Abstract

This paper discusses the monotone variational inequality over the solution set of the variational inequality problem and the fixed point set of a nonexpansive mapping. The strong convergence theorem for the proposed algorithm to the solution is guaranteed under some suitable assumptions.

1. Introduction

Let be a closed convex subset of a real Hilbert space with the inner product and the norm . We denote weak convergence and strong convergence by notations and , respectively.

A mapping is said to be monotone if is said to be -strongly monotone if there exists such that .    is said to be -inverse-strongly monotone if there exists such that .    is said to be -Lipschitz continuous if there exists such that . A linear bounded operator is said to be strongly positive on if there exists with the property .

Let be a -contraction if there exists such that

Let be nonexpansive such that 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 [1]). Let be a nonlinear mapping. The Hartmann-Stampacchia variational inequality [2] is to finding such that The set of solutions of (1.3) is denoted by . The variational inequality has been extensively studied in the literature [3, 4].

We discuss the following variational inequality problem over the fixed point set of a nonexpansive mapping (see [512]), which is called the hierarchical problem. Let a monotone, continuous mapping and a nonexpansive mapping . This solution set is denoted by .

We introduce the following variational inequality problem over solution set of variational inequality problem and the fixed point set of a nonexpansive mapping (see [1316]), which is called the triple hierarchical problem (or the triple hierarchical constrained optimization problem (see also [13])). Let an inverse-strongly monotone , a strongly monotone and Lipschitz continuous , and a nonexpansive mapping . where .

In 2009, Iiduka [13] introduced an iterative algorithm for the following triple hierarchical constrained optimization problem, the sequence defined by the iterative method below, with the initial guess is chosen arbitrarily, where and satisfies certain conditions. Let be an inverse-strongly monotone, be a strongly monotone and Lipschitz continuous, and be a nonexpansive mapping, then the sequence converges to strong analysis on (1.6).

In 2011, Ceng et al. [17] studied the new following algorithms. For is chosen arbitrarily, they defined a sequence iterative by where the mapping are nonexpansive mappings with . Let be a Lipschitzian and strongly monotone operator and be a contraction mapping satisfied some conditions. They proved that the proposed algorithms strongly converge to the minimum norm fixed point of .

Very recently, Yao et al. [18] studied the following algorithms. For is chosen arbitrarily, let the sequence be generated iteratively by where the sequences and are two sequences in . Then converges strongly to the unique solution of the variational inequality as follows. Find a point such that where is a strongly positive linear bounded operator, is a -contraction, and is a nonexpansive mapping satisfied some suitable conditions. The solution (1.9) is denoted by .

In this paper, we introduce a new iterative algorithm for solving the triple hierarchical problem, which contain algorithms (1.6) and (1.8) as follows: The strong convergence for the proposed algorithms to the solution is solved under some assumptions. Our results generalize and improve the results of Ceng et al. [17], Iiduka [13], Yao et al. [18], and some authors.

2. Preliminaries

Let be a real Hilbert space and be a nonempty closed convex subset of . The metric (or nearest point) projection from onto is the mapping which assigns to each point the unique point in satisfying the property The following properties of projection are useful and pertinent to our purposes.

Lemma 2.1. Given and ,(a), (b), (c) is a firmly nonexpansive mapping of onto and satisfies Consequently, is nonexpansive and monotone.

Lemma 2.2. There holds the following inequality in an inner product space

Lemma 2.3 (see [19]). Let be a closed convex subset of a real Hilbert space and let be a nonexpansive mapping. Then is demiclosed at zero, that is, implies .

Lemma 2.4 (see [20]). Each Hilbert space satisfies Opial’s condition, that is, for any sequence with , the inequality hold for each with .

Lemma 2.5 (see [21]). Let and be bounded sequences in a Banach space and let be a sequence in with . Suppose for all integers and . Then, .

Lemma 2.6 (see [10]). Let be -strongly monotone and -Lipschitz continuous and let . For , define by for all . Then, for all , hold, where .

Lemma 2.7 (see [22]). Assume is a sequence of nonnegative real numbers such that where and is a sequence in such that(i),(ii) or .Then .

Remark 2.8. If is a strongly positive linear bounded operator and is a -contraction, then for , the mapping is strongly monotone. In fact, we have

3. Main Results

In this section, we introduce a new iterative algorithm for solving monotone variational inequality problem (where is a strongly positive linear bounded operator, is a -contraction) over solution set of variational inequality problem over the fixed point set of a nonexpansive mapping.

Theorem 3.1. Let be a nonempty closed and convex subset of a real Hilbert space . Let be a strongly positive linear bounded operator, be a -contraction, and be a positive real number such that . Let be -Lipschitzian and -strongly monotone operators with constant and , respectively. Let be a nonexpansive mapping with . Let and , where . Assume that , where . Suppose is a sequence generated by the following algorithm arbitrarily and where satisfy the following conditions: and ;;;, and .
Then the sequence converges strongly to , which is the unique solution of another variational inequality

Proof. We will divide the proof into four steps.
Step  1. We will show is bounded. For any , we have From (3.1), we deduce that Substituting (3.3) into (3.4), we obtain By induction, it follows that Therefore, is bounded and so are , and .
Step  2. We will show that , and . From (3.1), we have It follows that where is a constant such that
By the conditions ()–() allow us to apply Lemma 2.7, we get On the other hand, we note that by (), it follows that From (3.7), we observe that It follows that From the conditions ()–() and the boundedness of , and , which implies that Hence, by Lemma 2.5, we have From (3.12) and (3.16), we obtain
Step  3. We will show that is proven. Choose a subsequence of such that The boundedness of implies the existences of a subsequence of and a point such that converges weakly to . We may assume without loss of generality that . Assume . Since with guarantee that which has a contradiction. Therefore, . Since , then , it follows that Setting and by (), we notice that Hence, we get Next we will show that is proven. Choose a subsequence of such that The boundedness of implies the existences of a subsequence of and a point such that converges weakly to . We may assume without loss of generality that . Assume . By with guarantee that which has a contradiction. Therefore, . From , we compute Using (3.10), we get
Step  4. Finally, we prove . We observe that From (3.1), we compute Since , and are all bounded, we can choose a constant such that It follows that where By the conditions (), (), (3.22), and (3.26), we get Now, applying Lemma 2.7 and (3.30), we conclude that . This completes the proof.

Next, the following example shows that all conditions of Theorem 3.1 are satisfied.

Example 3.2. For instance, let and . Then, clearly the sequences , satisfy the following condition (): We will show that the condition () is achieved. Indeed, we have The sequence satisfies the condition () by p-series. Next, we will show that the condition () is achieved. We compute The sequence satisfies the condition (). Finally, we will show that the condition () is achieved. We compute The sequence satisfies the condition ().

Corollary 3.3. Let be a nonempty closed and convex subset of a real Hilbert space . Let be a strongly positive linear bounded operator, be a -contraction, and be a positive real number such that . Let be a nonexpansive mapping with . Assume that . Suppose is a sequence generated by the following algorithm arbitrarily and where satisfy the following conditions ()–(). Then the sequence converges strongly to , which is the unique solution of variational inequality

Proof. Putting and in Theorem 3.1, we can obtain desired conclusion immediately.

Corollary 3.4. Let be a nonempty closed and convex subset of a real Hilbert space . Let be a nonexpansive mapping with . Suppose is a sequence generated by the following algorithm arbitrarily and where satisfy the following conditions ()–(). Then the sequence converges strongly to .

Proof. Putting and in Corollary 3.3, we can obtain desired conclusion immediately.

Remark 3.5. Our results generalize and improve the recent results of Iiduka [13] and Yao et al. [18].

Acknowledgment

The authors were supported by the Higher Education Research Promotion and National Research University Project of Thailand, Office of the Higher Education Commission (under Project NRU-CSEC no. 55000613) for financial support during the preparation of this paper.