Journal of Applied Mathematics

Volume 2014 (2014), Article ID 193749, 10 pages

http://dx.doi.org/10.1155/2014/193749

## New Mixed Equilibrium Problems and Iterative Algorithms for Fixed Point Problems in Banach Spaces

^{1}School of Mathematics and Sciences, Shijiazhuang University of Economics, Shijiazhuang 050031, China^{2}Department of Mathematics and Physics, North China Electric Power University, Baoding 071003, China

Received 30 October 2013; Accepted 16 December 2013; Published 9 January 2014

Academic Editor: Li Wei

Copyright © 2014 Minjiang Chen 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

We first introduce a new mixed equilibrium problem with a relaxed monotone mapping in Banach spaces and prove the existence of solutions of the equilibrium problem. Then we introduce a new iterative algorithm for finding a common element of the set of solutions of the equilibrium problem and the set of fixed points of a quasi--nonexpansive mapping and prove some strong convergence theorems of the iteration. Our results extend and improve the corresponding ones given by Wang et al., Takahashi and Zembayashi, and some others.

#### 1. Introduction

Let be a Banach space with the dual space and let be a nonempty closed convex subset of . Let be a bifunction from to , where denotes the set of numbers. The *equilibrium problem* is to find such that

The set of solutions of the above equilibrium problem is denoted by .

In order to solve the equilibrium problem, the bifunction is usually to be assumed that following conditions are satisfied:(A1) for all ;(A2) is monotone; that is, for all ;(A3)for all , (A4)for all , is convex and lower semicontinuous.

Recently, Takahashi and Zembayashi [1] extended the equilibrium problems and fixed point problems from Hilbert spaces to Banach spaces. More precisely, they gave the following iterative scheme: where is a relatively nonexpansive mapping from into itself such that , is the duality mapping on , and satisfies and for some . They proved that the sequence generated by (3) converges strongly to , where is the generalized projection of onto .

Very recently, Qin et al. [2] introduced the following hybrid algorithm to solve the equilibrium problems and fixed point problems for quasi--nonexpansive mappings in a uniformly convex and uniformly smooth Banach space with a nonempty closed convex subset of : where and are two quasi--nonexpansive mappings from into itself such that , , , and are three sequences in satisfying some certain conditions. They proved that the sequence generated by (4) converges strongly to , where is the generalized projection of onto .

In [3], Fang and Huang introduced a concept called a relaxed --monotone mapping.

A mapping is said to be *relaxed* *-**-monotone* if there exist a mapping and a function with for all and , where is a constant, such that

Especially, if for all and , where and are two constants, then is said to be *-monotone* (see, e.g., [4–6]). They proved that, under some suitable assumptions, the following variational inequality is solvable: find such that
where is a function from to . They also proved that the variational inequality (6) is equivalent to the following: find such that

In this paper, let us denote the set of solutions of the variational inequality (6) or (7) by .

It is worthy to notice that it is possible that is a singleton set, includes the finite elements or infinite elements. To show this, we see the following example.

*Example 1. *Let and . Define the mappings by for all , by for all , and by for all . Then the mapping is a relaxed --monotone mapping. In fact, for all , , we have

Hence, is a relaxed --monotone mapping. Now by defining the different function we find the solution set of (6). First we compute, for , ,
(i)Let the function for all . It is easy to see that the solution set . That is, only when , (6) holds for all .(ii)Let the function and for all . By (9), it is not hard to see that the solution set .(iii)Let the function for all where and for all . Then by (9) we see that . In this case, includes the unfinite elements.

Recently, Wang et al. [7] proposed the following equilibrium problem in a Hilbert space with a nonempty closed convex bounded subset : find such that where is a relaxed --monotone mapping. They introduced a new algorithm for solving the equilibrium problem in Hilbert spaces.

In [3], Fang and Huang did not give the algorithm to solve the variational inequality (6) or (7). In this paper, motivated and inspired by the above results, we introduced a new mixed equilibrium problem with the relaxed monotone mapping and prove the existence of solutions of the mixed equilibrium problem. Then we propose an 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 in Banach spaces. In particular, the variational inequality (6) or (7) may be solved by the algorithm proposed in this paper. Our results extend and improve the corresponding ones given by Takahashi and Zembayashi [1], Fang and Huang [3], and Wang et al. [7].

#### 2. Preliminaries

A Banach space is said to be *strictly convex* if for all with and . It is said to be *uniformly convex* if for any two sequences and in such that and .

Let be the unit sphere of . A Banach space is said to be smooth provided
exists for each . It is said to be *uniformly smooth* if the limit is attained uniformly for .

Let be the topological dual space of and be the normalized duality mapping from into given by

It is well known that if is uniformly convex, then is uniformly continuous on bounded subsets of and if is uniformly smooth, then is uniformly convex.

Recently, Alber [8] introduced a generalized projection operator in a Banach space with a nonempty closed convex subset , which is an analogue of the metric projection in Hilbert spaces:

let be a function defined by

The *generalized projection* is a mapping that assigns to an arbitrary point the minimum point of the functional ; that is, , where is a solution to the minimization problem

The existence and uniqueness of follows from the properties of the function and the strict monotonicity of the mapping (see, e.g., [8–12]).

Let be a smooth, strictly convex, and reflexive Banach space . Then, for all , we have the following [8, 9, 11]:

Let be a mapping from into itself. A point is said to be an *asymptotic fixed point* of [13] if contains a sequence which converges weakly to such that

The set of asymptotic fixed points of is denoted by . The mapping is said to be *relatively nonexpansive* [14–16] if

We recall that a mapping is said to be *nonexpansive* if for all , is said to be *-nonexpansive* if for all , and is said to be *quasi-**-nonexpansive* if

Notice that the class of quasi--nonexpansive mappings is more general than the class of relatively nonexpansive mappings [14–16] which requires the strong restriction that . Some examples of quasi--nonexpansive mappings may be found in [2].

A mapping is said to be closed if for any sequence such that and , then .

The following lemmas are needed for the proof of our main results in next section.

Lemma 2 (see [10]). *Let be a uniformly convex and smooth Banach space and let , be two sequences of . If and either or is bounded, then .*

Lemma 3 (see [8]). *Let be a nonempty closed convex subset of a smooth Banach space and . Then if and only if
*

*Lemma 4 (see [8]). Let be a reflexive, strictly convex, and smooth Banach space, let be a nonempty closed convex subset of and . Then
*

*The following lemma is contained implicitly in Step 1 of Theorem 3.1 of [17].*

*Lemma 5 (see [17]). Let be a uniformly convex and smooth Banach space, let be a nonempty closed convex subset of , and let be a closed quasi--nonexpansive mapping from into itself. Then is a closed convex subset of .*

*Lemma 6 (see [18]). Let be a uniformly convex Banach space and let be a closed ball of . Then there exists a continuous strictly increasing convex function with such that
for all and with .*

*3. Main Results*

*3. Main Results*

*For our main results, we introduce some definitions and lemmas as follows.*

*Definition 7 (see [3]). *Let be a Banach space with the dual space and let be a nonempty subset of . Let and be two mappings. The mapping is said to be *-hemicontinuous* if, for any fixed , the function defined by is continuous at .

*Definition 8. *Let be a Banach space with the dual space and let be a nonempty subset of . A mapping is called a KKM *mapping* if, for any ,
where denotes the family of all the nonempty subsets of .

*Lemma 9 (see [19]). Let be a nonempty subset of a Hausdorff topological vector space and let be a KKM mapping. If is closed in for all in and compact for some , then .*

*Lemma 10. Let be a reflexive Banach space with the dual space and let be a nonempty closed convex subset of . Let be an -hemicontinuous and relaxed --monotone mapping. Let be a bifunction from to satisfying and and let be a proper convex function from to . Let and . Assume that (i) for all ;(ii)for any fixed , the mapping is convex.*

Then the following problems (23) and (24) are equivalent:(I)find such that (II)find such that

*Proof. *Let be a solution of the problem (23). Since is relaxed - monotone, we have
Thus, is a solution of the problem (24).

Conversely, let be a solution of the problem (24) and let be any point with . From (24), it follows that . Letting
we have . Since is a solution of the problem (24), it follows that

The convexity of and conditions (i), (ii), (A1), and (A4) imply that

Thus, it follows from (27)-(28) that

Since is -hemicontinuous and , letting in (29), we get
for all with . When , the relation
is trivial. Therefore, is also a solution of the problem (23). This completes the proof.

*Lemma 11. Let be a reflexive Banach space with the dual space and let be a nonempty closed convex bounded subset of . Let be an -hemicontinuous and relaxed --monotone mapping, let be a bifunction from to satisfying and , and let be a proper convex function from to . Let and . Assume that (i) for all ;(ii)for any fixed , the mapping is convex and lower semicontinuous;(iii) is weakly lower semicontinuous; that is, for any net , converges to in implying that . Then the problem (23) is solvable.*

*Proof. *Define two set-valued mappings as follows:
respectively.

Now, we claim that is a KKM mapping. If is not a KKM mapping, then there exist and , , such that

By the definition of , we have

It follows that
which is a contradiction. This implies that is a KKM mapping.

Now, we prove that

For any given , letting , then

Since is relaxed --monotone, we have
which implies that and so

As in the proof of Theorem 2.2 in [3], also, we can obtain

Hence, there exists such that
This completes the proof.

*Next, let us denote the set of solutions of the following mixed equilibrium problem by : find such that
*

*It is easy to see that if for all , then the above mixed equilibrium problem reduces to the variational inequality problem (6) or (7).*

*Lemma 12. Let be a uniformly smooth, strictly convex Banach space with the dual space and let be a nonempty closed convex bounded subset of . Let be an -hemicontinuous and relaxed --monotone mapping, let be a bifunction from to satisfying (A1), (A2), and (A4), and let be a proper convex function from to . Let and define a mapping as follows:
for all . Assume that*

(i) , for all ;

(ii) for any fixed , the mapping is convex and lower semicontinuous;

(iii) is weakly lower semicontinuous; that is, for any net , converges to in implying that ;

(iv) for any , ;

(v) , for any and .

Then the following hold:(1) is single-valued;(2) is a firmly nonexpansive-type mapping; that is, for all , (3);(4) is quasi--nonexpansive satisfying for all and ;(5) is closed and convex.

*Proof. *We claim that is single-valued. Indeed, for any and , let . Then we have

Adding the two inequalities, it follows from (i) that

Thus, from (A2), we have
that is,

Since is relaxed --monotone and , one has

In (48), exchanging to , we get
that is,

Thus, it follows from (49), (51), and (iv) that
Since is strictly convex, we have .

Next, we claim that is a firmly nonexpansive-type mapping. Indeed, for any , we have

Adding the above two inequalities, by (i) and (A2), we get
that is,

Repeating the process from (48) to (52), we can obtain

Therefore, we have
which shows that is firmly nonexpansive.

Next, we claim that . Indeed, we have the following:

Finally, we prove (4) and (5). Indeed, we can obtain (4) directly from Lemma 2.8 and Lemma 2.9 of [1]. Moreover, since is quasi--nonexpansive, one sees that is closed and convex from Lemma 5. Therefore, is also closed and convex. This completes the proof.

*Remark 13. *In (42), if for all , then the equilibrium problem (42) is reduced to the equilibrium problem in [20] and if for all , then then the equilibrium problem (42) is reduced to the equilibrium problem in [21].

*In the following, we give the main result of this paper.*

*Theorem 14. Let be a nonempty closed convex bounded subset of a uniformly convex, uniformly smooth Banach space with the dual space . Let be an -hemicontinuous and relaxed - monotone mapping, let be a bifunction from to satisfying (A1)–(A4), and let be a proper convex and lower semicontinuous function from to . Let be a closed quasi--nonexpansive mapping such that . Assume that the conditions (i)–(v), Lemma 12, and the following condition hold:(vi) for all ,
Let be a sequence in generated by the following manner:
where is the duality mapping on , with , and with . Then the sequence generated by (52) converges strongly to .*

*Proof. *First, since the inequality is equivalent to the following:
is closed and convex for all . It follows from the definition of that is also closed and convex for all .

On the other hand, since every is quasi--nonexpansive, according to Lemma 12 (4), we have, for any ,
which shows that for all . This implies that for all . The definition of shows that for all . Hence, the sequence generated by the algorithm (60) is well defined.

Note that for all . Hence, by and , we have

Therefore, is nondecreasing. Since , is bounded. It follows that the limit of exists.

Let with . Since , from Lemma 4 it follows that

Letting in (64), it follows that . It follows from Lemma 2 that as . Hence is a Cauchy sequence. There exists a point such that as .

Next, we show that . By taking in (64), one arrives at

From Lemma 2, it follows that

Noticing that for all , we obtain

Thus, it follows from (65) that

It follows from Lemma 2 that

Combining (66) with (69), one has

Since as , we have

On the other hand, since is uniformly norm-to-norm continuous on bounded sets, one has

Since is a uniformly smooth Banach space, one knows that is a uniformly convex Banach space. Let . It follows from Lemma 6 that, for any ,

It follows that

On the other hand, one has

Thus, it follows from (70) and (72) that

Noting that and using (74) and (76), we get

It follows from the continuousness of that

Since is also uniformly norm-to-norm continuous on bounded sets, one gets
From the closeness of , we can conclude that .

Next, we show that . From (62), we arrived at

From and Lemma 12 (4), one has

Thus, it follows from (76) that

From Lemma 2, we get

Noting that , one obtains

From (A4) and (i), it follows that

Since is uniformly norm-to-norm continuous on bounded sets, one has

Noticing that for all , it follows from (A4), (ii), (85), and (86) that

For all and , define . Noticing that , one obtains , which yields that

It follows from (A1), (A4), (i), (ii), the convexity of , and (88) that
that is,

Letting , it follows from (A3), (vi), and the lower semicontinuity of that
This implies that .

Finally, we prove . From , we see that

Since for each , we have

By taking the limit in (93), one has
At this point, in view of Lemma 3, one sees that . This completes the proof.

*If in Theorem 14, then we have the following result.*

*Corollary 15. Let be a nonempty closed convex bounded subset of a uniformly convex, uniformly smooth Banach space with the dual space . Let be an -hemicontinuous and relaxed --monotone mapping and let be a proper convex and lower semicontinuous function from to . Let be a closed quasi--nonexpansive mapping such that . Assume that the conditions (i)–(v), Lemma 12, and the following condition hold:(vi) for all ,
Let be a sequence in generated by the following manner:
where is the duality mapping on , with , and with . Then the sequence generated by (96) converges strongly to .*

*If and in Theorem 14, then we have following.*

*Corollary 16. Let be a nonempty closed convex subset of a uniformly convex, uniformly smooth Banach space with the dual space . Let be a bifunction from to satisfying (A1)–(A4) and let be a closed quasi--nonexpansive mapping such that . Let be a sequence generated by the following manner:
where is the duality mapping on , with and with . Then the sequence generated by (97) converges strongly to .*

*Remark 17. *Theorem 14 improves the corresponding ones of Wang et al. [7] from Hilbert spaces to Banach spaces and those of Takahashi and Zembayashi [1] from relative nonexpansive mapping to quasi--nonexpansive mapping.

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that they have no conflict of interests.*

*Authors’ Contribution*

*Authors’ Contribution*

*The authors read and approved the final paper.*

*Acknowledgments*

*Acknowledgments*

*This work is supported by the Scientific Research Fund of Hebei Provincial Education Department (Grant no.: Z2013110), the Society Program Fund of Hebei Provincial Education Department (Grant no.: GH132068), and the Fundamental Research Fund for the Central Universities (Grant no.: 13MS109).*

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