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 $$E\ne\varnothing$$ (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 $$f(\partial U)=\{1\}$$.