Open Access

Iterative approximation of common element of solution sets of various nonlinear operator problems

Fixed Point Theory and Applications20132013:295

https://doi.org/10.1186/1687-1812-2013-295

Received: 28 March 2013

Accepted: 30 September 2013

Published: 9 November 2013

Abstract

In this paper, we prove strong convergence theorem for finding a common element of the set of fixed point of a finite family of nonexpansive mappings and a finite family of κ i -strictly pseudocontractive mappings and the set of a finite family of the set of solution of equilibrium problems by using the new mapping generated by a finite family of nonexpansive mappings and a finite family of κ i -strictly pseudocontractive mappings and a sequences of positive real numbers. Furthermore, by using our main result, we obtain two interesting theorems involving variational inequality problems and variational inclusion problems. In the last section, we give numerical examples to support our main results.

Keywords

nonexpansive mappingstrictly pseudocontractive mappingequilibrium problemvariational inequality problemvariational inclusion problem

1 Introduction

Let H be a real Hilbert space and C be a nonempty closed convex subset of H. A self mapping f : C C is a contraction on C if there exists a constant k [ 0 , 1 ) such that f ( x ) f ( y ) k x y , x , y C . Let T : C C be a mapping, a point x C is called a fixed point of T if and only if T x = x . In this paper, we use F ( T ) to denote the set of fixed point of T. Recall the following definitions.

Definition 1.1 A mapping T : C C is called nonexpansive if and only if for all x , y C ,
T x T y x y .
Definition 1.2 A mapping T : C C is called κ-strictly pseudocontractive [1] if and only if there exists a constant κ [ 0 , 1 ) such that for all x , y C ,
T x T y 2 x y 2 + κ ( I T ) x ( I T ) y 2 .
(1.1)

For such case, T is also said to be a κ-strictly pseudo contraction.

Note that the class of κ-strict pseudo-contractions strictly includes the class of nonexpansive mappings, that is T is nonexpansive if and only if T is 0-strict pseudo-contractive.

Let F : C × C R be a bifunction. The equilibrium problem for F is to determine its equilibrium points, i.e., the set
EP ( F ) = { x C : F ( x , y ) 0 , y C } .
(1.2)

Given T : C H , let F ( x , y ) = T x , y x for all x , y C . Then z EP ( F ) if and only if T z , y z 0 for all y C , that is, z is a solution of the variational inequality.

Equilibrium problems, which were introduced in [2] in 1994, have had a great impact and influence in the development of several branches of pure and applied sciences. Numerous problems in physics, minimization problems, Nash equilibria in noncooperative games, optimization and economics reduce to find a solution of EP ( F ) (see, for example, [24]). Some methods have been proposed to solve the equilibrium problem (see, for example, [57]).

In 2007, Takahashi and Takahashi [8] proved the following theorem.

Theorem 1.1 Let C be a nonempty closed convex subset of H. Let F be a bifunction from C × C to satisfying
  1. (A1)

    F ( x , x ) = 0 , x C ;

     
  2. (A2)

    F is monotone, i.e., F ( x , y ) + F ( y , x ) 0 , x , y C ;

     
  3. (A3)
    x , y , z C ,
    lim t 0 + F ( t z + ( 1 t ) x , y ) F ( x , y ) ;
     
  4. (A4)

    x C , y F ( x , y ) is convex and lower semicontinuous;

     
and let S be a nonexpansive mapping of C into H such that F ( S ) EP ( G ) . Let f be a contraction of H into itself, and let { x n } and { u n } be sequences generated by x 1 H and
F ( u n , y ) + 1 r n y u n , u n x n 0 , y C , x n + 1 = α n f ( x n ) + ( 1 α n ) S u n
for all n N , where { α n } [ 0 , 1 ] and { r n } ( 0 , 1 ) satisfy (C1)-(C3) as follows:
  1. (C1)

    α n 0 ;

     
  2. (C2)

    n = 0 α n = ;

     
  3. (C3)

    either n = 0 | α n + 1 α n | < or lim n α n + 1 α n = 1 ,

     

and lim inf n r n > 0 and n = 1 | r n + 1 r n | < .

Then { x n } and { u n } converge strongly to z F ( S ) EP ( F ) , where z = P F ( S ) EP ( F ) f ( z ) .

In 2010, Kangtunyakarn and Suantai [9] proved the strong convergence theorem by using the S-mapping generated by a finite family of strictly pseudocontractive mappings and a finite family of real number as follows.

Theorem 1.2 Let H be a Hilbert space, let f be an α-contraction on H, and let A be a strongly positive linear bounded self-adjoint operator with coefficient γ ¯ > 0 . Assume that 0 < γ < γ ¯ α . Let { T i } i = 1 N be a finite family of κ i -strictly pseudo contraction of H into itself for some κ i [ 0 , 1 ) and κ = max { κ i : i = 1 , 2 , , N } with i = 1 N F ( T i ) . Let S n be the S-mappings generated by T 1 , T 2 , , T N and α 1 ( n ) , α 2 ( n ) , , α N ( n ) , where α j ( n ) = ( α 1 n , j , α 2 n , j , α 3 n , j ) I × I × I , I = [ 0 , 1 ] , α 1 n , j + α 2 n , j + α 3 n , j = 1 and κ < a α 1 n , j , α 3 n , j b < 1 for all j = 1 , 2 , , N 1 , κ < c α 1 n , N 1 , κ α 3 n , N d < 1 , κ α 2 n , j e < 1 for all j = 1 , 2 , , N . For a point u H and x 1 H , let { x n } and { y n } be the sequences defined iteratively by
{ y n = β n x n + ( 1 β n ) S n x n , x n + 1 = α n γ ( a n u + ( 1 a n ) f ( x n ) ) + ( I α n A ) y n , n 1 ,
(1.3)
where { β n } , { α n } and { a n } are sequences in [ 0 , 1 ] . Assume that the following conditions hold:
  1. (i)

    lim n α n = 0 , n = 1 α n = and lim n a n = 0 ;

     
  2. (ii)

    n = 1 | α 1 n + 1 , j α 1 n , j | < , n = 1 | α 3 n + 1 , j α 3 n , j | < for all j { 1 , 2 , 3 , , N } and n = 1 | α n + 1 α n | < , n = 1 | β n + 1 β n | < and n = 1 | a n + 1 a n | < ;

     
  3. (iii)

    0 κ β n < θ < 1 for all n 1 for some θ ( 0 , 1 ) .

     
Then both { x n } and { y n } strongly converge to q i = 1 N F ( T i ) , which solves the following variational inequality
γ f ( q ) A q , p q 0 , p i = 1 N F ( T i ) .

Question Can we prove a strong convergence theorem for finding a common solution of the set of fixed point of a finite family of nonexpansive mappings and a finite family of strictly pseudocontractive mappings and a finite family of the set of solution of equilibrium problems?

Let C be a nonempty closed convex subset of Hilbert space H. Let { T i } i = 1 N be a finite family of κ i -strict pseudo-contractions of C into itself, and let { S i } i = 1 N be a finite family of nonexpansive mappings of C into itself. For each n N and j = 1 , 2 , , N , let α j ( n ) = ( α 1 n , j , α 2 n , j , α 3 n , j ) I × I × I , where I = [ 0 , 1 ] , α 1 n , j + α 2 n , j + α 3 n , j = 1 . We define the mapping S n A : C C as follows:
U n , 0 = I , U n , 1 = S 1 ( α 1 n , 1 T 1 U n , 0 + α 2 n , 1 U n , 0 + α 3 n , 1 I ) , U n , 2 = S 2 ( α 1 n , 2 T 2 U n , 1 + α 2 n , 2 U n , 1 + α 3 n , 2 I ) , U n , 3 = S 3 ( α 1 n , 3 T 3 U n , 2 + α 2 n , 3 U n , 2 + α 3 n , 3 I ) , U n , N 1 = S N 1 ( α 1 n , N 1 T N 1 U n , N 2 + α 2 n , N 1 U n , N 2 + α 3 n , N 1 I ) , S n A = U n , N = S N ( α 1 n , N T N U n , N 1 + α 2 n , N U n , N 1 + α 3 n , N I ) .
(1.4)

In Lemma 2.8, under suitable conditions of the real sequences { α 1 n , j } , { α 2 n , j } and { α 3 n , j } for every j = 1 , 2 , , N , we show that F ( S n A ) = i = 1 N F ( S i ) i = 1 N F ( T i ) and S n A is a nonexpansive mapping.

In this paper, motivated by the ongoing research and Theorems 1.1 and 1.2, we prove strong convergence theorem for finding a common solution of the set of fixed point of a finite family of nonexpansive mappings and a finite family of strictly pseudocontractive mappings and a finite family of the set of solution of equilibrium problems by using the mapping defined by (1.4). Furthermore, in the last section, we prove two interesting theorems involving a finite family of the set of solutions of variational inequality problem and variational inclusion problem. In the last section, we give numerical examples to support our main results.

2 Preliminaries

In this section, we need the following lemmas to prove our main result. Let C be a closed convex subset of a real Hilbert space H, 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 [10])

Given x H and y C . Then P C x = y if and only if the following inequality holds
x y , y z 0 , z C .

Lemma 2.2 (See [11])

Let { s n } be a sequence of nonnegative real numbers satisfying
s n + 1 = ( 1 α n ) s n + α n β n , n 0 ,
where { α n } , { β n } satisfy the conditions
( 1 ) { α n } [ 0 , 1 ] , n = 1 α n = ; ( 2 ) lim sup n β n 0 or n = 1 | α n β n | < .

Then lim n s n = 0 .

Lemma 2.3 (See [12])

Let { s n } be a sequence of nonnegative real numbers satisfying
s n + 1 = ( 1 α n ) s n + δ n , n 0 ,
where { α n } is a sequence in ( 0 , 1 ) and { δ n } is a sequence such that
( 1 ) n = 1 α n = ; ( 2 ) lim sup n δ n α n 0 or n = 1 | δ n | < .

Then lim n s n = 0 .

Lemma 2.4 (See [13])

Let C be a nonempty closed convex subset of a real Hilbert space H, and let S : C C be a self-mapping of C. If S is a κ-strict pseudo-contraction mapping, then S satisfies the Lipschitz condition
S x S y 1 + κ 1 κ x y , x , y C .
For solving the equilibrium problem for a bifunction F : C × C R , let us assume that F satisfies the following conditions:
  1. (A1)

    F ( x , x ) = 0 , x C ;

     
  2. (A2)

    F is monotone, i.e., F ( x , y ) + F ( y , x ) 0 , x , y C ;

     
  3. (A3)
    x , y , z C ,
    lim t 0 + F ( t z + ( 1 t ) x , y ) F ( x , y ) ;
     
  4. (A4)

    x C , y F ( x , y ) is convex and lower semicontinuous.

     

The following lemma appears implicitly in [2].

Lemma 2.5 (See [2])

Let C be a nonempty closed convex subset of H, and let F be a bifunction of C × C into satisfying (A1)-(A4). Let r > 0 and x H . Then there exists z C such that
F ( z , y ) + 1 r y z , z x
(2.1)

for all y C .

Lemma 2.6 (See [14])

Assume that F : C × C R satisfies (A1)-(A4). For r > 0 and x H , define a mapping T r : H C as follows:
T r ( x ) = { z C : F ( z , y ) + 1 r y z , z x 0 , y C }
for all z H . Then the following hold:
  1. (1)

    T r is single-valued;

     
  2. (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. (3)

    F ( T r ) = EP ( F ) ;

     
  4. (4)

    EP ( F ) is closed and convex.

     

Lemma 2.7 (See [15])

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 S is demi-closed at zero.

Definition 2.1 Let C be a nonempty convex subset of real Hilbert space. Let { T i } i = 1 N be a finite family of κ i -strict pseudo-contractions of C into itself, and let { S i } i = 1 N be a finite family of nonexpansive 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 . We define the mapping S A : C C as follows:
U 0 = I , U 1 = S 1 ( α 1 1 T 1 U 0 + α 2 1 U 0 + α 3 1 I ) , U 2 = S 2 ( α 1 2 T 2 U 1 + α 2 2 U 1 + α 3 2 I ) , U 3 = S 3 ( α 1 3 T 3 U 2 + α 2 3 U 2 + α 3 3 I ) , U N 1 = S N 1 ( α 1 N 1 T N 1 U N 2 + α 2 N 1 U N 2 + α 3 N 1 I ) , S A = U N = S N ( α 1 N T N U N 1 + α 2 N U N 1 + α 3 N I ) .

This mapping is called the S A -mapping generated by S 1 , S 2 , , S N , T 1 , T 2 , , T N and α 1 , α 2 , , α N .

Lemma 2.8 Let C be a nonempty closed convex subset of a real Hilbert space. Let { T i } i = 1 N be a finite family of κ i -strict pseudo-contractions of C into itself, and let { S i } i = 1 N be a finite family of nonexpansive mappings of C into itself with i = 1 N F ( S i ) i = 1 N F ( T i ) and κ = max { κ i : i = 1 , 2 , , N } , 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 , α 3 j ( κ , 1 ) for all j = 1 , 2 , , N 1 and α 1 N ( κ , 1 ] , α 3 N [ κ , 1 ) , α 2 j ( κ , 1 ) for all j = 1 , 2 , , N . Let S A be the S A -mapping generated by S 1 , S 2 , , S N , T 1 , T 2 , , T N and α 1 , α 2 , , α N . Then F ( S A ) = i = 1 N F ( S i ) i = 1 N F ( T i ) , and S A is a nonexpansive mapping.

Proof It is easy to see that i = 1 N F ( S i ) i = 1 N F ( T i ) F ( S A ) . Let x 0 F ( S A ) and x i = 1 N F ( S i ) i = 1 N F ( T i ) . Then we have
S A x 0 x 2 = S N ( α 1 N T N U N 1 x 0 + α 2 N U N 1 x 0 + α 3 N x 0 ) x 2 α 1 N T N U N 1 x 0 + α 2 N U N 1 x 0 + α 3 N x 0 x 2 = α 1 N ( T N U N 1 x 0 x ) + α 2 N ( U N 1 x 0 x ) + α 3 N ( x 0 x ) 2 = α 1 N T N U N 1 x 0 x 2 + α 2 N U N 1 x 0 x 2 + α 3 N x 0 x 2 α 1 N α 2 N T N U N 1 x 0 U N 1 x 0 2 α 1 N α 3 N T N U N 1 x 0 x 0 2 α 2 N α 3 N U N 1 x 0 x 0 2 α 1 N ( U N 1 x 0 x 2 + κ ( I T N ) U N 1 x 0 ( I T N ) x 2 ) + α 2 N U N 1 x 0 x 2 + α 3 N x 0 x 2 α 1 N α 2 N T N U N 1 x 0 U N 1 x 0 2 α 1 N α 3 N T N U N 1 x 0 x 0 2 α 2 N α 3 N U N 1 x 0 x 0 2 = ( 1 α 3 N ) U N 1 x 0 x 2 + α 1 N ( κ α 2 N ) ( I T N ) U N 1 x 0 2 + ( 1 ( 1 α 3 N ) ) x 0 x 2 α 1 N α 3 N T N U N 1 x 0 x 0 2 α 2 N α 3 N U N 1 x 0 x 0 2 ( 1 α 3 N ) U N 1 x 0 x 2 + α 1 N ( κ α 2 N ) ( I T N ) U N 1 x 0 2 + ( 1 ( 1 α 3 N ) ) x 0 x 2 α 2 N α 3 N U N 1 x 0 x 0 2 j = N 1 N ( 1 α 3 j ) U N 2 x 0 x 2 + ( 1 α 3 N ) α 1 N 1 ( κ α 2 N 1 ) ( I T N 1 ) U N 2 x 0 2 ( 1 α 3 N ) α 2 N 1 α 3 N 1 U N 2 x 0 x 0 2 + ( 1 j = N 1 N ( 1 α 3 j ) ) x 0 x 2 j = N 1 N ( 1 α 3 j ) U N 2 x 0 x 2 + ( 1 j = N 1 N ( 1 α 3 j ) ) x 0 x 2 j = N 2 N ( 1 α 3 j ) U N 3 x 0 x 2 + j = N 1 N ( 1 α 3 j ) α 1 N 2 ( κ α 2 N 2 ) ( I T N 2 ) U N 3 x 0 2 j = N 1 N ( 1 α 3 j ) α 2 N 2 α 3 N 2 U N 3 x 0 x 0 2 + ( 1 j = N 2 N ( 1 α 3 j ) ) x 0 x 2 j = N 2 N ( 1 α 3 j ) U N 3 x 0 x 2 + ( 1 j = N 2 N ( 1 α 3 j ) ) x 0 x 2 j = 3 N ( 1 α 3 j ) U 2 x 0 x 2 + j = 4 N ( 1 α 3 j ) α 1 3 ( κ α 2 3 ) ( I T 3 ) U 2 x 0 2 j = 4 N ( 1 α 3 j ) α 2 3 α 3 3 U 2 x 0 x 0 2 + ( 1 j = 3 N ( 1 α 3 j ) ) x 0 x 2
(2.2)
j = 3 N ( 1 α 3 j ) U 2 x 0 x 2 + ( 1 j = 3 N ( 1 α 3 j ) ) x 0 x 2 j = 2 N ( 1 α 3 j ) U 1 x 0 x 2 + j = 3 N ( 1 α 3 j ) α 1 2 ( κ α 2 2 ) ( I T 2 ) U 1 x 0 2 j = 3 N ( 1 α 3 j ) α 2 2 α 3 2 U 1 x 0 x 0 2 + ( 1 j = 2 N ( 1 α 3 j ) ) x 0 x 2
(2.3)
j = 2 N ( 1 α 3 j ) U 1 x 0 x 2 + ( 1 j = 2 N ( 1 α 3 j ) ) x 0 x 2 j = 1 N ( 1 α 3 j ) U 0 x 0 x 2 + j = 2 N ( 1 α 3 j ) α 1 1 ( κ α 2 1 ) ( I T 1 ) U 0 x 0 2 j = 2 N ( 1 α 3 j ) α 2 1 α 3 1 U 0 x 0 x 0 2 + ( 1 j = 1 N ( 1 α 3 j ) ) x 0 x 2 x 0 x 2 + j = 2 N ( 1 α 3 j ) α 1 1 ( κ α 2 1 ) ( I T 1 ) x 0 2 .
(2.4)
By (2.4), we have
j = 2 N ( 1 α 3 j ) α 1 1 ( α 2 1 κ ) ( I T 1 ) x 0 2 x 0 x 2 x 0 x 2 = 0 ,
which implies that T 1 x 0 = x 0 , that is, x 0 F ( T 1 ) . It implies that
U 1 x 0 = S 1 ( α 1 1 T 1 U 0 x 0 + α 2 1 U 0 x 0 + α 3 1 x 0 ) = S 1 x 0 .
(2.5)
By (2.3) and (2.5), we have
S A x 0 x 2 j = 2 N ( 1 α 3 j ) S 1 x 0 x 2 + j = 3 N ( 1 α 3 j ) α 1 2 ( κ α 2 2 ) ( I T 2 ) U 1 x 0 2 j = 3 N ( 1 α 3 j ) α 2 2 α 3 2 U 1 x 0 x 0 2 + ( 1 j = 2 N ( 1 α 3 j ) ) x 0 x 2 x 0 x 2 j = 3 N ( 1 α 3 j ) α 2 2 α 3 2 U 1 x 0 x 0 2 .
(2.6)
By (2.6), we have
j = 3 N ( 1 α 3 j ) α 2 2 α 3 2 U 1 x 0 x 0 2 0 .
It implies that
x 0 = U 1 x 0 .
(2.7)
By (2.5) and (2.7), we have x 0 F ( S 1 ) . Hence, we have
x 0 F ( S 1 ) F ( T 1 ) .
(2.8)
Since x 0 = U 1 x 0 and (2.3), we have
S A x 0 x 2 j = 2 N ( 1 α 3 j ) U 1 x 0 x 2 + j = 3 N ( 1 α 3 j ) α 1 2 ( κ α 2 2 ) ( I T 2 ) U 1 x 0 2 j = 3 N ( 1 α 3 j ) α 2 2 α 3 2 U 1 x 0 x 0 2 + ( 1 j = 2 N ( 1 α 3 j ) ) x 0 x 2 = j = 2 N ( 1 α 3 j ) x 0 x 2 + j = 3 N ( 1 α 3 j ) α 1 2 ( κ α 2 2 ) ( I T 2 ) x 0 2 + ( 1 j = 2 N ( 1 α 3 j ) ) x 0 x 2 .
It follows that
j = 3 N ( 1 α 3 j ) α 1 2 ( α 2 2 κ ) ( I T 2 ) x 0 2 0 ,
which implies that x 0 = T 2 x 0 , that is, x 0 F ( T 2 ) . Since x 0 = U 1 x 0 = T 2 x 0 , we have
U 2 x 0 = S 2 ( α 1 2 T 2 U 1 x 0 + α 2 2 U 1 x 0 + α 3 2 x 0 ) = S 2 x 0 .
(2.9)
By (2.2), we have
S A x 0 x 2 j = 3 N ( 1 α 3 j ) U 2 x 0 x 2 + j = 4 N ( 1 α 3 j ) α 1 3 ( κ α 2 3 ) ( I T 3 ) U 2 x 0 2 j = 4 N ( 1 α 3 j ) α 2 3 α 3 3 U 2 x 0 x 0 2 + ( 1 j = 3 N ( 1 α 3 j ) ) x 0 x 2 j = 3 N ( 1 α 3 j ) S 2 x 0 x 2 j = 4 N ( 1 α 3 j ) α 2 3 α 3 3 U 2 x 0 x 0 2 + ( 1 j = 3 N ( 1 α 3 j ) ) x 0 x 2 x 0 x 2 j = 4 N ( 1 α 3 j ) α 2 3 α 3 3 U 2 x 0 x 0 2 .
It follows that
j = 4 N ( 1 α 3 j ) α 2 3 α 3 3 U 2 x 0 x 0 2 0 .
It implies that
x 0 = U 2 x 0 .
(2.10)
By (2.9) and (2.10), we have x 0 F ( S 2 ) . Hence, we have
x 0 F ( S 2 ) F ( T 2 ) .
(2.11)
By continuing in this way, we can show that x 0 F ( S i ) F ( T i ) and x 0 = U i x 0 for all i = 1 , 2 , , N 1 . Finally, we shall show that x 0 F ( S N ) F ( T N ) . Since
S A x 0 x 2 ( 1 α 3 N ) U N 1 x 0 x 2 + α 1 N ( κ α 2 N ) ( I T N ) U N 1 x 0 2 + ( 1 ( 1 α 3 N ) ) x 0 x 2 = ( 1 α 3 N ) S N 1 x 0 x 2 + α 1 N ( κ α 2 N ) ( I T N ) x 0 2 + ( 1 ( 1 α 3 N ) ) x 0 x 2 x 0 x 2 + α 1 N ( κ α 2 N ) ( I T N ) x 0 2 .
It implies that
α 1 N ( α 2 N κ ) ( I T N ) x 0 2 0 ,
which implies that x 0 = T N x 0 , that is, x 0 F ( T N ) . It implies that
x 0 = S A x 0 = S N ( α 1 N T N U N 1 x 0 + α 2 N U N 1 x 0 + α 3 N x 0 ) = S N x 0 .
(2.12)

Then we have x 0 F ( S N ) F ( T N ) . Hence F ( S A ) i = 1 N F ( S i ) i = 1 N F ( T i ) .

Applying (2.4), we have that the mapping S A is a nonexpansive. □

Lemma 2.9 Let C be a nonempty closed convex subset of a real Hilbert space. Let { T i } i = 1 N be a finite family of κ i -strict pseudo-contractions of C into itself, and let { S i } i = 1 N be a finite family of nonexpansive mappings of C into itself with κ = max { κ i : i = 1 , 2 , , N } , and let 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 N , let S A and S n A be the S A -mapping generated by S 1 , S 2 , , S N , T 1 , T 2 , , T N and α 1 , α 2 , , α N and S 1 , S 2 , , S N , T 1 , T 2 , , T N and α 1 ( n ) , α 2 ( n ) , , α N ( n ) , respectively. Then lim n S n A x n S A x n = 0 for every bounded sequence { x n } in C.

Proof Let { x n } be a bounded sequence in C, U k and U n , k be generated by S 1 , S 2 , , S N , T 1 , T 2 , , T N and α 1 , α 2 , , α N and S 1 , S 2 , , S N , T 1 , T 2 , , T N and α 1 ( n ) , α 2 ( n ) , , α N ( n ) , respectively. For each n N , we have
U n , 1 x n U 1 x n = S 1 ( α 1 n , 1 T 1 x n + ( 1 α 1 n , 1 ) x n ) S 1 ( α 1 1 T 1 x n + ( 1 α 1 1 ) x n ) α 1 n , 1 T 1 x n + ( 1 α 1 n , 1 ) x n α 1 1 T 1 x n ( 1 α 1 1 ) x n = | α 1 n , 1 α 1 1 | T 1 x n x n ,
(2.13)
and for k { 2 , 3 , , N } , by using Lemma 2.4, we obtain
U n , k x n U k x n = S k ( α 1 n , k T k U n , k 1 x n + α 2 n , k U n , k 1 x n + α 3 n , k x n ) S k ( α 1 k T k U k 1 x n + α 2 k U k 1 x n + α 3 k x n ) α 1 n , k T k U n , k 1 x n + α 2 n , k U n , k 1 x n + α 3 n , k x n α 1 k T k U k 1 x n α 2 k U k 1 x n α 3 k x n = α 1 n , k ( T k U n , k 1 x n T k U k 1 x n ) + ( α 1 n , k α 1 k ) T k U k 1 x n + ( α 3 n , k α 3 k ) x n + α 2 n , k ( U n , k 1 x n U k 1 x n ) + ( α 2 n , k α 2 k ) U k 1 x n α 1 n , k T k U n , k 1 x n T k U k 1 x n + | α 1 n , k α 1 k | T k U k 1 x n + | α 3 n , k α 3 k | x n + α 2 n , k U n , k 1 x n U k 1 x n + | α 2 n , k α 2 k | U k 1 x n = α 1 n , k T k U n , k 1 x n T k U k 1 x n + | α 1 n , k α 1 k | T k U k 1 x n + α 2 n , k U n , k 1 x n U k 1 x n + | 1 α 1 n , k α 3 n , k 1 + α 1 k + α 3 k | U k 1 x n + | α 3 n , k α 3 k | x n α 1 n , k 1 + κ 1 κ U n , k 1 x n U k 1 x n + | α 1 n , k α 1 k | T k U k 1 x n + α 2 n , k U n , k 1 x n U k 1 x n + ( | α 1 k α 1 n , k | + | α 3 n , k α 3 k | ) U k 1 x n + | α 3 n , k α 3 k | x n 1 + κ 1 κ U n , k 1 x n U k 1 x n + | α 1 n , k α 1 k | T k U k 1 x n + 1 κ 1 κ U n , k 1 x n U k 1 x n + ( | α 1 k α 1 n , k | + | α 3 n , k α 3 k | ) U k 1 x n + | α 3 n , k α 3 k | x n 2 1 κ U n , k 1 x n U k 1 x n + | α 1 n , k α 1 k | ( T k U k 1 x n + U k 1 x n ) + | α 3 n , k α 3 k | ( U k 1 x n + x n ) .
(2.14)
By (2.13) and (2.14), we have
S n A x n S A x n = U n , N x n U N x n 2 1 κ U n , N 1 x n U N 1 x n + | α 1 n , N α 1 N | ( T N U N 1 x n + U N 1 x n ) + | α 3 n , N α 3 N | ( U N 1 x n + x n ) 2 1 κ ( 2 1 κ U n , N 2 x n U N 2 x n + | α 1 n , N 1 α 1 N 1 | ( T N 1 U N 2 x n + U N 2 x n ) + | α 3 n , N 1 α 3 N 1 | ( U N 2 x n + x n ) ) + | α 1 n , N α 1 N | ( T N U N 1 x n + U N 1 x n ) + | α 3 n , N α 3 N | ( U N 1 x n + x n ) = ( 2 1 κ ) 2 U n , N 2 x n U N 2 x n + j = N 1 N ( 2 1 κ ) N j | α 1 n , j α 1 j | ( T j U j 1 x n + U j 1 x n ) + j = N 1 N ( 2 1 κ ) N j | α 3 n , j α 3 j | ( U j 1 x n + x n ) ( 2 1 κ ) N 1 U n , 1 x n U 1 x n + j = 2 N ( 2 1 κ ) N j | α 1 n , j α 1 j | ( T j U j 1 x n + U j 1 x n ) + j = 2 N ( 2 1 κ ) N j | α 3 n , j α 3 j | ( U j 1 x n + x n ) = ( 2 1 κ ) N 1 | α 1 n , 1 α 1 1 | T 1 x n x n + j = 2 N ( 2 1 κ ) N j | α 1 n , j α 1 j | ( T j U j 1 x n + U j 1 x n ) + j = 2 N ( 2 1 κ ) N j | α 3 n , j α 3 j | ( U j 1 x n + x n ) .
(2.15)
This together with the assumption α i n , j α i j as n ( i = 1 , 3 , j = 1 , 2 , , N ), we can conclude that
lim n S n A x n S A x n = 0 .

 □

Lemma 2.10 Let C be a nonempty closed convex subset of a real Hilbert space. Let { T i } i = 1 N be a finite family of κ i -strict pseudo-contractions of C into itself, and let { S i } i = 1 N be a finite family of nonexpansive mappings of C into itself with κ = max { κ i : i = 1 , 2 , , N } , and let 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 n = 1 | α 1 n + 1 , j α 1 n , j | < , n = 1 | α 3 n + 1 , j α 3 n , j | < for all j { 1 , 2 , 3 , , N } . For every n N , let S n A be the S A -mapping generated by S 1 , S 2 , , S N , T 1 , T 2 , , T N and α 1 ( n ) , α 2 ( n ) , , α N ( n ) . Then n = 1 S n + 1 A z n S n A z n < for every bounded sequence { z n } in C.

Proof Let { z n } be a bounded sequence in C. For each n N and the definition of S A , we have
U n + 1 , 1 z n U n , 1 z n = S 1 ( α 1 n + 1 , 1 T 1 z n + ( 1 α 1 n + 1 , 1 ) z n ) S 1 ( α 1 n , 1 T 1 z n + ( 1 α 1 n , 1 ) z n ) α 1 n + 1 , 1 T 1 z n + ( 1 α 1 n + 1 , 1 ) z n α 1 n , 1 T 1 z n ( 1 α 1 n , 1 ) z n = | α 1 n + 1 , 1 α 1 n , 1 | T 1 z n z n .
(2.16)
For k { 2 , 3 , , N } , and using the same method as (2.14) in Lemma 2.9, we have
U n + 1 , k z n U n , k z n 2 1 κ U n + 1 , k 1 z n U n , k 1 z n + | α 1 n + 1 , k α 1 n , k | ( T k U n , k 1 z n + U n , k 1 z n ) + | α 3 n + 1 , k α 3 n , k | ( U n , k 1 z n + z n ) .
(2.17)
From (2.16), (2.17), and using the same method as (2.15) in Lemma 2.9, we have
S n + 1 A z n S n A z n ( 2 1 κ ) N 1 | α 1 n + 1 , 1 α 1 n , 1 | T 1 z n z n + j = 2 N ( 2 1 κ ) N j | α 1 n + 1 , j α 1 n , j | ( T j U n , j 1 z n + U n , j 1 z n ) + j = 2 N ( 2 1 κ ) N j | α 3 n + 1 , j α 3 n , j | ( U n , j 1 z n + z n ) .
It implies that
n = 1 S n + 1 A z n S n A z n < .

 □

3 Main result

Theorem 3.1 Let C be a nonempty closed convex subset of Hilbert spaces H, and let f be an α-contraction on H. Let F i be a bifunction from C × C into , for every i = 1 , 2 , , N satisfying (A1)-(A4). Let { T i } i = 1 N be a finite family of κ i -strict pseudo-contractions of C into itself, and let { S i } i = 1 N be a finite family of nonexpansive mappings of C into itself with F i = 1 N F ( S i ) i = 1 N F ( T i ) i = 1 N EP ( F i ) and κ = max { κ i : i = 1 , 2 , , N } , and let α j ( n ) = ( α 1 n , j , α 2 n , j , α 3 n , j ) I × I × I , j = 1 , 2 , 3 , , N , where I = [ 0 , 1 ] , α 1 n , j + α 2 n , j + α 3 n , j = 1 , α 1 n , j , α 2 n , j , α 3 n , j [ a , b ] ( κ , 1 ) for all j = 1 , 2 , , N . Let S n A be the S A -mapping generated by S 1 , S 2 , , S N , T 1 , T 2 , , T N and α 1 ( n ) , α 2 ( n ) , , α N ( n ) . Let { x n } and { z n } be the sequences generated by x 1 C and
{ F i ( u n i , y ) + 1 r n i y u n i , u n i x n 0 , y C  and  i = 1 , 2 , , N , z n = i = 1 N δ n i u n i , x n + 1 = α n f ( z n ) + ( 1 α n ) S n A z n , n 1 ,
(3.1)
where { α n } is a sequence in [ 0 , 1 ] . Assume that the following conditions hold:
  1. (i)

    lim n α n = 0 , n = 1 α n = ;

     
  2. (ii)

    n = 1 | α 1 n + 1 , j α 1 n , j | < , n = 1 | α 3 n + 1 , j α 3 n , j | < , for all j { 1 , 2 , 3 , , N } and n = 1 | α n + 1 α n | < ;

     
  3. (iii)

    i = 1 N δ n i = 1 , n = 1 | δ n + 1 i δ n i | < and lim n δ n i = δ i ( κ , 1 ) , for every i = 1 , 2 , , N ;

     
  4. (iv)

    κ < θ r n i η , for every i = 1 , 2 , , N and n = 1 | r n + 1 i r n i | < .

     

Then the sequence { x n } converges strongly to x = P F f ( x ) .

Proof Let p F , we have p i = 1 N EP ( F i ) from Lemma 2.6, we obtain p i = 1 N F ( T r n i ) . Since
F i ( u n i , y ) + 1 r n i y u n i , u n i x n 0 , y C  and  i = 1 , 2 , , N .
(3.2)
Again from Lemma 2.6, we have u n i = T r n i x n for every i = 1 , 2 , , N . By definition of x n , we have
x n + 1 p α n f ( z n ) p + ( 1 α n ) S n A z n p α n f ( z n ) f ( p ) + α n f ( p ) p + ( 1 α n ) S n A z n p α n α z n p + α n f ( p ) p + ( 1 α n ) z n p = α n f ( p ) p + ( 1 α n ( 1 α ) ) z n p = α n f ( p ) p + ( 1 α n ( 1 α ) ) i = 1 N δ n i ( u n i p ) α n f ( p ) p + ( 1 α n ( 1 α ) ) i = 1 N δ n i u n i p α n f ( p ) p + ( 1 α n ( 1 α ) ) x n p .
(3.3)

Put K = max { x 1 p , f ( p ) p 1 α } . By (3.3), we can show by induction that x n p K , n N . This implies that { x n } is bounded, and so are { u n i } , for every i = 1 , 2 , , N and { z n } .

Next, we will show that
lim n x n + 1 x n = 0 .
(3.4)
By nonexpansiveness of x n , we have
x n + 1 x n = α n f ( z n ) + ( 1 α n ) S n A z n α n 1 f ( z n 1 ) ( 1 α n 1 ) S n 1 A z n 1 = α n ( f ( z n ) f ( z n 1 ) ) + ( α n α n 1 ) f ( z n 1 ) + ( 1 α n ) ( S n A z n S n 1 A z n 1 ) + ( α n 1 α n ) S n 1 A z n 1 α n f ( z n ) f ( z n 1 ) + | α n α n 1 | f ( z n 1 ) + ( 1 α n ) S n A z n S n 1 A z n 1 + | α n 1 α n | S n 1 A z n 1 α n α z n z n 1 + | α n α n 1 | f ( z n 1 ) + ( 1 α n ) ( S n A z n S n A z n 1 + S n A z n 1 S n 1 A z n 1 ) + | α n 1 α n | S n 1 A z n 1 ( 1 α n ( 1 α ) ) z n z n 1 + | α n α n 1 | f ( z n 1 ) + ( 1 α n ) S n A z n 1 S n 1 A z n 1 + | α n 1 α n | S n 1 A z n 1 = ( 1 α n ( 1 α ) ) ( i = 1 N δ n i u n i i = 1 N δ n 1 i u n 1 i ) + | α n α n 1 | f ( z n 1 ) + ( 1 α n ) S n A z n 1 S n 1 A z n 1 + | α n 1 α n | S n 1 A z n 1 = ( 1 α n ( 1 α ) ) ( i = 1 N δ n i ( u n i u n 1 i ) + i = 1 N ( δ n i δ n 1 i ) u n 1 i ) + | α n α n 1 | f ( z n 1 ) + ( 1 α n ) S n A z n 1 S n 1 A z n 1 + | α n 1 α n | S n 1 A z n 1 ( 1 α n ( 1 α ) ) ( i = 1 N δ n i u n i u n 1 i + i = 1 N | δ n i δ n 1 i | u n 1 i ) + | α n α n 1 | f ( z n 1 ) + ( 1 α n ) S n A z n 1 S n 1 A z n 1 + | α n 1 α n | S n 1 A z n 1 .
(3.5)
From Lemma 2.10, we have
n = 1 S n + 1 A z n S n A z n < .
(3.6)
Since u n i = T r n i x n for every i = 1 , 2 , , N . By definition of T r n i , we have
F ( T r n i x n , y ) + 1 r n i y T r n i x n , T r n i x n x n 0 , y C ,
(3.7)
similarly,
F ( T r n + 1 i x n + 1 , y ) + 1 r n + 1 i y T r n + 1 i x n + 1 , T r n + 1 i x n + 1 x n + 1 0 , y C .
(3.8)
From (3.7) and (3.8), we obtain
F ( T r n i x n , T r n + 1 i x n + 1 ) + 1 r n i T r n + 1 i x n + 1 T r n i x n , T r n i x n x n 0
(3.9)
and
F ( T r n + 1 i x n + 1 , T r n i x n ) + 1 r n + 1 i T r n i x n T r n + 1 i x n + 1 , T r n + 1 i x n + 1 x n + 1 0 .
(3.10)
By (3.9) and (3.10), we have
1 r n i T r n + 1 i x n + 1 T r n i x n , T r n i x n x n + 1 r n + 1 i T r n i x n T r n + 1 i x n + 1 , T r n + 1 i x n + 1 x n + 1 0 .
It follows that
T r n i x n T r n + 1 i x n + 1 , T r n + 1 i x n + 1 x n + 1 r n + 1 i T r n i x n x n r n i 0 .
This implies that
0 T r n + 1 i x n + 1 T r n i x n , T r n i x n T r n + 1 i x n + 1 + T r n + 1 i x n + 1 x n r n i r n + 1 i ( T r n + 1 i x n + 1 x n + 1 ) .
It follows that
T r n + 1 i x n + 1 T r n i x n 2 T r n + 1 i x n + 1 T r n i x n , T r n + 1 i x n + 1 x n r n i r n + 1 i ( T r n + 1 i x n + 1 x n + 1 ) = T r n + 1 i x n + 1 T r n i x n , x n + 1 x n + ( 1 r n i r n + 1 i ) ( T r n + 1 i x n + 1 x n + 1 ) T r n + 1 i x n + 1 T r n i x n x n + 1 x n + ( 1 r n i r n + 1 i ) ( T r n + 1 i x n + 1 x n + 1 ) T r n + 1 i x n + 1 T r n i x n ( x n + 1 x n + | 1 r n i r n + 1 i | T r n + 1 i x n + 1 x n + 1 ) = T r n + 1 i x n + 1 T r n i x n ( x n + 1 x n + 1 r n + 1 i | r n + 1 i r n i | T r n + 1 i x n + 1 x n + 1 ) T r n + 1 i x n + 1 T r n i x n ( x n + 1 x n + 1 a | r n + 1 i r n i | T r n + 1 i x n + 1 x n + 1 ) .
It follows that
u n + 1 i u n i x n + 1 x n + 1 a | r n + 1 i r n i | u n + 1 i x n + 1
(3.11)

for every i = 1 , 2 , , N .

Substitute (3.11) into (3.5), we have
x n + 1 x n ( 1 α n ( 1 α ) ) ( i = 1 N δ n i u n i u n 1 i + i = 1 N | δ n i δ n 1 i | u n 1 i ) + | α n α n 1 | f ( z n 1 ) + ( 1 α n ) S n A z n 1 S n 1 A z n 1 + | α n 1 α n | S n 1 A z n 1 ( 1 α n ( 1 α ) ) ( i = 1 N δ n i ( x n + 1 x n + 1 a | r n + 1 i r n i | u n + 1 i x n + 1 ) + i = 1 N | δ n i δ n 1 i | u n 1 i ) + | α n α n 1 | f ( z n 1 ) + ( 1 α n ) S n A z n 1 S n 1 A z n 1 + | α n 1 α n | S n 1 A z n 1 = ( 1 α n ( 1 α ) ) ( x n + 1 x n + i = 1 N δ n i 1 a | r n + 1 i r n i | u n + 1 i x n + 1 + i = 1 N | δ n i δ n 1 i | u n 1 i ) + | α n α n 1 | f ( z n 1 ) + ( 1 α n ) S n A z n 1 S n 1 A z n 1 + | α n 1 α n | S n 1 A z n 1 ( 1 α n ( 1 α ) ) x n + 1 x n + i = 1 N δ n i 1 a | r n + 1 i r n i | u n + 1 i x n + 1 + i = 1 N | δ n i δ n 1 i | u n 1 i + | α n α n 1 | f ( z n 1 ) + S n A z n 1 S n 1 A z n 1 + | α n 1 α n | S n 1 A z n 1 .
(3.12)
By (3.12), (3.6), conditions (iii), (iv) and Lemma 2.3, we have
lim n x n + 1 x n = 0 .
(3.13)
From (3.11), (3.13) and condition (iv), we have
lim n u n + 1 i u n i = 0 , i = 1 , 2 , , N .
(3.14)
Let p F . From u n i = T r n i x n for every i = 1 , 2 , , N , we have
u n i p 2 = T r n i x n T r n i p 2 T r n i x n T r n i p , x n p = 1 2 ( u n i p 2 + x n p 2 u n i x n 2 ) .
It implies that
u n i p 2 x n p 2 u n i x n 2 .
(3.15)
By definition of { x n } and (3.15), we have
x n + 1 p 2 α n f ( z n ) p 2 + ( 1 α n ) S n A z n p 2 α n f ( z n ) p 2 + ( 1 α n ) z n p 2 = α n f ( z n ) p 2 + ( 1 α n ) i = 1 N δ n i ( u n i p ) 2 α n f ( z n ) p 2 + ( 1 α n ) i = 1 N δ n i u n i p 2 α n f ( z n ) p 2 + ( 1 α n ) i = 1 N δ n i ( x n p 2 u n i x n 2 ) α n f ( z n ) p 2 + x n p 2 ( 1 α n ) i = 1 N δ n i u n i x n 2 .
It implies that
( 1 α n ) i = 1 N δ n i u n i x n 2 α n f ( z n ) p 2 + x n p 2 x n + 1 p 2 α n f ( z n ) p 2 + ( x n p + x n + 1 p ) x n + 1 x n .
(3.16)
From conditions (i), (iii) and (3.13), we have
lim n u n i x n = 0 , i = 1 , 2 , , N .
(3.17)
Since
x n + 1 S n A z n = α n ( f ( z n ) S n A z n ) ,
from condition (i), we have
lim n x n + 1 S n A z n = 0 .
(3.18)
From the definition of z n , we have
z n x n = i = 1 N δ n i ( u n i x n ) i = 1 N δ n i u n i x n .
From condition (iii) and (3.17), we have
lim n z n x n = 0 .
(3.19)
Since
z n S n A z n z n x n + x n x n + 1 + x n + 1 S n A z n ,
by (3.13), (3.18) and (3.19), we have
lim n z n S n A z n = 0 .
(3.20)
Next, we show that
lim sup n f ( z ) z , x n z 0 ,
(3.21)
where z = P F f ( z ) . To show this inequality, take a subsequence { x n k } of { x n } such that
lim sup n f ( z ) z , x n z = lim k f ( z ) z , x n k z .
(3.22)

Without loss of generality, we may assume that a subsequence { x n k } of { x n } converges weakly to some q H . From (3.19), we have that { z n k } converges weakly to q.

Since κ < a α 1 n , j , α 2 n , j , α 3 n , j b < 1 for all j = 1 , 2 , , N . Without loss of generality, we may assume that
α 1 n k , j α 1 j ( κ , 1 ) , α 3 n k , j α 3 j ( κ , 1 ) and α 2 n k , j α 2 j ( κ , 1 ) as  k , j = 1 , 2 , , N .

Let S A be the S A -mapping generated by S 1 , S 2 , , S N , T 1 , T 2 , , T N and β 1 , β 2 , , β N , where β j = ( α 1 j , α 2 j , α 3 j ) , j = 1 , 2 , , N . By Lemma 2.8, S A is a nonexpansive mapping, and F ( S A ) = i = 1 N F ( S i ) i = 1 N F ( T i ) .

By Lemma 2.9, we have
lim k S n k A z n k S A z n k = 0 .
(3.23)
Since
z n k S A z n k z n k S n k A z n k + S n k A z n k S A z n k ,
by (3.20), (3.23), we have
lim k z n k S A z n k = 0 .
(3.24)
Since { z n k } converges weakly to q as k (3.24) and Lemma 2.7, we have
q F ( S A ) = i = 1 N F ( S i ) i = 1 N F ( T i ) .
(3.25)
Next, we show that q i = 1 N EP ( F i ) . To show this, we may assume that
lim k r n k i = r i [ θ , η ] , i = 1 , 2 , , N .
By Lemmas 2.5 and 2.6, for every i = 1 , 2 , , N , we define T r i : H C by
T r i ( x ) = { z C : F i ( z , y ) + 1 r i y z , z x 0 , y C } , x H  and  i = 1 , 2 , , N .
Then we have
F i ( T r i x n , y ) + 1 r i y T r i x n , T r i x n x n 0 , y C  and  i = 1 , 2 , , N .
From (3.1) and u n i = T r n i x n , we have
F i ( T r n i x n , y ) + 1 r n i y T r n i x n , T r n i x n x n 0 , y C  and  i = 1 , 2 , , N .
It implies that
F i ( T r i x n k , T r n k i x n k ) + 1 r i T r n k i x n k T r i x n k , T r i x n k x n k 0 , i = 1 , 2 , , N
and
F i ( T r n k i x n k , T r i x n k ) + 1 r n k i T r i x n k T r n k i x n k , T r n k i x n k x n k 0 , i = 1 , 2 , , N .
By (A2), we have
1 r i T r n k i x n k T r i x n k , T r i x n k x n k + 1 r n k i T r i x n k T r n k i x n k , T r n k i x n k x n k 0 .
It implies that
T r n k i x n k T r i x n k , T r i x n k x n k r i T r n k i x n k x n k r n k i 0 .
It follows that
T r n k i x n k T r i x n