Journal of Applied Mathematics

Volume 2012 (2012), Article ID 174318, 20 pages

doi:10.1155/2012/174318

Research Article

General Iterative Algorithms for Hierarchical Fixed Points Approach to Variational Inequalities

Nopparat Wairojjana and Poom Kumam

Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangmod, Bangkok 10140, Thailand

Received 24 March 2012; Accepted 16 May 2012

Academic Editor: Zhenyu Huang

Copyright © 2012 Nopparat Wairojjana and Poom Kumam. 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.

This paper deals with new methods for approximating a solution to the fixed point problem; find , where is a Hilbert space, is a closed convex subset of , is a -contraction from into , , is a strongly positive linear-bounded operator with coefficient , , is a nonexpansive mapping on and denotes the metric projection on the set of fixed point of . Under a suitable different parameter, we obtain strong convergence theorems by using the projection method which solves the variational inequality for , where . Our results generalize and improve the corresponding results of Yao et al. (2010) and some authors. Furthermore, we give an example which supports our main theorem in the last part.

1. Introduction

Throughout this paper, we assume that is a real Hilbert space where inner product and norm are denoted by and , respectively, and let be a nonempty closed convex subset of . A mapping is called nonexpansive if

We use to denote the set of fixed points of , that is, . It is assumed throughout the paper that is a nonexpansive mapping such that .

Recall that a mapping is a contraction on if there exists a constant such that

A mapping is called a strongly positive linear bounded operator on if there exists a constant with property

A mapping is called a strongly monotone operator with if

and is called a monotone operator if

We easily prove that the mapping is monotone operator, if is nonexpansive mapping.

The metric (or nearest point) projection from onto is mapping which assigns to each point the unique point satisfying the property

The variational inequality for a monotone operator, over , is to find a point in

A hierarchical fixed point problem is equivalent to the variational inequality for a monotone operator over the fixed point set. Moreover, to find a hierarchically fixed point of a nonexpansive mapping with respect to another nonexpansive mapping , namely, we find such that

Iterative methods for nonexpansive mappings have recently been applied to solve a convex minimization problem; see, for example, [1–5] and the references therein. A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space :

where is a given point in . In [5], it is proved that the sequence defined by the iterative method below, with the initial guess chosen arbitrarily,

converges strongly to the unique solution of the minimization problem (1.9) provided the sequence of parameters satisfies certain appropriate conditions.

On the other hand, Moudafi [6] introduced the viscosity approximation method for nonexpansive mappings (see [7] for further developments in both Hilbert and Banach spaces). Starting with an arbitrary initial , define a sequence recursively by

where is a sequence in . It is proved in [6, 7] that under certain appropriate conditions imposed on , the sequence generated by (1.11) strongly converges to the unique solution in of the variational inequality

In 2006, Marino and Xu [8] introduced a general iterative method for nonexpansive mapping. Starting with an arbitrary initial , define a sequence recursively by

They proved that if the sequence of parameters satisfies appropriate conditions, then the sequence generated by (1.13) strongly converges to the unique solution of the variational inequality

which is the optimality condition for the minimization problem

where is a potential function for (i.e., for ).

In 2010, Yao et al. [9] introduced an iterative algorithm for solving some hierarchical fixed point problem, starting with an arbitrary initial guess , define a sequence iteratively by

They proved that if the sequences and of parameters satisfies appropriate conditions, then the sequence generated by (1.16) strongly converges to the unique solution in of the variational inequality

In this paper we will combine the general iterative method (1.13) with the iterative algorithm (1.16) and consider the following iterative algorithm:

We will prove in Section 3 that if the sequences and of parameters satisfy appropriate conditions and then the sequence generated by (1.18) converges strongly to the unique solution in of the following variational inequality

In particular, if we take , under suitable difference assumptions on parameter, then the sequence generated by (1.18) converges strongly to the unique solution in of the following variational inequality

Our results improve and extend the recent results of Yao et al. [9] and some authors. Furthermore, we give an example which supports our main theorem in the last part.

2. Preliminaries

This section collects some lemma which can be used in the proofs for the main results in the next section. Some of them are known, others are not hard to derive.

Lemma 2.1 (Browder [10]). Let be a Hilbert space, be a closed convex subset of , and be a nonexpansive mapping with . If is a sequence in weakly converging to and if converges strongly to , then ; in particular, if then .

Lemma 2.2. Let and be any points. Then one has the following:

(1) That if and only if there holds the relation:

(2) That if and only if there holds the relation:

(3) There holds the relation:

Consequently, is nonexpansive and monotone.

Lemma 2.3 (Marino and Xu [8]). Let be a Hilbert space, be a closed convex subset of , be a contraction with coefficient , and be nonexpansive mapping. Let be a strongly positive linear bounded operator on a Hilbert space with coefficient . Then, for , for ,

(1)the mapping is strongly monotone with coefficient , that is,

(2)the mapping is strongly monotone with coefficient that is

Lemma 2.4 (Xu [4]). Assume that is a sequence of nonnegative numbers such that

where is a sequence in and is a sequence in such that

(1), (2) or . Then .

Lemma 2.5 (Marino and Xu [8]). Assume is a strongly positive linear bounded operator on a Hilbert space with coefficient and . Then .

Lemma 2.6 (Acedo and Xu [11]). Let be a closed convex subset of . Let be a bounded sequence in . Assume that

(1)The weak -limit set ,(2)For each , exists. Then is weakly convergent to a point in .

NotationWe use for strong convergence and for weak convergence.

3. Main Results

Theorem 3.1. Let be a nonempty closed convex subset of a real Hilbert space . Let be a -contraction with . Let be two nonexpansive mappings with . Let be a strongly positive linear bounded operator on with coefficient . and are two sequences in and . Starting with an arbitrary initial guess and is a sequence generated by

Suppose that the following conditions are satisfied:

(C1)   and , (C2)  , (C3)   and , or (C4)   and .

Then the sequence converges strongly to a point , which is the unique solution of the variational inequality:

Equivalently, one has .

Proof. We first show the uniqueness of a solution of the variational inequality (3.2), which is indeed a consequence of the strong monotonicity of . Suppose and both are solutions to (3.2), then and . It follows that

The strong monotonicity of (Lemma 2.3) implies that and the uniqueness is proved.

Next, we prove that the sequence is bounded. Since and by condition (C1) and (C2), respectively, we can assume, without loss of generality, that and for all . Take and from (3.1), we have

Since and by Lemma 2.5, we note that

By induction, we can obtain

which implies that the sequence is bounded and so are the sequences , , and .

Set . We get

It follows that

By (3.7) and (3.8), we get

From (3.1), we obtain

where is a constant such that

Substituting (3.10) into (3.8) to obtain

At the same time, we can write (3.12) as

From (3.12), (C4), and Lemma 2.5 or from (3.13), (C3), and Lemma 2.5, we can deduce that , respectively.

From (3.1), we have

Notice that , , and , so we obtain

Next, we prove

where . Since the sequence is bounded we can take a subsequence of such that

and . From (3.15) and by Lemma 2.1, it follows that . Hence, by Lemma 2.2(1) that

Now, by Lemma 2.2(1), we observe that

and so

Hence, it follows that

We observe that

Thus, by Lemma 2.4, as . This is completes.

From Theorem 3.1, we can deduce the following interesting corollary.

Corollary 3.2 (Yao et al. [9]). Let be a nonempty closed convex subset of a real Hilbert space . Let be a -contraction (possibly nonself) with . Let be two nonexpansive mappings with .   and are two sequences in . Starting with an arbitrary initial guess and is a sequence generated by

Suppose that the following conditions are satisfied:

(C1)   and , (C2)  , (C3)   and , or (C4)   and .

Then the sequence converges strongly to a point , which is the unique solution of the variational inequality:

Equivalently, one has . In particular, if one takes , then the sequence converges in norm to the Minimum norm fixed point of , namely, the point is the unique solution to the quadratic minimization problem:

Proof. As a matter of fact, if we take and in Theorem 3.1. This completes the proof.

Under different conditions on data we obtain the following result.

Theorem 3.3. Let be a nonempty closed convex subset of a real Hilbert space . Let be a -contraction (possibly nonself) with . Let be two nonexpansive mappings with . Let be a strongly positive linear bounded operator on a Hilbert space with coefficient and . and are two sequences in . Starting with an arbitrary initial guess and is a sequence generated by

Suppose that the following conditions are satisfied:

(C1)   and , (C2)  , (C5)  , (C6)  there exists a constant such that .

Then the sequence converges strongly to a point , which is the unique solution of the variational inequality:

Proof. First of all, we show that (3.27) has the unique solution. Indeed, let and be two solutions. Then

Analogously, we have

Adding (3.28) and (3.29), by Lemma 2.3, we obtain

and so . From (C2), we can assume, without loss of generality, that for all . By a similar argument in Theorem 3.1, we have

By induction, we obtain

which implies that the sequence is bounded. Since (C5) implies (C4) then, from Theorem 3.1, we can deduce .

From (3.1), we note that

Hence, it follows that

and so

Set . Then, we have

From (3.12) in Theorem 3.1 and (C6), we obtain

This together with Lemma 2.4 and (C2) imply that

From (3.36), for , we have

By Lemmas 2.2 and 2.3, we obtain

Now, we observe that

From (C1) and (C2), we have . Hence, from (3.1), we deduce and . Therefore,

Since , , , and , every weak cluster point of is also a strong cluster point. Note that the sequence is bounded, thus there exists a subsequence converging to a point . For all , it follows from (3.39) that

Letting , we obtain

By Lemma 2.6 and (3.27) having the unique solution, it follows that . Therefore, as . This completes the proof.

From Theorem 3.3, we can deduce the following interesting corollary.

Corollary 3.4 (Yao et al. [9]). Let be a nonempty closed convex subset of a real Hilbert space . Let be a -contraction (possibly nonself) with . Let be two nonexpansive mappings with . and are two sequences in Starting with an arbitrary initial guess and is a sequence generated by

Suppose that the following conditions are satisfied:

(C1)   and , (C2)  , (C5)  , (C6)  there exists a constant such that .

Then the sequence converges strongly to a point , which is the unique solution of the variational inequality:

Proof. As a matter of fact, if we take and in Theorem 3.3 then this completes the proof.

Corollary 3.5 (Yao et al. [9]). Let be a nonempty closed convex subset of a real Hilbert space . Let be two nonexpansive mappings with . and are two sequences in . Starting with an arbitrary initial guess and suppose is a sequence generated by

Suppose that the following conditions are satisfied:

(C1)   and , (C2)  , (C5)  , (C6)  there exists a constant such that .

Then the sequence converges strongly to a point , which is the unique solution of the variational inequality:

Proof. As a matter of fact, if we take , , and in Theorem 3.3 then this is completes the proof.

Remark 3.6. Prototypes for the iterative parameters are, for example, and (with ). Since and , it is not difficult to prove that (C5) is satisfied for and (C6) is satisfied if .

Remark 3.7. Our results improve and extend the results of Yao et al. [9] by taking and in Theorems 3.1 and 3.3.

The following is an example to support Theorem 3.3.

Example 3.8. Let ,,,, , ,, for every , we have and choose , and . Then is the sequence

and as , where is the unique solution of the variational inequality

Acknowledgments

The authors would like to thank the National Research University Project of Thailand’s Office of the Higher Education Commission under the Project NRU-CSEC no. 55000613 for financial support.

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