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Let $\Omega\subset {\mathbb R}^n$ be a domain. Let $a(x, \xi, y)$ be an "amplitude", i.e. a smooth function on $\Omega\times{\mathbb R}^n\times\Omega$ so that there is $m$, and for any compact set $B\subset\Omega$, one has $$ \sup_{x, y\in B}|D_x^\beta D_y^\gamma D_\xi^\alpha a(x, \xi, y)|\leq C_{\alpha, \beta,\gamma, B}(1+|\xi|)^{m-|\alpha|}. $$ Then one can define an operator $P_a$ on $C_0^\infty(\Omega)$: $$ P_au(x)=\int_{{\mathbb R}^n}\int_\Omega e^{(x-y)\xi i}a(x, \xi, y)u(y)dy d\xi. $$ Then in equation (8.23) Folland proceed to say that the Schwartz kernel of $P_a$ is $$ K(x, y)=a_2^\vee(x, x-y, y). $$
My question is, how to make sense of this? As I understand, $a_2^\vee$ is a distribution on $\Omega\times{\mathbb R}^n\times \Omega$ defined as $$\begin{aligned} \langle a_2^\vee, \varphi\rangle =&\int_\Omega\int_{{\mathbb R}^n}\int_\Omega a(x,\xi, y) \varphi_2^\vee(x, \xi, y)dxd\xi dy\\ =&\int_\Omega\int_{{\mathbb R}^n}\int_\Omega a(x,\xi, y) \Big[\int_{{\mathbb R}^n} e^{\eta\xi i}\varphi(x, \eta, y)d\eta\Big] dxd\xi dy. \end{aligned}$$ Thus the kernel $K$ seems to be restricting the above $a_2^\vee$ to the hyperplane $\xi=x-y$ in $\Omega\times{\mathbb R}^n\times\Omega$. So how to justify this? Is this way of writing $K$ is just a convenient notation?