Let $\mathbb{D}=\{z\in\mathbb{C}: |z|<1\}$ and suppose that $f:\mathbb{D}\to\mathbb{D}$ is a holomorphic function and that $f(\frac{1}{2})+f(-\frac{1}{2})=0$.
Prove that $f(0)\leq\frac{1}{4}$.
We can define the functions
$$\phi(z)=\frac{2z-1}{2-z},\ g(z)=\frac{f(z)+f(-z)}{2\phi(z)}$$
It is not so hard to obtain that $g(z)$ has a removable singularity at $\frac{1}{2}$ and that $\forall z\in\mathbb{D}:|g(z)|\leq 1$.
Finally, $|g(0)|\leq 1$ and so $|f(0)|\leq \frac{1}{2}$.
Can we get to $\frac{1}{4}$?
Edit:
Maybe we can use instead
$$\phi(z)=\frac{4z-1}{4-z},\ g(z)=\frac{f(2z)+f(-2z)}{2\phi(z)}$$