TITLE: Bound the derivative at zero of an analytic function with $f(0.5) = 0$
QUESTION [3 upvotes]: The following is a problem from the UW-Madison complex analysis qualifying exam.
Let $f:\mathbb{D}\to\mathbb{D}$ be analytic, where $\mathbb{D}=\{z : |z| < 1\}$. Assume that $f(\frac{1}{2}) = 0$. Show that $|f'(0)| \leq \frac{25}{32}$.
Using the Schwartz-Pick (S-P) lemma, we can say $$|f(0)|= \left|\frac{f(1/2)-f(0)}{1-\overline{f(1/2)}f(0)}\right| \leq \left|\frac{1/2-0}{1-\overline{\frac{1}{2}}0}\right| = 1/2.$$
Not sure if this will help.
There is a first part to this qual problem: Show that $|f'(0)| \leq 1-|f(0)|^2.$ It is a simple consequence of S-P lemma for $z=0$.

REPLY [2 votes]: We consider $B(z)=\frac{2z-1}{2-z}$ the disc automorphism st $B(1/2)=0, B(0)=-1/2$; then $B'(z)=\frac{3}{(2-z)^2}$ so $B'(0)=3/4$
Let $f(0)=a, f'(0)=b$ and note that $g=f/B$ is an analytic disc self map (analytic because $f(1/2)=0$ and $B$ has only one zero and disc self map because $|B|=1$ on the boundary) so applying schwarz pick one has $|g'(0)| \le 1-|g(0)|^2$ which means $$|-b/2-3a/4| \le 1/4-|a|^2$$
Now this implies $$3|a| \ge 2|b|-1+4|a|^2$$
But $3|a| \le 4|a|^2 + 9/16$ so $2|b|-1 \le 9/16$ and we are done!
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