Common fixed points for some generalized nonexpansive mappings and nonspreading-type mappings in uniformly convex Banach spaces

Abstract

In this article, we study the fixed point theorems for nonspreading mappings, defined by Kohsaka and Takahashi, in Banach spaces but using the sense of norm instead of using the function ϕ. Furthermore, we prove a weak convergence theorem for finding a common fixed point of two quasi-nonexpansive mappings having demiclosed property in a uniformly convex Banach space. Consequently, such theorem can be deduced to the case of the nonspreading type mappings and some generalized nonexpansive mappings.

MSC:49J40, 47J20.

1 Introduction

Let T be a mapping on a nonempty subset E of a Banach space X. The mapping T is said to be quasi-nonexpansive[1] if $F(T)\ne \mathrm{\varnothing }$ and $\parallel Tx-y\parallel \le \parallel x-y\parallel$ for all $x\in E$ and for all $y\in F(T)$, where $F(T)$ denoted the set of all fixed points of T.

In 2008, Suzuki [2] introduced a condition on T which is weaker than nonexpansiveness and stronger than quasi-nonexpansiveness, called condition (C) and obtained some fixed point theorems for such mappings.

Since then, Dhompongsa et al.[3] extended Suzuki’s main theorems to a wider class of Banach spaces. Furthermore, the fixed point theorems of such mappings have been studied by the authors of [4–6], etc.

During the same period, Kohsaka and Takahashi [7] introduced a nonlinear mapping called nonspreading mapping in a smooth, strictly convex, and reflexive Banach space X as follows:

Let E be a nonempty closed and convex subset of X. Then, a mapping $S:E\to E$ is said to be nonspreading if

$$\varphi (Sx,Sy)+\varphi (Sy,Sx)\le \varphi (Sx,y)+\varphi (Sy,x)$$

(1.1)

for all $x,y\in E$, where $\varphi (x,y)={\parallel x\parallel }^2-2〈x,Jy〉+{\parallel y\parallel }^2$ for all $x,y\in X$ and J is the duality mapping on E. When X is a Hilbert space, we know that $\varphi (x,y)={\parallel x-y\parallel }^2$ for all $x,y\in X$ so a mapping $S:E\to E$ is said to be nonspreading if

$$2{\parallel Sx-Sy\parallel }^2\le {\parallel Sx-y\parallel }^2+{\parallel x-Sy\parallel }^2$$

(1.2)

for all $x,y\in E$.

Since then, some fixed point theorems of such mapping has been studied by many researchers such as [8–10].

To discuss about weak convergence theorems for two nonexpansive mappings $T_1$$T_2$ on E to itself, Takahashi and Tamura [11] constructed the following iterative scheme:

In 2011, Dhompongsa et al.[12] showed, by giving examples, that the class of nonspreading mappings is different from the class of mappings satisfying condition (C) and proved weak convergence theorems for a common fixed point of such two mappings in Hilbert spaces by using Takahashi and Tamura’s iterative scheme.

In this article, motivated by Dhompongsa et al.[12], we prove some fixed point theorems for nonspreading mappings for a general Banach space, i.e., nonspreading mappings satisfying (1.2) instead of (1.1). Furthermore, we prove a weak convergence theorem for a common fixed point of any two quasi-nonexpansive mappings having demiclosed property in a uniformly convex Banach space. Consequently, such theorem can be deduced to the case of the nonspreading type mappings and some generalized nonexpansive mappings.

2 Preliminaries

Let E be a nonempty closed and convex subset of a Banach space X and $\{x_n\}$ be a bounded sequence in X. For $x\in X$, define the asymptotic radius of $\{x_n\}$ at x as the number

$$r(x,\{x_n\})=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x\parallel .$$

Let

$$r\equiv r(E,\{x_n\}):=inf\{r(x,\{x_n\}):x\in E\}$$

and

$$A\equiv A(E,\{x_n\}):=\{x\in E:r(x,\{x_n\})=r\}.$$

The number r and the set A are, respectively, called the asymptotic radius and asymptotic center of $\{x_n\}$ relative to E. It is known that $A(E,\{x_n\})$ is nonempty, weakly compact and convex as E is [13].

Definition 2.1[14]

A Banach space X is said to have the Opial property if for each sequence $\{x_n\}\subset X$ weakly converging to a point $x\in X$ (denote as $x_n\rightharpoonup x$) and for any $y\in X$ such that $y\ne x$ there holds

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

or equivalently

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

Definition 2.2 The modulus of convexity of a Banach space X is the function ${\delta }_X:[0,2]\to [0,1]$ defined by

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

for all $\epsilon \in [0,2]$. A Banach space X is said to be uniformly convex if ${\delta }_X(0)=0$ and ${\delta }_X(\epsilon )>0$ for all $0<\epsilon \le 2$.

In 2008, the following condition was defined by Suzuki [2]:

Definition 2.3[2]

Let T be a mapping on a subset E of Banach space X. Then T is said to be a satisfy condition (C) if

$$\frac{1}{2}\parallel x-Tx\parallel \le \parallel x-y\parallel \text{implies}\parallel Tx-Ty\parallel \le \parallel x-y\parallel$$

for all $x,y\in E$.

We further have the following from [2].

Theorem 2.4[2]

Let E be a weakly compact convex subset of a uniformly convex Banach space X. Let T be a mapping on E. Assume that T satisfies condition (C). Then T has a fixed point.

Proposition 2.5[2]

Assume that a mapping T satisfies condition (C) and has a fixed point. Then T is a quasi-nonexpansive mapping.

Lemma 2.6[2]

Let T be a mapping on a closed subset E of a Banach space X. Assume that T satisfies condition (C). Then$F(T)$is closed. Moreover, if X is strictly convex and E is convex, then$F(T)$is also convex.

Proposition 2.7[2]

Let T be a mapping on subset E of Banach space X with the Opial property. Assume that T satisfies condition (C). If$\{x_n\}$converges weakly to z and${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$, then$Tz=z$. That is$(I-T)$is demiclosed at 0.

In 2008, Kohsaka and Takahashi [7] introduced the following nonlinear mapping.

Definition 2.8[7]

Let X be a smooth, strictly convex, and reflexive Banach space, J be the duality mapping of X and let E be a nonempty closed convex subset of X. Then, a mapping $S:E\to E$ is said to be nonspreading if

$$\varphi (Sx,Sy)+\varphi (Sy,Sx)\le \varphi (Sx,y)+\varphi (Sy,x)$$

for all $x,y\in E$, where $\varphi (x,y)={\parallel x\parallel }^2-2〈x,Jy〉+{\parallel y\parallel }^2$ for all $x,y\in X$. In the case when X is a Hilbert space, S is said to be nonspreading if $2{\parallel Sx-Sy\parallel }^2\le {\parallel Sx-y\parallel }^2+{\parallel x-Sy\parallel }^2$ for all $x,y\in E$.

Theorem 2.9[7]

Let X be a smooth, strictly convex, and reflexive Banach space, E be a nonempty closed convex subset of X and let S be a nonspreading mapping of E into itself. Then the following are equivalent:

• there exists$x\in E$such that$\{S^{n}x\}$is bounded;

• $F(S)$is nonempty.

In 2011, Dhompongsa et al.[12] proved that, by giving the following examples, in Banach spaces, the class of nonspreading mappings for a general Banach space and the class of mappings satisfying condition (C) are different. For the sake of completeness, we give the proof.

Example 1[12]

Define a mapping T on $[0,3]$ by

$$Tx=\{\begin{array}{cc}0,\hfill & \text{if}x\ne 3;\hfill \\ 2,\hfill & \text{if}x=3.\hfill \end{array}$$

From [2], T does not satisfy condition (C). But T is nonspreading. Indeed if $x=3$ and $y\ne 3$, we have

$$2{\parallel Tx-Ty\parallel }^2=8<9={\parallel Ty-x\parallel }^2.$$

It is easy to see in the other cases that $2{\parallel Tx-Ty\parallel }^2\le {\parallel x-Ty\parallel }^2+{\parallel y-Tx\parallel }^2$.

Example 2[12] Define a mapping T on $[0,1]$ by

$$Tx=1-x\text{for all}x\in [0,1].$$

Thus, T is nonexpansive mapping and hence it satisfies condition (C). But T is not nonspreading. In fact, if $x=0$ and $y=1$, we have

$$2{\parallel Tx-Ty\parallel }^2=2>0={\parallel x-Ty\parallel }^2+{\parallel y-Tx\parallel }^2.$$

The authors also studied the iterative scheme of Takahashi and Tamura [11] for approximation a common fixed point of nonspreading mappings and Suzuki’s mappings in Hilbert spaces as follows:

Theorem 2.10[12]

Let E be a nonempty closed convex subset of a Hilbert space H, let S be a nonspreading mapping of E into itself and let T be a condition (C) mapping of E into itself such that$F(S)\cap F(T)\ne \mathrm{\varnothing }$. Define a sequence$\{x_n\}$and$\{z_n\}$as follows:

for all$n\in \mathbb{N}$, where$\{{\alpha }_n\}\subset (0,1]$and$\{{\beta }_n\}\subset [0,1]$. Then, the following hold.

• if${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$and${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n<\mathrm{\infty }$, then$\{x_n\}$generated by (A) and$\{z_n\}$generated by (B) converge weakly to$v\in F(S)$and$u\in F(T)$, respectively;

• if${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n(1-{\beta }_n)>0$, then$\{x_n\}$generated by (A) and$\{z_n\}$generated by (B) converge weakly to$u\in F(S)\cap F(T)$and$v\in F(S)\cap F(T)$, respectively, where$u={lim}_{n\to \mathrm{\infty }}{P}_{F(S)\cap F(T)}{x}_n$and$v={lim}_{n\to \mathrm{\infty }}{P}_{F(S)\cap F(T)}{z}_n$.

Since our purpose is to study fixed point theorems of mappings defined on uniformly convex Banach spaces, we need the following result.

Lemma 2.11[15]

Let E be a uniformly convex Banach space and$r>0$. Then there exists a strictly increasing, continuous, and convex function$g:[0,2r]\to \mathbb{R}$such that$g(0)=0$and

$${\parallel tx+(1-t)y\parallel }^2\le t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t)g(\parallel x-y\parallel )$$

for all$x,y\in B_r$and$t\in [0,1]$, where$B_r=\{z\in E:\parallel z\parallel \le r\}$.

3 Fixed point theorems for nonspreading mappings for a general Banach space

We recall that $S:E\to E$ is a nonspreading mapping for a general Banach space if

$$2{\parallel Sx-Sy\parallel }^2\le {\parallel Sx-y\parallel }^2+{\parallel x-Sy\parallel }^2\text{for all}x,y\in E.$$

First, we consider the existence of a fixed point for such mappings in Banach spaces.

Theorem 3.1 Let X be a Banach space and E be a nonempty weakly compact convex subset of X such that$A(E,\{x_n\})$is singleton for all bounded sequence$\{x_n\}$in X. If$S:E\to E$is a nonspreading mapping for a general Banach space, then$F(S)$is nonempty.

Proof Let $x\in E$. Since E is weakly compact, E is bounded and hence $\{S^{n}x\}$ is bounded $\mathrm{\forall }n\in \mathbb{N}$. Let $y\in A(E,\{S^{n}x\})$. By the definition of S, we have

$${\parallel S^{n}x-Sy\parallel }^2\le \frac{1}{2}{\parallel S^{n}x-y\parallel }^2+\frac{1}{2}{\parallel S^{n-1}x-Sy\parallel }^2.$$

Therefore,

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel S^{n}x-Sy\parallel }^2& \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\frac{1}{2}{\parallel S^{n}x-y\parallel }^2+\frac{1}{2}{\parallel S^{n-1}x-Sy\parallel }^2),\\ \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{1}{2}{\parallel S^{n}x-Sy\parallel }^2& \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{1}{2}{\parallel S^{n}x-y\parallel }^2\end{array}$$

thus, we have ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\parallel S^{n}x-Sy\parallel }^2\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\parallel S^{n}x-y\parallel }^2$. This implies that $Sy\in A(E,\{S^{n}x\})$. By the uniqueness of $A(E,\{S^{n}x\})$, we have $Sy=y$ and hence $F(S)$ is nonempty. □

It follows from the fact that, in a uniformly convex Banach space, the asymptotic center of a bounded sequence with respect to a bounded closed convex subset is singleton. So, we have the following.

Theorem 3.2 Let X be a uniformly convex Banach space and E be a nonempty weakly compact convex subset of X. If$S:E\to E$is a nonspreading mapping for a general Banach space, then$F(S)$is nonempty.

Proposition 3.3 Let X be a Banach space and E be a nonempty subset of X. If$S:E\to E$is a nonspreading mapping for a general Banach space and$F(S)\ne \mathrm{\varnothing }$. Then S is a quasi-nonexpansive mapping.

Proof Let $x\in E$ and $y\in F(S)$. By the definition of S, we have

$$2{\parallel Sx-y\parallel }^2=2{\parallel Sx-Sy\parallel }^2\le {\parallel Sx-y\parallel }^2+{\parallel x-Sy\parallel }^2.$$

Therefore, ${\parallel Sx-y\parallel }^2\le {\parallel x-Sy\parallel }^2={\parallel x-y\parallel }^2$ and hence the proof is complete. □

Theorem 3.4 Let X be a uniformly convex Banach space and E be a nonempty weakly compact convex subset of X. Assume that$S:E\to E$is a nonspreading mapping for a general Banach space and$T:E\to E$satisfies condition (C). If S and T are commutative, then$F(S)\cap F(T)\ne \mathrm{\varnothing }$.

Proof By Theorem 2.4 and Lemma 2.6, we have $F(T)$ is nonempty, closed, and convex. By the commutative of S and T, we have $Sx=S(Tx)=T(Sx)$, and hence $Sx\in F(T)$ for all $x\in F(T)$. Therefore, $S:F(T)\to F(T)$. Since E is weakly compact convex and $F(T)$ is a closed subset of E, $F(T)$ is weakly compact convex. By Theorem 3.2, we have $F(S)\ne \mathrm{\varnothing }$. So there exists $y\in F(S)$ such that $y=Sy\in F(T)$ which implies that $y\in F(S)\cap F(T)$. □

Open problem Can Theorem 3.4 be improved to a commutative family $\mathcal{F}$ of nonspreading mappings for a general Banach space when $\mathcal{F}$ generates a left reversible semigroup (i.e., any two right ideals have nonvoid intersection) (see [16, 17])?

We show the demiclosedness of a nonspreading mapping for a general Banach space as follows:

Theorem 3.5 Let X be a Banach space having Opial property and E be a nonempty closed convex subset of X. Assume that$S:E\to E$is a nonspreading mapping for a general Banach space. If$\{x_n\}$is a sequence in E such that$x_n\rightharpoonup x$and${lim}_{n\to \mathrm{\infty }}\parallel S{x}_n-x_n\parallel =0$, then$x\in F(S)$.

Proof Let $x_n\rightharpoonup x$ and ${lim}_{n\to \mathrm{\infty }}\parallel S{x}_n-x_n\parallel =0$. Assume that $Sx\ne x$. By Opial property of X, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-x\parallel }^2<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-Sx\parallel }^2.$$

By the definition of S, we have

$$\begin{array}{rcl}{\parallel x_n-Sx\parallel }^2& \le & {(\parallel x_n-S{x}_n\parallel +\parallel S{x}_n-Sx\parallel )}^2\\ \le & {\parallel x_n-S{x}_n\parallel }^2+2\parallel x_n-S{x}_n\parallel \parallel S{x}_n-Sx\parallel +\frac{1}{2}{\parallel S{x}_n-x\parallel }^2+\frac{1}{2}{\parallel x_n-Sx\parallel }^2.\end{array}$$

Since $x_n\rightharpoonup x$, $\{x_n\}$ is bounded and hence $\{S{x}_n-Sx\}$ is bounded. Thus ${lim}_{n\to \mathrm{\infty }}\parallel S{x}_n-x_n\parallel =0$ implies that

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-Sx\parallel }^2& \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel S{x}_n-x\parallel }^2\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{(\parallel S{x}_n-x_n\parallel +\parallel x_n-x\parallel )}^2.\end{array}$$

By the boundedness of $\{x_n-x\}$ and ${lim}_{n\to \mathrm{\infty }}\parallel S{x}_n-x_n\parallel =0$, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-Sx\parallel }^2\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-x\parallel }^2$$

which is a contradiction. Thus we have $x\in F(S)$. □

Lemma 3.6 Let X be a Banach space. Let E be a nonempty closed convex subset of X. If$S:E\to E$and$T:E\to E$are quasi-nonexpansive mappings such that$F(S)\cap F(T)\ne \mathrm{\varnothing }$. Let$\{x_n\}$be defined as

$$\{\begin{array}{c}{x}_1=x\in E,\hfill \\ x_{n+1}={\alpha }_{n}S\{{\beta }_{n}T{x}_n+(1-{\beta }_n)x_n\}+(1-{\alpha }_n)x_n\hfill \end{array}$$

for all$n\in \mathbb{N}$, where$\{{\alpha }_n\}\subset (0,1)$and$\{{\beta }_n\}\subset (0,1)$.

Then${lim}_{n\to \mathrm{\infty }}\parallel x_n-w\parallel$exists for all$w\in F(T)\cap F(S)$and$\{x_n\}$is bounded.

Proof Let $w\in F(S)\cap F(T)$ and $y_n={\beta }_{n}T{x}_n+(1-{\beta }_n)x_n$. By the quasi-nonexpansiveness of S and T, we have

$$\begin{array}{rcl}\parallel S{y}_n-w\parallel & \le & \parallel y_n-w\parallel \\ =& \parallel {\beta }_{n}T{x}_n+(1-{\beta }_n)x_n-w\parallel \\ \le & {\beta }_n\parallel x_n-w\parallel +(1-{\beta }_n)\parallel x_n-w\parallel \\ =& \parallel x_n-w\parallel .\end{array}$$

(3.1)

By (3.1) we have,

$$\begin{array}{rcl}\parallel x_{n+1}-w\parallel & =& \parallel {\alpha }_{n}S{y}_n+(1-{\alpha }_n)x_n-w\parallel \\ \le & {\alpha }_n\parallel S{y}_n-w\parallel +(1-{\alpha }_n)\parallel x_n-w\parallel \\ \le & {\alpha }_n\parallel x_n-w\parallel +(1-{\alpha }_n)\parallel x_n-w\parallel \\ =& \parallel x_n-w\parallel .\end{array}$$

We can conclude by induction that $\parallel x_n-w\parallel \le \parallel x-w\parallel$ for all $n\in \mathbb{N}$. This imply that $\{\parallel x_n-w\parallel \}$ is a decreasing and bounded sequence and hence ${lim}_{n\to \mathrm{\infty }}\parallel x_n-w\parallel$ exists. Furthermore, $\{x_n\}$ is bounded since $\parallel x_n\parallel \le \parallel x_n-w\parallel +\parallel w\parallel$. □

Now, we are in a position to prove our main result.

Theorem 3.7 Let X be a uniformly convex Banach space having Opial property. Let E be a nonempty closed convex subset of X. If$S:E\to E$and$T:E\to E$are quasi-nonexpansive mappings having demiclosed property. Assume that$F(S)\cap F(T)\ne \mathrm{\varnothing }$. Let$\{x_n\}$be defined as

$$\{\begin{array}{c}{x}_1=x\in E,\hfill \\ x_{n+1}={\alpha }_{n}S\{{\beta }_{n}T{x}_n+(1-{\beta }_n)x_n\}+(1-{\alpha }_n)x_n\hfill \end{array}$$

for all$n\in \mathbb{N}$, where$\{{\alpha }_n\}\subset (0,1)$and${\beta }_n\subset (0,1)$.

Then${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n(1-{\beta }_n)>0$imply that$x_n\rightharpoonup v\in F(S)\cap F(T)$.

Proof Let $w\in F(S)\cap F(T)$. As in the proof in Lemma 3.6, we have $\parallel x_n-w\parallel \le \parallel x-w\parallel$ for all $n\in \mathbb{N}$. Using Lemma 2.11, we put $r=\parallel x-w\parallel$ so that there exists a strictly increasing, continuous, and convex function $g:[0,2r]\to \mathbb{R}$ such that $g(0)=0$ and

$$\begin{array}{rcl}{\parallel y_n-w\parallel }^2& =& {\parallel {\beta }_n(T{x}_n-w)+(1-{\beta }_n)(x_n-w)\parallel }^2\\ \le & {\beta }_n{\parallel T{x}_n-w\parallel }^2+(1-{\beta }_n){\parallel x_n-w\parallel }^2-{\beta }_n(1-{\beta }_n)g(\parallel T{x}_n-x_n\parallel ).\end{array}$$

Hence, by the quasi-nonexpansiveness of T, we obtain

$${\parallel y_n-w\parallel }^2\le {\parallel x_n-w\parallel }^2-{\beta }_n(1-{\beta }_n)g(\parallel T{x}_n-x_n\parallel )$$

(3.2)

$$\le {\parallel x_n-w\parallel }^2.$$

(3.3)

From ${\parallel x_{n+1}-w\parallel }^2={\parallel {\alpha }_n(S{y}_n-w)+(1-{\alpha }_n)(x_n-w)\parallel }^2$ and (3.3), we put $r=\parallel x-w\parallel$ in Lemma 2.11 again to get a strictly increasing, continuous, and convex function $g:[0,2r]\to \mathbb{R}$ such that $g(0)=0$ and

$${\parallel x_{n+1}-w\parallel }^2\le {\alpha }_n{\parallel S{y}_n-w\parallel }^2+(1-{\alpha }_n){\parallel x_n-w\parallel }^2-{\alpha }_n(1-{\alpha }_n)g(\parallel S{y}_n-x_n\parallel ).$$

By the quasi-nonexpansiveness of S and from (3.3), we obtain

$${\parallel x_{n+1}-w\parallel }^2\le {\alpha }_n{\parallel y_n-w\parallel }^2+(1-{\alpha }_n){\parallel x_n-w\parallel }^2-{\alpha }_n(1-{\alpha }_n)g(\parallel S{y}_n-x_n\parallel )$$

(3.4)

$$\begin{array}{rc}\le & {\alpha }_n{\parallel x_n-w\parallel }^2+(1-{\alpha }_n){\parallel x_n-w\parallel }^2-{\alpha }_n(1-{\alpha }_n)g(\parallel S{y}_n-x_n\parallel )\\ =& {\parallel x_n-w\parallel }^2-{\alpha }_n(1-{\alpha }_n)g(\parallel S{y}_n-x_n\parallel ).\end{array}$$

(3.5)

Hence

$${\alpha }_n(1-{\alpha }_n)g(\parallel S{y}_n-x_n\parallel )\le {\parallel x_n-w\parallel }^2-{\parallel x_{n+1}-w\parallel }^2.$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, there exist $k_1>0$ and $N\in \mathbb{N}$ such that

$${\alpha }_n(1-{\alpha }_n)\ge k_1,\mathrm{\forall }n\ge N.$$

By Lemma 3.6, we have

$$0=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}({\parallel x_n-w\parallel }^2-{\parallel x_{n+1}-w\parallel }^2)$$

and hence

$$0\ge \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\alpha }_n(1-{\alpha }_n)g(\parallel S{y}_n-x_n\parallel )\ge k_1\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}g(\parallel S{y}_n-x_n\parallel ).$$

Since $k_1>0$, we have ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}g(\parallel S{y}_n-x_n\parallel )=0$ and hence ${lim}_{n\to \mathrm{\infty }}g(\parallel S{y}_n-x_n\parallel )=0$.

Since $\parallel S{y}_n-x_n\parallel \le 2\parallel x-w\parallel$ for all $n\in \mathbb{N}$, $\{\parallel S{y}_n-x_n\parallel \}$ is bounded and hence we can put $M={lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel S{y}_n-x_n\parallel$. So there exists $\{\parallel S{y}_{n_k}-x_{n_k}\parallel \}\subset \{\parallel S{y}_n-x_n\parallel \}$ such that

$$\underset{k\to \mathrm{\infty }}{lim}\parallel S{y}_{n_k}-x_{n_k}\parallel =M.$$

Since g is a continuous function, we have

$$0=\underset{k\to \mathrm{\infty }}{lim}g(\parallel S{y}_{n_k}-x_{n_k}\parallel )=g(M).$$

Since $g(0)=0$ and g is strictly increasing, $M=0$.

Therefore, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel S{y}_n-x_n\parallel =0$ and hence ${lim}_{n\to \mathrm{\infty }}\parallel S{y}_n-x_n\parallel =0$.

From (3.4), we have

$$\begin{array}{rcl}0& \le & {\alpha }_n(1-{\alpha }_n)g(\parallel S{y}_n-x_n\parallel )\\ \le & {\parallel x_n-w\parallel }^2-{\parallel x_{n+1}-w\parallel }^2+{\alpha }_n({\parallel y_n-w\parallel }^2-{\parallel x_n-w\parallel }^2).\end{array}$$

Hence,

$${\alpha }_n({\parallel x_n-w\parallel }^2-{\parallel y_n-w\parallel }^2)\le {\parallel x_n-w\parallel }^2-{\parallel x_{n+1}-w\parallel }^2.$$

(3.6)

Since ${\alpha }_n(1-{\alpha }_n)<{\alpha }_n$ for all $n\in \mathbb{N}$, $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n$.

Therefore, there exist $k_2>0$ and $N\in \mathbb{N}$ such that

$${\alpha }_n\ge k_2,\mathrm{\forall }n\ge N.$$

Then from (3.6) and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-w\parallel$ exists, we have ${lim}_{n\to \mathrm{\infty }}({\parallel x_n-w\parallel }^2-{\parallel y_n-w\parallel }^2)=0$.

On the other hand, we have from (3.2) that

$${\beta }_n(1-{\beta }_n)g(\parallel T{x}_n-x_n\parallel )\le {\parallel x_n-w\parallel }^2-{\parallel y_n-w\parallel }^2.$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n(1-{\beta }_n)>0$ so there exist $k_3>0$ and $N\in \mathbb{N}$ such that

$${\alpha }_n\ge k_3,\mathrm{\forall }n\ge N.$$

Therefore, we can conclude that ${lim}_{n\to \mathrm{\infty }}g(\parallel T{x}_n-x_n\parallel )=0$.

Similarly, the continuity and strictly convexity of g imply that ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$.

Since $\{x_n\}$ is bounded, there exists $\{x_{n_i}\}\subset \{x_n\}$ such that $x_{n_i}\rightharpoonup v$. From demiclosedness of T, we have $v\in F(T)$. Since

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel y_n-x_n\parallel & =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel {\beta }_{n}T{x}_n+(1-{\beta }_n)x_n-x_n\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\beta }_n\parallel T{x}_n-x_n\parallel ,\end{array}$$

where $\{{\beta }_n\}\subset (0,1)$ and ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$, we have ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$.

Using ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$ and $x_{n_i}\rightharpoonup v$, by passing through subsequences, if necessary, we can assume that there exists a weakly convergent subsequence $\{y_{n_i}\}$ of $\{y_n\}$ such that $y_{n_i}\rightharpoonup v$.

Furthermore, consider

$$\begin{array}{rcl}\parallel S{y}_n-y_n\parallel & =& \parallel S{y}_n-x_n+x_n-y_n\parallel \\ \le & \parallel S{y}_n-x_n\parallel +\parallel x_n-y_n\parallel .\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel S{y}_n-x_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel S{y}_n-y_n\parallel =0$.

By the demiclosedness of S, we have $v\in F(S)$ and hence $v\in F(S)\cap F(T)$.

Finally, we show that $x_n\rightharpoonup v$. Let $\{x_{n_k}\}$ be arbitrary subsequence of $\{x_n\}$. Since $\{x_{n_k}\}$ is bounded, there exists $\{x_{n_{k_i}}\}\subset \{x_{n_k}\}$ that $x_{n_{k_i}}\rightharpoonup u$. The same proof as v above, there exists $\{y_{n_{k_i}}\}\subset \{y_{n_k}\}$ such that $y_{n_{k_i}}\rightharpoonup u$ and $u\in F(S)\cap F(T)$.

Suppose that $v\ne u$. Using Lemma 3.6 to guarantee that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-v\parallel$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-u\parallel$ exist and hence we have from the Opial property that

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-v\parallel & =& \underset{i\to \mathrm{\infty }}{lim}\parallel x_{n_i}-v\parallel \\ <& \underset{i\to \mathrm{\infty }}{lim}\parallel x_{n_i}-u\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-u\parallel \\ =& \underset{i\to \mathrm{\infty }}{lim}\parallel x_{n_{k_i}}-u\parallel \\ <& \underset{i\to \mathrm{\infty }}{lim}\parallel x_{n_{k_i}}-v\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-v\parallel .\end{array}$$

This is a contradiction. So $x_n\rightharpoonup v\in F(T)\cap F(S)$. □

Since the class of nonspreading mappings for a general Banach space is different from the class of mappings satisfying condition (C), we can apply Proposition 2.5 and Proposition 3.3 to deduce Theorem 3.7 as follows:

Corollary 3.8 Let X be a uniformly convex Banach space having Opial property. Let E be a nonempty closed convex subset of X. Assume that$S:E\to E$is a nonspreading mapping for a general Banach space and$T:E\to E$satisfies condition (C) such that$F(S)\cap F(T)\ne \mathrm{\varnothing }$. Let$\{x_n\}$and$\{z_n\}$be defined as

for all$n\in \mathbb{N}$, where$\{{\alpha }_n\}\subset (0,1)$and${\beta }_n\subset (0,1)$.

If${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$and${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n(1-{\beta }_n)>0$, then$\{x_n\}$generated by (A) and$\{z_n\}$generated by (B) converge weakly to$u\in F(S)\cap F(T)$and$v\in F(S)\cap F(T)$, respectively.