Convergence theorems of a new iteration for two nonexpansive mappings

Abstract

The purpose of this paper is to introduce the following new general implicit iteration scheme for approximating the common fixed points of a pair of nonexpansive mappings in a uniformly convex Banach space: for any $x_0\in C$, the iterative process $\{x_n\}$ defined by $x_n=a_n{x}_{n-1}+b_{n}T{y}_n+c_{n}S{x}_n$, $y_n=a_n^{\prime }{x}_{n-1}+b_n^{\prime }{x}_n+c_n^{\prime }S{x}_{n-1}+d_n^{\prime }T{x}_n$, where $\{a_n\}$, $\{b_n\}$, $\{c_n\}$, $\{a_n^{\prime }\}$, $\{b_n^{\prime }\}$, $\{c_n^{\prime }\}$, $\{d_n^{\prime }\}$ are seven sequences of real numbers satisfying $a_n+b_n+c_n=1$, $a_n^{\prime }+b_n^{\prime }+c_n^{\prime }+d_n^{\prime }=1$, and $T,S:C\to C$ are two nonexpansive mappings. We approximate the common fixed points of these two mappings by weak and strong convergence of the scheme.

1 Introduction

Let C be a nonempty subset of a real Banach space E. A mapping T of C into itself is called nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ holds for all $x,y\in C$. We first recall the following two iterative processes due to Ishikawa [1] and Mann [2], respectively.

(I) Let C be a nonempty convex subset of E and let $T:C\to C$ be a mapping. For any given $x_0\in C$ the sequence $\{x_n\}$ defined by

$$\begin{array}{r}{x}_{n+1}=(1-a_n)x_n+a_{n}T{y}_n,\\ y_n=(1-b_n)x_n+b_{n}T{x}_n,n\ge 0,\end{array}$$

is called the Ishikawa iteration sequence, where $\{a_n\}$ and $\{b_n\}$ are two real sequences in $[0,1]$ satisfying some conditions. In particular, if $b_n=0$ for all $n\ge 0$, then $\{x_n\}$ defined by

$$x_0\in C,x_{n+1}=(1-a_n)x_n+a_{n}T{x}_n,n\ge 0,$$

is called the Mann iteration sequence.

In [3], Liu introduced the concepts of Ishikawa and Mann iterative processes with errors as follows.

(II) For a nonempty subset C of a Banach space E and a mapping $T:C\to C$, the sequence $\{x_n\}$ defined by

$$\begin{array}{r}{x}_0\in C,x_{n+1}=(1-a_n)x_n+a_{n}T{y}_n+u_n,\\ y_n=(1-b_n)x_n+b_{n}T{x}_n+v_n,n\ge 0,\end{array}$$

where $\{u_n\}$ and $\{v_n\}$ are two summable sequences in E. In particular, if $b_n=0$, $v_n=0$, the sequence $\{x_n\}$ is called the Mann iteration sequence with errors.

Unfortunately, the definitions of Liu, which depend on the convergence of the error terms, are against the randomness of errors. Xu [4] studied the following new iteration process.

(III) Let C be a nonempty convex subset of E and let $T:C\to C$ be a mapping. For any given $x_0\in C$ the sequence $\{x_n\}$ defined by

$$\begin{array}{r}{x}_{n+1}=a_n{x}_n+b_{n}T{y}_n+c_n{u}_n,\\ y_n=a_n^{\prime }{x}_n+b_n^{\prime }T{x}_n+c_n^{\prime }{v}_n,n\ge 0,\end{array}$$

is called the Ishikawa iteration sequence with errors. Here $\{u_n\}$ and $\{v_n\}$ are two bounded sequences in C and $\{a_n\}$, $\{b_n\}$, $\{c_n\}$, $\{a_n^{\prime }\}$, $\{b_n^{\prime }\}$, $\{c_n^{\prime }\}$ are six sequences in $[0,1]$ satisfying the following conditions: $a_n+b_n+c_n=1$, $a_n^{\prime }+b_n^{\prime }+c_n^{\prime }=1$, $n\ge 1$. In particular, if $b_n^{\prime }=0$, $c_n^{\prime }=0$, the sequence $\{x_n\}$ is called the Mann iteration sequence with errors. Chidume and Moore [5] studied the above schemes in 1999.

A generalization of Mann and Ishikawa iterative schemes was given by Das and Debata [6] and Takahashi and Tamura [7]. This scheme dealt with two mappings: $x_0\in C$

$$\begin{array}{r}{x}_n=a_n{x}_{n-1}+(1-a_n)T{y}_{n-1},\\ y_{n-1}=b_n{x}_{n-1}+(1-b_n)S{x}_{n-1},n\ge 1.\end{array}$$

Recently Khan and Fukhar [8] considered the above iterative process with bounded errors.

In the present paper, we consider the following scheme:

$$\begin{array}{r}{x}_0\in C,\\ x_n=a_n{x}_{n-1}+b_{n}T{y}_n+c_{n}S{x}_n,\\ y_n=a_n^{\prime }{x}_{n-1}+b_n^{\prime }{x}_n+c_n^{\prime }S{x}_{n-1}+d_n^{\prime }T{x}_n,n\ge 1,\end{array}$$

(1)

where $\{a_n\}$, $\{b_n\}$, $\{c_n\}$, $\{a_n^{\prime }\}$, $\{b_n^{\prime }\}$, $\{c_n^{\prime }\}$, $\{d_n^{\prime }\}$ are seven sequences of real numbers in $[0,1]$ satisfying $a_n+b_n+c_n=1$, $a_n^{\prime }+b_n^{\prime }+c_n^{\prime }+d_n^{\prime }=1$, and $T,S:C\to C$ are two nonexpansive mappings. Recently, some authors discuss similar issues (for example, please refer to [9–11]).

Approximating fixed points is an important subject in the theory of nonexpansive mappings and its applications in numerous applied areas. One is the convergence of iteration schemes constructed through nonexpansive mappings. In this paper, we study the iterative scheme given in (1) for weak and strong convergence for a pair of nonexpansive mappings in a uniformly convex Banach space. Before our discussions, we first recall the following definitions.

A Banach space E is said to satisfy Opial’s condition if whenever $\{x_n\}$ is a sequence in E which converges weakly to x, then

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-y\parallel \text{for all }y\in E,y\ne x.$$

It is well known that every Hilbert space satisfies the Opial condition (see for example [12]).

A mapping T is said to be semicompact (see, e.g., [13]) if for any sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ in C such that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$, there exists a subsequence ${\{x_{n_j}\}}_{j=1}^{\mathrm{\infty }}$ of ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ such that ${\{x_{n_j}\}}_{j=1}^{\mathrm{\infty }}$ converges strongly to some $u\in C$.

A mapping T with domain $D(T)$ and range $R(T)$ in E is said to be demiclosed at a point $p\in E$ if whenever $\{x_n\}$ is as sequence in $D(T)$ such that $\{x_n\}$ converges weakly to $x\in D(T)$ and $\{T{x}_n\}$ converges strongly to p, then $Tx=p$.

We shall make use of the following results.

Lemma 1.1 [14]

Let $\{s_n\}$, $\{t_n\}$ be two sequences of nonnegative real numbers satisfying

$$s_{n+1}\le s_n+t_n\mathrm{\forall }n\ge 1.$$

• (a) If ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n<\mathrm{\infty }$, then ${lim}_{n\to \mathrm{\infty }}{s}_n$ exists.

• (b) If ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n<\mathrm{\infty }$ and $\{s_n\}$ has a subsequence converging to zero, then ${lim}_{n\to \mathrm{\infty }}{s}_n=0$.

Lemma 1.2 [15]

Let E be a uniformly convex Banach space satisfying Opial’s condition and let C be a nonempty closed convex subset of E. Let T be a nonexpansive mapping of C into itself. Then $I-T$ is demiclosed with respect to zero.

Lemma 1.3 [16]

Suppose that E is a uniformly convex Banach space and $0<p\le t_n\le q<1$ for all positive integers n. Also suppose that $\{x_n\}$ and $\{y_n\}$ are two sequences of E such that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel x_n\parallel \le r$, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel y_n\parallel \le r$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel t_n{x}_n+(1-t_n)y_n\parallel =r$ hold for some $r\ge 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$.

2 Main results

We first give the following key lemma.

Lemma 2.1 Let E be a real uniformly convex Banach space and C its nonempty convex subset. Let $T,S:C\to C$ be nonexpansive mappings. Let $\{x_n\}$ be the sequence as defined in (1) with the following conditions:

• (1) $a_n\to 0$, $a_n^{\prime }\to 0$, $b_n^{\prime }\to 0$, as $n\to \mathrm{\infty }$;

• (2) $b_n,c_n,c_n^{\prime },d_n^{\prime }\in [\delta ,1-\delta ]$ for some $\delta \in (0,1)$;

• (3) $c_n^{\prime }+d_n^{\prime }\le \gamma$ for some $\gamma \in (0,1)$.

If $F:=F(T)\cap F(S)\ne \mathrm{\varnothing }$, then we have

• (i) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists for all $p\in F$ and $\{x_n\}$, $\{T{x}_n\}$ and $\{S{x}_n\}$ are all bounded;

• (ii) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0={lim}_{n\to \mathrm{\infty }}\parallel x_n-S{x}_n\parallel$.

Proof For any $p\in F$, we have

$$\begin{array}{rl}\parallel x_n-p\parallel & =\parallel a_n(x_{n-1}-p)+b_n(T{y}_n-p)+c_n(S{x}_n-p)\parallel \\ \le a_n\parallel x_{n-1}-p\parallel +b_n\parallel T{y}_n-p\parallel +c_n\parallel S{x}_n-p\parallel \\ \le a_n\parallel x_{n-1}-p\parallel +b_n\parallel y_n-p\parallel +c_n\parallel x_n-p\parallel .\end{array}$$

(2)

However, as the proof of the above inequality, it follows that

$$\begin{array}{rl}\parallel y_n-p\parallel & =\parallel a_n^{\prime }(x_{n-1}-p)+b_n^{\prime }(x_n-p)+c_n^{\prime }(S{x}_{n-1}-p)+d_n^{\prime }(T{x}_n-p)\parallel \\ \le a_n^{\prime }\parallel x_{n-1}-p\parallel +b_n^{\prime }\parallel x_n-p\parallel +c_n^{\prime }\parallel S{x}_{n-1}-p\parallel +d_n^{\prime }\parallel T{x}_n-p\parallel \\ \le a_n^{\prime }\parallel x_{n-1}-p\parallel +b_n^{\prime }\parallel x_n-p\parallel +c_n^{\prime }\parallel x_{n-1}-p\parallel +d_n^{\prime }\parallel x_n-p\parallel .\end{array}$$

(3)

Thus from (2) and (3), it is easy to check

$$\parallel x_n-p\parallel \le [a_n+b_n(a_n^{\prime }+c_n^{\prime })]\parallel x_{n-1}-p\parallel +[c_n+b_n(b_n^{\prime }+d_n^{\prime })]\parallel x_n-p\parallel ,$$

(4)

which implies that

$$\parallel x_n-p\parallel \le \parallel x_{n-1}-p\parallel ,$$

(5)

and the limit ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists for all $p\in F$. Furthermore $\{x_n\}$ is bounded and $\{S{x}_n\}$ and $\{T{x}_n\}$ are both bounded also. Now suppose ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel =c$ for some $c\ge 0$. By the inequalities (3) and (5), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel y_n-p\parallel \le c.$$

(6)

From the iterative process (1), we have

$$\parallel x_n-p\parallel =\parallel b_n[T{y}_n-p+a_n(x_{n-1}-S{x}_n)]+(1-b_n)[S{x}_n-p+a_n(x_{n-1}-S{x}_n)]\parallel .$$

(7)

Since $a_n\to 0$, $\{x_n\}$, $\{S{x}_n\}$ are both bounded and (6), it follows from Lemma 1.3 that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T{y}_n-S{x}_n\parallel =0.$$

(8)

Next, by the inequality (2) and (5), we have

$$\begin{array}{rl}\parallel x_n-p\parallel & \le a_n\parallel x_{n-1}-p\parallel +b_n\parallel y_n-p\parallel +c_n\parallel x_n-p\parallel \\ \le a_n\parallel x_{n-1}-p\parallel +b_n\parallel y_n-p\parallel +c_n\parallel x_{n-1}-p\parallel \\ =\parallel x_{n-1}-p\parallel +b_n(\parallel y_n-p\parallel -\parallel x_{n-1}-p\parallel ),\end{array}$$

(9)

which means

$$\frac{\parallel x_n-p\parallel -\parallel x_{n-1}-p\parallel }{b_n}+\parallel x_{n-1}-p\parallel \le \parallel y_n-p\parallel .$$

Taking liminf on both sides in the above inequality and by (6) and $b_n\in [\delta ,1-\delta ]$ for some $\delta \in (0,1)$, we have

$$c\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_n-p\parallel \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel y_n-p\parallel \le c,$$

which yields

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

(10)

It is easy to see that

$$\begin{array}{rl}\parallel y_n-p\parallel =& \parallel c_n^{\prime }[S{x}_{n-1}-p+a_n^{\prime }(x_{n-1}-T{x}_n)+b_n^{\prime }(x_n-T{x}_n)]\\ +(1-c_n^{\prime })[T{x}_n-p+a_n^{\prime }(x_{n-1}-T{x}_n)+b_n^{\prime }(x_n-T{x}_n)]\parallel .\end{array}$$

(11)

Since $a_n^{\prime }\to 0$, $b_n^{\prime }\to 0$, $\{S{x}_n\}$ and $\{T{x}_n\}$ are both bounded, then by Lemma 1.3 we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel S{x}_{n-1}-T{x}_n\parallel =0.$$

(12)

Moreover, by the iterative process (1) again, we get

$$\parallel x_n-T{y}_n\parallel \le a_n\parallel x_{n-1}-T{y}_n\parallel +c_n\parallel S{x}_n-T{y}_n\parallel ,$$

which, by $a_n\to 0$ as $n\to \mathrm{\infty }$ and (8), implies that

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

(13)

Hence from (8) and (13), we have

$$\parallel x_n-S{x}_n\parallel \le \parallel x_n-T{y}_n\parallel +\parallel T{y}_n-S{x}_n\parallel \to 0$$

(14)

as $n\to \mathrm{\infty }$. Note that

$$\begin{array}{rl}\parallel y_n-x_n\parallel & =\parallel a_n^{\prime }(x_{n-1}-x_n)+c_n^{\prime }(S{x}_{n-1}-x_n)+d_n^{\prime }(T{x}_n-x_n)\parallel \\ \le a_n^{\prime }\parallel x_{n-1}-x_n\parallel +c_n^{\prime }\parallel S{x}_{n-1}-T{x}_n\parallel +c_n^{\prime }\parallel T{x}_n-x_n\parallel +d_n^{\prime }\parallel T{x}_n-x_n\parallel ;\end{array}$$

(15)

then it follows from the above inequality that

$$\begin{array}{rl}\parallel x_n-T{x}_n\parallel \le & \parallel x_n-S{x}_n\parallel +\parallel S{x}_n-T{y}_n\parallel +\parallel T{y}_n-T{x}_n\parallel \\ \le & \parallel x_n-S{x}_n\parallel +\parallel S{x}_n-T{y}_n\parallel +\parallel y_n-x_n\parallel \\ \le & \parallel x_n-S{x}_n\parallel +\parallel S{x}_n-T{y}_n\parallel +a_n^{\prime }\parallel x_{n-1}-x_n\parallel \\ +c_n^{\prime }\parallel S{x}_{n-1}-T{x}_n\parallel +c_n^{\prime }\parallel T{x}_n-x_n\parallel +d_n^{\prime }\parallel T{x}_n-x_n\parallel .\end{array}$$

(16)

That is to say,

$$\begin{array}{r}\parallel x_n-T{x}_n\parallel \\ \le \frac{1}{1-\gamma }\{\parallel x_n-S{x}_n\parallel +\parallel S{x}_n-T{y}_n\parallel +a_n^{\prime }\parallel x_{n-1}-x_n\parallel +c_n^{\prime }\parallel S{x}_{n-1}-T{x}_n\parallel \},\end{array}$$

(17)

which, from (8), (12), (14), and $a_n^{\prime }\to 0$, means that

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

(18)

The proof of the lemma is completed. □

Now we give the weak convergence first.

Theorem 2.2 Let E be a uniformly convex Banach space satisfying Opial’s condition and C, S, T, and $\{x_n\}$ be as taken in Lemma  2.1. If $F:=F(T)\cap F(S)\ne \mathrm{\varnothing }$, then $\{x_n\}$ converges weakly to a common fixed point of S and T.

Proof Let $p\in F$. As the proof of Lemma 2.1, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists. Since E is uniformly convex, every bounded subset of E is weakly compact, so that there exists a subsequence $\{x_{n_k}\}$ of the bounded sequence $\{x_n\}$ such that $\{x_{n_k}\}$ converges weakly to a point $q\in C$. Therefore, it follows from (ii) in Lemma 2.1 that

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

By Lemma 1.2, we know $I-T$ is demiclosed, then it is easy to see that $q\in F(T)$. With a similar proof, it follows that $q\in F(S)$ also. Next we prove uniqueness. Suppose that this is not true, then there must exist a subsequence $\{x_{n_j}\}\subset \{x_n\}$ such that $\{x_{n_j}\}$ converges weakly to another $q^{\ast }\in C$ and $q^{\ast }\ne q$. Then by the same method given above, we can also prove $q^{\ast }\in F(T)\cap F(S)$.

Because we have proved that for any $p\in F$, the limit ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists, we can let

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-q\parallel =d_1,\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-q^{\ast }\parallel =d_2.$$

By the Opial condition of E, we have

$$\begin{array}{rl}{d}_1& =\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-q\parallel <\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-q^{\ast }\parallel =d_2\\ =\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_j}-q^{\ast }\parallel <\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_j}-q\parallel =d_1.\end{array}$$

This is a contradiction, hence $q=q^{\ast }$. This implies that $\{x_n\}$ converges weakly to a common fixed point of S and T. □

Next we give several strong convergence results.

Theorem 2.3 Let E be a uniformly convex Banach space and C, $\{x_n\}$ be as taken in Lemma  2.1. If one of the nonexpansive mappings T and S is semicompact and $F:=F(T)\cap F(S)\ne \mathrm{\varnothing }$, then $\{x_n\}$ converges strongly to a common fixed point of S and T.

Proof Since one of T and S is semicompact and by (ii) in Lemma 2.1, then there exists a subsequence ${\{x_{n_j}\}}_{j=1}^{\mathrm{\infty }}$ of the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ such that ${\{x_{n_j}\}}_{j=1}^{\mathrm{\infty }}$ converges strongly to u. Since C is closed, $u\in C$. Continuity of S and T gives $\parallel S{x}_{n_j}-Su\parallel \to 0$ and $\parallel T{x}_{n_j}-Tu\parallel \to 0$ as $n_j\to \mathrm{\infty }$. Then by Lemma 2.1,

$$\parallel Su-u\parallel =0=\parallel Tu-u\parallel .$$

This yields $u\in F$. By Lemma 2.1 again, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists for all $p\in F$, therefore $\{x_n\}$ must itself converge to $u\in F$. This completes the proof. □

Recall that a mapping $T:C\to C$, where C is a subset of E, is said to satisfy condition (A) if there exists a nondecreasing function $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $f(0)=0$, $f(r)>0$ for all $r\in (0,\mathrm{\infty })$ such that $\parallel x-Tx\parallel \ge f(d(x,F(T)))$ for all $x\in C$ where $d(x,F(T))=inf\{\parallel x-x^{\ast }\parallel :x^{\ast }\in F(T)\}$.

In [17], Senter and Dotson approximated fixed points of a nonexpansive mapping T by Mann iteration. Recently Maiti and Ghosh [18] and Tan and Xu [14] considered the approximation of fixed points of a nonexpansive mapping T by Ishikawa iteration under the same condition (A) which is weaker than the requirement that T is semicompact. Khan and Fukhar [8] modified this condition for two mappings $S,T:C\to C$ as follows.

Let C be a subset of a Banach space E. Two mappings $S,T:C\to C$ are said to satisfy condition ($A^{\prime }$) if there exists a nondecreasing function $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $f(0)=0$, $f(r)>0$ for all $r\in (0,\mathrm{\infty })$ such that $\frac{1}{2}(\parallel x-Tx\parallel +\parallel x-Sx\parallel )\ge f(d(x,F))$ for all $x\in K$, where $d(x,F)=inf\{\parallel x-x^{\ast }\parallel :x^{\ast }\in F=F(T)\cap F(S)\}$.

Note that condition ($A^{\prime }$) reduces to condition (A) when $S=T$. We use condition ($A^{\prime }$) to study the strong convergence of $\{x_n\}$ defined in (1).

Theorem 2.4 Let E be a uniformly convex Banach space and C, $\{x_n\}$ be as taken in Lemma  2.1. Let T, S satisfy the condition ($A^{\prime }$) and $F:=F(T)\cap F(S)\ne \mathrm{\varnothing }$, then $\{x_n\}$ converges strongly to a common fixed point of S and T.

Proof By Lemma 2.1, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel$ exists for all $x^{\ast }\in F=F(T)\cap F(S)$. Let it be c for some $c\ge 0$. If $c=0$, there is nothing to prove. Suppose $c>0$. By Lemma 2.1, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0={lim}_{n\to \mathrm{\infty }}\parallel x_n-S{x}_n\parallel$. Moreover, from (4), we have $\parallel x_n-x^{\ast }\parallel \le \parallel x_{n-1}-x^{\ast }\parallel$ which gives

$$\underset{x^{\ast }\in F}{inf}\parallel x_n-x^{\ast }\parallel \le \underset{x^{\ast }\in F}{inf}\parallel x_n-x^{\ast }\parallel .$$

That is to say, $d(x_n,F)\le d(x_{n-1},F)$ shows that ${lim}_{n\to \mathrm{\infty }}d(x_n,F)$ exists by virtue of Lemma 1.1. Now by condition ($A^{\prime }$), ${lim}_{n\to \mathrm{\infty }}f(d(x_n,F))=0$. By the properties of f, therefore ${lim}_{n\to \mathrm{\infty }}d(x_n,F)=0$. Next we can take a subsequence $\{x_{n_j}\}$ of $\{x_n\}$ and $\{y_j\}\subset F$ such that $\parallel x_{n_j}-y_j\parallel <2^{-j}$. Then following the method of proof of Tan and Xu [14], we find that $\{y_j\}$ is a Cauchy sequence in F and so it converges. Let $y_j\to y$. Since F is closed, therefore $y\in F$ and then $x_{n_j}\to y$. As ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel$ exists, $x_n\to y\in F$. This completes the proof. □

Remark 2.5 From the proof of the above results, it is easy to see that we can extend our theorems to the iterative process (1) with errors as follows: let C be a bounded closed convex subset of E and the sequence $\{x_n\}$ be defined by

$$\begin{array}{r}{x}_0\in C,\\ x_n=a_n{x}_{n-1}+b_{n}T{y}_n+c_{n}S{x}_n+e_n{u}_n,\\ y_n=a_n^{\prime }{x}_{n-1}+b_n^{\prime }{x}_n+c_n^{\prime }S{x}_{n-1}+d_n^{\prime }T{x}_n+e_n^{\prime }{v}_n\end{array}$$

(19)

where $\{a_n\}$, $\{b_n\}$, $\{c_n\}$, $\{e_n\}$, $\{a_n^{\prime }\}$, $\{b_n^{\prime }\}$, $\{c_n^{\prime }\}$, $\{d_n^{\prime }\}$, $\{e_n^{\prime }\}$, are sequences in $[0,1]$ with $a_n+b_n+c_n+e_n=1$, $a_n^{\prime }+b_n^{\prime }+c_n^{\prime }+d_n^{\prime }+e_n^{\prime }=1$ and $T,S:C\to C$ are both nonexpansive mappings, $\{u_n\},\{v_n\}\in C$.