Abstract

Let be a compact Hausdorff space, be a continuous involution on and denote the uniformly closed real subalgebra of consisting of all for which . Let be a compact metric space and let denote the complex Banach space of complex-valued Lipschitz functions of order on under the norm , where . For , the closed subalgebra of consisting of all for which as , denotes by . Let be a Lipschitz involution on and define for and for . In this paper, we give a characterization of extreme points of , where is a real linear subspace of or which contains 1, in particular, or .

1. Introduction and Preliminaries

We let , and denote the field of real numbers, complex numbers, and the unit circle, respectively. The symbol denotes a field that can be either or . The elements of are called scalars.

Let be a normed space over . We denote by and the dual space and the closed unit ball of , respectively. If is a subset , let denote the set of all extreme points of . Let be a subspace of and . A Hahn-Banach extension of to is a continuous linear functional such that and . The set of all Hahn-Banach extensions of to will be denoted by .

It is easy to see that if and are normed spaces over and is a linear isometry from onto , then is a bijection mapping between and .

For a complex normed space , we assume that denotes , regarded as a real normed space by restricting the scalar multiplication to real numbers.

Kulkarni and Limaye gave some conditions for to be an extreme point of in terms of the Hahn-Banach extension of to and the extreme points of as the following.

Theorem 1.1 (see [1, Theorem  2]). Let be a normed space over , be a nonzero linear subspace of and . (a)Let . Then, In particular, has an extension to some . Further, if such an extension is unique, then has a unique Hahn-Banach extension to .(b)Assume that whenever and for all with , one has , then .(c)If has a unique Hahn-Banach extension to and if , then .

Let be a compact Hausdorff space. We denote by the complex Banach algebra of all continuous complex-valued functions on under the uniform norm . For , consider the evaluation functional given by . Clearly, for all . It is well known [2, page 441] that

For and , we define the map by in fact, . Clearly, for all . Kulkarni and Limaye showed [1, Proposition  3] that and if and only if and .

Let be a continuous involution on ; that is, is continuous and is the identity map on . The map defined by , is an algebra involution on which is called the algebra involution induced by on . Define . Then, is a uniformly closed real subalgebra of which contains 1. The real algebras were first considered in [3]. For a detailed account of several properties of , we refer to [4].

Let . For each , let denote the restriction of to . Grzesiak obtained a characterization of the extreme points of in [5] and showed that if and only if for some . Further, if , then if and only if or .

Kulkarni and Limaye obtained [1, Theorem  4] a characterization of , where is a nonzero real linear subspace of .

Let be a compact metric space. For , we denote by the set of all complex-valued functions on for which is finite. Then, is a complex subalgebra of containing 1 and complex Banach space under the norm For , the complex subalgebra of consisting of all for which is denoted by . Clearly, is a closed linear subspace of and . These Banach spaces were first studied by Leeuw in [6].

Given a compact metric space , let , and let the compact Hausdorff space be the disjoint union of with , where is the Stone-Cech compactification of . For , consider the mapping defined for each by where and is the norm-preserving extension of to . Clearly, is a linear isometry from into , which is called the Leeuw’s linear isometry. Therefore, is a uniformly closed linear subspace of . It is well known (see [2, page 441]) that where is the evaluation functional at on .

For each and , define the linear functionals and in by and , respectively. Clearly, and for all . Therefore, . Moreover, for all   and for all . Thus, we have the following result.

Theorem 1.2. For , every extreme point of must be either of the form with or of the form with .

Roy proved the following result by using a result of Leeuw [6, Lemma  1.2].

Theorem 1.3 (see [7, Lemma  1.2]). For each , is an extreme point of .

Jimenez-Vargas and Villegas-Vallecillos used above results and obtained a characterization of linear isometries between and in [8].

A map is said to be Lipschitz map from the metric space to the metric space if there exists a constant such that for all .

Let be a compact metric space. The mapping is called a Lipschitz involution on , if is a Lipschitz map from to itself and an involution on . Clearly, every Lipschitz involution on is a continuous involution.

Let be a Lipschitz involution on the compact metric space and let be the algebra involution induced by on . Clearly, We define Then, the following statements hold.(i) (, resp.) is a real subalgebra of (, resp.).(ii) and .(iii) and .(iv) (, resp.) is a real subalgebra of which contains 1 and separates the points of .(v) (, resp.) is uniformly dense in (use (iv) and the Stone-Weierstrass theorem for real subalgebra of [3, Proposition  1.1].(vi)For , (vii)There exists a constant such that for all .(viii) is a real Banach space and is its closed real subspace.

The real Banach spaces and are called real Banach spaces of complex Lipschitz functions and first studied in [9].

We give a characterization of extreme points of the unit ball in the dual space of , and some its real linear subspaces for in Section 2. Next, we give a characterization of extreme points of the unit ball in the dual spaces of and some its real linear subspaces for in Section 3.

2. Real Linear Subspaces of Containing 1

In the remainder of this paper, we assume that , is a compact metric space, , is the Stone-Cech compactification of , is the compact Hausdorff space , is the Leeuw’s linear isometry from into , and is a Lipschitz involution on .

For each , we define the map by in fact, . Clearly, for all . Moreover, for all .

We first give a characterization of the extreme points of the unit ball in the as the following.

Proposition 2.1. By above notations, Further, for one has if and only if .

Proof. We define the map by . Clearly, is a real-linear mapping. For each , defining the map by . Clearly, and . It follows that and . Thus, is onto.
We claim that is an isometric. Let . Since for each , we have
Let be an arbitrary positive number. There exists with such that . Choose if and if . Then, , and . If , then , and so, It follows that
Thus, our claim is justified. The above arguments show that is a real-linear isometry from onto . Therefore, Since we conclude that by Theorems 1.2 and 1.3.
Clearly, for all . Therefore, It is obvious that if and , then . We now assume that , where . Letting and , we see that and ; that is, . If , there exists such that , but (define by , ); so that But this is not possible since . Thus, .

The next purpose is giving conditions for to be an extreme point of , where is a real subspace of Lip.

Theorem 2.2. Let be a real linear subspace of Lip containing 1. For , let . Let .
Let denote the set of such that(i)there is with and , (ii)for every , there is some with , and .Then, Further, if and , then if and only if either or .

Proof. Let . Letting in part (a) of Theorem 1.1, and using Proposition 2.1, we see that for some . To prove that , we consider with and show that . For every , we have that is, is a real number. Hence, This shows that . But since is an extreme point of , we must have . Thus, so that .
Next, let . We claim that the following statement hold.
For with and , there is such that , , but .
By condition (i), there is such that and .
Case 1 (). Let . Then, and so that . Also, If , then implies that that is, . Since , this shows that If , then implies that , that is, . Since , this shows that Case 2 (). By condition (ii), there is such that , and . Let . Now, and , so that . Also, Thus, our claim is justified. Let and in part (b) of Theorem 1.1. Consider such that for all with . By Proposition 2.1, for some . Thus, for all with . Since , and Re, we have . If , then Thus, it must be . Choose . Then, and for all with . By our claim, we must have or .
If , then clearly Now, let . Since , implies that Therefore, , again. Hence, we see that . This also shows that for all .
Conversely, our claim implies that if , and , then or . Thus, for and , we have if and only if or .