Let $\varphi$ be a function on a region $\Omega \in C^{n}$ which is locally integrable. Is it true that $\varphi = \psi_{1} -\psi_{2}$ where $\psi_{i}$ are psh?
$\begingroup$
$\endgroup$
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
No (if I decode your abbreviation "psh" correctly, as "plurisubharmonic"). Let $n=1$. Take a function $u(x,y)=0, x<0, u(x,u)=1, x>0$. It is locally integrable, but it is not a difference of subharmonic functions. In general, subharmonic functions are sort of "almost continuous".
answered Nov 23, 2012 at 15:58
-
$\begingroup$ Thank you. I want to change my question. What I want to know is that, given a singular hermitian metric, is it possible to characterize its singularity by the difference of two semipositive singular metric? $\endgroup$ Commented Nov 23, 2012 at 16:13
-
$\begingroup$ You have to define what a "singular Hermitian metric" means. $\endgroup$ Commented Nov 23, 2012 at 20:06
-
$\begingroup$ I just know a little from Demailly's book "analytic methods in algebraic geometry". A singular hermitian metric of a line bundle is a trivialization together with a locally integrable weight. A semipositive metric is a metric with a locally psh weight. $\endgroup$ Commented Nov 23, 2012 at 20:49