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Composite schemes for variational inequalities over equilibrium problems and variational inclusions
Journal of Inequalities and Applications volume 2013, Article number: 414 (2013)
Abstract
Let C be a nonempty closed convex subset of a Hilbert space H, and let be a nonlinear mapping. It is well known that the following classical variational inequality has been applied in many areas of applied mathematics, modern physical sciences, computerized tomography and many others. Find a point such that
In this paper, we consider the following variational inequality. Find a point such that
and, for solutions of the variational inequality (B) with the feasibility set C, which is the intersection of the set of solutions of an equilibrium problem and the set of a solutions of a variational inclusion, construct the two composite schemes, that is, the implicit and explicit schemes to converge strongly to the unique solution of the variational inequality (B).
Recently, many authors introduced some kinds of algorithms for solving the variational inequality problems, but, in fact, our two schemes are more simple for finding solutions of the variational inequality (B) than others.
MSC:49J30, 47H10, 47H17, 49M05.
1 Introduction
A very common problem in areas of mathematics and physical sciences consists of trying to find a point in a nonempty closed convex subset C of a Hilbert space H. This problem is related to the variational inequality problem (A). One frequently employed approach in solving the variational inequality problems is the approximation methods. Some approximation methods for solving variational inequality problems and the related optimization problems can be found in [1–16].
In this paper, we consider the following variational inequality. Find a point such that
where C is the intersection of the set of solutions of an equilibrium problem and the set of a variational inclusion. In fact, the reason that we focus on the set C in the equilibrium problems and the variational inclusion problems, plays a very important role in many practical applications.
For this purpose, we construct the following composite schemes, that is, the implicit scheme and the explicit scheme , respectively,
and
Our idea is to involve directly the operator to generate the two composite schemes (1.1) and (1.2) that converge strongly to solutions of the variational inequality problem (B). In fact, our two schemes are very simple.
2 Preliminaries
In this section, we introduce some notations and useful conclusions for our main results.
Let H be a real Hilbert space. Let be a nonlinear mapping, let be a function, and let be a bifunction.
Now, we consider the following equilibrium problem. Find a point such that
The set of solutions of problem (2.1) is denoted by EP. The equilibrium problems include fixed point problems, optimization problems and variational inequality problems as special cases. For the related works, see [17–30].
Let be a τ-contraction, that is, there exists a constant such that
and let be a nonexpansive mapping, that is,
Recall that a mapping is said to be α-inverse strongly monotone if there exists a constant such that
A mapping is said to be strongly positive if there exists a constant such that for all .
Let be a single-valued nonlinear mapping, and let be a set-valued mapping.
Now, we consider the following variational inclusion. Find a point such that
where θ is the zero element in H. The set of solutions of problem (2.2) is denoted by . The variational inclusion problems have been considered extensively in [31–38] and the references therein.
A set-valued mapping is said to be monotone if, for all , and imply . A monotone mapping is said to be maximal if its graph is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping T is maximal if and only if, for any , for all implies .
Let be a maximal monotone set-valued mapping. We define the resolvent operator associated with R and λ as follows:
where λ is a positive number. It is worth mentioning that the resolvent operator is single-valued, nonexpansive and 1-inverse strongly monotone and, further, a solution of problem (2.2) is a fixed point of the operator for all .
Throughout this paper, we assume that a bifunction and a convex function satisfy the following conditions:
(H1) for all ;
(H2) Θ is monotone, i.e., for all ;
(H3) for all , is weakly upper semi-continuous;
(H4) for all , is convex and lower semi-continuous;
(H5) for all and , there exists a bounded subset and such that, for any ,
Lemma 2.1 [39]
Let H be a real Hilbert space. Let be a bifunction, and let be a proper lower semi-continuous and convex function. For any and , define a mapping as follows:
Assume that conditions (H1)-(H5) hold. Then we have the following results:
-
(1)
For each , and is single-valued.
-
(2)
is firmly nonexpansive, i.e., for any ,
-
(3)
.
-
(4)
EP is closed and convex.
Lemma 2.2 [40]
Let be a maximal monotone mapping, and let be a Lipschitz-continuous mapping. Then the mapping is maximal monotone.
Lemma 2.3 [8]
Let H be a real Hilbert space. Let the mapping be α-inverse strongly monotone, and let be a constant. Then, we have
In particular, if , then is nonexpansive.
Lemma 2.4 [41]
Assume that is a sequence of nonnegative real numbers satisfying , where is a sequence in , and is a sequence such that
-
(a)
;
-
(b)
or .
Then .
3 Main results
In this section, we give our main results. In the sequel, we assume the following conditions are satisfied.
Condition 3.1 H is a real Hilbert space. is a lower semi-continuous and convex function, and is a bifunction satisfying conditions (H1)-(H5).
Condition 3.2 F is a strongly positive bounded linear operator with coefficient , is a τ-contraction satisfying , where is a constant, and is a maximal monotone mapping.
Condition 3.3 are an α-inverse strongly monotone operator and a β-inverse strongly monotone operator, respectively.
Condition 3.4 λ and μ are two constants such that and .
Condition 3.5 is nonempty.
Now, we first consider the following scheme.
Algorithm 3.1 For any , define a net as follows:
Remark 3.2 The net defined by (3.1) is well-defined. In fact, from Lemmas 2.1 and 2.3, we know that the mappings and and are nonexpansive. For any , we define a mapping . We note that is positive and . Hence we have
This shows that W is a contraction. Therefore, W has a unique fixed point, which is denoted by .
Theorem 3.3 The net defined by (3.1) converges strongly to the unique solution of the following variational inequality:
Remark 3.4 First, we can check easily that is strongly monotone with coefficient . Now, we show the uniqueness of the solution of the variational inequality (3.2). Suppose that and both are solutions to (3.2). Then we have
Adding up the last two inequalities gives
The strong monotonicity of implies that , and so, the uniqueness is proved.
Next, we give the detail proofs of Theorem 3.3.
Proof Pick up . It is clear that . Set and for all . It follows from Lemma 2.3 that
and
Therefore, we have
From (3.1), we get
and so,
Therefore, the net is bounded, and so , , and are all bounded. It follows from (3.3) and Lemma 2.3 that
By (3.1), we obtain
where is some constant satisfying
By (3.4) and (3.5), we have
and so,
which implies that
Since is firmly nonexpansive, we have
and so,
Since is 1-inverse strongly monotone, we have
which implies that
Thus, by (3.6) and (3.7), we obtain
Substituting (3.5) into (3.8), we get
Thus, we derive
and so,
By (3.1), we obtain
It follows that
Next, we show that the net is relatively norm-compact as . In fact, assume that is such that as . Put , and . From (3.9), we have
Since is bounded, without loss of generality, we may assume that converges weakly to a point .
Next, we prove that . We first show that . By , we know that
It follows from (H2) that
and so,
For any and , let . It follows from (3.11) that
Since , we have . Further, by the monotonicity of B, we have . Thus, from (H4) and the weakly lower semi-continuity of φ, and weakly, it follows that
From conditions (H1), (H4) and (3.12), we also have
and hence
Letting , we have
This implies that .
Next, we show that . In fact, since A is α-inverse strongly monotone, A is a Lipschitz continuous monotone mapping. It follows from Lemma 2.2 that is maximal monotone. Let , i.e., . Again, since , we have , i.e., . By virtue of the maximal monotonicity of R, we have
and so,
It follows from , and weakly that
It follows from the maximal monotonicity of that , i.e., . Hence . Therefore, if we can substitute for in (3.10), then we get
Consequently, the weak convergence of to actually implies that strongly. This shows the relative norm-compactness of the net as .
Now, we return to (3.10). If we take the limit as in (3.10), then we get
In particular, solves the following variational inequality
We know that the variational inequality (3.14) is equivalent to its dual variational inequality
Thus, by the uniqueness of the variational inequality, we deduce that the entire net converges in norm to as . This completes the proof. □
Next, we introduce an explicit scheme and prove its strong convergence to the unique solution of the variational inequality (3.2).
Algorithm 3.5 For any , define the sequence generated iteratively by
where is a real sequence in .
Theorem 3.6 Assume the following conditions are also satisfied:
(C1) and ;
(C2) .
Then the sequence generated by (3.15) converges strongly to the unique solution of the variational inequality (3.2).
Proof We write and for all . Then it follows from Lemma 2.3 that, for any ,
and
Hence we have
By induction, it follows from (3.15) that
Therefore, is bounded, and so, , , and are all bounded. It follows from (3.15) that
Note that
Substituting (3.18) into (3.17), we get
Notice that . This, together with the last inequality and Lemma 2.4, implies that
Again, using Lemma 2.3 and (3.16), we get
By (3.15), we obtain
where is a constant satisfying
From (3.19) and (3.20), we have
and so,
which implies that
Since is firmly nonexpansive, we have
and hence
Since is 1-inverse strongly monotone, we have
which implies that
Thus, by (3.21) and (3.22), we obtain
Substituting (3.20) into (3.23), we get
Thus, we derive
and so,
Next, we prove that
where is the unique solution of the variational inequality (3.2). To see this, we can take a subsequence of satisfying
and converges weakly to a point as . By the similar argument as in Theorem 3.3, we can deduce . Since solves the variational inequality (3.2), by combining (3.24) and (3.25), we get
Finally, we show that as . It follows from (3.15) that
that is,
where and . It is easy to see that and . Hence, by Lemma 2.4, we conclude that the sequence converges strongly to the point . This completes the proof. □
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Acknowledgements
Yonghong Yao was supported in part by NSFC 11071279 and NSFC 71161001-G0105. Yeol Je Cho was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant Number: 2012-0008170). Yeong-Cheng Liou was supported in part by NSC 101-2628-E-230-001-MY3 and NSC 101-2622-E-230-005-CC3.
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Yao, Y., Kang, J.I., Cho, Y.J. et al. Composite schemes for variational inequalities over equilibrium problems and variational inclusions. J Inequal Appl 2013, 414 (2013). https://doi.org/10.1186/1029-242X-2013-414
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DOI: https://doi.org/10.1186/1029-242X-2013-414