Convergence of perturbed composite implicit iteration process for a finite family of asymptotically nonexpansive mappings

Abstract

In this paper, we introduce a perturbed composite implicit iterative process with errors for a finite family of asymptotically nonexpansive mappings. Under Opial’s condition, semicompact and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}d(x_n,F(T))=0$ conditions, respectively, we prove that this iterative scheme converges weakly or strongly to a common fixed point of a finite family of asymptotically nonexpansive mappings in uniformly convex Banach spaces. The results presented in this paper generalize and improve the corresponding results of Sun (J. Math. Anal. Appl. 286:351-358, 2003), Chang (J. Math. Anal. Appl. 313:273-283, 2006), Gu (J. Math. Anal. Appl. 329:766-776, 2007), Thakur (Appl. Math. Comput. 190:965-973, 007), Rafiq (Rostock. Math. Kolloqu. 62:21-39, 2007) and some others.

MSC:47H9, 47H10.

1 Introduction

Let E be a real Banach space, K be a nonempty convex subset of E. Let $\{T_1,T_2,\dots ,T_N\}$ be a finite family of mappings from K into itself, and $F(T_i)$ be the set of fixed points of $T_i$ ($i\in I=\{1,2,\dots ,N\}$). $F(T)$ denotes the set of common fixed points of $\{T_1,T_2,\dots ,T_N\}$.

Recently, Xu and Ori [1] have introduced an implicit iteration process for a finite family of nonexpansive mappings as follows:

$$x_n={\alpha }_n{x}_{n-1}+(1-{\alpha }_n)T_n{x}_n,\mathrm{\forall }n\ge 1,$$

(1)

where $T_n=T_{n(modN)}$ (here the modN function takes values in I), $\{{\alpha }_n\}$ be a real sequence in $[0,1]$, $x_0$ be an initial point in K.

Sun [2] have extended this iterative process defined by Xu and Ori to a new iterative process for a finite family of asymptotically nonexpansive mappings, which is defined as follows:

$$x_n={\alpha }_n{x}_{n-1}+(1-{\alpha }_n)T_i^k{x}_n,n\ge 1,$$

(2)

where $n=(k-1)N+i$, $i\in I$.

Chang [3] have discussed the convergence of the implicit iteration process with errors for a finite family of asymptotically nonexpansive mappings as follows:

$$x_n={\alpha }_n{x}_{n-1}+(1-{\alpha }_n)T_{i(n)}^{k(n)}{x}_n+u_n,n\ge 1,$$

(3)

where $n=(k(n)-1)N+i(n)$, $i(n)\in I$, and $k(n)\ge 1$ with $k(n)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$. Under the hypotheses ${\sum }_{n=1}^{\mathrm{\infty }}\parallel u_n\parallel <\mathrm{\infty }$ and some appropriate conditions, they proved some results of weak and strong convergence for $\{x_n\}$ defined by (3). However, the condition ${\sum }_{n=1}^{\mathrm{\infty }}\parallel u_n\parallel <\mathrm{\infty }$ is not too reasonable, because this implies that $\{u_n\}$ are very small for n sufficiently big.

Gu [4] has extended the above implicit iteration processes. A composite implicit iteration process with random errors was introduced as follows:

$$\{\begin{array}{c}{x}_n=(1-{\alpha }_n-{\gamma }_n)x_{n-1}+{\alpha }_n{T}_n{y}_n+{\gamma }_n{u}_n,n\ge 1;\hfill \\ y_n=(1-{\beta }_n-{\delta }_n)x_n+{\beta }_n{T}_n{x}_n+{\delta }_n{v}_n,n\ge 1,\hfill \end{array}$$

(4)

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, $\{{\delta }_n\}$ are four real sequences in $[0,1]$ satisfying ${\alpha }_n+{\gamma }_n\le 1$ and ${\beta }_n+{\delta }_n\le 1$ for all $n\ge 1$, $\{u_n\}$, $\{v_n\}$ are two sequences in K and $x_0$ is an initial point. Some theorems were established on the strong convergence of the composite implicit iteration process defined by (4) for a finite family of mappings in real Banach spaces.

Thakur [5] has improved the composite implicit iteration process defined by (4) as follows:

$$\{\begin{array}{c}{x}_n=(1-{\alpha }_n)x_{n-1}+{\alpha }_n{T}_{i(n)}^{k(n)}{y}_n,n\ge 1;\hfill \\ y_n=(1-{\beta }_n)x_n+{\beta }_n{T}_{i(n)}^{k(n)}{x}_n,n\ge 1.\hfill \end{array}$$

(5)

Some theorems were proved on the weak and strong convergence of the composite implicit iteration process defined by (5) for a finite family of mappings in real uniformly convex Banach spaces.

Rafiq [6] have improved the implicit iterative process. The Mann type implicit iteration process was introduced in Hilbert spaces as follows:

$$x_n={\alpha }_n{x}_{n-1}+(1-{\alpha }_n)T{v}_n,n\ge 1,$$

(6)

where $v_n$ is a perturbation of $x_n$, and satisfy ${\sum }_{n\ge 1}\parallel x_n-v_n\parallel <\mathrm{\infty }$. Moreover, Ciric [7] also did some work in this respect.

Inspired and motivated by the above works, in this paper we will extend and improve the above iterative process to a perturbed composite implicit iterative process for a finite family of asymptotically nonexpansive mappings as follows:

$$\{\begin{array}{c}{x}_n=(1-{\alpha }_n-{\gamma }_n)x_{n-1}+{\alpha }_n{T}_{i(n)}^{k(n)}{y}_n+{\gamma }_n{u}_n,n\ge 1;\hfill \\ y_n=(1-{\beta }_n-{\delta }_n)x_{n-1}+{\beta }_n{T}_{i(n)}^{k(n)}{\tilde{x}}_n+{\delta }_n{v}_n,n\ge 1,\hfill \end{array}$$

(7)

where $n=(k(n)-1)N+i(n)$, $i(n)\in I$, $T_n=T_{n(modN)}$, $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, $\{{\delta }_n\}$ are four real sequences in $[0,1]$ satisfying ${\alpha }_n+{\gamma }_n\le 1$ and ${\beta }_n+{\delta }_n\le 1$ for all $n\ge 1$, $\{u_n\}$, $\{v_n\}$ are two sequences in K and $x_0$ is an initial point. $\{{\tilde{x}}_n\}$ be a sequence in K satisfying ${\sum }_{n\ge 1}\parallel x_n-{\tilde{x}}_n\parallel <\mathrm{\infty }$, which implies that $\parallel x_n-{\tilde{x}}_n\parallel \to 0$ ($n\to \mathrm{\infty }$). Therefore, ${\tilde{x}}_n$ is known as the perturbation of $x_n$, and $\{{\tilde{x}}_n\}$ is known as the perturbed sequence of $\{x_n\}$. This sequence $\{x_n\}$ defined by (7) is said to be the perturbed composite implicit iterative sequence with random errors.

Especially, (I) in the iterative process defined by (7), when ${\beta }_n=0$, ${\delta }_n=0$ for all $n\ge 1$, we have

$$x_n=(1-{\alpha }_n-{\gamma }_n)x_{n-1}+{\alpha }_n{T}_{i(n)}^{k(n)}{x}_{n-1}+{\gamma }_n{u}_n,n\ge 1.$$

(8)

At this time, the perturbed composite implicit iterative sequence generated by (7) becomes a Mann-type iterative sequence with random errors.

(II) In the iterative process defined by (7), when ${\beta }_n=1$, ${\delta }_n=0$ for all $n\ge 1$, we have

$$x_n=(1-{\alpha }_n-{\gamma }_n)x_{n-1}+{\alpha }_n{T}_{i(n)}^{2k(n)}{\tilde{x}}_n+{\gamma }_n{u}_n,n\ge 1.$$

(9)

At this time, the perturbed composite implicit iterative sequence generated by (7) becomes a perturbed implicit iterative sequence with random errors.

(III) In the iterative process defined by (7), when $x_{n-1}={\tilde{x}}_n$ for all $n\ge 1$, we have

$$\{\begin{array}{c}{x}_n=(1-{\alpha }_n-{\gamma }_n)x_{n-1}+{\alpha }_n{T}_{i(n)}^{k(n)}{y}_n+{\gamma }_n{u}_n,n\ge 1;\hfill \\ y_n=(1-{\beta }_n-{\delta }_n)x_{n-1}+{\beta }_n{T}_{i(n)}^{k(n)}{x}_{n-1}+{\delta }_n{v}_n,n\ge 1.\hfill \end{array}$$

(10)

At this time, the perturbed composite implicit iterative sequence generated by (7) becomes an Ishikawa-type iterative sequence with random errors for a finite family of asymptotically nonexpansive mappings $\{T_i,i\in I\}$.

From the above iterative processes defined by (1)-(6) and (8)-(10), we know that the iterative process (7) improves and extends some iterative process introduced by the recent literature. Moreover, we point out that the iterative process, defined by (7), in which it is not necessary to compute the value of the given operator at $x_n$, but compute an approximate point of $x_n$, are particularly useful in the numerical analysis. Therefore, the iterative sequence generated by (7) is better than some implicit iterative sequences at the existent aspect.

The main purpose of this paper is to study the convergence of the perturbed composite implicit iterative sequence $\{x_n\}$ defined by (7) for a finite family of asymptotically nonexpansive mappings under Opial’s condition, semicompact and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}d(x_n,F(T))=0$ conditions, respectively. The results presented in this paper generalized and improve the corresponding results of Sun [2], Chang [3], Gu [4], Thakur [5], Rafiq [6], and some others [1, 7–15].

2 Preliminaries

For the sake of convenience, we first recall some definitions and conclusions.

Definition 2.1 Let K be a closed subset of the real Banach space E and $T:K\to K$ be a mapping.

• 1. T is said to be semicompact, if for any bounded sequence $\{x_n\}$ in K such that $\parallel T{x}_n-x_n\parallel \to 0$ ($n\to \mathrm{\infty }$), then there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $x_{n_i}\to x^{\ast }\in E$;

• 2. T is said to be demiclosed at the origin, if for each sequence $\{x_n\}$ in K, the conditions $x_n\rightharpoonup x_0$ weakly and $T{x}_n\to 0$ strongly imply $T{x}_0=0$;

• 3. T is said to be asymptotically nonexpansive, if there exists a sequence $h_n\in [1,+\mathrm{\infty })$ with ${lim}_{n\to \mathrm{\infty }}{h}_n=1$ such that

  $$\parallel T^{n}x-T^{n}y\parallel \le h_n\parallel x-y\parallel ,\mathrm{\forall }x,y\in K,n\ge 1.$$

  (11)

• 4. Let T is said to be uniformly L-Lipschitizian if there exists a constant $L>0$ such that

  $$\parallel T^{n}x-T^{n}y\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }x,y\in E,n\ge 1.$$

Definition 2.2 [16]

A Banach space X is said to satisfy Opial’s condition if $x_n\rightharpoonup x$ weakly as $n\to \mathrm{\infty }$ and $x\ne y$ imply that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel x_n-x\parallel <{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel x_n-y\parallel$.

Lemma 2.1 Let K be a nonempty subset of E, $T_1,T_2,\dots ,T_N:K\to K$ be N asymptotically nonexpansive mappings. Then

• (i) there exists a sequence $\{h_n\}\subset [1,+\mathrm{\infty })$ with ${lim}_{n\to \mathrm{\infty }}{h}_n=1$ such that

  $$\parallel T_i^{n}x-T_i^{n}y\parallel \le h_n\parallel x-y\parallel ,\mathrm{\forall }x,y\in K,i\in I,n\ge 1;$$

  (12)

• (ii) $\{T_1,T_2,\dots ,T_N\}$ is uniformly Lipschitzian, i.e., there exists a constant L such that

  $$\parallel T_i^{n}x-T_i^{n}y\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }x,y\in K,i\in I,n\ge 1.$$

  (13)

Proof Since $T_1,T_2,\dots ,T_N:K\to K$ are N asymptotically nonexpansive mappings, then for every $i\in I$ and $n\in N$, there exists $h_n^{(i)}\in [1,+\mathrm{\infty })$ with ${lim}_{n\to \mathrm{\infty }}{h}_n^{(i)}=1$ such that

$$\parallel T_i^{n}x-T_i^{n}y\parallel \le h_n^{(i)}\parallel x-y\parallel ,\mathrm{\forall }x,y\in E.$$

Taking $h_n=max\{h_n^{(1)},h_n^{(2)},\dots ,h_n^{(N)}\}$, then $h_n\subset [1,+\mathrm{\infty })$, ${lim}_{n\to \mathrm{\infty }}{h}_n=1$ and (12) holds.

An asymptotically nonexpansive mapping must is a uniformly Lipschitzian mapping. Hence, for every $i\in I$ and $n\in N$, there exists $L_i$ such that

$$\parallel T_i^{n}x-T_i^{n}y\parallel \le L_i\parallel x-y\parallel ,\mathrm{\forall }x,y\in E.$$

Taking $L=max\{L_1,L_2,\dots ,L_N\}$, it is obvious that (13) holds. □

Lemma 2.2 [17]

Let E be a uniformly convex Banach space, K be a nonempty, closed and convex subset of E and $T:K\to K$ be an asymptotically nonexpansive mapping. Then $I-T$ is demi-closed at zero, i.e., for each sequence $\{x_n\}$ in K, if $\{x_n\}$ convergence weakly to $q\in E$ and $\{(I-T)x_n\}$ converges strongly to 0, then $(I-T)q=0$.

Lemma 2.3 [18]

Let E be a Banach space satisfying Opial’s condition, $\{x_n\}$ be a sequence in E. Let $u,v\in E$ be such that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-u\parallel$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-v\parallel$ exist. If $\{x_{n_k}\}$ and $\{x_{n_l}\}$ are two subsequences of $\{x_n\}$ which converge weakly to u and v, respectively, then $u=v$.

Lemma 2.4 [19]

Let E be a uniformly convex Banach space, b, c be two constants with $0<b<c<1$. Suppose that $\{t_n\}$ is a sequence in $[b,c]$ and $\{x_n\}$, $\{y_n\}$ are two sequences in E. Then the conditions ${lim}_{n\to \mathrm{\infty }}\parallel t_n{x}_n+(1-t_n)y_n\parallel =d$, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel x_n\parallel \le d$, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel y_n\parallel \le d$ imply that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$, where d is a nonnegative constant.

Lemma 2.5 [20]

Let $\{a_n\}$, $\{b_n\}$, $\{{\delta }_n\}$ are three sequences of nonnegative real numbers, if there exists $n_0$ such that

$$a_{n+1}\le (1+{\delta }_n)a_n+b_n,\mathrm{\forall }n>n_0,$$

where ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_n<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$. Then

• (i) ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists;

• (ii) ${lim}_{n\to \mathrm{\infty }}{a}_n=0$ whenever ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{a}_n=0$.

Lemma 2.6 Let E be a real Banach space and K be a nonempty closed convex subset of E. Let $T_1,T_2,\dots ,T_N:K\to K$ be N asymptotically nonexpansive mappings with $F(T)={\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Let $\{u_n\}$ and $\{v_n\}$ are two bounded sequences in K. If $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, $\{{\delta }_n\}$ be four real sequences in $[0,1]$ satisfying the following conditions:

• (i) ${\alpha }_n+{\gamma }_n\le 1$ and ${\beta }_n+{\delta }_n\le 1$ for all $n\ge 1$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n=\alpha <1$ or ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n=\beta <1$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}(h_n-1)<\mathrm{\infty }$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {\tilde{x}}_n-x_n\parallel <\mathrm{\infty }$.

Let $\{x_n\}$ be the perturbed composite implicit iterative sequence defined by (7), then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists for all $p\in F(T)$.

Proof Take $p\in F(T)$, it follows from (7) and Lemma 2.1 that

$$\begin{array}{rcl}\parallel x_n-p\parallel & \le & \parallel (1-{\alpha }_n-{\gamma }_n)x_{n-1}+{\alpha }_n{T}_{i(n)}^{k(n)}{y}_n+{\gamma }_n{u}_n-p\parallel \\ \le & (1-{\alpha }_n-{\gamma }_n)\parallel x_{n-1}-p\parallel +{\alpha }_n{h}_n\parallel y_n-p\parallel +{\gamma }_n\parallel u_n-p\parallel \end{array}$$

(14)

and

$$\begin{array}{rcl}\parallel y_n-p\parallel & \le & \parallel (1-{\beta }_n-{\delta }_n)x_{n-1}+{\beta }_n{T}_{i(n)}^{k(n)}{\tilde{x}}_n+{\delta }_n{v}_n-p\parallel \\ \le & (1-{\beta }_n-{\delta }_n)\parallel x_{n-1}-p\parallel +{\beta }_n{h}_n\parallel {\tilde{x}}_n-p\parallel +{\delta }_n\parallel v_n-p\parallel \\ \le & (1-{\beta }_n-{\delta }_n)\parallel x_{n-1}-p\parallel +{\beta }_n{h}_n\parallel {\tilde{x}}_n-x_n\parallel +{\beta }_n{h}_n\parallel x_n-p\parallel +{\delta }_n\parallel v_n-p\parallel .\end{array}$$

(15)

Substituting (15) into (14) and simplifying, we obtain

$$\begin{array}{rcl}(1-{\alpha }_n{\beta }_n{h}_n^2)\parallel x_n-p\parallel & \le & [1-{\alpha }_n-{\gamma }_n+{\alpha }_n{h}_n(1-{\beta }_n-{\delta }_n)]\parallel x_{n-1}-p\parallel \\ +{\alpha }_n{\beta }_n{h}_n^2\parallel {\tilde{x}}_n-x_n\parallel +{\alpha }_n{\delta }_n{h}_n\parallel v_n-p\parallel +{\gamma }_n\parallel u_n-p\parallel .\end{array}$$

(16)

We notice the hypotheses on $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{h_n\}$, by ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n=\alpha <1$, there exists $n_0\in N$ such that

$$1-{\alpha }_n{\beta }_n{h}_n^2\ge 1-{\alpha }_n{h}_n^2\ge \frac{1}{2}(1-\alpha )>0,n\ge n_0.$$

It follows from (16) that for $n\ge n_0$

$$\begin{array}{rcl}\parallel x_n-p\parallel & \le & \frac{1-{\alpha }_n-{\gamma }_n+{\alpha }_n{h}_n(1-{\beta }_n-{\delta }_n)}{1-{\alpha }_n{\beta }_n{h}_n^2}\parallel x_{n-1}-p\parallel \\ +\frac{1}{1-{\alpha }_n{\beta }_n{h}_n^2}({\alpha }_n{\beta }_n{h}_n^2\parallel {\tilde{x}}_n-x_n\parallel +{\alpha }_n{\delta }_n{h}_n\parallel v_n-p\parallel +{\gamma }_n\parallel u_n-p\parallel )\\ \le & [1+\frac{{\alpha }_n{\beta }_n{h}_n^2-{\alpha }_n+{\alpha }_n{h}_n(1-{\beta }_n)}{1-{\alpha }_n{\beta }_n{h}_n^2}]\parallel x_{n-1}-p\parallel \\ +\frac{2}{1-\alpha }({\alpha }_n{\beta }_n{h}_n^2\parallel {\tilde{x}}_n-x_n\parallel +{\alpha }_n{\delta }_n{h}_n\parallel v_n-p\parallel +{\gamma }_n\parallel u_n-p\parallel )\\ \le & \{1+\frac{2}{1-\alpha }[{\alpha }_n{\beta }_n{h}_n(h_n-1)+{\alpha }_n(h_n-1)]\}\parallel x_{n-1}-p\parallel \\ +\frac{2}{1-\alpha }({\alpha }_n{\beta }_n{h}_n^2\parallel {\tilde{x}}_n-x_n\parallel +{\alpha }_n{\delta }_n{h}_n\parallel v_n-p\parallel +{\gamma }_n\parallel u_n-p\parallel ).\end{array}$$

Hence, we have

$$\parallel x_n-p\parallel \le (1+{\theta }_n)\parallel x_{n-1}-p\parallel +{\eta }_n,n\ge n_0,$$

(17)

where

$${\theta }_n=\frac{2}{1-\alpha }[{\alpha }_n{\beta }_n{h}_n(h_n-1)+{\alpha }_n(h_n-1)],n\ge n_0$$

and

$${\eta }_n=\frac{2}{1-\alpha }({\alpha }_n{\beta }_n{h}_n^2\parallel {\tilde{x}}_n-x_n\parallel +{\alpha }_n{\delta }_n{h}_n\parallel v_n-p\parallel +{\gamma }_n\parallel u_n-p\parallel ),n\ge n_0.$$

From condition (iii), it is obvious that ${\sum }_{n=1}^{\mathrm{\infty }}{\theta }_n<\mathrm{\infty }$. In addition, since $\{\parallel u_n\parallel \}$, $\{\parallel v_n\parallel \}$ are all bounded, we deduce that ${\sum }_{n=1}^{\mathrm{\infty }}{\eta }_n<\mathrm{\infty }$ form (iii)-(iv). By virtue of (17) and Lemma 2.5, we obtain that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists. This completes the proof of Lemma 2.6. □

3 Main results and proofs

Theorem 3.1 Let E be a real Banach space and K be a nonempty, closed and convex subset of E. Let $T_1,T_2,\dots ,T_N:K\to K$ be N asymptotically nonexpansive mappings with $F(T)={\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Let $\{u_n\}$ and $\{v_n\}$ are two bounded sequences in K. If $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, $\{{\delta }_n\}$ be four real sequences in $[0,1]$ satisfying the following conditions:

• (i) ${\alpha }_n+{\gamma }_n\le 1$ and ${\beta }_n+{\delta }_n\le 1$ for all $n\ge 1$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$ or ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}(h_n-1)<\mathrm{\infty }$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {\tilde{x}}_n-x_n\parallel <\mathrm{\infty }$.

Then the perturbed composite implicit iterative sequence $\{x_n\}$ defined by (7) converges strongly to a common fixed point of $\{T_1,T_2,\dots ,T_N\}$ if and only if ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}d(x_n,F(T))=0$.

Proof The necessity of Theorem 3.1 is obvious. Now we prove the sufficiency of Theorem 3.1.

For arbitrary $p\in F(T)$, it follows from (17) in Lemma 2.6 that

$$\parallel x_n-p\parallel \le (1+{\theta }_n)\parallel x_{n-1}-p\parallel +{\eta }_n,\mathrm{\forall }n\ge n_0,$$

where ${\sum }_{n=1}^{\mathrm{\infty }}{\theta }_n<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\eta }_n<\mathrm{\infty }$. Hence, we have

$$d(x_n,F(T))\le (1+{\theta }_n)d(x_{n-1},F(T))+{\eta }_n,\mathrm{\forall }n\ge n_0.$$

(18)

It follows from (18) and Lemma 2.5 that limit ${lim}_{n\to \mathrm{\infty }}d(x_n,F(T))$ exists. By the assumption, we have ${lim}_{n\to \mathrm{\infty }}d(x_n,F(T))=0$. Consequently, for any given $\epsilon >0$, there exists a positive integer $N_1$ ($N_1>n_0$) such that

$$d(x_n,F(T))<\frac{\epsilon }{8},\sum _{k=n}^{\mathrm{\infty }}{\eta }_k<\frac{\epsilon }{8},\sum _{k=n}^{\mathrm{\infty }}{\theta }_k<1,\mathrm{\forall }n\ge N_1,$$

and there exists $p_1\in F(T)$ such that $\parallel x_n-p_1\parallel <\epsilon /8$, $\mathrm{\forall }n\ge N_1$. By (18) and the inequality $1+x\le {\text{e}}^x$ ($x\ge 0$), for any $n\ge N_1$ and all $m\ge 1$, we have

$$\begin{array}{rcl}\parallel x_{n+m}-x_n\parallel & \le & exp\{{\theta }_{n+m-1}\}\parallel x_{n+m-1}-p_1\parallel +{\eta }_{n+m-1}+\parallel x_n-p_1\parallel \\ \le & exp\{{\theta }_{n+m-1}+{\theta }_{n+m-2}\}\parallel x_{n+m-2}-p_1\parallel +exp\{{\theta }_{n+m-1}\}{\eta }_{n+m-2}\\ +{\eta }_{n+m-1}+\parallel x_n-p_1\parallel \le \cdots \\ \le & [exp\left\{\sum _{k=n}^{n+m-1}{\theta }_k\right\}+1]\parallel x_n-p_1\parallel +exp\left\{\sum _{k=n}^{n+m-1}{\theta }_k\right\}\sum _{k=n}^{n+m-1}{\eta }_k<\epsilon .\end{array}$$

Hence, $\{x_n\}$ is a Cauchy sequence in E. By the completeness of E, we can assume that $x_n\to x^{\ast }\in K$. Next we prove that $F(T)$ is a close subset of K. Let $\{p_n\}$ is a sequence in $F(T)$ which converges strongly to some p, then we have for any $i\in I$

$$\parallel p-T_{i}p\parallel \le \parallel p-p_n\parallel +\parallel p_n-T_{i}p\parallel \le (1+L)\parallel p-p_n\parallel \to 0(n\to \mathrm{\infty }).$$

Thus, $p\in F(T)$, and $F(T)$ is closed. Since ${lim}_{n\to \mathrm{\infty }}d(x_n,F(T))=0$, then $x^{\ast }\in F(T)$. Consequently, $\{x_n\}$ defined by (7) converges strongly to a common fixed point of $\{T_1,T_2,\dots ,T_N\}$ in K. This completes the proof of Theorem 3.1. □

Theorem 3.2 Let E be a real uniformly convex Banach space satisfying Opial’s condition and K be a nonempty closed convex subset of E. Let $T_1,T_2,\dots ,T_N:K\to K$ be N asymptotically nonexpansive mappings with $F(T)={\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Let $\{u_n\}$ and $\{v_n\}$ are two bounded sequences in K. If $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, $\{{\delta }_n\}$ be four real sequences in $[0,1]$ satisfying the following conditions:

• (i) ${\alpha }_n+{\gamma }_n\le 1$ and ${\beta }_n+{\delta }_n\le 1$ for all $n\ge 1$;

• (ii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1,{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}(h_n-1)<\mathrm{\infty }$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {\tilde{x}}_n-x_n\parallel <\mathrm{\infty }$.

Then the perturbed composite implicit iterative sequence $\{x_n\}$ defined by (7) converges weakly to a common fixed point of $\{T_1,T_2,\dots ,T_N\}$ in K.

Proof First, we prove that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_j{x}_n\parallel =0$ for all $j\in I$.

For any $p\in F(T)$, it follows from Lemma 2.6 that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists. Suppose that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel =d$, we have from (7)

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel & =& \underset{n\to \mathrm{\infty }}{lim}\parallel (1-{\alpha }_n)[x_{n-1}-p+{\gamma }_n(u_n-x_{n-1})]\\ +{\alpha }_n[T_{i(n)}^{k(n)}{y}_n-p+{\gamma }_n(u_n-x_{n-1})]\parallel =d.\end{array}$$

(19)

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel =d$, then $\{x_n\}$ be a bounded sequence. By virtue of the condition (iii) and the boundedness of sequences $\{x_n\}$ and $\{u_n\}$, we have

(20)

It follows from ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {\tilde{x}}_n-x_n\parallel <\mathrm{\infty }$ that ${lim}_{n\to \mathrm{\infty }}\parallel {\tilde{x}}_n-p\parallel ={lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel =d$. We have

(21)

Therefore, by (19), (20), (21), (ii) and Lemma 2.4, we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_{i(n)}^{k(n)}{y}_n-x_{n-1}\parallel =0.$$

Hence,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x_{n-1}\parallel \le \underset{n\to \mathrm{\infty }}{lim}[{\alpha }_n\parallel T_{i(n)}^{k(n)}{y}_n-x_{n-1}\parallel +{\gamma }_n\parallel u_n-x_{n-1}\parallel ]=0,$$

(22)

which implies that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_{n+j}\parallel =0$ for all $j\in I$. On the other hand, we also have

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\parallel T_{i(n)}^{k(n)}{x}_n-x_n\parallel & \le & \underset{n\to \mathrm{\infty }}{lim}[\parallel x_n-x_{n-1}\parallel +\parallel x_{n-1}-T_{i(n)}^{k(n)}{y}_n\parallel +\parallel T_{i(n)}^{k(n)}{y}_n-T_{i(n)}^{k(n)}{x}_n\parallel ]\\ \le & \underset{n\to \mathrm{\infty }}{lim}{h}_n\parallel y_n-x_n\parallel \le \underset{n\to \mathrm{\infty }}{lim}\parallel y_n-x_{n-1}\parallel +\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x_{n-1}\parallel \\ \le & \underset{n\to \mathrm{\infty }}{lim}[{\beta }_n\parallel T_{i(n)}^{k(n)}{\tilde{x}}_n-x_{n-1}\parallel +{\delta }_n\parallel v_n-x_{n-1}\parallel ]\\ \le & \underset{n\to \mathrm{\infty }}{lim}[{\beta }_n\parallel T_{i(n)}^{k(n)}{\tilde{x}}_n-T_{i(n)}^{k(n)}{x}_n\parallel +{\beta }_n\parallel T_{i(n)}^{k(n)}{x}_n-x_{n-1}\parallel ]\\ \le & \underset{n\to \mathrm{\infty }}{lim}[{\beta }_n{h}_n\parallel {\tilde{x}}_n-x_n\parallel +{\beta }_n\parallel T_{i(n)}^{k(n)}{x}_n-x_n\parallel +{\beta }_n\parallel x_{n-1}-x_n\parallel ].\end{array}$$

(23)

It follows from (22), (23), conditions (ii) and (iv) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_{i(n)}^{k(n)}{x}_n-x_n\parallel =0.$$

(24)

Since for each $n>N$, $n=(n-N)(modN)$, $n=(k(n)-1)N+i(n)$, hence $n-N=[(k(n)-1)-1]N+i(n-N)$, i.e. $k(n-N)=k(n)-1$ and $i(n-N)=i(n)$. Therefore, we have

$$\parallel T_n^{k(n)-1}{x}_n-T_{n-N}^{k(n)-1}{x}_{n-N}\parallel =\parallel T_n^{k(n)-1}{x}_n-T_n^{k(n)-1}{x}_{n-N}\parallel \le L\parallel x_n-x_{n-N}\parallel$$

(25)

and

$$\parallel T_{n-N}^{k(n)-1}{x}_{n-N}-x_{n-N}\parallel =\parallel T_{n-N}^{k(n-N)}{x}_{n-N}-x_{n-N}\parallel .$$

(26)

In view of (25) and (26), we have

$$\begin{array}{rcl}\parallel x_{n-1}-T_n{x}_n\parallel & \le & \parallel x_{n-1}-T_n^{k(n)}{x}_n\parallel +\parallel T_n{x}_n-T_n^{k(n)}{x}_n\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +\parallel x_n-T_n^{k(n)}{x}_n\parallel +L\parallel x_n-T_n^{k(n)-1}{x}_n\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +\parallel x_n-T_n^{k(n)}{x}_n\parallel \\ +L(\parallel T_n^{k(n)-1}{x}_n-T_{n-N}^{k(n)-1}{x}_{n-N}\parallel +\parallel T_{n-N}^{k(n)-1}{x}_{n-N}-x_n\parallel )\\ \le & \parallel x_n-x_{n-1}\parallel +\parallel x_n-T_n^{k(n)}{x}_n\parallel \\ +(L^2+L)\parallel x_n-x_{n-N}\parallel +L\parallel T_{n-N}^{k(n-N)}{x}_{n-N}-x_{n-N}\parallel .\end{array}$$

(27)

From (24) and (27), it is obviously that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n-1}-T_n{x}_n\parallel =0$, which implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_n{x}_n\parallel \le \underset{n\to \mathrm{\infty }}{lim}(\parallel x_{n-1}-T_n{x}_n\parallel +\parallel x_n-x_{n-1}\parallel )=0.$$

Consequently, we obtain that for all $i\in I$

$$\begin{array}{rcl}\parallel x_n-T_{n+i}{x}_n\parallel & \le & \parallel x_n-x_{n+i}\parallel +\parallel x_{n+i}-T_{n+i}{x}_{n+i}\parallel +\parallel T_{n+i}{x}_{n+i}-T_{n+i}{x}_n\parallel \\ \le & (1+L)\parallel x_n-x_{n+i}\parallel +\parallel x_{n+i}-T_{n+i}{x}_{n+i}\parallel \to 0(n\to \mathrm{\infty }).\end{array}$$

(28)

By virtue of (28), we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_i{x}_n\parallel =0$ for all $i\in I$.

Since E is uniformly convex, every bounded subset of E is weakly compact. Again since $\{x_n\}$ is a bounded subset in K, there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $\{x_{n_k}\}$ converges weakly to q in K, and ${lim}_{n_k\to \mathrm{\infty }}\parallel x_{n_k}-T_i{x}_{n_k}\parallel =0$ for all $i\in I$. By Lemma 2.2, we have that $(I-T_i)q=0$. Hence, $q\in F(T_i)$ for all $i\in I$. Therefore, $q\in F(T)$.

Next, we prove that $\{x_n\}$ converges weakly to q. Suppose that contrary, then there exists a subsequence $\{x_{n_j}\}$ of $\{x_n\}$ such that $\{x_{n_j}\}$ converges weakly to $q_1\in K$ and $q\ne q_1$. Using the same method, we can prove that $q_1\in F(T)$ and limit ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q_1\parallel$ exists. Without loss generality, we assume that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =d_1$, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q_1\parallel =d_2$, where $d_1$, $d_2$ are two nonnegative constants. By virtue of the Opial’s condition of E, we have

$$\begin{array}{rcl}{d}_1& =& \underset{n_k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-q\parallel <\underset{n_k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-q_1\parallel =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-q_1\parallel \\ =& \underset{n_j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_j}-q_1\parallel <\underset{n_j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_j}-q\parallel =d_1.\end{array}$$

This is contradictory. Hence, $q=q_1$, which implies that $\{x_n\}$ converges weakly to q. The proof of Theorem 3.2 is completed. □

Theorem 3.3 Let E be a real uniformly convex Banach space and K be a nonempty, closed and convex subset of E. Let $T_1,T_2,\dots ,T_N:K\to K$ be N asymptotically nonexpansive mappings with $F(T)={\bigcap }_{i=1}^{n}F(T_i)\ne \mathrm{\varnothing }$ and at least there exists $T_i$ ($i\in I$), it is semicompact. Let $\{u_n\}$ and $\{v_n\}$ are two bounded sequences in K. If $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, $\{{\delta }_n\}$ be four real sequences in $[0,1]$ satisfying the following conditions:

• (i) ${\alpha }_n+{\gamma }_n\le 1$ and ${\beta }_n+{\delta }_n\le 1$ for all $n\ge 1$;

• (ii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}(h_n-1)<\mathrm{\infty }$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {\tilde{x}}_n-x_n\parallel <\mathrm{\infty }$.

Then the perturbed composite implicit iterative sequence $\{x_n\}$ defined by (7) converges strongly to a common fixed point of $\{T_1,T_2,\dots ,T_N\}$ in K.

Proof Without loss of generality, we assume that $T_1$ is semicompact. By Theorem 3.2, we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_1{x}_n\parallel =0$. Hence, there exists a subsequence $\{x_{n_j}\}$ of $\{x_n\}$ such that $\{x_{n_j}\}\to x^{\ast }$ as $j\to \mathrm{\infty }$. Therefore, we have for all $i\in I$

$$\parallel T_i{x}^{\ast }-x^{\ast }\parallel \le \parallel T_i{x}^{\ast }-T_i{x}_{n_j}\parallel +\parallel T_i{x}_{n_j}-x_{n_j}\parallel +\parallel x_{n_j}-x^{\ast }\parallel .$$

(29)

It follows from (29) that $\parallel T_i{x}^{\ast }-x^{\ast }\parallel =0$ for all $i\in I$. This implies that $x^{\ast }\in F(T)$. Therefore, $x^{\ast }$ be a common fixed point of $\{T_i,i\in I\}$. By virtue of Lemma 2.6, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel$ exists. It follows from $x_{n_j}\to x^{\ast }\in E$ that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel =0$. Hence, the perturbed composites implicit iterative sequence $\{x_n\}$ generated by (7) strongly converges to a common fixed point of $\{T_i,i\in I\}$. This completes the proof of Theorem 3.3. □

Corollary 3.4 Let E be a real Banach space and K be a nonempty closed convex subset of E. Let $T_1,T_2,\dots ,T_N:K\to K$ be N asymptotically nonexpansive mappings with $F(T)={\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$ and let $\{u_n\}$ is a bounded sequence in K. If $\{{\alpha }_n\}$, $\{{\gamma }_n\}$ be two real sequences in $[0,1]$ satisfying the following conditions:

• (i) ${\alpha }_n+{\gamma }_n\le 1$ for all $n\ge 1$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}(h_n-1)<\mathrm{\infty }$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {\tilde{x}}_n-x_n\parallel <\mathrm{\infty }$.

Then the perturbed implicit iterative sequence $\{x_n\}$ defined by (9) converges strongly to a common fixed point of $\{T_1,T_2,\dots ,T_N\}$ if and only if ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}d(x_n,F(T))=0$.

Proof It is enough to take ${\beta }_n=1$, ${\delta }_n=0$ for all $n\in \mathbb{N}$ in Theorem 3.1. □

Corollary 3.5 Let E be a real uniformly convex Banach space satisfying Opial’s condition and K be a nonempty closed convex subset of E. Let $T_1,T_2,\dots ,T_N:K\to K$ be N asymptotically nonexpansive mappings with $F(T)={\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$ and let $\{u_n\}$ is a bounded sequence in K. If $\{{\alpha }_n\}$, $\{{\gamma }_n\}$ be two real sequences in $[0,1]$ satisfying the following conditions:

• (i) ${\alpha }_n+{\gamma }_n\le 1$ for all $n\ge 1$;

• (ii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}(h_n-1)<\mathrm{\infty }$.

Then the Mann type iterative sequence $\{x_n\}$ defined by (8) converges weakly to a common fixed point of $\{T_1,T_2,\dots ,T_N\}$ in K.

Proof It is sufficient to take ${\beta }_n={\delta }_n=0$ for all $n\in \mathbb{N}$ in Theorem 3.2. □

Corollary 3.6 Let E be a real uniformly convex Banach space and K be a nonempty closed convex subset of E. Let $T_1,T_2,\dots ,T_N:K\to K$ be N asymptotically nonexpansive mappings with $F(T)={\bigcap }_{i=1}^{n}F(T_i)\ne \mathrm{\varnothing }$ and at least there exists $T_i$ ($i\in I$), it is semicompact. Let $\{u_n\}$ is a bounded sequence in K. If $\{{\alpha }_n\}$, $\{{\gamma }_n\}$ be two real sequences in $[0,1]$ satisfying the following conditions:

• (i) ${\alpha }_n+{\gamma }_n\le 1$ for all $n\ge 1$;

• (ii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}(h_n-1)<\mathrm{\infty }$.

Then the Mann type iterative sequence $\{x_n\}$ defined by (8) converges strongly to a common fixed point of $\{T_1,T_2,\dots ,T_N\}$ in K.

Proof It is enough to take ${\beta }_n={\delta }_n=0$ for all $n\in \mathbb{N}$ in Theorem 3.3. □