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A Hybrid Extragradient Method for Solving

Ky Fan Inequalities, Variational Inequalities and

Fixed Point Problems

Supak Phiangsungnoen and Poom Kumam,

Member, IAENG.

Abstract—The purpose of this paper is to introduce a new hybrid extragradient iterative method for finding a common element of the set of points satisfying the Ky Fan inequalities, variational inequality and the set of fixed points of a strict pseudocontraction mapping in Hilbert spaces. Consequently, we obtain the strong convergence of an iterative algorithm generated by the hybrid extragradient projection method, under some suitable assumptions the function associated with Ky Fan inequality is pseudomonotone and weakly continuous.

Index Terms—Extragradient Method, Fixed point problems, Hybrid projection method,Variational inequality problems, ξ−strict pseudocontraction, Lipschitz continuity.

I. INTRODUCTION

T

HROUGHOUT this paper, we always assume that H be a real Hilbert space, whose inner product and norm denoted by h·,·i and k · k, respectively. Let C be a closed convex subset ofH andF :C×C→R, whereRdenotes

the set of real number, a bifunction. We concider the Ky Fan inequality which was first introduced by Ky Fan [25] as follows:

find x∈C such that F(x, y)≥0, ∀y∈C. (1) The set of such thatx∈C is denoted byEP(F), i.e.,

EP(F) ={x∈C:F(x, y)≥0, ∀y∈C}.

Numerous problems in physics, optimization, and economics reduce to find a solution of (1). Some methods have been proposed to solve the Ky Fan inequality (or some time called equilibrium problem, see [1,3,8,14]).

In 2005, Combettes and Hirstoaga [2] introduced an iter-ative scheme of finding the best approximation to the initial data whenEP(F)is nonempty and they also proved a strong convergence theorem. A mapping S :C →C is said to be

nonexpansiveif

kSx−Syk ≤ kx−yk,

for all x, y ∈ C. We denote by F ix(S) the set of fixed points of, S i.e., F ix(S) = {x ∈ C : Sx = x}. If C is bounded closed convex and S is a nonexpansive mapping of C into itself, thenF ix(S)is nonempty (see [6]).

This work was supported by the Higher Education Research Promotion and National Research University Project of Thailand, Office of the Higher Education Commission.

Department of Mathematics, Faculty of Science, King Mongkut’s Uni-versity of Technology Thonburi, Bangkok 10140, Thailand.

Emails: supuk [email protected] (S. Phiangsungnoen) and [email protected] (P. Kumam)

We write xn → x (xn * x, resp.) if {xn} converges

(weakly, resp.) to x. A mapping A of C into H is called

monotoneif

hAu−Av, u−vi ≥0.

The classicalvariational inequality problemis to findu∈C such that hv−u, Aui ≥ 0 for all v ∈ C. We denoted by V I(A, C) the set of solutions of this variational inequality problem. The variational inequality has been extensively studied in the literature. See, e.g. [19,20] and the references therein. A mapping A of C into H is called α- inverse-strongly-monotone if there exists a positive real number α such that

hAu−Av, u−vi ≥αkAu−Avk2, (2)

for all u, v ∈C.It is obvious that any α -inverse-strongly-monotone mappings A is monotone and Lipschitz continu-ous.

In 1953, Mann [7] introduced the iteration as follows: a sequence{xn} defined by

xn+1=αnxn+ (1−αn)Sxn, (3)

where the initial guess elementx0∈Cis arbitrary and{αn}

is a real sequence in [0,1]. The Mann iteration has been extensively investigated for nonexpansive mappings. One of the fundamental convergence results is proved by Reich [11].

In an infinite-dimensional Hilbert space, the Mann iter-ation can conclude only weak convergence [4]. Attempts to modify the Mann iteration method (3) so that strong convergence is guaranteed have recently been made. Nakajo and Takahashi [9] proposed the following modification of the Mann iteration method (3):

       

      

x0∈C is arbitrary,

yn=αnxn+ (1−αn)Sxn,

Cn={z∈C:kyn−zk ≤ kxn−zk},

Qn ={z∈C:hxn−z, x0−xni ≥0},

xn+1=PCn∩Qnx0, n= 0,1,2. . . ,

(4)

wherePC is metric projection on C.

For finding an element ofF ix(S)∩V I(A, C), Takahashi and Toyoda [18] introduced the following iterative scheme:

xn+1=αnxn+ (1−αn)SPC(xn−λnAxn) (5)

for everyn= 0,1,2, ..., where PC is the metric projection

on the set C, x0 = x ∈ C, {αn} is a sequence in (0,1)

and {λn} is a sequence in (0,2α). They proved a weak

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On the other hand, for finding an element of F ix(S), Takahashi et al., [17] introduced the following iteration procedure which is usually called the shrinking projection method. Let C be nonempty closed convex subset of a real Hilbert space H. Let {αn} be a sequence in (0,1). Let

x0 ∈H. For C1 =C and x1 =PC1x0, define a sequence

{xn} of C as follows: 

 

 

yn= (1−αn)xn+αnSnxn,

Cn+1={z∈Cn:kyn−zk ≤ kxn−zk},

xn+1=PCn+1x0, n≥1,

(6)

wherePCn is the metric projection ofH ontoCn and{Sn}

is a family of nonexpansive mappings. They proved that the sequence {xn} generated by (6) converges strongly to z =

PF ix(S)x0, whereF ix(S) =T ∞

n=1F ix(Sn).

Moreover, in 2012, Phan Tu Vuong, et. al., [26], they con-sidered the sequences{xn},{yn},{zn},and{tn}generated

byx0∈C and 

  

  

yn= arg miny∈C{λnF(xn, y) +12ky−xnk2},

zn= arg miny∈C{λnF(yn, y) +12ky−xnk2},

tn=αnxn+ (1−αn)[βnzn+ (1−βn)Szn],

xn+1=tn,

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for every n ∈ N, where {αn} ⊂ [0,1[, {βn} ⊂]0,1[, and

{λn} ⊂]0,1].

In this paper, we consider the sequences {xn},{yn},

{zn},{wn} and {tn} generated by x0 ∈ H, C1 = C,

x1=PC1x0 and

       

      

yn = arg miny∈C{λnF(xn, y) +12ky−xnk2},

zn= arg miny∈C{λnF(yn, y) +12ky−xnk2},

wn=PC(zn−λnAzn),

tn =αnxn+ (1−αn)[(1−µ)Swn+µPC(1−βn)wn],

Cn+1={z∈Cn:ktn−zk ≤ kxn−zk},

xn+1=PCn+1x0,

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for every n∈ N, where µ be a constant in(0,1),{αn} ⊂ [0,1),{βn} ⊂ (0,1),{λn} ⊂ (0,1]. We combine the

equations (5), (6) and (7) above for solving the EP(F)

with a fixed point and variational inequality problems. Consequencely, we obtained the strong convergent theorem for solving fixed point problems, Ky Fan inequality and variational inequality problems.

II. PRELIMINARIES

Let C be a closed convex subset of a Hilbert space H. Then the following hold:y

kx−yk2=kxk2− kyk22hxy, yi (9)

and

kλx+(1−λ)yk2=λkxk2+(1λ)kyk2λ(1λ)kxyk2

(10)

for all x, y ∈ H and λ ∈ [0,1]. It is also known that H satisfies theOpial’s condition [10], that is, for any sequence

{xn} withxn * x, the inequality lim inf

n→∞ kxn−xk<lim infn→∞ kxn−yk (11)

holds for everyy ∈ H withy 6=x. Also Hilbert space H satisfies the Kadec-Klee property [5, 15], that is, for any sequence{xn}withxn* x and kxnk → kxktogether

implykxn−xk →0.

Let C be a closed convex subset of H. For every point x∈H, there exists a unique nearest point in C, denoted by PCxsatisfying the property

kx−PCxk ≤ kx−yk. (12)

For a givenx∈H andz∈C,

z=PCx⇔ hx−z, z−yi ≥0, ∀y∈C. (13)

It is well known thatPC isfirmly nonexpansiveof H onto

C and satisfies

hx−y, PCx−PCyi ≥ kPCx−PCyk2. (14)

In the context of the variational inequality problem, this implies that

u∈V I(A, C)⇔u=PC(u−λAu), ∀λ >0. (15)

We also have that, for allu, v∈C andλ >0,

k(I−λA)u−(I−λA)vk2 = k(u−v)−λ(Au−Av)k2

≤ ku−vk2+λ(λ2α)kAuAvk2. (16)

So, if λ ≤ 2α, then I−λA is a nonexpansive mapping fromC toH.

Let us recall thatSis aξ-strict pseudocontraction mapping if there exists a scalarξ∈[0,1)such that

kSx−Syk2≤ kxyk2+ξk(xSx)(ySy)k2 (17)

for every x, y ∈ C. It is easy to see that a nonexpansive mapping onC is also a0-strict pseudocontraction mapping. Furthermore [24], if S is a ξ-strict pseudocontraction mapping, then the fixed point set F ix(S) is closed and convex and the mapping I−S is demiclosed at zero, i.e., satisfies the property

xn* x and Sxn−xn→0⇒Sx=x.

A set valued mappingT :H →2H is called monotone if

for allx, y∈H, f ∈T xandg∈T yimplyhx−y, f−gi ≥0. A monotone mappingT :H →2H is maximal if the graph

G(T)ofTis not properly contained in the graph of any other monotone mapping. It known that a monotone mappingT is maximal if and only if for(x, f)∈H×H,hx−y, f−hi ≥

0 for every (y, g) ∈ G(T) implies f ∈ T x. Let A be an inverse-strongly monotone mapping ofCintoHand letNCv

be the normal cone toC atv∈C, i.e.,

NCv={w∈H:hv−u, wi ≥0,∀u∈C}

and define

T v=

Av+NCv, v∈C,

∅, v /∈C.

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Proposition 1. ([21], Lemma 3.1) For every x∗ ∈EP(F), and everyn∈N, one has

(i) hxn−yn, y−yni ≤λnF(xn, y)−λnF(xn, yn),

∀y∈C;

(ii) kzn−x∗k2≤ kxn−x∗k2−(1−2λnc1)kyn−xnk2− (1−2λnc2)kzn−ynk2.

Lemma 2. [28] LetC be a closed convex subset of H and let{xn}be a bounded sequence in H. Assume that

1) the weakω−limitsetωw(xn)⊂C,

2) for eachz∈C,limn→∞kxn−zk exists. Then{xn} is weakly convergent to a point in C.

Proposition 3. [24] LetKbe a nonempty closed and convex subset ofH. Letu∈H and let{xn}be a sequence inH. If any weak limit point of{xn}belongs toK, andkxn−uk ≤

ku−PKuk for all n∈N, thenxn→PKu.

In order to apply this Proposition, we setK:=EP(F)∩

F ix(S)∩V I(A, C) and u= x0. Furthermore, we impose

that the sequence{xn}generated by our algorithms satisfies,

for alln∈N,kxn−x0k ≤ kxn+1−x0k ≤ kx˜0−x0k

wherex˜0=PKx0. In that case, the sequence{kxn−x0k}

is convergent and the sequence {xn} is bounded. These

properties will be useful to prove that any weak limit point of {xn} belongs toK.

For solving the Ky Fan inequality or equilibrium problem, let us assume that the following conditions (A1) - (A5) are satisfied on the bifunction F :C×C→R.

(A1) F(x, x) = 0for every x∈C;

(A2) F is pseudomonotone on C, i.e.,

F(x, y)≥0⇒F(y, x)≤0, ∀x, y∈C;

(A3) F is jointly weakly continuous onC×C in the sense that, ifx, y∈Cand{xn}and{yn}are two sequences

inC converging weakly toxandy, respectively, then F(xn, yn)→F(x, y);

(A4) F(x,·) is convex, lower semicontinuous, and subdif-ferentiable onC foe veryx∈C;

(A5) F satisfies the Lipschitz-type condition, there exist positive integerc1 andc2, for every x, y, z∈C,

F(x, y)+F(y, z)≥F(x, z)−c1ky−xk2−c2kz−yk2.

IfF satisfies the properties (A1)-(A4), then the setEP(F)

of solutions to the Ky Fan inequality is closed and convex.

Remark 4. A first example of functionF satisfying assump-tion (A5), is given by

F(x, y) =hG(x), y−xi ∀x, y∈C,

where G : C → H is Lipschitz continuous on C (with constantL >0)(see[22]). In that example,c1=c2=L\2.

Another example, related to the Cournot-Nash equilibrium model, is described in [23]. The functionF :C×C→Ris

defined, for every x, y∈C, by

F(x, y) =hG(x) +Q(y) +q, y−xi, with C = {x ∈Rn : Ax b}, F : C

Rn, Q ∈Rn×n,

a symmetric positive semidefinite matrix, and q ∈R. If F

is Lipschitz continuous on C (with constant L > 0) then F satisfies (A5) with c1 >0, c2 >0 such that 2

c1c2 ≥

L+kQk.

III. MAIN RESULT

A. An Algorithm

LetC be a nonempty, closed and convex subset of a real Hilbert space H, let F be a bifunction F : C× C into R satisfying conditions (A1)−(A5), let A be an α-inverse-strongly monotone mapping ofCintoH. Let S be aξ-strict pseudocontraction mapping fromC toC.

Algorithm 5. Choose the sequences{αn} ⊂[0,1),{βn} ⊂ (0,1),{λn} ⊂(0,1]andµbe a constant in(0,1).

Assume that the following 6 Steps S(1) - S(6) S(1) Letx0∈H,C1=C,x1=PC1x0. Setn= 0.

S(2) Solve successively the strongly convex programs

arg miny∈C{λnF(xn, y) +12ky−xnk2} and arg miny∈C{λnF(yn, y) +12ky−xnk2} to obtain

the unique optimal solution yn andzn, respectively.

S(3) Computewn=PC(zn−λnAzn).

S(4) Compute

tn=αnxn+ (1−αn)[(1−µ)Swn+µPC(1−βn)wn].

Ifyn=xn andtn=xn, then STOP:

xn∈EP(F)∩F ix(S)∩V I(A, C).Otherwise, go to

Step 5.

S(5) Computexn+1=PCn+1x0,

whereCn+1={z∈Cn:ktn−zk ≤ kxn−zk},

S(6) Setn:=n+ 1, and go to Step 2.

In the sequel, we also suppose that the sequences of parameters{αn},{βn},{λn}, ξ andµsatisfy the following

conditions: (1) - (4)

(1) {λn} ⊂ [λmin, λmax], where 0 < λmin ≤ λmax < min 1

2c1, 1 2c2 ;

(2) {αn} ⊂[0, c] for somec <1;

(3) limn→∞βn= 0 andP∞n=1βn =∞;

(4) ξ andµ be constant, where0≤ξ < µ <1.

Now, let {xn},{yn},{zn}, and {wn} be the sequences

generated by combination of the hybrid extragradient method, variational inequality by projection method and the fixed point method described at the beginning of this section.

B. A strong convergence theorem

Here we start our main theorem.

Theorem 6. LetCbe a nonempty, closed and convex subset of a real Hilbert space H, let F be a bifunction F : C×

C into R satisfying conditions (A1)−(A5), let A be an

α-inverse-strongly monotone mapping of C into H. Let S be aξ-strict pseudocontraction mapping from C to C and such thatΩ :=EP(F)∩F ix(S)∩V I(A, C)6=∅. Suppose that the sequences {αn},{βn},{λn} and µ satisfying the conditions(1)−(4). Then the sequence{xn} generated by Algorithm5converges strongly to the projection ofx0on to the setΩ.

Proof. Step 1.We show that{xn}is well defined.

First, we show thatCnis closed and convex for eachn≥1.

Indeed, it is obvious that C1 = C is closed and convex.

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Next, we show thatCk+1 is closed and convex for the same

k. Let z1, z2 ∈ Ck+1 and z = tz1+ (1−t)z2, where t ∈ (0,1). Notice that ktn−zk ≤ kxn−zk is equivalent to

ktn−xnk2+ 2htn−xn, xn−zi ≤0

Thus Ck+1 is closed and convex. Then, Cn is closed and

convex for anyn∈N. This implies that{xn}is well defined.

Step 2.Next, we show by induction thatΩ⊂Cnfor each

n≥1. Letx∗∈Ω. Thenx∗=PC(x∗−λnAx∗).

Since wn =PC(zn−λnAzn), we consider

kwn−x∗k2

= k[PC(zn−λnAzn)]−[PC(x∗−λnAx∗)]k2

≤ k(zn−λnAzn)−(x∗−λnAx∗)k2 = kzn−x∗k2−2λnhzn−x∗, Azn−Ax∗i

+λ2nkAzn−Ax∗k2

≤ kzn−x∗k2+λn(λn−2α)kAzn−Ax∗k2 (18)

≤ kzn−x∗k2.

By Proposition1 (ii), we have

kzn−x∗k2 ≤ kxn−x∗k2−(1−2λnc1)kyn−xnk2

−(1−2λnc2)kzn−ynk2 (19)

that is, we obtain kzn−x∗k ≤ kxn−x∗k

andkwn−x∗k ≤ kxn−x∗k.

Setun := (1−µ)Swn+µPC(1−βn)wn, for all n≥0.

Then, we havetn=αnxn+ (1−αn)un, It follows that

kun−x∗k2

= k(1−µ)Swn+µPC(1−βn)wn−x∗k2

≤ k(1−µ)(Swn−x∗) +µ

(1−βn)wn−x∗

k2

≤ kwn−x∗k2−(1−µ)(µ−ξ)kSwn−wnk2

−βnµ2kwnk2 (20)

≤ kwn−x∗k2≤ kxn−x∗k2

that is, kun−x∗k ≤ kxn−x∗k.

Since tn = αnxn + (1 −αn)un for every x∗ ∈ Ω,

we have

ktn−x∗k2 = kαnxn+ (1−αn)un−x∗k2

= kαn(xn−x∗) + (1−αn)(un−x∗)k2

≤ αnkxn−x∗k2+ (1−αn)kun−x∗k2

≤ αnkxn−x∗k2+ (1−αn)kxn−x∗k2

≤ kxn−x∗k2.

Hence ktn−x∗k ≤ kxn−x∗k. It follows thatx∗∈Cn+1.

This implies that

Ω :=EP(F)∩F ix(S)∩V I(A, C)⊂Cn, ∀n∈N.

Step 3. Next, we show that limn→∞kxn+1−xnk = 0

andlimn→∞ktn−xnk= 0.

From xn =PCnx0, we have

hx0−xn, xn−vi ≥0

for eachv∈Cn. UsingΩ⊂Cn for each we also have

hx0−xn, xn−zi ≥0, ∀z∈Ω, n∈N.

So, forz∈Ω, we have

0 ≤ hx0−xn, xn−zi

= hx0−xn, xn−x0+x0−zi

= −hx0−xn, x0−xni+hx0−xn, x0−zi

≤ −kx0−xnk2+kx0−xnkkx0−zk.

This implies that

kx0−xnk ≤ kx0−zk, ∀z∈Ω, n∈N.

From xn = PCnx0 and xn+1 = PCn+1x0 ∈ Cn+1 ⊂Cn,

we obtain

hx0−xn, xn−xn+1i ≥0. (21)

From (21), we have, forn∈N,

0 ≤ hx0−xn, xn−xn+1i

= hx0−xn, xn−x0+x0−xn+1i

= −hx0−xn, x0−xni+hx0−xn, x0−xn+1i

≤ −kx0−xnk2+kx0−xnkkx0−xn+1k.

It follows that

kx0−xnk ≤ kx0−xn+1k.

Thus the sequence{kxn−x0k}is bounded and nonincreasing

sequence, solimn→∞kxn−x0k exists, that is lim

n→∞kxn−x0k=m. (22)

Indeed, (21), we get

kxn−xn+1k2

= kxn−x0+x0−xn+1k2

= kxn−x0k2+ 2hxn−x0, x0−xn+1i +kx0−xn+1k2

= −kxn−x0k2+ 2hxn−x0, xn−xn+1i +kx0−xn+1k2

≤ −kxn−x0k2+kx0−xn+1k2.

From (22), we obtain

lim

n→∞kxn−xn+1k= 0. (23)

Sincexn+1∈Cn, we have

Cn ={z∈C:ktn−zk ≤ kxn−zk};

ktn−xn+1k ≤ kxn−xn+1k

and

kxn−tnk ≤ kxn−xn+1k+kxn+1−tnk

≤ 2kxn−xn+1k.

By (23), we obtain

lim

n→∞kxn−tnk= 0. (24)

Step 4. We will show that limn→∞kSwn −wnk = 0.

For x∗ ∈ Ω, from (18), (19) and (20), we can choose a constantM >0 such that,

sup n

(5)

We observe that

ktn−x∗k2

≤ αnkxn−x∗k2+ (1−αn)kun−x∗k2

≤ αnkxn−x∗k2+ (1−αn)kwn−x∗k2

−(1−µ)(µ−ξ)kSwn−wnk2−βnµ2kwnk2

= kxn−x∗k2−(1−αn)(1−2λnc1)kyn−xnk2

−(1−αn)(1−2λnc2)kzn−ynk2 +(1−αn)λn(λn−2α)kAz−Ax∗k2

−(1−αn)(1−µ)(µ−ξ)kSwn−wnk2

−(1−αn)βnµ2M.

Therefore, we get

(1−αn)(1−2λnc1)kyn−xnk2

≤ kxn−x∗k2− ktn−x∗k2−(1−αn)βnµ2M

=

kxn−x∗k+ktn−x∗k

kxn−tnk

−(1−αn)βnµ2M.

Since limn→∞kxn−tnk= 0,limn→∞βn= 0, 1−αn ≥1−c >0,1−2λnc1>1−2λmaxc1>0

and the sequence {xn},{tn}are bounded, we get lim

n→∞kyn−xnk= 0.

By similar way since limn→∞βn= 0,

1−αn ≥1−c >0and1−2λnc2>1−2λmaxc2>0,

we have

lim

n→∞kzn−ynk= 0.

Since limn→∞βn = 0,1−αn≥1−c >0

and−λn(λn−2α)>0,we obtain lim

n→∞kAz−Ax ∗k= 0.

By limn→∞βn= 0, 1−αn≥1−c >0

and(1−µ)(µ−ξ)>0,we have

lim

n→∞kSwn−wnk= 0.

Step 5. We will show that x˜∈Ω.

(5.1). We will show that x˜ ∈ EP(F). Since {xn} is

bounded, there exists a subsequence {xni} of {xn} which

xni *x˜andlimn→∞kxn−ynk= 0, we have thatyni*˜x.

On the other hand, by using Proposition 1(i), we have, for everyy∈C and for everyi∈N, that

hxni−yni, y−ynii ≤λniF(xni, y)−λniF(xni, yni).

Sincekxni−ynik →0 andy−yni* y−x˜ asi→ ∞and

since∀i∈N, 0< λmin≤λni ≤λmax,as i→ ∞, we get

F(˜x, y)≥0,∀y∈C. It means that x˜∈EP(F).

(5.2). We will show that x˜ ∈ F ix(S). Since {wn}

is bounded then there exists a subsequence {wni} of

{wn} which converges weakly to x. Since˜ S is a ξ-strict

pseudocontraction mapping, we know that the mapping I−S is demiclosed at zero. From kSwn−wnk → 0 as

n→ ∞ and{wni}*x. Thus, we obtain that˜ x˜∈F ix(S).

(5.3).Finally, we show thatx˜∈V I(A, C). Define T v=

Av+NCv, v∈C,

∅, v /∈C.

Then, we have thatT is maximal monotone operator. Since w ∈T v =Av+NC(v), we get w−Av ∈ NC(v).

Fromwn∈C, we have

hv−wn, w−Avi ≥0, (n= 1,2,3, ...).

We also havewn =PC(zn−λnAzn) and∀v∈C, we get

v−wn,

wn−zn

λn

+Azn

≥0. Therefore, we have

hv−wni, wi

≥ hv−wni, Avi

≥ hv−wni, Avi −

v−wni,

wni−zni

λni

+Azni

= hv−wni, Av−Awnii+hv−wni, Awni−Aznii

v−wni,

wni−zni

λni

.

Usingwni →x˜ andkwni−znik →0 which A is Lipshitz

continuous implies that

hv−x, w˜ i ≥0, as i→ ∞.

Since T is maximal monotone, we have x˜ ∈ T−1(0), and

hencex˜∈V I(A, C). Thus is clear thatx˜∈Ω.

Step 6. Finally, we show thatxn→x, where˜ x˜=PΩx0.

SinceΩis nonempty closed convex subset ofH, there exists a uniquex∗such thatx=P

Ωx0.Sincex∗∈Ω⊂Cn

andxn=PCnx0, we havekxn−x0k ≤ kx˜−x0k.

It follows fromx∗=PΩx0 and the lower semicontinuity of

norm that

kx∗−x0k ≤ kx˜−x0k ≤ lim

i→∞infkxni−x0k

≤ lim

i→∞supkxni−x0k ≤ kx ∗x

0k.

Thus, we obtain that

lim

i→∞kxni−x0k=kx˜−x0k=kx ∗x

0k.

Using the Kadec-Klee property ofH, we obtain that

lim

i→∞xni = ˜x=x ∗.

Since {xni} is an arbitrary subsequence of {xn}, we can

conclude that{xn}converges strongly toPΩx0. C. Reduced theorem

Corollary 7. LetCbe a nonempty, closed and convex subset of a real Hilbert space H, let F be a bifunction F : C×

C into R satisfying conditions (A1)−(A5), let A be an

α-inverse-strongly monotone mapping of C into H. Let S be aξ-strict pseudocontraction mapping from C to C and such thatΩ :=EP(F)∩F ix(S)∩V I(A, C)6=∅. Suppose that the sequences {αn},{βn},{λn} and µ satisfying the condition(i)−(iv). Then the sequences{xn} generated by Algorithm5converges strongly to the projection ofx0on to the setΩ.

(6)

ACKNOWLEDGMENTS

The first author would like to thank the Faculty of Sci-ence and Department of Mathematics, KMUTT for financial support. Also, the authors were supported by the Higher Education Research Promotion and National Research Uni-versity Project of Thailand, Office of the Higher Education Commission.

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