The mean ergodic theorem for nonexpansive mappings in multi-Banach spaces

Abstract

In this paper, we prove a mean ergodic theorem for nonexpansive mappings in multi-Banach spaces.

MSC:39A10, 39B72, 47H10, 46B03.

1 Introduction

Let X be a Banach space and C be a closed convex subset of X. For each $j\ge 1$, a mapping $T_j:C\to C$ is said to be nonexpansive on C if

$$\parallel T_{j}x-T_{j}y\parallel \le \parallel x-y\parallel$$

for all $x,y\in C$. For each $j\ge 1$, let $F(T_j)$ be the set of fixed points of $T_j$. If X is a strictly convex Banach space, then $F(T_j)$ is closed and convex.

In [1], Baillon proved the first nonlinear ergodic theorem such that, if X is a real Hilbert space and $F(T_j)\ne \mathrm{\varnothing }$ for each $j\ge 1$, then, for each $x\in C$, the sequence $\{S_{n,j}x\}$ defined by

$$S_{n,j}x=\frac{1}{n}(x+T_{j}x+\cdots +T_j^{n-1}x)$$

converges weakly to a fixed point of $T_j$. It was also shown by Pazy [2] that, if X is a real Hilbert space and $S_{n,j}x$ converges weakly to $y\in C$, then $y\in F(T)$. These results were extended by Baillon [3], Bruck [4] and Reich [5, 6] and [7].

2 Multi-Banach spaces

The notion of a multi-normed space was introduced by Dales and Polyakov in [8]. This concept is somewhat similar to an operator sequence space and has some connections with the operator spaces and Banach lattices. Observations on multi-normed spaces and examples are given in [8–10].

Let $(E,\parallel \cdot \parallel )$ be a complex normed space and let $k\in \mathbb{N}$. We denote by $E^k$ the linear space $E\oplus \cdots \oplus E$ consisting of k-tuples $(x_1,\dots ,x_k)$, where $x_1,\dots ,x_k\in E$. The linear operations on $E^k$ are defined coordinate-wise. The zero element of either E or $E^k$ is denoted by 0. We denote by ${\mathbb{N}}_k$ the set $\{1,2,\dots ,k\}$ and by ${\mathrm{\Sigma }}_k$ the group of permutations on k symbols.

Definition 2.1 A multi-norm on $\{E^k:k\in \mathbb{N}\}$ is a sequence ${\{{\parallel \cdot \parallel }_k\}}_{k\in \mathbb{N}}$ such that ${\parallel \cdot \parallel }_k$ is a norm on $E^k$ for each $k\in \mathbb{N}$ with $k\ge 2$ satisfying the following conditions:

(A1) ${\parallel (x_{\sigma (1)},\dots ,x_{\sigma (k)})\parallel }_k={\parallel (x_1,\dots ,x_k)\parallel }_k$ ($\sigma \in {\mathrm{\Sigma }}_k$, $x_1,\dots ,x_k\in E$);

(A2) ${\parallel ({\alpha }_1{x}_1,\dots ,{\alpha }_k{x}_k)\parallel }_k\le ({max}_{i\in {\mathbb{N}}_k}|{\alpha }_i|){\parallel (x_1,\dots ,x_k)\parallel }_k$ (${\alpha }_1,\dots ,{\alpha }_k\in \mathbb{C}$, $x_1,\dots ,x_k\in E$);

(A3) ${\parallel (x_1,\dots ,x_{k-1},0)\parallel }_k={\parallel (x_1,\dots ,x_{k-1})\parallel }_{k-1}$ ($x_1,\dots ,x_{k-1}\in E$);

(A4) ${\parallel (x_1,\dots ,x_{k-1},x_{k-1})\parallel }_k={\parallel (x_1,\dots ,x_{k-1})\parallel }_{k-1}$ ($x_1,\dots ,x_{k-1}\in E$).

In this case, we say that ${\{(E^k,{\parallel \cdot \parallel }_k)\}}_{k\in \mathbb{N}}$ is a multi-normed space.

Lemma 2.2 ([10])

Suppose that ${\{(E^k,{\parallel \cdot \parallel }_k)\}}_{k\in \mathbb{N}}$ is a multi-normed space and take $k\in \mathbb{N}$. Then we have the following:

• (1) ${\parallel (x,\dots ,x)\parallel }_k=\parallel x\parallel$ ($x\in E$);

• (2) ${max}_{i\in {\mathbb{N}}_k}\parallel x_i\parallel \le {\parallel x_1,\dots ,x_k\parallel }_k\le {\sum }_{i=1}^k\parallel x_i\parallel \le k{max}_{i\in {\mathbb{N}}_k}\parallel x_i\parallel$ ($x_1,\dots ,x_k\in E$).

It follows from (2) that, if $(E,\parallel \cdot \parallel )$ is a Banach space, then $(E^k,{\parallel \cdot \parallel }_k)$ is a Banach space for each $k\in \mathbb{N}$. In this case ${\{(E^k,{\parallel \cdot \parallel }_k)\}}_{k\in \mathbb{N}}$ is a multi-Banach space.

Now, we give two important examples of multi-norms for an arbitrary normed space E [8].

Example 2.3 The sequence ${\{{\parallel \cdot \parallel }_k\}}_{k\in \mathbb{N}}$ on $\{E^k:k\in \mathbb{N}\}$ defined by

$${\parallel (x_1,\dots ,x_k)\parallel }_k:=\underset{i\in {\mathbb{N}}_k}{max}\parallel x_i\parallel (x_1,\dots ,x_k\in E)$$

is a multi-norm, which is called the minimum multi-norm. The terminology ‘minimum’ is justified by the property (2).

Example 2.4 Let $\{({\parallel \cdot \parallel }_k^{\alpha }:k\in \mathbb{N}):\alpha \in A\}$ be the (nonempty) family of all multi-norms on $\{E^k:k\in \mathbb{N}\}$. For each $k\in \mathbb{N}$, set

$${\parallel (x_1,\dots ,x_k)\parallel }_k:=\underset{\alpha \in A}{sup}{\parallel (x_1,\dots ,x_k)\parallel }_k^{\alpha }(x_1,\dots ,x_k\in E).$$

Then ${\{{\parallel \cdot \parallel }_k\}}_{k\in \mathbb{N}}$ is a multi-norm on $\{E^k:k\in \mathbb{N}\}$, which is called the maximum multi-norm.

We need the following observation, which can easily be deduced from the triangle inequality for the norm ${\parallel \cdot \parallel }_k$ and the property (2) of multi-norms.

Lemma 2.5 Suppose that $k\in \mathbb{N}$ and $(x_1,\dots ,x_k)\in E^k$. For each $j\in \{1,\dots ,k\}$, let ${\{x_n^j\}}_{n\ge 1}$ be a sequence in E such that ${lim}_{n\to \mathrm{\infty }}{x}_n^j=x_j$. Then, for each $(y_1,\dots ,y_k)\in E^k$, we have

$$\underset{n\to \mathrm{\infty }}{lim}(x_n^1-y_1,\dots ,x_n^k-y_k)=(x_1-y_1,\dots ,x_k-y_k).$$

Definition 2.6 Let ${\{(E^k,{\parallel \cdots \parallel }_k)\}}_{k\in \mathbb{N}}$ be a multi-normed space. A sequence ${\{x_n\}}_{n\ge 1}$ in E is called a multi-null sequence if, for any $\epsilon >0$, there exists $n_0\in \mathbb{N}$ such that

$$\underset{k\in \mathbb{N}}{sup}{\parallel (x_n,\dots ,x_{n+k-1})\parallel }_k<\epsilon (n\ge n_0).$$

Let $x\in E$. We say that the sequence ${\{x_n\}}_{n\ge 1}$ is multi-convergent to a point $x\in E$ and write

$$\underset{n\to \mathrm{\infty }}{lim}{x}_n=x$$

if ${\{x_n-x\}}_n$ is a multi-null sequence.

3 Main results

To prove the main results in this paper, first, we introduce some lemmas.

Lemma 3.1 ([11])

Let ${\{(X^j,{\parallel \cdot \parallel }_j)\}}_{j\in \mathbb{N}}$ be a uniformly convex multi-Banach space with modulus of the convexity δ. Let $x_j,y_j\in X$. If ${\parallel (x_1,\dots ,x_j)\parallel }_j\le r$, ${\parallel (y_1,\dots ,y_j)\parallel }_j\le r$, $r\le R$ and ${\parallel (x_1-y_1,\dots ,x_j-y_j)\parallel }_j\ge ϵ>0$, then

$${\parallel (\lambda x_1+(1-\lambda )y_1,\dots ,\lambda x_j+(1-\lambda )y_j)\parallel }_j\le r(1-2\lambda (1-\lambda ){\delta }_R(ϵ))$$

for all $\lambda \in [0,1]$, where ${\delta }_R(ϵ)=\delta (\frac{ϵ}{R})$.

To proceed, let ${\{(X^j,{\parallel \cdot \parallel }_j)\}}_{j\in \mathbb{N}}$ denote a uniformly convex multi-Banach space with modulus of the convexity δ.

Lemma 3.2 Let C be a closed convex subset of X and for each $j\ge 1$, $T_j:C\to C$ be a nonexpansive mapping. Let $x\in C$, $f_j\in F(T_j)$ for each $j\ge 1$ and $0<\alpha \le \beta <1$. Then, for any $ϵ>0$, there exists $N>0$ such that, for all $n\ge N$,

$$\begin{array}{c}\parallel (T_1^k(\lambda T_1^{n}x+(1-\lambda )f_1)-(\lambda T_1^{n+k}x+(1-\lambda )f_1),\hfill \\ \dots ,T_j^k(\lambda T_j^{n}x+(1-\lambda )f_j)-(\lambda T_j^{n+k}x+(1-\lambda )f_j)){\parallel }_j\hfill \\ <ϵ\hfill \end{array}$$

for all $k>0$ and $\lambda \in [\alpha ,\beta ]$.

Proof Put

$$\begin{array}{c}r=\underset{n}{lim}{\parallel (T_1^{n}x-f_1,\dots ,T_j^{n}x-f_j)\parallel }_j,R={\parallel (x-f_1,\dots ,x-f_j)\parallel }_j,\hfill \\ c=min\{2\lambda (1-\lambda ):\alpha \le \lambda \le \beta \}.\hfill \end{array}$$

For given $ϵ>0$, choose $d>0$ such that $\frac{r}{r+d}>1-c{\delta }_R(ϵ)$. Then there exists $N>0$ such that, for all $n\ge N$,

$${\parallel (T_1^{n}x-f_1,\dots ,T_j^{n}x-f_j)\parallel }_j<r+d.$$

For each $n\ge N$, $k>0$ and $\alpha \le \lambda \le \beta$, we put

$$u_j=(1-\lambda )(T_j^{k}z-f_j),v_j=\lambda (T_j^{n+k}x-T_j^{k}z),$$

where $z_j=\lambda T_j^{n}x+(1-\lambda )f_j$. Then we have

$${\parallel (u_1,\dots ,u_j)\parallel }_j\le \lambda (1-\lambda ){\parallel (T_1^{n}x-f_1,\dots ,T_j^{n}x-f_j)\parallel }_j$$

and

$${\parallel (v_1,\dots ,v_j)\parallel }_j\le \lambda (1-\lambda ){\parallel (T_1^{n}x-f_1,\dots ,T_j^{n}x-f_j)\parallel }_j.$$

Suppose that

$$\begin{array}{c}{\parallel (u_1-v_1,\dots ,u_j-v_j)\parallel }_j\hfill \\ ={\parallel (T_1^{k}z-(\lambda T_1^{n+k}x+(1-\lambda )f_1),\dots ,T_1^{k}z-(\lambda T_j^{n+k}x+(1-\lambda )f_j))\parallel }_j\hfill \\ \ge ϵ.\hfill \end{array}$$

Then, by Lemma 3.1, we have

$$\begin{array}{c}{\parallel (\lambda u_1+(1-\lambda )v_1,\dots ,\lambda u_j+(1-\lambda )v_j)\parallel }_j\hfill \\ =\lambda (1-\lambda ){\parallel (T_1^{n+k}x-f_1,\dots ,T_j^{n+k}x-f_j)\parallel }_j\hfill \\ \le \lambda (1-\lambda ){\parallel (T_1^{n}x-f_1,\dots ,T_j^{n}x-f_j)\parallel }_j(1-2\lambda (1-\lambda ){\delta }_R(ϵ))\hfill \\ \le \lambda (1-\lambda ){\parallel (T_1^{n}x-f_1,\dots ,T_j^{n}x-f_j)\parallel }_j(1-c{\delta }_R(ϵ)).\hfill \end{array}$$

Hence we have

$$(r+d)(1-c{\delta }_R(ϵ))<r\le (r+d)(1-C{\delta }_R(ϵ)),$$

which is a contradiction. This completes the proof. □

Lemma 3.3 (Browder [12])

Let C be a closed convex subset of X and $T_j:C\to C$ be a nonexpansive mapping. If $\{u_i\}$ is a weakly convergent sequence in C with the weak limit $u_0$ and ${lim}_i\parallel u_i-T_j{u}_i\parallel =0$, then $u_0$ is a fixed point of $T_j$.

Lemma 3.4 Let C be a closed convex subset of X and, for each $j\ge 1$, $T_j:C\to C$ be a nonexpansive mapping. Then, for all $x\in C$ and $n>0$,

$$\underset{i\to \mathrm{\infty }}{lim}\underset{j\to \mathrm{\infty }}{sup}{\parallel (T_1^k{S}_{n,1}{T}_1^{i}x-S_{n,1}{T}_1^k{T}_1^{i}x,\dots ,T_j^k{S}_{n,j}{T}_j^{i}x-S_{n,j}{T}_j^k{T}_j^{i}x)\parallel }_j=0$$

(1)

uniformly for each $k\ge 1$.

Proof By induction on n, we prove this lemma. First, we prove the conclusion in the case $n=2$. Put

$$\begin{array}{c}r=\underset{n\to \mathrm{\infty }}{lim}\underset{j\ge 1}{sup}{\parallel (T_1^{n+1}x-T_1^{n}x,\dots ,T_j^{n+1}x-T_j^{n}x)\parallel }_j,\hfill \\ R={\parallel (x-T_{1}x,\dots ,x-T_{j}x)\parallel }_j,x_{i,j}=T_j^{i}x\hfill \end{array}$$

for each $i\ge 1$.

If $r\ne 0$, then, for any $ϵ>0$, choose $c>0$ such that $\frac{r}{r+c}>1-{\delta }_R(ϵ)/2$. Then there exists $N>0$ such that, for all $i\ge N$,

$${\parallel (T_1^k{x}_{i,1}-T_1^{k+1}{x}_{i,1},\dots ,T_j^k{x}_{i,j}-T_j^{k+1}{x}_{i,j})\parallel }_j\le r+c$$

for each $k\ge 1$. If we put

$$u_j=\frac{1}{2}(T_j^{k}z-T_j^k{x}_{i,j}),v_j=\frac{1}{2}(T_j^{k+1}{x}_{i,j}-T_j^k{z}_j),$$

where $i\ge N$, $k>0$ and $z_j=\frac{1}{2}(x_{i,j}+T_j{x}_{i,j})$, then we have

$$\begin{array}{rcl}{\parallel (u_1,\dots ,u_j)\parallel }_j& \le & \frac{1}{2}{\parallel (z_1-x_{i,1}-z_j-x_{i,j})\parallel }_j\\ =& \frac{1}{4}{\parallel (T_1{x}_{i,1}-x_{i,1},\dots ,T_j{x}_{i,j}-x_{i,j})\parallel }_j\\ \le & \frac{1}{4}(r+c).\end{array}$$

Similarly, we have ${\parallel (v_1,\dots ,v_j)\parallel }_j\le \frac{1}{4}(r+c)$. Suppose that

$$\begin{array}{c}{\parallel (u_1-v_1,\dots ,u_j-v_j)\parallel }_j\hfill \\ ={\parallel (T_1^k{z}_1-\frac{1}{2}(T_1^{k+1}{x}_{i,1}+T_1^k{x}_{i,1}),\dots ,T_j^k{z}_j-\frac{1}{2}(T_j^{k+1}{x}_{i,j}+T_j^k{x}_{i,j}))\parallel }_j\hfill \\ \ge ϵ.\hfill \end{array}$$

Then, by Lemma 3.1, we have

$$\begin{array}{rcl}{\parallel \frac{1}{2}(u_1+v_1,\dots ,u_j+v_j)\parallel }_j& =& \frac{1}{4}{\parallel (T_1^{k+1}{x}_{i,1}-T_1^k{x}_{i,1},\dots ,T_j^{k+1}{x}_{i,j}-T_j^k{x}_{i,j})\parallel }_j\\ \le & \frac{1}{4}(r+c)(1-\frac{1}{2}{\delta }_R(ϵ)),\end{array}$$

which contradicts $r>(r+c)(1-\frac{1}{2}{\delta }_R(ϵ))$.

If $r=0$, then, for any $ϵ>0$, choose $i>0$ so large that ${sup}_j{\parallel (u_1,\dots ,u_j)\parallel }_j<\frac{ϵ}{2}$. Hence we have

$$\begin{array}{c}\underset{j\ge 1}{sup}{\parallel (T_1^k{z}_1-\frac{1}{2}(T_1^{k+1}{x}_{i,1}+T_1^k{x}_{i,1}),\dots ,T_j^k{z}_j-\frac{1}{2}(T_j^{k+1}{x}_{i,j}+T_j^k{x}_{i,j}))\parallel }_j\hfill \\ =\underset{j\ge 1}{sup}{\parallel (u_1-v_1,\dots ,u_j-v_j)\parallel }_j\hfill \\ \le \underset{j\ge 1}{sup}{\parallel (u_1,\dots ,u_j)\parallel }_j+\underset{j\ge 1}{sup}{\parallel (v_1,\dots ,v_j)\parallel }_j\hfill \\ <ϵ.\hfill \end{array}$$

This completes the proof of the case $n=2$.

Now, suppose that

$$\underset{i\to \mathrm{\infty }}{lim}\underset{j\ge 1}{sup}{\parallel (T_1^k{S}_{n-1,1}{x}_{i,1}-S_{n-1,1}{T}_1^k{x}_{i,1},\dots ,T_j^k{S}_{n-1,j}{x}_{i,j}-S_{n-1,j}{T}_j^k{x}_{i,j})\parallel }_j=0$$

uniformly for each $k\ge 1$. We claim that

$$\underset{i\to \mathrm{\infty }}{lim}\underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1{x}_{i,1}-x_{i,1},\dots ,S_{n-1,j}{T}_j{x}_{i,j}-x_{i,j})\parallel }_j$$

exists. Put

$$r=\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1{x}_{i,1}-x_{i,1},\dots ,S_{n-1,j}{T}_j{x}_{i,j}-x_{i,j})\parallel }_j.$$

For any $ϵ>0$, choose $i>0$ such that

$$\underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1{x}_{i,1}-x_{i,1},\dots ,S_{n-1,j}{T}_j{x}_{i,j}-x_{i,j})\parallel }_j<r+\frac{ϵ}{2}$$

and

$$\underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1^k{x}_{i+1,1}-T_1^k{S}_{n-1,1}{x}_{i+1,1},\dots ,S_{n-1,j}{T}_j^k{x}_{i+1,j}-T_j^k{S}_{n-1,j}{x}_{i+1,j})\parallel }_j<\frac{ϵ}{2}.$$

Then we have

$$\begin{array}{c}\underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1{x}_{i+k,1}-x_{i+k,1},\dots ,S_{n-1,j}{T}_j{x}_{i+k,j}-x_{i+k,j})\parallel }_j\hfill \\ \le \underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1^k{x}_{i+1,1}-T_1^k{S}_{n-1,1}{x}_{i+1,1},\dots ,S_{n-1,j}{T}_j^k{x}_{i+1,j}-T_j^k{S}_{n-1,j}{x}_{i+1,j})\parallel }_j\hfill \\ +\underset{j\ge 1}{sup}{\parallel (T_1^k{S}_{n-1,1}{x}_{i+1,1}-T_1^k{x}_{i,1},\dots ,T_j^k{S}_{n-1,j}{x}_{i+1,j}-T_j^k{x}_{i,j})\parallel }_j\hfill \\ <\frac{ϵ}{2}+r+\frac{ϵ}{2}\hfill \\ =r+ϵ\hfill \end{array}$$

for all $k\ge 1$. Therefore, we have

$$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1{x}_{i,1}-x_{i,1},\dots ,S_{n-1,j}{T}_j{x}_{i,j}-x_{i,j})\parallel }_j\hfill \\ =\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1{x}_{i+k,1}-x_{i+k,1},\dots ,S_{n-1,j}{T}_j{x}_{i+k,j}-x_{i+k,j})\parallel }_j\hfill \\ <r+ϵ.\hfill \end{array}$$

Since $ϵ>0$ is arbitrary, we have

$$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1{x}_{i,1}-x_{i,1},\dots ,S_{n-1,j}{T}_j{x}_{i,1}-x_{i,1})\parallel }_j\hfill \\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1{x}_{i,1}-x_{i,1},\dots ,S_{n-1,j}{T}_j{x}_{i,j}-x_{i,j})\parallel }_j,\hfill \end{array}$$

i.e., ${lim}_{i\to \mathrm{\infty }}{sup}_{j\ge 1}{\parallel (S_{n-1,1}{T}_1{x}_{i,1}-x_{i,1},\dots ,S_{n-1,j}{T}_j{x}_{i,j}-x_{i,j})\parallel }_j$ exists.

Now, we put

$$r=\underset{i\to \mathrm{\infty }}{lim}\underset{j\ge 1}{sup}{\parallel (S_{n-1,1}{T}_1{x}_{i,1}-x_{i,1},\dots ,S_{n-1,j}{T}_j{x}_{i,j}-x_{i,j})\parallel }_j.$$

If $r\ne 0$, then, for any ϵ, choose $c>0$ such that

$$\frac{r-c}{r+2c}>1-\left(2\frac{(n-1)}{n^2}\right){\delta }_{3r}(ϵ).$$

Then there exists $N>0$ such that, if, for all $i\ge N$, we put

$$u_j=\frac{n}{(n-1)}(T_j^k{S}_{n,j}{x}_{i,j}-T_j^k{x}_{i,j}),v_j=n(S_{n-1,j}{T}_j^k{x}_{i+1,j}-T_j^k{S}_{n,j}{x}_{i,j}),$$

so

$$\begin{array}{c}{\parallel (u_1,\dots ,u_j)\parallel }_j\le {\parallel (S_{n-1,1}{T}_1{x}_{i,1}-x_{i,1},\dots ,S_{n-1,j}{T}_j{x}_{i,j}-x_{i,j})\parallel }_j\le r+c,\hfill \\ {\parallel (v_1,\dots ,v_2)\parallel }_j\hfill \\ \le n{\parallel (S_{n-1,1}{T}_1^k{x}_{i+1,1}-T_1^k{S}_{n-1,1}{x}_{i+1,1},\dots ,S_{n-1,j}{T}_j^k{x}_{i+1,j}-T_j^k{S}_{n-1,j}{x}_{i+1,j})\parallel }_j\hfill \\ +{\parallel (S_{n-1,1}{T}_1{x}_{i,1}-x_{i,1},\dots ,S_{n-1,j}{T}_j{x}_{i,j}-x_{i,j})\parallel }_j\hfill \\ \le r+2c\hfill \end{array}$$

and

$$\begin{array}{c}{\parallel (u_1-v_1,\dots ,u_j-v_j)\parallel }_j\hfill \\ =\frac{n}{n-1}{\parallel (T_1^k{S}_{n,1}{x}_{i,1}-S_{n,1}{T}_1^k{x}_{i,1},\dots ,T_j^k{S}_{n,j}{x}_{i,j}-S_{n,j}{T}_j^k{x}_{i,j})\parallel }_j.\hfill \end{array}$$

Hence, by the method in the proof of the case $n=2$, we have

$$\underset{j\ge 1}{sup}{\parallel (T_1^k{S}_{n,1}{x}_{i,1}-S_{n,1}{T}_1^k{x}_{i,1},\dots ,T_j^k{S}_{n,j}{x}_{i,j}-S_{n,j}{T}_j^k{x}_{i,j})\parallel }_j<ϵ$$

for all $k\ge 1$ and $i\ge N$.

If $r=0$, then, as in the proof of the case $n=2$, there exists $N^{\prime }$ such that, for each $i\ge N^{\prime }$,

$$\underset{j\ge 1}{sup}{\parallel (u_1,\dots ,u_j)\parallel }_j<\frac{ϵ}{2},\underset{j\ge 1}{sup}{\parallel (v_1,\dots ,v_j)\parallel }_j<\frac{ϵ}{2}.$$

Therefore, we have

$$\underset{j\ge 1}{sup}{\parallel (T_1^k{S}_{n,1}{x}_{i,1}-S_{n,1}{T}_1^k{x}_{i,1},\dots ,T_j^k{S}_{n,j}{x}_{i,j}-S_{n,j}{T}_j^k{x}_{i,j})\parallel }_j<ϵ.$$

This completes the proof. □

Now, assume that the norm of X is Frechet differentiable and then we have the following.

Proposition 3.5 ([4, 6, 13])

Let C be a closed convex subset of X and, for each $j\ge 1$, $T_j:C\to C$ be a nonexpansive mapping. If we put $W_j(x)={\cap }_m\overline{co}\{T_j^{k}x:k\ge m\}$ for all $x\in C$, then $W_j(x)\cap F(T_j)$ is at most one point.

In this paper, we give a new proof of the following theorem, which is due to Reich [6].

Theorem 3.6 Let ${\{(X^j,{\parallel \cdot \parallel }_j)\}}_{j\in \mathbb{N}}$ be a uniformly convex multi-Banach space which has the Fréchet differentiable norm. Let C be a closed convex subset of X and, for each $j\ge 1$, $T_j:C\to C$ be a nonexpansive mapping. Then the following statements are equivalent:

• (1) $F(T_j)\ne \mathrm{\varnothing }$.

• (2) $\{T_j^{n}x\}$ is bounded for all $x\in C$.

• (3) For all $x\in C$, $\{S_n{T}_j^{i}x\}$ converges weakly to a point $(y_1,\dots ,y_j)\in C^j$ uniformly for each $i\ge 1$.

Proof (1) ⟺ (2) is well known in [12].

(3) ⟺ (2) Suppose that, for some $x\in C$, there exists an unbounded subsequence $\{T_j^{n_i}x\}$ of $\{T_j^{n}x\}$. For each $j\ge 1$, since $T_j$ is a nonexpansive mapping, it follows that, for each $m>0$, the sequence $\{S_{m<j}{T}_j^{n_i}x\}$ is also unbounded, which contradicts the condition (3).

(2) ⟺ (3) Since $\{T_j^{n}x\}$ is bounded and

$$\begin{array}{c}{\parallel (T_1{S}_{n,1}{T}_1^{i}x-S_{n,1}{T}_1^{i}x,\dots ,T_j{S}_{n,j}{T}_j^{i}x-S_{n,j}{T}_j^{i}x)\parallel }_j\hfill \\ \le {\parallel (T_1{S}_{n,1}{T}_1^{i}x-S_{n,1}{T}_1{T}_1^{i}x,\dots ,T_j{S}_{n,j}{T}_j^{i}x-S_{n,j}{T}_j{T}_j^{i}x)\parallel }_j\hfill \\ +{\parallel (S_{n,1}{T}_1{T}_1^{i}x-S_{n,1}{T}_1^{i}x,\dots ,S_{n,j}{T}_j{T}_j^{i}x-S_{n,j}{T}_j^{i}x)\parallel }_j\hfill \\ \le {\parallel (T_1{S}_{n,1}{T}_1^{i}x-S_{n,1}{T}_1{T}_1^{i}x,\dots ,T_j{S}_{n,j}{T}_j^{i}x-S_{n,j}{T}_j{T}_j^{i}x)\parallel }_j\hfill \\ +\frac{1}{n}{\parallel (T_1^{i+1+n}x-T_1^{i}x,\dots ,T_j^{i+1+n}x-T_j^{i}x)\parallel }_j,\hfill \end{array}$$

there exists a sequence $\{S_{n,j}{T}_j^{i_n}x\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim}\underset{j\ge 1}{sup}{\parallel (T_1{S}_{n,1}{T}_1^{i_n}x-S_{n,1}{T}_1^{i_n}x,\dots ,T_j{S}_{n,j}{T}_j^{i_n}x-S_{n,j}{T}_j^{i_n}x)\parallel }_j=0.$$

Then, by Lemma 3.3 and Proposition 3.5, it follows that any weakly multi-convergent subsequence of $\{S_{n,j}{T}_j^{i_n}x\}$ multi-converges weakly to a point $y_j$, i.e., $S_{n,j}{T}_j^{i_n}x\rightharpoondown y_j$, where $y_j=W_j(x)\cap F(T_j)$. Also, by Lemma 3.4, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}\underset{j\ge 1}{sup}{\parallel (T_1{S}_{n,1}{T}_1^{i_n+kn+i}x-S_{n,1}{T}_1^{i_n+kn+i}x,\dots ,T_j{S}_{n,j}{T}_j^{i_n+kn+i}x-S_{n,j}{T}_j^{i_n+kn+i}x)\parallel }_j=0$$

for all $i,k\ge 1$. Therefore, $S_{n,j}{T}_j^{i_n+kn}{x}_i\rightharpoondown y_j$ uniformly for each $k\ge 1$.

On the other hand, for each $n\ge 1$ with $m\ge i_n$, we have

$$\begin{array}{rcl}{S}_{m,j}{T}_j^{i}x& =& \frac{1}{m}\sum _{k=0}^{m-1}{T}_j^k{x}_i\\ =& \frac{1}{m}(\sum _{k=i_n+tn}^{m-1}{T}_j^k{x}_i+n\left(\sum _{k=0}^t{S}_n{T}_j^{i_n+kn}{x}_i\right)+\sum _{k=0}^{i_n}{T}_j^k{x}_i),\end{array}$$

where $m=tn+i_n+r$, $r<n$. Since $\{S_{n,j}{T}_j^{i_n+kn}{x}_i\}$ multi-converges to $y_j$ uniformly for each $k\ge 1$, it follows that $\{S_{m,j}{T}_j^{i}x\}$ converges weakly to $y_j$ uniformly for each $i\ge 1$. This completes the proof. □