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`Abstract and Applied AnalysisVolume 2013 (2013), Article ID 631382, 19 pageshttp://dx.doi.org/10.1155/2013/631382`
Research Article

## Implicit and Explicit Iterative Methods for Systems of Variational Inequalities and Zeros of Accretive Operators

1Department of Mathematics, Shanghai Normal University, Shanghai 200234, China
2Scientific Computing Key Laboratory of Shanghai Universities, Shanghai 200234, China
3Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
4Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India

Received 10 September 2013; Accepted 22 September 2013

Copyright © 2013 Lu-Chuan Ceng 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

Based on Korpelevich's extragradient method, hybrid steepest-descent method, and viscosity approximation method, we propose implicit and explicit iterative schemes for computing a common element of the solution set of a system of variational inequalities and the set of zeros of an accretive operator, which is also a unique solution of a variational inequality. Under suitable assumptions, we study the strong convergence of the sequences generated by the proposed algorithms. The results of this paper improve and extend several known results in the literature.

#### 1. Introduction and Preliminaries

Throughout the paper, we use the notation to indicate the weak convergence and to indicate the strong convergence.

Let be a real Banach space whose dual space is denoted by . Let denote the unit sphere of . A Banach space is said to be uniformly convex if for each , there exists such that for all , It is known that a uniformly convex Banach space is reflexive and strictly convex.

A Banach space is said to be smooth if the limit exists for all ; in this case, is also said to have a Gâteaux differentiable norm. Moreover, it is said to be uniformly smooth if this limit is attained uniformly for all ; in this case, is also said to have a uniformly Fréchet differentiable norm. The norm of is said to be Fréchet differentiable if for each , the limit is attained uniformly for all . A function defined by is called the modulus of smoothness of . It is known that is uniformly smooth if and only if . If is a fixed number, then a Banach space is said to be -uniformly smooth if there exists a constant such that for all . As pointed out in [1], no Banach space is -uniformly smooth for .

The normalized duality mapping is defined by where denotes the generalized duality pairing. It is an immediate consequence of the Hahn-Banach theorem that is nonempty for all . It is also known that is single-valued if and only if is smooth, whereas if is uniformly smooth, then the mapping is norm-to-norm uniformly continuous on bounded subsets of . For further details on the geometry of Banach spaces, we refer to [2, 3] and the references therein.

Lemma 1 (see [4]). Let be a -uniformly smooth Banach space. Then where is the -uniformly smooth constant of and is the normalized duality mapping from into .

The following lemma is an immediate consequence of the subdifferential inequality of the function .

Lemma 2. Let be a smooth Banach space. Then, where is the normalized duality mapping of .

Proposition 3 (see [5]). Let be a real smooth and uniform convex Banach space and . Then, there exists a strictly increasing, continuous, and convex function , such that where .

Let be a nonempty closed convex subset of a real Banach space . A mapping is said to be -Lipschitzian (or Lipschtiz continuous) if there exists a constant such that for all . If , then is called -contraction. If , then is called nonexpansive.

The set of fixed points of is denoted by .

Lemma 4 (see [6]). Let be a nonempty closed convex subset of a strictly convex Banach space , and let be a sequence of nonexpansive mappings on . Suppose that is nonempty. Let be a sequence of positive numbers with . Then, a mapping on defined by for is defined well, nonexpansive, and .

Lemma 5 (see [7]). Let be a nonempty closed and convex subset of a Banach space and be a continuous and strong pseudocontraction mapping. Then, has a unique fixed point in .

Let be a nonempty subset of a Banach space . A mapping is said to be(a) accretive if for each , there exists such that where is the normalized duality mapping of ;(b)-strongly accretive if for each , there exists such that (c) pseudocontractive if for each , there exists such that (d)-strongly pseudocontractive if for each , there exists such that (e)-strictly pseudocontractive if for each , there exists such that

It is worth to mention that the definition of the inverse strongly accretive mapping is based on that of the inverse strongly monotone mapping, which was studied by so many authors; see, for example, [8, 9].

Lemma 6. Let be a nonempty closed convex subset of a real smooth Banach space , and let be a mapping.(a) If is -strongly accretive and -strictly pseudocontractive with , then is nonexpansive and is Lipschitz continuous with constant .(b) If is -strongly accretive and -strictly pseudocontractive with , then for any fixed is contractive with coefficient .

Proof. (a) From the -strictly pseudocontractivity of , we have for all , which implies that Hence, From the -strictly pseudocontractiveness and -strongly accretiveness of , we have for all , which implies that Since , is nonexpansive.
(b) Take a fixed arbitrarily. Observe that for all , Since , is contractive with coefficient .

Let be a real smooth Banach space . An operator is said to be strongly positive [10] if there exists a constant such that where is the identity mapping.

Lemma 7 (see [7]). Let be a strongly positive linear bounded operator on a smooth Banach space with coefficient and . Then, .

Recall that a possibly multivalued operator with domain and range in is accretive if for each and (), there exists such that . An accretive operator is said to satisfy the range condition if for all . An accretive operator is -accretive if for each . If is an accretive operator which satisfies the range condition, then we can define, for each , a mapping by , which is called the resolvent of . It is well known that is nonexpansive and for all . Hence, If , then the inclusion is solvable.

The following resolvent identity is well-known.

Lemma 8 (resolvent identity). For , , and ,

Recently, Aoyama et al. [11] studied an iterative scheme in the setting of a uniformly convex Banach space having a uniformly Gâteaux differentiable norm to compute the approximate zero of an accretive operator . They proved that the sequence generated by their scheme converges strongly to a zero of under some appropriate assumptions. Subsequently, Ceng et al. [12] introduced and analyzed a composite iterative scheme in the setting of a uniformly smooth Banach space or a reflexive Banach space having a weakly sequentially continuous duality mapping. Further, Jung [13] proposed and analyzed a composite iterative scheme by the viscosity approximation method for finding a zero of an accretive operator and established the strong convergence of the sequence generated by the proposed scheme.

Let be a closed and convex subset of a Banach space , and let be a nonempty subset of . A mapping is said to be sunny if whenever for all and . A mapping is called a retraction if . If a mapping is a retraction, then for all , where is the range of . A subset of is called a sunny nonexpansive retract of if there exists a sunny nonexpansive retraction from onto .

The following lemma concerns the sunny nonexpansive retraction.

Lemma 9 (see [14]). Let be a nonempty closed convex subset of a real smooth Banach space , let be a nonempty subset of , and let be a retraction of onto . Then, the following statements are equivalent:(a) is sunny and nonexpansive;(b);(c).

It is well known that if is a Hilbert space, then a sunny nonexpansive retraction is coincident with the metric projection from onto ; that is, . If is a nonempty closed convex subset of a strictly convex and uniformly smooth Banach space and if is a nonexpansive mapping with the fixed point set , then the set is a sunny nonexpansive retract of .

Let be a smooth Banach space, let be a nonempty closed convex set, and let be two nonlinear mappings. We consider the following system of variational inequalities (in short, SVI) which is earlier studied by Cai and Bu [15]. Find such that where and are two positive constants. When is a real Hilbert space, then SVI (23) is considered and studied by Ceng et al. [16]. In addition, if and , then the SVI reduces to the classical variational inequality problem (VIP) of finding such that It is worth to mention that the system of variational inequalities is used as a tool to study the Nash equilibrium problem; see, for example, [1719] and the references therein. We believe that the problem (23) could be used to study Nash equilibrium problem for two-player game. The theory of variational inequalities is well established because it has a wide range of applications in science, engineering, management, and social sciences. In 1976, Korpelevič [20] proposed an iterative algorithm for solving VIP (24) in Euclidean space, which is known as the extragradient method or Korpelevich’s extragradient method. For different generalizations of extragradient method for variational inequalities and fixed point problems, we refer to [2123] and the references therein. For further details on variational inequalities, we refer to [2, 19, 24, 25] and the references therein.

Whenever is a real smooth Banach space, , and , then SVI (23) reduces to the variational inequality problem (VIP) of finding such that which was considered by Aoyama et al. [26]. Very recently, Cai and Bu [15] constructed an iterative algorithm for solving SVI (23) and a common fixed point problem of an infinite family of nonexpansive mappings in a uniformly convex and -uniformly smooth Banach space. They proved the strong convergence of the sequence generated by the proposed algorithm.

Lemma 10. Let be a nonempty closed convex subset of a smooth Banach space . Let be a sunny nonexpansive retraction from onto , and let be nonlinear mappings. For given , is a solution of SVI (23) if and only if , where .

Proof. We can rewrite SVI (23) as which is equivalent to From Lemma 9, we get the conclusion.

Remark 11. In view of Lemma 10, we observe that which implies that is a fixed point of the mapping .

Lemma 12 (see [15, Lemma 2.8]). Let be a nonempty closed convex subset of a real -uniformly smooth Banach space , and for each , let be an -inverse-strongly accretive. Then, where . In particular, if , then is nonexpansive for .

Lemma 13 (see [15, Lemma 2.9]). Let be a nonempty closed convex subset of a real -uniformly smooth Banach space , and let be a sunny nonexpansive retraction from onto . For each , let be a -inverse-strongly accretive mapping and let be the mapping defined by If for , then is nonexpansive.

Throughout this paper, the set of fixed points of the mapping is denoted by .

Let be a real uniformly convex and -uniformly smooth Banach space, and let be a nonempty closed convex subset of such that . Let be a -strongly positive linear bounded operator with . The purpose of this paper is to introduce and analyze implicit and explicit iterative schemes for computing a common element of the solution set of system of variational inequalities and set of zeros of an accretive operator with domain and range in such that . Our methods are based on Korpelevich’s extragradient method, hybrid steepest-descent method, and viscosity approximation method. In particular, we propose the iterative methods for computing a point , which is also a unique solution of the following variational inequality: Under suitable assumptions, the strong convergence of the sequences generated by the proposed schemes is studied. Our results extend and improve several results presented in the recent past in the literature; see, for example, [10, 13, 15, 2729] and the references therein.

Now we present some known results and definitions which will be used in the sequel.

Let be a mean if is a continuous linear functional on satisfying . Then we know that is a mean on if and only if for every . According to time and circumstances, we use instead of . A mean on is called a Banach limit if and only if for every . We know that if is a Banach limit, then for every . So, if , and (resp., ), as , we have

Lemma 14 (see [30]). Let be a nonempty closed convex subset of a uniformly smooth Banach space , and let be a bounded sequence in . Let be a mean on and . Then, if and only if where is the normalized duality mapping of .

Recall that a Banach space is said to satisfy Opial’s condition if whenever is a sequence in which converges weakly to as , then

Lemma 15 (demiclosedness principle, see [31, Theorem 10.3]). Let be a reflexive Banach space satisfying Opial’s condition, let be a nonempty closed convex subset of , and let be a nonexpansive mapping. Then the mapping is demiclosed on , where is the identity mapping; that is, if is a sequence of such that and , then .

Remark 16. If the duality mapping from into is single-valued and weakly sequentially continuous, then satisfies Opial’s condition; see [32, Theorem 1]. Each Hilbert space and the sequence spaces with satisfy Opial’s condition; see [32]. Though an -space with does not usually satisfy Opial’s condition, each separable Banach space can be equivalently renormed, so that it satisfies Opial’s condition; see [33].

We close this section by mentioning a result on the sequence which will be used in the proof of the main results of this paper.

Lemma 17 (see [34]). Let be a sequence of nonnegative real numbers satisfying where , and satisfy the following conditions:(i) and ;(ii);(iii), and . Then .

#### 2. Implicit Iterative Scheme and Its Convergence Criteria

In this section, we propose an implicit hybrid steepest-descent viscosity scheme for finding the solution of SVI (23) such that such solution is also a zero of an accretive operator. We also study the strong convergence of the sequence generated by the proposed algorithm.

Theorem 18. Let be a nonempty closed convex subset of a uniformly convex and -uniformly smooth Banach space such that . Let be a sunny nonexpansive retraction from onto , and let be an accretive operator in such that . For each , let be an -inverse-strongly accretive mapping such that , where is the fixed point set of the mapping with for . Let be -strongly accretive and -strictly pseudocontractive with , and let be a fixed Lipschitzian strongly pseudocontractive mapping with pseudocontractive coefficient , , and Lipschitzian constant . Let be a -strongly positive linear bounded operator with . For each integer , let be defined by where for some and for all with . Then, as converges strongly to , which is a unique solution of the following variational inequality:

Proof. We first show that is defined well. Since , for all and , we have , for all . Define a mapping by Observe that Whenever and , is a continuous and strongly pseudocontractive mapping with pseudocontractive coefficient . By Lemma 5, we deduce that there exists a unique fixed point in , denoted by , which uniquely solves the fixed point equation
Let us show the uniqueness of a solution of VIP (41). Suppose that both and are solutions to VIP (41). Then, we have Adding up the above two inequalities, we obtain Note that Taking into account , we have , and hence the uniqueness is proved.
Next, we show that is bounded. Indeed, we note that , for all . Take a fixed arbitrarily. Utilizing Lemmas 6 and 7, we obtain which immediately yields It follows that Thus, is bounded and so are , , and .
Let us show that as .
Indeed, for simplicity we put and . Then, and . Hence from (48), we get From Lemma 12, we have From the last two inequalities, we obtain which together with (51) implies that Therefore, Since for , it follows from that Utilizing Proposition 3 and Lemma 9, we have which implies that Similarly, we derive which implies that Combining (58) and (60), we get which together with (51) implies that Therefore, Hence, from (56) and , we conclude that Utilizing the properties of and , we get which leads to that is, Note that for all and that is bounded and so are , and , where . Hence, we have Also, observe that This together with (67) and (68) implies that Since for all , utilizing Lemma 8, we have For any integer , for simplicity we put for all . Now let be a sequence in that converges to as , and define a function by where is a Banach limit. Define a set and a mapping where is a constant in . Then by Lemma 4, we have . We observe that So, from (67) and (71), we obtain