Relaxed Iterative Algorithms for Generalized Mixed Equilibrium Problems with Constraints of Variational Inequalities and Variational Inclusions

Ceng, Lu-Chuan; Chen, Chi-Ming; Wen, Ching-Feng; Pang, Chin-Tzong

doi:https://doi.org/10.1155/2014/345212

Abstract and Applied Analysis

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Variational Analysis, Optimization, and Fixed Point Theory 2014

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Research Article | Open Access

Volume 2014 | Article ID 345212 | https://doi.org/10.1155/2014/345212

Relaxed Iterative Algorithms for Generalized Mixed Equilibrium Problems with Constraints of Variational Inequalities and Variational Inclusions

Lu-Chuan Ceng,^1,2Chi-Ming Chen,³Ching-Feng Wen,⁴and Chin-Tzong Pang⁵

Academic Editor: Abdul Latif

Received27 Dec 2013

Accepted06 Jan 2014

Published16 Mar 2014

Abstract

We introduce and analyze a relaxed extragradient-like viscosity iterative algorithm for finding a solution of a generalized mixed equilibrium problem with constraints of several problems: a finite family of variational inequalities for inverse strongly monotone mappings, a finite family of variational inclusions for maximal monotone and inverse strongly monotone mappings, and a fixed point problem of infinitely many nonexpansive mappings in a real Hilbert space. Under some suitable conditions, we derive the strong convergence of the sequence generated by the proposed algorithm to a common solution of these problems which also solves a variational inequality problem.

1. Introduction

Let be a real Hilbert space with inner product and norm , a nonempty closed convex subset of , and the metric projection of onto . Let be a nonlinear mapping on . We denote by the set of fixed points of and by the set of all real numbers. A mapping is called strongly positive on if there exists a constant such that A mapping is called -Lipschitz continuous if there exists a constant such that In particular, if , then is called a nonexpansive mapping; if , then is called a contraction.

Let be a nonlinear mapping on . We consider the following variational inequality problem (VIP) [1]: find a point such that The solution set of VIP (3) is denoted by .

In 1976, Korpelevič [2] proposed an iterative algorithm for solving the VIP (3) in Euclidean space : with a given number, which is known as the extragradient method. The literature on the VIP is vast and Korpelevich’s extragradient method has received great attention given by many authors. See, for example, [3–10] and references therein.

Let be a real-valued function, a nonlinear mapping, and a bifunction. In 2008, Peng and Yao [11] introduced the following generalized mixed equilibrium problem (GMEP): finding such that We denote the set of solutions of (5) by .

Throughout this paper, it is assumed as in [11] that is a bifunction satisfying conditions (H1)–(H4) and is a lower semicontinuous and convex function with restriction (H5), where(H1) for all ,(H2) is monotone, that is, for any ,(H3) is upper hemicontinuous, that is, for each , (H4) is convex and lower semicontinuous for each ,(H5)for each and , there exists a bounded subset and such that, for any , given a positive number . Let be the solution set of the auxiliary mixed equilibrium problem; that is, for each , In particular, whenever , is rewritten as .

Let be an infinite family of nonexpansive mappings on and a sequence of nonnegative numbers in . For any , define a mapping on as follows: Such a mapping is called the -mapping generated by and .

In 2010, for the case where , Yao et al. [12] proposed the following hybrid iterative algorithm: where is a contraction, is differentiable and strongly convex, , , and , are given, for finding a common element of the set and the fixed point set of an infinite family of nonexpansive mappings on . They proved the strong convergence of the sequence generated by the hybrid iterative algorithm (10) to a point under some appropriate conditions. This point also solves the following optimization problem: where is the potential function of .

On the other hand, let be a single-valued mapping of into and a set-valued mapping with . Consider the following variational inclusion: find a point such that We denote by the solution set of the variational inclusion (11). In particular, if , then . If , then problem (11) becomes the inclusion problem introduced by Rockafellar [13]. Let a set-valued mapping be maximal monotone. We define the resolvent operator associated with and as follows: where is a positive number.

In 1998, Huang [14] studied problem (11) in the case where is maximal monotone and is strongly monotone and Lipschitz continuous with . Subsequently, Zeng et al. [15] further studied problem (11) in the case which is more general than Huang’s one [14].

Inspired by the above facts, we introduce and analyze an iterative algorithm by relaxed extragradient-like viscosity method for finding a solution of a generalized mixed equilibrium problem with constraints of several problems: a finite family of variational inequalities for inverse strongly monotone mappings, a finite family of variational inclusions for maximal monotone and inverse strongly monotone mappings, and a fixed point problem of infinitely many nonexpansive mappings in a real Hilbert space. Under some suitable conditions, we derive the strong convergence of the sequence generated by the proposed algorithm to a common solution of these problems. Such solution also solves a variational inequality problem. Several special cases are also discussed. The results presented in this paper are the supplement, extension, improvement, and generalization of the previously known results in this area.

2. Preliminaries

Throughout this paper, we assume that is a real Hilbert space whose inner product and norm are denoted by and , respectively. Let be a nonempty closed convex subset of . We write to indicate that the sequence converges weakly to and to indicate that the sequence converges strongly to . Moreover, we use to denote the weak -limit set of the sequence ; that is,

Definition 1. A mapping is called(i)monotone if (ii)-strongly monotone if there exists a constant such that (iii)-inverse strongly monotone if there exists a constant such that
It is easy to see that the projection is -ism. Inverse strongly monotone (also referred to as cocoercive) operators have been applied widely in solving practical problems in various fields.

Definition 2. A differentiable function is called(i)convex, if where is the Frechet derivative of at ;(ii)strongly convex, if there exists a constant such that
It is easy to see that if is a differentiable strongly convex function with constant then is strongly monotone with constant .
The metric (or nearest point) projection from onto is the mapping which assigns to each point the unique point satisfying the property
Some important properties of projections are gathered in the following proposition.

Proposition 3. For given and ,(i);(ii);(iii), for all . (This implies that is nonexpansive and monotone.)

By using the technique of [16], we can readily obtain the following elementary result.

Proposition 4 (see [17, Lemma 1 and Proposition 1]). Let be a nonempty closed convex subset of a real Hilbert space and let be a lower semicontinuous and convex function. Let be a bifunction satisfying the conditions (H1)–(H4). Assume that(i) is strongly convex with constant and the function is weakly upper semicontinuous for each ;(ii)for each and , there exists a bounded subset and such that, for any , Then the following hold:(a)for each ,(b) is single valued,(c) is nonexpansive if is Lipschitz continuous with constant and where for ,(d)for all and (e),(f) is closed and convex.In particular, whenever is a bifunction satisfying the conditions (H1)–(H4) and , for all , then that is, for any , ( is firmly nonexpansive) and In this case, is rewritten as . If, in addition, , then is rewritten as .

Remark 5. Suppose is strongly convex with constant and is Lipschitz continuous with constant . Then is -strongly monotone and -Lipschitz continuous with positive constants . Utilizing Proposition 4 () we obtain that, for all and ,

We need some facts and tools in a real Hilbert space which are listed as lemmas below.

Lemma 6. Let be a real inner product space. Then there holds the following inequality:

Lemma 7. Let be a real Hilbert space. Then the following hold:(a) for all ;(b) for all and with ;(c)if is a sequence in such that , it follows that
We have the following crucial lemmas concerning the -mappings defined by (9).

Lemma 8 (see [18, Lemma 3.2]). Let be a sequence of nonexpansive self-mappings on such that and let be a sequence in for some . Then, for every and the limit exists, where is defined by (9).

Lemma 9 (see [18, Lemma 3.3]). Let be a sequence of nonexpansive self-mappings on such that , and let be a sequence in for some . Then, .

Lemma 10 (see [19, Demiclosedness Principle]). Let be a nonempty closed convex subset of a real Hilbert space . Let be a nonexpansive self-mapping on . Then is demiclosed. That is, whenever is a sequence in weakly converging to some and the sequence strongly converges to some , it follows that . Here is the identity operator of .

Lemma 11. Let be a monotone mapping. In the context of the variational inequality problem the characterization of the projection (see Proposition 3 ()) implies

Lemma 12 (see [20]). Let and be bounded sequences in a real Banach space and a sequence in with . Suppose Then, .

Lemma 13 (see [21]). Assume that is a sequence of nonnegative real numbers such that where is a sequence in and is a real sequence such that(i);(ii) or .Then .

Recall that a set-valued mapping is called monotone if, for all , and imply A set-valued mapping is called maximal monotone if is monotone and for each , where is the identity mapping of . We denote by the graph of . It is known that a monotone mapping is maximal if and only if, for , for every implies . Next we provide an example to illustrate the concept of maximal monotone mapping.

Let be a monotone, -Lipschitz-continuous mapping and let be the normal cone to at ; that is, Define Then, is maximal monotone and if and only if ; see [13].

Assume that is a maximal monotone mapping. Let . In terms of Huang [14], there holds the following property for the resolvent operator .

Lemma 14. is single valued and firmly nonexpansive; that is,
Consequently, is nonexpansive and monotone.

Lemma 15 (see [9]). Let be a maximal monotone mapping with . Then, for any given , is a solution of problem (11) if and only if satisfies

Lemma 16 (see [15]). Let be a maximal monotone mapping with and let be a strongly monotone, continuous, and single-valued mapping. Then, for each , the equation has a unique solution for .

Lemma 17 (see [9]). Let be a maximal monotone mapping with and a monotone, continuous, and single-valued mapping. Then for each . In this case, is maximal monotone.

Lemma 18 (see [22, Lemma 2.8]). Let be a bounded sequence of nonnegative real numbers and a sequence of real numbers such that . Then, .

3. Main Results

We will introduce and analyze an iterative algorithm by relaxed extragradient-like viscosity method for finding a solution of a generalized equilibrium problem with constraints of several problems: a finite family of variational inclusions, a finite family of variational inequalities, and a fixed point problem in a real Hilbert space. Under appropriate conditions imposed on the parameter sequences we will prove strong convergence of the proposed algorithm.

Theorem 19. Let be a nonempty closed convex subset of a real Hilbert space . Let and be two integers. Let be a bifunction from to satisfying (H1)–(H4) and a lower semicontinuous and convex functional. Let be a maximal monotone mapping and let , , and be -inverse strongly monotone, -inverse strongly monotone, and -inverse strongly monotone, respectively, where and . Let be a sequence of nonexpansive mappings on and a sequence in for some . Let be a -strongly positive bounded linear operator with and a -contraction with . Let be the -mapping defined by (9). Assume that . Define − and for each and . Let , and be three sequences in . Assume that(i) is strongly convex with constant and its derivative is Lipschitz continuous with constant such that the function is weakly upper semicontinuous for each ,(ii)for each , there exist a bounded subset and such that, for any , (iii) and ,(iv) and ,(v) + ,(vi) and . Given arbitrarily, the sequence is generated iteratively by where and for each and . If and is firmly nonexpansive, then where is a unique solution in to the VIP

Proof. As , , and , we may assume, without loss of generality, that , and for all . Put for each and , and , where is the identity mapping on . Moreover, set and .
Since is a -strongly positive bounded linear operator on , we know that Taking into account that for all , we have That is, is positive. It follows that
In the meantime, it is not hard to find that and are nonexpansive. As a matter of fact, observe that, for all ,
In addition, note that That is, is strongly monotone and Lipschitz continuous. So, there exists a unique solution in to the VIP That is, .
We divide the rest of the proof into several steps.
Step 1. We show that is bounded. Indeed, take arbitrarily. Since , is -inverse strongly monotone, and , we have, for any , Since , , and is -inverse strongly monotone, where , , by Proposition 3 we obtain that for each Since , , and is -inverse strongly monotone, where , , by Lemma 14 we deduce that for each Hence from (37)–(49), we have Since is a -contraction with , from (37) and (50) we get By induction, we get Therefore, is bounded and so are the sequences , , , , and .
Step 2. We show that and . Indeed, put . Then it follows from conditions (iii) and (iv) that and hence Define Observe that and hence From (9), since , , and are all nonexpansive, we have where and for some .
On the other hand, we estimate and . First observe that Utilizing Remark 5 and Lemma 14, we have where for some . So, combining (57)–(60) we get where for some . Thus, from (61), , and conditions (v)-(vi) it follows that Since for all , by Lemma 12 we obtain from that which immediately yields Note that Consequently, it follows from (64) and that
Step 3. We prove .
Indeed, for any , we find that From (37), (48), (49), and (67), we obtain which immediately yields In particular, putting we have Since , , and , we obtain from (66) and the boundedness of and that Furthermore, from the firm nonexpansivity of , we have which implies that From (68) and (73), we have which immediately yields In particular, putting we have Since and , we deduce from (66) and (71) and the boundedness of , and that
Step 4. We prove that and for each and .
Indeed, let us show that and for each and . Observe that, for any , From (68) and (78) we have which leads to In particular, putting we have Since and , we obtain from (66), , , , and the boundedness of and that for each and .
Furthermore, by Proposition 3(iii) and Lemma 7(a), we obtain which implies that By Lemmas 7(a) and 14, we get which implies that From (68), (84), and (86) we have which hence implies that In particular, putting we have Since and , we obtain from (66) and (82) and the boundedness of , , and that for each and . Consequently, from (90) it follows that By (77), (91), and (92), we have
Step 5. We show that . Indeed, utilizing Lemma 7(b), from (37), (48), (49), and (67) we obtain that, for any , which implies that In particular, putting we have Since , , and , we deduce from (66) and the boundedness of and that Also, observe that Thus, from (77), (91), and (97) it follows that Moreover, note that From (99), [23, Remark 3.2], and the boundedness of we immediately obtain
Step 6. We show that .
Indeed, we observe that there exists a subsequence of such that Since is bounded, there exists a subsequence of which converges weakly to some . Without loss of generality, we may assume that . From (77), (90), and (93), we have that , , , and , where and . By (101) we have that as . Then, by Lemma 10 we obtain (due to Lemma 9). Next we prove that . Let where . Let . Since and , we have Also, from and , we have and hence Therefore we have From (90) and since is uniformly continuous, we obtain that . From , , and (90), we have Since is maximal monotone, we have and hence , , which implies . Next, we prove that . As a matter of fact, since is -inverse strongly monotone, is a monotone and Lipschitz continuous mapping. It follows from Lemma 17 that is maximal monotone. Let ; that is, . Again, since , , , we have That is, In terms of the monotonicity of , we get and hence In particular, Since (due to (90)) and (due to the Lipschitz continuity of ), we conclude from and , , that It follows from the maximal monotonicity of that ; that is, . Therefore, .
Next, we show that . In fact, from , we know that From (H2) it follows that Replacing by , we have Put for all and . Then, from (117) we have Since as , we deduce from the Lipschitz continuity of and that and as . Further, from the monotonicity of , we have . So, from (H4), the weakly lower semicontinuity of , and , we have From (H1), (H4), and (119) we also have and hence Letting , we have, for each , This implies that . Therefore, ,,),. This shows that . Consequently, from (102) and , we have
Step 7. Finally, we show that as .
Indeed, in terms of (68) we get which, together with (37), implies that where and Since , , and , we deduce that = and . Note that Hence from (123) and Lemma 18 it follows that Applying Lemma 13 to (125), we infer that the sequence converges strongly to . This completes the proof.

Corollary 20. Let be a nonempty closed convex subset of a real Hilbert space . Let be an integer. Let be a bifunction from to satisfying (H1)–(H4) and a lower semicontinuous and convex functional. Let be a maximal monotone mapping and let , , and be -inverse strongly monotone, -inverse strongly monotone, and -inverse strongly monotone, respectively, where . Let be a sequence of nonexpansive mappings on and a sequence in for some . Let be a -strongly positive bounded linear operator with and a -contraction with . Let be the -mapping defined by (9). Assume that ,,. Define for each . Let , , and be three sequences in . Assume that(i) is strongly convex with constant and its derivative is Lipschitz continuous with constant such that the function is weakly upper semicontinuous for each ,(ii)for each , there exist a bounded subset and such that, for any , (iii) and ,(iv) and ,(v),(vi), , and .Given arbitrarily, the sequence is generated iteratively by where and for each . If and is firmly nonexpansive, then where is a unique solution in to the VIP

Corollary 21. Let be a nonempty closed convex subset of a real Hilbert space . Let be an integer. Let be a bifunction from to satisfying (H1)–(H4) and a lower semicontinuous and convex functional. Let be a maximal monotone mapping and let and be -inverse strongly monotone and -inverse strongly monotone, respectively, where . Let be a sequence of nonexpansive mappings on and a sequence in for some . Let be a -strongly positive bounded linear operator with and a -contraction with . Let be the -mapping defined by (9). Assume that ,,. Define for each . Let , , and be three sequences in . Assume that(i) is strongly convex with constant and its derivative is Lipschitz continuous with constant such that the function is weakly upper semicontinuous for each ,(ii)for each , there exist a bounded subset and such that, for any , (iii) and ,(iv) and ,(v),(vi), , and .Given arbitrarily, the sequence is generated iteratively by where for each . If and is firmly nonexpansive, then where is a unique solution in to the VIP

Proof. In Theorem 19, putting for each , we know that the iterative scheme (37) reduces to (134). In this case, we get . Utilizing Theorem 19 we derive the desired result.

Corollary 22. Let be a nonempty closed convex subset of a real Hilbert space . Let be an integer. Let be a bifunction from to satisfying (H1)–(H4) and a lower semicontinuous and convex functional. Let be a maximal monotone mapping and let , , and be -inverse strongly monotone, -inverse strongly monotone, and -inverse strongly monotone, respectively, where . Let be a sequence of nonexpansive mappings on and a sequence in for some . Let be a -strongly positive bounded linear operator with and a -contraction with . Let be the -mapping defined by (9). Assume that . Define for each . Let , , and be three sequences in . Assume that(i) is strongly convex with constant and its derivative is Lipschitz continuous with constant such that the function is weakly upper semicontinuous for each ,(ii)for each , there exist a bounded subset and such that, for any , (iii) and ,(iv) and ,(v),(vi), , and .Given arbitrarily, the sequence is generated iteratively by where and for each . If and is firmly nonexpansive, then where is a unique solution in to the VIP

Corollary 23. Let be a nonempty closed convex subset of a real Hilbert space . Let and be two integers. Let be a bifunction from to satisfying (H1)–(H4) and a lower semicontinuous and convex functional. Let be a maximal monotone mapping and let and be -inverse strongly monotone and -inverse strongly monotone, respectively, where and . Let be a sequence of nonexpansive mappings on and a sequence in for some . Let be a -strongly positive bounded linear operator with and a -contraction with . Let be the -mapping defined by (9). Assume that . Define and for each and . Let , , and be three sequences in . Assume that(i) is strongly convex with constant and its derivative is Lipschitz continuous with constant such that the function is weakly upper semicontinuous for each ,(ii)for each , there exist a bounded subset and such that, for any , (iii) and ,(iv) and ,(v),(vi) is a bounded sequence in such that Given arbitrarily, the sequence is generated iteratively by where and for each and . If and is firmly nonexpansive, then where is a unique solution in to the VIP

Proof. In Theorem 19, for all is equivalent to Put . Then it follows that Observe that for all So, whenever for some , we obtain the desired result by using Theorem 19.

Let be a -strictly pseudocontractive mapping. For recent convergence result for strictly pseudocontractive mappings, we refer to [24]. Putting , we know that for all Note that Hence we have for all Consequently, if is a -strictly pseudocontractive mapping, then the mapping is -inverse strongly monotone.

Corollary 24. Let be a nonempty closed convex subset of a real Hilbert space . Let and be two integers. Let be a bifunction from to satisfying (H1)–(H4) and a lower semicontinuous and convex functional. Let be a maximal monotone mapping and let , and be -strictly pseudocontractive, -inverse strongly monotone, and -inverse strongly monotone, respectively, where and . Let be a sequence of nonexpansive mappings on and a sequence in for some . Let be a -strongly positive bounded linear operator with and a -contraction with . Let be the -mapping defined by (9). Assume that where . Define and for each and . Let , , and be three sequences in . Assume that(i) is strongly convex with constant and its derivative is Lipschitz continuous with constant such that the function is weakly upper semicontinuous for each ,(ii)for each , there exist a bounded subset and such that, for any , (iii) and ,(iv) and ,(v),(vi), , and . Given arbitrarily, the sequence is generated iteratively by where and for each and . If and is firmly nonexpansive, then where is a unique solution in to the VIP

Proof. Since is a -strictly pseudocontractive mapping, the mapping is -inverse strongly monotone. In this case, put . Moreover, we obtain that So, from Theorem 19, we obtain the desired result.

Corollary 25. Let be a nonempty closed convex subset of a real Hilbert space . Let and be two integers. Let be a bifunction from to satisfying (H1)–(H4) and a lower semicontinuous and convex functional. Let be a maximal monotone mapping and let , , and be -inverse strongly monotone, -inverse strongly monotone, and -inverse strongly monotone, respectively, where and . Let be a -strongly positive bounded linear operator with and a -contraction with . Assume that . Define and for each and . Let , , and be three sequences in . Assume that(i) is strongly convex with constant and its derivative is Lipschitz continuous with constant such that the function is weakly upper semicontinuous for each ,(ii)for each , there exist a bounded subset and such that, for any , (iii) and ,(iv) and ;(v),(vi), , and .Given arbitrarily, the sequence is generated iteratively by where and for each and . If and is firmly nonexpansive, then where is a unique solution in to the VIP

Proof. Put for all integers and all . Then, the desired result follows from Theorem 19.

Corollary 26. Let be a nonempty closed convex subset of a real Hilbert space . Let and be two integers. Let be a maximal monotone mapping and let and be -inverse strongly monotone and -inverse strongly monotone, respectively, where and . Let be a sequence of nonexpansive mappings on and a sequence in for some . Let be a -strongly positive bounded linear operator with and a -contraction with . Let be the -mapping defined by (9). Assume that . Define and for each and . Let , , and be three sequences in . Assume that(i) and ,(ii) and ,(iii).Given arbitrarily, the sequence is generated iteratively by where and for each and . If and is firmly nonexpansive, then where is a unique solution in to the VIP

Proof. Put , for all , for all , and . Take for all . Then we get in Theorem 19 and the conclusion follows.

Conflict of Interests

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

Acknowledgments

This research was partially supported by the National Science Foundation of China (11071169), Innovation Program of Shanghai Municipal Education Commission (09ZZ133), and Ph.D. Program Foundation of the Ministry of Education of China (20123127110002). This research was supported partly by the National Science Council of the Republic of China. This research was partially supported by a grant from NSC.

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Copyright © 2014 Lu-Chuan Ceng et al. 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.

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