# Given a $C_c^∞(G)$-valued random variable, is $C_c^∞(G)∋φ↦\text E[\langle\xi,φ\rangle]$ an element of the dual space of $C_c^∞(G)$?

Let

• $G\subseteq\mathbb R^d$ and $$\mathcal D:=C_c^\infty(G)$$ be equipped with some topology $\tau$
• $\mathcal D'$ be the dual space of $\mathcal D$ and $\langle\;\cdot\;,\;\cdot\;\rangle$ denote the dual pairing between $\mathcal D'$ and $\mathcal D$
• $\xi$ be a $\mathcal D'$-valued random variable

How can we show, that $$\operatorname E[\xi]:\mathcal D\to\mathbb R\;,\;\;\;\varphi\mapsto\operatorname E\left[\langle\xi,\varphi\rangle\right]\tag 1$$ is an element of $\mathcal D'$?

Clearly, the mapping $(1)$ is linear. But how can we show, that it is continuous and that $\operatorname E\left[\langle\xi,\varphi\rangle\right]$ is finite?

Maybe the answer depends on the choice of $\tau$. Maybe $\tau$ must be chosen in a way such that each linear and continuous mapping $\mathcal D\to\mathbb R$ is bounded.

In the Wikipedia article about distributions they equip $\mathcal D$ with a locally convex topology. Do we need this specific topology? Or can we use an arbitrary topology?

There is a "standard" locally convex topology on $\mathcal{D}$ generated by semi-norms (see the answer to your other question), and pretty much every user of generalized stochastic processes uses it (e.g. Gelfand & Vilenkin). It's always better to choose your topology before going on to work with continuity, dual space, etc. Local convexity is about the weakest assumption you would want to make, but unless you're driving to prove the most general result possible (not likely), just assume the standard topology. That being said, bounded linear functionals (i.e. $T:\mathcal{D}\rightarrow\Bbb{R}$) are always continuous; see Rudin (FA) Thm. 1.18.

As for showing that the mean of $\xi$ is an element of $\mathcal{D}^\prime$, you're on the right track conceptually, but you really need to be more specific about the process $\xi$ to make concrete statements. Typically one assumes things like "$\xi$ is a second-order process" which means that it has a well-defined mean $E[\xi]$ and correlation functional $B(\varphi,\psi) = E[\xi(\varphi)\xi(\psi)]$.

To explain a bit more, recall that for each fixed $\varphi\in\mathcal{D}$, $\langle \xi,\varphi\rangle$ is an $\Bbb{R}$-valued random variable with distribution $p_\varphi(x)$. This distribution might or might not have a well-defined expectation, so $E[\langle \xi,\varphi\rangle]$ might or might not be finite - this has everything to do with how the process $\xi$ behaves. For instance, if $\xi$ is Gaussian (a common but not necessary assumption), then $p_\varphi$ is Gaussian and hence has a finite first moment, etc.

As for continuity, I would prove boundedness instead. That is, show that $\varphi\mapsto E[\langle \xi,\varphi\rangle]$ is a bounded map from $\mathcal{D}$ to $\Bbb{R}$. Again, this requires assumptions on the process $\xi$!

Just out of curiosity, what source are you working from? I would highly recommend reading through Gelfand and Vilenkin "Generalized Functions Vol. 4".

• Thank you very much for your answer. I've read a paper where the authors define $\xi$ as I did above and use $\operatorname E[\xi]\in\mathcal D'$ without any further comment. This led me to the assumption that $\operatorname E[\xi]\in\mathcal D'$ would hold in general. I thought the boundedness of $\xi$ would immediately yield that $\operatorname E\left[\xi(\varphi)\right]$ is finite. But that's clearly wrong, since $\xi$ is in fact a function of $(\omega,\varphi)\in\Omega\times\mathcal D$ and while the boundedness is wrt $\varphi$ the expectation is taken wrt $\omega$. Commented Jan 27, 2016 at 18:06
• Let's conclude: 1. We can't assume $\operatorname E\left[\xi(\varphi)\right]$ without further assumptions. 2. In fact, we need to assume that the stochastic process $\left(\langle\xi,\varphi\rangle\right)_{\varphi\in\mathcal D}$ is $\mathcal L(\operatorname P)$-integrable, i.e. $\langle\xi,\varphi\rangle\in\mathcal L(\operatorname P)$ for all $\varphi\in\mathcal D$. Commented Jan 27, 2016 at 18:06
• Which assumption on $\xi$ do we need in order to show boundedness? Commented Jan 28, 2016 at 11:23
• @0xbadf00d I think the general assumption is exactly that $E[\xi]$ exists and is a bounded functional. At least, this is what I'm seeing in Gelfand & Vilenkin and a couple books by Zabczyk. Most "standard" stochastic process (Gaussian, Poisson, etc) will satisfy this assumption. Something more exotic? I'm not sure. Commented Jan 28, 2016 at 17:28