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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?

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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.

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