Journal of Applied MathematicsVolume 2012 (2012), Article ID 320421, 13 pagesdoi:10.1155/2012/320421

Research Article

An Iterative Algorithm for a Hierarchical Problem

Yonghong Yao,¹ Yeol Je Cho,² and Pei-Xia Yang¹

¹Department of Mathematics, Tianjin Polytechnic University, Tianjin 300387, China²Department of Mathematics Education and the RINS, Gyeongsang National University, Chinju 660-701, Republic of Korea

Received 29 September 2011; Accepted 11 November 2011

Academic Editor: Giuseppe Marino

Copyright © 2012 Yonghong Yao 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.

A general hierarchical problem has been considered, and an explicit algorithm has been presented for solving this hierarchical problem. Also, it is shown that the suggested algorithm converges strongly to a solution of the hierarchical problem.

1. Introduction

Let be a real Hilbert space with inner product and norm , respectively. Let be a nonempty closed convex subset of . The hierarchical problem is of finding such that where are two nonexpansive mappings and is the set of fixed points of . Recently, this problem has been studied by many authors (see, e.g., [1–15]). The main reason is that this problem is closely associated with some monotone variational inequalities and convex programming problems (see [16–19]).

Now, we briefly recall some historic results which relate to the problem (1.1).

For solving the problem (1.1), in 2006, Moudafi and Mainge [1] first introduced an implicit iterative algorithm: and proved that the net defined by (1.2) strongly converges to as , where satisfies , where is a mapping defined by or, equivalently, is the unique solution of the quasivariational inequality where the normal cone to , , is defined as follows:

Moreover, as , the net in turn weakly converges to the unique solution of the fixed point equation or, equivalently, is the unique solution of the variational inequality

Recently, Moudafi [2] constructed an explicit iterative algorithm: where and are two real numbers in . By using this iterative algorithm, Moudafi [2] only proved a weak convergence theorem for solving the problem (1.1).

In order to obtain a strong convergence result, Mainge and Moudafi [3] further introduced the following iterative algorithm: where and are two real numbers in , and proved that, under appropriate conditions, the iterative sequence generated by (1.8) has strong convergence.

Subsequently, some authors have studied some algorithms on hierarchical fixed problems (see, e.g., [4–15]).

Motivated and inspired by the results in the literature, in this paper, we consider a general hierarchical problem of finding such that, for any , where is the -mapping defined by (2.3) below and is a nonexpansive mapping, and introduce an explicit iterative algorithm which converges strongly to a solution of the hierarchical problem (1.9).

2. Preliminaries

Let a nonempty closed convex subset of a real Hilbert space . Recall that a mapping is said to be contractive if there exists a constant such that

A mapping is called nonexpansive if

Forward, we use to denote the fixed points set of .

Let be an infinite family of nonexpansive mappings and a real number sequence such that for each .

For each , define a mapping as follows:

Such is called the -mapping generated by and .

Lemma 2.1 (see [20]). Let be a nonempty closed convex subset of a real Hilbert space . Let be an infinite family of nonexpansive mappings of into itself with . Let be real numbers such that for each . Then one has the following results: (1)for any and , the limit exists; (2).

Using Lemma  3.1 in [21], we can define a mapping of into itself by for all . Thus we have the following.

Lemma 2.2 (see [21]). If is a bounded sequence in , then one has

Lemma 2.3 (see [22]). Let be a nonempty closed convex of a real Hilbert space and be nonexpansive mapping. Then is demiclosed on , that is, if and , then .

Lemma 2.4 (see [23]). Assume is a sequence of nonnegative real numbers such that where is a sequence in and are two sequences such that (i);(ii) or ;(iii). Then .

3. Main Results

In this section, we introduce our algorithm and give its convergence analysis.

Algorithm 3.1. Let be a nonempty closed convex subset of a real Hilbert space and be infinite family of nonexpansive mappings of into itself. Let be a contraction with coefficient . For any , let the sequence generated iteratively by where are two real numbers in and is the -mapping defined by (2.3).

Now, we give the convergence analysis of the algorithm.

Theorem 3.2. Let be a nonempty closed convex subset of a real Hilbert space and be an infinite family of nonexpansive mappings of into itself. Let be a contraction with coefficient . Assume that the set of solutions of the hierarchical problem (1.9) is nonempty. Let be two real numbers in and the sequence generated by (3.1). Assume that the sequence is bounded and (i) and ; (ii); (iii) and . Then and every weak cluster point of the sequence solves the following variational inequality

Proof. Set for each . Then we have It follows that From (3.1), we have Then we obtain From (2.3), since and are nonexpansive, we have where is a constant such that . Substituting (3.4) and (3.7) into (3.6), we get Therefore, it follows that where is a constant such that From (iii), we note that , which implies that Thus it follows from (iii) and (3.11) that Hence, applying Lemma 2.4 to (3.9), we immediately conclude that This implies that Thus, from (3.1) and (3.14), we have At the same time, we note that Hence we get Since the sequence is bounded, is also bounded. Thus there exists a subsequence of , which is still denoted by which converges weakly to a point . Therefore, by (3.17) and Lemma 2.3. By (3.1), we observe that that is, Set for each , that is, Using monotonicity of and , we derive that, for all , But, since , and (by Lemma 2.2), it follows from the above inequality that This suffices to guarantee that . As a matter of fact, if we take any , then there exists a subsequence of such that . Therefore, we have Note that . Hence solves the following problem: It is obvious that this is equivalent to the problem (1.9) since uniformly in any bounded set (by Lemma 2.2). Thus .Let be the unique solution of the variational inequality (3.2). Now, take a subsequence of such that Without loss of generality, we may further assume that . Then . Therefore, we have This completes the proof.

Theorem 3.3. Let be a nonempty closed convex subset of a real Hilbert space . Let be infinite family of nonexpansive mappings of into itself. Let be a contraction with coefficient . Assume that the set of solutions of the hierarchical problem (1.9) is nonempty. Let be two real numbers in and the sequence generated by (3.1). Assume that the sequence is bounded and (i), and ; (ii); (iii) and = = 0; (iv)there exists a constant such that , where Then the sequence defined by (3.1) converges strongly to a point , which solves the variational inequality problem (3.2).

Proof. From (3.1), we have Thus we have At the same time, we observe that Substituting (3.30) into (3.29), we get By Theorem 3.2, we note that every weak cluster point of the sequence is in . Since , then every weak cluster point of is also in . Consequently, since , we easily have On the other hand, we observe that Since is a solution of the problem (1.9) and , we have Thus it follows that We note that Hence we have From Theorem 3.2, we have . At the same time, we note that , , and are all bounded. Hence it follows from (i) and the above inequality that Finally, by (3.31)–(3.38) and Lemma 2.4, we conclude that the sequence converges strongly to a point . This completes the proof.

Remark 3.4. In the present paper, we consider the hierarchical problem (1.9) which includes the hierarchical problem (1.1) as a special case.From the above discussion, we can easily deduce the following result.

Algorithm 3.5. Let be a nonempty closed convex subset of a real Hilbert space and a nonexpansive mapping of into itself. Let be a contraction with coefficient . For any , let the sequence generated iteratively by where are two real numbers in .

Corollary 3.6. Let be a nonempty closed convex subset of a real Hilbert space . Let be a nonexpansive mapping. Let be a contraction with coefficient . Assume that the set of solutions of the hierarchical problem (1.1) is nonempty. Let be two real numbers in and the sequence generated by (3.1). Assume that the sequence is bounded and (i), and ; (ii); (iii) and ; (iv)there exists a constant such that , where Then the sequence defined by (3.39) converges strongly to a point , which solves the hierarchical problem (1.1).

Acknowledgment

This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant no. 2011-0021821).

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