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

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.

1 Introduction

Let H be a real Hilbert space and C be a nonempty closed convex subset of H. A self mapping f:CC is a contraction on C if there exists a constant k[0,1) such that f(x)f(y)kxy, x,yC. Let T:CC be a mapping, a point xC is called a fixed point of T if and only if Tx=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:CC is called nonexpansive if and only if for all x,yC,

TxTyxy.

Definition 1.2 A mapping T:CC is called κ-strictly pseudocontractive [1] if and only if there exists a constant κ[0,1) such that for all x,yC,

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×CR 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:CH, let F(x,y)=Tx,yx for all x,yC. Then zEP(F) if and only if Tz,yz0 for all yC, 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, xC;

  2. (A2)

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

  3. (A3)

    x,y,zC,

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

    xC, yF(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 nN, 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 zF(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,,N1, 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 uH 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)

    0κ β n <θ<1 for all n1 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 nN 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 :CC 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 xH, P C x satisfies the property

x P C x= min y C xy.

The following characterizes the projection P C .

Lemma 2.1 (See [10])

Given xH and yC. Then P C x=y if and only if the following inequality holds

xy,yz0,zC.

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 ,n0,

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 ,n0,

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:CC be a self-mapping of C. If S is a κ-strict pseudo-contraction mapping, then S satisfies the Lipschitz condition

SxSy 1 + κ 1 κ xy,x,yC.

For solving the equilibrium problem for a bifunction F:C×CR, let us assume that F satisfies the following conditions:

  1. (A1)

    F(x,x)=0, xC;

  2. (A2)

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

  3. (A3)

    x,y,zC,

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

    xC, yF(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 xH. Then there exists zC such that

F(z,y)+ 1 r yz,zx
(2.1)

for all yC.

Lemma 2.6 (See [14])

Assume that F:C×CR satisfies (A1)-(A4). For r>0 and xH, define a mapping T r :HC as follows:

T r (x)= { z C : F ( z , y ) + 1 r y z , z x 0 , y C }

for all zH. 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,yH;
  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:CC be a nonexpansive mapping. Then IS 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 :CC 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,,N1 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,,N1. 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 nN, 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 nN, 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 nN, 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 nN 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 pF, 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,yC 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 pK, nN. 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,yC,
(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,yC.
(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 pF. 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 qH. 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

qF ( 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 :HC by

T r i (x)= { z C : F i ( z , y ) + 1 r i y z , z x 0 , y C } ,xH 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,yC 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,yC 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 k , T r i x n k x n k r i r n k i ( T r n k i x n k x n k ) 0.

Then

0 T r n k i x n k T r i x n k , 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 r i r n k i ( T r n k i x n k x n k ) = T r n k i x n k T r i x n k , T r i x n k T r n k i x n k + ( 1 r i r n k i ) ( T r n k i x n k x n k ) .

It follows that

T r n k i x n k T r i x n k 2 T r n k i x n k T r i x n k , ( 1 r i r n k i ) ( T r n k i x n k x n k ) T r n k i x n k T r i x n k | 1 r i r n k i | T r n k i x n k x n k .

It implies that

T r n k i x n k T r i x n k 1 a | r n k i r i | T r n k i x n k x n k .

From lim k r n k i = r i and (3.17), we have

lim k T r n k i x n k T r i x n k =0,i=1,2,,N.
(3.26)

For every i=1,2,,N, we have

x n k T r i x n k x n k T r n k i x n k + T r n k i x n k T r i x n k = x n k u n k i + T r n k i x n k T r i x n k ,

by (3.17) and (3.26), we have

lim k x n k T r i x n k =0,i=1,2,,N.
(3.27)

Since a subsequence { x n k } of { x n } converges weakly to q as k, from (3.27) and Lemma 2.7, we have

qF( T r i ),i=1,2,,N.

Then

q i = 1 N F( T r i ).
(3.28)

From Lemma 2.6, we have EP( F i )=F( T r i ), i=1,2,,N. From (3.28), we have

q i = 1 N F( T r i )= i = 1 N EP( F i ).
(3.29)

By (3.25) and (3.29), we have

qF.
(3.30)

Since x n k q as k and qF and (3.22), we have

lim sup n f ( z ) z , x n z = lim k f ( z ) z , x n k z = f ( z ) z , q z 0.

Finally, we show that { x n } converges strongly to z= P F f(z). Putting z= P F f(z), by nonexpansiveness of S A , we have

x n + 1 z 2 = α n ( f ( z n ) z ) + ( 1 α n ) ( S n A z n z ) 2 ( 1 α n ) 2 S n A z n z 2 + 2 α n f ( z n ) z , x n + 1 z = ( 1 α n ) 2 S n A z n z 2 + 2 α n f ( z n ) f ( z ) , x n + 1 z + 2 α n f ( z ) z , x n + 1 z ( 1 2 α n + α n 2 ) z n z 2 + 2 α n α z n z x n + 1 z + 2 α n f ( z ) z , x n + 1 z ( 1 2 α n + α n 2 ) z n z 2 + α n α z n z 2 + α n α x n + 1 z 2 + 2 α n f ( z ) z , x n + 1 z ( 1 2 α n + α n α ) x n z 2 + α n 2 x n z 2 + α n α x n + 1 z 2 + 2 α n f ( z ) z , x n + 1 z = ( 1 α n α 2 α n ( 1 α ) ) x n z 2 + α n 2 x n z 2 + α n α x n + 1 z 2 + 2 α n f ( z ) z , x n + 1 z .

It implies that

x n + 1 z 2 ( 1 2 α n ( 1 α ) 1 α n α ) x n z 2 + α n 2 1 α n α x n z 2 + 2 α n 1 α n α f ( z ) z , x n + 1 z .

This implies that by condition (i), (3.21) and Lemma 2.2, we have that the sequence { x n } converges strongly to z= P F f(z). By (3.19), we have

z n z z n x n + x n z0as n.

This completes the proof. □

4 Applications

In this section, we apply our main result to prove strong convergence theorems involving variational inclusion problems and variational inequality problems. To prove these results, we need definition and lemmas as follows.

A set-valued mapping M:H 2 H is called monotone if for all x,yH, fMx and gMy imply that xy,fg0. A monotone mapping M:H 2 H is maximal if the graph 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, xy,fg0 for every (y,g)Graph(M) implies that fMx.

Next, we consider the following so-called variational inclusion problem: Find a uH such that

θBu+Mu,
(4.1)

where B:HH, M:H 2 H are two nonlinear mappings, and θ is zero vector in H (see, for instance, [1621]). The set of the solution of (4.1) is denoted by VI(H,B,M).

Definition 4.1 (See [16])

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),uH,

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

Lemma 4.1 (See [16])

uH is a solution of variational inclusion (4.1) if and only if u= J M , λ (uλBu), λ>0, i.e.,

VI(H,B,M)=F ( J M , λ ( I λ B ) ) ,λ>0.

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

Lemma 4.2 (See [6])

The resolvent operator J M , λ associated with M is single-valued, nonexpansive for all λ>0 and 1-inverse-strongly monotone.

A mapping A of C into H is called α-inverse strongly monotone, see [22], if there exists a positive real number α such that

xy,AxAyα A x A y 2

for all x,yC. The variational inequality problem is to find uC such that

Au,vu0
(4.2)

for all vC. The set of solutions of the variational inequality is denoted by VI(C,A). We need the following lemma to prove a strong convergence theorem in this section.

Lemma 4.3 (See [23])

Let C be a closed convex subset of Hilbert space H. Let A i :CH be mappings, and let G i :CC be defined by G i (y)= P C (I λ i A i )y with λ i >0, i=1,2,,N. Then x i = 1 N VI(C, A i ) if and only if x i = 1 N F( G i ).

Theorem 4.4 Let C be a nonempty closed convex subset of Hilbert spaces H, and let f be an α-contraction on H. For every i=1,2,,N, let F i be a bifunction from C×C into satisfying (A1)-(A4), let A i :CH be an α i -inverse strongly monotone, and let G i :CC be a mapping defined by G i (y)= P C (I λ i A i )y, yC with λ i (0,1](0,2 α i ). Let { T i } i = 1 N be a finite family of κ i -strict pseudo-contractions of C into itself with F i = 1 N VI(C, A 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 G 1 , G 2 ,, G 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 ,
(4.3)

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 First, we show that (I λ i A i ) is a nonexpansive mapping for every i=1,2,,N. For x,yC, we have

( I λ i A i ) x ( I λ i A i ) y 2 = x y λ i ( A i x A i y ) 2 = x y 2 2 λ i x y , A i x A i y + λ i 2 A i x A i y 2 x y 2 2 α i λ i A i x A i y 2 + λ i 2 A i x A i y 2 = x y 2 + λ i ( λ i 2 α i ) A i x A i y 2 x y 2 .
(4.4)

Thus, (I λ i A i ) is a nonexpansive mapping, and so is G i for all i=1,2,,N. Then we obtain the desired result from Lemma 4.3 and Theorem 3.1. □

Corollary 4.5 Let C be a nonempty closed convex subset of Hilbert spaces H, and let f be an α-contraction on H. For every i=1,2,,N, let F i be a bifunction from C×C into , satisfying (A1)-(A4), let A i :CH be an α i -inverse strongly monotone, and let G i :CC be a mapping defined by G i (y)= P C (I λ i A i )y, yC with λ i (0,1](0,2 α i ). Let { T i } i = 1 N be a finite family of nonexpansive mappings of C into itself with F i = 1 N VI(C, A i ) i = 1 N F( T i ) i = 1 N EP( F i ), 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](0,1) for all j=1,2,,N. Let S n A be the S A -mapping generated by G 1 , G 2 ,, G 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 ,
(4.5)

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 (0,1) for every i=1,2,,N;

  4. (iv)

    0<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 Since { T i } i = 1 N is a finite family of nonexpansive mappings, we have that { T i } i = 1 N is a finite family of κ i -strict pseudo-contractive mappings. From Theorem 4.4, we can draw the desired conclusion. □

Theorem 4.6 Let C be a nonempty closed convex subset of Hilbert spaces H, and let f be an α-contraction on H. For every i=1,2,,N, let F i be a bifunction from C×C into satisfying (A1)-(A4). Let M i :H 2 H be maximal monotone mappings for every i=1,2,,N, and let B i :HH be a δ i -inverse strongly monotone mapping for every i=1,2,,N. Let G i :HH be a mapping defined by J M i , η (Iη B i )x= G i x for every xH with η(0,2 δ i ) i=1,2,,N. Let { T i } i = 1 N be a finite family of κ i -strict pseudo-contractions of H into itself with F i = 1 N V(H, B i , M 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 G 1 , G 2 ,, G 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 H 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 ,
(4.6)

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 By using the same method as (4.4), we have that Iη B i is a nonexpansive mapping for every i=1,2,,N. By Lemma 4.2, we have J M i , η (Iη B i )= G i is a nonexpansive mapping for every i=1,2,,N. Then we obtain the desired result from Theorem 3.1. □

5 Example and numerical results

In the last section, we give numerical examples to support our main results.

Example 5.1 Let be the set of real numbers. For every i=1,2,,N, let the mappings F i :R×RR, T i :RR, S i :RR, f:RR defined by

F i ( x , y ) = i ( 4 y 2 + x y 5 x 2 ) , T i x = ( 1 ) 2 i + 1 3 2 x , S i x = 2 i 2 i + 1 x , f x = 1 3 x

for every x,yR.

Suppose that S n A is 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 ) , where α j ( n ) =( α 1 ( n , j ) , α 2 ( n , j ) , α 3 ( n , j ) ) and α 1 ( n , j ) = α 2 ( n , j ) = α 3 ( n , j ) = 1 3 for every n1 and j=1,2,,N. Let the sequences { x n } and { z n } be generated by (3.1), where α n = 1 5 n , δ n i =( 1 n 2 i + 1 n N × 2 N ) and r n i = i n n + 1 for every n1 and i=1,2,,N. Then the sequences { x n } and { z n } converge strongly to 0.

Solution. For every i=1,2,,N. It is easy to see that S i is nonexpansive and T i is 1 5 -strictly pseudo contractive mappings with {0}= i = 1 N F( S i ) i = 1 N F( T i ).

Since S n A is 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 ) , where α j ( n ) =( α 1 ( n , j ) α 2 ( n , j ) , α 3 ( n , j ) ) and α 1 ( n , j ) = α 2 ( n , j ) = α 3 ( n , j ) = 1 3 for every n1 and j=1,2,,N, then we have

U n , 0 x = x , U n , 1 x = 2 3 ( 1 3 × 3 2 U n , 0 + 1 3 × U n , 0 + 1 3 ) x , U n , 2 x = 4 5 ( 1 3 × 3 2 U n , 1 + 1 3 × U n , 1 + 1 3 ) x , U n , 3 x = 6 7 ( 1 3 × 3 2 U n , 2 + 1 3 × U n , 2 + 1 3 ) x , U n , N 1 x = 2 ( N 1 ) 2 ( N 1 ) + 1 ( 1 3 × 3 2 U n , N 2 + 1 3 × U n , N 2 + 1 3 ) x , S n A x = U n , N x = 2 N 2 N + 1 ( 1 3 × 3 2 U n , N 1 + 1 3 × U n , N 1 + 1 3 ) x

for every xR. From Lemma 2.8, we have {0}= i = 1 N F( S i ) i = 1 N F( T i )=F( S n A ). For every n1 and i=1,2,,N, we can see that i = 1 N δ n i = i = 1 N ( 1 n 2 i + 1 n N × 2 N )=1. From definition of F i , we have i = 1 N EP( F i )={0}. Then {0}= i = 1 N F( S i ) i = 1 N F( T i ) i = 1 N EP( F i )=F.

For every n1 and i=1,2,,N, the mappings F i , T i , S i and α n , r n i , δ n i satisfy conditions in Theorem 3.1. Then from Theorem 3.1, we have the sequences { x n } and { z n } converge to 0.

Next, we give numerical results to support this example. Let r>0 and zR. For every yR and i=1,2,,N, and from Lemma 2.5, there exist xR such that

F i ( x , y ) + 1 r y x , x z 0 i ( 4 y 2 + x y 5 x 2 ) + 1 r y x , x z 0 4 i r y 2 + i r x y 5 i r x 2 + ( y x ) ( x z ) 0 4 i r y 2 + i r x y 5 i r x 2 + x y x 2 z y + z x 0 4 i r y 2 + ( r i x + x z ) y ( 5 i r x 2 + x 2 z x ) 0 .

Put G(y)=4ir y 2 +(rix+xz)y(5ir x 2 + x 2 zx). Then G is a quadratic function of y with coefficient a=4ir, b=rix+xz, c=(5ir x 2 + x 2 zx). Next, we compute the discriminant Δ of G as follows:

Δ = b 2 4 a c = ( r i x + x z ) 2 + 4 ( 4 i r ) ( 5 i r x 2 + x 2 z x ) = ( r i x + x ) 2 2 z ( r i x + x ) + z 2 + 80 i 2 r 2 x 2 + 16 i r x 2 16 i r z x = 81 i 2 r 2 x 2 + 18 i r x 2 + x 2 18 i r z x 2 z x + z 2 = z 2 + ( 81 i 2 r 2 + 18 i r + 1 ) x 2 2 z x ( 9 i r + 1 ) = ( z x ( 9 i r + 1 ) ) 2 .

Since G(y)0 for all yR. If it has most one solution in , so Δ0. It implies that z=x(9ir+1). Then we have

x= T r z= z i ( 9 r ) + 1
(5.1)

for all r>0 and i=1,2,,N. From (3.1) and (5.1), we have

u n i = T r n i x n = x n i ( 9 r n i ) + 1
(5.2)

for every n1 and i=1,2,,N. Since α n = 1 5 n , δ n i =( 1 n 2 i + 1 n N × 2 N ), r n i = i n n + 1 and (5.2), we can rewrite (3.1) as follows:

{ z n = i = 1 N ( 1 n 2 i + 1 n N × 2 N ) x n i ( 9 i n n + 1 ) + 1 , x n + 1 = 1 5 n f ( z n ) + ( 1 1 5 n ) S n A z n , n 1
(5.3)

for every n1 and i=1,2,,N.

Put N=8 and initial points x 1 =700, x 1 =500 in (5.3) we have the following results respectively.

The numerical results for initial points x 1 =700 and x 1 =500 were shown in Tables 1 (Figure 1(b)) and 2 (Figure 1(a)), respectively. We observe that the sequences { x n } and { z n } converge to 0 i = 1 N F( S i ) i = 1 N F( T i ) i = 1 N EP( F i ).

Figure 1
figure 1

The convergence comparison with different initial values (a) x 1 =500 and (b) x 1 =700 .

Table 1 The values of { z n } and { x n } with initial points x 1 =700 , n=8 and N=8
Table 2 The values of { z n } and { x n } with initial points x 1 =500 , n=8 and N=8

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Acknowledgements

This research was supported by the Research Administration Division of King Mongkut’s Institute of Technology Ladkrabang.

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Kangtunyakarn, A. Iterative approximation of common element of solution sets of various nonlinear operator problems. Fixed Point Theory Appl 2013, 295 (2013). https://doi.org/10.1186/1687-1812-2013-295

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