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Let $f$ be an analytic function in the whole complex plane. If there exists $M>0$ such that $|f(z)|>M$ for all $z \in \Bbb Z$, show that $f$ is constant in the whole complex plane.

I tried to approach with contradiction supposing that $f$ is not constant. If so then there exists $z_1$ and $z_0$ such that $f(z_1) \ne f(z_0)$. Then $f(z_1)-f(z_0)\ne 0$ and $|f(z_1)-f(z_0)| \ne 0$. I then tried to use the triange inequality trick to break $|f(z_1)-f(z_0)| \ne 0$ apart as $$|(f(z_1)-f(z'))+(f(z')-f(z_0))| \le |f(z_1)-f(z')|+|f(z')-f(z_0)|$$

but I don't think I'm going anywhere with this approach. I think this resemble the Liouville theorem a bit, but it's like the converse of it?

Werner
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    What can you say about ${1 \over f}$? – copper.hat May 08 '22 at 20:43
  • Apply Liouville's theorem to $\frac{1}{f}$, which is entire and bounded above by $\frac{1}{M}$. – Reveillark May 08 '22 at 20:44
  • Should the condition be $|f(z)| > M$ for all $z\in \color{red}{\mathbb{C}}$ instead of just $z\in \mathbb{Z}$? Otherwise, the statement is false. For if $f(z) = \sin((2z + 1)\pi/2)$ and $M = 1/2$, then $f$ is analytic in $\mathbb{C}$ such that for all $z\in \mathbb{Z}$, $|f(z)| = 1 > M$. However, $f$ is non-constant. – kobe May 08 '22 at 20:51

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