---
tag: D6JTF3613D
kind: definition
name: Lelong number
msc: [32U25, 32U05]
defines: WVB60RKC3Z
uses: [FF2P7ER3QT, BYJHP6A906]
sources:
  - url: https://en.wikipedia.org/wiki/Lelong_number
---

## Definition

Let $\Omega \subseteq \C^n$ be open, let $u$ be a [[plurisubharmonic function|FF2P7ER3QT]] on $\Omega$ and let $x \in \Omega$.
The *Lelong number* of $u$ at $x$ is
$$\nu(u, x) = \lim_{r \to 0^+} \frac{\sup_{\abs{z - x} < r} u(z)}{\log r}.$$
Here $\abs{\cdot}$ is the Euclidean norm on $\C^n$.

## 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|BYJHP6A906]] 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 $\nu(u, x) = \nu(dd^c u, x)$.