Abstract
Let be a compact metric space. In 1987, Bade, Curtis, and Dales obtained a sufficient condition for density of a subspace of little Lipschitz algebra in this algebra and in particular showed that is dense in , whenever . Let be a compact subset of . We define new classes of Lipchitz algebras for and for , consisting of those continuous complex-valued functions on such that and , respectively. In this paper we obtain a sufficient condition for density of a linear subspace of extended little Lipschitz algebra in this algebra and in particular show that is dense in , whenever .
1. Introduction
Let be a locally compact Hausdorff space. The linear space of all continuous (bounded continuous) complex-valued functions on is denoted by (). It is known that under the uniform norm on , that is, is a commutative Banach algebra. The set of all in , which vanish at infinity, is denoted by , which is a closed linear subspace of . Clearly, , whenever is compact. The linear space of all complex regular Borel measures on is denoted by . It is known that under the norm () is a Banach space, where is the total variation of .
The Riesz representation theorem asserts that there exists a linear isometry from , the dual space onto . In fact, for each , there exists a unique measure with such that Let be a compact metric space and . The Lipschitz algebra is defined as the set of all complex-valued functions on such that Then is a subalgebra of . The subalgebra of is the set of all those complex-valued functions on for which as and is called little Lipschitz algebra of order .
We know that separates the points of , and , where . Also, if is infinite and , then . The algebras for and for are Banach function algebras on under the norm . Since these algebras are self-adjoint, they are uniformly dense in , by the Stone-Weierstrass theorem. We know that if is a Banach function algebra on a compact Hausdorff space such that is self-adjoint and whenever and for each , then is natural, that is, the maximal ideal space of coincides with . Hence, if is infinite, then the Lipschitz algebras for and for , are natural.
Extensive study of Lipschitz algebras started with Sherbert [1, 2]. Honary and Moradi introduced new classes of analytic Lipschitz algebras on compact plane sets and determined their maximal ideal spaces [3].
Bade et al. have obtained a sufficient condition for density of a linear subspace of in this algebra as follows.
Theorem 1.1 (see [4, Theorem 3.6]). Let be a compact metric space, and let be a linear subspace of . Suppose that there is a constant such that for each finite subset of and each , there exists with and with . Then is dense in .
They also showed that is dense in [4, Corollary 3.7]. We extend the above results for the more general classes of the Lipschitz algebras by generalizing and using some results that have been given by them.
Throughout this work we always assume that is a compact metric space, is nonempty compact subset of , and is a positive number.
Definition 1.2. The algebra of all continuous complex-valued functions on for which is denoted by , and the subalgebra of those for which as , when , is denoted by . The algebras and are called extended Lipschitz algebra and extended little Lipschitz algebra of order on with respect to , respectively.
It is easy to see that these extended Lipschitz algebras are both Banach algebras under the norm . In fact, is a Banach function algebra on for , and is a Banach function algebra on for . Note that if , then . We always assume that for and for . Note that is a proper subalgebra of when is infinite. Because if , then function defined by is an element of but does not belong to .
It is clear that whenever , the new classes of Lipschitz algebras coincide with the standard Lipschitz algebras. Also , whenever is finite. Hence, we may assume that is infinite.
By the Stone-Weierstrass theorem, and are both uniformly dense in .
Let be or . For each and for all , we have By the spectral radius theorem, where is the maximal ideal space of and is the Gelfand transform of on . Hence, by applying the main theorem in [5], we can show that is natural, that is, coincides with . We can prove this fact with another way. Since every self-adjoint inverse-closed Banach function algebra on a compact Hausdorff space is natural, and Banach function algebras and have the mentioned properties, they are natural.
In this paper we obtain a sufficient condition for density of a linear subspace of that is dense in this algebra by generalizing some results in [4]. In particular, we show that is dense in .
2. Representing Measure
We denote Obviously, is a locally compact Hausdorff space. We define the norm on by Then is a Banach space under the norm , since for all . Moreover, is a closed linear subspace of .
We define the norm on by Then is a Banach space under the norm , since for all . By applying the Riesz representation theorem, we obtain the following result which is a generalization of Theorem in [6], and one can prove it by the same method.
Theorem 2.1. For each , there exists a unique measure with such that where .
Definition 2.2. For , the function defined by is called Leeuw’s extension of on .
It is obvious that for each .
Theorem 2.3. Take .(i) is a linear isometry from the extended Lipschitz algebra into .(ii) is a closed linear subspace of .(iii)For each , there exists such that
Proof. (i) It is immediate.
(ii) Since is a linear subspace , is a linear subspace of by . Let , and let be an arbitrary positive number. There exists such that
for all with . Set . Clearly, is a compact subset of and
for all . It follows that . Therefore, is a subset . Since is a linear isometry and is a closed linear subspace of , we conclude that is a closed linear subspace of .
(iii) Let and define by
Then . By the Hahn-Banach extension theorem, there exists with such that
By Theorem 2.1, there exists with such that
Therefore,
On the other hand,
It follows that . This completes the proof.
Note that the map is not an algebra homomorphism and that its image is not a subalgebra of .
Definition 2.4. For , a measure and with is called a representing measure for on .
Note that a representing measure for on is not unique.
3. Main Results
In this section, by generalizing Theorem 1.1, we obtain a sufficient condition for which a linear subspace of is dense in this algebra. In particular, we show that is dense in .
Theorem 3.1. let be a linear subspace of which satisfies the following conditions:(a)if with , then , where is the closure of in .(b)there is a constant such that for each finite subset of and each , there exists with and with . Then is dense in .
Proof. We first show that if , then is dense in the little Lipschitz algebra . Clearly, is a linear subspace of . Let be a Finite subset of , and let . By Tietze’s extension theorem [7, Theorem 20.4], there exists such that . Clearly, . by (b), there exists such that . We define . Then , and
Thus is dense in by Theorem 1.1.
To prove the density of in , it is enough to show that if with for all , then for all .
Let such that for all . Continuity of implies that for all . By Theorem 2.3, there exists such that
where is Leeuw’s extension of on . We claim that
Let . We define the sequence of the subsets of by
Then is a compact subset of for each , , and . Let . By Urysohn’s lemma, there exists such that , , and . Define . Then and . Hence, by (a) and so . Thus
Let be the characteristic function of on . It is easy to see that
for all . Since for all and , we conclude that
by Lebesgue’s dominated convergence theorem. Thus
by (3.5) and (3.7). It follows that
Thus (3.3) is justified, by (3.2) and (3.9).
We now define the function , by
Clearly, is a linear functional on . Since
for all , we deduce that . We claim that for all . If , there exists such that , and so
Therefore, our claim is justified. It follows that for all , by the density of in and continuity of on . Let . If , then , and so . Therefore,
by (3.3). This completes the proof.
By applying the above result, we show that is dense in .
Bade et al. obtained the following result.
Lemma 3.2 (see [4, Lemma 3.3]). For each finite subset of and each , there exists with and with .
We now generalize the above lemma by applying it and Tietze’s extension theorem as follows.
Lemma 3.3. For each finite subset of and each , there exists with and with .
Proof. Let be a finite subset of , and let . Define . Then . By Lemma 3.2, there exists with and with . We now define the function by Clearly, is a compact subset of and . By Tietze’s extension theorem, there exists such that and . It follows that and . Furthermore, This completes the proof.
Theorem 3.4. is dense in .
Proof. Take . Then is a linear subspace of and for all with . Let be a finite subset of and . By Lemma 3.3, there exists with and with . Therefore, is dense in , by Theorem 3.1.
Corollary 3.5. is dense in .
Proof. It is enough to take in Theorem 3.4.