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.

  • $\begingroup$ I guess you meant $w(x)=(1+|x|^2)^{1/2}$ ... $\endgroup$ Sep 4, 2015 at 13:12
  • $\begingroup$ ... and $\omega$ instead of $x$ in the integral defining $H^s$ $\endgroup$ Sep 4, 2015 at 13:26
  • $\begingroup$ right, it is now edited. $\endgroup$
    – Goulifet
    Sep 4, 2015 at 17:07

1 Answer 1


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...

  • $\begingroup$ 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 $$\mu, \tau > d/2.$$ $\endgroup$
    – Goulifet
    Sep 4, 2015 at 18:22
  • $\begingroup$ 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. $\endgroup$
    – iolo
    May 20, 2021 at 8:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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