Proposition 1.7 (Properly supported smoothing operators). $L^{-\infty}=$ smoothing operators. Given $A \in L^{-\infty}$ with properly supported amplitude $a \in S^{-\infty}\left(\Omega \times \Omega \times \mathbb{R}^n\right)$ $$ A u(x)=\int_{\mathbb{R}^n} \int_{\Omega} e^{i\langle x-y, \xi\rangle} a(x, y, \xi) u(y) d y d \xi=\int_{\Omega}(\underbrace{\int_{\mathbb{R}^n} e^{i\langle x-y, \xi\rangle} a(x, y, \xi) d \xi}_{k(x, y)}) u(y) d y $$ Thus $$ A u(x)=\int_{\Omega} k(x, y) d y \quad \text { with } k \in C_{\text {proper }}^{\infty}(\Omega \times \Omega) $$ What about the converse? Given $k \in C_{\text {proper }}^{\infty}(\Omega \times \Omega)$ Is $A$ as above in $L^{-\infty}(\Omega)$ (1) Fix $\chi \in C_c^{\infty}\left(\mathbb{R}^n\right), \int_{\mathbb{R}^n} \chi=1, u \in \mathcal{E}(\Omega), x \in B \subset \subset \Omega$ (compact neighbourhood) $$ \begin{gathered} L:=\pi_2\left(\pi_1^{-1}(B) \cap \operatorname{supp} k\right) \quad \text { compact! }\\ \end{gathered} $$
$$ Au(x)=\int_L k(x, y) u(y) \int_{\mathbb{R}^n} \chi(\xi) e^{i\langle x-y, \xi\rangle} e^{-i\langle x-y, \xi\rangle} d \xi d y\\ =\int_{\mathbb{R}^n} \int_L e^{i\langle x-y, \xi\rangle} \underbrace{\left\{k(x, y) e^{-i\langle x-y, \xi\rangle} \chi(\xi)\right\}}_{a(x, y, \xi) \in S^{-\infty}\left(\Omega \times \Omega \times \mathbb{R}^n\right)} u(y) d y d \xi $$
Im going through this proof of that an operator whose kernel is properly supported is a smoothing operator. Here properly supported means that the projections to each component are proper maps.
Im not quite sure why we can restrict the area of integration to L, where a priori it should be over all of omega.
