AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation56268910.1155/2011/562689562689Research ArticleA New Iterative Algorithm for the Set of Fixed-Point Problems of Nonexpansive Mappings and the Set of Equilibrium Problem and Variational Inequality ProblemKangtunyakarnAtidHiranoNorimichiDepartment of MathematicsFaculty of ScienceKing Mongkut's Institute of Technology LadkrabangBangkok 10520Thailandkmitl.ac.th20110703201120110710201025012011110220112011Copyright © 2011 Atid Kangtunyakarn.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.

We introduce a new iterative scheme and a new mapping generated by infinite family of nonexpansive mappings and infinite real number. By using both of these ideas, we obtain strong convergence theorem for finding a common element of the set of solution of equilibrium problem and the set of variational inequality and the set of fixed-point problems of infinite family of nonexpansive mappings. Moreover, we apply our main result to obtain strong convergence theorems for finding a common element of the set of solution of equilibrium problem and the set of variational inequality and the set of common fixed point of pseudocontractive mappings.

1. Introduction

Let H be a real Hilbert space and let C be a nonempty closed convex subset of H. Let A:CH be a nonlinear mapping and let F:C×C be a bifunction. A mapping T of H into itself is called nonexpansive if Tx-Tyx-y  for  all  x,yH. We denote by F(T) the set of fixed points of T (i.e., F(T)={xH:Tx=x}). Goebel and Kirk  showed that F(T) is always closed convex, and also nonempty provided T has a bounded trajectory.

A bounded linear operator A on H is called strongly positive with coefficient γ¯ if there is a constant γ¯>0 with the propertyAx,xγ¯x2.

The equilibrium problem for F is to find xC, such thatF(x,y)0,yC. The set of solutions of (1.2) is denoted by EP(F). Many problems in physics, optimization, and economics are seeking some elements of EP(F), see [2, 3]. Several iterative methods have been proposed to solve the equilibrium problem, see, for instance, . In 2005, Combettes and Hirstoaga  introduced an iterative scheme for finding the best approximation to the initial data when EP(F) is nonempty and proved a strong convergence theorem.

The variational inequality problem is to find a point uC, such thatv-u,Au0,vC. The set of solutions of the variational inequality is denoted by VI(C,A). Numerous problems in physics, optimization, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games reduce to find element of (1.2) and (1.3).

A mapping A of C into H is called inverse-strongly monotone, see , if there exists a positive real number α, such thatx-y,Ax-AyαAx-Ay2 for all x,yC.

The problem of finding a common fixed point of a family of nonexpansive mappings has been studied by many authors. The well-known convex feasibility problem reduces to finding a point in the intersection of the fixed-point sets of a family of nonexpansive mapping (see [6, 7]).

The problem of finding a common element of EP(F) and the set of all common fixed points of a family of nonexpansive mappings is of wide interdisciplinary interest and importance. Many iterative methods are purposed for finding a common element of the solutions of the equilibrium problem and fixed-point problem of nonexpansive mappings, see .

In 2007, S. Takahashi and W. Takahashi  introduced a general iterative method for finding a common element of EP(F,A) and F(T). They defined {xn} in the following way:u,x1C,arbitrarily;F(zn,y)+1λny-zn,zn-xn0,yC,xn+1=βnf(xn)+(1-βn)Szn,nN, where {βn}[0,1], and proved strong convergence of the scheme (1.5) to zF(T)EP(F), where z=PF(S)EP(F)f(z) in the framework of a Hilbert space, under some suitable conditions on {βn}, {λn} and bifunction F.

In this paper, by motive of (1.5), we prove strong convergence theorem for finding a common element of the set of solution of equilibrium problem and the set of variational inequality and the set of fixed-point problems by using a new mapping generated by infinite family of nonexpansive mapping and infinite real number. Moreover, we apply our main result to obtain strong convergence theorems for finding a common element of the set of solution of equilibrium problem and the set of variational inequality and the set of common fixed point of pseudocontractive mappings.

2. Preliminaries

In this section, we collect and give some useful lemmas that will be used for our main result in the next section.

Let C be closed convex subset of a real Hilbert space H, and let PC be the metric projection of H onto C, that is, for xH, PCx satisfies the propertyx-PCx=minyCx-y. The following characterizes the projection PC.

Lemma 2.1 (see [<xref ref-type="bibr" rid="B11">11</xref>]).

Given xH and yC, then PCx=y if and only if there holds the inequality x-y,y-z0,zC.

Lemma 2.2 (see [<xref ref-type="bibr" rid="B4">12</xref>]).

Let E be a uniformly convex Banach space, let C be a nonempty closed convex subset of E, and let S:CC be a nonexpansive mapping, then I-S is demiclosed at zero.

Lemma 2.3 (see [<xref ref-type="bibr" rid="B20">13</xref>]).

Let {sn} be a sequence of nonnegative real numbers satisfying sn+1=(1-αn)sn+δn,n0, where {αn} is a sequence in (0,1) and {δn} is a sequence, such that

n=1αn=,

limsupnδn/αn0  or  n=1|δn|<

then limnsn=0.

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

F(x,x)=0,  for  all  xC,

F  is  monotone, that is, F(x,y)+F(y,x)0,  for  all  x,yC,

for all x,y,zC,

limt0+F(tz+(1-t)x,y)F(x,y),

for all xC,yF(x,y) is convex and lower semicontinuous.

The following lemma appears implicitly in .

Lemma 2.4 (see [<xref ref-type="bibr" rid="B1">2</xref>]).

Let C be a nonempty closed convex subset of H, and let F be a bifunction of C×C into satisfying (A1)–(A4). Let r>0 and xH, then there exists zC, such that F(z,y)+1ry-z,z-x, for all xC.

Lemma 2.5 (see [<xref ref-type="bibr" rid="B2">3</xref>]).

Assume that F:C×C satisfies (A1)–(A4). For r>0 and xH, define a mapping Tr:HC as follows: Tr(x)={zC:F(z,y)+1ry-z,z-x0,yC}, for all zH, then the following hold:

Tr is single valued,

Tr is firmly nonexpansive, that is,

Tr(x)-Tr(y)2Tr(x)-Tr(y),x-yx,yH,

F(Tr)=EP(F),

EP(F) is closed and convex.

Lemma 2.6 (see [<xref ref-type="bibr" rid="B21">14</xref>]).

Let H be a Hibert space, let C be a nonempty closed convex subset of H, and let A be a mapping of C into H. Let uC, then for λ>0, u=PC(I-λA)uuVI(C,A), where PC is the metric projection of H onto C.

Definition 2.7.

Let C be a nonempty convex subset of a real Hilbert space. Let Ti,  i=1,2, be mappings of C into itself. For each j=1,2,, let αj=(α1j,α2j,α3j)I×I×I where I[0,1] and α1j+α2j+α3j=1. For every n, we define the mapping Sn:CC as follows: Un,n+1=I,Un,n=α1nTnUn,n+1+α2nUn,n+1+α3nI,Un,n-1=α1n-1Tn-1Un,n+α2n-1Un,n+α3n-1I,  Un,k+1=α1k+1Tk+1Un,k+2+α2k+1Un,k+2+α3k+1I,Un,k=α1kTkUn,k+1+α2kUn,k+1+α3kI,  Un,2=α12T2Un,3+α22Un,3+α32I,Sn=Un,1=α11T1Un,2+α21Un,2+α31I. This mapping is called S-mapping generated by Tn,,T1 and αn,αn-1,,α1.

Lemma 2.8.

Let C be a nonempty closed convex subset of a real Hilbert space. Let {Ti}i=1 be nonexpansive mappings of C into itself with i=1F(Ti), and let αj=(α1j,α2j,α3j)I×I×I, where I=[0,1],  α1j+α2j+α3j=1,  α1j+α2jb<1, and α1j,α2j,α3j(0,1),  for  all  j=1,2,. For every n, let Sn be S-mapping generated by Tn,,T1 and αn,αn-1,,α1, then for every xC and k,  limnUn,kx exists.

Proof.

Let xC and yi=1F(Ti). Fix k, then for every n with nk, we have Un+1,kx-Un,kx2=α1kTkUn+1,k+1x+α2kUn+1,k+1x+α3kx-α1kTkUn,k+1x-α2kUn,k+1x-α3kx2=α1k(TkUn+1,k+1x-TkUn,k+1x)+α2k(Un+1,k+1x-Un,k+1x)2α1kTkUn+1,k+1x-TkUn,k+1x2+α2kUn+1,k+1x-Un,k+1x2α1kUn+1,k+1x-Un,k+1x2+α2kUn+1,k+1x-Un,k+1x2(1-α3k)Un+1,k+1x-Un,k+1x2  j=kn(1-α3j)Un+1,n+1x-Un,n+1x2=j=kn(1-α3j)α1n+1Tn+1Un+1,n+2x+α2n+1Un+1,n+2x+α3n+1x-x2=j=kn(1-α3j)α1n+1Tn+1x+(1-α1n+1)x-x2=j=kn(1-α3j)α1n+1(Tn+1x-x)2j=kn(1-α3j)(Tn+1x-y+y-x)2j=kn(1-α3j)(x-y+y-x)2j=kn(1-α3j)(2x-y)2bn-(k-1)(2x-y)2. It follows that Un+1,kx-Un,kxb(n-(k-1))/2(2x-y)=bn/2b(k-1)/2(2x-y)=anak-1M, where a=b1/2(0,1) and M=2x-y.

For any k,n,p,  p>0,  nk, we have Un+p,kx-Un,kxUn+p,kx-Un+p-1,kx+Un+p-1,kx-Un+p-2,kx++Un+1,kx-Un,kx=j=nn+p-1Uj+1,kx-Uj,kxj=nn+p-1ajak-1Man(1-a)ak-1M. Since a(0,1), we have limnan=0. From (2.12), we have that {Un,kx} is a Cauchy sequence. Hence, limnUn,kx exists.

For every k and xC, we define mapping U,k and S:CC as follows:limnUn,kx=U,kx,limnSnx=limnUn,1x=Sx. Such a mapping S is called S-mapping generated by Tn,Tn-1, and αn,αn-1,.

Remark 2.9.

For each n,Sn is nonexpansive and limnsupxDSnx-Sx=0 for every bounded subset D of C. To show this, let x,yC and D be a bounded subset of C, then we have Snx-Sny2=α11(T1Un,2x-T1Un,2y)+α21(Un,2x-Un,2y)+α31(x-y)2α11T1Un,2x-T1Un,2y2+α21Un,2x-Un,2y2+α31x-y2α11Un,2x-Un,2y2+α21Un,2x-Un,2y2+α31x-y2=(1-α31)Un,2x-Un,2y2+α31x-y2(1-α31)((1-α32)Un,3x-Un,3y2+α32x-y2)+α31x-y2=(1-α31)(1-α32)Un,3x-Un,3y2+α32(1-α31)x-y2+α31x-y2=j=12(1-α3j)Un,3x-Un,3y2+(1-j=12(1-α3j))x-y2  j=12(1-α3j)Un,n+1x-Un,n+1y2+(1-j=12(1-α3j))x-y2=x-y2. Then, we have that S:CC is also nonexpansive, Indeed, observe that for each x,yC, Sx-Sy=limnSnx-Snyx-y. By (2.11), we have Sn+1x-Snx=Un+1,1x-Un,1xanM. This implies that for m>n and xD, Smx-Snxj=nm-1Sj+1x-Sjxj=nm-1ajMan1-aM. By letting m, for any xD, we have Sx-Snxan1-aM. It follows that limnsupxDSnx-Sx=0.

Lemma 2.10.

Let C be a nonempty closed convex subset of a real Hilbert space. Let {Ti}i=1 be nonexpansive mappings of C into itself with i=1F(Ti), and let αj=(α1j,α2j,α3j)I×I×I, where I=[0,1],  α1j+α2j+α3j=1,  α1j+α2jb<1, and α1j,α2j,α3j(0,1)  for  all  j=1,2,. For every n, let Sn and S be S-mappings generated by Tn,,T1 and αn,αn-1,,α1 and Tn,Tn-1,, and αn,αn-1,, respectively, then F(S)=i=1F(Ti).

Proof.

It is easy to see that i=1F(Ti)F(S). For every n,k,  with  nk, let x0F(S) and x*i=1F(Ti), then we have Snx0-x*2=α11(T1Un,2x0-x*)+α21(Un,2x0-x*)+α31(x0-x*)2α11T1Un,2x0-x*2+α21Un,2x0-x*2+α31x0-x*2-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02α11Un,2x0-x*2+α21Un,2x0-x*2+α31x0-x*2-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02=(1-α31)Un,2x0-x*2+α31x0-x*2-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02(1-α31)((1-α32)Un,3x0-x*2+α32x0-x*2-α12α22T2Un,3x0-Un,3x02-α22α32Un,3x0-x02)+α31x0-x*2-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02=(1-α31)(1-α32)Un,3x0-x*2+α32(1-α31)x0-x*2+α31x0-x*2-α12α22(1-α31)T2Un,3x0-Un,3x02-α22α32(1-α31)Un,3x0-x02-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02=j=12(1-α3j)Un,3x0-x*2+(1-j=12(1-α3j))x0-x*2-α12α22(1-α31)T2Un,3x0-Un,3x02-α22α32(1-α31)Un,3x0-x02-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02j=12(1-α3j)((1-α33)Un,4x0-x*2+α33x0-x*2-α13α23T3Un,4x0-Un,4x02-α23α33Un,4x0-x02)+(1-j=12(1-α3j))x0-x*2-α12α22(1-α31)T2Un,3x0-Un,3x02-α22α32(1-α31)Un,3x0-x02-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02=j=12(1-α3j)(1-α33)Un,4x0-x*2+α33j=12(1-α3j)x0-x*2-α13α23j=12(1-α3j)T3Un,4x0-Un,4x02-α23α33j=12(1-α3j)Un,4x0-x02+(1-j=12(1-α3j))x0-x*2-α12α22(1-α31)T2Un,3x0-Un,3x02-α22α32(1-α31)Un,3x0-x02-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02=j=13(1-α3j)Un,4x0-x*2+(1-j=13(1-α3j))x0-x*2-α13α23j=12(1-α3j)T3Un,4x0-Un,4x02-α23α33j=12(1-α3j)Un,4x0-x02-α12α22(1-α31)T2Un,3x0-Un,3x02-α22α32(1-α31)Un,3x0-x02-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02  j=1k+1(1-α3j)Un,k+2x0-x*2+(1-j=1k+1(1-α3j))x0-x*2-α1k+1α2k+1j=1k(1-α3j)Tk+1Un,k+2x0-Un,k+2x02-α2k+1α3k+1j=1k(1-α3j)Un,k+2x0-x02-α1kα2kj=1k-1(1-α3j)TkUn,k+1x0-Un,k+1x02-α2kα3kj=1k-1(1-α3j)Un,k+1x0-x02-α1k-1α2k-1j=1k-2(1-α3j)Tk-1Un,kx0-Un,kx02-α2k-1α3k-1j=1k-2(1-α3j)Un,kx0-x02  -α13α23j=12(1-α3j)T3Un,4x0-Un,4x02-α23α33j=12(1-α3j)Un,4x0-x02-α12α22(1-α31)T2Un,3x0-Un,3x02-α22α32(1-α31)Un,3x0-x02-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02  j=1n(1-α3j)Un,n+1x0-x*2+(1-j=1n(1-α3j))x0-x*2-α1nα2nj=1n-1(1-α3j)TnUn,n+1x0-Un,n+1x02-α2nα3nj=1n-1(1-α3j)Un,n+1x0-x02  -α1k+1α2k+1j=1k(1-α3j)Tk+1Un,k+2x0-Un,k+2x02-α2k+1α3k+1j=1k(1-α3j)Un,k+2x0-x02-α1kα2kj=1k-1(1-α3j)TkUn,k+1x0-Un,k+1x02-α2kα3kj=1k-1(1-α3j)Un,k+1x0-x02-α1k-1α2k-1j=1k-2(1-α3j)Tk-1Un,kx0-Un,kx02-α2k-1α3k-1j=1k-2(1-α3j)Un,kx0-x02  -α13α23j=12(1-α3j)T3Un,4x0-Un,4x02-α23α33j=12(1-α3j)Un,4x0-x02-α12α22(1-α31)T2Un,3x0-Un,3x02-α22α32(1-α31)Un,3x0-x02-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02=x0-x*2-α1nα2nj=1n-1(1-α3j)TnUn,n+1x0-Un,n+1x02  -α1k+1α2k+1j=1k(1-α3j)Tk+1Un,k+2x0-Un,k+2x02-α2k+1α3k+1j=1k(1-α3j)Un,k+2x0-x02-α1kα2kj=1k-1(1-α3j)TkUn,k+1x0-Un,k+1x02-α2kα3kj=1k-1(1-α3j)Un,k+1x0-x02-α1k-1α2k-1j=1k-2(1-α3j)Tk-1Un,kx0-Un,kx02-α2k-1α3k-1j=1k-2(1-α3j)Un,kx0-x02  -α13α23j=12(1-α3j)T3Un,4x0-Un,4x02-α23α33j=12(1-α3j)Un,4x0-x02-α12α22(1-α31)T2Un,3x0-Un,3x02-α22α32(1-α31)Un,3x0-x02-α11α21T1Un,2x0-Un,2x02-α21α31Un,2x0-x02. For k and (2.20), we have α2k-1α3k-1j=1k-2(1-α3j)Un,kx0-x02x0-x*2-Snx0-x*2, as n. This implies that U,kx0=x0,  for  all  k.

Again by (2.20), we have α1kα2kj=1k-1(1-α3j)TkUn,k+1x0-Un,k+1x02x0-x*2-Snx0-x*2, as n. Hence, α1kα2kj=1k-1(1-α3j)TkU,k+1x0-U,k+1x020. From U,kx0=x0,  for  all  k, and (2.23), we obtain that Tkx0=x0,  for  all  k. This implies that x0i=1F(Ti).

3. Main ResultTheorem 3.1.

Let C be a nonempty closed convex subset of a Hilbert space H. Let F be bifunctions from C×C into satisfying (A1)–(A4). Let A:CH be a α-inverse-strongly monotone mapping. Let {Ti}i=1 be infinite family of nonexpansive mappings with 𝔉=i=1F(Ti)EP(F)VI(C,A), and let ρj=(α1j,α2j,α3j)I×I×I, where I=[0,1],  α1j+α2j+α3j=1,  α1j+α2jb<1, and α1j,α2j,α3j(0,1)  for  all  j=1,2,. For every n, let Sn and S be S-mappings generated by Tn,,T1 and ρn,ρn-1,,ρ1 and Tn,Tn-1,, and ρn,ρn-1,, respectively. Let {xn},{un} be sequences generated by x1,uC and F(un,y)+1rny-un,un-xn0,yC,xn+1=αnu+βnPC(I-λA)xn+γnSnPC(I-λA)un,n1, where {αn},{βn},{γn}(0,1), such that αn+βn+γn=1,βn[c,d](0,1)  rn[a,b](0,2α),  λ(0,2α). Assume that

limnαn=0  and  n=0αn=,

n=1α1n<,

n=1|rn+1-rn|,n=1|γn+1-γn|,n=1|αn+1-αn|,n=1|βn+1-βn|<,

then the sequence {xn},{un} converge strongly to z=P𝔉u.

Proof.

First, we show that (I-λA) is nonexpansive. Let x,yC. Since A is α-inverse-strongly monotone and λ<2α, we have (I-λA)x-(I-λA)y2=x-y-λ(Ax-Ay)2=x-y2-2λx-y,Ax-Ay+λ2Ax-Ay2x-y2-2αλAx-Ay2+λ2Ax-Ay2=x-y2+λ(λ-2α)Ax-Ay2x-y2. Thus, (I-λA) is nonexpansive. We will divide our proof into 5 steps.Step 1.

We shall show that the sequence {xn} is bounded. Since F(un,y)+1rny-un,un-xn0,yC. By Lemma 2.5, we have un=Trnxn and EP(F)=F(Trn).

Let z𝔉. By nonexpansiveness of I-λA and Trn, we have xn+1-z=αnu+βnPC(I-λA)xn+γnSnPC(I-λA)un-z=αn(u-z)+βn(PC(I-λA)xn-z)+γn(SnPC(I-λA)un-z)αnu-z+βnPC(I-λA)xn-z+γnSnPC(I-λA)un-zαnu-z+βnxn-z+γnun-z=αnu-z+βnxn-z+γnTrnxn-zαnu-z+(1-αn)xn-zmax{u-z,xn-z}. By induction, we can prove that {xn} is bounded and so is {un}.

Step 2.

We will show that limnxn+1-xn=0. By definition of xn, we have xn+1-xn=αnu+βnPC(I-λA)xn+γnSnPC(I-λA)un  -αn-1u-βn-1PC(I-λA)xn-1-γn-1Sn-1PC(I-λA)un-1=(αn-αn-1)u+βnPC(I-λA)xn-βn-1PC(I-λA)xn+βn-1PC(I-λA)xn+γnSnPC(I-λA)un-γn-1SnPC(I-λA)un+γn-1SnPC(I-λA)un-βn-1PC(I-λA)xn-1-γn-1Sn-1PC(I-λA)un-1=(αn-αn-1)u+(βn-βn-1)PC(I-λA)xn+βn-1(PC(I-λA)xn-PC(I-λA)xn-1)+(γn-γn-1)SnPC(I-λA)un+γn-1(SnPC(I-λA)un-Sn-1PC(I-λA)un-1)|αn-αn-1|u+|βn-βn-1|PC(I-λA)xn+βn-1PC(I-λA)xn-PC(I-λA)xn-1+|γn-γn-1|SnPC(I-λA)un+γn-1SnPC(I-λA)un-Sn-1PC(I-λA)un-1|αn-αn-1|u+|βn-βn-1|PC(I-λA)xn+βn-1xn-xn-1+|γn-γn-1|SnPC(I-λA)un+γn-1(SnPC(I-λA)un-SnPC(I-λA)un-1+SnPC(I-λA)un-1-Sn-1PC(I-λA)un-1)|αn-αn-1|u+|βn-βn-1|PC(I-λA)xn+βn-1xn-xn-1+|γn-γn-1|SnPC(I-λA)un+γn-1un-un-1+γn-1SnPC(I-λA)un-1-Sn-1PC(I-λA)un-1. Since un=Trnxn, by definition of Trn, we have F(Trnxn,y)+1rny-Trnxn,Trnxn-xn0,yC. Similarly, F(Trn+1xn+1,y)+1rn+1y-Trn+1xn+1,Trn+1xn+1-xn+10,yC. From (3.6) and (3.7), we obtain F(Trnxn,Trn+1xn+1)+1rnTrn+1xn+1-Trnxn,Trnxn-xn0,F(Trn+1xn+1,Trnxn)+1rn+1Trnxn-Trn+1xn+1,Trn+1xn+1-xn+10. By (3.8), we have 1rnTrn+1xn+1-Trnxn,Trnxn-xn+1rn+1Trnxn-Trn+1xn+1,Trn+1xn+1-xn+10. It follows that Trnxn-Trn+1xn+1,Trn+1xn+1-xn+1rn+1-Trnxn-xnrn0. This implies that 0Trn+1xn+1-Trnxn,Trnxn-Trn+1xn+1+Trn+1xn+1-xn-rnrn+1(Trn+1xn+1-xn+1). It follows that Trn+1xn+1-Trnxn2Trn+1xn+1-Trnxn,Trn+1xn+1-xn-rnrn+1(Trn+1xn+1-xn+1)=Trn+1xn+1-Trnxn,xn+1-xn+(1-rnrn+1)(Trn+1xn+1-xn+1)Trn+1xn+1-Trnxnxn+1-xn+(1-rnrn+1)(Trn+1xn+1-xn+1)Trn+1xn+1-Trnxn(xn+1-xn+|1-rnrn+1|Trn+1xn+1-xn+1)=Trn+1xn+1-Trnxn(xn+1-xn+1rn+1|rn+1-rn|Trn+1xn+1-xn+1)Trn+1xn+1-Trnxn(xn+1-xn+1a|rn+1-rn|Trn+1xn+1-xn+1). It follows that un+1-unxn+1-xn+1a|rn+1-rn|un+1-xn+1. Putting yn=PC(I-λA)un, then {yn} is bounded. By definition of Sn,for  all  n, we have Snyn-1-Sn-1yn-1=Un,1yn-1-Un-1,1yn-1=α11T1Un,2yn-1+α21Un,2yn-1+α31yn-1-α11T1Un-1,2yn-1-α21Un-1,2yn-1-α31yn-1(1-α31)Un,2yn-1-Un-1,2yn-1(1-α31)(1-α32)Un,3yn-1-Un-1,3yn-1=j=12(1-α3j)Un,3yn-1-Un-1,3yn-1  j=1n-1(1-α3j)Un,nyn-1-Un-1,nyn-1Un,nyn-1-yn-1=α1nTnyn-1+(1-α1n)yn-1-yn-1=α1nTnyn-1-yn-1α1n2yn-1-z. Substituting (3.13) and (3.14) into (3.5), we have xn+1-xn|αn-αn-1|u+|βn-βn-1|PC(I-λA)xn+βn-1xn-xn-1+|γn-γn-1|SnPC(I-λA)un+γn-1un-un-1+γn-1SnPC(I-λA)un-1-Sn-1PC(I-λA)un-1|αn-αn-1|u+|βn-βn-1|PC(I-λA)xn+βn-1xn-xn-1+|γn-γn-1|SnPC(I-λA)un+γn-1(xn+1-xn+1a|rn+1-rn|un+1-xn+1)+2γn-1α1nyn-1-z|αn-αn-1|u+|βn-βn-1|PC(I-λA)xn+(1-αn-1)xn-xn-1+|γn-γn-1|SnPC(I-λA)un+1a|rn+1-rn|un+1-xn+1+2γn-1α1nyn-1-z(1-αn-1)xn-xn-1+|βn-βn-1|M1+|αn-αn-1|M1+|γn-γn-1|M1+1a|rn+1-rn|M1+2α1nM1, where M1=maxn{u,PC(I-λA)xn,  SnPC(I-λA)un,  un-xn,  yn-z}. By (3.15), Lemma 2.3, and conditions (i)–(iii), we obtain limnxn+1-xn=0.

Step 3.

We shall show that limnxn-un=0.

Let v𝔉. Since un=Trnxn and Trn is firmly nonexpansive, we have v-Trnxn2=Trnv-Trnxn2Trnv-Trnxn,v-xn=12(Trnxn-v2+xn-v2-Trnxn-xn2). Hence, un-v2xn-v2-un-xn2. By (3.18), we have xn+1-v2=αn(u-v)+βn(PC(I-λA)xn-v)+γn(SnPC(I-λA)un-v)2αnu-v2+βnxn-v2+γnun-v2αnu-v2+βnxn-v2+γn(xn-v2-un-xn2)αnu-v2+xn-v2-γnun-xn2. it implies that γnun-xn2αnu-v2+xn-v2-xn+1-v2αnu-v2+(xn-v-xn+1-v)(xn-v+xn+1-v)αnu-v2+xn-xn+1(xn-v+xn+1-v). By (3.16) and condition (i), we have limnxn-un=0. Let z𝔉 and by nonexpansiveness of I-λA, we have xn+1-z2αnu-z2+βnPC(I-λA)xn-z2+γnSnPC(I-λA)un-z2-βnγnPC(I-λA)xn-SnPC(I-λA)unαnu-z2+(1-αn)xn-z2-βnγnPC(I-λA)xn-SnPC(I-λA)unαnu-z2+xn-z2-βnγnPC(I-λA)xn-SnPC(I-λA)un. It implies that βnγnPC(I-λA)xn-SnPC(I-λA)unαnu-z2+xn-z2-xn+1-z2=αnu-z2+(xn-z-xn+1-z)(xn-z+xn+1-z)=αnu-z2+xn-xn+1(xn-z+xn+1-z). By (3.16) and condition (i), we have limnPC(I-λA)xn-SnPC(I-λA)un=0. Since PC(I-λA)xn-SnPC(I-λA)xnPC(I-λA)xn-SnPC(I-λA)un+SnPC(I-λA)un-SnPC(I-λA)xnPC(I-λA)xn-SnPC(I-λA)un+un-xn, by (3.24) and (3.21), we have limnPC(I-λA)xn-SnPC(I-λA)xn=0. Since xn-PC(I-λA)xnxn-xn+1+xn+1-PC(I-λA)xnxn-xn+1+αnu-PC(I-λA)xn+γnSnPC(I-λA)un-PC(I-λA)xn, by (3.24), (3.16), and condition (i), we have limnxn-PC(I-λA)xn=0. Since xn-SnPC(I-λA)unxn-xn+1+xn+1-SnPC(I-λA)unxn-xn+1+αnu-SnPC(I-λA)un+βnPC(I-λA)xn-SnPC(I-λA)un, again by (3.24), (3.16), and condition (i), we have limnxn-SnPC(I-λA)un=0. Since xn-SnPC(I-λA)xnxn-SnPC(I-λA)un+SnPC(I-λA)un-SnPC(I-λA)xnxn-SnPC(I-λA)un+un-xn, by (3.21) and (3.30), we have limnxn-SnPC(I-λA)xn=0.

Step 4.

Putting z0=P𝔉u, we will show that limsupnu-z0,xn-z00. To show this inequality, take a subsequence {xnm} of {xn}, such that limsupnu-z0,xn-z0=limsupmu-z0,xnm-z0. Without loss of generality, we may assume that xnmω  as  m where ωC. By nonexpansiveness of PC(I-λA), (3.28), and Lemma 2.2, we have ωF(PC(I-λA)). By Lemma 2.6, we obtain that ωVI(C,A). Since unm-xnm0 as m, we have unmω  as  m. Since F(un,y)+1rny-un,un-xn0,yC. By (A2), we have 1rny-un,un-xnF(y,un),yC. In particular, y-unm,1rnm(unm-xnm)F(y,unm). By condition (A4), F(y,·) is lower semicontinuous and convex, and thus weakly semicontinuous. By (3.21) imply that (1/rnm)(unm-xnm)0 in norm. Therefore, letting m in (3.37), we have F(y,ω)limmF(y,unm)0,yC. Replacing y with yt:=ty+(1-t)ω, t(0,1], we have ytC, and using (A1), (A4), and (3.38), we obtain 0=F(yt,yt)tF(yt,y)+(1-t)F(yt,ω)tF(yt,y). Hence, F(ty+(1-t)ω,y)0,  for  all  t(0,1] and for all yC. Letting t0+ and using assumption (A3), we can conclude that F(ω,y)0,yC. Therefore, ωEP(F).

We will show that ωi=1F(Ti). By Lemma 2.10, we have F(S)=i=1F(Ti). Assume that ωSω. Using Opial s property, (3.32), ωF(PC(I-λA)), and Remark 2.9, we have liminfmxnm-ω<liminfmxnm-Sωliminfm(xnm-SnmPC(I-λA)xnm+SnmPC(I-λA)xnm-SnmPC(I-λA)ω+SnmPC(I-λA)ω-Sω)=liminfm(xnm-SnmPC(I-λA)xnm+SnmPC(I-λA)xnm-SnmPC(I-λA)ω+Snmω-Sω)liminfmxnm-ω. This is a contradiction, then ωi=1F(Ti). Hence, ω𝔉.

Since xnmω and ω𝔉, we have limsupnu-z0,xn-z0=limsupmu-z0,xnm-z0=u-z0,ω-z00.

Step 5.

Finally, we show that {xn} and {un} converse strongly to z0=P𝔉u. Putting z0=P𝔉u, by nonexpansiveness of PC(I-λA), Sn, and Trn, we have xn+1-z02=αn(u-z0)+βn(PC(I-λA)xn-z0)+γn(SnPC(I-λA)un-z0)2βn(PC(I-λA)xn-z0)+γn(SnPC(I-λA)un-z0)2+2αnu-z0,xn+1-z0βnPC(I-λA)xn-z02+γnSnPC(I-λA)un-z02+2αnu-z0,xn+1-z0βnxn-z02+γnTrnxn-z02+2αnu-z0,xn+1-z0(1-αn)xn-z02+2αnu-z0,xn+1-z0.

From Step 4 and Lemma 2.3, we obtain that {xn} converse strongly to z0=P𝔉u. By using (3.21), we have {un} converse strongly to z0=P𝔉u.

4. Application

Using our main theorem (Theorem 3.1), we obtain the following strong convergence theorems involving infinite family of κ-strict pseudocontractions.

To prove strong convergence theorem in this section, we need definition and lemma as follows.

Definition 4.1.

A mapping T:CC is said to be a κ-strongly pseudocontraction mapping, if there exists κ[0,1), such that Tx-Ty2x-y2+κ(I-T)x-(I-T)y2,x,yC.

Lemma 4.2 (see [<xref ref-type="bibr" rid="B19">15</xref>]).

Let C be a nonempty closed convex subset of a real Hilbert space H and T:CC a κ-strict pseudocontraction. Define S:CC by Sx=αx+(1-α)Tx for each xC. Then, as α[κ,1)  S is nonexpansive, such that F(S)=F(T).

Theorem 4.3.

Let C be a nonempty closed convex subset of a Hilbert space H. Let F be bifunctions from C×C into satisfying (A1)–(A4). Let A:CH be a α-inverse-strongly monotone mapping. Let {Ti}i=0 be infinite family of κi-pseudocontractions mappings with 𝔉=i=1F(Ti)EP(F)VI(C,A). Define a mapping Tκi by Tκi=κix+(1-κi)Tix, for  all  xC,  i, and let ρj=(α1j,α2j,α3j)I×I×I, where I=[0,1],α1j+α2j+α3j=1,α1j+α2jb<1, and α1j,α2j,α3j(0,1)  for  all  j=1,2,. For every n, let Sn and S be S-mappings generated by Tκn,,Tκ1 and ρn,ρn-1,,ρ1 and Tκn,Tκn-1,, and ρn,ρn-1,, respectively. Let {xn},{un} be sequences generated by x1,uC and F(un,y)+1rny-un,un-xn0,yC,xn+1=αnu+βnPC(I-λA)xn+γnSnPC(I-λA)un,n1, where {αn},{βn},{γn}(0,1), such that αn+βn+γn=1,  βn[c,d](0,1)rn[a,b](0,2α),  λ(0,2α). Assume that

limnαn=0  and  n=0αn=,

n=1α1n<,

n=1|rn+1-rn|,  n=1|γn+1-γn|,  n=1|αn+1-αn|,  n=1|βn+1-βn|<,

then the sequence {xn},{un} converges strongly to z=P𝔉u.

Proof.

For every i, by Lemma 4.2, we have that Tκi is nonexpansive mappings. From Theorem 3.1, we could have the desired conclusion.

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