- Open Access
Nonexpansive mappings on Abelian Banach algebras and their fixed points
© Fupinwong; licensee Springer 2012
- Received: 25 June 2012
- Accepted: 24 August 2012
- Published: 14 September 2012
A Banach space X is said to have the fixed point property if for each nonexpansive mapping 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) for each such that for each , (iii) 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).
- fixed point property
- nonexpansive mapping
- Abelian Banach algebra
A Banach space X is said to have the fixed point property (or weak fixed point property) if for each nonexpansive mapping 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.  showed that the space , 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  proved that each ω-almost weakly orthogonal closed subspace of , where is a metrizable compact space, has the weak fixed point property and , where is a compact set with , has the weak fixed point property.
As for the fixed point property, Dhompongsa et al.  showed that a -algebra has the fixed point property if and only if it is finite dimensional. Fupinwong and Dhompongsa  proved that each infinite dimensional unital Abelian Banach algebra X with satisfying: (i) (A) defined in , (ii) for each with for each , (iii) does not have the fixed point property. Alimohammadi and Moradi  used the above result to obtain sufficient conditions to show that some unital uniformly closed subalgebras of , 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  can be omitted.
We assume that the field of each vector space in this paper is complex.
We have is a unital Banach algebra with the unit and called the unitization of X. is also Abelian if X is Abelian.
for each .
If X is an Abelian Banach algebra, condition (A) is defined by:
(A) For each , there exists an element such that , for each .
It can be seen that if X satisfies (A), then so does the unitization of X.
for each .
The following lemma was proved in .
the Gelfand representation φ is a bounded isomorphism,
the inverse is also a bounded isomorphism.
Let X be an Abelian Banach algebra satisfying (A) and . It can be seen that X is embedded in as the closed subalgebra . Moreover, for each , x is in X if and only if .
is an infinite set.
If there exists a bounded sequence in X which contains no convergent subsequences and such that is finite for each , then there is an element such that is equal to or .
There is an element such that is an infinite set.
There exists a sequence in X such that , for each , and is a sequence of nonempty pairwise disjoint subsets of .
- (ii)Let be a bounded sequence in X which has no convergent subsequences and the set be finite for each . Consider a sequence in , so is finite for each . It follows from the proof of Lemma 2.10(ii) in  that
Assume to the contrary that is finite for each . Since X is infinite dimensional, so there is a bounded sequence in X which has no convergent subsequences. Thus is finite for each . It follows from (ii) that there exists such that is infinite. This leads to a contradiction.
- (iv)It follows from (iii) that there exists an element such that is infinite. We may assume that there exists a strictly decreasing sequence of real numbers such that
for each . Then , for each . Thus is a sequence in X. □
If is such that , for each , then ,
Then X does not have the fixed point property.
then 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 or the Fourier-Stieltjes algebra of a locally compact group G have property (A) when G is an infinite group?
Note that or 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 ). It is well known that is norm dense in with spectrum G.
This research was supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.
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