I am going to make an effort to understand the proof of the Riemann mapping theorem, which states that there exists a conformal map from any simply connected region that is not the entire plane to the unit disk. I learned of its significance that its combination with the Poisson integral formula can be used to solve basically any Dirichlet problem where the region in question in simply connected.
Definition A normal family of continuous functions is one for which every sequence in it has a uniformly convergent subsequence.
Montel’s theorem A family on domain of holomorphic functions which is locally uniformly bounded is a normal family.
Proof: Turns out holomorphic alongside local uniform boundness is enough for us to establish local equicontinuity via the Cauchy integral formula. On any compact set , we can find some for which for every point , . By local boundedness we have some such that in all of . Thus, for any , we can use Cauchy’s integral formula, for any . In that, the radius versus is used to bound the denominator with .
This shows it’s locally Lipschitz and thus locally equicontinuous. To choose the we can divide our by that Lipschitz constant alongside enforcing less than so as to stay inside the domain.
With this we can finish off with the Arzela-Ascoli theorem. ▢
Now take the family of analytic, injective functions from simply connected region onto the unit disk which take to . On this we have the following.
Proposition If is such that for all , , then surjects onto .
Proof: We prove the contrapositive. In order to do so, it suffices to find for any that hits not , , where are analytic with and is a self-map on that fixes and is not an automorphism. In that case, we can deduce from Schwarz lemma that and thereby from the chain rule that .
Recall that we have automorphisms on , , for all and that their inverses are also automorphisms. Let’s try to take to , then to via , and finally to . With this, we have a working . ▢
Nonemptiness of family
It is not difficult to construct an analytic injective self map on that sends to . The part of mapping to is in fact trivial with the s. To do that it suffices to map to as after that, we can invert.
Since is not the entire complex plane, there is some . By translation, we can assume that . Because the region is simply connected, there is a path from to outside the region, which means there is an analytic branch of the square root. For any that gets hits by that, does not. By the open mapping theorem, we can find a ball centered at that is entirely outside the region. With this, we can translate and dilate accordingly to shift that to the unit disk.
Construction of limit to surjection
We can see now that if we can construct a sequence of functions in our family that converges to an analytic one with the same zero at with maximal derivative (in absolute value) there, we are finished. Specifically, let be a sequence from such that
This can be done by taking functions with sufficiently increasing derivatives at . With Montel’s theorem on our obviously locally uniformly bounded family, we know that our family is normal, and thus by definition, we can extract some subsequence that is uniformly convergent on compact sets. Now it remains to show that the function converged to is analytic and injective.
The injective part follows from a corollary of Hurwitz’s theorem, which we now state.
Hurwitz’s theorem (corollary of) If is a sequence of injective analytic functions with converge uniformly on compact sets to , then is constant or injective.
Proof: Recall that Hurwitz’s theorem states that if has a zero of some multiplicity at some point , then for any , we will, past some in the index of the sequence, have zeros within for all , provided is not constantly . For any point to see that a non-constant can hit it only once, it suffices to do a translation by that point on all the s to turn it into a zero, so that the hypothesis of Hurwitz’s theorem, which in this case, bounds the number of zeros above by , with the s being injective, can be applied. ▢
To show analyticity, we can use Weierstrass’s theorem.
Weierstrass’s theorem Take and supposed it converges uniformly on compact sets to . Then the following hold:
a. is analytic.
b. converges to uniformly on compact sets.
Proof: This is a more standard theorem, so I will only sketch the proof. Recall the definition of compact as possessing the every cover has finite subcover property. This is so powerful, because we can for any collection of balls centered at every point of the cover, find a finite of them that covers the entire space, and finiteness allows us to take a maximum or minimum of finite s or s to uniformize some limit.
We can do the same here. For every on a compact set, express as integral of via Cauchy’s integral formula on some ball centered at . Uniform convergence of on the boundary to allows us to put the limit inside the integral to give us , as represented via Cauchy’s integral formula. The same can be done for the
Again we can use two radii as done in the proof of Montel’s theorem to impose uniform convergence on a smaller ball. ▢
Finally, our candidate conformal map to satisfies that . If not, convergence would be naught at since for all .
This gives us existence. There is also a uniqueness aspect of the Riemann mapping theorem that comes when one imposes . This is very elementary to prove and will be left to the reader.