Abstract

We suggest and study the convergence of some new iterative schemes for solving nonconvex equilibrium problems in Banach spaces. Many existing results have been obtained as particular cases.

1. Introduction

Let be a Banach space, and let be the dual space of . Let denote the duality pairing of and . Let be a nonempty closed subset of and let be a bifunction satisfying for all . A prototype of equilibrium problem associated with a closed convex set and a convex bifunction is given as follows:In this paper we introduce and study two appropriate extensions of from the convex case to the nonconvex case in Banach spaces setting. We consider the two following generalized equilibrium problems associated with , , and (resp., denoted by and ): where is the normalized duality mapping and is the functional defined byDue to the nonsymmetry of the terms and we can think about the symmetric functional . Thus, we can consider one more appropriate extension of equilibrium problem from Hilbert spaces to Banach spaces as follows:The reason for saying that the above three generalized equilibrium problems are the appropriate extensions from Hilbert spaces to Banach spaces and from convex cases to nonconvex cases is that in Hilbert spaces we have and the convex case is covered by taking . Obviously, any solution of and is a solution of . Thus, we are going to study the convergence of new iterative schemes to solutions of and .

We list some important properties of needed in our proofs, when is a reflexive smooth Banach space:(i), for all , ;(ii), for all , ;(iii), for all ;(iv) is continuous and is convex with respect to when is fixed and convex with respect to when is fixed;(v)whenever the space is smooth, the functional is differentiable and ;(vi) if and only if .

The proposed generalized equilibrium problems extend many existing equilibrium problems and variational inequalities from the convex case to the nonconvex case and from Hilbert spaces setting to Banach spaces setting.(1)If is a Hilbert space, the duality mapping is the identity operator and and so , , and become as follows:which has been introduced and studied in [1]. The same problem has been studied by Noor [2] and many authors (see, e.g., [35]).(2)If is a Hilbert space, is a convex closed set in , is a convex bifunction, and , all the generalized equilibrium problems , , and become as follows:which has been studied in various works (see, e.g., [3, 4] and the references therein).(3)If , with being a nonlinear operator, then and reduce, respectively, to the following:These inequalities are new even in Banach spaces. However, it has been studied, in Hilbert spaces, in [6], when is a uniformly prox-regular set (which is not necessarily a convex set) (see also [1, 2, 4]). By taking the last inequality becomes as follows:which is known as the classical variational inequality introduced and studied by Stampacchia in [7].

Our main aim of the present paper is to suggest and analyze some iterative schemes for solving the proposed generalized equilibrium problems and .

2. Main Results

We recall some definitions and results on -uniformly convex and -uniformly smooth Banach spaces (see, e.g., [8, 9]). The moduli of convexity and smoothness of are defined, respectively, byThe space is said to be uniformly convex whenever for all and is said to be uniformly smooth whenever . Let be real numbers. The space is said to be -uniformly convex (resp., -uniformly smooth) if there is a constant such thatIt is known (see, e.g., [8, 9]) that uniformly convex Banach spaces are reflexive, strictly convex and that uniformly smooth Banach spaces are reflexive. If is a -uniformly convex Banach space, then is a -uniformly smooth Banach space, where is the conjugate number of . If is a -uniformly smooth Banach space, then is a -uniformly convex Banach space, where .

The following lemma is needed in our proofs and for its proof we refer to [10].

Lemma 1. Let , be a -uniformly convex and -uniformly smooth Banach space, and let be a bounded set. Then there exist constants such that

Let be a reflexive smooth Banach space and let be a function and , where is finite. We recall from [10] that the proximal subdifferential is the set of all for which there are and so thatfor all around . We also recall ([10]) that the proximal normal cone of a nonempty closed subset in at is defined by , where is the indicator function of . The following proposition summarizes some properties of and that we need in our proofs. For their proofs we refer the reader to [10].

Proposition 2. Let be a reflexive smooth Banach space. Then the following assertions hold:(1)if is l.s.c. convex, then coincides with the subdifferential in the sense of convex analysis defined by(2)if is a closed convex set, then coincides with the normal cone in the sense of convex analysis defined by(3)if is a local minimum of , then ;(4), for any , where is the distance function associated with a nonempty closed set in ;(5)if is locally Lipschitz around with ratio , then ;(6);(7) is also characterized by the following global inequality:

We propose the following two iterative schemes:where is a given positive number. Under natural assumptions, we will prove the convergence of a subsequence of the sequence generated by (resp., ) to a solution of (resp., ).

To start our study we define two new classes of nonconvex sets and nonconvex functions as follows.

Definition 3. Let be a reflexive smooth Banach space. For a given , a subset is -uniformly prox-regular with respect to provided that for all and all nonzero we haveWe use the convention for .

Obviously, this class extends the class of uniformly prox-regular sets ([11, 12]) from Hilbert spaces to Banach spaces since in Hilbert spaces we have and the generalized proximal normal cone coincides with the usual proximal subdifferential (see [10] for more details on and ). We point out that a different extension of uniformly prox-regular sets to Banach spaces has been considered and studied recently in [13].

Definition 4. Let be a reflexive smooth Banach space. Let be a l.s.c. function and let be a nonempty closed set in . We will say that is -uniformly prox-regular over provided that for all and all we haveWe say that is -prox-regular around provided that is -uniformly prox-regular over some closed neighborhood of ; that is, there exists a closed neighborhood of such that for all , for all , inequality (14) holds for any .

Example 5. Consider the following.(1)Any l.s.c. proper convex function is -uniformly prox-regular over any nonempty closed set in its domain with .(2)The indicator function of -uniformly prox-regular set is -uniformly prox-regular over with respect to the same constant .(3)The distance function associated with a -uniformly prox-regular set is -uniformly prox-regular over with respect to the same constant . Indeed, for any and for all we have by Parts (4-5) in Proposition 2 that with and so by Definition 3: and hencethat is, is -uniformly prox-regular over with respect to the same constant .(4)Any lower- function over convex strongly compact in is -uniformly prox-regular over with some as the next propositions (Propositions 7 and 8) show.

Definition 6. Let be a reflexive smooth Banach space. A function is said to be lower- around a point if there exists an open convex neighborhood and such that is finite and convex over . We will say that is lower- around a subset if it is lower- around each point of .

Proposition 7. Let be a reflexive smooth Banach space and let be a convex strongly compact set in . If is a lower- function on , then is uniformly lower- over ; that is, for some the function is finite convex over .

Proof. Assume that is lower- on . For every there exist, by Definition 6, an open convex neighborhood of and such that is finite convex over . The family of open sets covers , so by the strong compactness of there exist points in such that , and sufficiently large such that is finite convex over and so over the convex set . This ensures the uniform lower- property of over and hence the proof is complete.

The next proposition proves the relationship between uniform lower- functions and -uniformly prox-regular functions.

Proposition 8. Let be a reflexive smooth Banach space, , let be a convex strongly compact subset of , and let be a l.s.c function on . If is uniformly lower- on with some ratio , then is -uniformly prox-regular over with respect to the constant .

Proof. Assume that is uniformly lower- on with some ratio ; that is, is convex over . Let and . Then by Part 6 in Proposition 2 we obtain Since is convex over we have by Part 1 in Proposition 2 that    and so by the definition of the convex subdifferential:This ensuresfor all and then the function is -uniformly prox-regular over with constant and hence the proof is complete.

Now, we are in position to state and prove our first proposition which extends the main result in [1] from Hilbert spaces to Banach spaces.

Proposition 9. Let be a reflexive smooth Banach space. If is a closed convex set and for all the function is convex Lipschitz with ratio over an open set containing , then is equivalent to the following subproblem:

Proof. Let be generated by , that is,with . Then there exists such thatBy the convexity of and Part 1 in Proposition 2 we have and so by the definition of the subdifferential for convex functions we haveSince , we obtainOn the other hand, by the convexity of and Part 2 in Proposition 2 we have which yields by the definition of convex normal cones the following inequality:Combining this inequality with (23) we obtainConversely, assume that is generated by , that is,Let and . Then the last inequality yieldsThis means that is a minimum of over . Thus, by Part 3 in Proposition 2, we have and consequently by the convexity of the set and the convexity of the function we can writeand soOn the other hand, since is Lipschitz continuous with ratio over an open set containing we have for all for small enoughLet . Then, taking and in the last inequality yieldsand so by we haveand hence , for all , which ensures that , that is, . Therefore, is generated by and the proof is complete.

The following proposition establishes an analogue result for and its proof follows the same lines of the previous proposition. So, its proof is omitted.

Proposition 10. Let be a reflexive smooth Banach space. If is a closed convex set and for all the function is convex Lipschitz with ratio over an open set containing , then is equivalent to the following subproblem:

The following proposition established a key tool of the proof of our main convergence result of the generalized equilibrium problem in the prox-regular setting.

Proposition 11. Let be a reflexive smooth Banach space and let . If is -uniformly -prox-regular and is -Lipschitz and -uniformly prox-regular over with ratio , then the sequence generated by satisfies the following inequality:where .

Proof. Let be generated by , that is,with . Then there exists such thatSince is -Lipschitz, then by Part 5 in Proposition 2 we have and hence . By definition of -uniform prox-regularity of we haveand henceOn the other hand, by the fact that and is -uniformly prox-regular over with ratio we haveCombining (37) and the last inequality we obtainwhich ensures thatThis completes the proof of the proposition.

Now, we are ready to state and prove our first main theorem of this paper.

Theorem 12. Let and be a -uniformly smooth Banach space. Let be a closed nonempty subset of and let be a bifunction. Let be a sequence generated by . Assume that(1) is -uniformly prox-regular with some ratio ;(2) is ball compact; that is, is compact for any ;(3)the solution set of is nonempty;(4) is -strongly monotone over , that is,for some ;(5) is upper semicontinuous with respect to the first variable over , that is,(6) is -uniformly prox-regular over with ratio ;(7)there exists such that for all ;(8)the positive number satisfies .Then, there exists subsequence of   that converges to which solves .

Proof. Let be a solution of . ThenBy the -strong monotonicity of over we haveBy setting in these two inequalities we getCombining these two inequalities we obtainUsing the 8th assumption of the theorem we have and henceThis combined with Proposition 11 giveswith and . Therefore,Define now the auxiliary real sequence . It is direct to check thatIndeed,It follows thatwhich ensures with (49) thatUsing the assumption yieldsTherefore, the sequence is a nonincreasing nonnegative sequence and so it is convergent to some limit and so it is bounded by some positive number . Thus, by the property (ii) of recalled in Section 1 we obtainand sothat is, is bounded and so by the -uniform convexity of we havefor some depending on and on the space . Here , where is the modulus of smoothness of . Using now (49) and (50) and the assumption we obtainTherefore, it follows from the 7th assumption of the theorem thatwhich ensures that . On the other hand, since is bounded in and is ball compact then there exists a subsequence which converges to some limit . Note that by Proposition 11 this subsequence satisfiesThus, by letting in inequality (60) and by taking into account the upper semicontinuity of and the continuity of and , we obtainTherefore, the assumption concludeswhich ensures that the limit is a solution of .

As a direct consequence of the previous theorem we have the following convex version which is new according to our modest knowledge.

Theorem 13. Let and be a -uniformly smooth Banach space. Let be a closed convex nonempty subset of and let be a convex bifunction. Let be a sequence generated by . Assume that(1) is ball compact; that is, is compact for any ;(2)the solution set of is nonempty;(3) is monotone over ; that is,(4) is upper semicontinuous with respect to the first variable over ;(5)there exists such that for all . Then, there exists subsequence of that converges to which solves .

Proof. It follows directly from the previous theorem with the constants and .

Now, we are going to prove a similar result for the second generalized equilibrium problem . To do that we need a different and more restrictive concept of monotonicity that we define as follows: a bifunction is said to be -relaxed strongly monotone with respect to for some and provided thatObserve that any -relaxed strong monotone with respect to is -strongly monotone with respect to . As a simple example of -relaxed strongly monotone bifunction we can take defined by . This bifunction is -relaxed strongly monotone with respect to (). The proof of the following proposition follows the same lines of the proof of Proposition 11. So its proof is omitted. It is needed in the proof of Theorem 15.

Proposition 14. Let be a reflexive smooth Banach space and let and . If is -uniformly prox-regular with ratio and is -Lipschitz and -uniformly prox-regular over with ratio , then the sequence generated by satisfies the following inequality:

Now, we prove our next main theorem concerning .

Theorem 15. Let and be a -uniformly smooth Banach space. Assume that Assumptions 1, 2, 3, 5, and 6 of Theorem 12 are satisfied and assume that is -relaxed strongly monotone with respect to some . If the constants , , , , , and satisfy the inequalitiesthen there exists a subsequence of the sequence generated by which converges to some solution of .

Proof. Let be a solution of . By setting in we getand by taking in the conclusion of Proposition 14 we obtainwith . The last two inequalities yieldUsing the -relaxed strong monotonicity of we writeTherefore, the two previous inequalities yieldObserve thatHence,We distinguish two cases.
Case 1. . In this case (73) ensures thatand since we obtain , for all , that is, , for all . This means that the sequence is constant and equals and hence we are done.
Case 2. Consider . In this case we use our assumptions on the constants , , , , , , and to ensure thatThus, (73) can be rewritten as follows:with . The conclusion of the theorem follows the same lines as in the proof of Theorem 12.

Remark 16. An inspection of the proof of the previous theorem shows that in the case when the ball compactness of is not needed and that all the sequence converges to the solution , whenever the space is assumed to satisfy the -uniform convexity instead of the -uniform smoothness. Indeed, with the assumption (73) ensures . This limit with the -uniform convexity of the space , and Lemma 1 ensure that and hence the proof is complete.

Corollary 17. Let , be a -uniformly smooth Banach space, be a closed convex nonempty subset of , and let be a convex bifunction. Let be a sequence generated by . Assume that(1) is ball compact; that is, is compact for any ;(2)the solution set of is nonempty;(3) satisfies for some (4) is upper semicontinuous with respect to the first variable over ;(5)there exists such that for all .Then, there exists a subsequence of that converges to which solves .

Proof. It follows directly from the previous theorem with the constants and .

It is a natural question to ask whether the additional assumption of -relaxed strong monotonicity in Theorem 15 can be replaced by the one used in Theorem 12. The answer is given in the next theorem with a different generalized subproblem given by the following: select such that

Theorem 18. Let and be a -uniformly smooth Banach space. Assume that the assumptions 2–5 of Theorem 12 are satisfied and assume that the constants , , , and satisfy the inequalitiesThen there exists a subsequence of the sequence generated by which converges to some solution of .

Proof. Let be a solution of . By setting in we getand by taking in we obtainThe last two inequalities yieldUsing the -strong monotonicity of we writeTherefore, the two previous inequalities yieldObserve thatHence,and soUsing our assumptions on the constants , , , and , we haveand hence we obtainSince by our assumptions we have , then the function is nondecreasing and henceUsing the equalityonce again yields . Therefore, (88) ensuresFinally, the conclusion of the theorem follows the same lines as in the proof of Theorem 12.

Remark 19. The same observation in Remark 16 holds for the previous theorem; that is, in the case when the ball compactness of is not needed and that the whole sequence converges to the solution , whenever the space is assumed to satisfy the -uniform convexity instead of the -uniform smoothness.

Corollary 20. Let , be a -uniformly smooth Banach space, let be a closed convex nonempty subset of , and let be a convex bifunction. Let be a sequence generated by . Assume that(1) is ball compact; that is, is compact for any ;(2)the solution set of is nonempty;(3) is monotone over , that is, (4) is upper semicontinuous with respect to the first variable over ;(5)there exists such that for all .Then, there exists subsequence of that converges to which solves .

Proof. It follows directly from the previous theorem with the constants and .

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The author would like to thank the referee for carefully reading the paper and for the well done report. This research was supported by the NSTIP strategic technologies program in Saudi Arabia, Award no. 11-MAT1916-02. Also, the author would like to thank Professor Hong-Kun Xu for the fruitful discussion on the subject of this research paper.