Proposition 4

Let E be a locally convex space. For every positive quadratic form Q on E', there exists one and only one promeasure $\Gamma_Q$ on E such that $\mathscr{F} \Gamma_Q = e^{-Q/2}$. The total mass of $\Gamma_Q$ is equal to 1.

The uniqueness of $\Gamma_Q$ follows from Prop. 3 of No. 3. The total mass of $\Gamma_Q$ is equal to $(\mathscr{F}\Gamma_Q)(0) = e^{-Q(0)/2} = 1$. We will prove existence in stages.

A) E of finite dimension n, and Q nondegenerate.

By Lemma 2 of No. 4, the measure $\gamma_1$ on $\mathbf{R}$ having density $t \mapsto (2\pi)^{-1/2}e^{-t^2/2}$ is bounded, of total mass 1. Set $\gamma = \gamma_1 \otimes \cdots \otimes \gamma_1$ (n factors). From Lemma 2 of No. 4, one deduces

Since Q is positive and nondegenerate, there exists a basis $(e'_1, \ldots, e'_n)$ of E' orthonormal for Q (Alg., Ch. IX, §7, No. 1). Let us denote by f the isomorphism $x \mapsto (e'_1(x), \ldots, e'_n(x))$ of E onto $\mathbf{R}^n$, and by $\Gamma_Q$ the measure $f^{-1}(\gamma)$ on E. Let $x' = a_1 e'_1 + \cdots + a_n e'_n$ be in E'; then $x'(f^{-1}(t_1, \ldots, t_n)) = \sum_{j=1}^n t_j a_j$ for $t_1, \ldots, t_n$ real, whence

Consequently, $\mathscr{F}\Gamma_Q = e^{-Q/2}$.

B) E finite-dimensional.

Let N be the linear subspace of E' formed by the $x'$ such that $Q(x') = 0$. Denote by M the orthogonal of N in E, and by j the canonical injection of M into E. The linear mapping ${}^t j : E' \to M'$ is surjective, with kernel N, therefore there exists on $M'$ a nondegenerate positive quadratic form q such that $Q = q \circ {}^t j$. By the foregoing, there exists a bounded measure $\Gamma$ on M such that $\mathscr{F}\Gamma = e^{-q/2}$. Setting $\Gamma_Q = j(\Gamma)$, we have

by formula (4) of No. 3.

C) The general case. Let $V \in \mathscr{F}(E)$. Denote by $p_V$ the canonical mapping of $E$ onto $E/V$, and by $Q_V$ the positive quadratic form $Q \circ {}^t p_V$ on $(E/V)'$; finally, let $\mu_V$ be the measure on $E/V$ with Fourier transform $e^{-Q_V/2}$ (cf. B)). If $W \in \mathscr{F}(E)$ is contained in $V$, then $p_V = p_{VW} \circ p_W$, whence $Q_V = Q_W \circ {}^t p_{VW}$; by formula (4) of No. 3, the measure $p_{VW}(\mu_W)$ has as Fourier transform the function $(e^{-Q_W/2}) \circ {}^t p_{VW} = e^{-Q_V/2}$, hence is equal to $\mu_V$. The family $(\mu_V)_{V \in \mathscr{F}(E)}$ is therefore a promeasure $\mu$ on $E$. Formula (5) of No. 3 shows that $\mathscr{F}\mu$ is equal to $e^{-Q/2}$.