A new multi-step iteration for solving a fixed point problem of nonexpansive mappings

Abstract

We introduce a new nonlinear mapping generated by a finite family of nonexpansive mappings. Weak and strong convergence theorems are also established in the setting of a Banach space.

MSC:47H09, 47H10.

1 Introduction

Let C be a nonempty, closed and convex subset of a real Banach space E. Let $T:C\to C$ be a nonlinear mapping. The fixed point set of T is denoted by $F(T)$, that is, $F(T)=\{x\in C:x=Tx\}$. Recall that a mapping T is said to be nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in C$, and a mapping $f:C\to C$ is called a contraction if there exists a constant $\alpha \in (0,1)$ such that $\parallel f(x)-f(y)\parallel \le \alpha \parallel x-y\parallel$ for all $x,y\in C$. We use ${\mathrm{\Sigma }}_C$ to denote a class of contractions with constant α.

Fixed point problems are now arising in a wide range of applications such as optimization, physics, engineering, economics and applied sciences. Many related problems can be cast as the problem of finding fixed points for nonlinear mappings. The interdisciplinary nature of fixed point problems is evident through a vast literature which includes a large body of mathematical and algorithmic developments.

In the literature, several types of iterations have been constructed and proposed in order to get convergence results for nonexpansive mappings in various settings. One classical iteration process is defined as follows: $x_1\in C$ and

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

where $\{{\alpha }_n\}\subset (0,1)$. This method was introduced in 1953 by Mann [1] and is known as the Mann iteration process. However, we note that it has only weak convergence in general; for instance, see [2].

In 1974, Ishikawa [3] proposed the following two-step iteration: $x_1\in C$ and

$$\begin{array}{c}{y}_n=(1-{\beta }_n)x_n+{\beta }_{n}T{x}_n,\hfill \\ x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}T{y}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are sequences in $(0,1)$. This method is often called the Ishikawa iteration process.

Very recently, Agarwal et al. [4] introduced a new iteration process as follows: $x_1\in C$ and

$$\begin{array}{c}{y}_n=(1-{\beta }_n)x_n+{\beta }_{n}T{x}_n,\hfill \\ x_{n+1}=(1-{\alpha }_n)T{x}_n+{\alpha }_{n}T{y}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are sequences in $(0,1)$. This method is called the S-iteration process. The weak convergence was studied in [5] for nonexpansive mappings. It was also shown in [5] that the convergence rate of the S-iteration process is faster than the Picard and Mann iteration processes for contractive mappings.

Firstly, motivated by Agarwal et al. [4], we have the aim to introduce and study a new mapping defined by the following definition.

Definition 1.1 Let C be a nonempty and convex subset of a real Banach space E. Let $T_1,T_2,\dots ,T_N$ be a finite family of nonexpansive mappings of C into itself, and let ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_N$ be real numbers such that $0\le {\lambda }_i\le 1$ for all $i=1,2,\dots ,N$. Define the mapping $B:C\to C$ as follows:

$$\begin{array}{r}{U}_1={\lambda }_1{T}_1+(1-{\lambda }_1)I,\\ U_2={\lambda }_2{T}_2{U}_1+(1-{\lambda }_2)T_1,\\ U_3={\lambda }_3{T}_3{U}_2+(1-{\lambda }_3)T_2,\\ ⋮\\ U_{N-1}={\lambda }_{N-1}{T}_{N-1}{U}_{N-2}+(1-{\lambda }_{N-1})T_{N-2},\\ B=U_N={\lambda }_N{T}_N{U}_{N-1}+(1-{\lambda }_N)T_{N-1}.\end{array}$$

(1.1)

Such a mapping B is called the B-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_N$. See [6–9] for the corresponding concept.

Secondly, using the definition above, we study weak convergence of the following Mann-type iteration process in a uniformly convex Banach space with a Fréchet differentiable norm or that satisfies Opial’s condition: $x_1\in C$ and

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

(1.2)

where $B_n$ is a B-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\lambda }_{n,1},{\lambda }_{n,2},\dots ,{\lambda }_{n,N}$ (see Section 3).

Finally, we discuss strong convergence of the iteration scheme involving the modified viscosity approximation method [10] defined as follows: $x_1\in C$ and

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{B}_n{x}_n,\mathrm{\forall }n\ge 1,$$

(1.3)

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are sequences in $(0,1)$, and $f\in {\mathrm{\Sigma }}_C$.

More references on earlier works promoting the theory of fixed points and common fixed points for nonexpansive mappings can be found in [11–21].

Throughout this paper, we use the notation:

• ⇀ for weak convergence and → for strong convergence.

• ${\omega }_{\omega }(x_n)=\{x:x_{n_i}\rightharpoonup x\}$ denotes the weak ω-limit set of $\{x_n\}$.

2 Preliminaries and lemmas

In this section, we begin by recalling some basic facts and lemmas which will be used in the sequel.

Let E be a real Banach space and let $U=\{x\in E:\parallel x\parallel =1\}$ be the unit sphere of E. A Banach space E is said to be strictly convex if for any $x,y\in U$,

$$x\ne y\text{implies}\parallel \frac{x+y}{2}\parallel <1.$$

It is also said to be uniformly convex if for each $\epsilon \in (0,2]$, there exists $\delta >0$ such that for any $x,y\in U$,

$$\parallel x-y\parallel \ge \epsilon \text{implies}\parallel \frac{x+y}{2}\parallel <1-\delta .$$

It is known that a uniformly convex Banach space is reflexive and strictly convex. Define a function $\delta :[0,2]\to [0,1]$ called the modulus of convexity of E as follows:

$$\delta (\epsilon )=inf\{1-\parallel \frac{x+y}{2}\parallel :x,y\in E,\parallel x\parallel =\parallel y\parallel =1,\parallel x-y\parallel \ge \epsilon \}.$$

Then E is uniformly convex if and only if $\delta (\epsilon )>0$ for all $\epsilon \in (0,2]$. A Banach space E is said to be smooth if the limit

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

(2.1)

exists for all $x,y\in U$. The norm is said to be uniformly Gâteaux differentiable if for $y\in U$, the limit is attained uniformly for $x\in U$. It is said to be Fréchet differentiable if for $x\in U$, the limit is attained uniformly for $y\in U$. It is said to be uniformly smooth or uniformly Fréchet differentiable if the limit (2.1) is attained uniformly for $x,y\in U$. The normalized duality mapping $J:E\to 2^{E^{\ast }}$ is defined by

$$J(x)=\{x^{\ast }\in E^{\ast }:〈x,x^{\ast }〉={\parallel x\parallel }^2={\parallel x^{\ast }\parallel }^2\}$$

for all $x\in E$. It is known that E is smooth if and only if the duality mapping J is single valued, and that if E has a uniformly Gâteaux differentiable norm, J is uniformly norm-to-weak^∗ continuous on each bounded subset of E. A Banach space E is said to satisfy Opial’s condition [22]. If $x\in E$ and $x_n\rightharpoonup x$, then

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

Let $T:C\to C$. Then $I-T$ is demiclosed at 0 if for all sequence $\{x_n\}$ in C, $x_n\rightharpoonup q$ and $\parallel x_n-T{x}_n\parallel \to 0$ imply $q=Tq$. It is known that if E is uniformly convex, C is nonempty closed and convex, and T is nonexpansive, then $I-T$ is demiclosed at 0 [23]. For more details, we refer the reader to [5, 24].

Lemma 2.1 [5]

Let E be a smooth Banach space. Then the following hold:

• (i) ${\parallel x+y\parallel }^2\ge {\parallel x\parallel }^2+2〈y,J(x)〉$ for all $x,y\in E$;

• (ii) ${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,J(x+y)〉$ for all $x,y\in E$.

Lemma 2.2 [24]

In a strictly convex Banach space E, if

$$\parallel x\parallel =\parallel y\parallel =\parallel \lambda x+(1-\lambda )y\parallel$$

for all $x,y\in E$ and $\lambda \in (0,1)$, then $x=y$.

Lemma 2.3 [25]

Let $\{x_n\}$ and $\{z_n\}$ be two sequences in a Banach space E such that

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n,n\ge 1,$$

where $\{{\beta }_n\}$ satisfies the conditions $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$.

If ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$, then $\parallel z_n-x_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Lemma 2.4 [26]

Let E be a uniformly convex Banach space with a Fréchet differentiable norm. Let C be a closed and convex subset of E, and let ${\{S_n\}}_{n=1}^{\mathrm{\infty }}$ be a family of $L_n$-Lipschitzian self-mappings on C such that ${\sum }_{n=1}^{\mathrm{\infty }}(L_n-1)<\mathrm{\infty }$ and $F={\bigcap }_{n=1}^{\mathrm{\infty }}F(S_n)\ne \mathrm{\varnothing }$. For arbitrary $x_1\in C$, define $x_{n+1}=S_n{x}_n$ for all $n\ge 1$. Then, for every $p,q\in F$, ${lim}_{n\to \mathrm{\infty }}〈x_n,J(p-q)〉$ exists, in particular, for all $u,v\in {\omega }_{\omega }(x_n)$ and $p,q\in F$, $〈u-v,J(p-q)〉=0$.

Lemma 2.5 [15]

Let E be a reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, let C be a nonempty closed convex subset of E, let $A:C\to C$ be a continuous strongly pseudocontractive mapping with constant $k\in [0,1)$, and let $T:C\to E$ be a continuous pseudocontractive mapping satisfying the weakly inward condition. If T has a fixed point in C, then the path $\{x_t\}$ defined by

$$x_t=tA{x}_t+(1-t)T{x}_t$$

converges strongly to a fixed point q of T as $t\to 0$, which is a unique solution of the variational inequality

$$〈(I-A)q,J(q-p)〉\le 0,\mathrm{\forall }p\in F(T).$$

Remark 2.6 Lemma 2.5 holds if $T:C\to C$ is a nonexpansive mapping and $A=f$ is a contraction.

The following lemma gives us a nice property of real sequences.

Lemma 2.7 [18]

Assume that $\{a_n\}$ is a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-c_n)a_n+b_n,\mathrm{\forall }n\ge 1,$$

where $\{c_n\}$ is a sequence in $(0,1)$ and $\{b_n\}$ is a sequence such that

• (a) ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n=\mathrm{\infty }$;

• (b) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{b_n}{c_n}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|b_n|<\mathrm{\infty }$.

Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

3 Weak convergence theorem

In this section, we give some properties concerning the B-mapping and then prove a weak convergence theorem for nonexpansive mappings.

Lemma 3.1 Let C be a nonempty, closed and convex subset of a strictly convex Banach space E. Let ${\{T_i\}}_{i=1}^N$ be a finite family of nonexpansive mappings of C into itself such that ${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$, and let ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_N$ be real numbers such that $0<{\lambda }_i<1$ for all $i=1,2,\dots ,N-1$ and $0<{\lambda }_N\le 1$. Let B be the B-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_N$. Then the following hold:

• (i) $F(B)={\bigcap }_{i=1}^{N}F(T_i)$;

• (ii) B is nonexpansive.

Proof (i) Since ${\bigcap }_{i=1}^{N}F(T_i)\subset F(B)$ is trivial, it suffices to show that $F(B)\subset {\bigcap }_{i=1}^{N}F(T_i)$. To this end, let $p\in F(B)$ and $x^{\ast }\in {\bigcap }_{i=1}^{N}F(T_i)$. Then we have

$$\begin{array}{rcl}\parallel p-x^{\ast }\parallel & =& \parallel Bp-x^{\ast }\parallel =\parallel {\lambda }_N(T_N{U}_{N-1}p-x^{\ast })+(1-{\lambda }_N)(T_{N-1}p-x^{\ast })\parallel \\ \le & {\lambda }_N\parallel U_{N-1}p-x^{\ast }\parallel +(1-{\lambda }_N)\parallel p-x^{\ast }\parallel \\ =& {\lambda }_N\parallel {\lambda }_{N-1}(T_{N-1}{U}_{N-2}p-x^{\ast })+(1-{\lambda }_{N-1})(T_{N-2}p-x^{\ast })\parallel +(1-{\lambda }_N)\parallel p-x^{\ast }\parallel \\ \le & {\lambda }_N{\lambda }_{N-1}\parallel U_{N-2}p-x^{\ast }\parallel +(1-{\lambda }_N{\lambda }_{N-1})\parallel p-x^{\ast }\parallel \\ =& {\lambda }_N{\lambda }_{N-1}\parallel {\lambda }_{N-2}(T_{N-2}{U}_{N-3}p-x^{\ast })+(1-{\lambda }_{N-2})(T_{N-3}p-x^{\ast })\parallel \\ +(1-{\lambda }_N{\lambda }_{N-1})\parallel p-x^{\ast }\parallel \\ \le & {\lambda }_N{\lambda }_{N-1}{\lambda }_{N-2}\parallel U_{N-3}p-x^{\ast }\parallel +(1-{\lambda }_N{\lambda }_{N-1}{\lambda }_{N-2})\parallel p-x^{\ast }\parallel \\ ⋮\\ =& {\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_3\parallel {\lambda }_2(T_2{U}_{1}p-x^{\ast })+(1-{\lambda }_2)(T_{1}p-x^{\ast })\parallel \\ +(1-{\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_3)\parallel p-x^{\ast }\parallel \\ \le & {\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2\parallel T_2{U}_{1}p-x^{\ast }\parallel +(1-{\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2)\parallel p-x^{\ast }\parallel \\ \le & {\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2\parallel U_{1}p-x^{\ast }\parallel +(1-{\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2)\parallel p-x^{\ast }\parallel \\ =& {\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2\parallel {\lambda }_1(T_{1}p-x^{\ast })+(1-{\lambda }_1)(p-x^{\ast })\parallel \\ +(1-{\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2)\parallel p-x^{\ast }\parallel \\ \le & {\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2{\lambda }_1\parallel T_{1}p-x^{\ast }\parallel +(1-{\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2{\lambda }_1)\parallel p-x^{\ast }\parallel \\ \le & {\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2{\lambda }_1\parallel p-x^{\ast }\parallel +(1-{\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2{\lambda }_1)\parallel p-x^{\ast }\parallel \\ =& \parallel p-x^{\ast }\parallel .\end{array}$$

(3.1)

This shows that

$$\parallel p-x^{\ast }\parallel ={\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2\parallel {\lambda }_1(T_{1}p-x^{\ast })+(1-{\lambda }_1)(p-x^{\ast })\parallel +(1-{\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_2)\parallel p-x^{\ast }\parallel ,$$

which turns out to be

$$\parallel p-x^{\ast }\parallel =\parallel {\lambda }_1(T_{1}p-x^{\ast })+(1-{\lambda }_1)(p-x^{\ast })\parallel .$$

Again by (3.1), we see that $\parallel p-x^{\ast }\parallel =\parallel T_{1}p-x^{\ast }\parallel$ and thus

$$\parallel p-x^{\ast }\parallel =\parallel T_{1}p-x^{\ast }\parallel =\parallel {\lambda }_1(T_{1}p-x^{\ast })+(1-{\lambda }_1)(p-x^{\ast })\parallel .$$

Using Lemma 2.2, we get that $T_{1}p=p$ and hence $U_{1}p=p$. Again by (3.1), we have

$$\begin{array}{rl}\parallel p-x^{\ast }\parallel =& {\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_3\parallel {\lambda }_2(T_2{U}_{1}p-x^{\ast })+(1-{\lambda }_2)(T_{1}p-x^{\ast })\parallel \\ +(1-{\lambda }_N{\lambda }_{N-1}\cdots {\lambda }_3)\parallel p-x^{\ast }\parallel ,\end{array}$$

which implies that

$$\parallel p-x^{\ast }\parallel =\parallel {\lambda }_2(T_2{U}_{1}p-x^{\ast })+(1-{\lambda }_2)(T_{1}p-x^{\ast })\parallel .$$

From (3.1) we see that $\parallel U_{1}p-x^{\ast }\parallel =\parallel T_2{U}_{1}p-x^{\ast }\parallel$. Since $U_{1}p=p$ and $T_{1}p=p$,

$$\parallel p-x^{\ast }\parallel =\parallel T_{2}p-x^{\ast }\parallel =\parallel {\lambda }_2(T_{2}p-x^{\ast })+(1-{\lambda }_2)(p-x^{\ast })\parallel .$$

Using Lemma 2.2, we get that $T_{2}p=p$ and hence $U_{2}p=p$.

By continuing this process, we can show that $T_{i}p=p$ and $U_{i}p=p$ for all $i=1,2,\dots ,N-1$. Finally, we obtain

$$\begin{array}{rcl}\parallel p-T_{N}p\parallel & \le & \parallel p-Bp\parallel +\parallel Bp-T_{N}p\parallel \\ =& \parallel p-Bp\parallel +(1-{\lambda }_N)\parallel p-T_{N}p\parallel ,\end{array}$$

which yields that $p=T_{N}p$ since $p\in F(B)$. Hence $p=T_{1}p=T_{2}p=\cdots =T_{N}p$ and thus $p\in {\bigcap }_{i=1}^{N}F(T_i)$.

(ii) The proof follows directly from (i). □

Lemma 3.2 Let C be a nonempty closed convex subset of a real Banach space E. Let ${\{T_i\}}_{i=1}^N$ be a finite family of nonexpansive mappings of C into itself such that ${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$, and let B be the B-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_N$. Let ${\{{\lambda }_{n,i}\}}_{i=1}^N$ be a real sequence in $(0,1)$. For every $n\in \mathbb{N}$, let $B_n$ be the B-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\lambda }_{n,1},{\lambda }_{n,2},\dots ,{\lambda }_{n,N}$ as follows:

$$\begin{array}{c}{U}_{n,1}={\lambda }_{n,1}{T}_1+(1-{\lambda }_{n,1})I,\hfill \\ U_{n,2}={\lambda }_{n,2}{T}_2{U}_1+(1-{\lambda }_{n,2})T_1,\hfill \\ U_{n,3}={\lambda }_{n,3}{T}_3{U}_2+(1-{\lambda }_{n,3})T_2,\hfill \\ ⋮\hfill \\ U_{n,N-1}={\lambda }_{n,N-1}{T}_{N-1}{U}_{N-2}+(1-{\lambda }_{n,N-1})T_{N-2},\hfill \\ B_n=U_{n,N}={\lambda }_{n,N}{T}_N{U}_{N-1}+(1-{\lambda }_{n,N})T_{N-1}.\hfill \end{array}$$

If ${\lambda }_{n,i}\to {\lambda }_i\in (0,1)$ for all $i=1,2,\dots ,N$, then ${lim}_{n\to \mathrm{\infty }}{B}_{n}x=Bx$ for all $x\in C$.

Proof Let $x\in C$ and $U_k$ and $U_{n,k}$ be generated by $T_1,T_2,\dots ,T_k$ and ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_k$, and $T_1,T_2,\dots ,T_k$ and ${\lambda }_{n,1},{\lambda }_{n,2},\dots ,{\lambda }_{n,k}$, respectively. Then

$$\parallel U_{n,1}x-U_{1}x\parallel =\parallel ({\lambda }_{n,1}-{\lambda }_1)(T_{1}x-x)\parallel \le |{\lambda }_{n,1}-{\lambda }_1|\parallel T_{1}x-x\parallel .$$

Let $k\in \{2,3,\dots ,N\}$ and $M=max\{\parallel T_k{U}_{k-1}x\parallel +\parallel T_{k-1}x\parallel :k=2,3,\dots ,N\}$. Then

$$\begin{array}{rcl}\parallel U_{n,k}x-U_{k}x\parallel & =& \parallel {\lambda }_{n,k}{T}_k{U}_{n,k-1}x+(1-{\lambda }_{n,k})T_{k-1}x-{\lambda }_k{T}_k{U}_{k-1}-(1-{\lambda }_k)T_{k-1}x\parallel \\ =& \parallel {\lambda }_{n,k}{T}_k{U}_{n,k-1}x-{\lambda }_{n,k}{T}_{k-1}x-{\lambda }_k{T}_k{U}_{k-1}+{\lambda }_k{T}_{k-1}x\parallel \\ \le & {\lambda }_{n,k}\parallel T_k{U}_{n,k-1}x-T_k{U}_{k-1}x\parallel +|{\lambda }_{n,k}-{\lambda }_k|\parallel T_k{U}_{k-1}x\parallel \\ +|{\lambda }_{n,k}-{\lambda }_k|\parallel T_{k-1}x\parallel \\ \le & \parallel U_{n,k-1}x-U_{k-1}x\parallel +|{\lambda }_{n,k}-{\lambda }_k|M.\end{array}$$

It follows that

$$\begin{array}{rcl}\parallel B_{n}x-Bx\parallel & =& \parallel U_{n,N}x-U_{N}x\parallel \\ \le & \parallel U_{n,N-1}x-U_{N-1}x\parallel +|{\lambda }_{n,N}-{\lambda }_N|M\\ \le & \parallel U_{n,N-2}x-U_{N-2}x\parallel +|{\lambda }_{n,N-1}-{\lambda }_{N-1}|M+|{\lambda }_{n,N}-{\lambda }_N|M\\ ⋮\\ \le & \parallel U_{n,1}x-U_{1}x\parallel +|{\lambda }_{n,2}-{\lambda }_2|M+\cdots +|{\lambda }_{n,N-1}-{\lambda }_{N-1}|M+|{\lambda }_{n,N}-{\lambda }_N|M\\ \le & |{\lambda }_{n,1}-{\lambda }_1|\parallel T_{1}x-x\parallel +|{\lambda }_{n,2}-{\lambda }_2|M+\cdots +|{\lambda }_{n,N-1}-{\lambda }_{N-1}|M\\ +|{\lambda }_{n,N}-{\lambda }_N|M.\end{array}$$

Since ${\lambda }_{n,i}\to {\lambda }_i$ as $n\to \mathrm{\infty }$ ($i=1,2,\dots ,N$), we thus complete the proof. □

Remark 3.3 It is easily seen that for all $n\in \mathbb{N}$, $B_n$ is nonexpansive.

Lemma 3.4 Let C be a nonempty closed convex subset of a real Banach space E. Let ${\{T_i\}}_{i=1}^N$ be a finite family of nonexpansive mappings of C into itself such that ${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Let ${\{{\lambda }_{n,i}\}}_{i=1}^N$ be a real sequence in $(0,1)$. For every $n\in \mathbb{N}$, let $B_n$ be the B-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\lambda }_{n,1},{\lambda }_{n,2},\dots ,{\lambda }_{n,N}$.

If ${lim}_{n\to \mathrm{\infty }}|{\lambda }_{n+1,i}-{\lambda }_{n,i}|=0$ for all $i=1,2,\dots ,N$, then

$$\underset{n\to \mathrm{\infty }}{lim}\parallel B_{n+1}{z}_n-B_n{z}_n\parallel =0$$

for each bounded sequence $\{z_n\}$ in C.

Proof Let $\{z_n\}$ be a bounded sequence in C. For $j\in \{0,1,\dots ,N-2\}$ and for some $M>0$, we have

$$\begin{array}{c}\parallel U_{n+1,N-j}{z}_n-U_{n,N-j}{z}_n\parallel \hfill \\ =\parallel {\lambda }_{n+1,N-j}{T}_{N-j}{U}_{n+1,N-j-1}{z}_n+(1-{\lambda }_{n+1,N-j})T_{N-j-1}{z}_n\hfill \\ -{\lambda }_{n,N-j}{T}_{N-j}{U}_{n,N-j-1}{z}_n-(1-{\lambda }_{n,N-j})T_{N-j-1}{z}_n\parallel \hfill \\ \le {\lambda }_{n+1,N-j}\parallel T_{N-j}{U}_{n+1,N-j-1}{z}_n-T_{N-j}{U}_{n,N-j-1}{z}_n\parallel \hfill \\ +|{\lambda }_{n+1,N-j}-{\lambda }_{n,N-j}|\parallel T_{N-j}{U}_{n,N-j-1}{z}_n\parallel \hfill \\ +|{\lambda }_{n+1,N-j}-{\lambda }_{n,N-j}|\parallel T_{N-j-1}{z}_n\parallel \hfill \\ \le \parallel U_{n+1,N-j-1}{z}_n-U_{n,N-j-1}{z}_n\parallel +|{\lambda }_{n+1,N-j}-{\lambda }_{n,N-j}|M.\hfill \end{array}$$

Using the relation above, we can show that

$$\begin{array}{rcl}\parallel B_{n+1}{z}_n-B_n{z}_n\parallel & =& \parallel U_{n+1,N}{z}_n-U_{n,N}{z}_n\parallel \\ \le & M\sum _{j=2}^N|{\lambda }_{n+1,j}-{\lambda }_{n,j}|+|{\lambda }_{n+1,1}-{\lambda }_{n,1}|(\parallel z_n\parallel +\parallel T_1{z}_n\parallel ).\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}|{\lambda }_{n+1,i}-{\lambda }_{n,i}|=0$ for all $i=1,2,\dots ,N$, we obtain the desired result. □

Using the concept of B-mapping, we study weak convergence of the sequence generated by Mann-type iteration process (1.2).

Theorem 3.5 Let E be a uniformly convex Banach space having a Fréchet differentiable norm or that satisfies Opial’s condition. Let C be a nonempty, closed and convex subset of E. Let ${\{T_i\}}_{i=1}^N$ be a finite family of nonexpansive mappings of C into itself such that ${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Let ${\{{\lambda }_{n,i}\}}_{i=1}^N$ be a real sequence in $(0,1)$ such that ${\lambda }_{n,i}\to {\lambda }_i$ ($i=1,2,\dots ,N$). For every $n\in \mathbb{N}$, let $B_n$ be the B-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\lambda }_{n,1},{\lambda }_{n,2},\dots ,{\lambda }_{n,N}$. Let $\{{\alpha }_n\}$ be a sequence in $(0,1)$ satisfying ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$. Let $\{x_n\}$ be generated by $x_1\in C$ and

$$x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n{B}_n{x}_n,\mathrm{\forall }n\ge 1.$$

Then $\{x_n\}$ converges weakly to $x^{\ast }\in {\bigcap }_{i=1}^{N}F(T_i)$.

Proof Let $p\in {\bigcap }_{i=1}^{N}F(T_i)$. Then $p=B_{n}p$ for all $n\ge 1$ and hence

$$\parallel x_{n+1}-p\parallel \le (1-{\alpha }_n)\parallel x_n-p\parallel +{\alpha }_n\parallel B_n{x}_n-p\parallel \le \parallel x_n-p\parallel .$$

It follows that $\{\parallel x_n-p\parallel \}$ is nonincreasing; consequently, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists. Assume $\parallel x_n-p\parallel >0$. Since E is uniformly convex, it follows (see, for example, [27]) that

$$\parallel x_{n+1}-p\parallel \le \parallel x_n-p\parallel \{1-2min\{{\alpha }_n,1-{\alpha }_n\}{\delta }_E\left(\frac{\parallel x_n-B_n{x}_n\parallel }{\parallel x_n-p\parallel }\right)\},$$

which implies that

$$\begin{array}{rcl}{\alpha }_n(1-{\alpha }_n)\parallel x_n-p\parallel {\delta }_E\left(\frac{\parallel x_n-B_n{x}_n\parallel }{\parallel x_n-p\parallel }\right)& \le & min\{{\alpha }_n,1-{\alpha }_n\}\parallel x_n-p\parallel {\delta }_E\left(\frac{\parallel x_n-B_n{x}_n\parallel }{\parallel x_n-p\parallel }\right)\\ \le & \frac{1}{2}\{\parallel x_n-p\parallel -\parallel x_{n+1}-p\parallel \}.\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, by the continuity of ${\delta }_E$, we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n-B_n{x}_n\parallel =0$. Since ${\lambda }_{n,i}\to {\lambda }_i$ ($i=1,2,\dots ,N$), let the mapping $B:C\to C$ be generated by $T_1,T_2,\dots ,T_N$ and ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_N$. Then, by Lemma 3.2, we have ${lim}_{n\to \mathrm{\infty }}\parallel B_{n}x-Bx\parallel =0$ for all $x\in C$. So we have

$$\begin{array}{rcl}\parallel x_n-B{x}_n\parallel & \le & \parallel x_n-B_n{x}_n\parallel +\parallel B_n{x}_n-B{x}_n\parallel \\ \le & \parallel x_n-B_n{x}_n\parallel +\underset{z\in \{x_n\}}{sup}\parallel B_{n}z-Bz\parallel \\ \to & 0.\end{array}$$

Since B is nonexpansive and E is uniformly convex, by the demiclosedness principle, ${\omega }_{\omega }(x_n)\subset F(B)$. Moreover, $F(B)={\bigcap }_{i=1}^{N}F(T_i)$ by Lemma 3.1(i).

We next show that ${\omega }_{\omega }(x_n)$ is a singleton. Indeed, suppose that $x^{\ast },y^{\ast }\in {\omega }_{\omega }(x_n)\subset {\bigcap }_{i=1}^{N}F(T_i)$. Define $S_n:C\to C$ by

$$S_{n}x=(1-{\alpha }_n)x+{\alpha }_n{B}_{n}x,x\in C.$$

Then $S_n$ is nonexpansive and $x^{\ast },y^{\ast }\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(S_n)$. Using Lemma 2.4, we have ${lim}_{n\to \mathrm{\infty }}〈x_n,J(x^{\ast }-y^{\ast })〉$ exists. Suppose that $\{x_{n_k}\}$ and $\{x_{m_k}\}$ are subsequences of $\{x_n\}$ such that $x_{n_k}\rightharpoonup x^{\ast }$ and $x_{m_k}\rightharpoonup y^{\ast }$. Then

$${\parallel x^{\ast }-y^{\ast }\parallel }^2=〈x^{\ast }-y^{\ast },J(x^{\ast }-y^{\ast })〉=\underset{k\to \mathrm{\infty }}{lim}〈x_{n_k}-x_{m_k},J(x^{\ast }-y^{\ast })〉=0.$$

This shows that $x^{\ast }=y^{\ast }$.

Assume that E satisfies Opial’s condition. Let $x^{\ast },y^{\ast }\in {\omega }_{\omega }(x_n)$ and $\{x_{n_k}\}$ and $\{x_{m_k}\}$ be subsequences of $\{x_n\}$ such that $x_{n_k}\rightharpoonup x^{\ast }$ and $x_{m_k}\rightharpoonup y^{\ast }$. If $x^{\ast }\ne y^{\ast }$, then

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x^{\ast }\parallel & =& \underset{k\to \mathrm{\infty }}{lim}\parallel x_{n_k}-x^{\ast }\parallel <\underset{k\to \mathrm{\infty }}{lim}\parallel x_{n_k}-y^{\ast }\parallel =\underset{k\to \mathrm{\infty }}{lim}\parallel x_{m_k}-y^{\ast }\parallel \\ <& \underset{k\to \mathrm{\infty }}{lim}\parallel x_{m_k}-x^{\ast }\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x^{\ast }\parallel ,\end{array}$$

which is a contradiction. It follows that $x^{\ast }=y^{\ast }$. Therefore $x_n\rightharpoonup x^{\ast }\in {\bigcap }_{i=1}^{N}F(T_i)$ as $n\to \mathrm{\infty }$. This completes the proof. □

4 Strong convergence theorem

In this section, we prove a strong convergence theorem for a finite family of nonexpansive mappings in Banach spaces.

Theorem 4.1 Let E be a strictly convex and reflexive Banach space having a uniformly Gâteaux differentiable norm. Let C be a nonempty, closed and convex subset of E. Let ${\{T_i\}}_{i=1}^N$ be a finite family of nonexpansive mappings of C into itself such that ${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Let ${\{{\lambda }_{n,i}\}}_{i=1}^N$ be a real sequence in $(0,1)$ such that ${\lambda }_{n,i}\to {\lambda }_i$ ($i=1,2,\dots ,N$). For every $n\in \mathbb{N}$, let $B_n$ be the B-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\lambda }_{n,1},{\lambda }_{n,2},\dots ,{\lambda }_{n,N}$. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $(0,1)$ which satisfy the conditions:

• (C1) ${\alpha }_n+{\beta }_n+{\gamma }_n=1$;

• (C2) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (C3) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$.

Let $f\in {\mathrm{\Sigma }}_C$ and define the sequence $\{x_n\}$ by $x_1\in C$ and

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{B}_n{x}_n,\mathrm{\forall }n\ge 1.$$

Then $\{x_n\}$ converges strongly to $q\in {\bigcap }_{i=1}^{N}F(T_i)$, where q is also the unique solution of the variational inequality

$$〈(I-f)(q),J(q-p)〉\le 0,\mathrm{\forall }p\in \bigcap _{i=1}^{N}F(T_i).$$

(4.1)

Proof We divide the proof into the following steps.

Step 1. We show that $\{x_n\}$ is bounded. Let $p\in {\bigcap }_{i=1}^{N}F(T_i)$. Then $p=B_{n}p$ for all $n\ge 1$ and hence, by the nonexpansiveness of ${\{B_n\}}_{n=1}^{\mathrm{\infty }}$, we have

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & =& \parallel {\alpha }_n(f(x_n)-p)+{\beta }_n(x_n-p)+{\gamma }_n(B_n{x}_n-p)\parallel \\ \le & {\alpha }_n\parallel f(x_n)-p\parallel +{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel \\ \le & {\alpha }_n\parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel \\ \le & {\alpha }_n\alpha \parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel \\ =& (1-{\alpha }_n(1-\alpha ))\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel \\ \le & max\{\parallel x_n-p\parallel ,\frac{1}{1-\alpha }\parallel f(p)-p\parallel \}.\end{array}$$

By induction, we can conclude that $\{x_n\}$ is bounded. So are $\{f(x_n)\}$ and $\{B_n{x}_n\}$.

Step 2. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$. To this end, we define $z_n=\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}$. From (1.3) we have

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & =& \parallel \frac{{\alpha }_{n+1}f(x_{n+1})+{\gamma }_{n+1}{B}_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_{n}f(x_n)+{\gamma }_n{B}_n{x}_n}{1-{\beta }_n}\parallel \\ =& \parallel \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(f(x_{n+1})-B_n{x}_n)+\frac{{\alpha }_n}{1-{\beta }_n}(B_n{x}_n-f(x_n))\\ +\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}(B_{n+1}{x}_{n+1}-B_n{x}_n)\parallel \\ \le & \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}M+\frac{{\alpha }_n}{1-{\beta }_n}M+\parallel B_{n+1}{x}_{n+1}-B_n{x}_n\parallel \\ \le & (\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}+\frac{{\alpha }_n}{1-{\beta }_n})M+\parallel B_{n+1}{x}_{n+1}-B_{n+1}{x}_n\parallel \\ +\parallel B_{n+1}{x}_n-B_n{x}_n\parallel \\ \le & (\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}+\frac{{\alpha }_n}{1-{\beta }_n})M+\parallel x_{n+1}-x_n\parallel +\parallel B_{n+1}{x}_n-B_n{x}_n\parallel \end{array}$$

for some $M>0$. It turns out that

$$\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel \le (\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}+\frac{{\alpha }_n}{1-{\beta }_n})M+\parallel B_{n+1}{x}_n-B_n{x}_n\parallel .$$

From conditions (C2), (C3) and Lemma 3.4, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

Lemma 2.3 yields that $\parallel z_n-x_n\parallel \to 0$ and hence

$$\parallel x_{n+1}-x_n\parallel =(1-{\beta }_n)\parallel z_n-x_n\parallel \to 0.$$

Step 3. We show that ${lim}_{n\to \mathrm{\infty }}\parallel B{x}_n-x_n\parallel =0$. Indeed, noting that

$$B_n{x}_n-x_n=\frac{1}{{\gamma }_n}\{(x_{n+1}-x_n)+{\alpha }_n(x_n-f(x_n))\},$$

we have, by (C2) and (C3),

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

Let $B:C\to C$ be the B-mapping generated by $T_1,T_2,\dots ,T_N$ and ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_N$. So, by Lemma 3.2, we have $B_{n}x\to Bx$ for all $x\in C$. It also follows that

$$\begin{array}{rcl}\parallel B{x}_n-x_n\parallel & \le & \parallel B{x}_n-B_n{x}_n\parallel +\parallel B_n{x}_n-x_n\parallel \\ \le & \underset{z\in \{x_n\}}{sup}\parallel Bz-B_{n}z\parallel +\parallel B_n{x}_n-x_n\parallel \\ \to & 0.\end{array}$$

For $t\in (0,1)$, we define a contraction as follows:

$$S_{t}x=tf(x)+(1-t)Bx.$$

Then there exists a unique path $x_t\in C$ such that

$$x_t=tf(x_t)+(1-t)B{x}_t.$$

From Lemma 2.5, we know that $x_t\to q$ as $t\to 0$, where $q\in F(B)$. Lemma 3.1(i) also yields that $q\in F(B)={\bigcap }_{i=1}^{N}F(T_i)$. Moreover, q is the unique solution of variational inequality (4.1).

Step 4. We show that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(q)-q,J(x_n-q)〉\le 0$. We see that

$$x_t-x_n=(1-t)(B{x}_t-x_n)+t(f(x_t)-x_n).$$

It follows, by Lemma 2.1(ii) that

$$\begin{array}{rcl}{\parallel x_t-x_n\parallel }^2& \le & {(1-t)}^2{\parallel B{x}_t-x_n\parallel }^2+2t〈f(x_t)-x_n,J(x_t-x_n)〉\\ \le & (1-2t+t^2){(\parallel x_t-x_n\parallel +\parallel B{x}_n-x_n\parallel )}^2\\ +2t〈f(x_t)-x_t,J(x_t-x_n)〉+2t{\parallel x_t-x_n\parallel }^2,\end{array}$$

which gives

$$〈f(x_t)-x_t,J(x_n-x_t)〉\le \frac{(1+t^2)\parallel x_n-B{x}_n\parallel }{2t}(2\parallel x_t-x_n\parallel +\parallel x_n-B{x}_n\parallel )+\frac{t}{2}{\parallel x_t-x_n\parallel }^2.$$

So we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(x_t)-x_t,J(x_n-x_t)〉\le \frac{t}{2}M$$

(4.2)

for some $M>0$. Since E has a uniformly Gâteaux differentiable norm, J is norm-to-weak^∗ uniformly continuous on bounded subsets of E. So we have

$$〈f(q)-q,J(x_n-q)-J(x_n-x_t)〉\to 0$$

(4.3)

and

$$〈f(q)-f(x_t)+x_t-q,J(x_n-x_t)〉\to 0$$

(4.4)

as $t\to 0$. On the other hand, we have

$$\begin{array}{rcl}〈f(q)-q,J(x_n-q)〉& =& 〈f(x_t)-x_t,J(x_n-x_t)〉\\ +〈f(q)-f(x_t)+x_t-q,J(x_n-x_t)〉\\ +〈f(q)-q,J(x_n-q)-J(x_n-x_t)〉.\end{array}$$

(4.5)

Since ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}$ and ${lim\hspace{0.17em}sup}_{t\to 0}$ are interchangeable, using (4.2)-(4.5), we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,J(x_n-q)〉\le 0.$$

Step 5. We show that $x_n\to q$ as $n\to \mathrm{\infty }$. In fact, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& =& {\alpha }_n〈f(x_n)-q,J(x_{n+1}-q)〉+{\beta }_n〈x_n-q,J(x_{n+1}-q)〉\\ +{\gamma }_n〈B_n{x}_n-q,J(x_{n+1}-q)〉\\ \le & {\alpha }_n\alpha \parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\\ +{\beta }_n\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +{\gamma }_n\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel \\ =& (1-{\alpha }_n(1-\alpha ))\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉\\ \le & \frac{1}{2}(1-{\alpha }_n(1-\alpha ))({\parallel x_n-q\parallel }^2+{\parallel x_{n+1}-q\parallel }^2)+{\alpha }_n〈f(q)-q,J(x_{n+1}-q)〉,\end{array}$$

which implies that

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2\le & \frac{1-{\alpha }_n(1-\alpha )}{1+{\alpha }_n(1-\alpha )}{\parallel x_n-q\parallel }^2+\frac{2{\alpha }_n}{1+{\alpha }_n(1-\alpha )}〈f(q)-q,J(x_{n+1}-q)〉\\ =& (1-\frac{2{\alpha }_n(1-\alpha )}{1+{\alpha }_n(1-\alpha )}){\parallel x_n-q\parallel }^2\\ +\frac{2{\alpha }_n}{1+{\alpha }_n(1-\alpha )}〈f(q)-q,J(x_{n+1}-q)〉.\end{array}$$

Put $c_n=\frac{2{\alpha }_n(1-\alpha )}{1+{\alpha }_n(1-\alpha )}$ and $b_n=\frac{2{\alpha }_n}{1+{\alpha }_n(1-\alpha )}〈f(q)-q,J(x_{n+1}-q)〉$. So it is easy to check that $\{c_n\}$ is a sequence in $(0,1)$ such that ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n=\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{b_n}{c_n}\le 0$. Hence, by Lemma 2.7, we conclude that $x_n\to q$ as $n\to \mathrm{\infty }$. This completes the proof. □