Theorem 4.1.7AA43DD93EFE4

Proof. Existence. Let $\mu _ j$ be probability measures on $K$ whose energies converge to the infimum $m$, which is finite by hypothesis. By theorem 2.6.5, after passing to a subsequence, $\mu _ j$ converges vaguely to a Radon measure $\mu $ on $\mathbb R^ d$. By proposition 2.6.2, applied to the open set $\mathbb R^ d\setminus K$, we have $\mu (\mathbb R^ d\setminus K)\leq \varliminf _ j\mu _ j(\mathbb R^ d\setminus K)=0$, so $\mu $ is carried by $K$; testing against a function in $C_ c(\mathbb R^ d)$ equal to one on $K$ then shows that $\mu (K)=1$. To prove that $I(\mu )\leq m$, we first show that

for every continuous function $F$ on $\mathbb R^ d\times \mathbb R^ d$.

Given $\delta >0$, cover $K$ by finitely many balls $B(x_ i,\delta )$ with $x_ i\in K$. By lemma 2.5.2, there are functions $0\leq \varphi _ i\in C_ c^\infty (B(x_ i,\delta ))$ with $\sum _ i\varphi _ i=1$ on $K$. For $x,y\in K$, a term $\varphi _ i(x)\varphi _ l(y)$ is nonzero only if $|x-x_ i|<\delta $ and $|y-x_ l|<\delta $, so

where $\omega $ is the modulus of continuity of $F$ on $K\times K$. The double integrals of the finite sums converge, since $\int \varphi _ i\, \mathrm d\mu _ j\to \int \varphi _ i\, \mathrm d\mu $. As all the measures are probability measures carried by $K$ and $\omega (2\delta )\to 0$ as $\delta \downarrow 0$, this proves the asserted convergence.

For $n>0$, the truncated kernel $\min \{ k,n\} $ is continuous on $\mathbb R^ d\times \mathbb R^ d$, including the diagonal, and it increases to $k$ as $n\to \infty $. Applying the convergence just proved to $F=\min \{ k,n\} $, we get

Since $k\geq 0$ on $K\times K$, monotone convergence in $n$ gives $I(\mu )\leq m$. Hence $\mu $ is a minimizer.

Uniqueness. Let $\mu $ and $\mu '$ be minimizers. The signed measure $\sigma =\mu -\mu '$ has total mass zero. Since $|\sigma |\leq \mu +\mu '$, its positive and negative parts satisfy $\sigma ^+=\frac12(|\sigma |+\sigma )\leq \mu $ and $\sigma ^-\leq \mu '$, so they have finite energy, the kernel being nonnegative on $K\times K$. All mutual energies are finite by proposition 4.1.5, so we may expand and obtain

Since $\frac12(\mu +\mu ')$ is a probability measure on $K$, this energy is at least $m$, so $I(\sigma )\leq 0$. By proposition 4.1.5, $\sigma =0$.

The inequality. Let $\nu $ be a probability measure on $K$ with finite energy. For $0<t\leq 1$ the probability measure $(1-t)\mu _ K+t\nu $ competes in the minimization, so

all terms being finite. Subtracting $V(K)$, dividing by $t$ and letting $t\downarrow 0$ gives $0\leq -2V(K)+2I(\mu _ K,\nu )$.

The capacities. Let $d\geq 3$. A point mass has infinite energy, so $K$ is not a single point and $\operatorname {diam}K>0$. The kernel is at least $((d-2)s_{d-1})^{-1}(\operatorname {diam}K)^{2-d}>0$ on $K\times K$, hence $V(K)>0$. Let $d=2$. Replacing $D$ by $D'>\operatorname {diam}K$ changes the energy of every probability measure on $K$ by the same constant $(2\pi )^{-1}\log (D'/D)$. Hence $\mu _ K$ does not depend on the scale, $V_{D'}(K)=V_ D(K)+(2\pi )^{-1}\log (D'/D)$, and $D'\mathrm e^{-2\pi V_{D'}(K)}=D\mathrm e^{-2\pi V_ D(K)}$. $\square$