Logarithmic potential

Remarks

Example: let $\mu$ be the uniform probability measure on the unit circle. Then $p_\mu(z) = 0$ for $\abs{z} \le 1$ and $p_\mu(z) = -\log\abs{z}$ for $\abs{z} \ge 1$, by the mean value property. For $\abs{z} > 1$ apply it to the harmonic function $w \mapsto \log\abs{z - w}$ on the disc $\abs{w} < \abs{z}$. For $\abs{z} < 1$ write $\abs{z - e^{it}} = \abs{1 - z e^{-it}}$ and apply it to the harmonic function $\zeta \mapsto \log\abs{1 - z\zeta}$ on the disc $\abs{\zeta} < 1/\abs{z}$; the case $\abs{z} = 1$ follows from $\int_0^{2\pi} \log\abs{1 - e^{it}} \, dt = 0$. Example: for $\mu = \sum_{j=1}^d \delta_{z_j}$ one gets $p_\mu(z) = -\log\abs{(z - z_1)\cdots(z - z_d)}$. Non-example: for Lebesgue measure on all of $\C$ the integral diverges at every point, which is why the definition asks for a finite measure with compact support. The function $p_\mu$ is harmonic on $\C \setminus \operatorname{supp}\mu$, and $p_\mu(z) = -\mu(\C) \log\abs{z} + o(1)$ as $\abs{z} \to \infty$. In the sense of distributions, $\Delta p_\mu = -2\pi\mu$. Some authors use the opposite sign, $\int \log\abs{z - w} \, d\mu(w)$, which is subharmonic.