# Rellich-Kondrachov compacteness Theorem for the Euclidean space with Gaussian measure

Let $\gamma_n: \mathbb{R}^n\to\mathbb{R}$ be the Gaussian distribution function defined by $$\gamma_n(x):=(2 \pi)^{-\frac{n}{2}} e^{-\frac{|x|^2}{2}}.$$ Let $d\gamma_n$ denote the following measure (weight) $\gamma_n(x) dx$ and consequently $L^2(\mathbb{R}^n,d\gamma_n)$ and $H^1(\mathbb{R}^n,d\gamma_n)$ be the weighted versions of the Sobolev spaces with respect to the measure $d\gamma_n$.

Does the following weighted version of the Rellich-Kondrachov Theorem hold?

$$H^1(\mathbb{R}^n,d\gamma_n) \text{ is } \textbf{compactly} \text{ embedded in } L^2(\mathbb{R}^n,d\gamma_n).$$

Proposition 1. Suppose that $E$ is a real Hilbert space with norm $\Vert -\Vert$ and $K: E\to E$ is a a compact, selfadjoint positive operator. Denote by $R(K)$ the range of $K$. Equip the range with the norm

$$\Vert x\Vert_K:=\Vert K^{-1} x\Vert,\;\;x\in R(K).$$

Then the inclusion $(R(K), \Vert-\Vert_K)\to (E,\Vert-\Vert)$ is compact.

$\DeclareMathOperator{\Tr}{TR}$

Proposition 2. Suppose that $E$ is a real Hilbert space and $K:E\to E$ is a positive selfadjoint operator such that for a sufficiently large positive integer $p$ the operator $K^p$ is trace class. Then the operator $K$ is compact.

Suppose now that $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bZ}{\mathbb{Z}}$ that $E=L^2(\bR^n,\gamma_n)$. Then $E$ admits a natural orthogonal basis consisting of the Hermite polynomials

$$H_\alpha(x_1,\dotsc, x_n)=H_{\alpha_1}(x_1)\dotsc H_{\alpha_n}(x_n),$$

$$\alpha=(\alpha_1,\dotsc,\alpha_n)\in\bZ^n_{\geq 0}.$$

Above, $H_k(x)$ denotes the degree $k$ Hermite polynomial in one variable $x$ defined by

$$H_k(x) =\delta^k 1,$$

where $\delta :C^\infty(\bR)\to C^\infty(\bR)$ is the creation operator $$\delta f(x)=-f'(x)+x f(x),\;\;\forall f\in C^\infty(\bR).$$ We have

$$\Vert H_\alpha\Vert^2 =\alpha!:=(\alpha_1!)\cdots (\alpha_n!),$$

where $\Vert-\Vert$ denotes the norm in $E=L^2(\bR^n,\gamma_n)$.

Denote by $E_m$ the subspace of $E$ spanned by the polynomials $H_\alpha$such that $|\alpha|=m$, where $|\alpha|=\alpha_1+\cdots +\alpha_n$. $\DeclareMathOperator{\Proj}{Proj}$ Note that

$$\dim E_m =\binom{m+n-1}{n-1} = O(m^{n-1})\;\;\mbox{as m\to \infty}.$$

Let $\Proj_m$ denote the orthogonal projection onto $E_m$. The norm $\Vert-\Vert_*$ on $H^1(\bR^n,\gamma_n)$ can be described in terms of the projectors $\Proj_m$. More precisely

$$\Vert f\Vert_*^2=\Vert f\Vert^2+\sum_{j=1}^n \Vert \partial_{x_j}\; f\Vert^2=\sum_{m\geq 0} (1+m)\Vert \Proj_m f\Vert^2 .$$

Now consider the operator

$$K : E\to E,\;\;K f= \sum_{m\geq 0}(1+m)^{-1/2} \Proj_m f.$$

This shows that $H^1(\bR^n,\gamma_n)$ can be identified with the range of $K$ equipped with the norm $\Vert -\Vert_K$ defined as in Proposition 1.

Clearly, for $p$ sufficiently large

$$\sum_{m\geq 0} (m+1)^{-p/2} \dim E_m <\infty.$$

This shows that $K^p$ is trace class for large $p$ and thus, by Proposition 2, $K$ is compact. Now conclude using Proposition 1.

• Interesting! And all the more so because $p$-Schatten-class embeddings are actually much stronger than compact embeddings. Dec 15, 2015 at 23:01
• I think the definition of the norm $\Vert - \Vert_*$ is missing. Maybe after "Then the norm on $H^1(\bR^n,\gamma_n)$ is equivalent with the norm..."? Dec 16, 2015 at 4:11
• @LorenzoCavallina Thanks for pointing this out. Dec 16, 2015 at 10:17