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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.

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 strong 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. (2021 Edit: see my article "A Second-Quantized Kolmogorov-Chentsov Theorem via the Operator Product Expansion" in CMP)

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.

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 $$ C(x,y)^2=\frac{\gamma^2}{|x-y|^{4[\phi]}}\ ? $$

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., 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.

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.

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 $Sq:S'(\mathbb{R}^d)\rightarrow S'(\mathbb{R}^d)$ 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., 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.

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 $$ C(x,y)^2=\frac{\gamma^2}{|x-y|^{4[\phi]}}\ ? $$

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., 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.