[proofplan] We localize the operator by inserting compactly supported cutoffs. Proper support ensures that, after localization, the operator has a smooth compactly supported kernel on $U \times U$. We then prove directly that any operator with such a kernel maps $H^s(\mathbb{R}^n)$ continuously into $H^t(\mathbb{R}^n)$ for arbitrary real $s,t$, using rapid decay of the [Fourier transform](/page/Fourier%20Transform) of the kernel. Finally, the local seminorm estimate gives the asserted continuity from $H^s_{\mathrm{loc}}(U)$ to $H^t_{\mathrm{loc}}(U)$. [/proofplan]

[step:Localize the properly supported smoothing operator by compact cutoffs] Fix $s,t \in \mathbb{R}$ and $\chi \in C_c^\infty(U)$. Let $K_R \in C^\infty(U \times U)$ denote the Schwartz kernel of the smoothing operator $R$. Proper support means that the projection of $\operatorname{supp} K_R \cap (\operatorname{supp}\chi \times U)$ onto the second factor is compact in $U$. Denote this compact set by \begin{align*} L_\chi := \{y \in U : \text{there exists } x \in \operatorname{supp}\chi \text{ with } (x,y) \in \operatorname{supp}K_R\}. \end{align*} Choose $\psi \in C_c^\infty(U)$ such that $\psi = 1$ on an open neighbourhood of $L_\chi$. Define the localized operator \begin{align*} A_{\chi,\psi}: C_c^\infty(U) \to C^\infty_c(U) \end{align*} by \begin{align*} A_{\chi,\psi}v := \chi R(\psi v). \end{align*} Its Schwartz kernel is the smooth compactly supported function \begin{align*} K_{\chi,\psi}: U \times U \to \mathbb{C} \end{align*} given by \begin{align*} K_{\chi,\psi}(x,y) := \chi(x)K_R(x,y)\psi(y). \end{align*} Since $\psi = 1$ on a neighbourhood of $L_\chi$, we have \begin{align*} \chi Ru = \chi R(\psi u) \end{align*} for every distribution $u$ for which the expression is locally defined. Thus it suffices to prove that every operator with a smooth compactly supported kernel maps $H^s(\mathbb{R}^n)$ continuously into $H^t(\mathbb{R}^n)$. [/step]

[step:Convert compact smoothness of the kernel into rapid Fourier decay]Extend $K_{\chi,\psi}$ by zero to a function \begin{align*} K: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C}. \end{align*} Then $K \in C_c^\infty(\mathbb{R}^n \times \mathbb{R}^n)$. Let $\widehat{K}: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C}$ denote its Fourier transform in both variables with the convention \begin{align*} \widehat{K}(\xi,\eta) := (2\pi)^{-n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} K(x,y)e^{-ix\cdot \xi}e^{-iy\cdot \eta}\,d\mathcal{L}^n(y)\,d\mathcal{L}^n(x). \end{align*} For $\zeta \in \mathbb{R}^n$, define \begin{align*} \langle \zeta \rangle := (1 + |\zeta|^2)^{1/2}. \end{align*} Since $K$ is smooth with compact support, repeated [integration by parts](/theorems/210) in the $x$ and $y$ variables gives the following rapid decay estimate: for every $M,N \in \mathbb{N}$ there exists $C_{M,N,K} > 0$ such that \begin{align*} |\widehat{K}(\xi,\eta)| \leq C_{M,N,K}\langle \xi \rangle^{-M}\langle \eta \rangle^{-N} \end{align*} for all $\xi,\eta \in \mathbb{R}^n$.[/step]

[guided]The point of passing to the Fourier side is that Sobolev norms are weighted $L^2$ norms of Fourier transforms. We first record exactly what compact smoothness gives. The function $K$ belongs to $C_c^\infty(\mathbb{R}^n \times \mathbb{R}^n)$, so every derivative $\partial_x^\alpha \partial_y^\beta K$ is integrable with respect to $\mathcal{L}^{2n}$. Let $M,N \in \mathbb{N}$. Applying [integration by parts](/theorems/2098) in the $x$ variables transfers powers of $\xi$ from the exponential $e^{-ix\cdot \xi}$ onto derivatives of $K$, and applying integration by parts in the $y$ variables transfers powers of $\eta$ from $e^{-iy\cdot \eta}$ onto derivatives of $K$. Because the derivatives of $K$ are compactly supported, no boundary terms appear. Hence there is a constant $C_{M,N,K} > 0$, depending only on finitely many $L^1(\mathbb{R}^{2n},\mathcal{L}^{2n})$ norms of derivatives of $K$, such that \begin{align*} |\widehat{K}(\xi,\eta)| \leq C_{M,N,K}\langle \xi \rangle^{-M}\langle \eta \rangle^{-N}. \end{align*} This is the analytic content of smoothing: after localization, the kernel has enough differentiability to produce arbitrarily many powers of decay in both Fourier variables.[/guided]

[step:Estimate the localized operator between arbitrary Sobolev orders] Let \begin{align*} A: \mathcal{S}(\mathbb{R}^n) \to C_c^\infty(\mathbb{R}^n) \end{align*} be the integral operator associated to $K$, namely \begin{align*} (Au)(x) := \int_{\mathbb{R}^n} K(x,y)u(y)\,d\mathcal{L}^n(y). \end{align*} For $u \in \mathcal{S}(\mathbb{R}^n)$, the Fourier transform of $Au$ has the form \begin{align*} \widehat{Au}(\xi) = \int_{\mathbb{R}^n} \widehat{K}_1(\xi,\eta)\widehat{u}(\eta)\,d\mathcal{L}^n(\eta), \end{align*} where $\widehat{K}_1$ differs from $\widehat{K}$ only by the harmless sign and normalization convention in the second variable. Therefore $\widehat{K}_1$ satisfies the same rapid decay estimates. Choose integers $M,N \in \mathbb{N}$ so large that \begin{align*} 2(M-t) > n \end{align*} and \begin{align*} 2(N+s) > n. \end{align*} Define the finite constant \begin{align*} B_{s,t,K}^2 := C_{M,N,K}^2\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\langle \xi \rangle^{2t-2M}\langle \eta \rangle^{-2N-2s}\,d\mathcal{L}^n(\eta)\,d\mathcal{L}^n(\xi). \end{align*} The chosen inequalities imply $B_{s,t,K}<\infty$. By the [Cauchy-Schwarz inequality](/theorems/432) in $L^2(\mathbb{R}^n,\mathcal{L}^n)$ applied in the $\eta$ variable, \begin{align*} \|Au\|_{H^t(\mathbb{R}^n)}^2 \leq B_{s,t,K}^2\|u\|_{H^s(\mathbb{R}^n)}^2. \end{align*} Thus \begin{align*} \|Au\|_{H^t(\mathbb{R}^n)} \leq B_{s,t,K}\|u\|_{H^s(\mathbb{R}^n)} \end{align*} for every $u \in \mathcal{S}(\mathbb{R}^n)$. Since $\mathcal{S}(\mathbb{R}^n)$ is dense in $H^s(\mathbb{R}^n)$, $A$ extends uniquely to a bounded [linear map](/page/Linear%20Map) \begin{align*} A: H^s(\mathbb{R}^n) \to H^t(\mathbb{R}^n). \end{align*} [/step]

[step:Pass the global localized estimate back to local Sobolev spaces] Apply the preceding step to the compactly supported smooth kernel $K_{\chi,\psi}$. There exists a constant $C_{\chi,\psi,s,t} > 0$ such that \begin{align*} \|\chi R(\psi v)\|_{H^t(\mathbb{R}^n)} \leq C_{\chi,\psi,s,t}\|\psi v\|_{H^s(\mathbb{R}^n)} \end{align*} for every $v \in H^s(\mathbb{R}^n)$. Now let $u \in H^s_{\mathrm{loc}}(U)$. Since $\psi \in C_c^\infty(U)$, the product $\psi u$, extended by zero outside $U$, belongs to $H^s(\mathbb{R}^n)$. By the support choice in the first step, \begin{align*} \chi Ru = \chi R(\psi u). \end{align*} Therefore \begin{align*} \|\chi Ru\|_{H^t(\mathbb{R}^n)} \leq C_{\chi,\psi,s,t}\|\psi u\|_{H^s(\mathbb{R}^n)}. \end{align*} This is precisely the seminorm estimate defining continuity of \begin{align*} R: H^s_{\mathrm{loc}}(U) \to H^t_{\mathrm{loc}}(U). \end{align*} Since $s,t \in \mathbb{R}$ and $\chi \in C_c^\infty(U)$ were arbitrary, the proof is complete. [/step]