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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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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
zhoraster, there does exist such a measure. This is Corollary 2 in Section III.2.2 of <i>Probability Distributions on Banach Spaces</i> by Vakhania, Tarieladze and Chobanyan. Furthermore, there does exist a Gaussian measure with covariance operator $I$. This is typical in the abstract Wiener space formalism for constructing Gaussian measures on Banach spaces. – Tom LaGatta Nov 15 '10 at 18:30
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 Answer 1

up vote 4 down vote accepted

The closure of $H^*$ in $L^2(H)$ does not contain the constants, or any other nonlinear function, since $L^2$ convergence preserves linearity.

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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

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