User:OdedSchramm/sb2

Ahlfors function

For each compact $E\subset \mathbb {C}$, there exists a unique extremal function, i.e. $f\in {\mathcal {H}}^{\infty }(\mathbb {C} \setminus K)$ such that $\|f\|\leq 1$, $f(\infty )=0$ and $f'\,(\infty )=\gamma (K)$. This function is called the Ahlfors function of K This can be proved by using a normal family argument involving Montel's theorem.

Proof of existence for a continuum

There is a relatively simple proof of the existence of an Ahlfors function, based on the Riemann mapping theorem, if we assume additionally that K is connected.

If K is compact and connected, we can assume $E\neq \varnothing$ (otherwise ${\mathcal {H}}^{\infty }(\mathbb {C} \setminus K)={\mathcal {H}}^{\infty }(\mathbb {C} )=\{{\text{constant functions}}\}$ by Liouville's theorem and hence $\gamma (K)=0$). Then there exists a unique connected component U of ${\hat {\mathbb {C} }}\setminus K$ that contains $\infty$, where ${\hat {\mathbb {C} }}=\mathbb {C} \cup \{\infty \}$ is the Riemann sphere.

The claim is that U is simply connected. To see this, consider first a smooth simple closed curve $\gamma$ in $U\cap \mathbb {C}$ and let $x_{0}$ be some point in $\gamma$. By the Jordan curve theorem (actually, since $\gamma$ is smooth, one only needs easy versions of the Jordan curve theorem), $\mathbb {C} \setminus \gamma$ contains a connected component, say $A$ that is disjoint from $K$. Then $A\subset U$. Moreover, since $\gamma$ is smooth, the union $\gamma \cup A$ is homeomorphic to

The Riemann mapping theorem now yields a biholomorphism $g\ :\ U\to B(0,1)$ such that $g(\infty )=0$ and $g'(\infty )>0$. (Here, $B(0,1)$ denotes the unit disk in $\mathbb {C}$.) Defining $g(z)=0$ for each $z\in \mathbb {C} \setminus (U\cup K)$, this defines a holomorphic map $g\ :\ \mathbb {C} \setminus K\to B(0,1)$. In particular, $g\in {\mathcal {H}}^{\infty }(\mathbb {C} \setminus E)$, so that $\gamma (E)\geq g'(\infty )$.

To prove the reverse inequality, let $f\in {\mathcal {H}}^{\infty }$ with $\|f\|_{\infty }\leq 1,\ f(\infty )=0$ and put $F:=f\circ g^{-1}$. Then $F\ :\ B(0,1)\to {\overline {B(0,1)}}$ is analytic (since f and g are),

$F(0)=f(g^{-1}(0))=f(\infty )=0$

and so we may apply the Schwarz lemma to F. Hence, $F'(0)\leq 1$. Thus,

$1\geq F'(0)={\frac {f'(g^{-1}(0))}{g'(g^{-1}(0))}}={\frac {f'(\infty )}{g'(\infty )}}$

which gives us $f'(\infty )\leq g'(\infty )$. Taking the supremum over all such f, we get $\gamma (E)\leq g'(\infty )$. This concludes the proof.

Additional properties assuming finite connectivity

Let $U:=\mathbb {C} \setminus E$. If $n\in \mathbb {N}$ and E has n components, then the Ahlfors function is analytic across $\partial U$. Moreover, if $\partial U$ is smooth, then $f(\partial U)=\{1\}$.