Ahlfors function
For each compact
, there exists a unique extremal function, i.e.
such that
,
and
. 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
(otherwise
by Liouville's theorem and hence
). Then there exists a unique connected component U of
that contains
, where
is the Riemann sphere.
The claim is that U is simply connected. To see this, consider first a smooth simple closed curve
in
and let
be some point in
. By the Jordan curve theorem (actually, since
is smooth, one only needs easy versions of the Jordan curve theorem),
contains a connected component, say
that is disjoint from
. Then
. Moreover, since
is smooth, the union
is homeomorphic to
The Riemann mapping theorem now yields a biholomorphism
such that
and
. (Here,
denotes the unit disk in
.) Defining
for each
, this defines a holomorphic map
. In particular,
, so that
.
To prove the reverse inequality, let
with
and put
. Then
is analytic (since f and g are),
![{\displaystyle F(0)=f(g^{-1}(0))=f(\infty )=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22f7a1210c8dce1de835d29ae9d881b007c1d64b)
and so we may apply the Schwarz lemma to F. Hence,
. Thus,
![{\displaystyle 1\geq F'(0)={\frac {f'(g^{-1}(0))}{g'(g^{-1}(0))}}={\frac {f'(\infty )}{g'(\infty )}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6885243d4aa0e913fc8e18b80e6f902a7bd2ecc)
which gives us
. Taking the supremum over all such f, we get
. This concludes the proof.
Additional properties assuming finite connectivity
Let
. If
and E has n components, then the Ahlfors function is analytic across
. Moreover, if
is smooth, then
.