# Riemann mapping theorem

Originally published at Inside the Mind of the G Machine. You can comment here or there.

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.

Involved in this is Montel’s theorem, which I will now state and prove.

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 $\mathcal{F}$ on domain $D$ 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 $K \subset D$, we can find some $r$ for which for every point $z_0 \in K$, $\overline{B(z_0, 2r)} \subset D$. By local boundedness we have some $M>0$ such that $|f(z)| \leq M$ in all of $B(z_0, 2r)$. Thus, for any $w \in K$, we can use Cauchy’s integral formula, for any $z \in B(w, r)$. In that, the radius $r$ versus $2r$ is used to bound the denominator with $2r^2$.

\begin{aligned} |f(z) - f(w)| &= \left| \oint_{\partial B(z_0, 2r)} \frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta)}{\zeta - w}d\zeta \right| \\ & \leq |z-w| \oint_{\partial B(z_0, 2r)} \left| \frac{f(\zeta)}{(\zeta - z)(\zeta - w)} \right| d\zeta \\ & < \frac{|z-w|2\pi r}{2\pi 2r^2} M. \end{aligned}

This shows it’s locally Lipschitz and thus locally equicontinuous. To choose the $\delta$ we can divide our $\epsilon$ by that Lipschitz constant alongside enforcing less than $2r$ so as to stay inside the domain.

With this we can finish off with the Arzela-Ascoli theorem.     ▢

Now take the family $\mathcal{F}$ of analytic, injective functions from simply connected region $\Omega$ onto $\mathbb{D}$ the unit disk which take $z_0$ to $0$. On this we have the following.

Proposition If $f \in \mathcal{F}$ is such that for all $g \in \mathcal{F}$, $|f'(z)| \geq g'(z)$, then $f$ surjects onto $\mathbb{D}$.

Proof:  We prove the contrapositive. In order to do so, it suffices to find for any $f$ that hits not $w \in \mathbb{D}$, $f = s \circ g$, where $s, g$ are analytic with $g(z_0) = 0$ and $s$ is a self-map on $\mathbb{D}$ that fixes $0$ and is not an automorphism. In that case, we can deduce from Schwarz lemma that $|s'(0)| < 1$ and thereby from the chain rule that $g'(z_0) > f'(z_0)$.

Recall that we have automorphisms on $\mathbb{D}$, $T_w = \frac{z-w}{1-wz}$, for all $w \in \mathbb{D}$ and that their inverses are also automorphisms. Let’s try to take $0$ to $w$, then $w$ to $w^2$ via $p(z) = z^2$, and finally $w^2$ to $0$. With this, we have a working $s = T_{w^2} \circ p \circ T_w^{-1}$.     ▢

Nonemptiness of family

It is not difficult to construct an analytic injective self map on $\mathbb{D}$ that sends $z_0$ to $0$. The part of mapping $z_0$ to $0$ is in fact trivial with the $T_w$s. To do that it suffices to map $\mathbb{D}$ to $\mathbb{C} \setminus \overline{\mathbb{D}}$ as after that, we can invert.

Since $D$ is not the entire complex plane, there is some $a \notin D$. By translation, we can assume that $a = 0$. Because the region is simply connected, there is a path from $0$ to $\infty$ outside the region, which means there is an analytic branch of the square root. For any $w$ that gets hits by that, $-w$ does not. By the open mapping theorem, we can find a ball centered at $-w$ 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 $z_0$ with maximal derivative (in absolute value) there, we are finished. Specifically, let $\{f_n\}$ be a sequence from $\mathcal{F}$ such that

$\lim_n |f'_n(z_0)| = \sup_{f \in \mathcal{F}} \{|f'(z_0)|\}$.

This can be done by taking functions with sufficiently increasing derivatives at $z_0$. 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 $f_n$ is a sequence of injective analytic functions with converge uniformly on compact sets to $f$, then $f$ is constant or injective.

Proof: Recall that Hurwitz’s theorem states that if $f$ has a zero of some multiplicity $m$ at some point $z_0$, then for any $\epsilon > 0$, we will, past some $N$ in the index of the sequence, have $m$ zeros within $B(z_0, \epsilon)$ for all $f_n, n > N$, provided $f$ is not constantly $0$. For any point to see that a non-constant $f$ can hit it only once, it suffices to do a translation by that point on all the $f_n$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 $1$, with the $f_n$s being injective, can be applied.     ▢

To show analyticity, we can use Weierstrass’s theorem.

Weierstrass’s theorem Take $\{f_n\}$ and supposed it converges uniformly on compact sets to $f$. Then the following hold:
a. $f$ is analytic.
b. $\{f'_n\}$ converges to $f'$ 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 $N$s or $\delta$s to uniformize some limit.

We can do the same here. For every $z$ on a compact set, express $f_n$ as integral of $\frac{f_n}{\zeta - z}$ via Cauchy’s integral formula on some ball centered at $z$. Uniform convergence of $\frac{f_n}{\zeta - z}$ on the boundary to $\frac{f}{\zeta-z}$ allows us to put the limit inside the integral to give us $f$, as represented via Cauchy’s integral formula. The same can be done for the $\{f'_n\}$

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 $\mathbb{D}$ satisfies that $f(z_0) = 0$. If not, convergence would be naught at $z_0$ since $f_n(z_0) = 0$ for all $n$.

This gives us existence. There is also a uniqueness aspect of the Riemann mapping theorem that comes when one imposes $f'(z_0) \in \mathbb{R}$. This is very elementary to prove and will be left to the reader.

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