55.3. RIEMANN MAPPING THEOREM 1743

Definition 55.3.2 Let F denote a collection of functions which are analytic on Ω, a region.Then F is normal if every sequence contained in F has a subsequence which convergesuniformly on compact subsets of Ω.

The following result is about a certain class of fractional linear transformations. RecallLemma 52.4.7 which is listed here for convenience.

Lemma 55.3.3 For α ∈ B(0,1) , let

φ α (z)≡ z−α

1−αz.

Then φ α maps B(0,1) one to one and onto B(0,1), φ−1α = φ−α , and

φ′α (α) =

1

1−|α|2.

The next lemma, known as Schwarz’s lemma is interesting for its own sake but willalso be an important part of the proof of the Riemann mapping theorem. It was stated andproved earlier but for convenience it is given again here.

Lemma 55.3.4 Suppose F : B(0,1)→ B(0,1) , F is analytic, and F (0) = 0. Then for allz ∈ B(0,1) ,

|F (z)| ≤ |z| , (55.3.3)

and ∣∣F ′ (0)∣∣≤ 1. (55.3.4)

If equality holds in 55.3.4 then there exists λ ∈ C with |λ |= 1 and

F (z) = λ z. (55.3.5)

Proof: First note that by assumption, F (z)/z has a removable singularity at 0 if itsvalue at 0 is defined to be F ′ (0) . By the maximum modulus theorem, if |z|< r < 1,∣∣∣∣F (z)

z

∣∣∣∣≤ maxt∈[0,2π]

∣∣F (reit)∣∣

r≤ 1

r.

Then letting r→ 1, ∣∣∣∣F (z)z

∣∣∣∣≤ 1

this shows 55.3.3 and it also verifies 55.3.4 on taking the limit as z→ 0. If equality holds in55.3.4, then |F (z)/z| achieves a maximum at an interior point so F (z)/z equals a constant,λ by the maximum modulus theorem. Since F (z) = λ z, it follows F ′ (0) = λ and so|λ |= 1. This proves the lemma.

Definition 55.3.5 A region, Ω has the square root property if whenever f , 1f : Ω→ C are

both analytic1, it follows there exists φ : Ω→ C such that φ is analytic and f (z) = φ2 (z) .

The next theorem will turn out to be equivalent to the Riemann mapping theorem.1This implies f has no zero on Ω.

55.3. RIEMANN MAPPING THEOREM 1743Definition 55.3.2 Let Y denote a collection of functions which are analytic on Q, a region.Then ¥ is normal if every sequence contained in ¥ has a subsequence which convergesuniformly on compact subsets of Q.The following result is about a certain class of fractional linear transformations. RecallLemma 52.4.7 which is listed here for convenience.Lemma 55.3.3 For a € B(0,1), letThen @ 4, maps B(0,1) one to one and onto B (0,1), 0g! = _q, and/a) = —,,.4(@) =The next lemma, known as Schwarz’s lemma is interesting for its own sake but willalso be an important part of the proof of the Riemann mapping theorem. It was stated andproved earlier but for convenience it is given again here.Lemma 55.3.4 Suppose F : B(0,1) — B(0,1), F is analytic, and F (0) = 0. Then for allz€B(0,1),IF (z)| < |zl, (55.3.3)and|F’(0)| <1. (55.3.4)If equality holds in 55.3.4 then there exists A € C with |A| = 1 andF(z) =Az. (55.3.5)Proof: First note that by assumption, F(z) /z has a removable singularity at 0 if itsvalue at 0 is defined to be F’ (0). By the maximum modulus theorem, if |z| <r < 1,F F itFe < max ve ebZz te [0,27] r rThen letting r > 1,FO! cy7 |=this shows 55.3.3 and it also verifies 55.3.4 on taking the limit as z > 0. If equality holds in55.3.4, then |F (z) /z| achieves a maximum at an interior point so F (z) /z equals a constant,A by the maximum modulus theorem. Since F (z) = Az, it follows F’(0) =A and so|A| = 1. This proves the lemma.Definition 55.3.5 A region, QO has the square root property if whenever f, ; :Q—- Careboth analytic', it follows there exists @ : Q—> C such that @ is analytic and f (z) = $7 (z).The next theorem will turn out to be equivalent to the Riemann mapping theorem.'This implies f has no zero on Q.