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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 320421, 13 pages
http://dx.doi.org/10.1155/2012/320421
Research Article

An Iterative Algorithm for a Hierarchical Problem

1Department of Mathematics, Tianjin Polytechnic University, Tianjin 300387, China
2Department 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 𝐻 be a real Hilbert space with inner product , and norm , respectively. Let 𝐶 be a nonempty closed convex subset of 𝐻. The hierarchical problem is of finding ̃𝑥Fix(𝑇) such that𝑆̃𝑥̃𝑥,𝑥̃𝑥0,𝑥Fix(𝑇),(1.1) where 𝑆,𝑇 are two nonexpansive mappings and Fix(𝑇) is the set of fixed points of 𝑇. Recently, this problem has been studied by many authors (see, e.g., [115]). The main reason is that this problem is closely associated with some monotone variational inequalities and convex programming problems (see [1619]).

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:𝑥𝑡,𝑠𝑥=𝑠𝑄𝑡,𝑠+𝑥(1𝑠)𝑡𝑆𝑡,𝑠+𝑥(1𝑡)𝑇𝑡,𝑠(1.2) and proved that the net {𝑥𝑡,𝑠} defined by (1.2) strongly converges to 𝑥𝑡 as 𝑠0, where 𝑥𝑡 satisfies 𝑥𝑡=projFix(𝑃𝑡)𝑄(𝑥𝑡), where 𝑃𝑡𝐶𝐶 is a mapping defined by 𝑃𝑡(𝑥)=𝑡𝑆(𝑥)+(1𝑡)𝑇(𝑥),𝑥𝐶,𝑡(0,1),(1.3) or, equivalently, 𝑥𝑡 is the unique solution of the quasivariational inequality 0(𝐼𝑄)𝑥𝑡+𝑁Fix(𝑃𝑡)𝑥𝑡,(1.4) where the normal cone to Fix(𝑃𝑡), 𝑁Fix(𝑃𝑡), is defined as follows: 𝑁Fix(𝑃𝑡)𝑃𝑥{𝑢𝐻𝑦𝑥,𝑢0},if𝑥Fix𝑡,,otherwise.(1.5)

Moreover, as 𝑡0, the net {𝑥𝑡} in turn weakly converges to the unique solution 𝑥 of the fixed point equation 𝑥=projΩ𝑄(𝑥) or, equivalently, 𝑥 is the unique solution of the variational inequality 0(𝐼𝑄)𝑥+𝑁Ω𝑥.(1.6)

Recently, Moudafi [2] constructed an explicit iterative algorithm:𝑥𝑛+1=1𝛿𝑛𝑥𝑛+𝛿𝑛𝜎𝑛𝑆𝑥𝑛+1𝜎𝑛𝑇𝑥𝑛,𝑛0,(1.7) where {𝛿𝑛} and {𝜎𝑛} are two real numbers in (0,1). 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:𝑥𝑛+1=1𝛿𝑛𝑄𝑥𝑛+𝛿𝑛𝜎𝑛𝑆𝑥𝑛+1𝜎𝑛𝑇𝑥𝑛,𝑛0,(1.8) where {𝛿𝑛} and {𝜎𝑛} are two real numbers in (0,1), and proved that, under appropriate conditions, the iterative sequence {𝑥𝑛} generated by (1.8) has strong convergence.

Subsequently, some authors have studied some algorithms on hierarchical fixed problems (see, e.g., [415]).

Motivated and inspired by the results in the literature, in this paper, we consider a general hierarchical problem of finding ̃𝑥Fix(𝑇) such that, for any 𝑛1,𝑊𝑛̃𝑥̃𝑥,𝑥̃𝑥0,𝑥Fix(𝑇),(1.9) where 𝑊𝑛 is the 𝑊-mapping defined by (2.3) below and 𝑇 is a nonexpansive mapping, and introduce an explicit iterative algorithm which converges strongly to a solution ̃𝑥 of the hierarchical problem (1.9).

2. Preliminaries

Let 𝐶 a nonempty closed convex subset of a real Hilbert space 𝐻. Recall that a mapping 𝑄𝐶𝐶 is said to be contractive if there exists a constant 𝛾(0,1) such that 𝑄𝑥𝑄𝑦𝛾𝑥𝑦,𝑥,𝑦𝐶.(2.1)

A mapping 𝑇𝐶𝐶 is called nonexpansive if𝑇𝑥𝑇𝑦𝑥𝑦,𝑥,𝑦𝐶.(2.2)

Forward, we use Fix(𝑇) to denote the fixed points set of 𝑇.

Let {𝑇𝑖}𝑖=1𝐶𝐶 be an infinite family of nonexpansive mappings and {𝜉𝑖}𝑖=1 a real number sequence such that 0𝜉𝑖1 for each 𝑖1.

For each 𝑛1, define a mapping 𝑊𝑛𝐶𝐶 as follows:𝑈𝑛,𝑛+1𝑈=𝐼,𝑛,𝑛=𝜉𝑛𝑇𝑛𝑈𝑛,𝑛+1+1𝜉𝑛𝑈𝐼,𝑛,𝑛1=𝜉𝑛1𝑇𝑛1𝑈𝑛,𝑛+1𝜉𝑛1𝑈𝐼,𝑛,𝑘=𝜉𝑘𝑇𝑘𝑈𝑛,𝑘+1+1𝜉𝑘𝑈𝐼,𝑛,𝑘1=𝜉𝑘1𝑇𝑘1𝑈𝑛,𝑘+1𝜉𝑘1𝑈𝐼,𝑛,2=𝜉2𝑇2𝑈𝑛,3+1𝜉2𝑊𝐼,𝑛=𝑈𝑛,1=𝜉1𝑇1𝑈𝑛,2+1𝜉1𝐼.(2.3)

Such 𝑊𝑛 is called the 𝑊-mapping generated by {𝑇𝑖}𝑖=1 and {𝜉𝑖}𝑖=1.

Lemma 2.1 (see [20]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let {𝑇𝑖}𝑖=1 be an infinite family of nonexpansive mappings of 𝐶 into itself with 𝑛=1Fix(𝑇𝑛). Let 𝜉1,𝜉2, be real numbers such that 0<𝜉𝑖𝑏<1 for each 𝑖1. Then one has the following results: (1)for any 𝑥𝐶 and 𝑘1, the limit lim𝑛𝑈𝑛,𝑘𝑥 exists; (2)Fix(𝑊)=𝑛=1Fix(𝑇𝑛).

Using Lemma  3.1 in [21], we can define a mapping 𝑊 of 𝐶 into itself by 𝑊𝑥=lim𝑛𝑊𝑛𝑥=lim𝑛𝑈𝑛,1𝑥 for all 𝑥𝐶. Thus we have the following.

Lemma 2.2 (see [21]). If {𝑥𝑛} is a bounded sequence in 𝐶, then one has lim𝑛𝑊𝑥𝑛𝑊𝑛𝑥𝑛=0.(2.4)

Lemma 2.3 (see [22]). Let 𝐶 be a nonempty closed convex of a real Hilbert space 𝐻 and 𝑇𝐶𝐶 be nonexpansive mapping. Then 𝑇 is demiclosed on 𝐶, that is, if 𝑥𝑛𝑥𝐶 and 𝑥𝑛𝑇𝑥𝑛0, then 𝑥=𝑇𝑥.

Lemma 2.4 (see [23]). Assume {𝑎𝑛} is a sequence of nonnegative real numbers such that 𝑎𝑛+11𝛾𝑛𝑎𝑛+𝛿𝑛𝛾𝑛+𝜂𝑛,𝑛1,(2.5) where {𝛾𝑛} is a sequence in (0,1) and {𝛿𝑛},{𝜂𝑛} are two sequences such that (i)𝑛=1𝛾𝑛=;(ii)limsup𝑛𝛿𝑛0 or 𝑛=1|𝛿𝑛𝛾𝑛|<;(iii)𝑛=1|𝜂𝑛|<.
Then lim𝑛𝑎𝑛=0.

3. Main Results

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

Algorithm 3.1. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻 and {𝑇𝑛}𝑛=1 be infinite family of nonexpansive mappings of 𝐶 into itself. Let 𝑄𝐶𝐶 be a contraction with coefficient 𝛾[0,1). For any 𝑥0𝐶, let {𝑥𝑛} the sequence generated iteratively by 𝑥𝑛+1=𝛼𝑛𝑊𝑛𝑥𝑛+1𝛼𝑛𝑇𝛽𝑛𝑄𝑥𝑛+1𝛽𝑛𝑥𝑛,𝑛0,(3.1) where {𝛼𝑛},{𝛽𝑛} are two real numbers in (0,1) and 𝑊𝑛 is the 𝑊-mapping defined by (2.3).

Now, we give the convergence analysis of the algorithm.

Theorem 3.2. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻 and {𝑇𝑛}𝑛=1 be an infinite family of nonexpansive mappings of 𝐶 into itself. Let 𝑄𝐶𝐶 be a contraction with coefficient 𝛾[0,1). Assume that the set Ω of solutions of the hierarchical problem (1.9) is nonempty. Let {𝛼𝑛},{𝛽𝑛} be two real numbers in (0,1) and {𝑥𝑛} the sequence generated by (3.1). Assume that the sequence {𝑥𝑛} is bounded and (i)lim𝑛𝛼𝑛=0 and lim𝑛(𝛽𝑛/𝛼𝑛)=0; (ii)𝑛=0𝛽𝑛=; (iii)lim𝑛(1/𝛽𝑛)|(1/𝛼𝑛)(1/𝛼𝑛1)|=0 and lim𝑛(𝑛1𝑖=1𝜉𝑖/𝛼𝑛𝛽𝑛)=lim𝑛(1/𝛼𝑛)|1(𝛽𝑛1/𝛽𝑛)|=0.
Then lim𝑛(𝑥𝑛+1𝑥𝑛/𝛼𝑛)=0 and every weak cluster point of the sequence {𝑥𝑛} solves the following variational inequalitỹ𝑥Ω,(𝐼𝑄)̃𝑥,𝑥̃𝑥0,𝑥Ω.(3.2)

Proof. Set 𝑦𝑛=𝛽𝑛𝑄𝑥𝑛+(1𝛽𝑛)𝑥𝑛 for each 𝑛0. Then we have 𝑦𝑛𝑦𝑛1=𝛽𝑛𝑄𝑥𝑛+1𝛽𝑛𝑥𝑛𝛽𝑛1𝑄𝑥𝑛11𝛽𝑛1𝑥𝑛1=𝛽𝑛𝑄𝑥𝑛𝑄𝑥𝑛1+𝛽𝑛𝛽𝑛1𝑄𝑥𝑛1+1𝛽𝑛𝑥𝑛𝑥𝑛1+𝛽𝑛1𝛽𝑛𝑥𝑛1.(3.3) It follows that 𝑦𝑛𝑦𝑛1𝛾𝛽𝑛𝑥𝑛𝑥𝑛1+1𝛽𝑛𝑥𝑛𝑥𝑛1+||𝛽𝑛𝛽𝑛1||𝑄𝑥𝑛1+𝑥𝑛1=1(1𝛾)𝛽𝑛𝑥𝑛𝑥𝑛1+||𝛽𝑛𝛽𝑛1||𝑄𝑥𝑛1+𝑥𝑛1.(3.4) From (3.1), we have 𝑥𝑛+1𝑥𝑛=𝛼𝑛𝑊𝑛𝑥𝑛+1𝛼𝑛𝑇𝑦𝑛𝛼𝑛1𝑊𝑛1𝑥𝑛11𝛼𝑛1𝑇𝑦𝑛1=𝛼𝑛𝑊𝑛𝑥𝑛𝑊𝑛𝑥𝑛1+𝛼𝑛𝛼𝑛1𝑊𝑛𝑥𝑛1+𝛼𝑛1𝑊𝑛𝑥𝑛1𝑊𝑛1𝑥𝑛1+1𝛼𝑛𝑇𝑦𝑛𝑇𝑦𝑛1+𝛼𝑛1𝛼𝑛𝑇𝑦𝑛1.(3.5) Then we obtain 𝑥𝑛+1𝑥𝑛𝛼𝑛𝑊𝑛𝑥𝑛𝑊𝑛𝑥𝑛1+1𝛼𝑛𝑇𝑦𝑛𝑇𝑦𝑛1+||𝛼𝑛𝛼𝑛1||𝑊𝑛𝑥𝑛1+𝑇𝑦𝑛1+𝛼𝑛1𝑊𝑛𝑥𝑛1𝑊𝑛1𝑥𝑛1𝛼𝑛𝑥𝑛𝑥𝑛1+1𝛼𝑛𝑦𝑛𝑦𝑛1+||𝛼𝑛𝛼𝑛1||𝑊𝑛𝑥𝑛1+𝑇𝑦𝑛1+𝛼𝑛1𝑊𝑛𝑥𝑛1𝑊𝑛1𝑥𝑛1.(3.6) From (2.3), since 𝑇𝑖 and 𝑈𝑛,𝑖 are nonexpansive, we have 𝑊𝑛𝑥𝑛1𝑊𝑛1𝑥𝑛1=𝜉1𝑇1U𝑛,2𝑥𝑛1𝜉1𝑇1𝑈𝑛1,2𝑥𝑛1𝜉1𝑈𝑛,2𝑥𝑛1𝑈𝑛1,2𝑥𝑛1=𝜉1𝜉2𝑇2𝑈𝑛,3𝑥𝑛1𝜉2𝑇2𝑈𝑛1,3𝑥𝑛1𝜉1𝜉2𝑈𝑛,3𝑥𝑛1𝑈𝑛1,3𝑥𝑛1𝜉1𝜉2𝜉𝑛1𝑈𝑛,𝑛𝑥𝑛1𝑈𝑛1,𝑛𝑥𝑛1𝑀1𝑛1𝑖=1𝜉𝑖,(3.7) where 𝑀1 is a constant such that sup𝑛1{𝑈𝑛,𝑛𝑥𝑛1𝑈𝑛1,𝑛𝑥𝑛1}𝑀1. Substituting (3.4) and (3.7) into (3.6), we get 𝑥𝑛+1𝑥𝑛𝛼𝑛𝑥𝑛𝑥𝑛1+1𝛼𝑛1(1𝛾)𝛽𝑛𝑥𝑛𝑥𝑛1+||𝛽𝑛𝛽𝑛1||𝑄𝑥𝑛1+𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑊𝑛𝑥𝑛1+𝑇𝑦𝑛1+𝛼𝑛1𝑀1𝑛1𝑖=1𝜉𝑖=1(1𝛾)𝛽𝑛1𝛼𝑛𝑥𝑛𝑥𝑛1+||𝛽𝑛𝛽𝑛1||𝑄𝑥𝑛1+𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑊𝑛𝑥𝑛1+𝑇𝑦𝑛1+𝛼𝑛1𝑀1𝑛1𝑖=1𝜉𝑖.(3.8) Therefore, it follows that 𝑥𝑛+1𝑥𝑛𝛼𝑛1(1𝛾)𝛽𝑛1𝛼𝑛𝑥𝑛𝑥𝑛1𝛼𝑛+||𝛽𝑛𝛽𝑛1||𝛼𝑛𝑄𝑥𝑛1+𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝛼𝑛𝑊𝑛𝑥𝑛1+𝑇𝑦𝑛1+𝛼𝑛1𝑀1𝑛1𝑖=1𝜉𝑖𝛼𝑛=1(1𝛾)𝛽𝑛1𝛼𝑛𝑥𝑛𝑥𝑛1𝛼𝑛1+1(1𝛾)𝛽𝑛1𝛼𝑛𝑥𝑛𝑥𝑛1𝛼𝑛𝑥𝑛𝑥𝑛1𝛼𝑛1+||𝛽𝑛𝛽𝑛1||𝛼𝑛𝑄𝑥𝑛1+𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝛼𝑛𝑊𝑛𝑥𝑛1+𝑇𝑦𝑛1+𝛼𝑛1𝑀1𝑛1𝑖=1𝜉𝑖𝛼𝑛1(1𝛾)𝛽𝑛1𝛼𝑛𝑥𝑛𝑥𝑛1𝛼𝑛1+||||1𝛼𝑛1𝛼𝑛1||||+||𝛼𝑛𝛼𝑛1||𝛼𝑛+||𝛽𝑛𝛽𝑛1||𝛼𝑛+𝑛1𝑖=1𝜉𝑖𝛼𝑛𝑀=1(1𝛾)𝛽𝑛1𝛼𝑛𝑥𝑛𝑥𝑛1𝛼𝑛1+(1𝛾)𝛽𝑛1𝛼𝑛×𝑀(1𝛾)1𝛼𝑛1𝛽𝑛||||1𝛼𝑛1𝛼𝑛1||||+1𝛽𝑛||𝛼𝑛𝛼𝑛1||𝛼𝑛+1𝛽𝑛||𝛽𝑛𝛽𝑛1||𝛼𝑛+𝑛1𝑖=1𝜉𝑖𝛼𝑛𝛽𝑛,(3.9) where 𝑀 is a constant such that sup𝑛1𝑀1,𝑥𝑛𝑥𝑛1,𝑊𝑛𝑥𝑛1+𝑇𝑦𝑛1,𝑄𝑥𝑛1+𝑥𝑛1𝑀.(3.10) From (iii), we note that lim𝑛(1/𝛼𝑛1)|𝛼𝑛𝛼𝑛1/𝛽𝑛𝛼𝑛|=0, which implies that lim𝑛1𝛽𝑛||𝛼𝑛𝛼𝑛1||𝛼𝑛=0.(3.11) Thus it follows from (iii) and (3.11) that lim𝑛1𝛽𝑛||||1𝛼𝑛1𝛼𝑛1||||+1𝛽𝑛||𝛼𝑛𝛼𝑛1||𝛼𝑛+1𝛽𝑛||𝛽𝑛𝛽𝑛1||𝛼𝑛+𝑛1𝑖=1𝜉𝑖𝛼𝑛𝛽𝑛=0.(3.12) Hence, applying Lemma 2.4 to (3.9), we immediately conclude that lim𝑛𝑥𝑛+1𝑥𝑛𝛼𝑛=0.(3.13) This implies that lim𝑛𝑥𝑛+1𝑥𝑛=0.(3.14) Thus, from (3.1) and (3.14), we have lim𝑛𝑥𝑛𝑇𝑦𝑛=0.(3.15) At the same time, we note that 𝑦𝑛𝑥𝑛=𝛽𝑛𝑄𝑥𝑛𝑥𝑛0.(3.16) Hence we get lim𝑛𝑦𝑛𝑇𝑦𝑛=0.(3.17) Since the sequence {𝑥𝑛} is bounded, {𝑦𝑛} is also bounded. Thus there exists a subsequence of {𝑦𝑛}, which is still denoted by {𝑦𝑛} which converges weakly to a point ̃𝑥𝐻. Therefore, ̃𝑥Fix(𝑇) by (3.17) and Lemma 2.3. By (3.1), we observe that 𝑥𝑛+1𝑥𝑛=𝛼𝑛𝑊𝑛𝑥𝑛𝑥𝑛+1𝛼𝑛𝑇𝑦𝑛𝑦𝑛+1𝛼𝑛𝛽𝑛𝑄𝑥𝑛𝑥𝑛,(3.18) that is, 𝑥𝑛𝑥𝑛+1𝛼𝑛=𝐼𝑊𝑛𝑥𝑛+1𝛼𝑛𝛼𝑛(𝐼𝑇)𝑦𝑛+𝛽𝑛1𝛼𝑛𝛼𝑛(𝐼𝑄)𝑥𝑛.(3.19) Set 𝑧𝑛=(𝑥𝑛𝑥𝑛+1)/𝛼𝑛 for each 𝑛1, that is, 𝑧𝑛=𝐼𝑊𝑛𝑥𝑛+1𝛼𝑛𝛼𝑛(𝐼𝑇)𝑦𝑛+𝛽𝑛1𝛼𝑛𝛼𝑛(𝐼𝑄)𝑥𝑛.(3.20) Using monotonicity of 𝐼𝑇 and 𝐼𝑊𝑛, we derive that, for all 𝑢Fix(𝑇), 𝑧𝑛,𝑥𝑛=𝑢𝐼𝑊𝑛𝑥𝑛,𝑥𝑛+𝑢1𝛼𝑛𝛼𝑛(𝐼𝑇)𝑦𝑛(𝐼𝑇)𝑢,𝑦𝑛+𝑢1𝛼𝑛𝛼𝑛(𝐼𝑇)𝑦𝑛,𝑥𝑛𝑦𝑛𝛽+𝑛1𝛼𝑛𝛼𝑛(𝐼𝑄)𝑥𝑛,𝑥𝑛𝑢𝐼𝑊𝑛𝑢,𝑥𝑛+𝛽𝑢𝑛1𝛼𝑛𝛼𝑛(𝐼𝑄)𝑥𝑛,𝑥𝑛𝑢+1𝛼𝑛𝛽𝑛𝛼𝑛(𝐼𝑇)𝑦𝑛,𝑥𝑛𝑄𝑥𝑛=(𝐼𝑊)𝑢,𝑥𝑛𝑢+𝑊𝑊𝑛𝑢,𝑥𝑛+𝛽𝑢𝑛1𝛼𝑛𝛼𝑛(𝐼𝑄)𝑥𝑛,𝑥𝑛+𝑢1𝛼𝑛𝛽𝑛𝛼𝑛(𝐼𝑇)𝑦𝑛,𝑥𝑛𝑄𝑥𝑛.(3.21) But, since 𝑧𝑛0,𝛽𝑛/𝛼𝑛0 and lim𝑛𝑊𝑛𝑢𝑊𝑢=0 (by Lemma 2.2), it follows from the above inequality that limsup𝑛(𝐼𝑊)𝑢,𝑥𝑛𝑢0,𝑢Fix(𝑇).(3.22) This suffices to guarantee that 𝜔𝑤(𝑥𝑛)Ω. As a matter of fact, if we take any 𝑥𝜔𝑤(𝑥𝑛), then there exists a subsequence {𝑥𝑛𝑗} of {𝑥𝑛} such that 𝑥𝑛𝑗𝑥. Therefore, we have (𝐼𝑊)𝑢,𝑥𝑢=lim𝑗(𝐼𝑊)𝑢,𝑥𝑛𝑗𝑢0,𝑢Fix(𝑇).(3.23) Note that 𝑥Fix(𝑇). Hence 𝑥 solves the following problem: 𝑥Fix(𝑇),(𝐼𝑊)𝑢,𝑥𝑢0,𝑢Fix(𝑇).(3.24) It is obvious that this is equivalent to the problem (1.9) since 𝑊𝑛𝑊 uniformly in any bounded set (by Lemma 2.2). Thus 𝑥Ω.
Let ̃𝑥 be the unique solution of the variational inequality (3.2). Now, take a subsequence {𝑥𝑛𝑖} of {𝑥𝑛} such thatlimsup𝑛(𝐼𝑄)̃𝑥,𝑥𝑛̃𝑥=lim𝑖(𝐼𝑄)̃𝑥,𝑥𝑛𝑖.̃𝑥(3.25) Without loss of generality, we may further assume that 𝑥𝑛𝑖𝑥. Then 𝑥Ω. Therefore, we have limsup𝑛(𝐼𝑄)̃𝑥,𝑥𝑛̃𝑥=(𝐼𝑄)̃𝑥,𝑥̃𝑥0.(3.26) This completes the proof.

Theorem 3.3. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let {𝑇𝑛}𝑛=1 be infinite family of nonexpansive mappings of 𝐶 into itself. Let 𝑄𝐶𝐶 be a contraction with coefficient 𝛾[0,1). Assume that the set Ω of solutions of the hierarchical problem (1.9) is nonempty. Let {𝛼𝑛},{𝛽𝑛} be two real numbers in (0,1) and {𝑥𝑛} the sequence generated by (3.1). Assume that the sequence {𝑥𝑛} is bounded and (i)lim𝑛𝛼𝑛=0, lim𝑛𝛽𝑛/𝛼𝑛=0 and lim𝑛𝛼2𝑛/𝛽𝑛=0; (ii)𝑛=0𝛽𝑛=; (iii)lim𝑛(1/𝛽𝑛)|(1/𝛼𝑛)(1/𝛼𝑛1)|=0 and lim𝑛𝑛1𝑖=1𝜉𝑖/𝛼𝑛𝛽𝑛 = lim𝑛(1/𝛼𝑛)|1(𝛽𝑛1/𝛽𝑛)| = 0; (iv)there exists a constant 𝑘>0 such that 𝑥𝑇𝑥𝑘Dist(𝑥,Fix(𝑇)), where Dist(𝑥,Fix(𝑇))=inf𝑦Fix(𝑇)𝑥𝑦.(3.27) Then the sequence {𝑥𝑛} defined by (3.1) converges strongly to a point ̃𝑥Fix(𝑇), which solves the variational inequality problem (3.2).

Proof. From (3.1), we have 𝑥𝑛+1̃𝑥=𝛼𝑛𝑊𝑛𝑥𝑛𝑊𝑛̃𝑥+𝛼𝑛𝑊𝑛+̃𝑥̃𝑥1𝛼𝑛𝑇𝑦𝑛̃𝑥.(3.28) Thus we have 𝑥𝑛+1̃𝑥2𝛼𝑛𝑊𝑛𝑥𝑛𝑊𝑛+̃𝑥1𝛼𝑛𝑇𝑦𝑛̃𝑥2+2𝛼𝑛𝑊𝑛̃𝑥̃𝑥,𝑥𝑛+1̃𝑥1𝛼𝑛𝑇𝑦𝑛̃𝑥2+𝛼𝑛𝑊𝑛𝑥𝑛𝑊𝑛̃𝑥2+2𝛼𝑛𝑊𝑛̃𝑥̃𝑥,𝑥𝑛+1̃𝑥1𝛼𝑛𝑦𝑛̃𝑥2+𝛼𝑛𝑥𝑛̃𝑥2+2𝛼𝑛𝑊𝑛̃𝑥̃𝑥,𝑥𝑛+1.̃𝑥(3.29) At the same time, we observe that 𝑦𝑛̃𝑥2=1𝛽𝑛𝑥𝑛̃𝑥+𝛽𝑛𝑄𝑥𝑛𝑄̃𝑥+𝛽𝑛(𝑄̃𝑥̃𝑥)21𝛽𝑛𝑥𝑛̃𝑥+𝛽𝑛𝑄𝑥𝑛𝑄̃𝑥2+2𝛽𝑛𝑄̃𝑥̃𝑥,𝑦𝑛̃𝑥1𝛽𝑛𝑥𝑛̃𝑥2+𝛽𝑛𝑄𝑥𝑛𝑄̃𝑥2+2𝛽𝑛𝑄̃𝑥̃𝑥,𝑦𝑛̃𝑥1𝛽𝑛𝑥𝑛̃𝑥2+𝛽𝑛𝛾2𝑥𝑛̃𝑥2+2𝛽𝑛𝑄̃𝑥̃𝑥,𝑦𝑛=̃𝑥11𝛾2𝛽𝑛𝑥𝑛̃𝑥2+2𝛽𝑛𝑄̃𝑥̃𝑥,𝑦𝑛̃𝑥.(3.30) Substituting (3.30) into (3.29), we get 𝑥𝑛+1̃𝑥2𝛼𝑛𝑥𝑛̃𝑥2+1𝛼𝑛11𝛾2𝛽𝑛𝑥𝑛̃𝑥2+2𝛽𝑛1𝛼𝑛𝑄̃𝑥̃𝑥,𝑦𝑛̃𝑥+2𝛼𝑛𝑊𝑛̃𝑥̃𝑥,𝑥𝑛+1=̃𝑥11𝛾2𝛽𝑛1𝛼𝑛𝑥𝑛̃𝑥2+2𝛽𝑛1𝛼𝑛𝑄̃𝑥̃𝑥,𝑦𝑛̃𝑥+2𝛼𝑛𝑊𝑛̃𝑥̃𝑥,𝑥𝑛+1=̃𝑥11𝛾2𝛽𝑛1𝛼𝑛𝑥𝑛̃𝑥2+1𝛾2𝛽𝑛1𝛼𝑛×21𝛾2𝑄̃𝑥̃𝑥,𝑦𝑛2̃𝑥+1𝛾21𝛼𝑛×𝛼𝑛𝛽𝑛𝑊𝑛̃𝑥̃𝑥,𝑥𝑛+1.̃𝑥(3.31) By Theorem 3.2, we note that every weak cluster point of the sequence {𝑥𝑛} is in Ω. Since 𝑦𝑛𝑥𝑛0, then every weak cluster point of {𝑦𝑛} is also in Ω. Consequently, since ̃𝑥=projΩ(𝑄̃𝑥), we easily have limsup𝑛𝑄̃𝑥̃𝑥,𝑦𝑛̃𝑥0.(3.32)
On the other hand, we observe that𝑊𝑛̃𝑥̃𝑥,𝑥𝑛+1=𝑊̃𝑥𝑛̃𝑥̃𝑥,projFix(𝑇)𝑥𝑛+1+𝑊̃𝑥𝑛̃𝑥̃𝑥,𝑥𝑛+1projFix(𝑇)𝑥𝑛+1.(3.33) Since ̃𝑥 is a solution of the problem (1.9) and projFix(𝑇)𝑥𝑛+1Fix(𝑇), we have 𝑊𝑛̃𝑥̃𝑥,projFix(𝑇)𝑥𝑛+1̃𝑥0.(3.34) Thus it follows that 𝑊𝑛̃𝑥̃𝑥,𝑥𝑛+1𝑊̃𝑥𝑛̃𝑥̃𝑥,𝑥𝑛+1projFix(𝑇)𝑥𝑛+1𝑊𝑛𝑥̃𝑥̃𝑥𝑛+1projFix(𝑇)𝑥𝑛+1=𝑊𝑛𝑥̃𝑥̃𝑥×Dist𝑛+11,Fix(𝑇)𝑘𝑊𝑛𝑥̃𝑥̃𝑥𝑛+1𝑇𝑥𝑛+1.(3.35) We note that 𝑥𝑛+1𝑇𝑥𝑛+1𝑥𝑛+1𝑇𝑥𝑛+𝑇𝑥𝑛𝑇𝑥𝑛+1𝛼𝑛𝑊𝑛𝑥𝑛𝑇𝑥𝑛+1𝛼𝑛𝑇𝑦𝑛𝑇𝑥𝑛+𝑥𝑛+1𝑥𝑛𝛼𝑛𝑊𝑛𝑥𝑛𝑇𝑥𝑛+𝑦𝑛𝑥𝑛+𝑥𝑛+1𝑥𝑛𝛼𝑛𝑊𝑛𝑥𝑛𝑇𝑥𝑛+𝛽𝑛𝑄𝑥𝑛𝑥𝑛+𝑥𝑛+1𝑥𝑛.(3.36) Hence we have 𝛼𝑛𝛽𝑛𝑊𝑛̃𝑥̃𝑥,𝑥𝑛+1𝛼̃𝑥2𝑛𝛽𝑛1𝑘𝑊𝑛𝑊̃𝑥̃𝑥𝑛𝑥𝑛𝑇𝑥𝑛+𝛼𝑛1𝑘𝑊𝑛̃𝑥̃𝑥𝑄𝑥𝑛𝑥𝑛+𝛼2𝑛𝛽𝑛𝑥𝑛+1𝑥𝑛𝛼𝑛1𝑘𝑊𝑛.̃𝑥̃𝑥(3.37) From Theorem 3.2, we have lim𝑛𝑥𝑛+1𝑥𝑛/𝛼𝑛=0. At the same time, we note that {(1/𝑘)𝑊𝑛̃𝑥̃𝑥𝑊𝑛𝑥𝑛𝑇𝑥𝑛}, {(1/𝑘)𝑊𝑛̃𝑥̃𝑥𝑄𝑥𝑛𝑥𝑛}, and {(1/𝑘)𝑊𝑛̃𝑥̃𝑥} are all bounded. Hence it follows from (i) and the above inequality that limsup𝑛𝛼𝑛𝛽𝑛𝑊𝑛̃𝑥̃𝑥,𝑥𝑛+1̃𝑥0.(3.38)
Finally, by (3.31)–(3.38) and Lemma 2.4, we conclude that the sequence {𝑥𝑛} converges strongly to a point ̃𝑥Fix(𝑇). 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 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻 and 𝑆 a nonexpansive mapping of 𝐶 into itself. Let 𝑄𝐶𝐶 be a contraction with coefficient 𝛾[0,1). For any 𝑥0𝐶, let{𝑥𝑛} the sequence generated iteratively by 𝑥𝑛+1=𝛼𝑛𝑆𝑥𝑛+1𝛼𝑛𝑇𝛽𝑛𝑄𝑥𝑛+1𝛽𝑛𝑥𝑛,𝑛0,(3.39) where {𝛼𝑛},{𝛽𝑛} are two real numbers in (0,1).

Corollary 3.6. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let S𝐶𝐶 be a nonexpansive mapping. Let 𝑄𝐶𝐶 be a contraction with coefficient 𝛾[0,1). Assume that the set Ω of solutions of the hierarchical problem (1.1) is nonempty. Let {𝛼𝑛},{𝛽𝑛} be two real numbers in (0,1) and {𝑥𝑛} the sequence generated by (3.1). Assume that the sequence {𝑥𝑛} is bounded and (i)lim𝑛𝛼𝑛=0, lim𝑛𝛽𝑛/𝛼𝑛=0 and lim𝑛𝛼2𝑛/𝛽𝑛=0; (ii)𝑛=0𝛽𝑛=; (iii)lim𝑛(1/𝛽𝑛)|(1/𝛼𝑛)(1/𝛼𝑛1)|=0 and lim𝑛(1/𝛼𝑛)|1(𝛽𝑛1/𝛽𝑛)|=0; (iv)there exists a constant 𝑘>0 such that 𝑥𝑇𝑥𝑘Dist(𝑥,Fix(𝑇)), where Dist(𝑥,Fix(𝑇))=inf𝑦Fix(𝑇)𝑥𝑦.(3.40) Then the sequence {𝑥𝑛} defined by (3.39) converges strongly to a point ̃𝑥Fix(𝑇), 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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