# Algorithmic Approach to the Equilibrium Points and Fixed Points

Authors: Lijin Guo; Shin Min Kang; Young Chel Kwun
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

The equilibrium and fixed point problems are considered. An iterative algorithm is presented. Convergence analysis of the algorithm is provided.


---

## Body

## 1. Introduction

LetH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥, respectively. Let C be a nonempty closed convex subset of H. Let Φ:C→H be a nonlinear operator and let Θ:C×C→R be a bifunction. The equilibrium problem is formulated as finding x†∈C such that
(1)Θ(x†,x)+〈Φ(x†),x-x†〉≥0,∀x∈C.
The solution set of (1) is denoted by EP. The problem (1) is very general in the sense that it includes, as special cases, optimization problems, variational inequalities, minimax problems, Nash equilibrium problem in noncooperative games, and others. For related work, please see, for example, [1–17]. Next, we recall several interesting results where Θ verifies the following usual conditions (C1)–(C4) which will be used in the sequel:(C1)
Θ(u,u)=0 for all u∈C;(C2)
Θ is monotone; that is, Θ(u,v)+Θ(v,u)≤0 for all u,v∈C;(C3)
for eachu,v,w∈C, lim⁡t↓0Θ(tw+(1-t)u,v)≤Θ(u,v);(C4)
for eachu∈C,  v↦Θ(u,v) is convex and lower semicontinuous.Theorem 1.
LetC be a nonempty closed convex subset of H. Let Θ:C×C→R be a bifunction which satisfies conditions (C1)–(C4). Let S:C→C be a nonexpansive mapping. Let ϕ:H→H be a contraction. For x0∈H arbitrarily, let the sequence {xn} be generated by
(2)Θ(yn,x†)+1λn〈x†-yn,yn-xn〉≥0,∀x†∈C,xn+1=αnϕ(xn)+(1-αn)Syn,n≥0,
where {αn}⊂[0,1] and {λn}⊂(0,∞) satisfy lim⁡n→∞αn=0, ∑n=1∞‍αn=∞,   ∑n=1∞‍|αn+1-αn|<∞,liminf⁡n→∞λn>0, and ∑n=1∞‍|λn+1-λn|<∞. Then, the sequence {xn} converges strongly to x‡=PF(S)∩EPϕ(x‡) provided F(S)∩EP≠∅.Chuang et al. [18] considered an iteration process of Halpern’s type for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points for a nonexpansive mapping with perturbation in a Hilbert space and they proved a strong convergence theorem for such iterations.Theorem 2.
LetC be a nonempty closed convex subset of H. Let Θ:C×C→R be a bifunction which satisfies conditions (C1)–(C4). Let S:C→H be a nonexpansive mapping. Let {un}⊂H be a sequence. For x0∈H arbitrarily, let the sequence {xn} be generated by
(3)Θ(yn,x†)+1λn〈x†-yn,yn-xn〉≥0,∀x†∈C,xn+1=αnun+(1-αn)(βnyn+(1-βn)Syn),n≥0,
where {αn}⊂[0,1] and {λn}⊂[a,∞) satisfy lim⁡n→∞αn=0,   ∑n=1∞‍αn=∞,lim⁡n→∞βn(1-βn)>0, and lim⁡n→∞un=u∈H. Then, the sequence {xn} converges strongly to x‡=PF(S)∩EP(u) provided F(S)∩EP≠∅.S. Takahashi and W. Takahashi [17] introduced the following iterative algorithm for finding an element of F(S)∩EP:
(4)Θ(yn,x†)+〈Φ(xn),x†-yn〉+1λn〈x†-yn,yn-xn〉≥0,∀x†∈C,xn+1=βnxn+S[αnu+(1-βn)yn],n≥0.And they proved that the sequence{xn} converges strongly to x‡=PF(S)∩EP⁡(u).Remark 3.
Algorithm (3) is involved in a variant anchor {un} and the parameters are also relaxed. In [17], the authors considered a general equilibrium problem.In this paper, our main purpose is to introduce a new iteration process for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points for a nonexpansive mapping in a Hilbert space and then we prove a strong convergence theorem for such iterations. Our iterations are very different from (2)–(4). As a special case, we can find the minimum norm solution of F(S)∩EP⁡.


## 2. Preliminaries

In the sequel, we assumeH is a real Hilbert space. Let C⊂H be a nonempty closed convex set. Recall that a mapping Φ:C→H is called α-inverse-strongly monotone if there exists a positive real number α>0 such that
(5)〈Φ(u)-Φ(v),u-v〉≥α∥Φ(u)-Φ(v)∥2,∀u,v∈C.
A mapping S:C→C is said to benonexpansive if
(6)∥Su-Sv∥≤∥u-v∥,∀u,v∈C.
We use F(T) to denote the set of fixed points of S.The following lemmas are useful for the next section.Lemma 4 (see [11]).
LetC be a nonempty closed convex subset of H. Let Θ:C×C→R be a bifunction which satisfies conditions (C1)–(C4). Let λ>0 and x∈H. Then, there exists y∈C such that
(7)Θ(y,x†)+1λ〈x†-y,y-x〉≥0,∀x†∈C.
Set Tλ(x)={y∈C:Θ(y,x†)+(1/λ)〈x†-y,y-x〉≥0,forallx†∈C}. Then the following hold:(i)
Tλ is single-valued and Tλ is firmly nonexpansive; that is for any x,y∈H,  ∥Tλx-Tλy∥2≤〈Tλx-Tλy,x-y〉;(ii)
EP is closed and convex and EP=F(Tλ).Lemma 5.
LetC, H, F, and Tλx be as in Lemma 4. Then the following holds:
(8)∥Tλx-Tγx∥2≤λ-γλ〈Tλx-Tγx,Tλx-x〉
for all λ,γ>0 and x∈H.Lemma 6 (see [19]).
LetC be a nonempty closed convex subset of H. Let the mapping Φ:C→H be α-inverse-strongly monotone and let λ>0 be a constant. Then, we have
(9)∥(I-λΦ)u-(I-λΦ)v∥2≤∥u-v∥2+λ(λ-2α)∥Φ(u)-Φ(v)∥2,∀u,v∈C.
In particular, if 0≤λ≤2α, then I-λΦ is nonexpansive.Lemma 7 (see [20]).
LetC be a closed convex subset of H and let S:C→C be a nonexpansive mapping with F(S)≠∅. Then, the mapping I-S is demiclosed. That is, if {xn} is a sequence in C such that xn→x* weakly and (I-S)xn→y strongly, then (I-S)x*=y.Lemma 8 (see [21]).
Let{Rn} and {Wn} be two bounded sequences in H. Let {βn} be a sequence in [0,1] satisfying 0<liminf⁡n→∞βn≤limsup⁡n→∞βn<1. Suppose that Rn+1=(1-βn)Wn+βnRn for all n≥0 and limsup⁡n→∞(∥Wn+1-Wn∥-∥Rn+1-Rn∥)≤0. Then, lim⁡n→∞∥Wn-Rn∥=0.Lemma 9 (see [22]).
Assume that{an} is a sequence of nonnegative real numbers such that
(10)an+1≤(1-γn)an+δnγn,
where {γn} is a sequence in (0,1) and {δn} is a sequence such that(1)
∑n=1∞‍γn=∞;(2)
limsup⁡n→∞δn≤0 or ∑n=1∞‍|δnγn|<∞. Then lim⁡n→∞an=0.


## 3. Main Results

In this section, we will prove our main results.Theorem 10.
LetC be a nonempty closed convex subset of H and let Θ:C×C→R be a bifunction satisfying conditions (C1)–(C4). Let Φ:C→H be an ς-inverse-strongly monotone mapping and let S:C→C be a nonexpansive mapping. Suppose that F(S)∩EP≠∅. Let {un} be a sequence in H. For x0∈C arbitrarily, let the sequence {xn} be generated by
(11)xn+1=ϖnxn+(1-ϖn)STρn[(1-ςn)xnvvvvvvvvvvvvvvvvvvv+ςnun-ρnΦ(xn)],n≥0,
where Tρn is defined as that in Lemma 4 and {ρn}⊂(0,2ς),    {ςn}⊂(0,1), and {ϖn}⊂(0,1) satisfy(i)
lim⁡n→∞ςn=0 and ∑n=1∞‍ςn=∞;(ii)
0<c≤ϖn≤d<1;(iii)
a(1-ςn)≤ρn≤b(1-ςn), where [a,b]⊂(0,2ς) and lim⁡n→∞(ρn+1-ρn)=0;(iv)
lim⁡n→∞un=u. Then {xn} generated by (11) converges strongly to PF(S)∩EP(u).We divide our proofs into several conclusions.Conclusion.
The sequence{xn} is bounded.Proof.
Letz∈F(S)∩EP⁡. We have z=Sz=Tρn(z-ρnΦ(z))=Tρn[ςnz+(1-ςn)(z-ρnΦ(z)/(1-ςn))] for all n≥0. Set zn=Tρn[(1-ςn)xn+ςnun-ρnΦ(xn)] for all n≥0. By Lemma 4, we know that Tρn is nonexpansive. By the convexity of ∥·∥, we derive
(12)∥zn-z∥2=∥Tρn[(1-ςn)xn+ςnun-ρnΦ(xn)]vvv-Tρn(z-ρnΦ(z))∥2=∥Tρn[ςnun+(1-ςn)(xn-ρn1-ςnΦ(xn))]xxx-Tρn[ςnz+(1-ςn)(z-ρn1-ςnΦ(z))]∥2≤∥[ςnun+(1-ςn)(xn-ρn1-ςnΦ(xn))]xxx-[ςnz+(1-ςn)(z-ρn1-ςnΦ(z))]∥2=∥(1-ςn)[(xn-ρn1-ςnΦ(xn))vvvvvvvvxxx-(z-ρn1-ςnΦ(z))]+ςn(un-z)∥2≤(1-ςn)∥(xn-ρn1-ςnΦ(xn))xxxxxxxxx-(z-ρn1-ςnΦ(z))∥2+ςn∥un-z∥2.
Since Φ is ς-inverse-strongly monotone, we know from Lemma 6 that
(13)∥(xn-ρn1-ςnΦ(xn))-(z-ρn1-ςnΦ(z))∥2≤∥xn-z∥2+ρn(ρn-2(1-ςn)ς)(1-ςn)2∥Φ(xn)-Φ(z)∥2.
It follows that
(14)∥zn-z∥2≤(1-ςn)(∥xn-z∥2+ρn(ρn-2(1-ςn)ς)(1-ςn)2hhhhhhhh×∥Φ(xn)-Φ(z)∥2ρn(ρn-2(1-ςn)ς)(1-ςn)2ρn(ρn-2(1-ςn)ς)(1-ςn)2)+ςn∥un-z∥2,≤(1-ςn)∥xn-z∥2+ςn∥un-z∥2.
So, we have that
(15)∥xn+1-z∥2=∥ϖn(xn-z)+(1-ϖn)(Szn-z)∥2≤ϖn∥xn-z∥2+(1-ϖn)∥zn-z∥2≤ϖn∥xn-z∥2+(1-ϖn)((1-ςn)∥xn-z∥2vvvvvvvvvvvvvvvvvvvvv+ςn∥un-z∥2)=[1-(1-ϖn)ςn]∥xn-z∥2+(1-ϖn)ςn∥un-z∥2≤max⁡{∥un-z∥2,∥un-z∥2}.
Note that lim⁡n→∞un=u∈H. Without loss of generality, we can assume that sup⁡n∥un-u∥≤M for some M>0.
By induction, we have(16)∥xn+1-z∥2≤max⁡{∥x0-z∥2,(∥un-u∥+∥u-z∥)2}≤max⁡{∥x0-z∥2,(M+∥u-z∥)2}.
Therefore, {xn} is bounded.Conclusion.
lim
⁡
n
→
∞
∥
x
n
+
1
-
x
n
∥
=
0 and lim⁡n→∞∥xn-Tρn[(1-ςn)xn+ςnun-ρnΦ(xn)]∥=0.Proof.
Puttingwn=(1-ςn)xn+ςnun-ρnΦ(xn) for all n≥0, we have
(17)zn+1-zn=Tρn+1wn+1-Tρn+1wn+Tρn+1wn-Tρnwn.
It follows that
(18)∥zn+1-zn∥≤∥Tρn+1wn+1-Tρn+1wn∥+∥Tρn+1wn-Tρnwn∥≤∥wn+1-wn∥+∥Tρn+1wn-Tρnwn∥.
From Lemma 6, we know that I-ρΦ is nonexpansive for all ρ∈(0,2ς). Thus, we have that I-((ρn+1)/(1-ςn+1))Φ is nonexpansive for all n due to the fact that (ρn+1)/(1-ςn+1)∈(0,2ς). Then, we get
(19)∥wn+1-wn∥=∥(1-ςn+1)xn+1+ςn+1un+1-ρn+1Φ(xn+1)v-((1-ςn)xn+ςnun-ρnΦ(xn))∥=∥(1-ςn+1)(xn+1-ρn+11-ςn+1Φ(xn+1))vv-(1-ςn)(xn-ρn1-ςnΦ(xn))vv+(ςn+1-ςn)un+1+ςn(un+1-un)ρn+11-ςn+1ρn+11-ςn+1Φ∥≤(1-ςn+1)∥(I-ρn+11-ςn+1Φ)xn+1vvvvvvvvv-(I-ρn+11-ςn+1Φ)xn∥+∥(1-ςn+1)xn-ρn+1Φ(xn)vvv-(1-ςn)xn+ρnΦ(xn)∥+|ςn+1-ςn|∥un+1∥+ςn∥un+1-un∥≤∥xn+1-xn∥+|ςn+1-ςn|(∥xn∥+∥un+1∥)+ςn∥un+1-un∥+|ρn+1-ρn|∥Φ(xn)∥.
By Lemma 5, we have
(20)∥Tρn+1wn-Tρnwn∥≤|ρn+1-ρn|ρn+1∥Tρn+1wn-wn∥.
From (18)–(20), we obtain
(21)∥zn+1-zn∥≤∥xn+1-xn∥+|ςn+1-ςn|(∥xn∥+∥un+1∥)+ςn∥un+1-un∥+|ρn+1-ρn|∥Φ(xn)∥+|ρn+1-ρn|ρn+1∥Tρn+1wn-wn∥.
Then,
(22)∥Szn+1-Szn∥≤∥zn+1-zn∥≤∥xn+1-xn∥+|ςn+1-ςn|×(∥xn∥+∥un+1∥)+ςn∥un+1-un∥+|ρn+1-ρn|∥Φ(xn)∥+|ρn+1-ρn|ρn+1∥Tρn+1wn-wn∥.
Therefore,
(23)∥Szn+1-Szn∥-∥xn+1-xn∥≤|ςn+1-ςn|(∥xn∥+∥un+1∥)+ςn∥un+1-un∥+|ρn+1-ρn|∥Φ(xn)∥+|ρn+1-ρn|ρn+1∥Tρn+1wn-wn∥.
Since ςn→0,  ρn+1-ρn→0 and liminf⁡n→∞ρn>0, we obtain
(24)limsup⁡n→∞(∥Szn+1-Szn∥-∥xn+1-xn∥)≤0.
By Lemma 8, we get
(25)lim⁡n→∞∥Szn-xn∥=0.
Consequently, we obtain
(26)lim⁡n→∞∥xn+1-xn∥=lim⁡n→∞(1-ϖn)∥Szn-xn∥=0.
From (11) and (14), we have
(27)∥xn+1-z∥2≤ϖn∥xn-z∥2+(1-ϖn)×∥STρn[(1-ςn)xn+ςnun-ρnΦ(xn)]-z∥2≤(1-ϖn){(1-ςn)(∥xn-z∥2+ρn(1-ςn)2vvvvvvvvvvvvvvvvvvv×(ρn-2(1-ςn)ς)vvvvvvvvvvvvvvvvvvv×∥Φ(xn)-Φ(z)∥2ρn(1-ςn)2)ρn(1-ςn)2vvvvvvvvv+ςn∥un-z∥2(∥xn-z∥2+ρn(1-ςn)2}+ϖn∥xn-z∥2=[1-(1-ϖn)ςn]∥xn-z∥2+(1-ϖn)ρn1-ςn×(ρn-2(1-ςn)ς)∥Φ(xn)-Φ(z)∥2+(1-ϖn)ςn∥un-z∥2≤∥xn-z∥2+(1-ϖn)ρn1-ςn(ρn-2(1-ςn)ς)×∥Φ(xn)-Φ(z)∥2+(1-ϖn)ςn∥un-z∥2.
Then, we obtain
(28)(1-ϖn)ρn1-ςn(2(1-ςn)ς-ρn)∥Φ(xn)-Φ(z)∥2≤∥xn-z∥2-∥xn+1-z∥2+(1-ϖn)ςn∥un-z∥2≤(∥xn-z∥-∥xn+1-z∥)∥xn+1-xn∥+(1-ϖn)ςn∥un-z∥2.
Since lim⁡n→∞ςn=0,    lim⁡n→∞∥xn+1-xn∥=0, and liminf⁡n→∞((1-ϖn)ρn/(1-ςn))(2(1-ςn)ς-ρn)>0, we have
(29)lim⁡n→∞∥Φ(xn)-Φ(z)∥=0.
Next, we show ∥xn-Tρnwn∥→0. By using the firm nonexpansivity of Tρn, we have
(30)∥Tρnwn-z∥2=∥Tρnwn-Tρn(z-ρnΦ(z))∥2≤〈wn-(z-ρnΦ(z)),Tρnwn-z〉=12(∥wn-(z-ρnΦ(z))∥2+∥Tρnwn-z∥2vvvvvv-∥Tρnwn(1-ςn)xn+ςnunvvvvvvvvvv-ρn(Φ(xn)-Φ(z))-Tρnwn∥2∥Tρnwn-z∥).
We note that
(31)∥wn-(z-ρnΦ(z))∥2≤(1-ςn)∥xn-z∥2+ςn∥un-z∥2.
Thus,
(32)∥Tρnwn-z∥2≤12(-ρn(Φ(xn)-Φ(z))Tρnwn∥2(1-ςn)∥xn-z∥2vv+ςn∥un-z∥2+∥Tρnwn-z∥2vv-∥(1-ςn)xn+ςnun-Tρnwnvvvvv-ρn(Φ(xn)-Φ(z))Tρnwn∥2);
that is,
(33)∥Tρnwn-z∥2≤(1-ςn)∥xn-z∥2+ςn∥un-z∥2-∥(1-ςn)xn+ςnun-Tρnwnccc-ρn(Φ(xn)-Φ(z))Tρnwn∥2=(1-ςn)∥xn-z∥2+ςn∥un-z∥2-∥(1-ςn)xn+ςnun-Tρnwn∥2+2ρn〈(1-ςn)xn+ςnun-Tρnwn,Φ(xn)vvvvvvv-Φ(z)(1-ςn)xn+ςnun-Tρnwn,Φ(xn)〉-ρn2∥Φ(xn)-Φ(z)∥2≤(1-ςn)∥xn-z∥2+ςn∥un-z∥2-∥(1-ςn)xn+ςnun-Tρnwn∥2+2ρn∥(1-ςn)xn+ςnun-Tρnwn∥×∥Φ(xn)-Φ(z)∥.
It follows that
(34)∥xn+1-z∥2≤ϖn∥xn-z∥2+(1-ϖn)(1-ςn)×∥xn-z∥2+(1-ϖn)ςn∥un-z∥2-(1-ϖn)∥(1-ςn)xn+ςnun-Tρnwn∥2+2ρn(1-ϖn)∥(1-ςn)xn+ςnun-Tρnwn∥×∥Φ(xn)-Φ(z)∥=[1-(1-ϖn)ςn]∥xn-z∥2+(1-ϖn)ςn∥un-z∥2-(1-ϖn)×∥(1-ςn)xn+ςnun-Tρnwn∥2+2ρn(1-ϖn)∥(1-ςn)xn+ςnun-Tρnwn∥vv×∥Φ(xn)-Φ(z)∥.
Hence,
(35)(1-ϖn)∥(1-ςn)xn+ςnun-Tρnwn∥2≤∥xn-z∥2-∥xn+1-z∥2-(1-ϖn)ςn∥xn-z∥2+(1-ϖn)ςn∥un-z∥2+2ρn(1-ϖn)xx×∥(1-ςn)xn+ςnun-Tρnwn∥∥Φ(xn)-Φ(z)∥≤(∥xn-z∥+∥xn+1-z∥)∥xn+1-xn∥+(1-ϖn)ςn∥un-z∥2+2ρn(1-ϖn)xx×∥(1-ςn)xn+ςnun-Tρnwn∥∥Φ(xn)-Φ(z)∥.
Since limsup⁡n→∞ϖn<1,  ∥xn+1-xn∥→0,  ςn→0, and ∥Φ(xn)-Φ(z)∥→0, we deduce
(36)lim⁡n→∞∥(1-ςn)xn+ςnun-Tρnwn∥=0.
This implies that
(37)lim⁡n→∞∥xn-Tρnwn∥=lim⁡n→∞∥xn-zn∥=0.Conclusion.
Consider(38)limsup⁡n→∞〈x~-u,xn-x~〉≥0,
where x~=PF(S)∩EP(u).Proof.
Since{xn} is bounded, there exists a subsequence {xni} of {xn} such that xni→w weakly and
(39)limsup⁡n→∞〈x~-u,xn-x~〉=lim⁡i→∞〈x~-u,xni-x~〉.
By (25) and (37), we deduce
(40)lim⁡n→∞∥xn-Sxn∥=0.
Hence,
(41)lim⁡i→∞∥xni-Sxni∥=0.
This together with Lemma 7 implies that w∈F(S).
Next we show thatx~∈EP⁡. Since zn=Tρn[(1-ςn)xn+ςnun-ρnΦ(xn)] for any y∈C, we have
(42)Θ(zn,y)+〈Φ(xn),y-zn〉+1ρn〈y-zn,zn-(1-ςn)xn-ςnun〉≥0.
From (C2), we have
(43)〈Φ(xn),y-zn〉+1ρn〈y-zn,zn-(1-ςn)xn-ςnun〉≥Θ(y,zn).
Put xt=ty+(1-t)x~ for all t∈(0,1) and y∈C. Then, we have xt∈C. So, from (43), we have
(44)〈xt-zn,Φ(xt)〉≥〈xt-zn,Φ(xt)〉-〈Φ(xn),xt-zn〉-1ρn〈xt-zn,zn-(1-ςn)xn-ςnun〉+Θ(xt,zn)=〈xt-zn,Φ(xt)-Φ(zn)〉+〈xt-zn,Φ(zn)-Φ(xn)〉-1ρn〈xt-zn,zn-(1-ςn)xn-ςnun〉+Θ(xt,zn).
Since Φ is ς-inverse-strongly monotone, Φ is (1/ς)-Lipschitzian. By (37), we derive that ∥Φ(zn)-Φ(xn)∥→0. Further, from monotonicity of Φ, we have 〈xt-zn,Φ(xt)-Φ(zn)〉≥0. Letting n→∞ in (45) and noting (C4), we have
(45)〈xt-x~,Φ(xt)〉≥Θ(xt,x~).
By (C1), (C4), and (45), we deduce
(46)0=Θ(xt,xt)≤tΘ(xt,y)+(1-t)Θ(xt,x~)≤tΘ(xt,y)+(1-t)〈xt-x~,Φ(xt)〉=tΘ(xt,y)+(1-t)t〈y-x~,Φ(xt)〉,
and hence
(47)0≤Θ(xt,y)+(1-t)〈y-x~,Φ(xt)〉.
Letting t→0, we have, for each y∈C,
(48)0≤Θ(x~,y)+〈y-x~,Φ(x~)〉.
This implies x~∈EP⁡. Therefore, we have w∈F(S)∩EP⁡. So,
(49)limsup⁡n→∞〈x~-u,xn-x~〉=lim⁡i→∞〈x~-u,xni-x~〉=〈x~-u,w-x~〉≥0.Setting vn=xn-(ρn/(1-ςn))(Φ(xn)-Φ(x~)) for all n and taking z=x~ in (29) to get ∥Φ(xn)-Φ(x~)∥→0, so, vn-xn→0. Since un→u, we have(50)limsup⁡n→∞〈un-x~,vn-x~〉=limsup⁡n→∞〈u-x~,vn-x~〉=limsup⁡n→∞〈u-x~,xn-x~〉≤0.Conclusion (xn→x~).Proof.
From (11), we have
(51)∥xn+1-x~∥2≤ϖn∥xn-x~∥2+(1-ϖn)∥STρnwn-x~∥2≤ϖn∥xn-x~∥2+(1-ϖn)∥Tρnwn-x~∥2=ϖn∥xn-x~∥2+(1-ϖn)×∥Tρnwn-Tρn(x~-ρnΦ(x~))∥2≤ϖn∥xn-x~∥2+(1-ϖn)∥wn-(x~-ρnΦ(x~))∥2=ϖn∥xn-x~∥2+(1-ϖn)×∥(1-ςn)xn+ςnun-ρnΦ(xn)-(x~-ρnΦ(x~))∥2=(1-ϖn)∥(1-ςn)((xn-ρn1-ςnΦ(xn))vvvvvvvvvvvvvvvvvv-(x~-ρn1-ςnΦ(x~)))vvvvvvvvvvvv+ςn(un-x~)ρn1-ςn∥2+ϖn∥xn-x~∥2=(1-ϖn)((1-ςn)2∥(xn-ρn1-ςnΦ(xn))vvvvvvvvvvvvvvvvvv-(x~-ρn1-ςnΦ(x~))∥2vvvvvvvvvv+2ςn(1-ςn)vvvvvvvvvv×〈ρn1-ςnun-x~,(xn-ρn1-ςnΦ(xn))vvvvvvvvvvvvvvvv-(x~-ρn1-ςnΦ(x~))〉vvvvvvvvvv+ςn2∥un-x~∥2∥(xn-ρn1-ςnΦ(xn)))+ϖn∥xn-x~∥2≤ϖn∥xn-x~∥2+(1-ϖn)×((1-ςn)2∥xn-x~∥2+2ςn(1-ςn)vvvvvvvvvv×〈un-x~,xn-ρn1-ςnvvvvvvvvvvvvvvvvv×(Φ(xn)-Φ(x~))vvvvvvvvvvvvvvvvvvvvvv-x~ρn1-ςn〉vvvvvvvv+ςn2∥un-x~∥2)≤[1-(1-ϖn)ςn]∥xn-x~∥2+(1-ϖn)ςn×{2(1-ςn)〈un-x~,vn-x~〉+ςn∥un-x~∥2}.
It is clear that ∑n=1∞‍(1-ϖn)ςn=∞ and limsup⁡n→∞(2(1-ςn)〈un-x~,vn-x~〉+ςn∥un-x~∥2)≤0. We can therefore apply Lemma 9 to conclude that xn→x~. This completes the proof.Corollary 11.
LetC be a nonempty closed convex subset of H and let Θ:C×C→R be a bifunction satisfying conditions (C1)–(C4). Let Φ:C→H be an ς-inverse-strongly monotone mapping and let S:C→C be a nonexpansive mapping. Suppose that F(S)∩EP≠∅. For x0∈C arbitrarily, let the sequence {xn} be generated by
(52)xn+1=ϖnxn+(1-ϖn)STρn[(1-ςn)xn-ρnΦ(xn)],n≥0,
where Tρn is defined as that in Lemma 4 and {ρn}⊂(0,2ς),  {ςn}⊂(0,1), and {ϖn}⊂(0,1) satisfy(i)
lim⁡n→∞ςn=0 and ∑n=1∞‍ςn=∞;(ii)
0<c≤ϖn≤d<1;(iii)
a(1-ςn)≤ρn≤b(1-ςn), where [a,b]⊂(0,2ς) and lim⁡n→∞(ρn+1-ρn)=0. Then {xn} generated by (52) converges strongly to PF(S)∩EP(0) which is the minimum norm element in F(S)∩EP.



---

*Source: 468593-2014-04-22.xml*