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?


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



Rellich-Kondrachov compacteness Theorem for the Euclidean space with Gaussian measure
I'll start with several known facts.



> 
> **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**.


