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I am trying to prove analyticity of a limit and I am at this situation.

I have a sequence of meromorphic functions $\{f_n\}$ and all of them have singularities at the same points. I have proved uniform convergence in an open disk around $0$, $U$. If I have pointwise convergence in an open set $V$, such that $U\subset V$, is this enough to get analyticity in $V$? If not what else do I need?

tst
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  • Don't quite understand some things: Do {$f_n$} have singularities of the same order at each of those points? And wouldn't f have a singularity at those points? So f could not be analytic at any of those points. Are any of them in U? – Betty Mock Nov 28 '13 at 05:37
  • I am not sure about the order yet, I still need to check that, but of course there are no singular points in $V$. – tst Nov 28 '13 at 17:45

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No, this is not enough even if all your functions are holomorphic.

A theorem of Osgood states that if $(f_n)$ is a sequence of holomorphic functions on a domain $\Omega$, that converges pointwise to some $f$, then there is a dense open subset $U\subset\Omega$ such that $f$ is holomorphic on $U$ and the convergence is lcally uniform on $U$. However, the set $\Omega\setminus U$ may be non-empty. See for example this question.

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mrf
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  • You are right, I checked the bibliography and the requirements are mild. If there exists an integrable function $g:\mathbb{C}\to\mathbb{R}_+$ such that $|f_n(z)|\le g(z)$, then the limit is analytic. – tst Nov 28 '13 at 17:54
  • @mrf glad to learn of this theorem of Osgood. – Betty Mock Nov 29 '13 at 18:51