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Let $f$ be a holomorphic function in $D(0,1)$ and $a\in D(0,1)$ such that $|f(z)|<1$ for all $z\in D(0,1)$, and $f(a)=f(-a)=0$. Show that $|f(0)|\le|a|^2$. What can we conclude if this holds with equality?

I was thinking of using the automorphism $(a-z)/(1- \bar{a}z)$ to show that $|f(0)|\le a\bar{a} = |a|^2$.

A.Γ.
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  • Use this result and set $z=0$. – A.Γ. Nov 02 '15 at 10:45
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    You can even relax the condition $f(a) = f(-a) = 0$ to $f(a) + f(-a) = 0$. The case $a=1/2$ is treated here http://math.stackexchange.com/questions/1177230/if-f-in-operatornamehold-f-frac12-f-frac12-0-prove, it can easily be generalised to arbitrary $a \in \mathbb D$. – Martin R Nov 02 '15 at 12:12

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As the comments indicate, one quick solution is obtained by introducing $$ g(z) = \frac{1- \bar az }{z-a } \frac{1+ \bar az }{z+a } \, f(z) $$ where the factors in front of $a$ are unimodular on the boundary. Since $g$ is holomorphic in the disk ($\pm a$ are removable singularities), It follows that $|g|\le 1$ in the disk, hence $|g(0)|\le 1$ which is the desired conclusion.