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Let $\Bbb D= \{z : |z|<1\}$ be the unit disc. Let $f:\Bbb D \to \Bbb D$ be a holomorphic function such that $|f(\frac{1}{2})|+|f(-\frac{1}{2})|=0$. Prove: $|f(0)| \leq \frac{1}{4}$.

The only thing I thought about was using Schwarz lemma, but I'm not sure how. I thought about defining $g(z)=\frac{f(z+\frac{1}{2})+f(z-\frac{1}{2})}{2}$, but now $g$ is not defined in all $\Bbb D$, and I'm pretty much stuck.

Did
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user401516
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  • Note that the weaker condition $f(\frac{1}{2}) + f(-\frac{1}{2}) = 0$ is sufficient to get the same result. – Martin R Feb 02 '18 at 08:26

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In the question's current form (i.e. there are zeroes at $\frac{1}{2}$ and $\frac{-1}{2}$), the claim immediately follows from Jensen's formula (applied with $r$ arbitrarily close to $1$):

$\log |f(0)| = \frac{1}{2\pi} \int_0^{2\pi} \log |f(re^{i\theta})|d\theta + \sum \log(\frac{|a_k|}{r})$ where $r$ is s.t. $f$ is analytic in an open set containing $|z| \le r$, and the sum is over all zeroes of $f$ in the open disk $|z| < r$.

mathworker21
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  • Unfortunately I didn't learn this formula, can you think of another way to solve it? – user401516 Feb 02 '18 at 08:15
  • @user401516 Luckily, the proof of Jensen is quite simple. So you can solve your problem merely by mimicing the easy proof of Jensen. – mathworker21 Feb 02 '18 at 08:17
  • @user401516 I should add though, that Jensen applies only if $f(0) \not = 0$. But of course we may assume this, for otherwise we would be immediately done. – mathworker21 Feb 02 '18 at 08:18