# Hybrid Iterative Scheme by a Relaxed Extragradient Method for Equilibrium Problems, a General System of Variational Inequalities and Fixed-Point Problems of a Countable Family of Nonexpansive Mappings

Authors: Qiao-Li Dong; Yan-Ni Guo; Fang Su
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

Based on the relaxed extragradient method and viscosity method, we introduce a new iterative method for finding a common element of solution of equilibrium problems, the solution set of a general system of variational inequalities, and the set of fixed points of a countable family of nonexpansive mappings in a real Hilbert space. Furthermore, we prove the strong convergence theorem of the studied iterative method. The results of this paper extend and improve the results of Ceng et al., (2008), W. Kumam and P. Kumam, (2009), Yao et al., (2010)  and many others.


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## Body

## 1. Introduction

LetH be a real Hilbert space with the inner product 〈·,·〉 and the norm ∥·∥. Let C be a closed convex subset of H. Let F be a bifunction of C×C into ℝ, where ℝ is the set of real numbers. The equilibrium problem for F:C×C→ℝ is to find x∈C such that (1.1)F(x,y)≥0,∀y∈C.
The set of solutions of (1.1) is denoted by EP(F). The equilibrium problems covers, as special cases, monotone inclusion problems, saddle point problems, minimization problems, optimization problems, variational inequality problems, Nash equilibria in noncooperative games, and various forms of feasibility problems (see [1–4] and the references therein).A mappingA:C→H is called α-inverse-strongly monotone if there exists a positive real number α such that
(1.2)〈Au-Av,u-v〉≥α‖Au-Av‖2,∀u,v∈C.
It is obvious that any α-inverse-strongly monotone mapping A is monotone and Lipschitz continuous. A mapping T:C→C is said to be  nonexpansive if
(1.3)‖Tx-Ty‖≤‖x-y‖,∀x,y∈C.
We denote by F(T) the set of fixed points of T. Recently, Wang and Guo [5] introduced an iterative scheme for a countable family of nonexpansive mappings.LetC be a nonempty closed convex subset of a real Hilbert space H. For a given nonlinear operator A:C→H, consider the following variational inequality problem of finding x*∈C such that
(1.4)〈Ax*,x-x*〉≥0,x∈C.
The set of solutions of the variational inequality (1.4) is denoted by VI(C,A) (see [6–9] and the references therein).LetA,B:C→H be two mappings. Consider the following problem of finding (x*,y*)∈C×C such that
(1.5)〈λAy*+x*-y*,x-x*〉≥0,∀x∈C,〈μBx*+y*-x*,x-y*〉≥0,∀x∈C,
which is called a general system of variational inequalities, where λ>0 and μ>0 are two constants. The set of solutions of (1.5) is denoted by GSVI(A,B,andC). In particular, if A=B, then problem (1.5) reduces to finding (x*,y*)∈C×C such that
(1.6)〈λAy*+x*-y*,x-x*〉≥0,∀x∈C,〈μAx*+y*-x*,x-y*〉≥0,∀x∈C,
which is defined by Verma [7] (see also [10]) and is called the new system of variational inequalities. Further, if we add up the requirement that x*=y*, then problem (1.6) reduces to the classical variational inequality problem (1.4). Recently, Yao et al. [11] presented system of variational inequalities in Banach space. For solving problem (1.5), recently, Ceng et al. [12] introduced and studied a relaxed extragradient method. Based on the relaxed extragradient method and the viscosity approximation method, W. Kumam and P. Kumam [13] constructed a new viscosity-relaxed extragradient approximation method. Very recently, based on the extragradient method, Yao et al. [14] proposed an iterative method for finding a common element of the set of a general system of variational inequalities and the set of fixed points of a strictly pseudocontractive mapping in a real Hilbert space.Motivated and inspired by the above works, in this paper, we introduce an iterative method based on the extragradient method and viscosity method for finding a common element of solution of equilibrium problems, the solution set of a general system of variational inequalities, and the set of fixed points of a countable family of nonexpansive mappings in a real Hilbert space. Furthermore, we prove the strong convergence theorem of the proposed iterative method.


## 2. Preliminaries

LetC be a closed convex subset of H, and let T:C→C be nonexpansive such that F(T)≠∅. For all x̂∈F(T) and all x∈C, we have
(2.1)‖x-x̂‖2≥‖Tx-Tx̂‖2=‖Tx-x̂‖2=‖(Tx-x)+(x-x̂)‖2=‖Tx-x‖2+‖x-x̂‖2+2〈Tx-x,x-x̂〉,
and hence
(2.2)‖Tx-x‖2≤2〈x-Tx,x-x̂〉,∀x̂∈F(T),∀x∈C.Remark 2.1.
LetA be α-inverse-strongly monotone. For all x,y∈C, λ>0,   we have
(2.3)‖(I-λA)x-(I-λA)y‖2=‖x-y‖2-2λ〈Ax-Ay,x-y〉+λ2‖Ax-Ay‖2≤‖x-y‖2+λ(λ-2α)‖Ax-Ay‖2.So,if λ≤2α, then I-λA    is a nonexpansive mapping from C to H.Recall that the (nearest point) projectionPC from H onto C assigns to each x∈H the unique point PCx∈C satisfying the property
(2.4)‖x-PCx‖=min⁡y∈C‖x-y‖.
The following characterizes the projection PC.Lemma 2.2.
Given thatx∈H and y∈C, then PCx=y if and only if there holds the inequality
(2.5)〈x-y,y-z〉≥0∀z∈C.Lemma 2.3 (see [15]).
LetH be a Hilbert space, C a closed convex subset of H, and T:C→C a nonexpansive mapping with F(T)≠∅. If {xn} is a sequence in C weakly converging to x∈C and if {(1-T)xn} converges strongly to y, then(1-T)x=y.Lemma 2.4 (see [16]).
Assume that{an} is a sequence of nonnegative real numbers such that
(2.6)αn+1≤(1-γn)αn+δn,∀n≥0,
where two sequences {γn}⊂(0,1) and {δn} satisfy(1)
∑n=1∞γn=∞;(2)
limsupn→∞δn/γn≤0or∑n=1∞|δn|<∞.Then  limn→∞αn=0.Lemma 2.5 (see [12]).
For givenx*,y*∈C, (x*,y*) is a solution of problem (1.5) if and only if x* is a fixed point of the mapping G:C→C defined by
(2.7)G(x)=PC[PC(x-μBx)-λAPC(x-μBx)],∀x∈C,
where  y*=PC(x*-μBx*).Note that the mappingG is nonexpansive provided that  λ∈(0,2α) and μ∈(0,2β).Throughout this paper, the set of fixed points of the mappingG is denoted by Γ.Lemma 2.6 (see [1]).
LetC be a nonempty closed convex subset of H and F:C×C→ℝ satisfy following conditions:  (A1)
F(x,x)=0,  ∀x∈C;(A2)
F is monotone, that is,  F(x,y)+F(y,x)≤0,  ∀x,y∈C;(A3)
limsupt→0+F(tz+(1-t)x,y)≤F(x,y),  ∀x,y,z∈C;(A4)
for eachx∈C, F(x,·)   is convex and lower semicontinuous.For  x∈C  and  r>0, set  TrF:H→C to be
(2.8)TrF(x)={z∈C:F(z,y)+1r〈y-z,z-x〉≥0,∀y∈C}.
Then TrF is well defined and the following holds:  (1)
TrF  is single valued;(2)
TrF  is firmly nonexpansive [17], that is, for any x,y∈E,
(2.9)‖TrFx-TrFy‖2≤〈TrFx-TrFy,x-y〉;(3)
F(TrF)=EP(F);(4)
EP(F) is closed and convex.By the proof of Lemma  5 in [2] (also see [3]), we have following lemma.Lemma 2.7.
LetC be a nonempty closed convex subset of a Hilbert space  H and  F:C×C→ℝ be a bifunction. Let  x∈C and  r1,r2∈(0,∞). Then
(2.10)‖Tr1x-Tr2x‖≤|1-r2r1|(‖Tr1x‖+‖x‖).


## 3. Main Results

Theorem 3.1.
LetC be a nonempty closed convex subset of a real Hilbert space  H. Let the mappings  A,B:C→H be  α-inverse strongly monotone and  β-inverse strongly monotone, respectively. Let  F be a bifunction from  C×C to  ℝ satisfying (A1)–(A4) and  {Tn}n=1∞:C→C be a countable family of nonexpansive mappings such that  Ω:=⋂n=1∞F(Tn)∩EP(F)∩Γ≠∅. Let  f:C→C be a contraction with coefficient  ρ∈(0,1/2). Set  β0=1. For given  x1∈C arbitrarily, let the sequences  {xn},  {yn},  {zn},  and  {un} be generated by
(3.1)F(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈C,zn=PC(un-μBun),yn=αnf(xn)+(1-αn)PC(zn-λAzn),xn+1=βnxn+σn∑i=1n(βi-1-βi)Tiyn+(1-βn)(1-σn)PC(zn-λAzn),
where λ∈(0,2α), μ∈(0,2β),  and sequences {αn}⊂[0,1], {βn}⊂[0,1], {σn}⊂[0,1],  and {rn}⊂(r,∞), r>0, are such that
(i)
{βn} is strictly decreasing,(ii)
0<liminfn→∞βn≤limsupn→∞βn<1,(iii)
limn→∞αn=0 and ∑n=1∞αn=∞,(iv)
σn>1/2(1-ρ),∑n=1∞|σn-σn-1|<∞,(v)
∑n=1∞|rn-rn-1|<∞.Then the sequence  {xn} generated by (3.1) converges strongly to  x*=PΩ·f(x*),  and  (x*,y*) is a solution of the general system of variational inequalities (1.5), where  y*=PC(x*-μBx*).Proof.
The proof is divided into several steps.
Step 1.
The sequence{xn} defined by (3.1) is bounded.
For eachx*∈Ω, from Lemma 2.6, we have un=Trnxn and hence
(3.2)‖un-x*‖=‖Trnxn-Trnx*‖≤‖xn-x*‖.
Since f is a ρ-contraction mapping, using (2.3) and (3.2), we have
(3.3)‖zn-y*‖2=‖PC(un-μBun)-PC(x*-μBx*)‖2≤‖(un-μBun)-(x*-μBx*)‖2≤‖un-x*‖2+μ(μ-2β)‖Bun-Bx*‖2≤‖xn-x*‖2+μ(μ-2β)‖Bun-Bx*‖2≤‖xn-x*‖2.
Set vn=PC(zn-λAzn). Since PC is nonexpansive, from (2.3), we have
(3.4)‖vn-x*‖2=‖PC(zn-λAzn)-PC(y*-λAy*)‖2≤‖zn-y*‖2+λ(λ-2α)‖Azn-Ay*‖2≤‖zn-y*‖2.
Hence we get
(3.5)‖yn-x*‖=‖αnf(xn)+(1-αn)vn-x*‖≤αn‖f(xn)-x*‖+(1-αn)‖zn-y*‖≤αn(‖f(xn)-f(x*)‖+‖f(x*)-x*‖)+(1-αn)‖xn-x*‖≤(1-αn(1-ρ))‖xn-x*‖+αn‖f(x*)-x*‖.
From (3.1) and (3.3)–(3.5), we get
(3.6)‖xn+1-x*‖=‖βnxn+σn∑i=1n(βi-1-βi)Tiyn+(1-βn)(1-σn)vn-x*‖≤βn‖xn-x*‖+σn∑i=1n(βi-1-βi)‖Tiyn-x*‖+(1-βn)(1-σn)‖vn-x*‖≤βn‖xn-x*‖+σn∑i=1n(βi-1-βi)‖yn-x*‖+(1-βn)(1-σn)‖zn-y*‖≤(1-σn(1-βn))‖xn-x*‖+σn(1-βn)‖yn-y*‖≤(1-σn(1-βn))‖xn-x*‖+σn(1-βn)[(1-αn(1-ρ))‖xn-x*‖+αn‖f(x*)-x*‖]≤[1-σn(1-βn)αn(1-ρ)]‖xn-x*‖+σn(1-βn)αn(1-ρ)11-ρ‖f(x*)-x*‖≤M,
where M=max{∥x1-x*∥,1/(1-ρ)∥f(x*)-x*∥}. Hence, {xn} is bounded and therefore {un}, {zn}, {vn},  and {yn} are also bounded.Step 2.
limn→∞∥xn+1-xn∥=0. Since un=Trnxn and un-1=Trn-1xn-1, using Lemma 2.7, we have
(3.7)‖un-un-1‖=‖Trnxn-Trn-1xn-1‖≤‖Trnxn-Trnxn-1‖+‖Trnxn-1-Trn-1xn-1‖≤‖xn-xn-1‖+|1-rnrn-1|(‖un-1‖+‖xn-1‖)≤‖xn-xn-1‖+1r|rn-1-rn|L,
where L=sup{∥un∥+∥xn∥:n=1,2,…}. From (3.1) and (3.7), it follows that
(3.8)‖zn-zn-1‖=‖PC(I-μB)un-PC(I-μB)un-1‖≤‖un-un-1‖≤‖xn-xn-1‖+1r|rn-1-rn|L.
From (3.1) and (3.8), we have
(3.9)‖yn-yn-1‖≤αn‖f(xn)-f(xn-1)‖+(1-αn)‖PC(I-λA)zn-PC(I-λA)zn-1‖≤αnρ‖xn-xn-1‖+(1-αn)‖zn-zn-1‖≤αnρ‖xn-xn-1‖+(1-αn)(‖xn-xn-1‖+1r|rn-1-rn|L)≤(1-αn(1-ρ))‖xn-xn-1‖+(1-αn)1r|rn-1-rn|L.
By definition of scheme (3.1), we have
(3.10)xn+1-xn=βn(xn-xn-1)+(βn-βn-1)xn-1+σn∑i=1n(βi-1-βi)(Tiyn-Tiyn-1)+(σn-σn-1)∑i=1n-1(βi-1-βi)Tiyn-1+σn(βn-1-βn)Tnyn-1+(1-βn)(1-σn)[PC(zn-λAzn)-PC(zn-1-λAzn-1)]-(1-βn-1)(1-σn-1)PC(zn-1-λAzn-1)+(1-βn)(1-σn)PC(zn-1-λAzn-1).
Thus, from (3.8)–(3.10),
(3.11)‖xn+1-xn‖≤βn‖xn-xn-1‖+(βn-1-βn)‖xn-1‖+σn∑i=1n(βi-1-βi)‖Tiyn-Tiyn-1‖+|σn-σn-1|∑i=1n-1(βi-1-βi)‖Tiyn-1‖+σn(βn-1-βn)‖Tnyn-1‖+(1-βn)(1-σn)‖PC(zn-λAzn)-PC(zn-1-λAzn-1)‖+|(1-βn)(1-σn)-(1-βn-1)(1-σn-1)|‖PC(zn-1-λAzn-1)‖≤βn‖xn-xn-1‖+(βn-1-βn)‖xn-1‖+σn∑i=1n(βi-1-βi)‖yn-yn-1‖+|σn-σn-1|∑i=1n-1(βi-1-βi)‖Tiyn-1‖+(βn-1-βn)‖Tnyn-1‖+(1-βn)(1-σn)‖zn-zn-1‖+(βn-1-βn)(1-σn)‖PC(I-λA)zn-1‖+(1-βn-1)|σn-σn-1|‖PC(I-λA)zn-1‖≤βn‖xn-xn-1‖+(βn-1-βn)‖xn-1‖+σn(1-βn)×[(1-αn(1-ρ))‖xn-xn-1‖+(1-αn)1r|rn-1-rn|L]+|σn-σn-1|∑i=1n-1(βi-1-βi)‖Tiyn-1‖+(βn-1-βn)‖Tnyn-1‖+(1-βn)(1-σn)(‖xn-xn-1‖+1r|rn-1-rn|L)+(βn-1-βn)‖PC(I-λA)zn-1‖+|σn-σn-1|‖PC(I-λA)zn-1‖≤[1-σn(1-βn)αn(1-ρ)]‖xn-xn-1‖+(βn-1-βn)M+1r|rn-1-rn|L+|σn-σn-1|∑i=1n-1(βi-1-βi)M+(βn-1-βn)M+((βn-1-βn)+|σn-σn-1|)M≤[1-σn(1-βn)αn(1-ρ)]‖xn-xn-1‖+1r|rn-1-rn|L+3(βn-1-βn)M+2|σn-σn-1|M,
where M=max{supn≥1∥xn∥,supi≥1,n≥1∥Tiyn∥,supn≥1∥PC(I-λA)zn∥}. Since {βn} is strictly decreasing, we have ∑n=2∞(βn-1-βn)=β1<∞. Further, by assumption conditions (iv)-(v), we have
(3.12)∑n=2∞{1r|rn+1-rn|L+3(βn-1-βn)M+2|σn-σn-1|M}<∞.
Thus, using Lemma 2.4, we have limn→∞∥xn+1-xn∥=0.Step 3.
limn→∞∥xn-un∥=0.
For anyx*∈Ω, it follows from Lemma 2.6 that
(3.13)‖un-x*‖2=‖Trnxn-Trnx*‖2≤〈Trnxn-Trnx*,xn-x*〉=〈un-x*,xn-x*〉=12(‖un-x*‖2+‖xn-x*‖2-‖un-xn‖2).
From (2.3), (3.1)–(3.4), we can get
(3.14)‖yn-x*‖2≤αn‖f(xn)-x*‖2+(1-αn)‖vn-x*‖2≤αn‖f(xn)-x*‖2+(1-αn)(‖zn-y*‖2+λ(λ-2α)‖Azn-Ay*‖2)≤αn‖f(xn)-x*‖2+(1-αn)‖zn-y*‖2≤αn‖f(xn)-x*‖2+(1-αn)‖un-x*‖2.
From (3.1), (3.3)-(3.4), and (3.13)-(3.14), it follows that
(3.15)‖xn+1-x*‖2=‖βnxn+σn∑i=1n(βi-1-βi)Tiyn+(1-βn)(1-σn)vn-x*‖2≤βn‖xn-x*‖2+(1-βn)‖σn11-βn∑i=1n(βi-1-βi)Tiyn+(1-σn)vn-x*‖2≤βn‖xn-x*‖2+(1-βn)σn‖11-βn∑i=1n(βi-1-βi)Tiyn-x*‖2+(1-βn)(1-σn)‖vn-x*‖2≤βn‖xn-x*‖2+(1-βn)σn11-βn∑i=1n(βi-1-βi)‖Tiyn-x*‖2+(1-βn)(1-σn)‖vn-x*‖2≤βn‖xn-x*‖2+(1-βn)σn‖yn-x*‖2+(1-βn)(1-σn)‖zn-y*‖2≤βn‖xn-x*‖2+(1-βn)σn[(1-αn)‖un-x*‖2+αn‖f(xn)-x*‖2]+(1-βn)(1-σn)‖un-x*‖2≤‖xn-x*‖2+(1-βn)σnαn(‖f(xn)-x*‖2-‖xn-x*‖2)-(1-βn)(1-σnαn)‖un-xn‖2
which implies that
(3.16)(1-βn)(1-σnαn)‖un-xn‖2≤(‖xn-x*‖+‖xn+1-x*‖)‖xn-xn+1‖+αn‖f(xn)-x*‖2.
Since {xn} is both bounded, using Step 2 and conditions (ii)–(iv), we conclude the result.Step 4.
limn→∞∥Azn-Ay*∥=0 and limn→∞∥Bxn-Bx*∥=0.Using (3.3), (3.14), and (3.15), we have
(3.17)‖xn+1-x*‖2≤βn‖xn-x*‖2+(1-βn)σn‖yn-x*‖2+(1-βn)(1-σn)‖zn-y*‖2≤βn‖xn-x*‖2+(1-βn)σn×[αn‖f(xn)-x*‖2+(1-αn)(‖zn-y*‖2+λ(λ-2α)‖Azn-Ay*‖2)]+(1-βn)(1-σn)‖zn-y*‖2≤βn‖xn-x*‖2+(1-βn)σn(1-αn)λ(λ-2α)‖Azn-Ay*‖2+αn‖f(xn)-x*‖2+(1-βn)‖zn-y*‖2≤βn‖xn-x*‖2+λ(λ-2α)‖Azn-Ay*‖2+αn‖f(xn)-x*‖2+(1-βn)[‖xn-x*‖2+μ(μ-2β)‖Bun-Bx*‖2]≤‖xn-x*‖2+λ(λ-2α)‖Azn-Ay*‖2+αn‖f(xn)-x*‖2+μ(μ-2β)‖Bun-Bx*‖2.
Therefore,
(3.18)λ(2α-λ)‖Azn-Ay*‖2+μ(2β-μ)‖Bun-Bx*‖2≤(‖xn-x*‖+‖xn+1-x*‖)‖xn-xn+1‖+αn‖f(xn)-y*‖2.
From Step 2, using condition (iii), we get limn→∞∥Azn-Ay*∥=0 and limn→∞∥Bun-Bx*∥=0. From the fact that the B is β-inverse strongly monotone operator, it follows that
(3.19)‖Bxn-Bx*‖≤‖Bxn-Bun‖+‖Bun-Bx*‖≤1β‖xn-un‖+‖Bun-Bx*‖.
Applying Step 3, we have limn→∞∥Bxn-Bx*∥=0.Step 5.
limn→∞∥xn-Tixn∥=0,foralli=1,2,….
Noting thatPC is firmly nonexpansive, we have(3.20)‖zn-y*‖2=‖PC(un-μBun)-PC(x*-μBx*)‖2≤〈(un-μBun)-(x*-μBx*),zn-y*〉=12[‖un-x*-μ(Bun-Bx*)‖2+‖zn-y*‖2-‖(un-zn)-μ(Bun-Bx*)-(x*-y*)‖2]≤12[‖un-x*‖2+‖zn-y*‖2-‖un-zn-(x*-y*)‖2+2μ〈un-zn-(x*-y*),Bun-Bx*〉-μ2‖Bun-Bx*‖2]≤12[11‖xn-x*‖2+‖zn-y*‖2-‖un-zn-(x*-y*)‖2+2μ〈un-zn-(x*-y*),Bun-Bx*〉11],‖vn-x*‖2=‖PC(zn-λAzn)-PC(y*-λAy*)‖2≤〈(zn-λAzn)-(y*-λAy*),vn-x*〉=12[‖zn-y*-λ(Azn-Ay*)‖2+‖vn-x*‖2-‖(zn-vn)-λ(Azn-Ay*)-(y*-x*)‖2]≤12[‖zn-y*‖2+‖vn-x*‖2-‖zn-vn-(y*-x*)‖2+2λ〈zn-vn-(y*-x*),Azn-Ay*〉-λ2‖Azn-Ay*‖2]≤12[11‖xn-x*‖2+‖vn-x*‖2-‖zn-vn-(y*-x*)‖2+2λ〈zn-vn-(y*-x*)Azn-Ay*〉11].
It follows that
(3.21)‖zn-y*‖2≤‖xn-x*‖2-‖un-zn-(x*-y*)‖2+2μ〈un-zn-(x*-y*),Bun-Bx*〉,(3.22)‖vn-x*‖2≤‖xn-x*‖2-‖zn-vn-(y*-x*)‖2+2λ〈zn-vn-(y*-x*),Azn-Ay*〉.
By (3.3), (3.14)-(3.15), and (3.22), we have
(3.23)‖xn+1-x*‖2≤(1-(1-βn)σn)‖xn-x*‖2+(1-βn)σn‖yn-x*‖2≤(1-(1-βn)σn)‖xn-x*‖2+(1-βn)σn[αn‖f(xn)-x*‖2+‖vn-x*‖2]≤(1-(1-βn)σn)‖xn-x*‖2+(1-βn)σn[12αn‖f(xn)-x*‖2+‖xn-x*‖2-‖zn-vn-(y*-x*)‖2+2λ〈zn-vn-(y*-x*)Azn-Ay*〉22]=‖xn-x*‖2+αn‖f(xn)-x*‖2+(1-βn)(1-σn)×(-‖zn-vn-(y*-x*)‖2+2λ〈zn-vn-(y*-x*),Azn-Ay*〉).
It follows that
(3.24)(1-βn)σn‖vn-z-(x*-y*)‖2≤(‖xn-x*‖+‖xn+1-x*‖)‖xn-xn+1‖+αn‖f(xn)-x*‖2+2λ(1-βn)σn〈zn-vn-(y*-x*),Azn-Ay*〉.
From conditions (ii)–(iv), Steps 2 and 4, we get the following:
(3.25)lim⁡n⟶∞‖vn-zn-(x*-y*)‖=0.
On the other hand, from (3.14)–(3.21), we have
(3.26)‖xn+1-x*‖2≤βn‖xn-x*‖2+(1-βn)σn(αn‖f(xn)-x*‖2+‖zn-y*‖2)+(1-βn)(1-σn)‖zn-y*‖2≤βn‖xn-x*‖2+αn‖f(xn)-x*‖2+(1-βn)‖zn-y*‖2≤βn‖xn-x*‖2+αn‖f(xn)-x*‖2+(1-βn)×[‖xn-x*‖2-‖un-zn-(x*-y*)‖2+2μ〈un-zn-(x*-y*),Bun-Bx*〉]=‖xn-x*‖2+αn‖f(xn)-x*‖2+(1-βn)(-‖un-zn-(x*-y*)‖2+2μ〈un-zn-(x*-y*),Bun-Bx*〉),
which implies that
(3.27)‖(1-βn)un-zn-(x*-y*)‖2≤(‖xn-x*‖+‖xn+1-x*‖)‖xn-xn+1‖+αn‖f(xn)-x*‖2+2μ(1-βn)×〈un-zn-(x*-y*),Bun-Bx*〉.
From (ii)-(iii), Steps 2 and 4, we get the following:
(3.28)lim⁡n⟶∞‖un-zn-(x*-y*)‖=0.
Combining (3.25) and (3.28), we get the following:
(3.29)lim⁡n⟶∞‖un-vn‖=0.
Using Step 3, we obtain the following:
(3.30)lim⁡n⟶∞‖xn-vn‖=0.
This together with ∥yn-vn∥=αn∥f(xn)-vn∥→0 implies that
(3.31)lim⁡n⟶∞‖xn-yn‖=0.
For any x*∈Ω, we have from (3.1) that
(3.32)σn∑i=1n(βi-1-βi)(〈xn-Tixn,xn-x*〉+〈Tixn-Tiyn,xn-x*〉)=σn∑i=1n(βi-1-βi)〈xn-Tiyn,xn-x*〉=〈xn-xn+1,xn-x*〉+(1-βn)(1-σn)〈vn-xn,xn-x*〉.
Since each Ti is nonexpansive, from (2.2), we have
(3.33)‖Tixn-xn‖2≤2〈xn-Tixn,xn-x*〉.
Hence, combining this inequality with (3.32), we get the following:
(3.34)12σn∑i=1n(βi-1-βi)‖Tixn-xn‖2≤-σn∑i=1n(βi-1-βi)〈Tixn-Tiyn,xn-x*〉+〈xn-xn+1,xn-x*〉+(1-βn)(1-σn)×〈vn-xn,xn-x*〉,
that is (noting that βn is strictly decreasing),
(3.35)‖Tixn-xn‖2≤2(1-βn)βi-1-βi‖xn-yn‖‖xn-x*‖+2σn(βi-1-βi)‖xn-xn+1‖‖xn-x*‖+2(1-βn)(1-σn)σn(βi-1-βi)‖vn-xn‖‖xn-x*‖.
Now from (3.30)-(3.31) and Step 2, we conclude that
(3.36)lim⁡n⟶∞‖Tixn-xn‖=0,∀i=1,2,…,
which completes the proof.Step 6.
limsupn→∞〈f(x*)-x*,xn-x*〉≤0,  where x*=PΩ·f(x*).
As{xn} is bounded, there exists a subsequence {xni} of {xn} such that {xni}→x̃ weakly. First, it is clear from Step 5 and Lemma 2.3 that x̃∈⋂n=1∞F(Tn). Next, we prove that x̃∈Γ. From (iii), Step 3 and (3.31), we note that
(3.37)‖yn-G(yn)‖≤αn‖f(xn)-G(yn)‖+(1-αn)‖PC[PC(un-μBun)-λAPC(un-μBun)]-G(yn)‖=αn‖f(xn)-G(yn)‖+(1-αn)‖G(un)-G(yn)‖≤αn‖f(xn)-G(yn)‖+(1-αn)‖un-yn‖⟶0.
According to (3.31) and Lemma 2.3, we obtain that  x̃∈Γ. Next, we show that  x̃∈EP(F). Indeed, by un=Trnxn, we have
(3.38)F(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈C.
From (A2), we have also
(3.39)1rn〈y-un,un-xn〉≥-F(un,y)≥F(y,un),
and hence,
(3.40)〈y-unk,unk-xnkrnk〉≥F(y,unk).
According to rn>r>0 and un-xn→0, we conclude that (uni-xni)/rni→0 and unk⇀x̃. From (A4), we obtain the following:
(3.41)F(y,x̃)≤0,∀y∈C.
For t with 0<t≤1 and y∈C, let yt=ty+(1-t)x̃. Since y∈C and x̃∈C (due to uni→x̃ as i→∞), we have yt∈C and hence F(yt,x̃)≤0. So, from (A1) and (A4), we have
(3.42)tF(yt,y)≥tF(yt,y)+(1-t)F(yt,x̃)≥tF(yt,yt)=0
and F(yt,y)≥0, for all t∈(0,1) and y∈C. From (A3), we obtain the following:
(3.43)F(x̃,y)≥0∀y∈C,
and hence x̃∈EP(F). Therefore, we obtain that  x̃∈Ω. Hence, it follows from Lemma 2.2 that
(3.44)limsup⁡n→∞〈f(x*)-x*,xn-x*〉=lim⁡i→∞〈f(x*)-x*,xni-x*〉=〈f(x*)-x*,x̃-x*〉≤0.Step 7.
limn→∞xn=x*.
From (3.1), (3.3)-(3.4), and the convexity of ∥·∥, we have
(3.45)‖yn-x*‖2=‖αnf(xn)+(1-αn)vn-x*‖2≤(1-αn)‖vn-x*‖2+2αn〈f(xn)-x*,yn-x*〉≤(1-αn)‖xn-x*‖2+2αn〈f(xn)-x*,yn-x*〉.
We can also get
(3.46)‖xn+1-x*‖2=‖βn(xn-x*)+σn∑i=1n(βi-1-βi)(Tiyn-x*)+(1-βn)(1-σn)(yn-x*)+(1-βn)(1-σn)αn(vn-f(xn))11‖2≤‖βn(xn-x*)+σn∑i=1n(βi-1-βi)(Tiyn-x*)+(1-βn)(1-σn)(yn-x*)‖2+2(1-βn)(1-σn)αn〈vn-f(xn),xn+1-x*〉≤βn‖xn-x*‖2+(1-βn)‖yn-x*‖2+2(1-βn)(1-σn)αn〈vn-x*,xn+1-x*〉+2(1-βn)(1-σn)αn〈x*-f(xn),xn+1-x*〉.
By (3.45), we have
(3.47)‖xn+1-x*‖2≤βn‖xn-x*‖2+(1-βn)[(1-αn)‖xn-x*‖2+2αn〈f(xn)-x*,yn-x*〉]+2(1-βn)(1-σn)αn‖vn-x*‖‖xn+1-x*‖+2(1-βn)αn(1-σn)〈x*-f(xn),xn+1-x*〉≤[1-(1-βn)αn]‖xn-x*‖2+2(1-βn)(1-σn)αn〈f(xn)-x*,yn-xn+1〉+2(1-βn)(1-σn)αn‖zn-y*‖‖xn+1-x*‖+2(1-βn)αnσn〈f(xn)-x*,yn-x*〉≤[1-(1-βn)αn]‖xn-x*‖2+2(1-βn)(1-σn)αn‖f(xn)-x*‖‖xn+1-yn‖+2(1-βn)(1-σn)αn‖xn-x*‖‖xn+1-x*‖+2(1-βn)αnσn〈f(xn)-x*,yn-xn〉+2(1-βn)αnσn〈f(xn)-f(x*),xn-x*〉+2(1-βn)αnσn〈f(x*)-x*,xn-x*〉≤[1-(1-βn)αn]‖xn-x*‖2+2(1-βn)(1-σn)αn‖f(xn)-x*‖‖xn+1-yn‖+(1-βn)(1-σn)αn(‖xn-x*‖2+‖xn+1-x*‖2)+2(1-βn)αnσn‖f(xn)-x*‖‖yn-xn‖+2(1-βn)αnσnρ‖xn-x*‖2+2(1-βn)αnσn〈f(x*)-x*,xn-x*〉,
which implies that
(3.48)‖xn+1-x*‖2≤[1-(2σn(1-ρ)-1)(1-βn)1-(1-βn)(1-σn)αnαn]‖xn-x*‖2+[(2σn(1-ρ)-1)(1-βn)]αn1-(1-βn)(1-σn)αn×[2(1-σn)2σn(1-ρ)-1‖f(xn)-x*‖‖xn+1-yn‖+2σn2σn(1-ρ)-1‖f(xn)-x*‖×‖yn-xn‖+2σn2σn(1-ρ)-1〈f(xn)-x*,xn-x*〉].
From liminfn→∞(2σn(1-ρ)-1)(1-βn)/(1-(1-βn)(1-σn))>0, it follows that ∑n=0∞((2σn(1-ρ)-1)(1-βn)/(1-(1-βn)(1-σn)αn))αn=∞. It is clear that
(3.49)limsup⁡n→∞[2(1-σn)2σn(1-ρ)-1‖f(xn)-x*‖‖xn+1-yn‖+2σn2σn(1-ρ)-1‖f(xn)-x*‖×‖yn-xn‖+2σn2σn(1-ρ)-1〈f(xn)-x*,xn-x*〉]≤0.
Therefore, all conditions of Lemma 2.4 are satisfied. Therefore, we immediately deduce that xn→x*. This completes the proof.Corollary 3.2.
LetC be a nonempty closed convex subset of a real Hilbert space H. Let the mappings A,B:C→H be α-inverse strongly monotone and β-inverse strongly monotone, respectively. Let F be a bifunction from C×C to ℝ satisfying (A1)–(A4) and T:C→C be a nonexpansive mapping such that Ω:=F(T)∩EP(F)∩Γ≠∅.Let f:C→C be a contraction with coefficient ρ∈(0,1/2). For given x0∈C arbitrarily, let the sequences {xn}, {yn}, {zn} and, {un} be generated By
(3.50)F(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈C,zn=PC(un-μBun),yn=αnf(xn)+(1-αn)PC(zn-λAzn),xn+1=βnxn+σn(1-βn)Tyn+(1-βn)(1-σn)PC(zn-λAzn),
where λ∈(0,2α), μ∈(0,2β),  and sequences {αn}⊂[0,1], {βn}⊂[0,1], {σn}⊂[0,1], and  {rn}⊂(r,∞), r>0, are such that  (i)
0<liminfn→∞βn≤limsupn→∞βn<1,(ii)
limn→∞αn=0and ∑n=1∞αn=∞,(iii)
σn>1/2(1-ρ),∑n=1∞|σn-σn-1|<∞,(iv)
∑n=1∞|rn-rn-1|<∞.Then the sequence {xn} generated by (3.50) converges strongly to x*=PΩ·f(x*),  and (x*,y*) is a solution of the general system of variational inequalities (1.5), where y*=PC(x*-μBx*).



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*Source: 572326-2012-02-28.xml*