Research Article | Open Access

# Strong Convergence Algorithms for Hierarchical Fixed Points Problems and Variational Inequalities

**Academic Editor:**Ya Ping Fang

#### Abstract

We introduce a new iterative scheme that converges strongly to a common fixed point of a countable family of nonexpansive mappings in a Hilbert space such that the common fixed point is a solution of a hierarchical fixed point problem. Our results extend the ones of Moudafi, Xu, Cianciaruso et al., and Yao et al.

#### 1. Introduction

Let be a real Hilbert space and a nonempty closed convex subset of . A mapping is called nonexpansive if one has If there exists a point such that , then is said to be a fixed point of . We denote the set of all fixed points of by . It is well known that is closed and convex if is nonexpansive.

Let be a mapping. The following problem is called a hierarchical fixed point problem: find such that

It is known that the hierarchical fixed point problem (1.2) links with some monotone variational inequalities and convex programming problems (see [1]).

In order to solve the hierarchical fixed point problem (1.2), Moudafi [2] introduced the following Krasnoselski-Mann algorithm: where and are two sequences in , and he proved that converges weakly to a fixed point of which is a solution of the problem (1.2).

Let be a mapping. The mapping is called a contraction if there exists a constant such that for all . For obtaining a strong convergence result, Mainge and Moudafi in [3] and Marino and Xu in [4] introduced the following algorithm: where is a nonexpansive mapping and and are two sequences in , and they proved that converges strongly to a fixed point of which is a solution of the problem (1.2). Recently, for solving the hierarchical fixed point problem (1.2), Cianciaruso et al. [5] also studied the following iterative scheme: where and are two sequences in . The authors proved some strong convergence results. Very recently, Yao et al. [1] introduced the following strong convergence iterative algorithm to solve the problem (1.2): where is a contraction and and are two sequences in . Under some certain restrictions on parameters, the authors proved that the sequence generated by (1.6) converges strongly to , which is the unique solution of the following variational inequality:

By changing the restrictions on parameters, the authors obtained another result on the iterative scheme (1.6), that is, the sequence generated by (1.6) converges strongly to a point , which is the unique solution of the following variational inequality: where is a constant.

Let be a nonexpansive mapping and a countable family of nonexpansive mappings. In this paper, motivated and inspired by the results of Yao et al. [1] and Marino and Xu [4], we introduce and study the following iterative scheme: where , is a strictly decreasing sequence in and is a sequence in . Under some certain conditions on parameters, we first prove that the sequence generated by (1.9) converges strongly to , which is the unique solution of the following variational inequality: By changing the restrictions on parameters, we also prove that the sequence converges strongly to , which is the unique solution of the following variational inequality: where is a constant. It is easy to see that, if for each and is a self-mapping of into itself, then our algorithm (1.9) is reduced to (1.6) of Yao et al. [1] Also, our results extend the corresponding ones of Moudafi [6], Xu [7], and Cianciaruso et al. [5].

#### 2. Preliminaries

Let be a Hilbert space and a nonempty closed convex subset of . Let be a nonexpansive mapping of into itself such that . For all and all , we have and hence

Let be an arbitrary point. There exists a unique nearest point in , denoted by , such that Moreover, we have the following:

Let denote the identity operator of , and let be a sequence in a Hilbert space and . Throughout this paper, denotes that strongly converges to and denotes that weakly converges to .

The following lemmas will be used in the next section.

Lemma 2.1 (see [8]). *Let be a nonexpansive mapping with . If is a sequence in weakly converging to a point and converges strongly to a point , then . In particular, if , then .*

Lemma 2.2 (see [9]). *Let be a contraction with coefficient and a nonexpansive mapping. Then one has the following. *(1)*The mapping is strongly monotone with coefficient , that is,
*(2)*The mapping is monotone, that is,
*

Lemma 2.3 (see [10]). *Let , be the sequences of nonnegative real numbers, and let . Suppose that is a real number sequence such that
**
Assume that . Then the following results hold. *(1)*If , where , then is a bounded sequence.*(2)*If one has
** then . *

#### 3. Main Results

Now, we give the main results in this paper.

Theorem 3.1. *Let be a nonempty closed convex subset of a real Hilbert space . Let be a -contraction with . Let be a nonexpansive mapping and a countable family of nonexpansive mappings of into itself such that . Set , and let be a strictly decreasing sequence and a sequence satisfying the following conditions: *(a)* and , *(b)*, *(c)* and . ** Then the sequence generated by (1.9) converges strongly to a point , which is the unique solution of the variational inequality
*

*Proof. *First, is a contraction from into itself with a constant and is complete, and there exists a unique such that . From (2.4), it follows that is the unique solution of the problem (3.1).

Now, we prove that converges strongly to . To this end, we first prove that is bounded. Take . Then it follows from (1.9) that
and hence (note that is strictly decreasing)
By condition (b), we can assume that for all . Hence, from above inequality, we get
For each , let , , and . Then , , and satisfy the condition of Lemma 2.3(1). Hence, it follows from Lemma 2.3(1) that is bounded and so are , , , and for all . Set for each . From (1.9), we have
where is a constant such that
From (1.9), we have
Substituting (3.7) into (3.5), we get that
Let , , and . Then conditions (a) and (c) imply that , , and satisfy the condition of Lemma 2.3(2). Thus, by Lemma 2.3(2), we can conclude that
Since for each and , we have
Now, fixing a , from (1.9) we have
Hence it follows that
where

Now, from (2.2) and (3.12), we get
From (a), (b), (3.9), and (3.14), we have
Since for each and is strictly decreasing, one has

Next, we prove that
Since is bounded, we can take a subsequence of such that and
From (3.16) and Lemma 2.1, we conclude that for each , that is, . Then
By (2.4), we have
Also,
Thus,
It follows that
Let , , and for all . Since
, and , we have
Therefore, it follows from Lemma 2.3(2) that
This completes the proof.

*Remark 3.2. *In (1.9), if , then it follows that . In this case, from (3.1), it follows that
that is,
Therefore, the point is the unique solution to the quadratic minimization problem

In Theorem 3.1, if for all , then we get the following result.

Corollary 3.3. *Let be a nonempty closed convex subset of a real Hilbert space . Let be a -contraction with . Let be a nonexpansive mapping and a nonexpansive mapping such that . Let and define a sequence by
**
where and are two sequences satisfying the following conditions: *(a)* and , *(b)*, *(c)* and .** Then the sequence converges strongly to a point , which is the unique solution of the variational inequality
*

In Corollary 3.3, if is a self-mapping of into itself, then we get the following result.

Corollary 3.4. *Let be a nonempty closed convex subset of a real Hilbert space . Let be a -contraction with . Let be a nonexpansive mapping and a nonexpansive mapping such that . Let and define a sequence by
**
where the sequences and are two sequences satisfying the following conditions: *(a)* and , *(b)*, *(c)* and .** Then the sequence converges strongly to a point , which is the unique solution of the variational inequality
*

By changing the restrictions on parameters in Theorem 3.1, we obtain the following.

Theorem 3.5. *Let be a nonempty closed convex subset of a real Hilbert space . Let be a -contraction with . Let be a nonexpansive mapping and a countable family of nonexpansive mappings such that . Set . Let and define a sequence by
**
where is a strictly decreasing sequence and is a sequence satisfying the following conditions: *(a)*, and , *(b)*, *(c)* and , *(d)*, *(e)*there exists a constant such that . ** Then the sequence generated by (3.34) converges strongly to a point , which is the unique solution of the variational inequality
*

*Proof. *First, the proof of Theoremâ€‰â€‰3.2 of [1] shows that (3.35) has the unique solution. By a similar argument as in that of Theorem 3.1, we can conclude that is bounded, and as . Note that conditions (a) and (b) imply that as . Hence we have
It follows that, for all ,
Now, it follows from (3.36) and (3.37) that, for all ,
From (3.8), we get
Note that
Hence, from (3.39), we have
Let and . From conditions (a) and (d), we have
By Lemma 2.3(2), we get
From (3.34), we have
Hence it follows that
and hence
Let . For any , we have
By Lemma 2.2, we have
By (2.4), we have
Now, from (3.47)â€“(3.51) it follows that
Observe that (3.52) implies that
Since , for all , and as , every weak cluster point of is also a strong cluster point. Since is bounded, there exists a subsequence of converging to a point . Note that as for all . By the demiclosed principle for a nonexpansive mapping, we have for all and so . From (3.47), (3.48), (3.50), and (3.51), it follows that, for all ,
Since , for all , and , letting in (3.54), we obtain