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Strong convergence of the hybrid method for a finite family of nonspreading mappings and variational inequality problems

Abstract

In this paper, we prove a strong convergence theorem by the hybrid method for finding a common element of the set of fixed points of a finite family of nonspreading mappings and the set of solutions of a finite family of variational inequality problems.

1 Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H. Then a mapping T:CC is said to be nonexpansive if TxTyxy for all x,yC. Recall that the mapping T:CC is said to be quasi-nonexpansive if Txpxp, xC and pF(T), where F(T) denotes the set of fixed points of T. In 2008, Kohsaka and Takahashi [1] introduced the mapping T called the nonspreading mapping in Hilbert spaces H and defined it as follows: 2 T x T y 2 T x y 2 + x T y 2 , x,yC.

Let A:CH. The variational inequality problem is to find a point uC such that

Au,vu0
(1.1)

for all vC. The set of solutions of (1.1) is denoted by VI(C,A).

The variational inequality has emerged as a fascinating and interesting branch of mathematical and engineering sciences with a wide range of applications in industry, finance, economics, social, ecology, regional, pure and applied sciences; see, e.g., [25].

A mapping A of C into H is called inverse-strongly monotone (see [6]) if there exists a positive real number α such that

xy,AxAyα A x A y 2

for all x,yC. Throughout this paper, we will use the following notation:

  1. 1.

    for weak convergence and → for strong convergence.

  2. 2.

    ω( x n )={x: x n i x} denotes the weak ω-limit set of { x n }.

In 2008, Takahashi, Takeuchi and Kubota [7] proved the following strong convergence theorems by using the hybrid method for nonexpansive mappings in Hilbert spaces.

Theorem 1.1 Let H be a Hilbert space and C be a nonempty closed convex subset of H. Let T be a nonexpansive mapping of C into H such that F(T) and let x 0 H. For C 1 =C and u 1 P C 1 x 0 , define a sequence { u n } of C as follows:

{ y n = α n u n + ( 1 α n ) u n , C n + 1 = { z C n : y n z u n z } , u n + 1 = P C n + 1 x 0 , n N ,

where 0 α n a<1 for all nN. Then { u n } converges strongly to z 0 = P F ( T ) x 0 .

In 2009, Iemoto and Takahashi [8] proved the convergence theorem of nonexpansive and nonspreading mappings as follows.

Theorem 1.2 Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let S be a nonspreading mapping of C into itself, and let T be a nonexpansive mapping of C into itself such that F(S)F(T). Define a sequence { x n } as follows.

{ x 1 C , x n + 1 = α n x n + ( 1 α n ) ( β n S x n + ( 1 β n ) T x n )

for all nN, where { α n },{ β n }[0,1]. Then the following hold:

  1. (i)

    If lim inf n α n (1 α n )>0 and n = 1 (1 β n )<, then { x n } converges weakly to vF(S).

  2. (ii)

    If n = 1 α n (1 α n )= and n = 1 β n <, then { x n } converges weakly to vF(T).

  3. (iii)

    If lim inf n α n (1 α n )>0 and lim inf n β n (1 β n )>0, then { x n } converges weakly to vF(S)F(T).

Inspired and motivated by these facts and the research in this direction, we prove the strong convergence theorem by the hybrid method for finding a common element of the set of fixed points of a finite family of nonspreading mappings and the set of solutions of a finite family of variational inequality problems.

2 Preliminaries

In this section, we collect and give 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, 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 [9])

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

xy,yz0zC.

Lemma 2.2 (See [8])

Let C be a nonempty closed convex subset of H. Then a mapping S:CC is nonspreading if and only if

S x S y 2 x y 2 +2xSx,ySy

for all x,yC.

Example 2.3 Let denote the reals with the usual norm. Let T:RR be defined by

Tx={ x 1 if  x ( , 0 ] , ( x + 1 ) if  x ( 0 , )

for all xR.

To see that T is a nonspreading mapping, if x,y(0,), then we have Tx=(x+1) and Ty=(y+1). From the definition of the mapping T, we have

| T x T y | 2 = | ( x + 1 ) ( ( y + 1 ) ) | 2 = | y x | 2 = | x y | 2

and

2 x T x , y T y = 2 x + x + 1 , y + y + 1 = 2 2 x + 1 , 2 y + 1 = 2 ( 2 x + 1 ) ( 2 y + 1 ) > 0 ( since  x , y > 0 ) .

The above implies that

| T x T y | 2 = | x y | 2 < | x y | 2 +2xTx,yTy.

For every x,y(,0], we have Tx=x1 and Ty=y1. From the definition of T, we have

| T x T y | 2 = | x 1 ( y 1 ) | 2 = | x y | 2 ,

and

2xTx,yTy=2 x ( x 1 ) , y ( y 1 ) =2.

From above, we have

| T x T y | 2 = | x y | 2 < | x y | 2 +2xTx,yTy.

Finally, for every x(,0] and y(0,), we have Tx=x1 and Ty=(y+1). From the definition of T, we have

and

2 x T x , y T y = 2 x ( x 1 ) , y + ( y + 1 ) = 2 1 , 2 y + 1 = 2 ( 2 y + 1 ) > 0 ( since  y > 0 ) .

From above, we have

| T x T y | 2 = | x + y | 2 = ( x + y ) 2 | x y | 2 < | x y | 2 + 2 x T x , y T y .

Hence, for all x,yR, we have

| T x T y | 2 < | x y | 2 +2xTx,yTy.

Then T is a nonspreading mapping.

Lemma 2.4 (See [1])

Let H be a Hilbert space, let C be a nonempty closed convex subset of H, and let S be a nonspreading mapping of C into itself. Then F(S) is closed and convex.

Lemma 2.5 (See [9])

Let H be a Hilbert space, let C be a nonempty closed convex subset of H, and let A be a mapping of C into H. Let uC. Then for λ>0,

u= P C (IλA)uuVI(C,A),

where P C is the metric projection of H onto C.

Lemma 2.6 (See [10])

Let C be a closed convex subset of a strictly convex Banach space E. Let { T n :nN} 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 n = 1 λ n =1. Then a mapping S on C defined by

S(x)= n = 1 λ n T n x

for xC is well defined, nonexpansive and F(S)= n = 1 F( T n ) holds.

Lemma 2.7 (See [11])

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.

Lemma 2.8 (See [12])

Let C be a closed convex subset of H. Let { x n } be a sequence in H and uH. Let q= P C u. If { x n } is such that ω( x n )C and satisfies the condition

x n uuq,nN,

then x n q, as n.

In 2009, Kangtunyakarn and Suantai [13] introduced an S-mapping generated by T 1 ,, T N and λ 1 ,, λ N as follows.

Definition 2.1 Let C be a nonempty convex subset of a real Banach space. Let { T 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. Define the mapping S:CC as follows:

(2.1)
(2.2)

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

The next lemma is very useful for our consideration.

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 nonspreading mappings of C into C 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 , α 3 j (0,1) for all j=1,2,,N1 and α 1 N (0,1], α 3 N [0,1), α 2 j [0,1) for all j=1,2,,N. Let S be the mapping generated by T 1 ,, T N and α 1 , α 2 ,, α N . Then F(S)= i = 1 N F( T i ) and S is a quasi-nonexpansive mapping.

Proof It easy to see that i = 1 N F( T i )F(S). Let x 0 F(S) and x i = 1 N F( T i ). Since { T i } i = 1 N is a finite family of nonspreading mappings of C into itself, for every yC, we have

T i y x 2 1 2 ( T i y x 2 + y x 2 ) .
(2.3)

This implies that

T i y x 2 y x 2 ,yC and i=1,2,,N.
(2.4)

From the definition of S and (2.4),

S 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 ( 1 α 3 N ) U N 1 x 0 x 2 + α 3 N x 0 x 2 = ( 1 α 3 N ) α 1 N 1 ( T N 1 U N 2 x 0 x ) + α 2 N 1 ( U N 2 x 0 x ) + α 3 N 1 ( x 0 x ) 2 + α 3 N x 0 x 2 ( 1 α 3 N ) ( α 1 N 1 T N 1 U N 2 x 0 x 2 + α 2 N 1 U N 2 x 0 x 2 + α 3 N 1 x 0 x 2 ) + α 3 N x 0 x 2 ( 1 α 3 N ) ( ( 1 α 3 N 1 ) U N 2 x 0 x 2 + α 3 N 1 x 0 x 2 ) + α 3 N x 0 x 2 = ( 1 α 3 N ) ( 1 α 3 N 1 ) U N 2 x 0 x 2 + α 3 N 1 ( 1 α 3 N ) x 0 x 2 + α 3 N 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 = 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 + ( 1 j = 2 N ( 1 α 3 j ) ) x 0 x 2 = j = 2 N ( 1 α 3 j ) α 1 1 ( T 1 x 0 x ) + ( 1 α 1 1 ) ( x 0 x ) 2 + ( 1 j = 2 N ( 1 α 3 j ) ) x 0 x 2 = j = 2 N ( 1 α 3 j ) ( α 1 1 T 1 x 0 x 2 + ( 1 α 1 1 ) x 0 x 2 α 1 1 ( 1 α 1 1 ) T 1 x 0 x 0 ) + ( 1 j = 2 N ( 1 α 3 j ) ) x 0 x 2 j = 2 N ( 1 α 3 j ) ( x 0 x 2 α 1 1 ( 1 α 1 1 ) T 1 x 0 x 0 2 ) + ( 1 j = 2 N ( 1 α 3 j ) ) x 0 x 2 .
(2.5)

From (2.5), we have

x 0 x 2 j = 2 N ( 1 α 3 j ) ( x 0 x 2 α 1 1 ( 1 α 1 1 ) T 1 x 0 x 0 2 ) + ( 1 j = 2 N ( 1 α 3 j ) ) x 0 x 2 ,

which implies that

x 0 x 2 x 0 x 2 α 1 1 ( 1 α 1 1 ) T 1 x 0 x 0 2 .
(2.6)

Since α 1 j (0,1) for all j=1,2,,N1 and (2.6), we have x 0 F( T 1 ). From x 0 = T 1 x 0 and the definition of S, we have

U 1 x 0 = α 1 1 T 1 x 0 + α 2 1 x 0 + α 3 1 x 0 = x 0 .

From (2.5) and x 0 F( U 1 ), we have

x 0 x 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 = 3 N ( 1 α 3 j ) α 1 2 T 2 U 1 x 0 + α 2 2 U 1 x 0 + α 3 2 x 0 x 2 + ( 1 j = 3 N ( 1 α 3 j ) ) x 0 x 2 = j = 3 N ( 1 α 3 j ) α 1 2 ( T 2 x 0 x ) + ( 1 α 1 2 ) ( x 0 x ) 2 + ( 1 j = 3 N ( 1 α 3 j ) ) x 0 x 2 = j = 3 N ( 1 α 3 j ) ( α 1 2 T 2 x 0 x 2 + ( 1 α 1 2 ) x 0 x 2 α 1 2 ( 1 α 1 2 ) T 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 ) ( x 0 x 2 α 1 2 ( 1 α 1 2 ) T 2 x 0 x 0 2 ) + ( 1 j = 3 N ( 1 α 3 j ) ) x 0 x 2 ,

which implies that

x 0 x 2 x 0 x 2 α 1 2 ( 1 α 1 2 ) T 2 x 0 x 0 2 .
(2.7)

Since α 1 j (0,1) for all j=1,2,,N1 and (2.7), we have x 0 F( T 2 ). From the definition of S and x 0 = T 2 x 0 , we have

U 2 x 0 = α 1 2 T 2 U 1 x 0 + α 2 2 U 1 x 0 + α 3 2 x 0 = x 0 .

By continuing in this way, we can show that x 0 F( T i ) and x 0 F( U i ) for all i=1,2,,N1.

Finally, we shall show that x 0 F( T N ).

Since

0 = S x 0 x 0 = α 1 N T N U N 1 x 0 + α 2 N U N 1 x 0 + α 3 N x 0 x 0 = α 1 N ( T N x 0 x 0 ) ,

and α 1 N (0,1], we obtain T N x 0 = x 0 so that x 0 F( T N ). Then we have x 0 i = 1 N F( T i ). Hence, F(S) i = 1 N F( T i ).

Next, we show that S is a quasi-nonexpansive mapping. Let xC and yF(S). From (2.5), we can imply that

S x y 2 j = 2 N ( 1 α 3 j ) ( x y 2 α 1 1 ( 1 α 1 1 ) T 1 x x ) + ( 1 j = 2 N ( 1 α 3 j ) ) x y 2 x y 2 .

Then we have the S-mapping is quasi-nonexpansive. □

Example 2.10 Let T 1 :[1,1][1,1] be a mapping defined by

T 1 x={ x + 1 2 if  x ( 0 , 1 ] , x + 1 2 if  x [ 1 , 0 ]

for all x[1,1].

Let T 2 :[1,1][1,1] be a mapping defined by

T 2 x={ x + 2 3 if  x ( 0 , 1 ] , x + 2 3 if  x [ 1 , 0 ]

for all x[1,1].

To see that T 1 is a nonspreading mapping, observe that if x,y(0,1], we have T 1 x= x + 1 2 and T 1 y= y + 1 2 . Then we have

| T 1 x T 1 y | 2 = | x + 1 2 y + 1 2 | 2 = 1 4 | x y | 2

and

2 x T 1 x , y T 1 y = 2 x ( x + 1 2 ) , y ( y + 1 2 ) = 2 x 1 2 , y 1 2 = 1 2 ( x 1 ) ( y 1 ) 0 ( since  x 1 , y 1 ,  then  ( x 1 ) ( y 1 ) 0 ) .

From above, we have

| x y | 2 + 2 x T 1 x , y T 1 y | x y | 2 1 4 | x y | 2 = | T 1 x T 1 y | 2 .

For every x,y[1,0], we have T 1 x= x + 1 2 and T 1 y= y + 1 2 . From the definition of T 1 , we have

| T 1 x T 1 y | 2 = | x + 1 2 ( y + 1 2 ) | 2 = | y x 2 | 2 = 1 4 | x y | 2

and

2 x T 1 x , y T 1 y = 2 x ( 1 x 2 ) , y ( 1 y 2 ) = 2 3 x 1 2 , 3 y 1 2 = 1 2 ( 3 x 1 ) ( 3 y 1 ) = 1 2 ( 3 x ( 3 y 1 ) ( 3 y 1 ) ) = 1 2 ( 9 x y 3 x 3 y + 1 ) > 0 ( since  1 x , y 0 ,  then  9 x y , 3 x , 3 y 0 ) .

From above, we have

| x y | 2 + 2 x T 1 x , y T 1 y > | x y | 2 1 4 | x y | 2 = | T 1 x T 1 y | 2 .

Finally, for every x(0,1] and y[1,0], we have T 1 x= x + 1 2 and T 1 y= y + 1 2 . From the definition of T 1 , we have

| T 1 x T 1 y | 2 = | x + 1 2 y + 1 2 | 2 = 1 4 | x + y | 2

and

2 x T 1 x , y T 1 y = 2 x ( x + 1 2 ) , y ( y + 1 2 ) = 2 x 1 2 , 3 y 1 2 = 1 2 ( x 1 ) ( 3 y 1 ) = 1 2 ( x ( 3 y 1 ) ( 3 y 1 ) ) = 1 2 ( 3 x y x 3 y + 1 ) = 1 2 ( 3 y ( x 1 ) + ( 1 x ) ) 0 ( since  0 < x 1  and  1 y 0 ,  then  3 y ( x 1 ) , ( 1 x ) 0 ) .

From above, we have

| x y | 2 + 2 x T 1 x , y T 1 y | x y | 2 = x 2 2 x y + y 2 = x 2 + 2 x y + y 2 4 x y x 2 + 2 x y + y 2 ( since  4 x y 0 ) = ( x + y ) 2 1 4 ( x + y ) 2 = | T 1 x T 1 y | 2 .

Then for all x,y[1,1], we have

| T 1 x T 1 y | 2 | x y | 2 +x T 1 x,y T 1 y.

Hence, we have T 1 is a nonspreading mapping.

Next, we show that T 2 is a nonspreading mapping. Let x,y(0,1], then we have T 2 x= x + 2 3 and T 2 y= y + 2 3 . From the definition of T 2 , we have

| T 2 x T 2 y | 2 = | x + 2 3 y + 2 3 | 2 = 1 9 | x y | 2

and

2 x T 2 x , y T 2 y = 2 x ( x + 2 3 ) , y ( y + 2 3 ) = 2 2 x 2 3 , 2 y 2 3 = 8 9 ( x 1 ) ( y 1 ) 0 ( since  0 < x , y 1 ,  then  ( x 1 ) ( y 1 ) 0 ) .

From above, we have

| x y | 2 + 2 x T 2 x , y T 2 y | x y | 2 1 9 | x y | 2 = | T 2 x T 2 y | 2 .

For every x,y[1,0], we have T 2 x= 2 x 3 and T 2 y= 2 y 3 . From the definition of T 2 , we have

| T 2 x T 2 y | 2 = | 2 x 3 2 y 3 | 2 = | y x 3 | 2 = 1 9 | x y | 2

and

2 x T 2 x , y T 2 y = 2 x ( 2 x 3 ) , y ( 2 y 3 ) = 2 4 x 2 3 , 4 y 2 3 = 8 9 ( 2 x 1 ) ( 2 y 1 ) = 8 9 ( 2 x ( 2 y 1 ) ( 2 y 1 ) ) = 8 9 ( 4 x y 2 x 2 y + 1 ) > 0 ( since  1 x , y 0 ,  then  4 x y , 2 x , 2 y 0 ) .

From above, we have

| x y | 2 + 2 x T 2 x , y T 2 y > | x y | 2 1 9 | x y | 2 = | T 2 x T 2 y | 2 .

Finally, for every x(0,1] and y[1,0], we have T 2 x= x + 2 3 and T 2 y= 2 y 3 . From the definition of T 2 , we have

| T 2 x T 2 y | 2 = | x + 2 3 2 y 3 | 2 = 1 9 | x + y | 2

and

2 x T 2 x , y T 2 y = 2 x ( x + 2 3 ) , y ( 2 y 3 ) = 2 2 x 2 3 , 4 y 2 3 = 8 9 ( x 1 ) ( 2 y 1 ) = 8 9 ( x ( 2 y 1 ) ( 2 y 1 ) ) = 8 9 ( 2 x y x 2 y + 1 ) = 8 9 ( 2 y ( x 1 ) + ( 1 x ) ) 0 ( since  0 < x 1  and  1 y 0 ,  then  2 y ( x 1 ) , ( 1 x ) 0 ) .

From above, we have

| x y | 2 + 2 x T 2 x , y T 2 y | x y | 2 = x 2 2 x y + y 2 = x 2 + 2 x y + y 2 4 x y ( x + y ) 2 ( since  4 x y 0 ) 1 9 | x + y | 2 = | T 2 x T 2 y | 2 .

Then for every x,y[1,1], we have

| T 2 x T 2 y | 2 | x y | 2 +2x T 2 x,y T 2 y.

Hence, we have T 2 is a nonspreading mapping. Observe that 1F( T 1 )F( T 2 ). Let the mapping S:[1,1][1,1] be the S-mapping generated by T 1 , T 2 and α 1 , α 2 , where α 1 =( 1 6 , 2 6 , 3 6 ) and ( 4 15 , 5 15 , 6 15 ). From Lemma 2.9, we have 1F(S).

3 Main result

Theorem 3.1 Let C be a nonempty closed convex subset of a Hilbert space H. For every i=1,2,,N, let A i :CH be an α i -inverse strongly monotone mapping, and let { T i } i = 1 N be a finite family of nonspreading mappings with F= i = 1 N F( T i ) i = 1 N VI(C, A i ). For every i=1,2,,N, define the mapping G i :CC by G i x= P C (Iλ A i )x xC and λ[c,d](0,2 α i ). Let ρ 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 (0,1) for all j=1,2,,N1 and α 1 N (0,1], α 3 N [0,1) α 2 j (0,1) for all j=1,2,,N, and let S be the S-mapping generated by T 1 , T 2 ,, T N and ρ 1 , ρ 2 ,, ρ N . Let { x n } be a sequence generated by x 1 C 1 =C and

{ z n = i = 1 N δ n i G i x n , y n = α n x n + β n S x n + γ n z n , C n + 1 = { z C n : y n z x n z } , x n + 1 = P C n + 1 x 1 , n 1 ,
(3.1)

where { α n },{ β n },{ γ n }[0,1], α n + β n + γ n =1 and suppose the following conditions hold:

Then the sequence { x n } converges strongly to P F x 1 .

Proof First, we show that (Iλ A i ) is a nonexpansive mapping for every i=1,2,,N. Let x,yC. Since A is an α i -inverse strongly monotone and λ<2 α i , we have

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

Thus (Iλ A i ) is a nonexpansive mapping for every i=1,2,,N. Since P C is a nonexpansive mapping, we have G i is a nonexpansive mapping for every i=1,2,,N. From Lemma 2.5, we have

F( G i )=F ( P C ( I λ A i ) ) =VI(C, A i ),i=1,2,,N.
(3.2)

From (3.2), VI(C, A i ) is closed and convex. Let zF. From (3.2), we have zF( P C (Iλ A i )) for every i=1,2,,N. By nonexpansiveness of G i , we have

z n z= i = 1 N δ n i ( G i x n z ) i = 1 N δ n i x n z= x n z.
(3.3)

Next, we show that C n is closed and convex for every nN. It is obvious that C n is closed. In fact, we know that for z C n ,

y n z x n zis equivalent to y n x n 2 +2 y n x n , x n z0.

So, for every z 1 , z 2 C n and t(0,1), it follows that

then, we have C n is convex. Since VI(C, A i ) is closed and convex for every i=1,2,,N, we have i = 1 N VI(C, A i ) is closed and convex. From Lemma 2.4, we have i = 1 N F( T i ) is closed and convex. Hence, we have F is closed and convex. This implies that P F is well defined. Next, we show that F C n for every nN. Let zF, then we have

y n z = α n ( x n z ) + β n ( S x n z ) + γ n ( z n z ) α n x n z + β n S x n z + γ n z n z x n z .

It follows that z C n . Hence, we have F C n for every nN. This implies that { x n } is well defined. Since x n = P C n x 1 , for every w C n , we have

x n x 1 w x 1 ,nN.
(3.4)

In particular, we have

x n x 1 P F x 1 x 1 .
(3.5)

By (3.4) we have { x n } is bounded, so are { G i x n }, { T i x n } for every i=1,2,,N, { z n }, { y n } and {S x n }. Since x n + 1 = P C n + 1 x 1 C n + 1 C n and x n = P C n x 1 , we have

0 x 1 x n , x n x n + 1 = x 1 x n , x n x 1 + x 1 x n + 1 x n x 1 2 + x n x 1 x 1 x n + 1 ,

which implies that

x n x 1 x n + 1 x 1 .

Hence, we have lim n x n x 1 exists. Since

x n x n + 1 2 = x n x 1 + x 1 x n + 1 2 = x n x 1 2 + 2 x n x 1 , x 1 x n + 1 + x 1 x n + 1 2 = x n x 1 2 + 2 x n x 1 , x 1 x n + x n x n + 1 + x 1 x n + 1 2 = x n x 1 2 2 x n x 1 2 + 2 x n x 1 , x n x n + 1 + x 1 x n + 1 2 x 1 x n + 1 2 x n x 1 2 ,
(3.6)

it implies that

lim n x n x n + 1 =0.
(3.7)

Since x n + 1 = P C n + 1 x 1 C n + 1 , we have

y n x n + 1 x n x n + 1 .

By (3.7) we have

lim n y n x n + 1 =0.
(3.8)

Since

y n x n y n x n + 1 + x n + 1 x n ,

by (3.7) and (3.8), we have

lim n y n x n =0.
(3.9)

Next, we will show that

lim n x n S x n =0.
(3.10)

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

(3.11)

From the definition of y n and (3.11), we have

y n z 2 α n x n z 2 + β n S x n z 2 + γ n z n z 2 α n x n z 2 + β n S x n z 2 + γ n i = 1 N δ n i P C ( I λ A i ) x n z 2 α n x n z 2 + β n S x n z 2 + γ n i = 1 N δ n i ( x n z 2 λ ( 2 α i λ ) A i x n A i z 2 ) = α n x n z 2 + β n S x n z 2 + γ n x n z 2 γ n i = 1 N δ n i λ ( 2 α i λ ) A i x n A i z 2 x n z 2 γ n i = 1 N δ n i λ ( 2 α i λ ) A i x n A i z 2 .

It follows that

γ n i = 1 N δ n i λ ( 2 α i λ ) A i x n A i z 2 x n z 2 y n z 2 ( x n z + y n z ) y n x n .

From conditions (i), (ii) and (3.9), it implies that

lim n A i x n A i z=0,i=1,2,,N.
(3.12)

Since

P C ( I λ A i ) x n z 2 ( I λ A i ) x n ( I λ A i ) z , P C ( I λ A i ) x n z = 1 2 ( ( I λ A i ) x n ( I λ A i ) z 2 + P C ( I λ A i ) x n z 2 ( I λ A i ) x n ( I λ A i ) z P C ( I λ A i ) x n + z 2 ) 1 2 ( x n z 2 + P C ( I λ A i ) x n z 2 x n P C ( I λ A i ) x n λ ( A i x n A i z ) 2 ) = 1 2 ( x n z 2 + P C ( I λ A i ) x n z 2 x n P C ( I λ A i ) x n 2 λ ( A i x n A i z ) 2 + 2 λ x n P C ( I λ A i ) x n , A i x n A i z ) ,

it implies that

P C ( I λ A i ) x n z 2 x n z 2 x n P C ( I λ A i ) x n 2 + 2 λ x n P C ( I λ A i ) x n A i x n A i z .
(3.13)

From the definition of y n and (3.13), we have

y n z 2 α n x n z 2 + β n S x n z 2 + γ n z n z 2 ( 1 γ n ) x n z 2 + γ n i = 1 N δ n i P C ( I λ A i ) x n z 2 ( 1 γ n ) x n z 2 + γ n i = 1 N δ n i ( x n z 2 x n P C ( I λ A i ) x n 2 + 2 λ x n P C ( I λ A i ) x n A i x n A i z ) = x n z 2 γ n i = 1 N δ n i x n P C ( I λ A i ) x n 2 + 2 γ n i = 1 N δ n i λ x n P C ( I λ A i ) x n A i x n A i z ,

which implies that

γ n i = 1 N δ n i x n P C ( I λ A i ) x n 2 x n z 2 y n z 2 + 2 γ n i = 1 N δ n i λ x n P C ( I λ A i ) x n A i x n A i z ( x n z + y n z ) y n x n + 2 γ n i = 1 N δ n i λ x n P C ( I λ A i ) x n A i x n A i z .

From conditions (i), (ii), (3.9) and (3.12), we have

lim n P C ( I λ A i ) x n x n =0,i=1,2,,N.
(3.14)

Since

z n x n i = 1 N δ n i P C ( I λ A i ) x n x n ,

from (3.14), we have

lim n z n x n =0.
(3.15)

Since

y n x n = β n (S x n x n )+ γ n ( z n x n )

from (3.9) and (3.15), we have

lim n S x n x n =0.

Next, we will show that

lim n T i U i 1 x n U i 1 x n =0,i=1,2,,N.
(3.16)

From the definition of y n , we have

y n z 2 α n x n z 2 + β n S x n z 2 + γ n z n z 2 ( 1 β n ) x n z 2 + β n α 1 N ( T N U N 1 x n z ) + α 2 N ( U N 1 x n z ) + α 3 N ( x n z ) ( 1 β n ) x n z 2 + β n ( α 1 N T N U N 1 x n z 2 + α 2 N U N 1 x n z 2 + α 3 N x n z 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) ( 1 β n ) x n z 2 + β n ( ( 1 α 3 N ) U N 1 x n z 2 + α 3 N x n z 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) = ( 1 β n ) x n z 2 + β n ( ( 1 α 3 N ) α 1 N 1 ( T N 1 U N 2 x n z ) + α 2 N 1 ( U N 2 x n z ) + α 3 N 1 ( x n z ) 2 + α 3 N x n z 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) ( 1 β n ) x n z 2 + β n ( ( 1 α 3 N ) ( α 1 N 1 T N 1 U N 2 x n z 2 + α 2 N 1 U N 2 x n z 2 + α 3 N 1 x n z 2 α 1 N 1 α 2 N 1 T N 1 U N 2 x n U N 2 x n 2 ) + α 3 N x n z 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) ( 1 β n ) x n z 2 + β n ( ( 1 α 3 N ) ( ( 1 α 3 N 1 ) U N 2 x n z 2 + α 3 N 1 x n z 2 α 1 N 1 α 2 N 1 T N 1 U N 2 x n U N 2 x n 2 ) + α 3 N x n z 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) = ( 1 β n ) x n z 2 + β n ( ( 1 α 3 N ) ( 1 α 3 N 1 ) U N 2 x n z 2 + ( 1 α 3 N ) α 3 N 1 x n z 2 α 1 N 1 α 2 N 1 ( 1 α 3 N ) T N 1 U N 2 x n U N 2 x n 2 + α 3 N x n z 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) = ( 1 β n ) x n z 2 + β n ( j = N 1 N ( 1 α 3 j ) U N 2 x n z 2 + ( 1 j = N 1 N ( 1 α 3 j ) ) x n z 2 α 1 N 1 α 2 N 1 ( 1 α 3 N ) T N 1 U N 2 x n U N 2 x n 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) = ( 1 β n ) x n z 2 + β n ( j = N 1 N ( 1 α 3 j ) α 1 N 2 ( T N 2 U N 3 x n z ) + α 2 N 2 ( U N 3 x n z ) + α 3 N 2 ( x n z ) 2 + ( 1 j = N 1 N ( 1 α 3 j ) ) x n z 2 α 1 N 1 α 2 N 1 ( 1 α 3 N ) T N 1 U N 2 x n U N 2 x n 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) ( 1 β n ) x n z 2 + β n ( j = N 1 N ( 1 α 3 j ) ( α 1 N 2 T N 2 U N 3 x n z 2 + α 2 N 2 U N 3 x n z 2 + α 3 N 2 x n z 2 α 1 N 2 α 2 N 2 T N 2 U N 3 x n U N 3 x n 2 ) + ( 1 j = N 1 N ( 1 α 3 j ) ) x n z 2 α 1 N 1 α 2 N 1 ( 1 α 3 N ) T N 1 U N 2 x n U N 2 x n 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) ( 1 β n ) x n z 2 + β n ( j = N 1 N ( 1 α 3 j ) ( ( 1 α 3 N 2 ) U N 3 x n z 2 + α 3 N 2 x n z 2 α 1 N 2 α 2 N 2 T N 2 U N 3 x n U N 3 x n 2 ) + ( 1 j = N 1 N ( 1 α 3 j ) ) x n z 2 α 1 N 1 α 2 N 1 ( 1 α 3 N ) T N 1 U N 2 x n U N 2 x n 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) = ( 1 β n ) x n z 2 + β n ( j = N 2 N ( 1 α 3 j ) U N 3 x n z 2 + ( 1 j = N 2 N ( 1 α 3 j ) ) x n z 2 α 1 N 2 α 2 N 2 j = N 1 N ( 1 α 3 j ) T N 2 U N 3 x n U N 3 x n 2 α 1 N 1 α 2 N 1 ( 1 α 3 N ) T N 1 U N 2 x n U N 2 x n 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) ( 1 β n ) x n z 2 + β n ( j = 1 N ( 1 α 3 j ) U 0 x n z 2 + ( 1 j = 1 N ( 1 α 3 j ) ) x n z 2 α 1 1 α 2 1 j = 2 N ( 1 α 3 j ) T 1 U 0 x n U 0 x n 2 α 1 N 2 α 2 N 2 j = N 1 N ( 1 α 3 j ) T N 2 U N 3 x n U N 3 x n 2 α 1 N 1 α 2 N 1 ( 1 α 3 N ) T N 1 U N 2 x n U N 2 x n 2 α 1 N α 2 N T N U N 1 x n U N 1 x n 2 ) = x n z 2 β n α 1 1 α 2 1 j = 2 N ( 1 α 3 j ) T 1 x n x n 2 β n α 1 N 2 α 2 N 2 j = N 1 N ( 1 α 3 j ) T N 2 U N 3 x n U N 3 x n 2 β n α 1 N 1 α 2 N 1 ( 1 α 3 N ) T N 1 U N 2 x n U N 2 x n 2 β n α 1 N α 2 N T N U N 1 x n U N 1 x n 2 .
(3.17)

From (3.17) and condition (ii), we have

β n α 1 1 α 2 1 j = 2 N ( 1 α 3 j ) T 1 x n x n 2 x n z 2 y n z 2 ( x n z + y n z ) y n x n .

Form (3.9), we have

lim n T 1 x n x n =0.
(3.18)

By using the same method as (3.18), we can conclude that

lim n T i U i 1 x n U i 1 x n =0,i=1,2,,N.

Let ω( x n ) be the set of all weakly ω-limit of { x n }. We shall show that ω( x n )F. Since { x n } is bounded, then ω( x n ). Let qω( x n ), there exists a subsequence { x n i } of { x n } which converges weakly to q.

Put Q:CC defined by

Qx= i = 1 N δ i G i x,xC.
(3.19)

Since G i = P C (Iλ A i ) is a nonexpansive mapping, for every i=1,2,,N, from Lemma 2.6 and 2.5, we have

F(Q)= i = 1 N F( G i )= i = 1 N VI(C, A i ).
(3.20)

Since

x n Q x n x n z n + z n Q x n = x n z n + i = 1 N δ n i G i x n i = 1 N δ i G i x n = x n z n + i = 1 N ( δ n i δ i ) G i x n x n z n + i = 1 N | δ n i δ i | G i x n ,

from the condition (i) and (3.15), we have

lim n x n Q x n =0.
(3.21)

From (3.21), we have

lim i x n i Q x n i =0.

From (3.19), it is easy to see that Q is a nonexpansive mapping. By Lemma 2.7 and x n i q as i, we have qF(Q)= i = 1 N F( G i ) From (3.2), we have

q i = 1 N VI(C, A i ).
(3.22)

Next, we will show that qF(S). Assume that qSq. From the Opial property, (3.10) and (3.16), we have

lim inf i x n i q 2 < lim inf i x n i S q 2 = lim inf i x n i S x n i + ( S x n i S q ) 2 = lim inf i ( x n i S x n i 2 + S x n i S q 2 + 2 x n i S x n i , S x n i S q ) = lim inf i S x n i S q 2 = lim inf i α 1 N T N U N 1 x n i + α 2 N U N 1 x n i + α 3 N x n i α 1 N T N U N 1 q α 2 N U N 1 q α 3 N q 2 = lim inf i α 1 N ( T N U N 1 x n i T N U N 1 q ) + α 2 N ( U N 1 x n i U N 1 q ) + α 3 N ( x n i q ) 2 lim inf i ( α 1 N T N U N 1 x n i T N U N 1 q 2 + α 2 N U N 1 x n i U N 1 q 2 + α 3 N x n i q 2 ) lim inf i ( α 1 N ( U N 1 x n i U N 1 q 2 + 2 U N 1 x n i T N U N 1 x n i , U N 1 q T N U N 1 q ) + α 2 N U N 1 x n i U N 1 q 2 + α 3 N x n i q 2 ) = lim inf i ( ( 1 α 3 N ) U N 1 x n i U N 1 q 2 + α 3 N x n i q 2 ) = lim inf i ( ( 1 α 3 N ) α 1 N 1 ( T N 1 U N 2 x n i T N 1 U N 2 q ) + α 2 N 1 ( U N 2 x n i U N 2 q ) + α 3 N 1 ( x n i q ) 2 + α 3 N x n i q 2 ) lim inf i ( ( 1 α 3 N ) ( α 1 N 1 T N 1 U N 2 x n i T N 1 U N 2 q 2 + α 2 N 1 U N 2 x n i U N 2 q 2 + α 3 N 1 x n i q 2 ) + α 3 N x n i q 2 ) lim inf i ( ( 1 α 3 N ) ( α 1 N 1 ( U N 2 x n i U N 2 q 2 + 2 U N 2 x n i T N 1 U N 2 x n i , U N 2 q T N 1 U N 2 q ) + α 2 N 1 U N 2 x n i U N 2 q 2 + α 3 N 1 x n i q 2 ) + α 3 N x n i q 2 ) = lim inf i ( ( 1 α 3 N ) ( ( 1 α 3 N 1 ) U N 2 x n i U N 2 q 2 + α 3 N 1 x n i q 2 ) + α 3 N x n i q 2 ) = lim inf i ( j = N 1 N ( 1 α 3 j ) U N 2 x n i U N 2 q 2 + ( 1 j = N 1 N ( 1 α 3 j ) ) x n i q 2 ) lim inf i ( j = 1 N ( 1 α 3 j ) U 0 x n i U 0 q 2 + ( 1 j = 1 N ( 1 α 3 j ) ) x n i q 2 ) = lim inf i ( j = 1 N ( 1 α 3 j ) x n i q 2 + ( 1 j = 1 N ( 1 α 3 j ) ) x n i q 2 ) = lim inf i x n i q 2 .

This is a contradiction. Then, we have qF(S). From Lemma 2.9, we have

q i = 1 N F( T i ).
(3.23)

From (3.22) and (3.23), we have qF. Hence, ω( x n )F. Therefore, by (3.5) and Lemma 2.8, we have { x n } converges strongly to P F x 1 . This completes the proof. □

The following result can be obtained from Theorem 3.1. We, therefore, omit the proof.

Corollary 3.2 Let C be a nonempty closed convex subset of a Hilbert space H. For every i=1,2,,N, let A i :CH be an α i -inverse strongly monotone mapping, and let T:CC be a nonspreading mapping with F=F(T) i = 1 N VI(C, A i ). For every i=1,2,,N, define the mapping G i :CC by G i x= P C (Iλ A i )x xC and λ[c,d](0,2 α i ). Let { x n } be a sequence generated by x 1 C 1 =C and

{ z n = i = 1 N δ n i G i x n , y n = α n x n + β n T x n + γ n z n , C n + 1 = { z C n : y n z x n z } , x n + 1 = P C n + 1 x 1 , n 1 ,
(3.24)

where { α n },{ β n },{ γ n }[0,1], α n + β n + γ n =1 and suppose the following conditions hold:

Then the sequence { x n } converges strongly to P F x 1 .

Corollary 3.3 Let C be a nonempty closed convex subset of a Hilbert space H. Let A:CH be an α-inverse strongly monotone mapping, and let { T i } i = 1 N be a finite family of nonspreading mappings with F= i = 1 N F( T i )VI(C,A). Let ρ 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 (0,1) for all j=1,2,,N1 and α 1 N (0,1], α 3 N [0,1), α 2 j (0,1) for all j=1,2,,N, and let S be the S-mapping generated by T 1 , T 2 ,, T N and ρ 1 , ρ 2 ,, ρ N . Let { x n } be a sequence generated by x 1 C 1 =C and

{ y n = α n x n + β n S x n + γ n P C ( I λ A ) x n , C n + 1 = { z C n : y n z x n z } , x n + 1 = P C n + 1 x 1 , n 1 ,
(3.25)

where { α n },{ β n },{ γ n }[a,b](0,1), α n + β n + γ n =1 and λ[c,d](0,2α). Then the sequence { x n } converges strongly to P F x 1 .

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