Let $\mu$ denote the centered Gaussian measure on $S'(\mathbb{R}^d)$ with covariance $$ \mathbb{E} [\phi(f)\phi(g)]=\int_{\mathbb{R}^d} \frac{\overline{\widehat{f}(\xi)} \widehat{g}(\xi)}{|\xi|^{d-2[\phi]}}\ d^d\xi $$ where $[\phi]>0$ is just a symbol for a real parameter (minus the Hurst exponent). The above random field is denoted by $FGF_{s}$ with $s=\frac{d}{2}-[\phi]$ in http://arxiv.org/abs/1407.5598

If $[\phi]<\frac{d}{2}$ then (via the nuclear theorem) the distribution $C(x,y)=\mathbb{E} [\phi(x)\phi(y)]$ in $S'(\mathbb{R}^{2d})$ is given up to a multiplicative constant $\gamma$ by $$ C(h)=\gamma \int_{\mathbb{R}^{2d}\backslash{\rm Diag}} h(x,y)\ \frac{1}{|x-y|^{2[\phi]}}\ d^dx\ d^d y $$ for all text functions $h\in S(\mathbb{R}^{2d})$. Here ${\rm Diag}$ denotes the diagonal.

The space $S'(\mathbb{R}^d)$ is equipped with the weak-$\ast$ topology and the corresponding Borel $\sigma$-algebra. For $U$ an open set in $\mathbb{R}^d$, let $\mathcal{F}(U)$ denote the $\sigma$-algebra generated by the maps $\phi\mapsto \phi(g)$, for $g$ a test function with support in $U$. Let me call a measurable map $\Xi:S'(\mathbb{R}^d)\rightarrow S'(\mathbb{R}^d)$ local if for any open set $U$, and any test function $f$ with support in $U$, the map $\phi\mapsto \Xi(\phi)(f)$ is measurable with respect to $\mathcal{F}(U)$.

If $[\phi]>\frac{d}{4}$, is it possible to have the existence of a local map $Sq:S'(\mathbb{R}^d)\rightarrow S'(\mathbb{R}^d)$ for which the random variable $\phi\mapsto Sq(\phi)(f)$ has finite variance and zero expectation for all $f\in S(\mathbb{R}^d)$, and such that the restriction of the covariance $$ \mathbb{E} [Sq(\phi)(f)\ Sq(\phi)(g)] $$ to the complement of the diagonal is given by integration against $$ 2 C(x,y)^2=\frac{2\gamma^2}{|x-y|^{4[\phi]}}\ ? $$

Perhaps a better reformulation of my question is: is there a no-go theorem which rules out the existence of such a "pointwise squaring" map $Sq$?

If $[\phi]<\frac{d}{4}$, I believe such a map can be constructed following, e.g., this article by Da Prato and Tubaro.

This article by Albeverio and Liang seem to point towards a negative answer as far as the existence of $Sq$. On the other hand this article by Magnen and Unterberger seems to point towards a positive answer.

Of course, I did not mean to say that these two articles are contradictory. They apply to questions which are different from (yet related to) the one asked here.

If finite variance for $\phi\mapsto Sq(\phi)(f)$ is too much to ask, I would settle for $\phi\mapsto Sq(\phi)(f)Sq(\phi)(g)$ being in $L^1$ if the supports of $f$ and $g$ are disjoint.

  • $\begingroup$ As far as I understand the article of Magnen and Unterberger, they renormalise the square not just by a usual Wick product, but by performing a change of measure. In some limiting procedure, the law of $\phi$ converges weakly to the original measure and the pair $(\phi, Sq(\phi))$ converges weakly to something non-trivial. The point here is that in the limit I don't think that $Sq(\phi)$ is actually a function of $\phi$ in their construction... $\endgroup$ – Martin Hairer Apr 21 '15 at 21:59
  • $\begingroup$ @Martin: yes this is also the impression I got. Complete dependence is not preserved under weak convergence of the joint probability distributions. So indeed their paper does not go as far as constructing a map $Sq$. I was wondering if 1) by working harder one could refine their result and construct such a map which would give the same join distribution or 2) there is some no-go theorem which would show that 1) is impossible. A related question is if there is a nonconstructive way of showing some $Sq$ map exists as in the theorem of Lyons and Victoir. $\endgroup$ – Abdelmalek Abdesselam Apr 21 '15 at 22:12
  • 2
    $\begingroup$ There is a completely constructive and general version of the Lyons-Victoir result, see Proposition 4.11 of the long paper arxiv.org/abs/1303.5113. Like LV, this however focuses on just getting the relevant algebraic relations and analytical bounds right. In your situation, there is no natural algebraic relation the product should satisfy, so the construction would simply return $0$... Regarding a possible no-go theorem, I suspect that it is indeed the case, but I don't recall having seen it stated in a precise form somewhere. $\endgroup$ – Martin Hairer Apr 22 '15 at 7:37
  • $\begingroup$ Thanks Martin. I didn't know that your magnum opus contained a generalization of the LV extension theorem... $\endgroup$ – Abdelmalek Abdesselam Apr 22 '15 at 14:37

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