# Density of linear functionals in $L^2$

Let $X$ be a locally convex topological linear space, and let $\mathbb P$ be a probability measure on $X$. Suppose that $\operatorname{var}(\varphi) < \infty$ for all continuous linear functionals $\varphi \in X^*$. By Theorem 3.ii of [Vakhania & Tarieladze 1978], the embedding $i : X^* \to L^2(X,\mathbb P)$ is continuous.

Let $C \subseteq L^2(X,\mathbb P)$ denote the subspace of almost-surely constant functions.

Is the direct sum $iX^* \oplus C$ dense in $L^2(X,\mathbb P)$?

• For instance if $X=\mathbb{R}$ and $\mathbb{P}$ is the Lebesgue measure on $[0,1]$, that is certainly not dense. Or am I missing something? – Pietro Majer Feb 4 '14 at 21:38
• The term "almost surely" sounds strange, as an outsider. "Surely" itself seems to imply some sort of "up to a null set", what would the almost add here, in that case? – Asaf Karagila Feb 4 '14 at 21:46
• @Asaf: a function $c : X \to \mathbb R$ is almost-surely constant if $\{ x : c(x) = \mathbb E[c] \}$ is a set of full measure. It is surely constant if that set equals $X$. The use of "almost surely" is common in the probability community. en.wikipedia.org/wiki/Almost_surely – Tom LaGatta Feb 4 '14 at 21:52
• Ah. That makes a lot of sense! (Assuming that $\Bbb E$ is the expected value.) Thank you, Tom! And thank you for the link. – Asaf Karagila Feb 4 '14 at 21:54
• Replacing my previous comment: if $X$ is reasonably nice (say separable Banach, or I think Frechet), and $\mathbb{P}$ is Gaussian, the set of all polynomials $p(f_1, \dots, f_k)$, where $f_1, \dots, f_k \in X^*$, is dense in $L^2(X,\mathbb{P})$. If you use Hermite polynomials, you can get an orthogonal decomposition $L^2 = \bigoplus_{n=0}^\infty \mathcal{H}_n$ which is the Wiener chaos decomposition. In particular $\mathcal{H}_0 = C$, and $\mathcal{H}_1$ is the closure of your $i X^*$. – Nate Eldredge Feb 5 '14 at 17:38

As requested, I'll elaborate here on the Gaussian case.

Suppose for instance that $X$ is a separable Banach space, and $\mathbb{P}$ is a Gaussian measure. Then each $f \in X^*$, when considered as a random variable on $(X, \mathbb{P})$ has a Gaussian distribution. Let $H$ denote the $L^2$-closure of $i X^*$. An $L^2$ limit of Gaussian random variables remains Gaussian, so every element of $H$ (and hence also $C \oplus H$) is Gaussian. But $L^2(X,\mathbb{P})$ clearly contains lots of non-Gaussian random variables (even when $X = \mathbb{R}^n$); as just one example, those of the form $f^2$ have a $\chi^2$ distribution.

However, since each $f \in X^*$ is Gaussian and hence has all moments, then for any polynomial $p$ in any number of variables $n$, we have that $p(f_1, \dots, f_n) \in L^2$. Then it can be shown by a Stone-Weierstrass type argument that the set of all such polynomials is indeed dense in $L^2$ (it's conveniently done with a functional version of the monotone class or $\pi$-$\lambda$ theorem). A key point is that because of the separability of $X$, one can show that $X^*$ generates the Borel $\sigma$-algebra of $X$. (Note that it is not necessary that $X^*$ be separable.)

This can be done in a more orderly way, too. We can find a countable sequence $e_i \in X^*$ which are $L^2$-orthonormal and span an $L^2$-dense subspace of $X^*$. Let $H_m(s)$ be the $m$th Hermite polynomial (e.g. $H_0(s) = 1$, $H_1(s) = s$, $H_2(s) = s^2 - 1$, etc) and consider the polynomials of the form $$F_{a_1, \dots, a_k}(x) = H_{a_1}(e_1(x)) \cdots H_{a_k}(e_k(x)).$$ These functions form an orthogonal basis of $L^2$. Moreover, if we let $\mathcal{H}_n$ be the closed linear span of $\{F_{a_1, \dots, a_k} : a_1 + \dots + a_k = n\}$, then we get an orthogonal decomposition $L^2 = \bigoplus_{n=0}^\infty \mathcal{H}_n$. This is the so-called Wiener chaos decomposition. Note that $\mathcal{H}_n$ is naturally isomorphic to the $n$-fold symmetric tensor of $H$ with itself, and so we get an isomorphism of $L^2$ with the bosonic Fock space over $H$.

In particular, $\mathcal{H}_0 = C$ and $\mathcal{H}_1 = H$. So you were on the right track, but you stopped too soon :)

I wrote a little bit about this in these lecture notes, which also has a few references.

• Nate, this is a great answer! Thanks much. What happens when $\mathbb P$ is non-Gaussian? Suppose that we only have a finite-variance assumption on continuous linear functionals: their higher moments may not exist. How big of a subspace of $L^2$ may we approximate by considering polynomials of linear functionals? – Tom LaGatta Feb 7 '14 at 16:23
• @TomLaGatta: This argument should still work as long as continuous linear functionals have moments of all orders. If higher moments don't exist, then you can't use any polynomials of degree higher than, say, 3, and you can't expect to get very much of $L^2$. For example, take $X = \mathbb{R}$ and $\mathbb{P}$ a measure with infinite third moment. Then the "polynomials in $X^*$" intersected with $L^2$ has dimension 2, whereas $L^2$ itself is still infinite dimensional. – Nate Eldredge Feb 7 '14 at 17:27
• @TomLaGatta: On the other hand, there are other approximating functions. For example, say we use trigonometric polynomials in $X^*$. These are bounded functions, so certainly in $L^2$, and by a similar argument they will be dense. – Nate Eldredge Feb 7 '14 at 17:29
• Thanks so much, Nate. This completely answers my question. – Tom LaGatta Feb 7 '14 at 17:58