TITLE: The Riemann mapping theorem via extremal problems
QUESTION [9 upvotes]: Let $X \subsetneq \mathbb{C}$ be a simply connected domain. The Riemann mapping theorem states that there exists a biholomorphism of $X$ onto the unit disk $\mathbb{D}$. A simple and elegant way to obtain such a map is to solve the following extremal problem :
$$\sup \{|h'(z_0)| : h:X \to \mathbb{D} \, \, \mbox{is holomorphic and one-to-one}, \,  h(z_0)=0 \},$$
where $z_0 \in X$ is some fixed point.
One first needs to show that the above class of functions is not-empty. Then, by an elementary normal families argument, the supremum is attained by some function $g$ with $g'(z_0)>0$, and it is not difficult to prove that $g$ is a biholomorphism of $X$ onto $\mathbb{D}$ which maps $z_0$ to $0$.
What happens if we drop the requirement that $h$ is one-to-one?
More precisely, consider the following extremal problem :
\begin{equation}
\sup \{|h'(z_0)| : h:X \to \mathbb{D} \, \, \mbox{is holomorphic},\, h(z_0)=0 \}
\end{equation}
Again by an elementary normal families argument, there exists an extremal holomorphic function $f:X \to \mathbb{D}$ such that $f'(z_0)$ equals the supremum.
Moreover, it is well-known that this function is unique; it is usually called the Ahlfors function. 
If we assume the existence of a biholomorphism $g:X \to \mathbb{D}$ with $g(z_0)=0$ and $g'(z_0)>0$, then it is easy to prove using the Schwarz's lemma and the uniqueness of the Ahlfors function that $h$ must be equal to the Ahlfors function. Therefore, my question is the following :
Is there a (simple?) proof of the Riemann mapping theorem using the second extremal problem?
Thank you and best regards,
Malik

REPLY [4 votes]: Ahlfors Mapping Theorem as stated in his 1946 paper Bounded Analytic Functions: Let $\Omega$ be a domain of connectivity $n \geq 1$, none of whose boundary components reduce to a point. Fix a point $a \in \Omega$ and consider the extremal problem $$ \mathrm{max} |f'(a)| $$
where $f$ is analytic (and single-valued) in $\Omega$ with $|f| < 1$. The solution of this problem is an $n : 1$ map onto $\mathbb{D}$.
So it would seem that the Riemann mapping theorem is just a special case of Ahlfors theorem. However, citing from  notes  by Brad Osgood on the Ahlfors mapping via the Szegö and Garabedian kernels
as presented in S. Bell’s book "The Cauchy Transform, Potential Theory, and Conformal Mapping": 
I don’t know other (easy) ways of seeing this, except that I think it can also be deduced through the connection
between the Riemann mapping theorem and the Bergman kernel, and the extremal properties of the Bergman kernel.
That may avoid some of the issues of smoothness at the boundary that come up in Bell’s approach. I suppose it also
follows from ‘the principle of the hyperbolic metric’, which is through potential theory. It’s all related.
...
There’s a slight catch here. Ahlfors needs a certain amount of boundary smoothness
to form integrals over ∂Ω, and he uses some preliminary conformal mapping to assume that ∂Ω
consists of analytic Jordan curves (a standard technique when working with ﬁnitely connected
domains). This requires the Riemann mapping theorem!  (The general methods in Bell’s book
assume $\mathcal{C}^\infty$ boundaries.)