Timeline for Extension of pluriharmonic functions
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 22, 2013 at 9:53 | comment | added | oydeis | Well, thanks again for you answer. What I was missing is the fact that $b_1=0$ implies that any pluriharmonic function is the real part of a holomorphic function. But now it's clear. | |
| Mar 20, 2013 at 15:03 | vote | accept | oydeis | ||
| Mar 20, 2013 at 13:37 | comment | added | Henri | Yes you are right of course; I just had in mind the case where $M$ is a domain in $\mathbb C^n$. My point was just that the assumptions on $\Omega$ should not be too restrictive like pseudoconvexity e.g.; but on $M surely! | |
| Mar 20, 2013 at 8:19 | comment | added | oydeis | I'm pretty sure one has to make an additional assumption about $\bar{\Omega}$ if $M\neq \mathbb{C}^n$ (or not Stein etc.). For example, if $M$ is compact and $\bar{\Omega}$ is the complement of some small (connected) open set in $M$ then there are plenty of non-constant holomorphic functions on $M\setminus \bar{\Omega}$ that do not extend to $M$. | |
| Mar 20, 2013 at 0:47 | comment | added | Henri | No, not even. You just need $\Omega$ relatively compact, and $M\setminus \overline{\Omega}$ connected. (btw, you should have defined $u$ on $M\setminus \overline{\Omega}$ only if you want to make sense of pluriharmonicity) | |
| Mar 19, 2013 at 15:50 | comment | added | oydeis | Thanks! And to apply Hartog's extension theorem I need to know that $\Omega$ is pseudoconvex, right? | |
| Mar 19, 2013 at 14:24 | comment | added | Henri | Oh, I just realized that I misunderstood your question; I thought that $u$ was only defined on $\Omega$. In fact, if $M\setminus \Omega$ has $b_1=0$, then every pluriharmonic function on $M\setminus \Omega$ is the real part of a holomorphic function; so $u=Re(f)$ for $f \in \mathcal O(M\setminus \Omega)$. Now, if $M\setminus \Omega$ is connected, Hartog's extension theorem tells you that $f$ extends (uniquely) to $M$, and then $Re(f)$ extends $u$ has a pluriharmonic function. | |
| Mar 19, 2013 at 14:00 | comment | added | Henri | 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. | |
| Mar 19, 2013 at 10:01 | comment | added | oydeis | 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? | |
| Mar 19, 2013 at 0:40 | comment | added | Henri | How about $\Omega=B(0,1)\subset \mathbb C^2$ and $u(z,w)=Re(e^{1/(z-1)})$? (cf remark of Alexandre) | |
| Mar 18, 2013 at 20:16 | comment | added | oydeis | And by pseudoconvex I mean strictly pseudoconvex. | |
| Mar 18, 2013 at 19:57 | comment | added | oydeis | Good point. I forgot to add that $\Omega$ should be a \textit{bounded} pseudoconvex domain. | |
| Mar 18, 2013 at 17:26 | history | answered | Alexandre Eremenko | CC BY-SA 3.0 |