# Existence and Iterative Approximation Methods for Generalized Mixed Vector Equilibrium Problems with Relaxed Monotone Mappings

Authors: Rabian Wangkeeree; Panu Yimmuang
Journal: Journal of Applied Mathematics
 (2013)

---

## Abstract

We first consider an auxiliary problem for the generalized mixed vector equilibrium problem with a relaxed monotone mapping and prove the existence and uniqueness of the solution for the auxiliary problem. We then introduce a new iterative scheme for approximating a common element of the set of solutions of a generalized mixed vector equilibrium problem with a relaxed monotone mapping and the set of common fixed points of a countable family of nonexpansive mappings. The results presented in this paper can be considered as a generalization of some known results due to Wang et al. (2010).


---

## Body

## 1. Introduction

LetH be a real Hilbert space with inner product〈·,·〉and norm∥·∥, respectively. LetXbe a nonempty closed convex subset ofH.Letφ:X×X→ℝ=(-∞,+∞)be a bifunction. The equilibrium problem EP(φ) is to findx∈Xsuch that
(1)φ(x,y)≥0,∀y∈X.
As pointed out by Blum and Oettli [1], EP(φ)provides a unified model of several problems, such as the optimization problem, fixed point problem, variational inequality, and complementarity problem.A mappingS:X→His called nonexpansive, if
(2)∥Sz-Sy∥≤∥z-y∥,∀z,y∈X.
We denote the set of all fixed points of S byF(S), that is,F(S)={z∈X:z=Sz}.It is well known that if X⊂H is bounded, closed, convex and S is a nonexpansive mapping of X onto itself, then F(S) is nonempty (see [2]). A mapping T:C→H is said to be relaxed η-α monotone if there exist a mapping η:C×C→H and a function α:H→ℝ positively homogeneous of degree p, that is, α(tz)=tpα(z) for allt>0andz∈Hsuch that
(3)〈Tx-Ty,η(x,y)〉≥α(x-y),∀x,y∈C,
wherep>1is a constant; see [3]. In the case ofη(x,y)=x-yfor allx,y∈C,  T is said to be relaxedα-monotone. In the case ofη(x,y)=x-yfor allx,y∈Candα(z)=k∥z∥p, wherep>1andk>0,  T is said to bep-monotone; see [4–6]. In fact, in this case, if p=2, then T is a k-strongly monotone mapping. Moreover, every monotone mapping is relaxedη-αmonotone withη(x,y)=x-yfor allx,y∈Candα=0.In 2000, Moudafi [7] introduced an iterative scheme of finding the solution of nonexpansive mappings and proved a strong convergence theorem. Recently, Huang et al. [8] introduced the approximate method for solving the equilibrium problem and proved the strong convergence theorem.Letφ:X×X→ℝbe a bifunction andT,A:X→Hnonlinear mappings. In 2010, Wang et al. [9] introduced the following generalized mixed equilibrium problem with a relaxed monotone mapping. 						(4)Findz∈Csuchthatφ(z,y)+〈Tz,η(y,z)〉+〈Az,y-z〉≥0,∀y∈C.
Problem (4) is very general setting, and it includes special cases of Nash equilibrium problems, complementarity problems, fixed point problems, optimization problems, and variational inequalities (see, e.g., [8, 10–13] and the references therein). Moreover, Wang et al. [9] studied the existence of solutions for the proposed problem and introduced a new iterative scheme for finding a common element of the set of solutions of a generalized equilibrium problem with a relaxed monotone mapping and the set of common fixed points of a countable family of nonexpansive mappings in a Hilbert space.It is well known that the vector equilibrium problem provides a unified model of several problems, for example, vector optimization, vector variational inequality, vector complementarity problem, and vector saddle point problem [14–16]. In recent years, the vector equilibrium problem has been intensively studied by many authors (see, e.g., [12, 14–19] and the references therein).Recently, Li and Wang [18] first studied the viscosity approximation methods for strong vector equilibrium problems and fixed point problems. Very recently, Shan and Huang [20] studied the problem of finding a common element of the set of fixed points of a nonexpansive mapping, the set of solutions of the generalized mixed vector equilibrium problem, and the solution set of a variational inequality problem with a monotone Lipschitz continuous mapping in Hilbert spaces. They first introduced an auxiliary problem for the generalized mixed vector equilibrium problem and proved the existence and uniqueness of the solution for the auxiliary problem. Furthermore, they introduced an iterative scheme for finding a common element of the set of fixed points of a nonexpansive mapping, the set of solutions of the generalized mixed vector equilibrium problem, and the solution set of a variational inequality problem with a monotone Lipschitz continuous mapping.LetYbe a Hausdorff topological vector space, and, let Cbe a closed, convex and pointed cone ofYwith intC≠∅.Letφ:X×X→Ybe a vector-valued bifunction. The strong vector equilibrium problem (for short, SVEP(φ)) is to findz∈Xsuch that
(5)φ(z,y)∈C,∀y∈X
and the weak vector equilibrium problem (for short, WVEP(φ)) is to findz∈Xsuch that
(6)φ(z,y)∉-intC,∀y∈X.In this paper, inspired and motivated by the works mentioned previously, we consider the following generalized mixed vector equilibrium problem with a relaxed monotone mapping (for short, GVEPR(φ,T)): findz∈Xsuch that
(7)φ(z,y)+e〈Tz,η(y,z)〉+e〈Az,y-z〉∈C,∀y∈X,
wheree∈intC,φ:X×X→Y, andT,A:X→Hare the mappings. The set of all solutions of the generalized mixed vector equilibrium problem with a relaxed monotone mapping is denoted by SGVEPR(φ,T), that is,
(8)SGVEPR(φ,T)={z∈X:φ(z,y)+e〈Tz,η(y,z)〉+e〈Az,y-z〉∈C,∀y∈X}.
IfA=0, we denote the set ASGVEPR(φ,T) by
(9)ASGVEPR(φ,T)={z∈X:φ(z,y)+e〈Tz,η(y,z)〉∈C,∀y∈X}.
Some special cases of the problem (7) are as follows.(1)
IfY=ℝ,  C=ℝ+, ande=1, then GVEPR(φ,T) (7) reduces to the generalized mixed equilibrium problem with a relaxed monotone mapping (4).(2)
IfT=0andA=0, then GVEPR(φ,T) (7) reduces to the classic vector equilibrium problem (5).We consider the auxiliary problem of GVEPR(φ,T) and prove the existence and uniqueness of the solutions of auxiliary problem of GVEPR(φ,T) under some proper conditions. By using the result for the auxiliary problem, we introduce a new iterative scheme for finding a common element of the set of solutions of a generalized mixed vector equilibrium problem with a relaxed monotone mapping and the set of common fixed points of a countable family of nonexpansive mappings and then obtain a strong convergence theorem. The results presented in this paper improve and generalize some known results of Wang et al. [9].


## 2. Preliminaries

LetA:X→Hbe aλ-inverse-strongly monotone mapping ofH.For allz,y∈Xandk>0, one has [21]
(10)∥(I-kA)z-(I-kA)y∥2≤∥z-y∥2+k(k-2λ)∥Az-Ay∥2.
Hence, if k∈(0,2λ), then I-kA is a nonexpansive mapping of X intoH.For each pointz∈H, there exists a unique nearest point of X, denoted by PXz, such that
(11)∥z-PXz∥≤∥z-y∥,
for ally∈X.Such aPXis called the metric projection fromHontoX.The well-known Browder’s characterization ofPXensures thatPXis a firmly nonexpansive mapping fromHontoX, that is,
(12)∥PXz-PXy∥2≤〈PXz-PXy,z-y〉,∀z,y∈H.
Further, we know that for anyz∈Handx∈X,  x=PXzif and only if
(13)〈z-x,x-y〉≥0,∀y∈X.
LetSbe a nonexpansive mapping ofXinto itself such thatF(S)≠∅.Then we have
(14)x^∈F(S)⟺∥Sx-x∥2≤2〈x-Sx,x-x^〉,∀x∈X,
which is obtained directly from the following:
(15)∥x-x^∥2≥∥Sx-Sx^∥2=∥Sx-x^∥2=∥Sx-x+(x-x^)∥2=∥Sx-x∥2+∥x-x^∥2+2〈Sx-x,x-x^〉.
This inequality is a very useful characterization ofF(S).Observe what is more that it immediately yields that Fix(S) is a convex closed set.Definition 1 (see [6, 22]).
LetX and Y be two Hausdorff topological vector spaces, E a nonempty, convex, subset of X and C a closed, convex and pointed cone of Y with intC≠∅.Let θ be the zero point of Y,  𝕌(θ) the neighborhood set of θ,  𝕌(x0) be the neighborhood set of x0, and f:E→Y a mapping. (1)
If for anyV∈𝕌(θ) inY, and there existsU∈𝕌(x0)such that
(16)f(x)∈f(x0)+V+C,∀x∈U∩E,
then f is called upperC-continuous on x0. If f is upper C-continuous for all x∈E,then f is called upper C-continuous on E.(2)
If for anyV∈𝕌(θ)in Y, and there exists U∈𝕌(x0) such that
(17)f(x)∈f(x0)+V-C,∀x∈U∩E,
thenfis called lowerC-continuous onx0.Iffis lowerC-continuous for allx∈E, then f is called lowerC-continuous onE.(3)
fis calledC-continuous iffis upperC-continuous and lowerC-continuous.(4)
If for anyx,y∈Eandt∈[0,1], and the mappingfsatisfies
(18)f(x)∈f(tx+(1-t)y)+Corf(y)∈f(tx+(1-t)y)+C,
thenfis called properC-quasiconvex.(5)
If for anyx1,x2∈Eandt∈[0,1], and the mappingfsatisfies
(19)tf(x1)+(1-t)f(x2)∈f(tx1-(1-t)x2)+C,
thenfis calledC-convex.Lemma 2 (see [19]).
LetXandYbe two real Hausdorff topological vector spaces,Eis a nonempty, compact, convex subset ofX, andCis a closed, convex, and pointed cone ofY.Assume thatf:E×E→Yandψ:E→Yare two vector valued mappings. Suppose thatfandψsatisfy the following:(i)
f(x,x)∈C, for allx∈E;(ii)
ψis upperC-continuous onE;(iii)
f(·,y)is lowerC-continuous for ally∈E;(iv)
f(x,·)+ψ(·)is properC-quasiconvex for allx∈E.
Then there exists a pointx∈Esatisfying
(20)F(x,y)∈C∖{0},∀y∈E,
where
(21)F(x,y)=f(x,y)+ψ(y)-ψ(x),∀x,y∈E.Definition 3 (see [3]).
LetEbe a Banach space with the dual spaceE*and letKbe a nonempty subset ofE.LetT:K→E*, andη:K×K→Ebe two mappings. The mappingT:K→E*is said to beη-hemicontinuous, if for any fixedx,y∈K, the functionf:[0,1]→(-∞,∞)defined by
(22)f(t)=〈T((1-t)x+ty),η(x,y)〉
is continuous at0+.


## 3. The Existence of Solutions for the Generalized Mixed Vector Equilibrium Problem with a Relaxed Monotone Mapping

For solving the generalized mixed vector equilibrium problem with a relaxed monotone mapping, we give the following assumptions. LetHbe a real Hilbert space with inner 〈·,·〉 and norm ∥·∥, respectively. Assume thatX⊆His nonempty, compact, convex subset, Yis real Hausdorff topological vector space, and C⊆Yis a closed, convex, and pointed cone. Let φ:X×X→Y,  T:X→H be two mappings. For any x∈H, define a mapping Φx:X×X→Y as follows:
(23)Φx(z,y)=φ(z,y)+e〈Tz,η(y,z)〉+er〈y-z,z-x〉,
where r is a positive number in ℝ ande∈C∖{0}.LetΦx,  φ, and T satisfy the following conditions: (A1)
for allz∈X,φ(z,z)=θ;(A2)
φis monotone, that is,φ(z,y)+φ(y,z)∈-Cfor all z,y∈X;(A3)
φ(·,y) is C-continuous for all y∈X;(A4)
φ(z,·) is C-convex, that is,
(24)tφ(z,y1)+(1-t)φ(z,y2)∈φ(z,ty1+(1-t)y2)+C,∀z,y1,y2∈X,∀t∈[0,1],(A5)
for ally∈X,z↦〈Tz,η(y,z)〉 is continuous, and for any u,v∈X,(25)y↦〈Tu,η(y,v)〉isconvexandlowersemicontinuous;(A6)
Φx(z,·) is proper C-quasiconvex for all z∈X and x∈H.Remark 4.
LetY=ℝ,  C=ℝ+, and e=1.For any y∈X, if φ(·,y) is upper semicontinuous and z↦〈Tz,η(y,z)〉 is continuous, then Φx(·,y) is lower C-continuous. In fact, since φ(·,y) is upper semicontinuous and z↦〈Tz,η(y,z)〉 is continuous, for any ϵ>0, there exists a δ>0 such that, for all z∈{z∈X,∥z-z0∥<δ}, we have
(26)Φx(z,y)<Φx(z0,y)+ϵ,
wherez0is a point inX.This means that Φx(·,y)is lowerC-continuous.Remark 5.
LetY=ℝ,  C=ℝ+ande=1.Assume thatφ(z,·)is a convex mapping for allz∈X. Then for anyy1,y2∈Xandt∈[0,1], we have
(27)Φx(z,ty1+(1-t)y2)=φ(z,ty1+(1-t)y2)+〈Tz,η(ty1+(1-t)y2,z)〉+1r〈ty1+(1-t)y2-z,z-x〉≤tφ(z,y1)+(1-t)φ(z,y2)+t〈Tz,η(y1,z)〉+(1-t)〈Tz,η(y2,z)〉+tr〈y1-z,z-x〉+1-tr〈y2-z,z-x〉=t(φ(z,y1)+〈Tz,η(y1,z)〉+1r〈y1-z,z-x〉)+(1-t)(φ(z,y2)+〈Tz,η(y2,z)〉1r+1r〈y2-z,z-x〉)=tΦx(z,y1)+(1-t)Φx(z,y2)≤max{Φx(z,y1),Φx(z,y2)},
which implies thatΦx(z,·)is properC-quasiconvex.Now we are in the position to state and prove the existence of solutions for the generalized mixed vector equilibrium problem with a relaxed monotone mapping.Theorem 6.
LetX be a nonempty, compact, convex subset of a real Hilbert space H. Let C be a closed, convex, and pointed cone of a Hausdorff topological vector space Y. Let T:X→H be an η-hemicontinuous and relaxed η-α-monotone mapping. Let φ:X×X→Y be a vector-valued bifunction. Suppose that all the conditions (A1)–(A6) are satisfied. Let r>0 and define a mapping Br:H→X as follows:
(28)Br(x)={z∈X:φ(z,y)+e〈Tz,η(y,z)〉er+er〈y-z,z-x〉∈C,∀y∈X}
for allx∈H.Assume that(i)
η(x,y)+η(y,x)=0, for allx,y∈X;(ii)
for anyx,y∈X,  α(x-y)+α(y-x)≥0.
Then, the following holds.(1)
Br(z)≠∅for allz∈X.(2)
Bris single-value.(3)
Bris a firmly nonexpansive mapping, that is, for allx,y∈X,(29)∥Brx-Bry∥2≤〈Brx-Bry,x-y〉,(4)
F(Br)=ASGVEPR(φ,T),(5)
ASGVEPR(φ,T) is closed and convex.Proof.
(1) In Lemma2, letf(z,y)=Φx(z,y), and, let ψ(z)=θfor allz,y∈Xandx∈H.Then it is easy to check thatfandΦsatisfy all the conditions of Lemma 2. Thus, there exists a pointz∈Xsuch that
(30)f(z,y)+ψ(z)-ψ(y)∈C,∀y∈X,x∈H,
which gives that, for any x∈H,
(31)φ(z,y)+e〈Tz,η(y,z)〉+er〈y-z,z-x〉∈C,∀y∈X.
Therefor we conclude that Br(x)≠∅ for all x∈H.
(2) Forx∈H and r>0, let z1,z2∈Br(x). Then
(32)φ(z1,y)+e〈Tz1,η(y,z1)〉+er〈y-z1,z1-x〉∈C,∀y∈X,(33)φ(z2,y)+e〈Tz2,η(y,z2)〉+er〈y-z2,z2-x〉∈C,∀y∈X.
Letting y=z2 in (32) and y=z1 in (33), adding (32) and (33), we have
(34)φ(z2,z1)+φ(z1,z2)+e〈Tz1-Tz2,η(z2,z1)〉+er〈z1-z2,z2-z1〉∈C.
By the monotonicity of φ, we have
(35)e〈Tz1-Tz2,η(z2,z1)〉+er〈z1-z2,z2-z1〉∈C.
Thus
(36)er〈z1-z2,z2-z1〉-e〈Tz2-Tz1,η(z2,z1)〉∈C.
Since T is relaxed η-α-monotone andr>0and the property ofC, one has
(37)e〈z1-z2,z2-z1〉-erα(z2-z1)∈C.
In (36) exchanging the position ofz1andz2, we get
(38)er〈z2-z1,z1-z2〉-eα(z1-z2)∈C,
that is,
(39)e〈z2-z1,z1-z2〉-erα(z1-z2)∈C.
Now, adding the inequalities (37) and (39),
(40)e〈z1-z2,z2-z1〉-erα(z2-z1)+e〈z2-z1,z1-z2〉-erα(z1-z2)∈C.
By using (iv), we have
(41)2e〈z2-z1,z1-z2〉∈C.
If〈z2-z1,z1-z2〉<0, then
(42)-2〈z2-z1,z1-z2〉>0.
This implies that
(43)-2e〈z2-z1,z1-z2〉∈C.
From (41) and (43), we have z1=z2 which is a contradiction. Thus
(44)〈z2-z1,z1-z2〉≥0,
so
(45)-∥z1-z2∥2=〈z1-z2,z2-z1〉≥0.
Hence z1=z2.Therefore Br is single value.
(3) For anyx1,x2∈H, let z1=Br(x1) and z2=Br(x2). Then
(46)φ(z1,y)+e〈Tz1,η(y,z1)〉+er〈y-z1,z1-x1〉∈C,∀y∈X,(47)φ(z2,y)+e〈Tz2,η(y,z2)〉+er〈y-z2,z2-x2〉∈C,∀y∈X.
Lettingy=z2in (46) andy=z1in (47), adding (46) and (47), we have
(48)φ(z1,z2)+φ(z2,z1)+e〈Tz1,η(z2,z1)〉+e〈Tz2,η(z1,z2)〉+er〈z2-z1,z1-z2-(x1-x2)〉∈C.
Sinceφis monotone andCis closed convex cone, we get
(49)〈Tz1-Tz2,η(z2,z1)〉+1r〈z2-z1,z1-z2-x1+x2〉≥0,
that is,
(50)1r〈z2-z1,z1-z2-x1+x2〉≥〈Tz2-Tz1,η(z2,z1)〉≥α(z2-z1).
In (50) exchanging the position of z1 and z2, we get
(51)1r〈z1-z2,z2-z1-x2+x1〉≥α(z1-z2).
Adding the inequalities (50) and (51), we have
(52)2〈z1-z2,z2-z1-x2+x1〉≥r(α(z1-z2)+α(z2-z1)).
It follows from (iv) that
(53)〈z1-z2,z2-z1-x2+x1〉≥0.
This implies that
(54)∥Brx1-Brx2∥2≤〈Brx1-Brx2,x1-x2〉.
This shows thatBris firmly nonexpansive.
(4) We claim thatF(Br)=ASGVEPR(φ,T).Indeed, we have the following:
(55)x∈F(Br)⟺x=Brx⟺φ(x,y)+e〈Tx,η(y,x)〉+er〈y-x,x-x〉∈C,∀y∈X⟺φ(x,y)+e〈Tx,η(y,x)〉∈C,∀y∈X⟺x∈ASGVEPR(φ,T).
(5) Since every firmly nonexpansive mapping is nonexpansive, we see thatBris nonexpansive. Since the set of fixed point of every nonexpansive mapping is closed and convex, we have that ASGVEPR(φ,T)is closed and convex. This completes the proof.


## 4. Convergence Analysis

In this section, we prove a strong convergence theorem which is one of our main results.Theorem 7.
LetXbe a nonempty, compact, convex subset of a real Hilbert spaceH.LetCbe a closed, convex cone of a real Hausdorff topological vector space Y ande∈C∖{0}. Letφ:X×X→Ysatisfy (A1)–(A6). LetT:X→Hbe an η-hemicontinuous and relaxed η-α-monotone mapping. Let A:X→H be aλ-inverse-strongly monotone mapping, and let{Sn}n=1∞be a countable family of nonexpansive mappings from X onto itself such that
(56)F:=∩n=1∞Fix(Sn)∩SGVEPR(φ,T)≠∅.
Assume that the conditions (i)-(ii) of Theorem 6 are satisfied. Putα0=1and assume that{αn}n=1∞⊂(0,1)is a strictly decreasing sequence. Assume that{βn}n=1∞⊂(c,d)with somec,d∈(0,1)and{λn}n=1∞⊂[a,b]with somea,b∈(0,2λ).Then, for anyx1∈X, the sequence {xn}, generated by
(57)φ(un,y)+e〈Tun,η(y,un)〉+e〈Axn,y-un〉+eλn〈y-un,un-xn〉∈C,∀y∈X,yn=αnxn+∑i=1n‍(αi-1-αi)βnSixn+(1-αn)(1-βn)un,Cn={z∈X:∥yn-z∥≤∥xn-z∥},Dn=∩j=1nCj,xn+1=PDnx1,n≥1,
converges strongly tox*∈PFx1.In particular, ifXcontains the origin 0 and takingx1=0, then the sequence{xn}generated by (57) converges strongly to the minimum norm element inF, that isx*=PF0.Proof.
We divide the proof into several steps.
Step
1. We will show that  Fis closed and convex, the sequence {xn} generated by (57) is well defined, and F⊂Dn, for alln≥1.
First, we prove thatF is closed and convex. It suffices to prove that SGVEPR(φ,T) is closed and convex. Indeed, it is easy to prove the conclusion by the following fact:
(58)∀p∈SGVEPR(φ,T)⟺φ(p,y)+e〈Tp,η(y,p)〉+eλn〈y-p,λnAp〉∈C,∀y∈X⟺φ(p,y)+e〈Tp,η(y,p)〉+eλn×〈y-p,p-(p-λnAp)〉∈C∀y∈X⟺p=Bλn(I-λnA)p.
This implies that
(59)SGVEPR(φ,T)=Fix[Bλn(I-λnA)].
SinceBλn(I-λnA)is a nonexpansive mapping forλn<2λand the set of fixed points of a nonexpansive mapping is closed and convex, we have that SGVEPR(φ,T)is closed and convex.
Next, we prove that the sequence{xn}generated by (57) is well defined andF⊂Dnfor alln≥1.By Definition ofCn, for allz∈X, the inequality
(60)∥yn-z∥≤∥xn-z∥
is equivalent to
(61)〈yn-xn,yn+xn〉-2〈yn-xn,z〉≤0.
It is easy to see that Cn is closed and convex for alln∈ℕ. Hence Dn is closed and convex for all n∈ℕ. For any p∈F, and since un=Bλn(xn-λnAxn) and I-λnA is nonexpansive, we have
(62)∥yn-p∥=∥αn(xn-p)+∑i=1n‍(αi-1-αi)βn(Sixn-p)∑i=1n‍+(1-αn)(1-βn)(un-p)∥≤αn∥xn-p∥+∑i=1n‍(αi-1-αi)βn∥Sixn-p∥+(1-αn)(1-βn)∥un-p∥≤αn∥xn-p∥+∑i=1n‍(αi-1-αi)βn∥xn-p∥+(1-αn)(1-βn)×∥Bλn(xn-λnAxn)-Bλn(p-λnAp)∥≤αn∥xn-p∥+(1-αn)βn∥xn-p∥+(1-αn)(1-βn)×∥(xn-λnAxn)-(p-λnAp)∥≤αn∥xn-p∥+(1-αn)βn∥xn-p∥+(1-αn)(1-βn)∥xn-p∥=∥xn-p∥.
This implies thatF⊂Cnfor alln∈ℕ. HenceF⊂∩j=1nCj.That is
(63)F⊂Dn,∀n∈ℕ.
SinceDnis nonempty closed convex, we get that the sequence{xn}is well defined. This completes the proof of Step 1.
Step
2. We shall show that ∥xn+1-xn∥→0asn→∞and there isx*∈Csuch that limn→∞∥xn-x*∥=0.
It easy to see thatDn+1⊂Dn for all n∈ℕ from the construction ofDn.Hence
(64)xn+2=PDn+1x1∈Dn+1⊂Dn.
Sincexn+1=PDnx1, we have
(65)∥xn+1-x1∥≤∥xn+2-x1∥,
for alln≥1. This implies that {∥xn-x1∥} is increasing. Note that C is bounded, we get that {∥xn-x1∥} is bounded. This shows that limn→∞∥xn-x1∥ exists.
Sincexn+1=PDnx1 and xm+1=PDmx1∈Dm⊂Dn for all m≥n, we have
(66)〈xn+1-x1,xm+1-xn+1〉≥0.
It follows from (66) that
(67)∥xm+1-xn+1∥2=∥xm+1-x1-(xn+1-x1)∥2=∥xm+1-x1∥2+∥xn+1-x1∥2-2〈xm+1-x1,xn+1-x1〉=∥xm+1-x1∥2+∥xn+1-x1∥2-2〈xn+1-x1,xm+1-xn+1+xn+1-x1〉=∥xm+1-x1∥2+∥xn+1-x1∥2-2〈xn+1-x1,xm+1-xn+1〉≤∥xm+1-x1∥2-∥xn+1-x1∥2.
By takingm=n+1in (67), we have
(68)∥xn+2-xn+1∥2≤∥xn+2-x1∥2-∥xn+1-x1∥2.
Since the limits of∥xn-x1∥exist, we get that
(69)∥xn+2-xn+1∥⟶0,asn⟶∞.
This implies that
(70)limn→∞∥xn+1-xn∥=0.
Moreover, from (67), we also have
(71)limm,n→∞∥xm+1-xn+1∥=0.
This shows that the sequence{xn}is a Cauchy sequence. Hence there isx*∈Csuch that
(72)xn⟶x*∈C,asn⟶∞.
Step
3. We shall show that ∥xn-un∥→0asn→∞.
Sincexn+1∈Cnandlimn→∞∥xn-xn+1∥=0, we have
(73)∥y-xn+1∥≤∥xn-xn+1∥⟶0asn⟶∞,
and hence
(74)∥yn-xn∥≤∥yn-xn+1∥+∥xn-xn+1∥⟶0asn⟶∞.
Note that un can be rewritten as un=Bλn(xn-λnAxn) for all n≥1. We take p∈F thus we have p=Bλn(p-λnAp). Since  Ais λ-inverse-strongly monotone, and 0<λn<2λ, we know that, for all n∈ℕ,(75)∥un-p∥2=∥Bλn(xn-λnAxn)-Bλn(p-λnAp)∥2≤∥xn-λnAxn-p+λnAp∥2=∥(xn-p)-λn(Axn-Ap)∥2=∥xn-p∥2-2λn〈xn-p,Axn-Ap〉+λn2∥Axn-Ap∥2≤∥xn-p∥2-2λnλ∥Axn-Ap∥2+λn2∥Axn-Ap∥2=∥xn-p∥2+λn(λn-2λ)∥Axn-Ap∥2≤∥xn-p∥2.
Using (57) and (75), we have
(76)∥yn-p∥2=∥αn(xn-p)∑i=1n‍+∑i=1n‍(αi-1-αi)βn(Sixn-p)∑i=1n‍+(1-αn)(1-βn)(un-p)∥2≤αn∥xn-p∥2+∑i=1n‍(αi-1-αi)βn∥Sixn-p∥2+(1-αn)(1-βn)∥un-p∥2≤αn∥xn-p∥2+∑i=1n‍(αi-1-αi)βn∥xn-p∥2+(1-αn)(1-βn)×(∥xn-p∥2+λn(λn-2λ)∥Axn-Ap∥2)=∥xn-p∥2+(1-αn)×(1-βn)λn(λn-2λ)∥Axn-Ap∥2,
and hence
(77)(1-αn)(1-d)a(2λ-b)∥Axn-Ap∥2≤(1-αn)(1-βn)λn(2λ-λn)∥Axn-Ap∥2≤∥xn-p∥2-∥yn-p∥2≤∥xn-yn∥(∥xn-p∥+∥yn-p∥).
Note that{xn}and{yn}are bounded,αn→0, andxn-ynconverges to0, we get that
(78)limn→∞∥Axn-Ap∥⟶0.
Using Theorem 6, we have
(79)∥un-p∥2=∥Bλn(xn-λnAxn)-Bλn(p-λnAp)∥2≤〈xn-λnAxn-(p-λnAp),un-p〉=12(∥xn-λnAxn-(p-λnAp)∥2+∥un-p∥2-∥xn-λnAxn-(p-λnAp)-(un-p)∥2)≤12(∥xn-p∥2+∥un-p∥2-∥xn-un-λn(Axn-Ap)∥2)=12(∥xn-p∥2+∥un-p∥2-∥xn-un∥2+2λn〈xn-un,Axn-Ap〉-λn2∥Axn-Ap∥2).
This implies that
(80)∥un-p∥2≤∥xn-p∥2-∥xn-un∥2+2λn〈xn-un,Axn-Ap〉-λn2∥Axn-Ap∥2.
From (80), we have
(81)∥yn-p∥2=∥αn(xn-p)+∑i=1n‍(αi-1-αi)βn(Sixn-p)∑i=1n‍+(1-αn)(1-βn)(un-p)∥2≤αn∥xn-p∥2+∑i=1n‍(αi-1-αi)βn∥Sixn-p∥2+(1-αn)(1-βn)∥un-p∥2≤αn∥xn-p∥2+(1-αn)βn∥xn-p∥2+(1-αn)(1-βn)×(∥xn-p∥2-∥xn-un∥2+2λn〈xn-un,Axn-Ap〉-λn2∥Axn-Ap∥2)≤∥xn-p∥2-(1-αn)(1-βn)∥xn-un∥2+2(1-αn)(1-βn)λn〈xn-un,Axn-Ap〉,
and hence
(82)(1-d)(1-αn)∥xn-un∥2≤(1-βn)(1-αn)∥xn-un∥2≤∥xn-yn∥(∥xn-p∥+∥yn-p∥)+2(1-αn)(1-βn)λn×∥xn-un∥∥Axn-Ap∥.
From (78) and limn→∞∥xn-yn∥=0, we have
(83)limn→∞∥xn-un∥=0.
Step
4. We show thatlimn→∞∥xn-Sixn∥=0, for all i=0,1,….
It follows from definition of scheme (57) that
(84)yn+∑i=1n‍(αi-1-αi)βn(xn-Sixn)-(1-αn)βnxn=αnxn+(1-αn)(1-βn)un,
that is,
(85)∑i=1n‍(αi-1-αi)βn(xn-Sixn)=xn-yn-xn+αnxn+(1-αn)βnxn+(1-αn)(1-βn)un=xn-yn+(1-αn)(βn-1)xn+(1-αn)(1-βn)un=xn-yn+(1-αn)(1-βn)(un-xn).
Hence, for anyp∈F, one has
(86)∑i=1n‍(αi-1-αi)βn〈xn-Sixn,xn-p〉=(1-αn)(1-βn)〈un-xn,xn-p〉(87)+〈xn-yn,xn-p〉.
Since eachSiis nonexpansive and by (36) we get that
(88)∥Sixn-xn∥2≤〈xn-Sixn,xn-p〉.
Hence, combining this inequality with (86), we have
(89)12∑i=1n‍(αi-1-αi)βn∥Sixn-xn∥2≤(1-αn)(1-βn)〈un-xn,xn-p〉+〈xn-yn,xn-p〉,
that is,
(90)∥Sixn-xn∥2≤2(1-αn)(1-βn)(αi-1-αi)βn〈un-xn,xn-p〉+2(αi-1-αi)βn〈xn-yn,xn-p〉≤2(1-αn)(1-βn)(αi-1-αi)βn∥un-xn∥∥xn-p∥+2(αi-1-αi)βn∥xn-yn∥∥xn-p∥.
Since ∥un-xn∥→0 and ∥xn-yn∥→0, we have
(91)limn→∞∥Sixn-xn∥=0,∀i=1,2,….
Step
5. We show thatxn→x*=PFx1.
First, we show thatx*∈∩i=1∞Fix(Si).Since
(92)limn→∞xn=x*,limn→∞∥Sixn-xn∥=0,
we have
(93)x*∈Fix(Si)foreachi=1,2,….
Hence x*∈∩i=1∞Fix(Si). Next, we show that x*∈SGVEPR(φ,T). Noting that un=Bλn(xn-λnAxn), one obtains
(94)φ(un,y)+e〈Tun,η(y,un)〉+e〈Axn,y-un〉+eλn〈y-un,un-xn〉∈C,∀y∈X,
which implies that
(95)0∈φ(y,un)-{e〈Tun,η(y,un)〉eλn+e〈Axn,y-un〉+eλn〈y-un,un-xn}+C,∀y∈X.
Putvt=ty+(1-t)x*, for allt∈(0,1)andy∈X.Then, we havevt∈X.So, from (95), we have
(96)e〈vt-un,Avt〉∈e〈vt-un,Avt〉-e〈vt-un,Axn〉-e〈vt-un,un-xnλn〉+φ(vt,un)+e〈Tun,η(un,vt)〉+C.=e〈vt-un,Avt-Aun〉+e〈vt-un,Aun-Axn〉-e〈vt-un,un-xnλn〉+φ(vt,un)+e〈Tun,η(un,vt)〉+C.
Since ∥xn-un∥→0 and the properties of T, we have
(97)∥Aun-Axn∥⟶0,un-xnλn⟶0,〈vt-un,Aun-Axn〉⟶0.
From the monotonicity of A, we have
(98)〈vt-un,Avt-Aun〉≥0.
Thus
(99)e〈vt-un,Avt-Aun〉∈C.
So, from (96)–(99) andη-hemicontinuity of T, we have
(100)e〈vt-x*,Avt〉∈φ(vt,x*)+e〈Tx*,η(x*,vt)〉+C.
Since φ is C-convex, we have
(101)tφ(vt,y)+(1-t)φ(vt,x*)∈φ(vt,vt)+C.
Since for any u,v∈X, and the mapping x↦〈Tv,η(x,u)〉 is convex, we have
(102)〈Tx*,η(vt,vt)〉≤t〈Tx*,η(y,vt)〉+(1-t)〈Tx*,η(x*,vt)〉.
This implies that
(103)et〈Tx*,η(y,vt)〉+e(1-t)〈Tx*,η(x*,vt)〉∈e〈Tx*,η(vt,vt)〉+C.
From (101) and (103), we get that
(104)tφ(vt,y)+(1-t)φ(vt,x*)+et〈Tx*,η(y,vt)〉+e(1-t)〈Tx*,η(x*,vt)〉∈e〈Tx*,η(vt,vt)〉+φ(vt,vt)+C=C,
which implies that
(105)-t(φ(vt,y)+e〈Tx*,η(y,vt)〉)-(1-t)(φ(vt,x*)+e〈Tx*,η(x*,vt)〉)∈-C.
From (100) and (105), we have
(106)-t(φ(vt,y)+e〈Tx*,η(y,vt)〉)∈(1-t)(φ(vt,x*)+e〈Tx*,η(x*,vt)〉)-C∈(1-t)e〈vt-x*,Avt〉-C.
This implies that
(107)-t(φ(vt,y)+e〈Tx*,η(y,vt)〉)-e(1-t)t〈y-x*,Avt〉∈-C.
It follows that
(108)(φ(vt,y)+e〈Tx*,η(y,vt)〉)+e(1-t)〈y-x*,Avt〉∈C.
As t→0, we obtain that for each y∈X,(109)(φ(x*,y)+e〈Tx*,η(y,x*)〉)+e(1-t)〈y-x*,Ax*〉∈C.
Hencex*∈SGVEPR(φ,T).Finally, we prove thatx*=PFx.Fromxn+1=PDnxandF⊂Dn, we have
(110)〈x-xn+1,xn+1-v〉≥0,∀v∈F.
Note that limn→∞xn=x*; we take the limit in (110), and then we have
(111)〈x-x*,x*-v〉≥0,∀v∈F.
We see that x*=PFx by (33). This completes the proof.Remark 8.
IfY=ℝ,  C=ℝ+, and e=1, then Theorem 7 extends and improves Theorem 3.1 of Wang et al. [9].



---

*Source: 973408-2013-07-25.xml*