Using Schwarz Lemma to find an upper bound for $f'(0)$

Let Ω denote the half-plane $\{z : Re z > −1\}$. Let $f : Ω → C$ be a holomorphic function such that $|f(z)| < 2$ for all $z ∈ Ω$ and $f(0) = 0$. Find an upper bound for $|f'(0)|$ and give an example to show that your upper bound is attained.

In my attempt at the solution I let $h = z/2$ be a map from $Ω$ to the unit circle and thus by the Schwarz Lemma, $h(f(z)) \le |z|$ and $h'(f(0)) \le 1$

I'm not sure how to get an inequality with $|f'(0)|$ from this.

• I think you should have $|h'(f(0))||f'(0)|\leq 1$ by chain rule. – Paul Mar 24 '15 at 2:32
• $h(z)=z/2$ doesn't map $\Omega$ to the unit circle! Schwarz lemma requires you have a holomorphic function $f:\mathbb E\to\mathbb E$ where $\mathbb E$ is the unit disk. – Pedro Tamaroff Mar 24 '15 at 2:48
• I think $h(z) = - {z \over 2+z}$ might be more appropriate. – copper.hat Mar 24 '15 at 2:49

Let $g:B(0,1) \to \Omega$ be given by $g(z) = - {2z \over 1+z}$.
Let $\phi = {1 \over 2} f \circ g$, and note that $\phi(0) = 0$ and $|\phi(z)| < 1$ for all $z \in B(0,1)$.
The Schwartz lemma gives $|\phi'(0)| \le 1$, and we have $\phi'(0) = {1 \over 2} f'(0) g'(0) = -f'(0)$, hence we have $|f'(0)| \le 1$.
If $|f'(0)| = 1$, then $|\phi'(0)| = 1$ and the Schwartz lemma tells us that $\phi(z) = c z$ for some $|c|=1$. Since $f = 2 \phi \circ g^{-1}$, we have $f(z) = -{ 2 c z \over z+2}$.
It is easy to check that $f'(0) = -c$, hence the bound is attained.