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Given that $f: \mathbb{C} \setminus \{0\} \rightarrow \mathbb{C}$ is a bounded holomorphic function. Prove that $f$ is constant.

There is a probably more generalized version of the problem here Bounded holomorphic functions on $\mathbb{C} \smallsetminus K$ are constant., but I want something simpler for this particular question. It mentions "If $K$ is finite, all its points are isolated, and since $f$ is bounded, removable singularities. Removing the singularities, we obtain a bounded entire function, which is constant by Liouville's theorem." I would like to make it more rigorous. Any help is appreciated.

The Limit Does Not Exist
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    Riemann’s removable singularity theorem + Liouville. Or if you know Morera’s theorem, then you can use that as well to first deduce $f$ is entire, and then use Liouvile. – peek-a-boo Mar 11 '23 at 04:13

2 Answers2

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This theorem justifies why $0$ is a removable singularity, i.e. there exists a holomorphic $g : \mathbb{C} \to \mathbb{C}$ that coincides with $f$ on $\mathbb{C} \setminus \{0\}$. Then apply Liouville's theorem.

angryavian
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$f: \mathbb{C} \setminus \{0\} \rightarrow \mathbb{C}$ is bounded holomorphic map.

$f$ is bounded in punctured disk around $0$, hence $0$ is removable singularity of $f$(by Riemann's theorem on removable singularity). Hence $f$ has a unique holomorphic extension to $\mathbb{C}$ say $F$.

$F$ is a bounded entire function, hence constant (by Liouville's theorem) and $F\mid_{\mathbb{C}\setminus \{0\}}=f=c$ For some $c\in\Bbb{C}$

Sourav Ghosh
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