#### 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, Theoremโโ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, Propositionโโ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, Theoremโโ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 Leeuwโ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 โโ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, Lemmaโโ1.2].

Theorem 1.3 (see [7, Lemmaโโ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, Propositionโโ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 Lip(๐,๐๐ผ) 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 Leeuwโ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 .

By using the above theorem, we give a characterization of extreme points of unit ball in the dual space of .

Corollary 2.3. Let be as in Theorem 2.2. For , letโโ denote the restriction of to . Then, Further, if and , then if and only if or .

Proof. Let . By Theorem 2.2, we have To prove , it is enough to show that for every ,(i)there is with and ,(ii)for every , there is with Let . We first define the function by where . Clearly, , , and . Let with . Then,
Therefore, and . Consequently, Hence, (i) holds.
We now assume that and define the function by Clearly, , , and there exists such that so that . We define the complex-valued function on by . Then, , , , and . Thus, we have , and so that . Since , . Thus, (ii) holds.

#### 3. Real Subspaces of lip(๐,๐๐ผ) Containing 1

Throughout this section, we assume that . Consider the mapping by . Then, is a linear isometric from into .

For each and , define the functionals and in lip by and , respectively. Clearly, for all and , and therefore, . Moreover, for all and for all .

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

Theorem 3.1. Every extreme point of must be either the form with or of the form with . Moreover, is an extreme point of for all .

Proof. Since is a Banach space and is a linear isometry from into , we conclude that is a uniformly closed subspace of . It is well known [2, page 441] that Let and define by . Then, is a linear isometry from onto , so , the adjoint of , is a linear isometry from onto . It is easily to show that for all . Let . Then, . Thus, there exists such that so that . It follows that
Now, let . Clearly, . Assume that and for all with . Since is a nonzero linear subspace of , we conclude that for some or for some by Theorem 1.1. Clearly, and . If for some , then we have Thus, it must be for some . Since , we have It follows that . We claim that . Let . We define the function by where . It is easy to show that , , and . Now, we define the function by and the function by . It is easy to see that , , , and . Therefore, we have , , , and so that . It must be that ; that is, . But