0
$\begingroup$

Let $f$ be a holomorphic, non-constant function on $D=\{0<|z|<10\}$. It is given that for every $n \in \mathbb{N}$, $|f(\frac{1}{n})| \le \frac{1}{n!}$.

Show that $f$ has an essential singularity at 0.

I have no idea how to even approach this. Why is it not possible for $f$ to have a pole at 0?

Note. In the original question it was written $n \in \mathbb{C}$ but I considered this a typo.

Thanks in advance!

  • 0
    See here http://math.stackexchange.com/questions/2104353/holomorphic-function-on-punctured-disk-satisfying-leftf-left-frac1n-righ – 2017-01-23
  • 0
    well, the explanation in that answer is really bad. he gives an example, but that does not explain the rule. – 2017-01-24

0 Answers 0