Let $\mathcal{S}(\mathbb{R})$ be the space of smooth and rapidly decaying functions and $\mathcal{S}'(\mathbb{R})$ its dual, the space of tempered distributions. Let $\mathscr{P}$ be a probability measure onto $\mathcal{S}'(\mathbb{R})$. Its characteristic functional is defined, for $f\in \mathcal{S}(\mathbb{R})$, by $$\widehat{\mathscr{P}}(f) = \int_{\mathcal{S}'(\mathbb{R})} \mathrm{e}^{\mathrm{i} \langle u ,f\rangle} \mathrm{d}\mathscr{P} (u).$$

There are strong connexions between the support of the measure $\mathscr{P}$ and the continuity of the characteristic functional of $\mathscr{P}$. For instance, if $\mathscr{P}$ is the measure associated with the Gaussian white noise with variance $1$, we have $\widehat{\mathscr{P}}(f) = \exp(-\frac{1}{2}\lVert f\rVert_2^2)$, that is continuous over $L_2(\mathbb{R})$. Applying for instance Theorem A.2 of [1], we deduce that $$\mathscr{P}\left( W_2^{-1/2-\epsilon}(\mathbb{R};(1+\lvert \cdot \rvert^2)^{(-1/2-\epsilon)/2})\right) =1$$ for every $\epsilon>0$, with $W_2^{s}(\mathbb{R};(1+\lvert \cdot \rvert^2)^{r/2})$ the weighted Sobolev space with regularity $s$ and weight function $(1+\lvert \cdot \rvert^2)^{r/2}$. This means that the Gaussian white noise is located in $W_2^{-1/2-\epsilon}(\mathbb{R};(1+\lvert \cdot \rvert^2)^{(-1/2-\epsilon)/2})$. This requires only to remark that the identity is an Hilbert-Schmidt operator from $W_2^{1/2+\epsilon}(\mathbb{R};(1+\lvert \cdot \rvert^2)^{(1/2+\epsilon)/2})$ to $L_2(\mathbb{R})$.

Question: Is there a similar result if we know that the characteristic functional $\widehat{\mathscr{P}}$ of $\mathscr{P}$ is continuous over $L^p$ for $1 \leq p < 2$?

NB. This is motivated by the fact that the characteristic functional of a S$\alpha$S white noise, of the form $\widehat{\mathscr{P}}(f) = \exp ( - \gamma^\alpha \lVert f \rVert^{\alpha}_{\alpha} )$, is continuous over $L^{\alpha}(\mathbb{R})$.

[1] T. Hida and Si Si, An Innovation Approach to Random Fields


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.