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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?

bill

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I removed the tag functional-analysis, which did not fit the question. Welcome to Math.SE! –  user53153 Jan 7 '13 at 14:53
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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 $p\in K$.)

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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
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@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
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@AD. Fine, but nobody uses $\mathbb{C}^n$ when they want to include $n=1$.... –  mrf Jan 7 '13 at 7:01
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