Lelong number

Remarks

The limit exists and lies in $[0, +\infty)$, because $\sup_{\abs{z - x} < r} u(z)$ is a convex increasing function of $\log r$. An equivalent form is $\nu(u, x) = \liminf_{z \to x} u(z) / \log \abs{z - x}$.

Example: let $f$ be holomorphic on $\Omega$ with order of vanishing $m$ at $x$, and $u = \log \abs{f}$. Near $x$ one has $c_1 r^m \le \sup_{\abs{z - x} < r} \abs{f(z)} \le c_2 r^m$ for constants $c_1, c_2 > 0$, so $\nu(u, x) = m$. In particular $\nu(\log \abs{z_1}, 0) = 1$.

If $u(x) > -\infty$, then $\nu(u, x) = 0$. The converse fails. On the unit ball of $\C^n$ let $u(z) = -\sqrt{-\log \abs{z}}$, a plurisubharmonic function because $t \mapsto -\sqrt{-t}$ is convex and increasing on $(-\infty, 0)$. Then $u(0) = -\infty$, but $\sup_{\abs{z} < r} u(z) / \log r = 1 / \sqrt{-\log r} \to 0$, so $\nu(u, 0) = 0$.

Let $T$ be a closed positive current of bidimension $(p, p)$. Its Lelong number $\nu(T, x)$ is the limit as $r \to 0$ of the trace mass of $T$ on the ball $B(x, r)$, divided by $\pi^p r^{2p} / p!$. The divisor $\pi^p r^{2p} / p!$ is the volume of a ball of radius $r$ in $\C^p$. With a suitable normalisation of $dd^c$, one has $\nu(u, x) = \nu(dd^c u, x)$.