Open Access

Iterative algorithms based on hybrid method and Cesàro mean of asymptotically nonexpansive mappings for equilibrium problems

Fixed Point Theory and Applications20142014:16

https://doi.org/10.1186/1687-1812-2014-16

Received: 29 September 2013

Accepted: 5 January 2014

Published: 21 January 2014

Abstract

Using Cesàro means of a mapping, we modify the progress of Mann’s iteration in hybrid method for asymptotically nonexpansive mappings in Hilbert spaces. Under suitable conditions, we prove that the iterative sequence converges strongly to a fixed point of an asymptotically nonexpansive mapping. We also introduce a new hybrid iterative scheme for finding a common element of the set of common fixed points of asymptotically nonexpansive mappings and the set of solutions of an equilibrium problem in Hilbert spaces.

MSC:47H10, 47J25, 90C33.

Keywords

strong convergence theoremasymptotically nonexpansive mappinghybrid methodequilibrium problemCesàro means

1 Introduction

Let H be a real Hilbert space with the inner product , and the norm . Let C be a nonempty closed convex subset of H. A mapping T : C C is said to be asymptotically nonexpansive if for each n 1 , there exists a nonnegative real number k n satisfying lim n k n = 1 such that
T n x T n y k n x y , x , y C ;

when k n 1 , T is called nonexpansive.

The concept of asymptotically nonexpansive mapping was introduced by Goebel and Kirk [1] in 1972. We denote by F ( T ) : = { x C : T x = x } the set of fixed points of T. It is well known that if T : H H is asymptotically nonexpansive, then F ( T ) is nonempty convex.

In 1953, Mann [2] introduced the iteration as follows: a sequence { x n } defined by
x n + 1 = α n x n + ( 1 α n ) T x n .
(1.1)
In an infinite-dimensional Hilbert space, Mann iteration could conclude only weak convergence [3]. Attempts to modify the Mann iteration method (1.1) so that strong convergence is guaranteed have recently been made. Nakajo and Takahashi [4] proposed the following modification of Mann iteration method for a nonexpansive mapping T in a Hilbert space:
{ x 0 C is arbitrary , y n = α n x n + ( 1 α n ) T x n , C n = { z C : y n z x n z } , Q n = { z C : x n z , x 0 x n 0 } , x n + 1 = P C n Q n ( x 0 ) , n = 0 , 1 , 2 ,
(1.2)

where P K denotes the metric projection from H onto a closed convex subset K of H. The above method is also called CQ method or hybrid method.

In 2006, Kim and Xu [5] adapted the iteration (1.2) in a Hilbert space. More precisely, they introduced the following iteration process for asymptotically nonexpansive mappings:
{ x 0 C is arbitrary, y n = α n x n + ( 1 α n ) T n x n , C n = { z C : y n z 2 x n z 2 + θ n } , Q n = { z C : x n z , x 0 x n 0 } , x n + 1 = P C n Q n ( x 0 ) , n = 0 , 1 , 2 ,
(1.3)
where
θ n = ( 1 α n ) ( k n 2 1 ) ( diam C ) 2 0 as  n .

They proved that { x n } converges in norm to P F ( T ) x 0 under some conditions. Several authors (see [6, 7]) have studied the convergence of hybrid method.

Baillon [8] first proved that the following Cesàro mean iterative sequence weakly converges to a fixed point of a nonexpansive mapping in Hilbert spaces:
T n x = 1 n + 1 i = 0 n T i x .

Shimizu and Takahashi [9] proved a strong convergence theorem of the above iteration for an asymptotically nonexpansive mapping in Hilbert spaces.

Let C be a nonempty closed convex subset of a real Hilbert space H, let f : C × C R be a functional, where is the set of real numbers. The equilibrium problem is to find x ˆ C such that
f ( x , y ) 0 , y C .
(1.4)

The set of solutions of (1.4) is denoted by EP ( f ) . Given a mapping T : C X , let f ( x , y ) = T x , y x , x , y C , then z EP ( f ) if and only if T z , y z 0 , y C , i.e., z is the solution of the variational inequality.

There are several other problems, for example, the complementarity problem, fixed point problem and optimization problem, which can also be written in the form of an equilibrium problem. So, equilibrium problems provide us with a systematic framework to study a wide class of problems arising in financial economics, optimization and operation research etc., which motivates the extensive concern. See, for example, [1014]. In recent years, equilibrium problems have been deeply and thoroughly researched. See, for example, [1520]. Some methods have been proposed to solve the equilibrium problem in a Hilbert space; see, for instance, [2123]. In 2011, Jitpeera, Katchang, and Kumam [24] found a common element of the set of solutions for mixed equilibrium problem, the set of solutions of the variational inequality for a β-inverse strongly monotone mapping, and the set of fixed points of a family of finitely nonexpansive mappings in a real Hilbert space by using the viscosity and Cesàro mean approximation method.

Motivated by the above-mentioned results, in this paper we introduce the following iteration process for asymptotically nonexpansive mappings T with C a closed convex bounded subset of a real Hilbert space:
{ x 0 C is arbitrary, y n = α n x n + ( 1 α n ) 1 n + 1 j = 0 n T j x n , C n = { z C : y n z 2 x n z 2 + θ n } , Q n = { z C : x n z , x 0 x n 0 } , x n + 1 = P C n Q n ( x 0 ) , n = 0 , 1 , 2 ,
(1.5)
where
θ n = ( 1 α n ) ( L n 2 1 ) ( diam C ) 2 0 as  n .

We shall prove that the above iterative sequence { x n } converges strongly to a fixed point of T under some proper conditions. In addition, we also introduce a new hybrid iterative scheme for finding a common element of the set of common fixed points of asymptotically nonexpansive mappings and the set of solutions of an equilibrium problem in Hilbert spaces.

We will use the notation for weak convergence and → for strong convergence.

2 Preliminaries

Let H be a real Hilbert space. Then
x y 2 = x 2 y 2 2 x y , y
(2.1)
and
λ x + ( 1 λ ) y 2 = λ x 2 + ( 1 λ ) y 2 λ ( 1 λ ) x y 2
(2.2)
for all x , y H and λ [ 0 , 1 ] . It is also known that H satisfies
  1. (1)
    Opial’s condition, that is, for any sequence { x n } with x n x , the inequality
    lim inf n x n x < lim inf n x n y

    holds for every y H with y x .

     
  2. (2)

    The Kadec-Klee property, that is, for any sequence { x n } with x n x and x n x together implies x n x 0 .

     
Let C be a nonempty closed convex subset of H. Then, for any x H , there exists a unique nearest point in C, denoted by P C ( x ) , such that
x P C ( x ) x y , y C .
Such a mapping P C is called the metric projection of H onto C. We know that P C is nonexpansive. Furthermore, for x H and z C ,
z = P C ( x ) if and only if x z , z y 0 , y C .

We also need the following lemmas.

Lemma 2.1 (See [25])

Let T be an asymptotically nonexpansive mapping defined on a bounded closed convex subset C of a Hilbert space H. Assume that { x n } is a sequence in C with the properties: (i)  x n z ; (ii)  T x n x n 0 . Then z F ( T ) .

Lemma 2.2 (See [9])

Let C be a nonempty bounded subset of a Hilbert space. Let T be an asymptotically nonexpansive mapping from C into itself such that F ( T ) is nonempty. Then, for any ε > 0 , there exists a positive integer l ε such that for any integer l l ε , there is a positive integer n l satisfying
1 n + 1 j = 0 n T j x T l ( 1 n + 1 j = 0 n T j x ) < ε , x C , n n l .
The equilibrium problem is to find x ˆ C such that
f ( x ˆ , y ) 0 , y C .

The set of solutions of the above inequality is denoted by EP ( f ) . For solving the equilibrium problem, let us assume that a bifunction f satisfies the following conditions:

(A1) f ( x , x ) = 0 for all x C ;

(A2) f is monotone, i.e., f ( x , y ) + f ( y , x ) 0 , for all x , y C ;

(A3) for each x , y , z C ,
lim sup t 0 + f ( t z + ( 1 t ) x , y ) f ( x , y ) ;

(A4) for each x C , f ( x , ) is convex and lower semi-continuous.

The following lemma appears implicitly in [21].

Lemma 2.3 ([21])

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

The following lemma was also given in [26].

Lemma 2.4 ([26])

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

    T r is single-valued;

     
  2. (2)
    T r is firmly nonexpansive, i.e.,
    T r x T r y 2 T r x T r y , x y , x , y H ;
     
  3. (3)

    F ( T r ) = EP ( f ) ;

     
  4. (4)

    EP ( f ) is closed and convex.

     

Lemma 2.5 ([26])

Let C be a nonempty closed convex subset of H, let f : C × C R be a functional, satisfying (A1)-(A4), and let r > 0 . Then, for x H and q F ( T r ) ,
q T r x 2 + T r x x 2 q x 2 .

3 Strong convergence theorem of modified Mann iteration based on hybrid method

Inspired by Kim and Xu’s results (see [5]), Mann-type iteration (1.3) is modified to obtain the strong convergence theorem as follows.

Theorem 3.1 Let C be a nonempty bounded closed convex subset of a real Hilbert space H and let T : C C be an asymptotically nonexpansive mapping with k n , denote L n = 1 n + 1 j = 0 n k j . Assume that { α n } n = 0 is a sequence in ( 0 , 1 ) such that α n a for all n and for some 0 < a < 1 . Define a sequence { x n } n = 0 in C by the following algorithm:
{ x 0 C is arbitrary , y n = α n x n + ( 1 α n ) 1 n + 1 j = 0 n T j x n , C n = { z C : y n z 2 x n z 2 + θ n } , Q n = { z C : x n z , x 0 x n 0 } , x n + 1 = P C n Q n ( x 0 ) , n = 0 , 1 , 2 ,
(3.1)
where
θ n = ( 1 α n ) ( L n 2 1 ) ( diam C ) 2 0 as  n .

Then { x n } converges strongly to P F ( T ) ( x 0 ) .

Proof First note that T has a fixed point in C (see [1]); that is, F ( T ) is nonempty. We divide the proof of this theorem into four steps as below.

Step 1. We show that C n and Q n are closed and convex for each n N { 0 } .

From the definition of C n and Q n , it is obvious that C n is closed and Q n is closed and convex for each n N { 0 } . We prove that C n is convex. Since y n z 2 x n z 2 + θ n is equivalent to
2 x n y n , z x n 2 y n 2 + θ n ,

it follows that C n is convex.

Step 2. We show that F ( T ) C n , n N { 0 } .

Let p F ( T ) and n N { 0 } . Then from
y n p 2 = α n x n + ( 1 α n ) 1 n + 1 j = 0 n T j x n p 2 α n x n p 2 + ( 1 α n ) ( 1 n + 1 j = 0 n T j x n p ) 2 α n x n p 2 + ( 1 α n ) ( 1 n + 1 j = 0 n k j x n p ) 2 α n x n p 2 + ( 1 α n ) L n 2 x n p 2 = x n p 2 + ( 1 α n ) ( L n 2 1 ) x n p 2 x n p 2 + θ n ,
we have p C n . Next, we show that F ( T ) Q n , n N { 0 } . We prove this by induction. For n = 0 , we have F ( T ) C = Q 0 . Suppose that F ( T ) Q n , then F ( T ) C n Q n and there exists a unique element x n + 1 C n Q n such that x n + 1 = P C n Q n ( x 0 ) . Then
x n + 1 z , x 0 x n + 1 0 , z C n Q n .
In particular,
x n + 1 p , x 0 x n + 1 0 , p F ( T ) .

It follows that F ( T ) Q n + 1 and hence F ( T ) Q n for each n. Thus we obtain F ( T ) C n Q n , n N { 0 } . This means that { x n } is well defined.

Step 3. We show that { x n } is bounded and lim n x n T x n = 0 .

It follows from the definition of Q n that x n = P Q n ( x 0 ) . Therefore
x n x 0 z x 0 for all  z Q n  and all  n N { 0 } .
Let z F ( T ) Q n for all n N { 0 } . Then
x n x 0 z x 0 for all  n N { 0 } .
On the other hand, from x n + 1 = P C n Q n ( x 0 ) Q n , we have
x n x 0 x n + 1 x 0 for all  n N { 0 } .
Therefore, the sequence { x n x 0 } is nondecreasing. Since C is bounded, we obtain that lim n x n x 0 exists. This implies that { x n } is bounded. Noticing again that x n + 1 = P C n Q n ( x 0 ) Q n and x n = P Q n ( x 0 ) , we have x n + 1 x n , x n x 0 0 . It follows from (2.1) that
x n + 1 x n 2 = ( x n + 1 x 0 ) ( x n x 0 ) 2 = x n + 1 x 0 2 x n x 0 2 2 x n + 1 x n , x n x 0 x n + 1 x 0 2 x n x 0 2
for all n N { 0 } . This implies that
lim n x n + 1 x n = 0 .
Since x n + 1 = P C n Q n C n , we have y n x n + 1 2 x n x n + 1 2 + θ n . Noticing that lim n k n = 1 , then L n = 1 n + 1 j = 0 n k j 1 as n . Hence
y n x n + 1 x n x n + 1 + θ n 0 .
Let A n = 1 n + 1 j = 0 n T j . Also since α n a < 1 for all n, then
x n A n x n = 1 1 α n y n x n 1 1 a ( y n x n + 1 + x n x n + 1 ) 0 .
From Lemma 2.2, we have
lim sup l lim sup n A n x n T l ( A n x n ) lim sup l lim sup n sup x C A n x T l ( A n x ) = 0 .
Therefore,
lim sup l lim sup n A n x n T l ( A n x n ) = 0 .
Put k = sup { k n : n 1 } < , then
x n T l x n x n A n x n + A n x n T l ( A n x n ) + T l ( A n x n ) T l x n x n A n x n + A n x n T l ( A n x n ) + k l x n A n x n ( 1 + k ) x n A n x n + A n x n T l ( A n x n ) 0
as n , l . Thus, we have
T x n x n T x n T l + 1 x n + T l + 1 x n T l + 1 x n + 1 + T l + 1 x n + 1 x n + 1 + x n + 1 x n k x n T l x n + x n + 1 + T l x n + 1 + ( k + 1 ) x n + 1 x n 0 .

Step 4. We show that { x n } converges strongly to P F ( T ) ( x 0 ) .

Put w = P F ( T ) ( x 0 ) . Since { x n } is bounded, let { x n k } be a subsequence of { x n } such that x n k w . By Lemma 2.1, we get w F ( T ) . Since x n = P Q n ( x 0 ) and w F ( T ) Q n , we have x n x 0 w x 0 . It follows from w = P F ( T ) ( x 0 ) and the weak lower semicontinuity of the norm that
w x 0 w x 0 lim inf k x n k x 0 lim sup k x n k x 0 w x 0 .

Thus, we obtain that lim k x n k x 0 = w x 0 = w x 0 . Using the Kadec-Klee property of H, we get lim k x n k = w = w . Since { x n k } is an arbitrary subsequence of { x n } , we can conclude that { x n } converges strongly to P F ( T ) ( x 0 ) . □

Remark 3.2 It is not difficult to see from the proof above that the boundedness of C can be discarded if T is a nonexpansive mapping.

4 Strong convergence theorem for equilibrium problems

In this section, we prove a strong convergence theorem for finding a common element of the set of zero points of an asymptotically nonexpansive mapping T and the set of solutions of an equilibrium problem in a Hilbert space.

Theorem 4.1 Let H be a real Hilbert space, and let C be a nonempty bounded closed convex subset of H, let f : C × C R be a functional, satisfying (A1)-(A4). Let T : C C be an asymptotically nonexpansive mapping with k n , denote L n = 1 n + 1 j = 0 n k j such that F ( T ) EP ( f ) . Let { x n } be a sequence generated by
{ x 0 C is arbitrary , y n = α n x n + ( 1 α n ) 1 n + 1 j = 0 n T j x n , u n C , such that  f ( u n , y ) + 1 r n y u n , u n y n 0 , y C , C n = { z C : u n z 2 x n z 2 + θ n } , Q n = { z C : x n z , x 0 x n 0 } , x n + 1 = P C n Q n ( x 0 ) , n = 0 , 1 , 2 , ,
(4.1)
where
θ n = ( 1 α n ) ( L n 2 1 ) ( diam C ) 2 0 , n .

Suppose that { α n } [ 0 , 1 ] and there exists a ( 0 , 1 ) such that α n a , n N , and { r n } ( 0 , ) such that lim inf n r n > 0 . Then { x n } converges strongly to P F ( T ) EP ( f ) ( x 0 ) .

Proof We divide the proof of this theorem into four steps as below.

Step 1. Similar to the proof of Step 1 in Theorem 3.1, it is easy to see that C n and Q n are closed convex sets for each n N { 0 } .

Step 2. We show that F ( T ) EP ( f ) C n Q n , n N { 0 } .

Let p F ( T ) EP ( f ) . Putting u n = T r n y n , n N { 0 } , by (2) of Lemma 2.4, we have T r n is relatively nonexpansive. Noticing that relatively nonexpansive mappings are nonexpansive in Hilbert spaces, then for any n N { 0 } ,
u n p 2 = T r n y n p 2 = T r n y n T r n p 2 y n p 2 .

From the proof of Step 2 in Theorem 3.1, we have u n p 2 x n p 2 + θ n . Thus p C n , hence F ( T ) EP ( f ) C n . Similar to the proof of Step 2 in Theorem 3.1, it is easy to see that F ( T ) EP ( f ) Q n . Therefore we have F ( T ) EP ( f ) C n Q n , n N { 0 } . This means that { x n } is well defined.

Step 3. We show that { x n } is bounded and lim n x n T x n = 0 .

Similar to the proof of Step 3 in Theorem 3.1, we may obtain that { x n } is bounded and
lim n x n + 1 x n = 0 .
(4.2)
Since x n + 1 = P C n Q n C n , then u n x n + 1 2 x n x n + 1 2 + θ n . Noticing that θ n 0 , we have
u n x n + 1 x n x n + 1 + θ n 0 .
Hence
lim n x n u n lim n x n x n + 1 + lim n x n + 1 u n 0 .
(4.3)
For p F ( T ) EP ( f ) C n , we have
u n p 2 x n p 2 + θ n .
(4.4)
Since u n = T r n y n , by (4.4) and Lemma 2.5, we get that
u n y n 2 = T r n y n y n 2 y n p 2 T r n y n p 2 x n p 2 + θ n u n p 2 0 .
(4.5)
Let A n = 1 n + 1 j = 0 n T j . Since α n a < 1 , n N , then by (4.3) and (4.5), we have
x n A n x n = 1 1 α n y n x n 1 1 a ( y n u n + u n x n ) 0 .
(4.6)
From Lemma 2.2, it follows that
lim sup l lim sup n A n x n T l ( A n x n ) lim sup l lim sup n sup x C A n x T l ( A n x ) = 0 .
Therefore
lim sup l lim sup n A n x n T l ( A n x n ) = 0 .
(4.7)
Put k = sup { k n : n 1 } < , then
y n T l x n = α n x n + ( 1 α n ) A n x n T l x n α n x n A n x n + A n x n T l ( A n x n ) + T l ( A n x n ) T l x n α n x n A n x n + A n x n T l ( A n x n ) + k l x n A n x n ( 1 + k ) x n A n x n + A n x n T l ( A n x n ) .
Let n , l , we get that y n T l x n 0 . Thus, by (4.3) and (4.5), we have
x n T l x n x n u n + u n y n + y n T l x n 0 .
(4.8)
Therefore, by (4.8) and (4.2), we obtain that
T x n x n T x n T l + 1 x n + T l + 1 x n T l + 1 x n + 1 + T l + 1 x n + 1 x n + 1 + x n + 1 x n k x n T l x n + ( k + 1 ) x n + 1 x n + T l + 1 x n + 1 x n + 1 0 .

Step 4. We show that { x n } converges strongly to P F ( T ) EP ( f ) ( x 0 ) .

Since { x n } is bounded, there exists a subsequence { x n k } of { x n } such that x n k w . By Lemma 2.1, we have w F ( T ) . Next we show w EP ( f ) .

From (4.3) and (4.5), we get that u n k w , y n k w . Since u n = T r n y n , then
f ( u n , y ) + 1 r n y u n , u n y n 0 , y C .
Replacing n by n k , we have from Condition (A2) that
1 r n k y u n k , u n k y n k f ( u n k , y ) f ( y , u n k ) , y C .
Let k , since lim inf n r n > 0 , by (4.5) and Condition (A4), we get that
f ( y , w ) 0 , y C .
For t ( 0 , 1 ) , y C , let y t = t y + ( 1 t ) w , then y t C , thus f ( y t , w ) 0 . By Condition (A1), we get that
0 = f ( y t , y t ) t f ( y t , y ) + ( 1 t ) f ( y t , w ) t f ( y t , y ) .
Dividing by t, we have
f ( y t , y ) 0 , y C .

Let t 0 . From Condition (A3), we obtain that f ( w , y ) 0 , y C . Therefore, w EP ( f ) .

Denote w = P F ( T ) EP ( f ) ( x 0 ) . Since x n + 1 = P C n Q n ( x 0 ) , w F ( T ) EP ( f ) C n Q n , then x n + 1 x 0 w x 0 . Since the norm is weakly lower semicontinuous, we have
w x 0 w x 0 lim inf k x n k x 0 lim sup k x n k x 0 w x 0 .

Hence lim k x n k x 0 = w x 0 = w x 0 . Using the Kadec-Klee property of H, we get lim k x n k = w = w . Since { x n k } is an arbitrary subsequence of { x n } , we can conclude that { x n } converges strongly to P F ( T ) EP ( f ) ( x 0 ) . □

Declarations

Acknowledgements

The authors would like to thank the anonymous referee for some suggestions to improve the manuscript. This work was supported by Research Fund for the Doctoral Program of Harbin University of Commerce (13DL002), Scientific Research Project Fund of the Education Department of Heilongjiang Province and Foundation of Heilongjiang Educational Committee (12521070).

Authors’ Affiliations

(1)
Mathematics and Applied Mathematics, Harbin University of Commerce
(2)
Department of Mathematics, Harbin University of Science and Technology

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© Zhang and Cui; licensee Springer. 2014

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