Uniformization of Riemann surfaces revisited

Abstract

We give an elementary and self-contained proof of the uniformization theorem for noncompact simply connected Riemann surfaces.

Appendices

Appendix A Proof of Perron’s principle

Since the family $\mathcal {F}$ is locally bounded above, we can define $u:X \longrightarrow \mathbb {R}$, $u=\sup _{f\in \mathcal {F}} f.$ Our aim is to show that u is harmonic. Since harmonicity is a local property, it suffices to prove that u is harmonic on a disk $D\subset X$. Let $A=\{ z_0, z_1,... \} \subset D$ be a dense subset. For each j, there exists a sequence $(v_{jk})_{k \in \mathbb N} \subset \mathcal F$ such that $u(z_j)=\lim _{k\rightarrow \infty } v_{jk}(z_j)$. Since $\mathcal F$ is a Perron family, the map $h_1=v_{11}^{(D)}$ belongs to $\mathcal F$, is harmonic on D, and $h_1 \ge v_{11}$. Suppose we constructed the functions $\{h_1,...,h_n\} \subset \mathcal F$ harmonic on D such that $h_k \ge h_{k-1}$ and $h_k \ge v_{ij}$ for all $k\in \{1,...,n\}$, $i,j\in \{1,...,k\}$. Pick $h_{n+1}=\max (v_{11},v_{12},...,h_n)^{(D)} \in \mathcal F$ harmonic on D, with $h_{n+1} \ge h_{n}$ and $h_{n+1} \ge v_{ij}$ for any $i,j \in \{1,...,n+1\}$. Then $h_n(z_j) \ge v_{jk}(z_j)$ for all $n \ge k \ge j$. Letting $k\rightarrow \infty$ in the previous inequality, we get:

$$\begin{aligned}\lim _{n\rightarrow \infty }h_n(z_j)=u(z_j).\end{aligned}$$

Since $(h_n)_{n \in \mathbb N}$ is an increasing sequence, we can consider the Lebesgue measurable function

$$\begin{aligned} w:D\longrightarrow \mathbb {R},&w=\lim _{n\rightarrow \infty } h_n\le \sup _{g\in \mathcal {F}}g_{\vert _{D}}=u_{\vert _D}. \end{aligned}$$

By the dominated convergence theorem, for every disk $D'$ centered at a point $x\in D$ we have

$$\begin{aligned} w(x)=\lim _{n\rightarrow \infty } h_n(x)=\lim _{n\rightarrow \infty } \frac{1}{\pi }\int _{D'} h_n(z)\mathrm{d}z=\frac{1}{\pi }\int _{D'} \lim _{n\rightarrow \infty } h_n(z)\mathrm{d}z=\frac{1}{\pi }\int _{D'} w(z)\mathrm{d}z. \end{aligned}$$

Here we used the mean property on disks for harmonic functions, which is an immediate consequence of the definition. This implies easily that w is $C^0$. Again by dominated convergence,

$$\begin{aligned} \frac{1}{2\pi } \int _{0}^{2\pi } w(e^{it}) \mathrm{d}t= & {} \frac{1}{2\pi } \int _{0}^{2\pi } \lim _{n\rightarrow \infty } h_n(e^{it}) \mathrm{d}t= \lim _{n\rightarrow \infty } \frac{1}{2\pi }\int _{0}^{2\pi } h_n(e^{it}) \mathrm{d}t\\= & {} \lim _{n\rightarrow \infty } h_n(x)=w(x) \end{aligned}$$

thus showing that w is harmonic. We claim that $w=u$.

Since $h_n \in \mathcal F$ for any $n \in \mathbb N$, then $h_n \le u$, so $w \le u$. But $w(z_j)=u(z_j)$ for all $j \in \mathbb N$, hence $w \ge v$ on the dense subset $A \subset D$ for every $v \in \mathcal F$. Since $\mathcal F$ is a family of continuous functions, it follows that $w \ge v$ for any $v \in \mathcal F$, therefore $w \ge u$, which ends the proof.

Appendix B Proof of Dirichlet’s principle

If f is constant, the conclusion is clear, otherwise let $m<M$ be the infimum, respectively, the supremum of f on $\partial Y$. Consider the family $\mathcal F$ of continuous functions $g:Y \longrightarrow [m,M]$ which are subharmonic on the interior of Y with $g_{\vert _{\partial Y}} \le f$. The first condition for $\mathcal F$ to be a Perron family is evident. Let $g \in \mathcal F$ and $D'$ a disk in Y. Using Remark 4, $g^{(D')}$ is subharmonic. By the maximum principle, $g^{(D')}$ still takes values in [m, M]. Thus $g^{(D')} \in \mathcal F$. Now $\mathcal {F}\ne \emptyset$ since it contains the constant function m. Let $F=\sup _{g \in \mathcal F} g$. By Perron’s principle, F is harmonic on .

We must prove that for every $x\in \partial Y$, $\lim _{\xi \rightarrow x}F(\xi )=f(x)$. Let us first show that $\liminf _{\xi \rightarrow x} F(\xi )\ge f(x)$. If $f(x)=m$, there is nothing to prove, hence let us suppose that $f(x)>m$. We work in a disk $D\subset X$ centered at x with local coordinate $\xi$.

Fix $\epsilon >0$ such that $f(0)-\epsilon >m$. There exists $1>\delta >0$ such that for every $\xi \in \partial Y$ with $|\xi |<\delta$ we have $f(0)-\epsilon <f(\xi )$. Consider the exterior normal to $\partial Y$ at $x=0$. Take a circle centered at a point p on this normal, of radius r small enough such that it touches Y only in 0. Choose $t>1$ such that $\log \frac{1}{t-1}+f(0)-\epsilon <m$ and set $R=rt$. By decreasing r if needed, we can assume that $R<\delta$. Clearly, $|p|\le |\xi -p|$ for every $\xi \in Y\cap D$. Define

$$\begin{aligned} u:D\rightarrow [m, \infty ),&u(\xi )=\max \left( m,\log \tfrac{|p|}{|\xi -p|} +f(0)-\epsilon \right) . \end{aligned}$$

Notice that u is the maximum of two harmonic functions, hence subharmonic. When $|\xi |$ is close to 0, $u(\xi )=\log \tfrac{|p|}{|\xi -p|}+f(0)-\epsilon$, while for $|\xi |\ge R$, $u(\xi )=m$. Hence u can be continuously extended (by the constant value m) to $Y\cup D$, and is subharmonic on . It is straightforward to check that $u_{\vert _{Y}}$ belongs to the family $\mathcal {F}$. Thus, for small $|\xi |$, we get:

$$\begin{aligned} F(\xi ) \ge u(\xi )=\log \tfrac{|p|}{|\xi -p|}+f(0)-\epsilon . \end{aligned}$$

Since $\epsilon$ was arbitrarily small, we obtain $\liminf _{\xi \rightarrow 0} F(\xi ) \ge f(0)$. We now prove that $\limsup _{\xi \rightarrow 0} F(\xi ) \le f(0)$. If $f(0)=M$, this is clear. Otherwise, take $\epsilon >0$ such that $f(0)+\epsilon < M$. As above, consider $t>1$ and $R=rt$ such that $-\log \frac{1}{t-1}+f(0)+ \epsilon >M$, and $f(\xi )<f(0)+ \epsilon$ for $|\xi | \le R$. Define

$$\begin{aligned} U:D\rightarrow (-\infty ,M],&U(\xi )=\min \left( M,-\log \tfrac{|p|}{|\xi -p|} + f(0)+\epsilon \right) . \end{aligned}$$

Notice that U is the minimum of two harmonic functions, thus $-U$ is subharmonic. Also $U(\xi )=M$ for $|\xi |>R$, $U(\xi )=-\log \tfrac{|p|}{|\xi -p|} + f(0)+\epsilon$ for small $|\xi |$, $U \ge m$, and $U(\xi )\ge f(\xi )$ on $\partial Y$. For every $g\in \mathcal {F}$, the continuous function $g-U$ is nonpositive on the boundary of the compact domain $\{|\xi |\le R \}\cap Y$ and subharmonic in the interior. By the maximum principle it follows that on this compact set $g\le U$, hence $F\le U$. Thus for small $|\xi |$,

$$\begin{aligned} F(\xi )\le -\log \tfrac{|p|}{|\xi -p|}+f(0)+\epsilon . \end{aligned}$$

Therefore $\limsup _{\xi \rightarrow 0} F(\xi ) \le f(0)$, showing that $\lim _{\xi \rightarrow 0} F(\xi )$ exists and equals f(0).

Appendix C Second countability in the presence of a holomorphic function

Let us prove that if there exists a nonconstant holomorphic function $f:X\longrightarrow {\mathbb {C}}$, then the Riemann surface X is second countable.

Let ${\mathcal {B}}$ be a countable basis of topology for ${\mathbb {C}}$. Let ${\mathcal {A}}$ be the set of those connected components of each $f^{-1}(U)$, $U \in {\mathcal {B}}$, which are second countable. We claim that A is a basis of topology for X. Indeed, let $D \subset X$ be an open set and $x \in D$. By the identity theorem for holomorphic functions, $f^{-1}\left( f(x) \right)$ is discrete, hence there exists an open neighborhood W of x, relatively compact in D, such that $W \cap f^{-1} \left( f(x) \right) = \lbrace x \rbrace$. We have that $f \left( \partial \overline{W} \right)$ is compact in ${\mathbb {C}}$. Since $f(x) \notin f \left( \partial \overline{W} \right)$, there exists $U \in {\mathcal {B}}$ which contains f(x) such that $U \cap f \left( \partial \overline{W} \right) =\emptyset$. Let V be the connected component of $f^{-1}(U)$ which contains x. Since $V \cap \partial \overline{W}=\emptyset$, we get that $V \subset W$, hence V is second countable. We found $x \in V \subset D$, with $V \in {\mathcal {A}}$, therefore the claim is proved. We next prove that ${\mathcal {A}}$ is countable.

Each $V \in {\mathcal {A}}$ intersects countably many other elements in ${\mathcal {A}}$. Otherwise, there would exist $U \in {\mathcal {B}}$ such that V intersects uncountably many connected components of $f^{-1}(U)$. It would follow that V contains uncountably many disjoint open subsets, which is a contradiction.

Consider $V_0 \in {\mathcal {A}}$. There exists a countable number of open sets from ${\mathcal {A}}$ which intersect $V_0$, denote their union with $V_0$ by $V_1$. For $k \ge 1$, define $V_{k+1}$ as the union of $V_k$ with those open sets in ${\mathcal {A}}$ which intersect $V_k$. By induction, $V_k$ is countable. It is clear that $\cup _{k \ge 1} V_k$ is both open and closed, hence $\cup _{k \ge 1} V_k=X$. Therefore all the open sets from ${\mathcal {A}}$ appear in this process, so ${\mathcal {A}}$ is a countable union of countable sets, hence countable.

Appendix D Compactness of families of injective holomorphic functions

Proof of Lemma 13

Denote by $r_n=\sup \{ r \in \mathbb {R}: r\mathbb {D}\subset f_n({\mathbb {D}}) \}$. Let $a_n \in \partial (r_n \overline{ \mathbb {D}}) {\setminus } f_n({\mathbb {D}})$. From the Schwarz lemma for the function ${f_n^{-1}}_{\vert _{r_n \mathbb {D}}}$, it follows that $|a_n|=r_n \le 1$. By extracting a subsequence, we can assume that $(a_n)_{n \in \mathbb {N}}$ is convergent. Consider the function $g_n: {\mathbb {D}} \longrightarrow \mathbb {C}$, $g_n=a_n^{-1} f_n$. It is easy to see that ${\mathbb {D}} \subset g_n({\mathbb {D}})$ and also $1 \notin g_n({\mathbb {D}})$. Since $g_n$ is holomorphic and injective, it follows that $g_n({\mathbb {D}})$ is simply connected; hence, we can construct the square root

$$\begin{aligned}&\psi _n:g_n({\mathbb {D}})\longrightarrow {\mathbb {C}}^{*}&\psi _n(z)=ie^{\frac{1}{2}\int _0^z \frac{dw}{w-1}}. \end{aligned}$$

Then $\psi _n(0)=i$ and $\psi _n^2(z)=z-1$ for $z \in g_n({\mathbb {D}})$. Since $\psi _n^2$ is injective, the image of $\psi _n$ cannot contain pairs of the form $(w,-w)$. Thus, there exists a disk D centered at $-i$ disjoint from the image of $\psi _n$ for every n. Define $h_n=\psi _n \circ g_n: {\mathbb {D}} \longrightarrow \mathbb {C}{\setminus } D$. There exists $0<r<\infty$ so that the homography $\alpha (z)=\frac{1}{z+i}$ maps $\mathbb {C}{\setminus } D$ into $r\mathbb {D}$. Using Montel’s Theorem and relabeling the sequence, we can assume that $\alpha \circ h_{n}$ converges to a holomorphic map $h:\mathbb {D}\longrightarrow r \overline{\mathbb {D}}$. If h is nonconstant, it is open by Sect. 2; hence, it maps $\mathbb {D}$ to $r\mathbb {D}$. Thus we can compose it with the inverse homography $\alpha ^{-1}$. Otherwise, h equals the constant $\frac{1}{2i}$. In both cases, $h_n$ converges to $\alpha ^{-1} \circ h$, thus $f_{n}=a_{n}\left( 1+ h_n^2 \right)$ converges in $\mathcal O (\mathbb {D})$. $\square$