[proofplan] The support condition is not needed for the boundedness of the already localized operator $\chi A\psi$, so we prove the stronger statement for arbitrary compactly supported cutoffs $\chi,\psi \in C_c^\infty(U)$. The compact supports of the cutoffs make $\chi A\psi$ a compactly supported local pseudodifferential operator of the same order $-r$, which can be represented as an operator on $\mathbb{R}^n$ after extending its kernel by zero. We then conjugate by Bessel potentials, use the pseudodifferential composition theorem to obtain an order-zero operator, apply $L^2$ boundedness for order-zero pseudodifferential operators, and finally pass back to the restriction definition of $H^s(U)$. [/proofplan]

[step:Replace the localized operator by a compactly supported operator on $\mathbb{R}^n$] Let $K_A \in \mathcal{D}'(U \times U)$ denote the Schwartz kernel of $A$, and let $\operatorname{Rel}(A) := \operatorname{supp} K_A \subset U \times U$ denote its support relation. Set \begin{align*} B := \chi A\psi. \end{align*} The Schwartz kernel of $B$ is \begin{align*} K_B(x,y) := \chi(x)K_A(x,y)\psi(y), \end{align*} where the products are the standard products of a distribution by smooth functions in the $x$ and $y$ variables. Hence \begin{align*} \operatorname{supp}K_B \subset \operatorname{Rel}(A) \cap (\operatorname{supp}\chi \times \operatorname{supp}\psi). \end{align*} Since $\chi,\psi \in C_c^\infty(U)$, the set $\operatorname{supp}\chi \times \operatorname{supp}\psi$ is a compact subset of $U \times U$. Therefore $K_B$ has compact support in $U \times U$. We use the following standard local pseudodifferential [extension theorem](/theorems/59): if $P \in \Psi^m(U)$ has Schwartz kernel compactly supported in $U \times U$, then the zero extension of its kernel defines an operator in $\Psi^m(\mathbb{R}^n)$ whose restriction to $U$ agrees with $P$. This theorem applies to $B$ because multiplication by compactly supported smooth cutoffs preserves the pseudodifferential order and because $K_B$ is compactly supported in $U \times U$. The compact containment of $\operatorname{supp} K_B$ in $U \times U$ ensures that the zero extension introduces no boundary singularity across $\partial(U \times U)$; the only pseudodifferential singularity remains the diagonal singularity already present inside $U \times U$. Thus $B$ extends by zero to an operator \begin{align*} \widetilde B : C_c^\infty(\mathbb{R}^n) \to C^\infty(\mathbb{R}^n) \end{align*} with \begin{align*} \widetilde B \in \Psi^{-r}(\mathbb{R}^n). \end{align*} Moreover, for every distribution $u \in \mathcal{D}'(U)$ and every extension $\widetilde u \in \mathcal{D}'(\mathbb{R}^n)$ of $u$, the restriction of $\widetilde B\widetilde u$ to $U$ equals $Bu$. This is because the $y$-support of the kernel of $B$ lies in $\operatorname{supp}\psi \subset U$, so $\widetilde B\widetilde u$ depends only on the restriction of $\widetilde u$ to $U$. [/step]

[step:Conjugate by Bessel potentials to obtain an order-zero operator]For each $t \in \mathbb{R}$, let \begin{align*} J_t : \mathcal{S}'(\mathbb{R}^n) \to \mathcal{S}'(\mathbb{R}^n) \end{align*} denote the Bessel potential operator with Fourier multiplier \begin{align*} \langle \xi\rangle^t := (1+|\xi|^2)^{t/2}. \end{align*} Thus $J_t \in \Psi^t(\mathbb{R}^n)$, and the Sobolev norm on $\mathbb{R}^n$ is \begin{align*} \|v\|_{H^t(\mathbb{R}^n)} := \|J_t v\|_{L^2(\mathbb{R}^n)}. \end{align*} Define \begin{align*} T := J_{s+r}\widetilde B J_{-s}. \end{align*} The standard composition theorem for pseudodifferential operators states that the composition of operators in $\Psi^{m_1}(\mathbb{R}^n)$ and $\Psi^{m_2}(\mathbb{R}^n)$ lies in $\Psi^{m_1+m_2}(\mathbb{R}^n)$, with the usual proper-support hypothesis for defining the composition. The compositions here are defined because $J_{s+r}$ and $J_{-s}$ are global Fourier multipliers and $\widetilde B$ has compactly supported kernel. Applying the theorem twice to the three factors of orders $s+r$, $-r$, and $-s$ gives \begin{align*} T \in \Psi^{(s+r)+(-r)+(-s)}(\mathbb{R}^n)=\Psi^0(\mathbb{R}^n). \end{align*}[/step]

[guided]We now translate the desired Sobolev gain into an $L^2$ statement. The operator $\widetilde B$ has order $-r$, so it should gain $r$ derivatives. Bessel potentials measure Sobolev regularity by converting $H^t$ norms into $L^2$ norms. For $t \in \mathbb{R}$, define the Bessel potential operator \begin{align*} J_t : \mathcal{S}'(\mathbb{R}^n) \to \mathcal{S}'(\mathbb{R}^n) \end{align*} to be the Fourier multiplier with symbol \begin{align*} \langle \xi\rangle^t := (1+|\xi|^2)^{t/2}. \end{align*} By definition of Sobolev spaces through Bessel potentials, \begin{align*} \|v\|_{H^t(\mathbb{R}^n)} := \|J_t v\|_{L^2(\mathbb{R}^n)}. \end{align*} Therefore, to estimate $\|\widetilde B F\|_{H^{s+r}(\mathbb{R}^n)}$, we need to estimate $\|J_{s+r}\widetilde B F\|_{L^2(\mathbb{R}^n)}$. If $F \in H^s(\mathbb{R}^n)$, then $F=J_{-s}g$ with $g:=J_sF \in L^2(\mathbb{R}^n)$. Thus \begin{align*} J_{s+r}\widetilde B F = J_{s+r}\widetilde B J_{-s}g. \end{align*} This motivates defining \begin{align*} T := J_{s+r}\widetilde B J_{-s}. \end{align*} Now we identify the order of $T$. The operator $J_{s+r}$ has order $s+r$, the operator $\widetilde B$ has order $-r$, and the operator $J_{-s}$ has order $-s$. The standard composition theorem for pseudodifferential operators says that composing an element of $\Psi^{m_1}(\mathbb{R}^n)$ with an element of $\Psi^{m_2}(\mathbb{R}^n)$ gives an element of $\Psi^{m_1+m_2}(\mathbb{R}^n)$, provided the composition is defined under the usual proper-support condition. Here the support condition is satisfied because $\widetilde B$ has compactly supported kernel, while $J_{s+r}$ and $J_{-s}$ are globally defined Fourier multipliers. Applying the theorem first to $\widetilde B J_{-s}$ and then to $J_{s+r}(\widetilde B J_{-s})$ gives \begin{align*} T \in \Psi^{(s+r)+(-r)+(-s)}(\mathbb{R}^n). \end{align*} The exponent simplifies to zero: \begin{align*} (s+r)+(-r)+(-s)=0. \end{align*} Thus \begin{align*} T \in \Psi^0(\mathbb{R}^n). \end{align*} This is the central point: the $r$ derivatives gained by $\widetilde B$ exactly cancel the $r$ extra derivatives demanded by the target space $H^{s+r}$.[/guided]

[step:Apply $L^2$ boundedness and undo the conjugation] By the Calderon-Vaillancourt $L^2$ [boundedness theorem](/theorems/181) for order-zero pseudodifferential operators, every operator in $\Psi^0(\mathbb{R}^n)$ with symbol in the standard order-zero class acts boundedly on $L^2(\mathbb{R}^n)$. Since $T \in \Psi^0(\mathbb{R}^n)$, there exists a constant $C_0>0$, depending on finitely many order-zero symbol seminorms of a full symbol of $T$, such that \begin{align*} \|Tg\|_{L^2(\mathbb{R}^n)} \le C_0\|g\|_{L^2(\mathbb{R}^n)} \end{align*} for every $g \in L^2(\mathbb{R}^n)$. Let $F \in H^s(\mathbb{R}^n)$ and define $g:=J_sF \in L^2(\mathbb{R}^n)$. Since $J_{-s}g=F$, the preceding estimate gives \begin{align*} \|\widetilde B F\|_{H^{s+r}(\mathbb{R}^n)} = \|J_{s+r}\widetilde B F\|_{L^2(\mathbb{R}^n)}. \end{align*} Using $F=J_{-s}g$, this becomes \begin{align*} \|J_{s+r}\widetilde B F\|_{L^2(\mathbb{R}^n)} = \|J_{s+r}\widetilde B J_{-s}g\|_{L^2(\mathbb{R}^n)}. \end{align*} By the definition of $T$, \begin{align*} \|J_{s+r}\widetilde B J_{-s}g\|_{L^2(\mathbb{R}^n)} = \|Tg\|_{L^2(\mathbb{R}^n)}. \end{align*} Therefore \begin{align*} \|\widetilde B F\|_{H^{s+r}(\mathbb{R}^n)} \le C_0\|g\|_{L^2(\mathbb{R}^n)} = C_0\|F\|_{H^s(\mathbb{R}^n)}. \end{align*} Thus \begin{align*} \widetilde B : H^s(\mathbb{R}^n) \to H^{s+r}(\mathbb{R}^n) \end{align*} is bounded. [/step]

[step:Pass from $\mathbb{R}^n$ back to the open set $U$] We use the restriction convention for Sobolev spaces on open sets: \begin{align*} H^t(U) := \{f|_U : f \in H^t(\mathbb{R}^n)\}, \end{align*} with quotient norm \begin{align*} \|u\|_{H^t(U)} := \inf\{\|F\|_{H^t(\mathbb{R}^n)} : F \in H^t(\mathbb{R}^n) \text{ and } F|_U=u\}. \end{align*} Let $u \in H^s(U)$, and let $F \in H^s(\mathbb{R}^n)$ be any extension of $u$. By the compatibility of $\widetilde B$ with restriction, \begin{align*} Bu = (\widetilde B F)|_U. \end{align*} Hence the quotient norm and the $\mathbb{R}^n$ estimate give \begin{align*} \|Bu\|_{H^{s+r}(U)} \le \|\widetilde B F\|_{H^{s+r}(\mathbb{R}^n)} \le C_0\|F\|_{H^s(\mathbb{R}^n)}. \end{align*} Taking the infimum over all such extensions $F$ yields \begin{align*} \|Bu\|_{H^{s+r}(U)} \le C_0\|u\|_{H^s(U)}. \end{align*} Since $B=\chi A\psi$, this proves that \begin{align*} \chi A\psi : H^s(U) \to H^{s+r}(U) \end{align*} is bounded. This is the desired Sobolev gain of order $r$ for the negative-order localized pseudodifferential operator. [/step]