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Find all holomorphic functions $f$ on $\mathbb{D}$ s.t the following condition holds for all $n>1$ $$\dfrac{1}{\sqrt{n}}<\left\vert f\left(\dfrac{1}{n}\right)\right\vert <\dfrac{2}{\sqrt{n}}.$$

My attempt:
From hypothesis, we have $f(0)=0$ and $\left\vert nf\left(\dfrac{1}{n}\right)\right\vert\geq \sqrt{n}\to\infty$ as $n\to\infty$. I'm trying to show that $\left\vert\dfrac{f\left(z\right)}{z}\right\vert$ is bounded in the neighbourhood of $z=0$ using $f(0)=0$. If that's true, then there is no such holomorphic function satisfying.
But I got stuck here. Could someone help me or have another way to deal with problem? Thanks in advance!

Alex Nguyen
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1 Answers1

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Let $$ f(z)=\sum_{k=0}^\infty a_kz^k $$ be the power series expansion of $f$. From $|f(1/n)| < 2/\sqrt{n}$ $(n>1)$ we get $f(0)=0$, hence $a_0=0$. Thus $$ f(z)= z \sum_{k=1}^\infty a_kz^{k-1} =: zg(z) $$ where $g$ is a holomorphic function on $\mathbb{D}$. Now $$ |g(1/n)| = |nf(1/n)| > \sqrt{n} \to \infty \quad (n \to \infty). $$ On the other hand $g(1/n) \to g(0)=a_1$ $(n \to \infty)$, a contradiction.

Gerd
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