# The Extreme and Support Points of a New Class of Analytic Functions with Positive Real Part

Authors: Zhigang Peng
Journal: Journal of Complex Analysis
 (2013)

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## Abstract

Suppose that0<α<β<+∞. Let 𝒫(α,β) denote the set of functions p(z) that are analytic in D={z:|z|<1} and satisfy Rep(z)>0(|z|<1) and α≤p(0)≤β. In this paper, we investigate the extreme points and support points of 𝒫(α,β).


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## Body

## 1. Introduction

By𝒜, we denote the space of functions analytic in the unit disk D={z:|z|<1}. Endowed with the topology of uniform convergence on compact subsets of the unit disk, 𝒜 is a locally convex topological vector space.LetX be a topological vector space and U a subset of X. If V⊂U, then V is called an extremal subset of U provided that whenever u=tx+(1-t)y, where u∈V, x,y∈U and 0<t<1, then x and y both belong to V. An extremal subset of U consisting of just one point is called an extreme point of U. Thus, an element x∈U is an extreme point of U if and only if x is not a proper convex combination of any two distinct points in U. The set of all extreme points of U is denoted by EU. It is apparent that if V is an extremal subset of U, then EV⊂EU. If X is a locally convex topological vector space and U is a nonempty compact subset of X, then EU is nonempty [1, page 44], [2, page 181]. For any subset U of X, we use HU to denote the closed convex hull of U. If U is a compact subset of the locally convex topological vector space X, then, by Krein-Milman theorem [1, page 44], [2, page 182], HU=HEU.Let𝒫 be the set of all functions p(z) which are analytic, have positive real part in D, and satisfy p(0)=1. Then 𝒫 is a compact subset of 𝒜 [1, page 39]. It is well known that E𝒫={(1+xz)/(1-xz):|x|=1} [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 thatℱ is a compact subset of 𝒜. A function f is called a support point of ℱ if f∈ℱ and there is a continuous linear functional J on 𝒜 such that ReJ is nonconstant on ℱ and
(1)ReJ(f)=max{ReJ(g):g∈ℱ}.
The set of all support points of ℱ is denoted by suppℱ. Hallenbeck and MacGregor [1, page 94], [8] proved that the set supp𝒫 consists of all functions which may be written as
(2)p(z)=∑k=1m‍λk1+xkz1-xkz,
where λk≥0, ∑k=1m‍λk=1, and |xk|=1  (m=1,2,…). The author [9] investigated the support points of a class of analytic functions with positive real part and a prescribed set of coefficients.Suppose that0<α<β<+∞. Let 𝒫(α,β) denote the set of functions p(z) that are analytic in D={z:|z|<1} and satisfy Rep(z)>0(|z|<1) and α≤p(0)≤β. It is apparent that p(z)∈𝒫(α,β) if and only if p(z)=μq(z) with some q(z)∈𝒫 and α≤μ≤β. Thus, it is easy to prove that 𝒫(α,β) is a compact subset of 𝒜. In this paper we investigate the extreme points and support points of 𝒫(α,β). In some ways, the results we obtained generalize the results of Holland, Hallenbeck, and MacGregor.


## 2. Main Results

Theorem 1.
p
(
z
) is an extreme point of 𝒫(α,β) if and only if
(3)p(z)=α1+xz1-xz,orβ1+xz1-xz,
where |x|=1.Proof.
Suppose that(4)α1+xz1-xz=tp1(z)+(1-t)p2(z),
where |x|=1,  0<t<1,  p1, p2∈𝒫(α,β). Then α=tp1(0)+(1-t)p2(0). Since α≤p1(0),  α≤p2(0), it follows that p1(0)=p2(0)=α. Consequently, p1(z)/α∈𝒫,  p2(z)/α∈𝒫. Notice that (4) is equivalent to
(5)1+xz1-xz=t1αp1(z)+(1-t)1αp2(z),
and that [1+xz]/[1-xz]∈E𝒫, and we have
(6)1αp1(z)=1αp2(z).
So, p1(z)=p2(z). This proves that [α(1+xz)]/[1-xz]∈E𝒫(α,β). Similarly, we can prove that [β(1+xz)]/[1-xz]∈E𝒫(α,β).
Conversely, Supposep(z)∈E𝒫(α,β). If α<p(0)<β, then p(0)=tα+(1-t)β for some 0<t<1. Since
(7)p(z)=tαp(0)p(z)+(1-t)βp(0)p(z),[αp(z)]p(0)∈𝒫(α,β),[βp(z)]p(0)∈𝒫(α,β),[αp(z)]p(0)≠[βp(z)]p(0),
it follows that p(z)∉E𝒫(α,β) which contradicts the assumption. Thus p(0)=α or p(0)=β.
Ifp(0)=α, then p(z)/α∈𝒫. By Herglotz formula [1, page 30], [10, page 22], we have
(8)p(z)=α∫|x|=1‍1+xz1-xzdμ(x),
where μ is a probability measure on the unit circle |x|=1. If μ is not a point mass, then there exist probability measures μ1 and μ2 on the unit circle |x|=1 such that μ1≠μ2 and μ=tμ1+(1-t)μ2 for some t with 0<t<1 [1, page 47]. Let
(9)p1(z)=α∫|x|=1‍1+xz1-xzdμ1(x),p2(z)=α∫|x|=1‍1+xz1-xzdμ2(x).
Then p1, p2∈𝒫(α,β), p1≠p2, and p=tp1+(1-t)p2. This implies that p is not an extreme point of 𝒫(α,β). We get a contradiction. So, μ is a point mass. Without loss of generality, we suppose μ({x})=1 for some unit complex number x. Then p(z)=α(1+xz)/(1-xz) with |x|=1. Similarly, if p(0)=β, then we can prove that p(z)=β(1+xz)/(1-xz), where |x|=1. The proof is completed.Theorem 2.
p is a support point of 𝒫(α,β) if and only if p satisfies one of the following conditions:(1)
p∈𝒫(α,β), p(0)=α or β,(2)
(10)p(z)=α∑k=1m‍λk1+xkz1-xkz+β∑k=m+1m+n‍λk1+xkz1-xkz,where λk≥0, ∑k=1m+n‍λk=1, |xk|=1, m and n are positive integers.Proof.
Letp0 be a support point of 𝒫(α,β). Then there is a continuous linear functional J defined on 𝒜 such that ReJ is not constant on 𝒫(α,β) and
(11)ReJ(p0)=max{ReJ(p):p∈𝒫(α,β)}.
Suppose that ReJ(p0)=M. Let 𝒬={q:q∈𝒫(α,β),ReJ(q)=M}. Since p0∈𝒬 and J is continuous, 𝒬 is a nonvacuous closed convex subset of 𝒫(α,β). As 𝒫(α,β) is a compact subset of 𝒜, so is 𝒬. By Krein-Milman theorem [1, page 44], [2, page 182], E𝒬 is nonempty, and 𝒬=HE𝒬. Suppose that q∈𝒬 and q=tq1+(1-t)q2, where q1, q2∈𝒫(α,β) and 0<t<1. Then
(12)M=ReJ(q)=tReJ(q1)+(1-t)ReJ(q2)≤tM+(1-t)M=M.
Since ReJ(q1)≤M,  ReJ(q2)≤M, it follows from (12) that
(13)ReJ(q1)=ReJ(q2)=M.
This implies that q1, q2∈𝒬. Thus 𝒬 is an extremal subset of 𝒫(α,β) and so E𝒬⊂E𝒫(α,β).
SinceJ is a continuous linear functional on 𝒜, there is a sequence {bn} of complex numbers satisfying
(14)lim-n→∞(|bn|)1/n<1,
such that
(15)J(f)=∑n=0∞‍bnan,
where f∈𝒜 and f(z)=∑n=0∞‍anzn  [1, page 42].
Let(16)G(x)=J(1+xz1-xz)=b0+∑n=1∞‍2bnxn.
Then G(x) is analytic in D-={x:|x|≤1}.
IfE𝒬 has infinitely many elements, then there must be infinitely many [α(1+xz)]/(1-xz) or infinitely many [β(1+xz)]/(1-xz) in E𝒬, where |x|=1. Without loss of generality, we assume that there are infinitely many [α(1+xz)]/(1-xz) in E𝒬 with |x|=1. Then ReG(x)=M/α has infinitely many solutions which implies that G(x) is constant in D-={z:|z|≤1}. So, E𝒬 contains all [α(1+xz)]/(1-xz) with |x|=1 but contains no elements such as [β(1+xz)]/(1-xz) with |x|=1. This prove that
(17)E𝒬={[α(1+xz)]1-xz:|x|=1},
and thus
(18)𝒬=HE𝒬={p:p∈𝒫(α,β),p(0)=α}.
In particular, p0∈𝒫(α,β) and p0(0)=α. Similarly, if we assume that there are infinitely many [β(1+xz)]/(1-xz) in E𝒬 with |x|=1, then we can prove that p0∈𝒫(α,β) and p0(0)=β.
In the case thatE𝒬 has only a finite number of elements, say, E𝒬={[α(1+xkz)]/(1-xkz), [β(1+xjz)]/(1-xjz):|xk|=|xj|=1, k=1,…,m;j=1,…,n}. Then 𝒬=HE𝒬 consists of functions given by (10). p0 especially must have the form given by (10).
Conversely, Suppose thatp0∈𝒫(α,β) and that p0(0)=α. Define a continuous linear functional J on 𝒜 by J(f)=-a0 where f(z)=∑n=0∞‍anzn. Since α<β, it is clear that ReJ is not constant on 𝒫(α,β) and ReJ(p0)=max{ReJ(p):p∈𝒫(α,β)}. So, p0∈supp𝒫(α,β). Similarly, we can prove that p0∈supp𝒫(α,β) if p∈𝒫(α,β) and p(0)=β.
Now suppose thatp0 has the form (10). Then, by Lemma  7.2 in [1], there is a function F analytic on D-={z:|z|≤1} such that ReF(z)≤0 when |z|≤1 and ReF(z)=0 if and only if z=xk  (k=1,2,…,m+n). Suppose that F(z)=∑n=0∞‍dnzn. Let
(19)b0=d0,bn=dn2(n=1,2,…).
Then lim¯n→∞|bn|1/n<1. Define a linear functional J on 𝒜 by
(20)J(f)=∑n=0∞‍bnan,
where
(21)f(z)=∑n=0∞‍anzn∈𝒜.
Then J is continuous [1, page 42]. Since
(22)J(α(1+xz)(1-xz))=αF(x),J(β(1+xz)(1-xz))=βF(x),
it follows that [1, page 44]
(23)max{ReJ(p):p∈𝒫(α,β)}=max{ReJ(p):p∈E𝒫(α,β)}=0.
Note that
(24)ReJ((1+xkz)(1-xkz))=ReF(xk)=0(k=1,2,…,m+n),
we have ReJ(p0)=0. If ReJ is constant on 𝒫(α,β), then ReF(x)=0 when |x|=1. But it is not the case. Therefore p0∈supp𝒫(α,β).Remark 3.
Though we assume thatα<β in Theorem 1, it is easy to see that Theorem 1 is valid for α=β=1, which is just the result of Holland. For α=β=1, Theorem 2 is invalid, since p∈𝒫(1,1)=𝒫 and p(0)=1 does not imply that p is a support point of 𝒫. In the case where α=β=1, Theorem 2 should be stated as follows.
The setsupp𝒫 consists of all functions which may be written as
(25)p(z)=∑k=1m‍λk1+xkz1-xkz,
where λk≥0, ∑k=1m‍λk=1, and |xk|=1  (m=1,2,…).
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.



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