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

Reading Demailly's Algebraic and Complex geometry in particular chapter 3 about positive currents, tha author defines the unbounded locus $L(u)$ of a plurisubharmonic function $u$ to be the set of points $x\in X$ such that $u$ is unbounded in every neighborhood of $x$. Then he shows that $L(u)$ contains the closure of $P(u)=u^{1}(-\infty)$, but in general this inclusion is strict: in fact, $u(z)=\sum k^{-2}\log(|z-1/k|+e^{-k^{3}})$ is everywhere finite in $\mathbb{C}$ but $L(u)=(0)$. what i don't understand is why is everywhere finite in $\mathbb{C}$, since if $z=1/k$ then $u(z)=-\infty$

share|cite|improve this question
No, when $k$ is fixed, $u(1/k)$ is not $-\infty$: it is the sum of $-k$ and some convergent series. The point is that near $0$ you can find arbitrary negative values of $u$ though $u$ has no pole. – Henri Dec 11 '11 at 20:51
thank if $u$ has logaritmic singularities then $u^{-1}(-\infty)=L(u)$? is it correct? – alike Dec 12 '11 at 16:57
Yes, it is correct! – Henri Dec 12 '11 at 19:46

Your Answer


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

Browse other questions tagged or ask your own question.