3

Suppose $V$ is a analytic variety of an open subset $U\subset ℂ^n$. Suppose that $f:U\setminus V\rightarrow C$ is holomorphic and that $f$ is $L^2$-bounded in $U$. Question: Is it true that there exists a unique holomorphic extension of $f$ to all of $U$?

Start wearing purple
  • 53,234
  • 13
  • 164
  • 223
user81500
  • 31

2 Answers2

1

Every analytic variety is pluripolar, according to page 2 here. Pluripolar sets are removable for $L^2$ holomorphic functions, according to page 20 here. (These are not the best references, but they are what I could find without going through paywalls.)

ˈjuː.zɚ79365
  • 3,004
  • Thank you very much for your help! It seems to me that it is possible to give a more direct proof of it using Laurent series as in the proof of (the classical) Riemann removable singularities theorem for bounded holomorphic functions. Any comments is welcome! – user81500 Jun 09 '13 at 08:27
  • @user81500 For $n=1$, of course: see this answer. I don't see how to extend this to higher dimensions. The idea of the Laurent series argument has more to do with Hartogs extension: it relies on being able to surround a singularity with a region on which $f$ is holomorphic. But Hartogs extension does not apply in your case. – ˈjuː.zɚ79365 Jun 09 '13 at 08:57
  • I was thinking to follow the book of Grauert-Remmert "Coherent analytic sheaves" at page 132 it takes the Laurent expansion – user81500 Jun 09 '13 at 10:55
  • Thanks for your answer. I was thinking to follow the book of Grauert-Remmert "Coherent analytic sheaves" at page 132 it takes the Laurent expansion $f(w)=\sum a_k(w^{'})w_1^k$, with $a_k(w^{'})$ holomorphic in $w^{'}=(w_2, \dots , w_n)$ and then to show that the L^2-boundness of $f$ implies that the coefficients $a_k(w^{'})=0$ for $k<0$. Can you please tell me where I need Hartogs extension? Thanks..... – user81500 Jun 09 '13 at 11:02
  • @user81500 I did not mean to confuse you by mentioning Hartogs. Forget about it. Since you have a book with a proof and I don't, there's nothing else I can add here. – ˈjuː.zɚ79365 Jun 09 '13 at 11:12
1

See Ohsawa's book Analysis of Several Complex Variables, Theorem 1.13 and Proposition 1.14. The proof relies on the fact that a $L^2$ function is holomorphic if and only if $\bar{\partial} f$ is $0$ in the sense of distribution. So we need to verify $\int f\bar{\partial}_i u=0$ for any smooth function $u$ with compact support. We can use a cut-off function supported in a $\varepsilon$-neighborhood of $V$ to separate the integration into two parts. Taking $\varepsilon\to 0$ will give us the desired identity. One property used when taking limits is that the volume growth of $\varepsilon$-neighborhood is of order $\varepsilon^2$. The conclusion can be applied to any subset with this property.
Also there is a proof following the discussion for $n=1$. We only need to deal with the case $V=\{h=0\}$ with $h$ holomorphic. Near the nonsingular points, a holomorphic transformation with $h=0$ can reduce the problem to one-dimensional case. So we have extended $f$ to the smooth part of $V$. Notice that the singular part has codimension at least $2$. A final use of Hartog-type extension theorem would give us the extension of $f$ to $U$.

user172654
  • 11