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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 218964, 7 pages
Algorithmic Approach to the Split Problems
Tianjin and Education Ministry, Key Laboratory of Advanced Composite Materials, Tianjin 300387, China
Received 6 June 2013; Accepted 26 June 2013
Academic Editor: Abdellah Bnouhachem
Copyright © 2013 Ming Ma. 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.
This paper deals with design algorithms for the split variational inequality and equilibrium problems. Strong convergence theorems are demonstrated.
Let be a real Hilbert space. Let and be two nonempty closed convex subsets of . Consider the following problem.
Problem 1. Find a point such that
This problem is called split feasibility problem when is a bounded linear operator. In this case, Problem 1 can be applied to many practical problems such as signal processing and image reconstruction. Specifically, we can find the prototype of Problem 1 in intensity-modulated radiation therapy; see, for example, [1–3]. Based on this relation, many mathematicians were devoted to study the split feasibility problem and develop its iterative algorithms. Related works can be found in [4–8] and the references therein.
Let be two mappings. Consider the variational inequality of finding such that
for all . We use to denote the set of solutions of (2). Variational inequality problems have important applications in many fields such as elasticity, optimization, economics, transportation, and structural analysis, and various numerical methods have been studied by many researchers; see, for instance, [9–17].
Let be an equilibrium bifunction; that is, for each . Consider the equilibrium problem which is to find such that
Denote the set of solutions of (3) by . The equilibrium problems include fixed point problems, optimization problems, and variational inequality problems as special cases. Some algorithms have been proposed to solve the equilibrium problems; see, for example, [18–22]. Thus it is an interesting topic associated with algorithmic approach to the variational inequality and equilibrium problems. In this paper, our main purpose is to study the following split problem involved in the variational inequality and equilibrium problems. Find a point such that
We are devoted to study (4) with operator being a nonlinear mapping. For this purpose, we develop an iterative algorithm for solving the split problem (4). We can compute iteratively by using our algorithm. Convergence analysis is given under some mild assumptions.
2. Basic Concepts
Let be a nonempty closed convex subset of a real Hilbert space . An operator is said to be (i)monotone for all ;(ii)strongly monotone for some constant and for all ;(iii)inverse-strongly monotone for some and for all ; in this case, is called -inverse strongly monotone;(iv)-inverse strongly -monotone for all and for some , where is a mapping.
A mapping is said to be(i)nonexpansive for all ;(ii)firmly nonexpansive for all ;(iii)-Lipschitz continuous for some constant and for all . In such a case, is said to be -Lipschitz continuous.
In the sequel, we use to denote the set of fixed points of .
Let be a multivalued mapping. The effective domain of is denoted by . is said to be (i)monotone for all , , and ;(ii)maximal monotone is monotone and its graph is not strictly contained in the graph of any other monotone operator on .
A function is said to be convex if for any and for any , .
Let be the metric projection from onto . It is known that satisfies the following inequality: for all and . From this characteristic inequality, we can deduce that is firmly nonexpansive.
3. Useful Lemmas
In this section, we present several lemmas which will be used in the next section.
Lemma 2 (see ). Let be a nonempty closed convex subset of a real Hilbert space . Let be a bifunction. Assume that satisfies the following conditions: for all ; is monotone, that is, for all ; for each , ; for each , is convex and lower semicontinuous.
Let and . Then there exists such that
Set for all . Then one have the following:(i) is single valued and is firmly nonexpansive,(ii) is closed and convex and .
Lemma 4 (see ). Let and be two bounded sequences in a Banach space , and let be a sequence in satisfying . Suppose for all and . Then, .
Lemma 5 (see ). Let be a nonempty closed convex subset of a real Hilbert space . Let be a nonexpansive mapping with . Then is demiclosed on .
Lemma 6 (see ). Let be a sequence. Assume that , where is a sequence in , and is a sequence satisfying and (or ). Then .
4. Main Results
In this section, we firstly present our problem and algorithm constructed. Consequently, we give the convergence analysis of the presented algorithm.
Problem 7. Let be a nonempty closed convex subset of a real Hilbert space . Assume that (1) is a weakly continuous and -strongly monotone mapping such that ; (2) is an -inverse strongly -monotone mapping; (3) is a bifunction satisfying conditions in Lemma 2. Our objective is to
We use to denote the set of solutions of (8). In the following, we assume that is nonempty. For solving Problem 7, we introduce the following algorithm.
Step 0 (initialization). Let
Step 1. For given , let the sequence be generated iteratively by where is the metric projection and is a real number sequence.
Step 2. For given , find such that
where and are two real number sequences.
Step 3. For the previous sequences and , let the sequence be generated by where is a real number sequence.
Theorem 9. Assume that the following conditions are satisfied: and ; ; , , and ; and . Then the sequence generated by Algorithm 8 converges strongly to .
Proof. Let . Hence and , noting that implies for all . Hence for all . Thus, from (10), we have
Condition and (13) imply that
From Lemma 2 and (11), we get for all . Since , from Lemma 2 we deduce that for all . So,
It follows that
Hence, is bounded. Since is -strongly monotone, we can get . So, . This implies that is bounded. Next, we show . From , we have
Using Lemma 3, we obtain
By condition , we have . So,
From (10), we have
It follows that
Since , , and and the sequences , , , and are bounded, we deduce that
Applying Lemma 4, we obtain
This together with the -strong monotonicity of implies that
From (13) and (16), we derive
Since , , and , we obtain
Set for all . By using the firm nonexpansivity of projection, we get
It follows that
From (29) and (32), we have
Then, we obtain
Since , , and , we deduce that
Next, we prove , where satisfies (GVI): , for all (note that is -strongly monotone; we can easily deduce that the solution of (GVI) is unique). We take a subsequence of such that
By the boundedness of , we can choose a subsequence of such that weakly. For the convenience, we may assume that . This implies that due to the weak continuity of . Now, we show . We firstly show .
Note that . Then we choose a subsequence of such that . From (26) and (36), we deduce that . Thus, . From Lemma 2, we know that is nonexpansive. By demiclosed principle (Lemma 5), we get immediately that .
Next we prove . Set
By , we know that is maximal -monotone. Let . Since and , we have . Noting that , we get It follows that Then, Since and , we deduce that by taking in (41). Thus, by the maximal -monotonicity of . Hence, . Therefore, . From (37), we obtain From (12), we have Using Lemma 6, we conclude that , and hence . This completes the proof.
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