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I have to prove that every bounded function holomorphic on $\mathbb{C}^2 \setminus K$ is constant, where $K$ is
$(a)$ a ball
$(b)$ a complex line
$(c)$ an arbitrary analytic subset

Now, I think the idea here is to show that the holomorphic function can be extended to all of $\mathbb{C}^2$ and thereafter using Liouville's theorem to show the bounded function as constant, but I am kind of lost on how to proceed to do that. Can I get some help?

Naweed G. Seldon
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  • Try and assume it isn't bounded, and then look at the coefficients (using the fact it is analytic, it has a Taylor series) and use Liouville's theorem to show they must all be $0$ (except for $a_0$). – NL1992 Apr 23 '19 at 12:55
  • I don't think I get the strategy completely, I'm not sure how I can sure that the coefficients are $0$. Can you show me an example of this calculation? – Naweed G. Seldon Apr 23 '19 at 12:58
  • For a ball centered at $z_0$, think of $\frac1{z-z_0}$. – robjohn Apr 23 '19 at 14:24

1 Answers1

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(i) If $K$ is a ball, we may use Hartogs' extension theorem.

(ii) Note a complex line is also an analytic subset. We may just consider the case $K$ is an analytic subset. But then $\mathbb{C}^2\setminus K$ is connected, and every locally bounded holomorphic functions on $\mathbb{C}^2\setminus K$ can be extended to $\mathbb{C}^2.$ See Demailly, complex analytic and differential geometry, P91 Remark 4.2.

qinxs
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