[proofplan] The negative order hypothesis means that $A$ gains $-m>0$ Sobolev derivatives locally. After multiplying the input by $\psi$, the pseudodifferential mapping theorem gives a bounded map from $H^s(U)$ into $H^{s-m}_{\mathrm{loc}}(U)$. Multiplication by $\chi$ localizes the output to a fixed compact subset of $U$, so the problem reduces to the compact embedding $H^{s-m}(V) \hookrightarrow H^s(V)$ on a bounded precompact smooth patch $V \Subset U$. Since $s-m>s$, Rellich compactness gives the desired compactness after composing the bounded Sobolev-gain map with the compact local inclusion. [/proofplan] [step:Choose a bounded smooth patch containing the output support] Let $K := \operatorname{supp}\chi$. Since $\chi \in C_c^\infty(U)$, the set $K$ is compact and $K \subset U$. Choose an [open set](/page/Open%20Set) $V \subset \mathbb{R}^n$ with smooth boundary such that $K \subset V \subset \overline{V} \subset U$ and $\overline{V}$ is compact. We write $V \Subset U$ for this compact containment. Define the multiplication operator \begin{align*} M_\chi : H^{s-m}(V) \to H^{s-m}(V) \end{align*} by $M_\chi v = \chi|_V\, v$. Since $\chi|_V \in C_c^\infty(V)$, multiplication by $\chi$ is bounded on every [Sobolev space](/page/Sobolev%20Space) $H^r(V)$ with $r \in \mathbb{R}$. Because $\operatorname{supp}(\chi v) \subset K \subset V$, the function $\chi v$ may be regarded as an element of $H^s(U)$ by extending it by zero outside $V$. Thus there is a bounded operator \begin{align*} E_\chi : H^s(V) \to H^s(U) \end{align*} given by $E_\chi w = w$ on $V$ and $E_\chi w = 0$ on $U \setminus V$ whenever $w$ is supported in $K$. [/step] [step:Use the pseudodifferential mapping theorem to gain Sobolev regularity locally] Define \begin{align*} B : H^s(U) \to H^{s-m}(V) \end{align*} by \begin{align*} B u := \left. A(\psi u) \right|_V. \end{align*} The input localization map $u \mapsto \psi u$ is bounded from $H^s(U)$ to the compactly supported Sobolev distributions in $U$. Since $A \in \Psi^m(U)$, the local Sobolev mapping theorem for pseudodifferential operators gives $A(\psi u) \in H^{s-m}_{\mathrm{loc}}(U)$ and, for the fixed precompact set $V \Subset U$, a constant $C_1 = C_1(A,\psi,V,s)>0$ such that $\|B u\|_{H^{s-m}(V)} \leq C_1 \|u\|_{H^s(U)}$ for all $u \in H^s(U)$. Hence $B : H^s(U) \to H^{s-m}(V)$ is bounded. Here we are using the standard compact-support-to-local form of the Sobolev mapping theorem for pseudodifferential operators, rather than a global mapping theorem on the possibly unbounded or nonsmooth open set $U$ (citing a result not yet in the wiki: Sobolev Mapping Theorem for Pseudodifferential Operators). [/step] [step:Apply Rellich compactness on the bounded patch] Since $m < 0$, we have $s-m > s$. The set $V$ is bounded with smooth boundary, so the Rellich-Kondrachov [compactness theorem](/theorems/2748) for Sobolev spaces on bounded smooth domains applies to the inclusion \begin{align*} J : H^{s-m}(V) \to H^s(V). \end{align*} Therefore $J$ is compact. [guided] The purpose of choosing $V$ is to place the problem on a bounded smooth domain, where Rellich compactness is available. The operator $A$ gives the regularity gain \begin{align*} H^s \to H^{s-m} \end{align*} locally, and because $m<0$, the exponent $s-m$ is strictly larger than $s$. Thus the output has positive extra Sobolev regularity. We now apply the Rellich-Kondrachov compactness theorem on the bounded smooth domain $V$ (citing a result not yet in the wiki: Rellich-Kondrachov Compactness Theorem). Its relevant hypothesis is that the source Sobolev order is strictly larger than the target Sobolev order. Here the source order is $s-m$ and the target order is $s$, and \begin{align*} s-m > s \end{align*} because $m<0$. Therefore the inclusion map \begin{align*} J : H^{s-m}(V) \to H^s(V) \end{align*} is compact. Concretely, every bounded sequence in $H^{s-m}(V)$ has a subsequence converging in $H^s(V)$. [/guided] [/step] [step:Factor the localized operator through the compact Sobolev inclusion] For $u \in H^s(U)$, the operator $\chi A\psi$ can be written as the composition $\chi A\psi u = E_\chi M_\chi J B u$. Indeed, $B u$ is the restriction of $A(\psi u)$ to $V$, $J$ regards it in the lower-order Sobolev space $H^s(V)$, $M_\chi$ multiplies by $\chi$, and $E_\chi$ extends the compactly supported result back to $U$. The operators $B$, $M_\chi$, and $E_\chi$ are bounded, while $J$ is compact. A composition of bounded operators with one [compact operator](/page/Compact%20Operator) is compact. Hence $\chi A\psi : H^s(U) \to H^s(U)$ is compact. This proves the theorem. [/step]