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

Fixed Point Theory and Applications20122012:188

https://doi.org/10.1186/1687-1812-2012-188

  • Received: 25 February 2012
  • Accepted: 5 October 2012
  • Published:

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.

Keywords

  • nonspreading mapping
  • quasi-nonexpansive mapping
  • S-mapping

1 Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H. Then a mapping T : C C is said to be nonexpansive if T x T y x y for all x , y C . Recall that the mapping T : C C is said to be quasi-nonexpansive if T x p x p , x C and p F ( 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 , y C .

Let A : C H . The variational inequality problem is to find a point u C such that
A u , v u 0
(1.1)

for all v C . The set of solutions of (1.1) is denoted by V I ( 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
x y , A x A y α A x A y 2
for all x , y C . 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 n N . 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 n N , 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 v F ( S ) .

     
  2. (ii)

    If n = 1 α n ( 1 α n ) = and n = 1 β n < , then { x n } converges weakly to v F ( 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 v F ( 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 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 [9])

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 [8])

Let C be a nonempty closed convex subset of H. Then a mapping S : C C is nonspreading if and only if
S x S y 2 x y 2 + 2 x S x , y S y

for all x , y C .

Example 2.3 Let denote the reals with the usual norm. Let T : R R be defined by
T x = { x 1 if  x ( , 0 ] , ( x + 1 ) if  x ( 0 , )

for all x R .

To see that T is a nonspreading mapping, if x , y ( 0 , ) , then we have T x = ( x + 1 ) and T y = ( 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 + 2 x T x , y T y .
For every x , y ( , 0 ] , we have T x = x 1 and T y = y 1 . From the definition of T, we have
| T x T y | 2 = | x 1 ( y 1 ) | 2 = | x y | 2 ,
and
2 x T x , y T y = 2 x ( x 1 ) , y ( y 1 ) = 2 .
From above, we have
| T x T y | 2 = | x y | 2 < | x y | 2 + 2 x T x , y T y .
Finally, for every x ( , 0 ] and y ( 0 , ) , we have T x = x 1 and T y = ( 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 , y R , we have
| T x T y | 2 < | x y | 2 + 2 x T x , y T y .

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 u C . Then for λ > 0 ,
u = P C ( I λ A ) u u V I ( 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 : n N } be a sequence of nonexpansive mappings on C. Suppose n = 1 F ( T n ) is nonempty. Let { λ n } be a sequence of positive numbers with n = 1 λ n = 1 . Then a mapping S on C defined by
S ( x ) = n = 1 λ n T n x

for x C 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 : C C be a nonexpansive mapping. Then I S 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 u H . Let q = P C u . If { x n } is such that ω ( x n ) C and satisfies the condition
x n u u q , n N ,

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 : C C 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 , , N 1 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 y C , 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 , y C  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 , , N 1 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 , , N 1 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 , , N 1 .

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 x C and y F ( 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 + 2 x T 2 x , y T 2 y .

Hence, we have T 2 is a nonspreading mapping. Observe that 1 F ( 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 1 F ( 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 : C H 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 V I ( C , A i ) . For every i = 1 , 2 , , N , define the mapping G i : C C by G i x = P C ( I λ A i ) x x C 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 , , N 1 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 ( I λ A i ) is a nonexpansive mapping for every i = 1 , 2 , , N . Let x , y C . 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 ( I λ 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 ) ) = V I ( C , A i ) , i = 1 , 2 , , N .
(3.2)
From (3.2), V I ( C , A i ) is closed and convex. Let z F . From (3.2), we have z F ( P C ( I λ 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 n N . It is obvious that C n is closed. In fact, we know that for z C n ,
y n z x n z is equivalent to y n x n 2 + 2 y n x n , x n z 0 .
So, for every z 1 , z 2 C n and t ( 0 , 1 ) , it follows that
then, we have C n is convex. Since V I ( C , A i ) is closed and convex for every i = 1 , 2 , , N , we have i = 1 N V I ( 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 n N . Let z F , 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 n N . 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 , n N .
(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 q ω ( x n ) , there exists a subsequence { x n i } of { x n } which converges weakly to q.

Put Q : C C defined by
Q x = i = 1 N δ i G i x , x C .
(3.19)
Since G i = P C ( I λ 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 V I ( 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 q F ( Q ) = i = 1 N F ( G i ) From (3.2), we have
q i = 1 N V I ( C , A i ) .
(3.22)
Next, we will show that q F ( S ) . Assume that q S q . 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 q F ( S ) . From Lemma 2.9, we have
q i = 1 N F ( T i ) .
(3.23)

From (3.22) and (3.23), we have q F . 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 : C H be an α i -inverse strongly monotone mapping, and let T : C C be a nonspreading mapping with F = F ( T ) i = 1 N V I ( C , A i ) . For every i = 1 , 2 , , N , define the mapping G i : C C by G i x = P C ( I λ A i ) x x C 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 : C H 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 ) V I ( 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 , , N 1 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 , 2 α ) . Then the sequence { x n } converges strongly to P F x 1 .

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, King Mongkut’s Institute of Technology Ladkrabang, Bangkok, 10520, Thailand

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