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RESEARC H Open Access
Strong convergence theorem for a generalized
equilibrium problem and system of variational
inequalities problem and infinite family of strict
pseudo-contractions
Atid Kangtunyakarn
Correspondence:

Department of Mathematics,
Faculty of Science, King Mongkut’s
Institute of Technology
Ladkrabang, Bangkok 10520,
Thailand
Abstract
In this article, we introduce a new mapping generated by an infinite family of 
i
-
strict pseudo-contractions and a sequence of positive real numbers. By using this
mapping, we consider an iterative method for finding a common element of the set
of a generalized equilibrium problem of the set of solution to a system of variational
inequalities, and of the set of fixed points of an infinite family of strict pseudo-
contractions. Strong convergence theorem of the purposed iteration is established in
the framework of Hilbert spaces.
Keywords: nonexpansive mappings, strongly positive operator, generalized equili-
brium problem, strict pseudo-contraction, fixed point
1 Introduction
Let C be a closed convex subset of a real Hilbert space H, and let G : C × C ® ℝ be a
bifunction. We know that the equilibrium problem for a bifunction G is to find x Î C
such that
G
(


x, y
)
≥ 0 ∀y ∈ C
.
(1:1)
The set of solutions of (1.1) is denoted by EP(G). Given a mapping T : C ® H,letG
(x, y)=〈Tx, y - x〉 fo r all x, y Î. Then, z Î EP(G) if and only if 〈Tz, y - z〉 ≥ 0 for all y
Î C,i.e.,z is a solution of the variational inequality. Let A : C ® H beanonlinear
mapping. The variational inequality problem is to find a u Î C such that

v − u,
A
u


0
(1:2)
for all v Î C. The set of solutions of the variational inequality is denoted by VI(C,
A). Now, we consider the following generalized equilibrium problem:
Find z ∈ C such that G
(
z, y
)
+ Az, y − z≥0, ∀y ∈ C
.
(1:3)
The set of such z Î C is denoted by EP(G, A), i.e.,
EP
(
G, A

)
= {z ∈ C : G
(
z, y
)
+ Az, y − z≥0, ∀y ∈ C
.
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>© 2011 Kangtunyakarn; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons
Attribution License ( which permits unrestricted use , distribution, and reproduction in
any medium, provided the original work is properly cited.
In the case of A ≡ 0, EP(G, A) is denoted by EP(G). In th e case of G ≡ 0, EP(G, A)is
also denoted by VI(C, A). Numerous problems in physics, optimization, variational
inequalities, minimax problems, the Nash equilibrium problem in noncooperative
games, and economics reduce to find a solution of (1.3) (see, for instance, [1]-[3]).
A mapping A of C into H is called inverse-strongly monot one (see [4]), if there exists
a positive real number a such that
x −
y
, Ax − A
y
≥α||Ax − A
y
||
2
for all x, y Î C.
A mapping T with domain D(T) and range R(T) is called nonexpansive if
||
Tx − T
y||


||
x −
y||
(1:4)
for all x, y Î D(T) and T is said to be - strict pseudo-contration if there exist  Î [0,
1) such that
||Tx − Ty||
2
≤||x − y||
2
+ κ||
(
I − T
)
x −
(
I − T
)
y||
2
, ∀x, y ∈ D
(
T
).
(1:5)
We know that the class of -strict pseudo-contractions includes class of nonexpan-
sive mappings. If  = 1, then T is said to be pseudo-contractive. T is strong pseudo-con-
traction if there exists a positive constant l Î (0, 1) such that T + lI is pseudo-
contractive. In a real Hilbert space H (1.5) is equivalent to

Tx − Ty, x − y≤ ||x − y||
2

1 − κ
2
||(I − T)x − (I − T)y||
2
∀x, y ∈ D(T)
.
(1:6)
T is pseudo-contractive if and only if
Tx − Ty , x − y≤ ||x − y||
2
∀x, y ∈ D
(
T
).
Then, T is strongly pseudo-contractive, if there exists a positive constant l Î (0, 1)
such that
Tx − Ty , x − y≤
(
1 − λ
)
x − y
2
, ∀x, y ∈ D
(
T
).
The class of -strict pseudo-contractions fall into the one between classes of nonex-

pansive mappings and pseudo-contractions, and the class of strong pseudo-contrac-
tions is independent of the class of -strict pseudo-contractions.
We denote by F(T) the set of fixed points of T.IfC ⊂ H is bounded, closed and con-
vex and T is a nonexpansive mapping of C into itself, then F(T) is nonem pty; for
instance, see [5]. Recently, Tada and Takahashi [6] and Takahashi and Takahashi [7]
considered iterative methods for finding an element of EP(G) ∩ F(T). Browder and Pet-
ryshyn [8] showed that if a -strict pseudo-contraction T has a fixed point in C, then
starting with an initial x
0
Î C, the sequence {x
n
} generated by the recursive formula:
x
n+1
= αx
n
+
(
1 − α
)
Tx
n
,
(1:7)
where a isaconstantsuchthat0< a <1, converges weakly to a fixed point of T.
Marino and Xu [9] extended Browder and Petryshyn’s above mentioned result by prov-
ing that the sequence {x
n
} generated by the following Manns algorithm [10]:
x

n+1
= α
n
x
n
+
(
1 − α
n
)
Tx
n
(1:8)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 2 of 16
converges weakly to a fixed point of T provided the control sequence

n
}

n
=
0
satisfies
the conditions that  < a
n
<1 for all n and


n

=
0

n
− κ)(1 − α
n
)=

.
Recently, in 2009, Qin et al. [11] introd uced a general iterative method for finding a
common element of EP(F, T), F(S), and F(D). They defined {x
n
} as follows:


















x
1
, u ∈ C,
Fu
n
, y + Tx
n
, y − u
n
 +
1
r
y − u
n
, u
n
− x
n
, ∀y ∈ C,
y
n
= P
C
(x
n
− ηBx
n
),
v
n

= P
C
(y
n
− λAy
n
),
x
n+1
= α
n
u + β
n
x
n
+ γ
n

1
S
k
x
n
+ μ
2
u
n
+ μ
3
v

n
), ∀n ∈ N
,
(1:9)
where the mapping D : C ® C is defined by D(x)=P
C
(P
C
(x - hBx)-lAP
C
(x - hBx)),
S
k
is the mapping defined by S
k
x = kx +(1-k)Sx, ∀x Î C, S : C ® C is a -strict
pseudo-contraction, and A, B : C Î H are a-inverse-strongly monotone mapping and
b-inverse-strongly monotone mappings, respectively. Under suitable conditions, they
proved strong convergence of {x
n
} defined by (1.9) to z = P
EP(F, T)∩F(S) ∩F(D)
u.
Let C be a nonempty convex subset of a real Hilbe rt sp ace. Let T
i
, i = 1, 2, be map-
pings of C into itself. For each j = 1, 2, , let
α
j
=(α

j
1
, α
j
2
, α
j
3
) ∈ I × I ×
I
where I = [0, 1]
and
α
j
1
+ α
j
2
+ α
j
3
=
1
.Foreveryn Î N, we define the mapping S
n
: C ® C as follows:
U
n,n+1
= I
U

n,n
= α
n
1
T
n
U
n,n+1
+ α
n
2
U
n,n+1
+ α
n
3
I
U
n,n−1
= α
n−1
1
T
n−1
U
n,n
+ α
n−1
2
U

n,n
+ α
n−1
3
I
.
.
.
U
n,k+1
= α
k+1
1
T
k+1
U
n,k+2
+ α
k+1
2
U
n,k+2
+ α
k+1
3
I
U
n,k
= α
k

1
T
k
U
n,k+1
+ α
k
2
U
n,k+1
+ α
k
3
I
.
.
.
U
n,2
= α
2
1
T
2
U
n,1
+ α
2
2
U

n,1
+ α
2
3
I
S
n
= U
n,1
= α
1
1
T
1
U
n,2
+ α
1
2
U
n,2
+ α
1
3
I.
This mapping is called S-mapping generated by T
n
, , T
1
and a

n
, a
n-1
, , a
1
.
Question. H ow can we define an iterative method for finding an element in
F =


i
=1
F( T
i
)

N
i
=1
EF(F
i
, A
i
)

N
i
=1
F( G
i

)
?
In this article, motivated by Qin et al. [11], by using S-mapping, we int roduce a new
iteration method for finding a common element of the set of a generalized equilibrium
problem of the set of solution to a system of variational inequalities, and of the set of
fixed points of an infinite family of strict pseudo-contractions. Our iteration scheme is
define as follows.
For u, x
1
Î C, let {x
n
} be generated by







F
i
v
i
n
, v + A
i
x
n
, v − v
i

n
 +
1
r
i
v − v
i
n
, v
i
n
− x
n
, ∀v ∈ C, i =1,2, , N
.
y
n
= 
N
i=1
δ
i
v
i
n
x
n+1
= α
n
u + β

n
x
n
+ γ
n
(a
n
S
n
x
n
+ b
n
Bx
n
+ c
n
y
n
), ∀n ∈ N.
For i = 1, 2, , N,letF
i
: C × C ® ℝ be bifunction, A
i
: C ® H be a
i
-inverse
strongly monotone and let G
i
: C ® C be defined by G

i
(y)=P
C
(I - l
i
A
i
)y, ∀y Î C with
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 3 of 16
(0, 1] ⊂ (0, 2 a
i
) such that
F =


i
=1
F( T
i
)

N
i
=1
EF(F
i
, A
i
)


N
i
=1
F( G
i
) =

, where B
is the K-mapping generated by G
1
, G
2
, , G
N
and b
1
, b
2
, , b
N
.
We prove a strong convergence theorem of purposed iterative sequence {x
n
}toa
point
z ∈ F
and z is a solution of (1.10)












x − z, A
1
z≥0
x − z, A
2
z≥0
.
.
.
x − z, A
N
z≥0, ∀x ∈ C and λ
i

(
0, 1] i =1,2, , N
.
(1:10)
2 Preliminaries
In this section, we coll ect and provide someusefullemmasthatwillbeusedforour
main result in the next section.

Let C be a closed convex subset of a real Hilbert space H,andletP
C
be the metric
projection of H onto C i.e., so that for x Î H, P
C
x satisfies the property:
|
x − P
C
x|| =m
i
n
y
∈C
||x − y||
.
The following characterizes the projection P
C
.
Lemma 2.1 [5]. Given x Î HandyÎ C. Then, P
C
x = y if and only if there holds the
inequality
x −
y
,
y
− z≥0 ∀z ∈ C
.
Lemma 2.2 [12]. Let {s

n
} be a sequence of nonnegative real number satisfying
s
n+1
=
(
1 − α
n
)
s
n
+ α
n
β
n
, ∀n ≥
0
where {a
n
}, {b
n
} satisfy the conditions
(1) {α
n
}⊂[0, 1],


n=1
α
n

= ∞;
(2) lim sup
n→∞
β
n
≤ 0 or


n
=1

n
β
n
| < ∞
.
Then lim
n®∞
s
n
=0.
Lemma 2.3 [13]. Let C be a closed convex subset of a strictly convex Banach space E.
Let {T
n
: n Î N} be a sequence of nonexpansive mappings on C. Suppose


n
=1
F( T

n
)
is
nonempty. Let {l
n
} be a sequence of positive numbers with


n
=1
λ
n
=
1
. Then, a mapping
S on C defined by
S(x)=

n
=1
λ
n
T
n
x
n
for x Î C is well defined, nonexpansive and
F( S)=



n
=1
F( T
n
)
hold.
Lemma 2.4 [14] . Let E be a uniformly convex Banach space, C be a nonempty closed
convex subset of E and S : C ® C be a nonexpansive mapping. Then, I - Sisdemi-
closed at zero.
Lemma 2.5 [15]. Let {x
n
} and {z
n
} be bounded sequences in a Banach space X and let
{b
n
} be a sequence in 0[1]with 0 <lim inf
n®∞
b
n
≤ lim sup
n®∞
b
n
<1.
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 4 of 16
Suppose
x
n+1

= β
n
x
n
+
(
1 − β
n
)
z
n
for all integer n ≥ 0 and
lim sup
n
→∞
(||z
n+1
− z
n
|| − ||x
n+1
− x
n
||) ≤ 0
.
Then lim
n®∞
||x
n
- z

n
|| = 0.
For solving the equilibrium problem for a bifunction F : C × C ® ℝ, let us assume
that F satisfies the following conditions:
(A1) F(x, x)=0∀x Î C;
(A2) F is monotone, i.e. F(x, y)+F(y, x) ≤ 0, ∀x, y Î C;
(A3) ∀x , y, z Î C,
lim
t

0
+
F( tz +(1− t)x, y) ≤ F(x, y)
;
(A4) ∀x Î C, y ↦ F(x, y) is convex and lower semicontinuous.
The following lemma appears implicitly in [1].
Lemma 2.6 [1]. LetCbeanonemptyclosedconvexsubsetofH,andletFbea
bifunction of C × C into ℝ satisfying (A1) - (A4). Let r >0 and x Î H. Then, there exists
z Î C such that
F( z , y)+
1
r
y − z, z − x

(2:1)
for all x Î C.
Lemma 2.7 [16]. Assume that F : C × C ® ℝ satisfies (A1) - (A4). For r >0 and x Î
H, define a mapping T
r
: H ® C as follows.

T
r
(x)={z ∈ C : F(z, y)+
1
r
y − z, z − x≥0, ∀y ∈ C}
.
for all z Î H. Then, the following hold.
(1) T
r
is single-valued,
(2) T
r
is firmly nonexpansive i.e
T
r
(
x
)
− T
r
(
y
)

2
≤T
r
(
x

)
− T
r
(
y
)
, x − y∀x, y ∈ H
;
(3) F(T
r
)=EP (F );
(4) EP (F) is closed and convex.
Definition 2.1 [17]. Let C be a nonempty convex subset of real Banach space. Let
{T
i
}
N
i
=
1
be a finite family of nonexpanxive mappings of C into itself, and let l
1
, ,l
N
be real num-
bers such that 0 ≤ l
i
≤ 1 for every i = 1, , N . We define a mapping K : C ® Casfollows.
U
1

= λ
1
T
1
+(1− λ
1
)I,
U
2
= λ
2
T
2
U
1
+(1− λ
2
)U
1
,
U
3
= λ
3
T
3
U
2
+(1− λ
3

)U
2
,
.
.
.
U
N−1
= λ
N−1
T
N−1
U
N−2
+(1− λ
N−1
)U
N−2
,
K = U
N
= λ
N
T
N
U
N−1
+
(
1 − λ

N
)
U
N−1
.
(2:3)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 5 of 16
Such a mapping K is called the K-mapping generated by T
1
, , T
N
and l
1
, , l
N
.
Lemma 2.8 [17]. LetCbeanonemptyclosedconvexsubsetofastrictlyconvex
Banach space. Let
{T
i
}
N
i
=
1
be a finite family of nonexpanxive mappings of C in to itself
with

N

i
=1
F( T
i
) =

and let l
1
, ,l
N
be real numbers such that 0 < l
i
<1 for every i =
1, , N -1and 0 < l
N
≤ 1. Let K be the K-mapping generated by T
1
, , T
N
and l
1
, ,
l
N
. Then
F( K)=

N
i
=1

F( T
i
)
.
Lemma 2.9 [9]. LetCbeanonemptyclosedconvexsubsetofarealHilbertspaceH
and S : C ® C be a self-mapping of C. If S is a -strict pseudo-contraction mapping,
then S satisfies the Lipschitz condition.
|
|Sx − Sy|| ≤
1+κ
1 −
κ
||x − y||, ∀x, y ∈ C
.
Lemma 2.10. Let C be a nonempty closed convex subset of a real Hilbert space. Let
{T
i
}
N
i
=
1
be 
i
-strict pseudo-contr action mappings of C into itself with


i
=1
F( T

i
) =

and 
=sup
i

i
and let
α
j
=(α
j
1
, α
j
2
, α
j
3
) ∈ I × I ×
I
, where I = [0, 1],
α
j
1
+ α
j
2
≤ b < 1

,
α
j
1
+ α
j
2
≤ b <
1
, and
α
j
1
, α
j
2
, α
j
3
∈ (κ,1
)
for all j = 1, 2, For every n Î
N, let S
n
be S-mapping generated by T
n
, ,T
1
and a
n

, a
n-1
, , a
1
. Then, for every x Î
C and k Î N, lim
n®∞
U
n
,
k
x exists.
Proof. Let x Î C and
y ∈


i
=1
F( T
i
)
. Fix k Î N, then for every n Î N with n ≥ k,
we have
U
n+1,k
x − U
n,k
x
2
= α

k
1
T
k
U
n+1,k+1
x + α
k
2
U
n+1,k+1
x + α
k
3
x − α
k
1
T
k
U
n,k+1
x
−α
k
2
U
n,k+1
x − α
k
3

x
2
= α
k
1
(T
k
U
n+1,k+1
x − T
k
U
n,k+1
x)+α
k
2
(U
n+1,k+1
x − U
n,k+1
x)
2
≤ α
k
1
T
k
U
n+1,k+1
x − T

k
U
n,k+1
x
2
+ α
k
2
U
n+1,k+1
x − U
n,k+1
x
2
−α
k
1
α
k
2
T
k
U
n+1,k+1
x − T
k
U
n,k+1
x − U
n+1,k+1

x + U
n,k+1
x
2
≤ α
k
1
(U
n+1,k+1
x − U
n,k+1
x
2
+ κ(I − T
k
)U
n+1,k+1
x
−(I − T
k
)U
n,k+1
x
2
)+α
k
2
U
n+1,k+1
x − U

n,k+1
x
2
−α
k
1
α
k
2
(I − T
k
)U
n,k+1
x − (I − T
k
)U
n+1,k+1
x
2
≤ (1 − α
k
3
)U
n+1,k+1
x − U
n,k+1
x
2
.
.

.
≤ 
n
j=k
(1 − α
j
3
)U
n+1,n+1
x − U
n,n+1
x
2
= 
n
j=k
(1 − α
j
3
)α
n+1
1
T
n+1
U
n+1,n+2
x + α
n+1
2
U

n+1,n+2
x + α
n+1
3
x − x
2
= 
n
j=k
(1 − α
j
3
)α
n+1
1
T
n+1
x +(1− α
n+1
1
)x − x
2
= 
n
j=k
(1 − α
j
3
)α
n+1

1
(T
n+1
x − x)
2
≤ 
n
j=k
(1 − α
j
3
)(T
n+1
x − y + y − x)
2
≤ 
n
j=k
(1 − α
j
3
)

1+κ
1 − κ
x − y + y − x

2
≤ 
n

j=k
(1 − α
j
3
)

2
1 − κ
x − y

2
≤ b
n−(k−1)

2
1 − κ
x − y 

2
.
It follows that
||U
n+1,k
x − U
n,k
x|| ≤ b
n − (k − 1)
2

2

1 − κ
||x − y||

Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 6 of 16
=
b
n
2
b
k−1
2

2
1 − κ
||x − y||

=
a
n
a
k−1
M,
(2:4)
where
a = b
1
2

(

0, 1
)
and
M =
2
1 −
κ
||x − y|
|
For any k, n, p Î N, p>0, n ≥ k, we have

U
n+p,k
x − U
n,k
x



U
n+p,k
x − U
n+p−1,k
x

+

U
n+p−1,k
x − U

n+p−2,k
x

+
.
+U
n+1,k
x − U
n,k
x
= 
n+p−1
j=n
U
j+1,k
x − U
j,k
x
≤ 
n+p−1
j=n
a
j
a
k−1
M

a
n
(

1 − a
)
a
k−1
M.
(2:5)
Since a Î (0, 1), we have lim
n®∞
a
n
=0.From(2.5),wehavethat{U
n
,
k
x}isaCau-
chy sequence. Hence lim
n®∞
U
n,k
x exists. □
For every k Î N and x Î C, we define mapping U
∞,k
and S : C Î C as follows:
lim
n


U
n,k
x = U

∞,k
x
(2:6)
and
lim
n


S
n
x = lim
n


U
n,1
x = S
x
(2:7)
Such a mapping S is called S-mapping generated by T
n
, T
n-1
, and a
n
, a
n-1
,
Remark 2.11. For each n Î N, S
n

is nonexpansive and lim
n®∞
sup
xÎD
||S
n
x - Sx|| = 0
for every bounded subset D of C. To show this, let x, y Î C and D be a bounded
subset of C. Then, we have
S
n
x − S
n
y
2
= α
1
1
(T
1
U
n,2
x − T
1
U
n,2
y)+α
1
2
(U

n,2
x − U
n,2
y)+α
1
3
(x − y)
2
≤ α
1
1
T
1
U
n,2
x − T
1
U
n,2
y
2
+ α
1
2
U
n,2
x − U
n,2
y
2

+ α
1
3
x − y
2
−α
1
1
α
1
2
T
1
U
n,2
x − T
1
U
n,2
y − U
n,2
x + U
n,2
y
2
≤ α
1
1
(U
n,2

x − U
n,2
y
2
+ κ(I − T
1
)U
n,2
x − (I − T
1
)U
n,2
y
2
)

1
2
U
n,2
x − U
n,2
y
2
+ α
1
3
x − y
2
− α

1
1
α
1
2
(I − T
1
)U
n,2
y − (I − T
1
)U
n,2
x
2
≤ (1 − α
1
3
)U
n,2
x − U
n,2
y
2
+ α
1
3
x − y
2
≤ (1 − α

1
3
)((1 − α
2
3
)U
n,3
x − U
n,3
y
2
+ α
2
3
x − y
2
)+α
1
3
x − y)
2
=(1− α
1
3
)(1 − α
2
3
)U
n,3
x − U

n,3
y
2
+ α
2
3
(1 − α
1
3
)x − y
2
+ α
1
3
x − y)
2
= 
2
j=1
(1 − α
j
3
)||U
n,3
x − U
n,3
y||
2
+(1− 
2

j=1
(1 − α
j
3
))||x − y||
2
.
.
.
≤ 
n
j=1
(1 − α
j
3
)||U
n,n+1
x − U
n,n+1
y||
2
+(1− 
n
j=1
(1 − α
j
3
))||x − y||
2
= ||x −

y
||
2
.
Then, we have that S : C ® C is also nonexpansive indeed, observe that for each x, y Î C
|
Sx − Sy|| = lim
n


||S
n
x − S
n
y|| ≤ ||x − y||
.
By (2.8), we have
||
S
n+1
x −
S
n
x|| = ||U
n+1,1
x − U
n,1
x|
|


a
n
M.
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 7 of 16
This implies that for m>nand x Î D,
|
|S
m
x − S
n
x|| ≤ 
m−1
j=n
||S
j+1
x − S
j
x|
|
≤ 
m−1
j=n
a
j
M

a
n
1 −

a
M.
By letting m ® ∞, for any x Î D, we have
|
|Sx − S
n
x|| ≤
a
n
1 −
a
M
.
(2:8)
It follows that
lim
n


sup
x∈D
||S
n
x − Sx|| =0
.
(2:9)
Lemma 2.12. Let C be a nonempty closed convex subset of a real Hilbert space. Let
{T
i
}


i
=
1
be 
i
-strict pseudo-contraction mappings of C into itself with


i
=1
F( T
i
) =

and
 =sup


i
and let
α
j
=(α
j
1
, α
j
2
, α

j
3
) ∈ I × I ×
I
, where I = [0, 1],
α
j
1
+ α
j
2
+ α
j
3
=
1
,
α
j
1
, α
j
2
, α
j
3
∈ (κ,1
)
and
α

j
1
, α
j
2
, α
j
3
∈ (κ,1
)
for all j = 1, For every n Î N, let S
n
and S
be S-mappings generated by T
n
, , T
1
and a
n
, a
n-1
, , a
1
and T
n
, T
n-1
, , and a
n
, a

n-
1
, , respectively. Then
F( S)=


i
=1
F( T
i
)
.
Proof. It is evident that


i
=1
F( T
i
) ⊆ F(S
)
. For every n, k Î N, with n ≥ k,letx
0
Î F
(S) and
x





i
=1
F( T
i
)
, we have
S
n
x
0
− x


2
= α
1
1
(T
1
U
n,2
x
0
− x

)+α
1
2
(U
n,2

x
0
− x

)+α
1
3
(x
0
− x

)
2
≤ α
1
1
T
1
U
n,2
x
0
− x


2
+ α
1
2
U

n,2
x
0
− x


2
+ α
1
3
x
0
− x


2
−α
1
1
α
1
2
T
1
U
n,2
x
0
− U
n,2

x
0

2
− α
1
2
α
1
3
U
n,2
x
0
− x
0

2
≤ α
1
1
(U
n,2
x
0
− x


2
+ κ(I − T

1
)U
n,2
x
0

2
)+α
1
2
U
n,2
x
0
− x


2

1
3
x
0
− x


2
− α
1
1

α
1
2
T
1
U
n,2
x
0
− U
n,2
x
0

2
− α
1
2
α
1
3
U
n,2
x
0
− x
0

2
=(1− α

1
3
)U
n,2
x
0
− x


2
+ α
1
3
x
0
− x


2
−α
1
1

1
2
− κ)T
1
U
n,2
x

0
− U
n,2
x
0

2
− α
1
2
α
1
3
U
n,2
x
0
− x
0

2
≤ (1 − α
1
3
)((1 − α
2
3
)U
n,3
x

0
− x


2
+ α
2
3
x
0
− x


2
−α
2
1

2
2
− κ)T
2
U
n,3
x
0
− U
n,3
x
0


2
− α
2
2
α
2
3
U
n,3
x
0
− x
0

2
)+α
1
3
x
0
− x


2
−α
1
1

1

2
− κ)T
1
U
n,2
x
0
− U
n,2
x
0

2
− α
1
2
α
1
3
U
n,2
x
0
− x
0

2
=(1− α
1
3

)(1 − α
2
3
)U
n,3
x
0
− x


2
+ α
2
3
(1 − α
1
3
)x
0
− x


2
+ α
1
3
x
0
− x



2
−α
2
1

2
2
− κ)(1 − α
1
3
)T
2
U
n,3
x
0
− U
n,3
x
0

2
− α
2
2
α
2
3
(1 − α

1
3
)U
n,3
x
0
− x
0

2
−α
1
1

1
2
− κ)T
1
U
n,2
x
0
− U
n,2
x
0

2
− α
1

2
α
1
3
U
n,2
x
0
− x
0

2
= 
2
j=1
(1 − α
j
3
)U
n,3
x
0
− x


2
+(1− 
2
j=1
(1 − α

j
3
)x
0
− x


2
−α
2
1

2
2
− κ)(1 − α
1
3
)T
2
U
n,3
x
0
− U
n,3
x
0

2
− α

2
2
α
2
3
(1 − α
1
3
)U
n,3
x
0
− x
0

2
−α
1
1

1
2
− κ)T
1
U
n,2
x
0
− U
n,2

x
0

2
− α
1
2
α
1
3
U
n,2
x
0
− x
0

2
≤ 
2
j=1
(1 − α
j
3
)((1 − α
3
3
)U
n,4
x

0
− x


2
+ α
3
3
x
0
− x


2
−α
3
1

3
2
− κ)T
3
U
n,4
x
0
− U
n,4
x
0


2
− α
3
2
α
3
3
U
n,4
x
0
− x
0

2
)
+(1 − 
2
j=1
(1 − α
j
3
)x
0
− x


2
− α

2
1

2
2
− κ)(1 − α
1
3
)T
2
U
n,3
x
0
− U
n,3
x
0

2
−α
2
2
α
2
3
(1 − α
1
3
)U

n,3
x
0
− x
0

2
− α
1
1

1
2
− κ)T
1
U
n,2
x
0
− U
n,2
x
0

2
−α
1
2
α
1

3
U
n,2
x
0
− x
0

2
= 
2
j=1
(1 − α
j
3
)(1 − α
3
3
)U
n,4
x
0
− x


2
+ α
3
3


2
j=1
(1 − α
j
3
)x
0
− x


2
−α
3
1

3
2
− κ)
2
j=1
(1 − α
j
3
)T
3
U
n,4
x
0
− U

n,4
x
0

2
−α
3
2
α
3
3

2
j=1
(1 − α
j
3
)U
n,4
x
0
− x
0

2
+(1− 
2
j=1
(1 − α
j

3
)x
0
− x


2
−α
2
1

2
2
− κ)(1 − α
1
3
)T
2
U
n,3
x
0
− U
n,3
x
0

2
− α
2

2
α
2
3
(1 − α
1
3
)U
n,3
x
0
− x
0

2
−α
1
1

1
2
− κ)T
1
U
n,2
x
0
− U
n,2
x

0

2
− α
1
2
α
1
3
U
n,2
x
0
− x
0

2
= 
3
j=1
(1 − α
j
3
)U
n,4
x
0
− x



2
+(1− 
3
j=1
(1 − α
j
3
)x
0
− x


2
−α
3
1

3
2
− κ)
2
j=1
(1 − α
j
3
)T
3
U
n,4
x

0
− U
n,4
x
0

2
−α
3
2
α
3
3

2
j=1
(1 − α
j
3
)U
n,4
x
0
− x
0

2
− α
2
1


2
2
− κ)(1 − α
1
3
)T
2
U
n,3
x
0
− U
n,3
x
0

2
−α
2
2
α
2
3
(1 − α
1
3
)U
n,3
x

0
− x
0

2
− α
1
1

1
2
− κ)T
1
U
n,2
x
0
− U
n,2
x
0

2
−α
1
2
α
1
3
U

n,2
x
0
− x
0

2
.
.
.
(2:10)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 8 of 16
.
.
.
≤ 
k+1
j=1
(1 − α
j
3
)||U
n,k+2
x
0
− x

||
2

+(1− 
k+1
j=1
(1 − α
j
3
)||x
0
− x

||
2
− α
k+1
1

k+1
2
− κ)
k
j=1
(1 − α
j
3
)||T
k+1
U
n,k+2
x
0

− U
n,k+2
x
0
||
2
− α
k+1
2
α
k+1
3

k
j=1
(1 − α
j
3
)||U
n,k+2
x
0
− x
0
||
2
− α
k
1


k
2
− κ)
k−1
j=1
(1 − α
j
3
)||T
k
U
n,k+1
x
0
− U
n,k+1
x
0
||
2
− α
k
2
α
k
3

k−1
j
=1

(1 − α
j
3
)||U
n,k+1
x
0
− x
0
||
2
(2:11)
.
.
.
−α
3
1

3
2
− κ)
2
j=1
(1 − α
j
3
)T
3
U

n,4
x
0
− U
n,4
x
0

2
− α
3
2
α
3
3

2
j=1
(1 − α
j
3
)U
n,4
x
0
− x
0

2
−α

2
1

2
2
− κ)(1 − α
1
3
)T
2
U
n,3
x
0
− U
n,3
x
0

2
− α
2
2
α
2
3
(1 − α
1
3
)U

n,3
x
0
− x
0

2
−α
1
1

1
2
− κ)T
1
U
n,2
x
0
− U
n,2
x
0

2
− α
1
2
α
1

3
U
n,2
x
0
− x
0

2
.
.
.
≤ 
n
j=1
(1 − α
j
3
)U
n,n+1
x
0
− x


2
+(1− 
n
j=1
(1 − α

j
3
)x
0
− x


2
−α
n
1

n
2
− κ)
n−1
j=1
(1 − α
j
3
)T
n
U
n,n+1
x
0
− U
n,n+1
x
0


2
−α
n
2
α
n
3

n−1
j=1
(1 − α
j
3
)U
n,n+1
x
0
− x
0

2
.
.
.
−α
k+1
1

k+1

2
− κ)
k
j=1
(1 − α
j
3
)T
k+1
U
n,k+2
x
0
− U
n,k+2
x
0

2
−α
k+1
2
α
k+1
3

k
j=1
(1 − α
j

3
)U
n,k+2
x
0
− x
0

2
−α
k
1

k
2
− κ)
k−1
j=1
(1 − α
j
3
)T
k
U
n,k+1
x
0
− U
n,k+1
x

0

2
−α
k
2
α
k
3

k−1
j=1
(1 − α
j
3
)U
n,k+1
x
0
− x
0

2
−α
k−1
1

k−1
2
− κ)

k−2
j=1
(1 − α
j
3
)T
k−1
U
n,k
x
0
− U
n,k
x
0

2
−α
k−1
2
α
k−1
3

k−2
j=1
(1 − α
j
3
)U

n,k
x
0
− x
0

2
.
.
.
−α
3
1

3
2
− κ)
2
j
=1
(1 − α
j
3
)T
3
U
n,4
x
0
− U

n,4
x
0

2
− α
3
2
α
3
3

2
j
=1
(1 − α
j
3
)U
n,4
x
0
− x
0

2
(2:12)
−α
2
1


2
2
− κ)(1 − α
1
3
)T
2
U
n,3
x
0
− U
n,3
x
0

2
− α
2
2
α
2
3
(1 − α
1
3
)U
n,3
x

0
− x
0

2
−α
1
1

1
2
− κ)T
1
U
n,2
x
0
− U
n,2
x
0

2
− α
1
2
α
1
3
U

n,2
x
0
− x
0

2
= x
0
− x


2
−α
n
1

n
2
− κ)
n−1
j=1
(1 − α
j
3
)T
n
U
n,n+1
x

0
− U
n,n+1
x
0

2
.
.
.
−α
k+1
1

k+1
2
− κ)
k
j=1
(1 − α
j
3
)T
k+1
U
n,k+2
x
0
− U
n,k+2

x
0

2
−α
k+1
2
α
k+1
3

k
j=1
(1 − α
j
3
)U
n,k+2
x
0
− x
0

2
−α
k
1

k
2

− κ)
k−1
j=1
(1 − α
j
3
)T
k
U
n,k+1
x
0
− U
n,k+1
x
0

2
−α
k
2
α
k
3

k−1
j=1
(1 − α
j
3

)U
n,k+1
x
0
− x
0

2
−α
k−1
1

k−1
2
− κ)
k−2
j=1
(1 − α
j
3
)T
k−1
U
n,k
x
0
− U
n,k
x
0


2
−α
k−1
2
α
k−1
3

k−2
j=1
(1 − α
j
3
)U
n,k
x
0
− x
0

2
.
.
.
−α
3
1

3

2
− κ)
2
j=1
(1 − α
j
3
)T
3
U
n,4
x
0
− U
n,4
x
0

2
− α
3
2
α
3
3

2
j=1
(1 − α
j

3
)U
n,4
x
0
− x
0

2
−α
2
1

2
2
− κ)(1 − α
1
3
)T
2
U
n,3
x
0
− U
n,3
x
0

2

− α
2
2
α
2
3
(1 − α
1
3
)U
n,3
x
0
− x
0

2
−α
1
1

1
2
− κ)T
1
U
n,2
x
0
− U

n,2
x
0

2
− α
1
2
α
1
3
U
n,2
x
0
− x
0

2
.
(2:13)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 9 of 16
For k Î N and (2.12), we have
α
k−1
2
α
k−1
3


k−2
j=1
(1 − α
j
3
)||U
n,k
x
0
− x
0
||
2
≤||x
0
− x

||
2
−||S
n
x
0
− x

||
2
,
(2:14)

as n ® ∞. This implies that U

,
k
x
0
= x
0
, ∀k Î N.
Again by (2.12), we have
α
k
1

k
2
− κ)
k−1
j
=1
(1 − α
j
3
)||T
k
U
n,k+1
x
0
− U

n,k+1
x
0
||
2
≤||x
0
− x

||
2
−||S
n
x
0
− x

||
2
,
(2:15)
as n ® ∞. Hence
α
k
1

k
2
− κ)
k−1

j
=1
(1 − α
j
3
)||T
k
U
∞,k+1
x
0
− U
∞,k+1
x
0
||
2
≤ 0
.
(2:16)
From U
∞,k
x
0
= x
0
, ∀k Î N, and (2.15), we obtain that T
k
x
0

= x
0
, ∀k Î N. This implies
that
x
0



i
=1
F( T
i
)
. □
Lemma 2.13. Let C be a closed convex subs et of Hilbert space H. Let A
i
: C ® Hbe
mappings and let G
i
: C ® CbedefinedbyG
i
(y)=P
C
(I - l
i
A
i
)ywithl
i

>0, ∀
i
=1,2,
N. Then
x



N
i
=1
VI(C, A
i
)
if and only if
x



N
i
=1
F( G
i
)
.
Proof. For given
x




N
i
=1
VI(C, A
i
)
,wehavex* Î VI(C, A
i
), ∀
i
=1,2, ,N.Since
〈A
i
x*, x - x*〉 ≥ 0, we have 〈l
i
A
i
x*, x - x*〉 ≥ 0, ∀l
i
>0, i = 1, 2, , N. It follows that
x


(
I − λ
i
A
i
)

x

, x − x

 = λ
i
A
i
x

, x − x

≥0, ∀x ∈ C, i =1,2, , N
.
(2:17)
Hence, x*=P
C
(I - l
i
A
i
)x*=G
i
(x*), ∀x Î C, i =1,2, ,N. Therefore, we have
x



N
i

=1
F( G
i
)
. For the converse, let
x



N
i
=1
F( G
i
)
; then, we have for every i = 1, ,
N, x*=G
i
(x*) = P
C
(I - l
i
A
i
)x*, ∀l
i
>0, i = 1, 2, , N. It implies that
x



(
I − λ
i
A
i
)
x

, x − x

 = λ
i
A
i
x

, x − x

≥0, ∀i =1,2, , N, ∀x ∈ C
.
(2:18)
Hence, 〈A
i
x*, x-x*〉 ≥ 0, ∀x Î C,sox* Î VI (C, A
i
), ∀i =1,2, ,N. Hence,
x




N
i
=1
VI(C, A
i
)
.

3 Main results
Theorem 3.1. Let C be a close d convex subset of Hilbert space H. For every i = 1, 2, ,
N, let F
i
: C × C ® ℝ be a bifunction satisfying (A
1
)-(A
4
), let A
i
: C ® Hbea
i
-inverse
strongly monotone and let G
i
: C ® C be defined by G
i
(y)=P
C
(I - l
i
A

i
)y, ∀y Î C with
l
i
Î (0, 1] ⊂ (0, 2a
i
). Let B : C ® C be the K-mapping generated by G
1
, G
2
, , G
N
and
b
1
, b
2
, , b
N
where b
i
Î (0, 1), ∀i = 1, 2, 3, , N-1, b
N
Î (0, 1] and let
{
T
i
}

i

=
1
be 
i
-
strict pseudo-contraction mappings of C into itself with  = sup
i

i
and let
ρ
j
=(α
j
1
, α
j
2
, α
j
3
) ∈ I × I ×
I
,whereI=[0,1],
α
j
1
+ α
j
2

+ α
j
3
=
1
,
α
j
1
+ α
j
2
≤ b < 1
, and
α
j
1
, α
j
2
, α
j
3
∈ (κ,1
)
for all j = 1, 2, . For every n Î N, let S
n
and S are S-mapping gener-
ated by T
n

, ,T
1
and r
n
, r
n-1
, , r
1
and T
n
, T
n-1
, , and r
n
, r
n-1
, ,respectively.
Assume that
F =


i
=1
F( T
i
)

N
i
=1

EF(F
i
, A
i
)

N
i
=1
F( G
i
) =

. For every n Î N, i =1,
2, , N, let {x
n
} and
{v
i
n
}
be generated by x
1
, u Î C and














F
i
v
i
n
, v + A
i
x
n
, v − v
i
n
 +
1
r
i
v − v
i
n
, v
i
n
− x

n
≥0, ∀v ∈ C
,
i =1,2, , N.
y
n
= 
N
i=1
δ
i
v
i
n
x
n+1
= α
n
u + β
n
x
n
+ γ
n
(a
n
S
n
x
n

+ b
n
Bx
n
+ c
n
y
n
), ∀n ∈ N,
(3:1)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 10 of 16
where {a
n
}, {b
n
}, {g
n
}, {a
n
}, {b
n
}, {c
n
} ⊂ (0, 1),
α
n
+ β
n
+ γ

n
= a
n
+ b
n
+ c
n
= 
N
i
=1
δ
i
=
1
,
and
{r
i
}
N
i
=1
⊂ (ς , τ) ⊂ (0, 2α
i
)
, satisfy the following conditions:
(i)
lim
n

→∞
α
n
=
0
and


n
=
0
α
n
=

,
(ii)
0 < lim inf
n→∞
β
n
≤ lim sup
n
→∞
β
n
<
1
,
(iii)

lim
n
→∞
a
n
=
a
,
lim
n
→∞
b
n
=
b
,
lim
n
→∞
c
n
=
c
, with a, b, c Î (0, 1).
Then, the sequence {x
n
}, {y
n
},
{v

i
n
}
, ∀i = 1, 2, , N, converge strongly to
z
= P

u
and z is
a solution of (1.10).
Proof. First, we show that (I - l
i
A
i
) is nonexpansive mapping for eve ry i = 1, 2, , N.
For x, y Î C, we have


(I − λ
i
A
i
)x − (I − λ
i
A
i
)y


2

= ||x − y − λ
i
(A
i
x − A
i
y)||
2
= ||x − y||
2
− 2λ
i
x − y, A
i
x − A
i
y + λ
2
i
||A
i
x − A
i
y||
2
≤||x − y||
2
− 2α
i
λ

i
||A
i
x − A
i
y||
2
+ λ
2
i
||A
i
x − A
i
y||
2
= ||x − y||
2
+ λ
i

i
− 2α
i
)||A
i
x − A
i
y||
2

≤||x −
y
||
2
.
(3:2)
Thus, (I-l
i
A
i
) is nonexpansive, and so are B and G
i
, for all i = 1, 2, , N.
Now, we shall divide our proof into five steps.
Step 1. We shall show that the sequence {x
n
} is bounded. Since
F( v
i
n
, v)+A
i
x
n
, v − v
i
n
 +
1
r

i
v − v
i
n
, v
i
n
− x
n
≥0, ∀v ∈ C, i =1,2, , N
,
(3:3)
we have
F( v
i
n
, v)+
1
r
i
v − v
i
n
, v
i
n
− (I − r
i
A
i

)x
n
≥0, ∀v ∈ C, i =1,2, , N
.
By Lemma 2.7, we have
v
i
n
= T
r
i
(I − r
i
A
i
)x
n
.
Let
z
=

. Then F(z, y)+〈y-z, A
i
z〉 ≥ 0 ∀y Î C, so we have
F( z , y)+
1
r
i
y − z, z − z + r

i
A
i
z≥0, ∀i =1,2, , N
.
Again by Lemma 2.7, we have
z
= T
r
i
(I − r
i
A
i
)
z
, ∀i = 1, 2, , N.SinceB is K-map-
ping generated by G
1
, G
2
, ,G
N
and b
1
, b
2
, ,b
N
and


N
i
=1
F( G
i
) =

. By Lemma 2.8,
we have

N
i
=1
F( G
i
)=F(B
)
.Since
z
=

,wehavez Î F(B). Setting e
n
= a
n
S
n
x
n

+ b
n
Bx
n
+ c
n
y
n
, ∀n Î N, we have

x
n+1
− z

= ||α
n
(u − z)+β
n
(x
n
− z)+γ
n
(e
n
− z)||
≤ α
n
||u − z|| + β
n
||x

n
− z|| + γ
n
||e
n
− z||
= α
n
||u − z|| + β
n
||x
n
− z|| + γ
n
||a
n
(S
n
x
n
− z)+b
n
(Bx
n
− z)+c
n
(y
n
− z)|
|

≤ α
n
||u − z|| + β
n
||x
n
− z|| + γ
n
((1 − c
n
)||x
n
− z|| + c
n
||y
n
− z||)
= α
n
||u − z|| + β
n
||x
n
− z|| + γ
n
((1 − c
n
)||x
n
− z|| + c

n
||
N
i=1
δ
i
(v
i
n
− z)||)
≤ α
n
||u − z|| + β
n
||x
n
− z|| + γ
n
((1 − c
n
)||x
n
− z|| + c
n

N
i=1
δ
i
||v

i
n
− z||)
≤ α
n
||u − z|| + β
n
||x
n
− z|| + γ
n
((1 − c
n
)||x
n
− z|| + c
n
||x
n
− z||)
= α
n
||u − z|| +(1− α
n
)||x
n
− z||,
≤ max
{||
u − z

||
,
||
x
n
− z
||}
.
(3:4)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 11 of 16
By induction, we can prove that {x
n
} is bounded, and so are
{
v
i
n
}
,{y
n
}, {Bx
n
}{S
n
x
n
},
{e
n

}.
Step 2. We will show that lim
n®∞
||x
n+1
- x
n
|| = 0. Let
d
n
=
x
n+1

β
n
x
n
1 −
β
n
,andthen
we have
x
n+1
=
(
1 − β
n
)

d
n
+ β
n
x
n
, ∀n ∈ N
.
(3:5)
From definition of d
n
, we have

d
n+1
− d
n

= ||
x
n+2

β
n+1
x
n+1
1 − β
n+1

x

n+1

β
n
x
n
1 − β
n
||
= ||
α
n+1
u + γ
n+1
e
n+1
1 − β
n+1

α
n
u + γ
n
e
n
1 − β
n
||
= ||
α

n+1
u +(1− β
n+1
− α
n+1
)e
n+1
1 − β
n+1

α
n
u +(1− β
n
− α
n
)e
n
1 − β
n
|
|
= ||
α
n+1
1 − β
n+1
(u − e
n+1
) −

α
n
1 − β
n
(u − e
n
)+e
n+1
− e
n
||

α
n+1
1 − β
n+1
||u − e
n+1
|| +
α
n
1 − β
n
||u − e
n
||
+
||
e
n+1

− e
n
||
.
(3:6)
By definition of e
n
, we have
e
n+1
− e
n
 = a
n+1
S
n+1
x
n+1
+ b
n+1
Bx
n+1
+ c
n+1
y
n+1
− a
n
S
n

x
n
− b
n
Bx
n
− c
n
y
n

= a
n+1
S
n+1
x
n+1
− a
n
S
n+1
x
n+1
+ a
n
S
n+1
x
n+1
+ b

n+1
Bx
n+1
− b
n
Bx
n+1
+b
n
Bx
n+1
+ c
n+1
y
n+1
− c
n
y
n+1
+ c
n
y
n+1
− a
n
S
n
x
n
− b

n
Bx
n
− c
n
y
n

= (a
n+1
− a
n
)S
n+1
x
n+1
+ a
n
(S
n+1
x
n+1
− S
n
x
n
)+(b
n+1
− b
n

)Bx
n+1
+b
n
(Bx
n+1
− Bx
n
)+(c
n+1
− c
n
)y
n+1
+ c
n
(y
n+1
− y
n
)
≤|a
n+1
− a
n
|S
n+1
x
n+1
 + a

n
S
n+1
x
n+1
− S
n
x
n
 + |b
n+1
− b
n
|Bx
n+1

+b
n
Bx
n+1
− Bx
n
 + |c
n+1
− c
n
|y
n+1
 + c
n

y
n+1
− y
n

≤|a
n+1
− a
n
|S
n+1
x
n+1
 + a
n
(S
n+1
x
n+1
− S
n+1
x
n
 + S
n+1
x
n
− S
n
x

n

)
+|b
n+1
− b
n
|Bx
n+1
 + b
n
Bx
n+1
− Bx
n
 + |c
n+1
− c
n
|y
n+1

+c
n

N
i=1
δ
i
T

r
i
(I − r
i
A
i
)x
n+1
− T
r
i
(I − r
i
A
i
)x
n

≤|a
n+1
− a
n
|S
n+1
x
n+1
 + a
n
(x
n+1

− x
n
 + S
n+1
x
n
− S
n
x
n
)
+|b
n+1
− b
n
|Bx
n+1
 + b
n
x
n+1
− x
n
 + |c
n+1
− c
n
|y
n+1


+c
n
x
n+1
− x
n

≤|a
n+1
− a
n
|S
n+1
x
n+1
 + a
n
x
n+1
− x
n
 + S
n+1
x
n
− S
n
x
n


+|b
n+1
− b
n
|Bx
n+1
 + b
n
x
n+1
− x
n
 + |c
n+1
− c
n
|y
n+1

+c
n
x
n+1
− x
n

= x
n+1
− x
n

 + |a
n+1
− a
n
|S
n+1
x
n+1
 + S
n+1
x
n
− S
n
x
n

+|b
n+1
− b
n
|Bx
n+1
 + |c
n+1
− c
n
|
y
n+1

.
(3:7)
By (3.6) and (3.7), we have
|
|d
n+1
− d
n
|| ≤
α
n+1
1 − β
n+1
||u − e
n+1
|| +
α
n
1 − β
n
||u − e
n
|
|
+
||
e
n+1
− e
n

||
(3:8)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 12 of 16

α
n+1
1 − β
n+1
||u − e
n+1
|| +
α
n
1 − β
n
||u − e
n
||
+ ||x
n+1
− x
n
|| + |a
n+1
− a
n
|||S
n+1
x

n+1
|| + ||S
n+1
x
n
− S
n
x
n
|
|
+ |b
n+1
− b
n
|||Bx
n+1
|| + |c
n+1
− c
n
|||
y
n+1
||.
(3:9)
It follows that
d
n+1
− d

n
−x
n+1
− x
n
≤
α
n+1
1 − β
n+1
u − e
n+1
 +
α
n
1 − β
n
u − e
n

+|a
n+1
− a
n
|S
n+1
x
n+1
 + S
n+1

x
n
− S
n
x
n

+|b
n+1
− b
n
|Bx
n+1
 + |c
n+1
− c
n
|y
n+1


α
n+1
1 −
β
n+1
u − e
n+1
 +
α

n
1 −
β
n
u − e
n

(3:10)
+|a
n+1
− a
n
|||S
n+1
x
n+1
|| + ||S
n+1
x
n
− Sx
n
|| + ||Sx
n
− S
n
x
n
|
|

+|b
n+1
− b
n
|||Bx
n+1
|| + |c
n+1
− c
n
|||
y
n+1
||.
(3:11)
From Remark 2.11 and conditions (i)-(iii), we have
lim sup
n
→∞
(||d
n+1
− d
n
|| − ||x
n+1
− x
n
||) ≤ 0
.
(3:12)

From (3.5), (3.12) and Lemma 2.5, we have
lim
n


||d
n
− x
n
|| =0
.
(3:13)
We can rewrite (3.5) as
x
n+1
− x
n
=
(
1 − β
n
)(
d
n
− x
n
)
, ∀n ∈ N
.
(3:14)

By (3.13) and (3.14), we have
lim
n
→∞
||x
n+1
− x
n
|| =0
.
(3:15)
Step. 3. Show that lim
n®∞
||x
n
-e
n
|| = 0. From (3.1), we have
x
n+1
− x
n
+ α
n
(
x
n
− u
)
= γ

n
(
e
n
− x
n
).
It implies that
γ
n
||e
n
− x
n
|| ≤ ||x
n+1
− x
n
|| + α
n
||x
n
− u||
.
By conditions (i), (ii), and (3.15), we have
lim
n
→∞
||e
n

− x
n
|| =0
.
(3:16)
Step. 4. We show that lim sup
n® ∞
〈u - z, x
n
-z〉 ≤ 0, where
z
= P

u
.Let
{x
n
j
}
be a
subsequence of {x
n
} such that
lim sup
n→∞
u − z, x
n
− z = lim sup
j
→∞

u − z, x
n
j
− z

(3:17)
Without loss of generality, we may assume that
{x
n
j
}
converges weakly to some q in
H. Next, we will show that
q ∈ F =


i
=1
F( T
i
)

N

i
=1
EF(F
i
, A
i

)

N

i
=1
F( G
i
)
.
(3:18)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 13 of 16
First, we define a mapping A : C ® C by
Ax = 
N
i
=1
δ
i
T
r
i
(I − r
i
A
i
)x, ∀x ∈ C
.
Since

F( T
r
i
(I − r
i
A
i
)) = EF(F
i
, A
i
), ∀i =1, 2, ,
N
,wehave

N
i
=1
F( T
r
i
(I − r
i
A
i
)) =

N
i
=1

EF(F
i
, A
i
) =

. By Lemma 2.3, we have
F( A)=

N
i
=1
F( T
r
i
(I − r
i
A
i
)
)
.
Next, we define Q : C ® C by
Q
x = aSx + bBx + cAx ∀x ∈ C
.
(3:19)
Again, by Lemma 2.3, we have
F( Q)=F(S)


F( B)

F( A)=


i
=1
F( T
i
)

N

i
=1
F( G
i
)

N

i
=1
EF(F
i
, A
i
)
.
By (3.19), we have

Qx
n
− e
n
 = aSx
n
+ bBx
n
+ cAx
n
− a
n
S
n
x
n
− b
n
Bx
n
− c
n
y
n

= aSx
n
− aS
n
x

n
+ aS
n
x
n
+ bBx
n
+ c
N
i=1
δ
i
T
r
i
(I − r
i
A
i
)x
n
− a
n
S
n
x
n
−b
n
Bx

n
− c
n

N
i=1
δ
i
T
r
i
(I − r
i
A
i
)x
n

= a(Sx
n
− S
n
x
n
)+(a − a
n
)S
n
x
n

+(b − b
n
)Bx
n
+(c − c
n
)
N
i=1
δ
i
T
r
i
(I − r
i
A
i
)x
n

≤ aSx
n
− S
n
x
n
 + |a − a
n
|S

n
x
n
 + |b − b
n
|Bx
n

+|c − c
n
|
N
i
=1
δ
i
T
r
i
(I − r
i
A
i
)x
n
.
(3:20)
By condition (iii), (3.20), and (2.11), we have
lim
n

→∞
||Qx
n
− e
n
|| =0
.
(3:21)
Since
||Q
x
n
− x
n
||

||Q
x
n
− e
n
||
+
||
e
n
− x
n
||.
by (3.16) and (3.21), we have

lim
n
→∞
||Qx
n
− x
n
|| =0
.
(3:22)
From, (3.22), we have
lim
j
→∞
||Qx
n
j
− x
n
j
|| =0
.
(3:23)
By Lemma 2.4, we obtain that
q ∈ F
(
Q
)
= F
.

(3:24)
From (3.17)
lim sup
n→∞
u − z, x
n
− z = lim sup
j
→∞
u − z, x
n
j
− z = u − z, q − z≤0
.
(3:25)
Step. 5. Finally, we show that lim
n®∞
x
n
= z, where
z
=
P

u
.
By nonexpansiveness of S
n
and B, we can show that ||e
n

-z|| ≤ ||x
n
-z||. Then,
x
n+1
− z
2
= α
n
(u − z)+β
n
(x
n
− z)+γ
n
(e
n
− z)
2
= α
n
u − z, x
n+1
− z + β
n
x
n
− z, x
n+1
− z + γ

n
e
n
− z, x
n+1
− z
≤ α
n
u − z, x
n+1
− z + β
n
x
n
− zx
n+1
− z + γ
n
e
n
− zx
n+1
− z
≤ α
n
u − z, x
n+1
− z + β
n
x

n
− zx
n+1
− z + γ
n
x
n
− zx
n+1
− z

≤ α
n
u − z, x
n+1
− z +(1− α
n
)x
n
− zx
n+1
− z
≤ α
n
u − z, x
n+1
− z +
(1 − α
n
)

2
(x
n
− z
2
+ x
n+1
− z
2
).
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 14 of 16
It follows that
x
n+1
− z
2
≤ 2α
n
u − z, x
n+1
− z +
(
1 − α
n
)
x
n
− z
2

.
(3:26)
From Step 4, (3.26), and Lemma 2.2, we have lim
n® ∞
x
n
= z,where
z
= P

u
.The
proof is complete. □
4 Applications
From Theorem 3.1, we obtain the following strong convergence theorems in a real
Hilbert space:
Theorem 4.1. Let C be a closed convex subset of Hilbe rt space H . For every i =1,2,
, N, let F
i
: C × C ® ℝ be a bif unction satisfying (A
1
)-(A
4
) and let
{T
i
}

i
=

1
be 
i
-strict
pseudo-contraction mappings of C into i tself with  =sup
i

i
and let
ρ
j
=(α
j
1
, α
j
2
, α
j
3
) ∈ I × I ×
I
, where I =[0,1],
α
j
1
+ α
j
2
+ α

j
3
=
1
,
α
j
1
+ α
j
2
≤ b < 1
, and
α
j
1
, α
j
2
, α
j
3
∈ (κ,1
)
for all j = 2, For every n Î N, let S
n
and S are S-mappings gener-
ated by T
n
, , T

1
and r
n
, r
n-1
, , r
1
and T
n
, T
n-1
, , and r
n
, r
n-1
, , respectively.
Assume that
F =


i
=1
F( T
i
)

N
i
=1
EF(F

i
) =

.ForeverynÎ N, i =1,2, ,N, let {x
n
}
and
{v
i
n
}
be generated by x
1
, u Î C and







F
i
v
i
n
, v +
1
r
i

v − v
i
n
, v
i
n
− x
n
≥0, ∀v ∈ C, i =1, 2, , N
.
y
n
= 
N
i=1
δ
i
v
i
n
x
n+1
= α
n
u + β
n
x
n
+ γ
n

(a
n
S
n
x
n
+ b
n
x
n
+ c
n
y
n
), ∀n ∈ N,
(4:1)
where {a
n
}, {b
n
}, {g
n
}, {a
n
}, {b
n
}, {c
n
} ⊂ (0, 1),
α

n
+ β
n
+ γ
n
= a
n
+ b
n
+ c
n
= 
N
i
=1
δ
i
=
1
,
and
{r
i
}
N
i
=1
⊂ (ς , τ) ⊂ (0, 2α
i
)

, satisfy the following conditions:
(i)
lim
n
→∞
α
n
=
0
and


n
=
0
α
n
=

,
(ii)
0 < lim inf
n→∞
β
n
≤ lim sup
n
→∞
β
n

<
1
,
(iii)
lim
n
→∞
a
n
=
a
,
lim
n
→∞
b
n
=
b
,
lim
n
→∞
c
n
=
c
, with a, b, c Î (0, 1),
Then, the sequence {x
n

}, {y
n
},
{v
i
n
}
, ∀i = 1, 2, , N, converge strongly to
z
= P

u
, and z
is solution of (1.10)
Proof. From Theorem 3.1, let A
i
≡ 0; then we have G
i
(y)=P
Cy
= y ∀y Î C. Then, we
get Bx
n
= x
n
∀n Î N. Then, from Theorem 3.1, we obtain the desired conclusion. □
Next theorem is derived from Theorem 3.1, and we modify the result of [11] as
follows:
Theorem 4.2. LetCbeaclosedconvexsubsetofHilbertspaceHandletF: C × C
® ℝ be a bifunction satisfying (A

1
)-(A
4
), let A : C ® Hbea-inverse strongly monotone
mapping, and let T be -strict pseudo-contraction mappings of C into itself. Define a
mapping T

by T

x = x +(1-)Tx, ∀x Î C. Assume that
F = F(T)

EF(F, A)

VI(C, A) =

. For every n Î N, let {x
n
} and {v
n
} be generated by
x
1
, u Î C and



Fv
n
, v + Ax

n
, v − v
n
 +
1
r
v − v
n
, v
n
− x
n
≥0, ∀v ∈ C
x
n+1
= α
n
u + β
n
x
n
+ γ
n
(aT
κ
x
n
+ bP
C
(I − λA)x

n
+ cv
n
), ∀n ∈ N
,
(4:2)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 15 of 16
where {a
n
}, {b
n
}, {g
n
}, {a, b, c} ⊂ (0, 1), a
n
+ b
n
+ g
n
= a + b + c =1,and {r, l} ⊂
(ς, τ) ⊂ (0, 2a) satisfy the following conditions:
(i)
lim
n
→∞
α
n
=
0

and


n
=
0
α
n
=

,
(ii)
0 < lim inf
n→∞
β
n
≤ lim sup
n


β
n
<
1
,
Then, the sequence {x
n
} and { v
n
} converge strongly to

z
= P

u
.
Proof. From Theorem 3.1, choose N =1andletA
1
= A, l
1
= l. Then, we have B(y)
= G
1
(y)=P
C
(I - lA)y, ∀y Î C. Choose
v
1
n
= v
n
, a = a
n
, b = b
n
, c = c
n
for all n Î N, and
let T

≡ S

1
: C ® C be S-mapping generated by T
1
and r
1
with T
1
= T and
α
1
1
= κ
, and
then we obtain the desired result from Theorem 3.1 □
Acknowledgements
The authors would like to thank Professor Dr. Suthep Suantai for his valuable suggestion in the preparation and
improvement of this article.
Competing interests
The author declares that they have no competing interests.
Received: 21 February 2011 Accepted: 29 July 2011 Published: 29 July 2011
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doi:10.1186/1687-1812-2011-23
Cite this article as: Kangtunyakarn: Strong convergence theorem for a generalized equilibrium problem and
system of variational inequalities problem and infinite family of strict pseudo-contractions. Fixed Point Theory and
Applications 2011 2011:23.
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
/>Page 16 of 16

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