# Liouville's Theorem. Example when the theorem does not work if $f$ is not entire function.

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?

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

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.