MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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".

share|cite|improve this answer
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? – Hu Zhengyu Nov 23 '12 at 16:13
You have to define what a "singular Hermitian metric" means. – Alexandre Eremenko Nov 23 '12 at 20:06
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. – Hu Zhengyu Nov 23 '12 at 20:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.