Theorem 4.5.2CC3D1AD9179B

Proof. The last assertion follows without Cauchy–Schwarz: $I(\lambda )\le I(\mu ,\lambda )=\int G\lambda \, d\mu \le I(\mu )$. To prove no polar mass, take a positive pole function $w$ on a Borel polar carrier $P$. On compact subsets $K$ of $P\cap \{ G\mu \le j\} $ the restricted measure has bounded potential on its support, so the everywhere-finite domination theorem, applied to $\varepsilon w$, gives $G(\mu |_ K)=0$. Thus $\mu (K)=0$. Inner regularity and $\mu (\{ G\mu =\infty \} )=0$ give $\mu (P)=0$. Every polar subset has a Borel polar carrier. Conversely a nonpolar Borel set has a compact subset of positive capacity, whose nonzero equilibrium measure has finite energy and gives it positive mass.

We establish Cauchy–Schwarz without assuming positivity for signed energies. For compact $K$ and continuous $f$ on $K$, the functional $J_ f(\eta )=I(\eta )-2\int f\, d\eta $ has a minimizer among positive measures on $K$ (the empty case is immediate). Positivity and compactness give a finite $\beta >0$ with $G\ge \beta $ on $K\times K$, hence $J_ f(\eta )\ge \beta \eta (K)^2-2\| f\| _\infty \eta (K)$, bounding the masses of a minimizing sequence. Compact measure selection and the continuous kernels $\min (G,m)$ give lower semicontinuity of energy: weak convergence of the product measures follows first for products of continuous functions, then for all continuous functions by their uniformly dense algebra. The zero measure gives a finite infimum, so the minimizer $\eta $ has finite energy. Variations $\eta -t\eta |_ U$, for $0<t<1$, give $G\eta \le f$ on its support. Variations adding the equilibrium measure of $K\cap \{ f-G\eta \ge \varepsilon \} $ give $G\eta \ge f$ quasi-everywhere on $K$. In these expansions all mutual integrals are finite: the adding measure has bounded potential, and the removing measure is a restriction of $\eta $.

For compactly supported $\mu ,\nu \in \mathcal E^+$ choose continuous $0\le f\le G\mu $ on their common compact support and let $\eta $ be this minimizer. The no-polar-mass assertion gives $I(\eta )=\int f\, d\eta $. The inequality $J_ f(t\nu )\ge J_ f(\eta )=-I(\eta )$ for all $t\ge 0$ therefore implies $\int f\, d\nu \le \| \nu \| \| \eta \| $ by its quadratic discriminant. Since $G\eta \le f$ on its support, it is finite there; off that compact support, kernel continuity makes it finite and harmonic. Thus domination applies and gives $G\eta \le G\mu $ and hence $\| \eta \| \le \| \mu \| $. Increase continuous minorants to $G\mu $ to obtain Cauchy–Schwarz. Compact restrictions and monotone convergence prove it for arbitrary finite-energy measures. If either energy is infinite, the extended inequality is immediate (zero energy implies the zero measure by the argument below).

Bilinear expansion now gives $I(\mu -\nu )\ge (\| \mu \| -\| \nu \| )^2\ge 0$. The discriminant of $I(\sigma +t\tau )$ proves signed Cauchy–Schwarz and the triangle inequality. To prove strict definiteness, take $q\in C_ c^\infty (\Omega )$ and $\lambda =(-\Delta q/c_ d)dx$. The kernel distributional identity directly gives $G\lambda =q$; both parts of $\lambda $ have bounded continuous potentials and compact supports, hence finite energy. If $I(\sigma )=0$, signed Cauchy–Schwarz gives $\int q\, d\sigma =I(\lambda ,\sigma )=0$. All smooth compact tests vanish, so the signed Radon measure $\sigma $ is zero. This also settles the zero-energy case of the extended positive inequality. $\square$