Theorem 3 (Prokhorov–Sazonov)

Let E be a Hilbert space and Es the space E equipped with the weakened topology. Let μ be a promeasure on E, and Φ its Fourier transform. The following conditions are equivalent:

a) The function Φ is continuous on E for the Sazonov topology.

b) For every ε > 0, there exists a nuclear positive quadratic form Q on E such that Φ(0) − $\Re \Phi$ ≤ ε + Q.

c) The promeasure μ is a measure on Es.

$b) \Rightarrow a)$: This follows from Prop. 8 of No. 8 (cf. the inequality (54)).

$a) \Rightarrow c)$: This follows from Theorem 2 of No. 10.

$c) \Rightarrow b)$: Suppose that μ is a measure on Es. Let ε > 0. For every integer $n \geq 1$, the set $B_n$ of $x \in E$ with norm $\leq n$ is a closed subset of $E_s$, and $E = \bigcup_{n \geq 1} B_n$. Therefore there exists an integer $n \geq 1$ such that $\mu(E - B_n) < \frac{\varepsilon}{2}$. The formula

defines a positive quadratic form Q on E. Set $C = \frac{n^2}{2} \mu(B_n)$. If $(e_1, \ldots, e_p)$ is a finite orthonormal sequence in E, then

for every $y \in B_n$ by Bessel’s inequality. It follows by integration that

therefore Q is nuclear.

Moreover, $1 - \cos t \leq \inf \left( 2, \frac{t^2}{2} \right)$ for every real number t, whence

for all $x \in E$. Thus b) is verified.

Q.E.D.