# When is the identity Hilbert-Schmidt between weighted Sobolev spaces?

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.

• I guess you meant $w(x)=(1+|x|^2)^{1/2}$ ... – Jean Duchon Sep 4 '15 at 13:12
• ... and $\omega$ instead of $x$ in the integral defining $H^s$ – Jean Duchon Sep 4 '15 at 13:26
• right, it is now edited. – Goulifet Sep 4 '15 at 17:07

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...
• This is enough to answer my question. Thanks a lot. Just to finish the argument, the integral you gave is equal to $$\lVert w^{-\mu} \rVert_2 \lVert \widehat{w^{-\tau}} \rVert_2 = \lVert w^{-\mu} \rVert_2 \lVert w^{-\tau} \rVert_2$$ thanks to Parceval, and therefore, the answer is $$\mu, \tau > d/2.$$ – Goulifet Sep 4 '15 at 18:22
• Do you happen to have a reference for $I_{s,\mu}$ being an isometry as well as for the Fourier transform of the multipliers? I am struggling to find any. – iolo May 20 at 8:56