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Given holomorphic non-constant function $f:D(0,1) \smallsetminus \{0\} \rightarrow \mathbb{C}$ so $\forall n=2,3,...:\ |f(\frac{1}{n})| \leq \frac{1}{n!}$ I need do show that $0$ is an essential singularity.

I tried expanding $f$ to it's Laurent series near $z_0 = 0$ and compute coefficients by:

$$a_{-k} = \frac{1}{2 \pi i} \int_{|z|=r} z^{n-1}f(z)dz$$ and take $r= \frac{1}{n}$ so $$a_{-k} = \frac{1}{2 \pi i} \int_{|z|=\frac{1}{n}} z^{n-1}f(z)dz $$

But I don't know how to if it's a fertile direction nor how to proceed if it is.

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2 Answers 2

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Suppose the singularity is removable. Then $f(0) = \lim_{n\to\infty} f(1/n) = 0$, so we may write $f(z) = z^k g(z)$ for some $k\in \mathbb{N}$ and $g$ holomorphic with $g(0) \neq 0$. In a small enough neighborhood around $0$, we may assume $|g| > C > 0$. Thus, in this neighborhood, we have $|f(z)| > C |z|^k$. For large enough $n$, it follows that $Cn^{-k} < 1/n!$, contradiction.

Similarly: suppose the singularity is a pole. Write $f(z) = g(z)/z^k$ where $g(0)\neq 0$. Then in a small enough neighborhood of $0$, we have $|f(z)| > C|z|^{-k}$ so for large enough $n$, it must be true that $1/n! > Cn^k$, contradiction.

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  • $\begingroup$ Why can you assume $|g|>C>0$ in the first case here? Is $\frac{1}{z^kg(z)} $bounded? $\endgroup$
    – Thesinus
    Sep 11, 2019 at 13:23
  • $\begingroup$ Similarly why can you state $|f(z)|>C|z|^{-k}$ in the second case? $\endgroup$
    – Thesinus
    Sep 11, 2019 at 13:39
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    $\begingroup$ Since $g(0)\neq 0$, by continuity of $g$ and local compactness of $\mathbb{C}$, we can find a neighborhood of $0$ on which $g$ is bounded away from $0$. But $1/f$ won't be bounded -- since $f(0)=0$. $\endgroup$ Sep 11, 2019 at 13:41
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    $\begingroup$ In the second case, we're again just using the fact that $g$ is bounded away from $0$, on account of $g(0)\neq 0$. $\endgroup$ Sep 11, 2019 at 13:43
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Hint:

Does $$\lim_{z\to0}f(z)=L$$ exist? (For finite or infinite $L$)

Suppose it does, and because $1/n\to0$ then $|L|\leq0\to L=0$. Now, you must check that this value is possible.

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  • $\begingroup$ Just to see if we are on the same page... At some point I'll have to mention Joseph Liouville? $\endgroup$
    – Uria Mor
    May 14, 2016 at 13:09

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