# Let $f$ be holomorphic with zeros at $\frac{1}{2}$ and $\frac{-1}{2}$ such that $|f(z)| < 1$ on $D$. Show $|f(i/2)| \leq \frac{8}{17}$.

Let $f$ be holomorphic on the unit disc $D$ with zeros at $\frac{1}{2}$ and $-\frac{1}{2}$ such that $|f(z)| < 1$ on $D$. Show $|f(i/2)| \leq \frac{8}{17}$.

I apply a Schwarz Lemma -esque argument.

Define $g(z)$ to be $\frac{f(z)}{(z-\frac{1}{2})(z+\frac{1}{2})}$ on $D\setminus\{\pm \frac{1}{2}\}$, and $\pm4f'(\pm\frac{1}{2})$ at $z = \pm\frac{1}{2}$. So $g$ is holomorphic. Now, we can apply maximum modulus principle on any smaller disc $D_r$ with radius $\frac{1}{2}<r<1$. This gives, for $z \in D_r$, $$|g(z)| \leq \sup_{|z|=r} |g(z)| \leq \sup_{|z|=r} \frac{|f(z)|}{|z^2-1/4|} \leq \frac{1}{r^2-1/4}.$$

The last step, I use the assumption on $f$, and the triangle inequality on the denominator: $|z^2-1/4| \geq |z^2| - 1/4 \ ( = r^2 - 1/4 \text{ on bdry of }D_r).$ From here, taking $r \to 1$ we get $|g(z)|\leq 4/3$. Thus $$|f(z)| = |g(z)||z^2-1/4| \implies |f(i/2)| \leq \frac{2}{3}.$$

This is the best bound I can get, but it's not good enough :(. Any tips appreciated! I'd prefer to keep a Schwarz Lemma type argument (as that was a hint given to me).

Try using Blaschke products. $$B_1(z)=\frac{z-1/2}{1-z/2}$$ has a zero at $1/2$ and has $|B_1(z)|=1$ on the unit circle. Also $$B_2(z)=\frac{z+1/2}{1+z/2}$$ has a zero at $-1/2$ and has $|B_2(z)|=1$ on the unit circle. Then $$g(z)=\frac{f(z)}{B_1(z)B_2(z)}$$ is bounded in absolute value by $1$ on the unit circle and so inside. Then $$|f(i/2)|=|g(i/2)||B_1(i/2)||B_2(i/2)| =|g(i/2)|\left|\frac{i^2/4-1/4}{1-i^2/16}\right|=\frac{8}{17}|g(i/2)| \le\frac8{17}.$$