Topics In Analysis

54.3.2 Regions With Square Root Property

Proof: Define ℱ to be the set of functions, f such that f : Ω → B

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 ≥ 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 are both analytic on Ω. Since Ω has the square root property, there exists an analytic function, ϕ : Ω → ℂ such that ϕ² = z − ξ for all z ∈ Ω,ϕ = . Since z − ξ is not constant, neither is ϕ and it follows from the open mapping theorem that ϕ is a region. Note also that ϕ is one to one because if ϕ = ϕ, then you can square both sides and conclude z₁ − ξ = z₂ − ξ implying z₁ = z₂.

Now pick a ∈ ϕ

. Thus = a. I claim there exists a positive lower bound to for z ∈ Ω. If not, there exists a sequence, ⊆ Ω such that

∘ ------ ∘ ------ ∘ ------ zn − ξ +a = zn − ξ + za − ξ ≡ εn → 0.

Then

∘ ------ ( ∘ -----) zn − ξ = εn − za − ξ (54.3.6)

(54.3.6)

and squaring both sides,

∘------ zn − ξ = ε2n + za − ξ − 2εn za − ξ.

Consequently,

= ε_n² − 2ε_n which converges to 0. Taking the limit in 54.3.6, it follows 2 = 0 and so ξ = z_a, a contradiction to ξΩ. Choose r > 0 such that for all z ∈ Ω, > r > 0. Then consider

ψ(z) ≡ √---r----. (54.3.7) z − ξ + a

(54.3.7)

This is one to one, analytic, and maps Ω into B

( > r). Thus ℱ is not empty and this proves the claim.

Claim 2: Let z₀ ∈ Ω. There exists a finite positive real number, η, defined by

η ≡ sup {|ψ ′(z )| : ψ ∈ ℱ } (54.3.8) 0

(54.3.8)

and an analytic function, h ∈ℱ such that

= η. Furthermore, h = 0.

Proof of Claim 2: First you show η < ∞. Let γ

= z₀ + re^it for t ∈ and r is small enough that B ⊆ Ω. Then for ψ ∈ℱ, the Cauchy integral formula for the derivative implies

′ 1 ∫ ψ(w) ψ (z0) = 2πi -------2dw γ (w − z0)

and so

≤2πr = 1∕r. Therefore, η < ∞ as desired. For ψ defined above in 54.3.7

′ − rϕ′(z0) − r(1∕2)(√z0-−-ξ)− 1 ψ (z0) =----------2 = -------------2----⁄= 0. (ϕ (z0)+ a) (ϕ (z0)+ a)

Therefore, η > 0. It remains to verify the existence of the function, h.

By Theorem 54.3.1, there exists a sequence,

, of functions in ℱ and an analytic function, h, such that

′ |ψ n(z0)| → η (54.3.9)

(54.3.9)

and

ψn → h,ψ′n → h′, (54.3.10)

(54.3.10)

uniformly on all compact subsets of Ω. It follows

|h′(z0)| = lim |ψ′n (z0)| = η (54.3.11) n→ ∞

(54.3.11)

and for all z ∈ Ω,

|h(z)| = lim |ψ (z)| ≤ 1. (54.3.12) n→∞ n

(54.3.12)

By 54.3.11, h is not a constant. Therefore, in fact,

< 1 for all z ∈ Ω in 54.3.12 by the open mapping theorem.

Next it must be shown that h is one to one in order to conclude h ∈ℱ. Pick z₁ ∈ Ω and suppose z₂ is another point of Ω. Since the zeros of h − h

have no limit point, there exists a circular contour bounding a circle which contains z₂ but not z₁ such that γ^∗ contains no zeros of h − h.

Using the theorem on counting zeros, Theorem 51.6.1, and the fact that ψ_n is one to one,

which shows that h − h has no zeros in B. In particular z₂ is not a zero of h−h. This shows that h is one to one since z₂≠z₁ was arbitrary. Therefore, h ∈ℱ. It only remains to verify that h = 0.

If h

≠0,consider ϕh

(z0)

∘h where ϕ_α is the fractional linear transformation defined in Lemma 54.3.3. By this lemma it follows ϕh

(z0)

∘ h ∈ℱ. Now using the chain rule,

Contradicting the definition of η. This proves Claim 2.

Claim 3: The function, h just obtained maps Ω onto B

.

Proof of Claim 3: To show h is onto, use the fractional linear transformation of Lemma 54.3.3. Suppose h is not onto. Then there exists α ∈ B

∖ h. Then 0≠ϕ_α ∘ h for all z ∈ Ω because

ϕ ∘ h(z) =-h(z)−-α- α 1 − αh(z)

and it is assumed α

h. Therefore, since Ω has the square root property, you can consider an analytic function z →. This function is one to one because both ϕ_α and h are. Also, the values of this function are in B by Lemma 54.3.3 so it is in ℱ.

Now let

------- ∘ ------ ψ ≡ ϕ√ ϕα∘h(z0) ∘ ϕα ∘ h. (54.3.13)

(54.3.13)

Thus

∘ --------- ψ (z0) = ϕ √ϕα∘h(z0) ∘ ϕα ∘h (z0) = 0

and ψ is a one to one mapping of Ω into B

so ψ is also in ℱ. Therefore,

| | |ψ′(z)| ≤ η,||(∘ ϕ-∘h)′(z )|| ≤ η. (54.3.14) 0 | α 0|

(54.3.14)

Define s

≡ w². Then using Lemma 54.3.3, in particular, the description of ϕ_α⁻¹ = ϕ_−α, you can solve 54.3.13 for h to obtain

Now

−1 F (0) = ϕ−α ∘s ∘ϕ−√ ϕα∘h(z0)(0) = ϕ α (ϕα ∘ h(z0)) = h(z0) = 0

and F maps B

into B. Also, F is not one to one because it maps B to B and has s in its definition. Thus there exists z₁ ∈ B such that ϕ−

√------- ϕα∘h(z0)

= − and another point z₂ ∈ B such that ϕ−

√ ------- ϕα∘h(z0)

= . However, thanks to s,F = F.

Since F

= h = 0, you can apply the Schwarz lemma to F. Since F is not one to one, it can’t be true that F = λz for = 1 and so by the Schwarz lemma it must be the case that < 1. But this implies from 54.3.15 and 54.3.14 that

a contradiction. This proves the theorem.

The following lemma yields the usual form of the Riemann mapping theorem.

Proof: Let f and

both be analytic on Ω. Then is analytic on Ω so by Corollary 50.7.23, there exists , analytic on Ω such that ^′ = on Ω. Then ^′ = 0 and so f = Ce

^F

= e^a+ibe

^F

. Now let F = + a + ib. Then F is still a primitive of f^′∕f and f = eF

(z)

. Now let ϕ ≡ e

12

F

(z)

. Then ϕ is the desired square root and so Ω has the square root property.

Proof: From Theorem 54.3.6 and Lemma 54.3.7 there exists a function, f : Ω → B

which is one to one, onto, and analytic such that f = 0. The assertion that f⁻¹ is analytic follows from the open mapping theorem.