Let f be analytic on ∆ The problem is: let $f$ be an analytic function on $\Delta$ and satisfy $|f|<1$. Prove that if $f(1/2)=f(−1/2)=0$, then $|f'(0)|\le 1/4$.
I tried to expand $f$ at $0$ and then plug in $1/2$ and $-1/2$ to evaluate the bound. It is quite straight forward if I use $f^{(n)}(0)/n! < 1$. But I got the bound to be $1/3$. Is there some key I am missing here?
 A: (One can proceed similarly as in If $f \in \operatorname{Hol}(D)$, $f(\frac{1}{2}) + f(-\frac{1}{2}) = 0$, prove that $|f(0)| \leq \frac{1}{4}$.)
It is actually sufficient to require that $f(\frac 12)=f(−\frac 12)$
instead of $f(\frac 12)=f(−\frac 12)=0$.
$(f(z) - f(-z))/2$ is an odd function in $\mathbb D$ (the unit disk). It follows  that there exists
a holomorphic function $g$ in $\mathbb D$ such that
$$
  z \, g(z^2) = \frac{f(z)-f(-z)}{2} \, . \tag 1
$$
Taking the derivates gives
$$
  g(z^2) + 2 z^2 g'(z^2) = \frac{f'(z)+f'(-z)}{2}
$$
and for $z= 0$ it follows that
$$
  g(0) = f'(0) \, .   
$$
$f(\frac 12)=f(−\frac 12)$ implies $g( \frac 14) = 0$, and 
 $|g(z)| < 1$ in $\mathbb D$ follows from the Schwarz lemma applied to
the right-hand side of $(1)$.
Then
$$
  h(z) = g \bigl(\frac{z + \frac 14}{1+ \frac 14 z} \bigr)
$$
satisfies $|h(z)| < 1$ in $\mathbb D$ and $h(0) = 0$. 
It follows from Schwarz lemma that $|h(z)| \le |z|$ in $\mathbb D$ and in particular
$$
 \frac 14 \ge |h(-\frac 14)| = |g(0)| = |f'(0)| \, .
$$
The example
$$
  f(z) = z \, \frac{z^2 - \frac 14}{1 - \frac 14 z^2} 
$$
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.
