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