Proposition 4.1.509A0D05A868F

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 $I(\sigma ,\tau )^2\leq I(\sigma )I(\tau )$. $\square$