Lemma 4.1.321858171114B

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, $\sigma =0$. $\square$