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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?

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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.

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  • $\begingroup$ 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 $\phi$. $\endgroup$ – IamWill Feb 14 at 23:04

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