[proofplan] We localize the smoothing kernel by the cutoffs $\chi$ and $\psi$, which produces a smooth compactly supported kernel on $U \times U$ and hence, after extension by zero, on $\mathbb{R}^n \times \mathbb{R}^n$. On $\mathbb{R}^n$ we prove the corresponding operator maps $H^s(\mathbb{R}^n)$ boundedly into $H^t(\mathbb{R}^n)$ by conjugating with Bessel potentials and reducing the problem to an $L^2$ Hilbert-Schmidt estimate. Finally, because the kernel is supported in the $y$-variable inside $U$, the operator depends only on the restriction class of the input in $H^s(U)$, so it descends to the quotient [Sobolev space](/page/Sobolev%20Space) with the same bound. [/proofplan]

[step:Extend the localized Schwartz kernel to a compactly supported smooth kernel on $\mathbb{R}^n \times \mathbb{R}^n$] Define \begin{align*} k: U \times U \to \mathbb{C}, \qquad k(x,y) = \chi(x)K_R(x,y)\psi(y). \end{align*} Since $K_R \in C^\infty(U \times U)$ and $\chi,\psi \in C_c^\infty(U)$, multiplication gives $k \in C^\infty(U \times U)$. Moreover, \begin{align*} \operatorname{supp} k \subset \operatorname{supp}\chi \times \operatorname{supp}\psi. \end{align*} Both $\operatorname{supp}\chi$ and $\operatorname{supp}\psi$ are compact subsets of $U$, so $\operatorname{supp} k$ is a compact subset of $U \times U$. Therefore the zero extension of $k$ defines a function \begin{align*} \widetilde{k}: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C} \end{align*} with $\widetilde{k} \in C_c^\infty(\mathbb{R}^n \times \mathbb{R}^n)$. Define the integral operator \begin{align*} \widetilde{S}: C_c^\infty(\mathbb{R}^n) \to C^\infty(\mathbb{R}^n) \end{align*} by \begin{align*} (\widetilde{S}f)(x) = \int_{\mathbb{R}^n} \widetilde{k}(x,y)f(y)\,d\mathcal{L}^n(y). \end{align*} This operator agrees on $U$ with the cutoff operator whose kernel is $\chi(x)K_R(x,y)\psi(y)$. [/step]

[step:Reduce the Sobolev estimate on $\mathbb{R}^n$ to an $L^2$ Hilbert-Schmidt estimate]For $r \in \mathbb{R}$, let \begin{align*} J^r: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}'(\mathbb{R}^n) \end{align*} denote the Bessel potential operator on $\mathbb{R}^n$, defined on Schwartz functions by the Fourier multiplier \begin{align*} \widehat{J^r f}(\xi) = (1+|\xi|^2)^{r/2}\hat{f}(\xi). \end{align*} It extends by completion to an isometric isomorphism $J^r:H^r(\mathbb{R}^n)\to L^2(\mathbb{R}^n)$. Here the [Fourier transform](/page/Fourier%20Transform) is taken with the symmetric normalization \begin{align*} \hat{f}(\xi) = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} f(x)e^{-ix\cdot \xi}\,d\mathcal{L}^n(x). \end{align*} With the Bessel-potential definition of Sobolev spaces, \begin{align*} \|f\|_{H^r(\mathbb{R}^n)} = \|J^r f\|_{L^2(\mathbb{R}^n)}. \end{align*} Define the mixed Fourier transform of $\widetilde{k}$ as the function $a: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C}$ given by \begin{align*} a(\xi,\eta) = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\widetilde{k}(x,y)e^{-ix\cdot \xi}e^{iy\cdot \eta}\,d\mathcal{L}^n(y)\,d\mathcal{L}^n(x). \end{align*} Since $\widetilde{k} \in C_c^\infty(\mathbb{R}^n \times \mathbb{R}^n)$, repeated [integration by parts](/theorems/210) in $x$ and $y$ shows that $a$ is a Schwartz function on $\mathbb{R}^n \times \mathbb{R}^n$. For $f \in C_c^\infty(\mathbb{R}^n)$, the Fourier transform of $\widetilde{S}f$ is \begin{align*} \widehat{\widetilde{S}f}(\xi) = \int_{\mathbb{R}^n} a(\xi,\eta)\hat{f}(\eta)\,d\mathcal{L}^n(\eta). \end{align*} Define $b: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C}$ by \begin{align*} b(\xi,\eta) = (1+|\xi|^2)^{t/2}a(\xi,\eta)(1+|\eta|^2)^{-s/2}. \end{align*} Because $a$ is Schwartz, multiplication by the polynomial weights $(1+|\xi|^2)^{t/2}$ and $(1+|\eta|^2)^{-s/2}$ still gives \begin{align*} b \in L^2(\mathbb{R}^n \times \mathbb{R}^n). \end{align*} Define the [linear map](/page/Linear%20Map) \begin{align*} A: L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n) \end{align*} initially for $g \in C_c^\infty(\mathbb{R}^n)$ by the Fourier-variable kernel formula \begin{align*} (Ag)(\xi) = \int_{\mathbb{R}^n} b(\xi,\eta)g(\eta)\,d\mathcal{L}^n(\eta). \end{align*} For each $g \in L^2(\mathbb{R}^n)$, Cauchy-Schwarz in the $\eta$ variable gives \begin{align*} |(Ag)(\xi)|^2 \leq \left(\int_{\mathbb{R}^n}|b(\xi,\eta)|^2\,d\mathcal{L}^n(\eta)\right)\|g\|_{L^2(\mathbb{R}^n)}^2. \end{align*} Integrating this inequality in $\xi$ with respect to $\mathcal{L}^n$ gives the Hilbert-Schmidt bound \begin{align*} \|Ag\|_{L^2(\mathbb{R}^n)} \leq \|b\|_{L^2(\mathbb{R}^n \times \mathbb{R}^n)}\|g\|_{L^2(\mathbb{R}^n)}. \end{align*} Applying this to $g=J^s f$ yields \begin{align*} \|\widetilde{S}f\|_{H^t(\mathbb{R}^n)} \leq \|b\|_{L^2(\mathbb{R}^n \times \mathbb{R}^n)}\|f\|_{H^s(\mathbb{R}^n)}. \end{align*} By density of $C_c^\infty(\mathbb{R}^n)$ in $H^s(\mathbb{R}^n)$, $\widetilde{S}$ extends uniquely to a bounded linear map \begin{align*} \widetilde{S}: H^s(\mathbb{R}^n) \to H^t(\mathbb{R}^n). \end{align*}[/step]

[guided]The purpose of the Bessel potentials is to convert an $H^s \to H^t$ estimate into an $L^2 \to L^2$ estimate. For $r \in \mathbb{R}$, the Bessel potential is the map \begin{align*} J^r: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}'(\mathbb{R}^n) \end{align*} defined on Schwartz functions by the Fourier multiplier \begin{align*} \widehat{J^r f}(\xi) = (1+|\xi|^2)^{r/2}\hat{f}(\xi). \end{align*} It extends by completion to an isometric isomorphism $J^r:H^r(\mathbb{R}^n)\to L^2(\mathbb{R}^n)$. Thus controlling $\widetilde{S}f$ in $H^t(\mathbb{R}^n)$ is the same as controlling $J^t\widetilde{S}f$ in $L^2(\mathbb{R}^n)$, while the input norm is $\|J^s f\|_{L^2(\mathbb{R}^n)}$. We now compute the kernel of the conjugated operator in Fourier variables. The mixed Fourier transform of the compactly supported smooth kernel $\widetilde{k}$ is the function $a: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C}$ given by \begin{align*} a(\xi,\eta) = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\widetilde{k}(x,y)e^{-ix\cdot \xi}e^{iy\cdot \eta}\,d\mathcal{L}^n(y)\,d\mathcal{L}^n(x). \end{align*} Because $\widetilde{k}$ is smooth and compactly supported in both variables, [integration by parts](/theorems/2098) in $x$ and $y$ gives rapid decay in both $\xi$ and $\eta$. Hence $a$ is a Schwartz function on $\mathbb{R}^n \times \mathbb{R}^n$. For $f \in C_c^\infty(\mathbb{R}^n)$, Fourier inversion and [Fubini's theorem](/theorems/2961) apply because all functions involved are smooth and compactly supported or rapidly decreasing. They give \begin{align*} \widehat{\widetilde{S}f}(\xi) = \int_{\mathbb{R}^n} a(\xi,\eta)\hat{f}(\eta)\,d\mathcal{L}^n(\eta). \end{align*} After applying $J^t$ to the output and writing $\hat{f}(\eta)=(1+|\eta|^2)^{-s/2}\widehat{J^s f}(\eta)$, the relevant kernel is the function $b: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C}$ defined by \begin{align*} b(\xi,\eta) = (1+|\xi|^2)^{t/2}a(\xi,\eta)(1+|\eta|^2)^{-s/2}. \end{align*} The factor $(1+|\xi|^2)^{t/2}(1+|\eta|^2)^{-s/2}$ grows at most polynomially in $(\xi,\eta)$, while $a$ decays faster than every polynomial. Therefore \begin{align*} b \in L^2(\mathbb{R}^n \times \mathbb{R}^n). \end{align*} Define the linear map \begin{align*} A: L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n) \end{align*} initially for $g \in C_c^\infty(\mathbb{R}^n)$ by \begin{align*} (Ag)(\xi) = \int_{\mathbb{R}^n} b(\xi,\eta)g(\eta)\,d\mathcal{L}^n(\eta). \end{align*} The $L^2$ hypothesis on $b$ is exactly the Hilbert-Schmidt condition. Indeed, Cauchy-Schwarz in the $\eta$ variable gives \begin{align*} |(Ag)(\xi)|^2 \leq \left(\int_{\mathbb{R}^n}|b(\xi,\eta)|^2\,d\mathcal{L}^n(\eta)\right)\left(\int_{\mathbb{R}^n}|g(\eta)|^2\,d\mathcal{L}^n(\eta)\right). \end{align*} Now integrate both sides over $\xi \in \mathbb{R}^n$ with respect to $\mathcal{L}^n$. This gives \begin{align*} \|Ag\|_{L^2(\mathbb{R}^n)}^2 \leq \|b\|_{L^2(\mathbb{R}^n \times \mathbb{R}^n)}^2\|g\|_{L^2(\mathbb{R}^n)}^2. \end{align*} Taking square roots proves that $A$ is bounded on $L^2(\mathbb{R}^n)$. Finally set $g=J^s f$. The previous estimate becomes \begin{align*} \|\widetilde{S}f\|_{H^t(\mathbb{R}^n)} = \|J^t\widetilde{S}f\|_{L^2(\mathbb{R}^n)} \leq \|b\|_{L^2(\mathbb{R}^n \times \mathbb{R}^n)}\|J^s f\|_{L^2(\mathbb{R}^n)}. \end{align*} Thus \begin{align*} \|\widetilde{S}f\|_{H^t(\mathbb{R}^n)} \leq \|b\|_{L^2(\mathbb{R}^n \times \mathbb{R}^n)}\|f\|_{H^s(\mathbb{R}^n)}. \end{align*} Since $C_c^\infty(\mathbb{R}^n)$ is dense in $H^s(\mathbb{R}^n)$ for every $s \in \mathbb{R}$, the estimate extends $\widetilde{S}$ uniquely to a bounded linear map from $H^s(\mathbb{R}^n)$ to $H^t(\mathbb{R}^n)$.[/guided]

[step:Descend the estimate to Sobolev spaces on $U$ with quotient norm] Let \begin{align*} \rho_U: H^s(\mathbb{R}^n) \to H^s(U) \end{align*} denote the restriction map. We define a candidate map \begin{align*} T: H^s(U) \to H^t(U) \end{align*} by choosing global extensions and then prove that the definition is independent of the choice. By definition of the quotient norm, \begin{align*} \|u\|_{H^s(U)} = \inf\{\|F\|_{H^s(\mathbb{R}^n)} : F \in H^s(\mathbb{R}^n), \rho_U F = u\}. \end{align*} For $u \in H^s(U)$, choose any $F \in H^s(\mathbb{R}^n)$ satisfying $\rho_U F=u$, and define \begin{align*} Tu = \rho_U(\widetilde{S}F). \end{align*} We verify that this definition is independent of the chosen extension. Suppose $F_1,F_2 \in H^s(\mathbb{R}^n)$ satisfy $\rho_U F_1=\rho_U F_2$. Define $G=F_1-F_2$. Then $G$ restricts to $0$ as a distribution on $U$. Let $\varphi \in C_c^\infty(\mathbb{R}^n)$ be an arbitrary [test function](/page/Test%20Function) in the $x$-variable, and define $\Phi_\varphi: \mathbb{R}^n \to \mathbb{C}$ by \begin{align*} \Phi_\varphi(y)=\int_{\mathbb{R}^n}\widetilde{k}(x,y)\varphi(x)\,d\mathcal{L}^n(x). \end{align*} Since $\widetilde{k} \in C_c^\infty(\mathbb{R}^n \times \mathbb{R}^n)$ and its $y$-support is contained in $\operatorname{supp}\psi \subset U$, we have $\Phi_\varphi \in C_c^\infty(U)$. Therefore the distributional pairing of $G$ with $\Phi_\varphi$ is zero. For smooth compactly supported $G$, Fubini's theorem gives the identity \begin{align*} \int_{\mathbb{R}^n}(\widetilde{S}G)(x)\varphi(x)\,d\mathcal{L}^n(x)=\int_{\mathbb{R}^n}G(y)\Phi_\varphi(y)\,d\mathcal{L}^n(y). \end{align*} For general $G \in H^s(\mathbb{R}^n)$, choose $G_j \in C_c^\infty(\mathbb{R}^n)$ with $G_j \to G$ in $H^s(\mathbb{R}^n)$. The already proved boundedness $\widetilde{S}:H^s(\mathbb{R}^n)\to H^t(\mathbb{R}^n)$ and the continuous embedding of $H^t(\mathbb{R}^n)$ into distributions allow passage to the limit on the left, while convergence in $H^s$ allows passage to the limit against the fixed test function $\Phi_\varphi$. Hence the same identity holds for $G$. Since $G$ restricts to $0$ as a distribution on $U$ and $\Phi_\varphi \in C_c^\infty(U)$, the right-hand side is zero. Therefore $\widetilde{S}G$ pairs to zero with every $\varphi \in C_c^\infty(\mathbb{R}^n)$, so $\widetilde{S}G=0$ and \begin{align*} \rho_U(\widetilde{S}F_1)=\rho_U(\widetilde{S}F_2). \end{align*} Thus $T$ is well-defined. The bound on $\widetilde{S}$ gives, for every extension $F$ of $u$, \begin{align*} \|Tu\|_{H^t(U)} \leq \|\widetilde{S}F\|_{H^t(\mathbb{R}^n)} \leq \|b\|_{L^2(\mathbb{R}^n \times \mathbb{R}^n)}\|F\|_{H^s(\mathbb{R}^n)}. \end{align*} Taking the infimum over all such extensions $F$ gives \begin{align*} \|Tu\|_{H^t(U)} \leq \|b\|_{L^2(\mathbb{R}^n \times \mathbb{R}^n)}\|u\|_{H^s(U)}. \end{align*} Thus $T:H^s(U)\to H^t(U)$ is bounded. [/step]

[step:Identify the descended operator with the localized smoothing operator and prove uniqueness] If $F \in C_c^\infty(\mathbb{R}^n)$ and $u=\rho_U F$, then for $x \in U$ the descended operator satisfies \begin{align*} (Tu)(x) = \int_{\mathbb{R}^n}\widetilde{k}(x,y)F(y)\,d\mathcal{L}^n(y). \end{align*} Since $\widetilde{k}(x,y)=\chi(x)K_R(x,y)\psi(y)$ for $(x,y)\in U\times U$ and $\widetilde{k}(x,y)=0$ when $y \notin U$, this becomes \begin{align*} (Tu)(x) = \int_U \chi(x)K_R(x,y)\psi(y)u(y)\,d\mathcal{L}^n(y). \end{align*} Hence $T$ is precisely the operator with kernel $\chi(x)K_R(x,y)\psi(y)$ on the restrictions to $U$ of global test functions. Let \begin{align*} \mathcal{D}_s(U) := \rho_U(C_c^\infty(\mathbb{R}^n)) \subset H^s(U). \end{align*} By the quotient construction of $H^s(U)$ and the density of $C_c^\infty(\mathbb{R}^n)$ in $H^s(\mathbb{R}^n)$, the subspace $\mathcal{D}_s(U)$ is dense in $H^s(U)$. Therefore any bounded linear map $H^s(U)\to H^t(U)$ agreeing with the above kernel operator on $\mathcal{D}_s(U)$ must agree with $T$ on all of $H^s(U)$. This proves uniqueness of the bounded extension, and the theorem follows. [/step]