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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 470354, 25 pages
http://dx.doi.org/10.1155/2012/470354
Research Article

Viscosity Iterative Schemes for Finding Split Common Solutions of Variational Inequalities and Fixed Point Problems

1Department of Mathematics, Honghe University, Yunnan 661100, China
2Department of Mathematics, National Kaohsiung Normal University, Kaohsiung 824, Taiwan

Received 12 August 2012; Accepted 25 September 2012

Academic Editor: Yonghong Yao

Copyright © 2012 Zhenhua He and Wei-Shih Du. 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 some new iterative schemes based on viscosity approximation method for finding a split common element of the solution set of a pair of simultaneous variational inequalities for inverse strongly monotone mappings in real Hilbert spaces with a family of infinitely nonexpansive mappings. Some strong convergence theorems are also given. Our results generalize and improve some well-known results in the literature and references therein.

1. Introduction

Throughout this paper, we denote by and , the sets of positive integers and real numbers, respectively. Let be a real Hilbert space, whose inner product and norm are denoted by and , respectively. Let be the identity mapping on and be a nonempty closed convex subset of . Let be a nonlinear operators. Then the canonical variational inequality problem for the operator ( or (VIP), for short) is to find such that We use the symbol to denote the solution set of (VIP), that is

(VIP) was extensively investigated and generalized to the vector variational inequality problems for single-valued or multivalued maps and contains optimization problems, quasi-variational inequality problems, equilibrium problems, fixed-point problems, complementary problems, bilevel problems, and semi-infinite problems as special cases and applications; see [16] and references therein.

Let be two nonlinear operators. In [7], some authors have considered the following pair of simultaneous variational inequality problems for operators and (, for short):

An element is a solution of if and only if . Clearly, reduces to (VIP) if .

Example 1.1. Let with usual inner product and let with . Define two real-valued functions by , . Then , and there exists such that and . If , then ; if , then and ; if , then and . So we have or which means that is the solution of .
Obviously, the problem is considered in the same subset of the same space. But many cases, two variational inequality problems often lie in different subset of spaces. So, as a further development of the problem , Censor et al. [8] presented a split variational inequality problem. Let be two real Hilbert spaces and and two closed convex sets. Let be a bounded linear operator. and are two nonlinear operators. The split variational inequality problem for and (, for short) is defined as follows:

It is well known to find a solution of (VIP) or a common element of the solution set of (VIP) and a fixed point of nonlinear operators, which has been studied by many authors (see [916]) using all kinds of auxiliary techniques and formulations. In 2005, Iiduka and Takahashi [9] established the following iteration scheme: let be arbitrary, define where is a nonexpansive mapping. They proved that the sequence defined by (1.6) strongly converge to , if the coefficient satisfy the following conditions:

In 2007, Chen et al. [10] studied the following iterative process: where is a nonexpansive mapping. If and , then they proved that converges strongly to , which solves the variational inequality:

For some split common solution problems, they have been studied by some authors; see [17, 18] and therein references. In this paper, we continue to study the (SVIP) and introduce some new iterative schemes based on viscosity approximation method for finding a common element of the fixed points set of nonexpansive mappings and the split solution set of a pair of variational inequalities for inverse strongly monotone mappings in real Hilbert spaces. Our results are new development of finding a common element of fixed point of nonlinear operators and variational inequality problems.

2. Preliminaries

In this paper, we use symbols and to denote strong and weak convergence, respectively. A Banach space , is said to satisfy Opial's condition, if for each sequence in with , we have It is well known that each Hilbert space satisfies Opial's condition; see, for example, [19]. Let be a mapping. In this paper, the set of fixed points of is denoted by .

A set-valued mapping is said to be monotone, if for all , and imply that . A monotone mapping is said to be maximal, if the graph of is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping is maximal, if and only if for , for every implies that . Let be a monotone mapping and let be the normal cone to at , that is, . Define Then is maximal monotone and if and only if , where is the zero vector of ; see, for example, [9, 20, 21] for more details.

For any , there exists a unique nearest point in , denoted by , such that for all . The mapping is called the projection operator (or metric projection) from onto .

Let and be two Hilbert spaces. Let and be two bounded linear operators. is called the adjoint operator (or adjoint) of , if for all , satisfies . It is known that the adjoint operator of a bounded linear operator on a Hilbert space always exists and is bounded linear and unique. Moreover, it is not hard to show that if is an adjoint operator of , then .

A mapping is said to be(1)expansive if there exists a constant such that for all . In particular, if , then is called expansive.(2)-strongly monotone if there exists a constant such that Clearly, any -strongly monotone mapping is -expansive.(3)-inverse strongly monotone if there exists a constant such that (4)Relaxed -cocoercive if there exists a constant such that (5)Relaxed -cocoercive if there exists constants such that Especially, if , then is -strongly monotone. So this class of mapping is more general than the class of strongly monotone mapping.(6)An -Lipschitz mapping if there exists a constant such that for all . In particular, if (, resp.), then is called a contraction (a nonexpansive mapping, resp.)

Remark ST (see [9]). If is -strongly monotone and -Lipschitz continuous, that is, for all , then is -inverse strongly monotone.

Example 2.1. Let , for all. Then it is easy to see that for any , Hence is a relaxed -cocoercive mapping, but is not a strongly monotone mapping.

Now, let be a family of infinitely nonexpansive mappings. In [22], a mapping is defined by the following: where . Such a mapping is called the -mapping generated by and .

The following properties for a -mapping are well known.

Theorem 2.2 (see [22, 23]). Let be a nonempty closed convex subset of a Hilbert space , let be a family of infinitely nonexpansive mappings from into itself such that is nonempty, and let be real numbers such that for any . Then the following statements hold:(1) is a nonexpansive mapping and .(2)For each and for each positive integer , the limit exists.(3)The mapping defined by , is a nonexpansive mapping satisfying and it is called the -mapping generated by and .

Theorem 2.3 (see [23]). Let be a nonempty closed convex subset of a Hilbert space , be nonexpansive mappings with , be a real sequence such that for any . If is any bounded subset of , then In particular, if is a bounded sequence in , then .

The following results are crucial in this paper.

Lemma 2.4 (see [19]). For a given , satisfies the inequality if and only if , where is a projection operator from onto .
It is well known that the projection operator is nonexpansive and satisfies

Lemma 2.5 (see [9]). The element is a solution of if and only if satisfies the relation , where is the projection operator, is a constant.

Lemma 2.6 (see [24]). Let be a nonnegative real sequence satisfying the following condition: where is some nonnegative integer, is a sequence in and is a sequence in such that (i); (ii) or is convergent.

Then .

Lemma 2.7 (see [25]). Let and be bounded sequences in a Banach space and let be a sequence in [0,1] with . Suppose for all integers and , then .

Lemma 2.8 (see [26]). Let be a real Banach space and be the normalized duality mapping, then for any the following inequality holds:
Especially, when , then . So, from Lemma 2.8, one has

The following result is simple, but it is very useful in this paper.

Lemma 2.9. Let , be two nonnegative real sequences. If , then .

3. Main Results

In this section, we construct an iteration scheme including a pair of mappings and which are -inverse strongly monotone to solve the split variational inequality problem. For the purpose we first give the following Lemmas.

Lemma 3.1 (see page 3 in [9]). Let be a -inverse strongly monotone mapping. Then is nonexpansive for any .

Example 3.2. Let for all and . Since is -inversely monotone. Let . It is easy to see that So is nonexpansive for all .

Applying Lemma 3.1, we have the following important result.

Lemma 3.3. Let be two -inverse strongly monotone mappings and be a nonexpansive mapping. Then for any given sequences and in , , , and are all nonexpansive for all .

The following conclusion is immediate from Lemma 2.5.

Lemma 3.4. The element is a solution of if and only if satisfies the relation where and are the projection operators, is a constant.

Theorem 3.5. Let , be two real Hilbert spaces and , two nonempty closed convex sets. Let and be -inversely monotone. Let be a bounded linear operator with adjoint operator . Let be a contraction with contraction constant . Let be a family of infinitely nonexpansive mappings of into itself and a nonexpansive mapping of into itself such that . Let be a real number and be a sequence of real numbers such that for every . For each , let be the -mapping of into itself generated by and . Let be a sequence generated by the following algorithm: where and are two constants and and are two sequences in . If and further satisfy the following conditions:

and ,

and , where ,

then the following statements hold:(a)there exists a unique such that ;(b) converges strongly to .

Proof. Let . By Lemma 3.3, , and are nonexpansive for all . For each , by (3.4) and Lemma 3.4, we obtain the following inequalities: Let for . Then for all . Next, we will show that the conclusion is true by several steps.
Step 1. We show that all , and are bounded.
To prove it, it suffices to show is bounded. Let . We claim that Indeed, it is obvious that (3.8) is true for . Assume that (3.7) is true for , . Since and by (3.5) and (3.7), it follows from (3.5) that which prove that (3.8) is true for . By induction, (3.8) holds for all . Hence, by (3.8), we know that is bounded and so are , , , and . This also means that there exists a bounded subset such that
Step 2. Prove .
For each , by Lemma 3.1, Similarly, Since , , we have where is a constant such that for any . Since , for each , we have where So, we have for any .
Choose a sequence such that , where , then we have It follows that where is a constant such that . From (3.20), (3.16), and (3.13) we have Applying the condition (C2), it follows from (3.21) that which implies
Applying Lemma 2.7, we obtain which implies that
Step 3. Prove .
For any , we have Similarly, From (3.5) again, we have where is a constant such that . It follows that which yields that (by the condition ).
For any , by (3.6), (3.7), and (3.27), we have So, From (3.26) and (3.30) again, we have which implies that (by the condition ).
On the other hand, since we get By (3.6) and (3.33), we have Using (3.7), (3.27), and (3.34), we obtain which implies According to (3.30) and (3.36), we derive that
Step 4. Prove .
Since we have . For any , since we get
It follows from (3.27), (3.7), and (3.40) that which yields that By Steps 1–3 and , it follows from (3.42) that .
Since for all , we have . Using it with and , we get .
Step 5. Prove and .
Indeed, since we have By Theorem 2.3, . Since we obtain .
Step 6. There exists a unique such that .
Indeed, for any , , Since , is a contraction on . Applying Banach contraction principle, there exists a unique such that .
Step 7. Prove .
For this purpose, we may choose subsequence of such that Since is a bounded sequence, there exists a subsequence of , which is still denoted by , such that . Therefore, we have Next we prove .
(a) . In fact, if , then we have This is a contradiction. Hence, , which implies that