Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers 3

up vote 7 down vote accepted

Nate, I once needed this result, so I proved it in Dirichlet forms with polynomial domain (Math. Japonica 37 (1992) 1015-1024). There may be better proofs out there, but you could start with this paper.

share|improve this answer
    
Hi Byron, I'm interested as well. I couldn't not find the paper using MathSciNet. Can I get a copy? –  Jonas Teuwen Nov 2 '11 at 21:34
    
Thanks Byron! I think my library has it, I'll head there right now. The MathSciNet link is ams.org/mathscinet-getitem?mr=1196376, for reference. –  Nate Eldredge Nov 2 '11 at 21:51
1  
@Jonas I have scanned the paper and added it to my "Publications" page. You want Proposition 1.3. I hope my argument holds up to scrutiny! It is not very profound, just a multi-dimensional version of Nate's argument. –  Byron Schmuland Nov 2 '11 at 22:16
    
@Byron: Thanks. I've got it. –  Jonas Teuwen Nov 2 '11 at 22:16
    
Thanks a lot for that! –  t.b. Nov 2 '11 at 22:28
show 1 more comment

Byron's paper, which he linked in his (accepted) answer, has a proof in a more general setting (where the Gaussian measure can be replaced by any measure with exponentially decaying tails). Here is a specialization of it to the Gaussian case, which I wrote up to include in some lecture notes. I guess I was on the right track, the trick was to differentiate more times.

$\newcommand{\R}{\mathbb{R}}$ For continuous $\psi : \R^k \to \R$, let \begin{equation*} I_i \psi(x_1, \dots, x_k) = \int_0^{x_i} \psi(x_1, \dots, x_{i-1}, y, x_{i+1}, \dots, x_k)dy. \end{equation*} By Fubini's theorem, all operators $I_i, 1 \le i \le k$ commute. If $\psi \in L^2(\mu)$ is continuous, then $I_i \psi$ is also continuous, and $\partial_i I_i \psi = \psi$. Moreover, $$\begin{align*} \int_{0}^\infty |I_i \psi (x_1, \dots, x_k)|^2 e^{-x_i^2/2} dx_i &= \int_{0}^\infty \big\lvert\int_0^{x_i} \psi(\dots, y,\dots)\,dy\big\rvert^2 e^{-x_i^2/2} dx_i \\ &\le \int_0^\infty \int_0^{x_i} |\psi(\dots, y, \dots)|^2 \,dy x_i e^{-x_i^2/2}\,dx_i && \text{Cauchy--Schwarz} \\ &= \int_0^\infty |\psi(\dots, x_i, \dots)|^2 e^{-x_i^2/2} dx_i \end{align*}$$ where in the last line we integrated by parts. We can make the same argument for the integral from $-\infty$ to $0$, adjusting signs as needed, so we have \begin{equation*} \int_\R |I_i \psi(x)|^2 e^{-x_i^2/2} dx_i \le \int_\R |\psi_i(x)|^2 e^{-x_i^2/2} dx_i. \end{equation*} Integrating out the remaining $x_j$ with respect to $e^{-x_j^2/2}$ shows $$ ||{I_i \psi}||_{L^2(\mu)}^2 \le ||{\psi}||_{L^2(\mu)}^2, $$ i.e. $I_i$ is a contraction on $L^2(\mu)$.

Now for $\phi \in C_c^\infty(\R^k)$, we can approximate $\partial_1 \dots \partial_k \phi$ in $L^2(\mu)$ norm by polynomials $q_n$. If we let $p_n = I_1 \dots I_k q_n$, then $p_n$ is again a polynomial, and $p_n \to I_1 \dots I_k \partial_1 \dots \partial_k \phi = \phi$ in $L^2(\mu)$. Moreover, $\partial_i p_n = I_1 \dots I_{i-1} I_{i+1} \dots I_k q_n \to I_1 \dots I_{i-1} I_{i+1} \dots I_k \partial_1 \dots \partial_k \phi = \partial_i \phi$ in $L^2(\mu)$ also.

share|improve this answer
add comment

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...

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.