If $f \in \operatorname{Hol}(D)$, $f(\frac{1}{2}) + f(-\frac{1}{2}) = 0$, prove that $|f(0)| \leq \frac{1}{4}$ If $f \in \operatorname{Hol}(D),f(\frac{1}{2}) + f(-\frac{1}{2}) = 0$, prove that $|f(0)| \leq \frac{1}{4}$
$D = \{ z \in \mathbb{C} : |z| < 1 \} $
My thoughts so far: Let's say $f(0) = a$. Define $g = \frac{z-a}{1-\bar{a}z}$ and $h(z) = (g(f(z))$
Now all the conditions for the Shwarz lemma are met, and I can conclude that $|h(\frac{1}{2})| \leq  \frac{1}{2}$ and $|h(-\frac{1}{2})| \leq  \frac{1}{2}$. The idea would then be to multiply the two inequalities together and try to somehow separate $a$, but the algebra gets really messy and I feel like I'm doing something wrong. Any help would be appreciated!
 A: $(f(z) + f(-z))/2$ is an even function in $D$. It follows (see below) that there exists
a holomorphic function $g$ in $D$ such that
$$
  g(z^2) = \frac{f(z)+f(-z)}{2} \, .
$$
$g$ satisfies $|g(z)| < 1$ in $D$ and $g( \frac 14) = 0$. Then
$$
  h(z) = g \bigl(\frac{z + \frac 14}{1+ \frac 14 z} \bigr)
$$
satisfies $|h(z)| < 1$ in $D$ and $h(0) = 0$. 
It follows from Schwarz lemma that $|h(z)| \le |z|$ in $D$ and in particular
$$
 \frac 14 \ge |h(-\frac 14)| = |g(0)| = |f(0)| \, .
$$
The example
$$
  f(z) = \frac{z^2 - \frac 14}{1 - \frac 14 z^2} 
$$
with $f(\frac 12) = f (-\frac 12) = 0$ and $f(0) = -\frac 14$ shows that the bound
$|f(0)| \le \frac 14$ is best possible.

Existence of $g$: If $F$ is an even holomorphic function in the unit disk $D$
then its power series has only terms with even exponents:
$$
   F(z) = \sum_{n = 0}^\infty a_{2n} z^{2n} \, . \tag 1
$$
Now define $g$ as
$$
   g(z) = \sum_{n = 0}^\infty a_{2n} z^n \, . \tag 2
$$
It is easy to see that if $R$ is the radius of convergence of $(1)$
then $R^2$ is the radius of convergence of $(2)$. Therefore $g$ is holomorphic
in $D$ and satisfies $g(z^2) = F(z) \, $.
