Lelong class

Remarks

Example: let $p$ be a polynomial on $\C^n$ of degree $d \ge 1$ and $u = \frac{1}{d} \log\abs{p}$. Then $u$ is plurisubharmonic, and $\abs{p(z)} \le M (1 + \abs{z})^d$ for some $M > 0$, so $u(z) \le \log(1 + \abs{z}) + \frac{1}{d} \log M$; hence $u \in \mathcal{L}$. It is not in $\mathcal{L}^+$ when $p$ has a zero, since $u = -\infty$ there. The function $\log(1 + \abs{z}^2)/2$ is in $\mathcal{L}^+$. Non-example: $u(z) = \abs{z_1}$ is plurisubharmonic, but $\abs{z_1} - \log(1 + \abs{z})$ is unbounded above, so $u \notin \mathcal{L}$. For $n = 1$, the logarithmic potentials $\int \log\abs{z - w} \, d\mu(w)$ for compactly supported probability measures $\mu$ are in $\mathcal{L}$. Some authors write $\log^+\abs{z} = \max(\log\abs{z}, 0)$ in place of $\log(1 + \abs{z})$; the class is the same, since the two functions differ by a bounded amount.