Here's the problem I'm having trouble with:
Let $f: \mathbb{D} \to \mathbb{D}$ be a holomorphic function with $f(\frac{1}{2}) + f(-\frac{1}{2}) = 0$. Prove $|f(0)|\leq \frac{1}{4}$.
I suspect that I somehow need to use some variant of Schwarz's lemma.
Here's my attempt:
Define $h(z) = \frac{f(z) + f(-z)}{2}$. Then, $h(0) = f(0)$ and $h(-\frac{1}{2}) = h(\frac{1}{2}) = 0$. Also, $h:\mathbb{D} \to \mathbb{D}$ is holomorphic, and $h(0) = f(0)$. By Schwarz-Pick's lemma, we have $$|\frac{h(\frac{1}{2}) - h(0)}{1-\overline{h(0)}h(\frac{1}{2})}| \leq |\frac{\frac{1}{2}-0}{1-\overline{0}\cdot \frac{1}{2}}|.$$
This shows $|h(0)| = |f(0)| \leq \frac{1}{2}$, however, it doesn't show what I need. Is there a way to improve the inequality and get the desired result?
$h \colon \mathbb{D} \to \mathbb{D}$ is holomorphic and has zeros at $\frac{1}{2}$ and $-\frac{1}{2}$. The same is true of $$g \colon z \mapsto \frac{z-\frac{1}{2}}{1 - \frac{1}{2}z}\cdot \frac{z + \frac{1}{2}}{1 + \frac{1}{2}z} = \frac{4z^2-1}{4-z^2}\,.$$ Moreover, we have $\lvert g(z)\rvert = 1$ for $\lvert z\rvert = 1$, and $g$ has no other zeros than the simple zeros at $\pm \frac{1}{2}$. It follows that $q \colon \mathbb{D} \to \overline{\mathbb{D}}$ is holomorphic, where $$q \colon z \mapsto \frac{h(z)}{g(z)}\,.$$ In particular, we have
$$1 \geqslant \lvert q(0)\rvert = \biggl\lvert \frac{h(0)}{g(0)}\biggr\rvert = 4\lvert f(0)\rvert\,.$$