Nonexpansive mappings on Abelian Banach algebras and their fixed points

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.

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.

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

$$(a,\lambda )(b,\mu )=(ab+\lambda b+\mu a,\lambda \mu ).$$

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.

If $\tilde{X}$ is the unitization of a Banach algebra 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

$$\{\tilde{\tau }:\tau \in \mathrm{\Omega }(X)\}\cup \{{\tau }_{\mathrm{\infty }}\},$$

where $\tilde{\tau }$ is defined from $\tau \in \mathrm{\Omega }(X)$ by

$$\tilde{\tau }((a,\lambda ))=\tau (a)+\lambda ,$$

for each $(a,\lambda )\in \tilde{X}$, and ${\tau }_{\mathrm{\infty }}$ is the canonical homomorphism defined by

$${\tau }_{\mathrm{\infty }}((a,\lambda ))=\lambda ,$$

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.

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

$$\hat{x}(\tau )=\tau (x),$$

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

$$inf\{r(x):x\in X,\parallel x\parallel =1\}>0.$$

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 $Y=\{\hat{x}\in C(\mathrm{\Omega }(\tilde{X})):\hat{x}({\tau }_{\mathrm{\infty }})=0\}$. 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

$$inf\{r(x):x\in X,\parallel x\parallel =1\}>0.$$

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 $\{{(\widehat{x_n})}^{-1}\{1\}\}$ 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

$$\mathrm{\Omega }(\tilde{X})=\{\tilde{\tau }:\tau \in \mathrm{\Omega }(X)\}\cup \{{\tau }_{\mathrm{\infty }}\},$$

where $\tilde{\tau }$ is defined from $\tau \in \mathrm{\Omega }(X)$ by $\tilde{\tau }((a,\lambda ))=\tau (a)+\lambda$, for each $(a,\lambda )\in \tilde{X}$, and ${\tau }_{\mathrm{\infty }}$ is the canonical homomorphism, so $\mathrm{\Omega }(X)$ is also infinite.

• (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

  $$\mathrm{\Omega }(\tilde{X})=\left(\bigcup _{n\in \mathbb{N}}{G}_n\right)\cup F,$$

where 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

$$\psi (\tau )=\{\begin{array}{cc}1,\hfill & \text{if }\tau \in G_1,\hfill \\ \frac{1}{n},\hfill & \text{if }\tau \in G_n,n\ge 2,\hfill \\ 0,\hfill & \text{if }\tau \in F.\hfill \end{array}$$

If ${\tau }_{\mathrm{\infty }}$ is in $G_{n_0}$, for some $n_0\in \mathbb{N}$, we may assume that $n_0=1$, defined $\psi :\mathrm{\Omega }(\tilde{X})\to \mathbb{R}$ by

$$\psi (\tau )=\{\begin{array}{cc}0,\hfill & \text{if }\tau \in G_1,\hfill \\ \frac{n-1}{n},\hfill & \text{if }\tau \in G_n,n\ge 2,\hfill \\ 1,\hfill & \text{if }\tau \in F.\hfill \end{array}$$

For each case, we have the inverse image of each closed set in $\psi (\mathrm{\Omega }(\tilde{X}))$ is closed, so $\psi \in C(\mathrm{\Omega }(\tilde{X}))$. Let $\phi :\tilde{X}\to C(\mathrm{\Omega }(\tilde{X}))$ be the Gelfand representation. Therefore, ${\phi }^{-1}(\psi )$ is an element in $\tilde{X}$, say $x_0$, 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 $\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \}$. We have $x_0\in X$ since ${\tau }_{\mathrm{\infty }}(x_0)=\psi ({\tau }_{\mathrm{\infty }})=0$.

• (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

  $$\{a_n\}\subset \widehat{x_1}(\mathrm{\Omega }(\tilde{X}))\subset [0,1],a_1<1,$$

and $\omega (x_1)=1$ for some $\omega \in \mathrm{\Omega }(\tilde{X})$. Define $g_1:[0,1]\to [0,1]$ by

$$g_1(t)=\{\begin{array}{cc}\frac{t}{a_1},\hfill & \text{if }t\in [0,a_1],\hfill \\ 1+\frac{(g_1(a_2)-1)(t-a_1)}{2(1-a_1)},\hfill & \text{if }t\in [a_1,1].\hfill \end{array}$$

So $g_1$ is a continuous function joining the points $(0,0)$ and $(a_1,1)$, and $g_1(1)\in (g_1(a_2),1)$. Let $\widehat{x_2}=g_1\circ \widehat{x_1}$, and define a continuous function $g_2:[0,1]\to [0,1]$ by

$$g_2(t)=\{\begin{array}{cc}\frac{t}{g_1(a_2)},\hfill & \text{if }t\in [0,g_1(a_2)],\hfill \\ 1+\frac{(g_2(g_1(a_3))-1)(t-g_1(a_2))}{2(1-g_1(a_2))},\hfill & \text{if }t\in [g_1(a_2),1].\hfill \end{array}$$

$g_2$ is joining the point $(0,0)$ and $(g_1(a_2),1)$ and $g_2(1)\in (g_2(g_1(a_3)),1)$. Let $\widehat{x_3}=g_2\circ \widehat{x_2}$. Continuing in this process, we obtain a sequence of points $\{x_n\}$ in $\tilde{X}$ with $\{\omega (x_n):\omega \in \mathrm{\Omega }(X)\}\subset [0,1]$, for each $n\in \mathbb{N}$, and $\{{(\widehat{x_n})}^{-1}\{1\}\}$ is a sequence of nonempty pairwise disjoint subsets of $\mathrm{\Omega }(\tilde{X})$. Since $g_n(0)=0$, for each $n\in \mathbb{N}$, so

$$\widehat{x_{i+1}}({\tau }_{\mathrm{\infty }})=(g_i\circ \cdots \circ g_1\circ \widehat{x_1})({\tau }_{\mathrm{\infty }})=(g_i\circ \cdots \circ g_1)(0)=0,$$

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 $\{{(\widehat{x_n})}^{-1}\{1\}\}$ is a sequence of nonempty pairwise disjoint subsets of $\mathrm{\Omega }(\tilde{X})$. Let $A_n={(\widehat{x_n})}^{-1}\{1\}$, and define $T_n:E_n\to E_n$ by

$$x\mapsto x_{n}x,$$

where

$$E_n=\{x\in X:0\le \omega (x)\le 1\text{ for each }\omega \in \mathrm{\Omega }(\tilde{X}),\text{ and }\omega (x)=1\text{ if }\omega \in A_n\}.$$

From (i) and (ii), $T_n:E_n\to E_n$ is a nonexpansive mapping on the bounded closed convex set $E_n$ for each $n\in \mathbb{N}$. Indeed, $E_n$ is bounded since

$$inf\{r(x):x\in X,\parallel x\parallel =1\}\le r\left(\frac{x}{\parallel x\parallel }\right)=\underset{\omega \in \mathrm{\Omega }(\tilde{X})}{sup}\left|\omega \left(\frac{x}{\parallel x\parallel }\right)\right|=\frac{1}{\parallel x\parallel }\underset{\omega \in \mathrm{\Omega }(\tilde{X})}{sup}|\omega (x)|$$

for each $x\in X$. So $T_n$ has a fixed point, say $y_n$, for each $n\in \mathbb{N}$. We have $y_n=x_n{y}_n$, hence $\widehat{y_n}=\widehat{x_n}\widehat{y_n}$, and then

$$\widehat{y_n}(\omega )=\{\begin{array}{cc}0,\hfill & \text{if }\omega \text{ is not in }{A}_n,\hfill \\ 1,\hfill & \text{if }\omega \text{ is in }{A}_n,\hfill \end{array}$$

for each $n\in \mathbb{N}$. We have $\parallel \widehat{y_m}-\widehat{y_n}\parallel =1$, if $m\ne n$, since $A_1,A_2,A_3,\dots$ are pairwise disjoint. Therefore, $\{\widehat{y_n}\}$ has no convergent subsequences. From Lemma 2.1, $\tilde{X}$ and $C(\mathrm{\Omega }(\tilde{X}))$ are homeomorphic. So $\{y_n\}$ has no convergent subsequences. From Lemma 2.2(ii), there exists 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 $\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \}$. Let $A_0={(\widehat{x_0})}^{-1}\{1\}$. Define $T_0:E_0\to E_0$ by

$$x\mapsto x_{0}x,$$

where

$$E_0=\{x\in X:0\le \omega (x)\le 1\text{ for each }\omega \in \mathrm{\Omega }(\tilde{X}),\text{ and }\omega (x)=1\text{ if }\omega \in A_0\}.$$

From (i) and (ii), $T_0$ is a nonexpansive mapping on the bounded closed convex set $E_0$. Hence $T_0$ has a fixed point, say $y_0$, i.e., $y_0=x_0{y}_0$. Therefore, $\widehat{y_0}=\widehat{x_0}\widehat{y_0}$. Then

$$\widehat{y_0}(\omega )=\{\begin{array}{cc}0,\hfill & \text{if }\omega \text{ is not in }{A}_0,\hfill \\ 1,\hfill & \text{if }\omega \text{ is in }{A}_0.\hfill \end{array}$$

Since $\widehat{y_0}=\widehat{x_0}\widehat{y_0}$, so we have $A_0={(\widehat{y_0})}^{-1}\{1\}$ and $\mathrm{\Omega }(\tilde{X})\mathrm{\setminus }{A}_0={(\widehat{y_0})}^{-1}\{0\}$. Then $\mathrm{\Omega }(\tilde{X})$ is a disjoint union of two compact sets $A_0$ and $\mathrm{\Omega }(\tilde{X})\mathrm{\setminus }{A}_0$. If

$$\{\omega (x_0):\omega \in \mathrm{\Omega }(\tilde{X})\}=\{0,1,\frac{1}{2},\frac{2}{3},\frac{3}{4},\dots \},$$

then $\{{(\widehat{x_0})}^{-1}\{\frac{n}{n+1}\}:n\in \mathbb{N}\}\cup \{{(\widehat{x_0})}^{-1}\{0\}\}$ is a pairwise disjoint open covering of the compact set $\mathrm{\Omega }(\tilde{X})\mathrm{\setminus }{A}_0$. This leads to a contradiction. Similarly, if

$$\{\omega (x_0):\omega \in \mathrm{\Omega }(\tilde{X})\}=\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \},$$

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.