Given a tempered distribution $s \in \mathcal{S}'(\mathbb{R}^{2d})$, define the Weyl pseudodifferential operator of symbol $s$ as the mapping $\mathcal{S}(\mathbb{R}^{d}) \rightarrow \mathcal{S}'(\mathbb{R}^{d})$ determined by $$\left(\textrm{Op}^{\textrm{w}}(s)u\right)(x) := \frac{1}{(2\pi)^{d}} \int_{\mathbb{R}^{d}} \int_{\mathbb{R}^{d}} s\left(\frac{x+x'}{2},\xi\right)e^{i(x-x')\cdot \xi} u(x') dx'd\xi, \quad x \in \mathbb{R}^{d},\ u \in \mathcal{S}(\mathbb{R}^{d}).$$

I understand that if the symbol $s$ belongs to $L^{1}(\mathbb{R}^{2d})$, then the above operator extends to a compact operator in $L^2(\mathbb{R}^{d})$. However, I haven't been able to prove it or to find the proof in any reference. Could you help me prove this statement?


For a Hörmander metric $g$ and a weight function $m$ on $\mathbb R^{2d}$, let $S(m,g)$ denote the corresponding class of pseudodifferential symbols $s=s(x,\xi)$, see Chapter XVIII of L. Hörmander, The Analysis of Linear Partial Differential Operators III. We additionally assume that $g$ satisfies the strong uncertainty principle (e.g., $\displaystyle g=|dx|^2 + \frac{|d\xi|^2}{1+|\xi|^2}$ is admissible).

In a paper of E. Buzano and F. Nicola , you'll find the following result: for $1\leq p< \infty$, $$ \text{$\operatorname{Op}^w\!S(m,g)$ is included in the Schatten class $S_p(L^2(\mathbb R^d))$ if and only if $m\in L^p(\mathbb R^{2d})$.} $$

It follows that if $s\in L^1(\mathbb R^{2d})$, then $\operatorname{Op}^w(s)$ is trace class in $L^2(\mathbb R^d)$. This latter result (including the formula $\displaystyle \operatorname{Tr}\operatorname{Op}^w(s)=(2\pi)^{-d}\int_{\mathbb R^{2d}}s(x,\xi)\,dxd\xi$ ) is already contained in Section 19.3. of L. Hörmander, op.cit.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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