[proofplan] We prove the result locally on an arbitrary compact subset of the separated region $W_r$. The separation condition gives uniform bounds for the coefficients of a differential operator in $\xi$ that fixes the phase $e^{i(x-y)\cdot \xi}$. Repeated [integration by parts](/theorems/210) transfers this operator from the phase to the amplitude, gaining one power of decay in $\xi$ at each application; choosing enough integrations by parts makes the resulting integral absolutely and locally uniformly convergent. Finally, $x$ and $y$ derivatives are handled by first differentiating the regularized integrals, observing that the differentiated amplitude is still a symbol of finite order, and repeating the same argument with a larger number of integrations by parts. [/proofplan]

[step:Fix a compact separated region and define the phase-adapted operator] Let $K \subset W_r$ be compact. Since $|x-y| \geq r$ on $W_r$, we have \begin{align*} |x-y|^{-1} \leq r^{-1} \end{align*} for every $(x,y) \in K$. For each $(x,y) \in K$, define the first-order differential operator in the $\xi$ variable \begin{align*} L_{x,y} := \frac{1}{|x-y|^2}\sum_{j=1}^n (x_j-y_j)D_{\xi_j} \end{align*} where $D_{\xi_j}:=-i\partial_{\xi_j}$. This operator acts on smooth functions of $\xi \in \mathbb{R}^n$. Since \begin{align*} D_{\xi_j}e^{i(x-y)\cdot \xi} = (x_j-y_j)e^{i(x-y)\cdot \xi} \end{align*} we obtain \begin{align*} L_{x,y}e^{i(x-y)\cdot \xi} = e^{i(x-y)\cdot \xi}. \end{align*} The formal transpose of $L_{x,y}$ with respect to integration against $\mathcal{L}^n$ in the $\xi$ variable is \begin{align*} L_{x,y}^t := -\frac{1}{|x-y|^2}\sum_{j=1}^n (x_j-y_j)D_{\xi_j}. \end{align*} The coefficients are independent of $\xi$, and on $K$ their absolute values are bounded by $r^{-1}$. [/step]

[step:Transfer derivatives from the phase to the amplitude]Let $N \in \mathbb{N}$ be chosen later. For $\varepsilon>0$, define the regularized map $F_\varepsilon:K\to\mathbb{C}$ by \begin{align*} F_\varepsilon(x,y) := \int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}\rho(\varepsilon \xi)A(x,y,\xi)\,d\mathcal{L}^n(\xi). \end{align*} The integrand is compactly supported in $\xi$, so repeated [integration by parts](/theorems/2098) in the $\xi$ variable gives \begin{align*} F_\varepsilon(x,y) = \int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}(L_{x,y}^t)^N\left[\rho(\varepsilon \xi)A(x,y,\xi)\right]\,d\mathcal{L}^n(\xi). \end{align*} Because $L_{x,y}^t$ is a linear combination of the operators $D_{\xi_j}$ with coefficients bounded on $K$, the Leibniz rule gives a finite expansion whose terms have the form \begin{align*} c_{\delta,\gamma}(x,y)\varepsilon^{|\delta|}(D^\delta \rho)(\varepsilon \xi)D_\xi^\gamma A(x,y,\xi) \end{align*} where $\delta,\gamma \in \mathbb{N}_0^n$, $|\delta|+|\gamma|=N$, and the coefficient functions $c_{\delta,\gamma}:K \to \mathbb{C}$ are uniformly bounded by constants depending only on $K,N,r$. Using the symbol estimate for $A$, each such term is bounded on $K$ by a constant times \begin{align*} \varepsilon^{|\delta|}|(D^\delta \rho)(\varepsilon \xi)|(1+|\xi|)^{m-|\gamma|}. \end{align*} If $\delta=0$, this is bounded by a constant times $(1+|\xi|)^{m-N}$. If $\delta \neq 0$, then $D^\delta \rho(\varepsilon \xi)$ is supported where $\varepsilon \xi$ lies in a fixed compact subset of $\mathbb{R}^n \setminus \{0\}$, hence $\varepsilon^{|\delta|}\leq C_\delta(1+|\xi|)^{-|\delta|}$ on that support. Therefore every term is bounded by \begin{align*} C_{K,N,\rho}(1+|\xi|)^{m-N}. \end{align*} Choose $N>m+n$. Then the function $\xi \mapsto (1+|\xi|)^{m-N}$ is integrable over $\mathbb{R}^n$ with respect to $\mathcal{L}^n$, so the family of regularized integrands is dominated by an integrable function uniformly for $(x,y)\in K$ and $\varepsilon \in (0,1]$.[/step]

[guided]The point of introducing $L_{x,y}$ is that it is tailored to the phase. The phase is $e^{i(x-y)\cdot \xi}$, and differentiating it in $\xi_j$ produces the factor $x_j-y_j$. Thus the normalized combination \begin{align*} L_{x,y} := \frac{1}{|x-y|^2}\sum_{j=1}^n (x_j-y_j)D_{\xi_j} \end{align*} satisfies \begin{align*} L_{x,y}e^{i(x-y)\cdot \xi}=e^{i(x-y)\cdot \xi}. \end{align*} This identity is useful because it lets us insert $L_{x,y}^N$ onto the phase without changing the integrand, and then integrate by parts to move those $N$ derivatives onto the amplitude. For the regularized map $F_\varepsilon:K\to\mathbb{C}$ defined by \begin{align*} F_\varepsilon(x,y) := \int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}\rho(\varepsilon \xi)A(x,y,\xi)\,d\mathcal{L}^n(\xi) \end{align*} the cutoff makes the integrand compactly supported in $\xi$, so integration by parts has no boundary term. Since the formal transpose of $D_{\xi_j}=-i\partial_{\xi_j}$ is $-D_{\xi_j}$ under integration with respect to $\mathcal{L}^n$, the formal transpose of $L_{x,y}$ is \begin{align*} L_{x,y}^t := -\frac{1}{|x-y|^2}\sum_{j=1}^n (x_j-y_j)D_{\xi_j}. \end{align*} Therefore \begin{align*} F_\varepsilon(x,y)=\int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}(L_{x,y}^t)^N\left[\rho(\varepsilon \xi)A(x,y,\xi)\right]\,d\mathcal{L}^n(\xi). \end{align*} Now we estimate the new amplitude. Each application of $L_{x,y}^t$ differentiates once in $\xi$ and has coefficients bounded by $r^{-1}$ on $K$, because $|x-y|\ge r$. After expanding with the Leibniz rule, every term has the form \begin{align*} c_{\delta,\gamma}(x,y)\varepsilon^{|\delta|}(D^\delta \rho)(\varepsilon \xi)D_\xi^\gamma A(x,y,\xi) \end{align*} with $|\delta|+|\gamma|=N$. The symbol estimate with $\alpha=0$ and $\beta=0$ gives \begin{align*} |D_\xi^\gamma A(x,y,\xi)|\leq C_{K,0,0,\gamma}(1+|\xi|)^{m-|\gamma|}. \end{align*} If no derivative lands on the cutoff, then $|\gamma|=N$ and we directly gain the decay $(1+|\xi|)^{m-N}$. If at least one derivative lands on the cutoff, then $D^\delta\rho(\varepsilon \xi)$ is supported where $\varepsilon \xi$ stays away from $0$, so the factor $\varepsilon^{|\delta|}$ can be converted into the decay factor $(1+|\xi|)^{-|\delta|}$ on that support. Since $|\delta|+|\gamma|=N$, the same bound follows: \begin{align*} \left|(L_{x,y}^t)^N\left[\rho(\varepsilon \xi)A(x,y,\xi)\right]\right|\leq C_{K,N,\rho}(1+|\xi|)^{m-N}. \end{align*} Choosing $N>m+n$ makes this majorant integrable over $\mathbb{R}^n$. This is the exact place where the repeated integration by parts creates enough decay to turn the oscillatory integral into an absolutely controlled integral.[/guided]

[step:Pass to the cutoff limit and obtain a locally uniform formula] For the same $N>m+n$, the previous estimate gives local uniform domination on $K$. Moreover, \begin{align*} (L_{x,y}^t)^N\left[\rho(\varepsilon \xi)A(x,y,\xi)\right]\to (L_{x,y}^t)^N A(x,y,\xi) \end{align*} for every $(x,y,\xi)\in K\times\mathbb{R}^n$ as $\varepsilon\downarrow 0$. By the [dominated convergence theorem](/theorems/4), applied with respect to $\mathcal{L}^n$ in the $\xi$ variable and with the same integrable majorant $C_{K,N,\rho}(1+|\xi|)^{m-N}$ uniformly for $(x,y)\in K$, the regularized integrals converge locally uniformly on $K$ to \begin{align*} F(x,y)=\int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}(L_{x,y}^t)^N A(x,y,\xi)\,d\mathcal{L}^n(\xi). \end{align*} This formula is independent of the cutoff $\rho$, because the right-hand side contains no cutoff. Since $K\subset W_r$ was arbitrary, the oscillatory integral defining $F$ exists on all of $W_r$ and is locally uniformly represented by the absolutely convergent integral above. [/step]

[step:Differentiate the phase and amplitude before repeating the argument] Let $\alpha,\beta \in \mathbb{N}_0^n$. Define the differentiated amplitude \begin{align*} B_{\alpha,\beta}(x,y,\xi):=e^{-i(x-y)\cdot \xi}D_x^\alpha D_y^\beta\left(e^{i(x-y)\cdot \xi}A(x,y,\xi)\right). \end{align*} This defines a smooth function $B_{\alpha,\beta}:W\times\mathbb{R}^n\to\mathbb{C}$. Expanding the derivatives by the product rule shows that $B_{\alpha,\beta}$ is a finite sum of terms of the form \begin{align*} p_{\alpha_1,\beta_1}(\xi)D_x^{\alpha_2}D_y^{\beta_2}A(x,y,\xi) \end{align*} where $\alpha_1+\alpha_2=\alpha$, $\beta_1+\beta_2=\beta$, and $p_{\alpha_1,\beta_1}$ is a polynomial in $\xi$ of degree at most $|\alpha_1|+|\beta_1|$. Therefore $B_{\alpha,\beta}$ satisfies local symbol estimates of order at most $m+|\alpha|+|\beta|$ and type $(1,0)$ in $\xi$. Choose $N_{\alpha,\beta}\in\mathbb{N}$ such that \begin{align*} N_{\alpha,\beta}>m+|\alpha|+|\beta|+n. \end{align*} Applying the preceding integration-by-parts argument to $B_{\alpha,\beta}$ gives locally [uniform convergence](/page/Uniform%20Convergence) on $K$ of \begin{align*} \int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}\rho(\varepsilon \xi)B_{\alpha,\beta}(x,y,\xi)\,d\mathcal{L}^n(\xi) \end{align*} to the absolutely convergent expression \begin{align*} \int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}(L_{x,y}^t)^{N_{\alpha,\beta}}B_{\alpha,\beta}(x,y,\xi)\,d\mathcal{L}^n(\xi). \end{align*} For each $\varepsilon>0$, the regularized function $F_\varepsilon$ is smooth in $(x,y)$, and differentiation under the integral is permitted because the cutoff makes the $\xi$ support compact. Thus \begin{align*} D_x^\alpha D_y^\beta F_\varepsilon(x,y)=\int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}\rho(\varepsilon \xi)B_{\alpha,\beta}(x,y,\xi)\,d\mathcal{L}^n(\xi). \end{align*} The standard theorem on convergence of derivatives applies locally on $K$: the functions $F_\varepsilon$ are smooth, $F_\varepsilon$ converges locally uniformly to $F$, and for every multi-index pair $\alpha,\beta$ the derivatives $D_x^\alpha D_y^\beta F_\varepsilon$ converge locally uniformly to the displayed oscillatory limit. Therefore $F$ is smooth on $K$ and \begin{align*} D_x^\alpha D_y^\beta F(x,y)=\operatorname{Os}\!\int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}B_{\alpha,\beta}(x,y,\xi)\,d\mathcal{L}^n(\xi). \end{align*} By the definition of $B_{\alpha,\beta}$, this is exactly \begin{align*} D_x^\alpha D_y^\beta F(x,y)=\operatorname{Os}\!\int_{\mathbb{R}^n}D_x^\alpha D_y^\beta\left(e^{i(x-y)\cdot \xi}A(x,y,\xi)\right)\,d\mathcal{L}^n(\xi). \end{align*} This proves both the smoothness of $F$ on $W_r$ and the asserted differentiation rule. [/step]