An iterative method for nonexpansive semigroups, variational inclusions and generalized equilibrium problems

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Adv. Fixed Point Theory, 4 (2014), No. 3, 325-343 ISSN: 1927-6303

AN ITERATIVE METHOD FOR NONEXPANSIVE SEMIGROUPS, VARIATIONAL INCLUSIONS AND GENERALIZED EQUILIBRIUM PROBLEMS

LIJUAN ZHANG∗, HUI TONG

College of Mathematics and Computer, Hebei University, Baoding 071002, China

Copyright c2014 Zhang and Tong. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. In this paper, we introduce a viscosity iterative scheme for finding a common element of the set of common fixed points of a one-parameter nonexpansive semigroup, the set of solutions to variational inclusions and the set of solutions to generalized equilibrium problems in a real Hilbert space. Strong convergence theorems for the common element are obtained.

Keywords: nonexpansive semigroup; variational inclusion; inverse strongly monotone mapping; generalized e-quilibrium problem.

2010 AMS Subject Classification:47H09, 47H10.

1. Introduction

LetH be a real Hilbert space with the inner producth·,·iand the induced normk · k. LetC be a nonempty closed convex subset ofH. Recall the following definitions.

A mappingT :C→H is said to be

∗_{Corresponding author}

E-mail address: [email protected] Received December 29, 2013

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(1) monotone if

hT x−Ty,x−yi ≥0, ∀x,y∈C.

(2)α-strongly monotone if there exists a constantα >0 such that

hT x−Ty,x−yi ≥αkx−yk2, ∀x,y∈C.

(3)α-inverse strongly monotone if there exists a constantα >0 such that

hT x−Ty,x−yi ≥αkT x−Tyk2, ∀x,y∈C.

(4)k-Lipschitz continuous if there exists a constantk>0 such that kT x−Tyk ≤kkx−yk, ∀x,y∈C.

IfT isα-inverse strongly monotone, thenT is _α1-Lipschitz continuous. In the case thatk=1,

T :C→His said to be nonexpansive.

A (one parameter) nonexpansive semigroup is a familyΓ={S(t):t≥0}of self-mapping of Cif the following conditions are satisfied:

(a)S(0)x=xfor allx∈C;

(b)S(s+t) =S(s)S(t)for alls,t≥0;

(c) for eacht>0,kS(t)x−S(t)yk ≤ kx−yk,x,y∈C; (d) for eachx∈C, the mappingS(·)xis continuous.

We useF(Γ)to denote the common fixed point set of the semigroupΓ, that is, F(Γ) ={x∈C:S(t)x=x,t ≥0}.

Let A:H →H be a single-valued nonlinear mapping and let M :H →2H be a set-valued mapping. The variational inclusion is to findx∈H such that

θ ∈A(x) +M(x), (1.1)

whereθ is a zero vector in H. The set of solutions to variational inclusion (1.1) is denoted by

I(A,M). WhenA=0, then (1.1) becomes the inclusion problem introduced by Rockafellar [1]. Letϕ :C→H be a nonlinear mapping. The variational inequality problem is to findx∈C

such that

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The set of solutions to variational inequality problem (1.2) is denoted byV I(C,ϕ). Finding a

common element of the set of fixed points of nonexpansive mappings and the set of solutions to a variational inequality problem has been studied extensively in the literature; see, for example, [2] and the references therein.

A set-valued mappingM:H →2H is called monotone if for allx,y∈H,f ∈Mxandg∈My imply hx−y,f −gi ≥0. A monotone mapping M :H →2H is maximal if the graph G(M)

is not properly contained in the graph of any other monotone mapping. It is known that a monotone mappingMis maximal if and only if for(x,f)∈H×H,hx−y,f−gi ≥0 for every

(y,g)∈G(M) implies f ∈Mx. The resolvent operator J_M_,_λ associated with M and λ is the

mappingJ_M_,_λ :H→H defined by

J_M_,_λ(u) = (I+λM)−1(u), u∈H,λ >0. (1.3)

It is known that the resolvent operator J_M_,_λ is single-valued, nonexpansive and 1-inverse-strongly monotone and that a solution of (1.1) is a fixed point of J_M_,_λ(I−λA),∀λ >0, see

[3]. If 0<λ <2α, it is easy to see thatJ_M,λ(I−λA)is nonexpansive andI(A,M)is closed and

convex.

LetGbe a bi-function ofC×CintoR, the set of reals andϕ:C→Hbe a nonlinear mapping.

The generalized equilibrium problem is to findx∈Csuch that

G(x,y) +hϕx,y−xi ≥0, ∀y∈C. (1.4)

The set of solutions to this generalized equilibrium problem (1.4) is denoted byEP. Thus

EP:={x∈C:G(x,y) +hϕx,y−xi ≥0,∀y∈C}.

In the case ofϕ≡0, the problem (1.4) reduces to an equilibrium problem, which is to findx∈C

such that

G(x,y)≥0,∀y∈C. (1.5)

and EP is then denoted by EP(G). In the case of G ≡0, the problem (1.4) reduces to the variational inequality problem (1.2) and EP is denoted by V I(C,ϕ). Numerous problems in

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(1.4). Some methods have been proposed to solve the generalized equilibrium problems and equilibrium problems.

For solving the equilibrium problem for a bifunction G:C×C→R, let us assume that F satisfies the following conditions:

(A1)G(x,x) =0 for allx∈C;

(A2)Gis monotone, i.e.,G(x,y) +G(y,x)≤0 for allx,y∈C; (A3) For eachx,y,z∈C, lim sup_t_→₀G(tz+ (1−t)x,y)≤G(x,y); (A4) For eachx∈C,y7→G(x,y)is convex and lower semicontinuous.

For finding an element ofF(T), whereT is a nonexpansive mapping. Moudafi [4] introduced the viscosity approximation method for nonexpansive mappings. Let f be a contraction on H, starting with an arbitrary initialx0∈H, define a sequence{xn}recursively by

x_n₊₁=αnf(xn) + (1−αn)T xn, n≥0,

where {αn} is a sequence in (0,1) satisfies certain conditions, the sequence {xn} converges strongly to the unique solutionqinF(T).

Tian [5] consider the following general iterative method

x_n+1=αnγf(xn) + (I−µ αnF)T xn, n≥0.

whereFis ak-Lipschitzian andη-strongly monotone operator withk>0,η>0,0<µ<2η

k2. If

the sequence{αn}satisfies appropriate conditions, then the sequence{xn}converges strongly to the unique solutionq∈F(T)of the variational inequality

h(γf −µF)q,p−qi ≤0, ∀p∈F(T).

For finding a common element in F(S)∩EP, Takahashi and Takahashi [6] introduced the following iterative scheme:



 

 

G(u_n,y) +hϕxn,y−uni+ 1

r_nhy−un,un−xni ≥0, ∀y∈C, x_n₊₁=βnxn+ (1−βn)S[αnu+ (1−αn)un].

(1.6)

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For finding a common element inF(S)∩EP∩I(A,M), Shehu [7] introduced the following iterative scheme:



 

 

G(u_n,y) +hϕxn,y−uni+ 1 rn

hy−u_n,u_n−x_ni ≥0, ∀y∈C,

x_n₊₁=βnxn+ (1−βn)S[αnf(xn) + (1−αn)JM,λ(un−λAun)].

(1.7)

Under the suitable conditions, some strong theorems are proved which extend the results of Takahashi and Takahashi [6].

For finding a common element inF(Γ)∩EP∩I(A,M), Shehu [8] introduced the following iterative scheme:



  

  

G(un,y) +hϕxn,y−uni+ 1 rn

hy−un,un−xni ≥0, ∀y∈C,

x_n+1=βnxn+ (1−βn)( 1 t_n

Z t_n

0

S(u)[αnf(xn) + (1−αn)JM,λ(un−λAun)]du).

(1.8)

Under the suitable conditions, some strong theorems are proved which extend the results of Shehu [7].

A nonexpansive semigroup is said to be uniformly asymptotically regular if for anyt≥0 and for any bounded subsetDofC,

lim

s→∞sup_x_∈_DkS(

t+s)x−S(s)xk=0.

LetCbe a nonempty closed convex sunset of a real Hilbert spaceH, andΓ={S(t):t>0} a nonexpansive semigroup onC such thatF(Γ) is nonempty. Let σ(t)x= 1_t R₀tS(u)xdu is an

uniformly asymptotically regular nonexpansive semigroup; see [9].

In this paper, motivated and inspired by the above results, we introduce an iterative scheme for finding a common element of the set of common fixed points of nonexpansive semigroups, the set of solutions to variational inclusions and the set of solutions to generalized equilibrium problems. Our results improved and extend many recent results in the literature.

2. Preliminaries

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to x. x_n→ximplies that {x_n}converges strongly to x. In a real Hilbert space, the following inequality holds:

kx+yk2≤ kxk2+2hy,x+yi, ∀x,y∈H. (2.1)

For every pointx∈H, there exists a unique nearest point inC, denoted byP_Cx, such that

kx−P_Cxk ≤ kx−yk, ∀y∈C.

PC is called the metric projection ofHontoC. PC is characterized by the following properties:

hx−P_Cx,P_Cx−yi ≥0, ∀y∈C. (2.2)

In the context of the variational inequality problem, this implies

p∈V I(C,ϕ)⇔p=PC(p−λ ϕp), ∀λ >0. (2.3)

It is well known that H satisfies the Opial’s condition [3], i.e., for any sequence {xn} with xn*x, the inequality

lim sup n→∞

kx_n−xk<lim sup n→∞

kx_n−yk,∀y∈H,y6=x.

In order to prove our main results, w shall make use of the following lemmas.

Lemma 2.1. [10]Let{xn}, {yn}be bounded sequences in a Banach space E and let{βn}be a sequence in[0,1]with0<lim infn→∞βn≤lim supn→∞βn<1. Suppose xn+1=βnxn+ (1− βn)yn,∀n≥0andlim supn→∞(kyn+1−ynk − kxn+1−xnk)≤0. Thenlimn→∞kyn−xnk=0. Lemma 2.2. [11]Let{sn}be a sequence of nonnegative real numbers such that:

s_n₊₁≤(1−λn)sn+βn, n≥0,

where{λn},{βn}satisfy the conditions:

(i){λn} ⊂(0,1)and∑∞_n=1λn=∞, (ii)lim sup_n_→_∞βn

λn ≤0or∑ ∞

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Lemma 2.3.[2]Let C be a nonempty closed subset of H and let G be a bifunction of C×C into R satisfying (A1)-(A4). Let r>0and x∈H. Then, there exists z∈C such that

G(z,y) +1

rhy−z,z−xi ≥0, ∀y∈C.

Lemma 2.4. [12]Assume that G:C×C→R satisfies (A1)-(A4). For r>0and x∈H, define a mapping T_r:H→C as follows:

T_rx={z∈C:G(z,y) +1

rhy−z,z−xi ≥0,∀y∈C} for all x∈H. Then the following hold:

(1) T_r is single-valued;

(2) T_r is firmly nonexpansive, i.e., for any x,y∈H,

kTrx−Tryk2≤ hTrx−Try,x−yi;

(3) F(Tr) =EP(G);

(4) EP(G)is closed and convex.

Lemma 2.5. [13] Let M :H →2H be a maximal monotone mapping and let A:H →H be a Lipschitz continuous mapping. Then the mapping M+A:H →2H is a maximal monotone mapping.

Lemma 2.6. [14] Let H be a real Hilbert space and let F :H →H be a k-Lipschitz and η

-strongly monotone operator with k>0, η >0. Let0<µ <2η/k2, B=I−tµF andµ(η−

µk2

2 ) =τ. Then for t ∈(0,min{1, 1

τ}), B is a contraction with a constant1−tτ.

3. Main results

Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let G: C×C →R be a bifunction satisfying conditions (A1)-(A4). Let ϕ :H →H be an θ

-inverse-strongly monotone mapping, A:C→H be anα-inverse-strongly monotone mapping and M:

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is Lipschitz mapping with the coefficient L, and F :H →H be a k-Lipschitz and η-strongly

monotone operator. Suppose that the sequences{xn},{un},{zn}are generated by x1∈H



     

     

G(un,y) +hϕxn,y−uni+ 1

r_nhy−un,un−xni ≥0, ∀y∈C, zn=βnun+ (1−βn)S(tn)un,

xn+1=αnγf(xn) +δnxn+ [(1−δn)I−αnµF]JM,λ(zn−λAzn),

where the sequences {αn},{βn},{δn} ⊆(0,1) and {rn} ⊆(0,∞), {tn} ⊆[0,∞) satisfying the following restrictions:

(C1) 0<a≤r_n≤b<2θ;

(C2) limn→∞|rn−rn+1|=0;

(C3) limn→∞αn=0,∑∞_n₌₁αn=∞; (C4) λ ∈(0,2α];

(C5) 0<lim infn→∞δn≤lim supn→∞δn<1;

(C6) limn→∞βn=0;

(C7) {t_n} ⊆[0,∞)be a real increasing sequence such thatlimn→∞tn=∞; (C8) 0<µ< 2η

k2, 0<γ <τ/L, µ(η− µk2

2 ) =τ.

Then the sequence {xn}converges strongly to q∈Ω, which is the unique solution in theΩto

the following variational inequality

hγf(q)−µFq,p−qi ≤0, ∀ p∈Ω. (3.1)

Equivalently, we have q=P_Ω(I+γf−µF)q.

Proof.We divide the proof into five steps. Step 1. Show that{xn}is bounded. For allx,y∈Candλ >0, we obtain

k(I−λA)x−(I−λA)yk2

=k(x−y)−λ(Ax−Ay)k2

=kx−yk2−2λhx−y,Ax−Ayi+λ2kAx−Ayk2

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So, I−λAis a nonexpansive mapping. Using (C1), we see that I−rnϕ is also nonexpansive. Letv_n=J_M_,_λ(z_n−λAzn)and let p∈Ω. It follows that

kvn−pk2=kJM,λ(zn−λAzn)−JM,λ(p−λAp)k 2

≤ k(z_n−λAzn)−(p−λAp)k2

≤ kz_n−pk2+λ(λ−2α)kAzn−Apk2 ≤ kzn−pk

2

andkzn−pk ≤(1−βn)kS(tn)un−pk+βnkun−pk ≤ kun−pk. From Lemma 2.4,un=Trn(xn−

rnϕxn)andTrn is nonexpansive. Hence, we have

kun−pk2=kTrn(xn−rnϕxn)−Trn(p−rnϕp)k 2

≤ k(x_n−r_nϕxn)−(p−rnϕp)k2

≤ kxn−pk2+rn(rn−2θ)kϕxn−ϕpk2

≤ kxn−pk2.

(3.2)

From (C3), (C8) and Lemma 2.6, we havek(1−δn)I−αnµFk ≤1−δn−αnτ. Further kx_n+1−pk

=kαn(γf(xn)−µF p) +δn(xn−p) + [(1−δn)I−αnµF](vn−p)k ≤αnkγf(xn)−µF pk+δnkxn−pk+ (1−δn−αnτ)kvn−pk

≤αnγkf(xn)−f(p)k+αnkγf(p)−µF pk+δnkxn−pk+ (1−δn−αnτ)kxn−pk ≤αnγLkxn−pk+αnkγf(p)−µF pk+δnkxn−pk+ (1−δn−αnτ)kxn−pk

= [1−αn(τ−γL)]kxn−pk+αnkγf(p)−µF pk. By induction, we have

kxn−pk ≤max{kx1−pk,

1

τ−γLkγf(p)−µF pk}.

Therefore {x_n} is bounded, we have {v_n},{u_n}, {S(t_n)u_n}, {Ax_n}, {Fv_n}, {f(x_n)} are also bounded.

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Putln=xn+₁1₋−_δδ_nnxn, this is,xn+1= (1−δn)ln+δnxn.Observing that kln+1−lnk ≤

αn+1

1−δn+1

(kγf(xn+1)k+kµFvn+1k) +

αn

1−δn

(kγf(xn)k+kµFvnk)

+kv_n+1−vnk.

(3.3)

SinceI−λAis a nonexpansive, we have

kvn+1−vnk=kJM,λ(zn+1−λAzn+1)−JM,λ(zn−λAzn)k

≤ k(z_n+1−λAzn+1)−(zn−λAzn)k

≤ kz_n+1−znk

≤(1−βn+1)kS(tn+1)un+1−S(tn)unk+βn+1kun+1−unk

+|βn+1−βn|kS(tn)un−unk

≤ kS(t_n₊₁)u_n₊₁−S(t_n₊₁)u_nk+kS[(t_n₊₁−t_n) +t_n]u_n−S(t_n)u_nk

+βn+1kun+1−unk+|βn+1−βn|kS(tn)un−unk

≤ kun+1−unk+ sup x∈{un},t≥0

kS(t+t_n)x−S(t_n)xk

+βn+1kun+1−unk+|βn+1−βn|kS(tn)un−unk.

(3.4)

On the other hand, fromu_n=T_r_n(x_n−r_nϕxn)andun+1=Trn+1(xn+1−rn+1ϕxn+1), we obtain

G(u_n,y) +hϕxn,y−uni+ 1 rn

hy−u_n,u_n−x_ni ≥0, ∀y∈C (3.5)

and

G(u_n₊₁,y) +hϕxn+1,y−un+1i+

1 rn+1

hy−u_n₊₁,u_n₊₁−x_n₊₁i ≥0, ∀y∈C. (3.6)

Substitutingy=u_n₊₁in (3.5) andy=u_nin (3.6), we have G(u_n,u_n₊₁) +hϕxn,un+1−uni+

1 rn

hun+1−un,un−xni ≥0 and

G(u_n₊₁,u_n) +hϕxn+1,un−un+1i+

1 r_n+1

hun−un+1,un+1−xn+1i ≥0.

So, from (A2), we have

0≤ hϕxn+1−ϕxn,un−un+1i+hun+1−un,

u_n−x_n

r_n −

u_n₊₁−x_n₊₁

r_n+1

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and hence,

0≤ hun+1−un,rn(ϕxn−ϕxn+1) + (un−xn)− r_n

r_n₊₁(un+1−xn+1)i

=hun+1−un,un−un+1+xn+1−rnϕxn+1−(xn−rnϕxn) + (1− r_n

rn+1

)(u_n₊₁−x_n₊₁)i.

It follows that

ku_n+1−unk2≤ kun+1−unk{kxn+1−xnk+ (1− rn

r_n₊₁)kun+1−xn+1k}.

From (C1), we have

kun+1−unk ≤ kxn+1−xnk+|1− r_n

rn+1

|kun+1−xn+1k

≤ kxn−xn+1k+

1

a|rn−rn+1|kun+1−xn+1k.

(3.7)

Substituting (3.4) and (3.7) into (3.3), we have kln+1−lnk − kxn+1−xnk

≤ αn+1 1−δn+1

(kγf(xn+1)k+kµFvn+1k) +

αn

1−δn

(kγf(xn)k+kµFvnk)

+βn+1kun+1−unk+ sup x∈{un},t≥0

kS(t+t_n)x−S(t_n)xk

+|βn+1−βn|kS(tn)un−unk+ 1

a|rn−rn+1|kun+1−xn+1k.

Since (C2), (C3), (C6), (C7) and the uniform asymptotic regularity of nonexpansive semigroup, we have

lim sup n→∞

(kl_n+1−lnk − kxn−xn+1k)≤0.

By Lemma 2.1, we have limn→∞kln−xnk=0. Consequently, we have lim

n→∞kxn+1−

x_nk= lim

n→∞(1−βn)kln−

x_nk=0. (3.8)

Step 3. Show that limn→∞kxn−S(t)xnk=0,∀t≥0.

Observing that

kxn−vnk ≤ kxn−xn+1k+kxn+1−vnk

≤ kxn−xn+1k+αnkγf(xn)−µFvnk+δnkxn−vnk.

It follows that

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From (C3), (C5) and (3.8), we have

lim n→∞kvn−

x_nk=0. (3.9)

LetM>0 be a constant such thatM>sup_n_≥₁max{kγf(xn)−µFvnk,kxn−pk}. From (2.1), (3.2) and the convexity ofk.k2_{, we obtain}

kxn+1−pk2

≤ kδn(xn−p) + (1−δn)(vn−p)k2+2αnhγf(xn)−µFvn,xn+1−pi

≤(1−δn)kvn−pk2+δnkxn−pk2+2αnM2 ≤(1−δn)kun−pk2+δnkxn−pk2+2αnM2

≤(1−δn)[kxn−pk2+rn(rn−2θ)kϕxn−ϕpk2] +δnkxn−pk2+2αnM2 ≤ kx_n−pk2+a(b−2θ)kϕxn−ϕpk2+2αnM2.

Hence,

a(2θ−b)kϕxn−ϕpk2≤2αnM2+ (kxn+1−pk+kxn−pk)kxn−xn+1k.

Using (C3) and (3.8), we havekϕxn−ϕpk →0,n→∞. Similarly, we also havekAzn−Apk → 0,asn→∞. SinceTrn is 1-inverse-strongly-monotone, we have

ku_n−pk2=kT_r_n(x_n−r_nϕxn)−Trn(p−rnϕp)k 2

≤ hxn−rnϕxn−(p−rnϕp),un−pi

= 1

2[kxn−rnϕxn−(p−rnϕp)k

2₊_ku

n−pk2− kxn−rnϕxn−(p−rnϕp)−(un−p)k2]

≤ 1

2[kxn−pk

2₊_ku

n−pk2− kxn−un−rn(ϕxn−ϕp)k2]

= 1

2[kxn−pk

2₊_ku

n−pk2− kxn−unk2+2rnhxn−un,ϕxn−ϕpi −r2nkϕxn−ϕpk2]. This implies that

kun−pk2≤ kxn−pk2− kxn−unk2+2rnhxn−un,ϕxn−ϕpi −r2nkϕxn−ϕpk2 ≤ kxn−pk2− kxn−unk2+2rnkxn−unkkϕxn−ϕpk.

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From (2.1) and (3.10), we obtain kxn+1−pk2

=kδn(xn−p) + (1−δn)(vn−p) +αn(γf(xn)−µFvn)k2

≤ kδn(xn−p) + [(1−δn)(vn−p)]k2+2αnhγf(xn)−µFvn,xn+1−pi

≤(1−δn)kvn−pk2+δnkxn−pk2+2αnM2 ≤(1−δn)kun−pk2+δnkxn−pk2+2αnM2

≤(1−δn)[kxn−pk2− kxn−unk2+2rnkxn−unkkϕxn−ϕpk] +δnkxn−pk2+2αnM2 ≤ kxn−pk2−(1−δn)kxn−unk2+2rnkxn−unkkϕxn−ϕpk+2αnM2.

This implies that

(1−δn)kxn−unk2≤ kxn−pk2− kxn+1−pk2+2rnkxn−unkkϕxn−ϕpk+2αnM2. Since (C3), (C5), (3.8) and limn→∞kϕxn−ϕpk=0, we have

lim n→∞

kxn−unk=0. (3.11)

SinceJ_M_,_λ is also 1-inverse-strongly-monotone, By the similar argument above, we also have lim

n→∞

kvn−znk=0. (3.12)

Fromzn=βnxn+ (1−βn)S(tn)un, we have from (C6) lim

n→∞

kz_n−S(t_n)u_nk= lim n→∞βn

ku_n−S(t_n)u_nk=0. (3.13)

From (3.9), (3.11), (3.12) and (3.13), we have

kxn−S(tn)xnk ≤ kxn−vnk+kvn−znk+kzn−S(tn)unk+kS(tn)un−S(tn)xnk

≤ kxn−vnk+kvn−znk+kzn−S(tn)unk+kun−xnk →0. Further, we have

kxn−S(t)xnk ≤ kxn−S(tn)xnk+kS(tn)xn−S(t)S(tn)xnk+kS(t)S(tn)xn−S(t)xnk

≤2kxn−S(tn)xnk+kS(tn)xn−S(t)S(tn)xnk ≤2kx_n−S(t_n)x_nk+ sup

x∈{xn},t≥0

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From (C7) and the uniform asymptotic regularity of the nonexpansive semigroup, we get

lim n→∞kxn−

S(t)x_nk=0, ∀t≥0. (3.14)

Step 4. We show that lim sup_n_→_∞hγf(q)−µFq,xn−qi ≤0.

From (C8), we obtainµF−γf is strongly monotone. Thenqis the uniqueness of a solution

of (3.1). Choose a subsequence{xni}of{xn}such that

lim sup n→∞

hγf(q)−µFq,xn−qi=lim i→∞

hγf(q)−µFq,xni−qi.

As {xni} is bounded, Without loss of generality that xni *z, We first show that z∈I(A,M).

SinceAis _α1 Lipschitz monotone andD(A) =H, we obtain from Lemma 2.5,M+Ais maximal monotone. Let(v,g)∈G(M+A), that is,g−Av∈M(v). Sincev_n_i=J_M_,_λ(z_n_i−λAzni), we get (I−λA)zni ∈(I+λM)vni, that is,

z_n_i−λAzni−vni

λ ∈M(vni).

Using the maximal monotonicity ofM+A, we obtain

hv−v_n_i,g−Av−zni−λAzni−vni

λ i ≥0. (3.15)

By the monotonicity ofAand (3.15), we have hv−vni,gi ≥ hv−vni,Av+

z_n_i−λAzni−vni

λ i

=hv−vni,Av−Avni+Avni−Azni+

z_n_i−v_n_i

λ i

≥ hv−vni,Avni−Aznii+hv−vni,

zni−vni

λ i.

It follow from (3.12), limi→∞kAvni−Aznik=0. From (3.9),vni*z, we have

lim n→∞

hv−v_n_i,gi=hv−z,gi ≥0.

Using the maximal monotonicity ofM+A, we obtainθ ∈(M+A)(v), this impliesz∈I(A,M).

Sinceu_n=T_r_n(x_n−r_nϕxn), for anyy∈C

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Replacenbyn_iand using (A2), we have

hϕxni,y−unii+hy−uni,

u_n_i−x_n_i

r_n_i i ≥G(y,uni). (3.16)

Lety_t =ty+ (1−t)zfor all 0<t≤1 and y∈C. Sincey∈Candz∈C, we havey_t ∈C. From (3.16), we have

hyt−uni,ϕyti ≥ hyt−uni,ϕyti − hyt−uni,ϕxnii − hyt−uni,

u_n_i−x_n_i

r_n_i i+G(yt,uni)

=hy_t−u_n_i,ϕyt−ϕunii+hyt−uni,ϕuni−ϕxnii

− hyt−uni,

u_n_i−x_n_i

r_n_i i+G(yt,uni).

From the monotonicity ofϕ, we havehyt−uni,ϕyt−ϕunii ≥0.From (3.11), we haveuni*z

andkϕxni−ϕunik →0. From (A4), we have

hyt−z,ϕyti ≥G(yt,z). (3.17) From (A1), (A4) and (3.17), we have

0=G(y_t,y_t)≤tG(y_t,y) + (1−t)G(y_t,z)

≤tG(y_t,y) + (1−t)hy_t−z,ϕyti

=tG(yt,y) + (1−t)thy−z,ϕyti

and hence G(y_t,y) + (1−t)hy−z,ϕyti ≥0. Lettingt →0 and (A3), we have G(z,y) +hy− z,ϕzi ≥0 for ally∈Cand hencez∈EP.

Finally, we show that z∈F(Γ). Assume the contrary that z6=S(t)zfor some t ∈[0,+∞). Then by the Opial’s condition, we obtain from (3.14) that

lim inf i→∞

kx_n_i−zk<lim inf i→∞

kx_n_i−S(t)zk

≤lim inf i→∞

(kxni−S(t)xnik+kS(t)xni−S(t)zk)

≤lim inf i→∞

kxni−zk.

This is a contradiction. Hencez∈F(Γ). Thusz∈Ω. This follows that lim sup

n→∞

hγf(q)−µFq,xn−qi=hγf(q)−µFq,z−qi ≤0. (3.18)

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We compute that kxn+1−qk2

=αnhγf(xn)−µFq,xn+1−qi+δnhxn−q,xn+1−qi

+h[(1−δn)I−αnµF](vn−q),xn+1−qi

≤αnγhf(xn)−f(q),xn+1−qi+αnhγf(q)−µFq,xn+1−qi

+δnkxn−qkkxn+1−qk+ (1−δn−αnτ)kvn−qkkxn+1−qk

≤αnγLkxn−qkkxn+1−qk+αnhγf(q)−µFq,xn+1−qi

+δnkxn−qkkxn+1−qk+ (1−δn−αnτ)kxn−qkkxn+1−qk

= (1−αn(τ−γL))kxn−qkkxn+1−qk+αnhγf(q)−µFq,xn+1−qi

≤1−αn(τ−γL)

2 kxn−qk

2₊1

2kxn+1−qk

2₊

αnhγf(q)−µFq,xn+1−qi,

which implies that

kxn+1−qk2≤[1−αn(τ−γL)]kxn−qk2+2αnhγf(q)−µFq,xn+1−qi.

By (3.18), (C3) and Lemma 2.2, we obtain limn→∞kxn−qk=0. This completes the proof.

Corollary 3.2. Let C be a nonempty closed convex subset of a real Hilbert space H. Let G:C×C →R be a bifunction satisfying conditions (A1)-(A4), ϕ :H →H be an θ

-inverse-strongly monotone mapping and let A:C →H be an α-inverse-strongly monotone mapping. Let Γ={S(t):t ≥0}be uniform asymptotically regular nonexpansive semigroup on C such thatΩ:=F(Γ)∩EP∩V I(C,A)6= /0.Let f :H→H is Lipschitz mapping with the coefficient L and let F :H→H be a k-Lipschitz andη-strongly monotone operator. Suppose the sequences

{xn},{un},{zn}are generated by x1∈H



     

     

r_nhy−un,un−xni ≥0,∀y∈C, zn=βnun+ (1−βn)S(tn)un,

xn+1=αnγf(xn) +δnxn+ [(1−δn)I−αnµF]PC(zn−λAzn).

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(C1) 0<a≤r_n≤b<2θ;

(C2) limn→∞|rn−rn+1|=0;

(C3) limn→∞αn=0,∑∞n=1αn=∞; (C4) λ ∈(0,2α];

(C7) {tn} ⊆[0,∞)be a real increasing sequence such thatlimn→∞tn=∞;

(C8) 0<µ< 2η

k2, 0<γ <τ/L, µ(η− µk2

2 ) =τ.

the variational inequality (3.1). Equivalently, we have q=P_Ω(I+γf−µF)q.

Proof. Take M =∂ δC :H →2H, where δC :H → [0,∞) is the indicator function of C, the subdifferential ∂ δC of δC is a maximal monotone operator. Then JM,λ =PC and I(A,M) =

V I(C,A). From the Theorem 3.1, we have the desired conclusion immediately.

Recall that mappingT :C→Cis calledα-strictly pseudocontractive if there existsα∈[0,1)

such that

kT x−Tyk2≤ kx−yk2+αk(T−I)x−(T−I)yk2, ∀x,y∈C.

Ifα =0, thenT is nonexpansive. PutA=I−T, Then, we have

k(I−A)x−(I−A)yk2≤ kx−yk2+αkAx−Ayk2, ∀x,y∈C.

Hence we have

hx−y,Ax−Ayi ≥1−α

2 kAx−Ayk

2_, _∀x_,_y_∈_C_.

ThenAis 1−α

2 -inverse strongly monotone.

Corollary 3.3. Let C be a nonempty closed convex subset of a real Hilbert space H. Let G:C×C →R be a bifunction satisfying conditions (A1)-(A4), ϕ :H →H be an θ

-inverse-strongly monotone mapping and let T :C→C be anα-strictly pseudocontractive maping. Let

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{u_n},{z_n}are generated by x₁∈H



     

     

G(un,y) +hϕxn,y−uni+ 1 rn

hy−un,un−xni ≥0,∀y∈C, zn=βnun+ (1−βn)S(tn)un,

xn+1=αnγf(xn) +δnxn+ [(1−δn)I−αnµF][(1−λ)zn+λT zn].

where the sequences {αn},{βn},{δn} ⊆(0,1) and {rn} ⊆(0,∞), {tn} ⊆[0,∞) satisfying the following restrictions:

(C1) 0<a≤rn≤b<2θ; (C2) limn→∞|rn−rn+1|=0;

(C3) limn→∞αn=0,∑∞n=1αn=∞; (C4) λ ∈(0,2α];

(C7) {tn} ⊆[0,∞)be a real increasing sequence such thatlimn→∞tn=∞; (C8) 0<µ< 2η_k2, 0<γ <τ/L, µ(η−

µk2

2 ) =τ.

the variational inequality (3.1). Equivalently, we have q=P_Ω(I+γf−µF)q.

Proof. Putting A=I−T, we have A is 1−α

2 -inverse strongly monotone. We have F(T) =

V I(C,A) and P_C(z_n−λAzn) = (1−λ)zn+λT zn. From Corollary 3.2, we have the desired conclusion immediately.

Conflict of Interests

The authors declare that there is no conflict of interests. Acknowledgements

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