0
$\begingroup$

I understand that if $f$ is holomorphic on $\Omega$ except possibly on $z_0$ and is bounded on $\Omega - \{z_0\}$ then $z_0$ is a removable singularity. But if we know that if $z_0$ is a removable singularity of $f$ does this imply that there is a neighborhood of $z_0$ on which $f$ is bounded? Can $f$ be holomorphic at a nearby point for which it tends to infinity?

$\endgroup$
1
$\begingroup$

Since the singularity is removable there is a neighborhood $U$ of $z_0$ and a holomorphic function $g$ on $U$ that coincides with $f$ on $U \setminus\{z_0\}$.

The function $g$ is in particular continuous. You can now, for instance, take a closed disk $D$ around $z_0$ with $D \subset U$ and note that $g$ is bounded on $D$ (a continuous function on a compact set is bounded) and thus $f$ is bounded on $D \setminus \{z_0\}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.