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Vietnam Journal of Mathematics

(will be inserted by the editor)

A Fixed Point Scheme for Nonexpansive Mappings, Variational

Inequalities and Equilibrium Problems

Pham N. Anh Le Q. Thuy Do D. Thanh

Received: 10 May 2013 / Accepted: 15 December 2013

Abstract The purpose of this paper is to introduce a new iteration scheme and prove a

strong convergence theorem for finding the common element of the fixed point set of a

nonexpansive mapping, the solution set of variational inequalities and the solution set of

equilibrium problems. Under certain conditions on parameters, we show that the iterative

sequences generated by the scheme strongly converge to the common element in a real

Hilbert space.

Keywords Nonexpansive Pseudomonotone Continuous Fixed point Variationalinequalities Equilibrium problems

Mathematics Subject Classification (2010) 65K10 90C33

1 Introduction

The equilibrium problems in the sense of Blum and Oettli (see [7]), shortly EP (f,C), arepresented as follows:

Find x C such that f(x,y) 0 for ally C, EP(f,C)

whereCis a nonempty closed convex subset of a real Hilbert space Hwith inner product

, and norm , f :C C R is a bifunction such that f(x,x) = 0 for all x Cand

This work is supported by the Vietnam Institute for Advanced Study in Mathematics.

P.N. Anh ( )Department of Scientific Fundamentals, Posts and Telecommunications Institute of Technology, Hanoi, Viet-

nam

E-mail: [email protected]

L.Q. Thuy

School of Applied Mathematics and Informatics, Ha Noi University of Science and Technology, No. 1, Dai

Co Viet Road, Hai Ba Trung, Hanoi, Vietnam

E-mail: [email protected]

D.D. Thanh

Department of Mathematics, Haiphong University, Vietnam

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2 Pham N. Anh et al.

f(x, ) is convex and subdifferentiable on Cfor every x C. The solution set of ProblemEP(f,C) is denoted by Sol(f,C). The bifunction f is called strongly monotone onCwith> 0, if

f(x,y) +f(y,x) x y2 x,y C;

monotone on C, if

f(x,y) +f(y,x) 0 x,y C;

pseudomonotoneonC, if

f(x,y) 0 f(y,x) 0 x,y C;

Lipschitz-type continuouson Cwith constantsc1> 0 and c2> 0 in the sense of Mastroeni(see [13]) if

f(x,y) +f(y,z) f(x,z) c1x y2 c2y z

2 x,y,z C.

The mappingB : C H is called-inverse strongly monotone, if

Bx By,x y Bx By2 x,y C.

The variational inequalities (shortly VI(C,B)) are to findx Csuch that

Bx,x x 0 x C. (1)

The solution set of Problem VI(C,B)is denoted by sol(C,B). The mappingg : C Cis saidto becontractivewith (0,1), if

g(x) g(y) x y x,y C.

The mappingS: C Cis called nonexpansive, if

Sx Sy x y x,y C.

Fix(S)denotes the set of fixed points ofS. A linear bounded operator A ofH into itself iscalledstrongly positiveif there is a constant> 0 with property that

Ax,x x2 x H.

In this paper, we are interested in the problem of finding a common element of the

solution set of the equilibrium problems Sol(f,C), the solution set of variational inequalitiessol(C,B)and the set of fixed points Fix(S), namely:

Find x Fix(S) sol(C,B) Sol(f,C). (2)

Problem (2) is very general and it includes the variational inequalities, the equilibrium prob-

lems, the fixed point problem, and the vector optimization problem as particular cases. It

is well-known that the problem of finding a common fixed point of mappings has been ex-

tensively investigated (see, for example, [3,6,12] and the references therein). In fact, x is

a solution to Problem VI(F,C) if and only if it is a fixed point of the mapping T, whereT(x):=PrC(x F(x))for allx C. Otherwise, for each> 0 andx C,x

Sol(f,C)if and only if it is a fixed point of the solution mapping S, whereS(x):=argmin{f(x,y) +x y2 : y C} (see [1]). So, we can say that, solving the problem of finding a commonelement of the solution set of the equilibrium problem EP(f,C), the variational inequality

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A Fixed Point Scheme for Nonexpansive Mappings 3

problem VI(F,C) and the set of fixed points of nonexpansive mappings, is an extended studyof the problem of finding a common fixed point of mappings.

Theory and methods for solving Problem (2) have been well developed by many re-

searchers (see [2, 46,8,14, 18,19] and the references therein). Combettes and Hirstoaga

[9] proposed an iterative scheme of finding the best approximation to the initial data when

Sol(f,C) = /0 and proved a strong convergence theorem. Motivated by this idea, Takahashiand Takahashi [16] introduced the viscosity approximation method for finding a common

element of the set of solutions of an equilibrium problem and the set of fixed points of a

nonexpansive mapping in a real Hilbert space. Recently, Anh [1] proposed a hybrid extra-

gradient algorithm for finding a point of the set Fix(S) Sol(f,C). The iteration sequence isgiven by

x0 C,

yk =argmin{kf(xk,y) + y xk2 :y C},

tk =argmin{kf(yk,y) + y xk2 :y C},

xk+1 = kx0 + (1 k)Sx

k.

(3)

Under certain conditions on parameters kandk, he showed that the sequences {xk} and

{yk} weakly converge to the point x Fix(S) Sol(f,C). Inspired and motivated by the ideaof Anh [1], Wangkeeree and Preechasilp [17] replaced xk+1 in (3) by

xk+1 :=PC

kg(x

k) + (1 kA)SPC(tkBtk)

,

wheretk =argmin{kf(y

k,y) + y xk2 :y C}

andg is contractive,A is a strongly positive linear bounded operator ofH into itself. Then

they showed that under some control conditions, the sequences {xk},{yk}, and {tk} stronglyconverge to an element of Fix(S) sol(B,C) Sol(f,C). Combining the iteration method in[17] and an improvement set of linesearch techniques for equilibrium problems, we intro-

duce a new iteration scheme for finding an element of Fix(S) sol(C,B) Sol(f,C). Then,we show that the iterative sequences generated by the scheme strongly converge to a solution

of Problem (2) in a real Hilbert space.

2 Preliminaries

LetCbe a nonempty closed convex subset of a real Hilbert space H. Let f :C C Rbe a bifunction. For solving the mixed equilibrium problem, let us assume the following

conditions for a bifunction f:

(A1) f(x,x) =0 for all x C;(A2) fis continuous onC;(A3) fis pseudomonotone onC;(A4) for eachx C, f(x, )is convex and subdifferentiable on C.

For each pointx H, there exists the unique nearest point in C, denoted byPC(x), such that

x PC(x) x y y C.

The mappingPCis called the metric projection on C.

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Lemma 1 Let C be a nonempty closed convex subset ofH, x H and y C. Then

(i) y=PC(x)if and only ifx y,y z 0for all z C;(ii) PCis a nonexpansive mapping on C;

(iii) x y,PC(x) PC(y) PC(x) PC(y)2 for all x,y H,

(iv) x PC(x),PC(x) y 0for all x H and y C.

Using Lemma 1, one can show that Problem (1) is equivalent to a fixed point problem.

Lemma 2 A point u C is a solution of the variational inequality(1) if and only if u satisfiesthe relation u=P

C(u Bu)for all> 0.

A set-valued mappingT : H 2H is called monotone if for all x,y H,u T x, andv Ty imply x y,u v 0. A monotone mappingT: H 2H ismaximalif the graphG(T) ofTis not properly contained in the graph of any other monotone mapping. It is knownthat a monotone mappingTis maximal if and only if for (x,u) H H, x y,u v 0for every(y,v) G(T) impliesu T x. LetB be an inverse-strongly monotone mapping ofCto H, letNC(v)be the normal cone to Catv C, that is,NC(v) = {w H : v u,w 0 for all u C}, and define

T v=

Bv +NC(v) if v C,

/0 if v/ C.(4)

ThenTis a maximal monotone and v T10 if and only ifv sol(C,B)(see [15]).

Now we collect some useful lemmas for proving the convergence results of this paper.

Lemma 3 (see [10]) Let C be a nonempty closed convex subset of a real Hilbert space H

and h: C R be convex and subdifferentiable on C. Then x is a solution to the followingconvex problem:

min{h(x): x C}

if and only if0 h(x) +NC(x), where NC(x

)is the out normal cone at x to C andh()denotes the subdifferential of g.

Lemma 4 (see [11]) Let C be a closed convex subset of a real Hilbert space H and let

S: C C be a nonexpansive mapping such thatFix(S) = /0. If a sequence {xn} C is suchthat xn z and xn Sxn 0, then z=Sz.

Lemma 5 (see [12]) Assume that A is a strongly positive linear bounded operator on a

Hilbert space Hwith coefficient> 0and0

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Algorithm 1 Choose positive sequences {n} (0,1); [a,b],0

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Proof(i) For anyx,y H, we have

P(IA +g)(x) P(IA +g)(y) (IA + g)(x) (IA + g)(y)

g(x) g(y) + IAx y

x y + (1 )x y

=

1 ( )

x y.

The Banachs contraction principle guarantees that P(IA +g)has a unique fixed point,saysq H. That is,q=P(IA +g)(q). By Lemma 1(i), we obtain

(g A)q,x q 0 x . (9)

(ii) If there existsn0 such thatd(xn) = 0, i.e.,xn =yn for alln n0, from Step 1, we have

tn =xn. By Rabian Wangkeeree and Pakkapon Preechasilp [17], the sequences {xn},{yn},and {tn} strongly converge to the same point q .

Now we considerd(xn) =0 and divide the proof into several steps.Step 1.Ifd(xn) =0, then there exists the smallest nonnegative integer mnsuch that

f(xn mn d(xn),yn) d(xn)2.

Indeed, for d(xn) =0 and (0,1), we suppose for contradiction that for everynonnegative integerm, we have

fxn md(xn),yn+d(xn)2 >0.Passing to the limit of the above inequality as m , by continuity of f, we obtain

fxn,yn

+d(xn)2 0. (10)

On the other hand, sinceyn is the unique solution of the strongly convex problem

min

nf(x

n,y) +1

2y xn2 : y C

,

we have

nf(xn,y) +

1

2y xn2 nf(x

n,yn) +1

2yn xn2 y C.

Withy=xn, the last inequality implies

nf(xn,yn) +12

d(xn)||2 0. (11)

Combining (10) with (11), we obtain

d(xn)2 1

2nd(xn)2.

Hence it must be either d(xn)= 0 or 12n

. The first case contradicts d(xn) =0,

while the second one contradicts the fact n 12>1.

Step 2.We claim that ifd(xn) =0 thenxn /Hn.Indeed, fromzn =xn mn d(xn), it follows that

yn zn =1 mn

mn(zn xn).

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Indeed, suppose x , this implies that x Sol(f,C). Using the definition of x,f(x,x) 0 for allx Cand fis pseudomonotone onC, we get

f(zn,x) 0. (14)

It follows fromvn 2f(zn,zn)that

f(zn,x) = f(zn,x) f(zn,zn)

vn,x zn. (15)

Combining (14) and (15), we have

vn,x zn 0.

By the definition ofHn, we havex Hn. Thus Sol(f,C) CHn, which means CHn.

Step 5.Letwn = PC(tn Btn)andx . We claim that ifd(xn) = 0 then the sequences

{xn},{wn} and {tn} are bounded.

Indeed, in the case d(xn) =0, by Step 3, we have tn =PCHn ( yn), i.e.,

yn tn,z tn 0 z CHn,

where yn =PHn (xn). Substitutingz=x Sol(f,C) CHnby Step 4, then we have

yn tn,x tn 0 yn tn,x yn +yn tn 0,

which implies that

tn yn2 tn yn,x yn.

Hence

tn x2 =tn yn +yn x2

=tn yn2 + yn x2 + 2tn yn,yn x

x yn, tn yn + yn x2 + 2tn yn,yn x

= yn x2 + tn yn,yn x

yn x2 tn yn2. (16)

Sincezn =xn mn d(xn)and

yn =PHn (xn) =xn

vn,xn zn

vn2 vn,

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we have

yn x2 =xn x2 +vn,xn zn2

vn4 vn2

2vn,xn zn

vn2 vn,xn x

=xn x2 +

mn vn,d(xn)

vn

2

2mn vn,d(xn)

vn2 vn,xn x

=xn x2

mn vn,d(xn)

vn

2

2mn vn,d(xn)

vn2 vn,xn x mn vn,d(xn)

vn2

=xn x2

mn vn,d(xn)

vn

2

2mn vn,d(xn)

vn2 (vn,xn x mn vn,d(xn))

=xn x2

mn vn,d(xn)

vn

2

2mn vn,d(xn)

vn2 vn,xn x mn d(xn)

=xn x2

mn vn,d(xn)

vn

2

2mn vn,d(xn)

vn2 vn,zn x. (17)

It follows fromvn 2f(zn,zn)that

f(zn,y) f(zn,zn) vn,y zn y C. (18)

Replacingy by yn and combining with assumptions f(zn,zn) = 0 and zn =xn mn d(xn),we have

f(zn,yn) vn,yn zn

=(1 mn )vn,d(xn).

Combining this inequality with (6) and assumption (0,1), we obtain

vn,d(xn)

1 mnd(xn)2. (19)

Substitutingy=x in (18) and using f(zn,zn) =0, we have

f(zn,x) vn,x zn. (20)

Since fis pseudomonotone onCand f(x,x) 0 x C, we have

f(zn,x) 0.

Combining this with (20), we get

0 vn,x zn. (21)

Using (17), (19) and (21), we have

yn x2 xn x2

mn vn,d(xn)

vn

2

xn x2

mn

vn(1 mn )2

d(xn)4. (22)

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Combining (16) with (22), we obtain

tn x2 xn x2 tn yn2

mn

vn(1 mn )

2d(xn)4.

On the other hand, we have IB is a nonexpansive mapping. Indeed, since B is a -strongly monotone mapping and 0

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we have

nf(xn,y) +

1

2y xn2 nf(x

n,yn) +1

2yn xn2 y C.

Withy=xn Cand f(xn,xn) =0, we have

0 nf(xn,yn) +

1

2yn xn2. (23)

By hypothesis, f(xn, )is convex and subdifferentiable on C, i.e.,

f(xn,y) f(xn,xn) un,y xn y C,

whereun 2f(xn,xn). Puttingy=yn, we have

f(xn,yn) un,yn xn.

Combining this and (23), we obtain

nun,yn xn +

1

2xn yn2 0,

and so

nunyn xn +12

xn yn2 0.

Hence

xn yn 2nun

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Thus, we have

xn+1 xn =PCng(xn) + (InA)Swn PCn1g(xn1) + (In1A)Swn1

(ng(xn) + (InA)Swn) n1g(xn1) + (In1A)Swn1

=ng(xn) ng(x

n1) +ng(xn1) n1g(x

n1) + (InA)Swn

(InA)Swn1 + (InA)Sw

n1 (In1A)Swn1

nxn xn1 + |n n1|g(x

n1)

+InASwn

Swn1

+ |n n1|ASwn1

nx

n xn1 + (1 n)wn wn1

+|n n1|(g(xn1) + ASwn1)

nxn xn1 + (1 n)x

n xn1

+|n n1|(g(xn1) + ASwn1)

= (1 n( ))xn xn1 + |n n1|M, (25)

whereM= supn0{g(xn1)+ASwn1}. Applying Lemma 6 for n:=n(),bn:=

|nn1|n

and using the assumptions (B1 B3), we have limn xn xn1= 0. Hence

limn xn+1 xn =0.

Step 8.We claim that limn xn tn =0.

Indeed, for eachx , we have

xn+1 x2 =PCng(xn) + (InA)Swnx2

ng(xn) + (InA)Sw

n x2

=n(g(xn) Ax) +I(Swn x) nA(Sw

n x)2

=n(g(xn) Ax) + (InA)(Sw

n x)2

(ng(xn) Ax + (1 n)Sw

n x)2

= 2n g(xn) Ax2 + (1 n)

2Swn x2

+2n(1 n)g(xn) AxSwn x

= 2n g(xn) Ax2 + (1 n)

2Swn x2


2

n (g

(xn

) g

(x

) + g

(x

) Ax

)

2

+ (1

n

)

2

Swn

x

2


= 2n g(xn) g(x)2 + 22n g(x

n) g(x)g(x) Ax

+2n g(x) Ax2 + (1 n)

2Swn x2


= 2n2g(xn) g(x)2 + 22n g(x

n) g(x)g(x) Ax

+2n g(x) Ax2 + (1 n)

2Swn Sx2


2n22xn x2 + (1 n)

2wn x2

+2n g(x) Ax2 + 22n g(x

n) g(x)g(x) Ax


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2n2xn x2 + (1 n)

2tn x2

+2n g(x) Ax2 + 22n g(x

n) g(x)g(x) Ax


2n2xn x2

+(1 n)2

xn x2 tn yn2

mnvn(1 mn )

2d(xn)4

+2n g(x

) Ax2 + 22n g(xn) g(x)g(x) Ax


(2n 2 + (1 n)

2)xn x2 (1 n)2tn yn2 +n

xn x2 (1 n)2tn yn2 +n,

where

n =2n g(x

) Ax2 + 22n g(xn) g(x)g(x) Ax

+2n(1 n)g(xn) AxSwn x (1 n)

2 mn

vn(1 mn )

2d(xn)4.

It then follows that

(1 n)2tn yn2 xn x2 xn+1 x +n.

Hence limn tn yn =0.

On the other hand xn yn2 = mn vn,d(xn)2

vn2 , which implies that limn xn yn =0.Since

xn tn xn yn + tn yn,

we obtain that limn xn tn =0.

Step 9.We claim that limn

xn Sxn =0.

Indeed, by Steps 7, 8, we have

xn+1 x2 2n g(xn) Ax2 + (1 n)

2wn x2

+2n(1 n)g(xn) Axwn x

2n g(xn) Ax2 + (1 n)

2(IB)tn (IB)x2


2n g(xn) Ax2 + (1 n)

2tn x2 +( 2)Btn Bx2+2n(1 n)g(x

n) Axwn x

2n g(xn) Ax2 + (1 n)

2tn x2

+(1 n)2( 2)Btn Bx2


2n g(xn) Ax2 + (1 n)

2tn x2

+(1 n)2a(b 2)Btn Bx2


2n g(xn) Ax2 + (1 n)

2xn x2

+(1 n)2a(b 2)Btn Bx2

+2n(1 n)g(xn) Axwn x.

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Hence

(1 n)2a(b 2)Btn Bx2 2n g(x

n) Ax2 + (1 n)2xn x2

xn+1 x2 + 2n(1 n)g(xn)

Axwn x.

Using(B1)and this inequality, we have limn Btn Bx =0.

On the other hand, by Lemma 1 and Step 7, we have

wn x2 =PC(tn Btn) PC(x Bx)2

(tn Btn) (x Bx),wn x

= 1

2((tn Btn) (x Bx)2 + wn x2

(tn Btn) (x Bx) (wn x)2)

1

2(tn x2 + wn x2 (tn wn) (Btn Bx)2)

1

2

tn x2 + wn x2 tn wn2

+

1

2(2tn wn,Btn Bx

2Btn Bx2).

Hence

wn x2 tn x2 tn wn2 + 2tn wn,Btn Bx 2

Btn Bx2.

From this inequality and Step 8, we have

xn+1 x2 2n g(xn) Ax2 + (1 n)

2wn x2


2n g(xn) Ax2 + (1 n)

2(tn x2 tn wn2

+2tn wn,Btn Bx 2

Btn Bx2)


2n g(xn) Ax2 + (1 n)

2xn x2 (1 n)2(tn wn2

+2(1 n)2tn wn,Btn Bx 2

(1 n)2Btn Bx2)+2n(1 n)g(x

n) Axwn x,

and hence

(1 n)2tn wn2 2n g(x

n) Ax2 + (1 n)2xn x2 xn+1 x2

+2(1 n)2tn wn,Btn Bx

2(1 n)

2Btn Bx2

+2n(1 n)g(xn) Axwn x.

Sincen 0 and Btn Bx 0, we obtain

limn

tn wn =0.

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From (8), we have

xn+1 Swn= PC(ng(xn) + (InA)Sw

n) Swn

ng(xn) + (InA)Sw

n Swn

= ng(xn) ASwn 0, asn .

Since

tn Stn tn xn + xn xn+1 + xn+1 Swn + Swn Stn

tn xn + xn xn+1 + xn+1 Swn + wn tn,

we obtain that tn Stn 0 as n . Moreover, we get

wn Swn wn tn + tn Stn + Stn Swn

2wn tn + tn Stn 0 as n .

Since

xn wn xn tn + tn wn,

it implies that

limn

xn wn =0.

Since

xn Sxn xn xn+1 + xn+1 Swn + Swn Sxn

xn xn+1 + xn+1 Swn + wn xn,

we obtain that

limn

xn Sxn =0.

Step 10.We claim that

limsupn

(g A)q,Swn q 0. (26)

Indeed, we choose a subsequence {wnk} of{wn} such that

limsupn

(g A)q,Swn q =limsupk

(g A)q,Swnk q.

Since {wn} is bounded, there exists a subsequence {wnki } of{wnk} which converges weaklytox. Next we show that x .

We prove that x Fix(S). We may assume without loss of generality that wnk x.Since wn Swn 0, we obtainSwnk p. Since xn Sxn 0, xn wn 0 and byLemma 4, we havex Fix(S).

We show thatx Sol(f,C). From Steps 7 and 10, we have

tnk x, xnk x, ynk x.

Sinceyn is the unique solution of the convex minimization problem

yn =argmin

nf(x

n,y) +1

2y xn2 : y C

,

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we have

0 2

nf(x

n,y) +1

2y xn2

(yn) +NC(y

n).

It follows that

0= nz +yn xn +zn,

wherez 2f(xn,yn)and zn NC(yn). By the definition of the normal cone NC(yn), we getthat

yn xn,y yn nz,yn y y C. (27)

On the other hand, since f(xn

, )is subdifferentiable onCand z 2f(xn

,yn

), we have

f(xn,y) f(xn,yn) z,y yn y C. (28)

Combining (27) with (27), we have

n(f(xn,y) f(xn,yn)) yn xn,yn y y C.

Hence

nk(f(xnk,y) f(xnk,ynk)) ynk xnk,ynk y y C.

Thus, using {n} [c,d], 0

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Step 11.We claim that xn q.Indeed, it follows from Step 2 that

xn+1 q2 = PC(ng(xn) + (InA)Sw

n) PC(q)2

ng(xn) + (InA)Sw

n q2

2n g(xn) Aq2 + InA

2Swn q2

+2n(InA)(Swn q),g(xn) Aq

2n g(xn) Aq2 + (1 n)

2Swn q2

+2nSwn

q,g(xn

) Aq 22

n A(Swn

q),g(xn

) Aq (1 n)

2wn q2 +2n g(xn) Aq2

+2nSwn q,g(xn) Aq 22n A(Sw

n q),g(xn) Aq

(1 n)2tn q2 +2n g(x

n) Aq2

+2nSwn q,g(xn) g(q) + 2nSw

n q,g(q) Aq

22n A(Swn q),g(xn) Aq

(1 n)2xn q2 +2n g(x

n) Aq2 + 2nwn qxn q

+2nSwn q,g(q) Aq 22n A(Sw

n q),g(xn) Aq

(1 n)2xn q2 +2n g(x

n) Aq2 + 2nxn q2

+2nSwn q,g(q) Aq 22n A(Sw

n q),g(xn) Aq

((1 n)2 + 2n)x

n q2 +2n g(xn) Aq2

+2nSwn q,g(q) Aq + 22n A(Swn q)g(xn) Aq

(1 2n())xn q2 +n(ng(x

n) Aq2 +n2xn q2

+2nA(Swn q)g(xn) Aq + 2Swn q,g(q) Aq).

Otherwise {xn}, {g(xn)}, and {Swn} are bounded, we can take a constant M>0 such that

M supn0

{ng(xn) Aq2 +n

2xn q2 + 2nA(Swn q)g(xn) Aq}.

This implies that

xn+1 q2 (1 2n())xn q2 +nn, (29)

wheren= 2Swn q,g(q) Aq +n M. From (26), it follows that limsupnn 0.

Applying Lemma 6 for (29), we obtain that xn

qasn . This completes the proof.

From Algorithm 1, we can easily deduce the algorithm for solving the equilibrium prob-

lem EP(f,C)as follows.

Algorithm 2 Choose positive sequences {n} (0,1), [a,b],0

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Step 2. Find the smallest positive integer number mnsuch that

f(xn mn d(xn),yn) d(xn)2.

Compute

tn =PCHn (xn),

where zn =xn mn d(xn),vn 2f(zk,zk) and Hn= {x H : v

n,x zn 0}.Increase n by1 and go back to Step1, and go to Step3.

Step 3. Compute

xn+1 =PC(ng(xn) + (InA)(tn)).

Increase n by1 and go back to Step1.

As in Theorem 1, we give a convergent result of Algorithm 2 as the following.

Theorem 2 LetH be a real Hilbert space and C be a closed convex subset of H. Let

f :C C R be a bifunction satisfying (A1)(A4) and A be a strongly positive linearbounded operator ofH into itself with coefficient> 0such thatA = 1. Let g : C C be

a contraction with coefficient (0,1). Assume that0

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