# Gaussian measure on function spaces

I'm reading this classic work and I'd like to get deeper inside some of its techniques. In particular, the authors state: "We construct a Gaussian measure $$d\mu_{0}(\phi)$$ on a measure space of continuous functions $$\phi(x), x\in \Lambda \subset \mathbb{R}^{3}$$ with covariance $$u$$: $$\begin{eqnarray} \int d\mu_{0}(\phi)e^{i\int f\phi} = e^{-\frac{1}{2}\int f u f} \tag{1}\label{1} \end{eqnarray}$$ It is then straightforward to show that: $$\begin{eqnarray} e^{-\beta U} = \int d\mu_{0}(\phi) e^{i\sqrt{\beta}\sum_{\alpha}e_{i(\alpha)}\phi(x_{\alpha})}" \tag{2}\label{2} \end{eqnarray}$$

First of all, how to construct such a Gaussian measure $$d\mu_{0}$$ on a space of continuous functions? Is it defined by condition (\ref{1}) or does (\ref{1}) follow as a consequence? Besides, how can we prove existence? Does anyone know any reference on this construction?

Second, equation (\ref{2}) seems to follow by taking $$f = \sum e_{i(\alpha)}\delta(x_{\alpha})$$. But how can we take such an $$f$$ is $$f$$ must be a continuous function rather than a distribution?

## 2 Answers

You should have a look at the book by Gelfand and Vilenkin

Generalized functions. Vol. 4: Applications of harmonic analysis

where they describe how to construct Gaussian measures on (duals) of nuclear spaces.

Thus, given an open set in $$\newcommand{\bR}{\mathbb{R}}$$ $$D\subset \bR^n$$ one begins by constructing a measure on the space $$C^{-\infty}(D)$$ of generalized functions on $$D$$. If the covariance kernel is sufficiently regular then this measure is concentrated one on a much smaller subspace.

Also, if you read French, I recommend this 1967 paper by Xavier Fernique. It is not the most comprehensive but I found it very helpful.

Finaly, there is V. Bogachev's book Gaussian Measures.

Just a quick answer for now. I would need to read carefully the definitions in the paper to be more precise.

In general you need the Bochner-Minlos Theorem which says there is a unique probability measure on Schwartz distribution for which (1) is satisfied. You can then convolve your random distribution $$\psi$$ by some nice continuous or smooth function to get a random disribution $$\phi$$ with law $$\mu_0$$. This relies on say $$u$$ being a convolution square. Then to prove (2) you can use (1) for the law of $$\psi$$ and not $$\phi$$. The mollifier then hits the $$\delta(x_{\alpha})$$'s.

Also, one may construct $$\phi$$ directly as $$\sum_{i} Z_i h_i$$ where the $$Z_i$$ are iid standard Gaussians and the $$h_i$$ are suitable functions like perhaps eigenfunctions for the Laplacian.

• Can you elaborate a little more, when you have a chance? It seems that this "space of continuous functions" must be, in fact, a schwartz space. I didn't follow you when you said to convolve $\psi$ with some function to get $\phi$. – IamWill Feb 14 '20 at 23:04