- Research Article
- Open Access

# The Fixed Point Property of Unital Abelian Banach Algebras

- W Fupinwong
^{1}and - S Dhompongsa
^{1}Email author

**2010**:362829

https://doi.org/10.1155/2010/362829

© W. Fupinwong and S. Dhompongsa. 2010

**Received:**28 October 2009**Accepted:**22 January 2010**Published:**28 January 2010

## Abstract

We give a general condition for infinite dimensional unital Abelian Banach algebras to fail the fixed point property. Examples of those algebras are given including the algebras of continuous functions on compact sets.

## Keywords

- Nonexpansive Mapping
- Pairwise Disjoint
- Banach Algebra
- Convergent Subsequence
- Point Property

## 1. Introduction

Let
be a Banach space. A mapping
is *nonexpansive* if

for each
The *fixed point set* of
is
We say that the space
has the *fixed point property* (or *weak* fixed point property) if for every nonempty bounded closed convex (or weakly compact convex, resp.) subset
of
and every nonexpansive mapping
we have
One of the central goals in fixed point theory is to solve the problem: which Banach spaces have the (weak) fixed point property?

For weak fixed point property, Alspach [1] exhibited a weakly compact convex subset of the Lebesgue space and an isometry without a fixed point, proving that the space does not have the weak fixed point property. Lau et al. [2] proved the following results.

Theorem 1.1.

Let be a locally compact Hausdorff space. If has the weak fixed point property, then X is dispersed.

Corollary.

Let be a locally compact group. Then the -algebra has the weak fixed point property if and only if is discrete.

Corollary.

A von Neumann algebra has the weak fixed point property if and only if is finite dimensional.

Continuing in this direction, Benavides and Pineda [3] developed the concept of -almost weak orthogonality in the Banach lattice and obtained the results.

Theorem.

Let be a -almost weakly orthogonal closed subspace of where is a metrizable compact space. Then has the weak fixed point property.

Theorem.

Let be a metrizable compact space. Then, the following conditions are all equivalent:

(1) is -almost weakly orthogonal,

Corollary.

Let be a compact set with Then has the weak fixed point property.

As for the fixed point property, Dhompongsa et al. [4] showed that a -algebra has the fixed point property if and only if it is finite dimensional. In this paper, we approach the question on the fixed point property from the opposite direction by identifying unital abelian Banach algebras which fail to have the fixed point property. As consequences, we obtain results on the algebra of continuous functions where is a compact set, and there is a unital abelian subalgebra of the algebra which does not have the fixed point property and does not contain the space

## 2. Preliminaries and Lemmas

The fields of real and complex numbers are denoted by
and
, respectively. The symbol
denotes a field that can be either
or
The elements of
are called *scalars*.

An element
in a unital algebra
is said to be *invertible* if there is an element
in
such that

In this case is unique and written

We define the *spectrum* of an element
of a unital algebra
over
to be the set

The *spectral radius* of
is defined to be

We note that a subalgbra of a normed algebra is itself a normed algebra with the norm got by restriction. The closure of a subalgebra is a subalgebra. A closed subalgebra of a Banach algebra is a Banach algebra. If
is a family of subalgebras of an algebra
then
is a subalgebra also. Hence, for any subset
of
there is the smallest subalgebra
of
containing
This algebra is called the subalgebra of
*generated* by
If
is the singleton
then
is the linear span of all powers
of
If
is a normed algebra, the closed algebra
*generated* by a set
is the smallest closed subalgebra containing
We can see that

We denote by the Banach algebra of continuous functions from a topological space to with the sup-norm

The following theorems are known as the Stone-Weierstrass approximation theorem for and respectively. For the details, the readers are referred to [5].

Theorem.

Let be a subalgebra of such that

Theorem.

Let be a compact space, a subalgebra of such that

(3°) implies that the conjugate of is in

A *character* on a unital algebra
over
is a nonzero homomorphism
We denote by
the set of characters on
Note that if
is a unital abelian complex Banach algebra, then

for each (see [6]).

Remark.

It is unknown if (2.5) is valid whenever Equation (2.5) obviously does not hold for a space with as the following example shows.

Example 2.4.

Thus so is not a real number, which is a contradiction.

We consider throughout this paper on Banach algebras for which and satisfy (2.5).

If
is a unital abelian Banach algebra, it follows from Proposition 2.5 that
is contained in the closed unit ball of
We endow
with the relative weak* topology and call the topological space
the *character space* of

Detailed proofs of the following propositions can be found in [6].

Proposition.

Let be a unital abelian Banach algebra. If then

Proposition.

If is a unital Banach algebra, then is compact.

If is a unital abelian Banach algebra, and we define a continuous function by

We call
the *Gelfand transform* of
and the homomorphism

is called the *Gelfand representation.*

The following two lemmas, Lemmas 2.7 and 2.10, will be used to prove our main theorem.

Lemma.

Then one has the following:

(i)the Gelfand representation is a bounded isomorphism,

(ii)the inverse is also a bounded isomorphism.

- (i)is injective since implies It is easily checked that is a bounded homomorphism, and is a subalgebra of separating the points of , and having the property that for any there is an element such that . In order to use the Stone-Weierstrass theorem to show that we shall show that is closed. We show that is closed by showing that is complete. Let be a Cauchy sequence in First, we show that the sequence is Cauchy. Assume on the contrary that is not Cauchy. Thus there exists and subsequences and of such that

- (ii)
follows from the open mapping theorem.

- (i)

Example.

Let denote the Banach algebra of complex-valued absolutely summable functions on the group of integers under convolution regarded as a real Banach algebra and let be the real subalgebra of consisting of those functions that satisfy Then the maximal ideal space of equals and the Gelfand transform is precisely the Fourier transform which maps into the real Banach algebra of continuous real-valued functions on under pointwise multiplication and maximum norm. Although the image of the Fourier transform is dense, it is clearly not all of since it is simply the real-valued functions in the Wiener space which consists of complex-valued functions whose Fourier series are absolutely summable. Therefore Lemma 2.7 shows that does not have Property (2.12).

Lemma.

Then one has the following:

(ii)if there exists a bounded sequence in which contains no convergent subsequences and such that is finite for each then there is an element with

Proof.

(i)If suffices to show that if is a finite set, for then the closed unit ball of is compact, and this will lead to us a contradiction. Let be a finite set, say , and let be a sequence in

Since the sequences are bounded, we can choose a subsequence of such that for each

Define by Thus there exists such that and consequently, since

First, we show that we can write

Note that are all closed and open. Since is a partition of for each is a partition of There are two cases to be considered.

Write

where is a nonempty closed set since is compact. And since has infinite elements, we can see that there exists a subsequence of such that and for each

Hence we have

Now we conclude that

and for some Define a continuous function to be linear on and on joining the points and and Put for some and define a continuous function similar to the way we construct The left part of is the line joining the point and and Then put for some Continuing in this process we obtain a sequence of points such that for each and is a sequence of nonempty pairwise disjoint subsets of We then obtain the required result.

## 3. Main Theorem

Now we prove our main theorem.

Theorem.

Let be an infinite dimensional unital abelian real Banach algebra satisfying each of the following:

(i)if is such that for each then

Then does not have fixed point property.

Proof.

is an open covering of This leads to a contradiction, since is compact.

From the above theorem we have the following.

Corollary 3.2.

Let be a compact Hausdorff topological space. If is infinite dimensional, then fails to have the fixed point property.

Proof.

Let denote the Banach algebra of all real bounded sequences with the sup-norm. The following two propositions tell us that there is a subalgebra of which does not contain but fails to have the fixed point property.

Proposition 3.3.

If is a subset of which contains an infinite bounded sequence and the identity, then the Banach subalgebra of generated by fails to have the fixed point property.

Proof.

Let be a subset of which contains an infinite bounded sequence and the identity. It follows that is unital and abelian. is infinite dimensional, since the set is a linearly independent subset of Next, we show that satisfies (i) and (ii) in Theorem 3.1.

Let be such that and for each Define by

so Now it follows from Theorem 3.1 that doesn't have the fixed point property.

Proposition.

Let with Then the Banach subalgebra of generated by the identity and does not contain the space

Proof.

for each From the above inequality, and since is arbitrary, we can see that the sequence does not lie in

## 4. Results on Complex Banach Algebras

Let be a unital abelian complex Banach algebra. Consider the following condition.

*(A)*
*For each*
*there exists an element*
*such that*
*for each*

If satisfies condition (A), then is a subspace of which is closed under the complex conjugation. By using the Stone-Weierstrass theorem for the complex Banach algebra and following the proof of Lemma 2.7, we obtain the following result.

Lemma.

Then one has the following:

(i)the Gelfand representation is a bounded isomorphism,

(ii)the inverse is also a bounded isomorphism.

Using Lemma 4.1 we obtain the complex counterpart of Lemma 2.10.

Lemma.

Then one has the following:

(ii)if there exists a bounded sequence in which contains no convergent subsequences and such that is finite for each then there is an element with

(iii)there is an element such that is an infinite set,

(iv)there exists a sequence in such that for each and is a sequence of nonempty pairwise disjoint subsets of

By using Lemmas 4.1 and 4.2, and by following the proof of Theorem 3.1, we get the following theorem.

Theorem.

## Declarations

### Acknowledgments

The authors would like to express their thanks to the referees for valuable comments, especially, to whom that provides them Remark 2.8(ii) and Example 2.9 for completeness. This work was supported by the Thailand Research Fund, grant BRG50800016.

## Authors’ Affiliations

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## Copyright

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