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## Existence and Uniqueness of Fixed Point in Various Abstract Spaces and Related Applications

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Research Article | Open Access

Volume 2014 |Article ID 678147 | 9 pages | https://doi.org/10.1155/2014/678147

# Strong Convergence of an Iterative Algorithm for Hierarchical Problems

Revised17 Jun 2014
Accepted27 Jun 2014
Published20 Jul 2014

#### Abstract

We introduce the triple hierarchical problem over the solution set of the variational inequality problem and the fixed point set of a nonexpansive mapping. The strong convergence of the algorithm is proved under some mild conditions. Our results extend those of Yao et al., Iiduka, Ceng et al., and other authors.

#### 1. Introduction

Let be a closed convex subset of a real Hilbert space with inner product and norm . We denote weak convergence and strong convergence by notations and , respectively. Let be a nonlinear mapping. The Hartman-Stampacchia variational inequality  is to find such that . The set of solutions is denoted by . is said to be a -contraction if there exists a constant such that . A mapping is said to be monotone if . A mapping is said to be - strongly monotone if there exists a positive real number such that . A mapping is said to be -inverse-strongly monotone if there exists a positive real number such that . A mapping is said to be -Lipschitz continuous if there exists a positive real number such that . A linear bounded operator is said to be strongly positive on if there exists a constant with the property . A mapping is said to be nonexpansive if .

A point is a fixed point of provided . Denote by the set of fixed points of ; that is, . If is bounded closed convex and is a nonexpansive mapping of into itself, then is nonempty (see ).

We discuss the following variational inequality problem over the fixed point set of a nonexpansive mapping (see ), which is said to be the hierarchical problem. Let a monotone, continuous mapping and a nonexpansive mapping . Find , where . This solution set is denoted by .

We introduce the following variational inequality problem over the solution set of variational inequality problem and the fixed point set of a nonexpansive mapping (see [17, 18]), which is said to be the triple hierarchical problem. Let an inverse-strongly monotone , a strongly monotone and Lipschitz continuous , and a nonexpansive mapping . Find , where .

In 2009, Yao et al.  considered the following two-step iterative algorithm with the initial guess which is chosen arbitrarily: where satisfies certain assumptions. Let be two nonexpansive mappings and let be a contraction mapping. Then, they proved that the above iterative sequence converges strongly to fixed point.

Next, Iiduka  introduced a monotone variational inequality with variational inequality constraint over the fixed point set of a nonexpansive mapping; the sequence defined by the iterative method below, with the initial guess , is chosen arbitrarily: where and satisfy certain conditions, is an inverse-strongly monotone, is a strongly monotone and Lipschitz continuous, and is a nonexpansive mapping; then the strongly convergence analysis of the sequence generated by (2) is proved under some appropriate conditions.

In 2011, Yao et al.  studied the hierarchical problem over the fixed point set. Let the sequences be generated by these two following algorithms:implicit algorithm explicit algorithm .They illustrated that these two algorithms converge strongly to the unique solution of the variational inequality which is to find such that where is a strongly positive linear bounded operator, is a -contraction, and is a nonexpansive mapping satisfying some conditions.

Very recently, Ceng et al.  studied the following new algorithms. For is chosen arbitrarily, they defined a sequence by where the mappings , are nonexpansive mappings with . Let be a Lipschitzian and strongly monotone operator and let be a contraction mapping satisfying some appropriate conditions. They proved that the proposed algorithms strongly converge to the minimum norm fixed point of .

In this paper, we consider a new iterative algorithm for solving the triple hierarchical problem over the solution set of the variational inequality problem and the fixed point set of a nonexpansive mapping which contain algorithms (1) and (4) as follows: where the mappings , are nonexpansive mappings with . Let be a Lipschitzian and strongly monotone operator, and let be a contraction mapping satisfying some mild conditions. Find a point such that This solution set of (6) is denoted by . The strong convergence for the proposed algorithms to the solution is solved under some appropriate assumptions. Our results improve the results of Ceng et al. , Iiduka , Yao et al. , Yao et al. , and some authors.

#### 2. Preliminaries

Let be a nonempty closed convex subset of . There holds the following inequality in an inner product space . For every point , there exists a unique nearest point in , denoted by , such that is called the metric projection of onto . It is well known that is a nonexpansive mapping of onto and satisfies for every . Moreover, is characterized by the following properties: and for all . Let be a monotone mapping of into . In the context of the variational inequality problem the characterization of projection (9) implies the following: It is also known that satisfies the Opial’s condition ; that is, for any sequence with , the inequality holds for every with .

Lemma 1 (see ). Let be a closed convex subset of a real Hilbert space and let be a nonexpansive mapping. Then is demiclosed at zero; that is,   and   imply .

Lemma 2 (see ). Let and be bounded sequences in a Banach space and let be a sequence in with . Suppose for all integers and . Then, .

Lemma 3 (see ). Let be -strongly monotone and -Lipschitz continuous and let . For , define by for all . Then, for all , hold, where .

Lemma 4 (see ). Assume that is a sequence of nonnegative real numbers such that where and is a sequence in such that(i);(ii) or .Then .

#### 3. Strong Convergence Theorem

In this section, we introduce an iterative algorithm of triple hierarchical for solving monotone variational inequality problems for -Lipschitzian and -strongly monotone operators over the solution set of variational inequality problems and the fixed point set of a nonexpansive mapping.

Theorem 5. Let be a nonempty closed and convex subset of a real Hilbert space . Let be -Lipschitzian and -strongly monotone operators with constant and , respectively, and let be a -contraction with coefficient . Let be a nonexpansive mapping with , and let be a nonexpansive mapping. Let and , where . Suppose that is a sequence generated by the following algorithm where is chosen arbitrarily: where satisfy the following conditions:(C1):;(C2):, , ;(C3):.Then converges strongly to , which is the unique solution of another variational inequality: where .

Proof. We will divide the proof into four steps.
Step  1. We will show that is bounded. Indeed, for any , we have From (13), we deduce that Substituting (15) into (16), we obtain By induction, it follows that Therefore, is bounded and so are , , , , and .
Step  2. We will show that . Setting , we obtain which implies that It follows from (13) that where is a constant such that Hence, conditions (C2) and (C3) allow us to apply Lemma 4; then we get By (21), we get Using the conditions (C2) and (C3), we can apply Lemma 4 to conclude that By (13), we compute From the condition (C2), we note that . At the same time, from (13), we also have By the conditions (C1) and (C2), we note that . Consider From (23), (26), and (27), we obtain We set ; then we get From (13), we have By the conditions (C1) and (C2) again, we note that . Consider From (29), , and , we obtain
Step  3. We will show that . Rewrite (13) as We observe that Set We note from (35) that This yields that, for each , In view of (38), is nonnegative due to the monotonicity of . From (38), we derive that Since (29) implies , as , from (25), then we get . Using (C1) and (30), , as and is bounded. We obtain from (39) that Since the sequence is bounded, we can take a subsequence of such that and . From (33), by the demiclosed principle of the nonexpansive mapping, it follows that . Then
Step  4. Finally, we will prove . From (13), we note that Using (43), we compute Since , , and are all bounded, we can choose a constant such that It follows that where Now, applying Lemma 4 and (35), we conclude that . This completes the proof.

Corollary 6. Let be a nonempty closed and convex subset of a real Hilbert space . Let be -Lipschitzian and -strongly monotone operators with constant and , respectively. Let be a nonexpansive mapping with , and let be a nonexpansive mapping. Let and , where . Suppose is a sequence generated by the following algorithm arbitrarily: where satisfy the following conditions (C1)–(C3). Then converges strongly to , which is the unique solution of variational inequality: where .

Proof. Putting in Theorem 5, we can obtain the desired conclusion immediately.

Corollary 7. Let be a nonempty closed and convex subset of a real Hilbert space . Let be a -contraction with coefficient , and let be a nonexpansive mapping with and a nonexpansive mapping. Suppose is a sequence generated by the following algorithm, , arbitrarily: where satisfy the following conditions (C1)–(C3). Then converges strongly to , which is the unique solution of variational inequality: where .

Proof. Putting , , and in Theorem 5, we can obtain the desired conclusion immediately.

Corollary 8. Let be a nonempty closed and convex subset of a real Hilbert space . Let be a nonexpansive mapping with and let be a nonexpansive mapping. Suppose is a sequence generated by the following algorithm, , arbitrarily: where satisfy the following conditions (C1)–(C3). Then converges strongly to , which is the unique solution of variational inequality:

Proof. Putting in Corollary 7, we can obtain the desired conclusion immediately.

Corollary 9. Let be a nonempty closed and convex subset of a real Hilbert space . Let be a -contraction with coefficient , and let be a nonexpansive mapping with and a nonexpansive mapping. Suppose is a sequence generated by the following algorithm, , arbitrarily: where satisfy the following conditions (C1)–(C3). Then converges strongly to , which is the unique solution of variational inequality:

Proof. Putting in Corollary 7, we can obtain the desired conclusion immediately.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The first author was supported by the Thailand Research Fund and the King Mongkut’s University of Technology Thonburi (Grant no. RSA5780059). The second author was supported by the Commission on Higher Education, the Thailand Research Fund, and Rajamangala University of Technology Lanna Chiangrai under Grant no. MRG5680157 during the preparation of this paper.

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