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I'm struggling with understanding Liouville's theorem, more precisely when does the theorem not hold. I'm looking for a non-holomorphic function to show that theorem doesn't work.

Liouville’s Theorem: Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be an bounded entire function. Then $f$ is constant.

My first guess is, I should take function $f(z)=\frac{1}{z}$ on set $|z|>1$, which is ofcourse bounded but not holomorphic in $z=0$.

I don't know how to move futher with the example or should I find another example of non-holomorphic function?

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    $\begingroup$ So you want a bounded but not holomorphic function that is not a constant? $\endgroup$
    – Randall
    Commented Feb 14, 2021 at 15:50
  • $\begingroup$ I'm into why entire function assumption is essential. $\endgroup$
    – Michal
    Commented Feb 14, 2021 at 15:57
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    $\begingroup$ The just let $f$ be $+1$ somewhere and $-1$ somewhere else. Bounded, not constant, and not entire because it's not even continuous. $\endgroup$
    – Randall
    Commented Feb 14, 2021 at 16:03

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What about $f(z)=\cos(\operatorname{Re}z)+\sin(\operatorname{Im}z)i$? It is bounded, non-holomorphic and non-constant. And, as an extra, it is continuous.

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