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Let $D$ be a non-empty connected open subset of $\mathbb{C}^n$. Let $f$ be a complex valued continuous function on $D$. Let $Z$ be the set of zeros of $f$. Suppose $f$ is holomoprphic on $D - Z$. Is $f$ holomorphic on $D$?

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2 Answers

I think the answer to your question is yes. There's probably a better answer than this, but I think the following argument should work.

For each $M\in \mathbb{R}$, let $\varphi_M\colon D\to \mathbb{R}$ be the function $\varphi_M(z) = \max(M, \log|f(z)|)$. This function is plurisubharmonic on $D$, and $\lim_{M\to -\infty} \varphi_M(z) = \log |f(z)|$ for each $z\in D$. Since decreasing limits of plurisubharmonic functions are plurisubharmonic, we conclude the function $\log|f(z)|$ is plurisubharmonic on $D$. In particular, $Z$ is a pluripolar set. One can then use the following result to conclude that $f$ is holomorphic in $D$.

Let $D\subset\mathbb{C}^n$ be open, and let $Z$ be a closed pluripolar subset of $D$. Suppose $f$ is holomorphic on $D\smallsetminus Z$. If $f$ is locally bounded at every point in $Z$, then $f$ extends to a holomorphic function on $D$.

For a reference, see Corollary 5 of Chapter Q of Volume 1 of Introduction to holomorphic functions of several variables by R.C. Gunning.

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Thanks. Unfortunately I don't have an easy access to the Gunning's book. –  Makoto Kato Oct 19 '12 at 6:35
@MakotoKato: This is also proved in Theorems 5.23 and 5.24 of Chapter 1 of the (free) book Complex Analytic and Differential Geometry by J.-P. Demailly. –  froggie Oct 19 '12 at 11:01
froggie, thanks. I'll see it. –  Makoto Kato Oct 19 '12 at 11:53
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This is a true and it is a Theorem due to Rado. There are some very nice proofs of Rado's Theorem in the case $n=1$, see e.g. Rudin's Real and Complex analysis 3rd edition Theorem 12.14.

For the general case, one can use pluripotential theory as in froggie's answer, but I believe the problem can be reduced to the one-dimensional case.

Unfortunately Rado's Theorem seems to be hard to find in textbooks. I suggest you google "Rado's Theorem several complex variables"; it gives some nice references.

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Do you use Hartogs' theorem to reduce it to the one-dimensional case? en.wikipedia.org/wiki/Hartogs'_theorem –  Makoto Kato Oct 19 '12 at 19:43
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