Now with Montel’s theorem the following will imply the Riemann mapping theorem. This approach is in
Rudin [109].
Theorem 17.6.4Let Ω≠ℂ for Ω a region and suppose Ω has the square root property. Then forz_{0}∈ Ω there exists h : Ω → B
(0,1)
such that h is one to one, onto, analytic, and h
(z0)
= 0.
Proof: Define ℱ to be the set of functions f such that f : Ω → B
(0,1)
is one to one and analytic. The
first task is to show ℱ is nonempty. Then, using Montel’s theorem it will be shown there is a function in ℱ,
h, such that
′
|h (z0)|
≥
′
|ψ (z0)|
for all ψ ∈ℱ. When this has been done it will be shown that h is actually
onto. This will prove the theorem.
Claim 1:ℱ is nonempty.
Proof of Claim 1: Since Ω≠ℂ it follows there exists ξ
∕∈
Ω. Then it follows z − ξ and
-1-
z−ξ
are both
analytic on Ω. Since Ω has the square root property, there exists an analytic function, ϕ : Ω → ℂ such that
ϕ^{2}
(z)
= z − ξ for all z ∈ Ω,ϕ
(z)
=
√-----
z − ξ
. Since z − ξ is not constant, neither is ϕ and it
follows from the open mapping theorem that ϕ
(Ω)
is a region. Note also that ϕ is one to one
because if ϕ
(z1)
= ϕ
(z2)
, then you can square both sides and conclude z_{1}− ξ = z_{2}− ξ implying
z_{1} = z_{2}.
Now pick a ∈ ϕ
(Ω)
. Thus
√ ------
za − ξ
= a. I claim there exists a positive lower bound to
, of functions in ℱ and an analytic function, h, such
that
|ψ′ (z)| → η (17.10)
n 0
(17.10)
and
ψn → h,ψ′n → h′, (17.11)
(17.11)
uniformly on all compact subsets of Ω. It follows
|h′(z0)| = lim |ψ′n (z0)| = η (17.12)
n→ ∞
(17.12)
and for all z ∈ Ω,
|h(z)| = lim |ψn (z)| ≤ 1. (17.13)
n→∞
(17.13)
By 17.12, h is not a constant. Therefore, in fact,
|h(z)|
< 1 for all z ∈ Ω in 17.13 by the open mapping
theorem.
It follows from Lemma 17.6.2 that h is one to one. It only remains to verify that h
(z0)
= 0.
If h
(z0)
≠0,consider ϕ_{h(z0)
}∘h where ϕ_{α} is the fractional linear transformation defined in Lemma 17.5.3.
By this lemma it follows ϕ_{h(z0)
}∘ h ∈ℱ. Now using the chain rule,