# Approximate Equilibrium Problems and Fixed Points

Authors: H. Mazaheri; S. A. M. Mohsenalhosseini
Journal: International Journal of Analysis
 (2013)

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## Abstract

We find a common element of the set of fixed points of a map and the set of solutions of an approximate equilibrium problem in a Hilbert space. Then, we show that one of the sequences weakly converges. Also we obtain some theorems about equilibrium problems and fixed points.


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## Body

## 1. Introduction

Equilibrium is “everywhere:” in economics, physics, engineering, chemistry, biology, and so forth. From the mathematical modelling point of view equilibrium can be described in fixed point theorems, optimisation problems, variational inequalities, complementarity problems, and so forth. Equilibrium systems can be studied from several points of view: existence of solutions; existence of nontrivial solutions; number of solutions; properties of the solution set; and the numerical approximation of solutions. In the first a short description of what is a mathematical equilibrium system in general, we will present several particular classes of approximate equilibrium systems and the relations between then.closed, bounded, and convex subset ofH.Condition 1.
The following condition appears implicitly in [1].
We assume that the mapφ:C×C→R satisfies the following conditions:(i)
φ(x,x)=0,forallx∈C;(ii)
φis monotone,that is,φ(x,y)+φ(y,x)≤0,forallx,y∈C;(iii)
forallx,y∈C,
(1)limt→0φ(xt+(1-t)x,y)≤φ(x,y);(iv)
foreachfixedx∈C, thefunctiony↦φ(x,y)isconvexandlowersemicontinuous.Condition 2.
Assume that the mapφ:C×C→R for ϵ>0 satisfies the following conditions.(A1)
φ(x,x)=ϵ,forallx∈C.(A2)
φisapproximatemonotone,that is,φ(x,y)+φ(y,x)≤ϵ,forallx,y∈C.(A3)
Forallx,y∈C,
(2)limt→0φ(xt+(1-t)x,y)≤φ(x,y).(A4)
Foreachfixedx∈C, thefunctiony↦φ(x,y)isconvexandlowersemicontinuous.Definition 1 (see [1]).
We say thatx*∈C is an equilibrium point of φ:C×C→R  if there exists a x*∈C, such that
(3)φ(x*,y)≥0∀y∈C;
the set of such x*∈C is denoted by EP(φ); that is,
(4)EP(φ)={x*∈C:φ(x*,y)≥0,∀y∈C}.Definition 2.
Supposeϵ>0, we say that x*∈C is an approximate equilibrium point of φ:C×C→R if there exists a x*∈C, such that
(5)φ(x*,y)≥ϵ∀y∈C.
In this paper, the set of such an x*∈C is denoted by AEP(φ), that is,
(6)AEP(φ)={x*∈C:φ(x*,y)≥ϵ,∀y∈C},
and we set(7)AF(T)={x∈C:d(x,Tx)<ϵfor  someϵ>0},F(T)={x∈C:Tx=x}.


## 2. Preliminaries

In the following we will present a known lemma which is needed in the proof of some results (see [2]).Lemma 3.
Letφ:C×C→R be a map satisfies Condition 1. Let r>0, and let  x∈H. Then there exists a z∈C, such that
(8)φ(z,y)+〈y-z,z-x〉r≥0∀y∈C.Lemma 4.
Letφ:C×C→R be a map satisfies Condition 2. Let r>0, ϵ>0, and let x∈H. Then there exists a z∈C, such that
(9)φ(z,y)+〈y-z,z-x〉r≥ϵ∀y∈C.Proof.
Consider the mapψ:C×C→R by ψ(x,y)=φ(x,y)-ϵ. The map ψ satisfies Condition 1. Then by Lemma 3, there exists a z∈C, such that
(10)ψ(z,y)+〈y-z,z-x〉r≥0∀y∈C.
Thus(11)φ(z,y)+〈y-z,z-x〉r≥ϵ∀y∈C.Lemma 5.
Letφ:C×C→R be a map satisfies Condition 2. For r>0, x∈H, ϵ>0 we defined Tr:H→C, such that
(12)Tr(x)={z∈C∣φ(z,y)+〈y-z,z-x〉r≥ϵ}.
Then(a)
Tr is single valued.(b)
Tr is firmly nonexpansive; that is,
(13)∥Tr(x)-Tr(y)∥2≤〈Tr(x)-Tr(y),x-y〉∀x,y∈H.(c)
F(Tr)=AEP(φ).(d)
AEP(φ) is nonempty, closed, and convex.Proof.
(a) Forx∈H and r>0, let z1,z2∈Tr(x). Then,
(14)φ(z1,y)+〈y-z1,z1-x〉r≥ϵ∀y∈C,φ(z2,y)+〈y-z2,z2-x〉r≥ϵ∀y∈C.
Then
(15)φ(z1,z2)≥〈z2-z1,x-z1〉r+ϵ,φ(z2,z1)≥〈z1-z2,x-z2〉r+ϵ.
Since φ is approximate monotone,
(16)ϵ≥φ(z1,z2)+φ(z2,z1)≥〈z2-z1,x-z1〉r+〈z1-z2,x-z2〉r+2ϵ≥〈z2-z1,z1-z2〉r+ϵ.
Now, since ϵ>0 and r>0, we have
(17)〈z2-z1,z1-z2〉≤0.
So, we have z1=z2.
(b) Now we claim thatTr is a firmly nonexpansive. Indeed, if x,y∈H,
(18)φ(Tr(x),Tr(y))+〈Tr(y)-Tr(x),Tr(x)-x〉r≥ϵ,φ(Tr(y),Tr(x))+〈Tr(x)-Tr(y),Tr(y)-y〉r≥ϵ.
Adding the two inequalities, we have
(19)φ(Tr(x),Tr(y))+φ(Tr(y),Tr(x))+〈Tr(y)-Tr(x),Tr(x)-Tr(y)-x+y〉r≥2ϵ.
With (A2), we have
(20)〈Tr(y)-Tr(x),Tr(x)-Tr(y)-(x-y)〉r≥ϵ.
Now since ϵ>0 and r>0, then
(21)〈Tr(y)-Tr(x),Tr(x)-Tr(y)-(x-y)〉≥0
so
(22)∥Tr(x)-Tr(y)∥2≤〈Tr(x)-Tr(y),x-y〉.
(c) Take x∈K. Then
(23)x∈F(Tr)→x=Tr(x)→φ(x,y)+〈x-x,y-x〉r≥ϵ∀y∈C→φ(x,y)≥ϵ∀y∈C→x∈AEP(φ).
(d) At last, we claim that AEP(φ) is closed and convex. Indeed, since Tr is firmly nonexpansive, Tr is also nonexpansive, and since the fixed-point set of a nonexpansive operator is closed and convex [3, proposition 1.5.3]. Therefore follows from (b), (c).In the following we will present a known theorem which is needed in the proof of some results (see [4]).Theorem 6.
Letφ:C×C→R be a map satisfies Condition 1, and let  S:C→C be a nonexpentive mapping such that F(S)∩EP(φ)≠∅. Let {xn} and {un} be sequences generated initially by an arbitrary element x1∈H and then by
(24)φ(un,y)+〈y-un,un-xn〉rn≥0∀y∈C,xn+1=αnun+(1-αn)Sun∀n≥1,
where {αn} and {rn} satisfy the following conditions:(i)
{αn}⊂[α,β] for some α,β∈(0,1);(ii)
{rn}⊂(0,∞) and liminfnrn>0.
Then, the sequences{xn} and {un} converge weakly to an element of F(S)∩EP(φ)≠∅.Lemma 7.
Letψ:C×C→R be a map by ψ(x,y)=φ(x,y)-ϵ that satisfies Condition 1, and let  S:C→C be a nonexpentive mapping, then F(S)∩AEP(φ)≠∅ if and only if F(S)∩EP(ψ)≠∅.In the following we will present a theorem which is extended Theorem6.Theorem 8.
Letφ:C×C→R be a map satisfies Condition 2, and let  S:C→C be a map such that F(S)∩AEP(φ)≠∅. Let {xn} and {un} be sequences generated initially by an arbitrary element x1∈H and then by
(25)φ(un,y)+〈y-un,un-xn〉rn≥ϵ∀y∈C,xn+1=αnun+(1-αn)Sun∀n≥1,
where {αn} and {rn} satisfy the following conditions:(i)
{αn}⊂[α,β] for some α,β∈(0,1);(ii)
{rn}⊂(0,∞) and liminfnrn>0.Then, the sequences {xn} and {un} converge weakly to an element of F(S)∩AEP(φ).Proof.
Suppose the napψ:C×C→R by ψ(x,y)=φ(x,y)-ϵ that satisfies Condition 1. Since by F(S)∩AEP(φ)≠∅, Lemma 9 implies that F(S)∩EP(ψ)≠∅ and holds conditions theorem (2.8), therefore the sequences {xn} and {un} converge weakly to an element of z∈F(S)∩EP(ψ).By Lemma 9 the sequences {xn} and {un} converge weakly to z∈F(S)∩AEP(φ).Lemma 9.
Letφ:C×C→R be a map satisfies Condition 1. For x∈C, define a mapping T:C→C such that is contraction. If φ(Tx,y)≥0  for all y∈C, then there exists x0∈EP(T).Proof.
SinceC is a nonempty closed, bounded, and convex subset of H, and T is continues, then by Browde’s theorem there exists a x0∈C:Tx0=x0. Since φ(Tx0,y)≥0 for all  y∈C, thus φ(x0,y)≥0  for all  y∈C.So ∃x0∈C:x0∈EP(T).Lemma 10.
Letφ:C×C→R be a map satisfies Condition 1. For x∈C, define a mapping T:C→C such that is contraction. If φ(Tnx,y)≥0,  for all y∈C, then x0∈EP(T).Proof.
SinceC is a nonempty closed, bounded, and convex subset of H, and T is contraction, now by Theorem 2.1 of [4], since {Tnx} converges to fixed point T, then there exists a x0∈C:Tnx0=x0. Since for all y∈C, φ(Tnx0,y)≥0, thus φ(x0,y)≥0 for all  y∈C. So there exists a x0∈C, such that  x0∈EP(T).Lemma 11.
Letφ:C→R be a real-valued function, let φ:C×C→R be a map satisfies Condition 2, and let T:C→C be a nonlinear onto mapping and satisfying
(26)∥x-Tx∥+φ(x,Tx)<ϵ∀x∈C.
If   EP(φ)≠∅,  then AF(T)≠∅.Proof.
Ifz∈EP(φ), then
(27)0≤φ(z,y)∀y∈C.
Since z∈C, there exists x∈C, such that Tx=z. Therefore φ(Tx,x)≥0, and so
(28)d(x,Tx)+φ(x,Tx)<ϵ.
It follows that d(x,Tx)<ϵ and x∈AF(T).Lemma 12.
Let the mapφ:C×C→R satisfy Condition 2. For x∈C, define a mapping T:C→C such that is contraction. If φ(x,Tx)<d(x,Tx),  for all  x∈C, then there exists a x0∈AEP(T).Proof.
SinceT is contraction by Theorem 2.1. of [1] AF(T)≠∅, forallϵ>0, and since for all x∈Cφ(x,Tx)<∥x-Tx∥, therefore for all x∈C and for some ϵ>0,φ(x,Tx)<ϵ,for allx∈C. Thusx0∈AEP(T).



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