How to compute $\sup_{f\in\mathcal F}|f(0)|$

Question

Consider $\mathcal F = \{f$ holomorphic in $\Bbb D$ with $f(\Bbb D)\subset\Bbb D,$ $f\left(\frac12\right)=f'\left(\frac12\right)=0\}$, where $\Bbb D$ is the unit disc. Compute $\sup_{f\in\mathcal F}|f(0)|$.

I was asked this on a exam, and I could only think about how every $f\in\mathcal F$ verifies that $|f(z)|<1$, for every $z\in B(0,1)$. So I said that $\sup_{f\in\mathcal F}|f(0)|=1$.

It felt too easy, so I don't think it is correct? In case it wasn't, what was I supossed to do?

Answer 1

Let $T(z) = \frac{1/2-z}{1-z/2}$ be the Möbius transformation which maps the unit disk onto itself with $T(0)=1/2$ and $T(1/2) = 0$, and define $g=f \circ T$.

• Show that $g: \Bbb D \to \Bbb D$ satisfies $g(0) = g'(0) = 0$.
• Conclude that $|g(z)| \le |z|^2$ for all $z \in \Bbb D$. (Hint: This is related to the Schwarz lemma.)
• Conclude that $|f(0)| \le 1/4$.
• Consider the function $T(z)^2$ to show that $1/4$ is not only an upper bound, but the supremum.

Answer 2

Consider the Blaschke factor $T(z)=(z+1/2)/(1+z/2)$. Let $f \in {\cal F}$ and set $g(z):=f(T(z))$. Then $g(\mathbb{D}) \subseteq \mathbb{D}$, $g(0)=f(1/2)=0$ and $g'(0)=f'(1/2)\cdot T'(0)=0$. Thus $g(z)=z^2h(z)$ for some $h \in H(\mathbb{D})$. By the maximum modulus principle $|h(z)| \le 1$ on $\mathbb{D}$. Thus $|f(0)|=|g(-1/2)| \le 1/4$. Setting $f_{max}(z)=(T^{-1}(z))^2$ we have $f_{max} \in {\cal F}$ and $|f_{max}(0)|=1/4$. Thus $\sup_{f \in \cal F} |f(0)|=1/4$ and the supremum is a maximum.