Set $w(x) = (1 + |x|^2)^{1/2}$ with $|\cdot|$ the Euclidian norm on $\mathbb{R}^n$. For $s,\mu \in \mathbb{R}$, we define the Sobolev space
$$H\_2^{s}(\mathbb{R}^n) = \left\{f : \lVert f \rVert\_{s} := \left( \int\_{\mathbb{R}^n} |\widehat{f}(\omega)|^2 w(\omega)^{2s} \mathrm{d} \omega \right)^{1/2} < \infty \right\}$$
with $\widehat{f}$ the Fourier transform of $f$, 
and the weighted Sobolev space 
$$H\_2^{s,\mu}(\mathbb{R}^n) = \left\{f : \lVert f \rVert\_{s,\mu} := \lVert w^\mu f \rVert\_{s} < \infty \right\}.$$


I am looking for sufficient conditions on $(s,\mu)$ such that the identity operator 
$$ \mathrm{I} : H\_2^{s,\mu}(\mathbb{R}^d) \rightarrow L\_2(\mathbb{R}^d) $$
is a Hilbert-Schmidt operator. 


First of all, we should have $s,u \geq 0$, such that $H\_2^{s,\mu}(\mathbb{R}^d) \subset L\_2(\mathbb{R}^d)$. If $(e\_k^{s,\mu})$ is an orthonormal basis of $H\_2^{s,\mu}(\mathbb{R}^d)$, then the identity is Hilbert-Schmidt if and only if
$$ \sum\_{k} \lVert e\_k^{s,\mu} \rVert < \infty$$
where $\lVert \cdot \rVert$ is the $L\_2$-norm. My problem is that I don't know a nice orthonormal bases on weighted Sobolev spaces such that the question becomes easy.


Any help would be appreciate.



When is the identity Hilbert-Schmidt between weighted Sobolev spaces?
There is an explicit operator that maps $L^2$ isometrically onto $H^{s,\mu}\_2$ :$$I\_{s,\mu}u(x)=(1+|x|^2)^{-\mu/2}(I-\Delta)^{-s/2}u(x)$$The (inverse) Fourier transform $k\_s(x)$ of $(1+|\omega|^2)^{-s/2}$ is also well documented (Bessel functions etc), and then $I\_{s,\mu}u(x)=\int (1+|x|^2)^{-\mu/2} k\_s(x-y)u(y)\ dy$ is Hilbert-Schmidt $L^2\to L^2$ iff its kernel is in $L^2(dx\ dy)$, i.e. iff $$\int\int (1+|x|^2)^{-\mu}k\_s(x-y)^2\ dx\ dy<\infty$$So, you don't really *need* an orthonormal basis...


