1. Introduction

By , we denote the space of functions analytic in the unit disk . Endowed with the topology of uniform convergence on compact subsets of the unit disk, is a locally convex topological vector space.

Let be a topological vector space and a subset of . If , then is called an extremal subset of provided that whenever , where , and , then and both belong to . An extremal subset of consisting of just one point is called an extreme point of . Thus, an element is an extreme point of if and only if is not a proper convex combination of any two distinct points in . The set of all extreme points of is denoted by . It is apparent that if is an extremal subset of , then . If is a locally convex topological vector space and is a nonempty compact subset of , then is nonempty [1, page 44], [2, page 181]. For any subset of , we use to denote the closed convex hull of . If is a compact subset of the locally convex topological vector space , then, by Krein-Milman theorem [1, page 44], [2, page 182], .

Let be the set of all functions which are analytic, have positive real part in , and satisfy . Then is a compact subset of [1, page 39]. It is well known that [1, page 48], [3–5]. Bellamy and Tkaczyńska [6] investigated the extreme points of some classes of analytic functions with positive real part and a prescribed set of coefficients. Peng [7] investigated the extreme points of a class of analytic functions with positive real part and a prescribed set of values.

Suppose that is a compact subset of . A function is called a support point of if and there is a continuous linear functional on such that is nonconstant on and The set of all support points of is denoted by . Hallenbeck and MacGregor [1, page 94], [8] proved that the set consists of all functions which may be written as where , , and   . The author [9] investigated the support points of a class of analytic functions with positive real part and a prescribed set of coefficients.

Suppose that . Let denote the set of functions that are analytic in and satisfy and . It is apparent that if and only if with some and . Thus, it is easy to prove that is a compact subset of . In this paper we investigate the extreme points and support points of . In some ways, the results we obtained generalize the results of Holland, Hallenbeck, and MacGregor.

2. Main Results

Theorem 1. is an extreme point of if and only if where .

Proof. Suppose that where ,  ,  , . Then . Since ,  , it follows that . Consequently, ,  . Notice that (4) is equivalent to and that , and we have So, . This proves that . Similarly, we can prove that .
Conversely, Suppose . If , then for some . Since it follows that which contradicts the assumption. Thus or .
If , then . By Herglotz formula [1, page 30], [10, page 22], we have where is a probability measure on the unit circle . If is not a point mass, then there exist probability measures and on the unit circle such that and for some with [1, page 47]. Let Then , , , and . This implies that is not an extreme point of . We get a contradiction. So, is a point mass. Without loss of generality, we suppose for some unit complex number . Then with . Similarly, if , then we can prove that , where . The proof is completed.

Theorem 2. is a support point of if and only if satisfies one of the following conditions:(1), or ,(2)   where , , , and are positive integers.

Proof. Let be a support point of . Then there is a continuous linear functional defined on such that is not constant on and Suppose that . Let . Since and is continuous, is a nonvacuous closed convex subset of . As is a compact subset of , so is . By Krein-Milman theorem [1, page 44], [2, page 182], is nonempty, and . Suppose that and , where , and . Then Since ,  , it follows from (12) that This implies that , . Thus is an extremal subset of and so .
Since is a continuous linear functional on , there is a sequence of complex numbers satisfying such that where and   [1, page 42].
Let Then is analytic in .
If has infinitely many elements, then there must be infinitely many or infinitely many in , where . Without loss of generality, we assume that there are infinitely many in with . Then has infinitely many solutions which implies that is constant in . So, contains all with but contains no elements such as with . This prove that and thus In particular, and . Similarly, if we assume that there are infinitely many in with , then we can prove that and .
In the case that has only a finite number of elements, say, , , . Then consists of functions given by (10). especially must have the form given by (10).
Conversely, Suppose that and that . Define a continuous linear functional on by where . Since , it is clear that is not constant on and . So, . Similarly, we can prove that if and .
Now suppose that has the form (10). Then, by Lemma  7.2 in [1], there is a function analytic on such that when and if and only if   . Suppose that . Let Then . Define a linear functional on by where Then is continuous [1, page 42]. Since it follows that [1, page 44] Note that we have . If is constant on , then when . But it is not the case. Therefore .

Remark 3. Though we assume that in Theorem 1, it is easy to see that Theorem 1 is valid for , which is just the result of Holland. For , Theorem 2 is invalid, since and does not imply that is a support point of . In the case where , Theorem 2 should be stated as follows.
The set consists of all functions which may be written as where , , and   .
This is the result of Hallenbeck and MacGregor [1, page 94], [8]. It is easy to see that Theorem 2 generalizes the result of Hallenbeck and MacGregor in some sense.

Acknowledgments

The author expresses his heartfelt thanks to the referee for his (or her) critical review and helpful suggestions for the improvement of this paper. This paper is supported by the Educational Commission of Hubei Province of China (D2011006).