Convergence theorems for mixed type asymptotically nonexpansive mappings

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.

1 Introduction

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

$$\parallel T^{n}x-T^{n}y\parallel \le k_n\parallel x-y\parallel$$

(1.1)

for all $x,y\in K$ and $n\ge 1$.

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 $P:E\to K$ be a nonexpansive retraction of E onto K. A nonself-mapping $T:K\to E$ is said to be asymptotically nonexpansive if there exists a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ with $k_n\to 1$ as $n\to \mathrm{\infty }$ such that

$$\parallel T{(PT)}^{n-1}x-T{(PT)}^{n-1}y\parallel \le k_n\parallel x-y\parallel$$

(1.2)

for all $x,y\in K$ and $n\ge 1$.

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:

$$\{\begin{array}{c}{x}_1\in K,\hfill \\ x_{n+1}=P((1-{\alpha }_n)x_n+{\alpha }_n{T}_1{(P{T}_1)}^{n-1}{x}_n)\hfill \end{array}$$

(1.3)

for each $n\ge 1$, where $\{{\alpha }_n\}$ 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:

$$\{\begin{array}{c}{x}_1\in K,\hfill \\ x_{n+1}=P((1-{\alpha }_n)x_n+{\alpha }_n{T}_1{(P{T}_1)}^{n-1}{y}_n),\hfill \\ y_n=P((1-{\beta }_n)x_n+{\beta }_n{T}_2{(P{T}_2)}^{n-1}{x}_n)\hfill \end{array}$$

(1.4)

for each $n\ge 1$, where $T_1,T_2:K\to E$ are two asymptotically nonexpansive nonself-mappings and $\{{\alpha }_n\}$, $\{{\beta }_n\}$ 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 $P:E\to K$ be a nonexpansive retraction of E onto K. Let $S_1,S_2:K\to K$ be two asymptotically nonexpansive self-mappings and $T_1,T_2:K\to E$ be two asymptotically nonexpansive nonself-mappings. Then we define the new iteration scheme of mixed type as follows:

$$\{\begin{array}{c}{x}_1\in K,\hfill \\ x_{n+1}=P((1-{\alpha }_n)S_1^n{x}_n+{\alpha }_n{T}_1{(P{T}_1)}^{n-1}{y}_n),\hfill \\ y_n=P((1-{\beta }_n)S_2^n{x}_n+{\beta }_n{T}_2{(P{T}_2)}^{n-1}{x}_n)\hfill \end{array}$$

(2.1)

for each $n\ge 1$, where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ 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 $F=F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)$ and denote the distance between a point z and a set A in E by $d(z,A)={inf}_{x\in A}\parallel z-x\parallel$.

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

Let E be a real Banach space, $E^{\ast }$ be the dual space of E and $J:E\to 2^{E^{\ast }}$ be the normalized duality mapping defined by

$$J(x)=\{f\in E^{\ast }:〈x,f〉=\parallel x\parallel \parallel f\parallel ,\parallel f\parallel =\parallel x\parallel \}$$

for all $x\in E$, where $〈\cdot ,\cdot 〉$ denotes duality pairing between E and $E^{\ast }$. 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 $P:E\to K$ such that $Px=x$ for all $x\in K$. Every closed convex subset of a uniformly convex Banach space is a retract. A mapping $P:E\to E$ is called a retraction if $P^2=P$. It follows that if a mapping P is a retraction, then $Py=y$ 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, $x_n\to x$ weakly as $n\to \mathrm{\infty }$ implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-y\parallel$$

for all $y\in E$ with $y\ne x$.

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

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

exists and is attained uniformly in $y\in U$.

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

Let K be a nonempty closed subset of a real Banach space E. A nonself-mapping $T:K\to E$ is said to be semi-compact [11] if, for any sequence $\{x_n\}$ in K such that $\parallel x_n-T{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$, there exists a subsequence $\{x_{n_j}\}$ of $\{x_n\}$ such that $\{x_{n_j}\}$ converges strongly to some $x^{\ast }\in K$.

Lemma 2.1 [15]

Let $\{a_n\}$, $\{b_n\}$ and $\{c_n\}$ be three nonnegative sequences satisfying the following condition:

$$a_{n+1}\le (1+b_n)a_n+c_n$$

for each $n\ge n_0$, where $n_0$ is some nonnegative integer, ${\sum }_{n=n_0}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$ and ${\sum }_{n=n_0}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$. Then ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists.

Lemma 2.2 [8]

Let E be a real uniformly convex Banach space and $0<p\le t_n\le q<1$ for each $n\ge 1$. Also, suppose that $\{x_n\}$ and $\{y_n\}$ are two sequences of E such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n\parallel \le r,\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel y_n\parallel \le r,\underset{n\to \mathrm{\infty }}{lim}\parallel t_n{x}_n+(1-t_n)y_n\parallel =r$$

hold for some $r\ge 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$.

Lemma 2.3 [10]

Let E be a real uniformly convex Banach space, K be a nonempty closed convex subset of E and $T:K\to E$ be an asymptotically nonexpansive mapping with a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ and $k_n\to 1$ as $n\to \mathrm{\infty }$. Then $I-T$ is demiclosed at zero, i.e., if $x_n\to x$ weakly and $x_n-T{x}_n\to 0$ strongly, then $x\in F(T)$, where $F(T)$ is the set of fixed points of T.

Lemma 2.4 [16]

Let X be a uniformly convex Banach space and C be a convex subset of X. Then there exists a strictly increasing continuous convex function $\gamma :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $\gamma (0)=0$ such that, for each mapping $S:C\to C$ with a Lipschitz constant $L>0$,

$$\parallel \alpha Sx+(1-\alpha )Sy-S[\alpha x+(1-\alpha )y]\parallel \le L{\gamma }^{-1}(\parallel x-y\parallel -\frac{1}{L}\parallel Sx-Sy\parallel )$$

for all $x,y\in C$ and $0<\alpha <1$.

Lemma 2.5 [16]

Let X be a uniformly convex Banach space such that its dual space $X^{\ast }$ has the Kadec-Klee property. Suppose $\{x_n\}$ is a bounded sequence and $f_1,f_2\in W_w(\{x_n\})$ such that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel \alpha x_n+(1-\alpha )f_1-f_2\parallel$$

exists for all $\alpha \in [0,1]$, 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 3.1 Let E be a real uniformly convex Banach space and K be a nonempty closed convex subset of E. Let $S_1,S_2:K\to K$ be two asymptotically nonexpansive self-mappings with $\{k_n^{(1)}\},\{k_n^{(2)}\}\subset [1,\mathrm{\infty })$ and $T_1,T_2:K\to E$ be two asymptotically nonexpansive nonself-mappings with $\{l_n^{(1)}\},\{l_n^{(2)}\}\subset [1,\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}(k_n^{(i)}-1)<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}(l_n^{(i)}-1)<\mathrm{\infty }$ for $i=1,2$, respectively, and $F=F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be the sequence defined by (2.1), where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two real sequences in $[0,1)$. Then

• (1) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel$ exists for any $q\in F$;

• (2) ${lim}_{n\to \mathrm{\infty }}d(x_n,F)$ exists.

Proof (1) Set $h_n=max\{k_n^{(1)},k_n^{(2)},l_n^{(1)},l_n^{(2)}\}$. For any $q\in F$, it follows from (2.1) that

$$\begin{array}{rcl}\parallel y_n-q\parallel & \le & \parallel (1-{\beta }_n)(S_2^n{x}_n-q)+{\beta }_n(T_2{(P{T}_2)}^{n-1}{x}_n-q)\parallel \\ \le & (1-{\beta }_n)h_n\parallel x_n-q\parallel +{\beta }_n{h}_n\parallel x_n-q\parallel \\ =& h_n\parallel x_n-q\parallel \end{array}$$

(3.1)

and so

$$\begin{array}{rcl}\parallel x_{n+1}-q\parallel & \le & \parallel (1-{\alpha }_n)(S_1^n{x}_n-q)+{\alpha }_n(T_1{(P{T}_1)}^{n-1}{y}_n-q)\parallel \\ \le & (1-{\alpha }_n)h_n\parallel x_n-q\parallel +{\alpha }_n{h}_n\parallel y_n-q\parallel \\ \le & (1-{\alpha }_n)h_n^2\parallel x_n-q\parallel +{\alpha }_n{h}_n^2\parallel x_n-q\parallel \\ =& [1+(h_n^2-1)]\parallel x_n-q\parallel .\end{array}$$

(3.2)

Since ${\sum }_{n=1}^{\mathrm{\infty }}(k_n^{(i)}-1)<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}(l_n^{(i)}-1)<\mathrm{\infty }$ for $i=1,2$, we have ${\sum }_{n=1}^{\mathrm{\infty }}(h_n^2-1)<\mathrm{\infty }$. It follows from Lemma 2.1 that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel$ exists.

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

  $$d(x_{n+1},F)\le [1+(h_n^2-1)]d(x_n,F)$$

for each $n\ge 1$. It follows from ${\sum }_{n=1}^{\mathrm{\infty }}(h_n^2-1)<\mathrm{\infty }$ and Lemma 2.1 that the conclusion (2) holds. This completes the proof. □

Lemma 3.2 Let E be a real uniformly convex Banach space and K be a nonempty closed convex subset of E. Let $S_1,S_2:K\to K$ be two asymptotically nonexpansive self-mappings with $\{k_n^{(1)}\},\{k_n^{(2)}\}\subset [1,\mathrm{\infty })$ and $T_1,T_2:K\to E$ be two asymptotically nonexpansive nonself-mappings with $\{l_n^{(1)}\},\{l_n^{(2)}\}\subset [1,\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}(k_n^{(i)}-1)<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}(l_n^{(i)}-1)<\mathrm{\infty }$ for $i=1,2$, respectively, and $F=F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be the sequence defined by (2.1) and the following conditions hold:

• (a) $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two real sequences in $[ϵ,1-ϵ]$ for some $ϵ\in (0,1)$;

• (b) $\parallel x-T_{i}y\parallel \le \parallel S_{i}x-T_{i}y\parallel$ for all $x,y\in K$ and $i=1,2$.

Then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-S_i{x}_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel x_n-T_i{x}_n\parallel =0$ for $i=1,2$.

Proof Set $h_n=max\{k_n^{(1)},k_n^{(2)},l_n^{(1)},l_n^{(2)}\}$. For any given $q\in F$, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel$ exists by Lemma 3.1. Now, we assume that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =c$. It follows from (3.2) and ${\sum }_{n=1}^{\mathrm{\infty }}(h_n^2-1)<\mathrm{\infty }$ that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel (1-{\alpha }_n)(S_1^n{x}_n-q)+{\alpha }_n(T_1{(P{T}_1)}^{n-1}{y}_n-q)\parallel =c$$

and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel S_1^n{x}_n-q\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{k}_n^{(1)}\parallel x_n-q\parallel =c.$$

Taking lim sup on both sides in (3.1), we obtain ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel y_n-q\parallel \le c$ and so

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel T_1{(P{T}_1)}^{n-1}{y}_n-q\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{l}_n^{(1)}\parallel y_n-q\parallel \le c.$$

Using Lemma 2.2, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel S_1^n{x}_n-T_1{(P{T}_1)}^{n-1}{y}_n\parallel =0.$$

(3.3)

By the condition (b), it follows that

$$\parallel x_n-T_1{(P{T}_1)}^{n-1}{y}_n\parallel \le \parallel S_1^n{x}_n-T_1{(P{T}_1)}^{n-1}{y}_n\parallel$$

and so, from (3.3), we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_1{(P{T}_1)}^{n-1}{y}_n\parallel =0.$$

(3.4)

Since

$$\begin{array}{rcl}\parallel x_n-q\parallel & \le & \parallel x_n-T_1{(P{T}_1)}^{n-1}{y}_n\parallel +\parallel T_1{(P{T}_1)}^{n-1}{y}_n-q\parallel \\ \le & \parallel x_n-T_1{(P{T}_1)}^{n-1}{y}_n\parallel +l_n^{(1)}\parallel y_n-q\parallel .\end{array}$$

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

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_n-q\parallel \ge c$$

by (3.4) and so

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-q\parallel =c.$$

Using (3.1), we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel (1-{\beta }_n)(S_2^n{x}_n-q)+{\beta }_n(T_2{(P{T}_2)}^{n-1}{x}_n-q)\parallel =c.$$

In addition, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel S_2^n{x}_n-q\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{k}_n^{(2)}\parallel x_n-q\parallel =c$$

and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel T_2{(P{T}_2)}^{n-1}{x}_n-q\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{l}_n^{(2)}\parallel x_n-q\parallel =c.$$

It follows from Lemma 2.2 that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel S_2^n{x}_n-T_2{(P{T}_2)}^{n-1}{x}_n\parallel =0.$$

(3.5)

Now, we prove that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_1{x}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_2{x}_n\parallel =0.$$

Indeed, since $\parallel x_n-T_2{(P{T}_2)}^{n-1}{x}_n\parallel \le \parallel S_2^n{x}_n-T_2{(P{T}_2)}^{n-1}{x}_n\parallel$ by the condition (b). It follows from (3.5) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_2{(P{T}_2)}^{n-1}{x}_n\parallel =0.$$

(3.6)

Since $S_2^n{x}_n=P(S_2^n{x}_n)$ and $P:E\to K$ is a nonexpansive retraction of E onto K, we have

$$\parallel y_n-S_2^n{x}_n\parallel \le {\beta }_n\parallel S_2^n{x}_n-T_2{(P{T}_2)}^{n-1}{x}_n\parallel$$

and so

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-S_2^n{x}_n\parallel =0.$$

(3.7)

Furthermore, we have

$$\parallel y_n-x_n\parallel \le \parallel y_n-S_2^n{x}_n\parallel +\parallel S_2^n{x}_n-T_2{(P{T}_2)}^{n-1}{x}_n\parallel +\parallel T_2{(P{T}_2)}^{n-1}{x}_n-x_n\parallel .$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-y_n\parallel =0.$$

(3.8)

Since $\parallel x_n-T_1{(P{T}_1)}^{n-1}{x}_n\parallel \le \parallel S_1^n{x}_n-T_1{(P{T}_1)}^{n-1}{x}_n\parallel$ by the condition (b) and

Using (3.3) and (3.8), we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel S_1^n{x}_n-T_1{(P{T}_1)}^{n-1}{x}_n\parallel =0$$

(3.9)

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_1{(P{T}_1)}^{n-1}{x}_n\parallel =0.$$

(3.10)

It follows from

$$\begin{array}{rcl}\parallel x_{n+1}-S_1^n{x}_n\parallel & =& \parallel P[(1-{\alpha }_n)S_1^n{x}_n+{\alpha }_n{T}_1{(P{T}_1)}^{n-1}{y}_n]-P\left(S_1^n{x}_n\right)\parallel \\ \le & {\alpha }_n\parallel S_1^n{x}_n-T_1{(P{T}_1)}^{n-1}{y}_n\parallel \end{array}$$

and (3.3) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-S_1^n{x}_n\parallel =0.$$

(3.11)

In addition, we have

$$\parallel x_{n+1}-T_1{(P{T}_1)}^{n-1}{y}_n\parallel \le \parallel x_{n+1}-S_1^n{x}_n\parallel +\parallel S_1^n{x}_n-T_1{(P{T}_1)}^{n-1}{y}_n\parallel .$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-T_1{(P{T}_1)}^{n-1}{y}_n\parallel =0.$$

(3.12)

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

$$\parallel S_1^n{x}_n-x_n\parallel \le \parallel S_1^n{x}_n-T_1{(P{T}_1)}^{n-1}{x}_n\parallel +\parallel T_1{(P{T}_1)}^{n-1}{x}_n-x_n\parallel ,$$

we have ${lim}_{n\to \mathrm{\infty }}\parallel S_1^n{x}_n-x_n\parallel =0$. It follows from (3.6) and the inequality

$$\parallel S_1^n{x}_n-T_2{(P{T}_2)}^{n-1}{x}_n\parallel \le \parallel S_1^n{x}_n-x_n\parallel +\parallel x_n-T_2{(P{T}_2)}^{n-1}{x}_n\parallel$$

that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel S_1^n{x}_n-T_2{(P{T}_2)}^{n-1}{x}_n\parallel =0.$$

(3.13)

Since

$$\parallel x_{n+1}-T_2{(P{T}_2)}^{n-1}{y}_n\parallel \le \parallel x_{n+1}-S_1^n{x}_n\parallel +\parallel S_1^n{x}_n-T_2{(P{T}_2)}^{n-1}{x}_n\parallel +l_n^{(2)}\parallel x_n-y_n\parallel ,$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-T_2{(P{T}_2)}^{n-1}{y}_n\parallel =0.$$

(3.14)

Again, since $(P{T}_i){(P{T}_i)}^{n-2}{y}_{n-1}$, $x_n\in K$ 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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_i{(P{T}_i)}^{n-1}{y}_{n-1}-T_i{x}_n\parallel =0$$

(3.16)

for $i=1,2$. Moreover, we have

$$\parallel x_{n+1}-y_n\parallel \le \parallel x_{n+1}-T_1{(P{T}_1)}^{n-1}{y}_n\parallel +\parallel T_1{(P{T}_1)}^{n-1}{y}_n-x_n\parallel +\parallel x_n-y_n\parallel .$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-y_n\parallel =0.$$

(3.17)

In addition, we have

$$\begin{array}{rcl}\parallel x_n-T_i{x}_n\parallel & \le & \parallel x_n-T_i{(P{T}_i)}^{n-1}{x}_n\parallel +\parallel T_i{(P{T}_i)}^{n-1}{x}_n-T_i{(P{T}_i)}^{n-1}{y}_{n-1}\parallel \\ +\parallel T_i{(P{T}_i)}^{n-1}{y}_{n-1}-T_i{x}_n\parallel \\ \le & \parallel x_n-T_i{(P{T}_i)}^{n-1}{x}_n\parallel +max\{\underset{n\ge 1}{sup}{l}_n^{(1)},\underset{n\ge 1}{sup}{l}_n^{(2)}\}\parallel x_n-y_{n-1}\parallel \\ +\parallel T_i{(P{T}_i)}^{n-1}{y}_{n-1}-T_i{x}_n\parallel \end{array}$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_1{x}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_2{x}_n\parallel =0.$$

Finally, we prove that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-S_1{x}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-S_2{x}_n\parallel =0.$$

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

$$\begin{array}{rcl}\parallel x_n-S_i{x}_n\parallel & \le & \parallel x_n-T_i{(P{T}_i)}^{n-1}{x}_n\parallel +\parallel S_i{x}_n-T_i{(P{T}_i)}^{n-1}{x}_n\parallel \\ \le & \parallel x_n-T_i{(P{T}_i)}^{n-1}{x}_n\parallel +\parallel S_i^n{x}_n-T_i{(P{T}_i)}^{n-1}{x}_n\parallel \end{array}$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-S_1{x}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-S_2{x}_n\parallel =0.$$

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 $|\cdot |$ and let $K=[-1,1]$. Define two mappings $S,T:K\to K$ by

$$Tx=\{\begin{array}{cc}-2sin\frac{x}{2},\hfill & \text{if }x\in [0,1],\hfill \\ 2sin\frac{x}{2},\hfill & \text{if }x\in [-1,0),\hfill \end{array}$$

and

$$Sx=\{\begin{array}{cc}x,\hfill & \text{if }x\in [0,1],\hfill \\ -x,\hfill & \text{if }x\in [-1,0).\hfill \end{array}$$

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

$$|Tx-Ty|=2|sin\frac{x}{2}-sin\frac{y}{2}|\le |x-y|.$$

If $x\in [0,1]$ and $y\in [-1,0)$ or $x\in [-1,0)$ and $y\in [0,1]$, then we have

$$\begin{array}{rcl}|Tx-Ty|& =& 2|sin\frac{x}{2}+sin\frac{y}{2}|\\ =& 4|sin\frac{x+y}{4}cos\frac{x-y}{4}|\\ \le & |x+y|\\ \le & |x-y|.\end{array}$$

This implies that T is nonexpansive and so T is an asymptotically nonexpansive mapping with $k_n=1$ for each $n\ge 1$. Similarly, we can show that S is an asymptotically nonexpansive mapping with $l_n=1$ for each $n\ge 1$.

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 $x,y\in [0,1]$. Then we have

$$|x-Ty|=|x+2sin\frac{y}{2}|=|Sx-Ty|.$$

Case 2. Let $x,y\in [-1,0)$. Then we have

$$|x-Ty|=|x-2sin\frac{y}{2}|\le |-x-2sin\frac{y}{2}|=|Sx-Ty|.$$

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

$$|x-Ty|=|x+2sin\frac{y}{2}|\le |-x+2sin\frac{y}{2}|=|Sx-Ty|.$$

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

$$|x-Ty|=|x-2sin\frac{y}{2}|=|Sx-Ty|.$$

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

Theorem 3.1 Under 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 $q^{\ast }$. Moreover, we know that

$$\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-S_1{x}_{n_j}\parallel =\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-S_2{x}_{n_j}\parallel =0$$

and

$$\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-T_1{x}_{n_j}\parallel =\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-T_2{x}_{n_j}\parallel =0$$

by Lemma 3.2, which imply that

$$\parallel x_{n_j}-q^{\ast }\parallel \le \parallel x_{n_j}-S_1{x}_{n_j}\parallel +\parallel S_1{x}_{n_j}-q^{\ast }\parallel \to 0$$

as $j\to \mathrm{\infty }$ and so $x_{n_j}\to q^{\ast }\in K$. Thus, by the continuity of $S_1$, $S_2$, $T_1$ and $T_2$, we have

$$\parallel q^{\ast }-S_i{q}^{\ast }\parallel =\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-S_i{x}_{n_j}\parallel =0$$

and

$$\parallel q^{\ast }-T_i{q}^{\ast }\parallel =\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-T_i{x}_{n_j}\parallel =0$$

for $i=1,2$. Thus it follows that $q^{\ast }\in F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)$. Furthermore, since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q^{\ast }\parallel$ exists by Lemma 3.1, we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q^{\ast }\parallel =0$. This completes the proof. □

Theorem 3.2 Under 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 ${lim}_{n\to \mathrm{\infty }}\parallel x_n-S_i{x}_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel x_n-T_i{x}_n\parallel =0$ 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 $q^{\ast }\in K$. Moreover, by the continuity of $S_1$, $S_2$, $T_1$ and $T_2$, we have $\parallel q^{\ast }-S_i{q}^{\ast }\parallel ={lim}_{j\to \mathrm{\infty }}\parallel x_{n_j}-S_i{x}_{n_j}\parallel =0$ and $\parallel q^{\ast }-T_i{q}^{\ast }\parallel ={lim}_{j\to \mathrm{\infty }}\parallel x_{n_j}-T_i{x}_{n_j}\parallel =0$ for $i=1,2$. Thus it follows that $q^{\ast }\in F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)$. Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q^{\ast }\parallel$ exists by Lemma 3.1, we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q^{\ast }\parallel =0$. This completes the proof. □

Theorem 3.3 Under the assumptions of Lemma  3.2, if there exists a nondecreasing function $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $f(0)=0$ and $f(r)>0$ for all $r\in (0,\mathrm{\infty })$ such that

$$f(d(x,F))\le \parallel x-S_{1}x\parallel +\parallel x-S_{2}x\parallel +\parallel x-T_{1}x\parallel +\parallel x-T_{2}x\parallel$$

for all $x\in K$, where $F=F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)$, 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 ${lim}_{n\to \mathrm{\infty }}\parallel x_n-S_i{x}_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel x_n-T_i{x}_n\parallel =0$ for $i=1,2$ by Lemma 3.2, we have ${lim}_{n\to \mathrm{\infty }}f(d(x_n,F))=0$. Since $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is a nondecreasing function satisfying $f(0)=0$, $f(r)>0$ for all $r\in (0,\mathrm{\infty })$ and ${lim}_{n\to \mathrm{\infty }}d(x_n,F)$ exists by Lemma 3.1, we have ${lim}_{n\to \mathrm{\infty }}d(x_n,F)=0$.

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

$$\parallel x_{n+1}-q\parallel \le [1+(h_n^2-1)]\parallel x_n-q\parallel$$

for each $n\ge 1$, where $h_n=max\{k_n^{(1)},k_n^{(2)},l_n^{(1)},l_n^{(2)}\}$ and $q\in F$. For any m, n, $m>n\ge 1$, we have

$$\begin{array}{rcl}\parallel x_m-q\parallel & \le & [1+(h_{m-1}^2-1)]\parallel x_{m-1}-q\parallel \\ \le & e^{h_{m-1}^2-1}\parallel x_{m-1}-q\parallel \\ \le & e^{h_{m-1}^2-1}{e}^{h_{m-2}^2-1}\parallel x_{m-2}-q\parallel \\ \le & \cdots \\ \le & e^{{\sum }_{i=n}^{m-1}(h_i^2-1)}\parallel x_n-q\parallel \\ \le & M\parallel x_n-q\parallel ,\end{array}$$

where $M=e^{{\sum }_{i=1}^{\mathrm{\infty }}(h_i^2-1)}$. Thus, for any $q\in F$, we have

$$\begin{array}{rcl}\parallel x_n-x_m\parallel & \le & \parallel x_n-q\parallel +\parallel x_m-q\parallel \\ \le & (1+M)\parallel x_n-q\parallel .\end{array}$$

Taking the infimum over all $q\in F$, we obtain

$$\parallel x_n-x_m\parallel \le (1+M)d(x_n,F).$$

Thus it follows from ${lim}_{n\to \mathrm{\infty }}d(x_n,F)=0$ that $\{x_n\}$ is a Cauchy sequence. Since K is a closed subset of E, the sequence $\{x_n\}$ converges strongly to some $q^{\ast }\in K$. 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 ${lim}_{n\to \mathrm{\infty }}d(x_n,F)=0$, $q^{\ast }\in F$, 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 4.1 Under the assumptions of Lemma  3.1, for all $q_1,q_2\in F=F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)$, the limit

$$\underset{n\to \mathrm{\infty }}{lim}\parallel t{x}_n+(1-t)q_1-q_2\parallel$$

exists for all $t\in [0,1]$, where $\{x_n\}$ is the sequence defined by (2.1).

Proof Set $a_n(t)=\parallel t{x}_n+(1-t)q_1-q_2\parallel$. Then ${lim}_{n\to \mathrm{\infty }}{a}_n(0)=\parallel q_1-q_2\parallel$ and, from Lemma 3.1, ${lim}_{n\to \mathrm{\infty }}{a}_n(1)={lim}_{n\to \mathrm{\infty }}\parallel x_n-q_2\parallel$ exists. Thus it remains to prove Lemma 4.1 for any $t\in (0,1)$.

Define the mapping $G_n:K\to K$ by

$$G_{n}x=P[(1-{\alpha }_n)S_1^{n}x+{\alpha }_n{T}_1{(P{T}_1)}^{n-1}P((1-{\beta }_n)S_2^{n}x+{\beta }_n{T}_2{(P{T}_2)}^{n-1}x)]$$

for all $x\in K$. It is easy to prove that

$$\parallel G_{n}x-G_{n}y\parallel \le h_n^4\parallel x-y\parallel$$

(4.1)

for all $x,y\in K$, 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 $1\le {\prod }_{j=n}^{\mathrm{\infty }}{h}_j^4\le e^{4{\sum }_{j=n}^{\mathrm{\infty }}{v}_j}$ and ${\sum }_{n=1}^{\mathrm{\infty }}{v}_n<\mathrm{\infty }$ that ${lim}_{n\to \mathrm{\infty }}{\prod }_{j=n}^{\mathrm{\infty }}{h}_j^4=1$. Setting

$$S_{n,m}=G_{n+m-1}{G}_{n+m-2}\cdots G_n$$

(4.2)

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

$$\parallel S_{n,m}x-S_{n,m}y\parallel \le \left(\prod _{j=n}^{n+m-1}{h}_j^4\right)\parallel x-y\parallel$$

for all $x,y\in K$ and $S_{n,m}{x}_n=x_{n+m}$, $S_{n,m}q=q$ for any $q\in F$. Let

$$b_{n,m}=\parallel t{S}_{n,m}{x}_n+(1-t)S_{n,m}{q}_1-S_{n,m}(t{x}_n+(1-t)q_1)\parallel .$$

(4.3)

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

$$\begin{array}{rcl}{b}_{n,m}& \le & \left(\prod _{j=n}^{n+m-1}{h}_j^4\right){\gamma }^{-1}(\parallel x_n-q_1\parallel -{\left(\prod _{j=n}^{n+m-1}{h}_j^4\right)}^{-1}\parallel S_{n,m}{x}_n-S_{n,m}{q}_1\parallel )\\ \le & \left(\prod _{j=n}^{\mathrm{\infty }}{h}_j^4\right){\gamma }^{-1}(\parallel x_n-q_1\parallel -{\left(\prod _{j=n}^{\mathrm{\infty }}{h}_j^4\right)}^{-1}\parallel x_{n+m}-q_1\parallel ).\end{array}$$

It follows from Lemma 3.1 and ${lim}_{n\to \mathrm{\infty }}{\prod }_{j=n}^{\mathrm{\infty }}{h}_j^4=1$ that ${lim}_{n\to \mathrm{\infty }}{b}_{n,m}=0$ uniformly for all m. Observe that

$$\begin{array}{rcl}{a}_{n+m}(t)& \le & \parallel S_{n,m}(t{x}_n+(1-t)q_1)-q_2\parallel +b_{n,m}\\ =& \parallel S_{n,m}(t{x}_n+(1-t)q_1)-S_{n,m}{q}_2\parallel +b_{n,m}\\ \le & \left(\prod _{j=n}^{n+m-1}{h}_j^4\right)\parallel t{x}_n+(1-t)q_1-q_2\parallel +b_{n,m}\\ \le & \left(\prod _{j=n}^{\mathrm{\infty }}{h}_j^4\right)a_n(t)+b_{n,m}.\end{array}$$

Thus we have ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{a}_n(t)\le {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{a}_n(t)$, that is, ${lim}_{n\to \mathrm{\infty }}\parallel t{x}_n+(1-t)q_1-q_2\parallel$ exists for all $t\in (0,1)$. This completes the proof. □

Lemma 4.2 Under the assumptions of Lemma  3.1, if E has a Fréchet differentiable norm, then, for all $q_1,q_2\in F=F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)$, the limit

$$\underset{n\to \mathrm{\infty }}{lim}〈x_n,j(q_1-q_2)〉$$

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 $〈x^{\ast }-y^{\ast },j(q_1-q_2)〉=0$ for all $q_1,q_2\in F$ and $x^{\ast },y^{\ast }\in W_w(\{x_n\})$.

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 4.1 Under the assumptions of Lemma  3.2, if E has 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 $q\in K$. By Lemma 3.2, we have

$$\underset{k\to \mathrm{\infty }}{lim}\parallel x_{n_k}-S_i{x}_{n_k}\parallel =\underset{k\to \mathrm{\infty }}{lim}\parallel x_{n_k}-T_i{x}_{n_k}\parallel =0$$

for $i=1,2$. It follows from Lemma 2.3 that $q\in F=F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)$.

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 $q_1\in K$. Then, by the same method given above, we can also prove that $q_1\in F$. So, $q,q_1\in F\cap W_w(\{x_n\})$. It follows from Lemma 4.2 that

$${\parallel q-q_1\parallel }^2=〈q-q_1,j(q-q_1)〉=0.$$

Therefore, $q_1=q$, which shows that the sequence $\{x_n\}$ converges weakly to q. This completes the proof. □

Theorem 4.2 Under the assumptions of Lemma  3.2, if the dual space $E^{\ast }$ of E has 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 $q\in F=F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)$.

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 $q^{\ast }\in K$. Then, as for q, we have $q^{\ast }\in F$. It follows from Lemma 4.1 that the limit

$$\underset{n\to \mathrm{\infty }}{lim}\parallel t{x}_n+(1-t)q-q^{\ast }\parallel$$

exists for all $t\in [0,1]$. Again, since $q,q^{\ast }\in W_w(\{x_n\})$, $q^{\ast }=q$ by Lemma 2.5. This shows that the sequence $\{x_n\}$ converges weakly to q. This completes the proof. □

Theorem 4.3 Under the assumptions of Lemma  3.2, if E satisfies 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 $q\in F=F(S_1)\cap F(S_2)\cap F(T_1)\cap F(T_2)$.

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 $\overline{q}\in K$ and $\overline{q}\ne q$. Then, as for q, we have $\overline{q}\in F$. Using Lemma 3.1, we have the following two limits exist:

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-q\parallel =c,\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-\overline{q}\parallel =c_1.$$

Thus, by Opial’s condition, we have

$$c=\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-q\parallel <\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-\overline{q}\parallel =\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{m_j}-\overline{q}\parallel <\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{m_j}-q\parallel =c,$$

which is a contradiction and so $q=\overline{q}$. This shows that the sequence $\{x_n\}$ converges weakly to q. This completes the proof. □