## Advance in Nonlinear Analysis: Algorithm, Convergence and Applications

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Shenghua Wang, Shin Min Kang, "Strong Convergence Iterative Algorithms for Equilibrium Problems and Fixed Point Problems in Banach Spaces", *Abstract and Applied Analysis*, vol. 2013, Article ID 619762, 9 pages, 2013. https://doi.org/10.1155/2013/619762

# Strong Convergence Iterative Algorithms for Equilibrium Problems and Fixed Point Problems in Banach Spaces

**Academic Editor:**Yisheng Song

#### Abstract

We first introduce the concept of Bregman asymptotically quasinonexpansive mappings and prove that the fixed point set of this kind of mappings is closed and convex. Then we construct an iterative scheme to find a common element of the set of solutions of an equilibrium problem and the set of common fixed points of a countable family of Bregman asymptotically quasinonexpansive mappings in reflexive Banach spaces and prove strong convergence theorems. Our results extend the recent ones of some others.

#### 1. Introduction

Let be a real reflexive Banach space with norm and the dual space of equipped with the inducted norm . Throughout this paper, is a proper, lower semicontinuous, and convex function and the Fenchel conjugate of is the function defined by

We denote by the domain of , that is, the set .

Let be a nonempty, closed, and convex subset of and a nonlinear mapping. The fixed points set of is denoted by

Recall that a mapping is said to be *nonexpansive* if, for each ,

Nakajo-Takahashi [1] introduced the following hybrid method which is the so-called * CQ-method* for a nonexpansive mapping in a Hilbert space :
where and is the metric projection from onto a closed and convex subset of . They proved that the sequence generated by (4) converges strongly to a fixed point of under suitable conditions.

Takahashi et al. [2] introduced a new hybrid iterative scheme called the * shrinking projection method* for a nonexpansive mapping in a Hilbert space as follows:
where , and they proved that the sequence generated by (5) converges strongly to a fixed point of under suitable conditions.

In 2010, Reich and Sabach [3] introduced the following two hybrid iterative schemes for Bregman strongly nonexpansive mappings â€‰ in a reflexive Banach space : where is the Bregman projection with respect to from onto a closed and convex subset of . They proved that the sequence generated by both (6) and (7) converges strongly to a common fixed point of .

The construction of fixed points for Bregman-type mappings via iterative processes has been investigated in, for example, [4â€“8].

In this paper, we design a new hybrid iterative scheme for finding a common element of the set of solutions of an equilibrium problem and the set of common fixed points of a countable family of Bregman asymptotically quasinonexpansive mappings in reflexive Banach spaces and prove some strong convergence theorems. Our results extend the recent one of Reich and Sabach [3].

#### 2. Preliminaries

Let be a real Banach space. For any and , we define the *right-hand derivative* of at in the direction by

The function is said to be *GĂ˘teaux differentiable* at if exists for any . In this case, coincides with , the value of the *gradient * of at . The function is said to be *GĂ˘teaux differentiable* if it is GĂ˘teaux differentiable for any int dom . The function is said to be FrĂ©chet differentiable at if this limit is attained uniformly in . Finally, is said to be *uniformly FrĂ©chet differentiable* on a subset of if the limit is attained uniformly for and .

Let be a reflexive Banach space. The Legendre function is defined from a general Banach space into (see [9]). According to [9], the function is *Legendre* if and only if it satisfies the following conditions(L1) The interior of the domain of (denoted by ) is nonempty; is GĂ˘teaux differentiable on and dom .(L2) The interior of the domain (denoted by int dom) is nonempty; is GĂ˘teaux differentiable on int dom and dom .

Since is reflexive, we always have (see [10]). This fact, when combined with conditions (L1) and (L2), implies the following equalities:

Also, conditions (L1) and (L2), in conjunction with [9], imply that the functions and are strictly convex on the interior of their respective domains. Several interesting examples of the Legendre functions are presented in [9, 11]. Especially, the functions with are Legendre, where the Banach space is smooth and strictly convex and, in particular, a Hilbert space.

Let be a convex and GĂ˘teaux differentiable function. The function defined as
is called the *Bregman distance* with respect to [12].

By the definition, we know the following property (the three point identity): for any and ,

Recall that the *Bregman projection* [13] of onto the nonempty, closed, and convex subset of is the necessarily unique vector satisfying

Let be a convex and GĂ˘teaux differentiable function. The function is said to be * totally convex* at if its modulus of total convexity at , that is, the function defined by
is positive whenever . The function is said to be *totally convex* when it is totally convex at every point . In addition, the function is said to be *totally convex on bounded sets* if is positive for any nonempty bounded subset of and , where the modulus of total convexity of the function on the set is the function defined by

Some examples of the totally convex functions can be found in [14, 15].

Recall that the function is said to be *sequentially consistent* [15] if, for any two sequences and in such that the first is bounded,

Let be a nonempty, closed, and convex subset of and a bifunction that satisfies the following conditions: (C1) for all ; (C2) is monotone, that is, for all ; (C3) for all ; (C4) for all is convex and lower semicontinuous.

The equilibrium problem with respect to is as follows: find such that

The set of all solutions of (16) is denoted by . The resolvent of a bifunction [16] is the operator denoted by For any , there exists such that ; see [3].

Let be a convex subset of and a mapping. A point in the closure of is said to be an *asymptotic fixed point* of [17, 18] if contains a sequence which converges weakly to such that the strong . The set of asymptotic fixed points of will be denoted by . The mapping is called * Bregman quasi-nonexpansive* if and
is said to be * Bregman (quasi)-strongly nonexpansive* [6] with respcet to a nonempty if
for all and , and if whenever is bounded, , and
it follows that
The mapping is called *Bregman firmly nonexpansive* if
for all .

Next, we introduce a new mapping that is called Bregman asymptotically quasinonexpansive mapping which is a natural extension of Bregman quasinonexpansive mapping introduced by Reich and Sabach [3]. The mapping is said to be * Bregman asymptotically quasi-nonexpansive* if there exists a sequence satisfying such that, for every ,
Obviously, every Bregman quasinonexpansive mapping is a Bregman asymptotically quasi-nonexpansive one with .

Let be a Banach space and a nonempty subset of . The mapping is said to be * uniformly asymptotically regular* on if
The mapping is said to be *closed* if, for any sequence in such that and , .

The following is an important result which will be used in the next section.

Lemma 1. *Let be a reflexive Banach space and a GĂ˘teaux differentiable and Legendre function which is totally convex on bounded sets. Let be a nonempty, closed and convex subset of and a closed Bergman asymptotically quasi-nonexpansive mapping with the sequence such that as . Then is closed and convex.*

* Proof. * The closedness of comes directly from the closedness of . Now, for arbitrary , , put . We prove that . Indeed, from the definition of , we see that
This implies that . It follows from Lemma 3 below that
that is, as . In view the closedness of , we can obtain the desired conclusion. This completes the proof.

Finally, we state some lemmas that will used in the proof of main results in next section.

Lemma 2 (see [7]). * If is uniformly FrĂ©chet 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 . *

Lemma 3 (see [14]). *The function is totally convex on bounded sets if and only if it is sequentially consistent. *

Lemma 4 (see [15]). *Suppose that is GĂ˘teaux differentiable and totally convex on . Let and a nonempty, closed, and convex subset of . If , then the following conditions are equivalent.*(i)* The vector is the Bregman projection of onto with respect to .*(ii)* The vector is the unique solution of the variational inequality.
*(iii)* The vector is the unique solution of the inequality
*

Lemma 5 (see [6]). * Let be a GĂ˘teaux differentiable and totally convex function. If and the sequence is bounded, then the sequence is also bounded. *

Lemma 6 (see [3]). *Let be a coercive i.e., and GĂ˘teaux differentiable function. Let be a closed and convex subset of . If the bifunction satisfies conditions â€“, then *(1)* is single-valued; *(2)* is a Bregman firmly nonexpansive mapping; *(3)*the set of fixed points of is the solution set of the equilibrium problem, that is, ; *(4)* is a closed and convex subset of ; *(5)*for all and , one has
*

#### 3. Main Results

Now, we give our main theorems.

Theorem 7. * Let be a reflexive Banach space and a coercive Legendre function which is bounded, uniformly FrĂ©chet differentiable, and totally convex on bounded subsets of . Let be a nonempty, closed, and convex subset of and a countable family of closed Bregman asymptotically quasi-nonexpansive mappings with the sequences such that for every . Let and suppose that . Let be a bifunction satisfying conditions (C1)â€“(C4). Assume that each is uniformly asymptotically regular and is nonempty and bounded. Let be a real sequence in with for every and for every . Let be a sequence generated by the following manner:
**
where for each . Then, defined by (30) converges strongly to as . *

*Proof. *First, we prove that the sequence is well defined. Note that
is
that is,
This shows that is closed and convex for every . From the definition of , it is easy to see that is closed and convex for every . For every and , Lemma 6 shows that and for any and . Hence,

Since for every , we have
This shows that for every . Thus for every . Further, we have for every . Thus the sequence is well defined.

From , by Lemma 4(iii) we have
for any . Hence the sequence is bounded. Therefore by Lemma 5 the sequence is bounded.

On the other hand, in view of and , from Lemma 4(iii) we have
that is,
Therefore the sequence is increasing, and since it is also bounded, exists. By the construction of , we have that and for any positive integer . It follows that
Letting in (39), we see that . It follows from Lemma 3 that as . Hence, is a Cauchy sequence. Since is a Banach space and is closed and convex, we can assume that
By taking in (39), we see that
Lemma 3 implies that
Since , we have
Then (41) implies that
Note that and , we have
for every . It follows from Lemma 3 that
for every . Note that
Combining (42) with (46), we see that
for every . This means that the sequence is bounded. Since is uniformly FrĂ©chet differentiable, it follows from Lemma 2 that
Since is uniformly FrĂ©chet differentiable, it is also uniformly continuous (see [19, Theorem 1.8, p.13]) and therefore
From the definition of the Bregman distance, we obtain that for every
for any . Since every sequence is bounded, is also bounded for every . Now from (48)â€“(50), we have
for any and for every .

In view of , by Lemma 6 (5) we have
Note that is bounded and as . It follows from (52) that
for every . Lemma 3 shows that
Note that . From (48) and (55) we get
for every . Note that

It follows from (40) and (56) that
for every . On the other hand, we have
Since every is uniformly asymptotically regular and (58), we obtain that, for every ,
that is, as . From the closedness of , we see that for every . Thus .

Next we prove that for every . Since is uniformly FrĂ©chet differentiable, is uniformly continuous. Thus, by (55) we have
Since , we have
We have from (C2) that
Letting , we have from (61) and (C4) that
For with and , let . Since and , we have and hence . So, from (C1) we have
Dividing by , we have
Letting , from (C3) we have
Therefore, . Thus .

Finally, we show that . Since for every , by Lemma 4(ii) we arrive at
Taking the limit as in (68), we obtain that
and hence by Lemma 4(ii). This completes the proof.

Corollary 8. *Let be a reflexive Banach space and a coercive Legendre function which is bounded, uniformly FrĂ©chet differentiable and totally convex on bounded subsets of . Let be a nonempty, closed, and convex subset of and a closed Bregman asymptotically quasi-nonexpansive mapping with the sequence such that . Let be a bifunction satisfying conditions (C1)â€“(C4). Assume that is uniformly asymptotically regular and is nonempty and bounded. Let be a sequence generated by the following manner:
**
where for each . Then, defined by (70) converges strongly to as . *

Since every Bregman quasi-nonexpansive mapping is Bregman quasi-asymptotically nonexpansive, we have the following results.

Corollary 9. * Let be a reflexive Banach space and let be a coercive Legendre function which is bounded, uniformly FrĂ©chet differentiable, and totally convex on bounded subsets of . Let be a nonempty, closed, and convex subset of . Let be a countable family of closed Bregman quasi-nonexpansive mappings and a bifunction satisfying conditions (C1)â€“(C4). Assume that . Let be a real sequence in with and for every . Let be a sequence generated by the following manner:
**
Then, defined by (71) converges strongly to as . *

Corollary 10. * Let be a reflexive Banach space and let be a coercive Legendre function which is bounded, uniformly FrĂ©chet differentiable, and totally convex on bounded subsets of . Let be a nonempty, closed, and convex subset of . Let be a closed Bregman quasi-nonexpansive mapping and a bifunction satisfying conditions (C1)â€“(C4). Assume that . Let be a sequence generated by the following manner:
**
Then, defined by (72) converges strongly to as *

*Remark 11. *Set for each and and for each and . Then and . Also, for every . Hence, and satisfy the conditions of Theorem 7.

*Remark 12. * It needs to notice that Corollaries 9 and 10 still hold if we replace the closedness of the mappings with .

In the equilibrium problem, the bifunction is usually required to satisfy conditions (C1)â€“(C4). But, if the condition (C3) is replaced with the following condition: for every fixed is continuous, then we have the following result:

Lemma 13. *Let be a coercive i.e., and GĂ˘teaux differentiable function. Let be a closed and convex subset of . If the bifunction satisfies conditions , , , and , then the mapping defined by (2.2) is closed. *

*Proof. * Let converge to and to . To end the conclusion, we need to prove that . Indeed, for each , Lemma 6 shows that there exists a unique such that , that is,

Since is uniformly FrĂ©chet differentiable, is uniformly continuous. So, taking the limit as in (73), by using we get
which implies that . This completes the proof.

If the bifunction satisfies conditions (C1), (C2), , and (C4) instead of (C1)â€“(C4), then we have a simple method to prove that in the proof of Theorem 7. Indeed, from the proof of Theorem 7, we see that

Note that as . This shows that as for every . It follows from the closedness of that . Lemma 6 shows that .

*Remark 14. *Obviously, the proof process of is simple if we replace condition (C3) with which is such that is closed. In fact, although condition is stronger than (C3), it is not easier to verify condition (C3) than to verify the condition . Hence, from this viewpoint, the condition is acceptable.

#### Acknowledgments

This work is supported by the Fundamental Research Fundsfor the Central Universities (Grant Number: 13MS109) and the HeBei Education 4 Department (Grant Number: 936101101).

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#### Copyright

Copyright © 2013 Shenghua Wang and Shin Min Kang. 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.