Strong Convergence Theorems for Mixed Equilibrium Problem and Asymptotically <svg xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" style="vertical-align:-0.06579971pt" id="M1" height="11.9431pt" version="1.1" viewBox="-0.0657574 -11.8773 7.81438 11.9431" width="7.81438pt"><g transform="matrix(.018,0,0,-0.018,0,0)"><path id="g113-74" d="M405 650H141L135 622C222 616 230 610 215 535L133 116C118 41 113 33 29 28L23 0H289L295 28C209 33 205 40 219 116L298 535C312 609 317 616 399 622L405 650Z"/></g></svg>-Nonexpansive Mapping in Banach Spaces

Deng, Bin-Chao; Chen, Tong; Yin, Yi-Lin

doi:https://doi.org/10.1155/2014/965737

Abstract and Applied Analysis

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Volume 2014 | Article ID 965737 | https://doi.org/10.1155/2014/965737

Strong Convergence Theorems for Mixed Equilibrium Problem and Asymptotically -Nonexpansive Mapping in Banach Spaces

Bin-Chao Deng,¹Tong Chen,²and Yi-Lin Yin¹

Academic Editor: Jong Kyu Kim

Received01 Jul 2014

Revised21 Aug 2014

Accepted21 Aug 2014

Published27 Oct 2014

Abstract

This paper aims to use a hybrid algorithm for finding a common element of a fixed point problem for a finite family of asymptotically nonexpansive mappings and the set solutions of mixed equilibrium problem in uniformly smooth and uniformly convex Banach space. Then, we prove some strong convergence theorems of the proposed hybrid algorithm to a common element of the above two sets under some suitable conditions.

1. Introduction

Let be a Banach space with norm . Let be a nonempty closed convex subset of and denoted the dual space of . Let be a nonlinear mapping and a bifunction from to , where denotes the set of numbers. The generalized equilibrium problem is to find such that The set of solution of (1) is denoted by , that is,

In this paper, we are interested in solving the generalized equilibrium problem with those given by where , are two bifunctions satisfying the following special properties , and : () , for all ; () is maximal monotone; () for all , , , we have ; () for all , the function is convex and weakly lower semicontinuous; () , for all ; () is monotone and maximal monotone, and weakly upper semicontinuous in the first variable; () is convex in the second variable; () for fixed and , there exist a bounded set and such that This is the well-know generalized mixed equilibrium problem, that is, to find an in such that The solution set of (5) is denoted by , that is, If , problem (5) reduces into mixed equilibrium problem for and , denoted by , which is to find such that (3).

If and , reduces into equilibrium problem for , denoted by , which is to find such that Mixed equilibrium problems are suitable and common format for investigation of various applied problems arising in economics, mathematical physics, transportation, communication systems, engineering, and other fields. Moreover, equilibrium problems are closely related with other general problems in nonlinear analysis, such as fixed points, game theory, variational inequality, and optimization problems. Recently, many authors studied a great number of iterative methods for solving a common element of the set of fixed points for a nonexpansive mapping and the set of solutions to a mixed equilibrium problem in the setting of Hilbert space and uniformly smooth and uniformly convex Banach space, respectively (please see, e.g., [1–11] and the references therein).

Let be a real Banach space with norm , let be a nonempty closed convex subset of , and let be the normalized duality mapping from into given by where denotes the dual space of and the generalized duality pairing between and . It is easily known that if is uniformly convex, then is uniformly continuous on bounded subsets of .

Consider the functional defined by It is obvious from the definition of that On the other hand, in a Hilbert space , (9) reduced to . Following Alber [12], the generalized projection is defined by where is a map that assigns to an arbitrary point the minimum point of the functional .

In 2011, Kim [13] considered the following shrinking projection methods to obtain a convergence theorem, and these methods were introduced in [14] for quasi--nonexpansive mappings in a uniformly convex and uniformly smooth Banach space.

Theorem 1 (see [13]). Let be a uniformly smooth and strictly convex Banach space which has the Kadec-Klee property and a nonempty closed convex subset of . Let be a bifunction from to satisfying and a closed and asymptotically quasi--nonexpansive mapping. Assume that T is asymptotically regular on and is nonempty and bounded. Let be a sequence generated in the following manner: where for each , is a real sequence in such that , and is a real sequence in , where is some positive real number and is the duality mapping on . Then the sequence converges strongly to , where is the generalized projection from onto .

Motivated and inspired by the researches going on in this direction (i.e., [4–11, 13–16]), the purpose of this paper is to use the following hybrid algorithm for finding a common element of the set of solutions to a mixed equilibrium problem and the set of the set of common fixed points for a finite family of asymptotically nonexpansive mappings in a uniformly smooth and uniformly convex Banach space.

Algorithm 2. Let
Consequently, under suitable conditions, we show that iterative algorithms converge strongly to a solution of some optimization problem. Note that our methods do not use any projection.

2. Preliminaries

Let be a mapping. Denote by the set of fixed points of , that is, . Throughout this paper, we always assume that . Now we need the following known definitions.

Definition 3. A mapping is said to be(1)nonexpansive, if , for all , ;(2)asymptotically nonexpansive, if there exists a sequence with such that , for all , and ;(3)quasi-nonexpansive, , for all and ;(4)asymptotically quasi-nonexpansive, if there exists a sequence with such that , for all , , and .

There are many concepts which generalize a notion of nonexpansive mapping. In 2004, Shahzad [17] introduced the following concepts about -nonexpansivity of a mapping .

Definition 4. Let and be two mappings of a nonempty subset , a real normal linear space . Then is said to be(i)-nonexpansive, if , for all , ;(ii)asymptotically -nonexpansive, if there exists a sequence with such that , for all , and ;(iii)asymptotically quasi--nonexpansive, if there exists a sequence with such that , for all , , and .

Lemma 5 (see [4]). Assume that is convex, , , and , for all . Then , for all .

Lemma 6 (see [18]). Let be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space , and let be a relatively nonexpansive mapping from into itself. Then is closed and convex.

Lemma 7 (see [19]). Let , and be sequences of nonnegative real sequences satisfying the following conditions: for all (1), (2), where and . Then exists.

Lemma 8 (see [20]). Let be a uniformly convex Banach space. Then, for each , there exists a strictly increasing, continuous, and convex function such that and for , , , where .

Lemma 9 (see [21]). Let be a uniformly convex Banach space and let , be two constants with . Suppose that is a sequence in and , are two sequence in such that holds some . Then .

Definition 10 (see [22]). The mappings , are said to be satisfying condition (A) if there is a nondecreasing function with , for each such that for all , where = = .

Lemma 11 (see [23]). Let be a uniformly convex Banach space satisfying the Opial’s condition, a nonempty closed subset of , and an asymptotically nonexpansive mapping. If the sequence is a weakly convergent sequence with the weak limit and if , then .

3. Main Results

Theorem 12. Let be a smooth, strictly convex, and reflexive Banach space, and let be a nonempty closed convex subset of . Let , be two bifunctions which satisfy the conditions , , and . Then for every , there exists a unique point such that

The proof goes over the following three steps.

Proof.
Step 1. There exists point such that
Consider the closed sets We will show that . Let , , be a finite subset of . Let be nonempty. Let for all . Then Assume, for contradiction, that By the convexity of and and the monotonicity of , we obtain that and that is absurd. Hence (20) cannot be true. and we have for some . Thus for some . Since for all , it follows that By the sets being closed, it follows form the standard version of the KKM-Theorem that In other words, any finite subfamily of the family has nonempty intersection. Since these sets are closed subsets of the compact set , it follows that the entire family has nonempty intersection. Hence
Step 2. For every , the following statement are equivalent:(i), , for all ,(ii), , for all .
Case 1. Let (ii) hold; since is monotone, one has Hence (i) follows.
Case 2. Let (i) hold, for with and , and let Then , and from (i), . By the properties of and , it follows then, for all , Let and thereby and using the hemicontinuity of we obtain in the limit
Step 3. Take . Then the function is convex and , for all . If , then set . If , then set , where is as in assumption for . In both cases , and . Hence it follows from the Lemma 5 that

Corollary 13. Let be a smooth, strictly convex, and reflexive Banach space, and let be a nonempty closed convex subset of . Let , be two bifunctions which satisfy the following conditions: , , and in Theorem 12. There for every and , there exists a unique point such that

Proof. Let and be given. Note that functions and also satisfy the conditions and . Therefore, for , there exists a unique point such that This completes the proof.

Under the same assumptions in Corollary 13, for every , we may define a single-valued mapping as follows: for , which is called the resolvent of and for .

Theorem 14. Let be a smooth, strictly convex, and reflexive Banach space, and let be a nonempty closed convex subset of . Let , be two bifunctions which satisfy conditions , , and . For and , define a mapping in (32). Then, the following hold:(a) is single-valued;(b) is a firmly nonexpansive mapping, that is, (c);(d) is closed and convex;(e).

Proof. We divide the proof into several steps.
Step 1 ( is single-valued). Indeed, for and , let . Then Adding the two inequalities, we obtain From , , and , we obtain Since is strictly convex, we obtain
Step 2 ( is a firmly nonexpansive mapping). For , , we obtain Adding the two inequalities, we obtain From , , and , we obtain Therefore, we have
Step 3 (). Indeed, we obtain the following equation:
Step 4 ( is closed and convex). From (c), we have , and from (b), we obtain Moreover, we obtain Hence, we obtain So we get Taking , we obtain
Next, we show that . Let . Then, there exists the sequence of such that and . Moreover, we obtain . Hence we have . Since is uniformly continuous on bounded sets, we obtain Form the definition of , we obtain Since the monotone of the , we have According to (48) and and form and , we obtain For with and , let ; then by the convexity of and we have Passing and by and , we have for all . Therefore, . So, we get . Therefore, we have that is a relatively nonexpansive mapping. From Lemma 6, then is closed and convex.
Step 5 (). From (b) and (45), for each , , we obtain Letting , we obtain

If and form Theorems 12 and 14, we obtain the following corollary.

Corollary 15 (see [24]). Let be a smooth, strictly convex, and reflexive Banach space, and be a nonempty closed convex subset of . Let be a bifunctions which satisfy conditions . Let be a lower semi-continuous and convex function. For and . Then, the following hold:(i), for all .(ii)If we define a mapping as follows: and the mapping has the following properties:(a) is single-valued;(b) is a firmly nonexpansive mapping, that is, (c);(d) is closed and convex;(e).

4. Strong Convergence Theorems

In this section, we introduce a new iterative scheme for finding a common element of the set of solutions of the mixed equilibrium problems and the set of fixed points for -asymptotically nonexpansive mapping in Banach spaces.

Theorem 16. Let be uniformly smooth and uniformly convex Banach space, and let be a nonempty closed convex subset of . Let , be two bifunctions which satisfy the conditions , , and , and let be -asymptotically nonexpansive self-mapping of with sequences such that , and let be asymptotically nonexpansive self-mapping of with sequences such that , and . For an initial point , generate a sequence by where is a sequence in , for some , and for . If the following conditions are satisfied:(i);(ii);(iii), then the sequence generated by (57) converges strongly to a fixed point in if and only if

Proof. We divide the proof into several steps.
Step 1 (The sequence is bounded). Let . Since is a -asymptotically nonexpansive mapping, it follows from and Theorem 14 that is nonempty closed convex subset and for each . Again from (57), we obtain that From (59) and (60), we obtain where Moreover since , for some , , , and , it follow that . Form (60) and, by Lemma 7, we obtain that the limit of exists for each . This implies that is bounded and so are , , , , and ; on the other hand, we obtain that . Then by Lemma 7, exists and, by assumption , we obtain Step 2 (). Taking lim sup on both sides in the above inequality, Since is asymptotically nonexpansive self-mappings of , we can get that , which on taking and using (64), we obtain Further, That means that It follows from Lemma 9 that Moveover, Thus, from (68), we have
Step 3 (). Use (57) again, and Lemma 8 that for , there exists a strictly increasing, continuous and convex function that and where From the discuss of the Step 1, we can easily know that . On the other hand, by (71) and the bounded sequence of , we obtain that From , (73) and the property of , we have The same as the proof of (74), we can easily obtain that From (57), we obtain that It follows that
Step 4 (). Let . Then, from (59) and (60), it follows that where Moreover since , for some , , and , we can easily claim that and . By Lemma 7, we obtain that exists and from Theorem 14(b) and (78), we have Thus, since converges, and and is bounded, it follows form Lemma 7 that
Step 5 (). By using the triangle inequality, we have Thus, from (74) and (81), we obtain that
Step 6 (). By using the triangle inequality again, we obtain From (74) and (81), we have From (57), we have From , , and (68), we obtain
Step 7 (). Since is bounded, there exists a subsequence of such that converges weakly to when for some . From (61), we have that converges weakly to and, by (77), we also have that converges weakly to . Also, by (85), (87), and Lemma 11, we obtain that .
Next, we show that ; that is, . Since is uniformly norm-to-norm continuous on bounded subset of , it follows from (61) that From the assumption , one sees Since is bounded and so is , there exists a subsequence of such that . Since is bounded, by (89), we also obtain . Noticing that , we obtain According to (89), we obtain . Then, by the conditions of () and (), we obtain Since and , we obtain For with and , let , we obtain So, from the conditions of (), (), (), and (), we have Consequently by () and (), as , and we obtain .
Step 8 (The sequence of converges strongly to a common ). From Step 1 and (61), for all , for with . This implies that . Then by Lemma 7, exists. Also by Step 6, , and by the condition (A) in Definition 10 which guarantees that . Since is a nondecreasing function and , it follows that . Form (81), we obtain We know that is Cauchy sequence in for all numbers , . This implies that converges strongly to . This completes the proof.

If is an asymptotically quasi-nonexpansive self-mapping in Theorem 16, we easily obtain the following corollary.

Corollary 17. Let be uniformly smooth and uniformly convex Banach space, and let be a nonempty closed convex subset of . Let , be two bifunctions which satisfy the conditions , , and , and let be asymptotically quasi-nonexpansive self-mapping of with sequences such that , and let be identity self-mapping of , and . For an initial point , generate a sequence by where is a sequence in , for some , and for . If the following conditions are satisfied:(i);(ii);(iii),then the sequence generated by (97) converges strongly to a fixed point in if and only if .

5. Numerical Example

In this section, we introduce an example of numerical test to illustrate the algorithms given in Corollary 17.

Example 1. Let , . The mixed equilibrium problem is to find such that where we define and .
Now, we can easily know that and satisfy the conditions , , and as follows: () for all ; () for all ; () for all , () for each , is convex and weakly lower semicontinuous. () for each ; () for all , and weakly upper semicontinuous in first variable; () for each , is convex.
Next, we find the formula of . From Theorem 14, we can claim that is single-valued, for any , , Let . Then is a quadratic function of with coefficients , , and . So its discriminant is According to for all , form , that is Therefore, it follows that and so
Now, let and , and define a mapping by for all . From the example in [25–27], we can easily know that is an asymptotically quasi-nonexpansive mapping; furthermore .
According to Theorem 14, we obtain and so . Therefore, all the assumptions in Corollary 17 are satisfied. we can obtain the following numerical algorithms.

Algorithm 18. Let , , and . It is claim to check that For an initial value and , let the sequences and generate by
Then, by the Corollary 17, the sequence converges to a solution of Example 1. Let and be the fixed point of the Algorithm 18. Using the software of MATLAB, we generated a sequence convergence to as shown in Figure 1.
Hence the sequence converges strongly to solve Example 1.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work is supported in part by National Natural Science Foundation of China (71272148, 71472135, and 71112175) and Humanity and Social Science Foundation for Youth of Ministry of Education of China (Project no. 12YJCZH095).

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Copyright © 2014 Bin-Chao Deng 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.

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