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If have got a $k$-fold connected surface $G$, which is bounded by n distinct, non-intersecting Jordan-curves. By Ahlfors it is known that there exists a unique function $\phi$ which maps G to the unit disc $\Delta$ and a previously picked point $p \in G$ is mapped to $0 \in \Delta$ with $\phi^{\prime}(p)$ is maximal.

I´m trying to show that the map $\phi$ is proper, by using that G is relatively compact.

Can someone give a hint how to show that?

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I assume that by $\Delta$ you mean the unit disk, not the unit cicle, as is written in the question.

There are several ways to prove that the Ahlfors map is a proper holomophic map of degree $n$ of $G$ onto $\Delta$. One of them is to use a variational argument (variation of zeros and boundary values), which was Ahlfors' original approach. For this, I suggest you take a look at the classical paper Bounded anaytic functions by Ahlfors.

Another way is to use the properties of the so-called Garabedian function on $G$ :

Suppose that $p=\infty$. Let $f$ be the Ahlfors map on $G$ for the point $\infty$, and let $\sigma$ be a measure on $\partial G$ of norm $\|\sigma\|=f'(\infty)$ such that $\int g d\sigma = g'(\infty)$ for all $g$ belonging to the class $A(G)$ of functions continuous on $\overline{G}$ and holomorphic in $G$. The measure $\mu=dz/(2\pi i) - \sigma$ is orthogonal to $A(G)$, so by a theorem of F. and M. Riesz we get that $\sigma = \psi(z) dz$ for some function $\psi$ that belongs to the Hardy space $H^1(G)$. This function $\psi$ is called the Garabedian function. It is not difficult then to see that the measure $f(z)\psi(z)dz$ is positive on $\partial G$, and the properness of the map $f$ can be deduced from this.

For more details on this approach, see the book Analytic capacity and measure by Garnett, Chapter 1, Theorem 4.1.

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  • $\begingroup$ Thanks for your answer, i hoped that there would be a way to show that the Ahlfors map is proper in a more directly sense, using the definition of properness and i hoped that this proof would allow a generalization to Riemann surfaces. $\endgroup$
    – Josh
    Jun 26, 2014 at 9:59
  • $\begingroup$ @Josh : You're welcome. I don't think there is a simple, easy way to prove properness of the Ahlfors map. $\endgroup$ Jun 26, 2014 at 12:12
  • $\begingroup$ Malik Younsi, regarding the above question about the properness of Ahlfors map, dont you think that we need to have knowledge about zeros of Ahlfors map? Thanks $\endgroup$
    – user59344
    Oct 10, 2014 at 15:01
  • $\begingroup$ I don't understand what you mean by that. In the case of planar domains bounded by $n$ disjoint analytic Jordan curves, the Ahlfors function is proper of degree $n$, so it has $n$ zeros. This is usually what is proved in order to obtain properness. $\endgroup$ Oct 10, 2014 at 18:01
  • $\begingroup$ Maybe I found a way to use the above mentioned surface G, in particular that $G$ is relatively compact. There is a theorem like "If $f: X \longrightarrow Y$ is a continuous mapping between topological spaces and $W$ is an open relatively compact subset of $X$, then the induced map $W \backslash f^{-1}(f(\partial W)) \longrightarrow Y \backslash f(\partial W)$ is proper". So it must be proved that there exists a map $\overline{\phi} : \overline{G} \longrightarrow \overline{\Delta}$ with $\overline{\phi} = \phi$ in $G$ $\endgroup$
    – Josh
    Dec 9, 2014 at 9:46

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