Kenneth Kuttler

55.4. ANALYTIC CONTINUATION 1747

and F maps B(0,1) into B(0,1). Also, F is not one to one because it maps B(0,1) toB(0,1) and has s in its definition. Thus there exists z1 ∈ B(0,1) such that

φ−√

φα◦h(z0)(z1) =−

and another point z2 ∈ B(0,1) such that φ−√

φα◦h(z0)(z2) =

12 . However, thanks to the

function s, which squares things, F (z1) = F (z2).Since F (0) = h(z0) = 0, you can apply the Schwarz lemma to F . Since F is not one to

one, 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 55.3.15 and 55.3.14 that

η =∣∣h′ (z0)

∣∣= ∣∣F ′ (ψ (z0))∣∣ ∣∣ψ ′ (z0)

∣∣=

∣∣F ′ (0)∣∣ ∣∣ψ ′ (z0)∣∣< ∣∣ψ ′ (z0)

∣∣≤ η ,

a contradiction. This proves the theorem.The following lemma yields the usual form of the Riemann mapping theorem.

Lemma 55.3.7 Let Ω be a simply connected region with Ω ̸= C. Then Ω has the squareroot property.

Proof: Let f and 1f both be analytic on Ω. Then f ′

f is analytic on Ω so by Corollary

51.7.23, there exists F̃ , analytic on Ω such that F̃ ′ = f ′f on Ω. Then

(f e−F̃

)′= 0 and so

f (z) = CeF̃ = ea+ibeF̃ . Now let F = F̃ + a+ ib. Then F is still a primitive of f ′/ f andf (z) = eF(z). Now let φ (z) ≡ e

12 F(z). Then φ is the desired square root and so Ω has the

square root property.

Corollary 55.3.8 (Riemann mapping theorem) Let Ω be a simply connected region withΩ ̸= C and let z0 ∈ Ω. Then there exists a function, f : Ω→ B(0,1) such that f is one toone, analytic, and onto with f (z0) = 0. Furthermore, f−1 is also analytic.

Proof: From Theorem 55.3.6 and Lemma 55.3.7 there exists a function, f : Ω→B(0,1)which is one to one, onto, and analytic such that f (z0) = 0. The assertion that f−1 isanalytic follows from the open mapping theorem.

55.4 Analytic Continuation55.4.1 Regular And Singular PointsGiven a function which is analytic on some set, can you extend it to an analytic functiondefined on a larger set? Sometimes you can do this. It was done in the proof of the Cauchyintegral formula. There are also reflection theorems like those discussed in the exercisesstarting with Problem 10 on Page 1649. Here I will give a systematic way of extending ananalytic function to a larger set. I will emphasize simply connected regions. The subjectof analytic continuation is much larger than the introduction given here. A good source formuch more on this is found in Alfors [3]. The approach given here is suggested by Rudin[113] and avoids many of the standard technicalities.

55.4. ANALYTIC CONTINUATION 1747and F maps B(0,1) into B(0,1). Also, F is not one to one because it maps B(0,1) toB(0, 1) and has s in its definition. Thus there exists z; € B (0,1) such that19 Vega) 1) = 3and another point z2 € B(0,1) such that ¢_ Joaahten) 2) = 5. However, thanks to thefunction s, which squares things, F (z;) = F (zz).Since F (0) =h(z9) =0, you can apply the Schwarz lemma to F. Since F is not one toone, it can’t be true that F (z) = Az for |A| = 1 and so by the Schwarz lemma it must be thecase that |F’(0)| < 1. But this implies from 55.3.15 and 55.3.14 thatIh’ (z0)| = |F (w(zo))| | ¥’ (z0)|IF’ (0)| |W’ (zo)| < |W’ (zo) | <n,a contradiction. This proves the theorem.The following lemma yields the usual form of the Riemann mapping theorem.ufLemma 55.3.7 Let Q be a simply connected region with Q #C. Then Q has the squareroot property.Proof: Let f and + both be analytic on Q. Then £ is analytic on © so by Corollary51.7.23, there exists F, analytic on Q such that Fl= £ on Q. Then (fe*) = 0 and sof(D= Ceh = ettibeF . Now let F = F +a+ib. Then F is still a primitive of f’/f andf(z) = eF ©, Now let @(z) = e2 (2). Then @ is the desired square root and so Q has thesquare root property.Corollary 55.3.8 (Riemann mapping theorem) Let Q be a simply connected region withQC and let z € Q. Then there exists a function, f :Q— B(0,1) such that f is one toone, analytic, and onto with f (z) =0. Furthermore, f~' is also analytic.Proof: From Theorem 55.3.6 and Lemma 55.3.7 there exists a function, f :Q — B(0,1)which is one to one, onto, and analytic such that f (zo) = 0. The assertion that fo! isanalytic follows from the open mapping theorem.55.4 Analytic Continuation55.4.1. Regular And Singular PointsGiven a function which is analytic on some set, can you extend it to an analytic functiondefined on a larger set? Sometimes you can do this. It was done in the proof of the Cauchyintegral formula. There are also reflection theorems like those discussed in the exercisesstarting with Problem 10 on Page 1649. Here I will give a systematic way of extending ananalytic function to a larger set. I will emphasize simply connected regions. The subjectof analytic continuation is much larger than the introduction given here. A good source formuch more on this is found in Alfors [3]. The approach given here is suggested by Rudin[113] and avoids many of the standard technicalities.