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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 812783, 10 pages
http://dx.doi.org/10.1155/2012/812783
Research Article

Approximation of Solutions of an Equilibrium Problem in a Banach Space

1School of Mathematics and Information Science, North China University of Water Resources and Electric Power, Zhengzhou 450011, China
2College of Science, Hebei University of Engineering, Handan 056038, China

Received 15 December 2011; Accepted 4 February 2012

Academic Editor: Rudong Chen

Copyright © 2012 Hecai Yuan and Guohong Shi. 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

An equilibrium problem is investigated based on a hybrid projection iterative algorithm. Strong convergence theorems for solutions of the equilibrium problem are established in a strictly convex and uniformly smooth Banach space which also enjoys the Kadec-Klee property.

1. Introduction

Equilibrium problems which were introduced by Fan [1] and Blum and Oettli [2] have had a great impact and influence on the development of several branches of pure and applied sciences. It has been shown that the equilibrium problem theory provides a novel and unified treatment of a wide class of problems which arise in economics, finance, image reconstruction, ecology, transportation, network, elasticity, and optimization. It has been shown [38] that equilibrium, problems include variational inequalities, fixed point, the Nash equilibrium, and game theory as special cases. A number of iterative algorithms have recently been studying for fixed point and equilibrium problems, see [926] and the references therein. However, there were few results established in the framework of the Banach spaces. In this paper, we suggest and analyze a projection iterative algorithm for finding solutions of equilibrium in a Banach space.

2. Preliminaries

In what follows, we always assume that 𝐸 is a Banach space with the dual space 𝐸. Let 𝐶 be a nonempty, closed, and convex subset of 𝐸. We use the symbol 𝐽 to stand for the normalized duality mapping from 𝐸 to 2𝐸 defined by𝑓𝐽𝑥=𝐸𝑥,𝑓=𝑥2=𝑓2,𝑥𝐸,(2.1)

where , denotes the generalized duality pairing of elements between 𝐸 and 𝐸.

Let 𝑈𝐸={𝑥𝐸𝑥=1} be the unit sphere of 𝐸. 𝐸 is said to be strictly convex if (𝑥+𝑦)/2<1 for all 𝑥,𝑦𝑈𝐸 with 𝑥𝑦. It is said to be uniformly convex if for any 𝜖(0,2] there exists 𝛿>0 such that for any 𝑥,𝑦𝑈𝐸,𝑥𝑦𝜖implies𝑥+𝑦21𝛿.(2.2)

It is known that a uniformly convex Banach space is reflexive and strictly convex; for details see [27] and the references therein.

Recall that a Banach space 𝐸 is said to have the Kadec-Klee property if a sequence {𝑥𝑛} of 𝐸 satisfies that 𝑥𝑛𝑥𝐶, where denotes the weak convergence, and 𝑥𝑛𝑥, where denotes the strong convergence, and then 𝑥𝑛𝑥. It is known that if 𝐸 is uniformly convex, then 𝐸 enjoys the Kadec-Klee property; for details see [26] and the references therein.

𝐸 is said to be smooth provided lim𝑡0(𝑥+𝑡𝑦𝑥)/𝑡 exists for all 𝑥,𝑦𝑈𝐸. It is also said to be uniformly smooth if the limit is attained uniformly for all 𝑥,𝑦𝑈𝐸.

It is well known that if 𝐸 is strictly convex, then 𝐽 is single valued; if 𝐸 is reflexive, and smooth, then 𝐽 is single valued and demicontinuous; for more details see [27, 28] and the references therein.

It is also well known that if 𝐷 is a nonempty, closed, and convex subset of a Hilbert space 𝐻, and 𝑃𝐷𝐻𝐷 is the metric projection from 𝐻 onto 𝐷, then 𝑃𝐷 is nonexpansive. This fact actually characterizes the Hilbert spaces, and consequently, it is not available in more general Banach spaces. In this connection, Alber [29] introduced a generalized projection operator Π𝐷 in the Banach spaces which is an analogue of the metric projection in the Hilbert spaces.

Let 𝐸 be a smooth Banach space. Consider the functional defined by𝜙(𝑥,𝑦)=𝑥22𝑥,𝐽𝑦+𝑦2,𝑥,𝑦𝐸.(2.3) Notice that, in a Hilbert space 𝐻, (2.3) is reduced to 𝜙(𝑥,𝑦)=𝑥𝑦2 for all 𝑥,𝑦𝐻. The generalized projection Π𝐶𝐸𝐶 is a mapping that is assigned to an arbitrary point 𝑥𝐸, the minimum point of the functional 𝜙(𝑥,𝑦); that is, Π𝐶𝑥=𝑥, where 𝑥 is the solution to the following minimization problem:𝜙𝑥,𝑥=min𝑦𝐶𝜙(𝑦,𝑥).(2.4)

The existence and uniqueness of the operator Π𝐶 follow from the properties of the functional 𝜙(𝑥,𝑦) and the strict monotonicity of the mapping 𝐽; see, for example, [27, 28]. In the Hilbert spaces, Π𝐶=𝑃𝐶. It is obvious from the definition of the function 𝜙 that()𝑦𝑥2)𝜙(𝑦,𝑥)(𝑦+𝑥2,𝑥,𝑦𝐸,(2.5)𝜙(𝑥,𝑦)=𝜙(𝑥,𝑧)+𝜙(𝑧,𝑦)+2𝑥𝑧,𝐽𝑧𝐽𝑦,𝑥,𝑦,𝑧𝐸.(2.6)

Let 𝑇𝐶𝐶 be a mapping. Recall that a point 𝑝 in 𝐶 is said to be an asymptotic fixed point of 𝑇 if 𝐶 contains a sequence {𝑥𝑛} which converges weakly to 𝑝 such that lim𝑛𝑥𝑛𝑇𝑥𝑛=0. The set of asymptotic fixed points of 𝑇 will be denoted by 𝐹(𝑇). 𝑇 is said to be relatively nonexpansive if𝐹(𝑇)=𝐹(𝑇),𝜙(𝑝,𝑇𝑥)𝜙(𝑝,𝑥),𝑥𝐶,𝑝𝐹(𝑇).(2.7)

The asymptotic behavior of a relatively nonexpansive mapping was studied in [27, 29, 30].

Let 𝑓 be a bifunction from 𝐶×𝐶 to , where denotes the set of real numbers. In this paper, we consider the following equilibrium problem. Find 𝑝𝐶 such that𝑓(𝑝,𝑦)0,𝑦𝐶.(2.8) We use EP(𝑓) to denote the solution set of the equilibrium problem (2.3). That is,EP(𝑓)={𝑝𝐶𝑓(𝑝,𝑦)0,𝑦𝐶}.(2.9)

Given a mapping 𝑄𝐶𝐸, let𝑓(𝑥,𝑦)=𝑄𝑥,𝑦𝑥,𝑥,𝑦𝐶.(2.10)

Then 𝑝EP(𝑓) if and only if 𝑝 is a solution of the following variational inequality. Find 𝑝 such that𝑄𝑝,𝑦𝑝0,𝑦𝐶.(2.11)

To study the equilibrium problem (2.8), we may assume that 𝑓 satisfies the following conditions:

(A1) 𝑓(𝑥,𝑥)=0,forall𝑥𝐶;

(A2) 𝑓 is monotone, that is, 𝑓(𝑥,𝑦)+𝑓(𝑦,𝑥)0,forall𝑥,𝑦𝐶;

(A3)limsup𝑡0𝑓(𝑡𝑧+(1𝑡)𝑥,𝑦)𝑓(𝑥,𝑦),𝑥,𝑦,𝑧𝐶;(2.12)

(A4) for each 𝑥𝐶, 𝑦𝑓(𝑥,𝑦) is convex and weakly lower semicontinuous.

In this paper, we study the problem of approximating solutions of equilibrium problem (2.8) based on a hybrid projection iterative algorithm in a strictly convex and uniformly smooth Banach space which also enjoys the Kadec-Klee property. To prove our main results, we need the following lemmas.

Lemma 2.1. Let 𝐸 be a strictly convex and uniformly smooth Banach space and 𝐶 a nonempty, closed, and convex subset of 𝐸. Let 𝑓 be a bifunction from 𝐶×𝐶 to satisfying (A1)–(A4). Let 𝑟>0 and 𝑥𝐸. Then (a)(see [2]). There exists 𝑧𝐶 such that1𝑓(𝑧,𝑦)+𝑟𝑦𝑧,𝐽𝑧𝐽𝑥0,𝑦𝐶.(2.13)(b)(see [31]). Define a mapping 𝑇𝑓𝑟𝐸𝐶 by𝑇𝑓𝑟1𝑥=𝑧𝐶𝑓(𝑧,𝑦)+𝑟.𝑦𝑧,𝐽𝑧𝐽𝑥,𝑦𝐶(2.14) Then the following conclusions hold:(1)𝑇𝑓𝑟 is single valued;(2)𝑇𝑓𝑟 is a firmly nonexpansive-type mapping; that is, for all 𝑥,𝑦𝐸,𝑇𝑓𝑟𝑥𝑇𝑓𝑟𝑦,𝐽𝑇𝑓𝑟𝑥𝐽𝑇𝑓𝑟𝑦𝑇𝑓𝑟𝑥𝑇𝑓𝑟𝑦,𝐽𝑥𝐽𝑦;(2.15)(3)𝐹(𝑇𝑓𝑟)=EP(𝑓); (4)EP(𝑓) is closed and convex;(5)𝑇𝑓𝑟 is relatively nonexpansive.

Lemma 2.2 (see [29]). Let 𝐸 be a reflexive, strictly convex, and smooth Banach space and 𝐶 a nonempty, closed, and convex subset of 𝐸. Let 𝑥𝐸, and 𝑥0𝐶. Then 𝑥0=Π𝐶𝑥 if and only if

𝑥0𝑦,𝐽𝑥𝐽𝑥00,𝑦𝐶.(2.16)

Lemma 2.3 (see [29]). Let 𝐸 be a reflexive, strictly convex, and smooth Banach space and 𝐶 a nonempty, closed, and convex subset of 𝐸, and 𝑥𝐸. Then

𝜙𝑦,Π𝐶𝑥Π+𝜙𝐶𝑥,𝑥𝜙(𝑦,𝑥),𝑦𝐶.(2.17)

Lemma 2.4 (see [27]). Let 𝐸 be a reflexive, strictly convex, and smooth Banach space. Then one has the following

𝜙(𝑥,𝑦)=0𝑥=𝑦,𝑥,𝑦𝐸.(2.18)

3. Main Results

Theorem 3.1. Let 𝐸 be a strictly convex and uniformly smooth Banach space which also enjoys the Kadec-Klee property and 𝐶 a nonempty, closed, and convex subset of 𝐸. Let 𝑓 be a bifunction from 𝐶×𝐶 to satisfying (A1)–(A4) such that EP(𝑓). Let {𝑥𝑛} be a sequence generated by the following manner: 𝑥0C𝐸chosenarbitrarily,1𝑥=𝐶,1=Π𝐶1𝑥0,𝑦𝑛𝑦𝐶,suchthat𝑓𝑛+1,𝑢𝑟𝑛𝑢𝑦𝑛,𝐽𝑦𝑛𝐽𝑥𝑛𝐶0,𝑢𝐶,𝑛+1=𝑢𝐶𝑛2𝑥𝑛𝑢,𝐽𝑥𝑛𝐽𝑦𝑛𝑥𝜙𝑛,𝑦𝑛,𝑥𝑛+1=Π𝐶𝑛+1𝑥0,𝑛1,(3.1) where {𝑟𝑛} is a real number sequence in [𝑟,), where 𝑟 is some positive real number. Then the sequence {𝑥𝑛} converges strongly to 𝑥=ΠEP(𝑓)𝑥0.

Proof. In view of Lemma 2.1, we see that EP(𝑓) is closed and convex. Next, we show that 𝐶𝑛 is closed and convex. It is not hard to see that 𝐶𝑛 is closed. Therefore, we only show that 𝐶𝑛 is convex. It is obvious that 𝐶1=𝐶 is convex. Suppose that 𝐶 is convex for some . Next, we show that 𝐶+1 is also convex for the same . Let 𝑎,𝑏𝐶+1 and 𝑐=𝑡𝑎+(1𝑡)𝑏, where 𝑡(0,1). It follows that 𝜙𝑥,𝑦2𝑥𝑎,𝐽𝑥𝐽𝑦𝑥,𝜙,𝑦2𝑥𝑏,𝐽𝑥𝐽𝑦,(3.2)where 𝑎,𝑏𝐶. From the above two inequalities, we can get that 𝜙𝑥,𝑦2𝑥𝑐,𝐽𝑥𝐽𝑦,(3.3)where 𝑐𝐶. It follows that 𝐶+1 is closed and convex. This completes the proof that 𝐶𝑛 is closed, and convex.
Next, we show that EP(𝑓)𝐶𝑛. It is obvious that EP(𝑓)𝐶=𝐶1. Suppose that EP(𝑓)𝐶 for some . For any 𝑧EP(𝑓)𝐶, we see from Lemma 2.1 that𝜙𝑧,𝑦𝜙𝑧,𝑥.(3.4) On the other hand, we obtain from (2.6) that 𝜙𝑧,𝑦=𝜙𝑧,𝑥𝑥+𝜙,𝑦+2𝑧𝑥,𝐽𝑥𝐽𝑦.(3.5) Combining (3.4) with (3.5), we arrive at 2𝑥𝑧,𝐽𝑥𝐽𝑦𝑥𝜙,𝑦(3.6)which implies that 𝑧𝐶+1. This shows that EP(𝑓)𝐶+1. This completes the proof that EP(𝑓)𝐶𝑛.
Next, we show that {𝑥𝑛} is a convergent sequence and strongly converges to 𝑥, where 𝑥EP(𝑓). Since 𝑥𝑛=Π𝐶𝑛𝑥0, we see from Lemma 2.2 that𝑥𝑛𝑧,𝐽𝑥0𝐽𝑥𝑛0,𝑧𝐶𝑛.(3.7) It follows from EP(𝑓)𝐶𝑛 that 𝑥𝑛𝑤,𝐽𝑥0𝐽𝑥𝑛0,𝑤EP(𝑓).(3.8) By virtue of Lemma 2.3, we obtain that 𝜙𝑥𝑛,𝑥0Π=𝜙𝐶𝑛𝑥0,𝑥0Π𝜙EP(𝑓)𝑥0,𝑥0Π𝜙EP(𝑓)𝑥0,𝑥𝑛Π𝜙EP(𝑓)𝑥0,𝑥0.(3.9) This implies that the sequence {𝜙(𝑥𝑛,𝑥0)} is bounded. It follows from (2.5) that the sequence {𝑥𝑛} is also bounded. Since the space is reflexive, we may assume that 𝑥𝑛𝑥. Since 𝐶𝑛 is closed and convex, we see that 𝑥𝐶𝑛. On the other hand, we see from the weakly lower semicontinuity of the norm that 𝜙𝑥,𝑥0=𝑥22𝑥,𝐽𝑥0+𝑥02liminf𝑛𝑥𝑛22𝑥𝑛,𝐽𝑥0𝑥+02=liminf𝑛𝜙𝑥𝑛,𝑥0limsup𝑛𝜙𝑥𝑛,𝑥0𝜙𝑥,𝑥0,(3.10)which implies that 𝜙(𝑥𝑛,𝑥0)𝜙(𝑥,𝑥0) as 𝑛. Hence, 𝑥𝑛𝑥 as 𝑛. In view of the Kadec-Klee property of 𝐸, we see that 𝑥𝑛𝑥 as 𝑛. Notice that 𝑥𝑛+1=ΠEP(𝑓)𝑥0𝐶𝑛+1𝐶𝑛. It follows that 𝜙𝑥𝑛+1,𝑥𝑛𝑥=𝜙𝑛+1,Π𝐶𝑛𝑥0𝑥𝜙𝑛+1,𝑥0Π𝜙𝐶𝑛𝑥0,𝑥0𝑥=𝜙𝑛+1,𝑥0𝑥𝜙𝑛,𝑥0.(3.11) Since 𝑥𝑛=Π𝐶𝑛𝑥0 and 𝑥𝑛+1=Π𝐶𝑛+1𝑥0𝐶𝑛+1𝐶𝑛, we arrive at 𝜙(𝑥𝑛,𝑥0)𝜙(𝑥𝑛+1,𝑥0). This shows that {𝜙(𝑥𝑛,𝑥0)} is nondecreasing. It follows from the boundedness that lim𝑛𝜙(𝑥,𝑥0) exists. It follows that lim𝑛𝜙𝑥𝑛+1,𝑥𝑛=0.(3.12) By virtue of 𝑥𝑛+1=Π𝐶𝑛+1𝑥0𝐶𝑛+1, we find that 𝜙𝑥𝑛,𝑦𝑛𝑥2𝑛𝑥𝑛+1,𝐽𝑥𝑛𝐽𝑦𝑛.(3.13) It follows that lim𝑛𝜙𝑥𝑛,𝑦𝑛=0.(3.14) In view of (2.5), we see that lim𝑛𝑥𝑛𝑦𝑛=0.(3.15) Since 𝑥𝑛𝑥, we find that lim𝑛𝑦𝑛=𝑥.(3.16) It follows that lim𝑛𝐽𝑦𝑛=𝐽𝑥.(3.17) This implies that {𝐽𝑦𝑛} is bounded. Note that both 𝐸 and 𝐸 are reflexive. We may assume that 𝐽𝑦𝑛𝑦𝐸. In view of the reflexivity of 𝐸, we see that there exists an element 𝑦𝐸 such that 𝐽𝑦=𝑦. It follows that 𝜙𝑥𝑛,𝑦𝑛=𝑥𝑛22𝑥𝑛,𝐽𝑦𝑛𝑦+𝑛2=𝑥𝑛22𝑥𝑛,𝐽𝑦𝑛+𝐽𝑦𝑛2.(3.18) Taking liminf𝑛 on the both sides of the equality above yields that 0𝑥22𝑥,𝑦+𝑦2=𝑥22𝑥,𝐽𝑦+𝐽𝑦2=𝑥22𝑥,𝐽𝑦+𝑦2=𝜙.𝑥,𝑦(3.19) That is, 𝑥=𝑦, which in turn implies that 𝑦=𝐽𝑥. It follows that 𝐽𝑦𝑛𝐽𝑥𝐸. Since 𝐸 enjoys the Kadec-Klee property, we obtain from (3.17) that lim𝑛𝐽𝑦𝑛=𝐽𝑥. Since 𝐽1𝐸𝐸 is demicontinuous, we find that 𝑦𝑛𝑥. This implies from (3.16) and the Kadec-Klee property of 𝐸 that lim𝑛𝑦𝑛=𝑥. This in turn implies that lim𝑛𝑦𝑛𝑥𝑛=0. Since 𝐽 is uniformly norm-to-norm continuous on any bounded sets, we find that lim𝑛𝐽𝑦𝑛𝐽𝑥𝑛=0.(3.20) Next, we show that 𝑥𝐸𝐹(𝑓). In view of Lemma 2.1, we find from 𝑦𝑛=𝑇𝑓𝑟𝑛𝑥𝑛 that 𝑓𝑦𝑛+1,𝑢𝑟𝑛𝑢𝑦𝑛,𝐽𝑦𝑛𝐽𝑥𝑛0,𝑢𝐶.(3.21) It follows from condition (A2) and (3.20) that 1𝑟𝑛𝑢𝑦𝑛𝐽𝑦𝑛𝐽𝑥𝑛𝑓𝑢,𝑦𝑛,𝑢𝐶.(3.22) In view of condition (A4), we obtain from (3.17) that 𝑓𝑢,𝑥0,𝑢𝐶.(3.23) For 0<𝑡<1 and 𝑢𝐶, define 𝑢𝑡=𝑡𝑢+(1𝑡)𝑥. It follows that 𝑢𝑡𝐶, which yields that 𝑓(𝑢𝑡,𝑥)0. It follows from conditions (A1) and (A4) that 𝑢0=𝑓𝑡,𝑢𝑡𝑢𝑡𝑓𝑡+𝑢,𝑢(1𝑡)𝑓𝑡,𝑥𝑢𝑡𝑓𝑡,𝑢.(3.24) That is, 𝑓𝑢𝑡,𝑢0.(3.25) Letting 𝑡0, we find from condition (A3) that 𝑓(𝑥,𝑢)0, forall𝑢𝐶. This implies that 𝑥EP(𝑓). This shows that 𝑥EP(𝑓).
Finally, we prove that 𝑥=ΠEP(𝑓)𝑥0. Letting 𝑛 in (3.8), we see that𝑥𝑤,𝐽𝑥0𝐽𝑥0,𝑤EP(𝑓).(3.26) In view of Lemma 2.2, we can obtain that 𝑥=ΠEP(𝑓)𝑥0. This completes the proof.

In the framework of the Hilbert spaces, we have the following.

Corollary 3.2. Let 𝐸 be a Hilbert space and 𝐶 a nonempty, closed, and convex subset of 𝐸. Let 𝑓 be a bifunction from 𝐶×𝐶 to satisfying (A1)–(A4) such that EP(𝑓). Let {𝑥𝑛} be a sequence generated by the following manner: 𝑥0𝐶𝐸chosenarbitrarily,1𝑥=𝐶,1=𝑃𝐶1𝑥0,𝑦𝑛𝑦𝐶,suchthat𝑓𝑛+1,𝑢𝑟𝑛𝑢𝑦𝑛,𝑦𝑛𝑥𝑛𝐶0,𝑢𝐶,𝑛+1=𝑢𝐶𝑛2𝑥𝑛𝑢,𝑥𝑛𝑦𝑛𝑥𝑛𝑦𝑛2,𝑥𝑛+1=𝑃𝐶𝑛+1𝑥0,𝑛1,(3.27) where {𝑟𝑛} is a real number sequence in [𝑟,), where 𝑟 is some positive real number. Then the sequence {𝑥𝑛} converges strongly to 𝑥=𝑃EP(𝑓)𝑥0.

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