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.
Fixed Point Theory and Applications volume 2010, Article number: 362829 (2010)
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.
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,
(2) is -weakly orthogonal,
(3)
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
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
(1°) separates the points of
(2°) annihilates no point of
Then is dense in
Theorem.
Let be a compact space, a subalgebra of such that
(1°) separates the points of
(2°) annihilates no point of
(3°) implies that the conjugate of is in
Then is dense 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.
Let be considered as a real unital abelian Banach algebra under ordinary complex multiplication and whose norm is the absolute value. We have Indeed, assume to the contrary that there is a non-zero homomorphism on and so
Thus so is not a real number, which is a contradiction.
Since so but
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.
Let be a unital abelian real Banach algebra with
Then one has the following:
(i)the Gelfand representation is a bounded isomorphism,
(ii)the inverse is also a bounded isomorphism.
Proof.
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
for each Write then for each But is Cauchy, and so we have Thus
which is a contradiction. Hence must be Cauchy and so for some Since for each so Thus is complete. The Stone-Weierstrass theorem can be applied to conclude that is surjective.
follows from the open mapping theorem.
Remark 2.8.
Lemma 2.7 tells us that if is a unital abelian real Banach algebra with property
then and are homeomorphic and isomorphic under Hence if we would like to consider the convergence of a sequence in we could look at the convergence of the corresponding sequence
Property (2.12) clearly implies the semisimplicity property () but the following example shows that it is stronger.
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.
Let be an infinite dimensional unital abelian real Banach algebra with
Then one has the following:
(i) is an infinite set,
(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,
there exists a sequence in such that for each and is a sequence of nonempty pairwise disjoint subsets of
Proof.
Let be an infinite dimensional unital abelian real Banach algebra with
(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
and for each
So is a subsequence of such that By Remark 2.8, where Thus is compact.
Let be a bounded sequence in which has no convergent subsequences and suppose that the set is finite for each By Remark 2.8, we will consider as a sequence of Gelfand transforms
First, we show that we can write
where is closed, are all closed and open, and is a partition of For each write
Define
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.
Case 1 ( is infinite).
Thus there exists such that
is an infinite set. Similarly, there exists such that
is an infinite set. Continuing in this process we obtain a sequence of the sets such that
is an infinite set, for each
Write
Thus are all closed and open, and
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
and is a partition of
Case 2 ().
To show that this case leads to a contradiction, we first observe that if are in the same then
for each Write if There exists a subsequence of such that for each
for some Define a Gelfand transform by if Since
so which is a contradiction.
Now we conclude that
where is closed, is closed and open for each and is a partition of Define a map by
We can check that the inverse image of each closed set in is closed. Therefore, is an element in say with
Assume to the contrary that is finite for each Since is infinite dimensional, so, as is noncompact, there exists a bounded sequence in which has no convergent subsequences. Hence is finite for each It follows from (ii) that
for some which is the contradiction.
From (iii), there is an element such that is an infinite set. We can choose so that there exists a strictly decreasing sequence such 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.
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
(ii)
Then does not have fixed point property.
Proof.
Let be an infinite dimensional unital abelian real Banach algebra satisfying (i) and (ii). Assume to the contrary that has fixed point property. From Lemma 2.10(iv), there exists a sequence in such that
for each and is a sequence of nonempty pairwise disjoint subsets of Write and define by
where
It follows from (i) that is a nonexpansive mapping on the bounded closed convex set for each Indeed, is bounded since
for each It follows that has a fixed point, say for each Since thus and then
for each Since are pairwise disjoint, so if Hence has no convergent subsequences. From Lemma 2.7, has no convergent subsequences too. It follows from the existence of and Lemma 2.10 (ii) that there exists an element in with
Write Define by
where
By (i), is a nonexpansive mapping on the bounded closed convex set Thus has a fixed point, say that is, Thus Consequently,
The set is open in since is continuous. Also the set is open in for each since is continuous. Thus,
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.
satisfies (i), (ii) in Theorem 3.1. Indeed, if is such that for each then for each Hence And since
so
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
for each Hence for each and thus
for each Clearly,
Since for each we have
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.
We have
If then for some and It follows that for each Write Hence
for each From the above inequality, and since is arbitrary, we can see that the sequence does not lie in
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.
Let be a unital abelian complex Banach algebra satisfying (A) and
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.
Let be an infinite dimensional unital abelian complex Banach algebra satisfying (A) and
Then one has the following:
(i) is an infinite set,
(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.
Let be an infinite dimensional unital abelian complex Banach algebra satisfying (A) and each of the following:
if is such that for each then ,
(ii)
Then does not have the fixed point property.
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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.
Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Fupinwong, W., Dhompongsa, S. The Fixed Point Property of Unital Abelian Banach Algebras. Fixed Point Theory Appl 2010, 362829 (2010). https://doi.org/10.1155/2010/362829
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DOI: https://doi.org/10.1155/2010/362829