Analytic functions in a punctured disk

Question

Let $f$ be continuous in the open disk $\{z:|z|<1\}$. If $f$ is analytic in $\{z:0<|z|<1\}$, ie: analytic in the punctured unit disk, then $f$ is analytic $\{z:|z|<1\}$.

How does one relate the continuity of the function to the analyticness of the function at the removed point?

I guess the Cauchy integral formula is not allowed? – Zardo Jun 10 '15 at 14:25 — Zardo, Jun 10 '15 at 14:25

score 6 · Accepted Answer · answered Jun 10 '15 at 14:27

6

Isolated singularities of analytic functions are of three types:

Removable singularities.
Poles.
Essential singularities.

In cases 2 and 3 the function is not bounded in any punctured neighbourhood of the singularity. Since $f$ is continuous, the only possible case is 1.

answered Jun 10 '15 at 14:27

user1337

24,381

Ah, that makes sense. Thank you! – user247211 Jun 10 '15 at 14:32

Analytic functions in a punctured disk

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