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Hindawi Publishing Corporation
Fixed Point Theory and Applications
Volume 2008, Article ID 134148, 17 pages
doi:10.1155/2008/134148
Research Article
An Extragradient Approximation Method for
Equilibrium Problems and Fixed Point Problems of
a Countable Family of Nonexpansive Mappings
Rabian Wangkeeree
Department of Mathematics, Faculty of Science, Naresuan University, Phitsanulok 65000, Thailand
Correspondence should be addressed to Rabian Wangkeeree,
Received 28 February 2008; Accepted 13 July 2008
Recommended by Huang Nanjing
We introduce a new iterative scheme for finding the common element of the set of common fixed
points of nonexpansive mappings, the set of solutions of an equilibrium problem, and the set of
solutions of the variational inequality. We show that the sequence converges strongly to a common
element of the above three sets under some parameters controlling conditions. Moreover, we apply
our result to the problem of finding a common fixed point of a countable family of nonexpansive
mappings, and the problem of finding a zero of a monotone operator. This main theorem extends
a recent result of Yao et al. 2007 and many others.
Copyright q 2008 Rabian Wangkeeree. 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.
1. Introduction
Let H be a real Hilbert space with inner product ·, · and norm ·,andletC be a closed
convex subset of H.LetF be a bifunction of C × C into R, where R is the set of real numbers.
The equilibrium problem for φ : C × C → R is to find x ∈ C such that
φx, y ≥ 0 ∀y ∈ C. 1.1
The set of solutions of 1.1 is denoted by EPφ. Given a mapping T : C → H,letφx, y
Tx,y − x for all x, y ∈ C. Then z ∈ EPφ if and only if Tz,y − z≥0 for all y ∈ C, that
is, z is a solution of the variational inequality. Numerous problems in physics, optimization,


and economics reduce to find a solution of 1.1. In 1997, Fl
˚
am and Antipin 1 introduced
an iterative scheme of finding the best approximation to initial data when EPφ is nonempty
and proved a strong convergence theorem.
Let A : C → H be a mapping. The classical variational inequality, denoted by VIA, C,
is to find x

∈ C such that

Ax

,v− x


≥ 0 1.2
2 Fixed Point Theory and Applications
for all v ∈ C. The variational inequality has been extensively studied in the literature. See,
for example, 2, 3 and the references therein. A mapping A of C into H is called α-inverse-
strongly monotone 4, 5 if there exists a positive real number α such that
Au − Av, u − v≥αAu − Av
2
1.3
for all u, v ∈ C. It is obvious that any α-inverse-strongly monotone mapping A is monotone
and Lipschitz continuous. A mapping S of C into itself is called nonexpansive if
Su − Sv≤u − v 1.4
for all u, v ∈ C. We denote by FS the set of fixed points of S. For finding an element of
FS ∩ VIA, C, under the assumption that a set C ⊆ H is nonempty, closed, and convex,
a mapping S : C → C is nonexpansive and a mapping A : C → H is α-inverse-strongly
monotone, Takahashi and Toyoda 6 introduced the following iterative scheme:

x
n1
 α
n
x
n


1 − α
n

SP
C

x
n
− λ
n
Ax
n

1.5
for every n  0, 1, 2, ,where x
0
 x ∈ C, {α
n
} is a sequence in 0, 1,and{λ
n
} is a sequence
in 0, 2α. They proved that if FS ∩ VIA, C

/
 ∅, then the sequence {x
n
} generated by 1.5
converges weakly to some z ∈ FS∩VIA, C. Recently, motivated by the idea of Korpelevi
ˇ
c’s
extragradient method 7, Nadezhkina and Takahashi 8 introduced an iterative scheme
for finding an element of FS ∩ VIA, C and the weak convergence theorem is presented.
Moreover, Zeng and Yao 9 proposed some new iterative schemes for finding elements
in FS ∩ VIA, C and obtained the weak convergence theorem for such schemes. Very
recently, Yao et al. 10 introduced the following iterative scheme for finding an element of
FS ∩VIA, C under some mild conditions. Let C be a closed convex subset of a real Hilbert
space H, A : C → H a monotone, L-Lipschitz continuous mapping, and S a nonexpansive
mapping of C into itself such that FS∩VIA, C
/
 ∅. Suppose that
x
1
 u ∈ C and {x
n
}, {y
n
}
are given by
y
n
 P
C


x
n
− λ
n
Ax
n

,
x
n1
 α
n
u  β
n
x
n
 γ
n
SP
C

x
n
− λ
n
Ay
n

∀n ∈ N,
1.6

where {α
n
}, {β
n
}, {γ
n
}⊆0, 1 and {λ
n
}⊆0, 1 satisfy some parameters controlling
conditions. They proved that the sequence {x
n
} defined by 1.6 converges strongly to a
common element of FS ∩ VIA, C.
On the other hand, S. Takahashi and W. Takahashi 11 introduced an iterative scheme
by the viscosity approximation method for finding a common element of the set of solution
1.1 and the set of fixed points of a nonexpansive mapping in a real Hilbert space. Let S :
C → C be a nonexpansive mapping. Starting with arbitrary initial x
1
∈ C, define sequences
{x
n
} and {u
n
} recursively by
φ

u
n
,y



1
r
n

y − u
n
,u
n
− x
n

≥ 0 ∀y ∈ C,
x
n1
 α
n
f

x
n



1 − α
n

Su
n
∀n ∈ N.

1.7
They proved that under certain appropriate conditions imposed on {α
n
} and {r
n
},the
sequences {x
n
} and {u
n
} converge strongly to z ∈ FS ∩ EPφ, where z  P
FS∩EPφ
fz.
Rabian Wangkeeree 3
Moreover, Aoyama et al. 12 introduced an iterative scheme for finding a common
fixed point of a countable family of nonexpansive mappings in Banach spaces and obtained
the strong convergence theorem for such scheme.
In this paper, motivated by Yao et al. 10, S. Takahashi and W. Takahashi 11 and
Aoyama et al. 12, we introduce a new extragradient method 4.2 which is mixed the
iterative schemes considered in 10–12 for finding a common element of the set of common
fixed points of nonexpansive mappings, the set of solutions of an equilibrium problem, and
the solution set of the classical variational inequality problem for a monotone L-Lipschitz
continuous mapping in a real Hilbert space. Then, the strong convergence theorem is proved
under some parameters controlling conditions. Further, we apply our result to the problem
of finding a common fixed point of a countable family of nonexpansive mappings, and the
problem of finding a zero of a monotone operator. The results obtained in this paper improve
and extend the recent ones announced by Yao et al. results 10 and many others.
2. Preliminaries
Let H be a real Hilbert space with norm · and inner product ·, · and let C be a closed
convex subset of H. For every point x ∈ H, there exists a unique nearest point in C, denoted

by P
C
x, such that


x − P
C
x


≤x − y∀y ∈ C. 2.1
P
C
is called the metric projection of H onto C. It is well known that P
C
is a nonexpansive
mapping of H onto C and satisfies

x − y, P
C
x − P
C
y




P
C
x − P

C
y


2
2.2
for every x, y ∈ H. Moreover, P
C
x is characterized by the following properties: P
C
x ∈ C and

x − P
C
x, y − P
C
x

≤ 0, 2.3
x − y
2



x − P
C
x


2




y − P
C
x


2
2.4
for all x ∈ H, y ∈ C. For more details, see 13. It is easy to see that the following is true:
u ∈ VIA, C ⇐⇒ u  P
C
u − λAu,λ>0. 2.5
A set-valued mapping T : H → 2
H
is called monotone if for all x, y ∈ H, f ∈ Tx, and
g ∈ Ty imply x − y, f − g≥0. A monotone mapping T : H → 2
H
is maximal if the graph of
GT of T is not properly contained in the graph of any other monotone mapping. It is known
that a monotone mapping T is maximal if and only if for x, f ∈ H × H, x − y, f − g≥0
for every y, g ∈ GT implies f ∈ Tx.LetB be a monotone map of C into H, L-Lipschitz
continuous mapping and let N
C
v be the normal cone to C at v ∈ C,thatis,N
C
v  {w ∈ H :
u − v, w≥0 for all u ∈ C}. Define
Tv 




Bv  N
C
v, v ∈ C;
∅,v
/
∈ C.
2.6
Then T is the maximal monotone and 0 ∈ Tv if and only if v ∈ VIC, B;see14.
4 Fixed Point Theory and Applications
The following lemmas will be useful for proving the convergence result of this paper.
Lemma 2.1 see 15. Let E, ·, · be an inner product space. Then for all x, y, z ∈ E and α, β, γ ∈
0, 1 with α  β  γ  1,one has
αx  βy  γz
2
 αx
2
 βy
2
 γz
2
− αβx − y
2
− αγx − z
2
− βγy − z
2
. 2.7

Lemma 2.2 see 16. Let {x
n
} and {z
n
} be bounded sequences in a Banach space E and let {β
n
} be
a sequence in 0, 1 with 0 < lim inf
n→∞
β
n
≤ lim sup
n→∞
β
n
< 1. Suppose x
n1
1 − β
n
z
n
 β
n
x
n
for all integers n ≥ 1 and lim sup
n→∞
z
n1
− z

n
−x
n1
− x
n
 ≤ 0. Then, lim
n→∞
z
n
− x
n
  0.
Lemma 2.3 see 17. Assume {a
n
} is a sequence of nonnegative real numbers such that
a
n1


1 − α
n

a
n
 δ
n
,n≥ 1, 2.8
where {α
n
} is a sequence in 0, 1 and {δ

n
} is a sequence in R suchthat
i


n1
α
n
 ∞ and
ii lim sup
n→∞
δ
n

n
 ≤ 0 or


n1

n
| < ∞.
Then lim
n→∞
a
n
 0.
Lemma 2.4 see 12, Lemma 3.2. Let C be a nonempty closed subset of a Banach space and let
{S
n

} be a sequence of mappings of C into itself. Suppose that


n1
sup{S
n1
z− S
n
z : z ∈ C} < ∞.
Then, for each y ∈ C, {S
n
y} converges strongly to some point of C. Moreover, let S be a mapping of
C into itself defined by
Sy  lim
n→∞
S
n
y ∀y ∈ C. 2.9
Then lim
n→∞
sup{Sz − S
n
z : z ∈ C}  0.
For solving the equilibrium problem for a bifunction φ : C × C → R, let us assume that
φ satisfies the following conditions:
A1 φx, x0 for all x ∈ C;
A2 φ is monotone, that is, φx, yφy, x ≤ 0 for all x, y ∈ C;
A3 for each x, y, z ∈ C, lim
t→0
φtz 1 − tx, y ≤ φx, y;

A4 for each x ∈ C, y → φx, y is convex and lower semicontinuous.
The following lemma appears implicitly in 18.
Lemma 2.5 see 18. Let C be a nonempty closed convex subset of H and let φ be a bifunction of
C × C into R satisfying (A1)–(A4). Let r>0 and x ∈ H. Then, there exists z ∈ C such that
φz, y
1
r
y − z, z − x≥0 ∀y ∈ C. 2.10
The following lemma was also given in 1.
Rabian Wangkeeree 5
Lemma 2.6 see 1. Assume that φ : C × C → R satisfies (A1)–(A4). For r>0 and x ∈ H, define
a mapping T
r
: H → C as follows:
T
r
x

z ∈ C : φz, y
1
r
y − z, z − x≥0 ∀y ∈ C

2.11
for all z ∈ H. Then, the following hold:
i T
r
is single-valued;
ii T
r

is firmly nonexpansive, that is, for any x, y ∈ H, T
r
x − T
r
y
2
≤T
r
x − T
r
y, x − y;
iii FT
r
EPφ;
iv EPφ is closed and convex.
3. Main results
In this section, we prove a strong convergence theorem.
Theorem 3.1. Let C be a closed convex subset of a real Hilbert space H.Letφ be a bifunction from C×
C to R satisfying (A1)–(A4), A : C → H a monotone L-Lipschitz continuous mapping and let {S
n
}
be a sequence of nonexpansive mappings of C into itself such that ∩

n1
FS
n
 ∩ VIA, C ∩ EPφ
/
 ∅.
Let the sequences {x

n
}, {u
n
}, and {y
n
} be generated by
x
1
 x ∈ C chosen arbitrarily,
φ

u
n
,y


1
r
n

y − u
n
,u
n
− x
n

≥ 0 ∀y ∈ C,
y
n

 P
C

u
n
− λ
n
Au
n

,
x
n1
 α
n
f

x
n

 β
n
x
n
 γ
n
S
n
P
C


u
n
− λ
n
Ay
n

∀n ≥ 1,
3.1
where {α
n
}, {β
n
}, {γ
n
}⊆0, 1, {λ
n
}⊆0, 1, and {r
n
}⊆0, ∞ satisfy the following conditions:
C1 α
n
 β
n
 γ
n
 1,
C2 lim
n→∞

α
n
 0,


n1
α
n
 ∞,
C3 0 < lim inf
n→∞
β
n
≤ lim sup
n→∞
β
n
< 1,
C4 lim
n→∞
λ
n
 0,
C5 lim inf
n→∞
r
n
> 0,



n1
|r
n1
− r
n
| < ∞.
Suppose that


n1
sup{S
n1
z − S
n
z : z ∈ B} < ∞ for any bounded subset B of C.Let
S be a mapping of C into itself defined by Sy  lim
n→∞
S
n
y for all y ∈ C and suppose that
FS∩

n1
FS
n
. Then the sequences {x
n
}, {u
n
}, and {y

n
} converge strongly to the same point
q ∈∩

n1
FS
n
 ∩ VIA, C ∩ EPφ,whereq  P


n1
FS
n
∩VIA,C∩EPφ
fq.
Proof. Let Q  P


n1
FS
n
∩VIA,C∩EPφ
. Since f is a contraction with α ∈ 0, 1,weobtain


Qfx − Qfy






fx − fy


≤ αx − y∀x, y ∈ C. 3.2
Therefore, Qf is a contraction of C into itself, which implies that there exists a unique element
q ∈ C such that q  Qfq. Then we divide the proof into several steps.
6 Fixed Point Theory and Applications
Step 1 {x
n
} is bounded. Indeed, put t
n
 P
C
u
n
− λ
n
Ay
n
 for all n ≥ 1. Let x

∈∩

n1
FS
n
 ∩
VIA, C ∩ EPφ.From2.5 we have x


 P
C
x

− λ
n
Ax

. Also it follows from 2.4 that


t
n
− x



2



u
n
− λ
n
Ay
n
− x




2



u
n
− λ
n
Ay
n
− t
n


2



u
n
− x



2
− 2λ
n

Ay

n
,u
n
− x


 λ
2
n


Ay
n


2



u
n
− t
n


2
 2λ
n

Ay

n
,u
n
− t
n

− λ
2
n


Ay
n


2



u
n
− x



2
 2λ
n

Ay

n
,x

− t
n




u
n
− t
n


2



u
n
− x



2



u

n
− t
n


2
 2λ
n

Ay
n
− Ax

,x

− y
n

 2λ
n

Ax

,x

− y
n

 2λ
n


Ay
n
,y
n
− t
n

.
3.3
Since A is monotone and x

is a solution of the variational inequality problem VIA, C,we
have

Ay
n
− Ax

,x

− y
n

≤ 0,

Ax

,x


− y
n

≤ 0. 3.4
This together with 3.3 implies that


t
n
− x



2



u
n
− x



2



u
n
− t

n


2
 2λ
n

Ay
n
,y
n
− t
n




u
n
− x



2



u
n
− y

n
y
n
− t
n



2
 2λ
n

Ay
n
,y
n
− t
n




u
n
− x



2




u
n
− y
n


2
− 2

u
n
− y
n
,y
n
− t
n




y
n
− t
n


2

 2λ
n

Ay
n
,y
n
− t
n




u
n
− x



2



u
n
− y
n


2




y
n
− t
n


2
 2

u
n
− λ
n
Ay
n
− y
n
,t
n
− y
n

.
3.5
From 2.3, we have

u

n
− λ
n
Au
n
− y
n
,t
n
− y
n

≤ 0, 3.6
so that

u
n
− λ
n
Ay
n
− y
n
,t
n
− y
n




u
n
− λ
n
Au
n
− y
n
,t
n
− y
n



λ
n
Au
n
− λ
n
Ay
n
,t
n
− y
n




λ
n
Au
n
− λ
n
Ay
n
,t
n
− y
n

≤ λ
n


Au
n
− Ay
n




t
n
− y
n



≤ λ
n
L


u
n
− y
n




t
n
− y
n


.
3.7
Hence it follows from 3.5 and 3.7 that


t
n
− x




2



u
n
− x



2



u
n
− y
n


2



y
n
− t
n



2
 2λ
n
L


u
n
− y
n




t
n
− y
n





u
n
− x




2



u
n
− y
n


2



y
n
− t
n


2
 λ
n
L



u
n
− y

n


2



y
n
− t
n


2

 u
n
− x


2


λ
n
L − 1



u

n
− y
n


2


λ
n
L − 1



y
n
− t
n


2
.
3.8
Rabian Wangkeeree 7
Since λ
n
→ 0asn →∞, there exists a positive integer N
0
such that λ
n

L − 1 ≤−1/3, when
n ≥ N
0
. Hence it follows from 3.8 that


t
n
− x






u
n
− x



. 3.9
Observe that


u
n
− x







T
r
n
x
n
− T
r
n
x






x
n
− x



, 3.10
and hence


t

n
− x






x
n
− x



. 3.11
Thus, we can calculate


x
n1
− x






α
n
f


x
n

 β
n
x
n
 γ
n
S
n
t
n
− x



≤ α
n


f

x
n

− x




 β
n


x
n
− x



 γ
n


t
n
− x



≤ α
n


f

x
n


− f

x




 α
n


f

x


− x



 β
n


x
n
− x




 γ
n


x
n
− x





1 − α
n
1 − α



x
n
− x



 α
n


f


x


− x





1 − α
n
1 − α



x
n
− x



 α
n
1 − α


f

x



− x



1 − α
.
3.12
It follows from induction that


x
n
− x



≤ max



x
1
− x



,



f

x


− x



1 − α

,n≥ N
0
. 3.13
Therefore, {x
n
} is bounded. Hence, so are {t
n
}, {S
n
t
n
}, {Au
n
}, {Ay
n
},and{fx
n
}.
Step 2 lim

n→∞
x
n1
− x
n
  0. Indeed, we observe that for any x, y ∈ C,



I − λ
n
A

x −

I − λ
n
A

y


2



x − y − λ
n
Ax − Ay



2
 x − y
2
− 2λ
n
x − y, Ax − Ay  λ
2
n
Ax − Ay
2
≤x − y
2
 λ
2
n
L
2
x − y
2


1  λ
2
n
L
2

x − y
2

,
3.14
which implies that



I − λ
n
A

x −

I − λ
n
A

y




1  λ
n
L

x − y. 3.15
Thus


t

n1
− t
n





P
C

u
n1
− λ
n1
Ay
n1

− P
C

u
n
− λ
n
Ay
n







u
n1
− λ
n1
Ay
n1


u
n
− λ
n
Ay
n







u
n1
− λ
n1
Au
n1




u
n
− λ
n1
Au
n

 λ
n1

Au
n1
− Ay
n1
− Au
n

 λ
n
Ay
n







u
n1
− λ
n1
Au
n1



u
n
− λ
n1
Au
n



 λ
n1



Au
n1






Ay
n1





Au
n



 λ
n


Ay
n




1  λ
n1
L



u
n1

− u
n


 λ
n1



Au
n1





Ay
n1





Au
n



 λ
n



Ay
n


.
3.16
8 Fixed Point Theory and Applications
On the other hand, from u
n
 T
r
n
x
n
and u
n1
 T
r
n1
x
n1
, we note that
φ

u
n
,y



1
r
n

y − u
n
,u
n
− x
n

≥ 0 ∀y ∈ C
, 3.17
φ

u
n1
,y


1
r
n1

y − u
n1
,u
n1
− x

n1

≥ 0 ∀y ∈ C.
3.18
Putting y  u
n1
in 3.17 and y  u
n
in 3.18, we have
φ

u
n
,u
n1


1
r
n

u
n1
− u
n
,u
n
− x
n


≥ 0,
φ

u
n1
,u
n


1
r
n1

u
n
− u
n1
,u
n1
− x
n1

≥ 0.
3.19
So, from A2, we have

u
n1
− u
n

,
u
n
− x
n
r
n

u
n1
− x
n1
r
n1

≥ 0 3.20
and hence

u
n1
− u
n
,u
n
− u
n1
 u
n1
− x
n


r
n
r
n1

u
n1
− x
n1


≥ 0. 3.21
Without loss of generality, let us assume that there exists a real number c such that r
n
>c>0
for all n ∈ N. Then, we have


u
n1
− u
n


2


u
n1

− u
n
,x
n1
− x
n


1 −
r
n
r
n1


u
n1
− x
n1





u
n1
− u
n






x
n1
− x
n







1 −
r
n
r
n1






u
n1
− x
n1




3.22
and hence


u
n1
− u
n





x
n1
− x
n



1
r
n1


r
n1
− r

n




u
n1
− x
n1





x
n1
− x
n



1
c


r
n1
− r
n



M,
3.23
where M  sup{u
n
− x
n
 : n ∈ N}. It follows from 3.16 and the last inequality that


t
n1
− t
n




1  λ
n1
L



x
n1
− x
n





1  λ
n1
L

1
c


r
n1
− r
n


M
 λ
n1



Au
n1





Ay

n1





Au
n



 λ
n
Ay
n
.
3.24
Rabian Wangkeeree 9
Setting z
n
α
n
fx
n
γ
n
S
n
t
n

/1 − β
n
,weobtainx
n1
1− β
n
z
n
 β
n
x
n
for all n ∈ N.Thus,
we have


z
n1
− z
n







α
n1
f


x
n1

 γ
n1
S
n1
t
n1
1 − β
n1

α
n
f

x
n

 γ
n
S
n
t
n
1 − β
n










α
n1
1 − β
n1
f

x
n1


γ
n1
1 − β
n1

S
n1
t
n1
− S
n
t
n



α
n
1 − β
n
f

x
n



1 −
α
n
1 − β
n

S
n
t
n


1 −
α
n1
1 − β
n1

S
n
t
n






α
n1
1 − β
n1


f

x
n1

− S
n
t
n



α
n

1 − β
n


S
n
t
n
− f

x
n




γ
n1
1 − β
n1


S
n1
t
n1
− S
n
t
n



.
3.25
It follows from 3.24 that


S
n1
t
n1
− S
n
t
n





S
n1
t
n1
− S
n1
t
n






S
n1
t
n
− S
n
t
n





t
n1
− t
n





S
n1
t
n
− S

n
t
n




1  λ
n1
L



x
n1
− x
n




1  λ
n1
L

1
c


r

n1
− r
n


M
 λ
n1



Au
n1





Ay
n1





Au
n




 λ
n


Ay
n





S
n1
t
n
− S
n
t
n


.
3.26
Combining 3.25 and 3.26, we have


z
n1
−z
n






x
n1
−x
n



α
n1
1 − β
n1


f

x
n1

− S
n
t
n




α
n
1 − β
n


S
n
t
n
− f

x
n




γ
n1
1 − β
n1

1λ
n1
L



x

n1
−x
n



γ
n1
1 − β
n1

1λ
n1
L

1
c
|r
n1
−r
n
|M

γ
n1
1 − β
n1
λ
n1




Au
n1





Ay
n1





Au
n




γ
n1
1 − β
n1
λ
n



Ay
n



γ
n1
1 − β
n1


S
n1
t
n
− S
n
t
n





x
n1
− x
n




α
n1
1 − β
n1


f

x
n1

− S
n
t
n



α
n
1 − β
n


S
n
t
n
− f


x
n




γ
n1
1 − β
n1
λ
n1
Lx
n1
− x
n
 
γ
n1
1 − β
n1

1  λ
n1
L

1
c
|r

n1
− r
n
|M

γ
n1
1 − β
n1
λ
n1



Au
n1





Ay
n1





Au
n





γ
n1
1 − β
n1
λ
n


Ay
n



γ
n1
1 − β
n1
sup



S
n1
t − S
n
t



: t ∈

t
n

.
3.27
10 Fixed Point Theory and Applications
This together with C1–C5 and lim
n→∞
sup{S
n1
t − S
n
t : t ∈{t
n
}}  0 implies that
lim sup
n→∞



z
n1
− z
n






x
n1
− x
n



≤ 0. 3.28
Hence, by Lemma 2.2,weobtainz
n
− x
n
→0asn →∞. It then follows that
lim
n→∞


x
n1
− x
n


 lim
n→∞

1 − β

n



z
n
− x
n


 0. 3.29
By 3.23 and 3.24, we also have
lim
n→∞


t
n1
− t
n


 lim
n→∞


u
n1
− u
n



 0. 3.30
Step 3 lim
n→∞
St
n
− t
n
  0. Indeed, pick any x

∈∩

n1
FS
n
 ∩ VIA, C ∩ EPφ,toobtain


u
n
− x



2



T

r
n
x
n
− T
r
n
x



2


T
r
n
x
n
− T
r
n
x

,x
n
− x





u
n
− x

,x
n
− x



1
2



u
n
− x



2



x
n
− x




2



x
n
− u
n


2

.
3.31
Therefore, u
n
− x


2
≤x
n
− x


2
−x
n

− u
n

2
.FromLemma 2.1 and 3.9, we obtain, when
n ≥ N
0
,that


x
n1
− x



2



α
n
f

x
n

 β
n
x

n
 γ
n
S
n
t
n
− x



2
≤ α
n


f

x
n

− x



2
 β
n



x
n
− x



2
 γ
n


S
n
t
n
− x



2
≤ α
n


f

x
n

− x




2
 β
n


x
n
− x



2
 γ
n


t
n
− x



2
≤ α
n



f

x
n

− x



2
 β
n


x
n
− x



2
 γ
n


u
n
− x




2
≤ α
n


f

x
n

− x



2
 β
n


x
n
− x



2
 γ
n




x
n
− x



2



x
n
− u
n


2

≤ α
n


f

x
n

− x




2


1 − α
n



x
n
− x



2
− γ
n


x
n
− u
n


2
3.32

and hence
γ
n


x
n
− u
n


2
≤ α
n


f

x
n

− x



2



x

n
− x



2



x
n1
− x



2
≤ α
n


f

x
n

− x



2




x
n
− x
n1





x
n
− x






x
n1
− x




.
3.33

It now follows from the last inequality, C1, C2, C3 and 3.29,that
lim
n→∞


x
n
− u
n


 0. 3.34
Noting that


y
n
− x
n





P
C

u
n
− λ

n
Au
n

− x
n





u
n
− x
n


 λ
n


Au
n


−→ 0asn −→ ∞ ,


y
n

− t
n





P
C

u
n
− λ
n
Au
n

− P
C

u
n
− λ
n
Ay
n



≤ λ

n


Au
n
− Ay
n


−→ 0asn −→ ∞ .
3.35
Rabian Wangkeeree 11
Thus


t
n
− x
n





t
n
− y
n






y
n
− x
n


−→ 0asn −→ ∞ . 3.36
We note that


S
n
y
n
− x
n1





S
n
y
n
− S
n

t
n





S
n
t
n
− x
n1





y
n
− t
n


 α
n


S
n

t
n
− f

x
n



 β
n


S
n
t
n
− x
n





y
n
− t
n



 α
n


S
n
t
n
− f

x
n



 β
n


S
n
t
n
− S
n
x
n


 β

n


S
n
x
n
− x
n





y
n
− t
n


 α
n


S
n
t
n
− f


x
n



 β
n


t
n
− x
n


 β
n


S
n
x
n
− x
n


.
3.37
Using 3.37, we have



S
n
x
n
− x
n





S
n
x
n
− S
n
y
n





S
n
y
n

− x
n1





x
n1
− x
n





x
n
− y
n





y
n
− t
n



 α
n


S
n
t
n
− f

x
n



 β
n


t
n
− x
n


 β
n



S
n
x
n
− x
n





x
n1
− x
n


,
3.38
so that

1−β
n



S
n
x
n

−x
n





x
n
−y
n





y
n
−t
n


α
n


S
n
t
n

−f

x
n



β
n


t
n
−x
n





x
n1
−x
n


.
3.39
This implies that
lim

n→∞


S
n
x
n
− x
n


 0. 3.40
It now follows from 3.36 and 3.40 that


S
n
t
n
− t
n





S
n
t
n

− S
n
x
n





S
n
x
n
− x
n





x
n
− t
n


≤ 2


t

n
− x
n





S
n
x
n
− x
n


−→ 0asn −→ ∞ .
3.41
Applying Lemma 2.4 and 3.41, we have


St
n
− t
n






St
n
− S
n
t
n





S
n
t
n
− t
n


≤ sup



St − S
n
t


: t ∈


t
n




S
n
t
n
− t
n


−→ 0.
3.42
It follows from the last inequality and 3.36 that


x
n
− St
n





x
n

− t
n





t
n
− St
n


−→ 0asn −→ ∞ . 3.43
Step 4 lim sup
n→∞
fq − q, x
n
− q≤0. Indeed, we choose a subsequence {t
n
i
} of {t
n
} such
that
lim sup
n→∞

fq − q, St
n

− q

 lim
i→∞

fq − q, St
n
i
− q

. 3.44
12 Fixed Point Theory and Applications
Without loss of generality, we may assume that {t
n
i
} converges weakly to z ∈ C.FromSt
n

t
n
→0, we obtain St
n
i
z.Now, we will show that z ∈ FS ∩ VIA, C ∩ EPφ. Firstly, we
will show z ∈ EPφ. Indeed, we observe that u
n
 T
r
n
x

n
,and
φ

u
n
,y


1
r
n

y − u
n
,u
n
− x
n

≥ 0 ∀y ∈ C. 3.45
From A2, we also have
1
r
n

y − u
n
,u
n

− x
n

≥ φ

y, u
n

, 3.46
and hence

y − u
n
i
,
u
n
i
− x
n
i
r
n
i

≥ φ

y, u
n
i


. 3.47
From u
n
−x
n
→0, x
n
−St
n
→0, and St
n
−t
n
→0, we get u
n
i
z. Since u
n
i
−x
n
i
/r
n
i

0, it follows by A4 that 0 ≥ φy, z for all y ∈ C. For t with 0 <t≤ 1andy ∈ C, let
y
t

 ty 1 − tz. Since y ∈ C and z ∈ C, we have y
t
∈ C and hence φy
t
,z ≤ 0. So, from A1
and A4, we have
0  φ

y
t
,y
t

≤ tφ

y
t
,y

1 − tφ

y
t
,z

≤ tφ

y
t
,y


3.48
and hence 0 ≤ φy
t
,y.FromA3, we have 0 ≤ φz, y for all y ∈ C, and hence z ∈ EPφ. By
the Opial’s condition, we can obtain that z ∈ FS. Next we will show that z ∈ VIA, C.Let
Tv 



Av  N
C
v, v ∈ C;
∅,v
/
∈ C.
3.49
Then T is maximal monotone see 14.Letv, w ∈ GT. Since w − Av ∈ N
C
v and t
n
∈ C,
we have v − t
n
,w− Av≥0. On the other hand, from t
n
 P
C
u
n

− λ
n
Ay
n
, we have

v − t
n
,t
n


u
n
− λ
n
Ay
n

≥ 0, 3.50
that is,

v − t
n
,
t
n
− u
n
λ

n
 Ay
n

≥ 0. 3.51
Therefore, we obtain

v − t
n
i
,w



v − t
n
i
,Av



v − t
n
i
,Av



v − t
n

i
,
t
n
i
− u
n
i
λ
n
i
 Ay
n
i



v − t
n
i
,Av− Ay
n
i

t
n
i
− u
n
i

λ
n
i



v − t
n
i
,Av− At
n
i



v − t
n
i
,At
n
i
− Ay
n
i
−

v − t
n
i
,

t
n
i
− u
n
i
λ
n
i



v − t
n
i
,At
n
i



v − t
n
i
,
t
n
i
− u
n

i
λ
n
i
 Ay
n
i



v − t
n
i
,At
n
i
− Ay
n
i



v − t
n
i
,
t
n
i
− u

n
i
λ
n
i

.
3.52
Rabian Wangkeeree 13
Noting that t
n
i
− y
n
i
→0, t
n
i
− u
n
i
→0asn →∞, A is Lipschitz continuous and 3.52,
we obtain
v − z, w≥0. 3.53
Since T is maximal monotone, we have z ∈ T
−1
0, and hence z ∈ VIA, C. Hence z ∈ FS ∩
VIA, C ∩ EPφ. The property of the metric projection implies that
lim sup
n→∞


fq − q, x
n
− q

 lim sup
n→∞

fq − q, St
n
− q

 lim
i→∞

fq − q, St
n
i
− q



fq − q, z − q

≤ 0.
3.54
Step 5 lim
n→∞
x
n

− q  0. Indeed, we observe that


x
n1
− q


2


α
n
f

x
n

 β
n
x
n
 γ
n
S
n
t
n
,x
n1

− q

 α
n

f

x
n

− q, x
n1
− q

 β
n

x
n
− q, x
n1
− q

 γ
n

S
n
t
n

− q, x
n1
− q


1
2
β
n



x
n
− q


2



x
n1
− q


2


1

2
γ
n



t
n
− q


2



x
n1
− q


2

 α
n
fx
n
 − fq,x
n1
− q  α
n

fq − q, x
n1
− q

1
2

1 − α
n




x
n
− q


2



x
n1
− q


2



1
2
α
n



f

x
n

− fq


2



x
n1
− q


2

 α
n

fq − q, x

n1
− q


1
2

1 − α
n

1 − α
2



x
n
− q


2

1
2

1 − α
n




x
n1
− q


2

1
2
α
n


x
n1
− q


2
 α
n

fq − q, x
n1
− q

,
3.55
which implies that



x
n1
− q


2


1 − α
n

1 − α
2



x
n
− q


2
 2α
n

fq − q, x
n1
− q


. 3.56
Setting δ
n
 2α
n
fq − q, x
n1
− q, we have lim sup
n→∞
δ
n

n
1 − α
2
 ≤ 0. Applying
Lemma 2.3 to 3.56, we conclude that {x
n
} converges strongly to q. Consequently, {u
n
} and
{y
n
} converge strongly to q. This completes the proof.
As in 12, Theorem 4.1, we can generate a sequence {S
n
} of nonexpansive mappings
satisfying condition



n1
sup{S
n1
z − S
n
z : z ∈ B} < ∞ for any bounded subset B of C
by using convex combination of a general sequence {T
k
} of nonexpansive mappings with a
common fixed point.
Corollary 3.2. Let C be a closed convex subset of a real Hilbert space H.Letφ be a bifunction from
C × C to R satisfying (A1)–(A4), A : C → H a monotone, L-Lipschitz continuous mapping and let
14 Fixed Point Theory and Applications

k
n
} be a family of nonnegative numbers with indices n, k ∈ N with k ≤ n such that
i

n
k1
β
k
n
 1 for all n ∈ N;
ii lim
n→∞
β
k
n

> 0 for every k ∈ N;
iii


n1

n
k1

k
n1
− β
k
n
| < ∞.
Let {T
k
} be a sequence of nonexpansive mappings of C into itself with ∩

k1
FT
k
 ∩ VIA, C ∩
EPφ
/
 ∅. Let x
1
 x ∈ C and {x
n
}, {y

n
} and {u
n
} be the sequences generated by
x
1
 x ∈ C chosen arbitrary,
φ

u
n
,y


1
r
n

y − u
n
,u
n
− x
n

≥ 0 ∀y ∈ C,
y
n
 P
C


u
n
− λ
n
Au
n

,
x
n1
 α
n
f

x
n

 β
n
x
n
 γ
n
n

k1
β
k
n

T
k
P
C

u
n
− λ
n
Ay
n

∀n ≥ 1,
3.57
where {α
n
}, {β
n
}, {γ
n
}⊆0, 1, {λ
n
}⊆0, 1, and {r
n
}⊆0, ∞ satisfy the following conditions:
C1 α
n
 β
n
 γ

n
 1,
C2 lim
n→∞
α
n
 0,


n1
α
n
 ∞,
C3 0 < lim inf
n→∞
β
n
≤ lim sup
n→∞
β
n
< 1,
C4 lim
n→∞
λ
n
 0,
C5 lim inf
n→∞
r

n
> 0,


n1
|r
n1
− r
n
| < ∞.
Then the sequences {x
n
}, {u
n
}, and {y
n
} converge strongly to the same point q ∈∩

k1
FT
k
 ∩
VIA, C ∩ EPφ,whereq  P


k1
FT
k
∩VIA,C∩EPφ
fq.

Setting S
n
≡ S and f : u in Theorem 3.1, we have the following result.
Corollary 3.3 see 10, Theorem 3.1. Let C be a closed convex subset of a real Hilbert space H.Let
A : C → H be a monotone, L-Lipschitz continuous mapping, and let S be a nonexpansive mapping of
C into itself such that FS ∩ VIA, C
/
 ∅. Suppose x
1
 u ∈ C and {x
n
}, {y
n
} are given by
y
n
 P
C

x
n
− λ
n
Ax
n

,
x
n1
 α

n
u  β
n
x
n
 γ
n
SP
C

x
n
− λ
n
Ay
n

,
3.58
where {λ
n
}, {α
n
}, {β
n
}, {γ
n
} are sequences in 0, 1 satisfying the following conditions:
i α
n

 β
n
 γ
n
 1,
ii lim
n→∞
α
n
 0,


n1
α
n
 ∞,
iii 0 < lim inf
n→∞
β
n
≤ lim sup
n→∞
β
n
< 1,
iv lim
n→∞
λ
n
 0.

Then {x
n
} converges strongly to P
FS∩VIA,C
u.
Proof. Put φx, y0 for all x, y ∈ C and {r
n
}  1inTheorem 3.1. Thus, we have u
n
 x
n
.
Then the sequence {x
n
} generated in Corallary 3.3 converges strongly to P
FS∩VIA,C
u.
Rabian Wangkeeree 15
4. Applications
In this section, we consider the problem of finding a zero of a monotone operator. A
multivalued operator S : H → 2
H
with domain DS{z ∈ H : Sz
/
 ∅} and range
RS{Sz : z ∈ DT} is said to be monotone if for each x
i
∈ DS and y
i
∈ Sx

i
,i 1, 2,
we have x
1
− x
2
,y
1
− y
2
≥0. A monotone operator S is said to be maximal if its graph
GS{x, y : y ∈ Sx} is not properly contained in the graph of any other monotone
operator. Let I denote the identity operator on H and let S : H → 2
H
be a maximal
monotone operator. Then we can define, for each r>0, a nonexpansive single-valued
mapping J
r
: H → H by J
r
I  rS
−1
. It is called the resolvent or the proximal mapping of
S. We also define the Yosida approximation A
r
by A
r
I − J
r
/r. We know that A

r
x ∈ SJ
r
x
and A
r
x≤inf{y : y ∈ Sx} for all x ∈ H.WealsoknowthatS
−1
0  FJ
r
 for all r>0; see,
for instance, Rockafellar 19 or Takahashi 20 .
Lemma 4.1 the resolvent identity. For λ, μ > 0, there holds the identity
J
λ
x  J
μ

μ
λ


1 −
μ
λ

J
λ
x


,x∈ H. 4.1
By using Theorem 3.1 and Lemma 4.1, we may obtain the following improvement.
Theorem 4.2. Let S : H → 2
H
be a maximal monotone operator. Let φ be a bifunction from C × C to
R satisfying (A1)–(A4), A : C → H a monotone L-Lipschitz continuous mapping of C into H such
that S
−1
0 ∩VIA, C ∩ EPφ
/
 ∅ and f a contraction of C into itself with coefficient α ∈ 0, 1.Let
the sequences {x
n
}, {u
n
} and {y
n
} be generated by
x
1
 x ∈ C chosen arbitrary,
φ

u
n
,y


1
r

n
y − u
n
,u
n
− x
n
≥0 ∀y ∈ C,
y
n
 P
C

u
n
− λ
n
Au
n

,
x
n1
 α
n
f

x
n


 β
n
x
n
 γ
n
J
r
n
P
C

u
n
− λ
n
Ay
n

∀n ≥ 1,
4.2
where {α
n
}, {β
n
}, {γ
n
}⊆0, 1, {λ
n
}⊆0, 1, and {r

n
}⊆0, ∞ satisfy the following conditions:
C1 α
n
 β
n
 γ
n
 1,
C2 lim
n→∞
α
n
 0,


n1
α
n
 ∞,
C3 0 < lim inf
n→∞
β
n
≤ lim sup
n→∞
β
n
< 1,
C4 lim

n→∞
λ
n
 0,
C5 lim inf
n→∞
r
n
> 0,


n1
|r
n1
− r
n
| < ∞.
Then {x
n
} converges strongly to q  P
S
−1
0∩VIA,C∩EPφ
fq.
Proof. We first verify that


n1
sup{J
r

n1
z − J
r
n
z : z ∈ B} < ∞ for any bounded subset B of
C.LetB be a bounded subset of C. Since S
−1
0FJ
r
n
 for each n ∈ N, {J
r
n
z : z ∈ B,n ∈ N}
is bounded. It follows from Lemma 4.1 that
J
r
n1
z  J
r
n

r
n
r
n1
z 

1 −
r

n
r
n1

J
r
n1
z

,z∈ H. 4.3
16 Fixed Point Theory and Applications
Thus


J
r
n1
z − J
r
n
z





r
n1
− r
n



r
n1


J
r
n1
z − z


≤ M


r
n1
− r
n


4.4
for each z ∈ B and n ∈ N, where M  sup{J
r
n1
z − z : z ∈ B, n ∈ N}/ inf{r
n
: n ∈ N}. Hence
we get



n1
sup



J
r
n1
z − J
r
n
z


: z ∈ B

≤ M


n1


r
n1
− r
n


< ∞. 4.5

By the assumption that


n1
|r
n1
−r
n
| < ∞,weobtainr
n
→ r for some r>0. Since J
r
z−J
r
n
z≤
|r − r
n
|/rz − J
r
z, we obtain that lim
n→∞
J
r
n
z  J
r
z for all z ∈ C. Since FJ
μ
S

−1
0 for
all μ>0, we have FJ
r
∩

n1
FJ
r
n
S
−1
0
/
 ∅. Therefore, by Theorem 3.1, {x
n
} converges
strongly to q  P
S
−1
0∩VIA,C∩EPφ
fq.
Acknowledgments
The author would like to thank the referees for reading this paper carefully, providing
valuable suggestions and comments, and pointing out a major error in the original version of
this paper. This research was partially supported by the Commission on Higher Education.
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