Abstract
An explicit algorithm is introduced to solve the monotone variational inequality over a triple hierarchical problem. The strong convergence for the proposed algorithm to the solution is guaranteed under some assumptions. Our results extend those of Iiduka (Nonlinear Anal. 71:e1292-e1297, 2009), Marino and Xu (J. Optim. Theory Appl. 149(1):61-78, 2011), Yao et al. (Fixed Point Theory Appl. 2011:794203, 2011) and other authors.
MSC:46C05, 47H06, 47H09, 47H10, 47J20, 47J25, 65J15.
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1 Introduction
Let C be a closed convex subset of a real Hilbert space H with the inner product and the norm . We denote weak convergence and strong convergence by notations ⇀ and →, respectively. A mapping A is a nonlinear mapping. The Hartmann-Stampacchia variational inequality [1] for finding such that
The set of solutions of (1.1) is denoted by . The variational inequality has been extensively studied in the literature [2, 3].
Let be called a ρ-contraction if there exists a constant such that
A mapping is said to be nonexpansive if
A point is a fixed point of T provided . Denote by the set of fixed points of T; that is, . If C is bounded closed convex and T is a nonexpansive mapping of C into itself, then is nonempty [4].
We discuss the following variational inequality problem over the fixed point set of a nonexpansive mapping (see [5–13]), which is called a hierarchical problem. Consider a monotone, continuous mapping and a nonexpansive mapping . Find , where . This solution set is denoted by Ξ.
We introduce the following variational inequality problem over the solution set of the variational inequality problem over the fixed point set of a nonexpansive mapping (see [14–18]), which is called a triple hierarchical problem. Consider an inverse-strongly monotone , a strongly monotone and Lipschitz continuous and a nonexpansive mapping . Find , where .
A mapping is said to be monotone if
A mapping is said to be α-strongly monotone if there exists a positive real number α such that
A mapping is said to be β-inverse-strongly monotone if there exists a positive real number β such that
A mapping is said to be L-Lipschitz continuous if there exists a positive real number L such that
A linear bounded operator A is said to be strongly positive on H if there exists a constant with the property
In 2009, Iiduka [14] 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 satisfy certain conditions. Let be an inverse-strongly monotone, be a strongly monotone and Lipschitz continuous and be a nonexpansive mapping, then the sequence converges strongly to the set solution of (1.2).
In 2011, Yao et al. [19] studied new algorithms. For chosen arbitrarily, let the sequence be generated iteratively by
where the sequences and are two sequences in . Then converges strongly to which is the unique solution of the variational inequality:
Let be a strongly positive linear bounded operator, be a ρ-contraction and be a nonexpansive mapping satisfying some conditions. The solution set of (1.3) is denoted by .
Very recently, Marino and Xu [20] generated a sequence through the recursive formula
where f is a contraction on C, the initial guess is arbitrary and , are two sequences in and are two nonexpansive self mappings. Strong convergence of the algorithm is proved under different circumstances of parameter selection.
In this paper, we introduce iterative algorithms and prove a strong convergence theorem for the following variational inequality over the triple hierarchical problem (1.4) below. Let be a β-strongly monotone and L-Lipschitz continuous. Find such that
where , T is a nonexpansive mapping, is a strongly positive linear bounded operator and is a ρ-contraction. This solution set of (1.4) is denoted by . Then the sequence strongly converges to the unique solution of (3.2) in Section 3.17 and we shall denote the set of such solutions by .
2 Preliminaries
Let H be a real Hilbert space and C be a nonempty closed convex subset of H. Recall that the (nearest point) projection from H onto C assigns, to each , the unique point in satisfying the property
The following characterizes the projection . We recall some lemmas which will be needed in the rest of this paper.
Lemma 2.1 The function is a solution of the variational inequality (1.1) if and only if satisfies the relation for all .
Lemma 2.2 For a given , , , .
It is well known that is a firmly nonexpansive mapping of H onto C and satisfies
Moreover, is characterized by the following properties: and for all , ,
Lemma 2.3 The following inequality holds in an inner product space H:
Lemma 2.4 [21]
Let C be a closed convex subset of a real Hilbert space H and let be a nonexpansive mapping. Then is demiclosed at zero, that is, and imply .
Lemma 2.5 [22]
Assume A is a strongly positive linear bounded operator on a Hilbert space H with the coefficient and , then .
Lemma 2.6 [23]
Each Hilbert space H satisfies Opial’s condition, that is, for any sequence with , the inequality
holds for each with .
Lemma 2.7 [24]
Let and be bounded sequences in a Banach space X and let be a sequence in with . Suppose for all integers and . Then .
Lemma 2.8 [10]
Let be β-strongly monotone and L-Lipschitz continuous and let . For , define by for all . Then, for all ,
holds, where .
Lemma 2.9 [25]
Assume is a sequence of nonnegative real numbers such that
where and is a sequence in ℛ such that
-
(i)
,
-
(ii)
or .
Then .
3 Strong convergence theorem
In this section, we introduce an iterative algorithm for solving the monotone variational inequality over a triple hierarchical problem.
Theorem 3.1 Let H be a real Hilbert space, C be a closed convex subset of H. Let be a strongly positive linear bounded operator, be a ρ-contraction, γ be a positive real number such that . Let be a nonexpansive mapping, be a β-strongly monotone and L-Lipschitz continuous. Let be a k-contraction mapping with . Assume that is nonempty set. Suppose is a sequence generated by the following algorithm arbitrarily:
where . If is used and if satisfies the following conditions:
(C1): , ;
(C2): ;
(C3): , ;
(C4): and .
Then converges strongly to , which is the unique solution of the variational inequality:
Proof We will divide the proof into four steps.
Step 1. We will show is bounded. For any , we have
By Lemma 2.8, it is found that
From (3.1), we get
where . Then the mathematical induction implies that
Therefore, is bounded and so are , , , , and .
Step 2. We claim that and . From (3.1), we have
and
Using (3.5) and (3.6), we get
where M is some constant such that
From (C1)-(C3) and the boundedness of , , , , and , by Lemma 2.9, we have
On the other hand, we note that
by (C3)-(C4) and it follows that
From (3.1), we compute
By (C3) and (C4), it follows that
Since
By (3.7), (3.8) and (3.9), we obtain
From (3.1), we compute
By (C3), it follows that
Since
From (3.7) and (3.12), we obtain
Step 3. First, 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 . with guarantees that
which is a contradiction. Therefore, . From , we find
Setting , by (C3)-(C4), we notice that
Hence, we get
Second, is proven. guarantees the existences of a subsequence of and a point such that and , . By the same discussion as in the proof of , we have . Let be fixed arbitrarily. Then it follows that is a nonexpansive mapping with , is a strongly positive linear bounded operator and is a contraction for all . From (3.1),
By (C3)-(C4), we observe that
Using (3.15) and (3.16),
which implies that
where for every . By the weak convergence of to , and (C3)-(C4), we get for all . A mapping A being a strongly positive linear bounded operator and f being a contraction ensure for all , that is, . Thus, , we have
From (3.13), we notice that
Third, is proven. Choose a subsequence of such that
The boundedness of implies the existence of a subsequence of and a point such that converges weakly to . By , we have , . We may assume, without loss of generality, that , . Assume . with guarantees that
This is a contradiction, that is, . From , we find
Step 4. Finally, we prove . By Lemma 2.8, we compute
Since , , , and are all bounded, we can choose a constant such that
It follows that
where
By (3.14), (3.17), (3.18) and (C3)-(C4), we get . Applying Lemma 2.9, we can conclude that . This completes the proof. □
4 An example
Next, the following example shows that all the conditions of Theorem 3.1 are satisfied.
Example 4.1 For instance, let , and . We will show that the condition (C1) is achieved. Then, clearly, the sequence
and
The sequence satisfies the condition (C1).
Next, we will show that the condition (C2) is achieved. We compute
The sequence satisfies the condition (C2).
Next, we will show that the condition (C3) is achieved. We compute
and
The sequence satisfies the condition (C3).
Finally, we will show that the condition (C4) is achieved.
Corollary 4.2 Let H be a real Hilbert space, C be a closed convex subset of H. Let be inverse-strongly monotone. Let be a nonexpansive mapping. Let be β-strongly monotone and L-Lipschitz continuous. Assume that is nonempty set. Suppose is a sequence generated by the following algorithm arbitrarily:
. If is used and if satisfies the following conditions:
(C1): , ;
(C2): ;
(C3): , ;
(C4): and .
Then converges strongly to , which is the unique solution of the variational inequality:
Proof Putting is the identity and in Theorem 3.1, we can obtain the desired conclusion immediately. □
Remark 4.3 Corollary 4.2 generalizes and improves the results of Iiduka [14].
Corollary 4.4 Let H be a real Hilbert space, C be a closed convex subset of H. Let be a strongly positive linear bounded operator, be a ρ-contraction, γ be a positive real number such that . Let be a nonexpansive mapping. Assume that Ω is nonempty set. Suppose is a sequence generated by the following algorithm arbitrarily:
where . If is used and if satisfies the following conditions:
(C1): , ;
(C2): ;
(C3): , ;
(C4): and .
Then converges strongly to , which is the unique solution of the variational inequality:
Proof Putting ϕ is the identity in Theorem 3.1, we can obtain the desired conclusion immediately. □
Remark 4.5 Corollary 4.4 generalizes and improves the results of Marino and Xu [20].
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The authors thank to the Higher Education Research Promotion and National Research University Project of Thailand, the Office of the Higher Education Commission for financial support during the preparation of this paper.
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Wairojjana, N., Jitpeera, T. & Kumam, P. The hybrid steepest descent method for solving variational inequality over triple hierarchical problems. J Inequal Appl 2012, 280 (2012). https://doi.org/10.1186/1029-242X-2012-280
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DOI: https://doi.org/10.1186/1029-242X-2012-280