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`Journal of Applied MathematicsVolume 2013 (2013), Article ID 574215, 8 pageshttp://dx.doi.org/10.1155/2013/574215`
Research Article

## Viscosity Method for Hierarchical Fixed Point Problems with an Infinite Family of Nonexpansive Nonself-Mappings

1Department of Mathematics, Shaoxing University, Shaoxing 312000, China
2Mathematical College, Sichuan University, Chengdu, Sichuan 610064, China

Received 4 January 2013; Accepted 22 March 2013

Copyright © 2013 Yaqin Wang. 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 viscosity method for hierarchical fixed point problems is presented to solve variational inequalities, where the involved mappings are nonexpansive nonself-mappings. Solutions are sought in the set of the common fixed points of an infinite family of nonexpansive nonself-mappings. The results generalize and improve the recent results announced by many other authors.

#### 1. Introduction and Preliminaries

Let a real Banach space and be the normalized duality mapping from into given by for all , where denotes the dual space of and the generalized duality pairing between and . If is a Hilbert space, then becomes the identity mapping on . A point is a fixed point of provided . Denote by the set of fixed points of ; that is, .

Let be a normed linear space with . The modulus of smoothness of is the function defined by The space is said to be smooth if . It is well known that if is smooth then is single valued. A Banach space is said to be strictly convex if ,  , implies .

Let be a nonempty closed convex subset of a real Banach space . Recall the following concepts.

Definition 1. (i) A mapping is a contraction if and if the following property is satisfied (ii) A mapping is nonexpansive provided (iii)A mapping is(a)accretive if for any there exists such that (b)-strongly accretive if for any there exists such that for some real constant .
Noting that if is nonexpansive, then is accretive; if is a contraction, then is -strongly accretive. particulary, if is a Hilbert space, then (strongly) accretive mappings become (strongly) monotone mappings.

Definition 2. Let and be nonempty subsets of a Banach space such that is nonempty closed convex and . (i)A mapping is called sunny, if for each and with .(ii)A mapping is called a retraction from to if is continuous and .(iii)A subset of is said to be a sunny nonexpansive retract of if there exists a sunny nonexpansive retraction of onto . For details, see [13].
Note that if is a Hilbert space, becomes the projection on , denoted by .
Let a nonexpansive self-mapping on and be a countable family of nonexpansive nonself-mappings of into such that . Then we consider the following problem: find hierarchically a common fixed point of the infinite family with respect to a nonexpansive mapping ; namely, find , such that

Particularly,if is a finite family of nonexpansive nonself-mappings, problem (7) has been studied by Ceng and Petruşel [4]. If and is an infinite family of nonexpansive self-mappings, Problem (7) reduces to the following problem: find hierarchically a common fixed point of with respect to a nonexpansive mapping , namely, find , such that which was studied by Zhang et al. [5]. If is a Hilbert space and , where is a nonexpansive mapping on , then problem (7) reduces to the following problem: finding hierarchically a fixed point of with respect to another nonexpansive mapping ; namely, find such that

Problem (7) includes many problems as special cases, so it is very important in the area of optimization and related fields, such as signal processing and image reconstruction (see [69]).

In 2007, Moudafi [10] introduced the following Krasnoselski-Mann's algorithm in Hilbert spaces: where and are two real sequences in (0,1) and and are two nonexpansive mappings of into itself. Furthermore, he established a weak convergence result for Algorithm (10) for solving problem (9).

Subsequently, Yao and Liou [11] derived a weak convergence result of algorithm (10) under the restrictions on parameters weaker than those in [10, Theorem 2.1].

Recently, Marino and Xu [12] introduced the following explicit hierarchical fixed point algorithm in Hilbert spaces: where is a contraction on and are two nonexpansive mappings of into itself and proved that the sequence generated by (11) converges strongly to a solution of problem (9).

Very recently, Zhang et al. [5] introduced the following iterative algorithm in order to find hierarchically a fixed point of Problem (8): where is a contraction, is a nonexpansive mapping, is a countable family of nonexpansive mappings, and is a mapping defined by Under suitable conditions on parameters and , they established some strong and weak convergence theorems. Note that, in [5], is an infinite family of self-mappings and is also a self-mapping. And they obtained the results in the setting of Hilbert spaces.

Motivated and inspired by the above researches, in a reflexive Banach space which admits a weakly sequentially continuous duality mapping , we propose and analyze an iteration process for a countable family of nonexpansive nonself-mappings and is a nonexpansive nonself-mapping as follows: where is a sunny nonexpansive retraction of onto and establishes a convergence theorem. particularly, if is a Hilbert space, we obtain some convergence results.

To prove the main results, we need the following lemmas.

Lemma 3 (see [1]). Let be a nonempty and convex subset of a smooth Banach space , , the normalized duality mapping of , and a retraction. Then the following conditions are equivalent:(i), for all and ;(ii) is both sunny and nonexpansive.

Lemma 4 (see [13, Lemma 3.1, 3.3]). Let be a real smooth and strictly convex Banach space and a nonempty closed and convex subset of which is also a sunny nonexpansive retract of . Assuming that is a nonexpansive mapping and is a sunny nonexpansive retraction of onto , then .

Lemma 5 (see [1]). Let be a real Banach space and the normalized duality mapping. Then for any , the following hold: (i),  for all;(ii),   for all .

Lemma 6 (see [14]). Let and be two sequences of nonnegative real numbers satisfying Then exists.

Lemma 7 (see [15]). Let be a sequence of nonnegative real numbers satisfying where and satisfy the following conditions: (i) or, equivalently, ;(ii); (iii). Then .

If Banach space admits sequentially continuous duality mapping from weak topology to weak * topology, then by [16, Lemma 1] we get that duality mapping is single-valued. In this case, duality mapping is also said to be weakly sequentially continuous, that is, for each with , then [16, 17].

Recall that a Banach space is said to be satisfying Opial’s condition if for any sequence in , implies that By [16, Lemma 1], we know that if admits a weakly sequentially continuous duality mapping, then satisfies Opial’s condition.

In the sequel, we also need the following lemmas.

Lemma 8 (see [17]). Let be a nonempty, closed and convex subset of a reflexive Banach space which satisfies Opial’s condition and a nonexpansive mapping. Then the mapping is demiclosed at zero, that is,
Let be a nonempty and convex subset of a Banach space . Then for , one defines the inward set as follows [2, 3]: A mapping is said to satisfy the inward condition if for all . is also said to satisfy the weakly inward condition if for each , is the closure of . Clearly and it is not hard to show that is a convex set if does.

Lemma 9 (see [18, Theorem 2.4]). Let be a reflexive Banach space which admits a weakly sequentially continuous duality mapping from to . Suppose is a nonempty closed convex subset of which is also a sunny nonexpansive retract of , and is a nonexpansive mapping satisfying the weakly inward condition and . Let be defined by where is a sunny nonexpansive retract of onto and satisfy the following conditions: (i); (ii); (iii)either or .Then converges strongly to a fixed point of such that is the unique solution in to the following variational inequality:

Remark 10. If a Banach space admits a sequentially continuous duality mapping from weak topology to weak star topology, from Lemma 1 of [16] it follows that is smooth. So for Lemma 9, if is a reflexive and strictly convex Banach space which admits a weakly sequentially continuous duality mapping , by Lemma 4, the weakly inward condition of can be removed.

#### 2. Main Results

Theorem 11. Let be a reflexive and strictly convex Banach space which admits a weakly sequentially continuous duality mapping and a nonempty, closed and convex subset of which is also a sunny nonexpansive retract of . Let be a nonexpansive nonself-mapping, a contractive mapping with a contractive constant and an infinite family of nonexpansive nonself-mappings such that . Let be defined by (13) and a sunny nonexpansive retraction of onto . Let be the sequence generated by (14), and and the sequences insatisfying the following conditions: (i); (ii); (iii). Then converges strongly to some point , which is the unique solution to the following variational inequality:

Proof. From condition (ii), without loss of generality, we can assume that .
First we prove that the sequence is bounded.
In fact, for any , we have By induction, Thus is bounded, so and are also bounded.
Next we prove that , where the sequence is defined by
By Lemma 9 and Remark 10, converges strongly to some point , which is the unique solution to the following variational inequality: Furthermore, we obtain where . It follows from conditions (i)-(ii) and Lemma 7 we have . Since as , , we get , which is the unique solution to the variational inequality (22).

Remark 12. Theorem 11 extends Theorem 2.1 in [5] from the following aspects: (i) from Hilbert spaces to reflexive and strictly convex Banach spaces which admits a weakly sequentially continuous duality mapping; (ii) for the infinite family of mappings from self-mappings to nonself-mappings. In addition, the existence of the sunny nonexpansive retraction has been proved in [19, Theorem 3.10].

Remark 13. If we take then since and (as ), it is not hard to find that the conditions (i)–(iii) are satisfied. For details, see [12, Remark 3.2].

In the sequel, we consider the result in the setting of Hilbert spaces.

Theorem 14. Let be a Hilbert space and a nonempty, closed and convex subset of . Let be a nonexpansive nonself-mapping, a contractive mapping with a contractive constant , and an infinite family of nonexpansive nonself-mappings such that . Let be the sequence generated by (14) and and the sequences insatisfying the following conditions: (i); (ii); (iii); (iv)there exists a constant such that for all .Then converges strongly to some point , which is the unique solution to the following variational inequality:

Proof. By condition (ii), without loss of generality, we can assume that . Similar to the proof of (24), for any , we have Thus is bounded. Furthermore, are all bounded. Put and . So and are also bounded.
Step 1. We prove that .
From (14), we obtain Substituting (32) into (31), we have By conditions (i), (iii), and Lemma 7, we have .
Step 2. We prove that , where is the -limit point set of in the weak topology: Noting that and , we have . Then from Step 1 we have . Furthermore, it follows from Lemmas 4 and 8 that , where .
Step 3. We show that .
It follows from (31) and (33) that By conditions (i) and (iii), , and Lemma 7, we have Thus from (35), we get
Step 4. We show that converges strongly to some point , which is the unique solution of (29).
Setting , we have Then Letting , from condition (i) and (36), we have . Noting that is monotone and is -strongly monotone, for any , from Lemma 3 we obtain Thus we have Since , by (37) we have Combining condition (ii), , (41), and (42), every weak cluster point of is also a strong cluster point. From (40), we obtain Note that the sequence is bounded; thus there exists a subsequence converging to a point . From Step 2, we have . Then it follows from the above inequality, (42), and that Replacing with , where and , we have Letting , we have If there exists another subsequence of converging to a point From Step 2, we also have . Then from (46) we obtain and, via interchanging and , Adding up these two inequalities yields which implies . Then converges strongly to , which is the solution to the following variational inequality: Since is -strongly monotone and is monotone, it is easy to see that the above variational inequality has a unique solution.

Remark 15. Theorem 14 extends Theorem 3.2 in [12] from the following aspects: (i) from a nonexpansive mapping to an infinite family of nonexpansive mappings ; (ii) from self-mappings to nonself-mappings.

#### Acknowledgments

The author is extremely grateful to the referees for their useful suggestions that improved the content of the paper. Supported by the China Postdoctoral Science Foundation Funded Project (no. 2012M511928).

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