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Let $\mathcal{F}$ be a family of holomorphic functions on the unit disk with $|f(z)|\leq 1, z\in \mathbb{D}$ and $f(1/4)=f(1/5)=0$. We need to find $\sup\{|f(0)|:f\in \mathcal{F}\}$.

I used the two automorphisms of the unit disk with zeros at $1/4$ and $1/5$. Their product, say $f$, is a function that belongs to this family, with $f = e^{i\alpha} \frac{4z-1}{z-4} \frac{5z-1}{z-5} $, $|f(0)| = 1/20$. I suspect this is the answer, but I do not know how to prove it. Any help would be appreciated.

Attila1177298
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  • You are on the right track. For the general case, consider $g(z) = f(z)/(\frac{4z-1}{z-4} \frac{5z-1}{z-5})$ and show that $g$ is holomorphic in $\Bbb D$ with $|g(z)| \le 1$. – Martin R May 17 '21 at 14:29
  • @MartinR $g$ could have poles at $1/4$ and $1/5$ no? – Attila1177298 May 17 '21 at 14:39
  • No, that are removable singularities because $f(z)=0$ at the zeros of the denominator. – See https://math.stackexchange.com/q/94122 for a more general result. – Martin R May 17 '21 at 14:41

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