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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 Applications20112011:38

https://doi.org/10.1186/1687-1812-2011-38

Received: 24 February 2011

Accepted: 18 August 2011

Published: 18 August 2011

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 mappingstrict pseudo contractionstrongly positive operatorvariational inclusion problemfixed 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 be a bifunction. A mapping T of H into itself is called nonexpansive if ||Tx - Ty|| ≤ ||x - y|| for all x, y H. We denote by F(T) the set of fixed points of T (i.e. F(T) = {x 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 mappings has been studied by many authors. The well-known convex feasibility problem reduces to finding a point in the intersection of the fixed point sets of a family of nonexpansive 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
A x , 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 α such that
x - y , A x - A y α A x - A y 2

for all x, y C.

The variational inequality problem is to find a point u C such that
v - u , A u 0 f o r a l l v C .
(1.1)

The set of solutions of (1.1) is denoted by V I(C, A). Many authors have studied methods for finding solution of variational inequality problems (see, e.g., [58]).

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 ,
(1.2)

where W n is the W-mapping generated by a finite family of nonexpansive mappings and real numbers, A : CH is relaxed (u,v) cocoercive and μ-Lipschitz continuous, and P C is a metric projection H onto C. Under suitable conditions of {s n }, {r n }{α n }, γ, they proved that {x n } converges strongly to an element of the set of variational inequality problem 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 ) S x n + α n γ f ( x n ) , n 0 ,
(1.3)
where S is a nonexpansive mapping, f is a contraction with the coefficient a (0, 1), A is a strongly positive bounded linear self-adjoint operator with the coefficient γ ̄ , and γ is a constant such that 0 < γ < γ a . They proved that {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 : HH is said to be monotone, if for each x, y H, we have
B x - B y , x - y 0 .

A set-valued mapping M : H → 2 H is called monotone if for all x, y H, f Mx and g 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) H × H, 〈x - y, f - g〉 ≥ 0 for every (y, g) Graph(M) implies f Mx.

Next, we consider the following so-called variational inclusion problem:

Find a u H such that
θ B u + M u
(1.4)

where B : HH, M : H → 2 H are two nonlinear mappings, and θ is zero vector in H (see, for instance, [1116]). The set of the solution of (1.4) is denoted by V I(H, B, M).

Let C be a nonempty closed convex subset of Banach space X. Let { T n } n = 1 be an infinite family of nonexpansive mappings of C into itself, and let λ1, λ2,..., be real numbers in [0, 1]; then we define the mapping K n : CC as follows:
U n , 0 = I (1) U n , 1 = λ 1 T 1 U n , 0 + ( 1 - λ 1 ) U n , 0 , (2) U n , 2 = λ 2 T 2 U n , 1 + ( 1 - λ 2 ) U n , 1 , (3) U n , 3 = λ 3 T 3 U n , 2 + ( 1 - λ 3 ) U n , 2 , (4) (5) U n , k = λ k T k U n , k - 1 + ( 1 - λ k ) U n , k - 1 (6) U n , k + 1 = λ k + 1 T k + 1 U n , k + ( 1 - λ k + 1 ) U n , k (7) (8) U n , n - 1 = λ n - 1 T n - 1 U n , n - 2 + ( 1 - λ n - 1 ) U n , n - 2 (9) K n = U n , n = λ n T n U n , n - 1 + ( 1 - λ n ) U n , n - 1 . (10) (11)

Such a mapping K n is called the K-mapping generated by T1, T2,..., T n and λ1, λ2,..., λ n .

Let x1 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 ) S x n ) ,
(1.5)

where A is a strongly positive linear-bounded self-adjoint operator with the coefficient 0 < γ < 1 , S : CC is the S - mapping generated by G1, G2,..., G N and ν1, ν2,..., ν N , where G i : HH is a mapping defined by J M i , η ( I - η B i ) x = G i x for every x H, and η (0, 2δ i ) for every i = 1, 2,..., N, f : HH is contractive mapping with coefficient θ (0, 1) and 0 < γ < γ θ , {α n }, {β n }, {γ n } are sequences in [0, 1].

In this article, by motivation of (1.3), we prove a strong convergence theorem of the proposed algorithm scheme (1.5) to an element z i = 1 F ( T i ) i = 1 N V ( H , B i , M i ) , under suitable conditions of {α n }, {β n }, {γ 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, and let P C be the metric projection of H onto C, i.e., for x H, P C x satisfies the property:
x - P C x = min y C x - y .

The following characterizes the projection P C .

Lemma 2.1. (see [17]) Given x H and y C. Then P C x = y if and only if there 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 {α n }, {β n } satisfy the conditions:
  1. (1)

    { α n } [ 0 , 1 ] , n = 1 α n = ;

     
  2. (2)

    lim sup n β n 0 o r n = 1 | α n β n | < .

     

Then limn→∞s n = 0.

Lemma 2.3. (see [19]) Let C be a closed convex subset of a strictly convex Banach space E. Let {T n : n } be a sequence of nonexpansive mappings on C. Suppose n = 1 F ( T n ) is nonempty. Let {λ n } be a sequence of positive numbers with Σ n = 1 λ n = 1 . Then a mapping S on C defined by
S ( x ) = Σ n = 1 λ n T n x n

for x 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 : CC be a nonexpansive mapping. Then I - S is demi-closed at zero.

Lemma 2.5. (see [21]) Let {x n } and {z n } be bounded sequences in a Banach space X and let {β n } be a sequence in [0,1] with 0 < lim inf n→∞ β n ≤ lim supn→∞β n < 1.

Suppose xn+1= β n x n +(1 - β n )z n for all integer n ≥ 0 and lim supn→∞(||zn+1- z n || - ||xn+1- x n ||) ≤ 0. Then limn→∞||x n - z n || = 0.

In 2009, Kangtunykarn and Suantai [5] introduced the S-mapping 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 } i = 1 N be a finite family of nonexpanxive mappings of C into itself. For each j = 1, 2,..., N, let α j = ( α 1 j , α 2 j , α 3 j ) I × I × I where I [0, 1] and α 1 j + α 2 j + α 3 j = 1 , define the mapping S : C → C as follows:
U 0 = I (1) U 1 = α 1 1 T 1 U 0 + α 2 1 U 0 + α 3 1 I (2) U 2 = α 1 2 T 2 U 1 + α 2 2 U 1 + α 3 2 I (3) U 3 = α 1 3 T 3 U 2 + α 2 3 U 2 + α 3 3 I (4) (5) U N - 1 = α 1 N - 1 T N - 1 U N - 2 + α 2 N - 1 U N - 2 + α 3 N - 1 I (6) S = U N = α 1 N T N U N - 1 + α 2 N U N - 1 + α 3 N I . (7) (8)
(2.1)

This mapping is called the S-mapping generated by T1,..., T N and α1, α2,..., α N .

Lemma 2.6. (see [5]) Let C be a nonempty closed convex subset of strictly convex. Let { T i } i = 1 N be a finite family of nonexpanxive mappings of C into itself with i = 1 N F ( T i ) and let α j = ( α 1 j , α 2 j , α 3 j ) I × I × I , j = 1,2,3,..., N, where I = [0, 1], α 1 j + α 2 j + α 3 j = 1 , α 1 j ( 0 , 1 ) for all j = 1,2,..., N-1, α 1 N ( 0 , 1 ] α 2 j , α 3 j [ 0 , 1 ) for all j = 1,2,..., N. Let S be the mapping generated by T1,..., T N and α1, α2,..., α N . Then F ( S ) = i = 1 N F ( T i ) .

Lemma 2.7. (see [5]) Let C be a nonempty closed convex subset of Banach space. Let { T i } i = 1 N be a finite family of nonexpansive mappings of C into itself and α j ( n ) = ( α 1 n , j , α 2 n , j , α 3 n , j ) , α j = ( α 1 j , α 2 j , α 3 j ) I × I × I , where I = [0,1], α 1 n , j + α 2 n , j + α 3 n , j = 1 and α 1 j + α 2 j + α 3 j = 1 such that α i n , j α i j [ 0 , 1 ] as n → ∞ for i = 1,3 and j = 1,2,3,..., N. Moreover, for every n , let S and S n be the S-mappings generated by T1, T2,..., T N and α1, α2,..., α N and T1, T2,..., T N and α 1 ( n ) , α 2 ( n ) , . . . , α N ( n ) , respectively. Then limn→∞||S n x - Sx|| = 0 for every x 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,λ : HH defined by
J M , λ ( u ) = ( I + λ M ) 1 ( u ) ,  u H ,

is called the resolvent operator associated with M, where λ is any positive number and I is identity mapping.

Lemma 2.8. (see [11]) u H is a solution of variational inclusion (1.4) if and only if u = JM, λ(u - λBu), λ > 0, i.e.,
V I ( H , B , M ) = F ( J M , λ ( I - λ B ) ) , λ > 0 .

Further, if λ (0, 2α], then V I(H, B, M) is closed convex subset in H.

Lemma 2.9. (see [22]) The resolvent operator J M,λ associated with M is single-valued, nonexpansive for all λ > 0 and 1-inverse-strongly monotone.

Lemma 2.10. In a strictly convex Banach space E, if
| | x | | = | | y | | = | | λ x + ( 1 - λ ) y | |

for all x, y E and λ (0, 1), then x = y.

Lemma 2.11. Let C be a nonempty closed convex subset of a strictly convex Banach 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 λ1, λ2,..., be real numbers such that 0 < λ i < 1 for every i = 1, 2,..., and i = 1 λ i < . For every n , let K n be the K-mapping generated by T1, T2,..., T n and λ1, λ2,..., λ n . Then for every x C and k , limn→∞K n x exits.

Proof. Let x C. Then for k, 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 (1) = 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 ) (2) 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 (3) 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 (4) = U n + 1 , k - 1 x - U n , k - 1 x (5) = 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 (6) 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 (7) U n + 1 , k - 2 x - U n , k - 2 x (8) (9) U n + 1 , 1 x - U n , 1 x (10) = 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 (11) = λ 1 T 1 x + ( 1 - λ 1 ) x - λ 1 T 1 x - ( 1 - λ 1 ) x (12) = 0 , (13) (14) 
(2.2)
which implies that Un+1,k= Un, kfor every k, n . Hence, K n = Un, n= Un+1,n. Since Kn+1x = Un+1,n+1x = λn+1Tn+1K n x + (1 - λn+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 C. For each 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 * (1) λ n T n U n , n - 1 x - x * + ( 1 - λ n ) U n , n - 1 x - x * (2) U n , n - 1 x - x * (3) = λ n - 1 T n - 1 U n , n - 2 x + ( 1 - λ n - 1 ) U n , n - 2 x - x * (4) λ n - 1 T n - 1 U n , n - 2 x - x * + ( 1 - λ n - 1 ) U n , n - 2 x - x * (5) U n , n - 2 x - x * (6) . (7) . (8) . (9) U n , 1 x - x * (10) = λ 1 T 1 U n , 0 x + ( 1 - λ 1 ) U n , 0 x - x * (11) λ 1 T 1 x - x * + ( 1 - λ 1 ) x - x * (12) = x - x * , (13) (14)
(2.4)
which implies that {K n x} is bounded, and so is {T n K n x}. For mn, 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 - + (1) - K n + 1 x + K n + 1 x - K n x (2) K m x - K m - 1 x + K m - 1 x - K m - 2 x + K m - 2 x - K m - 3 x + (3) + K n + 2 - K n + 1 x + K n + 1 x - K n x (4) = 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 + (5) + λ n + 1 T n + 1 K n x - K n x (6) M k = n + 1 m λ k , (7) (8) 
(2.5)

where M = supn{||Tn+1K n x - K n x||}. This implies that {K n x} is Cauchy sequence. Hence limn→∞K n x exists.

From Lemma 2.11, we can define a mapping K : CC as follows:
K x = lim n K n x , x C .

Such a mapping K is called the K-mapping generated by T1, T2,..., and λ1, λ2,.....

Remark 2.12. It is easy to see that for each n , K n is nonexpansive mappings. Let x, y C, then we have
K x - K y = lim n K n x - K n y x - y .
(2.6)
By (2.6), we have K : CC is nonexpansive mapping. Next, we will show that limn→∞supxD||K n x - Kx|| = 0 for every bounded subset D of C. To show this, let x, y C and D be a bounded subset of C. By (2.5), for mn, we have
K m x - K n x M k = n + 1 m λ k .
By letting m → ∞, for any x D, we have
K x - K n x M k = n + 1 λ k .
Since n = 1 λ n < , we have
lim n sup x D K x - K n x = 0 .

By the next lemma, we will show that F ( K ) = i = 1 F ( T i )

Lemma 2.13. Let C be a nonempty closed convex subset of a strictly convex Banach 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 λ1, λ2,..., be real numbers such that 0 < λ i < 1 for every i = 1, 2,... with i = 1 λ i < . Let K n and K be the K-mapping generated by T1, T2,... T n and λ1, λ2,... λ n and T1, T2,... and λ1, λ2,..., 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 x0 F (K) and x * i = 1 F ( T i ) . Let k be fixed. Since
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 * (1)  = λ n ( T n U n , n - 1 x 0 - x * ) + ( 1 - λ n ) ( U n , n - 1 x 0 - x * ) (2)  λ n T n U n , n - 1 x 0 - x * + ( 1 - λ n ) U n , n - 1 x 0 - x * (3)  U n , n - 1 x 0 - x * (4)  = λ n - 1 ( T n - 1 U n , n - 2 x 0 - x * ) + ( 1 - λ n - 1 ) U n , n - 2 ( x 0 - x * ) (5)  λ n - 1 T n - 1 U n , n - 2 x 0 - x * + ( 1 - λ n - 1 ) U n , n - 2 x 0 - x * (6)  U n , n - 2 x 0 - x * (7)  (8)  . (9)  . (10)  U n , k x 0 - x * (11)  = λ k ( T k U n , k - 1 x 0 - x * ) + ( 1 - λ k ) ( U n , k - 1 x 0 - x * ) (12)  λ k T k U n , k - 1 x 0 - x * + ( 1 - λ k ) U n , k - 1 x 0 - x * (13)  U n , k - 1 x 0 - x * (14)  . (15)  . (16)  . (17)  U n , 1 x 0 - x * (18)  = λ 1 ( T 1 x 0 - x * ) + ( 1 - λ 1 ) ( x 0 - x * ) (19)  λ 1 T 1 x 0 - x * + ( 1 - λ 1 ) x 0 - x * (20)  x 0 - x * , (21)  (22) 
(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 * ) (1)  λ 1 T 1 x 0 - x * + ( 1 - λ 1 ) x 0 - x * (2)  x 0 - x * , (3)  (4) 
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 T1x0 = x0, that is x0 F (T1). It follows that Un,1x0 = x0. 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 * ) (1) = λ 2 ( T 2 x 0 - x * ) + ( 1 - λ 2 ) ( x 0 - x * ) (2) λ 2 T 2 x 0 - x * + ( 1 - λ 2 ) x 0 - x * (3) x 0 - x * . (4) (5) 
It follows that
x 0 - x * = lim n K n x 0 - x * (1) λ 2 ( T 2 x 0 - x * ) + ( 1 - λ 2 ) ( x 0 - x * ) (2) λ 2 T 2 x 0 - x * + ( 1 - λ 2 ) x 0 - x * (3) x 0 - x * , (4) (5)
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 T2x0 = x0, that is x0 F (T2). It follows that Un,2x0 = x0. By using the same argument, we can conclude that T i x0 = x0 and U i x0 = x0 for i = 1, 2,..., k - 1. By (2.7), we have
K n x 0 - x * U n , k x 0 - x * (1) = λ k ( T k U n , k - 1 x 0 - x * ) + ( 1 - λ k ) ( U n , k - 1 x 0 - x * ) (2) = λ k ( T k x 0 - x * ) + ( 1 - λ k ) ( x 0 - x * ) (3) λ k T k x 0 - x * + ( 1 - λ k ) x 0 - x * (4) x 0 - x * . (5) (6)
It follows that
x 0 - x * = lim n K n x 0 - x * (1) = λ k ( T k x 0 - x * ) + ( 1 - λ k ) ( x 0 - x * ) (2) λ k T k x 0 - x * + ( 1 - λ k ) x 0 - x * (3) x 0 - x * , (4) (5)
(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 x0 = x0, that is x0 F (T k ). This implies that x 0 i = 1 F ( T i ) .

3 Main result

Theorem 3.1. 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 : HH be a δ i - inverse strongly monotone mapping for every i = 1, 2,..., N and { 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 : HH be defined by J M i , η ( I - η B i ) x = G i x for every x H and η (0, 2δ i ) for every i = 1, 2,..., N and let ν j = ( α 1 j , α 2 j , α 3 j ) I × I × I , j = 1, 2, 3,..., N, where I = [0, 1], α 1 j + α 2 j + α 3 j = 1 , α 1 j ( 0 , 1 ) for all j = 1,2,..., N-1, α 1 N ( 0 , 1 ] α 2 j , α 3 j [ 0 , 1 ) for all j = 1, 2,..., N.. Let S : CC be the S-mapping generated by G1, G2,..., G N and ν1, ν2,..., ν N . Let λ1, λ2,..., be real numbers such that 0 < λ i < 1 for every i = 1, 2,..., with i = 1 λ i < , and let K n be the K-mapping generated by T1, T2,..., T n and λ1, λ2,..., λ n , and let K be the K-mapping generated by T1, T2,..., and λ1, λ2,..., i.e.,
K x = lim n K n x
for every x C. Assume that F = i = 1 F ( T i ) i = 1 N V ( H , B i , M i ) . For every n , i = 1, 2,..., N, let x1 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 ) S x n ) ,
(3.1)

where f : HH is contractive mapping with coefficient θ (0, 1) and 0 < γ < γ θ . Let {α n }, {β n }, {γ n } be sequences in [0, 1], satisfying the following conditions:

(i) lim n α n = 0 and Σ n = 0 α n =

(ii) 0 < liminf n β n limsup 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. Let x, y H, since B i is δ i - inverse strongly monotone mapping and 0 < η < 2δ i , for every i = 1, 2,..., N, we have
( I - η B i ) x - ( I - η B i ) y 2 = x - y - η ( B i x - B i y ) 2 (1) = x - y 2 - 2 η x - y , B i x - B i y + η 2 B i x - B i y 2 (2) x - y 2 - 2 δ i η B i x - B i y 2 + η 2 B i x - B i y 2 (3) = x - y 2 + η ( η - 2 δ i ) B i x - B i y 2 (4) x - y 2 . (5) (6) 
(3.3)
Thus, (I - ηB i ) is a nonexpansive mapping for every i = 1, 2,..., N . By Lemma 2.9, we have G i = J M i , η ( I - η B 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 , η ( I - η B i ) x * , i = 1 , 2 , . . . N .
(3.4)
Let e n = γ n K n x n + (1 - γ n )Sx n . Since G i is a nonexpansive mapping for every 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 ) ( S x n - x * ) (1) γ n K n x n - x * + ( 1 - γ n ) S x n - x * (2) γ n x n - x * + ( 1 - γ n ) x n - x * (3) x n - x * . (4) (5)
(3.5)
Without loss of generality, by conditions (i) and (ii), we have α n ≤ (1 - β n )||A||-1. Since A is a strongly positive linear-bounded self-adjoint operator, we have
A = s u p { | A x , x | : x H , x = 1 } .
(3.6)
For each x C with ||x|| = 1, we have
( ( 1 - β n ) I - α n A ) x , x = 1 - β n - α n A x , x 1 - β n - α n A 0 ,
(3.7)
then (1 - β n )I -α 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 A x , x : x C , x = 1 } 1 β n α n A x , x 1 β n α n γ ¯ .
(3.8)

We shall divide our proof into six steps.

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 * (1) = α n γ f ( x n ) - α n A x * + α n A x * - β n x * + β n x * + β n x n (2) + ( ( 1 - β n ) I - α n A ) e n - x * (3) α n γ f ( x n ) - A x * + β n x n - x * + ( ( 1 - β n ) I - α n A ) ( e n - x * ) (4) α n γ f ( x n ) - A x * + β n x n - x * + ( ( 1 - β n ) I - α n γ ) x n - x * (5) α n ( γ f ( x n ) - γ f ( x * ) + γ f ( x * ) - A x * ) + ( 1 - α n γ ) x n - x * (6) α n γ θ x n - x * + α n γ f ( x * ) - A x * + ( 1 - α n γ ) x n - x * (7) = α n γ f ( x * ) - A x * + ( 1 - α n ( γ - γ θ ) ) x n - x * (8) m a x { x n - x * , γ f ( x * ) - A x * γ - γ θ } . (9) (10) 
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 , S x n , K n x n , G i x n D , n a n d i = 1 , 2 , , N .
(3.9)

Step 2. We will show that l i m 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 xn+1= β n x n + (1 - β 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 | | (1) = | | α n + 1 γ f ( x n + 1 ) + ( 1 - β n + 1 ) I - α n + 1 A e n + 1 1 - β n + 1 (2) - α n γ f ( x n ) + ( 1 - β n ) I - α n A e n 1 - β n | | (3) α n + 1 | | γ f ( x n + 1 ) - A e n + 1 1 - β n + 1 | | + | | e n + 1 - e n | | (4) + α n | | γ f ( x n ) - A e n 1 - β n | | . (5) (6)
(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 ) S x n + 1 - γ n K n x n - ( 1 - γ n ) S x n (1)  = γ n + 1 K n + 1 x n + 1 + ( 1 - γ n + 1 ) S x n + 1 - γ n + 1 K n x n + γ n + 1 K n x n (2)  - ( 1 - γ n + 1 ) S x n + ( 1 - γ n + 1 ) S x n - γ n K n x n - ( 1 - γ n ) S x n (3)  = γ n + 1 ( K n + 1 x n + 1 - K n x n ) + ( 1 - γ n + 1 ) ( S x n + 1 - S x n ) (4)  + ( γ n + 1 - γ n ) K n x n + ( γ n - γ n + 1 ) S x n (5)  γ n + 1 K n + 1 x n + 1 - K n x n + ( 1 - γ n + 1 ) x n + 1 - x n (6)  + γ n + 1 - γ n K n x n + γ n - γ n + 1 S x n (7)  γ n + 1 K n + 1 x n + 1 - K n x n + ( 1 - γ n + 1 ) x n + 1 - x n (8)  + 2 | γ n + 1 - γ n | M , (9) (10) 
where M = maxn{||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 ) - A e n + 1 1 - β n + 1 | | + α n | | γ f ( x n ) - A e n 1 - β n | | + e n + 1 - e n (1) α n + 1 | | γ f ( x n + 1 ) - A e n + 1 1 - β n + 1 | | + α n | | γ f ( x n ) - A e n 1 - β n | | (2) + γ n + 1 K n + 1 x n + 1 - K n x n + ( 1 - γ n + 1 ) x n + 1 - x n (3) + 2 | γ n + 1 - γ n | M (4) α n + 1 | | γ f ( x n + 1 ) - A e n + 1 1 - β n + 1 | | + α n | | γ f ( x n ) - A e n 1 - β n | | (5) + γ n + 1 K n + 1 x n + 1 - K n + 1 x n + K n + 1 x n - K n x n (6) + ( 1 - γ n + 1 ) x n + 1 - x n + 2 | γ n + 1 - γ n | M (7) α n + 1 | | γ f ( x n + 1 ) - A e n + 1 1 - β n + 1 | | + α n | | γ f ( x n ) - A e n 1 - β n | | + x n + 1 - x n (8) + K n + 1 x n - K n x n + 2 | γ n + 1 - γ n | M . (9) (10) 
(3.12)
It implies that
z n + 1 - z n - x n + 1 - x n α n + 1 | | γ f ( x n + 1 ) - A e n + 1 1 - β n + 1 | | + α n | | γ f ( x n ) - A e n 1 - β n | | (1) + K n + 1 x n - K n x n + 2 | γ n + 1 - γ n | M . (2) (3) 
(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 λ n 0 as n → ∞, 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
limsup 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 xn+1= α n γf + β n x n + ((1 - β n )I - α n A) e n , we have
x n + 1 - e n = α n ( γ f ( x n ) - A e n ) + β n ( x n - e n ) (1) α n γ f ( x n ) - A e n + β n x n - x n + 1 + x n + 1 - e n , (2) (3)
it implies that
( 1 - β n ) x n + 1 - e n α n γ f ( x n ) - A e n + β n x n - x n + 1 ,
and it follows that
x n + 1 - e n α n ( 1 - β n ) γ f ( x n ) - A e 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 - xn+1|| + ||xn+1- x n ||, by (3.17) and (3.19), we have
lim n e n - x n = 0 .
Step 4. Define a mapping Q : HH by
Q x = c K x + ( 1 - c ) S x , x H .
(3.20)
We will show that
lim n | | Q x n - x n | | = 0 .
(3.21)
Since
Q x n - e n = c K x n + ( 1 - c ) S x n - γ n K n x n - ( 1 - γ n ) S x n (1) c K x n - γ n K n x n + | γ n - c | S x n (2) = c K x n - γ n K x n + γ n K x n - γ n K n x n + | γ n - c | S x n (3) | c - γ n | K x n + γ n K x n - K n x n + | γ n - c | S x n (4) | c - γ n | K x n + sup> x D K x - K n x + | γ n - c | S x n . (5) (6)
By remark 2.12 and condition (iii), we have
lim n Q x 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 | | Q x 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 = P F ( I - ( A - γ f ) ) z . Let { x n j } be subsequence of {x n } such that
limsup 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 H. By nonexpansiveness of S and K, (3.20) and Lemma 2.3, we have that Q is nonexpansive mapping and
F ( Q ) = F ( K ) F ( S ) .
(3.25)
Since J M i , η ( I - η B i ) x = G i x for every x H and i = 1, 2,... N , by Lemma 2.8, we have
V I ( H , B i , M i ) = F ( J M i , η ( I - η B i ) ) = F ( G i ) , i = 1 , 2 , . . . N .
(3.26)
By Lemma 2.6 and Lemma 2.9, we have
F ( S ) = i = 1 N F ( G i ) = i = 1 N V I ( 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 ) i = 1 N V I ( 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 ) i = 1 N V I ( 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 = P F ( 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 (1) = α n ( γ f ( x n ) - A z ) + β n ( x n - z ) + ( 1 - β n ) I - α n A ( e n - z ) 2 (2) β n ( x n - z ) + ( 1 - β n ) I - α n A ( e n - z ) 2 + 2 α n γ f ( x n ) - A z , x n + 1 - z (3) = β n ( x n - z ) + ( 1 - β n ) I - α n A ( e n - z ) 2 + 2 α n γ f ( x n ) - A z , x n + 1 - z (4) β n ( x n - z ) + ( 1 - β n ) I - α n A e n - z 2 (5) + 2 α n γ f ( x n ) - γ f ( z ) , x n + 1 - z + 2 α n γ f ( z ) - A z , x n + 1 - z (6) β n ( x n - z ) + ( 1 - β n - α n γ ̄ ) e n - z 2 (7) + 2 α n γ θ x n - z x n + 1 - z + 2 α n γ f ( z ) - A z , x n + 1 - z (8) β n ( x n - z ) + ( 1 - β n - α n γ ̄ ) x n - z 2 (9) + 2 α n γ θ x n - z x n + 1 - z + 2 α n γ f ( z ) - A z , x n + 1 - z (10) ( 1 - α n γ ̄ ) x n - z 2 + α n γ θ x n - z 2 + x n + 1 - z 2 (11) + 2 α n γ f ( z ) - A z , x n + 1 - z (12) ( 1 - 2 α n γ ̄ + α n γ θ ) x n - z 2 + α n 2 γ ̄ 2 x n - z 2 + α n γ θ x n + 1 - z 2 (13) + 2 α n γ f ( z ) - A z , x n + 1 - z , (14) (15) 
it implies that
x n + 1 z 2 ( 1 2 α n γ ¯ + α n γ θ ) 1 α n γ θ x n z 2 + α n 2 γ ¯ 2 1 α n γ θ x n z 2 + 2 α n 1 α n γ θ γ f ( z ) A z , x n + 1 z = ( 1 α n γ θ + α n γ θ 2 α n γ ¯ + α n γ θ ) 1 α n γ θ x n z 2 + α n 2 γ ¯ 2 1 α n γ θ x n z 2 + 2 α n 1 α n γ θ γ f ( z ) A z , x n + 1 z = ( 1 α n γ θ 2 α n ( γ ¯ γ θ ) ) 1 α n γ θ x n z 2 + α n 2 γ ¯ 2 1 α n γ θ x n z 2 + 2 α n 1 α n γ θ γ f ( z ) A z , x n + 1 z = ( 1 2 α n ( γ ¯ γ θ ) ) 1 α n γ θ x n z 2 + α n 2 γ ¯ 2 1 α n γ θ x n z 2 + 2 α n 1 α n γ θ γ f ( z ) A z , x n + 1 z = ( 1 2 α n ( γ ¯ γ θ ) ) 1 α n γ θ x n z 2 + α n 1 α n γ θ ( α n γ ¯ 2 x n z 2 + 2 γ f ( z ) A z , x n + 1 z ) = ( 1 2 α n ( γ ¯ γ θ ) ) 1 α n γ θ x n z 2 + 2 ( γ ¯ γ θ ) 2 ( γ ¯ γ θ ) α n 1