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Nonexpansive mappings on Abelian Banach algebras and their fixed points
Fixed Point Theory and Applications volume 2012, Article number: 150 (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.
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 ωalmost 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
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
where \tilde{\tau} is defined from \tau \in \mathrm{\Omega}(X) by
for each (a,\lambda )\in \tilde{X}, and {\tau}_{\mathrm{\infty}} is the canonical homomorphism defined by
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 \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
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
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
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 \stackrel{\u02c6}{{x}_{1}}(\mathrm{\Omega}(\tilde{X}))\subset [0,1],\phantom{\rule{1em}{0ex}}{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
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 \stackrel{\u02c6}{{x}_{2}}={g}_{1}\circ \stackrel{\u02c6}{{x}_{1}}, and define a continuous function {g}_{2}:[0,1]\to [0,1] by
{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 \stackrel{\u02c6}{{x}_{3}}={g}_{2}\circ \stackrel{\u02c6}{{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 \{{(\stackrel{\u02c6}{{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
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
where
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
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 \stackrel{\u02c6}{{y}_{n}}=\stackrel{\u02c6}{{x}_{n}}\stackrel{\u02c6}{{y}_{n}}, and then
for each n\in \mathbb{N}. We have \parallel \stackrel{\u02c6}{{y}_{m}}\stackrel{\u02c6}{{y}_{n}}\parallel =1, if m\ne n, since {A}_{1},{A}_{2},{A}_{3},\dots are pairwise disjoint. Therefore, \{\stackrel{\u02c6}{{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}={(\stackrel{\u02c6}{{x}_{0}})}^{1}\{1\}. Define {T}_{0}:{E}_{0}\to {E}_{0} by
where
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, \stackrel{\u02c6}{{y}_{0}}=\stackrel{\u02c6}{{x}_{0}}\stackrel{\u02c6}{{y}_{0}}. Then
Since \stackrel{\u02c6}{{y}_{0}}=\stackrel{\u02c6}{{x}_{0}}\stackrel{\u02c6}{{y}_{0}}, so we have {A}_{0}={(\stackrel{\u02c6}{{y}_{0}})}^{1}\{1\} and \mathrm{\Omega}(\tilde{X})\mathrm{\setminus}{A}_{0}={(\stackrel{\u02c6}{{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
then \{{(\stackrel{\u02c6}{{x}_{0}})}^{1}\{\frac{n}{n+1}\}:n\in \mathbb{N}\}\cup \{{(\stackrel{\u02c6}{{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
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 FourierStieltjes 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.
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Acknowledgements
This research was supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.
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Fupinwong, W. Nonexpansive mappings on Abelian Banach algebras and their fixed points. Fixed Point Theory Appl 2012, 150 (2012). https://doi.org/10.1186/168718122012150
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DOI: https://doi.org/10.1186/168718122012150