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Let $(\xi_\phi)_{\phi \in L^2(\mathbb{R}_+ \times \mathbb{R}^d,\lambda_d)}$ be a collection of centered Gaussian processes on a probability space $(\Omega,\mathcal{F},P)$ such that $$\forall \phi \in L^2(\mathbb{R}_+\times \mathbb{R}^d),E[\xi_\phi^2]=\int_{\mathbb{R}_+ \times \mathbb{R}^d}(\phi(r,x))^2drdx.$$

We denote by $\mathcal{E}$ the collection of Schwartz functions and $\mathcal{E}'$ the collection of tempered (Schwartz) distributions.

  • How to construct $\overline{\xi}:\Omega \to \mathcal{E}'$ such that for all $\phi \in \mathcal{E},\overline{\xi}(\phi)=\xi_\phi$ a.s.?

  • Is it possible to have $\zeta:\Omega \to \mathcal{L}(L^2(\mathbb{R}_+ \times \mathbb{R}^d),\mathbb{R})$ such that such that for all $\phi \in L^2(\mathbb{R}_+ \times \mathbb{R}^d),\zeta(\phi)=\xi_\phi$ a.s.? ($\mathcal{L}(L^2(\mathbb{R}_+ \times \mathbb{R}^d),\mathbb{R})$ denotes the space of linear continuous functions).

We note that $$\forall \phi_1,\phi_2 \in L^2(\mathbb{R}_+\times \mathbb{R}^d),\xi_{\phi+\phi_2}=\xi_{\phi_1}+\xi_{\phi_2} \text{ a.s.}$$

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1 Answer 1

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Let me replace $\mathbb{R}_+\times \mathbb{R}^d$ by $\mathbb{R}$, the generalisation is an easy exercise. Write $\phi_n$ for the $n$th Hermite function, so that $\eta_n = \xi_{\phi_n}$ form a sequence of i.i.d. normal random variables. (I assume that you meant to say that the $\xi_\phi$'s are jointly Gaussian at the start of your question.)

Note then that for any fixed $\delta > 0$, $\sum_n |\eta_n|^2/n^{1+\delta} < \infty$ almost surely. Write $A$ for that event so that, for all $\omega \in A$, the series $\bar \xi(\omega) = \sum_{n} \eta_n(\omega) \phi_n$ converges in a weighted Sobolev space of order $-(1+\delta)/2$. These spaces are all included in $\mathcal{E}'$ and the fact that the required identity holds is a simple consequence of the fact that $\langle \phi,\phi_n\rangle \to 0$ faster than any power of $n$ for $\phi \in \mathcal{E}$.

The answer to the second question is no since it would imply that $\sup_n \eta_n(\omega) < \infty$ almost surely.

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  • $\begingroup$ Thank you Sir for your answer, for the second question, fixing $\omega \in \Omega,\overline{\xi}(\omega) \in\mathcal{E} '$ since $\overline{\mathcal{E}}=L^2$ (for the norm $\Vert\cdot \Vert_{L^2}$) then can't we extend $\overline{\xi}(\omega)$ to $L^2$? $\endgroup$
    – mathex
    Commented Mar 6 at 0:35
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    $\begingroup$ @mathex No. It is densely defined linear functional that's unbounded, so cannot be extended to a continuous (and therefore bounded) linear functional. $\endgroup$ Commented Mar 6 at 7:14

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