# General Iterative Methods for Equilibrium Problems and Infinitely Many Strict Pseudocontractions in Hilbert Spaces

Authors: Peichao Duan; Aihong Wang
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We propose an implicit iterative scheme and an explicit iterative scheme for finding a common element of the set of fixed point of infinitely many strict pseudocontractive mappings and the set of solutions of an equilibrium problem by the general iterative methods. In the setting of real Hilbert spaces, strong convergence theorems are proved. Our results improve and extend the corresponding results reported by many others.


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## Body

## 1. Introduction

LetH be a real Hilbert space and let C be a nonempty closed convex subset of H. Let F be a bifunction from C×C to ℝ, where ℝ is the set of real numbers.The equilibrium problem forF:C×C→ℝ is to find x∈C such that(1.1)F(x,y)≥0
for all y∈C. The set of such solutions is denoted by EP(F).A mappingS of C is said to be a κ-strict pseudocontraction if there exists a constant κ∈[0,1) such that(1.2)‖Sx-Sy‖2≤‖x-y‖2+κ‖(I-S)x-(I-S)y‖2
for all x,y∈C; see [1]. We denote the set of fixed points of S by F(S) (i.e., F(S)={x∈C:Sx=x}).Note that the class of strict pseudocontractions strictly includes the class of nonexpansive mappings which are mappingSonC such that(1.3)‖Sx-Sy‖≤‖x-y‖
for all x,y∈C. That is, S is nonexpansive if and only if S is a 0-strict pseudocontraction.Numerous problems in physics, optimization, and economics reduce to finding a solution of the equilibrium problem. Some methods have been proposed to solve the equilibrium problem (1.1); see, for instance, [2–4]. In particular, Combettes and Hirstoaga [5] proposed several methods for solving the equilibrium problem. On the other hand, Mann [6], Shimoji and Takahashi [7] considered iterative schemes for finding a fixed point of a nonexpansive mapping. Further, Acedo and Xu [8] projected new iterative methods for finding a fixed point of strict pseudocontractions.In 2006, Marino and Xu [3] introduced the general iterative method and proved that the algorithm converged strongly. Recently, Liu [2] considered a general iterative method for equilibrium problems and strict pseudocontractions. Tian [9] proposed a new general iterative algorithm combining an L-Lipschitzian and η-strong monotone operator. Very recently, Wang [10] considered a general composite iterative method for infinite family strict pseudocontractions.In this paper, motivated by the above facts, we introduce two iterative schemes and obtain strong convergence theorems for finding a common element of the set of fixed points of a infinite family of strict pseudocontractions and the set of solutions of the equilibrium problem (1.1).


## 2. Preliminaries

Throughout this paper, we always write⇀ for weak convergence and → for strong convergence. We need some facts and tools in a real Hilbert space H which are listed as below.Lemma 2.1.
LetH be a real Hilbert space. There hold the following identities:(i)
∥x-y∥2=∥x∥2-∥y∥2-2〈x-y,y〉, ∀x,y∈H;(ii)
∥tx+(1-t)y∥2=t∥x∥2+(1-t)∥y∥2-t(1-t)∥x-y∥2,∀t∈[0,1],∀x,y∈H.Lemma 2.2 (see [11]).
Assume that{αn} is a sequence of nonnegative real numbers such that
(2.1)αn+1≤(1-γn)αn+δn,
where {γn} is a sequence in (0,1) and {δn} is a sequence such that(i)
∑n=1∞γn=∞;(ii)
limn→∞sup(δn/γn)≤0or∑n=1∞|δn|<∞.Then, limn→∞αn=0.
Recall that given a nonempty closed convex subsetC of a real Hilbert space H, for any x∈H, there exists a unique nearest point in C, denoted by PCx, such that
(2.2)‖x-PCx‖≤‖x-y‖
for all y∈C. Such a PC is called the metric (or the nearest point) projection of H onto C. As known, y=PCx if and only if there holds the relation:
(2.3)〈x-y,y-z〉≥0∀z∈C.Lemma 2.3 (see [10]).
LetA:H→H be a L-Lipschitzian and η-strongly monotone operator on a Hilbert space H with L>0, η>0, 0<μ<2η/L2, and 0<t<1. Then, S=(I-tμA):H→H is a contraction with contractive coefficient 1-tτ and τ=(1/2)μ(2η-μL2).Lemma 2.4 (see [1]).
LetS:C→C be a κ-strict pseudocontraction. Define T:C→C by Tx=λx+(1-λ)Sx for each x∈C. Then, as λ∈[κ,1), T is a nonexpansive mapping such that F(T)=F(S).Lemma 2.5 (see [9]).
LetH be a Hilbert space and f:H→H be a contraction with coefficient 0<α<1, and A:H→H an L-Lipschitzian continuous operator and η-strongly monotone with L>0, η>0. Then for 0<γ<μη/α:
(2.4)〈x-y,(μA-γf)x-(μA-γf)y〉≥(μη-γα)‖x-y‖2,x,y∈H.
That is, μA-γf is strongly monotone with coefficient μη-γα.
Let{Sn} be a sequence of κn-strict pseudo-contractions. Define Sn′=θnI+(1-θn)Sn,θn∈[κn,1). Then, by Lemma 2.4, Sn′ is nonexpansive. In this paper, consider the mapping Wn defined by
(2.5)Un,n+1=I,Un,n=tnSn′Un,n+1+(1-tn)I,Un,n-1=tn-1Sn-1′Un,n+(1-tn-1)I,…,Un,i=tiSi′Un,i+1+(1-ti)I,…,Un,2=t2S2′Un,3+(1-t2)I,Wn=Un,1=t1S1′Un,2+(1-t1)I,
where t1,t2,… are real numbers such that 0≤tn<1. Such a mapping Wn is called a W-mapping generated by S1′,S2′,… and t1,t2,…. It is easy to see Wn is nonexpansive.Lemma 2.6 (see [7]).
LetC be a nonempty closed convex subset of a strictly convex Banach space E, let S1′,S2′,… be nonexpansive mappings of C into itself such that ∩i=1∞F(Si′)≠∅ and let t1,t2,… be real numbers such that 0<ti≤b<1, for every i=1,2,…. Then, for any x∈C and k∈N, the limit limn→∞Un,kx exists.
Using Lemma2.6, one can define the mapping W of C into itself as follows:
(2.6)Wx:=lim⁡n→∞Wnx=lim⁡n→∞Un,1x,x∈C.Lemma 2.7 (see [7]).
LetC be a nonempty closed convex subset of a strictly convex Banach space E. Let S1′,S2′,… be nonexpansive mappings of C into itself such that ∩i=1∞F(Si′)≠∅ and let t1,t2,… be real numbers such that 0<ti≤b<1,for all i≥1. If K is any bounded subset of C, then
(2.7)lim⁡n→∞sup⁡x∈K‖Wx-Wnx‖=0.Lemma 2.8 (see [12]).
LetC be a nonempty closed convex subset of a Hilbert space H,let{Si′:C→C} be a family of infinite nonexpansive mappings with ∩i=1∞F(Si′)≠∅, let t1,t2,… be real numbers such that 0<ti≤b<1, for every i=1,2,…. Then F(W)=∩i=1∞F(Si′).
For solving the equilibrium problem, assume that the bifunctionF satisfies the following conditions:(A1)
F(x,x)=0 for all x∈C;(A2)
F is monotone, that is, F(x,y)+F(y,x)≤0 for any x,y∈C;(A3)
for eachx,y,z∈C,limsupt→0F(tz+(1-t)x,y)≤F(x,y);(A4)
F(x,·) is convex and lower semicontinuous for each x∈C.
Recall some lemmas which will be needed in the rest of this paper.Lemma 2.9 (see [13]).
LetC be a nonempty closed convex subset of H, let F be bifunction from C×C to ℝ satisfying (A1)–(A4), and let r>0 and x∈H. Then, there exists z∈C such that
(2.8)F(z,y)+1r〈y-z,z-x〉≥0,∀y∈C.Lemma 2.10 (see [5]).
Forr>0,x∈H, define a mapping Tr:H→C as follows:
(2.9)Tr(x)={z∈C∣F(z,y)+1r〈y-z,z-x〉≥0,∀y∈C}
for all x∈H. Then, the following statements hold:(i)
Tr is single-valued;(ii)
Tr is firmly nonexpansive, that is, for any x,y∈H,(2.10)‖Trx-Try‖2≤〈Trx-Try,x-y〉;(iii)
F(Tr)=EP(F);(iv)
EP(F) is closed and convex.Lemma 2.11 (see [14]).
Let{xn} and {zn} be bounded sequences in a Banach space and let {βn} be a sequence of real numbers such that 0<liminfn→∞βn≤limsupn→∞βn<1 for all n=0,1,2,…. Suppose that xn+1=(1-βn)zn+βnxn for all n=0,1,2,… and limsupn→∞∥zn+1-zn∥-∥xn+1-xn∥≤0. Then limn→∞∥zn-xn∥=0.Lemma 2.12 (see [4]).
LetC,H,F, and Trx be as in Lemma 2.10. Then, the following holds:
(2.11)‖Tsx-Ttx‖2≤s-ts〈Tsx-Ttx,Tsx-x〉
for all s,t>0 and x∈H.Lemma 2.13 (see [10]).
LetH be a Hilbert space and let C be a nonempty 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 and if {(I-T)xn} converges strongly to y, then (I-T)x=y.


## 3. Main Result

Throughout the rest of this paper, we always assume thatf is a contraction of H into itself with coefficient α∈(0,1), and A is a L-Lipschitzian continuous operator and η-strongly monotone on H with L>0, η>0. Assume that 0<μ<2η/L2 and 0<γ<μ(η-(μL2/2))/α=τ/α.Define a mappingVn=βnI+(1-βn)WnTrn. Since both Wn and Trn are nonexpansive, it is easy to get Vn is also nonexpansive. Consider the following mapping Gn on H defined by(3.1)Gnx=αnγf(x)+(I-αnμA)Vnx,∀x∈H,n∈N,
where αn∈(0,1). By Lemmas 2.3 and 2.10, we have(3.2)‖Gnx-Gny‖≤αnγ‖f(x)-f(y)‖+(1-αnτ)‖Vnx-Vny‖≤αnγα‖x-y‖+(1-αnτ)‖x-y‖=(1-αn(τ-γα))‖x-y‖.
Since 0<1-αn(τ-γα)<1, it follows that Gn is a contraction. Therefore, by the Banach contraction principle, Gn has a unique fixed pointed xnf∈H such that(3.3)xnf=αnγf(xnf)+(I-αnμA)Vnxnf.For simplicity, we will writexn for xnf provided no confusion occurs. Next we prove the sequences {xn} converges strongly to a x*∈Ω=∩i=1∞F(Si)∩EP(F) which solves the variational inequality:(3.4)〈(γf-μA)x*,p-x*〉≤0,∀p∈Ω.
Equivalently, x*=PΩ(I-μA+γf)x*.Theorem 3.1.
LetC be a nonempty closed convex subset of a real Hilbert space H and F a bifunction from C×C to ℝ satisfying (A1)–(A4). Let Si:C→C be a family κi-strict pseudocontractions for some 0≤κi<1. Assume the set Ω=∩i=1∞F(Si)∩EP(F)≠∅. Let f be a contraction of H into itself with α∈(0,1) and let A be a L-Lipschitzian continuous operator and η-strongly monotone with L>0,η>0,0<μ<2η/L2 and 0<γ<μ(η-(μL2/2))/α=τ/α. For every n∈ℕ, let Wn be the mapping generated by Si′ and ti as in (2.5). Let {xn} and {un} be sequences generated by the following algorithm:
(3.5)un=Trnxn,yn=βnxn+(1-βn)Wnun,xn=αnγf(xn)+(I-μαnA)yn.
If {αn}, {βn}, and {rn} satisfy the following conditions:(i)
{αn}⊂(0,1), limn→∞αn=0;(ii)
0<liminfn→∞βn≤limsupn→∞βn<1;(iii)
{rn}⊂(0,∞), liminfn→∞rn>0.Then, {xn} converges strongly to a point x*∈Ω, which solves the variational inequality (3.4).Proof.
The proof is divided into several steps.
Step 1.
Show first that{xn} is bounded.
Take anyp∈Ω, by (3.5) and Lemma 2.3, we derive that
(3.6)‖xn-p‖=‖αn(γf(xn)-μAp)+(I-μαnA)yn-(I-μαnA)p‖≤αnαγ‖xn-p‖+αn‖γf(p)-μAp‖+(1-αnτ)‖yn-p‖≤(1-αn(τ-γα))‖xn-p‖+αn‖γf(p)-μAp‖.
It follows that ∥xn-p∥≤(∥γf(p)-μAp∥)/(τ-γα).
Hence,{xn} is bounded, so are {un} and {yn}. It follows from the Lipschitz continuity of A that {Axn} and {Aun} are also bounded. From the nonexpansivity of f and Wn, it follows that {f(xn)} and {Wnxn} are also bounded.Step 2.
Show that(3.7)lim⁡n→∞‖un-xn‖=0,lim⁡n→∞‖un-yn‖=0.
Notice that
(3.8)‖un-yn‖≤‖un-xn‖+‖xn-yn‖=‖un-xn‖+αn‖γf(xn)-μAyn‖.
By Lemma2.10, we have
(3.9)‖un-p‖2=‖Trnxn-Trnp‖2≤〈xn-p,un-p〉=12(‖un-p‖2+‖xn-p‖2-‖xn-un‖2).
It follows that
(3.10)‖un-p‖2≤‖xn-p‖2-‖xn-un‖2.
Thus, from Lemma2.1 and (3.10), we get
(3.11)‖xn-p‖2=‖αn(γf(xn)-μAp)+(I-μαnA)yn-(I-μαnA)p‖2≤(1-αnτ)2‖yn-p‖2+2αn〈γf(xn)-γf(p)+γf(p)-μAp,xn-p〉≤(1-αnτ)2‖un-p‖2+2αn〈γf(xn)-γf(p)+γf(p)-μAp,xn-p〉≤(1-αnτ)2(‖xn-p‖2-‖xn-un‖2)+2αnγα‖xn-p‖2+2αn‖γf(p)-μAp‖‖xn-p‖=(1-2αn(τ-γα)+(αnτ)2)‖xn-p‖2-(1-αnτ)2‖xn-un‖2+2αn‖γf(p)-μAp‖‖xn-p‖≤‖xn-p‖2+(αnτ)2‖xn-p‖2-(1-αnτ)2‖xn-un‖2+2αn‖γf(p)-μAp‖‖xn-p‖.
It follows that
(3.12)(1-αnτ)2‖xn-un‖2≤(αnτ)2‖xn-p‖2+2αn‖γf(p)-μAp‖‖xn-p‖.
Since αn→0, we have
(3.13)lim⁡n→∞‖un-xn‖=0.
From (3.8), it is easy to get
(3.14)lim⁡n→∞‖un-yn‖=0.Step 3.
Show that(3.15)lim⁡n→∞‖un-Wun‖=0,(3.16)‖un-Wnun‖≤‖un-yn‖+‖yn-Wnun‖=‖un-yn‖+βn(‖xn-un‖+‖un-Wnun‖).
This implies that
(3.17)(1-βn)‖un-Wnun‖≤‖un-yn‖+βn‖xn-un‖.
From condition (ii), (3.13), and (3.14), we have
(3.18)‖un-Wnun‖⟶0.
Notice that
(3.19)‖un-Wun‖≤‖un-Wnun‖+‖Wnun-Wun‖.
By Lemma 2.7 and (3.18), we get (3.15).
Since{un} is bounded, so there exists a subsequence {unj} which converges weakly to x*.Step 4.
Show thatx*∈Ω.
SinceC is closed and convex, C is weakly closed. So, we have x*∈C.
From (3.15), we obtain Wunj⇀x*. From Lemmas 2.8, 2.4, and 2.13, we have x*∈F(W)=∩i=1∞F(Si′)=∩i=1∞F(Si).
Byun=Trnxn, for all n≥1, we have
(3.20)F(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈C.
It follows from (A2) that
(3.21)1rn〈y-un,un-xn〉≥F(y,un),∀y∈C.
Hence, we get
(3.22)1rnj〈y-unj,unj-xnj〉≥F(y,unj),∀y∈C.
It follows from condition (iii), (3.13), and (A4) that
(3.23)0≥F(y,x*),∀y∈C.
For s with 0<s≤1 and y∈C, let ys=sy+(1-s)x*. Since y∈C and x*∈C, we obtain ys∈C and hence F(ys,x*)≤0. So, we have
(3.24)0=f(ys,ys)≤sF(ys,y)+(1-s)F(ys,x*)≤sF(ys,y).
Dividing by s, we get
(3.25)F(ys,y)≥0,∀y∈C.
Letting s→0 and from (A3), we get
(3.26)F(x*,y)≥0
for all y∈Candx*∈EP(F).Hencex*∈Ω.Step 5.
Show thatxn→x*,wherex*=PΩ(I-μA+γf)x*:
(3.27)xn-x*=αn(γf(xn)-μAx*)+(I-μαnA)yn-(I-μαnA)x*.
Hence, we obtain
(3.28)‖xn-x*‖2=αn〈γf(xn)-μAx*,xn-x*〉+〈(I-μαnA)yn-(I-μαnA)x*,xn-x*〉;≤αn〈γf(xn)-μAx*,xn-x*〉+(1-αnτ)‖xn-x*‖2.
It follows that
(3.29)‖xn-x*‖2≤1τ〈γf(xn)-μAx*,xn-x*〉=1τ(γ〈f(xn)-f(x*),xn-x*〉+〈γf(x*)-μAx*,xn-x*〉)≤1τ(γα‖xn-x*‖2+〈γf(x*)-μAx*,xn-x*〉).
This implies that
(3.30)‖xn-x*‖2≤〈γf(x*)-μAx*,xn-x*〉τ-γα.
In particular,
(3.31)‖xnj-x*‖2≤〈γf(x*)-μAx*,xnj-x*〉τ-γα.
Sincexnj⇀x*, it follows from (3.31) that xnj→x* as j→∞. Next, we show that x* solves the variational inequality (3.4).
By the iterative algorithm (3.5), we have
(3.32)xn=αnγf(xn)+(I-μαnA)yn=αnγf(xn)+(I-μαnA)Vnxn.
Therefore, we have
(3.33)μαnAxn-αnγf(xn)=μαnAxn-xn+(I-μαnA)Vnxn,
that is,
(3.34)(μA-γf)xn=-1αn((I-Vn)xn-μαn(Axn-AVn)xn).
Hence, for p∈Ω,
(3.35)〈(μA-γf)xn,xn-p〉=-1αn〈(I-Vn)xn-μαn(Axn-AVnxn),xn-p〉=-1αn〈(I-Vn)xn-(I-Vn)p,xn-p〉+μ〈Axn-AVnxn,xn-p〉≤μ〈Axn-AVnxn,xn-p〉.
SinceI-Vn is monotone (i.e., 〈x-y,(I-Vn)x-(I-Vn)y〉≥0, for all x,y∈H). This is due to the nonexpansivity of Vn.
Now replacingn in (3.35) with nj and letting j→∞, we obtain
(3.36)〈(μA-γf)x*,x*-p〉=lim⁡j→∞〈(μA-γf)xnj,xnj-p〉≤lim⁡j→∞μ〈Axnj-AVnxnj,xnj-p〉=0.
That is,x*∈Ω is a solution of (3.4). To show that the sequence {xn} converges strongly to x*, we assume that xnk→x̂. By the same processing as the proof above, we derive x̂∈Ω. Moreover, it follows from the inequality (3.36) that
(3.37)〈(μA-γf)x*,x*-x̂〉≤0.
Interchanging x* and x̂, we get
(3.38)〈(μA-γf)x̂,x̂-x*〉≤0.
By Lemma 2.5, adding up (3.37) and (3.38) yields
(3.39)(μη-γα)‖x*-x̂‖2≤〈(μA-γf)x*-(μA-γf)x̂,x*-x̂〉≤0.
Hence x*=x̂ and, therefore, xn→x* as n→∞,
(3.40)〈(I-μA+γf)x*-x*,x*-p〉≥0,∀p∈Ω.
This is equivalent to the fixed point equation:
(3.41)PΩ(I-μA+γf)x*=x*.Theorem 3.2.
LetC be a nonempty closed convex subset of a real Hilbert space H and F a bifunction from C×C to ℝ satisfying (A1)–(A4). Let Si:C→C be a family κi-strict pseudocontractions for some 0≤κi<1. Assume the set Ω=∩i=1∞F(Si)∩EP(F)≠∅. Let f be a contraction of H into itself with α∈(0,1) and let A be a L-Lipschitzian continuous operator and η-strongly monotone with L>0, η>0,0<μ<2η/L2, and 0<γ<μ(η-(μL2/2))/α=τ/α. For every n∈ℕ, let Wn be the mapping generated by Si′ and 0<ti≤b<1. Given x1∈H, let {xn} and {un} be sequences generated by the following algorithm:
(3.42)un=Trnxn,yn=βnxn+(1-βn)Wnun,xn+1=αnγf(xn)+(I-μαnA)yn.
If {αn}, {βn} and {rn} satisfy the following conditions:(i)
{αn}⊂(0,1), limn→∞αn=0 and ∑n=1∞αn=∞;(ii)
0<liminfn→∞βn≤limsupn→∞βn<1;(iii)
{rn}⊂(0,∞), liminfn→∞rn>0 and limn→∞|rn+1-rn|=0.Then, {xn} converges strongly to x*∈Ω, which solves the variational inequality (3.4).Proof.
The proof is divided into several steps.
Step 1.
Show first that{xn} is bounded.
Taking anyp∈Ω, we have
(3.43)‖xn+1-p‖=‖αn(γf(xn)-μAp)+(I-μαnA)yn-(I-μαnA)p‖≤αn(‖γf(xn)-γf(p)‖+‖γf(p)-μAp‖)+(1-αnτ)‖yn-p‖≤αnαγ‖xn-p‖+αn‖γf(p)-μAp‖+(1-αnτ)‖yn-p‖=(1-αn(τ-αγ))‖xn-p‖+αn(τ-αγ)‖γf(p)-μAp‖τ-αγ≤max⁡{‖xn-p‖,‖γf(p)-μAp‖τ-αγ}.
By induction, we obtain ∥xn-p∥≤max{∥x1-p∥,∥γf(p)-μAp)∥/(τ-αγ)},n≥1.Hence, {xn} is bounded, so are {un} and {yn}. It follows from the Lipschitz continuity of A that {Axn} and {Aun} are also bounded. From the nonexpansivity of f and Wn, it follows that {f(xn)} and {Wnxn} are also bounded.Step 2.
Show that(3.44)‖xn+1-xn‖⟶0.
Observe that(3.45)‖un+1-un‖=‖Trn+1xn+1-Trnxn‖≤‖Trn+1xn+1-Trn+1xn‖+‖Trn+1xn-Trnxn‖≤‖xn+1-xn‖+‖Trn+1xn-Trnxn‖,
and from (2.5), we have
(3.46)‖Wn+1un-Wnun‖=‖t1S1′Un+1,2un-t1S1′Un,2un‖≤t1‖Un+1,2un-Un,2un‖=t1‖t2S2′Un+1,3un-t2S2′Un,3un‖≤t1t2‖Un+1,3un-Un,3un‖≤⋯≤∏i=1nti‖Un+1,n+1un-Un,n+1un‖≤M1∏i=1nti,
where M1=supn{∥Un+1,n+1un-Un,n+1un∥}.
Supposexn+1=βnxn+(1-βn)zn, then zn=(xn+1-βnxn)/(1-βn)=(αnγf(xn)+(I-μαnA)yn-βnxn)/(1-βn).
Hence, we have(3.47)zn+1-zn=αn+1γf(xn+1)+(I-μαn+1A)yn+1-βn+1xn+11-βn+1-αnγf(xn)+(I-μαnA)yn-βnxn1-βn=αn+1(γf(xn+1)-μAyn+1)1-βn+1+yn+1-βn+1xn+11-βn+1-αn(γf(xn)-μAyn)1-βn-yn-βnxn1-βn=αn+1(γf(xn+1)-μAyn+1)1-βn+1+βn+1xn+1+(1-βn+1)Wn+1un+1-βn+1xn+11-βn+1-αn(γf(xn)-μAyn)1-βn-βnxn+(1-βn)Wnun-βnxn1-βn≤αn+1(γf(xn+1)-μAyn+1)1-βn+1-αn(γf(xn)-μAyn)1-βn+Wn+1un+1-Wnun.
It follows from (3.45), (3.46), and the above result that
(3.48)‖zn+1-zn‖≤αn+11-βn+1(‖γf(xn+1)‖+‖μAyn+1‖)+αn1-βn(‖γf(xn)‖+‖μAyn‖)+‖Wn+1un+1-Wnun‖≤(αn+11-βn+1+αn1-βn)M2+‖Wn+1un+1-Wn+1un‖+‖Wn+1un-Wnun‖≤(αn+11-βn+1+αn1-βn)M2+‖un+1-un‖+‖Wn+1un-Wnun‖≤‖xn+1-xn‖+‖Trn+1xn-Trnxn‖+(αn+11-βn+1+αn1-βn)M2+M1∏i=1nti,
where M2=supn{∥γf(xn)∥+∥μAyn∥}. Hence, we get
(3.49)‖zn+1-zn‖-‖xn+1-xn‖≤‖Trn+1xn-Trnxn‖+(αn+11-βn+1+αn1-βn)M2+M1∏i=1nti.
From condition (i), (iii), 0<tn≤b<1, and Lemma 2.12, we obtain
(3.50)limsup⁡n→∞(‖zn+1-zn‖-‖xn+1-xn‖)≤0.
By Lemma 2.11,we have limn→∞∥zn-xn∥=0. Thus,
(3.51)lim⁡n→∞‖xn+1-xn‖=lim⁡n→∞(1-βn)‖zn-xn‖=0.
By Lemma2.12, (3.45) and (3.44), we obtain
(3.52)‖un+1-un‖⟶0.Step 3.
Show that(3.53)‖xn-Wxn‖⟶0.
Observe that
(3.54)‖xn-Wnxn‖≤‖xn-Wnun‖+‖Wnun-Wnxn‖≤‖xn-Wnun‖+‖un-xn‖,‖xn-Wnun‖≤‖xn-xn+1‖+‖xn+1-yn‖+‖yn-Wnun‖=‖xn-xn+1‖+‖xn+1-yn‖+βn(‖un-xn‖+‖xn-Wnun‖).
From condition (i) and (3.5),we can obtain
(3.55)(1-βn)‖xn-Wnun‖≤‖xn-xn+1‖+‖xn+1-yn‖+βn‖un-xn‖≤‖xn-xn+1‖+αn‖γf(xn)-μAyn‖+βn‖un-xn‖.
By Lemma2.10, we get
(3.56)‖un-p‖2=‖Trnxn-Trnp‖2≤〈Trnxn-Trnp,xn-p〉=12(‖un-p‖2+‖xn-p‖2+‖xn-un‖2).
This implies that
(3.57)‖un-p‖2≤‖xn-p‖2-‖xn-un‖2.
By nonexpansivity of Wn, we have
(3.58)‖yn-p‖2≤βn‖xn-p‖2+(1-βn)‖un-p‖2≤‖xn-p‖2-(1-βn)‖xn-un‖2.
It follows from (3.42) that
(3.59)‖xn+1-p‖2=‖αn(γf(xn)-p)+(I-μαnA)yn-(I-μαnA)p+αn(p-μAp)‖2≤αn‖γf(xn)-p‖2+(1-αnτ)‖yn-p‖2+αn‖p-μAp‖2≤αn‖γf(xn)-p‖2+(1-αnτ)(‖xn-p‖2-(1-βn)‖xn-un‖2)+αn‖p-μAp‖2≤αn‖γf(xn)-p‖2+‖xn-p‖2-(1-βn)‖xn-un‖2+αn‖p-μAp‖2.
This implies that
(3.60)(1-βn)‖xn-un‖2≤αn(‖γf(xn)-p‖2+‖p-μAp‖2)+‖xn-p‖2-‖xn+1-p‖2≤αn(‖γf(xn)-p‖2+‖p-μAp‖2)+(‖xn-p‖+‖xn+1-p‖)‖xn+1-xn‖.
From condition (i), (ii), and (3.44), we have
(3.61)‖xn-un‖⟶0.
Further we have ∥xn-Wnun∥→0. Thus we get
(3.62)‖xn-Wnxn‖⟶0.
On the other hand, we have(3.63)‖xn-Wxn‖≤‖xn-Wnxn‖+‖Wnxn-Wxn‖≤‖xn-Wnxn‖+sup⁡xn∈C⁡‖Wnxn-Wxn‖.
Combining (3.62), the last inequality, and Lemma 2.7, we obtain (3.53).Step 4.
Show that(3.64)limsup⁡n→∞〈(γf-μA)x*,xn-x*〉≤0,
where x*=PΩ(I-μA+γf)x* is a unique solution of the variational inequality (3.4). Indeed, take a subsequence {xnj} of {xn} such that
(3.65)limsup⁡n→∞〈(γf-μA)x*,xn-x*〉=lim⁡j→∞〈(γf-μA)x*,xnj-x*〉.
Since{xnj} is bounded, there exists a subsequence {xnjk} of {xnj} which converges weakly to q. Without loss of generality, we can assume xnj⇀q. From (3.53), we obtain Wxnj⇀q.
By the same argument as in the proof of Theorem3.1, we have q∈Ω. Since x*=PΩ(I-μA+γf)x*, it follows that
(3.66)limsup⁡n→∞〈(γf-μA)x*,xn-x*〉=lim⁡j→∞〈(γf-μA)x*,xnj-x*〉=〈(γf-μA)x*,q-x*〉≤0.Step 5.
Show that(3.67)xn⟶x*.
Since
(3.68)〈(γf-μA)x*,xn+1-x*〉=〈(γf-μA)x*,xn+1-xn〉+〈(γf-μA)x*,xn-x*〉≤‖(γf-μA)x*‖‖xn+1-xn‖+〈(γf-μA)x*,xn-x*〉.
It follows from (3.44) and (3.66) that
(3.69)limsup⁡n→∞〈(γf-μA)x*,xn+1-x*〉≤0.‖xn+1-x*‖2=‖αnγf(xn)+(I-μαnA)yn-x*‖2=‖(I-μαnA)yn-(I-μαnA)x*+αn(γf(xn)-μAx*)‖2≤‖(I-μαnA)yn-(I-μαnA)x*‖2+2αn〈γf(xn)-μAx*,xn+1-x*〉≤(1-αnτ)2‖yn-x*‖2+2αn〈γf(xn)-γf(x*),xn+1-x*〉+2αn〈(γf-μA)x*,xn+1-x*〉≤(1-αnτ)2‖xn-x*‖2+αnαγ(‖xn-x*‖2+‖xn+1-x*‖2)+2αn〈(γf-μA)x*,xn+1-x*〉.
This implies that
(3.70)‖xn+1-x*‖2≤(1-αnτ)2+αnαγ1-αnαγ‖xn-x*‖2+2αn1-αnαγ〈(γf-μA)x*,xn+1-x*〉≤(1-2αn(τ-αγ)1-αnαγ)‖xn-x*‖2+2αn1-αnαγ〈(γf-μA)x*,xn+1-x*〉+(αnτ)21-αnαγM3,
where M3=supn∥xn-x*∥2, n≥1. It is easily to see that γn=2αn(τ-αγ)/(1-αnαγ). Hence, by Lemma 2.2, the sequence {xn} converges strongly to x*.Remark 3.3.
IfF≡0, then Theorem 3.2 reduces to Theorem 3.1 of Wang [10].



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*Source: 602513-2012-05-09.xml*