Convergence Theorems for Bregman K-Mappings and Mixed Equilibrium Problems in Reflexive Banach Spaces

Abstract

We introduce a new mixed equilibrium problem with a relaxed monotone mapping in a reflexive Banach space and prove the existence of solution of the equilibrium problem. Using Bregman distance, we introduce the concept of Bregman -mapping for a finite family of Bregman quasiasymptotically nonexpansive mappings and show the fixed point set of the Bregman -mapping is the set of common fixed points of . Using the Bregman -mapping, we introduce an iterative sequence for finding a common point in the set of a common fixed points of the finite family of Bregman quasiasymptotically nonexpansive mappings and the set of solutions of some mixed equilibrium problems. Strong convergence of the iterative sequence is proved. Our results generalise and improve many recent results in the literature.

1. Introduction

Let be a real Banach space and let be the dual of . Let be a nonempty closed and convex subset of . A mapping is called nonexpansive if . A point is a fixed point of if . We denote by the fixed point set of ; that is, . Let be a bifunction. The equilibrium problem with respect to and in the sense of Blum and Oettli [1] is to find such thatThe set of solutions of equilibrium problem is denoted by ; that is,In order to solve equilibrium problem (1), the bifunction is usually assumed to satisfy the following conditions: (C1) for all ;(C2) is monotone; that is, for all ;(C3)for all , ;(C4)for all is convex and lower semicontinuous.Fang and Huang [2] introduced the concept of relaxed monotone mapping for solving mixed equilibrium problems.

A mapping is said to be relaxed monotone (see also [3]) if there exist a mapping and a function with for all and , where such thatParticularly if and , where and are two constants, then is called monotone; see [4, 5].

Fang and Huang [2] proved that under some suitable assumptions, the following variational inequality is solvable: find such thatwhere is a function from to . They also proved that the following inequality is equivalent to variational inequality (4): find such thatThe mixed equilibrium problem (see [6, 7]) is to find such thatWe denote the set of solutions of mixed equilibrium problem (6) by . It is easily seen that if , then mixed equilibrium problem (6) reduces to variational inequality (4). In the case of and , then, coincides with .

Equilibrium problems and mixed equilibrium problems have been used as tools for solving problems arising from linear and nonlinear programming, optimization problems, variational inequalities, fixed point problems, and also problems in physics, economics, engineering, and so forth (see, e.g., [1, 6, 8–15] and the references therein).

Let be the normalised duality mapping defined by where denotes the generalised duality pairing. It is well known that if is smooth, strictly convex, and reflexive, then is single-valued, one-to-one, and onto.

Let be a convex function. We denote by the domain of ; that is, . The function is said to be coercive if . is said to be strongly coercive if . The Fenchel conjugate of is the function defined by The subdifferential of is a mapping defined by It is well known that (see [16]) if and only if for all . It is also known that if is a proper, convex, and lower semicontinuous function, then is a proper, convex, and lower semicontinuous function; see, for example, [17].

For any convex function , let and The right-hand derivative of at in the direction is defined byThe function is said to be Gâteaux differentiable at if exists for any . In this case coincides with , the value of the gradient of at . is said to be Gâteaux differentiable if it is Gâteaux differentiable at each . If the limit in (10) is attained uniformly in , then is said to be Fréchet differentiable at . is said to be uniformly Fréchet differentiable on a subset of if the limit in (10) is attained uniformly for every and . We know that if is uniformly Fréchet differentiable on bounded subset of , then is uniformly continuous on bounded set of (see, e.g., [18]).

The function is said to be essentially smooth if is both bounded and single-valued on its domain. It is called essentially strictly convex if is locally bounded on its domain and is strictly convex on every convex subset of . is said to be a Legendre function if it is both essentially smooth and essentially strictly convex. When the subdifferential of is single-valued, it coincides with the gradient; that is, ; see, for example, [19].

For a Legendre function , the following properties are well known:(i) is essentially smooth if and only if is essentially strictly convex; see [16];(ii); see [20];(iii) is Legendre if and only if is Legendre function; see [16];(iv)if is Legendre function, then is bijection satisfying , , and ; see [16].If is smooth and strictly convex, the function , , is Legendre function; see, for example, [21]. In this case , . In particular if is a Hilbert space we have , the identity mapping.

Let be a convex and Gâteaux differentiable function. The function defined byis called Bregman distance corresponding to ; see [22, 23]. It follows from the strict convexity of that and if and only if ; see [24].

Bregman projection with respect to of onto the nonempty closed convex subset of is the unique vector satisfying

Remark 1. If is smooth and strictly convex Banach space and , then we have and hence which is the Lyapunov function introduced by Alber [25] and the Bregman projection reduces to the generalised projection which is defined by If , a Hilbert space, then the Bregman projection reduces to the metric projection of onto .

Observing (11), we havewhich is called the three-point identity.

Let be a convex, Legendre, and Gâteaux differentiable function. Following [23, 25] we make use of the function associated with defined byThen is nonnegative and and Also from definition (15), it is obvious that and is convex in the second variable. Therefore for and , we haveMoreover by subdifferential inequality [26], we haveRecall that a mapping is said to be -quasi nonexpansive if and and . is -quasiasymptotically nonexpansive if and there exists a real sequence such that as and and . is called Bregman quasi nonexpansive if and and . is Bregman quasiasymptotically nonexpansive if and there exists a real sequence such that as and and . is said to be closed if for any sequence with and , .

It is worth mentioning that several iterative schemes have been constructed and proposed for finding points which solve fixed point problems and mixed equilibrium problems with relaxed monotone mappings in various settings. In 2010 Wang et al. [7] introduced the following iterative scheme for finding a common element of the set of solutions of a mixed equilibrium problem with relaxed monotone mapping and the set of fixed points of nonexpansive mappings in Hilbert spaces:where is a relaxed monotone mapping and is a nonexpansive mapping. Under some mild conditions on the three control sequences , , and , they obtained strong convergence of scheme (18) to common solution of mixed equilibrium problems and fixed point of nonexpansive mapping.

Recently, Chen et al. [27] introduced a new mixed equilibrium problem with the relaxed monotone mapping in uniformly convex and uniformly smooth Banach spaces and proved the existence of solutions of the mixed equilibrium problem. They also proposed the following iterative scheme to find the common element of the set of solutions of the mixed equilibrium problem and the set of fixed points of a quasi--nonexpansive mapping:where is a quasi--nonexpansive mapping from into itself and is the normalised duality mapping. Under some assumptions on the parameter sequences and , they obtained strong convergence of scheme (19) to common solution of mixed equilibrium problems and fixed point of nonexpansive mapping.

Motivated and inspired by the above results, in this paper we introduce and prove the existence of solutions of the mixed equilibrium problem with relaxed monotone mapping in reflexive Banach spaces. Using Bregman distance, we introduce the concept of Bregman -mapping of a finite family of Bregman quasiasymptotically nonexpansive mappings and propose an iterative sequence for finding a common element of the set of fixed points of a finite family of Bregman quasiasymptotically nonexpansive mappings and the set of solutions of mixed equilibrium problem.

2. Preliminaries

Let be a convex and Gâteaux differentiable function. The modulus of total convexity of at is the function defined by The function is totally convex at if for all and is totally convex if it is totally convex at each point . Let be a bounded subset of . For , define a functional on , defined by is totally convex on bounded set if for any bounded subset of and , where is the total convexity of the function on the set .

Let , for all and . The function is bounded if is bounded for all and is uniformly convex on bounded subsets of [28] if the function defined by satisfies where is called the gauge of uniform convexity of .

The gauge of uniform smoothness of is the function defined by The function is said to be uniformly smooth on if . We know that from [28, 29] is totally convex on bounded sets if and only if is uniformly convex on bounded sets.

Definition 2 (see [2]). Let be a Banach space with the dual and be a nonempty closed convex subset of . Let and be two mappings. Then is said to be -hemicontinuous if for any fixed , the function defined by is continuous at .

Definition 3 (see [30]). Let be a Banach space with the dual and be a nonempty subset of . A mapping is called a KKM mapping if for any finite set one has .

Remark 4. Observe that from Definition 3, if is a KKM mapping and such that for all , then is a KKM mapping.

In the sequel we will need the following lemmas.

Lemma 5 (see [31]). Let be a nonempty subset of a Hausdorff topological vector space and let be a mapping. If is closed in for all and compact for some , then .

Lemma 6 (see [32]). If , then the following statements are equivalent: (i)the function is totally convex at ;(ii)for any sequence

Recall that a function is called sequentially consistent [29] if for any two sequences and in such that is bounded,

Lemma 7 (see [24]). The function is totally convex on bounded sets if and only if is sequentially consistent.

Lemma 8 (see [26]). Let be a Legendre function such that is bounded on bounded subsets of . Let . If the sequence is bounded, then the sequence is bounded.

Lemma 9 (see [33]). Let be a constant and let be a uniformly convex function on bounded subsets of . Then for any and ,where is the gauge of the uniform convexity of .

Lemma 10 (see [34]). Let be a uniformly Fréchet differentiable function and bounded on bounded subsets of . Then is uniformly continuous on bounded subsets of from the strong topology of to strong topology of .

Lemma 11 (see [28]). Let be a strongly coercive function. If is uniformly continuous on bounded subsets of , then is uniformly convex on bounded subsets of .

Lemma 12 (see [28]). Let be a convex function which is bounded on bounded subsets of . Then the following assertions are equivalent: (i) is strongly coercive and uniformly convex on bounded subsets of ;(ii) is Fréchet differentiable and is uniformly norm-to-norm continuous on bounded subsets of .

Lemma 13 (see [29]). Let be a nonempty closed and convex subset of . Let be a Gâteaux differentiable and totally convex function. Let . Then (i) if and only if for all ;(ii) for all .

Lemma 14 (see [35]). Let be a reflexive Banach space and be a nonempty closed and convex subset of . Let be a Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex which is bounded on bounded subsets of . Let be a closed and Bregman quasiasymptotically nonexpansive mapping. If , then it is closed and convex.

Lemma 15. Let be a reflexive Banach space and be a nonempty closed and convex subset of . Let be a Legendre, uniformly Fréchet differentiable, strongly coercive, and totally convex function on bounded subsets of . Then the Bregman projection is continuous.

Proof. Let be a sequence in such that as . Let and . By Lemma 13(ii), we haveFrom inequality (28), we have Since converges, it is bounded and using the above inequality it follows that is bounded. The function is strongly coercive and totally convex which is bounded on bounded subsets of ; therefore in view of Lemma 12   is uniformly norm-to-norm continuous on bounded subsets of and consequently is bounded. Hence by Lemma 8 we obtain that is bounded.
Since we have . Therefore Since is uniformly Fréchet differentiable on bounded subsets of , it follows that is uniformly continuous on bounded subsets of (see, e.g., [18]). Thus, taking liminf as of both sides of the above inequality, we obtain Now let . Using (28), we have where is some natural number. As and , we obtain By total convexity of , we get as . This completes the proof.

3. Main Results

Lemma 16. Let be a nonempty, closed, and convex subset of a reflexive Banach space with the dual . Let be a convex and Gâteaux differentiable function. Let be -hemicontinuous and relaxed monotone mapping and be a bifunction satisfying and . Let be proper, convex, and lower semicontinuous. For and , suppose the following conditions hold: (i);(ii) is convex for fixed .Then problems (34) and (35) are equivalent.
Find such thatFind such that

Proof. Suppose (34) holds. Let be a solution of (34); then Since is relaxed monotone, we obtain This shows that is a solution of (35).
Conversely, suppose (35) holds; that is, is a solution of (35). Let such that ; then . Let , . Since is a solution of (35), we have Hence by (C1), (C4), (i), and (ii), we obtain Thus, Since is -hemicontinuous and , by allowing we obtain This shows that is a solution of (34)

Lemma 17. Let be a reflexive Banach space with the dual and let be a nonempty closed, convex, and bounded subset of . Let be a Legendre function and uniformly Fréchet differentiable on bounded subsets of . Let be -hemicontinuous and relaxed monotone mapping and be a bifunction satisfying and . Let be proper, convex lower semicontinuous. For and , suppose (i) for all ,(ii),(iii) is convex and lower semicontinuous for fixed ,(iv) is weakly lower semicontinuous.Then there exists such that

Proof. Define two set-valued mappings as follows: We claim is a KKM mapping. By contradiction suppose then there do not exist and such that and . This implies It follows that which is a contradiction. Thus is a KKM mapping.
Next we show that .
Let . Then Since is relaxed monotone, we have Showing that for all . By Remark 4 it follows that is a KKM mapping.
We claim also that is closed in the weak topology of . Let and be the weak closure point of . Since is reflexive, there exists a sequence such that as Observe that is equivalent toBy (iii) and (iv) and taking liminf as of both sides of (49) we obtain That is, . This implies is weakly closed for all . Since is weakly compact, then is weakly compact in for all .
It is clear that the solution sets of problem (34) and (35) are and . Using Lemmas 16 and 5 we obtain Hence there exists such that This completes the proof.

Lemma 18. Let be a reflexive Banach space with the dual and let be a nonempty closed, convex, and bounded subset of . Let be Legendre and Gâteaux differentiable function. Let be -hemicontinuous and relaxed monotone mapping and be a bifunction satisfying , , and . Let be proper, convex, and lower semicontinuous. For and , define a map by Assume that (i);(ii) is convex and lower semicontinuous for fixed ;(iii) is weakly lower semicontinuous;(iv).Then (1) is single-valued;(2) is a Bregman firmly nonexpansive type mapping; that is,(3);(4) is a Bregman quasi nonexpansive satisfying (5) is closed and convex.

Proof. First we show that is single-valued. Let ; thenBy using (C2), adding (56) yields By (i) we have Since is relaxed monotone, we obtain ThusInterchanging and in (60), we haveAdding (60) and (61), we have Hence By (iv), we have Thus,Since is convex and Gâteaux differentiable we haveBy (65) and (66) we obtain Since is Legendre function, then .
Next we show that is Bregman firmly nonexpansive type. Let ; thenAdding (68), using (i) and (C2), we obtain so that Since is relaxed monotone and , we haveAlso interchanging the roles of and in (71) and applying (iv), we have Hence,showing that is Bregman firmly nonexpansive type.
We now show that . Indeed Next we prove that is Bregman quasi nonexpansive mapping.
Sincewe have