Let $v: \mathbb{R}^{d} \to \mathbb{C}$ be positive-definite and write
$$ \langle \delta, v\delta \rangle := \int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\delta(x)v(x-y)\delta(y)dxdy$$
where $\delta$ stands for the Dirac delta distribution. I'm trying to prove that the following equality holds:
$$e^{-\frac{1}{2}\langle \delta, v \delta \rangle} = \int_{\mathcal{S}(\mathbb{R}^{d})}e^{i\delta(\phi)}d\mu_{v}(\phi) \hspace{1cm} (1)$$
where $\mu_{v}(\phi)$ is a gaussian measure on $\mathcal{S}(\mathbb{R}^{d})$ which is characterized by $v$. Note that (1) is a "functional version" of the "Fourier transform of a gaussian is a gaussian" result in $\mathbb{R}^{n}$.
My question
Is there any representation theorem analogous to Bochner's that allows me to prove (1)?
The Minlos-Bochner theorem is a functional version of Bochner's theorem on $\mathbb{R}^{n}$ but the measure given by it is defined on $\mathcal{S}'(\mathbb{R}^{d})$ rather than $\mathcal{S}(\mathbb{R}^{d})$.
In addition, is there a simpler way to prove (1) ?
Best Answer
Your question is ill-formulated as is. To have a well defined measure, you need to give a definition of your bilinear form $\langle \cdot,\cdot\rangle:\mathcal{S}'\times\mathcal{S}'\rightarrow\mathbb{R}$. For arbitrary temperate distributions $F$, $G$, what you wanted to define is, I presume, $$ \langle F,G\rangle=(F\otimes G)(\phi) $$ where $\phi(x,y)=v(x-y)$. The problem is that even if $v$ is in $\mathcal{S}$, $\phi$ is not a Schwartz function. In particular if $F=G=1$ (the constant function equal to 1), the bilinear form does not make sense, because you have decay in the $x-y$ direction but no decay in the $x+y$ direction.