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, Suppose p(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.
If p(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.
Proof.
Let p0 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).
Since J 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}.
If E𝒬 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 that E𝒬 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 that p0∈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 that p0 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 that α<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 set supp𝒫 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.