Common fixed points for some generalized nonexpansive mappings and nonspreadingtype mappings in uniformly convex Banach spaces
 Warunun Inthakon^{1, 2}Email author,
 Attapol Kaewkhao^{1} and
 Nutchari Niyamosot^{1}
https://doi.org/10.1186/168718122012110
© Inthakon et al.; licensee Springer 2012
Received: 24 April 2012
Accepted: 20 June 2012
Published: 10 July 2012
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 quasinonexpansive 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.
Keywords
1 Introduction
Let T be a mapping on a nonempty subset E of a Banach space X. The mapping T is said to be quasinonexpansive[1] if $F(T)\ne \mathrm{\varnothing}$ and $\parallel Txy\parallel \le \parallel xy\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 quasinonexpansiveness, 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:
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].
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 quasinonexpansive 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
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]
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]
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 quasinonexpansive 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$(IT)$is demiclosed at 0.
In 2008, Kohsaka and Takahashi [7] introduced the following nonlinear mapping.
Definition 2.8[7]
for all $x,y\in E$, where $\varphi (x,y)={\parallel x\parallel}^{2}2\u3008x,Jy\u3009+{\parallel y\parallel}^{2}$ for all $x,y\in X$. In the case when X is a Hilbert space, S is said to be nonspreading if $2{\parallel SxSy\parallel}^{2}\le {\parallel Sxy\parallel}^{2}+{\parallel xSy\parallel}^{2}$ 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]
It is easy to see in the other cases that $2{\parallel TxTy\parallel}^{2}\le {\parallel xTy\parallel}^{2}+{\parallel yTx\parallel}^{2}$.
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]
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]
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
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.
thus, we have ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty}}{\parallel {S}^{n}xSy\parallel}^{2}\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty}}{\parallel {S}^{n}xy\parallel}^{2}$. 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 quasinonexpansive mapping.
Therefore, ${\parallel Sxy\parallel}^{2}\le {\parallel xSy\parallel}^{2}={\parallel xy\parallel}^{2}$ 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$x\in F(S)$.
which is a contradiction. Thus we have $x\in F(S)$. □
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.
We can conclude by induction that $\parallel {x}_{n}w\parallel \le \parallel xw\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 $. □
Now, we are in a position to prove our main result.
for all$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)$.
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 $g(0)=0$ 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 ${lim}_{n\to \mathrm{\infty}}\parallel S{y}_{n}{x}_{n}\parallel =0$.
Since ${\alpha}_{n}(1{\alpha}_{n})<{\alpha}_{n}$ for all $n\in \mathbb{N}$, $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{\alpha}_{n}(1{\alpha}_{n})<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{\alpha}_{n}$.
Then from (3.6) and ${lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}w\parallel $ exists, we have ${lim}_{n\to \mathrm{\infty}}({\parallel {x}_{n}w\parallel}^{2}{\parallel {y}_{n}w\parallel}^{2})=0$.
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 ${lim}_{n\to \mathrm{\infty}}\parallel T{x}_{n}{x}_{n}\parallel =0$.
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 ${y}_{{n}_{i}}\rightharpoonup v$.
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 $u\in F(S)\cap F(T)$.
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:
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.
Declarations
Acknowledgements
This article is dedicated to Professor Anthony ToMing Lau for celebrating his great achievements in the development of fixed point theory and applications. The authors are indebted to the anonymous referee(s) for comments which lead to the improvement and for the kindness in providing us the open problem in the article. This research was (partially) supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.
Authors’ Affiliations
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