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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: $dim_{\mathbb{C}}(M)\geq 2$.

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2 Answers 2

up vote 5 down vote accepted

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.

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Good point. I forgot to add that $\Omega$ should be a \textit{bounded} pseudoconvex domain. –  oydeis Mar 18 '13 at 19:57
    
And by pseudoconvex I mean strictly pseudoconvex. –  oydeis Mar 18 '13 at 20:16
    
How about $\Omega=B(0,1)\subset \mathbb C^2$ and $u(z,w)=Re(e^{1/(z-1)})$? (cf remark of Alexandre) –  Henri Mar 19 '13 at 0:40
    
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? –  oydeis Mar 19 '13 at 10:01
    
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. –  Henri Mar 19 '13 at 14:00

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).

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Thanks for your reply. I'm not sure I understand the counterexample correctly. Isn't $u$ just plurisubharmonic on $\mathbb{C}^2\setminus \Omega$? –  oydeis Mar 21 '13 at 22:58
    
My previous example was confusing, so best thing is to forget it. Anyway, @Henri sketched an answer. –  Margaret Friedland Mar 21 '13 at 23:06
    
Thank you for the interesting references. –  oydeis Apr 1 '13 at 17:36

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