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:
[email protected]
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
(

)
= {z ∈ C : G
(
z, y
)
+ Az, y − z≥0, ∀y ∈ C
.
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
http://www.fixedpointtheoryandapplications.com/content/2011/1/23
© 2011 Kangtunyakarn; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons
Attribution License (http://creativecommons.org/licenses/by/2.0), 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

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
).

x
n+1
= α
n
x
n
+
(
1 − α
n
)
Tx
n
(1:8)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
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Page 2 of 16
converges weakly to a fixed point of T provided the control sequence
{α
n
}
∞
n
=
0
satisfies
the conditions that  < a
n
<1 for all n and


⎪
⎩
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
),

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

= 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

= 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

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

, 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

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
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Page 3 of 16
(0, 1] ⊂ (0, 2 a
i
) such that
F =

∞
i
=1
F( T
i
)

N
i
=1
EF(F
i

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)

− 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
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

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

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

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

N−1
+
(
1 − λ
N
)
U
N−1
.
(2:3)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:23
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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
=

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

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

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

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
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+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

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

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

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
, 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
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

+ 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

)(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

(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

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

=(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

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

− 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

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
∗

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

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

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
∗

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

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

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

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

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

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

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

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

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

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

)||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
(α
k+1
2
− κ)
k
j=1
(1 − α

− x
0
||
2
− α
k
1
(α
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
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

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

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

.
.
.
−α
k+1
1
(α
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

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


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

− x
0

2
(2:12)
−α
2
1
(α
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
− α
1
2
α
1
3
U
n,2
x
0
− x
0

2
= x
0
− x
∗

2
−α
n
1
(α
n
2
− κ)
n−1
j=1

)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

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

.
.
.
−α
3
1
(α
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

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
(α
1

http://www.fixedpointtheoryandapplications.com/content/2011/1/23
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

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)

∞,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

=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*=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

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

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

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 =

∞

be generated by x
1
, u Î C and
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
F
i
v
i
n
, v + A
i
x
n
, v − v
i
n
 +
1

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
http://www.fixedpointtheoryandapplications.com/content/2011/1/23
Page 10 of 16
where {a
n
}, {b
n
}, {g
n
}, {a
n

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
=
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

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

, 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

=

. 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

=

,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)+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

)||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||)

,{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 −

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
−

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
|| +

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

+ 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
)
≤|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

− 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

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
 + 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


− 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

||
+ ||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

+|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

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

(
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

|| + 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

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
http://www.fixedpointtheoryandapplications.com/content/2011/1/23
Page 13 of 16
First, we define a mapping A : C ® C by
Ax = 
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

)

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

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

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

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
= 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

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

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:

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

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
⎧
⎪

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)

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
=
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

. 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
+ 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

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

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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
http://www.fixedpointtheoryandapplications.com/content/2011/1/23
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