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RESEARC H Open Access
Iterative algorithms for finding a common solution
of system of the set of variational inclusion
problems and the set of fixed point problems
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 infinite family of
nonexpansive mapping and infinite real numbers. By means of the new mapping,
we prove a strong convergence theorem for finding a common element of the set
of fixed point problems of infinite family of nonexpansive mappings and the set of a
finite family of variational inclusion problems in Hilbert space. In the last section, we
apply our main result to prove a strong convergence theorem for finding a common
element of the set of fixed point problems of infinite family of strictly pseudo-
contractive mappings and the set of finite family of variational inclusion problems.
Keywords: nonexpansive mapping, strict pseudo contraction, strongly positive
operator, variational inclusion problem, fixed point
1 Introduction
Let H be a real Hilbert space and let C be a nonempty closed convex subset of H. Let A :
C ® H be a nonlinear mapping and let F : C × C ® ℝ be a bifunction. A mapping T of H
into itself is called nonexpansive if ||Tx - Ty|| ≤ ||x - y|| for all x, y Î H. We denote by F
(T) the set of fixed points of T (i.e. F(T)={x Î H : Tx = x}). Goebel and Kirk [1] showed
that F(T) is always closed convex and also nonempty provided T has a bounded trajectory.
The problem for finding a common fixed point of a family of nonexpansive map-
pings has been studied by many authors. The well-known convex f easibility problem


reduces to finding a point in the intersection of the fixed point sets of a family of non-
expansive mappings (see, e.g., [2,3]).
A bounded linear operator A on H is called strongly positive with coefficient
¯
γ
if
there exists a constant
¯
γ
>
0
with the property
Ax, x≥ ¯
γ
 x
2
.
A mapping A of C into H is called inverse-strongly monotone, 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.
The variational inequality problem is to find a point u Î C such that


v − u, Au

≥ 0forallv ∈ C
.
(1:1)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>© 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.
The set of solutions of (1.1) is denoted by VI( C, A). Many authors have studied
methods for finding solution of variational inequality problems (see, e.g., [5-8]).
In 2008, Qin et al. [9] introduced the following iterative scheme:

y
n
= P
C
(I − s
n
A)x
n
x
n+1
= α
n
γ f (W
n
x
n
)+(I − α

n
B)W
n
P
C
(I − r
n
A)y
n
, ∀n ∈ N
,
(1:2)
where W
n
is the W-mapping generated by a finite family o f nonexpansive mappings
and real numbers, A : C ® H is relaxed (u,v) cocoercive and μ-Lipschitz continuous, and
P
C
isametricprojectionH onto C. Under suitable conditions of {s
n
}, {r
n
}{a
n
}, g,they
proved that {x
n
} converges strongly to an element of the set of variational inequality pro-
blem and the set of a common fixed point of a finite family of nonexpansive mappings.
In 2006, Marino and Xu [10] introduced the iterative scheme as follows:

x
0
∈ H, x
n+1
=
(
I − α
n
A
)
Sx
n
+ α
n
γ f
(
x
n
)
, ∀n ≥ 0
,
(1:3)
where S is a nonexp ansive mapping, f is a contraction with the coefficient a Î (0, 1),
A is a stron gly positive bounded linear self-adjoint operator with the coefficient
¯
γ
, and
g is a constant such that
0 <γ <
γ

a
.Theyprovedthat{x
n
} generated by the above
iterative scheme converges strongly to the unique solution of the variational inequality:

(
A − γ f
)
x

, x − x

≥0, x ∈ F
(
S
)
.
We know that a mapping B : H ® H is said to be monotone, if for each x, y Î H,wehave

Bx − B
y
, x −
y
≥ 0
.
A set-valued mapping M : H ® 2
H
is called monotone if for all x, y Î H, f Î Mx and
g Î My imply 〈x - y, f - g〉 ≥ 0. A monotone mapping M : H ® 2

H
is maximal if the
graph of Graph(M) of M is not properly contained in the graph of any other monotone
mapping. It is known that a monotone mapping M is maximal if and only if for (x, f) Î
H × H, 〈x - y, f - g〉 ≥ 0 for every (y, g) Î Graph(M) implies f Î Mx.
Next, we consider the following so-called variational inclusion problem:
Find a u Î H such that
θ ∈
Bu + M
u
(1:4)
where B : H ® H, M : H ® 2
H
are two nonlinear mappings, and θ is zero vector in H
(see, for instance, [11-16]). The set of the solution of (1.4) is denoted by VI(H, B, M).
Let C be a nonempty closed convex subset of Banach space X. Let
{T
n
}

n
=
1
be an infi-
nite family of nonexpansive mappings of C into itself, and let l
1
, l
2
, , be real numbers
in [0, 1]; then we define the mapping K

n
: C ® C as follows:
U
n,0
= I
U
n,1
= λ
1
T
1
U
n,0
+(1− λ
1
)U
n,0
,
U
n,2
= λ
2
T
2
U
n,1
+(1− λ
2
)U
n,1

,
U
n,3
= λ
3
T
3
U
n,2
+(1− λ
3
)U
n,2
,
.
.
.
U
n,k
= λ
k
T
k
U
n,k−1
+(1− λ
k
)U
n,k−1
U

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

= λ
n
T
n
U
n,n−1
+
(
1 − λ
n
)
U
n,n−1
.
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 2 of 16
Such a mapping K
n
is called the K-mapping generated by T
1
, T
2
, , T
n
and l
1
, l
2
, , l
n

.
Let x
1
Î H and {x
n
} be the sequence generated by
x
n+1
= α
n
γ f
(
x
n
)
+ β
n
x
n
+
((
1 − β
n
)
I − α
n
A
)(
γ
n

K
n
x
n
+
(
1 − γ
n
)
Sx
n
),
(1:5)
where A is a strongly positive linear-bounded self-adjoint operator with the coeffi-
cient
0
<
γ
<
1
, S : C ® C is the S - mapping generated by G
1
, G
2
, , G
N
and ν
1
, ν
2

, ,
ν
N
,whereG
i
: H ® H is a mapping defined by
J
M
i,
η
(I − ηB
i
)x = G
i
x
for every x Î H,
and h Î (0, 2δ
i
) for every i = 1, 2, , N, f : H ® H is contractive mapping with coeffi-
cient θ Î (0, 1) and
0 <γ <
γ
θ
,{a
n
}, { b
n
}, {g
n
} are sequences in [0, 1].

In this article, by motivation of (1.3), we prove a s trong convergence theorem of the
proposed algorithm scheme (1.5) to an element
z



i
=1
F( T
i
)

N
i
=1
V(H, B
i
, M
i
)
,
under suitable conditions of {a
n
}, {b
n
}, { g
n
}.
2 Preliminaries
In this section, we provide some useful lemmas that will be used for our 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., 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. (see [17]) Given x Î HandyÎ C . Then P
C
x = yifandonlyifthere
holds the inequality
x −
y
,
y
− z≥0 ∀z ∈ C
.
Lemma 2.2. (see [18]) 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)

n
}⊂[0, 1],


n
=1

α
n
=

;
(2)
lim sup
n→∞
β
n
≤ 0or


n
=1

n
β
n
| <

.
Then lim
n®∞
s
n
=0.
Lemma 2.3.(see[19])LetCbeaclosedconvexsubsetofastrictlyconvexBanach
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. (see [20]) 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 - Sis
demi-closed at zero.
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 3 of 16
Lemma 2.5. (see [21]) 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.
Suppose x
n+1
= b
n
x
n
+(1 - b
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.

In 2009, Kangtunykarn and Suantai [5] introdu ced the S-mappin g generated by a
finite family of nonexpansive mappings and real numbers as follows:
Definition 2.1. 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. For each j = 1, 2, , N, let
α
j
=(α
j
1
, α
j
2
, α
j
3
) ∈ I × I ×
I
where I Î [0, 1] and
α
j
1
+ α

j
2
+ α
j
3
=
1
, define the mapping S :
C ® C as follows:
U
0
= I
U
1
= α
1
1
T
1
U
0
+ α
1
2
U
0
+ α
1
3
I

U
2
= α
2
1
T
2
U
1
+ α
2
2
U
1
+ α
2
3
I
U
3
= α
3
1
T
3
U
2
+ α
3
2

U
2
+ α
3
3
I
.
.
.
U
N−1
= α
N−1
1
T
N−1
U
N−2
+ α
N−1
2
U
N−2
+ α
N−1
3
I
S = U
N
= α

N
1
T
N
U
N−1
+ α
N
2
U
N−1
+ α
N
3
I.
(2:1)
This mapping is called the S-mapping generated by T
1
, , T
N
and a
1
, a
2
, , a
N
.
Lemma 2.6. (see [5]) Let C be a nonempty closed convex subset of strictly convex. Let
{T
i

}
N
i
=
1
be a finite family of nonexpanxive mappings of C into itself with

N
i
=1
F( T
i
) =

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

j
1
∈ (0, 1
)
,
α
j
1
∈ (0, 1
)
for all j = 1,2, , N-1,
α
N
1
∈ (0, 1]α
j
2
,
α
j
3
∈ [0, 1
)
for all j = 1,2, ,
N. Let S be the mapping generated by T
1
, , T
N
and a
1

, a
2
, , a
N
. Then
F( S)=

N
i
=1
F( T
i
)
.
Lemma 2.7. (see [5]) Let C be a nonempty closed convex subset of Banach space. Let
{T
i
}
N
i
=
1
be a finite family of nonexpansive mappings of C into itself and
α
j
=(α
j
1
, α
j

2
, α
j
3
) ∈ I × I ×
I
,
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
and
α
j
1
+ α
j
2
+ α
j
3
=
1
such that
α
n,j
i
→ α
j
i
∈ [0, 1
]
as n ® ∞ for i = 1,3
and j = 1,2,3, , N. Moreover, for every n Î N, let S and S
n
be the S-mappings generated
by T
1
, T

2
, , T
N
and a
1
, a
2
, , a
N
and T
1
, T
2
, , T
N
and
α
(
n
)
1
, α
(
n
)
2
, , α
(
n
)

N
, respectively.
Then lim
n®∞
||S
n
x - Sx || = 0 for every x Î C.
Definition 2.2.(see[11])Let M : H ® 2
H
be a multi-valued maximal monotone
mapping, then the single-valued mapping J
M,l
: H ® H defined by
J
M,λ
(
u
)
=
(
I + λM
)
−1
(
u
)
, ∀u ∈ H
,
is called the resolvent operator associated with M, where l is any positive number
and I is identity mapping.

Lemma 2.8. (see [11]) u Î H is a solution of variational inclusion (1.4) if and only if
u = J
M, l
(u - lBu), ∀l >0, i.e.,
VI
(
H, B, M
)
= F
(
J
M,λ
(
I − λB
))
, ∀λ>0
.
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 4 of 16
Further, if l Î (0, 2a], then V I(H, B, M) is closed convex subset in H.
Lemma 2.9. (see [22]) The resolvent operator J
M,l
associated with M is single-valued,
nonexpansive for all l >0 and 1-inverse-strongly monotone.
Lemma 2.10. In a strictly convex Banach space E, if
|
|x|| = ||y|| = ||λx +
(
1 − λ
)

y|
|
for all x, y Î E and l Î (0, 1), then x = y.
Lemma 2.11. LetCbeanonemptyclosedconvexsubsetofastrictlyconvexBanach
space. Let
{T
i
}

i=
1
be an infinite family of nonexpanxive mappings of C into itself with


i
=1
F( T
i
) =

and let l
1
, l
2
, , be real numbers such that 0 < l
i
<1 for every i = 1, 2, ,
and



i
=1
λ
i
<

. For every n Î N, let K
n
be the K-mapping generated by T
1
, T
2
, , T
n
and l
1
, l
2
, , l
n
. Then for every x Î C and k Î N, lim
n®∞
K
n
x exits.
Proof. Let x Î C. Then for k, n Î N, we have
 U
n+1,k
x − U
n,k

x  = λ
k
T
k
U
n+1,k−1
x +(1− λ
k
)U
n+1,k−1
x − λ
k
T
k
U
n,k−1
x − (1 − λ
k
)U
n,k−1
x 
= λ
k
(T
k
U
n+1,k−1
x − T
k
U

n,k−1
x)+(1− λ
k
)(U
n+1,k−1
x − U
n,k−1
x) 
≤ λ
k
 T
k
U
n+1,k−1
x − T
k
U
n,k−1
x  +(1 − λ
k
)  U
n+1,k−1
x − U
n,k−1
x 
≤ λ
k
 U
n+1,k−1
x − U

n,k−1
x  +(1 − λ
k
)  U
n+1,k−1
x − U
n,k−1
x 
= U
n+1,k−1
x − U
n,k−1
x 
= λ
k−1
T
k−1
U
n+1,k−2
x +(1− λ
k−1
)U
n+1,k−2
x − λ
k−1
T
k−1
U
n,k−2
x − (1 − λ

k−1
)U
n,k−2
x

≤ λ
k−1
 T
k−1
U
n+1,k−2
x − T
k−1
U
n,k−2
x  +(1 − λ
k−1
)  U
n+1,k−2
x − U
n,k−2
x 
≤ U
n+1,k−2
x − U
n,k−2
x 
.
.
.

≤ U
n+1,1
x − U
n,1
x 
= λ
1
T
1
U
n+1,0
x +(1− λ
1
)U
n+1,0
x − λ
1
T
1
U
n,0
x − (1 − λ
1
)U
n,0
x 
= λ
1
T
1

x +(1− λ
1
)x − λ
1
T
1
x − (1 − λ
1
)x 
=0
,
(2:2)
which implies that U
n+1,k
= U
n, k
for every k, n Î N.Hence,K
n
= U
n, n
= U
n+1,n
.
Since K
n+1
x = U
n+1,n+1
x = l
n+1
T

n+1
K
n
x +(1-l
n+1
)K
n
x, we have
K
n+1
x − K
n
x = λ
n+1
(
T
n+1
K
n
x − K
n
x
).
(2:3)
Let
x





i
=1
F( T
i
)
and x Î C. For each n Î N, we have
 K
n
x − x

 = λ
n
T
n
U
n,n−1
x +(1− λ
n
)U
n,n−1
x − x


≤ λ
n
 T
n
U
n,n−1
x − x


 +(1 − λ
n
)  U
n,n−1
x − x


≤ U
n,n−1
x − x


= λ
n−1
T
n−1
U
n,n−2
x +(1− λ
n−1
)U
n,n−2
x − x


≤ λ
n−1
 T
n−1

U
n,n−2
x − x

 +(1 − λ
n−1
)  U
n,n−2
x − x


≤ U
n,n−2
x − x


.
.
.
≤U
n,1
x − x


=  λ
1
T
1
U
n,0

x +(1− λ
1
)U
n,0
x − x


≤ λ
1
 T
1
x − x

 +(1 − λ
1
)  x − x


=

x − x


,
(2:4)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 5 of 16
which implies that {K
n
x} is bounded, and so is {T

n
K
n
x}. For m ≥ n, by (2.3) we have
 K
m
x − K
n
x  = K
m
x − K
m−1
x + K
m−1
x − K
m−2
x + K
m−2
x − + ···
− K
n+1
x + K
n+1
x − K
n
x 
≤ K
m
x − K
m−1

x  +  K
m−1
x − K
m−2
x  +  K
m−2
x − K
m−3
x  + ··
·
+  K
n+2
− K
n+1
x  +  K
n+1
x − K
n
x 
= λ
m
 T
m
K
m−1
x − K
m−1
x  +λ
m−1
 T

m−1
K
m−2
x − K
m−2
x  + ···
+ λ
n+1
 T
n+1
K
n
x − K
n
x 
≤ M
m

k
=
n
+1
λ
k
,
(2:5)
where M =sup
nÎN
{||T
n+1

K
n
x - K
n
x||}. This implies that {K
n
x} is Cauchy sequence.
Hence lim
n®∞
K
n
x exists.
From Lemma 2.11, we can define a mapping K : C ® C as follows:
Kx = lim
n


K
n
x, x ∈ C
.
Such a mapping K is called the K-mapping generated by T
1
, T
2
, , and l
1
, l
2
,

Remark 2.12. It is easy to see that for each n Î N, K
n
is nonexpansive mappings. Let
x, y Î C, then we have
 Kx − Ky = lim
n


 K
n
x − K
n
y ≤  x − y 
.
(2:6)
By (2.6), we have K : C ® C is nonexpansive mapping. Next, we will show that
lim
n®∞
sup
xÎD
||K
n
x - Kx|| = 0 for every bounded subset D of C. To show this, let x, y
Î C and D be a bounded subset of C. By (2.5), for m ≥ n, we have
 K
m
x − K
n
x ≤ M
m


k
=
n
+1
λ
k
.
By letting m ® ∞, for any x Î D, we have
 Kx − K
n
x ≤ M


k=n+1
λ
k
.
Since


n
=1
λ
n
<

, we have
lim
n→∞

sup
x

D
 Kx − K
n
x =0
.
By the next lemma, we will show that
F( K)=


i
=1
F( T
i
)
Lemma 2.13. LetCbeanonemptyclosedconvexsubsetofastrictlyconvexBanach
space. Let
{T
i
}

i
=
1
be an infinite family of nonexpansive mappings of C into itself with


i

=1
F( T
i
) =

, and let l
1
, l
2
, , be real numbers such that 0<l
i
<1for every i = 1, 2,
with


i
=1
λ
i
<

. Let K
n
and K be the K-mapping generated by T
1
, T
2
, T
n
and l

1
,
l
2
, l
n
and T
1
, T
2
, and l
1
, l
2
, , respectively. Then
F( K)=


i
=1
F( T
i
)
.
Proof. It is easy to see that


i
=1
F( T

i
) ⊆ F(K
)
. Next, we show that
F( K) ⊆


i
=1
F( T
i
)
.
Let x
0
Î F (K) and
x




i
=1
F( T
i
)
. Let k Î N be fixed. Since
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 6 of 16
 K

n
x
0
− x

 = λ
n
T
n
U
n,n−1
x
0
+(1− λ
n
)U
n,n−1
x
0
− x


= λ
n
(T
n
U
n,n−1
x
0

− x

)+(1− λ
n
)(U
n,n−1
x
0
− x

) 
≤ λ
n
 T
n
U
n,n−1
x
0
− x

 +(1 − λ
n
)  U
n,n−1
x
0
− x



≤ U
n,n−1
x
0
− x


= λ
n−1
(T
n−1
U
n,n−2
x
0
− x

)+(1− λ
n−1
)U
n,n−2
(x
0
− x

) 
≤ λ
n−1
 T
n−1

U
n,n−2
x
0
− x

 +(1 − λ
n−1
)  U
n,n−2
x
0
− x


≤ U
n,n−2
x
0
− x


.
.
.
.
.
≤ U
n,k
x

0
− x


= λ
k
(T
k
U
n,k−1
x
0
− x

)+(1− λ
k
)(U
n,k−1
x
0
− x

) 
≤ λ
k
 T
k
U
n,k−1
x

0
− x

 +(1 − λ
k
)  U
n,k−1
x
0
− x


≤ U
n,k−1
x
0
− x


.
.
.
≤ U
n,1
x
0
− x


= λ

1
(T
1
x
0
− x

)+(1− λ
1
)(x
0
− x

) 
≤ λ
1
 T
1
x
0
− x

 +(1 − λ
1
)  x
0
− x





x
0
− x


,
(2:7)
we have
 x
0
− x

 = lim
n→∞
 K
n
x
0
− x

≤λ
1
(T
1
x
0
− x

)+(1− λ

1
)(x
0
− x

)

≤ λ
1
 T
1
x
0
− x

 +(1 − λ
1
)  x
0
− x




x
0
− x


,

this implies that
 x
0
− x

 =  T
1
x
0
− x

 =  λ
1
(
T
1
x
0
− x

)
+
(
1 − λ
1
)(
x
0
− x


)

.
By Lemma 2.10, we have T
1
x
0
= x
0
, that is x
0
Î F (T
1
). It follows that U
n,1
x
0
= x
0
.By
(2.7), we have
 K
n
x
0
− x

≤U
n,2
x

0
− x

=  λ
2
(T
2
U
n,1
x
0
− x

)+(1− λ
2
)(U
n,1
x
0
− x

)

= λ
2
(T
2
x
0
− x


)+(1− λ
2
)(x
0
− x

) 
≤ λ
2
 T
2
x
0
− x

 +(1 − λ
2
)  x
0
− x




x
0
− x



.
It follows that
 x
0
− x

 = lim
n→∞
 K
n
x
0
− x


≤ λ
2
(T
2
x
0
− x

)+(1− λ
2
)(x
0
− x

) 

≤ λ
2
 T
2
x
0
− x

 +(1 − λ
2
)  x
0
− x




x
0
− x


,
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 7 of 16
which implies
 x
0
− x


=  T
2
x
0
− x

=  λ
2
(
T
2
x
0
− x

)
+
(
1 − λ
2
)(
x
0
− x

)

.
By Lemma 2.10, we obtain that T
2

x
0
= x
0
, that is x
0
Î F (T
2
). It follows that U
n,2
x
0
=
x
0
. By using the same argument, we can conclude that T
i
x
0
= x
0
and U
i
x
0
= x
0
for i =
1, 2, , k - 1. By (2.7), we have
 K

n
x
0
− x

≤U
n,k
x
0
− x


= λ
k
(T
k
U
n,k−1
x
0
− x

)+(1− λ
k
)(U
n,k−1
x
0
− x


)

= λ
k
(T
k
x
0
− x

)+(1− λ
k
)(x
0
− x

) 
≤ λ
k
 T
k
x
0
− x

 +(1 − λ
k
)  x
0
− x





x
0
− x


.
It follows that
 x
0
− x

 = lim
n→∞
 K
n
x
0
− x


= λ
k
(T
k
x
0

− x

)+(1− λ
k
)(x
0
− x

) 
≤ λ
k
 T
k
x
0
− x

 +(1 − λ
k
)  x
0
− x




x
0
− x



,
(2:8)
which implies
 x
0
− x

=  T
k
x
0
− x

=  λ
k
(
T
k
x
0
− x

)
+
(
1 − λ
k
)(
x

0
− x

)

.
(2:9)
By Lemma 2.10, we have T
k
x
0
= x
0
,thatisx
0
Î F (T
k
). This implies that
x
0



i
=1
F( T
i
)
.
3 Main result

Theorem 3.1. LetHbearealHilbertspace,andletM
i
: H ® 2
H
be maximal mono-
tone mappings for every i = 1, 2, , N. Let B
i
: H ® Hbeaδ
i
- inverse strongly mono-
tone mapping for every i = 1, 2, , Nand
{T
i
}

i=
1
an infinite family of nonexpansive
mappings from H into itself. Let A be a strongly positive linear-bounded self-adjoint
operator with the coefficient
0 <
γ
<
1
. Let G
i
: H ® Hbedefinedby
J
M
i,

η
(I − ηB
i
)x = G
i
x
for every x Î Handh Î (0, 2δ
i
) for every i = 1, 2, , Nandlet
ν
j
=(α
j
1
, α
j
2
, α
j
3
) ∈ I × I ×
I
, j = 1, 2, 3, , N, where I =[0,1],
α
j
1
+ α
j
2
+ α

j
3
=
1
,
α
j
1
∈ (0, 1
)
for all j = 1,2, , N-1,
α
N
1
∈ (0, 1]α
j
2
,
α
j
3
∈ [0, 1
)
for all j =1,2, ,N Let S : C
® C be the S-mapping generated by G
1
, G
2
, , G
N

and ν
1
, ν
2
, , ν
N
. Let l
1
, l
2
, , be
real numbers such that 0 < l
i
<1 for every i = 1, 2, , with


i
=1
λ
i
<

, and let K
n
be
the K-mapping generated by T
1
, T
2
, , T

n
and l
1
, l
2
, , l
n
, and let K be the K-mapping
generated by T
1
, T
2
, , and l
1
, l
2
, , i.e.,
Kx = lim
n


K
n
x
for every x Î C. Assume that
F =


i
=1

F( T
i
)

N
i
=1
V(H, B
i
, M
i
) =

. For every n Î
N, i = 1, 2, , N, let x
1
Î H and {x
n
} be the sequence generated by
x
n+1
= α
n
γ f
(
x
n
)
+ β
n

x
n
+
((
1 − β
n
)
I − α
n
A
)(
γ
n
K
n
x
n
+
(
1 − γ
n
)
Sx
n
),
(3:1)
where f : H ® H is contractive mapping with coefficient θ Î (0, 1) and
0 <γ <
γ
θ

.
Let {a
n
}, {b
n
}, {g
n
} be sequences in [0, 1], satisfying the following conditions:
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 8 of 16
(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
→∞
γ
n
= c ∈ (0, 1
)
Then {x
n
} converges strongly to
z

F
, which solves uniquely the following variational
inequality:

(
A − γ f
)

z, z − x

≤0, ∀x

∈ F
.
(3:2)
Equivalently, we have
P
F
(I − A + γ f)z =
z
.
Proof. Let z be the unique solution of (3.2). First, we will show that the mapping G
i
is
a nonexpansive mapping for every i = 1, 2, , N.Letx, y Î H,sinceB
i
is δ
i
-inverse
strongly monotone mapping and 0 < h <2δ
i
, for every i = 1, 2, , N, we have
 (I − ηB
i
)x − (I − ηB
i
)y
2

= x − y − η(B
i
x − B
i
y)
2
= x − y
2
− 2ηx − y, B
i
x − B
i
y + η
2
 B
i
x − B
i
y
2
≤ x − y
2
− 2δ
i
η  B
i
x − B
i
y
2

+ η
2
 B
i
x − B
i
y
2
= x − y
2
+ η(η − 2δ
i
)  B
i
x − B
i
y
2
≤ x −
y

2
.
(3:3)
Thus, (I - hB
i
) is a nonexpansive mapping for every i = 1, 2, , N . By Lemma 2.9, we
have
G
i

= J
M
i,
η
(I − ηB
i
)
is a nonexpansive mappings for every i = 1, 2, , N.Let
x


F
;
by Lemma 2.8, we have
x

= G
i
x

= J
M
i

(I − ηB
i
)x

, ∀i =1,2, N
.

(3:4)
Let e
n
= g
n
K
n
x
n
+(1-g
n
)Sx
n
.SinceG
i
is a nonexpansive mapping for ev ery i =1,
2, , N, we have that S is a nonexpansive mapping. By nonexpansiveness of K
n
we have
 e
n
− x

 = γ
n
(K
n
x
n
− x


)+(1− γ
n
)(Sx
n
− x

) 
≤ γ
n
 K
n
x
n
− x

 +(1 − γ
n
)  Sx
n
− x


≤ γ
n
 x
n
− x

 +(1 − γ

n
)  x
n
− x




x
n
− x


.
(3:5)
Without loss of generality, by conditions (i)and(ii), we have a
n
≤ (1 - b
n
)||A||
-1
.
Since A is a strongly positive linear-bounded self-adjoint operator, we have
 A  =su
p
{|Ax, x| : x ∈ H,  x  =1}
.
(3:6)
For each x Î C with ||x|| = 1, we have


((
1 − β
n
)
I − α
n
A
)
x, x =1− β
n
− α
n
Ax, x≥1 − β
n
− α
n
 A ≥0
,
(3:7)
then (1 - b
n
)I -a
n
A is positive. By (3.6) and (3.7), we have
 (1 − β
n
)I − α
n
A  =sup{((1 − β
n

)I − α
n
A)x, x : x ∈ C,  x  =1
}
=sup{1 − β
n
− α
n
Ax, x : x ∈ C,  x  =1}
≤ 1 − β
n
− α
n
Ax, x
≤ 1 − β
n
− α
n
γ
.
(3:8)
We shall divide our proof into six steps.
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 9 of 16
Step 1. We will show that the sequence {x
n
} is bounded. Let
x



F
,by(3.5)and
(3.8), we have
 x
n+1
− x

 = α
n
γ f (x
n
)+β
n
x
n
+ ((1 − β
n
)I − α
n
A)e
n
− x


= α
n
γ f (x
n
) − α
n

Ax

+ α
n
Ax

− β
n
x

+ β
n
x

+ β
n
x
n
+ ((1 − β
n
)I − α
n
A)e
n
− x


≤ α
n
 γ f (x

n
) − Ax

 +β
n
 x
n
− x

 +  ((1 − β
n
)I − α
n
A)(e
n
− x

)

≤ α
n
 γ f (x
n
) − Ax

 +β
n
 x
n
− x


 + ((1 − β
n
)I − α
n
γ )  x
n
− x


≤ α
n
( γ f (x
n
) − γ f (x

)  +  γ f (x

) − Ax

)+(1− α
n
γ )  x
n
− x


≤ α
n
γθ  x

n
− x

 + α
n
 γ f(x

) − Ax

 +(1− α
n
γ )  x
n
− x


= α
n
 γ f (x

) − Ax

 +(1− α
n
(γ − γθ))  x
n
− x


≤ max{ x

n
− x

,
 γ f (x

) − Ax


γ

γ
θ
}.
By induction, we can prove that {x
n
} is bounded, and so are {e
n
}, {K
n
x
n
}, {Sx
n
} and {G
i
(x
n
)} for every i = 1, 2, , N. Without loss of generality, we can assume that there exists
a bounded set D ⊂ H such that

e
n
, x
n
, Sx
n
, K
n
x
n
, G
i
x
n
∈ D, ∀n ∈ N andi =1,2, , N
.
(3:9)
Step 2. We will show that
lim
n→∞

x
n+1
− x
n

=
0
.
Define sequence {z

n
}by
z
n
=
1
1 −
β
n
(x
n+1
− β
n
x
n
)
.
Then x
n+1
= b
n
x
n
+(1- b
n
)z
n
.
Since {x
n

} is bounded, we have
|
|z
n+1
− z
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

γ f (x
n+1
)+

(1 − β
n+1
)I − α
n+1
A

e
n+1
1 − β
n+1


α
n
γ f (x
n
)+

(1 − β
n
)I − α
n
A

e
n

1 − β
n

||
≤ α
n+1
||
γ f (x
n+1
) − Ae
n+1
1 − β
n+1
|| + ||e
n+1
− e
n
||
+ α
n
||
γ f (x
n
) − Ae
n
1 −
β
n
||.
(3:10)

By definition of e
n
and nonexpansiveness of S, we have

e
n+1
− e
n

=


γ
n+1
K
n+1
x
n+1
+(1− γ
n+1
)Sx
n+1
− γ
n
K
n
x
n
− (1 − γ
n

)Sx
n


=


γ
n+1
K
n+1
x
n+1
+(1− γ
n+1
)Sx
n+1
− γ
n+1
K
n
x
n
+ γ
n+1
K
n
x
n
−(1 − γ

n+1
)Sx
n
+(1− γ
n+1
)Sx
n
− γ
n
K
n
x
n
− (1 − γ
n
)Sx
n


=


γ
n+1
(K
n+1
x
n+1
− K
n

x
n
)+(1− γ
n+1
)(Sx
n+1
− Sx
n
)
+(γ
n+1
− γ
n
)K
n
x
n
+(γ
n
− γ
n+1
)Sx
n


≤ γ
n+1

K
n+1

x
n+1
− K
n
x
n

+(1− γ
n+1
)

x
n+1
− x
n

+
|
γ
n+1
− γ
n
|
K
n
x
n

+
|

γ
n
− γ
n+1
|
Sx
n

≤ γ
n+1

K
n+1
x
n+1
− K
n
x
n

+(1− γ
n+1
)

x
n+1
− x
n

+2|

γ
n+1

γ
n
|M,
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 10 of 16
where M = max
nÎN
{||K
n
x
n
||, ||Sx
n
||}. Substituting (3.11) into (3.10), we have

z
n+1
− z
n

≤ α
n+1
||
γ f (x
n+1
) − Ae
n+1

1 − β
n+1
|| + α
n
||
γ f (x
n
) − Ae
n
1 − β
n
|| +

e
n+1
− e
n

≤ α
n+1
||
γ f (x
n+1
) − Ae
n+1
1 − β
n+1
|| + α
n
||

γ f (x
n
) − Ae
n
1 − β
n
||
+ γ
n+1

K
n+1
x
n+1
− K
n
x
n

+(1− γ
n+1
)

x
n+1
− x
n

+2|γ
n+1

− γ
n
|M
≤ α
n+1
||
γ f (x
n+1
) − Ae
n+1
1 − β
n+1
|| + α
n
||
γ f (x
n
) − Ae
n
1 − β
n
||
+ γ
n+1
(

K
n+1
x
n+1

− K
n+1
x
n

+

K
n+1
x
n
− K
n
x
n

)
+(1− γ
n+1
)

x
n+1
− x
n

+2|γ
n+1
− γ
n

|M
≤ α
n+1
||
γ f (x
n+1
) − Ae
n+1
1 − β
n+1
|| + α
n
||
γ f (x
n
) − Ae
n
1 − β
n
|| +

x
n+1
− x
n

+

K
n+1

x
n
− K
n
x
n

+2|
γ
n+1

γ
n
|M.
(3:12)
It implies that

z
n+1
− z
n



x
n+1
− x
n

≤ α

n+1
||
γ f (x
n+1
) − Ae
n+1
1 − β
n+1
|| + α
n
||
γ f (x
n
) − Ae
n
1 − β
n
||
+

K
n+1
x
n
− K
n
x
n

+2|

γ
n+1

γ
n
|M.
(3:13)
By (2.3), it implies that
K
n+1
x
n
− K
n
x
n
= λ
n+1
(
T
n+1
K
n
x
n
− K
n
x
n
),

since l
n
® 0asn ® ∞, we have
lim
n



K
n+1
x
n
− K
n
x
n

=0
.
(3:14)
By (3.13), (3.14) and conditions (i), (iii), we have
lim sup
n→∞
(

z
n+1
− z
n




x
n+1
− x
n

)
≤ 0
.
(3:15)
By Lemma 2.5, we have
lim
n



z
n
− x
n

=0
.
(3:16)
By condition (ii) and (3.16)
lim
n




x
n+1
− x
n

= lim
n


(1 − β
n
)

z
n
− x
n

=0
.
(3:17)
Step 3. We will show that
lim
n



e
n

− x
n

=0
.
(3:18)
Since x
n+1
= a
n
gf + b
n
x
n
+ ((1 - b
n
)I-a
n
A) e
n
, we have

x
n+1
− e
n

=



α
n
(γ f (x
n
) − Ae
n
)+β
n
(x
n
− e
n
)


≤ α
n


γ f (x
n
) − Ae
n


+ β
n
(

x

n
− x
n+1

+

x
n+1
− e
n

)
,
it implies that
(1 − β
n
)

x
n+1
− e
n

≤ α
n


γ f (x
n
) − Ae

n


+ β
n

x
n
− x
n+1

,
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 11 of 16
and it follows that

x
n+1
− e
n


α
n
(
1 − β
n
)



γ f (x
n
) − Ae
n


+
β
n
(
1 − β
n
)

x
n
− x
n+1

.
By conditions (i), (ii) and (3.17), we have
lim
n



x
n+1
− e
n


=0.
(3:19)
Since ||e
n
- x
n
|| ≤ ||e
n
- x
n+1
|| + ||x
n+1
- x
n
||, by (3.17) and (3.19), we have
lim
n



e
n
− x
n

=0
.
Step 4. Define a mapping Q : H ® H by
Qx = cKx +

(
1 − c
)
Sx, ∀x ∈ H
.
(3:20)
We will show that
lim
n


||Qx
n
− x
n
|| =0
.
(3:21)
Since

Qx
n
− e
n

=


cKx
n

+(1− c)Sx
n
− γ
n
K
n
x
n
− (1 − γ
n
)Sx
n




cKx
n
− γ
n
K
n
x
n

+ |γ
n
− c|

Sx

n

=

cKx
n
− γ
n
Kx
n
+ γ
n
Kx
n
− γ
n
K
n
x
n

+ |γ
n
− c|

Sx
n

≤|c − γ
n

|

Kx
n

+ γ
n

Kx
n
− K
n
x
n

+ |γ
n
− c|

Sx
n

≤|c − γ
n
|

Kx
n

+sup

x

D
{
Kx − K
n
x
}
+ |γ
n
− c|

Sx
n

.
By remark 2.12 and condition (iii), we have
lim
n



Qx
n
− e
n

=0
.
(3:22)

Since ||Qx
n
-x
n
|| ≤ ||Qx
n
-e
n
|| + ||e
n
-x
n
||, from (3.22) and (3.18), we have
lim
n


||Qx
n
− x
n
|| =0
.
Step 5. We will show that
lim sup
n


(γ f − A)z, x
n

− z≤0
,
(3:23)
where
z
= PF
(
I −
(
A − γ f
))z
. Let
{
x
n
j
}
be subsequence of {x
n
} such that
lim sup
n→∞
(γ f − A)z, x
n
− z = lim
j
→∞
(γ f − A)z, x
n
j

− z
.
(3:24)
Without loss of generality, we may assume that
{x
n
j
}
converses weakly to some q Î
H. By nonexpansive ness of S and K, (3.20) and Lemma 2.3, we have that Q is nonex-
pansive mapping and
F( Q)=F(K)

F( S)
.
(3:25)
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 12 of 16
Since
J
M
i,
η
(I − ηB
i
)x = G
i
x
for every x Î H and i = 1, 2, N , by Lemma 2.8, we have
VI(H, B

i
, M
i
)=F(J
M
i,
η
(I − ηB
i
)) = F(G
i
), ∀i =1,2, N
.
(3:26)
By Lemma 2.6 and Lemma 2.9, we have
F( S)=
N

i
=1
F( G
i
)=
N

i
=1
VI(H, B
i
, M

i
)
(3:27)
By Lemma 2.13, we have
F( K)=


i
=1
F( T
i
)
.
(3:28)
By (3.25), (3.27), and (3.28), we have
F( Q)=F(K)

F( S)=


i
=1
F( T
i
)

N

i
=1

VI(H, B
i
, M
i
)
.
(3:29)
Since
x
n
j

q
as j ® ∞, nonexpansiveness of Q, (3.21) and Lemma 2.4, we have
q ∈ F(Q)=


i
=1
F( T
i
)

N

i
=1
VI(H, B
i
, M

i
)=F
.
(3:30)
By (3.24) and (3.30), we have
lim sup
n
→∞
(γ f − A)z, x
n
− z = lim
n→∞
(γ f − A)z, x
n
j
− z = (γ f − A)z, q − z≤0
.
Step 6. Finally, we will show that x
n
® z as n ® ∞,where
z
= PF
(
I −
(
A − γ f
))z
.
Since


x
n+1
− z

2
=


α
n
γ f (x
n
)+β
n
x
n
+

(1 − β
n
)I − α
n
A

e
n
− z


2

=


α
n
(γ f (x
n
) − Az)+β
n
(x
n
− z)+

(1 − β
n
)I − α
n
A

(e
n
− z)


2



β
n

(x
n
− z)+

(1 − β
n
)I − α
n
A

(e
n
− z)


2
+2α
n
γ f (x
n
) − Az, x
n+1
− z
=



β
n
(x

n
− z)+

(1 − β
n
)I − α
n
A

(e
n
− z)



2
+2α
n
γ f (x
n
) − Az, x
n+1
− z





β
n

(x
n
− z)


+


(1 − β
n
)I − α
n
A



e
n
− z


2
+2α
n
γ f (x
n
) − γ f(z), x
n+1
− z +2α
n

γ f (z) − Az, x
n+1
− z




β
n
(x
n
− z)


+(1− β
n
− α
n
¯γ )

e
n
− z


2
+2α
n
γθ


x
n
− z

x
n+1
− z

+2α
n
γ f (z) − Az, x
n+1
− z




β
n
(x
n
− z)


+(1− β
n
− α
n
¯γ )


x
n
− z


2
+2α
n
γθ

x
n
− z

x
n+1
− z

+2α
n
γ f (z) − Az, x
n+1
− z


(1 − α
n
¯γ )

x

n
− z


2
+ α
n
γθ


x
n
− z

2
+

x
n+1
− z

2

+2α
n
γ f (z) − Az, x
n+1
− z
≤ (1 − 2α
n

¯γ + α
n
γθ)

x
n
− z

2
+ α
2
n
¯γ
2

x
n
− z

2
+ α
n
γθ

x
n+1
− z

2
+2α

n
γ f
(
z
)
− Az, x
n+1
− z,
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 13 of 16
it implies that

x
n+1
− z

2

(1 − 2α
n
¯γ + α
n
γθ)
1 − α
n
γθ

x
n
− z


2
+
α
2
n
¯γ
2
1 − α
n
γθ

x
n
− z

2
+

n
1 − α
n
γθ
γ f (z) − Az, x
n+1
− z
=
(1 − α
n
γθ + α

n
γθ − 2α
n
¯γ + α
n
γθ)
1 − α
n
γθ

x
n
− z

2
+
α
2
n
¯γ
2
1 − α
n
γθ

x
n
− z

2

+

n
1 − α
n
γθ
γ f (z) − Az, x
n+1
− z
=

1 − α
n
γθ − 2α
n
( ¯γ − γθ)

1 − α
n
γθ

x
n
− z

2
+
α
2
n

¯γ
2
1 − α
n
γθ

x
n
− z

2
+

n
1 − α
n
γθ
γ f (z) − Az, x
n+1
− z
=(1−

n
( ¯γ − γθ))
1 − α
n
γθ

x
n

− z

2
+
α
2
n
¯γ
2
1 − α
n
γθ

x
n
− z

2
+

n
1 − α
n
γθ
γ f (z) − Az, x
n+1
− z
=(1−

n

( ¯γ − γθ))
1 − α
n
γθ

x
n
− z

2
+
α
n
1 − α
n
γθ

α
n
¯γ
2

x
n
− z

2
+2γ f (z) − Az, x
n+1
− z


=(1−

n
( ¯γ − γθ))
1 − α
n
γθ

x
n
− z

2
+
2( ¯γ − γθ)
2( ¯γ − γθ)
α
n
1 − α
n
γθ

α
n
¯γ
2

x
n

− z

2
+2γ f (z) − Az, x
n+1
− z

=(1−

n
( ¯γ − γθ))
1 − α
n
γθ

x
n
− z

2
+

n
( ¯γ − γθ)
1 − α
n
γθ

α
n

¯γ
2
2( ¯γ − γθ)

x
n
− z

2
+
2
2
(
¯γ − γθ
)
γ f (z) − Az, x
n+1
− z

,
from condition i, step 5 and Lemma 2.2, we can conclude that x
n
® z as n ® ∞,
where z = P
F
(I-(A-gf))z. This completes the proof.
By means of our main result, we have the following results in the framework of Hilbert
space. To prove these results, we need definition and lemma as follows:
Definition 3.1. A mapping T : C ® Cissaidtobea-strict pseudo-contraction
mapping, if there exists  Î [0, 1) such that



Tx − Ty


2



x − y


2
+ κ


(I − T)x − (I − T)y


2
, ∀x, y ∈ C
.
Lemma 3.2. (see [23]) Let C be a nonempty closed convex subset of a real Hilbert space
H and T : C ® Ca-strict pseudo-contraction. Define S : C ® CbySx= ax +(1- a)Tx,
for each x Î C. Then, as a Î [, 1), S is nonexpansive such that F(S)=F(T).
Corollary 3.3. Let H be a real Hilbert space and let M
i
: H ® 2
H
be maximal monotone

mappings for every i = 1, 2, N. Let B
i
: H ® Hbeaδ
i
- inverse strongly monotone map-
ping for every i = 1, 2, Nand
{T
i
}

i
=
1
an infinite family of 
i
- strictly pseudo-contractive
mappings from H into itself. Define a mapping
T
κ
i
by
T
κ
i
= κ
i
x +(1− κ
i
)T
i

x
, ∀x Î H, i Î N.
Let A be a strongl y positive linear-bounded self-adjoint operator with t he coefficient
0 <
γ
<
1
. Let G
i
: H ® H be defined by
J
M
i,
η
(I − ηB
i
)x = G
i
x
for every x Î H and h Î (0,

i
) for every i = 1, 2, , N and let
ν
j
=(α
j
1
, α
j

2
, α
j
3
) ∈ I × I ×
I
, j = 1, 2, 3, , N, where I =[0,
1],
α
j
1
+ α
j
2
+ α
j
3
=
1
,
α
j
1
∈ (0, 1
)
for all j = 1, 2, , N-1,
α
N
1
∈ (0, 1]α

j
2
,
α
j
3
∈ [0, 1
)
for all j =
1,2, , N Let S : C ® C be the S - mapping generated by G
1
, G
2
, , G
N
and ν
1
, ν
2
, , ν
N
. Let
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 14 of 16
l
1
, l
2
, , be real numbers such that 0 < l
i

<1for every i = 1, 2, , with


i
=1
λ
i
<

, and
let K
n
be the K-mapping generated by
T
κ
1
, T
κ
2
, , T
κ
n
, and l
1
, l
2
, , l
n
, and let K be the
K-mapping generated by

T
κ
1
,
T
κ
2
,
,
and l
1
, l
2
, , i.e.,
Kx = lim
n
→∞
K
n
x
for every x Î C. Assume that
F =


i
=1
F( T
i
)


N
i
=1
V(H, B
i
, M
i
) =

. For every n Î
N, i = 1, 2, , N, let x
1
Î H and {x
n
} be the sequence generated by
x
n+1
= α
n
γ f (x
n
)+β
n
x
n
+

(1 − β
n
)I − α

n
A


n
K
n
x
n
+(1− γ
n
)Sx
n
)
,
(3:31)
where f : H ® H is contractive mapping with coefficient θ Î (0, 1) and
0 <γ <
γ
θ
.
Let {a
n
}, {b
n
}, {g
n
} be sequences in [0, 1], satisfying 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
→∞
γ
n
= c ∈ (0, 1
)
.
Then {x
n
} converges strongly to
z

F
, which solves uniquely the following variational
inequality:

(
A − γ f
)
z, z − x

≤0, ∀x

∈ F
.
(3:32)
Equivalently, we have
P
F

(I − A + γ f)z =
z
.
Proof.Foreveryi Î N, by L emma 3.2, we have that
T
κ
i
is a nonexpansive mapping
and


i
=1
F( T
κ
i
)=


i
=1
F( T
i
)
. From Theorem 3.1 and Lemma 2.13, we can reach the
desired conclusion.
Corollary 3.4. Let H be a real Hilbert space and let M : H ® 2
H
be maximal monotone
mappings. Let B : H ® Hbeaδ - inverse strongly monotone mapping and

{T
i
}

i
=
1
an infinite
family of 
i
- strictly pseudo-contractive m ap-pings from H into itself. Define a mapping
T
κ
i
= κ
i
x +(1− κ
i
)T
i
x
by
T
κ
i
= κ
i
x +(1− κ
i
)T

i
x
, ∀x Î H, i Î N. Let A be a strongly positive
linear-bounded self-adjoint operator with the coefficient
0 <
γ
<
1
. Let l
1
, l
2
, , be real
numbers such that 0 < l
i
<1 for every i = 1, 2, , with


i
=1
λ
i
<

, and let K
n
be the
K-mapping generated by
T
κ

1
, T
κ
2
, , T
κ
n
and l
1
, l
2
, , l
n
, and let K be the K-mapping
generated by
T
κ
1
, T
κ
2
, ,
and l
1
, l
2
, , i.e.,
Kx = lim
n



K
n
x
for every x Î C. Assume that
F =


i
=1
F( T
i
)

V(H, B, M) =

. For every n Î N, let x
1
Î H and {x
n
} be the sequence generated by
x
n+1
= α
n
γ f (x
n
)+β
n
x

n
+

(1 − β
n
)I − α
n
A


n
K
n
x
n
+(1− γ
n
)J
M,η
(I − ηB)x
n
)
,
(3:33)
where f : H ® H is contractive mapping with coefficient θ Î (0, 1) and
0 <γ <
γ
θ
, h
Î (0, 2δ), {a

n
}, {b
n
}, {g
n
} are sequences in [0, 1], satisfying the following conditions:
(i)
lim
n
→∞
α
n
=
0
and


n
=
0
α
n
=

,
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
/>Page 15 of 16
(ii)
0 < lim inf
n→∞

β
n
≤ lim sup
n
→∞
β
n
< 1,
,
(iii)
lim
n
→∞
γ
n
= c ∈ (0, 1
)
.
Then {x
n
} converges strongly to
z

F
, which solves uniquely the following variational
inequality

(
A − γ f
)

z, z − x

≤0, ∀x

∈ F
.
(3:34)
Equivalently, we have
P
F
(I − A + γ f)z =
z
.
Proof. Putting N = 1 in Corollary 3.3, we can reach the desired conclusion.
Competing interests
The authors declare that they have no competing interests.
Received: 24 February 2011 Accepted: 18 August 2011 Published: 18 August 2011
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doi:10.1186/1687-1812-2011-38
Cite this article as: Kangtunyakarn: Iterative algorithms for finding a common solution of system of the set of
variational inclusion problems and the set of fixed point problems. Fixed Point Theory and Applications 2011 2011:38.
Kangtunyakarn Fixed Point Theory and Applications 2011, 2011:38
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