I have been working on the following exercise: Let $\Omega \subsetneq \mathbb{C}$ simply connected, open set and $a,b \in \Omega$ fixed points. Let $C(\Omega, a ,b) = \sup \{|f'(a)| ; \quad f: \Omega \to \Omega \quad \text{holomorphic}, \quad f(a) = b\}$.
(1) if $\Omega = \{|z| < 1\} = \mathbb{D}$ and $a = b =0$ then compute $C(\Omega, 0, 0)$.
(2) Show that $C(\Omega, a, b) < \infty$ for any $\Omega \subsetneq \mathbb{C}$ and $a,b \in \Omega$.
(3) Does (2) still hold if $\Omega = \mathbb{C}$? Prove your assertion
My solution:
for (1) since $f$ is from unit disc to unit disc and $f(0) = 0 $ then Schwarz lemma says that $|f'(0)| \leq 1$. Taking supremum we get that $C(\mathbb{D}, 0, 0) = 1$.
for (2) since $\Omega$ is simply connected the Riemann Mapping Theorem states that there exist maps $g: \mathbb{D} \to \Omega$ such that $g$ is holomorphic, $g(0) = a$, and $g'(0) > 0$ and $h: \Omega \to \mathbb{D}$ such that $h$ is holomorphic, $h(b) = 0$, and $h'(b) > 0$. Then define map $F(z) = h\circ f \circ g (z)$. Note that $F:\mathbb{D} \to \mathbb{D}$ and $F(0) = 0$ by construction. Then by Schwarz lemma again we get that $F'(0) \leq 1$. But$$ F'(z) = h'(f(g(z)))f'(g(z))g'(z)$$$$ F'(0) = h'(b) f'(a) g'(0) \leq 1$$This implies that$$f'(a) \leq \frac{1}{h'(b) g'(0)} < \infty$$Since $h'(b) > 0 $ and $g'(0) > 0$. Taking supremum we get that $C(\Omega, a,b) < \infty$
I am stuck when it comes to (3). I believe that the supremum will not be finite but cannot come up with any counter examples. My guess comes from the fact that in the statement of Riemann Mapping theorem, $\Omega$ has to be a proper subset and not entire $\mathbb{C}$ but I don't know how to apply this to the problem since we are not mapping into $\mathbb{D}$, but rather $\Omega \to \Omega$ and the function is not even required to be conformal.