# Approximation of Solutions of an Equilibrium Problem in a Banach Space

Authors: Hecai Yuan; Guohong Shi
Journal: Journal of Applied Mathematics
 (2012)

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## 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.


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## Body

## 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 [3–8] 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 [9–26] 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 thatE 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 by(2.1)Jx={f*∈E*:〈x,f*〉=‖x‖2=‖f*‖2},∀x∈E,where〈·,·〉 denotes the generalized duality pairing of elements between E and E*.LetUE={x∈E:∥x∥=1} be the unit sphere of E. E is said to be strictly convex if ∥(x+y)/2∥<1 for all x,y∈UE with x≠y. It is said to be uniformly convex if for any ϵ∈(0,2] there exists δ>0 such that for any x,y∈UE,(2.2)‖x-y‖≥ϵimplies‖x+y2‖≤1-δ.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 spaceE is said to have the Kadec-Klee property if a sequence {xn} of E satisfies that xn⇀x∈C, where ⇀ denotes the weak convergence, and ∥xn∥→∥x∥, where → denotes the strong convergence, and then xn→x. 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 limt→0(∥x+ty∥-∥x∥)/t exists for all x,y∈UE. It is also said to be uniformly smooth if the limit is attained uniformly for all x,y∈UE.It is well known that ifE* 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 ifD is a nonempty, closed, and convex subset of a Hilbert space H, and PD:H→D 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.LetE be a smooth Banach space. Consider the functional defined by(2.3)ϕ(x,y)=‖x‖2-2〈x,Jy〉+‖y‖2,∀x,y∈E.
Notice that, in a Hilbert space H, (2.3) is reduced to ϕ(x,y)=∥x-y∥2 for all x,y∈H. The generalized projection ΠC:E→C is a mapping that is assigned to an arbitrary point x∈E, the minimum point of the functional ϕ(x,y); that is, ΠCx=x¯, where x¯ is the solution to the following minimization problem:(2.4)ϕ(x¯,x)=min⁡y∈C⁡ϕ(y,x).The existence and uniqueness of the operatorΠC 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(2.5)(‖y‖-‖x‖)2≤ϕ(y,x)≤(‖y‖+‖x‖)2,∀x,y∈E,(2.6)ϕ(x,y)=ϕ(x,z)+ϕ(z,y)+2〈x-z,Jz-Jy〉,∀x,y,z∈E.LetT:C→C 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 limn→∞∥xn-Txn∥=0. The set of asymptotic fixed points of T will be denoted by F̃(T). T is said to be relatively nonexpansive if(2.7)F̃(T)=F(T)≠∅,ϕ(p,Tx)≤ϕ(p,x),∀x∈C,∀p∈F(T).The asymptotic behavior of a relatively nonexpansive mapping was studied in [27, 29, 30].Letf be a bifunction from C×C to ℝ, where ℝ denotes the set of real numbers. In this paper, we consider the following equilibrium problem. Find p∈C such that(2.8)f(p,y)≥0,∀y∈C.
We use EP(f) to denote the solution set of the equilibrium problem (2.3). That is,(2.9)EP⁡(f)={p∈C:f(p,y)≥0,∀y∈C}.Given a mappingQ:C→E*, let(2.10)f(x,y)=〈Qx,y-x〉,∀x,y∈C.Thenp∈EP(f) if and only if p is a solution of the following variational inequality. Find p such that(2.11)〈Qp,y-p〉≥0,∀y∈C.To study the equilibrium problem (2.8), we may assume that f satisfies the following conditions:(A1)f(x,x)=0,forallx∈C;(A2)f is monotone, that is, f(x,y)+f(y,x)≤0,forallx,y∈C;(A3)(2.12)limsup⁡t↓0f(tz+(1-t)x,y)≤f(x,y),∀x,y,z∈C;(A4) for eachx∈C, y↦f(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.
LetE 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  ℝ satisfying  (A1)–(A4). Let  r>0 and  x∈E. Then (a)
(see [2]). There exists z∈C such that(2.13)f(z,y)+1r〈y-z,Jz-Jx〉≥0,∀y∈C.(b)
(see [31]). Define a mapping Trf:E→C by(2.14)Trfx={z∈C:f(z,y)+1r〈y-z,Jz-Jx〉,∀y∈C}.
Then the following conclusions hold:(1)
Trf is single valued;(2)
Trf is a firmly nonexpansive-type mapping; that is, for all x,y∈E,(2.15)〈Trfx-Trfy,JTrfx-JTrfy〉≤〈Trfx-Trfy,Jx-Jy〉;(3)
F(Trf)=EP(f);(4)
EP(f) is closed and convex;(5)
Trf is relatively nonexpansive.Lemma 2.2 (see [29]).
LetE be a reflexive, strictly convex, and smooth Banach space and  C a nonempty, closed, and convex subset of  E. Let  x∈E, and  x0∈C. Then  x0=ΠCx if and only if(2.16)〈x0-y,Jx-Jx0〉≥0,∀y∈C.Lemma 2.3 (see [29]).
LetE be a reflexive, strictly convex, and smooth Banach space and  C a nonempty, closed, and convex subset of  E, and  x∈E. Then(2.17)ϕ(y,ΠCx)+ϕ(ΠCx,x)≤ϕ(y,x),∀y∈C.Lemma 2.4 (see [27]).
LetE be a reflexive, strictly convex, and smooth Banach space. Then one has the following(2.18)ϕ(x,y)=0⟺x=y,∀x,y∈E.


## 3. Main Results

Theorem 3.1.
LetE 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  ℝ satisfying  (A1)–(A4) such that  EP(f)≠∅. Let  {xn} be a sequence generated by the following manner:
(3.1)x0∈Echosen arbitrarily,C1=C,x1=ΠC1x0,yn∈C,such thatf(yn,u)+1rn〈u-yn,Jyn-Jxn〉≥0,∀u∈C,Cn+1={u∈Cn:2〈xn-u,Jxn-Jyn〉≥ϕ(xn,yn)},xn+1=ΠCn+1x0,∀n≥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 Lemma2.1, 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 Ch is convex for some h∈ℕ. Next, we show that Ch+1 is also convex for the same h. Let a,b∈Ch+1 and c=ta+(1-t)b, where t∈(0,1). It follows that
(3.2)ϕ(xh,yh)≤2〈xh-a,Jxh-Jyh〉,ϕ(xh,yh)≤2〈xh-b,Jxh-Jyh〉,where a,b∈Ch. From the above two inequalities, we can get that
(3.3)ϕ(xh,yh)≤2〈xh-c,Jxh-Jyh〉,where c∈Ch. It follows that Ch+1 is closed and convex. This completes the proof that Cn is closed, and convex.
Next, we show thatEP(f)⊂Cn. It is obvious that EP(f)⊂C=C1. Suppose that EP(f)⊂Ch for some h∈ℕ. For any z∈EP(f)⊂Ch, we see from Lemma 2.1 that(3.4)ϕ(z,yh)≤ϕ(z,xh).
On the other hand, we obtain from (2.6) that
(3.5)ϕ(z,yh)=ϕ(z,xh)+ϕ(xh,yh)+2〈z-xh,Jxh-Jyh〉.
Combining (3.4) with (3.5), we arrive at
(3.6)2〈xh-z,Jxh-Jyh〉≥ϕ(xh,yh)which implies that z∈Ch+1. This shows that EP(f)⊂Ch+1. This completes the proof that EP(f)⊂Cn.
Next, we show that{xn} is a convergent sequence and strongly converges to x¯, where x¯∈EP(f). Since xn=ΠCnx0, we see from Lemma 2.2 that(3.7)〈xn-z,Jx0-Jxn〉≥0,∀z∈Cn.
It follows from EP(f)⊂Cn that
(3.8)〈xn-w,Jx0-Jxn〉≥0,∀w∈EP⁡(f).
By virtue of Lemma 2.3, we obtain that
(3.9)ϕ(xn,x0)=ϕ(ΠCnx0,x0)≤ϕ(ΠEP⁡(f)x0,x0)-ϕ(ΠEP⁡(f)x0,xn)≤ϕ(ΠEP⁡(f)x0,x0).
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 xn⇀x¯. Since Cn is closed and convex, we see that x¯∈Cn. On the other hand, we see from the weakly lower semicontinuity of the norm that
(3.10)ϕ(x¯,x0)=‖x¯‖2-2〈x¯,Jx0〉+‖x0‖2≤lim⁡inf⁡n→∞(‖xn‖2-2〈xn,Jx0〉+‖x0‖2)=lim⁡inf⁡n→∞ϕ(xn,x0)≤limsup⁡n→∞ϕ(xn,x0)≤ϕ(x¯,x0),which implies that ϕ(xn,x0)→ϕ(x¯,x0) as n→∞. Hence, ∥xn∥→∥x¯∥ as n→∞. In view of the Kadec-Klee property of E, we see that xn→x¯ as n→∞. Notice that xn+1=ΠEP(f)x0∈Cn+1⊂Cn. It follows that
(3.11)ϕ(xn+1,xn)=ϕ(xn+1,ΠCnx0)≤ϕ(xn+1,x0)-ϕ(ΠCnx0,x0)=ϕ(xn+1,x0)-ϕ(xn,x0).
Since xn=ΠCnx0 and xn+1=ΠCn+1x0∈Cn+1⊂Cn, we arrive at ϕ(xn,x0)≤ϕ(xn+1,x0). This shows that {ϕ(xn,x0)} is nondecreasing. It follows from the boundedness that limn→∞ϕ(x,x0) exists. It follows that
(3.12)lim⁡n→∞ϕ(xn+1,xn)=0.
By virtue of xn+1=ΠCn+1x0∈Cn+1, we find that
(3.13)ϕ(xn,yn)≤2〈xn-xn+1,Jxn-Jyn〉.
It follows that
(3.14)lim⁡n→∞ϕ(xn,yn)=0.
In view of (2.5), we see that
(3.15)lim⁡n→∞(‖xn‖-‖yn‖)=0.
Since xn→x¯, we find that
(3.16)lim⁡n→∞‖yn‖=‖x¯‖.
It follows that
(3.17)lim⁡n→∞‖Jyn‖=‖Jx¯‖.
This implies that {Jyn} is bounded. Note that both E and E* are reflexive. We may assume that Jyn⇀y*∈E*. In view of the reflexivity of E, we see that there exists an element y∈E such that Jy=y*. It follows that
(3.18)ϕ(xn,yn)=‖xn‖2-2〈xn,Jyn〉+‖yn‖2=‖xn‖2-2〈xn,Jyn〉+‖Jyn‖2.
Taking liminfn→∞ on the both sides of the equality above yields that
(3.19)0≥‖x¯‖2-2〈x¯,y*〉+‖y*‖2=‖x¯‖2-2〈x¯,Jy〉+‖Jy‖2=‖x¯‖2-2〈x¯,Jy〉+‖y‖2=ϕ(x¯,y).
That is, x¯=y, which in turn implies that y*=Jx¯. It follows that Jyn⇀Jx¯∈E*. Since E* enjoys the Kadec-Klee property, we obtain from (3.17) that limn→∞Jyn=Jx¯. Since J-1:E*→E is demicontinuous, we find that yn⇀x¯. This implies from (3.16) and the Kadec-Klee property of E that limn→∞yn=x¯. This in turn implies that limn→∞∥yn-xn∥=0. Since J is uniformly norm-to-norm continuous on any bounded sets, we find that
(3.20)lim⁡n→∞‖Jyn-Jxn‖=0.
Next, we show that x¯∈EF(f). In view of Lemma 2.1, we find from yn=Trnfxn that
(3.21)f(yn,u)+1rn〈u-yn,Jyn-Jxn〉≥0,∀u∈C.
It follows from condition (A2) and (3.20) that
(3.22)1rn‖u-yn‖‖Jyn-Jxn‖≥f(u,yn),∀u∈C.
In view of condition (A4), we obtain from (3.17) that
(3.23)f(u,x¯)≤0,∀u∈C.
For 0<t<1 and u∈C, define ut=tu+(1-t)x¯. It follows that ut∈C, which yields that f(ut,x¯)≤0. It follows from conditions (A1) and (A4) that
(3.24)0=f(ut,ut)≤tf(ut,u)+(1-t)f(ut,x¯)≤tf(ut,u).
That is,
(3.25)f(ut,u)≥0.
Letting t↓0, we find from condition (A3) that f(x¯,u)≥0, forallu∈C. This implies that x¯∈EP(f). This shows that x¯∈EP(f).
Finally, we prove thatx¯=ΠEP(f)x0. Letting n→∞ in (3.8), we see that(3.26)〈x¯-w,Jx0-Jx¯〉≥0,∀w∈EP⁡(f).
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.
LetE be a Hilbert space and  C a nonempty, closed, and convex subset of  E. Let  f be a bifunction from  C×C to  ℝ satisfying  (A1)–(A4) such that  EP(f)≠∅. Let  {xn} be a sequence generated by the following manner:
(3.27)x0∈Echosen arbitrarily,C1=C,x1=PC1x0,yn∈C,such thatf(yn,u)+1rn〈u-yn,yn-xn〉≥0,∀u∈C,Cn+1={u∈Cn:2〈xn-u,xn-yn〉≥‖xn-yn‖2},xn+1=PCn+1x0,∀n≥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¯=PEP(f)x0.



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*Source: 812783-2012-04-24.xml*