The Problem:

Suppose that $\displaystyle \Omega$ is a simply connected domain with $\displaystyle \Omega \ne \mathbb{C}$, $\displaystyle a \in \Omega$, and for some $\displaystyle r > 0, D(a;r) \subseteq \Omega$ (an open disk at a with radius r). For $\displaystyle \mathbb{D}$ a unit disk, let $\displaystyle f: \mathbb{D} \rightarrow \Omega$ be a conformal map with $\displaystyle f(0) = a$. Prove that $\displaystyle |f'(0)| \ge r$.

Ideas:

- There is a drawing (.pdf) attached that represents the set context.

- Application of the Riemann Ext Thm. provides a conformal map $\displaystyle h1$ from $\displaystyle \Omega$ onto $\displaystyle \mathbb{D}$ with $\displaystyle h1(a) = 0$ and $\displaystyle h1'(a) > 0$ (or another suitable rotation) and a map $\displaystyle h2$ from $\displaystyle f(\mathbb{D})$ onto $\displaystyle \mathbb{D}, h2(a) = 0$ and $\displaystyle h2'(a) > 0$.

- Seems like a setup for Schwarz's Lemma. For example, we can take $\displaystyle g = h1 \circ f$ with $\displaystyle g(0) = 0$ and $\displaystyle g(\mathbb{D}) \subseteq \mathbb{D}$. Then the lemma asserts that $\displaystyle |g(z)| \le |z|$ and $\displaystyle |g'(0)| \le 1$. So $\displaystyle |h1'(f(0))| |f'(0)| \le 1 \Rightarrow |h1'(a)| |f'(0)| \le 1$. I don't see where this helps with this proof.

Thank you in advance for your help in solving this.