Review of holomorphic functions of one variable - 2

In this post, we are going to prove the Riemann mapping theorem, which characterizes the simply connected open subsets in .

Theorem (Riemann mapping theorem). Suppose is a simply connected nonempty proper subset in and . Then there exists a unique bijective and holomorphic with nonvanishing derivative such that and .

Remark: A bijective and holomorphic function with nonvanishing derivative is usually called a conformal function. This is equivalent to that the function has a holomorphic inverse.

First let us consider the uniqueness statement.

Theorem (Schwarz lemma). Suppose is holomorphic and satisfies . Then for each and . If either there is a nonzero such that , or , then for some .

Proof: Since , there is a holomorphic function such that for each . For each , the maximal value of on can only be attained on , implying that It follows that on . Therefore for any and . If either of the equality holds, then must be constant on by the maximal module principle. Thus for some and .

The uniqueness statement of the Riemann mapping theorem then follows from the Schwarz lemma. In fact, suppose and both satisfies the conditions, then we can consider the holomorphic functions and . The Schwarz lemma then implies that and are only differ by a rotation. Since their derivatives are both positive real at , they are actually the same.

Before the proof of the existence statement, we introduce the concept of univalent functions. A function on a connected open set is called univalent if is injective and holomorphic.

Theorem. Suppose is a connected open set in and is univalent. Then has no zero on .

Proof: Assume for some . Let . Then is a zero of with order . Take a neighborhood of on which is the only zero of . By a similar proof to the open mapping theorem, for each sufficiently near , the total order of zeros of in the neighborhood is . Since is nonzero at these zeros, they must be different zeros of order , contradicting the injectivity of . We conclude that cannot have any zero on .

Since a bijective and holomorphic function is naturally univalent, it must have nonvanishing derivative, and hence become conformal.

Theorem (Montel theorem). Suppose is a connected open set in , and is a nonempty family of holomorphic functions on such that is uniformly bounded on each compact subset. Then is equicontinuous on each compact subset, and each sequence in has a subsequence that is uniformly convergent on each compact subset.

Proof: First we show that for each , there is such that and is equicontinuous on . Since is open, we can take . Let , and we see that . There is such that for each and each , we have . Consider any . Take . For each such that , and each , we have

Then we show that is equicontinuous on each compact subset. Consider any compact subset and . For each , there is a neighborhood such that is equicontinuous on . Since form an open cover of and is compact, there are such that By the Lebesgue number lemma, we have a such that for , implies that they are contained in the same for some . As shown in the previous part, for each there is such that Let . Then we can see that for each such that , and each , we have .

Now we show the final conclusion. Suppose is a sequence in . Take a sequence of compact substes of such that that for ech , and that for each compact , there is such that . By the previous part and the Arzela-Ascoli theorem, there is a subsequence of such that converges uniformly on . Then there is a subsequence of such that converges uniformly on . Doing this repeatedly, we have a list of sequence such that converges uniformly on for each . Let . For a compact , take such that contains . We see that is a subsequence of , implying that converges uniformly on .

Now we step into the proof of the existence statement of the Riemann mapping theorem.

STEP 1. Each simply connected nonempty proper subset of can be conformally mapped into a subset of , with the given maps to .

Take . Since is simply connected, we have take a simple-valued holomorphic branch of We can directly see that is injective. Fix a . We claim that has a positive lower bound on . Otherwise, there is a sequence in such that . Then implying that , a contradiction. Thus is a bounded univalent mapping on . By a scaling and a translation, we get a conformal mapping from to a subset of that maps to .

Therefore we can assume that is contained in and that .

Lemma. Suppose is a connected open set in and is a sequence of univalent functions on such that converges uniformly to on each compact subset. Then either is constant or is univalent.

Proof: The Cauchy integral formula and the uniform convergence on each compact subset implies that is holomorphic. Assume that is nonconstant and not injective. Then there are in such that . Since is nonconstant, there is such that is the only zero of in . It follows from the argument principle that Noting that each is injective, and that we have the uniform convergence we obtain a contradiction. The conclusion follows.

STEP 2. The “maximal” univalent function from to preserving the origin is the desired conformal mapping.

Let be the family of all univalent functions from to that maps to . Since the identity function belongs to , is nonempty. It is clear that is uniformly bounded. Take such that . By the Cauchy integral formula, we have Thus is finite. Meanwhile, as the identity function belongs to , we have . Suppose is a sequence in such that . By the Montel theorem, there is a subsequence of converging uniformly to a function on each compact subsets on . We can then see that is univalent and from the lemma and the fact that . Thus with .

It remains to show that is surjective onto . Assume that is not surjective. Then there is . Let Then is a conformal mapping from to itself that switches and . It follows that does not contain , and hence there is a simple-valued holomorphic branch of Consider It can be seen that is univalent with , i.e., . Let and Since is not injective, it cannot be a rotation, and then the Schwarz lemma suggests that . Noting that , we have If is positive, then , contracting to that is the supremum. Otherwise , a contradiction as well. In conclusion, must be surjective.

The proof of the Riemann mapping theorem is now completed.