An Explicit Viscosity Iterative Algorithm for Finding Fixed Points of Two Noncommutative Nonexpansive Mappings

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Vol. 11, No. 1 (2016), pp 69-83 DOI: 10.7508/ijmsi.2016.01.007

An Explicit Viscosity Iterative Algorithm for Finding Fixed

Points of Two Noncommutative Nonexpansive Mappings

H. R. Sahebi, A. Razani∗

Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran, Iran.

E-mail: [email protected] E-mail: [email protected]

Abstract. We suggest an explicit viscosity iterative algorithm for

find-ing a common element in the set of solutions of the general equilibrium problem system (GEPS) and the set of all common fixed points of two noncommuting nonexpansive self mappings in the real Hilbert space.

Keywords: General equilibrium problems, Strongly positive linear bounded operator,α−Inverse strongly monotone mapping, Fixed point, Hilbert space.

2000 Mathematics subject classification: 47H09, 47H10, 47J20. 1. Introduction

LetH be a real Hilbert space with inner producth., .iand norm||.||. Recall that a mappingT with domainD(T) and rangeR(T) inHis called nonexpan-sive iff for allx, y∈D(T),

kT x−T yk ≤ kx−yk.

F(T) denotes the set of fixed points of T. Moreover, H satisfies the Opial’s condition [6], if for any sequence{xn} withxn ⇀ x, the inequality

lim inf

n→∞kxn−xk<lim infn→∞kxn−yk, ∗_{Corresponding Author}

Received 06 June 2014; Accepted 17 January 2015 c

2016 Academic Center for Education, Culture and Research TMU 69

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holds for everyy∈H withx6=y.

Recall thatf is said to be weakly contractive [2] iff for allx, y∈D(T),

kf(x)−f(y)k ≤ kx−yk −φ(kx−yk),

for some φ : [0,∞) → [0,∞) is a continuous and strictly increasing function such that φ is positive on (0,∞) and φ(0) = 0. A mapping A is a strongly positive linear bounded operator on H if there exists a constant ¯γ > 0 such that

hAx, xi ≥γ¯kxk2_, _{for all}_x_∈_H.

Moreover,B :C →H is calledα−inverse strongly monotone if there exists a positive real numberα >0 such that for all x, y∈C

hBx−By, x−yi ≥αkBx−Byk2.

Let C be a nonempty closed convex subset of H. A : H →H be an inverse strongly monotone mapping andF :C×C→Rbe a bifunction. The general equilibrium problem is to find ˜x∈C such that for ally∈C,

F(˜x, y) +hAx, y−xi ≥0.

There are several other problems, for example, the complementarity problem, fixed point problem and optimization problem, which can also be written in the form of an EP. In other words, the general equilibrium problem system (GEPS) is an unifying model for several problems arising in physics, engineer-ing, science, optimization, economics, etc [1, 4].

To study the generalized equilibrium problem, we assume that the bifunction

F satisfies the following conditions: (A1) F(x, x) = 0, for allx∈C;

(A2) F is monotone, i.e.,F(x, y) +F(y, x)≤0 for allx, y∈C; (A3) for eachx, y, z∈C, lim supt→0−F(tz+ (1−t)x, y)≤F(x, y);

(A4) for eachx∈C y7→F(x, y) is convex and weakly lower semi-continuous. Recently, Yao and Chen [10] introduced a new iteration for two averaged self mappingsS andT on a closed convex subsetC as follows

(

x0=x∈C;

xn+1=αnx+ (1−αn)_(n+1)(n+2)2 P n i=0

Pn−i

j=0((ST)

j_Si−j_∨₍_ST₎i_Tj−i₎_x n,

wheren≥0 and

(ST)j_Si−j_∨₍_ST₎i_Tj−i₌

(ST)j_Si−j _if_i_≥_j

(ST)i_Tj−i _if_{i < j.} (1.1) By improving this idea, Jankaew et al. [5] considered the following iteration:

xn+1=αnf(xn)+βnxn+γn

2 (n+ 1)(n+ 2)

n

X

i=0 n−i

X

j=0

((ST)j_Si−j_∨₍_ST₎i_Tj−i₎_x n,

(1.2)

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where {αn}, {βn}, {γn} ⊂(0,1), αn+βn+γn = 1, f is a contraction map-ping on C. They proved that the iteration process (1.2) converges strongly to common fixed point of the mappingS andT which solves some variational inequality.

A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mappings on a real Hilbert spaceH:

min 1

2hAx, xi −h(x)

whereAis strongly positive linear bounded operator andhis a potential func-tion forγf, i. e., h′₍_x_{) =} _γf, _{for all}_x_∈ _H_{. In this paper, we consider and} analyze an iterative scheme for finding a common element of the set of solutions of the general equilibrium problem system (GEPS) and the set of all common fixed points of two noncommutative nonexpansive self mapping in the frame-work of a real Hilbert space. The results in this paper generalize and improve some well known results in Jankaew et al.[5] and others.

In order to prove our main results, we need the following lemmas.

Lemma 1.1. [3] Let C be a nonempty closed convex subset of H and F :

C×C → Rbe a bifunction satisfying (A1)−(A4). Then for any r > 0 and

x∈H there existsz∈C such that

F(z, y) +1

rhy−z, z−xi ≥0,∀y∈C.

Further, define

Trx={z∈C:F(z, y) + 1

rhy−z, z−xi ≥0,∀y∈C}

for allr >0 andx∈H. Then

(a) Tr is single-valued;

(b) Tr is firmly nonexpansive, i.e., for anyx, y∈H

kTrx−Tryk2≤ hTrx−Try, x−yi; (c) F(Tr) =GEP(F);

(d) kTsx−Trxk ≤ s−rs kTsx−xk; (e) GEP(F)is closed and convex.

Remark 1.2. It is clear that for anyx∈H andr >0, by Lemma 1.1(a), there existsz∈H such that

F(z, y) +1

rhy−z, z−xi ≥0,∀y∈H. (1.3)

Replacingxwithx−rψxin (1.3), we obtain

F(z, y) +hψx, y−zi+1

rhy−z, z−xi ≥0,∀y∈H.

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Lemma 1.3. [9]Assume{an} is a sequence of nonnegative numbers such that

an+1≤(1−αn)an+δn,

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

that

(i) lim

n→∞αn= 0, ∞

X

n=1

αn =∞,

(ii) lim sup n→∞

δn

αn

≤0 or

∞

X

n=1

|δn|<∞,

Then lim

n→∞an= 0.

Lemma 1.4. [5]LetCbe a nonempty bounded closed convex subset of a Hilbert spaceH, and letS, T be two nonexpansive mappings ofC into itself such that

F(ST) =F(S)T

F(T)6=∅. Let {xn} be a sequence defined as follows:

xn+1 = αnf(xn) +βnxn +γn

2 (n+ 1)(n+ 2)

n

X

i=0 n−i

X

j=0

((ST)j_Si−j_∨₍_ST₎i_Tj−i₎_x n,

and put

Λn= 2

(n+ 1)(n+ 2) n

X

i=0 n−i

X

j=0

((ST)j_Si−j_∨₍_ST₎i_Tj−i₎_x n.

Then,

lim sup n→∞ x∈C

kΛn(x)−STΛn(x)k= 0.

2. Explicit Viscosity Iterative Algorithm

In this section, we introduce an explicit viscosity iterative algorithm for finding a common element of the set of solution for an equilibrium problem system involving a bifunction defined on a closed convex subset and the set of fixed points for two noncommutative nonexpansive mappings.

Theorem 2.1. Let x0 ∈C, {un,i} ⊂C and C be a nonempty closed convex

subset of a real Hilbert space H, F1, F2, . . . , Fk be bifunctions from C ×C

toR satisfying (A1)−(A4),Ψ1,Ψ2, . . . ,Ψk be µi−inverse strongly monotone

mapping onC,f be a weakly contractive mapping with a functionφ onH,A

be a strongly positive linear bounded operator with coefficientγ¯ such thatγ¯ ≤ kAk ≤1,B be strongly positive linear bounded operator on H with coefficient

¯

β∈(0,1]such thatkBk= ¯β,S,T be nonexpansive mappings onC, such that

F(ST) =F(T S) =F(T)T

F(S)6=∅. Let {xn}be a sequence generated in the

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following manner:                                 

F1(un,1, y) +hΨ1xn, y−un,1i+r1nhy−un,1, un,1−xni ≥0, for ally∈C F2(un,2, y) +hΨ2xn, y−un,2i+_r1_nhy−un,2, un,2−xni ≥0, for ally∈C

.. .

Fk(un,k, y) +hΨkxn, y−un,ki+_r1_nhy−un,k, un,k−xni ≥0, for ally∈C

ωn= 1_kP k i=1un,i,

Λn= _(n+1)(n+2)2 Pn

i=0

Pn−i

j=0((ST)

j_Si−j_∨₍_ST₎i_Tj−i₎_ω n,

xn+1=αnγf(xn) +βnBxn+ ((1−εn)I−βnB−αnA)Λn.

where{αn} ⊂(0,1),{βn} and{εn} are the sequences in[0,1) such that εn ≤

αn and{rn} ⊂(0,∞)is a real sequence satisfying the following conditions: (C1) lim

n→∞αn= 0, ∞

X

n=1

αn=∞, (C2) lim

n→∞βn= 0,

P∞

n=1|βn+1−βn|<∞, (C3) P∞

n=1|rn+1−rn|<∞andlim inf

n→∞rn>0and0< b < rn < a <2µi for 1≤i≤k,

(C4) P∞

n=1|εn+1−εn|<∞, (C5) lim

n→∞

εn

αn = 0.

Then

(i) the sequence {xn}is bounded. (ii) lim

n→∞kxn+1−xnk= 0. (iii) lim

n→∞kΨixn−Ψix ∗_k

= 0. fori∈ {1,2, . . . , k}.

(iv) lim

n→∞kxn−Λnk= 0.

Proof. (i)Without loss of generality, we assume thatαn<(1−εn−βnkBk)kAk−1. SinceA,B are two strongly positive bounded linear operator onH, we have

kAk= sup{|hAx, xi|:x∈H,kxk= 1},

kBk= sup{|hBx, xi|:x∈H,kxk= 1}.

Also, (1−εn)I−βnB−αnAis positive. Indeed,

h((1−εn)I−βnB−αnA)x, xi = (1−εn)hx, xi −βnhBx, xi −αnhAx, xi

≥ 1−εn−βnkBk −αnkAk>0.

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Notice that

k(1−εn)I − βnB−αnAk

= sup{h((1−εn)I−βnB−αnA)x, xi:x∈H,kxk= 1}

= sup{(1−εn)hx, xi −βnhBx, xi −αnhAx, xi:x∈H,kxk= 1}

≤ 1−εn−βnβ¯−αn¯γ

≤ 1−βnβ¯−αnγ.¯

LetQ=PF(ST)TGEP(Fi,Ψi). It is clear that Q(I−A+γf) is a contraction.

Hence, there exists a unique elementz∈H such thatz=Q(I−A+γf)z. Let x∗ _∈ Tk

i=1F(ST)

T

GEP(Fi,Ψi). For any i = 1,2, . . . , k, I−rnΨi is a nonexpansive mapping andkun,i−x∗k ≤ kxn−x∗k. Alsokωn−x∗k ≤ kxn−x∗k. Thus

kxn+1−x∗k = kαnγf(xn) +βnBxn+ ((1−εn)I−βnB−αnA)Λn−x∗k

≤ αnkγf(xn)−Ax∗k+βnkBkkxn−x∗k

+k((1−εn)I−βnB−αnA)kkΛn−x∗k+εnkx∗k

≤ αnγkf(xn)−f(x∗)k+αnkγf(x∗)−Ax∗k+βnβ¯kxn−x∗k +(1−βnβ¯−αn¯γ)kxn−x∗k+αnkx∗k

≤ αnγkxn−x∗k −φ(kxn−x∗) +αnkγf(x∗)−Ax∗k +βnβ¯kxn−x∗k+ (1−βnβ¯−αn¯γ)kxn−x∗k+αnkx∗k

≤ (1−(¯γ−γ)αn)kxn−x∗k+αn(kγf(x∗)−Ax∗k+kx∗k)

≤ max{kxn−x∗k,

kγf(x∗₎₋_Ax∗_k ¯

γ−γ }.

By induction

kxn−x∗k ≤max{kx1−x∗k,

kγf(x∗₎₋_Ax∗_k ¯

γ−γ }.

and the sequence{xn}is bounded and also{f(xn)},{ωn}and{Λn}are bounded.

(ii)Note that un,i can be written as un,i =Trn,i(xn−rnψixn). It follows

from Lemma 1.1 that

kun+1,i−un,ik ≤ kxn+1−xnk+ 2Mi|rn+1−rn|, (2.1) where

Mi= max{sup{

kTrn+1,i(I−rnΨi)xn−Trn,i(I−rnΨi)xnk rn+1

},sup{kΨixnk}}.

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LetM = 1 k

k

X

i=1

2Mi<∞. Next, we estimate kωn+1−ωnk,

kωn+1−ωnk ≤ 1

k

X

i=1

kun+1,i−un,ik ≤ kxn+1−xnk+M|rn+1−rn|. (2.2)

Now, we prove that lim

n→∞kxn+1−xnk= 0. We observe that

kxn+2−xn+1k = kαn+1γf(xn+1) +βn+1Bxn+1

+((1−εn+1)I−βn+1B−αn+1A)Λn+1−αn+1γf(xn)

−βnBxn−((1−εn)I−βnB−αnA)Λnk = k((1−εn+1)I−βn+1B−αn+1A)(Λn+1−Λn)

+{(εn−εn+1)Λn+ (βn−βn+1)BΛn+ (αn−αn+1)AΛn} +αn+1γ(f(xn+1)−f(xn)) + (αn+1−αn)γf(xn)

+βn+1B(xn+1−xn) + (βn+1−βn)Bxnk

+αn+1γkxn+1−xnk −αn+1γφ(kxn+1−xnk) +|αn+1−αn|γkf(xn)k+βn+1β¯kxn+1−xnk +|βn+1−βn|β¯kxnk

+(αn+1γ+βn+1β¯)kxn+1−xnk −αn+1γφ(kxn+1−xnk)

whereK= sup{max{kΛnk+kxnk, γkf(xn)k+kAΛnk},∀n≥0}<∞. Let ∆n =kβn−βn+1kβK¯ +K|αn−αn+1|+|εn−εn+1|kΛnk, then

kxn+2−xn+1k ≤ (1−βn+1β¯−αn+1¯γ)kΛn+1−Λnk +(αn+1γ+βn+1β¯)kxn+1−xnk

−αn+1γφ(kxn+1−xnk) + ∆n,

(2.3)

From [5], we conclude

kΛn+1−Λnk ≤ kωn+1−ωnk+ 4

n+ 3kωn+1−x ∗

k+ 4

n+ 3kx ∗

k. (2.4)

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Substituting (2.2) and (2.4) into (2.3), thus

kxn+2−xn+1k ≤ (1−βn+1β¯−αn+1¯γ){kxn+1−xnk+M|rn+1−rn| + 4

n+ 3kωn+1−x ∗_k

+ 4

n+ 3kx ∗_k}

+(αn+1γ+βn+1β¯)kxn+1−xnk −αn+1γφ(kxn+1−xnk) +∆n,

for some positive constantM. It follows that

kxn+2−xn+1k ≤ (1−αn+1(¯γ−γρ))kxn+1−xnk +M(1−βn+1β¯−αn+1¯γ)|rn+1−rn| +(1−βn+1β¯−αn+1¯γ) 4

n+ 3kωn+1−x ∗

k

+(1−βn+1β¯−αn+1¯γ) 4

n+ 3kx ∗_k

+ ∆n.

By Lemma 1.3,

lim

n→∞kxn+1−xnk= 0. (2.5)

(iii)For anyi∈ {1,2, . . . , k},

kun,i−x ∗

k2 _{≤ k}₍_x n−x

∗

)−rn(Ψixn−Ψix ∗

)k2

= kxn−x∗k2−2rnhxn−x∗,Ψixn−Ψix∗i+rn2kΨixn−Ψix∗k2

≤ kxn−x∗k2−rn(2µi−rn)kΨixn−Ψix∗k2,

thus

kωn−x∗k2= k k

X

i=1 1

k(un,i−x

∗ )k2

≤ 1

k k

X

i=1

kun,i−x∗k2

≤ kxn−x∗k2−_k1 k

X

i=1

rn(2µi−rn)kΨixn−Ψix∗)k2.

(2.6)

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From (2.6),

kxn+1−x∗k2 = kαn(γf(xn)−Ax∗) +βnB(xn−x∗)

+((1−εn)I−βnB−αnA)(Λn−x∗)−εnx∗k2

≤ kαn(γf(xn)−Ax∗) +βnB(xn−x∗)

+((1−εn)I−βnB−αnA)(ωn−x∗) +εnx∗k2

≤ αnkγf(xn)−Ax∗k2+βnkBk2kxn−x∗k2 +(1−βnβ¯−αn¯γ)kΛn−x∗k2+ε2nkx

∗

k2

≤ αnkγf(xn)−Ax∗k2+βnβ¯2kxn−x∗k2 +(1−βnβ¯−αn¯γ)kωn−x∗k2+ε2nkx

∗_k2

≤ αnkγf(xn)−Ax∗k2+βnβ¯kxn−x∗k2 +(1−βnβ¯−αnγ¯){kxn−x∗k2

−1 k

k

X

i=1

rn(2µi−rn)kΨixn−Ψix ∗

)k2_}₊_ε2 nkx

∗

k2

≤ αnkγf(xn)−Ax∗k2+kxn−x∗k2

−(1−βnβ¯−αn¯γ) 1

k

X

i=1

rn(2µi−rn)kΨixn−Ψix∗)k2 +ε2nkx

∗

k2

and hence

(1 −βnβ¯−αnγ¯)1_k k

X

i=1

b(2µi−a)kΨixn−Ψix∗k2

≤αnkγf(xn)−Ax∗k2+kxn−x∗k2− kxn+1−x∗k2+ε2nkx ∗_k2

≤αnkγf(xn)−Ax∗k2+kxn+1−xnk(kxn+1−x∗k − kxn−x∗k) +ε2nkx

∗_k2_.

Since αn → 0 and εn ≤ αn then εn → 0 as n → ∞. The inequality (2.5) implies that

lim

n→∞kΨixn−Ψix ∗

k= 0,∀i= 1,2, . . . , k. (2.7)

(iv)By Lemma 1.1

kun,i−x∗k2≤ kxn−x∗k2− kxn−un,ik2+ 2rnkxn−un,ikkΨixn−Ψix∗k(2.8) and hence

kωn−x∗k2= kP k i=1

1

k(un,i−x ∗₎_k2

≤ 1

k

Pk

i=1kun,i−x∗k2

≤ kxn−x∗k2−1_kPk_i=1kun,i−xnk2 +1

k

Pk

i=12rnkxn−un,ikkΨixn−Ψix∗k.

(2.9)

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From (2.9),

kxn+1−x∗k2 ≤ αnkγf(xn)−Ax∗k2+βnβ¯kxn−x∗k2 +(1−βnβ¯−αn¯γ)kωn−x

∗

k2₊_ε2 nkx

∗

k2

≤ αnkγf(xn)−Ax∗k2+βnβ¯kxn−x∗k2 +(1−βnβ¯−αn¯γ){kxn−x∗k2−

1

k

X

i=1

kun,i−xnk2

+1

k

X

i=1

2rnkxn−un,ikkΨixn−Ψix∗k}+ε2nkx ∗_k2

thus

(1−βnβ¯−αn¯γ) 1

k

X

i=1

kun,i−xnk2

≤ αnkγf(xn)−Ax∗k2+kxn−x∗k2− kxn+1−x∗k2 +(1−βnβ¯−αn¯γ)

1

k

X

i=1

2rnkxn−un,ikkΨixn−Ψix∗k+ε2nkx ∗_k2

≤ αnkγf(xn)−Ax∗k2+kxn+1−xnk(kxn+1−x∗k − kxn−x∗k) +(1−βnβ¯−αn¯γ)

1

k

X

i=1

2rnkxn−un,ikkΨixn−Ψix∗k+ε2nkx ∗

k2

From the condition (C1), (2.5) and (2.7), we get lim

n→∞kun,i−xnk= 0. (2.10) It is easy to prove

lim

n→∞kωn−xnk= 0. (2.11)

By definition of the sequence{xn}, we obtain

kxn−Λnk ≤ kxn+1−xnk+kxn+1−Λnk

≤ kxn+1−xnk+kαnγf(xn) +βnBxn +((1−εn)I−βnB−αnA)Λn−Λnk

≤ kxn+1−xnk+αnkγf(xn)−AΛnk+βnβ¯kxn−Λnk+εnkΛnk. Then

kxn−Λnk ≤ 1 1−βnβ¯

kxn+1−xnk+

αn 1−βnβ¯n

kγf(xn)−AΛnk + εn

1−βnβ¯

kΛnk.

Thanks to the conditions (C1)−(C2) and (2.5), we conclude that lim

n→∞kxn−Λnk= 0. (2.12)

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Also

kωn−Λnk ≤ kωn−xnk+kxn−Λnk, hence

lim

n→∞kωn−Λnk= 0. (2.13)

Theorem 2.2. Suppose all assumptions of Theorem 2.1 are holds. Then the sequence {xn} is strongly convergent to a pointx¯, where

¯

x∈Tk

i=1F(ST)

T_GEP₍_F

i,Ψi) solves the variational inequality

h(A−γf)¯x,x¯−xi ≤0.

Equivalently,x¯=PTk i=1F(ST)

T_GEP

(Fi,Ψi)(I−A+γf)(¯x).

Proof. Observe thatPTk i=1F(ST)

T

GEP(Fi,Ψi)(I−A+γf) is a contraction ofH

into itself. SinceH is complete, there exists a unique element ¯x∈H such that ¯

x=PTk i=1F(ST)

T_GEP

(Fi,Ψi)(I−A+γf)(¯x).

Next, we prove

lim sup n→∞

h(A−γf)¯x,¯x−Λni ≤0 Let ˜x=PTk

i=1F(ST)

T

GEP(Fi,Ψi)x1, set

E={y¯∈H :ky¯−x˜k ≤ kx1−x˜k+

kγf(˜x)−A˜xk

¯

γ−γρ } \

C.

It is clear,E is nonempty closed bounded convex subset ofC and S(E)⊂E,

T(E)⊂E. Without loss of generality, we may assumeS andT are mappings ofE into itself. Since{Λn} ⊂E is bounded, there is a subsequence{Λnj} of {Λn}such that

lim sup n→∞

h(A−γf)¯x,x¯−Λni= lim

j→∞h(A−γf)¯x,x¯−Λnji. (2.14) As{Λnj}is also bounded, there exists a subsequence{Λnjl}of{Λnj}such that

Λn_jl ⇀ ξ. Without loss of generality, let Λnj ⇀ ξ. Now, we prove the following

items:

(i): ξ∈F(ST) =F(T)T F(S).

Assumeξ /∈F(ST). By Lemma 1.4 and Opial’s condition, lim inf

j→∞ kΛnj−ξk < lim infj→∞ kΛnj −ST(ξ)k

≤ lim inf

j→∞ (kΛnj −ST(Λnj)k+kST(Λnj)−ST(ξ)k)

≤ lim inf

j→∞ kΛnj −ξk. That is a contradiction. Henceξ=F(ST)ξ.

(ii): By the same argument as in the proof of [7, Theorem 3.2], we conclude that ξ ∈ GEP(Fi,Ψi), for all i = 1,2, . . . , k. Then ξ ∈ Tk_i=1GEP(Fi,Ψi). Now, in view of (2.14), we see

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lim sup n→∞

h(A−γf)¯x,x¯−Λni=h(A−γf)¯x,x¯−ξi ≤0,

Finally, we show that{xn} is strongly convergent to ¯x. As a matter of fact,

k xn+1−x¯k2=kαnγf(xn) +βnBxn+ ((1−εn)I−βnB−αnA)Λn−x¯k2 = kαn(γf(xn)−Ax¯) +βnB(xn−x¯)

+((1−εn)I−βnB−αnA)(Λn−x¯)−εnx¯k2

≤ kαn(γf(xn)−Ax¯) +βnB(xn−x¯)

+((1−εn)I−βnB−αnA)(Λn−x¯) +εnx¯k2 = kαn(γf(xn)−Ax¯) +βnB(xn−x¯)

+((1−εn)I−βnB−αnA)(Λn−x¯)k2 +2εnkαn(γf(xn)−Ax¯) +βnB(xn−x¯)

+((1−εn)I−βnB−αnA)(Λn−x¯)kkx¯k+ε2nkx¯k2 = kβnB(xn−x¯) + ((1−εn)I−βnB−αnA)(Λn−x¯)k2

+2αnh((1−εn)I−βnB−αnA)(Λn−x¯), γf(xn)−Ax¯i +2εnkαn(γf(xn)−Ax¯) +βnB(xn−x¯)

+((1−εn)I−βnB−αnA)(Λn−x¯)kkx¯k

+ε2nkx¯k2+α2nkγf(xn)−Ax¯k2+ 2αnβnhB(xn−x¯), γf(xn)−Ax¯i

≤ (βnkBkkxn−x¯k+k(1−εn)I−βnB−αnAkkΛn−x¯k)2

+2αnγhΛn−x, f¯ (xn)−f(¯x)i+ 2εnkαn(γf(xn)−Ax¯) +βnB(xn−x¯) +((1−εn)I−βnB−αnA)(Λn−x¯)kkx¯k+ε2nkx¯k2+α2nkγf(xn)−A¯xk2 +2αnβnhB(xn−x¯), γf(xn)−Ax¯i+ 2αnhΛn−x, γf¯ (xn)−Ax¯i

−2αnh(εnI+βnB+αnA)(Λn−x¯), γf(xn)−Ax¯i

≤ (βnβ¯kxn−x¯k+ (1−βnβ¯−αnγ¯)kxn−x¯k)2+ 2αnγkxn−x¯k2 +2εnkαn(γf(xn)−Ax¯) +βnB(xn−x¯)

+((1−εn)I−βnB−αnA)(Λn−x¯)kkx¯k+ε2nkx¯k 2

+α2nkγf(xn)−Ax¯k2+ 2αnβnhB(xn−x¯), γf(xn)−A¯xi +2αnhΛn−x, γf¯ (xn)−Ax¯i

= (1−2(¯γ−γ)αn)kxn−x¯k2+α2nγ¯2kxn−x¯k2+ 2εnkαn(γf(xn)−Ax¯) +βnB(xn−x¯) + ((1−εn)I−βnB−αnA)(Λn−x¯)kkx¯k+ε2nkx¯k2 +α2nkγf(xn)−Ax¯k2+ 2αnβnhB(xn−x¯), γf(xn)−A¯xi

+2αnhΛn−x, γf¯ (xn)−Ax¯i

≤ (1−2(¯γ−γ)αn)kxn−x¯k2+αnϑn,

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where

ζn = (¯γ−γρ)αn,

ϑn = αnγ¯2kxn−x¯k2+ 2kαn(γf(xn)−Ax¯)

+βnB(xn−¯x) + ((1−εn)I−βnB−αnA)(Λn−x¯)kkx¯k+εnkx¯k2 +αnkγf(xn)−Ax¯k2+ 2βnhB(xn−x¯), γf(xn)−A¯xi

+2αnhΛn−x, γf¯ (xn)−Ax¯i

−2αnh(εnI+βnB+αnA)(Λn−x¯), γf(xn)−A¯xi.

From conditions (C1)−(C2) and (C5) and Lemma 1.3, we obtain the sequence

{xn} strongly convergence to ¯x.

Using Theorems 2.1 and 2.2, we obtain the following corollaries.

Corollary 2.3. [5, Theorem 3.1] Let C be a nonempty closed convex subset of H. Suppose S andT are nonexpansive mappings of C into itself, such that

F(ST) =F(T S) =F(S)T

F(T)6=∅. Let f be a contraction mapping fromC

toC and{xn} be a sequence generated by x0=x∈C and

xn+1 = αnf(xn) +βn +(1−αn−βn)

2 (n+ 1)(n+ 2)

n

X

i=0 n−i

X

j=0

((ST)jSi−j∨(ST)iTj−i)xn

for alln∈NS

{0}, where {αn},{βn} ⊂[0,1]satisfy

lim

n→∞αn= 0, ∞

X

n=0

αn =∞.

If 0 < lim inf

n→∞βn ≤ lim sup_n→∞ βn < 1, then {xn} converges strongly to z ∈

F(S)T

F(T), where z = PF(S)TF(T)f(z) is the unique solution of the

vari-ational inequality

h(I−f)z, z−xi ≥0,∀x∈F(S)\F(T).

Proof. SettingFi= Ψi ≡0,∀i∈ {1,2, . . . , k}, A=B≡I, γ= ¯γ= 1,

εn= 0, and ωn=xn, φ(t) = (1−ρ)tin Theorems 2.1 and 2.2. Thus the proof

is straightforward.

Corollary 2.4. [5, Corollary 3.4] Let C be a nonempty closed convex subset of H. Suppose S and T are averaged mappings of C into itself, such that

F(S)T

F(T)6=∅. Letf be a contraction mapping fromC toC and{xn}be a

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sequence generated byx0=x∈C and

xn+1 = αnf(xn) +βn +(1−αn−βn)

2 (n+ 1)(n+ 2)

n

X

i=0 n−i

X

j=0

((ST)j_Si−j_∨₍_ST₎i_Tj−i₎_x n

for alln∈NS_{₀_}

, where {αn},{βn} ⊂[0,1]satisfy lim

n→∞αn= 0,n→∞limβn= 0, ∞

X

n=0

αn =∞.

If 0 < lim inf

n→∞βn ≤ lim sup_n→∞ βn < 1, then {xn} is strongly convergent to z ∈

F(S)T_F₍_T₎

, wherez =PF(S)TF(T)f(z)is the unique solution of the

varia-tional inequality

h(I−f)z, z−xi ≥0,∀x∈F(S)\F(T).

Proof. Set S and T averaged mappings of C into itself, Fi = Ψi ≡0, for all

i∈ {1,2, . . . , k}, A=B ≡I, γ = ¯γ= 1, εn = 0 and ωn =xn, φ(t) = (1−ρ)t in Theorems 2.1 and 2.2. Thus the proof is straightforward.

Corollary 2.5. [8, Theorem 1] Let C be a nonempty closed convex subset of H. Suppose S and T are nonexpansive mappings of C into itself, such that ST = T S and F(S)T

F(T) 6= ∅. Let {xn} be a sequence generated by

x0=x∈C and

xn+1 =αnx+ (1−αn)

2 (n+ 1)(n+ 2)

n

X

i=0 n−i

X

j=0

SiTjxn.

for alln∈N, where{αn} ⊂[0,1]satisfy lim

n→∞αn= 0, ∞

X

n=0

αn =∞.

Then{xn} is strongly convergent toz∈F(S)TF(T), wherePF(S)TF(T) is a

metric projection ofH ontoF(S)T F(T).

Proof. SettingST =T S,Fi= Ψi≡0,for alli∈ {1,2, . . . , k}, A=B≡I, γ= ¯

γ = 1, f(y) =x, for all y ∈C, εn =βn = 0 and ωn =xn, φ(t) = (1−ρ)t in Theorems 2.1 and 2.2. Thus the proof is straightforward.

Corollary 2.6. [10, Theorem 4] Let H be a Hilbert space and C a nonempty closed convex subset of H. SupposeS andT are averaged mappings ofC into itself such that F(S)T

F(T)is nonempty. Suppose that{αn}satisfies lim

n→∞αn= 0, ∞

X

n=0

αn =∞.

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For an arbitrary x∈C, the sequence {xn} generated by

(

x0=x,

xn+1=αnx+ (1−αn)_(n+1)(n+2)2 P n i=0

Pn−i

j=0((ST)jSi−j∨(ST)iTj−i)xn.

is strongly convergent to a common fixed pointP xof S andT, whereP is the metric projection ofH ontoF(S)T_F₍_T₎

.

Proof. SettingSandT be averaged mappings ofCinto itself andFi= Ψi≡0, for alli∈ {1,2, . . . , k}, A=B ≡I, γ = ¯γ= 1,f(y) =x,∀y ∈C,εn =βn = 0 and ωn = xn, φ(t) = (1−ρ)t in Theorems 2.1 and 2.2. Thus the proof is

straightforward.

Acknowledgments

We would like to thank the referees for their constructive comments and suggestions.

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