Common fixed-point results in uniformly convex Banach spaces

Abstract

We introduce a condition on mappings, namely condition $(K)$. In a uniformly convex Banach space, the condition is weaker than quasi-nonexpansiveness and weaker than asymptotic nonexpansiveness. We also present the existence theorem of common fixed points for a commuting pair consisting of a mapping satisfying condition $(K)$ and a multivalued mapping satisfying conditions $(E)$ and $(C_{\lambda })$ for some $\lambda \in (0,1)$.

1 Introduction

In 1978, Itoh and Takahashi [1] established the existence of common fixed points of a quasi-nonexpansive mapping and a multivalued nonexpansive mapping by an elementary constructive method in a Hilbert space. In 2006, Dhompongsa et al. [2] obtained a common fixed point result for a commuting pair of single-valued and multivalued nonexpansive mappings in uniformly convex Banach spaces. The analogy result in CAT(0) spaces was also proved by Dhompongsa et al. [3]. Since then, Shahzad and Markin [4] studied an invariant approximation problem and provided sufficient conditions for the existence of $z\in E\subseteq X$ such that $d(z,y)=dist(y,E)$ and $z=t(z)\in T(z)$, where $y\in X$, t and T are commuting nonexpansive mappings on E. In 2009, Shahzad [5] also obtained a common fixed point and invariant approximation result in a CAT(0) space in which t and T are weakly commuting.

Motivated by Suzuki’s result [6], Garcia-Falset et al. [7] introduced two kinds of generalizations for condition $(C)$, namely conditions $(E)$ and $(C_{\lambda })$ and studied both the existence of fixed points and their asymptotic behavior. Recently, Abkar and Eslamian [8] proved that if E is a nonempty closed convex and bounded subset of a complete CAT(0) space X, $t:E\to E$ is a single-valued quasi-nonexpansive mapping, and $T:E\to KC(E)$ is a multivalued mapping satisfying conditions $(E)$ and $(C_{\lambda })$ for some $\lambda \in (0,1)$ such that t and T are weakly commuting, then there exists a point $z\in E$ such that $z=t(z)\in T(z)$. This result was extended to the general setting of uniformly convex metric spaces by Laowang and Panyanak [9].

In this paper, we first introduce the following condition.

Definition 1.1 Let t be a mapping on a subset E of a Banach space X. Then t is said to satisfy condition $(K)$ if

• 1. the fixed point set $Fix(t)$ is nonempty closed and convex, and

• 2. for every $x\in Fix(t)$, a closed convex subset A with $t(A)\subseteq A$, and $y\in A$ such that $\parallel x-y\parallel =dist(x,A)$, we have $y\in Fix(t)$.

We show that, in a uniformly convex Banach space, condition $(K)$ is weaker than quasi-nonexpansiveness and weaker than asymptotic nonexpansiveness. We also present the existence theorem of common fixed points for a commuting pair consisting of a mapping satisfying condition $(K)$ and a multivalued mapping satisfying conditions $(E)$ and $(C_{\lambda })$ for some $\lambda \in (0,1)$. Consequently, such a theorem extends many results in the literature.

2 Preliminaries

In this section, we give some preliminaries.

A mapping t on a subset E of a Banach space X is called an asymptotically nonexpansive mapping if for each $n\in \mathbb{N}$, there exists a positive constant $k_n\ge 1$ with ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ such that

$$\parallel t^n(x)-t^n(y)\parallel \le k_n\parallel x-y\parallel \text{for all }x,y\in E.$$

If $k_n\equiv 1$ for all $n\in \mathbb{N}$, then t is called a nonexpansive mapping. We denote by $Fix(t)$ the set of fixed points of t, i.e., $Fix(t)=\{x\in E:x=t(x)\}$.

We shall denote by $FB(E)$ the family of nonempty bounded closed subsets of E and by $KC(E)$ the family of nonempty compact convex subsets of E. Let $H(\cdot ,\cdot )$ be the Hausdorff distance on $FB(X)$, i.e.,

$$H(A,B)=max\{\underset{a\in A}{sup}dist(a,B),\underset{b\in B}{sup}dist(b,A)\},A,B\in FB(X),$$

where $dist(a,B)=inf\{\parallel a-b\parallel :b\in B\}$ is the distance from the point a to the subset B.

A multivalued mapping $T:E\to FB(X)$ is said to be nonexpansive if

$$H(T(x),T(y))\le \parallel x-y\parallel \text{for all }x,y\in E.$$

Definition 2.1 A multivalued mapping $T:X\to FB(X)$ is said to satisfy condition $(E_{\mu })$ provided that

$$dist(x,T(y))\le \mu dist(x,T(x))+\parallel x-y\parallel ,\mathrm{\forall }x,y\in X.$$

We say that T satisfies condition $(E)$ whenever T satisfies $(E_{\mu })$ for some $\mu \ge 1$.

Definition 2.2 A multivalued mapping $T:X\to FB(X)$ is said to satisfy condition $(C_{\lambda })$ for some $\lambda \in (0,1)$ provided that

$$\lambda dist(x,T(x))\le \parallel x-y\parallel \Rightarrow H(T(x),T(y))\le \parallel x-y\parallel ,\mathrm{\forall }x,y\in X.$$

A point x is called a fixed point for a multivalued mapping T if $x\in T(x)$. A single valued mapping $t:E\to E$ and a multivalued mapping $T:E\to FB(E)$ are said to be commute if $t(T(x))\subseteq T(t(x))$ for all $x\in E$.

A Banach space X is said to be strictly convex if

$$\parallel x+y\parallel <2$$

for all $x,y\in X$ with $\parallel x\parallel =\parallel y\parallel =1$ and $x\ne y$. We recall that a Banach space X is called uniformly convex (Clarkson [10]) if, for each $\epsilon >0$, there is a $\delta >0$ such that if $\parallel x\parallel =\parallel y\parallel =1$ then

$$\parallel \frac{(x+y)}{2}\parallel \le 1-\delta .$$

It is obvious that uniform convexity implies strict convexity.

In 1991, Xu [11] proved the characterization of uniform convexity as follows.

Theorem 2.3 [11]

A Banach space X is uniformly convex if and only if for each fixed number $r>0$, there exists a continuous function $\phi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, $\phi (s)=0\iff s=0$, such that

$${\parallel \lambda x+(1-\lambda )y\parallel }^2\le \lambda {\parallel x\parallel }^2+(1-\lambda ){\parallel y\parallel }^2-\lambda (1-\lambda )\phi (\parallel x-y\parallel )$$

for all $\lambda \in [0,1]$ and all $x,y\in X$ such that $\parallel x\parallel \le r$ and $\parallel y\parallel \le r$.

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 as nonempty, weakly compact and convex as E is [12]. The sequence $\{x_n\}$ is called regular relative to E if $r(E,\{x_n\})=r(E,\{x_{n_k}\})$ for each subsequence $\{x_{n_k}\}$ of $\{x_n\}$.

Goebel [13] and Lim [14] proved the following lemma.

Lemma 2.4 Let $\{x_n\}$ be a bounded sequence in X, and let E be a nonempty closed convex subset of X. Then $\{x_n\}$ has a subsequence which is regular relative to E.

The following result was proved by Goebel and Kirk [15].

Lemma 2.5 Let $\{z_n\}$ and $\{w_n\}$ be bounded sequences in a Banach space X, and let $0<\lambda <1$. If, for every natural number n, we have $z_{n+1}=\lambda w_n+(1-\lambda )z_n$ and $\parallel w_{n+1}-w_n\parallel \le \parallel z_{n+1}-z_n\parallel$, then ${lim}_{n\to \mathrm{\infty }}\parallel w_n-z_n\parallel =0$.

3 Main results

We recall that a mapping t on a subset E of a Banach space X is called quasi-nonexpansive [16] if $\parallel x-t(y)\parallel \le \parallel x-y\parallel$ for all $y\in E$ and $x\in Fix(t)$. From the definition, we can see that nonexpansive mappings with a fixed point are quasi-nonexpansive.

We first show that a quasi-nonexpansive mapping defined on a strictly convex Banach space X satisfies condition $(K)$.

Proposition 3.1 Let E be a strictly convex subset of a Banach space. If $t:E\to E$ is a quasi-nonexpansive mapping, then t satisfies condition $(K)$.

Proof It is known that $Fix(t)$ is nonempty closed and convex [[17], Theorem 1]. Let $x\in Fix(t)$ and A be a closed convex subset with $t(A)\subseteq A$. Let $y\in A$ be such that $\parallel x-y\parallel =dist(x,A)$. Quasi-nonexpansiveness of t implies that

$$\parallel x-t(y)\parallel \le \parallel x-y\parallel .$$

Since E is strictly convex and $t(A)\subseteq A$, it must be the case that $t(y)=y$. Therefore, t satisfies condition $(K)$. □

An asymptotically nonexpansive mapping defined on a uniformly convex Banach space also satisfies condition $(K)$.

Proposition 3.2 Let X be a uniformly convex Banach space and E be a nonempty subset of X. If $t:E\to E$ is an asymptotically nonexpansive mapping with $Fix(t)\ne \mathrm{\varnothing }$, then t satisfies condition $(K)$.

Proof The set $Fix(t)$ is closed and convex [[18], Theorem 2]. Let $x\in Fix(t)$, A be a closed convex subset of E with $t(A)\subseteq A$, and $y\in A$ be such that $\parallel x-y\parallel =dist(x,A)$. By Theorem 2.3, there exists a continuous function ψ such that for all integers $l,m\ge 1$,

$$\begin{array}{rcl}{\parallel x-\left(\frac{t^l(y)+t^m(y)}{2}\right)\parallel }^2& \le & \frac{1}{2}{\parallel x-t^l(y)\parallel }^2+\frac{1}{2}{\parallel x-t^m(y)\parallel }^2-\frac{1}{4}\phi \left(\parallel t^l(y)-t^m(y)\parallel \right)\\ \le & \frac{1}{2}{k}_l^2{\parallel x-y\parallel }^2+\frac{1}{2}{k}_m^2{\parallel x-y\parallel }^2-\frac{1}{4}\phi \left(\parallel t^l(y)-t^m(y)\parallel \right).\end{array}$$

(3.1)

Since $\parallel x-y\parallel =dist(x,A)$ and A is convex, we have

$${\parallel x-y\parallel }^2\le {\parallel x-\left(\frac{t^l(y)+t^m(y)}{2}\right)\parallel }^2.$$

Thus,

$$\phi \left(\parallel t^l(y)-t^m(y)\parallel \right)\le 4((k_l^2+k_m^2)/2-1){\parallel x-y\parallel }^2.$$

Since t is asymptotically nonexpansive, the right-hand side of the inequality tends to zero as l, m tend to infinity. Hence, $\{t^i(y)\}$ is a Cauchy sequence. Let ${lim}_{i\to \mathrm{\infty }}{t}^i(y)=z\in A$. We have

$$\parallel t(z)-t^{i+1}(y)\parallel \le k_1\parallel z-t^i(y)\parallel .$$

By letting $i\to \mathrm{\infty }$, we can conclude that $\parallel t(z)-z\parallel \le 0$, that is $z\in Fix(t)$. Now, letting $l,m\to \mathrm{\infty }$ in (3.1) yields

$${\parallel x-z\parallel }^2\le {\parallel x-y\parallel }^2.$$

Since $z\in A$ and $\parallel x-y\parallel =dist(x,A)$, thus $y=z$. Therefore, $y\in Fix(t)$. □

The following example shows that the class of mappings satisfying condition $(K)$ is strictly wider than the class of quasi-nonexpansive mappings and asymptotically nonexpansive mappings.

Example 3.3 Let f be a function on $[0,1]$ defined by

$$f(x)=\{\begin{array}{cc}x,\hfill & x\le \frac{1}{2};\hfill \\ 0,\hfill & x>\frac{1}{2}.\hfill \end{array}$$

Then f is neither quasi-nonexpansive nor asymptotically nonexpansive. However, f satisfies condition $(K)$.

We are now in a position to state our main theorem.

Theorem 3.4 Let E be a nonempty bounded closed convex subset of a uniformly convex Banach space X. Let $t:E\to E$ be a mapping satisfying condition $(K)$, and let $T:E\to KC(E)$ be a multivalued mapping satisfying conditions $(E)$ and $(C_{\lambda })$ for some $\lambda \in (0,1)$. If t and T are commute, then they have a common fixed point, that is, there exists a point $z\in E$ such that $z=t(z)\in T(z)$.

Proof Commutative property of t and T implies that $t(T(x))\subseteq T(x)$ for all $x\in Fix(t)$. Then we have $Fix(t)\cap T(x)\ne \mathrm{\varnothing }$ for all $x\in Fix(t)$ since t satisfies condition $(K)$.

Now, we find an approximate fixed point sequence in $Fix(t)$ for T. Take $x_0\in Fix(t)$. Since $Fix(t)\cap T(x_0)\ne \mathrm{\varnothing }$, we choose $y_0\in Fix(t)\cap T(x_0)$. Define

$$x_1=(1-\lambda )x_0+\lambda y_0.$$

Since $Fix(t)$ is convex, we have $x_1\in Fix(t)$. Let $y_1\in T(x_1)$ such that $\parallel y_0-y_1\parallel =dist(y_0,T(x_1))$. We get $y_1\in Fix(t)$ since t satisfies condition $(K)$. Put

$$x_2=(1-\lambda )x_1+\lambda y_1.$$

Again, choose $y_2\in T(x_2)$ such that $\parallel y_1-y_2\parallel =dist(y_1,T(x_2))$. Similarly, we get $y_2\in Fix(t)$. We have a sequence $\{x_n\}\subseteq Fix(t)$ such that

$$x_{n+1}=(1-\lambda )x_n+\lambda y_n,$$

where $y_n\in T(x_n)\cap Fix(t)$ and $\parallel y_{n-1}-y_n\parallel =dist(y_{n-1},T(x_n))$. For every natural number $n\ge 1$, we have

$$\parallel x_{n+1}-x_n\parallel =\lambda \parallel x_n-y_n\parallel .$$

It follows that

$$\lambda dist(x_n,T(x_n))\le \lambda \parallel x_n-y_n\parallel =\parallel x_{n+1}-x_n\parallel .$$

Since T satisfies condition ($C_{\lambda }$), we have

$$H(T(x_n),T(x_{n+1}))\le \parallel x_n-x_{n+1}\parallel .$$

Hence, for each $n\ge 1$, we have

$$\parallel y_n-y_{n+1}\parallel =dist(y_n,T(x_{n+1}))\le H(T(x_n),T(x_{n+1}))\le \parallel x_n-x_{n+1}\parallel .$$

We now apply Lemma 2.5 to conclude that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$. That is, as $n\to \mathrm{\infty }$,

$$dist(x_n,T(x_n))\le \parallel x_n-y_n\parallel \to 0.$$

Passing through a subsequence, if necessary, we can assume that $\{x_n\}$ is regular. Let $A(Fix(t),\{x_n\})=\{z\}$. For each $n\ge 1$, we choose $z_n\in T(z)$ such that $\parallel x_n-z_n\parallel =dist(x_n,T(z))$. Since t satisfies condition $(K)$, we have $z_n\in Fix(t)$. The compactness of $T(z)$ implies that the sequence $\{z_n\}$ has a convergent subsequence $\{z_{n_k}\}$ with the limit point $w\in T(z)$. We also obtain $w\in Fix(t)$ since $Fix(t)$ is closed. By the condition $(E)$ of T, we have for some $\mu \ge 1$,

$$dist(x_{n_k},T(z))\le \mu dist(x_{n_k},T(x_{n_k}))+\parallel x_{n_k}-z\parallel .$$

Note that

$$\parallel x_{n_k}-w\parallel \le \parallel x_{n_k}-z_{n_k}\parallel +\parallel z_{n_k}-w\parallel \le \mu dist(x_{n_k},T(x_{n_k}))+\parallel x_{n_k}-z\parallel +\parallel z_{n_k}-w\parallel .$$

These entail

$$\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-w\parallel \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-z\parallel .$$

Since $\{x_n\}$ is regular, and an asymptotic center of a bounded sequence in a uniformly convex Banach space is a singleton set, these show that $z=w\in T(z)$. Hence, $z=t(z)\in T(z)$. □

As a consequence of Proposition 3.1, Proposition 3.2, and Theorem 3.4, we obtain the following corollaries.

Corollary 3.5 Let E be a nonempty bounded closed convex subset of a uniformly convex Banach space X. Let $t:E\to E$ be a quasi-nonexpansive mapping, and let $T:E\to KC(E)$ be a multivalued mapping satisfying conditions $(E)$ and $(C_{\lambda })$ for some $\lambda \in (0,1)$. If t and T are commute, then they have a common fixed point, that is, there exists a point $z\in E$ such that $z=t(z)\in T(z)$.

Corollary 3.6 Let E be a nonempty bounded closed convex subset of a uniformly convex Banach space X. Let $t:E\to E$ be an asymptotically nonexpansive mapping, and let $T:E\to KC(E)$ be a multivalued mapping satisfying conditions $(E)$ and $(C_{\lambda })$ for some $\lambda \in (0,1)$. If t and T are commute, then they have a common fixed point, that is, there exists a point $z\in E$ such that $z=t(z)\in T(z)$.