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Journal of Applied Mathematics

Volume 2013 (2013), Article ID 962691, 7 pages

http://dx.doi.org/10.1155/2013/962691

## Strong Convergence Theorem for Bregman Strongly Nonexpansive Mappings and Equilibrium Problems in Reflexive Banach Spaces

^{1}Department of Mathematics, Yibin University, Yibin, Sichuan 644007, China^{2}College of Statistics and Mathematics, Yunnan University of Finance and Economics, Kunming, Yunnan 650221, China

Received 30 April 2013; Accepted 10 June 2013

Academic Editor: Wei-Shih Du

Copyright © 2013 Jinhua Zhu 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.

#### Abstract

By using a new hybrid method, a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of Bregman strongly nonexpansive mappings in a reflexive Banach space is proved.

#### 1. Introduction

Throughout this paper, we denote by and the set of all real numbers and all nonnegative real numbers, respectively. We also assume that is a real reflexive Banach space, is the dual space of , is a nonempty closed convex subset of , and is the pairing between and . Let be a bifunction from . The equilibrium problem is to find The set of such solutions is denoted by .

Recall that a mapping is said to be nonexpansive, if We denote by the set of fixed points of .

Numerous problems in physics, optimization, and economics reduce to find a solution of the equilibrium problem. Some methods have been proposed to solve the equilibrium problem in a Hilbert spaces; see, for instance, Blum and Oettli [1], Combettes and Hirstoaga [2], and Moudafi [3]. Recently, Tada and Takahashi [4, 5] and S. Takahashi and W. Takahashi [6] obtained weak and strong convergence theorems for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping in a Hilbert space. In particular, Tada and Takahashi [4] established a strong convergence theorem for finding a common element of two sets by using the hybrid method introduced by Nakajo and Takahashi [7]. The authors also proved such a strong convergence theorem in a uniformly convex and uniformly smooth Banach space.

In this paper, motivated by Takahashi et al. [8], we prove a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a Bregman strongly nonexpansive mapping in a real reflexive Banach space by using the shrinking projection method. Using this theorem, we obtain two new strong convergence results for finding a solution of an equilibrium problem and a fixed point of Bregman strongly nonexpansive mappings in a real reflexive Banach space.

#### 2. Preliminaries and Lemmas

In the sequel, we begin by recalling some preliminaries and lemmas which will be used in the proof.

Let be a real reflexive Banach space with the norm and the dual space of . Throughout this paper, is a proper, lower semicontinuous, and convex function. We denote by the domain of , that is, the set .

Let . The subdifferential of at is the convex set defined by where the Fenchel conjugate of is the function defined by We know that the Young-Fenchel inequality holds:

A function on is coercive [9] if the sublevel set of is bounded; equivalently,

A function on is said to be strongly coercive [10] if

For any and , the right-hand derivative of at in the direction is defined by

The function is said to be Gâteaux differentiable at if exists for any . In this case, coincides with , the value of the gradient of at . The function is said to be Gâteaux differentiable if it is Gâteaux differentiable for any . The function is said to be Fréchet differentiable at if this limit is attained uniformly in . Finally, is said to be uniformly Fréchet differentiable on a subset of if the limit is attained uniformly for and . It is known that if is Gâteaux differentiable (resp., Fréchet differentiable) on , then is continuous and its Gâteaux derivative is norm-to-weak continuous (resp., continuous) on (see also [11, 12]). We will need the following result.

Lemma 1 (see [13]). *If is uniformly Fréchet differentiable and bounded on bounded subsets of , then is uniformly continuous on bounded subsets of from the strong topology of to the strong topology of .*

*Definition 2 (see [14]). *The function is said to be(i)essentially smooth, if is both locally bounded and single valued on its domain,(ii)essentially strictly convex, if is locally bounded on its domain and is strictly convex on every convex subset of ,(iii)Legendre if it is both essentially smooth and essentially strictly convex.

*Remark 3. *Let be a reflexive Banach space. Then we have the following.(i) is essentially smooth if and only if is essentially strictly convex (see [14, Theorem 5.4]).(ii) (see [12]).(iii) is Legendre if and only if is Legendre (see [14, Corollary 5.5]).(iv)If is Legendre, then is a bijection satisfying , ran = , and ran (see [14, Theorem 5.10]).

Examples of Legendre functions were given in [14, 15]. One important and interesting Legendre function is when is a smooth and strictly convex Banach space. In this case, the gradient of is coincident with the generalized duality mapping of ; that is, . In particular, the identity mapping in Hilbert spaces. In the rest of this paper, we always assume that is Legendre.

Let be a convex and Gâteaux differentiable function. The function defined as is called the Bregman distance with respect to [16].

Recall that the Bregman projection [17] of onto the nonempty closed and convex set is the necessarily unique vector satisfying Concerning the Bregman projection, the following are well known.

Lemma 4 (see [18]). *Let be a nonempty, closed, and convex subset of a reflexive Banach space . Let be a Gâteaux differentiable and totally convex function and let . Then*(a)* if and only if , for all .*(b)

Let be a convex and Gâteaux differentiable function. The modulus of total convexity of at is the function defined by

The function is called totally convex at if whenever . The function is called totally convex if it is totally convex at any point and is said to be totally convex on bounded sets if for any nonempty bounded subset of and , where the modulus of total convexity of the function on the set is the function defined by

The next lemma will be useful in the proof of our main results.

Lemma 5 (see [19]). *If , then the following statements are equivalent.*(i)*The function is totally convex at .*(ii)*For any sequence ,
*

Recall that the function is called sequentially consistent [18] if, for any two sequences and in such that the first one is bounded,

Lemma 6 (see [20]). *The function is totally convex on bounded sets if and only if the function is sequentially consistent.*

Lemma 7 (see [21]). *Let be a Gâteaux differentiable and totally convex function. If and the sequence is bounded, then the sequence is bounded too.*

Lemma 8 (see [21]). *Let be a Gâteaux differentiable and totally convex function, , and let be a nonempty, closed, and convex subset of . Suppose that the sequence is bounded and any weak subsequential limit of belongs to . If for any , then converges strongly to .*

Let be a convex subset of and let be a self-mapping of . A point is called an asymptotic fixed point of (see [22, 23]) if contains a sequence which converges weakly to such that . We denote by the set of asymptotic fixed points of .

*Definition 9. *A mapping with a nonempty asymptotic fixed point set is said to be(i)Bregman strongly nonexpansive (see [24, 25]) with respect to if
and if, whenever is bounded, and
it follows that
(ii)Bregman firmly nonexpansive [26] if, for all ,
or, equivalently,

The existence and approximation of Bregman firmly nonexpansive mappings were studied in [26]. It is also known that if is Bregman firmly nonexpansive and is Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of , then and is closed and convex (see [26]). It also follows that every Bregman firmly nonexpansive mapping is Bregman strongly nonexpansive with respect to .

Lemma 10 (see [27]). *Let be a real reflexive Banach space and a proper lower semicontinuous function; then is a proper weak lower semicontinuous and convex function. Thus, for all , we have
*

In order to solve the equilibrium problem, let us assume that a bifunction satisfies the following conditions [28]:(*A*_{1}), for all .(*A*_{2}) is monotone; that is, , for all .(*A*_{3}) lim for all .(*A*_{4}) The function is convex and lower semicontinuous.

The resolvent of a bifunction [29] is the operator defined by

From Lemma 1 in [24], if is a strongly coercive and Gâteaux differentiable function and satisfies conditions (*A*_{1}–*A*_{4}), then dom . We also know the following lemma which gives us some characterizations of the resolvent .

Lemma 11 (see [24]). *Let be a real reflexive Banach space and a nonempty closed convex subset of . Let be a Legendre function. If the bifunction satisfies the conditions ( A_{1})–(A_{4}), then the followings hold:*(i)

*is single-valued;*(ii)

*is a Bregman firmly nonexpansive operator;*(iii)

*;*(iv)

*is a closed and convex subset of ;*(v)

*for all and for all , we have*

#### 3. Strong Convergence Theorem

In this section, we proved a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and a fixed point of Bregman strongly nonexpansive mapping in a real reflexive Banach space by using the shrinking projection method.

Theorem 12. *Let be a nonempty, closed, and convex subset of a real reflexive Banach space and a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of . Let be a bifunction from to satisfying ( A_{1})–(A_{4}) and let be a Bregman strongly nonexpansive mapping from into itself such that and . Let be a sequence generated by and
*

*for every , where satisfies . Then, converges strongly to , where is the Bregman projection of onto .*

*Proof. *We divide the proof of Theorem 12 into five steps.

(I) We first prove that and both are closed and convex subset of for all . In fact, it follows from Lemma 11 and by Reich and Sabach [26] that and both are closed and convex. Therefore, is a closed and convex subset in . Furthermore, it is obvious that is closed and convex. Suppose that is closed and convex for some . Since the inequality is equivalent to

Therefore, we have

This implies that is closed and convex. The desired *conclusions* are proved. These in turn show that and are well defined.

(II) we prove that for all .

Indeed, it is obvious that . Suppose that for some . Let ; since , by Lemma 11 and (21), we have
Hence, we have . This implies that
So, is well defined.

(III) We prove that is a bounded sequence in .

By the definition of , we have for all . It follows from Lemma 4(b) that
This implies that is bounded. By Lemma 7, is bounded. Since is uniformly Fréchet differentiable and bounded on bounded subsets of , by Lemma 1 is uniformly continuous and bounded on bounded subsets of . This implies that is bounded.

(IV) Now we proved that .

From and , we have
Thus, is nondecreasing. So, the limit of exists. Since for all , we have . From , we have
Therefore, we have
From Lemma 5, we have
So, we have
This means that the sequence is bounded. Since is uniformly Fréchet differentiable, it follows from Lemma 1 that is uniformly continuous. Therefore, we have

Since is uniformly Fréchet differentiable on bounded subsets of , then is uniformly continuous on bounded subsets of (see [30, Theorem 1.8]). It follows that

From the definition of the Bregman distance, we obtain that
for any .

It follows from (34)–(37) that
On the other hand, from and Lemma 11(v), for any we have that
So, we have from (38) that
From Lemma 5, we have
So, from (34) and (41),we have
This means that the sequence is bounded. Since is uniformly Fréchet differentiable, it follows from Lemma 1 that
Since is uniformly Fréchet differentiable on bounded subsets of , then is uniformly continuous on bounded subsets of (see [30]). It follows that

From the definition of the Bregman distance, we obtain that
for any .

It follows from (42) to (45) that

On the other hand, for any we have
This together with (46), (16), and shows that
Since is Bregman strongly nonexpansive, it follows from (48) that

(V) Next, we prove that every weak subsequential limit of belongs to .

Since is bounded, there exists a subsequence of such that . Since is a Bregman strongly nonexpansive mapping with , we have .

From and (34), we have .

By , we have
Replacing by , we have from (*A*_{2}) that
Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. So, letting , we have from (35), (43), and (*A*_{4}) that
For and , letting , there are and . By condition (*A*_{1}) and (*A*_{4}), we have
Dividing both sides of the above equation by , we have , for all . Letting , from condition (*A*_{3}), we have
Therefore, .

(VI) Now, we prove .

Let . From , we have . Therefore, Lemma 8 implies that converges strongly to , as claimed. This completes the proof of Theorem 12.

Corollary 13. *Let be a nonempty, closed, and convex subset of a real reflexive Banach space and a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of . Let be a bifunction from to satisfying ( A_{1})–(A_{4}). Let be a sequence generated by , and
*

*for every . Then, converges strongly to , where is the Bregman projection of onto .*

*Proof. *Putting in Theorem 12, we obtain Corollary 13.

Corollary 14. *Let be a nonempty, closed, and convex subset of a real reflexive Banach space and a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of . let be a Bregman strongly nonexpansive mapping from into itself such that and . Let be a sequence generated by , and
**
for every , where satisfies . Then, converges strongly to , where is the Bregman projection of onto .*

*Proof. *Putting for all in Theorem 12, we obtain Corollary 14.

#### Acknowledgments

This work was supported by Scientific Research Fund of Sichuan Provincial Education Department (11ZB146) and Yunnan University of Finance and Economics.

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