2
$\begingroup$

Let $X_n$ in $S'$ and $\mu_n$, $\mu$ in $M(S')$. $S'$ is the space of tempered distributions. I'm looking for a reference that says if $< f, X_n >$ converges in distribution to $< f,X>$ for every $f$ in $S$, then $\mu_n$ weakly converges to $\mu$ where $\mu$ is the measure for the random variable $X$. Is anyone aware of such a result?

Also, same question for $X_n$ in $D'$ and $\mu_n$, $\mu$ in $M(D')$.

Thank you in advance for any insight, it's very much appreciated.

$\endgroup$

1 Answer 1

1
$\begingroup$

The only reference that I know which does all that is the thesis of Xavier Fernique. The paper that came out of it is: "Processus linéaires, processus généralisés". Annales de l'institut Fourier, 17 no. 1 (1967), p. 1-92.

However, it is in French and uses a lot of topological vector space machinery. Otherwise for $S'$ one can do that by hand. Use the isomorphism of $S$ with the space of sequences $$ \mathcal{s}=\{(x_n)_{n\ge 0}\in\mathbb{R}^{\mathbb{N}}\ |\ \forall k, ||x||_k<\infty\} $$ where $$ ||x||=\sup_{n\ge 0} \{(n+1)^k|x_n| \}\ . $$ The duality pairing between $y\in\mathcal{s}'$ and $x\in\mathcal{s}$ is given by $$ y(x)=\sum_{n\ge 0} y_n x_n\ . $$ Then follow similar steps for the proof of Levy's Convergence Theorem in finite dimension. The difficult step is in showing that if the characteristic functions of a sequence of probability measures $\mu_m$ in $\mathcal{s}'$ converge pointwise to a function which is continuous at the origin, then the sequence is tight. For this you can follow the discussion of Mitoma's Theorem (Thm 6.13) in the Saint Flour lectures notes by J. B. Walsh "An Introduction to Stochastic Partial Differential Equations" in Lect. Notes in Math. no. 1180. I suppose one can do the same for $D'$ using the Valdivia-Vogt isomorphism of $D$ with $\oplus_{\mathbb{N}}\mathcal{s}$ but I didn't try.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.