Modification of viscosity method for strictly pseudo-contractive mapping and equilibrium problem with some applications

Adv. Fixed Point Theory, 8 (2018), No. 2, 144-173 https://doi.org/10.28919/afpt/3619

ISSN: 1927-6303

MODIFICATION OF VISCOSITY METHOD FOR STRICTLY

PSEUDO-CONTRACTIVE MAPPING AND EQUILIBRIUM PROBLEM WITH SOME APPLICATIONS

KATANYOO THAPTHAS, ATID KANGTUNYAKARN∗

Department of Mathematics, King Mongkut’s Institute of Technology Ladkrabang, Bangkok, Thailand

Copyright c2018 Thapthas and 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.

Abstract. For the purpose of this article, we are using the concept of equilibrium problem and prove the strong convergence theorem by the viscosity approximation methods for finding a common element of the set of fixed points ofκi-strictly pseudo-contractive mappings and of a finite family of the set of solutions of equilibrium

prob-lems and variational inequality probprob-lems. Furthermore, we apply our main theorem for the numerical examples.

Keywords: viscosity approximation methods; strictly pseudo-contractive mapping; S-mapping; variational in-equality problems; the combination of equilibrium problems.

2010 AMS Subject Classification:47H09, 47H10, 47J20.

1.

Introduction

LetCbe a nonempty closed convex subset of a real Hilbert spaceHwith the inner producth·,·i

and the normk·k. A mappingT :C→Cis said to benonexpansiveifkT x−Tyk ≤ kx−yk, for

∗_{Corresponding author}

E-mail address: [email protected] Received November 19, 2017

allx,y∈C. Recall thatT is a κ-strictly pseudo-contractive mappingif there exists a constant κ∈[0,1)such that

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

Ifκ =0, then (1.1) reduces to nonexpansive mappings.

A pointx∈Cis called afixed pointofT ifT x=x. The set of fixed points ofT is denoted by

F(T) ={x∈C:T x=x}.

Recall that a mapping f :C→Cis said to becontractiveif there exists a constantη ∈(0,1)

such that, for allx,y∈H

kf(x)−f(y)k ≤ηkx−yk.

A mappingAofCintoH is calledα-inverse-strongly monotone if there exists a positive real

numberα such that

(1.2) hx−y,Ax−Ayi ≥αkAx−Ayk2, ∀x,y∈C.

LetA:C→H. The variational inequality problemsis to find a pointu∈Csuch that (1.3) hv−u,Aui ≥0, ∀v∈C.

The set of solutions of the variational inequality problems is denoted byV I(C,A).

Variational inequalities were introduced and investigated by Stampacchia [8] in 1964. It is well known that variational inequalities cover as diverse disciplines as partial differential equa-tions, optimal control, optimization, mathematical programming, mechanics and finance; see [9]−[11].

LetF :C×C→_Rbe a bifunction. The equilibrium problem for F is to determine its equi-librium point, that is to find a pointx∗∈Csuch thatF(x∗,y)≥0, for ally∈C.

The set of all solution of equilibrium problem is denoted by (1.4) EP(F) ={x∈C:F(x∗,y)≥0,∀y∈C}.

In 2013, Suwannaut and Kangtunyakarn [15] introduced the combination of equilibrium problemwhich is to findx∈Csuch that

(1.5)

N

∑

a_iF_i(x,y)≥0, ∀y∈C,

where F_i:C×C→_Rbe bifunction and a_i∈(0,1) with ∑N_i=1ai=1, for every i=1,2, ...,N. The set of solution (1.5) is denoted by

EP

N

∑

a_iF_i=nx∈C: N

∑

a_iF_i(x,y)≥0,∀y∈Co.

IfF_i=F,∀i=1,2, ...,N, then (1.5) reduces to (1.4).

In 2007, Takahashi and Takahashi [5] proved the following theorem.

Theorem 1.1. Let C be a nonempty closed convex subset of H. Let F be a bifunction from C×C to _Rsatisfying A1)−A4) and let S be a nonexpansive mapping of C into H such that F(S)∩EP(F)6= /0. Let f be a contraction of H into itself, let {x_n} and {u_n} be sequences generated by x₁∈H and

F(un,y) + 1

r_nhy−un,un−xni ≥0, ∀y∈C,

xn+1=αnf(xn) + (1−αn)Sun,

for all n∈_N,where{αn} ⊂[0,1]and{rn} ⊂[0,1]satisfy some control conditions. Then{xn}

and{u_n}converge strongly to z∈F(S)∩EP(F), where z=P_F(S)∩EP(F)f(z).

The explicit viscosity method for nonexpansive mappings generates a sequence{x_n}through the iteration process:

(1.6) x_n+1=αnf(xn) + (1−αn)T xn, n≥0,

whereIis the identity ofH and{αn}is a sequence in(0,1). It is well known [6, 7] that under certain conditions, the sequence{x_n}converges in norm to a fixed pointqof T which solves the variational inequality

whereSis the set of fixed points ofT, namely,S={x∈H:T x=x}.

Many authors proved a strong convergence theorem by using viscosity method; see, for in-stance, [5, 6].

In 2010, Kangtunyakarn [12] proved a strong convergence theorem of the iterative scheme (1.9) to a common fixed point ofq∈TN

i=1F(Ti).

Theorem 1.2. Let H be a Hilbert space, let f be anα-contraction on H and let A be a strongly

positive linear bounded self-adjoint operator with coefficientγ >0. Assume that 0<γ < γ

λ.

Let{T_i}N

i=1be a finite family ofκi-strict pseudo-contraction of H into itself, for someκi∈[0,1)

andκ=max{κi:i=1,2, ...,N}, withTNi=1F(Ti)6= /0. Let Sn be the S-mappings generated by

T₁,T₂, ...,T_Nandα₁(n),α₂(n), ...,α_N(n), whereα(_jn)= (α₁n,j,α₂n,j,α₃n,j)∈I×I×I,I= [0,1],α₁n,j+ α₂n,j+α₃n,j=1andκ<a≤α₁n,j,α₃n,j≤b<1, for all j=1,2, ...,N−1,κ<c≤α₁n,N≤1,κ≤ α₃n,N ≤d<1,κ ≤α₂n,j≤e<1, for all j=1,2, ...,N. For a point u∈H and x1∈H, let{xn}

and{yn}be the sequences defined iteratively by

(1.8)     

y_n=βnxn+ (1−βn)Snxn,

x_n+1=αnγ anu+ (1−an)f(xn)

+ (1−αnA)yn, n≥1,

where{βn},{αn}and{an}are sequences in[0,1]. Assume that the following conditions hold:

(i) lim

n→∞αn=0, ∞

∑

αn=∞and lim n→∞

a_n=0;

(ii) ∞

∑

α

n+1,j

1 −α

n,j 1

<∞,

∞

∑

α

n+1,j

3 −α

n,j 3

<∞, for all j∈ {1,2, ...,N}and

∞

∑

|λn+1−λn|<∞,

∞

∑

|βn+1−βn|<∞,

∞

∑

|a_n+1−an|<∞;

(iii) 0≤κ ≤βn<θ <1, for all n≥1, for someθ ∈(0,1).

Then both{x_n}and{y_n}strongly converges to q∈TN

i=1F(Ti)which solves the following

vari-ational inequality

(1.9) hγf(q)−Aq,p−qi ≤0, ∀p∈

N

\

i=1

F(Ti).

From Theorem 1.1 [5] and [15], we modify the viscosity methods as following:

For every i=1,2, ...,N, let F_i:C×C→_R be bifunction which satisfyA1)−A4)anda_i∈

andκ=max{κi:i=1,2, ...,N}. For each j=1,2, ...,N, letαj= (α₁j,α₂j,α₃j)∈I×I×I, where

I= [0,1]andα₁j+α₂j+α₃j=1, forx1∈Cand sequencexngenerated by

(1.10)



  

  

N

∑

aiFi(un,y) + 1

r_nhy−un,un−xni ≥0,∀y∈C, x_n+1=βn αnf(xn) + (1−αn)Sxn

+ (1−βn)PC(I−λA)un,∀n≥1,

whereA:C→H isα-inverse-strongly monotone mapping andS:C→CisS-mapping

gener-ated by a finite family of strictly pseudo-contractive mappings and a finite real numbers under suitable conditions of the parameters{βn},{αn},{rn} ∈[0,1]andλ ∈(0,2α).

Motivated by the above related literature, we prove a strong convergence theorem by modi-fying the viscosity methods for finding a common element of the set of solutions of equilibrium problems and variational inequality problems. Moreover, we apply our main result to obtain a strong convergence theorem for finding a common element of the set of fixed point ofκi-strictly pseudo-contractive mappings. Finally, we also give a numerical examples to support our main theorem.

2.

Preliminaries

In this section, we use some lemmas that will be used for our main result in the next section. LetC be a nonempty closed convex subset of a real Hilbert space H. We denote weak and strong convergence by 88*00 and88 →00, respectively, and let P_C be the metric projection of H

ontoC, that is, forx∈H,P_Cx∈Csatisfies the property

kx−P_Cxk=min

y∈Ckx−yk

and it is well-known that, for allx,y∈H andt∈[0,1],

ktx+ (1−t)yk2=tkxk2+ (1−t)kyk2−t(1−t)kx−yk2

andP_C is afirmly nonexpansive mappingofHontoC, that is,

Lemma 2.1. ([19]) For given z∈H and u∈C,

u=P_Cz⇔ hu−z,v−ui ≥0, ∀v∈C.

Lemma 2.2. ([16]) Each Hilbert space H satisfies Opial’s condition, i.e., for any sequence

{x_n} ⊂H with x_n*x, the inequality

lim inf n→∞ k

x_n−xk<lim inf n→∞ k

x_n−yk,

holds for every y∈H with y6=x.

Lemma 2.3. ([18]) Let{s_n}be a sequence of nonnegative real numbers satisfying s_n+1≤(1−αn)sn+δn,∀n≥0,

where{αn}is a sequence in(0,1)and{δn}is a sequence such that (1):

∞

∑

αn=∞,

(2): lim sup n→∞

δn

αn

≤0or ∞

∑

|δn|<∞.

Then, lim

n→∞

s_n=0.

Lemma 2.4. ([19]) Let H be a real Hilbert space, let C be a nonempty closed convex subset of H and let A be a mapping of C into H. Let u∈C. Then, forλ >0,

u=PC(I−λA)u⇔u∈V I(C,A),

where P_Cis the metric projection of H onto C.

Lemma 2.5. ([21]) Let C be a nonempty closed convex subset of a real Hilbert space H and S:C→C be a self-mapping of C. If S is aκ-strict pseudo-contractive mapping, then S satisfies

the Lipschitz condition

kSx−Syk ≤ 1+κ

1−κ kx−yk,∀x,y∈C.

For solving the equilibrium problem for a bifunction F :C×C →_R, let us assume that

(A2)F is monotone, i.e.,F(x,y) +F(y,x)≤0 for allx,y∈C; (A3) For eachx,y,z∈C,

lim

t↓0F tz+ (1−t)x,y

≤F(x,y); (A4) For eachx∈C,y7→F(x,y)is convex and lower semicontinuous.

Lemma 2.6. ([15]) Let C be a nonempty closed convex subset of a real Hilbert space H. For i=1,2, ...,N, let F_i:C×C→_R be bifunctions satisfying (A1)−(A4)with TN

i=1EP(Fi)6= /0.

Then

EP

N

∑

a_iF_i=

N

\

i=1

EP(F_i),

where a_i∈(0,1), for every i=1,2, ...,N and_∑N_i=1a_i=1.

Lemma 2.7. [14]) Let C be a nonempty close convex subset of H and F be a bifunction of C×C into_Rsatisfying(A1)−(A4). Let r>0and x∈H, then there exists z∈C such that

F(z,y) +1

rhy−z,z−xi ≥0,∀y∈C.

Lemma 2.8. ([17]) Assume that F :C×C →_R satisfies (A1)−(A4). For r > 0, define a mapping Tr:H→C as follows:

T_r(x) =nz∈C:F(z,y) +1

rhy−z,z−xi ≥0,∀y∈C

o ,

for all x∈H. Then, the following hold:

(i) T_r is single-valued;

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

kTr(x)−Tr(y)k2≤ hTr(p)−Tr(y),x−yi;

(iii) F(T_r) =EP(F);

(iv) EP(F)is closed and convex.

Remark2.9. From Lemma 2.6 and 2.8, ([15]) prove the following results; (i) ∑Ni=1aiFisatisfyingA1)−A4);

(ii) F(T_r) =TN

wherer>0 andai∈(0,1), for everyi=1,2, ...,N with∑N_i=1ai=1.

In 2009, Kangtunyakarn and Suantai ([20]) introduced theS-mapping generated by a finite family ofκi-strictly pseudo-contractions and a finite real numbers. The definition can be seen below:

Definition 2.1. Let C be a nonempty convex subset of a real Hilbert space. Let {Ti}N_i=1 be

a finite family of κi-strictly pseudo-contractions of C into itself. For each j =1,2, ...,N, let

αj = (α₁j,α₂j,α₃j)∈I×I×I, where I = [0,1] and α₁j+α₂j+α₃j = 1. Define the mapping

S:C→C as follows:

U₀=I,

U₁=α₁1T1U0+α21U0+α31I,

U2=α₁2T2U1+α₂2U1+α₃2I,

U3=α₁3T3U2+α₂3U2+α₃3I,

.

.

.

U_N−1=α₁N−1TN−1UN−2+α₂N−1UN−2+α₃N−1I,

S=U_N =α₁NTNUN−1+α₂NUN−1+α₃NI.

This mapping is called an S-mapping generated by T₁,T₂, ...,TN andα1,α2, ...,αN.

Lemma 2.10. ([22]) Let C be a nonempty closed convex subset of a real Hilbert space H. Let {T_i}N

i=1 be a finite family of κi-strictly pseudo-contractive mapping of C into itself with

TN

i=1F(Ti)6= /0 andκ =max{κi:i=1,2, ...,N}and let αj= (α₁j,α₂j,α₃j)∈I×I×I, where

I = [0,1],α₁j+α₂j+α₃j =1,α₁j,α₂j ∈(κ,1), for all i=1,2, ...,N−1 and α₁N ∈(κ,1],α₃N ∈

(κ,1],α₂j ∈(κ,1], for all j=1,2, ...,N, let S be the mapping generated by T1,T2, ...,TN and

3.

Main result

Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. For every i=1,2, ...,N, let Fi:C×C→Rbe bifunction with satisfy A1)−A4), Ti:C→C beκi-strictly

pseudo-contractive mapping and let A:C→H beα-inverse strongly monotone mapping with

F =TN

i=1EP(Fi)∩TNi=1F(Ti)∩V I(C,A)6= /0. Let S be S-mapping generated by T1,T2, ...,TN

andα1,α2, ...,αN, whereαj= (α₁j,α₂j,α₃j)∈I×I×I,I= [0,1]withα₁j+α₂j+α₃j=1andκ<

α₁j,α₃j<1, for all i=1,2, ...,N−1,κ<α₁N≤1,κ≤α₃N<1,κ≤α₂j<1, for all j=1,2, ...,N,

whereκ=max{κi:i=1,2, ...,N}. Let the sequence{xn}generated by x1∈C and

(3.1)

   

  

N

∑

a_iF_i(u_n,y) + 1

rn

hy−u_n,u_n−x_ni ≥0,∀y∈C,

x_n+1=βn αnf(xn) + (1−αn)Sxn

+ (1−βn)PC(I−λA)un,∀n≥1,

where{βn},{αn} ⊆[0,1]andλ ∈(0,2α). Suppose the following conditions hold:

(i) ∞

∑

αn=∞,lim

n→∞αn

=0,

(ii) 0<a≤βn,rn≤b<1, for all n≥1,

(iii) f :C→C beη-contraction,

(iv)

N

∑

ai=1, where ai>0, for all i=1,2, ...,N,

(v) ∞

∑

|αn+1−αn|<∞,

∞

∑

|βn+1−βn|<∞,

∞

∑

|r_n+1−r_n|<∞.

Then{x_n}converges strongly to z=P_F f(z).

Proof. First, we show that(I−λA) is a nonexpansive mapping. Let x,y∈C. Since A is α

-inverse strongly monotone andλ <2α, we have

k(I−λA)x−(I−λA)yk2=kx−yk2−2λhx−y,Ax−Ayi+λ2kAx−Ayk2 ≤ kx−yk2−2α λkAx−Ayk2+λ2kAx−Ayk2

=kx−yk2+λ(λ−2α)kAx−Ayk2

Then(I−λA)is a nonexpansive mapping. We will divide our proof into 5 steps.

Step 1: we show that the sequence{x_n}is bounded. Since N

∑

aiFi(un,y) + 1

r_nhy−un,un−xni ≥0,∀y∈C.

Form Remark 2.9, we haveun=Trnxnand TN

i=1EP(Fi) =F(Trn).

Letz∈F. By nonexpansiveness of(I−λA)andTrn, we obtain

kx_n+1−zk=β_n α_nf(x_n) + (1−α_n)Sx_n

+ (1−βn)PC(I−λA)un−z

=β_n α_nf(x_n) + (1−α_n)Sx_n−z

+ (1−βn) PC(I−λA)un−z

≤βn

α_n f(x_n)−z

+ (1−αn)(Sxn−z)

+ (1−βn)kPC(I−λA)un−zk

≤βn αnkf(xn)−f(z)k+αnkf(z)−zk+ (1−αn)kSxn−zk

+ (1−βn)kPC(I−λA)un−zk

≤βn αnηkxn−zk+αnkf(z)−zk+ (1−αn)kxn−zk

+ (1−βn)kun−zk

=βn

1−αn(1−η)

kx_n−zk+αnkf(z)−zk

+ (1−βn)kxn−zk

≤max

kx1−zk,

kf(z)−zk

1−η

.

By induction we can prove that{xn}is bounded and so is{un}.

Step 2: we will show that limn→∞kxn+1−xnk=0. By definition ofxn, we have

kx_n+1−xnk=

βn αnf(xn) + (1−αn)Sxn

+ (1−βn)PC(I−λA)un

−βn−1 αn−1f(xn−1) + (1−αn−1)Sxn−1

+ (1−βn−1)PC(I−λA)un−1

≤βn

α_nf(x_n) + (1−α_n)Sx_n

− αn−1f(xn−1) + (1−αn−1)Sxn−1 +|βn−βn−1| kαn−1f(xn−1) + (1−αn−1)Sxn−1k

+ (1−βn)kPC(I−λA)un−PC(I−λA)un−1k

+|βn−1−βn| kPC(I−λA)un−1k

≤βn αnkf(xn)−f(xn−1)k+|αn−αn−1| kf(xn−1)k

+ (1−αn)kSxn−Sxn−1k+|αn−1−αn| kSxn−1k

+|βn−βn−1| αn−1kf(xn−1)k+ (1−αn−1)kSxn−1k

+ (1−βn)kPC(I−λA)un−PC(I−λA)un−1k

+|βn−1−βn| kPC(I−λA)un−1k

≤βn αnηkxn−xn−1k+|αn−αn−1| kf(xn−1)k

+ (1−αn)kxn−xn−1k+|αn−1−αn| kSxn−1k

+|βn−βn−1| αn−1kf(xn−1)k+ (1−αn−1)kSxn−1k

+ (1−βn)kun−un−1k+|βn−1−βn| kPC(I−λA)un−1k

≤βn

|αn−αn−1|M+|αn−1−αn|M+ 1−αn(1−η)

kx_n−x_n−1k

+|βn−βn−1| αn−1M+ (1−αn−1)M+ (1−βn)kun−un−1k

+|βn−1−βn|M

=βn

2M|αn−αn−1|+ 1−αn(1−η)

kx_n−x_n−1k

+2M|βn−βn−1|+ (1−βn)kun−un−1k, (3.3)

whereM=maxn∈N

kf(x_n)k,kSx_nk,kP_C(I−λA)unk . Sinceun=Trnxnand definition ofTrn, we obtain

(3.4)

N

∑

a_iF_i(T_rnx_n,y) + 1

rn

hy−T_rnx_n,T_rnx_n−x_ni ≥0,∀y∈C

and (3.5)

N

∑

a_iF_i(T_rn+1x_n+1,y) + 1

rn+1

From (3.4) and (3.5). It follow that

(3.6)

N

∑

aiFi(Trnxn,Trn+1xn+1) +

1

r_n

Trn+1xn+1−Trnxn,Trnxn−xn ≥0 and (3.7) N

∑

aiFi(Trn+1xn+1,Trnxn) +

1

r_n+1

Trnxn−Trn+1xn+1,Trn+1xn+1−xn+1

≥0.

From (3.6),(3.7) and the fact that N

∑

a_iF_isatisfies (A2), we have

1

r_n

Trn+1xn+1−Trnxn,Trnxn−xn

+ 1

r_n+1

Trnxn−Trn+1xn+1,Trn+1xn+1−xn+1

≥0.

Which implies that

Trnxn−Trn+1xn+1,

Trn+1xn+1−xn+1

r_n+1 −

Trnxn−xn r_n

≥0.

It follows that

(3.8)

Trn+1xn+1−Trnxn,Trnxn−Trn+1xn+1+Trn+1xn+1−xn− rn

r_n+1(Trn+1xn+1−xn+1)

≥0.

From (3.8), we obtain

Trn+1xn+1−Trnxn

2

≤

Trn+1xn+1−Trnxn,Trn+1xn+1−xn− rn

r_n+1(Trn+1xn+1−xn+1)

=

Trn+1xn+1−Trnxn,xn+1−xn+ (1− rn

r_n+1

)(Trn+1xn+1−xn+1)

≤

Trn+1xn+1−Trnxn h

kx_n+1−xnk+|1−

rn

r_n+1|

×T_rn+1x_n+1−x_n+1 i

=T_rn+1x_n+1−T_rnx_n h

kx_n+1−xnk+ 1

r_n+1

|r_n+1−rn|

×T_rn+1x_n+1−xn+1

i

≤

T_rn+1x_n+1−T_rnx_n h

kxn+1−xnk+ 1

d|rn+1−rn|

×T_rn+1x_n+1−x_n+1 i

which yields

(3.9) kun+1−unk ≤ kxn+1−xnk+ 1

d|rn+1−rn| kun+1−xn+1k.

From (3.9), we have

(3.10) ku_n−u_n−1k ≤ kx_n−x_n−1k+1

d|rn−rn−1| kun−xnk.

By substituting (3.10) into (3.2), we have

kxn+1−xnk ≤βn

2M|αn−αn−1|+ 1−αn(1−η)kxn−xn−1k

+2M|βn−βn−1|+ (1−βn)kun−un−1k

≤βn

2M|αn−αn−1|+ 1−αn(1−η)

kx_n−x_n−1k

+2M|βn−βn−1|+ (1−βn) kxn−xn−1k+ 1

d|rn−rn−1| kun−xnk

= 1−βnαn(1−η)

kx_n−x_n−1k+2M|αn−αn−1|

+2M|βn−βn−1|+ (1−βn) 1

d|rn−rn−1| kun−xnk.

(3.11)

From (3.11), conditions (i),(v) and lemma 2.3, we obtain

(3.12) lim

n→∞

kx_n+1−xnk=0.

Step 3: We will show that lim n→∞k

u_n−x_nk = lim n→∞k

P_C(I−λA)un−xnk= lim n→∞k

Sx_n−x_nk=0. SinceT_rn is a firmly nonexpansive mapping, then we obtain

kz−Trnxnk

2

=kTrnz−Trnxnk

2

≤ hTrnz−Trnxn,z−xni

=1

2 kTrnxn−zk

2

+kx_n−zk2− kT_rnx_n−x_nk2 ,

which yields

By nonexpansiveness ofP_C(I−λA),(3.13) and definition ofxn, we have

kx_n+1−zk2=β_n α_nf(x_n) + (1−α_n)Sx_n−z

+ (1−βn) PC(I−λA)un−z

2

≤βn

α_n f(x_n)−z

+ (1−αn)(Sxn−z)

2

+ (1−βn)kPC(I−λA)un−zk2

≤βnαnkf(xn)−zk2+βn(1−αn)kxn−zk2+ (1−βn)kun−zk2

≤βnαnkf(xn)−zk2+βn(1−αn)kxn−zk2

+ (1−βn) kxn−zk2− kun−xnk2

≤βnαnkf(xn)−zk2+kxn−zk2−(1−βn)kun−xnk2,

which implies that

(1−βn)kun−xnk2≤βnαnkf(xn)−zk2+kxn−zk2− kxn+1−zk2

≤βnαnkf(xn)−zk2+ kxn−zk+kxn+1−zk

kxn+1−xnk. (3.14)

By (3.12),(3.14), conditions (i) and (ii), we have

(3.15) lim

n→∞k

u_n−x_nk=0.

Putw_n=αnf(xn) + (1−αn)Sxn. By definition ofxn, we have

kxn+1−zk2=kβnwn+ (1−βn)PC(I−λA)un−zk2

=β_n(w_n−z) + (1−β_n) P_C(I−λA)u_n−z

2

≤βnkwn−zk2+ (1−βn)kPC(I−λA)un−zk2−βn(1−βn)

=βnkαnf(xn) + (1−αn)Sxn−zk2+ (1−βn)kPC(I−λA)un−zk2

−βn(1−βn)kwn−PC(I−λA)unk2

≤βn αnkf(xn)−zk2+ (1−αn)kSxn−zk2+ (1−βn)kun−zk2

−βn(1−βn)kwn−PC(I−λA)unk2

=βnαnkf(xn)−zk2+βn(1−αn)kxn−zk2+ (1−βn)kun−zk2

−βn(1−βn)kwn−PC(I−λA)unk2

≤βnαnkf(xn)−zk2+βn(1−αn)kxn−zk2+ (1−βn)kxn−zk2

−βn(1−βn)kwn−PC(I−λA)unk2

=βnαnkf(xn)−zk2+ (1−βnαn)kxn−zk2−βn(1−βn)

× kwn−PC(I−λA)unk2

≤βnαnkf(xn)−zk2+kxn−zk2−βn(1−βn)kwn−PC(I−λA)unk2.

Which yields

βn(1−βn)kwn−PC(I−λA)unk2≤βnαnkf(xn)−zk2+kxn−zk2− kxn+1−zk2

≤βnαnkf(xn)−zk2

+ kx_n−zk+kx_n+1−zk

kx_n+1−x_nk.

(3.16)

By (3.12),(3.16), conditions (i) and (ii), we have

(3.17) lim

n→∞k

w_n−P_C(I−λA)unk=0.

By the definition ofx_n, we obtain

x_n+1−P_C(I−λA)un=βnwn−βnPC(I−λA)un

=βn wn−PC(I−λA)un

.

By (3.18), we have

kx_n−P_C(I−λA)xnk=kxn−xn+1+xn+1−PC(I−λA)un+PC(I−λA)un−PC(I−λA)xnk

≤ kx_n−x_n+1k+kx_n+1−P_C(I−λA)unk

+kP_C(I−λA)un−PC(I−λA)xnk

≤ kx_n−x_n+1k+βnkwn−PC(I−λA)unk+kun−xnk. Form (3.12),(3.15) and (3.17), we have

(3.19) lim

n→∞

kxn−PC(I−λA)xnk=0. Since

kxn−PC(I−λA)unk=kxn−PC(I−λA)xn+PC(I−λA)xn−PC(I−λA)unk

≤ kxn−PC(I−λA)xnk+kPC(I−λA)xn−PC(I−λA)unk

≤ kx_n−P_C(I−λA)xnk+kxn−unk.

From (3.15) and (3.19), we have

(3.20) lim

n→∞k

x_n−P_C(I−λA)unk=0. By the definition ofx_n, we obtain

x_n+1−x_n=βnαn f(xn)−xn

+βn(1−αn)(Sxn−xn)

+ (1−βn) PC(I−λA)un−xn

.

(3.21)

It follows that

βn(1−αn)kSxn−xnk ≤βnαnkf(xn)−xnk

+ (1−βn)kPC(I−λA)un−xnk+kxn+1−xnk. By (3.12),(3.20), conditions (i) and (ii), we have

(3.22) lim

n→∞

Step 4: We will show that lim sup n→∞

hf(z)−z,x_n−zi ≤0, wherez=P_F f(z).

To show this, choose a subsequence{x_nk}of{x_n}such that (3.23) lim sup

n→∞

hf(z)−z,x_n−zi=lim sup k→∞

hf(z)−z,x_nk−zi.

Without loss of generality, we can assume thatxnk*ω ask→∞, whereω ∈C.

From (3.15), we obtainu_nk*ω ask→∞.

Assume thatω∈/V I(C,A). SinceV I(C,A) =F PC(I−λA), we haveω 6=PC(I−λA)ω.

By nonexpansiveness ofP_C(I−λA),(3.19) and Opial’s condition, we have

lim inf k→∞

kxnk−ωk<lim inf

k→∞

kxnk−PC(I−λA)ωk

=lim inf k→∞

kx_nk−P_C(I−λA)xnk+PC(I−λA)xnk−PC(I−λA)ωk

≤lim inf k→∞

kx_nk−P_C(I−λA)xnkk

+lim inf k→∞

kP_C(I−λA)xnk−PC(I−λA)ωk

≤lim inf k→∞

kxnk−ωk.

This is a contradiction. Then we have

(3.24) ω∈V I(C,A).

Next, we will show thatω∈TN_i=1F(Ti). By Lemma 2.10, we haveF(S) =TN

i=1F(Ti). Assume that ω 6=Sω. Using Opial’s condition, (3.22), we obtain

lim inf k→∞

kx_nk−ωk<lim inf

k→∞

kx_nk−Sωk

=lim inf k→∞

kxnk−Sxnk+Sxnk−Sωk

≤lim inf k→∞

kx_nk−Sx_nkk+lim inf k→∞

kSx_nk−Sωk

≤lim inf k→∞

kx_nk−ωk.

This is a contradiction. Then we have

(3.25) ω ∈F(S) =

N

\

i=1

Next, we will show thatω∈TN_i=1EP(Fi). Since

N

∑

a_iF_i(u_n,y) + 1

r_nhy−un,un−xni ≥0,∀y∈Cand

N

∑

a_iF_isatisfies condition (A1)-(A4), we obtain

1

rn

hy−u_n,u_n−x_ni ≥

N

∑

a_iF_i(y,u_n),∀y∈C.

In particular, it follows that

(3.26)

y−u_nk,unk−xnk rnk

≥

N

∑

a_iF_i(y,u_nk),∀y∈C.

From (3.15),(3.26) and (A4), we have

(3.27)

N

∑

aiFi(y,ω)≤0,∀y∈C.

Put y_t :=ty+ (1−t)ω, for allt ∈(0,1], we haveyt ∈C. By using (A1),(A4) and (3.27), we have

0=

N

∑

a_iF_i(y_t,y_t)

=

N

∑

aiFi yt,ty+ (1−t)ω

≤t

N

∑

aiFi(yt,y) + (1−t) N

∑

aiFi(yt,ω)

≤t

N

∑

aiFi(yt,y).

It implies that

(3.28) 0≤

N

∑

aiFi ty+ (1−t)ω,y,

for allt ∈(0,1]andy∈C.

From (3.28), takingt→0+and using (A3), we can conclude that

0≤ lim t→0+

N

∑

a_iF_i ty+ (1−t)ω,y

≤

N

∑

Therefore,ω∈EP

N

∑

a_iF_i. By Lemma 2.6, we obtainEP

N

∑

a_iF_i=

N

\

i=1

EP(F_i). It follows that

(3.29) ω ∈

N

\

i=1

EP(Fi).

From (3.24),(3.25) and (3.29), we can deduce thatω ∈F.

Sincex_nk*ω ∈F and Lemma 2.1, we can conclude that

lim sup n→∞

hf(z)−z,x_n−zi=lim sup k→∞

hf(z)−z,x_nk−zi

=hf(z)−z,ω−zi

≤0,

(3.30)

wherez=P_Ff(z).

Step 5: Finally, we will show that the sequence{x_n}converges strongly toz=P_Ff(z).

By nonexpansive ofSandP_C(I−λA), we have

kx_n+1−zk2=

β_n α_nf(x_n) + (1−α_n)Sx_n

+ (1−βn)PC(I−λA)un−z

2

=β_nα_n f(x_n)−z

+βn(1−αn)(Sxn−z) + (1−βn) PC(I−λA)un−z

2

≤

β_n(1−α_n)(Sx_n−z) + (1−β_n) P_C(I−λA)u_n−z

2

+2βnαnhf(xn)−z,xn+1−zi

≤ βn(1−αn)kSxn−zk+ (1−βn)kPC(I−λA)un−zk

≤ (1−βnαn)kxn−zk

2

+2βnαnhf(xn)−f(z),xn+1−zi

+2βnαnhf(z)−z,xn+1−zi

≤ (1−βnαn)kxn−zk2+2βnαnkf(xn)−f(z)k kxn+1−zk

+2βnαnhf(z)−z,xn+1−zi

≤(1−βnαn)kxn−zk2+2βnαnηkxn−zk kxn+1−zk

+2βnαnhf(z)−z,xn+1−zi

≤(1−βnαn)kxn−zk2+βnαnηkxn−zk2+βnαnηkxn+1−zk2

+2βnαnhf(z)−z,xn+1−zi.

Which implies that

kx_n+1−zk2≤

1−βnαnη−βnαn+2βnαnη

1−βnαnη

kx_n−zk2

+ 2βnαn

1−βnαnη

hf(z)−z,xn+1−zi

=

1− βnαn

1−βnαnη

kx_n−zk2+ 2βnαnη

1−βnαnη

kx_n−zk2

+ 2βnαn

1−βnαnη

hf(z)−z,x_n+1−zi

=

1− βnαn

1−βnαnη

kx_n−zk2+ βnαn

1−βnαnη 2ηk

x_n−zk2

+2hf(z)−z,xn+1−zi

.

Applying the conditions (ii),(3.30) and Lemma 2.3, we have the sequence {x_n} converges strongly to z=P_F f(z). From (3.15), we obtain{u_n}converges strongly to z=P_Ff(z).This

completes the proof.

Corollary 3.2. Let C be a nonempty closed convex subset of a real Hilbert space H. Let F :

C×C→_Rbe a bifunction with satisfy A1)−A4), T_i:C→C beκi-strictly pseudo-contractive

mapping, for all i=1,2, ...,N and let A:C →H be α-inverse strongly monotone mapping

withF =EP(F)∩TN

i=1F(Ti)∩V I(C,A)6= /0. Let S be S-mapping generated by T1,T2, ...,TN

andα1,α2, ...,αN, whereαj= (α1j,α j 2,α

j

3)∈I×I×I,I= [0,1]withα j 1+α

j 2+α

j

α₁j,α₃j<1, for all i=1,2, ...,N−1,κ<α₁N≤1,κ≤α₃N<1,κ≤α₂j<1, for all j=1,2, ...,N,

whereκ=max{κi:i=1,2, ...,N}. Let the sequence{xn}generated by x1∈C and

(3.31)

   

  

F(u_n,y) + 1

r_nhy−un,un−xni ≥0,∀y∈C, x_n+1=βn αnf(xn) + (1−αn)Sxn

+ (1−βn)PC(I−λA)un,∀n≥1,

where{βn},{αn} ⊆[0,1]andλ ∈(0,2α). Suppose the following conditions hold:

(i) ∞

∑

αn=∞,lim n→∞αn

=0,

(ii) 0<a≤βn,rn≤b<1, for all n≥1,

(iii) f :C→C beη-contraction,

(iv) ∞

∑

|αn+1−αn|<∞,

∞

∑

|βn+1−βn|<∞,

∞

∑

|rn+1−rn|<∞.

Then{x_n}converges strongly to z=P_F f(z).

Proof. TakeF =F_i,∀i=1,2, ...,N.By Theorem 3.1, we obtain the desired conclusion.

Corollary 3.3. Let C be a nonempty closed convex subset of a real Hilbert space H. T_i:C→

C be κi-strictly pseudo-contractive mapping, for all i=1,2, ...,N and let A:C →H be α

-inverse strongly monotone mapping withF =EP(F)∩TN

i=1F(Ti)∩V I(C,A)6= /0. Let S be

S-mapping generated by T1,T2, ...,TN andα1,α2, ...,αN, whereαj= (α₁j,α₂j,α₃j)∈I×I×I,I=

[0,1] withα₁j+α₂j+α₃j =1 and κ <α₁j,α₃j<1, for all i=1,2, ...,N−1,κ <α₁N ≤1,κ ≤ α₃N<1,κ≤α₂j<1, for all j=1,2, ...,N, whereκ=max{κi:i=1,2, ...,N}. Let the sequence

{x_n}generated by x₁∈C and

(3.32) x_n+1=βn αnf(xn) + (1−αn)Sxn

+ (1−βn)PC(I−λA)xn,∀n≥1,

where{βn},{αn} ⊆[0,1]andλ ∈(0,2α). Suppose the following conditions hold:

(i) ∞

∑

αn=∞,lim

n→∞αn=0 ,

(ii) 0<a≤βn,rn≤b<1, for all n≥1,

(iv) ∞

∑

|αn+1−αn|<∞,

∞

∑

|βn+1−βn|<∞,

∞

∑

|r_n+1−rn|<∞.

Then{xn}converges strongly to z=PF f(z).

Proof. PutF_i=0,∀i=1,2, ...,N.Then we haveu_n=P_Cx_n=x_n,∀n∈_N. Therefore the conclu-sion of Corollary 3.3 can be obtained by Theorem 3.1.

4.

Application

In this section, we apply our main theorem to prove strong convergence theorems involving optimization problem.

Let us recall the standard constrained convex optimization problem as follows: (4.1) findx?∈Csuch thatg(x?) =min

x∈Cg(x),

whereg:C→_Ris a convex, Frechet differentiable function,C is closed-convex subset ofH. The set of all solutions of (4.1) is denoted byΩg.

The following lemmas is important to prove Theorem 4.2.

Lemma 4.1. ([23]) (Optimality condition) A necessary condition of optimality for a point x?∈C to be a solution of the minimization problem (4.1) is that x?solves the variational inequality

(4.2) h∇g(x?),x−x?i ≥0, ∀x∈C.

Equivalently, x?∈C solves the fixed point equation

x?=P_C x?−λ∇g(x?),

for every constantλ >0. if, in addition, g is convex, then the optimality condition (4.2) is also

sufficient.

all L≥0. Assume thatF =TN

i=1EP(Fi)∩TNi=1F(Ti)∩Ωg6= /0. Let S be S-mapping generated

by T₁,T₂, ...,T_Nandα1,α2, ...,αN, whereαj= (α₁j,α₂j,α₃j)∈I×I×I,I= [0,1]withα₁j+α₂j+

α₃j=1 andκ <α₁j,α₃j <1, for all i=1,2, ...,N−1,κ <α₁N ≤1,κ ≤α₃N <1,κ ≤α₂j<1,

for all j=1,2, ...,N, whereκ=max{κi:i=1,2, ...,N}. Let the sequence{xn}generated by

x₁∈C and

(4.3)

   

  

N

∑

a_iF_i(u_n,y) + 1

r_nhy−un,un−xni ≥0,∀y∈C, xn+1=βn αnf(xn) + (1−αn)Sxn

+ (1−βn)PC(I−λ∇g)un,∀n≥1,

where{βn},{αn} ⊆[0,1]andλ ∈(0,2_L). Suppose the following conditions hold:

(i) ∞

∑

αn=∞,lim

n→∞αn

=0,

(ii) 0<a≤βn,rn≤b<1, for all n≥1,

(iii) f :C→C beη-contraction, (iv)

N

∑

a_i=1, where a_i>0, for all i=1,2, ...,N,

(v) ∞

∑

|αn+1−αn|<∞,

∞

∑

|βn+1−βn|<∞,

∞

∑

|r_n+1−r_n|<∞.

Then{x_n}converges strongly to z=P_F f(z).

Proof. The conclusion of Theorem 4.2 can be obtained from Theorem 3.1 and Lemma 4.1.

5.

Example and Numerical Results

In this section, two examples are given to support Theorem 3.1 and Theorem 4.2, repectively.

Example 5.1. Let _R be the set of real numbers and let the mapping A: _R→_R defined by

Ax= 2x₃,∀x∈_R. For alli=1,2, ...,N, let the mappingT_i:_R→_Rdefined by

T_ix= _6i6i₊₁x, ∀x∈_R

and letF_i:_R×_R→_Rdefined by

F_i(x,y) =i(−7x2+xy+6y2),∀x,y∈_R. Furthermore, leta_i= ₇6i+

1

N

∑

aiFi(x,y) = N

∑

6

7i+ 1

N7N

i(−7x2+xy+6y2) =E(−7x2+xy+6y2),

where E =

N

∑

6

7i + 1

N7N

i, it is easy to check that ∑Ni=1aiFi satisfies all the conditions of Theorem 3.1. By the definition ofFi, we have

0≤

N

∑

aiFi(un,y) + 1

r_nhy−un,un−xni

=E(−7x2+xy+6y2) + 1

rn

(y−u_n)(u_n−x_n)

=E(−7x2+xy+6y2) + 1

rn

(yu_n−yx_n−u2_n−u_nx_n)

⇔

0≤Er_n(−7x2+xy+6y2) + (yu_n−yx_n−u2_n−u_nx_n) =6Er_ny2+Eu_nr_ny−7Eu2_nr_n+yu_n−yx_n−u2_n−u_nx_n

=6Er_ny2+ (u_n−x_n+Eu_nr_n)y+ (−7Eu2_nr_n−u2_n−u_nx_n).

LetG(y) =6Erny2+ un(1+Ern)−xn

y−7Eu2_nrn−un2−unxn. G(y)is a quadratic function of

ywith coefficienta=6Er_n,b=u_n(1+Er_n)−x_nandc=−7Eu2_nr_n−u2_n−u_nx_n. Determine the discriminant∆ofGas follows:

∆=b2−4ac

= u_n(1+Er_n)−x_n2−4(6Er_n)(−7Eu2_nr_n−u2_n−u_nx_n)

=u2_n(1+Er_n)2−2u_nx_n(1+Er_n) +x_n2+168E2u2_nr2_n+24Eu2_nr_n−24Eu_nx_nr_n

=u2_n+2Eu2_nrn+E2u2nrn2−2unxn−2Eunxnrn+x2n+168E2u2nr2n+24Eu2nrn

−24Eunxnrn

=u2_n+26Eu2_nr_n+169E2u2_nr_n2−2u_nx_n−26Eu_nx_nr_n+x2_n

=(u_n+13Eu_nr_n)2−2x_n(u_n+13Eu_nr_n) +x2_n

We know thatG(y)≥0,∀y∈_R. If it has at most one solution inR, then∆≤0, so we obtain

(5.1) u_n= xn

1+13∑N_i=1

6 7i+

1 N7N

irn

, f or all n∈_N.

For every j =1,2, ...,N, let α₁j = ₂1_j, α₂j = 3j−1_16j , α₃j = 13₁₆j−7_j . Thenαj=

1 2j,

3j−1 16j ,

13j−7 16j

, for all j=1,2, ...,N. LetS-mapping generated byT1,T2, ...,TN and α1,α2, ...,αN. From the definitionT_i,AandF_i, we have

{0}=

N

\

i=1

EP(Fi)∩ N

\

i=1

F(Ti)∩V I(C,A).

Putαn= _3n1,βn= 4n_17n+2,rn= _2nn₊₁, f(x) = 3x₅ andλ =1,∀n∈N. From (5.1) we rewrite (3.1) as follows:

x_n+1=

4n+2

17n

1

3nf(xn) +

1− 1

3n

Sx_n

+1−4n+2

17n

(I−A) xn

1+13 N

∑

6

7i+ 1

N7N

ir_n

,∀n≥1.

(5.2)

It is clear that the sequence{αn},{βn}and{rn}satisfy all the conditions of Theorem 3.1. From Theorem 3.1, we can conclude that the sequence{x_n}and{u_n}converges strongly to 0.

Table 1 shows that values of sequences{x_n} and{u_n}, wherex₁=−5 andx₁=5 and n=

x1=−5 x1=5

n un xn un xn

1 -0.825688 -5.000000 0.825688 5.000000

2 -0.241546 -1.706927 0.241546 1.706927

3 -0.070026 -0.525198 0.070026 0.525198

4 -0.019977 -0.154636 0.019977 0.154636

5 -0.005630 -0.044447 0.005630 0.044447 .

.

. ... ... ... ...

8 -0.000120 -0.000980 0.000120 0.000980 .

. .

. . .

. . .

. . .

. . .

11 -0.000002 -0.000020 0.000002 0.000020

12 -0.000001 -0.000006 0.000001 0.000006

13 -0.000000 -0.000002 0.000000 0.000002

14 -0.000000 -0.000000 0.000000 0.000000

TABLE1. The values of{u_n}and{x_n}wheren=14.

(A)x1=−5 (B)x1=5

FIGURE 1. The convergence comparison of the sequences{x_n} and{u_n} with different initial valuesx₁.

Example5.2. In this example, we consider the same mappings and parameters as in Example 5.1 except the following mappingg:_R→_Rbe defined bygx=2x2+1. It is clear that

{0}=

N

\

i=1

EP(F_i)∩

N

\

i=1

Putλ = 1₈. From (5.1), we rewrite (4.3) as follows:

x_n+1=

4n+2

17n

1

3nf(xn) +

1− 1

3n

Sx_n

+1−4n+2

17n

(I−1

8∇g)

xn

1+13 N

∑

6

7i+ 1

N7N

ir_n

,∀n≥1.

(5.3)

It is clear that the sequence{αn},{βn}and{rn}satisfy all the conditions of Theorem 4.2. From Theorem 4.2, we can conclude that the sequence{x_n}and{u_n}converges strongly to 0.

Table 2 shows that values of sequences{x_n} and{u_n}, wherex₁=−5 andx₁=5 and n=

N=14.

x1=−5 x1=5

n un xn un xn

1 -0.825688 -5.000000 0.825688 5.000000

2 -0.254147 -1.795972 0.254147 1.795972

3 -0.077666 -0.582496 0.077666 0.582496

4 -0.023370 -0.180898 0.023370 0.180898

5 -0.006949 -0.054859 0.006949 0.054859 . . . . . . . . . . . . . . .

8 -0.000175 -0.001422 0.000175 0.001422 . . . . . . . . . . . . . . .

11 -0.000004 -0.000035 0.000004 0.000035

12 -0.000001 -0.000010 0.000001 0.000010

13 -0.000000 -0.000003 0.000000 0.000003

14 -0.000000 -0.000001 0.000000 0.000001

(A)x1=−5 (B)x1=5

FIGURE 2. The convergence comparison of the sequences{x_n} and{u_n} with different initial valuesx₁.

Conclusion

(1) Table 1 and Figure 1 show that{x_n}and{u_n}converges to 0, where{0} ∈TN

i=1EP(Fi)∩

TN

i=1F(Ti)∩V I(C,A). The convergence of {xn}and{un}of Example 5.1 can be guar-anteed by Theorem 3.1.

(2) Table 2 and Figure 2 show that{xn}and{un}converges to 0, where{0} ∈TNi=1EP(Fi)∩

TN

i=1F(Ti)∩Ωg. The convergence of{xn}and{un}of Example 5.2 can be guaranteed by Theorem 4.2.

(3) From these Example, we obtain that the sequence{x_n}in Example 5.1 converges faster than the sequence{x_n}in Example 5.2.

Conflict of Interests

The authors declare that there is no conflict of interests.

AcknowledgementsThis paper was supported by the the Research and Innovation Services of King Mongkut’s Institute of Technology Ladkrabang.

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