# Nonexpansive mappings on Abelian Banach algebras and their fixed points

- W Fupinwong
^{1, 2}Email author

**2012**:150

https://doi.org/10.1186/1687-1812-2012-150

© Fupinwong; licensee Springer 2012

**Received: **25 June 2012

**Accepted: **24 August 2012

**Published: **14 September 2012

## Abstract

A Banach space *X* is said to have the fixed point property if for each nonexpansive mapping $T:E\to E$ on a bounded closed convex subset *E* of *X* has a fixed point. We show that each infinite dimensional Abelian complex Banach algebra *X* satisfying: (i) property (A) defined in (Fupinwong and Dhompongsa in Fixed Point Theory Appl. 2010:Article ID 34959, 2010), (ii) $\parallel x\parallel \le \parallel y\parallel $ for each $x,y\in X$ such that $|\tau (x)|\le |\tau (y)|$ for each $\tau \in \mathrm{\Omega}(X)$, (iii) $inf\{r(x):x\in X,\parallel x\parallel =1\}>0$ does not have the fixed point property. This result is a generalization of Theorem 4.3 in (Fupinwong and Dhompongsa in Fixed Point Theory Appl. 2010:Article ID 34959, 2010).

**MSC:**46B20, 46J99.

### Keywords

fixed point property nonexpansive mapping Abelian Banach algebra## 1 Introduction

A Banach space *X* is said to have the fixed point property (or weak fixed point property) if for each nonexpansive mapping $T:E\to E$ on a bounded closed convex (or weakly compact convex, resp.) subset *E* of *X* has a fixed point.

For the weak fixed point property of certain Banach algebras, Lau *et al.* [1] showed that the space ${C}_{0}(G)$, where *G* is a locally compact group, has the weak fixed point property if and only if *G* is discrete, and a von Neumann algebra has the weak fixed point property if and only if it is finite dimensional. Benavides and Pineda [2] proved that each *ω*-almost weakly orthogonal closed subspace of $C({K}_{1})$, where ${K}_{1}$ is a metrizable compact space, has the weak fixed point property and $C({K}_{2})$, where ${K}_{2}$ is a compact set with ${K}_{2}^{(\omega )}=\mathrm{\varnothing}$, has the weak fixed point property.

As for the fixed point property, Dhompongsa *et al.* [3] showed that a ${C}^{\ast}$-algebra has the fixed point property if and only if it is finite dimensional. Fupinwong and Dhompongsa [4] proved that each infinite dimensional unital Abelian Banach algebra *X* with $\mathrm{\Omega}(X)\ne \mathrm{\varnothing}$ satisfying: (i) (A) defined in [4], (ii) $\parallel x\parallel \le \parallel y\parallel $ for each $x,y\in X$ with $|\tau (x)|\le |\tau (y)|$ for each $\tau \in \mathrm{\Omega}(X)$, (iii) $inf\{r(x):x\in X,\parallel x\parallel =1\}>0$ does not have the fixed point property. Alimohammadi and Moradi [5] used the above result to obtain sufficient conditions to show that some unital uniformly closed subalgebras of $C(X)$, where *X* is a compact space, do not have the fixed point property.

In this paper, we show that the unitality in the result proved in [4] can be omitted.

## 2 Preliminaries and lemmas

We assume that the field of each vector space in this paper is complex.

*X*be a Banach algebra. Define $\tilde{X}=X\oplus \mathbb{C}$ and a multiplication on $\tilde{X}$ by

We have $\tilde{X}$ is a unital Banach algebra with the unit $(0,1)$ and called the unitization of *X*. $\tilde{X}$ is also Abelian if *X* is Abelian.

*X*and $\mathrm{\Omega}(X)$ is the set of characters on

*X*, then the set $\mathrm{\Omega}(\tilde{X})$ of characters on $\tilde{X}$ is equal to

for each $(a,\lambda )\in \tilde{X}$.

If *X* is an Abelian Banach algebra, condition (A) is defined by:

**(A)** *For each* $x\in X$, *there exists an element* $y\in X$ *such that* $\tau (y)=\overline{\tau (x)}$, *for each* $\tau \in \mathrm{\Omega}(X)$.

It can be seen that if *X* satisfies (A), then so does the unitization $\tilde{X}$ of *X*.

*X*be an Abelian Banach algebra. The Gelfand representation $\phi :X\to C(\mathrm{\Omega}(X))$ is defined by $x\mapsto \stackrel{\u02c6}{x}$, where $\stackrel{\u02c6}{x}$ is defined by

for each $\tau \in C(\mathrm{\Omega}(X))$.

The following lemma was proved in [4].

**Lemma 2.1**

*Let*

*X*

*be a unital Abelian Banach algebra satisfying*(

*A*)

*and*

*Then*:

- (i)
*the Gelfand representation**φ**is a bounded isomorphism*, - (ii)
*the inverse*${\phi}^{-1}$*is also a bounded isomorphism*.

Let *X* be an Abelian Banach algebra satisfying (A) and $inf\{r(x):x\in X,\parallel x\parallel =1\}>0$. It can be seen that *X* is embedded in $C(\mathrm{\Omega}(\tilde{X}))$ as the closed subalgebra $Y=\{\stackrel{\u02c6}{x}\in C(\mathrm{\Omega}(\tilde{X})):\stackrel{\u02c6}{x}({\tau}_{\mathrm{\infty}})=0\}$. Moreover, for each $x\in \tilde{X}$, *x* is in *X* if and only if ${\tau}_{\mathrm{\infty}}(x)=0$.

**Lemma 2.2**

*Let*

*X*

*be an infinite dimensional Abelian Banach algebra satisfying*(

*A*)

*and*

*Then we have*:

- (i)
$\mathrm{\Omega}(X)$

*is an infinite set*. - (ii)
*If there exists a bounded sequence*$\{{x}_{n}\}$*in**X**which contains no convergent subsequences and such that*$\{\tau ({x}_{n}):\tau \in \mathrm{\Omega}(X)\}$*is finite for each*$n\in \mathbb{N}$,*then there is an element*${x}_{0}\in X$*such that*$\{\omega ({x}_{0}):\omega \in \mathrm{\Omega}(\tilde{X})\}$*is equal to*$\{0,1,\frac{1}{2},\frac{2}{3},\frac{3}{4},\dots \}$*or*$\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \}$. - (iii)
*There is an element*${x}_{0}\in X$*such that*$\{\omega ({x}_{0}):\omega \in \mathrm{\Omega}(\tilde{X})\}$*is an infinite set*. - (iv)
*There exists a sequence*$\{{x}_{n}\}$*in**X**such that*$\{\omega ({x}_{n}):\omega \in \mathrm{\Omega}(\tilde{X})\}\subset [0,1]$,*for each*$n\in \mathbb{N}$,*and*$\{{(\stackrel{\u02c6}{{x}_{n}})}^{-1}\{1\}\}$*is a sequence of nonempty pairwise disjoint subsets of*$\mathrm{\Omega}(\tilde{X})$.

*Proof*(i) From Lemma 2.10(i) in [4], we have $\mathrm{\Omega}(\tilde{X})$ is infinite. Since

- (ii)Let $\{{x}_{n}\}$ be a bounded sequence in
*X*which has no convergent subsequences and the set $\{\tau ({x}_{n}):\tau \in \mathrm{\Omega}(X)\}$ be finite for each $n\in \mathbb{N}$. Consider $\{{x}_{n}\}$ a sequence in $\tilde{X}$, so $\{\omega ({x}_{n}):\omega \in \mathrm{\Omega}(\tilde{X})\}$ is finite for each $n\in \mathbb{N}$. It follows from the proof of Lemma 2.10(ii) in [4] that$\mathrm{\Omega}(\tilde{X})=\left(\bigcup _{n\in \mathbb{N}}{G}_{n}\right)\cup F,$

*F*is a closed set in $\mathrm{\Omega}(\tilde{X})$, ${G}_{n}$ is closed and open for each $n\in \mathbb{N}$, and $\{F,{G}_{1},{G}_{2},\dots \}$ is a partition of $\mathrm{\Omega}(\tilde{X})$. There are two cases to be considered. If ${\tau}_{\mathrm{\infty}}$ is in

*F*, defined $\psi :\mathrm{\Omega}(\tilde{X})\to \mathbb{R}$ by

- (iii)
Assume to the contrary that $\{\omega (x):\omega \in \mathrm{\Omega}(\tilde{X})\}$ is finite for each $x\in X$. Since

*X*is infinite dimensional, so there is a bounded sequence $\{{x}_{n}\}$ in*X*which has no convergent subsequences. Thus $\{\omega ({x}_{n}):\omega \in \mathrm{\Omega}(\tilde{X})\}$ is finite for each $n\in \mathbb{N}$. It follows from (ii) that there exists ${x}_{0}\in X$ such that $\{\omega ({x}_{0}):\omega \in \mathrm{\Omega}(\tilde{X})\}$ is infinite. This leads to a contradiction. - (iv)It follows from (iii) that there exists an element ${x}_{1}\in X$ such that $\{\omega ({x}_{1}):\omega \in \mathrm{\Omega}(\tilde{X})\}$ is infinite. We may assume that there exists a strictly decreasing sequence of real numbers $\{{a}_{n}\}$ such that$\{{a}_{n}\}\subset \stackrel{\u02c6}{{x}_{1}}(\mathrm{\Omega}(\tilde{X}))\subset [0,1],\phantom{\rule{1em}{0ex}}{a}_{1}<1,$

for each $i\in \mathbb{N}$. Then ${\tau}_{\mathrm{\infty}}({x}_{n})=0$, for each $n\in \mathbb{N}$. Thus $\{{x}_{n}\}$ is a sequence in *X*. □

## 3 Main theorem

**Theorem 3.1**

*Let*

*X*

*be an infinite dimensional Abelian Banach algebra satisfying*(

*A*)

*and each of the following*:

- (i)
*If*$x,y\in X$*is such that*$|\tau (x)|\le |\tau (y)|$,*for each*$\tau \in \mathrm{\Omega}(X)$,*then*$\parallel x\parallel \le \parallel y\parallel $, - (ii)
$inf\{r(x):x\in X,\parallel x\parallel =1\}>0$.

*Then* *X* *does not have the fixed point property*.

*Proof*Assume to the contrary that

*X*has the fixed point property. From Lemma 2.2(iv), there exists a sequence $\{{x}_{n}\}$ in

*X*such that $\{\omega ({x}_{n}):\omega \in \mathrm{\Omega}(\tilde{X})\}\subset [0,1]$ for each $n\in \mathbb{N}$, and $\{{(\stackrel{\u02c6}{{x}_{n}})}^{-1}\{1\}\}$ is a sequence of nonempty pairwise disjoint subsets of $\mathrm{\Omega}(\tilde{X})$. Let ${A}_{n}={(\stackrel{\u02c6}{{x}_{n}})}^{-1}\{1\}$, and define ${T}_{n}:{E}_{n}\to {E}_{n}$ by

*X*such that $\{\omega ({x}_{0}):\omega \in \mathrm{\Omega}(\tilde{X})\}$ is equal to $\{0,1,\frac{1}{2},\frac{2}{3},\frac{3}{4},\dots \}$ or $\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \}$. Let ${A}_{0}={(\stackrel{\u02c6}{{x}_{0}})}^{-1}\{1\}$. Define ${T}_{0}:{E}_{0}\to {E}_{0}$ by

*i.e.*, ${y}_{0}={x}_{0}{y}_{0}$. Therefore, $\stackrel{\u02c6}{{y}_{0}}=\stackrel{\u02c6}{{x}_{0}}\stackrel{\u02c6}{{y}_{0}}$. Then

then ${A}_{0}$ has a pairwise disjoint open covering, which is a contradiction. So we conclude that *X* does not have the fixed point property. □

The following question is interesting.

**Question 3.2** Does the Fourier algebra $A(G)$ or the Fourier-Stieltjes algebra $B(G)$ of a locally compact group G have property (A) when *G* is an infinite group?

Note that $A(G)$ or $B(G)$ are both commutative semigroup Banach algebras with the fixed point property if and only if *G* is finite (see Theorem 5.7 and Corollary 5.8 of [6]). It is well known that $A(G)$ is norm dense in ${C}_{0}(G)$ with spectrum *G*.

## Declarations

### Acknowledgements

This research was supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.

## Authors’ Affiliations

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