TITLE: Are polynomials dense in Gaussian Sobolev space?
QUESTION [18 upvotes]: Let $\mu$ be standard Gaussian measure on $\mathbb{R}^n$, i.e. $d\mu = (2\pi)^{-n/2} e^{-|x|^2/2} dx$, and define the Gaussian Sobolev space $H^1(\mu)$ to be the completion of $C_c^\infty(\mathbb{R}^n)$ under the inner product
$$\langle f,g \rangle_{H^1(\mu)} := \int f g\, d\mu + \int \nabla f \cdot \nabla g\, d\mu.$$
It is easy to see that polynomials are in $H^1(\mu)$.  Do they form a dense set?
I am quite sure the answer must be yes, but can't find or construct a proof in general.  I do have a proof for $n=1$, which I can post if anyone wants.  It may be useful to know that the polynomials are dense in $L^2(\mu)$.
Edit: Here is a proof for $n=1$.
It is sufficient to show that any $f \in C^\infty_c(\mathbb{R})$ can be approximated by polynomials.  We know polynomials are dense in $L^2(\mu)$, so choose a sequence of polynomials $q_n \to f'$ in $L^2(\mu)$.  Set $p_n(x) = \int_0^x q_n(y)\,dy + f(0)$; $p_n$ is also a polynomial.  By construction we have $p_n' \to f'$ in $L^2(\mu)$; it remains to show $p_n \to f$ in $L^2(\mu)$.  Now we have
$$ \begin{align*} \int_0^\infty |p_n(x) - f(x)|^2 e^{-x^2/2} dx &= \int_0^\infty \left(\int_0^x (q_n(y) - f'(y)) dy \right)^2 e^{-x^2/2} dx \\
&\le \int_0^\infty \int_0^x (q_n(y) - f'(y))^2\,dy \,x e^{-x^2/2} dx \\
&= \int_0^\infty (q_n(x) - f'(x))^2 e^{-x^2/2} dx \to 0 \end{align*}$$
where we used Cauchy-Schwarz in the second line and integration by parts in the third.  The $\int_{-\infty}^0$ term can be handled the same with appropriate minus signs.
The problem with $n > 1$ is I don't see how to use the fundamental theorem of calculus in the same way.

REPLY [8 votes]: I think I have a different proof. Let $\gamma$ be the Gaussian measure, that is, $\gamma$ is given by the Radon-Nikodym density,
$$\mathrm{d}\gamma(x) = \frac{\mathrm{e}^{-x^2}}{\sqrt{\pi}} \mathrm{d}x.$$
Also, consider the Ornstein-Uhlenbeck operator given as,
$$L := -\frac12 \Delta + x \cdot \nabla.$$
We can verify that for $u$ and $v$ in $C_{\mathrm{c}}^\infty(\mathbf R^d)$ we have the symmetricity
$$\int_{\mathbf{R}^d} u L v \, \gamma(\mathrm{d}x) = \int_{\mathbf{R}^d} \nabla u \cdot \nabla v \, \gamma(\textrm{d}x).$$
The nice thing about this is that the Hermite polynomials are an orthogonal basis for the Gaussian Hilbert space, that is, $L^2(\gamma)$. These can be define using the Rodrigues' formula, that is,
$$H_n(x) = (-1)^n \mathrm{e}^{x^2} \partial_x^n \mathrm{e}^{-x^2}.$$
Furthermore, we have,
$$L H_n = n H_n.$$
Also, we have,
$$H_n' = 2n H_{n - 1}.$$
Proving that the polynomials form a basis for $L^2(\gamma)$ is not hard, just consider the entire function
$$F(z) = \int_{-\infty}^\infty \mathrm e^{zx - x^2} \, \frac{\mathrm{d}x}{\sqrt\pi}.$$
Hence, so are the Hermite polynomials as they are linear combinations of the monomials $(x \mapsto x^n)_n$. Higher-order Hermite polynomials are the canonical tensor extensions. So, given an $f$ in $L^2$ we have that $f$ is given in the form
$$f = \lim_N f_N = \lim_N \sum_{n = 0}^N a_n \frac{a_n}{\sqrt{n! 2^n}}.$$
And $f_N$ gives the limit of functions that converges to $f$.
Also, recall the orthogonality (after scaling),
$$\int_{\mathbf{R}^d} h_n h_m \, \gamma(\mathrm{d}x) = \delta_{nm}.$$
where
$$h_n = \frac{H_n}{\sqrt{n! 2^n}}.$$
We can rewrite the inner product on the Sobolev space as
$$\langle u, v \rangle_{H^1} = \langle u, (L + 1) v \rangle_{L^2(\gamma)}.$$
We only have to care about the bilinear form
$$\mathcal E(u, v) = \int_{\mathbf{R}^d} u L v \, \gamma(\mathrm{d}x).$$
Or so, picking $u = v = f_N - f$ we have after noting that
$$g_N := f - f_N = \sum_{n = N + 1}^\infty a_n \frac{H_n}{\sqrt{n! 2^n}},$$
$$
\begin{align}
\mathcal E(u, v) &= \int_{\mathbf{R}^d} f_N L f_N \, \gamma(\mathrm{d}x)\\
&\quad + \int_{\mathbf{R}^d} f L f \, \gamma(\mathrm{d}x)\\
&\quad - 2\int_{\mathbf{R}^d} f_N L f \, \gamma(\mathrm{d}x)\\
&= \sum_{n = 0}^\infty n \frac{|a_n|^2}{n! 2^n} - \sum_{n = 0}^N n \frac{|a_n|^2}{n! 2^n}\\
&= \sum_{n = N + 1}^\infty n \frac{|a_n|^2}{n! 2^n}.
\end{align}.
$$
And as $\sum |a_n|^2$ converges due to the $L^2$ density, so should this.
I hope I did not make any mistakes, just occurred to me while I was biking...