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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

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  • $\begingroup$ Hi! I have just posted this question that is very related to yours. Essentially what I want to know is how to solve the first extremal problem you state when $X$ is not simply connected. Can you help me with this? $\endgroup$ – Leo Sera Apr 11 '16 at 3:59
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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.)

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  • $\begingroup$ Yes indeed, the Riemann mapping theorem is a direct consequence of Ahlfor's theorem on the properties of the Ahlfors function on $n$-connected domains. However, as you mention, every proof I know of this theorem uses analyticity of the boundary, which requires the Riemann mapping theorem. So this does not answer the question. The remark about the Bergman kernel is very interesting, I will look it up. Thank you. $\endgroup$ – Malik Younsi Mar 25 '14 at 16:42
  • $\begingroup$ I think the question should be rephrased as: "Is there a proof of Ahlfors theorem which does not need Riemann mapping theorem?" to make it clearer. From these notes it seems that the answer is affirmative at least for $\Omega$ with real analytic boundary. $\endgroup$ – Vít Tuček Mar 25 '14 at 17:09
  • $\begingroup$ What I am interested in here is just the particular case $n=1$ of Ahlfor's theorem, without any assumption on the boundary of the domain. $\endgroup$ – Malik Younsi Mar 26 '14 at 17:11
  • $\begingroup$ Hi @VítTuček ! I have just posted this question that is related to the one in this post . Essentially what I want to know is how to solve the extremal problem of the Riemann Mapping thm when $X$ is not simply connected. Can you help me with this? $\endgroup$ – Leo Sera Apr 11 '16 at 4:01

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