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Let $H$ be an infinite-dimensional, separable Hilbert space, and let $\gamma$ be a Radon probability measure on $H$ with mean zero and covariance operator the identity $I$.

Let $H^*$ denote the space of continuous linear functionals on $H$. By the Riesz representation theorem, $$H^* = \{ \langle k, \cdot \rangle : k \in H \}.$$ By the assumption that $\gamma$ has covariance operator $I$, for all $k \in H$, $$\int_H |\langle k, h \rangle|^2 \, \mathrm{d}\gamma(h) = \langle k, k \rangle < \infty,$$ so $H^*$ is contained in the separable Hilbert space $L^2(H)$.

My question: Is $H^*$ dense in $L^2(H)$?

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6  
Certainly not when H is finite dimensional. – Mark Meckes Nov 15 '10 at 17:41
    
Does there exist such a measure for an infinite-dimensional $H$? There is surely no such Gaussian measure, for example. – zhoraster Nov 15 '10 at 18:15
    
Good point, Mark. I've edited the question to include the assumption that $H$ is infinite-dimensional. – Tom LaGatta Nov 15 '10 at 18:29
2  
In general $|\langle \cdot, h\rangle|$ will not be in the closure of $H^*$. Maybe you want to ask whether the polynomials in elements of $H^*$ are dense? – Bill Johnson Nov 15 '10 at 18:48
1  
@Tom LaGatta: No, this ain't correct. A covariance operator of a Gaussian measure must be compact (even trace class). – zhoraster Nov 15 '10 at 23:32
up vote 6 down vote accepted

(Rewritten to give an answer more useful to future visitors.)

First of all, as noted in comments, there is no (countably additive) Gaussian measure on $H$ with covariance operator the identity.

However, if we take $\gamma$ to be some other Gaussian measure, the answer is no, $H^*$ is not dense in $L^2(H, \gamma)$. One way to see this is that every $f \in H^*$, considered as a random variable on $(H, \gamma)$, has a centered Gaussian distribution (in which we include the "degenerate" Gaussian distribution which is the constant 0). In particular $f$ has mean zero, and the mean-zero random variables are a proper closed subspace of $L^2(\gamma)$. (So for instance, the nonzero constants are not in the closure of $H^*$.)

Indeed, $L^2$ limit of centered Gaussian random variables is again centered Gaussian (this actually holds if you replace "$L^2$" by "in distribution"). Thus every random variable in the $L^2$ closure of $H^*$ is centered Gaussian, and hence non-Gaussian $L^2$ functions on $H$ are not in the closure of $H^*$ either.

It is true that if you allow polynomials $F(x) = p(f_1(x), \dots, f_n(x))$ where $f_1, \dots, f_n \in H^*$, then all such functions $F$ are in $L^2(H,\gamma)$, and they form a dense subspace. Indeed, if you let $p$ range over all Hermite polynomials of degree $n$, then the closed span of all corresponding $F$ gives you the space $\mathcal{H}_n$ which is the $n$th Wiener chaos, and we have an orthogonal decomposition $L^2(H, \gamma) = \bigoplus \mathcal{H}_n$. These spaces are also the eigenspaces of the Ornstein-Uhlenbeck operator $N$ (aka number operator).

Back to the "measure" with covariance operator $I$: we can consider such a "measure" as a finitely additive measure on the cylinder sets of $H$. I suppose it might be possible to study an $L^2$ space with respect to this finitely additive measure. I don't know much about such spaces, but I would guess that a similar argument would show that $H^*$ is still not dense.

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So, for example, might a constant agree a.e. with a linear functional? – Gerald Edgar Nov 15 '10 at 20:14
    
@GeraldEdgar As I have now explained more fully, no. (Except for the constant 0, of course.) – Nate Eldredge Jan 14 at 15:59

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