4.1.1 Kernel energies and equilibrium measuresFF9D3CA4D463

Proof. The function $h_ t^\sigma $ is bounded by $|\sigma |(\mathbb R^ d)$, and it is continuous by dominated convergence. Completing the square gives $|z-x|^2+|z-y|^2=2|z-\tfrac 12(x+y)|^2+\tfrac 12|x-y|^2$. Since $\int _{\mathbb R^ d}\mathrm e^{-4t|w|^2}\, \mathrm dw=(\pi /4t)^{d/2}$, it follows that

We integrate this identity against $\sigma \otimes \tau $. Fubini's theorem applies, because the integral of the absolute value of the integrand with respect to $\mathrm dz\otimes |\sigma |\otimes |\tau |$ is at most $(\pi /4t)^{d/2}|\sigma |(\mathbb R^ d)|\tau |(\mathbb R^ d)$, and integrating first in $x$ and in $y$ gives (4.1.4). For $\tau =\sigma $ it shows that $Q_ t(\sigma )$ is a positive multiple of $\int (h_ t^\sigma )^2\, \mathrm dz$. Hence $Q_ t(\sigma )\geq 0$, and if $Q_ t(\sigma )=0$, then $h_ t^\sigma =0$ almost everywhere, hence everywhere, because $h_ t^\sigma $ is continuous.

Now suppose that $Q_{t_ j}(\sigma )=0$ for a sequence $t_ j\to \infty $. The functions $g_ t(z)=(2t/\pi )^{d/2}\mathrm e^{-2t|z|^2}$ are nonnegative and have integral one, and $\int _{|z|\geq \delta }g_ t\, \mathrm d z\to 0$ as $t\to \infty $ for every $\delta >0$. Hence $g_ t*f\to f$ uniformly on $\mathbb R^ d$ for every $f\in C_ c(\mathbb R^ d)$: in the integral defining $g_ t*f-f$ we use the uniform continuity of $f$ on the region $|z|<\delta $ and the smallness of the Gaussian tail on the region $|z|\geq \delta $. Since $g_{t_ j}*\sigma =(2t_ j/\pi )^{d/2}h_{t_ j}^\sigma =0$, Fubini's theorem and the symmetry of $g_ t$ give

Letting $j\to \infty $, we get $\int f\, \mathrm d\sigma =0$ for every $f\in C_ c(\mathbb R^ d)$. Thus the positive and negative parts of $\sigma $ define the same positive linear functional on $C_ c(\mathbb R^ d)$, and the uniqueness assertion of theorem 2.3.3 shows that they are equal, that is, $\sigma =0$. $\square$

Proof. By (4.1.4), the expression $Q_ t(\sigma ,\tau )$ of lemma 4.1.3 is $(4t/\pi )^{d/2}$ times the $L^2$ inner product of $h_ t^\sigma $ and $h_ t^\tau $. Hence, by the Cauchy–Schwarz inequality in $z$,

for positive measures $\mu $ and $\nu $.

The case $d\geq 3$. Put $a=(d-2)/2>0$ and $c_ a=\int _0^\infty s^{a-1}\mathrm e^{-s}\, \mathrm d s\in (0,\infty )$. The substitution $s=tr^2$ gives

and both sides are $+\infty $ for $r=0$. By Tonelli's theorem,

Applying (4.1.6) and then the Cauchy–Schwarz inequality for the measure $t^{a-1}\, \mathrm dt$, we obtain $I(\mu ,\nu )\leq I(\mu )^{1/2}I(\nu )^{1/2}$. Now let $\sigma $ be a real linear combination of measures on $K$ with finite energy. Its total variation is dominated by a combination of the same measures with nonnegative coefficients, so the inequality just proved shows that $\iint k\, \mathrm d|\sigma |\, \mathrm d|\sigma |<\infty $. Consequently Fubini's theorem applies to the integrand $t^{a-1}\mathrm e^{-t|x-y|^2}$, and

by lemma 4.1.3. If $I(\sigma )=0$, then $Q_ t(\sigma )=0$ for almost every $t>0$, in particular for arbitrarily large $t$, and lemma 4.1.3 gives $\sigma =0$.

The case $d=2$. For $0<r\leq D$ we have

Indeed, the right side vanishes for $r=D$, and differentiation under the integral sign, which is legitimate for $r$ in a compact subinterval of $(0,\infty )$, shows that its derivative in $r$ is $-r\int _0^\infty \mathrm e^{-tr^2}\, \mathrm dt=-1/r$. The integrand is nonnegative, and since $|\mathrm e^{-t r^2}-\mathrm e^{-tD^2}|\leq tD^2$, its integral over $0<t<1$ is at most $D^2$. Since $|x-y|\leq \operatorname {diam}K<D$ for $x,y\in K$, Tonelli's theorem gives

with a nonnegative integrand. As $\int _1^\infty \mathrm e^{-tD^2}\, \mathrm dt/t<\infty $, it follows that $\int _1^\infty Q_ t(\mu )\, \mathrm dt/t<\infty $, and likewise for $\nu $. By (4.1.6) and the Cauchy–Schwarz inequality for the measure $\mathrm dt/t$ on $(1,\infty )$, also $\int _1^\infty Q_ t(\mu ,\nu )\, \mathrm dt/t<\infty $. Since the part of the integral over $0<t<1$ contributes at most $D^2\mu (K)\nu (K)$ to $4\pi I_ D(\mu ,\nu )$, the mutual energy $I_ D(\mu ,\nu )$ is finite.

Now let $\sigma $ be a signed measure on $K$ of total mass zero whose positive and negative parts $\sigma ^\pm $ have finite energy. Then

so Fubini's theorem applies to the integrand $(\mathrm e^{-t|x-y|^2}-\mathrm e^{-tD^2})/t$ with respect to $\mathrm dt\otimes \sigma \otimes \sigma $. Since $\sigma (K)=0$, the term $\mathrm e^{-tD^2}$ integrates to zero, and

In particular $I_ D(\sigma )$ does not depend on $D$. As in the case $d\geq 3$, $I_ D(\sigma )=0$ forces $\sigma =0$ by lemma 4.1.3.

It remains to prove the Cauchy–Schwarz inequality. Let $\mathcal E$ be the real linear span of the measures on $K$ with finite energy if $d\geq 3$, and the set of signed measures on $K$ of total mass zero whose positive and negative parts have finite energy if $d=2$. In the second case $\mathcal E$ is also a vector space. Indeed, let $\sigma ,\tau \in \mathcal E$ and $c\in \mathbb R$. Since $|\sigma +\tau |\leq |\sigma |+|\tau |$, we have $(\sigma +\tau )^\pm =\frac12\bigl (|\sigma +\tau |\pm (\sigma +\tau )\bigr )\leq \sigma ^\pm +\tau ^\pm $, while $(c\sigma )^\pm =c\sigma ^\pm $ if $c\geq 0$ and $(c\sigma )^\pm =|c|\sigma ^\mp $ if $c<0$. A positive measure on $K$ dominated by the sum of two measures with finite energy has finite energy, because the kernel is nonnegative on $K\times K$ and the mutual energy of two measures with finite energy is finite. Hence $\sigma +\tau $ and $c\sigma $ belong to $\mathcal E$.

In both cases, the total variation of an element of $\mathcal E$ is dominated by a sum of measures on $K$ with finite energy, so for $\sigma ,\tau \in \mathcal E$ the kernel is integrable with respect to $|\sigma |\otimes |\tau |$. Consequently $I(\sigma ,\tau )=\iint k\, \mathrm d\sigma \, \mathrm d\tau $ (with $k_ D$ in place of $k$ when $d=2$) is a symmetric bilinear form on $\mathcal E$, and $I(\sigma )=I(\sigma ,\sigma )\geq 0$ by what we have proved. Hence, for $\sigma ,\tau \in \mathcal E$, the quadratic polynomial $s\mapsto I(\sigma +s\tau )=I(\sigma )+2sI(\sigma ,\tau )+s^2I(\tau )$ is nonnegative on $\mathbb R$. Its discriminant is therefore nonpositive, which is the Cauchy–Schwarz inequality $I(\sigma ,\tau )^2\leq I(\sigma )I(\tau )$. $\square$

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 $D'\mathrm e^{-2\pi V_{D'}(K)}=D\mathrm e^{-2\pi V_ D(K)}$. $\square$