Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $D \subset \mathbb{C}^{n}$ be an open domain and $K \subset D$ be a compact subset such that $D - K$ is simply connected. Let furthermore $f$ be a analytic function defined on $D-K$. Is it possible to extend $f$ to the whole $D$? If yes, by which theorem. If not, what further assumption do I need in order to make it possible ($D$ beeing a polydisk, some convergence theorems etc....)? Or doesn't it work at all?


share|cite|improve this question
I removed the tag functional-analysis, which did not fit the question. Welcome to Math.SE! – user53153 Jan 7 '13 at 14:53

It's enough to assume that $D\setminus K$ is connected. Look up Hartogs' extension theorem, which you can find in any textbook on several complex variables. Wikipedia's version

Of course, you need to assume that $n > 1$. If $n=1$, it's very easy to find counterexamples, e.g. $f(z) = 1/(z-p)$, where $p\in K$.)

share|cite|improve this answer
yes but my function $f$ is not holomorphic? is the theorem also true for non-holomorphic but analytic functions? – bill Jan 7 '13 at 6:32
@bill What do you mean by 'analytic'? With the usual meaning of the words, they are synonymous. – mrf Jan 7 '13 at 6:39
that it can be locally written in a power-series – bill Jan 7 '13 at 6:51
@bill Then a function is holomorphic if and only if it is analytic (this can also be found in any textbook on the subject) – mrf Jan 7 '13 at 6:56
@AD. Fine, but nobody uses $\mathbb{C}^n$ when they want to include $n=1$.... – mrf Jan 7 '13 at 7:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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