Normalization of Gaussian w.r.t. Gaussian in a Banach space

Question

I would like to compute $$\int_X \exp\left(-\frac{1}{2}(Au)^2\right)\mathrm d\mu_0(u)$$ with a linear and continuous operator on a Banach space $A:X\to \mathbb R$ (in my case $X=C([0,1])$) and $\mu_0$ a centred Gaussian measure with covariance operator $Q$ on $X$. Is there any hope this can be calculated explicitly? It looks like this should be feasible as this is the normalization of a measure which is "Gaussian w.r.t another Gaussian". I have found something similar in Proposition 2 of this paper: http://projecteuclid.org/download/pdfview_1/euclid.ba/1440594948

But they need a Hilbert setting, which I don't have here.

Answer 1

Assume $\mu_0$ is centered. The covariance form (not operator) of $\mu_0$ is a bilinear form $Q$ on the dual $X^*$. $\newcommand{\bR}{\mathbb{R}}$ Regard the continuous linear functionals $\alpha:X\to\bR$ as random variables on the probability space $(X,\mu_0)$. Then $\newcommand{\bE}{\mathbb{E}}$

$$Q: X^*\times X^*\to\bR,\;\; Q (\alpha,\beta)=\bE[\alpha\cdot\beta]=\int_X\alpha(x)\beta(x) \mu_0(dx). $$

A continuous linear functional $A: X\to\bR$ is a Gaussian random variable with mean $0$ and variance $v_A:=Q(A,A)$. Assume first that $v_A\neq 0$. Then, for any $t\in\bR$, we have

$$ \int_X e^{- \frac{1}{2}(tA(x))^2} \mu_0(dx)=\bE\Big[e^{-\frac{1}{2}(tA)^2}\Big] $$

$$= \frac{1}{\sqrt{2\pi v_A}}\int_{\bR} e^{-\frac{1}{2}t^2a^2} e^{-\frac{a^2}{2v_A}} da =\frac{1}{\sqrt{2\pi v_A}}\int_{\bR}e^{-\frac{a^2}{2cA(t)}},$$

where

$$ \frac{a^2}{2c_A(t)}= \frac{a^2}{2}\Big( t^2+\frac{1}{v_A}\Big)\Rightarrow c_A(t)=\frac{v_A}{v_A t^2+1}. $$

Now observe that

$$ \int_{\bR}e^{-\frac{a^2}{2cA(t)}}=\sqrt{2\pi c_A(t)}, $$

so

$$ \int_X e^{- \frac{1}{2}(tA(x))^2} \mu_0(dx)=\frac{1}{v_At^2+1}. $$

This equality holds also when $v_A=0$.