Lemma 5

Tag 09SM, permanent. Cite it as [Bourbaki, Tag 09SM].

Integration, IX, p. 90, § 6. Promeasures and measures on a locally convex space

For every measure $\mu \in \mathscr{D}$ ,

\int_{\mathscr{C}} e^{i \langle f, \mu \rangle} \, dw'(f) = e^{-W(\mu)/2}.

Set $\mu = c_1 \varepsilon_{t_1} + c_2 \varepsilon_{t_2} + \cdots + c_n \varepsilon_{t_n}$ with $t_1, \ldots, t_n$ in $D$ and $c_1, \ldots, c_n$ in $\mathbf{R}$ . For every $g \in \Omega_0$ , the function $u(g)$ coincides with $g$ on $D$ ; therefore

\langle u(g), \mu \rangle = \sum_{j=1}^n c_j g(t_j) \qquad (g \in \Omega_0).

Also,

W(\mu) = \sum_{j,k=1}^n c_j c_k \inf(t_j, t_k),

and, since $m$ is the Gaussian measure on $\Omega$ with covariance $M$ , and $\Omega - \Omega_0$ is $m$ -negligible, we have

\int_{\Omega_0} e^{i \sum_{j=1}^n c_j g(t_j)} \, dm(g) = \exp \left( - \frac{1}{2} \sum_{j,k=1}^n c_j c_k \inf(t_j, t_k) \right).

Now, $\Omega - \Omega_0$ is $m$ -negligible and $w' = u(m)$ ; it follows that

\int_{\mathscr{C}} e^{i \langle f, \mu \rangle} \, dw'(f) = \int_{\Omega_0} e^{i \langle u(g), \mu \rangle} \, dm(g).

The formula (48) follows immediately from the formulas (49) to (52).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Lemma 6, § 6

Cites

Nothing.