigated by the authors in [11]. Generalized mixed variational inclusions involving -monotone operators have been discussed in [12]. Some results on -accretive operators and application for solving set-valued variational inclusions in Banach spaces have been proved in [7]. Some other related articles on the variational inclusion problems can be found in [1322].

Very recently, Ahmad et al. [23] introduced a new -cocoercive operator and its resolvent operator in the setting of Banach spaces. The authors proposed concrete examples in support of -cocoercive operators and they also proved the Lipschitz continuity of resolvent operator associated with -cocoercive operator. Motivated and inspired by the research works mentioned above, in this paper, we introduce and study a new system of -cocoercive mapping inclusions in Banach spaces. Using the resolvent operator associated with -cocoercive mapping, we suggest and analyze a new general algorithm and establish the existence and uniqueness of solutions for this system of -cocoercive mappings.

2. Preliminaries

Throughout this paper, we denote the set of positive integers by . Let be a Banach space with the norm and the dual space . For any , we denote the value of at by . When is a sequence in , we denote the strong convergence of to by as . We denote by the family of all nonempty subsets of . Let be the family of all nonempty, closed, and bounded subsets of . The Hausdörff metric on [24] is defined by where and .

Definition 1 (see [25]). A continuous and strictly increasing function such that and is called a gauge function.

Definition 2 (see [25]). Let be a Banach space. Given a gauge function , the mapping corresponding to defined by is called the duality mapping with gauge function .
In particular, if , the duality map is called the normalized duality mapping.

Lemma 3 (see [26]). Let be a real Banach space and be the normalized duality mapping. Then, for any , for all .

Definition 4. Let be a Banach space. Let and be two mappings and be the normalized duality mapping. Then, is called(i)-cocoercive, if there exists a constant such that (ii) -accretive, if (iii) -strongly accretive, if there exists a constant such that (iv) -relaxed cocoercive, if there exists a constant such that (v) Lipschitz continuous, if there exists a constant such that (vi) -expansive, if there exists a constant such that (vii) is said to be Lipschitz continuous, if there exists a constant such that

Definition 5. Let be a Banach space. Let , , be four single-valued mappings and be the normalized duality mapping. Then,(i) is said to be -cocoercive with respect to , if there exists a constant such that (ii) is said to be -relaxed cocoercive with respect to , if there exists a constant such that (iii) is said to be -Lipschitz continuous with respect to , if there exists a constant such that (iv) is said to be -Lipschitz continuous with respect to , if there exists a constant such that

Definition 6. Let be a Banach space. A set-valued mapping is said to be -cocoercive, if there exists a constant such that

Definition 7. Let be a Banach space. A mapping is said to be -Lipschitz continuous, if there exists a constant such that

Definition 8. Let be a Banach space. Let be the mappings. A mapping is said to be(i) Lipschitz continuous in the first argument with respect to , if there exists a constant such that (ii) Lipschitz continuous in the second argument with respect to , if there exists a constant such that (iii) -relaxed Lipschitz in the first argument with respect to , if there exists a constant such that (iv) -relaxed Lipschitz in the second argument with respect to , if there exists a constant such that

Definition 9. Let be a Banach space. Let , , be four single-valued mappings. Let be a set-valued mapping. is said to be -cocoercive operator with respect to and , if is -cocoercive and , for every .

Example 10. Let and be defined by Assume now that , are defined by
Let , where is the identity mapping. Then, is -cocoercive with respect to and .

Example 11. Let , the space of all real valued continuous functions defined on closed interval with the norm Let be defined by Let be defined by Suppose that , where for all . Then, for , we conclude that This proves that and is not -cocoercive with respect to and .

Proposition 12 (see [23]). Let be -cocoercive with respect to with constant and -relaxed cocoercive with respect to with constant , be -expansive and be -Lipschitz continuous and . Let be -cocoercive operator. Suppose that Then, , where .

Theorem 13 (see [23]). Let be -cocoercive with respect to with constant and -relaxed cocoercive with respect to with constant , be -expansive and be -Lipschitz continuous, and . Let be an -cocoercive operator with respect to and . Then, for each , the operator is single-valued.

Definition 14. Let be a Banach space. Let be -cocoercive with respect to with constant and -relaxed cocoercive with respect to with constant , be -expansive be -Lipschitz continuous and be -Lipschitz continuous, , and . Let be a -cocoercive operator with respect to and . Then, the resolvent is defined by

Theorem 15 (see [23]). Let be a Banach space. Let be -cocoercive with respect to with constant and -relaxed cocoercive with respect to with constant , be -expansive be -Lipschitz continuous, and be -Lipschitz continuous; and . Let be -cocoercive operator with respect to and . Then, the resolvent operator is -Lipschitz continuous, that is,

3. Strong Convergence Theorem

In this section, using the resolvent operator technique associated with -cocoercive operators, we propose a new generalized algorithm of nonlinear set-valued variational inclusions and establish strong convergence of iterative sequences produced by the method.

For , let be real Banach spaces with the norm . Let , , , , and be single-valued mappings, and , be set-valued mappings. Let , be -cocoercive and -cocoercive operators with respect to and , respectively. We consider the following problem.

Find , , and such that We call problem (30) a system of generalized -cocoercive operator inclusions.

Under the assumptions mentioned above, we have the following key and simple lemma.

Lemma 16. , , is a solution of problem (30) if and only if where , , and are constants.

Proof . This is an easy and direct consequence of Definition 14.

Algorithm 17. For , let be real Banach spaces with the norm . Let , , , , and be single-valued mappings, and , be set-valued mappings. Let , be such that, for each fixed , , and are -cocoercive and -cocoercive operators with respect to and , respectively. For any given constants , define the mappings and by For any given , , , let Since and , in view of Nadler's theorem [24], there exist and such that By induction, we define iterative sequences , , , and as follows: where , and are constants.

Theorem 18. For , let be real Banach spaces with the norm . Let , , , and be single-valued mappings, and , be set-valued mappings. Let and be such that, for each fixed , and are -cocoercive and -cocoercive operators with respect to and , respectively. Suppose that the following conditions are satisfied.(i) is -cocoercive with respect to with constant and -relaxed cocoercive with respect to with constant , .(ii) is -expansive and is -Lipschitz continuous, .(iii) is -Lipschitz continuous with respect to and -Lipschitz continuous with respect to , .(iv) is -Lipschitz continuous with constant and is -Lipschitz continuous with constant .(v) is -Lipschitz continuous with respect to in the first argument and -Lipschitz continuous with respect to in the second argument.(vi) is -Lipschitz continuous with respect to in the first argument and -Lipschitz continuous with respect to in the second argument.(vii) is -Lipschitz continuous, .(viii) is -relaxed Lipschitz continuous with respect to in the first argument and -relaxed Lipschitz continuous with respect to in the second argument with constants and , respectively.(ix) is -relaxed Lipschitz continuous with respect to in the first argument and -relaxed Lipschitz continuous with respect to in the second argument with constants and , respectively. Furthermore, assume that there exist constants such that and are constants satisfying the following conditions: Then, the iterative sequences , , , and generated by Algorithm 17 converge strongly to , , , and , respectively, and is a solution of problem (30).

Proof. In view of Theorem 13, the resolvent operator is -Lipschitz continuous. This, together with Algorithm 17 and (36), implies that Since is -Lipschitz continuous with respect to in the first argument and -Lipschitz continuous in the second argument, is -Lipschitz continuous, and is -Lipschitz continuous, by Algorithm 17, we get As is -Lipschitz continuous with respect to , we obtain Since is -Lipschitz continuous, we conclude that
Since is -relaxed Lipschitz continuous with respect to and -relaxed Lipschitz continuous with respect to in the first and second arguments with constants and , respectively, we have Employing Lemma 3 and taking into account (39)–(44), we obtain This implies that where
Using -Lipschitz continuity of with respect to , we deduce that In view of (41), (46), (48), (39) becomes Similarly, we have where
In view of (49) and (50), we obtain where , , and .
Letting , we obtain , where
Next, we define the norm on by One can easily check that is a Banach space.
Define . Then, we have
In view of (38), we conclude that . This implies that there exist and such that for all . It follows from (52) and (54) that In view of (56), we obtain This implies that for any , Since , it follows from (58) that and . This proves that is a Cauchy sequence in . Similarly, we conclude that is a Cauchy sequence in . Thus, there exist and such that and as .
Next, we prove that and . In view of Lipschitz continuity of and and Algorithm 17, we obtain From (59), we deduce that are Cauchy sequences in and , respectively. Thus, there exist and such that and as . Since is -Lipschitz continuous with constant , it is obvious that By the closedness of , we conclude that . Similarly, we have .
Assume now that Then, we have Since , , and as , it follows from (62) that and hence .
A similar argument shows that . Therefore, In view of Lemma 16, we conclude that is a solution of problem (30), which completes the proof.

At the end of this paper, we include the following simple example in support of Theorem 18.

Example 19. Let with the usual inner product. We define two mappings by Let a mapping be defined by By similar arguments, as in Example 4.1 of [27], we can prove the following.(1) is -cocoercive with respect to and -relaxed cocoercive with respect to .(2) is -expansive, for .(3) is -expansive, for .(4) is -Lipschitz continuous with constant with respect to and , for .(5) Let be defined by (6) Now, we define a mapping by Let be the identity mappings. It is obvious that these mappings are -Lipschitz continuous.(7)Assume that are defined by