16.6. RIEMANN MAPPING THEOREM 413

16.6.2 The Proof of Riemann Mapping TheoremThe existence part of the Riemann mapping theorem is from Montel’s theorem, Theorem16.6.1 and showing that what is obtaned is actually what is desired comes from the Schwarzlemma, Lemma 16.5.4 and the remarkable properties of the special fractional linear trans-formation of Lemma 16.5.3.

This approach is in Rudin [40] and Conway [10]. I will present it in a sequence oflemmas, each of which is interesting for its own sake. All that is needed for Ω⊊ C is thatit is a region with the square root property. Recall that simply connected Ω implies squareroot property, Lemma 16.5.6. In fact the other direction also holds but I won’t go into thathere.

Lemma 16.6.4 Let Ω have square root property, contain 0 and define F to be the set offunctions f such that f : Ω→ B(0,1) is one to one and analytic. Suppose F is nonempty.Letting η ≡ sup{|ψ ′ (0)| : ψ ∈F} , suppose there exists h ∈F with h′ (0) = η ,h(0) = 0.Then h is onto B(0,1).

Proof: Suppose α ∈ B(0,1) \ h(Ω). Then both φ α ◦ h,1/φ α ◦ h are analytic so sinceΩ has the square root property, there exists analytic

√φ α ◦h. Let φ α for |α| < 1 be from

Lemma 16.5.3, φ α (z)≡ z−α

1−αz and φ α (α) = 0. Let

ψ ≡ φ√φα◦h(0)

◦√

φ α ◦h (16.4)

Thus ψ (0) = φ√φα◦h(0)

◦√

φ α ◦h(0) = 0 and ψ is a one to one mapping of Ω into B(0,1)so ψ is also in F . Therefore,∣∣ψ ′ (0)∣∣≤ η ,

∣∣∣∣(√φ α ◦h)′(0)∣∣∣∣≤ η . (16.5)

Define s(w)≡ w2. Then using Lemma 16.5.3, in particular, the description of φ−1α = φ−α ,

you can solve 16.4 for h to obtain

h(z) = φ−α ◦ s◦φ−√

φα◦h(0)◦ψ =

 ≡F︷ ︸︸ ︷φ−α ◦ s◦φ−

√φα◦h(0)

◦ψ

(z) = (F ◦ψ)(z) (16.6)

Now F (0) = φ−α ◦ s◦φ−√

φα◦h(0)(0) = φ

−1α (φ α ◦h(0)) = h(0) = 0 and F maps B(0,1)

into B(0,1) because it is the composition of functions which map onto B(0,1). Also, Fis not one to one because, by Lemma 16.5.3, it maps B(0,1) onto B(0,1) and has s in itsdefinition. Indeed, there exists z1,z2 ∈ B(0,1) such that

φ−√

φα◦h(0)(z1) =−

12, φ−√

φα◦h(0)(z2) =

12.

Since φ−√

φα◦h(0)is one to one, z1 ̸= z2 but, since s(z) = z2, F (z1) = F (z2).

Since F (0) = h(0) = 0, you can apply the Schwarz lemma to F . Since F is not one toone, it can’t be true that F (z) = λ z for |λ |= 1 and so by the Schwarz lemma it must be thecase that |F ′ (0)|< 1. But this implies from 16.6 and 16.5 that

η =∣∣h′ (0)∣∣= ∣∣F ′ (ψ (0))

∣∣ ∣∣ψ ′ (0)∣∣= ∣∣F ′ (0)∣∣ ∣∣ψ ′ (0)∣∣< ∣∣ψ ′ (0)∣∣≤ η ,

a contradiction. Thus h is onto after all. ■