0
$\begingroup$

Im working out this proof needed for the caratheodory - koebe theory.

The idea is quite simply to understand but there is an argument using sqeunces which im questioning about.

The statement is the following:

Let $D \subset \mathbb{C}$ a domain with $0 \in D$. $f,g$ holomorphic in $D$ with $f,g$ not constant. $g$ is injective, $f(0) = 0$ and $\mid f(z) \mid \geq \mid g(z) \mid$ for all $z \in D$.

Then $\sup\lbrace r > 0 \mid B_r(0) \subset f(D) \rbrace \geq \sup\lbrace r > 0 \mid B_r(0) \subset f(D) \rbrace$

This seems quite confusing on the first look but is actually quite easy to understand (at least when drawing a picture).

(Note: $\sup\lbrace r > 0 \mid B_r(0) \subset f(D) \rbrace$ is called the inner radius of $f(D)$ and just represents the supremum of the largest disk you could fit into $f(D)$ around $0$.

Proof:

We proove this by taking a disk $B = B_r(0) \subset g(D)$. This is actually possible because $g(D)$ has to contain $0$ because of $f(0) = 0$ and $\mid f(z) \mid \geq \mid g(z) \mid$. Also $g(D)$ has to be a domain because of the open mapping theorem ($ g \neq $const). This means there is a disk fitting around $0$ in $g(D)$. We show the statement by showing $B \subset f(g^{-1}(B))$ (therefore $g$ has to be injective). This will follow from: Is $b \in \partial f(g^{-1}(B))$ then $\mid b \mid \geq r$.

After defining $B$ there has to be a sequence $a_n \in B$ with $a_n \rightarrow a \in \partial B \quad (n \rightarrow \infty)$ and $f(g^{-1}(a_n)) \rightarrow b \quad (n \rightarrow \infty)$.

That is my question. Why can we say that such sqequence has to exist?

Using this sequence we continue:

$\mid b \mid = \lim \mid f(g^{-1}(a_n)) \mid \geq \lim \mid g(g^{-1} ( a_n)) \mid = \mid a_n \mid = r$.

How can we say that there is such a sequence $a_n$ which converges in $B$ to $a \in \partial B$ AND $\lim f(g^{-1}(a_n)) = b \in \partial f(g^{-1} (B))$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.