55.3. RIEMANN MAPPING THEOREM 1741

maps the real line to the unit circle. It also takes the point, ξ to 0 and so it must map theupper half plane to the unit disk. You can verify the mapping is onto as well.

Example 55.2.6 Let z1 = 0,z2 = 1, and z3 = 2 and let w1 = 0,w2 = i, and w3 = 2i.

Then the equation to solve is

ww−2i

· −ii

=z

z−2· −1

1

Solving this yields w = iz which clearly works.

55.3 Riemann Mapping TheoremFrom the open mapping theorem analytic functions map regions to other regions or else tosingle points. The Riemann mapping theorem states that for every simply connected region,Ω which is not equal to all ofC there exists an analytic function, f such that f (Ω)=B(0,1)and in addition to this, f is one to one. The proof involves several ideas which have beendeveloped up to now. The proof is based on the following important theorem, a case ofMontel’s theorem. Before, beginning, note that the Riemann mapping theorem is a classicexample of a major existence theorem. In mathematics there are two sorts of questions,those related to whether something exists and those involving methods for finding it. Thereal questions are often related to questions of existence. There is a long and involvedhistory for proofs of this theorem. The first proofs were based on the Dirichlet principleand turned out to be incorrect, thanks to Weierstrass who pointed out the errors. For moreon the history of this theorem, see Hille [65].

The following theorem is really wonderful. It is about the existence of a subsequencehaving certain salubrious properties. It is this wonderful result which will give the existenceof the mapping desired. The other parts of the argument are technical details to set thingsup and use this theorem.

55.3.1 Montel’s TheoremTheorem 55.3.1 Let Ω be an open set in C and let F denote a set of analytic functionsmapping Ω to B(0,M)⊆C. Then there exists a sequence of functions from F , { fn}∞

n=1 andan analytic function, f such that f (k)n converges uniformly to f (k) on every compact subsetof Ω.

Proof: First note there exists a sequence of compact sets, Kn such that Kn ⊆ intKn+1 ⊆Ω for all n where here intK denotes the interior of the set K, the union of all open setscontained in K and ∪∞

n=1Kn = Ω. In fact, you can verify that

B(0,n)∩{

z ∈Ω : dist(z,ΩC)≤ 1

n

}works for Kn. Then there exist positive numbers, δ n such that if z ∈ Kn, then B(z,δ n) ⊆intKn+1. Now denote by Fn the set of restrictions of functions of F to Kn. Then let z ∈ Kn