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 $[\phi]$ is just a symbol for a real parameter (minus the Hurst exponent).
The above random field is called $FGF_{s}$ 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 function $h\in S(\mathbb{R}^{2d})$. 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 $Sq:S'(\mathbb{R}^d)\rightarrow S'(\mathbb{R}^d)$ 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
$$
C(x,y)^2=\frac{\gamma^2}{|x-y|^{4[\phi]}}\ .
$$

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.,
<a href="http://eprints.biblio.unitn.it/1189/1/UTM711.pdf">this article</a> by Da Prato and Tubaro.