Suppose $M$ is a complex manifold and $\Omega$ a (edit: bounded) pseudoconvex domain in $M$. Let $u:M\setminus\Omega\to\mathbb{R}$ be a pluriharmonic function. Is it true that $u$ has a pluriharmonic extension to $M$? edit: $dim_{\mathbb{C}}(M)\geq 2$.
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
11
Your condition $\dim M>2$ does not save the situation: you can have many counterexamples with $M=M'\times C^n$ where $\dim M'=1$ and your functions are independent of the second variable. And in dimension $1$ you certainly have plenty of pluriharmonic (=harmonic) functions which do not extend anywhere.
answered Mar 18, 2013 at 17:26
-
$\begingroup$ Good point. I forgot to add that $\Omega$ should be a \textit{bounded} pseudoconvex domain. $\endgroup$– oydeisCommented Mar 18, 2013 at 19:57
-
$\begingroup$ And by pseudoconvex I mean strictly pseudoconvex. $\endgroup$– oydeisCommented Mar 18, 2013 at 20:16
-
$\begingroup$ How about $\Omega=B(0,1)\subset \mathbb C^2$ and $u(z,w)=Re(e^{1/(z-1)})$? (cf remark of Alexandre) $\endgroup$– HenriCommented Mar 19, 2013 at 0:40
-
$\begingroup$ Yes, if $B(0,1)=\{z\in\mathbb{C}:|z|\leq 1\}\times \mathbb{C}$ then I agree. But I don't think $u$ is pluriharmonic on $M\setminus B(0,1)$ if $B(0,1)$ is the unit ball in $\mathbb{C}^2$. Or am I wrong? $\endgroup$– oydeisCommented Mar 19, 2013 at 10:01
-
$\begingroup$ Yes; if you want consider $\Omega = B(0, 1/2)$; by uniqueness of the pluriharmonic extension (ph functions are real analytic), $u$ won't extend to $\mathbb C^2$ either. $\endgroup$– HenriCommented Mar 19, 2013 at 14:00
$\begingroup$
$\endgroup$
2
In contrast, for plurisubharmonic functions and their subextensions the situation is more difficult: In: Fornæss, John Erik; Sibony, Nessim: Plurisubharmonic functions on ring domains. Complex analysis (University Park, Pa., 1986), 111–120, Lecture Notes in Math., 1268, Springer, Berlin, 1987 (MR0907057), the authors consider plurisubharmonic (psh) function on $B\setminus K$, where $B$ is a ball in $\mathbb{C}^n$ and $K$ is a polynomially convex set. They give the following counterexamples: one is a smooth psh function on $B\setminus K$, $K$ a polydisk, which does not have a psh subextension to $B$; another one is a discontinuous psh function on $B\setminus K$, $K$ a smaller ball, which does not have a psh subextension to $B$. There is also earlier work about psh functions that cannot be extended across a pseudoconvex set: Bedford, Eric; Burns, Dan Domains of existence for plurisubharmonic functions. Math. Ann. 238 (1978), no. 1, 67–69 (MR0510308).
-
$\begingroup$ Thanks for your reply. I'm not sure I understand the counterexample correctly. Isn't $u$ just plurisubharmonic on $\mathbb{C}^2\setminus \Omega$? $\endgroup$– oydeisCommented Mar 21, 2013 at 22:58
-
$\begingroup$ My previous example was confusing, so best thing is to forget it. Anyway, @Henri sketched an answer. $\endgroup$ Commented Mar 21, 2013 at 23:06