4.5.1 Green energy and the complete positive cone57DD989B07A4

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. $\square$

Proof. The minimum $q=\min (w,G\mu )$ is a nonnegative superharmonic minorant of a potential, hence is a potential $G\nu $ with $\nu \in \mathcal E^+$. The hypothesis gives $\int (G\mu -G\nu )d\mu =0$, whereas $\int (G\mu -G\nu )d\nu \ge 0$. Their difference is $I(\mu -\nu )\le 0$. Strict definiteness makes $\mu =\nu $, proving the assertion. The nonnegativity of $w$ is necessary, including when $\mu =0$. $\square$

Proof. In any inner product space weak lower semicontinuity of the norm follows from $\| z\| ^2=\lim \langle z_ j,z\rangle \le \| z\| \liminf \| z_ j\| $. Apply this to $z_ j=\mu _ j-\mu _ k$ to prove the strong-Cauchy/weak-limit assertion. If $p_ j=G\mu _ j\le G\mu _ k=p_ k$, then reciprocity gives

Thus monotone potentials give a strong Cauchy sequence: in the increasing case the norms are bounded by the limit norm, and in the decreasing case by the first norm, with the inequality direction reversed for the lower bound. Dominated or monotone integration of $p_ j$ against every finite-energy measure proves weak convergence to the prescribed limit.

For density, compact restrictions of $\mu $ converge strongly: their tail energies tend to zero by dominated convergence for the integrable kernel $G\, d\mu \, d\mu $. On each compact support, Lusin's theorem and the support-continuity argument split a measure of finite energy into disjoint compact restrictions with finite continuous potentials; finiteness is needed only almost everywhere for the original measure. Finite partial sums again converge strongly. For one such compact continuous potential $p$, let $D_ j$ be a regular exhaustion containing its support and put $q_ j=p$ outside $D_ j$, $q_ j=H^{D_ j}_ p$ inside. Sweeping makes $q_ j=G\nu _ j$ a potential, with compactly supported $\nu _ j$, and $q_ j\downarrow 0$. The preceding monotone assertion gives $\nu _ j\to 0$ strongly. Then $\mu -\nu _ j$ has compact support, and its Green potential $p-q_ j$ is continuous and compactly supported. Differences prove signed density.

Energy-bounded positive measures have uniformly bounded masses on each compact $K$: choose nonnegative $q\in C_ c^\infty $ equal to one on $K$, and use its signed finite-energy representing measure $\lambda $ above: $\mu _ j(K)\le \int q\, d\mu _ j\le \| \lambda \| \| \mu _ j\| $. Weak energy convergence therefore implies convergence on smooth compact tests and then on every continuous compact test by uniform approximation in a fixed compact neighborhood. Conversely a vague limit $\mu $ has $I(\mu )\le \liminf I(\mu _ j)$: insert compact cutoffs in both kernel variables, truncate the kernel, pass to product weak convergence, and increase cutoffs and truncations. Thus $\mu \in \mathcal E^+$. Vague convergence gives convergence against the dense compact-potential tests; Cauchy–Schwarz and the uniform energy bound extend it to every energy test, proving weak convergence.

Finally a strong Cauchy sequence in the positive cone has bounded energy and locally bounded masses. Diagonal compact measure selection gives a vaguely convergent subsequence and positive Radon limit. The preceding paragraph makes it an energy weak limit, and the strong-Cauchy assertion makes the whole sequence converge strongly. No completeness of the signed space was assumed. $\square$

Proof. The parallelogram identity bounds the distance between two almost minimizers by their errors above the infimum. Completeness of the positive cone and closedness of $A$ give the unique minimizing limit. Expanding the square along $\mu _ A+t(\nu -\mu _ A)$ and sending $t\downarrow 0$ gives the variational inequality; expanding the full square proves sufficiency. Testing zero and $2\mu _ A$ gives the cone characterization, with the converse immediate. Compact-support cones are closed because strong positive convergence implies vague convergence and preserves support.

The cone inequalities tested on restrictions of finite-energy measures on $K$ show $G\mu \le G\mu ^ K$ quasi-everywhere there; otherwise a positive-capacity compact subset of a level-set deficit supplies a finite-energy test with positive integral. Orthogonality then gives equality $\mu ^ K$-almost everywhere. Cartan domination gives $G\mu ^ K\le G\mu $ globally. Conversely these two pointwise conditions give all the cone inequalities, because energy measures charge no polar sets. The regularized reduction is a potential dominated by $G\mu $, has finite energy by potential order, and equals $G\mu $ quasi-everywhere on $K$; uniqueness identifies it with the projection. The affine functional differs from $\| \mu -\nu \| ^2$ by a constant. Finally $I(\mu _ K)=C(K)$ and $I(\mu _ K,\tau )=1$ for every finite-energy probability $\tau $ on $K$. Cauchy–Schwarz gives $I(\tau )\ge 1/C(K)$; equality and strict definiteness give uniquely $\tau =\mu _ K/C(K)$. Infinite-energy competitors cannot improve it. $\square$