Research

# Convergence theorems for mixed type asymptotically nonexpansive mappings

Weiping Guo1, Yeol Je Cho2* and Wei Guo3

Author Affiliations

1 School of Mathematics and Physics, Suzhou University of Science and Technology, Suzhou, Jiangsu, 215009, P.R. China

2 Department of Mathematics Education and the RINS College of Education, Gyeongsang National University, Chinju, 660-701, Korea

3 Department of Aerospace Engineering and Mechanics, University of Minnesota, Twin Cities, Minneapolis, MN, 55455, USA

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Fixed Point Theory and Applications 2012, 2012:224  doi:10.1186/1687-1812-2012-224

 Received: 27 April 2012 Accepted: 16 November 2012 Published: 11 December 2012

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In this paper, we introduce a new two-step iterative scheme of mixed type for two asymptotically nonexpansive self-mappings and two asymptotically nonexpansive nonself-mappings and prove strong and weak convergence theorems for the new two-step iterative scheme in uniformly convex Banach spaces.

##### Keywords:
mixed type asymptotically nonexpansive mapping; strong and weak convergence; common fixed point; uniformly convex Banach space

### 1 Introduction

Let K be a nonempty subset of a real normed linear space E. A mapping T : K K is said to be asymptotically nonexpansive if there exists a sequence { k n } [ 1 , ) with lim n k n = 1 such that

T n x T n y k n x y (1.1)

for all x , y K and n 1 .

In 1972, Goebel and Kirk [1] introduced the class of asymptotically nonexpansive self-mappings, which is an important generalization of the class of nonexpansive self-mappings, and proved that if K is a nonempty closed convex subset of a real uniformly convex Banach space E and T is an asymptotically nonexpansive self-mapping of K, then T has a fixed point.

Since then, some authors proved weak and strong convergence theorems for asymptotically nonexpansive self-mappings in Banach spaces (see [2-16]), which extend and improve the result of Goebel and Kirk in several ways.

Recently, Chidume et al.[10] introduced the concept of asymptotically nonexpansive nonself-mappings, which is a generalization of an asymptotically nonexpansive self-mapping, as follows.

Definition 1.1[10]

Let K be a nonempty subset of a real normed linear space E. Let P : E K be a nonexpansive retraction of E onto K. A nonself-mapping T : K E is said to be asymptotically nonexpansive if there exists a sequence { k n } [ 1 , ) with k n 1 as n such that

T ( P T ) n 1 x T ( P T ) n 1 y k n x y (1.2)

for all x , y K and n 1 .

Let K be a nonempty closed convex subset of a real uniformly convex Banach space E.

In 2003, also, Chidume et al.[10] studied the following iteration scheme:

{ x 1 K , x n + 1 = P ( ( 1 α n ) x n + α n T 1 ( P T 1 ) n 1 x n ) (1.3)

for each n 1 , where { α n } is a sequence in ( 0 , 1 ) and P is a nonexpansive retraction of E onto K, and proved some strong and weak convergence theorems for an asymptotically nonexpansive nonself-mapping.

In 2006, Wang [11] generalized the iteration process (1.3) as follows:

{ x 1 K , x n + 1 = P ( ( 1 α n ) x n + α n T 1 ( P T 1 ) n 1 y n ) , y n = P ( ( 1 β n ) x n + β n T 2 ( P T 2 ) n 1 x n ) (1.4)

for each n 1 , where T 1 , T 2 : K E are two asymptotically nonexpansive nonself-mappings and { α n } , { β n } are real sequences in [ 0 , 1 ) , and proved some strong and weak convergence theorems for two asymptotically nonexpansive nonself-mappings. Recently, Guo and Guo [12] proved some new weak convergence theorems for the iteration process (1.4).

The purpose of this paper is to construct a new iteration scheme of mixed type for two asymptotically nonexpansive self-mappings and two asymptotically nonexpansive nonself-mappings and to prove some strong and weak convergence theorems for the new iteration scheme in uniformly convex Banach spaces.

### 2 Preliminaries

Let E be a real Banach space, K be a nonempty closed convex subset of E and P : E K be a nonexpansive retraction of E onto K. Let S 1 , S 2 : K K be two asymptotically nonexpansive self-mappings and T 1 , T 2 : K E be two asymptotically nonexpansive nonself-mappings. Then we define the new iteration scheme of mixed type as follows:

{ x 1 K , x n + 1 = P ( ( 1 α n ) S 1 n x n + α n T 1 ( P T 1 ) n 1 y n ) , y n = P ( ( 1 β n ) S 2 n x n + β n T 2 ( P T 2 ) n 1 x n ) (2.1)

for each n 1 , where { α n } , { β n } are two sequences in [ 0 , 1 ) .

If S 1 and S 2 are the identity mappings, then the iterative scheme (2.1) reduces to the sequence (1.4).

We denote the set of common fixed points of S 1 , S 2 , T 1 and T 2 by F = F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) and denote the distance between a point z and a set A in E by d ( z , A ) = inf x A z x .

Now, we recall some well-known concepts and results.

Let E be a real Banach space, E be the dual space of E and J : E 2 E be the normalized duality mapping defined by

J ( x ) = { f E : x , f = x f , f = x }

for all x E , where , denotes duality pairing between E and E . A single-valued normalized duality mapping is denoted by j.

A subset K of a real Banach space E is called a retract of E[10] if there exists a continuous mapping P : E K such that P x = x for all x K . Every closed convex subset of a uniformly convex Banach space is a retract. A mapping P : E E is called a retraction if P 2 = P . It follows that if a mapping P is a retraction, then P y = y for all y in the range of P.

A Banach space E is said to satisfy Opial’s condition[17] if, for any sequence { x n } of E, x n x weakly as n implies that

lim sup n x n x < lim sup n x n y

for all y E with y x .

A Banach space E is said to have a Fréchet differentiable norm[18] if, for all x U = { x E : x = 1 } ,

lim t 0 x + t y x t

exists and is attained uniformly in y U .

A Banach space E is said to have the Kadec-Klee property[19] if for every sequence { x n } in E, x n x weakly and x n x , it follows that x n x strongly.

Let K be a nonempty closed subset of a real Banach space E. A nonself-mapping T : K E is said to be semi-compact[11] if, for any sequence { x n } in K such that x n T x n 0 as n , there exists a subsequence { x n j } of { x n } such that { x n j } converges strongly to some x K .

Lemma 2.1[15]

Let { a n } , { b n } and { c n } be three nonnegative sequences satisfying the following condition:

a n + 1 ( 1 + b n ) a n + c n

for each n n 0 , where n 0 is some nonnegative integer, n = n 0 b n < and n = n 0 c n < . Then lim n a n exists.

Lemma 2.2[8]

LetEbe a real uniformly convex Banach space and 0 < p t n q < 1 for each n 1 . Also, suppose that { x n } and { y n } are two sequences ofEsuch that

lim sup n x n r , lim sup n y n r , lim n t n x n + ( 1 t n ) y n = r

hold for some r 0 . Then lim n x n y n = 0 .

Lemma 2.3[10]

LetEbe a real uniformly convex Banach space, Kbe a nonempty closed convex subset ofEand T : K E be an asymptotically nonexpansive mapping with a sequence { k n } [ 1 , ) and k n 1 as n . Then I T is demiclosed at zero, i.e., if x n x weakly and x n T x n 0 strongly, then x F ( T ) , where F ( T ) is the set of fixed points ofT.

Lemma 2.4[16]

LetXbe a uniformly convex Banach space andCbe a convex subset ofX. Then there exists a strictly increasing continuous convex function γ : [ 0 , ) [ 0 , ) with γ ( 0 ) = 0 such that, for each mapping S : C C with a Lipschitz constant L > 0 ,

α S x + ( 1 α ) S y S [ α x + ( 1 α ) y ] L γ 1 ( x y 1 L S x S y )

for all x , y C and 0 < α < 1 .

Lemma 2.5[16]

LetXbe a uniformly convex Banach space such that its dual space X has the Kadec-Klee property. Suppose { x n } is a bounded sequence and f 1 , f 2 W w ( { x n } ) such that

lim n α x n + ( 1 α ) f 1 f 2

exists for all α [ 0 , 1 ] , where W w ( { x n } ) denotes the set of all weak subsequential limits of  { x n } . Then f 1 = f 2 .

### 3 Strong convergence theorems

In this section, we prove strong convergence theorems for the iterative scheme given in (2.1) in uniformly convex Banach spaces.

Lemma 3.1LetEbe a real uniformly convex Banach space andKbe a nonempty closed convex subset ofE. Let S 1 , S 2 : K K be two asymptotically nonexpansive self-mappings with { k n ( 1 ) } , { k n ( 2 ) } [ 1 , ) and T 1 , T 2 : K E be two asymptotically nonexpansive nonself-mappings with { l n ( 1 ) } , { l n ( 2 ) } [ 1 , ) such that n = 1 ( k n ( i ) 1 ) < and n = 1 ( l n ( i ) 1 ) < for i = 1 , 2 , respectively, and F = F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) . Let { x n } be the sequence defined by (2.1), where { α n } and { β n } are two real sequences in [ 0 , 1 ) . Then

(1) lim n x n q exists for any q F ;

(2) lim n d ( x n , F ) exists.

Proof (1) Set h n = max { k n ( 1 ) , k n ( 2 ) , l n ( 1 ) , l n ( 2 ) } . For any q F , it follows from (2.1) that

y n q ( 1 β n ) ( S 2 n x n q ) + β n ( T 2 ( P T 2 ) n 1 x n q ) ( 1 β n ) h n x n q + β n h n x n q = h n x n q (3.1)

and so

x n + 1 q ( 1 α n ) ( S 1 n x n q ) + α n ( T 1 ( P T 1 ) n 1 y n q ) ( 1 α n ) h n x n q + α n h n y n q ( 1 α n ) h n 2 x n q + α n h n 2 x n q = [ 1 + ( h n 2 1 ) ] x n q . (3.2)

Since n = 1 ( k n ( i ) 1 ) < and n = 1 ( l n ( i ) 1 ) < for i = 1 , 2 , we have n = 1 ( h n 2 1 ) < . It follows from Lemma 2.1 that lim n x n q exists.

(2) Taking the infimum over all q F in (3.2), we have

d ( x n + 1 , F ) [ 1 + ( h n 2 1 ) ] d ( x n , F )

for each n 1 . It follows from n = 1 ( h n 2 1 ) < and Lemma 2.1 that the conclusion (2) holds. This completes the proof. □

Lemma 3.2LetEbe a real uniformly convex Banach space andKbe a nonempty closed convex subset ofE. Let S 1 , S 2 : K K be two asymptotically nonexpansive self-mappings with { k n ( 1 ) } , { k n ( 2 ) } [ 1 , ) and T 1 , T 2 : K E be two asymptotically nonexpansive nonself-mappings with { l n ( 1 ) } , { l n ( 2 ) } [ 1 , ) such that n = 1 ( k n ( i ) 1 ) < and n = 1 ( l n ( i ) 1 ) < for i = 1 , 2 , respectively, and F = F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) . Let { x n } be the sequence defined by (2.1) and the following conditions hold:

(a) { α n } and { β n } are two real sequences in [ ϵ , 1 ϵ ] for some ϵ ( 0 , 1 ) ;

(b) x T i y S i x T i y for all x , y K and i = 1 , 2 .

Then lim n x n S i x n = lim n x n T i x n = 0 for i = 1 , 2 .

Proof Set h n = max { k n ( 1 ) , k n ( 2 ) , l n ( 1 ) , l n ( 2 ) } . For any given q F , lim n x n q exists by Lemma 3.1. Now, we assume that lim n x n q = c . It follows from (3.2) and n = 1 ( h n 2 1 ) < that

lim n ( 1 α n ) ( S 1 n x n q ) + α n ( T 1 ( P T 1 ) n 1 y n q ) = c

and

lim sup n S 1 n x n q lim sup n k n ( 1 ) x n q = c .

Taking lim sup on both sides in (3.1), we obtain lim sup n y n q c and so

lim sup n T 1 ( P T 1 ) n 1 y n q lim sup n l n ( 1 ) y n q c .

Using Lemma 2.2, we have

lim n S 1 n x n T 1 ( P T 1 ) n 1 y n = 0 . (3.3)

By the condition (b), it follows that

x n T 1 ( P T 1 ) n 1 y n S 1 n x n T 1 ( P T 1 ) n 1 y n

and so, from (3.3), we have

lim n x n T 1 ( P T 1 ) n 1 y n = 0 . (3.4)

Since

x n q x n T 1 ( P T 1 ) n 1 y n + T 1 ( P T 1 ) n 1 y n q x n T 1 ( P T 1 ) n 1 y n + l n ( 1 ) y n q .

Taking lim inf on both sides in the inequality above, we have

lim inf n y n q c

by (3.4) and so

lim n y n q = c .

Using (3.1), we have

lim n ( 1 β n ) ( S 2 n x n q ) + β n ( T 2 ( P T 2 ) n 1 x n q ) = c .

lim sup n S 2 n x n q lim sup n k n ( 2 ) x n q = c

and

lim sup n T 2 ( P T 2 ) n 1 x n q lim sup n l n ( 2 ) x n q = c .

It follows from Lemma 2.2 that

lim n S 2 n x n T 2 ( P T 2 ) n 1 x n = 0 . (3.5)

Now, we prove that

lim n x n T 1 x n = lim n x n T 2 x n = 0 .

Indeed, since x n T 2 ( P T 2 ) n 1 x n S 2 n x n T 2 ( P T 2 ) n 1 x n by the condition (b). It follows from (3.5) that

lim n x n T 2 ( P T 2 ) n 1 x n = 0 . (3.6)

Since S 2 n x n = P ( S 2 n x n ) and P : E K is a nonexpansive retraction of E onto K, we have

y n S 2 n x n β n S 2 n x n T 2 ( P T 2 ) n 1 x n

and so

lim n y n S 2 n x n = 0 . (3.7)

Furthermore, we have

y n x n y n S 2 n x n + S 2 n x n T 2 ( P T 2 ) n 1 x n + T 2 ( P T 2 ) n 1 x n x n .

Thus it follows from (3.5), (3.6) and (3.7) that

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

Since x n T 1 ( P T 1 ) n 1 x n S 1 n x n T 1 ( P T 1 ) n 1 x n by the condition (b) and

Using (3.3) and (3.8), we have

lim n S 1 n x n T 1 ( P T 1 ) n 1 x n = 0 (3.9)

and

lim n x n T 1 ( P T 1 ) n 1 x n = 0 . (3.10)

It follows from

x n + 1 S 1 n x n = P [ ( 1 α n ) S 1 n x n + α n T 1 ( P T 1 ) n 1 y n ] P ( S 1 n x n ) α n S 1 n x n T 1 ( P T 1 ) n 1 y n

and (3.3) that

lim n x n + 1 S 1 n x n = 0 . (3.11)

x n + 1 T 1 ( P T 1 ) n 1 y n x n + 1 S 1 n x n + S 1 n x n T 1 ( P T 1 ) n 1 y n .

Using (3.3) and (3.11), we obtain that

lim n x n + 1 T 1 ( P T 1 ) n 1 y n = 0 . (3.12)

Thus, using (3.9), (3.10) and the inequality

S 1 n x n x n S 1 n x n T 1 ( P T 1 ) n 1 x n + T 1 ( P T 1 ) n 1 x n x n ,

we have lim n S 1 n x n x n = 0 . It follows from (3.6) and the inequality

S 1 n x n T 2 ( P T 2 ) n 1 x n S 1 n x n x n + x n T 2 ( P T 2 ) n 1 x n

that

lim n S 1 n x n T 2 ( P T 2 ) n 1 x n = 0 . (3.13)

Since

x n + 1 T 2 ( P T 2 ) n 1 y n x n + 1 S 1 n x n + S 1 n x n T 2 ( P T 2 ) n 1 x n + l n ( 2 ) x n y n ,

from (3.8), (3.11) and (3.13), it follows that

lim n x n + 1 T 2 ( P T 2 ) n 1 y n = 0 . (3.14)

Again, since ( P T i ) ( P T i ) n 2 y n 1 , x n K for i = 1 , 2 and T 1 , T 2 are two asymptotically nonexpansive nonself-mappings, we have

(3.15)

for i = 1 , 2 . It follows from (3.12), (3.14) and (3.15) that

lim n T i ( P T i ) n 1 y n 1 T i x n = 0 (3.16)

for i = 1 , 2 . Moreover, we have

x n + 1 y n x n + 1 T 1 ( P T 1 ) n 1 y n + T 1 ( P T 1 ) n 1 y n x n + x n y n .

Using (3.4), (3.8) and (3.12), we have

lim n x n + 1 y n = 0 . (3.17)

x n T i x n x n T i ( P T i ) n 1 x n + T i ( P T i ) n 1 x n T i ( P T i ) n 1 y n 1 + T i ( P T i ) n 1 y n 1 T i x n x n T i ( P T i ) n 1 x n + max { sup n 1 l n ( 1 ) , sup n 1 l n ( 2 ) } x n y n 1 + T i ( P T i ) n 1 y n 1 T i x n

for i = 1 , 2 . Thus it follows from (3.6), (3.10), (3.16) and (3.17) that

lim n x n T 1 x n = lim n x n T 2 x n = 0 .

Finally, we prove that

lim n x n S 1 x n = lim n x n S 2 x n = 0 .

In fact, by the condition (b), we have

x n S i x n x n T i ( P T i ) n 1 x n + S i x n T i ( P T i ) n 1 x n x n T i ( P T i ) n 1 x n + S i n x n T i ( P T i ) n 1 x n

for i = 1 , 2 . Thus it follows from (3.5), (3.6), (3.9) and (3.10) that

lim n x n S 1 x n = lim n x n S 2 x n = 0 .

This completes the proof. □

Now, we find two mappings, S 1 = S 2 = S and T 1 = T 2 = T , satisfying the condition (b) in Lemma 3.2 as follows.

Example 3.1[20]

Let ℝ be the real line with the usual norm | | and let K = [ 1 , 1 ] . Define two mappings S , T : K K by

T x = { 2 sin x 2 , if  x [ 0 , 1 ] , 2 sin x 2 , if  x [ 1 , 0 ) ,

and

S x = { x , if  x [ 0 , 1 ] , x , if  x [ 1 , 0 ) .

Now, we show that T is nonexpansive. In fact, if x , y [ 0 , 1 ] or x , y [ 1 , 0 ) , then we have

| T x T y | = 2 | sin x 2 sin y 2 | | x y | .

If x [ 0 , 1 ] and y [ 1 , 0 ) or x [ 1 , 0 ) and y [ 0 , 1 ] , then we have

| T x T y | = 2 | sin x 2 + sin y 2 | = 4 | sin x + y 4 cos x y 4 | | x + y | | x y | .

This implies that T is nonexpansive and so T is an asymptotically nonexpansive mapping with k n = 1 for each n 1 . Similarly, we can show that S is an asymptotically nonexpansive mapping with l n = 1 for each n 1 .

Next, we show that two mappings S, T satisfy the condition (b) in Lemma 3.2. For this, we consider the following cases:

Case 1. Let x , y [ 0 , 1 ] . Then we have

| x T y | = | x + 2 sin y 2 | = | S x T y | .

Case 2. Let x , y [ 1 , 0 ) . Then we have

| x T y | = | x 2 sin y 2 | | x 2 sin y 2 | = | S x T y | .

Case 3. Let x [ 1 , 0 ) and y [ 0 , 1 ] . Then we have

| x T y | = | x + 2 sin y 2 | | x + 2 sin y 2 | = | S x T y | .

Case 4. Let x [ 0 , 1 ] and y [ 1 , 0 ) . Then we have

| x T y | = | x 2 sin y 2 | = | S x T y | .

Therefore, the condition (b) in Lemma 3.2 is satisfied.

Theorem 3.1Under the assumptions of Lemma 3.2, if one of S 1 , S 2 , T 1 and T 2 is completely continuous, then the sequence { x n } defined by (2.1) converges strongly to a common fixed point of S 1 , S 2 , T 1 and T 2 .

Proof Without loss of generality, we can assume that S 1 is completely continuous. Since { x n } is bounded by Lemma 3.1, there exists a subsequence { S 1 x n j } of { S 1 x n } such that { S 1 x n j } converges strongly to some q . Moreover, we know that

lim j x n j S 1 x n j = lim j x n j S 2 x n j = 0

and

lim j x n j T 1 x n j = lim j x n j T 2 x n j = 0

by Lemma 3.2, which imply that

x n j q x n j S 1 x n j + S 1 x n j q 0

as j and so x n j q K . Thus, by the continuity of S 1 , S 2 , T 1 and T 2 , we have

q S i q = lim j x n j S i x n j = 0

and

q T i q = lim j x n j T i x n j = 0

for i = 1 , 2 . Thus it follows that q F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) . Furthermore, since lim n x n q exists by Lemma 3.1, we have lim n x n q = 0 . This completes the proof. □

Theorem 3.2Under the assumptions of Lemma 3.2, if one of S 1 , S 2 , T 1 and T 2 is semi-compact, then the sequence { x n } defined by (2.1) converges strongly to a common fixed point of S 1 , S 2 , T 1 and T 2 .

Proof Since lim n x n S i x n = lim n x n T i x n = 0 for i = 1 , 2 by Lemma 3.2 and one of S 1 , S 2 , T 1 and T 2 is semi-compact, there exists a subsequence { x n j } of { x n } such that { x n j } converges strongly to some q K . Moreover, by the continuity of S 1 , S 2 , T 1 and T 2 , we have q S i q = lim j x n j S i x n j = 0 and q T i q = lim j x n j T i x n j = 0 for i = 1 , 2 . Thus it follows that q F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) . Since lim n x n q exists by Lemma 3.1, we have lim n x n q = 0 . This completes the proof. □

Theorem 3.3Under the assumptions of Lemma 3.2, if there exists a nondecreasing function f : [ 0 , ) [ 0 , ) with f ( 0 ) = 0 and f ( r ) > 0 for all r ( 0 , ) such that

f ( d ( x , F ) ) x S 1 x + x S 2 x + x T 1 x + x T 2 x

for all x K , where F = F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) , then the sequence { x n } defined by (2.1) converges strongly to a common fixed point of S 1 , S 2 , T 1 and T 2 .

Proof Since lim n x n S i x n = lim n x n T i x n = 0 for i = 1 , 2 by Lemma 3.2, we have lim n f ( d ( x n , F ) ) = 0 . Since f : [ 0 , ) [ 0 , ) is a nondecreasing function satisfying f ( 0 ) = 0 , f ( r ) > 0 for all r ( 0 , ) and lim n d ( x n , F ) exists by Lemma 3.1, we have lim n d ( x n , F ) = 0 .

Now, we show that { x n } is a Cauchy sequence in K. In fact, from (3.2), we have

x n + 1 q [ 1 + ( h n 2 1 ) ] x n q

for each n 1 , where h n = max { k n ( 1 ) , k n ( 2 ) , l n ( 1 ) , l n ( 2 ) } and q F . For any m, n, m > n 1 , we have

x m q [ 1 + ( h m 1 2 1 ) ] x m 1 q e h m 1 2 1 x m 1 q e h m 1 2 1 e h m 2 2 1 x m 2 q e i = n m 1 ( h i 2 1 ) x n q M x n q ,

where M = e i = 1 ( h i 2 1 ) . Thus, for any q F , we have

x n x m x n q + x m q ( 1 + M ) x n q .

Taking the infimum over all q F , we obtain

x n x m ( 1 + M ) d ( x n , F ) .

Thus it follows from lim n d ( x n , F ) = 0 that { x n } is a Cauchy sequence. Since K is a closed subset of E, the sequence { x n } converges strongly to some q K . It is easy to prove that F ( S 1 ) , F ( S 2 ) , F ( T 1 ) and F ( T 2 ) are all closed and so F is a closed subset of K. Since lim n d ( x n , F ) = 0 , q F , the sequence { x n } converges strongly to a common fixed point of S 1 , S 2 , T 1 and T 2 . This completes the proof. □

### 4 Weak convergence theorems

In this section, we prove weak convergence theorems for the iterative scheme defined by (2.1) in uniformly convex Banach spaces.

Lemma 4.1Under the assumptions of Lemma 3.1, for all q 1 , q 2 F = F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) , the limit

lim n t x n + ( 1 t ) q 1 q 2

exists for all t [ 0 , 1 ] , where { x n } is the sequence defined by (2.1).

Proof Set a n ( t ) = t x n + ( 1 t ) q 1 q 2 . Then lim n a n ( 0 ) = q 1 q 2 and, from Lemma 3.1, lim n a n ( 1 ) = lim n x n q 2 exists. Thus it remains to prove Lemma 4.1 for any t ( 0 , 1 ) .

Define the mapping G n : K K by

G n x = P [ ( 1 α n ) S 1 n x + α n T 1 ( P T 1 ) n 1 P ( ( 1 β n ) S 2 n x + β n T 2 ( P T 2 ) n 1 x ) ]

for all x K . It is easy to prove that

G n x G n y h n 4 x y (4.1)

for all x , y K , where h n = max { k n ( 1 ) , k n ( 2 ) , l n ( 1 ) , l n ( 2 ) } . Letting h n = 1 + v n , it follows from 1 j = n h j 4 e 4 j = n v j and n = 1 v n < that lim n j = n h j 4 = 1 . Setting

S n , m = G n + m 1 G n + m 2 G n (4.2)

for each m 1 , from (4.1) and (4.2), it follows that

S n , m x S n , m y ( j = n n + m 1 h j 4 ) x y

for all x , y K and S n , m x n = x n + m , S n , m q = q for any q F . Let

b n , m = t S n , m x n + ( 1 t ) S n , m q 1 S n , m ( t x n + ( 1 t ) q 1 ) . (4.3)

Then, using (4.3) and Lemma 2.4, we have

b n , m ( j = n n + m 1 h j 4 ) γ 1 ( x n q 1 ( j = n n + m 1 h j 4 ) 1 S n , m x n S n , m q 1 ) ( j = n h j 4 ) γ 1 ( x n q 1 ( j = n h j 4 ) 1 x n + m q 1 ) .

It follows from Lemma 3.1 and lim n j = n h j 4 = 1 that lim n b n , m = 0 uniformly for all m. Observe that

a n + m ( t ) S n , m ( t x n + ( 1 t ) q 1 ) q 2 + b n , m = S n , m ( t x n + ( 1 t ) q 1 ) S n , m q 2 + b n , m ( j = n n + m 1 h j 4 ) t x n + ( 1 t ) q 1 q 2 + b n , m ( j = n h j 4 ) a n ( t ) + b n , m .

Thus we have lim sup n a n ( t ) lim inf n a n ( t ) , that is, lim n t x n + ( 1 t ) q 1 q 2 exists for all t ( 0 , 1 ) . This completes the proof. □

Lemma 4.2Under the assumptions of Lemma 3.1, ifEhas a Fréchet differentiable norm, then, for all q 1 , q 2 F = F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) , the limit

lim n x n , j ( q 1 q 2 )

exists, where { x n } is the sequence defined by (2.1). Furthermore, if W w ( { x n } ) denotes the set of all weak subsequential limits of { x n } , then x y , j ( q 1 q 2 ) = 0 for all q 1 , q 2 F and x , y W w ( { x n } ) .

Proof This follows basically as in the proof of Lemma 3.2 of [12] using Lemma 4.1 instead of Lemma 3.1 of [12]. □

Theorem 4.1Under the assumptions of Lemma 3.2, ifEhas a Fréchet differentiable norm, then the sequence { x n } defined by (2.1) converges weakly to a common fixed point of S 1 , S 2 , T 1 and T 2 .

Proof Since E is a uniformly convex Banach space and the sequence { x n } is bounded by Lemma 3.1, there exists a subsequence { x n k } of { x n } which converges weakly to some q K . By Lemma 3.2, we have

lim k x n k S i x n k = lim k x n k T i x n k = 0

for i = 1 , 2 . It follows from Lemma 2.3 that q F = F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) .

Now, we prove that the sequence { x n } converges weakly to q. Suppose that there exists a subsequence { x m j } of { x n } such that { x m j } converges weakly to some q 1 K . Then, by the same method given above, we can also prove that q 1 F . So, q , q 1 F W w ( { x n } ) . It follows from Lemma 4.2 that

q q 1 2 = q q 1 , j ( q q 1 ) = 0 .

Therefore, q 1 = q , which shows that the sequence { x n } converges weakly to q. This completes the proof. □

Theorem 4.2Under the assumptions of Lemma 3.2, if the dual space E ofEhas the Kadec-Klee property, then the sequence { x n } defined by (2.1) converges weakly to a common fixed point of S 1 , S 2 , T 1 and T 2 .

Proof Using the same method given in Theorem 4.1, we can prove that there exists a subsequence { x n k } of { x n } which converges weakly to some q F = F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) .

Now, we prove that the sequence { x n } converges weakly to q. Suppose that there exists a subsequence { x m j } of { x n } such that { x m j } converges weakly to some q K . Then, as for q, we have q F . It follows from Lemma 4.1 that the limit

lim n t x n + ( 1 t ) q q

exists for all t [ 0 , 1 ] . Again, since q , q W w ( { x n } ) , q = q by Lemma 2.5. This shows that the sequence { x n } converges weakly to q. This completes the proof. □

Theorem 4.3Under the assumptions of Lemma 3.2, ifEsatisfies Opial’s condition, then the sequence { x n } defined by (2.1) converges weakly to a common fixed point of S 1 , S 2 , T 1 and T 2 .

Proof Using the same method as given in Theorem 4.1, we can prove that there exists a subsequence { x n k } of { x n } which converges weakly to some q F = F ( S 1 ) F ( S 2 ) F ( T 1 ) F ( T 2 ) .

Now, we prove that the sequence { x n } converges weakly to q. Suppose that there exists a subsequence { x m j } of { x n } such that { x m j } converges weakly to some q ¯ K and q ¯ q . Then, as for q, we have q ¯ F . Using Lemma 3.1, we have the following two limits exist:

lim n x n q = c , lim n x n q ¯ = c 1 .

Thus, by Opial’s condition, we have

c = lim sup k x n k q < lim sup k x n k q ¯ = lim sup j x m j q ¯ < lim sup j x m j q = c ,

which is a contradiction and so q = q ¯ . This shows that the sequence { x n } converges weakly to q. This completes the proof. □

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors read and approved the final manuscript.

### Acknowledgements

The project was supported by the National Natural Science Foundation of China (Grant Number: 11271282) and the second author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant Number: 2012-0008170).

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