Logarithmic Potential, Capacity, and Riesz Decomposition

Let $\mu$ be a finite positive Borel measure on $\mathbb{C}$ (Radon measure on an LCH space). Call an open set $V\subseteq \mathbb{C}$ $\mu$-null when $\mu (V)=0$. The support of $\mu$ is the complement of the union of all open $\mu$-null sets,

$\mathrm{supp}\mu :=\mathbb{C}\setminus \bigcup \{V\subseteq \mathbb{C}:V\text{ open and }\mu (V)=0\},$

which is a closed subset of $\mathbb{C}$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space). Equivalently, $x\in \mathrm{supp}\mu$ if and only if $\mu (V)>0$ for every open set $V\ni x$. A measure is carried by a Borel set $A$ when $\mu (\mathbb{C}\setminus A)=0$, the terminology of A positive, signed, or complex measure concentrated on a measurable set. We say $\mu$ has compact support when it is carried by some compact subset of $\mathbb{C}$.

The support is the smallest closed carrier, and the argument is choice-free. Let $\mathcal{U}$ be the union of all open $\mu$-null sets. The rational open squares $B$ of ${\mathbb{R}}^2\cong \mathbb{C}$ form a countable basis (${\mathbb{Q}}^n$ is a countable dense subset of ${\mathbb{R}}^n$, and rational open boxes form a countable basis). Let $\mathcal{N}$ be the countable family of those basis squares with $\mu (B)=0$. If $x\in \mathcal{U}$, then $x$ lies in some open $\mu$-null $V$, and the basis property supplies a square $B\in \mathcal{N}$ with $x\in B\subseteq V$; conversely every $B\in \mathcal{N}$ is contained in $\mathcal{U}$. Hence $\mathcal{U}$ is the countable union of the $\mu$-null sets $\mathcal{N}$ and is itself $\mu$-null by countable additivity (Measures on sigma-algebras). Therefore $\mu (\mathbb{C}\setminus \mathrm{supp}\mu )=0$: the support carries $\mu$. If $F$ is a closed set with $\mu (\mathbb{C}\setminus F)=0$, then $\mathbb{C}\setminus F$ is an open $\mu$-null set, so it is one of the sets in the defining union and $\mathrm{supp}\mu \subseteq F$; the support is thus the smallest closed set carrying $\mu$. In particular $\mu \mathrm{\ne }0$ if and only if $\mathrm{supp}\mu \mathrm{\ne }\mathrm{\varnothing }$, and if $\mu$ is carried by a compact $K$, then $\mathrm{supp}\mu \subseteq K$ is compact.

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 $z=w$.

Let $K\subseteq \mathbb{C}$ be compact and nonempty, and let $P(K)$ be the set of Borel probability measures on $K$ (Probability measures and probability spaces), viewed as measures on $\mathbb{C}$ carried by $K$. Every $\mu \in P(K)$ has compact support contained in $K$, so its logarithmic energy $I(\mu )\in (-\mathrm{\infty },+\mathrm{\infty }]$ is the quantity of Logarithmic potential and energy of a positive compactly supported measure; indeed the kernel $k(z,w)=\log (1\mathrm{/}\mathrm{\mid }z-w\mathrm{\mid })$ is bounded below on $K\times K$: it is $\ge -\log \mathrm{diam}K$ when $\mathrm{diam}K>0$, and $I(\mu )=+\mathrm{\infty }$ for the only measure of $P(K)$ when $K$ is a singleton.

The Robin constant of $K$ is

$V_K:={\inf }_{\mu \in P(K)}I(\mu )\in (-\mathrm{\infty },+\mathrm{\infty }]\mathrm{.}$

The set $P(K)$ is nonempty: it contains the Dirac measure at any point of $K$ (The Dirac set function at a point). Put $E_K:=\{I(\mu ):\mu \in P(K),I(\mu )<+\mathrm{\infty }\}$. If $E_K=\mathrm{\varnothing }$, all energies are $+\mathrm{\infty }$ and we set $V_K=+\mathrm{\infty }$. Otherwise $K$ is not a singleton, and $E_K$ is a nonempty subset of $\mathbb{R}$ bounded below by $-\log \mathrm{diam}K$, so Every nonempty set bounded below has an infimum gives its real infimum. Set $V_K=\inf E_K$ in this case. Adding the value $+\mathrm{\infty }$ does not change any real lower bound, so this is exactly the extended infimum displayed above. The logarithmic capacity of $K$ is

$\mathrm{cap}(K):=\{\begin{array}{ll}\exp (-V_K)& \text{if }{V}_K<+\mathrm{\infty },\\ 0& \text{if }{V}_K=+\mathrm{\infty },\end{array}$

with $\exp$ the real exponential function (The real exponential function and the number $e$ by a power series). Thus $\mathrm{cap}(K)\in [0,\mathrm{\infty })$ for every nonempty compact $K$, and for the empty set one uses the separate convention

$\mathrm{cap}(\mathrm{\varnothing }):=0.$

Zero capacity as total divergence of the energy. For nonempty compact $K$, $\mathrm{cap}(K)=0$ holds exactly when $V_K=+\mathrm{\infty }$, that is, exactly when $I(\mu )=+\mathrm{\infty }$ for every $\mu \in P(K)$: the set $\{I(\mu ):\mu \in P(K)\}$ lies in $(-\mathrm{\infty },+\mathrm{\infty }]$, so its infimum is $+\mathrm{\infty }$ precisely when all its members are $+\mathrm{\infty }$. Such a set is called polar; this is the defining dichotomy used throughout the page.

Monotonicity under inclusion. If $K\subseteq L$ are nonempty compact sets, then every Borel probability measure on $K$, extended by zero to the Borel subsets of $L$, is a Borel probability measure on $L$ with the same energy; hence $P(K)\subseteq P(L)$, the infimum over the larger set is no larger, $V_L\le V_K$, and $\mathrm{cap}(K)\le \mathrm{cap}(L)$. The explicit case $\mathrm{cap}(\mathrm{\varnothing })=0$ is consistent with this: the empty set is contained in every set.

Normalization. The sign $\exp (-V_K)$ rather than $\exp (V_K)$ is the convention of the cited sources: it makes $\mathrm{cap}(K)$ a length, for instance $\mathrm{cap}\overline{D(a,r)}=r$ for a disc. The constant $V_K$ and the capacity determine each other whenever $V_K<+\mathrm{\infty }$.

Choice. No choice principle is used in this definition. Choosing a point of a nonempty compact set to exhibit nonemptiness of $P(K)$ is a single selection from a nonempty set; the infimum is a set-theoretic construction on a fixed set of extended reals.

Let $K\subseteq \mathbb{C}$ be nonempty compact and let ${\mu }_n,\mu$ be Borel probability measures on $K$ with ${\mu }_n\Rightarrow \mu$. With the Borel kernel $k(z,w)=\log \frac{1}{\mathrm{\mid }z-w\mathrm{\mid }}$ assigned $+\mathrm{\infty }$ on the diagonal, the extended integrals $U^{\mu }(z)={\int }_{\mathbb{C}}k(z,w)d\mu (w)$ and $I(\mu )={\iint }_{\mathbb{C}\times \mathbb{C}}kd\mu \otimes d\mu$ of Logarithmic potential and energy of a positive compactly supported measure are unambiguous: for each fixed $z$, a Borel function of $w$ that equals $k(z,w)$ for $\mu$-almost every $w$ gives the same potential at $z$; a Borel kernel equal to $k$ for $(\mu \otimes \mu )$-almost every $(z,w)$ gives the same energy. Moreover

$U^{\mu }(z)\le {\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}{U}^{{\mu }_n}(z)(z\in \mathbb{C}),I(\mu )\le {\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}I({\mu }_n)\mathrm{.}$

Finally, if $\mu (\{z\})>0$ for some $z$ then $U^{\mu }(z)=+\mathrm{\infty }$, so for atomic measures the diagonal value is not a free convention. No choice principle is required.

Assume the Axiom of Countable Choice. Let $\mu ,\nu$ be finite positive Borel measures on $\mathbb{C}$ with compact support, equal total mass $\mu (\mathbb{C})=\nu (\mathbb{C})=M$, and finite logarithmic energy $I(\mu )<\mathrm{\infty }$, $I(\nu )<\mathrm{\infty }$ in the normalization of Logarithmic potential and energy of a positive compactly supported measure. Then:

1. the mixed energy $I(\mu ,\nu )$ is finite, and $I(\sigma ):=I(\mu )-2I(\mu ,\nu )+I(\nu )$ is a well-defined real number for $\sigma :=\mu -\nu$;
2. $I(\sigma )\ge 0$, and $I(\sigma )=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{Q}_t(\sigma )\frac{dt}{t}$ where $Q_t(\sigma ):=\iint e^{-t\mathrm{\mid }z-w{\mathrm{\mid }}^2}d\sigma (z)d\sigma (w)$ for $t>0$;
3. $I(\sigma )=0$ if and only if $\sigma =0$.

The case $M=0$ is included and settled separately: then $\mu =\nu =0$, hence $I(\mu )=I(\nu )=I(\mu ,\nu )=0$ by the zero clauses of Logarithmic potential and energy of a positive compactly supported measure, so $\sigma =0$, the number $I(\sigma )=0$ is well defined and the representation of (2) holds because $Q_t(0)=0$ for every $t>0$; assertion (3) is then a tautology. The proof below therefore assumes $M>0$.

Countable Choice enters at exactly one point, step 6.1, through the regularity of finite Borel measures on the second-countable space $\mathbb{C}$; the pointwise, Gaussian and convergence steps are choice-free.