Logarithmic potential and energy of a positive compactly supported measure

Identify $\mathbb{C}$ with ${\mathbb{R}}^2$ and let $dA$ be area Lebesgue measure. All measures below are positive Borel measures. The logarithmic kernel is

$k(z,w):=\log \frac{1}{\mathrm{\mid }z-w\mathrm{\mid }}(z,w\in \mathbb{C}),$

with the diagonal value $k(w,w):=+\mathrm{\infty }$; here $\log$ is the natural logarithm. The map $k$ is Borel on $\mathbb{C}\times \mathbb{C}$ and $k(z,w)=+\mathrm{\infty }$ exactly when $z=w$.

The kernel is used with the following two standing hypotheses, each stated separately where it is needed.

Potential. Let $\mu$ be a finite positive Borel measure of compact support. The logarithmic potential of $\mu$ is

$U^{\mu }(z):={\int }_{\mathbb{C}}k(z,w)d\mu (w)\in (-\mathrm{\infty },+\mathrm{\infty }](z\in \mathbb{C})\mathrm{.}$

The integral is the extended integral of the Borel function $w\mapsto k(z,w)$, which is bounded below on the fixed compact set $\mathrm{supp}\mu$ and takes the value $+\mathrm{\infty }$ only at $w=z$. Its negative is the subharmonic normalisation

$p_{\mu }(z):=-U^{\mu }(z)={\int }_{\mathbb{C}}\log \mathrm{\mid }z-w\mathrm{\mid }d\mu (w)\in [-\mathrm{\infty },+\mathrm{\infty })\mathrm{.}$

For the zero measure both extended integrals are empty sums; we record the clauses $U^0:=0$ and $p_0:=0$.

Energy. Let $\mu$ be a finite positive Borel measure of compact support. Choose $R>\mathrm{diam}(\mathrm{supp}\mu )$ and put $k_R(z,w):=k(z,w)+\log R$. Then $k_R\ge 0$ on $\mathrm{supp}\mu \times \mathrm{supp}\mu$, and the logarithmic energy of $\mu$ is

$I(\mu ):={\int }_{\mathbb{C}}{\int }_{\mathbb{C}}{k}_R(z,w)d\mu (z)d\mu (w)-\mu (\mathbb{C}{)}^2\log R\in (-\mathrm{\infty },+\mathrm{\infty }]\mathrm{.}$

Here the double integral of the nonnegative Borel function $k_R$ is the iterated extended integral, well defined by Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, and the value $I(\mu )$ does not depend on the choice of $R>\mathrm{diam}(\mathrm{supp}\mu )$. For the zero measure, whose support is empty and has no diameter, we record the separate clause $I(0):=0$; this is also the value of the displayed formula for every $R>0$ (both the double integral over the empty carrier and the subtracted term $\mu (\mathbb{C}{)}^2\log R$ are $0$), so the clause is the consistent extension of the definition and is used only to avoid mentioning $\mathrm{diam}\mathrm{\varnothing }$. For bounded Borel $B$ the finiteness $\mu (B)<\mathrm{\infty }$ is that of Radon measure on an LCH space.

Mixed energy. Let $\mu ,\nu$ be finite positive Borel measures of compact support and choose $R>\mathrm{diam}(\mathrm{supp}\mu \cup \mathrm{supp}\nu )$. The mixed energy of $\mu$ and $\nu$ is

$I(\mu ,\nu ):={\int }_{\mathbb{C}}{\int }_{\mathbb{C}}{k}_R(z,w)d\mu (z)d\nu (w)-\mu (\mathbb{C})\nu (\mathbb{C})\log R\in (-\mathrm{\infty },+\mathrm{\infty }],$

again independent of the choice of $R$. If $\mu =0$ or $\nu =0$ we record the clause $I(\mu ,\nu ):=0$; for $\mu =0$ the displayed formula gives $0$ for every $R>0$, and symmetrically for $\nu =0$, so this is again a consistent extension covering the case in which $\mathrm{supp}\mu \cup \mathrm{supp}\nu =\mathrm{\varnothing }$. When $I(\mu )=I(\mu ,\mu )$ the same symbol is used, and $I(\mu ,\nu )=I(\nu ,\mu )$ because $k$ is symmetric.

Why the shift is legitimate. Since $\mathrm{\mid }z-w\mathrm{\mid }\le \mathrm{diam}(\mathrm{supp}\mu )<R$ on the product of the supports, the two extensions of $I(\mu )$ obtained from two admissible radii $R<S$ agree by Tonelli: the nonnegative integrands $k+\log R$ and $k+\log S$ differ by the constant $\log (S\mathrm{/}R)$, whose integral against $\mu \otimes \mu$ is $\mu (\mathbb{C}{)}^2\log (S\mathrm{/}R)$.

Unbounded support. For a finite positive Borel measure $\mu$ whose support is not compact, $U^{\mu }(z)$ is defined by the extended integral when $\int {\log }^{+}\mathrm{\mid }z-w\mathrm{\mid }d\mu (w)<\mathrm{\infty }$, so its negative part has finite integral and $U^{\mu }(z)\in (-\mathrm{\infty },+\mathrm{\infty }]$. For energy, put $A:=\iint k^{+}d\mu d\mu$ and $B:=\iint k^{-}d\mu d\mu$, both nonnegative extended integrals defined by Tonelli. When at least one of $A,B$ is finite, define $I(\mu ):=A-B\in [-\mathrm{\infty },+\mathrm{\infty }]$. In particular, $A<\mathrm{\infty }$ and $B=+\mathrm{\infty }$ gives $I(\mu )=-\mathrm{\infty }$; if $B<\mathrm{\infty }$, the value lies in $(-\mathrm{\infty },+\mathrm{\infty }]$. If both parts are infinite, $I(\mu )$ is undefined. The shifted compact-support formulas above are not used on unbounded supports; the capacity theory of this page uses compactly supported measures only.

Finiteness on compact support. If $\mathrm{diam}(\mathrm{supp}\mu )>0$, then $k\ge -\log \mathrm{diam}(\mathrm{supp}\mu )$ on the product of the support with itself, so $I(\mu )>-\mathrm{\infty }$; if $\mathrm{supp}\mu =\{a\}$ then $I({\delta }_a)=+\mathrm{\infty }$. The diagonal value $+\mathrm{\infty }$ is not a removable convention: for every $\mu$ with $\mu (\{a\})>0$ the values $U^{\mu }(a)=+\mathrm{\infty }$ and $I(\mu )=+\mathrm{\infty }$ depend on which value is assigned at $z=w$.